CHAPTER 1
PROBLEM 1.1
Two solid cylindrical rods AB and BC are welded together at B and
loaded as shown. Knowing that the average normal stress must not
exceed 175 MPa in rod AB and 150 MPa in rod BC, determine the
smallest allowable values of d1 and d2.
SOLUTION
(a)
Rod AB
P = 40 + 30 = 70 kN = 70 × 103 N
σ AB =
d1 =
(b)
P
=
AAB
P
4P
=
2
d
π d12
4 1
π
4P
=
πσ AB
(4)(70 × 103 )
= 22.6 × 10−3 m
π (175 × 106 )
d1 = 22.6 mm 
Rod BC
P = 30 kN = 30 × 103 N
σ BC =
d2 =
P
=
ABC
4P
πσ BC
P
4P
=
2
d
π d 22
4 2
π
=
(4)(30 × 103 )
= 15.96 × 10−3 m
π (150 × 106 )
d 2 = 15.96 mm 
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PROBLEM 1.2
Two solid cylindrical rods AB and BC are welded together at B and loaded
as shown. Knowing that d1 = 50 mm and d 2 = 30 mm, find the average
normal stress at the midsection of (a) rod AB, (b) rod BC.
SOLUTION
(a)
Rod AB
P = 40 + 30 = 70 kN = 70 × 103 N
A=
σ AB =
(b)
π
4
d12 =
π
4
(50) 2 = 1.9635 × 103 mm 2 = 1.9635 × 10−3 m 2
P
70 × 103
=
= 35.7 × 106 Pa
−3
A 1.9635 × 10
σ AB = 35.7 MPa 
Rod BC
P = 30 kN = 30 × 103 N
A=
σ BC =
π
4
d 22 =
π
4
(30)2 = 706.86 mm 2 = 706.86 × 10−6 m 2
P
30 × 103
=
= 42.4 × 106 Pa
−6
A 706.86 × 10
σ BC = 42.4 MPa 
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PROBLEM 1.3
Two solid cylindrical rods AB and BC are welded together
at B and loaded as shown. Determine the magnitude of the
force P for which the tensile stress in rod AB is twice the
magnitude of the compressive stress in rod BC.
SOLUTION
σ AB
π
(2) 2 = 3.1416 in 2
4
P
P
=
=
AAB
3.1416
AAB =
= 0.31831 P
ABC =
σ BC
π
(3)2 = 7.0686 in 2
4
(2)(30) − P
=
AAB
=
60 − P
= 8.4883 − 0.14147 P
7.0686
Equating σ AB to 2σ BC
0.31831 P = 2(8.4883 − 0.14147 P)
P = 28.2 kips 
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PROBLEM 1.4
In Prob. 1.3, knowing that P = 40 kips, determine the
average normal stress at the midsection of (a) rod AB,
(b) rod BC.
PROBLEM 1.3 Two solid cylindrical rods AB and BC
are welded together at B and loaded as shown. Determine
the magnitude of the force P for which the tensile stress
in rod AB is twice the magnitude of the compressive
stress in rod BC.
SOLUTION
(a)
Rod AB
P = 40 kips (tension)
AAB =
σ AB
(b)
2
π d AB
=
π (2) 2
4
4
P
40
=
=
AAB
3.1416
= 3.1416 in 2
σ AB = 12.73 ksi 
Rod BC
F = 40 − (2)(30) = −20 kips, i.e., 20 kips compression.
ABC =
σ BC
2
π d BC
=
π (3) 2
4
4
−20
F
=
=
ABC
7.0686
= 7.0686 in 2
σ BC = −2.83 ksi 
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PROBLEM 1.5
Two steel plates are to be held together by means of 16-mmdiameter high-strength steel bolts fitting snugly inside cylindrical
brass spacers. Knowing that the average normal stress must not
exceed 200 MPa in the bolts and 130 MPa in the spacers,
determine the outer diameter of the spacers that yields the most
economical and safe design.
SOLUTION
At each bolt location the upper plate is pulled down by the tensile force Pb of the bolt. At the same time, the
spacer pushes that plate upward with a compressive force Ps in order to maintain equilibrium.
Pb = Ps
For the bolt,
σb =
Fb
4Pb
=
Ab
π db2
or
Pb =
For the spacer,
σs =
Ps
4Ps
=
As
π (d s2 − db2 )
or
Ps =
π
4
π
4
σ bdb2
σ s (d s2 − db2 )
Equating Pb and Ps ,
π
4
σ bdb2 =
ds =
π
4
σ s (d s2 − db2 )

σb 
1 +
d =
σs  b

200 

1 + 130  (16)


d s = 25.2 mm 
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PROBLEM 1.6
Two brass rods AB and BC, each of uniform diameter, will be brazed
together at B to form a nonuniform rod of total length 100 m, which will
be suspended from a support at A as shown. Knowing that the density of
brass is 8470 kg/m3, determine (a) the length of rod AB for which the
maximum normal stress in ABC is minimum, (b) the corresponding
value of the maximum normal stress.
SOLUTION
Areas:
AAB =
ABC =
From geometry,
Weights:
π
4
π
4
(15 mm)2 = 176.71 mm 2 = 176.71 × 10−6 m 2
(10 mm) 2 = 78.54 mm 2 = 78.54 × 10−6 m 2
b = 100 − a
WAB = ρ g AAB AB = (8470)(9.81)(176.71 × 10−6 ) a = 14.683 a
WBC = ρ g ABC  BC = (8470)(9.81)(78.54 × 10−6 )(100 − a) = 652.59 − 6.526 a
Normal stresses:
At A,
PA = WAB + WBC = 652.59 + 8.157a
σA =
At B,
(a)
PA
= 3.6930 × 106 + 46.160 × 103 a
AAB
PB = WBC = 652.59 − 6.526a
σB =
(1)
(2)
PB
= 8.3090 × 106 − 83.090 × 103 a
ABC
Length of rod AB. The maximum stress in ABC is minimum when σ A = σ B or
4.6160 × 106 − 129.25 × 103 a = 0
a = 35.71 m
(b)
 AB = a = 35.7 m 
Maximum normal stress.
σ A = 3.6930 × 106 + (46.160 × 103 )(35.71)
σ B = 8.3090 × 106 − (83.090 × 103 )(35.71)
σ A = σ B = 5.34 × 106 Pa
σ = 5.34 MPa 
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PROBLEM 1.7
Each of the four vertical links has an 8 × 36-mm uniform rectangular
cross section and each of the four pins has a 16-mm diameter.
Determine the maximum value of the average normal stress in the
links connecting (a) points B and D, (b) points C and E.
SOLUTION
Use bar ABC as a free body.
ΣM C = 0 :
(0.040) FBD − (0.025 + 0.040)(20 × 103 ) = 0
FBD = 32.5 × 103 N
Link BD is in tension.
3
ΣM B = 0 : − (0.040) FCE − (0.025)(20 × 10 ) = 0
FCE = −12.5 × 103 N
Link CE is in compression.
Net area of one link for tension = (0.008)(0.036 − 0.016) = 160 × 10−6 m 2.
For two parallel links,
(a)
σ BD =
A net = 320 × 10−6 m 2
FBD
32.5 × 103
=
= 101.56 × 106
Anet
320 × 10−6
σ BD = 101.6 MPa 
Area for one link in compression = (0.008)(0.036) = 288 × 10−6 m 2.
For two parallel links,
(b)
σ CE =
A = 576 × 10−6 m 2
FCE
−12.5 × 103
=
= −21.70 × 10−6
A
576 × 10−6
σ CE = −21.7 MPa 
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PROBLEM 1.8
Knowing that the link DE is 18 in. thick and 1 in. wide, determine
the normal stress in the central portion of that link when
(a) θ = 0°, (b) θ = 90°.
SOLUTION
Use member CEF as a free body.
Σ M C = 0 : − 12 FDE − (8)(60 sin θ ) − (16)(60 cos θ ) = 0
FDE = −40 sin θ − 80 cos θ lb.
1
ADE = (1)   = 0.125 in.2
8
F
σ DE = DE
ADE
(a)
θ = 0: FDE = −80 lb.
σ DE =
(b)
−80
0.125
σ DE = −640 psi 
θ = 90°: FDE = −40 lb.
σ DE =
−40
0.125
σ DE = −320 psi 
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PROBLEM 1.9
Link AC has a uniform rectangular cross section 161 in.
thick and 14 in. wide. Determine the normal stress in the
central portion of the link.
SOLUTION
Free Body Diagram of Plate
Note that the two 240-lb forces form a couple of moment
(240 lb)(6 in.) = 1440 lb ⋅ in.
Σ M B = 0 : 1440 lb ⋅ in − ( FAC cos 30°)(10 in.) = 0
FAC = 166.277 lb.
Area of link:
 1
 1

AAC = 
in.  in.  = 0.015625 in.2
16
4



Stress:
σ AC =
FAC
166.277
=
= 10640 psi
AAC
0.015625
σ AC = 10.64 ksi 
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PROBLEM 1.10
Three forces, each of magnitude P = 4 kN, are applied to
the mechanism shown. Determine the cross-sectional area
of the uniform portion of rod BE for which the normal
stress in that portion is +100 MPa.
SOLUTION
Draw free body diagrams of AC and CD.
Free Body CD:
ΣM D = 0: 0.150P − 0.250C = 0
C = 0.6P
Free Body AC:
Required area of BE:
M A = 0: 0.150 FBE − 0.350P − 0.450P − 0.450C = 0
FBE =
1.07
P = 7.1333 P = (7.133)(4 kN) = 28.533 kN
0.150
σ BE =
FBE
ABE
ABE =
FBE
σ BE
=
28.533 × 103
= 285.33 × 10−6 m 2
100 × 106
ABE = 285 mm 2 
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PROBLEM 1.11
The frame shown consists of four wooden members, ABC,
DEF, BE, and CF. Knowing that each member has a 2 × 4-in.
rectangular cross section and that each pin has a 1/2-in.
diameter, determine the maximum value of the average
normal stress (a) in member BE, (b) in member CF.
SOLUTION

Add support reactions to figure as shown.

Using entire frame as free body,
ΣM A = 0: 40 Dx − (45 + 30)(480) = 0
Dx = 900 lb.

Use member DEF as free body.
Reaction at D must be parallel to FBE and FCF .
Dy =
4
Dx = 1200 lb.
3
4

ΣM F = 0: − (30)  FBE  − (30 + 15) DY = 0
5


FBE = −2250 lb.
4

ΣM E = 0: (30)  FCE  − (15) DY = 0
5

FCE = 750 lb.
Stress in compression member BE
Area:
(a)
A = 2 in × 4 in = 8 in 2
σ BE =
FBE
−2250
=
A
8
σ BE = −281 psi 
Minimum section area occurs at pin.
Amin = (2)(4.0 − 0.5) = 7.0 in 2
Stress in tension member CF
(b)
σ CF =
FCF
750
=
Amin
7.0
σ CF =107.1 psi 
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PROBLEM 1.12
For the Pratt bridge truss and loading shown, determine the
average normal stress in member BE, knowing that the crosssectional area of that member is 5.87 in2.
SOLUTION

Use entire truss as free body.
ΣM H = 0: (9)(80) + (18)(80) + (27)(80) − 36 Ay = 0
Ay = 120 kips
Use portion of truss to the left of a section cutting members
BD, BE, and CE.


+ ↑ Σ Fy = 0: 120 − 80 −
σ BE =
12
FBE = 0
15
∴ FBE = 50 kips
FBE
50 kips
=
A
5.87 in 2
σ BE = 8.52 ksi 

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PROBLEM 1.13
An aircraft tow bar is positioned by means of a single
hydraulic cylinder connected by a 25-mm-diameter steel
rod to two identical arm-and-wheel units DEF. The mass
of the entire tow bar is 200 kg, and its center of gravity
is located at G. For the position shown, determine the
normal stress in the rod.
SOLUTION

FREE BODY – ENTIRE TOW BAR:
W = (200 kg)(9.81 m/s 2 ) = 1962.00 N
Σ M A = 0 : 850 R − 1150(1962.00 N) = 0
R = 2654.5 N


FREE BODY – BOTH ARM & WHEEL UNITS:
tan α =
100
675
α = 8.4270°
Σ M E = 0 : ( FCD cos α )(550) − R(500) = 0
FCD =
500
(2654.5 N)
550 cos 8.4270°
= 2439.5 N (comp.)

σ CD = −
2439.5 N
FCD
=−
ACD
π (0.0125 m) 2
= −4.9697 × 106 Pa
σ CD = −4.97 MPa 
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PROBLEM 1.14
A couple M of magnitude 1500 N ⋅ m is applied to the crank of an
engine. For the position shown, determine (a) the force P required to
hold the engine system in equilibrium, (b) the average normal stress in
the connecting rod BC, which has a 450-mm2 uniform cross section.
SOLUTION
Use piston, rod, and crank together as free body. Add wall reaction H
and bearing reactions Ax and Ay.
Σ M A = 0 : (0.280 m) H − 1500 N ⋅ m = 0
H = 5.3571 × 103 N
Use piston alone as free body. Note that rod is a two-force member;
hence the direction of force FBC is known. Draw the force triangle
and solve for P and FBE by proportions.
l =
2002 + 602 = 208.81 mm
P
200
=
H
60

∴
P = 17.86 × 103 N
(a)
P = 17.86 kN 
FBC
208.81
=
∴ FBC = 18.643 × 103 N
H
60
Rod BC is a compression member. Its area is
450 mm 2 = 450 × 10−6 m 2
Stress,
σ BC =

− FBC
−18.643 × 103
=
= −41.4 × 106 Pa
A
450 × 10−6
(b)
σ BC = −41.4 MPa 
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PROBLEM 1.15
When the force P reached 8 kN, the wooden specimen shown failed in
shear along the surface indicated by the dashed line. Determine the
average shearing stress along that surface at the time of failure.
SOLUTION
Area being sheared:
A = 90 mm × 15 mm = 1350 mm 2 = 1350 × 10−6 m 2
Force:
P = 8 × 103 N
Shearing stress:
τ =
P
8 × 103
−
= 5.93 × 106 Pa
A 1350 × 10−6
τ = 5.93 MPa 
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PROBLEM 1.16
The wooden members A and B are to be joined by plywood splice
plates, that will be fully glued on the surfaces in contact. As part of
the design of the joint, and knowing that the clearance between the
ends of the members is to be 14 in., determine the smallest allowable
length L if the average shearing stress in the glue is not to exceed
120 psi.
SOLUTION
There are four separate areas that are glued. Each of these areas transmits one half the 5.8 kip force. Thus
F =
1
1
P = (5.8) = 2.9 kips = 2900 lb.
2
2
Let l = length of one glued area and w = 4 in. be its width.
For each glued area,
A = lw
Average shearing stress:
τ =
F
F
=
A lw
The allowable shearing stress is τ = 120 psi
2900
F
=
= 6.0417 in.
τ w (120)(4)
Solving for l,
l =
Total length L:
L = l + (gap) + l = 6.0417 +
1
+ 6.0417
4
L = 12.33 in. 
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PROBLEM 1.17
A load P is applied to a steel rod supported as shown by an
aluminum plate into which a 0.6-in.-diameter hole has been
drilled. Knowing that the shearing stress must not exceed
18 ksi in the steel rod and 10 ksi in the aluminum plate,
determine the largest load P that can be applied to the rod.
SOLUTION
A1 = π dt = π (0.6)(0.4)
For steel:
= 0.7540 in 2
τ1 =
P
∴ P = A1τ1 = (0.7540)(18)
A
= 13.57 kips
A2 = π dt = π (1.6)(0.25) = 1.2566 in 2
For aluminum:
τ2 =
P
∴ P = A2τ 2 = (1.2566)(10) = 12.57 kips
A2
Limiting value of P is the smaller value, so
P = 12.57 kips 
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PROBLEM 1.18
Two wooden planks, each 22 mm thick and 160 mm
wide, are joined by the glued mortise joint shown.
Knowing that the joint will fail when the average
shearing stress in the glue reaches 820 kPa, determine
the smallest allowable length d of the cuts if the joint is
to withstand an axial load of magnitude P = 7.6 kN.
SOLUTION
Seven surfaces carry the total load P = 7.6 kN = 7.6 × 103.
Let t = 22 mm.
Each glue area is A = dt
τ =
P
7A
A=
P
7.6 × 103
=
= 1.32404 × 10−3 m 2
7τ
(7)(820 × 103 )
= 1.32404 × 103 mm 2
d =
A 1.32404 × 103
=
= 60.2
t
22
d = 60.2 mm 
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PROBLEM 1.19
The load P applied to a steel rod is distributed to a timber support by an annular
washer. The diameter of the rod is 22 mm and the inner diameter of the washer
is 25 mm, which is slightly larger than the diameter of the hole. Determine the
smallest allowable outer diameter d of the washer, knowing that the axial normal
stress in the steel rod is 35 MPa and that the average bearing stress between the
washer and the timber must not exceed 5 MPa.
SOLUTION
Steel rod: A =
π
4
(0.022) 2 = 380.13 × 10−6 m 2
σ = 35 × 106 Pa
P = σ A = (35 × 106 )(380.13 × 10−6 )
= 13.305 × 103 N
Washer: σ b = 5 × 106 Pa
Required bearing area:
Ab =
But, Ab =
π
4
P
σb
=
13.305 × 103
= 2.6609 × 10−3 m 2
5 × 106
(d 2 − di2 )
d 2 = di2 +
4 Ab
π
= (0.025)2 +
(4)(2.6609 × 10−3 )
π
−3
= 4.013 × 10 m
d = 63.3 × 10−3 m
2
d = 63.3 mm 
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PROBLEM 1.20
The axial force in the column supporting the timber beam shown is
P = 20 kips. Determine the smallest allowable length L of the bearing
plate if the bearing stress in the timber is not to exceed 400 psi.
SOLUTION
Bearing area: Ab = Lw
σb =
L=
P
P
=
Ab
Lw
20 × 103
P
=
= 8.33 in.
σ b w (400)(6)
L = 8.33 in. 
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without permission.
PROBLEM 1.21
An axial load P is supported by a short W8 × 40 column of crosssectional area A = 11.7 in.2 and is distributed to a concrete foundation
by a square plate as shown. Knowing that the average normal stress in
the column must not exceed 30 ksi and that the bearing stress on the
concrete foundation must not exceed 3.0 ksi, determine the side a of
the plate that will provide the most economical and safe design.
SOLUTION
For the column σ =
P
or
A
P = σ A = (30)(11.7) = 351 kips
For the a × a plate, σ = 3.0 ksi
A=
P
σ
=
351
= 117 in 2
3.0
Since the plate is square, A = a 2
a =
A = 117
a = 10.82 in. 
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PROBLEM 1.22
A 40-kN axial load is applied to a short wooden post that is
supported by a concrete footing resting on undisturbed soil.
Determine (a) the maximum bearing stress on the concrete
footing, (b) the size of the footing for which the average bearing
stress in the soil is 145 kPa.
SOLUTION
(a)
Bearing stress on concrete footing.
P = 40 kN = 40 × 103 N
A = (100)(120) = 12 × 103 mm 2 = 12 × 10−3 m 2
σ =
(b)
40 × 103
P
=
= 3.333 × 106 Pa
A 12 × 10−3
Footing area. P = 40 × 103 N
σ =
3.33 MPa 
σ = 145 kPa = 45 × 103 Pa
P
A
A=
P
σ
=
40 × 103
= 0.27586 m 2
145 × 103
Since the area is square, A = b 2
b=
A =
0.27586 = 0.525 m
b = 525 mm 
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PROBLEM 1.23
A 58 -in.-diameter steel rod AB is fitted to a round hole near end C of
the wooden member CD. For the loading shown, determine (a) the
maximum average normal stress in the wood, (b) the distance b for
which the average shearing stress is 100 psi on the surfaces indicated
by the dashed lines, (c) the average bearing stress on the wood.
SOLUTION
(a)
Maximum normal stress in the wood
5

Anet = (1)  4 −  = 3.375 in.2
8

1500
P
σ =
=
= 444 psi
3.375
Anet
(b)
σ = 444 psi 
Distance b for τ = 100 psi
For sheared area see dotted lines.
P
P
=
A 2bt
1500
P
b=
=
= 7.50 in.
2tτ
(2)(1)(100)
τ =
(c)
b = 7.50 in. 
Average bearing stress on the wood
σb =
P
P
1500
=
=
= 2400 psi
5
Ab
dt
  (1)
8
σ b = 2400 psi 
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PROBLEM 1.24
Knowing that θ = 40° and P = 9 kN, determine (a) the smallest
allowable diameter of the pin at B if the average shearing stress in
the pin is not to exceed 120 MPa, (b) the corresponding average
bearing stress in member AB at B, (c) the corresponding average
bearing stress in each of the support brackets at B.
SOLUTION
Geometry: Triangle ABC is an isoseles triangle with angles shown here.
Use joint A as a free body.
Law of sines applied to force triangle
P
FAB
FAC
=
=
sin 20° sin110° sin 50°
P sin110°
FAB =
sin 20°
(9)sin110°
=
= 24.73 kN
sin 20°
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PROBLEM 1.24 (Continued)
(a)
Allowable pin diameter.
τ =
FAB
F
2 FAB
= πAB 2 =
where FAB = 24.73 × 103 N
2
2 AP
πd
24d
d2 =
2 FAB
πτ
=
(2)(24.73 × 103 )
= 131.18 × 10−6 m 2
6
π (120 × 10 )
d = 11.45 × 10−3 m
(b)
11.45 mm 
Bearing stress in AB at A.
Ab = td = (0.016)(11.45 × 10−3 ) = 183.26 × 10−6 m 2
σb =
(c)
FAB
24.73 × 103
=
= 134.9 × 106
Ab
183.26 × 10−6
134.9 MPa 
Bearing stress in support brackets at B.
A = td = (0.012)(11.45 × 10−3 ) = 137.4 × 10−6 m 2
σb =
1
2
FAB
A
=
(0.5)(24.73 × 103 )
= 90.0 × 106
137.4 × 10−6
90.0 MPa 
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PROBLEM 1.25
Determine the largest load P which may be applied at A when θ = 60°,
knowing that the average shearing stress in the 10-mm-diameter pin at
B must not exceed 120 MPa and that the average bearing stress in
member AB and in the bracket at B must not exceed 90 MPa.
SOLUTION
Geometry: Triangle ABC is an isoseles triangle with angles shown here.
Use joint A as a free body.
Law of sines applied to force triangle
P
FAB
FAC
=
=
sin 30° sin 120° sin 30°
P=
FAB sin 30°
= 0.57735 FAB
sin 120°
P=
FAC sin 30°
= FAC
sin 30°
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PROBLEM 1.25 (Continued)
If shearing stress in pin at B is critical,
A=
π
4
d2 =
π
4
(0.010) 2 = 78.54 × 10−6 m 2
FAB = 2 Aτ = (2)(78.54 × 10−6 )(120 × 106 ) = 18.850 × 103 N
If bearing stress in member AB at bracket at A is critical,
Ab = td = (0.016)(0.010) = 160 × 10−6 m 2
FAB = Abσ b = (160 × 10−6 )(90 × 106 ) = 14.40 × 103 N
If bearing stress in the bracket at B is critical,
Ab = 2td = (2)(0.012)(0.010) = 240 × 10−6 m 2
FAB = Abσ b = (240 × 10−6 )(90 × 106 ) = 21.6 × 103 N
Allowable FAB is the smallest, i.e., 14.40 × 103N
Then from Statics
Pallow = (0.57735)(14.40 × 103 )
= 8.31 × 103 N
8.31 kN 
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PROBLEM 1.26
Link AB, of width b = 50 mm and thickness t = 6 mm, is used to support
the end of a horizontal beam. Knowing that the average normal stress in
the link is − 140 MPa, and that the average shearing stress in each of the
two pins is 80 MPa, determine (a) the diameter d of the pins, (b) the
average bearing stress in the link.
SOLUTION
Rod AB is in compression.
A = bt
where b = 50 mm and t = 6 mm
A = (0.050)(0.006) = 300 × 10−6 m 2
P = −σ A = −(−140 × 106 )(300 × 10−6 )
= 42 × 103 N
For the pin,
Ap =
π
Ap =
(a)
P
τ
=
and τ =
P
Ap
42 × 103
= 525 × 10−6 m 2
80 × 106
Diameter d
d =
(b)
4
d2
Bearing stress
σb =
4 Ap
π
=
(4)(525 × 10 −6 )
π
= 2.585 × 10−3 m
P
42 × 103
=
= 271 × 106 Pa
dt
(25.85 × 10−3 )(0.006)
d = 25.9 mm 
σ b = 271 MPa 
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PROBLEM 1.27
For the assembly and loading of Prob. 1.7, determine (a) the average
shearing stress in the pin at B, (b) the average bearing stress at B in
member BD, (c) the average bearing stress at B in member ABC, knowing
that this member has a 10 × 50-mm uniform rectangular cross section.
PROBLEM 1.7 Each of the four vertical links has an 8 × 36-mm uniform
rectangular cross section and each of the four pins has a 16-mm diameter.
Determine the maximum value of the average normal stress in the links
connecting (a) points B and D, (b) points C and E.
SOLUTION
Use bar ABC as a free body.
ΣM C = 0 : (0.040) FBD − (0.025 + 0.040)(20 × 103 ) = 0
FBD = 32.5 × 103 N
(a)
Shear pin at B
τ =
where
A=
τ =
(b)
Bearing: link BD
Bearing in ABC at B
π
4
d2 =
π
4
(0.016) 2 = 201.06 × 10−6 m 2
32.5 × 103
= 80.8 × 106
(2)(201.06 × 10−6 )
τ = 80.8 MPa 
A = dt = (0.016)(0.008) = 128 × 10−6 m 2
σb =
(c)
FBD
for double shear,
2A
1
2
FBD
A
=
(0.5)(32.5 × 103 )
= 126.95 × 106
−6
128 × 10
σ b = 127.0 MPa 
A = dt = (0.016)(0.010) = 160 × 10−6 m 2
σb =
FBD
32.5 × 103
=
= 203 × 106
−6
A
160 × 10
σ b = 203 MPa 
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PROBLEM 1.28
The hydraulic cylinder CF, which partially controls the position of rod
DE, has been locked in the position shown. Member BD is 58 in. thick
and is connected to the vertical rod by a 83 -in.-diameter bolt.
Determine (a) the average shearing stress in the bolt, (b) the bearing
stress at C in member BD.
SOLUTION
Use member BCD as a free body, and note that AB is a two force member.
l AB =
82 + 1.82 = 8.2 in.
 8

 1.8

ΣM C = 0: (4 cos 20°) 
FAB  − (4sin 20°) 
FAB 
 8.2

 8.2

−(7 cos 20°)(400sin 75°) − (7sin 20°)(400 cos 75°) = 0
3.36678FAB − 2789.35 = 0
∴ FAB = 828.49 lb
1.8
FAB + Cx + 400cos 75° = 0
8.2
(1.8)(828.49)
− 400cos 75° = 78.34 lb
Cx =
8.2
ΣFx = 0: −
8
FAB + C y − 400sin 75° = 0
8.2
(8)(828.49)
+ 400sin 75° = 1194.65 lb
Cy =
8.2
ΣFy = 0: −
C =
Cx2 + C y2 = 1197.2 lb
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PROBLEM 1.28 (Continued)
2
(a)
(b)
Shearing stress in the bolt: P = 1197.2 lb
A=
π 2 π 3
d =   = 0.11045 in 2
4
4 8
τ =
1197.2
P
=
= 10.84 × 103 psi =
A 0.11045
10.84 ksi 
 3  5 
Bearing stress at C in member BCD: P = 1197.2 lb Ab = dt =    = 0.234375 in 2
 8  8 
σb =
P
1197.2
=
= 5.11 × 103 psi = 5.11 ksi 
0.234375
Ab
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PROBLEM 1.29
The 1.4-kip load P is supported by two wooden members of uniform cross section
that are joined by the simple glued scarf splice shown. Determine the normal and
shearing stresses in the glued splice.
SOLUTION
P = 1400 lb
θ = 90° − 60° = 30°
A0 = (5.0)(3.0) = 15 in 2
σ =
P cos 2 θ
(1400)(cos 30°)2
=
A0
15
σ = 70.0 psi 
τ =
P sin 2θ
(1400)sin 60°
=
2 A0
(2)(15)
τ = 40.4 psi 
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PROBLEM 1.30
Two wooden members of uniform cross section are joined by the simple scarf
splice shown. Knowing that the maximum allowable tensile stress in the glued
splice is 75 psi, determine (a) the largest load P that can be safely supported, (b)
the corresponding shearing stress in the splice.
SOLUTION
A0 = (5.0)(3.0) = 15 in 2
θ = 90° − 60° = 30°
σ =
(a)
(b)
P=
P cos 2 θ
A0
σ A0
(75)(15)
=
= 1500 lb
2
cos θ
cos 2 30°
P = 1.500 kips 
P sin 2θ
(1500)sin 60°
=
2 A0
(2)(15)
τ = 43.3 psi 
τ =
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PROBLEM 1.31
Two wooden members of uniform rectangular cross section are
joined by the simple glued scarf splice shown. Knowing that
P = 11 kN, determine the normal and shearing stresses in the glued
splice.
SOLUTION
θ = 90° − 45° = 45°
P = 11 kN = 11 × 103 N
A0 = (150)(75) = 11.25 × 103 mm 2 = 11.25 × 10−3 m 2
σ =
P cos 2 θ
(11 × 103 ) cos 2 45°
=
= 489 × 103 Pa
−3
A0
11.25 × 10
σ = 489 kPa 
τ =
P sin 2θ
(11 × 103 )(sin 90°)
=
= 489 × 103 Pa
2 A0
(2)(11.25 × 10−3 )
τ = 489 kPa 
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PROBLEM 1.32
Two wooden members of uniform rectangular cross section are
joined by the simple glued scarf splice shown. Knowing that the
maximum allowable shearing stress in the glued splice is 620 kPa,
determine (a) the largest load P that can be safely applied, (b) the
corresponding tensile stress in the splice.
SOLUTION
θ = 90° − 45° = 45°
A0 = (150)(75) = 11.25 × 103 mm 2 = 11.25 × 10−3 m2
τ = 620 kPa = 620 × 103 Pa
P sin 2θ
τ =
2 A0
(a)
P=
2 A0τ
(2)(11.25 × 10−3 )(620 × 103 )
=
sin2θ
sin 90°
= 13.95 × 103 N
(b)
σ =
P = 13.95 kN 
P cos 2 θ
(13.95 × 103 )(cos 45°)2
=
A0
11.25 × 10−3
= 620 × 103 Pa
σ = 620 kPa 
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PROBLEM 1.33
A steel pipe of 12-in. outer diameter is fabricated from 14 -in.-thick plate by
welding along a helix that forms an angle of 25° with a plane perpendicular
to the axis of the pipe. Knowing that the maximum allowable normal and
shearing stresses in the directions respectively normal and tangential to the
weld are σ = 12 ksi and τ = 7.2 ksi, determine the magnitude P of the
largest axial force that can be applied to the pipe.
SOLUTION
1
d o = 6 in.
2
ri = ro − t = 6 − 0.25 = 5.75 in.
do = 12 in. ro =
A0 = π (ro2 − ri2 ) = π (62 − 5.752 ) = 9.228 in 2
θ = 25°
Based on σ = 12 ksi: σ =
P
cos 2 θ
A0
P=
Based on τ = 7.2 ksi: τ =
P=
A0σ
(9.228)(12 × 103 )
=
= 134.8 × 103 lb
cos 2 θ
cos 2 25°
P
sin 2θ
2 A0
2 A0τ
(2)(9.288)(7.2 × 103 )
=
= 174.5 × 103 lb
sin 2θ
sin 50°
The smaller calculated value of P is the allowable value.
P = 134.8 × 103 lb
P = 134.8 kips 
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PROBLEM 1.34
A steel pipe of 12-in. outer diameter is fabricated from 14 -in.-thick plate by
welding along a helix that forms an angle of 25° with a plane perpendicular
to the axis of the pipe. Knowing that a 66 kip axial force P is applied to the
pipe, determine the normal and shearing stresses in directions respectively
normal and tangential to the weld.
SOLUTION
1
d o = 6 in.
2
ri = ro − t = 6 − 0.25 = 5.75 in.
do = 12 in. ro =
A0 = π (ro2 − ri2 ) = π (62 − 5.752 ) = 9.228 in 2
θ = 25°
Normal stress:
Shearing stress:
σ =
τ =
P cos 2 θ
(66 × 103 ) cos 2 25°
=
= 5875 psi
A0
9.228
σ = 5.87 ksi 
P sin 2θ
(66 × 103 )sin 50°
=
= 2739 psi
2 A0
(2)(9.228)
τ = 2.74 ksi 
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PROBLEM 1.35
A 1060-kN load P is applied to the granite block shown. Determine the
resulting maximum value of (a) the normal stress, (b) the shearing stress.
Specify the orientation of the plane on which each of these maximum
values occurs.
SOLUTION
A0 = (140 mm)(140 mm) = 19.6 × 103 mm 2 = 19.6 × 10−3 m 2
P = 1060 × 103 N
σ =
(a)
P
1060 × 103
cos 2 θ =
cos 2 θ = 54.082 × 106 cos 2 θ
−3
A0
19.6 × 10
Maximum tensile stress = 0 at θ = 90°.
Maximum compressive stress = 54.1 × 106 at θ = 0°.
(b)
|σ |max = 54.1 MPa 
Maximum shearing stress:
τ max =
P
1060 × 103
=
= 27.0 × 106 Pa at θ = 45°.
2 A0
(2)(19.6 × 10−3 )
τ max = 27.0 MPa 
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PROBLEM 1.36
A centric load P is applied to the granite block shown. Knowing that the
resulting maximum value of the shearing stress in the block is 18 MPa,
determine (a) the magnitude of P, (b) the orientation of the surface on
which the maximum shearing stress occurs, (c) the normal stress exerted on
that surface, (d) the maximum value of the normal stress in the block.
SOLUTION
A0 = (140 mm)(140 mm) = 19.6 × 103 mm 2 = 19.6 × 10−3 m 2
τ max = 18 MPa = 18 × 106 Pa
θ = 45° for plane of τ max
(a)
Magnitude of P. τ max =
|P|
so P = 2 A0 τ max
2 A0
P = (2)(19.6 × 10−3 )(18 × 106 ) = 705.6 × 103 N
sin 2θ is maximum when 2θ = 90°
(b)
Orientation.
(c)
Normal stress at θ = 45°.
σ =
(d)
P = 706 kN 
θ = 45° 
P cos 2 θ
(705.8 × 103 ) cos 2 45°
=
= 18.00 × 106 Pa
A0
19.6 × 10−3
Maximum normal stress: σ max =
σ max =
σ = 18.00 MPa 
P
A0
705.8 × 103
= 36.0 × 106 Pa
19.6 × 10−3
σ max = 36.0 MPa (compression) 
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PROBLEM 1.37
Link BC is 6 mm thick, has a width w = 25 mm, and is made of a steel with
a 480-MPa ultimate strength in tension. What was the safety factor used if
the structure shown was designed to support a 16-kN load P?
SOLUTION
Use bar ACD as a free body and note that member BC is a two-force member.
ΣM A = 0:
(480) FBC − (600) P = 0
FBC =
Ultimate load for member BC:
600
(600)(16 × 103 )
= 20 × 103 N
P=
480
480
FU = σ U A
FU = (480 × 106 )(0.006)(0.025) = 72 × 103 N
Factor of safety:
F.S. =
FU
72 × 103
=
FBC
20 × 103
F.S. = 3.60 
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PROBLEM 1.38
Link BC is 6 mm thick and is made of a steel with a 450-MPa ultimate
strength in tension. What should be its width w if the structure shown is
being designed to support a 20-kN load P with a factor of safety of 3?
SOLUTION
Use bar ACD as a free body and note that member BC is a two-force member.
ΣM A = 0:
(480) FBC − 600 P = 0
FBC =
600 P (600)(20 × 103 )
=
= 25 × 103 N
480
480
For a factor of safety F.S. = 3, the ultimate load of member BC is
FU = (F.S.)( FBC ) = (3)(25 × 103 ) = 75 × 103 N
But FU = σ U A
∴A =
For a rectangular section
FU
σU
=
75 × 103
= 166.67 × 10−6 m 2
450 × 106
A = wt or w =
A 166.67 × 10−6
=
t
0.006
w = 27.8 × 10−3 m or 27.8 mm 
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PROBLEM 1.39
A 34 -in.-diameter rod made of the same material as rods AC and AD in
the truss shown was tested to failure and an ultimate load of 29 kips was
recorded. Using a factor of safety of 3.0, determine the required diameter
(a) of rod AC, (b) of rod AD.
SOLUTION
Forces in AC and AD.
Joint C:
ΣFy = 0:
Joint D:
Ultimate stress. From test on
ΣFy = 0:
3
4
σU =
-in. rod:
σ all =
Allowable stress:
(a)
Diameter of rod AC.
(b)
Diameter of rod AD.
σ all =
FAC
1 πd2
4
d2 =
d2 =
1
FAC − 10 kips = 0
5
FAC = 22.36 kips T
1
FAD − 10 kips = 0
17
FAD = 41.23 kips T
PU
29 kips
= 1 3 2 = 65.64 ksi
A
π (4)
4
σU
F .S .
4FAC
πσ all
4 FAD
πσ all
65.64 ksi
= 21.88 ksi
3.0
=
=
4(22.36)
= 1.301
π (21.88)
d = 1.141 in. 
=
4(41.23)
= 2.399
π (21.88)
d = 1.549 in. 
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PROBLEM 1.40
In the truss shown, members AC and AD consist of rods made of the
same metal alloy. Knowing that AC is of 1-in. diameter and that the
ultimate load for that rod is 75 kips, determine (a) the factor of safety
for AC, (b) the required diameter of AD if it is desired that both rods
have the same factor of safety.
SOLUTION
Forces in AC and AD.
Joint C:
ΣFy = 0:
Joint D:
ΣFy = 0:
F.S. =
PU
FAC
F.S. =
1
FAC − 10 kips = 0
5
FAC = 22.36 kips T
1
FAD − 10 kips = 0
17
FAD = 41.23 kips T
75 kips
22.36 kips
(a)
Factor of safety for AC.
(b)
For the same factor of safety in AC and AD, σ AD = σ AC .
F.S. = 3.35 
FAD
F
= AC
AAD
AAC
AAD =
Required diameter:
d AD =
41.23 π 2
FAD
AAC =
(1) = 1.4482 in 2
FAC
22.36 4
4 AAD
π
=
(4)(1.4482)
π
d AD = 1.358 in. 
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PROBLEM 1.41
Link AB is to be made of a steel for which the ultimate normal stress is
450 MPa. Determine the cross-sectional area for AB for which the
factor of safety will be 3.50. Assume that the link will be adequately
reinforced around the pins at A and B.
SOLUTION
P = (1.2)(8) = 9.6 kN
ΣM D = 0 :
−(0.8)( FAB sin 35°)
+ (0.2)(9.6) + (0.4)(20) = 0
FAB = 21.619 kN = 21.619 × 103 N
σ AB =
AAB =
FAB
σ
= ult
AAB
F. S .
( F. S.) FAB
σ ult
=
(3.50)(21.619 × 103 )
450 × 106
= 168.1 × 10−6 m 2
AAB = 168.1 mm 2 
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PROBLEM 1.42
A steel loop ABCD of length 1.2 m and of 10-mm diameter is placed as
shown around a 24-mm-diameter aluminum rod AC. Cables BE and DF,
each of 12-mm diameter, are used to apply the load Q. Knowing that the
ultimate strength of the steel used for the loop and the cables is 480 MPa
and that the ultimate strength of the aluminum used for the rod is
260 MPa, determine the largest load Q that can be applied if an overall
factor of safety of 3 is desired.
SOLUTION
Using joint B as a free body and considering symmetry,
2⋅
3
6
FAB − Q = 0 Q = FAB
5
5
Using joint A as a free body and considering symmetry,
4
FAB − FAC = 0
5
8 5
3
⋅ Q − FAC = 0 ∴ Q = FAC
5 6
4
2⋅
Based on strength of cable BF:
QU = σ U A = σ U
π
4
d 2 = (480 × 106 )
π
4
(0.012) 2 = 54.29 × 103 N
Based on strength of steel loop:
6
6
6
π
FAB, U = σ U A = σ U d 2
5
5
5
4
π
6
= (480 × 106 ) (0.010) 2 = 45.24 × 103 N
5
4
QU =
Based on strength of rod AC:
QU =
3
3
3
π
3
π
FAC , U = σ U A = σ U d 2 = (260 × 106 ) (0.024) 2 = 88.22 × 103 N
4
4
4
4
4
4
Actual ultimate load QU is the smallest, ∴ QU = 45.24 × 103 N
Allowable load:
Q=
QU
45.24 × 103
=
= 15.08 × 103 N
F. S .
3
Q = 15.08 kN 
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PROBLEM 1.43
Two wooden members shown, which support a 3.6 kip load, are joined
by plywood splices fully glued on the surfaces in contact. The ultimate
shearing stress in the glue is 360 psi and the clearance between the
members is 14 in. Determine the required length L of each splice if a
factor of safety of 2.75 is to be achieved.
SOLUTION
There are 4 separate areas of glue. Let l be the length of each area and w = 5 in. its width. Then the area
is A = lw.
Each glue area transmits one half of the total load.
1
F =   (3.6 kips) = 1.8 kips
2
Required ultimate load for each glue area:
FU = ( F . S .) F = (2.75)(1.8) = 4.95 kips
Required length of each glue area:
FU = τ U A = τ U lw
l =
Total length of splice:
FU
4.95 × 103
=
= 2.75 in.
τU w
(360)(5)
L=l+
1
in. + l
4
L = 2.75 + 0.25 + 2.75
L = 5.75 in. 
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PROBLEM 1.44
Two plates, each 18 in. thick, are used to splice a plastic strip as shown.
Knowing that the ultimate shearing stress of the bonding between the
surface is 130 psi, determine the factor of safety with respect to shear
when P = 325 lb.
SOLUTION
Bond area: (See figure)
1
(2.25)(0.75) + (2.25)(0.625) = 2.25 in 2
2
PU = 2 AτU = (2)(2.25)(130) = 585 lb.
A=
F. S . =
PU
585
=
= 1.800 
P
325
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PROBLEM 1.45
A load P is supported as shown by a steel pin that has been inserted in a
short wooden member hanging from the ceiling. The ultimate strength of
the wood used is 60 MPa in tension and 7.5 MPa in shear, while the
ultimate strength of the steel is 145 MPa in shear. Knowing that
b = 40 mm, c = 55 mm, and d = 12 mm, determine the load P if an
overall factor of safety of 3.2 is desired.
SOLUTION
Based on double shear in pin:
PU = 2 AτU = 2
=
π
4
π
4
d 2τ U
(2)(0.012)2 (145 × 106 ) = 32.80 × 103 N
Based on tension in wood:
PU = Aσ U = w (b − d )σ U
= (0.040)(0.040 − 0.012)(60 × 106 )
= 67.2 × 103 N
Based on double shear in the wood:
PU = 2 AτU = 2wcτU = (2)(0.040)(0.055)(7.5 × 106 )
= 33.0 × 103 N
Use smallest
Allowable:
PU = 32.8 × 103 N
P=
PU
32.8 × 103
=
= 10.25 × 103 N
F .S.
3.2
10.25 kN 
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PROBLEM 1.46
For the support of Prob. 1.45, knowing that the diameter of the pin is
d = 16 mm and that the magnitude of the load is P = 20 kN, determine
(a) the factor of safety for the pin, (b) the required values of b and c if the
factor of safety for the wooden members is the same as that found in part a
for the pin.
PROBLEM 1.45 A load P is supported as shown by a steel pin that has
been inserted in a short wooden member hanging from the ceiling. The
ultimate strength of the wood used is 60 MPa in tension and 7.5 MPa in
shear, while the ultimate strength of the steel is 145 MPa in shear.
Knowing that b = 40 mm, c = 55 mm, and d = 12 mm, determine the
load P if an overall factor of safety of 3.2 is desired.
SOLUTION
P = 20 kN = 20 × 103 N
(a)
Pin:
A=
Double shear:
τ =
π
4
d2 =
π
4
(0.016) 2 = 2.01.06 × 10−6 m 2
P
P
τU = U
2A
2A
PU = 2 AτU = (2)(201.16 × 10−6 )(145 × 106 ) = 58.336 × 103 N
F .S . =
(b)
Tension in wood:
where w = 40 mm = 0.040 m
b = 40.3 mm 
PU = 58.336 × 103 N for same F.S.
Double shear; each area is A = wc
c=
PU
PU
=
A
w(b − d )
PU
58.336 × 103
= 0.016 +
= 40.3 × 10−3 m
6
wσ U
(0.040)(60 × 10 )
Shear in wood:
F .S . = 2.92 
PU = 58.336 × 103 N for same F.S.
σU =
b=d +
PU
58.336 × 103
=
P
20 × 103
τU =
PU
P
= U
2 A 2wc
PU
58.336 × 103
=
= 97.2 × 10−3 m
2wτU
(2)(0.040)(7.5 × 106 )
c = 97.2 mm 
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PROBLEM 1.47
Three steel bolts are to be used to attach the steel plate shown to a wooden beam.
Knowing that the plate will support a 110-kN load, that the ultimate shearing
stress for the steel used is 360 MPa, and that a factor of safety of 3.35 is desired,
determine the required diameter of the bolts.
SOLUTION
110
= 36.667 kN
3
For each bolt,
P=
Required:
PU = ( F. S.) P = (3.35)(36.667) = 122.83 kN
τU =
d =
PU
P
4P
= π U 2 = U2
A
d
πd
4
4PU
πτU
=
(4)(122.83 × 103 )
= 20.8 × 10−3 m
π (360 × 106 )
d = 20.8 mm 
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PROBLEM 1.48
Three 18-mm-diameter steel bolts are to be used to attach the steel plate shown to
a wooden beam. Knowing that the plate will support a 110-kN load and that the
ultimate shearing stress for the steel used is 360 MPa, determine the factor of
safety for this design.
SOLUTION
For each bolt,
A=
π
4
d2 =
π
4
(18) 2 = 254.47 mm 2 = 254.47 × 10−6 m 2
PU = Aτ U = (254.47 × 10−6 )(360 × 106 )
= 91.609 × 103 N
For the three bolts,
PU = (3)(91.609 × 103 ) = 274.83 × 103 N
Factor of safety:
F. S . =
PU
274.83 × 103
=
P
110 × 103
F. S . = 2.50 
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PROBLEM 1.49
A steel plate 165 in. thick is embedded in a horizontal concrete slab and is
used to anchor a high-strength vertical cable as shown. The diameter of
the hole in the plate is 34 in., the ultimate strength of the steel used is 36 ksi,
and the ultimate bonding stress between plate and concrete is 300 psi.
Knowing that a factor of safety of 3.60 is desired when P = 2.5 kips,
determine (a) the required width a of the plate, (b) the minimum depth b
to which a plate of that width should be embedded in the concrete slab.
(Neglect the normal stresses between the concrete and the lower end of
the plate.)
SOLUTION
Based on tension in plate:
A = (a − d )t
PU = σ U A
F .S . =
σ (a − d )t
PU
= U
P
P
Solving for a,
a =d +
( F .S .) P
3 (3.60)(2.5)
= +
4
σU t
(36) 165
( )
(a) a = 1.550 in. 
Based on shear between plate and concrete slab,
τU = 0.300 ksi
A = perimeter × depth = 2(a + t )b
PU = τ U A = 2τ U (a + t )b
Solving for b,
b=
F .S. =
PU
P
( F .S.) P
(3.6)(2.5)
=
2(a + t )τU
(2) 1.550 + 165 (0.300)
(
)
(b) b = 8.05 in. 
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PROBLEM 1.50
Determine the factor of safety for the cable anchor in Prob. 1.49 when
P = 3 kips, knowing that a = 2 in. and b = 7.5 in.
PROBLEM 1.49 A steel plate 165 in, thick is embedded in a horizontal
concrete slab and is used to anchor a high-strength vertical cable as
shown. The diameter of the hole in the plate is 34 in., the ultimate
strength of the steel used is 36 ksi, and the ultimate bonding stress
between plate and concrete is 300 psi. Knowing that a factor of safety of
3.60 is desired when P = 2.5 kips, determine (a) the required width a
of the plate, (b) the minimum depth b to which a plate of that width
should be embedded in the concrete slab. (Neglect the normal stresses
between the concrete and the lower end of the plate.)
SOLUTION
Based on tension in plate:
A = (a − d )t
3  5 

=  2 −   = 0.3906 in 2
4  16 

PU = σ U A
= (36)(0.3906) = 14.06 kips
F .S . =
PU
14.06
=
= 4.69
P
3
Based on shear between plate and concrete slab:
5

A = perimeter × depth = 2(a + t )b = 2  2 +  (7.5)
16 

A = 34.69 in 2
τU = 0.300 ksi
PU = τ U A = (0.300)(34.69) = 10.41 kips
F .S . =
PU
10.41
=
= 3.47
P
3
Actual factor of safety is the smaller value.
F .S . = 3.47 
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PROBLEM 1.51
In the steel structure shown, a 6-mm-diameter pin
is used at C and 10-mm-diameter pins are used at B
and D. The ultimate shearing stress is 150 MPa at
all connections, and the ultimate normal stress is
400 MPa in link BD. Knowing that a factor of
safety of 3.0 is desired, determine the largest load P
that can be applied at A. Note that link BD is not
reinforced around the pin holes.
SOLUTION
Use free body ABC.
ΣM C = 0 : 0.280 P − 0.120 FBD = 0
P=
3
FBD
7
(1)
ΣM B = 0 : 0.160 P − 0.120 C = 0
P=
3
C
4
(2)
Tension on net section of link BD.
 400 × 106 
−3
−3
3
Anet = 
 (6 × 10 )(18 − 10)(10 ) = 6.40 × 10 N
3
F. S .


Shear in pins at B and D.
FBD = σ Anet =
FBD = τ Apin =
σU
 150 × 106   π 
−3 2
3
d 2 = 
   (10 × 10 ) = 3.9270 × 10 N
3
4
F. S. 4

 
τU π
Smaller value of FBD is 3.9270 × 103 N.
From (1)
3
P =   (3.9270 × 103 ) = 1.683 × 103 N
7
Shear in pin at C.
C = 2τ Apin = 2
From (2)
3
P =   (2.8274 × 103 ) = 2.12 × 103 N
4
Smaller value of P is allowable value.
 150 × 106   π 
−3 2
3
d 2 = (2) 
   (6 × 10 ) = 2.8274 × 10 N
3
4
F. S. 4

 
τU π
P = 1.683 × 103 N
P = 1.683 kN 
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PROBLEM 1.52
Solve Prob. 1.51, assuming that the structure has
been redesigned to use 12-mm-diameter pins at B
and D and no other change has been made.
PROBLEM 1.51 In the steel structure shown, a
6-mm-diameter pin is used at C and 10-mmdiameter pins are used at B and D. The ultimate
shearing stress is 150 MPa at all connections, and
the ultimate normal stress is 400 MPa in link BD.
Knowing that a factor of safety of 3.0 is desired,
determine the largest load P that can be applied
at A. Note that link BD is not reinforced around the
pin holes.
SOLUTION
Use free body ABC.
ΣM C = 0 : 0.280 P − 0.120 FBD = 0
P=
3
FBD
7
(1)
ΣM B = 0 : 0.160 P − 0.120 C = 0
P=
3
C
4
(2)
Tension on net section of link BD.
FBD = σ Anet =
 400 × 106 
−3
−3
3
Anet = 
 (6 × 10 )(18 − 12)(10 ) = 4.80 × 10 N
3
F. S.


σU
Shear in pins at B and D.
FBD = τ Apin =
 150 × 106   π 
−3 2
3
d 2 = 
   (12 × 10 ) = 5.6549 × 10 N
3
4
F. S. 4




τU π
Smaller value of FBD is 4.80 × 103 N.
From (1),
3
P =   (4.80 × 103 ) = 2.06 × 103 N
7
Shear in pin at C. C = 2τ Apin = 2
From (2),
 150 × 106   π 
−3 2
3
d 2 = (2) 
   (6 × 10 ) = 2.8274 × 10 N
3
4
F. S. 4




τU π
3
P =   (2.8274 × 103 ) = 2.12 × 103 N
4
Smaller value of P is the allowable value.
P = 2.06 × 103 N
P = 2.06 kN 
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PROBLEM 1.53
Each of the two vertical links CF connecting the two horizontal members AD
and EG has a uniform rectangular cross section 14 in. thick and 1 in. wide,
and is made of a steel with an ultimate strength in tension of 60 ksi. The pins
at C and F each have a 12 -in. diameter and are made of a steel with an
ultimate strength in shear of 25 ksi. Determine the overall factor of safety for
the links CF and the pins connecting them to the horizontal members.
SOLUTION
Use member EFG as free body.
ΣM E = 0 : 16 FCF − (26)(2) = 0
FCF = 3.25 kips
Failure by tension in links CF. (2 parallel links)
Net section area for 1 link: A = (b − d ) t = (1 − 12 )( 14 ) = 0.125 in 2
FU = 2 Aσ U = (2)(0.125)(60) = 15 kips
Failure by double shear in pins.
A=
π
4
2
d =
FU = 2 AτU
π 1
2
2
  = 0.196350 in
42
= (2)(0.196350)(25) = 9.8175 kips
Actual ultimate load is the smaller value. FU = 9.8175 kips
Factor of safety:
F. S . =
FU
9.8175
=
FCF
3.25
F. S . = 3.02 
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PROBLEM 1.54
Solve Prob. 1.53, assuming that the pins at C and F have been replaced by pins
with a 34 -in diameter.
PROBLEM 1.53 Each of the two vertical links CF connecting the two
horizontal members AD and EG has a uniform rectangular cross section 14 in.
thick and 1 in. wide, and is made of a steel with an ultimate strength in tension
of 60 ksi. The pins at C and F each have a 12 -in. diameter and are made of a
steel with an ultimate strength in shear of 25 ksi. Determine the overall factor
of safety for the links CF and the pins connecting them to the horizontal
members.
SOLUTION
Use member EFG as free body.
ΣM E = 0 : 16 FCF − (26)(2) = 0
FCF = 3.25 kips
Failure by tension in links CF. (2 parallel links)
A = (b − d ) t = (1 − 34 )( 14 ) = 0.0625 in 2
Net section area for 1 link:
FU = 2 Aσ U = (2)(0.0625)(60) = 7.5 kips
Failure by double shear in pins.
A=
π
4
d2 =
π 3
2
2
  = 0.44179 in
44
FU = 2 AτU = (2)(0.44179)(25) = 22.09 kips
Actual ultimate load is the smaller value. FU = 7.5 kips
Factor of safety:
F. S . =
FU
7.5
=
FCF
3.25
F. S . = 2.31 
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PROBLEM 1.55
In the structure shown, an 8-mm-diameter pin is
used at A, and 12-mm-diameter pins are used at B
and D. Knowing that the ultimate shearing stress is
100 MPa at all connections and that the ultimate
normal stress is 250 MPa in each of the two links
joining B and D, determine the allowable load P if
an overall factor of safety of 3.0 is desired.
SOLUTION
Statics: Use ABC as free body.
ΣM B = 0 : 0.20 FA − 0.18 P = 0
ΣM A = 0 : 0.20 FBD − 0.38 P = 0
Based on double shear in pin A: A =
FA =
2τ U A
=
π
4
10
FA
9
P=
d2 =
P=
π
10
FBD
19
(0.008) 2 = 50.266 × 10−6 m 2
4
(2)(100 × 106 )(50.266 × 10−6 )
= 3.351 × 103 N
3.0
F .S.
10
P=
FA = 3.72 × 103 N
9
Based on double shear in pins at B and D: A =
FBD =
2τ U A
=
π
4
d2 =
π
4
(0.012) 2 = 113.10 × 10−6 m 2
(2)(100 × 106 )(113.10 × 10−6 )
= 7.54 × 103 N
3.0
F .S.
10
P=
FBD = 3.97 × 103 N
19
Based on compression in links BD: For one link, A = (0.020)(0.008) = 160 × 10−6 m 2
2σ U A (2)(250 × 106 )(160 × 10−6 )
=
= 26.7 × 103 N
F .S.
3.0
10
P=
FBD = 14.04 × 103 N
19
FBD =
Allowable value of P is smallest, ∴ P = 3.72 × 103 N
P = 3.72 kN 
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PROBLEM 1.56
In an alternative design for the structure of Prob. 1.55, a
pin of 10-mm-diameter is to be used at A. Assuming that
all other specifications remain unchanged, determine the
allowable load P if an overall factor of safety of 3.0 is
desired.
PROBLEM 1.55 In the structure shown, an 8-mmdiameter pin is used at A, and 12-mm-diameter pins are
used at B and D. Knowing that the ultimate shearing
stress is 100 MPa at all connections and that the ultimate
normal stress is 250 MPa in each of the two links joining
B and D, determine the allowable load P if an overall
factor of safety of 3.0 is desired.
SOLUTION
Statics: Use ABC as free body.
ΣM B = 0: 0.20 FA − 0.18 P = 0
ΣM A = 0: 0.20 FBD − 0.38 P = 0
Based on double shear in pin A: A =
10
FA
9
10
P=
FBD
19
P=
π
4
d2 =
π
4
(0.010) 2 = 78.54 × 10−6 m 2
2τ U A (2)(100 × 106 )(78.54 × 10−6 )
=
= 5.236 × 103 N
F .S.
3.0
10
P=
FA = 5.82 × 103 N
9
FA =
Based on double shear in pins at B and D: A =
π
4
d2 =
π
4
(0.012) 2 = 113.10 × 10−6 m 2
2τ U A (2)(100 × 106 )(113.10 × 10−6 )
=
= 7.54 × 103 N
F .S.
3.0
10
P=
FBD = 3.97 × 103 N
19
FBD =
Based on compression in links BD: For one link, A = (0.020)(0.008) = 160 × 10−6 m 2
2σ U A (2)(250 × 106 )(160 × 10−6 )
=
= 26.7 × 103 N
F .S.
3.0
10
P=
FBD = 14.04 × 103 N
19
FBD =
Allowable value of P is smallest, ∴ P = 3.97 × 103 N
P = 3.97 kN 
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PROBLEM 1.57
The Load and Resistance Factor Design method is to be
used to select the two cables that will raise and lower a
platform supporting two window washers. The platform
weighs 160 lb and each of the window washers is
assumed to weigh 195 lb with equipment. Since these
workers are free to move on the platform, 75% of their
total weight and the weight of their equipment will be
used as the design live load of each cable. (a) Assuming
a resistance factor φ = 0.85 and load factors γ D = 1.2
and γ L = 1.5, determine the required minimum ultimate
load of one cable. (b) What is the conventional factor of
safety for the selected cables?
SOLUTION
γ D PD + γ L PL = φ PU
(a)
PU =
γ D PD + γ L PL
φ
1

3

(1.2)  × 160  + (1.5)  × 2 × 195 
2
4




=
0.85
PU = 629 lb 
Conventional factor of safety.
P = PD + PL =
(b)
F. S . =
PU
629
=
P
372.5
1
× 160 + 0.75 × 2 × 195 = 372.5 lb
2
F. S . = 1.689 
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PROBLEM 1.58
A 40-kg platform is attached to the end B of a 50-kg wooden
beam AB, which is supported as shown by a pin at A and by a
slender steel rod BC with a 12-kN ultimate load. (a) Using the
Load and Resistance Factor Design method with a resistance
factor φ = 0.90 and load factors γ D = 1.25 and γ L = 1.6,
determine the largest load that can be safely placed on the
platform. (b) What is the corresponding conventional factor
of safety for rod BC?
SOLUTION
3
ΣM A = 0 : (2.4) P − 2.4W1 − 1.2W2
5
5
5
∴ P = W1 + W2
3
6
For dead loading,
W1 = (40)(9.81) = 392.4 N, W2 = (50)(9.81) = 490.5 N
5
5
PD =   (392.4) +   (490.5) = 1.0628 × 103 N
3
6
For live loading,
W1 = mg W2 = 0
From which
m=
Design criterion.
PL =
5
mg
3
3 PL
5 g
γ D PD + γ L PL = φ PU
PL =
φ PU − γ D PD
(0.90)(12 × 103 ) − (1.25)(1.0628 × 10−3 )
=
γL
1.6
= 5.920 × 103 N
(a)
m=
Allowable load.
3 5.92 × 103
5
9.81
m = 362 kg 
Conventional factor of safety.
P = PD + PL = 1.0628 × 103 + 5.920 × 103 = 6.983 × 103 N
(b)
F. S . =
PU
12 × 103
=
P
6.983 × 103
F. S . = 1.718 
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PROBLEM 1.59
A strain gage located at C on the surface of bone AB indicates that the average
normal stress in the bone is 3.80 MPa when the bone is subjected to two 1200-N
forces as shown. Assuming the cross section of the bone at C to be annular and
knowing that its outer diameter is 25 mm, determine the inner diameter of the
bone’s cross section at C.
SOLUTION
σ =
Geometry:
A=
π
4
P
P
∴ A=
A
σ
(d12 − d 22 )
d 22 = d12 −
4A
π
= d12 −
d 22 = (25 × 10−3 ) 2 −
4P
πσ
(4)(1200)
π (3.80 × 106 )
= 222.9 × 10−6 m 2
d 2 = 14.93 × 10−3 m
d 2 = 14.93 mm 
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PROBLEM 1.60
Two horizontal 5-kip forces are applied to pin B of the assembly
shown. Knowing that a pin of 0.8-in. diameter is used at each
connection, determine the maximum value of the average normal
stress (a) in link AB, (b) in link BC.
SOLUTION
Use joint B as free body.
Law of Sines
FAB
FBC
10
=
=
sin 45° sin 60° sin 95°
FAB = 7.3205 kips
FBC = 8.9658 kips
Link AB is a tension member.
Minimum section at pin. Anet = (1.8 − 0.8)(0.5) = 0.5 in 2
(a)
Stress in AB :
σ AB =
FAB
7.3205
=
Anet
0.5
σ AB = 14.64 ksi 
Link BC is a compression member.
Cross sectional area is A = (1.8)(0.5) = 0.9 in 2
(b)
Stress in BC:
σ BC =
− FBC
−8.9658
=
A
0.9
σ BC = −9.96 ksi 
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PROBLEM 1.61
For the assembly and loading of Prob. 1.60, determine (a) the average
shearing stress in the pin at C, (b) the average bearing stress at C in
member BC, (c) the average bearing stress at B in member BC.
PROBLEM 1.60 Two horizontal 5-kip forces are applied to pin B of the
assembly shown. Knowing that a pin of 0.8-in. diameter is used at each
connection, determine the maximum value of the average normal stress
(a) in link AB, (b) in link BC.
SOLUTION
Use joint B as free body.
Law of Sines
FAB
FBC
10
=
=
sin 45° sin 60° sin 95°
(a)
Shearing stress in pin at C.
τ =
FBC = 8.9658 kips
FBC
2 AP
π
4
d2 =
π
(0.8) 2 = 0.5026 in 2
4
8.9658
= 8.92
τ =
(2)(0.5026)
AP =
τ = 8.92 ksi 
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PROBLEM 1.61 (Continued)
(b)
Bearing stress at C in member BC. σ b =
FBC
A
A = td = (0.5)(0.8) = 0.4 in 2
σb =
(c)
Bearing stress at B in member BC. σ b =
8.9658
= 22.4
0.4
σ b = 22.4 ksi 
FBC
A
A = 2td = 2(0.5)(0.8) = 0.8 in 2
σb =
8.9658
= 11.21
0.8
σ b = 11.21 ksi 
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PROBLEM 1.62
In the marine crane shown, link CD is known to have a
uniform cross section of 50 × 150 mm. For the loading
shown, determine the normal stress in the central portion
of that link.
SOLUTION
W = (80 Mg)(9.81 m/s 2 ) = 784.8 kN
Weight of loading:
Free Body: Portion ABC
 M A = 0: FCD (15 m) − W (28 m) = 0
28
28
(784.8 kN)
W =
15
15
= +1465 kN
FCD =
FCD
σ CD =
FCD
+1465 × 103 N
=
= +195.3 × 106 Pa
(0.050 m)(0.150 m)
A
σ CD = +195.3 MPa 
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PROBLEM 1.63
Two wooden planks, each 12 in. thick and 9 in. wide, are
joined by the dry mortise joint shown. Knowing that the wood
used shears off along its grain when the average shearing
stress reaches 1.20 ksi, determine the magnitude P of the axial
load that will cause the joint to fail.
SOLUTION
Six areas must be sheared off when the joint fails. Each of these areas has dimensions
being
A=
5
8
in. ×
1
2
in., its area
5 1
5 2
in = 0.3125 in 2
× =
8 2 16
At failure, the force carried by each area is
F = τ A = (1.20 ksi)(0.3125 in 2 ) = 0.375 kips
Since there are six failure areas,
P = 6F = (6)(0.375)
P = 2.25 kips 
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PROBLEM 1.64
Two wooden members of uniform rectangular cross section of sides
a = 100 mm and b = 60 mm are joined by a simple glued joint as shown.
Knowing that the ultimate stresses for the joint are σ U = 1.26 MPa in
tension and τU = 1.50 MPa in shear, and that P = 6 kN, determine the
factor of safety for the joint when (a) α = 20°, (b) α = 35°,
(c) α = 45°. For each of these values of α, also determine whether the
joint will fail in tension or in shear if P is increased until rupture
occurs.
SOLUTION
Let θ = 90° − α as shown.
From the text book:
or
σ =
P
cos 2 θ
A0
σ =
P
sin 2 α
A0
(1)
τ =
P
sin α cos α
A0
(2)
τ =
P
sin θ cos θ
A0
A0 = ab = (100 mm) (60 mm) = 6000 mm 2 = 6 × 10−3 m 2
σ U = 1.26 × 106 Pa
τU = 1.50 × 106 Pa
Ultimate load based on tension across the joint:
( PU )σ =
=
σ U A0
(1.26 × 106 )(6 × 10−3 )
=
sin 2 α
sin 2 α
7560
7.56
=
kN
2
sin α
sin 2 α
Ultimate load based on shear across the joint:
( PU )τ =
=
τU A0
(1.50 × 106 )(6 × 10−3 )
=
sin α cos α
sin α cos α
9000
9.00
kN
=
sin α cos α
sin α cos α
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PROBLEM 1.64 (Continued)
(a)
7.56
= 64.63 kN
sin 2 20°
9.00
= ( PU )τ =
= 28.00 kN
sin 20° cos 20°
α = 20° : ( PU )σ =
The smaller value governs. The joint will fail in shear and PU = 28.00 kN.
F .S . =
(b)
α = 35° :
PU
28.00
=
P
6
F .S . = 4.67 
7.56
= 22.98 kN
sin 2 35°
9.00
( PU )τ =
= 19.155 kN
sin 35° cos 35°
( PU )σ =
The joint will fail in shear and PU = 19.155 kN.
F .S . =
(c)
PU
19.155
=
P
6
F .S . = 3.19 
7.56
= 15.12 kN
sin 2 45°
9.00
( PU )τ =
= 18.00 kN
sin 45° cos 45°
α = 45°: ( PU )σ =
The joint will fail in tension and PU = 15.12 kN.
F .S . =
PU
15.12
=
P
6
F .S . = 2.52 
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PROBLEM 1.65
Member ABC, which is supported by a pin and bracket at C and a cable
BD, was designed to support the 16-kN load P as shown. Knowing that
the ultimate load for cable BD is 100 kN, determine the factor of safety
with respect to cable failure.
SOLUTION
Use member ABC as a free body, and note that member BD is a two-force member.
ΣM c = 0 : ( P cos 40°)(1.2) + ( P sin 40°)(0.6)
− ( FBD cos 30°)(0.6)
− ( FBD sin 30°)(0.4) = 0
1.30493P − 0.71962FBD = 0
FBD = 1.81335 P = (1.81335)(16 × 103 ) = 29.014 × 103 N
FU = 100 × 103 N
F. S . =
FU
100 × 103
=
FBD
29.014 × 103
F. S . = 3.45 
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PROBLEM 1.66
The 2000-lb load can be moved along the beam BD to any position
between stops at E and F. Knowing that σ all = 6 ksi for the steel
used in rods AB and CD, determine where the stops should be
placed if the permitted motion of the load is to be as large as
possible.
SOLUTION
Permitted member forces:
2
 π  1 
AB : ( FAB ) max = σ all AAB = (6)   
 4  2 
= 1.17810 kips
2
 π  5 
CD : ( FCD ) max = σ all ACD = (6)   
 4  8 
= 1.84078 kips
Use member BEFD as a free body.
P = 2000 lb = 2.000 kips
ΣM D = 0 : − (60) FAB + (60 − x E) P = 0
60 − xE =
60 FAB
(60)(1.17810)
=
P
2.000
= 35.343
x E = 24.7 in. 
ΣM B = 0 : 60 FCD − x F P = 0 
xF =
60FCD
(60)(1.84078)
=
P
2.000
x F = 55.2 in. 
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PROBLEM 1.67
Knowing that a force P of magnitude 750 N is applied to the pedal
shown, determine (a) the diameter of the pin at C for which the
average shearing stress in the pin is 40 MPa, (b) the corresponding
bearing stress in the pedal at C, (c) the corresponding bearing
stress in each support bracket at C.
SOLUTION
Draw free body diagram of BCD. Since BCD is a 3-force member, the reaction at C is directed toward
Point E, the intersection of the lines of action of the other two forces.
From geometry, CE =
3002 + 1252 = 325 mm
+ ↑ ΣFy = 0 :
(a)
τ pin =
(b)
σb =
(c)
σb =
1C
2
Apin
=
1C
2
π d2
4
d =
125
C − P = 0 C = 2.6 P = (2.6)(750) = 1950 N
325
2C
πτ pin
=
(2)(1950)
= 5.57 × 10−3 m
6
π (40 × 10 )
C
C
1950
=
=
= 38.9 × 106 Pa
Ab
dt
(5.57 × 10−3 )(9 × 10−3 )
1C
2
Ab
=
1950
C
=
= 35.0 × 106 Pa
2dt
(2)(5.57 × 10−3 )(5 × 10−3 )
d = 5.57 mm 
σ b = 38.9 MPa 
σ b = 35.0 MPa 
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PROBLEM 1.68
A force P is applied as shown to a steel reinforcing bar that has been
embedded in a block of concrete. Determine the smallest length L
for which the full allowable normal stress in the bar can be
developed. Express the result in terms of the diameter d of the bar,
the allowable normal stress σ all in the steel, and the average
allowable bond stress τ all between the concrete and the cylindrical
surface of the bar. (Neglect the normal stresses between the concrete
and the end of the bar.)
SOLUTION
For shear,
A = π dL
P = τ all A = τ allπ dL
For tension,
A=
π
4
d2
π

P = σ all A = σ all  d 2 
4


Equating,
Solving for L,
τ allπ dL = σ all
π
4
d2
Lmin = σ alld/4τ all 
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PROBLEM 1.69
The two portions of member AB are glued together along a plane forming
an angle θ with the horizontal. Knowing that the ultimate stress for the
glued joint is 2.5 ksi in tension and 1.3 ksi in shear, determine the range of
values of θ for which the factor of safety of the members is at least 3.0.
SOLUTION
A0 = (2.0)(1.25) = 2.50 in.2
P = 2.4 kips
PU = ( F. S .) P = 7.2 kips
Based on tensile stress:
σU =
cos 2 θ =
PU
cos 2 θ
A0
σ U A0
PU
cos θ = 0.93169
Based on shearing stress:
τU =
sin 2θ =
2θ = 64.52°
Hence,
=
(2.5)(2.50)
= 0.86806
7.2
θ = 21.3°
θ > 21.3°
PU
P
sin θ cos θ = U sin 2θ
A0
2 A0
2 A0τU
PU
=
(2)(2.50)(1.3)
= 0.90278
7.2
θ = 32.3°
θ < 32.3°
21.3° < θ < 32.3° 
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PROBLEM 1.70
The two portions of member AB are glued together along a plane forming
an angle θ with the horizontal. Knowing that the ultimate stress for the glued
joint is 2.5 ksi in tension and 1.3 ksi in shear, determine (a) the value of θ for
which the factor of safety of the member is maximum, (b) the corresponding
value of the factor of safety. (Hint: Equate the expressions obtained for the
factors of safety with respect to normal stress and shear.)
SOLUTION
A0 = (2.0)(1.25) = 2.50 in 2
At the optimum angle,
( F. S.)σ = ( F. S .)τ
σ =
σ U A0
P
cos 2 θ ∴ PU ,σ =
A0
cos 2 θ
Normal stress:
( F. S.)σ =
Shearing stress: τ =
σ U A0
P cos θ
Solving:
(b)
2
PU =
=
P
=
σ U A0
P cos 2 θ
τU A0
P
sin θ cos θ ∴ PU ,τ =
sin θ cos θ
A0
( F. S.)τ =
Equating:
PU ,σ
PU ,τ
P
=
τU A0
P sin θ cos θ
τ U A0
P sin θ cos θ
sin θ
τ
1.3
= tan θ = U =
= 0.520
cos θ
σU
2.5
(a)
θopt = 27.5° 
σ U A0
(12.5)(2.50)
=
= 7.94 kips
2
cos θ
cos 2 27.5°
F. S . =
PU
7.94
=
2.4
P
F. S . = 3.31 
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CHAPTER 2
PROBLEM 2.1
An 80-m-long wire of 5-mm diameter is made of a steel with E = 200 GPa and an ultimate tensile strength of
400 MPa. If a factor of safety of 3.2 is desired, determine (a) the largest allowable tension in the wire, (b) the
corresponding elongation of the wire.
SOLUTION
(a)
σ U = 400 × 106 Pa
A=
π
4
d2 =
π
4
(5) 2 = 19.635 mm 2 = 19.635 × 10−6 m 2
PU = σ U A = (400 × 106 ) (19.635 × 10−6 ) = 7854 N
Pall =
(b)
δ=
PU
7854
=
= 2454 N
F .S
3.2
PL
(2454) (80)
=
= 50.0 × 10−3 m
AE (19.635 × 10−6 )(200 × 109 )
Pall = 2.45 kN 
δ = 50.0 mm 
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PROBLEM 2.2
A steel control rod is 5.5 ft long and must not stretch more than 0.04 in. when a 2-kip tensile load is applied to it.
Knowing that E = 29 × 106 psi, determine (a) the smallest diameter rod that should be used, (b) the
corresponding normal stress caused by the load.
SOLUTION
(a)
δ=
PL
AE
: 0.04 in. =
(2000 lb) (5.5 × 12 in.)
6
A (29 × 10 psi)
1
A = π d 2 = 0.11379 in 2
4
d = 0.38063 in.
(b)
σ=
P
A
=
2000 lb
0.11379 in
2
= 17580 psi
d = 0.381 in. 
σ = 17.58 ksi 
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PROBLEM 2.3
Two gage marks are placed exactly 10 in. apart on a 12 -in.-diameter aluminum rod with E = 10.1 × 106 psi and
an ultimate strength of 16 ksi. Knowing that the distance between the gage marks is 10.009 in. after a load is
applied, determine (a) the stress in the rod, (b) the factor of safety.
SOLUTION
(a)
δ = 10.009 − 10.000 = 0.009 in.
ε=
(b)
F. S . =
σU
16
=
9.09
σ
δ
L
=
σ
E
σ=
Eδ (10.1 × 106 ) (0.009)
=
= 9.09 × 103 psi
L
10
σ = 9.09 ksi 
F. S . = 1.760 
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PROBLEM 2.4
An 18-m-long steel wire of 5-mm diameter is to be used in the manufacture of a prestressed concrete beam.
It is observed that the wire stretches 45 mm when a tensile force P is applied. Knowing that E = 200 GPa,
determine (a) the magnitude of the force P, (b) the corresponding normal stress in the wire.
SOLUTION
(a)
δ=
PL
, or
AE
P=
δ AE
L
1
1
with A = π d 2 = π (0.005) 2 = 19.6350 × 10−6 m 2
4
4
P=
(0.045 m)(19.6350 × 10−6 m 2 )(200 × 109 N/m 2 )
= 9817.5 N
18 m
P = 9.82 kN 
(b)
σ=
P
A
=
9817.5 N
19.6350 × 10
−6
6
m
2
= 500 × 10 Pa
σ = 500 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 2.5
A polystyrene rod of length 12 in. and diameter 0.5 in. is subjected to an 800-lb tensile load. Knowing that
E = 0.45 × 106 psi, determine (a) the elongation of the rod, (b) the normal stress in the rod.
SOLUTION
A=
(a)
δ=
(b)
σ=
PL
AE
P
A
=
=
(800) (12)
6
(0.19635) (0.45 ×10 )
800
0.19635
= 4074 psi
π
4
d2 =
π
4
= 0.1086
(0.5) 2 = 0.19635 in 2
δ = 0.1086 in. 
σ = 4.07 ksi 
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PROBLEM 2.6
A nylon thread is subjected to a 8.5-N tension force. Knowing that E = 3.3 GPa and that the length of the
thread increases by 1.1%, determine (a) the diameter of the thread, (b) the stress in the thread.
SOLUTION
(a)
1.1
= 0.011
100
ε=
Stress:
σ = Eε = (3.3 × 109 )(0.011) = 36.3 × 106 Pa
σ =
(b)
δ
Strain:
Area:
A=
Diameter:
d =
Stress:
=
L
P
A
P
σ
=
4A
π
8.5
= 234.16 × 10−9 m 2
36.3 × 106
=
(4)(234.16 × 10−9 )
π
= 546 × 10−6 m
d = 0.546 mm 
σ = 36.3 MPa 
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PROBLEM 2.7
Two gage marks are placed exactly 250 mm apart on a 12-mm-diameter aluminum rod. Knowing that, with an
axial load of 6000 N acting on the rod, the distance between the gage marks is 250.18 mm, determine the
modulus of elasticity of the aluminum used in the rod.
SOLUTION
δ = Δ L = L − L0 = 250.18 − 250.00 = 0.18 mm
δ 0.18 mm
ε=
=
= 0.00072
L0
π
250 mm
π
(12)2 = 113.097 mm 2 = 113.097 ×10−6 m 2
4
4
6000
P
σ= =

