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An Instructor’s Solutions Manual to Accompany
ISBN-13: 978-0-495-24458-5
ISBN-10: 0-495-24458-9
90000
9 780495 244585
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ISBN-13: 978-0-495-24458-5
ISBN-10: 0-495-24458-9
© 2009, 2004 Cengage Learning
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Page iii
Contents
1. Tension, Compression, and Shear
Normal Stress and Strain 1
Mechanical Properties of Materials 15
Elasticity, Plasticity, and Creep 21
Linear Elasticity, Hooke’s Law, and Poisson’s Ratio 25
Shear Stress and Strain 30
Allowable Stresses and Allowable Loads 51
Design for Axial Loads and Direct Shear 69
2. Axially Loaded Members
Changes in Lengths of Axially Loaded Members 89
Changes in Lengths under Nonuniform Conditions 105
Statically Indeterminate Structures 124
Thermal Effects 151
Stresses on Inclined Sections 178
Strain Energy 198
Impact Loading 212
Stress Concentrations 224
Nonlinear Behavior (Changes in Lengths of Bars) 231
Elastoplastic Analysis 237
3. Torsion
Torsional Deformations 249
Circular Bars and Tubes 252
Nonuniform Torsion 266
Pure Shear 287
Transmission of Power 294
Statically Indeterminate Torsional Members 302
Strain Energy in Torsion 319
Thin-Walled Tubes 328
Stress Concentrations in Torsion 338
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CONTENTS
4. Shear Forces and Bending Moments
Shear Forces and Bending Moments 343
Shear-Force and Bending-Moment Diagrams 355
5. Stresses in Beams (Basic Topics)
Longitudinal Strains in Beams 389
Normal Stresses in Beams 392
Design of Beams 412
Nonprismatic Beams 431
Fully Stressed Beams 440
Shear Stresses in Rectangular Beams 442
Shear Stresses in Circular Beams 453
Shear Stresses in Beams with Flanges 457
Built-Up Beams 466
Beams with Axial Loads 475
Stress Concentrations 492
6. Stresses in Beams (Advanced Topics)
Composite Beams 497
Transformed-Section Method 508
Beams with Inclined Loads 520
Bending of Unsymmetric Beams 529
Shear Stresses in Wide-Flange Beams 541
Shear Centers of Thin-Walled Open Sections 543
Elastoplastic Bending 558
7. Analysis of Stress and Strain
Plane Stress 571
Principal Stresses and Maximum Shear Stresses 582
Mohr’s Circle 595
Hooke’s Law for Plane Stress 608
Triaxial Stress 615
Plane Strain 622
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CONTENTS
8. Applications of Plane Stress (Pressure Vessels, Beams, and Combined Loadings)
Spherical Pressure Vessels 649
Cylindrical Pressure Vessels 655
Maximum Stresses in Beams 664
Combined Loadings 675
9. Deflections of Beams
Differential Equations of the Deflection Curve 707
Deflection Formulas 710
Deflections by Integration of the Bending-Moment Equation 714
Deflections by Integration of the Shear Force and Load Equations 722
Method of Superposition 730
Moment-Area Method 745
Nonprismatic Beams 754
Strain Energy 770
Castigliano’s Theorem 775
Deflections Produced by Impact 784
Temperature Effects 790
10. Statically Indeterminate Beams
Differential Equations of the Deflection Curve 795
Method of Superposition 809
Temperature Effects 839
Longitudinal Displacements at the Ends of Beams 843
11. Columns
Idealized Buckling Models 845
Critical Loads of Columns with Pinned Supports 851
Columns with Other Support Conditions 863
Columns with Eccentric Axial Loads 871
The Secant Formula 880
Design Formulas for Columns 889
Aluminum Columns 903
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CONTENTS
12. Review of Centroids and Moments of Inertia
Centroids of Plane Ares 913
Centroids of Composite Areas 915
Moment of Inertia of Plane Areas 919
Parallel-Axis Theorem 923
Polar Moments of Inertia 927
Products of Inertia 929
Rotation of Axes 932
Principal Axes, Principal Points, and Principal Moments of Inertia 936
Answers to Problems 944
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Page 1
1
Tension, Compression,
and Shear
Normal Stress and Strain
P1
Problem 1.2-1 A hollow circular post ABC (see figure) supports a load
P1 1700 lb acting at the top. A second load P2 is uniformly distributed
around the cap plate at B. The diameters and thicknesses of the upper and
lower parts of the post are dAB 1.25 in., tAB 0.5 in., dBC 2.25 in., and
tBC 0.375 in., respectively.
A
tAB
dAB
P2
(a) Calculate the normal stress AB in the upper part of the post.
(b) If it is desired that the lower part of the post have the same
compressive stress as the upper part, what should be the magnitude
of the load P2?
(c) If P1 remains at 1700 lb and P2 is now set at 2260 lb, what new
thickness of BC will result in the same compressive stress in both
parts?
B
dBC
tBC
C
Solution 1.2-1
PART (a)
PART (b)
P1 1700
dBC 2.25
AAB dAB 1.25
tAB 0.5
tBC 0.375
p [ dAB2 (dAB 2 tAB)2]
4
AAB 1.178
sAB 1443 psi
sAB P1
AAB
ABC p[ dBC2 1dBC 2tBC22]
ABC 2.209
4
P2 ABABC P1
P2 1488 lbs
CHECK:
;
P1 + P2
1443 psi
ABC
;
1
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CHAPTER 1 Tension, Compression, and Shear
Part (c)
P2 2260
dBC 2tBC P1 + P2
ABC
sAB
P1 + P2
2.744
sAB
dBC tBC A
dBC2 2
A
dBC tBC 0.499 inches
4 P1 + P2
a
b
p
sAB
Problem 1.2-2 A force P of 70 N is applied by
a rider to the front hand brake of a bicycle (P is
the resultant of an evenly distributed pressure). As
the hand brake pivots at A, a tension T develops in
the 460-mm long brake cable (Ae 1.075 mm2)
which elongates by 0.214 mm. Find normal
stress and strain in the brake cable.
Brake cable, L = 460 mm
Hand brake pivot A
37.5 mm A
P (Resultant
of distributed
pressure)
m
100
P 70 N
L 460 mm
Ae 1.075 mm2
0.214 mm
Statics: sum moments about A to get T 2P
s
T
Ae
s 103.2 MPa
â
d
L
â 4.65 * 10 4
E
s
1.4 * 105 MPa
â
;
;
NOTE: (E for cables is approx. 140 GPa)
;
T
50
m
Solution 1.2-2
4 P1 + P2
b
a
p
sAB
2
(dBC 2tBC)2
dBC2 4 P1 + P2
b
a
p
sAB
mm
Uniform hand
brake pressure
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3
SECTION 1.2 Normal Stress and Strain
Problem 1.2-3 A bicycle rider would like to compare the effectiveness of cantilever hand brakes [see figure part (a)] versus
V brakes [figure part (b)].
(a) Calculate the braking force RB at the wheel rims for each of the bicycle brake systems shown. Assume that all forces act in
the plane of the figure and that cable tension T 45 lbs. Also, what is the average compressive normal stress c on the
brake pad (A 0.625 in.2)?
(b) For each braking system, what is the stress in the brake cable
T
(assume effective cross-sectional area of 0.00167 in.2)?
(HINT: Because of symmetry, you only need to use the right
half of each figure in your analysis.)
4 in.
T
D
TDC = TDE
T
45°
TDE
4 in.
TDC
TDE
90°
E
C
TDCh
5 in.
4.25 in.
RB
A
G
Pivot points
anchored to frame
(a) Cantilever brakes
Solution 1.2-3
Apad 0.625 in.2
Acable 0.00167 in.2
(a) CANTILEVER BRAKES-BRAKING FORCE
RB & PAD PRESSURE
Statics: sum forces at D to get TDC T / 2
a MA 0
RB(1) TDCh(3) TDCv(1)
TDCh TDCv
TDCh T / 2
RB 90 lbs
;
so RB 2T vs 4.25T for V brakes (below)
HA
B
E
RB
1 in.
B
F
RB 2T
T
TDCv
2 in.
T 45 lbs
C
D
1 in.
F
1 in.
HA
Pivot points
anchored to frame
A
VA
VA
(b) V brakes
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CHAPTER 1 Tension, Compression, and Shear
spad RB
Apad
scable spad 144 psi
T
Acable
;
scable 26,946 psi
4.25
2.125
2
;
(same for
V-brakes (below))
(b) V BRAKES - BRAKING FORCE RB & PAD PRESSURE
a MA 0
RB 4.25T
spad RB
Apad
RB 191.3 lbs
spad 306 psi
;
;
Problem 1.2-4 A circular aluminum tube of length
L 400 mm is loaded in compression by forces P (see
figure). The outside and inside diameters are 60 mm and
50 mm, respectively. A strain gage is placed on the outside of
the bar to measure normal strains in the longitudinal direction.
(a) If the measured strain in 550 106, what is the
shortening of the bar?
(b) If the compressive stress in the bar is intended to be
40 MPa, what should be the load P?
Solution 1.2-4 Aluminum tube in compression
550
106
(b) COMPRESSIVE LOAD P
40 MPa
L 400 mm
d2 60 mm
A
d1 50 mm
(a) SHORTENING OF THE BAR
L (550
0.220 mm
106)(400 mm)
;
p 2
p
[d d21] [160 mm22 150 mm22]
4 2
4
P A (40 MPa)(863.9 mm2)
34.6 kN
;
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SECTION 1.2 Normal Stress and Strain
y
Problem 1.2-5
The cross section of a concrete corner column that
is loaded uniformly in compression is shown in the figure.
(a) Determine the average compressive stress c in the concrete if the
load is equal to 3200 k.
(b) Determine the coordinates xc and yc of the point where the resultant
load must act in order to produce uniform normal stress in the
column.
24 in.
16 in.
x
8 in.
Solution 1.2-5
P 3200 kips
1
A (24 + 20)(20 + 16 + 8) a 82 b 202
2
(a) sc P
A
103 in2
sc 2.13 ksi
;
24 8
b
2
1
2
+ (20)(16 + 8)(24 + 10) + 1822a 8b d
2
3
c(24)(20 + 16)(12) + (24 8)(8)a8 +
(b) xc A
xc 19.22 inches
;
20 + 16
b + (20)(16 + 8)
2
16 + 8
1
2
a
b + (24 8)(8)(4) + (82)a 8b d
2
2
3
c(24)(20 + 16)a8 +
yc yc 19.22 inches
20 in.
20 in.
8 in.
A 1.504
5
A
;
NOTE: xc & yc are the same as expected due to symmetry about a diagonal
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.2-6 A car weighing 130 kN when fully loaded is
pulled slowly up a steep inclined track by a steel cable (see figure).
The cable has an effective cross-sectional area of 490 mm2, and the
angle a of the incline is 30°.
Calculate the tensile stress st in the cable.
Solution 1.2-6 Car on inclined track
FREE-BODY DIAGRAM OF CAR
TENSILE STRESS IN THE CABLE
W Weight of car
T Tensile force
in cable
Angle of
incline
A Effective area
of cable
R1, R2 Wheel reactions (no friction force between
wheels and rails)
st Wsin a
T
A
A
SUBSTITUTE NUMERICAL VALUES:
W 130 kN
30°
A 490 mm
2
st (130 kN)(sin 30°)
490 mm2
133 MPa
;
EQUILIBRIUM IN THE INCLINED DIRECTION
©FT 0 Q + b T W sin a 0
T W sin
Problem 1.2-7 Two steel wires support a moveable overhead
camera weighing W 25 lb (see figure) used for close-up
viewing of field action at sporting events. At some instant, wire 1
is at on angle 20° to the horizontal and wire 2 is at an angle
48°. Both wires have a diameter of 30 mils. (Wire diameters
are often expressed in mils; one mil equals 0.001 in.)
Determine the tensile stresses 1 and 2 in the two wires.
T2
T1
b
a
W
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Page 7
SECTION 1.2 Normal Stress and Strain
Solution 1.2-7
NUMERICAL DATA
W 25 lb
a 20
d 30
p
180
b 48
T1 T2
103 in.
p
radians
180
T1cos( ) T2cos( )
T1 T2
a Fv 0
T2 a
cos(b)
cos (a)
T1sin( ) T2sin( ) W
cos (b)
sin(a) + sin (b) b W
cos(a)
TENSION IN WIRES
T2 T1 18.042 lb
TENSILE STRESSES IN WIRES
A wire EQUILIBRIUM EQUATIONS
a Fh 0
cos(b)
cos (a)
W
cos(b)
a
sin (a) + sin (b)b
cos(a)
T2 25.337 lb
Problem 1.2-8 A long retaining wall is braced by
wood shores set at an angle of 30° and supported by concrete
thrust blocks, as shown in the first part of the figure. The shores
are evenly spaced, 3 m apart.
For analysis purposes, the wall and shores are idealized as
shown in the second part of the figure. Note that the base of
the wall and both ends of the shores are assumed to be pinned.
The pressure of the soil against the wall is assumed to be
triangularly distributed, and the resultant force acting on a
3-meter length of the wall is F 190 kN.
If each shore has a 150 mm 150 mm square cross section,
what is the compressive stress c in the shores?
p 2
d
4
s1 T1
A wire
s1 25.5 ksi
;
s2 T2
A wire
s2 35.8 ksi
;
7
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.2-8 Retaining wall braced by wood shores
F 190 kN
A area of one shore
A (150 mm)(150 mm)
22,500 mm2
0.0225 m2
SUMMATION OF MOMENTS ABOUT POINT A
FREE-BODY DIAGRAM OF WALL AND SHORE
MA 0 哵哴
F(1.5 m) CV (4.0 m) CH (0.5 m) 0
or
(190 kN)(1.5 m) C(sin 30°)(4.0 m)
C(cos 30°)(0.5 m) 0
⬖ C 117.14 kN
COMPRESSIVE STRESS IN THE SHORES
C compressive force in wood shore
CH horizontal component of C
sc 117.14 kN
C
A
0.0225 m2
5.21 MPa
CV vertical component of C
;
CH C cos 30°
CV C sin 30°
Problem 1.2-9 A pickup truck tailgate supports a crate
(WC 150 lb), as shown in the figure. The tailgate weighs
WT 60 lb and is supported by two cables (only one is
shown in the figure). Each cable has an effective crosssectional area Ae 0.017 in2.
(a) Find the tensile force T and normal stress in each
cable.
(b) If each cable elongates 0.01 in. due to the weight
of both the crate and the tailgate, what is the average
strain in the cable?
WC = 150 lb
H = 12 in.
dc = 18 in.
Ca
ble
Crate
Truck
Tail gate
dT = 14 in.
L = 16 in.
WT = 60 lb
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Page 9
SECTION 1.2 Normal Stress and Strain
Solution 1.2-9
(a) T 2 Tv2 + T h2 T 184.4 lb
Wc 150 lb
Ae 0.017 in2
scable WT 60
(b) âcable 0.01
T
Ae
d
Lc
scable 10.8 ksi
cable 5
104
;
;
;
dc 18
dT 14
H 12
L 16
L c 2 L2 + H2
a Mhinge 0
Lc 20
2TvL Wcdc WT dT
Tv W cd c + W Td T
2L
Th L
T
H v
Tv 110.625 lb
T h 147.5
Problem 1.2-10 Solve the preceding problem if the
mass of the tail gate is MT 27 kg and that of the
crate is MC 68 kg. Use dimensions H 305 mm,
L 406 mm, dC 460 mm, and dT 350 mm. The
cable cross-sectional area is Ae 11.0 mm2.
(a) Find the tensile force T and normal stress in
each cable.
(b) If each cable elongates 0.25 mm due to the
weight of both the crate and the tailgate, what
is the average strain in the cable?
MC = 68 kg
dc = 460 mm
H = 305 mm
Ca
ble
Crate
Truck
Tail gate
dT = 350 mm
L = 406 mm
MT = 27 kg
9
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.2-10
(a) T 2 T2v + T2h
Mc 68
g 9.81
MT 27 kg
Wc Mcg
scable (b) âcable ;
T
Ae
scable 74.5 MPa
d
Lc
cable 4.92
;
104
;
WT 264.87
m
N kg
s2
0.25
Ae 11.0 mm2
dc 460
dT 350
H 305
L 406
L c 2 L2 + H2
a Mhinge 0
Lc 507.8 mm
2TvL Wcdc WT dT
Wc d c + WT d T
2L
Th s2
WT MTg
Wc 667.08
Tv m
T 819 N
L
T
H v
Tv 492.071 N
Th 655.019 N
Problem ★1.2-11 An L-shaped reinforced concrete slab
12 ft 12 ft (but with a 6 ft 6 ft cutout) and thickness
t 9.0 in. is lifted by three cables attached at O, B and D,
as shown in the figure. The cables are combined at point Q,
which is 7.0 ft above the top of the slab and directly above
the center of mass at C. Each cable has an effective crosssectional area of Ae 0.12 in2.
(a) Find the tensile force Ti (i 1, 2, 3) in each cable
due to the weight W of the concrete slab (ignore
weight of cables).
(b) Find the average stress i in each cable. (See
Table H-1 in Appendix H for the weight density
of reinforced concrete.)
F
Coordinates of D in ft
Q
(5, 5, 7)
T3
1
T1
7
D (5, 12, 0)
1
T2
5
5
z
O
(0, 0, 0)
y
x
6 ft
C (5, 5, 0) 5 7
7
6 ft
W
6 ft
B (12, 0, 0)
lb
Concrete slab g = 150 —3
ft
Thickness t, c.g at (5 ft, 5 ft, 0)
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SECTION 1.2 Normal Stress and Strain
11
Solution 1.2-11
CABLE LENGTHS
52 52 72 99
2
L1 299
L2 11.091
L2 25 + 7 + 7
2
2
5 7 7 123
2
2
L2 2123
2
L3 9.899
L3 272 + 72
7 7 98
2
T1
0.484
T2 0.385
P T Q P 0.589 Q
3
L1 9.95
L1 252 + 52 + 72
(a) SOLUTION FOR CABLE FORCES USING STATICS
(3 EQU, 3 UNKNOWNS)
T1 7 299
144
T1 0.484
5 2123
T2 144
d1 5 22
T3 12
T1L1
EA
T2L2
d2 EA
T2 0.385
T3L3
d3 EA
T3 0.589
a Tverti 0
T1
7
299
+ T2
7
2123
+ T3
1
22
1 CHECK
5
7
299
5
2123
5
299
7
2123
7
22
1
299
2123
22
q
7
0
1
7
2123
+ T3
1
22
1
1. sum about x-axis to get T3v, then T3
2. sum about y-axis to get T2v, then T2
3. sum vertical forces to get T1v, then T1 OR sum forces
in x-dir to get T1x in terms of T2x
SLAB WEIGHT & C.G.
W 150 1122 622
xcg 0
0
P 1Q
9
12
W 1.215
104
2 A 3 + A (6 + 3)
3A
xcg 5
same for ycg
ycg xcg
Multiply unit forces by W
-1
r
+ T2
299
T Tu W
For unit force in Z-direction
T1
T2 PT Q
3
T1
NOTE: preferred solution uses sum of moments about a
line as follows –
L3 722
2
check:
T1
Tu T2
PT Q
3
(b) s T
0.12
5877
T 4679 lb
P 7159 Q
s
;
49.0 ksi
39.0 ksi psi
P 60.0 ksi Q
;
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CHAPTER 1 Tension, Compression, and Shear
Problem *1.2-12
A round bar ACB of length 2L (see figure) rotates
about an axis through the midpoint C with constant angular speed v
(radians per second). The material of the bar has weight density g.
(a) Derive a formula for the tensile stress sx in the bar as a function
of the distance x from the midpoint C.
(b) What is the maximum tensile stress smax?
Solution 1.2-12
Rotating Bar
Consider an element of mass dM at distance from the
midpoint C. The variable ranges from x to L.
g
dM g A dj
angular speed (rad/s)
A cross-sectional area
weight density
g
mass density
g
We wish to find the axial force Fx in the bar at
Section D, distance x from the midpoint C.
The force Fx equals the inertia force of the part
of the rotating bar from D to B.
dF Inertia force (centrifugal force) of element of mass dM
g
dF (dM)(jv2) g Av2jdj
L
g 2
gAv2 2 2
Av jdj (L x )
2g
LD
Lx g
(a) TENSILE STRESS IN BAR AT DISTANCE x
Fx B
sx dF gv2 2
Fx
(L x2)
A
2g
(b) MAXIMUM TENSILE STRESS
x 0 smax Problem 1.2-13 Two gondolas on a ski lift are locked in
the position shown in the figure while repairs are being made
elsewhere. The distance between support towers is L 100 ft.
The length of each cable segment under gondola weights
WB 450 lb and WC 650 lb are DAB 12 ft, DBC 70 ft,
and DCD 20 ft. The cable sag at B is B 3.9 ft and that at
C(C) is 7.1 ft. The effective cross-sectional area of the cable
is Ae 0.12 in2.
;
gv2L2
2g
;
A
D
u1
DB
B
u2
DC
u3
C
WB
WC
(a) Find the tension force in each cable segment; neglect
the mass of the cable.
(b) Find the average stress (s ) in each cable segment.
L = 100 ft
Support
tower
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Page 13
13
SECTION 1.2 Normal Stress and Strain
Solution 1.2-13
WB 450
A
Wc 650 lb
D
u1
DB
u2
B
¢ B 3.9 ft
DC
u3
C
¢ C 7.1 ft
L 100 ft
WB
WC
DAB 12 ft
Support
tower
DBC 70 ft
DCD 20 ft
L = 100 ft
DAB DBC DCD 102 ft
Ae 0.12 in2
CONTRAINT EQUATIONS
COMPUTE INITIAL VALUES OF THETA ANGLES (RADIANS)
DAB cos(u1) + DBC cos (u2) + DCD cos(u3) L
u1 asin a
¢B
b
DAB
u1 0.331
u2 asin a
¢C ¢B
b
DBC
u2 0.046
u3 asin a
¢C
b
DCD
u3 0.363
DAB sin(u1) + DBC sin (u2) DCD sin(u3)
SOLVE SIMULTANEOUS EQUATIONS NUMERICALLY FOR TENSION
FORCE IN EACH CABLE SEGMENT
(a) STATICS AT B & C
TAB cos(u1) + TBC cos (u2) 0
TAB sin(u1) TBC sin(u2) WB
TBC cos(u2) + TCD cos (u3) 0
TBC sin(u2) TCD sin (u3) WC
TAB 1620 lb
TCB 1536 lb
TCD 1640 lb
;
CHECK EQUILIBRIUM AT B & C
TAB sin(u1) TBC sin (u2) 450
TBC sin(u2) TCD sin (u3) 650
(b) COMPUTE STRESSES IN CABLE SEGMENTS
sAB TAB
Ae
sBC TBC
Ae
sAB 13.5 ksi
sBC 12.8 ksi
sCD 13.67 ksi
;
sCD TCD
Ae
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Page 14
CHAPTER 1 Tension, Compression, and Shear
z
Problem 1.2-14 A crane boom of mass 450 kg with its
center of mass at C is stabilized by two cables AQ and BQ
(Ae 304 mm2 for each cable) as shown in the figure. A
load P 20 kN is supported at point D. The crane boom lies
in the y–z plane.
P
Q
y
C
om
(a) Find the tension forces in each cable: TAQ and
TBQ (kN); neglect the mass of the cables, but include
the mass of the boom in addition to load P.
(b) Find the average stress () in each cable.
D
bo
14
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2
1
B
2
Cr
an
e
01Ch01.qxd
2m
2m
55°
TBQ
O
5m
TAQ
5m
x
5m
A
3m
Solution 1.2-14
Data
Mboom 450 kg
g 9.81
m
s2
2TAQZ(3000) Wboom(5000) + P(9000)
Wboom Mboom g
TAQZ Wboom 4415 N
P 20 kN
Ae 304 mm2
TAQ A
22 + 22 + 12 T
AQz
2
TAQ 50.5 kN TBQ
(a) symmetry: TAQ = TBQ
(b) s a Mx 0
Wboom(5000) + P(9000)
2(3000)
TAQ
Ae
;
s 166.2 MPa
;
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SECTION 1.3 Mechanical Properties of Materials
Mechanical Properties of Materials
Problem 1.3-1 Imagine that a long steel wire hangs vertically from a high-altitude balloon.
(a) What is the greatest length (feet) it can have without yielding if the steel yields at 40 ksi?
(b) If the same wire hangs from a ship at sea, what is the greatest length? (Obtain the weight densities of steel and sea
water from Table H-1, Appendix H.)
Solution 1.3-1
Hanging wire of length L
W total weight of steel wire
S weight density of steel
490 lb/ft3
w weight density of sea water
(b) WIRE HANGING IN SEA WATER
F tensile force at top of wire
F (gS gW)AL smax 63.8 lb/ft3
A cross-sectional area of wire
max 40 ksi (yield strength)
(a) WIRE HANGING IN AIR
W SAL
smax
gS gW
40,000 psi
(490 63.8) lb/ft3
13,500 ft
smax W
gSL
A
Lmax smax 40,000 psi
(144 in.2/ft2)
gS
490 lb/ft3
11,800 ft
Lmax F
(gS gW)L
A
(144 in.2/ft2)
;
;
Problem 1.3-2 Imagine that a long wire of tungsten hangs vertically from a high-altitude balloon.
(a) What is the greatest length (meters) it can have without breaking if the ultimate strength (or breaking strength)
is 1500 MPa?
(b) If the same wire hangs from a ship at sea, what is
the greatest length? (Obtain the weight densities of tungsten and sea water from Table H-1, Appendix H.)
15
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.3-2
Hanging wire of length L
(b) WIRE HANGING IN SEA WATER
W total weight of tungsten wire
F tensile force at top of wire
T weight density of tungsten
F (T W)AL
190 kN/m3
W weight density of sea water
10.0 kN/m3
A cross-sectional area of wire
max 1500 MPa (breaking strength)
smax F
(gT gW)L
A
Lmax smax
gT gW
(a) WIRE HANGING IN AIR
W TAL
1500MPa
(190 10.0) kN/m3
8300 m
;
W
smax gTL
A
Lmax smax
1500MPa
gT
190 kN/m3
7900 m
;
Problem 1.3-3 Three different materials, designated A, B, and C,
are tested in tension using test specimens having diameters of
0.505 in. and gage lengths of 2.0 in. (see figure). At failure, the
distances between the gage marks are found to be 2.13, 2.48, and
2.78 in., respectively. Also, at the failure cross sections the diameters are found to be 0.484, 0.398, and 0.253 in., respectively.
Determine the percent elongation and percent reduction in
area of each specimen, and then, using your own judgment,
classify each material as brittle or ductile.
P
Gage
length
P
Solution 1.3-3 Tensile tests of three materials
where L1 is in inches.
Percent reduction in area
Percent elongation L1
L1 L0
(100) a 1b 100
L0
L0
L0 2.0 in.
Percent elongation a
L1
1b (100)
2.0
(Eq. 1)
d0 initial diameter
A0 A1
(100)
A0
A1
a1 b (100)
A0
d1 final diameter
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17
SECTION 1.3 Mechanical Properties of Materials
A1
d1 2
a b d0 0.505 in.
A0
d0
Percent reduction in area
c1 a
d1 2
b d(100)
0.505
(Eq. 2)
Material
L1
(in.)
d1
(in.)
% Elongation
(Eq. 1)
% Reduction
(Eq. 2)
Brittle or
Ductile?
A
2.13
0.484
6.5%
8.1%
Brittle
B
2.48
0.398
24.0%
37.9%
Ductile
C
2.78
0.253
39.0%
74.9%
Ductile
where d1 is in inches.
Problem 1.3-4 The strength-to-weight ratio of a structural material is defined as its load-carrying capacity divided by its weight.
For materials in tension, we may use a characteristic tensile stress (as obtained from a stress-strain curve) as a measure of
strength. For instance, either the yield stress or the ultimate stress could be used, depending upon the particular application. Thus,
the strength-to-weight ratio RS/W for a material in tension is defined as
RS/W s
g
in which is the characteristic stress and is the weight density. Note that the ratio has units of length.
Using the ultimate stress U as the strength parameter, calculate the strength-to-weight ratio (in units of meters) for
each of the following materials: aluminum alloy 6061-T6, Douglas fir (in bending), nylon, structural steel ASTM-A572, and
a titanium alloy. (Obtain the material properties from Tables H-1 and H-3 of Appendix H. When a range of values is given in
a table, use the average value.)
Solution 1.3-4 Strength-to-weight ratio
The ultimate stress U for each material is obtained
from Table H-3, Appendix H, and the weight density
is obtained from Table H-1.
The strength-to-weight ratio (meters) is
RS/W sU ( MPa)
g( kN/m3)
(103)
Values of U, , and RS/W are listed in the table.
U
(MPa)
(kN/m3)
RS/W
(m)
310
26.0
11.9
103
Douglas fir
65
5.1
12.7
103
Nylon
60
9.8
6.1
103
Structural steel
ASTM-A572
500
77.0
6.5
103
Titanium alloy
1050
44.0
23.9
103
Aluminum alloy
6061-T6
Titanium has a high strength-to-weight ratio, which is why
it is used in space vehicles and high-performance airplanes.
Aluminum is higher than steel, which makes it desirable for
commercial aircraft. Some woods are also higher than steel,
and nylon is about the same as steel.
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.3-5 A symmetrical framework consisting of three pin-connected
bars is loaded by a force P (see figure). The angle between the inclined bars
and the horizontal is 48°. The axial strain in the middle bar is measured
as 0.0713.
Determine the tensile stress in the outer bars if they are constructed of
aluminum alloy having the stress-strain diagram shown in Fig. 1-13.
(Express the stress in USCS units.)
A
B
C
a
D
P
Solution 1.3-5 Symmetrical framework
L length of bar BD
L1 distance BC
L cot
L(cot 48°) 0.9004 L
L2 length of bar CD
L csc
L(csc 48°) 1.3456 L
Elongation of bar BD distance DE BDL
BDL 0.0713 L
Aluminum alloy
48°
L3 distance CE
L3 2L21 (L âBDL)2
BD 0.0713
2(0.9004L)2 + L2(1 + 0.0713)2
Use stress-strain diagram of Figure 1-13
1.3994 L
elongation of bar CD
L3 L2 0.0538L
Strain in bar CD
0.0538L
d
0.0400
L2
1.3456L
From the stress-strain diagram of Figure 1-13:
⬇ 31 ksi
;
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Page 19
SECTION 1.3 Mechanical Properties of Materials
Problem 1.3-6 A specimen of a methacrylate plastic is tested in tension at room temperature (see figure), producing the stress-strain data
listed in the accompanying table.
Plot the stress-strain curve and determine the proportional limit,
modulus of elasticity (i.e., the slope of the initial part of the stress-strain
curve), and yield stress at 0.2% offset. Is the material ductile or brittle?
19
STRESS-STRAIN DATA FOR PROBLEM 1.3-6
P
P
Stress (MPa)
Strain
8.0
17.5
25.6
31.1
39.8
44.0
48.2
53.9
58.1
62.0
62.1
0.0032
0.0073
0.0111
0.0129
0.0163
0.0184
0.0209
0.0260
0.0331
0.0429
Fracture
Solution 1.3-6 Tensile test of a plastic
Using the stress-strain data given in the problem
statement, plot the stress-strain curve:
PL proportional limit
PL ⬇ 47 MPa
Modulus of elasticity (slope) ⬇ 2.4 GPa
;
;
Y yield stress at 0.2% offset
Y ⬇ 53 MPa
;
Material is brittle, because the strain after the proportional
;
limit is exceeded is relatively small.
Problem 1.3-7 The data shown in the accompanying table were
obtained from a tensile test of high-strength steel. The test specimen
had a diameter of 0.505 in. and a gage length of 2.00 in. (see figure for
Prob. 1.3-3). At fracture, the elongation between the gage marks was 0.12 in.
and the minimum diameter was 0.42 in.
Plot the conventional stress-strain curve for the steel and determine the
proportional limit, modulus of elasticity (i.e., the slope of the initial part of
the stress-strain curve), yield stress at 0.1% offset, ultimate stress, percent
elongation in 2.00 in., and percent reduction in area.
TENSILE-TEST DATA FOR PROBLEM 1.3-7
Load (lb)
1,000
2,000
6,000
10,000
12,000
12,900
13,400
13,600
13,800
14,000
14,400
15,200
16,800
18,400
20,000
22,400
22,600
Elongation (in.)
0.0002
0.0006
0.0019
0.0033
0.0039
0.0043
0.0047
0.0054
0.0063
0.0090
0.0102
0.0130
0.0230
0.0336
0.0507
0.1108
Fracture
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.3-7 Tensile test of high-strength steel
d0 0.505 in.
A0 pd20
4
L0 2.00 in.
ENLARGEMENT OF PART OF THE STRESS-STRAIN CURVE
0.200 in.2
CONVENTIONAL STRESS AND STRAIN
s
P
A0
Load P
(lb)
1,000
2,000
6,000
10,000
12,000
12,900
13,400
13,600
13,800
14,000
14,400
15,200
16,800
18,400
20,000
22,400
22,600
â
d
L0
Elongation (in.)
0.0002
0.0006
0.0019
0.0033
0.0039
0.0043
0.0047
0.0054
0.0063
0.0090
0.0102
0.0130
0.0230
0.0336
0.0507
0.1108
Fracture
STRESS-STRAIN DIAGRAM
Stress (psi)
5,000
10,000
30,000
50,000
60,000
64,500
67,000
68,000
69,000
70,000
72,000
76,000
84,000
92,000
100,000
112,000
113,000
Strain 0.00010
0.00030
0.00100
0.00165
0.00195
0.00215
0.00235
0.00270
0.00315
0.00450
0.00510
0.00650
0.01150
0.01680
0.02535
0.05540
RESULTS
Proportional limit ⬇ 65,000 psi
;
Modulus of elasticity (slope) ⬇ 30
106 psi
Yield stress at 0.1% offset ⬇ 69,000 psi
Ultimate stress (maximum stress)
⬇ 113,000 psi
;
Percent elongation in 2.00 in.
L1 L0
(100)
L0
0.12 in.
(100) 6%
2.00 in.
;
Percent reduction in area
A0 A1
(100)
A0
0.200 in.2 31%
p
4 (0.42
0.200 in.2
;
in.) 2
(100)
;
;
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Page 21
SECTION 1.4 Elasticity, Plasticity, and Creep
21
Elasticity, Plasticity, and Creep
Problem 1.4-1 A bar made of structural steel having the stressstrain diagram shown in the figure has a length of 48 in. The yield
stress of the steel is 42 ksi and the slope of the initial linear part of the
stress-strain curve (modulus of elasticity) is 30 103 ksi. The bar is
loaded axially until it elongates 0.20 in., and then the load is removed.
How does the final length of the bar compare with its original
length? (Hint: Use the concepts illustrated in Fig. 1-18b.)
Solution 1.4-1 Steel bar in tension
ELASTIC RECOVERY E
âE sB
42 ksi
0.00140
Slope
30 * 103 ksi
RESIDUAL STRAIN R
R B E 0.00417 0.00140
0.00277
PERMANENT SET
L 48 in.
Yield stress Y 42 ksi
Slope 30
103 ksi
0.20 in.
RL (0.00277)(48 in.)
0.13 in.
Final length of bar is 0.13 in. greater than its original
;
length.
STRESS AND STRAIN AT POINT B
sB sY 42 ksi
âB 0.20 in.
d
0.00417
L
48 in.
Problem 1.4-2 A bar of length 2.0 m is made of a structural steel
having the stress-strain diagram shown in the figure. The yield stress
of the steel is 250 MPa and the slope of the initial linear part of the
stress-strain curve (modulus of elasticity) is 200 GPa. The bar is
loaded axially until it elongates 6.5 mm, and then the load is
removed.
How does the final length of the bar compare with its original
length? (Hint: Use the concepts illustrated in Fig. 1-18b.)
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.4-2 Steel bar in tension
L 2.0 m 2000 mm
Yield stress Y 250 MPa
Slope 200 GPa
6.5 mm
ELASTIC RECOVERY E
âE sB
250 MPa
0.00125
Slope
200 GPa
RESIDUAL STRAIN R
R B E 0.00325 0.00125
0.00200
Permanent set RL (0.00200)(2000 mm)
4.0 mm
STRESS AND STRAIN AT POINT B
sB sY 250 MPa
âB 6.5 mm
d
0.00325
L
2000 mm
Final length of bar is 4.0 mm greater than its original
;
length.
Problem 1.4-3 An aluminum bar has length L 5 ft and diameter d 1.25 in. The stress-strain curve for the aluminum is
shown in Fig. 1-13 of Section 1.3. The initial straight-line part of the curve has a slope (modulus of elasticity) of 10
The bar is loaded by tensile forces P 39 k and then unloaded.
106 psi.
(a) What is the permanent set of the bar?
(b) If the bar is reloaded, what is the proportional limit? (Hint: Use the concepts illustrated in Figs. 1-18b and 1-19.)
Solution 1.4-3
RESIDUAL STRAIN
(a) PERMAMENT SET
Numerical data L 60 in
âE âB âE
d 1.25 in
PERMANENT SET
P 39 kips
STRESS AND STRAIN AT PT B
sB P
p 2
d
4
s B 31.8 ksi
From Figure 1-13 B 0.05
ELASTIC RECOVERY
sB
10(10)3
;
(b) PROPORTIONAL LIMIT WHEN RELOADED
B 31.78 ksi
âE âRL 2.81 in.
âR 0.047
âE 3.178 * 103
;
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Page 23
SECTION 1.4 Elasticity, Plasticity, and Creep
Problem 1.4-4 A circular bar of magnesium alloy is 750 mm long. The stressstrain diagram for the material is shown in the figure. The bar is loaded in
tension to an elongation of 6.0 mm, and then the load is removed.
(a) What is the permanent set of the bar?
(b) If the bar is reloaded, what is the proportional limit? (Hint: Use the
concepts illustrated in Figs. 1-18b and 1-19.)
23
200
s (MPa)
100
0
0
0.005
e
0.010
Solution 1.4-4
NUMERICAL DATA
L 750 mm
6 mm
STRESS AND STRAIN AT PT B
d
B 180 MPa
B 8 103
âB L
ELASTIC RECOVERY
178
slope slope 4.45 104
0.004
sB
âE slope
E 4.045 103
RESIDUAL STRAIN
R B E
R 3.955 103
(b) PROPORTIONAL LIMIT WHEN RELOADED
sB 180 MPa
;
(a) PERMANENT SET
âRL 2.97 mm
;
Problem 1.4-5 A wire of length L 4 ft and diameter d 0.125 in. is stretched by tensile forces P 600 lb. The wire is
made of a copper alloy having a stress-strain relationship that may be described mathematically by the following equation:
s
18,000P
1 + 300P
0 … P … 0.03 (s ksi)
in which P is nondimensional and has units of kips per square inch (ksi).
(a)
(b)
(c)
(d)
Construct a stress-strain diagram for the material.
Determine the elongation of the wire due to the forces P.
If the forces are removed, what is the permanent set of the bar?
If the forces are applied again, what is the proportional limit?
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.4-5 Wire stretched by forces P
L 4 ft 48 in.
d 0.125 in.
ALTERNATIVE FORM OF THE STRESS-STRAIN RELATIONSHIP
P 600 lb
Solve Eq. (1) for in terms of :
COPPER ALLOY
â
18,000â
s
1 + 300â
0 … â … 0.03 (s ksi) (Eq. 1)
(a) STRESS-STRAIN DIAGRAM (From Eq. 1)
s
0 … s … 54 ksi (s ksi)
18,000300s
(Eq. 2)
This equation may also be used when plotting the stressstrain diagram.
(b) ELONGATION OF THE WIRE
P
600 lb
48,900 psi 48.9 ksi
p
A
(0.125 in.)2
4
From Eq. (2) or from the stress-strain diagram:
s
0.0147
L (0.0147)(48 in.) 0.71 in.
STRESS AND STRAIN AT POINT B (see diagram)
B 48.9 ksi B 0.0147
ELASTIC RECOVERY E
âE sB
48.9 ksi
0.00272
Slope
18,000 ksi
INITIAL SLOPE OF STRESS-STRAIN CURVE
RESIDUAL STRAIN R
Take the derivative of with respect to :
R B E 0.0147 0.0027 0.0120
(1 + 300â)(18,000) (18,000)(300)s
ds
dâ
(1 + 300â)2
At â 0,
18,000
(1 + 300â)2
ds
18,00 ksi
dâ
⬖ Initial slope 18,000 ksi
(c) Permanent set RL (0.0120)(48 in.)
0.58 in.
;
(d) Proportional limit when reloaded B
B 49 ksi
;
;
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Page 25
SECTION 1.5 Linear Elasticity, Hooke’s Law, and Poisson’s Ratio
25
Linear Elasticity, Hooke’s Law, and Poisson’s Ratio
When solving the problems for Section 1.5, assume that the material behaves
linearly elastically.
Problem 1.5-1 A high-strength steel bar used in a large crane has diameter
d 2.00 in. (see figure). The steel has modulus of elasticity E 29 106 psi
and Poisson’s ratio v 0.29. Because of clearance requirements, the diameter
of the bar is limited to 2.001 in. when it is compressed by axial forces.
What is the largest compressive load Pmax that is permitted?
Solution 1.5-1 Steel bar in compression
STEEL BAR
d 2.00 in.
E 29
AXIAL STRESS
Max. d 0.001 in.
6
10 psi
v 0.29
Assume that the yield stress for the high-strength steel is
greater than 50 ksi. Therefore, Hooke’s law is valid.
0.001 in.
¢d
0.0005
d
2.00 in.
MAXIMUM COMPRESSIVE LOAD
AXIAL STRAIN
â 106 psi)(0.001724)
50.00 ksi (compression)
LATERAL STRAIN
â¿ E (29
0.0005
â¿
0.001724
v
0.29
p
Pmax sA (50.00 ksi)a b(2.00 in.)2
4
157 k
;
(shortening)
Problem 1.5-2 A round bar of 10 mm diameter is made of aluminum
alloy 7075-T6 (see figure). When the bar is stretched by axial forces P, its
diameter decreases by 0.016 mm.
Find the magnitude of the load P. (Obtain the material properties
from Appendix H.)
Solution 1.5-2
d 10 mm
Aluminum bar in tension
d 0.016 mm
AXIAL STRESS
E (72 GPa)(0.004848)
(Decrease in diameter)
349.1 MPa (Tension)
7075-T6
From Table H-2: E 72 GPa
v 0.33
From Table H-3: Yield stress Y 480 MPa
LATERAL STRAIN
â¿ 0.016mm
¢d
0.0016
d
10mm
AXIAL STRAIN
â¿
0.0016
v
0.33
0.004848 (Elongation)
â Because Y, Hooke’s law is valid.
LOAD P (TENSILE FORCE)
p
P sA (349.1 MPa)a b(10 mm)2
4
27.4 kN
;
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.5-3 A polyethylene bar having diameter d1 4.0 in. is placed inside a
Steel
tube
steel tube having inner diameter d2 4.01 in. (see figure). The polyethylene bar is
then compressed by an axial force P.
At what value of the force P will the space between the nylon bar and the steel
tube be closed? (For nylon, assume E 400 ksi and v 0.4.)
d1 d2
Polyethylene
bar
Solution 1.5-3
NORMAL STRAIN
NUMERICAL DATA
d1 4 in
d2 4.01 in.
v 0.4
p
A1 d12
4
d1 0.01 in
p
A2 d22
4
E 200 ksi
â1 A1 12.566 in2
103
AXIAL STRESS
1 1.25 ksi
COMPRESSION FORCE
LATERAL STRAIN
0.01
âp 4
1 6.25
v
1 E 1
A2 12.629 in2
¢d1
âp d1
â p
P EA11
p 2.5
3
10
P 15.71 kips
Problem 1.5-4 A prismatic bar with a circular cross section is loaded
by tensile forces P 65 kN (see figure). The bar has length L 1.75
m and diameter d 32 mm. It is made of aluminum alloy with modulus of elasticity E 75 GPa and Poisson’s ratio 1/3.
Find the increase in length of the bar and the percent decrease in
its cross-sectional area.
;
d
P
L
P
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SECTION 1.5 Linear Elasticity, Hooke’s Law, and Poisson’s Ratio
27
Solution 1.5-4
NUMERICAL DATA
P 65 kN
v
LATERAL STRAIN
p 1
3
DECREASE IN DIAMETER
d 32 mm L 1.75(1000) mm
d ƒpdƒ
E 75 GPa
p 2
d
4
Af Ai 804.248 mm2
p
1d ¢d22
4
Af 803.67 mm2
% decrease in x-sec area AXIAL STRAIN
â
d 0.011 mm
FINAL AREA OF CROSS SECTION
INITIAL AREA OF CROSS SECTION
Ai 104
p 3.592
P
EA i
1.078
103
Af Ai
(100)
Ai
0.072
;
;
INCREASE IN LENGTH
L L
¢ L 1.886 mm
;
Problem 1.5-5 A bar of monel metal as in the figure (length L 9 in., diameter d 0.225 in.) is loaded axially by a tensile
force P. If the bar elongates by 0.0195 in., what is the decrease in diameter d? What is the magnitude of the load P? Use the
data in Table H-2, Appendix H.
Solution 1.5-5
NUMERICAL DATA
DECREASE IN DIAMETER
E 25000 ksi
d pd
0.32
¢d 1.56 * 104 in.
L 9 in.
INITIAL CROSS SECTIONAL AREA
0.0195 in.
Ai d 0.225 in.
2.167
P EAi
3
10
P 2.15 kips
LATERAL STRAIN
p Ai 0.04 in.2
MAGNITUDE OF LOAD P
NORMAL STRAIN
d
â
L
p 2
d
4
p 6.933
104
;
;
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.5-6 A tensile test is peformed on a brass specimen
10 mm in diameter using a gage length of 50 mm (see figure).
When the tensile load P reaches a value of 20 kN, the distance
between the gage marks has increased by 0.122 mm.
(a) What is the modulus of elasticity E of the brass?
(b) If the diameter decreases by 0.00830 mm, what is Poisson’s ratio?
Solution 1.5-6 Brass specimen in tension
d 10 mm Gage length L 50 mm
P 20 kN 0.122 mm
d 0.00830 mm
AXIAL STRESS
P
20 k
s 254.6 MPa
p
A
(10 mm) 2
4
Assume is below the proportional limit so that
Hooke’s law is valid.
(a) MODULUS OF ELASTICITY
E
s
254.6 MPa
104 Gpa
â
0.002440
;
(b) POISSON’S RATIO
v
d d vd
v
¢d
0.00830 mm
0.34
âd
(0.002440)(10 mm)
;
AXIAL STRAIN
â
0.122 mm
d
0.002440
L
50 mm
Problem 1.5-7 A hollow, brass circular pipe ABC (see figure) supports a load
P1 26.5 kips acting at the top. A second load P2 22.0 kips is uniformly
distributed around the cap plate at B. The diameters and thicknesses of the upper
and lower parts of the pipe are dAB 1.25 in., tAB 0.5 in., dBC 2.25 in., and
tAB 0.375 in., respectively. The modulus of elasticity is 14,000 ksi. When both
loads are fully applied, the wall thickness of pipe BC increases by 200 3 106 in.
(a) Find the increase in the inner diameter of pipe segment BC.
(b) Find Poisson’s ratio for the brass.
(c) Find the increase in the wall thickness of pipe segment AB and the increase
in the inner diameter of AB.
P1
A
dAB
tAB
P2
B
Cap plate
dBC
tBC
C
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29
SECTION 1.5 Linear Elasticity, Hooke’s Law, and Poisson’s Ratio
Solution 1.5-7
NUMERICAL DATA
P1 26.5 kips
P2 22 kips
dAB 1.25 in.
tAB 0.5 in.
dBC 2.25 in.
tBC 0.375 in.
E 14000 ksi
tBC 200 106
(c) INCREASE IN THE WALL THICKNESS OF PIPE SEGMENT AB
AND THE INCREASE IN THE INNER DIAMETER OF AB
¢tBC
tBC
p
c dAB2 1 dAB 2tAB22 d
4
âAB P1
EAAB
tAB pABtAB
pBC 5.333
103
AB 1.607
pAB brassAB
pAB 5.464
104
¢tAB 2.73 * 104 in.
;
dABinner pAB(dAB 2tAB)
(a) INCREASE IN THE INNER DIAMETER OF PIPE
SEGMENT BC
âpBC AAB ¢dABinner 1.366 * 104 inches
104
dBCinner pBC(dBC 2tBC)
¢ dBCinner 8 * 104 inches
;
(b) POISSON’S RATIO FOR THE BRASS
ABC p
c d 2 1 dBC 2tBC22 d
4 BC
ABC 2.209 in.2
â BC 1P1 + P22
brass 1EABC2
âpBC
âBC
BC 1.568
103
brass 0.34
(agrees with App. H (Table H-2))
Problem 1.5-8 A brass bar of length 2.25 m with a square
cross section of 90 mm on each side is subjected to an axial
tensile force of 1500 kN (see figure). Assume that E 110
GPa and 0.34.
Determine the increase in volume of the bar.
90 mm
90 mm
1500 kN
1500 kN
2.25 m
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.5-8
CHANGE IN DIMENSIONS
NUMERICAL DATA
E 110 GPa 0.34 P 1500 kN
b 90 mm L 2250 mm
b pb
L L
INITIAL VOLUME
FINAL LENGTH AND WIDTH
Voli Lb2
Voli 1.822
7
Lf L L Lf 2.254
3
10 mm
bf b b
NORMAL STRESS AND STRAIN
P
185 MPa (less than yield so
s 2
b
Hooke’s Law applies)
s
â
1.684 103
E
LATERAL STRAIN
p p 5.724
b 0.052 mm
L 3.788 mm
103 mm
bf 89.948 mm
FINAL VOLUME
Volf Lfbf2 Volf 1.823
107 mm3
INCREASE IN VOLUME
V Volf Vol ¢V 9789 mm3
104
Shear Stress and Strain
Problem 1.6-1 An angle bracket having thickness t 0.75 in. is attached to the flange of a column by two 5/8-inch diameter
bolts (see figure). A uniformly distributed load from a floor joist acts on the top face of the bracket with a pressure
p 275 psi. The top face of the bracket has length L 8 in. and width b 3.0 in.
Determine the average bearing pressure b between the angle bracket and the bolts and the average shear stress aver in the
bolts. (Disregard friction between the bracket and the column.)
Distributed pressure on angle bracket
P
b
Floor slab
L
Floor joist
Angle bracket
Angle bracket
t
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SECTION 1.6 Shear Stress and Strain
31
Solution 1.6-1
NUMERICAL DATA
t 0.75 in.
L 8 in.
b 3. in.
p
275
ksi
1000
Ab dt
d
5
in.
8
Ab 0.469 in.2
BEARING STRESS
sb F
2Ab
s b 7.04 ksi
;
BEARING FORCE
F pbL
F 6.6 kips
SHEAR AND BEARING AREAS
AS p 2
d
4
SHEAR STRESS
tave F
2AS
tave 10.76 ksi
;
AS 0.307 in.2
Roof structure
Problem 1.6-2
Truss members supporting a roof are connected to a 26-mm-thick gusset plate by a 22 mm diameter
pin as shown in the figure and photo. The two end plates on
the truss members are each 14 mm thick.
Truss
member
(a) If the load P 80 kN, what is the largest bearing
stress acting on the pin?
(b) If the ultimate shear stress for the pin is 190 MPa, what
force Pult is required to cause the pin to fail in shear?
P
End
plates
(Disregard friction between the plates.)
P
Pin
t = 14 mm
Gusset
plate
26 mm
Solution 1.6-2
NUMERICAL DATA
(b) ULTIMATE FORCE IN SHEAR
tep 14 mm
Cross sectional area of pin
tgp 26 mm
Ap P 80 kN
dp 22 mm
Pult 2t ultAp
(a) BEARING STRESS ON PIN
P
gusset plate is thinner than
dptgp
(2 tep) so gusset plate controls
b 139.9 MPa
4
Ap 380.133 mm2
ult 190 MPa
sb p d2p
;
Pult 144.4 kN
;
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.6-3 The upper deck of a football stadium is supported by braces each of which transfers a load P 160 kips to
the base of a column [see figure part (a)]. A cap plate at the bottom of the brace distributes the load P to four flange plates
(tf 1 in.) through a pin (dp 2 in.) to two gusset plates (tg 1.5 in.) [see figure parts (b) and (c)].
Determine the following quantities.
(a) The average shear stress aver in the pin.
(b) The average bearing stress between the flange plates and the pin (bf), and also between the gusset
plates and the pin (bg).
(Disregard friction between the plates.)
Cap plate
Flange plate
(tf = 1 in.)
Pin (dp = 2 in.)
Gusset plate
(tg = 1.5 in.)
(b) Detail at bottom of brace
P
P = 160 k
Cap plate
(a) Stadium brace
Pin (dp = 2 in.)
P
Flange plate
(tf = 1 in.)
Gusset plate
(tg = 1.5 in.)
P/2
P/2
(c) Section through bottom of brace
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33
SECTION 1.6 Shear Stress and Strain
Solution 1.6-3
(b) BEARING STRESS ON PIN FROM FLANGE PLATE
NUMERICAL DATA
P 160 kips
dp 2 in.
tg 1.5 in.
P
4
sbf dp tf
tf 1 in.
s bf 20 ksi
;
(a) SHEAR STRESS ON PIN
t
V
a
p d2p
4
t
b
t 12.73 ksi
BEARING STRESS ON PIN FROM GUSSET PLATE
P
4
a
p d2p
4
P
2
sbg dp tg
b
sbg 26.7 ksi
;
;
Problem 1.6-4 The inclined ladder AB supports a house painter (82 kg) at
C and the self weight (q 36 N/m) of the ladder itself. Each ladder rail
(tr 4 mm) is supported by a shoe (ts 5 mm) which is attached to the
m)
ladder rail by a bolt of diameter dp 8 mm.
B
Bx
m)
(a) Find support reactions at A and B.
(b) Find the resultant force in the shoe bolt at A.
= 5 mm)
(c) Find maximum average shear () and bearing (b) stresses in the shoe
bolt at A.
C
H=7m
Typical rung
Shoe bolt at A
36
Ladder rail (tr = 4 mm)
N/
m
tr
q=
01Ch01.qxd
Shoe bolt (dp = 8 mm)
Ladder shoe (ts = 5 mm)
A
ts
Ax
A
—y
2
A
—y
2
a = 1.8 m
b = 0.7 m
Ay
Assume no slip at A
Shoe b
Section at base
Solution 1.6-4
(a) SUPPORT REACTIONS
NUMERICAL DATA
tr 4 mm
dp 8 mm
ts 5 mm
L 2( a + b)2 + H 2
P 82 kg (9.81 m/s )
2
P 804.42 N
a 1.8 m
b 0.7 m
H7m
q 36
N
m
L 7.433 m
LAC a
L
a + b
LAC 5.352 m
LCB b
L
a + b
LCB 2.081 m
LAC LCB 7.433 m
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CHAPTER 1 Tension, Compression, and Shear
SUM MOMENTS ABOUT A
Bx Pa + qL a
a + b
b
2
H
Bx 255 N
;
Ax Bx
A y 1072 N
As SHEAR AREA
SHEAR STRESS
Ay P qL
(b) RESULTANT FORCE IN SHOE BOLT AT A
BEARING STRESS
A resultant 2 A 2x + A 2y
ts 5 mm
s bshoe
t 5.48 MPa
tr 4 mm
;
Ab 80 mm2
A resultant
2
Ab
s bshoe 6.89 MPa
;
(c) MAXIMUM SHEAR AND BEARING STRESSES IN SHOE
BOLT AT A
dp 8 mm
As 50.265 mm2
Aresultant
2
t
2As
BEARING AREA Ab 2dpts
;
Aresultant 1102 N
p 2
d
4 p
;
CHECK BEARING STRESS IN LADDER RAIL
Aresultant
2
s brail dp tr
brail 17.22 MPa
T
Problem 1.6-5 The force in the brake cable of the V-brake
system shown in the figure is T 45 lb. The pivot pin at A has
diameter dp 0.25 in. and length Lp 5/8 in.
Use dimensions show in the figure. Neglect the weight of the
brake system.
(a) Find the average shear stress aver in the pivot pin where it
is anchored to the bicycle frame at B.
(b) Find the average bearing stress b,aver in the pivot pin over
segment AB.
Lower end of front brake cable
D
T
T
3.25 in.
Brake pads
C
HE
HC
1.0 in.
HB
B
HF
A
VF
Pivot pins
anchored to
frame (dP)
VB
LP
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35
SECTION 1.6 Shear Stress and Strain
Solution 1.6-5
NUMERICAL DATA
5
L in.
8
dp 0.25 in.
BC 1 in.
CD 3.25 in.
AS T 45 lb
EQUILIBRIUM - FIND HORIZONTAL FORCES
AT B AND C [VERTICAL REACTION VB 0]
a MB 0
HC HC 191.25 lb
HB T HC
(a) FIND THE AVE SHEAR STRESS ave IN THE PIVOT PIN WHERE
IT IS ANCHORED TO THE BICYCLE FRAME AT B:
T(BC + CD)
BC
a FH 0
HB 146.25 lb
tave pd2p
As 0.049 in.2
4
| H B|
AS
tave 2979 psi
(b) FIND THE AVE BEARING STRESS σb,ave IN THE PIVOT PIN
OVER SEGMENT AB.
Ab dpL
s b,ave Ab 0.156 in.2
| H B|
Ab
s b,ave 936 psi
Problem 1.6-6 A steel plate of dimensions 2.5 1.5 0.08 m
and weighing 23.1kN is hoisted by steel cables with lengths
L1 3.2 m and L2 3.9 m that are each attached to the plate by
a clevis and pin (see figure). The pins through the clevises are
18 mm in diameter and are located 2.0 m apart. The orientation
angles are measured to be 94.4° and 54.9°.
For these conditions, first determine the cable forces T1 and
T2, then find the average shear stress aver in both pin 1 and pin
2, and then the average bearing stress b between the steel plate
and each pin. Ignore the mass of the cables.
;
;
P
L1
Clevis and
pin 1
a = 0.6 m
b 1 b2
u
L2
a
2.0
Clevis
and pin 2
m
Center of mass
of plate
b=
1.0
Steel plate
(2.5 × 1.5 × 0.08 m)
m
Solution 1.6-6
SOLUTION APPROACH
NUMERICAL DATA
L1 3.2 m
u 94.4 a
a 0.6 m
L2 3.9 m
p
b rad.
a 54.9a
180
p
b rad.
180
d 1.166 m
STEP (1) d 2 a2 + b 2
STEP (2) u1 atan a
a
b
b
u1
180
30.964 degrees
p
STEP (3)-Law of cosines
H 2d2 + L12 2dL1cos(u + u1)
b1m
W 77.0(2.5 1.5 0.08)
W 23.1 kN
(77 wt density of steel, kN/m )
3
H 3.99 m
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CHAPTER 1 Tension, Compression, and Shear
STEP (4) b 1 acos a
b1
L22 + H2 d2
b
2L1H
180
13.789 degrees
p
STEP (5) b 2 acos a
L22 + H2 d2
b
2L2H
180
b2
16.95 degrees
p
STEP (6)
Check (b 1 + b 2 + u + a)
180
p
180.039 degrees
Statics
T1sin( 1) T2sin( 2)
T1 T2 a
sin(b 2)
b
sin(b 1)
T1 13.18 kN
T1cos( 1) T2cos( 2) 23.1
;
checks
SHEAR & BEARING STRESSES
dp 18 mm
AS p 2
dp
4
T1
2
t1ave AS
T2
2
t2ave AS
t 100 mm
Ab tdp
t1ave 25.9 MPa
;
t2ave 21.2 MPa
;
sin(b 2)
T1 T2 a
b
sin(b 1)
sb1 T1
Ab
sb1 7.32 MPa
;
T1cos( 1) T2cos( 2) W
s b2 T2
Ab
s b2 5.99 MPa
;
T2 W
sin(b 2)
cos(b 1)
+ cos(b 2)
sin(b 1)
T2 10.77 kN
;
Problem 1.6-7 A special-purpose eye bolt of shank diameter d 0.50 in. passes
y
through a hole in a steel plate of thickness tp 0.75 in. (see figure) and is secured by
a nut with thickness t 0.25 in. The hexagonal nut bears directly against the steel
plate. The radius of the circumscribed circle for the hexagon is r 0.40 in. (which
means that each side of the hexagon has length 0.40 in.). The tensile forces in three
cables attached to the eye bolt are T1 800 lb., T2 550 lb., and T3 1241 lb.
(a) Find the resultant force acting on the eye bolt.
(b) Determine the average bearing stress b between the hexagonal nut on the
eye bolt and the plate.
(c) Determine the average shear stress aver in the nut and also in the steel
plate.
T1
tp
T2
d
30°
2r
x
Cables
Nut
t
30°
Eye bolt
Steel plate
T3
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SECTION 1.6 Shear Stress and Strain
Solution 1.6-7
(c) AVE. SHEAR THROUGH NUT
CABLE FORCES
T1 800 lb
T2 550 lb
T3 1241 lb
(a) RESULTANT
P T2
23
+ T30.5
2
P 1097 lb
;
d 0.5 in.
t 0.25 in.
Asn dt
Asn 0
tnut 2793 psi
Ab 0.2194 in.2
P
sb Ab
Aspl 6rtp
hexagon (Case 25, App. D)
s b 4999 psi
tpl ;
P
A spl
tp 0.75
r 0.40
Aspl 2
tpl 609 psi
;
b
Problem 1.6-8 An elastomeric bearing pad consisting of two steel
plates bonded to a chloroprene elastomer (an artificial rubber) is
subjected to a shear force V during a static loading test (see figure).
The pad has dimensions a 125 mm and b 240 mm, and the
elastomer has thickness t 50 mm. When the force V equals 12 kN,
the top plate is found to have displaced laterally 8.0 mm with respect
to the bottom plate.
What is the shear modulus of elasticity G of the chloroprene?
P
A sn
;
SHEAR THROUGH PLATE
(b) AVE. BEARING STRESS
tnut a
V
t
Solution 1.6-8
NUMERICAL DATA
V 12 kN
b 240 mm
a 125 mm
t 50 mm
d 8 mm
AVERAGE SHEAR STRESS
tave V
ab
ave 0.4 MPa
AVERAGE SHEAR STRAIN
g ave SHEAR MODULUS G
d
t
ave
0.16
G
t ave
g ave
G 2.5 MPa
;
37
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.6-9 A joint between two concrete slabs A and B is
filled with a flexible epoxy that bonds securely to the concrete (see
figure). The height of the joint is h 4.0 in., its length is L 40
in., and its thickness is t 0.5 in. Under the action of shear forces
V, the slabs displace vertically through the distance d 0.002 in.
relative to each other.
(a) What is the average shear strain aver in the epoxy?
(b) What is the magnitude of the forces V if the shear modulus
of elasticity G for the epoxy is 140 ksi?
Solution 1.6-9 Epoxy joint between concrete slabs
(a) AVERAGE SHEAR STRAIN
gaver d
0.004
t
;
(b) SHEAR FORCES V
Average shear stress: aver G
h 4.0 in.
t 0.5 in.
L 40 in.
d 0.002 in.
G 140 ksi
Problem 1.6-10 A flexible connection consisting of rubber pads
(thickness t 9 mm) bonded to steel plates is shown in the figure.
The pads are 160 mm long and 80 mm wide.
(a) Find the average shear strain aver in the rubber if the
force P 16 kN and the shear modulus for the rubber
is G 1250 kPa.
(b) Find the relative horizontal displacement between the
interior plate and the outer plates.
V aver(hL) G
aver
aver(hL)
(140 ksi)(0.004)(4.0 in.)(40 in.)
89.6 k
;
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SECTION 1.6 Shear Stress and Strain
Solution 1.6-10 Rubber pads bonded to steel plates
(a) SHEAR STRESS AND STRAIN IN THE RUBBER PADS
taver gaver Rubber pads: t 9 mm
Length L 160 mm
P/2
8kN
625 kPa
bL
(80 mm)(160 mm)
taver
625 kPa
0.50
G
1250 kPa
;
(b) HORIZONTAL DISPLACEMENT
Width b 80 mm
avert
(0.50)(9 mm) 4.50 mm
;
G 1250 kPa
P 16 kN
Problem 1.6-11 A spherical fiberglass buoy used in an underwater experiment is anchored in shallow water by a chain [see
part (a) of the figure]. Because the buoy is positioned just below
the surface of the water, it is not expected to collapse from the
water pressure. The chain is attached to the buoy by a shackle
and pin [see part (b) of the figure]. The diameter of the pin is
0.5 in. and the thickness of the shackle is 0.25 in. The buoy has
a diameter of 60 in. and weighs 1800 lb on land (not including
the weight of the chain).
(a) Determine the average shear stress aver in the pin.
(b) Determine the average bearing stress b between the pin
and the shackle.
Solution 1.6-11
Submerged buoy
d diameter of buoy
60 in.
T tensile force in chain
dp diameter of pin
0.5 in.
t thickness of shackle
0.25 in.
W weight of buoy
1800 lb
W
weight density of sea water
63.8 lb/ft3
FREE-BODY DIAGRAM OF BUOY
FB buoyant force of water pressure
(equals the weight of the
displaced sea water)
V volume of buoy
pd 3
65.45 ft 3
6
FB W V 4176 lb
39
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CHAPTER 1 Tension, Compression, and Shear
EQUILIBRIUM
T FB W 2376 lb
(a) AVERAGE SHEAR STRESS IN PIN
Ap area of pin
Ap p 2
d 0.1963 in.2
4 p
taver T
6050 psi
2Ap
;
(b) BEARING STRESS BETWEEN PIN AND SHACKLE
Ab 2dpt 0.2500 in.2
sb T
9500 psi
Ab
;
Problem 1.6-12 The clamp shown in the figure is used to
support a load hanging from the lower flange of a steel beam.
The clamp consists of two arms (A and B) joined by a pin at C.
The pin has diameter d 12 mm. Because arm B straddles arm
A, the pin is in double shear.
Line 1 in the figure defines the line of action of the
resultant horizontal force H acting between the lower flange of
the beam and arm B. The vertical distance from this line to the
pin is h 250 mm. Line 2 defines the line of action of the
resultant vertical force V acting between the flange and arm B.
The horizontal distance from this line to the centerline of the
beam is c 100 mm. The force conditions between arm A and
the lower flange are symmetrical with those given for arm B.
Determine the average shear stress in the pin at C when the
load P 18 kN.
Solution 1.6-12
Clamp supporting a load P
FREE-BODY DIAGRAM OF CLAMP
h 250 mm
c 100 mm
P 18 kN
From vertical equilibrium:
V
P
9 kN
2
d diameter of pin at C 12 mm
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SECTION 1.6 Shear Stress and Strain
FREE-BODY DIAGRAMS OF ARMS A AND B
41
SHEAR FORCE F IN PIN
F
H 2
P 2
b
+
a
b
a
A 4
2
4.847 kN
AVERAGE SHEAR STRESS IN THE PIN
©MC 0 哵哴
VC Hh 0
H
taver F
F
42.9 MPa
Apin
pa 2
4
;
VC Pc
3.6 kN
h
2h
FREE-BODY DIAGRAM OF PIN
Problem ★1.6-13 A hitch-mounted bicycle rack is designed to carry up to four 30-lb. bikes mounted on and strapped to two
arms GH [see bike loads in the figure part (a)]. The rack is attached to the vehicle at A and is assumed to be like a cantilever beam
ABCDGH [figure part (b)]. The weight of fixed segment AB is W1 10 lb, centered 9 in. from A [see the figure part (b)] and the
rest of the rack weighs W2 40 lb, centered 19 in. from A. Segment ABCDG is a steel tube, 2 2 in., of thickness t 1/8 in.
Segment BCDGH pivots about a bolt at B of diameter dB 0.25 in. to allow access to the rear of the vehicle without removing the
hitch rack. When in use, the rack is secured in an upright position by a pin at C (diameter of pin dp 5/16 in.) [see photo and
figure part (c)]. The overturning effect of the bikes on the rack is resisted by a force couple Fh at BC.
(a)
(b)
(c)
(d)
Find the support reactions at A for the fully loaded rack;
Find forces in the bolt at B and the pin at C.
Find average shear stresses aver in both the bolt at B and the pin at C.
Find average bearing stresses b in the bolt at B and the pin at C.
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CHAPTER 1 Tension, Compression, and Shear
Bike loads
y
4 bike loads
19 in.
G
27 in.
G
Release pins
at C & G
5
(dp = — in.)
16
3 @ 4 in.
W2
H
a
1
2 in. 2 in. ( — in.)
8
C
Fixed
support
at A
MA
Ay
6 in.
D
C
D
A
H
F
2.125 in.
F
F
B
Bolt at B
1
(dB = — in.)
4
a
h = 7 in.
W1
Ax A
x
B
9 in.
h = 7 in.
F
8 in.
(a)
(b)
Pin at C
C
Pin at C
2.125 in.
D
Bolt at B
2 2 1/8 in.
tube
(c) Section a–a
Solution *1.6-13
A y 170 lb
NUMERICAL DATA
t
1
in.
8
h 7 in.
P 30 lb
L1 17 2.125 6
b 2 in.
W1 10 lb
W2 40 lb
5
in.
dp
16
dB 0.25 in.
(a) REACTIONS AT A
Ax 0
L1 25 in.
(dist from A to 1st bike)
MA W1(9) W2(19) P(4L1 4 8 12)
M A 4585 in.-lb
(b) FORCES IN BOLT AT B & PIN AT C
Fy 0
;
Ay W1 W2 4P
;
;
MB 0
By W2 4P
B y 160 lb
;
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SECTION 1.6 Shear Stress and Strain
RHFB
AsC 2
[W2(19 17) + P(6 + 2.125)
+ P(8.125 + 4) + P(8.125 + 8)
tC + P(8.125 + 12)]
Bx h
B x 254 lb
;
Bres 2Bx2 + By2
pd 2B
4
B res
tB A sB
AbB 2tdB
sbB B res
A bB
AbC 2tdp
AsB 0.098 in2
tB 3054 psi
Bx
A sC
AsC 0.153 in2
tC 1653 psi
sbC ;
Cx
AbC
AbB 0.063 in2
sbB 4797 psi
sbC 3246 psi
links, each 12 mm long between the centers of the pins (see
figure). You might wish to examine a bicycle chain and observe its
construction. Note particularly the pins, which we will assume to
have a diameter of 2.5 mm.
In order to solve this problem, you must now make two measurements on a bicycle (see figure): (1) the length L of the crank
arm from main axle to pedal axle, and (2) the radius R of the
sprocket (the toothed wheel, sometimes called the chainring).
(a) Using your measured dimensions, calculate the tensile force
T in the chain due to a force F 800 N applied to one of
the pedals.
(b) Calculate the average shear stress aver in the pins.
Bicycle chain
F force applied to pedal 800 N
L length of crank arm
;
AbC 0.078 in2
Problem 1.6-14 A bicycle chain consists of a series of small
Solution 1.6-14
;
t 0.125 in
;
(c) AVERAGE SHEAR STRESSES ave IN BOTH THE BOLT
AT B AND THE PIN AT C
AsB 2
4
(d) BEARING STRESSES B IN THE BOLT AT B AND THE PIN AT C
Cx Bx
B res 300 lb
pd 2p
R radius of sprocket
;
43
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CHAPTER 1 Tension, Compression, and Shear
(b) SHEAR STRESS IN PINS
MEASUREMENTS (FOR AUTHOR’S BICYCLE)
(1) L 162 mm
(2) R 90 mm
taver (a) TENSILE FORCE T IN CHAIN
©M axle 0
FL TR
T
FL
R
Substitute numerical values:
T
(800 N)(162 mm)
1440 N
90 mm
2T
T/ 2
T
2
Apin
pd2
pd
2
(4)
2FL
pd 2R
Substitute numerical values:
;
taver 2(800 N)(162 mm)
p(2.5 mm)2(90 mm)
147 MPa
Problem 1.6-15 A shock mount constructed as shown in the
figure is used to support a delicate instrument. The mount consists
of an outer steel tube with inside diameter b, a central steel bar of
diameter d that supports the load P, and a hollow rubber cylinder
(height h) bonded to the tube and bar.
(a) Obtain a formula for the shear in the rubber at a radial
distance r from the center of the shock mount.
(b) Obtain a formula for the downward displacement of the
central bar due to the load P, assuming that G is the shear
modulus of elasticity of the rubber and that the steel tube
and bar are rigid.
Solution 1.6-15
Shock mount
(a) SHEAR STRESS AT RADIAL DISTANCE r
As shear area at distance r
t
P
P
As
2prh
2prh
;
(b) DOWNWARD DISPLACEMENT g shear strain at distance r
g
t
P
G
2prhG
dd downward displacement for element dr
dd gdr d
r radial distance from center of shock mount to
element of thickness dr
L
dd Pdr
2prhG
b/2
Pdr
Ld/2 2prhG
d
b/2
P
dr
P
b/2
[In r]d/2
2phG Ld/2 r
2phG
d
P
b
ln
2phG
d
;
;
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SECTION 1.6 Shear Stress and Strain
45
Problem 1.6-16 The steel plane truss shown in the figure is loaded by three forces P, each of which is 490 kN. The truss mem-
bers each have a cross-sectional area of 3900 mm2 and are connected by pins each with a diameter of dp 18 mm. Members AC
and BC each consist of one bar with thickness of tAC tBC 19 mm. Member AB is composed of two bars [see figure part (b)]
each having thickness tAB/2 10 mm and length L 3 m. The roller support at B, is made up of two support plates, each having
thickness tsp/2 12 mm.
(a) Find support reactions at joints A and B and forces in members AB, BC, and AB.
(b) Calculate the largest average shear stress p,max in the pin at joint B, disregarding friction between the members;
see figures parts (b) and (c) for sectional views of the joint.
(c) Calculate the largest average bearing stress b,max acting against the pin at joint B.
P = 490 kN
P
C
a
b
A
45°
L=3m
Support
plate
and pin
Ax
b
B
45°
P
By
a
Ay
(a)
Member AB
FBC at 45°
Member AB
Member BC
Support
plate
Pin
By
—
2
By
—
2
(b) Section a–a at
joint B (Elevation view)
tAB
(2 bars, each — )
2
FAB
–––
2
Pin
P
—
2
FBC
FAB
–––
2
Support plate
tsp
(2 plates, each — )
2
P
—
2
Load P at joint B is applied
to the two support plates
(c) Section b–b at
joint B (Plan view)
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.6-16
(b) MAX. SHEAR STRESS IN PIN AT B
NUMERICAL DATA
L 3000 mm
P 490 kN
dp 18 mm
A 3900 mm
2
tAC 19 mm
tBC tAC
tAB 20 mm
tsp 24 mm
(a) SUPPORT REACTIONS AND MEMBER FORCES
Fx 0
Ax 0
1
L
L
By a P P b
L
2
2
a MA 0
By 0
;
;
Fy 0
Ay P
A y 490 kN
;
METHOD OF JOINTS
FAB P
FBC 0
FAC 22P
FAB 490 kN
FAC 693 kN
;
;
;
As tp max
pd2p
4
FAB
2
As
As 254.469 mm2
tp max 963 MPa
(c) MAX. BEARING STRESS IN PIN AT B (tab
STRESS ON AB WILL BE GREATER)
Ab dp
;
tsp SO BEARING
tAB
2
FAB
2
sb max Ab
sb max 1361 MPa
;
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SECTION 1.6 Shear Stress and Strain
Problem 1.6-17 A spray nozzle for a garden hose requires a force F 5 lb. to open the spring-loaded spray chamber AB.
47
The nozzle hand grip pivots about a pin through a flange at O. Each of the two flanges has thickness t 1/16 in., and the pin
has diameter dp 1/8 in. [see figure part (a)]. The spray nozzle is attached to the garden hose with a quick release fitting at B
[see figure part (b)]. Three brass balls (diameter db 3/16 in.) hold the spray head in place under water pressure force fp 30
lb. at C [see figure part (c)]. Use dimensions given in figure part (a).
(a) Find the force in the pin at O due to applied force F.
(b) Find average shear stress aver and bearing stress b in the pin at O.
Pin
Flange
t
dp
Pin at O
A
F
Top view at O
B
O
a = 0.75 in.
Spray
nozzle Flange
F
b = 1.5 in.
F
F
15°
c = 1.75 in.
F
Sprayer
hand grip
Water pressure
force on nozzle, f p
C
(b)
C
Quick
release
fittings
Garden hose
(c)
(a)
3 brass retaining
balls at 120°,
3
diameter db = — in.
16
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.6-17
Ox 12.68 lb
NUMERICAL DATA
t
F 5 lb
1
in.
16
dN fp 30 lb
a 0.75 in
dp 5
in.
8
b 1.5 in
1
in.
8
db u 15
3
in.
16
p
rad.
180
c 1.75 in
(a) FIND THE FORCE IN THE PIN AT O DUE TO APPLIED
FORCE F
Mo 0
FAB [F cos (u)( b a)] + F sin (u)(c)
a
FAB 7.849 lb
a FH 0
Ox FAB F cos ()
Oy 1.294 lb
Ores 2 O 2x + O 2y
Ores 12.74 lb
;
(b) FIND AVERAGE SHEAR STRESS tave AND BEARING STRESS
sb IN THE PIN AT O
As 2
pd2p
Ores
As
tO 4
Ab 2tdp
sbO tO 519 psi
Ores
Ab
;
sbO 816 psi
;
(c) FIND THE AVERAGE SHEAR STRESS ave IN THE BRASS
RETAINING BALLS AT B DUE TO WATER PRESSURE FORCE Fp
As 3
pd2b
4
tave fp
tave 362 psi
As
;
Oy F sin ()
y
Problem 1.6-18
A single steel strut AB with diameter
ds 8 mm. supports the vehicle engine hood of mass 20 kg which
pivots about hinges at C and D [see figures (a) and (b)]. The strut is
bent into a loop at its end and then attached to a bolt at A with
diameter db 10 mm. Strut AB lies in a vertical plane.
h = 660 mm
W hc = 490 mm
C
B
(a) Find the strut force Fs and average normal stress in the strut.
(b) Find the average shear stress aver in the bolt at A.
(c) Find the average bearing stress b on the bolt at A.
45°
C
x
A
30°
D
(a)
b = 254 mm c = 506 mm
y
a = 760 mm
d = 150 mm
h = 660 mm
Hood
C
Hinge
C
W
Fs
D
z
Strut
ds = 8 mm
(b)
A
H = 1041 mm
B
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49
SECTION 1.6 Shear Stress and Strain
Solution 1.6-18
NUMERICAL DATA
ds 8 mm
db 10 mm
Fsz m 20 kg
cd
2 H + ( c d)2
2
a 760 mm
b 254 mm
c 506 mm
d 150 mm
H
h 660 mm
hc 490 mm
2 H + ( c d)2
H h atan a30
Fs
where
2
0.946
cd
p
p
b + tan a45
bb
180
180
2 H + ( c d)2
2
0.324
H 1041 mm
W m (9.81m/s2)
(a) FIND THE STRUT
s IN THE STRUT
W 196.2 N
a + b + c
760 mm
2
M
lineDC
VECTOR rAB
UNIT VECTOR eAB
rAB
e AB | rAB|
0
W W
P
Q
0
hc
hc
rDC P b + c Q
D
0
rAB 1.041 * 103
P
Q
356
AND AVERAGE NORMAL STRESS
Fsy |W|hc
h
0
0.946
eAB P
0.324 Q
ƒeAB ƒ 1
rDC 2 H 2 + ( c d)2
2 H2 + ( c d)2
s
p 2
d
4 s
Fs
A strut
Astrut 50.265 mm2
s 3.06 MPa
;
db 10 mm
As 490
490
P
Q
760
H
Fs 153.9 N
H
(b) FIND THE AVERAGE SHEAR STRESS ave IN THE
BOLT AT A
0
W 196.2
P
Q
0
p 2
d
4 b
tave MD rDB Fs eAB W rDC
Fsy Fsy
Fs Astrut (ignore force at hinge C since it will vanish with
moment about line DC)
Fsx 0
0
FS
Fsy 145.664
0
H
rAB P
c dQ
M
FORCE
Fs
Fs
As
As 78.54 mm2
tave 1.96 Mpa
;
(c) FIND THE BEARING STRESS b ON THE BOLT AT A
Ab dsdb
sb Fs
Ab
Ab 80 mm2
s b 1.924 MPa
;
;
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.6-19
The top portion of a pole saw used to trim
small branches from trees is shown in the figure part (a).
The cutting blade BCD [see figure parts (a) and (c)] applies
a force P at point D. Ignore the effect of the weak return
spring attached to the cutting blade below B. Use properties
and dimensions given in the figure.
B
Rope, tension = T
a
T
Weak return spring
y
2T
x
(a) Find the force P on the cutting blade at D if the tension force in the rope is T 25 lb (see free body
diagram in part (b)].
(b) Find force in the pin at C.
(c) Find average shear stress ave and bearing stress b in
the support pin at C [see Section a–a through cutting
blade in figure part (c)].
C
Cutting
blade
Collar
Saw blade
D
a
P
(a) Top part of pole saw
B
T
20°
B
2T
50° BC = 6 in.
Cy
70°
C
in. D
C=1
D
P
Cx
x
20°
20°
Collar
3
(tc = — in.)
8
6 in. C
1 in.
D
70°
(b) Free-body diagram
Cutting blade
3
(tb = — in.)
32
Pin at C
1
(dp = — in.)
8
(c) Section a–a
Solution 1.6-19
NUMERICAL PROPERTIES
dp 1
in
8
T 25 lb
dCD 1 in
tb 3
in
32
SOLVE ABOVE EQUATION FOR P
tc 3
in
8
dBC 6 in
a
p
rad/deg
180
[T(6 sin (70a)) + 2T cos (20a)
P
6 sin (70a)) 2T sin (20a)(6 cos (70a))]
cos (20a)
P 395 lbs
;
(b) Find force in the pin at C
(a) Find the cutting force P on the cutting blade at D
if the tension force in the rope is T 25 lb:
Mc 0
MC T(6 sin(70 ))
2T cos (20 )(6 sin (70 ))
2T sin (20 )(6 cos (70 ))
P cos (20 )(1)
SOLVE FOR FORCES ON PIN AT C
Fx 0
Cx T 2T cos (20 ) P cos (40 )
Cx 374 lbs
Fy 0
;
Cy 2T sin (20 ) P sin (40 )
Cy 237 lbs
;
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SECTION 1.7 Allowable Stresses and Allowable Loads
RESULTANT AT C
BEARING STRESSES ON PIN ON EACH SIDE OF COLLAR
Cres 2 C 2x + C 2y
Cres 443 lbs
;
(c) Find maximum shear and bearing stresses in the
support pin at C (see section a-a through saw).
sbC
Cres
2
dp tc
sbC 4.72 ksi
;
BEARING STRESS ON PIN AT CUTTING BLADE
SHEAR STRESS - PIN IN DOUBLE SHEAR
p
As d2p
4
tave As 0.012 in
Cres
2As
2
sbcb Cres
dp tb
bcb 37.8 ksi
;
tave 18.04 ksi
Allowable Stresses and Allowable Loads
Problem 1.7.1 A bar of solid circular cross section is loaded in
tension by forces P (see figure). The bar has length L 16.0 in. and
diameter d 0.50 in. The material is a magnesium alloy having
modulus of elasticity E 6.4 106 psi. The allowable stress in
tension is allow 17,000 psi, and the elongation of the bar must not
exceed 0.04 in.
What is the allowable value of the forces P?
Solution 1.7-1
Magnesium bar in tension
p
Pmax smaxA (16.000 psi)a b (0.50 in.)2
4
3140 lb
L 16.0 in.
d 0.50 in.
E 6.4 106 psi
allow 17,000 psi
max 0.04 in.
MAXIMUM LOAD BASED UPON ELONGATION
emax dmax
0.04in.
0.00250
L
16 in.
smax Eâmax (6.4 * 106 psi)(0.00250)
16,000 psi
51
MAXIMUM LOAD BASED UPON TENSILE STRESS
p
Pmax sallowA (17,000 psi)a b(0.50 in.)2
4
3340 Ib
ALLOWABLE LOAD
Elongation governs.
Pallow 3140 lb
;
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.7-2 A torque T0 is transmitted between two flanged
T0
d
shafts by means of ten 20-mm bolts (see figure and photo). The
diameter of the bolt circle is d 250 mm.
If the allowable shear stress in the bolts is 90 MPa, what is the
maximum permissible torque? (Disregard friction between the
flanges.)
T0
T0
Solution 1.7-2 Shafts with flanges
NUMERICAL DATA
MAX. PERMISSIBLE TORQUE
r 10
^ bolts
Tmax ta As a r
d 250 mm
^ flange
d
b
2
Tmax 3.338 * 107 N # mm
As p r2
Tmax 33.4 kN # m
;
As 314.159 m
2
t a 85 MPa
Problem 1.7-3 A tie-down on the deck of a sailboat consists of a bent bar bolted at both ends, as shown in the figure. The
diameter dB of the bar is 1/4 in. , the diameter dW of the washers is 7/8 in. , and
the thickness t of the fiberglass deck is 3/8 in.
If the allowable shear stress in the fiberglass is 300 psi, and the allowable
bearing pressure between the washer and the fiberglass is 550 psi, what is the
allowable load Pallow on the tie-down?
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53
SECTION 1.7 Allowable Stresses and Allowable Loads
Solution 1.7-3 Bolts through fiberglass
dB 1
in.
4
P1
309.3 lb
2
dW 7
in.
8
P1 619 lb
t
3
in.
8
ALLOWABLE LOAD BASED UPON SHEAR STRESS IN
FIBERGLASS
allow 300 psi
Shear area As dWt
P1
t allow As t allow (pdWt)
2
7
3
(300 psi)(p)a in. b a in. b
8
8
ALLOWABLE LOAD BASED UPON BEARING PRESSURE
b 550 psi
Bearing area Ab 2
2
P2
p
7
1
sb Ab (550 psi) a b c a in.b a in. b d
2
4
8
4
303.7 lb
P2 607 lb
ALLOWABLE LOAD
Bearing pressure governs.
Pallow 607 lb
;
Problem 1.7-4
Two steel tubes are joined at B by four pins
(dp 11 mm), as shown in the cross section a–a in the
figure. The outer diameters of the tubes are dAB 40 mm and
dBC 28 mm. The wall thicknesses are tAB 6 mm and
tBC 7 mm. The yield stress in tension for the steel is
Y 200 MPa and the ultimate stress in tension is
U 340 MPa. The corresponding yield and ultimate values
in shear for the pin are 80 MPa and 140 MPa, respectively.
Finally, the yield and ultimate values in bearing between the
pins and the tubes are 260 MPa and 450 MPa, respectively.
Assume that the factors of safety with respect to yield stress
and ultimate stress are 4 and 5, respectively.
p 2
(d d2B)
4 W
a
Pin
tAB
dAB
A
tBC
B
dBC
C
P
a
(a) Calculate the allowable tensile force Pallow considering
tension in the tubes.
(b Recompute Pallow for shear in the pins.
(c) Finally, recompute Pallow for bearing between the pins
and the tubes. Which is the controlling value of P?
tAB
dp
tBC
dAB
dBC
Section a–a
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.7-4
Yield and ultimate stresses (all in MPa)
As TUBES:
Y 200
u 340
FSy 4
PIN (SHEAR):
Y 80
u 140
FSu 5
PIN (BEARING):
bY 260
bu 450
tubes and pin dimensions (mm)
dAB 40
tAB 6
dBC dAB 2tAB
tBC 7
dBC 28
dp 11
p
A netAB cd2AB (dAB 2tAB)2 4 dp tAB d
4
AnetAB 433.45 mm
2
p
A netBC cd 2BC (d BC 2t BC )2 4 dp t BC d
4
AnetAB 219.911 mm2 use smaller
sY
A
FSy netBC
PaT1 11.0 kN
PaT2 su
A
FSu netBC
p 2
dp
4
PaS1 14As2
As 95.033 mm2 (one pin)
tY
FSy
PaS1 7.60 kN
PaS2 14As2
;
tu
FSu
PaT1 1.1 104 N
;
PaT2 1.495 104
Problem 1.7-5 A steel pad supporting heavy machinery rests on
four short, hollow, cast iron piers (see figure). The ultimate
strength of the cast iron in compression is 50 ksi. The outer diameter of the piers is d 4.5 in. and the wall thickness is t 0.40 in.
Using a factor of safety of 3.5 with respect to the ultimate
strength, determine the total load P that may be supported by
the pad.
PaS2 10.64 kN
(c) Pallow CONSIDERING BEARING IN THE PINS
AbAB 4dptAB
AbAB 264 mm2
AbBC 4dpt BC
(a) Pallow CONSIDERING TENSION IN THE TUBES
PaT1 (b) Pallow CONSIDERING SHEAR IN THE PINS
Pab1 AbAB a
AbBC 308 mm2
sbY
b
FSy
Pab1 17.16 kN
Pab2 AbAB a
6 smaller controls
sbu
b
FSu
Pab1 1.716 * 104
;
Pab2 23.8 kN
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SECTION 1.7 Allowable Stresses and Allowable Loads
Solution 1.7-5 Cast iron piers in compression
Four piers
A
U 50 ksi
5.152 in.2
n 3.5
s allow
p 2
p
(d d20) [(4.5 in.)2 (3.7 in.)2]
4
4
sU
50 ksi
14.29 ksi
n
3.5
d 4.5 in.
P1 allowable load on one pier
sallow A (14.29 ksi)(5.152 in.2)
73.62 k
Total load P 4P1 294 k
t 0.4 in.
;
d0 d 2t 3.7 in.
Problem 1.7-6 The rear hatch of a van [BDCF in figure part (a)] is supported by two hinges at B1 and B2 and by two struts
A1B1 and A2B2 (diameter ds 10 mm) as shown in figure part (b). The struts are supported at A1 and A2 by pins, each with
diameter dp 9 mm and passing through an eyelet of thickness t 8 mm at the end of the strut [figure part (b)]. If a closing
force P 50 N is applied at G and the mass of the hatch Mh 43 kg is concentrated at C:
(a) What is the force F in each strut? [Use the free-body diagram of one half of the hatch in the figure part (c)]
(b) What is the maximum permissible force in the strut, Fallow, if the allowable stresses are as follows: compressive stress in
the strut, 70 MPa; shear stress in the pin, 45 MPa; and bearing stress between the pin and the end of the strut, 110 MPa.
127 mm
B2
B1
505 mm
505 mm
F
C
Mh
D
Bottom
part of
strut
G
P
F
A1
B
710 mm
Mh
—g
2
By
ds = 10 mm
A2
75 mm
Bx
10
460 mm
Eyelet
Pin support
A
F
t = 8 mm
(c)
(b)
(a)
Solution 1.7-6
(a) FORCE F IN EACH STRUT FROM STATICS (SUM
MOMENTS ABOUT B)
p
FV F cos1a2 FH F sin1a2
a 10
180
NUMERICAL DATA
Mh 43 kg
a 70 MPa
a 45 MPa
ba 110 MPa
ds 10 mm
dp 9 mm
P 50 N
g 9.81
t 8 mm
m
s2
G
C
D
g MB 0
FV(127) + FH(75)
P
—
2
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CHAPTER 1 Tension, Compression, and Shear
Mh
P
g (127 + 505) + [127 + 2(505)]
2
2
(b) MAX. PERMISSIBLE FORCE F IN EACH STRUT
Fmax IS SMALLEST OF THE FOLLOWING
p 2
d
4 s
p
ta dp2
4
F (127cos(a) + 75sin(a))
Fa1 sa
Mh
P
g (127 + 505) + [127 + 2(505)]
2
2
Fa2
Mh
P
g (127 + 505) + [127 + 2(505)]
2
2
F
(127 cos(a) + 75 sin(a))
F 1.171 kN
Fa1 5.50 kN
Fa2 2.86 kN
Fa3 sba dp t
;
Fa2
2.445
F
Fa3 7.92 kN
;
Problem 1.7-7 A lifeboat hangs from two ship’s davits, as shown in the
figure. A pin of diameter d 0.80 in. passes through each davit and supports two pulleys, one on each side of the davit.
Cables attached to the lifeboat pass over the pulleys and wind around
winches that raise and lower the lifeboat. The lower parts of the cables are
vertical and the upper parts make an angle 15° with the horizontal. The
allowable tensile force in each cable is 1800 lb, and the allowable shear
stress in the pins is 4000 psi.
If the lifeboat weighs 1500 lb, what is the maximum weight that should
be carried in the lifeboat?
Solution 1.7-7 Lifeboat supported by four cables
FREE-BODY DIAGRAM OF ONE PULLEY
Pin diameter d 0.80 in.
T tensile force in one cable
Tallow 1800 lb
allow 4000 psi
W weight of lifeboat
1500 lb
©Fhoriz 0
©Fvert 0
RH T cos 15° 0.9659T
RV T T sin 15° 0.7412T
V shear force in pin
V 2(RH)2 + (Rv)2 1.2175T
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SECTION 1.7 Allowable Stresses and Allowable Loads
ALLOWABLE TENSILE FORCE IN ONE CABLE BASED
MAXIMUM WEIGHT
UPON SHEAR IN THE PINS
Vallow tallow A pin
2011 lb
V 1.2175T
Shear in the pins governs.
p
(4000 psi)a b (0.80 in.)2
4
Tmax T1 1652 lb
Total tensile force in four cables
Vallow
1652 lb
T1 1.2175
4Tmax 6608 lb
Wmax 4Tmax W
6608 lb 1500 lb
ALLOWABLE FORCE IN ONE CABLE BASED UPON
TENSION IN THE CABLE
5110 lb
T2 Tallow 1800 lb
;
Problem 1.7-8 A cable and pulley system in figure part (a) supports a cage of mass 300 kg at B. Assume that this includes the
mass of the cables as well. The thickness of each the three steel pulleys is t 40 mm. The pin diameters are dpA 25 mm,
dpB 30 mm and dpC 22 mm [see figure, parts (a) and part (b)].
(a) Find expressions for the resultant forces acting on the pulleys at A, B, and C in terms of cable tension T.
(b) What is the maximum weight W that can be added to the cage at B based on the following allowable stresses? Shear
stress in the pins is 50 MPa; bearing stress between the pin and the pulley is 110 MPa.
a
C
dpA = 25 mm
L1
A
Cable
Cable
Pulley
a
t
L2
dpB
tB
Pin
dpC = 22 mm
dp
Support
bracket
B
dpB = 30 mm
Cage
W
(a)
Section a–a: pulley support
detail at A and C
Cage at B
Section a–a: pulley
support detail at B
(b)
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.7-8
OR check bearing stress
NUMERICAL DATA
g 9.81
M 300 kg
m
Wmax2 a 50 MPa
s2
ba 110 MPa
tA 40 mm
tB 40 mm
Wmax2 tC 50
dpA 25 mm
dpB 30
dpC 22 mm
(a) RESULTANT FORCES F ACTING ON PULLEYS A, B & C
FA 22 T
FC T
FB 2T
T
Mg
Wmax
+
2
2
Wmax 2T M g
(b) MAX. LOAD W THAT CAN BE ADDED AT B DUE TO a &
ba IN PINS AT A, B & C
PULLEY AT A
a sba Ab b M g
a sba tA dpA b M g
22
152.6 kN ( bearing at A)
2
Wmax3 2
1t A 2 M g
2 a s
p
Wmax3 cta a 2 dpB2 b d M g
4
Wmax4 DOUBLE SHEAR
FA aAs
22 T t aAs
Mg
Wmax
ta As
+
2
2
22
22
2
22
(shear at B)
2
(s A ) M g
2 ba b
Wmax4 sba t B dpB M g
PULLEY AT C
2
2T aAs
PULLEY AT B
Wmax4 129.1 kN
FA
tA As
Wmax1 22
Wmax3 67.7 kN
From statics at B
Wmax1 Wmax2
2
ataAs b M g
ata2
p
d A2 b M g
4 p
Wmax1
22.6
Mg
Wmax1 66.5 kN
;
(shear at A controls)
(bearing at B)
T ta As
Wmax5 21t a As2 M g
Wmax5 c2ta a 2
p 2
d b d Mg
4 pC
Wmax5 7.3 * 104
Wmax5 73.1 kN
Wmax6 2sbatC dp C M g
Wmax6 239.1 kN
(bearing at C)
(shear at C)
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SECTION 1.7 Allowable Stresses and Allowable Loads
Problem 1.7-9 A ship’s spar is attached at the base of a mast by a pin connection
Mast
(see figure). The spar is a steel tube of outer diameter d2 3.5 in. and inner diameter
d1 2.8 in. The steel pin has diameter d 1 in., and the two plates connecting the spar
to the pin have thickness t 0.5 in. The allowable stresses are as follows: compressive
stress in the spar, 10 ksi; shear stress in the pin, 6.5 ksi; and bearing stress between the
pin and the connecting plates, 16 ksi.
Determine the allowable compressive force Pallow in the spar.
P
Pin
Spar
Connecting
plate
Solution 1.7-9
COMPRESSIVE STRESS IN SPAR
p
Pa1 s a 1d22 d122
4
Pa1 34.636 kips
SHEAR STRESS IN PIN
Pa2 t a a 2
Pa2 10.21 kips
NUMERICAL DATA
d2 3.5 in.
dp 1 in.
a 10 ksi
d1 2.8 in.
controls
;
^double shear
t 0.5 in.
a 6.5 ksi
p 2
d b
4 p
ba 16 ksi
BEARING STRESS BETWEEN PIN & CONECTING PLATES
Pa3 ba(2dpt)
Problem 1.7-10 What is the maximum possible value of the
clamping force C in the jaws of the pliers shown in the figure if
the ultimate shear stress in the 5-mm diameter pin is 340 MPa?
What is the maximum permissible value of the applied load
P if a factor of safety of 3.0 with respect to failure of the pin is
to be maintained?
Pa3 16 kips
P
y
15
mm
90°
38
Rx
50°
x
C
50 mm
90°
P
C
Pin
10°
mm
Rx 140°
b=
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125 m
a
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.7-10
NUMERICAL DATA
u 340 MPa
FS 3
a 40
ta p
rad
180
ta tu
ta FS
As
Rx C cos ( )
a 50 cos ( )
Pmax pin at C in single shear
Ry P
125
Rx P
a 163.302 mm
Ac
2
a
a
cos (a) d + c1 +
sin(a) d
b
b
here
;
a
4.297
b
a/b mechanical advantage
FIND MAX. CLAMPING FORCE
P(a)
C
b
a Mpin 0
P(a)
cos(a)
b
R y Pc1 +
C ult PmaxFS a
a
sin(a) d
b
2
2
a
a
c cos1a2 d + c1 + sin(a) d ta A s
A
b
b
As ta A s
2
Pmax 445 N
C sin ( )
b 38 mm
STATICS
t a 113.333 MPa
Find Pmax
d 5 mm
2Rx2 + Ry2
tu
FS
a
b
b
Pult PmaxFS
C ult 5739 N
;
Pult 1335
C ult
4.297
P ult
p 2
d
4
Problem 1.7-11 A metal bar AB of weight W is suspended by a system of steel
2.0 ft
2.0 ft
7.0 ft
wires arranged as shown in the figure. The diameter of the wires is 5/64 in., and the
yield stress of the steel is 65 ksi.
Determine the maximum permissible weight Wmax for a factor of safety of
1.9 with respect to yielding.
5.0 ft
5.0 ft
W
A
B
Solution 1.7-11
NUMERICAL DATA
d
5
in.
64
sY
sa FSy
FORCES IN WIRES AC, EC, BD, FD
Y 65 ksi
FSy 1.9
a 34.211 ksi
a FV 0 at A, B, E or F
222 + 52
W
*
5
2
Wmax = 0.539 a A
FW 222 + 52
0.539
10
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SECTION 1.7 Allowable Stresses and Allowable Loads
Wmax 0.539 a
sY
p
b a d2 b
FS y
4
Wmax 0.305 kips
FCD 2a
;
FCD 2c
CHECK ALSO FORCE IN WIRE CD
a FH 0
FCD at C or D
2
22 + 52
2
2
222 + 52
2
W
5
61
F wb
a
222 + 52
W
*
bd
5
2
less than FAC so AC controls
Problem 1.7-12 A plane truss is subjected to loads 2P and P at joints B and C, respectively, as shown in the figure
part (a). The truss bars are made of two L102 76 6.4 steel angles [see Table E-5(b): cross sectional area of the two
angles, A 2180 mm2, figure part (b)] having an ultimate stress in tension equal to 390 MPa. The angles are connected
to an 12 mm-thick gusset plate at C [figure part (c)] with 16-mm diameter rivets; assume each rivet transfers an equal
share of the member force to the gusset plate. The ultimate stresses in shear and bearing for the rivet steel are 190 MPa
and 550 MPa, respectively.
Determine the allowable load Pallow if a safety factor of 2.5 is desired with respect to the ultimate load that can be
carried. (Consider tension in the bars, shear in the rivets, bearing between the rivets and the bars, and also bearing between
the rivets and the gusset plate. Disregard friction between the plates and the weight of the truss itself.)
F
FCF
G
a
a
A
B
a
FCG
Truss bars
C
a
a
D
Gusset
plate
Rivet
C
P
2P
a
FBC
FCD
(a)
P
(c)
Gusset plate
6.4 mm
12 mm
Rivet
(b) Section a–a
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.7-12
PCG 45.8 kN
NUMERICAL DATA
;
A 2180 mm
2
tg 12 mm
dr 16 mm
u 390 MPa
u 190 MPa
bu 550 MPa
su
FS
sa tang 6.4 mm
tu
FS
sba sbu
FS
MEMBER FORCES FROM TRUSS ANALYSIS
F BC
5
P
3
4
F CD P
3
22
0.471
3
22
P
FCF 3
4
F CG P
3
Pallow FOR TENSION ON NET SECTION IN TRUSS BARS
Anet A 2drtang
Anet 1975 mm2
A net
0.906
A
PCD 45.8 kN
;
Fmax
sba A b
N
3
PBC 3 a b(sba Ab)
5
PBC 81.101 kN
PCF 2 a
PCF 191.156 kN
3
22
b (sba Ab)
3
PCG 2 a b(sba Ab)
4
PCG 67.584 kN
3
PCD 2 a b(sba Ab)
4
PCD 67.584 kN
Finally, Pallow for bearing of rivets on gusset plate
Fallow aAnet
Pallow
3
PCD 2 a b(ta As)
4
Next, Pallow for bearing of rivets on truss bars
rivet bears on each angle in two angle
Ab 2drtang
pairs
FS 2.5
ta < so shear in rivets in CG & CD
controls Pallow here
allowable force in a member
so BC controls since it has the
largest member force for this loading
3
3
Pallow (sa A net)
F BCmax
5
5
Pallow 184.879 kN
Ab drtg
(bearing area for each rivert on gusset plate)
tg 12 mm
2tang 12.8 mm
so gusset will control over angles
Next, Pallow for shear in rivets (all are in double shear)
3
PBC 3 a b(sba Ab)
5
PBC 76.032 kN
p
A s 2 d r2
4
for one rivet in DOUBLE shear
PCF 2 a
PCF 179.209 kN
Fmax
t aA s
N
N number of rivets in a particular
member (see drawing of conn. detail)
3
PCG 2 a b(sba Ab)
4
PCG 63.36 kN
PCD 63.36 kN
3
22
b (sba Ab)
3
PBC 3 a b(ta As)
5
PBC 55.0 kN
3
PCD 2 a b(sba Ab)
4
PCF 2 a
PCF 129.7 kN
So, shear in rivets controls: Pallow = 45.8 kN
3
22
b (ta As)
3
PCG 2 a b(ta As)
4
;
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SECTION 1.7 Allowable Stresses and Allowable Loads
Problem 1.7-13
A solid bar of circular cross section (diameter d)
has a hole of diameter d/5 drilled laterally through the center of the
bar (see figure). The allowable average tensile stress on the net
cross section of the bar is allow.
d
d/5
P
d
(a) Obtain a formula for the allowable load Pallow that the bar
can carry in tension.
(b) Calculate the value of Pallow if the bar is made of brass with
diameter d 1.75 in. and allow 12 ksi.
(Hint: Use the formulas of Case 15 Appendix D.)
Solution 1.7-13
NUMERICAL DATA
1
1
2
26 b d
Pa sa c d2 a acosa b 2
5
25
a 12 ksi
d 1.75 in
1
2
acosa b 26
5
25
(a) FORMULA FOR PALLOW IN TENSION
From Case 15, Appendix D:
A 2r2 aa ab
r2
a
a acosa b
r
a
b
r
d
2
r 0.875 in.
180
78.463 degrees
p
b
d 2
d 2
c
a
b
a
b d
A 2
10
b
Aa
Pa aA
d
10
a 0.175 in.
2
d
b 26
5
0.587
Pa sa10.587 d22
;
0.587
0.748
0.785
(b) EVALUATE NUMERICAL RESULT
d 1.75 in.
b 2r2 a2
6 2
d b
25
a
d/5
P
Pa 21.6 kips
a 12 ksi
;
p
0.785
4
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.7-14 A solid steel bar of diameter d1 60 mm has a
hole of diameter d2 32 mm drilled through it (see figure). A steel
pin of diameter d2 passes through the hole and is attached to supports.
Determine the maximum permissible tensile load Pallow in the bar
if the yield stress for shear in the pin is Y 120 MPa, the yield stress
for tension in the bar is Y 250 MPa and a factor of safety of 2.0 with
respect to yielding is required. (Hint: Use the formulas of Case 15,
Appendix D.)
d2
d1
d1
P
Solution 1.7-14
SHEAR AREA (DOUBLE SHEAR)
NUMERICAL DATA
d1 60 mm
p
As 2a d22 b
4
d2 32 mm
Y 120 MPa
Y 250 MPa
As 1608 mm2
FSy 2
NET AREA IN TENSION (FROM CASE 15, APP. D)
ALLOWABLE STRESSES
Anet 2a
ta tY
FSy
a 60 MPa
sa sY
FSy
a 125 MPa
From Case 15, Appendix D:
A 2r 2 aa d2
a
2
b
2
ab
r
2
2
d2
c a d1 b a d2 b d
2 A 2
d2
2
≥acosa b ¥
d1
d1 2
a b
2
r
a arc cos
b 2r2 a2
d1 2
b
2
d1
2
d2
d2/2
arc cos
d1/2
d1
Anet 1003 mm2
Pallow in tension: smaller of values based on either shear
or tension allowable stress x appropriate area
Pa1 aAs
Pa2 aAnet
Pa1 96.5kN 6 shear governs
Pa2 125.4kN
;
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SECTION 1.7 Allowable Stresses and Allowable Loads
Resultant
of wind
pressure
Lh
2
C.P.
F
W
Pipe
column
z b
2
D
H
Lv
C
A
F
at each
4
bolt
h
y Overturning
moment
B about x axis
FH
x
W
at each
4
bolt
(a)
W
Pipe column
db
dw
FH
— = Rh
2
One half of over – turning
moment about x axis acts
on each bolt pair
Base
B
plate (tbp)
z
A
y
Footing
F/4
Tension
h
R
R
Compression
(b)
z
in.
FH
—
2
b=
1
F
4 R W
4
y
2 in
.
B
h
4
=1
D
FH
2
R
F
4
W
R
4
(c)
W
4
x
A
four bolts anchored in a concrete footing. Wind pressure p acts
normal to the surface of the sign; the resultant of the uniform
wind pressure is force F at the center of pressure. The wind force
is assumed to create equal shear forces F/4 in the y-direction at
each bolt [see figure parts (a) and (c)]. The overturning effect of
the wind force also causes an uplift force R at bolts A and C and
a downward force (R) at bolts B and D [see figure part (b)].
The resulting effects of the wind, and the associated ultimate
stresses for each stress condition, are: normal stress in each bolt
(u 60 ksi); shear through the base plate (u 17 ksi); horizontal shear and bearing on each bolt (hu 25 ksi and
bu 75 ksi); and bearing on the bottom washer at B (or D)
(bw 50 ksi).
Find the maximum wind pressure pmax (psf) that can be
carried by the bolted support system for the sign if a safety
factor of 2.5 is desired with respect to the ultimate wind load
that can be carried.
Use the following numerical data: bolt db 3⁄4 in.;
washer dw 1.5 in.; base plate tbp 1 in.; base plate
dimensions h 14 in. and b 12 in.; W 500 lb; H 17 ft;
sign dimensions (Lv 10 ft. Lh 12 ft.); pipe column
diameter d 6 in., and pipe column thickness t 3/8 in.
65
Sign (Lv Lh)
Problem 1.7-15 A sign of weight W is supported at its base by
C
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CHAPTER 1 Tension, Compression, and Shear
(1) COMPUTE pmax BASED ON NORMAL STRESS IN EACH BOLT
(GREATER AT B & D)
Solution 1.7-15
Numerical Data
u 60 ksi
u 17 ksi
bu 75 ksi
db 3
in.
4
h 14 in.
hu 25 ksi
bw 50 ksi
dw 1.5 in.
tbp 1 in.
b 12 in.
W 0.500 kips
Lv 10(12)
FSu 2.5
3
in.
8
H 204 in.
d 6 in.
H 17(12)
Lh 12(12)
Lv 120 in.
Lh 144 in.
su
FSu
a 24
a 6.8
sba tha sbu
FSu
ta thu
FSu
ba 30
tu
FSu
sbwa FORCES F AND R IN TERMS OF pmax
R pmax
LvLhH
2h
p
W
sa a d b2b 4
4
pmax1 LvLhH
2h
pmax1 11.98 psf
FH
2h
;
controls
W
4
t
p dw tbp
Rmax ta(p dw tbp) sbw
FSu
pmax2 R
p
W
Rmax sa a db2 b 4
4
p 2
d
4 b
R +
ha 10
bwa 20
F pmaxLvLh
s
W
4
(2) COMPUTE pmax BASED ON SHEAR THROUGH BASE PLATE
(GREATER AT B & D)
ALLOWABLE STRESSES (ksi)
sa t
R +
W
4
ta a p dw tbp b Lv Lh H
2h
pmax2 36.5 psf
W
4
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SECTION 1.7 Allowable Stresses and Allowable Loads
(3) COMPUTE pmax BASED ON HORIZONTAL SHEAR
ON EACH BOLT
th F
4
Fmax
p
a db2 b
4
pmax3 p
4tha a db2 b
4
tha(p db2)
Lv Lh
pmax3 147.3 psf
(5) COMPUTE pmax BASED ON BEARING UNDER THE TOP WASHER
AT A (OR C) AND THE BOTTOM WASHER AT B (OR D)
R +
sbw EACH BOLT
pmax5
F
4
sb (tbp db)
pmax4 Fmax 4ba(tbpdb)
4sba(tbpdb)
p
ad 2 db 2 b
4 w
Rmax sbwa c
(4) COMPUTE pmax BASED ON HORIZONTAL BEARING ON
W
4
p
W
adw2 db2b d 4
4
p
W
sbwa c (dw2 db2) d 4
4
LvLhH
2h
pmax5 30.2 psf
So, normal/stress in bolts controls; pmax 11.98 psf
LvLh
pmax4 750 psf
Problem 1.7-16 The piston in an engine is attached to a
connecting rod AB, which in turn is connected to a crank arm
BC (see figure). The piston slides without friction in a cylinder
and is subjected to a force P (assumed to be constant) while
moving to the right in the figure. The connecting rod, which has
diameter d and length L, is attached at both ends by pins. The
crank arm rotates about the axle at C with the pin at B moving
in a circle of radius R. The axle at C, which is supported by
bearings, exerts a resisting moment M against the crank arm.
(a) Obtain a formula for the maximum permissible force
Pallow based upon an allowable compressive stress c in
the connecting rod.
(b) Calculate the force Pallow for the following data:
c 160 MPa, d 9.00 mm, and R 0.28L.
Cylinder
P
Piston
Connecting rod
A
M
d
C
B
L
R
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.7-16
The maximun allowable force P occurs when cos has its
smallest value, which means that has its largest value.
LARGEST VALUE OF
d diameter of rod AB
FREE-BODY DIAGRAM OF PISTON
The largest value of occurs when point B is the farthest
distance from line AC. The farthest distance is the radius R
of the crank arm.
P applied force (constant)
C compressive force in connecting rod
RP resultant of reaction forces between cylinder
and piston (no friction)
:
a Fhoriz 0
P C cos
;
0
P C cos
MAXIMUM COMPRESSIVE FORCE C IN CONNECTING ROD
Cmax cAc
in which Ac area of connecting rod
pd2
Ac 4
MAXIMUM ALLOWABLE FORCE P
P Cmax cos
c Ac cos
Therefore,
—
BC R
—
Also, AC 2L2 R2
cos a R 2
2L2 R2
A 1 a b
L
L
(a) MAXIMUM ALLOWABLE FORCE P
Pallow c Ac cos
sc a
pd2
4
b
A
R 2
1 a b
L
(b) SUBSTITUTE NUMERICAL VALUES
c 160 MPa
R 0.28L
d 9.00 mm
R/L 0.28
Pallow 9.77 kN
;
;
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SECTION 1.8 Design for Axial Loads and Direct Shear
69
Design for Axial Loads and Direct Shear
Problem 1.8-1 An aluminum tube is required
to transmit an axial tensile force P 33 k
[see figure part (a)]. The thickness of the wall
of the tube is to be 0.25 in.
d
P
(a) What is the minimum required outer
diameter dmin if the allowable tensile
stress is 12,000 psi?
(b) Repeat part (a) if the tube will have a
hole of diameter d/10 at mid-length
[see figure parts (b) and (c)].
P
(a)
Hole of diameter d/10
d
d/10
P
P
d
(b)
(c)
Solution 1.8-1
(b) MIN. DIAMETER OF TUBE (WITH HOLES)
NUMERICAL DATA
P 33 kips
t 0.25 in.
a 12 ksi
p
d
A1 c 3d2 (d 2t)242 a bt d
4
10
(a) MIN. DIAMETER OF TUBE (NO HOLES)
p
A1 3d2 (d 2 t)24
4
P
A2 sa
A2 2.75 in2
t
b pt2
5
equating A1 & A2 and solving for d:
equating A1 & A2 and solving for d:
P
d
+ t
psat
A1 dapt d 3.75 in.
;
P
p t2
sa
d
t
p t
5
d 4.01 in.
;
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.8-2 A copper alloy pipe having yield stress Y 290
d
t =—
8
P
MPa is to carry an axial tensile load P 1500 kN [see figure
part (a)]. A factor of safety of 1.8 against yielding is to be used.
(a) If the thickness t of the pipe is to be one-eighth of its
outer diameter, what is the minimum required outer
diameter dmin?
(b) Repeat part (a) if the tube has a hole of diameter d/10
drilled through the entire tube as shown in the figure
[part (b)].
d
(a)
P
Hole of diameter d/10
d
t =—
8
d
(b)
Solution 1.8-2
NUMERICAL DATA
equate A1 & A2 and solve for d:
Y 290 MPa
d2 P 1500 kN
FSy 1.8
256
15p
(a) MIN. DIAMETER (NO HOLES)
A1 p 2
d 2
cd ad b d
4
8
p 15
A1 a d2 b
4 64
P
A2 sY
FSy
15
A1 p d2
256
A2 9.31 103 mm2
P
sY
P FS Q
y
256
15p
dmin Q
dmin 225mm
P
sY
P FSy Q
;
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71
SECTION 1.8 Design for Axial Loads and Direct Shear
(b) MIN. DIAMETER (WITH HOLES)
Redefine A1 - subtract area for two holes - then
equate to A2
d
p
d 2
d
A1 c cd2 ad b d 2a b a b d
4
8
10 8
A1 d2 a
P
sy
P FS Q
y
15
1
p b
256
40
15 2 1 2
pd d
256
40
A1 d2 a
15
1
p b
256
40
P
sy
15
1
p 0.159
256
40
y
dmin a
c
Problem 1.8-3 A horizontal beam AB with cross-sectional dimensions
(b 0.75 in.) (h 8.0 in.) is supported by an inclined strut CD and
carries a load P 2700 lb at joint B [see figure part (a)]. The strut,
which consists of two bars each of thickness 5b/8, is connected to the
beam by a bolt passing through the three bars meeting at joint C
[see figure part (b)].
(a) If the allowable shear stress in the bolt is 13,000 psi, what is the
minimum required diameter dmin of the bolt at C?
(b) If the allowable bearing stress in the bolt is 19,000 psi, what is
the minimum required diameter dmin of the bolt at C?
P FS Q
15
1
p b
256
40
dmin 242 mm
4 ft
;
5 ft
B
C
A
3 ft
P
D
(a)
b
Beam AB (b h)
h
—
2
Bolt (dmin)
h
—
2
5b
—
8
Strut CD
(b)
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.8-3
NUMERICAL DATA
P 2.7 kips
a 13 ksi
(b) dmin BASED ON ALLOWABLE BEARING AT JT C
b 0.75 in.
ba 19 ksi
h 8 in.
Bearing from beam ACB
(a) dmin BASED ON ALLOWABLE SHEAR - DOUBLE SHEAR
dmin IN STRUT
ta FDC
As
As 2 a
dmin
FDC 15 P/4
bd
dmin 0.711 inches
;
15
P
4
Bearing from strut DC sb 5
2 bd
8
15
P
4
p 2
d b
4
15
P
4
p
t a b
a a 2
15 P/4
b sba
sb sb 3
dmin 0.704 inches
P
bd
(lower than ACB)
;
Problem 1.8-4 Lateral bracing for an elevated pedestrian walkway is shown in the figure part (a). The thickness of the clevis
plate tc 16 mm and the thickness of the gusset plate tg 20 mm [see figure part (b)]. The maximum force in the diagonal
bracing is expected to be F 190 kN.
If the allowable shear stress in the pin is 90 MPa and the allowable bearing stress between the pin and both the clevis and
gusset plates is 150 MPa, what is the minimum required diameter dmin of the pin?
Clevis
Gusset plate
Gusset plate
tc
Pin
tg
Cl
ev
is
dmin
Diagonal brace
F
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SECTION 1.8 Design for Axial Loads and Direct Shear
Solution 1.8-4
NUMERICAL DATA
(2) dmin BASED ON ALLOW BEARING IN GUSSET & CLEVIS
F 190 kN
a 90 MPa
tg 20 mm
tc 16 mm
ba 150 MPa
(1) dmin BASED ON ALLOW SHEAR - DOUBLE SHEAR
IN STRUT
F
t
As
dmin p
As 2 a d2 b
4
F
p
t a b
Q a 2
dmin 36.7 mm
PLATES
Bearing on gusset plate
sb F
Ab
Ab tgd
dmin dmin 63.3 mm
6 controls
Bearing on clevis
Ab d(2tc)
dmin F
2t csba
dmin 39.6 mm
F
tgsba
;
73
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.8-5 Forces P1 1500 lb and P2 2500 lb are applied at joint C of plane truss ABC
P2
shown in the figure part (a). Member AC has thickness tAC 5/16 in. and member AB is composed of two bars each having thickness tAB/2 3/16 in. [see figure part (b)]. Ignore the effect
of the two plates which make up the pin support at A.
If the allowable shear stress in the pin is 12,000 psi and the allowable bearing stress in the
pin is 20,000 psi, what is the minimum required diameter dmin of the pin?
C
P1
L
A
a
B
L
a
Ax
By
Ay
(a)
tAC
Pin
support
plates
AC
tAB
—
2
AB
Pin
A
Ay
—
2
Ay
—
2
Section a–a
(b)
Solution 1.8-5
NUMERICAL DATA
P1 1.5 kips
P2 2.5 kips
5
tAC in.
16
3
tAB 2a b in.
16
a 12 ksi
ba 20 ksi
(1) dmin BASED ON ALLOWABLE SHEAR - DOUBLE SHEAR
IN STRUT; FIRST CHECK AB (SINGLE SHEAR IN EACH
BAR HALF)
Force in each bar of AB is P1/2
P1
2
t
AS
p
A s a d2 b
4
dmin
P1
2
p
t a b
a 2 4
dmin 0.282 in.
Next check double shear to AC; force in AC is (P1 + P2)/2
(P1 + P2)/ 2
dmin Q
ta a
dmin 0.461 inches
p
b
4
Finally check RESULTANT force on pin at A
R
P1 2
P1 + P2 2
b
Aa 2 b + a
2
R 2.136 kips
;
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SECTION 1.8 Design for Axial Loads and Direct Shear
dmin
R
2
p
t a b
a a 4
dmin 0.476 in.
(2) dmin BASED ON ALLOWABLE BEARING ON PIN
member AB bearing on pin
dmin P1
tABsba
P1 + P2
tACsba
P1
Ab
Ab tABd
dmin 0.2 in.
member AC bearing on pin
dmin sb Ab d(tAC)
dmin 0.64 in.
controls
;
Problem 1.8-6 A suspender on a suspension bridge consists of a cable that
passes over the main cable (see figure) and supports the bridge deck, which is
far below. The suspender is held in position by a metal tie that is prevented from
sliding downward by clamps around the suspender cable.
Let P represent the load in each part of the suspender cable, and let u
represent the angle of the suspender cable just above the tie. Finally, let sallow
represent the allowable tensile stress in the metal tie.
(a) Obtain a formula for the minimum required cross-sectional area of the tie.
(b) Calculate the minimum area if P 130 kN, u 75°, and sallow 80 MPa.
75
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.8-6 Suspender tie on a suspension bridge
F tensile force in cable
above tie
P tensile force in cable
below tie
allow allowable tensile
stress in the tie
(a) MINIMUM REQUIRED AREA OF TIE
Amin T
Pcotu
sallow
sallow
(b) SUBSTITUTE NUMERICAL VALUES:
P 130 kN
75°
allow 80 MPa
Amin 435 mm2
FREE-BODY DIAGRAM OF HALF THE TIE
Note: Include a small amount of the cable in the free-body
diagram
T tensile force in the tie
FORCE TRIANGLE
cotu T
P
T P cot Problem 1.8-7 A square steel tube of length L 20 ft and
width b2 10.0 in. is hoisted by a crane (see figure). The
tube hangs from a pin of diameter d that is held by the cables
at points A and B. The cross section is a hollow square with
inner dimension b1 8.5 in. and outer dimension b2 10.0
in. The allowable shear stress in the pin is 8,700 psi, and the
allowable bearing stress between the pin and the tube is
13,000 psi.
Determine the minimum diameter of the pin in order to
support the weight of the tube. (Note: Disregard the rounded
corners of the tube when calculating its weight.)
;
;
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77
SECTION 1.8 Design for Axial Loads and Direct Shear
Solution 1.8-7 Tube hoisted by a crane
T tensile force in cable
W weight of steel tube
d diameter of pin
b1 inner dimension of tube
8.5 in.
b2 outer dimension of tube
10.0 in.
W gs AL
(490 lb/ft3)(27.75 in.2)a
1,889 lb
DIAMETER OF PIN BASED UPON SHEAR
Double shear. 2allow Apin W
2(8,700 psi)a
L length of tube 20 ft
allow 8,700 psi
b 13,000 psi
WEIGHT OF TUBE
s
1 ft2
b (20 ft)
144 in.
p d2
b 1889 lb
4
d2 0.1382 in.2
d1 0.372 in.
DIAMETER OF PIN BASED UPON BEARING
b(b2 b1)d W
weight density of steel
(13,000 psi)(10.0 in. 8.5 in.) d 1,889 lb
490 lb/ft
d2 0.097 in.
3
A area of tube
b 22 b 21 (10.0 in.)2 (8.5 in.)2
27.75 in.
MINIMUM DIAMETER OF PIN
Shear governs.
Problem 1.8-8 A cable and pulley system at D is used to
bring a 230-kg pole (ACB) to a vertical position as shown in
the figure part (a). The cable has tensile force T and is attached
at C. The length L of the pole is 6.0 m, the outer diameter is
d 140 mm, and the wall thickness t 12 mm. The pole
pivots about a pin at A in figure part (b). The allowable shear
stress in the pin is 60 MPa and the allowable bearing stress
is 90 MPa.
Find the minimum diameter of the pin at A in order to support
the weight of the pole in the position shown in the figure part (a).
B
1.0 m
dmin 0.372 in.
Pole
C
Cable
30°
Pulley
5.0 m
a
T
A
D
4.0 m
a
(a)
d
ACB
Pin support
plates
A
Pin
Ay
—
2
Ay
—
2
(b)
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.8-8
ALLOWABLE SHEAR & BEARING STRESSES
CHECK SHEAR DUE TO RESULTANT FORCE ON PIN AT A
a 60 MPa
RA 2 A2x + A2y
ba 90 MPa
FIND INCLINATION OF & FORCE IN CABLE, T
let angle between pole & cable at C; use Law
of Cosines
DC A
52 + 42 2(5)(4)cos a120 p b
180
a acos c
DC 7.81 m
a
u
180
26.33 degrees
p
180
33.67
p
52 + DC2 42
d
2DC(5)
u 60a
p
b a
180
< ange between cable & horiz. at D
W 230 kg(9.81 m/s2)
W 2.256 103 N
STATICS TO FIND CABLE FORCE T
a MA 0
W(3 sin(30 deg)) TX(5 cos(30 deg))
Ty(5 sin(30 deg)) 0
substitute for Tx & Ty in terms of T & solve for T:
T
3
W
2
5
523
sin(u)
cos(u)
2
2
T 1.53 103 N
Ty T sin( )
Tx T cos( )
Tx 1.27 103 N
Ty 846.11 N
(1) dmin BASED ON ALLOWABLE SHEAR - DOUBLE SHEAR
AT A
Ax Tx
Ay Ty W
dmin
RA 3.35 103 N
RA
2
p
t a b
a a 4
dmin 5.96 mm 6controls
;
(2) dmin BASED ON ALLOWABLE BEARING ON PIN
dpole 140 mm
tpole 12 mm
Lpole 6000 mm
member AB BEARING ON PIN
sb RA
Ab
dmin Ab 2tpoled
RA
2tpole sba
dmin 1.55 mm
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SECTION 1.8 Design for Axial Loads and Direct Shear
Problem 1.8-9 A pressurized circular cylinder has a sealed cover plate fastened
with steel bolts (see figure). The pressure p of the gas in the cylinder is 290 psi,
the inside diameter D of the cylinder is 10.0 in., and the diameter dB of the bolts
is 0.50 in.
If the allowable tensile stress in the bolts is 10,000 psi, find the number n of
bolts needed to fasten the cover.
Solution 1.8-9 Pressurized cylinder
P
ppD 2
F
n
4n
Ab area of one bolt p 2
db
4
P allow Ab
sallow p 290 psi
D 10.0 in.
allow 10,000 psi
db 0.50 in.
n number of bolts
F total force acting on the cover plate from the
internal pressure
F pa
n
pD 2
b
4
NUMBER OF BOLTS
P tensile force in one bolt
ppD 2
pD 2
P
Ab
(4n)(p4 )d 2b
nd 2b
pD 2
d 2bsallow
SUBSTITUTE NUMERICAL VALUES:
n
(290 psi)(10 in.)2
(0.5 in.)2(10,000 psi)
Use 12 bolts
;
11.6
79
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.8-10 A tubular post of outer diameter d2 is guyed by two cables
fitted with turnbuckles (see figure). The cables are tightened by rotating the
turnbuckles, thus producing tension in the cables and compression in the post.
Both cables are tightened to a tensile force of 110 kN. Also, the angle between the
cables and the ground is 60°, and the allowable compressive stress in the post is
c 35 MPa.
If the wall thickness of the post is 15 mm, what is the minimum permissible
value of the outer diameter d2?
Solution 1.8-10
Tubular post with guy cables
d2 outer diameter
AREA OF POST
d1 inner diameter
A
t wall thickness
15 mm
T tensile force in a cable
110 kN
allow 35 MPa
P compressive force in post
2T cos 30°
REQUIRED AREA OF POST
A
P
s allow
2Tcos 30°
s allow
p
p 2
(d 2 d 21) [d 22 (d2 2t)2 ]
4
4
pt (d2 t)
EQUATE AREAS AND SOLVE FOR d2:
2T cos 30°
pt (d2 t)
sallow
d2 2T cos 30°
+ t
ptsallow
;
SUBSTITUTE NUMERICAL VALUES:
(d2)min 131 mm
;
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SECTION 1.8 Design for Axial Loads and Direct Shear
Problem 1.8-11 A large precast concrete panel for a warehouse is being raised to a vertical position using two sets of
81
cables at two lift lines as shown in the figure part (a). Cable 1 has length L1 22 ft and distances along the panel (see figure
part (b)) are a L1/2 and b L1/4. The cables are attached at lift points B and D and the panel is rotated about its base at
A. However, as a worst case, assume that the panel is momentarily lifted off the ground and its total weight must be
supported by the cables. Assuming the cable lift forces F at each lift line are about equal, use the simplified model of
one half of the panel in figure part (b) to perform your analysis for the lift position shown. The total weight of the panel
is W 85 kips. The orientation of the panel is defined by the following angles: 20° and 10°.
Find the required cross-sectional area AC of the cable if its breaking stress is 91 ksi and a factor of safety of 4 with respect
to failure is desired.
F
H
F
F
T2
b2
T1
B
a
b1
W
D
b
—
2
B
g
u
y
a
C
W
—
2
D
b
g
A
b
A
(a)
x
(b)
Solution 1.8-11
GEOMETRY
L1 22 ft
1
a L1
2
1
b L1
4
10 deg
a 2.5b 24.75 ft
20 deg
Using Law of cosines
L2 2(a + b)2 + L21 2(a + b)L1 cos(u )
L2 6.425 ft
b acos c
L21 + L22 ( a + b)2
d
2L1L2
b 26.484 degrees
1
( )
2
b 2 16.484 deg
1
b 1 10 deg
SOLUTION APPROACH: FIND T THEN AC T/(s U/FS)
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CHAPTER 1 Tension, Compression, and Shear
STATICS at point H
T1 27.042 kips
g Fx 0
T2 T1
H
SO
T2 T1
g FY 0
H
T1sin( 1) T2sin( 2)
sin(b 1)
sin(b 2)
T1cos(b 1) + T2 cos(b 2) F
and
F W/2,
W 85 kips
SO
T1 acos(b 1) +
sin(b 1)
sin(b 2)
T2 16.549 kips
COMPUTE REQUIRED CROSS-SECTIONAL AREA
u 91 ksi
Ac sin(b 1)
cos(b 2)b F
sin(b 2)
W
2
T1 sin(b 1)
acos(b 1) +
cos(b 2)b
sin(b 2)
Problem 1.8-12 A steel column of hollow circular cross section
is supported on a circular steel base plate and a concrete pedestal
(see figure). The column has outside diameter d 250 mm and
supports a load P 750 kN.
(a) If the allowable stress in the column is 55 MPa, what is the
minimum required thickness t? Based upon your result, select
a thickness for the column. (Select a thickness that is an even
integer, such as 10, 12, 14, . . . , in units of millimeters.)
(b) If the allowable bearing stress on the concrete pedestal is
11.5 MPa, what is the minimum required diameter D of the
base plate if it is designed for the allowable load Pallow that
the column with the selected thickness can support?
T1
su
FS
FS 4
su
22.75 ksi
FS
Ac 1.189 in2
;
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SECTION 1.8 Design for Axial Loads and Direct Shear
Solution 1.8-12 Hollow circular column
SUBSTITUTE NUMERICAL VALUES IN EQ. (1):
t 2 250 t +
(750 * 103 N)
p(55 N/mm2)
0
(Note: In this eq., t has units of mm.)
t2 250t 4,340.6 0
Solve the quadratic eq. for t:
t 18.77 mm
Use t 20 mm
d 250 mm
A t(d t) Pallow allow t(d t)
b 11.5 MPa (allowable pressure on concrete)
(a) THICKNESS t OF THE COLUMN
pt(d t) pt ptd +
2
t 2 td +
P
sallow
p
(4t)(d t) pt(d t)
4
D2 P
sallow
Pallow
pD 2
4
sb
4s allowt(d t)
sb
4(55 MPa)(20 mm)(230 mm)
11.5 MPa
D2 88,000 mm2
0
P
0
psallow
Area of base plate sallowpt(d t)
pD 2
4
sb
pd 2
p
A
(d 2t)2
4
4
Pallow allow A
where A is the area of the column with t 20 mm.
D diameter of base plate
s allow
;
For the column,
t thickness of column
A
;
(b) DIAMETER D OF THE BASE PLATE
P 750 kN
allow 55 MPa (compression in column)
P
tmin 18.8 mm
Dmin 297 mm
(Eq. 1)
D 296.6 mm
;
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.8-13 An elevated jogging track is supported at
intervals by a wood beam AB (L 7.5 ft) which is pinned at A
and supported by steel rod BC and a steel washer at B. Both
the rod (dBC 3/16 in.) and the washer (dB 1.0 in.) were
designed using a rod tension force of TBC 425 lb. The rod
was sized using a factor of safety of 3 against reaching the
ultimate stress u 60 ksi. An allowable bearing stress
ba 565 psi was used to size the washer at B.
Now, a small platform HF is to be suspended below a section
of the elevated track to support some mechanical and electrical
equipment. The equipment load is uniform load q 50 lb/ft and
concentrated load WE 175 lb at mid-span of beam HF. The
plan is to drill a hole through beam AB at D and install the same
rod (dBC) and washer (dB) at both D and F to support beam HF.
(a) Use u and ba to check the proposed design for rod DF
and washer dF; are they acceptable?
(b) Also re-check the normal tensile stress in rod BC and
bearing stress at B; if either is inadequate under the
additional load from platform HF, redesign them to
meet the original design criteria.
Original
structure
C
Steel rod,
3
dBC = — in.
16
TBC =
425 lb.
L
—
25
L = 7.5 ft
A
Wood beam supporting track
D
B
Washer
dB = 1.0 in.
3
New steel rod, dDF = — in.
16
WE = 175 lb
q = 50 lb/ft
New beam to support equipment
H
L
—
2
L
—
2
Hx
L
—
25
F
Washer, dF
(same at D
above)
Hy
Solution 1.8-13
NUMERICAL DATA
L 7.5(12)
u 60 ksi
q
50
12
dBC L 90 in.
ba 0.565 ksi
FSu 3
q 4.167
3
in.
16
TBC 425 lb
lb
in
WE 175 lb
dB 1.0 in
(a) FIND FORCE IN ROD DF AND FORCE ON WASHER AT F
L
L
qL
2
2
TDF L
aL
b
25
OK - less than a; rod is
;
acceptable
sDF 10.38 ksi
sa su
FSu
a 20 ksi
BEARING STRESS ON WASHER AT F:
sbF TDF
p 2
2
1d d BC2
4 B
OK - less than ba; washer is
;
acceptable
sbF 378 psi
WE
MH 0
TDF 286.458 lb
NORMAL STRESS IN ROD DF:
s DF TDF
p 2
d
4 BC
(b) FIND NEW FORCE IN ROD BC - SUM MOMENT ABOUT A FOR
UPPER FBD - THEN CHECK NORMAL STRESS IN BC &
BEARING STRESS AT B
MA 0
TBC2 TBCL + TDF a L TBC2 700 lb
L
L
b
25
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85
SECTION 1.8 Design for Axial Loads and Direct Shear
REVISED NORMAL STRESS IN ROD BC:
Original
structure
TBC2
s BC2 C
Steel rod,
3
dBC = — in.
16
p
a dBC2 b
4
sBC2 25.352 ksi
exceeds a = 20 ksi
L
—
25
L = 7.5 ft
SO RE-DESIGN ROD BC:
A
Wood beam supporting track
TBC2
dBCreqd p
sa
Q4
dBCreqd 0.211 in.
TBC2
B
Washer
dB = 1.0 in.
L
—
25
New beam to support equipment
H
RE-CHECK BEARING STRESS IN WASHER AT B:
p
c 1dB2 dBC22 d
4
D
3
New steel rod, dDF = — in.
16
WE = 175 lb
q = 50 lb/ft
dBCreqd . 16 3.38 in.
1
^say 4/16 1/4 in. dBC2 in.
4
s bB2 TBC =
425 lb.
bB2 924 psi
^ exceeds
ba = 565 psi
Washer, dF
(same at D
above)
L
—
2
L
—
2
F
Hx
Hy
SO RE-DESIGN WASHER AT B:
TBC2
dBC2
dBreqd 1.281 in.
p
sba
Q4
use 1 5/16 in washer at B: 1 + 5/16 1.312 in.
dBreqd ;
Problem 1.8-14 A flat bar of width b 60 mm and thickness t 10 mm
is loaded in tension by a force P (see figure). The bar is attached to a
support by a pin of diameter d that passes through a hole of the same size
in the bar. The allowable tensile stress on the net cross section of the bar is
sT 140 MPa, the allowable shear stress in the pin is tS 80 MPa, and
the allowable bearing stress between the pin and the bar is sB 200 MPa.
(a) Determine the pin diameter dm for which the load P will be a
maximum.
(b) Determine the corresponding value Pmax of the load.
d
P
b
t
P
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.8-14
Bar with a pin connection
SHEAR IN THE PIN
PS 2tS Apin 2tS a
pd 2
b
4
p
1
2(80 MPa)a b(d 2)a
b
4
1000
0.040 pd 2 0.12566d 2
(Eq. 2)
BEARING BETWEEN PIN AND BAR
PB B td
b 60 mm
(200 MPa)(10 mm)(d )a
t 10 mm
2.0 d
d diameter of hole and pin
1
b
1000
(Eq. 3)
GRAPH OF EQS. (1), (2), AND (3)
T 140 MPa
S 80 MPa
B 200 MPa
UNITS USED IN THE FOLLOWING CALCULATIONS:
P is in kN
and are in N/mm2 (same as MPa)
b, t, and d are in mm
TENSION IN THE BAR
PT T (Net area) t(t)(b d )
(140 MPa)(10 mm) (60 mm d) a
1.40 (60 d)
1
b
1000
(Eq. 1)
(a) PIN DIAMETER dm
PT PB or 1.40(60 d) 2.0 d
84.0
mm 24.7 mm
Solving, dm 3.4
(b) LOAD Pmax
Substitute dm into Eq. (1) or Eq. (3):
Pmax 49.4 kN
;
;
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Page 87
SECTION 1.8 Design for Axial Loads and Direct Shear
Problem 1.8-15 Two bars AC and BC of the same material support a vertical load P
(see figure). The length L of the horizontal bar is fixed, but the angle can be varied by
moving support A vertically and changing the length of bar AC to correspond with the
new position of support A. The allowable stresses in the bars are the same in tension and
compression.
We observe that when the angle is reduced, bar AC becomes shorter but the crosssectional areas of both bars increase (because the axial forces are larger). The opposite
effects occur if the angle is increased. Thus, we see that the weight of the structure
(which is proportional to the volume) depends upon the angle .
Determine the angle so that the structure has minimum weight without exceeding
the allowable stresses in the bars. (Note: The weights of the bars are very small compared to the force P and may be disregarded.)
87
A
θ
B
C
L
P
Solution 1.8-15 Two bars supporting a load P
LENGTHS OF BARS
L AC L
cos u
L BC L
WEIGHT OF TRUSS
g weight density of material
W g(AAC L AC + ABC L BC)
T tensile force in bar AC
C compressive force in bar BC
a Fvert 0
a Fhoriz 0
T
P
sin u
C
P
tan u
AREAS OF BARS
AAC T
P
sallow
sallow sin u
ABC C
P
sallow
sallow tan u
gPL
1
1
a
+
b
sallow sin u cos u
tan u
gPL 1 + cos2u
a
b
sallow sin u cos u
Eq. (1)
, P, L, and allow are constants
W varies only with Let k gPL
(k has unis of force)
sallow
W
1 + cos2u
(Nondimensional)
k
sin u cos u
Eq. (2)
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CHAPTER 1 Tension, Compression, and Shear
GRAPH OF EQ. (2):
df
du
(sin u cos u)(2)(cos u) ( sin u)
(1 + cos2u)( sin2u + cos2 u)
sin2u cos2u
sin2u cos2 u + sin2u cos2 u cos4u
sin2 u cos2 u
SET THE NUMERATOR 0 AND SOLVE FOR :
sin2 cos2 sin2 cos2 cos4 0
Replace sin2 by 1 cos2:
(1 cos2)(cos2) 1 cos2 cos2 cos4 0
Combine terms to simplify the equation:
ANGLE THAT MAKES WA MINIMUM
1 3 cos2 0
Use Eq. (2)
Let f 1 + cos2u
sin u cos u
df
0
du
54.7°
;
cos u 1
23
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Page 89
2
Axially Loaded
Members
Changes in Lengths of Axially Loaded Members
Problem 2.2-1 The L-shaped arm ABC shown in the figure
lies in a vertical plane and pivots about a horizontal pin at A.
The arm has constant cross-sectional area and total weight W.
A vertical spring of stiffness k supports the arm at point B.
Obtain a formula for the elongation of the spring due to the
weight of the arm.
k
A
B
C
b
b
b
—
2
Solution 2.2-1
Take first moments about A to find c.g.
x⫽
b
2b
2
W(b) + ≥
¥W(2 b)
5
5
P bQ
a bb
2
2
W
6
x⫽ b
5
Find force in spring due to weight of arm
a MA ⫽ 0
Fk ⫽
6
Wa bb
5
b
6
Fk ⫽ W
5
Find elongation of spring due to weight of arm
Fk
6W
d⫽
d⫽
;
k
5k
89
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CHAPTER 2
Page 90
Axially Loaded Members
Problem 2.2-2 A steel cable with nominal diameter
25 mm (see Table 2-1) is used in a construction yard
to lift a bridge section weighing 38 kN, as shown in
the figure. The cable has an effective modulus of
elasticity E ⫽ 140 GPa.
(a) If the cable is 14 m long, how much will it
stretch when the load is picked up?
(b) If the cable is rated for a maximum load of
70 kN, what is the factor of safety with respect
to failure of the cable?
Solution 2.2-2
Bridge section lifted by a cable
A ⫽ 304 mm2 (from
Table 2-1)
W ⫽ 38 kN
(b) FACTOR OF SAFETY
PULT ⫽ 406 kN (from Table 2-1)
Pmax ⫽ 70 kN
E ⫽ 140 GPa
L ⫽ 14 m
n⫽
PULT
406 kN
⫽
⫽ 5.8
Pmax
70 kN
(a) STRETCH OF CABLE
d⫽
(38 kN)(14 m)
WL
⫽
EA
(140 GPa)(304 mm2)
⫽ 12.5 mm
;
Problem 2.2-3 A steel wire and a copper wire have equal lengths and
support equal loads P (see figure). The moduli of elasticity for the steel and
copper are Es ⫽ 30,000 ksi and Ec ⫽ 18,000 ksi, respectively.
(a) If the wires have the same diameters, what is the ratio of the elongation
of the copper wire to the elongation of the steel wire?
(b) If the wires stretch the same amount, what is the ratio of the diameter of
the copper wire to the diameter of the steel wire?
;
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Page 91
SECTION 2.2
Solution 2.2-3
91
Changes in Lengths of Axially Loaded Members
Steel wire and copper wire
Equal lengths and equal
loads
Steel: Es ⫽ 30,000 ksi
dc
Es
30
⫽
⫽
⫽ 1.67
ds
Ec
18
(b) RATIO OF DIAMETERS (EQUAL ELONGATIONS)
Copper: Ec ⫽ 18,000 ksi
dc ⫽ ds
(a) RATIO OF
ELONGATIONS (EQUAL
DIAMETERS)
dc ⫽
PL
EcA
ds ⫽
;
PL
EsA
Ec a
d2c
d2s
Problem 2.2-4 By what distance h does the cage shown in the figure
move downward when the weight W is placed inside it?
Consider only the effects of the stretching of the cable, which
has axial rigidity EA ⫽ 10,700 kN. The pulley at A has diameter
dA ⫽ 300 mm and the pulley at B has diameter dB ⫽ 150 mm. Also,
the distance L1 ⫽ 4.6 m, the distance L2 ⫽ 10.5 m, and the weight
W ⫽ 22 kN. (Note: When calculating the length of the cable, include
the parts of the cable that go around the pulleys at A and B.)
PL
PL
⫽
or Ec Ac ⫽ Es As
Ec Ac
Es As
p 2
p
bdc ⫽ Es a bd2s
4
4
⫽
Es
Ec
dc
Es
30
⫽
⫽
⫽ ⫽ 1.29
ds
A Ec A 18
;
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CHAPTER 2
Page 92
Axially Loaded Members
Solution 2.2-4
Cage supported by a cable
dA ⫽ 300 mm
dB ⫽ 150 mm
LENGTH OF CABLE
L ⫽ L1 + 2L2 +
1
1
1pdA2 + (pdB)
4
2
L1 ⫽ 4.6 m
⫽ 4,600 mm + 21,000 mm + 236 mm + 236 mm
L2 ⫽ 10.5 m
⫽ 26,072 mm
EA ⫽ 10,700 kN
W ⫽ 22 kN
ELONGATION OF CABLE
d⫽
(11 kN)(26,072 mm)
TL
⫽
⫽ 26.8 mm
EA
(10,700 kN)
LOWERING OF THE CAGE
h ⫽ distance the cage moves downward
TENSILE FORCE IN CABLE
W
T⫽
⫽ 11 kN
2
h⫽
Problem 2.2-5 A safety valve on the top of a tank containing steam
under pressure p has a discharge hole of diameter d (see figure).
The valve is designed to release the steam when the pressure reaches the
value pmax.
If the natural length of the spring is L and its stiffness is k, what
should be the dimension h of the valve? (Express your result as a
formula for h.)
1
d ⫽ 13.4 mm
2
;
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Page 93
SECTION 2.2
Solution 2.2-5
Changes in Lengths of Axially Loaded Members
Safety valve
pmax ⫽ pressure when valve opens
L ⫽ natural length of spring (L ⬎ h)
k ⫽ stiffness of spring
FORCE IN COMPRESSED SPRING
F ⫽ k(L ⫺ h) (From Eq. 2-1a)
PRESSURE FORCE ON SPRING
P ⫽ pmax a
pd2
b
4
EQUATE FORCES AND SOLVE FOR h:
h ⫽ height of valve (compressed length of the spring)
d ⫽ diameter of discharge hole
p ⫽ pressure in tank
F ⫽ P k1L ⫺ h2 ⫽
h⫽L⫺
ppmax d2
4k
ppmaxd2
4
;
Problem 2.2-6 The device shown in the figure consists of
a pointer ABC supported by a spring of stiffness k ⫽ 800 N/m.
The spring is positioned at distance b ⫽ 150 mm from the pinned
end A of the pointer. The device is adjusted so that when there is no
load P, the pointer reads zero on the angular scale.
If the load P ⫽ 8 N, at what distance x should the load be placed
so that the pointer will read 3° on the scale?
Solution 2.2-6
Pointer supported by a spring
FREE-BODY DIAGRAM OF POINTER
⌺MA ⫽ 0 哵哴
Px
kb
Let ␣ ⫽ angle of rotation of pointer
Px
kb2
d
x⫽
tan a
tan a ⫽ ⫽ 2
b
P
kb
⫺ Px + (kd)b ⫽ 0
or d ⫽
SUBSTITUTE NUMERICAL VALUES:
P⫽8N
k ⫽ 800 N/m
b ⫽ 150 mm
␦ ⫽ displacement of spring
F ⫽ force in spring
⫽ k␦
a ⫽ 3°
(800 N/m)(150 mm)2
tan 3°
8N
⫽ 118 mm ;
x⫽
;
93
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CHAPTER 2
Page 94
Axially Loaded Members
Problem 2.2-7 Two rigid bars, AB and CD, rest on a
smooth horizontal surface (see figure). Bar AB is pivoted
end A, and bar CD is pivoted at end D. The bars are connected to each other by two linearly elastic springs of stiffness k. Before the load P is applied, the lengths of the
springs are such that the bars are parallel and the springs
are without stress.
Derive a formula for the displacement ␦C at point C when
the load P is acting near point B as shown. (Assume that the
bars rotate through very small angles under the action of the
load P.)
b
P
b
b
A
B
C
dC
D
Solution 2.2-7
(1) first sum moments about A for the entire structure
to get RD then sum vertical forces to get RA
a MA ⫽ 0
RD ⫽
DISPLACEMENT DIAGRAMS
1
[ P(2b)]
3b
A
2
RD ⫽ P
3
a FV ⫽ 0
RA ⫽ P ⫺ RD
B
2
UFBD
Fk2 ⫽ ⫺RA ⫺ P
⫺4
P
3
^ spring 2 is in
compression
P
UFBD
F k2 ⫽
b
RA
b
Fk1
Fk1 ⫽ RA ⫺ (P ⫹ Fk2)
UFBD
P
4
⫺ P + Pb
3
3
2
F k1 ⫽ P
3
^ spring 1 is
in tension
D
C
(P ⫹ Fk2)b ⫽ ⫺RAb
a Mk1 ⫽ 0
F k1 ⫽ a
dB
dB
P
RA ⫽
3
(2) next, cut through both springs & consider equilibrium of upper free body (UFBD) to find forces in
springs (assume initially that both springs are in
tension)
a FV ⫽ 0
(3) solve displacement equations to find ␦C
Fk2
dC
dC
2
dB
Fk1
2P
⫽
⫽
2
k
3 k
Fk2
dC
⫺4 P
elongation of spring 2 ⫽
⫺ dB ⫽
⫽
2
k
3 k
multiply 2nd equation above by (⫺1/2) and add to
first equation
3
4 P
16 P
16
d ⫽
dC ⫽
;
⫽ 1.778
4 C 3k
9 k
9
elongation of spring 1 ⫽ dC ⫺
(4) substitute dC into either equation to find ␦B
(not a required part of this problem)
4P
dB ⫽ 2dC ⫺
1st equ ⬎
3k
4P
16 P
b ⫺
d
dB ⫽ c2a
9 k
3k
20 P
20
dB ⫽
⫽ 2.222
9 k
9
2nd equ ⬎
dB ⫽
dC
4P
+
2
3k
4P
1 16 P
dB ⫽ c a
b +
d
2 9 k
3k
dB ⫽
20 P
9 k
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Page 95
SECTION 2.2
95
Changes in Lengths of Axially Loaded Members
Problem 2.2-8 The three-bar truss ABC shown in the figure has a span
L ⫽ 3 m and is constructed of steel pipes having cross-sectional
area A ⫽ 3900 mm2 and modulus of elasticity E ⫽ 200 GPa. Identical
loads P act both vertically and horizontally at joint C, as shown.
P
P
C
(a) If P ⫽ 650 kN, what is the horizontal displacement of joint B?
(b) What is the maximum permissible load value Pmax if the displacement of
joint B is limited to 1.5 mm?
45°
A
45°
B
L
Solution 2.2-8
P
P
By
a FV ⫽ 0
Ax
Ay
By
Ay ⫽ P ⫺ By
Method of Joints:
NUMERICAL DATA
A ⫽ 3900 mm2
E ⫽ 200 GPa
P ⫽ 650 kN
L ⫽ 3000 mm
␦Bmax ⫽ 1.5 mm
(a) FIND HORIZ. DISPL. OF JOINT B
a MA ⫽ 0
By ⫽
1
L
a2P b
L
2
By ⫽ P
a FH ⫽ 0
Ax ⫽ ⫺P
FAB ⫽ Ax
dB ⫽
F ABL
EA
Ay ⫽ 0
FACV ⫽ Ay
FAC ⫽ 0
FACV ⫽ 0
force in AB is P (tension) so elongation
of AB ⫽ horiz. displ. of jt B
dB ⫽
PL
EA
d B ⫽ 2.5 mm
;
(b) FIND Pmax IF DISPL. OF JOINT B ⫽ d Bmax ⫽ 1.5 mm
Pmax ⫽
EA
d
L Bmax
Pmax ⫽ 390 kN
;
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CHAPTER 2
Page 96
Axially Loaded Members
Problem 2.2-9 An aluminum wire having a diameter d ⫽ 1/10 in.
and length L ⫽ 12 ft is subjected to a tensile load P (see figure).
The aluminum has modulus of elasticity E ⫽ 10,600 ksi
If the maximum permissible elongation of the wire is 1/8 in. and the
allowable stress in tension is 10 ksi, what is the allowable load Pmax?
P
d
L
Solution 2.2-9
1
in
10
1
d a ⫽ in
8
d⫽
A⫽
pd2
4
L ⫽ 12(12) in
E ⫽ 10600 ⫻ (103) psi
s a ⫽ 10 * (103) psi
A ⫽ 7.854 ⫻ 10⫺3 in2
EA ⫽ 8.325 ⫻ 104 lb
Max. load based on elongation
EA
Pmax1 ⫽
d Pmax1 ⫽ 72.3 lb
L a
Max. load based on stress
Pmax2 ⫽ 78.5 lb
Pmax2 ⫽ ␴aA
; controls
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Page 97
SECTION 2.2
97
Changes in Lengths of Axially Loaded Members
Problem 2.2-10 A uniform bar AB of weight W ⫽ 25 N is
supported by two springs, as shown in the figure. The spring on the
left has stiffness k1 ⫽ 300 N/m and natural length L1 ⫽ 250 mm. The
corresponding quantities for the spring on the right are k2 ⫽ 400 N/m
and L2 ⫽ 200 mm. The distance between the springs is L ⫽ 350 mm,
and the spring on the right is suspended from a support that is
distance h ⫽ 80 mm below the point of support for the spring on the
left. Neglect the weight of the springs.
(a) At what distance x from the left-hand spring (figure part a)
should a load P ⫽ 18 N be placed in order to bring the bar to
a horizontal position?
(b) If P is now removed, what new value of k1 is required so that
the bar (figure part a) will hang in a horizontal position
under weight W?
(c) If P is removed and k1 ⫽ 300 N/m, what distance b should
spring k1 be moved to the right so that the bar (figure part a)
will hang in a horizontal position under weight W?
(d) If the spring on the left is now replaced by two springs in series
(k1 ⫽ 300N/m, k3) with overall natural length L1 ⫽ 250 mm
(see figure part b), what value of k3 is required so that the
bar will hang in a horizontal position under weight W?
New position of
k1 for part (c) only
k1
L1
b
k2
L2
W
A
B
P
x
Load P for
part (a) only
L
(a)
k3
L1
—
2
k1
L1
—
2
h
k2
L2
W
A
B
L
(b)
Solution 2.2-10
NUMERICAL DATA
W ⫽ 25 N
k1 ⫽ 0.300
N
mm
L1 ⫽ 250 mm
N
L2 ⫽ 200 mm
mm
L ⫽ 350 mm
h ⫽ 80 mm
P ⫽ 18 N
k2 ⫽ 0.400
(a) LOCATION OF LOAD P TO BRING BAR TO HORIZ.
POSITION
use statics to get forces in both springs
a MA ⫽ 0
F2 ⫽
F2 ⫽
1
L
aW + Pxb
L
2
W
x
+ P
2
L
h
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CHAPTER 2
Page 98
Axially Loaded Members
a FV ⫽ 0
F1 ⫽ W + P ⫺ F2
F1 ⫽
W
x
+ Pa1 ⫺ b
2
L
use constraint equation to define horiz. position,
then solve for location x
F1
F2
⫽ L2 + h +
k1
k2
L1 +
substitute expressions for F1 & F2 above into constraint equ. & solve for x
⫺ 2L1 L k1 k2 ⫺ k2WL ⫺ 2k2 P L + 2L2 L k1 k2 + 2 h L k1 k2 + k1W L
⫺2P1k1 + k22
x ⫽ 134.7 mm ;
x⫽
(b) NEXT REMOVE P AND FIND NEW VALUE OF SPRING
CONSTANT K1 SO THAT BAR IS HORIZ.
UNDER WEIGHT W
Now, F1 ⫽
W
2
F2 ⫽
W
since P ⫽ 0
2
same constraint equation as above but now P ⫽ 0:
Part (c) - continued
statics
a Mk1 ⫽ 0
F2 ⫽
wa
L
⫺ bb
2
L⫺b
a FV ⫽ 0
W
W
a b
2
2
L1 +
⫺ 1L2 + h2 ⫺
⫽0
k1
k2
F1 ⫽ W ⫺ F2
Wa
solve for k1
k1 ⫽
F1 ⫽ W ⫺
⫺ Wk2
[2k2[L1 ⫺ (L2 + h)]] ⫺ W
N
k1 ⫽ 0.204
mm
F1 ⫽
;
(c) USE K1 ⫽ 0.300 N/mm BUT RELOCATE
SPRING K1 (x ⫽ b) SO THAT BAR ENDS UP
IN HORIZ. POSITION UNDER WEIGHT W
L/2 – b
F1
b
F2
L/2
L/2
constraint equation - substitute above expressions
for F1 & F2 and solve for b
F1
F2
⫺ ( L2 + h) ⫺
⫽0
k1
k2
use the following data
L1 +
k1 ⫽ 0.300
N
mm
L2 ⫽ 200 mm
k2 ⫽ 0.4
L ⫽ 350 mm
L–b
FBD
2L1k1k2L + WLk2 ⫺ 2L2k1k2L ⫺ 2hk1k2L ⫺ Wk1L
(2L1k1k2) ⫺ 2L2k1k2 ⫺ 2hk1k2 ⫺ 2Wk1
L⫺b
WL
2( L ⫺ b)
W
b⫽
L
⫺ bb
2
b ⫽ 74.1 mm
;
N
mm
L1 ⫽ 250 mm
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Page 99
SECTION 2.2
(d) REPLACE SPRING K1 WITH SPRINGS IN SERIES:
K1 ⫽ 0.3N/mm, L1/2 AND K3, L1/2 - FIND K3
SO THAT BAR HANGS IN HORIZ. POSITION
statics
F1 ⫽
W
2
k3 ⫽
F2 ⫽
W
2
Changes in Lengths of Axially Loaded Members
new constraint equation; solve for k3
L1 +
F1
F1
F2
+
⫺ ( L2 + h) ⫺
⫽0
k1
k3
k2
W
W
W
2
2
2
L1 +
+
⫺ ( L2 + h) ⫺
⫽0
k1
k3
k2
Wk1k2
⫺2L1k1k2 ⫺ Wk2 + 2L2k1k2 + 2hk1k2 + Wk1
NOTE - equivalent spring constant for series springs
k1k3
ke ⫽
k1 + k3
Problem 2.2-11 A hollow, circular, cast-iron pipe (Ec ⫽ 12,000 ksi)
supports a brass rod (Eb ⫽ 14,000 ksi) and weight W ⫽ 2 kips, as
shown. The outside diameter of the pipe is dc ⫽ 6 in.
(a) If the allowable compressive stress in the pipe is 5000 psi and
the allowable shortening of the pipe is 0.02 in., what is the
minimum required wall thickness tc,min? (Include the weights
of the rod and steel cap in your calculations.)
(b) What is the elongation of the brass rod ␦r due to both load
W and its own weight?
(c) What is the minimum required clearance h?
99
k e ⫽ 0.204
k3 ⫽ 0.638
N
mm
;
N
mm
;
checks - same as (b) above
Nut & washer
3
dw = — in.
4
(
)
Steel cap
(ts = 1 in.)
Cast iron pipe
(dc = 6 in., tc)
Lr = 3.5 ft
Lc = 4 ft
(
Brass rod
1
dr = — in.
2
h
)
W
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CHAPTER 2
Page 100
Axially Loaded Members
Solution 2.2-11
The figure shows a section cut through the pipe, cap
and rod.
LET a ⫽
NUMERICAL DATA
tc2 ⫺ dctc ⫹ ␣ ⫽ 0
Ec ⫽ 12000 ksi
W ⫽ 2 kips
Eb ⫽ 14000 ksi
dc ⫽ 6 in
tc ⫽
1
dr ⫽ in.
2
g b ⫽ 3.009 * 10⫺4
Lc ⫽ 48 in
d c ⫺ 2 dc2 ⫺ 4a
2
g s ⫽ 2.836 * 10⫺4
in3
kips
dpipe ⫽
WtLc
EcAmin
Amin ⫽
␴a above
Lr ⫽ 42 in
ptc( dc ⫺ tc) ⫽
(a) MIN. REQ’D WALL THICKNESS OF CI PIPE, tcmin
first check allowable stress then allowable
shortening
Wcap ⫽ 8.018 ⫻ 10⫺3 kips
tc ⫽ 0.021 in.
Wt
sa
A pipe ⫽
p 2
[ d ⫺ (d c ⫺ 2 t c)2]
4 c
Amin ⫽ 0.402 in2
t c( d c ⫺ t c) ⫽
Wt
ps a
WtLc
pEc da
dc ⫺ 2 d2c ⫺ 4b
2
; min. based on da and sa
controls
(b) ELONGATION OF ROD DUE TO SELF WEIGHT &
ALSO WEIGHT W
Wrod ⫽ 2.482 ⫻ 10⫺3 kips
Amin ⫽
b⫽
␤ ⫽ 0.142
tc ⫽
p
⫽ gb a d2r L r b
4
Wt Lc
Ec da
tc2 ⫺ dctc ⫹ ␤ ⫽ 0
p
W cap ⫽ g s a d2c t s b
4
Wt ⫽ W ⫹ Wcap ⫹ Wrod
WtLc
Ec da
Amin ⫽ 0.447 in2 ⬍ larger than value based on
in3
Apipe ⫽ ␲ tc(dc ⫺ tc)
tc ⫽ 0.021 in
now check allowable shortening requirement
kips
ts ⫽ 1 in.
Wrod
␣ ⫽ 0.128
^ min. based on ␴a
␴a ⫽ 5 ksi ␦a ⫽ 0.02 in.
unit weights (see Table H-1)
Wt
ps a
Wt ⫽ 2.01 kips
dr ⫽
aW +
Wrod
b Lr
2
p
E b a dr 2 b
4
d r ⫽ 0.031 in
(c) MIN. CLEARANCE h
hmin ⫽ ␦a ⫹ ␦r
hmin ⫽ 0.051 in.
;
;
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Page 101
SECTION 2.2
Changes in Lengths of Axially Loaded Members
101
Problem 2.2-12 The horizontal rigid beam ABCD is supported
by vertical bars BE and CF and is loaded by vertical forces
P1 ⫽ 400 kN and P2 ⫽ 360 kN acting at points A and D,
respectively (see figure). Bars BE and CF are made of steel
(E ⫽ 200 GPa) and have cross-sectional areas ABE ⫽ 11,100 mm2
and ACF ⫽ 9,280 mm2. The distances between various points on
the bars are shown in the figure.
Determine the vertical displacements ␦A and ␦D of points A
and D, respectively.
Solution 2.2-12 Rigid beam supported by vertical bars
⌺MB ⫽ 0 哵哴
ABE ⫽ 11,100 mm2
ACF ⫽ 9,280 mm2
E ⫽ 200 GPa
LBE ⫽ 3.0 m
LCF ⫽ 2.4 m
P1 ⫽ 400 kN; P2 ⫽ 360 kN
(400 kN)(1.5 m) ⫹ FCF(1.5 m) ⫺ (360 kN)(3.6 m) ⫽ 0
FCF ⫽ 464 kN
⌺MC ⫽ 0 12
(400 kN)(3.0 m) ⫺ FBE(1.5 m) ⫺ (360 kN)(2.1 m) ⫽ 0
FBE ⫽ 296 kN
SHORTENING OF BAR BE
dBE ⫽
(296 kN)(3.0 m)
FBELBE
⫽
EABE
(200 GPa)(11,100 mm2)
⫽ 0.400 mm
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CHAPTER 2
Page 102
Axially Loaded Members
SHORTENING OF BAR CF
␦BE ⫺ ␦A ⫽ ␦CF ⫺ ␦BE or ␦A ⫽ 2␦BE ⫺ ␦CF
(464 kN)(2.4 m)
FCFLCF
⫽
dCF ⫽
EACF
(200 GPa)(9,280 mm2)
⫽ 0.600 mm
␦A ⫽ 2(0.400 mm) ⫺ 0.600 m
⫽ 0.200 mm ;
(Downward)
DISPLACEMENT DIAGRAM
2.1
(d ⫺ dBE)
1.5 CF
12
7
d D ⫽ dCF ⫺ dBE
5
5
12
7
⫽
(0.600 mm) ⫺ (0.400 mm)
5
5
⫽ 0.880 mm ;
(Downward)
dD ⫺ dCF ⫽
or
bars AB and BC, each having length b (see the first part of
the figure). The bars have pin connections at A, B, and C
and are joined by a spring of stiffness k. The spring is
attached at the midpoints of the bars. The framework has a
pin support at A and a roller support at C, and the bars are
at an angle ␣ to the hoizontal.
When a vertical load P is applied at joint B (see the
second part of the figure) the roller support C moves to the
right, the spring is stretched, and the angle of the bars
decreases from ␣ to the angle ␪.
Determine the angle ␪ and the increase ␦ in the distance
between points A and C. (Use the following data; b ⫽ 8.0
in., k ⫽ 16 lb/in., ␣ ⫽ 45°, and P ⫽ 10 lb.)
B
P
Problem 2.2-13 A framework ABC consists of two rigid
b
—
2
b
—
2
B
b
—
2
k
a
A
b
—
2
a
C
A
u
u
C
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Page 103
SECTION 2.2
Solution 2.2-13
103
Changes in Lengths of Axially Loaded Members
Framework with rigid bars and a spring
h ⫽ height from C to B ⫽ b sin ␪
L2
⫽ bcosu
2
F ⫽ force in spring due to load P
⌺MB ⫽ 0 哵 哴
h
P L2
a b ⫺ Fa b ⫽ 0 or Pcosu ⫽ Fsinu
2 2
2
WITH NO LOAD
DETERMINE THE ANGLE ␪
L2 ⫽ span from A to C
⌬S ⫽ elongation of spring
⫽ 2b cos ␪
S1 ⫽ length of spring
L1
⫽
⫽ bcosa
2
(Eq. 1)
⫽ S2 ⫺ S1 ⫽ b(cos ␪ ⫺ cos ␣)
For the spring: F ⫽ k(⌬S)
F ⫽ bk(cos ␪ ⫺ cos ␣)
Substitute F into Eq. (1):
P cos ␪ ⫽ bk(cos ␪ ⫺ cos ␣)(sin ␪)
or
P
cotu ⫺ cosu + cosa ⫽ 0
bk
;
(Eq. 2)
This equation must be solved numerically for the
angle ␪.
DETERMINE THE DISTANCE ␦
WITH LOAD P
␦ ⫽ L2 ⫺ L1 ⫽ 2b cos ␪ ⫺ 2b cos ␣
⫽ 2b(cos ␪ ⫺ cos ␣)
L1 ⫽ span from A to C
⫽ 2b cos ␣
S2 ⫽ length of spring
⫽
L2
⫽ bcosu
2
FREE-BODY DIAGRAM OF BC
From Eq. (2): cosa ⫽ cosu ⫺
Pcotu
bk
Therefore,
d ⫽ 2b acosu ⫺ cosu +
⫽
2P
cotu
k
Pcotu
b
bk
;
(Eq. 3)
NUMERICAL RESULTS
b ⫽ 8.0 in.
k ⫽ 16 lb/in.
␣ ⫽ 45°
P ⫽ 10 lb
Substitute into Eq. (2):
0.078125 cot ␪ ⫺ cos ␪ ⫹ 0.707107 ⫽ 0
Solve Eq. (4) numerically:
␪ ⫽ 35.1°
;
Substitute into Eq. (3):
␦ ⫽ 1.78 in.
;
(Eq. 4)
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CHAPTER 2
Page 104
Axially Loaded Members
Problem 2.2-14 Solve the preceding problem for the following data:
b ⫽ 200 mm, k ⫽ 3.2 kN/m, ␣ ⫽ 45°, and P ⫽ 50 N.
Solution 2.2-14
Framework with rigid bars and a spring
See the solution to the preceding problem.
Eq. (2):
P
cotu ⫺ cosu + cosa ⫽ 0
bk
Eq. (3):
2P
d⫽
cotu
k
k ⫽ 3.2 kN/m ␣ ⫽ 45°
0.078125 cot ␪ ⫺ cos ␪ ⫹ 0.707107 ⫽ 0
Solve Eq. (4) numerically:
␪ ⫽ 35.1°
;
Substitute into Eq. (3):
NUMERICAL RESULTS
b ⫽ 200 mm
Substitute into Eq. (2):
P ⫽ 50 N
␦ ⫽ 44.5 mm
;
(Eq. 4)
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Page 105
SECTION 2.3
105
Changes in Lengths under Nonuniform Conditions
Changes in Lengths under Nonuniform Conditions
Problem 2.3-1 Calculate the elongation of a copper bar of
solid circular cross section with tapered ends when it is
stretched by axial loads of magnitude 3.0 k (see figure).
The length of the end segments is 20 in. and the length of
the prismatic middle segment is 50 in. Also, the diameters at
cross sections A, B, C, and D are 0.5, 1.0, 1.0, and 0.5 in.,
respectively, and the modulus of elasticity is 18,000 ksi.
(Hint: Use the result of Example 2-4.)
Solution 2.3-1
A
B
C
3.0 k
20 in.
50 in.
Bar with tapered ends
MIDDLE SEGMENT (L 50 in.)
d2 (3.0 k)(50 in.)
PL
EA
(18,000 ksi) A p4 B (1.0 in.)2
0.0106 in.
dA dD 0.5 in.
P 3.0 k
dB dC 1.0 in.
E 18,000 ksi
END SEGMENT (L 20 in.)
ELONGATION OF BAR
d g
From Example 2-4:
d
d1 D
20 in.
4PL
pE dA dB
NL
2d1 + d2
EA
2(0.008488 in.) + (0.01061 in.)
0.0276 in.
;
4(3.0 k)(20 in.)
0.008488 in.
p(18,000 ksi)(0.5 in.)(1.0 in.)
Problem 2.3-2 A long, rectangular copper bar under a tensile load P
hangs from a pin that is supported by two steel posts (see figure). The
copper bar has a length of 2.0 m, a cross-sectional area of 4800 mm2,
and a modulus of elasticity Ec 120 GPa. Each steel post has a height
of 0.5 m, a cross-sectional area of 4500 mm2, and a modulus of elasticity
Es 200 GPa.
Steel
post
(a) Determine the downward displacement of the lower end of the
copper bar due to a load P 180 kN.
(b) What is the maximum permissible load Pmax if the displacement
is limited to 1.0 mm?
Copper
bar
P
3.0 k
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CHAPTER 2
Page 106
Axially Loaded Members
Solution 2.3-2
Copper bar with a tensile load
(a) DOWNWARD DISPLACEMENT (P 180 kN)
dc (180 kN)(2.0 m)
PLc
Ec Ac
(120 GPa)(4800 mm2)
0.625 mm
ds (P/2)Ls
(90 kN)(0.5 m)
EsAs
(200 GPa)(4500 mm2)
0.050 mm
d dc + ds 0.625 mm + 0.050 mm
0.675 mm
Lc 2.0 m
Ac 4800 mm2
;
(b) MAXIMUM LOAD Pmax (max 1.0 mm)
Ec 120 GPa
dmax
Pmax
P
d
Ls 0.5 m
As 4500 mm2
Pmax Pa
Pmax (180 kN)a
Es 200 GPa
dmax
b
d
1.0 mm
b 267 kN
0.675 mm
;
Problem 2.3-3 A steel bar AD (see figure) has a cross-sectional
area of 0.40 in.2 and is loaded by forces P1 2700 lb, P2 1800 lb,
and P3 1300 lb. The lengths of the segments of the bar are
a 60 in., b 24 in., and c 36 in.
P1
(a) Assuming that the modulus of elasticity E 30 10 psi,
calculate the change in length of the bar. Does the bar
elongate or shorten?
(b) By what amount P should the load P3 be increased so that the
bar does not change in length when the three loads are applied?
6
Solution 2.3-3
A
C
B
a
P2
b
D
c
Steel bar loaded by three forces
A 0.40 in.2
P1 2700 lb
P2 1800 lb
P3 1300 lb
E 30 106 psi
AXIAL FORCES
NAB P1 P2 P3 3200 lb
NBC P2 P3 500 lb
NCD P3 1300 lb
(a) CHANGE IN LENGTH
d g
NiLi
EiAi
1
(N L + NBCLBC + NCDLCD)
EA AB AB
P3
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Page 107
SECTION 2.3
1
6
2
The force P must produce a shortening equal to 0.0131 in.
in order to have no change in length.
[(3200 lb) (60 in.)
(30 * 10 psi)(0.40 in. )
+ (500 lb)(24 in.) (1300 lb) (36 in.)]
0.0131 in. (elongation)
107
Changes in Lengths under Nonuniform Conditions
‹ 0.0131 in. d ;
(b) INCREASE IN P3 FOR NO CHANGE IN LENGTH
PL
EA
P(120 in.)
(30 * 106 psi)(0.40 in.2)
P 1310 lb
;
P increase in force P3
Problem 2.3-4 A rectangular bar of length L has a slot in
the middle half of its length (see figure). The bar has width
b, thickness t, and modulus of elasticity E. The slot has
width b/4.
(a) Obtain a formula for the elongation of the bar due
to the axial loads P.
(b) Calculate the elongation of the bar if the material is
high-strength steel, the axial stress in the middle
region is 160 MPa, the length is 750 mm, and the
modulus of elasticity is 210 GPa.
Solution 2.3-4
b
—
4
P
t
b
L
—
4
P
L
—
2
L
—
4
Bar with a slot
STRESS IN MIDDLE REGION
s
P
A
P
4P
3bt
3
a btb
4
or
P
3s
bt
4
Substitute into the equation for :
d
t thickness L length of bar
(a) ELONGATION OF BAR
d g
P(L/4)
P(L/2)
P(L/4)
NiLi
+
+
3
EAi
E(bt)
E(bt)
E A 4 bt B
PL 1
4
1
7PL
a + + b
Ebt 4
6
4
6Ebt
;
7L P
7L 3s
7PL
a b a b
6Ebt
6E bt
6E 4
7sL
8E
(b) SUBSTITUTE NUMERICAL VALUES:
s 160 MPa L 750 mm E 210 GPa
d
7(160 MPa)(750 mm)
0.500 mm
8(210 GPa)
;
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CHAPTER 2
Page 108
Axially Loaded Members
Problem 2.3-5 Solve the preceding problem if the axial
stress in the middle region is 24,000 psi, the length is 30 in.,
and the modulus of elasticity is 30 106 psi.
b
—
4
P
L
—
4
Solution 2.3-5
t
b
P
L
—
2
L
—
4
Bar with a slot
STRESS IN MIDDLE REGION
P
A
s
P
4P
3bt
3
a btb
4
or
3s
P
bt
4
SUBSTITUTE INTO THE EQUATION FOR :
t thickness
L length of bar
d
(a) ELONGATION OF BAR
P(L/4)
P(L/4)
P(L/2)
NiLi
+
d g
+
3
EAi
E(bt)
E(bt)
E(4bt)
4
1
7PL
PL 1
a + + b
Ebt 4
6
4
6Ebt
;
7PL
7L P
7L 3s
a b a b
6Ebt
6E bt
6E 4
7sL
8E
(B) SUBSTITUTE NUMERICAL VALUES:
s 24,000 psi L 30 in.
E 30 * 106 psi
d
Problem 2.3-6 A two-story building has steel columns AB in the first floor
and BC in the second floor, as shown in the figure. The roof load P1 equals
400 kN and the second-floor load P2 equals 720 kN. Each column has length
L 3.75 m. The cross-sectional areas of the first- and second-floor columns
are 11,000 mm2 and 3,900 mm2, respectively.
(a) Assuming that E 206 GPa, determine the total shortening AC
of the two columns due to the combined action of the loads P1 and P2.
(b) How much additional load P0 can be placed at the top of the column
(point C) if the total shortening AC is not to exceed 4.0 mm?
7(24,000 psi)(30 in.)
8(30 * 106 psi)
P1 = 400 kN
0.0210 in.
;
C
L = 3.75 m
P2 = 720 kN
B
L = 3.75 m
A
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Page 109
SECTION 2.3
Solution 2.3-6
Changes in Lengths under Nonuniform Conditions
109
Steel columns in a building
(b) ADDITIONAL LOAD P0 AT POINT C
(dAC)max 4.0 mm
d0 additional shortening of the two columns
due to the load P0
d0 (dAC)max dAC 4.0 mm 3.7206 mm
0.2794 mm
Also, d0 Solve for P0:
(a) SHORTENING AC OF THE TWO COLUMNS
P0 NiLi
NABL
NBCL
dAC g
+
EiAi
EAAB
EABC
Ed0 AAB ABC
b
a
L AAB + ABC
SUBSTITUTE NUMERICAL VALUES:
(1120 kN)(3.75 m)
E 206 109 N/m2
(206 GPa)(11,000 mm2)
L 3.75 m
(400 kN)(3.75 m)
+
P0L
P0L
P0L 1
1
+
+
b
a
EAAB
EABC
E AAB
ABC
0 0.2794 103 m
AAB 11,000 106 m2
ABC 3,900 106 m2
2
(206 GPa)(3,900 mm )
P0 44,200 N 44.2 kN
1.8535 mm + 1.8671 mm 3.7206 mm
dAC 3.72 mm
;
;
Problem 2.3-7 A steel bar 8.0 ft long has a circular cross section
of diameter d1 0.75 in. over one-half of its length and diameter
d2 0.5 in. over the other half (see figure). The modulus of
elasticity E 30 106 psi.
(a) How much will the bar elongate under a tensile load
P 5000 lb?
(b) If the same volume of material is made into a bar of
constant diameter d and length 8.0 ft, what will be the
elongation under the same load P?
Solution 2.3-7
d1 = 0.75 in.
d2 = 0.50 in.
P = 5000 lb
P
4.0 ft
4.0 ft
Bar in tension
(a) ELONGATION OF NONPRISMATIC BAR
d g
P 5000 lb
E 30 106 psi
L 4 ft 48 in.
d
NiLi
PL 1
g
Ei Ai
E Ai
(5000 lb)(48 in.)
30 * 106 psi
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CHAPTER 2
Page 110
Axially Loaded Members
J 4 (0.75 in)
1
*
p
2
0.0589 in.
1
+
p
4 (0.50
;
in.)2 K
Ap d
(b) ELONGATION OF PRISMATIC BAR OF SAME VOLUME
Original bar: Vo A1L A2L L(A1 A2)
p
[(0.75 in.)2 + (0.50 in.)2] 0.3191 in.2
8
(5000 lb)(2)(48 in.)
P(2L)
EAp
(30 * 106 psi)(0.3191 in.2)
0.0501 in.
Prismatic bar: Vp Ap(2L)
L(A1 A2) Ap(2L)
Problem 2.3-8 A bar ABC of length L consists of two
parts of equal lengths but different diameters. Segment AB
has diameter d1 100 mm, and segment BC has diameter
d2 60 mm. Both segments have length L/2 0.6 m.
A longitudinal hole of diameter d is drilled through segment
AB for one-half of its length (distance L/4 0.3 m). The bar
is made of plastic having modulus of elasticity E 4.0 GPa.
Compressive loads P 110 kN act at the ends of the bar.
dmax
A
B
d2
C
d1
P
L
—
4
(a) If the shortening of the bar is limited to 8.0 mm, what
is the maximum allowable diameter dmax of the hole?
(See figure part a.)
(b) Now, if dmax is instead set at d2/2, at what distance b
from end C should load P be applied to limit the bar
shortening to 8.0 mm? (See figure part b.)
(c) Finally, if loads P are applied at the ends and
dmax d2/2, what is the permissible length x of the
hole if shortening is to be limited to 8.0 mm? (See
figure part c.)
;
NOTE: A prismatic bar of the same volume will always
have a smaller change in length than will a nonprismatic
bar, provided the constant axial load P, modulus E, and
total length L are the same.