= 53.052 × 106 Pa
A 113.097 × 10−6
A=
E=
d2 =
σ 53.052 × 106
=
= 73.683 × 109 Pa
ε
0.00072
E = 73.7 GPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 2.8
An aluminum pipe must not stretch more than 0.05 in. when it is subjected to a tensile load. Knowing that
E = 10.1 × 106 psi and that the maximum allowable normal stress is 14 ksi, determine (a) the maximum
allowable length of the pipe, (b) the required area of the pipe if the tensile load is 127.5 kips.
SOLUTION
(a)
δ=
PL
;
AE
L=
Thus,
EAδ
Eδ
(10.1 × 106 ) (0.05)
=
=
σ
P
14 × 103
L = 36.1 in. 
(b)
σ =
Thus,
P
;
A
A=
P
σ
=
127.5 × 103
14 × 103
A = 9.11 in 2 
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PROBLEM 2.9
An aluminum control rod must stretch 0.08 in. when a 500-lb tensile load is applied to it. Knowing that
σ all = 22 ksi and E = 10.1 × 106 psi, determine the smallest diameter and shortest length that can be selected for
the rod.
SOLUTION
σ all = 22 × 103 psi
P = 500 lb, δ = 0.08 in.
σ=
A=
P
< σ all
A
π
4
d2
A>
d=
P
σ all
=
4A
π
500
= 0.022727 in 2
22 × 103
=
(4)(0.022727)
π
Eδ
< σ all
L
Eδ (10.1 × 106 )(0.08)
L>
=
= 36.7 in.
σ all
22 × 103
d min = 0.1701 in. 
σ = Eε =
Lmin = 36.7 in. 
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PROBLEM 2.10
A square yellow-brass bar must not stretch more than 2.5 mm when it is subjected to a tensile load. Knowing
that E = 105 GPa and that the allowable tensile strength is 180 MPa, determine (a) the maximum allowable
length of the bar, (b) the required dimensions of the cross section if the tensile load is 40 kN.
SOLUTION
σ = 180 × 106 Pa P = 40 × 103 N
E = 105 × 109 Pa δ = 2.5 × 10−3 m
(a)
PL σ L
=
AE
E
Eδ (105 × 109 )(2.5 × 10−3 )
=
= 1.45833 m
L=
σ
180 × 106
δ=
L = 1.458 m 
(b)
σ =
A=
P
A
P
σ
A = a2
=
40 × 103
= 222.22 × 10−6 m 2 = 222.22 mm 2
180 × 106
a =
A =
222.22
a = 14.91 mm 
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PROBLEM 2.11
A 4-m-long steel rod must not stretch more than 3 mm and the normal stress must not exceed 150 MPa when
the rod is subjected to a 10-kN axial load. Knowing that E = 200 GPa, determine the required diameter of
the rod.
SOLUTION
L =4m
δ = 3 × 10−3 m,
σ = 150 × 106 Pa
E = 200 × 109 Pa, P = 10 × 103 N
Stress:
σ =
A=
Deformation:
P
A
P
σ
=
10 × 103
= 66.667 × 10−6 m 2 = 66.667 mm 2
6
150 × 10
δ =
PL
AE
A=
PL
(10 × 103 ) (4)
=
= 66.667 × 10−6 m 2 = 66.667 mm 2
Eδ
(200 × 109 ) (3 × 10−3 )
The larger value of A governs:
A = 66.667 mm 2
A=
π
4
d2
d=
4A
π
=
4(66.667)
π
d = 9.21 mm 
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PROBLEM 2.12
A nylon thread is to be subjected to a 10-N tension. Knowing that E = 3.2 GPa, that the maximum allowable
normal stress is 40 MPa, and that the length of the thread must not increase by more than 1%, determine the
required diameter of the thread.
SOLUTION
Stress criterion:
σ = 40 MPa = 40 × 106 Pa P = 10 N
σ =
A=
P
P
10 N
: A=
=
= 250 × 10−9 m 2
A
σ
40 × 106 Pa
π
4
d 2: d = 2
A
π
250 × 10−9
=2
π
= 564.19 × 10−6 m
d = 0.564 mm
Elongation criterion:
δ
L
= 1% = 0.01
δ =
PL
:
AE
A=
P /E 10 N/3.2 × 109 Pa
=
= 312.5 × 10−9 m 2
0.01
δ /L
d =2
A
π
=2
312.5 × 10−9
π
= 630.78 × 10−6 m 2
d = 0.631 mm
The required diameter is the larger value:
d = 0.631 mm 
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PROBLEM 2.13
The 4-mm-diameter cable BC is made of a steel with E = 200 GPa. Knowing
that the maximum stress in the cable must not exceed 190 MPa and that the
elongation of the cable must not exceed 6 mm, find the maximum load P that
can be applied as shown.
SOLUTION
LBC = 62 + 42 = 7.2111 m
Use bar AB as a free body.
 4

3.5 P − (6) 
FBC  = 0
7.2111


P = 0.9509 FBC
Σ M A = 0:
Considering allowable stress: σ = 190 × 106 Pa
A=
π
d2 =
4
FBC
σ=
A
π
4
(0.004) 2 = 12.566 × 10−6 m 2
∴ FBC = σ A = (190 × 106 ) (12.566 × 10−6 ) = 2.388 × 103 N
Considering allowable elongation: δ = 6 × 10−3 m
δ=
FBC LBC
AE
∴ FBC =
AEδ (12.566 × 10−6 )(200 × 109 )(6 × 10−3 )
=
= 2.091 × 103 N
LBC
7.2111
Smaller value governs. FBC = 2.091 × 103 N
P = 0.9509 FBC = (0.9509)(2.091 × 103 ) = 1.988 × 103 N
P = 1.988 kN 
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PROBLEM 2.14
The aluminum rod ABC ( E = 10.1 × 106 psi), which consists of two
cylindrical portions AB and BC, is to be replaced with a cylindrical steel
rod DE ( E = 29 × 106 psi) of the same overall length. Determine the
minimum required diameter d of the steel rod if its vertical deformation is
not to exceed the deformation of the aluminum rod under the same load
and if the allowable stress in the steel rod is not to exceed 24 ksi.
SOLUTION
Deformation of aluminum rod.
δA =
=
PLAB PLBC
+
AAB E ABC E
P  LAB LBC 
+


E  AAB ABC 
 12
+
π
 (1.5)2
4
= 0.031376 in.
=
Steel rod.
28 × 103
10.1 × 106
18
(2.25) 2
4
π




δ = 0.031376 in.
δ=
PL
PL
(28 × 103 )(30)
∴ A=
=
= 0.92317 in 2
6
EA
Eδ (29 × 10 )(0.031376)
σ=
P
A
∴ A=
P
σ
=
28 × 103
= 1.1667 in 2
24 × 103
Required area is the larger value.
A = 1.1667 in 2
Diameter:
d=
4A
π
=
(4)(1.6667)
π
d = 1.219 in. 
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PROBLEM 2.15
A 4-ft section of aluminum pipe of cross-sectional area 1.75 in2 rests on a
fixed support at A. The 58 -in.-diameter steel rod BC hangs from a rigid
bar that rests on the top of the pipe at B. Knowing that the modulus of
elasticity is 29 × 106 psi for steel, and 10.4 × 106 psi for aluminum,
determine the deflection of point C when a 15-kip force is applied at C.
SOLUTION
Rod BC:
LBC = 7 ft = 84 in. EBC = 29 × 106 psi
ABC =
δ C/B =
Pipe AB:
π
4
d2 =
π
4
(0.625) 2 = 0.30680 in 2
PLBC
(15 × 103 )(84)
=
= 0.141618 in.
EBC ABC
(29 × 106 )(0.30680)
LAB = 4 ft = 48 in. E AB = 10.4 × 106 psi
AAB = 1.75 in 2
δ B/A =
Total:
PLAB
(15 × 103 ) (48)
=
= 39.560 × 10−3 in.
6
E AB AAB (10.4 × 10 ) (1.75)
δ C = δ B/A + δ C/B = 39.560 × 10−3 + 0.141618 = 0.181178 in.
δ C = 0.1812 in. 
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PROBLEM 2.16
The brass tube AB ( E = 105 GPa) has a cross-sectional area of
140 mm2 and is fitted with a plug at A. The tube is attached at B to a
rigid plate that is itself attached at C to the bottom of an aluminum
cylinder ( E = 72 GPa) with a cross-sectional area of 250 mm2. The
cylinder is then hung from a support at D. In order to close the
cylinder, the plug must move down through 1 mm. Determine the force
P that must be applied to the cylinder.
SOLUTION
Shortening of brass tube AB:
LAB = 375 + 1 = 376 mm = 0.376 m
AAB = 140 mm 2 = 140 × 10−6 m 2
E AB = 105 × 109 Pa
δ AB =
PLAB
P(0.376)
=
= 25.578 × 10−9 P
−6
9
E AB AAB
(105 × 10 )(140 × 10 )
Lengthening of aluminum cylinder CD:
LCD = 0.375 m
δ CD =
Total deflection:
ACD = 250 mm 2 = 250 × 10−6 m 2
ECD = 72 × 109 Pa
PLCD
P(0.375)
=
= 20.833 × 10−9 P
9
−6
ECD ACD (72 × 10 )(250 × 10 )
δ A = δ AB + δ CD where δ A = 0.001 m
0.001 = (25.578 × 10−9 + 20.833 × 10−9 ) P
P = 21.547 × 103 N
P = 21.5 kN 
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PROBLEM 2.17
A 250-mm-long aluminum tube ( E = 70 GPa) of 36-mm outer
diameter and 28-mm inner diameter can be closed at both ends by
means of single-threaded screw-on covers of 1.5-mm pitch. With
one cover screwed on tight, a solid brass rod ( E = 105 GPa) of
25-mm diameter is placed inside the tube and the second cover is
screwed on. Since the rod is slightly longer than the tube, it is
observed that the cover must be forced against the rod by rotating
it one-quarter of a turn before it can be tightly closed. Determine
(a) the average normal stress in the tube and in the rod, (b) the
deformations of the tube and of the rod.
SOLUTION
Atube =
A rod =
π
4
π
4
(d o2 − di2 ) =
d2 =
π
4
π
4
(362 − 282 ) = 402.12 mm 2 = 402.12 × 10−6 m 2
(25)2 = 490.87 mm 2 = 490.87 × 10−6 m 2
PL
P(0.250)
=
= 8.8815 × 10−9 P
9
−6
Etube Atube (70 × 10 )(402.12 × 10 )
PL
P(0.250)
= −4.8505 × 10−9 P
=−
=
Erod Arod (105 × 106 )(490.87 × 10−6 )
δ tube =
δ rod
1



δ * =  turn  × 1.5 mm = 0.375 mm = 375 × 10−6 m
4
δ tube = δ * + δ rod
or δ tube − δ rod = δ *
8.8815 × 10−9 P + 4.8505 × 10−9 P = 375 × 10−6
P=
(a)
σ tube =
P
27.308 × 103
=
= 67.9 × 106 Pa
Atube 402.12 × 10−6
σ rod = −
(b)
0.375 × 10−3
= 27.308 × 103 N
−9
(8.8815 + 4.8505)(10 )
P
27.308 × 103
=−
= −55.6 × 106 Pa
Arod
490.87 × 10−6
δ tube = (8.8815 × 10−9 )(27.308 × 103 ) = 242.5 × 10−6 m
δ rod = −(4.8505 × 10−9 )(27.308 × 103 ) = −132.5 × 10−6 m
σ tube = 67.9 MPa 
σ rod = −55.6 MPa 
δ tube = 0.2425 mm 
δ rod = −0.1325 mm 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 2.18
The specimen shown is made from a 1-in.-diameter cylindrical steel rod
with two 1.5-in.-outer-diameter sleeves bonded to the rod as shown.
Knowing that E = 29 × 106 psi, determine (a) the load P so that the
total deformation is 0.002 in., (b) the corresponding deformation of the
central portion BC.
SOLUTION
(a)
δ =Σ
Pi Li P Li
= Σ
Ai Ei E Ai
−1
 L 
π
P = Eδ  Σ i  Ai = di2
4
 Ai 
L, in.
d, in.
A, in2
L/A, in–1
AB
2
1.5
1.7671
1.1318
BC
3
1.0
0.7854
3.8197
CD
2
1.5
1.7671
1.1318
6.083
P = (29 × 106 )(0.002)(6.083) −1 = 9.353 × 103 lb
(b)
δ BC =
PLBC P LBC 9.535 × 103
=
=
(3.8197)
ABC E E ABC
29 × 106
← sum
P = 9.53 kips 
δ = 1.254 × 10−3 in. 
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PROBLEM 2.19
Both portions of the rod ABC are made of an aluminum for which E = 70 GPa.
Knowing that the magnitude of P is 4 kN, determine (a) the value of Q so that
the deflection at A is zero, (b) the corresponding deflection of B.
SOLUTION
(a)
AAB =
ABC =
π
4
π
4
2
d AB
=
2
d BC
=
π
4
(0.020)2 = 314.16 × 10−6 m 2
π
(0.060)2 = 2.8274 × 10−3 m 2
4
Force in member AB is P tension.
Elongation:
δ AB =
PLAB
(4 × 103 )(0.4)
=
= 72.756 × 10−6 m
−6
9
EAAB (70 × 10 )(314.16 × 10 )
Force in member BC is Q − P compression.
Shortening:
δ BC =
(Q − P) LBC
(Q − P)(0.5)
=
= 2.5263 × 10−9 (Q − P )
EABC
(70 × 109 )(2.8274 × 10−3 )
For zero deflection at A, δ BC = δ AB
2.5263 × 10−9 (Q − P ) = 72.756 × 10−6 ∴ Q − P = 28.8 × 103 N
Q = 28.3 × 103 + 4 × 103 = 32.8 × 103 N
(b)
δ AB = δ BC = δ B = 72.756 × 10−6 m
Q = 32.8 kN 
δ AB = 0.0728 mm ↓ 
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PROBLEM 2.20
The rod ABC is made of an aluminum for which E = 70 GPa. Knowing that
P = 6 kN and Q = 42 kN, determine the deflection of (a) point A, (b) point B.
SOLUTION
AAB =
ABC =
π
4
π
4
2
d AB
=
2
d BC
=
π
4
π
4
(0.020) 2 = 314.16 × 10−6 m 2
(0.060)2 = 2.8274 × 10−3 m 2
PAB = P = 6 × 103 N
PBC = P − Q = 6 × 103 − 42 × 103 = −36 × 103 N
LAB = 0.4 m LBC = 0.5 m
δ AB =
PAB LAB
(6 × 103 )(0.4)
=
= 109.135 × 10−6 m
AAB E A (314.16 × 10−6 )(70 × 109 )
δ BC =
PBC LBC
(−36 × 103 )(0.5)
=
= −90.947 × 10−6 m
ABC E
(2.8274 × 10−3 )(70 × 109 )
(a)
δ A = δ AB + δ BC = 109.135 × 10−6 − 90.947 × 10−6 m = 18.19 × 10−6 m
(b)
δ B = δ BC = −90.9 × 10−6 m = −0.0909 mm
or
δ A = 0.01819 mm ↑ 
δ B = 0.0919 mm ↓ 
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PROBLEM 2.21
Members AB and BC are made of steel ( E = 29 × 106 psi) with crosssectional areas of 0.80 in2 and 0.64 in2, respectively. For the loading
shown, determine the elongation of (a) member AB, (b) member BC.
SOLUTION
(a)
LAB =
62 + 52 = 7.810 ft = 93.72 in.
Use joint A as a free body.
5
FAB − 28 = 0
7.810
= 43.74 kip = 43.74 × 103 lb
ΣFy = 0:
FAB
δ AB =
(b)
FAB LAB
(43.74 × 103 ) (93.72)
=
EAAB
(29 × 106 ) (0.80)
δ AB = 0.1767 in. 
Use joint B as a free body.
ΣFx = 0: FBC −
FBC =
δ BC =
6
FAB = 0
7.810
(6) (43.74)
= 33.60 kip = 33.60 × 103 lb.
7.810
FBC LBC
(33.60 × 103 ) (72)
=
EABC
(29 × 106 ) (0.64)
δ BC = 0.1304 in. 
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PROBLEM 2.22
The steel frame ( E = 200 GPa) shown has a diagonal brace BD with
an area of 1920 mm2. Determine the largest allowable load P if the
change in length of member BD is not to exceed 1.6 mm.
SOLUTION
δ BD = 1.6 × 10−3 m, ABD = 1920 mm 2 = 1920 × 10−6 m 2
LBD =
52 + 62 = 7.810 m, EBD = 200 × 109 Pa
δ BD =
FBD LBD
EBD ABD
FBD =
(200 × 109 ) (1920 × 10−6 )(1.6 × 10−3 )
EBD ABDδ BD
=
7.81
LBD
= 78.67 × 103 N
Use joint B as a free body.
ΣFx = 0:
5
FBD − P = 0
7.810
P=
5
(5)(78.67 × 103 )
FBD =
7.810
7.810
= 50.4 × 103 N
P = 50.4 kN 
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PROBLEM 2.23
For the steel truss ( E = 200 GPa) and loading shown, determine
the deformations of the members AB and AD, knowing that their
cross-sectional areas are 2400 mm2 and 1800 mm2, respectively.
SOLUTION
Statics: Reactions are 114 kN upward at A and C.
Member BD is a zero force member.
LAB = 4.02 + 2.52 = 4.717 m
Use joint A as a free body.
ΣFy = 0 : 114 +
2.5
FAB = 0
4.717
FAB = −215.10 kN
ΣFx = 0 : FAD +
FAD = −
4
FAB = 0
4.717
(4)(−215.10)
= 182.4 kN
4.717
Member AB:
δ AB =
FAB LAB
(−215.10 × 103 )(4.717)
=
EAAB
(200 × 109 )(2400 × 10−6 )
= −2.11 × 10−3 m
Member AD:
δ AD =
δ AB − 2.11 mm 
FAD LAD
(182.4 × 103 )(4.0)
=
EAAD
(200 × 109 )(1800 × 10−6 )
= 2.03 × 10−3 m
δ AD = 2.03 mm 
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PROBLEM 2.24
For the steel truss ( E = 29 × 106 psi) and loading shown, determine the
deformations of the members BD and DE, knowing that their crosssectional areas are 2 in2 and 3 in2, respectively.
SOLUTION
Free body: Portion ABC of truss
ΣM E = 0 : FBD (15 ft) − (30 kips)(8 ft) − (30 kips)(16 ft) = 0
FBD = + 48.0 kips
Free body: Portion ABEC of truss
ΣFx = 0 : 30 kips + 30 kips − FDE = 0 
FDE = + 60.0 kips

δ BD =
PL (+48.0 × 103 lb)(8 × 12 in.)
=
AE
(2 in 2 )(29 × 106 psi)
δ BD = +79.4 × 10−3 in. 
δ DE =
PL (+60.0 × 103 lb)(15 × 12 in.)
=
AE
(3 in 2 )(29 × 106 psi)
δ DE + 124.1 × 10−3 in. 
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PROBLEM 2.25
Each of the links AB and CD is made of aluminum ( E = 10.9 × 106 psi)
and has a cross-sectional area of 0.2 in.2. Knowing that they support the
rigid member BC, determine the deflection of point E.
SOLUTION
Free body BC:
ΣM C = 0 : − (32) FAB + (22) (1 × 103 ) = 0
FAB = 687.5 lb
ΣFy = 0 : 687.5 − 1 × 103 + FCD = 0
FCD = 312.5 lb
FAB LAB
(687.5) (18)
=
= 5.6766 × 10−3 in = δ B
6
EA
(10.9 × 10 ) (0.2)
F L
(312.5) (18)
= CD CD =
= 2.5803 × 10−3 in = δ C
6
EA
(10.9 × 10 ) (0.2)
δ AB =
δ CD
Deformation diagram:
Slope θ =
δ B − δC
LBC
=
3.0963 × 10−3
32
= 96.759 × 10−6 rad
δ E = δ C + LECθ
= 2.5803 × 10−3 + (22) (96.759 × 10−6 )
= 4.7090 × 10−3 in
δ E = 4.71 × 10−3 in ↓ 
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PROBLEM 2.26
3
The length of the 32
-in.-diameter steel wire CD has been adjusted so that
with no load applied, a gap of 161 in. exists between the end B of the rigid
beam ACB and a contact point E. Knowing that E = 29 × 106 psi,
determine where a 50-lb block should be placed on the beam in order to
cause contact between B and E.
SOLUTION
Rigid beam ACB rotates through angle θ to close gap.
θ=
1/16
= 3.125 × 10−3 rad
20
Point C moves downward.
δ C = 4θ = 4(3.125 × 10−3 ) = 12.5 × 10−3 in.
δ CD = δ C = 12.5 × 10−3 in.
ACD =
δ CD
π
d2 =
π 3 
d
F L
= CD CD
EACD
FCD =
2
= 6.9029 × 10−3 in 2


4  32 
EACDδ CD (29 × 106 ) (6.9029 × 10−3 ) (12.5 × 10−3 )
=
12.5
LCD
= 200.18 lb
Free body ACB:
ΣM A = 0: 4 FCD − (50) (20 − x) = 0
(4) (200.18)
= 16.0144
50
x = 3.9856 in.
20 − x =
For contact,
x < 3.99 in. 
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PROBLEM 2.27
Link BD is made of brass ( E = 105 GPa) and has a cross-sectional area of
240 mm2. Link CE is made of aluminum ( E = 72 GPa) and has a crosssectional area of 300 mm2. Knowing that they support rigid member ABC
determine the maximum force P that can be applied vertically at point A if
the deflection of A is not to exceed 0.35 mm.
SOLUTION
Free body member AC:
ΣM C = 0 : 0.350 P − 0.225 FBD = 0
FBD = 1.55556 P
ΣM B = 0 : 0.125 P − 0.225 FCE = 0
FCE = 0.55556 P
FBD LBD
(1.55556 P) (0.225)
=
= 13.8889 × 10−9 P
9
−6
EBD ABD (105 × 10 ) (240 × 10 )
F L
(0.55556 P) (0.150)
= CE CE =
= 3.8581 × 10−9 P
ECE ACE (72 × 109 ) (300 × 10−6 )
δ B = δ BD =
δ C = δ CE
Deformation Diagram:
From the deformation diagram,
Slope,
θ=
δ B + δC
LBC
=
17.7470 × 10−9 P
= 78.876 × 10−9 P
0.225
δ A = δ B + LABθ
= 13.8889 × 10−9 P + (0.125) (78.876 × 10−9 P)
= 23.748 × 10−9 P
Apply displacement limit. δ A = 0.35 × 10−3 m = 23.748 × 10−9 P
P = 14.7381 × 103 N
P = 14.74 kN 
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PROBLEM 2.28
Each of the four vertical links connecting the two rigid
horizontal members is made of aluminum ( E = 70 GPa)
and has a uniform rectangular cross section of 10 × 40 mm.
For the loading shown, determine the deflection of
(a) point E, (b) point F, (c) point G.
SOLUTION
Statics. Free body EFG.
ΣM F = 0 : − (400)(2 FBE ) − (250)(24) = 0
FBE = −7.5 kN = −7.5 × 103 N
ΣM E = 0 : (400)(2 FCF ) − (650)(24) = 0
FCF = 19.5 kN = 19.5 × 103 N
Area of one link:
A = (10)(40) = 400 mm 2
= 400 × 10−6 m 2
Length: L = 300 mm = 0.300 m
Deformations.
δ BE =
FBE L
(−7.5 × 103 )(0.300)
=
= −80.357 × 10−6 m
9
−6
EA
(70 × 10 )(400 × 10 )
δ CF =
FCF L
(19.5 × 103 )(0.300)
=
= 208.93 × 10−6 m
EA
(70 × 109 )(400 × 10−6 )
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PROBLEM 2.28 (Continued)
(a)
Deflection of Point E.
δ E = |δ BF |
δ E = 80.4 μ m ↑ 
(b)
Deflection of Point F.
δ F = δ CF
δ F = 209 μ m ↓ 
Geometry change.
Let θ be the small change in slope angle.
θ=
(c)
δE + δF
LEF
Deflection of Point G.
=
80.357 × 10−6 + 208.93 × 10−6
= 723.22 × 10−6 radians
0.400
δ G = δ F + LFG θ
δ G = δ F + LFG θ = 208.93 × 10−6 + (0.250)(723.22 × 10−6 )
= 389.73 × 10−6 m
δ G = 390 μ m ↓ 
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PROBLEM 2.29
A vertical load P is applied at the center A of the upper section of a
homogeneous frustum of a circular cone of height h, minimum radius a,
and maximum radius b. Denoting by E the modulus of elasticity of the
material and neglecting the effect of its weight, determine the deflection of
point A.
SOLUTION
Extend the slant sides of the cone to meet at a point O and place the origin of the coordinate system there.
tan α =
From geometry,
a1 =
b−a
h
a
b
, b1 =
,
tan α
tan α
r = y tan α
At coordinate point y, A = π r 2
Deformation of element of height dy:
dδ =
Pdy
AE
dδ =
P dy
P
dy
=
2
2
Eπ r
π E tan α y 2
Total deformation.
P
δA =
π E tan 2 α
=

b1
a1
 1
dy
P
=
− 
2
2
π E tan α  y 
y
b1 − a1 P(b1 − a1 )
P
=
2
π Eab
π E tan α a1b1
b1
=
a1
1 1
P
 − 
2
π E tan α  a1 b1 
δA =
Ph
↓ 
π Eab
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PROBLEM 2.30
A homogeneous cable of length L and uniform cross section is suspended from one end. (a) Denoting by ρ the
density (mass per unit volume) of the cable and by E its modulus of elasticity, determine the elongation of the
cable due to its own weight. (b) Show that the same elongation would be obtained if the cable were horizontal
and if a force equal to half of its weight were applied at each end.
SOLUTION
(a)
For element at point identified by coordinate y,
P = weight of portion below the point
= ρ g A(L − y )
Pdy ρ gA( L − y )dy ρ g ( L − y )
dδ =
=
=
dy
EA
EA
E
δ=

=
(b)
Total weight:
L
ρ g (L − y)
0
E
dy =
ρg 
L
1 2
 Ly − y 
E 
2 0
ρg 
2
L 
2
 L −

2 
E 
δ=
1 ρ gL2

2 E
W = ρ gAL
F=
EAδ EA 1 ρ gL2 1
=
⋅
= ρ gAL
L
L 2 E
2
1
F= W 
2
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PROBLEM 2.31
The volume of a tensile specimen is essentially constant while plastic deformation occurs. If the initial
diameter of the specimen is d1, show that when the diameter is d, the true strain is ε t = 2 ln(d1 /d ).
SOLUTION
If the volume is constant,
π
4
d 2L =
π
4
d12 L0
L d12  d1 
=
=
L0 d 2  d 
ε t = ln
2
L
d 
= ln  1 
L0
d 
2
ε t = 2ln
d1

d
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PROBLEM 2.32
Denoting by ε the “engineering strain” in a tensile specimen, show that the true strain is ε t = ln (1 + ε ).
SOLUTION
ε t = ln
Thus

L +δ
L
δ 
= ln 0
= ln 1 +  = ln (1 + ε )
L0
L0
L0 

ε t = ln (1 + ε ) 
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PROBLEM 2.33
An axial force of 200 kN is applied to the assembly shown by means of
rigid end plates. Determine (a) the normal stress in the aluminum shell,
(b) the corresponding deformation of the assembly.
SOLUTION
Let Pa = Portion of axial force carried by shell
Pb = Portion of axial force carried by core.
Thus,
with
δ=
Pa L
, or
Ea Aa
Pa =
Ea Aa
δ
L
δ=
Pb L
, or
Eb Ab
Pb =
Eb Ab
δ
L
P = Pa + Pb = ( Ea Aa + Eb Ab )
Aa =
Ab =
π
4
π
4
δ
L
[(0.060)2 − (0.025)2 ] = 2.3366 × 10−3 m 2
(0.025) 2 = 0.49087 × 10−3 m 2
P = [(70 × 109 ) (2.3366 × 10−3 ) + (105 × 109 ) (0.49087 × 10−3 )]
P = 215.10 × 106
Strain:
ε =
δ
L
=
δ
L
δ
L
200 × 103
P
=
= 0.92980 × 10−3
215.10 × 106
215.10 × 106
(a)
σ a = Ea ε = (70 × 109 ) (0.92980 × 10−3 ) = 65.1 × 106 Pa
(b)
δ = ε L = (0.92980 × 10−3 ) (300 mm)
σ a = 65.1 MPa 
δ = 0.279 mm 
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PROBLEM 2.34
The length of the assembly shown decreases by 0.40 mm when an axial
force is applied by means of rigid end plates. Determine (a) the magnitude
of the applied force, (b) the corresponding stress in the brass core.
SOLUTION
Let Pa = Portion of axial force carried by shell and Pb = Portion of axial force carried by core.
Thus,
with
δ=
Pa L
, or
Ea Aa
Pa =
Ea Aa
δ
L
δ=
Pb L
, or
Eb Ab
Pb =
Eb Ab
δ
L
P = Pa + Pb = ( Ea Aa + Eb Ab )
Aa =
Ab =
π
4
π
4
δ
L
[(0.060)2 − (0.025)2 ] = 2.3366 × 10−3 m 2
(0.025)2 = 0.49087 × 10−3 m 2
P = [(70 × 109 ) (2.3366 × 10−3 ) + (105 × 109 ) (0.49087 × 10−3 )]
with
δ
L
= 215.10 × 106
δ
L
δ = 0.40 mm, L = 300 mm
(a)
P = (215.10 × 106 )
(b)
σb =
0.40
= 286.8 × 103 N
300
Pb
Eδ
(105 × 109 )(0.40 × 10−3 )
= b =
= 140 × 106 Pa
−3
Ab
L
300 × 10
P = 287 kN 
σ b = 140.0 MPa 
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PROBLEM 2.35
A 4-ft concrete post is reinforced with four steel bars, each with a 34 -in. diameter.
Knowing that Es = 29 × 106 psi and Ec = 3.6 × 106 psi, determine the normal
stresses in the steel and in the concrete when a 150-kip axial centric force P is
applied to the post.
SOLUTION
 π  3 2 
As = 4     = 1.76715 in 2
 4  4  
Ac = 82 − As = 62.233 in 2
δs =
Ps L
Ps (48)
=
= 0.93663 × 10−6 Ps
As Es (1.76715)(29 × 106 )
δc =
Pc L
Pc (48)
=
= 0.21425 × 10−6 Pc
Ac Ec (62.233)(3.6 × 106 )
But δ s = δ c : 0.93663 × 10−6 Ps = 0.21425 × 10−6 Pc
Ps = 0.22875 Pc
Also:
Substituting (1) into (2):
Ps + Pc = P = 150 kips
(1)
(2)
1.22875Pc = 150 kips
Pc = 122.075 kips
From (1):
Ps = 0.22875(122.075) = 27.925 kips
σs = −
Ps
27.925
=−
As
1.76715
σ s = −15.80 ksi 
σc = −
Pc
122.075
=−
Ac
62.233
σ c = −1.962 ksi 
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PROBLEM 2.36
A 250-mm bar of 15 × 30-mm rectangular cross
section consists of two aluminum layers, 5-mm thick,
brazed to a center brass layer of the same thickness.
If it is subjected to centric forces of magnitude
P = 30 kN, and knowing that Ea = 70 GPa and
Eb = 105 GPa, determine the normal stress (a) in the
aluminum layers, (b) in the brass layer.
SOLUTION
For each layer,
A = (30)(5) = 150 mm 2 = 150 × 10−6 m 2
Let Pa = load on each aluminum layer
Pb = load on brass layer
Pa L Pb L
=
Ea A Eb A
Deformation.
δ=
Total force.
P = 2 Pa + Pb = 3.5 Pa
Solving for Pa and Pb,
Pa =
2
P
7
Pb =
Pb =
Eb
105
Pa =
Pa = 1.5 Pa
70
Ea
3
P
7
(a)
σa = −
Pa
2P
2 30 × 103
=−
=−
= −57.1 × 106 Pa
−6
A
7 A
7 150 × 10
σ a = −57.1 MPa 
(b)
σb = −
Pb
3P
3 30 × 103
=−
=−
= −85.7 × 106 Pa
A
7 A
7 150 × 10−6
σ b = −85.7 MPa 
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PROBLEM 2.37
Determine the deformation of the composite bar of
Prob. 2.36 if it is subjected to centric forces of
magnitude P = 45 kN.
PROBLEM 2.36 A 250-mm bar of 15 × 30-mm
rectangular cross section consists of two aluminum
layers, 5-mm thick, brazed to a center brass layer of
the same thickness. If it is subjected to centric forces
of magnitude P = 30 kN, and knowing that Ea = 70
GPa and Eb = 105 GPa, determine the normal stress
(a) in the aluminum layers, (b) in the brass layer.
SOLUTION
For each layer,
A = (30)(5) = 150 mm 2 = 150 × 10−6 m 2
Let Pa = load on each aluminum layer
Pb = load on brass layer
Pa L
PL
=− b
Ea A
Eb A
Deformation.
δ =−
Total force.
P = 2 Pa + Pb = 3.5 Pa
δ =−
=−
Pb =
Eb
105
Pa =
Pa = 1.5 Pa
70
Ea
Pa =
2
P
7
Pa L
2 PL
=−
Ea A
7 Ea A
2 (45 × 103 )(250 × 10−3 )
7 (70 × 109 )(150 × 10−6 )
= −306 × 10−6 m
δ = −0.306 mm 
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PROBLEM 2.38
Compressive centric forces of 40 kips are applied at both ends
of the assembly shown by means of rigid plates. Knowing
that Es = 29 × 106 psi and Ea = 10.1 × 106 psi, determine (a) the
normal stresses in the steel core and the aluminum shell,
(b) the deformation of the assembly.
SOLUTION
Let Pa = portion of axial force carried by shell
Ps = portion of axial force carried by core
Total force.
δ=
Pa L
Ea Aa
Pa =
Ea Aa
δ
L
δ=
Ps L
Es As
Ps =
Es As
δ
L
P = Pa + Ps = ( Ea Aa + Es As )
δ
L
Data:
δ
=ε =
L
P
Ea Aa + Es As
P = 40 × 103 lb
Aa =
As =
π
4
π
4
(d 02 − di2 ) =
2
d =
π
4
π
4
(2.52 − 1.0)2 = 4.1233 in 2
2
(1) = 0.7854 in 2
ε=
− 40 × 103
= −620.91 × 10−6
6
6
(10.1 × 10 )(4.1233) + (29 × 10 )(0.7854)
(a)
σ s = Es ε = (29 × 106 )(−620.91 × 10−6 ) = −18.01 × 103 psi

σ a = Ea ε = (10.1 × 106 )(620.91 × 10−6 ) = −6.27 × 103 psi 
(b)
δ = Lε = (10)(620.91 × 10−6 ) = −6.21 × 10−3
−18.01 ksi 
−6.27 ksi 
δ = −6.21 × 10−3 in. 
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PROBLEM 2.39
Three wires are used to suspend the plate shown. Aluminum wires of 18 -in.
diameter are used at A and B while a steel wire of 121 -in. diameter is used at
C. Knowing that the allowable stress for aluminum ( Ea = 10.4 × 106 psi) is
14 ksi and that the allowable stress for steel ( Es = 29 × 106 psi) is 18 ksi,
determine the maximum load P that can be applied.
SOLUTION
By symmetry,
PA = PB , and δ A = δ B
Also,
δC = δ A = δ B = δ
Strain in each wire:
δ
ε A = εB =
2L
, εC =
δ
L
= 2ε A
Determine allowable strain.
Wires A&B:
εA =
σA
EA
=
14 × 103
= 1.3462 × 10−3
10.4 × 106
ε C = 2 ε A = 2.6924 × 10−4
Wire C:
18 × 103
= 0.6207 × 10−3
EC 29 × 106
1
ε A = ε B = ε C = 0.3103 × 10−6
2
εC =
σC
=
σ A = E Aε A
σ C = 18 × 103 psi
∴
Allowable strain for wire C governs,
PA = AA E Aε A
π 1
2
(10.4 × 106 )(0.3103 × 10−6 ) = 39.61 lb
4  8 
PB = 39.61 lb
=
σ C = EC ε C
PC = ACσ C =
π1
2
(18 × 103 ) = 98.17 lb
4  12 
For equilibrium of the plate,
P = PA + PB + PC = 177.4 lb
P = 177.4 lb 
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PROBLEM 2.40
A polystyrene rod consisting of two cylindrical portions AB and BC is
restrained at both ends and supports two 6-kip loads as shown. Knowing
that E = 0.45 × 106 psi, determine (a) the reactions at A and C, (b) the
normal stress in each portion of the rod.
SOLUTION
(a)
We express that the elongation of the rod is zero:
δ=
But
PAB LAB
d2 E
4 AB
π
+
PAB = + RA
PBC LBC
π
4
2
d BC
E
=0
PBC = − RC
Substituting and simplifying:
RA LAB RC LBC
− 2
=0
2
d AB
d BC
RC =
LAB
LBC
2
2
 d BC 
25  2 

 RA = 
 RA
15
d
 1.25 
 AB 
RC = 4.2667 RA
From the free body diagram:
Substituting (1) into (2):
RA + RC = 12 kips
(2)
5.2667 RA = 12
RA = 2.2785 kips
From (1):
(1)
RA = 2.28 kips ↑ 
RC = 4.2667 (2.2785) = 9.7217 kips
RC = 9.72 kips ↑ 
(b)
σ AB =
PAB + RA
2.2785
=
=
AAB AAB π4 (1.25)2
σ AB = +1.857 ksi 
σ BC =
PBC − RC −9.7217
=
=
π
ABC
ABC
(2)2
4
σ BC = −3.09 ksi 
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PROBLEM 2.41
Two cylindrical rods, one of steel and the other of brass, are joined at
C and restrained by rigid supports at A and E. For the loading shown
and knowing that Es = 200 GPa and Eb = 105 GPa, determine
(a) the reactions at A and E, (b) the deflection of point C.
SOLUTION
A to C:
E = 200 × 109 Pa
π
(40) 2 = 1.25664 × 103 mm 2 = 1.25664 × 10−3 m 2
4
EA = 251.327 × 106 N
A=
C to E:
E = 105 × 109 Pa
π
(30) 2 = 706.86 mm 2 = 706.86 × 10−6 m 2
4
EA = 74.220 × 106 N
A=
A to B:
P = RA
L = 180 mm = 0.180 m
RA (0.180)
PL
=
δ AB =
EA 251.327 × 106
= 716.20 × 10−12 RA
B to C:
P = RA − 60 × 103
L = 120 mm = 0.120 m
δ BC =
PL ( RA − 60 × 103 )(0.120)
=
EA
251.327 × 106
= 447.47 × 10−12 RA − 26.848 × 10−6
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PROBLEM 2.41 (Continued)
C to D:
P = RA − 60 × 103
L = 100 mm = 0.100 m
PL ( RA − 60 × 103 )(0.100)
=
EA
74.220 × 106
= 1.34735 × 10−9 RA − 80.841 × 10−6
δ BC =
D to E:
P = RA − 100 × 103
L = 100 mm = 0.100 m
PL ( RA − 100 × 103 )(0.100)
=
EA
74.220 × 106
= 1.34735 × 10−9 RA − 134.735 × 10−6
δ DE =
A to E:
δ AE = δ AB + δ BC + δ CD + δ DE
= 3.85837 × 10−9 RA − 242.424 × 10−6
Since point E cannot move relative to A,
(a)
(b)
δ AE = 0
3.85837 × 10−9 RA − 242.424 × 10−6 = 0 RA = 62.831 × 103 N
RA = 62.8 kN ← 
RE = RA − 100 × 103 = 62.8 × 103 − 100 × 103 = −37.2 × 103 N
RE = 37.2 kN ← 
δ C = δ AB + δ BC = 1.16367 × 10−9 RA − 26.848 × 10−6
= (1.16369 × 10−9 )(62.831 × 103 ) − 26.848 × 10−6
= 46.3 × 10−6 m
δ C = 46.3 μ m → 
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PROBLEM 2.42
Solve Prob. 2.41, assuming that rod AC is made of brass and
rod CE is made of steel.
PROBLEM 2.41 Two cylindrical rods, one of steel and the
other of brass, are joined at C and restrained by rigid supports at A
and E. For the loading shown and knowing that Es = 200 GPa
and Eb = 105 GPa, determine (a) the reactions at A and E, (b) the
deflection of point C.
SOLUTION
A to C:
E = 105 × 109 Pa
π
(40) 2 = 1.25664 × 103 mm 2 = 1.25664 × 10−3 m 2
4
EA = 131.947 × 106 N
A=
C to E:
E = 200 × 109 Pa
π
(30) 2 = 706.86 mm 2 = 706.86 × 10−6 m 2
4
EA = 141.372 × 106 N
A=
A to B:
P = RA
L = 180 mm = 0.180 m
RA (0.180)
PL
=
δ AB =
EA 131.947 × 106
= 1.36418 × 10−9 RA
B to C:
P = RA − 60 × 103
L = 120 mm = 0.120 m
δ BC =
PL ( RA − 60 × 103 )(0.120)
=
EA
131.947 × 106
= 909.456 × 10−12 RA − 54.567 × 10−6
C to D:
P = RA − 60 × 103
L = 100 mm = 0.100 m
δ CD =
PL ( RA − 60 × 103 )(0.100)
=
EA
141.372 × 106
= 707.354 × 10−12 RA − 42.441 × 10−6
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PROBLEM 2.42 (Continued)
D to E:
P = RA − 100 × 103
L = 100 mm = 0.100 m
PL ( RA − 100 × 103 )(0.100)
=
EA
141.372 × 106
= 707.354 × 10−12 RA − 70.735 × 10−6
δ DE =
A to E:
δ AE = δ AB + δ BC + δ CD + δ DE
= 3.68834 × 10−9 RA − 167.743 × 10−6
Since point E cannot move relative to A,
(a)
(b)
δ AE = 0
3.68834 × 10−9 RA − 167.743 × 10−6 = 0 RA = 45.479 × 103 N
R A = 45.5 kN ← 
RE = RA − 100 × 103 = 45.479 × 103 − 100 × 103 = −54.521 × 103
RE = 54.5 kN ← 
δ C = δ AB + δ BC = 2.27364 × 10−9 RA − 54.567 × 10−6
= (2.27364 × 10−9 )(45.479 × 103 ) − 54.567 × 10−6
= 48.8 × 10−6 m
δ C = 48.8 μ m → 
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PROBLEM 2.43
The rigid bar ABCD is suspended from four identical wires. Determine the
tension in each wire caused by the load P shown.
SOLUTION
Deformations Let θ be the rotation of bar ABCD and δ A , δ B , δ C and δ D be the deformations of wires A, B,
C, and D.
From geometry,
θ=
δB − δ A
L
δ B = δ A + Lθ
δ C = δ A + 2 Lθ = 2δ B − δ A
(1)
δ D = δ A + 3Lθ = 3δ B − 2δ A
(2)
Since all wires are identical, the forces in the wires are proportional to the deformations.
TC = 2TB − TA
(1′)
TD = 3TB − 2TA
(2′)
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PROBLEM 2.43 (Continued)
Use bar ABCD as a free body.
ΣM C = 0 : − 2 LTA − LTB + LTD = 0
(3)
ΣFy = 0 : TA + TB + TC + TD − P = 0
(4)
Substituting (2′) into (3) and dividing by L,
−4TA + 2TB = 0
TB = 2TA
(3′)
Substituting (1′), (2′), and (3′) into (4),
TA + 2TA + 3TA + 4TA − P = 0
10TA = P
TA =
 1 
TB = 2TA = (2)   P
 10 
1
P 
10
TB =
1   1 
TC = (2)  P  −  P 
 5   10 
TC =
1 
 1 
TD = (3)  P  − (2)  P 
5


 10 
TD =
1
P 
5
3
P 
10
2
P 
5
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PROBLEM 2.44
The rigid bar AD is supported by two steel wires of 161 -in. diameter
( E = 29 × 106 psi) and a pin and bracket at D. Knowing that the
wires were initially taut, determine (a) the additional tension in each
wire when a 120-lb load P is applied at B, (b) the corresponding
deflection of point B.
SOLUTION
Let θ be the rotation of bar ABCD.
Then δ A = 24θ
δ C = 8θ 
δA =
PAE LAE
AE
PAE =
6
2
EAδ A (29 × 10 ) π4 ( 161 ) (24θ )
=
LAE
15
= 142.353 × 103θ
δC =
PCF LCF
AE
6 π 1
EAδ C (29 × 10 ) 4 ( 16 ) (8θ )
=
=
8
LCF
2
PCF
= 88.971 × 103θ
Using free body ABCD,
ΣM D = 0 :
−24PAE + 16P − 8PCF = 0
−24(142.353 × 103θ ) + 16(120) − 8(88.971 × 103θ ) = 0
θ = 0.46510 × 10−3 rad哷
(a)
(b)
PAE = (142.353 × 103 ) (0.46510 × 10−3 )
PAE = 66.2 lb 
PCF = (88.971 × 103 ) (0.46510 × 10−3 )
PCF = 41.4 lb 
δ B = 16θ = 16(0.46510 × 10−3 )
δ B = 7.44 × 10−3 in. ↓ 
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PROBLEM 2.45
The steel rods BE and CD each have a 16-mm diameter
( E = 200 GPa); the ends of the rods are single-threaded with a
pitch of 2.5 mm. Knowing that after being snugly fitted, the nut at
C is tightened one full turn, determine (a) the tension in rod CD,
(b) the deflection of point C of the rigid member ABC.
SOLUTION
Let θ be the rotation of bar ABC as shown.
Then
δ B = 0.15θ
But
δ C = δ turn −
PCD =
=
δ C = 0.25θ
PCD LCD
ECD ACD
ECD ACD
(δ turn − δ C )
LCD
(200 × 109 Pa) π4 (0.016 m) 2
2m
(0.0025m − 0.25θ )
= 50.265 × 103 − 5.0265 × 106 θ
δB =
PBE =
PBE LBE
EBE ABE
or
PBE =
EBE ABE
δB
LBE
(200 × 109 Pa) π4 (0.016 m) 2
3m
(0.15θ )
= 2.0106 × 106 θ
From free body of member ABC:
ΣM A = 0 : 0.15 PBE − 0.25PCD = 0
0.15(2.0106 × 106 θ ) − 0.25(50.265 × 103 − 5.0265 × 106 θ ) = 0
θ = 8.0645 × 10−3 rad
(a)
PCD = 50.265 × 103 − 5.0265 × 106 (8.0645 × 10−3 )
= 9.7288 × 103 N
(b)
PCD = 9.73 kN 
δ C = 0.25θ = 0.25(8.0645 × 10−3 )
= 2.0161 × 10−3 m
δ C = 2.02 mm ← 
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PROBLEM 2.46
Links BC and DE are both made of steel ( E = 29 × 106 psi) and are
1
in. wide and 14 in. thick. Determine (a) the force in each link
2
when a 600-lb force P is applied to the rigid member AF shown,
(b) the corresponding deflection of point A.
SOLUTION
Let the rigid member ACDF rotate through small angle θ clockwise about point F.
δ C = δ BC = 4θ in. →
δ D = −δ DE = 2θ in. →
Then
δ=
For links:
FL
EA
or
F=
EAδ
L
 1  1 
A =    = 0.125 in 2
 2  4 
LBC = 4 in.
LDE = 5 in.
FBC =
EA δ BC (29 × 106 )(0.125)(4θ )
=
= 3.625 × 106 θ
4
LBC
FDE =
EAδ DE (29 × 106 )(0.125)( −2θ )
=
= −1.45 × 106 θ
5
LDE
Use member ACDF as a free body.
ΣM F = 0 : 8 P − 4 FBC + 2 FDE = 0
1
1
FBC − FDE
2
4
1
1
600 = (3.625 × 106 )θ − ( −1.45 × 106 )θ = 2.175 × 106 θ
2
4
−3
+
θ = 0.27586 × 10 rad 哶
P=
(a)
(b)
FBC = (3.625 × 106 )(0.27586 × 10−3 )
FBC = 1000 lb 
FDE = −(1.45 × 106 )(0.27586 × 10−3 )
FDE = −400 lb 
Deflection at Point A.
δ A = 8θ = (8)(0.27586 × 10−3 )
δ A = 2.21 × 10−3 in → 
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PROBLEM 2.47
The concrete post ( Ec = 3.6 × 106 psi and α c = 5.5 × 10−6 / °F) is reinforced
with six steel bars, each of 78 -in. diameter ( Es = 29 × 106 psi and
α s = 6.5 × 10−6 / °F). Determine the normal stresses induced in the steel and
in the concrete by a temperature rise of 65°F.
SOLUTION
As = 6
π
4
d2 = 6
π 7
2
= 3.6079 in 2
4  8 
Ac = 102 − As = 102 − 3.6079 = 96.392 in 2
Let Pc = tensile force developed in the concrete.
For equilibrium with zero total force, the compressive force in the six steel rods equals Pc .
Strains:
εs = −
Matching: ε c = ε s
Pc
+ α s (ΔT )
Es As
εc =
Pc
+ α c ( ΔT )
Ec Ac
Pc
P
+ α c (ΔT ) = − c + α s (ΔT )
Ec Ac
Es As
 1
1 
+

 Pc = (α s − α c )(ΔT )
 Ec Ac Es As 


1
1
−6
+

 Pc = (1.0 × 10 )(65)
6
6
 (3.6 × 10 )(96.392) (29 × 10 )(3.6079) 
Pc = 5.2254 × 103 lb
σc =

Pc 5.2254 × 103
=
= 54.210 psi
Ac
96.392
σs = −
Pc
5.2254 × 103
=−
= −1448.32 psi 
As
3.6079
σ c = 54.2 psi
σ s = −1.448 ksi 
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PROBLEM 2.48
The assembly shown consists of an aluminum shell ( Ea = 10.6 × 106 psi,
αa = 12.9 × 10–6/°F) fully bonded to a steel core ( Es = 29 × 106 psi,
αs = 6.5 × 10–6/°F) and is unstressed. Determine (a) the largest allowable
change in temperature if the stress in the aluminum shell is not to exceed
6 ksi, (b) the corresponding change in length of the assembly.
SOLUTION
Since α a > α s , the shell is in compression for a positive temperature rise.
σ a = −6 ksi = −6 × 103 psi
Let
Aa =
As =
π
4
π
4
(d
)
2
o
− di2 =
d2 =
π
4
π
4
(1.252 − 0.752 ) = 0.78540 in 2
(0.75) 2 = 0.44179 in 2
P = −σ a Aa = σ s As
where P is the tensile force in the steel core.
σs = −
ε=
(α a − α s )(ΔT ) =
(6.4 × 10−6 )(ΔT ) =
(a)
ΔT = 145.91°F
(b)
ε=
σ a Aa
σs
Es
σs
Es
As
=
(6 × 103 )(0.78540)
= 10.667 × 103 psi
0.44179
+ α s ( ΔT ) =
−
σa
Ea
+ α a ( ΔT )
σa
Ea
10.667 × 103
6 × 103
+
= 0.93385 × 10−3
29 × 106
10.6 × 106
ΔT = 145.9°F 
10.667 × 103
+ (6.5 × 10−6 )(145.91) = 1.3163 × 10−3
29 × 106
or
ε=

−6 × 103
+ (12.9 × 10−6 )(145.91) = 1.3163 × 10−3
10.6 × 106
δ = Lε = (8.0)(1.3163 × 10−3 ) = 0.01053 in. 
δ = 0.01053 in. 
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PROBLEM 2.49
The aluminum shell is fully bonded to the brass core and the assembly is
unstressed at a temperature of 15 °C. Considering only axial deformations,
determine the stress in the aluminum when the temperature reaches 195 °C.
SOLUTION
Brass core:
E = 105 GPa
α = 20.9 × 10−6/ °C
Aluminum shell:
E = 70 GPa
α = 23.6 × 10−6 / °C
Let L be the length of the assembly.
Free thermal expansion:
ΔT = 195 − 15 = 180 °C
Brass core:
Aluminum shell:
(δT )b = Lα b ( ΔT )
(δT ) = Lα a (ΔT )
Net expansion of shell with respect to the core:
δ = L(α a − α b )(ΔT )
Let P be the tensile force in the core and the compressive force in the shell.
Brass core:
Eb = 105 × 109 Pa
π
(25) 2 = 490.87 mm 2
4
= 490.87 × 10−6 m 2
PL
(δ P )b =
Eb Ab
Ab =
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PROBLEM 2.49 (Continued)
Aluminum shell:
Ea = 70 × 109 Pa
π
(602 − 252 )
4
= 2.3366 × 103 mm 2
Aa =
= 2.3366 × 10−3 m 2
δ = (δ P )b + (δ P ) a
L(α b − α a )(ΔT ) =
PL
PL
+
= KPL
Eb Ab Ea Aa
where
K=
=
1
1
+
Eb Ab Ea Aa
1
1
+
−6
9
(105 × 10 )(490.87 × 10 ) (70 × 10 )(2.3366 × 10−3 )
9
= 25.516 × 10−9 N −1
Then
(α b − α a )(ΔT )
K
(23.6 × 10−6 − 20.9 × 10−6 )(180)
=
25.516 × 10−9
= 19.047 × 103 N
P=
Stress in aluminum:
σa = −
P
19.047 × 103
=−
= −8.15 × 106 Pa
−3
Aa
2.3366 × 10
σ a = −8.15 MPa 
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PROBLEM 2.50
Solve Prob. 2.49, assuming that the core is made of steel ( Es = 200 GPa,
α s = 11.7 × 10−6 / °C) instead of brass.
PROBLEM 2.49 The aluminum shell is fully bonded to the brass core
and the assembly is unstressed at a temperature of 15 °C. Considering
only axial deformations, determine the stress in the aluminum when the
temperature reaches 195 °C.
SOLUTION
E = 70 GPa α = 23.6 × 10−6 / °C
Aluminum shell:
Let L be the length of the assembly.
ΔT = 195 − 15 = 180 °C
Free thermal expansion:
Steel core:
(δT ) s = Lα s (ΔT )
Aluminum shell:
(δT ) a = Lα a (ΔT )
δ = L(α a − α s )( ΔT )
Net expansion of shell with respect to the core:
Let P be the tensile force in the core and the compressive force in the shell.
Es = 200 × 109 Pa, As =
Steel core:
(δ P ) s =
Aluminum shell:
π
4
(25) 2 = 490.87 mm 2 = 490.87 × 10−6 m 2
PL
Es As
Ea = 70 × 109 Pa
(δ P ) a =
PL
Ea Aa
π
(602 − 25) 2 = 2.3366 × 103 mm 2 = 2.3366 × 10−3 m 2
4
δ = (δ P ) s + (δ P )a
Aa =
L(α a − α s )(ΔT ) =
PL
PL
+
= KPL
Es As Ea Aa
where
K=
=
1
1
+
Es As Ea Aa
1
1
+
−6
9
(200 × 10 )(490.87 × 10 ) (70 × 10 )(2.3366 × 10−3 )
9
= 16.2999 × 10−9 N −1
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PROBLEM 2.50 (Continued)
Then
P=
(α a − α s )(ΔT ) (23.6 × 10−6 − 11.7 × 10−6 )(180)
=
= 131.41 × 103 N
K
16.2999 × 10−9
Stress in aluminum: σ a = −
P
131.19 × 103
=−
= −56.2 × 106 Pa
−3
Aa
2.3366 × 10
σ a = −56.2 MPa 
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PROBLEM 2.51
A rod consisting of two cylindrical portions AB and BC is restrained at both
ends. Portion AB is made of steel ( Es = 200 GPa, α s = 11.7 × 10−6 / °C) and
portion BC is made of brass ( Eb = 105 GPa, α b = 20.9 × 10−6 / °C). Knowing
that the rod is initially unstressed, determine the compressive force induced in
ABC when there is a temperature rise of 50 °C.
SOLUTION
AAB =
ABC =
π
4
π
4
2
d AB
=
2
d BC
=
π
4
π
4
(30)2 = 706.86 mm 2 = 706.86 × 10−6 m 2
(50)2 = 1.9635 × 103 mm 2 = 1.9635 × 10−3 m 2
Free thermal expansion:
δT = LABα s (ΔT ) + LBC α b (ΔT )
= (0.250)(11.7 × 10−6 )(50) + (0.300)(20.9 × 10−6 )(50)
= 459.75 × 10−6 m
Shortening due to induced compressive force P:
δP =
=
PL
PL
+
Es AAB Eb ABC
0.250 P
0.300 P
+
−6
9
(200 × 10 )(706.86 × 10 ) (105 × 10 )(1.9635 × 10−3 )
9
= 3.2235 × 10−9 P
For zero net deflection, δ P = δ T
3.2235 × 10−9 P = 459.75 × 10−6
P = 142.62 × 103 N
P = 142.6 kN 
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PROBLEM 2.52
A steel railroad track ( Es = 200 GPa, α s = 11.7 × 10−6 /°C) was laid out at a temperature of 6°C. Determine
the normal stress in the rails when the temperature reaches 48°C, assuming that the rails (a) are welded to
form a continuous track, (b) are 10 m long with 3-mm gaps between them.
SOLUTION
(a)
δT = α ( ΔT ) L = (11.7 × 10−6 )(48 − 6)(10) = 4.914 × 10−3 m
δP =
PL Lσ
(10)σ
=
=
= 50 × 10−12 σ
AE
E
200 × 109
δ = δT + δ P = 4.914 × 10−3 + 50 × 10−12 σ = 0
σ = −98.3 × 106 Pa
(b)
σ = −98.3 MPa
δ = δT + δ P = 4.914 × 10−3 + 50 × 10−12 σ = 3 × 10−3
3 × 10−3 − 4.914 × 10−3
50 × 10−12
= −38.3 × 106 Pa
σ=
σ = −38.3 MPa 
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PROBLEM 2.53
A rod consisting of two cylindrical portions AB and BC is restrained at
both ends. Portion AB is made of steel ( Es = 29 × 106 psi,
α s = 6.5 × 10−6 /°F) and portion BC is made of aluminum
( Ea = 10.4 × 106 psi, α a = 13.3 × 10−6 /°F). Knowing that the rod is
initially unstressed, determine (a) the normal stresses induced in portions
AB and BC by a temperature rise of 70°F, (b) the corresponding
deflection of point B.
SOLUTION
AAB =
Free thermal expansion.
π
4
(2.25) 2 = 3.9761 in 2 ABC =
π
4
(1.5) 2 = 1.76715 in 2
ΔT = 70°F
(δ T ) AB = LABα s (ΔT ) + (24)(6.5 × 10−6 )(70) = 10.92 × 10−3 in
(δT ) BC = LBC α a (ΔT ) = (32)(13.3 × 10−6 (70) = 29.792 × 10−3 in.
δT = (δT ) AB + (δT ) BC = 40.712 × 10−3 in.
Total:
Shortening due to induced compressive force P.
PLAB
24 P
=
= 208.14 × 10−9 P
Es AAB (29 × 106 )(3.9761)
PLBC
32 P
=
=
= 1741.18 × 10−9 P
Ea ABC (10.4 × 106 )(1.76715)
(δ P ) AB =
(δ P ) BC
δ P = (δ P ) AB + (δ P ) BC = 1949.32 × 10−9 P
Total:
For zero net deflection, δ P = δ T
(a)
(b)
1949.32 × 10−9 P = 40.712 × 10−3
P = 20.885 × 103 lb
σ AB = −
P
20.885 × 103
=−
= −5.25 × 103 psi
3.9761
AAB
σ AB = −5.25 ksi 
σ BC = −
P
20.885 × 103
=−
= −11.82 × 103 psi
1.76715
ABC
σ BC = −11.82 ksi 
(δ P ) AB = (208.14 × 10−9 )(20.885 × 103 ) = 4.3470 × 10−3 in.
δ B = (δT ) AB → + (δ P ) AB ← = 10.92 × 10−3 → + 4.3470 × 10−3 ←
δ B = 6.57 × 10−3 in. → 
or
(δ P ) BC = (1741.18 × 10−9 )(20.885 × 103 ) = 36.365 × 10−3 in.
δ B = (δT ) BC ← + (δ P ) BC → = 29.792 × 10−3 ← + 36.365 × 10−3 → = 6.57 × 10−3 in. →
(checks)
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PROBLEM 2.54
Solve Prob. 2.53, assuming that portion AB of the composite rod is made
of aluminum and portion BC is made of steel.
PROBLEM 2.53 A rod consisting of two cylindrical portions AB and BC
is restrained at both ends. Portion AB is made of steel ( Es = 29 × 106 psi,
α s = 6.5 × 10−6 /°F) and portion BC is made of aluminum
( Ea = 10.4 × 106 psi, α a = 13.3 × 10−6 /°F). Knowing that the rod is
initially unstressed, determine (a) the normal stresses induced in portions
AB and BC by a temperature rise of 70°F, (b) the corresponding
deflection of point B.
SOLUTION
AAB =
π
4
(2.25) 2 = 3.9761 in 2
Free thermal expansion.
ABC =
π
4
(1.5) 2 = 1.76715 in 2
ΔT = 70°F
(δ T ) AB = LABα a ( ΔT ) = (24)(13.3 × 10−6 )(70) = 22.344 × 10−3 in.
(δT ) BC = LBC α s (ΔT ) = (32)(6.5 × 10−6 )(70) = 14.56 × 10−3 in.
δT = (δT ) AB + (δT ) BC = 36.904 × 10−3 in.
Total:
Shortening due to induced compressive force P.
PLAB
24 P
=
= 580.39 × 10−9 P
Ea AAB (10.4 × 106 )(3.9761)
PLBC
32 P
=
=
= 624.42 × 10−9 P
6
Es ABC (29 × 10 )(1.76715)
(δ P ) AB =
(δ P ) BC
δ P = (δ P ) AB + (δ P ) BC = 1204.81 × 10−9 P
Total:
For zero net deflection, δ P = δ T
(a)
1204.81 × 10−9 P = 36.904 × 10−3
P = 30.631 × 103 lb
σ AB = −
P
30.631 × 103
=−
= −7.70 × 103 psi
3.9761
AAB
σ AB = −7.70 ksi 
σ BC = −
P
30.631 × 103
=−
= −17.33 × 103 psi
1.76715
ABC
σ BC = −17.33 ksi 
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PROBLEM 2.54 (Continued)
(b)
(δ P ) AB = (580.39 × 10−9 ) (30.631 × 103 ) = 17.7779 × 10−3 in.
δ B = (δT ) AB → + (δ P ) AB ← = 22.344 × 10−3 → + 17.7779 × 10−3 ←

or
δ B = 4.57 × 10−3 in. → 
(δ P ) BC = (624.42 × 10−9 )(30.631 × 103 ) = 19.1266 × 10−3 in.
δ B = (δT ) BC ← + (δ P ) BC → = 14.56 × 10−3 ← + 19.1266 × 10−3 → = 4.57 × 10−3 in. →
(checks)
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PROBLEM 2.55
A brass link ( Eb = 105 GPa, α b = 20.9 × 10−6 /°C) and a steel rod
(Es = 200 GPa, α s = 11.7 × 10−6 / °C) have the dimensions shown
at a temperature of 20 °C. The steel rod is cooled until it fits
freely into the link. The temperature of the whole assembly is
then raised to 45 °C. Determine (a) the final stress in the steel
rod, (b) the final length of the steel rod.
SOLUTION
Initial dimensions at T = 20 °C.
Final dimensions at T = 45 °C.
ΔT = 45 − 20 = 25 °C
Free thermal expansion of each part:
Brass link:
(δT )b = α b (ΔT ) L = (20.9 × 10−6 )(25)(0.250) = 130.625 × 10−6 m
Steel rod:
(δT ) s = α s (ΔT ) L = (11.7 × 10−6 )(25)(0.250) = 73.125 × 10−6 m
At the final temperature, the difference between the free length of the steel rod and the brass link is
δ = 120 × 10−6 + 73.125 × 10−6 − 130.625 × 10−6 = 62.5 × 10−6 m
Add equal but opposite forces P to elongate the brass link and contract the steel rod.
Brass link:
E = 105 × 109 Pa
Ab = (2)(50)(37.5) = 3750 mm 2 = 3.750 × 10−3 m 2
(δ P ) =
Steel rod:
PL
P(0.250)
=
= 634.92 × 10−12 P
EA (105 × 109 )(3.750 × 10−3 )
E = 200 × 109 Pa
(δ P ) s =
As =
π
4
(30) 2 = 706.86 mm 2 = 706.86 × 10−6 m 2
PL
P(0.250)
=
= 1.76838 × 10−9 P
Es As (200 × 109 )(706.86 × 10−6 )
(δ P )b + (δ P ) s = δ : 2.4033 × 10−9 P = 62.5 × 10−6 P = 26.006 × 103 N
σs = −
P
(26.006 × 103 )
=−
= −36.8 × 106 Pa
−6
As
706.86 × 10
(a)
Stress in steel rod:
(b)
Final length of steel rod: L f = L0 + (δ T ) s − (δ P ) s
σ s = −36.8 MPa 
L f = 0.250 + 120 × 10−6 + 73.125 × 10−6 − (1.76838 × 10−9 )(26.003 × 103 )
= 0.250147 m
L f = 250.147 mm 
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PROBLEM 2.56
Two steel bars ( Es = 200 GPa and α s = 11.7 × 10−6 /°C) are used to
reinforce a brass bar ( Eb = 105 GPa, α b = 20.9 × 10−6 / °C) that is
subjected to a load P = 25 kN. When the steel bars were fabricated, the
distance between the centers of the holes that were to fit on the pins
was made 0.5 mm smaller than the 2 m needed. The steel bars
were then placed in an oven to increase their length so that they
would just fit on the pins. Following fabrication, the temperature
in the steel bars dropped back to room temperature. Determine
(a) the increase in temperature that was required to fit the steel
bars on the pins, (b) the stress in the brass bar after the load is
applied to it.
SOLUTION
(a)
Required temperature change for fabrication:
δT = 0.5 mm = 0.5 × 10−3 m
Temperature change required to expand steel bar by this amount:
δT = Lα s ΔT , 0.5 × 10−3 = (2.00)(11.7 × 10−6 )(ΔT ),
ΔT = 0.5 × 10−3 = (2)(11.7 × 10−6 )(ΔT )
ΔT = 21.368 °C
(b)
21.4 °C 
Once assembled, a tensile force P* develops in the steel, and a compressive force P* develops in the
brass, in order to elongate the steel and contract the brass.
As = (2)(5)(40) = 400 mm2 = 400 × 10−6 m 2
Elongation of steel:
(δ P ) s =
F *L
P* (2.00)
=
= 25 × 10−9 P*
As Es (400 × 10−6 )(200 × 109 )
Contraction of brass: Ab = (40)(15) = 600 mm 2 = 600 × 10−6 m 2
(δ P )b =
P* L
P* (2.00)
=
= 31.746 × 10−9 P*
Ab Eb (600 × 10−6 )(105 × 109 )
But (δ P ) s + (δ P )b is equal to the initial amount of misfit:
(δ P ) s + (δ P )b = 0.5 × 10−3 , 56.746 × 10−9 P* = 0.5 × 10−3
P* = 8.811 × 103 N
Stresses due to fabrication:
Steel:
σ s* =
P* 8.811 × 103
=
= 22.03 × 106 Pa = 22.03 MPa
As 400 × 10−6
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PROBLEM 2.56 (Continued)
Brass:
σ b* = −
P*
8.811 × 103
=−
= −14.68 × 106 Pa = −14.68 MPa
−6
Ab
600 × 10
To these stresses must be added the stresses due to the 25 kN load.
For the added load, the additional deformation is the same for both the steel and the brass. Let δ ′ be the
additional displacement. Also, let Ps and Pb be the additional forces developed in the steel and brass,
respectively.
δ′ =
Ps L
PL
= b
As Es Ab Eb
As Es
(400 × 10−6 )(200 × 109 )
δ′ =
δ ′ = 40 × 106 δ ′
2.00
L
Ab Eb
(600 × 10−6 )(105 × 109 )
δ′ =
δ ′ = 31.5 × 106 δ ′
Pb =
2.00
L
Ps =
P = Ps + Pb = 25 × 103 N
Total
40 × 106 δ ′ + 31.5 × 106 δ ′ = 25 × 103
δ ′ = 349.65 × 10−6 m
Ps = (40 × 106 )(349.65 × 10−6 ) = 13.986 × 103 N
Pb = (31.5 × 106 )(349.65 × 10−6 ) = 11.140 × 103 N
σs =
Ps 13.986 × 103
=
= 34.97 × 106 Pa
As
400 × 10−6
σb =
Pb 11.140 × 103
=
= 18.36 × 106 Pa
Ab
600 × 10−6
Add stress due to fabrication.
Total stresses:
σ s = 34.97 × 106 + 22.03 × 106 = 57.0 × 106 Pa
σ s = 57.0 MPa
σ b = 18.36 × 106 − 14.68 × 106 = 3.68 × 106 Pa
σ b = 3.68 MPa 
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PROBLEM 2.57
Determine the maximum load P that may be applied to the brass bar of
Prob. 2.56 if the allowable stress in the steel bars is 30 MPa and the
allowable stress in the brass bar is 25 MPa.
PROBLEM 2.56 Two steel bars ( Es = 200 GPa and α s = 11.7 × 10–6/°C)
are used to reinforce a brass bar ( Eb = 105 GPa, α b = 20.9 × 10–6/°C)
that is subjected to a load P = 25 kN. When the steel bars were fabricated,
the distance between the centers of the holes that were to fit on the pins was
made 0.5 mm smaller than the 2 m needed. The steel bars were then placed
in an oven to increase their length so that they would just fit on the pins.
Following fabrication, the temperature in the steel bars dropped back to
room temperature. Determine (a) the increase in temperature that was
required to fit the steel bars on the pins, (b) the stress in the brass bar after
the load is applied to it.
SOLUTION
See solution to Problem 2.56 to obtain the fabrication stresses.
σ s* = 22.03 MPa
σ b* = 14.68 MPa
Allowable stresses:
σ s ,all = 30 MPa, σ b ,all = 25 MPa
Available stress increase from load.
σ s = 30 − 22.03 = 7.97 MPa
σ b = 25 + 14.68 = 39.68 MPa
Corresponding available strains.
εs =
εb =
σs
Es
σb
Eb
=
7.97 × 106
= 39.85 × 10−6
200 × 109
=
39.68 × 106
= 377.9 × 10−6
105 × 109
Smaller value governs ∴ ε = 39.85 × 10−6
Areas: As = (2)(5)(40) = 400 mm2 = 400 × 10−6 m 2
Ab = (15)(40) = 600 mm 2 = 600 × 10−6 m 2
Forces Ps = Es As ε = (200 × 109 )(400 × 10−6 )(39.85 × 10−6 ) = 3.188 × 103 N
Pb = Eb Abε = (105 × 109 )(600 × 10−6 )(39.85 × 10−6 ) = 2.511 × 10−3 N
Total allowable additional force:
P = Ps + Pb = 3.188 × 103 + 2.511 × 103 = 5.70 × 103 N
P = 5.70 kN 
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PROBLEM 2.58
Knowing that a 0.02-in. gap exists when the temperature is 75 °F,
determine (a) the temperature at which the normal stress in the
aluminum bar will be equal to −11 ksi, (b) the corresponding exact
length of the aluminum bar.
SOLUTION
σ a = −11 ksi = −11 × 103 psi
P = −σ a Aa = (11 × 103 )(2.8) = 30.8 × 103 lb
Shortening due to P:
δP =
=
PLb
PLa
+
Eb Ab Ea Aa
(30.8 × 103 )(14) (30.8 × 103 )(18)
+
(15 × 106 )(2.4) (10.6 × 106 )(2.8)
= 30.657 × 10−3 in.
Available elongation for thermal expansion:
δT = 0.02 + 30.657 × 10−3 = 50.657 × 10−3 in.
But δ T = Lbα b (ΔT ) + Laα a (ΔT )
= (14)(12 × 10−6 )( ΔT ) + (18)(12.9 × 10−6 )(ΔT ) = 400.2 × 10−6 ) ΔT
Equating, (400.2 × 10−6 )ΔT = 50.657 × 10−3
(a)
(b)
ΔT = 126.6 °F
Thot = Tcold + ΔT = 75 + 126.6 = 201.6 °F
δ a = Laα a (ΔT ) −
Thot = 201.6°F 
PLa
Ea Aa
= (18)(12.9 × 10−6 )(26.6) −
(30.8 × 103 )(18)
= 10.712 × 10−3 in.
6
(10.6 × 10 )(2.8)
Lexact = 18 + 10.712 × 10−3 = 18.0107 in.
L = 18.0107 in. 
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PROBLEM 2.59
Determine (a) the compressive force in the bars shown after a
temperature rise of 180 °F, (b) the corresponding change in length of
the bronze bar.
SOLUTION
Thermal expansion if free of constraint:
δT = Lbα b (ΔT ) + Laα a (ΔT )
= (14)(12 × 10−6 )(180) + (18)(12.9 × 10−6 )(180)
= 72.036 × 10−3 in.
Constrained expansion: δ = 0.02in.
Shortening due to induced compressive force P:
δ P = 72.036 × 10−3 − 0.02 = 52.036 × 10−3 in.
δP =
But
PLb
PLa  Lb
L
+
=
+ a
Eb Ab Ea Aa  Eb Ab Ea Aa

P



14
18
−9
=
+
 P = 995.36 × 10 P
6
6
 (15 × 10 )(2.4) (10.6 × 10 )(2.8) 
995.36 × 10−9 P = 52.036 × 10−3
Equating,
P = 52.279 × 103 lb
P = 52.3 kips 
(a)
(b)
δ b = Lbα b (ΔT ) −
PLb
Eb Ab
= (14)(12 × 10−6 )(180) −
(52.279 × 103 )(14)
= 9.91 × 10−3 in.
(15 × 106 )(2.4)
δ b = 9.91 × 10−3 in. 
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PROBLEM 2.60
At room temperature (20 °C) a 0.5-mm gap exists between the ends
of the rods shown. At a later time when the temperature has reached
140°C, determine (a) the normal stress in the aluminum rod, (b) the
change in length of the aluminum rod.
SOLUTION
ΔT = 140 − 20 = 120 °C
Free thermal expansion:
δT = Laα a (ΔT ) + Lsα s ( ΔT )
= (0.300)(23 × 10−6 )(120) + (0.250)(17.3 × 10−6 )(120)
= 1.347 × 10−3 m
Shortening due to P to meet constraint:
δ P = 1.347 × 10−3 − 0.5 × 10−3 = 0.847 × 10−3 m
PLa
PLs  La
L 
+
=
+ s P
Ea Aa Es As  Ea Aa Es As 


0.300
0.250
=
+
P
9
−6
9
−6 
 (75 × 10 )(2000 × 10 ) (190 × 10 )(800 × 10 ) 
= 3.6447 × 10−9 P
δP =
3.6447 × 10−9 P = 0.847 × 10−3
Equating,
P = 232.39 × 103 N
P
232.39 × 103
=−
= −116.2 × 106 Pa
Aa
2000 × 10−6
(a)
σa = −
(b)
δ a = Laα a (ΔT ) −
σ a = −116.2 MPa 
PLa
Ea Aa
= (0.300)(23 × 10−6 )(120) −
(232.39 × 103 )(0.300)
= 363 × 10−6 m
9
−6
(75 × 10 )(2000 × 10 )
δ a = 0.363 mm 
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PROBLEM 2.61
A 600-lb tensile load is applied to a test
coupon made from 161 -in. flat steel plate
(E = 29 × 106 psi and v = 0.30). Determine
the resulting change (a) in the 2-in. gage
length, (b) in the width of portion AB of the
test coupon, (c) in the thickness of
portion AB, (d) in the cross-sectional
area of portion AB.
SOLUTION
 1  1 
A =    = 0.03125 in 2
 2  16 
600
P
σ= =
= 19.2 × 103 psi
A 0.03125
σ 19.2 × 103
εx = =
= 662.07 × 10−6
E
29 × 106
(a)
δ x = L0ε x = (2.0)(662.07 × 10−6 )
δ l = 1.324 × 10−3 in. 
ε y = ε z = −vε x = −(0.30)(662.07 × 10−6 ) = −198.62 × 10−6
1
(b)
δ width = w0ε y =   ( −198.62 × 10−6 )
2
(c)
δ thickness = t0ε z =   (−198.62 × 10−6 )
 16 
(d)
 1 
δ w = −99.3 × 10−6 in. 
δ t = −12.41 × 10−6 in. 
A = wt = w0 (1 + ε y )t0 (1 + ε z )
= w0t0 (1 + ε y + ε z + ε y ε z )
ΔA = A − A0 = w0t0 (ε y + ε z + ε y ε z )
 1  1 
=    ( −198.62 × 10−6 − 198.62 × 10−6 + negligible term)
 2  16 
= −12.41 × 10−6 in 2
ΔA = −12.41 × 10−6 in 2 
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PROBLEM 2.62
In a standard tensile test, a steel rod of 22-mm diameter is subjected
to a tension force of 75 kN. Knowing that v = 0.3 and
E = 200 GPa, determine (a) the elongation of the rod in a 200-mm
gage length, (b) the change in diameter of the rod.
SOLUTION
P = 75 kN = 75 × 103 N
A=
σ =
εx =
π
4
d2 =
π
4
(0.022)2 = 380.13 × 10−6 m 2
P
75 × 103
=
= 197.301 × 106 Pa
A 380.13 × 10−6
σ
E
=
197.301 × 106
= 986.51 × 10−6
200 × 109
δ x = Lε x = (200 mm)(986.51 × 10−6 )
(a)
δ x = 0.1973 mm 

ε y = −vε x = −(0.3)(986.51 × 10−6 ) = −295.95 × 10−6 

δ y = d ε y = (22 mm)(−295.95 × 10−6 ) 

(b) δ y = −0.00651 mm 
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PROBLEM 2.63
A 20-mm-diameter rod made of an experimental plastic is subjected to a tensile force of magnitude P = 6 kN.
Knowing that an elongation of 14 mm and a decrease in diameter of 0.85 mm are observed in a 150-mm length,
determine the modulus of elasticity, the modulus of rigidity, and Poisson’s ratio for the material.
SOLUTION
Let the y-axis be along the length of the rod and the x-axis be transverse.
π
(20)2 = 314.16 mm 2 = 314.16 × 10−6 m 2
4
P
6 × 103
σy = =
= 19.0985 × 106 Pa
−6
A 314.16 × 10
δ y 14 mm
εy =
=
= 0.093333
L 150 mm
A=
Modulus of elasticity: E =
εx =
σy
εy
δx
d
Poisson’s ratio:
v=−
Modulus of rigidity:
G=
=
19.0985 × 106
= 204.63 × 106 Pa
0.093333
=−
P = 6 × 103 N
E = 205 MPa 
0.85
= −0.0425
20
εx
−0.0425
=−
0.093333
εy
E
204.63 × 106
=
= 70.31 × 106 Pa
2(1 + v)
(2)(1.455)
v = 0.455 
G = 70.3 MPa 
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PROBLEM 2.64
The change in diameter of a large steel bolt is carefully measured as the
nut is tightened. Knowing that E = 29 × 106 psi and v = 0.30,
determine the internal force in the bolt, if the diameter is observed to
decrease by 0.5 × 10−3 in.
SOLUTION
δ y = −0.5 × 10−3 in.
εy =
εy
d
v=−
=−
d = 2.5 in.
0.5 × 10−3
= −0.2 × 10−3
2.5
εy
:
εx
εx =
−ε y
v
=
0.2 × 10−3
= 0.66667 × 10−3
0.3
σ x = Eε x = (29 × 106 )(0.66667 × 10−3 ) = 19.3334 × 103 psi
A=
π
4
d2 =
π
4
(2.5) 2 = 4.9087 in 2
F = σ x A = (19.3334 × 103 )(4.9087) = 94.902 × 103 lb
F = 94.9 kips 
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PROBLEM 2.65
A 2.5-m length of a steel pipe of 300-mm outer diameter and 15-mm
wall thickness is used as a column to carry a 700-kN centric axial load.
Knowing that E = 200 GPa and v = 0.30, determine (a) the change in
length of the pipe, (b) the change in its outer diameter, (c) the change in
its wall thickness.
SOLUTION
d o = 0.3 m
t = 0.015 m
L = 2.5 m
di = do − 2t = 0.3 − 2(0.015) = 0.27 m
A=
(a)
δ =−
π
4
( do2 − di2 ) =
δ
L
4
(0.32 − 0.272 ) = 13.4303 × 10−3 m 2
PL
(700 × 103 )(2.5)
=−
EA
(200 × 109 )(13.4303 × 10−3 )
= −651.51 × 10−6 m
ε=
π
P = 700 × 103 N
=
δ = −0.652 mm 
−651.51 × 10−6
= −260.60 × 10−6
2.5
ε LAT = −vε = −(0.30)(−260.60 × 10−6 )
= 78.180 × 10−6
(b)
Δdo = do ε LAT = (300 mm)(78.180 × 10−6 )
Δd o = 0.0235 mm 
(c)