Equate volumes and solve for Ap:
Vo Vp
A1 + A2
1 p
a b(d21 + d22)
2
2 4
P
L
—
4
L
—
2
(a)
d
dmax = —2
2
A
B
P
d2
C
d1
P
L
—
4
L
—
4
L
—
2
b
(b)
d
dmax = —2
2
A
B
d2
C
d1
P
x
P
L
—
2
L
—
x
2
(c)
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Page 111
SECTION 2.3
Changes in Lengths under Nonuniform Conditions
Solution 2.3-8
b c
NUMERICAL DATA
d1 100 mm
d2 60 mm
L 1200 mm
E 4.0 GPa
P 110 kN
(a) find dmax if shortening is limited to a
d
p 2
d
4 1
A2 p 2
d
4 2
L
4
P
≥
E p
1d 2 dmax 22
4 1
L
L
4
2
+
¥
+
A1
A2
Eda p d1 2 d2 2 2PLd2 2 2PLd1 2
9 Edapd1 2d2 2 PLd2 2 2PLd1 2
dmax 23.9 mm
;
(b) Now, if dmax is instead set at d2 2, at what distance
b from end C should load P be applied to limit the bar
shortening to a 8.0 mm?
d2 2
p 2
c d1 a b d
4
2
p 2
p
A1 d1
A2 d2 2
4
4
A0 L
P L
+
+
d J
E 4A0
4A1
a
d
L
bb
2
A2
;
K
no axial force in segment at end of length b; set a &
solve for b
P x
J
+
E A0
a
a
L
xb
2
L
b
2
+
A1
A2
set a & solve for x
x
set to a and solve for dmax
dmax d1
b 4.16 mm
(c) Finally if loads P are applied at the ends and
dmax d2 2, what is the permissible length x
of the hole if shortening is to be limited to
a 8.0 mm?
a 8.0 mm
A1 Eda
L
L
L
+
bdd
A2 c
a
2
P
4A0
4A1
c A0 A1a
Ed a
L
1
b d A0 L
P
2 A2
2
x 183.3 mm
A1 A0
;
K
111
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CHAPTER 2
Page 112
Axially Loaded Members
Problem 2.3-9 A wood pile, driven into the earth, supports a load P entirely
P
by friction along its sides (see figure). The friction force f per unit length of pile
is assumed to be uniformly distributed over the surface of the pile. The pile has
length L, cross-sectional area A, and modulus of elasticity E.
(a) Derive a formula for the shortening of the pile in terms of P, L, E, and A.
(b) Draw a diagram showing how the compressive stress c varies throughout
the length of the pile.
Solution 2.3-9
Wood pile with friction
FROM FREE-BODY DIAGRAM OF PILE:
Fvert 0 *uarr* *darr* fL P 0 f P
(Eq. 1)
L
(a) SHORTENING OF PILE:
At distance y from the base:
N(y) axial force N(y) fy
dd f
N(y)dy
fy dy
EA
EA
f L
fL2
PL
L
ydy
d 10 dd 1
0
EA
2EA
2EA
d
PL
2EA
(b) COMPRESSIVE STRESS c IN PILE
sc (Eq. 2)
;
N(y)
fy
Py
A
A
AL
;
At the base (y 0): c 0
At the top(y L): sc See the diagram above.
P
A
L
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SECTION 2.3
Changes in Lengths under Nonuniform Conditions
113
Problem 2.3-10 Consider the copper tubes joined below using a “sweated” joint. Use the properties and dimensions
given.
(a) Find the total elongation of segment 2-3-4 (2-4) for an applied tensile force of P 5 kN. Use Ec 120 GPa.
(b) If the yield strength in shear of the tin-lead solder is y 30 MPa and the tensile yield strength of the copper is
y 200 MPa, what is the maximum load Pmax that can be applied to the joint if the desired factor of safety in
shear is FS 2 and in tension is FS 1.7?
(c) Find the value of L2 at which tube and solder capacities are equal.
Sweated
joint
P
Segment number
Solder joints
1
2
3
4
L2
L3
L4
5
P
d0 = 18.9 mm
t = 1.25 mm
d0 = 22.2 mm
t = 1.65 mm
L3 = 40 mm
L2 = L4 = 18 mm
Tin-lead solder in space
between copper tubes;
assume thickness of
solder equal zero
Solution 2.3-10
NUMERICAL DATA
P 5 kN
Ec 120 GPa
L2 18 mm
L4 L2
L3 40 mm
do3 22.2 mm
t3 1.65 mm
do5 18.9 mm
t5 1.25 mm
Y 200 MPa
Y 30 MPa
FS 2
tY
ta FSt
sa sY
FSs
FS 1.7
a
15 MPa
(a) ELONGATION OF SEGMENT 2-3-4
p
A2 [d2o3 (d o5 2 t5)2]
4
p 2
A3 [d o3 1d o3 2t322]
4
A2 175.835 mm2 A3 106.524 mm2
d24 L3
P L2 + L4
a
+
b
Ec
A2
A3
d24 0.024 mm
(b) MAXIMUM
LOAD
;
Pmax
THAT CAN BE APPLIED TO THE
JOINT
a 117.6 MPa
FIRST CHECK NORMAL STRESS
A1 p 2
[ d o5 1 d o5 2 t522]
4
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CHAPTER 2
Page 114
Axially Loaded Members
A1 69.311 mm2
smallest cross-sectional area
controls normal stress
Pmax aA1 Pmaxs 8.15 kN
; smaller than
Pmax based on shear below so normal stress controls
Pmax taAsh
(c) FIND THE VALUE OF L2 AT WHICH TUBE AND SOLDER
CAPACITIES ARE EQUAL
set Pmax based on shear strength equal to Pmax based
on tensile strength & solve for L2
next check shear stress in solder joint
Ash do5L2
Pmaxt 16.03 kN
Ash 1.069 103 mm2
L2 s aA1
ta1pd o52
L2 9.16 mm
Segment 1
;
Segment 2
Problem 2.3-11 The nonprismatic cantilever circular bar shown
has an internal cylindrical hole of diameter d/2 from 0 to x, so
the net area of the cross section for Segment 1 is (3/4)A. Load
P is applied at x, and load P/2 is applied at x L. Assume that
E is constant.
(a) Find reaction force R1.
(b) Find internal axial forces Ni in segments 1 and 2.
(c) Find x required to obtain axial displacement at joint 3 of
3 PL/EA.
(d) In (c), what is the displacement at joint 2, 2?
(e) If P acts at x 2L/3 and P/2 at joint 3 is replaced by P,
find so that 3 PL/EA.
(f) Draw the axial force (AFD: N(x), 0 x L) and axial
displacement (ADD: (x), 0 x L) diagrams using
results from (b) through (d) above.
3
—A
4
d
R1
A
d
—
2
x
3
2
L–x
3P
—
2
P
—
2
0
AFD 0
δ3
δ2
ADD 0
0
Solution 2.3-11
(a) STATICS a FH 0
R1 P R1 3
P
2
P
2
;
(b) DRAW FBD’S CUTTING THROUGH SEGMENT 1 & AGAIN
THROUGH SEGMENT 2
3P
P
N1 6 tension N2 6 tension
2
2
(c) FIND x REQUIRED TO OBTAIN AXIAL DISPLACEMENT AT
JOINT 3 OF 3 PL/EA
add axial deformations of segments 1 & 2 then set
to 3; solve for x
N2( L x)
N1x
PL
+
EA
EA
3
E A
4
P
—
2
P
3P
P
x
( L x)
2
2
PL
+
3
EA
EA
E A
4
3
L
L
x
x
;
2
2
3
(d) WHAT IS THE DISPLACEMENT AT JOINT 2, 2?
d2 N1x
3
E A
4
d2 2 PL
3 EA
d2 a
3P L
b
2 3
3
E A
4
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Page 115
SECTION 2.3
(e) IF x 2L/3 AND P/2 AT JOINT 3 IS REPLACED BY P,
FIND
SO THAT 3 PL/EA
2L
3
substitute in axial deformation expression above &
solve for
N1 (1 )P
[(1 + b)P]
2L
3
N2 P
bPaL x
2L
b
3
+
3
E A
4
EA
PL
EA
115
Changes in Lengths under Nonuniform Conditions
1 8 + 11b
PL
PL
9
EA
EA
(8 11 ) 9
1
;
11
0.091
b
(f) Draw AFD, ADD - see plots above for x L
3
Problem 2.3-12 A prismatic bar AB of length L, cross-sectional area A, modulus of elasticity E,
A
and weight W hangs vertically under its own weight (see figure).
(a) Derive a formula for the downward displacement C of point C, located at distance h from
the lower end of the bar.
C
(b) What is the elongation B of the entire bar?
(c) What is the ratio
half of the bar?
of the elongation of the upper half of the bar to the elongation of the lower
h
B
Solution 2.3-12 Prismatic bar hanging vertically
W Weight of bar
A
dy
C
y
L
(a) DOWNWARD DISPLACEMENT C
Consider an element at distance y from the lower end.
B
Wy
N(y)dy
Wydy
N(y) dd L
EA
EAL
W
L
L Wydy
dC 1h dd 1h
(L2 h2)
EAL
2EAL
W
(L2 h2)
2EAL
dB WL
2EA
;
(c) RATIO OF ELONGATIONS
Elongation of upper half of bar a h h
dC (b) ELONGATION OF BAR (h 0)
;
L
b:
2
3WL
8EA
Elongation of lower half of bar:
d upper d lower dB d upper b
d upper
d lower
3/8
3
1/8
WL
3WL
WL
2EA
8EA
8EA
;
L
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CHAPTER 2
Page 116
Axially Loaded Members
Problem 2.3-13 A flat bar of rectangular cross section,
b2
length L, and constant thickness t is subjected to tension by
forces P (see figure). The width of the bar varies linearly
from b1 at the smaller end to b2 at the larger end. Assume
that the angle of taper is small.
t
(a) Derive the following formula for the elongation of
the bar:
d
b2
PL
ln
Et(b2 b1)
b1
P
b1
L
P
(b) Calculate the elongation, assuming L 5 ft, t 1.0
in., P 25 k, b1 4.0 in., b2 6.0 in., and E 30
106 psi.
Solution 2.3-13 Tapered bar (rectangular cross section)
t thickness (constant)
L0 + L
x
b b1 a b b2 b1 a
b
L0
L0
x
A(x) bt b1 t a b
L0
From Eq. (1):
(Eq. 1)
Solve Eq. (3) for L0: L0 La
PL0 dx
Pdx
EA(x)
Eb1 tx
L0L
d
LL0
d
b1
b
b2 b1
b2
PL
ln
Et (b2 b1) b1
(Eq. 4)
(Eq. 5)
(b) SUBSTITUTE NUMERICAL VALUES:
PL0 L0L dx
dd Eb1 t LL0
x
L0L
PL0
L0 + L
PL0
ln x `
ln
Eb1 t
Eb1 t
L0
L0
(Eq. 3)
Substitute Eqs. (3) and (4) into Eq. (2):
(a) ELONGATION OF THE BAR
dd L0 + L
b2
L0
b1
L 5 ft 60 in.
(Eq. 2)
t 10 in.
P 25 k
b1 4.0 in.
b2 6.0 in.
E 30 106 psi
From Eq. (5): 0.010 in.
;
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Page 117
SECTION 2.3
Changes in Lengths under Nonuniform Conditions
Problem 2.3-14 A post AB supporting equipment in a laboratory
is tapered uniformly throughout its height H (see figure). The cross
sections of the post are square, with dimensions b b at the top
and 1.5b 1.5b at the base.
Derive a formula for the shortening of the post due to the
compressive load P acting at the top. (Assume that the angle of
taper is small and disregard the weight of the post itself.)
117
P
A
A
b
b
H
B
B
1.5b
Solution 2.3-14
Tapered post
Ay cross sectional area at distance y
1by22 b2
H2
(H + 0.5y)2
SHORTENING OF ELEMENT dy
dd Pdy
EAy
Pdy
2
Ea
b
H2
b 1H + 0.5y22
SHORTENING OF ENTIRE POST
d
Square cross sections
b width at A
1.5b width at B
by width at distance y
y
b + (1.5b b)
H
b
1H + 0.5y2
H
L
dd PH2
dy
Eb2 L0 (H + 0.5y)2
From Appendix C:
d
H
dx
L (a + bx)2
PH2
H
1
c
d
(0.5)(H + 0.5y) 0
Eb2
PH2
2
c
Eb
2PH
3Eb2
1
1
+
d
(0.5)(1.5H )
0.5H
;
1
b(a + bx)
1.5b
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CHAPTER 2
Page 118
Axially Loaded Members
Problem 2.3-15 A long, slender bar in the shape of a right circular cone
d
with length L and base diameter d hangs vertically under the action of its
own weight (see figure). The weight of the cone is W and the modulus of
elasticity of the material is E.
Derive a formula for the increase in the length of the bar due to
its own weight. (Assume that the angle of taper of the cone is small.)
L
Solution 2.3-15
Conical bar hanging vertically
ELEMENT OF BAR
W weight of cone
TERMINOLOGY
ELONGATION OF ELEMENT dy
Ny dy
Wy dy
4W
y dy
dd E Ay
E ABL
pd2 EL
Ny axial force acting on element dy
ELONGATION OF CONICAL BAR
Ay cross-sectional area at element dy
AB cross-sectional area at base of cone
pd2
4
1
ABL
3
d
L
dd 4W
pd2 EL L0
L
y dy 2WL
;
pd2 E
V volume of cone
Vy volume of cone below element dy
1
Ay y Wy weight of cone below element dy
3
Ay yW
Vy
Ny Wy
(W ) V
AB L
x
P
Problem 2.3-16 A uniformly tapered plastic tube AB of circular
dA
B
A
P
L
cross section and length L is shown in the figure. The average diameters at the ends are dA and dB 2dA. Assume E is constant. Find the
elongation of the tube when it is subjected to loads P acting at the
ends. Use the following numerial data: dA 35 mm, L 300 mm,
E 2.1 GPa, P 25 kN. Consider two cases as follows:
(a) A hole of constant diameter dA is drilled from B toward A to
form a hollow section of length x L/2 (see figure part a).
dA
dB
(a)
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Page 119
SECTION 2.3
119
Changes in Lengths under Nonuniform Conditions
(b) A hole of variable diameter d(x) is drilled from
B toward A to form a hollow section of length x L/2
and constant thickness t (see figure part b). (Assume
that t dA/20.)
x
P
B
A
P
dA
L
d(x)
t constant
dB
(b)
Solution 2.3-16
(a)
ELONGATION
FOR CASE OF CONSTANT DIAMETER HOLE
d() dA a1 +
d
1
P
a
d b
E L A()
P
d
E
d
d
b
L
J
p
d()2
solid portion of length L-x
4
p
hollow portion of length x
A() (d()2 dA 2)
4
A() d
P
c
E L0
2
p
c cdA a1 + b d d
4
L
L0
P
L2
4
+
E
( 2 x)pdA 2
J
L
4
pd()2
d +
JJ
4
d +
L
pdA2
LLx
L
+
LLx
4
LLx p1 d()2 d A 22
L
1
Lx
Lx
d d
1
c
2
p
c cdA a 1 + b d dA 2 d d
4
L
d
1
d
2
p
c c cdA a 1 + b d dA 2 d d K K K
4
L
ln(Lx) + ln(3Lx
ln(3)
L
L2
P
a4
2L
+ 2L
bd
c4
2 +
2
2
E ( 2 x)pdA
pdA
pdA
pdA 2
if x L/2
d
ln(3)
P 4 L
±
2L 2 2L
E 3 pd2A
pdA
Substitute numerical data
d 2.18 mm
;
K
5
1
lna Lb + lna Lb
2
2
p d2A
≤
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11:37 AM
CHAPTER 2
(b)
Axially Loaded Members
ELONGATION
FOR CASE OF VARIABLE DIAMETER HOLE BUT CONSTANT WALL THICKNESS t dA/20 OVER SEGMENT x
d() dA a1 +
d
d
d
Page 120
b
L
P
1
a
d b
E L A()
P
≥
E L0
Lx
A() p
d()2
4
A() dA 2
p
cd()2 a d() 2 b d
4
20
d
Lx
L
bd
d +
hollow portion of length x
L
4
pd()2
d +
LLx
L
4
pcdA a1 +
P
≥
E L0
solid portion of length L-x
4
dA 2
pc d() a d() 2
b d
20
2
4
LLx
pc c dA a 1 +
dA 2
b d cdA a 1 + b 2
d d
L
L
20
2
ln(3) + ln(13) + 2ln( dA) + ln( L)
L2
P
L
20L
c4
+ 4
2
2
E ( 2L + x)pdA
pdA
pdA 2
20L
2ln( dA) + ln (39L 20x)
pdA 2
d
if x L/2
d
ln(3) + ln(13) + 2ln( dA) + ln( L)
2ln( dA) + ln(29L)
P 4 L
+ 20L
b
a
20L
2
2
E 3 pdA
pdA
pdA 2
Substitute numerical data
d 6.74 mm
;
d¥
d ¥
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SECTION 2.3
121
Changes in Lengths under Nonuniform Conditions
Problem 2.3-17 The main cables of a suspension bridge
[see part (a) of the figure] follow a curve that is nearly parabolic
because the primary load on the cables is the weight of the
bridge deck, which is uniform in intensity along the horizontal.
Therefore, let us represent the central region AOB of one of the
main cables [see part (b) of the figure] as a parabolic cable supported at points A and B and carrying a uniform load of intensity
q along the horizontal. The span of the cable is L, the sag is h,
the axial rigidity is EA, and the origin of coordinates is at
midspan.
(a)
y
A
(a) Derive the following formula for the elongation of cable
AOB shown in part (b) of the figure:
d
L
—
2
L
—
2
B
h
qL3
16h2
(1 +
)
8hEA
3L2
O
q
(b) Calculate the elongation of the central span of one of
the main cables of the Golden Gate Bridge, for which the
dimensions and properties are L 4200 ft, h 470 ft,
q 12,700 lb/ft, and E 28,800,000 psi. The cable
consists of 27,572 parallel wires of diameter 0.196 in.
x
(b)
Hint: Determine the tensile force T at any point in the cable from a free-body diagram of part of the cable; then determine
the elongation of an element of the cable of length ds; finally, integrate along the curve of the cable to obtain an equation for
the elongation .
Solution 2.3-17
Cable of a suspension bridge
dy
8hx
2
dx
L
FREE-BODY DIAGRAM OF HALF OF CABLE
MB 0 哵哴
Hh +
H
qL L
a b 0
2 4
qL2
8h
Fhorizontal 0
HB H qL2
8h
(Eq. 1)
Fvertical 0
VB Equation of parabolic curve:
y
4hx2
L2
qL
2
(Eq. 2)
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CHAPTER 2
Page 122
Axially Loaded Members
FREE-BODY DIAGRAM OF SEGMENT DB OF CABLE
dd Tds
EA
ds 2(dx)2 + (dy)2 dx 1 + a
A
dx 1 + a
A
dx 1 +
8hx
L2
b
dy 2
b
dx
2
64h2x2
A
(Eq. 6)
L4
(a) ELONGATION OF CABLE AOB
d
©F horiz 0
TH HB
©F vert 0 VB Tv q a
Tv VB q a
qL2
8h
(Eq. 3)
L
xb 0
2
qL2
1
64h 2x 2
a1 +
b dx
EA L 8h
L4
For both halves of cable:
(Eq. 4)
qL2 2
b + (qx)2
A 8h
a
qL2
64h2x2
1 +
8h A
L4
ELONGATION d OF AN ELEMENT OF LENGTH ds
T ds
L EA
Substitute for T from Eq. (5) and for ds from
Eq. (6):
(Eq. 5)
L/2
qL2
64h2x2
a1 +
b dx
8h
L4
d
2
EA L0
d
qL3
16h2
a1 +
b
8hEA
3L4
TENSILE FORCE T IN CABLE
T 2T2H + T2v dd d
qL
qL
L
xb + qx
2
2
2
qx
L
;
(b) GOLDEN GATE BRIDGE CABLE
L 4200 ft
h 470 ft
q 12,700 lb/ft E 28,800,000 psi
27,572 wires of diameter d 0.196 in.
p
A (27,572)a b(0.196 in.)2 831.90 in.2
4
Substitute into Eq. (7):
133.7 in 11.14 ft
;
(Eq. 7)
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Page 123
SECTION 2.3
123
Changes in Lengths under Nonuniform Conditions
Problem 2.3-18 A bar ABC revolves in a horizontal plane about a
A
vertical axis at the midpoint C (see figure). The bar, which has length
2L and cross-sectional area A, revolves at constant angular speed .
Each half of the bar (AC and BC) has weight W1 and supports a
weight W2 at its end.
Derive the following formula for the elongation of one-half of the
bar (that is, the elongation of either AC or BC):
C
v
W1
W2
B
W1
L
W2
L
L22
(W + 3W2)
3gEA 1
in which E is the modulus of elasticity of the material of the bar
and g is the acceleration of gravity.
d
Solution 2.3-18 Rotating bar
Centrifugal force produced by weight W2
a
W2
b(L2)
g
AXIAL FORCE F(x)
F(x) angular speed
A cross-sectional area
E modulus of elasticity
g acceleration of gravity
F(x) axial force in bar at distance x from point C
Consider an element of length dx at distance x from
point C.
To find the force F(x) acting on this element, we must
find the inertia force of the part of the bar from distance
x to distance L, plus the inertia force of the weight W2.
Since the inertia force varies with distance from point C,
we now must consider an element of length d at distance , where varies from x to L.
d W1
b
Mass of element d a
L g
Acceleration of element 2
Centrifugal force produced by element
( mass)( acceleration) W12
d
gL
L
Lx
W12
W2L2
d +
gL
g
W12 2
W2L2
(L x2) +
2gL
g
ELONGATION OF BAR BC
L
F(x) dx
L0 EA
L
L
W12 2
W2L2dx
(L x2)dx +
gEA
L0
L0 2gL
L
L
2
W2L2dx L
W1L
2
2
c
L dx x dx d +
dx
2gLEA L0
gEA L0
L0
d
W2L22
W1L22
+
3gEA
gEA
L22
+ (W1 + 3W2)
3gEA
;
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CHAPTER 2
Page 124
Axially Loaded Members
Statically Indeterminate Structures
Problem 2.4-1 The assembly shown in the figure consists of a brass
core (diameter d1 0.25 in.) surrounded by a steel shell (inner diameter
d2 0.28 in., outer diameter d3 0.35 in.). A load P compresses the
core and shell, which have length L 4.0 in. The moduli of elasticity of
the brass and steel are Eb 15 106 psi and Es 30 106 psi,
respectively.
(a) What load P will compress the assembly by 0.003 in.?
(b) If the allowable stress in the steel is 22 ksi and the allowable stress
in the brass is 16 ksi, what is the allowable compressive load
Pallow? (Suggestion: Use the equations derived in Example 2-5.)
Solution 2.4-1
Cylindrical assembly in compression
Substitute numerical values:
Es As + Eb Ab (30 * 106 psi)(0.03464 in.2)
+ (15 * 106 psi)(0.04909 in.2)
1.776 * 106 lb
P (1.776 * 106 lb)a
1330 lb
0.003 in.
b
4.0 in.
;
(b) ALLOWABLE LOAD
d1 0.25 in.
Eb 15 106 psi
d2 0.28 in.
Es 30 106 psi
d3 0.35 in.
As L 4.0 in.
p 2
(d3 d22) 0.03464 in.2
4
p
Ab d21 0.04909 in.2
4
(a) DECREASE IN LENGTH ( 0.003 in.)
Use Eq. (2-13) of Example 2-5.
d
PL
Es As + Eb Ab
or
d
P (Es As + Es Ab)a b
L
s 22 ksi b 16 ksi
Use Eqs. (2-12a and b) of Example 2-5.
For steel:
ss PEs
Es As + Eb Ab
Ps (Es As + Eb Ab)
Ps (1.776 * 106 lb)a
22 ksi
30 * 106 psi
ss
Es
b 1300 lb
For brass:
sb PEb
Es As + Eb Ab
Ps (Es As + Eb Ab)
Ps (1.776 * 106 lb)a
Steel governs.
16 ksi
15 * 106 psi
Pallow 1300 lb
sb
Eb
b 1890 lb
;
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Page 125
SECTION 2.4
125
Statically Indeterminate Structures
Problem 2.4-2 A cylindrical assembly consisting of a brass core and
an aluminum collar is compressed by a load P (see figure). The length
of the aluminum collar and brass core is 350 mm, the diameter of the
core is 25 mm, and the outside diameter of the collar is 40 mm. Also, the
moduli of elasticity of the aluminum and brass are 72 GPa and 100 GPa,
respectively.
(a) If the length of the assembly decreases by 0.1% when the load
P is applied, what is the magnitude of the load?
(b) What is the maximum permissible load Pmax if the allowable
stresses in the aluminum and brass are 80 MPa and 120 MPa,
respectively? (Suggestion: Use the equations derived in
Example 2-5.)
Solution 2.4-2
Cylindrical assembly in compression
d
PL
Ea Aa + Eb Ab
or
d
P (Ea Aa + Eb Ab)a b
L
Substitute numerical values:
Ea Aa + Eb Ab (72 GPa)(765.8 mm2)
(100 GPa)(490.9 mm2)
55.135 MN + 49.090 MN
104.23 MN
P (104.23 MN)a
104.2 kN
A aluminum
0.350 mm
b
350 mm
;
B brass
(b) ALLOWABLE LOAD
L 350 mm
a 80 MPa b 120 MPa
da 40 mm
Use Eqs. (2-12a and b) of Example 2-5.
db 25 mm
For aluminum:
p
Aa (d2a d2b)
4
sa 765.8 mm2
Ea 72 GPa
Eb 100 GPa
490.9 mm2
p
Ab d2b
4
(a) DECREASE IN LENGTH
( 0.1% of L 0.350 mm)
Use Eq. (2-13) of Example 2-5.
PEa
Ea Aa + Eb Ab
Pa (104.23 MN)a
Pa (Ea Aa + Eb Ab) a
sa
b
Ea
80 MPa
b 115.8 kN
72 GPa
For brass:
sb PEb
Ea Aa + Eb Ab
Pb (104.23 MN)a
Pb (Ea Aa + Eb Ab) a
120 MPa
b 125.1 kN
100 GPa
Aluminum governs. Pmax 116 kN
;
sb
b
Eb
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CHAPTER 2
Page 126
Axially Loaded Members
Problem 2.4-3 Three prismatic bars, two of material A and one of material B,
transmit a tensile load P (see figure). The two outer bars (material A) are identical.
The cross-sectional area of the middle bar (material B) is 50% larger than the
cross-sectional area of one of the outer bars. Also, the modulus of elasticity of
material A is twice that of material B.
(a) What fraction of the load P is transmitted by the middle bar?
(b) What is the ratio of the stress in the middle bar to the stress in the outer bars?
(c) What is the ratio of the strain in the middle bar to the strain in the outer bars?
Solution 2.4-3
Prismatic bars in tension
FREE-BODY DIAGRAM OF END PLATE
STRESSES:
EQUATION OF EQUILIBRIUM
Fhoriz 0
PA PB P 0
PA
EA P
AA
EA AA + EB AB
sB PB
EB P
AB
EA AA + EB AB
(a) LOAD IN MIDDLE BAR
(2)
PB
EB AB
1
EA AA
P
EA AA + EB AB
+ 1
EB AB
Given:
FORCE-DISPLACEMENT RELATIONS
AA total area of both outer bars
‹
PA L
dA EA Ak
PB L
dB EB AB
(3)
Substitute into Eq. (2):
EA
2
EB
PB
P
AA
4
1 + 1
AB
1.5
3
1
3
1
8
11
EA AA
+ 1
a ba b + 1
3
EB AB
(b) RATIO OF STRESSES
PA L
PB L
EA AA
EB AB
(4)
sB
EB
1
sA
EA
2
;
SOLUTION OF THE EQUATIONS
(c) RATIO OF STRAINS
Solve simultaneously Eqs. (1) and (4):
EA AAP
PA EA AA + EB AB
EB AB P
PB EA AA + EB AB
All bars have the same strain
(5)
Substitute into Eq. (3):
d dA dB PL
EA AA + EB AB
(7)
(1)
EQUATION OF COMPATIBILITY
A B
sA (6)
Ratio 1
;
;
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Page 127
SECTION 2.4
Problem 2.4-4 A circular bar ACB of diameter d having a
cylindrical hole of length x and diameter d/2 from A to C is held
between rigid supports at A and B. A load P acts at L/2 from
ends A and B. Assume E is constant.
(a) Obtain formulas for the reactions RA and RB at supports
A and B, respectively, due to the load P (see figure part
a).
(b) Obtain a formula for the displacement at the point of
load application (see figure part a).
(c) For what value of x is RB (6/5) RA? (See figure part a.)
(d) Repeat (a) if the bar is now tapered linearly from A to B as
shown in figure part b and x L/2.
(e) Repeat (a) if the bar is now rotated to a vertical position,
load P is removed, and the bar is hanging under its own
weight (assume mass density ). (See figure part c.)
Assume that x L/2
127
Statically Indeterminate Structures
P, d
L
—
2
d
—
2
d
RA
RB
B
C
A
x
L–x
(a)
d
dB = —
2
d
—
2
dA = d
RA
RB
C
P, d
A
x
L–x
L
—
2
L
—
2
(b)
RB
B
L–x
C
d
—
2
x
d
A
RA
(c)
B
P applied
L
at —
2
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CHAPTER 2
Page 128
Axially Loaded Members
Solution 2.4-4
(a) reactions at A & B due to load P at L/2
AAC p 2
d 2
cd a b d
4
2
ACB p 2
d
4
AAC 3 2
pd
16
select RB as the redundant; use superposition and a compatibility equation at B
if x
Px
B1a +
EA AC
L/2
Pa
L
xb
2
L
x
P
x
2
≤
+
B1a ±
p 2
E 3
d
pd 2
4
16
ACB
2 2x + 3L
B1a P
3
Epd2
P
if x
B1b L/2
L
2
EA AC
P
B1b Ea
L
2
3
pd 2 b
16
B1b 8 PL
3 Epd2
the following expression for B2 is good for all x
B2 RB x
Lx
+
b
a
E AAC
ACB
B2 RB 16 x
Lx
+ 4
b
a
2
E 3 pd
pd 2
B2 RB
E
(a.1) solve for RB and RA assuming that x
compatibility:
B1a B2 0
x
Lx
+
p 2
3
P pd2
d Q
4
16
L/2
2 2x + 3L
a P
b
3
pd 2
RBa 16 x
Lx
a
+ 4
b
3 pd 2
pd 2
RBa 1 2x + 3L
P
2
x + 3L
;
^ check if x 0, RB P/2
statics:
RAa P RBa
RAa P 1 2x + 3L
P
2
x + 3L
RAa 3
L
P
2
x + 3L
;
^ check if x 0, RAa P/2
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Page 129
SECTION 2.4
(a.2) solve for RB and RA assuming that x
B1b B2 0
compatibility:
RBb
Statically Indeterminate Structures
129
L/2
8 PL
3 pd 2
16 x
Lx
a
+ 4
b
2
3 pd
pd 2
RBb 2PL
x + 3L
;
^ check if x L, RB P/2
RAb P a
RAb P RBb
statics:
da PL
L
x
2
+
ACB Q
2x + 3L
2
(x + 3L)Epd
(b.2) assume that x
b 1 RAb2
EAAC
L
2
x + L
x + 3L
;
L/2
(b.1) assume that x
RAa x
E P AAC
RAb P
axial force for segment 0 to L/2 RA & elongation of this segment
(b) find at point of load application;
da 2PL
b
x + 3L
da a
for x L/2
3
L
P
b
2
x + 3L
E
da P
8
L
7 Epd2
L
x
2
±
≤
+
p 2
3
d
pd2
4
16
x
;
L/2
b aP
x + L L
b
x + 3L 2
Ea
3
pd 2 b
16
b for x L/2
8
x + L
L
Pa
b
3
x + 3L Epd 2
8 L
b P
7 Epd 2
(c) For what value of x is R B (6/5) RA? Guess that x
1 2x + 3L
6 3
L
P
a
P
b 0
2
x + 3L
5 2
x + 3L
Now try RBb (6/5)RAb assuming that x
2PL
6
x + L
a P
b 0
x + 3L
5
x + 3L
So, there are two solutions for x.
;
same as a above (OK)
L/2 here & use RBa expression above to find x
1 10x 3L
P
0
10
x + 3L
x
L/2
2 2L + 3x
P
0
5
x + 3L
x
2
L
3
;
3L
10
;
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CHAPTER 2
Page 130
Axially Loaded Members
(d) repeat (a) above for tapered bar & x L/2
outer diameter
d(x) da1 AAC x
b
2L
p
d 2
c d(x)2 a b d
4
2
0 … x …
ACB p
d(x)2
4
ACB 1
d 2
p a b (4L2 4Lx + x2)
16
L
L
… x … L
2
L
2
AAC p
x 2
d 2
c c da 1 bd a b d
4
2L
2
AAC 1
d 2
pa b (3L2 4Lx + x2)
16
L
x 2
p
c da 1 bd
4
2L
ACB As in (a), use superposition and compatibility to find redundant RB & then RA
d B1 B1 P L2 1
d
E L0 AAC
8PL
Ep d
2
d B1 ( ln(5) + ln(3))
P L2
E L0
1
1
d 2
c pa b (3L2 4L + 2) d
16 L
d
PL
B1 1.301
Ed 2
L
d B2 L
RB
1
2 1
d +
d b
a
L
E L0 AAC
A
L CB
2
L
L
2
RB
1
1
≥
d d¥
d B2 2
2
E L
d
LL 1
2
2
0 1 p a d b 13L2 4L + 22
pa
b
14L
4L
+
2
2
16
L
16
L
B2 8L
RB
( 3 ln(5) + 3 ln(3) 2)
31p d 2 E
2
compatibility: B1 B2 0
statics:
RA P RB
RB B2 2.998
a1.301
PL
Ed2
L
a 2.998
b
Ed2
b
RA (P 0.434P)
RBL
Ed 2
RB 0.434P
;
RA 0.566P
;
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Page 131
SECTION 2.4
Statically Indeterminate Structures
131
(e) Find reactions if the bar is now rotated to a vertical position, load P is removed, and the bar is hanging under
its own weight (assume mass density ). Assume that x L/2.
AAC 3
pd2
16
ACB p 2
d
4
select RB as the redundant; use superposition and a compatibility equation at B
from (a) above:
dB2 compatibility: B1 B2 0
RB x
Lx
a
+
b
E AAC
ACB
B1 for x L/2, dB2 RB 14 L
a
b
E 3 pd 2
L
2
L
NAC
NCB
d d
L EA
EA
AC
CB
L0
L2
WAC rgAAC
where axial forces in bar due to self weight are:
(assume is measured upward from A)
NAC crgACB
L
L
+ rgAAC a b d
2
2
AAC 3
pd2
16
L
2
WCB rgACB
ACB p 2
d
4
NCB [ gACB(L )]
NAC B1 1
3
1
rgp d2 L rgp d2 a L b
8
16
2
L
2
1
3
1
rgpd2L rgpd2 a L b
8
16
2
L0
B1 a
3
E a pd2 b
16
L2
1
L2
11
rg
+
rg
b
24
E
8
E
B1 1
NCB c rgp d2( L ) d
4
L
d +
L2
7
rg
12
E
compatibility: B1 B2 0
RB statics:
a
7
L2
rg
b
12
E
1
rgpd 2L
8
;
RA (WAC WCB) RB
RA c crga
RA RB 14 L
a
b
3 Epd2
3
L
p
L
1
pd2 b + rga d2 b d rgpd2L d
16
2
4
2
8
3
rgp d 2 L
32
;
LL2
1
c rgpd2 (L) d
4
Ea p4 d2 b
7
0.583
12
d
L
2
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CHAPTER 2
Page 132
Axially Loaded Members
Problem 2.4-5 Three steel cables jointly support a load of 12 k (see figure). The diameter of
the middle cable is 3/4 in. and the diameter of each outer cable is 1/2 in. The tensions in the
cables are adjusted so that each cable carries one-third of the load (i.e., 4 k). Later, the load is
increased by 9 k to a total load of 21 k.
(a) What percent of the total load is now carried by the middle cable?
(b) What are the stresses M and O in the middle and outer cables, respectively?
(NOTE: See Table 2-1 in Section 2.2 for properties of cables.)
Solution 2.4-5
Three cables in tension
SECOND LOADING
P2 9 k (additional load)
AREAS OF CABLES (from Table 2-1)
Middle cable: AM 0.268 in.2
EQUATION OF EQUILIBRIUM
Fvert 0
2PO PM P2 0
Outer cables: AO 0.119 in.2
M O
(for each cable)
FORCE-DISPLACEMENT RELATIONS
FIRST LOADING
P1
P1 12 k aEach cable carries
or 4 k.b
3
(1)
EQUATION OF COMPATIBILITY
dM PML
EAM
(2)
dO Po L
EAo
(3, 4)
SUBSTITUTE INTO COMPATIBILITY EQUATION:
PML
POL
EAM
EAO
PM
PO
AM
AO
(5)
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Page 133
SECTION 2.4
SOLVE SIMULTANEOUSLY EQS. (1) AND (5):
PM P2 a
133
(a) PERCENT OF TOTAL LOAD CARRIED BY MIDDLE CABLE
2
AM
0.268 in.
b (9 k)a
b
AM + 2AO
0.506 in.2
4.767 k
Po P2 a
Statically Indeterminate Structures
AM
Percent 8.767 k
(100%) 41.7%
21 k
(b) STRESSES IN CABLES ( P/A)
Ao
0.119 in.2
b (9 k)a
b
+ 2AO
0.506 in.2
2.117 k
Middle cable: sM Outer cables: sO FORCES IN CABLES
8.767 k
0.268 in.2
6.117 k
0.119 in.2
32.7 ksi
51.4 ksi
Middle cable: Force 4 k 4.767 k 8.767 k
Outer cables: Force 4 k 2.117 k 6.117 k
(for each cable)
Problem 2.4-6 A plastic rod AB of length L 0.5 m has a
diameter d1 30 mm (see figure). A plastic sleeve CD of length
c 0.3 m and outer diameter d2 45 mm is securely bonded to the
rod so that no slippage can occur between the rod and the sleeve.
The rod is made of an acrylic with modulus of elasticity E1 3.1
GPa and the sleeve is made of a polyamide with E2 2.5 GPa.
(a) Calculate the elongation of the rod when it is pulled by
axial forces P 12 kN.
(b) If the sleeve is extended for the full length of the rod, what is
the elongation?
(c) If the sleeve is removed, what is the elongation?
Solution 2.4-6
;
Plastic rod with sleeve
P 12 kN
d1 30 mm
b 100 mm
L 500 mm
d2 45 mm
c 300 mm
Rod: E1 3.1 GPa
Sleeve: E2 2.5 GPa
Rod: A1 pd21
706.86 mm2
4
Sleeve: A2 p 2
(d d12) 883.57 mm2
4 2
E1A1 E2A2 4.400 MN
;
;
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CHAPTER 2
Page 134
Axially Loaded Members
(b) SLEEVE AT FULL LENGTH
(a) ELONGATION OF ROD
Part AC: dAC Pb
0.5476 mm
E1A1
Part CD: d CD L
500 mm
d dCD a b (0.81815 mm)a
b
c
300 mm
1.36 mm
PC
E1A1E2A2
(c) SLEEVE REMOVED
PL
d
2.74 mm
E1A1
0.81815 mm
(From Eq. 2-13 of Example 2-5)
2AC CD 1.91 mm
;
;
;
Problem 2.4-7 The axially loaded bar ABCD shown in the figure is held between
P
(a) Derive formulas for the reactions RA and RD at the ends of the bar.
(b) Determine the displacements B and C at points B and C, respectively.
(c) Draw a diagram in which the abscissa is the distance from the left-hand support to
any point in the bar and the ordinate is the horizontal displacement at that point.
Solution 2.4-7
2A1
A1
rigid supports. The bar has cross-sectional area A1 from A to C and 2A1 from C to D.
A
B
L
—
4
C
L
—
4
D
L
—
2
Bar with fixed ends
FREE-BODY DIAGRAM OF BAR
(a) REACTIONS
Solve simultaneously Eqs. (1) and (6):
RA 2P
3
RD P
3
;
(b) DISPLACEMENTS AT POINTS B AND C
dB dAB EQUATION OF EQUILIBRIUM
Fhoriz 0
RA RD P
(Eq. 1)
dC |dCD| EQUATION OF COMPATIBILITY
AB BC CD 0
FORCE-DISPLACEMENT EQUATIONS
dCD (RA P)(L / 4)
dBC EA1
RD(L/2)
E(2A1)
PL
(To the right)
12EA1
;
(c) AXIAL DISPLACEMENT DIAGRAM (ADD)
(Eqs. 3, 4)
(Eq. 5)
SOLUTION OF EQUATIONS
Substitute Eqs. (3), (4), and (5) into Eq. (2):
(RA P)(L)
RAL
RDL
+
0
4EA1
4EA1
4EA1
RDL
4EA1
(Eq. 2)
Positive means elongation.
RA(L/4)
dAB EA1
RAL
PL
(To the right)
4EA1
6EA1
(Eq. 6)
;
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Page 135
SECTION 2.4
135
Statically Indeterminate Structures
Problem 2.4-8 The fixed-end bar ABCD consists of three prismatic
segments, as shown in the figure. The end segments have cross-sectional
area A1 840 mm2 and length L1 200 mm. The middle segment has
cross-sectional area A2 1260 mm2 and length L2 250 mm. Loads PB
and PC are equal to 25.5 kN and 17.0 kN, respectively.
(a) Determine the reactions RA and RD at the fixed supports.
(b) Determine the compressive axial force FBC in the middle
segment of the bar.
Solution 2.4-8
Bar with three segments
PB 25.5 kN
PC 17.0 kN
L1 200 mm
L2 250 mm
A1 840 mm
A2 1260 mm2
2
m meter
SOLUTION OF EQUATIONS
Substitute Eqs. (3), (4), and (5) into Eq. (2):
FREE-BODY DIAGRAM
RA
RA
1
1
a 238.095 b +
a 198.413 b
E
m
E
m
EQUATION OF EQUILIBRIUM
Fhoriz 0 :
Simplify and substitute PB 25.5 kN:
;
RA a 436.508
PB RD PC RA 0 or
RA RD PB PC 8.5 kN
5,059.53
AD elongation of entire bar
(Eq. 2)
FORCE-DISPLACEMENT RELATIONS
dAB RAL1
RA
1
a238.05 b
EA1
E
m
dBC (RA PB)L2
EA2
RA
PB
1
1
a198.413 b a198.413 b
E
m
E
m
1
1
b + RD a 238.095 b
m
m
(Eq. 1)
EQUATION OF COMPATIBILITY
AD AB BC CD 0
RD
PB
1
1
a 198.413 b +
a 238.095 b 0
E
m
E
m
kN
m
(Eq. 6)
(a) REACTIONS RA AND RD
Solve simultaneously Eqs. (1) and (6).
From (1): RD RA 8.5 kN
(Eq. 3)
Substitute into (6) and solve for RA:
RA a 674.603
1
kN
b 7083.34
m
m
RA 10.5 kN
(Eq. 4)
;
RD RA 8.5 kN 2.0 kN
;
(b) COMPRESSIVE AXIAL FORCE FBC
dCD RDL1
RD
1
a238.095 b
EA1
E
m
(Eq. 5)
FBC PB RA PC RD 15.0 kN
;
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CHAPTER 2
Page 136
Axially Loaded Members
Problem 2.4-9 The aluminum and steel pipes shown in the figure are fastened to rigid
supports at ends A and B and to a rigid plate C at their junction. The aluminum pipe is
twice as long as the steel pipe. Two equal and symmetrically placed loads P act on the
plate at C.
(a) Obtain formulas for the axial stresses a and s in the aluminum and steel pipes,
respectively.
(b) Calculate the stresses for the following data: P 12 k, cross-sectional area of
aluminum pipe Aa 8.92 in.2, cross-sectional area of steel pipe As 1.03 in.2,
modulus of elasticity of aluminum Ea 10 106 psi, and modulus of elasticity
of steel Es 29 106 psi.
Solution 2.4-9
Pipes with intermediate loads
SOLUTION OF EQUATIONS
Substitute Eqs. (3) and (4) into Eq. (2):
RB(2L)
RAL
0
Es As
EaAa
(Eq. 5)
Solve simultaneously Eqs. (1) and (5):
RA 4Es As P
EaAa + 2EsAs
RB 2EaAaP
EaAa + 2EsAs
(Eqs. 6, 7)
(a) AXIAL STRESSES
Aluminum: sa RB
2EaP
Aa
EaAa + 2EsAs
;
(Eq. 8)
Pipe 1 is steel.
Pipe 2 is aluminum.
(compression)
Steel: ss EQUATION OF EQUILIBRIUM
Fvert 0
RA RB 2P
(Eq. 1)
EQUATION OF COMPATIBILITY
AB AC CB 0
dBC RB(2L)
EaAa
(Eq. 9)
(tension)
P 12 k
Aa 8.92 in.2
Ea 10 106 psi
FORCE-DISPLACEMENT RELATIONS
RAL
EsAs
;
(b) NUMERICAL RESULTS
(Eq. 2)
(A positive value of means elongation.)
dAC RA
4EsP
As
EaAa + 2EsAs
As 1.03 in.2
Es 29 106 psi
Substitute into Eqs. (8) and (9):
(Eqs. 3, 4))
a 1,610 psi (compression)
s 9,350 psi (tension)
;
;
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Page 137
SECTION 2.4
Statically Indeterminate Structures
137
Problem 2.4-10 A nonprismatic bar ABC is composed of two segments: AB of length
L1 and cross-sectional area A1; and BC of length L2 and cross-sectional area A2.
The modulus of elasticity E, mass density , and acceleration of gravity g are constants.
Initially, bar ABC is horizontal and then is restrained at A and C and rotated to a vertical
position. The bar then hangs vertically under its own weight (see figure). Let A1 2A2 A and L1 53 L, L2 25 L.
(a) Obtain formulas for the reactions RA and RC at supports A and C, respectively, due to
gravity.
(b) Derive a formula for the downward displacement B of point B.
(c) Find expressions for the axial stresses a small distance above points B and C,
respectively.
RA
A
A1
L1
B
Stress
elements
L2
A2
C
RC
Solution 2.4-10
(a) find reactions in 1-degree statically indeterminate structure
use superposition; select RA as the redundant
compatibility: A1 A2 0
segment weights:
WAB gA1L1
WBC gA2L2
find axial forces in each segment;
use variable measured from C toward A
NAB gA1(L1 L2 )
L2
NBC [WAB gA2(L2 )]
0
L1 L2
L2
displacement at A in released structure due to self weight
L2
d A1 L0
L2
d A1 L0
dA1 c
L L2
1
NBC
d +
E A2
LL2
NAB
d
E A1
L1 L2 rgA 1 L + L 2
[rgA1L1 + rgA21 L2 2]
1
1
2
d +
d
EA2
E
A
LL2
1
2L1 L2
L12 2L1L2 L22
2A1L1 A2L2
1
1
1
rgL2
rg
rgL2
bd
a
2
EA2
2
E
2
E
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CHAPTER 2
d A1 Page 138
Axially Loaded Members
rg
12L2A1L1 + A2L22 + A2L122
2( EA2)
Next, displacement at A in released structure due to redundant RA
A2 RA(fAB fBC)
d A2 RA a
L1
L2
+
b
EA1
EA2
enforce compatibility: A1 A2 0
solve for RA
rg
(2L2A1L1 + A2L22 + A2L12)
2(EA2)
RA fAB f B
2
RA A2L12 + 2A1L1L2 + A2L2
1
rgA1
2
L1A2 + L2A1
statics:
RC WAB WBC RA
RC crgA1L1 + rgA2L2 ;
A1
1
rg(2L2A1L1 + A2L22 + A2L12)
d
2
L1A2 + L2A1
2
A1L1 2 + 2A2L1L2 + A1L2
1
rgA2
2
L1A2 + L2A1
RC A1 A
For
RA 1
rgA
2
RC A2 A
2
L1 ;
3L
5
L2 A 3L 2
3L 2L
A 2L 2
a b + 2A
+ a b
2 5
5 5
2 5
3L A
2L
A
5 2
5
A
1
rg
2
2
Aa
3L 2
2L 2
A 3L 2L
b +2
+ Aa
b
5
2 5 5
5
3L A
2L
+
A
5 2
5
(b) use superposition to find displacement at point B
due to RA
L2
d B1 d B1 L
0
NBC
d
EA2
due to shortening of BC
rgL2
(2A1L1 + A2L2)
2(EA2)
B2 RA(fBC)
dB2 RA a
L2
b
EA 2
2L
5
RA 37
rgAL
70
RC 19
rgLA
70
B B1 B2
37
0.529
70
;
;
19
0.271
70
where B1 is due to gravity and B2 is
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Page 139
SECTION 2.4
dB c
Statically Indeterminate Structures
2
A2L12 + 2A1L1L2 + A2L2
L2
1
rgL2
(2A1L1 + A2L2) + rgA1
a
bd
2(EA2)
2
L1A2 + L2A1
EA2
A L A 2L 2
dB 1 r g L2 L1 1 1
2
(L1A2 L2A1) E
For
A1 A
A2 1
2L 3L
dB rg
2
5 5
A
2
L1 3L
5
3L A 2L
+
5
2 5
3L A
2L
a
+
Ab E
5 2
5
L2 A
dB 2L
5
L2
24
rg
175
E
;
(c) expressions for the average axial stresses a small distance above points B and C
NB axial force near B RA WAB
2
A2L12 + 2A1L1L2 + A2L2
1
b r gA1L1
NB a rgA1
2
L1A2 + L2A1
NB 3L
37
rgAL rgA
70
5
sB NB
A
NC RC
sB sC NB 1
rgL
14
a
;
19
rgLAb
70
A
2
1
rgAL
14
sC 19
rgL
35
;
139
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CHAPTER 2
Page 140
Axially Loaded Members
Problem 2.4-11 A bimetallic bar (or composite bar) of square cross
section with dimensions 2b 2b is constructed of two different metals
having moduli of elasticity E1 and E2 (see figure). The two parts of the
bar have the same cross-sectional dimensions. The bar is compressed by
forces P acting through rigid end plates. The line of action of the loads
has an eccentricity e of such magnitude that each part of the bar is
stressed uniformly in compression.
(a) Determine the axial forces P1 and P2 in the two parts of the bar.
(b) Determine the eccentricity e of the loads.
(c) Determine the ratio 1/2 of the stresses in the two parts of the bar.
Solution 2.4-11
Bimetallic bar in compression
FREE-BODY DIAGRAM
(a) AXIAL FORCES
(Plate at right-hand end)
Solve simultaneously Eqs. (1) and (3):
P1 PE1
E1 + E2
P2 PE2
E1 + E2
;
(b ECCENTRICITY OF LOAD P
Substitute P1 and P2 into Eq. (2) and solve for e:
e EQUATIONS OF EQUILIBRIUM
F 0 P1 P2 P
b
b
Pe + P1 a b P2 a b 0
2
2
M 0 哵哴
(c) RATIO OF STRESSES
(Eq. 2)
2 1
or
P2
P1
E2
E1
;
(Eq. 1)
EQUATION OF COMPATIBILITY
P1L
P2L
E2A
E1A
b(E2 E1)
2(E2 E1)
(Eq. 3)
s1 P1
A
s2 P2
A
s1
P1
E1
s2
P2
E2
;
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Page 141
SECTION 2.4
Statically Indeterminate Structures
141
Problem 2.4-12 A rigid bar of weight W 800 N hangs from three equally
spaced vertical wires (length L 150 mm, spacing a 50 mm): two of steel
and one of aluminum. The wires also support a load P acting on the bar. The
diameter of the steel wires is ds 2 mm, and the diameter of the aluminum wire
is da 4 mm. Assume Es 210 GPa and Ea 70 GPa.
a
L
S
a
A
S
Rigid bar
of weight W
(a) What load Pallow can be supported at the midpoint of the bar (x a) if the
allowable stress in the steel wires is 220 MPa and in the aluminum wire is
80 MPa? (See figure part a.)
(b) What is Pallow if the load is positioned at x a/2? (See figure part a.)
(c) Repeat (b) above if the second and third wires are switched as shown in figure
part b.
x
P
(a)
a
L
S
a
S
A
Rigid bar
of weight W
x
P
(b)
Solution 2.4-12
numerical data
W 800 N
a 50 mm
dA 4 mm
L 150 mm
dS 2 mm
ES 210 GPa
EA 70 GPa
Sa 220 MPa
AA p 2
d
4 A
AA 13 mm2
Aa 80 MPa
AS p 2
d
4 S
AS 3 mm2
(a) Pallow at center of bar
1-degree stat-indet - use reaction (RA) at top of aluminum bar as the redundant
compatibility: 1 2 0
d1 P +W
L
b
a
2
ESAS
statics:
2RS RA P W
downward displacement due to elongation of each steel wire under P W if alum.
wire is cut at top
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CHAPTER 2
d2 RA a
Page 142
Axially Loaded Members
L
L
+
b
2E SAS
EAAA
upward displ. due to shortening of steel wires & elongation of alum. wire
under redundant RA
enforce compatibility & then solve for RA
1 2
RA so
P +W
L
a
b
2
ESAS
RA ( P + W)
L
L
+
2ESAS
EAAA
EAAA
EAAA + 2ESAS
and
s Aa RA
AA
now use statics to find RS
P + W (P + W)
P + W RA
RS 2
RS EAAA
EAAA + 2ESAS
2
RS (P + W)
and
ESAS
EAAA + 2ESAS
RS
s Sa AS
compute stresses & apply allowable stress values
s Aa ( P + W)
EA
EAAA + 2ESAS
s Sa ( P + W)
ES
EAAA + 2ESAS
solve for allowable load P
PAa s Aa a
EAAA + 2ESAS
b W
EA
PAa 1713 N
PSa sSa a
EAAA + 2ESAS
b W
ES
PSa 1504 N
lower value of P controls
; Pallow is controlled by steel wires
(b) Pallow if load P at x a/2
again, cut aluminum wire at top, then compute elongations of left & right steel wires
d 1L a
d1 3P
W
L
b
+
ba
4
2
ESAS
d 1L + d 1R
2
d1 d 1R a
P
W
L
b
+
ba
4
2
ESAS
P +W
L
a
b where 1 displ. at x a
2
ESAS
Use 2 from (a) above
d2 RA a
L
L
+
b
2ESAS
EAAA
so equating 1 & 2, solve for RA RA ( P + W)
EAAA
EAAA + 2ESAS
same as in (a)
RSL RSL RA
3P
W
+
4
2
2
3P
W
+
4
2
(P + W)
stress in left steel wire exceeds that in right steel wire
EAAA
EAAA + 2ESAS
2
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Page 143
SECTION 2.4
RSL PEAAA + 6PESAS + 4WESAS
4EAAA + 8ESAS
s Sa Statically Indeterminate Structures
143
PEAAA + 6PESAS + 4WESAS 1
a b
4EAAA + 8ESAS
AS
solve for Pallow based on allowable stresses in steel & alum.
sSa(4ASEAAA + 8ESAS2 ) (4WESAS)
EAAA + 6ESAS
PSa PSa 820 N
;
PAa 1713 N
same as in (a)
steel controls
(c) Pallow if wires are switched as shown & x a/2
select RA as the redundant
statics on the two released structures
(1) cut alum. wire - apply P & W, compute forces in left & right steel wires, then compute displacements at each
steel wire
RSL d1L P
2
RSR L
P
a
b
2 ESAS
P
+W
2
d1R a
L
P
+ Wb a
b
2
ESAS
by geometry, at alum. wire location at far right is
d1 a
P
L
b
+ 2Wb a
2
ESAS
(2) next apply redundant RA at right wire, compute wire force & displ. at alum. wire
RSL RA
RSR 2RA
d2 RA a
5L
L
+
b
ESAS
EAAA
(3) compatibility equate 1, 2 and solve for RA then Pallow for alum. wire
RA a
P
L
+ 2Wb a
b
2
ES AS
5L
L
+
ESAS
EAAA
RA EAAAP + 4EAAAW
10EAAA + 2ESAS
sAa EAP + 4EAW
10EAAA + 2ESAS
PAa sAa(10EAAA + 2ESAS) 4EAW
EA
s Aa RA
AA
PAa 1713 N
(4) statics or superposition - find forces in steel wires then Pallow for steel wires
RSL P
+ RA
2
RSL EAAAP + 4EAAAW
P
+
2
10EAAA + 2ESAS
RSL 6EAAAP + PESAS + 4EAAAW
10EAAA + 2ESAS
larger than RSR below so use in allow. stress
calcs
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CHAPTER 2
RSR sSa Page 144
Axially Loaded Members
P
+ W 2RA
2
RSL
AS
RSR EAAAP + 4EAAAW
P
+ W
2
5EAAA + ES AS
RSR 3EAAAP + PESAS + 2EAAAW + 2WESAS
10EAAA + 2ESAS
PSa sSaAS a
10EAAA + 2ESAS
4EAAAW
b
6EAAA + ESAS
6EAAA + ESAS
2
10sSaASEAAA + 2sSaA S E S 4EAAAW
6EAAA + ESAS
PSa PSa 703 N
^ steel controls
;
Problem 2.4-13 A horizontal rigid bar of weight W 7200 lb is supported by
three slender circular rods that are equally spaced (see figure). The two outer
rods are made of aluminum (E1 10 106 psi) with diameter d1 0.4 in. and
length L1 40 in. The inner rod is magnesium (E2 6.5 106 psi) with diameter d2 and length L2. The allowable stresses in the aluminum and magnesium
are 24,000 psi and 13,000 psi, respectively.
If it is desired to have all three rods loaded to their maximum allowable
values, what should be the diameter d2 and length L2 of the middle rod?
Solution 2.4-13
Bar supported by three rods
BAR 1 ALUMINUM
E1 10 106 psi
FREE-BODY DIAGRAM OF RIGID BAR
EQUATION OF EQUILIBRIUM
Fvert 0
d1 0.4 in.
2F1 F2 W 0
L1 40 in.
(Eq. 1)
FULLY STRESSED RODS
1 24,000 psi
F1 1A1
F2 2A2
BAR 2 MAGNESIUM
E2 6.5 106 psi
d2 ?
L2 ?
2 13,000 psi
A1 Substitute into Eq. (1):
2s1 a
pd21
4
b + s2 a
pd22
4
b W
pd21
4
A2 pd22
4
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Page 145
SECTION 2.4
Diameter d1 is known; solve for d2:
d22
4W
ps2
2
2s1d1
s2
;
d2 (Eq. 2)
SUBSTITUTE NUMERICAL VALUES:
d22 2(24,000 psi)(0.4 in.)2
4(7200 lb)
p(13,000 psi)
13,000 psi
0.70518 in2. 0.59077 in.2 0.11441 in.2
d2 0.338 in.
;
FORCE-DISPLACEMENT RELATIONS
F1L1
L1
s1 a b
d1 E1A1
E1
F2L2
L2
s2 a b
E2A2
E2
(Eq. 5)
Substitute (4) and (5) into Eq. (3):
L1
L2
s1 a b s2 a b
E1
E2
Length L1 is known; solve for L2:
L2 L1 a
s1E2
b
s2E1
;
(Eq. 6)
SUBSTITUTE NUMERICAL VALUES:
EQUATION OF COMPATIBILITY
1 2
145
Statically Indeterminate Structures
(Eq. 3)
L2 (40 in.) a
24,000 psi 6.5 * 106 psi
ba
b
13,000 psi
10 * 106 psi
48.0 in.
(Eq. 4)
Problem 2.4-14 A circular steel bar ABC (E 200 GPa) has
cross-sectional area A1 from A to B and cross-sectional area A2 from B to C
(see figure). The bar is supported rigidly at end A and is subjected to a load P equal to 40 kN at
end C. A circular steel collar BD having cross-sectional area A3 supports the bar at B. The
collar fits snugly at B and D when there is no load.
Determine the elongation AC of the bar due to the load P. (Assume L1 2L3 250 mm,
L2 225 mm, A1 2A3 960 mm2, and A2 300 mm2.)
A
A1
L1
B
L3
A3
D
A2
C
P
L2
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CHAPTER 2
Page 146
Axially Loaded Members
Solution 2.4-14
Bar supported by a collar
FREE-BODY DIAGRAM OF BAR ABC AND COLLAR BD
SOLVE SIMULTANEOUSLY EQS. (1) AND (3):
PL3A1
PL1A3
RA RD L1A3 + L3A1
L1A3 + L3A1
CHANGES IN LENGTHS (Elongation is positive)
RAL1
PL1L3
PL2
dBC dAB EA1
E(L1A3 + L3A1)
EA2
ELONGATION OF BAR ABC
AC AB AC
SUBSTITUTE NUMERICAL VALUES:
P 40 kN
E 200 GPa
L1 250 mm
L2 225 mm
L3 125 mm
EQUILIBRIUM OF BAR ABC
Fvert 0
RA RD P 0
A1 960 mm2
(Eq. 1)
COMPATIBILITY (distance AD does not change)
AB(bar) BD(collar) 0
(Eq. 2)
A3 480 mm2
RESULTS:
(Elongation is positive.)
RA RD 20 kN
FORCE-DISPLACEMENT RELATIONS
RAL1
RDL3
dBD dAB EA1
EA3
Substitute into Eq. (2):
RDL3
RAL1
0
EA1
EA3
A2 300 mm2
AB 0.02604 mm
BC 0.15000 mm
AC AB AC 0.176 mm
;
(Eq. 3)
Problem 2.4-15 A rigid bar AB of length L 66 in. is hinged to a
support at A and supported by two vertical wires attached at points C and
D (see figure). Both wires have the same cross-sectional area (A 0.0272 in.2) and are made of the same material (modulus E 30 106
psi). The wire at C has length h 18 in. and the wire at D has length
twice that amount. The horizontal distances are c 20 in. and d 50 in.
(a) Determine the tensile stresses C and D in the wires due to the
load P 340 lb acting at end B of the bar.
(b) Find the downward displacement B at end B of the bar.
2h
h
A
C
D
B
c
d
P
L
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Page 147
SECTION 2.4
Solution 2.4-15
147
Statically Indeterminate Structures
Bar supported by two wires
EQUATION OF COMPATIBILITY
dc
dD
c
d
FORCE-DISPLACEMENT RELATIONS
dC TCh
EA
TD(2h)
EA
dD (Eq. 2)
(Eqs. 3, 4)
SOLUTION OF EQUATIONS
Substitute (3) and (4) into Eq. (2):
TD(2h)
TCh
TC
2TD
or
cEA
dEA
c
d
TENSILE FORCES IN THE WIRES
h 18 in.
(Eq. 5)
Solve simultaneously Eqs. (1) and (5):
2h 36 in.
TC c 20 in.
d 50 in.
2cPL
2
2
2c + d
dPL
TD 2
2c + d2
(Eqs. 6, 7)
TENSILE STRESSES IN THE WIRES
TC
2cPL
sC A
A(2c2 + d2)
L 66 in.
E 30 106 psi
A 0.0272 in.2
sD P 340 lb
FREE-BODY DIAGRAM
TD
dPL
A
A(2c2 + d2)
(Eq. 8)
(Eq. 9)
DISPLACEMENT AT END OF BAR
2hTD L
L
2hPL2
dB dD a b a b d
EA d
EA(2c2 + d 2)
(Eq. 10)
SUBSTITUTE NUMERICAL VALUES
2c2 d2 2(20 in.)2 (50 in.)2 3300 in.2
(a) sC 2cPL
2
2
A(2c + d )
10,000 psi
DISPLACEMENT DIAGRAM
sD 2
A(2c + d )
12,500 psi
(b) dB EQUATION OF EQUILIBRIUM
MA 0 哵 哴 TC (c) TD(d) PL
(Eq. 1)
(0.0272 in.2)(3300 in.2)
;
dPL
2
2(20 in.)(340 lb)(66 in.)
(50 in.)(340 lb)(66 in.)
(0.0272 in.2)(3300 in.2)
;
2hPL2
EA(2c2 + d2)
2(18 in.)(340 lb)(66 in.)2
(30 * 106 psi)(0.0272 in.2)(3300 in.2)
0.0198 in.
;
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CHAPTER 2
Page 148
Axially Loaded Members
Problem 2.4-16 A rigid bar ABCD is pinned at point B and supported by springs at A and D (see figure). The springs at A and D
have stiffnesses k1 10 kN/m and k2 25 kN/m, respectively, and
the dimensions a, b, and c are 250 mm, 500 mm, and 200 mm,
respectively. A load P acts at point C.
If the angle of rotation of the bar due to the action of the load P
is limited to 3°, what is the maximum permissible load Pmax?
a = 250 mm
A
b = 500 mm
B
C
D
P
c = 200 mm
k 2 = 25 kN/m
k1 = 10 kN/m
Solution 2.4-16
Rigid bar supported by springs
EQUATION OF EQUILIBRIUM
MB 0 FA(a) P(c) FD(b) 0
NUMERICAL DATA
a 250 mm
EQUATION OF COMPATIBILITY
dA
dD
a
b
FORCE-DISPLACEMENT RELATIONS
FA
FD
dD dA k1
k2
b 500 mm
SOLUTION OF EQUATIONS
c 200 mm
Substitute (3) and (4) into Eq. (2):
FA
FD
ak1
bk2
k1 10 kN/m
k2 25 kN/m
p
umax 3° rad
60
FREE-BODY DIAGRAM AND DISPLACEMENT DIAGRAM
(Eq. 1)
(Eq. 2)
(Eqs. 3, 4)
(Eq. 5)
SOLVE SIMULTANEOUSLY EQS. (1) AND (5):
ack1P
bck2P
FA 2
FD 2
2
a k1 + b k2
a k1 + b2k2
ANGLE OF ROTATION
FD
dD
bcP
cP
dD 2
2
u
k2
b
a k1 + b2k2
a k1 + b2k2
MAXIMUM LOAD
u
P (a2k1 + b2k2)
c
Pmax umax 2
(a k1 + b2k2)
c
;
SUBSTITUTE NUMERICAL VALUES:
Pmax p/60 rad
[(250 mm)2(10 kN/m)
200 mm
+ (500 mm)2(25 kN/m)]
1800 N
;
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Page 149
SECTION 2.4
Problem 2.4-17 A trimetallic bar is uniformly compressed by an
axial force P 9 kips applied through a rigid end plate (see figure).
The bar consists of a circular steel core surrounded by brass and
copper tubes. The steel core has diameter 1.25 in., the brass tube has
outer diameter 1.75 in., and the copper tube has outer diameter 2.25 in.
The corresponding moduli of elasticity are Es 30,0000 ksi, Eb 16,000 ksi, and Ec 18,000 ksi.
Calculate the compressive stresses ss, sb, and sc in the steel, brass,
and copper, respectively, due to the force P.
Statically Indeterminate Structures
P=9k
Copper tube
149
Brass tube
Steel core
1.25
in.
1.75
in.
2.25
in.
Solution 2.4-17
numerical properties (kips, inches)
dc 2.25 in.
db 1.75 in.
Ec 18000 ksi
ds 1.25 in.
Eb 16000 ksi
Es 30000 ksi
P 9 kips
EQUATION OF EQUILIBRIUM
Fvert 0
Ps Pb Pc P
(Eq. 1)
EQUATIONS OF COMPATIBILITY
s b
c s
(Eqs. 2)
FORCE-DISPLACEMENT RELATIONS
ds PsL
PbL
PcL
db dc EsAs
EbAb
EcAc
(Eqs. 3, 4, 5)
SOLUTION OF EQUATIONS
Substitute (3), (4), and (5) into Eqs. (2):
Pb Ps
EbAb
EcAc
P Ps
EsAs c
EsAs
(Eqs. 6, 7)
As p 2
d
4 s
Ab p 2
1d ds22
4 b
Ac p
1dc2 db22
4
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Axially Loaded Members
SOLVE SIMULTANEOUSLY EQS. (1), (6), AND (7):
Ps P
Es As
4 kips
Es As + Eb Ab + Ec Ac
Pb P
EbAb
2 kips
Es As + Eb Ab + Ec Ac
Pc P
Ec Ac
Es As + Eb Ab + Ec Ac
Ps Pb Pc 9
3 kips
statics check
COMPRESSIVE STRESSES
Let EA EsAs EbAb EcAc
ss Ps
PEs
As ©EA
sc Pc
PEc
Ac
©EA
sb Pb
PEb
Ab
©EA
compressive stresses
ss Ps
As
s 3 ksi
;
sb Pb
Ab
b 2 ksi
;
sc Pc
Ac
c 2 ksi
;
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SECTION 2.5
Thermal Effects
151
Thermal Effects
Problem 2.5-1 The rails of a railroad track are welded together at their ends (to form continuous rails and thus eliminate
the clacking sound of the wheels) when the temperature is 60°F.
What compressive stress is produced in the rails when they are heated by the sun to 120°F if the coefficient of thermal
expansion 6.5 106/°F and the modulus of elasticity E 30 106 psi?
Solution 2.5-1
Expansion of railroad rails
The rails are prevented from expanding because of
their great length and lack of expansion joints.
Therefore, each rail is in the same condition as a bar
with fixed ends (see Example 2-7).
The compressive stress in the rails may be calculated
from Eq. (2-18).
T 120°F 60°F 60°F
E(T)
(30 106 psi)(6.5 106/°F)(60°F)
11,700 psi
;
Problem 2.5-2 An aluminum pipe has a length of 60 m at a temperature of 10°C. An adjacent steel pipe at the same temperature is 5 mm longer than the aluminum pipe.
At what temperature (degrees Celsius) will the aluminum pipe be 15 mm longer than the steel pipe? (Assume that the coefficients of thermal expansion of aluminum and steel are a 23 106/°C and s 12 106/°C, respectively.)
Solution 2.5-2
Aluminum and steel pipes
INITIAL CONDITIONS
or, a(T )La
La 60 m
T0 10°C
Ls 60.005 m
T0 10°C
a 23 106/°C
s 12 106/°C
La L
s(T)Ls
Solve for T:
¢T ¢L + (Ls La)
aaLa asLs
;
FINAL CONDITIONS
Substitute numerical values:
Aluminum pipe is longer than the steel pipe by the
amount L 15 mm.
aLa sLs 659.9 106 m/°C
T increase in temperature
¢T a a(T )La
s s(T )Ls
15 mm + 5 mm
659.9 * 106 m/° C 30.31° C
T T0 + ¢T 10°C + 30.31°C
40.3°C
From the figure above:
a
La L
s
Ls
;
Ls
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Axially Loaded Members
Problem 2.5-3 A rigid bar of weight W 750 lb hangs from three equally spaced wires, two of steel and one of aluminum
(see figure). The diameter of the wires is 1/8 in . Before they were loaded, all three wires had the same length.
What temperature increase T in all three wires will result in the entire load being carried by the steel wires? (Assume
Es 30 106 psi, s 6.5 106/°F, and a 12 106/°F.)
Solution 2.5-3
Bar supported by three wires
1 increase in length of a steel wire due to temperature increase T
S
A
s (T)L
S
2 increase in length of a steel wire due to load
W/2
WL
2EsAs
3 increase in length of aluminum wire due to temperature increase T
W = 750 lb
a(T)L
S steel
A aluminum
W 750 lb
For no load in the aluminum wire:
1
2 3
d
1
in.
8
as(¢T)L +
As pd2
0.012272 in.2
4
or
Es 30 106 psi
EsAs 368,155 lb
¢T 6
a 12 10 /°F
L Initial length of wires
W
2EsAs(aa as)
;
Substitute numerical values:
6
s 6.5 10 /°F
WL
aa(¢T)L
2EsAs
¢T 750 lb
(2)(368,155 lb)(5.5 * 106/° F)
;
185°F
NOTE: If the temperature increase is larger than T,
the aluminum wire would be in compression, which is
not possible. Therefore, the steel wires continue to
carry all of the load. If the temperature increase is less
than T, the aluminum wire will be in tension and
carry part of the load.
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SECTION 2.5
Problem 2.5-4 A steel rod of 15-mm diameter is held snugly
(but without any initial stresses) between rigid walls by the
arrangement shown in the figure. (For the steel rod, use
12 106/°C and E 200 GPa.)
(a) Calculate the temperature drop T (degrees Celsius) at
which the average shear stress in the 12-mm diameter bolt
becomes 45 MPa.
Washer,
dw = 20 mm
12-mm diameter bolt
∆T
B
A
18 mm
15 mm
Clevis,
tc = 10 mm
(b) What are the average bearing stresses in the bolt and clevis
at A and the washer (dw 20 mm) and wall
(t wall 18mm) at B?
153
Thermal Effects
Solution 2.5-4
numerical properties
dr 15 mm
db 12 mm
dw 20 mm
tc 10 mm
twall 18 mm
b
6
12 (10 )
45 MPa
solve for T
E 200 GPa
¢T (a) TEMPERATURE DROP RESULTING IN BOLT SHEAR STRESS
T
ET
T 24°C
p
rod force P ( Ea¢T) d2r and bolt in double
4
P
2
shear with shear stress t AS
t
P
p 2
2 db
4
;
P ( E a ¢T)
p 2
d
4 r
P 10.18 kN
(b) BEARING STRESSES
p
so t b c( Ea¢T) dr2 d
2
4
pdb
2
tb 2t b db 2
a b
E a dr
Ea¢T dr 2
a b
2
db
P
2
bolt and clevis s bc bc 42.4 MPa
dbtc
washer at wall s bw bw 74.1 MPa
P
p
1d 2 d r2 2
4 w
;
(a) Derive a formula for the compressive stress c in the bar. (Assume that
the material has modulus of elasticity E and coefficient of thermal
expansion ).
∆TB
∆T
Problem 2.5-5 A bar AB of length L is held between rigid supports
and heated nonuniformly in such a manner that the temperature increase
T at distance x from end A is given by the expression T TBx3/L3,
where TB is the increase in temperature at end B of the bar
(see figure part a).
;
0
A
B
x
L
(a)
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Axially Loaded Members
(b) Now modify the formula in (a) if the rigid support at A is replaced by
an elastic support at A having a spring constant k (see figure part b).
Assume that only bar AB is subject to the temperature increase.
∆TB
∆T
0
k
A
B
x
L
(b)
Solution 2.5-5
(a) one degree statically indeterminate - use superposition
select reaction RB as the redundant; follow procedure below
Bar with nonuniform temperature change
COMPRESSIVE FORCE P REQUIRED TO SHORTEN THE BAR BY
THE AMOUNT EAd
1
EAa(¢TB)
L
4
P
COMPRESSIVE STRESS IN THE BAR
Ea(¢TB)
P
;
A
4
(b) one degree statically indeterminate - use
superposition
select reaction RB as the redundant then compute bar
elongations due to T & due to RB
L
due to temp. from above
dB1 a¢TB
4
ac At distance x:
¢T ¢TB a
x3
3
L
dB2 RB a
b
REMOVE THE SUPPORT AT THE END B OF THE BAR:
compatibility: solve for RB
RB Consider an element dx at a distance x from end A.
d Elongation of element dx
dd a(¢T)dx a(¢TB)a
x3
L3
bdx
d elongation of bar
L
L
1
dd a(¢TB) a 3 bdx a(¢TB)L
d
4
L
L0
L0
aa¢TB
a
B1
B2 0
L
b
4
1
L
+
b
k
EA
RB a¢TB
EA
EA
J 4a kL + 1 b K
so compressive stress in bar is:
sc x3
L
1
+
b
k
EA
RB
A
sc E a1¢TB2
4a
;
EA
+ 1b
kL
NOTE: sc in (b) is the same as in (a) if spring const. k
goes to infinity.
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Problem 2.5-6 A plastic bar ACB having two different solid circular
cross sections is held between rigid supports as shown in the figure.
The diameters in the left- and right-hand parts are 50 mm and
75 mm, respectively. The corresponding lengths are 225 mm and
300 mm. Also, the modulus of elasticity E is 6.0 GPa, and the coefficient of thermal expansion is 100 106/°C. The bar is subjected
to a uniform temperature increase of 30°C.
(a) Calculate the following quantities: (1) the compressive force
N in the bar; (2) the maximum compressive stress c; and
(3) the displacement C of point C.
(b) Repeat (a) if the rigid support at A is replaced by an elastic
support having spring constant k 50 MN/m (see figure
part b; assume that only the bar ACB is subject to the
temperature increase).
SECTION 2.5
Thermal Effects
50 mm C
75 mm
A
225 mm
155
B
300 mm
(a)
75 mm
50 mm C
A
k
225 mm
B
300 mm
(b)
Solution
NUMERICAL DATA
d1 50 mm
d2 75 mm
L1 225 mm
L2 300 mm
T 30°C
k 50 MN/m
(a) COMPRESSIVE
FORCE
N,
MAX. COMPRESSIVE STRESS
L1
E A1
C 0.314 mm ; () sign means jt C moves
left
100 (106/°C
E 6.0 GPa
DISPL. OF PT.
d C a ¢T( L1) RB
&
(b) COMPRESSIVE FORCE N, MAX. COMPRESSIVE STRESS &
DISPL. OF PT. C FOR ELASTIC SUPPORT CASE
Use RB as redundant as in (a)
C
p 2
p
d1
A2 d22
4
4
one-degree stat-indet - use RB as redundant
B1 T(L1
B1 T(L1
^ now add effect of elastic support; equate B1 and
B2 then solve for RB
A1 d B2 RB a
L2)
L1
L2
+
b
E A1
E A2
compatibility: B1 B2, solve for RB
RB a¢T( L1 + L2)
L1
L2
+
E A1
E A2
N RB
N 51.8 kN ;
max. compressive stress in AC since
it has the smaller area (A1 A2)
N
cmax 26.4 MPa
A1
displacement C of point C superposition of displacements in two released structures at C
scmax dB2 RB a
RB L2)
L1
L2
1
+
+ b
E A1
E A2
k
a¢T1 L1 + L22
L1
L2
1
+
+
E A1
EA2
k
N 31.2 kN
N RB
;
N
cmax 15.91 MPa
A1
super position
scmax d C a¢T( L1) RBa
C 0.546 mm
moves left
;
L1
1
+ b
E A1
k
; () sign means jt C
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Axially Loaded Members
Problem 2.5-7 A circular steel rod AB (diameter d1 1.0 in., length L1 3.0 ft) has a bronze sleeve (outer diameter d2 1.25 in., length L2 1.0 ft)
shrunk onto it so that the two parts are securely bonded (see figure).
Calculate the total elongation of the steel bar due to a temperature rise
T 500°F. (Material properties are as follows: for steel, Es 30 106 psi and
s 6.5 106/°F; for bronze, Eb 15 106 psi and b 11 106/°F.)
Solution 2.5-7
Steel rod with bronze sleeve
SUBSTITUTE NUMERICAL VALUES:
s 6.5 106/°F b 11 106/°F
Es 30 106 psi
Eb 15 106 psi
d1 1.0 in.
L1 36 in.
L2 12 in.
ELONGATION OF THE TWO OUTER PARTS OF THE BAR
1 s(T)(L1 L2)
6
(6.5 10 /°F)(500°F)(36 in. 12 in.)
0.07800 in.
ELONGATION OF THE MIDDLE PART OF THE BAR
The steel rod and bronze sleeve lengthen the same
amount, so they are in the same condition as the bolt
and sleeve of Example 2-8. Thus, we can calculate the
elongation from Eq. (2-21):
d2 As p 2
d 0.78540 in.2
4 1
d2 1.25 in.
Ab p
(d 2 d12) 0.44179 in.2
4 2
T 500°F
L2 12.0 in.
2 0.04493 in.
TOTAL ELONGATION
1
(as Es As + ab Eb Ab)(¢T)L2
Es As + Eb Ab
Problem 2.5-8 A brass sleeve S is fitted over a steel bolt B (see
figure), and the nut is tightened until it is just snug. The bolt has a
diameter dB 25 mm, and the sleeve has inside and outside diameters
d1 26 mm and d2 36 mm, respectively.
Calculate the temperature rise T that is required to produce a compressive stress of 25 MPa in the sleeve. (Use material properties as
follows: for the sleeve, S 21 106/°C and ES 100 GPa; for
the bolt, B 10 106/°C and EB 200 GPa.)
(Suggestion: Use the results of Example 2-8.)
2 0.123 in.
;
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SECTION 2.5
Solution 2.5-8
Thermal Effects
Brass sleeve fitted over a Steel bolt
sS
ES AS
b
a1 +
ES(aS aB)
EB AB
¢T ;
SUBSTITUTE NUMERICAL VALUES:
S 25 MPa
d2 36 mm
Subscript S means “sleeve”.
Subscript B means “bolt”.
ES 100 GPa
EQUATION (2-20A):
(aS aB)(¢T)ES EB AB
(Compression)
ES AS + EB AB
SOLVE FOR T:
sS ¢T sS(ES AS + EB AB)
(aS aB)ES EB AB
dB 25 mm
EB 200 GPa
6
S 21 10 /°C
Use the results of Example 2-8.
S compressive force in sleeve
d1 26 mm
B 10 106/°C
AS p 2
p
(d d21) (620 mm2)
4 2
4
AB ES AS
p
p
1.496
(dB)2 (625 mm2) 1 +
4
4
EB AB
¢T 25 MPa (1.496)
(100 GPa)(11 * 106/°C)
T 34°C
;
(Increase in temperature)
or
Problem 2.5-9 Rectangular bars of copper and aluminum are held by
pins at their ends, as shown in the figure. Thin spacers provide a separation
between the bars. The copper bars have cross-sectional dimensions
0.5 in. 2.0 in., and the aluminum bar has dimensions 1.0 in. 2.0 in.
Determine the shear stress in the 7/16 in. diameter pins if the temperature is raised by 100°F. (For copper, Ec 18,000 ksi and c 9.5 106/°F; for aluminum, Ea 10,000 ksi and a 13 106/°F.)
Suggestion: Use the results of Example 2-8.
Solution 2.5-9
Rectangular bars held by pins
Diameter of pin: dP Area of pin: AP 7
in. 0.4375 in.
16
p 2
d 0.15033 in.2
4 P
Area of two copper bars: Ac 2.0 in.2
Area of aluminum bar: Aa 2.0 in.2
T 100°F
157
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Axially Loaded Members
Copper: Ec 18,000 ksi c 9.5 106/°F
SUBSTITUTE NUMERICAL VALUES:
Aluminum: Ea 10,000 ksi
Pa Pc (3.5 * 106/° F)(100°F)(18,000 ksi)(2 in.2)
1 + a
a 13 106/°F
Use the results of Example 2-8.
Find the forces Pa and Pc in the aluminum bar and
copper bar, respectively, from Eq. (2-19).
18 2.0
ba
b
10 2.0
4,500 lb
FREE-BODY DIAGRAM OF PIN AT THE LEFT END
Replace the subscript “S” in that equation by “a” (for
aluminum) and replace the subscript “B” by “c” (for
copper), because for aluminum is larger than for
copper.
Pa Pc (aa ac)(¢T)Ea Aa Ec Ac
Ea Aa + Ec Ac
Note that Pa is the compressive force in the aluminum
bar and Pc is the combined tensile force in the two
copper bars.
Pa Pc (aa ac)(¢T)Ec Ac
Ec Ac
1 +
Ea Aa
V shear force in pin
Pc/2
2,250 lb
t average shear stress on cross section of pin
t
2,250 lb
V
AP
0.15033 in.2
t 15.0 ksi
;
Problem 2.5-10 A rigid bar ABCD is pinned at end A and supported by
two cables at points B and C (see figure). The cable at B has nominal
diameter dB 12 mm and the cable at C has nominal diameter
dC 20 mm. A load P acts at end D of the bar.
What is the allowable load P if the temperature rises by 60°C and each
cable is required to have a factor of safety of at least 5 against its ultimate
load?
(Note: The cables have effective modulus of elasticity E 140 GPa and
coefficient of thermal expansion 12 106/°C. Other properties of
the cables can be found in Table 2-1, Section 2.2.)
Solution 2.5-10
Rigid bar supported by two cables
FREE-BODY DIAGRAM OF BAR ABCD
From Table 2-1:
AB 76.7 mm2 E 140 GPa
T 60°C
AC 173 mm2
12 106/°C
EQUATION OF EQUILIBRIUM
MA 0 哵 哴 TB(2b)
or 2TB 4TC 5P
TB force in cable B
dB 12 mm
TC force in cable C
dC 20 mm
TC(4b) P(5b) 0
(Eq. 1)
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SECTION 2.5
DISPLACEMENT DIAGRAM
COMPATIBILITY:
C 2B
(Eq. 2)
FORCE-DISPLACEMENT AND TEMPERATURE-DISPLACEMENT
TBL
+ a(¢T)L
dB EAB
TCL
+ a(¢T)L
dC EAC
(Eq. 6)
SOLVE SIMULTANEOUSLY EQS. (1) AND (6):
TB 0.2494 P 3,480
TC 1.1253 P 1,740
in which P has units of newtons.
(Eq. 7)
(Eq. 8)
SOLVE EQS. (7) AND (8) FOR THE LOAD P:
PB 4.0096 TB 13,953
PC 0.8887 TC 1,546
(Eq. 3)
(Eq. 4)
Factor of safety 5
(TB)allow 20,400 N
TCL
2TBL
+ a(¢T)L + 2a(¢T)L
EAC
EAB
(Eq. 9)
(Eq. 10)
(TC)ULT 231,000 N
(TC)allow 46,200 N
From Eq. (9): PB (4.0096)(20,400 N)
95,700 N
SUBSTITUTE EQS. (3) AND (4) INTO EQ. (2):
13,953 N
From Eq. (10): PC (0.8887)(46,200 N) 1546 N
39,500 N
Cable C governs.
or
2TBAC TCAB E(T)AB AC
SUBSTITUTE NUMERICAL VALUES INTO EQ. (5):
TB(346) TC(76.7) 1,338,000
in which TB and TC have units of newtons.
ALLOWABLE LOADS
From Table 2-1:
(TB)ULT 102,000 N
RELATIONS
159
Thermal Effects
(Eq. 5)
Pallow 39.5 kN
Problem 2.5-11 A rigid triangular frame is pivoted at C and held by two identical
horizontal wires at points A and B (see figure). Each wire has axial rigidity EA 120 k
and coefficient of thermal expansion 12.5 106/°F.
(a) If a vertical load P 500 lb acts at point D, what are the tensile forces TA and
TB in the wires at A and B, respectively?
(b) If, while the load P is acting, both wires have their temperatures raised by
180°F, what are the forces TA and TB?
(c) What further increase in temperature will cause the wire at B to become slack?
;
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Axially Loaded Members
Solution 2.5-11
Triangular frame held by two wires
FREE-BODY DIAGRAM OF FRAME
(b) LOAD P AND TEMPERATURE INCREASE T
Force-displacement and temperature-displacement
relations:
TAL
+ a(¢T)L
(Eq. 8)
dA EA
TBL
+ a(¢T)L
EA
Substitute (8) and (9) into Eq. (2):
TAL
2TBL
+ a(¢T)L + 2a(¢T)L
EA
EA
dB or
EQUATION OF EQUILIBRIUM
MC 0 哵哴
(Eq. 9)
TA 2TB EA(T)
(Eq. 10)
Solve simultaneously Eqs. (1) and (10):
P(2b) TA(2b) TB(b) 0 or 2TA
TB 2P (Eq. 1)
DISPLACEMENT DIAGRAM
1
TA [4P + EAa(¢T)]
5
2
TB [P EAa(¢T)]
5
(Eq. 11)
(Eq. 12)
Substitute numerical values:
P 500 lb EA 120,000 lb
T 180°F
12.5 106/°F
1
TA (2000 lb + 270 lb) 454 lb
5
EQUATION OF COMPATIBILITY
A 2B
(Eq. 2)
(a) LOAD P ONLY
Force-displacement relations:
;
(c) WIRE B BECOMES SLACK
TAL
TBL
dB dA EA
EA
(L length of wires at A and B.)
Substitute (3) and (4) into Eq. (2):
(Eq. 3, 4)
Set TB 0 in Eq. (12):
P EA(T)
or
500 lb
P
EAa
(120,000 lb)(12.5 * 106/°F)
333.3°F
¢T TAL
2TBL
EA
EA
or TA 2TB
Solve simultaneously Eqs. (1) and (5):
4P
2P
TB 5
5
Numerical values:
P 500 lb
⬖TA 400 lb TB 200 lb
2
TB (500 lb 270 lb) 92 lb
5
;
TA (Eq. 5)
Further increase in temperature:
(Eqs. 6, 7)
;
T 333.3°F 180°F
153°F
;
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SECTION 2.5
161
Thermal Effects
Misfits and Prestrains
Problem 2.5-12 A steel wire AB is stretched between rigid supports (see
figure). The initial prestress in the wire is 42 MPa when the temperature is 20°C.
(a) What is the stress in the wire when the temperature drops to 0°C?
(b) At what temperature T will the stress in the wire become zero?
(Assume 14 106/°C and E 200 GPa.)
Solution 2.5-12
Steel wire with initial prestress
From Eq. (2-18): 2 E(T)
1
2 1
E(T)
Initial prestress: 1 42 MPa
42 MPa
(200 GPa)(14 106/°C)(20°C)
Initial temperature: T1 20°C
42 MPa
56 MPa 98 MPa
E 200 GPa
14 106/°C
(a) STRESS WHEN TEMPERATURE DROPS TO 0°C
T2 0°C
T 20°C
;
(b) TEMPERATURE WHEN STRESS EQUALS ZERO
1
2 0 1
E(T) 0
s1
¢T Ea
NOTE: Positive T means a decrease in temperature
and an increase in the stress in the wire.
(Negative means increase in temp.)
Negative T means an increase in temperature and a
decrease in the stress.
¢T Stress equals the initial stress 1 plus the additional
stress 2 due to the temperature drop.
42 MPa
(200 GPa)(14 * 106/°C
T 20°C 15°C 35°C
;
15°C
0.008 in.
Problem 2.5-13 A copper bar AB of length 25 in. and diameter 2 in. is placed in position at
A
room temperature with a gap of 0.008 in. between end A and a rigid restraint (see figure). The
bar is supported at end B by an elastic spring with spring constant k 1.2 106 lb/in.
(a) Calculate the axial compressive stress c in the bar if the temperature
rises 50°F. (For copper, use 9.6 106/°F and E 16 106 psi.)
(b) What is the force in the spring? (Neglect gravity effects.)
(C) Repeat (a) if k : 25 in.
d = 2 in.
B
k
C
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Axially Loaded Members
Solution 2.5-13
numerical data
compressive stress in bar
RA
s
957 psi
A
L 25 in. d 2 in. 0.008 in.
k 1.2 (106) lb/in.
E 16 (106) psi
9.6 (106)/°F T 50°F
p
A d2 A 3.14159 in2
4
(a) one-degree stat.-indet. if gap closes
TL
0.012 in.
exceeds gap
select RA as redundant & do superposition analysis
A1 d A2 RAa
A1
compatibility
RA d ¢
L
1
+
EA
k
L
1
+ b
EA
k
A2 A2 A1
(b) force in spring Fk RC
statics RA RC 0
RC RA
RC 3006 lb ;
(c) find compressive stress in bar if k goes to infinity
from expression for RA above, 1/k goes to zero, so
RA d¢
L
EA
2560 psi
RA
A
;
RA 3006 lb
2L
—
3
Problem 2.5-14 A bar AB having length L and axial rigidity EA is fixed
at end A (see figure). At the other end a small gap of dimension s exists
between the end of the bar and a rigid surface. A load P acts on the bar at
point C, which is two-thirds of the length from the fixed end.
If the support reactions produced by the load P are to be equal in
magnitude, what should be the size s of the gap?
Solution 2.5-14
s
RA 8042 lb
A
s
L
—
3
C
B
P
Bar with a gap (load P )
FORCE-DISPLACEMENT RELATIONS
d1 P A 2L
3B
EA
L length of bar
S size of gap
EA axial rigidity
Reactions must be equal; find S.
d2 RBL
EA
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SECTION 2.5
COMPATIBILITY EQUATION
Reactions must be equal.
1 2 S or
⬖ RA RB P 2RB
RBL
2PL
S
3EA
EA
(Eq. 1)
or S PL
6EA
;
NOTE: The gap closes when the load reaches the
value P/4. When the load reaches the value P, equal
to 6EAs/L, the reactions are equal (RA RB P/2).
When the load is between P/4 and P, RA is greater than
RB. If the load exceeds P, RB is greater than RA.
RA reaction at end A (to the left)
RB reaction at end B (to the left)
P RA
P
2
Substitute for RB in Eq. (1):
PL
2PL
S
3EA
2EA
EQUILIBRIUM EQUATION
RB 163
Thermal Effects
RB
Problem 2.5-15 Pipe 2 has been inserted snugly into Pipe 1,
but the holes for a connecting pin do not line up: there is a gap s.
The user decides to apply either force P1 to Pipe 1 or force P2 to
Pipe 2, whichever is smaller. Determine the following using the
numerical properties in the box.
Pipe 1 (steel)
Pipe 2 (brass)
Gap s
L1
RA
P1
L2
(a) If only P1 is applied, find P1 (kips) required to close gap s;
if a pin is then inserted and P1 removed, what are reaction
P2
P1 at L1
forces RA and RB for this load case?
(b) If only P2 is applied, find P2 (kips) required to close gap s;
L
P2 at —2
if a pin is inserted and P2 removed, what are reaction
2
forces RA and RB for this load case?
Numerical properties
(c) What is the maximum shear stress in the pipes, for the
E1 = 30,000 ksi, E2 = 14,000 ksi
loads in (a) and (b)?
a1 = 6.5 10–6/°F, a2 = 11 10–6/°F
(d) If a temperature increase T is to be applied to the entire
Gap s = 0.05 in.
structure to close gap s (instead of applying forces P1 and
L1 = 56 in., d1 = 6 in., t1 = 0.5 in., A1 = 8.64 in.2
P2), find the T required to close the gap. If a pin is inserted
L2 = 36 in., d2 = 5 in., t2 = 0.25 in., A2 = 3.73 in.2
after the gap has closed, what are reaction forces RA and RB
for this case?
(e) Finally, if the structure (with pin inserted) then cools to the
original ambient temperature, what are reaction forces RA and RB?
RB
Solution 2.5-15
(a) find reactions at A & B for applied force P1: first
compute P1, required to close gap
P1 E1A1
s
L1
P1 231.4 kips
;
stat-indet analysis with RB as the redundant
B1 s
L1
L2
+
b
d B2 RB a
E1A1
E2 A2
compatibility: B1
B2 0
RB s
a
L1
L2
+
b
E1A1
E2A2
RA RB
RB 55.2 k
;
(b) find reactions at A & B for applied force P2
E2A2
;
s P2 145.1 kips
L2
2
analysis after removing P2 is same as in (a) so
reaction forces are the same
P2 ;
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CHAPTER 2
Page 164
Axially Loaded Members
(c) max. shear stress in pipe 1 or 2 when either P1 or P2
P1
A1
;
is applied t maxa maxa 13.39 ksi
2
P2
A2
t maxb ;
maxb 19.44 ksi
2
(d) required ¢T and reactions at A & B
s
¢T reqd Treqd 65.8°F
a1L1 + a2L2
if pin is inserted but temperature remains at T
above ambient temp., reactions are zero
(e) if temp. returns to original ambient temperature, find
reactions at A & B
stat-indet analysis with RB as the redundant
compatibility: B1 B2 0
analysis is the same as in (a) & (b) above since gap
s is the same, so reactions are the same
;
Problem 2.5-16 A nonprismatic bar ABC made up of
a, T
RA
segments AB (length L1, cross-sectional area A1) and BC
(length L2, cross-sectional area A2) is fixed at end A and free at
A
L1, EA1 B
end C (see figure). The modulus of elasticity of the bar is E.
A small gap of dimension s exists between the end of the bar and
an elastic spring of length L3 and spring constant k3. If bar ABC only
(not the spring) is subjected to temperature increase T determine the following.
s
L2, EA2 C
D
L3, k3
(a) Write an expression for reaction forces RA and RD if the elongation of ABC exceeds gap length s.
(b) Find expressions for the displacements of points B and C if the elongation of ABC exceeds gap length s.
Solution 2.5-16
With gap s closed due to T, structure is one-degree
statically-indeterminate; select internal force (Q) at
juncture of bar & spring as the redundant; use superposition of two released structures in the solution
compatibility: rel1
rel2 s rel2 s rel1
rel2 s T(L1
L2)
rel1 relative displ. between end of bar at C & end of
spring due to T
rel1 T(L1 L2)
rel1 is greater than gap
length s
L1
L2
1
+
+
E A1
E A2
k3
E A1A2 k3
Q
L1A2 k3 + L2A1k3 + EA1A2
rel2 relative displ. between ends of bar & spring due
to pair of forces Q, one on end of bar at C &
the other on end of spring
Q
L1
L2
drel2 Q a
+
b +
E A1
E A2
k3
L1
L2
1
+
+
b
drel2 Q a
EA1
E A2
k3
Q
s a¢T1 L1 + L22
[ s a ¢T1 L1 + L22]
(a) REACTIONS AT A & D
statics: RA Q RD Q
RA s + a¢T1 L1 + L22
L1
L2
1
+
+
E A1
E A2
k3
RD RA
;
;
RD
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SECTION 2.5
(b) DISPLACEMENTS AT B & C
use superposition of displacements in the two
released structures
d B a¢T1 L12 RA a
L1
b
E A1
d B a ¢T 1 L12 [ s + a ¢T1 L1 + L22]
L1
L2
1
+
+
E A1
EA2
k3
;
Thermal Effects
d C a¢T1 L1 + L22 RAa
L1
L2
+
b
E A1
E A2
;
d C a ¢T1 L1 + L22 [ s + a ¢T1 L1 + L22]
a
165
L1
L2
1
+
+
E A1
EA2
k3
L1
b
EA1
a
L1
L2
+
b
EA1
EA2
Problem 2.5–17 Wires B and C are attached to a support at the left-hand
end and to a pin-supported rigid bar at the right-hand end (see figure).
Each wire has cross-sectional area A 0.03 in.2 and modulus of elasticity
E 30 106 psi. When the bar is in a vertical position, the length of each
wire is L 80 in. However, before being attached to the bar, the length of
wire B was 79.98 in. and of wire C was 79.95 in.
Find the tensile forces TB and TC in the wires under the action of a force
P 700 lb acting at the upper end of the bar.
700 lb
B
b
C
b
b
80 in.
Solution 2.5–17
Wires B and C attached to a bar
EQUILIBRIUM EQUATION
Mpin 0
TC(b)
2TB
P 700 lb
A 0.03 in.2
E 30 106 psi
LB 79.98 in.
LC 79.95 in.
;
TB(2b) P(3b)
TC 3P
(Eq. 1)
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CHAPTER 2
Page 166
Axially Loaded Members
DISPLACEMENT DIAGRAM
Combine Eqs. (3) and (5):
SB 80 in. LB 0.02 in.
TCL
SC + d
EA
SC 80 in. LC 0.05 in.
(Eq. 7)
Eliminate between Eqs. (6) and (7):
TB 2TC EASB
2EASC
L
L
(Eq. 8)
Solve simultaneously Eqs. (1) and (8):
Elongation of wires:
B SB
2
(Eq. 2)
C SC
(Eq. 3)
FORCE-DISPLACEMENT RELATIONS
TB L
dB EA
TC L
dC EA
EASB
2EASC
6P
+
5
5L
5L
TC 2EASB
4EASC
3P
+
5
5L
5L
(Eqs. 4, 5)
;
;
SUBSTITUTE NUMERICAL VALUES:
EA
2250 lb/in.
5L
TB 840 lb
SOLUTION OF EQUATIONS
45 lb 225 lb 660 lb
TC 420 lb 90 lb
;
;
450 lb 780 lb
(Both forces are positive, which means tension, as
required for wires.)
Combine Eqs. (2) and (4):
TBL
SB + 2d
EA
TB (Eq. 6)
P
Problem 2.5-18 A rigid steel plate is supported by three posts of
high-strength concrete each having an effective cross-sectional area
A 40,000 mm2 and length L 2 m (see figure). Before the load
P is applied, the middle post is shorter than the others by an amount
s 1.0 mm.
Determine the maximum allowable load Pallow if the allowable
compressive stress in the concrete is allow 20 MPa.
(Use E 30 GPa for concrete.)
S
s
C
C
C
L
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SECTION 2.5
Solution 2.5-18
Thermal Effects
167
Plate supported by three posts
EQUILIBRIUM EQUATION
P2 P
2P1
(Eq. 1)
COMPATIBILITY EQUATION
1 shortening of outer posts
2 shortening of inner post
1 2
s
(Eq. 2)
FORCE-DISPLACEMENT RELATIONS
d1 s size of gap 1.0 mm
L length of posts 2.0 m
A 40,000 mm2
allow 20 MPa
E 30 GPa
C concrete post
DOES THE GAP CLOSE?
Stress in the two outer posts when the gap is just
closed:
s
1.0 mm
s E Ea b (30 GPa) a
b
L
2.0 m
P1 L
EA
d2 (Eqs. 3, 4)
SOLUTION OF EQUATIONS
Substitute (3) and (4) into Eq. (2):
P1L
P2L
EAs
(Eq. 5)
+ s or P1 P2 EA
EA
L
Solve simultaneously Eqs. (1) and (5):
P 3P1 EAs
L
By inspection, we know that P1 is larger than P2.
Therefore, P1 will control and will be equal to allow A.
Pallow 3sallow A 15 MPa
Since this stress is less than the allowable stress, the
allowable force P will close the gap.
P2 L
EA
EAs
L
2400 kN 600 kN 1800 kN
1.8 MN
;
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Page 168
Axially Loaded Members
Problem 2.5-19 A capped cast-iron pipe is compressed by a brass rod,
Nut & washer
3
dw = — in.
4
as shown. The nut is turned until it is just snug, then add an additional
quarter turn to pre-compress the CI pipe. The pitch of the threads of the
bolt is p 52 mils (a mil is one-thousandth of an inch). Use the numerical
properties provided.
(
)
Steel cap
(tc = 1 in.)
(a) What stresses p and r will be produced in the cast-iron pipe and
brass rod, respectively, by the additional quarter turn of the nut?
(b) Find the bearing stress b beneath the washer and the shear stress
c in the steel cap.
Cast iron pipe
(do = 6 in.,
di = 5.625 in.)
Lci = 4 ft
Brass rod
1
dr = — in.
2
)
(
Modulus of elasticity, E:
Steel (30,000 ksi)
Brass (14,000 ksi)
Cast iron (12,000 ksi)
Solution 2.5-19
The figure shows a section through the pipe, cap and rod
Arod 0.196 in2
NUMERICAL PROPERTIES
compatibility equation
Lci 48 in.
Es 30000 ksi
Eb 14000 ksi
1
Ec 12000 ksi tc 1 in. p 52 (103) in. n 4
3
1
dr in. do 6 in. di 5.625 in.
dw in.
4
2
(a) FORCES & STRESSES IN PIPE & ROD
one degree stat-indet - cut rod at cap & use force in
rod (Q) as the redundant
rel1 relative displ. between cut ends of rod due to
1/4 turn of nut
rel1 np
ends of rod move apart, not
together, so this is ()
rel2 relative displ. between cut ends of rod due
pair of forces Q
L + 2tc
Lci
+
b
EbArod
EcApipe
p
p
Apipe (do2 di2)
Arod dr2
4
4
Q
Apipe 3.424 in2
rel1
rel2 0
np
Lci + 2tc
Lci
+
EbArod
EcApipe
Q 0.672 kips
Frod Q
Fpipe Q
statics
stresses
sc sb Fpipe
Apipe
Frod
Arod
c 0.196 ksi
b 3.42 ksi
;
;
(b) BEARING AND SHEAR STRESSES IN STEEL CAP
sb d rel2 Qa
tc Frod
p
(d 2 dr 2)
4 w
Frod
pdwtc
c
b 2.74 ksi
0.285 ksi
;
;
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SECTION 2.5
Thermal Effects
169
Problem 2.5-20 A plastic cylinder is held snugly between a rigid plate
and a foundation by two steel bolts (see figure).
Determine the compressive stress p in the plastic when the nuts on the
steel bolts are tightened by one complete turn.
Data for the assembly are as follows: length L 200 mm, pitch of the bolt
threads p 1.0 mm, modulus of elasticity for steel Es 200 GPa, modulus of
elasticity for the plastic Ep 7.5 GPa, cross-sectional area of one bolt As 36.0 mm2, and cross-sectional area of the plastic cylinder Ap 960 mm2.
Solution 2.5-20
Steel
bolt
L
Plastic cylinder and two steel bolts
COMPATIBILITY EQUATION
L 200 mm
P 1.0 mm
s elongation of steel bolt
Es 200 GPa
p shortening of plastic cylinder
As 36.0 mm2 (for one bolt)
s
Ep 7.5 GPa
p np
(Eq. 2)
FORCE-DISPLACEMENT RELATIONS
Ap 960 mm2
n 1 (See Eq. 2-22)
EQUILIBRIUM EQUATION
ds PsL
EsAs
dp PpL
Ep Ap
(Eq. 3, Eq. 4)
SOLUTION OF EQUATIONS
Substitute (3) and (4) into Eq. (2):
Pp L
PsL
+
np
Es As
Ep Ap
Solve simultaneously Eqs. (1) and (5):
Ps tensile force in one steel bolt
Pp compressive force in plastic cylinder
Pp 2Ps
Pp (Eq. 1)
2npEs AsEp Ap
L(Ep Ap + 2Es As)
(Eq. 5)
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CHAPTER 2
Page 170
Axially Loaded Members
STRESS IN THE PLASTIC CYLINDER
sp Pp
Ap
2EsAs 21.6 106 N
D EpAp
2np Es As Ep
sp L(Ep Ap + 2Es As)
2np N
2(1)(1.0 mm) N
a b a b
L D
200 mm
D
25.0 MPa
SUBSTITUTE NUMERICAL VALUES:
;
N Es As Ep 54.0 10 N /m
15
2
2
Problem 2.5-21 Solve the preceding problem if the data for the assembly are
as follows: length L 10 in., pitch of the bolt threads p 0.058 in., modulus
of elasticity for steel Es 30 106 psi, modulus of elasticity for the plastic
Ep 500 ksi, cross-sectional area of one bolt As 0.06 in.2, and crosssectional area of the plastic cylinder Ap 1.5 in.2
Solution 2.5-21
Steel
bolt
L
Plastic cylinder and two steel bolts
COMPATIBILITY EQUATION
L 10 in.
s elongation of steel bolt
p 0.058 in.
Es 30 106 psi
p shortening of plastic cylinder
s
p np
(Eq. 2)
As 0.06 in.2 (for one bolt)
Ep 500 ksi
Ap 1.5 in.2
n 1 (see Eq. 2-22)
FORCE-DISPLACEMENT RELATIONS
EQUILIBRIUM EQUATION
Ps tensile force in one steel bolt
ds Pp compressive force in plastic cylinder
Pp 2Ps
(Eq. 1)
Ps L
Es As
dp Pp L
Ep Ap
(Eq. 3, Eq. 4)
SOLUTION OF EQUATIONS
Substitute (3) and (4) into Eq. (2):
Pp L
Ps L
+
np
Es As
Ep Ap
(Eq. 5)
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Page 171
SECTION 2.5
SUBSTITUTE NUMERICAL VALUES:
Solve simultaneously Eqs. (1) and (5):
Pp N Es As Ep 900 109 lb2/in.2
2 np Es As Ep Ap
D Ep Ap
L(Ep Ap + 2Es As)
STRESS IN THE PLASTIC CYLINDER
sp Pp
Ap
sp 2 np Es As Ep
171
Thermal Effects
;
L(Ep Ap + 2Es As)
2Es As 4350 103 lb
2np N
2(1)(0.058in.) N
a b a b
L D
10 in.
D
2400 psi
;
d = np
Problem 2.5-22 Consider the sleeve made from two copper tubes joined by tin-lead
L1 = 40 mm,
d1 = 25 mm,
t1 = 4 mm
(a) Find the forces in the sleeve and bolt, Ps and PB, due to both the prestress
in the bolt and the temperature increase. For copper, use Ec 120 GPa and
c 17 106/°C; for steel, use Es 200 GPa and s 12 106/°C.
The pitch of the bolt threads is p 1.0 mm. Assume s 26 mm and bolt
diameter db 5 mm.
(b) Find the required length of the solder joint, s, if shear stress in the sweated
joint cannot exceed the allowable shear stress aj 18.5 MPa.
(c) What is the final elongation of the entire assemblage due to both temperature change T and the initial prestress in the bolt?
Brass
cap
T
S
T
L2 = 50 mm,
d2 = 17 mm,
t2 = 3 mm
solder over distance s. The sleeve has brass caps at both ends, which are held in
place by a steel bolt and washer with the nut turned just snug at the outset. Then, two
“loadings” are applied: n 1/2 turn applied to the nut; at the same time the internal
temperature is raised by T 30°C.
Copper
sleeve
Steel
bolt
Solution 2.5-22
p 2
d
4 b
The figure shows a section through the sleeve, cap and
bolt
Ab NUMERICAL PROPERTIES (SI UNITS)
Ab 19.635 mm2
n
1
2
p 1.0 mm
c 17 (106)/°C
Ec 120 GPa
6
s 12 (10 )/°C
Es 200 GPa
aj
18.5 MPa
L1 40 mm
d1 25 mm
A2 T 30°C
s 26 mm
t1 4 mm
db 5 mm
L2 50 mm
d1 2t1 17 mm
t2 3 mm
d2 17 mm
A1 p 2
[d 1 d1 2 t122]
4 1
A1 263.894 mm2
p
[ d 2 1 d2 2t222]
4 2
A2 131.947 mm2
(a) FORCES IN SLEEVE & BOLT
one-degree stat-indet - cut bolt & use force in bolt
(PB) as redundant (see sketches below)
B1 np
sT(L1
L2 s)
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CHAPTER 2
d B2 PB c
Axially Loaded Members
L1 + L2 s
L1 s
L2 s
s
+
+
+
d
Es Ab
EcA1
Ec A2
Ec( A1 + A2)
compatibility
PB Page 172
B1
B2 0
[ np + a s ¢T( L1 + L2 s)]
L1 + L2 s
L1 s
L2 s
s
c
+
+
+
d
EsAb
Ec A1
Ec A2
Ec ( A 1 + A2)
PB 25.4 kN
Sketches illustrating superposition procedure for statically-indeterminate analysis
δ = np
cap
∆T
L1
Actual
indeterminate
structure
under load(s)
=
S
∆T
L2
1° SI superposition analysis using
internal force in bolt as the
redundant
sleeve
bolt
Two released structures (see below) under:
(1)load(s); (2) redundant applied as a load
δ = np
∆T
+
Ps
S
S
∆T
cut
bolt
δB1
δB1
relative
displacement
across cut bolt, δB1
due to both δ and
∆T (positive if pieces
move together)
relative
displacement
across cut bolt, δB2
due to Pb (positive
if pieces move
together)
PB
δB2
PB
δB2
apply redundant
internal force Ps &
find relative
displacement
across cut bolt,
δB2
;
Ps PB
;
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Page 173
SECTION 2.5
(b) REQUIRED LENGTH OF SOLDER JOINT≈
P
t
As d2s
As
PB
sreqd pd2t aj
sreqd 25.7 mm
d s Ps c
Thermal Effects
173
L1 s
L2 s
s
+
+
d
Ec A1
Ec A2
Ec (A1 + A2)
s 0.064 mm
f b
s
f 0.35 mm
;
(c) FINAL ELONGATION
f net of elongation of bolt (b) & shortening of
sleeve (s)
d b PBa
L1 + L2 s
b
EsAb
b 0.413 mm
Problem 2.5-23 A polyethylene tube (length L) has a cap which when installed
compresses a spring (with undeformed length L1 L) by amount (L1 L).
Ignore deformations of the cap and base. Use the force at the base of the spring as
the redundant. Use numerical properties in the boxes given.
(a)
(b)
(c)
(d)
What is the resulting force in the spring, Fk?
What is the resulting force in the tube, Ft?
What is the final length of the tube, Lf?
What temperature change T inside the tube will result in zero force in
the spring?
d = L1 – L
Cap (assume rigid)
Tube
(d0, t, L, at, Et)
Spring (k, L1 > L)
Modulus of elasticity
Polyethylene tube (Et = 100 ksi)
Coefficients of thermal expansion
at = 80 10–6/°F, ak = 6.5 10–6/°F
Properties and dimensions
1
d0 = 6 in. t = — in.
8
kip
L1 = 12.125 in. > L = 12 in. k = 1.5 –––
in.
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CHAPTER 2
Page 174
Axially Loaded Members
Solution 2.5-23
solve for redundant Q
The figure shows a section through the tube, cap and
spring
Q
Properties & dimensions
Fk 0.174 kips
do 6 in.
At t
1
in.
8
At 2.307 in2
kip
k 1.5
in
L1 12.125 in. L 12 in.
L1 L
k 6.5(106)/F
T 0
(b) Ft force in tube Q
f
1
k
ft Lf L
Lf L Qft
c1
c2
t(T)L
i.e., add displacements
for the two released structures
to initial tube length L
Lf 12.01 in.
;
(d) Set Q 0 to find T required to reduce spring force
to zero
L
EtAt
2 rel. displ. across cut spring due to redundant
Q(f ft)
¢T reqd d
(a k L1 + a tL)
Treqd 141.9F
1 rel. displ. across cut spring due to precompression and T kTL1 tTL
compatibility: 1
;
(c) Final length of tube
t 80 (106)/F
(a) Force in spring Fk redundant Q
compressive force in
spring (Fk) & also
tensile force in tube
NOTE: if tube is rigid, Fk k 0.1875 kips
0.125 in.
note that Q result below is for
zero temp. (until part(d))
Flexibilities
;
Et 100 ksi
p
[ d 2 ( do 2 t)2]
4 o
spring is 1/8 in.
longer than tube
d + ¢T ( a kL1 + a tL)
Fk
f + ft
since t k, a temp.
increase is req’d to expand
tube so that spring force
goes to zero
2 0
Steel wires
Problem 2.5-24 Prestressed concrete beams are sometimes
manufactured in the following manner. High-strength steel wires
are stretched by a jacking mechanism that applies a force Q, as
represented schematically in part (a) of the figure. Concrete is then
poured around the wires to form a beam, as shown in part (b).
After the concrete sets properly, the jacks are released and the
force Q is removed [see part (c) of the figure]. Thus, the beam is left
in a prestressed condition, with the wires in tension and the concrete
in compression.
Let us assume that the prestressing force Q produces in the steel
wires an initial stress 0 620 MPa. If the moduli of elasticity of the
steel and concrete are in the ratio 12:1 and the cross-sectional areas
are in the ratio 1:50, what are the final stresses s and c in the two
materials?
Q
Q
(a)
Concrete
Q
Q
(b)
(c)
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Page 175
SECTION 2.5
Solution 2.5-24
Thermal Effects
Prestressed concrete beam
L length
0 initial stress in wires
Q
620 MPa
As
As total area of steel wires
Ac area of concrete
50 As
Es 12 Ec
Ps final tensile force in steel wires
Pc final compressive force in concrete
EQUILIBRIUM EQUATION
Ps Pc
COMPATIBILITY EQUATION AND
FORCE-DISPLACEMENT RELATIONS
STRESSES
(Eq. 1)
1 initial elongation of steel wires
sc QL
s0L
EsAs
Es
2 final elongation of steel wires
Ps
As
s0
EsAs
1 +
EcAc
Pc
s0
Ac
Es
Ac
+
As
Ec
0 620 MPa
3 shortening of concrete
Pc L
Ec Ac
1 2 3
or
s0L
PsL
PcL
Es
EsAs
EcAc
Solve simultaneously Eqs. (1) and (3):
s0As
EsAs
1 +
EcAc
;
;
SUBSTITUTE NUMERICAL VALUES:
PsL
EsAs
Ps Pc ss (Eq. 2, Eq. 3)
Es
12
Ec
As
1
Ac
50
ss 620 MPa
500 MPa (Tension)
12
1 +
50
sc 620 MPa
10 MPa (Compression)
50 + 12
;
;
175
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CHAPTER 2
Page 176
Axially Loaded Members
Problem 2.5-25 A polyethylene tube (length L) has a cap which is held in place
by a spring (with undeformed length L1 L). After installing the cap, the spring
is post-tensioned by turning an adjustment screw by amount . Ignore deformations of the cap and base. Use the force at the base of the spring as the redundant.
Use numerical properties in the boxes below.
(a)
(b)
(c)
(d)
What is the resulting force in the spring, Fk?
What is the resulting force in the tube, Ft?
What is the final length of the tube, Lf?
What temperature change T inside the tube will result in zero force in the
spring?
Cap (assume rigid)
Tube
(d0, t, L, at, Et)
Spring (k, L1 < L)
d = L – L1
Adjustment
screw
Modulus of elasticity
Polyethylene tube (Et = 100 ksi)
Coefficients of thermal expansion
at = 80 10–6/°F, ak = 6.5 10–6/°F
Properties and dimensions
1
d0 = 6 in. t = — in.
8
kip
L = 12 in. L1 = 11.875 in. k = 1.5 –––
in.
Solution 2.5-25
The figure shows a section through the tube, cap and spring
Properties & dimensions
do 6 in.
1
t in.
8
k 6.5(106)
k 1.5
At 2.307 in2
kip
in
t 80 (106)
p
At [ d2o 1 do 2t22]
4
spring is 1/8 in. shorter
than tube
L L1
0.125 in.
T 0
note that Q result below is for zero temp. (until part (d))
Et 100 ksi
L 12 in. L1 11.875 in.
Pretension & temperature
Flexibilities
f
1
k
ft L
EtAt
(a) Force in spring (Fk) redundant (Q)
follow solution procedure outlined in Prob. 2.5-23
solution
Q
d + ¢T 1a k L1 + a t L2
f + ft
Fk
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Page 177
SECTION 2.5
Fk 0.174 kips
;
also the compressive force in
the tube
(b) force in tube Ft Q 0.174 k
(c) Final length of tube & spring Lf L
Lf L Qft
t(T)L Lf 11.99 in.
c2
;
177
(d) Set Q 0 to find T required to reduce spring
force to zero
;
c1
Thermal Effects
¢ Treqd d
1a kL1 + a t L2
Treqd 141.6F
since t k, a temp.
drop is req’d to shrink
tube so that spring force
goes to zero
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CHAPTER 2
Page 178
Axially Loaded Members
Stresses on Inclined Sections
Problem 2.6-1 A steel bar of rectangular cross section
(1.5 in. 2.0 in.) carries a tensile load P (see figure). The allowable stresses in tension and shear are 14,500 psi and 7,100 psi,
respectively. Determine the maximum permissible load Pmax.
2.0 in.
P
P
1.5 in.
Solution 2.6-1
MAXIMUM LOAD - tension
2.0 in.
P
P
Pmax1 aA
Pmax1 43500 lbs
MAXIMUM LOAD - shear
Pmax2 2aA
1.5 in.
Because allow is less than one-half of allow, the shear
stress governs.
NUMERICAL DATA
A 3 in2
Pmax2 42,600 lbs
a 14500 psi
a 7100 psi
Problem 2.6-2 A circular steel rod of diameter d is subjected to a tensile
force P 3.5 kN (see figure). The allowable stresses in tension and shear
are 118 MPa and 48 MPa, respectively. What is the minimum permissible
diameter dmin of the rod?
d
P
Solution 2.6-2
P
d
P = 3.5 kN
Pmax 2t a a
dmin P 3.5 kN
a 118 MPa
a 48 MPa
Find Pmax then rod diameter
since a is less than 1/2 of a, shear governs
NUMERICAL DATA
p
dmin2 b
4
2
P
pt
A a
dmin 6.81 mm
;
P = 3.5 kN
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SECTION 2.6
Problem 2.6-3 A standard brick (dimensions 8 in. 4 in. 2.5 in.) is compressed
P
lengthwise by a force P, as shown in the figure. If the ultimate shear stress for brick is
1200 psi and the ultimate compressive stress is 3600 psi, what force Pmax is required to break
the brick?
8 in.
Solution 2.6-3
2.5 in.
Maximum shear stress:
t max 4 in.
4 in.
Standard brick in compression
P
8 in.
179
Stresses on Inclined Sections
2.5 in.
sx
P
2
2A
ult 3600 psi
ult 1200 psi
Because ult is less than one-half of ult, the shear stress
governs.
t max A 2.5 in. 4.0 in. 10.0 in.2
Maximum normal stress:
sx P
2A
or P max 2Atult
P max 2(10.0 in.2)(1200 psi) 24,000 lb
;
P
A
Problem 2.6-4 A brass wire of diameter d 2.42 mm is stretched tightly
between rigid supports so that the tensile force is T 98 N (see figure). The
coefficient of thermal expansion for the wire is 19.5 10-6/°C and the
modulus of elasticity is E = 110 GPa
(a) What is the maximum permissible temperature drop T if the allowable
shear stress in the wire is 60 MPa?
(b) At what temperature changes does the wire go slack?
T
d
T
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CHAPTER 2
Page 180
Axially Loaded Members
Solution 2.6-4
Brass wire in tension
d
T
a 60 MPa
T
A
p 2
d
4
T
2 ta
A
¢Tmax Ea
NUMERICAL DATA
d 2.42 mm
T 98 N
19.5 (106)/°C E 110 GPa
¢Tmax 46°C (drop)
(a) ¢Tmax (DROP IN TEMPERATURE)
(b) ¢T AT WHICH WIRE GOES SLACK
s
T
(E a ¢T)
A
ta E a ¢T
T
2A
2
max increase ¢T until s 0
s
2
T
E aA
¢T 9.93°C (increase)
¢T Problem 2.6-5 A brass wire of diameter d 1/16 in. is stretched between
rigid supports with an initial tension T of 37 lb (see figure). Assume that the
coefficient of thermal expansion is 10.6 106/°F and the modulus of
elasticity is 15 106 psi.)
d
T
T
(a) If the temperature is lowered by 60°F, what is the maximum
shear stress max in the wire?
(b) If the allowable shear stress is 10,000 psi, what is the
maximum permissible temperature drop?
(c) At what temperature change T does the wire go slack?
Solution 2.6-5
d
T
T
T
2ta
A
¢Tmax Ea
NUMERICAL DATA
d
1
in
16
T 37 lb
10.6 (106)/°F
Tmax 49.9°F
;
(c) T AT WHICH WIRE GOES SLACK
E 15 (10 ) psi
T 60°F
p 2
A d
4
(a) max (DUE TO DROP IN TEMPERATURE)
6
tmax (b) ¢Tmax FOR ALLOW. SHEAR STRESS
sx
2
T
(E a ¢T)
A
tmax 2
max 10800 psi
;
increase T until 0
¢T T
E aA
T 75.9°F (increase)
;
a 10000 psi
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SECTION 2.6
Stresses on Inclined Sections
Problem 2.6-6 A steel bar with diameter d 12 mm is subjected to
a tensile load P 9.5 kN (see figure).
(a) What is the maximum normal stress max in the bar?
(b) What is the maximum shear stress max?
(c) Draw a stress element oriented at 45° to the axis of the bar and show
all stresses acting on the faces of this element.
Solution 2.6-6
d = 12 mm
P
181
P = 9.5 kN
Steel bar in tension
(b) MAXIMUM SHEAR STRESS
The maximum shear stress is on a 45° plane and
equals x/2.
tmax P 9.5 kN
(a) MAXIMUM NORMAL STRESS
sx sx
42.0 MPa
2
(c) STRESS ELEMENT AT
;
45°
P
9.5 kN
84.0 MPa
p
A
(12 mm)2
4
max 84.0 MPa
;
NOTE: All stresses have units of MPa.
Problem 2.6-7 During a tension test of a mild-steel specimen
(see figure), the extensometer shows an elongation of 0.00120
in. with a gage length of 2 in. Assume that the steel is stressed
below the proportional limit and that the modulus of elasticity
E 30 106 psi.
(a) What is the maximum normal stress max in
the specimen?
(b) What is the maximum shear stress max?
(c) Draw a stress element oriented at an angle of 45° to the
axis of the bar and show all stresses acting on the
faces of this element.
2 in.
T
T
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CHAPTER 2
Solution 2.6-7
Page 182
Axially Loaded Members
Tension test
(b) MAXIMUM SHEAR STRESS
The maximum shear stress is on a 45° plane and
equals x/2.
tmax Elongation: 0.00120 in.
(2 in. gage length)
d
0.00120 in.
Strain: â 0.00060
L
2 in.
sx
9,000 psi
2
(c) STRESS ELEMENT AT
;
45°
Hooke’s law: x E (30 106 psi)(0.00060)
18,000 psi
(a) MAXIMUM NORMAL STRESS
x is the maximum normal stress.
max 18,000 psi
NOTE: All stresses have units of psi.
;
Problem 2.6-8 A copper bar with a rectangular cross section is held
45∞
without stress between rigid supports (see figure). Subsequently, the
temperature of the bar is raised 50°C.
Determine the stresses on all faces of the elements A and B, and show
these stresses on sketches of the elements.
(Assume 17.5 106/°C and E 120 GPa.)
Solution 2.6-8
A
B
Copper bar with rigid supports
MAXIMUM SHEAR STRESS
tmax T 50°C (Increase)
sx
52.5 MPa
2
STRESSES ON ELEMENTS A AND B
17.5 106/°C
E 120 GPa
STRESS DUE TO TEMPERATURE INCREASE
x E (T )
(See Eq. 2-18 of Section 2.5)
105 MPa (Compression)
NOTE: All stresses have units of MPa.
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SECTION 2.6
Problem 2.6-9 The bottom chord AB in a small truss ABC
(see figure) is fabricated from a W8 28 wide-flange steel
section. The cross-sectional area A 8.25 in.2 (Appendix E,
Table E-1 (a)) and each of the three applied loads P 45 k.
First, find member force NAB; then, determine the normal and
shear stresses acting on all faces of stress elements located in the
web of member AB and oriented at (a) an angle 0°, (b) an
angle 30°, and (c) an angle 45°. In each case, show the
stresses on a sketch of a properly oriented element.
Stresses on Inclined Sections
183
P
P = 45 kips
C
9 ft
B
12 ft
A
P
Ax
Ay
By
NAB
NAB
u
Solution 2.6-9
Statics
P 45 kips
MA 0
9
P
By 12
By 33.75 k
BCV By
FH 0 at B
degrees in web of AB
x 10.9 ksi
(a)
12
BCH BCV
9
BCH 45 k
NAB BCH P
NAB 90 kips (compression)
;
Normal and shear stresses on elements at 0, 30 & 45
(b)
0
sx NAB
A
A 8.25 in2
;
x 10.91 ksi
;
30°
on x face
xcos ( )2
8.18 ksi
xsin ( ) cos ( )
4.72 ksi
on y face
uu +
xcos( )2
xsin( ) cos( )
p
2
2.73 ksi
4.72 ksi
;
;
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CHAPTER 2
(c)
Page 184
Axially Loaded Members
45 degrees
xcos( )2
p
2
5.45 ksi
xsin( )cos( )
5.45 ksi
uu +
on y face
on x face
xcos( )
5.45 ksi
2
xsin ( ) cos ( )
5.45 ksi
;
;
Problem 2.6-10 A plastic bar of diameter d 32 mm is
compressed in a testing device by a force P 190 N applied as
shown in the figure.
P = 190 N
100 mm
(a) Determine the normal and shear stresses acting on all
faces of stress elements oriented at (1) an angle 0°,
(2) an angle 22.5°, and (3) an angle 45°. In each
case, show the stresses on a sketch of a properly oriented
element. What are max and max?
(b) Find max and max in the plastic bar if a re-centering
spring of stiffness k is inserted into the testing device, as
shown in the figure. The spring stiffness is 1/6 of the axial
stiffness of the plastic bar.
300 mm
200 mm
Re-centering
spring
(Part (b) only)
Plastic bar
u
d = 32 mm
k
Solution
NUMERICAL DATA
(2)
p 2
d
4
A 804.25 mm2
A
d 32 mm
P 190 N
on x face
xcos( )2
807 kPa
a 100 mm
(a) Statics - FIND COMPRESSIVE FORCE F & STRESSES IN
on y face
PLASTIC BAR
sx P( a + b)
a
F
A
x 0.945 MPa
or
(1)
0 degrees
xsin( ) cos( )
334.1 kPa
x 945 kPa
(3)
max 945 kPa
max 472 kPa
uu +
xcos( )2
138.39 kPa
F 760 N
from (1), (2) & (3) below
max x
;
xsin( ) cos( )
334 kPa ;
b 300 mm
F
22.50 degrees
on x face
xcos( )2
472 kPa
sx
472 kPa
2
x 945 kPa
45 degrees
;
;
xsin( ) cos( )
472 kPa ;
p
2
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Page 185
SECTION 2.6
on y face
xcos( )2
p
2
uu +
force in plastic bar
472.49 kPa
xsin( ) cos( )
472.49 kPa
δ/3
6k
k
2kδ
kδ
F
A
sx P
100 mm
185
4P
b
5k
8
P
5
F 304 N
normal and shear stresses in plastic bar
IN PLASTIC BAR
200 mm
F (2k) a
F
(b) ADD SPRING - FIND MAX. NORMAL & SHEAR STRESSES
100 mm
Stresses on Inclined Sections
δ
tmax x 0.38
max 378 kPa
sx
2
max 189 kPa
;
;
a Mpin 0
P(400) [2k (100) k (300)]
d
4 P
5k
Problem 2.6-11 A plastic bar of rectangular cross section (b 1.5 in.
L
—
2
and h 3 in.) fits snugly between rigid supports at room temperature (68°F)
but with no initial stress (see figure). When the temperature of the bar is
raised to 160°F, the compressive stress on an inclined plane pq at midspan
becomes 1700 psi.
L
—
2
L
—
4
p
6
u
P
(a) What is the shear stress on plane pq? (Assume 60 10 /°F
and E 450 103 psi.)
(b) Draw a stress element oriented to plane pq and show the stresses
Load P for part (c) only
acting on all faces of this element.
(c) If the allowable normal stress is 3400 psi and the allowable shear
stress is 1650 psi, what is the maximum load P (in x direction)
which can be added at the quarter point (in addition to
thermal effects above) without exceeding allowable stress values in the bar?
b
h
q
Solution 2.6-11
(a)
NUMERICAL DATA
b 1.5 in
h 3 in
A bh
T 92°F
T (160 68)°F
SHEAR STRESS ON PLANE PQ
STAT-INDET ANALYSIS GIVES, FOR REACTION AT RIGHT
SUPPORT:
R EA T
pq 1700 psi
A 4.5 in
2
6
60 (10 )/°F
E 450 (103) psi
sx R
A
R 11178 lb
x 2484 psi
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CHAPTER 2
Page 186
Axially Loaded Members
using xcos( )2
u acos a
s pq
A sx
cos1u22 b
s pq
from (a) (for temperature increase T):
sx
RR1 EA T
34.2°
Stresses in bar (0 to L/4)
pq xcos( )
stresses at
pq 1154 psi
;
pq 1700 psi
2
/2 (y face)
p 2
b
2
s y s xcosa u +
y 784 psi
tmax psi
4
115
)
t (b
E a¢T
3P
+
2
8A
4A
Pmax1 12ta + Ea¢ T2
3
Pmax1 34704 lb
ta 3 Pmax1
E a¢T
+
2
8A
max 1650 psi
check
(b) STRESS ELEMENT FOR PLANE PQ
784
sx
3P
tmax 4A
2
set max a & solve for Pmax1
s x Ea¢ T +
now with , can find shear stress on plane pq
pq xsin( )cos( )
RL1 EA T
psi
si
0 p θ = 34.2°
170
Par
s x Ea ¢ T +
x 3300 psi
3Pmax1
4A
less than a
Stresses in bar (L/4 to L)
sx
P
tmax 4A
2
set max a & solve for Pmax2
s x E a¢ T (c) MAX. LOAD AT QUARTER POINT
a 1650 psi
2 a 3300
a 3400 psi
less than a so
shear controls
stat-indet analysis for P at L/4 gives, for reactions:
RR2 P
4
RL2 3
P
4
(tension for 0 to L/4 & compression for rest of bar)
Pmax2 4A(2a E T)
Pmax2 14688 lb
tmax ;
Pmax2
E a¢ T
2
8A
s x Ea¢ T Pmax2
4A
shear in segment (L/4
to L) controls
max 1650 psi
x 3300 psi
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SECTION 2.6
L
—
2
Problem 2.6-12 A copper bar of rectangular cross section (b 18 mm
187
Stresses on Inclined Sections
and h 40 mm) is held snugly (but without any initial stress) between
rigid supports (see figure). The allowable stresses on the inclined plane pq
bL
at midspan, for which 55°, are specified as 60 MPa in compression
p
and 30 MPa in shear.
P
(a) What is the maximum permissible temperature rise T if the
allowable stresses on plane pq are not to be exceeded? (Assume A
17 106/°C and E 120 GPa.)
Load for part (c) only
(b) If the temperature increases by the maximum permissible amount,
what are the stresses on plane pq?
(c) If the temperature rise T 28°C, how far to the right of end A
(distance L, expressed as a fraction of length L) can load P 15 kN
be applied without exceeding allowable stress values in the bar? Assume
that a 75 MPa and a 35 MPa.
L
—
2
b
u
h
B
q
Solution 2.6-12
(b) STRESSES ON PLANE PQ FOR max. TEMP.
x E Tmax
pq xcos( )2
x 63.85 MPa
pq 21.0 MPa
pq xsin( )cos( )
NUMERICAL DATA
p
u 55a
b
180
(a) FIND Tmax BASED ON ALLOWABLE NORMAL & SHEAR
STRESS VALUES ON PLANE pq
s x
x E Tmax
¢ Tmax Ea
2
pq xcos( )
pq xsin( )cos( )
^ set each equal to corresponding allowable &
solve for x
s pqa
s x1 x1 182.38 MPa
cos1u22
sin1u2cos1u2
x2 63.85 MPa
lesser value controls so allowable shear stress governs
¢Tmax s x2
Ea
;
T 28 DEGREES C
P 15 kN from one-degree stat-indet analysis,
reactions RA & RB due to load P are:
b 18 mm
h 40 mm
A bh
A 720 mm2
pqa 60 MPa
pqa 30 Mpa
6
17 (10 )/°C
E 120 GPa
T 20°C
P 15 kN
s x2 pq 30 MPa
(c) ADD LOAD P IN X-DIRECTION TO TEMPERATURE
CHANGE & FIND LOCATION OF LOAD
radians
t pqa
;
Tmax 31.3°C
;
RA (1 )P
RB P
now add normal stresses due to P to thermal
stresses due to T (tension in segment 0 to L,
compression in segment L to L)
Stresses in bar (0 to L)
RA
sx
tmax A
2
shear controls so set max a & solve for s x Ea¢ T +
2t a E a¢ T +
b1
(1 b)P
A
A
[2 t a + Ea¢ T]
P
5.1
^ impossible so evaluate segment (L to L)
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CHAPTER 2
Page 188
Axially Loaded Members
Stresses in bar (L to L)
2t a Ea¢ T RB
sx
tmax A
2
set max a & solve for Pmax2
s x E a¢ T b
bP
A
A
[ 2 t a + E a ¢T]
P
0.62
;
NAC
Problem 2.6-13 A circular brass bar of diameter d is member AC
in truss ABC which has load P 5000 lb applied at joint C. Bar AC
is composed of two segments brazed together on a plane pq making
an angle 36° with the axis of the bar (see figure). The allowable
stresses in the brass are 13,500 psi in tension and 6500 psi in shear.
On the brazed joint, the allowable stresses are 6000 psi in tension
and 3000 psi in shear. What is the tensile force NAC in bar AC?
What is the minimum required diameter dmin of bar AC?
A
a
p
q
B
u = 60∞
d
C
P
NAC
Solution 2.6-13
NUMERICAL DATA
P 5 kips
36°
a 13.5 ksi
p
a
2
P
sin(60°)
NAC 5.77 kips
a 6.5 ksi
u
NAC (tension)
;
min. required diameter of bar AC
u 54°
ja 6.0 ksi
ja 3.0 ksi
tensile force NAC
Method of Joints at C
(1) check tension and shear in bars; a a/2 so shear
sx
controls tmax 2
2ta NAC
A
x 2a= 13 ksi
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Page 189
SECTION 2.6
Areqd NAC
2ta
dmin 4
Ap
Areqd 0.44 in2
sx dmin 0.75 in
Areqd
dreqd (2) check tension and shear on brazed joint
sx NAC
A
sX NAC
p 2
d
4
dreqd sja
x 17.37 ksi
cos(u)2
4 NAC
dreqd 0.65 in
A p sx
xsin( )cos( )
4 NAC
A p sX
sx `
set equal to ja & solve for x,
then dreqd
dreqd Problem 2.6-14 Two boards are joined by gluing along a scarf
joint, as shown in the figure. For purposes of cutting and gluing,
the angle between the plane of the joint and the faces of the
boards must be between 10° and 40°. Under a tensile load P,
the normal stress in the boards is 4.9 MPa.
tja
(sin(u) cos(u))
4 NAC
`
x 6.31 ksi
dreqd 1.08 in
A p sX
P
a
Two boards joined by a scarf joint
90° 70°
x cos
2
(4.9 MPa)(cos 70°)2
0.57 MPa
10° x sin
40°
(a) STRESSES ON JOINT WHEN
;
cos
(4.9 MPa)(sin 70°)(cos 70°)
Due to load P: x 4.9 MPa
20°
;
P
(a) What are the normal and shear stresses acting on the glued joint
if 20°?
(b) If the allowable shear stress on the joint is 2.25 MPa, what is the
largest permissible value of the angle ?
(c) For what angle will the shear stress on the glued joint be
numerically equal to twice the normal stress on the joint?
Solution 2.6-14
189
shear on brazed joint
tension on brazed joint
xcos( )2
Stresses on Inclined Sections
1.58 MPa
(b) LARGEST ANGLE
allow x sin
;
IF
allow 2.25 MPa
cos
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190
9/25/08
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CHAPTER 2
Page 190
Axially Loaded Members
The shear stress on the joint has a negative sign. Its
numerical value cannot exceed allow 2.25 MPa.
Therefore,
2.25 MPa (4.9 MPa)(sin )(cos ) or sin
0.4592
| | 2.25 MPa.
(c)
cos
1
From trigonometry: sin u cos u sin 2u
2
Therefore: sin 2 2(0.4592) 0.9184
Solving: 2 66.69°
33.34°
or
90° Since
or
113.31°
⬖ 56.66°
or
NOTE: If
;
is between 10° and 33.3°,
| | 2.25 MPa.
If
| | x sin
`
| | x cos2
cos
t0
` 2
s0
x sin
2xcos2
cos
2 cos
or
63.43°
33.34°
must be between 10° and 40°, we select
33.3°
Numerical values only:
sin
56.66°
if 2 ?
WHAT IS
26.6°
tan
2
90° ;
NOTE: For 26.6° and 63.4°, we find
0.98 MPa and 1.96 MPa.
Thus, `
is between 33.3° and 40°,
Problem 2.6-15 Acting on the sides of a stress element cut from a bar in
t0
` 2 as required.
s0
5000 psi
uniaxial stress are tensile stresses of 10,000 psi and 5,000 psi, as shown
in the figure.
tu tu
su = 10,000 psi
u
(a) Determine the angle and the shear stress and show all stresses on a
sketch of the element.
(b) Determine the maximum normal stress max and the maximum shear
stress max in the material.
10,000 psi
tu tu
5000 psi
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11:40 AM
Page 191
SECTION 2.6
Solution 2.6-15
Stresses on Inclined Sections
191
Bar in uniaxial stress
1
1
tanu 2
12
From Eq. (1) or (2):
tan2u u 35.26°
;
x 15,000 psi
x sin
cos
(15,000 psi)(sin 35.26°)(cos 35.26°)
7,070 psi
;
Minus sign means that acts clockwise on the plane
for which 35.26°.
(a) ANGLE
AND SHEAR STRESS
x cos2
10,000 psi
sx s0
2
cos u
10,000 psi
cos2u
PLANE AT ANGLE
(1)
90°
90° x[cos( 90°)]2 x[sin ]2
NOTE: All stresses have units of psi.
x sin2
(b) MAXIMUM NORMAL AND SHEAR STRESSES
90° 5,000 psi
sx s 090°
sin2u
5,000 psi
sin2u
max x 15,000 psi
(2)
tmax sx
7,500 psi
2
;
;
Equate (1) and (2):
10,000 psi
2
cos u
5,000 psi
sin2u
Problem 2.6-16 A prismatic bar is subjected to an axial force that produces a
tensile stress 65 MPa and a shear stress 23 MPa on a certain inclined plane
(see figure). Determine the stresses acting on all faces of a stress element oriented at
30° and show the stresses on a sketch of the element.
65 MPa
u
23 MPa
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11:40 AM
CHAPTER 2
Page 192
Axially Loaded Members
Solution 2.6-16
(4754 + 65s x)
0
s 2x
sx x
4754
65
u acos
find
& x for stress state shown above
xcos( )2
cos (u) sin (u) so
3
18.
Asx
A
PA s x Q
su
u 19.5°
a
su
1
65 MPa
73.1 MPa
MP
a
7
31.
su
MP
a
9
54.
sx
MP
θ = 30°
xsin( ) cos( )
tu
sx
a
tu
sx
2
b A
1
su
sx
su
su
sx Asx
a
su
sx
b
65
65 2
23 2
a b
a b sx
sx
sx
a
65 2
65
23 2
b a b + a b 0
sx
sx
sx
now find & for
1 xcos( )2
30°
1 54.9 MPa
xsin( )cos( )
su2 s xcosa u +
Problem 2.6-17 The normal stress on plane pq of a prismatic bar in
tension (see figure) is found to be 7500 psi. On plane rs, which makes
an angle 30° with plane pq, the stress is found to be 2500 psi.
Determine the maximum normal stress max and maximum shear
stress max in the bar.
p 2
b
2
;
31.7 MPa
2 18.3 MPa
;
;
p
r b
P
P
s
q
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11:40 AM
Page 193
SECTION 2.6
Solution 2.6-17
193
Stresses on Inclined Sections
Bar in tension
SUBSTITUTE NUMERICAL VALUES INTO EQ. (2):
cosu1
)
cos(u1 + 30°
7500 psi
A 2500 psi
23 1.7321
Solve by iteration or a computer program:
1
Eq. (2-29a):
MAXIMUM NORMAL STRESS (FROM EQ. 1)
xcos2
30°
smax sx PLANE pq: 1 xcos2
1
PLANE rs: 2 xcos2(
1
1 7500 psi
)
s1
cos2u1
s2
cos2(u1 + b)
s1
2
cos u1
10,000 psi
2 2500 psi
7500 psi
cos2 30°
;
MAXIMUM SHEAR STRESS
Equate x from 1 and 2:
sx 30°
(Eq. 1)
tmax sx
5,000 psi
2
;
or
cos2u1
2
cos (u1 + b)
cosu1
s1
s1
s2 cos(u1 + b) A s2
(Eq. 2)
Problem 2.6-18 A tension member is to be constructed of two
pieces of plastic glued along plane pq (see figure). For purposes of
cutting and gluing, the angle must be between 25° and 45°.
The allowable stresses on the glued joint in tension and shear are
5.0 MPa and 3.0 MPa, respectively.
u
p
P
P
q
(a) Determine the angle so that the bar will carry the largest
load P. (Assume that the strength of the glued joint
controls the design.)
(b) Determine the maximum allowable load Pmax if the
cross-sectional area of the bar is 225 mm2.
Solution 2.6-18
Bar in tension with glued joint
ALLOWABLE STRESS x IN TENSION
xcos2
xsin
25° 45°
sx su
2
cos u
5.0 MPa
cos2u
(1)
cos
Since the direction of is immaterial, we can write:
| xsin cos
A 225 mm2
On glued joint: allow 5.0 MPa
allow 3.0 MPa
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11:40 AM
CHAPTER 2
Page 194
Axially Loaded Members
or
sx (a) DETERMINE ANGLE FOR LARGEST LOAD
|tu|
sin u cosu
3.0 MPa
sin u cosu
Point A gives the largest value of x and hence the
largest load. To determine the angle corresponding to point A, we equate Eqs. (1) and (2).
(2)
GRAPH OF EQS. (1) AND (2)
5.0 MPa
2
cos u
tan u 3.0 MPa
sin u cos u
3.0
5.0
u 30.96°
;
(b) DETERMINE THE MAXIMUM LOAD
From Eq. (1) or Eq. (2):
sx 5.0 MPa
2
cos u
3.0 MPa
6.80 MPa
sin u cos u
Pmax xA (6.80 MPa)(225 mm2)
1.53 kN
Problem 2.6-19 A nonprismatic bar 1–2–3 of rectangular cross
section (cross sectional area A) and two materials is held snugly
(but without any initial stress) between rigid supports (see figure).
The allowable stresses in compression and in shear are specified as
a and a, respectively. Use the following numerical data: (Data:
b1 4b2/3 b; A1 2A2 A; E1 3E2/4 E; 1 5 2/4 ;
a1 4a2/3 a, a1 2a1/5, a2 3a2/5; let a 11 ksi,
P 12 kips, A 6 in.2, b 8 in. E 30,000 ksi, 6.5 10-6/°F;
1 52/3 490 lb/ft3)
(a) If load P is applied at joint 2 as shown, find an expression for
the maximum permissible temperature rise Tmax so that the
allowable stresses are not to be exceeded at either
location A or B.
(b) If load P is removed and the bar is now rotated to a vertical
position where it hangs under its own weight (load
intensity w1 in segment 1–2 and w2 in segment 2–3), find
an expression for the maximum permissible temperature
rise Tmax so that the allowable stresses are not exceeded at
either location 1 or 3. Locations 1 and 3 are each a short
distance from the supports at 1 and 3 respectively.
;
b1
b2
1
2
P
A
B
E2, A2, a2
E1, A1, a1
(a)
R1
1
W
w1 = —1
b1
E1, A1, b1
2
W
w2 = —2
b2
E2, A2, b2
3
R3
(b)
3
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Page 195
SECTION 2.6
Stresses on Inclined Sections
195
Solution 2.6-19
(a) STAT-INDET NONPRISMATIC BAR WITH LOAD P AT jt 2
apply load P and temp. change T - use R3 as
redundant & do superposition analysis
3a
3b
Pf12 ( 1b1 R3(f12 f23)
compatibility:
3a
85s a A + 45 P
64EAa
2b2)T
f12
b1
E1A1
3b
f23
max A b2
E2A2
f12 + f23
compression at Location B
due to both P and temp. increase
Pf23 + 1a1 b1 + a2b22¢T
f12 + f23
¢Tmax 3
sa
4
9
sa
ta2 20
2
sa
5
Numerical data
ta1 a 11 ksi
A 6 in2
P 12 kips
6.5 (106)/°F steel
E 30000 ksi
(1) check normal and shear stresses at element A
location & solve for Tmax using a1 & a1
sxA ¢Tmax f12
¢Tmax b1
E1 A1
f23
aa b +
4 3
a bb
5 4
Tmax 67.1°F
R3
A2
[s a2 A2 ( f12 + f23)] + P f12
¢Tmax (a1 b1 + a2 b2)
compression due to both temp. rise & load P
s xB ¢Tmax
3
b
3 A
b
4
b
≤¥ P
≥ sa ±
+
4 2 EA
4 A
EA
E
3 2
b2
E2 A2
3
3
b
b
b
4
4
±
≤
±
≤
≥saA
+
¥ + P
EA
4 A
4 A
E
E
3 2
3 2
68s aA + 45P
64EAa
(2) check normal and shear stresses at element B
location & solve for Tmax using a2 & a2
R1
A1
[s a1 A1 ( f12 + f23)] + P f23
(a1 b1 + a2 b2)
4 3
a bb
5 4
shear controls for Location A where
temp. rise causes compressive stress but
;
load P causes tensile stress
numerical data & allowable stresses
(normal & shear)
sa2 sxA
2
aa b +
compression due to temp. increase,
tension due to P, at Location A
a1 a
compression due to temp.
rise but tension due to P
3
3
b
b
b
4
4
2
≤ + P
+
2a s ab A ±
5
EA
4 A
4 A
E
E
3 2
3 2
R1 P R3
statics:
Tmax 82.1°F
¢Tmax
0
Pf12 1a1b1 + a2 b22¢T
R3 R1 ¢Tmax ¢Tmax a ab +
4 3
a bb
5 4
255s a A 320P
512EAa
¢ Tmax 21.7°F
;
normal stress controls for Location B where temp.
rise & load P both cause compressive stress;
as a result, permissible temp. rise is reduced at
B compared to Location A where temp. rise
effect is offset by load P effect
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9/25/08
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CHAPTER 2
tmax B ¢ Tmax Page 196
Axially Loaded Members
sxB
2
[2t a A2(f12 + f23)] Pf12
(a1b1 + a2b2)
Location A where temp. riseeffect is offset by load P effect
3
b
9
A b
4
b
≤¥ P
≥2 a sab ±
+
20
2 EA
4 A
EA
E
3 2
¢Tmax aa b +
¢ Tmax 4 3
a bb
5 4
153s a A 160 P
256 E Aa
Tmax 27.3°F
(b) STAT-INDET NONPRISMATIC BAR HANGING UNDER ITS OWN WEIGHT (GRAVITY)
apply gravity and temp. change T - use R3 as redundant & do superposition analysis
d3a 3b
W1
W2
f12 W2 f12 f23
2
2
(a1b1 + a2 b2) ¢T
R3(f12 f23)
f12 b1
E1A1
f23 compatibility:
3a
b2
E2 A2
3b
0
R3 a
W1
W2
f + W2 f12 +
f b + (a1b1 + a2 b2) ¢T
2 12
2 23
f12 + f23
^ compression at Location 3 due to both
P and temp. increase
statics:
R1 W1 W2 R3
R1 W1 + W2 a
W1
W2
f W f +
f b + (a1 b1 + a2 b2)¢T
2 12 2 12
2 23
f12 + f23
^ compression at Location 1 due to temp.
increase, tension due to W1 & W2
s x1 R1
A1
tmax1 s x1
2
s x3 R3
A2
tmax3 s x3
A2
numerical data & allowables stresses (normal & shear)
a1 a
a 11 ksi
b 8 in.
s a2 3
s
4 a
A 6 in2
g
0.490
123
t a1 2
s
5 a
E 30000 ksi
k/in3
t a2 9
s
20 a
6.5 (106)/°F
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Page 197
SECTION 2.6
Stresses on Inclined Sections
197
(1) check normal and shear stresses at element 1 location & solve for Tmax using a1 & a1 normal stress
sa1A1 W1 + W2 ¢Tmax a
W1
W2
f12 + W2 f12 +
f b + (a1b1 + a2 b2)¢T
2
2 23
f12 + f23
g1A1b1f12 + 2(g1A1b1)f23 + g2A2 b2f23 2 s a1A1( f12 + f23)
2(a1b1 + a2 b2)
3
3
3
b
b
b
b
4
3 A 3
4
b
4
≤
≥ gAb
+ 2(g Ab)
+ g a bb
¥ + 2 salA ±
+
EA
4 A
5 2 4
4 A
EA
4 A
E
E
E
3 2
3 2
3 2
¢Tmax 2a a b +
1121g b + 1360sa
¢Tmax 1024E a
4 3
a bb
5 4
^ sign difference because gravity
offsets effect of temp. rise
Tmax 74.9°F
Next, shear stress
1121g b + 1360a2
¢Tmax 2
s b
5 a
Tmax 59.9°F
1024E a
(2) check normal and shear stresses at element 3 location & solve for Tmax using a2 & a2 normal stress
¢ Tmax ¢Tmax s a2A2( f12 + f23) + a
W1
W2
f + W2 f12 +
f b
2 12
2 23
a1b1 + a2 b2
same sign because temp. rise & gravity
both produce compressive stress at element 3
3
3 A 3
3
b
g
a bb
b
A b
4
3
g Ab b
5 2 4
4
3 A 3
b
≥
+
¥ + ≥
a sab
+ g
a bb
+
+
¥
4
2 EA
4 A
2 EA
5 2 4
EA
2
4 A
E
E
3 2
3 2
ab +
¢Tmax 510sa + 545 g b
1024E a
4 3
a b
5 4
Tmax 28.1°F
shear stress
¢Tmax 510 a2
9
s b + 545g b
20 a
1024 E a
¢Tmax 25.3°F
;
shear at element 3 location controls
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198
9/25/08
11:42 AM
CHAPTER 2
Page 198
Axially Loaded Members
Strain Energy
When solving the problems for Section 2.7, assume that the
material behaves linearly elastically.
Problem 2.7-1 A prismatic bar AD of length L, cross-sectional
area A, and modulus of elasticity E is subjected to loads 5P, 3P,
and P acting at points B, C, and D, respectively (see figure).
Segments AB, BC, and CD have lengths L/6, L/2, and L/3,
respectively.
(a) Obtain a formula for the strain energy U of the bar.
(b) Calculate the strain energy if P ⫽ 6 k, L ⫽ 52 in.,
A ⫽ 2.76 in.2, and the material is aluminum with
E ⫽ 10.4 ⫻ 106 psi.
Solution 2.7-1
Bar with three loads
(a) STRAIN ENERGY OF THE BAR (EQ. 2-40)
P⫽6k
L ⫽ 52 in.
U⫽ g
E ⫽ 10.4 ⫻ 10 psi
6
A ⫽ 2.76 in.2
⫽
L
L
L
1
c(3P)2 a b + (⫺2P)2 a b + (P)2 a b d
2EA
6
2
3
⫽
23P2L
P2L 23
a b ⫽
2EA 6
12EA
INTERNAL AXIAL FORCES
NAB ⫽ 3P
NBC ⫽ ⫺2P
NCD ⫽ P
LENGTHS
LAB ⫽
L
6
LBC ⫽
L
2
LCD ⫽
L
3
N2i Li
2EiAi
;
(b) SUBSTITUTE NUMERICAL VALUES:
U⫽
23(6 k)2(52 in.)
12(10.4 * 106 psi)(2.76 in.2)
⫽ 125 in.-lb
;
Sec_2.7.qxd
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11:42 AM
Page 199
SECTION 2.7
Strain Energy
Problem 2.7-2 A bar of circular cross section having two different diameters
d and 2d is shown in the figure. The length of each segment of
the bar is L/2 and the modulus of elasticity of the material is E.
(a) Obtain a formula for the strain energy U of the bar due to the load P.
(b) Calculate the strain energy if the load P ⫽ 27 kN, the length L ⫽ 600 mm,
the diameter d ⫽ 40 mm, and the material is brass with E ⫽ 105 GPa.
Solution 2.7-2
Bar with two segments
(b) SUBSTITUTE NUMERICAL VALUES:
(a) STRAIN ENERGY OF THE BAR
Add the strain energies of the two segments of the
bar (see Eq. 2-40).
P2(L/2)
N2i Li
1
1
⫽
cp
⫹p 2 d
2
i⫽1 2 EiAi
2E
(2d)
(d
)
4
4
2
U⫽ g
1
5P2L
P2L 1
a 2 + 2b ⫽
⫽
pE 4d
d
4pEd2
P ⫽ 27 kN
L ⫽ 600 mm
d ⫽ 40 mm
E ⫽ 105 GPa
U⫽
5(27 kN2)(600 mm)
4p(105 GPa)(40 mm)2
;
⫽ 1.036 N # m ⫽ 1.036 J
Problem 2.7-3 A three-story steel column in a building supports roof
and floor loads as shown in the figure. The story height H is 10.5 ft, the
cross-sectional area A of the column is 15.5 in.2, and the modulus of elasticity
E of the steel is 30 ⫻ 106 psi.
Calculate the strain energy U of the column assuming P1 ⫽ 40 k and
P2 ⫽ P3 ⫽ 60 k.
;
199
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200
9/25/08
11:42 AM
CHAPTER 2
Solution 2.7-3
Page 200
Axially Loaded Members
Three-story column
Upper segment: N1 ⫽ ⫺P1
Middle segment: N2 ⫽ ⫺(P1 ⫹ P2)
Lower segment: N3 ⫽ ⫺(P1 ⫹ P2 ⫹ P3)
STRAIN ENERGY
U⫽ g
N2i Li
2EiAi
⫽
H 2
[P + (P1 + P2)2 + (P1 + P2 + P3)2]
2EA 1
⫽
H
[Q]
2EA
[Q] ⫽ (40 k)2 + (100 k)2 + (160 k)2 ⫽ 37,200 k2
H ⫽ 10.5 ft
E ⫽ 30 ⫻ 106 psi
A ⫽ 15.5 in.
2
2EA ⫽ 2(30 * 106 psi)(15.5 in.2) ⫽ 930 * 106 lb
P1 ⫽ 40 k
P2 ⫽ P3 ⫽ 60 k
To find the strain energy of the column, add the strain
energies of the three segments (see Eq. 2-40).
U⫽
(10.5 ft)(12 in./ft)
930 * 106 lb
⫽ 5040 in.-lb
Problem 2.7-4 The bar ABC shown in the figure is loaded by a
force P acting at end C and by a force Q acting at the midpoint B. The bar
has constant axial rigidity EA.
(a) Determine the strain energy U1 of the bar when the force P acts
alone (Q ⫽ 0).
(b) Determine the strain energy U2 when the force Q acts alone (P ⫽ 0).
(c) Determine the strain energy U3 when the forces P and Q act
simultaneously upon the bar.
;
[37,200 k2]
Sec_2.7.qxd
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11:42 AM
Page 201
SECTION 2.7
Solution 2.7-4
Strain Energy
201
Bar with two loads
(c) FORCES P AND Q ACT SIMULTANEOUSLY
(a) FORCE P ACTS ALONE (Q ⫽ 0)
U1 ⫽
P2L
2EA
Segment BC: UBC ⫽
P2(L/2)
P2L
⫽
2EA
4EA
Segment AB: UAB ⫽
(P + Q)2(L/2)
2EA
;
⫽
PQL
Q 2L
P2L
+
+
4EA
2EA
4EA
U3 ⫽ UBC + UAB ⫽
PQL
Q2L
P2L
+
+
2EA
2EA
4EA
(b) FORCE Q ACTS ALONE (P ⫽ 0)
U2 ⫽
Q2(L/2)
Q2L
⫽
2EA
4EA
;
;
(Note that U3 is not equal to U1 ⫹ U2. In this case,
U3 ⬎ U1 ⫹ U2. However, if Q is reversed in direction,
U3 ⬍ U1 ⫹ U2. Thus, U3 may be larger or smaller than
U1 ⫹ U2.)
Problem 2.7-5 Determine the strain energy per unit volume (units of psi) and the strain energy per unit weight (units of in.)
that can be stored in each of the materials listed in the accompanying table, assuming that the material is stressed to the proportional limit.
DATA FOR PROBLEM 2.7-5
Material
Weight
density
(lb/in.3)
Modulus of
elasticity
(ksi)
Proportional
limit
(psi)
Mild steel
Tool steel
Aluminum
Rubber (soft)
0.284
0.284
0.0984
0.0405
30,000
30,000
10,500
0.300
36,000
75,000
60,000
300
Solution 2.7-5
Strain-energy density
STRAIN ENERGY PER UNIT VOLUME
DATA:
Material
Weight
density
(lb/in.3)
Modulus of
elasticity
(ksi)
Proportional
limit
(psi)
Mild steel
Tool steel
Aluminum
Rubber (soft)
0.284
0.284
0.0984
0.0405
30,000
30,000
10,500
0.300
36,000
75,000
60,000
300
U⫽
P2L
2EA
Volume V ⫽ AL
Stress s ⫽
u⫽
s2PL
U
⫽
V
2E
P
A
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Axially Loaded Members
At the proportional limit:
At the proportional limit:
u ⫽ uR ⫽ modulus of resistance
uW ⫽
uR ⫽
s2PL
s2PL
2gE
(Eq. 2)
(Eq. 1)
2E
RESULTS
STRAIN ENERGY PER UNIT WEIGHT
U⫽
2
PL
2EA
Weight W ⫽ gAL
␥ ⫽ weight density
uW ⫽
Mild steel
Tool steel
Aluminum
Rubber (soft)
uR (psi)
uw (in.)
22
94
171
150
76
330
1740
3700
s2
U
⫽
W
2gE
Problem 2.7-6 The truss ABC shown in the figure is subjected to a horizontal
load P at joint B. The two bars are identical with cross-sectional area A and modulus
of elasticity E.
(a) Determine the strain energy U of the truss if the angle ␤ ⫽ 60°.
(b) Determine the horizontal displacement ␦B of joint B by equating the strain
energy of the truss to the work done by the load.
Truss subjected to a load P
↓
Solution 2.7-6
↓⫺
␤ ⫽ 60°
⌺Fvert ⫽ 0
LAB ⫽ LBC ⫽ L
⫺FAB sin ␤ ⫹ FBC sin ␤ ⫽ 0
sin b ⫽ 13/2
FAB ⫽ FBC
cos ␤ ⫽ 1/2
⌺Fhoriz ⫽ 0 : ←
FREE-BODY DIAGRAM OF JOINT B
⫺FAB cos ␤ ⫺ FBC cos ␤ ⫹ P ⫽ 0
FAB ⫽ FBC ⫽
⫹
(Eq. 1)
P
P
⫽
⫽P
2 cos b
2(1/2)
(Eq. 2)
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SECTION 2.7
NBC ⫽ ⫺P (compression)
dB ⫽
(a) STRAIN ENERGY OF TRUSS (EQ. 2-40)
(NBC)2L
(NAB)2L
N2i Li
P2L
⫽
+
⫽
2EiAi
2EA
2EA
EA
2 P2L
2PL
2U
⫽ a
b ⫽
P
P EA
EA
;
;
Problem 2.7-7 The truss ABC shown in the figure supports a
A
horizontal load P1 ⫽ 300 lb and a vertical load P2 ⫽ 900 lb. Both
bars have cross-sectional area A ⫽ 2.4 in.2 and are made of steel
with E ⫽ 30 ⫻ 106 psi.
(a) Determine the strain energy U1 of the truss when the load P1
acts alone (P2 ⫽ 0).
(b) Determine the strain energy U2 when the load P2 acts alone
(P1 ⫽ 0).
(c) Determine the strain energy U3 when both loads act
simultaneously.
Solution 2.7-7
203
(b) HORIZONTAL DISPLACEMENT OF JOINT B (EQ. 2-42)
Axial forces: NAB ⫽ P (tension)
U⫽ g
Strain Energy
30∞
C
B P1 = 300 lb
P2 = 900 lb
60 in.
Truss with two loads
LAB ⫽
LBC
120
in. ⫽ 69.282 in.
⫽
cos 30°
13
2EA ⫽ 2(30 ⫻ 106 psi)(2.4 in.2) ⫽ 144 ⫻ 106 lb
FORCES FAB AND FBC IN THE BARS
From equilibrium of joint B:
FAB ⫽ 2P2 ⫽ 1800 lb
FBC ⫽ P1 ⫺ P2 13 ⫽ 300 lb ⫺ 1558.8 lb
P1 ⫽ 300 lb
P2 ⫽ 900 lb
A ⫽ 2.4 in.2
E ⫽ 30 ⫻ 106 psi
LBC ⫽ 60 in.
Force
P1 alone
FAB
FBC
0
300 lb
13
cos b ⫽ cos 30° ⫽
2
P1 and P2
1800 lb
⫺1558.8 lb
1800 lb
⫺1258.8 lb
(a) LOAD P1 ACTS ALONE
U1 ⫽
␤ ⫽ 30°
1
sin b ⫽ sin 30° ⫽
2
P2 alone
(FBC)2LBC
(300 lb)2(60 in.)
⫽
2EA
144 * 106 lb
⫽ 0.0375 in.-lb
;
(b) LOAD P2 ACTS ALONE
U2 ⫽
1
c(F )2L + (FBC)2LBC d
2EA AB AB
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CHAPTER 2
⫽
Page 204
Axially Loaded Members
1
c(1800 lb)2(69.282 in.)
2EA
⫽
+ ( ⫺1558.8 lb)2(60 in.) d
⫽
370.265 * 106 lb2-in.
144 * 106 lb
+ (⫺ 1258.8 lb)2(60 in.) d
⫽ 2.57 in.-lb ;
⫽
(c) LOADS P1 AND P2 ACT SIMULTANEOUSLY
U3 ⫽
1
c(F )2L + (FBC)2LBC d
2EA AB AB
1
c(1800 lb)2(69.282 in.)
2EA
319.548 * 106 lb2-in.
144 * 106 lb
⫽ 2.22 i n.- lb
;
NOTE: The strain energy U3 is not equal to U1 ⫹ U2.
Problem 2.7-8 The statically indeterminate structure shown
in the figure consists of a horizontal rigid bar AB supported by
five equally spaced springs. Springs 1, 2, and 3 have stiffnesses
3k, 1.5k, and k, respectively. When unstressed, the lower ends of
all five springs lie along a horizontal line. Bar AB, which has
weight W, causes the springs to elongate by an amount ␦.
(a) Obtain a formula for the total strain energy U of the
springs in terms of the downward displacement
␦ of the bar.
(b) Obtain a formula for the displacement ␦ by equating the
strain energy of the springs to the work done by the
weight W.
(c) Determine the forces F1, F2, and F3 in the springs.
(d) Evaluate the strain energy U, the displacement ␦, and the
forces in the springs if W ⫽ 600 N and k ⫽ 7.5 N/mm.
1
3k
k
1.5k
2
3
1.5k
2
A
1
3k
B
W
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SECTION 2.7
Solution 2.7-8
205
Strain Energy
Rigid bar supported by springs
(c) FORCES IN THE SPRINGS
F1 ⫽ 3kd ⫽
F3 ⫽ kd ⫽
3W
3W
F2 ⫽ 1.5kd ⫽
10
20
W
10
;
(d) NUMERICAL VALUES
k1 ⫽ 3k
W ⫽ 600 N k ⫽ 7.5 N/mm ⫽ 7500 N/mm
k2 ⫽ 1.5k
k3 ⫽ k
U ⫽ 5kd2 ⫽ 5ka
␦ ⫽ downward displacement of rigid bar
kd2
Eq. (2-38b)
2
(a) STRAIN ENERGY U OF ALL SPRINGS
W 2
W2
b ⫽
10k
20k
⫽ 2.4 N # m ⫽ 2.4 J
For a spring: U ⫽
d⫽
W
⫽ 8.0 mm
10k
F1 ⫽
3W
⫽ 180 N
10
Wd
2
F2 ⫽
3W
⫽ 90 N
20
Strain energy of the springs equals 5k␦2
F3 ⫽
W
⫽ 60 N
10
U ⫽ 2a
2
2
3kd
1.5kd
kd
b + 2a
b +
2
2
2
2
⫽ 5kd2
;
(b) DISPLACEMENT ␦
Work done by the weight W equals
...
;
Wd
⫽ 5kd2
2
and d ⫽
W
10k
;
(a) Determine the strain energy U of the bar.
(b) Determine the elongation ␦ of the bar by equating the
strain energy to the work done by the force P.
;
;
;
;
NOTE: W ⫽ 2F1 ⫹ 2F2 ⫹ F3 ⫽ 600 N (Check)
Problem 2.7-9 A slightly tapered bar AB of rectangular cross
section and length L is acted upon by a force P (see figure). The
width of the bar varies uniformly from b2 at end A to b1 at end B.
The thickness t is constant.
;
A
B
b2
L
b1
P
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CHAPTER 2
Page 206
Axially Loaded Members
Solution 2.7-9
Tapered bar of rectangular cross section
Apply this integration formula to Eq. (1):
U⫽
b(x) ⫽ b2 ⫺
⫽
(b2 ⫺ b1)x
L
U⫽
A(x) ⫽ tb(x)
⫽ t cb2 ⫺
(b2 ⫺ b1)x
d
L
b2
PL
2U
⫽
ln
P
Et(b2 ⫺ b1) b1
;
NOTE: This result agrees with the formula derived in
Prob. 2.3-13.
(1)
1
dx
⫽ ln (a + bx)
b
L a + bx
Problem 2.7-10 A compressive load P is transmitted through a rigid plate to three
magnesium-alloy bars that are identical except that initially the middle bar is slightly
shorter than the other bars (see figure). The dimensions and properties of the assembly
are as follows: length L ⫽ 1.0 m, cross-sectional area of each bar A ⫽ 3000 mm2,
modulus of elasticity E ⫽ 45 GPa, and the gap s ⫽ 1.0 mm.
(a)
(b)
(c)
(d)
;
L
P2
P2dx
dx
⫽
⫽
x
2Etb(x)
2Et
b
⫺
(b
L0
L0 2
2 ⫺ b1)L
From Appendix C:
b2
P2L
ln
2Et(b2 ⫺ b1) b1
d⫽
[N(x)]2dx
( Eq. 2- 41)
L 2EA(x)
L
P2
⫺L
⫺L
c
ln b1 ⫺
ln b2 d
2Et (b2 ⫺ b1)
(b2 ⫺ b1)
(b) ELONGATION OF THE BAR (EQ. 2-42)
(a) STRAIN ENERGY OF THE BAR
U⫽
(b2 ⫺ b1)x L
P2
1
ln cb2 ⫺
c
dd
1
2Et ⫺(b2 ⫺ b1)1 2
L
0
L
Calculate the load P1 required to close the gap.
Calculate the downward displacement ␦ of the rigid plate when P ⫽ 400 kN.
Calculate the total strain energy U of the three bars when P ⫽ 400 kN.
Explain why the strain energy U is not equal to P␦/2.
(Hint: Draw a load-displacement diagram.)
P
s
L
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SECTION 2.7
Solution 2.7-10
Strain Energy
207
Three bars in compression
(c) STRAIN ENERGY U FOR P ⫽ 400 kN
U⫽ g
EAd2
2L
Outer bars:
␦ ⫽ 1.321 mm
Middle bar:
␦ ⫽ 1.321 mm ⫺ s
⫽ 0.321 mm
U⫽
s ⫽ 1.0 mm
L ⫽ 1.0 m
1
⫽ (135 * 106 N/m)(3.593 mm2)
2
For each bar:
⫽ 243 N # m ⫽ 243 J
A ⫽ 3000 mm2
E ⫽ 45 GPa
;
(d) LOAD-DISPLACEMENT DIAGRAM
EA
⫽ 135 * 106 N/m
L
U ⫽ 243 J ⫽ 243 N ⭈ m
(a) LOAD P1 REQUIRED TO CLOSE THE GAP
EAd
PL
In general, d ⫽
and P ⫽
EA
L
For two bars, we obtain:
P1 ⫽ 2 a
EA
[2(1.321 mm)2 + (0.321 mm)2]
2L
Pd
1
⫽ (400 kN)(1.321 mm) ⫽ 264 N # m
2
2
Pd
⫽ because the
2
load-displacement relation is not linear.
The strain energy U is not equal to
EAs
b ⫽ 2(135 * 106 N/m)(1.0 mm)
L
P1 ⫽ 270 kN
;
(b) DISPLACEMENT ␦ FOR P ⫽ 400 kN
Since P ⬎ P1, all three bars are compressed.
The force P equals P1 plus the additional force
required to compress all three bars by the amount
␦ ⫺ s.
P ⫽ P1 + 3 a
EA
b(d ⫺ s)
L
U ⫽ area under line OAB.
or 400 kN ⫽ 270 kN ⫹ 3(135 ⫻ 106 N/m)
(␦ ⫺ 0.001 m)
Solving, we get ␦ ⫽ 1.321 mm
;
Pd
⫽ area under a straight line from O to B, which is
2
larger than U.
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CHAPTER 2
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Axially Loaded Members
Problem 2.7-11 A block B is pushed against three springs by a force P
(see figure). The middle spring has stiffness k1 and the outer springs each
have stiffness k2. Initially, the springs are unstressed and the middle spring
is longer than the outer springs (the difference in length is denoted s).
(a) Draw a force-displacement diagram with the force P as ordinate
and the displacement x of the block as abscissa.
(b) From the diagram, determine the strain energy U1 of the springs
when x ⫽ 2s.
(c) Explain why the strain energy U1 is not equal to P␦/2, where ␦ ⫽ 2s.
Solution 2.7-11
Block pushed against three springs
Force P0 required to close the gap:
P0 ⫽ k1s
(a) FORCE-DISPLACEMENT DIAGRAM
(1)
FORCE-DISPLACEMENT RELATION BEFORE GAP IS CLOSED
P ⫽ k1x
(0 ⱕ x ⱕ s)(0 ⱕ P ⱕ P0)
(2)
FORCE-DISPLACEMENT RELATION AFTER GAP IS CLOSED
All three springs are compressed. Total stiffness equals
k1 ⫹ 2k2. Additional displacement equals x ⫺ s. Force
P equals P0 plus the force required to compress all three
springs by the amount x ⫺ s.
P ⫽ P0 ⫹ (k1 ⫹ 2k2)(x ⫺ s)
(b) STRAIN ENERGY U1 WHEN x ⫽ 2s
⫽ k1s ⫹ (k1 ⫹ 2k2)x ⫺ k1s ⫺ 2k2s
P ⫽ (k1 ⫹ 2k2)x ⫺ 2k2s
(x ⱖ s); (P ⱖ P0)
(3)
P1 ⫽ force P when x ⫽ 2s
⫽
Substitute x ⫽ 2s into Eq. (3):
P1 ⫽ 2(k1 ⫹ k2)s
U1 ⫽ Area below force - displacement curve
(4)
⫹
⫹
1
1
1
⫽ P0s + P0s + (P1 ⫺ P0)s ⫽ P0s + P1s
2
2
2
⫽ k1s2 + (k1 + k2)s2
U1 ⫽ (2k1 ⫹ k2)s2
;
(5)
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SECTION 2.7
(c) STRAIN ENERGY U1 IS NOT EQUAL TO
Pd
2
Pd
1
⫽ P1(2 s) ⫽ P1s ⫽ 2(k1 + k2)s2
2
2
(This quantity is greater than U1.)
For d ⫽ 2s:
209
Strain Energy
Pd
⫽ area under a straight line from O to B, which
2
is larger than U1.
Pd
Thus,
is not equal to the strain energy because
2
the force-displacement relation is not linear.
U1 ⫽ area under line OAB.
Problem 2.7-12 A bungee cord that behaves linearly
elastically has an unstressed length L0 ⫽ 760 mm and a
stiffness k ⫽ 140 N/m.The cord is attached to two pegs, distance b ⫽ 380 mm apart, and pulled at its midpoint by a
force P ⫽ 80 N (see figure).
b
A
B
(a) How much strain energy U is stored in the cord?
(b) What is the displacement ␦C of the point where the
load is applied?
(c) Compare the strain energy U with the quantity
P␦C/2.
(Note: The elongation of the cord is not small compared
to its original length.)
Solution 2.7-12
C
P
Bungee cord subjected to a load P.
DIMENSIONS BEFORE THE LOAD P IS APPLIED
From triangle ACD:
1
d ⫽ 2L20 ⫺ b2 ⫽ 329.09 mm
2
DIMENSIONS AFTER THE LOAD P IS APPLIED
L0 ⫽ 760 mm
L0
⫽ 380 mm
2
b ⫽ 380 mm
Let x ⫽ distance CD
k ⫽ 140 N/m
Let L1 ⫽ stretched length of bungee cord
(1)
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CHAPTER 2
Page 210
Axially Loaded Members
From triangle ACD:
L1 ⫽ L0 +
or
L1
b 2
⫽ a b + x2
2
A 2
(2)
L1 ⫽ 2b2 + 4x2
(3)
L0 ⫽ a1 ⫺
P 1 2
2
2
b + 4x2 ⫽ 1b + 4x
4kx
P 1 2
b b + 4x2
4kx
(7)
This equation can be solved for x.
EQUILIBRIUM AT POINT C
SUBSTITUTE NUMERICAL VALUES INTO EQ. (7):
Let F ⫽ tensile force in bungee cord
760 mm ⫽ c1 ⫺
(80 N)(1000 mm/m)
d
4(140 N/m)x
* 1(380 mm)2 + 4x2
760 ⫽ a1 ⫺
L1/2
F
P L1 1
⫽
F ⫽ a ba ba b
P/2
x
2
2
x
142.857 1
b 144,400 + 4x2
x
(4)
kd2
2
From Eq. (5):
Let ␦ ⫽ elongation of the entire bungee cord
F
P
b2
1 + 2
⫽
k
2k A
4x
(5)
Final length of bungee cord ⫽ original length ⫹ ␦
P
b2
L1 ⫽ L0 + d ⫽ L0 +
1 + 2
2k A
4x
SOLUTION OF EQUATIONS
Combine Eqs. (6) and (3):
L1 ⫽ L0 +
2
P
b
1 + 2 ⫽ 1b2 + 4x2
2k A
4x
(a) STRAIN ENERGY U OF THE BUNGEE CORD
U⫽
ELONGATION OF BUNGEE CORD
d⫽
(9)
Units: x is in millimeters
Solve for x (Use trial & error or a computer program):
x ⫽ 497.88 mm
P
b 2
⫽
1 + a b
2A
2x
(8)
(6)
d⫽
k ⫽ 140 N/m
P ⫽ 80 N
b2
P
1 + 2 ⫽ 305.81 mm
2k A
4x
1
U ⫽ (140 N/m)(305.81 mm)2 ⫽ 6.55 N.m
2
U ⫽ 6.55 J
;
(b) DISPLACEMENT ␦C OF POINT C
␦C ⫽ x ⫺ d ⫽ 497.88 mm ⫺ 329.09 mm
⫽ 168.8 mm
;
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Page 211
SECTION 2.7
(c) COMPARISON OF STRAIN ENERGY U WITH THE
QUANTITY P␦C/2
U ⫽ 6.55 J
PdC
1
⫽ (80 N)(168.8 mm) ⫽ 6.75 J
2
2
The two quantities are not the same. The work done by
the load P is not equal to P␦C/2 because the loaddisplacement relation (see below) is non-linear when
the displacements are large. (The work done by the
load P is equal to the strain energy because the bungee
cord behaves elastically and there are no energy
losses.)
U ⫽ area OAB under the curve OA.
PdC
⫽ area of triangle OAB, which is greater than U.
2
Strain Energy
211
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Axially Loaded Members
Impact Loading
The problems for Section 2.8 are to be solved on the basis of the assumptions
and idealizations described in the text. In particular, assume that the material
behaves linearly elastically and no energy is lost during the impact.
Collar
Problem 2.8-1 A sliding collar of weight W ⫽ 150 lb falls from a height
h ⫽ 2.0 in. onto a flange at the bottom of a slender vertical rod (see figure).
The rod has length L ⫽ 4.0 ft, cross-sectional area A ⫽ 0.75 in.2, and modulus
of elasticity E ⫽ 30 ⫻ 106 psi.
Calculate the following quantities: (a) the maximum downward
displacement of the flange, (b) the maximum tensile stress in the rod,
and (c) the impact factor.
L
Rod
h
Flange
Probs. 2.8-1, 2.8-2, 2.8-3
Solution 2.8-1
Collar falling onto a flange
(a) DOWNWARD DISPLACEMENT OF FLANGE
dst ⫽
WL
⫽ 0.00032 in.
EA
Eq. of (2-53):
dmax ⫽ dst c1 + a1 +
⫽ 0.0361 in.
2h 1/2
b d
dst
;
(b) MAXIMUM TENSILE STRESS (EQ. 2-55)
smax ⫽
Edmax
⫽ 22,600 psi
L
;
(c) IMPACT FACTOR (EQ. 2-61)
Impact factor ⫽
W ⫽ 150 lb
h ⫽ 2.0 in.
L ⫽ 4.0 ft ⫽ 48 in.
E ⫽ 30 ⫻ 10 psi
6
A ⫽ 0.75 in.
2
dmax
0.0361 in.
⫽
dst
0.00032 in.
⫽ 113
;
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SECTION 2.8
Impact Loading
Problem 2.8-2 Solve the preceding problem if the collar has mass
M ⫽ 80 kg, the height h ⫽ 0.5 m, the length L ⫽ 3.0 m, the cross-sectional
area A ⫽ 350 mm2, and the modulus of elasticity E ⫽ 170 GPa.
Solution 2.8-2
Collar falling onto a flange
(a) DOWNWARD DISPLACEMENT OF FLANGE
dst ⫽
WL
⫽ 0.03957 mm
EA
Eq. (2-53): dmax ⫽ dst c1 + a1 +
⫽ 6.33 mm
2h 1/2
b d
dst
;
(b) MAXIMUM TENSILE STRESS (EQ. 2-55)
smax ⫽
Edmax
⫽ 359 MPa
L
;
(c) IMPACT FACTOR (EQ. 2–61)
M ⫽ 80 kg
Impact factor ⫽
W ⫽ Mg ⫽ (80 kg)(9.81 m/s2)
⫽ 784.8 N
h ⫽ 0.5 m
L ⫽ 3.0 m
E ⫽ 170 GPa
A ⫽ 350 mm2
dmax
6.33 mm
⫽
dst
0.03957 mm
⫽ 160
;
Problem 2.8-3 Solve Problem 2.8-1 if the collar has weight W ⫽ 50 lb,
the height h ⫽ 2.0 in., the length L ⫽ 3.0 ft, the cross-sectional area
A ⫽ 0.25 in.2, and the modulus of elasticity E ⫽ 30,000 ksi.
Solution 2.8-3
Collar falling onto a flange
W ⫽ 50 lb
h ⫽ 2.0 in.
L ⫽ 3.0 ft ⫽ 36 in.
E ⫽ 30,000 psi
A ⫽ 0.25 in.2
(a) DOWNWARD DISPLACEMENT OF FLANGE
dst ⫽
WL
⫽ 0.00024 in.
EA
2h 1/2
b d
dst
;
Eq. (2 ⫺ 53): dmax ⫽ dst c1 + a1 +
⫽ 0.0312 in.
213
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11:43 AM
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Axially Loaded Members
(b) MAXIMUM TENSILE STRESS (EQ. 2–55)
smax ⫽
Edmax
⫽ 26,000 psi
L
(c) IMPACT FACTOR (EQ. 2-61)
Impact factor ⫽
;
dmax
0.0312 in.
⫽
dst
0.00024 in.
⫽ 130 ;
Problem 2.8-4 A block weighing W ⫽ 5.0 N drops inside a cylinder
from a height h ⫽ 200 mm onto a spring having stiffness k ⫽ 90 N/m
(see figure).
Block
(a) Determine the maximum shortening of the spring due to the
impact, and (b) determine the impact factor.
Cylinder
h
k
Prob. 2.8-4 and 2.8-5
Solution 2.8-4
W ⫽ 5.0 N
Block dropping onto a spring
h ⫽ 200 mm
k ⫽ 90 N/m
(a) MAXIMUM SHORTENING OF THE SPRING
dst ⫽
W
5.0 N
⫽
⫽ 55.56 mm
k
90 N/m
Eq. (2-53): dmax ⫽ dst c1 + a1 +
⫽ 215 mm
;
2h 1/2
b d
dst
(b) IMPACT FACTOR (EQ. 2-61)
Impact factor ⫽
dmax
215 mm
⫽
dst
55.56 mm
⫽ 3.9 ;
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SECTION 2.8
Impact Loading
Problem 2.8-5 Solve the preceding problem if the block weighs
W ⫽ 1.0 lb, h ⫽ 12 in., and k ⫽ 0.5 lb/in.
Solution 2.8-5
Block dropping onto a spring
(a) MAXIMUM SHORTENING OF THE SPRING
dst ⫽
W
1.0 lb
⫽
⫽ 2.0 in.
k
0.5 lb/in.
Eq. (2-53): dmax ⫽ dst c1 + a1 +
⫽ 9.21 in.
2h 1/2
b d
dst
;
(b) IMPACT FACTOR (EQ. 2-61)
dmax
9.21 in.
⫽
dst
2.0 in.
⫽ 4.6 ;
Impact factor ⫽
W ⫽ 1.0 lb
h ⫽ 12 in.
k ⫽ 0.5 lb/in.
Problem 2.8-6 A small rubber ball (weight W ⫽ 450 mN) is attached by a rubber cord to
a wood paddle (see figure). The natural length of the cord is L0 ⫽ 200 mm, its crosssectional area is A ⫽ 1.6 mm2, and its modulus of elasticity is E ⫽ 2.0 MPa. After being
struck by the paddle, the ball stretches the cord to a total length L1 ⫽ 900 mm.
What was the velocity v of the ball when it left the paddle? (Assume linearly elastic
behavior of the rubber cord, and disregard the potential energy due to any change in
elevation of the ball.)
Solution 2.8-6
Rubber ball attached to a paddle
WHEN THE RUBBER CORD IS FULLY STRETCHED:
U⫽
EAd2
EA
⫽
(L ⫺ L0)2
2L0
2L0 1
CONSERVATION OF ENERGY
KE ⫽ U
g ⫽ 9.81 m/s
E ⫽ 2.0 MPa
A ⫽ 1.6 mm
L0 ⫽ 200 mm
L1 ⫽ 900 mm
W ⫽ 450 mN
2
2
WHEN THE BALL LEAVES THE PADDLE
Wv2
KE ⫽
2g
v2 ⫽
Wv2
EA
(L1 ⫺ L0)2
⫽
2g
2L0
gEA
(L ⫺ L0)2
WL0 1
gEA
A WL0
v ⫽ (L1 ⫺ L0)
;
SUBSTITUTE NUMERICAL VALUES:
(9.81 m/s2) (2.0 MPa) (1.6 mm2)
A
(450 mN) (200 mm)
⫽ 13.1 m/s ;
v ⫽ (700 mm)
215
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Axially Loaded Members
Problem 2.8-7 A weight W ⫽ 4500 lb falls from a height h onto
a vertical wood pole having length L ⫽ 15 ft, diameter d ⫽ 12 in.,
and modulus of elasticity E ⫽ 1.6 ⫻ 106 psi (see figure).
If the allowable stress in the wood under an impact load is 2500 psi,
what is the maximum permissible height h?
W = 4,500 lb
h
d = 12 in.
L = 15 ft
Solution 2.8-7
Weight falling on a wood pole
E ⫽ 1.6 ⫻ 106 psi
␴allow ⫽ 2500 psi (⫽ ␴max)
Find hmax
STATIC STRESS
sst ⫽
W
4500 lb
⫽ 39.79 psi
⫽
A
113.10 in.2
MAXIMUM HEIGHT hmax
Eq. (2⫺59): smax ⫽ sst c1 + a1 +
2hE 1/2
b d
Lsst
or
smax
2hE 1/2
⫺ 1 ⫽ a1 +
b
sst
Lsst
Square both sides and solve for h:
h ⫽ hmax ⫽
W ⫽ 4500 lb
d ⫽ 12 in.
L ⫽ 15 ft ⫽ 180 in.
A⫽
2
pd
⫽ 113.10 in.2
4
Lsmax smax
a
⫺ 2b
2E
sst
;
SUBSTITUTE NUMERICAL VALUES:
hmax ⫽
(180 in.) (2500 psi) 2500 psi
⫺ 2b
a
2(1.6 * 106 psi) 39.79 psi
⫽ 8.55 in.
;
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SECTION 2.8
Impact Loading
Problem 2.8-8 A cable with a restrainer at the bottom hangs
vertically from its upper end (see figure). The cable has an
effective cross-sectional area A ⫽ 40 mm2 and an effective
modulus of elasticity E ⫽ 130 GPa. A slider of mass M ⫽ 35 kg
drops from a height h ⫽ 1.0 m onto the restrainer.
If the allowable stress in the cable under an impact load is
500 MPa, what is the minimum permissible length L of the cable?
Cable
Slider
L
h
Restrainer
Probs. 2.8-8, 2.8-2, 2.8-9
Solution 2.8-8
Slider on a cable
STATIC STRESS
sst ⫽
W
343.4 N
⫽ 8.585 MPa
⫽
A
40 mm2
MINIMUM LENGTH Lmin
Eq. (2⫺59): smax ⫽ sst c1 + a1 +
2hE 1/2
b d
Lsst
or
smax
2hE 1/2
⫺ 1 ⫽ a1 +
b
sst
Lsst
Square both sides and solve for L:
L ⫽ Lmin ⫽
2Ehsst
smax(smax ⫺ 2sst)
;
SUBSTITUTE NUMERICAL VALUES:
W ⫽ Mg ⫽ (35 kg)(9.81 m/s2) ⫽ 343.4 N
A ⫽ 40 mm2
h ⫽ 1.0 m
E ⫽ 130 GPa
␴allow ⫽ ␴max ⫽ 500 MPa
Find minimum length Lmin
Lmin ⫽
2(130 GPa) (1.0 m) (8.585 MPa)
(500 MPa) [500 MPa ⫺ 2(8.585 MPa)]
⫽ 9.25 mm
;
217
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CHAPTER 2
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Axially Loaded Members
Problem 2.8-9 Solve the preceding problem if the slider has
weight W ⫽ 100 lb, h ⫽ 45 in., A ⫽ 0.080 in.2, E ⫽ 21 ⫻ 106 psi,
and the allowable stress is 70 ksi.
Cable
Slider
L
h
Restrainer
Solution 2.8-9
Slider on a cable
STATIC STRESS
sst ⫽
100 lb
W
⫽
⫽ 1250 psi
A
0.080 in.2
MINIMUM LENGTH Lmin
Eq. (2⫺59): smax ⫽ sst c1 + a1 +
2hE 1/2
b d
Lsst
or
smax
2hE 1/2
⫺ 1 ⫽ a1 +
b
sst
Lsst
Square both sides and solve for L:
L ⫽ Lmin ⫽
2Ehsst
smax(smax ⫺ 2sst)
;
SUBSTITUTE NUMERICAL VALUES:
Lmin ⫽
W ⫽ 100 lb
A ⫽ 0.080 in.2
h ⫽ 45 in
E ⫽ 21 ⫻ 106 psi
␴allow ⫽ ␴max ⫽ 70 ksi
Find minimum length Lmin
2(21 * 106 psi) (45 in.) (1250 psi)
(70,000 psi) [70,000 psi ⫺ 2(1250 psi)]
⫽ 500 in.
;
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SECTION 2.8
Impact Loading
Problem 2.8-10 A bumping post at the end of a track in a railway
yard has a spring constant k ⫽ 8.0 MN/m (see figure). The maximum
possible displacement d of the end of the striking plate is 450 mm.
What is the maximum velocity ␯max that a railway car of weight
W ⫽ 545 kN can have without damaging the bumping post when it
strikes it?
Solution 2.8-10
Bumping post for a railway car
STRAIN ENERGY WHEN SPRING IS COMPRESSED TO THE
MAXIMUM ALLOWABLE AMOUNT
U⫽
kd2max
kd2
⫽
2
2
CONSERVATION OF ENERGY
KE ⫽ U
k ⫽ 8.0 MN/m
W ⫽ 545 kN
k
v ⫽ vmax ⫽ d
;
A W/g
d ⫽ maximum displacement of spring
d ⫽ ␦max ⫽ 450 mm
Wv2
kd2 2
kd2
v ⫽
⫽
2g
2
W/g
SUBSTITUTE NUMERICAL VALUES:
Find ␯max
KINETIC ENERGY BEFORE IMPACT
Mv2
Wv2
KE ⫽
⫽
2
2g
8.0 MN/m
vmax ⫽ (450 mm)
A (545 kN)/(9.81 m/s2)
⫽ 5400 mm/s ⫽ 5.4 m/s
Problem 2.8-11 A bumper for a mine car is constructed with
a spring of stiffness k ⫽ 1120 lb/in. (see figure). If a car weighing
3450 lb is traveling at velocity ␯ ⫽ 7 mph when it strikes the
spring, what is the maximum shortening of the spring?
;
v
k
219
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CHAPTER 2
Solution 2.8-11
11:44 AM
Page 220
Axially Loaded Members
Bumper for a mine car
k ⫽ 1120 lb/in.
W ⫽ 3450 lb
␯ ⫽ 7 mph ⫽ 123.2 in./sec
g ⫽ 32.2 ft/sec2 ⫽ 386.4 in./sec2
Find the shortening ␦max of the spring.
KINETIC ENERGY JUST BEFORE IMPACT
KE ⫽
Mv2
Wv2
⫽
2
2g
Conservation of energy
KE ⫽ U
kd2max
Wv2
⫽
2g
2
Solve for ␦max: dmax ⫽
Wv2
A gk
;
SUBSTITUTE NUMERICAL VALUES:
dmax ⫽
STRAIN ENERGY WHEN SPRING IS FULLY COMPRESSED
kd2max
U⫽
2
Problem 2.8-12 A bungee jumper having a mass of 55 kg leaps
from a bridge, braking her fall with a long elastic shock cord having
axial rigidity EA ⫽ 2.3 kN (see figure).
If the jumpoff point is 60 m above the water, and if it is desired to
maintain a clearance of 10 m between the jumper and the water, what
length L of cord should be used?
(3450 lb) (123.2 in./sec)2
A (386.4 in./sec2) (1120 lb/in.)
⫽ 11.0 in.
;
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SECTION 2.8
Solution 2.8-12
Impact Loading
Bungee jumper
SOLVE QUADRATIC EQUATION FOR ␦max:
dmax ⫽
⫽
WL
WL 2
WL 1/2
+ ca
b + 2L a
bd
EA
EA
EA
WL
2EA 1/2
c1 + a1 +
b d
EA
W
VERTICAL HEIGHT
h ⫽ C + L + dmax
h⫺C⫽L +
W ⫽ Mg ⫽ (55 kg)(9.81 m/s2)
⫽ 539.55 N
SOLVE FOR L:
h⫺C
L⫽
EA ⫽ 2.3 kN
1 +
Height: h ⫽ 60 m
2EA 1/2
WL
c1 + a 1 +
b d
EA
W
W
2EA 1/2
c1 + a 1 +
b d
EA
W
Clearance: C ⫽ 10 m
SUBSTITUTE NUMERICAL VALUES:
Find length L of the bungee cord.
W
539.55 N
⫽
⫽ 0.234587
EA
2.3 kN
P.E. ⫽ Potential energy of the jumper at the top of
bridge (with respect to lowest position)
U ⫽ strain energy of cord at lowest position
EAd2max
2L
or
W(L + dmax) ⫽
d2max ⫺
* c1 + a 1 +
⫽ 1.9586
50 m
⫽ 25.5 m
L⫽
1.9586
CONSERVATION OF ENERGY
P.E. ⫽ U
Numerator ⫽ h ⫺ C ⫽ 60 m ⫺ 10 m ⫽ 50 m
Denominator ⫽ 1 + (0.234587)
⫽ W(L ⫹ ␦max)
⫽
;
EAd2max
2L
2WL
2WL2
dmax ⫺
⫽0
EA
EA
;
1/2
2
b d
0.234587
221
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Axially Loaded Members
Problem 2.8-13 A weight W rests on top of a wall and is attached to one
end of a very flexible cord having cross-sectional area A and modulus of
elasticity E (see figure). The other end of the cord is attached securely
to the wall. The weight is then pushed off the wall and falls freely the full
length of the cord.
W
W
(a) Derive a formula for the impact factor.
(b) Evaluate the impact factor if the weight, when hanging statically,
elongates the band by 2.5% of its original length.
Solution 2.8-13
Weight falling off a wall
CONSERVATION OF ENERGY
P.E. ⫽ U
or
W(L + dmax) ⫽
d2max ⫺
EAd2max
2L
2WL
2WL2
dmax ⫺
⫽0
EA
EA
SOLVE QUADRATIC EQUATION FOR ␦max:
W ⫽ Weight
dmax ⫽
WL 2
WL
WL 1/2
+ ca
b + 2L a
bd
EA
EA
EA
Properties of elastic cord:
E ⫽ modulus of elasticity
STATIC ELONGATION
A ⫽ cross-sectional area
dst ⫽
L ⫽ original length
␦max ⫽ elongation of elastic cord
WL
EA
IMPACT FACTOR
P.E. ⫽ potential energy of weight before fall (with
respect to lowest position)
dmax
2EA 1/2
⫽ 1 + c1 +
d
dst
W
P.E. ⫽ W(L ⫹ ␦max)
NUMERICAL VALUES
Let U ⫽ strain energy of cord at lowest position
␦st ⫽ (2.5%)(L) ⫽ 0.025L
EAd2max
U⫽
2L
dst ⫽
WL
EA
W
⫽ 0.025
EA
;
EA
⫽ 40
W
Impact factor ⫽ 1 + [1 + 2(40)]1/2 ⫽ 10
;
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Page 223
SECTION 2.8
223
Impact Loading
Problem 2.8-14 A rigid bar AB having mass M ⫽ 1.0 kg and
length L ⫽ 0.5 m is hinged at end A and supported at end B by a nylon
cord BC (see figure). The cord has cross-sectional area A ⫽ 30 mm2,
length b ⫽ 0.25 m, and modulus of elasticity E ⫽ 2.1 GPa.
If the bar is raised to its maximum height and then released, what is
the maximum stress in the cord?
C
b
A
B
W
L
Solution 2.8-14
Falling bar AB
GEOMETRY OF BAR AB AND CORD BC
RIGID BAR:
W ⫽ Mg ⫽ (1.0 kg)(9.81 m/s2)
⫽ 9.81 N
L ⫽ 0.5 m
NYLON CORD:
A ⫽ 30 mm2
CD ⫽ CB ⫽ b
AD ⫽ AB ⫽ L
h ⫽ height of center of gravity of raised bar AD
␦max ⫽ elongation of cord
From triangle ABC:sin u ⫽
cos u ⫽
b
2b2 + L2
L
E ⫽ 2.1 GPa
2b2 + L2
2h
2h
⫽
From line AD: sin 2 u ⫽
AD
L
Find maximum stress ␴max in cord BC.
From Appendix C: sin 2 ␪ ⫽ 2 sin ␪ cos ␪
b ⫽ 0.25 m
L
2bL
b
2h
ba
b ⫽ 2
⫽ 2a
2
2
2
2
L
b + L2
2b + L
2b + L
bL2
and h ⫽ 2
(Eq. 1)
b + L2
‹
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CHAPTER 2
11:44 AM
Page 224
Axially Loaded Members
CONSERVATION OF ENERGY
P.E. ⫽ potential energy of raised bar AD
⫽ W ah +
Substitute from Eq. (1) into Eq. (3):
s2max ⫺
dmax
b
2
dmax
EAd2max
b⫽
2
2b
smaxb
For the cord: dmax ⫽
E
Substitute into Eq. (2) and rearrange:
s2max ⫺
W
2WhE
s
⫺
⫽0
A max
bA
(Eq. 4)
SOLVE FOR ␴max:
EAd2max
U ⫽ strain energy of stretched cord ⫽
2b
P.E. ⫽ U W a h +
W
2WL2E
⫽0
smax ⫺
A
A(b2 + L2)
(Eq. 2)
smax ⫽
W
8L2EA
c1 + 1 +
d
2A
A
W(b2 + L2)
;
SUBSTITUTE NUMERICAL VALUES:
␴max ⫽ 33.3 MPa
;
(Eq. 3)
Stress Concentrations
The problems for Section 2.10 are to be solved by considering
the stress-concentration factors and assuming linearly elastic behavior.
P
Problem 2.10-1 The flat bars shown in parts (a) and (b) of the figure are
P
d
b
subjected to tensile forces P ⫽ 3.0 k. Each bar has thickness t ⫽ 0.25 in.
(a) For the bar with a circular hole, determine the maximum stresses for
hole diameters d ⫽ 1 in. and d ⫽ 2 in. if the width b ⫽ 6.0 in.
(b) For the stepped bar with shoulder fillets, determine the maximum
stresses for fillet radii R ⫽ 0.25 in. and R ⫽ 0.5 in. if the bar
widths are b ⫽ 4.0 in. and c ⫽ 2.5 in.
(a)
R
P
c
b
(b)
Probs. 2.10-1 and 2.10-2
P
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Page 225
SECTION 2.10
Solution 2.10-1
P ⫽ 3.0 k
225
Flat bars in tension
(b) STEPPED BAR WITH SHOULDER FILLETS
t ⫽ 0.25 in.
(a) BAR WITH CIRCULAR HOLE (b ⫽ 6 in.)
FOR d ⫽ 1 in.:
c ⫽ b ⫺ d ⫽ 5 in.
3.0 k
P
⫽ 2.40 ksi
s nom ⫽ ⫽
ct
(5 in.) (0.25 in.)
1
K L 2.60
6
␴max ⫽ k␴nom ⬇ 6.2 ksi
b ⫽ 4.0 in.
s nom ⫽
Obtain K from Fig. 2-63
d/b ⫽
Stress Concentrations
c ⫽ 2.5 in.; Obtain k from Fig. 2-64
3.0 k
P
⫽
⫽ 4.80 ksi
ct
(2.5 in.) (0.25 in.)
FOR R ⫽ 0.25 in.: R/c ⫽ 0.1
b/c ⫽ 1.60
k ⬇ 2.30 ␴max ⫽ K␴nom ⬇ 11.0 ksi
FOR R ⫽ 0.5 in.: R/c ⫽ 0.2
K ⬇ 1.87
;
b/c ⫽ 1.60
␴max ⫽ K␴nom ⬇ 9.0 ksi
;
;
FOR d ⫽ 2 in.: c ⫽ b ⫺ d ⫽ 4 in.
s nom ⫽
d/b ⫽
P
3.0 k
⫽
⫽ 3.00 ksi
ct
(4 in.) (0.25 in.)
1
K L 2.31
3
␴max ⫽ K␴nom ⬇ 6.9 ksi
;
Problem 2.10-2 The flat bars shown in parts (a) and (b) of the figure
are subjected to tensile forces P ⫽ 2.5 kN. Each bar has thickness
t ⫽ 5.0 mm.
P
(a) For the bar with a circular hole, determine the maximum stresses for
hole diameters d ⫽ 12 mm and d ⫽ 20 mm if the width b ⫽ 60 mm.
(b) For the stepped bar with shoulder fillets, determine the maximum
stresses for fillet radii R ⫽ 6 mm and R ⫽ 10 mm if the bar widths are
b ⫽ 60 mm and c ⫽ 40 mm.
P
d
b
(a)
R
P
c
b
(b)
P
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CHAPTER 2
Axially Loaded Members
Solution 2.10-2
P ⫽ 2.5 kN
Page 226
Flat bars in tension
(b) STEPPED BAR WITH SHOULDER FILLETS
t ⫽ 5.0 mm
(a) BAR WITH CIRCULAR HOLE (b ⫽ 60 mm)
Obtain K from Fig. 2-63
FOR d ⫽ 12 mm: c ⫽ b ⫺ d ⫽ 48 mm
s nom ⫽
d/b ⫽
P
2.5 kN
⫽
⫽ 10.42 MPa
ct
(48 mm) (5 mm)
c ⫽ 40 mm;
Obtain K from Fig. 2-64
s nom ⫽
P
2.5 kN
⫽
⫽ 12.50 MPa
ct
(40 mm) (5 mm)
FOR R ⫽ 6 mm: R/c ⫽ 0.15
FOR R ⫽ 10 mm: R/c ⫽ 0.25
;
b/c ⫽ 1.5
␴max ⫽ K␴nom ⬇ 25 MPa
K ⬇ 2.00
1
K L 2.51
5
␴max ⫽ K␴nom ⬇ 26 MPa
b ⫽ 60 mm
b/c ⫽ 1.5
␴max ⫽ K␴nom ⬇ 22 MPa
K ⬇ 1.75
;
;
FOR d ⫽ 20 mm: c ⫽ b ⫺ d ⫽ 40 mm
s nom ⫽
d/b ⫽
1
3
2.5 kN
P
⫽
⫽ 12.50 MPa
ct
(40 mm) (5 mm)
K L 2.31
␴max ⫽ K␴nom ⬇ 29 MPa
;
Problem 2.10-3 A flat bar of width b and thickness t has a hole
of diameter d drilled through it (see figure). The hole may have
any diameter that will fit within the bar.
What is the maximum permissible tensile load Pmax if the allowable
tensile stress in the material is ␴t?
P
b
d
P
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Page 227
SECTION 2.10
Solution 2.10-3
␴t ⫽ allowable tensile stress
Find Pmax
Find K from Fig. 2-64
Pmax ⫽ s nom ct ⫽
smax
st
ct ⫽ (b ⫺ d)t
K
K
st
d
bt a1 ⫺ b
K
b
Because ␴t, b, and t are constants, we write:
⫽
P*⫽
stbt
⫽
227
Flat bar in tension
t ⫽ thickness
Pmax
Stress Concentrations
d
b
K
P*
0
0.1
0.2
0.3
0.4
3.00
2.73
2.50
2.35
2.24
0.333
0.330
0.320
0.298
0.268
We observe that Pmax decreases as d/b increases.
Therefore, the maximum load occurs when the hole
becomes very small.
d
a :0
b
Pmax ⫽
and K : 3b
stbt
3
;
1
d
a1 ⫺ b
K
b
Problem 2.10-4 A round brass bar of diameter d1 ⫽ 20 mm has
upset ends of diameter d2 ⫽ 26 mm (see figure). The lengths of
the segments of the bar are L1 ⫽ 0.3 m and L2 ⫽ 0.1 m.
Quarter-circular fillets are used at the shoulders of the bar, and
the modulus of elasticity of the brass is E ⫽ 100 GPa.
If the bar lengthens by 0.12 mm under a tensile load P, what is
the maximum stress ␴max in the bar?
Probs. 2.10-4 and 2.10-5
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CHAPTER 2
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Axially Loaded Members
Solution 2.10-4
Round brass bar with upset ends
Use Fig. 2-65 for the stress-concentration factor:
s nom ⫽
⫽
E ⫽ 100 GPa
␦ ⫽ 0.12 mm
dEA2
P
⫽
⫽
A1
2L2A1 + L1A2
dE
d1 2
2L2 a b + L1
d2
L2 ⫽ 0.1 m
SUBSTITUTE NUMERICAL VALUES:
L1 ⫽ 0.3 m
s nom ⫽
R ⫽ radius of fillets ⫽
26 mm ⫺ 20 mm
⫽ 3 mm
2
PL1
PL2
b +
d ⫽ 2a
EA2
EA1
Solve for P:
P⫽
dEA1A2
2L2A1 + L1A2
(0.12 mm) (100 GPa)
20 2
2(0.1 m) a b + 0.3 m
26
metal having the following properties: d1 ⫽ 1.0 in., d2 ⫽ 1.4 in.,
L1 ⫽ 20.0 in., L2 ⫽ 5.0 in., and E ⫽ 25 ⫻ 106 psi. Also, the bar
lengthens by 0.0040 in. when the tensile load is applied.
Solution 2.10-5
Use the dashed curve in Fig. 2-65. K ⬇ 1.6
␴max ⫽ K␴nom ⬇ (1.6) (28.68 MPa)
d2
P
;
d2
d1
L1
L2
L2
Round bar with upset ends
d ⫽ 2a
PL1
PL2
b +
EA2
EA1
Solve for P: P ⫽
s nom ⫽
␦ ⫽ 0.0040 in.
L1 ⫽ 20 in.
L2 ⫽ 5 in.
R ⫽ radius of fillets R ⫽
dEA1A2
2L2A1 + L1A2
Use Fig. 2-65 for the stress-concentration factor.
E ⫽ 25 ⫻ 106 psi
⫽ 0.2 in.
⫽ 28.68 MPa
3 mm
R
⫽
⫽ 0.15
D1
20 mm
⬇ 46 MPa
Problem 2.10-5 Solve the preceding problem for a bar of monel
dE
A1
2L2 a b + L1
A2
1.4 in. ⫺ 1.0 in.
2
⫽
dEA2
P
⫽
⫽
A1
2L2A1 + L1A2
dE
d1 2
2L2 a b + L1
d2
dE
A1
2L2 a b + L1
A2
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Page 229
SECTION 2.10
SUBSTITUTE NUMERICAL VALUES:
s nom ⫽
(0.0040 in.)(25 * 106 psi)
2(5 in.)a
1.0 2
b + 20 in.
1.4
Stress Concentrations
229
Use the dashed curve in Fig. 2-65. K ⬇ 1.53
⫽ 3,984 psi
␴max ⫽ K␴nom ⬇ (1.53)(3984 psi)
⬇ 6100 psi
;
0.2 in.
R
⫽
⫽ 0.2
D1
1.0 in.
Problem 2.10-6 A prismatic bar of diameter d0 ⫽ 20 mm is being compared
P1
with a stepped bar of the same diameter (d1 ⫽ 20 mm) that is enlarged
in the middle region to a diameter d2 ⫽ 25 mm (see figure).
The radius of the fillets in the stepped bar is 2.0 mm.
(a) Does enlarging the bar in the middle region make it stronger than the
prismatic bar? Demonstrate your answer by determining the maximum
permissible load P1 for the prismatic bar and the maximum permissible
load P2 for the enlarged bar, assuming that the allowable stress for the
material is 80 MPa.
(b) What should be the diameter d0 of the prismatic bar if it is to have the same
maximum permissible load as does the stepped bar?
P2
d0
d1
P1
d2
d1
Solution 2.10-6
P2
Prismatic bar and stepped bar
Fillet radius: R ⫽ 2 mm
Allowable stress: ␴t ⫽ 80 MPa
(a) COMPARISON OF BARS
Prismatic bar: P1 ⫽ stA0 ⫽ st a
pd20
b
4
p
⫽ (80 MPa)a b(20mm)2 ⫽ 25.1 kN
4
;
Stepped bar: See Fig. 2-65 for the stress-concentration
factor.
d0 ⫽ 20 mm
d1 ⫽ 20 mm
d2 ⫽ 25 mm
R ⫽ 2.0 mm
D1 ⫽ 20 mm
D2 ⫽ 25 mm
R/D1 ⫽ 0.10
D2/D1 ⫽ 1.25
K ⬇ 1.75
s nom ⫽
P2
P2
smax
⫽
s nom ⫽
p 2
A1
K
d
4 1
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CHAPTER 2
11:44 AM
Axially Loaded Members
P2 ⫽ s nom A1 ⫽
⫽a
Page 230
s max
st
A1 ⫽ A1
K
K
80 MPa p
b a b(20 mm)2
1.75
4
L 14.4 kN
(b) DIAMETER OF PRISMATIC BAR FOR THE SAME
ALLOWABLE LOAD
d0 ⫽
;
st pd21
pd20
b ⫽ a
b
4
K 4
d20 ⫽
20 mm
L 15.1 mm
11.75
;
P1 ⫽ P2 st a
d1
1K
L
d21
K
Enlarging the bar makes it weaker, not stronger. The
ratio of loads is P1/P2 ⫽ K ⫽ 1.75
Problem 2.10-7 A stepped bar with a hole (see figure) has widths
b ⫽ 2.4 in. and c ⫽ 1.6 in. The fillets have radii equal to 0.2 in.
What is the diameter dmax of the largest hole that can be drilled
through the bar without reducing the load-carrying capacity?
Solution 2.10-7
Stepped bar with a hole
b ⫽ 2.4 in.
BASED UPON HOLE (Use Fig. 2-63)
c ⫽ 1.6 in.
Fillet radius: R ⫽ 0.2 in.
b ⫽ 2.4 in.
c1 ⫽ b ⫺ d
Find dmax
Pmax ⫽ s nom c1t ⫽
smax
(b ⫺ d)t
K
d
1
⫽ a1 ⫺ bbtsmax
K
b
BASED UPON FILLETS (Use Fig. 2-64)
b ⫽ 2.4 in.
c ⫽ 1.6 in.
R/c ⫽ 0.125
b/c ⫽ 1.5
R ⫽ 0.2 in.
K ⬇ 2.10
smax c
smax
Pmax ⫽ s nomct ⫽
ct ⫽
a b(bt)
K
K
b
L 0.317 bt smax
d ⫽ diameter of the hole (in.)
d(in.)
0.3
0.4
0.5
0.6
0.7
d/b
K
Pmax/bt␴max
0.125
0.167
0.208
0.250
0.292
2.66
2.57
2.49
2.41
2.37
0.329
0.324
0.318
0.311
0.299
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Page 231
SECTION 2.11
Nonlinear Behavior (Changes in Lengths of Bars)
231
Nonlinear Behavior (Changes in Lengths of Bars)
A
Problem 2.11-1 A bar AB of length L and weight density ␥ hangs vertically
under its own weight (see figure). The stress-strain relation for the material is
given by the Ramberg-Osgood equation (Eq. 2-71):
P⫽
s0a s m
s
+
a b
E
E s0
L
Derive the following formula
d⫽
gL2
gL m
s0aL
+
a b
2E
(m + 1)E s0
B
for the elongation of the bar.
Solution 2.11-1
Bar hanging under its own weight
STRAIN AT DISTANCE x
Let A ⫽ cross-sectional area
Let N ⫽ axial force at distance x
N ⫽ ␥Ax
s⫽
N
⫽ gx
A
␧⫽
s0a s m gx
s0 gx m
s
+
a b ⫽
+
a b
E
E s0
E
aE s0
ELONGATION OF BAR
L
d⫽
⫽
L0
␧dx ⫽
L
L
gx
gx m
s0a
a b dx
dx +
E L0 s0
L0 E
gL2
gL m
s0aL
+
a b
2E
(m + 1)E s0
Q.E.D.
;
A
B
P1 C
Problem 2.11-2 A prismatic bar of length L ⫽ 1.8 m and cross-sectional
area A ⫽ 480 mm is loaded by forces P1 ⫽ 30 kN and P2 ⫽ 60 kN
(see figure). The bar is constructed of magnesium alloy having a stress-strain
curve described by the following Ramberg-Osgood equation:
2
P⫽
s
1
s 10
+
a
b
(s ⫽ MPa)
45,000
618 170
in which ␴ has units of megapascals.
(a) Calculate the displacement ␦C of the end of the bar when the load
P1 acts alone.
(b) Calculate the displacement when the load P2 acts alone.
(c) Calculate the displacement when both loads act simultaneously.
2L
—
3
L
—
3
P2
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CHAPTER 2
Axially Loaded Members
Solution 2.11-2
Axially loaded bar
(c) BOTH P1 AND P2 ARE ACTING
AB:s ⫽
L ⫽ 1.8 m
␧ ⫽ 0.008477
A ⫽ 480 mm2
P1 ⫽ 30 kN
dAB ⫽ ␧a
P2 ⫽ 60 kN
Ramberg–Osgood Equation:
1
s 10
s
+
a
b (s ⫽ MPa)
␧⫽
45,000
618 170
Find displacement at end of bar.
P1
30 kN
⫽ 62.5 MPa
⫽
A
480 mm2
␧ ⫽ 0.001389
dc ⫽ ␧ a
2L
b ⫽ 1.67 mm
3
BC:s ⫽
2L
b ⫽ 10.17 mm
3
P2
60 kN
⫽ 125 MPa
⫽
A
480 mm2
␧ ⫽ 0.002853
L
dBC ⫽ ␧a b ⫽ 1.71 mm
3
(a) P1 ACTS ALONE
AB: s ⫽
P1 + P2
90 kN
⫽ 187.5 MPa
⫽
A
480 mm2
;
dC ⫽ dAB + dBC ⫽ 11.88 mm
;
(Note that the displacement when both loads act
simultaneously is not equal to the sum of the displacements when the loads act separately.)
(b) P2 ACTS ALONE
P2
60 kN
⫽
⫽ 125 MPa
A
480 mm2
␧ ⫽ 0.002853
dc ⫽ ␧L ⫽ 5.13 mm ;
ABC:s ⫽
Problem 2.11-3 A circular bar of length L ⫽ 32 in. and diameter
d ⫽ 0.75 in. is subjected to tension by forces P (see figure).
The wire is made of a copper alloy having the following hyperbolic
stress-strain relationship:
s⫽
18,000P
0 … P … 0.03 (s ⫽ ksi)
1 + 300P
(a) Draw a stress-strain diagram for the material.
(b) If the elongation of the wire is limited to 0.25 in. and
the maximum stress is limited to 40 ksi, what is the allowable load P?
d
P
P
L
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Page 233
SECTION 2.11
Solution 2.11-3
Nonlinear Behavior (Changes in Lengths of Bars)
233
Copper bar in tension
(b) ALLOWABLE LOAD P
Max. elongation ␦max ⫽ 0.25 in.
Max. stress ␴max ⫽ 40 ksi
Based upon elongation:
L ⫽ 32 in.
A⫽
d ⫽ 0.75 in.
pd2
⫽ 0.4418 in.2
4
␧max ⫽
dmax
0.25 in.
⫽
⫽ 0.007813
L
32 in.
smax ⫽
18,000␧max
⫽ 42.06 ksi
1 + 300␧max
(a) STRESS-STRAIN DIAGRAM
s⫽
18,000␧
0 … ␧ … 0.03 (s ⫽ ksi)
1 + 300␧
BASED UPON STRESS:
␴max ⫽ 40 ksi
Stress governs. P ⫽ ␴max A ⫽ (40 ksi)(0.4418 in.2)
⫽ 17.7 k
Problem 2.11-4 A prismatic bar in tension has length L ⫽ 2.0 m
and cross-sectional area A ⫽ 249 mm2. The material of the bar has the stressstrain curve shown in the figure.
Determine the elongation ␦ of the bar for each of the following axial
loads: P ⫽ 10 kN, 20 kN, 30 kN, 40 kN, and 45 kN. From these results,
plot a diagram of load P versus elongation ␦ (load-displacement diagram).
;
200
s (MPa)
100
0
0
0.005
e
0.010
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9/25/08
CHAPTER 2
11:44 AM
Page 234
Axially Loaded Members
Solution 2.11-4
Bar in tension
L ⫽ 2.0 m
A ⫽ 249 mm2
STRESS-STRAIN DIAGRAM
(See the problem statement for the diagram)
LOAD-DISPLACEMENT DIAGRAM
P
(kN)
␴ ⫽ P/A
(MPa)
␧
(from diagram)
␦ ⫽ ␧L
(mm)
10
20
30
40
45
40
80
120
161
181
0.0009
0.0018
0.0031
0.0060
0.0081
1.8
3.6
6.2
12.0
16.2
NOTE: The load-displacement curve has the same
shape as the stress-strain curve.
Problem 2.11-5 An aluminum bar subjected to tensile forces P has length
L ⫽ 150 in. and cross-sectional area A ⫽ 2.0 in.2 The stress-strain behavior of the
aluminum may be represented approximately by the bilinear stress-strain diagram
shown in the figure.
Calculate the elongation ␦ of the bar for each of the following axial loads:
P ⫽ 8 k, 16 k, 24 k, 32 k, and 40 k. From these results, plot a diagram of load
P versus elongation ␦ (load-displacement diagram).
s
12,000
psi
E2 = 2.4 106 psi
E1 = 10 106 psi
0
e
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11:44 AM
Page 235
SECTION 2.11
Solution 2.11-5
Nonlinear Behavior (Changes in Lengths of Bars)
Aluminum bar in tension
LOAD-DISPLACEMENT DIAGRAM
L ⫽ 150 in.
A ⫽ 2.0 in.2
STRESS-STRAIN DIAGRAM
E1 ⫽ 10 ⫻ 106 psi
E2 ⫽ 2.4 ⫻ 106 psi
␴1 ⫽ 12,000 psi
␧1 ⫽
12,000 psi
s1
⫽
E1
10 * 106 psi
⫽ 0.0012
For 0 ⱕ ␴ ⱕ ␴1:
s
s
⫽
(s ⫽ psi)
E2
10 * 106psi
For ␴ ⱖ ␴1:
␧⫽
␧ ⫽ ␧1 +
⫽
s
Eq. (1)
s ⫺ 12,000
s ⫺ s1
⫽ 0.0012 +
E2
2.4 * 106
2.4 * 106
⫺ 0.0038 (s ⫽ psi)
Eq. (2)
P (k)
␴ ⫽ P/A
(psi)
␧ (from Eq.
1 or Eq. 2)
␦ ⫽ ␧L
(in.)
8
16
24
32
40
4,000
8,000
12,000
16,000
20,000
0.00040
0.00080
0.00120
0.00287
0.00453
0.060
0.120
0.180
0.430
0.680
235
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CHAPTER 2
11:44 AM
Page 236
Axially Loaded Members
Problem 2.11-6 A rigid bar AB, pinned at end A, is supported by a
wire CD and loaded by a force P at end B (see figure). The wire is
made of high-strength steel having modulus of elasticity E ⫽ 210 GPa
and yield stress ␴Y ⫽ 820 MPa. The length of the wire is L ⫽ 1.0 m
and its diameter is d ⫽ 3 mm. The stress-strain diagram for the steel is
defined by the modified power law, as follows:
C
L
A
D
B
s ⫽ EP 0 … s … sY
s ⫽ sY a
EP n
b s Ú sY
sY
P
2b
(a) Assuming n ⫽ 0.2, calculate the displacement ␦B at the end of
the bar due to the load P. Take values of P from 2.4 kN to
5.6 kN in increments of 0.8 kN.
b
(b) Plot a load-displacement diagram showing P versus ␦B.
Solution 2.11-6
Rigid bar supported by a wire
sY s 1/n
a b
E sY
3P
Axial force in wire: F ⫽
2
3P
F
Stress in wire: s ⫽ ⫽
A
2A
PROCEDURE: Assume a value of P
Calculate ␴ from Eq. (6)
Calculate ␧ from Eq. (4) or (5)
Calculate ␦B from Eq. (3)
From Eq. (2): ␧ ⫽
Wire: E ⫽ 210 GPa
␴Y ⫽ 820 MPa
L ⫽ 1.0 m
d ⫽ 3 mm
A⫽
pd2
⫽ 7.0686 mm2
4
STRESS-STRAIN DIAGRAM
␴ (MPa)
Eq. (6)
␧ Eq. (4)
or (5)
␦B (mm)
Eq. (3)
2.4
3.2
4.0
4.8
5.6
509.3
679.1
848.8
1018.6
1188.4
0.002425
0.003234
0.004640
0.01155
0.02497
3.64
4.85
6.96
17.3
37.5
For ␴ ⫽ ␴Y ⫽ 820 MPa:
␧ ⫽ 0.0039048
P ⫽ 3.864 kN
(n ⫽ 0.2)
(2)
(b) LOAD-DISPLACEMENT DIAGRAM
(a) DISPLACEMENT ␦B AT END OF BAR
3
3
␦ ⫽ elongation of wire dB ⫽ d ⫽ ␧L
2
2
Obtain ␧ from stress-strain equations:
(3)
(0 ⱕ ␴ ⱕ ␴Y)
E␧ n
s ⫽ sY a b
sY
(␴ ⱖ ␴Y)
From Eq. (1): ␧ ⫽
sE
(0 … s … sY)
(4)
(6)
P
(kN)
(1)
␴ ⫽ E␧
(5)
␦B ⫽ 5.86 mm
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Page 237
SECTION 2.12
Elastoplastic Analysis
237
u
C
Elastoplastic Analysis
The problems for Section 2.12 are to be solved assuming that the material is
elastoplastic with yield stress ␴Y, yield strain ⑀Y, and modulus of elasticity
E in the linearly elastic region (see Fig. 2-70).
A
u
Problem 2.12-1 Two identical bars AB and BC support a vertical load
P (see figure). The bars are made of steel having a stress-strain curve that
may be idealized as elastoplastic with yield stress ␴Y. Each bar has
cross-sectional area A.
Determine the yield load PY and the plastic load PP.
Solution 2.12-1
B
P
Two bars supporting a load P
JOINT B
⌺Fvert ⫽ 0
Structure is statically determinate. The yield load PY
and the plastic lead PP occur at the same time, namely,
when both bars reach the yield stress.
(2␴YA) sin ␪ ⫽ P
PY ⫽ PP ⫽ 2␴YA sin ␪
Problem 2.12-2 A stepped bar ACB with circular cross sections
is held between rigid supports and loaded by an axial force P at
midlength (see figure). The diameters for the two parts of the bar are
d1 ⫽ 20 mm and d2 ⫽ 25 mm, and the material is elastoplastic with
yield stress ␴Y ⫽ 250 MPa.
Determine the plastic load PP.
A
d1
;
C
L
—
2
d2
P
L
—
2
B
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CHAPTER 2
Solution 2.12-2
11:44 AM
Page 238
Axially Loaded Members
Bar between rigid supports
FAC ⫽ ␴YA1
FCB ⫽ ␴YA2
P ⫽ FAC ⫹ FCB
PP ⫽ ␴YA1 ⫹ ␴YA2 ⫽ ␴Y(A1 ⫹ A2)
;
SUBSTITUTE NUMERICAL VALUES:
d1 ⫽ 20 mm
d2 ⫽ 25 mm
␴Y ⫽ 250 MPa
DETERMINE THE PLASTIC LOAD PP:
At the plastic load, all parts of the bar are stressed to the
yield stress.
p
PP ⫽ (250 MPa)a b(d21 + d22)
4
p
⫽ (250 MPa)a b[(20 mm)2 + (25 mm)2]
4
⫽ 201 kN
Point C:
;
Problem 2.12-3 A horizontal rigid bar AB supporting a load P is hung
from five symmetrically placed wires, each of cross-sectional area A
(see figure). The wires are fastened to a curved surface of radius R.
R
(a) Determine the plastic load PP if the material of the wires is
elastoplastic with yield stress ␴Y.
(b) How is PP changed if bar AB is flexible instead of rigid?
(c) How is PP changed if the radius R is increased?
A
B
P
Solution 2.12-3
Rigid bar supported by five wires
(b) BAR AB IS FLEXIBLE
At the plastic load, each wire is stressed to the yield
stress, so the plastic load is not changed. ;
(a) PLASTIC LOAD PP
At the plastic load, each wire is stressed to the yield
stress. ⬖ PP ⫽ 5␴YA ;
F ⫽ ␴YA
(c) RADIUS R IS INCREASED
Again, the forces in the wires are not changed, so the
plastic load is not changed. ;
Sec_2.8-2.12.qxd
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Page 239
SECTION 2.12
Elastoplastic Analysis
239
Problem 2.12-4 A load P acts on a horizontal beam that is supported by four
rods arranged in the symmetrical pattern shown in the figure. Each rod has
cross-sectional area A and the material is elastoplastic with yield stress ␴Y.
Determine the plastic load PP.
a
a
P
Solution 2.12-4
Beam supported by four rods
F ⫽ ␴YA
Sum forces in the vertical direction and solve for the
load:
At the plastic load, all four rods are stressed to the yield
stress.
PP ⫽ 2F ⫹ 2F sin ␣
PP ⫽ 2␴YA (1 ⫹ sin ␣)
21 in.
Problem 2.12-5 The symmetric truss ABCDE shown in the figure
is constructed of four bars and supports a load P at joint E. Each of
the two outer bars has a cross-sectional area of 0.307 in.2, and each of
the two inner bars has an area of 0.601 in.2 The material is elastoplastic with yield stress ␴Y ⫽ 36 ksi.
Determine the plastic load PP.
A
;
54 in.
21 in.
C
B
D
36 in.
E
P
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CHAPTER 2
Solution 2.12-5
11:44 AM
Page 240
Axially Loaded Members
Truss with four bars
PLASTIC LOAD PP
At the plastic load, all bars are stressed to the yield
stress.
FAE ⫽ ␴YAAE
PP ⫽
FBE ⫽ ␴YABE
6
8
sY AAE + sY ABE
5
5
;
SUBSTITUTE NUMERICAL VALUES:
AAE ⫽ 0.307 in.2 ABE ⫽ 0.601 in.2
LAE ⫽ 60 in.
JOINT E
LBE ⫽ 45 in.
sY ⫽ 36 ksi
6
8
PP ⫽ (36 ksi) (0.307 in.2) + (36 ksi) (0.601 in.2)
5
5
Equilibrium:
3
4
2FAE a b + 2FBE a b ⫽ P
5
5
or
6
8
P ⫽ FAE + FBE
5
5
⫽ 13.26 k + 34.62 k ⫽ 47.9 k
Problem 2.12-6 Five bars, each having a diameter of 10 mm, support a
b
b
;
b
b
load P as shown in the figure. Determine the plastic load PP if the material is
elastoplastic with yield stress ␴Y ⫽ 250 MPa.
2b
P
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Page 241
SECTION 2.12
Solution 2.12-6
b
Elastoplastic Analysis
241
Truss consisting of five bars
b
b
At the plastic load, all five bars
are stressed to the yield stress
b
F ⫽ ␴YA
Sum forces in the vertical direction and solve for the load:
2b
PP ⫽ 2Fa
P
d ⫽ 10 mm
pd2
⫽ 78.54 mm2
A⫽
4
␴Y ⫽ 250 MPa
⫽
1
2
b + 2Fa
b + F
12
15
sYA
(512 + 415 + 5)
5
⫽ 4.2031sYA
;
Substitute numerical values:
PP ⫽ (4.2031)(250 MPa)(78.54 mm2)
⫽ 82.5 kN
Problem 2.12-7 A circular steel rod AB of diameter d ⫽ 0.60 in.
;
B
A
is stretched tightly between two supports so that initially the
tensile stress in the rod is 10 ksi (see figure). An axial force P is
then applied to the rod at an intermediate location C.
d
(a) Determine the plastic load PP if the material is elastoplastic
with yield stress ␴Y ⫽ 36 ksi.
(b) How is PP changed if the initial tensile stress is doubled
to 20 ksi?
Solution 2.12-7
A
P
C
Bar held between rigid supports
POINT C:
sYA
sYA
P
— C ¡ —
d ⫽ 0.6 in.
␴Y ⫽ 36 ksi
Initial tensile stress ⫽ 10 ksi
(a) PLASTIC LOAD PP
The presence of the initial tensile stress does not
affect the plastic load. Both parts of the bar must
yield in order to reach the plastic load.
p
PP ⫽ 2sYA ⫽ (2) (36 ksi)a b(0.60 in.)2
4
⫽ 20.4 k
;
(B) INITIAL TENSILE STRESS IS DOUBLED
PP is not changed.
;
B
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CHAPTER 2
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Axially Loaded Members
Problem 2.12-8 A rigid bar ACB is supported on a fulcrum
at C and loaded by a force P at end B (see figure). Three
identical wires made of an elastoplastic material
(yield stress ␴Y and modulus of elasticity E) resist the load P.
Each wire has cross-sectional area A and length L.
(a) Determine the yield load PY and the corresponding
yield displacement ␦Y at point B.
(b) Determine the plastic load PP and the corresponding
displacement ␦P at point B when the load just reaches
the value PP.
(c) Draw a load-displacement diagram with the load
P as ordinate and the displacement ␦B of point B
as abscissa.
Solution 2.12-8
L
A
C
B
P
L
a
a
a
a
Rigid bar supported by wires
(b) PLASTIC LOAD PP
(a) YIELD LOAD PY
Yielding occurs when the most highly stressed wire
reaches the yield stress ␴Y
At the plastic load, all wires reach the yield stress.
⌺MC ⫽ 0
PP ⫽
4sYA
3
;
At point A:
dA ⫽ (sYA)a
sYL
L
b ⫽
EA
E
At point B:
dB ⫽ 3dA ⫽ dP ⫽
⌺MC ⫽ 0
PY ⫽ ␴YA
At point A:
;
(c) LOAD-DISPLACEMENT DIAGRAM
;
sYA
sYL
L
dA ⫽ a
ba
b ⫽
2
EA
2E
At point B:
dB ⫽ 3dA ⫽ dY ⫽
3sYL
E
3sYL
2E
;
4
PP ⫽ PY
3
dP ⫽ 2dY
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SECTION 2.12
243
Elastoplastic Analysis
Problem 2.12-9 The structure shown in the figure consists of a
horizontal rigid bar ABCD supported by two steel wires, one of length L
and the other of length 3L/4. Both wires have cross-sectional area A and
are made of elastoplastic material with yield stress ␴Y and modulus of
elasticity E. A vertical load P acts at end D of the bar.
(a) Determine the yield load PY and the corresponding yield
displacement ␦Y at point D.
(b) Determine the plastic load PP and the corresponding displacement
␦P at point D when the load just reaches the value PP.
(c) Draw a load-displacement diagram with the load P as ordinate
and the displacement ␦D of point D as abscissa.
Solution 2.12-9
L
A
3L
4
B
C
D
P
2b
b
b
Rigid bar supported by two wires
FREE-BODY DIAGRAM
A ⫽ cross-sectional area
EQUILIBRIUM:
␴Y ⫽ yield stress
⌺MA ⫽ 0 哵哴
E ⫽ modulus of elasticity
FB(2b) ⫹ FC(3b) ⫽ P(4b)
2FB ⫹ 3FC ⫽ 4P
(3)
DISPLACEMENT DIAGRAM
FORCE-DISPLACEMENT RELATIONS
FBL
dC ⫽
dB ⫽
EA
3
FC a Lb
4
EA
(4, 5)
Substitute into Eq. (1):
COMPATIBILITY:
3
dC ⫽ dB
2
(1)
3FCL
3FBL
⫽
4EA
2EA
␦D ⫽ 2␦B
(2)
FC ⫽ 2FB
(6)
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CHAPTER 2
Page 244
Axially Loaded Members
STRESSES
From Eq. (3):
FB
FC
sC ⫽
sC ⫽ 2sB
(7)
A
A
Wire C has the larger stress. Therefore, it will yield first.
2(␴YA) ⫹ 3(␴YA) ⫽ 4P
sB ⫽
(a) YIELD LOAD
␴C ⫽ ␴Y
FB ⫽
(From Eq. 7)
1
s A
2 Y
From Eq. (3):
1
2a sYA b + 3(sYA) ⫽ 4P
2
P ⫽ PY ⫽ ␴YA
;
From Eq. (4):
sC
sY
⫽
sB ⫽
2
2
FC ⫽ ␴Y A
5
P ⫽ PP ⫽ sYA
4
FBL
sY L
⫽
EA
E
From Eq. (2):
dB ⫽
dD ⫽ dP ⫽ 2dB ⫽
2sYL
E
;
(c) LOAD-DISPLACEMENT DIAGRAM
5
PP ⫽ PY
4
;
From Eq. (4):
␦P ⫽ 2␦Y
FB L
sY L
⫽
dB ⫽
EA
2E
From Eq. (2):
dD ⫽ dY ⫽ 2dB ⫽
sY L
E
;
(b) PLASTIC LOAD
At the plastic load, both wires yield.
␴B ⫽ ␴Y ⫽ ␴C
FB ⫽ FC ⫽ ␴Y A
Problem 2.12-10 Two cables, each having a length L of approximately 40 m, support a l
oaded container of weight W (see figure). The cables, which have effective cross-sectional area
A ⫽ 48.0 mm2 and effective modulus of elasticity E ⫽ 160 GPa, are identical except that one
cable is longer than the other when they are hanging separately and unloaded. The difference
in lengths is d ⫽ 100 mm. The cables are made of steel having an elastoplastic stress-strain
diagram with ␴Y ⫽ 500 MPa. Assume that the weight W is initially zero and is slowly increased
by the addition of material to the container.
L
(a) Determine the weight WY that first produces yielding of the shorter cable. Also, determine
the corresponding elongation ␦Y of the shorter cable.
(b) Determine the weight WP that produces yielding of both cables. Also, determine the
elongation ␦P of the shorter cable when the weight W just reaches the value WP.
(c) Construct a load-displacement diagram showing the weight W as ordinate and the
elongation ␦ of the shorter cable as abscissa. (Hint: The load displacement diagram is
not a single straight line in the region 0 ⱕ W ⱕ WY.)
W
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Page 245
SECTION 2.12
Solution 2.12-10
Elastoplastic Analysis
Two cables supporting a load
L ⫽ 40 m
A ⫽ 48.0 mm2
(b) PLASTIC LOAD WP
E ⫽ 160 GPa
F1 ⫽ ␴YA
d ⫽ difference in length ⫽ 100 mm
WP ⫽ 2␴YA ⫽ 48 kN
␴Y ⫽ 500 MPa
INITIAL STRETCHING OF CABLE 1
Initially, cable 1 supports all of the load.
Let W1 ⫽ load required to stretch cable 1
to the same length as cable 2
EA
W1 ⫽
d ⫽ 19.2 kN
L
F2 ⫽ ␴YA
;
␦2P ⫽ elongation of cable 2
⫽ F2 a
sYL
L
b ⫽
⫽ 0.125 mm ⫽ 125 mm
EA
E
␦1P ⫽ ␦2P ⫹ d ⫽ 225 mm
␦P ⫽ ␦1P ⫽ 225 mm
;
(c) LOAD-DISPLACEMENT DIAGRAM
␦1 ⫽ 100 mm (elongation of cable 1)
s1 ⫽
W1
Ed
⫽
⫽ 400 MPa (s1 6 sY ‹ 7 OK)
A
L
(a) YIELD LOAD WY
Cable 1 yields first. F1 ⫽ ␴YA ⫽ 24 kN
␦1Y ⫽ total elongation of cable 1
d1Y ⫽ total elongation of cable 1
d1Y ⫽
F1L
sY L
⫽
⫽ 0.125 m ⫽ 125 mm
EA
E
dY ⫽ d1Y ⫽ 125 mm
;
d2Y ⫽ elongation of cable 2
⫽ d1Y ⫺ d ⫽ 25 mm
EA
F2 ⫽
d2Y ⫽ 4.8 kN
L
WY ⫽ F1 + F2 ⫽ 24 kN + 4.8 kN
⫽ 28.8 kN
;
dY
WY
⫽ 1.5
⫽ 1.25
W1
d1
dP
WP
⫽ 1.667
⫽ 1.8
WY
dY
0 ⬍ W ⬍ W1: slope ⫽ 192,000 N/m
W1 ⬍ W ⬍ WY: slope ⫽ 384,000 N/m
WY ⬍ W ⬍ WP: slope ⫽ 192,000 N/m
245
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11:44 AM
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Axially Loaded Members
Problem 2.12-11 A hollow circular tube T of length L ⫽ 15 in.
is uniformly compressed by a force P acting through a rigid plate
(see figure). The outside and inside diameters of the tube are 3.0 and
2.75 in., repectively. A concentric solid circular bar B of 1.5 in.
diameter is mounted inside the tube. When no load is present, there is
a clearance c ⫽ 0.010 in. between the bar B and the rigid plate. Both
bar and tube are made of steel having an elastoplastic stress-strain
diagram with E ⫽ 29 ⫻ 103 ksi and ␴Y ⫽ 36 ksi.
(a) Determine the yield load PY and the corresponding shortening
␦Y of the tube.
(b) Determine the plastic load PP and the corresponding shortening
␦P of the tube.
(c) Construct a load-displacement diagram showing the load P as
ordinate and the shortening ␦ of the tube as abscissa. (Hint: The
load-displacement diagram is not a single straight line in the
region 0 ⱕ P ⱕ PY.)
Solution 2.12-11
L ⫽ 15 in.
c ⫽ 0.010 in.
E ⫽ 29 ⫻ 103 ksi
␴Y ⫽ 36 ksi
P
c
T
T
B
T
L
Tube and bar supporting a load
TUBE:
d2 ⫽ 3.0 in.
d1 ⫽ 2.75 in.
AT ⫽
p 2
(d ⫺ d21) ⫽ 1.1290 in.2
4 2
B
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SECTION 2.12
BAR
Elastoplastic Analysis
(b) PLASTIC LOAD PP
FT ⫽ ␴YAT
d ⫽ 1.5 in.
AB ⫽
pd
⫽ 1.7671 in.2
4
⫽ 104,300 lb
;
␦BP ⫽ shortening of bar
INITIAL SHORTENING OF TUBE T
Initially, the tube supports all of the load.
Let P1 ⫽ load required to close the clearance
⫽ FB a
sYL
L
⫽ 0.018621 in.
b ⫽
EAB
E
␦TP ⫽ ␦BP ⫹ c ⫽ 0.028621 in.
EAT
c ⫽ 21,827 lb
L
Let ␦1 ⫽ shortening of tube
P1 ⫽
P1
⫽ 19,330 psi
s1 ⫽
AT
FB ⫽ ␴YAB
PP ⫽ FT ⫹ FB ⫽ ␴Y(AT ⫹ AB)
2
␦1 ⫽ c ⫽ 0.010 in.
␦P ⫽ ␦TP ⫽ 0.02862 in.
;
(c) LOAD-DISPLACEMENT DIAGRAM
(␴1 ⬍ ␴Y ⬖ OK)
(a) YIELD LOAD PY
Because the tube and bar are made of the same
material, and because the strain in the tube is larger
than the strain in the bar, the tube will yield first.
FT ⫽ ␴YAT ⫽ 40,644 lb
␦ TY ⫽ shortening of tube at the yield stress
s TY ⫽
FTL
sYL
⫽
⫽ 0.018621 in.
EAT
E
␦Y ⫽ ␦TY ⫽ 0.018621 in.
;
␦BY ⫽ shortening of bar
⫽ ␦TY ⫺ c ⫽ 0.008621 in.
dY
PY
⫽ 3.21
⫽ 1.86
P1
d1
EAB
d ⫽ 29,453 lb
L BY
dP
PP
⫽ 1.49
⫽ 1.54
PY
dY
FB ⫽
PY ⫽ FT ⫹ FB ⫽ 40,644 lb ⫹ 29,453 lb
⫽ 70,097 lb
PY ⫽ 70,100 lb
0 ⬍ P ⬍ P1: slope ⫽ 2180 k/in.
P1 ⬍ P ⬍ PY: slope ⫽ 5600 k/in.
;
PY ⬍ P ⬍ PP: slope ⫽ 3420 k/in.
247
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Page 249
3
Torsion
Torsional Deformations
Problem 3.2-1 A copper rod of length L ⫽ 18.0 in. is to be twisted
by torques T (see figure) until the angle of rotation between the ends
of the rod is 3.0°.
If the allowable shear strain in the copper is 0.0006 rad, what is
the maximum permissible diameter of the rod?
d
T
T
L
Probs. 3.2-1 and 3.2-2
Solution 3.2-1
Copper rod in torsion
d
T
T
L
L ⫽ 18.0 in.
From Eq. (3-3):
p
f ⫽ 3.0° ⫽ (3.0)a 180 b rad
␥max ⫽
⫽ 0.05236 rad
␥allow ⫽ 0.0006 rad
Find dmax
dmax ⫽
rf
df
⫽
L
2L
2L␥ allow
f
dmax ⫽ 0.413 in.
⫽
(2)(18.0 in.)(0.0006 rad)
0.05236 rad
;
Problem 3.2-2 A plastic bar of diameter d ⫽ 56 mm is to be twisted by torques T (see figure) until the angle of rotation
between the ends of the bar is 4.0°.
If the allowable shear strain in the plastic is 0.012 rad, what is the minimum permissible length of the bar?
249
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CHAPTER 3
Page 250
Torsion
Solution 3.2-2
NUMERICAL DATA
d ⫽ 56 mm
␥a ⫽ 0.012 radians
f ⫽ 4a
p
b radians
180
solution based on Equ. (3-3):
Lmin ⫽ 162.9 mm
Lmin ⫽
df
2␥a
;
Problem 3.2-3 A circular aluminum tube subjected to pure
torsion by torques T (see figure) has an outer radius r2 equal to
1.5 times the inner radius r1.
T
T
L
(a) If the maximum shear strain in the tube is measured as
400 ⫻ 10⫺6 rad, what is the shear strain ␥1 at the inner
surface?
(b) If the maximum allowable rate of twist is 0.125 degrees
per foot and the maximum shear strain is to be kept at
400 ⫻ 10⫺6 rad by adjusting the torque T, what is the
minimum required outer radius (r2)min?
r2
r1
Probs. 3.2-3, 3.2-4, and 3.2-5
Solution 3.2-3
NUMERICAL DATA
(b) MIN. REQUIRED OUTER RADIUS
r2 ⫽ 1.5r1 ␥max ⫽ 400 ⫻ (10⫺6) radians
u ⫽ 0.125a
p
1
ba b
180 12
r2min ⫽
␥max
u
r2min ⫽
r2min ⫽ 2.2 inches
␥max
u
;
␪ ⫽ 1.818 ⫻ 10⫺4 rad /m.
(a) SHEAR STRAIN AT INNER SURFACE AT RADIUS r1
␥1 ⫽
r1
␥
r2 max
␥1 ⫽
1
␥
1.5 max
␥1 ⫽ 267 ⫻ 10⫺6 radians
;
Problem 3.2-4 A circular steel tube of length L ⫽ 1.0 m is loaded in torsion by torques T (see figure).
(a) If the inner radius of the tube is r1 ⫽ 45 mm and the measured angle of twist between the ends is 0.5°, what is the
shear strain ␥1 (in radians) at the inner surface?
(b) If the maximum allowable shear strain is 0.0004 rad and the angle of twist is to be kept at 0.45° by adjusting the torque
T, what is the maximum permissible outer radius (r2)max?
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251
SECTION 3.2 Torsional Deformations
Solution 3.2-4
(b) MAX. PERMISSIBLE OUTER RADIUS
NUMERICAL DATA
L ⫽ 1000 mm
f ⫽ 0.45a
r1 ⫽ 45 mm
f ⫽ 0.5a
p
b radians
180
(a) SHEAR STRAIN AT INNER SURFACE
f
␥1 ⫽ r1
␥1 ⫽ 393 ⫻ 10⫺6 radians
L
p
b radians
180
␥max ⫽ 0.0004 radians
r2max ⫽ 50.9 mm
␥max ⫽ r2
f
L
r2max ⫽ ␥max
L
f
;
;
Problem 3.2-5 Solve the preceding problem if the length L ⫽ 56 in., the inner radius r1 ⫽ 1.25 in., the angle of twist is
0.5°, and the allowable shear strain is 0.0004 rad.
Solution 3.2-5
NUMERICAL DATA
(b) MAXIMUM PERMISSIBLE OUTER RADIUS (r2)max
L ⫽ 56 inches r1 ⫽ 1.25 inches
f ⫽ 0.5 a
f ⫽ 0.5 a
p
b radians
180
␥max
␥a ⫽ 0.0004 radians
(a) SHEAR STRAIN g1 (IN RADIANS) AT THE INNER
SURFACE
␥1 ⫽ r1
f
L
␥1 ⫽ 195 ⫻ 10⫺6 radians
;
p
b radians
180
f
⫽ r2
L
␥a ⫽ 0.0004 radians
L
r2max ⫽ ␥ a
f
r2max ⫽ 2.57 inches
;
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CHAPTER 3
Page 252
Torsion
Circular Bars and Tubes
P
Problem 3.3-1 A prospector uses a hand-powered winch
(see figure) to raise a bucket of ore in his mine shaft. The axle
of the winch is a steel rod of diameter d ⫽ 0.625 in. Also, the
distance from the center of the axle to the center of the lifting
rope is b ⫽ 4.0 in.
If the weight of the loaded bucket is W ⫽ 100 lb, what is
the maximum shear stress in the axle due to torsion?
d
W
b
W
Solution 3.3-1
Hand-powered winch
d ⫽ 0.625 in.
MAXIMUM SHEAR STRESS IN THE AXLE
b ⫽ 4.0 in.
From Eq. (3-12):
W ⫽ 100 lb
tmax ⫽
Torque T applied to the axle:
T ⫽ Wb ⫽ 400 lb-in.
tmax ⫽
16T
pd3
(16)(400 lb-in.)
p(0.625 in.)3
tmax ⫽ 8,340 psi
;
Problem 3.3-2 When drilling a hole in a table leg, a furniture
maker uses a hand-operated drill (see figure) with a bit of
diameter d ⫽ 4.0 mm.
(a) If the resisting torque supplied by the table leg is equal
to 0.3 N⭈m, what is the maximum shear stress in the drill bit?
(b) If the shear modulus of elasticity of the steel is G ⫽ 75 GPa,
what is the rate of twist of the drill bit (degrees per meter)?
d
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SECTION 3.3
Solution 3.3-2
253
Circular Bars and Tubes
Torsion of a drill bit
(b) RATE OF TWIST
From Eq. (3-14):
u⫽
d ⫽ 4.0 mm
T ⫽ 0.3 N⭈m G ⫽ 75 GPa
(a) MAXIMUM SHEAR STRESS
From Eq. (3-12):
tmax ⫽
tmax ⫽
16T
u⫽
T
GIP
0.3 N # m
p
(75 GPa)a
b(4.0 mm)4
32
u ⫽ 0.1592 rad/m ⫽ 9.12°/m
pd3
;
16(0.3 N # m)
p(4.0 mm)3
tmax ⫽ 23.8 MPa
;
Problem 3.3-3 While removing a wheel to change a tire,
a driver applies forces P ⫽ 25 lb at the ends of two of the arms
of a lug wrench (see figure). The wrench is made of steel with
shear modulus of elasticity G ⫽ 11.4 ⫻ 106 psi. Each arm of
the wrench is 9.0 in. long and has a solid circular cross section
of diameter d ⫽ 0.5 in.
(a) Determine the maximum shear stress in the arm
that is turning the lug nut (arm A).
(b) Determine the angle of twist (in degrees)
of this same arm.
P
9.0
in.
A
9.0
in.
d = 0.5 in.
P = 25 lb
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CHAPTER 3
Solution 3.3-3
Page 254
Torsion
Lug wrench
P ⫽ 25 lb
(a) MAXIMUM SHEAR STRESS
From Eq. (3-12):
(16)(450 l b - in.)
16T
⫽
tmax ⫽
3
pd
p(0.5 in.)3
L ⫽ 9.0 in.
d ⫽ 0.5 in.
G ⫽ 11.4 ⫻ 106 psi
T ⫽ torque acting on
arm A
tmax ⫽ 18,300 psi
;
(b) ANGLE OF TWIST
From Eq. (3-15):
(450 lb-in.)(9.0 in.)
TL
f⫽
⫽
GIP
p
(11.4 * 106 psi)a
b(0.5 in.)4
32
T ⫽ P(2L) ⫽ 2(25 lb)
(9.0 in.)
⫽ 450 lb-in.
f ⫽ 0.05790 rad ⫽ 3.32°
Problem 3.3-4 An aluminum bar of solid circular cross section
is twisted by torques T acting at the ends (see figure). The
dimensions and shear modulus of elasticity are as follows:
L ⫽ 1.4 m, d ⫽ 32 mm, and G ⫽ 28 GPa.
;
d
T
T
L
(a) Determine the torsional stiffness of the bar.
(b) If the angle of twist of the bar is 5°,
what is the maximum shear stress?
What is the maximum shear strain (in radians)?
Solution 3.3-4
(a) TORSIONAL STIFFNESS OF BAR
d ⫽ 32 mm
GI p
kT ⫽
L
G ⫽ 28 GPa
Ip ⫽
p 4
d
32
Ip ⫽ 1.029 ⫻ 105 mm4
kT ⫽
p
2811092a 0.0324 b
32
1.4
kT ⫽ 2059 N # m
(b) MAX SHEAR STRESS AND STRAIN
f ⫽ 5a
T ⫽ kT f
tmax ⫽
␶max ⫽ 27.9 MPa
gmax ⫽
;
p
b radians
180
d
Ta b
2
Ip
;
tmax
G
␥max ⫽ 997 ⫻ 10⫺6 radians
;
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SECTION 3.3
Circular Bars and Tubes
255
Problem 3.3-5 A high-strength steel drill rod used for boring
a hole in the earth has a diameter of 0.5 in. (see figure).
The allowable shear stress in the steel is 40 ksi and the shear
modulus of elasticity is 11,600 ksi.
What is the minimum required length of the rod so that
one end of the rod can be twisted 30° with respect to the other
end without exceeding the allowable stress?
Solution 3.3-5
d = 0.5 in.
T
T
L
Steel drill rod
T
d = 0.5 in.
T
T⫽
L
G ⫽ 11,600 psi
p
b rad ⫽ 0.52360 rad
180
Lmin ⫽
␶allow ⫽ 40 ksi
⫽
MINIMUM LENGTH
From Eq. (3-12): tmax ⫽
16T
pd3
TL
32TL
⫽
GIP
Gpd4
Gpd 4f
, substitute Tinto Eq. (1):
32L
tmax ⫽ a
d ⫽ 0.5 in.
f ⫽ 30° ⫽ (30°)a
From Eq. (3-15): f ⫽
ba
3
16
pd
Gpd4f
Gdf
b ⫽
32L
2L
Gdf
2t allow
(11,600 ksi)(0.5 in.)(0.52360 rad)
2(40 ksi)
Lmin ⫽ 38.0 in.
;
(1)
Problem 3.3-6 The steel shaft of a socket wrench has a
diameter of 8.0 mm. and a length of 200 mm (see figure).
If the allowable stress in shear is 60 MPa, what is the
maximum permissible torque Tmax that may be exerted with
the wrench?
Through what angle ␾ (in degrees) will the shaft twist
under the action of the maximum torque? (Assume G ⫽ 78 GPa
and disregard any bending of the shaft.)
d = 8.0 mm
T
L = 200 mm
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Torsion
Solution 3.3-6
Socket wrench
ANGLE OF TWIST
From Eq. (3-15): f ⫽
d ⫽ 8.0 mm
L ⫽ 200 mm
␶allow ⫽ 60 MPa
G ⫽ 78 GPa
MAXIMUM PERMISSIBLE TORQUE
16T
From Eq. (3-12): tmax ⫽
pd3
3
pd tmax
Tmax ⫽
16
3
Tmax ⫽
p(8.0 mm) (60 MPa)
16
Tmax ⫽ 6.03 N # m
;
Problem 3.3-7 A circular tube of aluminum is subjected to
torsion by torques T applied at the ends (see figure). The bar is
24 in. long, and the inside and outside diameters are 1.25 in. and
1.75 in., respectively. It is determined by measurement that the
angle of twist is 4° when the torque is 6200 lb-in.
Calculate the maximum shear stress ␶max in the tube, the
shear modulus of elasticity G, and the maximum shear strain
␥max (in radians).
TmaxL
GIP
From Eq. (3-12): Tmax ⫽
f⫽ a
f⫽
f⫽
pd3t max
L
ba
b
16
GIP
pd3tmaxL(32)
16G(pd4)
⫽
pd3tmax
16
IP ⫽
pd4
32
2tmaxL
Gd
2(60 MPa)(200 mm)
⫽ 0.03846 rad
(78 GPa)(8.0 mm)
f ⫽ 10.03846 rad2a
180
deg/radb ⫽ 2.20°
p
T
;
T
24 in.
1.25 in.
1.75 in.
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SECTION 3.3
Circular Bars and Tubes
257
Solution 3.3-7
NUMERICAL DATA
L ⫽ 24 in.
f ⫽ 4a
r2 ⫽
1.75
in.
2
p
b radians
180
MAX. SHEAR STRESS
p
Ip ⫽ 1 r24 ⫺ r142
2
tmax ⫽
Tr2
Ip
gmax ⫽
MAX. SHEAR STRAIN
1.25
in.
2
r1 ⫽
T ⫽ 6200 lb-in.
␥max ⫽ 0.00255 radians
r2
f
L
;
SHEAR MODULUS OF ELASTICITY
G
G⫽
G ⫽ 3.129 ⫻ 106 psi
tmax ⫽
Tr2
Ip
or
G⫽
Ip ⫽ 0.681 in.
4
␶max ⫽ 7965 psi
TL
fIp
G ⫽ 3.13 ⫻ 106 psi
tmax
gmax
;
;
Problem 3.3-8 A propeller shaft for a small yacht is made of a
solid steel bar 104 mm in diameter. The allowable stress in shear
is 48 MPa, and the allowable rate of twist is 2.0° in 3.5 meters.
Assuming that the shear modulus of elasticity is G ⫽ 80 GPa,
determine the maximum torque Tmax that can be applied to the
shaft.
d
T
T
L
Solution 3.3-8
NUMERICAL DATA
d ⫽ 104 mm
FIND MAX. TORQUE BASED ON ALLOWABLE RATE OF TWIST
␶a ⫽ 48 MPa
p
2a
b
180 rad
u⫽
3.5
m
Ip ⫽
p 4
d
32
f
u⫽
L
G ⫽ 80 GPa
Ip ⫽ 1.149 ⫻ 107 mm4
Tmax ⫽
GIpf
L
Tmax ⫽ GIp␪
Tmax ⫽ 9164 N # m
^ governs
;
FIND MAX. TORQUE BASED ON ALLOWABLE SHEAR STRESS
taIp
Tmax ⫽ 10,602 N # m
Tmax ⫽
d
2
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CHAPTER 3
Page 258
Torsion
Problem 3.3-9 Three identical circular disks A, B, and C are welded
P3
to the ends of three identical solid circular bars (see figure). The bars
lie in a common plane and the disks lie in planes perpendicular to the
axes of the bars. The bars are welded at their intersection D to form a
rigid connection. Each bar has diameter d1 ⫽ 0.5 in. and each disk has
diameter d2 ⫽ 3.0 in.
Forces P1, P2, and P3 act on disks A, B, and C, respectively, thus
subjecting the bars to torsion. If P1 ⫽ 28 lb, what is the maximum shear
stress ␶max in any of the three bars?
135∞
P1
P3
d1
A
D
135∞
P1
90∞
d2
P2
P2
Solution 3.3-9
B
Three circular bars
THE THREE TORQUES MUST BE IN EQUILIBRIUM
T3 is the largest torque
T3 ⫽ T1 12 ⫽ P1d2 12
MAXIMUM SHEAR STRESS (Eq. 3-12)
16T3
16P1d2 12
16T
tmax ⫽
⫽
⫽
3
3
pd
pd1
pd31
d1 ⫽ diameter of bars
⫽ 0.5 in.
tmax ⫽
d2 ⫽ diameter of disks
⫽ 3.0 in.
P1 ⫽ 28 lb
T1 ⫽ P1d2
T2 ⫽ P2d2
T3 ⫽ P3d2
C
16(28 lb)(3.0 in.)12
p(0.5 in.)3
⫽ 4840 psi
;
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SECTION 3.3
Problem 3.3-10 The steel axle of a large winch on an ocean liner is subjected to a
torque of 1.65 kN ⭈ m (see figure). What is the minimum required diameter dmin if the
allowable shear stress is 48 MPa and the allowable rate of twist is 0.75°/m? (Assume
that the shear modulus of elasticity is 80 GPa.)
Circular Bars and Tubes
T
259
d
T
Solution 3.3-10
NUMERICAL DATA
T ⫽ 1.65 kN # m
u a ⫽ 0.75 a
␶a ⫽ 48 MPa
G ⫽ 80 GPa
p
rad
b
180 m
32T
d ⫽
pGua
4
Ip ⫽
T
Gu
;
MIN. REQUIRED DIAMETER OF SHAFT BASED ON ALLOWABLE
MIN. REQUIRED DIAMETER OF SHAFT BASED ON
T
GIp
dmin ⫽ 63.3 mm
^ governs
SHEAR STRESS
ALLOWABLE RATE OF TWIST
u⫽
dmin ⫽ 0.063 m
p 4
T
d ⫽
32
Gu
32T
dmin ⫽ a
b
pGua
1
4
t⫽
Td
2Ip
t⫽
Td
p 4
2a
d b
32
1
16T 3
d
dmin ⫽ c
pta
dmin ⫽ 0.056 m
dmin ⫽ 55.9 mm
Problem 3.3-11 A hollow steel shaft used in a construction auger has
outer diameter d2 ⫽ 6.0 in. and inner diameter d1 ⫽ 4.5 in. (see figure).
The steel has shear modulus of elasticity G ⫽ 11.0 ⫻ 106 psi.
For an applied torque of 150 k-in., determine the following quantities:
(a) shear stress ␶2 at the outer surface of the shaft,
(b) shear stress ␶1 at the inner surface, and
(c) rate of twist ␪ (degrees per unit of length).
d2
Also, draw a diagram showing how the shear stresses vary in
magnitude along a radial line in the cross section.
d1
d2
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CHAPTER 3 Torsion
Solution 3.3-11 Construction auger
d2 ⫽ 6.0 in.
r2 ⫽ 3.0 in.
d1 ⫽ 4.5 in.
r1 ⫽ 2.25 in.
(c) RATE OF TWIST
u⫽
G ⫽ 11 ⫻ 106 psi
u ⫽ 157 * 10⫺6 rad/ in. ⫽ 0.00898°/ in.
T ⫽ 150 k-in.
IP ⫽
(150 k-in.)
T
⫽
GIP
(11 * 106 psi)(86.98 in.)4
p 4
(d ⫺ d14) ⫽ 86.98 in.4
32 2
;
(d) SHEAR STRESS DIAGRAM
(a) SHEAR STRESS AT OUTER SURFACE
t2 ⫽
(150 k-in.)(3.0 in.)
Tr2
⫽
IP
86.98 in.4
⫽ 5170 psi
;
(b) SHEAR STRESS AT INNER SURFACE
t1 ⫽
Tr1
r1
⫽
t ⫽ 3880 psi
IP
r2 2
;
Problem 3.3-12 Solve the preceding problem if the shaft has outer diameter d2 ⫽ 150 mm and inner diameter d1 ⫽ 100 mm.
Also, the steel has shear modulus of elasticity G ⫽ 75 GPa and the applied torque is 16 kN ⭈ m.
Solution 3.3-12 Construction auger
d2 ⫽ 150 mm
r2 ⫽ 75 mm
d1 ⫽ 100 mm
r1 ⫽ 50 mm
G ⫽ 75 GPa
T ⫽ 16 kN # m
p 4
IP ⫽
(d2 ⫺ d14) ⫽ 39.88 * 106 mm4
32
(a) SHEAR STRESS AT OUTER SURFACE
(16 kN # m)(75 mm)
Tr2
⫽
IP
39.88 * 106 mm4
⫽ 30.1 MPa
;
t2 ⫽
(b) SHEAR STRESS AT INNER SURFACE
t1 ⫽
Tr1
r1
⫽ t 2 ⫽ 20.1 MPa
IP
r2
;
(c) RATE OF TWIST
T
16 kN # m
u⫽
⫽
GIP
(75 GPa)(39.88 * 106 mm4)
␪ ⫽ 0.005349 rad/m ⫽ 0.306°/m
(d) SHEAR STRESS DIAGRAM
;
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SECTION 3.3
Circular Bars and Tubes
Problem 3.3-13 A vertical pole of solid circular cross section is twisted by
c
horizontal forces P ⫽ 1100 lb acting at the ends of a horizontal arm AB (see
figure). The distance from the outside of the pole to the line of action of each
force is c ⫽ 5.0 in.
If the allowable shear stress in the pole is 4500 psi, what is the minimum
required diameter dmin of the pole?
Solution 3.3-13
261
P
c
B
A
P
d
Vertical pole
tmax ⫽
P(2c + d)d
4
pd /16
⫽
16P(2c + d)
pd3
(␲␶max)d3 ⫺ (16P)d ⫺ 32Pc ⫽ 0
P ⫽ 1100 lb
SUBSTITUTE NUMERICAL VALUES:
c ⫽ 5.0 in.
UNITS: Pounds, Inches
␶allow ⫽ 4500 psi
(␲)(4500)d3 ⫺ (16)(1100)d ⫺ 32(1100)(5.0) ⫽ 0
Find dmin
or
d3 ⫺ 1.24495d ⫺ 12.4495 ⫽ 0
Solve numerically:
TORSION FORMULA
Tr
Td
⫽
tmax ⫽
IP
2IP
T ⫽ P12c + d2
d ⫽ 2.496 in.
dmin ⫽ 2.50 in.
IP ⫽
;
pd 4
32
Problem 3.3-14 Solve the preceding problem if the horizontal forces have magnitude P ⫽ 5.0 kN, the distance c ⫽ 125 mm,
and the allowable shear stress is 30 MPa.
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CHAPTER 3
Page 262
Torsion
Solution 3.3-14
Vertical pole
TORSION FORMULA
tmax ⫽
Tr
Td
⫽
IP
2IP
T ⫽ P(2c + d)
tmax ⫽
P(2c + d)d
4
pd /16
IP ⫽
⫽
pd 4
32
16P (2c + d)
pd 3
(␲␶max)d 3 ⫺ (16P)d ⫺ 32Pc ⫽ 0
SUBSTITUTE NUMERICAL VALUES:
P ⫽ 5.0 kN
UNITS: Newtons, Meters
c ⫽ 125 mm
(␲)(30 ⫻ 106)d3 ⫺ (16)(5000)d ⫺ 32(5000)(0.125) ⫽ 0
␶allow ⫽ 30 MPa
or
Find dmin
d3 ⫺ 848.826 ⫻ 10⫺6d ⫺ 212.207 ⫻ 10⫺6 ⫽ 0
d ⫽ 0.06438 m
Solve numerically:
dmin ⫽ 64.4 mm
Problem 3.3-15 A solid brass bar of diameter d ⫽ 1.25 in. is
subjected to torques T1, as shown in part (a) of the figure. The
allowable shear stress in the brass is 12 ksi.
(a) What is the maximum permissible value of the torques T1?
(b) If a hole of diameter 0.625 in. is drilled longitudinally through
the bar, as shown in part (b) of the figure, what is the maximum
permissible value of the torques T2?
(c) What is the percent decrease in torque and the percent decrease
in weight due to the hole?
T1
d
;
T1
(a)
d
T2
T2
(b)
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SECTION 3.3
263
Circular Bars and Tubes
Solution 3.3-15
␶a ⫽ 12 ksi
d2 ⫽ 1.25 in. d1 ⫽ 0.625 in.
p
1 d24 ⫺ d142
32
T2max ⫽
d2
2
d24 ⫺d12
1
T2max ⫽ tap
16
d2
ta
(a)
MAX. PERMISSIBILE VALUE OF TORQUE
T1max ⫽
taIp
T1 – SOLID BAR
p
ta d 4
32
T1max ⫽
d
2
d
2
1
T1max ⫽
tapd3
16
1
T1max ⫽
(12)p (1.25)3
16
;
T1max ⫽ 4.60 in.-k
(b) MAX. PERMISSIBILE VALUE OF TORQUE T2 –
HOLLOW BAR
T2max ⫽ 4.31 in.-k
(c) PERCENT
;
DECREASE IN TORQUE
IN WEIGHT DUE TO HOLE IN
&
PERCENT DECREASE
(b)
percent decrease in torque
T1max ⫺ T2max
(100) ⫽ 6.25%
T1max
;
percent decrease in weight (weight is proportional to
x-sec area)
A1 ⫽
d1
d2
p 2
d
4 2
A2 ⫽
p 2
1d ⫺ d122
4 2
A1 ⫺ A2
(100) ⫽ 25 %
A1
;
Problem 3.3-16 A hollow aluminum tube used in a roof structure has an outside
diameter d2 ⫽ 104 mm and an inside diameter d1 ⫽ 82 mm (see figure). The tube is
2.75 m long, and the aluminum has shear modulus G ⫽ 28 GPa.
(a) If the tube is twisted in pure torsion by torques acting at the ends, what is the
angle of twist (in degrees) when the maximum shear stress is 48 MPa?
(b) What diameter d is required for a solid shaft (see figure) to resist the same torque
with the same maximum stress?
(c) What is the ratio of the weight of the hollow tube to the weight of the solid shaft?
d1
d2
d
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CHAPTER 3 Torsion
Solution 3.3-16
Td22
NUMERICAL DATA
set tmax expression equal to
d2 ⫽ 104 mm
d1 ⫽ 82 mm
⫽
L ⫽ 2.75 ⫻ 103 mm
G ⫽ 28 GPa
I p ⫽ (p/32)(d 2 4 ⫺ d 14 )
32Td2
pa d24 ⫺ d14 b
d3 ⫽
Ip ⫽ 7.046 ⫻ 106 mm4
p
a d42 ⫺ d41 b
32
then solve for d
d24 ⫺ d14
d2
dreqd ⫽ a
1
d24 ⫺ d14 3
b dreqd⫽88.4 mm
d2
;
(c) RATIO OF WEIGHTS OF HOLLOW & SOLID SHAFTS
(a) FIND ANGLE OF TWIST ␶max ⫽ 48 MPa
f⫽
TL
GIp
f ⫽ (tmax)
f⫽ a
WEIGHT IS PROPORTIONAL TO CROSS SECTIONAL AREA
p 2
A d ⫺ d12 B
4 2
Ah
p
A s ⫽ d reqd 2
⫽ 0.524
4
As
Td2 2L
b
2Ip Gd2
Ah ⫽
2L
Gd2
So the weight of the tube is 52% of the solid shaft,
but they resist the same torque.
␾ ⫽ 0.091 radians
␾ ⫽ 5.19°
;
;
(b) REPLACE HOLLOW SHAFT WITH SOLID SHAFT - FIND
DIAMETER
d
2
p
T
tmax ⫽
32d 4
tmax ⫽
16T
d3p
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SECTION 3.3
265
Circular Bars and Tubes
Problem 3.3-17 A circular tube of inner radius r1 and outer radius r2 is
subjected to a torque produced by forces P ⫽ 900 lb (see figure). The
forces have their lines of action at a distance b ⫽ 5.5 in. from the outside
of the tube.
P
If the allowable shear stress in the tube is 6300 psi and the inner
radius r1 ⫽ 1.2 in., what is the minimum permissible outer radius r2?
P
P
r2
r1
P
b
Solution 3.3-17
2r2
b
Circular tube in torsion
SOLUTION OF EQUATION
UNITS: Pounds, Inches
Substitute numerical values:
6300 psi ⫽
P ⫽ 900 lb
b ⫽ 5.5 in.
␶allow ⫽ 6300 psi
r1 ⫽ 1.2 in.
4(900 lb)(5.5 in. + r2)(r2)
p[(r42) ⫺ (1.2 in.)4]
or
r42 ⫺ 2.07360
⫺ 0.181891 ⫽ 0
r2(r2 + 5.5)
or
r42 ⫺ 0.181891 r22 ⫺ 1.000402 r2 ⫺ 2.07360 ⫽ 0
Find minimum permissible radius r2
Solve numerically:
TORSION FORMULA
r2 ⫽ 1.3988 in.
T ⫽ 2P(b ⫹ r2)
IP ⫽
p 4
(r ⫺ r41)
2 2
2P(b + r2)r2
4P(b + r2)r2
Tr2
⫽
⫽
p 4
IP
p (r42 ⫺ r41)
(r2 ⫺ r41)
2
All terms in this equation are known except r2.
tmax ⫽
MINIMUM PERMISSIBLE RADIUS
r2 ⫽ 1.40 in.
;
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CHAPTER 3
Page 266
Torsion
Nonuniform Torsion
T1
Problem 3.4-1 A stepped shaft ABC consisting of two solid
d1
circular segments is subjected to torques T1 and T2 acting in
opposite directions, as shown in the figure. The larger segment
of the shaft has diameter d1 2.25 in. and length L1 30 in.;
the smaller segment has diameter d2 1.75 in. and length
L2 20 in. The material is steel with shear modulus
G 11 106 psi, and the torques are T1 20,000 lb-in.
and T2 8,000 lb-in.
Calculate the following quantities: (a) the maximum shear
stress max in the shaft, and (b) the angle of twist C (in degrees)
at end C.
Solution 3.4-1
d2
T2
B
A
C
L1
L2
Stepped shaft
SEGMENT BC
TBC T2 8,000 lb-in.
d1 2.25 in.
L1 30 in.
d2 1.75 in.
L2 20 in.
T2 8,000 lb-in.
` TABL1
G(Ip)AB
16(8,000 lb-in.)
p(1.75 in.)3
7602 psi
(8,000 lb-in.)(20 in.)
(11 * 106 psi)a
p
b(1.75 in.)4
32
16(12,000 lb-in.)
p(2.25 in.)3
C AB BC (0.013007 0.015797) rad
5365 psi
(12,000 lb-in.)(30 in.)
(11 * 106 psi)a
0.013007 rad
;
(b) ANGLE OF TWIST AT END C
TAB T2 T1 12,000 lb-in.
fAB TBCL2
G(Ip)BC
max 7600 psi
SEGMENT AB
pd31
fBC pd32
(a) MAXIMUM SHEAR STRESS
Segment BC has the maximum stress
T1 20,000 lb-in.
tAB `
16 TBC
0.015797 rad
G 11 10 psi
6
16 TAB
tBC p
b(2.25 in.)4
32
C 0.002790 rad 0.16°
;
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SECTION 3.4
Problem 3.4-2 A circular tube of outer diameter d3 70 mm and
inner diameter d2 60 mm is welded at the right-hand end to a fixed
plate and at the left-hand end to a rigid end plate (see figure). A solid
circular bar of diameter d1 40 mm is inside of, and concentric with,
the tube. The bar passes through a hole in the fixed plate and is
welded to the rigid end plate.
The bar is 1.0 m long and the tube is half as long as the bar. A
torque T 1000 N m acts at end A of the bar. Also, both the bar
and tube are made of an aluminum alloy with shear modulus of
elasticity G 27 GPa.
267
Nonuniform Torsion
Tube
Fixed
plate
End
plate
Bar
T
A
(a) Determine the maximum shear stresses in both the bar
and tube.
(b) Determine the angle of twist (in degrees) at end A of the bar.
Tube
Bar
d1
d2
d3
Solution 3.4-2
Bar and tube
TORQUE
T 1000 N m
(a) MAXIMUM SHEAR STRESSES
Bar: t bar 16T
79.6 MPa
;
pd31
T(d3/2)
32.3 MPa
Tube: t tube (Ip) tube
;
TUBE
d3 70 mm
Ltube 0.5 m
(Ip) tube (b) ANGLE OF TWIST AT END A
d2 60 mm
G 27 GPa
Bar: fbar p 4
(d d24)
32 3
Tube: ftube 1.0848 * 106 mm4
A 9.43°
(Ip) bar pd14
32
Lbar 1.0 m
TL tube
0.0171 rad
G(Ip) tube
A bar tube 0.1474 0.0171 0.1645 rad
BAR
d1 40 mm
TL bar
0.1474 rad
G(Ip) bar
G 27 GPa
251.3 * 103 mm4
;
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CHAPTER 3
Page 268
Torsion
Problem 3.4-3 A stepped shaft ABCD consisting of solid circular
segments is subjected to three torques, as shown in the figure. The
torques have magnitudes 12.5 k-in., 9.8 k-in., and 9.2 k-in. The
length of each segment is 25 in. and the diameters of the segments
are 3.5 in., 2.75 in., and 2.5 in. The material is steel with shear
modulus of elasticity G 11.6 103 ksi.
12.5 k-in.
3.5 in.
25 in.
D
C
B
A
(a) Calculate the maximum shear stress max in the shaft.
(b) Calculate the angle of twist D (in degrees) at end D.
9.8 k-in.
9.2 k-in.
2.75 in.
2.5 in.
25 in.
25 in.
Solution 3.4-3
NUMERICAL DATA (INCHES, KIPS)
TB 12.5 k-in.
TC 9.8k-in.
TD 9.2 k-in.
tBC Check BC:
1TC + TD2
p
d 4
32 BC
L 25 in.
dAB 3.5 in.
dBC 2.75 in.
dCD 2.5 in.
G 11.6 (103) ksi
dBC
2
BC 4.65 ksi
; controls
(b) ANGLE OF TWIST AT END D
(a) MAX. SHEAR STRESS IN SHAFT
T1 |RA|
torque reaction at A: RA (TB TC TD)
RA 31.5 in.-kip
|RA|
tAB dAB
2
p
d 4
32 AB
max 3.742 ksi
TD
Check CD:
tCD dCD
2
p
d 4
32 CD
CD 2.999 ksi
T2 TC TD
T3 TD
IP1 p
d 4
32 AB
p
IP3 d 4
32 CD
IP2 p
d 4
32 BC
TiLi
fD a
GIpi
fD T2
T3
L T1
a
+
+
b
G IP1
IP2
IP3
D 0.017 radians
fD 0.978 degrees
;
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SECTION 3.4
Problem 3.4-4 A solid circular bar ABC consists of two
segments, as shown in the figure. One segment has diameter
d1 56 mm and length L1 1.45 m; the other segment has
diameter d2 48 mm and length L2 1.2 m.
What is the allowable torque Tallow if the shear stress is not
to exceed 30 MPa and the angle of twist between the ends of the
bar is not to exceed 1.25°? (Assume G 80 GPa.)
269
Nonuniform Torsion
d1
d2
T
A
C
B
L1
T
L2
Solution 3.4-4
Tallow based on angle of twist
NUMERICAL DATA
d1 56 mm
d2 48 mm
L1 1450 mm
L2 1200 mm
f a 1.25 a
a 30 MPa
G 80 GPa
fmax p
b radians
180
Allowable torque
Tallow 16T
d32p
tapd23
Tallow 16
L1
J 32
a
p
d14 b
L2
+
a
Gf a
L1
a
Tallow based on shear stress
tmax T
G
p 4
d1 b
32
+
p 4
d b
32 2 K
L2
a
p 4
d2 b
32
Tallow 459 N # m
;
T1 =
T2 =
1000 lb-in. 500 lb-in.
T3 =
T4 =
800 lb-in. 500 lb-in.
governs
Tallow 651.441 N # m
Problem 3.4-5 A hollow tube ABCDE constructed of
monel metal is subjected to five torques acting in the directions
shown in the figure. The magnitudes of the torques are
T1 1000 lb-in., T2 T4 500 lb-in., and T3 T5 800 lb-in.
The tube has an outside diameter d2 1.0 in. The allowable
shear stress is 12,000 psi and the allowable rate of twist is
2.0°/ft.
Determine the maximum permissible inside diameter d1
of the tube.
A
B
C
D
d2 = 1.0 in.
T5 =
800 lb-in.
E
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CHAPTER 3
Page 270
Torsion
Solution 3.4-5
Hollow tube of monel metal
REQUIRED POLAR MOMENT OF INERTIA BASED UPON
ALLOWABLE SHEAR STRESS
tmax d2 1.0 in.
allow
allow 12,000 psi
Tmaxr
Ip
IP T max(d2/2)
0.05417 in.4
tallow
2°/ft 0.16667°/in.
REQUIRED
0.002909 rad/in.
ALLOWABLE ANGLE OF TWIST
From Table H-2, Appendix H: G 9500 ksi
TORQUES
u
Tmax
GIP
POLAR MOMENT OF INERTIA BASED UPON
IP Tmax
0.04704 in.4
Gu allow
SHEAR STRESS GOVERNS
Required IP 0.05417 in.4
IP T1 1000 lb-in.
T2 500 lb-in.
T4 500 lb-in.
T5 800 lb-in.
T3 800 lb-in.
INTERNAL TORQUES
p 4
(d d41)
32 2
d41 d43 32(0.05417 in.4)
32IP
(1.0 in.)4 p
p
0.4482 in.4
TAB T1 1000 lb-in.
d1 0.818 in.
TBC T1 T2 500 lb-in.
(Maximum permissible inside diameter)
;
TCD T1 T2 T3 1300 lb-in.
TDE T1 T2 T3 T4 800 lb-in.
Largest torque (absolute value only):
Tmax 1300 lb-in.
80 mm
Problem 3.4-6 A shaft of solid circular cross section consisting of two
segments is shown in the first part of the figure. The left-hand segment
has diameter 80 mm and length 1.2 m; the right-hand segment has
diameter 60 mm and length 0.9 m.
Shown in the second part of the figure is a hollow shaft made of the
same material and having the same length. The thickness t of the hollow
shaft is d/10, where d is the outer diameter. Both shafts are subjected to
the same torque.
If the hollow shaft is to have the same torsional stiffness as the solid
shaft, what should be its outer diameter d?
1.2 m
60 mm
0.9 m
d
2.1 m
d
t=—
10
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SECTION 3.4
Solution 3.4-6
Nonuniform Torsion
271
Solid and hollow shafts
SOLID SHAFT CONSISTING OF TWO SEGMENTS
TORSIONAL STIFFNESS
kT T
f
Torque T is the same for both shafts.
⬖ For equal stiffnesses, 1 2
98,741 m3 TLi
f1 g
GIPi
T(1.2 m)
p
Ga b(80 mm)4
32
T(0.9 m)
+
p
G a b (60 mm)4
32
32T
(29,297 m3 + 69,444 m3)
pG
d4 3.5569 m
d4
3.5569
36.023 * 106 m4
98,741
d 0.0775 m 77.5 mm
;
32T
(98,741 m3)
pG
HOLLOW SHAFT
d0 inner diameter 0.8d
f2 TL
GIp
T(2.1 m)
Ga
p
b[d4 (0.8d)4]
32
32T
2.1 m
32T 3.5569 m
a
b
a
b
4
pG 0.5904 d
pG
d4
UNITS: d meters
Problem 3.4-7 Four gears are attached to a circular shaft and transmit
the torques shown in the figure. The allowable shear stress in the shaft is
10,000 psi.
8,000 lb-in.
(a) What is the required diameter d of the shaft if it has a
solid cross section?
(b) What is the required outside diameter d if the shaft is
hollow with an inside diameter of 1.0 in.?
19,000 lb-in.
4,000 lb-in.
A
7,000 lb-in.
B
C
D
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CHAPTER 3
Page 272
Torsion
Solution 3.4-7
Shaft with four gears
(b) HOLLOW SHAFT
Inside diameter d0 1.0 in.
allow 10,000 psi
TBC 11,000 lb-in.
TAB 8000 lb-in.
TCD 7000 lb-in.
Tr
tmax Ip
10,000 psi (a) SOLID SHAFT
tmax d3 t allow 16T
pd3
d
Tmax a b
2
Ip
d
(11,000 lb-in.)a b
2
a
p
b[d4 (1.0 in.)4]
32
UNITS: d inches
16(11,000 lb-in.)
16Tmax
5.602 in.3
pt allow
p(10,000 psi)
Required d 1.78 in.
10,000 ;
56,023 d
d4 1
or
d 4 5.6023 d 1 0
Solving, d 1.832
Required d 1.83 in.
Problem 3.4-8 A tapered bar AB of solid circular cross section
is twisted by torques T (see figure). The diameter of the bar varies
linearly from dA at the left-hand end to dB at the right-hand end.
For what ratio dB/dA will the angle of twist of the tapered bar
be one-half the angle of twist of a prismatic bar of diameter dA?
(The prismatic bar is made of the same material, has the same length,
and is subjected to the same torque as the tapered bar.) Hint: Use the
results of Example 3-5.
;
B
A
T
T
L
dA
dB
Problems 3.4-8, 3.4-9 and 3.4-10
Solution 3.4-8
Tapered bar AB
ANGLE OF TWIST
TAPERED BAR (From Eq. 3-27)
b2 + b + 1
TL
f1 a
b
G(IP)A
3b 3
PRISMATIC BAR
TL
f2 G(IP)A
dB
b
dA
f1 1
f
2 2
or
3
3
b2 + b + 1
3b 3
2
2
dB
1.45
dA
1
2
2 20
SOLVE NUMERICALLY:
b
;
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SECTION 3.4 Nonuniform Torsion
273
Problem 3.4-9 A tapered bar AB of solid circular cross section is twisted
by torques T 36,000 lb-in. (see figure). The diameter of the bar varies
linearly from dA at the left-hand end to dB at the right-hand end. The bar
has length L 4.0 ft and is made of an aluminum alloy having shear
modulus of elasticity G 3.9 106 psi. The allowable shear stress in
the bar is 15,000 psi and the allowable angle of twist is 3.0°.
If the diameter at end B is 1.5 times the diameter at end A, what is
the minimum required diameter dA at end A? (Hint: Use the results of
Example 3–5).
Solution 3.4-9
Tapered bar
MINIMUM DIAMETER BASED
(From Eq. 3-27)
dB 1.5 dA
T 36,000 lb-in.
b dB/dA 1.5
L 4.0 ft 48 in.
b2 + b + 1
TL
TL
a
b (0.469136)
G(IP)A
G(IP)A
3b 3
(36,000 lb-in.)(48 in.)
(0.469136)
p
(3.9 * 106 psi)a bd4A
32
f
G 3.9 106 psi
allow 15,000 psi
allow 3.0°
0.0523599 rad
MINIMUM
DIAMETER BASED UPON ALLOWABLE SHEAR
STRESS
tmax 16T
pd3A
d3A UPON ALLOWABLE ANGLE OF
TWIST
16(36,000 lb-in.)
16 T
ptallow
p(15,000 psi)
12.2231 in.3
dA 2.30 in.
d4A 2.11728 in.4
d4A
2.11728 in.4
2.11728 in.4
0.0523599 rad
fallow
40.4370 in.4
dA 2.52 in.
ANGLE OF TWIST GOVERNS
Min. dA 2.52 in.
;
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Page 274
CHAPTER 3 Torsion
Problem 3.4-10 The bar shown in the figure is tapered linearly from
end A to end B and has a solid circular cross section. The diameter at the
smaller end of the bar is dA 25 mm and the length is L 300 mm.
The bar is made of steel with shear modulus of elasticity G 82 GPa.
If the torque T 180 N m and the allowable angle of twist is 0.3°,
what is the minimum allowable diameter dB at the larger end of the bar?
(Hint: Use the results of Example 3-5.)
Solution 3.4-10
Tapered bar
dA 25 mm
L 300 mm
G 82 GPa
T 180 N m
allow 0.3°
Find dB
DIAMETER BASED UPON ALLOWABLE ANGLE OF TWIST
p
(0.3°)a 180
rad
b
degrees
(180 N # m)(0.3 m)
p
(82 GPa)a b(25 mm)4
32
b2 + b + 1
0.304915 0.914745
3
3b 3
2
10
(From Eq. 3-27)
SOLVE NUMERICALLY:
dB
b
dA
f
b2 + b + 1
TL
p 4
a
b(IP)A d
G(IP)A
32 A
3b 3
1.94452
Min. dB dA 48.6 mm
;
a
b2 + b + 1
3b 3
b
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Page 275
SECTION 3.4
Nonuniform Torsion
Segment 1
Segment 2
275
Problem 3.4-11 The nonprismatic cantilever circular bar
shown has an internal cylindrical hole from 0 to x, so the net
polar moment of inertia of the cross section for segment 1 is
(7/8)Ip. Torque T is applied at x and torque T/2 is applied
at x L. Assume that G is constant.
(a)
(b)
(c)
(d)
(e)
Find reaction moment R1.
Find internal torsional moments Ti in segments 1 & 2.
Find x required to obtain twist at joint 3 of 3 TL/GIp
What is the rotation at joint 2, 2?
Draw the torsional moment (TMD: T(x), 0 x L) and
displacement (TDD: (x), 0 x L) diagrams.
x
7
—Ip
8
R1
Ip
T
1
2
T
—
2
3
x
L–x
T1
T2
TMD 0
φ2
TDD 0
0
φ3
0
Solution 3.4-11
(a) REACTION TORQUE R1
T
3
a Mx 0 R1 a T + 2 b R1 2 T ;
L
1
17
x + L
14
2
x
14 L
a b
17 2
(b) INTERNAL MOMENTS IN SEGMENTS 1 & 2
T1 R1
T1 1.5 T
T2 T
2
(c) FIND X REQUIRED TO OBTAIN TRWIST AT JT 3
TL
GIP
L
T 1x
7
Ga IP b
8
+
3
a Tbx
2
7
Ga IP b
8
3
a bx
2
7
a b
8
+
T2(L x)
GIP
T
a b(L x)
2
+
1
(L x)
2
GIP
;
(d) ROTATION AT JOINT 2 FOR X VALUE IN (C)
f2 TiLi
f3 a
GIPi
TL
GIP
7
L
17
x
f2 T1x
7
G a Ip b
8
12TL
17GIP
f2 3
7
a Tb a Lb
2
17
7
G a Ip b
8
;
(e) TMD & TDD – SEE PLOTS ABOVE
TMD is constant - T1 for 0 to x & T2 for x to L;
hence TDD is linear - zero at jt 1, 2 at jt 2 & 3
at jt 3
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CHAPTER 3
Page 276
Torsion
Problem 3.4-12 A uniformly tapered tube AB of hollow circular cross
section is shown in the figure. The tube has constant wall thickness t and
length L. The average diameters at the ends are dA and dB 2dA. The polar
moment of inertia may be represented by the approximate formula IP L d3t/4
(see Eq. 3-18).
Derive a formula for the angle of twist of the tube when it is subjected
to torques T acting at the ends.
B
A
T
T
L
t
t
dA
dB = 2dA
Solution 3.4-12
Tapered tube
t thickness (constant)
dA, dB average diameters at the ends
dB 2dA
Ip pd3t
(approximate formula)
4
ANGLE OF TWIST
Take the origin of coordinates at point O.
d(x) Lp(x) x
x
(d ) dA
2L B
L
p[d(x)]3t
ptd3A 3
x
4
4L3
For element of length dx:
df Tdx
GIP(x)
Tdx
ptd3A
G a 3 b x3
4TL3dx
pGtdA3 x3
4L
2L
f
LL
df 4TL3
2L
pGtd3A
LL x3
dx
3TL
2pGtd3A
;
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Page 277
SECTION 3.4
Problem 3.4-13 A uniformly tapered aluminum-alloy tube AB
of circular cross section and length L is shown in the figure. The
outside diameters at the ends are dA and dB 2dA. A hollow
section of length L/2 and constant thickness t dA/10 is cast
into the tube and extends from B halfway toward A.
(a) Find the angle of twist of the tube when it is subjected
to torques T acting at the ends. Use numerical values as
follows: dA 2.5 in., L 48 in., G 3.9 106 psi, and
T 40,000 in.-lb.
(b) Repeat (a) if the hollow section has constant diameter
dA. (See figure part b.)
L
dB
(a)
L
—
2
dA
A
from B to centerline
outer and inner diameters as function of x
0 … x …
L
2
d0(x) 2d A d0(x) dB a
xdA
L
di (x) (dB 2t)
di (x) dB dA
bx
L
[(2dA 2t) (dA 2t)]
x
L
1 9L + 5x
d
5 A
L
solid from centerline to A
L
… x … L
2
d0(x) 2dA L
x dA
L
L
T 32
1
1
2
f a b
dx + L 4 dx
4
4
G p P L0 d0 d i
L2 d0
Q
dA
B
T
L
(b)
PART (A) - CONSTANT THICKNESS
use x as integration variable measured from B toward A
B
T
dA
T
Solution 3.4-13
t constant
dB – 2t
L
—
2
A
T
277
Nonuniform Torsion
dB
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1:09 PM
Page 278
CHAPTER 3 Torsion
L
T 32
2
f a b≥
G p
L0
f 32
L
1
xdA 4
1
9L + 5x 4
a2dA b a
dA
b
L
5
L
dx +
L2
L
1
a 2dA xdA 4
b
L
T 125 3ln(2) + 2ln(7) ln(197)
125 2ln(19) + ln(181)
19
a
L
L
+
Lb
4
Gp
2
2
dA
dA 4
81dA 4
Simplifying: f 16TL
81GpdA4
a38 + 10125 ln a
71117
bb
70952
Use numerical properties as follows L 48 in.
a 0.049 radians
fa 2.79°
or
fa 3.868
G 3.9 106 psi
TL
GdA4
dA 2.5 in.
;
PART (B) - CONSTANT HOLE DIAMETER
0 … x …
d B dA
bx
L
d0( x) dB a
L
2
L
… x … L
2
d0(x) 2dA L
2
f
T 32
a b
G p
f
2
T 32
a b
G p
L0
1
4
P L0 d0 di
J
d0(x) 2 dA xdA
L
xdA
L
L
dx +
4
1
L d0 4
L
2
L
dx
Q
L
1
1
dx +
dx
1
xdA 4
xdA 4
L
2
4
K
a2dA b dA
a 2dA b
L
L
3
ln(5) + 2atana b
2
T 1
1 ln(3) + 2atan(2)
19
± L
fb 32
L
+
L≤
Gp 4
4
dA 4
dA 4
81dA 4
Simplifying, fb 3.057
TL
GdA4
Use numerical properties given above
b 0.039 radians
fa
fb
dx ¥
1.265
fb 2.21°
;
so tube (a) is more flexible than tube (b)
di (x) dA
t
dA
10
T 40000 in.-lb
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Page 279
SECTION 3.4
Problem 3.4-14 For the thin nonprismatic steel pipe of constant
thickness t and variable diameter d shown with applied torques at
joints 2 and 3, determine the following.
(a) Find reaction moment R1.
(b) Find an expression for twist rotation 3 at joint 3. Assume
that G is constant.
(c) Draw the torsional moment diagram (TMD: T(x),
0 x L).
2d
t
d
Nonuniform Torsion
t
d
T, f3
T/2
R1
2
1
279
L
—
2
x
3
L
—
2
T
T
—
2
0
TMD
Solution 3.4-14
(a) REACTION TORQUE R1
statics: T 0
T
R1 + T0
2
L
R1 T
2
2
2T
f3 Gpt L0
;
(b) ROTATION AT JOINT 3
p
3
Ga d12(x) t b
4
L
+
L
2
LL2
T
x 3
bd
L
dx
L
Gpd3t LL2
dx
L
2
2T
f3 Gpt L0
1
c2d a 1 x 3
bd
L
dx
2TL
+
T
2
L
2
L
0
0 … x …
L
… x … L
2
d23(x) d
f3 x
b
L
c2da 1 4T
+
d12( x) 2da1 1
dx
dx
p
Ga d23(x)3t b
4
use IP expression for thin walled tubes
f3 f3 Gpd3t
2TL
3TL
3
8Gp d t
19TL
8Gpd3t
+
Gpd3t
;
(c) TMD
TMD is piecewise constant: T(x) T/2 for
segment 1-2 & T(x) T for segment 2-3 (see plot
above)
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CHAPTER 3
Page 280
Torsion
Problem 3.4-15 A mountain-bike rider going uphill applies
Handlebar extension
d01, t01
torque T Fd (F 15 lb, d 4 in.) to the end of the handlebars
ABCD (by pulling on the handlebar extenders DE). Consider
the right half of the handlebar assembly only (assume the
bars are fixed at the fork at A). Segments AB and CD are
prismatic with lengths L1 2 in. and L3 8.5
in., and with outer diameters and thicknesses d01 1.25 in.,
t01 0.125 in., and d03 0.87 in., t03 0.115 in., respectively
as shown. Segment BC of length L2 1.2 in., however,
is tapered, and outer diameter and thickness vary linearly
between dimensions at B and C.
B
A
E
d03, t03
C
L1 L2
T = Fd
D
L3
d
Consider torsion effects only. Assume G 4000 ksi is
constant.
Derive an integral expression for the angle of twist D
of half of the handlebar tube when it is subjected to torque
T Fd acting at the end. Evaluate D for the given numerical
values.
45∞
Handlebar
extension
F
D
Handlebar
Solution 3.4-15
ASSUME THIN WALLED TUBES
Segments AB & CD
p
p
IP1 d01 3t01
IP3 d03 3t03
4
4
Segment BC
0
d02(x) d01 a1 d02(x) fD L2
x
x
b + d03 a b
L2
L2
d01L2 d01x + d03x
L2
t02(x) t01 a1 t02(x) x
x
x
b + t03 a b
L2
L2
t01L2 t01x + t03x
L2
L2
L3
1
Fd L1
+
dx +
p
G IP1
I
L0
P3
P
Q
d (x)3t02(x)
4 02
fD L2
L2 4
L1
4Fd
c 3 dx
Gp d01 t01 L0 (d01L2 d01x + d03x)3
* (t01L2 t01x + t03x)
+
L3
d03 3t03
d
;
NUMERICAL DATA
L1 2 in.
L2 1.2 in.
L3 8.5 in.
t03 0.115 in.
d01 1.25 in.
t01 0.125 in.
F 15 lb
d 4 in.
d03 0.87 in.
G 4 (106) psi
f D 0.142°
;
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SECTION 3.4
Nonuniform Torsion
281
Problem 3.4-16 A prismatic bar AB of length L and solid circular cross
section (diameter d) is loaded by a distributed torque of constant intensity
t per unit distance (see figure).
t
A
(a) Determine the maximum shear stress max in the bar.
(b) Determine the angle of twist between the ends of the bar.
B
L
Solution 3.4-16
Bar with distributed torque
(a) MAXIMUM SHEAR STRESS
Tmax tL
tmax 16Tmax
pd3
16tL
pd3
;
(b) ANGLE OF TWIST
T(x) tx
df t intensity of distributed torque
d diameter
IP T(x)dx
32 tx dx
GIP
pGd4
L
f
pd4
32
L0
df 32t
pGd4 L0
L
x dx 16tL2
pGd4
;
G shear modulus of elasticity
Problem 3.4-17 A prismatic bar AB of solid circular cross section
(diameter d) is loaded by a distributed torque (see figure). The intensity
of the torque, that is, the torque per unit distance, is denoted t(x) and
varies linearly from a maximum value tA at end A to zero at end B. Also,
the length of the bar is L and the shear modulus of elasticity of the
material is G.
(a) Determine the maximum shear stress max in the bar.
(b) Determine the angle of twist between the ends of the bar.
t(x)
A
L
B
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CHAPTER 3
Solution 3.4-17
Page 282
Torsion
Bar with linearly varying torque
(a) Maximum shear stress
tmax 16Tmax
pd
3
16TA
pd
3
8tAL
pd3
;
(b) ANGLE OF TWIST
T(x) torque at distance x from end B
T(x) t(x)x
tAx2
2
2L
IP pd4
32
T(x) dx
16tAx2 dx
GIP
pGLd4
L
L
16tA
16tA L2
2
df x
dx
;
f
pGLd4 L0
3pGd4
L0
df t(x) intensity of distributed torque
tA maximum intensity of torque
d diameter
G shear modulus
TA maximum torque
12 tAL
Problem 3.4-18 A nonprismatic bar ABC of solid circular
cross section is loaded by distributed torques (see figure).
The intensity of the torques, that is, the torque per unit
distance, is denoted t(x) and varies linearly from zero at A to a
maximum value T0/L at B. Segment BC has linearly distributed
torque of intensity t(x) T0/3L of opposite sign to that applied
along AB. Also, the polar moment of inertia of AB is twice that
of BC, and the shear modulus of elasticity of the material is G.
(a) Find reaction torque RA.
(b) Find internal torsional moments T(x) in segments AB
and BC.
(c) Find rotation C.
(d) Find the maximum shear stress tmax and its location
along the bar.
(e) Draw the torsional moment diagram (TMD: T(x),
0 x L).
T
—0
L
A
T0
—
6
Fc
IP
2Ip
RA
C
B
L
—
2
T0
—
3L
L
—
2
2°
2°
0
TMD
–T0
—
12
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SECTION 3.4 Nonuniform Torsion
283
Solution 3.4-18
(a) TORQUE REACTION RA
(d) MAXIMUM SHEAR STRESS ALONG BAR
T 0
STATICS:
RA +
1 T0 L
1 T0
L
a ba b a ba b 0
2 L
2
2 3L
2
RA +
1
T 0
6 0
RA T0
6
p
d 4
32 AB
p
For BC IP d 4
32 BC
For AB 2IP 1
;
dBC
1 4
a b dAB
2
(b) INTERNAL TORSIONAL MOMENTS IN AB & BC
T0
TAB (x) 6
TAB (x) a
TBC (x) T0 x
x
a b
L
L 2
P Q
2
T0
x2
2 T0 b
6
L
0 … x …
(L x) T0 (L x)
a b
L
3L
2
2
TBC (x) c a
x L 2 T0
b
d
L
3
L
… x … L
2
L
2
L
TAB(x)
TBC(x)
dx
dx +
L
GIP
L0 G(2IP)
L2
2
T0
x T0
L
2
2 6
3L
fC dx
G(2IP)
L0
L
+
c a
LL2
x L 2 T0
b
d
L
3
GIP
fC T0L
T0L
48GIP
72GIP
fC T0L
144GIP
;
L
2
;
tmax 8T0
3pdAB3
tmax
; controls
Just to right of B, T T0/12
T0 dBC
a
b
12 2
tmax p
d 4
32 BC
T0 0.841dAB
a
b
12
2
tmax p
(0.841dAB)4
32
;
(c) ROTATION AT C
fC At A, T T0/6
T0 dAB
6 2
p
d 4
32 AB
dx
tmax 2.243T0
pdAB 3
(e) TMD two 2nd degree curves: from T0/6 at A, to
T0/12 at B, to zero at C (with zero slopes at A & C
since slope on TMD is proportional to ordinate on
torsional loading) – see plot of T(x) above
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CHAPTER 3
Page 284
Torsion
Problem 3.4-19 A magnesium-alloy wire of diameter d 4 mm and
length L rotates inside a flexible tube in order to open or close
a switch from a remote location (see figure). A torque T is applied
manually (either clockwise or counterclockwise) at end B, thus
twisting the wire inside the tube. At the other end A, the rotation of
the wire operates a handle that opens or closes the switch.
A torque T0 0.2 N m is required to operate the switch.
The torsional stiffness of the tube, combined with friction between
the tube and the wire, induces a distributed torque of constant
intensity t 0.04 N m/m (torque per unit distance) acting along
the entire length of the wire.
T0 = torque
Flexible tube
B
d
A
t
(a) If the allowable shear stress in the wire is allow 30 MPa,
what is the longest permissible length Lmax of the wire?
(b) If the wire has length L 4.0 m and the shear modulus of
elasticity for the wire is G 15 GPa, what is the angle of twist (in degrees) between the ends of the wire?
Solution 3.4-19 Wire inside a flexible tube
(b) ANGLE OF TWIST d 4 mm
T0 0.2 N m
t 0.04 N m/m
(a) MAXIMUM LENGTH Lmax
allow 30 MPa
Equilibrium: T tL T0
16T
From Eq. (3-12): tmax pd3
tL + T0 L
L 4 m G 15 GPa
1 angle of twist due to distributed torque t
T
pGd4
(from problem 3.4-16)
2 angle of twist due to torque T0
pd3tmax
16
pd 3tmax
16
32 T0 L
T0 L
(from Eq. 3 -15)
GIP
pGd4
total angle of twist
1 2
1
(pd3tmax 16T0)
16t
1
Lmax (pd3tallow 16T0)
16t
16tL2
f
;
Substitute numerical values: Lmax 4.42 m
;
16L
pGd 4
(tL + 2T0)
;
Substitute numerical values:
2.971 rad 170°
;
T
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Page 285
SECTION 3.4
Problem 3.4-20 Two hollow tubes are connected by a pin
at B which is inserted into a hole drilled through both tubes at
B (see cross-section view at B). Tube BC fits snugly into tube
AB but neglect any friction on the interface. Tube inner and
outer diameters di (i 1, 2, 3) and pin diameter dp are labeled
in the figure. Torque T0 is applied at joint C. The shear
modulus of elasticity of the material is G.
Find expressions for the maximum torque T0,max which can
be applied at C for each of the following conditions.
B
d3
TA
d2
d1
d2
A
LA
(a) The shear in the connecting pin is less than some
allowable value (pin p,allow).
(b) The shear in tube AB or BC is less than some allowable
value (tube t,allow).
(c) What is the maximum rotation C for each of cases
(a) and (b) above?
Solution 3.4-20
Pin at B is in shear at interface between the two
tubes; force couple V # d2 T0
V
T0
d2
tpin tpin T0
d2
a
pdp 2
4
b
V
AS
tpin T0,max tp,allow a
ttubeAB pd2dp 2
b
p
ad3 4 d2 4 b
32
p
IPAC ad 4 d1 4 b
32 2
ttubeAB d3
T0 a b
2
IPAB
p
(d 4 d2 4)
32 3
16T0d3
p(d3 4 d2 4)
T0,max tt,allow c
p(d3 4 d2 4)
d
16d3
;
and based on tube BC:
;
(b) T0,max BASED ON ALLOWABLE SHEAR IN TUBES
AB & BC
IPAB d3
b
2
so based on tube AB:
4T0
p d2 d2p
4
ttubeAB T0 a
ttubeBC ttubeBC T0 a
d2
b
2
p
(d 4 d1 4)
32 2
16T0d2
p(d2 4 d1 4)
T0,max tt,allow
J
p(d2 4 d1 4)
16d2
K
T0, Fc
C
Pin dp LB
Cross-section at B
(a) T0,max BASED ON ALLOWABLE SHEAR IN PIN AT B
285
Nonuniform Torsion
;
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CHAPTER 3 Torsion
(c)
MAX. ROTATION AT
SHEAR IN PIN AT
C
B
BASED ON EITHER ALLOWABLE
OR ALLOWABLE SHEAR STRESS IN
TUBES
MAX. ROTATION BASED ON ALLOWABLE SHEAR IN
PIN AT B
fC T0,max
fCmax G
a
LA
+
IPAB
t p,allow a
LB
IPBC
p d2d2p
4
b
4
4
J 32 (d3 d2 )
p
fCmax tp, allow a
J
8d2dp 2
G
MAX.
+
LB
p
(d 4 d1 4) K
32 2
(d3 d2 )
4
+
LB
(d2 d1 4)
4
K
;
AB
fCmax tt, allow c
J
2(d3 4 d2 4)
d
Gd3
LA
(d3 d2 )
4
4
+
LB
K
;
K
;
(d2 d1 4)
4
ROTATION BASED ON ALLOWABLE SHEAR STRESS
IN TUBE
b
LA
4
ROTATION BASED ON ALLOWABLE SHEAR STRESS
IN TUBE
b
G
LA
MAX.
BC
fCmax tt, allow c
J
2(d2 4 d1 4)
d
Gd2
LA
(d3 d2 )
4
4
+
LB
(d2 d1 4)
4
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Page 287
SECTION 3.5
Pure Shear
287
Pure Shear
Problem 3.5-1 A hollow aluminum shaft (see figure) has outside
diameter d2 4.0 in. and inside diameter d1 2.0 in. When twisted
by torques T, the shaft has an angle of twist per unit distance equal
to 0.54°/ft. The shear modulus of elasticity of the aluminum is
G 4.0 106 psi.
d2
T
T
L
(a) Determine the maximum tensile stress max in the shaft.
(b) Determine the magnitude of the applied torques T.
d1 d2
Probs. 3.5-1, 3.5-2, and 3.5-3
Solution 3.5-1
d2 4.0 in.
Hollow aluminum shaft
d1 2.0 in. 0.54°/ft
(a) MAXIMUM TENSILE STRESS
G 4.0 106 psi
max occurs on a 45° plane and is equal to max.
MAXIMUM SHEAR STRESS
max max 6280 psi
max Gr (from Eq. 3-7a)
r d2 /2 2.0 in.
1 ft
prad
u (0.54°/ft)a
ba
b
12 in. 180 degree
785.40 106 rad/in.
max (4.0 106 psi)(2.0 in.)(785.40 106 rad/in.)
6283.2 psi
;
(b) APPLIED TORQUE
Use the torsion formula tmax T
tmaxIP
r
IP p
[(4.0 in.)4 (2.0 in.)4]
32
23.562 in.4
T
(6283.2 psi) (23.562 in.4)
2.0 in.
74,000 lb-in.
Tr
IP
;
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CHAPTER 3 Torsion
Problem 3.5-2 A hollow steel bar (G 80 GPa) is twisted by torques T (see figure). The twisting of the bar produces
a maximum shear strain max 640 106 rad. The bar has outside and inside diameters of 150 mm and 120 mm,
respectively.
(a) Determine the maximum tensile strain in the bar.
(b) Determine the maximum tensile stress in the bar.
(c) What is the magnitude of the applied torques T ?
Solution 3.5-2 Hollow steel bar
G 80 GPa max 640 106 rad
d2 150 mm
IP max Gmax (80 GPa)(640 106)
d1 120 mm
51.2 MPa
p 4
(d d 41)
32 2
max max 51.2 MPa
p
[(150 mm)4 (120 mm)4]
32
Torsion formula: tmax (a) MAXIMUM TENSILE STRAIN
gmax
320 * 106
2
;
(c) APPLIED TORQUES
29.343 * 106 mm4
âmax (b) MAXIMUM TENSILE STRESS
T
;
Td2
Tr
IP
2IP
2(29.343 * 106 mm4)(51.2 MPa)
2IPtmax
d2
150 mm
20,030 N # m
20.0 kN # m
;
Problem 3.5-3 A tubular bar with outside diameter d2 4.0 in. is twisted by torques T 70.0 k-in. (see figure). Under the
action of these torques, the maximum tensile stress in the bar is found to be 6400 psi.
(a) Determine the inside diameter d1 of the bar.
(b) If the bar has length L 48.0 in. and is made of aluminum with shear modulus G 4.0 106 psi, what is the angle of
twist (in degrees) between the ends of the bar?
(c) Determine the maximum shear strain max (in radians)?
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SECTION 3.5
Solution 3.5-3
d2 4.0 in.
T 70.0 k-in. 70,000 lb-in.
f
Torsion formula: tmax TL
GIp
From torsion formula, T (a) INSIDE DIAMETER d1
Td2
Tr
IP
2IP
‹ f
(70.0 k-in.)(4.0 in.)
Td2
2tmax
2(6400 psi)
21.875 in.4
Also, Ip 289
Tubular bar
max 6400 psi max max 6400 psi
IP Pure Shear
2(48 in.)(6400 psi)
(4.0 * 106 psi)(4.0 in.)
0.03840 rad
;
(c) MAXIMUM SHEAR STRAIN
gmax Equate formulas:
p
[256 in.4 d14] 21.875 in.4
32
Solve for d1: d1 2.40 in.
2IPtmax
2Ltmax
L
a
b d2
GIP
Gd2
f 2.20°
p 4
p
(d d 14) [(4.0 in.)4 d14]
32 2
32
2IP tmax
d2
6400 psi
tmax
G
4.0 * 106 psi
1600 * 106 rad
;
;
(b) ANGLE OF TWIST
L 48 in.
G 4.0 106 psi
Problem 3.5-4 A solid circular bar of diameter d 50 mm
(see figure) is twisted in a testing machine until the applied
torque reaches the value T 500 N m. At this value of torque,
a strain gage oriented at 45° to the axis of the bar gives a reading
P 339 106.
What is the shear modulus G of the material?
d = 50 mm
Strain gage
T
45°
T = 500 N·m
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CHAPTER 3 Torsion
Solution 3.5-4 Bar in a testing machine
Strain gage at 45°:
max
SHEAR STRESS (FROM EQ. 3-12)
6
339 10
tmax d 50 mm
G
SHEAR STRAIN (FROM EQ. 3-33)
max
pd
3
16(500 N # m)
p(0.050 m)3
20.372 MPa
SHEAR MODULUS
T 500 N m
max 2
16T
6
678 10
tmax
20.372 MPa
30.0 GPa
gmax
678 * 106
;
Problem 3.5-5 A steel tube (G 11.5 106 psi) has an outer diameter d2 2.0 in. and an inner diameter d1 1.5 in.
When twisted by a torque T, the tube develops a maximum normal strain of 170 106.
What is the magnitude of the applied torque T ?
Solution 3.5-5
Steel tube
G 11.5 106 psi
max
d2 2.0 in.
d1 1.5 in.
170 106
IP p 2
p
1d d142 [(2.0 in.)4 (1.5 in.)4]
32 2
32
1.07379 in.
Equate expressions:
Td2
Ggmax
2IP
SOLVE FOR TORQUE
4
T
SHEAR STRAIN (FROM EQ. 3-33)
max 2
max
340 106
SHEAR STRESS (FROM TORSION FORMULA)
Td2
Tr
tmax IP
2IP
Also, max Gmax
2GIPgmax
d2
2(11.5 * 106 psi)(1.07379 in.4)(340 * 106)
2.0 in.
4200 lb-in.
;
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Page 291
SECTION 3.5
Pure Shear
291
Problem 3.5-6 A solid circular bar of steel (G 78 GPa) transmits a torque T 360 N m. The allowable stresses
in tension, compression, and shear are 90 MPa, 70 MPa, and 40 MPa, respectively. Also, the allowable tensile strain is
220 106.
Determine the minimum required diameter d of the bar.
Solution 3.5-6
Solid circular bar of steel
T 360 N m
G 78 GPa
DIAMETER BASED UPON ALLOWABLE TENSILE STRAIN
ALLOWABLE STRESSES
Tension: 90 MPa Compression: 70 MPa
Shear: 40 MPa
Allowable tensile strain: max 220 106
max 2
tmax DIAMETER BASED UPON ALLOWABLE STRESS
The maximum tensile, compressive, and shear stresses
in a bar in pure torsion are numerically equal.
Therefore, the lowest allowable stress (shear stress)
governs.
allow 40 MPa
tmax 16T
3
pd
d3 max;
16T
3
pd
max Gmax 2G
d3 max
16T
16T
ptmax
2pGâmax
16(360 N # m)
2p(78 GPa)(220 * 106)
53.423 * 106 m3
d 0.0377 m 37.7 mm
TENSILE STRAIN GOVERNS
d3 16(360 N # m)
16T
pt allow
p(40 MPa)
dmin 37.7 mm
;
d 45.837 106 m3
3
d 0.0358 m 35.8 mm
Problem 3.5-7 The normal strain in the 45° direction on the
surface of a circular tube (see figure) is 880 106 when the
torque T 750 lb-in. The tube is made of copper alloy with
G 6.2 106 psi.
Strain gage
T
45°
If the outside diameter d2 of the tube is 0.8 in., what is the
inside diameter d1?
Solution 3.5-7
d2 0.80 in.
Circular tube with strain gage
T 750 lb-in. G 6.2 106 psi
Strain gage at 45°:
max
880 106
T = 750 lb-in.
d 2 = 0.8 in.
MAXIMUM SHEAR STRAIN
max 2
max
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CHAPTER 3 Torsion
MAXIMUM SHEAR STRESS
INSIDE DIAMETER
tmax Ggmax 2Gâmax
Substitute numerical values:
tmax IP T(d2/2)
IP
IP Td2
Td2
2tmax
4Gâmax
8Td2
pGâmax
d14 d24 8(750 lb-in.) (0.80 in.)
p (6.2 * 106 psi) (880 * 10 6)
0.4096 in.4 0.2800 in.4 0.12956 in.4
Td2
p 4
(d d14) 32 2
4Gâmax
d24 d14 d42 (0.8 in.)4 d1 0.60 in.
;
8Td2
pGâmax
Problem 3.5-8 An aluminium tube has inside diameter d1 50 mm, shear modulus of elasticity G 27 GPa, and torque
T 4.0 kN m. The allowable shear stress in the aluminum is 50 MPa and the allowable normal strain is 900 106.
Determine the required outside diameter d2.
Solution 3.5-8
d1 50 mm
Aluminum tube
G 27 GPa
T 4.0 kN m allow 50 MPa
NORMAL STRAIN GOVERNS
allow
900 106
allow 48.60 MPa
Determine the required diameter d2.
REQUIRED DIAMETER
ALLOWABLE SHEAR STRESS
t
(allow)1 50 MPa
Tr
IP
48.6 MPa ALLOWABLE SHEAR STRESS BASED ON NORMAL STRAIN
âmax Rearrange and simplify:
g
t
2
2G
(allow)2 2G
allow
(4000 N # m)(d2/2)
p 4
[d2 (0.050 m)4]
32
t 2Gâmax
2(27 GPa)(900 106)
48.6 MPa
d42 (419.174 * 106)d2 6.25 * 106 0
Solve numerically:
d2 0.07927 m
d2 79.3 mm
;
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SECTION 3.5
293
Pure Shear
Problem 3.5-9 A solid steel bar (G 11.8 106 psi) of
diameter d 2.0 in. is subjected to torques T 8.0 k-in.
acting in the directions shown in the figure.
(a) Determine the maximum shear, tensile, and compressive
stresses in the bar and show these stresses on sketches of
properly oriented stress elements.
(b) Determine the corresponding maximum strains (shear, tensile, and compressive) in the bar and show these
strains on sketches of the deformed elements.
Solution 3.5-9
Solid steel bar
T 8.0 k-in.
(b) MAXIMUM STRAINS
G 11.8 10 psi
6
gmax (a) MAXIMUM STRESSES
tmax T = 8.0 k-in.
d = 2.0 in.
T
16T
3
pd
432 * 106 rad
16(8000 lb-in.)
5093 psi
3
p(2.0 in.)
âmax ;
t 5090 psi c 5090 psi
gmax
5093 psi
G
11.8 * 106 psi
;
t
Problem 3.5-10 A solid aluminum bar (G 27 GPa) of
diameter d 40 mm is subjected to torques T 300 N m
acting in the directions shown in the figure.
(a) Determine the maximum shear, tensile, and compressive
stresses in the bar and show these stresses on sketches of
properly oriented stress elements.
(b) Determine the corresponding maximum strains (shear,
tensile, and compressive) in the bar and show these
strains on sketches of the deformed elements.
gmax
216 * 106
2
216 106
d = 40 mm
T
;
c
216 106
;
T = 300 N·m
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CHAPTER 3
Page 294
Torsion
Solution 3.5-10
Solid aluminum bar
(b) MAXIMUM STRAINS
(a) MAXIMUM STRESSES
tmax 16T
3
pd
16(300 N #
m)
gmax 3
p(0.040 m)
23.87 MPa
tmax
23.87 MPa
G
27 GPa
884 * 106 rad
;
t 23.9 MPa c 23.9 MPa
;
âmax t
;
gmax
442 * 106
2
442 106
c
442 106
;
Transmission of Power
Problem 3.7-1 A generator shaft in a small hydroelectric plant turns
120 rpm
at 120 rpm and delivers 50 hp (see figure).
d
(a) If the diameter of the shaft is d 3.0 in., what is the maximum
shear stress max in the shaft?
(b) If the shear stress is limited to 4000 psi, what is the minimum
permissible diameter dmin of the shaft?
Solution 3.7-1
n 120 rpm
TORQUE
H
Generator shaft
H 50 hp
d diameter
2pnT
H hp n rpm T 1b-ft
33,000
33,000 H
(33,000)(50 hp)
T
2pn
2p(120 rpm)
2188 1b-ft 26,260 1b-in.
(a) MAXIMUM SHEAR STRESS max
d 3.0 in.
50 hp
tmax 16T
3
pd
16(26,260 1b-in.)
tmax 4950 psi
p (3.0 in.)3
;
(b) MINIMUM DIAMETER dmin
allow 4000 psi
d3 16(26,260 1b-in.)
16T
33.44 in.3
ptallow
p (4000 psi)
dmin 3.22 in.
;
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Page 295
SECTION 3.7
Transmission of Power
295
Problem 3.7-2 A motor drives a shaft at 12 Hz and delivers 20 kW
of power (see figure).
12 Hz
d
(a) If the shaft has a diameter of 30 mm, what is the maximum shear
stress max in the shaft?
(b) If the maximum allowable shear stress is 40 MPa, what is the
minimum permissible diameter dmin of the shaft?
Solution 3.7-2
f 12 Hz
20 kW
Motor-driven shaft
P 20 kW 20,000 N m/s
16T
tmax pd3
TORQUE
P 2 fT
P watts
T
20,000 W
P
265.3 N # m
2pf
2p(12 Hz)
(a) MAXIMUM SHEAR STRESS max
16(265.3 N # m)
p(0.030 m)3
50.0 MPa
f Hz s1
T Newton meters
;
(b) MINIMUM DIAMETER dmin
allow 40 MPa
d3 16(265.3 N # m)
16T
pt allow
p(40 MPa)
33.78 106 m3
d 30 mm
dmin 0.0323 m 32.3 mm
Problem 3.7-3 The propeller shaft of a large ship has outside
;
100 rpm
18 in.
diameter 18 in. and inside diameter 12 in., as shown in the figure.
The shaft is rated for a maximum shear stress of 4500 psi.
(a) If the shaft is turning at 100 rpm, what is the maximum
horsepower that can be transmitted without exceeding the
allowable stress?
12 in.
18 in.
(b) If the rotational speed of the shaft is doubled but the power
requirements remain unchanged, what happens to the shear
stress in the shaft?
Solution 3.7-3
Hollow propeller shaft
d2 18 in. d1 12 in. allow 4500 psi
p 4
(d d42) 8270.2 in.4
IP 32 2
TORQUE
T(d2/2)
tmax IP
T
2t allowIP
T
d2
2(4500 psi)(8270.2 in.4)
18 in.
4.1351 * 106 1b-in.
344,590 1b-ft.
(a) HORSEPOWER
n 100 rpm
n rpm
H
H
2pnT
33,000
T lb-ft H hp
2p(100 rpm)(344,590 lb-ft)
33,000
6560 hp
;
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Torsion
(b) ROTATIONAL SPEED IS DOUBLED
H
2pnT
33,000
If n is doubled but H remains the same, then T is
halved.
If T is halved, so is the maximum shear stress.
⬖ Shear stress is halved
;
Problem 3.7-4 The drive shaft for a truck (outer diameter
60 mm and inner diameter 40 mm) is running at 2500 rpm
(see figure).
2500 rpm
60 mm
(a) If the shaft transmits 150 kW, what is the maximum shear
stress in the shaft?
(b) If the allowable shear stress is 30 MPa, what is the
maximum power that can be transmitted?
40 mm
60 mm
Solution 3.7-4
d2 60 mm
IP Drive shaft for a truck
d1 40 mm
n 2500 rpm
p 4
(d d14) 1.0210 * 106 m4
32 2
(a) MAXIMUM SHEAR STRESS max
P power (watts) P 150 kW 150,000 W
T torque (newton meters) n rpm
P
2pnT
60
T
60(150,000 W)
572.96 N # m
2p(2500 rpm)
T
60P
2pn
tmax (572.96 N # m)(0.060 m)
Td2
2 IP
2(1.0210 * 106 m4)
16.835 MPa
tmax 16.8 MPa
;
(b) MAXIMUM POWER Pmax
allow 30 MPa
Pmax P
tallow
30 MPa
(150 kW) a
b
tmax
16.835 MPa
267 kW
;
Problem 3.7-5 A hollow circular shaft for use in a pumping station is being designed with an inside diameter equal to
0.75 times the outside diameter. The shaft must transmit 400 hp at 400 rpm without exceeding the allowable shear stress
of 6000 psi.
Determine the minimum required outside diameter d.
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Page 297
SECTION 3.7
Solution 3.7-5
H hp
d0 inside diameter
T
0.75 d
n 400 rpm
allow 6000 psi
IP n rpm
T lb-ft
(33,000)(400 hp)
33,000 H
2pn
2p(400 rpm)
5252.1 lb-ft 63,025 lb-in.
MINIMUM OUTSIDE DIAMETER
p 4
[d (0.75 d)4] 0.067112 d 4
32
TORQUE
H
297
Hollow shaft
d outside diameter
H 400 hp
Transmission of Power
tmax Td
2IP
IP 0.067112 d 4 2pnT
33,000
Td
Td
2tmax
2t allow
(63,025 lb-in.)(d)
2(6000 psi)
d3 78.259 in.3
dmin 4.28 in.
;
Problem 3.7-6 A tubular shaft being designed for use on a construction site must transmit 120 kW at 1.75 Hz. The inside
diameter of the shaft is to be one-half of the outside diameter.
If the allowable shear stress in the shaft is 45 MPa, what is the minimum required outside diameter d?
Solution 3.7-6
Tubular shaft
d outside diameter
T
d0 inside diameter
MINIMUM OUTSIDE DIAMETER
0.5 d
P 120 kW 120,000 W
f 1.75 Hz
allow 45 MPa
IP p 4
[d (0.5 d)4] 0.092039 d 4
32
tmax Td
2IP
IP 0.092039 d4 Td
Td
2tmax
2t allow
(10,913.5 N # m)(d)
2(45 MPa)
d3 0.0013175 m3
TORQUE
P 2 fT
120,000 W
P
10,913.5 N # m
2pf
2p(1.75 Hz)
P watts
f Hz
dmin 110 mm
d 0.1096 m
;
T newton meters
Problem 3.7-7 A propeller shaft of solid circular cross section
and diameter d is spliced by a collar of the same material
(see figure). The collar is securely bonded to both parts
of the shaft.
What should be the minimum outer diameter d1 of the collar in
order that the splice can transmit the same power as the solid shaft?
d1
d
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CHAPTER 3 Torsion
Solution 3.7-7 Splice in a propeller shaft
EQUATE TORQUES
SOLID SHAFT
tmax 16 T1
pd3
T1 pd3tmax
16
For the same power, the torques must be the same.
For the same material, both parts can be stressed to the
same maximum stress.
HOLLOW COLLAR
IP T2(d1/2)
T2r
p 4
(d1 d 4) tmax 32
IP
IP
2tmaxIP
2tmax p
a
b(d14 d 4)
T2 d1
d1 32
ptmax 4
(d1 d 4)
16 d1
‹ T1 T2
or a
pd3tmax
ptmax 4
(d d 4)
16
16d1 1
d1 4
d1
b 10
d
d
(Eq. 1)
MINIMUM OUTER DIAMETER
Solve Eq. (1) numerically:
Min. d1 1.221 d
;
Problem 3.7-8 What is the maximum power that can be delivered by a hollow propeller shaft (outside diameter 50 mm,
inside diameter 40 mm, and shear modulus of elasticity 80 GPa) turning at 600 rpm if the allowable shear stress is 100 MPa
and the allowable rate of twist is 3.0°/m?
Solution 3.7-8 Hollow propeller shaft
d2 50 mm
d1 40 mm
G 80 GPa
n 600 rpm
allow 100 MPa allow 3.0°/m
IP p 4
(d d41) 362.3 * 109 m4
32 2
BASED UPON ALLOWABLE SHEAR STRESS
tmax T1(d2/2)
IP
T1 2t allowIP
d2
2(100 MPa)(362.3 * 109 m4)
T1 0.050 m
1449 N # m
BASED UPON ALLOWABLE RATE OF TWIST
T2
T2 GIPallow
u
GIP
T (80 GPa) (362.3 * 10
2
* a
9 4
m )(3.0°/m)
p
rad /degreeb
180
T2 1517 N m
SHEAR STRESS GOVERNS
Tallow T1 1449 N m
MAXIMUM POWER
2p(600 rpm)(1449 N # m)
2pnT
P
60
60
P 91,047 W
Pmax 91.0 kW
;
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SECTION 3.7
Transmission of Power
299
Problem 3.7-9 A motor delivers 275 hp at 1000 rpm to the end of a shaft (see figure). The gears at B and C take out 125
and 150 hp, respectively.
Determine the required diameter d of the shaft if the allowable shear stress is 7500 psi and the angle of twist between the
motor and gear C is limited to 1.5°. (Assume G 11.5 106 psi, L1 6 ft, and L2 4 ft.)
Motor
C
A
d
B
L2
L1
PROBS. 3.7-9 and 3.7-10
Solution 3.7-9
Motor-driven shaft
FREE-BODY DIAGRAM
L1 6 ft
L2 4 ft
TA 17,332 lb-in.
d diameter
TC 9454 lb-in.
n 1000 rpm
d diameter
allow 7500 psi
(
AC)allow
1.5° 0.02618 rad
G 11.5 106 psi
TORQUES ACTING ON THE SHAFT
2pnT
H
33,000
T
TB 7878 lb-in.
INTERNAL TORQUES
TAB 17,332 lb-in.
TBC 9454 lb-in.
H hp n rpm T lb-ft
DIAMETER BASED UPON ALLOWABLE SHEAR STRESS
The larger torque occurs in segment AB
33,000 H
2pn
At point A: TA 33,000(275 hp)
2p(1000 rpm)
1444 lb-ft
17,332 lb-in.
At point B: TB At point C: TC tmax 16TAB
d3 3
pd
16TAB
pt allow
16(17,332 lb-in.)
11.77 in.3
p(7500 psi)
d 2.27 in.
DIAMETER BASED UPON ALLOWABLE ANGLE OF TWIST
125
275
TA 7878 lb-in.
150
275
TA 9454 lb-in.
IP pd4
32
f
TL
32TL
GIP
pGd4
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CHAPTER 3 Torsion
Segment AB:
fAB fAB fBC 32TAB LAB
pGd 4
32(17,330 lb in.)(6 ft)(12 in./ft)
p(11.5 * 106 psi)d 4
1.1052
d
AC)allow
d4
0.02618 rad
0.02618 1.5070
d4
d 2.75 in.
32 TBCLBC
pGd
(
1.5070
and
d 2.75 in.
Angle of twist governs
Segment BC:
fBC d4
From A to C: fAC fAB + fBC ⬖
4
0.4018
;
4
32(9450 lb-in.)(4 ft)(12 in./ft)
p(11.5 * 106 psi)d 4
Problem 3.7-10 The shaft ABC shown in the figure is driven by a motor that delivers 300 kW at a rotational speed of
32 Hz. The gears at B and C take out 120 and 180 kW, respectively. The lengths of the two parts of the shaft are L1 1.5 m
and L2 0.9 m.
Determine the required diameter d of the shaft if the allowable shear stress is 50 MPa, the allowable angle of twist
between points A and C is 4.0°, and G 75 GPa.
Solution 3.7-10 Motor-driven shaft
L1 1.5 m
L2 0.9 m
d diameter
f 32 Hz
At point A: TA 300,000 W
1492 N # m
2p(32 Hz)
At point B: TB 120
T 596.8 N # m
300 A
At point C: TC 180
T 895.3 N # m
300 A
FREE-BODY DIAGRAM
allow 50 MPa
G 75 GPa
(
AC)allow
4° 0.06981 rad
TORQUES ACTING ON THE SHAFT
P 2 fT P watts
T newton meters
T
P
2pf
f Hz
TA 1492 N m
TB 596.8 N m
TC 895.3 N m
d diameter
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Page 301
SECTION 3.7 Transmission of Power
INTERNAL TORQUES
Segment BC:
TAB 1492 N m
fBC TBC 895.3 N m
DIAMETER BASED UPON ALLOWABLE SHEAR STRESS
fBC 32 TBCLBC
pGd 4
tmax 16(1492 N # m)
16 TAB
d pt allow
p(50 MPa)
3
pd 3
d3 0.0001520 m3
d 0.0534 m 53.4 mm
DIAMETER BASED UPON ALLOWABLE ANGLE OF TWIST
IP pd 4
32
f
TL
32TL
GIP
pGd 4
fAB From A to C: fAC fAB + fBC (
AC)allow
0.06981 rad
⬖ 0.06981 0.1094 * 106
d4
49.3mm
SHEAR STRESS GOVERNS
32 TABLAB
pGd
4
0.3039 * 10
d4
32(1492 N #
6
m)(1.5 m)
p(75 GPa)d
4
p(75 GPa)d 4
d4
and d 0.04933 m
Segment AB:
fAB 32(895.3 N # m)(0.9 m)
0.1094 * 106
The larger torque occurs in segment AB
16 TAB
d 53.4mm
;
0.4133 * 106
d4
301
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CHAPTER 3
Page 302
Torsion
Statically Indeterminate Torsional Members
Problem 3.8-1 A solid circular bar ABCD with fixed supports is
acted upon by torques T0 and 2T0 at the locations shown in the figure.
Obtain a formula for the maximum angle of twist ␾max of the bar.
(Hint: Use Eqs. 3-46a and b of Example 3-9 to obtain the reactive
torques.)
TA
A
T0
2T0
B
C
3L
—
10
3L
—
10
D
TD
D
TD
4L
—
10
L
Solution 3.8-1
Circular bar with fixed ends
ANGLE OF TWIST AT SECTION B
From Eqs. (3-46a and b):
TA(3L/10)
9T0 L
⫽
GIP
20GIP
TA ⫽
T0 LB
L
fB ⫽ fAB ⫽
TB ⫽
T0 LA
L
ANGLE OF TWIST AT SECTION C
APPLY THE ABOVE FORMULAS TO THE GIVEN BAR:
TA ⫽ T0 a
15T0
7
4
b + 2T0 a b ⫽
10
10
10
15T0
3
6
TD ⫽ T0 a b + 2T0 a b ⫽
10
10
10
fC ⫽ fCD ⫽
TD(4L/10)
3T0 L
⫽
GIP
5GIP
MAXIMUM ANGLE OF TWIST
fmax ⫽ fC ⫽
3T0 L
5GIP
Problem 3.8-2 A solid circular bar ABCD with fixed supports at
ends A and D is acted upon by two equal and oppositely directed
torques T0, as shown in the figure. The torques are applied at points
B and C, each of which is located at distance x from one end of the
bar. (The distance x may vary from zero to L/2.)
(a) For what distance x will the angle of twist at points B and C be
a maximum?
(b) What is the corresponding angle of twist ␾max?
(Hint: Use Eqs. 3-46a and b of Example 3-9 to obtain the
reactive torques.)
TA
A
;
T0
T0
B
C
x
x
L
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SECTION 3.8
Solution 3.8-2
303
Statically Indeterminate Torsional Members
Circular bar with fixed ends
(a) ANGLE OF TWIST AT SECTIONS B AND C
φB
φ max
0
From Eqs. (3-46a and b):
fB ⫽ fAB ⫽
L/A
T0 LB
L
dfB
T0
⫽
(L ⫺ 4x)
dx
GIPL
TB ⫽
T0 LA
L
dfB
⫽ 0; L ⫺ 4x ⫽ 0
dx
or x ⫽
L
4
x
TAx
T0
⫽
(L ⫺ 2x)(x)
GIP
GIPL
TA ⫽
APPLY THE ABOVE FORMULAS TO THE GIVEN BAR:
L/2
;
(b) MAXIMUM ANGLE OF TWIST
fmax ⫽ (fB)max ⫽ (fB)x⫽ L4 ⫽
TA ⫽
T0L
8GIP
;
T0(L ⫺ x)
T0x
T0
⫺
⫽ (L ⫺ 2x) TD ⫽ TA
L
L
L
Problem 3.8-3 A solid circular shaft AB of diameter d is fixed against rotation at
both ends (see figure). A circular disk is attached to the shaft at the location shown.
What is the largest permissible angle of rotation ␾max of the disk if the allowable
shear stress in the shaft is ␶allow? (Assume that a ⬎ b. Also, use Eqs. 3-46a and b of
Example 3-9 to obtain the reactive torques.)
Disk
A
d
B
a
Solution 3.8-3
b
Shaft fixed at both ends
Assume that a torque T0 acts at the disk.
The reactive torques can be obtained from Eqs. (3-46a
and b):
T0b
T0a
TB ⫽
L
L
Since a ⬎ b, the larger torque (and hence the larger
stress) is in the right hand segment.
TA ⫽
L⫽a⫹b
a⬎b
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CHAPTER 3
tmax ⫽
T0 ⫽
Page 304
Torsion
ANGLE OF ROTATION OF THE DISK (FROM Eq. 3-49)
TB(d/2)
T0 ad
⫽
IP
2LIP
2LIPtmax
ad
(T0)max ⫽
2L IPt allow
ad
f⫽
T0 ab
GLIP
(T0)maxab
2bt allow
⫽
GLIp
Gd
fmax ⫽
;
Problem 3.8-4 A hollow steel shaft ACB of outside diameter 50 mm
and inside diameter 40 mm is held against rotation at ends A and B
(see figure). Horizontal forces P are applied at the ends of a vertical
arm that is welded to the shaft at point C.
Determine the allowable value of the forces P if the maximum
permissible shear stress in the shaft is 45 MPa. (Hint: Use Eqs. 3-46a
and b of Example 3-9 to obtain the reactive torques.)
200 mm
A
P
200 mm
C
B
P
600 mm
400 mm
Solution 3.8-4
Hollow shaft with fixed ends
GENERAL FORMULAS:
T0 ⫽ P(400 mm)
LB ⫽ 400 mm
LA ⫽ 600 mm
L ⫽ LA ⫹ LB ⫽ 1000 mm
d2 ⫽ 50 mm d1 ⫽ 40 mm
␶allow ⫽ 45 MPa
APPLY THE ABOVE FORMULAS TO THE GIVEN SHAFT
TA ⫽
P(0.4 m)(400 mm)
T0 LB
⫽
⫽ 0.16 P
L
1000 mm
TB ⫽
P(0.4 m)(600 mm)
T0 LA
⫽
⫽ 0.24 P
L
1000 mm
UNITS: P ⫽ Newtons T ⫽ Newton meters
From Eqs. (3-46a and b):
TA ⫽
T0 LB
L
T0 LA
L
The larger torque, and hence the larger shear
stress, occurs in part CB of the shaft.
TB ⫽
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Page 305
SECTION 3.8
Tmax ⫽ TB ⫽ 0.24 P
Tmax(d/2)
IP
Tmax ⫽
2tmaxIP
d
0.24P ⫽
(Eq. 1)
P⫽
␶max ⫽ 45 ⫻ 10 N/m
Ip ⫽
2(45 * 106 N/m2)(362.26 * 10⫺9 m4)
0.05 m
⫽ 652.07 N # m
UNITS: Newtons and meters
6
305
Substitute numerical values into (Eq. 1):
SHEAR STRESS IN PART CB
tmax ⫽
Statically Indeterminate Torsional Members
2
652.07 N # m
⫽ 2717 N
0.24 m
Pallow ⫽ 2720 N
p 4
(d ⫺d 4) ⫽ 362.26 * 10⫺9m4
32 2 1
;
d ⫽ d2 ⫽ 0.05 mm
Problem 3.8-5 A stepped shaft ACB having solid circular cross
Solution 3.8-5
B
T0
6.0 in.
15.0 in.
Combine Eqs. (1) and (3) and solve for T0:
(T0)AC ⫽
LAIPB
1
pd3 t
a1 +
b
16 A allow
LBIPA
⫽
LAdB4
1
pd3At allow a 1 +
b
16
LBdA4
Find (T0)max
REACTIVE TORQUES (from Eqs. 3-45a and b)
LBIPA
b
TA ⫽ T0 a
LBIPA + LAIPB
LAIPB
b
LBIPA + LAIPB
(1)
ALLOWABLE TORQUE BASED UPON SHEAR STRESS
AC
Substitute numerical values:
ALLOWABLE TORQUE BASED UPON SHEAR STRESS
IN SEGMENT CB
tCB ⫽
16TA
pdA3
1
1
pd3 t ⫽
pd3 t
TA ⫽
16 A AC 16 A allow
(4)
(T0)AC ⫽ 3678 lb-in.
(2)
IN SEGMENT
tAC ⫽
C
A
Stepped shaft ACB
dA ⫽ 0.75 in. dB ⫽ 1.50 in.
LA ⫽ 6.0 in. LB ⫽ 15.0 in.
␶allow ⫽ 6000 psi
TB ⫽ T0 a
1.50 in.
0.75 in.
sections with two different diameters is held against rotation at
the ends (see figure).
If the allowable shear stress in the shaft is 6000 psi, what is
the maximum torque (T0)max that may be applied at section C?
(Hint: Use Eqs. 3-45a and b of Example 3-9 to obtain the reactive
torques.)
TB ⫽
(3)
16TB
pdB3
1
1
pd 3 t ⫽
pd 3 t
16 B CB 16 B allow
(5)
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CHAPTER 3
Page 306
Torsion
Combine Eqs. (2) and (5) and solve for T0:
(T0)CB ⫽
SEGMENT AC GOVERNS
(T0)max ⫽ 3680 lb-in.
LBIPA
1
pd3 t
a1 +
b
16 B allow
LAIPB
LBdA4
1
pd3Bt allow a1 +
b
⫽
16
LAdB4
Substitute numerical values:
(T0)CB ⫽ 4597 lb-in.
NOTE: From Eqs. (4) and (6) we find that
(6)
(T0)AC
LA dB
⫽ a ba b
(T0)CB
LB dA
which can be used as a partial check on the results.
20 mm
Problem 3.8-6 A stepped shaft ACB having solid circular cross
sections with two different diameters is held against rotation at
the ends (see figure).
If the allowable shear stress in the shaft is 43 MPa, what is the
maximum torque (T0)max that may be applied at section C?
(Hint: Use Eqs. 3-45a and b of Example 3-9 to obtain the reactive
torques.)
Solution 3.8-6
;
25 mm
B
C
A
T0
225 mm
450 mm
Stepped shaft ACB
1
1
pd 3 t ⫽
pd 3 t
16 A AC 16 A allow
dA
⫽ 20 mm
dB
⫽ 25 mm
LA
⫽ 225 mm
Combine Eqs. (1) and (3) and solve for T0:
LB
⫽ 450 mm
(T0)AC ⫽
TA ⫽
␶allow ⫽ 43 MPa
⫽
Find (T0)max
REACTIVE TORQUES (from Eqs. 3-45a and b)
LAIPB
1
pdA3 t allow a 1 +
b
16
LBIPA
LAdB4
1
pdA3t allow a 1 +
b
16
LBdA4
(1)
(T0)AC ⫽ 150.0 N ⭈ m
TB ⫽ T0 a
(2)
IN SEGMENT
ALLOWABLE TORQUE BASED UPON SHEAR STRESS
IN SEGMENT AC
tAC ⫽
16TA
pd3A
(4)
Substitute numerical values:
LBIPA
b
TA ⫽ T0 a
LBIPA + LAIPB
LAIPB
b
LBIPA + LAIPB
(3)
ALLOWABLE TORQUE BASED UPON SHEAR STRESS
CB
tCB ⫽
TB ⫽
16TB
pd3B
1
1
pd3 t ⫽
pd3 t
16 B CB 16 B allow
(5)
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Page 307
SECTION 3.8
Combine Eqs. (2) and (5) and solve for T0:
(T0)CB ⫽
(T0)max ⫽ 150 N ⭈ m
(T0)AC
LA dB
⫽ a ba b
(T0)CB
LB dA
(6)
which can be used as a partial check on the results.
Problem 3.8-7 A stepped shaft ACB is held against rotation at ends
dA
A and B and subjected to a torque T0 acting at section C (see figure).
The two segments of the shaft (AC and CB) have diameters dA and dB,
respectively, and polar moments of inertia IPA and IPB, respectively.
The shaft has length L and segment AC has length a.
C
T0
a
L
Stepped shaft
SEGMENT AC: dA, IPA LA ⫽ a
or
SEGMENT CB: dB, IPB LB ⫽ L ⫺ a
REACTIVE TORQUES (from Eqs. 3-45a and b)
Solve for a/L:
(a)
dB
IPA
A
(a) For what ratio a/L will the maximum shear stresses be the same
in both segments of the shaft?
(b) For what ratio a/L will the internal torques be the same in
both segments of the shaft? (Hint: Use Eqs. 3-45a and b of
Example 3-9 to obtain the reactive torques.)
TA ⫽ T0 a
;
NOTE: From Eqs. (4) and (6) we find that
Substitute numerical values:
(T0)CB ⫽ 240.0 N ⭈ m
Solution 3.8-7
LBIPA
LAIPB
b; TB ⫽ T0 a
b
LBIPA + LAIPB
LBIPA + LAIPB
EQUAL SHEAR STRESSES
TA(dA/2)
TB(dB/2)
tCB ⫽
tAC ⫽
IPA
IPB
␶AC ⫽ ␶CB or
TAdA
TBdB
⫽
IPA
IPB
Substitute TA and TB into Eq. (1):
LBIPAdA
LAIPBdB
⫽
IPA
IPB
or
307
SEGMENT AC GOVERNS
LBIPA
1
pd 3 t
a1 +
b
16 B allow
LAIPB
LBdA4
1
b
⫽ pdB3t allow a1 +
16
LAdB4
Statically Indeterminate Torsional Members
LBdA ⫽ LAdB
(L ⫺ a)dA ⫽ adB
(b) EQUAL TORQUES
TA ⫽ TB or LBIPA ⫽ LAIPB
or
(L ⫺ a)IPA ⫽ aIPB
Solve for a/L:
(Eq. 1)
dA
a
⫽
L
dA + dB
or
IPA
a
⫽
L
IPA + IPB
dA4
a
⫽ 4
L
dA + dB4
;
;
IPB
B
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CHAPTER 3
Page 308
Torsion
Problem 3.8-8 A circular bar AB of length L is fixed against
rotation at the ends and loaded by a distributed torque t(x) that
varies linearly in intensity from zero at end A to t0 at end B
(see figure).
Obtain formulas for the fixed-end torques TA and TB.
t0
t(x)
TA
TB
A
B
x
L
Solution 3.8-8
Fixed-end bar with triangular load
ELEMENT OF DISTRIBUTED LOAD
dTA ⫽ Elemental reactive torque
t(x) ⫽
dTB ⫽ Elemental reactive torque
t0x
L
From Eqs. (3-46a and b):
T0 ⫽ Resultant of distributed torque
L
L
t0 x
t0 L
T0 ⫽
t(x)dx ⫽
dx ⫽
2
L0
L0 L
EQUILIBRIUM
t0L
TA + TB ⫽ T0 ⫽
2
L⫺x
x
b dTB ⫽ t(x)dxa b
L
L
REACTIVE TORQUES (FIXED-END TORQUES)
dTA ⫽ t(x)dxa
L
t0L
x
L⫺x
TA ⫽ dTA ⫽
a t0 b a
bdx ⫽
L
L
6
L
L0
L
t0 L
x
x
TB ⫽ dTB ⫽ a t0 b a b dx ⫽
;
L L
3
L
L0
NOTE: TA + TB ⫽
Problem 3.8-9 A circular bar AB with ends fixed against rotation has
a hole extending for half of its length (see figure). The outer diameter
of the bar is d2 ⫽ 3.0 in. and the diameter of the hole is d1 ⫽ 2.4 in.
The total length of the bar is L ⫽ 50 in.
At what distance x from the left-hand end of the bar should a torque
T0 be applied so that the reactive torques at the supports will be equal?
t0L
(check)
2
25 in.
A
;
25 in.
T0
3.0 in.
B
x
2.4
in.
3.0
in.
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SECTION 3.8
Solution 3.8-9
Statically Indeterminate Torsional Members
309
Bar with a hole
IPA ⫽ Polar moment of inertia at left-hand end
IPB ⫽ Polar moment of inertia at right-hand end
fB ⫽
T0(L/2)
GIPA
⫺
L ⫽ 50 in.
L/2 ⫽ 25 in.
(2)
Substitute Eq. (1) into Eq. (2) and simplify:
d2 ⫽ outer diameter
fB ⫽
⫽ 3.0 in.
d1 ⫽ diameter of hole
T0 L
L
x
L
L
c
+
⫺
+
⫺
d
G 4IPB
4IPA
IPB
2IPB
2IPA
COMPATIBILITY ␾B ⫽ 0
⫽ 2.4 in.
x
T0 ⫽ Torque applied at distance x
‹
IPB
Find x so that TA ⫽ TB
⫽
3L
L
⫺
4IPB 4IPA
soLVE FOR x:
EQUILIBRIUM
TA ⫹ TB ⫽ T0 ‹ TA ⫽ TB ⫽
TB(L/2)
TB(L/2)
T0(x ⫺ L/2)
+
⫺
GIPB
GIPA
GIPB
T0
2
(1)
REMOVE THE SUPPORT AT END B
x⫽
IPB
L
a3 ⫺
b
4
IPA
IPB
d24 ⫺ d14
d1 4
⫽
⫽
1
⫺
a
b
IPA
d2
d24
d1 4
L
x ⫽ c2 + a b d ;
4
d2
SUBSTITUTE NUMERICAL VALUES:
x⫽
2.4 in. 4
50 in.
c2 + a
b d ⫽ 30.12 in.
4
3.0 in.
;
␾B ⫽ Angle of twist at B
Problem 3.8-10 A solid steel bar of diameter d1 ⫽ 25.0 mm is
enclosed by a steel tube of outer diameter d3 ⫽ 37.5 mm and inner
diameter d2 ⫽ 30.0 mm (see figure). Both bar and tube are held
rigidly by a support at end A and joined securely to a rigid plate
at end B. The composite bar, which has a length L ⫽ 550 mm,
is twisted by a torque T ⫽ 400 N ⭈ m acting on the end plate.
(a) Determine the maximum shear stresses ␶1 and ␶2 in the bar and
tube, respectively.
(b) Determine the angle of rotation ␾ (in degrees) of the end plate,
assuming that the shear modulus of the steel is G ⫽ 80 GPa.
(c) Determine the torsional stiffness kT of the composite bar.
(Hint: Use Eqs. 3-44a and b to find the torques in the bar and tube.)
Tube
A
B
T
Bar
End
plate
L
d1
d2
d3
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CHAPTER 3
Page 310
Torsion
Solution 3.8-10
Bar enclosed in a tube
TORQUES IN THE BAR (1) AND TUBE (2)
FROM EQS. (3-44A AND B)
Bar: T1 ⫽ T a
IP1
b ⫽ 100.2783 N # m
IP1 + IP2
Tube: T2 ⫽ T a
IP2
b ⫽ 299.7217 N # m
IP1 + IP2
(a) MAXIMUM SHEAR STRESSES
Bar: t1 ⫽
T1(d1/2)
⫽ 32.7 MPa
IP1
Tube: t2 ⫽
d1 ⫽ 25.0 mm
d2 ⫽ 30.0 mm
d3 ⫽ 37.5 mm
T 2L
T1L
⫽
⫽ 0.017977 rad
GIP1
GIP2
f⫽
POLAR MOMENTS OF INERTIA
␾ ⫽ 1.03°
Tube: IP2 ⫽
;
(b) ANGLE OF ROTATION OF END PLATE
G ⫽ 80 GPa
p 4
Bar: IP1 ⫽
d ⫽ 38.3495 * 10⫺9 m4
32 1
T2(d3/2)
⫽ 49.0 MPa
IP2
;
;
(c) TORSIONAL STIFFNESS
p
1d 4 ⫺ d242 ⫽ 114.6229 * 10⫺9 m4
32 3
kT ⫽
T
⫽ 22.3 kN # m
f
Problem 3.8-11 A solid steel bar of diameter d1 ⫽ 1.50 in. is enclosed
by a steel tube of outer diameter d3 ⫽ 2.25 in. and inner diameter
d2 ⫽ 1.75 in. (see figure). Both bar and tube are held rigidly by a support
at end A and joined securely to a rigid plate at end B. The composite bar,
which has length L ⫽ 30.0 in., is twisted by a torque T ⫽ 5000 lb-in.
acting on the end plate.
(a) Determine the maximum shear stresses ␶1 and ␶2 in the bar and
tube, respectively.
(b) Determine the angle of rotation ␾ (in degrees) of the end plate,
assuming that the shear modulus of the steel is G ⫽ 11.6 ⫻ 106 psi.
(c) Determine the torsional stiffness kT of the composite bar. (Hint: Use
Eqs. 3-44a and b to find the torques in the bar and tube.)
;
Tube
A
B
T
Bar
End
plate
L
d1
d2
d3
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Page 311
SECTION 3.8
Solution 3.8-11
Statically Indeterminate Torsional Members
311
Bar enclosed in a tube
TORQUES IN THE BAR (1) AND TUBE (2)
FROM EQS. (3-44A AND B)
Bar: T1 ⫽ T a
IPI
b ⫽ 1187.68 lb-in.
IPI + IPI
Tube: T2 ⫽ T a
IP2
b ⫽ 3812.32 lb-in.
IP1 + IP2
(a) MAXIMUM SHEAR STRESSES
Bar: t1 ⫽
T1(d1/2)
⫽ 1790 psi
IP1
T2(d3/2)
⫽ 2690 psi
IP2
Tube: t2 ⫽
;
;
(b) ANGLE OF ROTATION OF END PLATE
d1 ⫽ 1.50 in.
d2 ⫽ 1.75 in.
d3 ⫽ 2.25 in.
f⫽
G ⫽ 11.6 ⫻ 10 psi
6
␾ ⫽ 0.354°
POLAR MOMENTS OF INERTIA
Bar: IP1 ⫽
p 4
d ⫽ 0.497010 in.4
32 1
T 2L
T1L
⫽
⫽ 0.006180015 rad
GIP1
GIP2
;
(c) TORSIONAL STIFFNESS
kT ⫽
p
Tube: IP2 ⫽ 1d34 ⫺ d242 ⫽ 1.595340 in.4
32
T
⫽ 809 k - in.
f
T
Problem 3.8-12 The composite shaft shown in the figure is
manufactured by shrink-fitting a steel sleeve over a brass core so that
the two parts act as a single solid bar in torsion. The outer diameters
of the two parts are d1 ⫽ 40 mm for the brass core and d2 ⫽ 50 mm for
the steel sleeve. The shear moduli of elasticity are Gb ⫽ 36 GPa for the
brass and Gs ⫽ 80 GPa for the steel.
;
Steel sleeve
Brass core
T
Assuming that the allowable shear stresses in the brass and steel are
␶b ⫽ 48 MPa and ␶s ⫽ 80 MPa, respectively, determine the maximum
permissible torque Tmax that may be applied to the shaft. (Hint: Use
Eqs. 3-44a and b to find the torques.)
d1 d2
Probs. 3.8-12 and 3.8-13
Solution 3.8-12
Composite shaft shrink fit
d1 ⫽ 40 mm
d2 ⫽ 50 mm
GB ⫽ 36 GPa
GS ⫽ 80 GPa
Allowable stresses:
␶B ⫽ 48 MPa ␶S ⫽ 80 MPa
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CHAPTER 3 Torsion
BRASS CORE (ONLY)
Eq. (3-44b): TS ⫽ T a
GSIPS
b
GBIPB + GSIPS
⫽ 0.762082 T
T ⫽ TB ⫹ TS (CHECK)
ALLOWABLE TORQUE T BASED UPON BRASS CORE
IPB ⫽
p 4
d ⫽ 251.327 * 10⫺9 m4
32 1
GBIPB ⫽ 9047.79 N ⭈ m2
TB(d1/2)
2tBIPB
TB ⫽
IPB
d1
Substitute numerical values:
tB ⫽
TB ⫽ 0.237918 T
STEEL SLEEVE (ONLY)
⫽
2(48 MPa)(251.327 * 10⫺9 m4)
40 mm
T ⫽ 2535 N ⭈ m
ALLOWABLE TORQUE T BASED UPON STEEL SLEEVE
IPS
p 4
⫽
(d ⫺ d14) ⫽ 362.265 * 10⫺9 m4
32 2
GSIPS ⫽ 28,981.2 N ⭈ m
2
TORQUES
Total torque: T ⫽ TB ⫹ TS
Eq. (3-44a): TB ⫽ T a
GBIPB
b
GBIPB + GS IPS
tS ⫽
TS(d2/2)
IPS
TS ⫽
2tSIPS
d2
SUBSTITUTE NUMERICAL VALUES:
TS ⫽ 0.762082 T
⫽
2(80 MPa)(362.265 * 10⫺9 m4)
50 mm
T ⫽ 1521 N ⭈ m
STEEL SLEEVE GOVERNS
Tmax ⫽ 1520 N ⭈ m
;
⫽ 0.237918 T
Problem 3.8-13 The composite shaft shown in the figure is manufactured by shrink-fitting a steel sleeve over a brass core
so that the two parts act as a single solid bar in torsion. The outer diameters of the two parts are d1 ⫽ 1.6 in. for the brass
core and d2 ⫽ 2.0 in. for the steel sleeve. The shear moduli of elasticity are Gb ⫽ 5400 ksi for the brass and Gs ⫽ 12,000 ksi
for the steel.
Assuming that the allowable shear stresses in the brass and steel are ␶b ⫽ 4500 psi and ␶s ⫽ 7500 psi, respectively,
determine the maximum permissible torque Tmax that may be applied to the shaft. (Hint: Use Eqs. 3-44a and b to find
the torques.)
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SECTION 3.8 Statically Indeterminate Torsional Members
Solution 3.8-13 Composite shaft shrink fit
TORQUES
Total torque: T ⫽ TB ⫹ TS
Eq. (3-44 a): TB ⫽ T a
GBIPB
b
GBIPB + GSIPS
⫽ 0.237918 T
d1 ⫽ 1.6 in.
Eq. (3-44 b): TS ⫽ T a
d2 ⫽ 2.0 in.
GB ⫽ 5,400 psi
GS ⫽ 12,000 psi
GSIPS
b
GBIPB + GSIPS
⫽ 0.762082 T
Allowable stresses:
T ⫽ TB ⫹ TS (CHECK)
␶B ⫽ 4500 psi ␶S ⫽ 7500 psi
ALLOWABLE TORQUE T BASED UPON BRASS CORE
BRASS CORE (ONLY)
TB(d1/2)
2tBIPB
TB ⫽
IPB
d1
Substitute numerical values:
tB ⫽
TB ⫽ 0.237918 T
⫽
IPB ⫽
p 4
d1 ⫽ 0.643398 in.4
32
GBIPB ⫽ 3.47435 ⫻ 106 lb-in.2
STEEL SLEEVE (ONLY)
2(4500 psi)(0.643398 in.4)
1.6 in.
T ⫽ 15.21 k-in.
ALLOWABLE TORQUE T BASED UPON STEEL SLEEVE
TS(d2/2)
2tSIPS
TS ⫽
IPS
d2
Substitute numerical values:
tS ⫽
tS ⫽ 0.762082 T ⫽
2(7500 psi)(0.927398 in.4)
2.0 in.
T ⫽ 9.13 k-in.
STEEL SLEEVE GOVERNS
IPS ⫽
Tmax ⫽ 9.13 k-in.
;
p
(d 4 ⫺ d14) ⫽ 0.927398 in.4
32 2
GSIPS ⫽ 11.1288 ⫻ 106 lb-in.2
Problem 3.8-14 A steel shaft (Gs ⫽ 80 GPa) of total length L ⫽ 3.0 m is encased for one-third of its length by a brass
sleeve (Gb ⫽ 40 GPa) that is securely bonded to the steel (see figure). The outer diameters of the shaft and sleeve are
d1 ⫽ 70 mm and d2 ⫽ 90 mm. respectively.
313
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CHAPTER 3
Page 314
Torsion
(a) Determine the allowable torque T1 that may be applied
to the ends of the shaft if the angle of twist between the
ends is limited to 8.0°.
(b) Determine the allowable torque T2 if the shear stress in
the brass is limited to ␶b ⫽ 70 MPa.
(c) Determine the allowable torque T3 if the shear stress
in the steel is limited to ␶s ⫽ 110 MPa.
(d) What is the maximum allowable torque Tmax if all three
of the preceding conditions must be satisfied?
Brass
sleeve
Steel
shaft
d2 = 90 mm
d1 = 70 mm
T
T
A
B
1.0 m
L
= 2.0 m
2
d1
C
L
= 2.0 m
2
d1
Brass
sleeve
d2
d1
Steel
shaft
d2
Solution 3.8-14
(a) ALLOWABLE TORQUE T1 BASED ON TWIST AT ENDS OF
8 DEGREES
First find torques in steel (Ts) & brass (Tb) in segment
in which they are joined - 1 degree stat-indet; use Ts
as the internal redundant; see equ. 3-44a in text
example
statics
Gb IPb
b
Gs IPs + GbIPb
now find twist of 3 segments:
Tb ⫽ T1 ⫺ Ts
Tb ⫽ T1 a
L
L
L
Ts
T1
4
4
2
␾⫽
+
+
Gb IPb
Gs IPs
Gs IPs
T1
Gs IPs
Ts ⫽ T1 a
b
Gs IPs + Gb IPb
For middle term, brass sleeve & steel shaft twist the same so could use Tb(L/4)/(Gb IPb) instead
Let ␾a = ␾ allow; substitute expression for Ts then simplifiy; finally, solve for T1, allow
Gs IPs
L
L
L
T1 a
b
T1
Gs IPs + Gb IPb 4
4
2
fa ⫽
+
+
Gb IPb
Gs IPs
Gs IPs
T1
L
L
L
T1
T1
4
4
2
fa ⫽
+
+
Gb IPb
Gs IPs + Gb IPb
Gs IPs
T1
L
1
2
1
fa ⫽ T1 a
+
b
+
4 Gb IPb
GsIPs + Gb IPb
Gs IPs
T1, allow ⫽
4␾a
Gb IPb(GsIPs + Gb IPb)Gs IPs
c 2 2
d
2
L Gs IPs + 4 Gb IPb Gs IPs + 2 Gb2 IPb
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SECTION 3.8 Statically Indeterminate Torsional Members
f a ⫽ 8a
NUMERICAL VALUES
Gs ⫽ 80 GPa
Gb ⫽ 40 GPa
d1 ⫽ 70 mm
d2 ⫽ 90 mm
IPs ⫽
p 4
d1
32
L ⫽ 3.0 m
IPs ⫽ 2.357 ⫻ 10⫺6 m4
p
A d 4 ⫺ d14 B
32 2
T1, allow ⫽ 9.51 kN⭈ m
TORQUE
STRESS IN BRASS,
␶b
T2
;
BASED ON ALLOWABLE SHEAR
␶b ⫽ 70 MPa
First check hollow segment 1 (brass sleeve only)
T2
t⫽
d2
2
IPb
T2, allow ⫽
T2, allow ⫽ 6.35 kN⭈m
2tbIPb
d2
;
controls over T2 below also check segment 2 with
brass sleeve over steel shaft
Tb
t⫽
d2
2
IPb
Tb ⫽ T2 a
T2, allow
T2, allow ⫽ 13.69 kN⭈m
so T2 for hollow segment controls
(c) ALLOWABLE
TORQUE
STRESS IN STEEL,
␶s
T3
Ts
t⫽
d1
2
where from stat-indet analysis above
IPS
Ts ⫽ T3 a
GsIPS
b
Gs IPS + Gb IPb
T3, allow ⫽
2ts (Gs IPs + Gb IPb)
d1Gs
T3, allow ⫽ 13.83 kN⭈m
also check segment 3 with steel shaft alone
d1
2
t⫽
IPs
T3
T3, allow ⫽
T3, allow ⫽ 7.41 kN⭈m
where from stat-indet analysis above
GbIPb
b
Gs IPS + Gb IPb
2tb(Gs IPS + Gb IPb)
⫽
d2 Gb
BASED ON ALLOWABLE SHEAR
␶s ⫽ 110 MPa
First check segment 2 with brass sleeve over steel
shaft
IPb ⫽ 4.084 ⫻ 10⫺6 m4
IPb ⫽
(b) ALLOWABLE
p
b rad
180
315
2ts IPs
d1
; controls over T3 above
(d) Tmax IF ALL PRECEDING CONDITIONS MUST BE
CONSIDERED
from (b) above
Tmax ⫽ 6.35 kN⭈m
; max. shear stress in
hollow brass sleeve in
segment 1 controls overall
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CHAPTER 3
Page 316
Torsion
Problem 3.8-15 A uniformly tapered aluminum-alloy tube AB of circular cross section and length L is fixed against rota-
tion at A and B, as shown in the figure. The outside diameters at the ends are dA and dB ⫽ 2dA. A hollow section of length
L/2 and constant thickness t ⫽ dA/10 is cast into the tube and extends from B halfway toward A. Torque To is applied at L/2.
(a) Find the reactive torques at the supports, TA and TB. Use numerical values as follows: dA ⫽ 2.5 in., L ⫽ 48 in.,
G ⫽ 3.9 ⫻ 106 psi, T0 ⫽ 40,000 in.-lb.