Δt = tε LAT = (15 mm)(78.180 × 10−6 )
Δt = 0.001173 mm 
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PROBLEM 2.66
An aluminum plate ( E = 74 GPa and v = 0.33) is subjected to a
centric axial load that causes a normal stress σ. Knowing that, before
loading, a line of slope 2:1 is scribed on the plate, determine the slope
of the line when σ = 125 MPa.
SOLUTION
The slope after deformation is
tan θ =
2(1 + ε y )
1+ εx
125 × 106
= 1.6892 × 10−3
E
74 × 109
ε y = −vε x = −(0.33)(1.6892 × 10−3 ) = −0.5574 × 10−3
εx =
tan θ =
σx
=
2(1 − 0.0005574)
= 1.99551
1 + 0.0016892
tan θ = 1.99551 
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PROBLEM 2.67
The block shown is made of a magnesium alloy, for which
E = 45 GPa and v = 0.35. Knowing that σ x = −180 MPa,
determine (a) the magnitude of σ y for which the change in the
height of the block will be zero, (b) the corresponding change in
the area of the face ABCD, (c) the corresponding change in the
volume of the block.
SOLUTION
(a)
δy = 0
εy = 0
σz = 0
1
(σ x − vσ y − vσ z )
E
σ y = vσ x = (0.35)(−180 × 106 )
εy =
= −63 × 106 Pa
σ y = −63 MPa 
1
v
(0.35)( −243 × 106 )
(σ z − vσ x − vσ y ) = − (σ x + σ y ) =
= −1.89 × 10−3
E
E
45 × 109
σ x − vσ y
1
157.95 × 106
ε x = (σ x − vσ y − vσ Z ) =
=−
= −3.51 × 10−3
E
E
45 × 109
εz =
(b)
A0 = Lx Lz
A = Lx (1 + ε x ) Lz (1 + ε z ) = Lx Lz (1 + ε x + ε z + ε xε z )
Δ A = A − A0 = Lx Lz (ε x + ε z + ε x ε z ) ≈ Lx Lz (ε x + ε z )
Δ A = (100 mm)(25 mm)(−3.51 × 10−3 − 1.89 × 10−3 )
(c)
Δ A = −13.50 mm 2 
V0 = Lx Ly Lz
V = Lx (1 + ε x ) Ly (1 + ε y ) Lz (1 + ε z )
= Lx Ly Lz (1 + ε x + ε y + ε z + ε xε y + ε y ε z + ε z ε x + ε x ε y ε z )
ΔV = V − V0 = Lx Ly Lz (ε x + ε y + ε z + small terms)
ΔV = (100)(40)(25)(−3.51 × 10−3 + 0 − 1.89 × 10−3 )
ΔV = −540 mm3 
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PROBLEM 2.68
A 30-mm square was scribed on the side of a large steel pressure
vessel. After pressurization, the biaxial stress condition at the square
is as shown. For E = 200 GPa and v = 0.30, determine the change
in length of (a) side AB, (b) side BC, (c) diagnonal AC.
SOLUTION
Given:
σ x = 80 MPa
σ y = 40 MPa
Using Eq’s (2.28):
εx =
1
80 − 0.3(40)
(σ x − vσ y ) =
= 340 × 10−6
200 × 103
E
εy =
1
40 − 0.3(80)
(σ y − vσ x ) =
= 80 × 10−6
200 × 103
E
(a) Change in length of AB.
δ AB = ( AB)ε x = (30 mm)(340 × 10−6 ) = 10.20 × 10−3 mm
δ AB = 10.20 μ m 
(b) Change in length of BC.
δ BC = ( BC )ε y = (30 mm)(80 × 10−6 ) = 2.40 × 10−3 mm
δ BC = 2.40 μ m 
(c) Change in length of diagonal AC.
From geometry,
Differentiate:
But
Thus,
( AC )2 = ( AB)2 + ( BC ) 2
2( AC ) Δ( AC ) = 2( AB)Δ( AB) + 2( BC )Δ( BC )
Δ( AC ) = δ AC
Δ( AB) = δ AB
Δ( BC ) = δ BC
2( AC )δ AC = 2( AB)δ AB + 2( BC )δ BC
δ AC =
AB
BC
1
1
(10.20 μ m)+
(2.40 μ m)
δ AB +
δ BC =
AC
AC
2
2
δ AC = 8.91 μ m 
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PROBLEM 2.69
The aluminum rod AD is fitted with a jacket that is used to apply a
hydrostatic pressure of 6000 psi to the 12-in. portion BC of the rod.
Knowing that E = 10.1 × 106 psi and v = 0.36, determine (a) the change in
the total length AD, (b) the change in diameter at the middle of the rod.
SOLUTION
σ x = σ z = − P = −6000 psi
σy = 0
1
(σ x − vσ y − vσ z )
E
1
=
[−6000 − (0.36)(0) − (0.36)( −6000)]
10.1 × 106
= −380.198 × 10−6
1
ε y = (−vσ x + σ y − vσ z )
E
1
[−(0.36)(−6000) + 0 − (0.36)( −6000)]
=
10.1 × 106
= 427.72 × 10−6
εx =
Length subjected to strain ε x: L = 12 in.
(a)
δ y = Lε y = (12)(427.72 × 10−6 )
(b)
δ x = d ε x = (1.5)(−380.198 × 10−6 )
δ l = 5.13 × 10−3 in. 
δ d = −0.570 × 10−3 in. 
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PROBLEM 2.70
For the rod of Prob. 2.69, determine the forces that should be applied to
the ends A and D of the rod (a) if the axial strain in portion BC of the rod
is to remain zero as the hydrostatic pressure is applied, (b) if the total
length AD of the rod is to remain unchanged.
PROBLEM 2.69 The aluminum rod AD is fitted with a jacket that is
used to apply a hydrostatic pressure of 6000 psi to the 12-in. portion BC
of the rod. Knowing that E = 10.1 × 106 psi and v = 0.36, determine
(a) the change in the total length AD, (b) the change in diameter at the
middle of the rod.
SOLUTION
Over the pressurized portion BC,
σx =σz = −p
σy =σy
1
(−vσ x + σ y − vσ z )
E
1
= (2vp + σ y )
E
(ε y ) BC =
(a)
(ε y ) BC = 0
2vp + σ y = 0
σ y = −2vp = −(2)(0.36)(6000)
= −4320 psi
A=
π
d2 =
π
(1.5) 2 = 1.76715 in 2
4
4
F = Aσ y = (1.76715)(−4320) = −7630 lb
i.e.,
(b)
Over unpressurized portions AB and CD,
(ε y ) AB
7630 lb compression 
σx =σz = 0
σy
= (ε y )CD =
E
For no change in length,
δ = LAB (ε y ) AB + LBC (ε y ) BC + LCD (ε y )CD = 0
( LAB + LCD )(ε y ) AB + LBC (ε y ) BC = 0
σy
12
(2vp + σ y ) = 0
E
E
24vp
(24)(0.36)(6000)
σy = −
=−
= −2592 psi
20
20
P = Aσ y = (1.76715)(−2592) = −4580 lb
P = 4580 lb compression 
(20 − 12)
+
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PROBLEM 2.71
In many situations, physical constraints prevent
strain from occurring in a given direction.
For example, ε z = 0 in the case shown, where
longitudinal movement of the long prism is
prevented at every point. Plane sections
perpendicular to the longitudinal axis remain
plane and the same distance apart. Show that for
this situation, which is known as plane strain,
we can express σ z , ε x , and ε y as follows:
σ z = v(σ x + σ y )
1
[(1 − v 2 )σ x − v(1 + v)σ y ]
E
1
ε y = [(1 − v 2 )σ y − v(1 + v)σ x ]
E
εx =
SOLUTION
εz = 0 =
1
(−vσ x − vσ y + σ z ) or σ z = v(σ x + σ y )
E
1
(σ x − vσ y − vσ z )
E
1
= [σ x − vσ y − v 2 (σ x + σ y )]
E
1
= [(1 − v 2 )σ x − v(1 + v)σ y ]
E

εx =
1
( −vσ x + σ y − vσ z )
E
1
= [−vσ x + σ y − v 2 (σ x + σ y )]
E
1
= [(1 − v 2 )σ y − v(1 + v)σ x ]
E

εy =

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PROBLEM 2.72
In many situations, it is known that the normal stress in a given direction is zero,
for example, σ z = 0 in the case of the thin plate shown. For this case, which is
known as plane stress, show that if the strains εx and εy have been determined
experimentally, we can express σ x , σ y , and ε z as follows:
σx = E
ε x + vε y
1− v
2
σy = E
ε y + vε x
1− v
2
εz = −
v
(ε x + ε y )
1− v
SOLUTION
σz = 0
εx =
1
(σ x − vσ y )
E
(1)
εy =
1
( −vσ x + σ y )
E
(2)
Multiplying (2) by v and adding to (1),
ε x + vε y =
1 − v2
σx
E
or
σx =
E
(ε x + vε y )
1 − v2

or
σy =
E
(ε y + vε x )
1 − v2

Multiplying (1) by v and adding to (2),
ε y + vε x =
1 − v2
σy
E
1
v
E
(−vσ x − vσ y ) = −
⋅
(ε x + vε y + ε y + vε x )
E
E 1 − v2
v(1 + v)
v
=−
(ε x + ε y ) = −
(ε x + ε y )
2
1− v
1− v
εz =

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PROBLEM 2.73
For a member under axial loading, express the normal strain ε′ in a
direction forming an angle of 45° with the axis of the load in terms of the
axial strain εx by (a) comparing the hypotenuses of the triangles shown in
Fig. 2.49, which represent, respectively, an element before and after
deformation, (b) using the values of the corresponding stresses of σ ′ and
σx shown in Fig. 1.38, and the generalized Hooke’s law.
SOLUTION
Figure 2.49
(a)
[ 2(1 + ε ′)]2 = (1 + ε x ) 2 + (1 − vε x ) 2
2(1 + 2ε ′ + ε ′2 ) = 1 + 2ε x + ε x2 + 1 − 2vε x + v 2ε x2
4ε ′ + 2ε ′2 = 2ε x + ε x2 − 2vε x + v 2ε x2
4ε ′ = 2ε x − 2vε x
Neglect squares as small
(A)
ε′ =
1− v
εx 
2
(B)
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PROBLEM 2.73 (Continued)
(b)
vσ ′
E
E
1− v P
=
⋅
E 2A
1− v
σx
=
2E
ε′ =
=
σ′
−
1− v
εx
2

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PROBLEM 2.74
The homogeneous plate ABCD is subjected to a biaxial loading as
shown. It is known that σ z = σ 0 and that the change in length of
the plate in the x direction must be zero, that is, ε x = 0. Denoting
by E the modulus of elasticity and by v Poisson’s ratio, determine
(a) the required magnitude of σ x , (b) the ratio σ 0 / ε z .
SOLUTION
σ z = σ 0 , σ y = 0, ε x = 0
εx =
1
1
(σ x − vσ y − vσ z ) = (σ x − vσ 0 )
E
E
σ x = vσ 0 
(a)
(b)
εz =
1
1
1 − v2
(−vσ x − vσ y + σ z ) = (−v 2σ 0 − 0 + σ 0 ) =
σ0
E
E
E
σ0
E
=

ε z 1 − v2
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PROBLEM 2.75
A vibration isolation unit consists of two blocks of hard rubber
bonded to a plate AB and to rigid supports as shown. Knowing
that a force of magnitude P = 25 kN causes a deflection
δ = 1.5 mm of plate AB, determine the modulus of rigidity of the
rubber used.
SOLUTION
F =
1
1
P = (25 × 103 N) = 12.5 × 103 N
2
2
τ =
F
12.5 × 103 N)
=
= 833.33 × 103 Pa
A (0.15 m)(0.1 m)
δ = 1.5 × 10−3 m
h = 0.03 m
1.5 × 10−3
= 0.05
h
0.03
τ
833.33 × 103
G = =
= 16.67 × 106 Pa
γ
0.05
γ=
δ
=
G = 16.67 MPa 
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PROBLEM 2.76
A vibration isolation unit consists of two blocks of hard rubber with a
modulus of rigidity G = 19 MPa bonded to a plate AB and to rigid
supports as shown. Denoting by P the magnitude of the force applied
to the plate and by δ the corresponding deflection, determine the
effective spring constant, k = P/δ , of the system.
SOLUTION
δ
h
Shearing strain:
γ =
Shearing stress:
τ = Gγ =
Force:
Gδ
h
1
GAδ
P = Aτ =
2
h
P
P=
k =
with
A = (0.15)(0.1) = 0.015 m 2
k =
2GAδ
h
2GA
h
Effective spring constant:
δ
=
or
h = 0.03 m
2(19 × 106 Pa)(0.015 m 2 )
= 19.00 × 106 N/m
0.03 m
k = 19.00 × 103 kN/m 
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PROBLEM 2.77
The plastic block shown is bonded to a fixed base and to a horizontal rigid
plate to which a force P is applied. Knowing that for the plastic used
G = 55 ksi, determine the deflection of the plate when P = 9 kips.
SOLUTION
Consider the plastic block. The shearing force carried is P = 9 × 103 lb
The area is A = (3.5)(5.5) = 19.25 in 2
Shearing stress:
τ =
Shearing strain:
γ =
But
γ =
P 9 × 103
=
= 467.52 psi
A
19.25
τ
G
=
467.52
= 0.0085006
55 × 103
δ
∴ δ = hγ = (2.2)(0.0085006)
h
δ = 0.0187 in. 
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PROBLEM 2.78
A vibration isolation unit consists of two blocks of hard rubber bonded to
plate AB and to rigid supports as shown. For the type and grade of rubber
used τ all = 220 psi and G = 1800 psi. Knowing that a centric vertical force
of magnitude P = 3.2 kips must cause a 0.1-in. vertical deflection of
the plate AB, determine the smallest allowable dimensions a and b of the
block.
SOLUTION
Consider the rubber block on the right. It carries a shearing force equal to
1
2
1
2
P.
P
The shearing stress is
τ=
or required area
A=
But
A = (3.0)b
Hence,
b=
Use
b = 2.42 in
Shearing strain.
γ=
But
γ=
δ
a
Hence,
a=
δ
0.1
=
= 0.818 in.
γ 0.12222
A
P 3.2 × 103
=
= 7.2727 in 2
2τ (2)(220)
A
= 2.42 in.
3.0
τ
G
=
and
bmin = 2.42 in. 
τ = 220 psi
220
= 0.12222
1800
amin = 0.818 in. 
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PROBLEM 2.79
The plastic block shown is bonded to a rigid support and to a vertical plate to
which a 55-kip load P is applied. Knowing that for the plastic used G = 150 ksi,
determine the deflection of the plate.

SOLUTION
A = (3.2)(4.8) = 15.36 in 2
P = 55 × 103 lb
τ =
P 55 × 103
=
= 3580.7 psi
A
15.36
G = 150 × 103 psi
γ =

τ
=
G
h = 2 in.
3580.7
= 23.871 × 10−3
150 × 103
δ = hγ = (2)(23.871 × 10−3 )
= 47.7 × 10−3 in.
δ = 0.0477 in. ↓ 
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PROBLEM 2.80
What load P should be applied to the plate of Prob. 2.79 to produce a
1
16
-in. deflection?
PROBLEM 2.79 The plastic block shown is bonded to a rigid support and to a vertical plate to which a
55-kip load P is applied. Knowing that for the plastic used G = 150 ksi, determine the deflection of the plate.
SOLUTION
1
in. = 0.0625 in.
16
h = 2 in.
δ =
γ =
δ
h
=
0.0625
= 0.03125
2
G = 150 × 103 psi
τ = Gγ = (150 × 103 )(0.03125)
= 4687.5 psi
A = (3.2)(4.8) = 15.36 in 2
P = τ A = (4687.5)(15.36)
= 72 × 103 lb
72 kips 
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PROBLEM 2.81
Two blocks of rubber with a modulus of rigidity G = 12 MPa are
bonded to rigid supports and to a plate AB. Knowing that c = 100 mm
and P = 45 kN, determine the smallest allowable dimensions a and b of
the blocks if the shearing stress in the rubber is not to exceed 1.4 MPa
and the deflection of the plate is to be at least 5 mm.
SOLUTION
Shearing strain:
γ=
a=
Shearing stress:
τ=
b=
δ
=
a
Gδ
τ
1
2
P
A
τ
G
=
(12 × 106 Pa)(0.005 m)
= 0.0429 m
1.4 × 106 Pa
=
P
2bc
P
45 × 103 N
=
= 0.1607 m
2cτ 2(0.1 m) (1.4 × 106 Pa)
a = 42.9 mm 
b = 160.7 mm 
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PROBLEM 2.82
Two blocks of rubber with a modulus of rigidity G = 10 MPa are bonded
to rigid supports and to a plate AB. Knowing that b = 200 mm and
c = 125 mm, determine the largest allowable load P and the smallest
allowable thickness a of the blocks if the shearing stress in the rubber
is not to exceed 1.5 MPa and the deflection of the plate is to be at least
6 mm.
SOLUTION
Shearing stress:
τ=
1
2
P
A
=
P
2bc
P = 2bcτ = 2(0.2 m)(0.125 m)(1.5 × 103 kPa)
Shearing strain:
γ=
a=
δ
a
=
Gδ
τ
P = 75.0 kN 
τ
G
=
(10 × 106 Pa)(0.006 m)
= 0.04 m
1.5 × 106 Pa
a = 40.0 mm 
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PROBLEM 2.83∗
Determine the dilatation e and the change in volume of the 200-mm
length of the rod shown if (a) the rod is made of steel with E = 200 GPa
and v = 0.30, (b) the rod is made of aluminum with E = 70 GPa and
v = 0.35.
SOLUTION
π
π
d 2 = (22) 2 = 380.13 mm 2 = 380.13 × 10−6 m 2
4
4
3
P = 46 × 10 N
P
σ x = = 121.01 × 106 Pa
A
σy =σz = 0
A=
εx =
σ
1
(σ x − vσ y − vσ z ) = x
E
E
ε y = ε z = −vε x = −v
σx
E
(1 − 2v)σ x
1
e = ε x + ε y + ε z = (σ x − vσ x − vσ x ) =
E
E
Volume:
(a)
V = AL = (380.13 mm 2 )(200 mm) = 76.026 × 103 mm3
ΔV = Ve
Steel:
e=
(1 − 0.60)(121.01 × 106 )
= 242 × 10−6
9
200 × 10
ΔV = (76.026 × 103 )(242 × 10−6 ) = 18.40 mm3
(b)
Aluminum:
e=
(1 − 0.70)(121.01 × 106 )
= 519 × 10−6
9
70 × 10
ΔV = (76.026 × 103 )(519 × 10−6 ) = 39.4 mm3
e = 242 × 10−6 
ΔV = 18.40 mm3 
e = 519 × 10−6 
ΔV = 39.4 mm3 
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PROBLEM 2.84∗
Determine the change in volume of
the 2-in. gage length segment AB in
Prob. 2.61 (a) by computing the
dilatation of the material, (b) by
subtracting the original volume of
portion AB from its final volume.
SOLUTION
From Problem 2.61, thickness =
(a)
1
in.,
16
E = 29 × 106 psi, v = 0.30.
 1  1 
A =    = 0.03125 in 2
 2  16 
Volume: V0 = AL 0 = (0.03125)(2.00) = 0.0625 in 3
P
600
=
= 19.2 × 103 psi
σy =σz = 0
A 0.03125
σ
1
19.2 × 103
= 662.07 × 10−6
ε x = (σ x − vσ y − vσ z ) = x =
E
E
29 × 106
σx =
ε y = ε z = −v ε x = −(0.30)(662.07 × 10−6 ) = −198.62 × 10−6
e = ε x + ε y + ε z = 264.83 × 10−6
ΔV = V0 e = (0.0625) (264.83 × 10−6 ) = 16.55 × 10−6 in 3
(b)

From the solution to Problem 2.61,
δ x = 1.324 × 10−3 in.,
δ y = −99.3 × 10−6 in.,
δ z = −12.41 × 10−6 in.
The dimensions when under a 600-lb tensile load are:
Length:
L = L0 + δ x = 2 + 1.324 × 10−3 = 2.001324 in.
Width:
w = w0 + δ y =
Thickness:
Volume:
1
− 99.3 × 10−6 = 0.4999007 in.
2
1
t = t0 + δ z = − 12.41 × 10−6 = 0.06248759 in.
16
V = Lwt = 0.062516539 in 3
ΔV = V − V0 = 0.062516539 − 0.0625 = 16.54 × 10−6 in 3

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PROBLEM 2.85∗
A 6-in.-diameter solid steel sphere is lowered into the ocean to a point where the pressure is 7.1 ksi (about
3 miles below the surface). Knowing that E = 29 × 106 psi and v = 0.30, determine (a) the decrease in
diameter of the sphere, (b) the decrease in volume of the sphere, (c) the percent increase in the density of the
sphere.
SOLUTION
For a solid sphere,
V0 =
π
6
d 03
π
(6.00)3
6
= 113.097 in.3
=
σ x = σ y = σ z = −p
= −7.1 × 103 psi
1
(σ x − vσ y − vσ z )
E
(1 − 2v) p
(0.4)(7.1 × 103 )
=−
=−
E
29 × 106
εx =
= −97.93 × 10−6
Likewise,
ε y = ε z = −97.93 × 10−6
e = ε x + ε y + ε z = −293.79 × 10−6
(a)
−Δd = −d 0ε x = −(6.00)(−97.93 × 10−6 ) = 588 × 10−6 in.
(b)
−ΔV = −V0 e = −(113.097)(−293.79 × 10−6 ) = 33.2 × 10−3 in 3
(c)
Let m = mass of sphere.
−Δd = 588 × 10−6 in. 
− ΔV = 33.2 × 10−3 in 3 
m = constant.
m = ρ0V0 = ρV = ρV0 (1 + e)
ρ − ρ0 ρ
V
m
1
=
−1 =
× 0 −1 =
−1
V0 (1 + e) m
1+ e
ρ0
ρ0
= (1 − e + e 2 − e3 + ) − 1 = −e + e 2 − e3 + 
≈ −e = 293.79 × 10−6
ρ − ρ0
× 100% = (293.79 × 10−6 )(100%)
ρ0
0.0294% 
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PROBLEM 2.86∗
(a) For the axial loading shown, determine the change in height and the
change in volume of the brass cylinder shown. (b) Solve part a,
assuming that the loading is hydrostatic with σ x = σ y = σ z = −70 MPa.
SOLUTION
h0 = 135 mm = 0.135 m
π
π
(85) 2 = 5.6745 × 103 mm 2 = 5.6745 × 10−3 m 2
4
4
V0 = A0 h0 = 766.06 × 103 mm3 = 766.06 × 10−6 m3
A 0=
(a)
d 02 =
σ x = 0, σ y = −58 × 106 Pa, σ z = 0
σy
1
58 × 106
(−vσ x + σ y − vσ z ) =
=−
= −552.38 × 10−6
E
E
105 × 109
εy =
Δh = h 0ε y = (135 mm)( − 552.38 × 10−6 )
e=
(1 − 2v)σ y (0.34)(−58 × 106 )
1 − 2v
(σ x + σ y + σ z ) =
=
= −187.81 × 10−6
9
E
E
105 × 10
ΔV = V0 e = (766.06 × 103 mm3 )(−187.81 × 10−6 )
(b)
Δh = −0.0746 mm 
σ x = σ y = σ z = −70 × 106 Pa
εy =
σ x + σ y + σ z = −210 × 106 Pa
1
1 − 2v
(0.34)(−70 × 106 )
(−vσ x + σ y − vσ z ) =
= −226.67 × 10−6
σy =
E
E
105 × 109
Δh = h 0ε y = (135 mm)( −226.67 × 10−6 )
e=
ΔV = −143.9 mm3 
Δh = −0.0306 mm 
1 − 2v
(0.34)( −210 × 106 )
(σ x + σ y + σ z ) =
= −680 × 10−6
9
E
105 × 10
ΔV = V0 e = (766.06 × 103 mm3 )(−680 × 10−6 )
ΔV = −521 mm3 
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PROBLEM 2.87∗
A vibration isolation support consists of a rod A of radius R1 = 10 mm and
a tube B of inner radius R 2 = 25 mm bonded to an 80-mm-long hollow
rubber cylinder with a modulus of rigidity G = 12 MPa. Determine the
largest allowable force P that can be applied to rod A if its deflection is not
to exceed 2.50 mm.
SOLUTION
Let r be a radial coordinate. Over the hollow rubber cylinder, R1 ≤ r ≤ R2 .
Shearing stress τ acting on a cylindrical surface of radius r is
τ=
P
P
=
A 2π rh
The shearing strain is
γ=
τ
=
G
P
2π Ghr
Shearing deformation over radial length dr,
dδ
=γ
dr
d δ = γ dr =
P dr
2π Gh r
Total deformation.
δ=

R2
R1
dδ =
P
2π Gh
R2
P
ln r
R1
2π Gh
R
P
ln 2
=
R1
2π Gh
=
Data:
R1 = 10 mm = 0.010 m,
R1
R2 = 25 mm = 0.025 m, h = 80 mm = 0.080 m
G = 12 × 106 Pa
P=
dr
r
P
(ln R2 − ln R1 )
=
2π Gh
2π Ghδ
or P =
ln( R2 / R1 )

R2
δ = 2.50 × 10−3 m
(2π )(12 × 106 ) (0.080) (2.50 × 10−3 )
= 16.46 × 103 N
ln (0.025/0.010)
16.46 kN 
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PROBLEM 2.88
A vibration isolation support consists of a rod A of radius R1 and a tube B of
inner radius R2 bonded to a 80-mm-long hollow rubber cylinder with a
modulus of rigidity G = 10.93 MPa. Determine the required value of the ratio
R2/R1 if a 10-kN force P is to cause a 2-mm deflection of rod A.
SOLUTION
Let r be a radial coordinate. Over the hollow rubber cylinder, R1 ≤ r ≤ R2 .
Shearing stress τ acting on a cylindrical surface of radius r is
τ=
P
P
=
A 2π rh
The shearing strain is
γ=
τ
G
=
P
2π Ghr
Shearing deformation over radial length dr,
dδ
=γ
dr
d δ = γ dr
P dr
drδ =
2π Gh r
Total deformation.
δ=

R2
R1
dδ =
P
2π Gh
R2
P
ln r
R1
2π Gh
R
P
=
ln 2
R1
2π Gh
=
ln
dr
R1 r
P
(ln R2 − ln R1 )
=
2π Gh

R2
R2 2π Ghδ (2π ) (10.93 × 106 ) (0.080) (0.002)
=
=
= 1.0988
R1
P
10.103
R2
= exp (1.0988) = 3.00
R1
R2 /R1 = 3.00 
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without permission.
PROBLEM 2.89∗
The material constants E, G, k, and v are related by Eqs. (2.33) and (2.43). Show that any one of these
constants may be expressed in terms of any other two constants. For example, show that
(a) k = GE/(9G − 3E) and (b) v = (3k – 2G)/(6k + 2G).
SOLUTION
k=
(a)
1+ v =
and
G=
E
2(1 + v)
E
E
or v =
−1
2G
2G
k=
(b)
E
3(1 − 2v)
E
2 EG
2 EG
=
=

 E
  3[2G − 2 E + 4G ] 18G − 6 E
− 1 
3 1 − 2 
 2G  

k=
EG

9G − 6 E
v=
3k − 2G

6k + 2G
k
2(1 + v)
=
G 3(1 − 2v)
3k − 6kv = 2G + 2Gv
3k − 2G = 2G + 6k
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PROBLEM 2.90∗
Show that for any given material, the ratio G/E of the modulus of rigidity over the modulus of elasticity is
always less than 12 but more than 13 . [Hint: Refer to Eq. (2.43) and to Sec. 2.13.]
SOLUTION
G=
E
2(1 + v)
Assume v > 0 for almost all materials, and v <
or
1
2
E
= 2(1 + v)
G
for a positive bulk modulus.
E
 1
< 2 1 +  = 3
G
2

Applying the bounds,
2≤
Taking the reciprocals,
1
G 1
>
>
2
E 3
or
1 G 1
< < 
3
E 2
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PROBLEM 2.91∗
A composite cube with 40-mm sides and the properties shown is
made with glass polymer fibers aligned in the x direction. The cube
is constrained against deformations in the y and z directions and is
subjected to a tensile load of 65 kN in the x direction. Determine
(a) the change in the length of the cube in the x direction, (b) the
stresses σ x , σ y , and σ z .
Ex = 50 GPa
vxz = 0.254
E y = 15.2 GPa vxy = 0.254
Ez = 15.2 GPa vzy = 0.428
SOLUTION
Stress-to-strain equations are
εx =
σx
Ex
εy = −
εz = −
−
v yxσ y
Ey
vxyσ x
Ex
+
−
σy
Ey
vzxσ z
Ez
−
vzyσ z
Ez
vxzσ x v yzσ y σ z
−
+
Ex
Ey
Ez
vxy
Ex
v yz
Ey
=
=
v yx
Ey
vzy
Ez
vzx vxz
=
Ez Ex
(1)
(2)
(3)
(4)
(5)
(6)
ε y = 0 and ε z = 0.
The constraint conditions are
Using (2) and (3) with the constraint conditions gives
vzy
vxy
1
σy −
σz =
σx
Ey
Ez
Ex
−
v yz
Ey
σy +
V
1
σ z = xz σ x
Ez
Ex
(7)
(8)
1
0.428
0.254
σy −
σz =
σ x or σ y − 0.428σ z = 0.077216 σ x
15.2
15.2
50
0.428
1
0.254
−
σy +
σz =
σ x or − 0.428 σ y + σ z = 0.077216 σ x
15.2
15.2
50
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PROBLEM 2.91∗ (Continued)
Solving simultaneously,
σ y = σ z = 0.134993 σ x
Using (4) and (5) in (1),
εx =
Ex =
=
vxy
v
1
σx −
σ y − xz σ z
Ex
Ex
E
1
[1 − (0.254) (0.134993) − (0.254)(0.134993)]σ x
Ex
0.93142 σ x
Ex
A = (40)(40) = 1600 mm 2 = 1600 × 10−6 m 2
P
65 × 103
=
= 40.625 × 106 Pa
A 1600 × 10−6
(0.93142) (40.625 × 103 )
εx =
= 756.78 × 10−6
9
50 × 10
σx =
(a)
δ x = Lx ε x = (40 mm) (756.78 × 10−6 )
δ x = 0.0303 mm 
(b)
σ x = 40.625 × 106 Pa
σ x = 40.6 MPa 

σ y = σ z = (0.134993) (40.625 × 106 ) = 5.48 × 106 Pa 
σ y = σ z = 5.48 MPa 
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PROBLEM 2.92∗
The composite cube of Prob. 2.91 is constrained against
deformation in the z direction and elongated in the x direction by
0.035 mm due to a tensile load in the x direction. Determine (a) the
stresses σ x , σ y , and σ z , (b) the change in the dimension in the
y direction.
Ex = 50 GPa
vxz = 0.254
E y = 15.2 GPa vxy = 0.254
Ez = 15.2 GPa vzy = 0.428
SOLUTION
εx =
σx
Ex
εy = −
εz = −
−
v yxσ y
vxyσ x
Ex
Ey
+
−
σy
Ey
vzxσ z
Ez
−
vzyσ z
Ez
vxzσ x v yzσ y σ z
−
+
Ex
Ey
Ez
vxy
Ex
v yz
Ey
=
=
v yx
Ey
vzy
Ez
vzx vxz
=
Ez Ex
Constraint condition:
Load condition :
(1)
(2)
(3)
(4)
(5)
(6)
εz = 0
σy = 0
0=−
From Equation (3),
σz =
vxz
1
σx + σz
Ex
Ez
vxz Ez
(0.254) (15.2)
σx =
= 0.077216 σ x
Ex
50
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PROBLEM 2.92∗ (Continued)
From Equation (1) with σ y = 0,
εx =
=
1
1
[σ x − 0.254 σ z ] =
[1 − (0.254) (0.077216)]σ x
Ex
Ex
=
0.98039
σx
Ex
σx =
But,
εx =
(a)
σx =
δx
Lx
=
v
v
1
1
σ x − zx σ z = σ x − xz σ z
Ex
Ez
Ex
Ex
Exε x
0.98039
0.035 mm
= 875 × 10−6
40 mm
(50 × 109 ) (875 × 10−6 )
= 44.625 × 103 Pa
0.98039
σ x = 44.6 MPa 
σy = 0
σ z = (0.077216) (44.625 × 106 ) = 3.446 × 106 Pa
From (2),
εy =
vxy
Ex
σx +
σ z = 3.45 MPa 
vzy
1
σy −
σz
Ey
Ez
(0.254) (44.625 × 106 )
(0.428) (3.446 × 106 )
0
+
−
50 × 109
15.2 × 109
−6
= −323.73 × 10
=−
(b)
δ y = Ly ε y = (40 mm) ( − 323.73 × 10−6 )
δ y = −0.0129 mm 
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PROBLEM 2.93
Two holes have been drilled through a long steel bar that is subjected to a centric
axial load as shown. For P = 6.5 kips, determine the maximum value of the
stress (a) at A, (b) at B.
SOLUTION
(a)
At hole A:
r =
1 1 1
⋅ = in.
2 2 4
d =3−
1
= 2.50 in.
2
1
Anet = dt = (2.50)   = 1.25 in 2
2
σ non =
P
6.5
=
= 5.2 ksi
Anet 1.25
2 ( 14 )
2r
=
= 0.1667
D
3
From Fig. 2.60a,
K = 2.56
σ max = Kσ non = (2.56)(5.2)
(b)
r =
At hole B:
1
(1.5) = 0.75,
2
σ max = 13.31 ksi 
d = 3 − 1.5 = 1.5 in.
1
Anet = dt = (1.5)   = 0.75 in 2 ,
2
σ non =
P
6.5
=
= 8.667 ksi
Anet
0.75
2r
2(0.75)
=
= 0.5
D
3
From Fig. 2.60a,
K = 2.16
σ max = Kσ non = (2.16)(8.667)
σ max = 18.72 ksi 
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PROBLEM 2.94
Knowing that σ all = 16 ksi, determine the maximum allowable value of the
centric axial load P.
SOLUTION
At hole A:
r =
1 1 1
⋅ = in.
2 2 4
d =3−
1
= 2.50 in.
2
1
Anet = dt = (2.50)   = 1.25 in 2
2
2 ( 14 )
2r
=
= 0.1667
D
3
From Fig. 2.60a,
K = 2.56
σ max =
At hole B:
r =
KP
Anet
∴ P=
Anetσ max
(1.25)(16)
=
= 7.81 kips
K
2.56
1
(1.5) = 0.75 in,
2
d = 3 − 1.5 = 1.5 in.
1
Anet = dt = (1.5)   = 0.75 in 2 ,
2
2r
2(0.75)
=
= 0.5
D
3
From Fig. 2.60a,
K = 2.16
P=
Smaller value for P controls.
Anetσ max
(0.75)(16)
=
= 5.56 kips
K
2.16
P = 5.56 kips 
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PROBLEM 2.95
Knowing that the hole has a diameter of 9 mm, determine (a) the radius
rf of the fillets for which the same maximum stress occurs at the hole A
and at the fillets, (b) the corresponding maximum allowable load P if
the allowable stress is 100 MPa.
SOLUTION
1
r =   (9) = 4.5 mm
2
For the circular hole,
d = 96 − 9 = 87 mm
2r
2(4.5)
=
= 0.09375
D
96
Anet = dt = (0.087 m)(0.009 m) = 783 × 10−6 m 2
K hole = 2.72
From Fig. 2.60a,
σ max =
P=
(a)
For fillet,
K hole P
Anet
Anetσ max
(783 × 10−6 )(100 × 106 )
=
= 28.787 × 103 N
K hole
2.72
D = 96 mm, d = 60 mm
D 96
=
= 1.60
d
60
Amin = dt = (0.060 m)(0.009 m) = 540 × 10−6 m 2
σ max =
(5.40 × 10−6 )(100 × 106 )
K fillet P
A σ
∴ K fillet = min max =
Amin
P
28.787 × 103
= 1.876
From Fig. 2.60b,
rf
d
≈ 0.19 ∴ rf ≈ 0.19d = 0.19(60)
r f = 11.4 mm 
(b) P = 28.8 kN 
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PROBLEM 2.96
For P = 100 kN, determine the minimum plate thickness t required if
the allowable stress is 125 MPa.
SOLUTION
At the hole:
rA = 20 mm
d A = 88 − 40 = 48 mm
2rA
2(20)
=
= 0.455
DA
88
From Fig. 2.60a,
K = 2.20
σ max =
t =
At the fillet:
From Fig. 2.60b,
KP
KP
KP
=
∴ t =
Anet
d At
d Aσ max
(2.20)(100 × 103 N)
= 36.7 × 10−3 m = 36.7 mm
6
(0.048 m)(125 × 10 Pa)
D
88
=
= 1.375
dB
64
D = 88 mm,
d B = 64 mm
rB = 15 mm
rB
15
=
= 0.2344
dB
64
K = 1.70
σ max =
t =
KP
KP
=
Amin
d Bt
KP
d Bσ max
=
(1.70)(100 × 103 N)
= 21.25 × 10−3 m = 21.25 mm
(0.064 m)(125 × 106 Pa)
The larger value is the required minimum plate thickness.

t = 36.7 mm 
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PROBLEM 2.97
The aluminum test specimen shown is subjected to two equal and opposite
centric axial forces of magnitude P. (a) Knowing that E = 70 GPa
and σ all = 200 MPa, determine the maximum allowable value of P and the
corresponding total elongation of the specimen. (b) Solve part a, assuming that
the specimen has been replaced by an aluminum bar of the same length and a
uniform 60 × 15-mm rectangular cross section.
SOLUTION
σ all = 200 × 106 Pa E = 70 × 109 Pa
Amin = (60 mm) (15 mm) = 900 mm 2 = 900 × 10−6 m 2
(a)
Test specimen.
D = 75 mm, d = 60 mm, r = 6 mm
D 75
=
= 1.25
d 60
From Fig. 2.60b
K = 1.95
P=
r
6
=
= 0.10
d 60
σ max = K
P
A
Aσ max (900 × 10−6 ) (200 × 106 )
=
= 92.308 × 103 N
K
1.95
P = 92.3 kN 
Wide area A* = (75 mm) (15 mm) = 1125 mm 2 = 1.125 × 10−3 m 2
δ =Σ
Pi Li
P L 92.308 × 103  0.150
0.300
0.150 
= Σ i =
+
+

−3
−6
9
Ai Ei E Ai
70 × 10
900 × 10
1.125 × 10−3 
1.125 × 10
= 7.91 × 10−6 m
(b)
Uniform bar.
P = Aσ all = (900 × 10−6 )(200 × 106 ) = 180 × 103 N

δ = 0.791 mm 
δ=
PL
(180 × 103 )(0.600)
=
= 1.714 × 10−3 m 
AE (900 × 10−6 )(70 × 109 )
P = 180.0 kN 
δ = 1.714 mm 
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PROBLEM 2.98
For the test specimen of Prob. 2.97, determine the maximum value of the
normal stress corresponding to a total elongation of 0.75 mm.
PROBLEM 2.97 The aluminum test specimen shown is subjected to two equal
and opposite centric axial forces of magnitude P. (a) Knowing that E = 70 GPa
and σ all = 200 MPa, determine the maximum allowable value of P and the
corresponding total elongation of the specimen. (b) Solve part a, assuming that
the specimen has been replaced by an aluminum bar of the same length and a
uniform 60 × 15-mm rectangular cross section.
SOLUTION
δ =Σ
Pi Li P Li
= Σ
Ei Ai E Ai
δ = 0.75 × 10−3 m
L1 = L3 = 150 mm = 0.150 m,
L2 = 300 mm = 0.300 m
A1 = A3 = (75 mm) (15 mm) = 1125 mm 2 = 1.125 × 10−3 m 2
A2 = (60 mm) (15 mm) = 900 mm2 = 900 × 10−6 m 2
Σ
Li
0.150
0.300
0.150
=
+
+
= 600 m −1
−3
−6
−3
Ai 1.125 × 10
900 × 10
1.125 × 10
P=
Stress concentration.
(70 × 109 ) (0.75 × 10−3 )
Eδ
=
= 87.5 × 103 N
Li
600
Σ
Ai
D = 75 mm, d = 60 mm, r = 6 mm
D 75
=
= 1.25
d 60
From Fig. 2.60b
6
r
=
= 0.10
d 60
K = 1.95
σ max = K
P
(1.95) (87.5 × 103 )
=
= 189.6 × 106 Pa
−6
Amin
900 × 10
σ max = 189.6 MPa 
Note that σ max < σ all . 
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PROBLEM 2.99
A hole is to be drilled in the plate at A. The diameters of the bits
available to drill the hole range from 12 to 11/2 in. in 14 -in.
increments. If the allowable stress in the plate is 21 ksi,
determine (a) the diameter d of the largest bit that can be used if
the allowable load P at the hole is to exceed that at the fillets, (b)
the corresponding allowable load P.
SOLUTION
At the fillets:
D
4.6875
=
= 1.5
d
3.125
From Fig. 2.60b,
K = 2.10
r
0.375
=
= 0.12
d
3.125
Amin = (3.125)(0.5) = 1.5625 in 2
σ max = K
Pall =
Aminσ all
(1.5625)(21)
=
= 15.625 kips
K
2.10
Anet = ( D − 2r )t , K from Fig. 2.60a
At the hole:
σ max = K
with
Hole diam.
0.5 in.
Pall
= σ all
Amin
r
0.25 in.
P
= σ all
Anet
∴
Pall =
D = 4.6875 in.
t = 0.5 in.
d = D − 2r
2r/D
4.1875 in.
0.107
Anetσ all
K
σ all = 21 ksi
K
2.68
Anet
Pall
2.0938 in2
16.41 kips
2
0.75 in.
0.375 in.
3.9375 in.
0.16
2.58
1.96875 in
16.02 kips
1 in.
0.5 in.
3.6875 in.
0.213
2.49
1.84375 in2
15.55 kips
2
1.25 in.
0.625 in.
3.4375 in.
0.267
2.41
1.71875 in
14.98 kips
1.5 in.
0.75 in.
3.1875 in.
0.32
2.34
1.59375 in2
14.30 kips
(a)
Largest hole with Pall > 15.625 kips is the
(b)
Allowable load Pall = 15.63 kips
3
4
-in. diameter hole.


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PROBLEM 2.100
(a) For P = 13 kips and d = 12 in., determine the maximum stress
in the plate shown. (b) Solve part a, assuming that the hole at A is
not drilled.
SOLUTION
Maximum stress at hole:
Use Fig. 2.60a for values of K.
2r
0.5
=
= 0.017,
D
4.6875
K = 2.68
Anet = (0.5)(4.6875 − 0.5) = 2.0938 in 2
σ max = K
P
(2.68)(13)
=
= 16.64 ksi,
Anet
2.0938
Maximum stress at fillets:
Use Fig. 2.60b for values of K.
r
0.375
=
= 0.12
d
3.125
D
4.6875
=
= 1.5
d
3.125
K = 2.10
Amin = (0.5)(3.125) = 1.5625 in 2
σ max = K
(2.10)(13)
P
=
= 17.47 ksi
Amin
1.5625
(a)
With hole and fillets:
17.47 ksi 
(b)
Without hole:
17.47 ksi 
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PROBLEM 2.101
Rod ABC consists of two cylindrical portions AB and BC; it is made of a mild
steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 250 MPa.
A force P is applied to the rod and then removed to give it a permanent set
δ p = 2 mm. Determine the maximum value of the force P and the maximum
amount δ m by which the rod should be stretched to give it the desired permanent
set.
SOLUTION
AAB =
ABC =
π
4
π
(30)2 = 706.86 mm 2 = 706.86 × 10−6 m 2
(40) 2 = 1.25664 × 103 mm 2 = 1.25664 × 10−3 m 2
4
Pmax = Aminσ Y = (706.86 × 10−6 )(250 × 106 ) = 176.715 × 103 N
Pmax = 176.7 kN 
δ′ =
P′LAB P ′LBC
(176.715 × 103 )(0.8)
(176.715 × 103 )(1.2)
+
=
+
EAAB
EABC (200 × 109 )(706.86 × 10−6 ) (200 × 109 )(1.25664 × 10−3 )
= 1.84375 × 10−3 m = 1.84375 mm
δ p = δ m − δ ′ or δ m = δ p + δ ′ = 2 + 1.84375
δ m = 3.84 mm 
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PROBLEM 2.102
Rod ABC consists of two cylindrical portions AB and BC; it is made of a mild
steel that is assumed to be elastoplastic with E = 200 GPa and σ Y = 250 MPa.
A force P is applied to the rod until its end A has moved down by an amount
δ m = 5 mm. Determine the maximum value of the force P and the permanent
set of the rod after the force has been removed.
SOLUTION
AAB =
4
(30)2 = 706.86 mm 2 = 706.86 × 10−6 m 2
π
(40) 2 = 1.25664 × 103 mm 2 = 1.25644 × 10−3 m 2
4
= Aminσ Y = (706.86 × 10−6 )(250 × 106 ) = 176.715 × 103 N
ABC =
Pmax
π
Pmax = 176.7 kN 
δ′ =
P′LAB P ′LBC
(176.715 × 103 )(0.8)
(176.715 × 103 )(1.2)
+
=
+
EAAB
EABC (200 × 109 )(706.68 × 10−6 ) (200 × 109 )(1.25664 × 10−3 )
= 1.84375 × 10−3 m = 1.84375 mm
δ p = δ m − δ ′ = 5 − 1.84375 = 3.16 mm
δ p = 3.16 mm 
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PROBLEM 2.103
The 30-mm square bar AB has a length L = 2.2 m; it is made of a mild steel that is
assumed to be elastoplastic with E = 200 GPa and σ Y = 345 MPa. A force P is applied
to the bar until end A has moved down by an amount δ m . Determine the maximum
value of the force P and the permanent set of the bar after the force has been removed,
knowing that (a) δ m = 4.5 mm, (b) δ m = 8 mm.
SOLUTION
A = (30)(30) = 900 mm2 = 900 × 10−6 m 2
δY = Lε Y =
Lσ Y (2.2)(345 × 106 )
=
= 3.795 × 10−3 = 3.795 mm
E
200 × 109
If δ m ≥ δ Y , Pm = Aσ Y = (900 × 10−6 )(345 × 106 ) = 310.5 × 103 N
Unloading: δ ′ =
Pm L σ Y L
=
= δY = 3.795 mm
AE
E
δ P = δm − δ ′
(a)
δ m = 4.5 mm > δY Pm = 310.5 × 103 N
δ perm = 4.5 mm − 3.795 mm
(b)
δ m = 8 mm > δY Pm = 310.5 × 103 N

δ perm = 8.0 mm − 3.795 mm
δ m = 310.5 kN 
δ perm = 0.705 mm 
δ m = 310.5 kN 
δ perm = 4.205 mm 
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PROBLEM 2.104
The 30-mm square bar AB has a length L = 2.5 m; it is made of mild steel that is assumed
to be elastoplastic with E = 200 GPa and σ Y = 345 MPa. A force P is applied to the bar
and then removed to give it a permanent set δ p . Determine the maximum value of the
force P and the maximum amount δ m by which the bar should be stretched if the desired
value of δ p is (a) 3.5 mm, (b) 6.5 mm.
SOLUTION
A = (30)(30) = 900 mm 2 = 900 × 10−6 m 2
δY = Lε Y =
Lσ Y (2.5)(345 × 106 )
=
= 4.3125 × 103 m = 4.3125 mm
9
E
200 × 10
When δ m exceeds δ Y , thus producing a permanent stretch of δ p , the maximum force is
Pm = Aσ Y = (900 × 10−6 )(345 × 106 ) = 310.5 × 103 N
= 310.5 kN 
δ p = δ m − δ ′ = δ m − δY ∴ δ m = δ p + δY
(a)
δ p = 3.5 mm δ m = 3.5 mm + 4.3125 mm
= 7.81 mm 
(b)
δ p = 6.5 mm δ m = 6.5 mm + 4.3125 mm
= 10.81 mm 
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PROBLEM 2.105
Rod AB is made of a mild steel that is assumed to be elastoplastic
with E = 29 × 106 psi and σ Y = 36 ksi. After the rod has been attached
to the rigid lever CD, it is found that end C is 83 in. too high. A vertical
force Q is then applied at C until this point has moved to position C ′.
Determine the required magnitude of Q and the deflection δ1 if the
lever is to snap back to a horizontal position after Q is removed.
SOLUTION
Since the rod AB is to be stretched permanently, the peak force in the rod is P = PY , where
PY = Aσ Y =
π 3
2
(36) = 3.976 kips
4  8 
Referring to the free body diagram of lever CD,
ΣM D = 0: 33 Q − 22 P = 0
22
(22)(3.976)
Q=
P=
= 2.65 kips
33
33
Q = 2.65 kips 
During unloading, the spring back at B is
δ B = LABε Y =
LABσ Y (60)(36 × 103 )
=
= 0.0745 in.
E
29 × 106
From the deformation diagram,
Slope:
θ=
δB
22
=
δC
33
∴ δC =
33
δ B = 0.1117 in.
−22
δ C = 0.1117 in. 
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PROBLEM 2.106
Solve Prob. 2.105, assuming that the yield point of the mild steel is
50 ksi.
PROBLEM 2.105 Rod AB is made of a mild steel that is assumed to
be elastoplastic with E = 29 × 106 psi and σ Y = 36 ksi. After the rod
has been attached to the rigid lever CD, it is found that end C is 83 in.
too high. A vertical force Q is then applied at C until this point has
moved to position C′. Determine the required magnitude of Q and the
deflection δ1 if the lever is to snap back to a horizontal position after
Q is removed.
SOLUTION
Since the rod AB is to be stretched permanently, the peak force in the rod is P = PY , where
PY = Aσ Y =
π 3
2
(50) = 5.522 kips
4  8 
Referring to the free body diagram of lever CD,
ΣM D = 0: 33Q − 22 P = 0
Q=
22
(22)(5.522)
P=
= 3.68 kips
33
33
Q = 3.68 kips 
During unloading, the spring back at B is
δ B = LAB ε Y =
LABσ Y (60)(50 × 103 )
=
= 0.1034 in.
E
29 × 106
From the deformation diagram,
Slope:
θ=
δB
22
=
δC
33
∴
δC =
33
δB
22
δ C = 0.1552 in. 
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PROBLEM 2.107
Each cable has a cross-sectional area of 100 mm2 and is made of an
elastoplastic material for which σ Y = 345 MPa and E = 200 GPa. A force
Q is applied at C to the rigid bar ABC and is gradually increased from 0 to
50 kN and then reduced to zero. Knowing that the cables were initially taut,
determine (a) the maximum stress that occurs in cable BD, (b) the
maximum deflection of point C, (c) the final displacement of point C.
(Hint: In Part c, cable CE is not taut.)
SOLUTION
Elongation constraints for taut cables.
Let θ = rotation angle of rigid bar ABC.
θ=
δ BD
LAB
δ BD =
=
δ CE
LAC
LAB
1
δ CE = δ CE
LAC
2
(1)
Equilibrium of bar ABC.
M A = 0 : LAB FBD + LAC FCE − LAC Q = 0
Q = FCE +
LAB
1
FBD = FCE + FBD
2
LAC
Assume cable CE is yielded.
FCE = Aσ Y = (100 × 10−6 )(345 × 106 ) = 34.5 × 103 N
From (2),
FBD = 2(Q − FCE ) = (2)(50 × 103 − 34.5 × 103 ) = 31.0 × 103 N
(2)
Since FBD < Aσ Y = 34.5 × 103 N, cable BD is elastic when Q = 50 kN.
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PROBLEM 2.107 (Continued)
(a)
σ CE = σ Y = 345 MPa
Maximum stresses.
σ BD =
(b)
FBD 31.0 × 103
=
= 310 × 106 Pa
−6
A
100 × 10
σ BD = 310 MPa 
Maximum of deflection of point C.
δ BD =
From (1),
FBD LBD
(31.0 × 103 )(2)
=
= 3.1 × 10−3 m
EA
(200 × 109 )(100 × 10−6 )
δ C = δ CE = 2δ BD = 6.2 × 10−3 m
Permanent elongation of cable CE: (δ CE ) p = (δ CE ) −
6.20 mm ↓ 
σ Y LCE
E
FCE LCE
σ L
= (δ CE ) max − Y CE
EA
E
6
(345 × 10 )(2)
= 6.20 × 10−3 −
= 2.75 × 10−3 m
9
200 × 10
(δ CE ) P = (δ CE ) max −
(c)
Unloading. Cable CE is slack ( FCE = 0) at Q = 0.
From (2),
FBD = 2(Q − FCE ) = 2(0 − 0) = 0
Since cable BD remained elastic, δ BD =
FBD LBD
= 0.
EA

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PROBLEM 2.108
Solve Prob. 2.107, assuming that the cables are replaced by rods of the
same cross-sectional area and material. Further assume that the rods are
braced so that they can carry compressive forces.
PROBLEM 2.107 Each cable has a cross-sectional area of 100 mm2
and is made of an elastoplastic material for which σ Y = 345 MPa and
E = 200 GPa. A force Q is applied at C to the rigid bar ABC and is
gradually increased from 0 to 50 kN and then reduced to zero. Knowing
that the cables were initially taut, determine (a) the maximum stress
that occurs in cable BD, (b) the maximum deflection of point C, (c) the
final displacement of point C. (Hint: In Part c, cable CE is not taut.)
SOLUTION
Elongation constraints.
Let θ = rotation angle of rigid bar ABC.
θ=
δ BD =
δ BC
LAB
=
δ CE
LAC
LAB
1
δ CE = δ CE
LAC
2
(1)
Equilibrium of bar ABC.
M A = 0: LAB FBD + LAC FCE − LAC Q = 0
Q = FCE +
LAB
1
FBD = FCE + FBD
LAC
2
(2)
Assume cable CE is yielded. FCE = Aσ Y = (100 × 10−6 )(345 × 106 ) = 34.5 × 103 N
From (2),
FBD = 2(Q − FCE ) = (2)(50 × 103 − 34.5 × 103 ) = 31.0 × 103 N
Since FBD < Aσ Y = 34.5 × 103 N, cable BD is elastic when Q = 50 kN.
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PROBLEM 2.108 (Continued)
(a)
σ CE = σ Y = 345 MPa
Maximum stresses.
σ BD =
(b)
FBD 31.0 × 103
=
= 310 × 106 Pa
−6
A
100 × 10
σ BD = 310 MPa 
Maximum of deflection of point C.
δ BD =
FBD LBD
(31.0 × 103 )(2)
=
= 3.1 × 10−3 m
EA
(200 × 109 )(100 × 10−6 )
δ C = δ CE = 2δ BD = 6.2 × 10−3 m
From (1),
6.20 mm ↓ 
′ = δ C′
Unloading. Q′ = 50 × 103 N, δ CE
′ = 12 δ C′
δ BD
From (1),
′′ =
Elastic FBD
′ =
FCE
(200 × 109 )(100 × 10−6 )( 12 δ C′ )
′
EAδ BD
=
= 5 × 106 δ C′
LBD
2
′
EAδ CE
(200 × 109 )(100 × 10−6 )(δ C′ )
=
= 10 × 106 δ C′
LCE
2
From (2),
′ + 12 FBD
′ = 12.5 × 106 δ C′
Q′ = FCE
Equating expressions for Q′,
12.5 × 106 δ C′ = 50 × 103
δ C′ = 4 × 10−3 m
(c)
Final displacement.
δ C = (δ C ) m − δ C′ = 6.2 × 10−3 − 4 × 10−3 = 2.2 × 10−3 m
2.20 mm ↓ 
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PROBLEM 2.109
Rod AB consists of two cylindrical portions AC and BC, each with a cross-sectional
area of 1750 mm2. Portion AC is made of a mild steel with E = 200 GPa and
σ Y = 250 MPa, and portion CB is made of a high-strength steel with E = 200 GPa
and σ Y = 345 MPa. A load P is applied at C as shown. Assuming both steels to be
elastoplastic, determine (a) the maximum deflection of C if P is gradually increased
from zero to 975 kN and then reduced back to zero, (b) the maximum stress in each
portion of the rod, (c) the permanent deflection of C.
SOLUTION
Displacement at C to cause yielding of AC.
L σ
(0.190)(250 × 106 )
δ C ,Y = LAC ε Y , AC = AC Y , AC =
= 0.2375 × 10−3 m
9
E
200 × 10
FAC = Aσ Y , AC = (1750 × 10−6 )(250 × 106 ) = 437.5 × 103 N
Corresponding force.
FCB = −
EAδ C
(200 × 109 )(1750 × 10−6 )(0.2375 × 10−3 )
=−
= −437.5 × 103 N
0.190
LCB
For equilibrium of element at C,
FAC − ( FCB + PY ) = 0
PY = FAC − FCB = 875 × 103 N
Since applied load P = 975 × 103 N > 875 × 103 N, portion AC yields.
FCB = FAC − P = 437.5 × 103 − 975 × 103 N = −537.5 × 103 N
(a)
δC = −
FCB LCD
(537.5 × 103 )(0.190)
=
= 0.29179 × 10−3 m
EA
(200 × 109 )(1750 × 10−6 )
0.292 mm 
(b)
Maximum stresses: σ AC = σ Y , AC = 250 MPa
(c)
FBC
537.5 × 103
=−
= −307.14 × 106 Pa = −307 MPa
−6
A
1750 × 10
Deflection and forces for unloading.
P′ L
P′ L
L
′ = − PAC
′ AC = − PAC
′
δ ′ = AC AC = − CB CB ∴ PCB
EA
EA
LAB
σ BC =

250 MPa 
−307 MPa 
′ − PCB
′ = 2 PAC
′ PAC
′ = 487.5 × 10−3 N
P′ = 975 × 103 = PAC
δ′ =
(487.5 × 103 )(0.190)
= 0.26464 × 103 m
−6
9
(200 × 10 )(1750 × 10 )
δ p = δ m − δ ′ = 0.29179 × 10−3 − 0.26464 × 10−3
= 0.02715 × 10−3 m
0.0272 mm 
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PROBLEM 2.110
For the composite rod of Prob. 2.109, if P is gradually increased from zero until the
deflection of point C reaches a maximum value of δ m = 0.3 mm and then decreased
back to zero, determine (a) the maximum value of P, (b) the maximum stress in each
portion of the rod, (c) the permanent deflection of C after the load is removed.
PROBLEM 2.109 Rod AB consists of two cylindrical portions AC and BC, each with a
cross-sectional area of 1750 mm2. Portion AC is made of a mild steel with E = 200 GPa
and σ Y = 250 MPa, and portion CB is made of a high-strength steel with E = 200 GPa
and σ Y = 345 MPa. A load P is applied at C as shown. Assuming both steels to be
elastoplastic, determine (a) the maximum deflection of C if P is gradually increased
from zero to 975 kN and then reduced back to zero, (b) the maximum stress in each
portion of the rod, (c) the permanent deflection of C.
SOLUTION
Displacement at C is δ m = 0.30 mm. The corresponding strains are
ε AC =
δm
LAC
ε CB = −
=
δm
LCB
0.30 mm
= 1.5789 × 10−3
190 mm
=−
0.30 mm
= −1.5789 × 10−3
190 mm
Strains at initial yielding:
ε Y, AC =
ε Y, CB =
(a)
σ Y, AC
E
σ Y, BC
E
250 × 106
= 1.25 × 10−3
(yielding)
200 × 109
345 × 106
=−
= −1.725 × 10−3
(elastic)
200 × 109
=
Forces: FAC = Aσ Y = (1750 × 10−6 )(250 × 106 ) = 437.5 × 10−3 N
FCB = EAε CB = (200 × 109 )(1750 × 10−6 )(−1.5789 × 10−3 ) = −552.6 × 10−3 N
For equilibrium of element at C, FAC − FCB − P = 0
P = FAC − FCD = 437.5 × 103 + 552.6 × 103 = 990.1 × 103 N
(b)
Stresses: AC : σ AC = σ Y, AC
CB : σ CB =
FCB
552.6 × 103
=−
= −316 × 106 Pa
−6
A
1750 × 10
= 990 kN 
= 250 MPa 
−316 MPa 
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PROBLEM 2.110 (Continued)
(c) Deflection and forces for unloading.
δ′ =
′ LAC
PAC
P′ L
L
′ = − PAC
′ AC = − PAC
= − CB CB ∴ PCB
EA
EA
LAB
′ − PCB
′ = 2 PAC
′ = 990.1 × 103 N ∴ PAC
′ = 495.05 × 103 N
P′ = PAC
δ′ =
(495.05 × 103 )(0.190)
= 0.26874 × 10−3 m = 0.26874 mm
(200 × 109 )(1750 × 10−6 )
δ p = δ m − δ ′ = 0.30 mm − 0.26874 mm
0.031mm 
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PROBLEM 2.111
Two tempered-steel bars, each 163 -in. thick, are bonded to a 12 -in. mild-steel
bar. This composite bar is subjected as shown to a centric axial load of
magnitude P. Both steels are elastoplastic with E = 29 × 106 and with yield
strengths equal to 100 ksi and 50 ksi, respectively, for the tempered and mild
steel. The load P is gradually increased from zero until the deformation of
the bar reaches a maximum value δ m = 0.04 in. and then decreased back to
zero. Determine (a) the maximum value of P, (b) the maximum stress in the
tempered-steel bars, (c) the permanent set after the load is removed.
SOLUTION
1
For the mild steel, A1 =   (2) = 1.00 in 2
2
δY1 =
Lσ Y 1 (14)(50 × 103 )
=
= 0.024138 in.
E
29 × 106
 3
For the tempered steel, A2 = 2   (2) = 0.75 in 2
 16 
δY 2 =
Lσ Y 2 (14)(100 × 103 )
=
= 0.048276 in.
E
29 × 103
Total area: A = A1 + A2 = 1.75 in 2
δ Y 1 < δ m < δY 2 . The mild steel yields. Tempered steel is elastic.
(a)
Forces: P1 = A1σ Y 1 = (1.00)(50 × 103 ) = 50 × 103 lb
P2 =
EA2δ m (29 × 103 )(0.75)(0.04)
=
= 62.14 × 103 lb
14
L
P = P1 + P2 = 112.14 × 103 lb = 112.1kips
(b)
Stresses: σ1 =
σ2 =
Unloading:
(c)
P = 112.1 kips 
P1
= σ Y 1 = 50 × 103 psi = 50 ksi
A1
P2 62.14 × 103
=
= 82.86 × 103 psi = 82.86 ksi
A2
0.75
δ′ =
82.86 ksi 
PL (112.14 × 103 )(14)
=
= 0.03094 in.
EA
(29 × 106 )(1.75)
Permanent set: δ p = δ m − δ ′ = 0.04 − 0.03094 = 0.00906 in.
0.00906in. 
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PROBLEM 2.112
For the composite bar of Prob. 2.111, if P is gradually increased from zero to
98 kips and then decreased back to zero, determine (a) the maximum
deformation of the bar, (b) the maximum stress in the tempered-steel bars,
(c) the permanent set after the load is removed.
PROBLEM 2.111 Two tempered-steel bars, each 163 -in. thick, are bonded to
a 12 -in. mild-steel bar. This composite bar is subjected as shown to a centric
axial load of magnitude P. Both steels are elastoplastic with E = 29 × 106 psi
and with yield strengths equal to 100 ksi and 50 ksi, respectively, for the
tempered and mild steel.
SOLUTION
Areas:
Mild steel:
1
A1=   (2) = 1.00 in 2
2
Tempered steel:
 3
A2 = 2   (2) = 0.75 in 2
 16 
A = A1+ A2 = 1.75 in 2
Total:
Total force to yield the mild steel:
σY1 =
PY
∴ PY = Aσ Y 1 = (1.75)(50 × 103 ) = 87.50 × 103 lb
A
P > PY , therefore, mild steel yields.
Let P1 = force carried by mild steel.
P2 = force carried by tempered steel.
P1 = A1σ1 = (1.00)(50 × 103 ) = 50 × 103 lb
P1 + P2 = P, P2 = P − P1 = 98 × 103 − 50 × 103 = 48 × 103 lb
(a)
δm =
P2 L
(48 × 103 )(14)
=
EA2 (29 × 106 )(0.75)
(b)
σ2 =
P2 48 × 103
=
= 64 × 103 psi
A2
0.75
Unloading: δ ′ =
(c)
= 0.03090in. 
= 64 ksi 
PL
(98 × 103 )(14)
=
= 0.02703 in.
EA (29 × 106 )(1.75)
δ P = δ m − δ ′ = 0.03090 − 0.02703
= 0.00387 in. 
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PROBLEM 2.113
The rigid bar ABC is supported by two links, AD and BE, of uniform
37.5 × 6-mm rectangular cross section and made of a mild steel that is
assumed to be elastoplastic with E = 200 GPa and σ Y = 250 MPa.
The magnitude of the force Q applied at B is gradually increased from
zero to 260 kN. Knowing that a = 0.640 m, determine (a) the value of
the normal stress in each link, (b) the maximum deflection of point B.
SOLUTION
ΣM C = 0 : 0.640(Q − PBE ) − 2.64 PAD = 0
Statics:
δ A = 2.64θ , δ B = aθ = 0.640θ
Deformation:
Elastic analysis:
A = (37.5)(6) = 225 mm 2 = 225 × 10−6 m 2
PAD =
σ AD
PBE
σ BE
EA
(200 × 109 )(225 × 10−6 )
δA =
δ A = 26.47 × 106 δ A
LAD
1.7
= (26.47 × 106 )(2.64θ ) = 69.88 × 106 θ
P
= AD = 310.6 × 109 θ
A
(200 × 109 )(225 × 10−6 )
EA
=
δB =
δ B = 45 × 106 δ B
1.0
LBE
= (45 × 106 )(0.640θ ) = 28.80 × 106 θ
P
= BE = 128 × 109 θ
A
From Statics, Q = PBE +
2.64
PAD = PBE + 4.125PAD
0.640
= [28.80 × 106 + (4.125)(69.88 × 106 ] θ = 317.06 × 106 θ
θY at yielding of link AD:
σ AD = σ Y = 250 × 106 = 310.6 × 109 θ
θY = 804.89 × 10−6
QY = (317.06 × 106 )(804.89 × 10−6 ) = 255.2 × 103 N
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PROBLEM 2.113 (Continued)
(a)
Since Q = 260 × 103 > QY , link AD yields.
σ AD = 250 MPa 
PAD = Aσ Y = (225 × 10−6 )(250 × 10−6 ) = 56.25 × 103 N
From Statics, PBE = Q − 4.125PAD = 260 × 103 − (4.125)(56.25 × 103 )
PBE = 27.97 × 103 N
σ BE =
(b)
δB =
PBE 27.97 × 103
=
= 124.3 × 106 Pa
−6
A
225 × 10
PBE LBE
(27.97 × 103 )(1.0)
=
= 621.53 × 10−6 m
EA
(200 × 109 )(225 × 10−6 )
σ BE = 124.3 MPa 
δ B = 0.622 mm ↓ 
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PROBLEM 2.114
Solve Prob. 2.113, knowing that a = 1.76 m and that the magnitude of
the force Q applied at B is gradually increased from zero to 135 kN.
PROBLEM 2.113 The rigid bar ABC is supported by two links, AD
and BE, of uniform 37.5 × 6-mm rectangular cross section and made
of a mild steel that is assumed to be elastoplastic with E = 200 GPa
and σ Y = 250 MPa. The magnitude of the force Q applied at B is
gradually increased from zero to 260 kN. Knowing that a = 0.640 m,
determine (a) the value of the normal stress in each link, (b) the
maximum deflection of point B.
SOLUTION
ΣM C = 0 : 1.76(Q − PBE ) − 2.64 PAD = 0
Statics:
δ A = 2.64θ , δ B = 1.76θ
Deformation:
Elastic Analysis:
A = (37.5)(6) = 225 mm 2 = 225 × 10−6 m 2
PAD =
σ AD
PBE
σ BE
EA
(200 × 109 )(225 × 10−6 )
δA =
δ A = 26.47 × 106 δ A
LAD
1.7
= (26.47 × 106 )(2.64θ ) = 69.88 × 106 θ
P
= AD = 310.6 × 109 θ
A
(200 × 109 )(225 × 10−6 )
EA
=
δB =
δ B = 45 × 106 δ B
1.0
LBE
= (45 × 106 )(1.76θ ) = 79.2 × 106 θ
P
= BE = 352 × 109 θ
A
From Statics, Q = PBE +
2.64
PAD = PBE + 1.500 PAD
1.76
= [73.8 × 106 + (1.500)(69.88 × 106 ] θ = 178.62 × 106 θ
θY at yielding of link BE:
σ BE = σ Y = 250 × 106 = 352 × 109 θY
θY = 710.23 × 10−6
QY = (178.62 × 106 )(710.23 × 10−6 ) = 126.86 × 103 N
Since Q = 135 × 103 N > QY , link BE yields.
σ BE = σ Y = 250 MPa 
PBE = Aσ Y = (225 × 10−6 )(250 × 106 ) = 56.25 × 103 N

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PROBLEM 2.114 (Continued)
From Statics, PAD =
1
(Q − PBE ) = 52.5 × 103 N
1.500
(a)
From elastic analysis of AD,
(b)
σ AD =
PAD 52.5 × 103
=
= 233.3 × 106
−6
A
225 × 10
θ=
PAD
= 751.29 × 10−3 rad
69.88 × 106
δ B = 1.76θ = 1.322 × 10−3 m
σ AD = 233 MPa 
δ B = 1.322 mm ↓ 
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PROBLEM 2.115∗
Solve Prob. 2.113, assuming that the magnitude of the force Q
applied at B is gradually increased from zero to 260 kN and then
decreased back to zero. Knowing that a = 0.640 m, determine (a)
the residual stress in each link, (b) the final deflection of point B.
Assume that the links are braced so that they can carry compressive
forces without buckling.
PROBLEM 2.113 The rigid bar ABC is supported by two links,
AD and BE, of uniform 37.5 × 6-mm rectangular cross section and
made of a mild steel that is assumed to be elastoplastic with
E = 200 GPa and σ Y = 250 MPa. The magnitude of the force Q
applied at B is gradually increased from zero to 260 kN. Knowing
that a = 0.640 m, determine (a) the value of the normal stress in
each link, (b) the maximum deflection of point B.
SOLUTION
See solution to Problem 2.113 for the normal stresses in each link and the deflection of Point B after loading.
σ AD = 250 × 106 Pa
σ BE = 124.3 × 106 Pa
δ B = 621.53 × 10−6 m
The elastic analysis given in the solution to Problem 2.113 applies to the unloading.
Q′ = 317.06 × 106 θ ′
Q′ =
Q
260 × 103
=
= 820.03 × 10−6
317.06 × 106 317.06 × 106
σ ′AD = 310.6 × 109 θ = (310.6 × 109 )(820.03 × 10−6 ) = 254.70 × 106 Pa
′ = 128 × 109 θ = (128 × 109 )(820.03 × 10−6 ) = 104.96 × 106 Pa
σ BE
δ B′ = 0.640θ ′ = 524.82 × 10−6 m
(a)
(b)
Residual stresses.
′ = 250 × 106 − 254.70 × 10−6 = −4.70 × 106 Pa
σ AD , res = σ AD − σ AD
= −4.70 MPa 
′ = 124.3 × 106 − 104.96 × 106 = 19.34 × 106 Pa
σ BE , res = σ BE − σ BE
= 19.34 MPa 
δ B , P = δ B − δ B′ = 621.53 × 10−6 − 524.82 × 10−6 = 96.71 × 10−6 m
= 0.0967 mm ↓ 
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PROBLEM 2.116
A uniform steel rod of cross-sectional area A is attached to rigid supports
and is unstressed at a temperature of 45 °F. The steel is assumed to be
elastoplastic with σ Y = 36 ksi and E = 29 × 106 psi. Knowing that
α = 6.5 × 10−6 /°F, determine the stress in the bar (a) when the temperature
is raised to 320 °F, (b) after the temperature has returned to 45 °F .
SOLUTION
Let P be the compressive force in the rod.
Determine temperature change to cause yielding.
σ L
PL
+ L α (ΔT ) = − Y + Lα ( ΔT )Y = 0
AE
E
3
σ
36 × 10
(ΔT )Y = Y =
= 190.98 °F
Eα (29 × 106 )(6.5 × 10−6 )
δ =−
But ΔT = 320 − 45 = 275 °F > (ΔTY )
(a)
σ = −σ Y = −36 ksi 
Yielding occurs.
Cooling:
(ΔT)′ = 275 °F
δ ′ = δ P′ = δ T′ = −
P′L
+ Lα (ΔT )′ = 0
AE
P′
= − Eα (ΔT )′
A
= −(29 × 106 )(6.5 × 10−6 )(275) = −51.8375 × 103 psi
σ′=
(b) Residual stress:
σ res = −σ Y − σ ′ = −36 × 103 + 51.8375 × 103 = 15.84 × 10 psi
15.84 ksi 
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PROBLEM 2.117
The steel rod ABC is attached to rigid supports and is unstressed at a
temperature of 25 °C. The steel is assumed elastoplastic, with E = 200
GPa and σ y = 250 MPa. The temperature of both portions of the rod is
then raised to 150 °C . Knowing that α = 11.7 × 10−6 / °C, determine
(a) the stress in both portions of the rod, (b) the deflection of point C.
SOLUTION
AAC = 500 × 10−6 m 2
LAC = 0.150 m
ACB = 300 × 10−6 m 2
LCB = 0.250 m
δ P + δT = 0
Constraint:
Determine ΔT to cause yielding in portion CB.
−
PLAC PLCB
−
= LABα ( ΔT )
EAAC EACB
ΔT =
P  LAC LCB 
+


LAB Eα  AAC ACB 
At yielding, P = PY = ACBσ Y = (300 × 10−6 )(2.50 × 106 ) = 75 × 103 N
(ΔT )Y =
=
PY
LAB Eα
 LAC LCB 
+


 AAC ACB 
75 × 103
0.250
 0.150
+
9
−6 
−6
(0.400)(200 × 10 )(11.7 × 10 )  500 × 10
300 × 10−6
Actual ΔT:
150 °C − 25 °C = 125 °C > (ΔT )Y
Yielding occurs. For ΔT > (ΔT )Y ,
(a)
(b)

 = 90.812 °C

P = PY = 75 × 103 N
σ AC = −
PY
75 × 103
=−
= −150 × 10−6 Pa
AAC
500 × 10−6
σ AC = −150 MPa 
σ CB = −
PY
= −σ Y
ACB
σ CB = −250 MPa 
For ΔT > (ΔT )Y , portion AC remains elastic.
δ C /A = −
=−
PY LAC
+ LACα (ΔT )
EAAC
(75 × 103 )(0.150)
+ (0.150)(11.7 × 10−6 )(125) = 106.9 × 10−6 m
(200 × 109 )(500 × 10−6 )
Since Point A is stationary, δ C = δ C /A = 106.9 × 10−6 m
δ C = 0.1069 mm → 
\
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PROBLEM 2.118∗
Solve Prob. 2.117, assuming that the temperature of the rod is raised
to 150°C and then returned to 25 °C.
PROBLEM 2.117 The steel rod ABC is attached to rigid supports
and is unstressed at a temperature of 25 °C. The steel is assumed
elastoplastic, with E = 200 GPa and σ Y = 250 MPa. The
temperature of both portions of the rod is then raised to 150 °C.
Knowing that α = 11.7 × 10−6 / °C, determine (a) the stress in both
portions of the rod, (b) the deflection of point C.
SOLUTION
AAC = 500 × 10−6 m 2
Constraint:
LAC = 0.150 m
ACB = 300 × 10−6 m 2
LCB = 0.250 m
δ P + δT = 0
Determine ΔT to cause yielding in portion CB.
−
PLAC PLCB
−
= LABα ( ΔT )
EAAC EACB
ΔT =
P
LAB Eα
 LAC LCB 
+


 AAC ACB 
At yielding, P = PY = ACBσ Y = (300 × 10−6 )(250 × 106 ) = 75 × 103 N
PY  LAC LCB 
75 × 103
0.250 
 0.150
+
+