(b) Repeat (a) if the hollow section has constant diameter dA.
Fixed against
rotation
TA
L
—
2
A
d(x)
t constant
TB
T0
dA
Fixed against
rotation
B
dB
L
(a)
TA
A
Fixed against
rotation
B
L
—
2
dA
TB
T0
dA
Fixed against
rotation
dB
L
(b)
Solution 3.8-15
Solution approach-superposition: select TB as the redundant (1° SI )
L
—
2
TA1
A
B
ϕ1(same results for parts a & b)
T0
f1 ⫽
L
+
TA2
L
—
2
A
81GpdA
4
f1 ⫽ 2.389
See Prob. 3.4-13 for results
for w2 for Parts a & b
B
ϕ2
TB
L
608T0L
f2a ⫽ 3.868
f2a ⫽ 3.057
T0 L
Gd A 4
T0L
Gd A 4
T0L
GdA 4
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SECTION 3.8
CONSTANT THICKNESS OF HOLLOW SECTION OF TUBE
CONSTANT DIAMETER OF HOLE
␾1 ⫺ ␾2 ⫽ 0
TB ⫽ a
TB ⫽ 24708 in-lb
TA ⫽ T0 ⫺ TB TA ⫽ 15292 in-1b
ba
4
608T0 L
81GpdA
TB ⫽ 2.45560
GdA4
b
b
a
TB ⫽ a
81G pdA4 3.86804L
608T0 L
T0
TB ⫽ 1.94056
p
317
(b) REACTIVE TORQUES, TA & TB, FOR CASE OF
(a) REACTIVE TORQUES, TA & TB, FOR CASE OF
compatibility equation:
TB ⫽ redundant
T0 ⫽ 40000 in.-lb
Statically Indeterminate Torsional Members
T0
p
GdA4
b
3.05676L
TB ⫽ 31266 in.-lb
TA ⫽ T0 ⫺ TB TA ⫽ 8734 in.-lb
;
;
TA ⫹ TB ⫽ 40,000 in.-1b (check)
;
;
TA ⫹ TB ⫽ 40,000 in.-lb (check)
Problem 3.8-16 A hollow circular tube A (outer diameter dA, wall
thickness tA) fits over the end of a circular tube B (dB, tB), as shown
in the figure. The far ends of both tubes are fixed. Initially, a hole
through tube B makes an angle ␤ with a line through two holes in
tube A. Then tube B is twisted until the holes are aligned, and a pin
(diameter dp) is placed through the holes. When tube B is released,
the system returns to equilibrium. Assume that G is constant.
(a) Use superposition to find the reactive torques TA and TB at the
supports.
(b) Find an expression for the maximum value of ␤ if the shear
stress in the pin, ␶p, cannot exceed ␶p,allow.
(c) Find an expression for the maximum value of ␤ if the shear
stress in the tubes, ␶t, cannot exceed ␶t,allow.
(d) Find an expression for the maximum value of ␤ if the bearing
stress in the pin at C cannot exceed ␴b,allow.
IPA
IPB
TA
Tube A
A
TB
Tube B
B
C
L
L
b Pin at C
Tube A
Tube B
Cross-section at C
Solution 3.8-16
(a) SUPERPOSITION
TO FIND TORQUE REACTIONS
-
USE
TB
AS THE REDUNDANT
compatibility: ␾B1 ⫹ ␾B2 ⫽ 0
␾B1 ⫽ ⫺␤ ⬍ joint tubes by pin then release end B
fB2 ⫽
TBL 1
1
a
+
b
G IPA
IPB
fB2 ⫽
TBL IPB + IPA
a
b
G
IPAIPB
TB ⫽
Gb IPAIPB
a
b
L IPA ⫹IPB
TA ⫽ ⫺TB
; statics
(b) ALLOWABLE SHEAR IN PIN RESTRICTS MAGNITUDE OF ␤
TB ⫽ FORCE COUPLE VdB WITH V ⫽ SHEAR IN
C
TB
V
V⫽
tp ⫽
dB
As
TORQUE
PIN AT
TB
dB
tp, allow ⫽
p 2
d
4 P
;
b max ⫽ tp, allow
ca
Gb IPAIPB
a
b
L IPA ⫹ IPB
tp, allow ⫽
p
dB dp2
4
L
4G
IPB + IPA
b dBpdP 2 d
IPAIPB
;
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CHAPTER 3 Torsion
(c) ALLOWABLE SHEAR IN TUBES RESTRICTS MAGNITUDE
OF ␤
dA
TB
2
tmax ⫽
IPA
tmax ⫽
or
tmax
Bearing stresses from tubes A & B are:
sbA ⫽
dB
TB
2
⫽
IPB
TB
dA ⫺ tA
sbA ⫽
dPtA
Gb IPAIPB dA
a
b
L IPA ⫹IPB 2
Gb
IPAIPB
a
b
L IPA + IPB
Gb IPAIPB dB
a
b
L IPA ⫹IPB 2
sbA ⫽
IPB
simplifying these two equ., then solving for ␤
gives:
tmax ⫽
Gb
IPB
dA
a
b
L IPA + IPB 2
sbA ⫽ b
Gb
dB
IPA
a
b
L IPA + IPB 2
b max ⫽ tt, allow a
IPA + IPB
2L
ba
b
GdA
IPB
sbB ⫽ b
;
or
b max
dA ⫺tA
dPtA
Gb
IPAIPB
a
b
L IPA + IPB
IPA + IPB
2L
⫽ tt, allow a
ba
b
GdB
IPA
dB ⫺ tB
dPtB
sbB ⫽
or
tmax ⫽
TB
dB ⫺ tB
sbB ⫽
dP tB
substitute TB expression from part (a), then simplify
␧ solve for b
IPA
or
tmax ⫽
FA
FB
sbB ⫽
dPtA
dPtB
;
where lesser value of ␤ controls
(d) ALLOWABLE BEARING STRESS IN PIN RESTRICTS
MAGNITUDE OF ␤
Torque TB ⫽ force couple FB (dB ⫺ tB) or
FA(dA ⫺ tA), with F ⫽ ave. bearing force on
pin at C
L(IPB
IPAIPB
G
L (IPB + IPA)(dB ⫺ tB)dPtB
b max ⫽ sb, allow
c
L
G
(IPB + IPA)(dA ⫺ tA)dPtA
d
IPAIPB
b max ⫽ s b, allow
c
GIPAIPB
+ IPA)(dA ⫺ tA) dP tA
;
L
G
(IPB + IPA)(dB ⫺ tB)dPtB
d
IPAIPB
where lesser value controls
;
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SECTION 3.9
Strain Energy in Torsion
319
Strain Energy in Torsion
Problem 3.9-1 A solid circular bar of steel (G ⫽ 11.4 ⫻ 106 psi)
with length L ⫽ 30 in. and diameter d ⫽ 1.75 in. is subjected to
pure torsion by torques T acting at the ends (see figure).
d
T
T
(a) Calculate the amount of strain energy U stored in the bar
when the maximum shear stress is 4500 psi.
(b) From the strain energy, calculate the angle of twist ␾
(in degrees).
Solution 3.9-1
L
Steel bar
⫽
pd2Lt2max
16G
(Eq. 2)
Substitute numerical values:
U ⫽ 32.0 in.-lb
G ⫽ 11.4 ⫻ 106 psi
(b) ANGLE OF TWIST
L ⫽ 30 in.
U⫽
d ⫽ 1.75 in.
␶max ⫽ 4500 psi
tmax ⫽
IP ⫽
16 T
pd3
Tf
2
f⫽
2U
T
Substitute for T and U from Eqs. (1) and (2):
pd3tmax
T⫽
16
pd 4
32
;
f⫽
(Eq. 1)
2Ltmax
Gd
(Eq. 3)
Substitute numerical values:
␾ ⫽ 0.013534 rad ⫽ 0.775°
;
(a) STRAIN ENERGY
U⫽
pd3tmax 2 L
T2L
32
⫽ a
b a
b a 4b
2GIP
16
2G pd
Problem 3.9-2 A solid circular bar of copper (G ⫽ 45 GPa) with length L ⫽ 0.75 m and diameter d ⫽ 40 mm is subjected
to pure torsion by torques T acting at the ends (see figure).
(a) Calculate the amount of strain energy U stored in the bar when the maximum shear stress is 32 MPa.
(b) From the strain energy, calculate the angle of twist ␾ (in degrees)
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Torsion
Solution 3.9-2
Copper bar
(a) STRAIN ENERGY
U⫽
pd3tmax 2 L
32
T2L
⫽ a
b a
ba
b
2GIP
16
2G pd 4
L ⫽ 0.75 m
pd2Lt2max
16G
Substitute numerical values:
d ⫽ 40 mm
U ⫽ 5.36 J
⫽
G ⫽ 45 GPa
␶max ⫽ 32 MPa
tmax ⫽
IP ⫽
T⫽
3
pd
;
(b) ANGLE OF TWIST
3
16T
pd tmax
16
pd4
32
Tf
2U
f⫽
2
T
Substitute for T and U from Eqs. (1) and (2):
2Ltmax
f⫽
(Eq. 3)
Gd
Substitute numerical values:
U⫽
(Eq. 1)
␾ ⫽ 0.026667 rad ⫽ 1.53°
Problem 3.9-3 A stepped shaft of solid circular cross sections
(see figure) has length L ⫽ 45 in., diameter d2 ⫽ 1.2 in., and
diameter d1 ⫽ 1.0 in. The material is brass with G ⫽ 5.6 ⫻ 106 psi.
Determine the strain energy U of the shaft if the angle of
twist is 3.0°.
d2
;
d1
T
T
L
—
2
L
—
2
Solution 3.9-3
(Eq. 2)
Stepped shaft
⫽
8T2L 1
1
a 4 + 4b
pG d2
d1
Also, U ⫽
(Eq. 1)
Tf
2
(Eq. 2)
Equate U from Eqs. (1) and (2) and solve for T:
d1 ⫽ 1.0 in.
T⫽
d2 ⫽ 1.2 in.
L ⫽ 45 in.
pGd14 d24f
16L(d41 + d42)
Tf
pGf2 d14 d24
⫽
a
b
2
32L d41 + d42
G ⫽ 5.6 ⫻ 10 psi (brass)
U⫽
␾ ⫽ 3.0° ⫽ 0.0523599 rad
SUBSTITUTE NUMERICAL VALUES:
STRAIN ENERGY
U ⫽ 22.6 in.-lb
6
16 T2(L/2)
16 T2(L/2)
T2L
U⫽ a
⫽
+
4
2GIP
pGd2
pGd41
;
f ⫽ radians
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SECTION 3.9
Strain Energy in Torsion
321
Problem 3.9-4 A stepped shaft of solid circular cross sections (see figure) has length L ⫽ 0.80 m, diameter d2 ⫽ 40 mm,
and diameter d1 ⫽ 30 mm. The material is steel with G ⫽ 80 GPa.
Determine the strain energy U of the shaft if the angle of twist is 1.0°.
Solution 3.9-4
Stepped shaft
Also, U ⫽
Tf
2
(Eq. 2)
Equate U from Eqs. (1) and (2) and solve for T:
T⫽
d1 ⫽ 30 mm
d2 ⫽ 40 mm
L ⫽ 0.80 m
G ⫽ 80 GPa (steel)
U⫽
pG d14 d24f
16L(d14 + d24)
Tf
pGf2 d14 d24
⫽
a
b
2
32L d14 + d24
f ⫽ radians
␾ ⫽ 1.0° ⫽ 0.0174533 rad
SUBSTITUTE NUMERICAL VALUES:
STRAIN ENERGY
2
TL
⫽
U⫽ a
2GIP
⫽
U ⫽ 1.84 J
16T2(L/2)
pGd24
8T2L 1
1
a
+ 4b
pG d24
d1
;
16T2(L/2)
+
pGd14
(Eq. 1)
Problem 3.9-5 A cantilever bar of circular cross section and length L is
fixed at one end and free at the other (see figure). The bar is loaded by a
torque T at the free end and by a distributed torque of constant intensity
t per unit distance along the length of the bar.
(a) What is the strain energy U1 of the bar when the load T acts alone?
(b) What is the strain energy U2 when the load t acts alone?
(c) What is the strain energy U3 when both loads act simultaneously?
Solution 3.9-5
Cantilever bar with distributed torque
G ⫽ shear modulus
IP ⫽ polar moment of inertia
T ⫽ torque acting at free end
t ⫽ torque per unit distance
t
L
T
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Torsion
(a) LOAD T ACTS ALONE (Eq. 3-51a)
U1 ⫽
T2L
2GIP
(c) BOTH LOADS ACT SIMULTANEOUSLY
;
(b) LOAD t ACTS ALONE
From Eq. (3-56) of Example 3-11:
At distance x from the free end:
t2L3
U2 ⫽
6GIP
T(x) ⫽ T + tx
;
L
L
[T(x)]2
1
dx ⫽
(T + tx)2dx
2GI
2GI
P
PL
L0
0
T2L
TtL2
t2L3
⫽
+
+
;
2GIP
2GIP
6GIP
U3 ⫽
NOTE: U3 is not the sum of U1 and U2.
Problem 3.9-6 Obtain a formula for the strain energy U of the statically
2T0
indeterminate circular bar shown in the figure. The bar has fixed supports
at ends A and B and is loaded by torques 2T0 and T0 at points C and D,
respectively.
Hint: Use Eqs. 3-46a and b of Example 3-9, Section 3.8, to obtain the
reactive torques.
Solution 3.9-6
3L
b
4
L
T0 a b
4
+
L
L
⫽
7T0
4
L
—
4
D
L
—
2
⫽
L
L
L
1
2
2
cT 2 a b + TCD
a b + TDB
a bd
2GIp AC 4
2
4
⫽
7T0 2 L
1
ca⫺
b a b
2GIP
4
4
+ a
INTERNAL TORQUES
7T0
4
C
n
Ti2Li
U⫽ a
i⫽1 2GiIPi
5T0
TB ⫽ 3T0 ⫺ TA ⫽
4
TAC ⫽ ⫺
B
STRAIN ENERGY (from Eq. 3-53)
From Eq. (3-46a):
TA ⫽
A
Statically indeterminate bar
REACTIVE TORQUES
(2T0)a
T0
TCD ⫽
T0
4
TDB ⫽
5T0
4
U⫽
5T0 2 L
T0 2 L
b a b + a
b a bd
4
2
4
4
19T02L
32GIP
;
L
—
4
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SECTION 3.9
Strain Energy in Torsion
323
Problem 3.9-7 A statically indeterminate stepped shaft ACB is fixed at
ends A and B and loaded by a torque T0 at point C (see figure). The two
segments of the bar are made of the same material, have lengths LA and LB,
and have polar moments of inertia IPA and IPB.
Determine the angle of rotation ␾ of the cross section at C by using
strain energy.
Hint: Use Eq. 3-51b to determine the strain energy U in terms of the
angle ␾. Then equate the strain energy to the work done by the torque T0.
Compare your result with Eq. 3-48 of Example 3-9, Section 3.8.
A
IPA
T0
C
IPB
LA
B
LB
Solution 3.9-7
Statically indeterminate bar
WORK DONE BY THE TORQUE T0
W⫽
T0f
2
EQUATE U AND W AND SOLVE FOR ␾
T0f
Gf2 IPA
IPB
a
+
b ⫽
2
LA
LB
2
f⫽
STRAIN ENERGY (FROM EQ. 3-51B)
;
(This result agrees with Eq. (3-48) of Example 3-9,
Section 3.8.)
GIPif2i
GIPAf2
GIPBf2
⫽
+
U⫽ a
2LA
2LB
i⫽1 2Li
n
⫽
T0LALB
G(LBIPA + LAIPB)
Gf2 IPA
IPB
+
b
a
2
LA
LB
Problem 3.9-8 Derive a formula for the strain energy U of the cantilever bar shown in the figure.
The bar has circular cross sections and length L. It is subjected to a distributed torque of intensity t per unit distance. The
intensity varies linearly from t ⫽ 0 at the free end to a maximum value t ⫽ t0 at the support.
t0
t
L
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CHAPTER 3 Torsion
Solution 3.9-8 Cantilever bar with distributed torque
x ⫽ distance from right-hand end of the bar
ELEMENT d␰
STRAIN ENERGY OF ELEMENT dx
Consider a differential element dj at distance j from the
right-hand end.
dU ⫽
[T(x)]2dx
t0 2
1
⫽
a b x4dx
2GIP
2GIP 2L
⫽
t20
8L2GIP
x4 dx
STRAIN ENERGY OF ENTIRE BAR
L
U⫽
L0
dT ⫽ external torque acting on this element
dT ⫽ t(j)dj
j
⫽ t0 a bdj
L
U⫽
ELEMENT dx AT DISTANCE x
T(x) ⫽ internal torque acting on this element
T(x) ⫽ total torque from x ⫽ 0 to x ⫽ x
x
T(x) ⫽
⫽
L0
t0x2
2L
dT ⫽
x
L0
t0 a
j
bdj
L
t20
dU ⫽
t20L3
40GIP
L
x4 dx
8L2GIP L0
t20
L5
⫽ 2
a b
8L GIP 5
;
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SECTION 3.9
Problem 3.9-9 A thin-walled hollow tube AB of conical shape has
constant thickness t and average diameters dA and dB at the ends
(see figure).
B
A
T
T
(a) Determine the strain energy U of the tube when it is subjected
to pure torsion by torques T.
(b) Determine the angle of twist ␾ of the tube.
L
t
Note: Use the approximate formula IP ⬇ ␲d3t/4 for a thin circular
ring; see Case 22 of Appendix D.
t
dB
dA
Solution 3.9-9
325
Strain Energy in Torsion
Thin-walled, hollow tube
Therefore,
L
dx
3
dB ⫺ dA
L0
cdA + a
bx d
L
⫽⫺
t ⫽ thickness
dA ⫽ average diameter at end A
dB ⫽ average diameter at end B
⫽⫺
d(x) ⫽ average diameter at distance x from end A
d(x) ⫽ dA + a
dB ⫺ dA
bx
L
⫽
pd3t
4
U⫽
3
p[d(x)]3t
dB ⫺ dA
pt
IP(x) ⫽
⫽ cdA + a
bx d
4
4
L
L
T2dx
L0 2GIP(x)
L
dx
2T2
⫽
3
pGt L0
dB ⫺ dA
cdA + a
bx d
L
From Appendix C:
⫽ ⫺
L
2(dB ⫺ dA)(dB)
2
+
2(dB ⫺ dA)(dA)2
L(dA + dB)
2dA2 dB2
1
2b(a + bx)2
2T2 L(dA + dB)
T2L dA + dB
⫽
a 2 2 b
pGt 2dA2dB2
pGt
dA dB
Work of the torque T: W ⫽
U⫽
dx
L
(b) ANGLE OF TWIST
(a) STRAIN ENERGY (FROM EQ. 3-54)
L (a + bx)3
2 †
2(dB ⫺ dA)
dB ⫺ dA
cdA + a
bx d
L
L
0
Substitute this expression for the integral into the equation
for U (Eq. 1):
POLAR MOMENT OF INERTIA
IP ⫽
L
1
W⫽U
(Eq. 1)
Tf
T2L(dA + dB)
⫽
2
pGt d2Ad2B
Solve for ␾:
f⫽
Tf
2
2TL(dA + dB)
pGt d2A d2B
;
;
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Torsion
Problem 3.9-10 A hollow circular tube A fits over the end of
IPA
a solid circular bar B, as shown in the figure. The far ends of both
bars are fixed. Initially, a hole through bar B makes an angle ␤ with
a line through two holes in tube A. Then bar B is twisted until the
holes are aligned, and a pin is placed through the holes.
When bar B is released and the system returns to equilibrium,
what is the total strain energy U of the two bars? (Let IPA and IPB
represent the polar moments of inertia of bars A and B, respectively.
The length L and shear modulus of elasticity G are the same for
both bars.)
IPB
Tube A
Bar B
L
L
b
Tube A
Bar B
Solution 3.9-10
Circular tube and bar
TUBE A
COMPATIBILITY
␾A ⫹ ␾B ⫽ ␤
FORCE-DISPLACEMENT RELATIONS
fA ⫽
T ⫽ torque acting on the tube
␾A ⫽ angle of twist
BAR B
TL
GIPA
fB ⫽
TL
GIPB
Substitute into the equation of compatibility and solve
for T:
T⫽
bG
IPAIPB
a
b
L IPA + IPB
STRAIN ENERGY
U⫽ g
⫽
T 2L
T 2L
T 2L
⫽
+
2GIP
2GIPA
2GIPB
T2L 1
1
a
+
b
2G IPA
IPB
Substitute for T and simplify:
U⫽
T ⫽ torque acting on the bar
␾B ⫽ angle of twist
b 2G IPA IPB
a
b
2L IPA + IPB
;
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SECTION 3.9
Strain Energy in Torsion
Problem 3.9-11 A heavy flywheel rotating at n revolutions per minute is
rigidly attached to the end of a shaft of diameter d (see figure). If the
bearing at A suddenly freezes, what will be the maximum angle of twist ␾
of the shaft? What is the corresponding maximum shear stress in the shaft?
(Let L ⫽ length of the shaft, G ⫽ shear modulus of elasticity, and
Im ⫽ mass moment of inertia of the flywheel about the axis of the shaft.
Also, disregard friction in the bearings at B and C and disregard the mass
of the shaft.)
A
d
n (rpm)
B
C
Hint: Equate the kinetic energy of the rotating flywheel to the strain
energy of the shaft.
Solution 3.9-11
Rotating flywheel
p 4
d
32
IP ⫽
d ⫽ diameter of shaft
U⫽
pGd 4f2
64L
UNITS:
d ⫽ diameter
n ⫽ rpm
IP ⫽ (length)4
␾ ⫽ radians
KINETIC ENERGY OF FLYWHEEL
K.E. ⫽
G ⫽ (force)/(length)2
1
Imv2
2
2pn
v⫽
60
n ⫽ rpm
2pn 2
1
b
K.E. ⫽ Im a
2
60
L ⫽ length
U ⫽ (length)(force)
EQUATE KINETIC ENERGY AND STRAIN ENERGY
Solve for ␾:
f⫽
2 2
⫽
p n Im
1800
Im ⫽ (force)(length)(second)2
␻ ⫽ radians per second
K.E. ⫽ (length)(force)
STRAIN ENERGY OF SHAFT (FROM EQ. 3-51b)
U⫽
GIPf
2L
2
2n
2A
15d
2pImL
G
;
MAXIMUM SHEAR STRESS
t⫽
UNITS:
pGd 4f2
p2n2Im
⫽
1800
64 L
K.E. ⫽ U
T(d/2)
IP
f⫽
TL
GIP
Eliminate T:
t⫽
Gdf
2L
2pImL
2L15d A G
2pGIm
n
tmax ⫽
15d A L
tmax ⫽
Gd2n
2
;
327
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Torsion
Thin-Walled Tubes
Problem 3.10-1 A hollow circular tube having an inside diameter of 10.0 in.
and a wall thickness of 1.0 in. (see figure) is subjected to a torque T ⫽ 1200 k-in.
Determine the maximum shear stress in the tube using (a) the approximate
theory of thin-walled tubes, and (b) the exact torsion theory. Does the approximate
theory give conservative or nonconservative results?
10.0 in.
1.0 in.
Solution 3.10-1
Hollow circular tube
APPROXIMATE THEORY (EQ. 3-63)
t1 ⫽
T
2
2pr t
⫽
1200 k-in.
2p(5.5 in.)2(1.0 in.)
␶approx ⫽ 6310 psi
⫽ 6314 psi
;
EXACT THEORY (EQ. 3-11)
T ⫽ 1200 k-in.
t ⫽ 1.0 in.
t2 ⫽
T(d2/2)
⫽
IP
r ⫽ radius to median line
r ⫽ 5.5 in.
d2 ⫽ outside diameter ⫽ 12.0 in.
d1 ⫽ inside diameter ⫽ 10.0 in.
⫽
Td2
p
2a
b 1d24 ⫺ d142
32
16(1200k-in.)(12.0 in.)
p[(12.0 in.)4 ⫺ (10.0 in.)4]
⫽ 6831 psi
t exact ⫽ 6830 psi
;
Because the approximate theory gives stresses that are
too low, it is nonconservative. Therefore, the approximate theory should only be used for very thin tubes.
Problem 3.10-2 A solid circular bar having diameter d is to be
replaced by a rectangular tube having cross-sectional dimensions
d ⫻ 2d to the median line of the cross section (see figure).
Determine the required thickness tmin of the tube so that the
maximum shear stress in the tube will not exceed the maximum
shear stress in the solid bar.
t
t
d
d
2d
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SECTION 3.10
Solution 3.10-2
Thin-Walled Tubes
329
Bar and tube
SOLID BAR
tmax ⫽
16T
pd3
(Eq. 3-12)
Am ⫽ (d)(2d) ⫽ 2d2
(Eq. 3-64)
T
T
⫽
2tAm
4td2
(Eq. 3-61)
tmax ⫽
EQUATE THE MAXIMUM SHEAR STRESSES AND SOLVE FOR t
RECTANGULAR TUBE
16T
pd3
⫽
T
4td2
tmin ⫽
pd
64
;
If t ⬎ tmin, the shear stress in the tube is less than the
shear stress in the bar.
Problem 3.10-3 A thin-walled aluminum tube of rectangular
cross section (see figure) has a centerline dimensions b ⫽ 6.0 in.
and h ⫽ 4.0 in. The wall thickness t is constant and equal to 0.25 in.
t
h
(a) Determine the shear stress in the tube due to a torque
T ⫽ 15 k-in.
(b) Determine the angle of twist (in degrees) if the length L of
the tube is 50 in. and the shear modulus G is 4.0 ⫻ 106 psi.
b
Probs. 3.10-3 and 3.10-4
Solution 3.10-3
Thin-walled tube
Eq. (3-64): Am ⫽ bh ⫽ 24.0 in.2
J⫽
Eq. (3-71) with t1 ⫽ t2 ⫽ t:
J ⫽ 28.8 in.4
(a) SHEAR STRESS (EQ. 3-61)
t⫽
b ⫽ 6.0 in.
h ⫽ 4.0 in.
t ⫽ 0.25 in.
T ⫽ 15 k-in.
L ⫽ 50 in.
G ⫽ 4.0 ⫻ 106 psi
T
⫽ 1250 psi
2tAm
;
(b) ANGLE OF TWIST (EQ. 3-72)
f⫽
TL
⫽ 0.0065104 rad
GJ
⫽ 0.373°
;
2b2h2t
b + h
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Torsion
Problem 3.10-4 A thin-walled steel tube of rectangular cross section (see figure) has centerline dimensions b ⫽ 150 mm
and h ⫽ 100 mm. The wall thickness t is constant and equal to 6.0 mm.
(a) Determine the shear stress in the tube due to a torque T ⫽ 1650 N ⭈ m.
(b) Determine the angle of twist (in degrees) if the length L of the tube is 1.2 m and the shear modulus G is 75 GPa.
Solution 3.10-4
Thin-walled tube
b ⫽ 150 mm
(a) SHEAR STRESS (Eq. 3-61)
h ⫽ 100 mm
t⫽
t ⫽ 6.0 mm
L ⫽ 1.2 m
f⫽
G ⫽ 75 GPa
TL
⫽ 0.002444 rad
GJ
⫽ 0.140°
Eq. (3-64): Am ⫽ bh ⫽ 0.015 m2
J⫽
;
(b) ANGLE OF TWIST (Eq. 3-72)
T ⫽ 1650 N ⭈ m
Eq. (3-71) with t1 ⫽ t2 ⫽ t:
T
⫽ 9.17 MPa
2tAm
;
2b2h2t
b + h
J ⫽ 10.8 ⫻ 10⫺6 m4
Tube (1)
Problem 3.10-5 A thin-walled circular tube and a solid circular bar of
Bar (2)
the same material (see figure) are subjected to torsion. The tube and bar
have the same cross-sectional area and the same length.
What is the ratio of the strain energy U1 in the tube to the strain
energy U2 in the solid bar if the maximum shear stresses are the same in
both cases? (For the tube, use the approximate theory for thin-walled bars.)
Solution 3.10-5 THIN-WALLED TUBE (1)
Am ⫽ ␲r2
tmax ⫽
J ⫽ 2␲r3t
12pr2ttmax22L
T 2L
⫽
2GJ
2G(2pr3t)
prtt2maxL
G
A
But rt ⫽
2p
⫽
At2max L
‹ U1 ⫽
2G
SOLID BAR (2)
T
T
⫽
2tAm
2pr 2t
T ⫽ 2␲r 2t␶max
U1 ⫽
A ⫽ 2␲rt
A ⫽ pr22
IP ⫽
p 4
r
2 2
Tr2
pr23tmax
2T
⫽ 3 T⫽
IP
2
pr2
3
2
2 2
2
(pr2 tmax) L
pr2 tmaxL
TL
⫽
U2 ⫽
⫽
2GIP
4G
p 4
8G a r2 b
2
tmax ⫽
But pr22 ⫽ A
RATIO
U1
⫽2
U2
;
‹ U2 ⫽
2
L
Atmax
4G
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SECTION 3.10
Thin-Walled Tubes
t = 8 mm
Problem 3.10-6 Calculate the shear stress ␶ and the angle of twist ␾ (in
r = 50 mm
degrees) for a steel tube (G ⫽ 76 GPa) having the cross section shown
in the figure. The tube has length L ⫽ 1.5 m and is subjected to a torque
T ⫽ 10 kN ⭈ m.
r = 50 mm
b = 100 mm
Solution 3.10-6
Steel tube
SHEAR STRESS
G ⫽ 76 GPa.
t⫽
10 kN # m
T
⫽
2tAm
2(8 mm)(17,850 mm2)
L ⫽ 1.5 m
T ⫽ 10 kN ⭈ m
Am ⫽ ␲r2 ⫹ 2br
Am ⫽ ␲ (50 mm)2 ⫹ 2(100 mm)(50 mm)
⫽ 17,850 mm2
Lm ⫽ 2b ⫹ 2␲r
⫽ 35.0 MPa
;
ANGLE OF TWIST
f⫽
(10 kN # m)(1.5 m)
TL
⫽
GJ
(76 GPa)(19.83 * 106 mm4)
⫽ 0.00995 rad
⫽ 0.570°
;
⫽ 2(100 mm) ⫹ 2␲(50 mm)
⫽ 514.2 mm
J⫽
4(8 mm)(17,850 mm2)2
4tA2m
⫽
Lm
514.2 mm
⫽ 19.83 * 106 mm4
Problem 3.10-7 A thin-walled steel tube having an elliptical cross
331
t
section with constant thickness t (see figure) is subjected to a torque
T ⫽ 18 k-in.
Determine the shear stress ␶ and the rate of twist ␪ (in degrees
per inch) if G ⫽ 12 ⫻ 106 psi, t ⫽ 0.2 in., a ⫽ 3 in., and b ⫽ 2 in.
(Note: See Appendix D, Case 16, for the properties of an ellipse.)
2b
2a
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CHAPTER 3
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Torsion
Solution 3.10-7
Elliptical tube
FROM APPENDIX D, CASE 16:
Am ⫽ pab ⫽ p(3.0 in.)(2.0 in.) ⫽ 18.850 in.2
Lm L p[1.5(a + b) ⫺ 1ab]
⫽ p[1.5(5.0 in.) ⫺ 26.0 in.2] ⫽ 15.867 in.
J⫽
⫽ 17.92 in.4
T ⫽ 18 k-in.
SHEAR STRESS
G ⫽ 12 ⫻ 106 psi
t⫽
t ⫽ constant
t ⫽ 0.2 in
4(0.2 in.)(18.850 in.2)2
4tA2m
⫽
Lm
15.867 in.
a ⫽ 3.0 in.
b ⫽ 2.0 in.
18 k-in.
T
⫽
2tAm
2(0.2 in.)(18.850 in.2)
⫽ 2390 psi
;
ANGLE OF TWIST PER UNIT LENGTH (RATE OF TWIST)
u⫽
f
T
18 k-in.
⫽
⫽
L
GJ
(12 * 106 psi)(17.92 in.)4
u ⫽ 83.73 * 10⫺6 rad/in. ⫽ 0.0048°/in.
Problem 3.10-8 A torque T is applied to a thin-walled tube having
a cross section in the shape of a regular hexagon with constant wall
thickness t and side length b (see figure).
Obtain formulas for the shear stress ␶ and the rate of twist ␪.
;
t
b
Solution 3.10-8
Regular hexagon
b ⫽ Length of side
t ⫽ Thickness
Lm ⫽ 6b
FROM APPENDIX D, CASE 25:
␤ ⫽ 60° n ⫽ 6
Am ⫽
⫽
b
nb2
6b2
cot ⫽
cot 30°
4
2
4
3 13b2
2
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SECTION 3.10
SHEAR STRESS
T
T13
t⫽
⫽
2tAm
9b2t
u⫽
;
Thin-Walled Tubes
2T
T
2T
⫽
⫽
3
GJ
G(9b t)
9Gb3t
333
;
(radians per unit length)
ANGLE OF TWIST PER UNIT LENGTH (RATE OF TWIST)
4A2mt
J⫽
Lm
L0
ds
t
⫽
4A2mt
9b3t
⫽
Lm
2
Problem 3.10-9 Compare the angle of twist ␾1 for a thin-walled circular tube
(see figure) calculated from the approximate theory for thin-walled bars with the
angle of twist ␾2 calculated from the exact theory of torsion for circular bars.
t
r
(a) Express the ratio ␾1/␾2 in terms of the nondimensional ratio ␤ ⫽ r/t.
(b) Calculate the ratio of angles of twist for ␤ ⫽ 5, 10, and 20. What conclusion
about the accuracy of the approximate theory do you draw from these results?
Solution 3.10-9
C
Thin-walled tube
(a) RATIO
f1
f2
⫽
4r 2 + t 2
Let b ⫽
APPROXIMATE THEORY
TL
f1 ⫽
GJ
J ⫽ 2␲r 3t
(b)
f1 ⫽
TL
GIP
2pGr3t
From Eq. (3-17): Ip ⫽
TL
2TL
f2 ⫽
⫽
GIP
pGrt(4r 2 + t 2)
r
t
f1
f2
t2
4r 2
1
⫽1 +
;
4b 2
␤
␾1/␾2
5
10
20
1.0100
1.0025
1.0006
TL
EXACT THEORY
f2 ⫽
4r 2
⫽1 +
prt
(4r 2 + t 2)
2
As the tube becomes thinner and ␤ becomes larger, the
ratio ␾1/␾2 approaches unity. Thus, the thinner the tube,
the more accurate the approximate theory becomes.
Problem 3.10-10 A thin-walled rectangular tube has uniform thickness t
and dimensions a ⫻ b to the median line of the cross section (see figure).
How does the shear stress in the tube vary with the ratio ␤ ⫽ a/b if
the total length Lm of the median line of the cross section and the torque T
remain constant?
From your results, show that the shear stress is smallest when the
tube is square (␤ ⫽ 1).
t
b
a
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Torsion
Solution 3.10-10
Rectangular tube
T, t, and Lm are constants.
Let k ⫽
2T
tL2m
⫽ constant t ⫽ k
(1 + b)2
b
t ⫽ thickness (constant)
a, b ⫽ dimensions of the tube
b⫽
a
b
t
a b
⫽4
k min
Lm ⫽ 2(a ⫹ b) ⫽ constant
T ⫽ constant
T
2tAm
tL2m
ALTERNATE SOLUTION
Am ⫽ ab ⫽ ␤b2
t⫽
Lm ⫽ 2b(1 ⫹ ␤) ⫽ constant
2T (1 + b)2
c
d
b
tL2m
2T b(2)(1 + b) ⫺ (1 + b)2(1)
dt
⫽ 2c
d ⫽0
db
tLm
b2
2
Lm
d
Am ⫽ b c
2(1 + b)
Lm
b⫽
2(1 + b)
Am ⫽
8T
From the graph, we see that ␶ is minimum when ␤ ⫽ 1
and the tube is square.
SHEAR STRESS
t⫽
tmin ⫽
or 2␤ (1 ⫹ ␤) ⫺ (1 ⫹ ␤)2 ⫽ 0
bL2m
⬖␤ ⫽ 1
2
4(1 + b)
T(4)(1 + b)2
2T(1 + b)2
T
⫽
⫽
t⫽
2
2tAm
2tbLm
tL2m b
;
Thus, the tube is square and ␶ is either a minimum or a
maximum. From the graph, we see that ␶ is a minimum.
Problem 3.10-11 A tubular aluminum bar (G ⫽ 4 ⫻ 106 psi) of square
cross section (see figure) with outer dimensions 2 in. ⫻ 2 in. must resist a
torque T ⫽ 3000 lb-in.
Calculate the minimum required wall thickness tmin if the allowable
shear stress is 4500 psi and the allowable rate of twist is 0.01 rad/ft.
t
2 in.
2 in.
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SECTION 3.10
Solution 3.10-11
335
Thin-Walled Tubes
Square aluminum tube
THICKNESS t BASED UPON SHEAR STRESS
t⫽
T
2tAm
tAm ⫽
T
2t
T
2t
t(b ⫺ t)2 ⫽
b ⫽ in. T ⫽ lb-in. ␶ ⫽ psi
UNITS: t ⫽ in.
t(2.0 in. ⫺ t)2 ⫽
3000 lb-in.
1
⫽ in.3
2(4500 psi)
3
3t(2 ⫺ t)2 ⫺ 1 ⫽ 0
Solve for t: t ⫽ 0.0915 in.
Outer dimensions:
2.0 in. ⫻ 2.0 in.
THICKNESS t BASED UPON RATE OF TWIST
G ⫽ 4 ⫻ 106 psi
T ⫽ 3000 lb-in.
u⫽
␶allow ⫽ 4500 psi
u allow ⫽ 0.01 rad/ft ⫽
0.01
rad/in.
12
T
T
⫽
GJ
Gt(b ⫺ t)3
3000 lb-in
t(2.0 in. ⫺ t)3 ⫽
6
(4 * 10 psi)(0.01/12 rad/in.)
9
⫽
10
⫽ 2.0 in.
Centerline dimension ⫽ b ⫺ t
10t(2 ⫺ t)3 ⫺ 9 ⫽ 0
Am ⫽ (b ⫺ t)2
Solve for t:
J⫽
Lm
Lm ⫽ 4(b ⫺ t)
4t(b ⫺ t)
⫽ t(b ⫺ t)3
4(b ⫺ t)
4
⫽
T
Gu
G ⫽ psi ␪ ⫽ rad/in.
UNITS: t ⫽ in.
Let b ⫽ outer dimension
4tA2m
t(b ⫺ t)3 ⫽
t ⫽ 0.140 in.
ANGLE OF TWIST GOVERNS
tmin ⫽ 0.140 in.
Problem 3.10-12 A thin tubular shaft of circular cross section
(see figure) with inside diameter 100 mm is subjected to a torque
of 5000 N ⭈ m.
If the allowable shear stress is 42 MPa, determine the required
wall thickness t by using (a) the approximate theory for a thin-walled
tube, and (b) the exact torsion theory for a circular bar.
;
100 mm
t
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CHAPTER 3 Torsion
Solution 3.10-12
Thin tube
(b) EXACT THEORY
T ⫽ 5,000 N ⭈ m
d1 ⫽ inner diameter ⫽ 100 mm
t⫽
Tr2
Ip
Ip ⫽
p 4
p
(r ⫺ r41) ⫽ [(50 + t)4 ⫺(50)4]
2 2
2
42 MPa ⫽
␶allow ⫽ 42 MPa
t is in millimeters.
(50 + t)4 ⫺ (50)4
(5000 N # m)(2)
⫽
50 + t
(p)(42 MPa)
r ⫽ Average radius
⫽ 50 mm +
t
2
⫽
r1 ⫽ Inner radius
t ⫽ 7.02 mm
r2 ⫽ Outer radius
⫽ 50 mm ⫹ t Am ⫽ ␲r2
(a) APPROXIMATE THEORY
T
T
T
⫽
⫽
2tAm
2t(pr 2)
2pr 2 t
5,000 N # m
42 MPa ⫽
t 2
2pa50 + b t
2
t⫽
or
5,000 N # m
t 2
5 * 106
b ⫽
⫽
mm3
2
2p(42 MPa)
84p
Solve for t:
t ⫽ 6.66 mm
5 * 106
mm3
21p
Solve for t:
⫽ 50 mm
t a50 +
(5,000 N # m)(50 + t)
p
[(50 + t)4 ⫺ (50)4]
2
;
;
The approximate result is 5% less than the exact
result. Thus, the approximate theory is nonconservative and should only be used for thin-walled tubes.
Sec_3.9-3.11.qxd
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SECTION 3.10
Problem 3.10-13 A long, thin-walled tapered tube AB of circular cross
section (see figure) is subjected to a torque T. The tube has length L and
constant wall thickness t. The diameter to the median lines of the cross
sections at the ends A and B are dA and dB, respectively.
Derive the following formula for the angle of twist of the tube:
f⫽
T
T
L
t
t
Hint: If the angle of taper is small, we may obtain approximate
results by applying the formulas for a thin-walled prismatic tube to a
differential element of the tapered tube and then integrating along the
axis of the tube.
Solution 3.10-13
B
A
2TL dA + dB
a 2 2 b
pGt
dAdB
337
Thin-Walled Tubes
dB
dA
Thin-walled tapered tube
For entire tube:
f⫽
4T
pGT L0
L
dx
3
dB ⫺ dA
cdA + a
bx d
L
From table of integrals (see Appendix C):
1
t ⫽ thickness
dx
(a + bx)3
⫽ ⫺
1
2b(a + bx)2
dA ⫽ average diameter at end A
dB ⫽ average diameter at end B
T ⫽ torque
d(x) ⫽ average diameter at distance x from end A.
d(x) ⫽ dA + a
J ⫽ 2pr 3t ⫽
dB ⫺ dA
bx
L
3
pd t
4
3
dB ⫺ dA
pt
pt
J(x) ⫽ [d(x)]3 ⫽ cdA + a
bx d
4
4
L
For element of length dx:
df ⫽
Tdx
⫽
GJ(x)
4Tdx
3
dB ⫺ dA
GptcdA + a
bx d
L
L
4T
f⫽
pGt
⫽
f⫽
J
1
⫺
2a
2
dB ⫺ dA
dB ⫺ dA
# xb K0
b adA +
L
L
4T
L
L
c⫺
+
d
pGt
2(dB ⫺ dA)d2B
2(dB ⫺ dA)d2A
2TL dA + dB
a 2 2 b
pGt
dAdB
;
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Torsion
Stress Concentrations in Torsion
D2
The problems for Section 3.11 are to be solved by considering the
stress-concentration factors.
Problem 3.11-1 A stepped shaft consisting of solid circular
D1
T
T
segments having diameters D1 ⫽ 2.0 in. and D2 ⫽ 2.4 in. (see figure)
is subjected to torques T. The radius of the fillet is R ⫽ 0.1 in.
If the allowable shear stress at the stress concentration is 6000 psi,
what is the maximum permissible torque Tmax?
Solution 3.11-1
R
Probs. 3.11-1 through 3.11-5
Stepped shaft in torsion
USE FIG. 3-48 FOR THE STRESS-CONCENTRATION FACTOR
D2
2.4 in.
⫽
⫽ 1.2
D1
2.0 in.
R
0.1 in.
⫽
⫽ 0.05
D1
2.0 in.
K ⬇ 1.52
tmax ⫽ Kt nom ⫽ K a
16 Tmax
pD31
b
pD31tmax
16K
p(2.0 in.)3(6000 psi)
⫽
⫽ 6200 lb-in.
16(1.52)
D1 ⫽ 2.0 in.
Tmax ⫽
D2 ⫽ 2.4 in.
R ⫽ 0.1 in.
␶allow ⫽ 6000 psi
⬖ Tmax ⬇ 6200 lb-in.
;
Problem 3.11-2 A stepped shaft with diameters D1 ⫽ 40 mm and D2 ⫽ 60 mm is loaded by torques T ⫽ 1100 N ⭈ m
(see figure).
If the allowable shear stress at the stress concentration is 120 MPa, what is the smallest radius Rmin that may be used for
the fillet?
Solution 3.11-2
Stepped shaft in torsion
USE FIG. 3-48 FOR THE STRESS-CONCENTRATION FACTOR
tmax ⫽ Kt nom ⫽ Ka
16T
pD31
b
p(40 mm)3(120 MPa)
pD31tmax
⫽
⫽ 1.37
16 T
16(1100 N # m)
D2
60 mm
⫽
⫽ 1.5
D1
40 mm
K⫽
D1 ⫽ 40 mm
D2 ⫽ 60 mm
T ⫽ 1100 N ⭈ m
From Fig. (3-48) with
␶allow ⫽ 120 MPa
we get
D2
⫽ 1.5 and K ⫽ 1.37,
D1
R
L 0.10
D1
⬖ Rmin ⬇ 0.10(40 mm) ⫽ 4.0 mm
;
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339
SECTION 3.11 Stress Concentrations in Torsion
Problem 3.11-3 A full quarter-circular fillet is used at the shoulder of a stepped shaft having diameter D2 ⫽ 1.0 in.
(see figure). A torque T ⫽ 500 lb-in. acts on the shaft.
Determine the shear stress ␶max at the stress concentration for values as follows: D1 5 0.7, 0.8, and 0.9 in. Plot a graph
showing ␶max versus D1.
Solution 3.11-3 Stepped shaft in torsion
D1 (in.)
D2/D1
R(in.)
R/D1
K
␶max(psi)
0.7
0.8
0.9
1.43
1.25
1.11
0.15
0.10
0.05
0.214
0.125
0.056
1.20
1.29
1.41
8900
6400
4900
D2 ⫽ 1.0 in.
T ⫽ 500 lb-in.
D1 ⫽ 0.7, 0.8, and 0.9 in.
Full quarter-circular fillet (D2 ⫽ D1 ⫹ 2R)
R⫽
D2 ⫺ D1
D1
⫽ 0.5 in. ⫺
2
2
USE FIG. 3-48 FOR THE STRESS-CONCENTRATION FACTOR
tmax ⫽ Kt nom ⫽ K a
⫽K
16 T
pD31
16(500 lb-in.)
pD31
b
⫽ 2546
K
D31
NOTE that ␶max gets smaller as D1 gets larger, even
though K is increasing.
Problem 3.11-4 The stepped shaft shown in the figure is required to transmit 600 kW of power at 400 rpm.
The shaft has a full quarter-circular fillet, and the smaller diameter D1 ⫽ 100 mm.
If the allowable shear stress at the stress concentration is 100 MPa, at what diameter D2 will this stress be reached?
Is this diameter an upper or a lower limit on the value of D2?
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CHAPTER 3 Torsion
Solution 3.11-4 Stepped shaft in torsion
P ⫽ 600 kW
D1 ⫽ 100 mm
n ⫽ 400 rpm ␶allow ⫽ 100 MPa
Use the dashed line for a full quarter-circular fillet.
R
L 0.075
D1
R ⬇ 0.075 D1 ⫽ 0.075 (100 mm)
Full quarter-circular fillet
⫽ 7.5 mm
2pnT
( Eq. 3-42 of Section 3.7)
POWER P ⫽
60
P ⫽ watts
n ⫽ rpm
T ⫽ Newton meters
60(600 * 103 W)
60P
⫽
⫽ 14,320 N # m
T⫽
2pn
2p(400 rpm)
D2 ⫽ D1 ⫹ 2R ⫽ 100 mm ⫹ 2(7.5 mm) ⫽ 115 mm
⬖ D2 ⬇ 115 mm
;
This value of D2 is a lower limit
;
(If D2 is less than 115 mm, R/D1 is smaller, K is larger,
and ␶max is larger, which means that the allowable stress
is exceeded.)
USE FIG. 3-48 FOR THE STRESS-CONCENTRATION FACTOR
tmax ⫽ Kt nom ⫽ K a
K⫽
⫽
16T
pD31
b
tmax(pD31)
16T
(100 MPa)(p)(100 mm)3
⫽ 1.37
16(14,320 N # m)
Problem 3.11-5 A stepped shaft (see figure) has diameter D2 ⫽ 1.5 in. and a full quarter-circular fillet. The allowable shear
stress is 15,000 psi and the load T ⫽ 4800 lb-in.
What is the smallest permissible diameter D1?
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Page 341
SECTION 3.11 Stress Concentrations in Torsion
341
Solution 3.11-5 Stepped shaft in torsion
Use trial-and-error. Select trial values of D1
D2 ⫽ 1.5 in.
␶allow ⫽ 15,000 psi
T ⫽ 4800 lb-in.
Full quarter-circular fillet D2 ⫽ D1 ⫹ 2R
R⫽
D1
D2 ⫺ D1
⫽ 0.75 in. ⫺
2
2
D1 (in.)
R (in.)
R/D1
K
␶max(psi)
1.30
1.35
1.40
0.100
0.075
0.050
0.077
0.056
0.036
1.38
1.41
1.46
15,400
14,000
13,000
USE FIG. 3-48 FOR THE STRESS-CONCENTRATION FACTOR
tmax ⫽ Kt nom ⫽ K a
16T
pD31
K 16(4800 lb-in.)
⫽ 3c
d
p
D1
⫽ 24,450
b
K
D31
From the graph, minimum D1 ⬇ 1.31 in.
;
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04Ch04.qxd
9/24/08
4:19 AM
Page 343
4
Shear Forces and
Bending Moments
Shear Forces and Bending Moments
800 lb
1600 lb
Problem 4.3-1 Calculate the shear force V and bending moment M at a cross
section just to the left of the 1600-1b load acting on the simple beam AB shown
in the figure.
A
B
30 in.
50 in.
120 in.
40 in.
Solution 4.3-1
gMA ⫽ 0: RB ⫽
3800
⫽ 1267 lb
3
3400
gMB ⫽ 0: RA ⫽
⫽ 1133 lb
3
FREE-BODY DIAGRAM OF SEGMENT DB
gFVERT ⫽ 0: V ⫽ 1600 lb ⫺ 1267 lb
⫽ 333 lb
;
g MD ⫽ 0: M ⫽ 11267 lb2(40 in.)
⫽
152000 #
lb in ⫽ 50667 lb # in.
3
;
1600 lb
D
B
40 in
RB
343
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4:19 AM
CHAPTER 4
Page 344
Shear Forces and Bending Moments
Problem 4.3-2 Determine the shear force V and bending moment M at the
6.0 kN
midpoint C of the simple beam AB shown in the figure.
2.0 kN/m
C
A
B
0.5 m
1.0 m
2.0 m
4.0 m
1.0 m
Solution 4.3-2
FREE-BODY DIAGRAM OF SEGMENT AC
g MA ⫽ 0:
RB ⫽ 3.9375 kN
g MB ⫽ 0:
RA ⫽ 5.0625 kN
g FVERT ⫽ 0:
V ⫽ RA ⫺ 6 ⫽ ⫺0.938 kN
g MC ⫽ 0:
M ⫽ RA ⭈ 2 m ⫺ 6 kN ⭈ 1 m
⫽ 4.12 kN⭈m
;
Problem 4.3-3 Determine the shear force V and bending moment M at the
Pb
P
midpoint of the beam with overhangs (see figure). Note that one load acts
downward and the other upward. Also clockwise moments Pb are applied
at each support.
;
Pb
b
L
P
b
Solution 4.3-3
Pb
Pb
Pb
FREE-BODY DIAGRAM (C IS THE MIDPOINT)
1
(2Pb ⫺ (b + L)P ⫺ Pb)
L
⫽ P (upward)
gMB ⫽ 0: RA ⫽
g MA ⫽ 0:
g FVERT ⫽ 0:
V ⫽ RA ⫺ P ⫽ 0
g MC ⫽ 0: M ⫽ ⫺Pa b +
RB ⫽ P (downward)
+ RA
;
L
b
2
L
+ Pb ⫽ 0
2
;
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SECTION 4.3
Problem 4.3-4 Calculate the shear force V and bending moment M at a cross
4.0 kN
section located 0.5 m from the fixed support of the cantilever beam AB shown
in the figure.
B
1.0 m
2.0 m
Cantilever beam
4.0 kN
g FVERT ⫽ 0:
1.5 kN/m
A
B
V ⫽ 4.0 kN ⫹ (1.5 kN/m)(2.0 m)
⫽ 4.0 kN ⫹ 3.0 kN ⫽ 7.0 kN
1.0 m
1.5 kN/m
A
1.0 m
Solution 4.3-4
345
Shear Forces and Bending Moments
1.0 m
2.0 m
g MD ⫽ 0:
;
M ⫽ ⫺(4.0 kN)(0.5 m)
FREE-BODY DIAGRAM OF SEGMENT DB
⫺ (1.5 kN/m)(2.0 m)(2.5 m)
⫽ ⫺2.0 kN ⭈ m ⫺ 7.5 kN ⭈ m
Point D is 0.5 m from support A.
⫽ ⫺9.5 kN ⭈ m
Problem 4.3-5 Determine the shear force V and bending moment M at a cross
section located 18 ft from the left-hand end A of the beam with an overhang
shown in the figure.
;
400 lb/ft
300 lb/ft
B
A
10 ft
10 ft
C
6 ft
6 ft
Solution 4.3-5
FREE-BODY DIAGRAM OF SEGMENT AD
g MB ⫽ 0:
RA ⫽ 2190 lb
g MA ⫽ 0:
RB ⫽ 3610 lb
Point D is 18ft from support A.
g FVERT ⫽ 0:
V ⫽ 2190 lb ⫺ (400 lb/ft)(10 ft)
;
⫽ ⫺1810 lb
g Mc ⫽ 0: M ⫽ (2190 lb)(18 ft)
⫺ (400 lb/ft)(10 ft)(13 ft)
⫽ ⫺12580 lb⭈ ft
;
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Shear Forces and Bending Moments
Problem 4.3-6 The beam ABC shown in the figure is simply
supported at A and B and has an overhang from B to C. The
loads consist of a horizontal force P1 ⫽ 4.0 kN acting at the
end of a vertical arm and a vertical force P2 ⫽ 8.0 kN acting
at the end of the overhang.
Determine the shear force V and bending moment M at a
cross section located 3.0 m from the left-hand support.
(Note: Disregard the widths of the beam and vertical arm and
use centerline dimensions when making calculations.)
Solution 4.3-6
P1 = 4.0 kN
P2 = 8.0 kN
1.0 m
A
B
4.0 m
C
1.0 m
Beam with vertical arm
FREE-BODY DIAGRAM OF SEGMENT AD
Point D is 3.0 m from support A.
g MB ⫽ 0:
RA ⫽ 1.0 kN (downward)
g MA ⫽ 0:
RB ⫽ 9.0 kN (upward)
g FVERT ⫽ 0:
V ⫽ ⫺RA ⫽ ⫺1.0 kN
g MD
M ⫽ ⫺RA(3.0 m) ⫺ 4.0 kN ⭈ m
⫽ ⫺7.0 kN ⭈ m
;
⫽ 0:
Problem 4.3-7 The beam ABCD shown in the figure has overhangs at each
end and carries a uniform load of intensity q.
For what ratio b/L will the bending moment at the midpoint of the beam
be zero?
;
q
A
D
B
b
C
L
b
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SECTION 4.3
Solution 4.3-7
347
Shear Forces and Bending Moments
Beam with overhangs
FREE-BODY DIAGRAM OF LEFT-HAND HALF OF BEAM:
Point E is at the midpoint of the beam.
From symmetry and equilibrium of vertical forces:
L
RB ⫽ RC ⫽ q ab + b
2
gME ⫽ 0 哵 哴
1
L 2
L
⫺ RB a b + qa b ab + b ⫽ 0
2
2
2
L L
1
L 2
b a b + qa b ab + b ⫽ 0
2
2
2
2
⫺ qa b +
Solve for b/L:
b
1
⫽
L
2
;
Problem 4.3-8 At full draw, an archer applies a pull of 130 N to the bowstring
of the bow shown in the figure. Determine the bending moment at the midpoint
of the bow.
70°
1400 mm
350 mm
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Solution 4.3-8
Page 348
Shear Forces and Bending Moments
Archer’s bow
FREE-BODY DIAGRAM OF SEGMENT BC
g MC ⫽ 0
T(cos b)a
M⫽Ta
P ⫽ 130 N
␤ ⫽ 70°
⫽
H ⫽ 1400 mm
H
b + T(sin b)(b) ⫺ M ⫽ 0
2
H
cos b + b sin b b
2
P H
a + b tan b b
2 2
SUBSTITUTE NUMERICAL VALUES:
⫽ 1.4 m
b ⫽ 350 mm
M⫽
⫽ 0.35 m
FREE-BODY DIAGRAM OF POINT A
T ⫽ tensile force in the bowstring
g FHORIZ ⫽ 0:
哵哴
2T cos ␤ ⫺ P ⫽ 0
T⫽
P
2 cos b
130 N 1.4 m
c
+ (0.35 m)(tan 70°) d
2
2
M ⫽ 108 N # m
;
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SECTION 4.3
Problem 4.3-9 A curved bar ABC is subjected to loads in the form of
two equal and opposite forces P, as shown in the figure. The axis of the
bar forms a semicircle of radius r.
Determine the axial force N, shear force V, and bending moment M
acting at a cross section defined by the angle ␪.
Solution 4.3-9
349
Shear Forces and Bending Moments
M
B
P
A
V
r
u
O
N
P
P
C
u
A
Curved bar
g FN ⫽ 0
Q⫹ b⫺
N ⫺ P sin ␪ ⫽ 0
N ⫽ P sin ␪
g FV ⫽ 0
⫹
R a
⫺
V ⫺ P cos ␪ ⫽ 0
V ⫽ P cos ␪
g MO ⫽ 0
哵哴
;
;
M ⫺ Nr ⫽ 0
M ⫽ Nr ⫽ Pr sin ␪
;
Problem 4.3-10 Under cruising conditions the distributed
load acting on the wing of a small airplane has the idealized
variation shown in the figure.
Calculate the shear force V and bending moment M at the
inboard end of the wing.
1600 N/m
2.6 m
Solution 4.3-10
900 N/m
2.6 m
1.0 m
Airplane wing
(Minus means the shear force acts opposite to the direction shown in the figure.)
LOADING (IN THREE PARTS)
SHEAR FORCE
g FVERT ⫽ 0
V +
+
c⫹ T⫺
1
(700 N/m)(2.6 m) + (900 N/m)(5.2 m)
2
1
1900 N/m2(1.0 m) ⫽ 0
2
V ⫽ ⫺6040 N ⫽ ⫺6.04 kN
;
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Shear Forces and Bending Moments
BENDING MOMENT
M ⫽ 788.67 N ⭈ m ⫹ 12,168 N ⭈ m ⫹ 2490 N ⭈ m
g MA ⫽ 0 哵哴
⫽ 15,450 N ⭈ m
1
2.6 m
(700 N/m) (2.6 m) a
b
2
3
+ (900 N/m) (5.2 m) (2.6 m)
1
1.0 m
+ (900 N/m) (1.0 m) a 5.2 m +
b ⫽0
2
3
⫽ 15.45 kN ⭈ m
;
⫺M +
Problem 4.3-11 A beam ABCD with a vertical arm CE is supported as
E
a simple beam at A and D (see figure). A cable passes over a small pulley
that is attached to the arm at E. One end of the cable is attached to the
beam at point B.
What is the force P in the cable if the bending moment in the
beam just to the left of point C is equal numerically to 640 lb-ft?
(Note: Disregard the widths of the beam and vertical arm and use
centerline dimensions when making calculations.)
Cable
A
B
8 ft
C
6 ft
Solution 4.3-11
6 ft
Beam with a cable
FREE-BODY DIAGRAM OF SECTION AC
g MC ⫽ 0
UNITS:
P in lb
M in lb-ft
P
哵哴
4P
4P
(6 ft) +
(12 ft) ⫽ 0
5
9
8P
M⫽ ⫺
lb-ft
15
M⫺
Numerical value of M equals 640 lb-ft.
8P
lb-ft
15
and P ⫽ 1200 lb
‹ 640 lb-ft ⫽
;
D
6 ft
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SECTION 4.3
351
Shear Forces and Bending Moments
Problem 4.3-12 A simply supported beam AB supports a trapezoidally
distributed load (see figure). The intensity of the load varies linearly from
50 kN/m at support A to 25 kN/m at support B.
Calculate the shear force V and bending moment M at the midpoint of
the beam.
50 kN/m
25 kN/m
A
B
4m
Solution 4.3-12
FREE-BODY DIAGRAM OF SECTION CB
Point C is at the midpoint of the beam.
g MB ⫽ 0:
⫺RA (4m) + (25 kN/m) (4m) (2m)
2
1
⫹(25 kN/m)(4 m)a b a 4 m b ⫽ 0
2
3
RA ⫽ 83.33 kN
g FVERT ⫽ 0: RA + RB
1
⫺ (50 kN/m + 25 kN/m)(4 m) ⫽ 0
2
RB ⫽ 66.67 kN
g FVERT ⫽ 0: V ⫺ (25 kN/m)(2 m)
⫺ (12.5 kN/m)(2 m)
V ⫽ ⫺4.17 kN
gMC ⫽ 0:
1
+ RB ⫽ 0
2
;
⫺ M ⫺ (25 kN/m)(2 m)(1 m)
⫺ (12.5 kN/m)(2 m)
1
1
a2 m b
2
3
+ RB (2 m) ⫽ 0
M ⫽ 75 kN ⭈ m
;
Problem 4.3-13 Beam ABCD represents a reinforced-concrete foundation beam
that supports a uniform load of intensity q1 ⫽ 3500 lb/ft (see figure). Assume that
the soil pressure on the underside of the beam is uniformly distributed with
intensity q2.
(a) Find the shear force VB and bending moment MB at point B.
(b) Find the shear force Vm and bending moment Mm at the midpoint
of the beam.
q1 = 3500 lb/ft
B
C
A
D
3.0 ft
q2
8.0 ft
3.0 ft
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CHAPTER 4
Solution 4.3-13
Page 352
Shear Forces and Bending Moments
Foundation beam
(b) V AND M AT MIDPOINT E
g FVERT ⫽ 0:
‹ q2 ⫽
q2(14 ft) ⫽ q1(8 ft)
8
q ⫽ 2000 lb/ft
14 1
g FVERT ⫽ 0:
(a) V AND M AT POINT B
Vm ⫽ 0
a FVERT
;
g ME ⫽ 0:
⫽ 0:
VB ⫽ 6000 lb
g MB ⫽ 0:
Vm ⫽ (2000 lb/ft)(7 ft)
⫺ (3500 lb/ft)(4 ft)
Mm ⫽ (2000 lb/ft)(7 ft)(3.5 ft)
⫺ (3500 lb/ft)(4 ft)(2 ft)
;
Mm ⫽ 21,000 lb-ft
MB ⫽ 9000 lb-ft
;
;
Problem 4.3-14 The simply-supported beam ABCD is loaded by
a weight W ⫽ 27 kN through the arrangement shown in the figure.
The cable passes over a small frictionless pulley at B and is attached
at E to the end of the vertical arm.
Calculate the axial force N, shear force V, and bending moment M
at section C, which is just to the left of the vertical arm.
(Note: Disregard the widths of the beam and vertical arm and use
centerline dimensions when making calculations.)
E
Cable
1.5 m
A
B
2.0 m
C
2.0 m
W = 27 kN
D
2.0 m
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SECTION 4.3
Solution 4.3-14
Shear Forces and Bending Moments
353
Beam with cable and weight
FREE-BODY DIAGRAM OF PULLEY AT B
RA ⫽ 18 kN
RD ⫽ 9 kN
FREE-BODY DIAGRAM OF SEGMENT ABC OF BEAM
g FHORIZ ⫽ 0:
g FVERT ⫽ 0:
g MC ⫽ 0:
N ⫽ 21.6 kN (compression)
V ⫽ 7.2 kN
;
;
M ⫽ 50.4 kN ⭈ m
;
y
Problem 4.3-15 The centrifuge shown in the figure rotates in a horizontal plane
(the xy plane) on a smooth surface about the z axis (which is vertical) with an
angular acceleration ␣. Each of the two arms has weight w per unit length and
supports a weight W ⫽ 2.0 wL at its end.
Derive formulas for the maximum shear force and maximum bending moment
in the arms, assuming b ⫽ L/9 and c ⫽ L/10.
c
L
b
W
x
W
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Shear Forces and Bending Moments
Solution 4.3-15
Rotating centrifuge
SUBSTITUTE NUMERICAL DATA:
Tangential acceleration ⫽ r␣
Inertial force Mr a ⫽
W ⫽ 2.0 wL b ⫽
W
ra
g
Maximum V and M occur at x ⫽ b.
W
(L + b + c)a +
g
Lb
Wa
(L + b + c)
⫽
g
L⫹b
Vmax ⫽
+
Mmax ⫽
wLa
(L + 2b)
2g
;
Wa
(L + b + c)(L + c)
g
L⫹b
+
wa
x(x ⫺ b)dx
g
Lb
Wa
(L + b + c)(L + c)
⫽
g
+
wL2a
(2L + 3b)
6g
;
wa
x dx
g
Vmax ⫽
91wL2a
30g
Mmax ⫽
229wL3a
75g
L
L
c⫽
9
10
;
;
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SECTION 4.5
355
Shear-Force and Bending-Moment Diagrams
Shear-Force and Bending-Moment Diagrams
When solving the problems for Section 4.5, draw the shear-force
and bending-moment diagrams approximately to scale and label all
critical ordinates, including the maximum and minimum values.
Probs. 4.5-1 through 4.5-10 are symbolic problems and
Probs. 4.5-11 through 4.5-24 are numerical problems. The remaining
problems (4.5-25 through 4.5-40) involve specialized topics, such
as optimization, beams with hinges, and moving loads.
P
a
P
a
A
B
Problem 4.5-1 Draw the shear-force and bending-moment diagrams for
a simple beam AB supporting two equal concentrated loads P (see figure).
L
Solution 4.5-1
Simple beam
Problem 4.5-2 A simple beam AB is subjected to a counterclockwise
couple of moment M0 acting at distance a from the left-hand support
(see figure).
Draw the shear-force and bending-moment diagrams for this beam.
M0
A
B
a
L
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CHAPTER 4
Solution 4.5-2
Page 356
Shear Forces and Bending Moments
Simple beam
q
Problem 4.5-3 Draw the shear-force and bending-moment diagrams for
A
a cantilever beam AB carrying a uniform load of intensity q over one-half of
its length (see figure).
B
L
—
2
Solution 4.5-3
Cantilever beam
3qL2
MA = —
8
q
A
B
RA =
qL
2
L
—
2
—
L
—
2
qL
2
—
V
0
M
0
qL2
–—
8
3qL2
–—
8
qL
2
—
L
—
2
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SECTION 4.5
357
Shear-Force and Bending-Moment Diagrams
Problem 4.5-4 The cantilever beam AB shown in the figure
is subjected to a concentrated load P at the midpoint and
a counterclockwise couple of moment M1 PL/4 at
the free end.
Draw the shear-force and bending-moment diagrams for this
beam.
Solution 4.5-4
PL
M1 = —–
4
P
A
B
L
—
2
L
—
2
Cantilever beam
RA P
MA to a concentrated load P and a clockwise couple M1 PL/3 acting at the
third points.
Draw the shear-force and bending-moment diagrams for this beam.
A
B
L
—
3
Solution 4.5-5
PL
M1 = —–
3
P
A
B
L
—
3
P
RA= —–
3
L
—
3
L
—
3
2P
RB= —–
3
P/3
V
0
Vmax = –2P/3
PL/9
Mmax = 2PL/9
M 0
–PL/9
PL
M1 = —–
3
P
Problem 4.5-5 The simple beam AB shown in the figure is subjected
PL
4
L
—
3
L
—
3
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CHAPTER 4
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Shear Forces and Bending Moments
Problem 4.5-6 A simple beam AB subjected to couples M1 and 3M1
M1
acting at the third points is shown in the figure.
Draw the shear-force and bending-moment diagrams for this beam.
3M1
A
B
L
—
3
L
—
3
L
—
3
Solution 4.5-6
M1
3M1
A
B
L
—
3
L
—
3
L
—
3
RA
V
RB
2M 1
L
0
7M1
3
5M 1
3
M
0
2M 1
3
2M 1
3
Problem 4.5-7 A simply supported beam ABC is loaded by a vertical
load P acting at the end of a bracket BDE (see figure).
Draw the shear-force and bending-moment diagrams for beam ABC.
B
A
C
D
E
P
L
—
4
L
—
4
L
—
2
L
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SECTION 4.5
Solution 4.5-7
Beam with bracket
P
Problem 4.5-8 A beam ABC is simply supported at A and B
and has an overhang BC (see figure). The beam is loaded by two
forces P and a clockwise couple of moment Pa that act through
the arrangement shown.
Draw the shear-force and bending-moment diagrams for
beam ABC.
Solution 4.5-8
359
Shear-Force and Bending-Moment Diagrams
Beam with overhang
P
A
Pa
C
B
a
a
a
a
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Shear Forces and Bending Moments
Problem 4.5-9 Beam ABCD is simply supported at B and C and has
overhangs at each end (see figure). The span length is L and each overhang
has length L/3. A uniform load of intensity q acts along the entire length of
the beam.
Draw the shear-force and bending-moment diagrams for this beam.
q
A
D
B
L
3
Solution 4.5-9
C
L
3
L
Beam with overhangs
x1 L
15
0.3727L
6
q0
Problem 4.5-10 Draw the shear-force and bending-moment diagrams
for a cantilever beam AB supporting a linearly varying load of maximum
intensity q0 (see figure).
A
B
L
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SECTION 4.5
Solution 4.5-10
Shear-Force and Bending-Moment Diagrams
Cantilever beam
q0 = 10 lb/in.
Problem 4.5-11 The simple beam AB supports a triangular load
of maximum intensity q0 10 lb/in. acting over one-half of the
span and a concentrated load P 80 lb acting at midspan
(see figure).
Draw the shear-force and bending-moment diagrams for this
beam.
Solution 4.5-11
361
P = 80 lb
A
B
L =
—
40 in.
2
L =
—
40 in.
2
Simple beam
q0 = 10 lb/in.
P = 80 lb
MA 0: RB (80 in.) (80 lb)(40 in.)
A
1
2
(10 lb/in. )140 in.2(40 + 40 in.) 0
2
3
L =
—
40 in.
2
RB 206.7 lb
1
g FVERT 0: RA + RB 80 lb a10 lb/in. b(40 in.) 0
2
RA 73.3 lb
B
L =
—
40 in.
2
RA
RB
73.3 lb
V
0
–6.67 lb
–207 lb
2933 lb-in
M 0
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Shear Forces and Bending Moments
Problem 4.5-12 The beam AB shown in the figure supports a uniform load
3000 N/m
of intensity 3000 N/m acting over half the length of the beam. The beam rests on
a foundation that produces a uniformly distributed load over the entire length.
Draw the shear-force and bending-moment diagrams for this beam.
A
B
0.8 m
Solution 4.5-12
1.6 m
0.8 m
Beam with distributed loads
200 lb
Problem 4.5-13 A cantilever beam AB supports a couple and a concentrated
load, as shown in the figure.
Draw the shear-force and bending-moment diagrams for this beam.
400 lb-ft
A
B
5 ft
5 ft
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SECTION 4.5
Solution 4.5-13
Shear-Force and Bending-Moment Diagrams
363
Cantilever beam
Problem 4.5-14 The cantilever beam AB shown in the figure is
2.0 kN/m
subjected to a triangular load acting throughout one-half of its length
and a concentrated load acting at the free end.
Draw the shear-force and bending-moment diagrams for this beam.
2.5 kN
B
A
2m
Solution 4.5-14
4.5 kN
V
2.5 kN
2.5 kN
0
M 0
0
–5 kN • m
–11.33 kN • m
2m
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Shear Forces and Bending Moments
Problem 4.5-15 The uniformly loaded beam ABC has simple supports
25 lb/in.
at A and B and an overhang BC (see figure).
A
Draw the shear-force and bending-moment diagrams for this beam.
C
B
72 in.
Solution 4.5-15
Beam with an overhang
a uniform load of intensity 12 kN/m and a concentrated moment of
magnitude 3 kN # m at C (see figure).
Draw the shear-force and bending-moment diagrams for this beam.
A
Beam with an overhang
3 kN • m
12 kN/m
C
B
1.6 m
1.6 m
RA
1.6 m
RB
15.34 kN
V
0 kN
0
–3.86 kN
max 9.80 kN • m 9.18 kN • m
3 kN • m
M 0
1.28 m
C
B
1.6 m
A
3 kN • m
12 kN/m
Problem 4.5-16 A beam ABC with an overhang at one end supports
Solution 4.5-16
48 in.
1.6 m
1.6 m
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SECTION 4.5
365
Shear-Force and Bending-Moment Diagrams
Problem 4.5-17 Consider the two beams below; they are loaded the same but have different support conditions. Which
beam has the larger maximum moment?
First, find support reactions, then plot axial force (N), shear (V) and moment (M) diagrams for both beams. Label all
critical N, V & M values and also the distance to points where N, V &/or M is zero.
PL
L
—
2
A
B
L
—
2
L
—
4
C
P
4 L
—
4
3
D
PL
Ay
Ax
Cy
(a)
PL
A
L
—
2
B
L
—
2
L
—
4
P
4 L
—
4
3
PL
Cy
Dy
Dx
(b)
Solution 4.5-17
BEAM (a):
g MA 0: Cy 0
N 0
1 4 5
a P Lb P (upward)
L 5 4
g FV 0: Ay 4
P
P Cy (downward)
5
5
g FH 0: Ax 3
P (right)
5
–3P/5(compression)
4P/5
0
V 0
–P/5
0
M 0
–PL/10
–11PL/10
–PL/5
–6PL/5
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Shear Forces and Bending Moments
BEAM (b):
3P/5
g MD 0: Cy 2
2 4 1
a P Lb P (upward)
L 5 4
5
g FV 0: Dy 4
2
P Cy P (upward)
5
5
N
0
2P/5
V
0
3
g FH 0: Dx P (right)
5
–2P/5
⬖ The first case has the larger maximum moment
6
a PL b
5
PL/10
M
0
;
–PL
Problem 4.5-18 The three beams below are loaded the same and have the same support conditions. However, one has a
moment release just to the left of C, the second has a shear release just to the right of C, and the third has an axial release
just to the left of C. Which beam has the largest maximum moment?
First, find support reactions, then plot axial force (N), shear (V) and moment (M) diagrams for all three beams. Label all
critical N, V & M values and also the distance to points where N, V &/or M is zero.
PL at C
A
L
—
2
B
L
—
2
L
—
4
C
P
3
4 L
—
4 D
PL at B
Moment
release
Ax
Ay
Cy
Dy
(a)
PL at C
A
L
—
2
B
L
—
2
L
—
4
C
PL at B
Ax
Ay
P
4 L
—
4 D
3
Shear
release
Cy
Dv
(b)
PL at C
A
L
—
2
B
PL at B
Ay
Ax
L
—
2
Axial force
release
(c)
C
L P 4 L
—
—
4
4
3
Cx
Cy
0
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SECTION 4.5
367
Shear-Force and Bending-Moment Diagrams
Solution 4.5-18
BEAM (a): MOMENT RELEASE
N
0
0
Ay P (upward)
Cy Dy –3P/5 (compression)
13
P (downward)
5
12
P (upward)
5
P
V
0
–8P/5
3
Ax P (right)
5
PL/2
M
PL
–12P/5
3PL/5
0
–PL/2
BEAM (b): SHEAR RELEASE
Ay N
0
0
1
P (upward)
5
1
Cy P (downward)
5
Dy 4
P (upward)
5
Ax 3
P (right)
5
–3P/5 (compression)
P/5
V
0
–4P/5
M
PL/10
0
–9PL/10
BEAM (c): AXIAL RELEASE
N
–3P/5 (compression)
4P/5
V
0
M
0
Ax 0
Cx 3
P (right)
5
⬖ The third case has the largest maximum moment
6
a PLb
5
;
–4PL/5
0
1
Ay P (downward)
5
Cy P (upward)
PL/5
–P/5
–PL/10
–11PL/10
–PL/5
–6PL/5
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Shear Forces and Bending Moments
Problem 4.5-19 A beam ABCD shown in the figure is simply
supported at A and B and has an overhang from B to C. The loads
consist of a horizonatal force P1 400 lb acting at the end of the
vertical arm and a vertical force P2 900 lb acting at the end of
the overhang.
Draw the shear-force and bending-moment diagrams for this
beam. (Note: Disregard the widths of the beam and vertical arm
and use centerline dimensions when making calculations.)
Solution 4.5-19
Beam with vertical arm
Problem 4.5-20 A simple beam AB is loaded by two segments
of uniform load and two horizontal forces acting at the ends of a vertical arm
(see figure).
Draw the shear-force and bending-moment diagrams for this beam.
Solution 4.5-20
Simple beam
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SECTION 4.5
369
Shear-Force and Bending-Moment Diagrams
Problem 4.5-21 The two beams below are loaded the same and have the same support conditions. However, the location of
internal axial, shear and moment releases is different for each beam (see figures). Which beam has the larger maximum
moment?
First, find support reactions, then plot axial force (N), shear (V) and moment (M) diagrams for both beams. Label all
critical N, V & M values and also the distance to points where N, V &/or M is zero.
MAz
PL
A
L
—
2
B
L
—
2
L
—
4
C
P
3
4 L
—
4 D
Ax
PL
Axial force
release
Ay
Shear
release
Moment
release
Cy
Dy
Dx
(a)
MAz
PL
A
L
—
2
B
L
—
2
L
—
4
C
P
3
4 L
—
4 D
Ax
PL
Ay
Shear
release
Axial force
release
Moment
release
Cy
Dy
Dx
(b)
Solution 4.5-21
Support reactions for both beams:
MAz 0, Ax 0, Ay 0
Cy Dx 3P/5(tension)
N
0
2
2
P ( upward), Dy P ( upward)
5
5
3
P ( rightward)
5
2P/ 5
V 0
–2P/ 5
⬖ These two cases have the same maximum
moment (PL) ;
(Both beams have the same N, V and M diagrams)
–PL/10
M 0
–PL
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CHAPTER 4
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Shear Forces and Bending Moments
Problem 4.5-22 The beam ABCD shown in the figure has
overhangs that extend in both directions for a distance of 4.2 m
from the supports at B and C, whiich are 1.2 m apart.
Draw the shear-force and bending-moment diagrams for this
overhanging beam.
10.6 kN/m
5.1 kN/m
5.1 kN/m
A
D
B
C
4.2 m
4.2 m
1.2 m
Solution 4.5-22
Beam with overhangs
Problem 4.5-23 A beam ABCD with a vertical arm CE is supported as a
simple beam at A and B (see figure). A cable passes over a small pulley
that is attached to the arm at E. One end of the cable is attached to the
beam at point B. The tensile force in the cable is 1800 lb.
Draw the shear-force and bending-moment diagrams for beam ABCD.
(Note: Disregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.)
E
1800 lb
Cable
A
B
6 ft
8 ft
C
6 ft
D
6 ft
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SECTION 4.5
Solution 4.5-23
Shear-Force and Bending-Moment Diagrams
371
Beam with a cable
Note: All forces have units of pounds.
Problem 4.5-24 Beams ABC and CD are supported
MAz
at A, C and D, and are joined by a hinge (or moment release)
just to the left of C and a shear release just to the right of C.
The support at A is a sliding support (hence reaction
Ay 0 for the loading shown below). Find all support
reactions then plot shear (V) and moment (M) diagrams.
Label all critical V & M values and also the distance to
points where either V &/or M is zero.
W0 = P/L
A
L
—
2
L
—
B 2
C
Ax
L
—
2
PL
Ay
Sliding
support
Moment
release
Cy
Dy
Solution 4.5-24
MAz PL (clockwise), Ax 0, Ay 0
;
1
1
P (upward), Dy P (upward)
12
6
;
Cy VMAX P
MMAX PL
6
P/12
V 0
0.289L
–P/6
PL
;
M 0
0.016PL
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CHAPTER 4
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Shear Forces and Bending Moments
Problem 4.5-25 The simple beam AB shown in the figure supports a
5k
concentrated load and a segment of uniform load.
Draw the shear-force and bending-moment diagrams for this beam.
2.0 k/ft
C
A
5 ft
B
10 ft
20 ft
Solution 4.5-25
Simple beam
5k
2.0 k/ft
C
A
B
5 ft
10 ft
20 ft
6.25
1.25
V 0
(Kips)
3.125
–13.75
37.5
M 0
(k-ft)
max 47.3
6.25
Problem 4.5-26 The cantilever beam shown in the figure supports a concentrated
load and a segment of uniform load.
Draw the shear-force and bending-moment diagrams for this cantilever beam.
3 kN
1.0 kN/m
A
0.8 m
B
0.8 m
1.6 m
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SECTION 4.5
Solution 4.5-26
Shear-Force and Bending-Moment Diagrams
373
Cantilever beam
Problem 4.5-27 The simple beam ACB shown in the figure is subjected to
180 lb/ft
a triangular load of maximum intensity 180 lb/ft and a concentrated moment of
300 lb-ft at A.
300 lb-ft
Draw the shear-force and bending-moment diagrams for this beam.
A
B
C
6.0 ft
7.0 ft
Solution 4.5-27
Simple beam
180 lb/ft
300 lb-ft
A
B
C
6.0 ft
7.0 ft
197.1
V 0
(lb)
0 lb
3.625 ft
max 776
300
M
0
(lb-ft)
–343
–433
403
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Shear Forces and Bending Moments
Problem 4.5-28 A beam with simple supports is subjected to a trapezoidally
3.0 kN/m
1.0 kN/m
distributed load (see figure). The intensity of the load varies from 1.0 kN/m
at support A to 3.0 kN/m at support B.
Draw the shear-force and bending-moment diagrams for this beam.
A
B
2.4 m
Solution 4.5-28
V 2.0 x Set V 0:
Simple beam
x2
2.4
(x meters; V kN)
x1 1.2980 m
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SECTION 4.5
Problem 4.5-29 A beam of length L is being designed to support a uniform load
q
of intensity q (see figure). If the supports of the beam are placed at the ends,
creating a simple beam, the maximum bending moment in the beam is ql 2/8.
However, if the supports of the beam are moved symmetrically toward the middle
of the beam (as pictured), the maximum bending moment is reduced.
Determine the distance a between the supports so that the maximum bending
moment in the beam has the smallest possible numerical value.
Draw the shear-force and bending-moment diagrams for this condition.
Solution 4.5-29
A
Beam with overhangs
M1 M2 a (2 22) L 0.5858L
q
(L a)2
8
qL2
(3 222) 0.02145qL2
8
The maximum bending moment is smallest when
M1 M2 (numerically).
q(L a)2
8
qL2
qL
a
(2a L)
M2 RA a b 2
8
8
M1 M2
B
a
L
Solve for a:
M1 375
Shear-Force and Bending-Moment Diagrams
(L a)2 L(2a L)
x1 0.3536 a
0.2071 L
;
;
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CHAPTER 4
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Shear Forces and Bending Moments
Problem 4.5-30 The compound beam ABCDE shown in the figure
consists of two beams (AD and DE) joined by a hinged connection at D.
The hinge can transmit a shear force but not a bending moment. The loads
on the beam consist of a 4-kN force at the end of a bracket attached at
point B and a 2-kN force at the midpoint of beam DE.
Draw the shear-force and bending-moment diagrams for this
compound beam.
4 kN
1m
B
C
1m
A
E
2m
Solution 4.5-30
2 kN
D
2m
2m
2m
Compound beam
Problem 4.5-31 The beam shown below has a sliding support
at A and an elastic support with spring constant k at B.
A distributed load q(x) is applied over the entire beam. Find all
support reactions, then plot shear (V) and moment (M) diagrams
for beam AB; label all critical V & M values and also the
distance to points where any critical ordinates are zero.
MA
y
)
A
Ax
q(x
Linear
q0
B x
L
k
By
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SECTION 4.5
Shear-Force and Bending-Moment Diagrams
377
Solution 4.5-31
MA By q0 2
L (clockwise), Ax 0
6
q0
L (upward)
2
V
;
0
–q0L/2
;
L2/6
q0
M
Problem 4.5-32 The shear-force diagram for a simple beam
is shown in the figure.
Determine the loading on the beam and draw the bendingmoment diagram, assuming that no couples act as loads on
the beam.
0
12 kN
V
0
–12 kN
2.0 m
Solution 4.5-32
Simple beam (V is given)
1.0 m
1.0 m
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CHAPTER 4
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Shear Forces and Bending Moments
Problem 4.5-33 The shear-force diagram for a beam is shown
in the figure. Assuming that no couples act as loads on the beam,
determine the forces acting on the beam and draw the bendingmoment diagram.
652 lb
580 lb
572 lb
500 lb
V
0
–128 lb
–448 lb
4 ft
Solution 4.5-33
16 ft
4 ft
Forces on a beam (V is given)
FORCE DIAGRAM
Problem 4.5-34 The compound beam below has an internal
moment release just to the left of B and a shear release just to
the right of C. Reactions have been computed at A, C and D
and are shown in the figure.
First, confirm the reaction expressions using statics, then
plot shear (V) and moment (M) diagrams. Label all critical
V and M values and also the distance to points where either
V and/or M is zero.
w0 L2
MA = ––––
12
w0
w0
A
B
L
L
Ax = 0
—
—
2 Moment
2
release
w0 L
w0 L
Ay = ––––
Cy = ––––
6
3
C
D
L
—
2
Shear
release
–w0 L
Dy = ––––
4
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SECTION 4.5
Shear-Force and Bending-Moment Diagrams
379
Solution 4.5-34
FREE-BODY DIAGRAM
w 0 L2
––––
12
w0
0
w0 L
––––
6
w0 L
––––
6
V
w0 L
––––
6
w0 L
––––
6
w0
w0 L2
––––
24
w0 L2
––––
24
w0 L
––––
3
w0 L
––––
4
–w0 L
––––
4
0
–w0 L
––––
3
L
6
––––
w0 L2
––––
72
M 0
L2
–w0
––––
12
L
––––
3
–w0 L2
––––
24
Problem 4.5-35 The compound beam below has an shear
MA
release just to the left of C and a moment release just to the
right of C. A plot of the moment diagram is provided below
for applied load P at B and triangular distributed loads w(x) on
segments BC and CD.
Ax
First, solve for reactions using statics, then plot axial
Ay
force (N) and shear (V) diagrams. Confirm that the moment
diagram is that shown below. Label all critical N and V & M
values and also the distance to points where N, V &/or M is zero.
w0
A
w0
B
L
—
2
w0 L
P = ––––
2
4
3
C
D
L
—
2
Shear
release
Cy
L
—
2
Moment
release Dy
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CHAPTER 4
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Shear Forces and Bending Moments
Solution 4.5-35
Solve for reactions using statics
MA 7w0 2
L (clockwise),
30
Ax 3
w L (left)
10 0
Ay 3
w0L (downward)
20
;
;
Cy w0
L (upward)
12
;
Dy w0
L (upward)
6
;
;
FREE-BODY DIAGRAM
–7w0L2/60
3w0L/10
3w0L/20
w0L2/24 w0L2/24
w0L/2
w0L/4
w0
w0
w0L/4
w0L/12
3w0L/10 (tension)
N 0
w0L/4
w0L/12
V
0
0.289L
–3w0L/20
7w0L2/60
2w0L2/125
M 0
–w0L2/24
–w0L/6
w0L/6
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SECTION 4.5
Shear-Force and Bending-Moment Diagrams
Problem 4.5-36 A simple beam AB supports two connected wheel loads
P and 2P that are distance d apart (see figure). The wheels may be placed
at any distance x from the left-hand support of the beam.
P
Solution 4.5-36
2P
x
(a) Determine the distance x that will produce the maximum shear force
in the beam, and also determine the maximum shear force Vmax.
(b) Determine the distance x that will produce the maximum bending
moment in the beam, and also draw the corresponding bendingmoment diagram. (Assume P 10 kN, d 2.4 m, and L 12 m.)
381
d
A
B
L
Moving loads on a beam
P 10 kN
d 2.4 m
L 12 m
Reaction at support B:
2P
P
P
x +
(x + d) (2d + 3x)
L
L
L
Bending moment at D:
RB MD RB (L x d)
(a) MAXIMUM SHEAR FORCE
By inspection, the maximum shear force occurs at support B when the larger load is placed close to, but not
directly over, that support.
P
(2d + 3x) (L x d)
L
P
[3x2 + (3L 5d)x + 2d(L d)]
L
dMD
P
(6x + 3L 5d) 0
dx
L
x
Solve for x:
L
5d
a3 b 4.0 m
6
L
;
Substitute x into Eq (1):
L 2
P
5d 2
c 3a b a 3 b + (3L 5d)
L
6
L
Mmax x L d 9.6 m
;
d
Vmax RB P a3 b 28 kN
L
5d
L
b + 2d(L d) d
* a b a3 6
L
;
(b) MAXIMUM BENDING MOMENT
By inspection, the maximum bending moment occurs at
point D, under the larger load 2P.
Note:
PL
d 2
a3 b 78.4 kN # m
12
L
RA P
d
a 3 + b 16 kN
2
L
RB P
d
a 3 b 14 kN
2
L
;
Eq.(1)
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CHAPTER 4
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Shear Forces and Bending Moments
Problem 4.5-37 The inclined beam below represents the loads
applied to a ladder: the weight (W) of the house painter and the
self weight (w) of the ladder itself. Find support reactions at A
and B, then plot axial force (N), shear (V) and moment (M)
diagrams. Label all critical N, V & M values and also the
distance to points where any critical ordinates are zero. Plot
N, V & M diagrams normal to the inclined ladder.
θ
B
Bx
6f
t
W = 150 lb
=2
.5
lb/
ft
w
18
ft
θ
θ
θ
A
8 ft
Ax
θ
Ay
θ
Bx sin θ = 47.5 lb
sin
–47.5 lb
–16.79 lb
–30.98 lb
W
W cos θ = 50 lb
Bx cos θ = –16.79 lb
B
θ=
14
1.4
lb
Solution 4.5-37
7.5 lb
–42.5 lb
Ax cos θ
+ Ay sin θ = 214.8 lb
=0
os
θ
wc
ws
in
θ=
2.3
57
.83
lb/
ft
3l
b/f
t
–172.4 lb
270 lb·ft
N
A
Ay cos θ – Ax sin θ = 22.5 lb
22.5 lb
V
–214.8 lb
M
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Page 383
SECTION 4.5
cos u 8
1
2 12
, sin u 18 + 6
3
3
383
Shear-Force and Bending-Moment Diagrams
(2) Use u to find forces at ends A & B which are along
and perpendicular to member AB (see free-body diagram); also resolve forces W and w into components
along & perpendicular to member AB
Solution procedure:
(3) Starting at end A, plot N, V and M diagrams (see
plots)
(1) Use statics to find reaction forces at A & B
g FV 0: Ay 150 2.5 (18 6) 210 lb
Ay 210 lb (upward)
g MA 0: Bx # 24 sin 150
#
;
6 2.5
Bx 50.38 lb (left)
g FH 0; Ax 50.38 lb (right)
# 24 # 4 0
;
;
Problem 4.5-38 Beam ABC is supported by
MD
a tie rod CD as shown (see Prob. 10.4-9).
Two configurations are possible: pin support
at A and downward triangular load on AB, or
pin at B and upward load on AB. Which has
the larger maximum moment?
First, find all support reactions, then plot
axial force (N), shear (V) and moment (M)
diagrams for ABC only and label all critical
N, V & M values. Label the distance to
points where any critical ordinates are zero.
Dy
Dx
D
Moment
releases
q0 at B
y
L
—
4
x)
ear q(
Lin
Ax
A
L
B
PL
(a)
Solution 4.5-38
4q0L/9
q0L/2
7q0L2/9
7q0L2/9
q0L/2
q0L/2
q0L/2
q0L/2
4q0L/9
q0L
q0L2
17q0L/18
4q0L/9
q0L/2
4q0L/9
x
C
L
—
2
Ay
FREE-BODY DIAGRAM—BEAM (a)
L
—
4P=q L
0
4q0L/9
12:43 PM
Shear Forces and Bending Moments
Use statics to find reactions at A and D for Beam (a)
;
17
Ay q L (upward)
18 0
4
Dy q0L (downward)
9
;
MD 0
;
–4q0L/9
D
q0L/2(tension)
N
0
A
C 0
B
17q0L/18
V
4q0L/9
0
0
q0L2
7q0L2/9
M
;
(compression)
1
Ax q0L (left)
2
1
Dx q0L (left)
2
–q0L/2
CHAPTER 4
Page 384
q0L/2
384
9/25/08
q0L2/4
04Ch04.qxd
0
0
MD
Dy
Dx
D
Moment
releases
q0 at B
y
x)
ear q(
Lin
A
L
—
4 P=q L
0
L
—
4
B
L
L
—
2
By
(b)
Bx
x
C
PL
;
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Page 385
SECTION 4.5
Shear-Force and Bending-Moment Diagrams
FREE-BODY DIAGRAM—BEAM (b)
5q0L/3
q0L/2
q0L2/6
q0L2/6
q0L2
q0L/2 q0L/2
q0L/2
5q0L/3
q0L/2
q0L
5q0L/3
5q0L/3
Use statics to find reactions at B and D for Beam (b)
1
q L (right)
2 0
;
1
5
7
By q0L + q0L q0L (upward)
2
3
6
1
q L (right)
2 0
;
5
Dy q0L (downward)
3
;
D
N
0
B
A
C
0
q0L/2
(compression)
5q0L/3
q0L/2
V
0
0
q0L2
q0L2/6
M 0
–5q0L/3
(compression)
MD 0
;
q0L/2
Dx ;
–q0L/2
Bx –q0L2/4
04Ch04.qxd
0
385
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CHAPTER 4
Page 386
Shear Forces and Bending Moments
Problem 4.5-39 The plane frame below consists of column AB and beam BC
which carries a triangular distributed load. Support A is fixed and there is a
roller support at C. Column AB has a moment release just below joint B.
Find support reactions at A and C, then plot axial force (N), shear (V) and
moment (M) diagrams for both members. Label all critical N, V & M values
and also the distance to points where any critical ordinates are zero.
q0
B
C
L
Moment
release
RCy
2L
A
RAx
RAy
MA
Solution 4.5-39
Use statics to find reactions at A and C
MA 0
;
q0
L (upward)
6
q0
L (upward)
3
RAx 0
;
B
B
;
;
B
0
B
N
–3w0L/20 (compression)
RAy RCy 0
C
w0L/6
0
0
V
0
0.5774L
8w0L2/125
A
A
0
0
A
0
N
V
M
M 0
–w0L/3
9/25/08
12:43 PM
Page 387
SECTION 4.5
Shear-Force and Bending-Moment Diagrams
Problem 4.5-40 The plane frame shown below is part of an
elevated freeway system. Supports at A and D are fixed but
there are moment releases at the base of both columns (AB and
DE), as well near in column BC and at the end of beam BE.
Find all support reactions, then plot axial force (N), shear (V)
and moment (M) diagrams for all beam and column members.
Label all critical N, V & M values and also the distance to
points where any critical ordinates are zero.
387
750 N/m
C
F
45 kN
Moment
release
7m
1500 N/m
E
B
18 kN
7m
19 m
A
D
Dx
Ax
MA
Ay
MD
Dy
Solution 4.5-40
Solution procedure:
(4) g MB 0 for AB: Ax 0
(1) MA MD 0 due to moment releases
(2) g MA 0: Dy 61,164 N 61.2 kN
(5) g FH 0: Dx 63 kN
(6) Draw separate FBD’s of each member (see below) to
find N, V and M for each member; plot diagrams (see
below)
(3) g Fy 0: Ay 18,414 N 18.41kN
756 kN·m
756 kN·m
FREE-BODY DIAGRAM
750 N/m
C C
32.7 kN
1500 N/m
B
32.7 kN
B
18.41 kN
B
A
18.41 kN
14.25 kN
E
45 kN
45 kN
46.9 kN
61.2 kN
E
14.25 kN
46.9 kN
F
F
46.9 kN
441kN·m
32.7 kN
441kN·m
04Ch04.qxd
E
63 kN
D
63 kN
61.2 kN
12:43 PM
Shear Forces and Bending Moments
0
C
–46.9 kN
F
B
E
A
D
–61.2 kN
0
AXIAL FORCE DIAGRAM.
() COMPRESSION
F
C
–32.7 kN
–46.9 kN
45 kN
0
14.25 kN
E
–14.25 kN
B
63 kN
0
A
D
SHEAR FORCE DIAGRAM.
756 kN·m F
C
0
0
756 kN·m
67.7 kN·m
B
E
0
CHAPTER 4
Page 388
32.7 kN
388
9/25/08
18.41 kN
04Ch04.qxd
D
A
BENDING MOMENT DIAGRAM
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4:59 AM
Page 389
5
Stresses in Beams
(Basic Topics)
d
Longitudinal Strains in Beams
Problem 5.4-1 Determine the maximum normal strain âmax produced in
a steel wire of diameter d 1/16 in. when it is bent around a cylindrical
drum of radius R 24 in. (see figure).
Solution 5.4-1
R
Steel wire
R 24 in.
d
1
in.
16
From Eq. (5-4):
y
âmax r
Substitute numerical values:
âmax 1/16 in.
1300 * 106
2(24 in.) + 1/16 in.
d/2
d
R + d/2
2R + d
d = diameter
Problem 5.4-2 A copper wire having diameter d 3 mm is bent into
a circle and held with the ends just touching (see figure). If the maximum
permissible strain in the copper is âmax 0.0024, what is the shortest
length L of wire that can be used?
Solution 5.4-2
;
L = length
Copper wire
d 3 mm âmax 0.0024
L 2pr r L
2p
From Eq. (5-4):
âmax y
d/2
pd
r
L/2p
L
Lmin p(3 mm)
pd
3.93 m
âmax
0.0024
;
389
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CHAPTER 5
Page 390
Stresses in Beams (Basic Topics)
Problem 5.4-3 A 4.5 in. outside diameter polyethylene pipe
designed to carry chemical wastes is placed in a trench and bent
around a quarter-circular 90° bend (see figure). The bent section of
the pipe is 46 ft long.
Determine the maximum compressive strain âmax in the pipe.
90°
Solution 5.4-3
Polyethylene pipe
Angle equals 90° or p/2 radians,
L length of 90° bend
L 46 ft 552 in.
d 4.5 in.
2pr
pr
L
4
2
r r radius of curvature
r
2L
L
p
p/2
âmax y
d/2
r
2L/p
âmax pd
p 4.5 in.
a
b 6400 * 106
4L
4 552 in.
Problem 5.4-4 A cantilever beam AB is loaded by a couple M0 at
its free end (see figure.) The length of the beam is L 1.5 m and the
longitudinal normal strain at the top surface is 0.001. The distance
from the top surface of the beam to the neutral surface is 75 mm.
Calculate the radius of curvature r, the curvature k, and the vertical
deflection d at the end of the beam.
;
d
A
B
M0
L
Solution 5.4-4
Deflection: constant curvature for pure bending so
gives a circular arc; assume flat deflection curve
(small defl.) so BC L
NUMERICAL DATA
L 2.0 m
âmax 0.0012
c 82.5 mm
RADIUS OF CURVATURE
c
r
r 68.8 m
âmax
sin(u) 1
r
L
u asina b
r
;
u 0.029 radians
CURVATURE
k
L
r
k 1.455 * 105 m1
;
L
0.029
r
1 cos(u) 4.232 * 104
d r (1 cos(u))
r 6.875 * 104 mm
d 29.1 mm
;
05Ch05.qxd
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Page 391
SECTION 5.4
Problem 5.4-5 A thin strip of steel of length L 20 in. and thickness
391
Longitudinal Strains in Beams
M0
t 0.2 in. is bent by couples M0 (see figure). The deflection at the
midpoint of the strip (measured from a line joining its end points) is found
to be 0.20 in.
Determine the longitudinal normal strain â at the top surface of the strip.
M0
t
d
L
—
2
L
—
2
Solution 5.4-5
NUMERICAL DATA
L 28 inches
solving for r:
t 0.25 inches
r
d 0.20 inches
LONGITUDINAL NORMAL STRAIN AT TOP SURFACE
t
2
t
â
â
r
2r
d r (1 cos(u))
L
2
sin(u) r
L
sin(u) 2r
L
2r
d r a 1 cosa
1 cos a
r
L
b
2r
0.20
1cos a
14
b
r
numerical solution for radius of curvature r gives
r 489.719 inches
strain at top (compressive):
t
â 2.552 * 104
â
2r
â 255 11062
assume angle is small so that
u
insert numerical data:
d
;
L
bb
2r
Problem 5.4-6 A bar of rectangular cross section is loaded and
supported as shown in the figure. The distance between supports is
L 1.5 m and the height of the bar is h 120 mm. The deflection
at the midpoint is measured as 3.0 mm.
What is the maximum normal strain â at the top and bottom of
the bar?
h
P
d
P
a
L
—
2
L
—
2
a
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CHAPTER 5
Page 392
Stresses in Beams (Basic Topics)
Solution 5.4-6
NUMERICAL DATA
d r a1 cosa
h 120 mm
L 1.5 m
d 3.0 mm
‹ r a 1 cos a
NORMAL STRAIN AT TOP OF BAR:
h
2
â
r
h
2r
â
L
2
sin(u) r
u
L
b b d 0
2r
numerical solution for radius of curvature r gives
r 93.749 m
tensile strain, r radius of
curvature
SMALL DEFLECTION SO SMALL ANGLE
L
bb
2r
strain at top (compressive):
h
â
â 640 * 106
2r
;
u
L
2r
Normal Stresses in Beams
Problem 5.5-1 A thin strip of hard copper (E 16,000 ksi) having length
L 90 in. and thickness t 3/32 in. is bent into a circle and held with the
ends just touching (see figure).
3
t = — in.
32
(a) Calculate the maximum bending stress smax in the strip.
(b) By what percent does the stress increase or decrease if the thickness of
the strip is increased by 1/32 in.?
Solution 5.5-1
smaxnew 69.813 ksi
(a) MAXIMUM BENDING STREES
E 16000 ksi
L 90 inches
t
2
sEP Q
r
L
r
2p
r 14.324 inches
smax 52.4 ksi
smax ;
(b) % CHANGE IN STRESS
tnew 4
32
smaxnew Etnew
2r
Et
2r
t
3
inches
32
smaxnew smax
(100) 33.3
smax
;
33% increase (linear) in max.stress due to increase
in t; same as % increase in thickness t
3
4
32
32
(100) 33.3
3
32
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Page 393
SECTION 5.5
393
Normal Stresses in Beams
Problem 5.5-2 A steel wire (E 200 GPa) of diameter d 1.25 mm
is bent around a pulley of radius R0 500 mm (see figure).
(a) What is the maximum stress smax in the wire?
(b) By what percent does the stress increase or decrease if the radius of the pulley is
increased by 25%?
R0
d
Solution 5.5-2
(a) MAX. NORMAL STRESS IN WIRE
d 1.25 mm
E 200 GPa
d
2
s
r
E
E
smax smax 250 MPa
R0 500 mm
(b) % CHANGE IN MAX. STRESS DUE TO INCREASE
IN PULLEY RADIUS BY 25%
d
2
E
snew d
R0 +
2
d
2
d
2
1.25 R0 +
snew 199.8 MPa
snew smax
(100) 20%
smax
;
;
L = length
Problem 5.5-3 A thin, high-strength steel rule (E 30 106 psi)
having thickness t 0.175 in. and length L 48 in. is bent by couples
M0 into a circular are subtending a central angle a 40° (see figure).
t
M0
M0
(a) What is the maximum bending stress smax in the rule?
(b) By what percent does the stress increase or decrease if the central angle is
increased by 10%?
a
Solution 5.5-3
(b) % CHANGE IN STRESS DUE TO 10% INCREASE
IN ANGLE a
(a) MAX. BENDING STRESS
a 40 a
p
b
180
L 48 inches
r
L
a
a 0.698 radians
t 0.175 in.
E 30 (106) psi
r 68.755 inches
t
2
Et
smax smax r
2r
smax 38.2 ksi
;
E
snew E t (1.1a)
2L
snew smax
(100) 10%
smax
linear increase (%)
Eta
smax 2L
snew 41997 psi
;
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4:59 AM
CHAPTER 5
Page 394
Stresses in Beams (Basic Topics)
Problem 5.5-4 A simply supported wood beam AB with span length
q
L 4 m carries a uniform load of intensity q 5.8 kN/m (see figure).
(a) Calculate the maximum bending stress smax due to the load q if the
beam has a rectangular cross section with width b 140 mm and
height h 240 mm.
(b) Repeat (a) but use the trapezoidal distuibuted load shown in the
figure part (b).
A
h
B
b
L
(a)
q
—
2
q
A
B
L
(b)
Solution 5.5-4
(a) MAX. BENDING STRESS DUE TO UNIFORM LOAD q
2
qL
8
Mmax S
(b)
MAX. BENDING STRESS DUE TO TRAPEZOIDAL LOAD
RA c
I
h
2
uniform load (q/2) & triang. load (q/2)
3
bh
12
S
h
2
smax RA 1
S bh2
6
L4m
qL2
8
Mmax 11.6 kN # m
smax 8.63 MPa
b 140 mm
q
1 x q
x a
bx 0
2
2 L2
3x2 + 6Lx 4L2 0
6 L 1184 L22
2(3)
x
1
a 1 + 184b
L
6
xmax 0.52753 L
h 240 mm
Mmax RA x
3 L2
smax q 2
4 bh
kN
q 5.8
m
1
qL
3
find x location of zero shear
qL2
8
smax 1
a bh2 b
6
Mmax
S
1 q
1 q1
a bL + a
b Ld
2 2
3 22
;
q
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Page 395
SECTION 5.5
Mmax RAxmax q xmax2
1 xmax q xmax2
a
b
2 2
2 L 2
3
Mmax 9.40376 * 102 qL2
Mmax 8.727 kN # m
smax Normal Stresses in Beams
395
Mmax
S
smax 6.493 * 103
N
m2
smax 6.49 MPa
;
Problem 5.5-5 Each girder of the lift bridge (see figure) is
180 ft long and simply supported at the ends. The design load
for each girder is a uniform load of intensity 1.6 k/ft. The girders
are fabricated by welding tree steel plates so as to form an
I-shaped cross section (see figure) having section modulus
S 3600 in.3.
What is the maximum bending stress smax in a girder due
to the uniform load?
Solution 5.5-5
Bridge girder
L 180 ft
q 1.6 k/ft
S 3600 in.
3
Mmax qL2
8
smax qL2
Mmax
S
8S
smax (1.6 k/ft)(180 ft)2(12 in./ft)
8(3600 in.3)
21.6 ksi
;
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CHAPTER 5
Page 396
Stresses in Beams (Basic Topics)
Problem 5.5-6 A freight-car axle AB is loaded approximately as shown in
the figure, with the forces P representing the car loads (transmitted to the axle
through the axle boxes) and the forces R representing the rail loads (transmitted
to the axle through the wheels). The diameter of the axle is d 80 mm, the
distance between centers of the rails is L, and the distance between the forces
P and is R is b 200 mm.
Calculate the maximum bending stress smax in the axle if P 47 kN.
P
P
B
A
d
d
R
b
R
L
b
Solution 5.5-6
NUMERICAL DATA
d 82 mm
MAX. BENDING STRESS
b 220 mm
P 50 kN
I
pd4
64
I 2.219 * 106 m4
Mmax Pb
smax Md
2I
smax 203 MPa
;
Mmax 11 kN # m
Problem 5.5-7 A seesaw weighing 3 lb/ft of length is occupied by two
children, each weighing 90 lb (see figure). The center of gravity of each
child is 8 ft from the fulcrum. The board is 19 ft long, 8 in. wide, and
1.5 in. thick.
What is the maximum bending stress in the board?
Solution 5.5-7
Seesaw
b 8 in.
h 1.5 in.
q 3 lb/ft
P 90 lb
d 8.0 ft
L 9.5 ft
2
Mmax Pd +
qL
720 lb-ft + 135.4 lb-ft
2
855.4 lb-ft 10,264 lb-in.
S
2
bh
3.0 in.3.
6
smax 10,264 lb-in.
M
3420 psi
S
3.0 in.3
;
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Page 397
SECTION 5.5
397
Normal Stresses in Beams
Problem 5.5-8 During construction of a highway bridge,
the main girders are cantilevered outward from one pier toward
the next (see figure). Each girder has a cantilever length of
48 m and an I-shaped cross section with dimensions shown
in the figure. The load on each girder (during construction)
is assumed to be 9.5 kN/m, which includes the weight of
the girder.
Determine the maximum bending stress in a girder due to
this load.
52 mm
2600 mm
28 mm
620 mm
Solution 5.5-8
NUMERICAL DATA
tf 52 mm
h 2600 mm
L 48 m
I
tw 28 mm
bf 620 mm
q 9.5
kN
m
L
Mmax qL a b
2
Mmax h
smax 2I
Mmax 1.094 * 104 kN m
smax 101 MPa
;
1
1
(b ) h3 (b tw) [ h 2 (tf)]3
12 f
12 f
I 1.41 * 1011 mm4
Problem 5.5-9 The horizontal beam ABC of an oil-well pump
has the cross section shown in the figure. If the vertical pumping force acting at
end C is 9 k and if the distance from the line of action of that force to point B is
16 ft, what is the maximum bending stress in the beam due to the pumping
force?
Horizontal beam transfers loads as part of oil well pump
C
B
A
0.875 in.
22 in.
0.625
in.
8.0 in.
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CHAPTER 5
Page 398
Stresses in Beams (Basic Topics)
Solution 5.5-9
NUMERICAL DATA
FC 9 k
MAX. BENDING STRESS AT B
BC 16 ft
Mmax F C (BC)
Mmax 144 k-ft
smax 1
1
I
(8) (22)3 (8 0.625)
12
12
* [22 2 (0.875)]
;
4
P
a
P
a
L
b
h
q
Railroad tie (or sleeper)
P 175 kN
b 300 mm
L 1500 mm
q
I
smax 9.53 ksi
3
Problem 5.5-10 A railroad tie (or sleeper) is subjected to two rail
loads, each of magnitude P 175 kN, acting as shown in the figure.
The reaction q of the ballast is assumed to be uniformly distributed
over the length of the tie, which has cross-sectional dimensions
b 300 mm and h 250 mm.
Calculate the maximum bending stress smax in the tie due to
the loads P, assuming the distance L 1500 mm and the overhang
length a 500 mm.
DATA
22
b
2
I 1.995 * 10 in.
3
Solution 5.5-10
Mmax (12) a
2P
L + 2a
h 250 mm
a 500 mm
S
bh2
3.125 * 103 m3
6
Substitute numerical values:
M1 17,500 N # m
M2 21,875 N # m
Mmax 21,875 N # m
MAXIMUM BENDING STRESS
BENDING-MOMENT DIAGRAM
smax 21,875 N # m
Mmax
7.0 MPa
5
3.125 * 103 m3
(Tension on top; compression on bottom)
M1 qa2
Pa2
2
L + 2a
M2 2
q L
PL
a + ab 2 2
2
2
L
PL
P
a + ab L + 2a 2
2
P
(2a L)
4
;
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Page 399
SECTION 5.5
Normal Stresses in Beams
399
Problem 5.5-11 A fiberglass pipe is lifted by a sling, as shown in the
figure. The outer diameter of the pipe is 6.0 in., its thickness is 0.25 in.,
and its weight density is 0.053 lb/in.3 The length of the pipe is L 36 ft
and the distance between lifting points is s 11 ft.
Determine the maximum bending stress in the pipe due to its own
weight.
s
L
Solution 5.5-11
Pipe lifted by a sling
d2 6.0 in.
L 36 ft 432 in.
s 11 ft 132 in.
g 0.053 lb/in.3
t 0.25 in.
d1 d2 2t 5.5 in.
p
A (d22 d21) 4.5160 in.2
4
a (L s)/2 150 in.
BENDING-MOMENT DIAGRAM
I
p 4
(d d14) 18.699 in.4
64 2
q gA (0.053 lb/in.3)(4.5160 in.2) 0.23935 lb/in.
MAXIMUM BENDING STRESS
smax smax Mmax c
I
M2 qL L
a sb 2,171.4 lb-in.
4 2
Mmax 2,692.7 lb-in.
d2
3.0 in.
2
(2,692.7 lb-in.)(3.0 in.)
18.699 in.4
(Tension on top)
qa2
2,692.7 lb-in.
M1 2
c
432 psi
;
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CHAPTER 5
Page 400
Stresses in Beams (Basic Topics)
Problem 5.5-12 A small dam of height h 2.0 m is constructed of
vertical wood beams AB of thickness t 120 mm, as shown in the figure.
Consider the beams to be simply supported at the top and bottom.
Determine the maximum bending stress smax in the beams, assuming
that the weight density of water is g 9.81 kN.m3
A
h
t
B
Solution 5.5-12
Vertical wood beam
MAXIMUM BENDING MOMENT
RA q0L
6
q0 x 3
6L
q0 Lx
q0 x 3
6
6L
q0L
q0x 2
dM
L
0 x
dx
6
2L
13
M RAx h 2.0 m
t 120 mm
g 9.81kN/ m3(water)
Let b = width of beam perpendicular to the plane
of the figure
Substitute x L/13 into the equation for M:
Let q0 = maximum intensity of distributed load
Mmax q0 gbh S bt2
6
q0L
q0
q0 L2
L
L3
a
b a
b 6
6L 313
13
913
For the vertical wood beam: L h; Mmax q0 h 2
9 13
Maximum bending stress
smax 2q0 h2
2gh3
Mmax
S
313 bt2
313 t2
SUBSTITUTE NUMERICAL VALUES:
smax 2.10 MPa
;
NOTE: For b 1.0 m, we obtain q0 19,620 N/m,
S 0.0024 m3,
Mmax 5,034.5 N # m, and smax Mmax/S 2.10 MPa
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Page 401
SECTION 5.5
Normal Stresses in Beams
401
y
Problem 5.5-13 Determine the maximum tensile stress st
(due to pure bending about a horizontal axis through C
by positive bending moments M ) for beams having cross
sections as follows (see figure).
x
b1
xc
C
(a) A semicircle of diameter d
(b) An isosceles trapezoid with bases b1 b and b2 4b/3,
and altitude h
(c) A circular sector with p/3 and r d/2
x
xc
C
h
y
a
xc
C
a
r
d
b2
O
(a)
(b)
(c)
x
Solution 5.5-13
MAX. TENSILE STRESS DUE TO POSITIVE BENDING MOMENT
IS ON BOTTOM OF BEAM CROSS-SECTION
r4
(a + sin (a) cos(a))
4
Ix (a) SEMICIRCLE
ybar From Appendix D, Case 10:
(9p2 64)r4
(9p2 64)d4
72p
1152p
Ic c
;
A d2 a
c
2a
p
b
12
d
b
2
3
From Appendix D, Case 8:
h3(b21 + 4b1b2 + b22)
36(b1 + b2)
73bh3
756
c
Mc
360M
Ic
73bh2
Ix a
d 4
b
2
4
A 0.2618 d2
p
sina b
3
±
≤
p
3
a
d 2 p
b a b
2
3
c 0.276 d
p
p
p
+ sina b cos a b b
3
3
3
Ix 0.02313 d 4
h(2b1 + b2)
10h
3(b1 + b2)
21
st A a
For a p/3, r d/2:
(b) ISOSCELES TRAPEZOID
IC c ybar
d1
2d
4r
3p
3p
Mc
768M
M
30.93 3
st 2
3
Ic
(9p 64)d
d
2r sin (a)
a
b
3
a
;
(c) CIRCULAR SECTOR WITH a p/3, r d/2
IC Ix A y2bar
IC cd 4
(4p313)
p
d 13 2
d 2 a 12 b c a
bd d
768
2 p
IC 3.234 * 103 d 4
max. tensile stress st From Appendix D, Case 13:
Mc
IC
A r 2 (a)
Problem 5.5-14 Determine the maximum bending stress smax
(due to pure bending by a moment M) for a beam having a cross
section in the form of a circular core (see figure). The circle has
diameter d and the angle b 60°. (Hint: Use the formulas given
in Appendix D, Cases 9 and 15.)
C
b
b
d
st 85.24
M
d3
;
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CHAPTER 5
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Stresses in Beams (Basic Topics)
Solution 5.5-14
Circular core
From Appendix D, Cases 9 and 15:
Iy r
4
4
r
ab
pr
2ab
aa 2 + 4 b
4
2
r
r
d
2
a
p
b
2
b radians a radians a r sin b
Iy 4
4
4
4
3
d4 p
1
pd 4
a b + sin 4b b
64
32 2
4
d4
(4b sin4b)
128
MAXIMUM BENDING STRESS
b r cos b
pd
d p
a b sin b cos b + 2 sin b cos3 b b
64
32 2
pd
d p
a b (sin b cos b)(1 2 cos2 b) b
64
32 2
d4 p
1
pd 4
a b a sin 2b b (cos 2b) b
64
32 2
2
smax smax Mc
Iy
c r sin b 64M sin b
;
d (4b sin 4b)
3
For b 60° p/3 rad:
576M
M
smax 10.96 3
3
(8p13 + 9)d
d
P
Problem 5.5-15 A simple beam AB of span length L 24 ft
is subjected to two wheel loads acting at distance d = 5 ft apart
(see figure). Each wheel transmits a load P = 3.0 k, and the
carriage may occupy any position on the beam.
Determine the maximum bending stress smax due to the wheel
loads if the beam is an I-beam having section modulus S 16.2 in.3
Solution 5.5-15
d
sin b
2
d
;
P
A
B
C
L
Wheel loads on a beam
Substitute x into the equation for M:
L 24 ft 288 in.
d 5 ft 60 in.
P3k
S 16.2 in.
3
Mmax P
d 2
aL b
2L
2
MAXIMUM BENDING STRESS
smax Mmax
d 2
P
aL b
S
2LS
2
MAXIMUM BENDING MOMENT
Substitute numerical values:
P
P
P
L x + (L x d) (2L d 2x)
L
L
L
P
2
M RA x (2L x dx 2x )
L
P
d
dM
L
(2L d 4x) 0 x dx
L
2
4
smax RA 3k
2(288 in.)(16.2 in.3)
21.4 ksi
;
;
(288 in. 30 in.)2
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SECTION 5.5
Problem 5.5-16 Determine the maximum tensile stress st and
maximum compressive stress sc due to the load P acting on the
simple beam AB (see figure).
Data are as follows: P 6.2 kN, L 3.2 m, d 1.25 m,
b 80 mm, t 25 mm, h 120 mm, and h1 90 mm.
403
Normal Stresses in Beams
t
P
d
A
B
h1
h
L
b
Solution 5.5-16
NUMERICAL DATA
MAX. MOMENT & NORMAL STRESSES
P 6.2 kN
L 3.2 m
d 1.25 m
b 80 mm
t 25 mm
h 120 mm
Mmax Beam cross section properties: centroid and moment
of inertia
Aw
c1 Af + Aw
c2 h c1
I
sc Mmax c1
I
sc 61.0 MPa
;
MAX. TENSILE STRESS AT BOTTOM (c c2)
Aw th1
(h h1)
h1
+ Af c h d
2
2
Mmax 4.7 kN # m
MAX. COMPRESSIVE STRESS AT TOP (c c1)
h1 90 mm
Af b (h h1)
Pd (L d)
L
st c1 76 mm
Mmax c2
I
st 35.4 MPa
;
c2 44 mm dist. to C from bottom
1
1
t h31 +
b (h h1)3
12
12
(h h1) 2
h1 2
+ Af cc2 d + Aw a c1 b
2
2
I 5879395.2 mm4
250 lb
Problem 5.5-17 A cantilever beam AB, loaded by a uniform load and a
concentrated load (see figure), is constructed of a channel section.
Find the maximum tensile stress st and maximum compressive stress sc
if the cross section has the dimensions indicated and the moment of inertia
about the z axis (the neutal axis) is I 3.36 in.4 (Note: The uniform load
represents the weight of the beam.)
22.5 lb/ft
B
A
5.0 ft
3.0 ft
y
z
C
0.617 in.
2.269 in.
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CHAPTER 5
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Stresses in Beams (Basic Topics)
Solution 5.5-17
NUMERICAL DATA
MAXIMUM STRESSES
c1 0.617 in.
I 3.36 in.
4
c2 2.269 in.
MAmax 22.5 (8)2
+ 250 (5) ft-lb
2
st MAmax c1
I
st 4341 psi
sc MAmax c2
I
sc 15964 psi
;
;
MAmax 1970 ft-lb
MAmax (12) 23640 in.-lb
q
Problem 5.5-18 A cantilever beam AB of isosceles trapezoidal
cross section has length L 0.8 m, dimensions b1 80 mm,
b2 90 mm, and height h 110 mm (see figure). The beam is
made of brass weighing 85 kN/m3.
b1
C
h
L
b2
(a) Determine the maximum tensile stress st and maximum
compressive stress sc due to the beam’s own weight.
(b) If the width b1 is doubled, what happens to the stresses?
(c) If the height h is doubled, what happens to the stresses?
Solution 5.5-18
NUMERICAL DATA
MAX. TENSILE STRESS AT SUPPORT (TOP)
g 85
L 0.8 m
b1 80 mm
kN
st 3
m
b 2 90 mm
(a) MAX. STRESSES DUE TO BEAM’S OWN WEIGHT
q L2
2
q gA
A
1
(b + b 2) h
2 1
A 9.35 * 103 mm2
q 7.9475 * 102
I h3
h (2b1 b2)
3 (b1 b2)
ybar 53.922 mm
36 (b1 b2)
I 9.417 * 10 mm
4
Mmax ybar
I
sc 1.456 MPa
;
(b) DOUBLE b1& RECOMPUTE STRESSES
b1 160 mm
1
(b + b2) h
2 1
q gA
1 b21 4 b1 b2 b222
6
sc A
N
m
Mmax 254.32 N # m
ybar st 1.514 MPa
;
MAX. COMPRESSIVE STRESS AT SUPPORT (BOTTOM)
h 110 mm
Mmax Mmax (h ybar)
I
A 1.375 * 104 mm2
q 1.169 * 103
N
m
qL2
2
Mmax 374 N # m
Mmax ybar h (2 b1 + b2)
3 (b1 + b2)
ybar 60.133 mm
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Page 405
SECTION 5.5
I h3
1b21 + 4 b1 b2 + b222
Mmax 36 (b1 + b2)
I 1.35 * 107 mm4
ybar MAX. TENSILE STRESS AT SUPPORT (TOP)
st Mmax (h ybar)
I
st 1.381 MPa
I h3
qL2
2
405
Normal Stresses in Beams
Mmax 508.64 N # m
h (2b1 + b2)
3 (b1 + b2)
ybar 107.843 mm
(b12 + 4b1 b2 + b22)
;
36 1b1 + b22
I 7.534 * 107 mm4
MAX. COMPRESSIVE STRESS AT SUPPORT (BOTTOM)
sc MAX. TENSILE STRESS AT SUPPORT (TOP)
Mmax ybar
2
sc 1.666 MPa
;
st Mmax (h ybar)
I
st 0.757 MPa
;
(c) DOUBLE h & RECOMPUTE STRESSES
MAX. COMPRESSIVE STRESS AT SUPPORT (BOTTOM)
b1 80 mm h 220 mm
A
1
(b + b2) h
2 1
q gA
sc A 1.87 * 104 mm2
q 1.589 * 103
Mmax ybar
I
sc 0.728 MPa
;
N
m
200 lb/ft
Problem 5.5-19 A beam ABC with an overhang from B to C supports
a uniform load of 200 lb/ft throughout its length (see figure). The beam is
a channel section with dimensions as shown in the figure. The moment of
inertia about the z axis (the neutral axis) equals 8.13 in.4
Calculate the maximum tensile stress st and maximum compressive
stress sc due to the uniform load.
A
C
B
12 ft
6 ft
y
0.787 in.
z
C
2.613 in.
Solution 5.5-19
NUMERICAL DATA
LOCATON OF ZERO SHEAR IN SPAN AB & MAX. (+)
MOMENT IN SPAN AB
lb
I 8.13 in.4
ft
c2 2.613 in.
c1 0.787 in.
q 200
xmax a MA 0
a Fv 0
RB (18)2
2
12
RA q (18) RB
xmax 4.5 ft
MmaxAB RA xmax q
COMPUTE SUPPORT REACTIONS
q
RA
q
xmax2
2
MmaxAB 2025 ft-lb
RB 2700 lb
R A 900 lb
max. () moment at B
MB q
(6)2
2
MB 3600 ft-lb
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Stresses in Beams (Basic Topics)
MAX. STRESSES IN SPAN AB
MmaxAB (12) c1
I
MmaxAB (12) c2
st I
st 7810 psi
;
sC MAX. STRESSES IN SPAN BC
MB (12) c2
I
sc 13885 psi
sc 2352 psi
sc st max. tensile stress
MB (12) c1
I
;
st 4182 psi
A
Problem 5.5-20 A frame ABC travels horizontally with an acceleration
a0 (see figure). Obtain a formula for the maximum stress smax in the
vertical arm AB, which had length L, thickness t , and mass density r.
max. compressive stress
t
a0 = acceleration
L
B
Solution 5.5-20
Accelerating frame
L length of vertical arm
t thickness of vertical arm
r mass density
a0 acceleration
Let b width of arm perpendicular to the plane of the figure
Let q inertia force per unit distance along vertical arm
TYPICAL UNITS FOR USE
VERTICAL ARM
t meters (m)
IN THE PRECEDING EQUATION
SI units: r kg/m3 N # s2/m4
L meters (m)
a0 m/s2
smax N/m2 (pascals)
q rbta0
S
bt2
6
USCS units: r slug/ft3 lb-s2/ft4
qL2
rbta0L2
Mmax 2
2
smax 3rL2a0
Mmax
S
t
L ft a0 ft/s2
t ft
smax lb/ft (Divide by 144 to obtain psi)
2
;
C
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Page 407
SECTION 5.5
Problem 5.5-21 A beam of T-section is supported and
loaded as shown in the figure. The cross section has width
b 2 1/2 in., height h 3 in., and thickness t 3/8 in.
Determine the maximum tensile and compressive stresses
in the beam.
407
Normal Stresses in Beams
3
t=—
8 in.
P = 700 lb
q = 100 lb/ft
L1 = 3 ft
3
t=—
8 in.
L2 = 8 ft
h=
3 in.
1
b = 2—
2 in.
L3 = 5 ft
Solution 5.5-21
NUMERICAL DATA
L1 3 ft
L2 8 ft
q 100
P 700 lb
t
3
in.
8
a Fv 0
L3 5 ft
R lf 281 lb
Moment diagram (843.75 ft-lb at load P, 1250 ft-lb
at right support)
lb
ft
h 3 in.
Rlf P + qL3 Rrt
8.438E+02
b 2.5 in.
Find centroid of cross section (c2 from bottom,
Aw t (h t)
c1 from top)
Af t b
Af
c2 t
ht
+ Aw at +
b
2
2
c2 1 in.
Af + Aw
c1 h c2
check c1 c1 2
–1.250E+03
MP 843.75 ft-lb
c1 2 in.
Aw a
Mrt 1250 ft-lb
ht
t
b + Af ah b
2
2
MAX. STRESSES IN BEAM
at load P
Af + Aw
c1 + c2 3
MP (12) c1
sc 12494 psi
I
(max. compressive stress)
equals h
sc MOMENT OF INERTIA
I
1
1
t 2
t (h t)3 +
b t 3 + Af ac2 b
12
12
2
+ Aw cc1 (h t) 2
d
2
FIND SUPPORT REACTIONS-SUM MOMENTS ABOUT LEFT
SUPPORT
a Mlf 0
Rrt 919 lb
Rrt L2
MP (12) c2
I
st 5842 psi
at right support
I 2 in.4
PL1 + qL3 aL2 +
st ;
L3
b
2
Mrt (12) c2
sc 8654 psi
I
Mrt (12) c1
st st 18509 psi
I
(max. tensile stress)
sc ;
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Stresses in Beams (Basic Topics)
Problem 5.5-22 A cantilever beam AB with a rectangular
cross section has a longitudinal hole drilled throughout its length
(see figure). The beam supports a load P 600 N. The cross
section is 25 mm wide and 50 mm high, and the hole has a
diameter of 10 mm.
Find the bending stresses at the top of the beam, at the top
of the hole, and at the bottom of the beam.
10 mm
50 mm
A
B
12.5 mm
37.5 mm
P = 600 N
L = 0.4 m
25 mm
Solution 5.5-22
Rectangular beam with a hole
MOMENT OF INERTIA ABOUT THE NEUTRAL AXIS
(THE z AXIS)
All dimensions in millimeters.
Rectangle:
Iz Ic + Ad2
1
(25)(50)3 + (25)(50)(25 24.162)2
12
MAXIMUM BENDING MOMENT
M PL (600 N)(0.4 m) 240 N # m
260,420 + 878 261,300 mm4
PROPERTIES OF THE CROSS SECTION
Hole:
A1 area of rectangle
Iz Ic + Ad2 (25 mm)(50 mm) 1250 mm2
A2 area of hole
p
(10 mm)2 78.54 mm2
4
A area of cross section
A1 A2 1171.5 mm
Using line B B as reference axis:
©Aiyi A1(25 mm) A2(37.5 mm) 28,305 mm3
Aiyi
28,305 mm3
y a
24.162 mm
A
1171.5 mm2
Distances to the centroid C:
c2 y 24.162 mm
c1 50 mm c2 25.838 mm
p
(10)4 + (78.54)(37.5 24.162)2
64
490.87 + 13,972 14,460 mm4
Cross-section:
I 261,300 14,460 246,800 mm4
STRESS AT THE TOP OF THE BEAM
(240 N # m)(25.838 mm)
Mc1
s1 I
246,800 mm4
25.1 MPa
(tension)
;
STRESS AT THE TOP OF THE HOLE
My
s2 y c1 7.5 mm 18.338 mm
I
(240 N # m)(18.338 mm)
s2 17.8 MPa
246,800 mm4
(tension)
STRESS AT THE BOTTOM OF THE BEAM
(240 N # m)(24.162 mm)
Mc2
I
246,800 mm4
23.5 MPa
;
(compression)
s3 ;
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SECTION 5.5
Problem 5.5-23 A small dam of height h 6 ft is constructed of
Normal Stresses in Beams
Steel beam
vertical wood beams AB, as shown in the figure. The wood beams,
which have thickness t 2.5 in., are simply supported by horizontal
steel beams at A and B.
Construct a graph showing the maximum bending stress smax in
the wood beams versus the depth d of the water above the lower
support at B. Plot the stress smax (psi) as the ordinate and the
depth d (ft) as the abscissa. (Note: The weight density g of water
equals 62.4 lb/ft3.)
A Wood beam
t
t
Wood beam
Steel beam
h
d
B
Side view
Solution 5.5-23
Vertical wood beam in a dam
h 6 ft
t 2.5 in.
g 62.4 lb/ft3
Let b width of beam
(perpendicular to the
figure)
Let q0 intensity of
load at depth d
q0 gbd
ANALYSIS OF BEAM
L h 6 ft
q0d2
RA 6L
q0d
d
RB a3 b
6
L
MAXIMUM BENDING STRESS
1
Section modulus: S bt2
6
Mmax
6 q0d2
d
2d d
bd
2c
a1 +
S
6
L
3L
A 3L
bt
q0 g bd
smax smax smax q0d2
d
2d d
a1 +
b
6
L
3L A 3L
t2
a1 d
2d d
b
+
L
3L A 3L
(62.4)d3
(2.5)2
a1 ;
d
d d
b
+
6
9 A 18
0.1849d3(54 9d + d12d )
d
A 3L
Mc RA(L d) gd3
SUBSTITUTE NUMERICAL VALUES:
d depth of water (ft) (Max. d h 6 ft)
L h 6 ft g 62.4 lb/ ft3 t 2.5 in.
smax psi
x0 d
Mmax Top view
q0d2
d
a1 b
6
L
d(ft)
0
1
2
3
4
5
6
smax(psi)
0
9
59
171
347
573
830
;
409
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CHAPTER 5
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Stresses in Beams (Basic Topics)
Problem 5.5-24 Consider the nonprismatic cantilever beam
of circular cross section shown. The beam has an internal cylindrical
hole in segment 1; the bar is solid (radius r) in segment 2. The beam
is loaded by a downward triangular load with maximum intensity q0
as shown.
Find expressions for maximum tensile and compressive flexural
stresses at joint 1.
q0
y
Linea
P = q0L/2
r q(x
)
x
M1
1
R1
2
2L
—
3
Segment 1
0.5 EI
3
L
—
3
Segment 2
EI
Solution 5.5-24
STATICS
a Fv 0
R1 MAX. STRESSES AT JOINT 1
R1 q0L
1
2L
q0 a b 2
3
2
1
qL
6 0
g M1 0
q0 L
1
2L
1 2L
b Ld
M1 c q0 a b a
2
3
3 3
2
M1 23
q0 L2
54
23
0.426
54
MAX. COMPRESSION AT TOP (RADIUS r)
23
q L2 (r)
M1 r
54 0
sc sc 0.5 EI
EI
2
sc 23 q0 L2 r
27 EI
;
23
0.852
27
Max. tensile stress at bottom same magnitude as
compressive stress at top
Problem 5.5-25 A steel post (E 30 106 psi) having thickness t 1/8 in.
and height L 72 in. supports a stop sign (see figure: s 12.5 in.). The height
of the post L is measured from the base to the centroid of the sign. The stop sign
is subjected to wind pressure p 20 lb/ft2 normal to its surface. Assume that
the post is fixed at its base.
(a) What is the resultant load on the sign? [See Appendix D, Case 25, for
properties of an octagon, n 8].
(b) What is the maximum bending stress smax in the post?
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SECTION 5.5
Normal Stresses in Beams
s
L
y
5/8 in.
Section A–A
z
Circular cutout, d = 0.375 in.
Post, t = 0.125 in.
c1
1.5 in.
C
c2
Stop sign
0.5 in. 1.0 in.
1.0 in. 0.5 in.
Wind load
Numerical properties of post
A = 0.578 in.2, c1 = 0.769 in., c2 = 0.731 in.,
Iy = 0.44867 in.4, Iz = 0.16101 in.4
A
A
Elevation
view of post
411
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CHAPTER 5
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Stresses in Beams (Basic Topics)
Solution 5.5-25
(a) RESULTANT LOAD F ON SIGN
s 12.5 in.
p 20 psf
b
(b) MAX. BENDING STRESS IN POST
360 p
a
b
n
180
b
ns2
A
cot a b
4
2
A 754.442 in.2
F 104.8 lb
;
I Z 0.16101 in.4
c1 0.769 in.
b 0.785 rad
or A 5.239 ft2
F pA
L 72 in.
n8
c2 0.731 in.
Mmax
628.701 ft-lb
12
Mmax FL
sc Mmax c1
Iz
sc 36.0 ksi
;
(max. bending stress at base of post)
st Mmax c2
Iz
st 34.2 ksi
Design of Beams
P
Problem 5.6-1 The cross section of a narrow-gage railway
Wood
tie
d
b
Steel
girder
(b)
s1
(a)
Railway cross tie
Mmax S
P(s1 s2)
15,000 lb-in.
2
1
5d 2
bd 2
(50 in.)(d 2) 6
6
6
Mmax s allow S
s1 50 in.
P
Steel rail
bridge is shown in part (a) of the figure. The bridge is
constructed with longitudinal steel girders that support the
wood cross ties. The girders are restrained against lateral
buckling by diagonal bracing, as indicated by the dashed lines.
The spacing of the girders is s1 50 in. and the spacing
of the rails is s2 30 in. The load transmitted by each rail to a
single tie is P 1500 lb. The cross section of a tie, shown in
part (b) of the figure, has width b 5.0 in. and depth d.
Determine the minimum value of d based upon an
allowable bending stress of 1125 psi in the wood tie.
(Disregard the weight of the tie itself.)
Solution 5.6-1
s2
b 5.0 in.
d depth of tie
s2 30 in.
P 1500 lb
sallow 1125 psi
d inches
15,000 (1125) a
Solving, d 2 16.0 in.
5d 2
b
6
dmin 4.0 in.
3P(s1 s2)
NOTE: Symbolic solution: d 2 bsallow
;
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Page 413
SECTION 5.6
Problem 5.6-2 A fiberglass bracket ABCD of solid circular cross section has
the shape and dimensions shown in the figure. A vertical load p 40 N acts at
the free end D.
Determine the minimum permissible diameter dmin of the bracket if the
allowable bending stress in the material is 30 MPa and b 37 mm.
(Note: Disregard the weight of the bracket itself.)
Design of Beams
413
6b
A
B
2b
D
C
2b
P
Solution 5.6-2
sa (3Pb) a
a
dmin
b
2
pdmin4
64
b
1
dmin3 96Pb
psa
1
96Pb 3
dmin a
b
psa
96 (40) (37) 3
dmin c
d
p (30)
dmin 11.47 mm
;
P 2750 lb
Problem 5.6-3 A cantilever beam of length L 7.5 ft supports a uniform
load of intensity q 225 lb/ft and a concentrated load P 2750 lb (see figure).
Calculate the required section modulus S if sallow 17,000 psi. Then select a
suitable wide-flange beam (W shape) from Table E-1(a), Appendix E, and recalculate S taking into account the weight of beam. Select a new beam size if necessary.
q 225 lb/ft
L = 7.5 ft
Solution 5.6-3
sa 17000 psi
q 225
lb
ft
Mmax1 PL +
P 2750 lb
Mmax2 2.774 * 104 lb-ft
smax qL2
2
below allowable -OK
Mmax1 2.695 * 104 lb-ft
Mmax1 (12)
Sreqd sa
w 26
Sreqd 19.026 in.
3
try W 8 * 28 (S 24.3 in. )
3
Check - add weight per ft for beam
lb
ft
Mmax2 PL +
Sact 24.3 in.3
(q + w) L2
2
smax 13699 psi
Repeat for W14 * 26 which is lighter than W8 * 28
Find Sreqd without beam weight
W 28
Mmax2 (12)
Sact
L 7.5 ft
lb
ft
Mmax3 PL +
Sact 35.3 in.3
(q + w) L2
2
Mmax3 2.768 * 104 lb-ft
smax M max3 (12)
Sact
smax 9411 psi
well below allowable - OK
use W 14 * 26
;
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Stresses in Beams (Basic Topics)
Problem 5.6-4 A simple beam of length L 5 m carries a uniform load
P = 22.5 kN 1.5 m
kN
of intensity q 5.8
and a concentrated load 22.5 kN (see figure).
m
Assuming sallow 110 MPa, calculate the required section modulus S.
Then select an 200 mm wide-flange beam (W shape) from Table E-1(b)
Appendix E, and recalculate S taking into account the weight of beam. Select a
new 200 mm beam if necessary.
q = 5.8 kN/m
L=5m
Solution 5.6-4
NUMERICAL DATA
RECOMPUTE MAX. MOMENT WITH BEAM MASS INCLUDED &
L 5 m q 5.8
THEN CHECK ALLOWABLE STRESS
kN
m
w a41.7
b 1.5 m
P 22.5 kN
a L b a 3.5 m
RA RB qL
Pb
+
2
L
qL
Pa
+
2
L
qL + P 51.5 kN
N
m
w 409.077
sallow 110 MPa
statics
kg
M
b a 9.81 2 b
m
s
RA 21.25 kN
RA aq +
RB 30.25 kN
RA + RB 51.5 kN
Sact 398 * 103 mm3
W
bL
1000
+
2
RA 22.273 kN
Pd
L
xm RA
q + W
xm 3.587 m greater than a so max. moment at load pt
LOCATE POINT OF ZERO SHEAR
xm RA
q
Mmax RA a xm 3.664 m
Mmax 39.924 kN # m
greater than dist. a to load P so zero shear is at load
point
Mmax RA a q a2
2
(q + W ) a2
2
Mmax 38.85 kN # m
smax Mmax
S act
smax 100.311 MPa
OK, less than 110 MPa
FIND REQUIRED SECTION MODULUS
Sreqd Mmax
sallow
Sreqd 353.182 * 103 mm3
select W 200 * 41.7
;
(Sact 398 * 103 mm3)
Calculate the required section modulus S if sallow 17,000 psi, L 28 ft,
P 2200 lb, and q 425 lb/ft. Then select a suitable I-beam (S shape) from
Table E-2(a), Appendix E, and recalculate S taking into account the weight of
the beam. Select a new beam size if necessary.
P
q
Problem 5.6-5 A simple beam AB is loaded as shown in the figure.
q
B
A
L
—
4
L
—
4
L
—
4
L
—
4
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SECTION 5.6
Design of Beams
415
Solution 5.6-5
NUMERICAL DATA
sa 17000 psi
L 28 ft
P 2200 lb
q 425
lb
ft
FIND REACTIONS (EQUAL DUE TO SYMMETRY) THEN MAX.
MOMENT AT CENTER OF BEAM
RA P
L
+ q
2
4
Mmax RA
RA 4.075 * 103 lb
qL L
L
1L
a +
b
2
4 4
24
Mmax 2.581 * 104 ft-lb
RECOMPUTE REACTIONS AND MAX. MOMENT THEN CHECK
lb
MAX. STRESS
w 25.4
ft
RA P
L
L
+ q + w
2
4
2
Mmax RA
RA 4.431 * 103 lb
qL L
L
1L
L 1L
a +
b w a
b
2
4 4
24
2 22
Mmax 2.83 * 104 ft-lb
smax Mmax (12)
Sact
smax 13,806 psi less than allowable so OK
Compute Sreqd & then select S shape
Sreqd Mmax (12)
sa
select S 10 * 25.4
Sreqd 18.221 in.3
;
(Sact 24.6 in. , w 25.4 lb/ft)
3
Problem 5.6-6 A pontoon bridge (see figure) is constructed of two
longitudinal wood beams, known an balks, that span between adjacent
pontoons and support the transverse floor beams, which are called
chesses.
For purposes of design, assume that a uniform floor load of
8.0 kPa acts over the chesses. (This load includes an allowance for
the weights of the chesses and balks.) Also, assume that the chesses
are 2.0 m long and that the balks are simply supported with a span
of 3.0 m. The allowable bending stress in the wood is 16 MPa.
If the balks have a square cross section, what is their
minimum required width bmin?
Solution 5.6-6
Chess
Pontoon
Balk
Pontoon bridge
FLOOR LOAD: W 8.0 kPa
ALLOWABLE STRESS: sallow 16 MPa
Lc length of chesses
2.0 m
Lb length of balks
3.0 m
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Stresses in Beams (Basic Topics)
LOADING DIAGRAM FOR ONE BALK
Section modulus S Mmax S
W total load
‹
wLbLc
q
b3
6
qL2b
(8.0 kN/m)(3.0 m)2
9,000 N # m
8
8
9,000 N # m
Mmax
562.5 * 106 m3
sallow
16 MPa
b3
562.5 * 106 m3 and b3 3375 * 106 m3
6
Solving, bmin 0.150 m 150 mm
wLc
W
2Lb
2
;
(8.0 kPa)(2.0 m)
2
8.0 kN/m
Problem 5.6-7 A floor system in a small building consists of
wood planks supported by 2 in. (nominal width) joists spaced at
distance s, measured from center to center (see figure). The span
length L of each joist is 10.5 ft, the spacing s of the joists is 16 in.,
and the allowable bending stress in the wood is 1350 psi. The uniform floor load is 120 lb/ft2, which includes an
allowance for the weight of the floor system itself.
Calculate the required section modulus S for the joists, and
then select a suitable joist size (surfaced lumber) from Appendix F,
assuming that each joist may be represented as a simple beam
carrying a uniform load.
Planks
s
Joists
Solution 5.6-7
s
L
s
Floor joists
Mmax qL2
1
(13.333 lb/in.)(126 in.)2 26,460 lb-in.
8
8
Required S 26,460 lb/in.
Mmax
19.6 in.3
sallow
1350 psi
From Appendix F: Select 2 * 10 in. joists
sallow 1350 psi
L 10.5 ft 126 in.
w floor load 120 lb/ft2 0.8333 lb/in.2
s spacing of joists 16 in.
q ws 13.333 lb/in.
;
;
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SECTION 5.6
Design of Beams
Problem 5.6-8 The wood joists supporting a plank floor (see figure) are
40 mm * 180 mm in cross section (actual dimensions) and have a span
length L 4.0 m. The floor load is 3.6 kPa, which includes the weight
of the joists and the floor.
Calculate the maximum permissible spacing s of the joists if the
allowable bending stress is 15 MPa. (Assume that each joist may be
represented as a simple beam carrying a uniform load.)
Solution 5.6-8
Spacing of floor joists
L 4.0 m
w floor load 3.6 kPa
sallow 15 MPa
s spacing of joists
q ws
S
SPACING OF JOISTS
2
bh
6
4 bh2sallow
3wL2
Substitute numerical values:
qL2
wsL2
Mmax 8
8
S
smax 2
smax 2
Mmax
wsL
bh
sallow
8sallow
6
4(40 mm)(180 mm)2(15 MPa)
3(3.6 kPa)(4.0 m)2
0.450 m 450 mm
;
;
417
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Stresses in Beams (Basic Topics)
q0
Problem 5.6-9 A beam ABC with an overhang from B to C is
constructed of a C 10 30 channel section (see figure). The beam
supports its own weight (30 lb/ft) plus a triangular load of maximum
intensity q0 acting on the overhang. The allowable stresses in tension
and compression are 20 ksi and 11 ksi, respectively.
Determine the allowable triangular load intensity q0,allow if the
distance L equals 3.5 ft.
A
C
B
L
L
2.384 in.
0.649 in.
C
3.033 in.
10.0 in.
Solution 5.6-9
NUMERICAL DATA
w 30
lb
ft
check tension on top
sat 20 ksi
sac 11 ksi
L 3.5 ft
c1 2.384 in.
from Table E-3(a)
MB MB c1
I22
q0allow c2 0.649 in.
I22 3.93 in.
4
MAX. MOMENT IS AT B (TENSION TOP, COMPRESSION BOTTOM)
MB wL
st L
1
2
+ q0L a Lb
2
2
3
1
1
wL2 + q0L2
2
3
Problem 5.6-10 A so-called “trapeze bar” in a hospital
room provides a means for patients to exercise while in bed
(see figure). The bar is 2.1 m long and has a cross section in
the shape of a regular octagon. The design load is 1.2 kN applied
at the midpoint of the bar, and the allowable bending stress is
200 MPa.
Determine the minimum height h of the bar. (Assume
that the ends of the bar are simply supported and that the
weight of the bar is negligible.)
3
L
2
MB s at
csat a
q0allow 628 lb/ft
I22
c1
I22
1
b wL2 d
c1
2
;
governs
check compression on bottom
q0allow 3
L
2
csac a
q0allow 1314
I22
1
b wL2 d
c2
2
lb
ft
C
h
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SECTION 5.6
Solution 5.6-10
Trapeze bar (regular octagon)
P 1.2 kN L 2.1 m
sallow 200 MPa
b 0.41421h
‹ Ic 1.85948(0.41421h)4 0.054738h4
Determine minimum height h.
SECTION MODULUS
MAXIMUM BENDING MOMENT
S
Mmax 419
Design of Beams
(1.2 kN)(2.1 m)
PL
630 N # m
4
4
PROPERTIES OF THE CROSS SECTION
Use Appendix D, Case 25, with n 8
b length of one side
b
360°
360°
45°
n
8
b
b
tan
(from triangle)
2
h
b
h
cot 2
b
Ic
0.054738h4
0.109476h3
h/2
h/2
MINIMUM HEIGHT h
M
s
630 N # m
3.15 * 106 m3
0.109476h3 200 MPa
s
M
S
S
h3 28.7735 * 106 m3 h 0.030643 m
‹ hmin 30.6 mm
;
ALTERNATIVE SOLUTION (n 8)
M
PL
4
b 45° tan
b
b
12 1 cot 12 1
2
2
b (12 1)h h (12 + 1)b
b
45°
For b 45°: tan
0.41421
h
2
45°
h
cot
2.41421
b
2
Ic a
S a
11 + 812 4
412 5 4
bb a
bh
12
12
412 5 3
bh
6
3PL
2(412 5)sallow
;
h3 28.7735 * 106 m3 hmin 30.643 mm
;
h3 Substitute numerical values:
MOMENT OF INERTIA
4
Ic b
b
nb
acot b a3 cot2 1 b
192
2
2
Ic 8b4
(2.41421)[3(2.41421)2 1] 1.85948b4
192
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Stresses in Beams (Basic Topics)
Problem 5.6-11 A two-axle carriage that is part of an overhead traveling
crane in a testing laboratory moves slowly across a simple beam AB
(see figure). The load transmitted to the beam from the front axle is 2200 lb
and from the rear axle is 3800 lb. The weight of the beam itself may be
disregarded.
3800 lb
5 ft
2200 lb
A
B
(a) Determine the minimum required section modulus S for the beam
if the allowable bending stress is 17.0 ksi, the length of the beam is
18 ft, and the wheelbase of the carriage is 5 ft.
(b) Select the most economical I-beam (S shape) from Table E-2(a), Appendix E.
18 ft
Solution 5.6-11
NUMERICAL DATA
P1 2200 lb
L 18 ft
P2 3800 lb
d 5 ft
sa 17 ksi
(a) FIND REACTION RA THEN AN EXPRESSION FOR MOMENT
UNDER LARGER LOAD P2; LET X DIST. FROM A
TO LOAD P2
RA P2 a
L (x + d )
Lx
b + P1 c
d
L
L
M2 RA x
M2 x cP2 a
L (x + d )
Lx
b + P1 c
dd
L
L
xP2 LP2 x2 xP1L P1x2 xP1d
L
Take derivative of MA & set to zero to find max.
bending moment at x x m
M2 xm (P1 + P2) L P1d
2 (P1 + P2)
xm 8.083 ft
L (xm + d)
L xm
b + P1 c
d
L
L
RA 2694 lb
RA P2 a
Mmax xm cP2 a
L(xm d)
L xm
b P1 c
dd
L
L
Mmax 21780 ft-lb
Sreqd Mmax
sa
Sreqd 15.37 in.3
;
(b) SELECT MOST ECONOMICAL S SHAPE FROM
TABLE E-2(A)
select S8 * 23
;
Sact 16.2 in.3
d xP2LP2x2 xP1LP1x2 xP1d
a
b
dx
L
P2L 2P2x + P1L 2P1x P1d
L
P2L 2P2x + P1L 2P1x P1d 0
Problem 5.6-12 A cantilever beam AB of circular cross section and length
L 450 mm supports a load P 400 N acting at the free end (see figure).
The beam is made of steel with an allowable bending stress of 60 MPa.
Determine the required diameter dmin of the beam, considering the
effect of the beam’s own weight.
A
B
d
P
L
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SECTION 5.6
Solution 5.6-12
MINIMUM DIAMETER
Mmax sallow S
PL +
77.0 kN/m
3
pgd 2L2
pd 3
sallow a
b
8
32
Rearrange the equation:
WEIGHT OF BEAM PER UNIT LENGTH
q ga
421
Cantilever beam
L 450 mm
P 400 N
sallow 60 MPa
g weight density of steel
DATA
Design of Beams
sallow d 3 4gL2 d 2 pd 2
b
4
32 PL
0
p
(Cubic equation with diameter d as unknown.)
MAXIMUM BENDING MOMENT
Substitute numerical values (d meters):
q L2
pgd3L2
PL +
Mmax PL +
2
8
(60 * 106 N/m2)d3 4(77,000 N/m3)(0.45 m)2d2
SECTION MODULUS
pd 3
S
32
32
(400 N)(0.45 m) 0
p
60,000d 3 62.37d 2 1.833465 0
Solve the equation numerically:
d 0.031614 m
;
q
Problem 5.6-13 A compound beam ABCD (see figure) is
supported at points A, B, and D and has a splice at point C.
The distance a 6.25 ft, and the beam is a S 18 70
wide-flange shape with an allowable bending stress of 12,800 psi.
dmin 31.61 mm
A
B
C
D
Splice
(a) If the splice is a moment release, find the allowable
4a
uniform load qallow that may be placed on top of the
beam, taking into account the weight of the beam itself.
[See figure part (a).]
(b) Repeat assuming now that the splice is a shear release, as in figure part (b).
a
4a
(a)
(b)
Moment Shear
release release
Solution 5.6-13
NUMERICAL DATA
lb
S 103 in.3
w 70
ft
a 6.25 ft
sa 12800 psi
MZ 2.000E+00
@ 2.000E+00
MZ 9.453E–01
@ 1.375E+00
×
×
(a) MOMENT RELEASE AT C-GIVES MAX. MOMENT AT B
(SEE MOMENT DIAGRAM) 2.5 q a2
Mmax
Mmax [1qallow + w2 a2 (2.5)]
S
and Mmax s a S
sa lb
S 103 in.3
ft
sa 12800 psi
a 6.25 ft
w 70
MZ –2.500E+00
@ 4.000E+00
×
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sa S
12 in./ft
qallow Page 422
2.5 a2
lb
qallow 1055
ft
MZ 8.00E+00
w
;
for moment release
(b) SHEAR RELEASE AT C-GIVES MAX. MOMENT AT C
(SEE MOMENT DIAGRAM) 8 q a2
sa S
12 in./ft
qallow 8a2
qallow 282
;
for shear release
w
Problem 5.6-14 A small balcony constructed of wood is
supported by three identical cantilever beams (see figure).
Each beam has length L1 2.1 m, width b, and height h 4b/3.
The dimensions of the balcon floor are L1 * L2,
with L2 2.5 m. The design load is 5.5 kPa acting over the
entire floor area. (This load accounts for all loads except
the weights of the cantilever beams, which have a weight
density g 5.5 kN/m3.) The allowable bending stress in the
cantilevers is 15 MPa.
Assuming that the middle cantilever supports 50% of the
load and each outer cantilever supports 25% of the load,
determine the required dimensions b and h.
Solution 5.6-14
lb
ft
4b
h= —
3
L2
b
L1
Compound beam
MAXIMUM BENDING MOMENT
(q q0)L21
1
(6875 N/m7333b2)(2.1 m)2
Mmax 2
2
15,159 + 16,170b2 (N # m)
bh2
8b3
6
27
Mmax sallow S
L1 2.1 m L2 2.5 m Floor dimensions: L1 * L2
Design load w 5.5 kPa
g 5.5 kN/m3 (weight density of wood beam)
sallow 15 MPa
S
MIDDLE BEAM SUPPORTS 50% OF THE LOAD.
Rearrange the equation:
‹ q wa
L2
2.5 m
b (5.5 kPA)a
b 6875 N/m
2
2
WEIGHT OF BEAM
q0 gbh 4gb2
4
(5.5 kN/m2) b2
3
3
7333b2 (N/m)
(b meters)
15,159 + 16,170b2 (15 * 106 N/m2)a
8b3
b
27
(120 * 106)b3 436,590b2 409,300 0
SOLVE NUMERICALLY FOR DIMENSION b
4b
0.2023 m
h
b 0.1517 m
3
REQUIRED DIMENSIONS
b 152 mm
h 202 mm
;
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SECTION 5.6
Problem 5.6-15 A beam having a cross section in the form of an
unsymmetric wide-flange shape (see figure) is subjected to a negative
bending moment acting about the z axis.
Determine the width b of the top flange in order that the stresses at
the top and bottom of the beam will be in the ratio 4:3, respectively.
423
Design of Beams
y
b
1.5 in.
1.25 in.
z
12 in.
C
1.5 in.
16 in.
Solution 5.6-15
Unsymmetric wide-flange beam
AREAS OF THE CROSS SECTION (in.2)
A1 1.5b A2 (12)(1.25) 15 in.2
A3 (16)(1.5) 24 in.2
A A1 + A2 + A3 39 + 1.5b (in.2)
FIRST MOMENT OF THE CROSS-SECTIONAL
AREA ABOUT THE LOWER EDGE B-B
QBB gyi Ai (14.25)(1.5b) + (7.5)(15) + (0.75)(24)
Stresses at top and bottom are in the ratio 4:3.
Find b (inches)
h height of beam 15 in.
LOCATE CENTROID
stop
c1
4
sbottom
c2
3
4
60
8.57143 in.
c1 h 7
7
3
45
c2 h 6.42857 in.
7
7
130.5 + 21.375b (in.3)
DISTANCE c2 FROM LINE B-B TO THE CENTROID C
c2 QBB
130.5 + 21.375b
45
in.
A
39 + 1.5b
7
SOLVE FOR b
(39 + 1.5b)(45) (130.5 + 21.375b)(7)
82.125b 841.5 b 10.25 in.
;
y
Problem 5.6-16 A beam having a cross section in the form of a channel
(see figure) is subjected to a bending moment acting about the z axis.
Calculate the thickness t of the channel in order that the bending stresses
at the top and bottom of the beam will be in the ratio 7:3, respectively.
t
z
t
C
152 mm
t
55 mm
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Stresses in Beams (Basic Topics)
Solution 5.6-16
ratio of top to bottom stresses c1/c2 7/3
NUMERICAL DATA
h 152 mm
b 55 mm
1 11385 186 t + t2
2
131 + t
take 1st moments to find distances c1 & c2
1st moments about base
c2 J
t
b
(h 2t) (t) + 2bt a b
2
2
2bt + t (h 2t)
c1 b c2
c2 t
55
(152 2t) (t) + 2.55t a b
2
2
t
55
(152 2t) (t) + 2.55t a b
2
2
2.55t + t(152 2t)
1 11385 186 t + t
2
131 + t
2
111385 186 t + t22
76 t + t2 3025
7/3
7 a 76 tt2 3025b d 0
t2 109 t + 1298 0
t
109 11092 4 (1298)
2
Problem 5.6-17 Determine the ratios of the weights of three beams that
have the same length, are made of the same material, are subjected to the
same maximum bending moment, and have the same maximum bending
stress if their cross sections are (1) a rectangle with height equal to twice
the width, (2) a square, and (3) a circle (see figures).
Solution 5.6-17
K
2.55 t + t (152 2 t)
c3 c a 11385186 tt2 b d
2.55t + t (152 2t)
c1 55 c1 55
t
(152 2t) (t) + 2.55 t a b
2
2
t 13.61 mm
h = 2b
b
;
a
a
d
Ratio of weights of three beams
Beam 1: Rectangle (h 2b)
Beam 2: Square (a side dimension)
Beam 3: Circle (d diameter)
L, g, Mmax, and smax are the same in all three beams.
M
S section modulus S s
Since M and s are the same, the section moduli must
be the same.
bh2
2b3
(1) RECTANGLE: S 6
3
A1 2b2 2 a
3S 1/3
b a b
2
3S 2/3
b 2.6207S2/3
2
(2) SQUARE: S a3
6
a (6S)1/3
A2 a2 (6S)2/3 3.3019 S2/3
(3) CIRCLE: S pd 3
32
A3 d a
32S 1/3
b
p
pd 2
p 32S 2/3
a
b 3.6905 S2/3
4
4 p
Weights are proportional to the cross-sectional areas
(since L and g are the same in all 3 cases).
W1 : W2 : W3 A1 : A2 : A3
A1 : A2 : A3 2.6207 : 3.3019 : 3.6905
W1 : W2 : W3 1 : 1.260 : 1.408
;
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SECTION 5.6
425
Design of Beams
t
Problem 5.6-18 A horizontal shelf AD of length L 915 mm, width
b 305 mm, and thickness t 22 mm is supported by brackets at B and C
[see part (a) of the figure]. The brackets are adjustable and may be placed
in any desired positions between the ends of the shelf. A uniform load of
intensity q, which includes the weight of the shelf itself, acts on the shelf
[see part (b) of the figure].
Determine the maximum permissible value of the load q if the allowable
bending stress in the shelf is sallow 7.5 MPa and the position of the supports is
adjusted for maximum load-carrying capacity.
A
B
D
C
b
L
(a)
q
A
D
B
C
L
(b)
Solution 5.6-18
NUMERICAL DATA
Substitute x into the equation for either M1 or |M2|:
b 305 mm
L 915 mm
t 22 mm
sallow 7.5 MPa
Mmax qL2
(3 2 12)
8
Mmax sallow S sallow a
MOMENT DIAGRAM
Eq. (1)
bt 2
b
6
Eq. (2)
Equate Mmax from Eqs. (1) and (2) and solve for q:
qmax 4bt2sallow
3L2(3 2 12)
Substitute numerical values:
For maximum load-carrying capacity, place the
supports so that M1 |M2|.
Let x length of overhang
M1 ‹
qL
(L 4x)
8
|M2| qL
qx2
(L 4x) 8
2
Solve for x: x L
(12 1)
2
qx2
2
qmax 10.28 kN/m
;
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Stresses in Beams (Basic Topics)
Problem 5.6-19 A steel plate (called a cover plate) having cross-sectional
dimensions 6.0 in. * 0.5 in. is welded along the full length of the bottom flange
of a W 12 * 50 wide-flange beam (see figure, which shows the beam cross
section).
What is the percent increase in the smaller section modulus (as compared to
the wide-flange beam alone)?
W 12 50
6.0 0.5 in. cover plate
Solution 5.6-19
FIND I ABOUT HORIZ. CENTROIDAL AXIS
NUMERICAL PROPERTIES FOR W 12 * 50
(FROM TABEL E-1(a))
2
A 14.6 in.
d 12.2 in.
c1 c2
c1 d
2
I 391 in.4
Ih I + A a c1 + (6) (0.5) a c2 c1 d
0.5
+ (6) (0.5) ad +
b
2
2
A + (6) (0.5)
c2 (d + 0.5) c1
0.5 2
b
2
Ih 491.411in.4
S 64.2 in.3
FIND SMALLER SECTION MODULUS
Ih
Stop Stop 68.419 in.3
c1
% increase in smaller section modulus
Stop S
;
(100) 6.57%
S
FIND CENTROID OF BEAM WITH COVER PLATE (TAKE
1ST MOMENTS ABOUT TOP TO FIND c1 7 c2)
A
d 2
1
b +
(6) (0.5)3
2
12
c1 7.182 in.
c2 5.518 in.
Problem 5.6-20 A steel beam ABC is simply supported
at A and B and has an overhang BC of length L 150 mm
(see figure). The beam supports a uniform load of intensity
q 4.0 kN/m over its entire span AB and 1.5q over BC.
The cross section of the beam is rectangular with width b
and height 2b. The allowable bending stress in the steel is
sallow 60 MPa, and its weight density is 77.0 kN/m3.
1.5 q
q
C
A
2b
B
2L
L
(a) Disregarding the weight of the beam, calculate the required width b of the rectangular cross section.
(b) Taking into account the weight of the beam, calculate the required width b.
b
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SECTION 5.6
Design of Beams
427
Solution 5.6-20
NUMERICAL DATA
L 150 mm
sa 60 MPa
(b) NOW MODIFY-INCLUDE BEAM WEIGHT
kN
q4
m
kN
g 77 3
m
w gA
and
and
L2
2
Equate Mmax1 to Mmax2 & solve for bmin
at B
Mmax2 s a S
2
3
a sa b b3 1gL22 b2 qL2 0
3
4
2
S b3
3
Insert numerical values, then solve for b
Equate Mmax1 to Mmax2 & solve for bmin
bmin 11.92 mm
1
9 qL2 3
bmin a
b
8 sa
bmin 11.91 mm
L2
2
2
Mmax s a a b3 b
3
Mmax (1.5q + w)
(a) IGNORE BEAM SELF WEIGHT-FIND bmin
Mmax1 1.5 q
w g 12b22
;
;
Problem 5.6-21 A retaining wall 5 ft high is constructed of
horizontal wood planks 3 in. thick (actual dimension) that are
supported by vertical wood piles of 12 in. diameter (actual
dimension), as shown in the figure. The lateral earth pressure
is p1 100 lb/ft2 at the top of the wall and p2 400 lb/ft2 at
the bottom.
Assuming that the allowable stress in the wood is 1200 psi,
calculate the maximum permissible spacing s of the piles.
(Hint: Observe that the spacing of the piles may be
governed by the load-carrying capacity of either the planks
or the piles. Consider the piles to act as cantilever beams
subjected to a trapezoidal distribution of load, and consider
the planks to act as simple beams between the piles. To be on
the safe side, assume that the pressure on the bottom plank is
uniform and equal to the maximum pressure.)
3 in.
p1 = 100 lb/ft2
12 in.
diam.
12 in.
diam.
s
5 ft
3 in.
Top view
p2 = 400 lb/ft2
Side view
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CHAPTER 5
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Stresses in Beams (Basic Topics)
Solution 5.6-21
Retaining wall
(1) PLANK AT THE BOTTOM OF THE DAM
t thickness of plank 3 in.
b width of plank (perpendicular to the plane
of the figure)
p2 maximum soil pressure
400 lb/ft2 2.778 lb/in.2
s spacing of piles
q p2b sallow 1200 psi
S section modulus
2
Mmax qs
p2bs
8
8
Mmax s allow S
2
or
S
bt
6
p2bs 2
bt 2
sallow a
b
8
6
Solve for s:
s
4sallow t 2
A
3p 2
2
72.0 in.
q1 p1s
q2 p2s
d diameter of pile 12 in.
Divide the trapezoidal load into two triangles
(see dashed line).
Mmax S
pd 3
32
Mmax sallow S
or
pd 3
sh 2
(2p1 + p2) sallow a
b
6
32
Solve for s:
s
(2) VERTICAL PILE
h 5 ft 60 in.
p1 soil pressure at the top
100 lb/ft 2 0.6944 lb/in.2
1
sh2
1
2h
h
(q1) (h)a b (q2)(h)a b (2p1 p2)
2
3
2
3
6
3psallow d 3
16h2 (2p1 + p2)
PLANK GOVERNS
81.4 in.
smax 72.0 in.
Problem 5.6-22 A beam of square cross section (a length of each side) is bent in
the plane of a diagonal (see figure). By removing a small amount of material at the top
and bottom corners, as shown by the shaded triangles in the figure, we can increase the
section modulus and obtain a stronger beam, even though the area of the cross section
is reduced.
(a) Determine the ratio b defining the areas that should be removed in order to obtain
the strongest cross section in bending.
(b) By what percent is the section modulus increased when the areas are removed?
;
y
a
z
ba
C
a
ba
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SECTION 5.6
Solution 5.6-22
removed
429
Design of Beams
Beam of square cross section with corners
RATIO OF SECTION MODULI
S
(1 + 3b)(1 b)2
S0
Eq. (1)
GRAPH OF EQ. (1)
a length of each side
ba amount removed
Beam is bent about the z axis.
ENTIRE CROSS SECTION (AREA 0)
I0 a4
12
c0 a
12
S0 I0
a3 12
c0
12
(a) VALUE OF b
S/S0
d S
a b 0
db S0
SQUARE mnpq (AREA 1)
I1 FOR A MAXIMUM VALUE OF
(1 b)4a4
12
Take the derivative and solve this equation for b .
b
PARALLELOGRAM mm, n, n (AREA 2)
1
I2 (base)(height)3
3
1
9
;
(b) MAXIMUM VALUE OF S/S0
(1 b)a 3
ba4
1
I2 (ba12)c
d (1 b)3
3
6
12
Substitute b 1/9 into Eq. (1). (S/S0)max 1.0535
The section modulus is increased by 5.35% when
;
the triangular areas are removed.
REDUCED CROSS SECTION (AREA qmm, n, p, pq)
a4
I I1 + 2I2 (1 + 3b)(1 b)3
12
c
(1 b) a
12
S
I
12 a3
(1 + 3b)(1 b)2
c
12
b
—
9
Problem 5.6-23 The cross section of a rectangular beam having
width b and height h is shown in part (a) of the figure. For reasons
unknown to the beam designer, it is planned to add structural projections
of width b/9 and height d to the top and bottom of the beam [see part
(b) of the figure].
For what values of d is the bending-moment capacity of the beam
increased? For what values is it decreased?
d
h
b
(a)
h
d
b
—
9
(b)
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Page 430
Stresses in Beams (Basic Topics)
Solution 5.6-23
Beam with projections
Graph of
S2
d
versus
S1
h
d
h
0
0.25
0.50
0.75
1.00
(1) ORIGINAL BEAM
I1 bh3
12
c1 h
2
S1 S2
S1
1.000
0.8426
0.8889
1.0500
1.2963
I1
bh2
c1
6
(2) BEAM WITH PROJECTIONS
I2 1 8b 3
1 b
a bh +
a b(h + 2d)3
12 9
12 9
b
[8h3 + (h + 2d)3]
108
1
h
c2 + d (h + 2d)
2
2
S2 b[8h3 + (h + 2d)3]
I2
c2
54(h + 2d)
RATIO OF SECTION MODULI
3
3
b [8h + (h + 2d) ]
S2
S1
9(h + 2d)(bh2)
8 + a1 +
2d
9a 1 +
b
h
EQUAL SECTION MODULI
Set
S2
d
1 and solve numerically for .
S1
h
d
0.6861 and
h
d
0
h
2d 3
b
h
Moment capacity is increased when
d
7 0.6861
;
h
Moment capacity is decreased when
d
6 0.6861
;
h
NOTES:
S2
2d 3
2d
1 when a1 +
b 9a 1 +
b + 80
S1
h
h
or
d
0.6861 and 0
h
3 1
S2
d
14
is minimum when 0.2937
S1
h
2
a
S2
b 0.8399
S1 min
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SECTION 5.7
431
Nonprismatic Beams
Nonprismatic Beams
Problem 5.7-1 A tapered cantilever beam AB of length
L has square cross sections and supports a concentrated load
P at the free end [see figure part (a)]. The width and height of
the beam vary linearly from hA at the free end to hB at the
fixed end.
Determine the distance x from the free end A to the cross
section of maximum bending stress if hB 3hA.
(a) What is the magnitude smax of the maximum bending
stress? What is the ratio of the maximum stress to the
largest stress B at the support?
(b) Repeat (a) if load P is now applied as a uniform load
of intensity q P/L over the entire beam, A is restrained
by a roller support and B is a sliding support [see figure,
part (b)].
q = P/L
B
hA
A
B
A
hB
x
P
Sliding
support
x
L
L
(a)
(b)
Solution 5.7-1
(a) FIND MAX. BENDING STRESS FOR TAPERED
sB s(L)
CANTILEVER
h(x) hA a1 +
2x
b
L
M(x)
s(x) S(x)
s(x) s(x) S(x) h(x)3
6
2x
bd
L
3
6PxL3
hA3 (L
2PL
9hA 3
4PL
6(P)(x)
chA a1 +
sB smax
9hA 3
sB
2PL
smax
2
sB
9hA 3
(b) REPEAT (A) BUT NOW FOR DISTRIBUTED UNIFORM
LOAD OF P/L OVER ENTIRE BEAM
+ 2x)3
d
s(x) 0 then solve for xmax
dx
a Fv 0
6PxL3
d
c 3
d 0
dx hA (L + 2x)3
M(x) c c RA x c6P
L3
hA3 (L 2x)3
L + 4x
hA3 (L + 2x)4
0
L
smax s a b
4
smax 4PL
9hA 3
36Px
so
L3
hA3 (L2x)4
x
smax ;
;
d 0
L
4
L
6P L3
4
L 3
hA3 a L + 2 b
4
M(x) Px M(x)
s(x) S(x)
RA P
P
x
xa b d d
L
2
1 2P
x
2 L
Px s(x) 1 2P
x
2 L
c hA a 1 +
s(x) 3xP (2L + x)
2x 3
bd
L
6
L2
hA3 (L + 2x)3
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Stresses in Beams (Basic Topics)
xmax 0.20871 L
d
s(x) 0 then solve for xmax
dx
smax s(0.20871 L)
PL
smax 0.394 3
;
hA
2
d
L
c 3xP ( 2L x) 3
d 0
dx
hA (L2x)3
c 3P ( 2L + x)
3xP
L2
hA3(L
sB s(L) So
3
+ 2x)
L2
sB hA3 (L + 2x)3
2
+ 18xP (2L + x)
L
hA3 (L
4
+ 2x)
d 0
smax
sB
PL
9hA3
a 0.39385
PL
hA3
b
PL
9hA3
smax
3.54
sB
Simplifying
;
L2 5xL + x 2 0 so
xmax
5 152 4
L
2
Problem 5.7-2 A tall signboard is supported by two vertical beams consisting of thin-walled, tapered circular tubes
[see figure]. For purposes of this analysis, each beam may be represented as a cantilever AB of length L 8.0 m subjected to
a lateral load P 2.4 kN at the free end. The tubes have constant thickness t 10.0 mm and average diameters dA 90 mm
and dB 270 mm at ends A and B, respectively.
Because the thickness is small compared to the diameters, the moment of inertia at any cross section may be obtained
from the formula I pd3t/8 (see Case 22, Appendix D), and therefore, the section modulus may be obtained from the formula
S pd2t/4.
(a) At what distance x from the free end does the maximum bending stress occur? What is the magnitude smax of the
maximum bending stress? What is the ratio of the maximum stress to the largest stress sB at the support?
(b) Repeat (a) if concentrated load P is applied upward at A and downward uniform load q(x) 2P/L is applied over the
entire beam as shown. What is the ratio of the maximum stress to the stress at the location of maximum moment?
2P
q(x) = —
L
P = 2.4 kN
Wind
load
B
A
t
B
A
x
P
d
L = 8.0 m
t = 10.0 mm
x
L = 8.0 m
(b)
dA = 90 mm
(a)
dB = 270 mm
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SECTION 5.7
433
Nonprismatic Beams
Solution 5.7-2
(a) FIND MAX. BENDING STRESS FOR TAPERED CANTILEVER
d(x) d A a1 +
2x
b
L
2
S(x) dA 90 mm
M(x)
s(x) S(x)
4P
s(x) pt
4P
xL2
s(x) d
c 2
pt dA (L + 2x)2
J
c dA a 1 +
xL2
d 4P
c
c 2
dd 0
dx pt dA (L + 2x)2
xL2
P
d 0
pt dA2 (L + 2x)3
L + 2x
ptdA2 (L + 2x)3
L
4m
2
d 0
smax L 2
aL + 2 b
2
PL
4 P L
9 pt dA2
P x
x b
L 2
M(x) a Px 2
M(x) Pxa
L + x
b
L
M(x)
S(x)
s(x) Px a
L + x
b
L
pt
2x 2
cdA a 1 +
bd
4
L
L
ptdA2 (L 2x)2
tension on top, compression on bottom of beam
d
L
c 4Px (L x)
d 0
2
dx
ptdA (L 2x)2
¥
c 4P ( L + x)
4Px
L
ptdA2 (L + 2x)2
L
ptdA2 (L
+ 2x)2
16Px (L x)
2ptdA2
Stress at support sB s(L)
sB (b) REPEAT (A) BUT NOW ADD DISTRIBUTED LOAD
d
s(x) 0 then solve for xmax
dx
;
L 2
L
2
dA2
2p (0.010) (0.090)2
;
smax 37.7 MPa
s(x) 4Px (L x)
L
smax s a b
2
4P
≥
pt
2x
bd
L
K
2
s(x) L2
P
c4
2
pt dA (L + 2x)2
smax ;
(2400) (8)
smax x
d
s(x) 0 then solve for xmax
dx
so xmax 4 P L
b
9 pt dA2
Evaluate using numerical data
dB 270 mm
c4PL2
a
smax
9
sB
8
L 8 m t 10 mm
or
2ptdA2
smax
sB
pd(x) t
4
P 2.4 kN
16
PL
OR simplifying
L
4
xmax 2 m
L
ptdA2 (L 2x)3
c 4PL2
so xmax ;
d 0
L + 4x
ptdA2 (L + 2x)3
d 0
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Stresses in Beams (Basic Topics)
stress at support
L
smax s a b
4
sB s(L)
L
L
smax 4P aL b
4
4
smax J
L
p t dA2 aL 2
PL
ptdA2 (L + 2L2)
sB 0 so no ratio of smax/sB is possible
L 2
b K
4
MAX. MOMENT AT L/2 SO COMPARE
3 p t dA2
evaluate using numerical data
Stress at location of max. moment
L
L
L
sa b 4P a L b
2
2
2
L8m
P 2.4 kN
d A 90 mm
t 10 mm
dB 270 mm
smax L
sB 4PL ( L + L)
L
ptdA2 aL 2
L 2
b
2
1
L
L
sa b P
2
4 ptdA2
(2400) (8)
3p (0.010) (0.090)2
smax 25.2 MPa
;
PL
smax/s(L/2) 3ptdA2
L
1
a P
b
4 ptdA2
4
3
;
Problem 5.7-3 A tapered cantilever beam AB having rectangular cross sections is subjected to a concentrated
load P 50 lb and a couple M0 800 lb-in. acting at the free end [see figure part (a)]. The width b of the beam is
constant and equal to 1.0 in., but the height varies linearly from hA 2.0 in. at the loaded end to hB 3.0 in.
at the support.
(a) At what distance x from the free end does the maximum bending stress smax occur? What is the magnitude
smax of the maximum bending stress? What is the ratio of the maximum stress to the largest stress sB at
the support?
(b) Repeat (a) if, in addition to P and M0, a triangular distributed load with peak intensity q0 3P/L acts upward
over the entire beam as shown. What is the ratio of the maximum stress to the stress at the location of maximum
moment?
P = 50 lb
P = 50 lb
A M0 = 800 lb-in.
hA =
2.0 in.
B
hB =
3.0 in.
x
b = 1.0 in.
3P
q0 = —
L
A M0 = 800 lb-in.
x
L = 20 in.
(a)
b = 1.0 in.
L = 20 in.
(b)
B
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SECTION 5.7
Nonprismatic Beams
435
Solution 5.7-3
(a) FIND MAX. BENDING STRESS FOR TAPERED CANTILEVER
FIG. (A)
x
b
h(x) hA a1 +
2L
numerical data
6
24P
+ x)2
L2
bhA2 (2L
+ x)2
L2
0
bhA2 (2L x)3
2PL + Px + 2M0
d 0
OR simplifying c24L2
bhA2 (2L + x)3
2 1PL M02
so x P
xmax 8 in.
;
agrees with plot at left
M(x)
S(x)
Evaluate max. stress & stress at B using
numerical data
2000
smax s(8)
1500
sB s(20)
smax
1.042
sB
M(x)
(in.-lb)
1000
0
10
x (in.)
20
1260
1240
σ (x)
(psi)
1220
20
;
sB 1200 psi
;
h(x) hA a 1 +
x
b
2L
4
PL
5
P
q0 3
L
M0 800 in.-lb
I(x) 10
x (in.)
smax 1250 psi
(b) FIND MAX. BENDING STRESS FOR TAPERED
CANTILEVER, FIG. (B)
M0 0
L2
bhA2 (2L
2 124Px 24M02
6
M(x) Px + M0
1200
2
x
bd
2L
d
L2
c24 1Px + M02
d 0
dx
bhA2 (2L + x)2
2
x
b chA a1 +
bd
2L
500
b c hA a 1 +
d
s(x) 0 then solve for xmax
dx
4
M0 PL M0 800 in.-lb
5
bh(x)3
I(x)
I(x) S(x) h(x)
12
2
bh(x)2
S(x) 6
s(x) Px + M0
s(x) 24 1Px + M02
P 50 lb L 20 in.
hA 2 in. hB 3 in. b 1 in.
S(x) s(x) bh(x)3
12
S(x) I(x)
h(x)
2
S(x) bh(x)2
6
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S(x) Page 436
Stresses in Beams (Basic Topics)
b chA a1 +
2
x
bd
2L
then solve for xmax
d
s(x) c124PL 12x2 q02
dx
6
1 x
x
a q0 b x
2
L
3
M(x) Px + M0 +
s(x) d
s(x) 0
dx
L
bhA2 (2L + x)2
M(x)
S(x)
4x3q0) *
1500
2 (24PxL + 24M0L
L
bhA2 (2L
+ x)3
d 0
Simplifying
12PL2 + 6PxL + 6x2 q0L
+ x3 q0 + 12M0 L 0
M(x)
1000
(in.-lb)
Solve for xmax
xmax 4.642 in.
;
Max. stress & stress at B
500
0
10
x (in.)
20
smax s (xmax)
smax 1235 psi
1400
sB s (20)
;
sB 867 psi
FIND MAX. MOMENT AND STRESS AT LOCATION OF MAX.
1200
MOMENT
1000
d
M(x) 0
dx
σ (x)
(psi)
xm 800
0
10
x (in.)