=

9
−6 
−6
LAB Eα  AAC ACB  (0.400)(200 × 10 )(11.7 × 10 )  500 × 10
300 × 10−6 
= 90.812 °C
(ΔT )Y =
Actual
ΔT : 150 °C − 25 °C = 125 °C > (ΔT )Y
Yielding occurs. For
ΔT > (ΔT )Y
Cooling: (ΔT )′ = 125 °C P′ =
P = PY = 75 × 103 N
ELABα (ΔT )′
(
LAC
AAC
+
LCB
ACB
)
=
(200 × 109 )(0.400)(11.7 × 10−6 )(125)
= 103.235 × 103 N
0.150
0.250
+
−6
−6
500 × 10
300 × 10
Residual force: Pres = P′ − PY = 103.235 × 103 − 75 × 103 = 28.235 × 103 N (tension)
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PROBLEM 2.118∗ (Continued)
(a)
(b)
Residual stresses.
σ AC =
Pres 28.235 × 103
=
AAC
500 × 10−6
σ AC = 56.5 MPa 
σ CB =
Pres 28.235 × 103
=
ACB
300 × 10−6
σ CB = 9.41 MPa 
Permanent deflection of point C. δ C =
Pres LAC
EAAC
δ C = 0.0424 mm → 
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PROBLEM 2.119∗
For the composite bar of Prob. 2.111, determine the residual stresses in
the tempered-steel bars if P is gradually increased from zero to
98 kips and then decreased back to zero.
PROBLEM 2.111 Two tempered-steel bars, each 163 -in. thick, are
bonded to a 12 -in. mild-steel bar. This composite bar is subjected as
shown to a centric axial load of magnitude P. Both steels are
elastoplastic with E = 29 × 106 psi and with yield strengths equal to
100 ksi and 50 ksi, respectively, for the tempered and mild steel. The
load P is gradually increased from zero until the deformation of the bar
reaches a maximum value δ m = 0.04 in. and then decreased back to
zero. Determine (a) the maximum value of P, (b) the maximum stress
in the tempered-steel bars, (c) the permanent set after the load is
removed.
SOLUTION
Areas:
Mild steel:
1
A1 =   (2) = 1.00 in 2
2
 3
Tempered steel: A2 = (2)   (2) = 0.75 in 2
 16 
Total: A = A1+ A2 = 1.75 in 2
Total force to yield the mild steel: σ Y 1 =
PY
A
∴ PY = Aσ Y 1 = (1.75)(50 × 103 ) = 87.50 × 103 lb
P > PY; therefore, mild steel yields.
P1 = force carried by mild steel.
Let
P2 = force carried by tempered steel.
P1 = A1σ Y 1 = (1.00)(50 × 103 ) = 50 × 103 lb
P1 + P2 = P, P2 = P − P1 = 98 × 103 − 50 × 103 = 48 × 103 lb
σ2 =
Unloading: σ ′ =
P2 48 × 103
=
= 64 × 103 psi
A2
0.75
P 98 × 103
=
= 56 × 103 psi
A
1.75
Residual stresses.
Mild steel:
σ1,res = σ 1 − σ ′ = 50 × 103 − 56 × 103 = −6 × 10−3 psi = −6 ksi
Tempered steel:
σ 2,res = σ 2 − σ 1 = 64 × 103 − 56 × 103 = 8 × 103 psi
8.00 ksi 
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PROBLEM 2.120∗
For the composite bar in Prob. 2.111, determine the residual stresses in
the tempered-steel bars if P is gradually increased from zero until the
deformation of the bar reaches a maximum value δ m = 0.04 in. and is
then decreased back to zero.
PROBLEM 2.111 Two tempered-steel bars, each 163 -in. thick, are bonded
to a 12 -in. mild-steel bar. This composite bar is subjected as shown to a
centric axial load of magnitude P. Both steels are elastoplastic
with E = 29 × 106 psi and with yield strengths equal to 100 ksi and 50 ksi,
respectively, for the tempered and mild steel. The load P is gradually
increased from zero until the deformation of the bar reaches a maximum
value δ m = 0.04 in. and then decreased back to zero. Determine (a) the
maximum value of P, (b) the maximum stress in the tempered-steel bars,
(c) the permanent set after the load is removed.
SOLUTION
Lδ
(14)(50 × 103 )
1
= 0.024138 in
A1 =   (2) = 1.00 in 2 δ Y 1 = Y 1 =
E
29 × 106
2
For the mild steel,
Lδ
(14)(100 × 103 )
 3
For the tempered steel, A2 = 2   (2) = 0.75 in 2 δ Y 2 = Y 2 =
= 0.048276 in.
E
29 × 106
 16 
A = A1 + A2 = 1.75 in 2
Total area:
δY 1 < δ m < δ Y 2
The mild steel yields. Tempered steel is elastic.
P1 = A1δY 1 = (1.00)(50 × 103 ) = 50 × 103 lb
Forces:
P2 =
Stresses:
Unloading:
σ1 =
EA2δ m (29 × 106 )(0.75)(0.04)
=
= 62.14 × 103 lb
14
L
P1
P
62.14 × 103
= δY 1 = 50 × 103 psi σ 2 = 2 =
= 82.86 × 103 psi
0.75
A1
A2
σ′ =
P 112.14
=
= 64.08 × 103 psi
A
1.75
Residual stresses. σ1,res = σ 1 − σ ′ = 50 × 103 − 64.08 × 103 = −14.08 × 103 psi = −14.08 ksi
σ 2,res = σ 2 − σ ′ = 82.86 × 103 − 64.08 × 103 = 18.78 × 103 psi = 18.78 ksi

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PROBLEM 2.121∗
Narrow bars of aluminum are bonded to the two sides of a thick steel plate as
shown. Initially, at T1 = 70°F, all stresses are zero. Knowing that the
temperature will be slowly raised to T2 and then reduced to T1, determine
(a) the highest temperature T2 that does not result in residual stresses, (b) the
temperature T2 that will result in a residual stress in the aluminum equal to
58 ksi. Assume α a = 12.8 × 10−6 / °F for the aluminum and α s = 6.5 × 10−6 / °F
for the steel. Further assume that the aluminum is elastoplastic, with
E = 10.9 × 106 psi and σ Y = 58 ksi. (Hint: Neglect the small stresses in the
plate.)
SOLUTION
Determine temperature change to cause yielding.
δ=
PL
+ Lα a ( ΔT )Y = Lα s ( ΔT )Y
EA
P
= σ = − E (α a − α s )(ΔT )Y = −σ Y
A
σY
58 × 103
(ΔT )Y =
=
= 844.62 °F
6
E (α a − α s ) (10.9 × 10 )(12.8 − 6.5)(10−6 )
(a)
T2Y = T1 + (ΔT )Y = 70 + 844.62 = 915 °F
915 °F 
After yielding,
δ=
σY L
E
+ Lα a (ΔT ) = Lα s (ΔT )
Cooling:
δ′ =
P′L
+ Lα a (ΔT )′ = Lα s (ΔT )′
AE
The residual stress is
σ res = σ Y −
Set
σ res = −σ Y
−σ Y = σ Y − E (α a − α s )(ΔT )
ΔT =
(b)
P′
= σ Y − E (α a − α s )(ΔT )
A
2σ Y
(2)(58 × 103 )
=
= 1689 °F
E (α a − α s ) (10.9 × 106 )(12.8 − 6.5)(10−6 )
T2 = T1 + ΔT = 70 + 1689 = 1759 °F
1759 °F 
If T2 > 1759 °F, the aluminum bar will most likely yield in compression.
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PROBLEM 2.122∗
Bar AB has a cross-sectional area of 1200 mm2 and is made of a steel
that is assumed to be elastoplastic with E = 200 GPa and
σ Y = 250 MPa. Knowing that the force F increases from 0 to 520 kN
and then decreases to zero, determine (a) the permanent deflection of
point C, (b) the residual stress in the bar.
SOLUTION
A = 1200 mm 2 = 1200 × 10−6 m 2
Force to yield portion AC:
PAC = Aσ Y = (1200 × 10−6 )(250 × 106 )
= 300 × 103 N
For equilibrium, F + PCB − PAC = 0.
PCB = PAC − F = 300 × 103 − 520 × 103
= −220 × 103 N
δC = −
PCB LCB (220 × 103 )(0.440 − 0.120)
=
EA
(200 × 109 )(1200 × 10−6 )
= 0.293333 × 10−3 m
PCB
220 × 103
=
A 1200 × 10−6
= −183.333 × 106 Pa
σ CB =
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PROBLEM 2.122∗ (Continued)
Unloading:
δ C′ =
′ =
PAC
′ LAC
′ ) LCB
PAC
P′ L
( F − PAC
= − CB CB =
EA
EA
EA
L  FL
L
′  AC + BC  = CB
PAC
EA 
EA
 EA
FLCB
(520 × 103 )(0.440 − 0.120)
=
= 378.182 × 103 N
LAC + LCB
0.440
′ = PAC
′ − F = 378.182 × 103 − 520 × 103 = −141.818 × 103 N
PCB
′
PAC
378.182 × 103
=
= 315.152 × 106 Pa
−6
A
1200 × 10
P′
141.818 × 103
′ = BC = −
σ BC
= −118.182 × 106 Pa
−6
A
1200 × 10
(378.182)(0.120)
= 0.189091 × 10−3 m
δ C′ =
9
6
(200 × 10 )(1200 × 10 )
′ =
σ AC
(a)
δ C , p = δ C − δ C′ = 0.293333 × 10−3 − 0.189091 × 10−3 = 0.1042 × 10−3 m
= 0.1042 mm 
(b)
σ AC , res = σ Y − σ ′AC = 250 × 106 − 315.152 × 106 = −65.2 × 106 Pa
= −65.2 MPa 
′ = −183.333 × 106 + 118.182 × 106 = −65.2 × 106 Pa
σ CB , res = σ CB − σ CB
= −65.2 MPa 
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PROBLEM 2.123∗
Solve Prob. 2.122, assuming that a = 180 mm.
PROBLEM 2.122 Bar AB has a cross-sectional area of 1200 mm2 and
is made of a steel that is assumed to be elastoplastic with E = 200 GPa
and σ Y = 250 MPa. Knowing that the force F increases from 0 to
520 kN and then decreases to zero, determine (a) the permanent
deflection of point C, (b) the residual stress in the bar.
SOLUTION
A = 1200 mm 2 = 1200 × 10−6 m 2
Force to yield portion AC:
PAC = Aσ Y = (1200 × 10−6 )(250 × 106 )
= 300 × 103 N
For equilibrium, F + PCB − PAC = 0.
PCB = PAC − F = 300 × 103 − 520 × 103
= −220 × 103 N
δC = −
PCB LCB (220 × 103 )(0.440 − 0.180)
=
EA
(200 × 109 )(1200 × 10−6 )
= 0.238333 × 10−3 m
PCB
220 × 103
=−
A
1200 × 10−6
= −183.333 × 106 Pa
σ CB =
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PROBLEM 2.123∗ (Continued)
Unloading:
′ LAC
′ ) LCB
PAC
P′ L
( F − PAC
= − CB CB =
EA
EA
EA
L  FL
L
′  AC + BC  = CB
= PAC
EA 
EA
 EA
δ C′ =
′ =
PAC
FLCB
(520 × 103 )(0.440 − 0.180)
=
= 307.273 × 103 N
0.440
LAC + LCB
′ = PAC
′ − F = 307.273 × 103 − 520 × 103 = −212.727 × 103 N
PCB
δ C′ =
(307.273 × 103 )(0.180)
= 0.230455 × 10−3 m
(200 × 109 )(1200 × 10−6 )
′
PAC
307.273 × 103
=
= 256.061 × 106 Pa
−6
A
1200 × 10
′
PCB
−212.727 × 103
′ =
σ CB
=
= −177.273 × 106 P
A
1200 × 10−6
′ =
σ AC
(a)
δ C , p = δ C − δ C′ = 0.238333 × 10−3 − 0.230455 × 10−3 = 0.00788 × 10−3 m
(b)
σ AC ,res = σ AC − σ ′AC = 250 × 106 − 256.061 × 106 = −6.06 × 106 Pa
= −6.06 MPa 
′ = −183.333 × 106 + 177.273 × 106 = −6.06 × 106 Pa
σ CB ,res = σ CB − σ CB
= −6.06 MPa 
= 0.00788 mm 

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PROBLEM 2.124
Rod BD is made of steel ( E = 29 × 106 psi) and is used to brace the axially
compressed member ABC. The maximum force that can be developed in member
BD is 0.02P. If the stress must not exceed 18 ksi and the maximum change in
length of BD must not exceed 0.001 times the length of ABC, determine the
smallest-diameter rod that can be used for member BD.
SOLUTION
FBD = 0.02 P = (0.02) (130) = 2.6 kips = 2.6 × 103lb
Considering stress: σ = 18 ksi = 18 × 103 psi
σ=
FBD
A
∴ A=
FBD
σ
=
2.6
= 0.14444 in 2
18
Considering deformation: δ = (0.001)(144) = 0.144 in.
δ=
FBD LBD
AE
∴
A=
FBD LBD
(2.6 × 103 ) (54)
=
= 0.03362 in 2
6
Eδ
(29 × 10 ) (0.144)
Larger area governs. A = 0.14444 in 2
A=
π
4
d2 ∴
d=
4A
π
=
(4) (0.14444)
π
d = 0.429in. 
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PROBLEM 2.125
Two solid cylindrical rods are joined at B and loaded as shown. Rod AB is
made of steel ( E = 200 GPa) and rod BC of brass ( E = 105 GPa). Determine
(a) the total deformation of the composite rod ABC, (b) the deflection of
point B.
SOLUTION
Rod AB:
FAB = − P = −30 × 103 N
LAB = 0.250 m
E AB = 200 × 109 GPa
δ AB
Rod BC:
π
(30)2 = 706.85 mm 2 = 706.85 × 10−6 m 2
4
F L
(30 × 103 )(0.250)
= AB AB = −
= −53.052 × 10−6 m
−6
9
E AB AAB
(200 × 10 )(706.85 × 10 )
AAB =
FBC = 30 + 40 = 70 kN = 70 × 103 N
LBC = 0.300 m
EBC = 105 × 109 Pa
δ BC
π
(50) 2 = 1.9635 × 103 mm 2 = 1.9635 × 10−3 m 2
4
F L
(70 × 103 )(0.300)
= BC BC = −
= −101.859 × 10−6 m
9
−3
EBC ABC
(105 × 10 )(1.9635 × 10 )
ABC =
δ tot = δ AB + δ BC = −154.9 × 10−6 m
(a)
Total deformation:
(b)
Deflection of Point B. δ B = δ BC
= −0.1549 mm 
δ B = 0.1019 mm ↓ 
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PROBLEM 2.126
Two solid cylindrical rods are joined at B and loaded as shown. Rod AB is
made of steel ( E = 29 × 106 psi), and rod BC of brass ( E = 15 × 106 psi).
Determine (a) the total deformation of the composite rod ABC, (b) the
deflection of point B.
SOLUTION
Portion AB:
PAB = 40 × 103 lb
LAB = 40 in.
d = 2 in.
E AB
δ AB =
Portion BC:
π
d2 =
π
(2) 2 = 3.1416 in 2
4
4
= 29 × 106 psi
AAB =
PAB LAB
(40 × 103 )(40)
=
= 17.5619 × 10−3 in.
E AB AAB (29 × 106 )(3.1416)
PBC = −20 × 103 lb
LBC = 30 in.
d = 3 in.
ABC =
π
d2 =
π
(3) 2 = 7.0686 in 2
4
4
EBC = 15 × 106 psi
δ BC =
PBC LBC
(−20 × 103 )(30)
=
= −5.6588 × 10−3 in.
6
EBC ABC (15 × 10 )(7.0686)
(a)
δ = δ AB + δ BC = 17.5619 × 10−6 − 5.6588 × 10−6
δ = 11.90 × 10−3 in. ↓ 
(b)
δ B = −δ BC
δ B = 5.66 × 10−3 in. ↑ 
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PROBLEM 2.127
The uniform wire ABC, of unstretched length 2l, is attached to the
supports shown and a vertical load P is applied at the midpoint B.
Denoting by A the cross-sectional area of the wire and by E the
modulus of elasticity, show that, for δ  l , the deflection at the
midpoint B is
δ =l3
P
AE
SOLUTION
Use approximation.
sin θ ≈ tan θ ≈
Statics:
l
ΣFY = 0 : 2 PAB sin θ − P = 0
PAB =
Elongation:
δ
P
Pl
≈
2sin θ 2δ
δ AB =
PAB l
Pl 2
=
AE 2 AEδ
Deflection:
From the right triangle,
(l + δ AB ) 2 = l 2 + δ 2
2
δ 2 = l 2 + 2l δ AB + δ AB
− l2
 1 δ AB
= 2l δ AB 1 +
2 l

≈
δ3 ≈

 ≈ 2l δ AB

Pl 3
AEδ
Pl 3
P
∴ δ ≈ l3
AE
AE

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PROBLEM 2.128
The brass strip AB has been attached to a fixed support
at A and rests on a rough support at B. Knowing that
the coefficient of friction is 0.60 between the strip and
the support at B, determine the decrease in temperature
for which slipping will impend.
SOLUTION
Brass strip:
E = 105 GPa
α = 20 × 10−6 / °C
ΣFy = 0 : N − W = 0
N =W
ΣFx = 0 : P − μ N = 0
δ =−
Data:
PL
+ Lα ( ΔT ) = 0
EA
P = μW = μ mg
ΔT =
μ mg
P
=
EAα EAα
μ = 0.60
A = (20)(3) = 60 mm 2 = 60 × 10−6 m 2
m = 100 kg
g = 9.81 m/s 2
E = 105 × 109 Pa
ΔT =
(0.60)(100)(9.81)
(105 × 109 )(60 × 10−6 )(20 × 106 )
ΔT = 4.67 °C 
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PROBLEM 2.129
Members AB and CD are 1 18 -in.-diameter steel rods, and members BC and AD
are 78 -in.-diameter steel rods. When the turnbuckle is tightened, the diagonal
member AC is put in tension. Knowing that E = 29 × 106 psi and h = 4 ft,
determine the largest allowable tension in AC so that the deformations in
members AB and CD do not exceed 0.04 in.
SOLUTION
δ AB = δ CD = 0.04 in.
h = 4ft = 48 in. = LCD
ACD =
δ CD
π
d2 =
π
4
4
FCD LCD
=
EACD
FCD =
(1.125) 2 = 0.99402 in 2
EACDδ CD (29 × 106 )(0.99402)(0.04)
=
= 24.022 × 103 lb
48
LCD
Use joint C as a free body.
ΣFy = 0 : FCD −
FAC =
4
5
FAC = 0 ∴ FAC = FCD
5
4
5
(24.022 × 103 ) = 30.0 × 103 lb
4
FAC = 30.0 kips 
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PROBLEM 2.130
The 1.5-m concrete post is reinforced with six steel bars, each with a 28-mm
diameter. Knowing that Es = 200 GPa and Ec = 25 GPa, determine the
normal stresses in the steel and in the concrete when a 1550-kN axial centric
force P is applied to the post.
SOLUTION
Let Pc = portion of axial force carried by concrete.
Ps = portion carried by the six steel rods.
δ=
Pc L
Ec Ac
Pc =
Ec Acδ
L
δ=
Ps L
Es As
Ps =
Es Asδ
L
P = Pc + Ps = ( Ec Ac + Es As )
δ
L
δ
−P
ε= =
L Ec Ac + Es As
6π
(28) 2 = 3.6945 × 103 mm 2
4
4
= 3.6945 × 10−3 m 2
As = 6
Ac =
π
π
d s2 =
d c2 − As =
π
(450) 2 − 3.6945 × 103
4
4
= 155.349 × 103 mm 2
= 153.349 × 10−3 m 2
L = 1.5 m
ε=
1550 × 103
= 335.31 × 10−6
(25 × 109 )(155.349 × 10−3 ) + (200 × 109 )(3.6945 × 10−3 )
σ s = Es ε = (200 × 109 )(335.31 × 10−6 ) = 67.1 × 106 Pa
σ s = 67.1 MPa 
σ c = Ecε = (25 × 109 )( −335.31 × 10−6 ) = 8.38 × 106 Pa
σ c = 8.38 MPa 
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PROBLEM 2.131
The brass shell (α b = 11.6 × 10−6 /°F) is fully bonded to the steel core
(α s = 6.5 × 10−6 /°F). Determine the largest allowable increase in
temperature if the stress in the steel core is not to exceed 8 ksi.
SOLUTION
Let Ps = axial force developed in the steel core.
For equilibrium with zero total force, the compressive force in the brass shell is Ps .
Strains:
εs =
Ps
+ α s (ΔT )
Es As
εb = −
Matching:
Ps
+ α b (ΔT )
Eb Ab
ε s = εb
Ps
P
+ α s (ΔT ) = − s + α b (ΔT )
Es As
Eb Ab
 1
1 
+

 Ps = (α b − α s )(ΔT )
 Es As Eb Ab 
(1)
Ab = (1.5) (1.5) − (1.0) (1.0) = 1.25 in 2
As = (1.0) (1.0) = 1.0 in 2
αb − α s = 5.1 × 10−6 /°F
Ps = σ s As = (8 × 103 ) (1.0) = 8 × 103 lb
1
1
1
1
+
=
+
= 87.816 × 10−9 lb −1
Es As Eb Ab (29 × 106 ) (1.0) (15 × 106 ) (1.25)
From (1),
(87.816 × 10−9 ) (8 × 103 ) = (5.1 × 10−6 ) (ΔT )
ΔT = 137.8 °F 
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PROBLEM 2.132
A fabric used in air-inflated structures is subjected to a biaxial loading
that results in normal stresses σ x = 120 MPa and σ z = 160 MPa.
Knowing that the properties of the fabric can be approximated as
E = 87 GPa and v = 0.34, determine the change in length of (a) side AB,
(b) side BC, (c) diagonal AC.
SOLUTION
σ x = 120 × 106 Pa,
σ y = 0,
σ z = 160 × 106 Pa
1
1
(σ x − vσ y − vσ z ) =
[120 × 106 − (0.34)(160 × 106 )] = 754.02 × 10−6
E
87 × 109
1
1
ε z = ( −vσ x − vσ y + σ z ) =
[−(0.34)(120 × 106 ) + 160 × 106 ] = 1.3701 × 10−3
E
87 × 109
εx =
(a)
δ AB = ( AB)ε x = (100 mm)(754.02 × 10−6 )
= 0.0754 mm 
(b)
δ BC = ( BC )ε z = (75 mm)(1.3701 × 10−6 )
= 0.1028 mm 
Label sides of right triangle ABC as a, b, and c.
c2 = a 2 + b2
Obtain differentials by calculus.
2c dc = 2a da + 2b db
dc =
a
b
da + db
c
c
But a = 100 mm,
b = 75 mm,
c = (1002 + 752 ) = 125 mm
da = δ AB = 0.0754 mm
db = δ BC = 0.1370 mm
(c)
δ AC = dc =
100
75
(0.0754) +
(0.1028)
125
125
= 0.1220 mm 
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PROBLEM 2.133
An elastomeric bearing (G = 0.9 MPa) is used to support a bridge
girder as shown to provide flexibility during earthquakes. The beam
must not displace more than 10 mm when a 22-kN lateral load is
applied as shown. Knowing that the maximum allowable shearing
stress is 420 kPa, determine (a) the smallest allowable dimension b,
(b) the smallest required thickness a.
SOLUTION
Shearing force:
P = 22 × 103 N
Shearing stress:
τ = 420 × 103 Pa
P
P
∴ A=
τ
A
22 × 103
=
= 52.381 × 10−3 m 2
420 × 103
= 52.381 × 103 mm 2
A = (200 mm)(b)
τ=
(a)
b=
γ=
(b)
A
52.381 × 103
=
= 262 mm
200
200
τ
G
=
But γ =
b = 262 mm 
420 × 103
= 466.67 × 10−3
6
0.9 × 10
δ
a
∴ a=
δ
10 mm
=
= 21.4 mm
γ 466.67 × 10−3
a = 21.4 mm 
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PROBLEM 2.134
Knowing that P = 10 kips, determine the maximum stress when (a) r = 0.50 in.,
(b) r = 0.625 in.
SOLUTION
P = 10 × 103 lb D = 5.0 in. d = 2.50 in.
5.0
D
=
= 2.00
2.50
d
3
Amin = dt = (2.50)   = 1.875 in 2
4
(a)
r = 0.50 in.
From Fig. 2.60b,
σ max =
0.50
r
=
= 0.20
2.50
d
K = 1.94
KP
(1.94) (10 × 103 )
=
= 10.35 × 103 psi
Amin
1.875
10.35 ksi 
(b)
r = 0.625 in.
σ max =
r
0.625
=
= 0.25 K = 1.82
d
2.50
KP
(1.82) (10 × 103 )
=
= 9.71 × 103 psi
Amin
1.875
9.71 ksi 
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PROBLEM 2.135
The uniform rod BC has a cross-sectional area A and is made of a
mild steel that can be assumed to be elastoplastic with a modulus of
elasticity E and a yield strength σ y . Using the block-and-spring
system shown, it is desired to simulate the deflection of end C of the
rod as the axial force P is gradually applied and removed, that is, the
deflection of points C and C′ should be the same for all values of P.
Denoting by μ the coefficient of friction between the block and the
horizontal surface, derive an expression for (a) the required mass m
of the block, (b) the required constant k of the spring.
SOLUTION
Force-deflection diagram for Point C or rod BC.
P < PY = Aσ Y
For
PL
EA
= PY = Aσ Y
δC =
Pmax
P=
EA
δC
L
Force-deflection diagram for Point C′ of block-and-spring system.
ΣFy = 0 : N − mg = 0
N = mg
ΣFx = 0 : P − F f = 0
P = Ff
If block does not move, i.e., F f < μ N = μ mg or P < μ mg ,
δ c′ =
then
P
K
or
P = kδ c′
If P = μ mg, then slip at P = Fm = μ mg occurs.
If the force P is the removed, the spring returns to its initial length.
(a)
Equating PY and Fmax,
(b)
Equating slopes,
Aσ Y = μ mg
k=
EA
L
m=
Aσ Y
μg


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CHAPTER 3
PROBLEM 3.1
(a) Determine the maximum shearing stress caused by a 4.6-kN ⋅ m torque T in
the 76-mm-diameter shaft shown. (b) Solve part a, assuming that the solid shaft
has been replaced by a hollow shaft of the same outer diameter and of 24-mm
inner diameter.
SOLUTION
(a)
Solid shaft:
c=
J =
τ =
d
= 38 mm = 0.038 m
2
π
2
c4 =
π
2
(0.038) 4 = 3.2753 × 10−6 m 4
Tc (4.6 × 103 )(0.038)
=
= 53.4 × 106 Pa
J
3.2753 × 10−6
τ = 53.4 MPa 
(b)
Hollow shaft:
do
= 0.038 m
2
1
c1 = di = 12 mm = 0.012 m
2
c2 =
J =
τ =
π
(c
2
4
2
)
− c14 =
π
2
(0.0384 − 0.0124 ) = 3.2428 × 10−6 m 4
Tc (4.6 × 103 )(0.038)
=
= 53.9 × 106 Pa
−6
J
3.2428 × 10
τ = 53.9 MPa 
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PROBLEM 3.2
(a) Determine the torque T that causes a maximum shearing stress of 45
MPa in the hollow cylindrical steel shaft shown. (b) Determine the
maximum shearing stress caused by the same torque T in a solid cylindrical
shaft of the same cross-sectional area.
SOLUTION
(a)
Given shaft:
J =
J =
π
2
π
(c
4
2
− c14
)
(454 − 304 ) = 5.1689 × 106 mm 4 = 5.1689 × 10−6 m 4
2
Tc
τ =
J
T =
T =
Jτ
c
(5.1689 × 10−6 )(45 × 106 )
= 5.1689 × 103 N ⋅ m
−3
45 × 10
T = 5.17 kN ⋅ m 
(b)
Solid shaft of same area:
(
)
A = π c22 − c12 = π (452 − 302 ) = 3.5343 × 103 mm 2
π c 2 = A or c =
J =
τ =
π
2
c4, τ =
A
π
= 33.541 mm
2T
Tc
=
J
π c3
(2)(5.1689 × 103 )
= 87.2 × 106 Pa
π (0.033541)3
τ = 87.2 MPa 
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PROBLEM 3.3
Knowing that d = 1.2 in., determine the torque T that causes a
maximum shearing stress of 7.5 ksi in the hollow shaft shown.
SOLUTION
c2 =
1
1
d 2 =   (1.6) = 0.8 in.
2
2
c1 =
1
1
d1 =   (1.2) = 0.6 in.
2
2
J =
π
2
(c
4
2
)
− c14 =
π
2
c = 0.8 in.
(0.84 − 0.64 ) = 0.4398 in 4
Tc
J
Jτ max
(0.4398)(7.5)
=
T =
c
0.8
τ max =
T = 4.12 kip ⋅ in 
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PROBLEM 3.4
Knowing that the internal diameter of the hollow shaft shown is d = 0.9in.,
determine the maximum shearing stress caused by a torque of magnitude
T = 9 kip ⋅ in.
SOLUTION
c2 =
1
1
d 2 =   (1.6) = 0.8 in.
2
2
c1 =
1
1
d1 =   (0.9) = 0.45 in.
2
2
(
)
π
c24 − c14 = (0.84 − 0.454 ) = 0.5790 in 4
2
2
(9)(0.8)
Tc
=
=
0.5790
J
J =
τ max
π
c = 0.8 in.
τ max = 12.44 ksi 
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PROBLEM 3.5
A torque T = 3 kN ⋅ m is applied to the solid bronze cylinder shown.
Determine (a) the maximum shearing stress, (b) the shearing stress at
point D which lies on a 15-mm-radius circle drawn on the end of the
cylinder, (c) the percent of the torque carried by the portion of the cylinder
within the 15 mm radius.
SOLUTION
(a)
c=
J =
1
d = 30 mm = 30 × 10−3 m
2
π
c4 =
2
π
2
(30 × 10−3 ) 4 = 1.27235 × 10−6 m 4
T = 3 kN = 3 × 103 N
τm =
Tc (3 × 103 )(30 × 10−3 )
=
= 70.736 × 106 Pa
J
1.27235 × 10−6
τ m = 70.7 MPa 
(b)
ρ D = 15 mm = 15 × 10−3 m
τD =
(c)
τD =
TD =
ρD
c
τ =
TD ρ D
JD
π
2
(15 × 10−3 )(70.736 × 10−6 )
(30 × 10−3 )
TD =
J Dτ D
ρD
=
π
2
τ D = 35.4 MPa 
ρ D3 τ D
(15 × 10−3 )3 (35.368 × 106 ) = 187.5 N ⋅ m
TD
187.5
× 100% =
(100%) = 6.25%
T
3 × 103
6.25%
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PROBLEM 3.6
(a) Determine the torque that can be applied to a solid shaft of 20-mm diameter without exceeding an
allowable shearing stress of 80 MPa. (b) Solve Part a, assuming that the solid shaft has been replaced by a
hollow shaft of the same cross-sectional area and with an inner diameter equal to half of its own outer
diameter.
SOLUTION
(a)
Solid shaft:
c=
J =
T =
(b)
Hollow shaft:
1
1
d = (0.020) = 0.010 m
2
2
π
c4 −
2
π
2
(0.10) 4 = 15.7080 × 10−9 m 4
Jτ max
(15.7080 × 10−9 )(80 × 106 )
=
= 125.664
0.010
c
T = 125.7 N ⋅ m 
Same area as solid shaft.
2

1   3
A = π c22 − c12 = π c22 −  c2   = π c22 = π c 2
 2   4

2
2
(0.010) = 0.0115470 m
c2 =
c=
3
3
1
c1 = c2 = 0.0057735 m
2
(
J =
T =
π
2
)
(c
4
2
τ max J
c2
)
− c14 =
=
π
2
(0.01154704 − 0.00577354 ) = 26.180 × 10−9 m 4
(80 × 106 )(26.180 × 10−9 )
= 181.38
0.0115470
T = 181.4 N ⋅ m 
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PROBLEM 3.7
The solid spindle AB has a diameter ds= 1.5 in. and is made of a steel with
an allowable shearing stress of 12 ksi, while sleeve CD is made of a brass
with an allowable shearing stress of 7 ksi. Determine the largest torque T
that can be applied at A.
SOLUTION
Analysis of solid spindle AB:
c=
1
d s = 0.75 in.
2
τ=
Tc
J
T =
Analysis of sleeve CD:
c2 =
π
2
T=
π
Jτ
= τ c3
c
2
(12 × 103 )(0.75)3 = 7.95 × 103 lb ⋅ in
1
1
d o = (3) = 1.5 in.
2
2
c1 = c2 − t = 1.5 − 0.25 = 1.25 in.
J=
T=
The smaller torque governs:

π
(c
2
4
2
)
− c14 =
π
2
(1.54 − 1.254 ) = 4.1172 in 4
Jτ
(4.1172)(7 × 103 )
=
= 19.21 × 103 lb ⋅ in
c2
1.5
T = 7.95 × 103 lb ⋅ in
T = 7.95 kip ⋅ in 
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PROBLEM 3.8
The solid spindle AB is made of a steel with an allowable shearing stress of
12 ksi, and sleeve CD is made of a brass with an allowable shearing stress
of 7 ksi. Determine (a) the largest torque T that can be applied at A if the
allowable shearing stress is not to be exceeded in sleeve CD, (b) the
corresponding required value of the diameter d s of spindle AB.
SOLUTION
(a)
Analysis of sleeve CD:
1
1
do = (3) = 1.5 in.
2
2
c1 = c2 − t = 1.5 − 0.25 = 1.25 in.
c2 =
J=
T=
π
(c
2
4
2
)
− c14 =
π
2
(1.54 − 1.254 ) = 4.1172 in 4
(4.1172)(7 × 103 )
Jτ
=
= 19.21 × 103 lb ⋅ in
1.5
c2
T = 19.21 kip ⋅ in 
(b)
Analysis of solid spindle AB:
τ=
Tc
J
π
19.21 × 103
J
T
= c3 = =
= 1.601 in 3
τ
2
c
12 × 103
c=3
(2)(1.601)
π
= 1.006 in.
d s = 2c
d = 2.01in. 
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PROBLEM 3.9
The torques shown are exerted on pulleys A and B. Knowing that both
shafts are solid, determine the maximum shearing stress (a) in shaft AB,
(b) in shaft BC.
SOLUTION
(a)
Shaft AB:
TAB = 300 N ⋅ m, d = 0.030 m, c = 0.015 m
τ max =
Tc
2T
(2)(300)
=
=
3
J
πc
π (0.015)3
= 56.588 × 106 Pa
(b)
Shaft BC:
τ max = 56.6 MPa 
TBC = 300 + 400 = 700 N ⋅ m
d = 0.046 m, c = 0.023 m
τ max =
Tc
2T
(2)(700)
=
=
3
J
πc
π (0.023)3
= 36.626 × 106 Pa
τ max = 36.6 MPa 
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PROBLEM 3.10
In order to reduce the total mass of the assembly of Prob. 3.9, a new design
is being considered in which the diameter of shaft BC will be smaller.
Determine the smallest diameter of shaft BC for which the maximum value
of the shearing stress in the assembly will not increase.
SOLUTION
Shaft AB:
TAB = 300 N ⋅ m, d = 0.030 m, c = 0.015 m
τ max =
2T
(2)(300)
Tc
=
=
3
J
πc
π (0.015)3
= 56.588 × 106 Pa = 56.6 MPa
Shaft BC:
TBC = 300 + 400 = 700 N ⋅ m
d = 0.046 m, c = 0.023 m
τ max =
2T
(2)(700)
Tc
=
=
3
J
πc
π (0.023)3
= 36.626 × 106 Pa = 36.6 MPa
The largest stress (56.588 × 106 Pa) occurs in portion AB.
Reduce the diameter of BC to provide the same stress.
Tc
2T
=
J
π c3
(2)(700)
=
= 7.875 × 10−6 m3
π (56.588 × 106 )
TBC = 700N ⋅ m
c3 =
2T
πτ max
τ max =
c = 19.895 × 10−3 m
d = 2c = 39.79 × 10−3 m
d = 39.8 mm 
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PROBLEM 3.11
Knowing that each portion of the shafts AB, BC,
and CD consist of a solid circular rod, determine
(a) the shaft in which the maximum shearing
stress occurs, (b) the magnitude of that stress.
SOLUTION
Shaft AB:
T = 48 N ⋅ m
1
d = 7.5 mm = 0.0075 m
2
Tc
2T
=
= 3
J
πc
(2) (48)
=
= 72.433MPa
π (0.0075)3
c=
τ max
τ max
Shaft BC:
T = − 48 + 144 = 96 N ⋅ m
c=
Shaft CD:
τ max =
Tc
2T
(2) (96)
= 3 =
= 83.835 MPa
J
πc
π (0.009)3
T = − 48 + 144 + 60 = 156 N⋅ m
c=
Answers:
1
d = 9 mm
2
1
d = 10.5 mm
2
τ max =
Tc
2T
(2 × 156)
= 3 =
= 85.79 MPa
J
πc
π (0.0105)3
(a) shaft CD
(b) 85.8 MPa 
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PROBLEM 3.12
Knowing that an 8-mm-diameter hole has been
drilled through each of the shafts AB, BC, and
CD, determine (a) the shaft in which the
maximum shearing stress occurs, (b) the
magnitude of that stress.
SOLUTION
1
d1 = 4 mm
2
Hole:
c1 =
Shaft AB:
T = 48 N ⋅ m
c2 =
J =
τ max =
Shaft BC:
τ max
2
(c
4
2
)
− c14 =
π
2
(0.00754 − 0.0044 ) = 4.5679 × 10−9 m 4
Tc2
(48)(0.0075)
=
= 78.810 MPa
J
4.5679 × 10−9
π
(c
2
)
c2 =
1
d 2 = 9 mm
2
π
(0.0094 − 0.0044 ) = 9.904 × 10−9 m 4
2
Tc
(96)(0.009)
= 2 =
= 87.239 MPa
J
9.904 × 10−9
4
2
− c14 =
T = − 48 + 144 + 60 = 156 N ⋅ m
J =
τ max =
Answers:
π
T = − 48 + 144 = 96 N ⋅ m
J =
Shaft CD:
1
d 2 = 7.5 mm
2
π
(c
2
4
2
)
− c14 =
π
2
c2 =
1
d 2 = 10.5 mm
2
(0.01054 − 0.0044 ) = 18.691 × 10−9 m 4
Tc2 (156)(0.0105)
=
= 87.636 MPa
J
18.691 × 10−9
(a) shaft CD
(b) 87.6 MPa 
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PROBLEM 3.13
Under normal operating conditions, the
electric motor exerts a 12-kip ⋅ in.
torque at E. Knowing that each shaft is
solid, determine the maximum shearing
in (a) shaft BC, (b) shaft CD, (c) shaft
DE.
SOLUTION
(a)
Shaft BC:
From free body shown:
TBC = 3 kip ⋅ in
τ =
τ =
(b)
Shaft CD:
Tc
Tc
2 T
=
=
3
π
J
c4 π c
2
3 kip ⋅ in
2
π 1

 × 1.75 in . 
2

3
(1)
τ = 2.85 ksi 
From free body shown:
TCD = 3 + 4 = 7 kip ⋅ in
From Eq. (1):
τ =
(c)
Shaft DE:
2 T
2 7 kip ⋅ in
=
3
π c π (1 in.)3
τ = 4.46 ksi 
From free body shown:
TDE = 12 kip ⋅ in
From Eq. (1):
τ =
2 T
2 12 kip ⋅ in
=
3
3
π c π 1

 × 2.25 in. 
2

τ = 5.37 ksi 
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PROBLEM 3.14
Solve Prob.3.13, assuming that a 1in.-diameter hole has been drilled
into each shaft.
PROBLEM 3.13 Under normal
operating conditions, the electric
motor exerts a 12-kip ⋅ in. torque at
E. Knowing that each shaft is solid,
determine the maximum shearing in
(a) shaft BC, (b) shaft CD, (c) shaft
DE.
SOLUTION
(a)
Shaft BC:
From free body shown:
c2 =
τ =
(b)
Shaft CD:
(c
2
)
− c14 =
π
2
2
(c
4
2
)
− c14 =
π
2
(1.04 − 0.54 ) = 1.47262 in 4
From free body shown:
τ =
TCD = 3 + 4 = 7 kip ⋅ in
Tc (7 kip ⋅ in)(1.0 in.)
=
J
1.47262 in 4
c2 =
τ = 3.19 ksi 
1
(2.0) = 1.0 in.
2
π
τ =
J =
1
(1) = 0.5 in.
2
(0.8754 − 0.54 ) = 0.82260 in 4
From free body shown:
J =
Shaft DE:
4
2
c1 =
Tc (3 kip ⋅ in)(0.875 in.)
=
J
0.82260 in 4
c2 =
(c)
1
(1.75) = 0.875 in.
2
π
J =
TBC = 3 kip ⋅ in
τ = 4.75 ksi 
TDE = 12 kip ⋅ in
2.25
= 1.125 in.
2
π
(c
2
4
2
)
− c14 =
π
2
(1.1254 − 0.54 ) = 2.4179 in 4
Tc (12 kip ⋅ in)(1.125 in.)
=
J
2.4179 in 4
τ = 5.58 ksi 
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PROBLEM 3.15
The allowable shearing stress is 15 ksi in the 1.5-in.-diameter steel rod AB and
8 ksi in the 1.8-in.-diameter brass rod BC. Neglecting the effect of stress
concentrations, determine the largest torque that can be applied at A.
SOLUTION
τ max =
Rod AB:
τ max = 15 ksi
T =
Rod BC:
π
π
Tc
, J = c 4 , T = c3τ max
J
2
2
π
2
π
2
The allowable torque is the smaller value.
1
d = 0.75 in.
2
(0.75)3 (15) = 9.94 kip ⋅ in
τ max = 8 ksi
T =
c=
c=
1
d = 0.90 in.
2
(0.90)3 (8) = 9.16 kip ⋅ in
T = 9.16 kip ⋅ in 
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PROBLEM 3.16
The allowable shearing stress is 15 ksi in the steel rod AB and 8 ksi in the
brass rod BC. Knowing that a torque of magnitude T = 10 kip ⋅ in. is applied
at A, determine the required diameter of (a) rod AB, (b) rod BC.
SOLUTION
τ max =
(a)
Rod AB:
Tc
π
2T
, J = , c3 =
J
2
πτ max
T = 10 kip ⋅ in
c3 =
τ max = 15 ksi
(2)(10)
= 0.4244 in 3
π (15)
c = 0.7515 in.
(b)
Rod BC:
T = 10 kip ⋅ in
c3 =
d = 2c = 1.503 in. 
τ max = 8 ksi
(2)(10)
= 0.79577 in 2
π (8)
c = 0.9267 in.
d = 2c = 1.853 in. 
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PROBLEM 3.17
The allowable stress is 50 MPa in the brass rod AB and
25 MPa in the aluminum rod BC. Knowing that a torque of
magnitude T = 1250 N ⋅ m is applied at A, determine the
required diameter of (a) rod AB, (b) rod BC.
SOLUTION
τ max =
(a)
Rod AB:
c3 =
Tc
J
J =
π
2
c4
c3 =
2T
πτ max
(2)(1250)
= 15.915 × 10−6 m3
π (50 × 106 )
c = 25.15 × 10−3 m = 25.15 mm
d AB = 2c = 50.3 mm 
(b)
Rod BC:
c3 =
(2)(1250)
= 31.831 × 10−6 m3
6
π (25 × 10 )
c = 31.69 × 10−3 m = 31.69 mm
d BC = 2c = 63.4 mm 
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PROBLEM 3.18
The solid rod BC has a diameter of 30 mm and is made of an
aluminum for which the allowable shearing stress is 25 MPa.
Rod AB is hollow and has an outer diameter of 25 mm; it is
made of a brass for which the allowable shearing stress is 50
MPa. Determine (a) the largest inner diameter of rod AB for
which the factor of safety is the same for each rod, (b) the
largest torque that can be applied at A.
SOLUTION
Solid rod BC:
τ =
Tc
J
J =
π
2
c4
τ all = 25 × 106 Pa
c=
Tall =
Hollow rod AB:
1
d = 0.015 m
2
π
2
c3τ all =
π
2
(0.015)3 (25 × 106 ) = 132.536 N ⋅ m
τ all = 50 × 106 Pa
Tall = 132.536 N ⋅ m
1
1
d 2 = (0.025) = 0.0125 m
2
2
c1 = ?
c2 =
Tall =
Jτ all
π 4
τ
c2 − c14 all
=
c2
2
c2
c14 = c24 −
(
2Tallc2
πτ all
= 0.01254 −
(2)(132.536)(0.0125)
= 3.3203 × 10−9 m 4
6
π (50 × 10 )
c1 = 7.59 × 10−3 m = 7.59 mm
(a)
(b)
)
Allowable torque.
d1 = 2c1 = 15.18 mm 
Tall = 132.5 N ⋅ m 
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PROBLEM 3.19
The solid rod AB has a diameter dAB = 60 mm. The pipe CD has
an outer diameter of 90 mm and a wall thickness of 6 mm.
Knowing that both the rod and the pipe are made of a steel for
which the allowable shearing stress is 75 Mpa, determine the
largest torque T that can be applied at A.
SOLUTION
τ all = 75 × 106 Pa Tall =
Rod AB:
c=
Tall =
1
d = 0.030 m
2
π
2
c3τ all =
π
2
Jτ all
c
J=
π
2
c4
(0.030)3 (75 × 106 )
= 3.181 × 103 N ⋅ m
Pipe CD:
c2 =
J =
Tall =
1
d 2 = 0.045 m
2
π
(c
2
4
2
)
− c14 =
c1 = c2 − t = 0.045 − 0.006 = 0.039 m
π
2
(0.0454 − 0.0394 ) = 2.8073 × 10−6 m 4
(2.8073 × 10−6 )(75 × 106 )
= 4.679 × 103 N ⋅ m
0.045
Allowable torque is the smaller value.
Tall = 3.18 × 103 N ⋅ m
3.18 kN ⋅ m 
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PROBLEM 3.20
The solid rod AB has a diameter dAB = 60 mm and is made of a
steel for which the allowable shearing stress is 85 Mpa. The
pipe CD, which has an outer diameter of 90 mm and a wall
thickness of 6 mm, is made of an aluminum for which the
allowable shearing stress is 54 MPa. Determine the largest
torque T that can be applied at A.
SOLUTION
Rod AB:
τ all = 85 × 106 Pa
Tall =
=
Pipe CD:
c=
1
d = 0.030 m
2
π
Jτ all
= c3 τ all
2
c
π
2
(0.030)3 (85 × 106 ) = 3.605 × 103 N ⋅ m
τ all = 54 × 106 Pa
c2 =
1
d 2 = 0.045 m
2
c1 = c2 − t = 0.045 − 0.006 = 0.039 m
J =
Tall =
π
(c
2
4
2
)
− c14 =
π
2
(0.0454 − 0.0394 ) = 2.8073 × 10−6 m 4
Jτ all (2.8073 × 10−6 )(54 × 106 )
=
= 3.369 × 103 N ⋅ m
0.045
c2
Allowable torque is the smaller value.
Tall = 3.369 × 103 N ⋅ m
3.37 kN ⋅ m 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 3.21
A torque of magnitude T = 1000 N ⋅ m is applied at
D as shown. Knowing that the diameter of shaft AB
is 56 mm and that the diameter of shaft CD is
42 mm, determine the maximum shearing stress in
(a) shaft AB, (b) shaft CD.
SOLUTION
TCD = 1000 N ⋅ m
TAB =
(a)
(b)
Shaft AB:
Shaft CD:
rB
100
TCD =
(1000) = 2500 N ⋅ m
rC
40
c=
1
d = 0.028 m
2
τ=
Tc
2T
(2) (2500)
= 3=
= 72.50 × 106
J
π c π (0.028)3
c=
1
d = 0.020 m
2
τ=
Tc
2T
(2) (1000)
= 3=
= 68.7 × 106
J
π c π (0.020)3
72.5 MPa 
68.7 MPa 
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PROBLEM 3.22
A torque of magnitude T = 1000 N ⋅ m is applied
at D as shown. Knowing that the allowable
shearing stress is 60 MPa in each shaft, determine
the required diameter of (a) shaft AB, (b) shaft CD.
SOLUTION
TCD = 1000 N ⋅ m
TAB =
(a)
Shaft AB:
rB
100
TCD =
(1000) = 2500 N ⋅ m
rC
40
τ all = 60 × 106 Pa
τ =
Tc
2T
= 3
J
πc
c3 =
2T
πτ
=
(2) (2500)
= 26.526 × 10−6 m3
π (60 × 106 )
c = 29.82 × 10−3 = 29.82 mm
(b)
Shaft CD:
d = 2c = 59.6 mm 
τ all = 60 × 106 Pa
τ =
Tc
2T
= 3
J
πc
c3 =
2T
πτ
=
(2) (1000)
= 10.610 × 10−6 m3
π (60 × 106 )
c = 21.97 × 10−3 m = 21.97 mm
d = 2c = 43.9 mm 
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PROBLEM 3.23
Under normal operating conditions, a motor exerts a
torque of magnitude TF = 1200 lb ⋅ in. at F.
Knowing that rD = 8 in., rG = 3 in., and the
allowable shearing stress is 10.5 ksi ⋅ in each shaft,
determine the required diameter of (a) shaft CDE,
(b) shaft FGH.
SOLUTION
TF = 1200 lb ⋅ in
TE =
rD
8
TF = (1200) = 3200 lb ⋅ in
rG
3
τ all = 10.5 ksi = 10500 psi
τ =
(a)
2T
Tc
= 3
J
πc
c3 =
2T
πτ
Shaft CDE:
c3 =
(2) (3200)
= 0.194012 in 3
π (10500)
c = 0.5789 in.
(b)
or
d DE = 2c
d DE = 1.158 in. 
Shaft FGH:
c3 =
(2) (1200)
π (10500)
c = 0.4174 in.
= 0.012757 in 3
d FG = 2c
d FG = 0.835 in. 
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PROBLEM 3.24
Under normal operating conditions, a motor exerts a
torque of magnitude TF at F. The shafts are made of a
steel for which the allowable shearing stress is 12 ksi
and have diameters dCDE = 0.900 in. and
dFGH = 0.800 in. Knowing that rD = 6.5 in. and
rG = 4.5 in., determine the largest allowable value of
TF.
SOLUTION
τ all = 12 ksi
Shaft FG:
1
d = 0.400 in.
2
π
Jτ
= all = c3τ all
c
2
c=
TF , all
=
Shaft DE:
c=
TE , all =
π
2
(0.400)3 (12) = 1.206 kip ⋅ in
1
d = 0.450 in.
2
π
2
c3τ all
π
(0.450)3 (12) = 1.7177 kip ⋅ in
2
r
4.5
TF = G TE TF , all =
(1.7177) = 1.189 kip ⋅ in
rD
6.5
=
Allowable value of TF is the smaller.
TF = 1.189 kip ⋅ in 
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PROBLEM 3.25
The two solid shafts are connected by gears as shown and are made
of a steel for which the allowable shearing stress is 8500 psi.
Knowing that a torque of magnitude TC = 5 kip ⋅ in. is applied at C
and that the assembly is in equilibrium, determine the required
diameter of (a) shaft BC, (b) shaft EF.
SOLUTION
τ max = 8500 psi = 8.5 ksi
(a)
Shaft BC:
TC = 5 kip ⋅ in
τ max =
c=
Tc
2T
=
J
π c3
3
c=
3
2T
πτ max
(2)(5)
= 0.7208 in.
π (8.5)
d BC = 2c = 1.442 in. 
(b)
Shaft EF:
TF =
c=
rD
2.5
TC =
(5) = 3.125 kip ⋅ in
rA
4
3
2T
πτ max
=
3
(2)(3.125)
= 0.6163 in.
π (8.5)
d EF = 2c = 1.233 in. 
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PROBLEM 3.26
The two solid shafts are connected by gears as shown and are
made of a steel for which the allowable shearing stress is
7000 psi. Knowing the diameters of the two shafts are,
respectively, d BC = 1.6 in. and d EF = 1.25 in., determine the
largest torque TC that can be applied at C.
SOLUTION
τ max = 7000 psi = 7.0 ksi
Shaft BC:
d BC = 1.6 in.
1
d = 0.8 in.
2
π
Jτ max
TC =
= τ max c3
2
c
c=
=
Shaft EF:
π
2
(7.0)(0.8)3 = 5.63 kip ⋅ in
d EF = 1.25 in.
1
d = 0.625 in.
2
π
Jτ max
TF =
= τ max c3
c
2
c=
=
By statics,
TC =
π
2
(7.0)(0.625)3 = 2.684 kip ⋅ in
rA
4
TF =
(2.684) = 4.30 kip ⋅ in
rD
2.5
Allowable value of TC is the smaller.
TC = 4.30 kip ⋅ in 
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PROBLEM 3.27
A torque of magnitude T = 100 N ⋅ m is applied to shaft AB of the
gear train shown. Knowing that the diameters of the three solid
shafts are, respectively, d AB = 21 mm, dCD = 30 mm, and
d EF = 40 mm, determine the maximum shearing stress in
(a) shaft AB, (b) shaft CD, (c) shaft EF.
SOLUTION
Statics:
Shaft AB:
Gears B and C:
Force on gear circles.
TAB = TA = TB = T
rB = 25 mm, rC = 60 mm
FBC =
TC =
Shaft CD:
Gears D and E:
Force on gear circles.
rC
60
TB =
T = 2.4 T
rB
25
TCD = TC = TD = 2.4 T
rD = 30 mm, rE = 75 mm
FDE =
TE =
Shaft EF:
TB
T
= C
rB
rC
TD
T
= E
rD
rE
rE
75
TD =
(2.4 T ) = 6 T
rD
30
TEF = TE = TF = 6T
Maximum Shearing Stresses. τ max =
Tc
2T
=
J
π c3
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PROBLEM 3.27 (Continued)
(a)
(b)
Shaft AB:
Shaft CD:
T = 100 N ⋅ m
c=
1
d = 10.5 mm = 10.5 × 10−3 m
2
τ max =
(2)(100)
= 55.0 × 106 Pa
π (10.5 × 10−3 )3
T = (2.4)(100) = 240 N ⋅ m
c=
τ max =
(c)
Shaft EF:
τ max = 55.0 MPa 
1
d = 15 mm = 15 × 10−3 m
2
(2)(240)
= 45.3 × 106 Pa
−3 3
π (15 × 10 )
τ max = 45.3 MPa 
T = (6)(100) = 600 N ⋅ m
c=
τ max =
1
d = 20 mm = 20 × 10−3 m
2
(2)(600)
π (20 × 10−3 )3
= 47.7 × 106 Pa
τ max = 47.7 MPa 
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PROBLEM 3.28
A torque of magnitude T = 120 N ⋅ m is applied to shaft AB of
the gear train shown. Knowing that the allowable shearing stress
is 75 MPa in each of the three solid shafts, determine the
required diameter of (a) shaft AB, (b) shaft CD, (c) shaft EF.
SOLUTION
Statics:
Shaft AB:
Gears B and C:
Force on gear circles.
TAB = TA = TB = T
rB = 25 mm, rC = 60 mm
FBC =
TC =
Shaft CD:
Gears D and E:
Force on gear circles.
rC
60
TB =
T = 2.4 T
rB
25
TCD = TC = TD = 2.4 T
rD = 30 mm, rE = 75 mm
FDE =
TE =
Shaft EF:
TB
T
= C
rB
rC
TD
T
= E
rD
rE
rE
75
TD =
(2.4 T ) = 6 T
rD
30
TEF = TE = TF = 6 T
Required Diameters.
τ max =
c=
2T
Tc
=
J
π c3
3
2T
πτ
d = 2c = 2 3
2T
πτ max
τ max = 75 × 106 Pa
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PROBLEM 3.28 (Continued)
(a)
Shaft AB:
TAB = T = 120 N ⋅ m
d AB = 2 3
(b)
Shaft CD:
Shaft EF:
d AB = 20.1 mm 
TCD = (2.4)(120) = 288 N ⋅ m
dCD = 2 3
(c)
2(120)
= 20.1 × 10−3 m
π (75 × 106 )
(2)(288)
π (75 × 106 )
= 26.9 × 10−3 m
dCD = 26.9 mm 
TEF = (6)(120) = 720 N ⋅ m
d EF = 2 3
(2)(720)
= 36.6 × 10−3 m
3
π (75 × 10 )
d EF = 36.6 mm 
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PROBLEM 3.29
(a) For a given allowable shearing stress, determine the ratio T/w of the
maximum allowable torque T and the weight per unit length w for the
hollow shaft shown. (b) Denoting by (T /w)0 the value of this ratio for a
solid shaft of the same radius c2 , express the ratio T/w for the hollow shaft
in terms of (T /w)0 and c1 /c2.
SOLUTION
w = weight per unit length,
ρ g = specific weight,
W = total weight,
L = length
w=
Tall =
(
)
(a)
T
= c12 + c22 τ all
W

c1 = 0 for solid shaft:
(b)
(T /w) h
c2
= 1 + 12
(T /w)0
c2
W
ρ gLA
=
= ρ gA = ρ gπ c22 − c12
L
L
(
(
)(
)
)
2
2
2
2
π c24 − c14
π c2 + c1 c2 − c1
Jτ all
τ all =
τ all
=
2 c2
2
c2
c2
(
)
c12 + c22 τ all
T
(hollow shaft) 
=
2ρ gc2
w
c2τ all
T 
(solid shaft)
  =
2 pg
 w 0
c12 
T  T  
=
1
+

 
    
c22 
 w   w 0 
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PROBLEM 3.30
While the exact distribution of the shearing stresses in a
hollow-cylindrical shaft is as shown in Fig. a, an approximate
value can be obtained for τmax by assuming that the stresses
are uniformly distributed over the area A of the cross section,
as shown in Fig. b, and then further assuming that all of the
elementary shearing forces act at a distance from O equal to
the mean radius ½(c1 + c2 ) of the cross section. This
approximate value τ 0 = T /Arm , where T is the applied torque.
Determine the ratio τ max /τ 0 of the true value of the maximum
shearing stress and its approximate value τ 0 for values
of c1 /c2 , respectively equal to 1.00, 0.95, 0.75, 0.50, and 0.
SOLUTION
For a hollow shaft:
τ max =
τ0 =
By definition,
Tc2
2Tc2
2Tc2
2Tc2
=
=
=
4
4
2
2
2
2
J
π c2 − c1
π c2 − c1 c2 + c1
A c22 + c12
(
)
(
)(
)
(
T
2T
=
Arm
A(c2 + c1)
τ max
c (c + c )
1 + (c1 /c2 )
= 2 2 2 21 =
τ0
c2 + c1
1 + (c1 /c2 ) 2
Dividing,
)

c1 /c2
1.0
0.95
0.75
0.5
0.0
τ max /τ 0
1.0
1.025
1.120
1.200
1.0

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PROBLEM 3.31
(a) For the solid steel shaft shown (G = 77 GPa), determine the angle
of twist at A. (b) Solve part a, assuming that the steel shaft is hollow
with a 30-mm-outer diameter and a 20-mm-inner diameter.
SOLUTION
(a)
c=
1
d = 0.015 m,
2
−9
J =
π
2
c4 =
π
2
(0.015) 4
J = 79.522 × 10 m , L = 1.8 m, G = 77 × 109 Pa
T = 250 N ⋅ m
ϕ =
ϕ =
(b)
4
ϕ =
TL
GJ
(250)(1.8)
= 73.49 × 10−3 rad
9
−9
(77 × 10 )(79.522 × 10 )
(73.49 × 10−3 )180
ϕ = 4.21° 
π
1
π 4
c2 = 0.015 m, c1 d1 = 0.010 m, J =
c2 − c14
2
2
π
TL
J = (0.0154 − 0.0104 ) = 63.814 × 10−9 m 4 ϕ
2
GJ
(
ϕ =
)
(250)(1.8)
180
= 91.58 × 10−3 rad =
(91.58 × 10−3 )
9
−9
π
(77 × 10 )(63.814 × 10 )
ϕ = 5.25° 

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PROBLEM 3.32
For the aluminum shaft shown (G = 27 GPa), determine (a) the torque T
that causes an angle of twist of 4°, (b) the angle of twist caused by the same
torque T in a solid cylindrical shaft of the same length and cross-sectional
area.
SOLUTION
TL
GJ
ϕ =
(a)
T =
GJ ϕ
L
ϕ = 4° = 69.813 × 10−3 rad, L = 1.25 m
G = 27 GPa = 27 × 109 Pa
J =
π
(c
2
4
2
)
− c14 =
π
2
(0.0184 − 0.0124 ) = 132.324 × 10−9 m 4
(27 × 109 )(132.324 × 109 )(69.813 × 10−3 )
1.25
= 199.539 N⋅ m
T =
(b)
Matching areas:
(
A = π c 2 = π c22 − c12
T = 199.5 N ⋅ m 
)
c = c22 − c12 = 0.0182 − 0.0122 = 0.013416 m
J =
ϕ =
π
2
c4 =
π
2
(0.013416) 4 = 50.894 × 10−9 m 4
TL
(195.539)(1.25)
=
= 181.514 × 10−3 rad
GJ (27 × 109 )(50.894 × 109 )
ϕ = 10.40° 
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PROBLEM 3.33
Determine the largest allowable diameter of a 10-ft-long steel rod (G = 11.2 × 106 psi) if the rod is to be twisted
through 30° without exceeding a shearing stress of 12 ksi.
SOLUTION
L = 10 ft = 120 in.
ϕ = 30° =
30π
= 0.52360 rad
180
τ = 12 ksi = 12 × 103 psi
TL
GJ ϕ
Tc GJ ϕ c Gϕ c
τL
, T =
, τ =
, c=
=
=
ϕ =
GJ
L
J
JL
L
Gϕ
c=
(12 × 103 )(120)
= 0.24555 in.
(11.2 × 106 )(0.52360)
d = 2c = 0.491 in. 
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PROBLEM 3.34
While an oil well is being drilled at a depth of 6000 ft, it is observed that the top of the 8-in.-diameter steel
drill pipe rotates though two complete revolutions before the drilling bit starts to rotate. Using
G = 11.2 × 106 psi, determine the maximum shearing stress in the pipe caused by torsion.
SOLUTION
For outside diameter of 8 in., c = 4 in.
For two revolutions, ϕ = 2(2π ) = 4π radians.
G = 11.2 × 106 psi
L = 6000 ft = 72000 in.
From text book,
TL
GJ
Tc
τm =
J
ϕ =
Divide (2) by (1).
(1)
(2)
τ m Gc
=
L
ϕ
τm =
Gcϕ (11 × 106 )(4)(4π )
=
= 7679 psi
72000
L
τ m = 7.68 ksi 
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PROBLEM 3.35
The electric motor exerts a 500 N ⋅ m torque on
the aluminum shaft ABCD when it is rotating at a
constant speed. Knowing that G = 27 GPa and that
the torques exerted on pulleys B and C are as shown,
determine the angle of twist between (a) B and C,
(b) B and D.
SOLUTION
(a)
Angle of twist between B and C.
TBC = 200 N ⋅ m, LBC = 1.2 m
c=
J BC =
ϕ B/C =
(b)
1
d = 0.022 m, G = 27 × 109 Pa
2
π
2
c 4 = 367.97 × 10−9 m
TL
(200)(1.2)
=
= 24.157 × 10−3 rad
GJ
(27 × 109 )(367.97 × 109 )
ϕ B /C = 1.384° 
Angle of twist between B and D.
TCD = 500 N ⋅ m, LCD = 0.9 m, c =
J CD =
ϕC /D
π
2
c4 =
π
1
d = 0.024 m, G = 27 × 109 Pa
2
(0.024) 4 = 521.153 × 10−9 m 4
2
(500)(0.9)
=
= 31.980 × 10−3 rad
(27 × 109 )(521.153 × 109 )
ϕ B /D = ϕ B /C + ϕC /D = 24.157 × 10−3 + 31.980 × 10−3 = 56.137 × 10−3 rad
ϕ B /D = 3.22° 
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PROBLEM 3.36
The torques shown are exerted on pulleys B, C,
and D. Knowing that the entire shaft is made of
steel (G = 27 GPa), determine the angle of
twist between (a) C and B, (b) D and B.
SOLUTION
(a)
Shaft BC:
c=
J BC =
1
d = 0.015 m
2
π
4
c 4 = 79.522 × 10−9 m 4
LBC = 0.8 m, G = 27 × 109 Pa
ϕ BC =
(b)
Shaft CD:
TL
(400)(0.8)
=
= 0.149904 rad
GJ
(27 × 109 )(79.522 × 10−9 )
ϕ BC = 8.54° 
1
π
d = 0.018 m
J CD = c 4 = 164.896 × 10−9 m 4
2
4
= 1.0 m TCD = 400 − 900 = −500 N ⋅ m
c=
LCD
ϕCD =
(−500)(1.0)
TL
=
= −0.11230 rad
GJ
(27 × 109 )(164.896 × 10−9 )
ϕ BD = ϕ BC + ϕCD = 0.14904 − 0.11230 = 0.03674 rad
ϕ BD = 2.11° 
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PROBLEM 3.37
The aluminum rod BC (G = 26 GPa) is bonded to the brass rod AB
(G = 39 GPa). Knowing that each rod is solid and has a diameter of
12 mm, determine the angle of twist (a) at B, (b) at C.
SOLUTION
Both portions:
c=
1
d = 6 mm = 6 × 10−3 m
2
π
π
c 4 = (6 × 10−3 )4 = 2.03575 × 10−9 m 4
2
2
T = 100 N ⋅ m
J =
Rod AB:
(a)
ϕ B = ϕ AB =
Rod BC:
GAB = 39 × 109 Pa, LAB = 0.200 m
TLAB
(100)(0.200)
=
= 0.25191 rad
G AB J
(39 × 109 )(2.03575 × 10−9 )
GBC = 26 × 109 Pa, LBC = 0.300 m
ϕ BC =
(b)
ϕ B = 14.43° 
TLBC
(100)(0.300)
=
= 0.56679 rad
GBC J
(26 × 109 )(2.03575 × 10−9 )
ϕC = ϕ B + ϕ BC = 0.25191 + 0.56679 = 0.81870 rad
ϕC = 46.9° 
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PROBLEM 3.38
The aluminum rod AB (G = 27 GPa) is bonded to the brass
rod BD (G = 39GPa). Knowing that portion CD of the
brass rod is hollow and has an inner diameter of 40 mm,
determine the angle of twist at A.
SOLUTION
G = 27 × 109 Pa, L = 0.400 m
Rod AB:
T = 800 N ⋅ m c =
Part BC:
π
(0.018) 4 = 164.896 × 10−9 m
2
2
TL
(800)(0.400)
=
=
= 71.875 × 10−3 rad
9
−9
GJ
(27 × 10 )(164.896 × 10 )
J =
ϕ A/B
π
1
d = 0.018 m
2
G = 39 × 109 Pa
c4 =
L = 0.375 m, c =
T = 800 + 1600 = 2400 N ⋅ m, J =
ϕ B /C
1
d = 0.030 m
2
π
c4 =
π
(0.030) 4 = 1.27234 × 10−6 m 4
2
2
(2400)(0.375)
TL
=
=
= 18.137 × 10−3 rad
GJ
(39 × 109 )(1.27234 × 10−6 )
1
d1 = 0.020 m
2
1
c2 = d 2 = 0.030 m, L = 0.250 m
2
c1 =
Part CD:
ϕC / D
Angle of twist at A.
π
(
)
π
c24 − c14 = (0.0304 − 0.0204 ) = 1.02102 × 10−6 m 4
2
2
(2400)(0.250)
TL
=
=
= 15.068 × 10−3 rad
9
−6
GJ
(39 × 10 )(1.02102 × 10 )
J =
ϕ A = ϕ A/B + ϕ B/C + ϕC/D
= 105.080 × 10−3 rad
ϕ A = 6.02° 
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PROBLEM 3.39
The solid spindle AB has a diameter d s =1.5 in. and is made of a steel with
G = 11.2 × 106 psi and τ all = 12 ksi, while sleeve CD is made of a brass
with G = 5.6 × 106 psi and τ all = 7 ksi. Determine the angle through end
A can be rotated.
SOLUTION
Stress analysis of solid spindle AB:
c=
1
ds = 0.75 in.
2
τ =
Tc
J
(12 × 103 )(0.75)3 = 7.95 × 103 lb ⋅ in
2
1
1
c2 = do = (3) = 1.5in.
2
2
c1 = c2 − t = 1.5 − 0.25 = 1.25in.
J =
π
T =
The smaller torque governs.
Deformation of spindle AB:
(c
2
)
− c14 =
π
2
(1.54 − 1.254 ) = 4.1172in 4
T = 7.95 × 103 lb ⋅ in
c = 0.75 in.
ϕ AB =
π
2
c 4 = 0.49701 in 4 , L = 12 in., G = 11.2 × 106 psi
TL
(7.95 × 103 ) (12)
=
= 0.017138 radians
GJ
(11.2 × 106 )(0.49701)
J = 4.1172 in 4 , L = 8 in., G = 5.6 × 106 psi
ϕCD =
Total angle of twist:
4
2
Jτ
(4.1172)(7 × 10−3 )
=
= 19.21 × 103 lb ⋅ in
c2
1.5
J =
Deformation of sleeve CD:
Jτ
π
= τ c3
c
2
π
T =
Stress analysis of sleeve CD:
T =
TL
(7.95 × 103 ) (8)
=
= 0.002758 radians
GJ
(5.6 × 106 ) (4.1172)
ϕ AD = ϕ AB + ϕCD = 0.019896 radians
ϕ AD = 1.140° 
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PROBLEM 3.40
The solid spindle AB has a diameter d s = 1.75in. and is made of
a steel with G = 11.2 × 106 psi and τ all = 12 ksi, while sleeve
CD is made of a brass with G = 5.6 × 106 psi and τ all = 7 ksi.
Determine (a) the largest torque T that can be applied at A if the
given allowable stresses are not to be exceeded and if the angle of
twist of sleeve CD is not to exceed 0.375°, (b) the corresponding
angle through which end A rotates.
SOLUTION
Spindle AB:
c=
J =
Sleeve CD:
π
2
c4 =
π
2
L = 12 in., τ all = 12 ksi, G = 11.2 × 106 psi
(0.875)4 = 0.92077in 4
c1 = 1.25 in., c2 = 1.5 in., L = 8 in., τ all = 7 ksi
J =
(a)
1
(1.75in.) = 0.875 in.
2
π
2
(c
4
2
)
− c14 = 4.1172 in 4 ,
G = 5.6 × 106 psi
Largest allowable torque T.
Ciriterion: Stress in spindle AB.
Critrion: Stress in sleeve CD.
Jτ
c
τ =
Tc
J
T=
(0.92077)(12)
= 12.63 kip ⋅ in
0.875
T=
Criterion: Angle of twist of sleeve CD
T =
Jτ 4.1172 in 4
=
(7 ksi)
c2
1.5 in.
T = 19.21 kip ⋅ in
φ = 0.375° = 6.545 × 10−3 rad
φ =
TL
JG
T =
JG
(4.1172)(5.6 × 106 )
φ =
(6.545 × 10−3 )
L
8
T = 18.86 kip ⋅ in
T = 12.63 kip ⋅ in 
The largest allowable torque is
(b)
Angle of rotation of end A.
φ A = φ A / D = φ A / B + φC / D = 
Ti Li
=T
J i Gi

Li
J i Gi


12
8
= (12.63 × 103 ) 
+
6
6 
 (0.92077)(11.2 × 10 ) (4.1172)(5.6 × 10 ) 
= 0.01908 radians
ϕ A = 1.093° 
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PROBLEM 3.41
Two shafts, each of 78 -in. diameter, are
connected by the gears shown. Knowing
that G = 11.2 × 106 psi and that the shaft
at F is fixed, determine the angle through
which end A rotates when a 1.2 kip ⋅ in.
torque is applied at A.
SOLUTION
Calculation of torques.
Circumferential contact force between gears B and E.
F =
TAB
T
= EF
rB
rE
TEF =
rE
TAB
rB
TAB = 1.2 kip ⋅ in = 1200 lb ⋅ in
TEF =
6
(1200) = 1600 lb ⋅ in
4.5
Twist in shaft FE.
L = 12 in., c =
J =
ϕ E /F =
π
2
4
c =
ϕB =
−3 4
  = 57.548 × 10 in
2  16 
δ = rEϕ E = rBϕ B
rE
6
(29.789 × 10−3 ) = 39.718 × 10−3 rad
ϕE =
4.5
rB
L = 8 + 6 = 14 in.
ϕ A/B =
Rotation at A.
4
ϕ E = ϕ E/F = 29.789 × 10−3 rad
Tangential displacement at gear circle.
Twist in shaft BA.
π7
TL
(1600)(12)
=
= 29.789 × 10−3 rad
−3
6
GJ
(11.2 × 10 )(57.548 × 10 )
Rotation at E.
Rotation at B.
1
7
d =
in., G = 11.2 × 106 psi
2
16
J = 57.548 × 10−3 in 4
(1200)(14)
TL
=
= 26.065 × 10−3 rad
GJ
(11.2 × 106 )(57.548 × 10−3 )
ϕ A = ϕ B + ϕ A/B = 65.783 × 10−3 rad
ϕ A = 3.77° 
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PROBLEM 3.42
Two solid shafts are connected by gears as shown. Knowing
that G = 77.2 GPa for each shaft, determine the angle
through which end A rotates when TA = 1200 N ⋅ m.
SOLUTION
Calculation of torques:
Circumferential contact force between gears B and C.
TAB = 1200 N ⋅ m
Twist in shaft CD:
c=
J =
ϕC/D
Rotation angle at C.
TCD =
1
d = 0.030 m,
2
π
c4 =
π
F =
TAB
T
= CD
rB
rC
Twist in shaft AB:
L = 1.2 m, G = 77.2 × 109 Pa
(0.030) 4 = 1.27234 × 10−6 m 4
2
2
(3600)(1.2)
TL
=
=
= 43.981 × 10−3 rad
GJ
(77.2 × 109 )(1.27234 × 10−9 )
ϕC = ϕC /D = 43.981 × 10−3 rad
ϕB =
c=
J =
ϕ A/B
Rotation angle at A.
rC
TAB
rB
240
(1200) = 3600 N ⋅ m
80
Circumferential displacement at contact points of gears B and C.
Rotation angle at B.
TCD =
δ = rCϕC = rBϕ B
rC
240
(43.981 × 10−3 ) = 131.942 × 10−3 rad
ϕC =
80
rB
1
d = 0.021m, L = 1.6 m, G = 77.2 × 109 Pa
2
π
c4 =
π
(0.021)4 = 305.49 × 109 m 4
2
2
(1200)(1.6)
TL
=
=
= 81.412 × 10−3 rad
GJ
(77.2 × 109 )(305.49 × 10−9 )
ϕ A = ϕ B + ϕ A/B = 213.354 × 10−3 rad
ϕ A = 12.22° 
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PROBLEM 3.43
A coder F, used to record in digital form the rotation
of shaft A, is connected to the shaft by means of the
gear train shown, which consists of four gears and
three solid steel shafts each of diameter d. Two of the
gears have a radius r and the other two a radius nr. If
the rotation of the coder F is prevented, determine in
terms of T, l, G, J, and n the angle through which end
A rotates.
SOLUTION
TAB = TA
TCD =
rC
T
T
TAB = AB = A
rB
n
n
TEF =
rE
T
T
TCD = CD = A2
rD
n
n
TEF lEF
T l
= 2A
GJ
n GJ
ϕ
r
Tl
= E ϕE = E = 3 A
rD
n
n GJ
ϕ E = ϕ EF =
ϕD
TCDlCD
T l
= A
GJ
nGJ
1
Tl
T l
T l 1
ϕC = ϕ D + ϕCD = 3 A + A = A  3 + 
GJ  n
n
n GJ nGJ
ϕCD =
1 
ϕ
rC
Tl  1
ϕC = C =
 4 + 2
rB
n
GJ  n
n 
T l
T l
= AB AB = A
GJ
GJ
ϕB =
ϕ AB
ϕ A = ϕ B + ϕ AB =
TAl  1
1

+ 2 + 1 

4
GJ  n
n

 
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PROBLEM 3.44
For the gear train described in Prob. 3.43, determine
the angle through which end A rotates when
T = 5lb ⋅ in.,
l = 2.4in.,
d = 1/16in.,
G = 11.2 × 106 psi, and n = 2.
PROBLEM 3.43 A coder F, used to record in digital
form the rotation of shaft A, is connected to the shaft
by means of the gear train shown, which consists of
four gears and three solid steel shafts each of diameter
d. Two of the gears have a radius r and the other two a
radius nr. If the rotation of the coder F is prevented,
determine in terms of T, l, G, J, and n the angle
through which end A rotates.
SOLUTION
See solution to Prob. 3.43 for development of equation for ϕ A.
ϕA =
Data:
Tl 
1
1 
1+ 2 + 4

GJ 
n
n 
T = 5 lb ⋅ in, l = 2.4 in., c =
n = 2, J =
ϕA =
π
2
c4 =
π 1 
1
1
d =
in., G = 11.2 × 106 psi
2
32
4
−6 4
  = 1.49803 × 10 in
2  32 
(5)(2.4)
1
1 

1 + +  = 938.73 × 10−3 rad 
−6 
4 16 
(11.2 × 10 )(1.49803 × 10 ) 
6
ϕ A = 53.8° 
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PROBLEM 3.45
The design of the gear-and-shaft system shown
requires that steel shafts of the same diameter be
used for both AB and CD. It is further required
that τ max ≤ 60 MPa, and that the angle φD through
which end D of shaft CD rotates not exceed 1.5°.
Knowing that G = 77 GPa, determine the required
diameter of the shafts.
SOLUTION
TCD = TD = 1000 N ⋅ m
For design based on stress, use larger torque.
TAB =
rB
100
TCD =
(1000) = 2500 N ⋅ m
rC
40
TAB = 2500 N ⋅ m
Tc
2T
=
J
π c3
2T
(2)(2500)
c3 =
=
= 26.526 × 10−6 m3
πτ
π (60 × 106 )
τ =
c = 29.82 × 10−3 m = 29.82 mm , d = 2c = 59.6 mm
Design based on rotation angle.
Shaft AB:
ϕ D = 1.5° = 26.18 × 10−3 rad
TAB = 2500 N ⋅ m,
L = 0.4 m
TL
(2500)(0.4) 1000
=
=
GJ
GJ
GJ
1000

ϕ B = ϕ AB = GJ

Gears 
ϕC = rB ϕ B =  100  1000  = 2500




rC
GJ
 40  GJ 
ϕ AB =
Shaft CD:
TCD = 1000 N ⋅ m,
L = 0.6 m
TL
(1000) (0.6) 600
=
=
GJ
GJ
GJ
2500 600 3100
3100
= ϕC + ϕCD =
+
=
= π 4
GJ
GJ
GJ
G2c
ϕCD =
ϕD
c4 =
(2) (3100)
(2)(3100)
=
= 979.06 × 10−9 m 4
π Gϕ D
π (77 × 109 )(26.18 × 10−3 )
c = 31.46 × 10−3 m = 31.46 mm,
Design must use larger value for d.
d = 2c = 62.9 mm
d = 62.9 mm 
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PROBLEM 3.46
The electric motor exerts a torque of
800 N ⋅ m on the steel shaft ABCD when it
is rotating at constant speed. Design
specifications require that the diameter of
the shaft be uniform from A to D and that
the angle of twist between A to D not
exceed 1.5°. Knowing that τ max ≤ 60 MPa
and G = 77 GPa, determine the minimum
diameter shaft that can be used.
SOLUTION
Torques:
TAB = 300 + 500 = 800 N ⋅ m
TBC = 500 N ⋅ m,
TCD = 0
τ = 60 × 106 Pa
Design based on stress.
τ =
Tc
2T
=
J
π c3
c3 =
2T
πτ
=
(2)(800)
= 8.488 × 10−6 m3
6
π (60 × 10 )
−3
c = 20.40 × 10 m = 20.40 mm,
Design based on deformation.
d = 2c = 40.8 mm
ϕ D/ A = 1.5° = 26.18 × 10−3 rad
ϕD / C = 0
(500)(0.6) 300
TBC LBC
=
=
GJ
GJ
GJ
(800) (0.4) 320
T L
= AB AB =
=
GJ
GJ
GJ
620
620
(2)(620)
= ϕ D / C + ϕC / B + ϕ B / A =
= π 4 =
GJ
π Gc 4
G2c
ϕ C/ B =
ϕ B/ A
ϕ D/ A
c4 =
(2)(620)
(2)(620)
=
= 195.80 × 10−9 m 4
π Gϕ D/ A π (77 × 109 )(26.18 × 10−3 )
c = 21.04 × 10−3 m = 21.04 mm,
Design must use larger value of d.
d = 2c = 42.1 mm
d = 42.1 mm 
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PROBLEM 3.47
The design specifications of a 2-m-long solid circular transmission shaft require that the angle of twist of the
shaft not exceed 3° when a torque of 9 kN ⋅ m is applied. Determine the required diameter of the shaft,
knowing that the shaft is made of (a) a steel with an allowable shearing stress of 90 MPa and a modulus of
rigidity of 77 GPa, (b) a bronze with an allowable shearing of 35 MPa and a modulus of rigidity of 42 GPa.
SOLUTION
ϕ = 3° = 52.360 × 10−3 rad, T = 9 × 103 N ⋅ m
L = 2.0 m
2TL
2TL
TL
=
∴ c4 =
based on twist angle.
4
π Gϕ
GJ
πc G
2T
2T
Tc
τ =
= 2 ∴ c3 =
based on shearing stress.
πτ
J
πc
ϕ =
(a)
Steel shaft:
Based on twist angle,
τ = 90 × 106 Pa, G = 77 × 109 Pa
c4 =
(2)(9 × 103 )(2.0)
= 2.842 × 10−6 m 4
π (77 × 109 )(52.360 × 10−3 )
c = 41.06 × 10−3 m = 41.06 mm
Based on shearing stress,
c3 =
d = 2c = 82.1 mm
(2)(9 × 103 )
= 63.662 × 10−6 m3
π (90 × 106 )
c = 39.93 × 10−3 m = 39.93 mm
d = 2c = 79.9 mm
d = 82.1 mm 
Required value of d is the larger.
(b)
Bronze shaft:
Based on twist angle,
τ = 35 × 106 Pa, G = 42 × 109 Pa
c4 =
(2)(9 × 103 )(2.0)
= 5.2103 × 10−6 m 4
9
−3
π (42 × 10 )(52.360 × 10 )
c = 47.78 × 10−3 m = 47.78 mm
Based on shearing stress,
c3 =
(2)(9 × 103 )
= 163.702 × 10−6 m3
π (35 × 106 )
c = 54.70 × 10−3 m = 54.70 mm
Required value of d is the larger.
d = 2c = 95.6 mm
d = 2c = 109.4 mm
d = 109.4 mm 
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PROBLEM 3.48
A hole is punched at A in a plastic sheet by applying a 600-N
force P to end D of lever CD, which is rigidly attached to the
solid cylindrical shaft BC. Design specifications require that
the displacement of D should not exceed 15 mm from the
time the punch first touches the plastic sheet to the time it
actually penetrates it. Determine the required diameter of
shaft BC if the shaft is made of a steel with G = 77 GPa and
τ all = 80 MPa.
SOLUTION
T = rP = (0.300 m)(600 N) = 180 N ⋅ m
Torque
Shaft diameter based on displacement limit.
ϕ =
δ
r
=
15 mm
= 0.005 rad
300 mm
ϕ =
TL
2TL
=
GJ
π Gc 4
c4 =
2TL
(2)(180)(0.500)
=
= 14.882 × 10−9 m 4
π Gϕ π (77 × 109 )(0.05)
c = 11.045 × 10−3 m = 11.045 m
d = 2c = 22.1 mm
Shaft diameter based on stress.
τ = 80 × 106 Pa
c3 =
2T
πτ
=
τ =
Tc
2T
=
J
π c3
(2)(180)
= 1.43239 × 10−6 m3
π (80 × 106 )
c = 11.273 × 10−3 m = 11.273 mm
Use the larger value to meet both limits.
d = 2c = 22.5 mm
d = 22.5 mm 
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PROBLEM 3.49
The design specifications for the gear-and-shaft system
shown require that the same diameter be used for both
shafts, and that the angle through which pulley A will
rotate when subjected to a 2-kip ⋅ in. torque TA while
pulley D is held fixed will not exceed 7.5°. Determine
the required diameter of the shafts if both shafts are made
of a steel with G = 11.2 × 106 psi and τ all = 12 ksi.
SOLUTION
Statics:
Gear B.
ΣM B = 0:
rB F − TA = 0 F = TB /rB
ΣM C = 0:
Gear C.
rC F − TD = 0
TD = rC F =
n=
rC
5
= = 2.5
rB
2
Torques in shafts.
TAB = TA = TB
Deformations:
ϕC /D =
ϕ A/B
Kinematics:
ϕD = 0
rC
TA = nTB
rB
TCD L
GJ
T L
= AB
GJ
TCD = TC = nTB = nTA
nTAL
GJ
T L
= A
GJ
=
ϕC = ϕ D + ϕC/D = 0 +
rBϕ B = −rCϕ B
ϕB = −
ϕ A = ϕC + ϕ B /C =
nTAL
GJ
rC
ϕC = −nϕC
rB
ϕB
n 2TA L
GJ
n 2TA L TA L (n2 + 1)TA L
+
=
GJ
GJ
GJ
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PROBLEM 3.49 (Continued)
Diameter based on stress.
Tm = TCD = nTA
Largest torque:
τm =
c=
Tmc
2nTA
=
J
π c3
3
2nTA
πτ m
=
3
τ m = τ all = 12 × 103 psi, TA = 2 × 103 lb ⋅ in
(2)(2.5)(2 × 103 )
= 0.6425 in., d = 2c = 1.285 in.
π (12 × 103 )
Diameter based on rotation limit.
ϕ = 7.5° = 0.1309 rad
ϕ =
c=
(n 2 + 1)TA L (2)(7.25)TAL
=
GJ
π c 4G
4
(2)(7.25)TA L
=
π Gϕ
Choose the larger diameter.
4
L = 8 + 16 = 24 in.
(2)(7.25)(2 × 103 )(24)
= 0.62348 in., d = 2c = 1.247 in.
π (11.2 × 106 )(0.1309)
d = 1.285 in. 
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PROBLEM 3.50
Solve Prob. 3.49, assuming that both shafts are made of a
brass with G = 5.6 × 106 psi and τ all = 8 ksi.
PROBLEM 3.49 The design specifications for the gearand-shaft system shown require that the same diameter
be used for both shafts, and that the angle through which
pulley A will rotate when subjected to a 2-kip ⋅ in.
torque TA while pulley D is held fixed will not exceed
7.5°. Determine the required diameter of the shafts if
both shafts are made of a steel with G = 11.2 × 106 psi
and τ all = 12 ksi.
SOLUTION
Statics:
Gear B.
ΣM B = 0:
rB F − TA = 0 F = TB /rB
ΣM C = 0:
Gear C.
rC F − TD = 0
TD = rC F =
n=
rC
5
= = 2.5
rB
2
Torques in shafts.
TAB = TA = TB
Deformations:
ϕC /D =
ϕ A/B
Kinematics:
ϕD = 0
rC
TA = nTB
rB
TCD L
GJ
T L
= AB
GJ
TCD = TC = nTB = nTA
nTAL
GJ
T L
= A
GJ
=
ϕC = ϕ D + ϕC/D = 0 +
rBϕ B = −rCϕ B
ϕB = −
ϕ A = ϕ C + ϕ B /C =
nTAL
GJ
rC
ϕC = −nϕC
rB
ϕB
n 2TA L
GJ
n 2TA L TA L (n2 + 1)TA L
+
=
GJ
GJ
GJ
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PROBLEM 3.50 (Continued)
Diameter based on stress.
Tm = TCD = nTA
Largest torque:
τm =
c=
Tmc
2nTA
=
J
π c3
3
2nTA
πτ m
=
3
τ m = τ all = 8 × 103 psi, TA = 2 × 103 lb ⋅ in
(2)(2.5)(2 × 103 )
= 0.7355 in., d = 2c = 1.471 in.
π (8 × 103 )
Diameter based on rotation limit.
ϕ = 7.5° = 0.1309 rad
ϕ =
c=
(n 2 + 1)TA L (2)(7.25)TAL
=
GJ
π c 4G
4
(2)(7.25)TA L
=
π Gϕ
Choose the larger diameter.
4
L = 8 + 16 = 24 in.
(2)(7.25)(2 × 103 )(24)
= 0.7415 in., d = 2c = 1.483 in.
π (5.6 × 106 )(0.1309)
d = 1.483 in. 
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PROBLEM 3.51
A torque of magnitude T = 4 kN⋅ m is applied at end A of the
composite shaft shown. Knowing that the modulus of rigidity
is 77 GPa for the steel and 27 GPa for the aluminum,
determine (a) the maximum shearing stress in the steel core,
(b) the maximum shearing stress in the aluminum jacket, (c)
the angle of twist at A.
SOLUTION
c1 =
Steel core:
1
d1 = 0.027 m
2
J1 =
π
2
c14 =
π
2
(0.027) 4 = 834.79 × 10−9
−9
9
G1J1 = (77 × 10 )(834.79 × 10 ) = 64.28 × 103 N ⋅ m 2
Torque carried by steel core.
T1 = G1J1ϕ /L
Aluminum jacket:
c1 =
J2 =
1
d1 = 0.027 m,
2
π
(c
2
4
2
)
− c14 =
π
2
c2 =
1
d 2 = 0.036 m
2
(0.0364 − 0.027 4 ) = 1.80355 × 10−6 m 4
G2 J 2 = (27 × 10 )(1.80355 × 10−6 ) = 48.70 × 103 N ⋅ m 2
9
Torque carried by aluminum jacket. T2 = G2 J 2ϕ /L
T = T1 + T2 = (G1J1 + G2 J 2 ) ϕ /L
Total torque:
ϕ
L
(a)
=
T
4 × 103
=
= 35.406 × 10−3 rad/m
G1 J1 + G2 J 2 64.28 × 103 + 48.70 × 103
Maximum shearing stress in steel core.
τ = G1γ = G1c1
(b)
ϕ
= 73.6 × 106 Pa
73.6 MPa 
= 34.4 × 106 Pa
34.4 MPa 
Maximum shearing stress in aluminum jacket.
τ = G2γ = G2c2
(c)
= (77 × 109 )(0.027)(35.406 × 10−3 )
L
Angle of twist.
ϕ
L
= (27 × 109 )(0.036)(35.406 × 10−3 )
ϕ =L
ϕ
L
= (2.5)(35.406 × 10−3 ) = 88.5 × 10−3 rad
ϕ = 5.07° 
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PROBLEM 3.52
The composite shaft shown is to be twisted by applying a
torque T at end A. Knowing that the modulus of rigidity is 77
GPa for the steel and 27 GPa for the aluminum, determine the
largest angle through which end A can be rotated if the
following allowable stresses are not to be exceeded:
τ steel = 60 MPa and τ aluminum = 45 MPa .
SOLUTION
τ max = Gγ max = Gcmax
ϕall
L
Steel core:
τ all
Gcmax
L
for each material.
τ all = 60 × 106 Pa , cmax =
ϕall
L
Aluminum Jacket:
=
ϕ
=
60 × 106
= 28.860 × 10−3 rad/m
(77 × 109 )(0.027)
τ all = 45 × 106 Pa , cmax =
ϕall
L
=
1
d = 0.036 m, G = 27 × 109 Pa
2
45 × 106
= 46.296 × 10−3 rad/m
(27 × 109 )(0.036)
Smaller value governs:
ϕall
Allowable angle of twist:
ϕall = L
L
1
d = 0.027 m, G = 77 × 109 Pa
2
= 28.860 × 10−3 rad/m
ϕall
L
= (2.5) (28.860 × 10−3 ) = 72.15 × 10−3 rad
ϕall = 4.13° 
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PROBLEM 3.53
The solid cylinders AB and BC are bonded together at B and are
attached to fixed supports at A and C. Knowing that the modulus of
rigidity is 3.7 × 106 psi for aluminum and 5.6 × 106 psi for brass,
determine the maximum shearing stress (a) in cylinder AB, (b) in
cylinder BC.
SOLUTION
The torques in cylinders AB and BC are statically indeterminate. Match the rotation ϕ B for each cylinder.
Cylinder AB:
c=
1
d = 0.75 in.
2
L = 12 in.
π
c 4 = 0.49701 in 4
2
T L
TAB (12)
= 6.5255 × 10−6TAB
ϕ B = AB =
6
GJ
(3.7 × 10 )(0.49701)
J =
Cylinder BC:
c=
J =
1
d = 1.0 in.
2
π
c4 =
π
L = 18 in.
(1.0) 4 = 1.5708 in 4
2
2
TBC L
TBC (18)
ϕB =
=
= 2.0463 × 10−6TBC
GJ
(5.6 × 106 )(1.5708)
Matching expressions for ϕΒ
6.5255 × 10−6TAB = 2.0463 × 10−6TBC
TBC = 3.1889 TAB
Equilibrium of connection at B :
TAB + TBC − T = 0
(1)
T = 12.5 × 103 lb ⋅ in
TAB + TBC = 12.5 × 103
(2)
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PROBLEM 3.53 (Continued)
4.1889 TAB = 12.5 × 103
Substituting (1) into (2),
TAB = 2.9841 × 103 lb ⋅ in
(a)
Maximum stress in cylinder AB.
τ AB =
(b)
TBC = 9.5159 × 103 lb ⋅ in
TABc (2.9841 × 103 ) (0.75)
=
= 4.50 × 103 psi
J
0.49701
τ AB = 4.50 ksi 
Maximum stress in cylinder BC.
τ BC =
TBC c (9.5159 × 103 ) (1.0)
=
= 6.06 × 103 psi
J
1.5708
τ BC = 6.06 ksi 
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PROBLEM 3.54
Solve Prob. 3.53, assuming that cylinder AB is made of steel, for
which G = 11.2 × 106 psi.
PROBLEM 3.53 The solid cylinders AB and BC are bonded together
at B and are attached to fixed supports at A and C. Knowing that the
modulus of rigidity is 3.7 × 106 psi for aluminum and 5.6 × 106 psi for
brass, determine the maximum shearing stress (a) in cylinder AB,
(b) in cylinder BC.
SOLUTION
The torques in cylinders AB and BC are statically indeterminate. Match the rotation ϕ B for each cylinder.
Cylinder AB:
Cylinder BC.
1
π
d = 0.75 in.
L = 12 in. J = c 4 = 0.49701 in 4
2
2
TAB L
TAB (12)
=
= 2.1557 × 10−6 TAB
ϕB =
GJ
(11.2 × 106 )(0.49701)
c=
1
π
π
d = 1.0 in. L = 18 in. J = c 4 = (1.0)4 = 1.5708 in 4
2
2
2
T L
TBC (18)
= 2.0463 × 10−6TBC
ϕ B = BC =
6
GJ
(5.6 × 10 )(1.5708)
c=
Matching expressions for ϕB
2.1557 × 10−6TAB = 2.0463 × 10−6TBC
Equilibrium of connection at B : TAB + TBC − T = 0
TAB = 6.0872 × 103 lb ⋅ in
(2)
TBC = 6.4128 × 103 lb ⋅ in
Maximum stress in cylinder AB.
τ AB =
(b)
TAB + TBC = 12.5 × 103
(1)
2.0535 TAB = 12.5 × 103
Substituting (1) into (2),
(a)
TBC = 1.0535 TAB
TABc (6.0872 × 103 ) (0.75)
=
= 9.19 × 103 psi
J
0.49701
τ AB = 9.19 ksi 
Maximum stress in cylinder BC.
τ BC =
TBC c (6.4128 × 103 )(1.0)
=
= 4.08 × 103 psi
J
1.5708
τ BC = 4.08 ksi 
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PROBLEM 3.55
Two solid steel shafts are fitted with flanges that are then
connected by bolts as shown. The bolts are slightly undersized
and permit a 1.5° rotation of one flange with respect to the
other before the flanges begin to rotate as a single unit.
Knowing that G = 11.2 × 106 psi, determine the maximum
shearing stress in each shaft when a torque of T of magnitude
420 kip ⋅ ft is applied to the flange indicated.
PROBLEM 3.55 The torque T is applied to flange B.
SOLUTION
Shaft AB:
T = TAB , LAB = 2 ft = 24 in., c =
π
1
d = 0.625 in.
2
π
c 4 = (0.625) 4 = 0.23968 in 4
2
2
T L
ϕ B = AB AB
GJ AB
J AB =
TAB =
GJ ABϕ B (11.2 × 106 ) (0.23968)
ϕB
−
LAB
24
= 111.853 × 103ϕ B
Shaft CD:
Applied torque:
T = 420 kip ⋅ ft = 5040 lb ⋅ in
T = TCD , LCD = 3ft = 36 in., c =
J CD =
π
c4 =
π
2
2
T L
ϕC = CD CD
GJ CD
TCD =
1
d = 0.75 in.
2
(0.75) 4 = 0.49701 in 4
(11.2 × 106 )(0.49701)
GJ CD ϕC
ϕC = 154.625 × 103ϕC
=
36
LCD
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PROBLEM 3.55 (Continued)
Clearance rotation for flange B:
ϕ ′B = 1.5° = 26.18 × 10−3 rad
Torque to remove clearance:
′ = (111.853 × 103 )(26.18 × 10−3 ) = 2928.3 lb ⋅ in
TΑΒ
Torque T ′′ to cause additional rotation ϕ ′′:
T ′′ = 5040 − 2928.3 = 2111.7 lb ⋅ in
′′ + T CD
′′
T ′′ = TΑΒ
2111.7 = (111.853 × 103 )ϕ ′′ + (154.625 × 103 )ϕ ′′ ∴
ϕ ′′ = 7.923 × 10−3 rad
′′ = (111.853 × 103 ) (7.923 × 10−3 ) = 886.21 lb ⋅ in
TΑΒ
′′ = (154.625 × 103 ) (7.923 × 10−3 ) = 1225.09 lb ⋅ in
TCD
Maximum shearing stress in AB.
τ AB =
TAB c (2928.3 + 886.21)(0.625)
=
= 9950 psi
J AB
0.23968
τ AB = 9.95 ksi 
TCDc
(1225.09)(0.75)
=
= 1849 psi
J CD
0.49701
τ CD = 1.849 ksi 
Maximum shearing stress in CD.
τ CD =
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PROBLEM 3.56
Two solid steel shafts are fitted with flanges that are then
connected by bolts as shown. The bolts are slightly undersized
and permit a 1.5° rotation of one flange with respect to the
other before the flanges begin to rotate as a single unit.
Knowing that G = 11.2 × 106 psi, determine the maximum
shearing stress in each shaft when a torque of T of magnitude
420 kip ⋅ ft is applied to the flange indicated.
PROBLEM 3.56 The torque T is applied to flange C.
SOLUTION
Shaft AB:
T = TAB , LAB = 2ft = 24 in., c =
π
1
d = 0.625 in.
2
π
c 4 = (0.625)4 = 0.23968 in 4
2
2
T L
ϕ B = AB AB
GJ AB
J AB =
TAB =
GJ ABϕ B (11.2 × 106 ) (0.23968)
ϕB
−
LAB
24
= 111.853 × 103ϕ B
Shaft CD:
Applied torque:
T = 420 kip ⋅ ft = 5040 lb ⋅ in
T = TCD ,
J CD =
π
c4 =
LCD = 3 ft = 36 in., c =
π
2
2
T L
ϕC = CD CD
GJ CD
TCD =
Clearance rotation for flange C:
1
d = 0.75 in.
2
(0.75) 4 = 0.49701 in 4
GJ CD ϕC
(11.2 × 106 )(0.49701)
ϕC = 154.625 × 103 ϕC
=
LCD
36
ϕC′ = 1.5° = 26.18 × 10−3 rad
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PROBLEM 3.56 (Continued)
Torque to remove clearance:
′ = (154.625 × 103 )(26.18 × 10−3 ) = 4048.1 lb ⋅ in
TCD
Torque T′′ to cause additional rotation ϕ ′′:
T ′′ = 5040 − 4048.1 = 991.9 lb ⋅ in
′′ + T CD
′′
T ′′ = TΑΒ
991.9 = (111.853 × 103 )ϕ ′′ + (154.625 × 103 )ϕ ′′ ∴
ϕ ′′ = 3.7223 × 10 −3 rad
′′ = (111.853 × 10−3 ) (3.7223 × 10−3 ) = 416.35 lb ⋅ in
TAB
′′ = (154.625 × 10−3 ) (3.7223 × 10−3 ) = 575.56 lb ⋅ in
TCD
Maximum shearing stress in AB.
τ AB =
TAB c (416.35) (0.625)
=
= 1086 psi
J AB
0.23968
τ AB = 1.086 ksi 
Maximum shearing stress in CD.
τ CD =
TCDc
(4048.1 + 575.56)(0.75)
=
= 6980 psi
J CD
0.49701
τ CD = 6.98 ksi 
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PROBLEM 3.57
Ends A and D of two solid steel shafts AB and CD are fixed, while
ends B and C are connected to gears as shown. Knowing that a
4-kN ⋅ m torque T is applied to gear B, determine the maximum
shearing stress (a) in shaft AB, (b) in shaft CD.
SOLUTION
Gears B and C:
φB =
rC
40
φC =
φC
rB
100
φB = 0.4 φC
(1)
 M C = 0 : TCD = rC F
(2)
 M B = 0 : T − TAB = rB F
(3)
Solve (2) for F and substitute into (3):
T − TAB =
rB
TCD
rC
T = TAB +
100
TCD
40
(4)
T = TAB + 2.5 TCD
L = 0.3 m, c = 0.030 m
Shaft AB:
φB = φB
A
=
TAB L
TAB (0.3)
T
=
= 235.79 × 103 AB
π
JG
G
4
(0.030) 6
2
(5)
L = 0.5 m, c = 0.0225 m
Shaft CD:
φC = φC
D
=
TCD L
JG
π
2
TCD (0.5)
(0.0225) 4 G
= 1242 × 103
TCD
G
(6)
Substitute from (5) and (6) into (1):
φB = 0.4φC :
235.79 × 103
TCD = 0.47462 = TAB
TAB
T
= 0.4 × 1242 × 103 CD
G
G
(7)
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PROBLEM 3.57 (Continued)
Substitute for TCD from (7) into (4):
T = TAB + 2.5 (0.47462 TAB )
T = 2.1865 TAB (8)
4000 N ⋅ m = 2.1865TAB
TAB = 1829.4 N ⋅ m
For T = 4 kN ⋅ m, Eq. (8) yields
TCD = 0.47462(1829.4) = 868.3 N ⋅ m
Substitute into (7):
(a)
Stress in AB:
τ AB =
(b)
TAB c 2 TAB
2 1829.4
=
=
= 43.1 × 106
3
J
π c
π (0.030)3
τ AB = 43.1 MPa 
Stress in CD:
τ CD =
TCDc
2 TCD
2 868.3
=
=
= 48.5 × 106
3
3
J
π c
π (0.0225)
τ CD = 48.5 MPa 
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PROBLEM 3.58
Ends A and D of the two solid steel shafts AB and CD are fixed,
while ends B and C are connected to gears as shown. Knowing that
the allowable shearing stress is 50 MPa in each shaft, determine the
largest torque T that may be applied to gear B.
SOLUTION
Gears B and C:
φB =
rC
40
φC =
φC
rB
100
φB = 0.4 φC
(1)
ΣM C = 0: TCD = rC F
(2)
ΣM B = 0: T − TAB = rB F
(3)
Solve (2) for F and substitute into (3):
T − TAB =
rB
TCD
rC
T = TAB +
100
TCD
40
(4)
T = TAB + 2.5 TCD
L = 0.3 m, c = 0.030 m
Shaft AB:
φB = φB / A =
TAB L
TAB (0.3)
T
=
= 235.77 × 103 AB
π
JG
G
(0.030) 4 G
2
(5)
L = 0.5 m, c = 0.0225 m
Shaft CD:
φC = φC / D =
TCD L
TCD (0.5)
T
=
= 1242 × 103 CD
π
JG
G
(0.0225) 4 G
2
(6)
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PROBLEM 3.58 (Continued)
Substitute from (5) and (6) into (1) :
φB = 0.4 φC :
235.79 × 103
TAB
T
= 0.4 × 1242 × 103 CD
G
G
TCD = 0.47462 TAB
(7)
Substitute for TCD from (7) into (4):
T = TAB + 2.5 (0.47462 TAB )
T = 2.1865 TAB
(8)
T = 4.6068 TCD
(9)
Solving (7) for TAB and substituting into (8),
 TCD 
T = 2.1865 

 0.47462 
Stress criterion for shaft AB:
τ AB = τ all = 50 MPa:
τ AB =
=
TABc
J
π
TAB = τ AB = c3τ AB
2
J
c
π
2
(0.030 m)3 (50 × 106 Pa) = 2120.6 N ⋅ m
T = 2.1865(2120.6 N ⋅ m) = 4.64 kN ⋅ m 
From (8):
Stress criterion for shaft CD:
τ CD = τ all = 50 MPa:
τ CD =
From (7):
TCD c
π
π
TCD = c3τ CD = (0.0225 m)3 (50 × 106 Pa)
2
2
J
= 894.62 N ⋅ m
T = 4.6068(894.62 N ⋅ m) = 4.12 kN ⋅ m
The smaller value for T governs.
T = 4.12 kN ⋅ m 
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PROBLEM 3.59
The steel jacket CD has been attached to the 40-mm-diameter steel
shaft AE by means of rigid flanges welded to the jacket and to the
rod. The outer diameter of the jacket is 80 mm and its wall thickness
is 4 mm. If 500 N ⋅ m torques are applied as shown, determine the
maximum shearing stress in the jacket.
SOLUTION
c=
Solid shaft:
JS =
1
d = 0.020 m
2
π
2
c4 =
π
2
(0.020) 4 = 251.33 × 10−9 m 4
1
d = 0.040 m
2
c1 = c2 − t = 0.040 − 0.004 = 0.036 m
c2 =
Jacket:
JJ =
π
(c
2
4
2
)
− c14 =
π
2
(0.0404 − 0.0364 )
−6
= 1.3829 × 10 m 4
Torque carried by shaft.
TS = GJ S ϕ /L
Torque carried by jacket.
TJ = GJ J ϕ /L
Total torque.
T = TS + TJ = ( J S + J J ) G ϕ /L
TJ =
∴
Gϕ
T
=
L
JS + JJ
(1.3829 × 10−6 ) (500)
JJ
= 423.1 N ⋅ m
T =
JS + JJ
1.3829 × 10−6 + 251.33 × 10−6
Maximum shearing stress in jacket.
τ =
TJ c2
(423.1) (0.040)
=
= 12.24 × 106 Pa
−6
JJ
1.3829 × 10
12.24 MPa 
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PROBLEM 3.60
A solid shaft and a hollow shaft are made of the same material and
are of the same weight and length. Denoting by n the ratio c1 / c2 ,
show that the ratio Ts /Th of the torque Ts in the solid shaft to the
torque
Th
in
the
hollow
shaft
is
(a)
(1 − n 2 ) /(1 + n 2 ) if the maximum shearing stress is the same in
each shaft, (b) (1 − n)/ (1 + n 2 ) if the angle of twist is the same for
each shaft.
SOLUTION
For equal weight and length, the areas are equal.
(
)
π c02 = π c22 − c12 = π c22 (1 − n 2 )
Js =
(a)
π
2
c04 =
2
c24 (1 − n 2 ) 2
Jh =
π
(c
2
4
2
)
− c14 =
π
2
c24 (1 − n 4 )
For equal stresses.
τ =
Ts c0
Tc
= h 2
Js
Jh
Ts
Jc
= s 2 =
Th
J hc0
(b)
π
c0 = c2 (1 − n 2 )1/2
∴
c 4 (1 − n 2 ) 2 c2
2 2
π c 4 (1 − n 4 )c (1 − n 2 )1/2
2
2 2
π
=
1 − n2
(1 − n 2 )1/2
=
(1 + n 2 ) (1 − n 2 )1/2
1 + n2

For equal angles of twist.
ϕ =
Ts L
TL
= h
GJ s
GJ h
Ts
J
= s =
Th
Jh
c 4 (1 − n 2 )2
2 2
π c 4 (1 − n 4 )
2 2
π
=
(1 − n 2 ) 2 1 − n2
=
1 − n4
1 + n2

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PROBLEM 3.61
A torque T is applied as shown to a solid tapered shaft AB. Show by
integration that the angle of twist at A is
φ =
7TL
12π Gc 4
SOLUTION
Introduce coordinate y as shown.
r =
cy
L
Twist in length dy:
dϕ =
Tdy
Tdy
2TL4dy
=
=
π
GJ
π Gc 4 y 4
G r4
2
2L
ϕ = L
2TL4 dy
2TL 2L dy
=

4 4
π Gc y
π Gc 4 L y 4
2L
2TL4  1 
2TL4
=
−
=


π Gc 4  3 y 3 L
π Gc 4
=
1 
 1
+ 3
−
3
3L 
 24L
2TL4  7 
7TL
=
4 
3
π Gc  24L  12π Gc 4

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PROBLEM 3.62
The mass moment of inertia of a gear is to be determined experimentally by
using a torsional pendulum consisting of a 6-ft steel wire. Knowing that
G = 11.2 × 106 psi, determine the diameter of the wire for which the torsional
spring constant will be 4.27 lb ⋅ ft/rad.
SOLUTION
Torsion spring constant K = 4.27 lb ⋅ ft/rad = 51.24 lb ⋅ in/rad
K =
c4 =
T
ϕ
=
π Gc 4
T
GJ
=
=
2L
TL /GJ
L
2 LK
(2) (72) (51.24)
=
= 209.7 × 10−6 in 4
6
πG
π (11.2 × 10 )
c = 0.1203 in.
d = 2c = 0.241 in. 
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PROBLEM 3.63
An annular plate of thickness t and modulus G is used to
connect shaft AB of radius r1 to tube CD of radius r2. Knowing
that a torque T is applied to end A of shaft AB and that end D
of tube CD is fixed, (a) determine the magnitude and location
of the maximum shearing stress in the annular plate, (b) show
that the angle through which end B of the shaft rotates with
respect to end C of the tube is
ϕ BC =
T  1
1 
 2 − 2 

4π Gt  r1
r2 
SOLUTION
Use a free body consisting of shaft AB and an inner portion of the plate BC, the outer radius of this portion
being ρ .
The force per unit length of circumference is τ t.
ΣM = 0
τ t (2πρ ) ρ − T = 0
T
2π t ρ 2
τ =
(a)
Maximum shearing stress occurs at ρ = r1
γ =
Shearing strain:
τ
G
τ max =
=
T
2π tr12

T
2π GT ρ 2
The relative circumferential displacement in radial length d ρ is
d δ = γ d ρ = ρ dϕ
dϕ = γ
dϕ =
(b)
ϕ B/C =
=

r2
r1
T dρ
T
=
3
2π Gt
2π Gt ρ

r2
r1
dρ
ρ3
dρ
ρ
T
dρ
T dρ
=
2
2π Gt ρ ρ
2π GT ρ 3
=
T  1  r2
−

2π Gt  2 ρ 2  r1
1 
1 
T  1
T  1
− 2 + 2  =
 2 − 2
2π Gt  2r2
2r1  4π Gt  r1
r2 

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PROBLEM 3.64
Determine the maximum shearing stress in a solid shaft of 12-mm diameter as it transmits 2.5 kW at a
frequency of (a) 25 Hz, (b) 50 Hz.
SOLUTION
c=
(a)
(b)
f = 25 Hz
f = 50 Hz
T =
1
d = 6 mm = 0.006 m
2
P
2π f
=
P = 2.5 kW = 2500 W
2500
= 15.9155 N ⋅ m
2π (25)
τ =
Tc
2T
2(15.9155)
=
=
= 46.9 × 106 Pa
3
J
πc
π (0.006)3
T =
2500
= 7.9577 N ⋅ m
2π (50)
τ =
2(7.9577)
= 23.5 × 106 Pa
π (0.006)3
τ = 46.9 MPa 
τ = 23.5 MPa 
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PROBLEM 3.65
Determine the maximum shearing stress in a solid shaft of 1.5-in. diameter as it transmits 75 hp at a speed of
(a) 750 rpm, (b) 1500 rpm.
SOLUTION
c=
(a)
f =
T =
(b)
1
d = 0.75 in.
2
P = 75 hp = (75)(6600) = 495 × 103 lb ⋅ in/s
750
= 12.5 Hz
60
P
2π f
=
495 × 103
= 6.3025 × 103 lb ⋅ in
2π (12.5)
τ =
Tc
2T
(2) (6.3025 × 103 )
=
=
= 9.51 × 103 psi
J
π c3
π (0.75)3
f =
1500
= 25 Hz
60
T =
495 × 103
= 3.1513 × 103 lb ⋅ in
2π (25)
τ =
(2) (3.1513 × 103 )
= 4.76 × 103 psi
3
π (0.75)
τ = 9.51 ksi 
τ = 4.76 ksi 
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PROBLEM 3.66
Design a solid steel shaft to transmit 0.375 kW at a frequency of 29 Hz, if the shearing stress in the shaft is not
to exceed 35 MPa.
SOLUTION
τ all = 35 × 106 Pa P = 0.375 × 103 W
T=
τ =
P
2π f
=
f = 29 Hz
0.375 × 103
= 2.0580 N ⋅ m
2π (29)
Tc
2T
=
J
π c3
∴
c3 =
2T
πτ
=
(2) (2.0580)
π (35 × 106 )
= 37.43 × 10−9 m3
c = 3.345 × 10−3 m = 3.345 mm
d = 2c
d = 6.69 mm 
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PROBLEM 3.67
Design a solid steel shaft to transmit 100 hp at a speed of 1200 rpm, if the maximum shearing stress is not to
exceed 7500 psi.
SOLUTION
τ all = 7500 psi P = 100 hp = 660 × 103 lb ⋅ in/s
f =
τ =
1200
= 20 Hz
60
Tc
2T
=
J
π c3
c = 0.7639in.
T =
∴ c3 =
2T
πτ
d = 2c
P
2π f
=
=
660 × 103
= 5.2521 × 103 lb ⋅ in
2π (20)
(2) (5.2521 × 103 )
= 0.4458 in 3
π (7500)
d = 1.528 in. 
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PROBLEM 3.68
Determine the required thickness of the 50-mm tubular shaft of Example 3.07 if is to transmit the same power
while rotating at a frequency of 30 Hz.
SOLUTION
P = 100 kW = 100 × 103 W
From Example 3.07,
τ all = 60 MPa = 60 × 106 Pa
c2 =
1
d = 25 mm = 0.025 m
2
f = 30 Hz
T =
P
= 530.52 N ⋅ m
2π f
π
(c
2
− c14
)
c14 = c24 −
2Tc2
= 0.0254 −
J =
4
2
πτ
τ =
Tc2
2Tc2
=
J
π c24 − c14
(
)
(2)(530.52) (0.025)
= 249.90 × 10−9 m 4
π (60 × 106 )
c1 = 22.358 × 10−3 m = 22.358 mm
t = c2 − c1 = 25 mm − 22.358 mm = 2642 mm
t = 2.64 mm 
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PROBLEM 3.69
While a steel shaft of the cross section shown rotates at 120 rpm, a
stroboscopic measurement indicates that the angle of twist is 2°
in a 12-ft length. Using G = 11.2 × 106 psi, determine the power
being transmitted.
SOLUTION
ϕ = 2° = 34.907 × 10−3 rad
c2 =
J =
1
d o = 1.5in.
2
π
(c
2
4
2
)
− c14 =
c1 =
π
2
L = 12ft =144 in.
1
di = 0.6in.
2
(1.54 − 0.64 ) = 7.7486 in 4
120
f =
= 2 Hz
60
T =
GJ ϕ
(11.2 × 106 ) (7.7486) (34.907 × 10−3 )
=
= 21.037 × 103 lb ⋅ in
L
144
P = 2π fT = 2π (2) (21.037 × 103 ) = 264.36 × 103 lb ⋅ in/s
Since 1 hp = 6600 lb ⋅ in/s,
P = 40.1 hp 
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PROBLEM 3.70
The hollow steel shaft shown (G = 77.2 GPa, τ all = 50 MPa)
rotates at 240 rpm. Determine (a) the maximum power that can
be transmitted, (b) the corresponding angle of twist of the shaft.
SOLUTION
1
d 2 = 30 mm
2
1
c1 = d1 = 12.5 mm
2
c2 =
J =
π
(c
2
4
2
)
− c14 =
π
2
[(30) 4 − (12.5) 4 ]
= 1.234 × 106 mm 4 = 1.234 × 10−6 m 4
τ m = 50 × 106 Pa
τm =
(a)
(50 × 106 )(1.234 × 10−6 )
τ J
Tc
T = m =
= 2056.7 N ⋅ m
J
c
30 × 10−3
Angular speed.
f = 240 rpm = 4 rev/sec = 4 Hz
Power being transmitted.
P = 2π f T = 2π (4)(2056.7) = 51.7 × 103 W
P = 51.7 kW 
(b)
Angle of twist.
ϕ =
TL
(2056.7)(5)
=
= 0.1078 rad
GJ
(77.2 × 109 )(1.234 × 10−6 )
ϕ = 6.17° 
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PROBLEM 3.71
As the hollow steel shaft shown rotates at 180 rpm, a
stroboscopic measurement indicates that the angle of twist of
the shaft is 3°. Knowing that G = 77.2 GPa, determine (a) the
power being transmitted, (b) the maximum shearing stress in the
shaft.
SOLUTION
1
d 2 = 30 mm
2
1
c1 = d1 = 12.5 mm
2
c2 =
J =
π
(c
2
4
2
)
− c14 =
π
2
[(30)4 − (12.5)4 )]
= 1.234 × 10 mm 4 = 1.234 × 10−6 m 4
6
ϕ = 3° = 0.05236 rad
(a)
ϕ =
TL
GJ
T =
GJ ϕ
(77.2 × 109 )(1.234 × 10−6 )(0.0536)
=
= 997.61 N ⋅ m
L
5
Angular speed:
f = 180 rpm = 3 rev/sec = 3 Hz
Power being transmitted.
P = 2π f T = 2π (3)(997.61) = 18.80 × 103 W
P = 18.80 kW 
(b)
Maximum shearing stress.
τm =
Tc2
(997.61)(30 × 10−3 )
=
J
1.234 × 10−6
= 24.3 × 106 Pa
τ m = 24.3 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 3.72
The design of a machine element calls for a 40-mm-outer-diameter
shaft to transmit 45 kW. (a) If the speed of rotation is 720 rpm,
determine the maximum shearing stress in shaft a. (b) If the speed of
rotation can be increased 50% to 1080 rpm, determine the largest inner
diameter of shaft b for which the maximum shearing stress will be the
same in each shaft.
SOLUTION
(a)
f =
720
= 12 Hz
60
P = 45 kW = 45 × 103 W
T =
(b)
P
2π f
=
45 × 103
= 596.83 N ⋅ m
2π (12)
c=
1
d = 20 mm = 0.020 m
2
τ =
Tc
2T
(2)(596.83)
=
=
= 47.494 × 106 Pa
3
J
πc
π (0.020)3
f =
1080
= 18 Hz
60
T =
45 × 103
= 397.89 N ⋅ m
2π (18)
τ =
Tc2
2Tc2
=
J
π c24 − c14
c14 = c24 −
(
τ max = 47.5 MPa 
)
2Tc2
πτ
c14 = 0.0204 −
(2)(397.89)(0.020)
= 53.333 × 10−9
π (47.494 × 106 )
c1 = 15.20 × 10−3 m = 15.20 mm
d 2 = 2c1 = 30.4 mm 
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PROBLEM 3.73
A steel pipe of 3.5-in. outer diameter is to be used to transmit a torque
of 3000 lb ⋅ ft without exceeding an allowable shearing stress of 8 ksi.
A series of 3.5-in.-outer-diameter pipes is available for use. Knowing
that the wall thickness of the available pipes varies from 0.25 in. to
0.50 in. in 0.0625-in. increments, choose the lightest pipe that can be
used.
SOLUTION
T = 3000 lb ⋅ ft = 36 × 103 lb ⋅ in
1
do = 1.75 in.
2
Tc
2Tc2
τ= 2=
J
π c24 − c14
c2 =
(
c14 = c24 −
)
2Tc2
(2)(36 × 103 )(1.75)
= 1.754 −
= 4.3655 in 4
πτ
π(8 × 103 )
c1 = 1.4455 in.
Required minimum thickness: t = c2 − c1
t = 1.75 − 1.4455 = 0.3045 in.
Available thicknesses: 0.25 in., 0.3125 in., 0.375 in., etc.
Use
t = 0.3125 in. 
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PROBLEM 3.74
The two solid shafts and gears shown are used to
transmit 16 hp from the motor at A, operating at a speed
of 1260 rpm, to a machine tool at D. Knowing that the
maximum allowable shearing stress is 8 ksi, determine
the required diameter (a) of shaft AB, (b) of shaft CD.
SOLUTION
(a)
Shaft AB:
P = 16 hp = (16)(6600) = 105.6 × 103 lb ⋅ in/sec
f =
1260
= 21 Hz
60
τ = 8 ksi = 8 × 103 psi
TAB =
P
2π f
=
105.6 × 103
= 800.32 lb ⋅ in
2π (21)
τ =
2T
Tc
=
J
π c3
c=
3
c=
3
2T
πτ
(2)(800.32)
= 0.399 in.
π (8 × 103 )
d AB = 2c = 0.799 in.
(b)
Shaft CD:
TCD =
c=
d AB = 0.799 in. 
rC
5
TAB = (800.32) = 1.33387 × 103 lb ⋅ in
rB
3
3
2T
πτ
=
3
(2)(1.33387 × 103 )
= 0.473 in.
π (8 × 103 )
dCD = 2c = 0.947 in.
dCD = 0.947 in. 
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PROBLEM 3.75
The two solid shafts and gears shown are used to
transmit 16 hp from the motor at A operating at a speed
of 1260 rpm to a machine tool at D. Knowing that each
shaft has a diameter of 1 in., determine the maximum
shearing stress (a) in shaft AB, (b) in shaft CD.
SOLUTION
(a)
Shaft AB:
P = 16 hp = (16)(6600) = 105.6 × 103 lb ⋅ in/sec
f =
TAB =
1260
= 21 Hz
60
P
2π f
=
105.6 × 103
= 800.32 lb ⋅ in
2π (21)
1
d = 0.5 in.
2
2T
Tc
τ =
=
J
π c3
(2)(800.32)
=
= 4.08 × 103 psi
3
π (0.5)
c=
(b)
Shaft CD:
TCD =
τ =
τ AB = 4.08 ksi 
rC
5
TAB = (800.32) = 1.33387 × 103 lb ⋅ in
rB
3
2T
(2)(1.33387 × 103 )
=
= 6.79 × 103 psi
π c3
π (0.5)3
τ CD = 6.79 ksi 
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PROBLEM 3.76
Three shafts and four gears are used to form a gear train
that will transmit 7.5 kW from the motor at A to a
machine tool at F. (Bearings for the shafts are omitted in
the sketch.) Knowing that the frequency of the motor is
30 Hz and that the allowable stress for each shaft is
60 MPa, determine the required diameter of each shaft.
SOLUTION
P = 7.5 kW = 7.5 × 103 W
Shaft AB:
f AB = 30 Hz
τ=
c3AB =
TAB =
τ all = 60 MPa = 60 × 106 Pa
P
7.5 × 103
=
= 39.789 N ⋅ m
2π f AB
2π(30)
Tc AB
2T
= 3
∴
J AB
πc AB
c3AB =
2T
πτ
(2)(39.789)
= 422.17 × 10−9 m3
π(60 × 106 )
c AB = 7.50 × 10−3 m = 7.50 mm
Shaft CD:
d AB = 2c AB = 15.00 mm 
P
7.5 × 103
=
= 99.472 N ⋅ m
2πfCD
2π(12)
fCD =
rB
60
f AB =
(30) = 12 Hz
rC
150
τ CD =
TcCD
2T
2T
2(99.472)
3
= 3
∴ cCD
= CD =
= 1.05543 × 10−6 m3
πτ
J CD
πcCD
π(60 × 106 )
TCD =
cCD = 10.18 × 10−3 m = 10.18 mm
Shaft EF:
f EF =
τ=
dCD = 2cCD = 20.4 mm 
rD
60
fCD =
(12) = 4.8 Hz
rE
150
TcEF
2T
= 3
J EF
πcEF
3
∴ cEF
=
TEF =
P
7.5 × 103
=
= 248.68 N ⋅ m
2πf EF 2 π (4.8)
(2)(248.68)
= 2.6886 × 10−6 m3
6
π(60 × 10 )
cEF = 13.82 × 10−3 m = 13.82 mm
d EF = 2cEF = 27.6 mm 
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PROBLEM 3.77
Three shafts and four gears are used to form a gear train that will
transmit power from the motor at A to a machine tool at F. (Bearings
for the shafts are omitted in the sketch.) The diameter of each shaft is
as follows: d AB =16 mm, dCD = 20 mm, d EF = 28 mm. Knowing
that the frequency of the motor is 24 Hz and that the allowable
shearing stress for each shaft is 75 MPa, determine the maximum
power that can be transmitted.
SOLUTION
τ all = 75 MPa = 75 × 106 Pa
Shaft AB:
c AB =
Tall =
1
d AB = 0.008 m
2
π
2
c3AB τ all =
π
2
τ =
Tc AB
2T
=
J AB
π c3AB
(0.008)3 (75 × 106 ) = 60.319 N ⋅ m
f AB = 24 Hz Pall = 2π f ABTall = 2π (24) (60.319) = 9.10 × 103 W
Shaft CD:
1
dCD = 0.010 m
2
π 3
π
2T
Tc
τ = CD = 3 ∴ Tall = cCD
τ all = (0.010)3 (75 × 106 ) = 117.81 N ⋅ m
2
2
J CD
π cCD
cCD =
fCD =
Shaft EF:
60
rB
(24) = 9.6 Hz
f AB =
150
rC
cEF =
Pall = 2π fCDTall = 2π (9.6) (117.81) = 7.11 × 103 W
1
d EF = 0.014 m
2
π
3
cEF
τ all =
π
(0.014)3 (75 × 106 ) = 323.27 N ⋅ m
2
2
r
60
f EF = D fCD =
(9.6) = 3.84 Hz
rE
150
Tall =
Pall = 2π f EFTall = 2π (3.84) (323.27) = 7.80 ×103 W
Maximum allowable power is the smallest value.
Pall = 7.11× 103 W = 7.11 kW 
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PROBLEM 3.78
A 1.5-m-long solid steel of 48 mm diameter is to transmit 36 kW between a motor and a machine tool.
Determine the lowest speed at which the shaft can rotate, knowing that G = 77.2 GPa, that the maximum
shearing stress must not exceed 60 MPa, and the angle of twist must not exceed 2.5°.
SOLUTION
P = 36 × 103 W,
c=
1
d = 0.024 m, L = 1.5 m, G = 77.2 × 109 Pa
2
Torque based on maximum stress:
τ =
Tc
J
T =
Jτ
π
π
= c3 τ = (0.024)3 (60 × 106 ) = 1.30288 × 103 N ⋅ m
c
2
2
Torque based on twist angle:
ϕ=
TL
GJ
T =
τ = 60 MPa = 60 × 106 Pa
ϕ = 2.5° = 43.633 × 10−3 rad
GJ ϕ π c 4Gϕ
π (0.024)4 (77 × 109 ) (43.633 × 10−3 )
=
=
L
2L
(2) (1.5)
= 1.17033 × 103 N ⋅ m
Smaller torque governs, so T =1.17033× 103 N ⋅ m
P = 2π fT
f =
P
36 × 103
=
2π T
2π (1.17033 × 103 )
f = 4.90 Hz 
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PROBLEM 3.79
A 2.5-m-long steel shaft of 30-mm diameter rotates at a frequency of 30 Hz. Determine the maximum power
that the shaft can transmit, knowing that G = 77.2 GPa, that the allowable shearing stress is 50 MPa, and that
the angle of twist must not exceed 7.5°.
SOLUTION
c=
1
d = 15 mm = 0.015 m L = 2.5 m
2
τ = 50 × 106 Pa τ =
Stress requirement.
T =
τJ
c
=
π
2
τ c3 =
π
2
Tc
J
(50 × 106 )(0.015)3 = 265.07 N ⋅ m
ϕ = 7.5° = 130.90 × 10−3 rad G = 77.2 × 109 Pa
Twist angle requirement.
ϕ =
T =
TL
2TL
=
GJ
π Gc 4
π
2
Gc 4ϕ =
π
2
(77.2 × 109 )(0.015)4 (130.90 × 10−3 ) = 803.60 N ⋅ m
Smaller value of T is the maximum allowable torque.
T = 265.07 N ⋅ m
Power transmitted at f = 30 Hz.
P = 2π f T = 2π (30)(265.07) = 49.96 × 103 W
P = 50.0 kW 
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PROBLEM 3.80
A steel shaft must transmit 210 hp at a speed of 360 rpm. Knowing that G = 11.2 × 106 psi, design a solid
shaft so that the maximum shearing stress will not exceed 12 ksi, and the angle of twist in a 8.2-ft length must
not exceed 3°.
SOLUTION
Power:
P = (210 hp)(6600 in ⋅ lb/s/hp) = 1.336 × 106 in ⋅ lb/s
Angular speed:
f = (360 rpm)
Torque:
T =
Stress requirement:
τ = 12 ksi, τ =
c=
Angle of twist requirement:
P
2π f
3
2T
πτ
=
=
1 min.
= 6 Hz
60 sec
1.386 × 106
= 36.765 × 103 lb ⋅ in
(2π )(6)
3
Tc
2T
= 3
J
πc
(2)(36.765 × 103 )
= 1.2494 in.
π (12 × 103 )
ϕ = 3° = 52.36 × 10−3 rad
L = 8.2 ft = 98.4 in.
ϕ =
TL
2TL
=
GJ
π Gc 4
c=
4
2TL
=
π Gϕ
4
(2)(36.765 × 103 )(98.4)
= 1.4077 in.
π (11.2 × 106 )(52.36 × 10−3 )
The larger value is the required radius. c = 1.408 in.
d = 2c
d = 2.82 in. 
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PROBLEM 3.81
The shaft-disk-belt arrangement shown is used to transmit 3 hp from
point A to point D. (a) Using an allowable shearing stress of 9500 psi,
determine the required speed of shaft AB. (b) Solve part a, assuming that
the diameters of shafts AB and CD are, respectively, 0.75 in. and 0.625 in.
SOLUTION
τ = 9500 psi
τ =
Tc
2T
=
J
π c3
P = 3 hp = (3)(6600) = 19800 lb ⋅ in/s
T =
π
2
c3τ
Allowable torques.
3
in. diameter shaft:
c=
3
−
4
in. diameter shaft:
c = 83 in., Tall =
Statics:
Allowable torques.
3
(9500) = 786.9 lb ⋅ in
2  8 
TC = rC ( F1 − F2 )
rB
1.125
TC =
TC = 0.25TC
rC
4.5
TB,all = 455.4 lb ⋅ in
TC ,all = 786.9 lb ⋅ in
Assume
TC = 786.9 lb ⋅ in
Then
TB = (0.25)(786.9) = 196.73 lb ⋅ in < 455.4 lb ⋅ in (okay)
P = 2π fT
(b)
π  3
TB = rB ( F1 − F2 )
TB =
(a)
5
π5
in., Tall =   (9500) = 455.4 lb ⋅ in
16
2  16 
5
−
8
Allowable torques.
P
19800
=
2π TB
2π (196.73)
f AB =
TB,all = 786.9 lb ⋅ in
TC ,all = 455.4 lb ⋅ in
Assume
TC = 455.4 lb ⋅ in
Then
TB = (0.25)(455.4) = 113.85 lb ⋅ in < 786.9 lb ⋅ in
P = 2π fT
f AB = 16.02 Hz 
f AB =
P
19800
=
2π TB
2π (113.85)
f AB = 27.2 Hz 
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PROBLEM 3.82
A 1.6-m-long tubular steel shaft of 42-mm outer diameter d1 is to be
made of a steel for which τ all = 75 MPa and G = 77.2 GPa. Knowing
that the angle of twist must not exceed 4° when the shaft is subjected to a
torque of 900 N ⋅ m, determine the largest inner diameter d2 that can be
specified in the design.
SOLUTION
c1 =
1
d1 = 0.021 m L = 1.6 m
2
Based on stress limit: τ = 75 MPa = 75 × 106 Pa
τ =
Tc1
∴
J
J =
Tc1
τ
=
(900)(0.021)
= 252 × 10−9 m 4
75 × 106
Based on angle of twist limit: ϕ = 4° = 69.813 × 10−3 rad
ϕ =
TL
GJ
∴ J =
TL
(900)(1.6)
=
= 267.88 × 10−9 m 4
Gϕ
(77 × 109 )(69.813 × 10−3 )
Larger value for J governs.
J =
π
2
(c
4
1
c24 = c14 −
− c24
2J
π
J = 267.88 × 10−9 m 4
)
= 0.0214 −
(2)(267.88 × 10−9 )
c2 = 12.44 × 10−3 m = 12.44 mm
π
= 23.943 × 10−9 m 4
d 2 = 2c2 = 24.9 mm 
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PROBLEM 3.83
A 1.6-m-long tubular steel shaft ( G = 77.2 GPa ) of 42-mm outer
diameter d1 and 30-mm inner diameter d2 is to transmit 120 kW between
a turbine and a generator. Knowing that the allowable shearing stress is
65 MPa and that the angle of twist must not exceed 3°, determine the
minimum frequency at which the shaft can rotate.
SOLUTION
c1 =
J =
1
1
d1 = 0.021 m, c2 = d 2 = 0.015 m
2
2
π
2
(c
4
1
)
− c24 =
π
2
(0.0214 − 0.0154 ) = 225.97 × 10−9 m 4
Based on stress limit: τ = 65 MPa = 65 × 106 Pa
τ =
Tc1
Jτ
(225.97 × 10−9 )(65 × 106 )
or T =
=
= 699.43 N ⋅ m
J
G
0.021
Based on angle of twist limit: ϕ = 3° = 52.36×10−3 rad
ϕ =
TL
GJ ϕ
(77 × 109 )(225.97 × 10−9 )(52.36 × 10−3 )
or T =
=
GJ
L
1.6
= 569.40 N ⋅ m
T = 569.40 N ⋅ m
Smaller torque governs.
P = 120 kW = 120 × 103 W
P = 2π fT
so
f =
P
120 × 103
=
2π T
2π (569.40)
f = 33.5 Hz 
or 2010 rpm 
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PROBLEM 3.84
Knowing that the stepped shaft shown transmits a torque of magnitude
T = 2.50 kip ⋅ in., determine the maximum shearing stress in the shaft
3
when the radius of the fillet is (a) r = 18 in., (b) r = 16
in.
SOLUTION
D
2
=
= 1.33
d
1.5
D = 2 in. d = 1.5 in.
1
d = 0.75 in.
T = 2.5 kip ⋅ in
2
2T
(2)(2.5)
Tc
=
=
= 3.773 ksi
3
J
πc
π (0.75)3
c=
(a)
r =
1
in. r = 0.125 in.
8
r
0.125
=
= 0.0833
d
1.5
K = 1.42
From Fig. 3.32,
τ max = K
(b)
r =
Tc
= (1.42)(3.773)
J
3
in.
16
τ max = 5.36 ksi 
r = 0.1875 in.
r
0.1875
=
= 0.125
d
1.5
From Fig. 3.32,
τ max = K
K = 1.33
Tc
= (1.33)(3.773)
J
τ max = 5.02 ksi 
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PROBLEM 3.85
Knowing that the allowable shearing stress is 8 ksi for the stepped shaft
shown, determine the magnitude T of the largest torque that can be
3
transmitted by the shaft when the radius of the fillet is (a) r = 16
in.,
(b) r = 14 in.
SOLUTION
D = 2 in.
c=
1
d = 0.75 in.
2
τ max = K
(a)
r =
D
= 1.33
d
d = 1.5 in.
Tc
J
3
in.
16
τ max = 8 ksi
or
T =
Jτ max
πτ c3
= max
Kc
2K
r = 0.1875 in.
r
0.1875
=
= 0.125
d
1.5
From Fig. 3.32,
T =
(b)
r =
K = 1.33
π (8)(0.75)3
T = 3.99 kip ⋅ in 
(2)(1.33)
1
in.
4
r = 0.25 in.
r
0.25
=
= 0.1667
d
1.5
From Fig. 3.32,
T =
π (8)(0.75)3
(2)(1.27)
K = 1.27
T = 4.17 kip ⋅ in 
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PROBLEM 3.86
The stepped shaft shown must transmit 40 kW at a speed of 720 rpm.
Determine the minimum radius r of the fillet if an allowable stress of
36 MPa is not to be exceeded.
SOLUTION
Angular speed:
 1 Hz 
f = (720 rpm) 
 = 12 Hz
 60 rpm 
Power:
P = 40 × 103 W
Torque:
T =
P
2π f
=
40 × 103
= 530.52 N ⋅ m
2π (12)
In the smaller shaft, d = 45 mm, c = 22.5 mm = 0.0225 m
τ =
Tc
2T
(2)(530.52)
=
=
= 29.65 × 106 Pa
3
3
J
πc
π (0.0225)
Using τ max = 36 MPa = 36 × 106 Pa results in
K =
τ max
36 × 106
=
= 1.214
τ
29.65 × 106
From Fig 3.32 with
D 90 mm
r
=
= 2,
= 0.24
d
45 mm
d
r = 0.24 d = (0.24)(45 mm)
r = 10.8 mm 
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PROBLEM 3.87
The stepped shaft shown must transmit 45 kW. Knowing that the allowable
shearing stress in the shaft is 40 MPa and that the radius of the fillet is
r = 6 mm, determine the smallest permissible speed of the shaft.
SOLUTION
r
6
=
= 0.2
d
30
D
60
=
=2
d
30
From Fig. 3.32,
For smaller side,
K = 1.26
c=
1
d = 15 mm = 0.015 m
2
τ =
KTc 2 KT
=
J
π c3
T =
π c3τ
2K
=
π (0.015)3 (40 × 106 )
(2)(1.26)
P = 45 kW = 45 × 103
f =
= 168.30 N ⋅ m
P = 2π fT
P
45 × 103
=
= 42.6 Hz
2π T
2π (168.30 × 103 )
f = 42.6 Hz 
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PROBLEM 3.88
The stepped shaft shown must rotate at a frequency of 50 Hz. Knowing that
the radius of the fillet is r = 8 mm and the allowable shearing stress is 45
MPa, determine the maximum power that can be transmitted.
SOLUTION
τ =
KTc
2 KT
=
J
π c3
T =
π c3τ
2K
1
d = 15 mm = 15 × 10−3 m
2
D = 60 mm, r = 8 mm
d = 30 mm c =
60
D
=
= 2,
30
d
8
r
=
= 0.26667
30
d
From Fig. 3.32,
K = 1.18
Allowable torque.
T =
Maximum power.
P = 2π f T = (2π )(50)(202.17) = 63.5 × 103 W
π (15 × 10−3 )3 (45 × 106 )
(2)(1.18)
= 202.17 N ⋅ m
P = 63.5 kW 
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PROBLEM 3.89
In the stepped shaft shown, which has a full quarter-circular fillet, D = 1.25 in.
d = 1 in.
Knowing that the speed of the shaft is
and
2400 rpm and that the allowable shearing stress is 7500 psi, determine the
maximum power that can be transmitted by the shaft.
SOLUTION
D 1.25
=
= 1.25
d
1.0
1
r = ( D − d ) = 0.15 in.
2
r
0.15
=
= 0.15
1.0
d
From Fig. 3.32,
For smaller side,
K = 1.31
c=
1
d = 0.5 in.
2
τ =
π c3τ
KTc
Jτ
=
T =
2K
J
Kc
T =
π (0.5)3 (7500)
(2)(1.31)
= 1.1241 × 103 lb ⋅ in
f = 2400 rpm = 40 Hz
P = 2π f T = 2π (40)(1.1241 × 103 )
= 282.5 × 103 lb ⋅ in/s
P = 42.8 hp 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 3.90
A torque of magnitude T = 200 lb ⋅ in. is applied to the stepped shaft shown,
which has a full quarter-circular fillet. Knowing that D = 1 in., determine the
maximum shearing stress in the shaft when (a) d = 0.8 in., (b) d = 0.9 in.
SOLUTION
(a)
D 1.0
=
= 1.25
0.8
d
1
r = ( D − d ) = 0.1 in.
2
0.1
r
=
= 0.125
d
0.8
From Fig. 3.32,
K = 1.31
For smaller side,
c=
1
d = 0.4 in.
2
KTc
2KT
=
J
π c3
(2)(1.31)(200)
=
= 2.61 × 103 psi
π (0.4)3
τ =
(b)
τ = 2.61 ksi 
D 1.0
=
= 1.111
0.9
d
1
r = ( D − d ) = 0.05
2
r
0.05
=
= 0.05
d
1.0
From Fig. 3.32,
K = 1.44
For smaller side,
c=
1
d = 0.45 in.
2
τ =
2 KT
(2)(1.44)(200)
=
= 2.01 × 103 psi
π c3
π (0.45)3
τ = 2.01 ksi 
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PROBLEM 3.91
In the stepped shaft shown, which has a full quarter-circular fillet, the
allowable shearing stress is 80 MPa. Knowing that D = 30 mm, determine
the largest allowable torque that can be applied to the shaft if
(a) d = 26 mm, (b) d = 24 mm.
SOLUTION
τ = 80 × 106 Pa
(a)
D
30
=
= 1.154
d
26
r =
1
( D − d ) = 2 mm
2
From Fig. 3.32,
K = 1.36
Smaller side,
c=
1
d = 13 mm = 0.013 m
2
τ =
2KT
KTc
=
J
π c3
T =
(b)
r
2
=
= 0.0768
d
26
D
30
=
= 1.25
d
24
From Fig. 3.32,
r =
π c3τ
2K
=
1
( D − d ) = 3 mm
2
K = 1.31
c=
T =
π (0.013)3 (80 × 106 )
(2)(1.36)
= 203 N ⋅ m
T = 203 N ⋅ m 
r
3
=
= 0.125
d
24
1
d = 12 mm = 0.012 m
2
π c3τ
2K
=
π (0.012)3 (80 × 106 )
(2)(1.31)
= 165.8 N ⋅ m
T = 165.8 N ⋅ m 
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PROBLEM 3.92
A 30-mm diameter solid rod is made of an elastoplastic material with τ Y = 3.5 MPa. Knowing that the elastic
core of the rod is 25 mm in diameter, determine the magnitude of the applied torque T.
SOLUTION
τ Y = 3.5 × 106 Pa, c =
1
(30 mm) = 15 mm = 0.015 m
2
1
(25 mm) = 12.5 mm = 0.0125 m
2
J
π
π
TY = τ Y = c3τ Y = (0.015)3 (3.5 × 106 ) = 18.555 N ⋅ m
c
2
2
ρY =
T =

ρ3  4
4 
(0.0125)3 
TY 1 − Y3  = (18.555) 1 −

3 
c  3
(0.015)3 

= 21.2 N ⋅ m
T = 21.2 N ⋅ m 
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PROBLEM 3.93
The solid circular shaft shown is made of a steel that is assumed to be
elastoplastic with τ Y = 21 ksi. Determine the magnitude T of the
applied torques when the plastic zone is (a) 0.8 in. deep, (b) 1.2 in. deep.
SOLUTION
τ Y = 21 ksi
TY =
τY J
c
= τY
π
2
c3
π
(1.5 in.)3
2
TY = 111.3 kip ⋅ in
= (21 ksi)
(a)
For t = 0.8 in. ρY = 1.5 − 0.8 = 0.7 in.
Eq. (3.32)
T =

4 
1 ρY3  4
1 (0.7 in.)3 
TY 1 −
=
⋅
−
(111.3
kip
in)
1



3 
4 c3  3
4 (1.5 in.)3 

T = 144.7 kip ⋅ in 
(b)
For t = 1.2 in.
T =
ρY = 1.5 − 1.2 = 0.3 in.

4 
1 ρY3  4
1 (0.3 in.)3 
TY 1 −
=
⋅
−
(111.3
kip
in)
1



3 
4 c3  3
4 (1.5 in.)3 

T = 148.1 kip ⋅ in 
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PROBLEM 3.94
The solid circular shaft shown is made of a steel that is assumed to be
elastoplastic with τ Y = 145 MPa. Determine the magnitude T of the
applied torques when the plastic zone is (a) 16 mm deep, (b) 24 mm
deep.
SOLUTION
c = 32 mm = 0.032 m
τ Y = 145 × 106 Pa
TY =
π
π
Jτ Y
= c3τ Y = (0.032)3 (145 × 106 )
c
2
2
= 7.4634 × 103 N ⋅ m
(a)
t P = 16 mm = 0.016 m
ρY = c − tP = 0.032 − 0.016 = 0.016 m
4  1 ρY3  4
1 0.0163 
3 
=
×
−
T = TY 1 −
(7.4634
10
)
1

 4 0.0323 
3  4 c3  3


= 9.6402 × 103 N ⋅ m
(b)
T = 9.64 kN ⋅ m 
t P = 24 mm = 0.024 m
ρY = c − tP = 0.032 − 0.024 = 0.008 m
T =
4 
1 ρY 3  4
1 0.0083 
3 
=
×
−
TY 1 −
(7.4634
10
)
1




3 
4 c3  3
4 0.0323 

= 9.9123 × 103 N ⋅ m
T = 9.91 kN ⋅ m 
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PROBLEM 3.95
The solid shaft shown is made of a mild steel that is assumed to be
elastoplastic with G = 11.2 × 106 psi and τ Y = 21 ksi. Determine the
maximum shearing stress and the radius of the elastic core caused by the
T = 100 kip ⋅ in.,
application
of
torque
of
magnitude
(a)
(b) T = 140 kip ⋅ in.
SOLUTION
c = 1.5 in., J =
(a)
π
2
c 4 = 7.9522 in 4 , τ Y = 21 ksi
T = 100 kip ⋅ in
τm =
Tc (100 kip ⋅ in)(1.5 in.)
=
J
7.9522 in 4
τ m = 18.86 ksi 
Since τ m < τ Y , shaft remains elastic.
c = 1.500 in. 
Radius of elastic core:
(b)
T = 140 kip ⋅ in
τm =
(140)(1.5)
= 26.4 ksi.
7.9522
Impossible: τ m = τ Y = 21.0 ksi  
Plastic zone has developed. Torque at onset of yield is TY =
Eq. (3.32):
T =
J
7.9522
(21 ksi) = 111.33 kip ⋅ in
τY =
c
1.5
4 
1 ρY3 
TY 1 −

3 
4 c3 
3
T
140
 ρY 
 c  = 4 − 3 T = 4 − 3 111.33 = 0.22743


Y
ρY = 0.6104c = 0.6104(1.5 in.)
ρY
c
= 0.6104
ρY = 0.916 in. 
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PROBLEM 3.96
It is observed that a straightened paper clip can be twisted through several revolutions by the application of a
torque of approximately 60 mN ⋅ m. Knowing that the diameter of the wire in the paper clip is 0.9 mm,
determine the approximate value of the yield stress of the steel.
SOLUTION
c=
1
d = 0.45 mm = 0.45 × 10−3 m
2
TP = 60 mN ⋅ m = 60 × 10−3 N ⋅ m
TP =
4
4 Jτ Y
4 π
2π 3
= ⋅ c3 TY =
TY =
c τY
3
3 c
3 2
3
τY =
3TP
(3)(60 × 10−3 )
=
= 314 × 106 Pa
2π c3
2π (0.45 × 10−3 )3
τ Y = 314 MPa 
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PROBLEM 3.97
The solid shaft shown is made of a mild steel that is assumed to be
elastoplastic with τ Y = 145MPa. Determine the radius of the elastic core
caused by the application of a torque equal to 1.1 TY, where TY is the
magnitude of the torque at the onset of yield.
SOLUTION
c=
ρY
c
=
1
d = 15 mm
2
3
4−3
T
=
TY
T =
3
4   ρY  
TY 1 − 
 
3   c  
4 − (3)(1.1) = 0.88790
ρY = 0.88790 c = (0.88790)(15 mm)
ρY = 13.32 mm 
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PROBLEM 3.98
For the solid circular shaft of Prob. 3.95, determine the angle of twist caused
by the application of a torque of magnitude (a) T = 80 kip ⋅ in.,
(b) T = 130 kip ⋅ in.
SOLUTION
1
1
d = (3) = 1.5 in. τ Y = 21 × 103 psi
2
2
c=
L = 4 ft = 48 in. J =
Torque at onset of yielding:
Tc
J
τ =
τY J
TY =
(a)
T =
c
=
2
c4 =
π
2
(1.5)4 = 7.9522 in 4
τJ
c
(21 × 103 )(7.9522)
= 111.330 × 103 lb ⋅ in
1.5
T = 80 kip ⋅ in = 80 × 103 lb ⋅ in
Since T < TY , the shaft is fully elastic. ϕ =
TL
GJ
(80 × 103 )(48)
= 43.115 × 10−3 rad
(11.2 × 106 )(7.9522)
ϕ =
(b)
π
3
T = 130 kip ⋅ in = 130 × 10 lb ⋅ in
ϕY =
T > TY
3
4   ϕY  
T = TY 1 − 
 
3  ϕ  


TY L
(111.330 × 103 )(48)
=
= 60.000 × 10−3 rad
6
GJ
(11.2 × 10 )(7.9522)
T
ϕY
= 34−3
=
TY
ϕ
ϕ =
ϕ = 2.47° 
ϕY
0.79205
=
3
4−
(3)(130 × 103 )
= 0.79205
111.330 × 103
60.000 × 10−3
= 75.75 × 10−3 rad
0.79205
ϕ = 4.34° 
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PROBLEM 3.99
The solid shaft shown is made of a mild steel that is assumed to be
elastoplastic with G = 77.2 GPa and τ Y = 145 MPa. Determine the angle
of twist caused by the application of a torque of magnitude (a)
T = 600 N ⋅ m, (b) T = 1000 N ⋅ m.
SOLUTION
1
d = 15 mm = 15 × 10−3 m
2
c=
Torque at onset of yielding:
τ =
TY =
(a)
π c3τ Y
2
=
π (15 × 10−3 )3 (145 × 106 )
2
= 768.71 N ⋅ m
T = 600 N ⋅ m. Since T ≤ TY , the shaft is elastic.
ϕ=
(b)
Tc
2T
=
J
π c3
T = 1000 N ⋅ m.
TL
2TL
(2)(600)(1.2)
=
=
= 0.11728 rad
GJ
π c 4G π (15 × 10−3 ) 4 (77.2 × 109 )
ϕ = 6.72° 
T > TY A plastic zone has developed.
3
T 
4   ϕY  
ϕY
= 3 4 − 3 
TY 1 − 
 
3  ϕ  
ϕ
 TY 


T L
2T L
(2)(768.71)(2.1)
ϕY = Y = Y4 =
= 0.15026 rad
GJ
π c G π (15 × 10−3 ) 4 (77.2 × 109 )
T =
ϕY
(3)(1000)
= 34−
= 0.46003
ϕ
768.71
ϕY
0.15026
=
= 0.32663 rad
ϕ =
0.46003
0.46003
ϕ = 18.71° 
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PROBLEM 3.100
A 3-ft-long solid shaft has a diameter of 2.5 in. and is made of a mild steel that is assumed to be elastoplastic
with τ Y = 21 ksi and G = 11.2 × 106 psi. Determine the torque required to twist the shaft through an angle of
(a) 2.5°, (b) 5°.
SOLUTION
L = 3 ft = 36 in., c =
J =
(a)
π
2
π
c4 =
2
1
d = 1.25 in., τ Y = 21 × 103 psi
2
(1.25) 4 = 3.835 in4
(3.835)(21 × 103 )
Jτ Y
=
= 64.427 × 103 lb ⋅ in
1.25
c
τY =
TY c
J
ϕY =
(64.427 × 103 )(36)
TY L
=
= 53.999 × 10−3 rad = 3.0939°
GJ
(11.2 × 106 )(3.835)
TY =
ϕ = 2.5° = 43.633 × 10−3 rad
ϕ < ϕY
ϕ =
T =
The shaft remains elastic.
TL
GJ
GJ ϕ
(11.2 × 106 )(3.835)(43.633 × 10−3 )
=
L
36
= 52.059 × 103 lb ⋅ in
(b)
ϕ = 5° = 87.266 × 10−3 rad
ϕ > ϕY
T =
T = 52.1 kip ⋅ in 
A plastic zone occurs.
3
4 
1ϕ  
TY 1 −  Y  
3 
4 ϕ  


3

4
1  53.999 × 10−3  
3 
= (64.427 × 10 ) 1 − 


3
4  87.266 × 10−3  


= 80.814 × 103 lb ⋅ in
T = 80.8 kip ⋅ in 
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PROBLEM 3.101
For the solid shaft of Prob. 3.99, determine (a) the magnitude of the torque T
required to twist the shaft through an angle of 15°, (b) the radius of the
corresponding elastic core.
PROBLEM 3.99 The solid shaft shown is made of a mild steel that is
assumed to be elastoplastic with G = 77.2 GPa and τ Y = 145 MPa.
Determine the angle of twist caused by the application of a torque of
magnitude (a) T = 600 N ⋅ m, (b) T = 1000 N ⋅ m.
SOLUTION
1
d = 15 mm = 15 × 10−3 m
2
ϕ = 15° = 0.2618 rad
c=
ϕY =
(a)
Lγ Y
Lτ
(1.2)(145 × 106 )
= Y =
= 0.15026 rad
c
cG
(15 × 10−3 )(77.2 × 109 )
Since ϕ > ϕY , there is a plastic zone.
τY =
TY =
TY c
2T
=
J
π c3
π c3τ Y
2
=
π (15 × 10−3 )3 (145 × 106 )
2
= 768.71 N ⋅ m
3
3

4 
1ϕ   4
1  0.15026  
T = TY 1 −  Y   = (768.71) 1 − 
 
3 
4 ϕ   3
4  0.2618  



= 976.5 N ⋅ m
(b)
T = 977 N ⋅ m 
Lγ Y = ρY ϕ = cϕY
ρY =
cϕY
ϕ
=
(15 × 10−3 )(0.15026)
= 8.61 × 10−3 m
0.2618
ρY = 8.61 mm 
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PROBLEM 3.102
The shaft AB is made of a material that is elastoplastic with
τ Y = 12 ksi and G = 4.5 × 106 psi. For the loading shown, determine
(a) the radius of the elastic core of the shaft, (b) the angle of twist at
end B.
SOLUTION
(a)
Radius of elastic core.
τ Y = 12 × 103 psi
c = 0.5in.
TY =
Jτ Y
π
π
= c3τ Y = (0.5)3 (12 × 103 )
2
2
c
= 2356.2 lb ⋅ in
T = 2560 lb ⋅ in > TY (plastic region with elastic core)
T =
ρY 3
c
c
(b)
=4−
3
ρY
3
4 
1 ρY 

TY 1 −
3 
4 c3 


3T
(3)(2560)
= 4−
= 0.74051
2356.2
TY
ρY = (0.9047)(0.5)
= 0.9047
ρY = 0.452 in. 
Angle of twist.
L = 6.4 ft = 76.8 in.
ϕY =
G = 4.5 × 106 psi
TY L
2T L
(2)(2356.2)(76.8)
= Y4 =
= 0.4096 radians
JG
π c G π (0.5) 4 (4.5 × 106 )
ϕY
ρ
= Y
ϕ
c
ϕ =
ϕY c 0.4096
=
= 0.4527 radians
ρY
0.9047
ϕ = 25.9° 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 3.103
A 1.25-in.-diameter solid circular shaft is made of a material that is assumed to be elastoplastic with
τ Y = 18 ksi and G = 11.2 × 106 psi. For an 8-ft length of the shaft, determine the maximum shearing stress
and the angle of twist caused by a 7.5 kip ⋅ in. torque.
SOLUTION
c=
1
d = 0.625 in., G = 11.2 × 106 psi, τ Y = 18 ksi = 18000 psi
2
L = 8 ft = 96 in. T = 7.5 kip ⋅ in = 7.5 × 103 lb ⋅ in
TY =
Jτ Y
π
π
= c3τ Y = (0.625)3 (18000) = 6.9029 × 103 lb ⋅ in
c
2
2
T > TY : plastic region with elastic core
γY =
cϕY
L
∴ τ max = τ Y = 18 ksi
∴ ϕY =
τ max = 18 ksi 
Lγ Y
Lτ
(96)(18000)
= Y =
= 246.86 × 10−3 rad
c
cG
(0.625)(11.2 × 106 )
4 
1 ϕ Y3 
TY 1 −

3 
4 ϕ 3 
1
1
ϕ
=
=
= 1.10533
3
ϕY
3T
(3)(7.5
10
)
×
34 −
34 −
TY
6.9029 × 103
T =
ϕ = 1.10533ϕY = (1.10533)(246.86 × 10−3 ) = 272.86 × 10−3 rad
ϕ = 15.63° 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 3.104
A 18-mm-diameter solid circular shaft is made of a material that is assumed to be elastoplastic with
τ Y = 145 MPa and G = 77 GPa. For a 1.2-m length of the shaft, determine the maximum shearing stress and
the angle of twist caused by a 200 N ⋅ m torque.
SOLUTION
τ Y = 145 × 106 Pa, c =
TY =
1
d = 0.009 m, L = 1.2 m, T = 200 N ⋅ m
2
Jτ Y
π
π
= c3τ Y = (0.009)3 (145 × 106 ) = 166.04 N ⋅ m
c
2
2
τ max = τ Y = 145MPa 
T > TY (plastic region with elastic core)
ϕY =
T =
TY L
2T L
(2)(166.04)(1.2)
= Y4 =
= 251.08 × 10−3 radians
GJ
π c G π (0.009)4 (77 × 109 )
4 
1 ϕ3 
TY 1 −

3 
4 ϕY3 
3
 ϕY 
3T
(3)(200)
=4−
= 0.38641
  = 4−
166.04
ϕ
T
 
Y
ϕ =
ϕ
0.72837
=
ϕY
= 0.72837
ϕ
251.08 × 10−3
= 344.7 × 10−3 radians
0.72837

ϕ = 19.75° 
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PROBLEM 3.105
A solid circular rod is made of a material that is assumed to be elastoplastic. Denoting by TY and φY,
respectively, the torque and the angle of twist at the onset of yield, determine the angle of twist if the torque is
increased to (a) T = 1.1 TY , (b) T = 1.25 TY , (c) T = 1.3 TY .
SOLUTION
T =