Px + M0 s(x) b chA a1 +
q0 x3
6L
2
x
bd
2L
6
s(x) 41 6PxL 6 M0 L + x3 q02
L
*
bhA2 (2L
+ x)2
20
P (2L)
A q0
sm s(xm)
smax
1.215
sm
q0x3
d
aPx + M0 b 0
dx
6L
xm 16.33 in.
sm 1017 psi
;
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SECTION 5.7
Nonprismatic Beams
437
Problem 5.7-4 The spokes in a large flywheel are modeled as beams fixed at one end and loaded by a force P and a
couple M0 at the other (see figure). The cross sections of the spokes are elliptical with major and minor axes (height and width,
respectively) having the lengths shown in the figure part (a). The cross-sectional dimensions vary linearly from end A to end B.
Considering only the effects of bending due to the loads P and M0, determine the following quantities.
(a)
(b)
(c)
(d)
(e)
The largest bending stress sA at end A
The largest bending stress sB at end B
The distance x to the cross section of maximum bending stress
The magnitude smax of the maximum bending stress
Repeat (d) if uniform load q(x) 10P/3L is added to loadings P and M0, as shown in the figure part (b).
P = 12 kN
M0 = 10 kN•m
10P
q(x) = —
3L
B
A
P
x
M0
L = 1.25 m
A
B
x
L = 1.25 m
hA = 90 mm
hB = 120 mm
(b)
bA = 60 mm
bB = 80 mm
(a)
Solution 5.7-4
(a-d) FIND MAX. BENDING STRESS FOR TAPERED
30
CANTILEVER
numerical data
L 1.25 m bA 60 mm hA 90 mm
bB 80 mm hB 120 mm
M(x)
20
(kN•m)
P 12 kn M0 10 kN # m
h(x) hA a1 I(x) x
b
3L
p b(x) h(x)3
64
b(x) bA a 1 +
S(x) p b(x) h(x)2
S(x) 32
S(x) p bA hA2 a1 +
32
x 3
b
3L
I(x)
h(x)
2
x
b
3L
10
0
0.5
x (m)
1
0
0.5
x (m)
1
240
230
σ (x)
220
(MPa)
210
200
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CHAPTER 5
Page 438
Stresses in Beams (Basic Topics)
M(x) Px + M0
s(x) s(x) M(x)
S(x)
I(x) p b(x) h(x)3
64
S(x) p b(x) h(x)2
32
Px + M0
p bA hA2 a1 +
x 3
b
3L
32
s(x) 864 a
Px + M0
d
s (x) 0
dx
p bA hA2
b a
L3
(3L + x)3
b
S(x) s(x) p bAhA2 (3L + x)3
Px + M0
L3
0
2592
2
p bAhA (3L + x)4
M(x)
S(x)
10
5
3PL + 2Px + 3M0
p bAhA2 (3L + x)4
d 0
0
3(PL M0)
so xmax 2P
xmax 0.625 m
;
0
σ (x)
(MPa)
Evaluate using numerical data
smax 231 MPa
;
100
sA s(0)
sB s(L)
smax
1.045
sB
sA 210 MPa
sB 221 MPa
;
;
0
(e) FIND MAX. BENDING STRESS INCLUDING
UNIFORM LOAD
bB 80 mm
P 12 kN
hB 120 mm
M0 10 kN # m
x
b
h(x) hA a1 +
3L
b(x) bA a1 +
x
b
3L
1
0.5
x (m)
1
200
smax s(xmax)
bA 60 mm
0.5
x (m)
300
agrees with plot above
L 1.25 m
10 P x2
3 L 2
M(x)
(kN•m)
OR simplfying
c864L3
32
15
L3
P
I(x)
h(x)
2
x 3
b
3L
p bA hA2 a 1 +
M(x) P x + M0 then solve for xmax
Px + M0
d
L3
c864
d 0
2
dx
p bAhA (3L + x)3
864
S(x) hA 90 mm
0
P x + M0 s(x) ≥
10 P x2
3 L 2
p bA hA2 a 1 +
32
x 3
b
3L
¥
s(x) 288 13 P x L 3 M0 L
L2
+ 5 P x22
p bA hA2 (3 L + x)3
d
s(x) 0 then solve for xmax
dx
05Ch05.qxd
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Page 439
SECTION 5.7
L
*
pbAhA2 (3L
+ x)3
9PL2 36PxL + 5Px2 9M0L 0
d 0
Solving for x max: xmax 0.105m
L2
d
s (x) c(864 PL 2880 P x)
dx
p bA hA2 (3 L x)3
3 1864 P x L 864 M0 L 1440Px22
L2
*
pbAhA2 (3L
+ x)4
439
OR
d
c 288 13 Px L 3 M0 L + 5 P x22
dx
2
Nonprismatic Beams
d 0
solution agrees with plot above, evaluate using
numerical data
smax s(xmax)
sA s(0)
sB s(L)
smax 214 MPa
;
sA 210 MPa
;
sB 0 MPa
;
OR simplifying
(288 L2)
c9PL2 36PxL + 5Px2 9M0L d
cpbAhA2 (3L + x)4 d
0
Problem 5.7-5 Refer to the tapered cantilever beam of solid circular cross section shown in Fig. 5-24 of Example 5-9.
(a) Considering only the bending stresses due to the load P, determine the range of values of the ratio dB/dA for which
the maximum normal stress occurs at the support.
(b) What is the maximum stress for this range of values?
Solution 5.7-5
Tapered cantilever beam
FROM EQ. (5-32), EXAMPLE 5-9
s1 32Px
Eq. (1)
x 3
pcdA + (dB dA)a b d
L
After simplification:
FIND THE VALUE OF x THAT MAKES s1 A MAXIMUM
Let s1 u
v
ds1
dx
va
du
dv
b ua b
dx
dx
2
v
x 3
N p cdA + (dB dA)a b d [32P]
L
x 2 1
[32Px][p] [3]cdA + (dB dA)a b d c (dB dA) d
L
L
N
D
x 2
x
N 32pPcdA + (dB dA)a b d cdA 2(dB dA) d
L
L
x 6
D p 2 cdA + (dB dA) d
L
ds1
N
dx
D
x
32PcdA 2(dB dA) d
L
x 4
pcdA + (dB dA) a b d
L
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CHAPTER 5
Page 440
Stresses in Beams (Basic Topics)
ds1
x
0 dA 2(dB dA)a b 0
dx
L
dA
x
‹ L
2(dB dA)
1
2a
dB
1b
dA
(a) GRAPH OF x/L VERSUS dB/dA (EQ. 2)
Maximum bending stress occurs at the support when
1 …
Eq. (2)
dB
… 1.5
dA
;
(b) MAXIMUM STRESS (AT SUPPORT B)
Substitute x/L 1 into Eq. (1):
smax 32PL
;
pdB3
Fully Stressed Beams
q
Problems 5.7-6 to 5.7-8 pertain to fully stressed beams of rectangular
cross section. Consider only the bending stresses obtained from the
flexure formula and disregard the weights of the beams.
B
Problem 5.7-6 A cantilever beam AB having rectangular cross sections
A
hx
with constant width b and varying height hx is subjected to a uniform load
of intensity q (see figure).
How should the height hx vary as a function of x (measured from the
free end of the beam) in order to have a fully stressed beam? (Express hx
in terms of the height hB at the fixed end of the beam.)
hB
x
L
hx
hB
b
b
Solution 5.7-6
Fully stressed beam with constant width and varying height
hx height at distance x
hB height at end B
b width (constant)
AT DISTANCE x: M 2
3qx
M
S
bhx2
3q
hx x
A bsallow
sallow qx 2
2
AT THE FIXED END (x L):
hB L
S
bhx2
6
3q
A bsallow
Therefore,
hx
x
hB
L
hx hB x
L
;
05Ch05.qxd
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Page 441
SECTION 5.7
Problem 5.7-7 A simple beam ABC having rectangular cross sections
with constant height h and varying width bx supports a concentrated load
P acting at the midpoint (see figure).
How should the width bx vary as a function of x in order to have a
fully stressed beam? (Express bx in terms of the width bB at the midpoint
of the beam.)
Fully Stressed Beams
P
A
h
B
C
x
L
—
2
L
—
2
h
h
bx
Solution 5.7-7
441
bB
Fully stressed beam with constant height and varying width
h height of beam (constant)
L
bx width at distance x from end Aa 0 … x … b
2
bB width at midpoint B (x L/2)
Px
1
AT DISTANCE x M S b x h2
2
6
3Px
M
3Px
sallow bx 2
S
bx h
sallow h2
AT MIDPOINT B (x L/2)
bB 3PL
2sallowh2
Therefore,
bx
2bB x
2x
and bx bb
L
L
;
NOTE: The equation is valid for 0 … x …
L
and the
2
beam is symmetrical about the midpoint.
q
Problem 5.7-8 A cantilever beam AB having rectangular cross sections
with varying width bx and varying height hx is subjected to a uniform
load of intensity q (see figure). If the width varies linearly with x
according to the equation bx bB x/L, how should the height hx vary as
a function of x in order to have a fully stressed beam? (Express hx in
terms of the height hB at the fixed end of the beam.)
B
hB
hx
A
x
L
hx
hB
bx
bB
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CHAPTER 5
Page 442
Stresses in Beams (Basic Topics)
Solution 5.7-8
Fully stressed beam with varying width and varying height
hx height at distance x
hB height at end B
bx width at distance x
bB width at end B
3qLx
hx A bB sallow
AT THE FIXED END (x L)
x
bx bB a b
L
hB 3qL2
A bB sallow
AT DISTANCE x
M
qx 2
2
sallow bx h2x
bB x
(hx)2
6
6L
3qLx
S
Therefore,
hx
x
hB A L
x
AL
hx hB
;
M
S
bB h2x
Shear Stresses in Rectangular Beams
Problem 5.8-1 The shear stresses t in a rectangular beam are given by
Eq. (5-39):
t
V h2
a y21 b
2I 4
in which V is the shear force, I is the moment of inertia of the cross-sectional
area, h is the height of the beam, and y1 is the distance from the neutral axis to
the point where the shear stress is being determined (Fig. 5-30).
By integrating over the cross-sectional area, show that the resultant
of the shear stresses is equal to the shear force V.
Solution 5.8-1
Resultant of the shear stresses
V shear force acting on the cross section
R resultant of shear stresses t
h/2
R
Lh/2
12V
h/2
tbdy1 2
h/2
(b)
a
L0
V h2
a y21 b bdy1
2I 4
2
h
y21 b dy1
4
bh
L0
12V 2h3
b V
3 a
24
h
I
bh3
12
t
V h2
a y21 b
2I 4
3
‹ R V Q.E.D.
;
05Ch05.qxd
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Page 443
SECTION 5.8
443
Shear Stresses in Rectangular Beams
Problem 5.8-2 Calculate the maximum shear stress tmax
22.5 kN/m
and the maximum bending stress smax in a wood beam
(see figure) carrying a uniform load of 22.5 kN/m (which
includes the weight of the beam) if the length is 1.95 m and the
cross section is rectangular with width 150 mm and height
300 mm, and the beam is (a) simply supported as in the figure
part (a) and (b) has a sliding support at right as in the figure
part (b).
300 mm
150 mm
1.95 m
(a)
22.5 kN/m
1.95 m
(b)
Solution 5.8-2
q 22
kN
m
smax b 150 mm
h 300 mm
L 1.95 m
tmax tmax 715 kPa
MAXIMUM BENDING STRESS
M
qL2
8
S
bh2
6
;
V qL
A bh
3V
2A
smax 4.65 MPa
(b) MAXIMUM SHEAR STRESS
(a) MAXIMUM SHEAR STRESS
qL
V
2
M
S
tmax ;
3V
2A
tmax 1430 kPa
;
MAXIMUM BENDING STRESS
M
qL2
2
smax M
S
Problem 5.8-3 Two wood beams, each of rectangular cross section
(3.0 in. 4.0 in., actual dimensions) are glued together to form a
4.0 in.
solid beam of dimensions 6.0 in. 4.0 in. (see figure). The beam
is simply supported with a span of 8 ft.
What is the maximum moment Mmax that may be
6.0 in.
applied at the left support if the allowable shear stress in the
glued joint is 200 psi? (Include the effects of the beam’s
own weight, assuming that the wood weighs 35 lb/ft3.)
smax 18.59 MPa
;
M
8 ft
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CHAPTER 5
Page 444
Stresses in Beams (Basic Topics)
Solution 5.8-3
L 8 ft
b 4 in.
h 6 in.
g 35
t allow 200 psi
Ab#h
lb
tmax ft3
q g A weight of beam per unit distance
q 5.833
V
1b
ft
Maximum load Mmax
qL
3V
3 M
a +
b
2A
2A L
2
qL2
2 AL
tmax 3
2
qL2
2 AL
tallow Mmax 3
2
M
Mmax 25.4 k-ft
Problem 5.8-4 A cantilever beam of length L 2 m supports a
load P 8.0 kN (see figure). The beam is made of wood with
cross-sectional dimensions 120 mm * 200 mm.
Calculate the shear stresses due to the load P at points located
25 mm, 75 mm, and 100 mm from the top surface of the
beam. from these results, plot a graph showing the distribution of
shear stresses from top to bottom of the beam.
Solution 5.8-4
qL
M
+
L
2
;
P = 8.0 kN
200 mm
L=2m
120 mm
Shear stresses in a cantilever beam
Distance from the
top surface (mm)
Eq. (5-39): t 2
V h
a y21 b
2I 4
h 200 mm (y1 mm)
(200)2
y21 d
4
2(80 * 106)
8,000
c
(t N/mm2 MPa)
t 50 * 106(10,000 y21) (y1 mm; t MPa)
0
t
(kPa)
100
25
75
0.219
219
50
50
0.375
375
75
25
0.469
469
0
0.500
500
GRAPH OF SHEAR STRESS t
bh3
I
80 * 106 mm4
12
t
(MPa)
0
100 (N.A.)
V P 8.0 kN 8,000 N
t
y1
(mm)
0
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Page 445
SECTION 5.8
Problem 5.8-5 A steel beam of length L 16 in. and cross-
q = 240 lb/in.
sectional dimensions b 0.6 in. and h 2 in. (see figure) supports a
uniform load of intensity q 240 lb/in., which includes the weight
of the beam.
Calculate the shear stresses in the beam (at the cross
section of maximum shear force) at points located 1/4 in., 1/2 in.,
3/4 in., and 1 in. from the top surface of the beam. From these
calculations, plot a graph showing the distribution of shear stresses
from top to bottom of the beam.
Solution 5.8-5
h = 2 in.
Shear stresses in a simple beam
V h2
Eq. (5-39): t a y21 b
2I 4
y1
(in.)
t
(psi)
0
1.00
0
0.25
0.75
1050
0.50
0.50
1800
0.75
0.25
2250
0
2400
qL
bh3
1920 lb I 0.4 in.4
2
12
1.00 (N.A.)
GRAPH OF SHEAR STRESS t
UNITS: POUNDS AND INCHES
t
b = 0.6 in.
L = 16 in.
Distance from the
top surface (in.)
V
445
Shear Stresses in Rectangular Beams
1920 (2)2
c
y21 (2400)(1 y21) d
2(0.4) 4
(t psi; y1 in.)
Problem 5.8-6 A beam of rectangular cross section (width b and height h)
supports a uniformly distributed load along its entire length L. The
allowable stresses in bending and shear are sallow and tallow, respectively.
(a) If the beam is simply supported, what is the span length L0 below
which the shear stress governs the allowable load and above which the
bending stress governs?
(b) If the beam is supported as a cantilever, what is the length L0
below which the shear stress governs the allowable load and above which
the bending stress governs?
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CHAPTER 5
Stresses in Beams (Basic Topics)
Solution 5.8-6
b width
Page 446
Beam of rectangular cross section
h height
L length
(b) CANTILEVER BEAM
q intensity of load
Uniform load
ALLOWABLE STRESSES
BENDING
sallow and tallow
Mmax (a) SIMPLE BEAM
smax BENDING
Mmax qL2
8
S
bh2
6
qallow qL2
2
3qL
Mmax
S
4bh2
4sallow bh2
qallow 2
3L
qL
2
tmax 3qL
3V
2A
4bh
Vmax qL A bh
(1)
A bh
tmax 3qL
3V
2A
2bh
qallow 2tallow bh
3L
h sallow
L0 a
b
2 tallow
;
(2)
Equate (1) and (2) and solve for L0:
sallow
b
tallow
(4)
Equate (3) and (4) and solve for L0:
4tallowbh
qallow 3L
L 0 ha
(3)
3L2
SHEAR
SHEAR
Vmax bh2
6
3qL2
Mmax
S
bh2
sallowbh2
3
smax S
NOTE: If the actual length is less than L 0, the shear
stress governs the design. If the length is greater than
L0, the bending stress governs.
;
Problem 5.8-7 A laminated wood beam on simple supports is built up
by gluing together four 2 in. 4 in. boards (actual dimensions) to form
a solid beam 4 in. 8 in. in cross section, as shown in the figure.
The allowable shear stress in the glued joints is 65 psi, and the allowable
bending stress in the wood is 1800 psi.
If the beam is 9 ft long, what is the allowable load P acting at the
one-third point along the beam as shown? (Include the effects of the
beam’s own weight, assuming that the wood weighs 35 lb/ft3.)
3 ft
P
2 in.
2 in.
2 in.
2 in.
L 9 ft
4 in.
Solution 5.8-7
L 9 ft
b 4 in.
h 8 in.
A bh
t allow 65 psi
s allow 1800 psi
WEIGHT OF BEAM PER UNIT DISTANCE
g 35
lb
ft3
q gA
q 7.778
1b
ft
ALLOWABLE LOAD BASED UPON SHEAR STRESS IN THE
GLUED JOINTS; MAX. SHEAR STRESS AT NEUTRAL AXIS
t
VQ
Ib
tmax 3V
2A
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Page 447
SECTION 5.8
VP
tmax MP
qL
3V
3
2
aP +
b
2A
2A
3
2
Pmax A t allow
447
ALLOWABLE LOAD BASED UPON BENDING STRESS
qL
2
+
3
2
P aA tmax Shear Stresses in Rectangular Beams
S
3 qL
b
4
qL
q
2
3 ft +
3 ft (3 ft)2
3
2
2
b h2
6
qL
q
2
3 ft +
3 ft (3 ft)2
3
2
2
M
smax S
S
q
sallow S3
3 qL
a
(3ft)b
Pmax (3 ft) 2
2 2
2
P
3 qL
4
Pmax 2.03 k (governs)
Pmax 3.165 k
P allow 2.03 k
Problem 5.8-8 A laminated plastic beam of square cross
section is built up by gluing together three strips, each
10 mm 30 mm in cross section (see figure). The beam has
a total weight of 3.6 N and is simply supported with span
length L 360 mm.
Considering the weight of the beam (q) calculate the
maximum permissible CCW moment M that may be placed
at the right support.
;
M
q
10 mm
10 mm 30 mm
10 mm
L
30 mm
(a) If the allowable shear stress in the glued joints is
0.3 MPa.
(b) If the allowable bending stress in the plastic is 8 MPa.
Solution 5.8-8
(a) FIND M BASED ON ALLOWABLE SHEAR STRESS IN GLUED
JOINT
b 30 mm
h 30 mm
W 3.6 N
L 360 mm
q
ta 0.3 MPa
N
m
beam distributed weight
MAX. SHEAR ST LEFT SUPPORT
Vm ta bh h
3 3
Q
b h2
9
Q
4
Ib
3bh
W
L
q 10
Q
qL
M
+
2
L
Vm Q
Ib
I
and Vm t a a
3
bh
12
Ib 2 3
b h
12
Ib
b
Q
M L cta a
qL
Ib
b d
Q
2
M L cta a
qL
3bh
b d
4
2
Mmax 72.2 N # M
;
b h2
9
Q
2 3
Ib
b h
12
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CHAPTER 5
Page 448
Stresses in Beams (Basic Topics)
(b) FIND M BASED ON ALLOWABLE BENDING STRESS AT
h/2 FROM NA AT LOCATION (xm) OF MAX. BENDING
MOMENT, Mm
qL
qx2
M
M(x) a
+ bx
2
L
2
Mm a
d
M(x) 0
dx
qL
M
L
M
+ b a +
b
2
L
2
qL
qa
2
use to find location of zero shear where
max. moment occurs
simplifying
qx2
M
d qL
ca
+ bx
d
dx
2
L
2
Mm xm L
M 2
+
b
2
qL
2
2
1 1qL + 2 M2
8q
L2
bh2
b
6
M
1
qL +
qx 0
2
L
also Mm sa S
L
M
+
2
qL
Equating both Mm expressions & solving for M where
sa 8 MPa
MAX. MOMENT Mm
Mm a
qL
qxm
M
+ b xm 2
L
2
2
M
A
sa a
Mm sa a
bh2
b a8 qL2 b qL2
6
2
Mmax 9.01 N # m
Problem 5.8-9 A wood beam AB on simple supports with span length
equal to 10 ft is subjected to a uniform load of intensity 125 lb/ft acting
along the entire length of the beam, a concentrated load of magnitude
7500 lb acting at a point 3 ft from the right-hand support, and a moment at
A of 18,500 ft-lb (see figure). The allowable stresses in bending and shear,
respectively, are 2250 psi and 160 psi.
;
7500 lb
18,500 ft-lb
125 lb/ft
3 ft
A
B
(a) From the table in Appendix F, select the lightest beam that will
support the loads (disregard the weight of the beam).
(b) Taking into account the weight of the beam (weight density 5 35 lb/ft3),
verify that the selected beam is satisfactory, or if it is not, select a new beam.
10 ft
Solution 5.8-9
(a) q 125
1b
ft
L 10 ft
P 75001b M 18500 ft-b
d 3 ft
sAllow 2250 psi
t allow 160 psi
RB 7.725 * 103 1b
Vmax RB
Vmax 7.725 * 103 1b
qd2
Mmax RB d 2
qL
d
M
+ P RA 2
L
L
Mmax 2.261 * 104 1b-ft
RA 1.025 * 103 1b
tmax RB qL
Ld
M
+ P
+
2
L
L
3V
2A
Areq Areq 72.422 in.2
3Vmax
2tallow
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SECTION 5.8
smax M
S
Sreq Mmax
sallow
Sreq 120.6 in.3
From Appendix F: Select 8 * 12 in. beam (nominal
;
dimensions)
S 165.3 in.3
A 86.25 in.2
Vmax RB
Areq g 35
ft3
8 * 12 beam is still satisfactory for shear.
Mmax RB d qbeam g A
qd2
2
Mmax 2.293 * 104 1b-ft
Sreq RB 7.725 * 103 1b +
qbeam L
2
Mmax
sallow
;
Use 8 * 12 in. beam
Problem 5.8-10 A simply supported wood beam of rectangular
cross section and span length 1.2 m carries a concentrated load
P at midspan in addition to its own weight (see figure). The cross
section has width 140 mm and height 240 mm. The weight density
of the wood is 5.4 kN/m3.
Calculate the maximum permissible value of the load P if
(a) the allowable bending stress is 8.5 MPa, and (b) the allowable
shear stress is 0.8 MPa.
P
240 mm
0.6 m
0.6 m
140 mm
Simply supported wood beam
(a) ALLOWABLE P BASED UPON BENDING STRESS
P
240 mm
0.6 m
0.6 m
h 240 mm
A bh 33,600 mm2
S
Sreq 122.3 in.3 < S
8 * 12 beam is still satisfactory for moment.
RB 7.83 * 103 1b
b 140 mm
1b
ft
qtotal q + qbeam q total 145.964
1b
q beam 20.964
ft
Solution 5.8-10
3Vmax
2 tallow
Areq 73.405 in.2 < A
(b) REPEAT (A) CONSIDERING THE WEIGHT OF THE BEAM
1b
449
Shear Stresses in Rectangular Beams
bh2
1344 * 103 mm3
6
140 mm
sallow 8.5 MPa s Mmax +
Mmax
S
qL2
P(1.2 m)
PL
+
4
8
4
(181.44 N/m)(1.2 m)2
8
0.3 P + 32.66 N # m
(P newtons; M N # m)
g 5.4 kN/m3
Mmax Ssallow (1344 * 103 mm3)(8.5 MPa)
L 1.2 m q gbh 181.44 N/m
11,424 N # m
Equate values of Mmax and solve for P:
0.3P + 32.66 11,424 P 37,970 N
or P 38.0 kN
;
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CHAPTER 5
Page 450
Stresses in Beams (Basic Topics)
(b) ALLOWABLE LOAD P BASED UPON SHEAR STRESS
tallow 0.8 MPa t V
3V
2A
qL
(181.44 N/m)(1.2 m)
P
P
+
+
2
2
2
2
P
+ 108.86 (N)
2
2At
2
V
(33,600 mm2)(0.8 MPa) 17,920 N
3
3
Equate values of V and solve for P:
P
+ 108.86 17,920 P 35,622 N
2
or P 35.6 kN
;
NOTE: The shear stress governs and
Pallow 35.6 kN
Problem 5.8-11 A square wood platform, 8 ft * 8 ft in area,
rests on masonry walls (see figure). The deck of the platform is
constructed of 2 in. nominal thickness tongue-and-groove planks
(actual thickness 1.5 in.; see Appendix F) supported on two
8-ft long beams. The beams have 4 in. * 6 in. nominal
dimensions (actual dimensions 3.5 in. * 5.5 in.).
The planks are designed to support a uniformly distributed
load w (lb/ft2) acting over the entire top surface of the platform.
The allowable bending stress for the planks is 2400 psi and
the allowable shear stress is 100 psi. When analyzing the
planks, disregard their weights and assume that their reactions
are uniformly distributed over the top surfaces of the supporting
beams.
(a) Determine the allowable platform load w1 (lb/ft2) based
upon the bending stress in the planks.
(b) Determine the allowable platform load w2 (lb/ft2) based
upon the shear stress in the planks.
(c) Which of the preceding values becomes the allowable
load wallow on the platform?
(Hints: Use care in constructing the loading diagram for the
planks, noting especially that the reactions are distributed loads
instead of concentrated loads. Also, note that the maximum shear
forces occur at the inside faces of the supporting beams.)
8 ft
8 ft
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Page 451
SECTION 5.8
Solution 5.8-11
Shear Stresses in Rectangular Beams
451
Wood platform with a plank deck
Load on one plank:
q c
w(lb/ft2)
2
2
144 in. / ft
d(b in.) Reaction R qa
wb
(lb/in.)
144
96 in.
wb
wb
b a
b (48) 2
144
3
(R lb; w lb/ft2; b in.)
Mmax occurs at midspan.
Mmax Ra
q(48 in.)2
3.5 in.
89 in.
+
b 2
2
3
wb
wb
89
(46.25) (1152) wb
3
144
12
(M lb-in.; w lb/ft2; b in.)
Platform: 8 ft * 8 ft
t thickness of planks
Allowable bending moment:
1.5 in.
Mallow s allow S (2400 psi)(0.375 b)
w uniform load on the deck (lb/ft )
900 b (lb-in.)
2
sallow 2400 psi
Equate Mmax and Mallow and solve for w:
tallow 100 psi
89
wb 900 b w1 121 lb/ft2
12
2
Find wallow (lb/ft )
(a) ALLOWABLE LOAD BASED UPON BENDING STRESS IN THE
(b) ALLOWABLE
;
LOAD BASED UPON SHEAR STRESS IN THE
PLANKS
PLANKS
Let b width of one plank (in.)
See the free-body diagram in part (a).
A 1.5b (in.2)
S
b
(1.5 in.)2
6
Vmax occurs at the inside face of the support.
Vmax qa
89 in.
b 44.5q
2
(44.5)a
0.375b (in.3)
Free-body diagram of one plank supported on the
beams:
89 wb
wb
b 144
288
(V lb; w lb/ft2; b in.)
Allowable shear force:
t
3V
2A
Vallow 2Atallow
3
2(1.5 b)(100 psi)
100 b (lb)
3
Equate Vmax and Vallow and solve for w:
89wb
100b w2 324 lb/ft2
288
;
(c) ALLOWABLE LOAD
Bending stress governs. wallow 121 lb/ft2
;
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CHAPTER 5
Page 452
Stresses in Beams (Basic Topics)
Problem 5.8-12 A wood beam ABC with simple supports
at A and B and an overhang BC has height h 300 mm
(see figure). The length of the main span of the beam is
L 3.6 m and the length of the overhang is L/3 1.2 m.
The beam supports a concentrated load 3P 18 kN at the
midpoint of the main span and a moment PL/2 10.8 kN . m
at the free end of the overhang. The wood has weight
density g 5.5 kN/m3.
L
—
2
3P
PL
M = –––
2
A
h=
300 mm
C
B
L
—
3
L
b
(a) Determine the required width b of the beam
based upon an allowable bending stress of 8.2 MPa.
(b) Determine the required width based upon an allowable
shear stress of 0.7 MPa.
Solution 5.8-12
Numerical data:
L 3.6 m
A bh
g 5.5
s
h 300 mm
P 6 kN
kN
m3
M
PL
2
qbeam g A
Reactions, max. shear and moment equations
RA M
4
3P
4
+ qbeam L P qbeam L
2
L
9
9
RB M
8
3P
8
+
+ qbeam L 2 P + qbeam L
2
L
9
9
Vmax RB 2 P +
8
L
q
9 beam
L
L2
PL
17
q
L2
MD RA qbeam
2
2
2
18 beam
MB 3PL
b 87.8 mm
sallow h2
Vmax 2 P +
t
b
8
L
q
9 beam
3 Vmax
3 Vmax
2A
2 bh
8
4
3P
3
a2 P + qbeam L b + gL
2 bh
9
bh
3
3P
ha t allow
b 89.1 mm
(a) REQUIRED WIDTH b BASED UPON BENDING STRESS
sallow 8.2 MPa
PL
2
;
(b) REQUIRED WIDTH b BASED UPON SHEAR STRESS
tallow 0.7 MPa
4
gLb
3
b 89.074 mm
Shear stress governs
PL
2
Mmax MB b
6 Mmax
Mmax
S
bh2
; (governs)
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Page 453
SECTION 5.9
Shear Stresses in Circular Beams
453
Shear Stresses in Circular Beams
Problem 5.9-1 A wood pole of solid circular cross section (d diameter)
q0 = 20 lb/in.
is subjected to a horizontal force P 450 lb (see figure). The length of the pole is
L 6 ft, and the allowable stresses in the wood are 1900 psi in bending and
120 psi in shear.
Determine the minimum required diameter of the pole based upon
(a) the allowable bending stress, and (b) the allowable shear stress.
d
d
L
Solution 5.9-1
q 20
1b
in
3
L 6 ft
dmin s allow 1900 psi
dmin 5.701 in.
t allow 120 psi
Vmax Mmax qL
2
Vmax 720 1b
qL 2 L
2 3
(b) BASED UPON SHEAR STRESS
t
Mmax 2.88 * 103 1b-ft
(a) BASED UPON BENDING STRESS
s
32 Mmax
A p sallow
32 M
M
S
pd3
4V
16V
3A
3pd2
dmin 16 Vmax
A 3p tallow
dmin 3.192 in.
Bending stress governs
dmin 5.70 in.
;
Problem 5.9-2 A simple log bridge in a remote area consists of two
parallel logs with planks across them (see figure). The logs
are Douglas fir with average diameter 300 mm. A truck moves
slowly across the bridge, which spans 2.5 m. Assume that the weight
of the truck is equally distributed between the two logs.
Because the wheelbase of the truck is greater than 2.5 m, only
one set of wheels is on the bridge at a time. Thus, the wheel load on
one log is equivalent to a concentrated load W acting at any position
along the span. In addition, the weight of one log and the planks it
supports is equivalent to a uniform load of 850 N/m acting on the log.
Determine the maximum permissible wheel load W based upon
(a) an allowable bending stress of 7.0 MPa, and (b) an allowable
shear stress of 0.75 MPa.
x
W
850 N/m
300 mm
2.5 m
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CHAPTER 5
Solution 5.9-2
Page 454
Stresses in Beams (Basic Topics)
Log bridge
Diameter d 300 mm
sallow 7.0 MPa
tallow 0.75 MPa
Find allowable load W
(b) BASED UPON SHEAR STRESS
Maximum shear force occurs when wheel is adjacent to support (x 0).
Vmax W +
(a) BASED UPON BENDING STRESS
Maximum moment occurs when wheel is at midspan
(x L/2).
Mmax qL
WL
+
4
8
A
0.625W + 664.1 (N # m) (W newtons)
S
W + 1062.5 N (W newtons)
2
W
1
(2.5 m) + (850 N/m)(2.5 m)2
4
8
pd3
2.651 * 103m3
32
qL
1
W + (850 N/m)(2.5 m)
2
2
pd2
0.070686 m2
4
tmax 4Vmax
3A
Vmax 3Atallow
3
(0.070686 m2)(0.75 MPa)
4
4
39,760 N
Mmax Ssallow (2.651 * 103 m3)(7.0 MPa)
‹ W + 1062.5 N 39,760 N
18,560 N # m
W 38,700 N 38.7 kN
;
‹ 0.625W + 664.1 18,560
W 28,600 N 28.6 kN
;
b
Problem 5.9-3 A sign for an automobile service station is supported by
two aluminum poles of hollow circular cross section, as shown in the
figure. The poles are being designed to resist a wind pressure of 75 lb/ft2
against the full area of the sign. The dimensions of the poles and sign
are h1 20 ft, h2 5 ft, and b 10 ft. To prevent buckling of the walls of
the poles, the thickness t is specified as one-tenth the outside diameter d.
(a) Determine the minimum required diameter of the poles based
upon an allowable bending stress of 7500 psi in the aluminum.
(b) Determine the minimum required diameter based upon an
allowable shear stress of 2000 psi.
h2
d
t=—
10
Wind
load
d
h1
Probs. 5.9.3 and 5.9.4
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Page 455
SECTION 5.9
Solution 5.9-3
Wind load on a sign
b width of sign
b 10 ft
p 75 lb/ft2
sallow 7500 psi
tallow 2000 psi
d diameter
t
(b) REQUIRED DIAMETER BASED UPON SHEAR STRESS
Vmax W 1875 lb
t
W wind force on one pole
b
W ph2 a b 1875 lb
2
d
10
(a) REQUIRED DIAMETER BASED UPON BENDING STRESS
Mmax W ah1 +
h2
b 506,250 lb-in.
2
4
p
(d24 d24) d2 d d1 d 2t d
64
5
I
4d 4
pd 4 369
p
a
b
I cd 4 a b d 64
5
64 625
369pd 4
(in.4)
40,000
c
d
2
Shear Stresses in Circular Beams
M(d/2)
17.253 M
Mc
I
369pd 4/40,000
d3
(17.253)(506,250 lb-in.)
17.253 Mmax
d3 sallow
7500 psi
s
1164.6 in.
r1 r2 d
2
d
d
d
2d
t 2
2
10
5
r22 + r2r1 + r21
r2 2 + r1 2
2d 2
d 2d
d 2
a b + a ba b + a b
2
2
5
5
61
2
2
41
d
2d
a b + a b
2
5
A
p 2
p
4d 2
9pd2
(d2 d21) cd2 a b d 4
4
5
100
100
4V 61
V
a ba
b 7.0160 2
3 41 9pd2
d
7.0160 Vmax
d2 tallow
t
(d inches)
3
4V r22 + r2r1 + r12
b
a
3A
r2 2 + r1 2
d 10.52 in.
;
Problem 5.9-4 Solve the preceding problem for a sign and poles
having the following dimensions: h1 6.0 m, h2 1.5 m, b 3.0 m,
and t d/10. The design wind pressure is 3.6 kPa, and the allowable
stresses in the aluminum are 50 MPa in bending and 14 MPa in shear.
(7.0160)(1875 lb)
6.5775 in.2
2000 psi
d 2.56 in.
;
(Bending stress governs.)
455
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CHAPTER 5
Page 456
Stresses in Beams (Basic Topics)
Solution 5.9-4
Wind load on a sign
b width of sign
(b) REQUIRED DIAMETER BASED UPON SHEAR STRESS
Vmax W 8.1 kN
b 3.0 m
p 3.6 kPa
sallow 50 MPa
tallow 16 MPa
d diameter
t
4V r2 2 + r1 r2 + r1 2
a
b
3A
r2 2 + r1 2
r1 d
d
d
2d
t 2
2
10
5
W wind force on one pole
(a) REQUIRED DIAMETER BASED UPON BENDING STRESS
h2
b 54.675 kN # m
Mmax W ah1 +
2
s
Mc
I
I
A
4
d
5
4d 4
p 4
cd a b d
64
5
I
d
2
d 2
2d 2
a b + a b
2
5
M(d/2)
17.253 M
Mc
s
4
I
369pd /40,000
d3
(17.253)(54.675 kN # m)
17.253Mmax
d3
sallow
50 MPa
0.018866 m3
;
p
(d 2 d1 2)
4 2
p 2
4d 2
9pd2
cd a b d 4
5
100
100
V
4V 61
a ba
b 7.0160 2
3 41 9pd2
d
(7.0160)(8.1 kN)
7.0160 Vmax
d2 tallow
14 MPa
0.004059 m2
(d meters)
d 0.266 m 266 m
r2 2 + r1 2
d 2d
2d 2
d 2
a b + a ba b + a b
2
5
5
5
t
369pd 4
pd 4 369
a
b
(m 4)
64 625
40,000
c
p 4
(d d41)
64 2
d2 d d1 d 2t d
2
r2 2 + r1r2 + r1 2
b
W ph2 a b 8.1 kN
2
d
t
10
r2 d 0.06371 m 63.7 mm
Bending stress governs
;
61
41
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Page 457
SECTION 5.10
Shear Stresses in Beams with Flanges
Shear Stresses in Beams with Flanges
Problem 5.10-1 through 5.10-6 A wide-flange beam (see figure) having
y
the cross section described below is subjected to a shear force V. Using the
dimensions of the cross section, calculate the moment of inertia and then
determine the following quantites:
(a) The maximum shear stress tmax in the web.
(b) The minimum shear stress tmin in the web.
(c) The average shear stress taver (obtained by dividing the shear
force by the area of the web) and the ratio tmax/taver.
(d) The shear force Vweb carried in the web and the ratio Vweb /V.
z
O h1
h
t
NOTE: Disregard the fillets at the junctions of the web and flanges and
determine all quantities, including the moment of inertia, by considering
the cross section to consist of three rectangles.
b
Probs 5.10.1through 5.-10.6
Problem 5.10-1 Dimensions of cross section: b 6 in., t 0.5 in.,
h 12 in., h1 10.5 in., and V 30 k.
Solution 5.10-1
Wide-flange beam
(b) MINIMUM SHEAR STRESS IN THE WEB (Eq. 5-48b)
b 6.0 in.
tmin t 0.5 in.
h 12.0 in.
taver V 30 k
1
(bh3 bh31 + th31) 333.4 in.4
12
;
;
(d) SHEAR FORCE IN THE WEB (Eq. 5-49)
(a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-48a)
V
(bh2 bh21 + th21) 5795 psi
8It
V
5714 psi
th1
tmax
1.014
taver
MOMENT OF INERTIA (Eq.5-47)
tmax ;
(c) AVERAGE SHEAR STREAR IN THE WEB (Eq. 5-50)
h1 10.5 in.
I
Vb 2
(h h12) 4555 psi
8It
;
Vweb th1
(2tmax + tmin) 28.25 k
3
Vweb
0.942
V
;
;
457
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CHAPTER 5
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Stresses in Beams (Basic Topics)
Problem 5.10-2 Dimensions of cross section: b 180 mm, t 12 mm,
h 420 mm, h1 380 mm, and V 125 kN.
Solution 5.10-2
Wide-flange beam
b 180 mm
(b) MINIMUM SHEAR STRESS IN THE WEB (Eq. 5-48b)
t 12 mm
tmin h 420 mm
taver V 125 kN
V
27.41 MPa
th1
tmax
1.037
taver
MOMENT OF INERTIA (Eq. 5-47)
1
(bh3 bh31 + th31) 343.1 * 106 mm4
12
;
;
(d) SHEAR FORCE IN THE WEB (Eq. 5-49)
Vweb (a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-48a)
tmax ;
(c) AVERAGE SHEAR STRESS IN THE WEB (Eq. 5-50)
h1 380 mm
I
Vb 2
(h h21) 21.86 MPa
8It
V
(bh2 bh21 + th21) 28.43 MPa
8It
;
th1
(2tmax + tmin) 119.7 kN
3
Vweb
0.957
V
;
;
Problem 5.10-3 Wide-flange shape, W 8 * 28 (see Table E-1(a),
Appendix E); V 10 k.
Solution 5.10-3
Wide-flange beam
(b) MINIMUM SHEAR STRESS IN THE WEB (Eq. 5-48b)
W 8 * 28
b 6.535 in.
tmin t 0.285 in.
h 8.06 in.
taver V 10 k
1
(bh3 bh31 + th31) 96.36 in.4
12
;
;
(d) SHEAR FORCE IN THE WEB (EQ. 5-49)
(a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-48a)
V
(bh2 bh21 + th21) 4861 psi
8It
V
4921 psi
th1
tmax
0.988
taver
MOMENT OF INERTIA (Eq. 5-47)
tmax ;
(c) AVERAGE SHEAR STRESS IN THE WEB (Eq. 5-50)
h1 7.13 in.
I
Vb 2
(h h21) 4202 psi
8It
;
Vweb th1
(2tmax + tmin) 9.432 k
3
Vweb
0.943
V
;
;
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Page 459
SECTION 5.10
459
Shear Stresses in Beams with Flanges
Problem 5.10-4 Dimensions of cross section: b 220 mm, t 12 mm,
h 600 mm, h1 570 mm, and V 200 kN .
Solution 5.10-4
Wide-flange beam
b 220 mm
(c) AVERAGE SHEAR STRESS IN THE WEB (EQ. 5-50)
t 12 mm
taver h 600 mm
V
29.24 MPa
th1
;
tmax
1.104
taver
h1 570 mm
V 200 kN
(d) SHEAR FORCE IN THE WEB (Eq. 5-49)
MOMENT OF INERTIA (Eq. 5-47)
Vweb 1
I
(bh3 bh31 + th31) 750.0 * 106 mm4
12
(a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-48a)
tmax V
(bh2 bh21 + th21) 32.28 MPa
8It
th1
(2tmax + tmin) 196.1 kN
3
Vweb
0.981
V
;
;
;
(b) MINIMUM SHEAR STRESS IN THE WEB (EQ. 5-48b)
tmin Vb 2
(h h21) 21.45 MPa
8It
;
Problem 5.10-5 Wide-flange shape, W 18 * 71
(see Table E-1(a), Appendix E); V 21 k.
Solution 5.10-5
Wide-flange beam
(b) MINIMUM SHEAR STRESS IN THE WEB (EQ. 5-48b)
W 18 * 71
b 7.635 in.
tmin t 0.495 in.
h 18.47 in.
taver V 21 k
1
(bh3 bh31 + th31) 1162 in.4
12
;
;
(d) SHEAR FORCE IN THE WEB (EQ. 5-49)
(a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-48a)
V
(bh2 bh21 + th21) 2634 psi
8It
V
2518 psi
th1
tmax
1.046
taver
MOMENT OF INERTIA (Eq. 5-47)
tmax ;
(c) AVERAGE SHEAR STRESS IN THE WEB (EQ. 5-50)
h1 16.85 in.
I
Vb 2
(h h21) 1993 psi
8It
;
Vweb th1
(2tmax + tmin) 20.19 k
3
Vweb
0.961
V
;
;
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CHAPTER 5
Page 460
Stresses in Beams (Basic Topics)
Problem 5.10-6 Dimensions of cross section: b 120 mm, t 7 mm,
h 350 mm, h1 330 mm, and V 60 kN
Solution 5.10-6
Wide-flange beam
(b) MINIMUM SHEAR STRESS IN THE WEB (Eq. 5-48)
b 120 mm
tmin t 7 mm
h 350 mm
taver V 60 kN
V
25.97 MPa
th1
tmax
1.093
taver
MOMENT OF INERTIA (Eq. 5-47)
1
(bh3 bh31 + th31) 90.34 * 106 mm4
12
;
;
(d) SHEAR FORCE IN THE WEB (Eq. 5-49)
Vweb (a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-48a)
tmax ;
(c) AVERAGE SHEAR STRESS IN THE WEB (Eq. 5-50)
h1 330 mm
I
Vb 2
(h h21) 19.35 MPa
8It
V
(bh2 bh21 + th21) 28.40 MPa
8It
th1
(2tmax + tmin) 58.63 kN
3
Vweb
0.977
V
;
;
;
Problem 5.10-7 A cantilever beam AB of length L 6.5 ft supports a
trapezoidal distributed load of peak intensity q, and minimum intensity
q/2, that includes the weight of the beam (see figure). The beam is a
steel W 12 14 wide-flange shape (see Table E-1(a), Appendix E).
Calculate the maximum permissible load q based upon
(a) an allowable bending stress sallow 18 ksi and (b) an allowable
shear stress tallow 7.5 ksi. (Note: Obtain the moment of inertia
and section modulus of the beam from Table E-1(a))
q
—
2
q
B
A
L = 6.5 ft
Solution 5.10-7
b 3.97 in.
I 88.6 # in.4
t 0.2 in.
Vmax t f 0.225 in.
S 14.9 in.3
h 11.9 in.
h1 h 2 tf
h1 11.45 in.
L 6.5 ft
s allow 18 ksi
t allow 7.5 ksi
a
q
+ qb L
2
2
Vmax Mmax 1q 2
1 q 2L
L +
L
22
22
3
Mmax 5
qL2
12
3
qL
4
W 12 14
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Page 461
SECTION 5.10
q
12S sallow
5L2
q 1270 lb/ft
q 3210
(b) MAXIMUM LOAD UPON SHEAR STRESS
tmax Vmax
1 bh2 bh21 + th212
8It
Problem 5.10-8 A bridge girder AB on a simple span of
length L 14 m supports a distributed load of maximum
intensity q at midspan and minimum intensity q/2 at supports
A and B that includes the weight of the girder (see figure).
The girder is constructed of three plates welded to form the
cross section shown.
Determine the maximum permissible load q based upon
(a) an allowable bending stress sallow 110 MPa and
(b) an allowable shear stress tallow 50 MPa.
461
3 qL
1 bh2 bh21 + th212
32It
tallow32It
q
3 L1 bh2 bh21 + th212
(a) MAXIMUM LOAD BASED UPON BENDING STRESS
5 2
qL
12
M
s
S
S
Shear Stresses in Beams with Flanges
lb
ft
Shear stress governs
q 1270 lb/ft
q
q
—
2
q
—
2
A
B
L = 14 m
;
450 mm
32 mm
16 mm
1800 mm
32 mm
450 mm
Solution 5.10-8
L 14 m
(a) MAXIMUM LOAD BASED UPON BENDING STRESS
h 1864 mm h1 1800 mm
b 450 mm
I
tf 32 mm tw 16 mm
1
1 bh3 bh31 + tw h312
12
I 3.194 * 10 mm
10
S
2I
h
RA RB 4
S 3.427 * 107 mm3
qL
qL
3
+
qL
22
42
8
sallow 110 MPa
qLL
qLL
3
L
qL 8
2
22 4
24 6
Mmax 5
qL2
48
5
qL2
Mmax
48
s
S
S
qmax sallow S
5 2
L
48
qmax 184.7
kN
m
;
;
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CHAPTER 5
Page 462
Stresses in Beams (Basic Topics)
(b) MAXIMUM LOAD BASED UPON SHEAR STRESS
tallow 50 MPa
Vmax R A tmax 3
qL
8
qmax 3 qL
1 bh2 bh21 + th212
64It
64 tallow Itw
3 L 1bh2 bh12 tw h122
qmax 247 kN/m
Vmax
1 bh2 bh21 + th212
8It
;
‹ Bending stress governs: qmax 184.7 kN/m
Problem 5.10-9 A simple beam with an overhang supports a uniform
load of intensity q 1200 lb/ft and a concentrated load P 3000 lb
(see figure). The uniform load includes an allowance for the wight
of the beam. The allowable stresses in bending and shear are 18 ksi
and 11 ksi, respectively.
Select from Table E-2 (a), Appendix E, the lightest I-beam (S shape) A
that will support the given loads.
8 ft
q = 1200 lb/ft
C
12 ft
4 ft
Beam with an overhand
sallow 18 ksi
q 1200
P = 3000 lb
B
(Hint: Select a beam based upon the bending stress and then
calculate the maximum shear stress. If the beam is overstressed in
shear, select a heavier beam and repeat.)
Solution 5.10-9
P = 3000 lb
t allow 11 ksi
lb
ft
L 12 ft
P 3000 lb
Sum moments about A & Solve for RB
Find moment at D (at Load P between A and B)
MD R A 8 ft q
(8 ft)2
2
MD 1.28 * 104 lb-ft
Mmax | MB|
Mmax 2.16 * 104 lb-ft
2
RB ;
4
1
q a Lb
+ P(8 ft + 16 ft)
3
2
12 ft
RB 1.88 * 104 lb
Sum forces in vertical direction
RA q (16 ft) + 2P R B
R A 6.4 * 103 lb
Vmax R B (P + q4 ft)
Vmax 1.1 * 104 lb at B
(4 ft)2
MB P (4 ft) q
2
MB 2.16 * 10 lb-ft
4
Required section modulus:
S
Mmax
sallow
S 14.4 in.3
Lightest beam is S 8 * 23 (from Table E-2(a))
I 64.7 in.4
S 16.2 in.3
b 4.17 in.
t 0.441 in.
t f 0.425 in.
h 8 in.
h1 h 2 tf h1 7.15 in.
Check max. shear stress
tmax Vmax
1 bh2 bh21 + th212
8 It
tmax 3674 6 11,000 psi so ok for shear
Select S 8 * 23 beam
;
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Page 463
SECTION 5.10
Shear Stresses in Beams with Flanges
Problem 5.10-10 A hollow steel box beam has the rectangular
cross section shown in the figure. Determine the maximum allowable
shear force V that may act on the beam if the allowable shear
stress in 36 Mpa.
463
20
mm
450
10 mm mm
10 mm
20
mm
200 mm
Solution 5.10-10
Rectangular box beam
tallow 36 MPa
Find Vallow
t
Vallow I
VQ
It
tallowIt
Q
1
1
(200)(450)3 (180)(410)3
12
12
484.9* 106mm4
Q (200)a
410 410
450 450
ba
b (180)a
ba
b
2
4
2
4
1.280 * 106 mm3
Vallow tallow It
Q
(36 MPa)(484.9 * 106 mm4)(20 mm)
273 kN
1.280 * 106 mm3
;
t 2(10 mm) 20 mm
Problem 5.10-11 A hollow aluminum box beam has the square cross section
shown in the figure. Calculate the maximum and minimum shear stresses tmax
and tmin in the webs of the beam due to a shear force V 28 k.
1.0 in.
1.0 in.
12 in.
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CHAPTER 5
Page 464
Stresses in Beams (Basic Topics)
Solution 5.10-11
Square box beam
Q a
1
(b3 b31) 91.0 in.3
8
V 28 k 28,000 lb
t1 1.0 in .
b 12 in.
b21 b1
b2 b
ba b a ba b
2
4
2
4
tmax VQ
(28,000 lb)(91.0 in.3)
1424 psi
It
(894.67 in.4)(2.0 in.)
1.42 ksi
b1 10 in.
;
MINIMUM SHEAR STRESS IN THE WEB (AT LEVEL A.A)
VQ
t
It
t 2t1 2.0 in .
bt1
b t1
Q Ay (bt1)a b a b (bt1)
2 2
2
MOMENT OF INERTIA
t1 b b1
2
MAXIMUM SHEAR STRESS IN THE WEB (AT NEUTRAL AXIS)
Q
(12 in.)
[(12 in.)2 (10 in.)2] 66.0 in.3
8
b
b2
Q A1y1 A2y2 A1 ba b 2
2
tmin 1 4
I
(b b41) 894.67 in.4
12
A2 b1 a
b1
b21
b
2
2
b
1 b
y1 a b 2 2
4
b
Q (b2 b21)
8
VQ
(28,000 lb)(66.0 in.3)
1033 psi
It
(894.67 in4)(2.0 in.)
1.03 ksi
;
b1
1 b1
y2 a b 2 2
4
y
Problem 5.10-12 The T-beam shown in the figure has cross-sectional dimensions
as follows: b 220 mm, t 15 mm, h 300 mm, and h1 275 mm . The beam
is subjected to a shear force V 60 kN.
Determine the maximum shear stress tmax in the web of the beam.
t
h1
z
C
c
b
Probs 5.10.12 and 5.-10.13
Solution 5.10-12
h 300 mm
h1 280 mm
b 210 mm
t 16 mm
t f h h1
V 68 kN
t f 20 mm
LOCATION OF NEUTRAL AXIS
c
b1 h h12 a
c 87.419 mm
c1 c
h h1
h1
b + t h1 a h b
2
2
b1 h h12 + t h1
c1 87.419 mm
c2 h c c2 212.581 mm
h
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Page 465
SECTION 5.10
MOMENT OF INERTIA ABOUT THE z-AXIS
Iweb 1 3
1
t c + t1 c1 tf 23
3 2
3
Iweb 5.287 * 107 mm4
Iflange tf 2
1
b tf3 + b tf ac1 b
12
2
Shear Stresses in Beams with Flanges
Iflange 2.531 * 107 mm4
I Iweb + Iflange
I 7.818 * 107 mm4
FIRST MOMENT OF AREA ABOVE THE z AXIS
c2
Q tc2
2
VQ
tmax tmax 19.7 MPa
;
It
Problem 5.10-13 Calculate the maximum shear stress tmax in the web
of the T-beam shown in the figure if b 10 in., t 0.5 in., h 7 in.,
h1 6.2 in., and the shear force V 5300 lb.
Solution 5.10-13
T-beam
h 7 in.
h1 6.2 in.
b 10 in.
t 0.5 in.
tf h h1
tf 0.8 in.
LOCATION OF NEUTRAL AXIS
b 1 h h12 a
c 1.377 in.
c1 c
h h1
h1
b + t h1 a h b
2
2
b 1 h h12 + t h1
c1 1.377 in.
c2 h c
Iweb 1
1
t c 3 + t1 c1 tf23
3 2
3
Iweb 29.656 in.4
V 5300 lb
c
MOMENT OF INERTIA ABOUT THE z-AXIS
c2 5.623 in.
Iflange tf 2
1
btf3 + btf a c1 b
12
2
Iflange 8.07 in.4
I Iweb + Iflange
I 37.726 in.4
FIRST MOMENT OF AREA ABOVE THE z AXIS
c2
Q tc2
2
VQ
tmax tmax 2221 psi
;
It
465
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CHAPTER 5
Page 466
Stresses in Beams (Basic Topics)
Built-Up Beams
Problem 5.11-1 A prefabricated wood I-beam serving as a floor joist
y
has the cross section shown in the figure. The allowable load in shear
for the glued joints between the web and the flanges is 65 lb/in. in the
longitudinal direction.
Determine the maximum allowable shear force Vmax for the beam.
0.75 in.
z
0.625 in.
8 in.
O
0.75 in.
5 in.
Solution 5.11-1
Wood I-beam
All dimensions in inches.
Find Vmax based upon shear in the glued joints.
Allowable load in shear for the glued joints is 65 lb/in.
‹ fallow 65 lb/in.
fallow I
Q
f
VQ
I
I
(b t)h31
bh3
12
12
Vmax ;
1
1
(5) (9.5)3 (4.375)(8)3 170.57 in.4
12
12
Q Qflange Af df
(5)(0.75)(4.375) 16.406 in.3
Vmax fallowI
Q
(65 lb/in.)(170.57 in.4)
16.406 in.3
676 lb
;
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Page 467
SECTION 5.11
Built-Up Beams
Problem 5.11-2 A welded steel girder having the cross section shown in the figure
y
is fabricated of two 300 mm * 25 mm flange plates and a 800 mm * 16 mm
web plate. The plates are joined by four fillet welds that run continuously for the
length of the girder. Each weld has an allowable load in shear of 920 kN/m.
Calculate the maximum allowable shear force Vmax for the girder.
25 mm
z
16 mm
O
800 mm
25 mm
300 mm
Solution 5.11-2
h 850 mm
h1 800 mm
b 300 mm
t 16 mm
Qflange 3.094 * 106 mm3
f allow 920
tf 25 mm
I
(b t)h13
b h3
12
12
f
4
Qflange A f df Qflange b t f a
f 2 fallow
(2 welds, one either side of web)
I 3.236 * 10 mm
9
kN
m
h tf
2
b
VQ
I
Vmax Vmax 1.924 MN
fI
Qflange
;
y
Problem 5.11-3 A welded steel girder having the cross section shown in the figure
is fabricated of two 20 in. * 1 in. flange plates and a 60 in. * 5/16 in. web plate.
The plates are joined by four longitudinal fillet welds that run continuously throughout
the length of the girder.
If the girder is subjected to a shear force of 280 kips, what force F (per inch of
length of weld) must be resisted by each weld?
1 in.
z
O
60 in.
5
— in.
16
1 in.
20 in.
467
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CHAPTER 5
Page 468
Stresses in Beams (Basic Topics)
Solution 5.11-3
h 62 in.
h1 60 in.
b 20 in.
5
t
in.
16
Qflange btf a
2
b
Qflange 610 in3
tf 1 in.
V 280 k
(b t)h13
bh3
I
12
12
F
I 4.284 * 10 in.
4
h tf
4
f 2F VQflange
F 1994 * 103 lb.in.
21
F 1994 lb/in.
Qflange Af df
VQ
I
;
Problem 5.11-4 A box beam of wood is constructed of two
y
25 mm
260 mm * 50 mm boards and two 260 mm * 25 mm boards (see figure).
The boards are nailed at a longitudinal spacing s 100 mm.
If each nail has a allowable shear force F 1200 N, what is the
maximum allowable shear force Vmax?
z
O
50
mm
50
mm
260 mm
260 mm
Solution 5.11-4
25 mm
Wood box beam
All dimensions in millimeters.
b 260 b1 260 2(50) 160
h 310 h1 260
s nail spacing 100 mm
F allowable shear force for one nail 1200 N
f shear flow between one
flange and both webs
2(1200 N)
2F
24 kN/ m
fallow s
100 mm
fallow I
Q
f
VQ
I
I
1
(bh3 b1h31) 411.125 * 106 mm4
12
Vmax Q Qflange Afdf (260)(25)(142.5)
926.25 * 103 mm4
Vmax fallowI
(24 kN/ m)(411.25 * 106 mm4)
.
Q
926.25 * 103 mm3
10.7 kN
;
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Page 469
SECTION 5.11
Problem 5.11-5 A box beam is constructed of four wood
boards as shown in the figure part (a). The webs are 8 in. 1 in.
and the flanges are 6 in. 1 in. boards (actual dimensions),
joined by screws for which the allowable load in shear is
F 250 lb per screw.
(a) Calculate the maximum permissible longitudinal
spacing smax of the screws if the shear
force V is 1200 lb.
(b) Repeat (a) if the flanges are attached to the webs
using a horizontal arrangement of screws as shown
in the figure part (b).
Solution 5.11-5
V 1200 lb
y
z
1 in.
8 in.
1 in.
1 in.
1 in.
6 in.
6 in.
1 in.
(a)
F 250 lb
1 in.
(b)
(b) Horizontal screws
h1 8 in.
t 1 in.
(b 2t) h13
bh
I
12
12
3
Qa bt (4.5 in.)
I 329.333 in.4
h1 6 in.
b 8 in.
t 1 in.
(b 2t) h13
bh
12
12
3
I 233.333 in.4
Qb (b 2 t) t (3.5 in.)
Qa 27 in.3
f
Qb 21 in.3
VQ
2F
I
s
smax 2FI
VQa
smax 5.08 in.
h 8 in.
I
VQ
2F
I
S
smax 1 in.
Wood box beam
h 10 in.
f
Web
8 in.
O
469
Flange
Flange
1 in. Web
(a) Vertical screws
b 6 in.
Built-Up Beams
2FI
VQb
smax 4.63 in.
;
;
y
Problem 5.11-6 Two wood box beams (beams A and B)
have the same outside dimensions (200 mm * 360 mm)
and the same thickness (t 20 mm) throughout, as shown
|in the figure on the next page. Both beams are formed by
nailing, with each nail having an allowable shear load of
250 N. The beams are designed for a shear force V 3.2 kN.
y
A
z
(a) What is the maximum longitudinal spacing SA for the
t=
nails in beam A?
20 mm
(b) What is the maximum longitudinal spacing sB for the
nails in beam B?
(c) Which beam is more efficient in resisting the shear force?
B
O
360
mm
z
O
t=
20 mm
200 mm
200 mm
360
mm
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CHAPTER 5
Solution 5.11-6
Page 470
Stresses in Beams (Basic Topics)
Two wood box beams
Cross-sectional dimensions are the same.
(a) BEAM A
All dimensions in millimeters.
Q Af df (bt)a
b 200 b1 200 2(20) 160
680 * 103 mm3
h 360 h1 360 2(20) 320
t 20
sA F allowable load per nail 250 N
V shear force 3.2 kN
I
1
(bh3 b1 h31) 340.69 * 106 mm4
12
‹ smax ;
(b) BEAM B
Q Afdf (b 2t)(t)a
f shear flow between one flange and both webs
VQ
2F
s
I
(2)(250 N)(340.7 * 106 mm4)
2FI
VQ
(3.2 kN)(680 * 103 mm3)
78.3 mm
s longitudinal spacing of the nails
f
ht
1
b (200)(20)a b (340)
2
2
ht
b
2
1
(160)(20) (340)
2
2FI
VQ
544 * 103 mm3
sB (2)(250 N)(340.7 * 106 mm4)
2FI
VQ
(3.2 kN)(544 * 103 mm3)
97.9 mm
;
(c) BEAM B IS MORE EFFICIENT because the shear flow on
the contact surfaces is smaller and therefore fewer
;
nails are needed.
3
— in.
16
Problem 5.11-7 A hollow wood beam with plywood webs has the
cross-sectional dimensions shown in the figure. The plywood is attached
to the flanges by means of small nails. Each nail has an allowable load
in shear of 30 lb.
Find the maximum allowable spacing s of the nails at cross sections
where the shear force V is equal to (a) 200 lb and (b) 300 lb.
3
— in.
16
3 in.
y
z
3
in.
4
8 in.
O
3
in.
4
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Page 471
SECTION 5.11
Solution 5.11-7
471
Built-Up Beams
Wood beam with plywood webs
(a) V 200 lb
All dimensions in inches.
b 3.375 b1 3.0
smax h 8.0 h1 6.5
2.77 in.
F allowable shear force for one nail 30 lb
s longitudinal spacing of the nails
f shear flow between one flange and both webs
f
VQ
2F
I
s
I
1
(bh3 b1h31) 75.3438 in.4
12
‹ smax 2FI
VQ
2(30 lb)(75.344 in.4)
2FI
VQ
(200 lb)(8.1563 in.3)
;
(b) V 300 lb
By proportion,
smax (2.77 in.)a
200
b 1.85 in.
300
;
Q Qflange Afdf (3.0)(0.75)(3.625) 8.1563 in.3
y
Problem 5.11-8 A beam of T cross section is formed by nailing together two boards
having the dimensions shown in the figure.
If the total shear force V acting on the cross section is 1500 N and each nail may
carry 760 N in shear, what is the maximum allowable nail spacing s?
240 mm
60 mm
z
C
200 mm
60 mm
Solution 5.11-8
V 1500 N
F allow 760 N
h1 200 mm
b 240 mm
t 60 mm
h 260 mm
MOMENT OF INERTIA ABOUT THE NEUTRAL AXIS
I
A bt + h1t A 2.64 * 104 mm2
LOCATION OF NEUTRAL AXIS (z AXIS)
c2 h1
t
bt ah1 b + th1
2
2
A
1 3
1
tc + t1 h1 c223
3 2
3
+
1 3
t 2
bt + bt a c1 b
12
2
I 1.549 * 108 mm4
FIRST MOMENT OF AREA OF FLANGE
c2 170.909 mm
t
Q bt a c1 b
2
c1 h c2
Q 8.509 * 105 mm3
c1 89.091 mm
MAXIMUM ALLOWABLE SPACING OF NAILS
f
smax VQ
F
I
s
F allowI
VQ
smax 92.3 mm
;
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Page 472
CHAPTER 5
Stresses in Beams (Basic Topics)
Problem 5.11-9 The T-beam shown in the figure is fabricated by welding
together two steel plates. If the allowable load for each weld is 1.8 k/in. in the
longitudinal direction, what is the maximum allowable shear force V?
y
0.6 in.
5.5 in.
z
C
0.5 in.
4.5 in.
Solution 5.11-9
F allow 1.8
T-beam (welded)
MOMENT OF INERTIA ABOUT THE NEUTRAL AXIS
k
in.
h1 5.5 in.
b 4.5 in.
t1 0.6 in.
t2 0.5 in.
I
+
h 6 in.
A bt2 + h1t1 A 5.55 in.2
LOCATION OF NEUTRAL AXIS (z AXIS)
c2 t2
h1
bt2 + t1 h1 a
+ t2 b
2
2
c2 2.034 in.
c1 3.966 in.
A
c1 h c 2
1
1
t c 3 + t1 1 c2 t223
3 1 1
3
t2 2
1
b t23 + bt2 ac2 b
12
2
I 20.406 in.4
FIRST MOMENT OF AREA OF FLANGE
Q b t2 ac2 t2
b
2
Q 4.014 in.3
MAXIMUM ALLOWABLE SHEAR FORCE
f
VQ
2F
I
Vmax 2 Fallow I
Q
Vmax 18.30 k
Problem 5.11-10 A steel beam is built up from a W 410 * 85 wide-flange beam
and two 180 mm * 9 mm cover plates (see figure). The allowable
load in shear on each bolt is 9.8 kN.
What is the required bolt spacing s in the longitudinal direction if the shear
force V = 110kN (Note: Obtain the dimensions and moment of inertia of the
W shape from Table E-1(b).)
;
y
z
180 mm 9 mm
cover plates
W 410 85
O
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Page 473
SECTION 5.11
Built-Up Beams
473
Solution 5.11-10
F allow 9.8 kN
V 110 kN
+ Acp a c W 410 * 85
I 4.57 * 108 mm4
Aw 10800 mm2 hw 417 mm
Iw 310 * 106 mm4
First moment of area of one flange
Acp (180) (9) (2) mm2 for two plates
Q 180 mm (9 mm)a c h hw + (9 mm) (2)
Maximum allowable spacing of nails
LOCATION OF NEUTRAL AXIS (z AXIS)
h
2
f
c 217.5 mm
Moment of inertia about the neutral axis
smax 3
I Iw +
9 mm
b
2
Q 3.451 * 105 mm3
A Aw + Acp A 1.404 * 104 mm2
c
9 mm 2
b
2
180 mm (9 mm)
(2)
12
VQ
2F
I
s
2 Fallow I
VQ
smax 236 mm
;
Problem 5.11-11 The three beams shown have approximately the same cross-sectional area. Beam 1 is a W 14 82 with
flange plates; Beam 2 consists of a web plate with four angles; and Beam 3 is constructed of 2 C shapes with flange plates.
(a)
(b)
(c)
(d)
Which design has the largest moment capacity?
Which has the largest shear capacity?
Which is the most economical in bending?
Which is the most economical in shear?
Assume allowable stress values are: sa 18 ksi and ta 11 ksi. The most economical beam is that having the largest capacityto-weight ratio. Neglect fabrication costs in answering (c) and (d) above. (Note: Obtain the dimensions and properties of all rolled
shapes from tables in Appendix E.)
8 0.52
4 0.375
Four angles
1
66—
2
W 14 82
Beam 1
Solution 5.11-11
8 0.52
14 0.675
4 0.375
Beam 2
b1 8 in.
Beam 3
Built-up steel beam
Beam 1: properties and dimensions for W14 * 82 with
flange plates
AW 24 in.2
C 15 50
hw 14.3 in.
t1 0.52 in.
Iw 88l in.4
h1 hw + 2t1
bf1 10.1 in.
tf1 0.855 in.
tw1 0.51 in.
AI AW + 2b1t1
AI 32.32 in.2
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CHAPTER 5
I1 Iw +
Page 474
Stresses in Beams (Basic Topics)
Q1 b1 t1 a
t1 2
b1 + t31
hw
2 + b1 t1 a
+ b 2
12
2
2
I1 1.338 * 103 in4
Beam 2: properties and dimensions for L6 * 6 * 1/2
angles with web plate
Aa 5.77 in.2
ca 1.67 in.
Ia 19.9 in.4
b2 14 in.
t2 0.675 in.
h2 b2
A2 4Aa + b2t2
I2 4Ia + Aa a
+ tw1
tf1
h1
t1
hw
b + bf1 tf1 a
b
2
2
2
2
a
Q2 2 Aa a
h2
ca b + t2
2
Q3 b3 t3 a
bf3 3.72 in.
A3 2Ac + 2b3 t3
I3 Ic 2 +
h3 hc + 2t3
tw3 0.716 in.
A3 32.4 in.2
I3 985.328 in.4
(a) Beam with largest moment capacity; largest section
modulus controls
Mmax sallow S
2I1
h1
S1 174.449 in.3
S2 2I2
h2
S2 127.09 in.3
S3 2I3
h3
S3 125.121 in.3
largest value
BEAM WITH LARGEST SHEAR CAPACITY: LARGEST
Q3 79.826 in.3
I2 t2
4.964 * 103 mm2
Q2
largest value
Itw
Case (3) with maximum
has the largest shear
Q
capacity ;
(c) MOST ECONOMICAL BEAM IN BENDING HAS LARGEST
BENDING CAPACITY-TO-WEIGHT RATIO
S3
3.862 in.
A3
6
S2
3.907 in.
A2
6
;
Case (1) is the most economical in bending.
Itw/Q
(d) MOST ECONOMICAL BEAM IN SHEAR HAS LARGEST
SHEAR CAPACITY-TO-WEIGHT RATIO
I1 tw1
0.213
Q1 A1
RATIO CONTROLS
tallow I tw
Vmax O
2
S1
5.398 in.
A1
case (1) with maximum S has the largest moment
capacity ;
(b)
2
hc
tf3 b
2
I2 2tw3
1.14 * 104 mm2
Q3
b3t33
hc
t3 2
2 + b3 t3 a + b 2
12
2
2
S1 a
I1 tw1
4.448 * 103 mm2
Q1
Ic 404 in.4
tf3 0.65 in.
2
tf3
h3
t3
hc
b + 2bf3 tf3 a b
2
2
2
2
+ 2tw3
Beam 3: properties and dimensions for C15 * 50 with
flange plates
t3 0.375 in.
b2 2
b
2
Q2 78.046 in.
2
b2
t2 b32
ca b 4 +
2
12
hc 15 in.
a
3
I2 889.627 in.4
b3 4 in.
Q1 98.983 in.3
2
ha 6 in.
A2 32.53 in.2
Ac 14.7 in.2
2
hw
tf1 b
2
6
6
I2 t2
0.237
Q2 A2
I3 tw3
0.273
Q3 A3
Case (3) is the most economical in shear.
;
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Page 475
SECTION 5.12
Beams with Axial Loads
Problem 5.11-12 Two W 310 * 74 steel wide-flange beams are bolted together to form
a built-up beam as shown in the figure. What is the maximum permissible bolt spacing s
if the shear force V 80 kN and the allowable load in shear on each bolt is F 13.5 kN
(Note: Obtain the dimensions and properties of the W shapes from Table E-1(b).)
W 310 74
W 310 74
Solution 5.11-12
V 80 kN
FIRST MOMENT OF AREA OF FLANGE
W 310 * 74
F allow 13.5 kN
hw 310 mm
A w 9420 mm
2
4
Location of neutral axis (z axis)
c hw
Q Aw
I w 163 * 10 mm
6
c 310 mm
hw 2
b d (2)
2
Q 1.46 * 106 mm3
MAXIMUM ALLOWABLE SPACING OF NAILS
f
MOMENT OF INERTIA ABOUT THE NEUTRAL AXIS
I c Iw + Aw a
hw
2
VQ
2F
I
s
smax 2Fallow I
VQ
smax 180 mm
;
I 7.786 * 108 mm4
Beams with Axial Loads
When solving the problems for Section 5.12, assume that the bending
moments are not affected by the presence of lateral deflections.
P = 25 lb
Problem 5.12-1 While drilling a hole with a brace and bit, you exert
a downward force P 25 lb on the handle of the brace (see figure).
The diameter of the crank arm is d 7/16 in. and its lateral offset
is b 4-7/8 in.
Determine the maximum tensile and compressive stresses st and
sc, respectively, in the crank.
7
d= —
16 in.
7
b = 4—
8 in.
475
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CHAPTER 5
Page 476
Stresses in Beams (Basic Topics)
Solution 5.12-1
Brace and bit
P 25 lb (compression)
M Pb (25 lb)(4 7/8 in.)
121.9 lb-in.
MAXIMUM STRESSES
st 166 psi + 14,828 psi 14,660 psi
d diameter
d 7/16 in.
S
sc pd2
0.1503 in.2
4
A
P
121.9 lb-in.
M
25 lb
+
+
2
A
S
0.1503 in.
0.008221 in.3
;
P
M
166 psi 14,828 psi
A
S
14,990 psi
;
3
pd
0.008221 in.3
32
Problem 5.12-2 An aluminum pole for a street light weights 4600 N
and supports an arm that weights 660 N (see figure). The center of gravity
of the arm is 1.2 m from the axis of the pole. A wind force of 300 N also acts
in the (y) direction at 9 m above the base. The outside diameter of the
pole (at its base) is 225 mm, and its thickness is 18 mm.
Determine the maximum tensile and compressive stresses st and sc,
respectively, in the pole (at its base) due to the weights and the wind force.
W2 = 660 N
1.2 m
P1 = 300 N
W1 = 4600 N
18 mm
9m
z
y
x
y
x
Solution 5.12-2
W1 4600 N
b 1.2 m
Mx W2 b + P1h
W2 660 N
h9m
Mx 3.492 * 103 N # m
P1 300 N
d1 225 mm t 18 mm
MAXIMUM STRESS
d2 d1 2 t
A
p 2
1d1 d222
4
I
A 1.171 * 104 mm2
p
1d1 4 d2 42
64
I 6.317 * 107 mm4
AT BASE OF POLE
Pz W1 + W2
Pz 5.26 * 10 N
3
V y P1
st a Pz
Mx d1
+
b
A
I 2
st 5.77 * 103 kPa
5770 kPa
sc a ;
Pz Mx d1
b
A
I 2
sc 6.668 * 103
(Axial force )
V y 300 N
(Shear force)
(Moment)
6668 kPa
;
;
225 mm
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Page 477
SECTION 5.12
Problem 5.12-3 A curved bar ABC having a circular axis (radius
h
B
r 12 in.) is loaded by forces P 400 lb (see figure). The cross
section of the bar is rectangular with height h and thickness t.
If the allowable tensile stress in the bar is 12,000 psi and the
height h 1.25 in., what is the minimum required thickness tmin ?
477
Beams with Axial Loads
C
A
P
P
45°
45°
r
h
t
Solution 5.12-3
Curved bar
TENSILE STRESS
st r radius of curved bar
e r r cos 45°
tmin Pr
(2 12)
2
P
r
c1 + 3(2 12) d
hsallow
h
SUBSTITUE NUMERICAL VALUES:
P 400 lb s allow 12,000 psi
CROSS SECTION
h height t thickness A ht S P
r
c1 + 3(2 12) d
ht
h
MINIMUM THICKNESS
1
b
r a1 12
M Pe 3Pr(2 12)
P
M
P
+
+
A
S
ht
th2
1 2
th
6
r 12 in. h 1.25 in.
tmin 0.477 in.
;
B
Problem 5.12-4 A rigid frame ABC is formed by welding two
steel pipes at B (see figure). Each pipe has cross-sectional area
A 11.31 * 103 mm2, moment of inertia I 46.37 * 106 mm4,
and outside diameter d 200 mm.
Find the maximum tensile and compressive stresses st and sc,
respectively, in the frame due to the load P 8.0 kN if L H 1.4 m.
d
d
P
H
A
C
d
L
L
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CHAPTER 5
Solution 5.12-4
Page 478
Stresses in Beams (Basic Topics)
Rigid frame
AXIAL FORCE: N RA sin a P
sin a
2
PL
2
BENDING MOMENT: M RAL TENSILE STRESS
st Load P at midpoint B
P
REACTIONS: RA RC 2
BAR AB:
SUBSTITUTE NUMERICAL VALUES
P 8.0 kN L H 1.4 m a 45°
sina 1/12 d 200 mm
A 11.31 * 103 mm2 I 46.37 * 106 mm4
H
tan a L
sin a N
Mc
P sin a
PLd
+
+
A
I
2A
4I
st H
1H2 + L2
2(11.31 * 103 mm2)
(8.0 kN)(1.4 m)(200 mm)
+
4(46.37 * 106 mm4)
d diameter
c d/2
(8.0 kN)(1/12)
0.250 MPa + 12.08 MPa
11.83 MPa (tension)
sc ;
N
Mc
0.250 MPa 12.08 MPa
A
I
12.33 MPa (compression)
Problem 5.12-5 A palm tree weighing 1000 lb is inclined
at an angle of 60° (see figure). The weight of the tree may be
resolved into two resultant forces, a force P1 900 lb acting at
a point 12 ft from the base and a force P2 100 lb acting at
the top of the tree, which is 30 ft long. The diameter at the base
of the tree is 14 in.
Calculate the maximum tensile and compressive stresses
st and sc, respectively, at the base of the tree due to its weight.
;
P2 = 100 lb
30 ft
12 ft
P1 = 900 lb
60°
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Page 479
SECTION 5.12
Solution 5.12-5
479
Beams with Axial Loads
Palm tree
M P1L1 cos 60°P2 L2 cos 60°
[(900 lb)(144 in.) + (100 lb)(360 in.)] cos 60°
82,800 lb-in.
N (P1 + P2) sin 60° (1000 lb) sin 60° 866 lb
FREE-BODY DIAGRAM
MAXIMUM TENSILE STRESS
P1 900 lb
st P2 100 lb
L1 12 ft 144 in.
L2 30 ft 360 in.
d 14 in.
A
pd2
153.94 in.2
4
S
pd3
269.39 in.3
32
82,800 lb-in.
N
M
866 lb
+
+
2
A
S
153.94 in.
269.39 in.3
5.6 psi + 307.4 psi 302 psi
MAXIMUM COMPRESSIVE STRESS
sc 5.6 psi 307.4 psi 313 psi
Problem 5.12-6 A vertical pole of aluminum is fixed at the base and pulled
at the top by a cable having a tensile force T (see figure). The cable is attached
at the outer edge of a stiffened cover plate on top of the pole and makes an
angle a 20° at the point of attachment. The pole has length L 2.5 m and a
hollow circular cross section with outer diameter d2 280 mm and inner
diameter d1 220 mm. The circular cover plate has diameter 1.5d2.
Determine the allowable tensile force Tallow in the cable if the allowable
compressive stress in the aluminum pole is 90 MPa.
T
a
L
Solution 5.12-6
d1 220 mm
d2 280 mm
t
A
d2 d1
2
PN T cos (a)
V T sin (a)
a 20°
p
1d 2 d1 22
4 2
A 2.356 * 104 mm2
L 2.5 m
I
p
1d 4 d1 42
64 2
I 1.867 * 108 mm4
M VL + PN a
;
1.5 d2
d2
sallow 90 MPa
;
(Axial force)
(Shear force)
1.5 d2
b
2
(Moment).
d1
d2
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CHAPTER 5
Page 480
Stresses in Beams (Basic Topics)
Allowable Tensile Force
sc T cos (a)
PN
M d2
A
I 2
A
T sin (a) L T cos (a) a
I
sallow
Tallow 1.5 d2
b
2
d2
2
cos (a)
+
A
sin (a) L + cos (a) a
Tallow 108.6 kN
I
1.5 d2
b
2
d2
2
;
Problem 5.12-7 Because of foundation settlement,
a circular tower is leaning at an angle a to the vertical
(see figure). The structural core of the tower is a circular
cylinder of height h, outer diameter d2, and inner diameter d1.
For simplicity in the analysis, assume that the weight
of the tower is uniformly distributed along the height.
Obtain a formula for the maximum permissible angle
a if there is to be no tensile stress in the tower.
h
d1
d2
a
Solution 5.12-7
Leaning tower
CROSS SECTION
W weight of tower
a angle of tilt
A
p 2
(d d21)
4 2
I
p 4
(d d41)
64 2
p 2
(d d21)(d22 + d21)
64 2
d22 + d21
I
A
16
c
d2
2
AT THE BASE OF THE TOWER
h
N W cos a M Wa bsin a
2
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Page 481
SECTION 5.12
TENSILE STRESS (EQUAL TO ZERO)
st ‹
Mc
Wcosa
N
+
A
I
A
d2
W h
+ a sinab a b 0
I 2
2
hd2 sin a
cos a
A
4I
MAXIMUM ANGLE a
d22 + d12
a arctan
4hd2
Problem 5.12-8 A steel bar of solid circular cross section and length
Beams with Axial Loads
tan a 481
d22 + d12
4I
hd2A
4hd2
;
y
L 2.5 m is subjected to an axial tensile force T 24 kN and a
bending moment M 3.5 kN m (see figure).
d
(a) Based upon an allowable stress in tension of 110 MPa, determine
the required diameter d of the bar; disregard the weight of
the bar itself.
(b) Repeat (a) including the weight of the bar.
M
–z-direction
z
T
L
x
Solution 5.12-8
M 3.5 kN # m
g steel 77
T 24 kN
kN
L 2.5 m
3
m
p 2
d
4
c
d
2
I
p 4
d
64
(a) DISREGARD WEIGHT OF BAR
MAX. TENSILE STRESS AT TOP OF BEAM AT SUPPORT
smax Md
T
T
M d
+
+
p 2
p 42
A
I 2
d
d
4
64
sallow 4T
2
pd
d 70 mm
d (SUBSTITUTE sallow)
;
(b) INCLUDE WEIGHT OF BAR
sallow 110 MPa
A
SOLVE NUMERICALLY FOR
32 M
+
p d3
Mmax M +
Agsteel L2
2
AT TOP OF BEAM AT SUPPORT
st sallow Mmax d
T
+
A
I 2
SUBSTITUTE MMAX FROM ABOVE, SOLVE FOR d NUMERICALLY
d 76.5 mm
;
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CHAPTER 5
Page 482
Stresses in Beams (Basic Topics)
Problem 5.12-9 A cylindrical brick chimney of height H weighs
w 825 lb/ft of height (see figure). The inner and outer diameters are
d1 3 ft and d2 4 ft, respectively. The wind pressure against the
side of the chimney is p = 10 lb/ft2 of projected area.
Determine the maximum height H if there is to be no tension in
the brickwork
p
w
H
d1
d2
Solution 5.12-9
Brick Chimney
I
d2
H
q
w
p 2
p 4
(d2 d41) (d d21) (d22 d21)
64
64 2
I
1 2
(d2 + d21)
A
16
d2
2
c
AT BASE OF CHIMNEY
M qH a
N W wH
V
M
TENSILE STRESS (EQUAL TO ZERO)
s1 N
Md2
N
+
0
A
2I
p wind pressure
pd2 H2
d22 + d12
2wH
8d2
q intensity of load pd2
SOLVE FOR H
d2 outer diameter
1
H
b pd2 H2
2
2
H
2I
M
N
Ad2
or
w(d22 + d21)
d1 inner diameter
SUBSTITUTE NUMERICAL VALUES
W total weight of chimney wH
w 825 lb/ft
d2 4 ft
CROSS SECTION
q 10 lb/ft
Hmax 32.2 ft
A
p 2
(d2 d21)
4
2
;
4pd22
d1 3 ft
;
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Page 483
SECTION 5.12
483
Beams with Axial Loads
Problem 5.12-10 A flying buttress transmits a load P 25 kN, acting
at an angle of 60° to the horizontal, to the top of a vertical buttress AB
(see figure). The vertical buttress has height h 5.0 m and rectangular
cross section of thickness t 1.5 m and width b 1.0 m (perpendicular
to the plane of the figure). The stone used in the construction weighs
y 26 kN/m3.
What is the required weight W of the pedestal and statue above the
vertical buttress (that is, above section A) to avoid any tensile stresses
in the vertical buttress?
Flying
buttress
P
W
60°
A
A
—t
2
h
t
B
Solution 5.12-10
h
t
B
Flying buttress
FREE-BODY DIAGRAM OF VERTICAL BUTTRESS
CROSS SECTION
A bt (1.0 m)(1.5 m) 1.5 m2
1
1
S bt2 (1.0 m)(1.5 m)2 0.375 m3
6
6
AT THE BASE
N W + WB + P sin 60°
W + 195 kN + (25 kN) sin 60°
W + 216.651 kN
M (Pcos 60°) h (25 kN) (cos 60°) (5.0 m)
62.5 kN # m
TENSILE STRESS (EQUAL TO ZERO)
P 25 kN
st h 5.0 m
t 1.5 m
b width of buttress perpendicular to the figure
g 26 kN/m
3
WB weight of vertical buttress
195 kN
62.5 kN # m
W + 216.651 kN
2
1.5 m
+
0.375 m3
or W 216.651 kN + 250 kN 0
b 1.0 m
bthg
N
M
+
A
S
W 33.3 kN
;
0
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CHAPTER 5
Page 484
Stresses in Beams (Basic Topics)
Problem 5.12-11 A plain concrete wall (i.e., a wall with no steel
t
reinforcement) rests on a secure foundation and serves as a small
dam on a creek (see figure). The height of the wall is h 6.0 ft
and the thickness of the wall is t 1.0 ft.
(a) Determine the maximum tensile and compressive stresses st and
sc, respectively, at the base of the wall when the water level reaches the
top (d h). Assume plain concrete has weight density gc 145 Ib/ft3.
(b) Determine the maximum permissible depth dmax of the water if
there is to be no tension in the concrete.
Solution 5.12-11
h
d
Concrete wall
h height of wall
t thickness of wall
b width of wall (perpendicular to the figure)
gc width density of concrete
gw weight density of water
d depth of water
W weight of wall
STRESSES AT THE BASE OF THE WALL
(d DEPTH OF WATER)
d 3gw
W
M
+
hgc +
A
S
t2
d 3gw
W
M
sc hgc 2
A
S
t
st (a) STRESSES AT THE BASE WHEN d h
W bhtgc
h 6.0 ft 72 in. d 72 in.
F resultant force for the water pressure
t 1.0 ft 12 in.
MAXIMUM WATER PRESSURE = gw d
gc 145 lb/ft3 F
1
1
(d)(gw d) (b) bd2gw
2
2
d
1
M F a b bd3gw
3
6
1
A bt S bt2
6
145
lb/in.3
1728
gw 62.4 Ib/ft3 62.4
lb/in.3
1728
Eq. (1)
Eq.(2)
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Page 485
SECTION 5.12
Substitute numerical values into Eqs. (1) and (2):
st 6.042 psi + 93.600 psi 87.6 psi
d3 (72 in.)(12 in.)2 a
;
sc 6.042 psi 93.600 psi 99.6 psi
dmax 28.9 in.
;
485
Beams with Axial Loads
145
b 24,092 in.3
62.4
;
(b) MAXIMUM DEPTH FOR NO TENSION
Set st = 0 in Eq. (1):
hgc +
d3gw
2
t
0 d3 ht2 a
gc
b
gw
Problem 5.12-12 A circular post, a rectangular post, and a post of cruciform
cross section are each compressed by loads that produce a resultant force P acting
at the edge of the cross section (see figure). The diameter of the circular post and the
depths of the rectangular and cruciform posts are the same.
(a) For what width b of the rectangular post will the maximum tensile
stresses be the same in the circular and rectangular posts?
(b) Repeat (a) for the post with cruciform cross section.
(c) Under the conditions described in parts (a) and (b), which post has the
largest compressive stress?
P
P
P
b
4 — = b
4
x
b
d
d
d
Load P here
d
4 — = d
4
Solution 5.12-12
(a) EQUAL MAXIMUM TENSILE STRESSES
COMPRESSION sc CIRCULAR POST
A
p 2
d
4
S
p 3
d
32
M
Pd
2
Tension
st P
M
A
S
4P
pd
2
16 P
pd
2
20P
pd 2
RECTANGULAR POST
P
M
4P
16 P
12 P
+
2 +
2
A
S
pd
pd
pd2
A bd
TENSION
st S
bd2
6
M
Pd
2
P
M
P 3P
2P
+
A
S
bd bd
bd
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CHAPTER 5
Page 486
Stresses in Beams (Basic Topics)
COMPRESSION s c P
M
P
3P
4P
A
S
bd
bd
bd
Equate tensile stress expressions, solve for b
12 P
2
pd
2P
bd
6
1
pd
b
b
pd
6
;
(b) CRUCIFORM CROSS SECTION
A cbd a
S c
3
bd
b d
1 2
3
+ a b
d bd 2
2 12
2 2 12 d
32
Pd
16P
M
3bd
3
2 a bd2 b
32
TENSION
st COMPRESSION
12 P
2
pd
2P
3bd
1
3
pd
b
;
substitute expressions for b above & compare
compressive stresses
16 P
12 P
4P
+
3bd
3bd
3bd
sc sc M
P
A
S
20 P
pd2
RECTANGULAR POST
4P
24 P
pd
pd 2
a bd
6
CRUCIFORM POST
20 P
20 P
sc pd
pd 2
3
d
3
Rectangular post has the largest compressive
stress ;
4P
16 P
20 P
3bd
3bd
3bd
Problem 5.12-13 Two cables, each carrying a tensile force
P 1200 lb, are bolted to a block of steel (see figure). The
block has thickness t 1 in. and width b 3 in.
Steel block loaded by cables
P 1200 lb d 0.25 in.
t
d
+ 0.625 in.
2
2
b
P
(a) If the diameter d of the cable is 0.25 in., what are the maximum tensile
and compressive stresses st and sc, respectively, in the block?
(b) If the diameter of the cable is increased (without changing
the force P), what happens to the maximum tensile and compressive stresses?
t 1.0 in. e pd
3
(c) THE LARGEST COMPRESSIVE STRESS
sc M
P
+
A
S
Solution 5.12-13
b
CIRCULAR POST
bd
bd
22
3
Equate compressive stresses & solve for b
b width of block
3.0 in.
t
P
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Page 487
SECTION 5.12
MAXIMUM COMPRESSIVE STRESS (AT BOTTOM OF BLOCK)
CROSS SECTION OF BLOCK
A bt 30 in.2
I
1 3
bt 0.25 in.4
12
t
y 0.5 in.
2
sc (a) MAMIMUM TENSILE STRESS (AT TOP OF BLOCK)
y
t
0.5 in.
2
Pey
P
st +
A
I
Pey
P
+
A
I
(1200 lb)(0.625 in.)( 0.5 in.)
1200 lb
3 in.2
+
0.25 in.4
400 psi 1500 psi 1100 psi
;
(1200 lb)(0.625 in.)(0.5 in.)
1200 lb
3 in.2
+
(b) IF d IS INCREASED, increase the eccentricity e
increases and both stresses in magnitude.
0.25 in.4
400 psi + 1500 psi 1900 psi
;
Problem 5.12-14 A bar AB supports a load P acting at the centroid
b
—
2
of the end cross section (see figure). In the middle region of the bar
the cross-sectional area is reduced by removing one-half of the bar.
A
(a) If the end cross sections of the bar are square with sides of
length b, what are the maximum tensile and compressive stresses st
and sc, respectively, at cross section mn within the reduced region?
(b) If the end cross sections are circular with diameter b, what
are the maximum stresses st and sc?
b
b
b
b
—
2
m
(a)
b
—
2
n
B
P
b
(b)
Solution 5.12-14
Bar with reduced cross section
(a) SQUARE BAR
(b) CIRCULAR BAR
Cross section mn is a rectangle.
Cross section mn is a semicircle
b2
b
A (b)a b 2
2
1 pb2
pb2
A a
b 0.3927 b2
2 4
8
b
M Pa b
4
c
I
1
b 3
b4
(b) a b 12
2
96
b
4
STRESSES
P
Mc
2P
6P
8P
+
2 + 2 2
;
A
I
b
b
b
P
Mc
2P
6P
4P
sc 2 2 2
;
A
I
b
b
b
st 487
Beams with Axial Loads
From Appendix D, Case 10:
b 4
I 0.1098a b 0.006860 b4
2
M Pa
2b
b 0.2122 Pb
3p
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CHAPTER 5
Page 488
Stresses in Beams (Basic Topics)
FOR TENSION
2.546
FOR COMPRESSION:
b
2b
0.2878 b
2
3p
P
(0.2122 Pb)(0.2122 b)
Mct
P
P
+
+
A
I
0.3927 b2
0.006860 b4
Problem 5.12-15 A short column constructed
of a W 12 * 35 wide-flange shape is subjected to a resultant
compressive load P 12 k having its line of action at the
midpoint of one flange (see figure).
(a) Determine the maximum tensile and compressive
stresses st and sc, respectively, in the column.
(b) Locate the neutral axis under this loading condition.
(c) Recompute maximum tensile and compressive stresses
if a C 10 15.3 is attached to one flange, as shown.
2
0.3927 b
2.546
STRESSES
st + 6.564
2
b
Mc
P
c
sc A
I
2b
4r
ct 0.2122 b
3p
3p
cc r ct P
P
2
b
P
2
b
9.11
P
b2
;
(0.2122 Pb)(0.2878 b)
8.903
0.006860 b4
P
2
b
6.36
P
b2
;
y
P = 25 k
C 10 15.3
(Part c only)
z
C
2
W 12 35
1
1
2
Solution 5.12-15
Column of wide-flange shape
PROPERTIES OF EACH SHAPE:
sc W 12 * 35
C 10 * 15.3
Aw 10.3 in.3
Ac 4.48 in.2
hw 12.5 in.
twc 0.24 in.
tf 0.52 in.
Iw 285 in.
y0 Ic 2.27 in. (2-2 axis)
4
(a) THE MAXIMUM TENSILE AND COMPRESSIVE STRESSES
LOCATION OF CENTROID FOR W 12 35 ALONE
hw
cw 2
cw 6.25 in.
P 25 k
hw tf
ew 2 2
st Pew
P
+
c
Aw
Iw w
sc 5711 psi
Iw
Aw ew
y0 4.62 in.
h hw + twc
A Aw + Ac
ew 5.99 in.
;
;
(C) COMBINED COLUMN, W 12 * 35 with C 10 * 15.3
h 12.74 in.
st 857 psi
;
(b) NEUTRAL AXIS (W SHAPE ALONE)
xp 0.634 in.
4
P Pew
c
Aw Iw w
A 14.78 in.2
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SECTION 5.12
LOCATION OF CENTROID OF COMBINED SHAPE
c
hw
Aw a b + Ac (h xp)
2
A
I Iw + Aw ac c 8.025 in.
hw 2
b
2
+ Ic + Ac (h xp c)2
I 394.334 in.4
(a) Determine the maximum tensile and compressive
stresses st and sc, respectively, in the column.
(b) Locate the neutral axis under this loading condition.
(c) Recompute maximum tensile and compressive stresses
if a 120 mm 10 mm cover plate is added to one flange
as shown.
2
st P
Pe
+
c
A
I
sc P
Pe
(h c)
A
I
y0 I
Ae
Problem 5.12-16 A short column of wide-flange shape is
subjected to a compressive load that produces a resultant force
P 55 kN acting at the midpoint of one flange (see figure).
tf
e hw P 25 k
489
Beams with Axial Loads
e 4.215 in.
c
st 453 psi
;
sc 2951 psi
;
y0 6.33 in. (from centroid)
;
y
P = 55 kN
z
Cover plate
(120 mm 10 mm)
(Part c only)
y
P
C
8 mm
z
200
mm
C
12 mm
160
mm
Solution 5.12-16
P 55 kN
(a) MAXIMUM TENSILE AND COMPRESSIVE STRESSES FOR W
SHAPE ALONE
PROPERTIES AND DIMENSIONS FOR W SHAPE
b 160 mm
tf 12 mm
d 200 mm
tw 8 mm
Aw bd (b tw) (d 2 tf)
Aw 5.248 * 103 mm2
(b tw) (d 2 tf)3
bd3
Iw 12
12
Iw 3.761 * 107 mm4
e
tf
d
2
2
e 94 mm
st P
Pe d
+
Aw
Iw 2
st 3.27 MPa
sc P
Pe d
Aw
Iw 2
sc 24.2 MPa
(b) NEUTRAL AXIS (W SHAPE ALONE)
y0 Iw
Aw e
y0 76.2 mm
;
(c) COMBINED COLUMN-W SHAPE & COVER PLATE
bp 120 mm tp 10 mm
;
;
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Stresses in Beams (Basic Topics)
h dtp
I 4.839 * 107 mm4
tf
e 74.459 mm
ed c
2
h 210 mm
A Aw + bp tp
A 6.448 * 103 mm2
CENTROID OF COMPOSITE SECTION
tp
d
Aw + bp tp ad + b
2
2
c
A
st sc c 119.541 mm
I Iw + Aw ac bp t3p
+
12
d
b
2
+ bp tp a d +
y0 tp
2
cb
P Pe
(h c)
Aw Iw
sc 20.3 MPa
;
I
Ae
y0 100.8 mm (from centrioid)
2
1
(a) Determine the maximum tensile stress st in the
angle section.
(b) Recompute the maximum tensile stress if two angels
are used and P is applied as shown in the figure part (b).
1
L44—
2
C
1
3
1
1
2L44—
2
C
P 2
3
P
(a)
(b)
Angle section in tension
(b) TWO ANGLES: L 4 * 4 * 1/2
(a) ONE ANGLE: L 4 * 4 * 1/2
AL 3.75 in.2
A 2AL
rmin 0.776 in.
t 0.5 in.
t 0.5 in
c 1.18 in.
c 1.18 in
e ac
t
b 12
2
e 1.315 in
P 12.5 k
c1 c 12
AL rmin2
M Pe
Mc1
P
+
AL
I3
IL 5.52 in.4 (2-2 axis)
e ac
t
b
2
e 0.93 in.
P 12.5 k
c1 1.699 in.
I 2IL
I3 2.258 in.
4
M Pe
M 16.44 k-in.
MAXIMUM TENSILE STRESS OCCURS AT CORNER
st ;
2
L 4 4 2 inch angle section (see Table E-4(a) in
Appendix E) is subjected to a tensile load P 12.5 kips
that acts through the point where the midlines of the legs
intersect [see figure part (a)].
I3 st 1.587 MPa
NEUTRAL AXIS
2
Problem 5.12-17 A tension member constructed of an
Solution 5.12-17
P
Pe
+
c
A
I
st 15.48 ksi
;
I 11.04 in.4
M 11.625 k-in.
MAXIMUM TENSILE STRESS OCCURS AT THE LOWER EDGE
st P
Mc
+
A
I
st 2.91 ksi
;
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Page 491
SECTION 5.12
Beams with Axial Loads
Two L 76 76 6.4 angles
Problem 5.12-18 A short length of a 200 * 17.1 channel
is subjected to an axial compressive force P that has its line
of action through the midpoint of the web of the channel
[(see figure(a)].
y
C 200 × 17.1
P
(a) Determine the equation of the neutral axis under this
z
z
C
loading condition.
(b) If the allowable stresses in tension and compression
(a)
are 76 MPa and 52 MPa respectively, find the
maximum permissible load Pmax.
(c) Repeat (a) and (b) if two L 76 76 6.4 angles are
added to the channel as shown in the figure part (b).
See Table E-3(b) in Appendix E for channel properties and Table E-4(b) for angle properties.
y
P
C
C 200 × 17.1
(b)
Solution 5.12-18
sc
P
tw = 5.59mm
bf = 57.4
Ac 2170 mm2
1
e
c1
Ac
Ic
Pmax 67.3 kN
dc 203 mm c1 14.5 mm
L 76 * 76 * 6.4
Ic 0.545 * 106 mm4 (z-axis)
AL 929 mm2
c2 bf c1 c2 42.9 mm
cL 21.2 mm
A 4.028 * 103 mm2
st 76 MPa s c 52 MPa
A Ac + 2 AL
ECCENTRICITY OF THE LOAD
h bf + 76 mm
tw
2
e 11.705 mm
(a) LOCATION OF THE NEUTRAL AXIS (CHANNEL ALONE)
Ic
y0 Ac # e
y0 21.5 mm
;
Pe
P
+
c
A
I 2
P 165.025 kN
P
Pe
sc c
A
I 1
h 133.4 mm
CENTROID OF COMPOSITE SECTION
Ac 1bf c12 + 2 AL 1bf + cL2
c
A
c 59.367 mm
I Ic + Ac 1bf c1 c22
+ 2 IL + 2 AL 1bf + cL c22
I 2.845 * 106 mm4
(b) FIND PMAX
st IL 0.512 * 106 mm4
COMPOSITE SECTION
ALLOWABLE STRESSES
e c1 ;
(c) COMBINED COLUMN WITH 2-ANGLES
C 200 * 17.1
P
st
1
e
+ c2
Ac
Ic
491
e bf tw
c
2
e 4.762 mm
bf 57.4 mm
LOCATION OF THE NEUTRAL AXIS
y0 I
Ae
y0 148.3 mm
;
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Page 492
Stresses in Beams (Basic Topics)
y0 148.3 mm 7 h 133.4 mm
;
P
Thus, this composite section has no tensile stress
sc P
Pe
+
c
A
I
sc
1
e
+ c
A
I
Pmax 149.6 kN
;
Stress Concentrations
The problems for Section 5.13 are to be solved considering the
stress-concentration factors.
M
M
h
Problem 5.13-1 The beams shown in the figure are subjected
to bending moments M 2100 lb-in. Each beam has a rectangular
cross section with height h 1.5 in. and width b 0.375 in.
(perpendicular to the plane of the figure).
d
(a)
(a) For the beam with a hole at midheight, determine the
maximum stresses for hole diameters d 0.25, 0.50, 0.75,
and 1.00 in.
(b) For the beam with two identical notches (inside height
h1 1.25 in .), determine the maximum stresses for notch radii
R 0.05, 0.10, 0.15, and 0.20 in.
2R
M
M
h
Probs. 5.13.1 through 5.13-4
h1
(b)
Solution 5.13-1
M 2100 lb-in. h 1.5 in. b 0.375 in.
(b) BEAM WITH NOTCHES
(a) BEAM WITH A HOLE
d
1
…
h
2
h1 1.25 in.
Eq.(5-57): sc 1
d
Ú
h
2
Eq.(5-56): sB d (in.)
d
h
sc Eq. (1)
(psi)
0.25
0.50
0.75
1.00
0.1667
0.3333
0.5000
0.6667
15,000
15,500
17,100
—
6Mh
Eq. (5-58)
b(h3 d3)
50,400
3.375 d
3
(1)
snom 12Md
b(h3 d3)
67,200 d
3.375 d3
(2)
sB
Eq. (2) (psi)
—
—
17,100
28,300
h
1.5 in.
1.2
h1
1.25 in.
sm ax
(psi)
15,000
15,500
17,100
28,300
NOTE: The larger the hole, the larger the stress.
6M
bh21
21,500 psi
R (in)
R
h1
K
(Fig. 5-50)
sm ax Ks nom
sm ax (psi)
0.05
0.10
0.15
0.20
0.04
0.08
0.12
0.16
3.0
2.3
2.1
1.9
65,000
49,000
45,000
41,000
NOTE: The larger the notch radius, the smaller the stress.
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Page 493
SECTION 5.13
Stress Concentrations
493
Problem 5.13-2 The beams shown in the figure are subjected to bending
moments M 250 N # m. Each beam has a rectangular cross section with height
h 44 mm and width b 10 mm (perpendicular to the plane of the figure).
(a) For the beam with a hole at midheight, determine the maximum stresses
for hole diameters d 10, 16, 22 and 28 mm.
(b) For the beam with two identical notches (inside height h1 40 mm ),
determine the maximum stresses for notch radii R 2, 4, 6, and 8 mm.
Solution 5.13-2
M 250 N # m h 44 mm b 10 mm
(b) BEAM WITH NOTCHES
(a) BEAM WITH A HOLE
1
d
…
h
2
sc sB d (mm)
10
16
22
28
Eq. (5-57):
6Mh
b(h3 d3)
d
1
Ú
h
2
h1 40 mm
Eq. (5-58): snom 6
6.6 * 10
85,180 d3
MPa
b(h3 d3)
d
h
0.227
0.364
0.500
0.636
R
(mm)
300 * 103d
MPa
85,180 d3
sB
sc
Eq. (2)
Eq. (1) (MPa) (MPa)
78
81
89
—
—
—
89
133
6M
bh21
93.8 MPa
(1)
Eq. (5-56):
12Md
h
44 mm
1.1
h1
40 mm
(2)
sm ax
(MPa)
2
4
6
8
R
h1
K
(Fig. 5-50)
smax Ks nom
smax (MPa)
0.05
0.10
0.15
0.20
2.6
2.1
1.8
1.7
240
200
170
160
NOTE: The larger the notch radius, the smaller the stress.
78
81
89
133
NOTE: The larger the hole, the larger the stress.
Problem 5.13-3 A rectangular beam with semicircular notches, as shown
in part (b) of the figure, has dimensions h 0.88 in. and h1 0.80 in. The
maximum allowable bending stress in the metal beam is smax 60 ksi, and
the bending moment is M 600 lb-in.
Determine the minimum permissible width bmin of the beam.
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CHAPTER 5
Solution 5.13-3
h 0.88 in.
Page 494
Stresses in Beams (Basic Topics)
Beam with semicircular notches
h1 0.80 in.
smax 60 ksi M 600 lb-in.
1
h h1 + 2R R (h h1) 0.04 in.
2
0.04 in.
R
0.05
h1
0.80 in.
smax Ksnom Ka
60 ksi 2.57c
6M
bh21
b
6(600 lb-in.)
b(0.80 in.)2
d
Solve for b:
;
bmin L 0.24 in.
From Fig. 5-50: K L 2.57
Problem 5.13-4 A rectangular beam with semicircular notches,
as shown in part (b) of the figure, has dimension h 120 mm and
h1 100 mm . The maximum allowable bending stress in the plastic
beam is smax 6 MPa, and the bending moment is M 150 N # m.
Determine the minimum permissible width bmin of the beam.
Solution 5.13-4
Beam with semicircular notches
h 120 mm
h1 100 mm
smax 6 MPa
M 150 N # m
smax Ksnom Ka
1
h h1 + 2R R (h h1) 10 mm
2
6 MPa 2.20c
R
10 mm
0.10
h1
100 mm
Solve for b:
From Fig.5-50: K L 2.20
Problem 5.13-5 A rectangular beam with notches and a
hole (see figure) has dimensions h 5.5 in., h1 5 in., and width
b 1.6 in. The beam is subjected to a bending moment
M 130 k-in., and the maximum allowable bending stress
in the material (steel) is smax 42,000 psi.
(a) What is the smallest radius Rmin that should be
used in the notches?
(b) What is the diameter dmax of the largest hole that should
be drilled at the midheight of the beam?
6M
bh21
b
6(150 N # m)
b(100 mm)2
bmin L 33 mm
d
;
2R
M
M
h1
h
d
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Page 495
SECTION 5.13
Solution 5.13-5
Beam with notches and a hole
h 5.5 in. h1 5 in. b 1.6 in.
M 130 k-in. smax 42,000 psi
(b) LARGEST HOLE DIAMETER
Assume
(a) MINIMUM NOTCH RADIUS
sB 5.5 in.
h
1.1
h1
5 in.
snom K
6M
bh21
Stress Concentrations
12Md
b(h3 d3)
42,000 psi 19,500 psi
12(130 k-in.)d
(1.6 in.)[(5.5 in.)3 d3]
d3 + 23.21d 166.4 0
42,000 psi
smax
2.15
snom
19,500 psi
h
From Fig. 5-50, with K 2.15 and
1.1, we get
h1
R
L 0.090
h1
‹ Rmin L 0.090h1 0.45 in.
1
d
7 and use Eq. (5-56).
h
2
;
Solve numerically:
dmax 4.13 in.
;
or
495
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5:32 AM
Page 497
6
Stresses in Beams
(Advanced Topics)
Composite Beams
When solving the problems for Section 6.2, assume that the
component parts of the beams are securely bonded by
adhesives or connected by fasteners. Also, be sure to use the
general theory for composite beams described in Sect. 6.2.
y
Problem 6.2-1 A composite beam consisting of fiberglass
faces and a core of particle board has the cross section
shown in the figure. The width of the beam is 2.0 in., the
thickness of the faces is 0.10 in., and the thickness of the
core is 0.50 in. The beam is subjected to a bending moment
of 250 lb-in. acting about the z axis.
Find the maximum bending stresses sface and score
in the faces and the core, respectively, if their respective
moduli of elasticity are 4 ⫻ 106 psi and 1.5 ⫻ 106 psi.
0.10 in.
z
0.50 in.
C
0.10 in.
2.0 in.
Solution 6.2-1 Composite beam
b ⫽ 2 in.
h ⫽ 0.7 in.
hc ⫽ 0.5 in.
M ⫽ 250 lb-in.
E1 ⫽ 4 ⫻ 10 psi
6
E2 ⫽ 1.5 ⫻ 106 psi
I1 ⫽
b 3
(h ⫺ h3c) ⫽ 0.03633 in.4
12
I2 ⫽
bh3c
⫽ 0.02083 in.4
12
From Eq. (6-6b): score ⫽ ;
E1I1 + E2I2 ⫽ 176,600 lb-in.2
From Eq. (6-6a): sface ⫽ ;
M(h/2)E1
E1I1 + E2I2
⫽ ; 1980 psi
M(hc / 2)E2
E1I1 + E2I2
⫽ ;531psi
;
;
497
9/24/08
Page 498
CHAPTER 6 Stresses in Beams (Advanced Topics)
y
Problem 6.2-2 A wood beam with cross-sectional dimensions
200 mm ⫻ 300 mm is reinforced on its sides by steel plates 12 mm
thick (see figure). The moduli of elasticity for the steel and wood
are Es ⫽ 190 GPa and Ew ⫽ 11 GPa, respectively. Also, the corresponding allowable stresses are ss ⫽ 110 MPa and sw ⫽ 7.5 MPa.
(a) Calculate the maximum permissible bending moment
Mmax when the beam is bent about the z axis.
(b) Repeat part a if the beam is now bent about its y axis.
z
12 mm
z
C
200 mm
498
5:32 AM
300 mm
06Ch06.qxd
200 mm
12 mm
300 mm
12 mm
12 mm
y
C
(a)
(b)
Solution 6.2-2
MAXIMUM MOMENT BASED UPON THE WOOD
y
Mmax_w ⫽ sallow_w
1
2
J
Ew Iw + Es Is
h
a b Ew
2
Mmax_w ⫽ 69.1 kN # m
z
300 mm
C
MAXIMUM MOMENT BASED UPON THE STEEL
Mmax_s ⫽ sallow_s
200 mm
12 mm
K
J
Ew Iw + Es Is
h
a b Es
2
Mmax_s ⫽ 58.7 kN # m
12 mm
K
Mmax ⫽ min (Mmax_w, Mmax_s)
(a) BENT ABOUT THE Z AXIS
b ⫽ 200 mm
Ew ⫽ 11 GPa
t ⫽ 12 mm
h ⫽ 300 mm
Es ⫽ 190 GPa
sallow_w ⫽ 7.5 MPa
sallow_s ⫽ 110 MPa
Iw ⫽
bh3
12
Iw ⫽ 4.50 * 108 mm4
Is ⫽
2th3
12
Is ⫽ 5.40 * 107 mm4
EwIw ⫹ EsIs ⫽ 1.52 ⫻ 107 N⭈m2
STEEL GOVERNS.
Mmax ⫽ 58.7 kN # m
(b) BENT ABOUT THE Y AXIS
Iw ⫽
b3 h
12
Is ⫽ 2c
Iw ⫽ 2.00 * 108 mm4
t3h
b + t 2
+ th a
b d
12
2
Is ⫽ 8.10 * 107 mm4
Ew Iw + Es Is ⫽ 1.76 * 107 N # m2
;
9/24/08
5:32 AM
Page 499
499
SECTION 6.2 Composite Beams
MAXIMUM MOMENT BASED UPON THE STEEL
MAXIMUM MOMENT BASED UPON THE WOOD
Mmax_w ⫽ sallow_w
J
Ew Iw + Es Is
b
a bEw
2
Mmax_w ⫽ 119.9 kN # m
Mmax_s ⫽ sallow_s
K
Ew Iw + Es Is
J a
Mmax_s ⫽ 90.9 kN # m
b
+ tb Es K
2
;
Mmax ⫽ min (Mmax_w, Mmax_s)
Mmax ⫽ 90.9 kN # m
STEEL GOVERNS.
y
Problem 6.2-3 A hollow box beam is constructed with webs
(a) If the allowable stresses are 2000 psi for the plywood
and 1750 psi for the pine, find the allowable bending
moment Mmax when the beam is bent about the z axis.
(b) Repeat part a if the b.eam is now bent about its y axis.
1.5 in.
z
C
;
z
1.5 in.
1.5 in.
1 in.
3.5 in.
of Douglas-fir plywood and flanges of pine, as shown in the
figure in a cross-sectional view. The plywood is 1 in. thick
and 12 in. wide; the flanges are 2 in. ⫻ 4 in. (nominal size).
The modulus of elasticity for the plywood is 1,800,000 psi
and for the pine is 1,400,000 psi.
12 in.
06Ch06.qxd
C
1 in.
1.5 in.
1 in.
12 in.
3.5 in.
1 in.
(b)
(a)
Solution 6.2-3
(a) BENT ABOUT THE Z AXIS
t
t
1
I1 ⫽
b(h3 ⫺ h31)
12
I2 ⫽
2th3
12
I1 ⫽ 291 in.4
I2 ⫽ 288 in.4
E1I1 + E2I 2 ⫽ 9.26 * 108 lb # in.2
h1
2
MAXIMUM MOMENT BASED UPON THE WOOD
h
Mmax_1 ⫽ sallow_1
1
2
b ⫽ 3.5 in.
h
a b E1
2
Mmax_1 ⫽ 193 k # in.
b
t ⫽ 1 in.
J
E1 I1 + E2 I2
;
K
MAXIMUM MOMENT BASED UPON THE PLYWOOD
h ⫽ 12 in.
h1 ⫽ 9 in.
E1 ⫽ 1.4 * 106 psi
E2 ⫽ 1.8 * 106 psi
sallow_1 ⫽ 1750 psi
sallow_2 ⫽ 2000 psi
Mmax_2 ⫽ sallow_2
J
E1 I1 + E2 I2
Mmax_2 ⫽ 172 k # in.
h
a b E2
2
;
K
y
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Mmax ⫽ min (Mmax_1, Mmax_2)
MAXIMUM MOMENT BASED UPON THE PLYWOOD
Mmax ⫽ 172 k # in.
PLYWOOD GOVERNS.
;
Mmax_2 ⫽sallow_2
(b) BENT ABOUT THE Y AXIS
Ja
Mmax_2 ⫽ 96 k # in.
b3 (h ⫺ h1)
I1 ⫽
12
I2 ⫽ 2 c
E1 I1 + E2 I2
I1 ⫽ 11 in.
4
3
2
t h
b + t
+ th a
b d
12
2
I2 ⫽ 123 in.4
b
+ tb E2 K
2
Mmax ⫽ min (Mmax_1, Mmax_1)
Mmax ⫽ 96 k # in.
PLYWOOD GOVERNS.
;
E1 I1 + E2 I2 ⫽ 2.37 * 108 lb # in.2
MAXIMUM MOMENT BASED UPON THE WOOD
Mmax_1 ⫽ sallow_1
J
E1 I1 + E2 I2
b
a b E1
2
Mmax_1 ⫽ 170 k # in.
K
Problem 6.2-4 A round steel tube of outside diameter d and an brass core of diameter
S
2d/3 are bonded to form a composite beam, as shown in the figure.
Derive a formula for the allowable bending moment M that can be carried by the
beam based upon an allowable stress ss in the steel. (Assume that the moduli of
elasticity for the steel and brass are Es and Eb, respectively.)
B
2d/3
d
Solution 6.2-4
Core (2): I2 ⫽
B
E1 I1 + E2 I2 ⫽ Es I1 + Eb I2
⫽
2d/3
Mallow ⫽ ss
d
Tube (1): I1 ⫽
p 2d 4 pd 4
a b ⫽
64 3
324
S
p
2d 4
65
cd 4 ⫺ a b d ⫽
pd 4
64
3
5184
Mallow ⫽
J
pd 4
(65Es + 16 Eb)
5184
E1I1 + E2 I2
d
a b Es
2
K
sspd 3
Eb
a 65 + 16 b
2592
Es
;
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SECTION 6.2 Composite Beams
Problem 6.2-5 A beam with a guided support and 10 ft span
supports a distributed load of intensity q ⫽ 660 lb/ft over its first
half (see figure part a) and a moment M0 ⫽ 300 ft-lb at joint B.
The beam consists of a wood member (nominal dimensions 6 in.
⫻ 12 in., actual dimensions 5.5 in. ⫻ 11.5 in. in cross
section, as shown in the figure part b) that is reinforced by
0.25-in.-thick steel plates on top and bottom. The moduli
of elasticity for the steel and wood are Es ⫽ 30 ⫻ 106 psi
and Ew ⫽ 1.5 ⫻ 106 psi, respectively.
y
0.25 in.
q
11.5 in.
M0
z
A
5 ft
C
C
B
5 ft
0.25 in.
(a) Calculate the maximum bending stresses ss in the steel
(a)
plates and sw in the wood member due to the applied loads.
(b) If the allowable bending stress in the steel plates is
sas ⫽ 14,000 psi and that in the wood is saw ⫽ 900 psi,
find qmax. (Assume that the moment at B, M0, remains
at 300 ft-lb.)
(c) If q ⫽ 660 lb/ft and allowable stress values in (b) apply, what is M0,max at B?
5.5 in.
(b)
Solution 6.2-5
q ⫽ 660 lb/it
M0 ⫽ 300 lb # ft
L ⫽ 10 ft
(b) MAXIMUM UNIFORM DISTRIBUTED LOAD
MAXIMUM MOMENT BASED UPON WOOD
(a) MAXIMUM BENDING STRESSES
sallow_w ⫽ 900 psi
L 3L
Mmax ⫽ q a b a b + M0
2
4
Mmax ⫽ 25050 lb # ft
b ⫽ 5.5 in.
Wood (1):
From sallow_w ⫽
h1 ⫽ 11.5 in.
I1 ⫽
b 3
1h ⫺ h312
12
sw ⫽
ss ⫽
h1
b Ew
2
Ew I1 + EsI2
h
Mmax a b Es
2
Ew I1 + Es I2
Ew I1 + Es I2
sallow_s ⫽ 14000 psi
t ⫽ 0.25 in.
Es ⫽ 30 * 106 psi
From sallow_s ⫽
I2 ⫽ 94.93 in.4
h
Mallow_s a b Es
2
Ew I1 + Es I2
Mallow_s ⫽ 25236 lb-ft
Ew I1 + Es I 2 ⫽ 3.894 * 109 lb # in.2
Mmax a
h1
b Ew
2
MAXIMUM MOMENT BASED UPON STEEL PLATE
I1 ⫽ 697.07 in.4
b ⫽ 5.5 in.
h ⫽ 12 in.
Plate (2):
I2 ⫽
bh31
12
Mallow_w a
Mallow_w ⫽ 33857 lb-ft
Ew ⫽ 1.5 * 106 psi
MAXIMUM ALLOWABLE MOMENT
sw ⫽ 666 psi
;
Mallow ⫽ min (Mallow_s, Mallow_w)
STEEL PLATES GOVERN
ss ⫽ 13897 psi
;
501
Mallow ⫽ 25236 lb-ft
;
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CHAPTER 6 Stresses in Beams (Advanced Topics)
MAXIMUM UNIFORM DISTRIBUTED LOAD
L 3L
From Mallow ⫽ qmax a b a b + M0
2
4
qmax ⫽ 665lb/ft
;
(c) MAXIMUM APPLIED MOMENT
L 3L
From Mallow ⫽ q a b a b + Mo_max
2
4
M0_max ⫽ 486 lb-ft
;
y
Problem 6.2-6 A plastic-lined steel pipe has the cross-sectional shape shown in
the figure. The steel pipe has outer diameter d3 ⫽ 100 mm and inner diameter d2 ⫽
94 mm. The plastic liner has inner diameter d1 ⫽ 82 mm. The modulus of elasticity
of the steel is 75 times the modulus of the plastic.
Determine the allowable bending moment Mallow if the allowable stress in the
steel is 35 MPa and in the plastic is 600 kPa.
z
C
d1
Solution 6.2-6 Steel pipe with plastic liner
MAXIMUM MOMENT BASED UPON THE STEEL (1)
From Eq. (6-6a):
Mmax ⫽ (s1)allow c
⫽ (s1)allow
(1) Pipe: ds ⫽ 100 mm
d2 ⫽ 94 mm
Es ⫽ E1 ⫽ modulus of elasticity
(s1)allow ⫽ 35 MPa
(2) Liner: d2 ⫽ 94 mm
d1 ⫽ 32 mm
Ep ⫽ E2 ⫽ modulus of elasticity
(s2)allow ⫽ 600 kPa
E1 ⫽ 75E2
E1/E2 ⫽ 75
I1 ⫽
p 4
(d ⫺ d 42) ⫽ 1.076 * 10⫺6 m4
64 3
I2 ⫽
p 4
(d ⫺ d 41) ⫽ 1.613 * 10⫺6 m4
64 2
E1I1 + E2I2
d
(d3/2)E1
(E1/E2)I1 + I2
⫽ 768 N # m
(d3/2)(E1/E2)
MAXIMUM MOMENT BASED UPON THE PLASTIC (2)
From Eq. (6-6b):
Mmax ⫽ (s2)allow c
⫽ (s2)allow c
STEEL GOVERNS.
E1I1 + E2I2
d
(d2/2)E2
(E1/E2)I1 + I2
d ⫽ 1051 N # m
(d2/2)
Mallow ⫽ 768 N # m
;
d2 d3
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503
SECTION 6.2 Composite Beams
Problem 6.2-7 The cross section of a sandwich beam consisting of aluminum
y
alloy faces and a foam core is shown in the figure. The width b of the beam
is 8.0 in., the thickness t of the faces is 0.25 in., and the height hc of the core
is 5.5 in. (total height h ⫽ 6.0 in.). The moduli of elasticity are 10.5 ⫻ 106 psi for
the aluminum faces and 12,000 psi for the foam core. A bending moment
M ⫽ 40 k-in. acts about the z axis.
Determine the maximum stresses in the faces and the core using (a) the
general theory for composite beams, and (b) the approximate theory for sand wich
beams.
t
z
Probs. 6.2-7 and 6.2-8
Solution 6.2-7
C
hc
b
t
h
Sandwich beam
I2 ⫽
bh3c
⫽ 110.92 in.4
12
M ⫽ 40 k.in. E1I1 + E2I2 ⫽ 348.7 * 106 lb-in.2
(a) GENERAL THEORY (EQS. 6-6a AND b)
sface ⫽ s1 ⫽
M(h/2)E1
⫽ 3610 psi
E1I1 + E2I2
score ⫽ s2 ⫽
M(hc / 2)E2
⫽ 4 psi
E1I1 + E2I2
;
;
(1) ALUMINUM FACES:
b ⫽ 8.0 in.
t ⫽ 0.25 in.
h ⫽ 6.0 in.
E1 ⫽ 10.5 * 106 psi
I1 ⫽
I1 ⫽
b 3
(h ⫺ h3c ) ⫽ 33.08 in.4
12
b 3
(h ⫺ h3c ) ⫽ 33.08 in.4
12
sface ⫽
Mh
⫽ 3630 psi
2I1
score ⫽ 0
(2) Foam core:
b ⫽ 8.0 in.
(b) APPROXIMATE THEORY (EQS. 6-8 AND 6-9)
hc ⫽ 5.5 in.
;
;
E2 ⫽ 12,000 psi
Problem 6.2-8 The cross section of a sandwich beam consisting of fiberglass faces and a lightweight plastic core is shown in
the figure. The width b of the beam is 50 mm, the thickness t of the faces is 4 mm, and the height hc of the core is 92 mm (total
height h ⫽ 100 mm). The moduli of elasticity are 75 GPa for the fiberglass and 1.2 GPa for the plastic. A bending moment M ⫽
275 N # m acts about the z axis.
Determine the maximum stresses in the faces and the core using (a) the general theory for composite beams, and (b) the
approximate theory for sandwich beams.
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Solution 6.2-8
Sandwich beam
(a) GENERAL THEORY (EQS. 6-6a AND b)
sface ⫽ s1 ⫽
M(h/2)E1
⫽ 14.1 MPa
E1I1 + E2I2
;
score ⫽ s2 ⫽
M(hc / 2)E2
⫽ 0.21 MPa
E1I1 + E2I2
;
(b) APPROXIMATE THEORY (EQS. 6-8 AND 6-9)
I1 ⫽
(1) Fiber glass faces:
b ⫽ 50 mm
t ⫽ 4 mm
h ⫽ 100 mm
sface ⫽
E1 ⫽ 75 GPa
I1 ⫽
b 3
(h ⫺ h3c ) ⫽ 0.9221 * 106 m4
12
Mh
⫽ 14.9 MPa
2I1
score ⫽ 0
b 3
(h ⫺ h3c ) ⫽ 0.9221 * 10⫺6 m4
12
;
;
(2) Plastic core:
hc ⫽ 92 mm
b ⫽ 50 mm
I2 ⫽
bh3c
12
E2 ⫽ 1.2 GPa
⫽ 3.245 * 10⫺6 m4
M ⫽ 275 N # m
E1I1 + E2I2 ⫽ 73,050 N # m2
Problem 6.2-9 A bimetallic beam used in a temperature-control switch
consists of strips of aluminum and copper bonded together as shown in the
figure, which is a cross-sectional view. The width of the beam is 1.0 in., and
each strip has a thickness of 1/16 in.
Under the action of a bending moment M ⫽ 12 lb-in. acting about the z
axis, what are the maximum stresses sa and sc in the aluminum and copper,
respectively? (Assume Ea ⫽ 10.5 ⫻ 106 psi and Ec ⫽ 16.8 ⫻ 106 psi.)
y
A
z
O
1.0 in.
Solution 6.2-9 Bimetallic beam
NEUTRAL AXIS (EQ. 6-3)
CROSS SECTION
L1
ydA ⫽ y1A1 ⫽ (h1 ⫺ t/2)(bt)
⫽ (h1 ⫺ 1/32)(1)(1/16) in.3
(1) Aluminum E1 ⫽ Ea ⫽ 10.5 ⫻ 106 psi
(2) Copper
E2 ⫽ Ec ⫽ 16.8 ⫻ 10 psi
6
M ⫽ 12 lb-in.
L2
1
— in.
16
ydA ⫽ y2A2 ⫽ (h1 ⫺ t ⫺ t/2)(bt)
⫽ (h1 ⫺ 3/32)(1)(1/16) in.3
Eq. (6-3): E1 11 ydA + E2 12 ydA ⫽ 0
C
1
— in.
16
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SECTION 6.2 Composite Beams
(10.5 ⫻ 106)(h1 ⫺ 1/32)(1/16)
⫹ (16.8 ⫻ 106)(h1 ⫺ 3/32)(1/16) ⫽ 0
MAXIMUM STRESSES (EQS. 6-6a AND b)
sa ⫽ s1 ⫽
Mh1E1
⫽ 4120 psi
E1I1 + E2I2
;
sc ⫽ s2 ⫽
Mh2E2
⫽ 5230 psi
E1I1 + E2I2
;
Solve for h1: h1 ⫽ 0.06971 in.
h2 ⫽ 2(1/16 in.) ⫺ h1 ⫽ 0.05529 in.
505
MOMENTS OF INERTIA (FROM PARALLEL-AXIS THEOREM)
I1 ⫽
bt3
+ bt(h1 ⫺ t/2)2 ⫽ 0.0001128 in.4
12
I2 ⫽
bt3
+ bt(h2 ⫺ t/2)2 ⫽ 0.00005647 in.4
12
E1I1 + E2I2 ⫽ 2133 lb-in.2
Problem 6.2-10 A simply supported composite beam
y
3 m long carries a uniformly distributed load of intensity
q ⫽ 3.0 kN/m (see figure). The beam is constructed of a wood
member, 100 mm wide by 150 mm deep, reinforced on its
lower side by a steel plate 8 mm thick and 100 mm wide.
Find the maximum bending stresses sw and ss in the wood
and steel, respectively, due to the uniform load if the moduli of
elasticity are Ew ⫽ 10 GPa for the wood and Es ⫽ 210 GPa for
the steel.
q = 3.0 kN/m
150 mm
z
O
8
mm
3m
100 mm
Solution 6.2-10 Simply supported composite beam
BEAM: L ⫽ 3 m
2
Mmax ⫽
q ⫽ 3.0 kN/m
qL
⫽ 3375 N # m
8
CROSS SECTION
b ⫽ 100 mm
h ⫽ 150 mm
t ⫽ 8 mm
(1) Wood: E1 ⫽ Ew ⫽ 10 GPa
(2) Steel: E2 ⫽ Es ⫽ 210 GPa
NEUTRAL AXIS
L1
ydA ⫽ y1A1 ⫽ (h1 ⫺ h/2)(bh)
⫽ (h1 ⫺ 75)(100)(150) mm3
L2
ydA ⫽ y2A2 ⫽ ⫺ (h + t/2 ⫺ h1)(bt)
⫽ ⫺ (154 ⫺ h1)(100)(18) mm3
Eq. (6-3): E1 11ydA + E2 12ydA ⫽ 0
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CHAPTER 6 Stresses in Beams (Advanced Topics)
MAXIMUM STRESSES (EQS. 6-6a AND b)
(10 GPa)(h1 ⫺ 75)(100)(150)(10⫺9)
⫹ (210 GPa)(h1 ⫺ 154)(100)(8)(10⫺9) ⫽ 0
sw ⫽ s1 ⫽
Solve for h1: h1 ⫽ 116.74 mm
Mh1E1
E1I1 + E2I2
⫽ 5.1MPa (Compression)
h2 ⫽ h ⫹ t ⫺ h1 ⫽ 41.26 mm
MOMENTS OF INERTIA (FROM PARALLEL-AXIS THEOREM)
I1 ⫽
bh3
+ bh(h1 ⫺ h/2)2 ⫽ 54.26 * 106 mm4
12
I2 ⫽
bt2
+ bt(h2 ⫺ t/2)2 ⫽ 1.115 * 106 mm4
12
ss ⫽ s2 ⫽
;
Mh2E2
E1I1 + E2I2
⫽ 37.6MPa (Tension)
;
E1I1 + E2I2 ⫽ 776,750 N # m2
Problem 6.2-11 A simply supported wooden I-beam with a 12 ft span supports
a distributed load of intensity q ⫽ 90 lb/ft over its length (see figure part a). The
beam is constructed with a web of Douglas-fir plywood and flanges of pine glued
2 in.
to the web as shown in the figure part b. The plywood is 3/8 in. thick; the flanges are
2 in. ⫻ 2 in. (actual size). The modulus of
z
q
8 in.
elasticity for the plywood is 1,600,000 psi
and for the pine is 1,200,000 psi.
(a) Calculate the maximum bending
stresses in the pine flanges and in
the plywood web.
(b) What is qmax if allowable stresses
are 1600 psi in the flanges and
1200 psi in the web?
A
3
— in. plywood
8
(Douglas fir)
2 in.
B
12 ft
y 2 in. ⫻ 2 in. pine flange
1
— in.
2
C
2 in.
(b)
(a)
Solution 6.2-11
q ⫽ 90 lb/f
L ⫽ 12ft
I2 ⫽2c
(a) MAXIMUM BENDING STRESSES
Mmax
qL2
⫽
8
Plywood (1):
+ (b ⫺ t) (h2 ⫺ a) a
Mmax ⫽ 1620 lb # ft
t⫽
3
in.
8
I2 ⫽ 65.95 in.4
h1 ⫽7 in.
Eplywood ⫽ 1.6 * 10 psi
Pine (2): b ⫽ 2 in.
I1 ⫽ 10.72 in.4
h2 ⫽ 2 in.
Epine ⫽ 1.2 * 106 psi
h1
h2 ⫺ a 2
⫺
b d
2
2
;
Eplywood I1 ⫹ EpineI2 ⫽ 96.287 ⫻ 106 lbin.2
6
t h13
I1 ⫽
12
h1 + a 2 (b ⫺ t)(h2 ⫺ a)3
ba3
+ ba a
b +
12
2
12
splywood ⫽
a⫽
1
in.
2
Mmax a
h1
b Eplywood
2
Eplywood I1 + Epine I2
splywood ⫽ 1131 psi
;
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507
SECTION 6.2 Composite Beams
spine ⫽
Mmax a
h1
+ a b Epine
2
From sallow_pine ⫽
Eplywood I1 + Epine I2
spine ⫽ 969 psi
h1
+ ab Epine
2
Eplywood I1 + Epine I2
Mallow_pine ⫽ 2675 lb # ft
;
(b) MAXIMUM UNIFORM DISTRIBUTED LOAD
MAXIMUM MOMENT BASED UPON PLYWOOD
MAXIMUM ALLOWABLE MOMENT
Mallow ⫽ min (Mallow_plywood, Mallow_pine)
Mallow ⫽ 1719 lb⭈ft
PLYWOOD GOVERNS.
sallow_plywood ⫽ 1200 psi
From sallow_plywood ⫽
Mallow_pine a
Mallow_plywood a
h1
bEplywood
2
Eplywood I1 + Epine I2
;
MAXIMUM UNIFORM DISTRIBUTED LOAD
From Mallow ⫽
qmax L2
8
qmax ⫽ 95.5 lb/ft
Mallow_plywood ⫽ 1719 lb # ft
;
MAXIMUM MOMENT BASED UPON PINE
sallow_pine ⫽ 1600 psi
6 mm ⫻ 80 mm
steel plate
Problem 6.2-12 A simply supported composite beam with a
3.6 m span supports a triangularly distributed load of peak
intensity q0 at midspan (see figure part a). The beam is constructed
of two wood joists, each 50 mm ⫻ 280 mm, fastened to two steel
plates, one of dimensions 6 mm ⫻ 80 mm and the lower plate of
dimensions 6 mm ⫻ 120 mm (see figure part b). The modulus of
elasticity for the wood is 11 GPa and for the steel is 210 GPa.
If the allowable stresses are 7 MPa for the wood and 120 MPa
for the steel, find the allowable peak load intensity q0,max when the
beam is bent about the z axis. Neglect the weight of the beam.
50 mm ⫻ 280 mm
wood joist
y
280 mm
C
z
6 mm ⫻ 120 mm
steel plate
q0
A
1.8 m
1.8 m
B
(a)
(b)
Solution 6.2-12
L ⫽ 3.6 m
Steel (2):
t1 ⫽ 50 mm
b1 ⫽ 80 mm
b2 ⫽ 120 mm
DETERMINE NEUTRAL AXIS
WOOD (1):
t2 ⫽ 6 mm
h ⫽ 280 mm
Ew ⫽ 11 GPa
h
y1 dA ⫽ y1 A1 ⫽ a ⫺ h1b (2t1 h)
2
L
L
Es ⫽ 210 GPa
y2 dA ⫽ y2 A 2 ⫽ ah ⫺ h1 ⫺
⫺ ah1 ⫺
b1
b (t2 b1)
2
b2
b (t 2 b 2)
2
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CHAPTER 6 Stresses in Beams (Advanced Topics)
From E1
Ew a
L
y1 dA + E2
L
MAXIMUM MOMENT BASED UPON WOOD
y2 dA ⫽ 0
sallow_w ⫽ 7 MPa
b1
h
⫺ h1 b (2t1 h) + Es c a h ⫺ h1 ⫺ b
2
2
(t2 b1) ⫺ ah1 ⫺
From
sallow_w ⫽
Mallow_w (h ⫺ h1)Ew
Ew I1 + Es I2
Mallow_w ⫽ 18.68 kN # m
b2
b (t2 b 2) d ⫽ 0
2
MAXIMUM MOMENT BASED UPON STEEL
h1 ⫽ 136.4 mm
sallow_s ⫽ 120 MPa
MOMENT OF INERTIA
From sallow_s ⫽
2
t1 h 3
h
Wood (1): I1 ⫽ 2 c
+ (t1 h)a ⫺ h1 b d
12
2
Mallow_s (h ⫺ h1)Es
Ew I1 + Es I2
Mallow_s ⫽ 16.78 kN # m
I1 ⫽ 183.30 * 10 mm
6
Steel (2):
I2 ⫽
4
MAXIMUM ALLOWABLE MOMENT
t2 b31
b1 2
+ t2 b1 ah ⫺ h1 ⫺ b
12
2
+
t2 b23
12
+ t2 b2 ah1 ⫺
Mallow ⫽ min(Mallow_w,Mallow_s)
b2 2
b
2
Mallow ⫽ 16.78 kN # m
STEEL GOVERNS.
;
MAXIMUM UNIFORM DISTRIBUTED LOAD
I2 ⫽ 10.47 ⫻ 106 mm4
Ew I1 ⫹ Es I2 ⫽ 4.22 ⫻ 1012 N # mm2
From
Mallow ⫽
qomax L2
12
qomax ⫽ 15.53 kN/m
;
Transformed-Section Method
When solving the problems for Section 6.3, assume that
the component parts of the beams are securely bonded by
adhesives or connected by fasteners. Also, be sure to use
the transformed-section method in the solutions.
y
Problem 6.3-1 A wood beam 8 in. wide and 12 in. deep
3.5 in.
(nominal dimensions) is reinforced on top and bottom by
0.25-in.-thick steel plates (see figure part a).
z
C
z
11.25 in.
0.25 in
11.5 in.
(a) Find the allowable bending moment Mmax about the
z axis if the allowable stress in the wood is 1,100 psi
and in the steel is 15,000 psi. (Assume that the ratio
of the moduli of elasticity of steel and wood is 20.)
(b) Compare the moment capacity of the beam in part a with
that shown in the figure part b which has two 4 in. ⫻ 12 in.
joists (nominal dimensions) attached to a 1/4 in. ⫻ 11.0 in.
steel plate.
1
— in. ⫻ 11.0 in.
4
steel plate
y
C
4 ⫻ 12
joists
0.25 in
7.5 in.
(a)
(b)
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SECTION 6.3 Transformed-Section Method
Solution 6.3-1
Mmax ⫽min(M1, M2)
(a) FIND Mmax
b ⫽ 7.5 in.
(1) Wood beam
h1 ⫽ 11.5 in.
Mmax ⫽ 422 k-in.
STELL GOVERNS
;
sallow_w ⫽ 1100 psi
b ⫽ 7.5 in.
(2) Steel plates
h2 ⫽ 12 in.
t ⫽ 0.25 in.
sallow_s ⫽ 15000 psi
TRANSFORMED SECTION (WOOD)
n ⫽ 20
IT ⫽
bh31
12
bT ⫽ 150 in.
+ 2c
b ⫽ 3.5 in.
(1) Wood beam
(2) Steel plates
h1 ⫽ 11.25 in.
h2 ⫽ 11 in.
t ⫽ 0.25 in.
WIDTH OF STEEL PLATES
bT ⫽ nt bT ⫽ 5 in.
WIDTH OF STEEL PLATES
bT ⫽ nb
(b) COMPARE MOMENT CAPACITIES
t3 bT
h2 ⫺ t 2
+ t bT a
b d
12
2
IT ⫽ 2
bh13 bT h23
+
12
12
MAXIMUM MOMENT BASED UPON THE WOOD (1)
M1 ⫽
IT ⫽ 3540 in.
4
IT ⫽1385 in.4
sallow_w IT
h1
2
M1 ⫽ 271 k # in.
MAXIMUM MOMENT BASED UPON THE WOOD (1)
M1 ⫽
sallow_w IT
h1
2
M1 ⫽ 677 k # in.
M2 ⫽
MAXIMUM MOMENT BASED UPON THE STEEL (2)
M2 ⫽
sallow_s I T
h2n
2
MAXIMUM MOMENT BASED UPON THE STEEL (2)
M2 ⫽ 442 k # in.
sallow_s IT
h2 n
2
M2 ⫽ 189 k # in.
Mmax ⫽min(M1, M2)
STELL GOVERNS.
Mmax ⫽189 k-in.
;
THE MOMENT CAPACITY OF THE BEAM IN (a) IS 2.3
(b)
TIMES MORE THAN THE BEAM IN
y
Problem 6.3-2 A simple beam of span length 3.2 m carries a uniform load of intensity
48 kN/m. The cross section of the beam is a hollow box with wood flanges and steel side
plates, as shown in the figure. The wood flanges are 75 mm by 100 mm in cross section,
and the steel plates are 300 mm deep.
What is the required thickness t of the steel plates if the allowable stresses are
120 MPa for the steel and 6.5 MPa for the wood? (Assume that the moduli of elasticity
for the steel and wood are 210 GPa and 10 GPa, respectively, and disregard the weight
of the beam.)
75 mm
z
300 mm
C
75 mm
100 mm
t
t
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Solution 6.3-2
Mmax ⫽
Box beam
Width of steel plates
qL2
⫽ 61.44 kN # m
8
SIMPLE BEAM:
L ⫽ 3.2 m
(1) Wood flanges: b ⫽ 100 mm
⫽ nt ⫽ 21t
q ⫽ 48 kN/m
All dimensions in millimeters.
h ⫽ 300 mm
IT ⫽
h1 ⫽ 150 mm
(s1)allow ⫽ 6.5 MPa
1
1
(100 + 42t)(300)3 ⫺
(100)(150)3
12
12
⫽ 196.9 * 106 mm4 + 94.5t * 106 mm4
Ew ⫽ 10 GPa
(2) Steel plates:
t ⫽ thickness
h ⫽ 300 mm
(s2)allow ⫽ 120 MPa
Es ⫽ 210 GPa
TRANSFORMED SECTION (WOOD)
REQUIRED THICKNESS BASED UPON THE WOOD (1)
(EQ. 6-15)
s1 ⫽
M(h/2)
IT
(IT)1 ⫽
Mmax(h/2)
(s1)allow
⫽ 1.418 * 109 mm4
Equate IT and (IT)1 and solve for t : t1 ⫽ 12.92 mm
REQUIRED THICKNESS BASED UPON THE STEEL (2) (EQ. 6-17)
s2 ⫽
M(h/2)n
IT
(IT)2 ⫽
Mmax(h/2)n
(s2)allow
⫽ 1.612 * 109 mm4
Equate IT and (IT)2 and solve for t : t2 ⫽ 14.97 mm
tmin ⫽ 15.0 mm
STEEL GOVERNS.
;
Wood flanges are not changed
n⫽
Es
⫽ 21
Ew
y
Problem 6.3-3 A simple beam that is 18 ft long supports a
uniform load of intensity q. The beam is constructed of two
C 8 ⫻ 11.5 sections (channel sections or C shapes) on either side
of a 4 ⫻ 8 (actual dimensions) wood beam (see the cross section
shown in the figure part a). The modulus of elasticity of the steel
(Es ⫽ 30,000 ksi) is 20 times that of the wood (Ew).
z
C 8 ⫻ 11.5
C
z
y
(a) If the allowable stresses in the steel and wood are 12,000
C
psi and 900 psi, respectively, what is the allowable load
Wood beam
C 8 ⫻ 11.5 Wood beam
qallow? (Note: Disregard the weight of the beam, and see
Table E-3a of Appendix E for the dimensions and proper(a)
(b)
ties of the C-shape beam.)
(b) If the beam is rotated 90° to bend about its y axis (see figure part b), and uniform load q ⫽ 250 lb/ft is applied, find
the maximum stresses ss and sw in the steel and wood, respectively. Include the weight of the beam. (Assume weight
densities of 35 lb/ft3 and 490 lb/ft3 for the wood and steel, respectively.)
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SECTION 6.3 Transformed-Section Method
Solution 6.3-3
(b) BENT ABOUT THE Y AXIS (INCLUDING THE WEIGHT OF
THE BEAM) q ⫽ 250 lb/ft.
L ⫽ 18 ft
(1) Wood beam
rw ⫽ 35 lb/ft
(1) Wood beam
(a) BENT ABOUT THE Z AXIS
b ⫽ 4 in.
qw ⫽ 7.778 Ib/ft.
h ⫽ 8 in.
sallow_w ⫽ 900 psi
qs ⫽ 11.5 lb/ft.
(2) Steel Channels
Iz ⫽ 32.5 in.4
(2) Steel Channel h ⫽ 8.0 in.
Iy ⫽ 1.31 in.
c ⫽ 0.572 in.
4
qw ⫽ bhrw
sallow_s ⫽ 12000 psi
As ⫽ 3.37 in.2
qtotal ⫽ q + qw + 2qs
bs ⫽ 2.26 in.
qtotal ⫽ 281 lb/ft.
2
Mmax ⫽
qtotal L
8
Mmax ⫽ 11.4 k # ft.
TRANSFORMED SECTION (WOOD)
TRANSFORMED SECTION (WOOD)
n ⫽ 20
bh3
IT ⫽
+ 2 Iz n
12
IT ⫽
IT ⫽ 1471 in.
4
b3h
b 2
+ 2n cIy + As a c + b d
12
2
IT ⫽ 987 in.4
MAXIMUM MOMENT BASED UPON THE WOOD (1)
M1 ⫽
sallow_w IT
h/2
MAXIMUM STRESS IN THE WOOD (1)
M1 ⫽ 331 k # in.
sw_max ⫽
MAXIMUM MOMENT BASED UPON THE STEEL (2)
M2 ⫽
sallow_s IT
hn/ 2
M2 ⫽ 221 k # in.
ss_max ⫽
Mmax ⫽ 221 k # in
qallow L2
8
qallow ⫽ 454 lb/ft.
nMmax a
sw_max ⫽ 277 psi
;
b
+ bsb
2
IT
ss_max ⫽ 11782 psi
ALLOWABLE LOAD ON a 18-FT-LONG SIMPLE BEAM
From Mmax ⫽
IT
MAXIMUM MOMENT BASED UPON THE STEEL (2)
Mmax ⫽ min(M1, M2)
STEEL GOVERNS.
b
Mmax a b
2
;
;
Problem 6.3-4 The composite beam shown in the figure is simply
supported and carries a total uniform load of 50 kN/m on a span length
of 4.0 m. The beam is built of a wood member having cross-sectional
dimensions 150 mm ⫻ 250 mm and two steel plates of cross-sectional
dimensions 50 mm ⫻ 150 mm.
Determine the maximum stresses ss and sw in the steel and wood,
respectively, if the moduli of elasticity are Es ⫽ 209 GPa and
Ew ⫽ 11 GPa. (Disregard the weight of the beam.)
y
50 kN/m
50 mm
z
C
250 mm
50 mm
4.0 m
150 mm
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Solution 6.3-4 Composite beam
SIMPLE BEAM:
L ⫽ 4.0 m
qL2
⫽ 100 kN # m
8
(1) Wood beam: b ⫽ 150 mm
q ⫽ 50 kN/m
Mmax ⫽
h2 ⫽ 350 mm
TRANSFORMED SECTION (WOOD)
⫽ nb ⫽ (19) (150 mm) ⫽ 2850 mm
All dimensions in millimeters.
h1 ⫽ 250 mm
Ew ⫽ 11 GPa
(2) Steel plates: b ⫽ 150 mm
Width of steel plates
t ⫽ 50 mm
Es ⫽ 209 GPa
IT ⫽
1
1
(2850)(350)3 ⫺ (2850 ⫺ 150)(250)3
12
12
⫽ 6.667 * 109 mm4
MAXIMUM STRESS IN THE WOOD (1) (EQ. 6-15)
sw ⫽ s1 ⫽
Mmax(h1/2)
⫽ 1.9 MPa
IT
;
MAXIMUM STRESS IN THE STEEL (2) (EQ. 6-17)
ss ⫽ s2 ⫽
Mmax(h2/2)n
⫽ 49.9 MPa
IT
;
Wood beam is not changed.
n⫽
Es
209
⫽
⫽ 19
Ew
11
Problem 6.3-5 The cross section of a beam made of thin strips of aluminum
separated by a lightweight plastic is shown in the figure. The beam has width
b ⫽ 3.0 in., the aluminum strips have thickness t ⫽ 0.1 in., and the plastic segments
have heights d ⫽ 1.2 in. and 3d ⫽ 3.6 in. The total height of the beam is h ⫽ 6.4 in.
The moduli of elasticity for the aluminum and plastic are Ea ⫽ 11 ⫻ 106 psi and
Ep ⫽ 440 ⫻ 103 psi, respectively.
Determine the maximum stresses sa and sp in the aluminum and plastic,
respectively, due to a bending moment of 6.0 k-in.
y
t
z
d
C
3d
d
Probs. 6.3-5 and 6.3-6
b
h ⫽ 4t ⫹ 5d
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SECTION 6.3 Transformed-Section Method
513
Solution 6.3-5 Plastic beam with aluminum strips
(1) Plastic segments: b ⫽ 3.0 in.
d ⫽ 1.2 in.
3d ⫽ 3.6 in.
Ep ⫽ 440 ⫻ 103 psi
(2) Aluminum strips: b ⫽ 3.0 in.
All dimensions in inches.
Plastic: I1 ⫽ 2c
t ⫽ 0.1 in.
+
Ea ⫽ 11 ⫻ 106 psi
h ⫽ 4t ⫹ 5d ⫽ 6.4 in.
M ⫽ 6.0 k-in.
1
(3.0)(1.2)3 + (3.0)(1.2)(2.50)2 d
12
1
(3.0)(3.6)3 ⫽ 57.528 in.4
12
Aluminum:
I2 ⫽ 2c
TRANSFORMED SECTION (PLASTIC)
+
1
(75)(0.1)3 + (75)(0.1)(3.15)2
12
1
(75)(0.1)3 + (75)(0.1)(1.85)2 d
12
⫽ 200.2 in.4
IT ⫽ I1 + I2 ⫽ 257.73 in.4
MAXIMUM STRESS IN THE PLASTIC (1) (EQ. 6-15)
sp ⫽ s1 ⫽
M(h/2 ⫺ t)
⫽ 72 psi
IT
;
MAXIMUM STRESS IN THE ALUMINUM (2) (EQ. 6-17)
Plastic segments are not changed.
Ea
n⫽
⫽ 25
Ep
sa ⫽ s2 ⫽
M(h/2)n
⫽ 1860 psi
IT
;
Width of aluminum strips
⫽ nb ⫽ (25)(3.0 in.) ⫽ 75 in.
Problem 6.3-6 Consider the preceding problem if the beam has width b ⫽ 75 mm, the aluminum strips have thickness t ⫽ 3 mm, the
plastic segments have heights d ⫽ 40 mm and 3d ⫽ 120 mm, and the total height of the beam is h ⫽ 212 mm. Also, the moduli of
elasticity are Ea ⫽ 75 GPa and Ep ⫽ 3 GPa, respectively.
Determine the maximum stresses sa and sp in the aluminum and plastic, respectively, due to a bending moment of 1.0 kN ⭈ m.
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Solution 6.3-6 Plastic beam with aluminum strips
(1) Plastic segments: b ⫽ 75 mm
3d ⫽ 120 mm
(2) Aluminum strips: b ⫽ 75 mm
All dimensions in millimeters.
d ⫽ 40 mm
Ep ⫽ 3 GPa
Plastic: I1 ⫽ 2c
t ⫽ 3 mm
Ea ⫽ 75 GPa
+
h ⫽ 4t ⫹ 5d ⫽ 212 mm
1
(75)(40)3 + (75)(40)(83)2 d
12
1
(75)(120)3
12
⫽ 52.934 * 106 mm4
M ⫽ 1.0 kN # m
ALUMINUM:
TRANSFORMED SECTION (PLASTIC)
I2 ⫽ 2c
+
1
(1875)(3)3 + (1875)(3)(104.5)2
12
1
(1875)(3)3 + (1875)(3)(61.5)2 d
12
⫽ 165.420 * 106 mm4
IT ⫽ I1 + I2 ⫽ 218.35 * 106 mm4
MAXIMUM STRESS IN THE PLASTIC (1) (EQ. 6-15)
sp ⫽ s1 ⫽
Plastic segments are not changed.
M(h/2 ⫺ t)
⫽ 0.47 MPa
IT
;
MAXIMUM STRESS IN THE ALUMINUM (2) (EQ. 6-17)
Ea
n⫽
⫽ 25
Ep
sa ⫽ s2 ⫽
Width of aluminum strips
M(h/2)n
⫽ 12.14 MPa
IT
;
⫽ nb ⫽ (25)(75 mm) ⫽ 1875 mm
Problem 6.3-7 A simple beam that is 18 ft long supports a uniform load of intensity q. The beam is constructed of two angle
sections, each L 6 ⫻ 4 ⫻ 1/2, on either side of a 2 in. ⫻ 8 in. (actual dimensions) wood beam (see the cross section shown in
the figure part a). The modulus of elasticity of the steel is 20 times that of the wood.
(a) If the allowable stresses in the
4 in.
steel and wood are 12,000 psi
and 900 psi, respectively, what
is the allowable load qallow?
(Note: Disregard the weight of
z
the beam, and see Table E-5a 8 in. 6 in.
of Appendix E for the dimensions and properties of the
Steel angle
angles.)
(b) Repeat part a if a 1 in. ⫻ 10 in.
wood flange (actual dimensions)
is added (see figure part b).
(a)
4 in.
yC
y
Wood flange
C
z
Wood beam
2 in.
Wood beam
Steel angle
2 in.
(b)
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SECTION 6.3 Transformed-Section Method
Solution 6.3-7
L ⫽ 18 ft
bf ⫽ 1 in.
(b) ADDITIONAL WOOD FLANGE
hf ⫽ 10 in.
(a) WOOD BEAM AND STEEL ANGLES
b ⫽ 2 in.
(1) Wood beam
h ⫽ 8 in.
TRANSFORMED SECTION (WOOD)
sallow_w ⫽ 900 psi
NEUTRAL AXIS
(2) Steel Channel Iz ⫽ 17.3 in.
4
As ⫽ 4.75 in.2
d ⫽ 1.98 in.
hs ⫽ 6 in.
sallow_s ⫽ 12000 psi
TRANSFORMED SECTION (WOOD)
n ⫽ 20
h1_b ⫽ 3.015 in.
IT_b ⫽ [IT + (bh + 2nAs)
NEUTRAL AXIS
From
bh a
L
y1 dA +
L
(h1 + bf ⫺ h1_b)2]
ny2 dA ⫽ 0
+ c
h
⫺ h1 b ⫺ 2nAs(h1 ⫺ d) ⫽ 0
2
12
+ bf hf a h1_b ⫺
bf
2
2
b d
IT_b ⫽ 905 in.
MAXIMUM MOMENT BASED UPON THE WOOD (1)
2
bh3
h
+ bh a ⫺ h1 b d
12
2
M1 ⫽
+ 2n[Iz + As(h1 ⫺ d) ] IT ⫽ 838 in.
2
b3f hf
4
h1 ⫽ 2.137 in.
IT ⫽ c
y1 dA +
ny2 dA ⫽ 0
L
L
(bh + 2nAs) (h1 + bf ⫺ h1_b)
bf
⫺ bf hf a h1_b ⫺ b ⫽ 0
2
From
4
sallow_wIT_b
h + bf ⫺ h1_b
M1 ⫽ 136.0 k # in.
MAXIMUM MOMENT BASED UPON THE STEEL (2)
MAXIMUM MOMENT BASED UPON THE WOOD (1)
M1 ⫽
sallow_wIT
M2 ⫽
M1 ⫽ 128.6 k # in.
h ⫺ h1
sallow_sIT
WOOD GOVERNS
M2 ⫽ 130.1 k # in.
(hs ⫺ h1)n
M2 ⫽ 136.2 k # in.
From Mmax ⫽
Mmax ⫽ 128.6 k # in.
WOOD GOVERNS
;
ALLOWABLE LOAD ON A 18-FT-LONG SIMPLE BEAM
qallowL2
8
qallow ⫽ 264 lb/ft.
;
Mmax ⫽136.0 k # in
;
ALLOWABLE LOAD ON A 18-FT-LONG SIMPLE BEAM
Mmax ⫽ min (M1, M2)
From Mmax ⫽
(hs + bf ⫺ h1_b)n
Mmax ⫽ min (M1, M2)
MAXIMUM MOMENT BASED UPON THE STEEL (2)
M2 ⫽
sallow_sIT_b
qallowL2
8
qallow ⫽ 280 lb/ft
;
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Problem 6.3-8 The cross section of a composite beam made of aluminum
y
and steel is shown in the figure. The moduli of elasticity are Ea ⫽ 75 GPa and
Es ⫽ 200 GPa.
Under the action of a bending moment that produces a maximum stress
of 50 MPa in the aluminum, what is the maximum stress ss in the steel?
Aluminum
40 mm
Steel
z
O 80 mm
30 mm
Solution 6.3-8 Composite beam of aluminum and steel
(1) Aluminum:
b ⫽ 30 mm
ha ⫽ 40 mm
All dimensions in millimeters.
sa ⫽ 50 MPa
Use the base of the cross section as a reference line.
Ea ⫽ 75 GPa
(2) Steel: b ⫽ 30 mm
Es ⫽ 200 GPa
hs ⫽ 80 mm
ss ⫽ ?
TRANSFORMED SECTION (ALUMINUM)
h2 ⫽
©yi Ai
(40)(80)(80) + (100)(30)(40)
⫽
©Ai
(80)(80) + (30)(40)
⫽ 49.474 mm
h1 ⫽ 120 ⫺ h2 ⫽ 70.526 mm
MAXIMUM STRESS IN THE ALUMINUM (1) (EQ. 6-15)
sa ⫽ s1 ⫽
Mh1
IT
MAXIMUM STRESS IN THE STEEL (2) (EQ. 6-17)
ss ⫽ s2 ⫽
Aluminum part is not changed.
Es
200
⫽
⫽ 2.667
n⫽
Ea
75
Mh2n
IT
(49.474)(2.667)
ss
h2n
⫽ 1.8707
⫽
⫽
sa
h1
70.526
ss ⫽ 1.8707 (50 MPa) ⫽ 93.5 MPa
;
Width of steel part
⫽ nb ⫽ (2.667)(30 mm) ⫽ 80 mm
Problem 6.3-9 A beam is constructed of two angle sections, each
L 5 ⫻ 3 ⫻ 1/2, which reinforce a 2 ⫻ 8 (actual dimensions) wood plank
(see the cross section shown in the figure). The modulus of elasticity for
the wood is Ew ⫽ 1.2 ⫻ 106 psi and for the steel is Es ⫽ 30 ⫻ 106 psi.
Find the allowable bending moment Mallow for the beam if the
allowable stress in the wood is sw ⫽ 1100 psi and in the steel is
ss ⫽ 12,000 psi. (Note: Disregard the weight of the beam, and see Table
E-5a of Appendix E for the dimensions and properties of the angles.)
2 ⫻ 8 wood plank
C
y
z
5 in.
3 in.
Steel angles
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SECTION 6.3 Transformed-Section Method
Solution 6.3-9
b ⫽ 2 in.
(1) Wood beam
h ⫽ 8 in.
sallow_w ⫽ 1100 psi
Iz ⫽ 9.43 in.4
(2) Steel Angle
As ⫽ 3.75 in.
2
sallow_s ⫽ 12000 psi
Ew ⫽ 1.2⭈106 psi
IT ⫽ c
d ⫽ 1.74 in.
hs ⫽ 5 in.
Es ⫽ 30⭈10 psi
6
b3h
b 2
+ bha h1 ⫺ b d
12
2
+ 2n[Iz + As(b + d ⫺ h1)2]
IT ⫽ 588 in.4
MAXIMUM MOMENT BASED UPON THE WOOD (1)
TRANSFORMED SECTION (WOOD)
Es
n⫽
Ew
n ⫽ 25
M1 ⫽
sallow_wIT
M1 ⫽ 183.4 k # in.
h1
MAXIMUM MOMENT BASED UPON THE STEEL (2)
NEUTRAL AXIS
From
1 y1 dA + 1 ny2 dA ⫽ 0
bha h1 ⫺
b
b ⫺ 2nAs(b + d ⫺ h1) ⫽ 0
2
h1 ⫽ 3.525 in.
M2 ⫽
sallow_sIT
(hs + b ⫺ h1)n
M2 ⫽ 81.1 k # in.
Mmax ⫽ min (M1, M2)
Mmax ⫽ 81.1 k-in.
STEEL GOVERNS
;
y
Problem 6.3-10 The cross section of a bimetallic strip is shown in the
figure. Assuming that the moduli of elasticity for metals A and B are
EA ⫽ 168 GPa and EB ⫽ 90 GPa, respectively, determine the smaller of
the two section moduli for the beam. (Recall that section modulus is equal
to bending moment divided by maximum bending stress.) In which material does the maximum stress occur?
z
A
O
B
3 mm
3 mm
10 mm
Solution 6.3-10 Bimetallic strip
Metal A: b ⫽ 10 mm
hA ⫽ 3 mm
Metal B does not change.
EA ⫽ 168 GPa
Metal B: b ⫽ 10 mm
hB ⫽ 3 mm
TRANSFORMED SECTION
TRANSFORMED SECTION (METAL B)
n⫽
EB ⫽ 90 GPa
EA
168
⫽
⫽ 1.8667
EB
90
Width of metal A
⫽ nb ⫽ (1.8667)(10 mm) ⫽ 18.667 mm
All dimensions in millimeters.
Use the base of the cross section as a reference line.
h2 ⫽
©yi Ai
(1.5)(10)(13) + (4.5)(18.667)(3)
⫽
©Ai
(10)(3) + (18.667)(3)
⫽ 3.4535 mm
h1 ⫽ 6 ⫺ h2 ⫽ 2.5465 mm
517
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CHAPTER 6 Stresses in Beams (Advanced Topics)
IT ⫽
1
(10)(3)3 + (10)(3)(h2 ⫺ 1.5)2
12
+
1
(18.667)(3)3 + (18.667)(3)(h1 ⫺ 1.5)2
12
MAXIMUM STRESS IN MATERIAL A (EQ. 6-17)
sA ⫽ s2 ⫽
Mh1n
IT
SA ⫽
IT
M
⫽
sA
h1n
⫽ 50.6 mm3
⫽ 240.31mm
4
SMALLER SECTION MODULUS
MAXIMUM STRESS IN MATERIAL B (EQ. 6-15)
Mh2
sB ⫽ s1 ⫽
IT
IT
M
SB ⫽
⫽
⫽ 69.6 mm3
sB
h2
SA ⫽ 50.6 mm3
;
‹ Maximum stress occurs in metal A.
;
y
Problem 6.3-11 A W 12 ⫻ 50 steel wide-flange beam and a segment of a
4-inch thick concrete slab (see figure) jointly resist a positive bending
moment of 95 k-ft. The beam and slab are joined by shear connectors that
are welded to the steel beam. (These connectors resist the horizontal shear
at the contact surface.) The moduli of elasticity of the steel and the concrete
are in the ratio 12 to 1.
Determine the maximum stresses ss and sc in the steel and concrete,
respectively. (Note: See Table E-1a of Appendix E for the dimensions and
properties of the steel beam.)
30 in.
4 in.
z
O
W 12 ⫻ 50
Solution 6.3-11 Steel beam and concrete slab
(1) Concrete: b ⫽ 30 in.
t ⫽ 4 in.
(2) Wide-flange beam: W 12 ⫻ 50
d ⫽ 12.19 in.
I ⫽ 394 in.4
A ⫽ 14.7 in.2
M ⫽ 95 k-ft ⫽ 1140 k-in.
TRANSFORMED SECTION (CONCRETE)
All dimensions in inches.
Use the base of the cross section as a reference line.
nI ⫽ 4728 in.4
h2 ⫽
nA ⫽ 176.4 in.2
©yiAi
(12.19/2)(176.4) + (14.19)(30)(4)
⫽
©Ai
176.4 + (30)(4)
⫽ 9.372 in.
h1 ⫽ 16.19 ⫺ h2 ⫽ 6.818 in.
IT ⫽
1
(30)(4)3 + (30)(4)(h1 ⫺ 2)2 + 4728
12
+ (176.4)(h2 ⫺ 12.19/2)2 ⫽ 9568 in.4
MAXIMUM STRESS IN THE CONCRETE (1) (EQ. 6-15)
sc ⫽ s1 ⫽
No change in dimensions of the concrete.
n⫽
Es
E2
⫽
⫽ 12
Ec
E1
Width of steel beam is increased by the factor n to
transform to concrete.
Mh1
⫽ 812 psi (Compression)
IT
;
MAXIMUM STRESS IN THE STEEL (2) (EQ. 6-17)
ss ⫽ s2 ⫽
Mh2n
⫽ 13,400 psi (Tension)
IT
;
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519
SECTION 6.3 Transformed-Section Method
Problem 6.3-12 A wood beam reinforced by an aluminum channel section is
150 mm
shown in the figure. The beam has a cross section of dimensions 150 mm by
250 mm, and the channel has a uniform thickness of 6 mm.
If the allowable stresses in the wood and aluminum are 8.0 MPa and 38 MPa,
respectively, and if their moduli of elasticity are in the ratio 1 to 6, what is the
maximum allowable bending moment for the beam?
y
216 mm
250 mm
z
O
40 mm
6 mm
162 mm
Solution 6.3-12
Wood beam and aluminum channel
(1) Wood beam: bw ⫽ 150 mm
hw ⫽ 250 mm
(sw)allow ⫽ 8.0 MPa
(2) Aluminum channel: t ⫽ 6 mm
ba ⫽ 162 mm
ha ⫽ 40 mm
(sa)allow ⫽ 38 MPa
Use the base of the cross section as a reference line.
h2 ⫽
©yiAi
©Ai
Area A1: y1 ⫽ 3
A1 ⫽ (972)(6) ⫽ 5832
y1A1 ⫽ 17,496 mm3
Area A2: y2 ⫽ 23
TRANSFORMED SECTION (WOOD)
A2 ⫽ (36)(34) ⫽ 1224
y2A2 ⫽ 28,152 mm3
Area A3: y3 ⫽ 131
A3 ⫽ (150)(250) ⫽ 37,500
y3A3 ⫽ 4,912,500 mm3
h2 ⫽
y1A1 + 2y2A2 + y3A3
4,986,300 mm3
⫽
A1 + 2A2 + A3
45,780 mm2
⫽ 108.92 mm
h1 ⫽ 256 ⫺ h2 ⫽ 147.08 mm
MOMENT OF INERTIA
Area A1: I1 ⫽
Wood beam is not changed.
Ea
n⫽
⫽6
Ew
Width of aluminum channel is increased.
nb ⫽ (6)(162 mm) ⫽ 972 mm
nt ⫽ (6)(6 mm) ⫽ 36 mm
All dimensions in millimeters.
1
(972)(6)3 + (972)(6)(h2 ⫺ 3)2
12
⫽ 65,445,000 mm4
Area A2: I2 ⫽
1
(36)(34)3
12
+ (36)(34)(h2 ⫺ 6 ⫺ 17)2
⫽ 9,153,500 mm4
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Area A3: I3 ⫽
MAXIMUM MOMENT BASED UPON ALUMINUM (2) (EQ. 6-17)
1
(150)(250)3
12
+ (150)(250)(h1 ⫺ 125)
2
⫽ 213,597,000 mm4
sa ⫽ s2 ⫽
Mh2n
IT
WOOD GOVERNS.
M2 ⫽
(sa)allowIT
⫽ 17.3 kN # m
h2n
Mallow ⫽ 16.2 kN # m
;
IT ⫽ I1 + 2I2 + I3 ⫽ 297.35 * 106 mm4
MAXIMUM MOMENT BASED UPON THE WOOD (1) (EQ. 6-15)
sw ⫽ s1 ⫽
Mh1
IT
M1 ⫽
(sw)allowIT
⫽ 16.2 kN # m
h1
Beams with Inclined Loads
y
When solving the problems for Section 6.4, be sure to draw a sketch of
he cross section showing the orientation of the neutral axis and the
locations of the points where the stresses are being found.
Problem 6.4-1 A beam of rectangular cross section supports an
z
inclined load P having its line of action along a diagonal of the cross
section (see figure). Show that the neutral axis lies along the other
diagonal.
h
C
b
P
Solution 6.4-1 Location of neutral axis
Iy ⫽
hb3
12
Iz
h2
Iy
⫽
b2
See Figure 6-15b.
b ⫽ angle between the z axis and the neutral axis nn
u ⫽ angle between the y axis and the load P
u ⫽ a ⫹ 180°
tan u ⫽ tan (a ⫹ 180°) ⫽ tan a
Load P acts along a diagonal.
b
b/2
⫽
tan a ⫽
h/2
h
Iz ⫽
3
bh
12
(Eq. 6-23):
tan b ⫽
Iz
tan u ⫽
Iy
⫽ a
h2
b2
b2
tan u
b
h
ba b ⫽
h
b
‹ The neutral axis lies along
the other diagonal. QED
h2
;
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SECTION 6.4 Beams with Inclined Loads
Problem 6.4-2 A wood beam of rectangular cross section
y
(see figure) is simply supported on a span of length L.
The longitudinal axis of the beam is horizontal, and the cross
section is tilted at an angle a. The load on the beam is a vertical
uniform load of intensity q acting through the centroid C.
Determine the orientation of the neutral axis and
calculate the maximum tensile stress smax if b 80 mm,
h 140 mm, L 1.75 m, a 22.5°, and q 7.5 kN/m.
b
h
C
q
z
a
Probs. 6.4-2 and 6.4-3
Solution 6.4-2
q 7.5 kN/m
L 1.75 m
h 140 mm
b 80 mm
a 22.5 deg
Iz bh3
12
Iz 18.293 * 106 mm4
NEUTRAL AXIS nn
BENDING MOMENTS
2
My qsin(a)L
8
My 1099 N # m
Iz
b a tan a tan(a)b
Iy
Mz qcos(a)L2
8
Mz 2653 N # m
MAXIMUM TENSILE STRESS (AT POINT A)
MOMENT OF INERTIA
hb3
Iy 12
smax Iy 5.973 * 10 mm
6
b
My a b
2
Iy
4
b 51.8°
Mz a
smax 17.5 MPa
;
h
b
2
Iz
;
Problem 6.4-3 Solve the preceding problem for the following data: b 6 in., h 10 in., L 12.0 ft, tan a 1/3, and
q 325 lb/ft.
Solution 6.4-3
L 12 ft
q 325 lb/ft
b 6 in.
1
a a tan a b
3
h 10 in.
BENDING MOMENTS
Mz qsin(a)L
8
My 22199 lb-in.
Mz 66598 lb-in.
MOMENT OF INERTIA
Iy hb3
12
Iy 180 in.4
Iz bh3
12
Iz 500 in.4
2
My qcos(a)L2
8
521
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CHAPTER 6 Stresses in Beams (Advanced Topics)
MAXIMUM TENSILE STRESS (AT POINT A)
NEUTRAL AXIS nn
Iz
b a tan a tan(a)b
Iy
b 42.8°
;
smax b
My a b
2
Iy
h
b
2
Mz a
Iz
smax 1036 psi
;
y
Problem 6.4-4 A simply supported wide-flange beam of span length L carries
a vertical concentrated load P acting through the centroid C at the midpoint of the
span (see figure). The beam is attached to supports inclined at an angle a to the
horizontal.
Determine the orientation of the neutral axis and calculate the maximum
stresses at the outside corners of the cross section (points A, B, D, and E ) due to
the load P. Data for the beam are as follows: W 250 44.8 section, L 3.5 m,
P 18 kN, and a 26.57°. (Note: See Table E-1b of Appendix E for the
dimensions and properties of the beam.)
P
n
E
D
b
C
B
z
A
n
a
Probs. 6.4-4 and 6.4-5
Solution 6.4-4
L 3.5 m
P 18 kN
W 250 44.8
Wide-flange beam:
Iy 6.95 106 mm4
d 267 mm
a 26.57 deg
Iz 70.8 106 mm4
b 148 mm
BENDING STRESSES
sx(z,y) POINT A:
Myz
Iy
zA Mzy
Iz
b
2
yA sA 102 MPa
BENDING MOMENTS
My Psin(a)L
4
My 7045 N # m
Mz Pcos(a)L
4
Mz 14087 N # m
POINT B:
NEUTRAL AXIS NN
Iz
b a tan a tan(a)b
Iy
b 78.9°
;
d
2
sA sx(zA, yA)
;
b
d
yB 2
2
sB 48 MPa
sB sx(zB, yB)
zB POINT D:
sD sB
sD 48 MPa
POINT E:
sE sA
sE 102 MPa
;
;
;
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523
SECTION 6.4 Beams with Inclined Loads
Problem 6.4-5 Solve the preceding problem using the following data: W 8 21 section, L 84 in., P 4.5 k,
and a 22.5°.
Solution 6.4-5
L 84 in.
P 4.5 k
BENDING STRESSES
a 22.5°
sx(z, y) WIDE-FLANGE BEAM:
W 8 21
Iy 9.77 in.4
d 8.28 in.
Mz Iy
Iz 75.3 in.4
b 5.270 in.
POINT A:
zA Mz y
Iz
b
2
yA d
2
sA sx(zA, yA)
BENDING MOMENTS
My My z
Psin(a)L
4
My 36164 lb-in.
Pcos(a)L
4
Mz 87307 lb-in.
sA 14554 psi
POINT B:
zB b
2
;
yB sB sx(zB, yB)
sB 4953 psi
NEUTRAL AXIS nn
b a tan a tan(a)b
Iy
Iz
b 72.6°
d
2
;
;
POINT D:
sD sB
sD 4953 psi
POINT E:
sE sA
sE 14554 psi
Problem 6.4-6 A wood cantilever beam of rectangular cross section and length
;
;
y
L supports an inclined load P at its free end (see figure).
Determine the orientation of the neutral axis and calculate the maximum
tensile stress smax due to the load Pt Data for the beam are as follows:
b 80 mm, h 140 mm, L 2.0 m, P 575 N, and a 30°.
h
z
C
a
P
b
Probs. 6.4-6 and 6.4-7
Solution 6.4-6
L 2.0 m
P 575 N
h 140 mm
b 80 mm
a 30°
BENDING MOMENTS
My Pcos(a)L
Mz Psin(a)L
My 996 N # m
Mz 575 N # m
MOMENT OF INERTIA
Iy hb3
12
Iy 5.973 * 106 mm4
Iz bh3
12
Iz 18.293 * 106 mm4
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CHAPTER 6 Stresses in Beams (Advanced Topics)
MAXIMUM TENSILE STRESS (AT POINT A)
NEUTRAL AXIS nn
Iz
b a tan a tan(a + 90°)b
Iy
b 79.3°
smax ;
b
My a b
2
Iy
smax 8.87 MPa
h
Mz a b
2
Iz
;
Problem 6.4-7 Solve the preceding problem for a cantilever beam with data as follows: b 4 in., h 9 in., L 10.0 ft,
P 325 lb, and a 45°.
Solution 6.4-7
L 10.0 ft
h 9 in.
P 325 lb
b 4 in.
Iz
b a tan a tan(a + 90°)b
Iy
a 45°
b 78.8 deg
BENDING MOMENTS
My Pcos(a)L
Mz Psin(a)L
My 27577 lb # in.
Mz 27577 lb # in.
MOMENT OF INERTIA
hb3
Iy 12
Iz 3
bh
12
NEUTRAL AXIS nn
MAXIMUM TENSILE STRESS (AT POINT A)
smax Iy 48.000 in.
;
4
b
My a b
2
Iy
smax 1660 psi
h
Mz a b
2
Iz
;
Iz 243.000 in.4
Problem 6.4-8 A steel beam of I-section (see figure) is simply supported at
the ends. Two equal and oppositely directed bending moments M0 act at the
ends of the beam, so that the beam is in pure bending. The moments act in
plane mm, which is oriented at an angle a to the xy plane.
Determine the orientation of the neutral axis and calculate the maximum tensile stress smax due to the moments M0. Data for the beam are as
follows: S 200 27.4 section, M0 4 kN # m., and a 24°. (Note: See
Table E-2b of Appendix E for the dimensions and properties of the beam.)
m
y
C
z
M0
a
m
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525
SECTION 6.4 Beams with Inclined Loads
Solution 6.4-8
Mo 4.0 kN # m
S 200 27.4
NEUTRAL AXIS nn
a 24°
Iy 1.54 106 mm4
Iz 23.9 106 mm4
d 203 mm
b 102 mm
My 1627 N # m
Mz M0 cos(a)
b 81.8°
;
MAXIMUM TENSILE STRESS (AT POINT A)
BENDING MOMENTS
My M0 sin(a)
Iz
b a tan a tan(a)b
Iy
Mz 3654 N # m
smax b
My a b
2
Iy
d
Mz a b
2
Iz
smax 69.4 MPa
;
Problem 6.4-9 A cantilever beam of wide-flange cross section and length
y
L supports an inclined load P at its free end (see figure).
Determine the orientation of the neutral axis and calculate the maximum
tensile stress smax due to the load P.
Data for the beam are as follows: W 10 45 section, L 8.0 ft, P 1.5 k,
and a 55°. (Note: See Table E-1a of Appendix E for the dimensions and properties of the beam.)
P
a
z
C
Probs. 6.4-9 and 6.4-10
Solution 6.4-9 Cantilever beam with inclined load
BENDING MOMENTS
My (P cos a)L 82,595 lb-in.
Mz (P sin a)L 117,960 lb-in.
NEUTRAL AXIS nn (EQ. 6-23)
u 90° a 35°
tan b Iz
Iy
tan u (see Fig. 6-15)
248
tan 35° 3.2519
53.4
b 72.91°
P 1.5 k 1500 lb
L 8.0 ft 96 in.
a 55°
d 10.10 in.
MAXIMUM TENSILE STRESS (POINT A) (EQ. 6-18)
zA b/2 4.01 in.
yA d/2 5.05 in.
W 10 45
Iy 53.4 in.4
;
Iz 248 in.4
b 8.02 in.
smax sA My zA
Iy
Mz yA
Iz
8600 psi
;
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Problem 6.4-10 Solve the preceding problem using the following data: W 310 129 section, L 1.8 m, P 9.5 kN, and
a 60°. (Note: See Table E-1b of Appendix E for the dimensions and properties of the beam.)
Solution 6.4-10
P 9.5 kN
L 1.8 m
W 310 129
MAXIMUM TENSILE STRESS (AT POINT A)
a 60°
Iy 100 106 mm4
Iz 308 106 mm4
d 318 mm
b 307 mm
smax b
My a b
2
Iy
d
Mz a b
2
Iz
smax 20.8 MPa
BENDING MOMENTS
My Pcos(a)L
My 8550 N # m
Mz Psin(a)L
Mz 14809 N # m
;
NEUTRAL AXIS nn
Iz
b a tan a tan(90° a) b
Iy
b 60.6°
b 60.6°
;
Problem 6.4-11 A cantilever beam of W 12 14 section and
y
length L 9 ft supports a slightly inclined load P 500 lb at the
free end (see figure).
(a) Plot a graph of the stress sA at point A as a function of the
angle of inclination a.
(b) Plot a graph of the angle b, which locates the neutral axis
nn, as a function of the angle a. (When plotting the
graphs, let a vary from 0 to 10°.) (Note: See Table E-1a of
Appendix E for the dimensions and properties of the
beam.)
A
n
b
C
z
n
P
a
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SECTION 6.4 Beams with Inclined Loads
Solution 6.4-11
Cantilever beam with inclined load
(b) NEUTRAL AXIS nn (EQ. 6-23)
u 180° + a
tan b Iz
Iy
(see Fig. 6-15)
tan u Iz
Iy
tan(180° + a)
88.6
tan(180° + a) 37.54 tan a
2.36
b arctan(37.54 tan a)
P 500 lb
L 9 ft 108 in.
Iy 2.36 in.
Iz 88.6 in.
d 11.91 in.
b 3.970 in.
4
;
W 12 14
4
BENDING MOMENTS
My (P sin a)L 54,000 sin a
Mz (P cos a)L 54,000 cos a
(a) STRESS AT POINT A (EQ. 6-18)
zA b/2 1.985 in.
yA d/2 5.955 in.
sA MyzA
Iy
Mz yA
Iz
45,420 sin a
+ 3629 cos a (psi)
;
y
Problem 6.4-12 A cantilever beam built up from two channel shapes, each
C 200 17.1, and of length L supports an inclined load P at its free end
(see figure).
Determine the orientation of the neutral axis and calculate the maximum
tensile stress smax due to the load P. Data for the beam are as follows:
L 4.5 m, P 500 N, and a 30°.
z
C
a
P
C 200 17.1
527
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Solution 6.4-12
L 4.5 m
P 500 N
Iz
b a tan a tan (90 ° a)b
Iy
Double C 200 17.1
BUILT UP BEAM:
Icy 0.545 106 mm4
Icz 13.5 106 mm4
c 14.5 mm
bc 57.4 mm
[
]
Iy 2 Icy Ac(bc c)2
Iz 2Icz
NEUTRAL AXIS nn
a 30°
Iy 9.08 106 mm4
Iz 27.0 106 mm4
d 203 mm
b 2bc
b 79.0°
;
Ac 2170 mm2
MAXIMUM TEMSILE STRESS
sx(z, y) b 114.8 mm
POINT A:
BENDING MOMENTS
My Pcos(a)L
My 1949 N # m
Mz Psin(a)L
Mz 1125 N # m
Myz
Iy
Mzy
zA Iz
b
2
yA sA sx(zA, yA)
d
2
sA 16.6 MPa
;
Problem 6.4-13 A built-up steel beam of I-section with channels attached to the flanges (see figure part a) is simply
supported at the ends. Two equal and oppositely directed bending moments M0 act at the ends of the beam, so that the beam
is in pure bending. The moments act in plane mm, which is oriented at an angle a to the xy plane.
(a) Determine the orientation of the neutral
axis and calculate the maximum tensile
stress smax due to the moments M0.
(b) Repeat part a if the channels are now
with their flanges pointing away from the
beam flange, as shown in figure part b.
Data for the beam are as follows:
S 6 12.5 section with C 4 5.4 sections attached to the flanges,
M0 45 k-in., and a 40°. (Note:
See Tables E-2a and E-3a of Appendix E
for the dimensions and properties of the
S and C shapes.)
m
y
m
C 4 5.4
y
C 4 5.4
z
C
M0
S 6 12.5
a
C 4 5.4
C
z
S 6 12.5
a
M0
C 4 5.4
m
(a)
m
(b)
Solution 6.4-13
Mo 45 k # in.
a 40°
S 6 12.5:
Isy 1.80 in.4
Isz 22.0 in.4
ds 6.0 in.
bs 3.33 in.
As 3.66 in.2
Iz Isz + 2cIcy + Ac a
C 4 5.4:
Icy 0.312 in.4
Icz 3.85 in.4
Iz 46.1 in.4
dc 4.0 in.
bc 1.58 in.
Ac 1.58 in.2
d ds + 2twc
twc 0.184 in.
c 0.457 in.
(a) BUILT UP SECTION:
b dc
Iy Isy 2Icz Iy 9.50 in.4
2
ds
+ twc cb d
2
d 6.368 in.
b 4.0 in.
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SECTION 6.5 Bending of Unsymmetric Beams
NEUTRAL AXIS nn
BENDING MOMENTS
My Mosin(a)
My 28.9 k # in.
Mz Mocos(a)
Iz
b a tan a tan(a)b
Iy
Mz 34.5 k # in.
b 79.4°
NEUTRAL AXIS nn
MAXIMUM TENSILE STRESS
Iz
b a tan a tan( a) b
Iy
b 76.2°
sx(z, y) ;
POINT A:
MAXIMUM TENSILE STRESS
sx(z, y) POINT A:
My z
Iy
Mz y
zA s A sx(zA, yA)
;
My z
Iy
Mz y
Iz
zA b
2
sA sx(zA, yA)
yA sA 8704 psi
Iz
b
2
yA d
2
sA 8469 psi
(b) BUILT UP SECTION: Iy Isy 2Icz
Iz Isz + 2 cIcy + Ac a
;
Iy 9.50 in.4
2
ds
+ cb d
2
Iz 60.4 in.4
d ds + 2bc
b dc
d 9.160 in.
b 4.000 in.
Bending of Unsymmetric Beams
y
When solving the problems for Section 6.5, be sure to draw a
sketch of the cross section showing the orientation of the neutral
axis and the locations of the points where the stresses are being
found.
Problem 6.5-1 A beam of channel section is subjected to a
bending moment M having its vector at an angle u to the z axis
(see figure).
Determine the orientation of the neutral axis and calculate
the maximum tensile stress st and maximum compressive stress
sc in the beam.
Use the following data: C 8 11.5 section, M 20 k-in.,
tan u 1/3. (Note: See Table E-3a of Appendix E for the
dimensions and properties of the channel section.)
M
z
d
2
u
Probs. 6.5-1 and 6.5-2
C
;
529
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Solution 6.5-1
Channel section
NEUTRAL AXIS nn (EQ. 6-40)
tan b Iz
Iy
tan u 32.6
(1/3) 8.2323
1.32
b 83.07°
;
MAXIMUM TENSILE STRESS (POINT A) (EQ. 6-38)
zA c 0.571 in.
st sA yA d/2 4.00 in.
(M sin u)zA
(M cos u)yA
Iy
Iz
5060 psi
M 20 k-in.
tan u 1/3
u 18.435°
C 8 11.5
c 0.571 in.
Iy 1.32 in.4
d 8.00 in.
Iz 32.6 in.4
b 2.260 in.
;
MAXIMUM COMPRESSIVE STRESS (POINT B) (EQ. 6-38)
zB (b c) (2.260 0.571) 1.689 in.
yz d/2 4.00 in.
sc sB (M sin u)zB
(M cos u)yB
Iy
Iz
10,420 psi
;
Problem 6.5-2 A beam of channel section is subjected to a bending moment M having its vector at an angle u to the z axis
(see figure).
Determine the orientation of the neutral axis and calculate the maximum tensile stress st and maximum compressive
stress sc in the beam. Use a C 200 20.5 channel section with M 0.75 kN # m and u 20°.
Solution 6.5-2
M 0.75 kN # m
C 200 20.5
MAXIMUM TENSILE STRESS (AT POINT A)
u 20°
Iy 0.633 # 106 mm4
Iz 15.0 106 mm4
d 203 mm
b 59.4 mm
c 14.1 mm
Iy
d
Mz a b
2
smax 10.5 MPa
BENDING MOMENTS
My Msin(u)
My 257 N # m
Mz Mcos(u)
Mz 705 N # m
NEUTRAL AXIS
smax My(c)
smax My(b + c)
Iy
smax 23.1 MPa
b 83.4°
;
MAXIMUM TENSILE STRESS (AT POINT B)
nn
Iz
b atan a tan(u)b
Iy
Iz
;
d
Mz a b
2
Iz
;
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SECTION 6.5 Bending of Unsymmetric Beams
Problem 6.5-3 An angle section with equal legs is subjected to a
2
bending moment M having its vector directed along the 1-1 axis, as
shown in the figure.
Determine the orientation of the neutral axis and calculate the
maximum tensile stress st and maximum compressive stress sc if the
angle is an L 6 6 3/4 section and M 20 k-in. (Note: See Table
E-4a of Appendix E for the dimensions and properties of the
angle section.)
M
1
Probs. 6.5-3 and 6.5-4
Solution 6.5-3
C
1
2
Angle section with equal legs
NEUTRAL AXIS nn (EQ. 6-40)
tan b Iz
Iy
tan u 44.85
tan 45° 3.8831
11.55
b 75.56°
;
MAXIMUM TENSILE STRESS (POINT A) (EQ. 6-38)
zA c12 2.517 in.
st sA (M sin u)zA
(M cos u)yA
Iy
Iz
3080 psi
M 20 k-in.
L 6 6 3/4 in.
h b 6 in.
c 1.78 in.
A 8.44 in.2
Iy Ar2min 11.55 in.4
Iz I1 I2 Iy 44.85 in.4
;
MAXIMUM COMPRESSIVE STRESS (POINT B) (EQ. 6-38)
zB c12 h/12 1.725 in.
I1 I2 28.2 in.4
u 45°
yA 0
rmin 1.17 in.
yB h/12 4.243 in.
sc sB (M sin u)zB
(M cos u)yB
Iy
Iz
3450 psi
;
Problem 6.5-4 An angle section with equal legs is subjected to a bending moment M having its vector directed along the
1–1 axis, as shown in the figure.
Determine the orientation of the neutral axis and calculate the maximum tensile stress st and maximum compressive
stress sc if the section is an L 152 152 12.7 section and M 2.5 kN # m. (Note: See Table E-4b of Appendix E for the
dimensions and properties of the angle section.)
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Solution 6.5-4
M 2.5 kN # m
L 152 152 12.7
rmin 30.0 mm
Iy Armin2
MAXIMUM TENSILE STRESS (AT POINT A)
u 45°
I1 8.28 106 mm4
I2 I1
A 3720 mm2
h 152 mm
Iz 13.212 106 mm4
bh
Mz Mcos(u)
NEUTRAL AXIS
Iy
smax My 1768 N # m
Mz 1768 N # m
Mz(0)
Iz
;
MAXIMUM TENSILE STRESS (AT POINT B)
c 42.4 mm4
BENDING MOMENTS
My Msin(u)
My(c12)
smax 31.7 MPa
Iy 3.348 106 mm4
Iz I1 I2 Iy
smax My a c12 h
b
12
Iy
smax 39.5 MPa
Mz a
h
b
12
Iz
;
nn
Iz
b atan a tan(u)b
Iy
b 75.8°
;
Problem 6.5-5 A beam made up of two unequal leg angles is subjected to a bending moment M having its vector at an
angle u to the z axis (see figure part a).
(a) For the position shown in the figure, determine the orientation of the neutral axis and calculate the maximum tensile
stress st and maximum compressive stress sc in the beam. Assume that u 30° and M 30 k-in.
(b) The two angles are now inverted and attached back-to-back to form a lintel beam which supports two courses of brick
façade (see figure part b). Find the new orientation of the neutral axis and calculate the maximum tensile stress st and
maximum compressive stress sc in the beam using u 30° and M 30 k-in.
y
1
1
L53— —
2
2
1
1
L53— —
2
2
y
Lintel beam supporting
brick facade
z
u
C
M
z
u
3
— in.
4
(a)
M
C
(b)
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SECTION 6.5 Bending of Unsymmetric Beams
533
Solution 6.5-5
M 30 k # in.
u 30°
IL1 10.0 in.4
L5 * 3q1/2 * 1/2
d 1.65 in.
c 0.901 in.
IL2 4.02 in.4
hL1 5 in.
AL 4 in.2
hL2 3.5 in.
3
gap in.
4
Iz 20.000 in.4
2
gap
+ cb d
2
Iy 2 cIL2 + AL a
Iy 21.065 in.
4
h hL1
h 5.000 in.
b gap + 2hL2
b 7.750 in.
h1 d
h1 1.650 in.
BENDING MOMENTS
My Msin (u)
Mz Mcos (u)
zA st sx(zA, yA)
gap
+ t
2
yA h + h1
st 4263 psi
;
MAXIMUM COMPRESSIVE STRESS
1
t in.
2
Iz 2IL1
(a) BUILT UP SECTION:
POINT A:
My 1.250 k # ft
Mz 2.165 k # ft
POINT B:
zB b
2
yB h1
sc sx(zB, yB)
sc 4903 psi
Iz 2IL1
(b) BUILT UP SECTION:
Iy 2(IL2 ALc2)
h hL1
Iz 20.000 in.4
Iy 14.534 in.4
h 5.000 in.
b 7.000 in.
;
b 2hL2
h1 h d
h1 3.350 in.
NEUTRAL AXIS nn
Iz
b a tan a tan(u)b
Iy
b 38.5 deg
;
MAXIMUM TENSILE STRESS
NEUTRAL AXIS nn
b atan a tan( u)b
Iy
sx(z, y) b 28.7°
POINT A:
Iz
;
MAXIMUM TENSILE STRESS
sx(z, y) Myz
Iy
Mzy
Iz
Myz
Iy
zA st sx(zA, yA)
Iz
b
2
yA h + h1
st 5756 psi
;
MAXIMUM COMPRESSIVE STRESS
POINT B:
zB t
sc sx(zB,yB)
yB h1
sc 4868 psi
y1
Problem 6.5-6 The Z-section of Example 12-7 is subjected to
M 5 kN # m, as shown.
Determine the orientation of the neutral axis and calculate
the maximum tensile stress st and maximum compressive stress
sc in the beam. Use the following numerical data: height
h 200 mm, width b 90 mm, constant thickness t 15 mm,
and up 19.2°. Use I1 32.6 106 mm4 and
I2 2.4 106 mm4 from Example 12-7.
Mzy
;
y
b
h
—
2
M
t
up
x
C
h
—
2
x1
t
t
b
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Solution 6.5-6
M 5 kN # m
ypA 91.97 mm
u 19.2°
st Z-SECTION
Izp I1
Izp 32.6 10 mm
Iyp I2
Iyp 2.4 10 mm
6
6
h 200 mm
b 90 mm
4
Myp(zpA)
Iyp
st 40.7 MPa
Mzp(ypA)
Izp
;
4
t 15 mm
BENDING MOMENTS
MAXIMUM COMPRESSIVE STRESS (AT POINT B)
t
h
zpB a b cos(u) a bsin(u)
2
2
Myp Msin(u)
Myp 1644 N # m
zpB 39.97 mm
Mzp Mcos(u)
Mzp 4722 N # m
h
t
ypB a bsin(u) + a bcos(u)
2
2
NEUTRAL AXIS nn
b atan a
Izp
Iyp
tan(u)b
ypB 91.97 mm
b 78.1°
;
Myp(zpB)
Iyp
Mzp(ypB)
Izp
sc 40.7 MPa
MAXIMUM TENSILE STRESS (AT POINT A)
h
t
zpA a bcos(u) a
bsin(u)
2
2
sc ;
zpA 39.97 mm
h
t
bcos(u)
ypA a bsin(u) + a
2
2
Problem 6.5-7 The cross section of a steel beam is
constructed of a W 18 71 wide-flange section with a
6 in 1/2 in. cover plate welded to the top flange and a
C 10 30 channel section welded to the bottom flange. This
beam is subjected to a bending moment M having its vector at
an angle u to the z axis (see figure).
Determine the orientation of the neutral axis and calculate the maximum tensile stress st and maximum compressive
stress sc in the beam. Assume that u 30° and M 75 k-in.
(Note: The cross sectional properties of this beam were computed in Examples 12-2 and 12-5.)
Plate
1
6 in. — in.
2
M
y
C1
W 18 71
u
z
y1
C2
c
C
y3
C 10 30
C3
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SECTION 6.5 Bending of Unsymmetric Beams
535
Solution 6.5-7
M 75 k # in.
PLATE:
Iplate NEUTRAL AXIS nn
u 30°
1
in.
2
bp bph3p
hp 6 in.
Iplate 9.00 in.4
12
hw 18.47 in.
W SECTION:
bw 7.635 in.
Iz
b atan a tan(u)b
Iy
b 82.3°
;
MAXIMUM TENSILE STRESS
sx(z, y) Myz
Iy
Iwy 60.3 in.
Mzy
Iz
4
POINT A:
hc 10.0 in.
C SECTION:
Icz 103 in.
bc 3.033 in.
zA hc
2
yA st sx(zA, yA)
4
st 1397 psi
BUILT-UP SECTION:
MAXIMUM COMPRESSIVE STRESS
cbar 1.80 in.
POINT B:
Iz 2200 in.4
Iy Iwy Icz Iplate
zB bw
2
yB sc sx(zB, yB)
Iy 172.3 in.
4
hw
bc + cbar
2
;
hw
+ cbar
2
sc 1157 psi
;
BENDING MOMENTS
My Msin(u)
My 3.125 k # ft
Mz Mcos(u)
Mz 5.413 k # ft
Problem 6.5-8 The cross section of a steel beam is shown in the
figure. This beam is subjected to a bending moment M having its
vector at an angle u to the z axis.
Determine the orientation of the neutral axis and calculate the
maximum tensile stress st and maximum compressive stress
sc in the beam. Assume that u 22.5° and M 4.5 kN # m.
Use cross sectional properties Ix1 93.14 106 mm4,
Iy1 152.7 106 mm4, and up 27.3°.
y1
y
180mm
180 mm
x1
30 mm
105 mm
15 mm
30 mm
z
up
C
u
M
y = 52.5 mm
90 mm
30 mm
90 mm
O
30 mm
120 mm
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Solution 6.5-8
M 4.5 kN # m
u 22.5°
MAXIMUM TENSILE STRESS (AT POINT A)
zA BUILT-UP SECTION:
t
2
yA h
ybar
2
b 120 mm
t 30 mm
zpA (zA) cos (uP)(yA) sin (uP)
h 180 mm
b1 360 mm
ypA (zA) sin (uP)(yA) cos (uP)
t + h
b
2
ypA 133.51 mm
ybar tb1 a
tb1 + [2tb + (h 2t)t]
ybar 52.5 mm
Iyp 152.7 * 106 mm4
BENDING MOMENTS
Myp Msin(uP u)
Myp 377 N # m
Mzp Mcos(uP u)
Mzp 4484 N # m
NEUTRAL AXIS nn
b 2.93°
Izp
Iyp
Iyp
Mzp(ypA)
Izp
;
MAXIMUM COMPRESSIVE STRESS (AT POINT B)
Iyp Iy1
Izp 93.14 * 106 mm4
b atan a
Myp(zpA)
st 6.56 MPa
uP 27.3 deg
Izp Ix1
st zpA 52.03 mm
tan(uP u) b
zB b1
2
yB h
+ t ybar
2
zpB (zB) cos (uP)(yB) sin (uP)
zpB 128.99 mm
ypB (zB) sin (uP)(yB) cos (uP)
ypB 142.5 mm
sc Myp(zpB)
Iyp
sc 6.54 MPa
Mzp(ypB)
Izp
;
;
Problem 6.5-9 A beam of semicircular cross section of radius r is subjected
to a bending moment M having its vector at an angle u to the z axis (see figure).
Derive formulas for the maximum tensile s, and the maximum compressive stress sc in the beam for u 0, 45°, and 90°. (Note: Express the results in
the from M/r3, where is a numerical value.)
y
M
z
u
O
C
r
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SECTION 6.5 Bending of Unsymmetric Beams
Soultion 6.5.-9
537
Semicircle
sc sD M(r c)
Iy
5.244
M
;
r3
FOR u 45°: Eq. (6q40): tan b tan b 9p2
9p2 64
Iz
Iy
tan u
(1) 3.577897
b 74.3847°
90° b 15.6153°
r radius
c
4r
0.42441r
3p
Iy (9p2 64) 4
r
72p
MAXIMUM TENSILE STRESS for u 45° occurs at point A.
z A c 0.42441r
From (Eq. 6-38):
st sA 0.109757r4
Iz pr4
8
(M sin u)zA
(M cos u)yA
Iy
Iz
4.535
st maximum tensile stress
2.546
Mr
8M
Iz
pr 3
z E c r cos (90° ) 0.53868r
;
r3
From (Eq. 6-38):
sc sB sA M
Mc
Iy
3.867
M
8M
pr
;
r3
FOR u 90°: st so r3
;
r3
y E r sin (90° ) 0.26918r
M
2.546
M
MAXIMUM COMPRESSIVE STRESS for u 45° occurs at
point E. where the tangent to the circle is parallel to the
neutral axis nn.
sc maximum compressive stress
FOR u 0°: st sA yA r
;
3
sC sE (M sin u)zE
(M cos u)yE
Iy
Iz
3.955
M
r3
;
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Problem 6.5-10 A built-up beam supporting a condominium balcony is made up of a structural T (one half of a
W 200 31.3) for the top flange and web and two angles (2L 102 76 6.4, long legs back-to-back) for the bottom
flange and web, as shown. The beam is subjected to a bending moment M having its vector at an angle u to the
z axis (see figure).
Determine the orientation of the neutral axis and calculate the maximum tensile stress st and maximum compressive
stress sc in the beam. Assume that u 30° and M 15 kN # m.
Use the following numerical properties: c1 4.111 mm, c2 4.169 mm, bf 134 mm, Ls 76 mm,
A 4144 mm2, Iy 3.88 106 mm4, and Iz 34.18 106 mm4.
y
bf /2
u, b
bf /2
c1
C
z
u
c2
M
Ls
Ls
Solution 6.5-10
M 15 kN # m
MAXIMUM TENSILE STRESS (AT POINT A)
u 30°
zA BUILT-UP SECTION:
c1 4.111 mm
Ls 76 mm
c2 4.169 mm
bf 134 mm
A 4144 mm
2
Iy 3.88 106 mm4
Iz 34.18 106 mm4
BENDING MOMENTS
st My 7500 N # m
Mz Mcos(180° u)
Mz 12990 N # m
My(zA)
Iy
Mz(yA)
Iz
st 131.1 MPa
zB Ls
sc NEUTRAL AXIS nn
Iz
b atan a tan(180° u) b
Iy
;
yA c1
2
;
MAXIMUM COMPRESSIVE STRESS (AT POINT B)
My Msin(180° u)
b 78.9°
bf
My(zB)
Iy
yB c2
Mz(yB)
Iz
sc 148.5 MPa
;
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SECTION 6.5 Bending of Unsymmetric Beams
Problem 6.5-11 A steel post (E 30 106 psi) having
thickness t 1/8 in. and height L 72 in. supports a stop
sign (see figure). The stop sign post is subjected to a bending
moment M having its vector at an angle u to the z axis.
Determine the orientation of the neutral axis and
calculate the maximum tensile stress st and maximum
compressive stress sc in the beam. Assume that u 30° and
M 5.0 k-in.
Use the following numerical properties for the post: A 0.578 in2, c1 0.769 in., c2 0.731 in., Iy 0.44867 in4, and
Iz 0.16101 in4.
A
Section A–A
5/8 in.
Circular cutout,
d = 0.375 in.
y
Post, t = 0.125 in.
c1
z
1.5 in.
C
u
Stop
sign
c2
M
1.0 in.
0.5 in.
1.0 in.
0.5 in.
A
L
Elevation
view of post
Solution 6.5-11
M 5 k # in
MAXIMUM TENSILE STRESS
u 30°
Post: A 0.578 in.2
c1 0.769 in.
sx(z, y) c2 0.731 in.
Iy 0.44867 in.4
POINT A:
Iz 0.16101 in.4
My z
Iy
Mz y
Iz
zA 1.5 in.
st sx(zA, yA)
yA c2
st 28.0 ksi
;
BENDING MOMENTS
My Msin(u)
My 0.208 k # ft
MAXIMUM COMPRESSIVE STRESS
Mz Mcos(u)
Mz 0.361 k # ft
POINT B:
Iz
b atan a tan(u)b
Iy
zB sc sx(zB, yB)
NEUTRAL AXIS nn
b 11.7°
;
539
5
in.
8
yB c1
sc 24.2 ksi
;
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Problem 6.5-12 A C 200 17.1 channel section has an angle
with equal legs attached as shown; the angle serves as a lintel
beam. The combined steel section is subjected to a bending
moment M having its vector directed along the z axis, as shown
in the figure. The centroid C of the combined section is located
at distances xc and yc from the centroid (C1) of the channel alone.
Principal axes x1 and y1 are also shown in the figure and properties Ix1, Iy1 and up are given below.
Find the orientation of the neutral axis and calculate the
maximum tensile stress st and maximum compressive stress sc if
the angle is an L 76 76 6.4 section and M 3.5 kN # m. Use
the following properties for principal axes for the combined
section: Ix1 18.49 106 mm4 Iy1 1.602 106 mm4,
up 7.448° (CW), xc 10.70 mm, yc 24.07 mm.
y1
y
C
C1
M
z
yc
up
xc
x1
L 76 76 6.4
lintel
C 200 17.1
Solution 6.5-12
M 3.5 kN # m
ANGLE:
ca 21.2 mm
CHANNEL:
zpA (zA) cos (up) (yA) sin (up)
up 7.448°
cc 14.5 mm
zpA 63.18 mm
La 76 mm
ypA (zA) sin (up) + (yA) cos (up)
dc 203 mm
ypA 69.83 mm
bc 57.4 mm
st BUILT-UP SECTION:
ybar 24.07 mm
Izp Ix1
xbar 10.70 mm
Iyp
st 31.0 MPa
Iyp Iy1
Izp 18.49 106 mm4
Myp(zpA)
Iyp 1.602 106 mm4
Izp
;
MAXIMUM COMPRESSIVE STRESS (AT POINT B)
zB xbar cc
BENDING MOMENTS
Mzp(ypA)
yB dc
+ ybar
2
Myp Msin(up)
Myp 454 N # m
zpB (zB) cos (up) (yB) sin (up)
Mzp Mcos(up)
Mzp 3470 N # m
zpB 20.05 mm
ypB (zB) sin (up) + (yB) cos (up)
NEUTRAL AXIS nn
b a tan a
Izp
Iyp
tan( up)b
ypB 124.0 mm
b 56.5°
MAXIMUM TENSILE STRESS (AT POINT A)
zA xbar cc bc
dc
yA + ybar
2
;
sc Myp(zpB)
Iyp
sc 29.0 MPa
Mzp(ypB)
Izp
;
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SECTION 6.8 Shear Stresses in Wide-Flange Beams
541
Shear Stresses in Wide-Flange Beams
When solving the problems for Section 6.8, assume the cross sections
are thin-walled. Use centerline dimensions for all calculations and
derivations, unless otherwise specified
Problem 6.8-1 A simple beam of W 10 30 wide-flange cross
section supports a uniform load of intensity q 3.0 k/ft on a span
of length L 12 ft (see figure). The dimensions of the cross
section are h 10.5 in., b 5.81 in., tf 0.510 in., and
tw 0.300 in.
y
b
—
2
q
A
(a) Calculate the maximum shear stress tmax on cross section
A–A located at distance d 2.5 ft from the end of the beam.
(b) Calculate the shear stress t at point B on the cross
section. Point B is located at a distance a 1.5 in.
from the edge of the lower flange.
A
Probs. 6.8-1 and 6.8-2
L
b
—
2
h
—
2
tw
z
C
tf
h
—
2
B
a
d
Solution 6.8-1
(a) MAXIMUM SHEAR STRESS
SIMPLE BEAM:
q 3.0 k/ft
R
qL
2
L 12 ft
R 18.0 k
V |R qd|
d 2.5 ft
V 10.5 k
b 5.81 in.
tf 0.510 in.
tw 0.30 in.
3
tmax 3584 psi
a 1.5 in.
t1 2
twh
btfh
+
Iz 12
2
btf
h Vh
+ b
tw
4 2Iz
;
(b) SHEAR STRESS AT POINT B
CROSS SECTION:
h 10.5 in.
tmax a
Iz 192.28 in.4
bhV
4Iz
a
tB (t1)
b
2
b
2.9 in.
2
t1 832.8 psi
tB 430 psi
;
Problem 6.8-2 Solve the preceding problem for a W 250 44.8 wide-flange shape with the following data: L 3.5 m,
q 45 kN/m, h 267 mm, b 148 mm, tf 13 mm, tw 7.62 mm, d 0.5 m, and a 50 mm.
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Solution 6.8-2
(a) MAXIMUM SHEAR STRESS
SIMPLE BEAM:
q 45 kN/m
R
qL
2
L 3.5 m
tmax a
R 78.8 kN d 0.5 m
tmax 29.7 MPa
V 56.3 kN
V |R qd|
b/2 74.0 mm
a 50 mm
h 267 mm
b 148 mm
tf 13 mm
tw 7.62 mm
;
(b) SHEAR STRESS AT POINT B
CROSS SECTION:
Iz btf
h Vh
+ b
tw
4 2Iz
twh3
btfh2
+
12
2
t1 bhV
4Iz
tB a
(t )
b/2 1
t1 999.1 psi
tB 4.65 MPa
;
Iz 80.667 * 106 mm4
Problem 6.8-3 A beam of wide-flange shape, W 8 28, has the cross section
y
shown in the figure. The dimensions are b 6.54 in., h 8.06 in., tw 0.285
in., and tf 0.465 in. The loads on the beam produce a shear force V 7.5 k
at the cross section under consideration.
tf
(a) Using centerline dimensions, calculate the maximum shear stress tmax
in the web of the beam.
(b) Using the more exact analysis of Section 5.10 in Chapter 5, calculate
the maximum shear stress in the web of the beam and compare it with
the stress obtained in part a.
z
h
C
tw
tf
Probs. 6.8-3 and 6.8-4
b
Solution 6.8-3
b 6.54 in.
h 8.06 in.
tf 0.465 in.
V 7.5 k
tw 0.285 in.
(a) CALCULATIONS BASED ON CENTERLINE DIMENSIONS
Maximum shear stress in the web:
tmax a
btf
h Vh
+ b
tw
4 2Iz
tmax 3448 psi
;
Moment of inertia:
Iz twh3
btfh2
+
12
2
Iz 111.216 in.4
(b) CALCULATIONS BASED ON MORE EXACT ANALYSIS
h2 h tf
h1 7.6 in.
h2 8.5 in.
h1 h tf
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SECTION 6.9 Shear Centers of Thin-Walled Open Sections
Moment of inertia:
I
Maximum shear stress in the web:
1
(bh32 bh31 + twh31)
12
tmax V
(bh22 bh21 + twh21)
8Itw
tmax 3446 psi
I 109.295 in.4
;
Problem 6.8-4 Solve the preceding problem for a W 200 41.7 shape with the following data: b 166 mm,
h 205 mm, tw 7.24 mm, tf 11.8 mm, and V 38 kN.
Solution 6.8-4
b 166 mm
h 205 mm
tf 11.8 mm
V 38 kN
tw 7.24 mm
(b) CALCULATIONS BASED ON MORE EXACT ANALYSIS
h2 h tf
h2 216.8 mm
h1 h tf
h1 193.2 mm
(a) CALCULATIONS BASED ON CENTERLINE DIMENSIONS
Moment of inertia:
Moment of inertia:
Iz twh3
btfh2
+
12
2
I
I 45.556 * 106 mm4
Iz 46.357 * 106 mm4
Maximum shear stress in the web:
tmax a
btf
h Vh
+ b
tw
4 2Iz
tmax 27.04 MPa
1
(bh32 bh31 + twh31)
12
Maximum shear stress in the web:
tmax V
(bh22 bh21 + twh21)
8Itw
tmax 27.02 MPa
;
;
Shear Centers of Thin-Walled Open Sections
When locating the shear centers in the problems for Section 6.9,
assume that the cross sections are thin-walled and use centerline
dimensions for all calculations and derivations.
y
Problem 6.9-1 Calculate the distance e from the centerline of
the web of a C 15 40 channel section to the shear center S
(see figure). (Note: For purposes of analysis, consider the
flanges to be rectangles with thickness tf equal to the average
flange thickness given in Table E-3a in Appendix E.)
S
z
e
Probs. 6.9-1 and 6.9-2
C
543
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Solution 6.9-1
C 15 * 40
bf 3.520 in.
d 15.0 in.
tw 0.520 in.
tf 0.650 in.
b bf h d tf
tw
2
e
h 14.350 in.
3b2 tf
h tw + 6btf
e 1.027 in.
;
b 3.260 in.
Problem 6.9-2 Calculate the distance e from the centerline of the web of a C 310 45 channel section to the shear
center S (see figure). (Note: For purposes of analysis, consider the flanges to be rectangles with thickness tf equal to the
average flange thickness given in Table E-3b in Appendix E.)
Solution 6.9-2
C 310 * 45
bf 80.5 mm
d 305 mm
tf 12.7 mm
tw 13.0 mm
b bf tw
2
b 74.0 mm
h d tf
h 292.3 mm
2
e
3b tf
htw + 6btf
e 22.1 mm
Problem 6.9-3 The cross section of an unbalanced wide-flange
y
beam is shown in the figure. Derive the following formula for
the distance h1 from the centerline of one flange to the shear
center S:
h1 ;
t2
t2b32h
z
t1b31 + t2b32
b1
t1
S
Also, check the formula for the special cases of a T-beam
(b2 t2 0) and a balanced wide-flange beam (t2 t1 and
b2 b1).
h1
h2
h
Solution 6.9-3 Unbalanced wide-flange beam
FLANGE 1:
t1 VQ
Izt1
Q (b1/2)(t1)(b1/4) t1 b2
C
t1b21
8
Vb21
8Iz
Vt1b31
2
F1 (t1)(b1)(t1) 3
12Iz
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SECTION 6.9 Shear Centers of Thin-Walled Open Sections
Solve Eqs. (1) and (2): h1 FLANGE 2:
F2 Vt2b32
12Iz
t2b32h
t1b31 + t2b32
T-BEAM
b2 t2 0;
‹ h1 0
Shear force V acts through the shear center S.
‹ a MS F1h1 F2h2 0
;
WIDE-FLANGE BEAM
or (t1b31 ) h1 (t2b 32) h2
(1)
h1 h2 h
(2)
t2 t1 and b2 b1;
‹ h1 h/2
;
Problem 6.9-4 The cross section of an unbalanced wide-flange beam is
y
tf
shown in the figure. Derive the following formula for the distance e from
the centerline of the web to the shear center S:
e
;
tw
3tf (b22 b21)
htw + 6tf (b1 + b2)
Also, check the formula for the special cases of a channel section (b1 0
and b2 b) and a doubly symmetric beam (b1 b2 b/2).
h
—
2
S
z
C
e
tf
b1
h
—
2
b2
Solution 6.9-4 Unbalanced wide-flange beam
t1 VQ
b1hV
Itf
2Iz
F1 b1t1tf
b21htfV
2
4Iz
Shear force V acts through the shear center S.
b22htfV
‹ a MS F3e F1h + F2h 0
F2 4Iz
t2 F3 V
b2hV
2Iz
e
h2tf 2
F2h F1h
(b b21)
F3
4Iz 2
545
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CHAPTER 6 Stresses in Beams (Advanced Topics)
twh3
h 2
+ 2(b1 + b2)(tf)a b
12
2
Iz e
CHANNEL SECTION (b1 0, b2 b)
e
2
h
[htw + 6 tf (b1 + b2)]
12
3tf (b22 b21)
htw + 6tf (b1 + b2)
3b2tf
htw + 6btf
(Eq. 6 - 65)
DOUBLY SYMMETRIC BEAM (b1 b2 b/2)
e 0 (Shear center coincides with the centroid)
;
y
Problem 6.9-5 The cross section of a channel beam with double
flanges and constant thickness throughout the section
is shown in the figure.
Derive the following formula for the distance e from the
centerline of the web of the shear center S:
e
3b2(h21 + h22)
h32
+
6b(h21
+
S
z
h22)
C
h1 h2
e
b
Solution 6.9-5 Channel beam with double flanges
Shear force V acts through the shear center S.
‹ a MS F3e + F1h2 + F2h1 0
e
F2h1 + F1h2
b2t 2
(h + h22)
F3
4Iz 1
Iz th32
+ 2 [bt(h2/2)2 + bt(h1/2)2]
12
e
t thickness
V(bt)a
h2
b
2
tA VQA
Izt
F1 b2h2tV
1
tAbt 2
4Iz
tB bh1V
2Iz
F3 V
Izt
F2 b2h1tV
4Iz
bh2V
2Iz
t 3
[h + 6b(h21 + h22)]
12 2
3b2(h21 + h22)
h32 + 6b(h21 + h22)
;
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547
SECTION 6.9 Shear Centers of Thin-Walled Open Sections
Problem 6.9-6 The cross section of a slit circular tube of
y
constant thickness is shown in the figure.
(a) Show that the distance e from the center of the circle
to the shear center S is equal to 2r in the figure part a.
(b) Find an expression for e if flanges with the same
thickness as that of the tube are added, as shown in
the figure part b.
Flange
(r/2 t)
y
z
z
r
S
r/2
r
S
C
e
C
e
Flange
(r/2 t)
(b)
(a)
Solution 6.9-6
u
(a) QA L
y dA L0
for 0 … u 6
(r tsin(f)) df
u
QA QA r t(1 cos(u))
2
tA VQA
Vr2(1 cos(u))
Izt
Iz
Iz pr t
V(1 cos(u))
prt
TC moment of shear stresses about center C.
2p
tA VQA
Vr2(1 cos(u))
Izt
Iz
for
3p
p
… u 6
2
2
Vr
p
QA Shear force V acts through the shear center S.
Moment of the shear force V about any point must
be equal to the moment of the shear stresses about
that same point.
TC
2r
e
V
(b) Iz pr3t + 2 J
Iz pr3t +
r 3
ta b
2
12
r 5r 2
+ t a b K
2 4
19 3
19
tr tr3 ap +
b
12
12
L
y dA L0
(rtsin(f)) df +
QA r2ta
13
cos(u)b
8
Vr2 a
13
cos(u)b
8
(1 cos(u)) du 2Vr
©MC Ve TC
(rtsin(f)) df
L0
u
At point A: dA rtdu
TC tAr dA L
L0
L
y dA QA r2t(1 cos(u))
3
tA p
2
tA r 5r
t
2 4
Iz
At point A: dA rtdu
TC moment of shear stresses about center C.
;
TC L
tAr dA 2
p
2
Vr4t
Iz
L
0
(1 cos(u)) du +
a
13
cos(u)b du
8
3p
2
Vr4t
Iz
Lp2
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CHAPTER 6 Stresses in Beams (Advanced Topics)
TC L
tAr dA Vr4t
p 2
Iz
3p
2
+
TC Vr4t
Vr4t 13
a
cos(u)b du
Iz
8
Lp2
p2
1
13p + 16
+ Vr4t
Iz
8
Iz
21Vr4tp
8Iz
Shear force V acts through the shear center S.
Moment of the shear force V about any point must
be equal to the moment of the shear stresses about
that same point.
©MC Ve TC
e
e
TC
V
21r4tp
8Iz
21r4tp
19
8ctr3 a p +
bd
12
63pr
1.745r
24p + 38
;
Problem 6.9-7 The cross section of a slit square tube of constant thickness is
y
shown in the figure. Derive the following formula for the distance e from the
corner of the cross section to the shear center S:
e
b
212
b
z
S
e
Solution 6.9-7 Slit square tube
b length of each side
t thickness
t
VQ
Iz t
FROM A TO B:
Q
ts2
212
At A: Q 0 tA 0
C
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SECTION 6.9 Shear Centers of Thin-Walled Open Sections
At B: Q tB F1 tb2
212
At B: tB b2V
212Iz
F2 tB bt +
tB bt
b3tV
3
612Iz
b2V
12Iz
2
5tb3V
(tD tB) bt 3
612Iz
‹ g Ms 0
2(F1/12)(b 12 + e) + 2(F2/ 12) (e) 0
b
S
b
b + st a
b
Q bt a
212
12
212
t
At D: tD Shear force V acts through the shear center S.
FROM B TO D:
b2V
212Iz
549
Substitute for F1 and F2 and solve for e:
b
212
e
ts
tb2
+
(2b s)
212
212
;
VQ
V b2
s
c
+
(2b s) d
Izt
Iz 212
212
Problem 6.9-8 The cross section of a slit rectangular tube of
y
constant thickness is shown in the figures.
(a) Derive the following formula for the distance e from the
centerline of the wall of the tube in figure part (a) to the
b(2h + 3b)
shear center S: e 2(h + 3b)
(b) Find an expression for e if flanges with the same
thickness as that of the tube are added as shown in
figure part (b).
Flange
(h/4 t)
y
h
—
2
z
S
z
C
e
h
—
2
S
C
e
h
—
2
b
—
2
h
—
2
b
—
2
b
—
2
(a)
b
—
2
(b)
Solution 6.9-8
(a) FROM A TO B: Q tA 0
F1 tB ts2
2
h2V
8Iz
tBt h
th3V
a b 3 2
48Iz
FROM B TO C: tB h2V
8Iz
t
VQ
s2V
Izt
2Iz
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CHAPTER 6 Stresses in Beams (Advanced Topics)
th h
h
th
a b + bta b (h + 4b)
2 4
2
8
QC h(h + 4b)V
8Iz
tC bht(h + 2b)V
1
F2 (tB + tC)bt 2
8Iz
©FVERT V
F3 V a1 +
F3 2 F1 V
In flange:
QC_ flange t sa
Fflange h
4
s
tV
h3Vt
c (2s h) d ds Iz
96Iz
L0 4
FROM C TO D:
Q
th3
b
24Iz
s
h
ts
+ b (2s + h)
2
4
4
FCD b h
h 3h
h
th2
+ t a b + t a b + t s
8
2 2
4 8
2
b
2
L0
c
th2 b h
h 3h
t a b t a b 8
2 2
4 8
Shear force V acts through the shear center S.
©Ms 0
(b + e) 0
solve for
Iz 2 c
ts2
2
VQ
s2V
Izt
2Iz
F1 tB h2V
8Iz
tBt h
th3V
a b
3 2
48Iz
FROM B TO C: tB QC tC h2 # V
8Iz
th h
b h
th
a b + t a b (h + 2b)
2 4
2 2
8
h(h + 2b)V
8Iz
bht(h + b)V
1
b
FBC (tB + tC) t 2
2
16Iz
th3
b
16Iz
Shear force V acts through the shear center S.
©Ms 0
;
t
F3 2F1 2Fflange V
F3 Va 1 +
b 2h + 3b
e a
b
2 h + 3b
(b) FROM A TO B: Q tA 0
©FVERT V
bh2t(2h + 3b)
e
12Iz
1 3
h 2
th2
th + bt a b d (h + 3b)
12
2
6
Therefore
Vthb
h Vt
(7h 12b)
t s d ds 2 Izt
64Iz
F3e + F2h + 2F1
F3e + (FBC + FCD)h + 2F1
(b + e) + 2Fflange a
e
b2h2t
61bth3
+
192Iz
4Iz
e
th2b
(43h + 48b)
192Iz
Iz 2c
b
+ eb 0
2
1 3
h 2
1 h 3
th + bta b +
ta b +
12
2
12 4
th2
h 3h 2
(23h + 48b)
ta b a b d 4
8
96
b 43h + 48b
b
e a
2 23h + 48b
;
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SECTION 6.9 Shear Centers of Thin-Walled Open Sections
Problem 6.9-9 A U-shaped cross section of constant thickness
y
is shown in the figure. Derive the following formula for the
distance e from the center of the semicircle to the shear
center S:
e
551
b
r
2(2r2 + b2 + pbr)
4b + pr
S
z
O
Also, plot a graph showing how the distance e (expressed
as the nondimensional ratio e/r) varies as a function of the
ratio b/r. (Let b/r range from 0 to 2.)
C
e
Solution 6.9-9 U-shaped cross section
At angle u: dA rtdu
r radius
F1 force in AB
t thickness
F2 force in EF
p
T0 trdA tr2tdu
L
L0
T0 moment in BDE
FROM A TO B: tA 0
F1 p
tB VQ
V(btr)
Vbr
Izt
Izt
Iz
2
bttB
Vb rt
2
2Iz
u
FROM B TO E: Q1 ydA (r cos f) rtdf
L
L0
r2t sin u
QB btr
Vr3t (b + r sin u)du
Iz
L0
Vr3t
(pb + 2r)
Iz
Shear force V acts through the shear center S. Moment
of the shear force V about any point must be equal to
the moment of the shear stresses about that same point.
‹ a M0 Ve T0 + F1(2r)
e
T0 + 2F1r
r2t
(pbr + 2r2 + b2)
V
Iz
Iz pr3t
+ 2(btr2)
2
Qu QB + Q1 btr + r2t sin u
VQB
Vr (b + r sin u)
tu Iz t
Iz
e
2(2r 2 + b2 + pbr)
4b + pr
;
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CHAPTER 6 Stresses in Beams (Advanced Topics)
GRAPH
NOTE: When b/r 0,
2(2 + b2/r2 + pb/r)
e
r
4b/r + p
e/r 4
(Eq. 6-73)
p
Problem 6.9-10 Derive the following formula for the distance e from the centerline
y
of the wall to the shear center S for the C-section of constant thickness shown in the
figure:
e
a
3bh2(b + 2a) 8ba3
h
—
2
h2(h + 6b + 6a) + 4a2(2a 3h)
S
z
Also, check the formula for the special cases of a channel section (a 0) and a slit
rectangular tube (a h/2).
e
C
h
—
2
a
b
Solution 6.9-10 C-section of constant thickness
a
t thickness
F1 FROM A TO B:
Q st a
h
s
a+ b
2
2
tA 0 tB t
a
V
(h a)
2
Iz
VQ
h
s V
sa a + b
Iz t
2
2 Iz
L0
a
ttds tV
h
s
s a a + b ds
Iz L0
2
2
a2t(3h 4a)V
12Iz
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SECTION 6.9 Shear Centers of Thin-Walled Open Sections
FROM B TO C:
tB Substitute for F1, F2, and F3 and solve for e:
V
a
(h a)
2
Iz
QC at a
a
h
h
b + bta b
2
2
2
at
bht
(h a) +
2
2
a
bh V
tC c (h a) +
d
2
2 Iz
e bt [3h2(b + 2a) 8a3]
12 Iz
Iz 2a
1 3
h 2
6
th b + 2bta b (h 2a)3
12
2
12
t 2
[h (h + 6b + 6a) + 4a2(2a 3h)]
12
3bh2(b + 2a) 8ba3
1
bt
V
F2 (tB + tC)bt [2a(h a) + bh]
2
4
Iz
e FROM C TO E:
CHANNEL SECTION (a 0)
g FVERT V F3 2F1 V
e
a 2 t(3h 4a)
F3 V c1 +
d
6Iz
h (h + 6b + 6a) + 4a2(2a 3h)
2
3b2
h + 6b
;
(agrees with Eq. 6-65 when tf tw)
SLIT RECTANGULAR TUBE (a h/2)
Shear force V acts through the shear center S.
‹ a Ms 0
553
F3(e) + F2h + 2F1(b + e) 0
e
b(2h + 3b)
(agrees with the result of Prob. 6.9-8)
2(h + 3b)
Problem 6.9-11 Derive the following formula for the distance e from the centerline of the
y
wall to the shear center S for the hat section of constant thickness shown in the figure:
e
a
3bh2(b + 2a) 8ba3
h2(h + 6b + 6a) + 4a2(2a + 3h)
h
—
2
Also, check the formula for the special case of a channel section (a 0).
S
z
e
C
h
—
2
a
b
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Solution 6.9-11 Hat section of constant thickness
t thickness
FROM A TO B
t FROM C TO E:
Q st a
s
h
+ a b
2
2
VQ
h
s V
sa + a b
Izt
2
2 Iz
V
a
tA 0 tB (h + a)
2
Iz
a
F1 L0
a
ttds s
tV
h
sa + a bds
Iz L0
2
2
a2t (3h + 4a)V
12Iz
FROM B TO C
QC at a
tB V
a
(h + a)
2
Iz
a
h
at
bht
h
+ b + bta b (h + a) +
2
2
2
2
2
g FVERT V
F3 V c1 F3 + 2F1 V
2
a t(3h + 4a)
d
6Iz
Shear force V acts through the shear center S.
‹ g MS 0
F3e F2h 2F1(b e) 0
Substitute for F1, F2, and F3 and solve for e:
e
bt[3h2(b + 2a) 8a3]
12Iz
Iz 1 3
h 2
t
1
th + 2bta b +
(h + 2a)3 th3
12
2
12
12
e
t 2
[h (h + 6b + 6a) + 4a2(2a + 3h)]
12
3bh2(b + 2a) 8ba3
2
h (h + 6b + 6a) + 4a2(2a + 3h)
bh V
a
tc c (h + a) +
d
2
2 Iz
CHANNEL SECTION (a 0)
1
bt
V
F2 (tB + tc) bt [2a(h + a) + bh]
2
4
Iz
e
3b2
h + 6b
;
(agrees with Eq. 6-65 when tf tw)
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555
SECTION 6.9 Shear Centers of Thin-Walled Open Sections
Problem 6.9-12 The cross section of a sign post of constant thickness is shown in the
figure. Derive the formula below for the distance e from the centerline of the wall of the post
to the shear center S. Also, compare this formula with that given in Prob. 6.9-11 for the
special case of b 0 here and a h/2 in both formulas.
y
a
b
b
b sin(b )
a
S
C
z
a
e
b
b
b sin(b )
a
Solution 6.9-12
FROM A TO B
s
QAB st a2a + bsin (b ) b
2
tAB FAB FAB VQAB
Iz t
L
tdA a V cst a2a bsin (b)
L0
Izt
s
bd
2
t ds
Vta2(5a 3bsin (b))
6Iz
FROM B TO C
a
s
QBC at a2a + bsin (b) b + ts a a + bsin (b) sin (b)b
2
2
3
1
QBC a2t + atbsin (b) + sta + stbsin (b) s 2tsin (b)
2
2
tBC VQBC
Iz t
b
FBC FBC L
tdA 3
1
V a a2t + atbsin (b) + sta + stbsin (b) s2tsin (b)b
2
2
L0
Izt
Vtb a9a2 + 6absin (b) + 3ba + 2b2sin (b)b
6Iz
tds
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CHAPTER 6 Stresses in Beams (Advanced Topics)
SHEAR FORCE V ACTS THROUGH THE SHEAR CENTER S.
a ME 0
e 2cos (b)
e
Ve + 2FABbcos (b) FBCcos (b)(2a) 0
(FBCa FABb)
V
tbacos (b)
a4a2 + 3absin (b) + 3ab + 2b2sin (b)b
3Iz
;
Now, compare this formula with that given in Prob. 6.9-11 for the special case of b 0 here and a = h/2 in both formulas.
FIRST MODIFY ABOVE FORMULA FOR b 0 & a = h/2:
h
tb cos (0)
2
h 2
h
h
c4a b 3 bsin (0) 3 b2b2sin (0) d
e
3Iz
2
2
2
e
bht ah2 3bh
b
2
6Iz
where Iz for the hat section of #6.9-11 is as follows:
3
h 3
ta b
2
2
Iz h
th
+ 2bt a b + 2
12
2
12
Iz h2t (3b + 4h)
6
h h
h 2
+ 2t a + b
2 2
4
substituting expression for Iz & simplifying gives:
bhtah2 e
6
e
3bh
b
2
h2t(3b + 4h)
6
b(3b 2h)
6b 8h
for
b 0 and
a
h
2
;
NOW MODIFY FORMULA FOR e FROM #6.9-11 AND COMPARE TO ABOVE
e
e
e
3bh2(b 2a) 8ba3
h2(h 6b 6a)4a2(2a 3h)
h
h 3
3bh2 ab 2 b 8ba b
2
2
h
h 2 h
h2 a h 6b 6 b 4a b a2 3hb
2
2
2
b(3b 2h)
6b 8h
same expressions as that above from sign post solution
;
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557
SECTION 6.9 Shear Centers of Thin-Walled Open Sections
Problem 6.9-13 A cross section in the shape of a circular arc of constant thickness is shown
y
in the figure. Derive the following formula for the distance e from the center of the arc to the
shear center S:
e
2r(sin b b cos b)
b sin b cos b
r
in which b is in radians. Also, plot a graph showing how the distance e varies as b varies
from 0 to p.
z
b
S
O
C
b
e
Solution 6.9-13
Circular arc
Shear force V acts through the shear center S. Moment
of the shear force V about any point must be equal to the
moment of the shear stresses about that same point.
‹ a M0 Ve T0 e T0/V
e
t thickness
r radius
2r(sin b b cos b)
b sin b cos b
;
GRAPH
At angle u:
b
Q
L
ydA Lu
(r sin f) rtdf
r2t(cos u cos b)
t
Vr2(cos u cos b)
VQ
Izt
Iz
b
Iz L
y2 dA L b
(r sin f)2rtdf
r3t(b sin b cos b)
V(cos u cos b)
t
rt(b sin b cos b)
2(sin b b cos b)
e
r
b sin b cos b
SEMICIRCULAR ARC (b p/2):
4
e
(Eq. 6-73)
p
r
T0 moment of shear stresses
SLIT CIRCULAR ARC (b p):
At angle u, dA rtdu
e
2 (Prob. 6.9-6)
r
b
T0 V(cos u cos b)
trdA rtdu
L
L b t (b sin b cos b
2 Vr (sin b b cos b)
b sin b cos b
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Elastoplastic Bending
y
The problems for Section 6.10 are to be solved using the assumption that
the material is elastoplastic with yield stress sY.
b1
Problem 6.10-1 Determine the shape factor f for a cross section in the shape of
a double trapezoid having the dimensions shown in the figure.
Also, check your result for the special cases of a rhombus (b1 0) and a
rectangle (b1 b2).
h
—
2
z
C
b1
b2
Solution 6.10-1
Double trapezoid
SHAPE FACTOR f
f
(EQ. 6-79)
2(2b1 + b2)
Z
S
3b1 + b2
;
SPECIAL CASE – RHOMBUS
Neutral axis passes through the centroid C.
Use case 8, Appendix D.
SECTION MODULUS S
h 3
Iz 2 a b (3b1 + b2)/12
2
h3
(3b1 + b2)
48
c h/2
S
b1 0
f2
SPECIAL CASE – RECTANGLE
I
h2
(3b1 + b2)
c
24
PLASTIC MODULUS Z (EQ. 6-78)
h
h
A 2a b(b1 + b2)/2 (b1 + b2)
2
2
1 h 2b1 + b2
b
y1 y2 a b a
3 2
b1 + b2
z
h2
A
(y1 + y2) (2b1 + b2)
2
12
b1 b2
f
3
2
h
—
2
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559
SECTION 6.10 Elastoplastic Bending
Problem 6.10-2 (a) Determine the shape factor f for a
y
hollow circular cross section having inner radius r1 and outer
radius r2 (see figure).
(b) If the section is very thin, what is the shape factor?
r1
z
C
r2
Solution 6.10-2 Hollow circular cross sections
(a) SHAPE FACTOR f (EQ. 6-79)
f
16r2(r32 r31)
Z
S
3p(r42 r41)
;
(b) THIN SECTION (r1 : r2)
Rewrite the expression for the shape factor f.
(r23 r13) (r2 r1)(r22 r1r2 r12)
(r24 r14) (r2 r1)(r2 r1)(r22 r12)
Neutral axis passes through the centroid C.
f
Use cases 9 and 10, Appendix D.
SECTION MODULUS S
Iz p 4
(r r41 ) c r2
4 2
S
Iz
p 4
(r r41)
c
4r2 2
gyi Ai
g Ai
Let r1 0 f y
4r
3p
4r2 pr22
4r1 pr21
ba
b a
ba
b
3p
2
3p
2
p/2(r22 r21)
4 r32 r31
a
b
3p r22 r21
y1 y2 Z 16 3
4
a b L 1.27
p
3p 4
;
SPECIAL CASE OF A SOLID CIRCULAR CROSS SECTION
A p(r22 r21 ) For a semicircle,
y1 16 1 + r1/r2 + (r1/r2)2
d
c
3p (1 + r1/r2)(1 + r21/r22)
Let r1/r2 : 1 f PLASTIC MODULUS Z (EQ. 6-78)
a
16r2 r22 + r1r2 + r21
d
c
3p (r2 + r1)(r22 + r21)
A
4
(y + y2) (r32 r31)
2 1
3
16 1
16
a b 3p 1
3p
(Eq. 6-90)
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Problem 6.10-3 A propped cantilever beam of length L 54 in.
y
q
with a sliding support supports a uniform load of intensity q (see
figure). The beam is made of steel (sY 36 ksi) and has a rectangular cross section of width b 4.5 in. and height h 6.0 in.
What load intensity q will produce a fully plastic condition
in the beam?
z
C
Sliding support
L = 54 in.
h = 6.0 in.
b = 4.5 in.
Solution 6.10-3
L 54 in.
sy 36 ksi
b 4.5 in.
Mmax MAXIMUM BENDING MOMENT:
h 6.0 in.
qL2
2
Let
Mmax MP
q
gives
Therefore q 1000 lb/in.
sybh2
2L2
;
2
PLASTIC MOMENT:
MP sybh
4
y
Problem 6.10-4 A steel beam of rectangular cross section is 40 mm wide and 80 mm high
(see figure). The yield stress of the steel is 210 MPa.
(a) What percent of the cross-sectional area is occupied by the elastic core if the beam is
subjected to a bending moment of 12.0 kNm acting about the z axis?
(b) What is the magnitude of the bending moment that will cause 50% of the cross section
to yield?
z
C
40 mm
Solution 6.10-4
sy 210 MPa
b 40 mm
h 80 mm
2e
56.7%
h
(a) ELASTIC CORE
M 12.0 kN # m
My MP 6
M
is between
(b) ELASTIC CORE
e
2
4
MP 13.4 k N # m
My
1 3
M
eh
a b
A2 2
My
and
;
sybh2
My 9.0 kN # m
sybh
Percent of cross-sectional area is
MP
e 22.678 mm
h
4
M My a
e 20 mm
3
2e2
2b
2
h
M 12.3 kN # m
;
80 mm
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SECTION 6.10 Elastoplastic Bending
Problem 6.10-5 Calculate the shape factor f for the wideflange beam shown in the figure if
h 12.2 in., b 8.08 in., tf 0.64 in., and tw 0.37 in.
y
tf
z
h
C
tw
tf
b
Probs. 6.10-5 and 6.10-6
Solution 6.10-5
h 12.2 in.
b 8.08 in.
PLASTIC MODULUS
tf 0.64 in.
tw 0.37 in.
1
Z [bh2 (b tw)(h 2tf)2]
4
Z 70.8 in.3
SECTION MODULUS
I
1 3
1
bh (b tw)(h 2tf)3
12
12
SHAPE FACTOR
I 386.0 in.4
c
h
2
f
c 6.1 in.
S
I
c
Z
S
f 1.12
;
S 63.3 in.3
Problem 6.10-6 Solve the preceding problem for a wide-flange beam with h 404 mm, b 140 mm, tf 11.2 mm,
and tw 6.99 mm.
Solution 6.10-6
h 404 mm
b 140 mm
PLASTIC MODULUS
tf 11.2 mm
tw 6.99 mm
1
Z [bh2 (b tw)(h 2t f)2]
4
Z 870.4 * 103 mm3
SECTION MODULUS
1 3
1
bh (b tw)(h 2tf)3
12
12
I 153.4 * 106 mm4
I
h
c
c 202.0 mm
2
S 759.2 * 103 mm3
I
S
c
SHAPE FACTOR
f
Z
S
f 1.15
;
561
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Problem 6.10-7 Determine the plastic modulus Z and shape factor f for a W 12 ⫻ 14 wide-flange beam. (Note: Obtain the
cross-sectional dimensions and section modulus of the beam from Table E-1a in Appendix E.)
Solution 6.10-7
W 12 ⫻ 14
SHAPE FACTOR
h ⫽ 11.9 in.
b ⫽ 3.97 in.
tw ⫽ 0.200 in.
S ⫽ 14.9 in.
tf ⫽ 0.225 in.
3
f⫽
Z
S
f ⫽ 1.14
;
PLASTIC MODULUS
1
Z ⫽ [bh2 ⫺ (b ⫺ t w)(h ⫺ 2t f)2]
4
Z ⫽ 16.98 in.3
;
Problem 6.10-8 Solve the preceding problem for a W 250 ⫻ 89 wide-flange beam. (Note: Obtain the cross-sectional
dimensions and section modulus of the beam from Table E-1b in Appendix E.)
Solution 6.10-8
SHAPE FACTOR
W 250 ⫻ 89
h ⫽ 259 mm
b ⫽ 257 mm
tw ⫽ 10.7 mm
S ⫽ 1090 ⫻ 103 mm3
tf ⫽ 17.3 mm
f⫽
Z
S
f ⫽ 1.11
;
PLASTIC MODULUS
Z⫽
1
[bh 2 ⫺ (b ⫺ tw)(h ⫺ 2tf)2]
4
Z ⫽ 1.209 * 106 mm3
;
Problem 6.10-9 Determine the yield moment MY, plastic moment MP, and shape factor f for a W 16 ⫻ 100 wide-flange
beam if sY ⫽ 36 ksi. (Note: Obtain the cross-sectional dimensions and section modulus of the beam from Table E-1a in
Appendix E.)
Solution 6.10-9
W 16 ⫻ 100
YIELD MOMENT
h ⫽ 17.0 in.
b ⫽ 10.4 in.
tf ⫽ 0.985 in.
tw ⫽ 0.585 in.
S ⫽ 175 in.
sy ⫽ 36 ksi
3
My ⫽ syS
My ⫽ 525 k-f t
;
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SECTION 6.10 Elastoplastic Bending
PLASTIC MODULUS
563
SHAPE FACTOR
1
Z ⫽ [bh2 ⫺(b ⫺ tw)(h ⫺2 tf)2]
4
Z ⫽ 197.1 in.3
f⫽
Z
S
f ⫽ 1.13
;
PLASTIC MOMENT
MP ⫽ syZ
MP ⫽ 591 k-ft
;
Problem 6.10-10 Solve the preceding problem for a W 410 ⫻ 85 wide-flange beam. Assume that sY ⫽ 250 MPa.
(Note: Obtain the cross-sectional dimensions and section modulus of the beam from Table E-1b in Appendix E.)
Solution 6.10-10
W 410 ⫻ 85
PLASTIC MOMENT
h ⫽ 417 mm
b ⫽ 181 mm
tw ⫽ 10.9 mm
S ⫽ 1510⭈10 mm
tf ⫽ 18.2 mm
3
MP ⫽ syZ
SHAPE FACTOR
f⫽
YIELD MOMENT
My ⫽ 378 kN # m
;
3
sy ⫽ 250 MPa
My ⫽ syS
MP ⫽ 427 kN # m
Z
S
f ⫽ 1.13
;
;
PLASTIC MODULUS
1
Z ⫽ cbh 2 ⫺(b ⫺ tw) (h ⫺ 2tf )2 d
4
Z ⫽ 1.708 * 106 mm3
Problem 6.10-11 A hollow box beam with height h ⫽ 16 in., width b ⫽ 8 in., and constant
wall thickness t ⫽ 0.75 in. is shown in the figure. The beam is constructed of steel with
yield stress sY ⫽ 32 ksi.
Determine the yield moment MY, plastic moment MP, and shape factor f.
y
t
z
C
t
Probs. 6.10-11 and 6.10-12
b
h
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Solution 6.10-11
h ⫽ 16 in.
Hollow box beam
b ⫽ 8 in.
t ⫽ 0.75 in.
PLASTIC MODULUS
Use (Eq. 6-86) with tw ⫽ 2t and tf ⫽ t:
sY ⫽ 32 ksi
SECTION MODULUS (S ⫽ I/c)
Z⫽
1
1
bh3 ⫺
(b ⫺ 2t)(h ⫺ 2t)3
12
12
⫽ 1079 in.4
⫽ 170.3 in.3
I⫽
c⫽
h
⫽ 8.0 in.
2
S⫽
1
[bh 2 ⫺(b ⫺2t)(h ⫺ 2t)2]
4
PLASTIC MOMENT (EQ. 6-77)
I
⫽ 134.9 in.3
c
Mp ⫽ sYZ ⫽ 5450 k-in.
;
SHAPE FACTOR (EQ. 6-79)
YIELD MOMENT (EQ. 6-74)
MY ⫽ sYS ⫽ 4320 k-in.
f⫽
;
MP
Z
⫽ ⫽ 1.26
MY
S
;
Problem 6.10-12 Solve the preceding problem for a box beam with dimensions h ⫽ 0.5 m, b ⫽ 0.18 m, and t ⫽ 22 mm.
The yield stress of the steel is 210 MPa.
Solution 6.10-12
h ⫽ 0.5 m
b ⫽ 0.18 m
t ⫽ 22 mm
PLASTIC MODULUS
1
[bh2 ⫺ (b ⫺ 2t)(h ⫺ 2t)2]
4
sy ⫽ 210 MPa
Z⫽
SECTION MODULUS
Z ⫽ 4.180 * 106 mm3
I⫽
1 3
1
bh ⫺
(b ⫺ 2t)(h ⫺ 2t)3
12
12
PLASTIC MOMENT
MP ⫽ syZ
I ⫽ 800.4 * 106 mm4
c⫽
h
2
I
S⫽c
SHAPE FACTOR
c ⫽ 250 mm
f⫽
S ⫽ 3.202 * 10 mm
6
3
YIELD MOMENT
My ⫽ syS
MP ⫽ 878 kN # m
My ⫽ 672 kN # m
;
Z
S
f ⫽ 1.31
;
;
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SECTION 6.10 Elastoplastic Bending
565
y
Problem 6.10-13 A hollow box beam with height h ⫽ 9.5 in., inside height h1 ⫽ 8.0 in.,
width b ⫽ 5.25 in., and inside width b1 ⫽ 4.5 in. is shown in the figure.
Assuming that the beam is constructed of steel with yield stress sY ⫽ 42 ksi,
calculate the yield moment MY, plastic moment MP, and shape factor f.
h1
z
C
h
b1
b
Probs. 6.10-13 through 6.10-16
Solution 6.10-13
h ⫽ 9.5 in.
b ⫽ 5.25 in.
b1 ⫽ 4.5 in.
sy ⫽ 42 ksi
h1 ⫽ 8.0 in.
PLASTIC MODULUS
1
Z ⫽ (bh2 ⫺ b1h12)
4
Z ⫽ 46.5 in.3
SECTION MODULUS
PLASTIC MOMENT
1
I ⫽ (bh3 ⫺ b1h13)
12
c⫽
h
2
I ⫽ 183.10 in.
MP ⫽ syZ
I
S⫽ c
SHAPE FACTOR
4
c ⫽ 4.75 in.
S ⫽ 38.55 in.3
f⫽
YIELD MOMENT
My ⫽ syS
My ⫽ 1619 k-in.
Z
S
MP ⫽ 1951 k-in.
f ⫽ 1.21
;
;
;
Problem 6.10-14 Solve the preceding problem for a box beam with dimensions h ⫽ 200 mm, h1 ⫽ 160 mm, b ⫽ 150 mm,
and b1 ⫽ 130 mm. Assume that the beam is constructed of steel with yield stress sY ⫽ 220 MPa.
Solution 6.10-14
h ⫽ 200 mm
Hollow box beam
b ⫽ 150 mm
h1 ⫽ 160 mm
b1 ⫽ 130 mm
PLASTIC MODULUS
sY ⫽ 220 MPa
Use (Eq. 6-86) with b ⫺ tw ⫽ b1 and h ⫺ 2tf ⫽ h1
Z⫽
SECTION MODULUS (S ⫽ I/c)
I⫽
1
(bh3 ⫺ b1h31) ⫽ 55.63 * 106 mm4
12
c⫽
h
⫽ 100 mm
2
I
S ⫽ c ⫽ 556.3 * 103 mm3
1
(bh 2 ⫺ b1h21) ⫽ 668.0 * 103 mm3
4
PLASTIC MOMENT (EQ. 6-77)
MP ⫽ sYZ ⫽ 147 kN # m
;
SHAPE FACTOR (EQ. 6-79)
YIELD MOMENT (EQ. 6-74)
MY ⫽ sYS ⫽ 122 kN # m
;
f⫽
MP
Z
⫽ ⫽ 1.20
MY
S
;
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Problem 6.10-15 The hollow box beam shown in the figure is subjected to a bending moment M of such magnitude that
the flanges yield but the webs remain linearly elastic.
(a) Calculate the magnitude of the moment M if the dimensions of the cross section are h ⫽ 15 in., h1 ⫽ 12.75 in.,
b ⫽ 9 in., and b1 ⫽ 7.5 in. Also, the yield stress is sY ⫽ 33 ksi.
Solution 6.10-15
h ⫽ 15 in.
b1 ⫽ 7.5 in.
b ⫽ 9 in.
h1 ⫽ 12.75 in.
sy ⫽ 33 ksi
1
(b ⫺ b1)h21
6
M1 ⫽ syS1
h + h1
b
2
M2 ⫽ 4636 k-in.
(a) BENDING MOMENT
ELASTIC CORE
S1 ⫽
M2 ⫽ Fa
S1 ⫽ 40.64 in.3
M ⫽ M1 + M2
M ⫽ 5977 k-in.
;
(b) PERCENT DUE TO ELASTIC CORE
M1 ⫽ 1341 k-in.
M1
⫽ 22.4%
M
PLASTIC FLANGES
;
F ⫽ force in one flange
1
F ⫽ syba b(h ⫺ h1)
2
F ⫽ 334.1 k
Problem 6.10-16 Solve the preceding problem for a box beam with dimensions h ⫽ 400 mm, h1 ⫽ 360 mm, b ⫽ 200 mm,
and b1 ⫽ 160 mm, and with yield stress sY ⫽ 220 MPa.
Solution 6.10-16
h ⫽ 400 mm
h1 ⫽ 360 mm
Hollow box beam
b ⫽ 200 mm
b1 ⫽ 160 mm
(a) BENDING MOMENT
sY ⫽ 220 MPa
(see Figure 6-47, Example 6-9)
ELASTIC CORE
1
S1 ⫽ (b ⫺ b1)h21 ⫽ 864 * 103 mm3
6
M1 ⫽ sYS1 ⫽ 190.1 kN # m
PLASTIC FLANGES
F ⫽ force in one flange
1
F ⫽ sYba b(h ⫺h1) ⫽ 880.0 kN
2
M2 ⫽ F a
h + h1
b ⫽ 334.4 kN # m
2
M ⫽ M1 ⫹ M2 ⫽ 524 kN⭈m
;
(b) PERCENT DUE TO ELASTIC CORE
Percent ⫽
M1
(100) ⫽ 36%
M
;
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SECTION 6.10 Elastoplastic Bending
567
Problem 6.10-17 A W 10 ⫻ 60 wide-flange beam is subjected to a bending moment M of such magnitude that the flanges
yield but the web remains linearly elastic.
(a) Calculate the magnitude of the moment M if the yield stress is sY ⫽ 36 ksi.
(b) What percent of the moment M is produced by the elastic core?
Solution 6.10-17
W 10 ⫻ 60
(a) BENDING MOMENT
h ⫽ 10.2 in.
b ⫽ 10.1 in.
tw ⫽ 0.420 in.
tf ⫽ 0.680 in.
sy ⫽ 36 ksi
M1 ⫽ syS1
;
(b) PERCENT DUE TO ELASTIC CORE
ELASTIC CORE
1
S1 ⫽ (h ⫺ 2tf ) 2tw
6
M ⫽ 2551 k-in.
M ⫽ M1 + M2
S1 ⫽ 5.47 in.
3
M1
⫽ 7.7%
M
;
M1 ⫽ 196.9 k-in.
PLASTIC FLANGES
F ⫽ force in one flange
F ⫽ 247.2 k
F ⫽ sybtf
M2 ⫽ 2354 k-in.
M2 ⫽ F(h ⫺ tf)
y
Problem 6.10-18 A singly symmetric beam of T-section (see figure) has cross-sectional
dimensions b ⫽ 140 mm, a ⫽ 190.8 mm, tw ⫽ 6.99 mm, and tf ⫽ 11.2 mm.
Calculate the plastic modulus Z and the shape factor f.
tw
a
z
O
tf
b
Solution 6.10-18
b ⫽ 140 mm
a ⫽ 190.8 mm
tw ⫽ 6.99 mm
tf ⫽ 11.2 mm
c2 ⫽
btf + atw
c1 ⫽ 149.98 mm
1
1
1
t c 3 + bc23 ⫺ (b ⫺ tw)(c2 ⫺ tf)3
3 w 1
3
3
Iz ⫽ 11.41 * 106 mm4
Iz ⫽
ELASTIC BENDING
tf
a
a b btf + a + tfbatw
2
2
c1 ⫽ a + t f ⫺ c2
c2 ⫽ 52.02 mm
S⫽
Iz
c1
S ⫽ 76.1 * 103 mm3
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CHAPTER 6 Stresses in Beams (Advanced Topics)
PLASTIC BENDING
A ⫽ btf + atw
h2 ⫽
A ⫽ 2902 mm
2
A
2b
h2 ⫽ 10.4 mm
h1 ⫽ a + tf ⫺ h2
y2
⫺bar
⫽ h2 /2
h1 ⫽ 191.6 mm
y2
⫺bar
⫽ 5.18 mm
1
1
(b ⫺ tw)(tf ⫺ h 2)2 + twh12
2
2
y1 bar ⫽
⫺
A /2
y1 bar ⫽ 88.50 mm
⫺
Z⫽
A
(y
+ y2 bar)
⫺
2 1⫺bar
Z ⫽ 136 * 103 mm3
f⫽
Z
S
;
f ⫽ 1.79
;
Problem 6.10-19 A wide-flange beam of unbalanced cross section has the dimensions
shown in the figure.
Determine the plastic moment MP if sY ⫽ 36 ksi.
y
10 in.
0.5 in.
z
O
7 in.
0.5 in.
0.5 in.
5 in.
Solution 6.10-19 Unbalanced wide-flange beam
NEUTRAL AXIS UNDER FULLY PLASTIC CONDITIONS
A
⫽ h1tw + (b1 ⫺ tw)tf
2
from which we get h1 ⫽ 1.50 in.
h2 ⫽ d ⫺ h1 ⫽ 8.50 in.
PLASTIC MODULUS
y1 ⫽
sY ⫽ 36 ksi
b1 ⫽ 10 in.
b2 ⫽ 5 in.
tw ⫽ 0.5 in.
d ⫽ 8 in.
d1 ⫽ 7 in.
tf ⫽ 0.5 in.
A ⫽ b1tf ⫹ b2tf ⫹ d1tw ⫽ 11.0 in.2
⫽
g y i Ai
A/2
(h1/ 2)(tw)(h1) + (h1 ⫺ tf / 2)(b1 ⫺ tw)(tf)
A/2
⫽ 1.182 in.
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SECTION 6.10 Elastoplastic Bending
y2 ⫽
⫽
g yi Ai
A /2
569
PLASTIC MOMENT
MP ⫽ sy Z ⫽ 1120 k-in.
;
(h2/2)(tw)(h2) + (h2 ⫺ tf/2)(b2 ⫺ tw)(tf)
A/2
⫽ 4.477 in.
Z⫽
A
(y + y2) ⫽ 31.12 in.3
2 1
Problem 6.10-20 Determine the plastic moment MP for a beam having
y
the cross section shown in the figure if sY ⫽ 210 MPa.
120
mm
z
150
mm
O
250 mm
30 mm
Solution 6.10-20 Cross section of beam
sY ⫽ 210 MPa
d2 ⫽ 150 mm
d1 ⫽ 120 mm
NEUTRAL AXIS FOR FULLY PLASTIC CONDITIONS
Cross section is divided into two equal areas.
A⫽
p
[(150 mm) 2 ⫺ (120 mm) 2]
4
+ (250 mm) (30 mm) ⫽ 13,862 mm2
A
⫽ 6931 mm2
2
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CHAPTER 6 Stresses in Beams (Advanced Topics)
(h2)(30 mm) ⫽
A
⫽ 6931 mm2
2
h2 ⫽ 231.0 mm
(Dimensions are in millimeters)
p
Z ⫽ (h1 ⫺ 75)a b(d22 ⫺ d12)
4
h1 ⫽ 150 mm + 250 mm ⫺ h2 ⫽ 169.0 mm
+ ca
PLASTIC MODULUS
+ a
g yi Ai
for upper half of cross section
y1 ⫽
A/ 2
g yi Ai
y2 ⫽
for lower half of cross section
A/2
Z⫽
A
( y + y 2) ⫽ (g yi A i)upper + ( g yi Ai)lower
2 1
h1 ⫺ 150
b(30)(h1 ⫺150) d
2
h2
b(30)(h2)
2
⫽ 598,000 + 5,400 + 800,400
⫽ 1404 * 103 mm3
PLASTIC MOMENT
MP ⫽ sPZ ⫽ (210 MPa)(1404 ⫻ 103 mm3)
⫽ 295 kN # m
;
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Page 571
7
Analysis of Stress
and Strain
Plane Stress
1200 psi
Problem 7.2-1 An element in plane stress is subjected to stresses sx ⫽ 4750 psi,
sy ⫽ 1200 psi, and txy ⫽ 950 psi, as shown in the figure.
Determine the stresses acting on an element oriented at an angle u ⫽ 60˚ from the
x axis, where the angle u is positive when counterclockwise. Show these stresses on a
sketch of an element oriented at the angle u.
950 psi
4750 psi
Solution 7.2-1
sx ⫽ 4750 psi
sy ⫽ 1200 psi
txy ⫽ 950 psi
u ⫽ 60°
sx1 ⫽
sx ⫺ sy
sx + sy
+
2
sx1 ⫽ 2910 psi
2
;
tx1y1 ⫽ ⫺
sx ⫺ sy
2
sin(2u) + txy cos(2u)
tx1y1 ⫽ ⫺2012 psi
cos(2u) + txy sin(2u)
;
sy1 ⫽ sx + sy ⫺ sx1
sy1 ⫽ 3040 psi
Problem 7.2-2 Solve the preceding problem for an element in plane stress
subjected to stresses sx ⫽ 100 MPa, sy ⫽ 80 MPa, and txy ⫽ 28 MPa, as shown in
the figure.
Determine the stresses acting on an element oriented at an angle u ⫽ 30˚ from
the x axis, where the angle u is positive when counterclockwise. Show these stresses
on a sketch of an element oriented at the angle u.
;
80 MPa
28 MPa
100 MPa
571
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CHAPTER 7
Page 572
Analysis of Stress and Strain
Solution 7.2-2
sx ⫽ 100 MPa
sy ⫽ 80 MPa
txy ⫽ 28 MPa
u ⫽ 30°
sx1 ⫽
sx ⫺ sy
sx + sy
+
2
2
sx1 ⫽ 119.2 MPa
tx1y1 ⫽ ⫺
sx ⫺ sy
2
sin(2u) + txy cos(2u)
tx1y1 ⫽ 5.30 MPa
cos (2u) + txy sin (2u)
;
sy1 ⫽ sx + sy ⫺ sx1
sy1 ⫽ 60.8 MPa
;
;
Problem 7.2-3 Solve Problem 7.2-1 for an element in plane stress subjected to
2300 psi
stresses sx ⫽ ⫺5700 psi, sy ⫽ ⫺2300 psi, and txy ⫽ 2500 psi, as shown in the figure.
Determine the stresses acting on an element oriented at an angle u ⫽ 50˚ from the x
axis, where the angle u is positive when counterclockwise. Show these stresses on a sketch
of an element oriented at the angle u.
2500 psi
5700 psi
Solution 7.2-3
sx ⫽ ⫺5700 psi
sy ⫽ ⫺2300 psi
txy ⫽ 2500 psi
u ⫽ 50°
sx1 ⫽
sx ⫺ sy
sx + sy
+
2
sx1 ⫽ ⫺1243 psi
2
tx1y1 ⫽ ⫺
sx ⫺ sy
2
sin (2u) + txy cos (2u)
tx1y1 ⫽ 1240 psi
cos (2u) + txy sin (2u)
;
;
sy1 ⫽ sx + sy ⫺ sx1
sy1 ⫽ ⫺6757 psi
;
Problem 7.2-4 The stresses acting on element A in the web of
a train rail are found to be 40 MPa tension in the horizontal
direction and 160 MPa compression in the vertical direction
(see figure). Also, shear stresses of magnitude 54 MPa act in the
directions shown.
Determine the stresses acting on an element oriented at a
counterclockwise angle of 52˚ from the horizontal. Show these
stresses on a sketch of an element oriented at this angle.
160 MPa
A
A
Side
View
40 MPa
54 MPa
Cross
Section
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Page 573
SECTION 7.2
Plane Stress
573
Solution 7.2-4
sx ⫽ 40 MPa
sy ⫽ ⫺160 MPa
txy ⫽ ⫺54 MPa
u ⫽ 52°
sx1 ⫽
sx ⫺ sy
sx + sy
+
2
2
sx1 ⫽ ⫺136.6 MPa
tx1y1 ⫽ ⫺
sx ⫺ sy
2
sin (2u) + txy cos (2u)
tx1y1 ⫽ ⫺84.0 MPa
cos(2u) + txy sin(2u)
;
sy1 ⫽ sx + sy ⫺ sx1
sy1 ⫽ 16.6 MPa
;
;
Problem 7.2-5 Solve the preceding problem if the normal
and shear stresses acting on element A are 6500 psi, 18,500 psi,
and 3800 psi (in the directions shown in the figure).
Determine the stresses acting on an element oriented at a
counterclockwise angle of 30˚ from the horizontal. Show these
stresses on a sketch of an element oriented at this angle.
18,500 psi
6500 psi
A
A
3800 psi
Side
View
Cross
Section
Solution 7.2-5
sx ⫽ 6500 psi
sy ⫽⫺18500 psi
txy ⫽ ⫺3800 psi
u ⫽ 30°
sx1 ⫽
sx ⫺ sy
sx + sy
+
2
sx1 ⫽ ⫺3041 psi
2
;
tx1y1 ⫽ ⫺
sx ⫺ sy
2
sin (2u) + txy cos (2u)
tx1y1 ⫽ ⫺12725 psi
cos (2u) + txy sin (2u)
;
sy1 ⫽ sx + sy ⫺ sx1
sy1 ⫽ ⫺8959 psi
Problem 7.2-6 An element in plane stress from the fuselage of an airplane is subjected
to compressive stresses of magnitude 27 MPa in the horizontal direction and tensile
stresses of magnitude 5.5 MPa in the vertical direction (see figure). Also, shear stresses
of magnitude 10.5 MPa act in the directions shown.
Determine the stresses acting on an element oriented at a clockwise angle of 35˚
from the horizontal. Show these stresses on a sketch of an element oriented at this
angle.
;
5.5 MPa
27 MPa
10.5 MPa
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CHAPTER 7
Page 574
Analysis of Stress and Strain
Solution 7.2-6
sx ⫽ ⫺27 MPa
sy ⫽ 5.5 MPa
txy ⫽ ⫺10.5 MPa
u ⫽ ⫺35°
sx1 ⫽
sx ⫺ sy
sx + sy
+
2
2
sx1 ⫽ ⫺6.4 MPa
tx1y1 ⫽ ⫺
sx ⫺ sy
2
sin (2u) + txy cos (2u)
tx1y1 ⫽ ⫺18.9 MPa
cos (2u) + txy sin (2u)
;
sy1 ⫽ sx ⫹sy ⫺ sx1
sy1 ⫽ ⫺15.1 MPa
;
;
Problem 7.2-7 The stresses acting on element B in the
web of a wide-flange beam are found to be 14,000 psi compression in the horizontal direction and 2600 psi compression in the vertical direction (see figure). Also, shear stresses
of magnitude 3800 psi act in the directions shown.
Determine the stresses acting on an element oriented at a
counterclockwise angle of 40˚ from the horizontal. Show
these stresses on a sketch of an element oriented at this angle.
2600 psi
B
14,000 psi
B
3800 psi
Side
View
Cross
Section
Solution 7.2-7
sx ⫽ ⫺14000 psi
txy ⫽ ⫺3800 psi
sy ⫽ ⫺2600 psi
u ⫽ 40°
sx1 ⫽
tx1y1 ⫽ ⫺
sx ⫺ sy
2
tx1y1 ⫽ 4954 psi
sx ⫺ sy
sx + sy
+
2
sx1 ⫽ ⫺13032 psi
2
cos (2u) + txy sin (2u)
sin (2u) + txy cos (2u)
;
sy1 ⫽ sx + sy ⫺ sx1
sy1 ⫽ ⫺3568 psi
;
;
Problem 7.2-8 Solve the preceding problem if the normal and shear stresses acting on
13 MPa
element B are 46 MPa, 13 MPa, and 21 MPa (in the directions shown in the figure) and
the angle is 42.5˚ (clockwise).
21 MPa
B
46 MPa
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Page 575
SECTION 7.2
Plane Stress
Solution 7.2-8
sx ⫽ ⫺46 MPa
sy ⫽ ⫺13 MPa
txy ⫽ 21 MPa
u ⫽ ⫺42.5°
sx1 ⫽
sx ⫺ sy
sx + sy
+
2
2
sx1 ⫽ ⫺51.9 MPa
tx1y1 ⫽ ⫺
;
;
y
112 psi
30°
350 psi
O
x
120 psi
Seam
Plane stress (angle ␪ )
sx ⫽ 350 psi
sy ⫽112 psi
txy ⫽ ⫺120 psi
u ⫽ 30°
sx ⫺ sy
sx + sy
+
2
2
⫽ 187 psi
tx1y1 ⫽ ⫺
sin (2u) + txy cos (2u)
sy1 ⫽ sx + sy ⫺ sx1
sy1 ⫽ ⫺7.1 MPa
;
pond is subjected to stresses sx ⫽ 350 psi, sy ⫽112 psi,
and txy ⫽ ⫺120 psi, as shown by the plane-stress
element in the first part of the figure.
Determine the normal and shear stresses acting on a
seam oriented at an angle of 30° to the element, as
shown in the second part of the figure. Show these
stresses on a sketch of an element having its sides
parallel and perpendicular to the seam.
sx1 ⫽
2
tx1y1 ⫽ ⫺14.6 MPa
cos (2u) + txy sin (2u)
Problem 7.2-9 The polyethylene liner of a settling
Solution 7.2-9
sx ⫺ sy
sx ⫺ sy
2
⫽ ⫺163 psi
cos 2u + txy sin 2u
;
sin 2u + txy cos 2u
;
sy1 ⫽ sx + sy ⫺ sx1 ⫽ 275 psi
;
The normal stress on the seam equals 187 psi
;
tension.
The shear stress on the seam equals 163 psi, acting
clockwise against the seam.
;
575
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CHAPTER 7
Page 576
Analysis of Stress and Strain
Problem 7.2-10 Solve the preceding problem if the normal
y
and shear stresses acting on the element are sx ⫽ 2100 kPa,
sy ⫽ 300 kPa, and txy ⫽ ⫺560 kPa, and the seam is oriented
at an angle of 22.5° to the element (see figure).
300 kPa
22.5°
2100 kPa
x
O
Seam
560 kPa
Solution 7.2-10
Plane stress (angle ␪ )
sx1 ⫽
sx ⫺sy
sx + sy
+
2
2
⫽ 1440 kPa
tx1 y1 ⫽ ⫺
cos 2u + txy sin 2u
;
sx ⫺ sy
2
⫽ ⫺1030 kPa
sin 2u + txy cos 2u
;
sy1 ⫽ sx + sy ⫺ sx1 ⫽ 960 kPa
sx ⫽ 2100 kPa
u ⫽ 22.5°
sy ⫽ 300 kPa
txy ⫽⫺560 kPa
;
The normal stress on the seam equals 1440 kPa
tension. ;
The shear stress on the seam equals 1030 kPa, acting
clockwise against the seam. ;
Problem 7.2-11 A rectangular plate of dimensions 3.0 in. * 5.0 in. is formed
by welding two triangular plates (see figure). The plate is subjected to a tensile
stress of 500 psi in the long direction and a compressive stress of 350 psi in the
short direction.
Determine the normal stress sw acting perpendicular to the line of the weld
and the shear tw acting parallel to the weld. (Assume that the normal stress sw
is positive when it acts in tension against the weld and the shear stress tw is
positive when it acts counterclockwise against the weld.)
350 psi
ld
We
3 in.
5 in.
500 psi
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Page 577
SECTION 7.2
Solution 7.2-11
Plane Stress
577
Biaxial stress (welded joint)
tx1y1 ⫽ ⫺
sx ⫺ sy
2
sin 2u + txy cos 2u ⫽ ⫺ 375 psi
sy1 ⫽ sx + sy ⫺ sx1 ⫽ ⫺125 psi
STRESSES ACTING ON THE WELD
sx ⫽ 500 psi
u ⫽ arctan
sx1 ⫽
sy ⫽ ⫺350 psi
txy ⫽ 0
sw ⫽ ⫺125 psi
3 in.
⫽ arctan 0.6 ⫽ 30.96°
5 in.
sx ⫺ sy
sx + sy
+
2
2
tw ⫽ 375 psi
;
;
cos 2u + txy sin 2u
⫽ 275 psi
12.0 MPa
Problem 7.2-12 Solve the preceding problem for a plate of dimensions
100 mm * 250 mm subjected to a compressive stress of 2.5 MPa in the
long direction and a tensile stress of 12.0 MPa in the short direction
(see figure).
Solution 7.2-12
ld
We
100 mm
250 mm
Biaxial stress (welded joint)
tx1y1 ⫽ ⫺
sx ⫺ sy
2
sin 2u + txy cos 2u ⫽ 5.0 MPa
sy1 ⫽ sx + sy ⫺ sx1 ⫽ 10.0 MPa
STRESSES ACTING ON THE WELD
sx ⫽ ⫺2.5 MPa
u ⫽ arctan
sx1 ⫽
sy ⫽ 12.0 MPa
txy ⫽ 0
100 mm
⫽ arctan 0.4 ⫽ 21.80°
250 mm
sx ⫺ sy
sx + sy
+
2
⫽ ⫺0.5 MPa
2.5 MPa
2
cos 2u + txy sin 2u
sw ⫽ 10.0 MPa
;
tw ⫽ ⫺5.0 MPa
;
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CHAPTER 7
Page 578
Analysis of Stress and Strain
Problem 7.2-13 At a point on the surface of a machine the material
y
is in biaxial stress with sx ⫽ 3600 psi, and sy ⫽ ⫺1600 psi, as
shown in the first part of the figure. The second part of the figure
shows an inclined plane aa cut through the same point in the material
but oriented at an angle u.
Determine the value of the angle u between zero and 90° such
that no normal stress acts on plane aa. Sketch a stress element
having plane aa as one of its sides and show all stresses acting on
the element.
1600 psi
a
u
3600 psi
O
x
a
Solution 7.2-13
Biaxial stress
STRESS ELEMENT
sx1 ⫽ 0
u ⫽ 56.31°
sy1 ⫽ sx + sy ⫺ sx1 ⫽ 2000 psi
sx ⫽ 3600 psi
sy ⫽ ⫺1600 psi
txy ⫽ 0
tx1y1 ⫽ ⫺
sx ⫺ sy
2
;
sin 2u + txy cos 2u
⫽ ⫺2400 psi
Find angle u for s ⫽ 0.
s ⫽ normal stress on plane a-a
sx1 ⫽
sx ⫺ sy
sx + sy
+
2
2
cos 2u + txy sin 2u
⫽ 1000 + 2600 cos 2u(psi)
For sx1 ⫽ 0, we obtain cos 2u ⫽ ⫺
‹
2u ⫽112.62° and
1000
2600
u ⫽ 56.31°
Problem 7.2-14 Solve the preceding problem for sx ⫽ 32 MPa
y
and sy ⫽ ⫺50 MPa (see figure).
50 MPa
a
u
32 MPa
O
x
a
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Page 579
SECTION 7.2
Solution 7.2-14
Plane Stress
579
Biaxial stress
STRESS ELEMENT
sx1 ⫽ 0
u ⫽ 38.66°
sy1 ⫽ sx + sy ⫺ sx1 ⫽ ⫺18 MPa
sx ⫽ 32 MPa
tx1y1 ⫽ ⫺
sy ⫽ ⫺50 MPa
txy ⫽ 0
s x ⫺ sy
2
sin 2u + txy cos 2u
⫽ ⫺40 MPa
Find angles u for s ⫽ 0.
;
;
s ⫽ normal stress on plane a-a
sx1 ⫽
sx ⫺ sy
sx + sy
+
2
2
cos 2u + txy sin 2u
⫽ ⫺9 + 41 cos 2u ( MPa)
For sx1 ⫽ 0, we obtain cos 2u ⫽
‹
9
41
2u ⫽ 77.32° and u ⫽ 38.66°
;
Problem 7.2-15 An element in plane stress from the frame of a racing car is
y
oriented at a known angle u (see figure). On this inclined element, the normal and
shear stresses have the magnitudes and directions shown in the figure.
Determine the normal and shear stresses acting on an element whose sides
are parallel to the xy axes, that is, determine sx, sy, and txy. Show the results on
a sketch of an element oriented at u ⫽ 0˚.
2475 psi
3950 psi
u = 40°
14,900 psi
O
Solution 7.2-15
Transform from u ⫽ 40°
sx ⫽ ⫺14900 psi
txy ⫽ 2475 psi
to
u ⫽ 0°
sy ⫽ ⫺3950 psi
sx ⫺ sy
2
sin (2u) + txy cos (2u)
tx1y1 ⫽ ⫺4962 psi
u ⫽ ⫺40°
sx1 ⫽
tx1y1 ⫽ ⫺
;
sy1 ⫽ sx + sy ⫺ sx1
sx ⫺ sy
sx + sy
+
2
sx1 ⫽ ⫺12813 psi
2
;
cos (2u) + txy sin (2u)
sy1 ⫽ ⫺6037 psi
;
x
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CHAPTER 7
Page 580
Analysis of Stress and Strain
Problem 7.2-16 Solve the preceding problem for the element shown in the
y
figure.
24.3 MPa
62.5 MPa
u = 55°
O
x
24.0 MPa
Solution 7.2-16
Transform from u ⫽ 55°
sx ⫽ ⫺24.3 MPa
txy ⫽ ⫺24 MPa
to
u ⫽ 0°
sy ⫽ 62.5 MPa
sx ⫺ sy
2
sin (2u) + txy cos (2u)
tx1y1 ⫽ ⫺32.6 MPa
u ⫽ ⫺55°
sx1 ⫽
tx1y1 ⫽ ⫺
;
sy1 ⫽ sx + sy ⫺ sx1
sx ⫺ sy
sx + sy
+
2
2
sx1 ⫽ 56.5 MPa
cos (2u) + txy sin (2u)
sy1 ⫽ ⫺18.3 MPa
;
;
Problem 7.2-17 A plate in plane stress is subjected to normal stresses
sx and sy and shear stress txy, as shown in the figure. At counterclockwise angles
u ⫽ 35˚ and u ⫽ 75˚ from the x axis, the normal stress is 4800 psi tension.
If the stress sx equals 2200 psi tension, what are the stresses sy and txy?
y
sy
txy
O
sx = 2200 psi
x
Solution 7.2-17
sx ⫽ 2200 psi
At u ⫽35°
sy unknown
txy unknown
and u ⫽ 75°, sx1 ⫽ 4800 psi
Find sy and txy
sx1 ⫽
sx ⫺ sy
sx + sy
+
2
2
For
u ⫽ 35°
sx1 ⫽ 4800 psi
4800 psi ⫽
cos (2u) + txy sin (2u)
or
2200 psi ⫺ sy
2200 psi + sy
+
2
2
* cos (70°) + txy sin (70°)
0.32899 sy + 0.93969 txy ⫽ 3323.8 psi
(1)
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Page 581
Plane Stress
581
0.93301sy + 0.50000 txy ⫽ 4652.6 psi
(2)
SECTION 7.2
For
u ⫽ 75°:
or
sx1 ⫽ 4800 psi
4800 psi ⫽
Solve Eqs. (1) and (2):
2200 psi ⫺ sy
2200 psi + sy
sy ⫽ 3805 psi
txy ⫽ 2205 psi
;
+
2
2
* cos (150°) + txy sin (150°)
Problem 7.2-18 The surface of an airplane wing is subjected to plane stress
y
with normal stresses sx and sy and shear stress txy, as shown in the figure. At a
counterclockwise angle u ⫽ 32˚ from the x axis, the normal stress is 37 MPa
tension, and at an angle u ⫽ 48˚, it is 12 MPa compression.
If the stress sx equals 110 MPa tension, what are the stresses sy and txy?
sy
txy
O
sx = 110 MPa
x
Solution 7.2-18
sx ⫽ 110 MPa
sy unknown
At u ⫽ 32°, sx1 ⫽ 37 MPa
sxy unknown
(tension)
At u ⫽ 48°, sx1 ⫽ ⫺12 MPa
(compression)
Find sy and txy
sx1 ⫽
For
sx ⫺sy
sx + sy
+
2
2
cos(2u) + txy sin (2u)
u ⫽ 32°
37 MPa ⫽
For
0.28081sy + 0.89879txy ⫽ ⫺42.11041 MPa (1)
u ⫽ 48°:
sx1 ⫽ ⫺12 MPa
110 MPa + sy
110 MPa ⫺ sy
⫺12 MPa ⫽
+
2
2
* cos (96°) + txy sin (96°)
or
0.55226sy + 0.99452txy ⫽ ⫺61.25093 MPa (2)
Solve Eqs. (1) and (2):
sx1 ⫽ 37 MPa
110 MPa + sy
or
sy ⫽ ⫺60.7 MPa
110 MPa ⫺ sy
+
2
2
* cos (64°) + txy sin (64°)
txy ⫽ ⫺27.9 MPa
;
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Analysis of Stress and Strain
Problem 7.2-19 At a point in a structure subjected to plane stress, the stresses are
y
sx ⫽ ⫺4100 psi, sy ⫽ 2200 psi, and txy ⫽ 2900 psi (the sign convention for these
stresses is shown in Fig. 7-1). A stress element located at the same point in the structure
(but oriented at a counterclockwise angle u1 with respect to the x axis) is subjected to
the stresses shown in the figure (sb, tb, and 1800 psi).
Assuming that the angle u1 is between zero and 90˚, calculate the normal stress sb, the
shear stress tb, and the angle u1
Solution 7.2-19
sx ⫽ ⫺4100 psi
txy ⫽ 2900 psi
For
u ⫽ u1:
Find
Stress
sy1 ⫽ sb
1800 psi u1
O
x
SOLVE NUMERICALLY:
tx1y1 ⫽ tb
2u1 ⫽ 87.32°
sb, tb, and u1
sb
u1 ⫽ 43.7°
;
Shear Stress tb
sb ⫽ sx + sy ⫺1800 psi
sb ⫽ ⫺3700 psi
;
tb ⫽ ⫺
Angle u1
sx1 ⫽
tb
1800 psi ⫽ ⫺950 psi ⫺ 3150 psi cos12u12
+ 2900 psi sin 12u12
sy ⫽ 2200 psi
sx1 ⫽ 1800 psi
sb
sx ⫺ sy
sx + sy
+
2
2
cos ( 2u) + txy sin ( 2u)
sx ⫺ sy
2
sin 12u12 + txy cos 12u12
tb ⫽ 3282 psi
;
Principal Stresses and Maximum Shear Stresses
When solving the problems for Section 7.3, consider only the in-plane stresses (the stresses in the xy plane).
Problem 7.3-1 An element in plane stress is subjected to stresses sx ⫽ 4750 psi, sy ⫽ 1200 psi, and txy ⫽ 950 psi (see the
figure for Problem 7.2-1).
Determine the principal stresses and show them on a sketch of a properly oriented element.
Solution 7.3-1
sx ⫽ 4750 psi
sy ⫽ 1200 psi
atana
up1 ⫽
sx ⫺ sy
up1 ⫽ 14.08°
2
up2 ⫽ up1 + 90°
s1 ⫽
PRINCIPAL STRESSES
2 txy
txy ⫽ 950 psi
b
s2 ⫽
up2 ⫽ 104.08°
sx ⫺ sy
sx + sy
+
2
2
sx ⫺ sy
sx + sy
+
2
s1 ⫽ 4988 psi
s2 ⫽ 962 psi
2
;
;
cos 12up12 + txy sin 12up12
cos 12up22 + txy sin 12up22
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SECTION 7.3 Principal Stresses and Maximum Shear Stresses
583
Problem 7.3-2 An element in plane stress is subjected to stresses sx ⫽ 100 MPa, sy ⫽ 80 MPa, and txy ⫽ 28 MPa (see the
figure for Problem 7.2-2).
Determine the principal stresses and show them on a sketch of a properly oriented element.
Solution 7.3-2
sx ⫽ 100 MPa
sy ⫽ 80 MPa
txy ⫽ 28 MPa
s1 ⫽
PRINCIPAL STRESSES
up1 ⫽
atan a
2 txy
sx ⫺ sy
s2 ⫽
b
+
2
2
sx ⫺ sy
sx + sy
+
2
2
s1 ⫽ 120 MPa
2
s2 ⫽ 60 MPa
up1 ⫽ 35.2°
up2 ⫽ up1 + 90°
sx ⫺ sy
sx + sy
cos12up12 + txy sin12up12
cos12up22 + txy sin12up22
;
;
up2 ⫽ 125.17°
Problem 7.3-3 An element in plane stress is subjected to stresses sx ⫽ ⫺5700 psi, sy ⫽ ⫺2300 psi, and txy ⫽ 2500 psi
(see the figure for Problem 7.2-3).
Determine the principal stresses and show them on a sketch of a properly oriented element.
Solution 7.3-3
sx ⫽ ⫺5700 psi
sy ⫽ ⫺2300 psi
txy ⫽ 2500 psi
s1 ⫽
PRINCIPAL STRESSES
up2 ⫽
atan a
2txy
sx ⫺ sy
s2 ⫽
b
+
2
2
sx ⫺ sy
sx + sy
+
2
s1 ⫽ ⫺ 977 psi
2
s2 ⫽ ⫺ 7023 psi
up2 ⫽ ⫺27.89°
up1 ⫽ up2 + 90°
sx ⫺ sy
sx + sy
2
cos12up12 + txy sin12up12
cos12up22 + txy sin12up22
;
;
up1 ⫽ 62.1°
Problem 7.3-4 The stresses acting on element A in the web of a train rail are found to be 40 MPa tension in the horizontal
direction and 160 MPa compression in the vertical direction (see figure). Also, shear stresses of magnitude 54 MPa act in the
directions shown (see the figure for Problem 7.2-4).
Determine the principal stresses and show them on a sketch of a properly oriented element.
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CHAPTER 7 Analysis of Stress and Strain
Solution 7.3-4
sx ⫽ 40 MPa
sy ⫽ ⫺160 MPa
txy ⫽ ⫺54 MPa
s1 ⫽
PRINCIPAL STRESSES
up1 ⫽
atan a
2txy
sx ⫺ sy
s2 ⫽
b
sx ⫺ sy
sx + sy
+
2
2
sx ⫺ sy
sx + sy
+
2
2
s1 ⫽ 53.6 MPa
2
s2 ⫽ ⫺ 173.6 MPa
up1 ⫽ ⫺ 14.2°
up2 ⫽ up1 + 90°
cos 12up12 + txy sin 12up12
cos 12up22 + txy sin 12up22
;
;
up2 ⫽ 75.8°
Problem 7.3-5 The normal and shear stresses acting on element A are 6500 psi, 18,500 psi, and 3800 psi (in the directions
shown in the figure) (see the figure for Problem 7.2-5).
Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented
element.
Solution 7.3-5
sx ⫽ 6500 psi
sy ⫽ ⫺18500 psi
txy ⫽ ⫺3800 psi
s2 ⫽ ⫺19065 psi
PRINCIPAL ANGLES
up1 ⫽
atan a
2txy
sx ⫺ sy
MAXIMUM SHEAR STRESSES
b
tmax ⫽
2
up1 ⫽ ⫺8.45°
s2 ⫽
up2 ⫽ 81.55°
sx ⫺ sy
sx + sy
+
2
2
sx ⫺ sy
sx + sy
+
2
sx ⫺ sy 2
a
b + txy2
A
2
us1 ⫽ up1 ⫺ 45°
up2 ⫽ up1 + 90°
s1 ⫽
s1 ⫽ 7065 psi
2
cos 12up12 + txy sin 12up12
saver ⫽
sx + sy
2
tmax ⫽ 13065 psi
us1 ⫽ ⫺53.4°
;
;
saver ⫽ ⫺6000 psi
;
cos 12up22 + txy sin 12up22
Problem 7.3-6 An element in plane stress from the fuselage of an airplane is subjected to compressive stresses of
magnitude 27 MPa in the horizontal direction and tensile stresses of magnitude 5.5 MPa in the vertical direction. Also,
shear stresses of magnitude 10.5 MPa act in the directions shown (see the figure for Problem 7.2-6).
Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented
element.
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SECTION 7.3 Principal Stresses and Maximum Shear Stresses
585
Solution 7.3-6
sx ⫽ ⫺27 MPa
sy ⫽ 5.5 MPa
txy ⫽ ⫺10.5 MPa
s2 ⫽ ⫺30.1 MPa
PRINCIPAL ANGLES
2txy
b
atan a
sx ⫺ sy
up2 ⫽
2
MAXIMUM SHEAR STRESSES
tmax ⫽
up2 ⫽ 16.43°
up1 ⫽ up2 + 90°
s1 ⫽
s2 ⫽
up1 ⫽ 106.43°
sx ⫺ sy
sx + sy
+
2
2
sx ⫺ sy
sx + sy
+
2
s1 ⫽ 8.6 MPa
2
cos 12up12 + txy sin 12up12
sx ⫺ sy 2
b + txy2
A
2
a
tmax ⫽ 19.3 MPa
;
us1 ⫽ up1 ⫺ 45°
saver ⫽
us1 ⫽ 61.4°
sx + sy
saver ⫽ ⫺10.8 MPa
2
;
cos 12up22 + txy sin 12up22
Problem 7.3-7 The stresses acting on element B in the web of a wide-flange beam are found to be 14,000 psi compression
in the horizontal direction and 2600 psi compression in the vertical direction. Also, shear stresses of magnitude 3800 psi act
in the directions shown (see the figure for Problem 7.2-7).
Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented
element.
Solution 7.3-7
sx ⫽ ⫺14000 psi
txy ⫽ ⫺3800 psi
sy ⫽ ⫺2600 psi
up2 ⫽
2txy
sx ⫺ sy
b
s1 ⫽
tmax ⫽
up1 ⫽ 106.85°
sx ⫺ sy
+
2
2
2
cos12up22 + txy sin12up22
MAXIMUM SHEAR STRESSES
up2 ⫽ 16.85°
sx + sy
+
s2 ⫽ ⫺15151 psi
2
up1 ⫽ up2 + 90°
sx ⫺ sy
sx + sy
s1 ⫽ ⫺1449 psi
PRINCIPAL ANGLES
atan a
s2 ⫽
2
cos12up12 + txy sin12up12
sx ⫺ sy 2
a
b + txy2
A
2
us1 ⫽ up1 ⫺ 45°
saver ⫽
sx + sy
2
tmax ⫽ 6851 psi
us1 ⫽ 61.8°
saver ⫽ ⫺ 8300 psi
;
;
;
Problem 7.3-8 The normal and shear stresses acting on element B are sx ⫽ ⫺46 MPa, sy ⫽ ⫺13 MPa, and txy ⫽ 21 MPa
(see figure for Problem 7.2-8).
Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly oriented
element.
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Page 586
Analysis of Stress and Strain
Solution 7.3-8
sx ⫽ ⫺46 MPa
sy ⫽ ⫺13 MPa
txy ⫽ 21 MPa
s2 ⫽ ⫺56.2 MPa
PRINCIPAL ANGLES
up2 ⫽
atan a
2txy
sx ⫺ sy
MAXIMUM SHEAR STRESSES
b
tmax ⫽
2
up1 ⫽ up2 + 90°
s1 ⫽
s2 ⫽
up1 ⫽ 64.08°
sx ⫺ sy
+
2
2
sx ⫺ sy
sx + sy
a
sx ⫺ sy
A
2
us1 ⫽ up1 ⫺ 45°
up2 ⫽ ⫺25.92°
sx + sy
s1 ⫽ ⫺2.8 MPa
+
2
2
cos 12up12 + txy sin 12up12
saver ⫽
sx ⫺ sy
2
2
b + txy2
tmax ⫽ 26.7 MPa
us1 ⫽ 19.08°
;
;
saver ⫽ ⫺29.5 MPa
;
cos 12up22 + txy sin 12up22
Problem 7.3-9 A shear wall in a reinforced concrete building is
subjected to a vertical uniform load of intensity q and a horizontal
force H, as shown in the first part of the figure. (The force H
represents the effects of wind and earthquake loads.) As a
consequence of these loads, the stresses at point A on the surface
of the wall have the values shown in the second part of the
figure (compressive stress equal to 1100 psi and shear stress equal
to 480 psi).
q
1100 psi
H
480 psi
A
A
(a) Determine the principal stresses and show them on a sketch
of a properly oriented element.
(b) Determine the maximum shear stresses and associated
normal stresses and show them on a sketch of a properly oriented element.
Solution 7.3-9
sx ⫽ 0
Shear wall
sy ⫽ ⫺1100 psi
txy ⫽ ⫺480 psi
2txy
sx ⫺ sy
⫽ ⫺0.87273
2up ⫽ ⫺41.11° and up ⫽ ⫺20.56°
2up ⫽ 138.89° and up ⫽ 69.44°
sx1 ⫽
f
s2 ⫽ ⫺1280 psi and up2 ⫽ 69.44°
(a) PRINCIPAL STRESSES
tan 2up ⫽
Therefore, s1 ⫽180 psi and up1 ⫽ ⫺20.56°
sx ⫺ sy
sx + sy
+
2
2
cos 2u + txy sin 2u
For 2up ⫽ ⫺41.11°:
sx1 ⫽ 180 psi
For 2up ⫽ 138.89°:
sx1 ⫽ ⫺1280 psi
;
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SECTION 7.3
587
Principal Stresses and Maximum Shear Stresses
(b) MAXIMUM SHEAR STRESSES
tmax ⫽
sx ⫺ sy 2
b + t2xy ⫽ 730 psi
A
2
a
us1 ⫽ up1⫺ 45° ⫽ ⫺ 65.56° and t ⫽ 730 psi
us2 ⫽ up1+ 45° ⫽ 24.44°
saver ⫽
sx + sy
and t ⫽ ⫺730 psi
⫽ ⫺550 psi
2
f
;
;
Problem 7.3-10 A propeller shaft subjected to combined torsion and axial thrust
is designed to resist a shear stress of 56 MPa and a compressive stress of 85 MPa
(see figure).
(a) Determine the principal stresses and show them on a sketch of a properly
oriented element.
(b) Determine the maximum shear stresses and associated normal stresses
and show them on a sketch of a properly oriented element.
85 MPa
56 MPa
Solution 7.3-10
sx ⫽ ⫺85 MPa
sy ⫽ 0 MPa
Txy ⫽ ⫺56 MPa
up2 ⫽
2txy
sx ⫺ sy
tmax ⫽
s2 ⫽
up1 ⫽ 116.4°
sx ⫺ sy
sx + sy
+
2
2
sx ⫺ sy
sx + sy
+
2
a
sx ⫺ sy
A
2
tmax ⫽ 70.3 MPa
2
up1 ⫽ up2 + 90°
;
(b) MAXIMUM SHEAR STRESSES
b
up2 ⫽ 26.4°
s1 ⫽
;
s2 ⫽ ⫺112.8 MPa
(a) PRINCIPAL STRESSES
atan a
s1 ⫽ 27.8 MPa
2
;
cos 12up12 + txy sin 12up12
cos 12up22 + txy sin 12up22
us1 ⫽ up1 ⫺ 45°
saver ⫽
sx + sy
2
2
b + txy2
;
us1 ⫽ 71.4°
;
saver ⫽ ⫺42.5 MPa
;
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CHAPTER 7
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Analysis of Stress and Strain
y
Problems 7.3-11 sx ⫽ 2500 psi, sy ⫽ 1020 psi, txy ⫽ ⫺900 psi
(a) Determine the principal stresses and show them on a sketch of a properly
oriented element.
(b) Determine the maximum shear stresses and associated normal stresses and
show them on a sketch of a properly oriented element.
sy
txy
sx
O
x
Probs. 7.3-11 through 7.3-16
Solution 7.3-11
sx ⫽ 2500 psi
sy ⫽ 1020 psi
txy ⫽ ⫺900 psi
(a) PRINCIPAL STRESSES
tan(2up) ⫽
2txy
up1 ⫽
sx ⫺ sy
atan a
2txy
sx ⫺ sy
b
s1 ⫽
2
s2 ⫽
+
2
2
sx ⫺ sy
sx + sy
+
2
2
s1 ⫽ 2925 psi
For up2 ⫽ 64.7°:
s2 ⫽ 595 psi
tmax ⫽
up2 ⫽ 64.71°
sx ⫺ sy
sx + sy
For up1 ⫽ ⫺25.3°:
;
;
(b) MAXIMUM SHEAR STRESSES
up1 ⫽ 25.29°
up2 ⫽ 90 ° + up1
Therefore,
A
a
sx ⫺ sy 2
b + txy2
2
tmax ⫽ 1165 psi
us1 ⫽ ⫺70.3° and
us1 ⫽ up1 ⫺ 45°
;
t1 ⫽ 1165 psi
cos 12up12 + txy sin 12up12
us2 ⫽ up1 ⫺ 45°
t2 ⫽ ⫺1165 psi
cos 12up22 + txy sin 12up22
saver ⫽
sx + sy
us2 ⫽ 19.71° and
;
saver ⫽ 1760 psi
2
;
Problems 7.3-12 sx ⫽ 2150 kPa, sy ⫽ 375 kPa, txy ⫽ ⫺460 kPa
(a) Determine the principal stresses and show them on a sketch of a properly oriented element.
(b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly
oriented element.
Solution 7.3-12
sx ⫽ 2150 kPa
sy ⫽ 375 kPa
txy ⫽ ⫺460 kPa
up1 ⫽ ⫺13.70°
up2 ⫽ 90° + up1
(a) PRINCIPAL STRESSES
tan(2up) ⫽
2txy
sx ⫺ sy
up1 ⫽
atan a
2txy
sx ⫺sy
2
b
s1 ⫽
s2 ⫽
sx ⫺ sy
sx + sy
+
2
2
sx ⫺ sy
sx + sy
+
2
2
up2 ⫽ 76.30°
cos 12up12 + txy sin 12up12
cos 12up22 + txy sin 12up22
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SECTION 7.3 Principal Stresses and Maximum Shear Stresses
tmax ⫽ 145 psi
Therefore,
For up1 ⫽ ⫺ 13.70°
s1 ⫽ 2262 kPa
For up2 ⫽ 76.3°
s2 ⫽ 263 kPa
us1 ⫽ ⫺ 58.7°
us1 ⫽ up1 ⫺ 45°
and t1 ⫽ 1000 kPa
;
;
us2 ⫽ up1 + 45° us2 ⫽ 31.3° ;
and t2 ⫽⫺1000 kPa
sx + sy
saver ⫽
saver ⫽ 1263 kPa
2
(b) MAXIMUM SHEAR STRESSES
tmax ⫽
Problems 7.3-13
A
a
sx ⫺ sy
2
;
2
b + txy2
sx ⫽ 14,500 psi, sy ⫽ 1070 psi, txy ⫽ 1900 psi
(a) Determine the principal stresses and show them on a sketch of a properly oriented element.
(b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly
oriented element.
Solution 7.3-13
sx ⫽ 14500 psi
sy ⫽ 1070 psi
txy ⫽ 1900 psi
(a) PRINCIPAL STRESSES
tan(2up) ⫽
2txy
up1 ⫽
sx ⫺ sy
atan a
2txy
sx ⫺sy
b
2
up1 ⫽ 7.90°
up2 ⫽ 90° + up1
s1 ⫽
s2 ⫽
up2 ⫽ 97.90°
sx ⫺ sy
sx + sy
+
2
2
sx + sy
sx ⫺ sy
+
2
2
cos 12up12 + txy sin 12up12
cos 12up22 + txy sin 12up22
Therefore,
For up1 ⫽ 7.90°
s1 ⫽ 14764 psi
;
For up2 ⫽ 97.9°
s2 ⫽ 806 psi
;
(b) MAXIMUM SHEAR STRESSES
tmax ⫽
sx ⫺ sy
2
2
b + txy2
tmax ⫽ 6979 psi
us1 ⫽ u p1 ⫺ 45°
and t1 ⫽ 6979 psi
us1 ⫽ ⫺37.1°
us2 ⫽ 52.9°
us2 ⫽ up1 + 45°
and t2 ⫽⫺6979 psi
saver ⫽
Problems 7.3-14
A
a
sx + sy
2
;
;
saver ⫽ 7785 psi
sx ⫽ 16.5 MPa, sv ⫽ ⫺91 MPa, txy ⫽ ⫺39 MPa
(a) Determine the principal stresses and show them on a sketch of a properly oriented element.
(b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly
oriented element.
589
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CHAPTER 7 Analysis of Stress and Strain
Solution 7.3-14
sx ⫽ 16.5 MPa
sy ⫽ ⫺91 MPa
txy ⫽ ⫺39 MPa
(b) MAXIMUM SHEAR STRESSES
(a) PRINCIPAL STRESSES
2txy
tan(2up) ⫽
up1 ⫽
sx ⫺ sy
atan a
2txy
sx ⫺sy
tmax ⫽
b
s1 ⫽
us1 ⫽ ⫺63.0°
us1 ⫽ up1 ⫺ 45°
and t1 ⫽ 66.4 MPa
2
s2 ⫽
+
2
2
sx ⫺ sy
sx + sy
+
2
us2 ⫽ 27.0°
us2 ⫽ up1 + 45°
and t2 ⫽ ⫺ 66.4 MPa
up2 ⫽72.02°
sx ⫺ sy
sx + sy
2
2
2
b + txy2
tmax ⫽ 9631.7 psi
up1 ⫽ ⫺ 17.98°
up2 ⫽ 90° + up1
A
sx ⫺ sy
a
cos 12up12 + txy sin 12up12
saver ⫽
sx + sy
;
;
saver ⫽ ⫺37.3 MPa
2
cos 12up22 + txy sin 12up22
Therefore,
For up1 ⫽ ⫺ 17.98°
For up2 ⫽ 72.0°
Problems 7.3-15
s1 ⫽ 29.2 MPa
s2 ⫽ ⫺ 103.7 MPa
sx ⫽ ⫺3300 psi, sy ⫽ ⫺11,000 psi, txy ⫽ 4500 psi
(a) Determine the principal stresses and show them on a sketch of a properly oriented element.
(b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly
oriented element.
Solution 7.3-15
sy ⫽⫺11000 psi
s ⫽ ⫺3300 psi
txy ⫽ 4500 psi
(a) PRINCIPAL STRESSES
tan(2up) ⫽
2txy
sx ⫺ sy
up1 ⫽
atan a
2txy
sx ⫺sy
b
2
up1 ⫽ 24.73°
up2 ⫽ 90° + up1
s1 ⫽
s2 ⫽
up2 ⫽ 114.73°
sx ⫺ sy
sx + sy
+
2
2
sx + sy
sx ⫺ sy
+
2
2
cos 12up12 + txy sin 12up12
cos 12up22 + txy sin 12up22
Therefore,
For up1 ⫽ 24.7°
s1 ⫽ ⫺1228 psi
For up2 ⫽ 114.7°
s2 ⫽ ⫺13072 psi
(b) MAXIMUM SHEAR STRESSES
tmax ⫽
A
a
sx ⫺ sy
2
2
b + txy2
tmax ⫽ 5922 psi
us1 ⫽ up1 ⫺ 45°
and t1 ⫽ 5922 psi
us1 ⫽ ⫺20.3°
us2 ⫽ up1 + 45°
and t2 ⫽ ⫺5922 psi
us2 ⫽ 69.7°
saver ⫽
sx + sy
2
;
;
saver ⫽ ⫺7150 psi
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SECTION 7.3
Principal Stresses and Maximum Shear Stresses
591
Problems 7.3-16 sx ⫽ ⫺108 MPa, sy ⫽ 58 MPa, txy ⫽ ⫺58 MPa
(a) Determine the principal stresses and show them on a sketch of a properly oriented element.
(b) Determine the maximum shear stresses and associated normal stresses and show them on a sketch of a properly
oriented element.
Solution 7.3-16
sx ⫽ ⫺108 MPa
sy ⫽ 58 MPa
txy ⫽ ⫺58 MPa
(a) PRINCIPAL STRESSES
tan(2up) ⫽
atana
2txy
up2 ⫽
sx ⫺ sy
tmax ⫽
2txy
sx ⫺sy
b
s1 ⫽
s2 ⫽
up1 ⫽ 107.47°
sx ⫺ sy
sx + sy
+
2
2
sx ⫺ sy
sx + sy
+
2
2
A
a
sx ⫺ sy
2
2
b + txy2
tmax ⫽ 14686.1 psi
us1 ⫽ 62.47°
us1 ⫽ up1 ⫺ 45°
and t1 ⫽ 101.3 MPa
2
up2 ⫽ 17.47°
up1 ⫽ 90° + up2
(b) MAXIMUM SHEAR STRESSES
cos 12up12 + txy sin 12up12
;
us2 ⫽ 152.47° ;
us2 ⫽ up1 + 45°
and t2 ⫽⫺101.3 MPa
sx + sy
saver ⫽
saver ⫽ ⫺25.0 MPa
2
cos 12up22 + txy sin 12up22
Therefore,
For up1 ⫽ 107.47°
s1 ⫽ 76.3 MPa
For up2 ⫽ 17.47°
s2 ⫽ ⫺126.3 MPa
;
;
Problem 7.3-17 At a point on the surface of a machine component, the stresses
y
acting on the x face of a stress element are sx ⫽ 5900 psi and txy ⫽ 1950 psi
(see figure).
What is the allowable range of values for the stress sy if the maximum shear stress
is limited to t0 ⫽ 2500 psi?
sy
txy = 1950 psi
O
sx = 5900 psi
x
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Analysis of Stress and Strain
Solution 7.3-17
sx ⫽ 5900 psi
sy
unknown
txy ⫽ 1950 psi
Therefore, 2771 psi … sy … 9029 psi
Find the allowable range of values for sy if the
From Eq. (1):
maximum allowable shear stresses is tmax ⫽ 2500 psi
sx ⫺ sy 2
(1)
tmax ⫽ a
b + txy2
A
2
t max (sy1) ⫽
A
a
sx ⫺ sy1
2
2
b + txy2
Solve for sy
sy ⫽ sx ⫹
J
22tmax 2 ⫺txy2
⫺ a22tmax2 ⫺txy2 b
K
sy ⫽ a
9029
b psi
2771
2.771 ksi
9.029 ksi
tmax(σy1)
2.5 ksi
σy1
Problem 7.3-18 At a point on the surface of a machine component the stresses
y
acting on the x face of a stress element are sx ⫽ 42 MPa and txy ⫽ 33 MPa (see figure).
What is the allowable range of values for the stress sy if the maximum shear
stress is limited to t0 ⫽ 35 MPa?
sy
txy = 33 MPa
O
sx = 42 MPa
x
Solution 7.3-18
sx ⫽ 42 MPa
sy
unknown
txy ⫽ 33 MPa
Find the allowable range of values for sy if the
maximum allowable shear stresses is tmax ⫽ 35 MPa
tmax ⫽
A
a
sx ⫺ sy
2
2
b + txy2
(1)
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Page 593
SECTION 7.3
Solve for sy
sy ⫽ sx ⫹
593
Principal Stresses and Maximum Shear Stresses
From Eq. (1):
22tmax 2 ⫺txy2
2
2
J ⫺ a22tmax ⫺txy b K
65.3
b MPa
sy ⫽ a
18.7
tmax (sy1) ⫽
A
a
sx ⫺ sy1
2
2
b + txy2
Therefore, 18.7 MPa … sy … 65.3 MPa
65.3 MPa
18.7 MPa
35 MPa
tmax (σy1)
σy1
An element in plane stress is subjected to stresses sx ⫽ 5700 psi
and txy ⫽ ⫺2300 psi (see figure). It is known that one of the principal stresses equals
6700 psi in tension.
Problem 7.3-19
y
sy
(a) Determine the stress sy.
(b) Determine the other principal stress and the orientation of the principal planes,
then show the principal stresses on a sketch of a properly oriented element.
5700 psi
O
x
2300 psi
Solution 7.3-19
sx ⫽ 5700 psi
sy unknown txy ⫽ ⫺2300 psi
(a) STRESS sy
s1 ⫽ 6700 psi
s1 ⫽
2
;
(b) PRINCIPAL STRESSES
Because sy is smaller than a given principal stress,
we know that the given stress is the larger principal
stress.
sx + sy
sy ⫽ 1410 psi
Solve for sy
sx ⫺ sy 2
+
a
b + txy2
A
2
tan (2up) ⫽
2txy
sx ⫺ sy
atana
up1 ⫽
up1 ⫽ ⫺23.50°
up2 ⫽ 90° + up1
up2 ⫽ 66.50°
2txy
sx ⫺sy
2
b
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CHAPTER 7
Page 594
Analysis of Stress and Strain
s1 ⫽
s2 ⫽
sx ⫺ sy
sx + sy
+
2
2
+ txy sin12up12
For up1 ⫽ ⫺23.5° : s1 ⫽ 6700 psi
For up2 ⫽ 66.5°
sx ⫺ sy
sx + sy
Therefore,
cos 12up12
+
2
2
+ txy sin 12up22
:
s2 ⫽ 410 psi
;
;
cos 12up22
Problem 7.3-20 An element in plane stress is subjected to stresses sx ⫽ ⫺50 MPa
and txy ⫽ 42 MPa (see figure). It is known that one of the principal stresses equals
33 MPa in tension.
y
sy
(a) Determine the stress sy.
(b) Determine the other principal stress and the orientation of the principal planes,
then show the principal stresses on a sketch of a properly oriented element.
42 MPa
50 MPa
O
x
Solution 7.3-20
sx ⫽ ⫺50 MPa
sy unknown
txy ⫽ 42 MPa
up2 ⫽ ⫺26.85°
up1 ⫽ 90° + up2
(a) STRESS sy
s1 ⫽
Because sy is smaller than a given principal stress,
we know that the given stress is the larger principal
stress.
s1 ⫽ 33 MPa
s1 ⫽
sx + sy
+
A
2
Solve for sy
a
s x ⫺ sy 2
2
b +
s2 ⫽
txy2
sy ⫽ 11.7 MPa
;
tan(2up) ⫽
2txy
For up1 ⫽ 63.2°
sx ⫺ sy
up2 ⫽
atan a
sx ⫺sy
2
sx ⫺ sy
sx + sy
+
2
2
+ txy sin 12up12
sx ⫺ sy
sx + sy
+
2
2
+ txy sin 12up22
cos 12up12
cos 12up22
Therefore,
(b) PRINCIPAL STRESSES
2txy
up1 ⫽ 63.15°
b
:
s1 ⫽ 33.0 MPa
For up2 ⫽ ⫺26.8° : s 2 ⫽ ⫺71.3 MPa
;
;
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SECTION 7.4
595
Mohr’s Circle
Mohr’s Circle
y
The problems for Section 7.4 are to be solved using Mohr’s circle. Consider only the
in-plane stresses (the stresses in the xy plane).
Problem 7.4-1 An element in uniaxial stress is subjected to tensile stresses
sx ⫽ 11,375 psi, as shown in the figure. Using Mohr’s circle, determine:
11,375 psi
O
(a) The stresses acting on an element oriented at a counterclockwise angle u ⫽ 24°
from the x axis.
(b) The maximum shear stresses and associated normal stresses.
Show all results on sketches of properly oriented elements.
x
Solution 7.4-1
sx ⫽ 11375 psi sy ⫽ 0 psi
(a) ELEMENT AT
u ⫽ 24°
2u ⫽ 48° R ⫽
sx
2
txy ⫽ 0 psi
;
sy1 ⫽ 1882 psi
(b) MAXIMUM SHEAR STRESSES
R ⫽ 5688 psi
Point C: sc ⫽ R sc ⫽ 5688 psi
Point D: sx1 ⫽ R + R cos(2u)
sx1 ⫽ 9493 psi
;
;
tx1y1 ⫽ ⫺R sin (2u)
tx1y1 ⫽ ⫺4227 psi
Point S1: us1 ⫽
tmax ⫽ R
us1 ⫽⫺45°
tmax ⫽ 5688 psi
Point S2: us2 ⫽
tmax ⫽ ⫺R
;
⫺ 90°
2
saver ⫽ R
90°
2
;
;
us2 ⫽ 45°
;
tmax ⫽ ⫺5688 psi
;
saver ⫽ 5688 psi
;
œ
PointD : sy1 ⫽ R ⫺ R cos (2u)
y
Problem 7.4-2 An element in uniaxial stress is subjected to tensile stresses
sx ⫽ 49 MPa, as shown in the figure Using Mohr’s circle, determine:
(a) The stresses acting on an element oriented at an angle u ⫽ ⫺27° from the x
axis (minus means clockwise).
(b) The maximum shear stresses and associated normal stresses.
Show all results on sketches of properly oriented elements.
49 MPa
O
Solution 7.4-2
sx ⫽ 49 MPa sy ⫽ 0 MPa
(a) ELEMENT AT
u ⫽ ⫺27°
2u ⫽ ⫺54.0° R ⫽
Point C:
txy ⫽ 0 MPa
sc ⫽ R
sx
2
R ⫽ 24.5 MPa
sc ⫽ 24.5 MPa
Point D:
sx1 ⫽ R + R cos (|2u|)
sx1 ⫽ 38.9 MPa
;
tx1y1 ⫽ ⫺R sin (2u)
tx1y1 ⫽ 19.8 MPa
;
œ
Point D sy1 ⫽ R ⫺ R cos ( |2u| )
;
sy1 ⫽ 10.1 MPa
x
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Analysis of Stress and Strain
(b) MAXIMUM SHEAR STRESSES
Point S1:
tmax ⫽ R
90°
2
Point S2: us2 ⫽
⫺90°
us1 ⫽
2
tmax ⫽ ⫺R
us1 ⫽ ⫺45.0°
;
tmax ⫽ 24.5 MPa
saver ⫽ R
us2 ⫽ 45.0°
;
tmax ⫽ ⫺24.5 MPa
saver ⫽ 24.5 MPa
;
;
;
Problem 7.4-3 An element in uniaxial stress is subjected to compressive stresses
y
of magnitude 6100 psi, as shown in the figure. Using Mohr’s circle, determine:
1
(a) The stresses acting on an element oriented at a slope of 1 on 2 (see figure).
(b) The maximum shear stresses and associated normal stresses.
Show all results on sketches of properly oriented elements.
2
O
6100 psi
x
Solution 7.4-3
sx ⫽ ⫺6100 psi sy ⫽ 0 psi
txy ⫽ 0 psi
œ
Point D : sy1 ⫽ R ⫺ R cos (2u)
sy1 ⫽ ⫺1220 psi
(a) ELEMENT AT A SLOPE OF 1 ON 2
1
u ⫽ atan a b
2
u ⫽ 26.565°
Point S1:
sx
2u ⫽ 53.130° R ⫽
2
Point C:
Point D:
sc ⫽ R
(b) MAXIMUM SHEAR STRESSES
;
R ⫽ ⫺3050 psi
tmax ⫽ ⫺R
sc ⫽ ⫺3050 psi
Point S2:
sx1 ⫽ R + R cos (2u)
sx1 ⫽ ⫺4880 psi
tx1y1 ⫽ ⫺R sin (2u)
;
tx1y1 ⫽ 2440 psi
us1 ⫽
;
90°
2
us1 ⫽ 45°
tmax ⫽ 3050 psi
us2 ⫽
⫺ 90°
2
;
;
us2 ⫽ ⫺45°
tmax ⫽ R
tmax ⫽ ⫺3050 psi
saver ⫽ R
saver ⫽ ⫺3050 psi
;
;
;
Problem 7.4-4 An element in biaxial stress is subjected to stresses sx ⫽ ⫺48 MPa and
sy ⫽ 19 MPa, as shown in the figure. Using Mohr’s circle, determine:
y
(a) The stresses acting on an element oriented at a counterclockwise angle u ⫽ 25°
from the x axis.
(b) The maximum shear stresses and associated normal stresses.
19 MPa
Show all results on sketches of properly oriented elements.
48 MPa
O
x
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SECTION 7.4
597
Mohr’s Circle
Solution 7.4-4
sx ⫽ ⫺48 MPa
(a) ELEMENT AT
sy ⫽ 19 MPa
u ⫽ 25°
2u ⫽ 50.0 deg R ⫽
txy ⫽ 0 MPa
Point S1: us1 ⫽
;
|sx| + |sy|
2
Point C: sx ⫽ sx + R
(b) MAXIMUM SHEAR STRESSES
us1 ⫽ 45.0°
R ⫽ 33.5 MPa
sc ⫽ ⫺14.5 MPa
tmax ⫽ R
sx1 ⫽ ⫺36.0 MPa
;
tmax ⫽ ⫺R
;
;
tmax ⫽ ⫺33.5 MPa
saver ⫽ sc
Point D œ : sy1 ⫽ sc + R cos(2u)
;
⫺ 90°
2
us2 ⫽ ⫺45.0°
tx1y1 ⫽ ⫺R sin(2u)
tx1y1 ⫽ 25.7 MPa
;
tmax ⫽ 33.5 MPa
Point S2: us2 ⫽
Point D: sx1 ⫽ sc ⫺ R cos(2u)
90°
2
saver ⫽ ⫺14.5 MPa
;
;
sy1 ⫽ 7.0 MPa
Problem 7.4-5 An element in biaxial stress is subjected to stresses sx ⫽ 6250 psi
y
and sy ⫽ ⫺1750 psi, as shown in the figure. Using Mohr’s circle, determine:
(a) The stresses acting on an element oriented at a counterclockwise angle
u ⫽ 55° from the x axis.
(b) The maximum shear stresses and associated normal stresses.
Show all results on sketches of properly oriented elements.
1750 psi
6250 psi
O
x
Solution 7.4-5
sx ⫽ 6250 psi
(a) ELEMENT AT
sy ⫽ ⫺1750 psi
txy ⫽ 0 psi
u ⫽ 60°
2u ⫽ 120° R ⫽
|sx| + |sy|
2
Point C: sc ⫽ sx ⫺ R
Point S1: us1 ⫽
R ⫽ 4000 psi
sc ⫽ 2250 psi
Point D: sx1 ⫽ sc + R cos(2u)
sx1 ⫽ 250 psi
(b) MAXIMUM SHEAR STRESSES
us1 ⫽ ⫺45°
tmax ⫽ R
;
tx1y1 ⫽ ⫺3464 psi
Point D œ : sy1 ⫽ sc ⫺ R cos(2u)
;
sy1 ⫽ 4250 psi
;
tmax ⫽ 4000 psi
Point S2: us2 ⫽
tx1y1 ⫽ ⫺R sin (2u)
⫺ 90°
2
90°
2
;
us2 ⫽ 45°
;
tmax ⫽ R
tmax ⫽ 4000 psi
;
saver ⫽ sc
saver ⫽ 2250 psi
;
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CHAPTER 7
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Analysis of Stress and Strain
y
Problem 7.4-6 An element in biaxial stress is subjected to stresses sx ⫽ ⫺29 MPa
and sy ⫽ 57 MPa, as shown in the figure. Using Mohr’s circle, determine:
57 MPa
(a) The stresses acting on an element oriented at a slope of 1 on 2.5 (see figure).
(b) The maximum shear stresses and associated normal stresses.
Show all results on sketches of properly oriented elements.
1
2.5
29 MPa x
O
Solution 7.4-6
sx ⫽ ⫺29 MPa
sy ⫽ 57 MPa
txy ⫽ 0 MPa
(b) MAXIMUM SHEAR STRESSES
Point S1: us1 ⫽
(a) ELEMENT AT A SLOPE OF 1 ON 2.5
u ⫽ atan a
1
b
2.5
2u ⫽ 43.603° R ⫽
u ⫽ 21.801°
|sx| + |sy|
2
Point C: sc ⫽ sx + R
tmax ⫽ R
;
tmax ⫽ ⫺R
Point D: sx1 ⫽ sc ⫺ R cos (2u)
sx1 ⫽ ⫺17.1 MPa
tx1y1 ⫽ R sin (2u)
Point
Dœ:
saver ⫽ sc
;
tx1y1 ⫽ 29.7 MPa
;
;
⫺ 90°
2
us2 ⫽ ⫺45.0°
sc ⫽ 14.0 MPa
us1 ⫽ 45.0°
tmax ⫽ 43.0 MPa
Point S2: us2 ⫽
R ⫽ 43.0 MPa
90°
2
;
tmax ⫽ ⫺43.0 MPa
saver ⫽ 14.0 MPa
;
;
;
sy1 ⫽ sc + R cos (2u)
sy1 ⫽ 45.1 MPa
Problem 7.4-7 An element in pure shear is subjected to stresses txy ⫽ 2700 psi, as shown
y
in the figure. Using Mohr’s circle, determine:
(a) The stresses acting on an element oriented at a counterclockwise angle u ⫽ 52° from
the x axis.
(b) The principal stresses.
Show all results on sketches of properly oriented elements.
2700 psi
O
x
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Page 599
SECTION 7.4
599
Mohr’s Circle
Solution 7.4-7
sx ⫽ 0 psi
sy ⫽ 0 psi
(a) ELEMENT AT
txy ⫽ 2700 psi
Point P1: up1 ⫽
u ⫽ 52°
2u ⫽ 104.0° R ⫽ txy
R ⫽ 2700 psi
Point D: sx1 ⫽ R cos (2u ⫺ 90°)
sx1 ⫽ 2620 psi
tx1y1 ⫽ ⫺653 psi
Point
90°
2
up1 ⫽ 45°
s1 ⫽ R
Point P2: up2 ⫽
;
tx1y1 ⫽ ⫺R sin (2u ⫺ 90°)
Dœ
(b) PRINCIPAL STRESSES
;
s1 ⫽ 2700 psi
;
⫺ 90°
2
up2 ⫽ ⫺45°
;
;
: sy1 ⫽ ⫺R cos (2u ⫺ 90°)
sy1 ⫽ ⫺2620 psi
s2 ⫽ ⫺R
s2 ⫽ ⫺2700 psi
;
;
Problem 7.4-8 An element in pure shear is subjected to stresses txy ⫽ ⫺14.5 MPa, as
shown in the figure. Using Mohr’s circle, determine:
y
(a) The stresses acting on an element oriented at a counterclockwise angle u ⫽ 22.5°
from the x axis
(b) The principal stresses.
Show all results on sketches of properly oriented elements.
O
x
14.5 MPa
Solution 7.4-8
sx ⫽ 0 MPa
sy ⫽ 0 MPa
(a) ELEMENT AT
txy ⫽ ⫺14.5 MPa
u ⫽ 22.5°
Point P1: up1 ⫽
2u ⫽ 45.00°
R ⫽ |txy|
270°
u ⫽ 135.0°
2 p1
s1 ⫽ R
R ⫽ 14.50 MPa
Point D: sx1 ⫽ ⫺R cos (2u ⫺ 90°)
sx1 ⫽ ⫺10.25 MPa
;
tx1y1 ⫽ R sin (2u ⫺ 90°)
Point P2: up2 ⫽
;
s1 ⫽ 14.50 MPa
⫺ 270°
2
up2 ⫽ ⫺135.0°
;
s2 ⫽ ⫺R
tx1y1 ⫽ ⫺10.25 MPa
;
Point D œ : sy1 ⫽ R cos (2u ⫺ 90°)
sy1 ⫽ 10.25 MPa
(b) PRINCIPAL STRESSES
;
s2 ⫽ ⫺14.50 MPa
;
;
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Analysis of Stress and Strain
Problem 7.4-9 An element in pure shear is subjected to stresses txy ⫽ 3750 psi, as
shown in the figure. Using Mohr’s circle, determine:
y
3
4
(a) The stresses acting on an element oriented at a slope of 3 on 4 (see figure).
(b) The principal stresses.
Show all results on sketches of properly oriented elements.
x
O
3750 psi
Solution 7.4-9
sx ⫽ 0 psi
sy ⫽ 0 psi
txy ⫽ 3750 psi
(b) PRINCIPAL STRESSES
3
u ⫽ atan a b
4
u ⫽ 36.870°
2u ⫽ 73.740°
R ⫽ txy
s1 ⫽ R
R ⫽ 3750 psi
Point D: sx1 ⫽ R cos (2u ⫺ 90°)
sx1 ⫽ 3600 psi
;
tx1y1 ⫽ ⫺R sin (2u ⫺ 90°)
tx1y1 ⫽ 1050 psi
90°
2
Poin P1: up1 ⫽
(a) ELEMENT AT A SLOPE OF 3 ON 4
Point P2: up2 ⫽
up1 ⫽ 45°
s1 ⫽ 3750 psi
;
⫺ 90°
2
up2 ⫽ ⫺45°
s2 ⫽ ⫺R
;
;
s2 ⫽ ⫺3750 psi
;
;
Point D sy1 ⫽ ⫺R cos (2u ⫺ 90°)
œ:
sy1 ⫽ ⫺3600 psi
;
Problem 7.4-10 sx ⫽ 27 MPa, sy ⫽ 14 MPa, txy ⫽ 6 MPa, u ⫽ 40°
y
Using Mohr’s circle, determine the stresses acting on an element oriented at an
angle u from the x axis. Show these stresses on a sketch of an element oriented at the
angle u. (Note: The angle u is positive when counterclockwise and negative when
clockwise.)
sy
txy
sx
O
Probs. 7.4-10 through 7.4-15
x
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Page 601
SECTION 7.4 Mohr’s Circle
601
Solution 7.4-10
sx ⫽ 27 MPa
sy ⫽ 14 MPa
txy ⫽ 6 MPa
sx + sy
saver ⫽ 20.50 MPa
2
R ⫽ 2(sx ⫺ saver)2 + txy 2
a ⫽ atan a
b ⫽ 37.29°
Point D: sx1 ⫽ saver + R cos (b)
u ⫽ 40°
saver ⫽
b ⫽ 2u ⫺ a
txy
sx ⫺ saver
R ⫽ 8.8459 MPa
sx1 ⫽ 27.5 MPa
;
tx1y1 ⫽ ⫺ R sin (b)
tx1y1 ⫽ ⫺ 5.36 MPa
;
Point D œ : sy1 ⫽ saver ⫺ R cos (b)
b
a ⫽ 42.71°
sy1 ⫽ 13.46 MPa
;
Problem 7.4-11 sx ⫽ 3500 psi, sy ⫽ 12,200 psi, txy ⫽ ⫺3300 psi, u ⫽ ⫺51°
Using Mohr’s circle, determine the stresses acting on an element oriented at an angle u from the x axis. Show these stresses on a
sketch of an element oriented at the angle u. (Note: The angle u is positive when counterclockwise and negative when clockwise.)
Solution 7.4-11
sx ⫽ 3500 psi
u ⫽ ⫺ 51°
saver ⫽
sy ⫽ 12200 psi
b ⫽ 180° + 2u ⫺ a
b ⫽ 40.82°
Point D: sx1 ⫽ saver + R cos (b)
;
sx + sy
sx1 ⫽ 11982 psi
saver ⫽ 7850 psi
2
R ⫽ 2(sx ⫺ saver) + txy
2
a ⫽ atan a
txy ⫽ ⫺ 3300 psi
txy
sx ⫺ saver
b
tx1y1 ⫽ ⫺ R sin (b)
R ⫽ 5460 psi
2
;
tx1y1 ⫽ ⫺ 3569 psi
;
Point D sy1 ⫽ saver ⫺ R cos (b)
œ:
a ⫽ 37.18°
sy1 ⫽ 3718 psi
;
Problem 7.4-12 sx ⫽ ⫺47 MPa, sy ⫽ ⫺186 MPa, txy ⫽ ⫺29 MPa, u ⫽ ⫺33°
Using Mohr’s circle, determine the stresses acting on an element oriented at an angle u from the x axis. Show these stresses on a
sketch of an element oriented at the angle u. (Note: The angle u is positive when counterclockwise and negative when clockwise.)
Solution 7.4-12
sx ⫽⫺47MPa
sy ⫽⫺186MPa
txy ⫽⫺29MPa
u ⫽ ⫺ 33°
saver ⫽
sx1 ⫽ ⫺ 61.7 MPa
sx + sy
saver ⫽ ⫺ 116.50 MPa
2
R ⫽ 2(sx ⫺ saver) + txy
2
a ⫽ atan a `
Point D: sx1 ⫽ saver + R cos (b)
txy
sx ⫺ saver
b ⫽ ⫺ 2u ⫺ a
`b
R ⫽ 75.3077 MPa
2
a ⫽ 22.65°
b ⫽ 43.35°
;
tx1y1 ⫽⫺R sin (b)
tx1y1 ⫽⫺51.7 MPa
;
Point D œ : sy1 ⫽ saver ⫺ R cos (b)
sy1 ⫽ ⫺ 171.3 MPa
;
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Page 602
CHAPTER 7 Analysis of Stress and Strain
Problem 7.4-13 sx ⫽ ⫺1720 psi, sy ⫽ ⫺680 psi, txy ⫽ 320 psi, u ⫽ 14°
Using Mohr’s circle, determine the stresses acting on an element oriented at an angle u from the x axis. Show these
stresses on a sketch of an element oriented at the angle u. (Note: The angle u is positive when counterclockwise and negative
when clockwise.)
Solution 7.4-13
sx ⫽ ⫺ 1720 psi
sy ⫽ ⫺ 680 psi
txy ⫽ 380 psi
Point D: sx1 ⫽ saver ⫺ R cos(b)
u ⫽ 14°
saver ⫽
sx1 ⫽ ⫺ 1481 psi
sx + sy
saver ⫽ ⫺ 1200 psi
2
tx1y1 ⫽ 580 psi
;
Point D : sy1 ⫽ saver + R cos (b)
œ
R ⫽ 2(sx ⫺ saver) + txy
2
a ⫽ atan a
tx1y1 ⫽ R sin (b)
;
txy
|sx ⫺ saver|
b ⫽ 2u + a
b
sy1 ⫽ ⫺ 919 psi
R ⫽ 644.0 psi
2
;
a ⫽ 36.16°
b ⫽ 64.16°
Problem 7.4-14 sx ⫽ 33 MPa, sy ⫽ ⫺9 MPa, txy ⫽ 29 MPa, u ⫽ 35°
Using Mohr’s circle, determine the stresses acting on an element oriented at an angle u from the x axis. Show these
stresses on a sketch of an element oriented at the angle u. (Note: The angle u is positive when counterclockwise and negative
when clockwise.)
Solution 7.4-14
sx ⫽ 33 MPa sy ⫽ ⫺ 9 MPa txy ⫽ 29 MPa
Point D: sx1 ⫽ saver + R cos (b)
u ⫽ 35°
saver ⫽
sx1 ⫽ 46.4 MPa
sx + sy
R ⫽ 2(sx ⫺ saver) +
2
a ⫽ atan a `
txy
sx ⫺ saver
b ⫽ 2u ⫺ a
tx1y1 ⫽ ⫺ R sin (b)
saver ⫽ 12.00 MPa
2
txy2
`b
R ⫽ 35.8050 MPa
;
tx1y1 ⫽ ⫺ 9.81 MPa
;
Point D : sy1 ⫽ saver ⫺ R cos (b)
a ⫽ 54.09°
œ
sy1 ⫽ ⫺22.4 MPa
;
b ⫽ 15.91°
Problem 7.4-15 sx ⫽ ⫺5700 psi, sy ⫽ 950 psi, txy ⫽ ⫺2100 psi, u ⫽ 65°
Using Mohr’s circle, determine the stresses acting on an element oriented at an angle u from the x axis. Show these
stresses on a sketch of an element oriented at the angle u. (Note: The angle u is positive when counterclockwise and negative
when clockwise.)
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Page 603
SECTION 7.4
603
Mohr’s Circle
Solution 7.4-15
sx ⫽ ⫺5700 psi sy ⫽ 950 psi txy ⫽ ⫺2100 psi
u ⫽ 65°
saver ⫽
Point D: sx1 ⫽ saver + R cos (b)
sx1 ⫽ ⫺1846 psi
sx + sy
R ⫽ 2(sx ⫺ saver) + txy
2
a ⫽ atan a
tx1y1 ⫽ R sin (b)
saver ⫽ ⫺2375 psi
2
|txy|
|sx ⫺ saver|
2
b
b ⫽ 180° ⫺ 2u + a
;
tx1y1 ⫽ 3897 psi
;
Point D sy1 ⫽ saver ⫺ R cos (b)
œ:
sy1 ⫽ ⫺2904 psi
R ⫽ 3933 psi
;
a ⫽ 32.28°
b ⫽ 82.28°
Problems 7.4-16 sx ⫽ ⫺29.5 MPa, sy ⫽ 29.5 MPa, txy ⫽ 27 MPa
y
Using Mohr’s circle, determine (a) the principal stresses and (b) the maximum
shear stresses and associated normal stresses. Show all results on sketches of properly
oriented elements.
sy
txy
sx
O
x
Probs. 7.4-16 through 7.4-23
Solution 7.4-16
sx ⫽ ⫺29.5 MPa sy ⫽ 29.5 MPa txy ⫽ 27 MPa
saver ⫽
sx + sy
2
a ⫽ atan a `
txy
sx ⫺ saver
`b
R ⫽ 39.9906 MPa
a ⫽ 42.47°
180° ⫺ a
2
up2 ⫽ up1 ⫺ 90°
90°⫺a
2
s2 ⫽ ⫺40.0 MPa
us1 ⫽ 23.8°
us2 ⫽ 90° + us1
us2 ⫽ 113.8°
Point S1: saver ⫽ 0 MPa
up1 ⫽ 68.8°
up2 ⫽ ⫺21.2°
tmax ⫽ R
;
;
;
(b) MAXIMUM SHEAR STRESSES
us1 ⫽
(a) PRINCIPAL STRESSES
up1 ⫽
s1 ⫽ 40.0 MPa
Point P2: s2 ⫽ ⫺R
saver ⫽ 0 MPa
R ⫽ 2(sx ⫺ saver)2 + txy2
Point P1: s1 ⫽ R
;
;
;
tmax ⫽ 40.0 MPa
;
;
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Page 604
CHAPTER 7 Analysis of Stress and Strain
Problems 7.4-17 sx ⫽ 7300 psi, sy ⫽ 0 psi, txy ⫽ 1300 psi
Using Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal
stresses. Show all results on sketches of properly oriented elements.
Solution 7.4-17
sx ⫽ 7300 psi sy ⫽ 0 psi txy ⫽ 1300 psi
saver ⫽
sx + sy
s1 ⫽ 7525 psi
saver ⫽ 3650 psi
2
R ⫽ 2(sx ⫺ saver) +
2
a ⫽ atan a `
Point P1: s1 ⫽ R + saver
txy
sx ⫺ saver
txy2
`b
Point P2: s2 ⫽ ⫺ R + saver
a
2
up2 ⫽
a + 180°
2
s2 ⫽ ⫺ 225 psi
R ⫽ 3875 psi
a ⫽ 19.60°
(b) MAXIMUM SHEAR STRESSES
us1 ⫽
(a) PRINCIPAL STRESSES
up1 ⫽
up1 ⫽ 9.80°
;
⫺ 90° + a
2
us2 ⫽ 90° + us1
;
us1 ⫽ ⫺ 35.2°
us2 ⫽ 54.8°
Point S1: saver ⫽ 3650 psi
tmax ⫽ R
up2 ⫽ 99.8°
;
tmax ⫽ 3875 psi
;
Problems 7.4-18 sx ⫽ 0 MPa, sy ⫽ ⫺23.4 MPa, txy ⫽ ⫺9.6 MPa
Using Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal
stresses. Show all results on sketches of properly oriented elements.
Solution 7.4-18
sx ⫽ 0 MPa sy ⫽ ⫺ 23.4 MPa txy ⫽ ⫺ 9.6 MPa
saver ⫽
sx + sy
R ⫽ 2(sx ⫺ saver)2 + txy2
a ⫽ atan a `
s1 ⫽ 3.43 MPa
saver ⫽ ⫺ 11.70 MPa
2
txy
sx ⫺ saver
`b
Point P1: s1 ⫽ R + saver
Point P2: s2 ⫽ ⫺ R + saver
s2 ⫽ ⫺ 26.8 MPa
R ⫽ 15.1344 MPa
(a) PRINCIPAL STRESSES
up1
up2 ⫽ up1 + 90°
⫺ 90° ⫺ a
2
us2 ⫽ 90 ° + us1
up1 ⫽ ⫺ 19.68°
up2 ⫽ 70.32°
;
(b) MAXIMUM SHEAR STRESSES
a ⫽ 39.37°
us1 ⫽
⫺a
⫽
2
;
;
us1 ⫽ ⫺ 64.7°
;
us2 ⫽ 25.3°
Point S1: saver ⫽ ⫺ 11.70 MPa
;
tmax ⫽ R
tmax ⫽ 15.13 MPa
;
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Page 605
SECTION 7.4 Mohr’s Circle
605
Problems 7.4-19 sx ⫽ 2050 psi, sy ⫽ 6100 psi, txy ⫽ 2750 psi
Using Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal
stresses. Show all results on sketches of properly oriented elements.
Solution 7.4-19
sx ⫽ 2050 psi sy ⫽ 6100 psi txy ⫽ 2750 psi
saver ⫽
sx + sy
s1 ⫽ 7490 psi
saver ⫽ 4075 psi
2
R ⫽ 2(sx ⫺ saver) +
2
a ⫽ atan a `
Point P1: s1 ⫽ R + saver
txy
sx ⫺ saver
txy2
`b
Point P2: s2 ⫽ ⫺ R + saver
R ⫽ 3415 psi
a ⫽ 53.63°
s2 ⫽ 660 psi
;
(b) MAXIMUM SHEAR STRESSES
us1 ⫽
(a) PRINCIPAL STRESSES
⫺ 90° + a
2
us2 ⫽ 90° + us1
180° ⫺ a
up1 ⫽ 63.2° ;
2
⫺a
⫽
up2 ⫽ ⫺ 26.8°
;
2
up1 ⫽
up2
;
us1 ⫽ ⫺18.2°
us2 ⫽ 71.8°
Point S1: saver ⫽ 4075 psi
tmax ⫽ R
;
;
tmax ⫽ 3415 psi
;
Problems 7.4-20 sx ⫽ 2900 kPa, sy ⫽ 9100 kPa, txy ⫽ ⫺3750 kPa
Using Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal
stresses. Show all results on sketches of properly oriented elements.
Solution 7.4-20
sx ⫽ 2900 kPa sy ⫽ 9100 kPa txy ⫽ ⫺ 3750 kPa
saver ⫽
sx + sy
R ⫽ 2(sx ⫺ saver)2 + txy2
a ⫽ atan a `
s1 ⫽ 10865 KPa
saver ⫽ 6000 kPa
2
txy
sx ⫺ saver
`b
Point P1: s1 ⫽ R + saver
Point P2: s2 ⫽ ⫺ R + saver
s2 ⫽ 1135 kPa
R ⫽ 4865.4393 kPa
(a) PRINCIPAL STRESSES
a + 180°
2
up2 ⫽
a
2
90° + a
2
us2 ⫽ 90° + us1
up1 ⫽ 115.2°
up2 ⫽ 25.2°
;
;
(b) MAXIMUM SHEAR STRESSES
a ⫽ 50.42°
us1 ⫽
up1 ⫽
;
;
us1 ⫽ 70.2°
us2 ⫽ 160.2°
Point S1: saver ⫽ 6000 kPa
tmax ⫽ R
;
;
;
tmax ⫽ 4865 kPa
;
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CHAPTER 7 Analysis of Stress and Strain
Problems 7.4-21 sx ⫽ ⫺11,500 psi, sy ⫽ ⫺18,250 psi, txy ⫽ ⫺7200 psi
Using Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal
stresses. Show all results on sketches of properly oriented elements.
Solution 7.4-21
sx ⫽ ⫺ 11500 psi
sy ⫽ ⫺ 18250 psi
Point P1: s1 ⫽ R + saver
txy ⫽ ⫺ 7200 psi
saver ⫽
s1 ⫽ ⫺ 6923 psi
sx + sy
R ⫽ 2(sx ⫺ saver)2 + txy2
a ⫽ atan a `
Point P2: s2 ⫽ ⫺ R + saver
saver ⫽ ⫺ 14875 psi
2
txy
sx ⫺ saver
`b
s2 ⫽ ⫺ 22827 psi
R ⫽ 7952 psi
⫺a
2
a ⫽ 64.89°
us1 ⫽
up2 ⫽
180° ⫺ a
2
270° ⫺ a
2
us2 ⫽ 90° + us1
up1 ⫽ ⫺ 32.4°
us1 ⫽ 102.6°
tmax ⫽ R
;
us2 ⫽ 192.6°
Point S1: saver ⫽ ⫺ 14875 psi
;
up2 ⫽ 57.6°
;
(b) MAXIMUM SHEAR STRESSES
(a) PRINCIPAL STRESSES
up1 ⫽
;
;
tmax ⫽ 7952 psi
;
Problems 7.4-22 sx ⫽ ⫺3.3 MPa, sy ⫽ 8.9 MPa, txy ⫽ ⫺14.1 MPa
Using Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal
stresses. Show all results on sketches of properly oriented elements.
Solution 7.4-22
sx ⫽⫺ 3.3 MPa
sy ⫽ 8.9 MPa
txy ⫽⫺ 14.1 MPa
saver ⫽
sx + sy
up1 ⫽
a + 180°
2
up2 ⫽
a
2
saver ⫽ 2.8 MPa
2
R ⫽ 2(sx ⫺ saver) +
2
a ⫽ atan a `
(a) PRINCIPAL STRESSES
txy
sx ⫺ saver
txy2
`b
R ⫽ 15.4 MPa
a ⫽ 66.6°
up1 ⫽ 123.3°
up2 ⫽ 33.3°
Point P1: s1 ⫽ R + saver
s1 ⫽ 18.2 MPa
Point P2: s2 ⫽ ⫺ R + saver
;
;
;
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Page 607
SECTION 7.4 Mohr’s Circle
s2 ⫽ ⫺ 12.6 MPa
us2 ⫽ 90° + us1
;
Point S1: saver ⫽ 2.8 MPa
(b) MAXIMUM SHEAR STRESSES
90° + a
us1 ⫽
2
us2 ⫽ 168.3°
tmax ⫽ R
;
tmax ⫽ 15.4 MPa
;
us1 ⫽ 78.3°
Problems 7.4-23 sx ⫽ 800 psi, sy ⫽ ⫺2200 psi, txy ⫽ 2900 psi
Using Mohr’s circle, determine (a) the principal stresses and (b) the maximum shear stresses and associated normal
stresses. Show all results on sketches of properly oriented elements.
Solution 7.4-23
sy ⫽ ⫺ 2200 psi
txy ⫽ 2900 psi
sx ⫽ 800 psi
sx + sy
saver ⫽
saver ⫽ ⫺ 700 psi
2
R ⫽ 2(sx ⫺ saver)2 + txy2
a ⫽ atan a `
txy
sx ⫺ saver
`b
(a) PRINCIPAL STRESSES
a
up1 ⫽
up1 ⫽ 31.3°
2
up2 ⫽
180° + a
2
R ⫽ 3265 psi
a ⫽ 62.65°
Point P1: s1 ⫽ R + saver
s1 ⫽ 2565 psi
Point P2: s2 ⫽ ⫺ R + saver
s2 ⫽ ⫺ 3965 psi
up2 ⫽ 121.3°
⫺ 90° + a
2
us2 ⫽ 90° + us1
us1 ⫽ ⫺ 13.7°
us2 ⫽ 76.3°
Point S1: saver ⫽ ⫺ 700 psi
;
;
(b) MAXIMUM SHEAR STRESSES
us1 ⫽
;
;
tmax ⫽ R
;
;
;
tmax ⫽ 3265 psi
607
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CHAPTER 7
Page 608
Analysis of Stress and Strain
Hooke’s Law for Plane Stress
When solving the problems for Section 7.5, assume that the material
is linearly elastic with modulus of elasticity E and Poisson’s ratio n.
sy
y
Problem 7.5-1 A rectangular steel plate with thickness t 0.25 in. is
subjected to uniform normal stresses x and y, as shown in the figure.
Strain gages A and B, oriented in the x and y directions, respectively, are
attached to the plate. The gage readings give normal strains ex 0.0010
(elongation) and ey 0.0007 (shortening).
Knowing that E 30 106 psi and 0.3, determine the stresses
x and y and the change t in the thickness of the plate.
Solution 7.5-1
sx
x
Probs. 7.5-1 and 7.5-2
âz SUBSTITUTE NUMERICAL VALUES:
(s + sy) 128.5 * 106
E x
¢t âzt 32.1 * 106 in.
Eq. (7-40a):
(1 )2
O
Eq. (7-39c):
E 30 * 106 psi 0.3
sx A
Rectangular plate in biaxial stress
t 0.25 in. âx 0.0010 ây 0.0007
E
B
;
(Decrease in thickness)
(âx + ây) 26,040 psi
;
Eq. (7-40b):
sy E
(1 )2
(ây + âx) 13,190 psi
;
Problem 7.5-2 Solve the preceding problem if the thickness of the steel plate is t 10 mm, the gage readings are
ex 480 106 (elongation) and ey 130 106 (elongation), the modulus is E 200 GPa, and Poisson’s ratio is 0.30.
Solution 7.5-2
Rectangular plate in biaxial stress
t 10 mm âx 480 * 106
ây 130 * 10
Eq. (7-40b):
6
sy E 200 GPa 0.3
E
(1 )2
SUBSTITUTE NUMERICAL VALUES:
Eq. (7-39c):
Eq. (7-40a):
âz sx E
(1 )2
(âx + ây) 114.1 MPa
;
(ây + âx) 60.2 MPa
(s + sy) 261.4 * 106
E x
¢t âz t 2610 * 106 mm
(Decrease in thickness)
;
;
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Page 609
SECTION 7.5
Problem 7.5-3 Assume that the normal strains Px and Py for an
Hooke’s Law for Plane Stress
y
element in plane stress (see figure) are measured with strain gages.
(a) Obtain a formula for the normal strain Pz in the z direction
in terms of Px, Py, and Poisson’s ratio .
(b) Obtain a formula for the dilatation e in terms of Px, Py,
and Poisson’s ratio .
sy
txy
sx
O
x
z
Solution 7.5-3
Plane stress
Given: âx, ây, (b) DILATATION
Eq. (7-47): e (a) NORMAL STRAIN âz
Eq. (7-34c): âz Eq. (7-36a): sx Eq. (7-36b): sy (s + sy)
E x
E
(1 2)
E
(1 2)
(âx + ây)
1 2
(sx + sy)
E
Substitute sx and sy from above and simplify:
e
1 2
(â + ây)
1 x
;
(ây + âx)
Substitute sx and sy into the first equation and simplify:
âz (â + ây)
1 x
;
sy
Problem 7.5-4 A magnesium plate in biaxial stress is subjected
to tensile stresses sx 24 MPa and sy 12 MPa (see figure).
The corresponding strains in the plate are x 440 106 and
y 80 106.
Determine Poisson’s ratio and the modulus of elasticity E
for the material.
y
O
x
sx
Probs. 7.5-4 through 7.5-7
Solution 7.5-4
Biaxial stress
sx 24 MPa sy 12 MPa
âx 440 * 106
ây 80 * 106
POISSON’S RATION AND MODULUS OF ELASTICITY
Eq. (7-39a): âx 1
(s sy)
E x
Eq. (7-39b): ây 1
(s sx)
E y
Substitute numerical values:
E (440 * 106) 24 MPa (12 MPa)
E (80 * 106) 12 MPa (24 MPa)
Solve simultaneously:
0.35 E 45 GPa
;
609
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Page 610
CHAPTER 7 Analysis of Stress and Strain
Problem 7.5-5 Solve the preceding problem for a steel plate with sx 10,800 psi (tension), sy 5400 psi (compression), ex 420 106 (elongation), and ey 300 106 (shortening).
Solution 7.5-5 Biaxial stress
sx 10,800 psi sy 5400 psi
âx 420 * 10
6
ây 300 * 10
Substitute numerical values:
6
E (420 * 106) 10,800 psi ( 5400 psi)
E (300 * 106) 5400 psi (10,800 psi)
POISSON’S RATIO AND MODULUS OF ELASTICITY
Solve simultaneously:
1/3 E 30 * 106 psi
1
Eq. (7-39a): âx (sx sy)
E
Eq. (7-39b): ây ;
1
(s sx)
E y
Problem 7.5-6 A rectangular plate in biaxial stress (see figure) is subjected to normal stresses x 90 MPa (tension) and
y 20 MPa (compression). The plate has dimensions 400 * 800 * 20 mm and is made of steel with E 200 GPa and
0.30.
(a) Determine the maximum in-plane shear strain gmax in the plate.
(b) Determine the change ¢t in the thickness of the plate.
(c) Determine the change ¢V in the volume of the plate.
Solution 7.5-6 Biaxial stress
sx 90 MPa sy 20 MPa
E 200 GPa 0.30
(b) CHANGE IN THICKNESS
Dimensions of Plate: 400 mm * 800 mm * 20 mm
Shear Modulus (Eq. 7-38):
E
76.923 GPa
G
2(1 + )
Principal stresses: s1 90 MPa s2 20 MPa
s1 s2
55.0 MPa
2
Eq. (7-35): gmax tmax
715 * 106
G
(sx + sy) 105 * 106
E
¢t âz t 2100 * 106 mm
;
(Decrease in thickness)
(c) CHANGE IN VOLUME
(a) MAXIMUM IN-PLANE SHEAR STRAIN
Eq. (7-26): tmax Eq. (7-39c): âz From Eq. (7-47): ¢V V0 a
1 2
b (sx + sy)
E
V0 (400)(800)(20) 6.4 * 106 mm3
;
Also, a
1 2
b (sx + sy) 140 * 106
E
‹ ¢V (6.4 * 106 mm3)(140 * 106)
896 mm3
(Increase in volume)
;
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Page 611
SECTION 7.5
Hooke’s Law for Plane Stress
611
Problem 7.5-7 Solve the preceding problem for an aluminum plate with sx 12,000 psi (tension), sy 3,000 psi
(compression), dimensions 20 30 0.5 in., E 10.5 106 psi, and 0.33.
Solution 7.5-7
Biaxial stress
sx 12,000 psi sy 3,000 psi
(b) CHANGE IN THICKNESS
E 10.5 * 10 psi 0.33
6
Eq. (7-39c): âz Dimensions of Plate: 20 in. * 30 in. * 0.5 in.
Shear Modulus (Eq. 7-38):
G
282.9 * 106
¢t âz t 141 * 106 in.
E
3.9474 * 106 psi
2(1 + )
(c) CHANGE IN VOLUME
Principal stresses: s1 12,000 psi
From Eq. (7-47): ¢V V0 a
s2 3,000 psi
s1 s2
7,500 psi
2
tmax
Eq. (7-35): gmax 1,900 * 106
G
;
(Decrease in thickness)
(a) MAXIMUM IN-PLANE SHEAR STRAIN
Eq. (7-26): tmax (s + sy)
E x
1 2
b (sx + sy)
E
V0 (20)(30)(0.5) 300 in.3
;
Also, a
1 2
b (sx + sy) 291.4 * 106
E
‹ ¢V (300 in.3)(291.4 * 106)
0.0874 in.3
;
(Increase in volume)
Problem 7.5-8 A brass cube 50 mm on each edge is compressed
P = 175 kN
in two perpendicular directions by forces P 175 kN (see figure).
Calculate the change V in the volume of the cube and the
strain energy U stored in the cube, assuming E 100 GPa and
0.34.
Solution 7.5-8
P = 175 kN
Biaxial stress-cube
sx sy P
b
2
(175 kN)
(50 mm)2
70.0 MPa
CHANGE IN VOLUME
Eq. (7-47): e 1 2
(sx + sy) 448 * 106
E
V0 b 3 (50 mm)3 125 * 103mm3
¢V eV0 56 mm3
Side b 50 mm P 175 kN
E 100 GPa 0.34 ( Brass)
(Decrease in volume)
;
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Analysis of Stress and Strain
U uV0 (0.03234 MPa)(125 * 103 mm3)
STRAIN ENERGY
Eq. (7-50): u 1
(s2 + s2y 2sxsy)
2E x
4.04 J
;
0.03234 MPa
Problem 7.5-9 A 4.0-inch cube of concrete (E 3.0 * 106 psi, 0.1)
is compressed in biaxial stress by means of a framework that is loaded as
shown in the figure.
Assuming that each load F equals 20 k, determine the change ¢V in
the volume of the cube and the strain energy U stored in the cube.
F
F
Solution 7.5-9
Biaxial stress – concrete cube
CHANGE IN VOLUME
b 4 in.
Eq. (7-47): e E 3.0 * 106 psi
V0 b 3 (4 in.)3 64 in.3
0.1
F 20 kips
1 2
(sx + sy) 0.0009429
E
¢V eV0 0.0603 in.3
;
(Decrease in volume)
STRAIN ENERGY
Joint A:
Eq. (7-50): u P F12
28.28 kips
sx sy P
1768 psi
b2
1
(s2 + s2y 2sxsy)
2E x
0.9377 psi
U uV0 60.0 in.-lb
;
Py
Problem 7.5-10 A square plate of width b and thickness t is loaded by
normal forces Px and Py, and by shear forces V, as shown in the figure.
These forces produce uniformly distributed stresses acting on the side
faces of the place.
Calculate the change V in the volume of the plate and the strain
energy U stored in the plate if the dimensions are b 600 mm and
t 40 mm, the plate is made of magnesium with E 45 GPa and
v 0.35, and the forces are Px 480 kN, Py 180 kN, and
V 120 kN.
t
V
y
Px
V
V
b O
b
x
V
Py
Probs. 7.5-10 and 7.5-11
Px
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SECTION 7.5 Hooke’s Law for Plane Stress
613
Solution 7.5-10 Square plate in plane stress
b 600 mm
t 40 mm
E 45 GPa
v 0.35 (magnesium)
Px
20.0 MPa
bt
Py
sy 7.5 MPa
bt
V
txy 5.0 MPa
bt
Px 480 kN
sx Py 180 kN
V 120 kN
CHANGE IN VOLUME
1 2
Eq. (7-47): e (sx + sy) 183.33 * 106
E
V0 b 2t 14.4 * 106 mm3
¢V eV0 2640 mm3
;
(Increase in volume)
STRAIN ENERGY
Eq. (7-50): u G
t2xy
1 2
(sx + s2y 2sxsy) +
2E
2G
E
16.667 GPa
2(1 + )
Substitute numerical values:
u 4653 Pa
U uV0 67.0 N # m 67.0 J
;
Problem 7.5-11 Solve the preceding problem for an aluminum plate with b 12 in., t 1.0 in., E 10,600 ksi, 0.33,
Px 90 k, Py 20 k, and V 15 k.
Solution 7.5-11 Square plate in plane stress
b 12.0 in.
t 1.0 in.
STRAIN ENERGY
E 10,600 ksi 0.33 (aluminum)
Px
7500 psi
bt
Py
sy 1667 psi
bt
Px 90 k
sx Py 20 k
V 15 k
txy G
u 2.591 psi
V
1250 psi
bt
U uV0 373 in.-lb
1 2
(sx + sy) 294 * 106
E
V0 b 2t 144 in.3
¢V eV0 0.0423 in.3
(Increase in volume)
E
3985 ksi
2(1 + )
Substitute numerical values:
CHANGE IN VOLUME
Eq. (7-47): e t2xy
1
2
2
Eq. (7-50): u (s + sy 2sxsy) +
2E x
2G
;
;
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Analysis of Stress and Strain
Problem 7.5-12 A circle of diameter d 200 mm is etched on a
z
brass plate (see figure). The plate has dimensions 400 400 20 mm.
Forces are applied to the plate, producing uniformly distributed normal
stresses x 42 MPa and y 14 MPa.
Calculate the following quantities: (a) the change in length ac
of diameter ac; (b) the change in length bd of diameter bd; (c) the
change t in the thickness of the plate; (d) the change V in the
volume of the plate, and (e) the strain energy U stored in the plate.
(Assume E 100 GPa and v 0.34.)
y
sy
d
sx
a
c
sx
b
x
sy
Solution 7.5-12
Plate in biaxial stress
sx 42 MPa sy 14 MPa
(c) CHANGE IN THICKNESS
(s + sy)
E x
190.4 * 106
Eq. (7-39c): âz Dimensions: 400 * 400 * 20 (mm)
Diameter of circle: d 200 mm
E 100 GPa 0.34 (Brass)
¢t âz t 0.00381 mm
(decrease)
;
(a) CHANGE IN LENGTH OF DIAMETER IN x DIRECTION
Eq. (7-39a): âx 1
(s sy) 372.4 * 106
E x
¢ac âx d 0.0745 mm
(increase)
;
(b) CHANGE IN LENGTH OF DIAMETER IN y DIRECTION
Eq. (7-39b): ây 1
(s sx) 2.80 * 106
E y
¢bd ây d
560 * 106 mm
(decrease)
;
(d) CHANGE IN VOLUME
Eq. (7-47):
e
1 2
(sx + sy) 179.2 * 106
E
V0 (400)(400)(20) 3.2 * 106 mm3
¢V eV0 573 mm3 ;
(increase)
(e) STRAIN ENERGY
Eq. (7-50): u 1 2
(s + s2y 2sx sy)
2E x
7.801 * 103 MPa
U uV0 25.0 N # m 25.0 J
;
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SECTION 7.6
615
Triaxial Stress
Triaxial Stress
When solving the problems for Section 7.6, assume that the material is
linearly elastic with modulus of elasticity E and Poisson’s ratio n.
y
a
c
Problem 7.6-1 An element of aluminum in the form of a rectangular
parallelepiped (see figure) of dimensions a ⫽ 6.0 in., b ⫽ 4.0 in, and
c ⫽ 3.0 in. is subjected to triaxial stresses sx ⫽ 12,000 psi,
sy ⫽ ⫺ 4,000 psi, and sz ⫽ ⫺1,000 psi acting on the x, y, and z faces,
respectively.
Determine the following quantities: (a) the maximum shear stress tmax
in the material; (b) the changes ¢a, ¢b, and ¢c in the dimensions of the
element; (c) the change ¢V in the volume; and (d) the strain energy U
stored in the element. (Assume E ⫽ 10,400 ksi and ␯ ⫽ 0.33.)
b
O
x
z
Probs. 7.6-1 and 7.6-2
Solution 7.6-1
Triaxial stress
sx ⫽ 12,000 psi sy ⫽ ⫺4,000 psi
¢a ⫽ aâx ⫽ 0.0079 in. ( increase)
sz ⫽ ⫺1,000 psi
¢b ⫽ bây ⫽ ⫺0.0029 in. ( decrease)
a ⫽ 6.0 in. b ⫽ 4.0 in. c ⫽ 3.0 in.
E ⫽ 10,400 ksi ␯ ⫽ 0.33 ( aluminum)
(a) MAXIMUM SHEAR STRESS
s1 ⫽ 12,000 psi s2 ⫽ ⫺1,000 psi
(c) CHANGE IN VOLUME
Eq. (7-56):
e⫽
s3 ⫽ ⫺4,000 psi
tmax ⫽
¢c ⫽ câz ⫽ ⫺0.0011 in. ( decrease)
s1 ⫺ s3
⫽ 8,000 psi
2
;
M
;
1 ⫺ 2␯
(sx + sy + sz) ⫽ 228.8 * 10⫺6
E
V ⫽ abc
¢V ⫽ e (abc) ⫽ 0.0165 in.3 ( increase)
;
(b) CHANGES IN DIMENSIONS
Eq. (7-53 a): âx ⫽
sx
␯
⫺ (sy + sz)
E
E
⫽1312.5 * 10⫺6
Eq. (7- 53 b): ây ⫽
sy
E
⫺
␯
(s + sx)
E z
⫽⫺733.7 * 10⫺6
Eq. (7-53 c): âz ⫽
sz
E
⫺
␯
(s + sy)
E x
⫽⫺350.0 * 10⫺6
(d) STRAIN ENERGY
Eq. (7-57a): u ⫽
1
(s â + sy ây + sz âz)
2 x x
⫽ 9.517 psi
U ⫽ u (abc) ⫽ 685 in.-lb
;
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Analysis of Stress and Strain
Problem 7.6-2 Solve the preceding problem if the element is steel (E = 200 GPA, ␯ ⫽ 0.30) with dimensions a = 300 mm,
b = 150 mm, and c = 150 mm and the stresses are sx ⫽ ⫺60 MPa, sy ⫽ ⫺40 MPa, and sz ⫽ ⫺40 MPa.
Solution 7.6-2
Triaxial stress
sx ⫽ ⫺60 MPa sy ⫽ ⫺40 MPa
¢a ⫽ aâx ⫽ ⫺0.0540 mm (decrease)
sz ⫽ ⫺40 MPa
¢b ⫽ bây ⫽ ⫺0.0075 mm (decrease)
a ⫽ 300 mm
b ⫽ 150 mm
E ⫽ 200 GPa
␯ ⫽ 0.30
c ⫽ 150 mm
(steel)
(a) MAXIMUM SHEAR STRESS
s1 ⫽ ⫺40 MPa s2 ⫽ ⫺40 MPa
s3 ⫽ ⫺60 MPa
tmax ⫽
s1 ⫺ s3
⫽ 10.0 MPa
2
Eq. (7-53 b): ây ⫽
Eq. (7-53 c): âz ⫽
sx
␯
⫺ (sy + sz) ⫽ ⫺ 180.0 * 10⫺6
E
E
sy
E
sz
E
(c) CHANGE IN VOLUME
Eq. (7-56):
e⫽
M
;
1 ⫺ 2␯
(sx + sy + sz) ⫽ ⫺280.0 * 10⫺6
E
V ⫽ abc
;
(b) CHANGES IN DIMENSIONS
Eq. (7-53 a): âx ⫽
¢c ⫽ câz ⫽ ⫺0.0075 mm. (decrease)
⫺
␯
(s + sx) ⫽ ⫺50.0 * 10⫺6
E z
⫺
␯
(s + sy) ⫽ ⫺ 50.0 * 10⫺6
E x
¢V ⫽ e(abc) ⫽ ⫺ 1890 mm3 (decrease)
;
(d) STRAIN ENERGY
1
(s â + sy ây + sz âz)
2 x x
⫽ 0.00740 MPa
Eq. (7-57 a): u ⫽
U ⫽ u (abc) ⫽ 50.0 N # m ⫽ 50.0 J
;
y
Problem 7.6-3 A cube of cast iron with sides of length a = 4.0 in.
(see figure) is tested in a laboratory under triaxial stress. Gages mounted
on the testing machine show that the compressive strains in the material
are Px ⫽ ⫺ 225 * 10⫺6 and Py ⫽ Pz ⫽ ⫺37.5 * 10⫺6.
Determine the following quantities: (a) the normal stresses sx, sy, and
sz acting on the x, y, and z faces of the cube; (b) the maximum shear
stress tmax in the material; (c) the change ¢V in the volume of the
cube; and (d) the strain energy U stored in the cube. (Assume E = 14,000
ksi and ␯ ⫽ 0.25.)
a
a
a
O
z
Probs. 7.6-3 and 7.6-4
x
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SECTION 7.6 Triaxial Stress
617
Solution 7.6-3 Triaxial stress (cube)
âx ⫽ ⫺ 225 * 10⫺6
ây ⫽ ⫺ 37.5 * 10⫺6
âz ⫽ ⫺ 37.5 * 10⫺6
a ⫽ 4.0 in.
(c) CHANGE IN VOLUME
Eq. (7-55): e ⫽ âx + ây + âz ⫽ ⫺ 0.000300
V ⫽ a3
E ⫽ 14,000 ksi ␯ ⫽ 0.25 (cast iron)
¢V ⫽ ea 3 ⫽ ⫺ 0.0192 in.3 ( decrease)
(a) NORMAL STRESSES
Eq. (7-54a):
(d) STRAIN ENERGY
E
[(1 ⫺ ␯)âx + ␯(ây + âz)]
sx ⫽
(1 + ␯)(1 ⫺ 2␯)
⫽ ⫺ 4200 psi
;
Eq. (7-57a): u ⫽
;
1
(sx âx + sy ây + sz âz)
2
⫽ 0.55125 psi
U ⫽ ua ⫽ 35.3 in.-lb
3
In a similar manner, Eqs. (7-54 b and c) give
sy ⫽ ⫺ 2100 psi sz ⫽ ⫺ 2100 psi ;
;
(b) MAXIMUM SHEAR STRESS
s1 ⫽ ⫺ 2100 psi s2 ⫽ ⫺ 2100 psi
s3 ⫽ ⫺ 4200 psi
tmax ⫽
s1 ⫺ s3
⫽ 1050 psi
2
;
Problem 7.6-4 Solve the preceding problem if the cube is granite (E ⫽ 60 GPa, ␯ ⫽ 0.25) with dimensions a = 75 mm and
compressive strains Px ⫽ ⫺ 720 * 10⫺6 and Py ⫽ Pz ⫽ ⫺ 270 * 10⫺6.