4
1 ϕY 3 
TY 1 −

3
4 ϕ 3 

ϕY
3T
= 34−
TY
ϕ
(a)
T
= 1.10
TY
ϕ
=
ϕY
(b)
T
= 1.25
TY
ϕ
=
ϕY
(c)
T
= 1.3
TY
ϕ
=
ϕY
3
3
3
or
ϕ
=
ϕY
1
3
4−
3T
TY
1
= 1.126
4 − (3)(1.10)
ϕ = 1.126 ϕY 
1
= 1.587
4 − (3)(1.25)
ϕ = 1.587 ϕY 
1
= 2.15
4 − (3)(1.3)
ϕ = 2.15 ϕY 
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PROBLEM 3.106
The hollow shaft shown is made of a steel that is assumed to be
elastoplastic with τ Y = 145 MPa and G = 77.2 GPa. Determine
the magnitude T of the torque and the corresponding angle of twist
(a) at the onset of yield, (b) when the plastic zone is 10 mm deep.
SOLUTION
(a)
At the onset of yield, the stress distribution is the elastic distribution with τ max = τ Y .
c2 =
J =
1
1
d 2 = 0.030 m, c1 = d1 = 0.0125 m
2
2
π
(c
2
4
2
τ max = τ Y =
)
− c14 =
π
2
(0.0304 − 0.01254 ) = 1.2340 × 10−6 m 4
TY c2
Jτ
(1.2340 × 10−6 )(145 × 106 )
∴ TY = Y =
= 5.9648 × 103 N ⋅ m
J
c2
0.030
TY = 5.96 kN ⋅ m 
TY L
(5.9643 × 103 )(5)
=
= 313.04 × 10−3 rad
GJ
(77.2 × 109 )(2.234010−6 )
ϕY =
(b)
ρϕ
L
ϕ =

ρY = c2 − t = 0.030 − 0.010 = 0.020 m
t = 0.010 m
γ =
=
ϕY = 17.94° 
ρY ϕ
L
= γY =
τY
G
τY L
(145 × 106 )(5)
=
= 469.56 × 10−3 rad
G ρY
(77.2 × 109 )(0.020)
Torque T1 carried by elastic portion:
τ = τ Y at ρ = ρY .
τY =
T1ρY
J1
ϕ = 26.9° 
c1 ≤ ρ ≤ ρY
where
J1 =
π
2
(ρ
4
Y
− c14
)
π
(0.0204 − 0.01254 ) = 212.978 × 10−9 m 4
2
Jτ
(212.978 × 10−9 )(145 × 106 )
T1 = 1 Y =
= 1.5441 × 103 N ⋅ m
ρY
0.020
J1 =
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PROBLEM 3.106 (Continued)
Torque T2 carried by plastic portion:
c2
2
T2 = 2π ρ τ Y ρ d ρ = 2π τ Y
Y
=
ρ3
3
c2
=
ρY
2π
τ Y c23 − ρY3
3
(
)
2π
(145 × 106 )(0.0303 − 0.0203 ) = 5.7701 × 103 N ⋅ m
3
Total torque:
T = T1 + T2 = 1.5541 × 103 + 5.7701 × 103 = 7.3142 × 103 N ⋅ m

T = 7.31 kN ⋅ m 
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PROBLEM 3.107
For the shaft of Prob. 3.106, determine (a) angle of twist at which
the section first becomes fully plastic, (b) the corresponding
magnitude T of the applied torque. Sketch the T − φ curve for the
shaft.
PROBLEM 3.106 The hollow shaft shown is made of a steel that
is assumed to be elastoplastic with τ Y = 145 MPa and
G = 77.2 GPa. Determine the magnitude T of the torque and the
corresponding angle of twist (a) at the onset of yield, (b) when the
plastic zone is 10 mm deep.
SOLUTION
1
d1 = 0.0125 m
2
c1 =
(a)
1
d 2 = 0.030 m
2
For onset of fully plastic yielding, ρY = c1
τ = τY ∴ γ =
TP = 2π
=
τY
G
=
ρY ϕ
L
=
c1ϕ
L
Lτ Y
(25)(145 × 106 )
=
= 751.295 × 10−3 rad
c1G
(0.0125)(77.2 × 109)
ϕf =
(b)
c2 =

c2
c1
2
τ Y ρ d ρ = 2π τ Y
ρ3
3
c2
=
c1
2π
τ Y c23 − c13
3
(
)
2π
(145 × 106 )(0.0303 − 0.01253 ) = 7.606 × 103 N ⋅ m
3
From Prob. 3.101,
ϕY = 17.94°
Also from Prob. 3.101,
ϕ f = 43.0° 
TP = 7.61 kN ⋅ m 
TY = 5.96 kN ⋅ m
ϕ = 26.9°
T = 7.31 kN ⋅ m
Plot T vs ϕ using the following data.
ϕ , deg
T kN ⋅ m
0
17.94
26.9
43.0
>43.0
0
5.96
7.31
7.61
7.61
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PROBLEM 3.108
A steel rod is machined to the shape shown to form a tapered solid shaft to which
torques of magnitude T = 75 kip ⋅ in. are applied. Assuming the steel to be
elastoplastic with τ Y = 21 ksi and G = 11.2 × 106 psi, determine (a) the radius of the
elastic core in portion AB of the shaft, (b) the length of portion CD that remains fully
elastic.
SOLUTION
(a)
c=
In portion AB.
TY =
T =
ρY
c
=
1
d = 1.25 in.
2
JABτ Y
π
π
= c3τ Y = (1.25)3 (21 × 103 ) = 64.427 × 103 lb ⋅ in
2
2
c
ρ3 
4 
TY 1 − Y3 
3 
c 
3
4−
3T
=
TY
3
4−
(3)(75 × 103 )
= 0.79775
64.427 × 103
ρY = 0.79775c = (0.79775)(1.25) = 0.99718 in.
(b)
For yielding at point C.
ρY = 0.997 in. 
τ = τ Y , c = cx , T = 75 × 103 lb ⋅ in
T =
J Cτ Y
π
= cx3τ Y
cx
2
cx =
3
2T
πτ Y
=
3
(2)(75 × 103 )
= 1.31494 in.
π (21 × 103 )
Using proportions from the sketch,
1.50 − 1.31494
x
=
1.50 − 1.25
5
x = 3.70 in. 
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PROBLEM 3.109
If the torques applied to the tapered shaft of Prob. 3.108 are slowly increased,
determine (a) the magnitude T of the largest torques that can be applied to the shaft,
(b) the length of the portion CD that remains fully elastic.
PROBLEM 3.108 A steel rod is machined to the shape shown to form a tapered solid
shaft to which torques of magnitude T = 75 kip ⋅ in. are applied. Assuming the steel
to be elastoplastic with τ Y = 21 ksi and G = 11.2 × 106 psi, determine
(a) the radius of the elastic core in portion AB of the shaft, (b) the length of portion
CD that remains fully elastic.
SOLUTION
(a)
The largest torque that may be applied is that which makes portion AB fully plastic.
c=
In portion AB,
TY =
1
d = 1.25 in.
2
Jτ Y
π
π
= c3τ Y = (1.25)3 (21 × 103 ) = 64.427 × 103 lb ⋅ in
2
2
c
For fully plastic shaft, ρY = 0
T =
T =
(b)
4 
1 ρY3  4
TY 1 −
= T
3 
4 c3  3
4
(64.427 × 103 ) = 85.903 × 103 lb ⋅ in
3
For yielding at point C, τ = τ Y ,
c = cx ,
τY =
Tcx
2T
=
Jx
π cx3
cx =
3
2T
πτ Y
=
3
T = 85.9 kip ⋅ in 
T = 85.903 × 103 lb ⋅ in
(2)(85.903 × 103 )
= 1.37580 in.
π (21 × 103 )
Using proportions from the sketch,
1.50 − 1.37580
x
=
1.50 − 1.25
5
x = 2.48 in. 
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PROBLEM 3.110
A hollow shaft of outer and inner diameters respectively equal to 0.6 in.
and 0.2 in. is fabricated from an aluminum alloy for which the stressstrain diagram is given in the diagram shown. Determine the torque
required to twist a 9-in. length of the shaft through 10°.
SOLUTION
ϕ = 10° = 174.53 × 10−3 rad
c1 =
1
1
d1 = 0.100 in, c2 = d 2 = 0.300 in.
2
2
γ max =
c2ϕ
(0.300)(174.53 × 10−3 )
=
= 0.00582
L
9
γ min =
c1ϕ
(0.100)(174.53 × 10−3 )
=
= 0.00194
L
9
z=
Let
γ
=
γ max
T = 2π

c2
c1
ρ
z1 =
c2
ρ 2τ d ρ = 2π c23
I=
where the integral I is given by
c1 1
=
c2 3

1
1/3

1
z1
z 2τ dz = 2π c23 I
z 2τ dz
Evaluate I using a method of numerical integration. If Simpson’s rule is used, the integration formula is
I =
Δz
Σ w z2 τ
3
where w is a weighting factor. Using Δ z = 16 , we get the values given in the table below.
z
1/3
1/2
2/3
5/6
1
γ
0.00194
0.00291
0.00383
0.00485
0.00582
I =
τ, ksi
8.0
10.0
11.5
13.0
14.0
z 2τ , ksi
0.89
2.50
5.11
9.03
14.0
w
1
4
2
4
1
wz 2τ , ksi
0.89
10.00
10.22
36.11
14.00
71.22
←⎯
⎯ Σwz 2τ
(1/6)(71.22)
= 3.96 ksi
3
T = 2π c23I = 2π (0.300)3 (3.96) = 0.671 kip ⋅ in
T = 671 lb ⋅ in 
Note: Answer may differ slightly due to differences of opinion in reading the stress-strain curve.
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PROBLEM 3.111
Using the stress-strain diagram shown, determine (a) the torque that
causes a maximum shearing stress of 15 ksi in a 0.8-in.-diameter solid
rod, (b) the corresponding angle of twist in a 20-in. length of the rod.
SOLUTION
(a)
τ max = 15 ksi
c=
1
d = 0.400 in.
2
γ max = 0.008
From the stress-strain diagram,
z =
Let
γ
γ max
T = 2π
I=
where the integral I is given by
1

c
0
=
ρ
c
ρ 2τ d ρ = 2π c3
1
 z τ dz = 2π c I
2
3
0
 z τ dz
2
0
Evaluate I using a method of numerical integration. If Simpson’s rule is used, the integration formula is
I =
Δz
Σ w z2 τ
3
where w is a weighting factor. Using Δ z = 0.25, we get the values given in the table below.
wz 2τ , ksi
γ
τ, ksi
z 2τ , ksi
w
0
0.000
0
0.000
1
0.00
0.25
0.002
8
0.500
4
2.00
0.5
0.004
12
3.000
2
6.00
0.75
0.006
14
7.875
4
31.50
1.0
0.008
15
15.000
1
15.00
z
54.50
I =
(0.25)(54.50)
= 4.54 ksi
3
T = 2π c3I = 2π (0.400)3 (4.54)
(b)
γ max =
ϕ =
←⎯
⎯ Σwz 2τ
T = 1.826 kip ⋅ in 
cϕ
L
Lγ m
(20)(0.008)
=
= 400 × 10−3 rad
c
0.400
ϕ = 22.9° 
Note: Answers may differ slightly due to differences of opinion in reading the stress-strain curve.
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PROBLEM 3.112
A 50-mm-diameter cylinder is made of a brass for which the stressstrain diagram is as shown. Knowing that the angle of twist is 5° in a
725-mm length, determine by approximate means the magnitude T of
torque applied to the shaft
SOLUTION
ϕ = 5° = 87.266 × 10−3 rad
γ max
1
d = 0.025m, L = 0.725 m
2
cϕ
(0.025)(87.266 × 10−3 )
=
=
= 0.00301
L
0.725
z =
Let
c=
γ
γ max
T = 2π

c
0
=

ρ
c
ρ 2τ d ρ = 2π c3
1
 z τ dz = 2π c I
2
3
0
where the integral I is given by I =
1
 z τ dz
2
0
Evaluate I using a method of numerical integration. If Simpson’s rule is used, the integration formula is
I =
Δz
 wz 2τ
3
where w is a weighting factor. Using Δz = 0.25, we get the values given in the table below.
γ
z
τ , MPa
z 2τ , MPa
w
wz 2τ , MPa
0
0
0
0
1
0
0.25
0.00075
30
1.875
4
7.5
0.5
0.0015
55
13.75
2
27.5
0.75
0.00226
75
42.19
4
168.75
1.0
0.00301
80
80
1
80
←  wz 2τ
= 283.75 × 106 Pa
283.75
I =
(0.25)(283.75 × 106 )
= 23.65 × 106 Pa
3
T = 2π c3I = 2π (0.025)3 (23.65 × 106 ) = 2.32 × 103 N ⋅ m
T = 2.32 kN ⋅ m 
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PROBLEM 3.113
Three points on the nonlinear stress-strain diagram used in Prob. 3.112 are (0,0), (0.0015,55 MPa), and
(0.003,80 MPa). By fitting the polynomial τ = A + Bγ + Cγ 2 through these points, the following
approximate relation has been obtained.
T = 46.7 × 109 γ − 6.67 × 1012 γ 2
Solve Prob. 3.113 using this relation, Eq. (3.2), and Eq. (3.26).
PROBLEM 3.112 A 50-mm diameter cylinder is made of a brass for which the stress-strain diagram is as
shown. Knowing that the angle of twist is 5° in a 725-mm length, determine by approximate means the
magnitude T of torque applied to the shaft.
SOLUTION
ϕ = 5° = 87.266 × 10−3 rad, c =
γ max =
Let
z=
1
d = 0.025 m, L = 0.725 m
2
cϕ
(0.025)(87.266 × 10−3 )
=
= 3.009 × 10−3
0.725
L
γ
ρ
=
γ max c
T = 2π

c
0
ρ 2τ d ρ = 2π c3
1
 z τ dz
2
0
The given stress-strain curve is
2
τ = A + Bγ + Cγ 2 = A + Bγ max z + Cγ max
z2
T = 2π c3
 z ( A + Bγ
1
2
0
max z
)
2
+ Cγ max
z 2 dz
1
 1
2
= 2π c3 A z 2 dz + Bγ max z 3 dz + Cγ max
0
 0
1
1 2 
1
= 2π c 2  A + Bγ max + Cγ max

4
5
3


Data:
A = 0,
1
A = 0,
3

1

 z dz
4
0
B = 46.7 × 109 , C = −6.67 × 1012
1
1
Bγ max = (46.7 × 109 )(3.009 × 10−3 ) = 35.13 × 103
4
4
1 2
1
Cγ max = − (6.67 × 1012 )(3.009 × 10−3 )2 = −12.08 × 103
5
5
T = 2π (0.025)3(0 + (35.13 × 103 − 12.08 × 103 ) = 2.26 × 103 N ⋅ m
T = 2.26 kN ⋅ m 
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PROBLEM 3.114
The solid circular drill rod AB is made of a steel that is assumed to be elastoplastic
with τ Y = 22 ksi and G = 11.2 × 106 psi. Knowing that a torque T = 75 kip ⋅ in. is
applied to the rod and then removed, determine the maximum residual shearing stress
in the rod.
SOLUTION
c = 1.2 in.
π
L = 35 ft = 420 in.
π
c 4 = (1.2) 4 = 3.2572 in 4
2
2
(3.2572)(22)
Jτ
= 59.715 kip ⋅ in
TY = Y =
1.2
c
J =
Loading:
T = 75 kip ⋅ in
T =
ρY3
c
3
ρY
c
4 
1 ρY3 
TY 1 −

3 
4 c3 
=4−
3T
(3)(75)
= 4−
= 0.23213
59.715
TY
= 0.61458,
ρY = 0.61458c = 0.73749 in.
Unloading:
τ′ =
Tρ
J
At ρ = c
τ′ =
(75)(1.2)
= 27.63 ksi
3.2572
At ρ = ρY
τ′ =
(75)(0.73749)
= 16.98 ksi
3.2572
where
T = 75 kip ⋅ in
Residual:
τ res = τ load − τ ′
At ρ = c
τ res = 22 − 27.63 = −5.63 ksi
At ρ = ρY
τ res = 22 − 16.98 = 5.02 ksi
maximum τ res = 5.63 ksi 
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PROBLEM 3.115
In Prob. 3.114, determine the permanent angle of twist of the rod.
PROBLEM 3.114 The solid circular drill rod AB is made of steel that is assumed to
be elastoplastic with τ Y = 22 ksi and G = 11.2 × 106 psi. Knowing that a torque
T = 75kip ⋅ in. is applied to the rod and then removed, determine the maximum
residual shearing stress in the rod.
SOLUTION
From the solution to Prob. 3.114,
c = 1.2 in.
J = 3.2572 in 4
ρY
c
= 0.61458
ρY = 0.73749 in.
After loading,
γ =
ρϕ
L
∴ ϕ =
ϕload =
During unloading,
Lγ
ρ
=
Lγ Y
ρY
=
Lτ Y
ρY G
L = 35 ft = 420 in.
(420)(22 × 103 )
= 1.11865 rad = 64.09°
(0.73749)(11.2 × 106 )
ϕ′ =
TL
GJ
ϕ′ =
(75 × 103 )(420)
= 0.86347 rad = 49.47°
(11.2 × 106 )(3.2572)
(elastic)
T = 5 × 103 N ⋅ m
Permanent angle of twist.
ϕperm = ϕload − ϕ ′ = 1.11865 − 0.86347 = 0.25518
ϕ = 14.62° 
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PROBLEM 3.116
The solid shaft shown is made of a steel that is assumed to be
elastoplastic with τ Y = 145 MPa and G = 77.2 GPa. The torque is
increased in magnitude until the shaft has been twisted through 6°; the
torque is then removed. Determine (a) the magnitude and location of the
maximum residual shearing stress, (b) the permanent angle of twist.
SOLUTION
ϕ = 6° = 104.72 × 10−3 rad
c = 0.016 m
γ max =
γY =
ρY
c
=
(0.016)(104.72 × 10−3 )
cϕ
=
= 0.0027925
0.6
L
τY
G
=
γY
γ max
145 × 106
= 0.0018782
77.2 × 109
0.0018
=
= 0.67260
0.0027925
π
π
c 4 = (0.016)4 = 102.944 × 10−9 m 4
2
2
Jτ
π
π
TY = Y = c3τ Y = (0.016)3 (145 × 106 ) = 932.93 N ⋅ m
2
2
c
J =
At end of loading.
Unloading:
At ρ = c
At ρ = ρY
Residual:
(a)
Tload =
4 
1 ρ Y3  4
1


TY 1 −
= (932.93) 1 − (0.67433)3  = 1.14855 × 103 N ⋅ m
3 

3 
4 c  3
4


T ′ = 1.14855 × 103 N ⋅ m
elastic
τ′ =
T ′c
(1.14855 × 103 )(0.016)
=
= 178.52 × 106 Pa
J
102.944 × 10−9
τ′ =
T ′c ρY
= (178.52 × 106 )(0.67433) = 120.38 × 106 Pa
J c
ϕ′ =
(1.14855 × 103 )(0.6)
T ′L
=
= 86.71 × 10−3 rad = 4.97°
GJ
(77.2 × 109 )(102.944 × 10−9 )
τ res = τ load − τ ′
ϕ perm = ϕload − ϕ ′
At ρ = c
τ res = 145 × 106 − 178.52 × 106 = −33.52 × 106 Pa
τ res = −33.5 MPa
At ρ = ρY
τ res = 145 × 106 − 120.38 × 106 = 24.62 × 106 Pa
τ res = 24.6 MPa
Maximum residual stress:
(b)
ϕperm = 104.72 × 10−3 − 86.71 × 10−3 = 17.78 × 10−3 rad
33.5 MPa at ρ = 16 mm 
ϕperm = 1.032° 
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PROBLEM 3.117
After the solid shaft of Prob. 3.116 has been loaded and unloaded as
described in that problem, a torque T1 of sense opposite to the original
torque T is applied to the shaft. Assuming no change in the value of ϕY ,
determine the angle of twist ϕ1 for which yield is initiated in this second
loading and compare it with the angle ϕY for which the shaft started to
yield in the original loading.
PROBLEM 3.116 The solid shaft shown is made of a steel that is
assumed to be elastoplastic with τ Y = 145 MPa and G = 77.2 GPa.
The torque is increased in magnitude until the shaft has been twisted
through 6°; the torque is then removed. Determine (a) the magnitude
and location of the maximum residual shearing stress, (b) the permanent
angle of twist.
SOLUTION
From the solution to Prob. 3.116,
c = 0.016 m, L = 0.6 m
τ Y = 145 × 106 Pa,
J = 102.944 × 10−9 m 4
The residual stress at ρ = c is
τ res = 33.5 MPa
For loading in the opposite sense, the change in stress to produce reversed yielding is
τ1 = τ Y − τ res = 145 × 106 − 33.5 × 106 = 111.5 × 106 Pa
(102.944 × 10−9 )(111.5 × 106 )
T1c
Jτ
∴ T1 = 1 =
0.016
J
c
= 717 N ⋅ m
τ1 =
Angle of twist at yielding under reversed torque.
ϕ1 =
T1L
(717 × 103 )(0.6)
=
= 54.16 × 10−3 rad
GJ
(77.2 × 109 )(102.944 × 10−9 )
ϕ1 = 3.10° 
Angle of twist for yielding in original loading.
γ =
ϕY =
τY
G
=
cϕY
L
Lτ Y
(0.6)(145 × 106 )
=
= 70.434 × 10−3 rad
cG
(0.016)(77.2 × 109 )
ϕY = 4.04° 
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PROBLEM 3.118
The hollow shaft shown is made of a steel that is assumed to be
elastoplastic with τ Y = 145 MPa and G = 77.2 GPa. The
magnitude T of the torques is slowly increased until the plastic
zone first reaches the inner surface of the shaft; the torques are
then removed. Determine the magnitude and location of the
maximum residual shearing stress in the rod.
SOLUTION
1
d1 = 12.5 mm
2
1
c2 = d 2 = 30 mm
2
c1 =
When the plastic zone reaches the inner surface, the stress is equal to τ Y . The corresponding torque is
calculated by integration.
dT = ρτ dA = ρτ Y (2πρ d ρ ) = 2π τ Y ρ 2 d ρ
T = 2π τ Y

C2
ρ2 dρ =
C1
=
Unloading.
2π
τ Y c23 − c13
3
(
)
2π
(145 × 106 )[(30 × 10−3 )3 − (12.5 × 10−3 )3 ] = 7.6064 × 103 N ⋅ m
3
T ′ = 7.6064 × 103 N ⋅ m
J =
π
(c
2
4
2
)
− c14 =
π
2
[(30)4 − (12.5)4 ] = 1.234 × 106 mm 4 = 1.234 × 10−6 m 4
τ1′ =
(7.6064 × 103 )(12.5 × 10−3 )
T ′c1
=
= 77.050 × 106 Pa = 77.05 MPa
J
1.234 × 10−6
τ 2′ =
(7.6064 × 103 )(30 × 10−3 )
T ′c2
=
= 192.63 × 106 Pa = 192.63 MPa
J
1.234 × 10−6
Residual stress.
Inner surface:
τ res = τ Y − τ1′ = 145 − 77.05 = 67.95 MPa
Outer surface:
τ res = τ Y − τ 2′ = 145 − 192.63 = −47.63 MPa
Maximum residual stress:
68.0 MPa at inner surface. 
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PROBLEM 3.119
In Prob. 3.118, determine the permanent angle of twist of the rod.
PROBLEM 3.118 The hollow shaft shown is made of a steel that is
and
assumed to be elastoplastic with τ Y = 145 MPa
G = 77.2 GPa. The magnitude T of the torques is slowly increased
until the plastic zone first reaches the inner surface of the shaft; the
torques are then removed. Determine the magnitude and location of
the maximum residual shearing stress in the rod.
SOLUTION
1
d1 = 12.5 mm
2
1
c2 = d 2 = 30 mm
2
c1 =
When the plastic zone reaches the inner surface, the stress is equal to τ Y . The corresponding torque is
calculated by integration.
dT = ρτ dA = ρτ Y (2πρ d ρ ) = 2πτ Y ρ 2 d ρ
T = 2πτ Y
=

c2
c1
ρ2 dρ =
2π
τ Y c23 − c13
3
(
)
2π
(145 × 106 )[(30 × 10−3 )3 − (12.5 × 10−3 )3 ] = 7.6064 × 103 N ⋅ m
3
Rotation angle at maximum torque.
c1ϕmax
τ
= γY = Y
L
G
ϕmax =
Unloading.
τY L
Gc1
=
(145 × 106 )(5)
= 0.75130 rad
(77.2 × 109 )(12.5 × 10−3 )
T ′ = 7.6064 × 103 N ⋅ m
J =
ϕ′ =
π
(c
2
4
2
)
− c14 =
π
2
[(30)4 − (12.5)4 ] = 1.234 × 106 mm 4 = 1.234 × 10−6 m 4
(7.6064 × 103 )(5)
T ′L
=
= 0.39922 rad
GJ
(77.2 × 109 )(1.234 × 10−6 )
Permanent angle of twist.
ϕperm = ϕmax − ϕ ′ = 0.75130 − 0.39922 = 0.35208 rad
ϕperm = 20.2° 
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PROBLEM 3.120
A torque T applied to a solid rod made of an elastoplastic material is increased until
the rod is fully plastic and them removed. (a) Show that the distribution of residual
shearing stresses is as represented in the figure. (b) Determine the magnitude of the
torque due to the stresses acting on the portion of the rod located within a circle of
radius c0.
SOLUTION
(a)
ρY = 0, Tload =
After loading:
Tc
2T
2(Tload ) 4
=
=
= τY
3
3
J
πc
π c3
4 ρ
τ ′ = τY
3 c
τ′ =
Unloading:
4
3
4ρ 

= τ Y 1 −
3c 
c

τ res = τ Y − τ Y
To find c0 set,
τ res = 0 and ρ = c0
T0 = 2π

c0
0
ρ 2τ d ρ = 2π
 ρ3 4 ρ4 
= 2πτ Y 
−

 3 3 4c 

(3/4) c
0
(3/4) c
0
at ρ = c
ρ
Residual:
0 =1−
(b)
4
4π 3
2π 3
TY =
c τY =
c τY
3
32
3
4c0
3
∴ c0 = c
3c
4
c0 = 0.150c 
4ρ
 dρ
3 c
 1  3 3  4  1  3 4 
= 2πτ Y c3    −     
 3  4   3  4  4  


ρ 2τ Y 1 −
27  9π
9
τ Y c3 = 0.2209 τ Y c3
= 2πτ Y c3  −
=
 64 256  128
T0 = 0.221τ Y c3 
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PROBLEM 3.121
Determine the largest torque T that can be applied to each of the two
brass bars shown and the corresponding angle of twist at B, knowing that
τ all = 12 ksi and G = 5.6 × 106 psi.
SOLUTION
L = 25 in., G = 5.6 × 106 psi, τ all = 12 × 103 psi
τ max =
ϕ =
(a)
(b)
T
c1ab 2
or
T = c1ab 2τ max
TL
c2ab3G
or
ϕ =
a = 4 in., b = 1 in.,
a
= 4.0
b
c1Lτ max
c2bG
From Table 3.1: c1 = 0.282,
From (1):
T = (0.282)(4)(1)2 (12 × 103 ) = 13.54 × 103
From (2):
ϕ =
a = 2.4in.,
(1)
(2)
c2 = 0.281
T = 13.54 kip ⋅ in 
(0.282)(25)(12 × 103 )
= 0.05376 radians
(0.281)(1)(5.6 × 106 )
b = 1.6in.,
a
= 1.5
b
From Table 3.1: c1 = 0.231,
ϕ = 3.08° 
c2 = 0.1958
From (1):
T = (0.231)(2.4)(1.6)2 (12 × 103 ) = 17.03 × 103
T = 17.03 kip ⋅ in 
From (2):
ϕ =
(0.231)(25)(12 × 103 )
= 0.0395 radians
(0.1958)(1.6)(5.6 × 106 )
ϕ = 2.26° 
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PROBLEM 3.122
Each of the two brass bars shown is subjected to a torque of magnitude
T = 12.5 kip · in. Knowing that G = 5.6 × 106 psi, determine for each
bar the maximum shearing stress and the angle of twist at B.
SOLUTION
(a)
L = 25 in.,
G = 5.6 × 106 psi,
a = 4 in,
b = 1 in,
From Table 3.1:
τ max =
ϕ =
T = 12.5 × 103 lb ⋅ in
a
= 4.0
b
c1 = 0.282,
c2 = 0.281
T
12.5 × 103
=
= 11.08 × 103
(0.282)(4)(1) 2
c1ab 2
τ max = 11.08 ksi 
TL
(12.5 × 103 )(25)
=
= 0.04965 radians
(0.282)(4)(1)3 (5.6 × 106 )
c2ab3G
ϕ = 2.84° 
(b)
a = 2.4 in.,
From Table 3.1:
τ max =
ϕ =
b = 1.6 in.,
a
= 1.5
b
c1 = 0.231,
c2 = 0.1958
T
12.5 × 103
=
= 8.81 × 103
2
2
(0.231)(2.4)(1.6)
c1ab
τ max = 8.81 ksi 
TL
(12.5 × 106 )(25)
=
= 0.02899 radians
3
(0.1958)(2.4)(1.6)3 (5.6 × 106 )
c2ab G
ϕ = 1.661° 
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PROBLEM 3.123
Each of the two aluminium bars shown is subjected to a torque of
magnitude T = 1800 N ⋅ m. Knowing that G = 26 GPa, determine
for each bar the maximum shearing stress and the angle of twist at
B.
SOLUTION
T = 1800 N ⋅ m
(a)
τ max =
ϕ =
c1 = 0.208,
c2 = 0.1406
T
1800
=
= 40.1 × 106 Pa
2
2
c1ab
(0.208)(0.060)(0.060)
TL
(1800)(0.300)
=
= 0.011398 radians
3
c2ab G
(0.1406)(0.060)(0.060)3 (26 × 109 )
a = 95 mm = 0.095 m,
From Table 3.1:
τ max =
ϕ =
G = 26 × 109 Pa
a
= 1.0
b
a = b = 60 mm = 0.060 m
From Table 3.1:
(b)
L = 0.300 m
b = 38 mm = 0.038 m,
c1 = 0.258,
τ max = 40.1 MPa 
ϕ = 0.653° 
a
= 2.5
b
c2 = 0.249
T
1800
=
= 50.9 × 106 Pa
2
2
c1ab
(0.258)(0.095)(0.038)
TL
(1800)(0.300)
=
= 0.01600 radians
3
c2ab G
(0.249)(0.095)(0.038)3 (26 × 109 )
τ max = 50.9 MPa 
ϕ = 0.917° 
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PROBLEM 3.124
Determine the largest torque T that can be applied to each of the
two aluminium bars shown and the corresponding angle of twist at
B, knowing τ all = 50 MPa and G = 26 GPa.
SOLUTION
L = 0.300 m, G = 26 × 109 Pa, τ all = 50 × 106 Pa
τ max =
ϕ =
(a)
T
c1ab 2
or
TL
c2ab 4G
or
a = b = 60 mm = 0.060 m,
From Table 3.1:
(b)
c1 = 0.208,
T = c1ab 2τ max
ϕ =
(1)
c1Lτ max
c2bG
(2)
a
= 1.0
b
c2 = 0.1406
From (1):
T = (0.208)(0.060)(0.060)2 (50 × 106 ) = 2246 N ⋅ m
From (2):
ϕ =
(0.208)(0.300)(50 × 106 )
= 0.01422 radians
(0.1406)(0.060)(26 × 109 )
a = 95 mm = 0.095 m, b = 38 mm = 0.038 m,
From Table 3.1:
T = 2.25 kN ⋅ m 
ϕ = 0.815° 
a
= 2.5
b
c1 = 0.258, c2 = 0.249
From (1):
T = (0.258)(0.095)(0.038) 2 (50 × 106 ) = 1770 N ⋅ m
From (2):
ϕ =
(0.258)(0.300)(50 × 106 )
= 0.01573 radians
(0.249)(0.038)(26 × 109 )
T = 1.770 kN ⋅ m 
ϕ = 0.901° 
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PROBLEM 3.125
Determine the largest allowable square cross section of a steel shaft of length 20 ft if the maximum shearing
stress is not to exceed 10 ksi when the shaft is twisted through one complete revolution. Use
G = 11.2 × 106 psi.
SOLUTION
L = 20 ft = 240 in.
τ max = 10 ksi = 10 × 103 psi
ϕ = 1 rev = 2π radians
τ max =
T
c1ab 2
(1)
TL
c2ab3G
(2)
ϕ =
Divide (2) by (1) to eliminate T.
ϕ
τ max
=
c1ab 2 L
cL
= 1
3
c2bG
c2ab G
c1Lτ max
c2Gϕ
Solve for b.
b=
For a square section,
a
= 1.0
b
From Table 3.1,
c1 = 0.208,
c2 = 0.1406
b=
(0.208)(240)(10 × 103 )
(0.1406)(11.2 × 106 )(2π )
b = 0.0505 in. 
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PROBLEM 3.126
Determine the largest allowable length of a stainless steel shaft of
3 3
×
8 4
-in. cross section if the shearing stress
is not to exceed 15 ksi when the shaft is twisted through 15°. Use G = 11.2 × 106 psi.
SOLUTION
3
in. = 0.75 in.
4
3
b = in. = 0.375 in.
8
a=
τ max = 15 ksi = 15 × 103 psi
ϕ = 15° =
τ max =
ϕ =
Divide (2) by (1) to eliminate T.
Solve for L.
ϕ
τ max
=
L=
15π
rad = 0.26180 rad
180
T
c1ab 2
(1)
TL
c2ab3G
(2)
c1ab 2 L
cL
= 1
3
c2bG
c2ab G
c2bGϕ
c1τ max
a
0.75
=
=2
0.375
b
Table 3.1 gives
c1 = 0.246, c2 = 0.229
L=
(0.229)(0.375)(11.2 × 106 )(0.26180)
= 68.2 in.
(0.246)(15 × 103 )
L = 68.2 in. 
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PROBLEM 3.127
The torque T causes a rotation of 2° at end B of the stainless steel bar
shown. Knowing that b = 20 mm and G = 75GPa, determine the
maximum shearing stress in the bar.
SOLUTION
a = 30 mm = 0.030 m
b = 20 mm = 0.020 m
ϕ = 2° = 34.907 × 10−3 rad
ϕ =
TL
c2ab3Gϕ
T
∴
=
L
c2ab3G
τ max =
T
c2ab3Gϕ
c bGϕ
=
= 2
c1L
c1ab 2
c1ab 2 L
30
a
=
= 1.5.
20
b
From Table 3.1,
c1 = 0.231
c2 = 0.1958
τ max =
(0.1958)(20 × 10−3 )(75 × 109 )(34.907 × 10−3 )
= 59.2 × 106 Pa
(0.231)(750 × 10−3 )
τ max = 59.2 MPa 
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PROBLEM 3.128
The torque T causes a rotation of 0.6° at end B of the aluminum bar
shown. Knowing that b = 15 mm and G = 26 GPa, determine the
maximum shearing stress in the bar.
SOLUTION
a = 30 mm = 0.030 m
b = 15 mm = 0.015 m
ϕ = 0.6° = 10.472 × 10−3 rad
ϕ=
c2 ab3Gϕ
TL
∴
=
T
c1 L
c2 ab3G
τ max =
c2 ab3Gϕ c2 bGϕ
T
=
=
c1 L
c1ab 2
c1ab 2 L
a 30
=
= 2.0
b 15
From Table 3.1,
c1 = 0.246
c2 = 0.229
τ max =
(0.229)(15 × 10−3 )(26 × 109 )(10.472 × 10−3 )
(0.246)(750 × 10−3 )
= 5.07 × 106 Pa
τ max = 5.07 MPa 
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PROBLEM 3.129
Two shafts are made of the same material. The cross section of shaft A is a
square of side b and that of shaft B is a circle of diameter b. Knowing that the
shafts are subjected to the same torque, determine the ratio τ A /τ B of maximum
shearing stresses occurring in the shafts.
SOLUTION
a
= 1, c1 = 0.208 (Table 3.1)
b
T
T
=
τA =
2
0.208b3
c1ab
A.
Square:
B.
Circle:
c=
Ratio:
τA
1
π
=
⋅
= 0.3005π
τ B 0.208 16
1
Tc
2T
16T
b τB =
=
=
2
J
π c3 π b3
τA
= 0.944 
τB
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PROBLEM 3.130
Shafts A and B are made of the same material and have the same
cross-sectional area, but A has a circular cross section and B has a
square cross section. Determine the ratio of the maximum shearing
stresses occurring in A and B, respectively, when the two shafts are
subjected to the same torque (TA = TB ). Assume both deformations
to be elastic.
SOLUTION
Let c be the radius of circular section A and b be the side square section B.
For equal areas,
π c2 = b2
TAC
2T
= A3
J
πc
Circle:
τA =
Square:
a
=1
b
τB =
b=c π
c1 = 0.208
from Table 3.1
TB
T
= B3
c1ab 2
c1b
Ratio:
τ A 2TA c1b3 2c1b3 TA
T
=
=
= 2c1 π A
3
3
TB
τ B π c TB
π c TB
For TA = TB ,
τA
= (2)(0.208) π
τB
τA
= 0.737 
τB
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PROBLEM 3.131
Shafts A and B are made of the same material and have the same crosssectional area, but A has a circular cross section and B has a square
cross section. Determine the ratio of the maximum torques TA and TB
that can be safely applied to A and B, respectively.
SOLUTION
Let c = radius of circular section A and b = side of square section B.
For equal areas π c 2 = b 2 ,
c=
Circle:
τA =
b
π
TAc
2TA
=
J
π c3
∴
TA =
π
2
c3τ A
Square:
From Table 3.1,
c1 = 0.208
τB =
TA
T
= B3
c1ab 2
c1b
π c3τ
B
Ratio:
TA
= 2 3
TB
c1b τ B
For the same stresses,
τB = τ A ∴
∴
TB = c1b3τ B
π ⋅ b3 τ
1 τA
2 π 3/2 B
=
=
3
c1b τ B
2c1 π τ B
TA
1
=
TB
(2) (0.208) π
TA
= 1.356 
TB
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PROBLEM 3.132
Shafts A and B are made of the same material and have the same length
and cross-sectional area, but A has a circular cross section and B has a
square cross section. Determine the ratio of the maximum values of the
angles ϕ A and ϕ B through which shafts A and B, respectively, can be
twisted.
SOLUTION
Let c = radius of circular section A and b = side of square section B.
For equal areas,
π c2 = b2 ∴ b = π c
Circle:
γ max =
Square:
From Table 3.1,
Ratio:
For equal stresses, τ A = τ B
τA
G
=
cϕ A
L
c1 = 0.208,
∴
ϕA =
Lτ A
cG
c2 = 0.1406
τB =
TB
TB
=
2
c1ab
0.208 b3
ϕB =
0.208 b3τ B L 1.4794 Lτ B
TB L
=
=
bG
0.1406 b 4G
c2ab3G
∴
TB = 0.208 b3τ B
ϕ A Lτ A
bG
bτ
τ
=
⋅
= 0.676 A = 0.676 π A
cG 1.4794Lτ B
cτ B
ϕB
τB
ϕB
= 0.676 π
ϕA
ϕB
= 1.198 
ϕA
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PROBLEM 3.133
Each of the three aluminum bars shown is to be twisted through
an angle of 2°. Knowing that b = 30 mm, τ all = 50 MPa, and
G = 27 GPa, determine the shortest allowable length of each
bar.
SOLUTION
ϕ = 2° = 34.907 × 10−3 rad, τ = 50 × 106 Pa
τ =
For square and rectangle,
T
c1ab2
Divide to eliminate T; then solve for L.
(a)
Square:
L=
(b)
a
= 1.0
b
From Table 3.1,
c=
cGϕ
τ
Rectangle:
L=
TL
c2ab3G
ϕ
c ab2 L
= 1 3
τ
c2ab G
L=
=
c2bGϕ
c1τ
c1 = 0.208, c2 = 0.1406
L = 382 mm 
1
Tc
TL
b = 0.015 m τ =
ϕ =
2
J
GJ
Divide to eliminate T; then solve for L.
(c)
ϕ =
(0.1406)(0.030)(27 × 109 )(34.907 × 10−3 )
= 382 × 10−3 m
6
(0.208)(50 × 10 )
Circle:
L=
G = 27 × 109 Pa, b = 30 mm = 0.030 m
ϕ
JL
L
=
=
cGJ
cG
τ
(0.015)(27 × 109 )(34.907 × 10−3 )
= 283 × 10−3 m
50 × 106
a = 1.2b
a
= 1.2
b
From Table 3.1,
(0.1661)(0.030)(27 × 109 )(34.907 × 10−3 )
= 429 × 10−3 m
(0.219)(50 × 106 )
L = 283 mm 
c1 = 0.219, c2 = 0.1661
L = 429 mm 
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PROBLEM 3.134
Each of the three steel bars is subjected to a torque as shown.
Knowing that the allowable shearing stress is 8 ksi and that
b = 1.4 in., determine the maximum torque T that can be applied
to each bar.
SOLUTION
τ max = 8 ksi, b = 1.4 in.
(a)
a = b = 1.4 in.
Square:
a
= 1.0
b
c1 = 0.208
From Table 3.1,
τ max =
T
c1ab 2
T = c1ab 2τ max
T = (0.208)(1.4)(1.4)(1.4) 2 (8)
(b)
Circle:
1
b = 0.7 in.
2
2T
π
Tc
=
=
T = c3τ max
3
J
2
πc
c=
τ max
T =
(c)
T = 4.57 kip ⋅ in 
π
2
(0.7)3 (8)
Rectangle:
a = (1.2)(1.4) = 1.68 in.
From Table 3.1,
c1 = 0.219
T = 4.31 kip ⋅ in 
a
= 1.2
b
T = c1ab 2τ max = (0.219)(1.68)(1.4)2 (8)
T = 5.77 kip ⋅ in 
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PROBLEM 3.135
A 36-kip ⋅ in. torque is applied to a 10-ft-long steel angle with an L8 × 8 × 1 cross section.
From Appendix C, we find that the thickness of the section is 1 in. and that its area is,
15.00 in2. Knowing that G = 11.2 × 106 psi, determine (a) the maximum shearing stress
along line a-a, (b) the angle of twist.
SOLUTION
a =
(a)
A 15 in 2
=
= 15 in., b = 1 in.,
t
1 in.
Since
a
1
b
> 5, c1 = c2 = 1 − 0.630 
b
a
3
or
c1 = c2 =
1
0.630 
= 0.3193
1−
3 
15 
T = 36 × 103 lb ⋅ in; L = 120 in.; G = 11.2 × 106 psi
T
Maximum shearing stress:
τ max =
c1ab 2
τ max =
(b)
a
= 15
b
Angle of twist:
36 × 103
= 7.52 × 103 psi
2
(0.3193)(15)(1)
ϕ =
TL
c2ab 2G
ϕ =
(36 × 103 )(120)
= 0.08052 radians
(0.3193)(15)(1)3 (11.2 × 106 )
τ max = 7.52 ksi 
ϕ = 4.61° 
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PROBLEM 3.136
A 3-m-long steel angle has an L203 × 152 × 12.7 cross section. From
Appendix C, we find that the thickness of the section is 12.7 mm and
that its area is 4350 mm2. Knowing that τ all = 50 MPa and that
G = 77.2 GPa, and ignoring the effect of stress concentration,
determine (a) the largest torque T that can be applied, (b) the
corresponding angle of twist.
SOLUTION
A = 4350 mm 2 b = 12.7 mm a = ?
a =
Equivalent rectangle.
A 4350
=
= 342.52 mm
b
12.7
a
= 26.97
b
b
1
c1 = c2 = 1 − 0.630  = 0.32555
a
3
(a)
τ max =
T
c1ab 2
τ max = 50 × 106 Pa
T = c1ab2τ max = (0.32555) (26.97 × 10−3 ) (12.7 × 10−3 ) 2 (50 × 106 )
= 70.807 N ⋅ m
(b)
ϕ =
T = 70.8 N ⋅ m 
TL
(70.807) (3)
=
3
c2ab G
(0.32555)(26.97 × 10−3 )(12.7 × 10−3 ) (77.2 × 109 )
= 0.15299 rad
ϕ = 8.77° 
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PROBLEM 3.137
An 8-ft-long steel member with a W8 × 31 cross section is subjected to a 5-kip ⋅ in. torque.
The properties of the rolled-steel section are given in Appendix C. Knowing that
G = 11.2 × 106 psi, determine (a) the maximum shearing stress along line a-a, (b) the
maximum shearing stress along line b-b, (c) the angle of twist. (Hint: consider the web and
flanges separately and obtain a relation between the torques exerted on the web and a
flange, respectively, by expressing that the resulting angles of twist are equal.)
SOLUTION
a 7.995
=
= 18.38
b 0.435
Tf L
1
c1 = c2 = 1 − 0.630 b = 0.3219
ϕf =
a
3
c2ab3G
Gϕ f
Gϕ
T f = c2ab3
= Kf
where K f = c2ab3
L
L
K f = (0.3219) (7.995) (0.435)3 = 0.2138 in 3
a = 7.995 in., b = 0.435 in.,
Flange:
)
(
a = 8.0 − (2)(0.435) = 7.13 in., b = 0.285 in.,
Web:
a
7.13
=
= 25.02
b
0.285
b
Tw L

c1 = c2 = 1 1 − 0.630  = 0.3249
ϕw =
3
a
c2ab3G
Gϕ w
Gϕ
Tw = c2ab3
K w = c2ab3
= Kw
where
L
L
K w = (0.3249) (7.13) (0.285)3 = 0.0563 in 4
ϕ f = ϕw = ϕ
For matching twist angles:
T = 2T f + Tw = (2 K f + K w )
Total torque.
Gϕ
T
=
,
2K p + K w
L
Tf =
Tf =
KfT
2K f + K w
,
K wT
2K f + K w
(0.2138)(5000)
(0.0563) (5000)
= 2221 lb ⋅ in; Tw =
= 557 lb ⋅ in
(2) (0.2138) + 0.0563
(2) (0.2138) + 0.0563
Tf
2221
= 4570 psi
(0.3219)(7.995)(0.435) 2
τ f = 4.57 ksi 
Tw
557
=
= 2960 psi
2
c1ab
(0.3249)(7.13)(0.285)2
τ w = 2.96 ksi 
(a)
τf =
(b)
τw =
(c)
Gϕ
T
TL
=
∴ ϕ =
L
2K f + K w
G(2 K f + K w )
ϕ =
Tw =
Gϕ
L
c1ab
2
=
where L = 8 ft = 96 in.
(5000)(96)
= 88.6 × 10−3 rad
(11.2 × 10 )[(2)(0.2138) + 0.563]
6
ϕ = 5.08° 
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PROBLEM 3.138
A 4-m-long steel member has a W310 × 60 cross section. Knowing that
G = 77.2 GPa and that the allowable shearing stress is 40 MPa, determine
(a) the largest torque T that can be applied, (b) the corresponding angle of twist.
Refer to Appendix C for the dimensions of the cross section and neglect the
effect of stress concentrations. (See hint of Prob. 3.137.)
SOLUTION
W310 × 60, L = 4 m, G = 77.2 G Pa, τ all = 40 MPa
For one flange:
From App. C,
Eq. (3.45):
c1 = c2 =
Eq. (3.44):
φf =
Tf L
3
c2ab G
=
a = 203 mm, b = 13.1 mm, a/b = 15.50
1
0.630 
= 0.320
1−

3
15.50 
T f (4)
0.320(0.203)(0.0131)3 (77.2 × 109 )
(1)
φ f = 355.04 × 10−6T f
For web:
From App. C,
Eq. (3.45):
c1 = c2 =
Eq. (3.44):
φw =
a = 303 − 2(13.1) = 276.8 mm, b = 7.5 mm, a/b = 36.9
1
0.630 
= 0.328
1−

3
36.9 
Tw (4)
0.328(0.2768)(0.0075)3 (77.2 × 109 )
φw = 1.354.2 × 10−6Tw
(2)
Since angle of twist is the same for flanges and web:
φ f = φ w:
355.04 × 10−6T f = 1354.2 × 10−6Tw
T f = 3.814Tw
(3)
But the sum of the torques exerted on the two flanges and on the web is equal to the torque T applied to the
member:
2T f + Tw = T
(4)
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PROBLEM 3.138 (Continued)
Substituting for T f from (3) into (4):
2(3.814Tw ) + Tw = T
From (3):
T f = 3.814(0.11589T )
Tw = 0.11589T
T f = 0.44205T
(5)
(6)
For one flange:
From Eq. (3.43):
T f = c1ab 2τ max = 0.320(0.203)(0.0131) 2 (40 × 106 )
= 445.91 N ⋅ m
Eq. (6):
445.91 = 0.44205T
For web:
Tw = c1ab 2τ max = 0.328(0.2768)(0.0075)2 (40 × 106 )
T = 1009 N ⋅ m 
= 204.28 N ⋅ m
Eq. (5):
204.28 = 0.11589T
(a)
Largest allowable torque:
(b)
Angle of twist: Use T f , which is critical.
Eq. (1):
Use the smaller value.
φ = φ f = (355.04 × 10−6 )(445.91) = 0.15831 rad
T = 1763 N ⋅ m 
T = 1009 N ⋅ m 
φ = 9.07° 
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PROBLEM 3.139
A torque T = 750 kN ⋅ m is applied to a hollow shaft shown that has a uniform 8-mm
wall thickness. Neglecting the effect of stress concentrations, determine the shearing
stress at points a and b.
SOLUTION
Detail of corner.
1
t = e tan 30°
2
t
e=
2 tan 30°
=
8
= 6.928 mm
2 tan 30°
b = 90 − 2e = 76.144 mm
Area bounded by centerline.
a =
1
3
3 2
3
(76.144)2
b
b=
b =
2 2
4
4
= 2510.6 mm 2 = 2510.6 × 10−6 m 2
t = 0.008 m
τ =
T
750
=
= 18.67 × 106 Pa
−6
2ta (2)(0.008) (2510 × 10 )
τ = 18.67 MPa 
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PROBLEM 3.140
A torque T = 5 kN ⋅ m is applied to a hollow shaft having the cross section
shown. Neglecting the effect of stress concentrations, determine the shearing
stress at points a and b.
SOLUTION
T = 5 × 103 N ⋅ m
Area bounded by centerline.
a = bh = (69) (115) = 7.935 × 103 mm 2
= 7.935 × 10−3 m 2
At point a:
t = 6 mm = 0.006 m
τ =
T
5 × 103
=
2ta (2) (0.006) (7.935 × 10−3 )
= 52.5 × 106 Pa
At point b:
τ = 52.5 MPa 
t = 10 mm = 0.010 m
τ =
T
5 × 103
=
= 31.5 × 106 Pa
−3
2ta (2)(0.010)(7.935 × 10 )
τ = 31.5 MPa 
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PROBLEM 3.141
A 90-N ⋅ m torque is applied to a hollow shaft having the cross section
shown. Neglecting the effect of stress concentrations, determine the shearing
stress at points a and b.
SOLUTION
Area bounded by centerline.
a = 52 × 52 − 39 × 39 +
π
4
(39)2 = 2378 mm 2 = 2.378 × 10−3 m 2
T = 90 N ⋅ m

τa =
T
90 N ⋅ m
=
−3
2ta
2(4 × 10 m)(2.378 × 10−3 m 2 )
τ a = 4.73 MPa 
τb =
T
90 N ⋅ m
=
−3
2ta
2(2 × 10 m)(2.378 × 10−3m 2 )
τ b = 9.46 MPa 
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PROBLEM 3.142
A 5.6 kN ⋅m torque is applied to a hollow shaft having the cross section
shown. Neglecting the effect of stress concentrations, determine the
shearing stress at points a and b.
SOLUTION
Area bounded by centerline.
a = (96 mm)(95mm) +
π
2
(47.5mm)2 = 12.664 × 103 mm2
= 12.664 × 10−3 m 2
t = 5 mm = 0.005m 
At point a,
τ =
5.6 × 103
T
=
= 44.2 × 106 Pa
−3
2at
(2)(12.664 × 10 )(0.005)
τ = 44.2 MPa 
t = 8 mm = 0.008m
At point b,
τ =
T
5.6 × 103
=
= 27.6 × 106 Pa
2at
(2)(12.664 × 10−3 )(0.008)
τ = 27.6 MPa 
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PROBLEM 3.143
A hollow member having the cross section shown is formed from sheet
metal of 2-mm thickness. Knowing that the shearing stress must not
exceed 3 MPa, determine the largest torque that can be applied to the
member.
SOLUTION
Area bounded by centerline.
a = (48)(18) + (30)(18)
= 1404 mm 2 = 1404 × 10−6 m 2
t = 0.002 m
τ =
T
2ta
or
T = 2taτ = (2)(0.002)(1404 × 10−6 )(3 × 106 )
T = 16.85 N ⋅ m 
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PROBLEM 3.144
A hollow brass shaft has the cross section shown. Knowing that the
shearing stress must not exceed 12 ksi and neglecting the effect of stress
concentrations, determine the largest torque that can be applied to the
shaft.
SOLUTION
Calculate the area bounded by the center line of the wall cross section. The area is a rectangle with two semicircular cutouts.
b = 5 − 0.2 = 4.8 in.
h = 6 − 0.5 = 5.5 in.
r = 1.5 + 0.1 = 1.6 in.
π 
a = bh − 2  r 2  = (4.8)(5.5) − π (1.6) 2 = 18.3575 in 2
2 
T
τ max =
τ max = 12 × 103 psi
tmin = 0.2 in.
2atmin
T = 2atmin τ max = (2)(18.3575) (0.2)(12 × 103 ) = 88.116 × 103 lb ⋅ in
T = 88.1kip ⋅ in = 7.34 kip ⋅ ft 
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PROBLEM 3.145
A hollow member having the cross section shown is to be formed from sheet metal of
0.06 in. thickness. Knowing that a 1250 lb ⋅ in.-torque will be applied to the member,
determine the smallest dimension d that can be used if the shearing stress is not to
exceed 750 psi.
SOLUTION
Area bounded by centerline.
a = (5.94)(2.94 − d ) + 1.94 d = 17.4636 − 4.00 d
t = 0.06 in., τ = 750 psi, T = 1250 lb ⋅ in
T
τ=
2ta
a=
17.4636 − 4.00d =
d =
T
2tτ
1250
= 13.8889
(2) (0.06) (750)
3.5747
= 0.894 in.
4.00
d = 0.894 in. 
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PROBLEM 3.146
A hollow member having the cross section shown is to be formed from sheet metal of
0.06 in. thickness. Knowing that a 1250 lb ⋅ in.-torque will be applied to the member,
determine the smallest dimension d that can be used if the shearing stress is not to
exceed 750 psi.
SOLUTION
Area bounded by centerline.
a = (5.94)(2.94) − 2.06 d = 17.4636 − 2.06 d
t = 0.06 in., τ = 750 psi, T = 1250 lb ⋅ in
τ =
T
2ta
a=
17.4636 − 2.06d =
d =
T
2tτ
1250
= 13.8889
(2) (0.06) (750)
3.5747
= 1.735 in.
2.06
d = 1.735 in. 
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PROBLEM 3.147
A hollow cylindrical shaft was designed to have a uniform wall thickness of
0.1 in. Defective fabrication, however, resulted in the shaft having the cross
section shown. Knowing that a 15 kip ⋅ in.-torque is applied to the shaft,
determine the shearing stresses at points a and b.
SOLUTION
Radius of outer circle = 1.2 in.
Radius of inner circle = 1.1 in.
Mean radius = 1.15 in.
Area bounded by centerline.
a = π rm2 = π (1.15) 2 = 4.155 in 2
t = 0.08 in.
At point a,
τ =
T
15
=
2ta (2)(0.08)(4.155)
τ = 22.6 ksi 
t = 0.12 in.
At point b,
τ =
T
15
=
2ta (2)(0.12)(4.155)
τ = 15.04 ksi 
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PROBLEM 3.148
A cooling tube having the cross section shown is formed from a sheet of stainless
steel of 3-mm thickness. The radii c1 = 150 mm and c2 = 100 mm are measured to
the center line of the sheet metal. Knowing that a torque of magnitude T = 3 kN ⋅ m
is applied to the tube, determine (a) the maximum shearing stress in the tube, (b) the
magnitude of the torque carried by the outer circular shell. Neglect the dimension of
the small opening where the outer and inner shells are connected.
SOLUTION
Area bounded by centerline.
(
)
a = π c12 − c22 = π (1502 − 1002 ) = 39.27 × 103 mm 2
= 39.27 × 10−3 m 2
t = 0.003 m
T
3 × 103
=
= 12.73 × 106 Pa
2ta
(2)(0.003)(39.27 × 10−3 )
(a)
τ =
(b)
T1 = (2π c1t τ c1) = 2π c12 t τ
= 2π (0.150) 2 (0.003) (12.73 × 106 ) = 5.40 × 103 N ⋅ m
τ = 12.76 MPa 
T1 = 5.40 kN ⋅ m 
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PROBLEM 3.149
A hollow cylindrical shaft of length L, mean radius cm , and uniform
thickness t is subjected to a torque of magnitude T. Consider, on the
one hand, the values of the average shearing stress τ ave and the angle
of twist ϕ obtained from the elastic torsion formulas developed in
Sections 3.4 and 3.5 and, on the other hand, the corresponding values
obtained from the formulas developed in Sec. 3.13 for thin-walled
shafts. (a) Show that the relative error introduced by using the thinwalled-shaft formulas rather than the elastic torsion formulas is the
same for τ ave and ϕ and that the relative error is positive and
proportional to the ratio t/cm . (b) Compare the percent error
corresponding to values of the ratio t /cm of 0.1, 0.2, and 0.4.
SOLUTION
Let c2 = outer radius = cm +
1
2
t and c1 = inner radius = cm − 12 t
π
(c
2
4
2
)
− c14 =
π
(
)
c22 + c12 (c2 + c1)(c2 − c1)
2
1
1 
π
=  cm2 + cmt + t 2 + cm2 − cmt + t 2  (2cm ) t
2
4
4 
J =
1 

= 2π  cm2 + t 2  cmt
4 

Tc
T
τm = m =
 2 1 2
J
2π  cm + t  t
4 

TL
TL
=
ϕ1 =
 2 1 2
JG
2π  cm + t  cmt G
4 

Area bounded by centerline.
a = π cm2
τ ave =
ϕ2 =
(a)

Ratios:
T
T
=
2ta
2π cm2 t
TL
ds TL(2π cm /t )
TL
=
=

2 
2 2
4a G t
4(π cm ) G
2π cm3 t G
)
(
2π cm2 + 1 t 2 t
1 t2
T
τ ave
4
1
=
×
=
+
T
τm
4 cm2
2π cm2 t
(
)
2π cm2 + 1 t 2 cmtG
1 t2
TL
ϕ2
4
1

=
×
=
+
TL
ϕ1 2π cm3 tG
4 cm


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PROBLEM 3.149 (Continued)
2
(b)
τ ave
ϕ
1 t
−1 = 2 −1 =
τm
ϕ1
4 cm2

t
cm
0.1
0.2
0.4
1 t2
4 cm2
0.0025
0.01
0.04
%
0.25%
1%
4%

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PROBLEM 3.150
Equal torques are applied to thin-walled tubes of the same length L,
same thickness t, and same radius c. One of the tubes has been slit
lengthwise as shown. Determine (a) the ratio τ b / τ a of the maximum
shearing stresses in the tubes, (b) the ratio ϕb /ϕ a of the angles of twist
of the shafts.
SOLUTION
Without slit:
Area bounded by centerline. a = π c 2
τa =
T
T
=
2ta
2π c 2t
J ≈ 2π c3t
With slit:
ϕa =
a = 2π c, b = t ,
c1 = c2 =
TL
TL
=
GJ
2π c3t G
a
2π c
=
>> 1
b
t
1
3
τb =
3T
T
=
2
2π ct 2
c1ab
ϕb =
3TL
T
=
3
2π ct 3G
c2ab G
(a)
Stress ratio:
τb
3T
2π c 2t
3c
=
⋅
=
2
τa
T
t
2π ct
(b)
Twist ratio:
ϕb
3TL
2π c3t G 3c 2
=
⋅
= 2
ϕa
TL
2π ct 3G
t
τ b 3c

=
t
τa
ϕb 3c 2
= 2 
ϕa
t
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PROBLEM 3.151
The ship at A has just started to drill for oil on the ocean floor at a depth of 5000 ft.
Knowing that the top of the 8-in.-diameter steel drill pipe (G = 11.2 × 106 psi)
rotates through two complete revolutions before the drill bit at B starts to operate,
determine the maximum shearing stress caused in the pipe by torsion.
SOLUTION
TL
GJ ϕ
T =
GJ
L
Tc GJ ϕ c Gϕ c
τ =
=
=
J
JL
L
ϕ =
ϕ = 2 rev = (2)(2π ) = 12.566 rad,
c=
1
d = 4.0 in.
2
L = 5000 ft = 60000 in.
τ =
(11.2 × 106 )(12.566)(4.0)
= 9.3826 × 103 psi
60000
τ = 9.38 ksi 
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PROBLEM 3.152
The shafts of the pulley assembly shown are to be
redesigned. Knowing that the allowable shearing
stress in each shaft is 8.5 ksi, determine the smallest
allowable diameter of (a) shaft AB, (b) shaft BC.
SOLUTION
(a)
Shaft AB:
TAB = 3.6 × 103 lb ⋅ in
τ max = 8.5 ksi = 8.5 × 103 psi
J =
c=
(b)
Shaft BC:
π
2
3
c4
2TAB
πτ max
τ max =
=
3
2T
Tc
=
J
π c3
(2)(3.6 × 103 )
= 0.646 in.
π (8.5 × 103 )
d AB = 2c = 1.292 in. 
TBC = 6.8 × 103 lb ⋅ in
τ max = 8.5 × 103 psi
c=
3
2TBC
πτ max
=
3
(2)(6.8 × 103 )
= 0.7985 in.
π (8.5 × 103 )
d BC = 2c = 1.597 in. 
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PROBLEM 3.153
A steel pipe of 12-in. outer diameter is fabricated from
1
4 -in. -thick plate by welding along a helix which forms an angle
of 45° with a plane perpendicular to the axis of the pipe.
Knowing that the maximum allowable tensile stress in the weld
is 12 ksi, determine the largest torque that can be applied to the
pipe.
SOLUTION
From Eq. (3.14) of the textbook,
σ 45 = τ max
hence,
τ max = 12 ksi = 12 × 103 psi
1
1
d0 = (12) = 6.00 in.
2
2
c1 = c2 − t = 6.00 − 0.25 = 5.75 in.
c2 =
J =
τ max =
T =
π
(c
2
Tc
J
4
2
)
− c14 =
T =
π
2
[(6.00)4 − (5.75)4 ] = 318.67 in.
τ max J
c
3
(12 × 10 )(318.67)
= 637 × 103 lb ⋅ in
6.00
T = 637 kip ⋅ in 
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PROBLEM 3.154
For the gear train shown, the diameters of the three solid shafts
are:
d AB = 20 mm dCD = 25mm d EF = 40 mm
Knowing that for each shaft the allowable shearing stress is
60 MPa, determine the largest torque T that can be applied.
SOLUTION
Statics:
TAB = T
TCD
T
= AB
rC
rB
TCD =
rC
75
TAB =
T = 2.5T
rB
30
TEF
T
= CD
rF
rD
TEF =
rF
90
TCD =
(2.5T ) = 7.5T
rD
30
Determine the magnitude of T so that the stress is 60 MPa = 60 × 106 Pa.
Shaft AB:
Tc
J
c=
1
d AB = 10 mm = 0.010 m
2
TAB = T =
Shaft CD:
c=
Tshaft =
π
2
c=
(60 × 106 )(0.010)3
T = 94.2 N ⋅ m
1
dCD = 12.5 mm = 0.0125 m
2
TCD = 2.5T =
Shaft EF:
π
Jτ
= τ c3
2
c
τ =
π
2
(60 × 106 )(0.0125)3
T = 73.6 N ⋅ m
1
d EF = 20 mm = 0.020 m
2
TEF = 7.5T =
π
2
(60 × 106 )(0.020)3
T = 100.5 N ⋅ m
The smallest value of T is the largest torque that can be applied.
T = 73.6 N ⋅ m 
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PROBLEM 3.155
Two solid steel shafts (G = 77.2GPa) are connected to a coupling disk B
and to fixed supports at A and C. For the loading shown, determine (a) the
reaction at each support, (b) the maximum shearing stress in shaft AB, (c) the
maximum shearing stress in shaft BC.
SOLUTION
Shaft AB:
1
d = 25 mm = 0.025 m
2
π
π
T L
= c 4 = (0.025) 4 = 613.59 × 10−9 m 4 ϕ B = AB AB
2
2
GJ AB
T = TAB , LAB = 0.200 m, c =
J AB
TAB =
Shaft BC:
(77.2 × 109 )(613.59 × 10−9 )
GJ AB
ϕB =
ϕ B = 236.847 × 103 ϕ B
0.200
LAB
1
d = 19 mm = 0.019 m
2
π
π
T L
ϕ B = BC BC
= c 4 = (0.019) 4 = 204.71 × 10−9 m 4
2
2
GJ BC
T = TBC , LBC = 0.250 m, c =
J BC
TBC =
GJ BC
(77.2 × 109 )(204.71 × 10−9 )
= 63.214 × 103ϕ B
ϕB =
0.250
LBC
Equilibrium of coupling disk.
T = TAB + TBC
1.4 × 103 = 236.847 × 103ϕ B + 63.214 × 103ϕ B
ϕ B = 4.6657 × 10−3 rad.
TAB = (236.847 × 103 )(4.6657 × 10−3 ) = 1.10506 × 103 N ⋅ m
TBC = (63.214 × 103 )(4.6657 × 10−3 ) = 294.94 N ⋅ m
(a)
TA = TAB = 1105 N ⋅ m 
Reactions at supports.
TC = TBC = 295 N ⋅ m 
(b)
Maximum shearing stress in AB.
τ AB =
(c)
TABc (1.10506 × 103 )(0.025)
=
= 45.0 × 106 Pa
J AB
613.59 × 10−9
τ AB = 45.0 MPa 
Maximum shearing stress in BC.
τ BC =
TBC c
(294.94)(0.019)
=
= 27.4 × 106 Pa
J BC
204.71 × 10−9
τ BC = 27.4 MPa 
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PROBLEM 3.156
In the bevel-gear system shown, α = 18.43°. Knowing that the
allowable shearing stress is 8 ksi in each shaft and that the system is in
equilibrium, determine the largest torque TA that can be applied at A.
SOLUTION
Using stress limit for shaft A:
τ = 8 ksi,
TA =
c=
1
d = 0.25 in.
2
π
π
Jτ
= τ c3 = (8)(0.25)3 = 0.1963 kip ⋅ in
2
2
c
Using stress limit for shaft B:
τ = 8 ksi, c =
From statics,
1
d = 0.3125in.
2
TB =
π
π
Jτ
= τ c3 = (8)(0.3125)3 = 0.3835 kip ⋅ in
2
2
c
TA =
rA
TB = (tan α )TB
rB
TA = (tan18.43°)(0.3835) = 0.1278 kip ⋅ in
The allowable value of TA is the smaller.
TA = 0.1278 kip ⋅ in
TA = 127.8 lb ⋅ in 
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PROBLEM 3.157
Three solid shafts, each of
3
4
-in. diameter, are
connected by the gears shown. Knowing that
G = 11.2 × 106 psi, determine (a) the angle
through which end A of shaft AB rotates, (b) the
angle through which end E of shaft EF rotates.
SOLUTION
Geometry:
rB = 1.5 in., rC = 6 in., rF = 2in.
LAB = 48 in., LCD = 36 in., LEF = 48 in.
Statics:
TA = 100 lb ⋅ in
TE = 200 lb ⋅ in
Gear B.
ΣM B = 0:
−rB F1 + TA = 0
−1.5F1 + 100 = 0
F1 = 67.667 lb
Gear F.
ΣM F = 0:
−rF F2 + TE = 0
−2F2 + 200 = 0
F2 = 100 lb
Gear C.
ΣM C = 0:
−rC F1 − rC F2 + TC = 0
−(6)(66.667) − (6)(100) + TC = 0
TC = 1000 lb ⋅ in
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PROBLEM 3.157 (Continued)
Deformations:
c=
For all shafts,
J =
Kinematics:
1
d = 0.375 in.
2
π
2
c 4 = 0.031063 in 4
ϕ A/B =
TAB LAB
(100)(48)
=
= 0.013797 rad
GJ
(11.2 × 106 )(0.031063)
ϕ E /F =
TEF LEF
(200)(48)
=
= 0.027594 rad
GJ
(11.2 × 106 )(0.031063)
ϕ C /D =
TCD LCD
(1000)(36)
=
= 0.103476 rad
GJ
(11.2 × 106 )(0.031063)
ϕC = ϕC/D = 0.103476 rad
rBϕ B = rCϕC
(a)
rC
6
(0.103476) = 0.41390 rad
ϕC
rB
1.5
ϕ A = ϕ B + ϕ A/B = 0.41390 + 0.01397 = 0.42788 rad
rFϕ F = rCϕC
(b)
ϕB =
ϕF =
ϕ A = 24.5°

rC
6
ϕC = (0.103476) = 0.31043 rad
rF
2
ϕ E = ϕ F + ϕ E/F = 0.31043 + 0.027594 = 0.33802 rad
ϕ E = 19.37°

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PROBLEM 3.158
The design specifications of a 1.2-m-long solid transmission shaft require that the angle of twist of the shaft
not exceed 4° when a torque of 750 N ⋅ m is applied. Determine the required diameter of the shaft, knowing
that the shaft is made of a steel with an allowable shearing stress of 90 MPa and a modulus of rigidity of
77.2 GPa.
SOLUTION
T = 750 N ⋅ m, ϕ = 4° = 69.813 × 10−3 rad,
L = 1.2 m, J =
π
2
c4
τ = 90 MPa = 90 × 106 Pa G = 77.2 GPa = 77.2 × 109 Pa
Based on angle of twist. ϕ =
c=
Based on shearing stress. τ =
c=
Use larger value.
TL
2TL
=
GJ
π Gc 4
4
2TL
=
π Gϕ
4
(2)(750)(1.2)
= 18.06 × 10−3 m
π (77.2 × 109 )(69.813 × 10−3 )
Tc
2T
=
J
π c3
3
2T
πτ
=
3
(2)(750)
= 17.44 × 10−3 m
π (90 × 106 )
c = 18.06 × 10−3 m = 18.06 mm
d = 2c = 36.1 mm 
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PROBLEM 3.159
The stepped shaft shown rotates at 450 rpm. Knowing that r = 0.5in.,
determine the maximum power that can be transmitted without
exceeding an allowable shearing stress of 7500 psi.
SOLUTION
d = 5 in.
D = 6 in.
r = 0.5 in.
D
6
= = 1.20
d
5
r
0.5
=
= 0.10
d
5
K = 1.33
From Fig. 3.32,
1
d = 2.5 in.
2
KTc
2KT
τ =
=
J
π c3
c=
For smaller side,
T =
π c3τ
2K
=
π (2.5)3 (7500)
(2)(1.33)
= 138.404 × 103 lb ⋅ in
f = 450 rpm = 7.5 Hz
Power.
P = 2π f T = 2π (7.5)(138.404 × 103 ) = 6.52 × 106 in ⋅ lb/s
Recalling that 1 hp = 6600 in ⋅ lb/s,
P = 988 hp 
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PROBLEM 3.160
A 750-N ⋅ m torque is applied to a hollow shaft having the cross section shown
and a uniform 6-mm wall thickness. Neglecting the effect of stress concentrations,
determine the shearing stress at points a and b.
SOLUTION
Area bounded by centerline.
a =2
π
2
(33) 2 + (60)(66) = 7381 mm 2
= 7381 × 10−6 m 2
t = 0.006 m at both a and b,
Then at points a and b,
τ =
T
750
=
= 8.47 × 106 Pa
2ta
(2) (0.006) (7381 × 10−6 )
τ = 8.47 MPa 
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PROBLEM 3.161
The composite shaft shown is twisted by applying a torque T at end A.
Knowing that the maximum shearing stress in the steel shell is
150 MPa, determine the corresponding maximum shearing stress in the
aluminum core. Use G = 77.2 GPa for steel and G = 27 GPa for
aluminum.
SOLUTION
Let G1, J1, and τ1 refer to the aluminum core and G2 , J 2 , and τ 2 refer to the steel shell.
At the outer surface on the steel shell,
γ2 =
c2ϕ
ϕ γ2
τ
∴
=
= 2
L
L
c2
c2G2
At the outer surface of the aluminum core,
γ1 =
Matching
ϕ
L
c1ϕ
ϕ γ1
τ
∴
=
= 1
L
L
c1
c1G
for both components,
τ2
c2G2
Solving for τ 2 ,
τ2 =
=
τ1
c1G1
0.030 27 × 109
c2 G2
⋅
⋅
⋅ 150 × 106 = 39.3 × 106 Pa
τ1 =
9
0.040 77.2 × 10
c1 G1
τ 2 = 39.3 MPa 
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PROBLEM 3.162
Two solid brass rods AB and CD are brazed to a brass sleeve EF.
Determine the ratio d 2 /d1 for which the same maximum shearing stress
occurs in the rods and in the sleeve.
SOLUTION
Let
c1 =
1
d1
2
Shaft AB:
τ1 =
Tc1
2T
=
J1
π c13
Sleeve EF.
τ2 =
Tc2
2Tc2
=
J2
π c24 − c14
and
(
2T
2Tc2
=
3
π c1
π c24 − c14
For equal stresses,
(
c2 =
1
d2
2
)
)
c24 − c14 = c13c2
Let x =
c2
c1
x 4 − 1 = x or
x = 41 + x
Solve by successive approximations starting with x0 = 1.0.
x1 =
4
2 = 1.189, x2 =
x4 =
4
2.220 = 1.221, x5 =
x = 1.221
4
2.189 = 1.216, x3 =
c2
= 1.221
c1
4
4
2.216 = 1.220
2.221 = 1.221 (converged).
d2
= 1.221 
d1
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CHAPTER 4
PROBLEM 4.1
Knowing that the couple shown acts in a vertical plane, determine the
stress at (a) point A, (b) point B.
SOLUTION
For rectangle:
I=
1 3
bh
12
For cross sectional area:
I = I1 + I 2 + I 3 =
1
1
1
(2)(1.5)3 +
(2)(5.5)3 +
(2)(1.5)3 = 28.854 in 4
12
12
12
(a)
y A = 2.75 in.
σA = −
My A
(25)(2.75)
=−
I
28.854
(b)
yB = 0.75 in.
σB = −
MyB
(25)(0.75)
=−
I
28.854
σ A = −2.38 ksi 
σ B = −0.650 ksi 
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PROBLEM 4.2
Knowing that the couple shown acts in a vertical plane, determine
the stress at (a) point A, (b) point B.
SOLUTION
For rectangle:
I =
1 3
bh
12
Outside rectangle:
I1 =
1
(80)(120)3
12
I1 = 11.52 × 106 mm 4 = 11.52 × 10−6 m 4
Cutout:
I2 =
1
(40)(80)3
12
I 2 = 1.70667 × 106 mm 4 = 1.70667 × 10−6 m 4
Section:
(a)
I = I1 − I 2 = 9.81333 × 10−6 m 4
y A = 40 mm = 0.040 m
σA = −
My A
(15 × 103 )(0.040)
=−
= −61.6 × 106 Pa
I
9.81333 × 10−6
σ A = −61.6 MPa 
(b)
yB = −60 mm = −0.060 m
σB = −
MyB
(15 × 103 )(−0.060)
=−
= 91.7 × 106 Pa
−6
I
9.81333 × 10
σ B = 91.7 MPa 
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PROBLEM 4.3
Using an allowable stress of 16 ksi, determine the largest couple that can be
applied to each pipe.
SOLUTION
(a)
I =
π
(r
4
4
o
)
− ri4 =
π
4
(0.64 − 0.54 ) = 52.7 × 10−3 in 4
c = 0.6 in.
σ =
Mc
:
I
M =
σI
c
=
(16)(52.7 × 10−3 )
0.6
M = 1.405 kip ⋅ in 
(b)
π
(0.7 4 − 0.54 ) = 139.49 × 10−3 in 4
4
c = 0.7 in.
I =
σ =
Mc
:
I
M =
σI
c
=
(16)(139.49 × 10−3 )
0.7
M = 3.19 kip ⋅ in 
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PROBLEM 4.4
A nylon spacing bar has the cross section shown. Knowing that the
allowable stress for the grade of nylon used is 24 MPa, determine the
largest couple Mz that can be applied to the bar.
SOLUTION
I = I rect − I circle =
=
1 3 π 4
bh − r
12
4
1
π
(100)(80)3 − (25) 4 = 3.9599 × 106 mm 4
12
4
= 3.9599 × 10−6 m
c=
80
= 40 mm = 0.040 m
2
σ =
Mc
:
I
Mz =
σI
c
=
(24 × 106 )(3.9599 × 10−6 )
= 2.38 × 103 N ⋅ m
0.040
M z = 2.38 kN ⋅ m 
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PROBLEM 4.5
A beam of the cross section shown is extruded from an aluminum
alloy for which σ Y = 250 MPa and σ U = 450 MPa. Using a factor
of safety of 3.00, determine the largest couple that can be applied to
the beam when it is bent about the z-axis.
SOLUTION
Allowable stress.
=
σU
F.S .
=
450
= 150 MPa
3
= 150 × 106 Pa
Moment of inertia about z-axis.
1
(16)(80)3 = 682.67 × 103 mm 4
12
1
I 2 = (16)(32)3 = 43.69 × 103 mm 4
12
I 3 = I1 = 682.67 × 103 mm 4
I1 =
I = I1 + I 2 + I 3 = 1.40902 × 106 mm 4 = 1.40902 × 10−6 m 4
1
Mc
with c = (80) = 40 mm = 0.040 m
2
I
I σ (1.40902 × 10−6 )(150 × 106 )
=
= 5.28 × 103 N ⋅ m
M=
0.040
c
σ=
M = 5.28 kN ⋅ m 
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PROBLEM 4.6
Solve Prob. 4.5, assuming that the beam is bent about the y-axis.
PROBLEM 4.5 A beam of the cross section shown is extruded from an
aluminum alloy for which σ Y = 250 MPa and σ U = 450 MPa. Using a
factor of safety of 3.00, determine the largest couple that can be applied to
the beam when it is bent about the z-axis.
SOLUTION
=
Allowable stress:
σU
F.S .
=
450
= 150 MPa
3.00
= 150 × 106 Pa
Moment of inertia about y-axis.
1
(80)(16)3 + (80)(16)(16) 2 = 354.987 × 103 mm 4
12
1
I 2 = (32)(16)3 = 10.923 × 103 mm 4
12
I 3 = I1 = 354.987 × 103 mm 4
I1 =
I = I1 + I 2 + I 3 = 720.9 × 103 mm 4 = 720.9 × 10−9 m 4
1
Mc
with c = (48) = 24 mm = 0.024 m
2
I
I σ (720.9 × 10−9 )(150 × 106 )
=
= 4.51 × 103 N ⋅ m
M=
0.024
c
σ=
M = 4.51 kN ⋅ m 
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PROBLEM 4.7
Two W4 × 13 rolled sections are welded together as shown. Knowing that for the steel
alloy used σ Y = 36 ksi and σ U = 58 ksi and using a factor of safety of 3.0, determine
the largest couple that can be applied when the assembly is bent about the z axis.
SOLUTION
Properties of W4 × 13 rolled section.
(See Appendix C.)
Area = 3.83 in 2
Depth = 4.16 in.
I x = 11.3 in 4
For one rolled section, moment of inertia about axis a-a is
I a = I x + Ad 2 = 11.3 + (3.83)(2.08) 2 = 27.87 in 4
For both sections,
I z = 2 I a = 55.74 in 4
c = depth = 4.16 in.
M all
σU
58
= 19.333 ksi
F .S . 3.0
σ I (19.333) (55.74)
= all =
c
4.16
σ all =
=
σ=
Mc
I
M all = 259 kip ⋅ in 
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PROBLEM 4.8
Two W4 × 13 rolled sections are welded together as shown. Knowing that for the steel
alloy used σ Y = 36 ksi and σ U = 58 ksi and using a factor of safety of 3.0, determine the
largest couple that can be applied when the assembly is bent about the z axis.
SOLUTION
Properties of W4 × 13 rolled section.
(See Appendix C.)
Area = 3.83 in 2
Width = 4.060 in.
I y = 3.86 in 4
For one rolled section, moment of inertia about axis b-b is
I b = I y + Ad 2 = 3.86 + (3.83)(2.030) 2 = 19.643 in 4
For both sections,
I z = 2 I b = 39.286 in 4
c = width = 4.060 in.
σU
58
=
= 19.333 ksi
F .S . 3.0
σ I (19.333) (39.286)
= all =
4.060
c
σ all =
M all
σ=
Mc
I
M all = 187.1 kip ⋅ in 
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PROBLEM 4.9
Two vertical forces are applied to a beam of the cross section shown. Determine
the maximum tensile and compressive stresses in portion BC of the beam.
SOLUTION
A1 =
π
2
r2 =
π
2
(25) 2 = 981.7 mm 2
A2 = bh = (50)(25) = 1250 mm 2
y =
4r
(4)(25)
=
= 10.610 mm
3π
3π
h
25
y2 = − = −
= −12.5 mm
2
2
y1 =
A1 y1 + A2 y2
(981.7)(10.610) + (1250)(−12.5)
=
= −2.334 mm
A1 + A2
981.7 + 1250
I1 = I x1 − A1 y12 =
π
r 4 − A1 y12 =
π
(25) 4 − (981.7)(10.610) 2 = 42.886 × 106 mm 4
8
8
d1 = y1 − y = 10.610 − ( −2.334) = 12.944 mm
I1 = I1 + A1d12 = 42.866 × 103 + (981.7)(12.944)2 = 207.35 × 103 mm 4
1 3 1
bh = (50)(25)3 = 65.104 × 103 mm 4
12
12
d 2 = y2 − y = −12.5 − (−2.334) = 10.166 mm
I2 =
I 2 = I 2 + A2 d 22 = 65.104 × 103 + (1250)(10.166)2 = 194.288 × 103 mm 4
I = I1 + I 2 = 401.16 × 103 mm 4 = 401.16 × 10−9 m 4
ytop = 25 + 2.334 = 27.334 mm = 0.027334 m
ybot = −25 + 2.334 = −22.666 mm = −0.022666 m
M − Pa = 0 :
σ top =
σ bot =
−Mytop
M = Pa = (4 × 103 )(300 × 10−3 ) = 1200 N ⋅ m
(1200)(0.027334)
= −81.76 × 106 Pa
401.16 × 10−9
σ top = −81.8 MPa 
−Mybot
(1200)(−0.022666)
=−
= 67.80 × 106 Pa
I
401.16 × 10−9
σ bot = 67.8 MPa 
I
=−
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PROBLEM 4.10
Two vertical forces are applied to a beam of the cross
section shown. Determine the maximum tensile and
compressive stresses in portion BC of the beam.
SOLUTION
A, mm 2
y0 , mm
A y0 , mm3

600
30
18 × 103

600
30
18 × 103

300
5
1.5 × 103
1500
Y0 =
37.5 × 103
37.5 × 103
= 25 mm
1500
Neutral axis lies 25 mm above the base.
1
(10)(60)3 + (600)(5)2 = 195 × 103 mm 4 I 2 = I1 = 195 mm 4
12
1
I 3 = (30)(10)3 + (300)(20) 2 = 122.5 × 103 mm 4
12
I = I1 + I 2 + I 3 = 512.5 × 103 mm 4 = 512.5 × 10−9 m 4
I1 =
ytop = 35 mm = 0.035 m
ybot = −25 mm = −0.025 m
a = 150 mm = 0.150 m P = 10 × 103 N
M = Pa = (10 × 103 )(0.150) = 1.5 ×103 N ⋅ m
σ top = −
σ bot = −
M ytop
I
=−
(1.5 × 103 )(0.035)
= −102.4 × 106 Pa
512.5 × 10−9
M ybot
(1.5 × 103 )(−0.025)
=−
= 73.2 × 106 Pa
I
512.5 × 10−9
σ top = −102.4 MPa (compression) 
σ bot = 73.2 MPa (tension) 
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PROBLEM 4.11
Two vertical forces are applied to a beam of the cross section shown.
Determine the maximum tensile and compressive stresses in portion
BC of the beam.
SOLUTION
A
y0
A y0

8
7.5
60

6
4
24

4
0.5
Σ
18
Yo =
2
86
86
= 4.778 in.
18
Neutral axis lies 4.778 in. above the base.
1
1
b1h13 + A1d12 = (8)(1)3 + (8)(2.772)2 = 59.94 in 4
12
12
1
1
I 2 = b2 h23 + A2 d 22 = (1)(6)3 + (6)(0.778)2 = 21.63 in 4
12
12
1
1
3
2
I 3 = b3 h3 + A3 d3 = (4)(1)3 + (4)(4.278) 2 = 73.54 in 4
12
12
I = I1 + I 2 + I 3 = 59.94 + 21.63 + 73.57 = 155.16 in 4
ytop = 3.222 in. ybot = −4.778 in.
I1 =
M − Pa = 0
M = Pa = (25)(20) = 500 kip ⋅ in
σ top = −
σ bot = −
Mytop
I
=−
(500)(3.222)
155.16
Mybot
(500)(−4.778)
=−
I
155.16
σ top = −10.38 ksi (compression) 
σ bot = 15.40 ksi (tension) 
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PROBLEM 4.12
Knowing that a beam of the cross section shown is bent about a horizontal
axis and that the bending moment is 6 kN ⋅ m, determine the total force
acting on the top flange.
SOLUTION
The stress distribution over the entire cross section is given by the bending stress formula:
σx = −
My
I
where y is a coordinate with its origin on the neutral axis and I is the moment of inertia of the entire cross
sectional area. The force on the shaded portion is calculated from this stress distribution. Over an area
element dA, the force is
dF = σ x dA = −
My
dA
I
The total force on the shaded area is then

F = dF = −

My
M
dA = −
I
I
 ydA = −
M * *
y A
I
where y * is the centroidal coordinate of the shaded portion and A* is its area.
d1 = 54 − 18 = 36 mm
d 2 = 54 + 36 − 54 = 36 mm
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PROBLEM 4.12 (Continued)
Moment of inertia of entire cross section:
1
1
b1h13 + A1d12 = (216)(36)3 + (216)(36)(36)2 = 10.9175 × 106 mm4
12
12
1
1
3
2
I 2 = b2 h2 + A2 d 2 = (72)(108)3 + (72)(108)(36)2 = 17.6360 × 106 mm 4
12
12
I = I1 + I 2 = 28.5535 × 106 mm4 = 28.5535 × 10−6 m 4
I1 =
For the shaded area,
A* = (216)(36) = 7776 mm 2
y * = 36 mm
A* y * = 279.936 × 103 mm3 = 279.936 × 10−6 m3
F =−
MA* y * (6 × 103 )(279.936 × 10−6 )
=
I
28.5535 × 10−6
= 58.8 × 103 N
F = 58.8 kN 
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PROBLEM 4.13
Knowing that a beam of the cross section shown is bent about a horizontal
axis and that the bending moment is 6 kN ⋅ m, determine the total force
acting on the shaded portion of the web.
SOLUTION
The stress distribution over the entire cross section is given by the bending stress formula:
σx = −
My
I
where y is a coordinate with its origin on the neutral axis and I is the moment of inertia of the entire cross
sectional area. The force on the shaded portion is calculated from this stress distribution. Over an area
element dA, the force is
dF = σ x dA = −
My
dA
I
The total force on the shaded area is then

F = dF = −

My
M
dA = −
I
I
 ydA = −
M * *
y A
I
where y * is the centroidal coordinate of the shaded portion and A* is its area.
d1 = 54 − 18 = 36 mm
d 2 = 54 + 36 − 54 = 36 mm
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PROBLEM 4.13 (Continued)
Moment of inertia of entire cross section:
1
1
b1h13 + A1d12 = (216)(36)3 + (216)(36)(36)2 = 10.9175 × 106 mm 4
12
12
1
1
I 2 = b2 h23 + A2 d 22 = (72)(108)3 + (72)(108)(36) 2 = 17.6360 × 106 mm 4
12
12
I = I1 + I 2 = 28.5535 × 106 mm 4 = 28.5535 × 10−6 m 4
I1 =
For the shaded area,
A* = (72)(90) = 6480 mm 2
y * = 45 mm
A* y * = 291.6 × 103 mm3 = 291.6 × 10−6 m
F=
MA* y * (6 × 103 )(291.6 × 10−6 )
=
I
28.5535 × 10−6
= 61.3 × 103 N
F = 61.3 kN 
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PROBLEM 4.14
Knowing that a beam of the cross section shown is bent about a horizontal axis
and that the bending moment is 50 kip ⋅ in., determine the total force acting
(a) on the top flange, (b) on the shaded portion of the web.
SOLUTION
The stress distribution over the entire cross-section is given by the bending stress formula:
σx = −
My
I
where y is a coordinate with its origin on the neutral axis and I is the moment of
inertia of the entire cross sectional area. The force on the shaded portion is
calculated from this stress distribution. Over an area element dA, the force is
dF = σ x dA = −
My
dA
I
The total force on the shaded area is then

F = dF = −

My
M
dA = −
I
I
 ydA = −
M * *
y A
I
where y * is the centroidal coordinate of the shaded portion and A* is its area.
Calculate the moment of inertia.
1
1
(6 in.)(7 in.)3 − (4 in.)(4 in.)3 = 150.17 in 4
12
12
M = 50 kip ⋅ in
I =
(a)
Top flange:
A* = (6 in.)(1.5 in.) = 9 in 2
F =
(b)
Half web:
50 kip ⋅ in
(9 in 2 )(2.75 in.) = 8.24 kips
150.17 in 4
A* = (2 in.)(2 in.) = 4 in 2
F =
y * = 2 in. + 0.75 in. = 2.75 in.
F = 8.24 kips 
y * = 1 in.
50 kip ⋅ in
(4 in 2 )(1 in.) = 1.332 kips
150.17 in 4
F = 1.332 kips 
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PROBLEM 4.15
The beam shown is made of a nylon for which the allowable stress is
24 MPa in tension and 30 MPa in compression. Determine the largest
couple M that can be applied to the beam.
SOLUTION
A, mm 2
y0 , mm
A y0 , mm3

600
22.5
13.5 × 103

300
7.5
2.25 × 103
Σ
900
Y0 =
15.75 × 103
15.5 × 103
= 17.5 mm
900
The neutral axis lies 17.5 mm above the bottom.
ytop = 30 − 17.5 = 12.5 mm = 0.0125 m
ybot = −17.5 mm = −0.0175 m
1
1
b1h13 + A1d12 = (40)(15)3 + (600)(5)2 = 26.25 × 103 mm 4
12
12
1
1
I 2 = b2 h23 + A2 d 22 = (20)(15)3 + (300)(10)2 = 35.625 × 103 mm 4
12
12
I = I1 + I 2 = 61.875 × 103 mm 4 = 61.875 × 10−9 m 4
I1 =
|σ | =
My
I
M=
σI
y
Top: (tension side)
M=
(24 × 106 )(61.875 × 10−9 )
= 118.8 N ⋅ m
0.0125
Bottom: (compression)
M=
(30 × 106 )(61.875 × 10−9 )
= 106.1 N ⋅ m
0.0175
Choose smaller value.
M = 106.1 N ⋅ m 
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PROBLEM 4.16
Solve Prob. 4.15, assuming that d = 40 mm.
PROBLEM 4.15 The beam shown is made of a nylon for which the
allowable stress is 24 MPa in tension and 30 MPa in compression.
Determine the largest couple M that can be applied to the beam.
SOLUTION
A, mm 2
y0 , mm
A y0 , mm3

600
32.5
19.5 × 103

500
12.5
6.25 × 103
Σ
1100
Y0 =
25.75 × 103
25.75 × 103
= 23.41 mm
1100
The neutral axis lies 23.41 mm above the bottom.
ytop = 40 − 23.41 = 16.59 mm = 0.01659 m
ybot = −23.41 mm = −0.02341 m
1
1
b1h13 + A1d12 = (40)(15)3 + (600)(9.09)2 = 60.827 × 103 mm 4
12
12
1
1
I 2 = b2 h22 + A2 d 22 = (20)(25)3 + (500)(10.91)2 = 85.556 × 103 mm 4
12
12
I = I1 + I 2 = 146.383 × 103 mm 4 = 146.383 × 10−9 m 4
I1 =
|σ | =
My
I
M=
σI
y
Top: (tension side)
M=
(24 × 106 )(146.383 × 10−9 )
= 212 N ⋅ m
0.01659
Bottom: (compression)
M=
(30 × 106 )(146.383 × 10−9 )
= 187.6 N ⋅ m
0.02341
Choose smaller value.
M = 187.6 N ⋅ m 
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PROBLEM 4.17
Knowing that for the extruded beam shown the allowable stress is 12 ksi
in tension and 16 ksi in compression, determine the largest couple M that
can be applied.
SOLUTION
A
y0

2.25
1.25
2.8125

2.25
0.25
0.5625
4.50
Y=
A y0
3.375
3.375
= 0.75 in.
4.50
The neutral axis lies 0.75 in. above bottom.
ytop = 2.0 − 0.75 = 1.25 in.,
ybot = −0.75 in.
1
1
b1h13 + A1d12 = (1.5)(1.5)3 + (2.25)(0.5) 2 = 0.984375 in 4
12
12
1
1
I 2 = b2 h22 + A2 d 22 = (4.5)(0.5)3 + (2.25)(0.5)2 = 0.609375 in 4
12
12
I = I1 + I 2 = 1.59375 in 4
I1 =
σ =
My
I
M =
σI
y
Top: (compression)
M =
(16)(1.59375)
= 20.4 kip ⋅ in
1.25
Bottom: (tension)
M =
(12)(1.59375)
= 25.5 kip ⋅ in
0.75
Choose the smaller as Mall.
M all = 20.4 kip ⋅ in 
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PROBLEM 4.18
Knowing that for the casting shown the allowable stress is 5 ksi in
tension and 18 ksi in compression, determine the largest couple M
that can be applied.
SOLUTION
Locate the neutral axis and compute the moment of inertia.
Y =
 Ai yi
 Ai
I =
1
bi hi3 for rectangle
12
I = ( Ai di2 + I )
di = yi − Y
I , in 4
Part
A, in 2
yi , in.
Ai yi , in 3
1
1.5
1.25
1.875
0.3333
0.1667
0.03125
2
1.0
0.75
0.75
0.1667
0.0277
0.02083
3
0.5
0.25
0.125
0.6667
0.2222
0.01042
Σ
3.0
0.4166
0.0625
Y =
 Ay 2.75
=
= 0.9167 in.
A
3.0
Ai di2 , in 4
di , in.
2.75
I = ( I + Ad 2 ) = 0.4166 + 0.0625 = 0.479 in 4
Allowable bending moment.
σ =
Tension at A:
Mc
I
or
M =
σI
c
σ A ≤ 5 ksi
c A = 0.583 in.
M ≤
Compression at B:
(5)(0.479)
= 4.11 kip ⋅ in
0.583
σ B ≤ 18 ksi
M ≤
The smaller value is the allowable value of M.
cB = 0.917 in.
(18)(0.479)
= 9.40 kip ⋅ in
0.917
M = 4.11 kip ⋅ in 
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PROBLEM 4.19
Knowing that for the extruded beam shown the allowable stress
is 120 MPa in tension and 150 MPa in compression, determine
the largest couple M that can be applied.
SOLUTION
 rectangle
 circular cutout
A1 = (150)(250) = 37.5 × 103 mm 2
A2 = −π (50) 2 = − 7.85398 × 103 mm 2
A = A1 + A2 = 29.64602 × 103 mm 2
y1 = 0 mm
y2 = −50 mm
Y =
ΣA y
ΣA
(37.5 × 103 )(0) + (−7.85393 × 103 )(−50)
29.64602 × 103
= 13.2463 mm
Y =
I X ′ = Σ( I + Ad 2 ) = I1 − I 2
1

=  (150)(250)3 + (37.5 × 103 )(13.2463) 2 
12

π

−  (50) 4 + (7.85398 × 103 )(50 + 13.2463) 2 
4

= 201.892 × 106 − 36.3254 × 106 = 165.567 × 106 mm 4
= 165.567 × 10−6 m 4
Top: (tension side)
c = 125 − 13.2463 = 111.7537 mm = 0.11175 m
σ=
Bottom: (compression side)
M=
I σ (165.567 × 10−6 )(120 × 106 )
=
c
0.11175
= 177.79 × 103 N ⋅ m
c = 125 + 13.2463 = 138.2463 mm
= 0.13825 m
σ=
Choose the smaller.
Mc
I
Mc
I
M=
I σ (165.567 × 10−6 )(150 × 106 )
=
c
0.13825
= 179.64 × 103 N ⋅ m
M = 177.8 × 103 N ⋅ m
M = 177.8 kN ⋅ m 
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PROBLEM 4.20
Knowing that for the extruded beam shown the allowable stress is
120 MPa in tension and 150 MPa in compression, determine the largest
couple M that can be applied.
SOLUTION
A, mm 2
y0 , mm
A y0 , mm3
d, mm

2160
27
58320
3

1080
36
38880
3
Σ
3240
Y =
97200
= 30 mm
3240
97200
The neutral axis lies 30 mm above the bottom.
ytop = 54 − 30 = 24 mm = 0.024 m
ybot = −30 mm = −0.030 m
1
1
b1h13 + A1d12 = (40)(54)3 + (40)(54)(3)2 = 544.32 × 103 mm 4
12
12
1
1
1
I 2 = b2 h22 + A2 d 22 = (40)(54)3 + (40)(54)(6)2 = 213.84 × 103 mm 4
36
36
2
3
4
I = I1 + I 2 = 758.16 × 10 mm = 758.16 × 10−9 m 4
I1 =
|σ | =
My
I
|M | =
σI
y
Top: (tension side)
M=
(120 × 106 )(758.16 × 10−9 )
= 3.7908 × 103 N ⋅ m
0.024
Bottom: (compression)
M=
(150 × 106 )(758.16 × 10−9 )
= 3.7908 × 103 N ⋅ m
0.030
Choose the smaller as Mall.
M all = 3.7908 × 103 N ⋅ m
M all = 3.79 kN ⋅ m 
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PROBLEM 4.21
A steel band saw blade, that was originally straight, passes over 8-in.-diameter pulleys
when mounted on a band saw. Determine the maximum stress in the blade, knowing that
it is 0.018 in. thick and 0.625 in. wide. Use E = 29 × 106 psi.
SOLUTION
Band blade thickness:
t = 0.018 in.
Radius of pulley:
r=
1
d = 4.000 in.
2
Radius of curvature of centerline of blade:
1
2
ρ = r + t = 4.009 in.
c=
1
t = 0.009 in.
2
c
0.009
= 0.002245
4.009
Maximum strain:
εm =
Maximum stress:
σ m = Eε m = (29 × 106 )(0.002245)
ρ
=
σ m = 65.1 × 103 psi
σ m = 65.1 ksi 
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PROBLEM 4.22
Straight rods of 0.30-in. diameter and 200-ft length are sometimes
used to clear underground conduits of obstructions or to thread wires
through a new conduit. The rods are made of high-strength steel and,
for storage and transportation, are wrapped on spools of 5-ft
diameter. Assuming that the yield strength is not exceeded,
determine (a) the maximum stress in a rod, when the rod, which is
initially straight, is wrapped on a spool, (b) the corresponding
bending moment in the rod. Use E = 29 × 106 psi .
SOLUTION
Radius of cross section:
r =
Moment of inertia:
I =
D = 5 ft = 60 in.
ρ =
1
1
d = (0.30) = 0.15 in.
2
2
π
4
r4 =
π
4
(0.15) 4 = 397.61 × 10−6 in 4
1
D = 30 in.
2
c = r = 0.15 in.
(a)
σ max =
(b)
M =
EI
ρ
Ec
ρ
=
=
(29 × 106 )(0.15)
= 145 × 103 psi
30
(29 × 106 )(397.61 × 10−6 )
30
σ max = 145 ksi 
M = 384 lb ⋅ in 
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PROBLEM 4.23
A 900-mm strip of steel is bent into a full circle by two couples applied as
shown. Determine (a) the maximum thickness t of the strip if the allowable
stress of the steel is 420 MPa, (b) the corresponding moment M of the
couples. Use E = 200 GPa .
SOLUTION
When the rod is bent into a full circle, the circumference is 900 mm. Since the circumference is equal to
2π times ρ, the radius of curvature, we get
ρ =
900 mm
= 143.24 mm = 0.14324 m
2π
σ = Eε =
Stress:
Ec
or
ρ
c=
ρσ
E
For σ = 420 MPa and E = 200 GPa,
c=
(a)
(0.14324)(420 × 106 )
= 0.3008 × 10−3 m
200 × 109
Maximum thickness:
t = 2c = 0.6016 × 10−3 m
t = 0.602 mm 
Moment of inertia for a rectangular section.
I =
(b)
bt 3
(8 × 10−3 )(0.6016 × 10−3 )3
=
= 145.16 × 10−15 m 4
12
12
Bending moment:
M=
M =
EI
ρ
(200 × 109 )(145.16 × 10−15 )
= 0.203 N ⋅ m
0.14324
M = 0.203 N ⋅ m 
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PROBLEM 4.24
A 60 N ⋅ m couple is applied to the steel bar shown. (a) Assuming that
the couple is applied about the z axis as shown, determine the maximum
stress and the radius of curvature of the bar. (b) Solve part a, assuming
that the couple is applied about the y axis. Use E = 200 GPa.
SOLUTION
(a)
Bending about z-axis.
1 3 1
bh = (12)(20)3 = 8 × 103 mm4 = 8 × 10−9 m 4
12
12
20
c=
= 10 mm = 0.010 m
2
I=
σ=
1
ρ
(b)
=
Mc (60)(0.010)
=
= 75.0 × 106 Pa
I
8 × 10−9
M
60
=
= 37.5 × 10−3 m −1
EI (200 × 109 )(8 × 10 −9 )
σ = 75.0 MPa 
ρ = 26.7 m 
Bending about y-axis.
1 3 1
bh = (20)(12)3 = 2.88 × 103 mm4 = 2.88 × 10−9 m4
12
12
12
c=
= 6 mm = 0.006 m
2
Mc (60)(0.006)
=
= 125.0 × 106 Pa
σ=
I
2.88 × 10−9
I=
1
ρ
=
M
60
=
= 104.17 × 10−3 m −1
−9
9
EI (200 × 10 )(2.88 × 10 )
σ = 125.0 MPa 
ρ = 9.60 m 
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PROBLEM 4.25
A couple of magnitude M is applied to a square bar of side a. For
each of the orientations shown, determine the maximum stress and
the curvature of the bar.
SOLUTION
1 3 1 3 a4
bh = aa =
12
12
12
a
c=
2
I=
σ max
1
ρ
a
Mc M 2
=
= 4
I
a
12
=
M
M
=
EI E a 4
12
σ max =
1
ρ
=
6M

a3
12M

Ea 4
For one triangle, the moment of inertia about its base is
I1 =
1 3 1
bh =
12
12
I 2 = I1 =

a
2
)
a4
24
I = I1 + I 2 =
c=
(
3
 a 
a4
2a 
=

24
 2
a4
12
σ max =
1
ρ
=
Mc Ma/ 2 6 2 M
= 4
=
I
a /12
a3
M
M
=

EI E a 4
12
σ max =
1
ρ
8.49M

a3
=
12M

Ea 4
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PROBLEM 4.26
A portion of a square bar is removed by milling so that its cross section
is as shown. The bar is then bent about its horizontal axis by a couple M.
Considering the case where h = 0.9h0, express the maximum stress in the
bar in the form σ m = kσ 0 , where σ 0 is the maximum stress that would
have occurred if the original square bar had been bent by the same
couple M, and determine the value of k.
SOLUTION
I = 4 I1 + 2 I 2
 1
1
= (4)   h h3 + (2)   (2h0 − 2h) (h3 )
 12 
3
1
4
4
4
= h 4 + h 0 h3 − h h3 = h 0 h3 − h 4
3
3
3
3
c=h
σ=
For the original square,
3M
Mc
Mh
=
=
3
4
4
I
(4h 0 − 3h) h 2
h h −h
3 0
h = h0 , c = h0 .
σ0 =
3M
3M
= 3
2
(4h0 − 3h0 )h0
h0
h03
h03
σ
=
=
= 0.950
σ 0 (4h0 − 3h) h 2 (4h0 − (3)(0.9)h0 )(0.9 h02 )
σ = 0.950 σ 0
k = 0.950 
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PROBLEM 4.27
In Prob. 4.26, determine (a) the value of h for which the maximum stress
σ m is as small as possible, (b) the corresponding value of k.
PROBLEM 4.26 A portion of a square bar is removed by milling so that
its cross section is as shown. The bar is then bent about its horizontal
axis by a couple M. Considering the case where h = 0.9h0, express the
maximum stress in the bar in the form σ m = kσ 0 , where σ 0 is the
maximum stress that would have occurred if the original square bar had
been bent by the same couple M, and determine the value of k.
SOLUTION
I = 4 I1 + 2 I 2
 1 
1
= (4)   hh3 + (2)   (2h0 − 2h) h3
 12 
3
1 4 4
4
4
= h − h 0 h3 − h3 = h 0 h3 − h 4
3
3
3
3
I 4
c=h
= h 0 h 2 − h3
c 3
I
d 4

is maximum at
h 0 h 2 − h3  = 0.

c
dh  3

8
h 0 h − 3h 2 = 0
3
2

3
I 4 8  8 
256 3
= h0  h0  −  h0  =
h0
729
c 3 9  9 
For the original square,
h = h0
c = h0
σ0 =
h=
σ=
8
h 0 
9
Mc 729M
=

I
256h03
I0 1 3
= h0
c0 3
Mc0 3M
= 2
I0
h0
σ 729 1 729
=
⋅ =
= 0.949
σ 0 256 3 768
k = 0.949 
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PROBLEM 4.28
A couple M will be applied to a beam of rectangular cross section that
is to be sawed from a log of circular cross section. Determine the ratio
d/b, for which (a) the maximum stress σm will be as small as possible,
(b) the radius of curvature of the beam will be maximum.
SOLUTION
Let D be the diameter of the log.
D2 = b2 + d 2
I =
(a)
1 3
bd
12
d 2 = D 2 − b2
c=
σm is the minimum when
1
d
2
I
is maximum.
c
I
1
= b( D 2 − b 2 )
c
6
d I 1 2
 = D −
db  c  6
d =
(b)
ρ =
I
1
= bd 2
c
6
D2 −
=
1 2
1
D b − b3
6
6
3 2
b =0
6
1 2
D =
3
b=
1
D
3
2
D
3
d
=
b
2 
d
=
b
3 
EI
M
ρ is maximum when I is maximum,
1 3
bd is maximum, or b 2d 6 is maximum.
12
( D 2 − d 2 )d 6 is maximum.
6D 2d 5 − 8d 7 = 0
b=
D2 −
3 2
1
D = D
4
2
d =
3
D
2
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PROBLEM 4.29
For the aluminum bar and loading of Sample Prob. 4.1, determine (a) the radius of curvature ρ ′ of a
transverse cross section, (b) the angle between the sides of the bar that were originally vertical. Use
E = 10.6 × 106 psi and v = 0.33.
SOLUTION
From Sample Prob. 4.1,
I = 12.97 in 4
1
ρ
(a)
=
M
103.8 × 103
=
= 755 × 10−6 in −1
EI (10.6 × 106 )(12.97)
1
1
= v = (0.33) (755 × 10−6 ) = 249 × 10−6 in −1
ρ′
ρ
ρ ′ = 4010 in.
(b)
M = 103.8 kip ⋅ in
θ=
length of arc b
3.25
=
=
= 810 × 10−6 rad
ρ ′ 4010
radius
ρ ′ = 334 ft 
θ = 0.0464° 
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PROBLEM 4.30
For the bar and loading of Example 4.01, determine (a) the radius of curvature ρ , (b) the radius of curvature
ρ ′ of a transverse cross section, (c) the angle between the sides of the bar that were originally vertical. Use
E = 29 × 106 psi and v = 0.29.
SOLUTION
M = 30 kip ⋅ in,
From Example 4.01,
(a)
(b)
(c)
1
ρ
=
I = 1.042 in 4
M
(30 × 103 )
=
= 993 × 10−6 in −1
EI (29 × 106 )(1.042)
ε 1 = vε =
vc
ρ
=v
ρ = 1007 in. 
c
ρ′
1
1
= v = (0.29)(993 × 10−6 )in.−1 = 288 × 10−6 in.−1
ρ′
ρ
ρ ′ = 3470 in. 
length of arc b
0.8
=
=
= 230 × 10−6 rad
ρ ′ 3470
radius
θ = 0.01320° 
θ=
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PROBLEM 4.31
A W200 × 31.3 rolled-steel beam is subjected to a couple M of moment
45 kN ⋅ m. Knowing that E = 200 GPa and v = 0.29, determine (a) the
radius of curvature ρ , (b) the radius of curvature ρ ′ of a transverse cross
section.
SOLUTION
For W 200 × 31.3 rolled steel section,
I = 31.4 × 106 mm4
= 31.4 × 10−6 m 4
(a)
(b)
1
ρ
=
M
45 × 103
=
= 7.17 × 10−3 m −1
EI (200 × 109 ) (31.4 × 10−6 )
1
1
= v = (0.29) (7.17 × 10−3 ) = 2.07 × 10−3 m −1
ρ′
ρ
ρ = 139.6 m 
ρ ′ = 481 m 
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PROBLEM 4.32
It was assumed in Sec. 4.3 that the normal stresses σ y in a
member in pure bending are negligible. For an initially straight
elastic member of rectangular cross section, (a) derive an
approximate expression for σ y as a function of y, (b) show that
( σ y ) max = −(c/2 ρ )(σ x )max and, thus, that σ y can be neglected in
all practical situations. (Hint: Consider the free-body diagram of
the portion of beam located below the surface of ordinate y and
assume that the distribution of the stress σ x is still linear.)
SOLUTION
Denote the width of the beam by b and the length by L.
θ=
Using the free body diagram above, with
Σ Fy = 0 :
σy = −
(a)
σ y bL + 2
θ
2
sin
L
2
σ x = −(σ x )max
But,
(σ )
σ y = x max
ρc

y
−c
cos
(σ )
y2
ydy = x max
ρc
2

y
−c

θ
2
y
−c
L
ρ
≈1
σ x bdy sin
σ x dy ≈ −
θ
L

θ
2
y
−c
=0
σ x dy = −
1
ρ

y
−c
σ x dy
y
c
y
σy =
−c
(σ x )max 2
( y − c 2 ) 
2ρ c
The maximum value σ y occurs at y = 0 .
(b)
(σ y ) max = −
(σ x )max c 2
(σ ) c
= − x max 
2ρ c
2ρ
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PROBLEM 4.33
A bar having the cross section shown has been formed by securely
bonding brass and aluminum stock. Using the data given below,
determine the largest permissible bending moment when the composite
bar is bent about a horizontal axis.
Aluminum
Modulus of elasticity
Allowable stress
Brass
70 GPa
105 GPa
100 MPa
160 MPa
SOLUTION
Use aluminum as the reference material.
For aluminum,
n = 1.0
For brass,
n = Eb /Ea = 105 / 70 = 1.5
Values of n are shown on the figure.
For the transformed section,
n1
1.0
b1 h13 =
(8) (32)3 = 21.8453 × 103 mm 4
12
12
n
1.5
I 2 = 2 b2 h23 =
(32)(32)3 = 131.072 × 103 mm 4
12
12
I 3 = I1 = 21.8453 × 103 mm 4
I1 =
I = I1 + I 2 + I 3 = 174.7626 × 103 mm 4 = 174.7626 × 10−9 m 4
|σ | =
Aluminum:
σI
ny
σ = 100 × 106 Pa
(100 × 106 )(174.7626 × 10−9 )
= 1.0923 × 103 N ⋅ m
(1.0)(0.016)
n = 1.5, | y | = 16 mm = 0.016 m,
M=
Choose the smaller value.
M=
n = 1.0, | y | = 16 mm = 0.016 m,
M=
Brass:
nM y
I
σ = 160 × 106 Pa
(160 × 106 )(174.7626 × 10−9 )
= 1.1651 × 103 N ⋅ m
(1.5)(0.016)
M = 1.092 × 103 N ⋅ m
M = 1.092 kN ⋅ m 
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PROBLEM 4.34
A bar having the cross section shown has been formed by securely
bonding brass and aluminum stock. Using the data given below,
determine the largest permissible bending moment when the
composite bar is bent about a horizontal axis.
Aluminum
Brass
70 GPa
105 GPa
100 MPa
160 MPa
Modulus of elasticity
Allowable stress
SOLUTION
Use aluminum as the reference material.
For aluminum, n = 1.0
For brass, n = Eb /Ea = 105/70 = 1.5
Values of n are shown on the sketch.
For the transformed section,
n1
1.5
(8)(32)3 = 32.768 × 103 mm 4
b1h13 =
12
12
n
1.0
I 2 = 2 b2 H 23 − h23 =
(32)(323 − 163 ) = 76.459 × 103 mm 4
12
12
I 3 = I1 = 32.768 × 103 mm 4
I1 =
(
)
I = I1 + I 2 + I 3 = 141.995 × 103 mm 4 = 141.995 × 10−9 m 4
|σ | =
Aluminum:
M=
σI
ny
n = 1.0, | y | = 16 mm = 0.016 m, σ = 100 × 106 Pa
M=
Brass:
nMy
I
(100 × 106 )(141.995 × 10−9 )
= 887.47 N ⋅ m
(1.0)(0.016)
n = 1.5, | y | = 16 mm = 0.016 m, σ = 160 × 106 Pa
M=
Choose the smaller value.
(160 × 106 )(141.995 × 10−9 )
= 946.63 N ⋅ m
(1.5)(0.016)
M = 887 N ⋅ m 
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PROBLEM 4.35
For the composite bar indicated, determine the largest permissible bending
moment when the bar is bent about a vertical axis.
PROBLEM 4.35 Bar of Prob. 4.33.
Modulus of elasticity
Allowable stress
Aluminum
Brass
70 GPa
105 GPa
100 MPa
160 MPa
SOLUTION
Use aluminum as the reference material.
For aluminum,
n = 1.0
For brass,
n = Eb /Ea = 105/70 = 1.5
Values of n are shown on the figure.
For the transformed section,
n1
1.0
(32) (8)3 + (1.0)[(32)(8)](20) 2 = 103.7653 × 103 mm 4
h1 b13 + n1 A1d12 =
12
12
n
1.5
(32)(32)3 = 131.072 × 103 mm 4
I 2 = 2 h2b23 =
12
12
I 3 = I1 = 103.7653 × 103 mm 4
I1 =
I = I1 + I 2 + I 3 = 338.58 × 103 mm 4 = 338.58 × 10−9 m 4
|σ | =
n My
I
Aluminum:
M=
σ = 100 × 106 Pa
(100 × 106 )(338.58 × 10−9 )
= 1.411 × 103 N ⋅ m
(1.0)(0.024)
n = 1.5, | y | = 16 mm = 0.016 m,
M=
Choose the smaller value.
ny
n = 1.0, | y | = 24 mm = 0.024 m,
M=
Brass:
σI
σ = 160 × 106 Pa
(160 × 106 )(338.58 × 10−9 )
= 2.257 × 103 N ⋅ m
(1.5)(0.016)
M = 1.411 × 103 N ⋅ m
M = 1.411 kN ⋅ m 
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PROBLEM 4.36
For the composite bar indicated, determine the largest permissible
bending moment when the bar is bent about a vertical axis.
PROBLEM 4.36 Bar of Prob. 4.34.
Modulus of elasticity
Allowable stress
Aluminum
70 GPa
100 MPa
Brass
105 GPa
160 MPa
SOLUTION
Use aluminum as the reference material.
For aluminum, n = 1.0
For brass, n = Eb /Ea = 105/70 = 1.5
Values of n are shown on the sketch.
For the transformed section,
n1
1.5
(32)(483 − 323 ) = 311.296 × 103 mm 4
h1 B13 − b13 =
12
12
n2
1.0
I 2 = h2b23 =
(8)(32)3 = 21.8453 × 103 mm 4
12
12
I 3 = I 2 = 21.8453 × 103 mm 4
(
I1 =
)
I = I1 + I 2 + I 3 = 354.99 × 103 mm 4 = 354.99 × 10−9 m 4
|σ | =
Aluminum:
M=
σI
ny
n = 1.0, | y | = 16 mm = 0.016 m, σ = 100 × 106 Pa
M=
Brass:
nMy
I
(100 × 106 )(354.99 × 10−9 )
= 2.2187 × 103 N ⋅ m
(1.0)(0.016)
n = 1.5 | y | = 24 mm = 0.024 m σ = 160 × 106 Pa
M=
(160 × 106 )(354.99 × 10−9 )
= 1.57773 × 103 N ⋅ m
(1.5)(0.024)
Choose the smaller value.
M = 1.57773 × 103 N ⋅ m
M = 1.578 kN ⋅ m 
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PROBLEM 4.37
Wooden beams and steel plates are securely bolted together to form the composite
member shown. Using the data given below, determine the largest permissible bending
moment when the member is bent about a horizontal axis.
Modulus of elasticity:
Allowable stress:
Wood
Steel
2 × 106 psi
29 × 106 psi
2000 psi
22 ksi
SOLUTION
Use wood as the reference material.
n = 1.0 in wood
n = Es /Ew = 29/2 = 14.5 in steel
For the transformed section,
n1
1.0
b1h13 =
(3)(10)3 = 250 in 4
12
12
n
14.5  1 
3
4
I 2 = 2 b2 h23 =
  (10) = 604.17 in
12
12  2 
I1 =
I 3 = I1 = 250 in 4
I = I1 + I 2 + I 3 = 1104.2 in 4
σ =
Wood:
nMy
I
∴ M =
n = 1.0,
M =
Steel:
ny
y = 5 in., σ = 2000 psi
(2000)(1104.2)
= 441.7 × 103 lb ⋅ in
(1.0)(5)
n = 14.5,
M =
σI
y = 5 in., σ = 22 ksi = 22 × 103 psi
(22 × 103 )(1104.2)
= 335.1 × 103 lb ⋅ in
(14.5)(5)
Choose the smaller value.
M = 335 × 103 lb ⋅ in
M = 335 kip ⋅ in 
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PROBLEM 4.38
Wooden beams and steel plates are securely bolted together to form the composite
member shown. Using the data given below, determine the largest permissible bending
moment when the member is bent about a horizontal axis.
Modulus of elasticity:
Allowable stress:
Wood
Steel
2 × 106 psi
29 × 106 psi
2000 psi
22 ksi
SOLUTION
Use wood as the reference material.
n = 1.0 in wood
n = Es /Ew = 29/2 = 14.5 in steel
For the transformed section,
I1 =
n1
b1h13 + n1 A1d12
12
3
14.5  1 
1
(5)   + (14.5)(5)   (5.25) 2 = 999.36 in 4
=
12
2
2
n
1.0
(6)(10)3 = 500 in 4
I 2 = 2 b2 h22 =
12
12
I 3 = I1 = 999.36 in 4
I = I1 + I 2 + I 3 = 2498.7 in 4
σ =
Wood:
nMy
I
∴ M =
n = 1.0,
M =
Steel:
ny
y = 5 in., σ = 2000 psi
(2000)(2499)
= 999.5 × 103 lb ⋅ in
(1.0)(5)
n = 14.5,
M =
σI
y = 5.5 in., σ = 22 ksi = 22 × 103 psi
(22 × 103 )(2499)
= 689.3 × 103 lb ⋅ in
(14.5)(5.5)
Choose the smaller value.
M = 689 × 103 lb ⋅ in
M = 689 kip ⋅ in 
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PROBLEM 4.39
A steel bar and an aluminum bar are bonded together to form the composite
beam shown. The modulus of elasticity for aluminum is 70 GPa and for steel is
200 GPa. Knowing that the beam is bent about a horizontal axis by a couple of
moment M = 1500 N ⋅ m, determine the maximum stress in (a) the aluminum,
(b) the steel.
SOLUTION
Use aluminum as the reference material.
For aluminum,
n =1
For steel,
n = Es /Ea = 200/70 = 2.8571
Transformed section:
Part
1
2
Σ
Y0 =
A, mm2
600
1200
nA, mm2
1714.3
1200
2914.3
109714
= 37.65 mm
2914.3
yo , mm
50
20
nAyo , mm3
85714
24000
109714
d, mm
12.35
17.65
d = y0 − Y0
n1
2.8571
(30)(20)3 + (1714.3)(12.35) 2 = 318.61 × 103 mm 4
b1 h13 + n1 A1 d12 =
12
12
n
1
I 2 = 2 b2 h23 + n2 A2 d 22 = (30)(40)3 + (1200)(17.65) 2 = 533.83 × 103 mm 4
12
12
3
I = I1 + I 2 = 852.44 × 10 mm 4 = 852.44 × 10−9 m 4
I1 =
M = 1500 N ⋅ m
Stress:
(a)
σ =−
Aluminum:
n My
I
n = 1,
σa = −
(b)
Steel:
y = −37.65 mm = −0.03765 m
(1)(1500)(−0.03765)
= 66.2 × 106 Pa
852.44 × 10−9
n = 2.8571,
σs = −
σ a = 66.2 MPa 
y = 60 − 37.65 = 22.35 mm = 0.02235 m
n My
(2.8571)(1500) (0.02235)
=−
= −112.4 × 106 Pa
I
852.44 × 10−9
σ s = −112.4 MPa 
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PROBLEM 4.40
A steel bar and an aluminum bar are bonded together to form the composite
beam shown. The modulus of elasticity for aluminum is 70 GPa and for steel is
200 GPa. Knowing that the beam is bent about a horizontal axis by a couple of
moment M = 1500 N ⋅ m, determine the maximum stress in (a) the aluminum,
(b) the steel.
SOLUTION
Use aluminum as the reference material.
For aluminum,
n =1
For steel,
n = Es /Ea = 200/70 = 2.8571
Transformed section:
A, mm2
Part
nA, mm2
yo , mm
nAyo , mm3
d, mm
1
600
600
50
30000
25.53
2
1200
3428.5
20
68570
4.47
4028.5
Σ
Y0 =
98570
= 24.47 mm
4028.5
98570
d = y0 − Y0
n1
1
b1 h13 + n1 A1 d12 = (30)(20)3 + (600)(25.53)2 = 411.07 × 103 mm 4
12
12
n
2.8571
I 2 = 2 b2 h23 + n2 A2 d 22 =
(30)(40)3 + (3428.5)(4.47) 2 = 525.64 × 103 mm 4
12
12
I = I1 + I 2 = 936.71 × 103 mm 4 = 936.71 × 10−9 m 4
I1 =
M = 1500 N ⋅ m
Stress:
(a)
Aluminum:
σ =−
nM y
I
n = 1,
σa = −
(b)
Steel:
y = 60 − 24.47 = 35.53 mm = 0.03553 m
(1)(1500)(0.03553)
= −56.9 × 106 Pa
−9
936.71 × 10
n = 2.8571,
σs = −
σ a = −56.9 MPa 
y = −24.47 mm = −0.02447 m
(2.8571)(1500) (−0.02447)
= 111.9 × 106 Pa
−9
936.71 × 10
σ s = 111.9 MPa 
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PROBLEM 4.41
The 6 × 12-in. timber beam has been strengthened by bolting to it the steel
reinforcement shown. The modulus of elasticity for wood is 1.8 × 106 psi and for
steel, 29 × 106 psi. Knowing that the beam is bent about a horizontal axis by a
couple of moment M = 450 kip ⋅ in., determine the maximum stress in (a) the
wood, (b) the steel.
SOLUTION
Use wood as the reference material.
For wood,
n =1
For steel,
n = Es / Ew = 29 /1.8 = 16.1111
Transformed section:
421.931
112.278
= 3.758 in.
Yo =
 = wood


 = steel
A, in 2
nA, in 2
72
72
2.5
40.278
yo
6
−0.25
112.278
nA yo , in 3
432
−10.069
421.931
The neutral axis lies 3.758 in. above the wood-steel interface.
n1
1
b1h13 + n1 A1d12 = (6)(12)3 + (72)(6 − 3.758)2 = 1225.91 in 4
12
12
n2
16.1111
I 2 = b2 h23 + n2 A2 d 22 =
(5) (0.5)3 + (40.278)(3.578 + 0.25)2 = 647.87 in 4
12
12
I = I1 + I 2 = 1873.77 in 4
I1 =
σ =−
M = 450 kip ⋅ in
(a)
Wood:
n = 1,
y = 12 − 3.758 = 8.242 in
σw = −
(b)
Steel:
(1) (450) (8.242)
= −1.979 ksi
1873.77
n = 16.1111,
σs = −
nMy
I
σ w = −1.979 ksi 
y = −3.758 − 0.5 = −4.258 in
(16.1111) (450) (−4.258)
= 16.48 ksi
1873.77
σ s = 16.48 ksi 
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PROBLEM 4.42
The 6 × 12-in. timber beam has been strengthened by bolting to it the steel
reinforcement shown. The modulus of elasticity for wood is 1.8 × 106 psi
and for steel, 29 × 106 psi. Knowing that the beam is bent about a
horizontal axis by a couple of moment M = 450 kip ⋅ in., determine the
maximum stress in (a) the wood, (b) the steel.
SOLUTION
Use wood as the reference material.
For wood,
n =1
For steel,
n=
Es 29 × 106
=
= 16.1111
Ew 1.8 × 106
For C8 × 11.5 channel section,
A = 3.38 in 2 , t w = 0.220 in., x = 0.571 in., I y = 1.32 in 4
For the composite section, the centroid of the channel (part 1) lies 0.571 in. above the bottom of the section.
The centroid of the wood (part 2) lies 0.220 + 6.00 = 6.22 in. above the bottom.
Transformed section:
A, in2
3.38
Part
1
72
2
y , in.
0.571
nAy , in 3
31.091
d, in.
3.216
72
6.22
447.84
2.433
478.93
126.456
Σ
Y0 =
nA, in2
54.456
478.93 in 3
= 3.787 in.
126.456 in 2
d = y0 − Y0
The neutral axis lies 3.787 in. above the bottom of the section.
I1 = n1 I1 + n1 A1d12 = (16.1111)(1.32) + (54.456)(3.216) 2 = 584.49 in 4
n2
1
b2 h23 + n2 A2 d 22 = (6)(12)3 + (72)(2.433)2 = 1290.20 in 4
12
12
4
I = I1 + I 2 = 1874.69 in
I2 =
M = 450 kip ⋅ in
(a)
Wood:
n = 1,
σw = −
(b)
Steel:
n My
I
y = 12 + 0.220 − 3.787 = 8.433 in.
σ =−
(1)(450)(8.433)
= −2.02 ksi
1874.69
n = 16.1111,
σs = −
σ w = −2.02 ksi 
y = −3.787 in.
(16.1111) (450) (−3.787)
= 14.65 ksi
1874.67
σ s = 14.65 ksi 
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PROBLEM 4.43
For the composite beam indicated, determine the radius of curvature caused by
the couple of moment 1500 N ⋅ m.
Beam of Prob. 4.39.
SOLUTION
See solution to Prob. 4.39 for the calculation of I.
I = 852.44 × 10−9 m 4
1
ρ
=
Ea = 70 × 109 Pa
M
1500
=
= 0.02513 m −1
9
EI (70 × 10 )(852.44 × 10−9 )
ρ = 39.8 m 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.44
For the composite bar indicated, determine the radius of curvature caused by the
couple of moment 1500 N ⋅ m.
Beam of Prob. 4.40.
SOLUTION
See solution to Prob. 4.40 for calculation of I.
I = 936.71 × 10−9 m 4
1
ρ
=
Ea = 70 × 109 Pa
M
1500
=
= 0.02288 m −1
9
EI (70 × 10 )(936.71 × 10−9 )
ρ = 43.7 m 
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PROBLEM 4.45
For the composite beam indicated, determine the radius of curvature caused by the
couple of moment 450 kip ⋅ in.
Beam of Prob. 4.41.
SOLUTION
See solution to Prob. 4.41 for calculation of I.
I = 1873.77 in 4
Ew = 1.8 × 106 psi
M = 450 kip ⋅ in = 450 × 103 lb ⋅ in
1
ρ
=
M
450 × 103
=
= 133.42 × 10−6 in −1
6
EI (1.8 × 10 )(1873.77)
ρ = 7495 in. = 625 ft 
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PROBLEM 4.46
For the composite beam indicated, determine the radius of curvature caused by the
couple of moment 450 kip ⋅ in.
Beam of Prob. 4.42.
SOLUTION
See solution to Prob. 4.42 for calculation of I.
I = 1874.69 in 4
Ew = 1.8 × 106 psi
M = 450 kip ⋅ in = 450 × 103 lb ⋅ in
1
ρ
=
M
450 × 103
=
= 133.36 × 10−6 in −1
6
EI (1.8 × 10 )(1874.69)
ρ = 7499 in. = 625 ft 
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PROBLEM 4.47
The reinforced concrete beam shown is subjected to a positive bending
moment of 175 kN ⋅ m. Knowing that the modulus of elasticity is 25 GPa
for the concrete and 200 GPa for the steel, determine (a) the stress in the
steel, (b) the maximum stress in the concrete.
SOLUTION
n=
Es 200 GPa
=
= 8.0
25 GPa
Ec
As = 4 ⋅
π
π 
d 2 = (4)   (25) 2 = 1.9635 × 103 mm 2
4
4
nAs = 15.708 × 103 mm 2
Locate the neutral axis:
x
− (15.708 × 103 )(480 − x) = 0
2
150 x 2 + 15.708 × 103 x − 7.5398 × 106 = 0
300 x
Solve for x:
−15.708 × 103 + (15.708 × 103 ) 2 + (4)(150)(7.5398 × 106 )
(2)(150)
x = 177.87 mm,
480 − x = 302.13 mm
x=
1
I = (300) x3 + (15.708 × 103 )(480 − x) 2
3
1
= (300)(177.87)3 + (15.708 × 103 )(302.13)2
3
= 1.9966 × 109 mm 4 = 1.9966 × 10−3 m 4
nMy
σ =−
I
(a)
Steel:
y = −302.45 mm = −0.30245 m
σ =−
(b)
Concrete:
(8.0)(175 × 103 )(−0.30245)
= 212 × 106 Pa
−3
1.9966 × 10
σ = 212 MPa 
y = 177.87 mm = 0.17787 m
σ =−
(1.0)(175 × 103 )(0.17787)
= −15.59 × 106 Pa
1.9966 × 10−3
σ = −15.59 MPa 
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PROBLEM 4.48
Solve Prob. 4.47, assuming that the 300-mm width is increased to
350 mm.
PROBLEM 4.47 The reinforced concrete beam shown is subjected to a
positive bending moment of 175 kN ⋅ m. Knowing that the modulus of
elasticity is 25 GPa for the concrete and 200 GPa for the steel, determine
(a) the stress in the steel, (b) the maximum stress in the concrete.
SOLUTION
n=
Es 200 GPa
=
= 8.0
Ec
25 GPa
As = 4
π
π 
d 2 = (4)   (25)2
4
4
= 1.9635 × 103 mm 2
nAs = 15.708 × 103 mm 2
Locate the neutral axis:
x
− (15.708 × 103 )(480 − x) = 0
2
175 x 2 + 15.708 × 103 x − 7.5398 × 106 = 0
350 x
Solve for x:
−15.708 × 103 + (15.708 × 103 ) 2 + (4)(175)(7.5398 × 106 )
(2)(175)
x = 167.48 mm, 480 − x = 312.52 mm
x=
1
I = (350) x3 + (15.708 × 103 )(480 − x) 2
3
1
= (350)(167.48)3 + (15.708 × 103 )(312.52) 2
3
= 2.0823 × 109 mm 4 = 2.0823 × 10−3 m 4
nMy
σ =−
I
(a)
Steel:
y = −312.52 mm = −0.31252 m
σ =−
(b)
Concrete:
(8.0)(175 × 103 )(−0.31252)
= 210 × 106 Pa
2.0823 × 10−3
σ = 210 MPa 
y = 167.48 mm = 0.16748 m
σ =−
(1.0)(175 × 103 )(0.16748)
= −14.08 × 106 Pa
−3
2.0823 × 10
σ = −14.08 MPa 
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PROBLEM 4.49
A concrete slab is reinforced by 16-mm-diameter steel rods placed on
180-mm centers as shown. The modulus of elasticity is 20 GPa for
concrete and 200 GPa for steel. Using an allowable stress of 9 MPa for
the concrete and of 120 MPa for the steel, determine the largest bending
moment in a portion of slab 1 m wide.
SOLUTION
n=
Es 200 GPa
=
= 10
20 GPa
Ec
Consider a section 180-mm wide with one steel rod.
As =
π
d2 =
π
(16)2 = 201.06 mm 2
4
4
nAs = 2.0106 × 103 mm 2
Locate the neutral axis:
x
− (100 − x)(2.0106 × 103 ) = 0
2
90 x 2 + 2.0106 × 103 x − 201.06 × 103 = 0
180 x
Solving for x:
−2.0106 × 103 + (2.0106 × 103 ) 2 + (4)(90)(201.06 × 103 )
(2)(90)
x = 37.397 mm 100 − x = 62.603 mm
x=
1
I = (180) x3 + (2.0106 × 103 )(100 − x) 2
3
1
= (180)(37.397)3 + (2.0106 × 103 )(62.603) 2
3
= 11.018 × 106 mm 4 = 11.018 × 10−6 m 4
Concrete:
σI
σ =−
nMy
I
n = 1,
y = 37.397 mm = 0.037397 m,
M=
∴ M=
ny
σ = 9 × 106 Pa
(9 × 106 )(11.018 × 10−6 )
= 2.6516 × 103 N ⋅ m
(1.0)(0.037397)
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PROBLEM 4.49 (Continued)
n = 10,
Steel:
M=
Choose the smaller value.
y = 62.603 mm = 0.062603 m, σ = 120 × 106 Pa
(120 × 106 )(11.018 × 10−6 )
= 2.1120 × 103 N ⋅ m
(10)(0.062603)
M = 2.1120 × 103 N ⋅ m
The above is the allowable positive moment for a 180-mm wide section. For a 1-m = 1000-mm width,
1000
= 5.556
mutiply by
180
M = (5.556)(2.1120 × 103 ) = 11.73 × 103 N ⋅ m
M = 11.73 kN ⋅ m 
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PROBLEM 4.50
A concrete slab is reinforced by 16-mm-diameter steel rods placed on
180-mm centers as shown. The modulus of elasticity is 20 GPa for
concrete and 200 GPa for steel. Using an allowable stress of 9 MPa
for the concrete and of 120 MPa for the steel, determine the largest
allowable positive bending moment in a portion of slab 1 m wide.
Solve Prob. 4.49, assuming that the spacing of the 16-mm-diameter
rods is increased to 225 mm on centers.
SOLUTION
n=
Es 200 GPa
=
= 10
20 GPa
Ea
Consider a section 225-mm wide with one steel rod.
As =
π
d2 =
π
(16)2 = 201.06 mm 2
4
4
nAs = 2.0106 × 103 mm 2
Locate the neutral axis:
x
225 x − (100 − x)(2.0106 × 103 ) = 0
2
112.5 x 2 + 2.0106 x − 201.06 × 103 = 0
Solving for x:
−2.0106 × 103 + (2.0106 × 103 ) 2 + (4)(112.5)(201.06 × 103 )
(2)(112.5)
x = 34.273 mm
100 − x = 65.727
1
I = (225) x3 + 2.0106 × 103 (100 − x) 2
3
1
= (225)(34.273)3 + (2.0106 × 103 )(65.727) 2
3
= 11.705 × 106 mm 4 = 11.705 × 10−6 m 4
x=
|σ | =
Concrete:
nMy
I
n = 1,
M=
∴ M =
σI
ny
y = 34.273 mm = 0.034273 m, σ = 9 × 106 Pa
(9 × 106 )(11.705 × 10−6 )
= 3.0738 × 103 N ⋅ m
(1)(0.034273)
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PROBLEM 4.50 (Continued)
n = 10,
Steel:
M=
Choose the smaller value.
y = 65.727 mm = 0.065727 m, σ = 120 × 106 Pa
(120 × 106 )(11.705 × 10−6 )
= 2.1370 × 103 N ⋅ m
(10)(0.065727)
M = 2.1370 × 103 N ⋅ m
The above is the allowable positive moment for a 225-mm-wide section. For a 1-m = 1000-mm section,
1000
= 4.4444
multiply by
225
M = (4.4444)(2.1370 × 103 ) = 9.50 × 103 N ⋅ m
M = 9.50 kN ⋅ m 
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PROBLEM 4.51
A concrete beam is reinforced by three steel rods placed as shown. The
modulus of elasticity is 3 × 106 psi for the concrete and 29 × 106 psi for the
steel. Using an allowable stress of 1350 psi for the concrete and 20 ksi for the
steel, determine the largest allowable positive bending moment in the beam.
SOLUTION
n=
Es 29 × 10 6
=
= 9.67
Ec
3 × 106
As = 3
2
π
 π  7 
d 2 = (3)    = 1.8040 in 2
4
 4  8 
nAs = 17.438 in 2
x
− (17.438)(14 − x) = 0
2
4 x 2 + 17.438 x − 244.14 = 0
Locate the neutral axis:
8x
Solve for x.
x=
−17.438 + 17.4382 + (4)(4)(244.14)
= 5.6326 in.
(2)(4)
14 − x = 8.3674 in.
I =
1 3
1
8x + nAs (14 − x)2 = (8)(5.6326)3 + (17.438)(8.3674) 2 = 1697.45 in 4
3
3
σ =
nMy
I
∴ M=
σI
ny
n = 1.0,
Concrete:
M =
M =
Choose the smaller value.
σ = 1350 psi
(1350)(1697.45)
= 406.835 × 103 lb ⋅ in = 407 kip ⋅ in
(1.0)(5.6326)
n = 9.67,
Steel:
y = 5.6326 in.,
y = 8.3674 in., σ = 20 × 103 psi
(20 × 103 )(1697.45)
= 419.72 lb ⋅ in = 420 kip ⋅ in
(9.67)(8.3674)
M = 407 kip ⋅ in
M = 33.9 kip ⋅ ft 
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PROBLEM 4.52
Knowing that the bending moment in the reinforced concrete beam is
+100 kip ⋅ ft and that the modulus of elasticity is 3.625 × 106 psi for the
concrete and 29 × 106 psi for the steel, determine (a) the stress in the
steel, (b) the maximum stress in the concrete.
SOLUTION
n=
Es
29 × 106
=
= 8.0
Ec
3.625 × 106

π 
As = (4)   (1) 2 = 3.1416 in 2
4
nAs = 25.133 in
2
Locate the neutral axis.
x
(24)(4)( x + 2) + (12 x)   − (25.133)(17.5 − 4 − x) = 0
2
96 x + 192 + 6 x 2 − 339.3 + 25.133x = 0
x=
Solve for x.
or
6 x 2 + 121.133x − 147.3 = 0
−121.133 + (121.133) 2 + (4)(6)(147.3)
= 1.150 in.
(2)(6)
d3 = 17.5 − 4 − x = 12.350 in.
1
b1h13 +
12
1
I 2 = b2 x3 =
3
I1 =
A1d12 =
1
(24)(4)3 + (24)(4)(3.150) 2 = 1080.6 in 4
12
1
(12)(1.150)3 = 6.1 in 4
3
I 3 = nA3d32 = (25.133)(12.350) 2 = 3833.3 in 4
I = I1 + I 2 + I 3 = 4920 in 4
σ =−
(a)
Steel:
nMy
I
n = 8.0
y = −12.350 in.
σs = −
(b)
Concrete:
n = 1.0,
σc = −
where M = 100 kip ⋅ ft = 1200 kip ⋅ in.
(8.0)(1200)(−12.350)
4920
σ s = 24.1 ksi 
y = 4 + 1.150 = 5.150 in.
(1.0)(1200)(5.150)
4920
σ c = −1.256 ksi 
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PROBLEM 4.53
The design of a reinforced concrete beam is said to be balanced if the maximum stresses
in the steel and concrete are equal, respectively, to the allowable stresses σ s and σ c .
Show that to achieve a balanced design the distance x from the top of the beam to the
neutral axis must be
d
x=
1+
σ s Ec
σ c Es
where Ec and Es are the moduli of elasticity of concrete and steel, respectively, and d
is the distance from the top of the beam to the reinforcing steel.
SOLUTION
nM (d − x)
Mx
σc =
I
I
σ s n( d − x )
d
=
=n −n
σc
x
x
σs =
Eσ
d
1 σs
=1+
=1+ c s
x
n σc
E sσ c
x=

d
Ecσ s
1+
E sσ c
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PROBLEM 4.54
For the concrete beam shown, the modulus of elasticity is 3.5 × 106 psi for the concrete and
29 × 106 psi for the steel. Knowing that b = 8 in. and d = 22 in., and using an allowable
stress of 1800 psi for the concrete and 20 ksi for the steel, determine (a) the required area As
of the steel reinforcement if the beam is to be balanced, (b) the largest allowable bending
moment. (See Prob. 4.53 for definition of a balanced beam.)
The design of a reinforced concrete beam is said to be balanced if the maximum stresses in
the steel and concrete are equal, respectively, to the allowable stresses σ s and σ c .
SOLUTION
Es 29 × 10 6
=
= 8.2857
Ec 3.5 × 106
nM ( d − x)
Mx
σs =
σc =
I
I
σ s n( d − x )
d
=
=n −n
σc
x
x
n=
1 σs
1
20 × 103
d
=1+
=1+
⋅
= 2.3410
8.2857 1800
x
n σc
x = 0.42717 d = (0.42717)(22) = 9.398 in.
Locate neutral axis:
(a)
As =
bx
d − x = 22 − 9.398 = 12.602 in .
x
− nAs ( d − x) = 0
2
bx 2
(8)(9.398) 2
=
= 3.3835 in 2
2n(d − x) (2)(8.2857)(12.602)
As = 3.38 in 2 
1
1
I = bx3 + nAs ( d − x) 2 = (8)(9.398)3 + (8.2857)(3.3835)(12.602)2 = 6665.6 in 4
3
3
n My
σI
M=
σ=
I
ny
(b)
Steel:
Concrete:
n = 1.0
y = 9.398 in. σ c = 1800 psi
M=
(1800)(6665.6)
= 1.277 × 106 lb ⋅ in
(1.0)(9.398)
n = 8.2857 | y | = 12.602 in. σ s = 20 × 103 psi
M=
(20 × 103 )(6665.6)
= 1.277 × 106 lb ⋅ in
(8.2857)(12.602)
Note that both values are the same for balanced design.
M = 106.4 kip ⋅ ft 
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PROBLEM 4.55
Five metal strips, each 40 mm wide, are bonded together to form the
composite beam shown. The modulus of elasticity is 210 GPa for the
steel, 105 GPa for the brass, and 70 GPa for the aluminum. Knowing that
the beam is bent about a horizontal axis by a couple of moment 1800 N ⋅
m, determine (a) the maximum stress in each of the three metals,
(b) the radius of curvature of the composite beam.
SOLUTION
Use aluminum as the reference material.
n = 1 in aluminum.
n = Es / Ea = 210 / 70 = 3 in steel.
n = Eb / Ea = 105 / 70 = 1.5 in brass.
Due to symmetry of both the material arrangement and the geometry, the
neutral axis passes through the center of the steel portion.
For the transformed section,
n1
1
b1h13 + n1 A1d12 = (40)(10)3 + (40)(10)(25) 2 = 253.33 × 103 mm 4
12
12
n2
1.5
3
2
(40)(10)3 + (1.5)(40)(10)(15)2 = 140 × 103 mm 4
I 2 = b2 h2 + n2 A2 d 2 =
12
12
n
3.0
(40)(20)3 = 80 × 103 mm 4
I 3 = 3 b3 h33 =
12
12
I 4 = I 2 = 140 × 103 mm 4
I 5 = I1 = 253.33 × 103 mm 4
I1 =
I=
(a)
= 866.66 × 103 mm 4 = 866.66 × 10−9 m 4
nMy
where M = 1800 N ⋅ m
I
n = 1.0
y = −30 mm = 0.030 m
Aluminum:
σ =−
σa =
(1.0)(1800)(0.030)
= 62.3 × 106 Pa
−9
866.66 × 10
Brass:
σb =
Steel:
σs =
(b)
I
n = 1.5
y = −20 mm = −0.020 m
(1.5)(1800)(0.020)
= 62.3 × 106 Pa
−9
866.66 × 10
n = 3.0
σ b = 62.3 MPa 
y = −10 mm = −0.010 m
(3.0)(1800)(0.010)
= 62.3 × 106 Pa
−9
866.66 × 10
Radius of curvature.
σ a = 62.3 MPa 
1
ρ
=
M
1800
=
= 0.02967 m −1
9
Ea I (70 × 10 )(866.66 × 10−9 )
σ s = 62.3 MPa 
ρ = 33.7 m 
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PROBLEM 4.56
Five metal strips, each of 40 mm wide, are bonded together to form the
composite beam shown. The modulus of elasticity is 210 GPa for the
steel, 105 GPa for the brass, and 70 GPa for the aluminum. Knowing
that the beam is bent about a horizontal axis by a couple of moment
1800 N ⋅ m, determine (a) the maximum stress in each of the three
metals, (b) the radius of curvature of the composite beam.
SOLUTION
Use aluminum as the reference material.
n = 1.0 in aluminum.
n = Es / Ea = 210 / 70 = 3 in steel.
n = Eb / Ea = 105 / 70 = 1.5 in brass.
Due to symmetry of both the material arrangement and the geometry, the neutral axis
passes through the center of the aluminum portion.
For the transformed section,
n1
1.5
b1h13 + n1 A1d12 =
(40)(10)3 + (1.5)(40)(10)(25) 2 = 380 × 103 mm 4
12
12
n2
3.0
3
2
I 2 = b2 h2 + n2 A2 d 2 =
(40)(10)3 + (3.0)(40)(10)(15)2 = 280 × 103 mm 4
12
12
n3
1.0
I 3 = b3 h33 =
(40)(20)3 = 26.67 × 103 mm 4
12
12
I 4 = I 2 = 280 × 103 mm 4
I 5 = I1 = 380 × 103 mm 4
I1 =
I=
(a)
σ =−
nMy
I
Aluminum:
σa =
Steel:
σs =
(b)
where
6
mm 4 = 1.3467 × 10−6 m 4
M = 1800 N ⋅ m
n = 1,
y = −10 mm = −0.010 m
(1.0)(1800)(0.010)
= 13.37 × 106 Pa
1.3467 × 10−6
Brass:
σb =
 I = 1.3467 × 10
n = 1.5,
y = −30 mm = −0.030 m
(1.5)(1800)(0.030)
= 60.1 × 106 Pa
1.3467 × 10−6
n = 3.0,
σ b = 60.1 MPa 
y = −20 mm = −0.020 m
(3.0)(1800)(0.020)
= 80.1 × 106 Pa
1.3467 × 10−6
Radius of curvature.
σ a = 13.37 MPa 
1
ρ
=
M
1800
=
= 0.01909 m −1
9
−6
Ea I (70 × 10 )(1.3467 × 10 )
σ s = 80.1 MPa 
ρ = 52.4 m 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.57
The composite beam shown is formed by bonding together a brass rod and an
aluminum rod of semicircular cross sections. The modulus of elasticity is
15 × 106 psi for the brass and 10 × 106 psi for the aluminum. Knowing that the
composite beam is bent about a horizontal axis by couples of moment
8 kip ⋅ in., determine the maximum stress (a) in the brass, (b) in the aluminum.
SOLUTION
For each semicircle,
yo =
r = 0.8 in.
π
A=
2
4r
(4)(0.8)
=
= 0.33953 in.
3π
3π
r 2 = 1.00531 in 2 ,
I base =
π
8
r 4 = 0.160850 in 4
I = I base − Ayo2 = 0.160850 − (1.00531)(0.33953)2 = 0.044953 in 4
Use aluminum as the reference material.
n = 1.0 in aluminum
n=
Eb 15 × 106
=
= 1.5 in brass
Ea 10 × 106
Locate the neutral axis.
A, in2
nA, in2
yo , in.