Solution 7.6-4 Triaxial stress (cube)
âx ⫽ ⫺ 720 * 10⫺6
ây ⫽ ⫺ 270 * 10⫺6
âz ⫽ ⫺ 270 * 10⫺6
a ⫽ 75 mm
␯ ⫽ 0.25
E ⫽ 60 GPa
(Granite)
;
Eq. (7-55): e ⫽ âx + ây + âz ⫽ ⫺ 1260 * 10⫺6
V ⫽ a3
E
[(1 ⫺ ␯)âx + ␯(âx + âz)]
sx ⫽
(1 + ␯)(1 ⫺ 2␯)
;
In a similar manner, Eqs. (7-54 b and c) give
sy ⫽ ⫺ 43.2 MPa sz ⫽ ⫺ 43.2 MPa ;
(b) MAXIMUM SHEAR STESS
s1 ⫽ ⫺ 43.2 MPa s2 ⫽ ⫺ 43.2 MPa
s3 ⫽ ⫺ 64.8 MPa
s1 ⫺ s3
⫽ 10.8 MPa
2
(c) CHANGE IN VOLUME
(a) NORMAL STRESSES
Eq.(7-54a):
⫽ ⫺ 64.8 MPa
tmax ⫽
¢V ⫽ ea 3 ⫽ ⫺ 532 mm3 ( decrease)
;
(d) STRAIN ENERGY
1
Eq. (7-57 a): u ⫽ (sxâx + syây + szâz)
2
⫽ 0.03499 MPa ⫽ 34.99 kPa
U ⫽ ua ⫽ 14.8 N # m ⫽ 14.8 J
3
;
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Analysis of Stress and Strain
Problem 7.6-5 An element of aluminum in triaxial stress (see figure) is
subjected to stresses sx ⫽ 5200 psi (tension), sy ⫽ ⫺4750 psi
(compression), and sz ⫽ ⫺3090 psi (compression). It is also known that the normal strains in the x and y directions
are Px ⫽ 7138.8 * 10⫺6 (elongation) and Py ⫽ ⫺502.3 * 10⫺6 (shortsx
ening).
What is the bulk modulus K for the aluminum?
y
sy
sz
sx
O
x
sz
sy
z
Probs. 7.6-5 and 7.6-6
Solution 7.6-5
Triaxial stress (bulk modulus)
sx ⫽ 5200 psi sy ⫽ ⫺4750 psi
sz ⫽ ⫺3090 psi âx ⫽ 713.8 * 10
ây ⫽ ⫺502.3 * 10
Substitute numerical values and rearrange:
⫺6
⫺6
(713.8 * 10⫺6) E ⫽ 5200 + 7840 ␯
(⫺502.3 * 10
⫺6
) E ⫽ ⫺4750 ⫺ 2110 ␯
Find K.
Units: E = psi
sx
␯
⫺ (sy + sz)
E
E
sy
␯
Eq. (7-53 b): ây ⫽
⫺ (sx + sy)
E
E
Solve simultaneously Eqs. (1) and (2):
Eq. (7-53 a): âx ⫽
(1)
(2)
E ⫽ 10.801 * 106 psi ␯ ⫽ 0.3202
Eq. (7-16): K ⫽
E
⫽ 10.0 * 10⫺6 psi
3(1 ⫺ 2␯)
;
Problem 7.6-6
Solve the preceding problem if the material is nylon subjected to compressive stresses
sx ⫽ ⫺ 4.5 MPa, sy ⫽ ⫺3.6 MPa, and sz ⫽ ⫺2.1 MPa, and the normal strains are Px ⫽ ⫺740 * 10⫺6 and
Py ⫽ ⫺ 320 * 10⫺6 (shortenings).
Solution 7.6-6
Triaxial stress (bulk modulus)
sx ⫽ ⫺4.5 MPa sy ⫽ ⫺3.6 MPa
sz ⫽ ⫺2.1 MPa âx ⫽ ⫺740 * 10
ây ⫽ ⫺320 * 10
Substitute numerical values and rearrange:
⫺6
⫺6
Find K.
sx
␯
⫺ (sy + sz)
E
E
sy
␯
Eq. (7-53 b): ây ⫽
⫺ (sz + sx)
E
E
Eq. (7-53 a): âx ⫽
(⫺ 740 * 10⫺6) E ⫽ ⫺4.5 + 5.7 ␯
(⫺ 320 * 10
⫺6
(1)
) E ⫽ ⫺3.6 + 6.6 ␯
(2)
Units: E = MPa
Solve simultaneously Eqs. (1) and (2):
E ⫽ 3,000 MPa ⫽ 3.0 GPa ␯ ⫽ 0.40
Eq. (7-16): K ⫽
E
⫽ 5.0 GPa
3(1 ⫺ 2␯)
;
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SECTION 7.6
Problem 7.6-7 A rubber cylinder R of length L and cross-sectional
F
F
area A is compressed inside a steel cylinder S by a force F that
applies a uniformly distributed pressure to the rubber (see figure).
(a) Derive a formula for the lateral pressure p between the
rubber and the steel. (Disregard friction between the rubber
and the steel, and assume that the steel cylinder is rigid when
compared to the rubber.)
(b) Derive a formula for the shortening d of the rubber cylinder.
Solution 7.6-7
619
Triaxial Stress
R
S
L
S
Rubber cylinder
Solve for p: p ⫽
F
v
a b
1⫺v A
;
(b) SHORTENING
Eq. (7-53 b): ây ⫽
sx ⫽ ⫺p sy ⫽ ⫺
F
A
sz ⫽ ⫺p
âx ⫽ âz ⫽ 0
⫽ ⫺
E
⫺
␯
(s + sx)
E z
␯
F
⫺ ( ⫺2p)
EA
E
Substitute for p and simplify:
ây ⫽
F (1 + ␯)(⫺1 + 2␯)
EA
1⫺␯
(Positive ây represents an increase in strain, that is,
elongation.)
(a) LATERAL PRESSURE
Eq. (7-53 a): âx ⫽
sy
sx
␯
⫺ (sy + sz)
E
E
F
or 0 ⫽ ⫺p ⫺ ␯ a ⫺ ⫺ pb
A
d ⫽ ⫺âyL
d⫽
(1 + ␯)(1 ⫺ 2␯) FL
a
b
(1 ⫺ ␯)
EA
;
(Positive d represents a shortening of the rubber
cylinder.)
Problem 7.6-8 A block R of rubber is confined between plane
F
F
parallel walls of a steel block S (see figure). A uniformly distributed pressure p0 is applied to the top of the rubber block by
a force F.
(a) Derive a formula for the lateral pressure p between the
rubber and the steel. (Disregard friction between the
rubber and the steel, and assume that the steel block is
rigid when compared to the rubber.)
(b) Derive a formula for the dilatation e of the rubber.
(c) Derive a formula for the strain-energy density u of the
rubber.
S
S
R
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CHAPTER 7 Analysis of Stress and Strain
Solution 7.6-8 Block of rubber
(b) DILATATION
Eq. (7-56): e ⫽
1 ⫺ 2␯
(sx + sy + sz)
E
⫽
1 ⫺ 2␯
(⫺ p ⫺ p0)
E
Substitute for p:
e⫽ ⫺
sx ⫽ ⫺ p
sy ⫽ ⫺ p0
(a) LATERAL PRESSURE
OR
sx
␯
⫺ (sy + sz)
E
E
0 ⫽ ⫺ p ⫺ ␯ ( ⫺p0)
;
sz ⫽ 0
âx ⫽ 0 ây Z 0 âz Z 0
Eq. (7-53 a): âx ⫽
(1 + ␯)(1 ⫺ 2␯)p0
E
(c) STRAIN ENERGY DENSITY
Eq. (7-57b):
u⫽
‹ p ⫽ ␯p0
;
1
v
(s2x + s2y + s2z ) ⫺ (sx sy + sx sz + sy sz)
2E
E
Substitute for sx, sy, sz, and p:
u⫽
(1 ⫺ ␯2)p 20
2E
;
Problem 7.6-9 A solid spherical ball of brass (E ⫽ 15 * 106 psi ␯ ⫽ 0.34) is lowered into the ocean to a depth of 10,000 ft.
The diameter of the ball is 11.0 in.
Determine the decrease ¢d in diameter, the decrease ¢V in volume, and the strain energy U of the ball.
Solution 7.6-9
E ⫽ 15 * 10
⫺6
Brass sphere
psi ␯ ⫽ 0.34
DECREASE IN VOLUME
Lowered in the ocean to depth h ⫽ 10,000 ft
Eq. (7-60): e ⫽ 3â0 ⫽ 283.6 * 10⫺6
4
4
11.0 in. 3
b ⫽ 696.9 in.3
V0 ⫽ pr 3 ⫽ (p)a
3
3
2
Diameter d ⫽ 11.0 in.
Sea water: g ⫽ 63.8 lb/ft3
Pressure: s0 ⫽ gh ⫽ 638,000 lb/ft2 ⫽ 4431 psi
DECREASE IN DIAMETER
s0
Eq. (7-59): â0 ⫽ (1 ⫺ 2␯) ⫽ 94.53 * 10⫺6
E
¢d ⫽ â0d ⫽ 1.04 * 10⫺3 in.
(decrease)
;
¢V ⫽ eV0 ⫽ 0.198 in.3
(decrease)
;
STRAIN ENERGY
Use Eq. (7-57 b) with sx ⫽ sy ⫽ sz ⫽ s0:
u⫽
3(1 ⫺ 2␯)s20
⫽ 0.6283 psi
2E
U ⫽ uV0 ⫽ 438 in.-lb
;
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SECTION 7.6 Triaxial Stress
621
Problem 7.6-10 A solid steel sphere (E ⫽ 210 GPa, ␯ = 0.3) is subjected to hydrostatic pressure p such that its volume is
reduced by 0.4%.
(a) Calculate the pressure p.
(b) Calculate the volume modulus of elasticity K for the steel.
(c) Calculate the strain energy U stored in the sphere if its diameter is d ⫽ 150 mm
Solution 7.6-10
Steel sphere
E ⫽ 210 GPa ␯ ⫽ 0.3
(b) VOLUME MODULUS OF ELASTICITY
Hydrostatic Pressure. V0 = Initial volume
¢V ⫽ 0.004 V0
Dilatation: e ⫽
¢V
⫽ 0.004
V0
3s0(1 ⫺ 2␯)
Eq.(7-60): e ⫽
E
s0 ⫽
s0
700 MPa
⫽
⫽ 175 GPa
E
0.004
;
(c) STRAIN ENERGY (d = diameter)
d = 150 mm r = 75 mm
From Eq. (7-57b) with sx ⫽ sy ⫽ sz ⫽ s0:
(a) PRESSURE
or
Eq. (7-63): K ⫽
Ee
⫽ 700 MPa
3(1 ⫺ 2␯)
Pressure p ⫽ s0 ⫽ 700 MPa
u⫽
3(1 ⫺ 2␯)s20
⫽ 1.40 MPa
2E
V0 ⫽
4pr 3
⫽ 1767 * 10⫺6 m3
3
U ⫽ uV0 ⫽ 2470 N # m ⫽ 2470 J
;
;
Problem 7.6-11 A solid bronze sphere (volume modulus of elasticity K ⫽ 14.5 ⫻ 106 psi) is suddenly heated around its
outer surface. The tendency of the heated part of the sphere to expand produces uniform tension in all directions at the center
of the sphere.
If the stress at the center is 12,000 psi, what is the strain? Also, calculate the unit volume change e and the strain-energy
density u at the center.
Solution 7.6-11 Bronze sphere (heated)
K ⫽ 14.5 * 106 psi
s0 ⫽ 12,000 psi (tension at the center)
STRAIN AT THE CENTER OF THE SPHERE
s0
Eq. (7-59): â0 ⫽ (1 ⫺ 2␯)
E
E
Eq. (7-61): K ⫽
3(1 ⫺ 2␯)
Combine the two equations:
â0 ⫽
s0
⫽ 276 * 10 ⫺6
3K
;
UNIT VOLUME CHANGE AT THE CENTER
Eq. (7-62): e ⫽
s0
⫽ 828 * 10 ⫺6
K
;
STRAIN ENERGY DENSITY AT THE CENTER
Eq. (7-57b) with sx ⫽ sy ⫽ sz ⫽ s0:
3(1 ⫺ 2␯)s20
s02
⫽
2E
2K
u ⫽ 4.97 psi ;
u⫽
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CHAPTER 7
Page 622
Analysis of Stress and Strain
Plane Strain
y
sy
When solving the problems for Section 7.7, consider only the in-plane strains (the
strains in the xy plane) unless stated otherwise. Use the transformation equations
of plane strain except when Mohr’s circle is specified (Problems 7.7-23 through
7.7-28).
sx
h
Problem 7.7-1 A thin rectangular plate in biaxial stress is subjected to
stresses sx and sy, as shown in part (a) of the figure on the next page. The
width and height of the plate are b 8.0 in. and h 4.0 in., respectively.
Measurements show that the normal strains in the x and y directions are
Px 195 * 106 and Py 125 * 106, respectively.
With reference to part (b) of the figure, which shows a two-dimensional
view of the plate, determine the following quantities: (a) the increase ¢d in the
length of diagonal Od; (b) the change ¢f in the angle f between diagonal Od
and the x axis; and (c) the change ¢c in the angle c between diagonal Od and
the y axis.
b
x
z
(a)
y
d
c
h
f
O
x
b
(b)
Probs. 7.7-1 and 7.7-2
Solution 7.7-1
Plate in biaxial stress
For u f 26.57°, âx1 130.98 * 106
¢d âx1L d 0.00117 in.
;
(b) CHANGE IN ANGLE f
Eq. (7-68): a (âx ây) sinu cosu gxy sin2u
For u f 26.57°: a 128.0 * 106 rad
b 8.0 in.
h 4.0 in.
ây 125 * 106
âx 195 * 106
gxy 0
Minus sign means line Od rotates clockwise
(angle f decreases).
¢f 128 * 106 rad (decrease)
h
f arctan 26.57°
b
;
(c) CHANGE IN ANGLE c
L d 1b 2 + h2 8.944 in.
Angle c increases the same amount that f
decreases.
(a) INCREASE IN LENGTH OF DIAGONAL
¢c 128 * 106 rad (increase)
âx1 âx ây
âx + ây
+
2
2
cos 2u +
gxy
2
sin 2u
;
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Page 623
SECTION 7.7
Plane Strain
623
Problem 7.7-2 Solve the preceding problem if b 160 mm, h 60 mm, Px 410 * 106, and Py 320 * 106.
Solution 7.7-2
Plate in biaxial stress
For u f 20.56°: âx1 319.97 * 10 6
¢d âx1 L d 0.0547 mm
;
(b) CHANGE IN ANGLE f
Eq. (7-68): a (âx ây) sin u cos u gxy sin2u
For u f 20.56°: a 240.0 * 106 rad
b 160 mm
h 60 mm
ây 320 * 106
Minus sign means line Od rotates clockwise (angle f
decreases.)
âx 410 * 106
gxy 0
¢f 240 * 106 rad (decrease)
h
f arctan 20.56°
b
2
1
L d b + h2 170.88 mm
(c) CHANGE IN ANGLE c
Angle c increases the same amount that f
decreases.
¢c 240 * 106 rad (increase)
(a) INCREASE IN LENGTH OF DIAGONAL
âx1 âx ây
âx + ây
+
2
2
cos 2u +
;
gxy
2
;
sin 2u
Problem 7.7-3 A thin square plate in biaxial stress is subjected
to stresses sx and sy, as shown in part (a) of the figure . The
width of the plate is b 12.0 in. Measurements show that the
normal strains in the x and y directions are Px 427 * 106
and Py 113 * 106, respectively.
With reference to part (b) of the figure, which shows a
two-dimensional view of the plate, determine the following
quantities: (a) the increase ¢d in the length of diagonal Od;
(b) the change ¢f in the angle f between diagonal Od and
the x axis; and (c) the shear strain g associated with diagonals
Od and cf (that is, find the decrease in angle ced).
y
sy
y
c
sx
b
e
b
b
f
x
z
d
O
b
(b)
(a)
PROBS. 7.7-3 and 7.7-4
f
x
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CHAPTER 7 Analysis of Stress and Strain
Solution 7.7-3 Square plate in biaxial stress
(b) CHANGE IN ANGLE f
Eq. (7-68): a (âx ây) sin u cos u gxy sin2u
For u f 45°: a 157 * 106 rad
Minus sign means line Od rotates clockwise (angle f
decreases.)
¢f 157 * 106 rad (decrease)
(c) SHEAR STRAIN BETWEEN DIAGONALS
âx 427 * 106
b 12.0 in.
ây 113 * 106
Eq. (7-71b):
gxy 0
f 45°
;
gx1y1
2
âx ây
2
sin 2u +
gxy
2
cos 2u
For u f 45°: gx1y1 314 * 106 rad
L d b12 16.97 in.
(Negative strain means angle ced increases)
(a) INCREASE IN LENGTH OF DIAGONAL
g 314 * 106 rad
âx1 âx ây
âx + ây
+
2
2
cos 2u +
gxy
2
;
sin 2u
For u f 45°: âx1 270 * 106
¢d âx1L d 0.00458 in.
;
Problem 7.7-4 Solve the preceding problem if b 225 mm, Px 845 * 106 , and Py 211 * 106.
Solution 7.7-4 Square plate in biaxial stress
(a) INCREASE IN LENGTH OF DIAGONAL
âx1 âx ây
âx + ây
+
2
2
cos 2u +
gxy
2
sin 2u
For u f 45°: âx1 528 * 106
¢d âx1L d 0.168 mm
;
(b) CHANGE IN ANGLE f
Eq. (7-68): a (âx ây) sin u cos u gxy sin2u
b 225 mm
ây 211 * 106
âx 845 * 106
f 45°
L d b12 318.2 mm
gxy 0
For u f 45°: a 317 * 106 rad
Minus sign means line Od rotates clockwise (angle f
decreases.)
¢f 317 * 106 rad (decrease)
;
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Page 625
SECTION 7.7
g x1y1
2
âx ây
2
625
For u f 45°: gx1y1 634 * 106 rad
(c) SHEAR STRAIN BETWEEN DIAGONALS
Eq. (7- 71b):
Plane Strain
sin 2u +
gxy
2
cos 2u
(Negative strain means angle ced increases)
g 634 * 106 rad
y
Problem 7.7-5 An element of material subjected to plane strain (see
figure) has strains as follows: Px 220 * 106, Py 480 * 106, and
gxy 180 * 106.
Calculate the strains for an element oriented at an angle u 50° and show these
strains on a sketch of a properly oriented element.
ey
gxy
1
O
1
ex
x
Probs. 7.7-5 through 7.7-10
Solution 7.7-5
Element in plane strain
âx 220 * 106
ây 480 * 106
gxy 180 * 106
âx1 gx1 y1
2
âx ây
âx + ây
+
2
2
âx ây
2
cos 2u +
sin 2u +
g xy
2
gxy
2
sin 2u
cos 2u
ây1 âx + ây âx1
For u 50°:
âx1 461 * 106
gx1y1 225 * 106
ây1 239 * 106
Problem 7.7-6 Solve the preceding problem for the following data: Px 420 * 106, Py 170 * 106, gxy 310 * 106,
and u 37.5°.
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Page 626
CHAPTER 7 Analysis of Stress and Strain
Solution 7.7-6 Element in plane strain
âx 420 * 106
ây 170 * 106
gxy 310 * 106
âx1 gx1y1
2
âx ây
âx + ây
+
2
2
âx ây
sin 2u +
2
cos 2u +
gxy
2
gxy
2
sin 2u
cos 2u
ây1 âx + ây âx1
For u 37.5°:
âx1 351 * 106
gx1y1 490 * 106
ây1 101 * 106
Problem 7.7-7 The strains for an element of material in plane strain (see figure) are as follows: Px 480 * 106,
Py 140 * 106, and gxy 350 * 106.
Determine the principal strains and maximum shear strains, and show these strains on sketches of properly oriented
elements.
Solution 7.7-7 Element in plane strain
âx 480 * 106
ây 140 * 106
gxy 350 * 106
PRINCIPAL STRAINS
â1, 2 âx + ây
;
2
âx ây 2
gxy 2
b + a b
A
2
2
a
310 * 106 ; 244 * 106
â2 66 * 106
â1 554 * 106
gxy
1.0294
tan 2up âx ây
2up 45.8° and
up 22.9° and
MAXIMUM SHEAR STRAINS
âx ây 2
gxy 2
gmax
a
b + a
b
2
A
2
2
134.2°
244 * 106
67.1°
gmax 488 * 106
For up 22.9°:
âx1 âx ây
âx + ây
+
2
2
cos 2u +
gxy
2
us1 up1 45° 67.9° or 112.1°
sin 2u
up2 67.1°
â1 554 * 106
â2 66 * 106
;
;
;
us2 us1 + 90° 22.1°
554 * 106
‹ up1 22.9°
gmax 488 * 106
gmin 488 * 106
âaver âx + ây
2
;
310 * 106
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Page 627
SECTION 7.7 Plane Strain
Problem 7.7-8 Solve the preceding problem for the following strains: Px 120 * 106, Py 450 * 106, and
gxy 360 * 106.
Solution 7.7-8 Element in plane strain
âx 120 * 106
ây 450 * 106
MAXIMUM SHEAR STRAINS
âx ây 2
gxy 2
gmax
a
b + a
b
2
A
2
2
gxy 360 * 106
PRINCIPAL STRAINS
â1,2 âx + ây
2
337 * 106
âx ây 2
gxy 2
a
;
b + a b
A
2
2
gmax 674 * 106
us1 up1 45° 118.9°
165 * 106 ; 377 * 106
â1 172 * 106
tan 2up gxy
âx ây
â2 502 * 106
0.6316
âaver up 163.9° and 73.9°
For up 163.9°:
âx ây
âx + ây
+
2
2
cos 2u +
gxy
2
sin 2u
172 * 106
‹ up1 163.9°
up2 73.9°
â1 172 * 106
â2 502 * 10
;
6
;
;
us2 us1 90° 28.9°
gmin 674 * 106
2up 327.7° and 147.7°
âx1 gmax 674 * 106
âx + ây
2
;
165 * 106
627
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Page 628
CHAPTER 7 Analysis of Stress and Strain
Problem 7.7-9 A element of material in plane strain (see figure) is subjected to strains Px 480 * 106, Py 70 * 106,
and gxy 420 * 106.
Determine the following quantities: (a) the strains for an element oriented at an angle u 75°, (b) the principal strains,
and (c) the maximum shear strains. Show the results on sketches of properly oriented element.
Solution 7.7-9 Element in plane strain
âx 480 * 106
ây 70 * 106
gxy 420 * 106
âx1 gx1y1
2
âx ây
âx + ây
+
2
2
âx ây
2
cos 2u +
sin 2u +
gxy
2
gxy
2
For up 22.85°:
âx1 âx ây
âx + ây
+
2
2
gxy
2
sin 2u
568 * 106
sin 2u
â1 568 * 106
‹ up1 22.8°
cos 2u
cos 2u +
up2 112.8
°
â2 18 * 10
6
;
;
ây1 âx + ây âx1
For u 75°:
âx1 202 * 106
gx1y1 569 * 106
ây1 348 * 106
MAXIMUM SHEAR STRAINS
âx ây 2
gxy 2
gmax
a
b + a
b 293 * 106
2
A
2
2
gmax 587 * 106
us1 up1 45° 22.2° or 157.8°
gmax 587 * 106
us2 us1 + 90° 67.8°
PRINCIPAL STRAINS
â1,2 ;
âx + ây
;
2
275 * 10
âx ây 2
gxy 2
b + a
b
A
2
2
6
a
; 293 * 10
6
â2 18 * 106
â1 568 * 106
gxy
1.0244
tan 2up âx ây
2up 45.69° and 225.69°
up 22.85° and 112.85°
gmin 587 * 106
âaver âx + ây
2
;
275 * 106
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Page 629
629
SECTION 7.7 Plane Strain
Problem 7.7-10 Solve the preceding problem for the following data: Px 1120 * 106, Py 430 * 106,
gxy 780 * 106, and u 45°.
Solution 7.7-10 Element in plane strain
âx 1120 * 106
gxy 780 * 10
âx1 gx1y1
2
ây 430 * 106
6
up2 155.7°
âx ây
âx + ây
+
2
2
âx ây
2
â1 254 * 106
‹ up1 65.7°
sin 2u +
cos 2u +
gxy
2
gxy
2
â2 1296 * 10
sin 2u
cos 2u
ây1 âx + ây âx1
For u 45°:
âx1 385 * 106
gx1y1 690 * 106
ây1 1165 * 106
MAXIMUM SHEAR STRAINS
âx ây 2
gxy 2
gmax
a
b + a
b
2
A
2
2
521 * 106
gmax 1041 * 106
us1 up1 45° 20.7°
gmax 1041 * 106
PRINCIPAL STRAINS
âx + ây
âx ây 2
gxy 2
â1, 2 ;
a
b + a
b
2
A
2
2
775 * 10
6
; 521 * 10
us2 us1 + 90° 110.7°
gmin 1041 * 106
6
6
â2 1296 * 10
â1 254 * 10
gxy
1.1304
tan 2up âx ây
6
2up 131.5° and 311.5°
up 65.7° and 155.7°
For up 65.7°:
âx1 âx ây
âx + ây
+
2
254 * 106
2
cos 2u +
gxy
2
;
sin 2u
âaver âx + ây
2
;
775 * 106
6
;
;
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CHAPTER 7
Page 630
Analysis of Stress and Strain
Problem 7.7-11 A steel plate with modulus of elasticity E 30 * 106 psi and
Poisson’s ratio 0.30 is loaded in biaxial stress by normal stresses sx and sy
(see figure). A strain gage is bonded to the plate at an angle f 30°.
If the stress sx is 18,000 psi and the strain measured by the gage is P 407 * 106,
what is the maximum in-plane shear stress (tmax)xy and shear strain (gmax)xy? What is the
maximum shear strain (gmax)xz in the xz
plane? What is the maximum shear strain (gmax)yz in the yz plane?
y
sy
sx
f
x
z
Probs. 7.7-11 and 7.7-12
Solution 7.7-11
Steel plate in biaxial stress
sx 18,000 psi
gxy 0
E 30 * 106 psi
sy ?
MAXIMUM IN-PLANE SHEAR STRESS
0.30
Strain gage: f 30°
â 407 * 10
(tmax)xy 6
sx sy
2
7800 psi
;
UNITS: All stresses in psi.
STRAINS FROM EQS. (1), (2), AND (3)
STRAIN IN BIAXIAL STRESS (EQS. 7-39)
âx 576 * 106
âx 1
1
(18,000 0.3sy)
(s sy) E x
30 * 106
(1)
ây 1
1
(sy 5400)
(sy sx ) E
30 * 106
(2)
âz 0.3
(s sy) (18,000sy)
E x
30 * 106
âx ây
âx + ây
+
2
2
cos 2u +
gxy
2
âz 204 * 106
MAXIMUM SHEAR STRAINS (EQ. 7-75)
xy plane:
(3)
2
(gmax)xz
2
yz plane:
(gmax) yz
2
gyz 0
(4)
A
a
âx ây
2
gxy
2
b + a
2
A
a
âx âz
2
2
b + a
b
2
6
gxz
2
;
b
2
(gmax) xz 780 * 106
gxz 0
sin 2u
1
1
407 * 106 a b a
b(12,600 + 0.7sy)
2 30 * 106
1
1
+ a ba
b (23,400 1.3sy) cos 60°
2 30 * 106
sy 2400 psi
Solve for sy :
(gmax) xy
gxy 0 (gmax) xy 676 * 10
xz plane:
STRAINS AT ANGLE f 30° (EQ. 7-71a)
âx1 ây 100 * 106
A
a
ây 2
âz 2
b + a
;
gyz 2
2
b
(gmax) yz 104 * 106
;
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Page 631
SECTION 7.7 Plane Strain
631
Problem 7.7-12 Solve the preceding problem if the plate is made of aluminum with E 72 GPa and 1/3, the stress
sx is 86.4 MPa, the angle f is 21° , and the strain P is 946 * 106.
Solution 7.7-12 Aluminum plate in biaxial stress
sx 86.4 MPa
gxy 0
E 72 GPa
sy ?
MAXIMUM IN-PLANE SHEAR STRESS
1/3
â 946 * 10
Strain gage: f 21°
(tmax) xy 6
(1)
1
1
(sy 28.8)
ây (sy sx) E
72,000
(2)
1/ 3
(86.4 sy)
âz (sx sy) E
72,000
946 * 10
2
6
2
cos 2u +
gxy
2
(gmax) xy
2
(gmax)xz
2
âx ây
2
2
b + a
gxy
2
sin 2u
yz plane:
(gmax) yz
2
gyz 0
A
a
âx âz
2
2
b + a
b
2
6
gxz
2
b
;
2
(gmax) xz 1600 * 106
gxz 0
(4)
A
a
gxy 0 (gmax) xy 1200 * 10
xz plane:
1
4
1
b a115.2 sy b cos 42°
+ a ba
2 72,000
3
sy 21.55 MPa
xy plane:
(3)
1
1
2
a ba
b a 57.6 + sy b
2 72,000
3
Solve for sy:
ây 101 * 106
MAXIMUM SHEAR STRAINS (EQ. 7-75)
STRAINS AT ANGLE f 21° (EQ. 7-71a)
+
;
âz 500 * 106
1
1
1
(86.4 sy)
âx (sx sy) E
72,000
3
âx1 32.4 MPa
âx 1100 * 106
STRAIN IN BIAXIAL STRESS (EQS. 7-39)
âx ây
2
STRAINS FROM EQS. (1), (2), AND (3)
UNITS: All stresses in MPa.
âx + ây
sx sy
A
a
ây 2
âz 2
b + a
;
gyz 2
2
b
(gmax) yz 399 * 106
;
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CHAPTER 7
Page 632
Analysis of Stress and Strain
sy
Problem 7.7-13 An element in plane stress is subjected to stresses sx 8400 psi,
sy 1100 psi, and txy 1700 psi (see figure). The material is aluminum with modulus
of elasticity E 10,000 ksi and Poisson’s ratio 0.33.
Determine the following quantities: (a) the strains for an element oriented at an angle
u 30°, (b) the principal strains, and (c) the maximum shear strains. Show the results on
sketches of properly oriented elements.
txy
y
O
x
Probs. 7.7-13 and 7.7-14
Solution 7.7-13
Element in plane strain
sx 8400 psi
sy 1100 psi
txy 1700 psi
E 10,000 ksi
PRINCIPAL STRAINS
0.33
HOOKE’S LAW (EQS. 7-34 AND 7-35)
âx 1
(s sy) 876.3 * 106
E x
ây 1
(s sx) 387.2 * 106
E y
gxy txy
2txy(1 + )
G
FOR u 30°:
âx1 E
+
2
2
cos 2u +
2
âx ây
2
sin 2u +
gxy
2
;
2
â1 426 * 10
tan 2up gxy
2
756 * 106
gx1y1
âx + ây
A
245 * 10
452.2 * 106
âx ây
âx + ây
â1,2 âx ây
a
6
2
; 671 * 10
6
gxy
âx ây
2
b + a
gxy
2
434 * 106
gx1y1 868 * 106
ây1 âx + ây âx1 267 * 106
2
6
â2 961 * 106
0.3579
2up 19.7° and 199.7°
up 9.8° and 99.8°
FOR up 9.8°:
sin 2u
âx1 âx ây
âx + ây
+
2
2
cos 2u +
gxy
2
916 * 106
cos 2u
b
‹ up1 99.8° â1 426 * 106
up2 9.8° â2 916 * 10
6
;
;
sin 2u
sx
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Page 633
SECTION 7.7 Plane Strain
MAXIMUM SHEAR STRAINS
âx ây 2
gxy 2
gmax
a
b + a
b
2
A
2
2
671 * 106
gmax 1342 * 106
us1 up1 45° 54.8°
gmax 1342 * 106
;
us2 us1 + 90° 144.8°
gmin 1342 * 106
;
âx + ây
âaver 245 * 106
2
Problem 7.7-14 Solve the preceding problem for the following data: sx 150 MPa, sy 210 MPa,
txy 16 MPa, and u 50°. The material is brass with E 100 GPa and 0.34.
Solution 7.7-14 Element in plane strain
sx 150 MPa sy 210 MPa
txy 16 MPa E 100 GPa 0.34
HOOKE’S LAW (EQS. 7-34 AND 7-35)
âx 1
(s sy) 786 * 106
E x
ây 1
(s sx) 1590 * 106
E y
gxy txy
G
2txy(1 + )
E
429 * 106
FOR u 50°:
âx1 âx ây
âx + ây
+
2
2
cos 2u +
gxy
2
1469 * 106
gx1y1
2
âx ây
2
sin 2u +
gxy
2
cos 2u
358.5 * 106
gx1y1 717 * 106
ây1 âx + ây âx1 907 * 106
sin 2u
PRINCIPAL STRAINS
âx + ây
âx ây 2
gxy 2
;
b + a
b
â1,2 a
2
A
2
2
1188 * 106 ; 456 * 106
â1 732 * 106 â2 1644 * 106
gxy
0.5333
tan 2up âx ây
2up 151.9° and 331.9°
up 76.0° and 166.0°
633
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CHAPTER 7
Page 634
Analysis of Stress and Strain
FOR up 76.0°:
âx1 MAXIMUM SHEAR STRAINS
âx ây
âx + ây
+
2
1644 * 10
2
cos 2u +
gxy
2
sin 2u
456 * 106
6
‹ up1 166.0° â1 732 * 106
up2 76.0° â2 1644 * 106
âx ây 2
gxy 2
gmax
a
b + a
b
2
A
2
2
;
;
gmax 911 * 106
us1 up1 45° 121.0°
gmax 911 * 106
;
us2 us1 90° 31.0°
gmin 911 * 106
âaver âx + ây
2
;
1190 * 106
Problem 7.7-15 During a test of an airplane wing, the strain gage readings from a 45° rosette
(see figure) are as follows: gage A, 520 * 106; gage B, 360 * 106; and gage C, 80 * 106 .
Determine the principal strains and maximum shear strains, and show them on sketches of
properly oriented elements.
y
45°
B
C
45°
A
O
Probs. 7.7-15 and 7.7-16
x
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Page 635
SECTION 7.7 Plane Strain
Solution 7.7-15
45° strain rosette
âA 520 * 106
âC 80 * 10
âB 360 * 106
6
âx âA 520 * 10
6
ây âC 80 * 10
gxy 2âB âA âC 280 * 10
6
âx + ây
;
;
6
;
MAXIMUM SHEAR STRAINS
6
âx ây 2
gxy 2
gmax
a
b + a
b
2
A
2
2
331 * 106
PRINCIPAL STRAINS
A
2
‹ up1 12.5° â1 551 * 106
up2 102.5° â2 111 * 10
FROM EQS. (7-77) AND (7-78) OF EXAMPLE 7-8:
â1,2 a
âx ây
2
2
b + a
gxy
2
b
2
220 * 106 ; 331 * 106
â1 551 * 106
â2 111 * 106
gmax 662 * 106
us1 up1 45° 32.5°or 147.5°
gmax 662 * 106
gxy
âx ây
;
us2 us1 + 90° 57.5°
gmin 662 * 106
âaver tan 2up 635
âx + ây
2
;
220 * 106
0.4667
2up 25.0° and 205.0°
up 12.5° and 102.5°
For up 12.5°:
âx1 âx ây
âx + ây
+
2
2
cos 2u +
gxy
2
sin 2u
551 * 106
Problem 7.7-16 A 45°strain rosette (see figure) mounted on the surface of an automobile frame gives the following
readings: gage A, 310 * 106; gage B, 180 * 106; and gage C, 160 * 106.
Determine the principal strains and maximum shear strains, and show them on sketches of properly oriented elements.
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Page 636
CHAPTER 7 Analysis of Stress and Strain
Solution 7.7-16
45° strain rosette
âA 310 * 106
âB 180 * 106
MAXIMUM SHEAR STRAINS
âx ây 2
gxy 2
gmax
a
b + a
b
2
A
2
2
âC 160 * 106
FROM EQS. (7-77) AND (7-78) OF EXAMPLE 7-8:
âx âA 310 * 10
6
ây âC 160 * 10
gxy 2âB âA âC 210 * 10
6
2
A
a
âx ây
2
2
b + a
gxy
2
b
2
â1 332 * 106
gxy
âx ây
âaver â2 182 * 106
0.4468
2up 24.1° and 204.1°
up 12.0° and 102.0°
FOR up 12.0°:
âx + ây
âx ây
gxy
âx1 +
cos 2u +
sin 2u
2
2
2
332 * 106
‹ up1 12.0° â1 332 * 106
up2 102.0° â2 182 * 10
;
6
;
;
us2 us1 + 90° 57.0°
gmin 515 * 106
75 * 106 ; 257 * 106
tan 2up gmax 515 * 106
gmax 515 * 106
âx + ây
;
257 * 106
us1 up1 45° 33.0° or 147.0°
PRINCIPAL STRAINS
â1,2 6
âx + ây
2
;
75 * 106
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Page 637
SECTION 7.7
Problem 7.7-17 A solid circular bar of diameter d 1.5 in. is
Plane Strain
d
subjected to an axial force P and a torque T (see figure). Strain gages
A and B mounted on the surface of the bar give reading
Pa 100 * 106 and Pb 55 * 106. The bar is made of steel
having E 30 * 106 psi and 0.29.
637
T
P
C
(a) Determine the axial force P and the torque T.
(b) Determine the maximum shear strain gmax and the maximum
shear stress tmax in the bar.
B
45∞
A
C
Solution 7.7-17
Circular bar (plane stress)
Bar is subjected to a torque T and an axial force P.
E 30 * 10 psi 0.29
STRAIN AT u 45°
6
âx1 Diameter d 1.5 in.
+
2
2
cos 2u +
âx1 âB 55 * 106
STRAIN GAGES
At u 0°:
âx ây
âx + ây
gxy
2
sin 2u
2u 90°
Substitute numerical values into Eq. (1):
âA âx 100 * 106
55 * 106 35.5 * 106 (0.0649 * 106)T
At u 45°: âB 55 * 106
Solve for T:
T 1390 lb-in
;
ELEMENT IN PLANE STRESS
4P
P
sx A
pd 2
âx 100 * 106
sy 0
txy MAXIMUM SHEAR STRAIN AND MAXIMUM SHEAR STRESS
16T
gxy (0.1298 * 106)T 180.4 * 106 rad
pd 3
ây âx 29 * 106
sx
4P
E
pd 2E
âx ây 2
gxy 2
gmax
a
b + a
b
2
A
2
2
111 * 106 rad
AXIAL FORCE P
âx Eq.( 7-75):
2
P
pd Eâx
5300 lb
4
SHEAR STRAIN
txy
2txy(1 + )
32T(1 + )
gxy G
E
pd 3E
(0.1298 * 106)T (T lb-in.)
;
gmax 222 * 106 rad
tmax Ggmax 2580 psi
;
;
(1)
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CHAPTER 7
Page 638
Analysis of Stress and Strain
Problem 7.7-18 A cantilever beam of rectangular cross
h
section (width b 25 mm, height h 100 mm) is loaded
by a force P that acts at the midheight of the beam and is
inclined at an angle a to the vertical (see figure). Two strain
gages are placed at point C, which also is at the midheight
of the beam. Gage A measures the strain in the horizontal
direction and gage B measures the strain at an angle b = 60°
to the horizontal. The measured strains are Pa 125 * 106
and Pb 375 * 106.
Determine the force P and the angle a, assuming the
material is steel with E 200 GPa and 1/3.
b
h
C
a
P
b
B
b
A
C
Probs. 7.7-18 and 7.7-19
Solution 7.7-18
Cantilever beam (plane stress)
Beam loaded by a force P acting at an angle a.
1/3
E 200 GPa
b 25 mm
h 100 mm
Axial force F P sin a
Shear force V P cos a
(At the neutral axis, the bending moment produces no
stresses.)
At u 60°:
âx sx
P sin a
E
bhE
P sin a bhEâx 62,500 N
txy
3(1 + ) P cos a
3P cos a
gxy G
2bhG
bhE
(8.0 * 109) P cos a
(1)
(2)
FOR u 60°:
STRAIN GAGES
At u 0°:
HOOKE’S LAW
âA âx 125 * 10
âB 375 * 10
6
6
ELEMENT IN PLANE STRESS
P sin a
F
sx A
bh
sy 0
âx1 âx ây
âx + ây
+
2
2
âx1 âB 375 * 106
gxy
2
sin 2u
(3)
2u 120°
Substitute into Eq. (3):
375 * 106 41.67 * 106 41.67 * 106
(3.464 * 109)P cos a
3V
3P cos a
txy 2A
2bh
or P cos a 108,260 N
âx 125 * 106
SOLVE EQS. (1) AND (4):
ây âx 41.67 * 106
cos 2u +
a 30°
tan a 0.5773
P 125 kN
(4)
;
;
Problem 7.7-19 Solve the preceding problem if the cross-sectional dimensions are b 1.0 in. and h 3.0 in., the gage
angle is b = 75°, the measure strains are Pa 171 * 106 and Pb 266 * 106, and the material is a magnesium alloy
with modulus E 6.0 * 106 psi and Poisson’s ratio 0.35.
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Page 639
Plane Strain
639
P sin a bhEâx 3078 lb
txy
3(1 + v)P cos a
3P cos a
gxy G
2bhG
bhE
(1)
SECTION 7.7
Solution 7.7-19
Cantilever beam (plane stress)
Beam loaded by a force P acting at an angle a.
E 6.0 * 10 psi
6
0.35
b 1.0 in.
h 3.0 in.
Axial force F P sin a Shear foce V P cos a
(At the neutral axis, the bending moment produces
no stresses.)
HOOKE’S LAW
âx (225.0 * 109)P cos a
STRAIN GAGES
At u 0°:
At u 75°:
sx
P sin a
E
bhE
âA âx 171 * 106
âB 266 * 10
6
ELEMENT IN PLANE STRESS
F
P sin a
sx A
bh
(3)
2u 150°
Substitute into Eq. (3):
266 * 106 55.575 * 106 99.961 * 106
3P cos a
3V
txy 2A
2bh
âx 171 * 106
FOR u 75°:
âx + ây
âx ây
gxy
+
cos 2u +
sin 2u
âx1 2
2
2
âx1 âB 266 * 106
sy 0
(2)
(56.25 * 109)P cos a
ây âx 59.85 * 106
or P cos a 3939.8 lb
(4)
SOLVE EQS. (1) AND (4):
tan a 0.7813
P 5000 lb
a 38°
;
;
y
Problem 7.7-20 A 60° strain rosette, or delta rosette, consists of three
electrical-resistance strain gages arranged as shown in the figure. Gage A measures
the normal strain Pa in the direction of the x axis. Gages B and C measure the
strains Pb and Pc in the inclined directions shown.
Obtain the equations for the strains Px, Py, and gxy associated with the xy axis.
B
60°
O
60°
A
C
60°
x
Solution 7.7-20 Delta rosette (60° strain rosette)
STRAIN GAGES
Gage A at u 0°
Gage B at u 60°
Strain âA
Strain âB
Gage C at u 120°
FOR u 0°:
âx âA
Strain âC
;
FOR u 60°:
âx + ây
âx ây
gxy
+
cos 2u +
sin 2u
âx1 2
2
2
âA + ây
âA ây
gxy
+
(cos 120°) + 2 (sin 120°)
âB 2
2
3ây
gxy 13
âA
+
+
âB (1)
4
4
4
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CHAPTER 7
Page 640
Analysis of Stress and Strain
FOR u 120°:
âx + ây
âx ây
gxy
+
cos 2u +
sin 2u
âx1 2
2
2
âA + ây
âA ây
gxy
+
(cos 240°)
(sin 240°)
âC 2
2
2
3ây
gxy 13
âA
+
âC (2)
4
4
4
SOLVE EQS. (1) AND (2):
ây 1
(2âB + 2âC âA)
3
gxy 2
(âB âC)
13
;
;
Problem 7.7-21 On the surface of a structural component in a space vehicle, the
y
strainsare monitored by means of three strain gages arranged as shown in the figure.
During a certain maneuver, the following strains were recorded: Pa 1100 * 106,
Pb 200 * 106, and Pc 200 * 106.
Determine the principal strains and principal stresses in the material, which is a
magnesium alloy for which E 6000 ksi and 0.35. (Show the principal strains
and principal stresses on sketches of properly oriented element.)
B
30°
O
Solution 7.7-21
C
x
A
30-60-90° strain rosette
Magnesium alloy: E 6000 ksi 0.35
STRAIN GAGES
Gage A at u 0°
PA 1100 * 106
Gage B at u 90°
âB 200 * 106
Gage C at u 150°
âC 200 * 10
6
FOR u 0°:
âx âA 1100 * 106
FOR u 90°:
ây âB 200 * 106
FOR u 150°:
âx + ây
âx ây
gxy
âx1 âC +
cos 2u +
sin 2u
2
2
2
200 * 106 650 * 106 + 225 * 106
0.43301gxy
Solve for gxy: gxy 1558.9 * 106
PRINCIPAL STRAINS
âx + ây
âx ây 2
gxy 2
â1,2 ;
a
b + a
b
2
A
2
2
650 * 106 ; 900 * 106
â1 1550 * 106
â2 250 * 106
tan2 up gxy
13 1.7321
âx ây
2up 60°
up 30°
FOR up 30°:
âx + ây
âx ây
gxy
âx1 +
cos 2u +
sin 2u
2
2
2
1550 * 106
‹ up1 30°
up2 120°
â1 1550 * 106
â2 250 * 10
6
;
;
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Page 641
SECTION 7.7
Plane Strain
641
PRINCIPAL STRESSES (see Eqs. 7-36)
s1 E
1
2
(â1 + â2)
s2 E
1 2
(â2 + â1)
Substitute numerical values:
s1 10,000 psi
s2 2,000 psi
;
y
Problem 7.7-22 The strains on the surface of an experimental device made of pure
aluminum (E 70 Gpa, 0.33) and tested in a space shuttle were measured by
means of strain gages. The gages were oriented as shown in the figure, and the measured
strains were Pa 1100 * 106, Pb 1496 * 106, and Pc 39.44 * 10 * 6.
What is the stress sx in the x direction?
B
O
Solution 7.7-22
0.33
40°
x
FOR u 140°:
âx + ây
âx ây
gxy
+
cos 2u +
sin 2u
âx1 2
2
2
STRAIN GAGES
âA 1100 * 106
Gage A at u 0°
Gage B at u 40°
âB 1496 * 10
Gage C at u 140°
Substitute âx1 âc 39.44 * 106 and
6
âC 39.44 * 10
âx 1100 * 106; then simplify and rearrange:
6
0.41318ây 0.49240gxy 684.95 * 106
âx âA 1100 * 106
SOLVE EQS. (1) AND (2):
ây 200.3 * 106
FOR u 40°:
âx + ây
âx ây
gxy
+
cos 2u +
sin 2u
âx1 2
2
2
6
sx ; then simplify and rearrange:
0.41318ây + 0.49240gxy 850.49 * 106
gxy 1559.2 * 106
HOOKE’S LAW
Substitute âx1 âB 1496 * 106 and
âx 1100 * 10
A
40-40-100° strain rosette
Pure aluminum: E 70 GPa
FOR u 0°:
40°
C
(1)
E
1 2
(âx + ây) 91.6 MPa
;
(2)
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Page 642
CHAPTER 7 Analysis of Stress and Strain
Problem 7.7-23 Solve Problem 7.7-5 by using Mohr’s circle for plane strain.
Solution 7.7-23 Element in plane strain
âx 220 * 106
gxy 180 * 106
R 2(130 * 10
158.11 * 10
a arctan
ây 480 * 106
gxy
90 * 106 u 50°
2
6 2
) + (90 * 10
6
90
34.70°
130
b 180° a 2u 45.30°
6 2
)
POINT C: âx1 350 * 106
POINT D (u 50° ):
âx1 350 * 106 + R cos b 461 * 106
gx1y1
R sin b 112.4 * 106
2
gx1y1 225 * 106
POINT D ¿ (u 140°):
âx1 350 * 106 R cos b 239 * 106
gx1y1
R sin b 112.4 * 106
2
gx1y1 225 * 106
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Page 643
SECTION 7.7 Plane Strain
Problem 7.7-24 Solve Problem 7.7-6 by using Mohr’s circle for plane strain.
Solution 7.7-24 Element in plane strain
âx 420 * 106
gxy 310 * 106
ây 170 * 106
gxy
155 * 106
u 37.5°
2
POINT C: âx1 125 * 106
POINT D (u 37.5°):
âx1 125 * 106 + R cos b 351 * 106
gx1y1
R sin b 244.8 * 106
2
gx1y1 490 * 106
POINT D œ (u 127.5°):
âx1 125 * 106 R cos b 101 * 106
gx1y1
R sin b 244.8 * 106
2
gx1y1 490 * 106
R 2(295 * 106)2 + (155 * 106)2
333.24 * 106
a arctan
155
27.72°
295
b 2u a 47.28°
643
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Page 644
CHAPTER 7 Analysis of Stress and Strain
Problem 7.7-25 Solve Problem 7.7-7 by using Mohr’s circle for plane strain.
Solution 7.7-25 Element in plane strain
âx 480 * 106
ây 140 * 106
gxy
gxy 350 * 106
175 * 106
2
MAXIMUM SHEAR STRAINS
2us2 90°a 44.17°
us2 22.1°
2us1 2us2 + 180° 224.17°
Point S1: âaver 310 * 10
6
gmax 2R 488 * 106
Point S2: âaver 310 * 106
gmin 488 * 106
R 2(175 * 10
243.98 * 10
6 2
) + (170 * 10
6 2
)
6
a arctan
175
45.83°
170
POINT C:
âx1 310 * 106
PRINCIPAL STRAINS
2up2 180° a 134.2°
up2 67.1°
2up1 2up2 + 180° 314.2° up1 157.1°
Point P1: â1 310 * 106 + R 554 * 106
Point P2: â2 310 * 106 R 66 * 106
us1 112.1°
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Page 645
SECTION 7.7 Plane Strain
Problem 7.7-26 Solve Problem 7.7-8 by using Mohr’s circle for plane strain.
Solution 7.7-26 Element in plane strain
âx 120 * 106
ây 450 * 106
gxy
180 * 106
gxy 360 * 106
2
MAXIMUM SHEAR STRAINS
2us2 90° a 57.72°
us2 28.9°
2us1 2us2 + 180° 237.72°
Point S1: âaver 165 * 10
6
gmax 2R 674 * 106
Point S2: âaver 165 * 106
R 2(285 * 10
337.08 * 10
a arctan
6 2
) + (180 * 10
6 2
)
6
180
32.28°
285
Point C: âx1 165 * 106
PRINCIPAL STRAINS
2up2 180° a 147.72°
up2 73.9°
2up1 2up2 + 180° 327.72°
Point P1: â1 R 165 * 10
6
up1 163.9°
172 * 106
Point P2: â2 165 * 106 R 502 * 106
gmin 674 * 106
us1 118.9°
645
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Page 646
CHAPTER 7 Analysis of Stress and Strain
Problem 7.7-27 Solve Problem 7.7-9 by using Mohr’s circle for plane strain.
Solution 7.7-27 Element in plane strain
âx 480 * 106
gxy 420 * 106
ây 70 * 106
gxy
210 * 106
2
u 75°
PRINCIPAL STRAINS
2up1 a 45.69° up1 22.8°
2up2 2up1 + 180° 225.69°
Point P1: â1 275 * 10
6
up2 112.8°
+ R 568 * 106
Point P2: â2 275 * 106 R 18 * 106
R 2(205 * 106)2 + (210 * 106)2
293.47 * 106
a arctan
210
45.69°
205
b a + 180° 2u 75.69°
Point C: âx1 275 * 106
Point D (u 75°):
âx1 275 * 106 R cos b 202 * 106
gx1y1
R sin b 284.36 * 106
2
gx1y1 569 * 106
Point
Dœ
(u 165°):
âx1 275 * 106 + R cos b 348 * 106
gx1y1
R sin b 284.36 * 106
2
gx1y1 569 * 106
MAXIMUM SHEAR STRAINS
2us2 90° + a 135.69°
2us1 2us2 + 180° 315.69°
Point S1: âaver 275 * 10
6
gmax 2R 587 * 106
Point S2: âaver 275 * 106
gmin 587 * 106
us2 67.8°
us1 157.8°
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Page 647
SECTION 7.7 Plane Strain
647
Problem 7.7-28 Solve Problem 7.7-10 by using Mohr’s circle for plane strain.
Solution 7.7-28 Element in plane strain
âx 1120 * 106
gxy 780 * 106
ây 430 * 106
gxy
2
390 * 106
u 45°
PRINCIPAL STRAINS
2up1 180° a 131.50°
up1 65.7°
2up2 2up1 + 180° 311.50°
Point P1: â1 775 * 10
6
up2 155.7°
+ R 254 * 106
Point P2: â2 775 * 106 R 1296 * 106
R 2(345 * 106)2 + (390 * 106)2
520.70 * 106
a arctan
390
48.50°
345
b 180° a 2u 41.50°
Point C: âx1 775 * 106
MAXIMUM SHEAR STRAINS
Point D: (u 45°):
2us2 2us1 + 180° 221.50°
âx1 775 * 106 + R cos b 385 * 106
gx1y1
R sin b 345 * 106 gx1y1 690 * 106
2
Point D ¿ : (u 135°)
âx1 775 * 106 R cos b 1165 * 106
gx1y1
R sin b 345 * 106
2
gx1y1 690 * 106
2us1 90° a 41.50°
Point S1: âaver 775 * 10
us1 20.7°
6
gmax 2R 1041 * 106
Point S2: âaver 775 * 106
gmin 1041 * 106
us2 110.7°
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Page 648
08Ch08.qxd
9/18/08
11:03 AM
Page 649
8
Applications of Plane Stress
(Pressure Vessels, Beams,
and Combined Loadings)
Spherical Pressure Vessels
When solving the problems for Section 8.2, assume that the given radius or
diameter is an inside dimension and that all internal pressures are gage pressures.
Problem 8.2-1 A large spherical tank (see figure) contains gas at a
pressure of 450 psi. The tank is 42 ft in diameter and is constructed of high-strength
steel having a yield stress in tension of 80 ksi.
Determine the required thickness (to the nearest 1/4 inch) of the wall of the tank
if a factor of safety of 3.5 with respect to yielding is required.
Probs. 8.2-1 and 8.2-2
Solution 8.2-1
Radius:
r
1
42 * 12
2
r 252 in.
Internal Pressure: p 450 psi
Yield stress:
t
prn
2sY
t 2.481 in.
to nearest 1/4 inch, t min 2.5 in.
;
sY 80 ksi (steel)
Factor of safety: n 3.5
MINIMUM WALL THICKNESS tmin
From Eq. (8-1): smax pr
pr
sY
or
2t
n
2t
649
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9/18/08
11:03 AM
CHAPTER 8
Page 650
Applications of Plane Stress
Problem 8.2-2 Solve the preceding problem if the internal pressure is 3.75 MPa, the diameter is 19 m, the yield stress is
570 MPa, and the factor of safety is 3.0.
Determine the required thickness to the nearest millimeter.
Solution 8.2-2
Radius:
r
1
(19 m)
2
r 9.5 103 mm
Internal Pressure: p 3.75 MPa
Yield stress:
t
prn
2sY
t 93.8 mm
Use the next higher millimeter t min 94 mm
sY 570 MPa
Factor of safety: n 3
MINIMUM WALL THICKNESS tmin
From Eq. (8-1): smax pr
pr
sY
or
2t
n
2t
Problem 8.2-3 A hemispherical window (or viewport) in a decompression chamber
(see figure) is subjected to an internal air pressure of 80 psi. The port is attached to
the wall of the chamber by 18 bolts.
Find the tensile force F in each bolt and the tensile stress s in the viewport if the
radius of the hemisphere is 7.0 in. and its thickness is 1.0 in.
Solution 8.2-3
Hemispherical viewport
FREE-BODY DIAGRAM
T total tensile force in 18 bolts
a FHORIZ T pA 0
T pA p(pr 2)
F force in one bolt
F
Radius:
r 7.0 in.
Internal pressure: p 80 psi
Wall thickness:
18 bolts
t 1.0 in.
1
T
(ppr 2) 684 lb
18
18
;
TENSILE STRESS IN VIEWPORT (EQ. 8-1)
s
pr
280 psi
2t
;
;
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SECTION 8.2
651
Spherical Pressure Vessels
Problem 8.2-4 A rubber ball (see figure) is inflated to a pressure of 60 kPa. At that pressure
the diameter of the ball is 230 mm and the wall thickness is 1.2 mm. The rubber has modulus of
elasticity E 3.5 MPa and Poisson’s ratio v 0.45.
Determine the maximum stress and strain in the ball.
Prob. 8.2-4, 8.2-5
Solution 8.2-4
Rubber ball
CROSS-SECTION
MAXIMUM STRESS (EQ. 8-1)
smax (60 kPa)(115 mm)
pr
2t
2(1.2 mm)
2.88 MPa
;
MAXIMUM STRAIN (EQ. 8-4)
Radius:
r (230 mm)/2 115 mm
Internal pressure:
p 60 kPa
Wall thickness:
t 1.2 mm
âmax pr
(60 kPa)(115 mm)
(1 v) (0.55)
2tE
2(1.2 mm)(3.5 MPa)
0.452
;
Modulus of elasticity: E 3.5 MPa (rubber)
v 0.45 (rubber)
Poisson’s ratio:
Problem 8.2-5 Solve the preceding problem if the pressure is 9.0 psi, the diameter is 9.0 in.,
the wall thickness is 0.05 in., the modulus of elasticity is 500 psi, and Poisson’s ratio is 0.45.
Solution 8.2-5
Rubber ball
Modulus of elasticity: E 500 psi (rubber)
CROSS-SECTION
v 0.45 (rubber)
Poisson’s ratio:
MAXIMUM STRESS (EQ. 8-1)
smax 1
(9.0 in.) 4.5 in.
2
Radius:
r
Internal pressure:
p 9.0 psi
Wall thickness:
t 0.05 in.
(9.0 psi)(4.5 in.)
pr
405 psi
2t
2(0.05 in.)
;
MAXIMUM STRAIN (EQ. 8-4)
âmax (9.0 psi)(4.5 in.)
pr
(1 v) (0.55)
2tE
2(0.05 in.)(500 psi)
0.446
;
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Page 652
Applications of Plane Stress
Problem 8.2-6 A spherical steel pressure vessel (diameter 480 mm,
Cracks in
coating
thickness 8.0 mm) is coated with brittle lacquer that cracks when the strain
reaches 150 106 (see figure).
What internal pressure p will cause the lacquer to develop cracks? (Assume
E 205 GPa and v 0.30.)
Solution 8.2-6
Spherical vessel with brittle coating
Cracks occur when max 150 106
pr
(1 v)
From Eq. (8-4): âmax 2tE
CROSS-SECTION
‹ P
r 240 mm
t 8.0 mm
E 205 GPa (steel)
v 0.30
P
2tE âmax
r(1 v)
2(8.0 mm)(205 GPa)(150 * 106)
(240 mm)(0.70)
2.93 MPa
;
Problem 8.2-7 A spherical tank of diameter 48 in. and wall thickness 1.75 in. contains
compressed air at a pressure of 2200 psi. The tank is constructed of two hemispheres
joined by a welded seam (see figure).
Weld
(a) What is the tensile load f (lb per in. of length of weld) carried by the weld?
(b) What is the maximum shear stress tmax in the wall of the tank?
(c) What is the maximum normal strain in the wall? (For steel, assume
E 30 106 psi and v 0.29.)
Probs. 8.2-7 and 8.2-8
Solution 8.2-7
r 24 in.
t 1.75 in.
E 30 106 psi
v 0.29 (steel)
(b) MAXIMUM SHEAR STRESS IN WALL (EQ. 8-3)
tmax (2200 psi)(24 in.)
pr
7543 psi
4t
4 (1.75 in.)
(a) TENSILE LOAD CARRIED BY WELD
T Total load
f load per inch
T pA ppr 2
c Circumference of tank 2pr
T p1pr 2 pr (2200 psi)(24 in.)
f c
2pr
2
2
2
26.4 k/in.
;
(c) MAXIMUM NORMAL STRAIN IN WALL (EQ. 8-4)
âmax (2200 psi)(24 in.)(0.71)
pr (1 v)
2 tE
2(1.75 in.)130106 psi2
3.57 * 104
;
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SECTION 8.2
653
Spherical Pressure Vessels
Problem 8.2-8 Solve the preceding problem for the following data: diameter 1.0 m, thickness 48 mm, pressure 22 MPa,
modulus 210 GPa, and Poisson’s ratio 0.29.
Solution 8.2-8
r 0.5 m
E 210 GPa
t 48 mm
(b) MAXIMUM SHEAR STRESS IN WALL (EQ. 8-3)
v 0.29 (steel)
tmax (a) TENSILE LOAD CARRIED BY WELD
T Total load
57.3 MPa
f load per inch
T pA ppr2
c Circumference of tank 2pr
p1pr 2
pr
(22 MPa)(0.5 m)
T
c
2pr
2
2
5.5 MN/m
;
(c) MAXIMUM NORMAL STRAIN IN WALL (EQ. 8-4)
2
f
(22 MPa)(0.5 m)
pr
4t
4 (48 mm)
âmax pr (1 v)
(22 MPa)(0.5 m) 0.71
2 tE
2(48 mm)(210 GPa)
3.87 * 104
;
;
Problem 8.2-9 A spherical stainless-steel tank having a diameter of 22 in. is used to store propane gas at a pressure of
2450 psi. The properties of the steel are as follows: yield stress in tension, 140,000 psi; yield stress in shear, 65,000 psi;
modulus of elasticity, 30 106 psi; and Poisson’s ratio, 0.28. The desired factor of safety with respect to yielding is 2.8.
Also, the normal strain must not exceed 1100 106.
Determine the minimum permissible thickness tmin of the tank.
Solution 8.2-9
r 11 in.
E 30 106 psi
p 2450 psi
v 0.28 (steel)
sY 140000 psi
t2 n 2.8
max 1100 106
tY 65000 psi
MIMIMUM WALL THICKNESS t
(1) TENSION (EQ. 8-1)
pr
sY
2a b
n
(2450 psi) (11 in.)
t1 smax pr
2 t1
2
2.8
t3 0.269 in.
pr
4t2
(2450 psi)(11 in.)
pr
0.29 in.
65000 psi
tY
4n
4
2.8
(3) STRAIN (EQ. 8-4)
pr
2smax
140000 psi
tmax (2) SHEAR (EQ. 8-3)
âmax pr
(1 v)
2t 3 E
pr
(1 v)
2âmax E
(2450 psi)(11 in.)
211100 * 1062130 * 106 psi2
0.72
0.294 in.
t3 t2 t1
Thus, tmin 0.294 in.
;
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Page 654
Applications of Plane Stress
Problem 8.2-10 Solve the preceding problem if the diameter is 500 mm, the pressure is 18 MPa, the yield stress in tension
is 975 MPa, the yield stress in shear is 460 MPa, the factor of safety is 2.5, the modulus of elasticity is 200 GPa, Poisson’s
ratio is 0.28, and the normal strain must not exceed 1210 106.
Solution 8.2-10
r 250 mm
E 200 GPa
p 18 MPa
v 0.28 (steel)
(2) SHEAR (EQ. 8-3)
n 2.5
tY 460 MPa
max 1210 106
MINIMUM WALL THICKNESS t
t1 pr
2smax
smax pr
4t 2
(18 MPa)(250 mm)
pr
6.114 mm
tY
460 MPa
4
4
n
2.5
pr
(3) STRAIN (EQ. 8-4) âmax (1 v)
2t 3E
t2 sY 975 MPa
(1) TENSION (EQ. 8-1)
tmax pr
2t 1
pr
sY
2a b
n
(18 MPa)(250 mm)
5.769 mm
975 MPa
2
2.5
t3 pr
2âmax E
(1 v)
(18 MPa) (250 mm)
211210 * 1062(200 GPa)
0.72
6.694 mm
t3 t2 t1
Thus, tmin 6.69 mm
Problem 8.2-11 A hollow pressurized sphere having radius r 4.8 in. and
wall thickness t 0.4 in. is lowered into a lake (see figure). The compressed
air in the tank is at a pressure of 24 psi (gage pressure when the tank is out
of the water).
At what depth D0 will the wall of the tank be subjected to a compressive
stress of 90 psi?
D0
;
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SECTION 8.3
Solution 8.2-11
Cylindrical Pressure Vessels
655
Pressurized sphere under water
CROSS-SECTION
D0 depth of water (in.)
r 4.8 in.
p1 24 psi
t 0.4 in.
g density of water 62.4 lb/ft
3
(1) IN AIR: p1 24 psi
p2 gD0 a
( p1 p2)r
pr
2t
2t
90 psi (2) UNDER WATER: p1 24 psi
1728 in.3/ ft3
b D0 0.036111 D0 ( psi)
Compressive stress in tank wall equals 90 psi.
(Note: s is positive in tension.)
s
(1) IN AIR
62.4 lb/ft3
s 90 psi
(24 psi 0.03611 D0)(4.8 in.)
2(0.4 in.)
144 0.21667 D0
Solve for D0: D0 234
0.21667
1080 in. 90 ft
(2) UNDER WATER
Cylindrical Pressure Vessels
When solving the problems for Section 8.3, assume that the given radius or diameter is an
inside dimension and that all internal pressures are gage pressures.
Problem 8.3-1 A scuba tank (see figure) is being designed for an internal pressure of
1600 psi with a factor of safety of 2.0 with respect to yielding. The yield stress of the steel is
35,000 psi in tension and 16,000 psi in shear.
If the diameter of the tank is 7.0 in., what is the minimum required wall thickness?
;
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CHAPTER 8
Solution 8.3-1
Page 656
Applications of Plane Stress
Scuba tank
sallow sY
17,500 psi
n
t allow tY
8,000 psi
n
Find required wall thickness t.
(1) BASED ON TENSION (EQ. 8-5)
t1 pr
sallow
(1600 psi)(3.5 in.)
17,500 psi
Cylindrical pressure vessel
p 1600 psi
r 3.5 in.
n 2.0
d 7.0 in.
sY 35,000 psi
Y 16,000 psi
pr
t
0.320 in.
tmax (2) BASED ON SHEAR (EQ. 8-10)
t2 smax pr
2t
(1600 psi)(3.5 in.)
pr
0.350 in.
2tallow
2(8,000 psi)
Shear governs since t2 t1
tmin 0.350 in.
;
Problem 8.3-2 A tall standpipe with an open top (see figure) has diameter d 2.2 m
d
and wall thickness t 20 mm.
(a) What height h of water will produce a circumferential stress of 12 MPa in the wall of
the standpipe?
(b) What is the axial stress in the wall of the tank due to the water pressure?
h
Solution 8.3-2
d 2.2 m
r 1.1 m
t 20 mm
weight density of water g 9.81 kN/m3
height of water
h
water pressure
p h
(a) HEIGHT OF WATER
s1 pr
0.00981h (1.1 m)
12 MPa t
20 mm
h
12 (20)
22.2 m
0.00981 (1.1)
(b) AXIAL
STRESS
IN
THE
WALL
DUE
TO
WATER
PRESSURE ALONE
Because the top of the tank is open, the internal
pressure of the water produces no axial (longitudinal) stresses in the wall of the tank. Axial stress
equals zero.
;
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SECTION 8.3
Problem 8.3-3 An inflatable structure used by a traveling
Cylindrical Pressure Vessels
657
Longitudinal seam
circus has the shape of a half-circular cylinder with closed ends
(see figure). The fabric and plastic structure is inflated by a
small blower and has a radius of 40 ft when fully inflated. A
longitudinal seam runs the entire length of the “ridge” of the
structure.
If the longitudinal seam along the ridge tears open when it
is subjected to a tensile load of 540 pounds per inch of seam,
what is the factor of safety n against tearing when the internal
pressure is 0.5 psi and the structure is fully inflated?
Solution 8.3-3
Inflatable structure
Half-circular cylinder
r 40 ft 480 in.
Internal pressure p 0.5 psi
T tensile force per unit length of longitudinal seam
Seam tears when T Tmax 540 lb/in.
Find factor of safety against tearing.
CIRCUMFERENTIAL STRESS (EQ. 8-5)
s1 pr
where t thickness of fabric
t
Actual value of T due to internal pressure s1t
T s1t pr (0.5 psi)(480 in.) 240 lb/in.
FACTOR OF SAFETY
n
Tmax
540 lb/in.
2.25
T
240 lb/in.
;
Problem 8.3-4 A thin-walled cylindrical pressure vessel of radius r is subjected
simultaneously to internal gas pressure p and a compressive force F acting at the
ends (see figure).
What should be the magnitude of the force F in order to produce pure shear
in the wall of the cylinder?
Solution 8.3-4
F
Cylindrical pressure vessel
STRESSES (SEE EQ. 8-5 AND 8-6):
r Radius
p Internal pressure
s1 pr
t
s2 pr
pr
F
F
2t
A
2t
2prt
F
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Applications of Plane Stress
FOR PURE SHEAR, the stresses s1 and s2 must be equal in
magnitude and opposite in sign (see, e.g., Fig. 7-11 in
Section 7.3).
s1 s2
OR
pr
pr
F
a b
t
2t
2prt
Solve for F: F 3ppr2
;
Problem 8.3-5 A strain gage is installed in the longitudinal direction on the surface
of an aluminum beverage can (see figure). The radius-to-thickness ratio of the can is
200. When the lid of the can is popped open, the strain changes by 僆0 170 106.
What was the internal pressure p in the can? (Assume E 10 106 psi
and v 0.33.)
12 FL OZ (355 mL)
Solution 8.3-5
Aluminum can
STRAIN IN LONGITUDINAL DIRECTION (EQ. 8-11a)
â2 pr
(1 2v)
2tE
â2 â0
‹
or p p
2tEâ2
r(1 2v)
2tEâ0
2Eâ0
(r)(1 2v)
(r/t) (1 2v)
Substitute numerical values:
r
200
t
E 10 106 psi
v 0.33
0 change in strain when pressure is released
170 106
Find internal pressure p.
p
2(10 * 106 psi)(170 * 106)
(200)(1 0.66)
50 psi
;
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SECTION 8.3
Problem 8.3-6 A circular cylindrical steel tank (see figure) contains
a volatile fuel under pressure. A strain gage at point A records the
longitudinal strain in the tank and transmits this information to a control
room. The ultimate shear stress in the wall of the tank is 84 MPa, and
a factor of safety of 2.5 is required.
At what value of the strain should the operators take action to reduce
the pressure in the tank? (Data for the steel are as follows: modulus of
elasticity E 205 GPa and Poisson’s ratio v 0.30.)
659
Cylindrical Pressure Vessels
Pressure relief
valve
Cylindrical tank
A
Solution 8.3-6
tULT 84 MPa
E 205 GPa
n 2.5
tULT
n
tmax v 0.3
tmax 33.6 MPa
Find maximum allowable strain reading at the gage
s1 pr
t
s2 pr
2t
From Eq. (8-11a)
â2 pr
(1 2v)
2tE
â2max âmax From Eq. (8-10)
tmax pr
s1
2
2t
Pmax pmax r
tmax
(1 2v) (1 2v)
2tE
E
tmax
(1 2v)
E
max 6.56 105
2ttmax
r
Cylinder
Problem 8.3-7 A cylinder filled with oil is under pressure from a
piston, as shown in the figure. The diameter d of the piston is 1.80 in.
and the compressive force F is 3500 lb. The maximum allowable shear
stress tallow in the wall of the cylinder is 5500 psi.
What is the minimum permissible thickness tmin of the cylinder
wall? (See the figure on the next page.)
F
p
Piston
Probs. 8.3-7 and 8.3-8
Solution 8.3-7
Cylinder with internal pressure
Maximum shear stress (Eq. 8-10):
tmax d 1.80 in.
r 0.90 in.
F 3500 lb
tallow 5500 psi
Find minimum thickness tmin.
F
F
Pressure in cylinder: p A
pr 2
pr
F
2t
2prt
Minimum thickness:
t min F
2prtallow
Substitute numerical values:
t min 3500 lb
0.113 in.
2p(0.90 in.)(5500 psi)
;
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CHAPTER 8
Page 660
Applications of Plane Stress
Problem 8.3-8 Solve the preceding problem if d 90 mm, F 42 kN, and tallow 40 MPa.
Solution 8.3-8
Cylinder with internal pressure
Maximum shear stress (Eq. 8-10):
pr
F
2t
2prt
tmax Minimum thickness:
d 90 mm
F 42.0 kN
r 45 mm
t min tallow 40 MPa
Find minimum thickness tmin.
Pressure in cylinder: p F
2pr tallow
Substitute numerical values:
F
F
A
pr 2
t min 42.0 kN
3.71 mm
2p(45 mm)(40 MPa)
;
Problem 8.3-9 A standpipe in a water-supply system (see figure) is 12 ft in
diameter and 6 inches thick. Two horizontal pipes carry water out of the
standpipe; each is 2 ft in diameter and 1 inch thick. When the system is
shut down and water fills the pipes but is not moving, the hoop stress at the
bottom of the standpipe is 130 psi.
(a) What is the height h of the water in the standpipe?
(b) If the bottoms of the pipes are at the same elevation as the bottom of the
standpipe, what is the hoop stress in the pipes?
Solution 8.3-9
Vertical standpipe
(a) FIND HEIGHT h OF WATER IN THE STANDPIPE
p pressure at bottom of standpipe gh
From Eq. (8-5): s1 pr
ghr
t
t
or h Substitute numerical values:
h
d 12 ft 144 in.
g 62.4 lb/ft3 r 72 in.
t 6 in.
62.4
lb/in.3
1728
s1 hoop stress at bottom of standpipe 130 psi
(130 psi)(6 in.)
62.4
a
lb/in.3 b (72 in.)
1728
25 ft
;
300 in.
s1t
gr
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SECTION 8.3
HORIZONTAL PIPES
d1 2 ft 24 in.
r1 12 in.
t1 1.0 in.
s1 (b) FIND HOOP STRESS s1 IN THE PIPES
Since the pipes are 2 ft in diameter, the depth of
water to the center of the pipes is about 24 ft.
h1 ⬇ 24 ft 288 in.
p1 gh1
Cylindrical Pressure Vessels
661
p1r1
gh 1r1
t1
t1
a
62.4
lb/in.3 b (288 in.)(12 in.)
1728
1.0 in.
125 psi
Based on the average pressure in the pipes:
s1 ⬇ 125 psi
;
Problem 8.3-10 A cylindrical tank with hemispherical heads is constructed of
steel sections that are welded circumferentially (see figure). The tank diameter is
1.25 m, the wall thickness is 22 mm, and the internal pressure is 1750 kPa.
(a) Determine the maximum tensile stress sh in the heads of the tank.
(b) Determine the maximum tensile stress sc in the cylindrical part
of the tank.
(c) Determine the tensile stress sw acting perpendicular to the welded joints.
(d) Determine the maximum shear stress th in the heads of the tank.
(e) Determine the maximum shear stress tc in the cylindrical part of the tank.
Welded seams
Probs. 8.3-10 and 8.3-11
Solution 8.3-10
d 1.25 m
r
d
2
t 22 mm
p 1750 kPa
(c) TENSILE STRESS IN WELDS (EQ. 8-6)
sw pr
2t
sw 24.9 MPa
;
(a) MAXIMUM TENSILE STRESS IN HEMISPHERES (EQ. 8-1)
sh pr
2t
sh 24.9 MPa
;
(d) MAXIMUM SHEAR STRESS IN HEMISPHERES (EQ. 8-3)
th pr
4t
th 12.43 MPa
;
(b) MAXIMUM STRESS IN CYLINDER (EQ. 8-5)
sc pr
t
sc 49.7 MPa
;
(e) MAXIMUM SHEAR STRESS IN CYLINDER (EQ. 8-10)
tc pr
2t
tc 24.9 MPa
;
Problem 8.3-11 A cylindrical tank with diameter d 18 in. is subjected to internal gas pressure p 450 psi. The tank is
constructed of steel sections that are welded circumferentially (see figure). The heads of the tank are hemispherical. The
allowable tensile and shear stresses are 8200 psi and 3000 psi, respectively. Also, the allowable tensile stress perpendicular
to a weld is 6250 psi.
Determine the minimum required thickness tmin of (a) the cylindrical part of the tank and (b) the hemispherical heads.
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Page 662
Applications of Plane Stress
Solution 8.3-11
d 18 in.
r
d
2
sallow 8200 psi
(tension)
tallow 3000 psi
(shear)
sa 6250 psi
WELD
t min p 450 psi
tmax pr
t min 2tallow
s
WELD
tmin 0.324 in.
t min 0.675 in.
TENSION
t min smax pr
2sallow
pr
2t
tmin 0.247 in.
tmax SHEAR
pr
2t
t min tmin 0.675 in.
t min 0.338 in.
pr
4tallow
pr
4t
tmin 0.338 in.
;
pr
2t
Problem *8.3-12 A pressurized steel tank is constructed with a helical weld
that makes an angle a 55° with the longitudinal axis (see figure). The tank
has radius r 0.6 m, wall thickness t 18 mm, and internal pressure p 2.8 MPa. Also, the steel has modulus of elasticity E 200 GPa and Poisson’s
ratio v 0.30.
Determine the following quantities for the cylindrical part of the tank.
(a)
(b)
(c)
(d)
;
(b) FIND MINIMUM THICKNESS OF HEMISPHERES
(tension)
(a) FIND MINIMUM THICKNESS OF CYLINDER
pr
TENSION
smax t
pr
t min tmin 0.494 in.
sallow
SHEAR
pr
2sa
Helical weld
a
The circumferential and longitudinal stresses.
The maximum in-plane and out-of-plane shear stresses.
The circumferential and longitudinal strains.
The normal and shear stresses acting on planes parallel and perpendicular to
the weld (show these stresses on a properly oriented stress element).
Probs. 8.3-12 and 8.3-13
Solution 8.3-12
a 55 deg
p 2.8 MPa
r 0.6 m
t 18 mm
E 200 GPa
v 0.3
(a) CIRCUMFERENTIAL STRESS
pr
s1 t
s1 93.3 MPa
pr
2t
s2 46.7 MPa
t1 s1 s2
2
t1 23.3 MPa
OUT-OF-PLANE SHEAR STRESS
;
LONGITUDIAL STRESS
s2 (b) IN-PLANE SHEAR STRESS
;
t2 s1
2
t2 46.7 MPa
;
;
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SECTION 8.3
(c) CIRCUMFERENTIAL STRAIN
s1
â1 (2 v)
2E
1 3.97 104
;
For u 35 deg
sx sy
sx + sy
+
cos (2u)
sx1 2
2
LONGITUDINAL STRAIN
s2
(1 2v)
â2 E
+ txy sin (2u)
5
2 9.33 10
;
(d) STRESS ON WELD
u 90 deg a
sx s2
sx1 62.0 MPa
;
sx sy
tx1y1 sin (2u) + txy cos (2u)
2
tx1y1 21.9 MPa
u 35 deg
sx 46.667 MPa
sy 93.333 MPa
663
Cylindrical Pressure Vessels
sy1 sx
sy s1
;
sy sx1
sy1 78.0 MPa
;
txy 0
Problem *8.3-13 Solve the preceding problem for a welded tank with a 62°, r 19 in., t 0.65 in., p 240 psi,
E 30 106 psi, and v 0.30.
Solution 8.3-13
a 62 deg
r 19 in.
p 240 psi
E 30 106 psi
LONGITUDINAL STRAIN
t 0.65 in.
v 0.3
â2 (a) CIRCUMFERENTIAL STRESS
s1 pr
t
s1 7015 psi
s2
(1 2v)
E
2 4.68 105
(d) STRESS ON WELD
;
u 90 deg a
u 28 deg
LONGITUDIAL STRESS
sx s2
pr
s2 2t
sy 7.015 10 psi
;
s1 s2
2
t1 1754 psi
;
s1
2
t2 3508 psi
sx1 4281 psi
;
sx sy
sin (2u) + txy cos (2u)
tx1y1 2
;
tx1y1 1454 psi
(c) CIRCUMFERENTIAL STRAIN
â1 s1
(2 v)
2E
txy 0
+ txy sin (2u)
OUT-OF-PLANE SHEAR STRESS
t2 sy s1
For u 28 deg
sx + sy
sx sy
+
cos (2u)
sx1 2
2
(b) IN-PLANE SHEAR STRESS
t1 sx 3.508 103 psi
3
s2 3508 psi
;
1 1.988 104
;
sy1 sx
;
sy sx1
sy1 6242 psi
;
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Page 664
Applications of Plane Stress
Maximum Stresses in Beams
When solving the problems for Section 8.4, consider only the in-plane
stresses and disregard the weights of the beams
Problem 8.4-1 A cantilever beam of rectangular cross section is subjected
P
to a concentrated load P ⫽ 17 k acting at the free end (see figure). The beam
has width b ⫽ 3 in. and height h ⫽ 12 in. Point A is located at distance
c ⫽ 2.5 ft from the free end and distance d ⫽ 9 in. from the bottom of
the beam.
Calculate the principal stresses s1 and s2 and the maximum shear stress
tmax at point A. Show these stresses on sketches of properly oriented elements.
A
h
c
b
d
Probs. 8.4-1 and 8.4-2
Solution 8.4-1
P ⫽ 17 k
c ⫽ 2.5 ft
h ⫽ 12 in.
v ⫽ 0.3
b ⫽ 3 in.
d ⫽ 9 in.
For u1 ⫽ up
sx1 ⫽
STRESS AT POINT A
bh3
I⫽
12
V⫽P
I ⫽ 432 in.
sx ⫽ ⫺
cos (2 u1)
sx ⫺ sy
2
txy ⫽ t
sy ⫽ 0
2txy
1
b
up ⫽ atan a
2
sx ⫺ sy
cos (2 u2)
sx2 ⫽ ⫺77.971 psi
Therefore
2txy
sx ⫺ sy
up1 ⫽ u2
txy ⫽ 531.25 psi
s2 ⫽ 3620 psi
⫽ 0.3
up ⫽ 8.35 deg
s2 ⫽ sx1
up2 ⫽ u1
up1 ⫽ 98.4 deg
;
;
up2 ⫽ 8.35 deg
;
;
MAXIMUM SHEAR STRESSES
tmax ⫽
⫽
2
s1 ⫽ ⫺78.0 psi
PRINCIPAL STRESSES
2txy
sx ⫺ sy
+
s1 ⫽ sx2
Q ⫽ 40.5 in.3
t ⫽ 531.25 psi
u2 ⫽ 98.35 deg
+ txy sin (2 u2)
sx ⫽ 3.542 ⫻ 103 psi
sx ⫽ 3.542 ⫻ 103 psi
tan12up2 ⫽
2
sx + sy
sx2 ⫽
yA ⫽ 3 in.
d
h
Q ⫽ bd a ⫺ b
2
2
VQ
t⫽
Ib
2
For u2 ⫽ 90 deg ⫹ up
V ⫽ 1.7 ⫻ 104 lb
MyA
I
+
sx1 ⫽ 60.306 Mpa
M ⫽ ⫺5.1 ⫻ 105 lb-in.
h
yA ⫽ ⫺ + d
2
sx ⫺ sy
sx + sy
+ txy sin (2u1)
4
M ⫽ ⫺Pc
u1 ⫽ 8.35 deg
A
a
sx ⫺ sy
2
tmax ⫽ 1849 psi
2
2
b + txy
;
us1 ⫽ up1 ⫺ 45 deg
us1 ⫽ 53.4 deg
us2 ⫽ us1 ⫹ 90 deg
us2 ⫽ 143.4 deg
savg ⫽
sx + sy
2
savg ⫽ 1771 psi
;
;
;
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SECTION 8.4
665
Maximum Stresses in Beams
Problem 8.4-2 Solve the preceding problem for the following data: P ⫽ 130 kN, b ⫽ 80 mm, h ⫽ 260 mm, c ⫽ 0.6 m,
and d ⫽ 220 mm.
Solution 8.4-2
P ⫽ 130 kN
c ⫽ 0.6 m
b ⫽ 80 mm
d ⫽ 220 mm
h ⫽ 260 mm
v ⫽ 0.3
For u1 ⫽ up
sx1 ⫽
STRESS AT POINT A
I⫽
bh3
12
M ⫽ ⫺7.8 ⫻ 104 N # m
V⫽P
V ⫽ 1.3 ⫻ 105 N
h
yA ⫽ ⫺ + d
2
sx ⫽ ⫺
MyA
I
Q ⫽ bd a
2
2
cos (2u1)
Q ⫽ 3.52 ⫻ 105 mm3
t ⫽ 4.882 MPa
txy ⫽ t
sy ⫽ 0
txy ⫽ 4.882 MPa
PRINCIPAL STRESSES
2txy
sx ⫺ sy
⫽
2 txy
1
b
up ⫽ a tan a
2
sx ⫺ sy
For u2 ⫽ 90 deg ⫹ up
2 txy
sx ⫺ sy
u2 ⫽ 94.628 deg
sx ⫺ sy
sx + sy
+
2
2
cos (2u2)
+ txy sin (2u 2)
sx ⫽ 59.911 MPa
sx ⫽ 59.911 MPa
tan(2up) ⫽
+
sx1 ⫽ 60.306 MPa
sx2 ⫽
yA ⫽ 90 mm
d
h
⫺ b
2
2
VQ
t⫽
Ib
sx ⫺ sy
sx + sy
+ txy sin (2u1)
I ⫽ 1.172 ⫻ 108 mm4
M ⫽ ⫺Pc
u1 ⫽ 4.628 deg
⫽ 0.163
up ⫽ 4.628 deg
sx2 ⫽ ⫺0.395 MPa
Therefore
s1 ⫽ sx1
s1 ⫽ 60.3 MPa
s2 ⫽ sx2
up2 ⫽ u2
up1 ⫽ 4.63 deg
;
s2 ⫽ ⫺0.395 MPa
;
;
up2 ⫽ 94.6 deg
;
MAXIMUM SHEAR STRESSES
sx ⫺ sy 2
2
b + t xy
tmax ⫽ a
A
2
tmax ⫽ 30.4 MPa
;
us1 ⫽ up1 ⫺ 45 deg
us1 ⫽ ⫺40.4 deg
us2 ⫽ us1 ⫹ 90 deg
us2 ⫽ 49.6 deg
savg ⫽
Problem 8.4-3 A simple beam of rectangular cross section
(width 4 in., height 10 in.) carries a uniform load of 1200 lb/ft
on a span of 12 ft (see figure).
Find the principal stresses s1 and s2 and the maximum shear
stress tmax at a cross section 2 ft from the left-hand support at
each of the following locations: (a) the neutral axis, (b) 2 in. above
the neutral axis and (c) the top of the beam. (Disregard the direct
compressive stresses produced by the uniform load bearing against
the top of the beam.)
up1 ⫽ u1
sx + sy
2
savg ⫽ 30.0 deg
;
;
;
1200 lb/ft
10 in.
4 in.
2 ft
12 ft
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CHAPTER 8
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Applications of Plane Stress
Solution 8.4-3
b ⫽ 4 in.
h ⫽ 10 in.
bh3
I⫽
12
I ⫽ 333.333 in.4
c ⫽ 2 ft
q ⫽ 1200 lb/ft
RA ⫽
qL
2
A ⫽ bh
txy ⫽ ⫺
s1 ⫽
L ⫽ 12 ft
s2 ⫽
c2
2
M ⫽ 1.44 ⫻ 105 lb # in.
V ⫽ RA ⫺ qc
sx
sx 2
2
+
a b + txy
2
A 2
s2 ⫽ ⫺890 psi
V ⫽ 4.8 ⫻ 10 lb
sy ⫽ 0
txy ⫽ ⫺
tmax ⫽
s1 ⫽ 180 psi
sx 2
2
b + txy
A 2
tmax ⫽ 458 psi
3V
2A
s2 ⫽ ⫺s1
tmax ⫽ s1
My
sx ⫽ ⫺
I
sx ⫽ ⫺2.16 ⫻ 103 psi
I
txy ⫽ 0
Uniaxial stress: s1 ⫽ sy
s2 ⫽ sx
d ⫽ 3 in.
sx ⫽ ⫺864 psi
h
d
Q ⫽ bd a ⫺ b
2
2
sx ⫽ ⫺
h
Ma b
2
sy ⫽ 0
;
(b) 2 IN. ABOVE THE NEUTRAL AXIS
y ⫽ 2 in.
;
(c) TOP OF THE BEAM
s2 ⫽ ⫺180 psi
tmax ⫽ 180 psi
;
a
txy ⫽ ⫺180 psi
Pure shear: s1 ⫽ ⫺txy
;
sx
sx 2
2
⫺ a b + txy
2
A 2
3
(a) NEUTRAL AXIS
sx ⫽ 0
txy ⫽ ⫺151.2 psi
s1 ⫽ 25.7 psi
RA ⫽ 7.2 ⫻ 103 lb
M ⫽ RA c ⫺ q
VQ
Ib
sy ⫽ 0
tmax ⫽
s1 ⫽ 0 psi
s2 ⫽ ⫺2.16 ⫻ 103 psi
sx 2
2
b + txy
A 2
a
tmax ⫽ 1080 psi
;
Q ⫽ 42 in.
3
Problem 8.4-4 An overhanging beam ABC with a guided support
at A is of rectangular cross section and supports concentrated
loads P both at A and at the free end C (see figure). The span length
from A to B is L, and the length of the overhang is L/2. The cross
section has width b and height h. Point D is located midway
between the supports at a distance d from the top face of the beam.
Knowing that the maximum tensile stress (principal stress) at
point D is s1 ⫽ 35 MPa. determine the magnitude of the load P.
Data for the beam are as follows: L ⫽ 1.75 m, b ⫽ 50 mm,
h ⫽ 200 mm, and d ⫽ 40 mm.
P
P
d
A
D
C
B
L
—
2
L
—
2
L
—
2
Probs. 8.4-4 and 8.4-5
h
b
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Page 667
SECTION 8.4
Maximum Stresses in Beams
Solution 8.4-4
L ⫽ 1.75 m
b ⫽ 50 mm
Q ⫽ bd a
h ⫽ 200 mm
d ⫽ 40 mm
txy ⫽
Maximum principal stress at point D:
RB ⫽ 0
M A ⫽ PL + P
L
2
MA ⫽
L
3
M D ⫽ ⫺ PL + P ⫽ ⫺PL
2
2
3
PL
2
bh3
12
Q ⫽ 1.6 ⫻ 105 mm3
VQ
⫽ (96P) N>m2
Ib
PRINCIPAL STRESSES
VD ⫽ P
s1 ⫽
STRESS AT POINT D
I⫽
h
d
⫺ b
2
2
⫽
I ⫽ 3.333 ⫻ 107 mm4
sx + sy
+
2
A
a
y ⫽ 60 mm
P ⫽ 11.10 kN
My
⫺(⫺ PL) y
sx ⫽ ⫺
⫽
⫽ (3150P) N>m2
I
I
2
2
2
b + txy
sx
sx 2
2
a b + txy
⫽ (3.153 * 103 ) P
+
2
A 2
With s1 ⫽ 35 MPa
h
y⫽ ⫺d
2
sx ⫺ sy
P⫽
s1
3.153 * 103
;
sy ⫽ 0
Problem 8.4-5 Solve the preceding problem if the stress and dimensions are as follows: s1 ⫽ 2450 psi, L ⫽ 80 in.,
b ⫽ 2.5 in., h ⫽ 10 in., and d ⫽ 2.5 in.
Solution 8.4-5
L ⫽ 80 in.
b ⫽ 2.5 in.
h ⫽ 10 in.
d ⫽ 2.5 in.
Maximum principal stress at point D:
RB ⫽ 0
M A ⫽ PL + P
L
2
3
L
M D ⫽ ⫺ PL + P ⫽ ⫺PL
2
2
MA ⫽
3
PL
2
VD ⫽ P
STRESS AT POINT D
I⫽
bh3
12
y⫽
h
⫺d
2
sx ⫽ ⫺
sy ⫽ 0
I ⫽ 208.333 in.4
y ⫽ 2.5 in.
My
⫺(⫺ PL)y
lb
⫽
⫽ (0.96P)
I
I
in.2
Q ⫽ bd a
txy ⫽
h
d
⫺ b
2
2
Q ⫽ 23.438 in.3
VQ
⫽ (0.045P) lb/in.2
Ib
PRINCIPAL STRESSES
s1 ⫽
⫽
sx + sy
+
2
A
a
sx ⫺ sy
2
2
2
b + txy
sx 2
sx
1b
2
+
a b + txy
⫽ (0.962P) 2
2
A 2
in.
With s1 ⫽ 2450 psi
P ⫽ 2.55 k
;
P⫽
s1
0.962
667
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Page 668
Applications of Plane Stress
Problem 8.4-6 A beam of wide-flange cross section (see figure) has the following dimensions:
b
b ⫽ 120 mm, t ⫽ 10 mm, h ⫽ 300 mm, and h1 ⫽ 260 mm. The beam is simply supported with
span length L ⫽ 3.0 m. A concentrated load P ⫽ 120 kN acts at the midpoint of the span.
At a cross section located 1.0 m from the left-hand support, determine the principal stresses s1
and s2 and the maximum shear stress tmax at each of the following locations: (a) the top of the beam,
(b) the top of the web, and (c) the neutral axis.
t
h1 h
Probs. 8.4-6 and 8.4-7
Solution 8.4-6
Simply supported beam
sy ⫽ 0 txy ⫽ 0
Uniaxial stress: s1 ⫽ 0
s2 ⫽ ⫺82.7 MPa
tmax ⫽ 41.3 MPa
;
K
(b) TOP OF THE WEB (POINT B)
sx ⫽ ⫺
P ⫽ 120 kN
M⫽
L ⫽ 3.0 m
Pc
⫽ 60 kN # m
2
V⫽
b ⫽ 120 mm
t ⫽ 10 mm
h ⫽ 300 mm
h1 ⫽ 260 mm
I⫽
c ⫽ 1.0 m
P
⫽ 60 kN
2
(b ⫺ t)h 31
bh3
⫺
⫽ 108.89 * 106 mm4
12
12
(a) TOP OF THE BEAM (POINT A)
My
(60 kN # m)(130 mm)
⫽ ⫺
I
108.89 * 106 mm4
⫽ ⫺71.63 MPa
sy ⫽ 0
Q ⫽ (b)a
txy ⫽ ⫺
h ⫺ h1 h + h1
ba
b ⫽ 336 * 103 mm3
2
4
VQ
(60 kN)(336 * 103 mm3)
⫽ ⫺
It
(108.89 * 106 mm4)(10 mm)
⫽ ⫺18.51 MPa
s1,2 ⫽
sx + sy
;
2
A
a
sx ⫺ sy
2
2
2
b + txy
⫽ ⫺35.82 ; 40.32 MPa
s1 ⫽ 4.5 MPa, s2 ⫽ ⫺76.1 MPa
tmax ⫽
A
a
sx ⫺ sy
2
;
2
2
b + txy
⫽ 40.3 MPa
(c) NEUTRAL AXIS (POINT C)
sx ⫽ 0
sx ⫽
(60 kN # m)(150 mm)
Mc
⫽ ⫺
I
108.89 * 106 mm4
⫽ ⫺82.652 MPa
sy ⫽ 0
h1 h1
h h
Q ⫽ ba b a b ⫺ (b ⫺ t)a b a b
2 4
2
4
⫽ 420.5 * 103 mm3
;
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SECTION 8.4
txy ⫽ ⫺
VQ
(60 kN)(420.5 * 103 mm3)
⫽ ⫺
It
(108.89 * 106 mm4)(10 mm)
⫽ ⫺23.17 MPa
Maximum Stresses in Beams
Pure shear: s1 ⫽ 23.2 MPa,
s2 ⫽ ⫺23.2 MPa
K
tmax ⫽ 23.2 MPa
669
;
Problem 8.4-7 A beam of wide-flange cross section (see figure) has the following dimensions: b ⫽ 5 in., t ⫽ 0.5 in.,
h ⫽ 12 in., and h1 ⫽ 10.5 in. The beam is simply supported with span length L ⫽ 10 ft and supports a uniform load q ⫽ 6 k/ft.
Calculate the principal stresses s1 and s2 and the maximum shear stress tmax at a cross section located 3 ft from
the left-hand support at each of the following locations: (a) the bottom of the beam, (b) the bottom of the web, and
(c) the neutral axis.
Solution 8.4-7
Simply supported beam
q ⫽ 6.0 k/ft
L ⫽ 10 ft ⫽ 120 in.
M⫽
c ⫽ 3 ft ⫽ 36 in.
V⫽
qL
⫺ qc ⫽ 12,000 lb
2
b ⫽ 5.0 in.
I⫽
qc2
qLc
⫺
⫽ 756,000 lb-in.
2
2
t ⫽ 0.5 in.
(b ⫺
bh
⫺
12
12
3
t)h 31
sx ⫽ ⫺
h ⫽ 12 in.
h1 ⫽ 10.5 in.
⫽ 285.89 in.4
⫽ 13,883 psi
txy ⫽ ⫺
(756,000 lb-in.)(⫺ 6.0 in.)
Mc
⫽
I
285.89 in.4
⫽ 15,866 psi
h ⫺ h1 h + h1
ba
b ⫽ 21.094 in.3
2
4
VQ
(12,000 lb)(21.094 in.3)
⫽ ⫺
It
(285.89 in.4)(0.5 in.)
⫽ ⫺1771 psi
s1,2 ⫽
sy ⫽ 0 txy ⫽ 0
Uniaxial stress: s1 ⫽ 15,870 psi,
d
s2 ⫽ 0 tmax ⫽ 7930 psi
My
(756,000 lb-in.)(⫺5.25 in.)
⫽ ⫺
I
285.89 in.4
sy ⫽ 0 Q ⫽ ba
(a) BOTTOM OF THE BEAM (POINT A)
sx ⫽ ⫺
(b) BOTTOM OF THE WEB (POINT B)
sx + sy
;
2
A
a
sx ⫺ sy
2
⫽ 6941.5 ; 7163.9 psi
;
2
2
b + txy
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Applications of Plane Stress
s1 ⫽ 14,100 psi, s2 ⫽ ⫺220 psi
tmax ⫽
A
a
sx ⫺ sy
2
;
txy ⫽ ⫺
2
b + txy2 ⫽ 7160 psi
;
⫽ ⫺2349 psi
Pure shear: s1 ⫽ 2350 psi,
s2 ⫽ ⫺2350 psi,
K
tmax ⫽ 2350 psi
(c) NEUTRAL AXIS (POINT C)
sx ⫽ 0
VQ
(12,000 lb)(27.984 in.3)
⫽ ⫺
It
(285.89 in.4)(0.5 in.)
sy ⫽ 0
h1 h1
h h
Q ⫽ ba b a b ⫺ (b ⫺ t)a b a b
2 4
2
4
;
⫽ 27.984 in.3
Problem 8.4-8 A W 200 ⫻ 41.7 wide-flange beam (see Table E-1(b).
100 kN
Appendix E) is simply supported with a span length of 2.5 m (see figure).
The beam supports a concentrated load of 100 kN at 0.9 m from support B.
At a cross section located 0.7 m from the left-hand support, determine
the principal stresses s1 and s2 and the maximum shear stress tmax at each
of the following locations: (a) the top of the beam, (b) the top of the web, and
(c) the neutral axis.
W 200 × 41.7
A
B
D
0.7 m
0.9 m
2.5 m
0.9 m
Solution 8.4-8
RB ⫽ 100 kN a
0.7 + 0.9
b
2.5
RA ⫽ 100 kN ⫺ RB
RB ⫽ 64 kN (upward)
RA ⫽ 36 kN
(upward)
(a) TOP OF THE BEAM (POINT A)
sx ⫽ ⫺
At the point D
M ⫽ RA(0.7 m)
V ⫽ RA
M ⫽ 25.2 kN # m
V ⫽ 36 kN
sy ⫽ 0
h
Ma b
2
I
sx ⫽ ⫺63.309 MPa
txy ⫽ 0
Uniaxial stress: s1 ⫽ sy
W200 ⫻ 41.7
I ⫽ 40.8 ⫻ 106 mm4
b ⫽ 166 mm
s1 ⫽ 0
;
s2 ⫽ sx
s2 ⫽ ⫺63.3 MPa
tmax ⫽ 31.7 MPa
;
tw ⫽ 7.24 mm
(b) TOP OF THE WEB (POINT B)
tf ⫽ 11.8 mm
Ma
h ⫽ 205 mm
h1 ⫽ h ⫺ 2t f
h1 ⫽ 181.4 mm
sx ⫽ ⫺
h1
b
2
I
sx ⫽ ⫺56.021 MPa
sy ⫽ 0
Q ⫽ ba
tmax ⫽ `
h ⫺ h1 h + h1
ba
b
2
4
Q ⫽ 1.892 ⫻ 105 mm3
;
sx
`
2
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SECTION 8.4
txy ⫽ ⫺
s1 ⫽
VQ
I tw
txy ⫽ ⫺23.061 MPa
sx + sy
+
2
A
s1 ⫽ 8.27 MPa
s2 ⫽
sx + sy
2
sx ⫺
a
2
A
a
2
b + txy
⫺
A
sx ⫺ sy
a
sx ⫺ sy
2
2
2
2
b + txy
sx ⫽ 0
sy ⫽ 0
h1
h1
h
h
Q ⫽ b a b a b ⫺ 1 b ⫺ t w2 a b a b
2
4
2
4
txy ⫽ ⫺
VQ
I tw
txy ⫽ ⫺26.69 MPa
Pure shear: s1 ⫽ ƒ txy ƒ
;
2
2
b + txy
tmax ⫽ 36.3 MPa
(c) NEUTRAL AXIS (POINT C)
Q ⫽ 2.19 ⫻ 105 mm3
;
s2 ⫽ ⫺64.3 MPa
tmax ⫽
sy 2
671
Maximum Stresses in Beams
s1 ⫽ 26.7 Mpa
s2 ⫽ ⫺s1
s2 ⫽ ⫺26.7 Mpa
tmax ⫽ txy
tmax ⫽ ⫺26.7 Mpa
;
;
;
;
Problem 8.4-9 A W 12 ⫻ 14 wide-flange beam (see Table E-1(a), Appendix E)
7.5 k
is simply supported with a span length of 120 in. (see figure). The beam supports
two anti-symmetrically placed concentrated loads of 7.5 k each.
At a cross section located 20 in. from the right-hand support, determine the
principal stresses s1 and s2 and the maximum shear stress tmax at each of the
following locations: (a) the top of the beam, (b) the top of the web, and
(c) the neutral axis.
7.5 k
W 12 × 14
D
40 in.
40 in.
20 in. 20 in.
120 in.
Solution 8.4-9
RB ⫽
7.5 # 80 ⫺ 7.5 # 40
k
120
RA ⫽ ⫺RB
RA ⫽ ⫺2.5 k
RB ⫽ 2.5 k (upward)
(downward)
At Section C-C
M ⫽ RB # 20 in.
V ⫽ ⫺RB
M ⫽ 50 k # in.
V ⫽ ⫺2.5 k
(a) TOP OF THE BEAM (POINT A)
sx ⫽ ⫺
sy ⫽ 0
h
Ma b
2
I
sx ⫽ ⫺3.361 ⫻ 103 psi
txy ⫽ 0
Uniaxial stress: s1 ⫽ sy
W12 ⫻ 14
I ⫽ 88.6 in.4
b ⫽ 3.970 in.
s1 ⫽ 0
;
tmax ⫽ 1680 psi
s2 ⫽ ⫺3361 psi
(b) TOP OF THE WEB (POINT B)
tf ⫽ 0.225 in.
Ma
h ⫽ 11.91 in.
h1 ⫽ h ⫺ 2tf
h1 ⫽ 11.46 in.
sx ⫽ ⫺
sy ⫽ 0
I
tmax ⫽ `
;
;
tw ⫽ 0.200 in.
h1
b
2
s2 ⫽ sx
sx ⫽ ⫺3.234 ⫻ 103 psi
sx
`
2
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CHAPTER 8
Q ⫽ ba
txy ⫽ ⫺
s1 ⫽
Page 672
Applications of Plane Stress
h ⫺ h1
h + h1
ba
b
2
4
VQ
I tw
sx + sy
+
A
2
sx + sy
2
A
a
sx ⫺ sy
a
2
⫺
A
a
sx ⫺ sy
2
Q ⫽ 8.502 in3
2
2
b + txy
txy ⫽ ⫺
2
sy 2
2
b + txy
2
tmax ⫽ 1777 psi
VQ
I tw
txy ⫽ 1.2 ⫻ 103 psi
Pure shear:
2
b + txy
;
sx ⫺
sy ⫽ 0
h1
h1
h
h
Q ⫽ ba b a b ⫺ 1 b ⫺ t w2a b a b
2
4
2
4
;
s2 ⫽ ⫺3393 psi
tmax ⫽
sx ⫽ 0
txy ⫽ 736.289 psi
s1 ⫽ 159.8 psi
s2 ⫽
(C) NEUTRAL AXIS (POINT C)
Q ⫽ 5.219 in.3
s1 ⫽ ƒ txy ƒ
s1 ⫽ 1200 psi
s2 ⫽ ⫺s1
s2 ⫽ ⫺1200 psi
;
tmax ⫽ txy
tmax ⫽ 1200 psi
;
;
;
Problem *8.4-10 A cantilever beam of T-section is loaded
by an inclined force of magnitude 6.5 kN (see figure). The line
of action of the force is inclined at an angle of 60° to the
horizontal and intersects the top of the beam at the end cross
section. The beam is 2.5 m long and the cross section has the
dimensions shown.
Determine the principal stresses s1 and s2 and the
maximum shear stress tmax at points A and B in the web
of the beam near the support.
60°
B
6.5 kN
A
y
2.5 m
80 80
mm mm
25 mm
z
C
160 mm
25 mm
Solution 8.4-10
P ⫽ 6.5 kN
L ⫽ 2.5 m
A ⫽ 8 ⫻ 10 mm
3
Location of centroid C
c2 ⫽
©( yi Ai)
A
c2 ⫽
c2 ⫽ 126.25 mm
A ⫽ 2(160 mm)(25 mm)
b ⫽ 160 mm
2
t ⫽ 25 mm
From Eq. (12-7b) in Chapter 12:
(160 mm) (25 mm) a160⫹
c1 ⫽ 185 mm ⫺ c2
25
b mm + (160 mm) (25 mm) (80 mm)
2
A
c1 ⫽ 58.75 mm
MOMENT OF INTERTIA
IZ ⫽
1 3
1
1
t c2 + b c31 ⫺ ( b ⫺ t) ( c1 ⫺ t)3
3
3
3
IZ ⫽ 2.585 ⫻ 107 mm4
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Page 673
SECTION 8.4
EQUIVALENT LOADS AT FREE END OF BEAM
PH ⫽ ⫺P cos (60 deg)
Pv ⫽ 5.629 kN
PH ⫽ ⫺3.25 kN
Mc ⫽ PH c1
Pv ⫽ P sin (60 deg)
Mc ⫽ ⫺190.938 N # m
STRESS RESULTANTS AT FIXED END OF BEAM
N0 ⫽ PH
N0 ⫽ ⫺3.25 kN
V ⫽ Pv
V ⫽ 5.629 kN
M ⫽ ⫺Mc ⫺ Pv L
M ⫽ ⫺1.388 ⫻ 104 N # m
Stress at point A (bottom of web)
sx ⫽
N0
Mc1
+
A
IZ
sx ⫽ ⫺31.951 MPa
Uniaxial stress: s1 ⫽ sy
s1 ⫽ 0
s2 ⫽ sx
sy ⫽ 0
tmax ⫽ `
s2 ⫽ ⫺32.0 MPa
;
;
txy ⫽ 0
sx
`
2
tmax ⫽ 15.98 MPa
Stress at point B (top of web)
sx ⫽
M1c1 ⫺ t2
N0
⫺
A
Iz
Q ⫽ bt ac1 ⫺
txy ⫽ ⫺
s1 ⫽
s2 ⫽
t
b
2
VQ
Iz t
txy ⫽ ⫺1.611 MPa
+
2
tmax ⫽
2
A
a
sy ⫽ 0
Q ⫽ 1.85 ⫻ 105 mm3
sx ⫺ sy
A
a
sx ⫺ sy
A
a
sx + sy
sx + sy
sx ⫽ 17.715 MPa
⫺
sx ⫺ sy
2
2
2
2
b + txy2
2
b + t2xy
2
b + t2xy
s1 ⫽ 17.86 MPa
s2 ⫽ ⫺0.145 MPa
tmax ⫽ 9.00 MPa
;
;
;
;
Maximum Stresses in Beams
673
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CHAPTER 8
Page 674
Applications of Plane Stress
Problem *8.4-11 A simple beam of rectangular cross section has
span length L ⫽ 62 in. and supports a concentrated moment
M ⫽ 560 k-in at midspan (see figure). The height of the beam
is h ⫽ 6 in. and the width is b ⫽ 2.5 in.
Plot graphs of the principal stresses s1 and s2 and the
maximum shear stress tmax, showing how they vary over the
height of the beam at cross section mn, which is located 24 in.
from the left-hand support.
L
M at —
2
m
h
n
b
L
Probs. 8.4-11 and 8.4-12
Solution 8.4-11
2 × 103
sx(y) ⫽ ⫺
Prin. stresses (psi)
1 × 103
Q(y) ⫽ ba
σ.1(y)
σ.2(y)
0
txy(y) ⫽
τmax (y)
–1 × 103
–2.5
–3
M ⫽ 560 k # in.
b ⫽ 2.5 in.
bh
12
M
RA ⫽
L
–2
–1.5
–1
–0.5
0
y
0.5
1
1.5
2
2.5
3
s1(y) ⫽
sx(y)
sx(y) 2
a
+
b + txy(y)2
2
A
2
s2(y) ⫽
sx(y)
sx(y) 2
⫺ a
b + txy(y)2
2
A
2
3
L ⫽ 62 in.
c ⫽ 24 in.
h ⫽ 6 in.
RA ⫽ 9.032 k
sx(y) 2
b + txy(y)2
A
2
a
s2(3 in.) ⫽ ⫺14,452 psi
(upward)
RB ⫽ ⫺9.032 k
tmax(3 in.) ⫽ 7226 psi
s1(0) ⫽ 903 psi
(downward)
At section m-n
V ⫽ RA
tmax(y) ⫽
s1(3 in.) ⫽ 0 psi
I ⫽ 45 in4
RB ⫽ ⫺RA
M ⫽ RAc
1
h
h
⫺ yb a b a + yb
2
2
2
VQ(y)
Ib
dist. y (in.)
I⫽
sy ⫽ 0
PRINCIPAL STRESSES
–2 × 103
–3
3
My
I
M ⫽ 216.774 k # in.
V ⫽ 9.032 k
s2(0) ⫽ ⫺903 psi
tmax(0) ⫽ 903 psi
s1(⫺3 in.) ⫽ 14,452 psi
s2(⫺3 in.) ⫽ 0 psi
tmax(⫺3 in.) ⫽ 7226 psi
Problem *8.4-12 Solve the preceding problem for a cross section mn located 0.18 m from the support if L ⫽ 0.75 m,
M ⫽ 65 kN # m, h ⫽ 120 mm, and b ⫽ 20 mm.
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Page 675
SECTION 8.5
675
Combined Loadings
Solution 8.4-12
sx( y) ⫽
Prin. stresses (MPa)
08Ch08.qxd
Q( y) ⫽ b a
txy( y) ⫽
sy ⫽ 0
1
h
h
⫺ yb a b a + yb
2
2
2
VQ( y)
Ib
PRINCIPAL STRESSES
dist.y(mm)
M ⫽ 65 kN # m
b ⫽ 20 mm
3
bh
I⫽
12
RA ⫽
My
I
M
L
RB ⫽ ⫺RA
L ⫽ 0.75 m
c ⫽ 0.18 m
h ⫽ 120 mm
I ⫽ 2.88 * 10 mm
4
V ⫽ RA
s2( y) ⫽
sx( y)
sx( y) 2
⫺ a
b + txy( y)2
2
A
2
tmax( y) ⫽
sx( y) 2
b + txy( y)2
A
2
a
s1(0) ⫽ 54.2 MPa
(upward)
RB ⫽ ⫺86.667 kN
s2(60 mm) ⫽ ⫺325 MPa
tmax(60 mm) ⫽ 162.5 MPa
(downward)
At section m-n
M ⫽ RA c
sx( y)
sx( y) 2
a
+
b + txy( y)2
2
A
2
s1(60 mm) ⫽ 0 MPa
6
RA ⫽ 86.667 kN
s1( y) ⫽
s2(0) ⫽ ⫺54.2 MPa
tmax(0) ⫽ 54.2 MPa
s1(⫺60 mm) ⫽ 325 MPa
s2(⫺60 mm) ⫽ 0 MPa
tmax(⫺60 mm) ⫽ 162.5 MPa
M ⫽ 15.6 kN # m
V ⫽ 86.667 kN
Combined Loadings
y0
The problems for Section 8.5 are to be solved assuming that the
structures behave linearly elastically and that the stresses caused by
two or more loads may be superimposed to obtain the resultant stresses
acting at a point. Consider both in-plane and out-of-plane shear
stresses unless otherwise specified.
b2
b1
D
B
C
P
Problem 8.5-1 A bracket ABCD having a hollow circular cross
section consists of a vertical arm AB, a horizontal arm BC parallel to
the x0 axis, and a horizontal arm CD parallel to the z0 axis (see figure).
The arms BC and CD have lengths b1 ⫽ 3.6 ft and b2 ⫽ 2.2 ft, respectively. The outer and inner diameters of the bracket are d2 ⫽ 7.5 in. and
d1 ⫽ 6.8 in. A vertical load P ⫽ 1400 lb acts at point D. Determine the
maximum tensile, compressive, and shear stresses in the vertical arm.
A
z0
x0
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CHAPTER 8
Page 676
Applications of Plane Stress
Solution 8.5-1
b1 ⫽ 3.6 ft
b2 ⫽ 2.2 ft
d2 ⫽ 7.5 in.
P ⫽ 1400 lb
MAXIMUM COMPRESSIVE STRESS
d1 ⫽ 6.8 in.
A⫽
p
ad 22 ⫺ d 21 b
4
A ⫽ 7.862 in.2
I⫽
p
ad 24 ⫺ d 14 b
64
I ⫽ 50.36 in.4
sc ⫽ ⫺
P
⫺
A
Ma
d2
b
2
I
sc ⫽ ⫺5456 psi
;
MAXIMUM SHEAR STRESS
VERTICAL ARM AB
M⫽
P 2b 21
+
b 22
Uniaxial stress
M ⫽ 7.088 * 10
4
lb # in.
tmax ⫽ |sc|
tmax ⫽ 5456 psi
;
MAXIMUM TENSILE STRESS
st ⫽ ⫺
P
+
A
Ma
d2
b
2
I
st ⫽ 5100 psi
;
Problem 8.5-2 A gondola on a ski lift is supported by two bent arms, as
W
shown in the figure. Each arm is offset by the distance b ⫽ 180 mm
from the line of action of the weight force W. The allowable stresses in
the arms are 100 MPa in tension and 50 MPa in shear.
If the loaded gondola weighs 12 kN, what is the minimum diameter
d of the arms?
d
b
W
(a)
(b)
08Ch08.qxd
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Page 677
SECTION 8.5
Solution 8.5-2
Gondola on a ski lift
b ⫽ 180 mm
W⫽
12 kN
⫽ 6 kN
2
sallow ⫽ 100 MPa (tension)
A⫽
pd
4
S⫽
3
pd
32
MAXIMUM TENSILE STRESS
W
M
4W
32 Wb
+
+
⫽
2
A
S
pd
pd 3
pst 3
or a
bd ⫺ d ⫺ 8b ⫽ 0
4W
st ⫽
677
SUBSTITUTE NUMERICAL VALUES:
p(100 Mpa)
pst
psallow
1
⫽
⫽
⫽ 13,089.97 2
4W
4W
4(6 kN)
m
8b ⫽ 1.44 m
Find dmin
2
tallow ⫽ 50 MPa
Combined Loadings
13,090 d3 ⫺ d ⫺ 1.44 ⫽ 0
(d ⫽ meters)
Solve numerically: d ⫽ 0.04845 m
⬖ dmin ⫽ 48.4 mm
;
MAXIMUM SHEAR STRESS
st
(uniaxial stress)
2
Since tallow is one-half of sallow, the minimum diameter
for shear is the same as for tension.
tmax ⫽
Problem 8.5-3 The hollow drill pipe for an oil well (see figure) is 6.2 in. in outer
diameter and 0.75 in. in thickness. Just above the bit, the compressive force in the
pipe (due to the weight of the pipe) is 62 k and the torque (due to drilling) is 185 k-in.
Determine the maximum tensile, compressive, and shear stresses in the
drill pipe.
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CHAPTER 8
Page 678
Applications of Plane Stress
Solution 8.5-3
P ⫽ compressive force
T ⫽ Torque
txy ⫽
d2 ⫽ outer diameter
d1 ⫽ inner diameter
P ⫽ 62 k
t ⫽ 0.75 in.
T ⫽ 185 k # in.
d1 ⫽ d2 ⫺ 2t
p
A ⫽ a d 22 ⫺ d 2b
1
4
p
Ip ⫽
ad 24 ⫺ d 14 b
32
d2 ⫽ 6.2 in.
d1 ⫽ 4.7 in.
Ta
d2
b
2
txy ⫽ 5903 psi
Ip
PRINCIPAL STRESSES
s1 ⫽
sx ⫺ sy
A
a
sx ⫺ sy
A
a
sx + sy
+
2
2
b + txy2
2
s1 ⫽ 3963 psi
A ⫽ 12.841 in.2
Ip ⫽ 97.16 in.
4
s2 ⫽
sx + sy
2
⫺
2
b + txy2
2
s2 ⫽ ⫺8791 psi
STRESSES AT THE OUTER SURFACE
st ⫽ s1
MAXIMUM TENSILE STRESS
st ⫽ 3963 psi
;
MAXIMUM COMPRESSIVE STRESS sc ⫽ s2
sc ⫽ ⫺8791 psi
;
MAXIMUM IN-PLANESHEAR STRESS
tmax ⫽
sy ⫽ ⫺
sx ⫽ 0
P
A
A
a
sx ⫺ sy
2
tmax ⫽ 6377 psi
sy ⫽ ⫺4828 psi
b + txy2
;
NOTE: Since the principal stresses have opposite signs,
the maximum in-plane shear is larger than the maximum
out-of-plane shear stress.
P
Problem 8.5-4 A segment of a generator shaft is subjected to a torque T and an axial
force P, as shown in the figure. The shaft is hollow (outer diameter d2 ⫽ 300 mm
and inner diameter d1 ⫽ 250 mm) and delivers 1800 kW at 4.0 Hz.
If the compressive force P ⫽ 540 kN, what are the maximum tensile, compressive, and shear stresses in the shaft?
T
T
Probs. 8.5-4 and 8.5-5
P
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Page 679
SECTION 8.5
Combined Loadings
679
Solution 8.5-4
P ⫽ Compressive_force
P0 ⫽ Power
f ⫽ frequency
T ⫽ torgue ⫽
P0
2p f
d2 ⫽ outer diameter
sy ⫽ ⫺
P ⫽ 540 kN
txy ⫽
P0 ⫽ 1800 kW
P0
2p f
T ⫽ 7.162 ⫻ 104 N # m
d2 ⫽ 300 mm
d1 ⫽ 250 mm
T⫽
A⫽
p
ad 22 ⫺ d 21 b
4
A ⫽ 2.16 ⫻ 104 mm2
Ip ⫽
p
ad 42 ⫺ d 14 b
32
sy ⫽ ⫺25.002 MPa
sx ⫽ 0
Ta
d2
b
2
txy ⫽ 26.093 MPa
Ip
d1 ⫽ inner diameter
f ⫽ 4.0 Hz
P
A
PRINCIPAL STRESSES
sx + sy
sx ⫺ sy 2
2
s1 ⫽
a
+
b + txy
2
A
2
s1 ⫽ 16.432 MPa
s2 ⫽
sx + sy
2
⫺
A
a
sx ⫺ sy
2
s2 ⫽ ⫺41.434 MPa
Ip ⫽ 4.117 ⫻ 108 mm4
STRESSES AT THE OUTER SURFACE
2
2
b + txy
MAXIMUM TENSILE STRESS
st ⫽ 16.43 MPa
st ⫽ s1
;
MAXIMUM COMPRESSIVE STRESS
sc ⫽ ⫺41.4 MPa
sc ⫽ s2
;
MAXIMUM IN-PLANESHEAR STRESS
tmax ⫽
A
a
sx ⫺ sy
2
tmax ⫽ 28.9 MPa
2
2
b + txy
;
NOTE: Since the principal stresses have opposite signs,
the maximum in-plane shear is larger than the maximum
out-of-plane shear stress.
Problem 8.5-5 A segment of a generator shaft of hollow circular cross section is subjected to a torque T ⫽ 240 k-in.
(see figure). The outer and inner diameters of the shaft are 8.0 in. and 6.25 in., respectively.
What is the maximum permissible compressive load P that can be applied to the shaft if the allowable in-plane shear
stress is tallow ⫽ 6250 psi?
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CHAPTER 8
Page 680
Applications of Plane Stress
Solution 8.5-5
P ⫽ compressive force
d2 ⫽ outer diameter
T ⫽ Torque
d1 ⫽ inner diameter
A⫽
P
A
sx ⫽ 0
T ⫽ 240 k # in.
d2 ⫽ 8 in.
sy ⫽ ⫺
d1 ⫽ 6.25 in.
p
ad 22 ⫺ d 21 b
4
tallow ⫽ 6250 psi
A ⫽ 19.586 in.2
Ip ⫽
p
ad 42 ⫺ d 14 b
32
txy ⫽
d2
Ta b
2
txy ⫽ 3805 psi
Ip
MAXIMUM IN-PLANESHEAR STRESS
a
sx ⫺ sy
2
b + t 2xy
Ip ⫽ 252.321 in.4
tmax ⫽
STRESSES AT THE OUTER SURFACE
2 ⫺ t22 4
sy ⫽ 11tmax
xy
A
P ⫽ sy A
2
tmax ⫽ tallow
sy ⫽ 9917 psi
P ⫽ 194.2 k
;
NOTE: The maximum in-plane shear is larger than the
maximum out-of-plane shear stress.
Problem 8.5-6 A cylindrical tank subjected to internal pressure p is
simultaneously compressed by an axial force F ⫽ 72 kN (see figure). The
cylinder has diameter d ⫽ 100 mm and wall thickness t ⫽ 4 mm.
Calculate the maximum allowable internal pressure pmax based upon an
allowable shear stress in the wall of the tank of 60 MPa.
Solution 8.5-6
F
F
Cylindrical tank with compressive force
CIRCUMFERENTIAL STRESS (TENSION)
s1 ⫽
F ⫽ 72 kN
Units: s1 ⫽ MPa
p ⫽ internal pressure
d ⫽ 100 mm
pr
p(50 mm)
⫽
⫽ 12.5 p
t
4 mm
t ⫽ 4 mm
tallow ⫽ 60 MPa
p ⫽ MPa
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Page 681
SECTION 8.5
LONGITUDINAL STRESS (TENSION)
s2 ⫽
pr
pr
F
F
⫺ ⫽
⫺
2t
A
2t
2prt
⫽ 6.25p ⫺
72,000 N
2p(50 mm)(4 mm)
⫽ 6.25p ⫺ 57.296 Mpa
Units: s2 ⫽ MPa
p ⫽ MPa
Combined Loadings
681
OUT-OF-PLANE SHEAR STRESSES
Case 2: tmax ⫽
s1
⫽ 6.25 p; 60 MPa ⫽ 6.25 p
2
Solving, p2 ⫽ 9.60 MPa
Case 3: tmax ⫽
s2
⫽ 3.125 p ⫺ 28.648 MPa
2
60 MPa ⫽ 3.125 p ⫺ 28.648 MPa
Solving, p3 ⫽ 28.37 MPa
BIAXIAL STRESS
IN-PLANE SHEAR STRESS (CASE 1)
CASE 2, OUT-OF-PLANE SHEAR STRESS GOVERNS
s1 ⫺ s2
⫽ 3.125 p + 28.648 Mpa
tmax ⫽
2
pmax ⫽ 9.60 MPa
;
60 MPa ⫽ 3.125 p ⫹ 28.648 MPa
Solving, p1 ⫽ 10.03 MPa
Problem 8.5-7 A cylindrical tank having diameter d ⫽ 2.5 in. is
subjected to internal gas pressure p ⫽ 600 psi and an external tensile load T ⫽ 1000 lb (see figure).
Determine the minimum thickness t of the wall of the tank
based upon an allowable shear stress of 3000 psi.
Solution 8.5-7
T
Cylindrical tank with tensile load
CIRCUMFERENTIAL STRESS (TENSION)
(600 psi)(1.25 in.)
pr
750
⫽
⫽
t
t
t
Units: s1 ⫽ psi
t ⫽ inches
s2 ⫽ psi
s1 ⫽
T ⫽ 1000 lb
t ⫽ thickness
p ⫽ 600 psi
d ⫽ 2.5 in.
tallow ⫽ 3000 psi
T
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Page 682
Applications of Plane Stress
LONGITUDINAL STRESS (TENSION)
s2 ⫽
pr
pr
T
T
+ ⫽
+
2t
A
2t
2prt
375
1000 lb
375
127.32
502.32
⫽
+
⫽
+
+
t
2p(1.25 in.)t
t
t
t
BIAXIAL STRESS
3000 psi ⫽
123.84
t
Solving, t 1 ⫽ 0.0413 in.
OUT-OF-PLANE SHEAR STRESSES
Case 2: tmax ⫽
s1
375
375
⫽
; 3000 ⫽
2
t
t
Solving, t2 ⫽ 0.125 in.
Case 3: tmax ⫽
s2
251.16
251.16
⫽
; 3000 ⫽
2
t
t
Solving, t3 ⫽ 0.0837 in.
CASE 2, OUT-OF-PLANE SHEAR STRESS GOVERNS
tmin ⫽ 0.125 in.
;
IN-PLANE SHEAR STRESS (CASE 1)
tmax ⫽
s1 ⫺ s2
247.68
123.84
⫽
⫽
2
2t
t
Problem 8.5-8 The torsional pendulum shown in the figure consists of a horizontal
circular disk of mass M ⫽ 60 kg suspended by a vertical steel wire (G ⫽ 80 GPa) of
length L ⫽ 2 m and diameter d ⫽ 4 mm.
Calculate the maximum permissible angle of rotation fmax of the disk (that is, the
maximum amplitude of torsional vibrations) so that the stresses in the wire do not
exceed 100 MPa in tension or 50 MPa in shear.
d = 4 mm
L=2m
fmax
M = 60 kg
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SECTION 8.5
Solution 8.5-8
Combined Loadings
683
Torsional pendulum
sx ⫽
TENSILE STRESS
W
⫽ 46.839 MPa
A
sx ⫽ 0
sy ⫽ st ⫽ 46.839 MPa
txy ⫽ ⫺80 fmax (MPa)
L ⫽ 2.0 m
d ⫽ 4.0 mm
M ⫽ 60 kg
G ⫽ 80 GPa
sallow ⫽ 100 MPa
A⫽
tallow ⫽ 50 MPa
2
pd
⫽ 12.5664 mm2
4
PRINCIPAL STRESSES
W ⫽ Mg ⫽ (60 kg)(9.81 m/s ) ⫽ 588.6 N
2
s1, 2 ⫽
s1, 2
sx + sy
;
2
A
a
sx ⫺ sy
2
2
b + t 2xy
⫽ 23.420 ; 1123.42022 + 6400f2max ( MPa)
Note that s1 is positive and s2 is negative. Therefore,
the maximum in-plane shear stress is greater than the
maximum out-of-plane shear stress.
MAXIMUM ANGLE OF ROTATION BASED ON TENSILE STRESS
s1 ⫽ maximum tensile stress
sallow ⫽ 100 MPa
‹ 100 MPa ⫽ 23.420 ; 1123.42022 + 6400f2max
(100 ⫺ 23.420)2 ⫽ 123.42022 + 6400f2max
5316 ⫽ 6400f2max
TORQUE: T ⫽
GIp fmax
L
(EQ. 3-15)
fmax ⫽ 0.9114 rad ⫽ 52.2°
MAXIMUM ANGLE OF ROTATION BASED ON IN-PLANE SHEAR
STRESS
SHEAR STRESS: t ⫽
t⫽ a
GIpfmax
L
t ⫽ 80 fmax
ba
Tr
Ip
(EQ. 3-11)
tmax ⫽
Gr fmax
r
b⫽
⫽ (80 * 106 Pa)fmax
IP
L
Units: t ⫽ MPa
fmax ⫽ radians
sx ⫺ sy 2
b + t2xy ⫽ 1(23.420)2 + 6400f2max
A
2
a
tallow ⫽ 50 MPa
50 ⫽ 1(23.420)2 + 6400f2max
(50)2 ⫽ (23.420)2 + 6400f2max
Solving, fmax ⫽ 0.5522 rad ⫽ 31.6°
SHEAR STRESS GOVERNS
fmax ⫽ 0.552 rad ⫽ 31.6°
;
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Applications of Plane Stress
Problem 8.5-9 Determine the maximum tensile, compressive,
P = 160 lb
and shear stresses at points A and B on the bicycle pedal crank
shown in the figure.
The pedal and crank are in a horizontal plane and points A
and B are located on the top of the crank. The load P ⫽ 160 lb
acts in the vertical direction and the distances (in the horizontal
plane) between the line of action of the load and points A and B
are b1 ⫽ 5.0 in., b2 ⫽ 2.5 in. and b3 ⫽ 1.0 in. Assume that the
crank has a solid circular cross section with diameter d ⫽ 0.6 in.
Crank
d = 0.6 in.
A
B
b3
b1 = 5.0 in.
b3 = 1.0 in.
b2 = 2.5 in.
A
B
b3
b3
b2
⫻
P
b1
Top view
Solution 8.5-9
P ⫽ 160 lb
d ⫽ 0.6 in.
STRESS AT POINT A:
b1 ⫽ 5.0 in.
b2 ⫽ 2.5 in.
t⫽
16 TA
s⫽
MA
S
b3 ⫽ 1.0 in.
S⫽
pd
32
3
STRESS RESULTANTS on cross section at point A:
Torque:
TA ⫽ Pb2
Moment:
MA ⫽ Pb1
Shear Force:
VA ⫽ P
t ⫽ 9.431 ⫻ 103 psi
p d3
s ⫽ 3.773 ⫻ 104 psi
(The shear force V produces no shear stresses at point A)
PRINCIPAL STRESSES AND MAXIMUM SHEAR STRESS
sx ⫽ 0
sy ⫽ s
txy ⫽ ⫺t
sx + sy
a
STRESS RESULTANTS at point B:
s1 ⫽
TB ⫽ P(b1 ⫹ b3)
s1 ⫽ 3.995 ⫻ 10 psi
MB ⫽ P(b2 ⫹ b3)
VB ⫽ P
+
A
2
sx ⫺ sy
2
2
b + t 2xy
4
s2 ⫽
sx + sy
2
⫺
A
a
sx ⫺ sy
s2 ⫽ ⫺2.226 ⫻ 10 psi
3
2
2
b + t 2xy
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SECTION 8.5
MAXIMUM TENSILE STRESS
;
s2 ⫽
sx + sy
2
MAXIMUM COMPRESSIVE STRESS
sc ⫽ ⫺2226 psi
sc ⫽ s2
;
MAXIMUM IN-PLANE SHEAR STRESS
tmax ⫽
A
a
sx ⫺ sy
2
s⫽
MB
S
pd
2
2
b + t2xy
MAXIMUM TENSILE STRESS
st ⫽ 39,410 psi
;
s ⫽ 2.641 ⫻ 104 psi
sc ⫽ ⫺13,000 psi
;
MAXIMUM IN-PLANE SHEAR STRESS
tmax ⫽
t ⫽ 2.264 ⫻ 104 psi
3
sx ⫺ sy
s2 ⫽ ⫺1.3 ⫻ 10
sc ⫽ s2
STRESS AT POINT B
16TB
A
a
MAXIMUM COMPRESSIVE STRESS
;
NOTE: Since the principal stresses have opposite signs,
the maximum in-plane shear is larger than the maximum
out-of-plane shear stress.
t⫽
⫺
4
st ⫽ s1
2
b + t 2xy
tmax ⫽ 21,090 psi
A
a
sx ⫺ sy
2
tmax ⫽ 26,210 psi
2
b + txy2
;
NOTE: Since the principal stresses have opposite signs,
the maximum in-plane shear is larger than the maximum
out-of-plane shear stress.
(The shear force V produces no shear stresses at point A)
PRINCIPAL STRESSES AND MIXIMUM SHEAR STRESS
sx ⫽ 0
s1 ⫽
sy ⫽ s
sx + sy
+
2
A
txy ⫽ ⫺t
a
sx ⫺ sy
2
685
s1 ⫽ 3.941 ⫻ 104 psi
st ⫽ 39,950 psi
st ⫽ s1
Combined Loadings
2
b + t 2xy
Problem 8.5-10 A cylindrical pressure vessel having radius r ⫽ 300 mm
and wall thickness t ⫽ 15 mm is subjected to internal pressure
p ⫽ 2.5 MPa. In addition, a torque T ⫽ 120 kN # m acts at each end of
the cylinder (see figure).
(a) Determine the maximum tensile stress smax and the maximum inplane shear stress tmax in the wall of the cylinder.
(b) If the allowable in-plane shear stress is 30 MPa, what is the maximum allowable torque T?
T
T
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Applications of Plane Stress
Solution 8.5-10
Cylindrical pressure vessel
s1 ⫽ 56.4 MPa
s2 ⫽ 18.6 MPa
‹ smax ⫽ 56.4 MPa
;
MAXIMUM IN-PLANE SHEAR STRESS
tmax ⫽
T ⫽ 120 kN # m
t ⫽ 15 mm
A
r ⫽ 300 mm
P ⫽ 2.5 MPa
a
sx ⫺ sy
2
2
b + t 2xy ⫽ 18.9 MPa
;
(b) MAXIMUM ALLOWABLE TORQUE T
tallow ⫽ 30 MPa (in-plane shear stress)
STRESSES IN THE WALL OF THE VESSEL
pr
⫽ 25 MPa
sx ⫽
2t
Tr
txy ⫽ ⫺
Ip
T
2pr 2t
tmax ⫽
sx ⫽
(EQ. 3-11)
Ip ⫽ 2pr 3t
txy ⫽ ⫺
pr
sy ⫽
⫽ 50 MPa
t
A
a
sx ⫺ sy
2
2
b + t2xy
(1)
pr
pr
⫽ 25 MPa sy ⫽
⫽ 50 MPa
2t
t
T
⫽ ⫺117.893 * 10⫺6 T
(EQ. 3-18)
txy ⫽ ⫺
⫽ ⫺ 14.147 MPa
Units: txy ⫽ MPa
2pr 2t
T⫽N#m
Substitute into Eq. (1):
tmax ⫽ tallow ⫽ 30 MPa
⫽ 2(⫺ 12.5 MPa)2 + (⫺117.893 * 10⫺6 T)2
Square both sides, rearrange, and solve for T:
(30)2 ⫽ (12.5)2 ⫹ (117.893 ⫻ 10⫺6)2 T2
T2 ⫽
743.750
13,899 * 10⫺12
⫽ 53,512 * 106 ( N # m)2
T ⫽ 231.3 * 103 N # m
Tmax ⫽ 231 kN # m
(a) PRINCIPAL STRESSES
s1, 2 ⫽
sx + sy
;
2
A
a
sx ⫺ sy
⫽ 37.5 ; 18.878 MPa
2
2
b + t2xy
;
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SECTION 8.5
Problem 8.5-11 An L-shaped bracket lying in a horizontal plane
Combined Loadings
A
supports a load P ⫽ 150 lb (see figure). The bracket has a hollow
rectangular cross section with thickness t ⫽ 0.125 in. and outer
dimensions b ⫽ 2.0 in. and h ⫽ 3.5 in. The centerline lengths of
the arms are b1 ⫽ 20 in. and b2 ⫽ 30 in.
Considering only the load P, calculate the maximum tensile
stress st, maximum compressive stress sc, and maximum shear
stress tmax at point A, which is located on the top of the bracket
at the support.
687
t = 0.125 in.
b1 = 20 in.
h = 3.5 in.
b2 = 30 in.
b = 2.0 in.
P = 150 lb
Solution 8.5-11
P ⫽ 150 lb
L-shaped bracket
b1 ⫽ 20 in.
t ⫽ 0.125 in.
b2 ⫽ 30 in.
h ⫽ 3.5 in.
b ⫽ 2.0 in.
STRESSES AT POINT A ON THE TOP OF THE BRACKET
t⫽
T
4500 lb-in.
⫽ 2844 psi
⫽
2tAm
2(0.125 in.)(6.3281 in.2)
s⫽
(3000 lb-in.)(1.75 in.)
Mc
⫽ 2454 psi
⫽
I
2.1396 in.4
FREE-BODY DIAGRAM OF BRACKET
(The shear force V produces no stresses at point A.)
STRESS ELEMENT AT POINT A
(This view is looking downward at the top of the bracket.)
STRESS RESULTANTS AT THE SUPPORT
Torque:
Moment:
T ⫽ Pb2 ⫽ (150 lb)(30 in.) ⫽ 4500 lb-in.
M ⫽ Pb1 ⫽ (150 lb)(20 in.) ⫽ 3000 lb-in.
Shear force:
V ⫽ P ⫽ 150 lb
PROPERTIES OF THE CROSS SECTION
For torsion:
Am ⫽ (b ⫺ t)(h ⫺ t) ⫽ (1.875 in.)(3.375 in.) ⫽ 6.3281 in.2
For bending: c ⫽
I⫽
⫽
h
⫽ 1.75 in.
2
1
1
(bh3) ⫺ (b ⫺ 2t)(h ⫺ 2t)3
12
12
1
1
(2.0 in.)(3.5 in.)3 ⫺ (1.75 in.)(3.25 in.)3
12
12
⫽ 2.1396 in.4
sx ⫽ 0
sy ⫽ s ⫽ 2454 psi
txy ⫽ ⫺t ⫽ ⫺2844 psi
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Applications of Plane Stress
PRINCIPAL STRESSES AND MAXIMUM SHEAR STRESS
s1, 2 ⫽
sx + sy
;
A
2
a
sx ⫺ sy
2
2
b + t 2xy
⫽ 1227 psi ; 3097 psi
s1 ⫽ 4324 psi
tmax ⫽
A
a
s2 ⫽ ⫺1870 psi
sx ⫺ sy
2
2
b + t 2xy ⫽ 3097 psi
MAXIMUM COMPRESSIVE STRESS:
sc ⫽ ⫺1870 psi
;
MAXIMUM SHEAR STRESS:
tmax ⫽ 3100 psi
;
NOTE: Since the principal stresses have opposite signs,
the maximum in-plane shear stress is larger than the
maximum out-of-plane shear stress.
MAXIMUM TENSILE STRESS:
st ⫽ 4320 psi
;
Problem 8.5-12 A semicircular bar AB lying in a horizontal plane is
supported at B (see figure). The bar has centerline radius R and
weight q per unit of length (total weight of the bar equals pqR). The
cross section of the bar is circular with diameter d.
Obtain formulas for the maximum tensile stress st, maximum compressive stress sc, and maximum in-plane shear stress tmax at the top of
the bar at the support due to the weight of the bar.
Solution 8.5-12
O
A
B
R
Semicircular bar
STRESSES AT THE TOP OF THE BAR AT B
64qR2
M B(d/2)
(2qR2)(d/2)
⫽
⫽
I
pd 4/64
pd 3
2
16qR2
TB(d/2)
(pqR )(d/2)
tB ⫽
⫽
⫽
IP
pd 4/32
d3
sB ⫽
d ⫽ diameter of bar
R ⫽ radius of bar
q ⫽ weight of bar per unit length
W ⫽ weight of bar ⫽ pqR
Weight of bar acts at the center of gravity
From Case 23, Appendix D, with b ⫽ p/2, we get
y⫽
2R
p
‹ c⫽
2R
p
Bending moment at B: MB ⫽ Wc ⫽ 2qR2
Torque at B: TB ⫽ WR ⫽ pqR2
(Shear force at B produces no shear stress at the top of
the bar.)
d
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SECTION 8.5
STRESS ELEMENT AT THE TOP OF THE BAR AT B
sx ⫽ 0
MAXIMUM TENSILE STRESS
st ⫽ s1 ⫽
sy ⫽ sB
txy ⫽ tB
689
Combined Loadings
⫽ 29.15
16qR2
pd 3
(2 + 24 + p2)
qR2
;
d3
PRINCIPAL STRESSES:
s1, 2 ⫽
sx + sy
;
2
A
a
sx ⫺ sy
2
MAXIMUM COMPRESSIVE STRESS
2
2
b + txy
sc ⫽ s2 ⫽
sB
sB 2
⫽
;
a ⫺ b + tB2
2
A
2
⫽
⫽
32qR 2
pd 3
16qR 2
pd 3
;
A
a
32qR2
pd 3
2
b + a
16qR2
d3
b
pd 3
(2 ⫺ 24 + p2)
⫽ ⫺8.78
2
(2 ; 24 + p2 )
16qR2
qR2
d3
;
MAXIMUM IN-PLANE SHEAR STRESS (EQ. 7-26)
tmax ⫽
16qR2
1
24 + p 2
(s1 ⫺ s2) ⫽
2
pd 3
;
z
Problem 8.5-13 An arm ABC lying in a horizontal plane and
supported at A (see figure) is made of two identical solid steel
bars AB and BC welded together at a right angle. Each bar is
20 in. long.
Knowing that the maximum tensile stress (principal stress)
at the top of the bar at support A due solely to the weights of the
bars is 932 psi, determine the diameter d of the bars.
y
A
x
B
C
Solution 8.5-13
Horizontal arm ABC
L ⫽ length of AB and BC
d ⫽ diameter of AB and BC
A ⫽ cross-sectional area
⫽ pd2/4
g ⫽ weight density of steel
q ⫽ weight per unit length of bars
⫽ gA ⫽ pgd2/4
(1)
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Applications of Plane Stress
RESULTANT FORCES ACTING ON AB
P ⫽ weight of AB and BC
P ⫽ qL ⫽ pgLd2/4
(2)
T ⫽ torque due to weight of BC
sx ⫽ 0
sy ⫽ sA
txy ⫽ ⫺tA
(8)
Substitute (8) into (7):
qL2
pgL2d 2
L
T ⫽ (qL)a b ⫽
⫽
2
2
8
(3)
MA ⫽ bending moment at A
MA ⫽ PL ⫹ PL /2 ⫽ 3PL /2 ⫽ 3pgL2d2/8
s1 ⫽
sA
sA 2
+
a b + t 2A
2
A 2
Substitute from (5) and (6) and simplify:
(4)
s1 ⫽
STRESSES AT THE TOP OF THE BAR AT A
gL2
2gL2
(6 + 140) ⫽
(3 + 110)
d
d
sA ⫽ normal stress due to MA
SOLVE FOR d
M(d/2)
M(d/2)
12gL2
32M
⫽
⫽
⫽
sA ⫽
4
3
I
d
pd /64
pd
d⫽
(5)
2gL2
(3 + 110)
s1
;
(10)
(11)
SUBSTITUTE NUMERICAL VALUES INTO EQ. (11):
tA ⫽ shear stress due to torque T
T(d/2)
T(d/2)
2gL2
16T
⫽
tA ⫽
⫽
⫽
Ip
d
pd 4/32
pd 3
(9)
g ⫽ 490 lb/ft3 ⫽ 0.28356 lb/in.3
(6)
L ⫽ 20 in.
d ⫽ 1.50 in.
s1 ⫽ 932 psi
;
STRESS ELEMENT ON TOP OF THE BAR AT A
s1 ⫽ principal tensile stress (maximum tensile stress)
s1 ⫽
sx + sy
+
2
A
a
sx ⫺ sy
2
2
b + t 2xy
(7)
Problem 8.5-14 A pressurized cylindrical tank with flat ends is loaded
by torques T and tensile forces P (see figure). The tank has radius
r ⫽ 50 mm and wall thickness t ⫽ 3 mm. The internal pressure
p ⫽ 3.5 MPa and the torque T ⫽ 450 N # m.
What is the maximum permissible value of the forces P if the
allowable tensile stress in the wall of the cylinder is 72 MPa?
T
P
T
P
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Page 691
SECTION 8.5
Solution 8.5-14
r ⫽ 50 mm
T ⫽ 450 N # m
Combined Loadings
691
Cylindrical tank
t ⫽ 3.0 mm
p ⫽ 3.5 MPa
sallow ⫽ 72 MPa
Units: sx ⫽ MPa, P ⫽ newtons
sy ⫽
pr
⫽ 58.333 MPa
t
CROSS SECTION
A ⫽ 2prt ⫽ 2p(50 mm)(3.0 mm) ⫽ 942.48 mm2
txy ⫽ ⫺
IP ⫽ 2pr3t ⫽ 2p(50 mm)3(3.0 mm)
(450 N # m)(50 mm)
Tr
⫽ ⫺
IP
2.3562 * 106 mm4
⫽ ⫺9.5493 MPa
⫽ 2.3562 ⫻ 106 mm4
MAXIMUM TENSILE STRESS
STRESSES IN THE WALL OF THE TANK
smax ⫽ sallow ⫽ 72 MPa
⫽
sx + sy
+
2
A
72 ⫽ 43.750 + (530.52 * 10
a
⫺6
sx ⫺ sy
2
2
b + t 2xy
)P
+ 2[⫺ 14.583 + (530.52 * 10⫺6)P]2 + ( ⫺9.5493)2
or
28.250 ⫺ 0.00053052P
⫽ 2( ⫺14.583 + 0.00053052 P)2 + 91.189
Square both sides and simplify:
sx ⫽
pr
P
+
2t
A
(3.5 MPa)(50 mm)
P
+
⫽
2(3.0 mm)
942.48 mm2
494.21 ⫽ 0.014501 P
SOLVE FOR P
P ⫽ 34,080 N
Pmax ⫽ 34.1 kN
OR
;
⫽ 29.167 MPa + 1.0610 * 10⫺3P
z
Problem 8.5-15 A post having a hollow circular cross section supports
a horizontal load P ⫽ 240 lb acting at the end of an arm that is 5 ft long (see figure).
The height of the post is 27 ft, and its section modulus is S ⫽ 15 in.3 Assume that outer
radius of the post, r2 ⫽ 4.5 in., and inner radius r1 ⫽ 4.243 in.
(a) Calculate the maximum tensile stress smax and maximum in-plane shear stress
tmax at point A on the outer surface of the post along the x-axis due to the load P.
Load P acts in a horizontal plane at an angle of 30° from a line which is parallel
to the (⫺x) axis.
(b) If the maximum tensile stress and maximum in-plane shear stress at point A are limited to 16,000 psi and 6000 psi, respectively, what is the largest permissible value of
the load P?
5 ft
30°
P = 240 lb
27 ft
x
A
y
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Applications of Plane Stress
Solution 8.5-15
(a) MAXIMUM
P ⫽ 240 lb
b ⫽ 5 ft
length of arm
h ⫽ 27 ft
height of post
r2 ⫽ 4.5 in.
A⫽
⫺
s2 ⫽
I ⫽ 67.507 in.4
Vx ⫽ P cos(30deg)
Vy ⫽ ⫺P sin(30deg)
+
2
a
A
sx ⫺
b + t 2xy
2
sx + sy
2
⫺
A
a
sx ⫺ sy
2
2
b + t 2xy
MAXIMUM TENSILE STRESS
Q ⫽ 9.825 in.3
smax ⫽ 4534 psi
smax ⫽ s1
REACTIONS AT THE SUPPORT
T ⫽ ⫺P cos(30deg)b
sx + sy
txy ⫽ t
sy 2
s2 ⫽ ⫺44.593 psi
Ip ⫽ 135.014 in.4
M ⫽ P cos(30deg)h
sy ⫽ 4.489 ⫻ 103 psi
s1 ⫽ 4.534 ⫻ 103 psi
A ⫽ 7.059 in.2
2
1 r 32 ⫺ r 312
3
Q⫽
s1 ⫽
t ⫽ 0.257 in.
r12 )
p
I ⫽ 1 r 42 ⫺ r 412
4
Ip ⫽ 2I
sx ⫽ 0
r1 ⫽ 4.243 in.
t ⫽ r2 ⫺ r1
p(r22
TENSILE STRESS AND MAXIMUM SHEAR
STRESS
;
MAXIMUM SHEAR STRESS
M ⫽ 6.734 ⫻ 104 lb # in
T ⫽ ⫺1.247 ⫻ 10
4
lb # in
Vx ⫽ 207.846 lb
tmax ⫽
A
a
sx ⫺ sy
2
tmax ⫽ 2289 psi
2
b + t 2xy
;
Vy ⫽ ⫺120 lb
(b) ALLOWABLE LOAD P
STRESSES AT POINT A
sallow ⫽ 16000 psi
tallow ⫽ 6000 psi
The stresses at point A are proportional to the load P.
Based on tensile stress:
Pallow sallow
⫽
P
smax
Pallow ⫽
sallow P
smax
Pallow ⫽ 847.01lb
Based on shear stress:
Pallow tallow
⫽
tmax
P
Pallow ⫽ 629.07 lb
sx ⫽ 0
sy ⫽
t⫽
Mr2
I
Pallow ⫽
sy ⫽ 4.489 ⫻ 103 psi
Vy Q
Tr2
+
Ip
I 2t
t ⫽ ⫺449.628 psi
(The shear force Vx produces no stress at point A)
Pallow ⫽ 629 lb
;
tallow P
tmax
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SECTION 8.5
Problem 8.5-16 A sign is supported by a pipe (see figure) having outer
2.0 m
diameter 110 mm and inner diameter 90 mm. The dimensions of the
sign are 2.0 m ⫻ 1.0 m, and its lower edge is 3.0 m above the base.
Note that the center of gravity of the sign is 1.05 m from the axis of the
pipe. The wind pressure against the sign is 1.5 kPa.
Determine the maximum in-plane shear stresses due to the wind
pressure on the sign at points A, B, and C, located on the outer surface
at the base of the pipe.
Rose’s
Editing Co.
1.0 m
1.05 m
to c.g.
Pipe
110 mm
3.0 m
X
B
X
C
Section X-X
Solution 8.5-16
d2 ⫽ 110 mm
I⫽
d1 ⫽ 90 mm
p 4
(d ⫺ d 14)
64 2
Ip ⫽ 2I
t ⫽ 10 mm
MAXIMUM SHEAR STRESS AT POINT A
I ⫽ 3.966 ⫻ 106 mm4
Ip ⫽ 7.933 ⫻ 106 mm4
1
1d 3 ⫺ d 312
12 2
Q⫽
Q ⫽ 5.017 * 104 mm3
SIGN:
A ⫽ 2m2
h ⫽ a3 +
Area
1
bm
2
b ⫽ 1.05 m
WIND PRESSURE
horizontal distance
from the center of
gravity of the sign to
the axis of the pipe
pw ⫽ 1.5 kPa
P ⫽ pw A
STRESS RESULTANTS AT THE BASE
M ⫽ Ph
M ⫽ 10.5 kN # m
T ⫽ Pb
T ⫽ 3.15 kN # m
V⫽P
Height from the base to
the center of gravity of
the sign
V ⫽ 3 kN
P ⫽ 3 kN
sx ⫽ 0
sy ⫽
Md 2
2I
sy ⫽ 145.603 MPa
txy ⫽
Td 2
2Ip
txy ⫽ 21.84 MPa
tmax ⫽
A
a
sx ⫺ sy
2
tmax ⫽ 76.0 MPa
C
A
A
B
PIPE:
693
Combined Loadings
2
b + t 2xy
;
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Page 694
Applications of Plane Stress
MAXIMUM SHEAR STRESS AT POINT B
MAXIMUM SHEAR STRESS AT POINT C
sx ⫽ 0
sx ⫽ 0
sy ⫽ 0
sy ⫽ 0
txy ⫽
VQ
Td2
⫺
2 Ip
I(2t)
tmax ⫽
A
Pure shear
a
sx ⫺ sy
2
txy ⫽ 19.943 MPa
2
b + t2xy
tmax ⫽ 19.94 MPa
;
VQ
Td2
+
txy ⫽ 23.738 MPa
2 Ip
I(2t)
sx ⫺ sy 2
tmax ⫽ a
b + t 2xy
A
2
txy ⫽
Pure shear
tmax ⫽ 23.7 MPa
Problem 8.5-17 A sign is supported by a pole of hollow circular
8 ft
cross section, as shown in the figure. The outer and inner diameters of the pole are 10.5 in. and 8.5 in., respectively. The pole is
42 ft high and weighs 4.0 k. The sign has dimensions 8 ft ⫻ 3 ft
and weighs 500 lb. Note that its center of gravity is 53.25 in.
from the axis of the pole. The wind pressure against the sign is
35 lb/ft2.
(a) Determine the stresses acting on a stress element at point
A, which is on the outer surface of the pole at the “front”
of the pole, that is, the part of the pole nearest to the
viewer.
(b) Determine the maximum tensile, compressive, and shear
stresses at point A.
;
Hilda’s Office
3 ft
10.5 in.
8.5 in.
42 ft
X
X
A
A
Section X-X
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Page 695
SECTION 8.5
695
Combined Loadings
Solution 8.5-17
PIPE:
d2 ⫽ 10.5 in.
d1 ⫽ 8.5 in.
c⫽
d2
2
(a) STRESSES AT POINT A
p 2
1d ⫺ d 212
A ⫽ 29.845 in.2
4 2
p 4
I ⫽ 340.421 in.4
1d 2 ⫺ d 412
I⫽
64
Ip ⫽ 2I
Ip ⫽ 680.842 in.4
1 3
Q ⫽ 45.292 in.3
1d ⫺ d 31 2
Q⫽
12 2
A⫽
W1 ⫽ 4000 lb
SIGN:
sx ⫽ 0
As ⫽ (8)(3) ft
2
As ⫽ 24 ft
Area
Height from the base to the center of gravity
of the sign
sy ⫽ ⫺
sxy ⫽
3
h ⫽ a42 ⫺ b ft
2
Horizontal distance from the center of gravity
of the sign to the axis of the pipe
b ⫽ a(4)12 +
10.5
b in.
2
s1 ⫽
sy ⫽ 6145 psi
txy ⫽ 345 psi
sx + sy
+
2
A
sx ⫺ sy
a
2
sx + sy
2
⫺
A
a
sx ⫺ sy
2
P ⫽ pw As
s2 ⫽ ⫺19.30 psi
P ⫽ 840 lb
M ⫽ 4.082 ⫻ 105 lb # in.
T ⫽ Pb
T ⫽ 4.473 ⫻ 104 lb # in.
V ⫽ 840 lb
Nz ⫽ 4.5 ⫻ 103 lb
;
;
tmax ⫽
A
a
sx ⫺ sy
2
Max. shear stress
2
b + t 2xy
s1 ⫽ 6164 psi
Max. compressive stress
M ⫽ Ph
Nz ⫽ W1 ⫹ W2
Td2
2Ip
Md2
2I
pw ⫽ 35 lb/ft2
STRESS RESULTANTS AT THE BASE
V⫽P
+
A
Max. tensile stress
s2 ⫽
W2 ⫽ 500 lb
Nz
(b) MAXIMUM STRESSES AT POINT A
b ⫽ 53.25 in.
WIND PRESSURE
;
2
;
2
b + t 2xy
;
2
b + t2xy
tmax ⫽ 3092 psi
;
NOTE: Since the principal stresses have opposite signs,
the maximum in-plane shear is larger than the maximum
out-of-plane shear stress.
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Applications of Plane Stress
Problem 8.5-18 A horizontal bracket ABC consists of two
perpendicular arms AB of length 0.5 m, and BC of length of
0.75 m. The bracket has a solid circular cross section with
diameter equal to 65 mm. The bracket is inserted in a frictionless
sleeve at A (which is slightly larger in diameter) so is free to
rotate about the z0 axis at A, and is supported by a pin at C.
Moments are applied at point C as follows: M1 ⫽ 1.5 kN # m in
the x-direction and M2 ⫽ 1.0 kN # m acts in the (⫺z)
direction.
Considering only the moments M1 and M2, calculate the
maximum tensile stress st, the maximum compressive stress sc,
and the maximum in-plane shear stress tmax at point p, which is
located at support A on the side of the bracket at midheight.
y0
A
0.75 m
Frictionless sleeve
embedded in
support
p
x0
B
z0
y0
C
M2
0.5 m
p
M1
x0
O
65 mm
Cross section at A
Solution 8.5-18
d ⫽ 65 mm
T⫽0
b1 ⫽ 0.5 m
length of arm BC
b2 ⫽ 0.75 m
length of arm AB
Torsional frictionless sleeve at support A (MZ)
Vy ⫽ ⫺
M1 ⫽ 1.5 kN # m
sx ⫽ 0
M2 ⫽ 1.0 kN # m
sy ⫽ 0
PROPERTIES OF THE CROSS SECTION
txy ⫽
d ⫽ 65 mm
r⫽
d
2
M2
b1
Vy Q
tmax ⫽
txy ⫽ ⫺0.804 MPa
Id
A
a
sx ⫺ sy
2
2
b + t 2xy
A⫽
p 2
d
4
A ⫽ 3.318 ⫻ 103 mm2
Pure Shear
I⫽
p 4
d
64
I ⫽ 8.762 ⫻ 105 mm4
STRESSES AT POINT p ON THE SIDE OF THE BRACKET
Ip ⫽ 2 I
Q⫽
Ip ⫽ 1.752 ⫻ 106 mm4
2 3
r
3
Q ⫽ 2.289 ⫻ 104 mm3
STRESS RESULTANTS AT SUPPORT A
Nz ⫽ 0
Axial force
My ⫽ 0
Mx ⫽ ⫺ M2
b2
+ M1
b1
Mx ⫽ 0
tmax ⫽ 0.804 MPa
;
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Page 697
SECTION 8.5
Problem 8.5-19 A cylindrical pressure vessel with flat ends is
697
Combined Loadings
y0
subjected to a torque T and a bending moment M (see figure). The
outer radius is 12.0 in. and the wall thickness is 1.0 in. The loads are
as follows: T ⫽ 800 k-in., M ⫽ 1000 k-in., and the internal pressure
p ⫽ 900 psi.
Determine the maximum tensile stress st, maximum compressive
stress sc, and maximum shear stress tmax in the wall of the cylinder.
T
M
M
T
x0
z0
Solution 8.5-19
Cylindrical pressure vessal
Internal pressure:
Bending moment:
p ⫽ 900 psi
M ⫽ 1000 k-in.
Torque:
T ⫽ 800 k-in.
Outer radius:
r2 ⫽ 12 in.
Wall thickness:
t ⫽ 1.0 in.
Mean radius:
r ⫽ r2 ⫺ t/2 ⫽ 11.5 in.
Outer diameter:
d2 ⫽ 24 in.
Inner diameter:
d1 ⫽ 22 in.
MOMENT OF INERTIA
sx ⫽
⫽ 2668.2 psi
sy ⫽
pr
⫽ 10,350 psi
t
txy ⫽
Tr2
⫽ ⫺1002.7 psi
Ip
PRINCIPAL STRESSES
s1, 2 ⫽
p
I ⫽ (d 42 ⫺ d 41) ⫽ 4787.0 in.4
64
Ip ⫽ 2I ⫽ 9574.0 in.4
pr
Mr2
⫺
⫽ 5175.0 psi ⫺ 2506.8 psi
2t
I
sx + sy
;
2
A
a
sx ⫺ sy
2
2
b + t2xy
⫽ 6509.1 psi ; 3969.6 psi
s1 ⫽ 10,479 psi
s2 ⫽ 2540 psi
NOTE: Since the stresses due to T and p are the same
everywhere in the cylinder, the maximum stresses
occur at the top and bottom of the cylinder where
the bending stresses are the largest.
MAXIMUM SHEAR STRESSES
PART (a). TOP OF THE CYLINDER
t⫽
Stress element on the top of the cylinder as seen from
above.
⬖ tmax ⫽ 5240 psi
In-plane:
t ⫽ 3970 psi
Out-of-plane:
s1
2
or
s2
2
t⫽
s1
⫽ 5240 psi
2
MAXIMUM STRESSES FOR THE TOP OF THE CYLINDER
st ⫽ 10,480 psi
tmax ⫽ 5240 psi
sc ⫽ 0 (No compressive stresses)
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Applications of Plane Stress
PART (b). BOTTOM OF THE CYLINDER
MAXIMUM SHEAR STRESSES
Stress element on the bottom of the cylinder as seen
from below.
In-plane:
t ⫽ 1669 psi
Out-of-plane:
t⫽
s1
2
or
s2
2
t⫽
s1
⫽ 5340 psi
2
⬖ tmax ⫽ 5340 psi
MAXIMUM STRESSES FOR THE BOTTOM OF THE CYLINDER
st ⫽ 10,680 psi
sc ⫽ 0 (No compressive stresses)
tmax ⫽ 5340 psi
PART (c). ENTIRE CYLINDER
pr
Mr2
sx ⫽
+
⫽ 5175.0 psi + 2506.8 psi
2t
I
⫽ 7681.8 psi
st ⫽ 10,680 psi
;
sc ⫽ 0 (No compressive stresses)
pr
sy ⫽
⫽ 10,350 psi
t
txy ⫽ ⫺
The largest stresses are at the bottom of the cylinder.
tmax ⫽ 5340 psi
;
;
Tr2
⫽ ⫺1002.7 psi
Ip
PRINCIPAL STRESSES
s1, 2 ⫽
sx + sy
;
2
A
a
sx ⫺ sy
2
2
b + t2xy
⫽ 9015.9 psi ; 1668.9 psi
s1 ⫽ 10,685 psi
s2 ⫽ 7347 psi
y0
Problem 8.5-20 For purposes of analysis, a segment of the crankshaft in a
vehicle is represented as shown in the figure. Two loads P act as shown, one
parallel to (⫺x0) and another parallel to z0; each load P equals 1.0 kN. The
crankshaft dimensions are b1 ⫽ 80 mm, b2 ⫽ 120 mm, and b3 ⫽ 40 mm.
The diameter of the upper shaft is d ⫽ 20 mm.
(a) Determine the maximum tensile, compressive, and shear stresses at
point A, which is located on the surface of the upper shaft at the z0 axis.
(b) Determine the maximum tensile, compressive, and shear stresses at
point B, which is located on the surface of the shaft at the y0 axis.
b1 = 80 mm
B
A
x0
z0
d = 20 mm
b2 = 120 mm
P
b3 = 40 mm
P = 1.0 kN
08Ch08.qxd
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Page 699
SECTION 8.5
Combined Loadings
699
Solution 8.5-20
P ⫽ 1.0 kN
sx ⫽ ⫺155.972 MPa (compressive)
b1 ⫽ 80 mm
sy ⫽ 0
b2 ⫽ 120 mm
txy ⫽
b3 ⫽ 40 mm
s1 ⫽
PROPERTIES OF THE CROSS SECTION
d ⫽ 20 mm
d
r⫽
2
p
A ⫽ d2
4
A ⫽ 314.159 mm
I⫽
p 4
d
64
+
2
sx + sy
2
tmax ⫽
I ⫽ 7.854 ⫻ 103 mm4
4
2 3
r Q ⫽ 666.667 mm3
3
STRESS RESULTANTS AT THE SUPPORT
Vx ⫽ P
(Axial force in X-dir.)
Vy ⫽ 0
(Shear force in Y-dir.)
Vz ⫽ P
(Shear force in Z-dir.)
Mx ⫽ Pb2
A
a
⫺
A
A
sx ⫺ sy
2
a
sx ⫺ sy
a
sx ⫺ sy
2
2
2
b + t2xy
2
b + t2xy
2
b + t2xy
s1 ⫽ 31.2 MPa
MAX. TENSILE STRESS
Ip ⫽ 1.571 ⫻ 10 mm
4
txy ⫽ 76.394 MPa
sx ⫺ sy
2
Ip ⫽ 2I
Q⫽
s2 ⫽
M xd
2Ip
MAX. COMPRESSIVE STRESS
s2 ⫽ ⫺187.2 MPa
;
MAX. SHEAR STRESS
tmax ⫽ 109.2 MPa
(b) STRESSES AT POINT B
sx ⫽ ⫺
Mz
Vx
+
r
A
I
(Torsional Moment)
Mx ⫽ 120 kN # mm
My ⫽ P(b1 ⫹ b3)
(Bending Moment)
My ⫽ 120 kN # mm
Mz ⫽ Pb2 (Bending Moment)
Mz ⫽ 120 kN # mm
(a) STRESSES AT POINT A
sx ⫽ 149.606 MPa (tensile)
sy ⫽ 0
txy ⫽
s1 ⫽
s2 ⫽
sx ⫽ ⫺
My
Vx
⫺
r
A
I
;
Vz Q
Mxd
+
2Ip
Id
sx ⫺ sy
A
a
sx ⫺ sy
A
a
sx + sy
+
2
sx + sy
tmax ⫽
2
A
a
⫺
sx ⫺ sy
2
txy ⫽ 80.639 MPa
2
2
2
b + t2xy
2
b + t2xy
2
b + t2xy
;
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CHAPTER 8
Page 700
Applications of Plane Stress
s1 ⫽ 184.8 MPa
MAX. TENSILE STRESS
;
MAX. COMPRESSIVE STRESS
s2 ⫽ ⫺ 35.2 MPa
tmax ⫽ 110.0 MPa
MAX. SHEAR STRESS
;
NOTE: Since the principal stresses have opposite signs,
the maximum in-plane shear is larger than the
maximum out-of-plane shear stress.
;
Problem 8.5-21 A moveable steel stand supports an automobile engine weighing W ⫽ 750 lb as shown in figure part (a).
The stand is constructed of 2.5 in. ⫻ 2.5 in. ⫻ 1/8 in. thick steel tubing. Once in position the stand is restrained by pin
supports at B and C. Of interest are stresses at point A at the base of the vertical post; point A has coordinates (x ⫽ 1.25, y ⫽
0, z ⫽ 1.25) (inches). Neglect the weight of the stand.
(a) Initially, the engine weight acts in the (⫺z) direction through point Q which has coordinates (24, 0, 1.25); find the
maximum tensile, compressive, and shear stresses at point A.
(b) Repeat (a) assuming now that, during repair, the engine is rotated about its own longitudinal axis (which is parallel to
the x axis) so that W acts through Q⬘ (with coordinates (24, 6, 1.25)) and force Fy ⫽ 200 lb is applied parallel to the y
axis at distance d ⫽ 30 in.
17 in.
z
17 in.
1.25 in.
O
B
W
C
y
A
24 in.
d=
.
0 in
Q
3
Q'
12 in.
6 in.
D
x
Fy
(b) Top view
2.5 in. ⫻ 2.5 in. ⫻ 1/8 in.
A
B
Q
C
Cx
y
24 in.
D
x
17 in.
36 in.
Dz
(a)
Cz
Cy
08Ch08.qxd
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Page 701
SECTION 8.5
701
Combined Loadings
Solution 8.5-21
W ⫽ 750 1b
Fy ⫽ 200 1b
A ⫽ b2 ⫺ (b ⫺ 2t)2
Am ⫽ 6 in.
(a)
A ⫽ 1in.2
1
[ b 4 ⫺ ( b ⫺ 2 t)4]
12
SHEAR
s⫽
I ⫽ 1 in.4
tt ⫽
sx ⫽ 0
sy ⫽ s
+
2
sx + sy
s2 ⫽
2
tmax ⫽
s1 ⫽ 0
A
a
⫺
txy ⫽ t
sx ⫺ sy
A
a
sx ⫺ sy
A
a
sx ⫺ sy
2
t⫽0
& MAX SHEAR STRESS
sx + sy
s1 ⫽
1
1 b 3 ⫺ b 312
8
Fy Q
SHEAR DUE TO TORSIONAL MOMENT
2
2
t ⫽ tt ⫹ tT
t ⫽ 4633 psi
PRINCIPAL STRESSES
s1 ⫽
2
s2 ⫽
+
2
sx + sy
2
A
a
⫺
sx ⫺ sy
A
sx ⫺ sy
A
a
sx ⫺ sy
2
s1 ⫽ 988 psi
txy ⫽ t
a
sx + sy
tmax ⫽
;
& MAX SHEAR STRESS
sy ⫽ s
2
b + t2xy
2
2
2
b + t2xy
;
s2 ⫽ ⫺20730 psi
;
s2 ⫽ ⫺21719 psi
;
tmax ⫽ 10365 psi
;
tmax ⫽ 11354 psi
;
ENGINE WEIGHT ACTS THROUGH
POINT Q ¿ & FORCE FY
ACTS IN Y-DIR
My ⫽ 18000 in.-1b
SHEAR
s⫽
Mx ⫽ Wy1
& NORMAL STRESSES AT A
My c
⫺W
⫺
A
I
s ⫽ ⫺20730 psi
same since element A lies on NA for bending about
x-axis
tT ⫽
tT ⫽ 4255 psi
2
b + t2xy
MZ USING APPROX.
3-65)
Mz ⫽ 6000 in.-1b
sx ⫽ 0
b + t2xy
Q ⫽ 1 in.3
tt ⫽ 378 psi
I (2 t)
Mz ⫽ Fy d
s ⫽ ⫺20730 psi
TRANSVERSE SHEAR
b1 ⫽ 2 in.
THEORY (EQU.
& NORMAL STRESSES AT A
Myc
⫺W
⫺
A
I
b1 ⫽ b ⫺ 2t
Q⫽
(POINT Q)
MZ (MAX
PROB. #5.10-11)
TORSION DUE TO
IN WEB-SEE
b
c⫽ .
2
My ⫽ Wx1
PRINCIPAL STRESSES
(b)
Am ⫽ (b ⫺ t)2
ENGINE WEIGHT ACTS THROUGH X-AXIS
Mx ⫽ 0
x1 ⫽ 24 in.
d ⫽ 30 in.
y1 ⫽ 6 in.
2
I⫽
FY &
1
in.
8
t⫽
b ⫽ 2.5 in.
SHEAR STRESS DEPENDS ON TRANSVERSE SHEAR DUE TO
2
b + t2xy
2
b + t2xy
Mz
2 tAm
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CHAPTER 8
Page 702
Applications of Plane Stress
Problem 8.5-22 A mountain bike rider going uphill applies force P ⫽ 65 N to each end of the handlebars ABCD, made of
aluminum alloy 7075-T6, by pulling on the handlebar extenders (DF on right handlebar segment). Consider the right half
of the handlebar assembly only (assume the bars are fixed at the fork at A). Segments AB and CD are prismatic with lengths
L1 and L3 and with outer diameters and thicknesses d01, t01 and d03, t03, respectively, as shown. Segment BC of length L2,
however, is tapered and, outer diameter and thickness vary linearly between dimensions at B and C. Consider shear, torsion,
and bending effects only for segment AD; assume DF is rigid.
Find maximum tensile, compressive, and shear stresses adjacent to support A. Show where each maximum stress value
occurs.
y
Handlebar extension
d01 = 32 mm
F
t01 = 3.15 mm
d03 = 22 mm
t03 = 2.95 mm
B
A
x
C
F Handlebar
extension
d = 100 mm
z
D
d03
L3 = 220 mm
P
45°
D
L1 = 50 mm L2 = 30 mm
y
Handlebar
(a)
(b) Section D–F
Solution 8.5-22
z
P ⫽ 65 N
L1 ⫽ 50 mm
45°
L2 ⫽ 30 mm
A2
45
A1
L3 ⫽ 220 mm
°
d01 ⫽ 32 mm
ax
y
Vm
d03 ⫽ 22 mm
d ⫽ 100 mm
T
M
A3
PROPERTIES OF THE CROSS SECTION AT POINT A
r⫽
d01
2
t01 ⫽ 3.15 mm
STRESS RESULTANTS AT POINT A1
p
A ⫽ [d 201 ⫺ 1d01 ⫺ t 0122]
4
A ⫽ 150.543 mm2
p 4
[d ⫺ 1d01 ⫺ t 0124]
I⫽
64 01
I ⫽ 1.747 ⫻ 10 mm
Ip ⫽ 2I
Ip ⫽ 3.493 ⫻ 104 mm4
1
Q⫽
[ d 3 ⫺ 1 d01 ⫺ t 0123]
12 01
x
Section at point A
Nx ⫽ 0
Vmax. ⫽ P
4
4
Q ⫽ 729.625 mm
3
(Axial force in X-dir.)
(Max. Shear force)
Vmax. ⫽ 0.065 kN
T ⫽ Pd
(Torsional Moment)
M ⫽ P(L1 ⫹ L2 ⫹ L3)
M ⫽ 19.5 kN # mm
T ⫽ 6.5 kN # mm
(Bending Moment)
08Ch08.qxd
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Page 703
SECTION 8.5
703
Combined Loadings
sx ⫽ 0
sy ⫽ 0
txy ⫽
s1 ⫽
s2 ⫽
sx ⫽ ⫺
M
r
I
2
A
sx + sy
2
A
a
⫺
sx ⫺ sy
A
a
sx ⫺
2
2
2
b + t2xy
2
b + t2xy
sy 2
b + t2xy
2
s1 ⫽ 3.41 MPa
;
MAX. COMPRESSIVE STRESS
T d01
txy ⫽
2Ip
+
2
sx + sy
2
A
a
s2 ⫽ ⫺3.41 MPa
txy ⫽ 2.977 MPa
A
a
sx ⫺ sy
sx ⫺ sy
A
a
sx + sy
tmax ⫽
sx ⫺ sy
MAX. TENSILE STRESS
sy ⫽ 0
s2 ⫽
+
txy ⫽ 3.408 MPa
a
sx + sy
tmax ⫽
sx ⫽ ⫺17.863 MPa (compressive stress)
s1 ⫽
Vmax. Q
T d01
+
2Ip
I 2 t 01
⫺
sx ⫺ sy
2
2
2
;
tmax ⫽ 3.41 MPa
MAX. SHEAR STRESS
2
b +
t2xy
NOTE: Since the principal stresses have opposite signs,
the maximum in-plane shear is larger than the
maximum out-of-plane shear stress.
2
b + t2xy
STRESS RESULTANTS AT POINT A3
2
b + t2xy
MAX. TENSILE STRESS
s1 ⫽ 0.483 MPa
;
MAX. COMPRESSIVE STRESS
s2 ⫽ ⫺18.35 MPa
;
MAX. SHEAR STRESS
tmax ⫽ 9.42 MPa
;
NOTE: Since the principal stresses have opposite signs,
the maximum in-plane shear is larger than the
maximum out-of-plane shear stress.
STRESS RESULTANTS AT POINT A2
;
sx ⫽
M
r
I
sx ⫽ 17.863 MPa (tensile stress)
sy ⫽ 0
txy ⫽
s1 ⫽
s2 ⫽
Td01
2 Ip
txy ⫽ 2.977 MPa
sx ⫺ sy
A
a
sx ⫺ sy
A
a
sx + sy
+
2
sx + sy
2
⫺
2
2
2
b + t2xy
2
b + t2xy
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CHAPTER 8
tmax ⫽
A
a
Page 704
Applications of Plane Stress
sx ⫺ sy
2
MAX. COMPRESSIVE STRESS
2
b + t2xy
MAX. TENSILE STRESS
s1 ⫽ 18.35 MPa
;
s2 ⫽ ⫺0.483 MPa
;
MAX. SHEAR STRESS
tmax ⫽ 9.42 MPa
;
NOTE: Since the principal stresses have opposite signs,
the maximum in-plane shear is larger than the
maximum out-of-plane shear stress.
Problem 8.5-23 Determine the maximum tensile, compressive, and shear stresses acting on the cross section of the tube at
point A of the hitch bicycle rack shown in the figure.
The rack is made up of 2 in. ⫻ 2 in. steel tubing which is 1/8 in. thick. Assume that the weight of each of four bicycles
is distributed evenly between the two support arms so that the rack can be represented as a cantilever beam (ABCDEF) in
the x-y plane. The overall weight of the rack alone is W ⫽ 60 lb. directed through C, and the weight of each bicycle is B ⫽
30 lb.
B
—
2
y
6 in.
Bike loads B
3 @ 4 in.
F
E
4 loads, each B
E
F
33 in.
B
— at each tie down point
2
W
MAZ
Ay
C
C
A
Fixed
support B
D
1
2 in. ⫻ 2 in. ⫻ — in. steel tube
8
x
Ax
B
17 in.
1
— in.
8
2 in.
2 in.
A
D
7 in.
2 in.
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SECTION 8.5
Combined Loadings
705
Solution 8.5-23
TOP OF THE CROSS SECTION (AT POINT A)
t ⫽ 0.125 in.
Cross section
d2 ⫽ 2 in.
Outer width
d1 ⫽ d2 ⫺ 2t
Inner width
A ⫽ d22 ⫺ d12
A ⫽ 0.938 in.2
I⫽
Thickness
d 42
d 41
⫺
12
12
I ⫽ 0.552 in.4
Q ⫽ (d2 ⫺ 2t) t a
d2
d2 d2
t
⫺ b + 2
t
2
2
2
4
Q ⫽ 0.33 in.3
The distance between point A and the center of
load W
sx ⫽
M Az d2
2I
sx ⫽ 8.591 ⫻ 103 psi (tensile stress)
sy ⫽ 0
txy ⫽ 0
s1 ⫽
s2 ⫽
sx ⫺ sy
A
a
sx ⫺ sy
A
a
sx + sy
+
2
sx + sy
tmax ⫽
2
A
a
⫺
sx ⫺
2
2
2
b + t2xy
2
b + t2xy
sy 2
2
b + t2xy
b1 ⫽ 17 in.
The distance between the point A and the center of a
bike load B
MAX. TENSILE STRESS
3.4
b2 ⫽ a17 + 2 + 6 +
b in.
2
MAX. COMPRESSIVE STRESS
W ⫽ 60 lb
tmax ⫽ 4295 psi
B ⫽ 30 lb
REACTIONS AT SUPPORT A
MAz ⫽ Wb1 ⫹ 4Bb2
Ay ⫽ W ⫹ 4B
Ax ⫽ 0
MAz ⫽ 4.74 ⫻ 103 lb # in.
Ay ⫽ 180 lb
s1 ⫽ 8591 psi
;
s2 ⫽ 0
;
MAX. SHEAR STRESS
;
NOTE: Since the principal stresses have opposite signs,
the maximum in-plane shear is larger than the
maximum out-of-plane shear stress.
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CHAPTER 8
Page 706
Applications of Plane Stress
SIDE OF THE CROSS SECTION (AT POINT A)
BOTTOM OF THE CROSS SECTION (AT POINT A)
sx ⫽ 0
sx ⫽ ⫺
sy ⫽ 0
txy ⫽
s1 ⫽
s2 ⫽
sx ⫽ ⫺8.591 ⫻ 103 psi (compressive stress)
Ay Q
txy ⫽ 430.726 psi
I2 t
sx ⫺ sy
A
a
sx ⫺ sy
A
a
sx + sy
+
2
sx + sy
tmax ⫽
2
A
a
M Az d2
2I
⫺
sx ⫺ sy
2
2
2
sy ⫽ 0
txy ⫽ 0
2
b + txy2
s1 ⫽
+
2
2
b + txy2
s2 ⫽
sx + sy
2
b + txy2
s1 ⫽ 431 psi
MAX. TENSILE STRESS
sx ⫺ sy
A
a
sx ⫺ sy
A
a
sx + sy
tmax ⫽
2
A
a
⫺
sx ⫺
2
2
2
b + txy2
;
s1 ⫽ 0
MAX. TENSILE STRESS
s2 ⫽ ⫺431 psi
MAX. COMPRESSIVE STRESS
;
tmax ⫽ 431 psi
2
b + txy2
sy 2
MAX. COMPRESSIVE STRESS
MAX. SHEAR STRESS
2
b + txy2
;
NOTE: Since the principal stresses have opposite signs,
the maximum in-plane shear is larger than the
maximum out-of-plane shear stress.
s2 ⫽ ⫺8591 psi
;
;
MAX. SHEAR STRESS
tmax ⫽ 4295 psi
;
NOTE: Since the principal stresses have opposite signs,
the maximum in-plane shear is larger than the
maximum out-of-plane shear stress.
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Page 707
9
Deflections of Beams
Differential Equations of the Deflection Curve
y
The beams described in the problems for Section 9.2 have constant flexural
rigidity EI.
Problem 9.2-1 The deflection curve for a simple beam AB (see figure) is
B
A
given by the following equation:
␯⫽ ⫺
q0 x
(7L4 ⫺ 10L2x2 + 3x4)
360LEI
x
L
Describe the load acting on the beam.
Probs. 9.2-1 and 9.2-2
Solution 9.2-1
Simple beam
q0 x
␯⫽ ⫺
(7L4 ⫺ 10L2x2 + 3x4)
360LEI
Take four consecutive derivatives and obtain:
␯–– ⫽ ⫺
q0 x
LEI
From Eq. (9-12c): q ⫽ ⫺ EI␯–– ⫽
q0x
L
;
The load is a downward triangular load of maximum
;
intensity q0.
707
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CHAPTER 9
Page 708
Deflections of Beams
Problem 9.2-2 The deflection curve for a simple beam AB (see figure) is given by the following equation:
␯⫽ ⫺
q0 L4
4
p EI
sin
px
L
(a) Describe the load acting on the beam.
(b) Determine the reactions RA and RB at the supports.
(c) Determine the maximum bending moment Mmax.
Solution 9.2-2
Simple beam
4
␯⫽ ⫺
␯¿ ⫽ ⫺
q0 L
4
p EI
q0 L3
3
p EI
sin
(b) REACTIONS (EQ. 9-12b)
px
L
cos
V ⫽ EI␯–¿ ⫽
px
L
q0 L
px
cos
p
L
At x ⫽ 0: V ⫽ RA ⫽
q0 L2
px
␯– ⫽ 2 sin
L
p EI
␯–¿ ⫽
At x ⫽ L: V ⫽ ⫺RB ⫽ ⫺
q0 L
px
cos
pEI
L
␯–– ⫽ ⫺
q0 L
p
RB ⫽
q0
px
sin
EI
L
;
q0 L
;
p
q0 L
p
;
(c) MAXIMUM BENDING MOMENT (EQ. 9-12a)
M ⫽ EI␯– ⫽
(a) LOAD (EQ. 9-12c)
px
q ⫽ ⫺EI␯–– ⫽ q0 sin
;
L
The load has the shape of a sine curve, acts downward,
;
and has maximum intensity q0.
q0 L2
2
p
sin
px
L
L
For maximum moment, x ⫽ ;
2
Problem 9.2-3 The deflection curve for a cantilever beam AB (see figure) is
Mmax ⫽
q0 L2
p2
;
y
given by the following equation:
␯⫽ ⫺
q0 x2
(10L3 ⫺ 10L2x + 5Lx2 ⫺ x3)
120LEI
A
B
x
Describe the load acting on the beam.
Probs. 9.2-3 and 9.2.-4
L
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SECTION 9.2 Differential Equations of the Deflection Curve
709
Solution 9.2-3 Cantilever beam
␯⫽ ⫺
q0x2
(10L3 ⫺ 10L2x + 5Lx2 ⫺ x3)
120LEI
Take four consecutive derivatives and obtain:
␯–– ⫽ ⫺
q0
(L ⫺ x)
LEI
From Eq. (9-12c):
q ⫽ ⫺ EI␯–– ⫽ q0 a1 ⫺
x
b
L
;
The load is a downward triangular load of maximum
;
intensity q0.
Problem 9.2-4 The deflection curve for a cantilever beam AB (see figure) is given by the following equation:
␯⫽ ⫺
q0 x2
360L2EI
(45L4 ⫺ 40L3x + 15L2x2 ⫺ x4)
(a) Describe the load acting on the beam.
(b) Determine the reactions RA and MA at the support.
Solution 9.2-4 Cantilever beam
␯– ⫽
␯– ⫽
␯–¿ ⫽
␯–– ⫽
q0x2
(45L4 ⫺ 40L3x + 15L2x2 ⫺ x4)
360L2EI
q0
(15L4x ⫺ 20L3x2 + 10L2x2 ⫺ x5)
⫺
60L2EI
q0
(3L4 ⫺ 8 L3x + 6 L2x2 ⫺ x4)
⫺
12L2EI
q0
⫺ 2 (⫺ 2L3 + 3L2x ⫺ x3)
3L EI
q0
⫺ 2 (L2 ⫺ x2)
L EI
␯⫽ ⫺
(a) LOAD (EQ. 9-12c)
q ⫽ ⫺ EI␯–– ⫽ q0 a1 ⫺
(b) REACTIONS RA AND MA (EQ. 9-12b AND EQ. 9-12a)
V ⫽ EI␯–¿ ⫽ ⫺
x
L2
b
3L2
(⫺2L3 + 3L2x ⫺ x3)
At x ⫽ 0: V ⫽ RA ⫽
M ⫽ EI␯– ⫽ ⫺
2
q0
q0
12L2
2q0 L
3
(3L4 ⫺ 8L3x + 6L2x2 ⫺ x4)
;
The load is a downward parabolic load of maximum
intensity q0.
;
;
At x ⫽ 0: M ⫽ MA ⫽ ⫺
q0 L2
4
;
NOTE: Reaction RA is positive upward.
Reaction MA is positive clockwise (minus means MA is
counterclockwise).
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CHAPTER 9
Page 710
Deflections of Beams
Deflection Formulas
q
Problems 9.3-1 through 9.3-7 require the calculation of deflections using
the formulas derived in Examples 9-1, 9-2, and 9-3. All beams have
constant flexural rigidity EI.
h
Problem 9.3-1 A wide-flange beam (W 12 ⫻ 35) supports a uniform
load on a simple span of length L ⫽ 14 ft (see figure).
Calculate the maximum deflection dmax at the midpoint and the
angles of rotation u at the supports if q ⫽ 1.8 k/ft and E ⫽ 30 ⫻ 106 psi.
Use the formulas of Example 9-1.
Solution 9.3-1
Simple beam (uniform load)
W 12 ⫻ 35
L ⫽ 14 ft ⫽ 168 in.
q ⫽ 1.8 k/ft ⫽ 150 lb/in.
I ⫽ 285 in.
L
Probs. 9.3-1 through 9.3-3
ANGLE OF ROTATION AT THE SUPPORTS
(EQs. 9-19 AND 9-20)
E ⫽ 30 ⫻ 10 psi
6
4
u ⫽ uA ⫽ uB ⫽
MAXIMUM DEFLECTION (EQ. 9-18)
⫽
5 qL4
5(150 lb/in.)(168 in.)4
⫽
384 EI
384(30 * 106 psi)(285 in.4)
⫽ 0.182 in.
;
dmax ⫽
qL3
24 EI
(150 lb/in.)(168 in.)3
24(30 * 106 psi)(285 in.4)
⫽ 0.003466 rad ⫽ 0.199°
;
Problem 9.3-2 A uniformly loaded steel wide-flange beam with simple supports (see figure) has a downward deflection of
10 mm at the midpoint and angles of rotation equal to 0.01 radians at the ends.
Calculate the height h of the beam if the maximum bending stress is 90 MPa and the modulus of elasticity is 200 GPa.
(Hint: Use the formulas of Example 9-1.)
Solution 9.3-2
Simple beam (uniform load)
d ⫽ dmax ⫽ 10 mm
u ⫽ uA ⫽ uB ⫽ 0.01 rad
s ⫽ smax ⫽ 90 MPa
Maximum bending moment:
E ⫽ 200 GPa
M⫽
Calculate the height h of the beam.
5qL4
384EId
or q ⫽
384EI
5L4
qL3
24EIu
or q ⫽
Eq. (9-19): u ⫽ uA ⫽
24EI
L3
Eq. (9-18): d ⫽ dmax ⫽
16d
Equate (1) and (2) and solve for L: L ⫽
5u
Flexure formula: s ⫽
Mc
Mh
⫽
I
2I
(1)
qL2
8
‹ s⫽
qL2h
16 I
Solve Eq. (4) for h: h ⫽
(4)
16Is
(5)
qL2
Substitute for q from (2) and for L from (3):
(2)
h⫽
(3)
32sd
15Eu2
;
Substitute numerical values:
h⫽
32(90 MPa)(10 mm)
15(200 GPa)(0.01 rad)2
⫽ 96 mm
;
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Page 711
SECTION 9.3
Deflection Formulas
711
Problem 9.3-3 What is the span length L of a uniformly loaded simple beam of wide-flange cross section (see figure) if the
maximum bending stress is 12,000 psi, the maximum deflection is 0.1 in., the height of the beam is 12 in., and the modulus
of elasticity is 30 ⫻ 106 psi? (Use the formulas of Example 9-1.)
Solution 9.3-3
Simple beam (uniform load)
s ⫽ smax ⫽ 12,000 psi
h ⫽ 12 in.
d ⫽ dmax ⫽ 0.1 in.
Solve Eq. (2) for q: q ⫽
E ⫽ 30 ⫻ 106 psi
Flexure formula: s ⫽
5qL4
384EId
or q ⫽
384EI
5L4
(1)
qL2
8
‹ s⫽
L2 ⫽
24 Ehd
5s
L⫽
24 Ehd
;
A 5s
Substitute numerical values:
Mh
Mc
⫽
I
2I
L2 ⫽
Maximum bending moment:
M⫽
(3)
L2h
Equate (1) and (2) and solve for L:
Calculate the span length L.
Eq. (9-18): d ⫽ dmax ⫽
16Is
qL2h
16I
(2)
24(30 * 106 psi)(12 in.)(0.1 in.)
⫽ 14,400 in.2
5(12,000 psi)
L ⫽ 120 in. ⫽ 10 ft
Problem 9.3-4 Calculate the maximum deflection dmax of a uniformly loaded
simple beam (see figure) if the span length L ⫽ 2.0 m, the intensity of the
uniform load q ⫽ 2.0 kN/m, and the maximum bending stress s ⫽ 60 MPa.
The cross section of the beam is square, and the material is aluminum
having modulus of elasticity E ⫽ 70 GPa. (Use the formulas of Example 9-1.)
;
q = 2.0 kN/m
L = 2.0 m
Solution 9.3-4
L ⫽ 2.0 m
Simple beam (uniform load)
q ⫽ 2.0 kN/m
s ⫽ smax ⫽ 60 MPa
Solve for b3: b3 ⫽
E ⫽ 70 GPa
Substitute b into Eq. (2): dmax ⫽
CROSS SECTION (square; b ⫽ width)
I⫽
b4
12
S⫽
b3
6
5qL4
32 Eb4
qL2
qL2
M
:s⫽
⫽
Flexure formula with M ⫽
8
S
8S
5Ls 4Ls 1/3
a
b
24E 3q
Substitute numerical values:
(1)
(2)
3qL
4b3
5(2.0 m)(60 MPa)
5Ls
1
1
⫽
⫽
m⫽
mm
24E
24(70 GPa)
2800
2.8
a
4(2.0 m)(60 MPa) 1/3
4 Ls 1/3
b ⫽ c
d ⫽ 10(80)1/3
3q
3(2000 N/m)
dmax ⫽
2
Substitute for S: s ⫽
(4)
(The term in parentheses is nondimensional.)
5qL4
Maximum deflection (Eq. 9-18): d ⫽
384 EI
Substitute for I: d ⫽
3qL2
4s
(3)
10(80)1/3
mm ⫽ 15.4 mm
2.8
;
;
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CHAPTER 9
Page 712
Deflections of Beams
q
Problem 9.3-5 A cantilever beam with a uniform load (see figure) has a
height h equal to 1/8 of the length L. The beam is a steel wideflange
section with E ⫽ 28 ⫻ 106 psi and an allowable bending stress of
17,500 psi in both tension and compression.
Calculate the ratio d/L of the deflection at the free end to the length,
assuming that the beam carries the maximum allowable load. (Use the
formulas of Example 9-2.)
Solution 9.3-5
1
h
⫽
L
8
L
Cantilever beam (uniform load)
E ⫽ 28 * 106 psi
Solve for q:
s ⫽ 17,500 psi
q⫽
Calculate the ratio d/L.
Maximum deflection (Eq. 9-26): dmax ⫽
qL4
8EI
qL3
d
‹ ⫽
L
8EI
(1)
(2)
2
4Is
Substitute q from (3) into (2):
s L
d
⫽
a b
L
2E h
2
Problem 9.3-6 A gold-alloy microbeam attached to a silicon
L ⫽ 27.5 ␮m
b ⫽ 4.0 ␮m
qL ⫽ 17.2 ␮N
t ⫽ 0.88 ␮m
dmax ⫽ 2.46 ␮m
b
L
Substitute numerical values:
Eg ⫽
3(17.2 mN)(27.5 mm)3
2(4.0 mm)(0.88 mm)3(2.46 mm)
⫽ 80.02 * 109 N/m2 or
Determine Eg.
4
bt3
12
t
Gold-alloy microbeam
Cantilever beam with a uniform load.
I⫽
;
q
wafer behaves like a cantilever beam subjected to a uniform load
(see figure). The beam has length L ⫽ 27.5 ␮m and rectangular
cross section of width b ⫽ 4.0 ␮m and thickness t ⫽ 0.88 ␮m.
The total load on the beam is 17.2 ␮N.
If the deflection at the end of the beam is 2.46 ␮m, what is
the modulus of elasticity Eg of the gold alloy? (Use the formulas
of Example 9-2.)
Eq. (9-26): d ⫽
;
17,500 psi
1
d
⫽
(8) ⫽
6
L
400
2(28 * 10 psi)
qL
qL h
Mc
h
⫽ a
ba b ⫽
I
2
2I
4I
Solution 9.3-6
(3)
L2h
Substitute numerical values:
qL2
:
Flexure formula with M ⫽
2
s⫽
h
4
qL
qL
or Eg ⫽
8EgI
8Idmax
Eg ⫽
3qL4
2bt3dmax
;
Eg ⫽ 80.0 GPa
;
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Page 713
SECTION 9.3
Deflection Formulas
Problem 9.3-7 Obtain a formula for the ratio dC /dmax of the deflection at the
midpoint to the maximum deflection for a simple beam supporting a concentrated load P (see figure).
From the formula, plot a graph of dC/dmax versus the ratio a/L that
defines the position of the load (0.5 ⬍ a/L ⬍ 1). What conclusion do you
draw from the graph? (Use the formulas of Example 9-3.)
713
P
A
B
a
b
L
Solution 9.3-7
Simple beam (concentrated load)
Eq. (9-35): dC ⫽
Pb(3L2 ⫺ 4b2)
48EI
Eq. (9-34): dmax ⫽
(a Ú b)
Pb(L2 ⫺ b2)3/2
913LEI
(3 13L)(3L2 ⫺ 4b2)
dc
⫽
dmax
16(L2 ⫺ b2)3/2
GRAPH OF dc/dmax VERSUS b ⫽ a/L
Because a ⱖ b, the ratio b versus from 0.5 to 1.0.
(a Ú b)
dc
b
dmax
0.5
0.6
0.7
0.8
0.9
1.0
1.0
0.996
0.988
0.981
0.976
0.974
(a Ú b)
Replace the distance b by the distance a by substituting
L ⫺ a for b:
(3 13L)(⫺ L2 + 8aL ⫺ 4a2)
dc
⫽
dmax
16(2aL ⫺ a2)3/2
Divide numerator and denominator by L2:
dc
⫽
dmax
dc
⫽
dmax
(3 13L)a⫺1 + 8
16L a2
a2 3/2
a
⫺ 2b
L
L
(3 13L)a⫺ 1 + 8
16a 2
a2
a
⫺ 4 2b
L
L
a2
a
⫺ 4 2b
L
L
a2 3/2
a
⫺ 2b
L
L
;
NOTE: The deflection dc at the midpoint of the beam is
almost as large as the maximum deflection dmax. The
greatest difference is only 2.6% and occurs when the
load reaches the end of the beam (b ⫽ 1).
ALTERNATIVE FORM OF THE RATIO
Let b ⫽
a
L
(3 13)(⫺ 1 + 8b ⫺ 4b 2)
dc
⫽
dmax
16(2b ⫺ b 2)3/2
;
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CHAPTER 9
Page 714
Deflections of Beams
Deflections by Integration of the Bending-Moment Equation
Problems 9.3-8 through 9.3-16 are to be solved by integrating the second-order
differential equation of the deflection curve (the bending-moment equation).
The origin of coordinates is at the left-hand end of each beam, and all beams
have constant flexural rigidity EI.
y
P
A
Problem 9.3-8 Derive the equation of the deflection curve for a cantilever
B
beam AB supporting a load P at the free end (see figure). Also, determine
the deflection dB and angle of rotation uB at the free end. (Note: Use the
second-order differential equation of the deflection curve.)
Solution 9.3-8
L
Cantilever beam (concentrated load)
BENDING-MOMENT EQUATION (EQ. 9-12a)
EI␯– ⫽ M ⫽ ⫺P(L ⫺ x)
␯¿ ⫽ ⫺
Px
(2L ⫺ x)
2EI
EI␯¿ ⫽ ⫺PLx +
Px2
+ C1
2
dB ⫽ ⫺␯(L) ⫽
n⬘(0) ⫽ 0
⬖ C1 ⫽ 0
uB ⫽ ⫺␯¿(L) ⫽
B.C.
PLx2
Px3
EI␯ ⫽ ⫺
+
+ C2
2
6
B.C. n (0) ⫽ 0
⬖ C2 ⫽ 0
␯⫽ ⫺
Px2
(3L ⫺ x)
6EI
PL3
3EI
;
PL2
2EI
;
(These results agree with Case 4, Table G-1.)
;
y
Problem 9.3-9 Derive the equation of the deflection curve
for a simple beam AB loaded by a couple M0 at the left-hand
support (see figure). Also, determine the maximum deflection
dmax. (Note: Use the second-order differential equation of the
deflection curve.)
M0
B
A
L
Solution 9.3-9
Simple beam (couple M0)
BENDING-MOMENT EQUATION (EQ. 9-12a)
x
EI␯– ⫽ M ⫽ M0 a1 ⫺ b
L
EI␯¿ ⫽ M0 ax ⫺
EI␯ ⫽ M0 a
x2
b + C1
2L
x2
x3
⫺
b + C1x + C2
2
6L
B.C.
n(0) ⫽ 0
⬖ C2 ⫽ 0
B.C.
n(L) ⫽ 0
‹ C1 ⫽ ⫺
␯⫽ ⫺
M0 x
(2L2 ⫺ 3Lx + x2)
6LEI
M0 L
3
;
x
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SECTION 9.3
MAXIMUM DEFLECTION
␯¿ ⫽ ⫺
dmax ⫽ ⫺(␯)x⫽x1
⫽
Set ␯¿ ⫽ 0 and solve for x:
13
b
3
715
Substitute x1 into the equation for n:
M0
(2L2 ⫺ 6Lx + 3x2)
6LEI
x1 ⫽ La 1 ⫺
Deflections by Integration of the Bending-Moment Equation
M0 L2
;
913EI
(These results agree with Case 7, Table G-2.)
;
Problem 9.3-10 A cantilever beam AB supporting a triangularly distributed
y
load of maximum intensity q0 is shown in the figure.
Derive the equation of the deflection curve and then obtain formulas
for the deflection dB and angle of rotation uB at the free end. (Note: Use the
second-order differential equation of the deflection curve.)
q0
x
B
A
L
Solution 9.3-10
Cantilever beam (triangular load)
BENDING-MOMENT EQUATION (EQ. 9-12a)
EI␯– ⫽ M ⫽ ⫺
q0
(L ⫺ x)3
6L
q0
EI␯¿ ⫽
(L⫺ x)4 + C1
24L
3
B.C.
n⬘(0) ⫽ 0
‹ C1 ⫽ ⫺
q0 L
24
3
EI␯ ⫽ ⫺
B.C.
q0
q0 L x
(L ⫺ x)5 ⫺
+ C2
120L
24
n(0) ⫽ 0
␯⫽ ⫺
q0 x2
(10L3 ⫺ 10L2x + 5Lx2 ⫺ x3)
120LEI
␯¿ ⫽ ⫺
q0 x
(4L3 ⫺ 6L2x + 4Lx2 ⫺ x3)
24LEI
dB ⫽ ⫺␯(L) ⫽
uB ⫽ ⫺␯¿(L) ⫽
q0 L4
30EI
q0 L3
24EI
;
;
;
(These results agree with Case 8, Table G-1.)
q0 L4
‹ C2 ⫽
120
Problem 9.3-11 A cantilever beam AB is acted upon by a uniformly distributed
moment (bending moment, not torque) of intensity m per unit distance along the
axis of the beam (see figure).
Derive the equation of the deflection curve and then obtain formulas for the
deflection dB and angle of rotation uB at the free end. (Note: Use the second-order
differential equation of the deflection curve.)
y
m
B
A
L
x
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CHAPTER 9
Page 716
Deflections of Beams
Solution 9.3-11
Cantilever beam (distributed moment)
BENDING-MOMENT EQUATION (EQ. 9-12a)
EI␯– ⫽ M ⫽ ⫺m(L ⫺ x)
EI␯¿ ⫽ ⫺m aLx ⫺
B.C.
n⬘(0) ⫽ 0
x2
b + C1
2
⬖ C1 ⫽ 0
x3
Lx2
⫺ b + C2
EI␯ ⫽ ⫺ma
2
6
B.C.
n(0) ⫽ 0
⬖ C2 ⫽ 0
mx2
(3L ⫺ x)
6 EI
mx
␯¿ ⫽ ⫺
(2L ⫺ x)
2EI
␯⫽ ⫺
mL3
3 EI
mL2
uB ⫽ ⫺␯¿(L) ⫽
2EI
dB ⫽ ⫺␯(L) ⫽
Problem 9.3-12 The beam shown in the figure has a guided support
;
;
;
y
MA
at A and a spring support at B. The guided support permits vertical
movement but no rotation.
Derive the equation of the deflection curve and determine
the deflection dB at end B due to the uniform load of intensity q.
(Note: Use the second-order differential equation of the deflection
curve.)
q
A
L
x
B
k = 48EI/L3
RB = kdB
Solution 9.3-12
BENDING-MOMENT EQUATION
2
EI ␯– ⫽ M(x) ⫽
2
qL
qx
⫺
2
2
q L2 x
q x3
EI␯¿ ⫽
⫺
+ C1
2
24
2 2
4
qL x
qx
EI ␯¿ ⫽
⫺
+ C1 x + C2
2
24
B.C.
n⬘(0) ⫽ 0
C1 ⫽ 0
B.C.
␯(L) ⫽
␯(x) ⫽ ⫺
qL
qL4
⫽⫺
k
48EI
C2 ⫽⫺
11qL4
48
q
(2 x4 ⫺ 12x2L2 + 11L4)
48EI
dB ⫽ ⫺v(L) ⫽
qL4
48EI
;
;
Note that RB ⫽ kdB ⫽ qL which agrees with a Fvert ⫽ 0
09Ch09.qxd
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Page 717
SECTION 9.3
717
Deflections by Integration of the Bending-Moment Equation
Problem 9.3-13 Derive the equations of the deflection curve for a
y
simple beam AB loaded by a couple M0 acting at distance a from the
left-hand support (see figure). Also, determine the deflection d0 at
the point where the load is applied. (Note: Use the second-order
differential equation of the deflection curve.)
M0
B
A
a
x
b
L
Simple beam (couple M0)
Solution 9.3-13
BENDING-MOMENT EQUATION (EQ. 9-12a)
M0x
EI␯– ⫽ M ⫽
L
EI␯¿ ⫽
4 (n)Left ⫽ (n)Right
(0 … x … a)
‹ C4 ⫽ ⫺
M0 x2
+ C1 (0 … x … a)
2L
EI␯– ⫽ M ⫽ ⫺
EI␯¿ ⫽ ⫺
B.C.
B.C.
M0
(L ⫺ x)
L
(a … x … L)
M0
x2
aLx ⫺ b + C2
L
2
1 (n⬘)Left ⫽ (n⬘)Right
C1 ⫽
␯⫽ ⫺
(a … x … L)
␯⫽ ⫺
at x ⫽ a
at x ⫽ a
M0a2
2
M0 2
(2L ⫺ 6aL + 3a2)
6L
M0 x
(6aL ⫺ 3a2 ⫺ 2L2 ⫺ x2)
6 LEI
(0 … x … a)
M0
(3a2L ⫺ 3a2x ⫺ 2L2x + 3Lx2 ⫺ x3)
6 LEI
‹ C2 ⫽ C1 + M0a
(a … x … L)
3
EI␯ ⫽
B.C.
M0x
+ C1x + C3 (0 … x … a)
6L
2 n(0) ⫽ 0
2
d0 ⫽ ⫺␯(a) ⫽
⬖ C3 ⫽ 0
3
M0x
M0x
EI␯ ⫽ ⫺
+
+ C1x + M0ax + C4
2
6L
(a … x … L)
B.C.
3 n(L) ⫽ 0
‹ C4 ⫽ ⫺M0L aa ⫺
⫽
;
;
M0a(L ⫺ a)(2a ⫺ L)
3LEI
M0ab(2a ⫺ L)
3LEI
;
NOTE: d0 is positive downward. The preceding results
agree with Case 9, Table G-2.
L
b ⫺ C1L
3
Problem 9.3-14 Derive the equations of the deflection curve for a
y
cantilever beam AB carrying a uniform load of intensity q over part of the
span (see figure). Also, determine the deflection dB at the end of the beam.
(Note: Use the second-order differential equation of the deflection curve.)
q
B
A
a
b
L
x
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CHAPTER 9
Page 718
Deflections of Beams
Solution 9.3-14
Cantilever beam (partial uniform load)
BENDING-MOMENT EQUATION (EQ. 9-12a)
B.C.
q
q
EI␯– ⫽ M ⫽ ⫺ (a ⫺ x)2 ⫽ ⫺ (a2 ⫺ 2ax + x2)
2
2
B.C.
q 2
x3
aa x ⫺ ax2 +
b + C1
2
3
1 n⬘(0) ⫽ 0
EI␯¿¿ ⫽ M ⫽ 0
(a ⱕ x ⱕ L)
EI␯¿ ⫽ C2
(a ⱕ x ⱕ L)
B.C.
B.C.
␯ ⫽⫺
2 (v¿)Left ⫽ (v¿)Right at x ⫽ a
q a2x2
ax3
x4
EI␯ ⫽ ⫺ a
⫺
+
b + C3
2 2
3
12
‹ C4 ⫽
qx2
(6a2 ⫺ 4ax + x2) (0 … x … a)
24EI
␯⫽ ⫺
qa3
‹ C2 ⫽ ⫺
6
qa3x
+ C4 (a … x … L)
6
4 (n)Left ⫽ (n)Right at x ⫽ a
(0 … x … a)
⬖ C1 ⫽ 0
⬖ C3 ⫽ 0
EI␯ ⫽ C2x + C4 ⫽ ⫺
(0 … x … a)
EI␯¿ ⫽ ⫺
3 n(0) ⫽ 0
qa3
(4x ⫺ a)
24EI
dB ⫽ ⫺␯(L) ⫽
(a … x … L)
qa3
(4L ⫺ a)
24EI
;
;
(0 … x … a)
y
q0
MA
beam AB supporting a distributed load of peak intensity q0 acting over one-half
of the length (see figure). Also, obtain formulas for the deflections dB and dC at
points B and C, respectively. (Note: Use the second-order differential equation of
the deflection curve.)
x
A
L/2
C
L/2
RA
Solution 9.3-15
L
For 0 … x …
2
q0 L x
q0 L2
EI␯– ⫽ M(x) ⫽
⫺
4
6
q0 L x2
q0L2 x
EI␯¿ ⫽
⫺
+ C1
8
6
EI␯ ⫽
q0 L x3
q0 L2 x2
⫺
+ C1 x + C2
24
12
B.C.
n⬘(0) ⫽ 0
B.C.
n(0) ⫽ 0
C1 ⫽ 0
C2 ⫽ 0
5q0 L3
L
␯¿ a b ⫽ ⫺
2
96EI
;
(These results agree with Case 2, Table G-1.)
Problem 9.3-15 Derive the equations of the deflection curve for a cantilever
BENDING-MOMENT EQUATION
qa4
24
q0 L 3
(x ⫺ 2L x2)
24EI
;
q0 L4
L
dC ⫽ ⫺␯a b ⫽
2
64EI
;
␯(x) ⫽
For
L
… x … L
2
EI␯– ⫽ M(x) ⫽
q0L2
q0
q0 L x
⫺
⫺ (L ⫺ x)
4
6
L
ax ⫺
2 q0
L 2 1
b ⫺ cq0 ⫺
2
2
L
(L ⫺ x)d a x ⫺
L 22
b
2 3
B
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Page 719
SECTION 9.3
EI␯– ⫽ M(x) ⫽
⫺ q0
(⫺3L2 x + L3
3L
B.C.
5q0 L3
L
␯¿ a b ⫽ ⫺
2
96EI
B.C.
q0 L4
L
␯a b ⫽ ⫺
2
64 EI
+ 3L x2 ⫺ x3)
q0 ⫺3 2 2
EI␯¿ ⫽⫺
a
L x + L3x
3L
2
+ L x3 ⫺
␯(x) ⫽
x4
b + C3
4
C3 ⫽
C4 ⫽
5
q L3
192 0
⫺1
q L4
320 0
⫺q0
(⫺160L2 x3 + 160L3 x2
960LEI
+ 80L x4 ⫺ 16 x5 ⫺ 25L4 x
q0 ⫺ 1 2 3
1
1
EI ␯ ⫽⫺
a
L x + L3 x2 + L x4
3L 2
2
4
⫺
719
Deflections by Integration of the Bending-Moment Equation
+ 3 L5)
;
7q0L4
dB ⫽ ⫺␯(L) ⫽
160EI
1 5
x b + C3 x + C4
20
Problem 9.3-16 Derive the equations of the deflection curve for a simple beam
;
y
AB with a distributed load of peak intensity q0 acting over the left-hand half of the
span (see figure). Also, determine the deflection dC at the midpoint of the beam.
(Note: Use the second-order differential equation of the deflection curve.)
q0
A
L/2
C
B
L/2
RA
RB
Solution 9.3-16
BENDING-MOMENT EQUATION
For 0 … x …
EI␯ ⫽
L
2
+ C 1x + C 2
5q0 L x
2q0 L
EI ␯– ⫽ M(x) ⫽
⫺
a ⫺ xb
24
L 2
a
B.C.
EI␯¿ ⫽
2q0
x
1
b ⫺ cq0 ⫺
2
2
L
C2 ⫽ 0
2 3
q0 5L x
2x5
a
⫺ x4 L +
b
24L
6
5
(2)
+ C 1x
For
q0
(5L2 x ⫺12 x2 L + 8 x3)
24L
L
… x … L
2
EI␯– ⫽ M(x) ⫽
2 2
q0 5L x
a
⫺ 4 x3 L + 2 x4 b
24L
2
+ C1
n(0) ⫽ 0
EI␯ ⫽
2
L
2
a ⫺ xb d x x
2
3
EI␯– ⫽
q0 5L2 x3
2 x5
a
⫺ x4 L +
b
24L
6
5
(1)
5q0 L x
1 L
L
⫺ q0 a x ⫺ b
24
2 2
6
EI␯– ⫽
Lq0
( ⫺x + L)
24
EI␯¿ ⫽
Lq0 ⫺x2
a
+ L xb + C3
24
2
(3)
x
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CHAPTER 9
EI␯ ⫽
Page 720
Deflections of Beams
Lq0 ⫺x3
L x2
a
+
b + C 3x + C 4
24
6
2
(4)
q0 L4
+ C3 L + C4
72
(5)
0⫽
B.C.
␯(L) ⫽ 0
B.C.
L
L
␯¿ L a b ⫽ ␯¿ R a b
2
2
+ 96 x4 ⫺ 53L4)
1
1
q L3 + C1 ⫽
q L3 + C3
96 0
64 0
B.C.
1
q L4
1920 0
L
For 0 … x …
2
q0 x
␯(x) ⫽
(200 x2 L2 ⫺ 240 x3 L
5760LEI
C4 ⫽
(6)
L
L
␯L a b ⫽ ␯R a b
2
2
For
L
… x … L
2
␯(x) ⫽
13
1
5
1
q0 L4 + C1 L ⫽
qo L4 + C 3 L
5760
2
1152
2
⫺ Lq0
(40 x3 ⫺ 120L x2
5760EI
+ 83L2 x ⫺ 3L3)
;
4
(7)
+ C4
;
3q0 L
L
dC ⫽ ⫺␯a b ⫽
2
1280EI
;
From (5)–(7)
C1 ⫽
⫺ 53
q L3
5760 0
C3 ⫽
⫺83
q L3
5760 0
y
Problem 9.3-17 The beam shown in the figure has a guided
support at A and a roller support at B. The guided support permits
vertical movement but no rotation. Derive the equation of the
deflection curve and determine the deflection dA at end A and also
dC at point C due to the uniform load of intensity q ⫽ P/L applied
over segment CB and load P at x ⫽ L/3. (Note: Use the secondorder differential equation of the deflection curve.)
MA
L
—
3
B
A
C
L
—
2
Solution 9.3-17
BENDING-MOMENT EQUATION
For 0 … x …
L
3
EI␯– ⫽ M(x) ⫽
EI ␯¿ ⫽
EI␯ ⫽
B.C. ␯¿(0)
For
⫽0
L
L
… x …
3
2
C1 ⫽ 0
19
PL
24
19
PL x + C1
24
19
PL x2 + C 1 x + C2
48
19
19
EI␯ ⫽ PL x2 + C2
EI ␯¿ ⫽ PL x
24
48
EI␯– ⫽ M(x) ⫽
L
19
PL ⫺ P a x ⫺ b
24
3
P
q=—
L
P
L
—
2
x
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Page 721
SECTION 9.3 Deflections by Integration of the Bending-Moment Equation
19
PL
PL ⫺ P x +
24
3
EI␯– ⫽ M(x) ⫽
EI␯¿ ⫽
EI␯ ⫽
For
B.C.
B.C.
B.C.
B.C.
B.C.
19
Px2
PL x
PLx⫺
+
+ C3
24
2
3
P x3 PL x2
19
PL x2 ⫺
+
+ C3 x + C4
48
6
6
L
… x … L
2
␯(L) ⫽ 0
EI ␯– ⫽ M(x) ⫽
19
P
L 21
PL
PL ⫺ Px +
⫺ ax ⫺ b
24
3
L
2 2
EI␯– ⫽ M(x) ⫽
19
PL
Px
PL
Px2
PL ⫺ Px +
+
⫺
⫺
24
3
2L
2
8
EI ␯¿ ⫽
Px2
PL x
19
PL x
Px2
P x3
PL x ⫺
+
+
⫺
+ C5
⫺
24
2
3
6L
4
8
EI ␯ ⫽
PL x2
P x4
Px3
PL x2
19
Px3
+
PL x2 ⫺
+
⫺
⫺
+ C5 x + C6
48
6
6
24L
12
16
PL4
PL3
PLL2
19
PL3
PLL2
PL L2 ⫺
+
⫺
+
⫺
+ C5 L + C6
48
6
6
24L
12
16
0⫽
L
L
␯¿L a b ⫽ ␯¿R a b
3
3
0⫽ ⫺
L
L
␯L a b ⫽ ␯R a b
3
3
␯L(a) ⫽ ␯R(a)
L 2
Pa b
3
C2 ⫽ ⫺
L
L
␯¿L a b ⫽ ␯¿R a b
2
2
+
2
3
L 3
Pa b
3
C3 ⫽ ⫺
C3
L
PL a b
3
+
6
L 3
Pa b
2
L 2
Pa b
2
+
L
+ C3 a b + C4
3
⫺
6L
4
L 4
Pa b
2
L 3
Pa b
2
L
+ C4 ⫽ ⫺
2
24L
(2)
+ C3
L 2
PL a b
3
6
(1)
+
12
L
PL a b
2
8
⫺
(4)
+ C5
L 2
PL a b
2
16
(3)
L
+ C5 a b + C6
2
(5)
From (1)–(5)
C2 ⫽
721
⫺3565 3
PL
10368
C3 ⫽
⫺1 2
PL
18
C4 ⫽
⫺ 389
PL3
1152
C5 ⫽
⫺5
PL2
144
L
3
␯(x) ⫽
⫺PL
(⫺4104 x2 + 3565 L2)
10368 EI
For
L
L
… x …
3
2
␯(x) ⫽
⫺P
(⫺648 L x2 + 192 x3 + 64 L2 x + 389L3)
1152EI
For
L
… x … L
2
␯(x) ⫽
⫺P
( ⫺72L2 x2 + 12 x3L + 6 x4 + 5L3 x + 49L4)
144EIL
For 0 … x …
dA ⫽ ⫺␯(0) ⫽
3565PL3
10368EI
;
L
3109PL3
dC ⫽ ⫺␯ a b ⫽
3
10368EI
C6 ⫽
;
;
;
;
⫺ 49
PL3
144
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CHAPTER 9
Page 722
Deflections of Beams
Deflections by Integration of the Shear Force and Load Equations
The beams described in the problems for Section 9.4 have constant flexural
rigidity EI. Also, the origin of coordinates is at the left-hand end of each beam.
y
Problem 9.4-1 Derive the equation of the deflection curve for a cantilever
A
M0
B
x
beam AB when a couple M0 acts counterclockwise at the free end (see figure).
Also, determine the deflection dB and slope uB at the free end. Use the
third-order differential equation of the deflection curve (the shear-force
equation).
Cantilever beam (couple M0)
Solution 9.4-1
SHEAR-FORCE EQUATION (EQ. 9-12b).
EIv–¿ ⫽ V ⫽ 0
EIv– ⫽ M ⫽ M0 ⫽ C1
1 M ⫽ M0
EIv¿ ⫽ C1x + C2 ⫽ M0x + C2
B.C.
B.C.
3 n(0) ⫽ 0
M0 x2
␯⫽
2EI
EIv¿¿ ⫽ C1
B.C.
L
2 n⬘(0) ⫽ 0
‹ C2 ⫽ 0
M0x2
EI␯ ⫽
+ C3
2
␯¿ ⫽
⬖ C3 ⫽ 0
;
M0 x
EI
M0 L2
(upward)
2EI
dB ⫽ ␯(L) ⫽
;
M0 L
(counterclockwise)
EI
uB ⫽ ␯¿(L) ⫽
;
(These results agree with Case 6, Table G-1.)
px
q = q0 sin —
L
Problem 9.4-2 A simple beam AB is subjected to a distributed load of
intensity q ⫽ q0 sin px/L, where q0 is the maximum intensity of the load
(see figure).
Derive the equation of the deflection curve, and then determine the
deflection dmax at the midpoint of the beam. Use the fourth-order
differential equation of the deflection curve (the load equation).
y
B
A
L
Solution 9.4-2
Simple beam (sine load)
LOAD EQUATION (EQ. 9-12c).
EI␯–– ⫽ ⫺q ⫽ ⫺q0 sin
px
L
L
px
EI␯–¿ ⫽ q0 a b cos
+ C1
p
L
2
L
px
+ C1x + C2
EI␯– ⫽ q0 a b sin
p
L
EIv–(0) ⫽ 0
B.C.
1 EIv– ⫽ M
B.C.
2 EIv–(L) ⫽ 0
EI␯¿ ⫽ ⫺q0 a
EI␯ ⫽ ⫺q0 a
‹ C1 ⫽ 0
3
L
px
+ C3
b cos
p
L
L 4 px
+ C3x + C4
b sin
p
L
‹ C2 ⫽ 0
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Page 723
SECTION 9.4
B.C.
3 n(0) ⫽ 0
⬖ C4 ⫽ 0
B.C.
4 n(L) ⫽ 0
⬖ C3 ⫽ 0
␯⫽ ⫺
q0 L4
p4EI
q0L4
L
dmax ⫽ ⫺␯a b ⫽ 4
2
p EI
;
(These results agree with Case 13, Table G-2.)
px
L
sin
723
Deflections by Integration of the Shear Force and Load Equations
;
Problem 9.4-3 The simple beam AB shown in the figure has
moments 2M0 and M0 acting at the ends.
Derive the equation of the deflection curve, and then determine
the maximum deflection dmax. Use the third-order differential
equation of the deflection curve (the shear-force equation).
y
2M0
B
A
M0
x
L
Solution 9.4-3
Simple beam with two couples
Reaction at support A: RA ⫽
3M0
L
(downward)
Shear force in beam: V ⫽ ⫺RA ⫽ ⫺
3M0
L
⫽ ⫺
SHEAR-FORCE EQUATION (EQ. 9-12b)
EI␯–¿ ⫽ V ⫽ ⫺
EI␯– ⫽ ⫺
B.C.
EI␯¿ ⫽ ⫺
EI␯ ⫽ ⫺
␯¿ ⫽ ⫺
3M0
L
3M0 x2
+ 2M0 x + C2
2L
M0 x3
+ M0 x2 + C2 x + C3
2L
2 n(0) ⫽ 0
‹ C3 ⫽ 0
B.C.
3 n(L) ⫽ 0
‹ C2 ⫽ ⫺
M0 x
(L ⫺ x)2
2LEI
;
M0
(L ⫺ x) (L ⫺ 3x)
2LEI
Set n⬘ ⫽ 0 and solve for x:
EIn⬘⬘ (0) ⫽ 2M0
B.C.
M0 x 2
(L ⫺ 2 Lx + x2)
2LEI
MAXIMUM DEFLECTION
3M0 x
+ C1
L
1 EIn⬘⬘ ⫽ M
␯⫽ ⫺
M0 L
2
⬖ C1 ⫽ 2M0
x1 ⫽ L and x2 ⫽
L
3
Maximum deflection occurs at x2 ⫽
2M0 L2
L
dmax ⫽ ⫺␯a b ⫽
3
27EI
L
.
3
(downward)
;
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CHAPTER 9
Page 724
Deflections of Beams
Problem 9.4-4 A beam with a uniform load has a guided support at one
y
end and spring support at the other. The spring has stiffness k ⫽ 48EI/L3.
Derive the equation of the deflection curve by starting with the third-order
differential equation (the shear-force equation). Also, determine the angle of
rotation uB at support B.
MA
q
A
L
x
B
k = 48EI/L3
RB = kdB
Solution 9.4-4
SHEAR-FORCE EQUATION
B.C.
n⬘(0) ⫽ 0
B.C.
␯(L) ⫽
EI␯–¿ ⫽ V ⫽ ⫺qx
2
EI␯– ⫽ ⫺
qx
+ C1
2
2
B.C.
n⬘⬘(L) ⫽ M(L) ⫽ 0
2
EI ␯– ⫽
C1 ⫽
qL
2
2
qL
qx
⫺
2
2
qL2 x
q x3
EI␯¿ ⫽
⫺
+ C2
2
6
EI ␯ ⫽
C3 ⫽ ⫺
C2 ⫽ 0
qL4
qL
⫽⫺
k
48EI
11qL4
48
q
(2 x4 ⫺12 x2 L2 ⫹11L4)
;
48EI
qL3
uB ⫽⫺␯¿(L) ⫽⫺
(Counterclockwise)
3EI
␯(x) ⫽ ⫺
;
q x4
qL2 x2
⫺
+ C2 x + C3
4
24
Problem 9.4-5 The distributed load acting on a cantilever beam AB has an
intensity q given by the expression q0 cos px/2L, where q0 is the maximum
intensity of the load (see figure).
Derive the equation of the deflection curve, and then determine the
deflection dB at the free end. Use the fourth-order differential equation of the
deflection curve (the load equation).
y
q0
px
q = q0 cos —
2L
B
A
L
x
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SECTION 9.4
Solution 9.4-5
Cantilever beam (cosine load)
LOAD EQUATION (EQ. 9-12 c)
B.C.
px
EI␯–– ⫽ ⫺q ⫽ ⫺ q0 cos
2L
EI␯–¿ ⫽ ⫺q0 a
B.C.
3 n⬘(0) ⫽ 0
EI␯ ⫽ ⫺q0 a
2L
px
+ C1
b sin
p
2L
1 EIn⬘⬘⬘ ⫽ V
EI␯– ⫽ q0 a
B.C.
2q0L
‹ C1 ⫽
p
EIn⬘⬘⬘ (L) ⫽ 0
2q0 Lx
2L
px
+
b cos
+ C2
p
p
2L
q0Lx3
q0L2x2
2L
px
+
⫺
b cos
+ C4
p
p
2L
3p
4 n(0) ⫽ 0
␯⫽ ⫺
2
⬖ C3 ⫽ 0
4
‹ C4 ⫽
2 EIn⬘⬘ ⫽ M
EI␯¿ ⫽ q0 a
EIn⬘⬘(L) ⫽ 0
2
2q0L
p
px
⫺ 48L3 + 3p3Lx2 ⫺ p3x3 b
2L
dB ⫽ ⫺␯(L) ⫽
2
q0Lx
2q0L x
2L 3 px
+
b sin
⫺
+ C3
p
p
p
2L
p4
3p4EI
a48L3 cos
‹ C2 ⫽ ⫺
16q0L4
q0L
2
B.C.
725
Deflections by Integration of the Shear Force and Load Equations
2q0L4
3p4EI
(p3 ⫺ 24)
;
;
(These results agree with Case 10, Table G-1.)
Problem 9.4-6 A cantilever beam AB is subjected to a parabolically
varying load of intensity q ⫽ q0(L2 ⫺ x2)/L2, where q0 is the maximum
intensity of the load (see figure).
Derive the equation of the deflection curve, and then determine the
deflection dB and angle of rotation uB at the free end. Use the fourth-order
differential equation of the deflection curve (the load equation).
y
L2 ⫺x2
q = q0 —
L2
q0
x
B
A
L
Solution 9.4-6
Cantilever beam (parabolic load)
LOAD EQUATION (EQ. 9-12 c)
EI␯–– ⫽ ⫺q ⫽ ⫺
EI␯–¿ ⫽ ⫺
B.C.
q0
2
L
L2
2 EIn⬘⬘ ⫽ M
(L2 ⫺ x2)
aL2x ⫺
1 EIn⬘⬘⬘ ⫽ V
EI␯– ⫽ ⫺
q0
B.C.
EI␯¿ ⫽ ⫺
3
x
b + C1
3
EIn⬘⬘⬘(L) ⫽ 0
‹ C1 ⫽
q0 L2x2
2q0L
x4
a
⫺
b +
x + C2
2
2
12
3
L
2q0L
3
B.C.
EIn⬘⬘(L) ⫽ 0
‹ C2 ⫽ ⫺
q0L2
4
q0 L2x3
q0Lx2
q0L2x
x5
a
⫺
b +
⫺
+ C3
2
6
60
3
4
L
3 n⬘(0) ⫽ 0
⬖ C3 ⫽ 0
q0 L2x4
q0Lx3
q0L2x2
x6
EI␯ ⫽ ⫺ 2 a
⫺
b +
⫺
+ C4
24
360
9
8
L
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CHAPTER 9
B.C.
Page 726
Deflections of Beams
4 n(0) ⫽ 0
⬖ C4 ⫽ 0
␯¿ ⫽ ⫺
2
␯ ⫽⫺
q0 x
360 L2EI
(45L4 ⫺ 40L3x + 15L2x2 ⫺x4)
19q0L4
dB ⫽ ⫺␯(L) ⫽
360 EI
;
q0 x
60L2EI
(15L4 ⫺ 20L3x + 10L2x2 ⫺ x4)
uB ⫽ ⫺␯¿(L) ⫽
q0L3
15EI
;
;
Problem 9.4-7 A beam on simple supports is subjected to a parabolically
4q0 x
(L ⫺ x)
q= —
L2
distributed load of intensity q ⫽ 4q0 x(L ⫺ x)/L2, where q0 is the maximum intensity of the load (see figure).
Derive the equation of the deflection curve, and then determine the
maximum deflection dmax. Use the fourthorder differential equation of the
deflection curve (the load equation).
y
B
A
x
L
Solution 9.4-7
Simple beam (parabolic load)
LOAD EQUATION (EQ. 9-12 c)
EI␯–– ⫽ ⫺q ⫽ ⫺
EI␯–¿ ⫽ ⫺
EI␯– ⫽ ⫺
B.C.
B.C.
2q0
3L2
q0
3L2
30L2
L
(L ⫺ x) ⫽ ⫺
4q0
L2
(Lx ⫺ x2)
EI␯ ⫽ ⫺
B.C.
(2Lx ⫺ x ) + C1x + C2
3
2 EIn⬘⬘(L) ⫽ 0
q0
2
3 (Symmetry)
(3Lx2 ⫺ 2x3) + C1
1 EIn⬘⬘ ⫽ M
EI␯¿ ⫽ ⫺
4q0 x
B.C.
⬖ C2 ⫽ 0
q0L
‹ C1 ⫽
3
2
30L
a L5x ⫺
4 n(0) ⫽ 0
4
EIn⬘⬘(0) ⫽ 0
q0
␯⫽ ⫺
L
␯¿ a b ⫽ 0
2
q0 x
90L2EI
‹ C3 ⫽ ⫺
qoL3
30
5L3x3
x6
+ Lx5 ⫺ b + C4
3
3
⬖ C4 ⫽ 0
(3L5 ⫺ 5L3x2 + 3Lx4 ⫺ x5)
61q0L4
L
dmax ⫽ ⫺␯a b ⫽
2
5760 EI
;
;
( ⫺5L3x2 + 5Lx4 ⫺ 2x5) + C3
Problem 9.4-8 Derive the equation of the deflection curve for beam
AB, with guided support at A and roller at B, carrying a triangularly
distributed load of maximum intensity q0 (see figure). Also, determine
the maximum deflection dmax of the beam. Use the fourth-order differential equation of the deflection curve (the load equation).
MA
y
q0
x
A
L
B
RB
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SECTION 9.4
Deflections by Integration of the Shear Force and Load Equations
727
Solution 9.4-8
LOAD EQUATION
EI ␯–– ⫽ ⫺q ⫽ ⫺q0 +
EI ␯¿ ⫽ ⫺
q0 x
L
EI ␯ ⫽ ⫺
2
EI ␯–¿ ⫽ ⫺q0 x +
B.C.
q0 x
+ C1
2L
n⬘⬘⬘(0) ⫽ V(0) ⫽ 0
B.C.
C1 ⫽ 0
q0 x3
q0 x2
+
+ C2
2
6L
n⬘⬘(L) ⫽ M(L) ⫽ 0
EI␯– ⫽ ⫺
q0 x5
q0 L2 x2
q0 x4
+
+
24
120L
6
+ C3 x + C4
q0 x2
EI␯–¿ ⫽ ⫺q0 x +
2L
EI ␯– ⫽ ⫺
q0 L2 x
q0 x4
q0 x3
+
+ C3
+
6
24L
3
B.C.
n⬘(0) ⫽ 0
C3 ⫽ 0
B.C.
n(L) ⫽ 0
C4 ⫽ ⫺
␯(x) ⫽
C2 ⫽
2q0 L4
15
q0
1⫺ 5x4 L + x5
120EIL
+ 20L3 x2 ⫺ 16L52
q0 L2
3
q0 x2
q0 x3
q0 L2
+
+
2
6L
3
;
MAXIMUM DEFLECTION
dmax ⫽ ⫺␯(0) ⫽
Problem 9.4-9 Derive the equations of the deflection curve for beam
ABC, with guided support at A and roller support at B, supporting a
uniform load of intensity q acting on the over-hang portion of the
beam (see figure). Also, determine deflection dC and angle of rotation
uC. Use the fourth-order differential equation of the deflection curve
(the load equation).
2q0 L4
15EI
;
y
MA
q
x
A
L
B
L/2
C
RB
Solution 9.4-9
LOAD EQUATION
B.C.
EI␯–– ⫽ ⫺q ⫽ 0 (0 … x … L)
EI ␯ ⫽ ⫺
EI ␯– ⫽ C1 x + C2 (0 … x … L)
n⬘⬘⬘(0) ⫽ V(0) ⫽ 0
B.C.
␯–(0) ⫽ M(0) ⫽ ⫺
EI ␯– ⫽ ⫺
EI ␯¿ ⫽ ⫺
qL2
8
qL2x
+ C3
8
B.C.
C1 ⫽ 0
2
qL
8
C2 ⫽ ⫺
C3 ⫽ 0
2 2
EI␯–¿ ⫽ C1 (0 … x … L)
B.C.
n⬘(0) ⫽ 0
qL2
8
qL x
+ C4
16
n(L) ⫽ 0
␯(x) ⫽ ⫺
C4 ⫽
qL4
16
qL2 2
1x ⫺ L22
16E I
(0 … x … L)
LOAD EQUATION
EI ␯–– ⫽ ⫺q
aL … x …
3L
b
2
;
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CHAPTER 9
Page 728
Deflections of Beams
EI ␯–¿ ⫽ ⫺ qx + C5 aL … x …
3L
b
2
⫺ q x2
3L
+ C5 x + C6 aL … x …
b
2
2
EI ␯– ⫽
3L
3L
b ⫽ Va b ⫽ 0
2
2
C5 ⫽
3qL
2
3L
3L
b ⫽ Ma b ⫽ 0
2
2
C6 ⫽
9qL2
8
B.C.
␯–¿ a
B.C.
␯– a
+
EI␯ ⫽
⫺q x2
3qL x
9qL2
+
⫺
2
2
8
B.C.
EI ␯¿ ⫽
⫺q x3
3qL x2
9qL2 x
+
⫺
+ C7
6
4
8
␯(x) ⫽
n⬘L(L) ⫽ n⬘R(L)
qL3
⫺ qL3
3qL3
9qL3
⫽
+
⫺
+ C7
⫺
8
6
4
8
5
qL3
C7 ⫽
12
5
qL3
12
3qL x3
9 qL2 x2
⫺q x 4
+
⫺
24
12
16
5
qL3 x + C8
12
+
EI␯– ⫽
B.C.
⫺qx3
3qL x2
9qL2 x
+
⫺
6
4
8
EI␯¿ ⫽
n(L) ⫽ 0
⫺1 4
qL
16
C8 ⫽
⫺q
1⫺20xL3 + 27L2 x2
48EI
aL … x …
⫺12L x3 + 2 x4 + 3L42
dC ⫽ ⫺␯a
9qL4
3L
b ⫽
2
128EI
uC ⫽ ⫺␯¿ a
Problem 9.4-10 Derive the equations of the deflection curve for beam AB,
with guided support at A and roller support at B, supporting a distributed load
of maximum intensity q0 acting on the right-hand half of the beam (see
figure). Also, determine deflection dA, angle of rotation uB, and deflection dC
at the midpoint. Use the fourth-order differential equation of the deflection
curve (the load equation).
;
;
7qL3
3L
b ⫽
2
48EI
MA
3L
b
2
;
(Clockwise)
y
q0
x
A
L/2
C
L/2
B
RB
Solution 9.4-10
LOAD EQUATION
EI ␯–– ⫽ ⫺q ⫽ 0
B.C.
L
a0 … x … b
2
EI ␯–¿ ⫽ C1 a0 … x …
L
b
2
EI ␯– ⫽ C1 x + C2 a0 … x …
L
b
2
n⬘⬘⬘(0) ⫽ V(0) ⫽ 0
C1 ⫽ 0
2
B.C.
␯–(0) ⫽ M(0) ⫽
EI ␯– ⫽
q0 L2
12
EI ␯¿ ⫽
q0 L2 x
+ C3
12
q0 L
12
a0 … x …
C2 ⫽
q0 L2
12
L
b
2
a0 … x …
L
b
2
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Page 729
729
SECTION 9.4 Deflections by Integration of the Shear Force and Load Equations
B.C.
n⬘(0) ⫽ 0
EI␯ ⫽
C3 ⫽ 0
q0 L2 x2
+ C4
24
EI ␯ ⫽ ⫺
a0 … x …
L
b
2
q0 x4
q0 x5
q0 L x3
q0 L2 x2
+
+
⫺
12
60L
8
24
+
5q0 L3 x
41
⫺
q L4
192
960 0
LOAD EQUATION
a
2q0
L
L
EI ␯–– ⫽ ⫺q0 +
ax ⫺ b a … x … Lb
L
2
2
B.C.
2 q0 x
EI ␯–– ⫽ ⫺ 2 q0 +
L
2
EI ␯–¿ ⫽ ⫺ 2q0 x +
a
q0 x
+ C5
L
L
… x … Lb
2
L 2
q0 L2 a b
2
24
+ C4 ⫽ ⫺
L 4
q0 a b
2
B.C.
L
L
␯–¿ a b ⫽ V a b ⫽ 0
2
2
+
60L
B.C.
q0 L2
L
L
␯– a b ⫽ Ma b ⫽
2
2
12
q0 L2
12
q0 x3
3q0 L x
q0 L2
EI␯– ⫽ ⫺ q0 x +
+
⫺
3L
4
12
2
EI ␯¿ ⫽⫺
q0 x3 q0 x4 3q0 L x2
+
+
3
12 L
8
EI ␯¿ ⫽ ⫺
C7 ⫽
q0 x3 q0 x4 3q0 L x2
+
+
3
12L
8
q0 L2 x 5q0 L3
+
+
12
192
q0 x4
q0 x5
q0 L x3
EI ␯ ⫽ ⫺
+
+
12
60L
8
2 2
B.C.
192
C4 ⫽
3
q0 L x
5q0 L x
+
+ C8
24
192
n(L) ⫽ 0
L
2
⫺
⫺
L 2
q0 L2 a b
2
⫺ 19
q L4
480 0
EI ␯ ⫽
q0 L2 x2
19
⫺
q L4
24
480 0
␯(x) ⫽ ⫺
⫺41
C8 ⫽
q L4
960 0
5
q L3
192 0
24
41
q L4
960 0
a0 … x …
L
b
2
q0 L2
(⫺20x2 + 19L2)
480 EI
a 0 … x … Lb
L
L
␯¿ L a b ⫽ ␯¿ R a b
2
2
⫺
+
q0 L2 x
+ C7
12
⫺
B.C.
8
5q0 L3
3q0 L
4
C6 ⫽
L 3
q0 L a b
2
⫹
L
… x … Lb
2
C5 ⫽
L 5
q0 a b
2
12
q0 x
+ C5 x + C6
3L
a
;
L
L
␯L a b ⫽ ␯R a b
2
2
3
EI␯– ⫽ ⫺ q0 x2 +
L
… x … Lb
2
␯(x) ⫽ ⫺
;
q0
(80x4 L ⫺ 16 x5 ⫺120L2
960LEI
x3 + 40 L3 x2 ⫺25L4 x⫹41L5
a
dA ⫽ ⫺␯(0) ⫽
19
q L4
480EI 0
uB ⫽ ⫺␯¿(L) ⫽ ⫺
13
q L3
192EI 0
L
7
dC ⫽ ⫺␯ a b ⫽
q L4
2
240EI 0
L
… x … Lb
2
;
;
;
;
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CHAPTER 9
Page 730
Deflections of Beams
Method of Superposition
P
The problems for Section 9.5 are to be solved by the method of
superposition. All beams have constant flexural rigidity EI.
P
P
A
B
Problem 9.5-1 A cantilever beam AB carries three equally spaced
L
—
3
concentrated loads, as shown in the figure. Obtain formulas for the
angle of rotation uB and deflection dB at the free end of the beam.
Solution 9.5-1
L 2
Pa b
3
Pa
2L 2
b
3
+
2EI
2EI
dB PL2
7PL2
+
2EI
9EI
L 2
Pa b
3
6EI
;
+
P
(a) Determine the deflection d1 at the midpoint of the beam.
(b) If the same total load (5P) is distributed as a uniform load
on the beam, what is the deflection d2 at the midpoint?
(c) Calculate the ratio of d1 to d2.
Solution 9.5-2
2L 2
b
3
6EI
a 3L P
P
P
P
A
B
L
—
6
d2 L 2
d1 c3L2 4 a b d
24 EI
6
(c)
L
Pa b
3
2
3
L
PL
c3L2 4 a b d +
24EI
3
48EI
;
2L
b
3
;
L
—
6
L
—
6
L
—
6
L
—
6
L
—
6
(b) Table G-2, Case 1 qL 5P
L
Pa b
6
11PL3
144EI
L
b +
3
Pa
Simple beam with 5 loads
(a) Table G-2, Cases 4 and 6
a3L 5PL3
PL3
3EI
9EI
Problem 9.5-2 A simple beam AB supports five equally spaced
loads P (see figure).
+
L
—
3
Cantilever beam with 3 loads
Table G-1, Cases 4 and 5
uB L
—
3
5qL4
25PL3
384EI
384EI
;
d1
11 384
88
a
b 1.173
d2
144 25
75
;
09Ch09.qxd
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Page 731
SECTION 9.5
Problem 9.5-3 The cantilever beam AB shown in the figure has an extension
L
BCD attached to its free end. A force P acts at the end of the extension.
A
B
(a) Find the ratio a/L so that the vertical deflection of point B will be zero.
(b) Find the ratio a/L so that the angle of rotation at point B will be zero.
D
C
a
P
Solution 9.5-3
Cantilever beam with extension
Table G-1, Cases 4 and 6
PL3
PaL2
0
3EI
2EI
2
PL
PaL
0
(b) uB 2EI
EI
(a) dB a
2
L
3
a
1
L
2
;
;
Problem 9.5-4 Beam ACB hangs from two springs, as shown
in the figure. The springs have stiffnesses k1 and k2 and the beam
has flexural rigidity EI.
(a) What is the downward displacement of point C, which is
at the midpoint of the beam, when the moment M0 is
applied? Data for the structure are as follows:
M0 10.0 kNm, L 1.8 m, EI 216 kNm2,
k1 250 kN/m, and k2 160 kN/m.
(b) Repeat (a) but remove M0 and apply uniform load
q 3.5 kN/m to the entire beam.
RA = k1 dA
k1
A
RB = k2 dB
k2
M0
L/2
C
L/2
B
q = 3.5 kN/m (for Part (b) only)
Solution 9.5-4
M0 10.0 kN # m
k1 250 kN /m
L 1.8 m
k2 160 kN / m
q 3.5 kN/ m
EI 216 kN # m2
dA 22.22 mm
dB 34.72 mm
Downward
Upward
Table G-2, Case 8
(a) RA M0
L
RB dA RA
k1
dB RB
k2
M0
L
731
Method of Superposition
dC 0 +
1#
(dA + dB)
2
dC 6.25 mm Upward
;
09Ch09.qxd
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CHAPTER 9
(b) RA qL
2
RA
dA k1
Page 732
Deflections of Beams
dB 19.69 mm
RB RA
Table G-2, Case 1
RB
dB k2
dC dA 12.60 mm
5qL4
1
+ (dA + dB)
384EI
2
dC 18.36 mm Downward
Problem 9.5-5 What must be the equation y f(x) of the axis of the
slightly curved beam AB (see figure) before the load is applied in order that
the load P, moving along the bar, always stays at the same level?
;
P
y
B
A
L
Solution 9.5-5
Slightly curved beam
Let x distance to load P
d downward deflection at load P
Table G-2, Case 5:
d
Initial upward displacement of the beam must equal d.
‹ y
Px2(L x)2
3LEI
;
Px2(L x)2
P(L x)x 2
[L (L x)2 x2] 6LEI
3LEI
Problem 9.5-6 Determine the angle of rotation uB and
deflection dB at the free end of a cantilever beam AB having a
uniform load of intensity q acting over the middle third of its
length (see figure).
q
B
A
L
—
3
Solution 9.5-6
Cantilever beam (partial uniform load)
q intensity of uniform load
Original load on the beam:
Load No. 1:
L
—
3
L
—
3
x
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Page 733
SECTION 9.5
Load No. 2:
Method of Superposition
Table G-1, Case 2
uB q 2L 3
q L 3
7qL3
a b a b 6EI 3
6EI 3
162EI
dB q
q
2L 3
2L
L 3
L
a b a 4L b
a b a 4L b
24EI 3
3
24EI 3
3
SUPERPOSITION:
23qL4
648EI
;
;
Original load Load No. 1 minus Load No. 2
Problem 9.5-7 The cantilever beam ACB shown in the figure has
A
48 in.
Cantilever beam (two loads)
dB M0(L/2)
L
PL3
a 2L b +
2EI
2
3EI
3M0L2
PL3
+
( downward deflection)
8EI
3EI
SUBSTITUTE NUMERICAL VALUES:
EI 2.1 106 k-in.2
M0 35 k-in.
P 2.5 k
L 96 in.
Table G-1, Cases 4, 6, and 7
B
C
48 in.
dC 2.5 k
35 k-in.
flexural rigidity EI 2.1 106 k-in.2 Calculate the downward deflections dC and dB at points C and B, respectively, due to the simultaneous
action of the moment of 35 k-in. applied at point C and the concentrated load of 2.5 k applied at the free end B.
Solution 9.5-7
733
P(L/2)2
M0(L/2)2
L
+
a3L b
2EI
6EI
2
M0L2
5PL3
+
( downward deflection)
8EI
48EI
dC 0.01920 in. 0.10971 in.
0.0905 in.
;
dB 0.05760 in. 0.35109 in.
0.293 in.
;
09Ch09.qxd
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1:32 PM
CHAPTER 9
Page 734
Deflections of Beams
Problem 9.5-8 A beam ABCD consisting of a simple span BD and an
L
—
3
L
—
2
overhang AB is loaded by a force P acting at the end of the bracket CEF
(see figure).
A
(a) Determine the deflection dA at the end of the overhang.
(b) Under what conditions is this deflection upward? Under what
conditions is it downward?
2L
—
3
B
C
F
E
D
P
a
Solution 9.5-8
Beam with bracket and overhang
Consider part BD of the beam.
(a) DEFLECTION AT THE END OF THE OVERHANG
PL2
L
(10L 9a)
dA uB a b 2
324 EI
;
( upward deflection)
a
10
(b) Deflection is upward when 6
and downward
L
9
10
a
;
when 7
L
9
M0 Pa
Table G-2, Cases 5 and 9
uB P(L/3)(2L/3)(5L/3)
6LEI
+
L2
L2
Pa
c6a b 3a b 2L2 d
6LEI
3
9
PL
(10L 9a) ( clockwise angle)
162EI
Problem 9.5-9 A horizontal load P acts at end C of the bracket
C
ABC shown in the figure.
(a) Determine the deflection dC of point C.
(b) Determine the maximum upward deflection dmax of member AB.
Note: Assume that the flexural rigidity EI is constant throughout the
frame. Also, disregard the effects of axial deformations and consider
only the effects of bending due to the load P.
P
H
B
A
L
09Ch09.qxd
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Page 735
SECTION 9.5
Method of Superposition
735
Bracket ABC
Solution 9.5-9
BEAM AB
(a) ARM BC
M0 PH
Table G-1, Case 4
3
PH
PH3
PH2L
+ uBH +
3EI
3EI
3EI
dC PH2
(L + H)
3EI
;
(b) MAXIMUM DEFLECTION OF BEAM AB
Table G-2,
Table G-2, Case 7: uB PHL
M0L
3EI
3EI
Problem 9.5-10 A beam ABC having flexural rigidity
EI 75 kNm2 is loaded by a force P 800 N at end
C and tied down at end A by a wire having axial rigidity
EA 900 kN (see figure).
What is the deflection at point C when the load P is applied?
Solution 9.5-10
Case 7: dmax M0L2
913EI
PHL2
913EI
B
A
;
C
P = 800 N
0.5 m
0.5 m
0.75 m
D
Beam tied down by a wire
CONSIDER AB AS A SIMPLE BEAM
M0 PL2
Table G-2, Case 7: u¿B M0 L1
PL1L2
3EI
3EI
CONSIDER THE STRETCHING OF WIRE AD
d¿A (Force in AD) a
EI 75 kN # m2
P 800 N
DEFLECTION dC OF POINT C
EA 900 kN
H 0.5 m
PL2
PL2H
H
H
b a
ba
b EA
L1
EA
EAL1
L1 0.5 m
L2 0.75 m
CONSIDER BC AS A CANTILEVER BEAM
Table G-1, Case 4: dC¿ dc dc¿ uB¿ (L2)dB¿ a
PL32
3EI
L2
b
L1
PL1L22
PL22H
PL32
+
+
3EI
3EI
EAL21
;
SUBSTITUTE NUMERICAL VALUES:
dC 1.50 mm 1.00 mm 1.00 mm 3.50 mm
;
09Ch09.qxd
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1:32 PM
CHAPTER 9
Page 736
Deflections of Beams
Problem 9.5-11 Determine the angle of rotation uB and deflection dB at the
q0
y
free end of a cantilever beam AB supporting a parabolic load defined by the
equation q q0x2/L2 (see figure).
A
B
x
L
Solution 9.5-11
LOAD: q Cantilever beam (parabolic load)
q0x2
qdx element of load
2
L
L
q0
q0L3
10EI
x4dx 2EIL2 L0
;
L
(qdx)(x2)
(3L x)
6EI
L0
L
q0x2
1
a 2 b (x2)(3L x) dx
6EI L0
L
dB TABLE G-1, CASE 5
(Set a equal to x)
L
L
uB 2
2
(qdx)(x )
q0x
1
a 2 b x2dx
2EI
2EI
L
L0
L0
q0
6EIL2 L0
Problem 9.5-12 A simple beam AB supports a uniform load of intensity q
acting over the middle region of the span (see figure).
Determine the angle of rotation uA at the left-hand support and the
deflection dmax at the midpoint.
L
(x4)(3Lx) dx 13q0L4
180EI
;
q
B
A
a
a
L
Solution 9.5-12
Simple beam (partial uniform load)
LOAD: qdx element of load
Replace P by qdx
Replace a by x
Integrate x from a to L/2
L/3 qdx
L/2
q
uA (x)(L x) (xL x2)dx
2EI La
La 2EI
q
(L3 6a2L + 4a3)
;
24EI
TABLE G-2, CASE 6
TABLE G-2, CASE 6
uA Pa(L a)
2EI
Replace P by qdx
Pa
(3L2 4a2)
24EI
Replace a by x
dmax Integrate x from a to L/2
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1:32 PM
Page 737
SECTION 9.5
L/2
dmax qdx
(x)(3L2 4x2)
24EI
dmax La
L/2
q
(3L2x 4x3)dx
24EI La
q
(5L4 24a2L2 + 16a4)
384EI
L 2
L 3
4L(L a) a b + L a b d
2
2
;
q(L a)2
qa2
[2L (L a)]2 (2L a)2
24LEI
24LEI
qa2
L
L
c La2 + 4L2 a b + a2 a b
24LEI
2
2
L 3
L 2
6La b + 2a b d
2
2
Table G-2, Case 3
q
(L3 6La2 + 4a3)
24EI
q(L/2)
c(L a)4 4L(L a)3
24LEI
L 2
+ 4L2(L a)2 + 2(L a)2 a b
2
ALTERNATE SOLUTION (not recommended; algebra is
extremely lengthy)
uA dmax ;
Problem 9.5-13 The overhanging beam ABCD supports two
concentrated loads P and Q (see figure).
q
(5L4 24L2a2 + 16a4)
384EI
;
y
P
MA
(a) For what ratio P/Q will the deflection at point B be zero?
(b) For what ratio will the deflection at point D be zero?
(c) If Q is replaced by uniform load with intensity q (on the
overhang), repeat (a) and (b) but find ratio P/(qa)
Q
A
C
x
B
L
—
2
D
L
—
2
RC
q
Solution 9.5-13
(a) DEFLECTION AT POINT B
Table G-2 Cases 6 and 10
dB L
Pa b
2
737
Method of Superposition
L
L 2
c3 a b (2L) 3 a b
6EI
2
2
L
Qa a b
2
L 2
L
a b d c(2L) a b d
2
2EI
2
dB 0
9a
P
Q
4L
;
a
(for Part (c))
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CHAPTER 9
Page 738
Deflections of Beams
(b) DEFLECTION AT POINT D
Table G-2 Case 6; Table G-1 Case 4; Table G-2
Case 10
dD L
L
P a b c(2L) d
2
2
2E I
Table G-2 Case 6; Table G-1 Case 1; Table G-2
Case 10
(a)
8a (3L + a)
P
Q
9L2
dD L
L
P a b c(2L) d
2
2
;
2EI
4
+
(c.1) DEFLECTION AT POINT B
dD 0
Table G-2 Cases 6 and 10
dB ;
(c.2) DEFLECTION AT POINT D
Qa (2L)
Q a3
+
(a)
+
3EI
2EI
dD 0
P
9a
qa
8L
dB 0
L
Pa b
2
qa
+
8EI
a
qa2
b (2L)
2
2EI
(a)
(a)
a (4L + a)
P
qa
3L2
;
L
L 2
c3 a b (2L) 3 a b
6EI
2
2
2
L
a b d 2
a
qa2
L
ba b
2
2
2 EI
L
c(2 L) a b d
2
Problem 9.5-14 A thin metal strip of total weight W and length L is placed
d
across the top of a flat table of width L/3 as shown in the figure.
What is the clearance d between the strip and the middle of the table?
(The strip of metal has flexural rigidity EI.)
L
—
3
Solution 9.5-14
L
—
6
L
—
6
Thin metal strip
W total weight q W
L
EI flexural rigidity
FREE BODY DIAGRAM (the part of the strip above the table)
TABLE G-2, CASES 1 AND 10
d 5q
M0 L 2
L 4
a b +
a b
384EI 3
8EI 3
5qL4
qL4
+
31,104EI
1296EI
19qL4
31,104EI
But q W
:
L
‹ d
19WL3
31,104EI
;
L
—
3
09Ch09.qxd
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1:32 PM
Page 739
SECTION 9.5
Problem 9.5-15 An overhanging beam ABC with flexural rigidity
739
Method of Superposition
y
EI 15 k-in.2 is supported by a guided support at A and by a spring
of stiffness k at point B (see figure). Span AB has length L 30 in. and
carries a uniform load. The over-hang BC has length b 15 in. For
what stiffness k of the spring will the uniform load produce no
deflection at the free end C?
q
MA
C
A
L
B
k
x
b
RB
Solution 9.5-15
EI 15kip # in2.
L 30 in.
b 15 in.
RB qL
Table G-2, Case 1
3EI
for dC 0
k
Therefore
k 3.33lb/in
bL2
;
3
dC uB b dB q (2L)
qL
(b) 24EI
k
Problem 9.5-16 A beam ABCD rests on simple supports at
B and C (see figure). The beam has a slight initial curvature so that
end A is 18 mm above the elevation of the supports and end D is
12 mm above. What moments M1 and M2, acting at points A and D,
respectively, will move points A and D downward to the level of the
supports? (The flexural rigidity EI of the beam is 2.5 106 N # m2
and L 2.5 m).
M1
M2
B
18 mm
A
C
L
L
12 mm
L
D
Solution 9.5-16
EI 2.5106 N # m2
L 2.5 m
dA 18 mm
dD 12 mm
Table G-2, Case 7
M2L
M 1L
+
uB 3EI
6 EI
M 2L
M 1L
+
uC 6 EI
3EI
dD M 2 L2
M 1L
M 2L
+ a
+
bL
2EI
6EI
3EI
5M 1 + M 2 6dA EI
5M 2 + M 1 6dD EI
(1)
L2
(2)
L2
DEFLECTION AT POINT A AND D
SOLVE EQUATION (1) AND (2)
Table G-1, Case 6
dA M 1 L2
+ uB L
2EI
dA M 1 L2
M 1L
M 2L
+ a
+
bL
2EI
3EI
6EI
dD M 2 L2
+ uC L
2EI
M1 EI(5dA dD)
M2 2
4L
Therefore
M 1 7800 N # m
;
M 2 4200 N # m
;
EI(5dD dA)
4L2
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CHAPTER 9
Page 740
Deflections of Beams
Problem 9.5-17 The compound beam ABC shown in the figure
has a guided support at A and a fixed support at C. The beam
consists of two members joined by a pin connection (i.e., moment
release) at B. Find the deflection d under the load P.
P
MA
MC
A
B
3b
C
2b
b
RC
Solution 9.5-17
Table G-1, Case 4
dB P(2b)3
3EI
DEFLECTION UNDER THE LOAD P
Table G-2, Case 6
d
P(b)
c3(b) (8b) 3(b)2 (b)2 d + dB
6EI
d
P (2b)3
P(b)
c3(b) (8 b) 3b2 b2 d +
6EI
3EI
d
6 Pb3
EI
Problem 9.5-18 A compound beam ABCDE (see figure) consists of two
parts (ABC and CDE) connected by a hinge (i.e., moment release) at C.
The elastic support at B has stiffness k EI/b3 Determine the deflection
dE at the free end E due to the load P acting at that point.
;
P
D
C
B
A
2b
b
b
Solution 9.5-18
CONSIDER BEAM CDE
CONSIDER BEAM ABC
RB 3P
2
dB RB
3P
k
2k
Table G-2, Case 7; Table G-1, Case 4
Upward
Table G-2, Case 7; Table G-1, Case 4
dC dC Pb(2 b)
P b3
2b + b
b +
+ dB a
b
3EI
3EI
2b
Pb(2 b)
Pb3
3P 2b + b
b +
+
a
b
3EI
3EI
2k
2b
P(4b3 k + 9E I)
4EIk
Upward
dE E
EI
k= —
b3
(Pb) (b)
(Pb)(b)
Pb3
b +
+ dC b
3EI
3EI
3EI
+
for k dE P(4b3 k + 9EI )
Pb3
+
3EI
4EIk
EI
b3
47Pb3
12EI
;
b
09Ch09.qxd
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1:32 PM
Page 741
SECTION 9.5
Method of Superposition
741
Problem 9.5-19 A stekel beam ABC is simply supported at A and held
by a high-strength steel wire at B (see figure). A load P 240 lb acts at
the free end C. The wire has axial rigidity EA 1500 103 lb, and the
beam has flexural rigidity EI 36 106 lb-in.2
What is the deflection dC of point C due to the load P?
Wire
20 in.
A
C
20 in.
Solution 9.5-19
P = 240 lb
Beam
B
30 in.
Beam supported by a wire
(2) ASSUME THAT THE WIRE STRETCHES
T tensile force in the wire
P
(b + c)
b
dB P 240 lb
b 20 in.
c 30 in.
h 20 in.
Beam: EI 36 106 lb-in.2
Wire: EA 1500 103 lb
Ph(b + c)
Th
EA
EAb
d– C dB a
Ph(b + c)2
b + c
b b
EAb2
(downward)
(3) DEFLECTION AT POINT C
dC dc¿ dc¿¿ P(bc)c
(1) ASSUME THAT POINT B IS ON A SIMPLE SUPPORT
c2 h(b c)
d
3EI
EAb2
;
Substitute numerical values:
dC 0.10 in. + 0.02 in. 0.12 in.
;
Pc3
Pc3
b
uB¿ c (Pc)a
bc
3EI
3EI
3EI
Pc2
(b + c) (downward)
3EI
dC¿ Problem 9.5-20 The compound beam shown in the figure
consists of a cantilever beam AB (length L) that is pin-connected to a simple
beam BD (length 2L). After the beam is constructed, a clearance c exists
between the beam and a support at C, midway between points B and D.
Subsequently, a uniform load is placed along the entire length of the beam.
What intensity q of the load is needed to close the gap at C and bring
the beam into contact with the support?
q
D
A
C
B
L
Moment
release
L
c
L
09Ch09.qxd
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9/27/08
1:32 PM
CHAPTER 9
Solution 9.5-20
Page 742
Deflections of Beams
Compound beam
dc¿¿ downward displacement of point C due to dB
BEAM BCD WITH A SUPPORT AT B
dc¿ 5q(2L)4
384EI
11qL4
1
dc¿¿ dB 2
48EI
5qL4
24EI
DOWNWARD DISPLACEMENT OF POINT C
dc dc¿ dc¿¿ BEAM BCD WITH A SUPPORT AT B
5qL4 11qL4 7qL4
24EI 48EI
16EI
c clearance
CANTILEVER BEAM AB
dB qL4
(qL)L3
+
8EI
3EI
4
11qL
24EI
(downward)
c dC 7qL4
16EI
INTENSITY OF LOAD TO CLOSE THE GAP
q
16EIc
7L4
;
Problem 9.5-21 Find the horizontal deflection dh and vertical deflection dn at the free end C of the
frame ABC shown in the figure. (The flexural rigidity EI is constant throughout the frame.)
Note: Disregard the effects of axial deformations and consider only the effects of bending
due to the load P.
P
B
C
c
b
A
Solution 9.5-21
Frame ABC
MEMBER BC WITH B FIXED AGAINST ROTATION:
MEMBER AB:
Table G-1, Case 4:
dh horizontal deflection
of point B
dc¿ Table G-1, Case 6:
dh (Pc)b2
Pcb2
2EI
2EI
Pcb
uB EI
Since member BC does not change in length,
dh is also the horizontal displacement of point C.
‹ dh Pcb2
2EI
;
Pc3
3EI
VERTICAL DEFLECTION OF POINT C
dc dv dc¿ uBC d Pc3 Pcb
(c)
3EI EI
Pc2
(c + 3b)
3EI
Pc2
(c3b)
3EI
;
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Page 743
SECTION 9.5
743
Method of Superposition
Problem 9.5-22 The frame ABCD shown in the figure is squeezed by two collinear
forces P acting at points A and D. What is the decrease d in the distance between
points A and D when the loads P are applied? (The flexural rigidity EI is constant
throughout the frame.)
Note: Disregard the effects of axial deformations and consider only the effects
of bending due to the loads P.
P
B
A
a
D
C
L
Solution 9.5-22
P
Frame ABCD
MEMBER BC:
MEMBER BA:
Table G-1, Case 4: dA Table G-2, Case 10: uB (PL)a
PLa
2EI
2EI
PL3
+ uBL
3EI
PLa
PL3
+
(L)
3EI
2EI
PL2
(2L + 3a)
6EI
DECREASE IN DISTANCE BETWEEN POINTS A AND D
d 2dA PL2
(2L + 3a)
3EI
Problem 9.5-23 A beam ABCDE has simple supports at B and D and symmetrical
overhangs at each end (see figure). The center span has length L and each overhang
has length b. A uniform load of intensity q acts on the beam.
(a) Determine the ratio b/L so that the deflection dC at the midpoint of
the beam is equal to the deflections dA and dE at the ends.
(b) For this value of b/L, what is the deflection dC at the midpoint?
Solution 9.5-23
BEAM BCD:
;
q
A
E
B
b
C
D
L
Beam with overhangs
Table G-2, Case 1 and Case 10:
qL3
qb2 L
a
b
24EI
2 2EI
qL 2
(L 6b2) (clockwise is positive)
24EI
uB b
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CHAPTER 9
dC ⫽
Page 744
Deflections of Beams
qL2
5qL4
qb2 L2
⫺
a
b ⫽
(5L2 ⫺ 24b2)
384EI
2 8EI
384EI
(downward is positive)
(1)
Rearrange and simplify the equation:
48b4 ⫹ 96b3L ⫹ 24b2L2 ⫺ 16bL3 ⫺ 5L4 ⫽ 0
or
b 3
b 2
b
b 4
48a b + 96a b + 24a b ⫺ 16 a b ⫺ 5 ⫽ 0
L
L
L
L
BEAM AB:
(a) RATIO
b
L
Solve the preceding equation numerically:
Table G-1, Case 1:
qb4
qb4
qL 2
⫺ uBb ⫽
⫺
(L ⫺ 6b2)b
dA ⫽
8EI
8EI
24EI
qb
(3b3 + 6b2L ⫺ L3)
⫽
24EI
(downward is positive)
DEFLECTION dC EQUALS DEFLECTION dA
qL2
qb
(5L2 ⫺ 24b2) ⫽
(3b3 + 6b2L ⫺ L3)
384EI
24EI
b
⫽ 0.40301
L
b
⫽ 0.4030
L
Say,
;
(b) DEFLECTION dC (EQ. 1)
qL2
(5L2 ⫺ 24b2)
384EI
qL2
[5L2 ⫺ 24 (0.40301 L)2]
⫽
384EI
qL4
⫽ 0.002870
EI
dC ⫽
(downward deflection)
;
Problem 9.5-24 A frame ABC is loaded at point C by a force P acting at an angle
a to the horizontal (see figure). Both members of the frame have the same length and
the same flexural rigidity.
Determine the angle a so that the deflection of point C is in the same direction
as the load. (Disregard the effects of axial deformations and consider only the effects
of bending due to the load P.)
Note: A direction of loading such that the resulting deflection is in the
same direction as the load is called a principal direction. For a given load
on a planar structure, there are two principal directions, perpendicular to each other.
P
L
a
B
C
L
A
Solution 9.5-24
Principal directions for a frame
P1 and P2 are the components of the load P
P1 ⫽ P cos a
P2 ⫽ P sin a
If P1 ACTS ALONE
¿
dH
⫽
P1L3
3EI
dv¿ ⫽ uBL ⫽ a
(to the right)
P1L2
P1L3
bL ⫽
2EI
2EI
(downward)
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Page 745
SECTION 9.6
¿¿
dH
⫽
If P2 ACTS ALONE
dv¿¿ ⫽
P2L3
2EI
(to the left)
P2L3
P2L3
P2L2
4P2L3
⫹ uBL ⫽
⫹a
bL ⫽
3EI
3EI
EI
3EI
(upward)
DEFLECTIONS DUE TO THE LOAD P
dH ⫽
P1L3
P2L3
L3
⫺
⫽
(2P1 ⫺ 3P2)
3EI
2EI
6EI
(to the right)
P1L3 4P2L3
L3
⫹
⫽
(⫺3P1 + 8P2)
dv ⫽ ⫺
2EI
3EI
6EI
(upward)
745
Moment-Area Method
dv
⫺3P1 ⫹8P2
⫽
dH
2P1 ⫺3P2
⫽
⫺ 3Pcosa + 8Psina
⫺3 + 8 tana
⫽
2Pcosa ⫺ 3Psina
2 ⫺ 3 tana
PRINCIPAL DIRECTIONS
The deflection of point C is in the same direction as the
load P.
‹ tan a ⫽
P2
d␯
⫽
P1
dH
or tan a ⫽
⫺ 3 + 8 tan a
2 ⫺ 3 tan a
Rearrange and simplity: tan2a ⫹ 2 tan a ⫺ 1 ⫽ 0
(quadratic equation)
Solving, tan a ⫽ ⫺1 ; 12
a ⫽ 22.5°,
112.5°,
⫺67.5°,
⫺157.5°
;
Moment-Area Method
q
The problems for Section 9.6 are to be solved by the moment-area method. All beams have
constant flexural rigidity EI.
Problem 9.6-1 A cantilever beam AB is subjected to a uniform load of intensity q
Solution 9.6-1
M
EI
B
A
acting throughout its length (see figure).
Determine the angle of rotation uB and the deflection dB at the free end.
L
Cantilever beam (uniform load)
DIAGRAM:
uB/A ⫽ uB ⫺ uA ⫽ A1 ⫽
uA ⫽ 0
uB ⫽
qL3
6EI
qL3
6EI
(clockwise)
;
DEFLECTION
ANGLE OF ROTATION
Use absolute values of areas.
qL2
qL3
1
b ⫽
Appendix D, Case 18: A1 ⫽ (L)a
3
2EI
6EI
3L
x⫽
4
Q1 ⫽ First moment of area A1 with respect to B
qL4
qL3 3L
ba b ⫽
Q1 ⫽ A1x ⫽ a
6EI
4
8EI
4
qL
(Downward)
;
dB ⫽ Q1 ⫽
8EI
(These results agree with Case 1, Table G-1.)
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1:33 PM
CHAPTER 9
Page 746
Deflections of Beams
Problem 9.6-2 The load on a cantilever beam AB has a triangular distribution
q0
with maximum intensity q0 (see figure).
Determine the angle of rotation uB and the deflection dB at the free end.
B
A
L
Solution 9.6-2
M
EI
Cantilever beam (triangular load)
x⫽
DIAGRAM
b(n + 1)
4L
⫽
n + 2
5
uB/A ⫽ uB ⫺ uA ⫽ A1 ⫽
uA ⫽ 0
uB ⫽
q0 L3
24EI
q0 L3
24EI
;
(clockwise)
DEFLECTION
ANGLE OF ROTATION
Q1 ⫽ First moment of area A1 with respect to B
Use absolute values of areas.
Q1 ⫽ A1x ⫽ a
Appendix D, Case 20:
A1 ⫽
q0 L2
q0 L3
bh
1
⫽ (L)a
b ⫽
n + 1
4
6EI
24EI
dB ⫽ Q1 ⫽
q0 L3
q0 L4
4L
ba b ⫽
24EI
5
30EI
q0 L4
30EI
;
(Downward)
(These results agree with Case 8, Table G-1.)
Problem 9.6-3 A cantilever beam AB is subjected to a concentrated load P and
P
a couple M0 acting at the free end (see figure).
Obtain formulas for the angle of rotation uB and the deflection dB at end B.
A
B
M0
L
Solution 9.6-3
M
EI
DIAGRAM
Cantilever beam (force P and couple M0)
NOTE: A1 is the M/EI diagram for M0 (rectangle). A2 is
the M/EI diagram for P (triangle).
ANGLE OF ROTATION
Use the sign conventions for the moment-area theorems
(page 713 of textbook).
A1 ⫽
M0 L
EI
x1 ⫽
L
2
A2 ⫽ ⫺
PL2
2EI
x2 ⫽
2L
3
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1:33 PM
Page 747
SECTION 9.6
A0 ⫽ A1 + A2 ⫽
M0 L
PL2
⫺
EI
2EI
uB/A ⫽ uB ⫺ uA ⫽ A0
uB ⫽ A0 ⫽
tB/A ⫽ Q ⫽ dB
uA ⫽ 0
747
Moment-Area Method
dB ⫽
M0 L2
PL3
⫺
2EI
3EI
(dB is positive when upward)
2
M0 L
PL
⫺
EI
2EI
FINAL RESULTS
(uB is positive when counterclockwise)
To match the sign conventions for uB and dB used in
Appendix G, change the signs as follows.
DEFLECTION
uB ⫽
M0 L
PL2
⫺
2EI
EI
dB ⫽
M0 L2
PL3
⫺
3EI
2EI
Q ⫽ first moment of areas A1 and A2 with respect to
point B
M0 L2
PL3
Q ⫽ A1x1 + A2x2 ⫽
⫺
2EI
3EI
(These results agree with Cases 4 and 6, Table G-1.)
q
of a cantilever beam AB with a uniform load of intensity q acting over the middle third
of the length (see figure).
A
B
L
—
3
M
EI
;
(positive downward)
Problem 9.6-4 Determine the angle of rotation uB and the deflection dB at the free end
Solution 9.6-4
;
(positive clockwise)
L
—
3
L
—
3
Cantilever beam with partial uniform load
qL3
1 L qL2
b ⫽
A3 ⫽ a b a
2 3
9EI
54EI
DIAGRAM
x3 ⫽
2 L
8L
2L
+ a b ⫽
3
3 3
9
A0 ⫽ A1 + A2 + A3 ⫽
7qL3
162EI
uB/A ⫽ uB ⫺ uA ⫽ A0
uA ⫽ 0
ANGLE OF ROTATION
Use absolute values of areas. Appendix D, Cases 1, 6,
and 18:
(clockwise)
;
DEFLECTION
Q ⫽ A1x1 + A2x2 + A3x3 ⫽
L
3 L
7L
+ a b⫽
3
4 3
12
qL2
qL3
L
b⫽
A2 ⫽ a b a
3
18EI
54EI
7qL3
162EI
Q ⫽ first moment of area A0 with respect to point B
qL2
qL3
1 L
A1 ⫽ a b a
b⫽
3 3
18EI
162EI
x1 ⫽
uB ⫽
2L
L
5L
x2 ⫽
+ ⫽
3
6
6
dB ⫽ Q ⫽
23qL4
648EI
23qL4
648EI
(Downward)
;
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1:33 PM
CHAPTER 9
Page 748
Deflections of Beams
Problem 9.6-5 Calculate the deflections dB and dC at points B and C, respectively,
of the cantilever beam ACB shown in the figure. Assume M0 ⫽ 36 k-in., P ⫽ 3.8 k,
L ⫽ 8 ft, and EI ⫽ 2.25 ⫻ 109 lb-in.2
A
M0
P
C
B
L
—
2
Solution 9.6-5
M
EI
L
—
2
Cantilever beam (force P and couple M0)
DEFLECTION dC
DIAGRAM
QC ⫽ first moment of areas A1 and left-hand part of A2
with respect to point C
⫽ a
M0 L L
PL
L L
ba ba b ⫺ a
ba ba b
EI
2
4
2EI
2
4
⫺
⫽
L L
1 PL
a
ba ba b
2 2EI
2
3
L2
(6M0 ⫺ 5PL)
48EI
NOTE: A1 is the M/EI diagram for M0 (rectangle). A2 is
the M/EI diagram for P (triangle).
tC/A ⫽ QC ⫽ dC
Use the sign conventions for the moment-area theorems
(page 713 of textbook).
(dC is positive when upward)
DEFLECTION dB
ASSUME DOWNWARD DEFLECTIONS
(change the signs of dB and dC)
QB ⫽ first moment of areas A1 and A2 with respect to
point B
M0 L 3L
ba ba b
⫽ A1x1 + A2x2 ⫽ a
EI
2
4
2L
1 PL
⫺ a b(L)a b
2 EI
3
L2
(9M0 ⫺ 8PL)
⫽
24EI
tB/A ⫽ QB ⫽ dB
L2
dB ⫽
(9M0 ⫺ 8PL)
24EI
(dB is positive when upward)
dC ⫽
L2
(6M0 ⫺ 5PL)
48EI
dB ⫽
L2
(8PL ⫺ 9M0)
24EI
;
dC ⫽
L2
(5PL ⫺ 6M0)
48EI
;
ARE POSITIVE
SUBSTITUTE NUMERICAL VALUES:
M0 ⫽ 36 k-in.
L ⫽ 8 ft ⫽ 96 in.
P ⫽ 3.8 k
EI ⫽ 2.25 ⫻ 106 k-in.2
dB ⫽ 0.4981 in. ⫺ 0.0553 in. ⫽ 0.443 in.
;
dC ⫽ 0.1556 in. ⫺ 0.0184 in. ⫽ 0.137 in.
;
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Page 749
SECTION 9.6
Problem 9.6-6 A cantilever beam ACB supports two concentrated loads P1 and P2
as shown in the figure.
Calculate the deflections dB and dC at points B and C, respectively. Assume
P1 ⫽ 10 kN, P2 ⫽ 5 kN, L ⫽ 2.6 m, E ⫽ 200 GPa, and I ⫽ 20.1 ⫻ 106 mm4.
Solution 9.6-6
M
EI
749
Moment-Area Method
A
P1
P2
C
B
L
—
2
L
—
2
Cantilever beam (forces P1 and P2)
1 P1L L L
L
1 P2L
2L
dB ⫽ a
ba ba + b + a
b(L)a b
2 2EI
2
2
3
2 EI
3
DIAGRAMS
⫽
P2L3
5P1L3
+
48EI
3EI
;
(downward)
DEFLECTION dC
dC ⫽ tC/A ⫽ QC ⫽ first moment of areas to the left
of point C with respect to point C
P2 L L L
1 P1L L L
dc ⫽ a
ba ba b + a
ba ba b
2 2EI
2
3
2EI
2
4
+
⫽
P1 ⫽ 10 kN
E ⫽ 200 GPa
P2 ⫽ 5 kN
L ⫽ 2.6 m
I ⫽ 20.1 ⫻ 106 mm4
Use absolute values of areas.
DEFLECTION dB
dB ⫽ tB/A ⫽ QB ⫽ first moment of areas with respect
1 P2L L L
a
ba ba b
2 2EI
2
3
5P2L3
P1L3
+
24EI
48EI
;
(downward)
SUBSTITUTE NUMERICAL VALUES:
dB ⫽ 4.554 mm + 7.287 mm ⫽ 11.84 mm
dC ⫽ 1.822 mm + 2.277 mm ⫽ 4.10 mm
;
;
(deflections are downward)
to point B
Problem 9.6-7 Obtain formulas for the angle of rotation uA at support A and the
deflection dmax at the midpoint for a simple beam AB with a uniform load of
intensity q (see figure).
q
A
B
L
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CHAPTER 9
Solution 9.6-7
Page 750
Deflections of Beams
Simple beam with a uniform load
DEFLECTION CURVE AND
M
EI
DIAGRAM
tB/A ⫽ BB1 ⫽ first moment of areas A1 and A2with
respect to point B
qL4
L
⫽ (A1 + A2)a b ⫽
2
24EI
uA ⫽
qL3
BB1
⫽
L
24EI
;
(clockwise)
DEFLECTION dmax AT THE MIDPOINT C
Distance CC1 ⫽
qL4
1
(BB1) ⫽
2
48EI
tC2/A ⫽ C2C1 ⫽ first moment of area A1 with respect
to point C
⫽ A1x1 ⫽ a
dmax ⫽ maximum deflection (distance CC2)
dmax ⫽ CC2 ⫽ CC1 ⫺ C2C1 ⫽
Use absolute values of areas.
ANGLE OF ROTATION AT END A
Appendix D, Case 17:
qL3
qL4
3L
ba b ⫽
24EI
16
128EI
⫽
5qL4
384EI
qL4
qL4
⫺
48EI
128EI
;
(downward)
(These results agree with Case 1 of Table G-2.)
qL3
2 L qL2
b⫽
A1 ⫽ A2 ⫽ a b a
3 2
8EI
24EI
3 L
3L
x1 ⫽ a b ⫽
8 2
16
Problem 9.6-8 A simple beam AB supports two concentrated loads P at
the positions shown in the figure. A support C at the midpoint of the beam is
positioned at distance d below the beam before the loads are applied.
Assuming that d ⫽ 10 mm, L ⫽ 6 m, E ⫽ 200 GPa, and
I ⫽ 198 ⫻ 106 mm4, calculate the magnitude of the loads P so that the beam
just touches the support at C.
Solution 9.6-8
Simple beam with two equal loads
DEFLECTION CURVE AND
M
EI
DIAGRAM
dC ⫽ deflection at the midpoint C
P
d
A
P
B
C
L
—
4
L
—
4
L
—
4
L
—
4
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Page 751
SECTION 9.6
A1 ⫽
PL2
16EI
x1 ⫽
3L
8
A2 ⫽
PL2
32EI
x2 ⫽
L
6
⫽ A1x1 + A2x2 ⫽
⫽
PL2 3L
PL2 L
a b +
a b
16EI 8
32EI 6
11PL3
384EI
Use absolute values of areas.
d ⫽ gap between the beam and the support at C
DEFLECTION dC AT MIDPOINT OF BEAM
MAGNITUDE OF LOAD TO CLOSE THE GAP
At point C, the deflection curve is horizontal.
d⫽d⫽
dC ⫽ tB/C ⫽ first moment of area between B and C
with respect to B
751
Moment-Area Method
11PL3
384EI
P⫽
384EId
;
11L3
SUBSTITUTE NUMERICAL VALUES:
d ⫽ 10 mm
L⫽6m
I ⫽ 198 ⫻ 10 mm
6
Problem 9.6-9 A simple beam AB is subjected to a load in the form
of a couple M0 acting at end B (see figure).
Determine the angles of rotation uA and uB at the supports and
the deflection d at the midpoint.
E ⫽ 200 GPa
P ⫽ 64 kN
4
;
M0
A
B
L
Solution 9.6-9
Simple beam with a couple M0
DEFLECTION CURVE AND
M
EI
DIAGRAM
ANGLE OF ROTATION uA
tB/A ⫽ BB1 ⫽ first moment of area between A and B
with respect to B
M0 L2
1 M0
L
⫽ a
b (L)a b ⫽
2 EI
3
6EI
uA ⫽
BB1
M0 L
⫽
L
6EI
(clockwise)
;
ANGLE OF ROTATION uB
tA/B ⫽ AA1 ⫽ first moment of area between A and B
with respect to A
⫽
d ⫽ deflection at the midpoint C
d ⫽ distance CC2
Use absolute values of areas.
uB ⫽
M0 L2
2L
1 M0
a
b (L)a b ⫽
2 EI
3
3EI
AA1
M0 L
⫽
L
3EI
(Counterclockwise)
;
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CHAPTER 9
Page 752
Deflections of Beams
DEFLECTION d AT THE MIDPOINT C
2
Distance CC1 ⫽
M0 L
1
(BB1) ⫽
2
12EI
tC2/A ⫽ C2C1 ⫽ first moment of area between A and C
with respect to C
d ⫽ CC1 ⫺ C2C1 ⫽
⫽
M0 L2
16EI
M0 L2
M0 L2
⫺
12EI
48EI
(Downward)
;
(These results agree with Case 7 of Table G-2.)
M0 L2
1 M0
L L
⫽ a
ba ba b ⫽
2 2EI
2
6
48EI
Problem 9.6-10 The simple beam AB shown in the figure supports two
P
P
equal concentrated loads P, one acting downward and the other upward.
Determine the angle of rotation uA at the left-hand end, the deflection
d1 under the downward load, and the deflection d2 at the midpoint
of the beam.
A
B
a
a
L
Solution 9.6-10
Simple beam with two loads
Because the beam is symmetric and the load is
antisymmetric, the deflection at the midpoint is zero.
‹ d2 ⫽ 0
;
ANGLE OF ROTATION uA AT END A
tC/A ⫽ CC1 ⫽ first moment of area between A and C
with respect to C
⫽ A1 a
⫽
uA ⫽
a
L
2 L
⫺ a + b + A2 a b a ⫺ ab
2
3
3
2
Pa(L ⫺ a)(L ⫺ 2a)
12EI
Pa(L ⫺ a)(L ⫺ 2a)
CC1
⫽
L/2
6LEI
(clockwise)
DEFLECTION d1 UNDER THE DOWNWARD LOAD
Distance
DD1 ⫽ a
⫽
Pa(L ⫺ 2a)
M1
⫽
EI
LEI
Pa2(L ⫺ 2a)
1 M1
A1 ⫽ a
b (a) ⫽
2 EI
2LEI
Pa(L ⫺ 2a)2
1 M1 L
A2 ⫽ a
b a ⫺ ab ⫽
2 EI
2
4LEI
a
b(CC1)
L/2
Pa2(L ⫺ a)(L ⫺ 2a)
6LEI
tD2/A ⫽ D2D1 ⫽ first moment of area between A
and D with respect to D
Pa3(L ⫺ 2a)
a
⫽ A1 a b ⫽
3
6LEI
d1 ⫽ DD1 ⫺ D2D1
⫽
Pa2(L ⫺ 2a)2
6LEI
(Downward)
;
;
09Ch09.qxd
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Page 753
SECTION 9.6
Moment-Area Method
Problem 9.6-11 A simple beam AB is subjected to couples M0 and 2M0 as
M0
shown in the figure. Determine the angles of rotation uA and uB at the beam
and the deflection d at point D where the load M0 is applied.
2M0
A
B
D
L
—
3
Solution 9.6-11
753
E
L
—
3
L
—
3
Simple beam with two couples
DEFLECTION CURVE AND
M
EI
ANGLE OF ROTATION uB AT END B
DIAGRAM
tA/B ⫽ AA1 ⫽ first moment of area between A and B
with respect to A
⫽ A1 a
uB ⫽
2L
L
2L
L
2L
b + A2 a +
b + A3 a
+ b ⫽0
9
3
9
3
9
AA1
⫽0
L
;
DEFLECTION d AT POINT D
M0 L2
1
Distance DD1 ⫽ (BB1) ⫽
3
18EI
tD2/A ⫽ D2D1 ⫽ first moment of area between A
and D with respect to D
M0L2
L
⫽ A1 a b ⫽
9
54EI
d ⫽ DD1 ⫺ D2D1 ⫽
(downward)
M0 L
1 M0 L
A1 ⫽ A2 ⫽ a
ba b ⫽
2 EI
3
6EI
A3 ⫽ ⫺
M0 L
6EI
ANGLE OF ROTATION uA AT END A
tB/A ⫽ BB1 ⫽ first moment of area between A and B
with respect to B
⫽ A1 a
⫽
uA ⫽
L
L
2L
L
2L
+ b + A2 a + b ⫹ A3 a b
3
9
3
9
9
M0 L2
6EI
BB1
M0 L
⫽
L
6EI
(clockwise)
;
M0 L2
27EI
;
NOTE: This deflection is also the maximum deflection.
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1:35 PM
CHAPTER 9
Page 754
Deflections of Beams
Nonprismatic Beams
Problem 9.7-1 The cantilever beam ACB shown in the figure has moments of
A
inertia I2 and I1 in parts AC and CB, respectively.
(a) Using the method of superposition, determine the deflection dB at the free
end due to the load P.
(b) Determine the ratio r of the deflection dB to the deflection d1 at the free end
of a prismatic cantilever with moment of inertia I1 carrying the same load.
(c) Plot a graph of the deflection ratio r versus the ratio I2/I1 of the moments of
inertia. (Let I2 /I1 vary from 1 to 5.)
Solution 9.7-1
I2
L
—
2
P
C
I1
B
L
—
2
Cantilever beam (nonprismatic)
Use the method of superposition.
(3) Total deflection at point B
(a) DEFLECTION dB AT THE FREE END
dB ⫽ (dB)1 + (dB)2 ⫽
(1) Part CB of the beam:
(b) PRISMATIC BEAM d1 ⫽
(dB)1 ⫽
P L 3
PL3
a b ⫽
3EI1 2
24EI1
(2) Part AC of the beam:
3
Ratio: r ⫽
7I1
PL3
a1 +
b
24EI1
I2
;
PL3
3 EI1
dB
7I1
1
⫽ a1 +
b
d1
8
I2
;
(c) GRAPH OF RATIO
2
3
dC ⫽
P(L/2)
(PL/2)(L/2)
5PL
+
⫽
3EI2
2EI2
48EI2
uC ⫽
P(L/2)2
(PL/2)(L/2)
3PL2
+
⫽
2EI2
EI2
8EI2
L
7PL3
(dB)2 ⫽ dC + uC a b ⫽
2
24EI2
I2
I1
r
1
2
3
4
5
1.00
0.56
0.42
0.34
0.30
09Ch09.qxd
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Page 755
SECTION 9.7
755
Nonprismatic Beams
Problem 9.7-2 The cantilever beam ACB shown in the figure supports a uniform
q
load of intensity q throughout its length. The beam has moments of inertia I2 and I1
in parts AC and CB, respectively.
(a) Using the method of superposition, determine the deflection dB at the free
end due to the uniform load.
(b) Determine the ratio r of the deflection dB to the deflection d1 at the free end
of a prismatic cantilever with moment of inertia I1 carrying the same load.
(c) Plot a graph of the deflection ratio r versus the ratio I2/I1 of the moments of
inertia. (Let I2 /I1 vary from 1 to 5.)
Solution 9.7-2
A
I2
L
—
2
dB ⫽ (dB)1 + (dB)2 ⫽
(1) Part CB of the beam:
(b) PRISMATIC BEAM d1 ⫽
(dB)1 ⫽
q L 4
qL4
a b ⫽
8EI1 2
128EI1
Ratio: r ⫽
(c) GRAPH OF RATIO
4
q(L/2)
+
8EI2
a
qL
b(L/2)3
2
3EI2
2
qL
L 2
ba b
8
2
+
2EI2
17qL4
⫽
384EI2
3
⫽
L
—
2
q(L/2)
(qL/2)(L/2)2
(qL2/8)(L/2)
+
+
6EI2
2EI2
EI2
7qL3
48EI2
15qL4
L
(dB)2 ⫽ dC + uC a b ⫽
2
128EI2
qL4
15I1
a1 +
b
128EI1
I2
qL4
8EI1
dB
15I1
1
⫽ a1 +
b
d1
16
I2
(2) Part AC of the beam:
uC ⫽
I1
(3) Total deflection at point B
(a) DEFLECTION dB AT THE FREE END
a
B
Cantilever beam (nonprismatic)
Use the method of superposition
dC ⫽
C
I2
I1
r
1
2
3
4
5
1.00
0.53
0.38
0.30
0.25
;
;
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1:35 PM
CHAPTER 9
Page 756
Deflections of Beams
*Problem 9.7-3 Beam ACB hangs from two springs, as shown in the
RA = k1dA
figure. The springs have stiffnesses k1 and k2 and the beam has flexural
rigidity EI.
(a) What is the downward displacement of point C, which is at the
midpoint of the beam, when the moment M0 is applied? Data for
the structure are as follows: M0 ⫽ 7.5 k-ft, L ⫽ 6 ft, EI ⫽ 520 k-ft2,
k1 ⫽ 17 k/ft, and k2 ⫽ 11 k/ft.
(b) Repeat (a) but remove M0 and, instead, apply uniform load q
over the entire beam.
k1
2EI
RB = k2dB
M0
k2
EI
B
A
L/2
C
L/2
B
q = 250 lb/ft (for Part (b) only)
*Solution 9.7-3
M0 ⫽ 7.5 kip/ft
L ⫽ 6 ft
EI ⫽ 520 kip/ft2
k1 ⫽ 17 kip/ft
k2 ⫽ 11 kip/ft
q ⫽ 250 lb/ft
(a) BENDING-MOMENT EQUATIONS-MOMENT M0 AT C
2EI␯– ⫽ M ⫽
M0x
L
a0 … x …
2EI␯¿ ⫽
M0x2
⫹C1
2L
2EI␯ ⫽
M0x3
+ C1x + C 2
6L
B.C.
␯(0) ⫽ 0
a0 … x …
M0 ax ⫺
EI␯¿ ⫽ ⫺ M0x +
EI␯ ⫽ ⫺
B.C.
B.C.
B.C.
2EI␯ ⫽
L
b
2
L
2
L
b
2
a0 … x …
C2 ⫽ 0
M0
+
EI␯– ⫽ ⫺
2
L
b
2
M0x
+ C3
2L
⫺
M0x3
+ C 1x
6L
⫽ ⫺M0 +
a
M 0x
L
a
a0 … x …
L
b
2
L
… x … Lb
2
L
… x … Lb
2
M0x2
M0x3
+
+ C3x + C4
2
6L
␯(L) ⫽ 0
L
b
2
a
L
… x … Lb
2
M0L3
M0L2
+
+ C3L + C4 ⫽ 0
2
6L
L
L
␯¿L a b ⫽ ␯¿R a b
2
2
L
L
␯L a b ⫽ ␯R a b
2
2
L 2
L 2
M0 a b
M0 a b
2
2
1
L
J
+ C1 K ⫽ ⫺M0 +
+ C3
2
2L
2
2L
L 3
L 2
L 3
M0 a b
M0 a b
M0 a b
2
2
2
1
L
L
J
+ C1 K ⫽ ⫺
+
+ C3 + C4
2
6L
2
2
6L
2
(1)
(2)
(3)
09Ch09.qxd
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1:35 PM
Page 757
SECTION 9.7 Nonprismatic Beams
From (1), (2), and (3)
C1 ⫽ 0
C3 ⫽
7
ML
16 0
⫺5
M L2
48 0
C4 ⫽
;
Therefore
␯(x) ⫽
M0x3
12EIL
␯(x) ⫽
M0
(⫺ 24x2L + 8x3 + 21L2x ⫺ 5L3)
48EIL
a0 … x …
L
b
2
a
L
… x … Lb
2
DEFLECTION AT A AND B
RA ⫽
M0
L
dA ⫽
RA
k1
RB ⫽ ⫺
M0
L
RB
k2
dB ⫽
dA ⫽ 0.88 in.
dB ⫽ ⫺1.36 in.
Downward
Upward
DEFLECTION AT POINT C
L
1
dC ⫽ ⫺␯ a b + (dA + dB)
2
2
dC ⫽ ⫺
L 3
M0 a b
2
+
12EIL
1
(d + dB)
2 A
dC ⫽ ⫺ 0.31 in. Upward
;
(b) BENDING-MOMENT EQUATIONS-UNIFORM LOAD q
qLx
qx2
L
2EI␯– ⫽ M ⫽
⫺
a0 … x … b
2
2
2
2
3
qLx
qx
L
2EI␯¿ ⫽
⫺
+ C1 a 0 … x … b
4
6
2
2EI␯ ⫽
B.C.
qLx3
qx4
⫺
+ C 1x + C 2
12
24
␯(0) ⫽ 0
C2 ⫽ 0
EI␯– ⫽
qLx
qx2
⫺
2
2
EI␯¿ ⫽
qLx2
qx3
⫺
⫹ C3
4
6
EI␯ ⫽
a
a0 … x …
2EI␯ ⫽
L
b
2
qLx3
qx4
⫺
+ C1x
12
24
L
… x … Lb
2
a
L
… x … Lb
2
qx4
qLx3
⫺
+ C3x + C4
12
24
a
L
… x … Lb
2
a0 … x …
L
b
2
757
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Page 758
CHAPTER 9 Deflections of Beams
B.C.
B.C.
B.C.
␯(L) ⫽ 0
qLL3
qL4
+ C3L + C4 ⫽ 0
⫺
12
24
L
L
␯¿L a b ⫽ ␯¿R a b
2
2
L
L
vL a b ⫽ ␯R a b
2
2
⫽
L 3
qL a b
2
12
(1)
L 2
L 3
L 3
L 2
qL a b
qa b
qa b
qLa b
2
2
2
2
1
J
⫺
+ C1 K ⫽
⫺
+ C3
2
4
6
4
6
1
J
2
⫺
L 3
qLa b
2
12
L 4
qa b
2
24
⫺
+ C3
L 4
qa b
2
24
L
+ C1 K
2
L
+ C4
2
(3)
From (1), (2), and (3)
⫺7 3
qL
128
C1 ⫽
C3 ⫽
⫺37 3
qL
768
C4 ⫽
5
qL4
768
Therefore
␯(x) ⫽ ⫺
␯(x) ⫽
qx
(⫺ 32Lx2 + 16x3 + 21L3)
768EI
a0 … x …
q
(64Lx3 ⫺ 32x4 ⫺ 37L3x + 5L4)
768EI
a
L
b
2
L
… x … Lb
2
DEFLECTION AT A AND B
RA ⫽
dA ⫽
qL
2
RB ⫽
qL
2
RA
k1
dA ⫽ 0.53 in.
dB ⫽
Downward
RB
k2
dB ⫽ 0.82 in.
Downward
DEFLECTION AT POINT C
L
1
dC ⫽ ⫺␯ a b + (dA + dB)
2
2
L
2
L 2
L 3
1
dC ⫽
c ⫺ 32La b + 16 a b + 21L3 d + (dA + dB)
768EI
2
2
2
q
dC ⫽ 0.75 in. Downward
;
(2)
09Ch09.qxd
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1:35 PM
Page 759
SECTION 9.7
759
Nonprismatic Beams
Problem 9.7-4 A simple beam ABCD has moment of inertia I near the supports
q
and moment of inertia 2I in the middle region, as shown in the figure. A uniform
load of intensity q acts over the entire length of the beam.
Determine the equations of the deflection curve for the left-hand half of the
beam. Also, find the angle of rotation uA at the left-hand support and the deflection
dmax at the midpoint.
A
B
C
I
D
I
2I
L
—
4
L
—
4
L
Solution 9.7-4
Simple beam (nonprismatic)
Use the bending-moment equation (Eq. 9-12a).
B.C.
L
1 Symmetry: ␯¿ a b ⫽ 0
2
REACTIONS, BENDING MOMENT, AND DEFLECTION CURVE
From Eq. (4): C2 ⫽ ⫺
2EI␯¿ ⫽
qL3
24
qL x2 qx3 qL3
⫺
⫺
4
6
24
a
L
L
… x … b
4
2
(5)
SLOPE AT POINT B (FROM THE RIGHT)
RA ⫽ RB ⫽
qL
2
2
M ⫽ Rx ⫺
Substitute x ⫽
2
qx
qLx
qx
⫽
⫺
2
2
2
EI␯¿B ⫽ ⫺
B.C.
L
into Eq. (5):
4
11qL3
768
(6)
2 CONTINUITY OF SLOPES AT POINT B
(vB¿ )Left ⫽ (vB¿ )Right
From Eqs. (3) and (6):
BENDING-MOMENT
EQUATIONS FOR THE LEFT-HAND HALF
11qL3
qL L 2 q L 3
a b ⫺ a b + C1 ⫽⫺
4 4
6 4
768
‹C1 ⫽ ⫺
7qL3
256
OF THE BEAM
EI␯– ⫽ M ⫽
qx2
qLx
⫺
2
2
a0 … x …
2
qLx qx
⫺
E(2I )␯– ⫽ M ⫽
2
2
L
b
4
L
L
a … x … b
4
2
(1)
qL x2 qx3
⫺
+ C1
4
6
2
2EI␯¿ ⫽
3
qL x
qx
⫺
+ C2
4
6
a0 … x …
a
L
b
4
L
L
… x … b
4
2
EI␯¿ ⫽
7qL3
qL x2 qx3
⫺
⫺
4
6
256
EI␯¿ ⫽
qL3
qL x2 qx3
⫺
⫺
8
12
48
(2)
INTEGRATE EACH EQUATION
EI␯¿ ⫽
SLOPE OF THE BEAM (FROM EQS. 3 AND 5)
(3)
a
L
b
4
(7)
L
L
… x … b
4
3
(8)
ANGLE OF ROTATION uA (FROM EQ. 7)
uA ⫽ ⫺v¿(0) ⫽
(4)
a0 … x …
7qL3
(positive clockwise)
256EI
;
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CHAPTER 9
Page 760
Deflections of Beams
INTEGRATE EQS. (7) AND (8)
3
4
3
EI␯ ⫽
7qL x
qL x qx
⫺
⫺
+ C3
12
24
256
EI␯ ⫽
qL x3 qx4 qL3x
⫺
⫺
+ C4
24
48
48
B.C.
DEFECTION OF THE BEAM (FROM EQS. 9 AND 10)
a0 … x …
a
L
b
4
(9)
␯⫽ ⫺
qx
121L3 ⫺ 64Lx2 + 32x32
768EI
a0 … x …
L
L
… x … b (10)
4
2
␯⫽ ⫺
3 Deflection at support A
a
DEFECTION AT POINT B (FROM THE LEFT)
L
into Eq. (9) with C3 ⫽ 0
4
35 qL4
EI␯B ⫽ ⫺
6144
B.C.
;
q
(13L4 + 256L3x ⫺ 512Lx3 + 256x4)
12,288EI
n(0) ⫽ 0 From Eq. (9): C3 ⫽ 0
Substitute x ⫽
L
b
4
L
L
… x … b
4
2
;
MAXIMUM DEFLECTION (AT THE MIDPOINT E)
(From the preceding equation for v.)
(11)
dmax ⫽ ⫺va
31qL4
L
b ⫽
2
4096EI
(positive downward)
;
4 Continuity of deflections at point B
(nB)Right ⫽ (nB)Left
From Eqs. (10) and (11):
q L 4
qL3 L
35qL4
qL L 3
a b ⫺
a b ⫺
a b + C4 ⫽ ⫺
24 4
48 4
48 4
6144
‹C4 ⫽ ⫺
13qL4
12,288
Problem 9.7-5 A beam ABC has a rigid segment from A to B and a flexible segment
with moment of inertia I from B to C (see figure). A concentrated load P acts at point B.
Determine the angle of rotation uA of the rigid segment, the deflection dB at point B,
and the maximum deflection dmax.
Rigid
P
I
A
B
L
—
3
2L
—
3
C
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Page 761
761
SECTION 9.7 Nonprismatic Beams
Solution 9.7-5 Simple beam with a rigid segment
EI␯ ⫽
‹ dB ⫽
uA ⫽
3dB x
L
a0 … x …
L
b
3
(1)
3dB
L
a0 … x …
L
b
3
(2)
␯¿ ⫽ ⫺
PL2
+ 3EIdB
54
8PL3
729EI
dB
8PL2
⫽
L/3
243EI
EI␯¿
B.C.
(3)
Substitute for dB in Eq. (5) and simplify:
␯⫽
P
(7L3 ⫺ 61L2x + 81L x2 ⫺ 27x3)
486EI
1 At x ⫽ L/3,
‹ C1 ⫽ ⫺
a
L
… x … Lb
3
(7)
MAXIMUM DEFLECTION
v¿ ⫽ 0 gives x1 ⫽
3EIdB
Px2
5PL2
PLx
⫺
⫺
⫺
EI␯¿ ⫽
3
6
54
L
a
EI␯ ⫽
(6)
P
(⫺ 61L2 + 162Lx ⫺ 81x2)
486EI
3dB
␯¿ ⫽ ⫺
L
3EIdB
5PL2
⫺
54
L
L
… x … Lb
3
Also,
␯¿ ⫽
Px2
PLx
⫽
⫺
+ C1
3
6
(5)
;
a
Px
PL
⫺
3
3
L
… x … Lb
3
;
FROM B TO C
EI␯– ⫽ M ⫽
a
L
3 At x ⫽ , (nB)Left ⫽ (nB)Right (Eqs. 1 and 5)
3
B.C.
␯⫽ ⫺
PL3
+ 3EIdB
54
3EIdBx
Px3
5PL2x
PLx2
⫺
⫺
⫺
6
18
54
L
⫺
FROM A TO B
‹ C2 ⫽ ⫺
2 n(L) ⫽ 0
B.C.
L
… x … Lb
3
3EIdB x
Px3
5PL2x
PLx2
⫺
⫺
⫺
+ C2
6
18
54
L
a
L
… x … Lb
3
L
(9 ⫺ 225) ⫽ 0.5031L
9
Substitute x1 in Eq. (6) and simplify:
(4)
␯max ⫽ ⫺
4015PL3
6561EI
dmax ⫽ ⫺␯max ⫽
PL3
4015PL3
⫽ 0.01363
6561EI
EI
;
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CHAPTER 9
Page 762
Deflections of Beams
Problem 9.7-6 A simple beam ABC has moment of inertia 1.5I from A to B and I from
P
B to C (see figure). A concentrated load P acts at point B.
Obtain the equations of the deflection curves for both parts of the beam. From the
equations, determine the angles of rotation uA and uC at the supports and the deflection
dB at point B.
Solution 9.7-6
1.5I
EI␯ ⫽
DEFLECTION CURVE
B.C.
L
—
3
2L
—
3
PLx2
Px3
⫺
+ C2x + C4
6
18
B.C.
B.C.
2Px
3I
b␯– ⫽ M ⫽
2
3
a0 … x …
Px
PL
⫺
EI␯– ⫽ M ⫽
3
3
4Px2
+ C1
18
EI␯¿ ⫽
Px2
PLx
⫺
+ C2
3
2
(8)
C4 ⫽ ⫺
PL3
⫺ C2L
9
L
a … x … Lb
3
4 Continuity of deflections at point B
(1)
From Eqs. (6), (8), and (7):
(2)
4P L 3
PL L 2
L
P L 3
a b + C1 a b ⫽
a b ⫺
a b
54 3
3
6 3
18 3
L
+ C2 a b + C4
3
a0 … x …
a
L
b
3
L
… x … Lb
3
10PL3
+ C2L + 3C4
243
(3)
C1L ⫽
(4)
SOLVE EQS. (5), (8), (9), AND (10)
C1 ⫽ ⫺
(n⬘B)Left ⫽ (n⬘B)Right
38PL2
729
C2 ⫽ ⫺
175PL2
1458
(10)
C3 ⫽ 0
13PL3
1458
From Eqs. (3) and (4):
C4 ⫽
PL L
P L 2
4P L 2
a b + C1 ⫽
a b ⫺ a b + C2
18 3
3 3
6 3
SLOPES OF THE BEAM (FROM EQS. 3 AND 4)
11PL2
C2 ⫽ C1 ⫺
162
(5)
INTEGRATE EQS. (3) AND (4)
4Px3
+ C1x + C3
54
(9)
(nB)Left ⫽ (nB)Right
L
b
3
1 Continuity of slopes at point B
EI␯ ⫽
(7)
3 Deflection at support C
INTEGRATE EACH EQUATION
EI␯¿ ⫽
L
… x … Lb
3
C3 ⫽ 0
From Eq. (6):
v(L) ⫽ 0 From Eq. (7):
BENDING-MOMENT EQUATIONS
a
2 Deflection at support A
n (0) ⫽ 0
B.C.
C
B
Simple beam (nonprismatic)
Use the bending-moment equation (Eq. 9-12a).
Ea
I
A
a0 … x …
L
b
3
(6)
␯¿ ⫽ ⫺
2P
L
(19L2 ⫺ 81x2) a 0 … x … b
729EI
3
(11)
09Ch09.qxd
9/27/08
1:35 PM
Page 763
SECTION 9.7
␯¿ ⫽ ⫺
DEFLECTIONS OF THE BEAM
P
(175L2 ⫺ 486Lx + 243x2)
1458EI
a
Substitute C1, C2, C3, and C4 into Eqs. (6) and (7):
L
… x … Lb
3
(12)
ANGLE OF ROTATION uA (FROM EQ. 11)
uA ⫽ ⫺␯¿(0) ⫽
38PL2
729EI
(positive counterclockwise)
a0 … x …
2Px
(19L2 ⫺ 27x2)
729EI
␯⫽ ⫺
P
(⫺ 13L3 + 175L2x ⫺ 243Lx2 + 81x3)
1458EI
a
;
L
b
3
␯⫽ ⫺
;
(positive clockwise)
ANGLE OF ROTATION uC (FROM EQ. 12)
34PL2
uC ⫽ ␯¿(L) ⫽
729EI
763
Nonprismatic Beams
DEFLECTION AT POINT B a x ⫽
L
32PL3
dB ⫽⫺␯a b ⫽
3
2187EI
;
L
… x … Lb
3
;
L
b
3
;
(positive downward)
Problem 9.7-7 The tapered cantilever beam AB shown in the
figure has thin-walled, hollow circular cross sections of constant
thickness t. The diameters at the ends A and B are dA and
dB ⫽ 2dA, respectively. Thus, the diameter d and moment of
inertia I at distance x from the free end are, respectively,
dA
d⫽
(L + x)
L
I⫽
P
B
A
dA
t
dB = 2dA
d
x
ptd3A
IA
ptd3
⫽
(L + x)3 ⫽ 3 (L + x)3
3
8
8L
L
L
in which IA is the moment of inertia at end A of the beam.
Determine the equation of the deflection curve and the
deflection dA at the free end of the beam due to the load P.
Solution 9.7-7
M ⫽ ⫺Px
Tapered cantilever beam
EI␯– ⫽ ⫺ Px
I⫽
IA
3
L
(L + x)3
␯¿ ⫽
Px
PL3
x
␯– ⫽ ⫺
⫽ ⫺
c
d
EI
EIA (L + x)3
(1)
From Appendix C:
␯¿ ⫽
B.C.
xdx
L (L + x)
3
⫽⫺
3
PL
L + 2x
c
d + C1
EIA 2(L + x)2
1 ␯¿ (L) ⫽ 0
‹ C1 ⫽ ⫺
or
␯¿ ⫽
INTEGRATE EQ. (1)
L + 2x
PL3 L + 2x
3PL2
c
d ⫺
2
EIA 2(L + x)
8EIA
PL3
L
PL3
x
3PL2
c
d +
c
d ⫺
2
2
EIA 2(L + x)
EIA (L + x)
8EIA
INTEGRATE EQ. (2)
2
2(L + x)
From Appendix C:
dx
L (L + x)
2
2
3PL
8EIA
xdx
L (L + x)2
⫽ ⫺
⫽
1
L + x
L
+ ln (L + x)
L + x
(2)
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764
1:35 PM
CHAPTER 9
Deflections of Beams
⫺
Substitute C2 into Eq. (3).
2
3PL
x + C2
8EIA
3x
1
L + x
L
PL3
⫺
+ + lna
bd
c
EIA 2(L + x)
8L
8
2L
␯⫽
L
3x
PL3
c
+ ln (L + x) ⫺
d + C2
EIA 2(L + x)
8L
B.C.
DEFLECTION OF THE BEAM
PL3 L
1
PL3
L
a b a⫺
b+
c
+ ln (L + x)d
EIA 2
L + x
EIA L + x
␯⫽
⫽
Page 764
(3)
PL3 1
‹ C2 ⫽
c ⫺ ln (2L) d
EIA 8
2 ␯ (L) ⫽ 0
;
DEFLECTION dA AT END A OF THE BEAM
dA ⫽ ⫺␯(0) ⫽
⫽ 0.06815
Note: ln
PL3
(8 ln 2 ⫺ 5)
8EIA
PL3
(positive downward)
EIA
;
1
⫽ ⫺ln 2
2
Problem 9.7-8 The tapered cantilever beam AB shown in the
figure has a solid circular cross section. The diameters at the ends
A and B are dA and dB ⫽ 2dA, respectively. Thus, the diameter d
and moment of inertia I at distance x from the free end are,
respectively,
P
dB = 2dA
dA
dA
d⫽
(L + x)
L
pdA4
(L
4
4
I⫽
pd
⫽
64
64L
+ x)4 ⫽
B
A
x
d
L
IA
L4
(L + x)4
in which IA is the moment of inertia at end A of the beam.
Determine the equation of the deflection curve and the
deflection dA at the free end of the beam due to the load P.
Solution 9.7-8
M ⫽ ⫺Px
Tapered cantilever beam
EI␯– ⫽ ⫺Px
I⫽
IA
(L + x)4
B.C.
4
L
4
␯– ⫽ ⫺
Px
PL
x
⫽ ⫺
c
d
EI
EIA (L + x)4
(1)
␯¿ ⫽
‹ C1 ⫽ ⫺
PL2
12EIA
PL2
PL4 L⫹ 3x
c
d
⫺
EIA 6(L ⫹x)3
12EIA
or
INTEGRATE EQ. (1)
xdx
L + 3x
˚L (L + x) ⫽ ⫺ 6(L + x)
From A ppendix C:
␯¿ ⫽
1 n⬘(L) ⫽ 0
PL4 L⫹3x
c
d ⫹ C1
EIA 6(L⫹ x)3
4
3
␯¿ ⫽
L
PL4
x
PL4
c
d
⫹
c
d
3
EIA 6(L ⫹x)
EIA 2(L ⫹x)3
⫺
PL2
12EIA
(2)
09Ch09.qxd
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1:36 PM
Page 765
SECTION 9.7
INTEGRATE EQ. (2)
dx
From Appendix C:
L(L + x)
3
xdx
L(L + x)3
␯⫽
⫽ ⫺
⫽
B.C.
1
‹ C2 ⫽
PL3 7
a b
EIA 24
2(L + x) 2
⫺(L + 2x)
DEFLECTION OF THE BEAM
2(L + x)2
Substitute C2 into Eq. (3)
PL4 L
1
1 2 PL4 1
L + 2x
a b a⫺ b a
b +
a b c⫺
d
EIA 6
2 L+x
EIA 2
2(L + x)2
⫺
2 n(L) ⫽ 0
765
Nonprismatic Beams
PL2
x + C2
12EIA
4L (2L + 3x)
2x
PL3
c7 ⫺
⫺
d
24EIA
L
(L + x)2
␯⫽
;
DEFLECTION dA AT END A OF THE BEAM
L(L + 2x)
PL3
L2
x
⫽
c⫺
⫺
⫺
d + C2
2
2
EIA
12L
12(L + x)
4(L + x)
(3)
dA ⫽ ⫺␯(0) ⫽
PL3
24EIA
(positive downward)
;
Problem 9.7-9 A tapered cantilever beam AB supports a
concentrated load P at the free end (see figure). The cross
sections of the beam are rectangular with constant width b,
depth dA at support A, and depth dB ⫽ 3dA/2 at the support.
Thus, the depth d and moment of inertia I at distance x from the
free end are, respectively,
dA
d⫽
(2L + x)
2L
I⫽
P
B
A
3dA
dB = —
2
dA
x
IA
bd A3
bd 3
(2L + x)3 ⫽ 3 (2L + x) 3
⫽
3
12
96L
8L
d
b
L
in which IA is the moment of inertia at end A of the beam.
Determine the equation of the deflection curve and the
deflection dA at the free end of the beam due to the load P.
Solution 9.7-9
M ⫽ ⫺P x
␯– ⫽ ⫺
Tapered cantilever beam
EI␯– ⫽ ⫺P x
3
I⫽
IA
3
8L
(2L + x) 3
x
Px
8PL
⫽⫺
c
d
EI
EIA (2L + x)3
(1)
1 n⬘(L) ⫽ 0
␯¿ ⫽
‹ C1 ⫽ ⫺
16PL2
9EIA
16PL2
8PL3 L⫹ x
c
d
⫺
EIA (2L ⫹x)2
9EIA
or
INTEGRATE EQ. (1)
From Appendix C:
B.C.
xdx
L(2L + x)3
8PL3 L⫹x
d ⫹ C1
␯¿ ⫽ EI c
3
A (2L⫹x)
⫽ ⫺
2L + 2x
2(2L + x)2
L
8PL3
x
8PL3
d
⫹
c
d
␯¿ ⫽ EI c
2
EIA (2L ⫹x)2
A (2L ⫹x)
⫺
16PL2
9EIA
(2)
09Ch09.qxd
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CHAPTER 9
Page 766
Deflections of Beams
DEFLECTION OF THE BEAM
INTEGRATE EQ. (2)
dx
From Appendix C:
L (2L + x)
2
xdx
⫽
L(2L + x)
2
3
⫽ ⫺
Substitute C2 into Eq. (3).
1
2L + x
2L
+ ln (2L + x)
2L + x
3
8PL
L
8PL
2L
␯⫽
a⫺
b +
c
EIA
2L + x
EIA 2L + x
⫽
B.C.
+ lna
2L + x
bd
3L
dA ⫽ ⫺␯(0) ⫽
16x
PL3 8L
c
+ 8 ln(2L + x) ⫺
d + C2
EIA 2L + x
9L
(3)
8PL3 1
‹ C2 ⫽ ⫺
c + ln (3L)d
EIA 9
2 n(L) ⫽ 0
2x
1
L
8PL3
⫺
⫺
c
EIA 2L + x
9L
9
;
DEFLECTION dA AT END A OF THE BEAM
16PL2
x + C2
9EIA
+ ln (2L + x) d ⫺
␯⫽
⫽ 0.1326
NOTE: ln
8PL2
3
7
clna b ⫺ d
EIA
2
18
PL3
EIA
(positive downward)
3
2
⫽ ⫺ln
3
2
Problem 9.7-10 A tapered cantilever beam AB supports a concentrated
load P at the free end (see figure). The cross sections of the beam are
rectangular tubes with constant width b and outer tube depth dA at A, and
outer tube depth dB ⫽ 3dA/2 at support B. The tube thickness is constant,
t ⫽ dA/20. IA is the moment of inertia of the outer tube at end A of the beam.
If the moment of inertia of the tube is approximated as Ia(x) as
defined, find the equation of the deflection curve and the deflection dA at
the free end of the beam due to the load P.
(
10x
3
Ia(x) = IA 4 + 27 L
Solution 9.7-10
EI␯ – ⫽ M ⫽ ⫺Px
␯– ⫽
⫺Px
⫽
EIa (x)
⫺ Px
EIA a
From Appendix C:
⫺P
␯¿ ⫽
≥
EIA
␯¿ ⫽
3
10x
+
b
4 27L
3
x
L (a + bx)3
⫽
dx ⫽ ⫺
3
10
⫹2
x
4
27L
2a
⫺P
EIA
2
10 2 3 10
b a ⫹
xb
27L
4 27L
x
a
3 10x 3
+
b
4 27L
a + 2bx
2b 2 (a + bx)2
¥ ⫹C1
PL3 19683
81L
80x
c
⫹
d ⫹C1
EIA 50 (81L⫹40x)2 (81L ⫹40x)2
3
(
IA =
bd 3
12
d
b
P
dA
dB = 3dA/2
x
BENDING-MOMENT EQUATIONS
;
L
t
09Ch09.qxd
9/27/08
1:36 PM
Page 767
SECTION 9.7
From Appendix C:
1
L (a + bx)2
x
L (a + bx)
2
dx ⫽
dx ⫽
Nonprismatic Beams
⫺1
b(a + bx)
1
b
2
a
a
+ ln(a + bx)b
a + bx
␯⫽
PL 19683
⫺81L
80
81L
c
⫹
a
+ ln (81L + 40x)bd + C1x + C2
EIA 50 40(81L + 40x) 40 2 81L + 40x
␯⫽
19683PL3 81L + 162ln (81L + 40x) L + 80ln (81L + 40x) x
a
b + C1x + C2
2000EIA
81L + 40 x
B.C.
␯¿(L) ⫽ 0
C1 ⫽
⫺ 3168963 PL2
732050 EIA
B.C.
␯(L) ⫽ 0
C2 ⫽
⫺ 19683PL3
(3361+ 29282 ln (121L))
29282000EIA
3
␯(x) ⫽
19683PL3
40x
81L
81
6440x
3361
a
+ 2 ln a
+
b⫺
⫺
b
2000EIA 81L + 40x
121
121L
14641L 14641
dA ⫽ ⫺␯(0) ⫽
PL3
19683PL3
11
a⫺2820 + 14641 lna bb ⫽ 0.317
7320500EIA
9
EIA
;
;
P
**Problem 9.7-11 Repeat Problem 9.7-10 but now use the tapered propped
cantilever tube AB, with guided support at B, shown in the figure which
supports a concentrated load P at the guided end.
Find the equation of the deflection curve and the deflection dB at the
guided end of the beam due to the load P.
dA
dB = 3dA/2
x
Solution 9.7-11
BENDING-MOMENT EQUATIONS
EI␯– ⫽ M ⫽ Px
␯– ⫽
Px
⫽
EIa(x)
Px
EIA a
From Appendix C:
␯¿ ⫽
P
≥
EIA
⫽
3
3 10x
+
b
4 27L
x
L (a + bx)3
P
EIA
dx ⫽ ⫺
3
10
⫹2
x
4
27L
10 2 3
10 2
2a
xb
b a ⫹
27L
4 27L
¥ ⫹ C1
x
a
3 10x 3
+
b
4 27L
a + 2bx
2b 2(a + bx)2
L
767
09Ch09.qxd
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768
1:36 PM
CHAPTER 9
␯¿ ⫽ ⫺
Page 768
Deflections of Beams
80x
PL3 19683
81L
⫹
c
d ⫹C1
EIA 50
(81L⫹40x)2 (81L⫹40x)2
From Appendix C:
1
2
dx ⫽
2
dx ⫽
L (a + bx)
x
L (a + bx)
⫺1
b(a + bx)
1
b
2
a
a
+ ln(a + bx)b
a + bx
␯⫽ ⫺
PL 19683
⫺ 81L
80
81L
c
⫹
a
+ ln (81L + 40x)bd + C1x + C2
EIA 50
40 (81L + 40x) 40 2 81L + 40x
␯⫽ ⫺
19683PL3 81L + 162 ln(81L + 40x)L + 80 ln(81L + 40x)x
a
b + C1x + C2
2000EIA
81L + 40 x
3
B.C.
␯¿(L) ⫽ 0
C1 ⫽
3168963 PL2
732050 EIA
B.C.
␯(L) ⫽ 0
C2 ⫽
19683PL3
(1+ 2 ln(81L))
2000EIA
␯(x) ⫽ ⫺
19683PL3
81L
40x
6440x
a
⫺1b
+ 2 ln a1 +
b⫺
2000EIA 81L + 40x
81L
14641L
;
PL3
19683PL3
11
a⫺2820 + 14641 lna bb ⫽ 0.317
7320500EIA
9
EIA
;
dB ⫽ ⫺ ␯(L) ⫽
q
Problem 9.7-12 A simple beam ACB is constructed with square
cross sections and a double taper (see figure). The depth of the
beam at the supports is dA and at the midpoint is dC ⫽ 2dA. Each
half of the beam has length L. Thus, the depth d and moment of
inertia I at distance x from the left-hand end are, respectively,
dA
d ⫽ (L + x)
L
I⫽
d A4
IA
d4
⫽
(L + x)4 ⫽ 4 (L + x)4
12
12 L4
L
in which IA is the moment of inertia at end A of the beam. (These
equations are valid for x between 0 and L, that is, for the left-hand
half of the beam.)
(a) Obtain equations for the slope and deflection of the
left-hand half of the beam due to the uniform load.
(b) From those equations obtain formulas for the angle of
rotation uA at support A and the deflection dC at the midpoint.
C
A
B
d
d
x
L
L
09Ch09.qxd
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1:36 PM
Page 769
SECTION 9.7 Nonprismatic Beams
769
Solution 9.7-12 Simple beam with a double taper
L ⫽ length of one-half of the beam
I⫽
IA
(L + x) 4
L4
ANGLE OF ROTATION AT SUPPORT A
uA ⫽ ⫺␯¿(0) ⫽
(0 … x … L)
(x is measured from the left-hand support A)
qL3
16EIA
qx 2
qx 2
⫽ qL x ⫺
Bending moment: M ⫽ RAx ⫺
2
2
From Appendix C:
From Eq. (9-12a):
␯ ⫽⫺
␯– ⫽
qL5x
⫺
EIA(L + x)4
qx 2
2
x 2dx
L(L + x)
3
⫽
L(3L + 4 x)
B.C.
+ ln (L + x)
(0 … x … L)
(0 … x … L) (1)
2EIA(L + x)4
2(L + x)2
qL3
8L2(3L + 4x)
cx ⫺
16EIA
2(L + x)2
⫺8L ln (L + x)d + C2
qL4x 2
;
INTEGRATE EQ. (3)
Reactions: RA ⫽ RB ⫽ qL
EI␯– ⫽ M ⫽ qL x ⫺
(positive clockwise)
2 n(0) ⫽ 0
‹ C2 ⫽ ⫺
(4)
qL4 3
a + ln Lb
2EIA 2
INTEGRATE EQ. (1)
xdx
From Appendix C:
L (L + x)
4
2
x dx
L (L + x)
4
␯¿ ⫽
⫽⫺
⫽⫺
DEFLECTION OF THE BEAM
L + 3x
3
6(L + x)
2
L + 3L x + 3x
2
3(L + x)3
␯ ⫽⫺
qL5
L⫹3x
c⫺
d
EIA
6(L⫹x)3
⫺
qL4
L2 + 3L x + 3x 2
c⫺
d + C1
2EIA
3(L + x)3
B.C.
qL x
2EIA(L + x) 3
+ C1
1 (symmetry) n⬘(L) ⫽ 0
(0 … x … L)
dC ⫽ ⫺␯(L) ⫽
(2)
3
2EIA(L⫹x)
⫽⫺
⫺
qL3
16EIA
qL3
8L x 2
c1 ⫺
d
16EIA
(L⫹x)3
qL4
qL4
(3 ⫺ 4 ln 2) ⫽ 0.02843
8EIA
EIA
(positive downward)
qL3
‹ C1 ⫽ ⫺
16EIA
Substitute C1 into Eq. (2).
qL4x 2
(0 … x … L)
;
DEFLECTION AT THE MIDPOINT C OF THE BEAM
SLOPE OF THE BEAM
␯¿ ⫽
qL4 (9L2 + 14L x + x 2) x
x
c
⫺ln a1 + bd
2EIA
L
8L(L + x)2
(0 … x … L)
4 2
⫽
Substitute C2 into Eq. (4) and simplify. (The algebra is
lengthy.)
(3)
;
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CHAPTER 9
Page 770
Deflections of Beams
Strain Energy
The beams described in the problems for Section 9.8 have constant flexural
rigidity EI.
A
Problem 9.8-1 A uniformly loaded simple beam AB (see figure) of span
length L and rectangular cross section (b ⫽ width, h ⫽ height) has a maximum bending stress smax due to the uniform load.
Determine the strain energy U stored in the beam.
Solution 9.8-1
b
L
Simple beam with a uniform load
Given: L, b, h, smax
Bending moment: M ⫽
Find: U (strain energy)
qx
qLx
⫺
2
2
Solve for q: q ⫽
2
M 2dx
Strain energy (Eq. 9-80a): U ⫽
L0 2EI
U⫽
q 2L5
⫽
240EI
smax ⫽
L2h
(1)
16Is2max L
15h2E
Substitute I ⫽
M max c
M maxh
Maximum stress: smax ⫽
⫽
I
2I
qL2
8
16Ismax
Substitute q into Eq. (1):
L
M max ⫽
bh 3
:
12
U⫽
P
M⫽
Px
2
a0 … x …
Strain energy (Eq. 9-80a):
L0
B
L
—
2
L
—
2
Simple beam with a concentrated load
(a) BENDING MOMENT
L/2
;
A
(a) Evaluate the strain energy of the beam from the bending moment in the beam.
(b) Evaluate the strain energy of the beam from the equation of the deflection
curve.
(c) From the strain energy, determine the deflection d under the load P.
U⫽2
2
4bhLs max
45E
qL2h
16I
Problem 9.8-2 A simple beam AB of length L supports a concentrated load P
at the midpoint (see figure).
Solution 9.8-2
h
B
M 2dx
P 2L3
⫽
2 EI
96EI
L
b
2
(b) DEFLECTION CURVE
From Table G-2, Case 4:
␯ ⫽⫺
;
Px
(3L2 ⫺ 4x 2)
48EI
d␯
P
⫽ ⫺
(L2 ⫺4x 2)
dx
16EI
a0 … x …
d 2␯
dx
2
⫽
Px
2EI
L
b
2
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Page 771
SECTION 9.8
Strain energy (Eq. 9-80b):
L/2
U⫽2
L0
EI d 2␯ 2
a
b dx ⫽ EI
2 dx 2
L0
P 2L3
⫽
96EI
Strain Energy
771
(c) DEFLECTION d UNDER THE LOAD P
L/2
a
From Eq. (9-82a):
Px 2
b dx
2EI
d⫽
PL3
2U
⫽
P
48EI
;
;
y
Problem 9.8-3 A propped cantilever beam AB of length L, and with guided support
at A, supports a uniform load of intensity q (see figure).
q
(a) Evaluate the strain energy of the beam from the bending moment in the beam.
(b) Evaluate the strain energy of the beam from the equation of the deflection curve.
x
A
B
L
Solution 9.8-3
q
d
␯⫽ ⫺
(8L3 ⫺12Lx 2 + 4x 3)
dx
24EI
(a) BENDING-MOMENT EQUATIONS
Measure x from end B
d2
q
( ⫺2Lx + x 2)
2EI
qx 2
M ⫽ qLx ⫺
2
dx
Strain Energy (Eq. 9-80a):
Strain energy (Eq. 9-80b):
M2
dx
L0 2EI
qx 2 2
q 2L5
1
aqLx ⫺
b dx ⫽
2
15EI
L0 2EI
(b) DEFLECTION CURVE
2
EI d 2
a 2 ␯b dx
L0 2 dx
L
2
q
EI
U⫽
c⫺
(⫺2Lx + x 2 )d dx
2EI
L0 2
U⫽
L
⫽
␯⫽ ⫺
L
L
U⫽
2
;
U⫽
q 2L5
15EI
;
Measure x from end B
␯⫽ ⫺
qx
(8L3 ⫺ 4Lx 2 + x 3)
24EI
Problem 9.8-4 A simple beam AB of length L is subjected to loads that produce a
symmetric deflection curve with maximum deflection d at the midpoint of the span (see
figure).
How much strain energy U is stored in the beam if the deflection curve is
(a) a parabola, and (b) a half wave of a sine curve?
d
A
L
—
2
B
L
—
2
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CHAPTER 9
Deflections of Beams
Solution 9.8-4
Simple beam (symmetric deflection curve)
d ⫽ maximum deflection
at midpoint
L, EI, d
GIVEN:
Page 772
Determine the strain energy U.
Assume the deflection n is positive downward.
(b) DEFLECTION CURVE IS A SINE CURVE
␯ ⫽ d sin
px
L
d 2␯
p 2d
dx
(a) DEFLECTION CURVE IS A PARABOLA
4dx
␯⫽
d ␯
2
dx 2
2
L
d␯
4d
⫽ 2 (L ⫺2x)
dx
L
(L ⫺x)
⫽⫺
8d
2
⫽⫺
L
sin
px
L
Strain energy (Eq. 9-80b):
L
U⫽
⫽
L2
2
d␯
pd
px
⫽
cos
dx
L
L
L
EI d 2␯ 2
EI
p 2d 2
px
a 2 b dx ⫽
a⫺ 2 b sin2
dx
2
2
L
dx
L
L0
L0
p4EId2
4L3
;
Strain energy (Eq. 9-80b):
L
U⫽
⫽
L
EI d 2␯ 2
EI
8d 2
a⫺ 2 b dx
a 2 b dx ⫽
2 L0
L
L0 2 dx
32EId2
L3
;
Problem 9.8-5 A beam ABC with simple supports at A and B and an overhang
P
BC supports a concentrated load P at the free end C (see figure).
B
A
(a) Determine the strain energy U stored in the beam due to the load P.
(b) From the strain energy, find the deflection dC under the load P.
(c) Calculate the numerical values of U and dC if the length L is 8 ft, the
overhang length a is 3 ft, the beam is a W 10 ⫻ 12 steel wide-flange section, and the load P produces a maximum stress of 12,000 psi in the
beam. (Use E ⫽ 29 ⫻ 106 psi.)
Solution 9.8-5
a
L
Simple beam with an overhang
(a) STRAIN ENERGY (use Eq. 9-80a)
FROM B TO C: M ⫽ ⫺Px
a
UBC ⫽
1
P 2a 3
(⫺Px)2 dx ⫽
6EI
L0 2EI
TOTAL STRAIN ENERGY:
U ⫽ UAB + UBC ⫽
P 2a 2
(L + a)
6EI
(b) DEFLECTION dC UNDER THE LOAD P
FROM A TO B: M ⫽ ⫺
Pax
L
L
UAB ⫽
M 2dx
Pax 2
P 2a 2L
1
⫽
a⫺
b dx ⫽
L
6EI
L 2EI
L0 2EI
From Eq. (9-82a):
dC ⫽
2U
Pa 2
⫽
(L + a)
P
3EI
C
;
;
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Page 773
SECTION 9.8
(c) CALCULATE U AND dc
Data: L ⫽ 8 ft ⫽ 96 in.
W 10 ⫻ 12
U⫽
a ⫽ 3 ft ⫽ 36 in.
smax ⫽ 12,000 psi
I ⫽ 53.8 in.4
c⫽
s2max I(L + a)
P 2a 2(L + a)
⫽
6EI
6c 2E
⫽ 241 in.-lb
E ⫽ 29 ⫻ 106 psi
dc ⫽
9.87
d
⫽
⫽ 4.935 in.
2
2
773
Strain Energy
;
Pa 2(L + a)
smaxa(L + a)
⫽
3EI
3cE
⫽ 0.133 in.
;
Express load P in terms of maximum stress:
smax ⫽
M maxc
smax I
Mc
Pac
⫽
⫽
! ! ! ! ‹ P⫽
I
I
I
ac
y
Problem 9.8-6 A simple beam ACB supporting a uniform load q over
the first half of the beam and a couple of moment M0 at end B is shown in the
figure.
Determine the strain energy U stored in the beam due to the load q and
the couple M0 acting simultaneously.
q
M0
A
B
L
—
2
L
—
2
Solution 9.8-6
FROM A TO MID-SPAN
STRAIN ENERGY (EQ. 9-80A):
Bending-Moment Equations
U2 ⫽
M⫽ a
L
M0
qx 2
3qL
+
bx ⫺
8
L
2
2
L
3qL
M0
M
1
dx ⫽
ca
+
bx
8
L
LL 2EI
LL 2EI
⫺
STRAIN ENERGY (EQ. 9-80A):
U1 ⫽
L
2
L0
U2 ⫽
2
M
dx
2 EI
L
2
2 2
3qL M 0
qx
1
ca
⫹
bx ⫺
d dx
⫽
2EI
8
L
2
L0
U1 ⫽
L
a3L4q 2 + 30qL2M 0 + 80M 02b
3840EI
FROM MID-SPAN TO B
Bending-Moment Equations
M⫽ a
3qL M 0
qL
L
+
b x ⫺ ax ⫺ b
8
L
2
4
2
2
qL
L 2
a x⫺ b d dx
2
4
L
a L4q 2 + 32qL2M 0 + 448M 20 b
3072EI
STRAIN ENERGY OF THE ENTIRE BEAM
U ⫽ U1 + U2 ⫽
L
a 17L4q 2
15360EI
+ 280qL2M 0 + 2560M 20 b
;
x
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CHAPTER 9
Page 774
Deflections of Beams
Problem 9.8-7 The frame shown in the figure consists of a beam ACB
L
supported by a struct CD. The beam has length 2L and is continuous
through joint C. A concentrated load P acts at the free end B.
Determine the vertical deflection dB at point B due to the load P.
Note: Let EI denote the flexural rigidity of the beam, and let EA denote
the axial rigidity of the strut. Disregard axial and shearing effects in the
beam, and disregard any bending effects in the strut.
L
B
A
C
P
L
D
Solution 9.8-7
Frame with beam and strut
L CD ⫽ length of strut
BEAM ACB
⫽ 12L
F ⫽ axial force in strut
⫽ 2 12P
For part AC of the beam: M ⫽ ⫺Px
USTRUT ⫽
F 2L CD
2EA
USTRUT ⫽
(212P)2(12L)
412P 2L
⫽
2EA
EA
FRAME
U ⫽ UBEAM + USTRUT ⫽
L
1
M 2dx
P 2L3
⫽
(⫺P x) 2dx ⫽
2EI L0
6EI
L 2EI
P 2L3
For part CB of the beam: UCB ⫽ UAC ⫽
6EI
UAC ⫽
Entire beam: UBEAM ⫽ UAC
SRUCT CD
P 2L3
+ UCB ⫽
3EI
(Eq. 2-37a)
DEFLECTION dB AT POINT B
From Eq. (9-82a):
dB ⫽
2U
2PL3
812PL
⫽
+
P
3EI
EA
;
P 2L3
4 12P 2L
+
3EI
EA
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Page 775
SECTION 9.9
775
Castigliano’s Theorem
Castigliano’s Theorem
M0
The beams described in the problems for Section 9.9 have constant flexural
rigidity EI.
A
B
Problem 9.9-1 A simple beam AB of length L is loaded at the left-hand end
by a couple of moment M0 (see figure).
Determine the angle of rotation uA at support A. (Obtain the solution by
determining the strain energy of the beam and then using Castigliano’s theorem.)
Solution 9.9-1
L
Simple beam with couple M0
STRAIN ENERGY
U⫽
M 20 L
M 20 L
x 2
M 2dx
⫽
a 1 ⫺ b dx ⫽
2EI L0
L
6EI
L 2EI
CASTIGLIANO’S THEOREM
M0
RA ⫽
L
uA ⫽
(downward)
M ⫽ M 0 ⫺ RAx ⫽ M 0 ⫺
⫽ M 0 a1 ⫺
M 0x
L
M 0L
dU
⫽
dM 0
3EI
;
(clockwise)
(This result agrees with Case 7, Table G-2)
x
b
L
Problem 9.9-2 The simple beam shown in the figure supports a concentrated
P
load P acting at distance a from the left-hand support and distance b from the
right-hand support.
Determine the deflection dD at point D where the load is applied. (Obtain
the solution by determining the strain energy of the beam and then using
Castigliano’s theorem.)
A
a
b
L
Solution 9.9-2
Simple beam with load P
M AD ⫽ RAx ⫽
Pbx
L
M DB ⫽ RBx ⫽
Pax
L
STRAIN ENERGY
RA ⫽
Pb
L
RB ⫽
Pa
L
a
UAD ⫽
B
D
U⫽
M 2dx
L 2EI
1
Pbx 2
P 2a 3b 2
a
b dx ⫽
2EI L0
L
6EIL2
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CHAPTER 9
Page 776
Deflections of Beams
b
UDB ⫽
1
Pax 2
P 2a 2b 3
a
b dx ⫽
2EI L0
L
6EIL2
U ⫽ UAD
P 2a 2b 2
+ UDB ⫽
6LEI
CASTIGLIANO’S THEOREM
dD ⫽
Pa 2b 2
dU
⫽
dP
3LEI
;
(downward)
Problem 9.9-3 An overhanging beam ABC supports a concentrated load P at
the end of the overhang (see figure). Span AB has length L and the overhang has
length a.
Determine the deflection dC at the end of the overhang. (Obtain the
solution by determining the strain energy of the beam and then using
Castigliano’s theorem.)
Solution 9.9-3
Pa
L
A
L
U⫽
C
a
M AB ⫽ ⫺RAx ⫽ ⫺
L
1
Pax 2
P 2a 2L
a⫺
b dx ⫽
2EI L0
L
6EI
UCB ⫽
a
1
P 2a 3
(⫺ Px)2dx ⫽
2EI L0
6EI
U ⫽ UAB + UCB ⫽
Pax
L
M 2dx
L 2EI
UAB ⫽
(downward)
M CB ⫽ ⫺Px
B
Overhanging beam
STRAIN ENERGY
RA ⫽
P
P 2a 2
(L + a)
6EI
CASTIGLIANO’S THEOREM
dC ⫽
dU
Pa 2
⫽
(L + a)
dP
3EI
Problem 9.9-4 The cantilever beam shown in the figure supports a triangularly
(downward)
;
q0
distributed load of maximum intensity q0.
Determine the deflection dB at the free end B. (Obtain the solution by
determining the strain energy of the beam and then using Castigliano’s theorem.)
B
A
L
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Page 777
SECTION 9.9
Solution 9.9-4
Castigliano’s Theorem
777
Cantilever beam with triangular load
CASTIGLIANO’S THEOREM
dB ⫽
q0L4
0U
PL3
⫽
+
0P
3EI
30EI
(downward)
(This result agrees with Cases 1 and 8 of Table G-1.)
SET P ⫽ 0: dB ⫽
q0L4
30EI
;
P ⫽ fictitious load corresponding to deflection dB
M ⫽ ⫺Px ⫺
q0x 3
6L
STRAIN ENERGY
L
q0x 3 2
1
M 2dx
a⫺Px ⫺
⫽
b dx
2EI L0
6L
L 2EI
Pq0L4
q 20L5
P 2L3
⫽
+
+
6EI
30EI
42EI
U⫽
q
Problem 9.9-5 A simple beam ACB supports a uniform load of intensity q on
the left-hand half of the span (see figure).
Determine the angle of rotation uB at support B. (Obtain the solution by using
the modified form of Castigliano’s theorem.)
C
A
B
L
—
2
Solution 9.9-5
L
—
2
Simple beam with partial uniform load
BENDING MOMENT AND PARTIAL DERIVATIVE FOR
AC
SEGMENT
M AC ⫽ RAx ⫺
qx 2
3qL
qx 2
M0
⫽ a
+
bx ⫺
2
8
L
2
a0 … x …
M0 ⫽ fictitious load corresponding to angle
of rotation uB
RA ⫽
3qL
M0
+
8
L
RB ⫽
qL
M0
⫺
8
L
L
b
2
0M AC
x
⫽
0M 0
L
BENDING MOMENT AND PARTIAL DERIVATIVE FOR
SEGMENT CB
M CB ⫽ RBx + M 0 ⫽ a
qL
M0
⫺
b x + M0
8
L
a0 … x …
L
b
2
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CHAPTER 9
Page 778
Deflections of Beams
SET FICTITIOUS LOAD M0 EQUAL TO ZERO
0M CB
x
⫽ ⫺ + 1
0M 0
L
uB ⫽
MODIFIED CASTIGLIANO’S THEOREM (EQ. 9-88)
uB ⫽
⫽
a
M
0M
ba
bdx
EI
0M
L
0
1
EI L0
L/2
1
+
EI L0
1
EI L0
L/2
a
qx 2
3qLx
x
⫺
b a b dx
8
2
L
L/2
qLx
x
1
a
b a1 ⫺ b dx
EI L0
8
L
3
3
qL
qL
⫽
+
128EI
96EI
+
ca
3qL
qx 2 x
M0
+
bx⫺
d c d dx
8
L
2
L
L/2
qL
M0
x
ca
⫺
b x + M 0 d c1 ⫺ d dx
8
L
L
⫽
7qL3
384EI
;
(counterclockwise)
(This result agrees with Case 2, Table G-2.)
P1
Problem 9.9-6 A cantilever beam ACB supports two concentrated loads
P1 and P2, as shown in the figure.
Determine the deflections dC and dB at points C and B, respectively.
(Obtain the solution by using the modified form of Castigliano’s theorem.)
A
P2
C
L
—
2
B
L
—
2
Cantilever beam with loads P1 and P2
Solution 9.9-6
MODIFIED CASTIGLIANO’S THEOREM FOR DEFLECTION dC
dC ⫽
1
EI L0
L/2
(M CB)a
0M CB
b dx
0P1
L
+
BENDING MOMENT AND PARTIAL DERIVATIVES
CB
0M AC
1
(M AC)a
b dx
EI LL/2
0P1
FOR SEGMENT
M CB ⫽ ⫺P2x
0M CB
⫽0
0P1
L
⫽0 +
L
a0 … x … b
2
0M CB
⫽ ⫺x
0P2
⫽
1
L
L
c ⫺P1 a x ⫺ b ⫺ P2x d a ⫺ x b dx
EI LL/2
2
2
L3
(2P1 + 5P2)
48EI
;
BENDING MOMENT AND PARTIAL DERIVATIVES FOR
AC
MODIFIED CASTIGLIANO’S THEOREM FOR DEFLECTION dB
L
M AC ⫽ ⫺P1 ax ⫺ b ⫺ P2x
2
dB ⫽
SEGMENT
0M AC
L
⫽ ⫺x
0P1
2
L
a … x … Lb
2
0M AC
⫽ ⫺x
0P2
1
EI L0
L/2
(M CB)a
L
+
0M CB
b dx
0P2
0M AC
1
(M )a
b dx
EI LL/2 AC
0P2
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Page 779
SECTION 9.9
⫽
1
EI L0
L/2
(⫺P2x) (⫺x)dx
L
L
1
c ⫺P1 ax ⫺ b ⫺ P2x d(⫺x)dx
+
EI LL/2
2
Castigliano’s Theorem
⫽
P2L3
L3
+
(5P1 + 14P2)
24EI
48EI
⫽
L3
(5P1 + 16P2)
48EI
779
;
(These results can be verified with the aid of Cases 4
and 5, Table G-1.)
q
Problem 9.9-7 The cantilever beam ACB shown in the figure is subjected to
a uniform load of intensity q acting between points A and C.
Determine the angle of rotation uA at the free end A. (Obtain the solution
by using the modified form of Castigliano’s theorem.)
C
A
L
—
2
Solution 9.9-7
B
L
—
2
Cantilever beam with partial uniform load
MODIFIED CASTIGLIANO’S THEOREM (EQ. 9-88)
uA ⫽
⫽
M0 ⫽ fictitious load corresponding to the angle
of rotation uA
BENDING MOMENT AND PARTIAL DERIVATIVE FOR
AC
SEGMENT
M AC ⫽ ⫺M 0 ⫺
qx 2
2
a0 … x …
L
b
2
M
0M
a ba
b dx
L EI 0M 0
1
EI L0
+
L/2
a⫺M 0 ⫺
L
qL
L
1
c ⫺ M0 ⫺
a x ⫺ b d ( ⫺1)dx
EI LL/2
2
4
SET FICTITIOUS LOAD M0 EQUAL TO ZERO
uA ⫽
1
EI L0
L/2
L
qx 2
qL
1
L
a b a x ⫺ b dx
dx +
2
EI LL/2 2
4
0M AC
⫽ ⫺1
0M 0
⫽
qL3
qL3
+
48EI
8EI
BENDING MOMENT AND PARTIAL DERIVATIVE FOR
SEGMENT CB
⫽
7qL3
48EI
M CB ⫽ ⫺M 0 ⫺
0M CB
⫽ ⫺1
0M 0
qL
L
ax ⫺ b
2
4
a
L
… x … Lb
2
qx 2
b( ⫺1)dx
2
(counterclockwise)
;
(This result can be verified with the aid of Case 3,
Table G-1.)
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CHAPTER 9
Page 780
Deflections of Beams
Problem 9.9-8 The frame ABC supports a concentrated load P at point C (see figure).
Members AB and BC have lengths h and b, respectively.
Determine the vertical deflection dC and angle of rotation uC at end C of the frame.
(Obtain the solution by using the modified form of Castigliano’s theorem.)
b
B
C
P
h
A
Solution 9.9-8
Frame with concentrated load
MODIFIED CASTIGLIANO’S THEOREM FOR DEFLECTION dC
dC ⫽
a
0M
M
ba
bdx
EI
0P
L
h
⫽
b
1
1
(Pb + M 0)(b) dx +
(Px + M 0)(x) dx
EI L0
EI L0
Set M0 ⫽ 0:
h
dC ⫽
P ⫽ concentrated load acting at point C
(corresponding to the deflection dC)
M0 ⫽ fictitious moment corresponding to the
angle of rotation uC
BENDING MOMENT AND PARTIAL DERIVATIVES FOR
MEMBER AB
M AB ⫽ Pb + M 0 (0 … x … h)
0M AB
⫽b
0P
0M AB
⫽1
M0
BENDING MOMENT AND PARTIAL DERIVATIVES FOR
BC
MEMBER
⫽
0M BC
⫽1
0M 0
Pb 2
(3h + b)
3EI
;
(downward)
MODIFIED CASTIGLIANO’S THEOREM FOR ANGLE OF
uC
ROTATION
uC ⫽
0M
M
ba
b dx
L EI 0M 0
a
h
⫽
b
1
1
(Pb + M 0)(1)dx +
(Px + M 0)(1) dx
EI L0
EI L0
Set M0 ⫽ 0:
h
uC ⫽
M BC ⫽ Px + M 0 (0 … x … b)
0M BC
⫽x
0P
b
1
1
Pb 2dx +
Px 2dx
EI L0
EI L0
⫽
b
1
1
Pb dx +
Pxdx
EI L0
EI L0
Pb
(2h + b)
2EI
(clockwise)
;
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Page 781
SECTION 9.9
Problem 9.9-9 A simple beam ABCDE supports a uniform load of
q
intensity q (see figure). The moment of inertia in the central part of the
beam (BCD) is twice the moment of inertia in the end parts (AB and DE).
Find the deflection dC at the midpoint C of the beam. (Obtain the
solution by using the modified form of Castigliano’s theorem.)
B
A
C
I
D
L
—
4
E
I
2I
L
—
4
Solution 9.9-9
781
Castigliano’s Theorem
L
—
4
L
—
4
Nonprismatic beam
MODIFIED CASTIGLIANO’S THEOREM (EQ. 9-88)
Integrate from A to C and multiply by 2.
dC ⫽ 2
a
M AC 0M AC
ba
b dx
0P
L EI
⫽ 2a
1
b
EI L0
+ 2a
P ⫽ fictitiuous load corresponding to the deflection ␦C
at the midpoint
qL
P
+
RA ⫽
2
2
dC ⫽
qLx
qx 2
Px
⫺
+
2
2
2
0M AC
x
⫽
0P
2
a0 … x …
L
b
2
a0 … x …
L
b
2
a
qLx
qx 2
Px
x
⫺
+
b a b dx
2
2
2
2
L/2
qLx
qx 2
Px
x
1
b
a
⫺
+
b a b dx
2EI LL/4
2
2
2
2
SET FICTITIOUS LOAD P EQUAL TO ZERO
BENDING MOMENT AND PARTIAL DERIVATIVE FOR THE
LEFT-HAND HALF OF THE BEAM (A TO C)
M AC ⫽
L/4
2
EI L0
+
⫽
dC ⫽
L/4
a
qLx
qx 2
x
⫺
b a bdx
2
2
2
L/2
qLx
qx 2
1
x
a
⫺
b a bdx
EI LL/4
2
2
2
67qL4
13qL4
+
6,144EI
12,288EI
31qL4
4096EI
Problem 9.9-10 An overhanging beam ABC is subjected to a couple MA
MA
at the free end (see figure). The lengths of the overhang and the main span
are a and L, respectively.
Determine the angle of rotation uA and deflection dA at end A. (Obtain the
solution by using the modified form of Castigliano’s theorem.)
;
(downward)
A
B
a
C
L
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CHAPTER 9
Solution 9.9-10
Page 782
Deflections of Beams
Overhanging beam ABC
Set P ⫽ 0:
uA ⫽
⫽
MA ⫽ couple acting at the free end A (corresponding to
the angle of rotation uA)
P ⫽ fictitious load corresponding to the deflection dA
BENDING MOMENT AND PARTIAL DERIVATIVES
AB
FOR SEGMENT
M AB ⫽ ⫺M A ⫺ Px (0 … x … a)
0M AB
⫽ ⫺1
0M A
MA
(L + 3a) (counterclockwise)
3EI
;
MODIFIED CASTIGLIANO’S THEOREM FOR DEFLECTION dA
dA ⫽
a
M
0M
ba
bdx
0P
L EI
a
⫽
0M AB
⫽ ⫺x
0P
BENDING MOMENT AND PARTIAL DERIVATIVES
BC
MA
Pa
+
(downward)
Reaction at support C: RC ⫽
L
L
M Ax
Pax
⫺
(0 … x … L)
M BC ⫽ ⫺RC x ⫽ ⫺
L
L
a
L
M Ax
1
1
x
M Adx +
a
b a b dx
EI L0
EI L0
L
L
1
(⫺M A ⫺ Px)(⫺x)dx
EI L0
L
+
M Ax
Pax
ax
1
a⫺
⫺
b a⫺ bdx
EI L0
L
L
L
Set P ⫽ 0:
FOR SEGMENT
0M BC
x
⫽ ⫺
0M A
L
0M BC
ax
⫽ ⫺
0P
L
dA ⫽
⫽
a
L
M Ax ax
1
1
M Axdx +
a
b a b dx
EI L0
EI L0
L
L
M Aa
(2L + 3a) (downward)
6EI
;
MODIFIED CASTIGLIANO’S THEOREM FOR ANGLE OF
uA
ROTATION
a
M
0M
bdx
ba
EI
0M
L
A
a
1
(⫺ M A ⫺ Px)(⫺1) dx
⫽
EI L0
uA ⫽
+
L
M Ax
Pax
x
1
a⫺
⫺
b a ⫺ bdx
EI L0
L
L
L
Problem 9.9-11 An overhanging beam ABC rests on a simple support at A and
a spring support at B (see figure). A concentrated load P acts at the end of the
overhang. Span AB has length L, the overhang has length a, and the spring
has stiffness k.
Determine the downward displacement dC of the end of the overhang. (Obtain
the solution by using the modified form of Castigliano’s theorem.)
P
B
A
C
k
L
a
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Page 783
SECTION 9.9
Solution 9.9-11
783
Castigliano’s Theorem
Beam with spring support
TOTAL STRAIN ENERGY U
U ⫽ UB + US ⫽
P 2(L + a)2
M 2dx
+
2kL2
L 2EI
APPLY CASTIGLIANO’S THEOREM (EQ. 9-87)
dC ⫽
RA ⫽
RB ⫽
Pa
L
(downward)
⫽
P
(L + a) (upward)
L
BENDING MOMENT AND PARTIAL DERIVATIVE FOR
SEGMENT AB
M AB ⫽ ⫺RAx ⫽ ⫺
Pax
L
dM AB
ax
⫽ ⫺
dP
L
dC ⫽
ax
1
Pax
a⫺
b a⫺ bdx
EI L0
L
L
+
dM BC
⫽ ⫺x (0 … x … a)
dP
P 2(L + a)2
R2B
⫽
US ⫽
2k
2kL2
P(L + a)2
M
dM
ba
b dx +
dP
kL2
L EI
a
L
⫽
BENDING MOMENT AND PARTIAL DERIVATIVE FOR
SEGMENT BC
STRAIN ENERGY OF THE SPRING (EQ. 2-38a)
P(L + a)2
d
M 2dx
+
dP L 2EI
kL2
DIFFERENTIATE UNDER THE INTEGRAL SIGN (MODIFIED
CASTIGLIANO’S THEOREM)
(0 … x … L)
M BC ⫽ ⫺Px
dU
d
M 2dx
d P 2(L + a)2
d
⫽
+
c
dP
dP L 2EI
dP
2kL2
⫽
dC ⫽
a
P(L + a)2
1
(⫺Px)(⫺x) dx +
EI L0
kL2
P(L + a)2
Pa 3
Pa 2L
+
+
3EI
3EI
kL2
Pa 2(L + a)
P(L + a)2
+
3EI
kL2
;
STRAIN ENERGY OF THE BEAM (EQ. 9-80a)
UB ⫽
M 2dx
L 2EI
q
Problem 9.9-12 A symmetric beam ABCD with overhangs at both ends
supports a uniform load of intensity q (see figure).
Determine the deflection dD at the end of the overhang. (Obtain the
solution by using the modified form of Castigliano’s theorem.)
A
B
L
—
4
D
C
L
L
—
4
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CHAPTER 9
Solution 9.9-12
Page 784
Deflections of Beams
Beam with overhangs
SEGMENT CD M CD ⫽ ⫺
qx 2
⫺ Px
2
a0 … x …
L
b
4
0M CD
⫽ ⫺x
0P
MODIFIED CASTIGLIANO’S THEOREM FOR DEFLECTION dD
dD ⫽
q ⫽ intensity of uniform load
P ⫽ fictitious load corresponding to the deflection dD
⫽
L
⫽ length of segments AB and CD
4
L ⫽ length of span BC
RB ⫽
3qL
P
⫺
4
4
RC ⫽
M
0M
ba
bdx
EI
0P
L
1
EI L0
+
3qL
5P
+
4
4
BENDING MOMENTS AND PARTIAL DERIVATIVES
SEGMENT AB
qx 2
0M AB
L
M AB ⫽ ⫺
⫽ 0 a0 … x … b
2
0P
4
SEGMENT BC
M BC ⫽ ⫺ cq ax +
a
3 qL
q
L 2
P
⫺ bx
⫽ ⫺ ax + b + a
2
4
4
4
(0 … x … L)
0M BC
x
⫽ ⫺
0P
4
a⫺
qx 2
b(0)dx
2
L
q
3qL
L 2
P
1
c ⫺ ax + b + a
⫺ bx d
EI L0
2
4
4
4
1
x
* c⫺ ddx +
4
EI L0
SET P ⫽ 0:
dD ⫽
L/4
a⫺
qx 2
⫺ Pxb ( ⫺x)dx
2
L
q
3qL
L 2
x
1
c ⫺ ax + b +
x d c ⫺ d dx
EI L0
2
4
4
4
+
L
1
L
b d c ax + b d + RBx
4
2
4
L/4
⫽ ⫺
1
EI L0
L/4
a⫺
qx 2
b( ⫺x)dx
2
qL4
37qL4
5qL4
+
⫽ ⫺
768EI
2048EI
6144EI
(Minus means the deflection is opposite in direction to
the fictitious load P.)
‹ dD ⫽
37qL4
6144EI
(upward)
;
Deflections Produced by Impact
The beams described in the problems for Section 9.10 have constant
flexural rigidity EI. Disregard the weights of the beams themselves,
and consider only the effects of the given loads.
W
h
A
Problem 9.10-1 A heavy object of weight W is dropped onto the midpoint of a
simple beam AB from a height h (see figure).
Obtain a formula for the maximum bending stress smax due to the falling weight
in terms of h, sst, and dst, where sst is the maximum bending stressand dst is the deflection at the midpoint when the weight W acts on the beam as a statically applied load.
Plot a graph of the ratio smax/sst (that is, the ratio of the dynamic stress
to the static stress) versus the ratio h/dst. (Let h/dst vary from 0 to 10.)
B
L
—
2
L
—
2
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Page 785
SECTION 9.10
Solution 9.10-1
785
Weight W dropping onto a simple beam
MAXIMUM DEFLECTION (EQ. 9-94)
dmax ⫽ dst ⫹ (d2st ⫹ 2hdst)1/2
MAXIMUM BENDING STRESS
For a linearly elastic beam, the bending stress s is
proportional to the deflection d
‹
Deflections Produced by Impact
dmax
smax
2h 1/2
⫽
⫽ 1 + a1 +
b
sst
dst
dst
2h 1/2
smax ⫽ sst c1 + a1 +
b d
dst
;
NOTE: dst ⫽
h
dst
smax
sst
0
2.5
5.0
7.5
10.0
2.00
3.45
4.33
5.00
5.58
WL3
for a simple beam with a load at the
48EI
midpoint.
GRAPH OF RATIO smax/sst
W
Problem 9.10-2 An object of weight W is dropped onto the midpoint of a
simple beam AB from a height h (see figure). The beam has a rectangular cross
section of area A.
Assuming that h is very large compared to the deflection of the beam when
the weight W is applied statically, obtain a formula for the maximum bending
stress smax in the beam due to the falling weight.
h
A
B
L
—
2
L
—
2
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1:37 PM
CHAPTER 9
Page 786
Deflections of Beams
Solution 9.10-2
Weight W dropping onto a simple beam
Height h is very large.
sst ⫽
M
WL
⫽
S
4S
dst ⫽
WL3
48EI
MAXIMUM DEFLECTION (EQ. 9-95)
dmax ⫽ 12hdst
MAXIMUM BENDING STRESS
I⫽
smax
dmax
2h
⫽
⫽
s st
dst
A dst
s2st
3WEI
⫽ 2
dst
S L
bd 3
12
S⫽
bd 2
6
smax ⫽
I
S
(1)
18WhE
(2)
2
⫽
3
3
⫽
bd
A
(3)
constructed of a W 8 ⫻ 21 wide-flange section (see figure). A
weight W ⫽ 1500 lb falls through a height h ⫽ 0.25 in. onto
the end of the beam.
Calculate the maximum deflection dmax of the end of the beam
and the maximum bending stress smax due to the falling weight.
(Assume E ⫽ 30 ⫻ 106 psi.)
;
A AL
W = 1500 lb
Problem 9.10-3 A cantilever beam AB of length L ⫽ 6 ft is
W 8 ⫻ 21
h = 0.25 in.
A
B
L = 6 ft
Cantilever beam
DATA: L ⫽ 6 ft ⫽ 72 in.
h ⫽ 0.25 in.
W 8 ⫻ 21
16S 2
Substitute (2) and (3) into (1):
2hsst2
smax ⫽
A dst
Solution 9.10-3
W 2L2
For a RECTANGULAR BEAM (with b, depth d):
For a linearly elastic beam, the bending stress s is
proportional to the deflection d.
‹
s2st ⫽
Equation (9-94):
W ⫽ 1500 lb
dmax ⫽ dst + (d2st + 2hdst)1/2 ⫽ 0.302 in.
E ⫽ 30 ⫻ 106 psi
I ⫽ 75.3 in.4
S ⫽ 18.2 in.3
;
MAXIMUM BENDING STRESS
MAXIMUM DEFLECTION (EQ. 9-94)
Consider a cantilever beam with load P at the free end:
Equation (9-94) may be used for any linearly elastic structure by substituting dst ⫽ W/k, where k is the stiffness of
the particular structure being considered. For instance:
Simple beam with load at midpoint:
smax ⫽
k⫽
Ratio:
M max
PL
⫽
S
S
Cantilever beam with load at the free end: k ⫽
For the cantilever beam in this problem:
dst ⫽
‹ smax ⫽
(1500 lb)(72 in.)3
WL3
⫽
3EI
3(30 * 106 psi)(75.3 in.4)
⫽ 0.08261 in.
3 EI
L3
PL3
3EI
smax
3 EI
⫽
dmax
SL2
48EI
L3
dmax ⫽
3EI
SL2
dmax ⫽ 21,700 psi
;
09Ch09.qxd
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1:37 PM
Page 787
SECTION 9.10
W
Problem 9.10-4 A weight W ⫽ 20 kN falls through a
height h ⫽ 1.0 mm onto the midpoint of a simple beam of
A
length L ⫽ 3 m (see figure). The beam is made of wood with square
cross section (dimension d on each side) and E ⫽ 12 GPa.
If the allowable bending stress in the wood is sallow ⫽ 10 MPa, what
is the minimum required dimension d?
Solution 9.10-4
787
Deflections Produced by Impact
h
d
B
d
L
—
2
L
—
2
Simple beam with falling weight W
DATA: W ⫽ 20 kN
E ⫽ 12 GPa
h ⫽ 1.0 mm
SUBSTITUTE (2) AND (3) INTO EQ. (1)
L ⫽ 3.0 m
2smaxd 3
8hEd 4 1/2
b
⫽ 1 + a1 +
3WL
WL3
sallow ⫽ 10 MPa
CROSS SECTION OF BEAM (SQUARE)
SUBSTITUTE NUMERICAL VALUES:
d ⫽ dimension of each side
d4
I⫽
12
2(10 MPa)d 3
8(1.0 mm)(12 GPa)d 4 1/2
d
⫽ 1 + c1 +
3(20 kN)(3.0 m)
(20 kN)(3.0 m)3
d3
S⫽
6
1000 3
1600 4 1/2
(d ⫽ meters)
d ⫺ 1 ⫽ c1 +
d d
9
9
MAXIMUM DEFLECTION (EQ. 9-94)
dmax ⫽ dst ⫹ (d 2st ⫹ 2hdst)1/2
SQUARE BOTH SIDES, REARRANGE, AND SIMPLIFY
a
MAXIMUM BENDING STRESS
For a linearly elastic beam, the bending stress s is
proportional to the deflection d.
‹
smax
dmax
2h
⫽
⫽ 1 + a1 +
b
s st
dst
dst
2500d 3 ⫺ 36d ⫺ 45 ⫽ 0 (d ⫽ meters)
1/2
(1)
WL
6
M
3WL
s st ⫽
⫽a
b a 3b ⫽
S
4
d
2d 3
WL3
WL3 12
WL3
⫽
a 4b ⫽
48 EI
48 E d
4 Ed 4
SOLVE NUMERICALLY
d ⫽ 0.2804 m ⫽ 280.4 mm
STATIC TERMS sst AND dst
dst ⫽
1600
2000
1000 2 3
b d ⫺
d⫺
⫽0
9
9
9
For minimum value, round upwa