1.00531
1.50796
0.33953
0.51200

1.00531
1.00531
−0.33953
−0.34133
2.51327
Σ
nAyo , in 3
Yo =
0.17067
= 0.06791 in.
2.51327
The neutral axis lies 0.06791 in.
above the material interface.
0.17067
d1 = 0.33953 − 0.06791 = 0.27162 in., d 2 = 0.33953 + 0.06791 = 0.40744 in.
I1 = n1I + n1 Ad12 = (1.5)(0.044957) + (1.5)(1.00531)(0.27162)2 = 0.17869 in 4
I 2 = n2 I + n2 Ad 22 = (1.0)(0.044957) + (1.0)(1.00531)(0.40744) 2 = 0.21185 in 4
I = I1 + I 2 = 0.39054 in 4
(a)
Brass:
n = 1.5,
y = 0.8 − 0.06791 = 0.73209 in.
σ =−
(b)
Aluminium:
n = 1.0,
nMy
(1.5)(8)(0.73209)
=−
I
0.39054
σ = −22.5 ksi 
y = −0.8 − 0.06791 = −0.86791 in.
σ =−
nMy
(1.0)(8)(−0.86791)
=−
I
0.39054
σ = 17.78 ksi 
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PROBLEM 4.58
A steel pipe and an aluminum pipe are securely bonded together to form
the composite beam shown. The modulus of elasticity is 200 GPa for the
steel and 70 GPa for the aluminum. Knowing that the composite beam is
bent by a couple of moment 500 N ⋅ m, determine the maximum stress
(a) in the aluminum, (b) in the steel.
SOLUTION
Use aluminum as the reference material.
n = 1.0 in aluminum
n = Es / Ea = 200 / 70 = 2.857 in steel
For the transformed section,
Steel:
I s = ns
Aluminium:
I a = na
π
4
π
(r
4
o
(r
4
4
o
π 
− ri4 = (2.857)   (164 − 104 ) = 124.62 × 103 mm 4
4
)
π 
− ri4 = (1.0)   (194 − 164 ) = 50.88 × 103 mm 4
4
)
I = I s + I a = 175.50 × 103 mm 4 = 175.5 × 10−9 m 4
(a)
Aluminum:
c = 19 mm = 0.019 m
σa =
na Mc (1.0)(500)(0.019)
=
= 54.1 × 106 Pa
−9
I
175.5 × 10
σ a = 54.1 MPa 
(b)
Steel:
c = 16 mm = 0.016 m
σs =
ns Mc (2.857)(500)(0.016)
=
= 130.2 × 106 Pa
−9
I
175.5 × 10
σ s = 130.2 MPa 
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PROBLEM 4.59
The rectangular beam shown is made of a plastic for which the
value of the modulus of elasticity in tension is one-half of its value
in compression. For a bending moment M = 600 N ⋅ m, determine
the maximum (a) tensile stress, (b) compressive stress.
SOLUTION
n=
1
on the tension side of neutral axis.
2
n = 1 on the compression side.
Locate neutral axis.
n1bx
x
h−x
− n2 b(h − x)
=0
2
2
1 2 1
bx − b(h − x) 2 = 0
2
4
1
1
(h − x) 2 x =
( h − x)
2
2
1
x=
h = 0.41421 h = 41.421 mm
2 +1
h − x = 58.579 mm
x2 =
1
1
I1 = n1 bx3 = (1)   (50)(41.421)3 = 1.1844 × 106 mm 4
3
3
1
 1  1 
I 2 = n2 b(h − x)3 =    (50)(58.579)3 = 1.6751 × 106 mm 4
3
 2  3 
I = I1 + I 2 = 2.8595 × 106 mm 4 = 2.8595 × 10−6 m 4
(a)
Tensile stress:
n=
1
,
2
σ =−
(b)
Compressive stress:
n = 1,
σ =−
y = −58.579 mm = −0.058579 m
nMy
(0.5)(600)(−0.058579)
=−
= 6.15 × 106 Pa
I
2.8595 × 10−6
σ t = 6.15 MPa 
y = 41.421 mm = 0.041421 m
nMy
(1.0)(600)(0.041421)
=−
= −8.69 × 106 Pa
I
2.8595 × 10−6
σ c = −8.69 MPa 
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PROBLEM 4.60*
A rectangular beam is made of material for which the modulus of elasticity is Et in tension and Ec in
compression. Show that the curvature of the beam in pure bending is
1
ρ
=
M
Er I
where
Er =
4 Et Ec
( Et + Ec ) 2
SOLUTION
Use Et as the reference modulus.
Then Ec = nEt .
Locate neutral axis.
x
h−x
− b( h − x )
=0
2
2
nx 2 − ( h − x)2 = 0
nx = (h − x)
nbx
h
x=
I trans
1
ρ
n +1
nh
h−x=
n +1
3
 h  1 3 
n 3 1
n   3
3

= bx + b(h − x) = 
 bh
 +
 3  n + 1   n + 1  
3
3


(
)
=
1 n + n3/ 2
1 n 1+ n
1
3
bh
=
bh3 = ×
3
3
3 n +1
3 n +1
3
=
M
M
=
Et I trans Er I
(
)
(
)
where I =
n
(
)
n +1
2
bh3
1 3
bh
12
Er I = Et I trans
Er =
=
Et I trans 12
= 3 × Et ×
I
bh
3
(
4 Et Ec /Et
=
) (
Ec /Et + 1
2
n
(
)
n +1
4 Et Ec
Ec + Et
)
2
bh3
2
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PROBLEM 4.61
Semicircular grooves of radius r must be milled as shown in the sides
of a steel member. Using an allowable stress of 60 MPa, determine the
largest bending moment that can be applied to the member when
(a) r = 9 mm, (b) r = 18 mm.
SOLUTION
(a)
d = D − 2r = 108 − (2)(9) = 90 mm
D 108
=
= 1.20
d
90
From Fig. 4.32,
r
9
=
= 0.1
d 90
K = 2.07
1
(18)(90)3
12
= 1.0935 × 106 mm 4
I=
= 1.0935 × 10−6 m 4
1
d = 45 mm = 0.045 m
2
KMc
σ=
I
c=
M=
(b)
σI
Kc
=
(60 × 106 )(1.0935 × 10−6 )
= 704
(2.07)(0.045)
M = 704 N ⋅ m 
d = 108 − (2)(18) = 72 mm
D 108
r 18
=
= 1.5
=
= 0.25
d
72
d 72
1
c = d = 36 mm = 0.036 m
2
From Fig. 4.32,
K = 1.61
1
I = (18)(72)3 = 559.87 × 103 mm 4 = 559.87 × 10−9 m 4
12
M=
σI
Kc
=
(60 × 106 )(559.87 × 10−9 )
= 580
(1.61)(0.036)
M = 580 N ⋅ m 
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PROBLEM 4.62
Semicircular grooves of radius r must be milled as shown in the sides
of a steel member. Knowing that M = 450 N ⋅ m, determine the
maximum stress in the member when the radius r of the semicircular
grooves
is
(a) r = 9 mm, (b) r = 18 mm.
SOLUTION
(a)
d = D − 2r = 108 − (2)(9) = 90 mm
D 108
=
= 1.20
d
90
From Fig. 4.32,
r
9
=
= 0.1
d 90
K = 2.07
1 3 1
bh = (18)(90)3
12
12
= 1.0935 × 106 mm 4
I=
= 1.0935 × 10−6 m 4
1
c = d = 45 mm = 0.045 m
2
KMc (2.07)(450)(0.045)
=
I
1.0935 × 10−6
= 38.3 × 106 Pa
σ max =
(b)
σ max = 38.3 MPa 
d = D − 2r = 108 − (2)(18) = 72 mm
D 108
=
= 1.5
d
72
From Fig. 4.32,
r 18
=
= 0.25
d 72
K = 1.61
1
d = 72 mm = 0.036 m
2
1
I = (18)(72)3 = 559.87 × 103 mm 4 = 559.87 × 10−9 m 4
12
c=
σ max =
KMc (1.61)(450)(0.036)
=
= 46.6 × 106 Pa
I
559.87 × 10−9
σ max = 46.6 MPa 
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PROBLEM 4.63
Knowing that the allowable stress for the beam shown is 90 MPa, determine
the allowable bending moment M when the radius r of the fillets is (a) 8 mm,
(b) 12 mm.
SOLUTION
1 3
1
bh =
(8)(40)3 = 42.667 × 103 mm 4 = 42.667 × 10−9 m 4
12
12
c = 20 mm = 0.020 m
I =
D 80 mm
=
= 2.00
40 mm
d
(a)
r
8 mm
=
= 0.2
d
40 mm
σ max = K
Mc
I
From Fig. 4.31,
M =
σ max I
Kc
=
K = 1.50
(90 × 106 )(42.667 × 10−9 )
(1.50)(0.020)
M = 128 N ⋅ m 
(b)
r
12 mm
=
= 0.3
d
40 mm
From Fig. 4.31,
M =
K = 1.35
(90 × 106 )(42.667 × 10−9 )
(1.35)(0.020)
M = 142 N ⋅ m 
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PROBLEM 4.64
Knowing that M = 250 N ⋅ m, determine the maximum stress in the beam
shown when the radius r of the fillets is (a) 4 mm, (b) 8 mm.
SOLUTION
1 3
1
bh =
(8)(40)3 = 42.667 × 103 mm 4 = 42.667 × 10−9 m 4
12
12
c = 20 mm = 0.020 m
I =
D 80 mm
=
= 2.00
40 mm
d
(a)
r
4 mm
=
= 0.10
d
40 mm
σ max = K
(b)
K = 1.87
Mc (1.87)(250)(0.020)
=
= 219 × 106 Pa
I
42.667 × 10−9
r
8 mm
=
= 0.20
d
40 mm
σ max = K
From Fig. 4.31,
From Fig. 4.31,
σ max = 219 MPa 
K = 1.50
Mc
(1.50)(250)(0.020)
=
= 176 × 106 Pa
I
42.667 × 10−9
σ max = 176 MPa 
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PROBLEM 4.65
The allowable stress used in the design of a steel bar is
12 ksi. Determine the largest couple M that can be applied
to the bar (a) if the bar is designed with grooves having
semicircular portions of radius r = 34 in. as shown in
Fig. a, (b) if the bar is redesigned by removing the
material above and below the dashed lines as shown in
Fig. b.
SOLUTION
7
in. = 0.875 in.
8
D = 7.5 in.
3
in. = 0.75 in.
4
d = 5 in.
r
0.75
=
= 0.15
d
5
D 7.5
=
= 1.5
d
5
t =
Dimensions:
Stress concentration factors:
Figs. 4.32 and 4.31
Configuration (a):
K = 1.92
Configuration (b):
K = 1.58
I =
Moment of inertia:
c=
1
d = 2.5 in.
2
σm =
KMc
I
r =
1 3
1
td =
(0.875)(5)3 = 9.115 in 4
12
12
σ m = 12 ksi
M =
Iσ m
Kc
(a)
M =
(9.115)(12)
(1.92)(2.5)
M = 22.8 kip ⋅ in 
(b)
M =
(9.115)(12)
(1.58)(2.5)
M = 27.7 kip ⋅ in 
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PROBLEM 4.66
A couple of moment M = 20 kip ⋅ in. is to be applied to the
end of a steel bar. Determine the maximum stress in the bar
(a) if the bar is designed with grooves having semicircular
portions of radius r = 12 in., as shown in Fig. a, (b) if the bar
is redesigned by removing the material above and below the
dashed line as shown in Fig. b.
SOLUTION
7
in. = 0.875 in.
8
D = 7.5 in.
1
in. = 0.5 in.
2
d = 5 in.
r
0.5
=
= 0.10
d
5
D 7.5
=
= 1.5
d
5
t =
Dimensions:
Stress concentration factors:
Figs. 4.32 and 4.31
Configuration (a):
K = 2.22
Configuration (b):
K = 1.80
I =
Moment of inertia:
c=
1
d = 2.5 in.
2
Maximum stress:
r =
1 3
1
td =
(0.875)(5)3 = 9.115 in 4
12
12
M = 20 kip ⋅ in
σm =
KMc
I
(a)
σm =
(2.22)(20)(2.5)
9.115
σ m = 12.2 ksi 
(b)
σm =
(1.80)(20)(2.5)
9.115
σ m = 9.9 ksi 
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PROBLEM 4.67
The prismatic bar shown is made of a steel that is assumed to be
elastoplastic with σ Y = 300 MPa and is subjected to a couple M
parallel to the x axis. Determine the moment M of the couple for which
(a) yield first occurs, (b) the elastic core of the bar is 4 mm thick.
SOLUTION
(a)
I=
1 3 1
bh = (12)(8)3 = 512 mm 4
12
12
= 512 × 10−12 m 4
c=
1
h = 4 mm = 0.004 m
2
(300 × 106 )(512 × 10−12 )
0.004
c
= 38.4 N ⋅ m
MY =
(b)
yY =
1
(4) = 2 mm
2
σY I
=
yY 2
= = 0.5
c
4
 1  y 2 
3
M = M Y 1 −  Y  
2
 3  c  
3
 1

= (38.4) 1 − (0.5) 2 
2
 3

= 52.8 N ⋅ m
M Y = 38.4 N ⋅ m 
M = 52.8 N ⋅ m 
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PROBLEM 4.68
Solve Prob. 4.67, assuming that the couple M is parallel to the z axis.
PROBLEM 4.67 The prismatic bar shown is made of a steel that is
assumed to be elastoplastic with σ Y = 300 MPa and is subjected to a
couple M parallel to the x axis. Determine the moment M of the couple
for which (a) yield first occurs, (b) the elastic core of the bar is 4 mm
thick.
SOLUTION
(a)
I=
1 3 1
bh = (8)(12)3 = 1.152 × 103 mm 4
12
12
= 1.152 × 10−9 m 4
1
c = h = 6 mm = 0.006 m
2
(300 × 106 )(1.152 × 10−9 )
c
0.006
= 57.6 N ⋅ m
MY =
(b)
yY =
1
(4) = 2 mm
2
σY I
=
M Y = 57.6 N ⋅ m 
yY 2 1
= =
c
6 3
M=
 1  y 2 
3
M Y 1 −  Y  
2
 3  c  
 1  1 2 
3
= (57.6) 1 −   
2
 3  3  
= 83.2 N ⋅ m
M = 83.2 N ⋅ m 
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PROBLEM 4.69
The prismatic bar shown, made of a steel that is assumed to be elastoplastic
with E = 29 × 106 psi and σ Y = 36 ksi, is subjected to a couple of
1350 lb ⋅ in. parallel to the z axis. Determine (a) the thickness of the elastic
core, (b) the radius of curvature of the bar.
SOLUTION
3
(a)
I=
1  1  5 
4
−3
   = 10.1725 × 10 in
12  2  8 
c=
15
  = 0.3125 in.
28
(36 × 103 )(10.1725 × 10−3 )
0.3125
c
= 1171.872 lb ⋅ in
MY =
σY I
=
3
M = MY
2
 1  y 3 
1 −  Y  
 3  c  
yY
(2)(1350)
M
= 3−2
= 3−
= 0.83426
1171.872
c
MY
yY = (0.83426)(0.3125) = 0.26071 in.
Thickness of elastic core:
(b)
yY = ε Y ρ =
σY
2 yY = 0.521 in. 
ρ
E
y E (0.26071)(29 × 106 )
= 210.02 in.
ρ= Y =
σY
36 × 103
ρ = 17.50 ft 
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PROBLEM 4.70
Solve Prob. 4.69, assuming that the 1350-lb ⋅ in. couple is parallel to the
y axis.
PROBLEM 4.69 The prismatic bar shown, made of a steel that is assumed
to be elastoplastic with E = 29 × 106 psi and σ Y = 36 ksi, is subjected to a
couple of 1350 lb ⋅ in parallel to the z axis. Determine (a) the thickness of
the elastic core, (b) the radius of curvature of the bar.
SOLUTION
3
(a)
I=
1  5  1 
−3 4
   = 6.5104 × 10 in
12  8  2 
 1  1 
c =    = 0.25 in.
 2  2 
(36 × 103 )(6.5104 × 10−3 )
0.25
c
= 937.5 lb ⋅ in
MY =
σY I
=
2
3  1  yY  
M = 1 − 

2  3  c  
yY
(2)(1350)
M
= 3−2
= 3−
= 0.34641
937.5
c
MY
yY = (0.34641)(0.25) = 0.086603 in.
Thickness of elastic core:
(b)
yY = ε Y ρ =
σY
2 yY = 0.1732 in. 
ρ
E
y E (0.086603)(29 × 106 )
= 69.763 in.
ρ= Y =
σY
36 × 103
ρ = 5.81 ft 
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PROBLEM 4.71
A bar of rectangular cross section shown is made of a steel that is
assumed to be elastoplastic with E = 200 GPa and σ Y = 300 MPa.
Determine the bending moment M for which (a) yield first occurs,
(b) the plastic zones at the top and bottom of the bar are 12 mm thick.
SOLUTION
1
(30)(40)3 = 160 × 103 mm 4 = 160 × 10−9 m 4
12
1
c = (40) = 20 mm = 0.020 m
2
I =
(a)
First yielding:
MY =
σ =
Mc
= σY
I
Iσ Y
(160)(10−9 )(300 × 106 )
=
= 2400 N ⋅ m
c
0.020
M Y = 2.40 kN ⋅ m 
(b)
Plastic zones are 12 mm thick:
yY = 20 mm − 12 mm = 8 mm
yY
8 mm
=
= 0.4
c
20 mm
M =
=
2

3
1 y  
M Y 1 −  Y  
2
3  c  

3
1


(2400) 1 − (0.4)2  = 3408 N ⋅ m
2
3


M = 3.41 kN ⋅ m 
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PROBLEM 4.72
Bar AB is made of a steel that is assumed to be elastoplastic with
E = 200 GPa and σ Y = 240 MPa. Determine the bending moment M
for which the radius of curvature of the bar will be (a) 18 m, (b) 9 m.
SOLUTION
1
(30)(40)3 = 160 × 103 mm 4 = 160 × 10−9 m 4
12
1
c = (40) = 20 mm = 0.020 m
2
I =
Iσ Y
(160 × 10−9 )(240 × 106 )
=
= 1920 N ⋅ m
c
0.020
M
1
1920
= Y =
= 0.0600 m −1
EI
ρY
(200 × 109 )(160 × 10−9 )
MY =
(a)
1
ρ = 18 m:
ρ
= 0.05556 m −1 < 0.0600 m −1
The bar is fully elastic.
M =
EI
ρ
=
(200 × 109 )(160 × 10−9 )
= 1778 N ⋅ m
18
M = 1.778 kN ⋅ m 
(b)
1
ρ = 9 m:
ρ
= 0.11111 m −1 > 0.0600 m −1
The bar is partially plastic.
σY =
EyY
ρ
yY =
ρσ Y
E
(9)(240 × 106 )
= 0.0108 m = 10.8 mm
200 × 109
yY
10.8 mm
=
= 0.54
c
20 mm
yY =
M =
3
MY
2
2

1 y   3
1


1 −  Y   = (1920) 1 − (0.54)2  = 2600 N ⋅ m
c
3
2
3
  



M = 2.60 kN ⋅ m 
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PROBLEM 4.73
A beam of the cross section shown is made of a steel that is assumed to be
elastoplastic with E = 200 GPa and σ Y = 240 MPa. For bending about the zaxis, determine the bending moment at which (a) yield first occurs, (b) the
plastic zones at the top and bottom of the bar are 30-mm thick.
SOLUTION
(a)
1 3 1
bh = (60)(90)3 = 3.645 × 106 mm 4 = 3.645 × 10−6 m 4
12
12
1
c = h = 45 mm = 0.045 m
2
σ I (240 × 106 )(3.645 × 10−6 )
MY = Y =
= 19.44 × 10 N ⋅ m
c
0.045
I=
M Y = 19.44 kN ⋅ m 
R1 = σ Y A1 = (240 × 106 )(0.060)(0.030)
= 432 × 103 N
y1 = 15 mm + 15 mm = 0.030 m
1
1
R2 = σ Y A2 =   (240 × 106 )(0.060)(0.015)
2
2
= 108 × 103 N
2
y2 = (15 mm) = 10 mm = 0.010 m
3
(b)
M = 2( R1 y1 + R2 y2 ) = 2[(432 × 103 )(0.030) + (108 × 103 )(0.010)]
= 28.08 × 103 N ⋅ m
M = 28.1 kN ⋅ m 
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PROBLEM 4.74
A beam of the cross section shown is made of a steel that is assumed to be
elastoplastic with E = 200 GPa and σ Y = 240 MPa. For bending about the
z axis, determine the bending moment at which (a) yield first occurs,
(b) the plastic zones at the top and bottom of the bar are 30-mm thick.
SOLUTION
(a)
1 3 1
bh = (60)(90)3 = 3.645 × 106 mm 4
12
12
1
1
I cutout = bh3 = (30)(30)3 = 67.5 × 103 mm 4
12
12
I = 3.645 × 106 − 67.5 × 103 = 3.5775 × 106 mm 4
I rect =
= 3.5775 × 10−6 mm 4
1
h = 45 mm = 0.045 m
2
σ I (240 × 106 )(3.5775 × 10−6 )
MY = Y =
0.045
c
3
= 19.08 × 10 N ⋅ m
c=
M Y = 19.08 kN ⋅ m 
R1 = σ Y A1 = (240 × 106 )(0.060)(0.030) = 432 × 103 N
y1 = 15 mm + 15 mm = 30 mm = 0.030 m
1
1
σ Y A2 = (240 × 106 )(0.030)(0.015) = 54 × 103 N
2
2
2
y2 = (15 mm) = 10 mm = 0.010 m
3
R2 =
(b)
M = 2( R1 y1 + R2 y2 )
= 2[(432 × 103 )(0.030) + (54 × 103 )(0.010)]
= 27.00 × 103 N ⋅ m
M = 27.0 kN ⋅ m 
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PROBLEM 4.75
A beam of the cross section shown is made of a steel that is assumed to be
elastoplastic with E = 29 × 106 psi and σ Y = 42 ksi. For bending about the
z axis, determine the bending moment at which (a) yield first occurs, (b) the
plastic zones at the top and bottom of the bar are 3 in. thick.
SOLUTION
(a)
1
1
b1h13 + A1d12 = (6)(3)3 + (6)(3)(3)2 = 175.5 in 4
12
12
1
1
3
I 2 = b2 h2 = (3)(3)3 = 6.75 in 4
12
12
I 3 = I1 = 175.5 in 4
I1 =
I = I1 + I 2 + I 3 = 357.75 in 4
c = 4.5 in.
MY =
σY I
c
=
(42)(357.75)
4.5
M Y = 3339 kip ⋅ in 
R1 = σ Y A1 = (42)(6)(3) = 756 kip
y1 = 1.5 + 1.5 = 3 in.
1
1
R2 = σ Y A2 = (42)(3)(1.5)
2
2
= 94.5 kip
2
y2 = (1.5) = 1.0 in.
3
(b)
M = 2( R1 y1 + R2 y2 ) = 2[(756)(3) + (94.5)(1.0)]
M = 4725 kip ⋅ in 
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PROBLEM 4.76
A beam of the cross section shown is made of a steel that is assumed to be
elastoplastic with E = 29 × 106 psi and σ Y = 42 ksi. For bending about the
z axis, determine the bending moment at which (a) yield first occurs, (b) the
plastic zones at the top and bottom of the bar are 3 in. thick.
SOLUTION
(a)
1
1
b1h13 + A1d12 = (3)(3)3 + (3)(3)(3) 2 = 87.75 in 4
12
12
1
1
I 2 = b2 h23 = (6)(3)3 = 13.5 in 4
12
12
I 3 = I1 = 87.75 in 4
I1 =
I = I1 + I 2 + I 3 = 188.5 in 4
c = 4.5 in.
σ I (42)(188.5)
MY = Y =
c
4.5
M Y = 1759 kip ⋅ in 
R1 = σ Y A1 = (42)(3)(3) = 378 kip
y1 = 1.5 + 1.5 = 3.0 in.
1
1
R2 = σ Y A2 = (42)(6)(1.5)
2
2
= 189 kip
2
y2 = (1.5) = 1.0 in.
3
(b)
M = 2( R1 y1 + R2 y2 ) = 2[(378)(3.0) + (189)(1.0)]
M = 2646 kip ⋅ in 
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PROBLEM 4.77
For the beam indicated (of Prob. 4.73), determine (a) the fully plastic
moment M p , (b) the shape factor of the cross section.
SOLUTION
From Problem 4.73,
E = 200 GPa and σ Y = 240 MPa.
A1 = (60)(45) = 2700 mm 2
= 2700 × 10−6 m 2
R = σ Y A1
= (240 × 106 )(2700 × 10−6 )
= 648 × 103 N
d = 45 mm = 0.045 m
(a)
(b)
M p = Rd = (648 × 103 )(0.045) = 29.16 × 103 N ⋅ m
M p = 29.2 kN ⋅ m 
1 3 1
bh = (60)(90)3 = 3.645 × 106 mm 4 = 3.645 × 10−6 m 4
12
12
c = 45 mm = 0.045 m
I=
MY =
σY I
c
=
(240 × 106 )(3.645 × 10−6 )
= 19.44 × 103 N ⋅ m
0.045
k=
Mp
MY
=
29.16
19.44
k = 1.500 
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PROBLEM 4.78
For the beam indicated (of Prob. 4.74), determine (a) the fully plastic moment
M p , (b) the shape factor of the cross section.
SOLUTION
From Problem 4.74,
(a)
E = 200 GPa and σ Y = 240 MPa.
R1 = σ Y A1
= (240 × 106 )(0.060)(0.030)
= 432 × 103 N
y1 = 15 mm + 15 mm = 30 mm
= 0.030 m
R2 = σ Y A2
= (240 × 106 )(0.030)(0.015)
= 108 × 103 N
1
y2 = (15) = 7.5 mm = 0.0075 m
2
M p = 2( R1 y1 + R2 y2 ) = 2[( 432× 103 )(0.030) + (108× 103 )(0.0075)]
= 27.54 × 103 N ⋅ m
(b)
M p = 27.5 kN ⋅ m 
1 3 1
bh = (60)(90)3 = 3.645 × 106 mm 4
12
12
1 3 1
I cutout = bh = (30)(30)3 = 67.5 × 103 mm 4
12
12
I = I rect − I cutout = 3.645 × 106 − 67.5 × 103 = 3.5775 × 103 mm 4
I rect =
= 3.5775 × 10−9 m 4
1
h = 45 mm = 0.045 m
2
σ I (240 × 106 )(3.5775 × 10−9 )
= 19.08 × 103 N ⋅ m
MY = Y =
0.045
c
M p 27.54
k=
=
MY 19.08
c=
k = 1.443 
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PROBLEM 4.79
For the beam indicated (of Prob. 4.75), determine (a) the fully plastic moment
M p , (b) the shape factor of the cross section.
SOLUTION
From Problem 4.75,
(a)
E = 29 × 106 and σ Y = 42 ksi.
R1 = σ Y A1 = (42)(6)(3) = 756 kip
y1 = 1.5 + 1.5 = 3.0 in.
R2 = σ Y A2 = (42)(3)(1.5) = 189 kip
y2 =
1
(1.5) = 0.75 in.
2
M p = 2( R1 y1 + R2 y2 ) = 2[( 756)( 3.0) + (189)(0. 75)]
(b)
M p = 4819.5 kip ⋅ in 
1
1
b1h13 + A1d12 = (6)(3)3 + (6)(3)(3)2 = 175.5 in 4
12
12
1
1
I 2 = b2 h23 = (3)(3)3 = 6.75 in 4
12
12
I 3 = I1 = 175.5 in 4
I1 =
I = I1 + I 2 + I 3 = 357.75 in 4
c = 4.5 in.
σY I
(42)(357.75)
= 3339 kip ⋅ in
4.5
4819.5
=
k=
MY
3339
MY =
c
Mp
=
k = 1.443 
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PROBLEM 4.80
For the beam indicated (of Prob. 4.76), determine (a) the fully plastic moment
M p , (b) the shape factor of the cross section.
SOLUTION
From Problem 4.76,
(a)
E = 29 × 106 psi and σ Y = 42 ksi.
R1 = σ Y A1 = (42)(3)(3) = 378 kip
y1 = 1.5 + 1.5 = 3.0 in.
R2 = σ Y A2 = (42)(6)(1.5) = 378 kip
y2 =
1
(1.5) = 0.75 in.
2
M p = 2( R1 y1 + R2 y2 ) = 2[( 378)( 3.0) + ( 378)(0. 75)]
(b)
M p = 2835 kip ⋅ in 
1
1
b1h13 + A1d12 = (3)(3)3 + (3)(3)(3)2 = 87.75 in 4
12
12
1
1
3
I 2 = b2 h2 = (6)(3)3 = 13.5 in 4
12
12
I 3 = I1 = 87.75 in 4
I1 =
I = I1 + I 2 + I 3 = 188.5 in 4
c = 4.5 in.
σY I
(42)(188.5)
= 1759.3 kip ⋅ in
c
4.5
Mp
2835
k=
=
MY 1759.3
MY =
=
k = 1.611 
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PROBLEM 4.81
Determine the plastic moment M p of a steel beam of the cross section shown,
assuming the steel to be elastoplastic with a yield strength of 240 MPa.
SOLUTION
A=
For a semicircle:
π
2
r 2;
r =
4r
3π
R = σY A
Resultant force on semicircular section:
Resultant moment on entire cross section:
M p = 2 Rr =
Data:
4
σY r3
3
σ Y = 240 MPa = 240 × 106 Pa,
Mp =
r = 18 mm = 0.018 m
4
(240 × 106 )(0.018)3 = 1866 N ⋅ m
3
M p = 1.866 kN ⋅ m 
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PROBLEM 4.82
Determine the plastic moment M p of a steel beam of the cross section shown,
assuming the steel to be elastoplastic with a yield strength of 240 MPa.
SOLUTION
Total area:
A = (50)(90) − (30)(30) = 3600 mm 2
1
A = 1800 mm 2
2
1 A
1800
x= 2 =
= 36 mm
b
50
A1 = (50)(36) = 1800 mm 2 ,
y1 = 18 mm,
A1 y1 = 32.4 × 103 mm3
A2 = (50)(14) = 700 mm 2 ,
y2 = 7 mm,
A2 y2 = 4.9 × 103 mm3
A3 = (20)(30) = 600 mm 2 ,
y3 = 29 mm,
A3 y3 = 17.4 × 103 mm3
A4 = (50)(10) = 500 mm 2 ,
y4 = 49 mm,
A4 y4 = 24.5 × 103 mm3
A1 y1 + A2 y2 + A3 y3 + A4 y4 = 79.2 × 103 mm3 = 79.2 × 10−6 m3
M p = σ Y ΣAi yi = (240 × 106 )(79.2 × 10−6 ) = 19.008 × 103 N ⋅ m
M p = 19.01 kN ⋅ m 
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PROBLEM 4.83
Determine the plastic moment M p of a steel beam of the cross section shown, assuming
the steel to be elastoplastic with a yield strength of 240 MPa.
SOLUTION
Total area:
A=
1
(30)(36) = 540 mm 2
2
By similar triangles,
b 30
=
y 36
Since
A1 =
b=
Half area:
1
A = 270 mm 2 = A1
2
5
y
6
1
5 2
by =
y ,
2
12
y2 =
12
A1
5
12
(270) = 25.4558 mm
5
b = 21.2132 mm
1
A1 = (21.2132)(25.4558) = 270 mm 2 = 270 × 10−6 m 2
2
A2 = (21.2132)(36 − 25.4558) = 223.676 mm 2 = 223.676 × 10−6 m 2
y=
A3 = A − A1 − A2 = 46.324 mm 2 = 46.324 × 10−6 m 2
Ri = σ Y Ai = 240 × 106 Ai
R1 = 64.8 × 103 N, R2 = 53.6822 × 103 N, R3 = 11.1178 × 103 N
1
y1 = y = 8.4853 mm = 8.4853 × 10−3 m
3
1
y2 = (36 − 25.4558) = 5.2721 mm = 5.2721 × 10−3 m
2
2
y3 = (36 − 25.4558) = 7.0295 mm = 7.0295 × 10−3 m
3
M p = R1 y1 + R2 y2 + R3 y3 = 911 N ⋅ m
M p = 911 N ⋅ m 
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PROBLEM 4.84
Determine the plastic moment M p of a steel beam of the cross section shown, assuming
the steel to be elastoplastic with a yield strength of 240 MPa.
SOLUTION
Let c1 be the outer radius and c2 the inner radius.
A1 y1 = Aa ya − Ab yb
 π  4c   π  4c 
=  c12  1  −  c22  2 
 2  3π   2  3π 
3
2 3
=
c1 − c2
3
(
A2 y2 = A1 y1 =
)
(
2 3
c1 − c 22
3
)
M p = σ Y ( A1 y1 + A2 y2 ) =
Data:
(
4
σ Y c13 − c32
3
)
σ Y = 240 MPa = 240 × 106 Pa
c1 = 60 mm = 0.060 m
c2 = 40 mm = 0.040 m
4
(240 × 106 )(0.0603 − 0.0403 )
3
= 48.64 × 103 N ⋅ m
Mp =
M p = 48.6 kN ⋅ m 
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PROBLEM 4.85
Determine the plastic moment M p of the cross section shown, assuming the steel
to be elastoplastic with a yield strength of 36 ksi.
SOLUTION
A = (1.8)(0.4) + (0.6)(1.2) = 1.44 in 2
Total area:
1
A = 0.72 in 2
2
1
A 0.72
x= 2 =
= 1.2 in.
b
0.6
Neutral axis lies 1.2 in. below the top.
1
1
A = 0.72 in 2 , y1 = (1.2) = 0.6 in.
2
2
1
1
2
A2 = A = 0.72 in , y2 = (0.4) = 0.2 in.
2
2
M p = σ Y ( A1 y1 + A2 y2 )
A1 =
= (36)[(0.72)(0.6) + (0.72)(0.2)] = 20.7 kip ⋅ in
M p = 20.7 kip ⋅ in 
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PROBLEM 4.86
Determine the plastic moment M p of the cross section shown, assuming the steel to
be elastoplastic with a yield strength of 36 ksi.
SOLUTION
Total area:
1 1
1
A = (4)   +   (3) + (2)   = 4.5 in 2
2
2
   
2
1
A = 2.25 in 2
2
A1 = 2.00 in 2 ,
y1 = 0.75,
A1 y1 = 1.50 in 3
A2 = 0.25 in 2 ,
y2 = 0.25,
A2 y2 = 0.0625 in 3
A3 = 1.25 in 2 ,
y3 = 1.25,
A3 y3 = 1.5625 in 3
A4 = 1.00 in 2 ,
y4 = 2.75,
A4 y4 = 2.75 in 3
M p = σ Y ( A1 y1 + A2 y2 + A3 y3 + A4 y4 )
= (36)(1.50 + 0.0625 + 1.5625 + 2.75)
M p = 212 kip ⋅ in 
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PROBLEM 4.87
For the beam indicated (of Prob. 4.73), a couple of moment equal to the full
plastic moment M p is applied and then removed. Using a yield strength of
240 MPa, determine the residual stress at y = 45 mm .
SOLUTION
M p = 29.16 × 103 N ⋅ m
See solutions to Problems 4.73 and 4.77.
I = 3.645 × 10−6 m 4
c = 0.045 m
σ′ =
Mp c
M max y
=
I
I
UNLOADING
LOADING
σ′ =
at y = c = 45 mm
RESIDUAL STRESSES
(29.16 × 103 )(0.045)
= 360 × 106 Pa
−6
3.645 × 10
σ res = σ ′ − σ Y = 360 × 106 − 240 × 106
= 120 × 106 Pa
σ res = 120.0 MPa 
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PROBLEM 4.88
For the beam indicated (of Prob. 4.74), a couple of moment equal to the
full plastic moment M p is applied and then removed. Using a yield
strength of 240 MPa, determine the residual stress at y = 45 mm .
SOLUTION
M p = 27.54 × 103 N ⋅ m (See solutions to Problems 4.74 and 4.78.)
I = 3.5775 × 10−6 m 4 ,
LOADING
c = 0.045 m
σ′ =
Mp c
M max y
=
I
I
σ′ =
(27.54 × 103 )(0.045)
= 346.4 × 106 Pa
−6
3.5775 × 10
UNLOADING
at
y =c
RESIDUAL STRESSES
σ res = σ ′ − σ Y = 346.4 × 106 − 240 × 106 = 106.4 × 106 Pa
σ res = 106.4 MPa 
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PROBLEM 4.89
A bending couple is applied to the bar indicated, causing plastic zones 3 in.
thick to develop at the top and bottom of the bar. After the couple has been
removed, determine (a) the residual stress at y = 4.5in. , (b) the points
where the residual stress is zero, (c) the radius of curvature corresponding
to the permanent deformation of the bar.
SOLUTION
See solution to Problem 4.75 for bending couple and stress distribution.
M = 4725 kip ⋅ in
σ Y = 42 ksi
(a)
yY = 1.5 in.
I = 357.75 in 4
c = 4.5 in.
Mc (4725)(4.5)
=
= 59.43 ksi
I
357.75
MyY
(4725)(1.5)
σ ′′ =
=
= 19.81 ksi
I
357.75
At y = c, σ res = σ ′ − σ Y = 59.43 − 42 = 17.43 ksi
σ′ =
At
σ res = 0 ∴
y0 =
(c)
At

y = yY , σ res = σ ′′ − σ Y = 19.81 − 42 = −22.19 ksi
UNLOADING
LOADING
(b)
E = 29 × 106 psi = 29 × 103 ksi
RESIDUAL STRESSES
My0
− σY = 0
I
Iσ Y
(357.75)(42)
=
= 3.18 in.
M
4725
Answer:
y0 = −3.18in., 0, 3.18 in. 
y = yY , σ res = −22.19 ksi
σ =−
Ey
ρ
∴ ρ =−
Ey
σ
=
(29 × 103 )(1.5)
= 1960 in.
22.19
ρ = 163.4 ft. 
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PROBLEM 4.90
A bending couple is applied to the bar indicated, causing plastic zones 3 in.
thick to develop at the top and bottom of the bar. After the couple has been
removed, determine (a) the residual stress at y = 4.5in., (b) the points
where the residual stress is zero, (c) the radius of curvature corresponding
to the permanent deformation of the bar.
SOLUTION
See solution to Problem 4.76 for bending couple and stress distribution during loading.
M = 2646 kip ⋅ in
I = 188.5 in 4
σ Y = 42 ksi
(a)
(c)
E = 29 × 106 psi = 29 × 103 ksi
c = 4.5 in.
Mc (2646)(4.5)
=
= 63.17 ksi
I
188.5
MyY
(2646)(1.5)
σ ′′ =
=
= 21.06 ksi
I
188.5
σ′ =
At
y = c, σ res = σ ′ − σ Y = 63.17 − 42 = 21.17 ksi
At
y = yY , σ res = σ ′′ − σ Y = 21.06 − 42
σ res = −20.94 ksi 
UNLOADING
LOADING
(b)
yY = 1.5in.
My0
= σY
I
Iσ Y
(188.5)(42)
y0 =
=
= 2.992 in.
M
2646
RESIDUAL STRESSES
σ res = 0 ∴
At
y = yY ,
σ =−
Ey
ρ
Answer:
y0 = −2.992 in.,
0,
2.992 in. 
σ res = −20.94 ksi
∴ ρ =−
Ey
σ
=
(29 × 103 )(1.5)
= 2077 in.
20.94
ρ = 173.1 ft 
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PROBLEM 4.91
A bending couple is applied to the beam of Prob. 4.73, causing plastic zones
30 mm thick to develop at the top and bottom of the beam. After the couple has
been removed, determine (a) the residual stress at y = 45 mm, (b) the points
where the residual stress is zero, (c) the radius of curvature corresponding to
the permanent deformation of the beam.
SOLUTION
See solution to Problem 4.73 for bending couple and stress distribution during loading:
M = 28.08 × 103 N ⋅ m
I = 3.645 × 10−6 m 4
σ Y = 240 MPa
(a)
E = 200 GPa
c = 0.045 m
σ′ =
Mc (28.08 × 103 )(0.045)
=
= 346.7 × 106 Pa = 346.7 MPa
I
3.645 × 10−6
σ ′′ =
M yY
(28.08 × 103 )(0.015)
=
= 115.6 × 106 Pa = 115.6 MPa
−6
I
3.645 × 10
At y = c, σ res = σ ′ − σ Y = 346.7 − 240
σ res = 106.7 MPa 
At y = yY , σ res = σ ′′ − σ Y = 115.6 − 240
σ res = −124.4 MPa
UNLOADING
LOADING
(b)
yY = 15 mm = 0.015 m
σ res = 0 ∴
y0 =
RESIDUAL STRESSES
M y0
− σY = 0
I
Iσ Y
(3.645 × 10−6 )(240 × 106 )
=
= 31.15 × 10−3 m = 31.15 mm
M
28.08 × 103
Answer: y0 = −31.15 mm, 0, 31.15 mm 
(c)
At
y = yY , σ res = −124.4 × 106 Pa
σ =−
Ey
ρ
∴ ρ =−
Ey
σ
=
(200 × 109 )(0.015)
−124.4 × 106
ρ = 24.1 m 
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PROBLEM 4.92
A beam of the cross section shown is made of a steel that is assumed to be
elastoplastic with E = 29 × 106 psi and σ Y = 42 ksi . A bending couple is
applied to the beam about z axis, causing plastic zones 2 in. thick to develop at
the top and bottom of the beam. After the couple has been removed, determine
(a) the residual stress at y = 2 in., (b) the points where the residual stress is
zero, (c) the radius of curvature corresponding to the permanent deformation of
the beam.
SOLUTION
See solution to Problem 4.76 for bending couple and stress distribution during loading.
M = 406 kip ⋅ in
σ Y = 42 ksi
(a)
yY = 1.0 in.
I = 14.6667 in 4
E = 29 × 106 psi = 29 × 103 ksi
c = 2 in.
Mc (406)(2)
=
= 55.36 ksi
I
14.6667
MyY
(406)(1.0)
σ ′′ =
=
= 27.68 ksi
I
14.6667
σ′ =
σ res = σ ′ − σ Y = 55.36 − 42
At y = c,
At y = yY , σ res = σ ′′ − σ Y = 27.68 − 42
UNLOADING
LOADING
(b)
σ res = 0 ∴
y0 =
(c)
σ =−
Ey
ρ
σ res = −14.32 ksi 
RESIDUAL STRESSES
M y0
− σY = 0
I
Iσ Y
(14.6667)(42)
=
= 1.517 in.
M
406
At y = yY ,
σ res = 13.36 ksi 
Answer: y0 = −1.517 in., 0, 1.517 in. 
σ res = −14.32 ksi
∴ ρ =−
Ey
σ
=
(29 × 103 )(1.0)
= 2025 in.
14.32
ρ = 168.8 ft 
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PROBLEM 4.93*
A rectangular bar that is straight and unstressed is bent into an arc of circle of radius by two couples of
moment M. After the couples are removed, it is observed that the radius of curvature of the bar is ρ R .
Denoting by ρY the radius of curvature of the bar at the onset of yield, show that the radii of curvature satisfy
the following relation:
1
ρR
2 

1
3 ρ 
1  ρ  
1 − 
= 1 −
 
ρ
2 ρY 
3  ρY   



SOLUTION
1
ρY
Let m denote
=
MY
,
EI
M =

3
1 ρ2 
M Y 1 −
,
2
3 ρY2 

M
:
MY
M
3
ρ2 
ρ2
= 1 − 2  ∴
= 3−2m
MY
2
ρY 
ρY2
1
1 M
1 mM Y
1
m
=
−
=
−
=
−
ρR
ρ EI
ρ
EI
ρ ρY
m=
=
1
ρ  1  3 ρ 
1 ρ 2  
m
−
=
−
−
1
1
1





ρ
ρY  ρ  2 ρY 
3 ρY2  

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PROBLEM 4.94
A solid bar of rectangular cross section is made of a material that is assumed to be elastoplastic. Denoting by
M Y and ρY , respectively, the bending moment and radius of curvature at the onset of yield, determine (a) the
radius of curvature when a couple of moment M = 1.25 M Y is applied to the bar, (b) the radius of curvature
after the couple is removed. Check the results obtained by using the relation derived in Prob. 4.93.
SOLUTION
(a)
1
ρY
=
m=
MY
,
EI
M =

3
1 ρ2 
M Y 1 −

2
3 ρY2 

M
3
1 ρ2 
= 1 −

MY
2
3 ρY2 
Let m =
M
= 1.25
MY
ρ
= 3 − 2m = 0.70711
ρY
ρ = 0.70711ρY 
(b)
1
ρR
=
=
1
ρ
−
M
1 mM Y
1
m
1
1.25
=
−
=
−
=
−
EI
EI
0.70711ρY
ρ
ρ ρY
ρY
0.16421
ρY
ρ R = 6.09ρY 
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PROBLEM 4.95
The prismatic bar AB is made of a steel that is assumed to be elastoplastic and
for which E = 200 GPa . Knowing that the radius of curvature of the bar is
2.4 m when a couple of moment M = 350 N ⋅ m is applied as shown,
determine (a) the yield strength of the steel, (b) the thickness of the elastic
core of the bar.
SOLUTION
M =

3
1 ρ2 
M Y 1 −

2
3 ρY2 


1 ρ 2σ Y2 
1 −

3 E 2c 2 

=
3 σY I
2 c
=
3 σ Y b(2c)3 
1 ρ 2σ Y2 
1 −

2 12c 
3 E 2c 2 

1 ρ 2σ Y2 
= σ Y bc 2 1 −

3 E 2c 2 

(a)

ρ 2σ Y2 
bc 2σ y 1 −
=M
3E 2c 2 

Cubic equation for σ Y
Data:
E = 200 × 109 Pa
M = 420 N ⋅ m
ρ = 2.4 m
b = 20 mm = 0.020 m
c=
1
h = 8 mm = 0.008 m
2
(1.28 × 10−6 ) σ Y 1 − 750 × 10−21σ Y2  = 350
σ Y 1 − 750 × 10−21σ Y2  = 273.44 × 106
Solving by trial,
(b)
yY =
σY ρ
E
=
σ Y = 292 × 106 Pa
σ Y = 292 MPa 
(292 × 106 )(2.4)
= 3.504 × 10−3 m = 3.504 mm
200 × 109
thickness of elastic core = 2 yY = 7.01 mm 
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PROBLEM 4.96
The prismatic bar AB is made of an aluminum alloy for
which the tensile stress-strain diagram is as shown.
Assuming that the σ -ε diagram is the same in
compression as in tension, determine (a) the radius of
curvature of the bar when the maximum stress is 250 MPa,
(b) the corresponding value of the bending moment. (Hint:
For part b, plot σ versus y and use an approximate method
of integration.)
SOLUTION
(a)
σ m = 250 MPa = 250 × 106 Pa
ε m = 0.0064 from curve
1
h = 30 mm = 0.030 m
2
b = 40 mm = 0.040 m
c=
1
ρ
(b)
=
ε m 0.0064
=
= 0.21333 m −1
c
0.030
ρ = 4.69 m 
ε = −ε m
Strain distribution:
M = −
Bending couple:
y
= −ε mu
c
c
−c
u =
where
c
0
y
ε
1
yσ bdy = 2b  y |σ | d y = 2bc 2  0 u |σ | du = 2bc 2 J
1
where the integral J is given by  0 u |σ | du
Evaluate J using a method of numerial integration. If Simpson’s rule is used, the integration formula is
J =
Δu
Σwu |σ |
3
where w is a weighting factor.
Using Δu = 0.25, we get the values given in the table below:
u
0
0.25
0.5
0.75
1.00
|ε |
0
0.0016
0.0032
0.0048
0.0064
|σ |, (MPa)
0
110
180
225
250
u |σ |, (MPa)
0
27.5
90
168.75
250
w
1
4
2
4
1
wu |σ |, (MPa)
0
110
180
675
250
1215
← Σwu |σ |
(0.25)(1215)
= 101.25 MPa = 101.25 × 106 Pa
3
M = (2)(0.040)(0.030)2 (101.25 × 106 ) = 7.29 × 103 N ⋅ m
J =
M = 7.29 kN ⋅ m 
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PROBLEM 4.97
The prismatic bar AB is made of a bronze alloy for which the tensile
stress-strain diagram is as shown. Assuming that the σ − ε diagram
is the same in compression as in tension, determine (a) the
maximum stress in the bar when the radius of curvature of the bar is
100 in., (b) the corresponding value of the bending moment. (See
hint given in Prob. 4.96.)
SOLUTION
(a)
ρ = 100 in., b = 0.8 in., c = 0.6 in.
εm =
(b)
c
ρ
=
0.6
= 0.006
100
Strain distribution:
Bending couple:
From the curve,
ε = −ε m
M = −
y
= −ε mu
c
c
−c
where u =
c
0
σ m = 43 ksi 
y
ε
1
y σ bdy = 2b  y |σ | dy = 2bc 2  0 u |σ | du = 2bc 2 J
1
where the integral J is given by  0 u |σ | du
Evaluate J using a method of numerial integration. If Simpson’s rule is used, the integration formula is
J =
Δu
Σwu|σ |
3
where w is a weighting factor.
Using Δu = 0.25, we get the values given the table below:
u
|ε |
0
0.25
0.5
0.75
1.00
0
0.0015
0.003
0.0045
0.006
|σ |, ksi
0
25
36
40
43
u |σ |, ksi
0
6.25
18
30
43
J =
w
1
4
2
4
1
wu |σ |, ksi
0
25
36
120
43
224
← Σwu |σ |
(0.25) (224)
= 18.67 ksi
3
M = (2)(0.8)(0.6)2 (18.67)
M = 10.75 kip ⋅ in 
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PROBLEM 4.98
A prismatic bar of rectangular cross section is made of an alloy for
which the stress-strain diagram can be represented by the relation
ε = kσ n for σ > 0 and ε = − kσ n for σ < 0 . If a couple M is
applied to the bar, show that the maximum stress is
σm =
1 + 2n Mc
3n
I
SOLUTION
Strain distribution:
ε = −ε m
y
= −ε mu where
c
u =
y
c
Bending couple:
c
c
c
M = −  − c y σ bdy = 2b  0 y |σ | dy = 2bc 2  0
y
dy
|σ |
c
c
1
= 2 bc 2  0 u |σ | du
For
ε = Kσ n ,
ε m = Kσ m
 σ 
ε
=u =

εm
 σm 
Then
n
∴
1
| σ | = σ mu n
1
1
1 1+ 1n
du
M = 2bc 2  0 uσ mu n du = 2bc 2σ m  0 u
2+ 1
u n
= 2bc σ m
2 + 1n
2
σm =
1
2n
2
 0 = 2n + 1 bc σ m
2n + 1 M
2 bc 2
Recall: that
I
1 b(2c)3
2
1
2 c
=
= bc 2 ∴
=
2
3
3 I
c 12 c
bc
Then
σm =
2n + 1 Mc
3n
I
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PROBLEM 4.99
A short wooden post supports a 6-kip axial load as shown. Determine the
stress at point A when (a) b = 0, (b) b = 1.5in., (c) b = 3in.
SOLUTION
A = π r 2 = π (3) 2 = 28.27 in 2
π
π
r 4 = (3) 4 = 63.62 in 4
4
4
I
63.62
S = =
= 21.206 in 3
c
3
P = 6 kips
M = Pb
I =
(a)
b=0
σ =−
(b)
M =0
P
6
=−
= −0.212 ksi
A
28.27
σ = −212 psi 
b = 1.5 in. M = (6)(1.5) = 9 kip ⋅ in
σ =−
P M
6
9
−
=−
−
= −0.637 ksi
28.27 21.206
A
S
σ = −637 psi 
(c)
b = 3 in.
σ =−
M = (6)(3) = 18 kip ⋅ in
P M
6
18
−
=−
−
= −1.061 ksi
28.27 21.206
A
S
σ = −1061 psi 
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PROBLEM 4.100
As many as three axial loads, each of magnitude P = 10 kips, can be
applied to the end of a W8 × 21 rolled-steel shape. Determine the stress at
point A, (a) for the loading shown, (b) if loads are applied at points 1 and 2
only.
SOLUTION
For W8 × 21 Appendix C gives
A = 6.16 in 2 , d = 8.28 in., I x = 75.3 in 4
At point A,
(a)
Centric loading:
1
d = 4.14 in.
2
F My
−
σ =
A
I
y =
F = 30 kips, M = 0
σ =
(b)
Eccentric loading:
30
6.16
σ = 4.87 ksi 
F = 2 P = 20 kips
M = −(10)(3.5) = −35 kip ⋅ in
σ =
20
(−35)(4.14)
−
6.16
75.3
σ = 5.17 ksi 
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PROBLEM 4.101
Knowing that the magnitude of the horizontal force P is 8 kN,
determine the stress at (a) point A, (b) point B.
SOLUTION
A = (30)(24) = 720 mm 2 = 720 × 10−6 m 2
e = 45 − 12 = 33mm = 0.033 m
1 3
1
(30)(24)3 = 34.56 × 103 mm 4 = 34.56 × 10−9 m 4
bh =
12
12
1
c = (24 mm) = 12 mm = 0.012 m
P = 8 × 103 N
2
I =
M = Pe = (8 × 103 )(0.033) = 264 N ⋅ m
(a)
σA = −
P Mc
8 × 103
(264)(0.012)
−
=−
=
= −102.8 × 106 Pa
A
I
720 × 10−6
34.56 × 10−9
σ A = −102.8 MPa 
(b)
σB = −
P Mc
8 × 103
(264)(0.012)
+
=−
+
= 80.6 × 106 Pa
−6
A
I
720 × 10
34.56 × 10−9
σ B = 80.6 MPa 
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PROBLEM 4.102
The vertical portion of the press shown consists of a
rectangular tube of wall thickness t = 10 mm. Knowing that the
press has been tightened on wooden planks being glued
together until P = 20 kN, determine the stress at (a) point A,
(b) point B.
SOLUTION
Rectangular cutout is 60 mm × 40 mm.
A = (80) (60) − (60)(40) = 2.4 × 103 mm 2 = 2.4 × 10−3 m 2
I =
1
1
(60)(80)3 − (40)(60)3 = 1.84 × 106 mm 4
12
12
= 1.84 × 10−6 m 4
c = 40 mm = 0.040 m e = 200 + 40 = 240 mm = 0.240 m
P = 20 × 103 N
M = Pe = (20 × 103 )(0.240) = 4.8 × 103 N ⋅ m
(a)
σA =
P Mc
20 × 103
(4.8 × 103 )(0.040)
+
=
+
= 112.7 × 106 Pa
A
I
2.4 × 10−3
1.84 × 10−6
σ A = 112.7 MPa 
(b)
σB =
P Mc
20 × 103
(4.8 × 103 ) (0.040)
−
=
−
= −96.0 × 106 Pa
A
I
2.4 × 10−3
1.84 × 10−6
σ B = −96.0 MPa 
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PROBLEM 4.103
Solve Prob. 4.102, assuming that t = 8 mm.
PROBLEM 4.102 The vertical portion of the press shown
consists of a rectangular tube of wall thickness t = 10 mm.
Knowing that the press has been tightened on wooden planks
being glued together until P = 20 kN, determine the stress at
(a) point A, (b) point B.
SOLUTION
Rectangular cutout is 64 mm × 44 mm.
A = (80) (60) − (64)(44) = 1.984 × 103 mm 2
= 1.984 × 10−3 mm 2
I =
1
1
(60)(80)3 − (44)(64)3 = 1.59881 × 106 mm 2
12
12
= 1.59881 × 10−6 m 4
c = 40 mm = 0.004
e = 200 + 40 = 240 mm = 0.240 m
3
P = 20 × 10 N
M = Pe = (20 × 103 )(0.240) = 4.8 × 103 N ⋅ m
(a)
σA =
P Mc
20 × 103
(4.8 × 103 )(0.040)
+
=
+
= 130.2 × 106 Pa
A
I
1.984 × 10−3
1.59881 × 10−6
(b)
σB =
P Mc
20 × 103
(4.8 × 103 ) (0.040)
−
=
−
= −110.0 × 106 Pa
A
I
1.984 × 10−3
1.59881 × 10−6
σ A = 130.2 MPa 
σ B = −110.0 MPa 
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PROBLEM 4.104
Determine the stress at points A and B, (a) for the loading
shown, (b) if the 60-kN loads are applied at points 1 and 2 only.
SOLUTION
(a)
Loading is centric.
P = 180 kN = 180 × 103 N
A = (90)(240) = 21.6 × 103 mm 2 = 21.6 × 10−6 m 2
σ =−
At A and B:
P
180 × 103
=
= −8.33 × 106 Pa
−3
A 21.6 × 10
σ A = σ B = −8.33 MPa 
(b) Eccentric loading.
P = 120 kN = 120 × 103 N
M = (60 × 103 )(150 × 10−3 ) = 9.0 × 103 N ⋅ m
1 3
1
bh =
(90)(240)3 = 103.68 × 106 mm 4 = 103.68 × 10−6 m 4
12
12
c = 120 mm = 0.120 m
I =
At A:
σA = −
At B:
σB −
P Mc
120 × 103
(9.0 × 103 )(0.120)
−
=−
−
= −15.97 × 106 Pa 
A
I
21.6 × 10−3
103.68 × 10−6
P Mc
120 × 103
(9.0 × 103 )(0.120)
+
=−
+
= 4.86 × 106 Pa
A
I
21.6 × 10−3
103.68 × 10−6
σ A = −15.97 MPa 
σ B = 4.86 MPa 
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PROBLEM 4.105
Knowing that the allowable stress in section ABD is 10 ksi,
determine the largest force P that can be applied to the bracket
shown.
SOLUTION
Statics: M = 2.45 P
Cross section: A = (0.9)(1.2) = 1.08 in 2
c=
1
(0.9) = 0.45 in.
2
I =
1
(1.2)(0.9)3 = 0.0729 in 4
12
At point B: σ = −10 ksi
P Mc
−
A
I
P
(2.45P)(0.45)
−10 = −
−
= −16.049P
1.08
0.0729
σ =−
P = 0.623 kips
P = 623 lb 

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PROBLEM 4.106
Portions of a 12 × 12 -in. square bar have been bent to
form the two machine components shown. Knowing
that the allowable stress is 15 ksi, determine the
maximum load that can be applied to each
component.
SOLUTION
The maximum stress occurs at point B.
σ B = −15 ksi = −15 × 103 psi
σB = −
1 ec
+
A
I
K =
where
P Mc
P Pec
−
=− −
= − KP
A
I
A
I
e = 1.0 in.
A = (0.5) (0.5) = 0.25 in 2
I =
1
(0.5)(0.5)3 = 5.2083 × 10−3 in 4 for all centroidal axes.
12
(a)
(a)
c = 0.25 in.
K =
1
(1.0) (0.25)
+
= 52 in −2
−3
0.25 5.2083 × 10
P=−
(b)
(b)
c=
σB
K
=−
(−15 × 103 )
52
P = 288 lb 
0.5
= 0.35355 in.
2
K =
1
(1.0)(0.35355)
+
= 71.882 in −2
0.25 5.2083 × 10−3
P=−
σB
K
=−
(−15 × 103 )
71.882
P = 209 lb 
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PROBLEM 4.107
The four forces shown are applied to a rigid plate supported by a solid
steel post of radius a. Knowing that P = 100 kN and a = 40 mm,
determine the maximum stress in the post when (a) the force at D is
removed, (b) the forces at C and D are removed.
SOLUTION
For a solid circular section of radius a,
A = π a2
σ =−
a4
M x = − Pa,
σ =−
F
4P
=− 2
A
πa
Mz = 0
F M xz
3P
(− Pa)(−a)
7P
−
=− 2 −
=− 2
2
π
A
I
a
πa
πa
4
Forces at C and D are removed:
F = 2 P,
Resultant bending couple:
σ =−
M =
M x = − Pa,
M x2 + M z2 =
F Mc
2P
−
=− 2 −
A
I
πa
M z = − Pa
2 Pa
2 Pa a
2+4 2 P
=−
= −2.437 P/a 2
2
π a2
π
a
4
P = 100 × 103 N,
Numerical data:

4
Force at D is removed:
F = 3P,
(b)
π
F = 4 P, M x = M z = 0
Centric force:
(a)
I =
a = 0.040 m
Answers:
(a)
σ =−
(7) (100 × 103 )
= −139.3 × 106 Pa
2
π (0.040)
σ = −139.3 MPa 

(b)
σ =−
(2.437)(100 × 103 )
= −152.3 × 106 Pa
(0.040) 2
σ = −152.3 MPa 
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PROBLEM 4.108
A milling operation was used to remove a portion of a solid bar
of square cross section. Knowing that a = 30 mm, d = 20 mm,
and σ all = 60 MPa, determine the magnitude P of the largest
forces that can be safely applied at the centers of the ends of the
bar.
SOLUTION
A = ad ,
I =
1
ad 3 ,
12
c=
a d
−
2 2
P Mc
P
6Ped
σ = +
=
+
A
I
ad
ad 3
3 P (a − d )
P
σ =
+
= KP
ad
ad 2
1
d
2
e=
Data:
a = 30 mm = 0.030 m
K =
P=
where
K =
1
3(a − d )
+
ad
ad 2
d = 20 mm = 0.020 m
1
(3) (0.010)
+
= 4.1667 × 103 m −2
(0.030) (0.020) (0.030) (0.020) 2
σ
K
=
60 × 106
= 14.40 × 103 N
3
4.1667 × 10
P = 14.40 kN 
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PROBLEM 4.109
A milling operation was used to remove a portion of a solid bar of
square cross section. Forces of magnitude P = 18 kN are applied at
the centers of the ends of the bar. Knowing that a = 30 mm and
σ all = 135 MPa, determine the smallest allowable depth d of the
milled portion of the bar.
SOLUTION
A = ad ,
e=
I =
1
ad 3 ,
12
c=
1
d
2
a d
−
2 2
P 1 (a − d ) 1 d
P Mc
P
Pec
P
2 = P + 3P ( a − d )
=
+
=
+ 2
σ = +
1
A
I
ad
I
ad
ad
ad 2
ad 3
12
3P 2P
2P
d − 3P = 0
σ = 2 −
or σ d 2 +
ad
a
d
2


1   2P 
2P 
d =
 

 + 12 Pσ −
2σ   a 
a 


Solving for d,
a = 0.030 m,
Data:
d =
P = 18 × 103 N, σ = 135 × 106 Pa
2


1
(2) (18 × 103 ) 
  (2) (18 × 103 ) 
3
6
+
×
×
−
12(18
10
)
(135
10
)
 


0.030
0.030
(2) (135 × 106 )  




= 16.04 × 10−3 m
d = 16.04 mm 
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PROBLEM 4.110
A short column is made by nailing two 1 × 4-in. planks to a 2 × 4-in.
timber. Determine the largest compressive stress created in the column by a
12-kip load applied as shown in the center of the top section of the timber if
(a) the column is as described, (b) plank 1 is removed, (c) both planks are
removed.
SOLUTION
(a)
Centric loading: 4 in. × 4 in. cross section
σ =−
(b)
P
12
=−
A
16
Eccentric loading: 4 in. × 3 in. cross section
1
c =   (3) = 1.5 in.
2
I =
σ = −0.75 ksi 
A = (4) (3) = 12 in 2
e = 1.5 − 1.0 = 0.5 in.
1 3
1
bh =
(4) (3)3 = 9 in 4
12
12
σ =−
(c)
A = (4)(4) = 16 in 2
P Pec
12 (12) (0.5) (1.5)
−
=− −
A
I
12
9
Centric loading: 4 in. × 2 in. cross section
σ =−
P
12
=−
A
8
σ = −2.00 ksi 
A = (4) (2) = 8 in 2
σ = −1.50 ksi 
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PROBLEM 4.111
An offset h must be introduced into a solid circular rod of
diameter d. Knowing that the maximum stress after the offset
is introduced must not exceed 5 times the stress in the rod
when it is straight, determine the largest offset that can be
used.
SOLUTION
For centric loading,
σc =
P
A
For eccentric loading,
σe =
P Phc
+
A
I
Given
σe = 5 σc
P Phc
P
+
=5
A
I
A
Phc
P
=4
I
A
∴
π

(4)  d 4 
4I
64

 = 1d
h=
=
cA  d  π 2  2
  d 
 2  4 
h = 0.500 d 
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PROBLEM 4.112
An offset h must be introduced into a metal tube of 0.75-in. outer
diameter and 0.08-in. wall thickness. Knowing that the maximum
stress after the offset is introduced must not exceed 4 times the
stress in the tube when it is straight, determine the largest offset
that can be used.
SOLUTION
1
d = 0.375 in.
2
c1 = c − t = 0.375 − 0.08 = 0.295 in.
c=
(
)
A = π c 2 − c12 = π (0.3752 − 0.2952 )
= 0.168389 in 2
I =
π
(c
4
4
)
− c14 =
π
4
(0.3754 − 0.2954 )
−3
= 9.5835 × 10 in 4
For centric loading,
σ cen =
P
A
For eccentric loading,
σ ecc =
P Phc
+
A
I
σ ecc = 4 σ cen
hc
3
=
I
A
or
h=
P Phc
P
+
=4
A
I
A
3I
(3)(9.5835 × 10−3 )
=
Ac (0.168389)(0.375)
h = 0.455 in. 
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PROBLEM 4.113
A steel rod is welded to a steel plate to form the machine
element shown. Knowing that the allowable stress is 135 MPa,
determine (a) the largest force P that can be applied to the
element, (b) the corresponding location of the neutral axis.
Given: The centroid of the cross section is at C and
I z = 4195 mm 4 .
SOLUTION
(a)
A = (3)(18) +
π
4
(6)2 = 82.27 mm 2 = 82.27 × 10−6 m 2
I = 4195 mm 4 = 4195 × 10−12 m 4
e = 13.12 mm = 0.01312 m
Based on tensile stress at y = −13.12 mm = −0.01312 m
σ =
P Pec  1 ec 
+
=  +  P = KP
A
I
I 
A
K =
1 ec
1
(0.01312)(0.01312)
+
=
+
= 53.188 × 103 m −2
−6
−12
A
I
82.27 × 10
4195 × 10
P=
(b)
σ
K
=
135 × 106
= 2.538 × 103 N
3
53.188 × 10
Location of neutral axis.
P = 2.54 kN 
σ =0
σ =
P My
P Pey
−
=
−
=0
A
I
A
I
ey
1
=
I
A
y =
I
4195 × 10−12
=
= 3.89 × 10−3 m
−6
Ae (82.27 × 10 )(0.01312)
y = 3.89 mm 
The neutral axis lies 3.89 mm to the right of the centroid or 17.01 mm to the right of the line of action
of the loads.
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PROBLEM 4.114
A vertical rod is attached at point A to the cast iron hanger
shown. Knowing that the allowable stresses in the hanger
are σ all = +5 ksi and σ all = −12 ksi , determine the
largest downward force and the largest upward force that
can be exerted by the rod.
SOLUTION
Ay
(1 × 3)(0.5) + 2(3 × 0.25)(2.5)
X = 
=
A
(1 × 3) + 2(3 × 0.75)

X =
12.75 in 3
= 1.700 in.
7.5 in 2
A = 7.5 in 2
σ all = +5 ksi
σ all = −12 ksi
 1

I c =   bh3 + Ad 2 
12


1
1
(3)(1)3 + (3 × 1)(1.70 − 0.5) 2 + (1.5)(3)3 + (1.5 × 3)(2.5 − 1.70) 2
=
12
12
I c = 10.825 in 4
Downward Force.
M = P(1.5 in.+1.70 in.) = (3.20 in.) P
At D: σ D = +
P Mc
+
A
I
P
(3.20) P(1.70)
+
7.5
10.825
+ 5 = P(+0.6359)
+ 5 ksi =
At E: σ E = +
P = 7.86 kips ↓ 
P Mc
−
A
I
P
(3.20) P(2.30)
−
7.5
10.825
− 12 = P(−0.5466)
− 12 ksi =
We choose the smaller value.
P = 21.95 kips ↓
P = 7.96 kips ↓ 
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PROBLEM 4.114 (Continued)
Upward Force.
M = P(1.5 in. + 1.70 in.) = (3.20 in.) P
At D: σ D = +
P Mc
−
A
I
P
(3.20) P(1.70)
−
7.5
10.825
−12 = P(−0.6359)
−12 ksi = −
At E: σ E = +
P = 18.87 kips ↑ 
P Mc
+
A
I
P
(3.20) P(2.30)
+
7.5
10.825
+5 = P(+0.5466)
+5 ksi =
We choose the smaller value.
P = 9.15 kips ↑ 
P = 9.15 kips ↑ 
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PROBLEM 4.115
Solve Prob. 4.114, assuming that the vertical rod is attached
at point B instead of point A.
PROBLEM 4.114 A vertical rod is attached at point A to the
cast iron hanger shown. Knowing that the allowable stresses
in the hanger are σ all = +5 ksi and σ all = −12 ksi ,
determine the largest downward force and the largest upward
force that can be exerted by the rod.
SOLUTION
Ay
(1 × 3)(0.5) + 2(3 × 0.25)(2.5)
=
X = 
A
(1 × 3) + 2(3 × 0.75)

X =
12.75 in 3
= 1.700 in.
7.5 in 2
A = 7.5 in 2
σ all = +5 ksi
σ all = −12 ksi
 1

I c =   bh3 + Ad 2 
12


1
1
=
(3)(1)3 + (3 × 1)(1.70 − 0.5) 2 + (1.5)(3)3 + (1.5 × 3)(2.5 − 1.70) 2
12
12
I c = 10.825 in 4
Downward Force.
σ all = +5 ksi
σ all = −12 ksi
M = (2.30 in. + 1.5 in.) = (3.80 in.) P
At D: σ D = +
P Mc
−
A
I
P
(3.80) P(1.70)
−
7.5
10.825
− 12 = P(−0.4634)
−12 ksi = +
At E: σ E = +
P = 25.9 kips ↓
P Mc
+
A
I
P
(3.80) P(2.30)
+
7.5
10.825
+5 = P(+0.9407)
+5 ksi = +
We choose the smaller value.
P = 5.32 kips ↓ 
P = 5.32 kips ↓ 
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PROBLEM 4.115 (Continued)
Upward Force.

σ all = +5 ksi

M = (2.30 in. + 1.5 in.)P = (3.80 in.)P 


σ all = −12 ksi 
At D: σ D = −
P Mc
+

A
I
P
(3.80) P(1.70)
+
7.5
10.825
5 = P(+0.4634)

5 ksi = −

At E: σ E = −
P = 10.79 kips ↑ 
P Mc
−
A
I
P
(3.80) P(2.30)
−
7.5
10.825
−12 = P(−0.9407)
−12 ksi = −
We choose the smaller value.
P = 12.76 kips ↑ 
P = 10.79 kips ↑ 
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PROBLEM 4.116
Three steel plates, each of 25 × 150-mm cross section, are welded
together to form a short H-shaped column. Later, for architectural
reasons, a 25-mm strip is removed from each side of one of the
flanges. Knowing that the load remains centric with respect to the
original cross section, and that the allowable stress is 100 MPa,
determine the largest force P (a) that could be applied to the original
column, (b) that can be applied to the modified column.
SOLUTION
(a)
Centric loading:
σ =−
P
A
A = (3)(150)(25) = 11.25 × 103 mm2 = 11.25 × 10−3 m 2
P = −σ A = −(−100 × 106 )(11.25 × 10−3 )
= 1.125 × 106 N
(b)
P = 1125 kN 
Eccentric loading (reduced cross section):
A, 103 mm 2
y , mm
A y (103 mm3 )

3.75
87.5
328.125
76.5625

3.75
0
0
10.9375

2.50
−87.5
−218.75
98.4375
Σ
10.00
d , mm
109.375
Y =
ΣAy 109.375 × 103
=
= 10.9375 mm
ΣA
10.00 × 103
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PROBLEM 4.116 (Continued)
The centroid lies 10.9375 mm from the midpoint of the web.
1
1
b1h13 + A1d12 =
(150)(25)3 + (3.75 × 103 )(76.5625) 2 = 22.177 × 106 mm 4
12
12
1
1
I2 =
b2h23 + A2d 22 =
(25) (150)3 + (3.75 × 103 )(10.9375) 2 = 7.480 × 106 mm 4
12
12
1
1
I3 =
b3h33 + A3d32 =
(100)(25)3 + (2.50 × 103 ) (98.4375)2 = 24.355 × 106 mm 4
12
12
I1 =
I = I1 + I 2 + I 3 = 54.012 × 106 mm 4 = 54.012 × 10−6 m 4
c = 10.9375 + 75 + 25 = 110.9375 mm = 0.1109375 m
M = Pe
σ =−
K =
where
e = 10.4375 mm = 10.4375 × 10−3 m
P Mc
P Pec
−
=− −
= − KP
A
I
A
I
A = 10.00 × 10−3 m 2
1 ec
1
(101.9375 × 10−3 )(0.1109375)
= 122.465 m −2
+
=
+
−6
−3
A
I
10.00 × 10
54.012 × 10
P=−
σ
K
=−
(−100 × 106 )
= 817 × 103 N
122.465
P = 817 kN 
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PROBLEM 4.117
A vertical force P of magnitude 20 kips is applied at
point C located on the axis of symmetry of the cross
section of a short column. Knowing that y = 5 in.,
determine (a) the stress at point A, (b) the stress at
point B, (c) the location of the neutral axis.
SOLUTION
Locate centroid.
Part
A, in 2
y , in.

12
5

8
2
Σ
20
Eccentricity of load: e = 5 − 3.8 = 1.2 in.
I1 =
Ay , in 3
60
16
76
1
(6)(2)3 + (12)(1.2)2 = 21.28 in 4
12
y =
I2 =
ΣAy
76
=
= 3.8 in.
ΣA
20
1
(2)(4)3 + (8)(1.8) 2 = 36.587 in 4
12
I = I1 + I 2 = 57.867 in 4
(a)
Stress at A: c A = 3.8 in.
σA = −
(b)
σ A = 0.576 ksi 
Stress at B: cB = 6 − 3.8 = 2.2 in.
σB = −
(c)
P Pec A
20 20(1.2)(3.8)
+
=−
+
20
57.867
A
I
P PecB
20 20(1.2)(2.2)
+
=−
−
20
57.867
A
I
σ B = −1.912 ksi 
P Pea
ea
1
+
=0 ∴
=
A
I
I
A
I
57.867
a =
=
= 2.411 in.
Ae (20)(1.2)
Location of neutral axis: σ = 0
σ =−
Neutral axis lies 2.411 in. below centroid or 3.8 − 2.411 = 1.389 in. above point A.
Answer: 1.389 in. from point A. 
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PROBLEM 4.118
A vertical force P is applied at point C located
on the axis of symmetry of the cross section of
a short column. Determine the range of values
of y for which tensile stresses do not occur in
the column.
SOLUTION
Locate centroid.
A, in 2


Σ
Eccentricity of load:
12
8
20
e = y − 3.8 in.
I1 =
y , in.
Ay , in 3
5
2
60
16
76
y =
ΣAi yi
76
=
= 3.8 in.
ΣAi
20
y = e + 3.8 in.
1
(6)(2)3 + (12)(1.2)2 = 21.28 in 4
12
1
(2)(4)3 + (8)(1.8)2 = 36.587 in 4
12
I2 =
I = I1 + I 2 = 57.867 in 4
If stress at A equa ls zero, c A = 3.8 in.
σA = −
e=
P Pec A
+
=0
A
I
∴
ec A
1
=
I
A
I
57.867
=
= 0.761 in.
Ac A
(20)(3.8)
y = 0.761 + 3.8 = 4.561 in.
If stress at B equa ls zero. cB = 6 − 3.8 = 2.2 in.
P PecB
ecB
1
−
=0
∴
=−
A
I
I
A
I
57.867
e=−
=−
= −1.315 in.
AcB
(20)(2.2)
σB = −
y = −1.315 + 3.8 = 2.485 in.
Answer: 2.485 in. < y < 4.561 in. 
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PROBLEM 4.119
Knowing that the clamp shown has been tightened until
P = 400 N, determine (a) the stress at point A,
(b) the stress at point B, (c) the location of the neutral
axis of section a − a.
SOLUTION
Cross section: Rectangle  + Circle 
A1 = (20 mm)(4 mm) = 80 mm 2
y1 =
1
(20 mm) = 10 mm
2
A2 = π (2 mm)2 = 4π mm 2
y2 = 20 − 2 = 18 mm
Ay
(80)(10) + (4π )(18)
cB = y = 
=
= 11.086 mm
80 + 4π
A
c A = 20 − y = 8.914 mm
d1 = 11.086 − 10 = 1.086 mm
d 2 = 18 − 11.086 = 6.914 mm
I1 = I1 + A1d12 =
I 2 = I 2 + A2d 22 =
1
(4)(20)3 + (80)(1.086) 2 = 2.761 × 103 mm 4
12
π
4
(2) 4 + (4π )(6.914)2 = 0.613 × 103 mm 4
I = I1 + I 2 = 3.374 × 103 mm 4 = 3.374 × 10−9 m 4
A = A1 + A2 = 92.566 mm 2 = 92.566 × 10−6 m 2
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PROBLEM 4.119 (Continued)
e = 32 + 8.914 = 40.914 mm = 0.040914 m
M = Pe = (400 N)(0.040914 m) = 16.3656 N ⋅ m
(a)
Point A:
σA =
P Mc
400
(16.3656)(8.914 × 10−3 )
+
=
+
−6
A
I
92.566 × 10
3.374 × 10−9
= 4.321 × 106 + 43.23 × 106 = 47.55 × 106 Pa
(b)
Point B:
σB =
P Mc
400
(16.3656)(11.086)
−
=
−
A
I
92.566 × 10−6
3.374 × 10−9
= 4.321 × 106 − 53.72 × 106 = −49.45 × 106 Pa
(c)
σ A = 47.6 MPa
Neutral axis:
σ B = −49.4 MPa
By proportions,
a
20
=
47.55 47.55 + 49.45
a = 9.80 mm
9.80 mm below top of section
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PROBLEM 4.120
The four bars shown have the same cross-sectional area. For the given
loadings, show that (a) the maximum compressive stresses are in the
ratio 4:5:7:9, (b) the maximum tensile stresses are in the ratio 2:3:5:3.
(Note: the cross section of the triangular bar is an equilateral triangle.)
SOLUTION
Stresses:
At A,
σA = −
Aec A 
P Pec A
P
−
= − 1 +
A
I
A
I 
At B,
σB = −
P PecB P  AecB

+
= 
− 1
A
I
A I

1 4
1
1

2
 A1 = a , I1 = 12 a , c A = cB = 2 a, e = 2 a



 1  1  
(a 2 )  a  a  


P
 2  2  
σ A = − 1 +
1 2


A

a



12




 2  1  1  

 (a )  a  a  
 2  2  − 1
σ = P 
 B A
1 2

a



12



a
π

2
2
, I 2 = c4 , e = c
 A2 = π c = a ∴ c =
4
π





P  (π c 2 )(c)(c) 
σ A = − 1 +


π 4
A2 

c


4







P  (π c 2 )(c)(c) 
σ B =
− 1

π 4
A2 


c


4


σ A = −4
P

A1
σB = 2
P

A1
σ A = −5
P

A2
σB = 3
P

A2
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PROBLEM 4.120 (Continued)

2
1
2
a I3 = a4 e = c
 A3 = a c =
2
12



 2  2  

a 
a 
 ( a 2 ) 
2 
2  

P


σ A = − 1 +

1 4
A3

a


12








 2  2  

a 
 ( a 2 ) 
 2 a  

2 
P

 −1
σ B =

1 4
A3 

a



12





σ A = −7
P

A3

σB = 5
P

A3
  3 2   s  s  
s  
 


P   4
 3  3  

σ A = − 1 +

A4
3 4


s


96


σ A = −9
P

A4
  3 2   s  s  
s  


 
P   4
 3  2 3  

σB = 
− 1
A4
3
4


s


96


σB = 3
P

A4
A4 =
1  3 
3 2
(s) 
s =
s
2  2  4
3
1  3 
3 4
I4 =
s
s =
s


36  2 
96
cA =
2 3
s
=e
s=
3 2
3
cB = s
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PROBLEM 4.121
The C-shaped steel bar is used as a dynamometer to determine the magnitude P of
the forces shown. Knowing that the cross section of the bar is a square of side
40 mm and that strain on the inner edge was measured and found to be 450 μ,
determine the magnitude P of the forces. Use E = 200 GPa.
SOLUTION
At the strain gage location,
σ = Eε = (200 × 109 )(450 × 10−6 ) = 90 × 106 Pa
A = (40)(40) = 1600 mm 2 = 1600 × 10−6 m 2
1
(40)(40)3 = 213.33 × 103 mm 4 = 213.33 × 10−9 m 4
12
e = 80 + 20 = 100 mm = 0.100 m
I =
c = 20 mm = 0.020 m
σ =
P Mc
P Pec
+
=
+
= KP
A
I
A
I
K =
1 ec
1
(0.100)(0.020)
+
=
+
= 10.00 × 103 m −2
−6
A
I
1600 × 10
213.33 × 10−9
P=
σ
K
=
90 × 106
= 9.00 × 103 N
10.00 × 103
P = 9.00 kN 
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PROBLEM 4.122
An eccentric force P is applied as shown to a steel bar of
25 × 90-mm cross section. The strains at A and B have been
measured and found to be
ε A = +350 μ
ε B = −70 μ
Knowing that E = 200 GPa, determine (a) the distance d, (b) the
magnitude of the force P.
SOLUTION
1
h = 45 mm = 0.045 m
2
A = bh − (25)(90) = 2.25 × 103 mm 2 = 2.25 × 10−3 m 2
h = 15 + 45 + 30 = 90 mm
b = 25 mm
c=
1 3 1
bh = (25)(90)3 = 1.51875 × 106 mm 4 = 1.51875 × 10−6 m 4
12
12
y A = 60 − 45 = 15 mm = 0.015 m
yB = 15 − 45 = −30 mm = −0.030 m
I=
Stresses from strain gages at A and B:
σ A = Eε A = (200 × 109 )(350 × 10−6 ) = 70 × 106 Pa
σ B = Eε B = (200 × 109 )( −70 × 10−6 ) = −14 × 106 Pa
Subtracting,
σA −σB = −
M =−
σA =
P MyA
−
A
I
(1)
σB =
P MyB
−
A
I
(2)
M ( y A − yB )
I
I (σ A − σ B )
(1.51875 × 10−6 )(84 × 106 )
=−
= −2835 N ⋅ m
0.045
y A − yB
Multiplying (2) by y A and (1) by yB and subtracting,
P=
(a)
(b)
y Aσ B − yBσ A = ( y A − yB )
P
A
A( y Aσ B − yBσ A ) (2.25 × 10−3 )[(0.015)(−14 × 106 ) − (−0.030)(70 × 106 )]
=
= 94.5 × 103 N
0.045
y A − yB
M = − Pd ∴ d = −
−2835
M
=−
= 0.030 m
P
94.5 × 103
d = 30.0 mm 
P = 94.5 kN 
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PROBLEM 4.123
Solve Prob. 4.122, assuming that the measured strains are
ε A = +600 μ
ε B = +420 μ
PROBLEM 4.122 An eccentric force P is applied as shown to a
steel bar of 25 × 90-mm cross section. The strains at A and B have
been measured and found to be
ε A = +350μ
ε B = −70 μ
Knowing that E = 200 GPa, determine (a) the distance d, (b) the
magnitude of the force P.
SOLUTION
1
h = 45 mm = 0.045 m
2
A = bh = (25)(90) = 2.25 × 103 mm 2 = 2.25 × 10−3 m 2
h = 15 + 45 + 30 = 90 mm
b = 25 mm
c=
1 3 1
bh = (25)(90)3 = 1.51875 × 106 mm 4 = 1.51875 × 10−6 m 4
12
12
y A = 60 − 45 = 15 mm = 0.015 m yB = 15 − 45 = −30 mm = −0.030 m
I=
Stresses from strain gages at A and B:
σ A = Eε A = (200 × 109 )(600 × 10−6 ) = 120 × 106 Pa
σ B = Eε B = (200 × 109 )(420 × 10−6 ) = 84 × 106 Pa
Subtracting,
σA =
P My A
−
A
I
(1)
σB =
P MyB
−
A
I
(2)
σA −σB = −
M =−
M ( y A − yB )
I
I (σ A − σ B )
(1.51875 × 10−6 )(36 × 106 )
=−
= −1215 N ⋅ m
0.045
y A − yB
Multiplying (2) by y A and (1) by yB and subtracting,
P=
y Aσ B − yBσ A = ( y A − yB )
P
A
A( y Aσ B − yBσ A ) (2.25 × 10−3 )[(0.015)(84 × 106 ) − ( −0.030)(120 × 106 )]
=
= 243 × 103 N
y A − yB
0.045
M = − Pd
(a)
(b)
∴ d =−
−1215
M
=−
= 5 × 10−3 m
3
P
243 × 10
d = 5.00 mm 
P = 243 kN 
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PROBLEM 4.124
A short length of a W8 × 31 rolled-steel shape supports a rigid plate on which two
loads P and Q are applied as shown. The strains at two points A and B on the
centerline of the outer faces of the flanges have been measured and found to be
ε A = −550 × 10−6 in./in.
ε B = −680 × 10−6 in./in.
Knowing that E = 29 × 106 psi, determine the magnitude of each load.
SOLUTION
Strains:
ε A = −550 × 10−6 in./in.
εC =
Stresses:
ε B = −680 × 10−6 in./in.
1
1
(ε A + ε B ) = (−550 − 680)10−6 = −615 × 10−6 in./in.
2
2
σ A = Eε A = (29 × 106 psi)(−550 × 10−6 in./in.) = −15.95 ksi
σ C = Eε C = (29 × 106 psi)(−615 × 10−6 in./in.) = −17.935 ksi
A = 9.13 in 2
W8 × 31
S = 27.5 in 3
M = (4.5 in.)( P − Q)
At point C:
σC = −
P+Q
P+Q
; − 17.835 ksi = −
A
9.13 in 2
P + Q = 162.83 kips (1)
At point A:
σA = −
P+Q M
−
A
S
−15.95 ksi = −17.835 ksi −
Solve simultaneously,
(4.5 in.)( P − Q)
;
27.5 in 3
P − Q = −11.52 kips (2)
P = 25.7 kips Q = 87.2 kips
P = 25.7 kips ↓ 
Q = 87.2 kips ↓ 
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PROBLEM 4.125
Solve Prob. 4.124, assuming that the measured strains are
ε A = +35 × 10−6 in./in. ε B = −450 × 10−6 in./in.
PROBLEM 4.124 A short length of a W8 × 31 rolled-steel shape supports a rigid
plate on which two loads P and Q are applied as shown. The strains at two points A
and B on the centerline of the outer faces of the flanges have been measured and
found to be
ε A = −550 × 10−6 in./in.
ε B = −680 × 10−6 in./in.
Knowing that E = 29 × 106 psi, determine the magnitude of each load.
SOLUTION
See solution and figures of Prob. 4.124.
ε A = +35 × 10−6 in./in.;
εC =
ε B = −450 × 10−6 in./in.
1
1
(ε A + ε B ) = (35 − 450)10−6 in./in. = −207.5 × 10−6 in./in.
2
2
Stresses:
σ A = Eε A = (29 × 106 psi)(+35 × 10−6 in./in.) = +1.015 ksi
σ C = Eε C = (29 × 106 psi)(−207.5 × 10−6 in./in.) = −6.0175 ksi
At point C:
σC = −
P+Q
P+Q
; − 6.0175 ksi = −
A
9.13 in 2
P + Q = 54.94 kips (1)
At point A:
σA = −
P+Q M
−
A
S
+1.015 ksi = −6.0175 −
(4.5 in.)( P − Q)
27.5 in 3
P − Q = −42.98 kips (2)
Solve simultaneously,
P = 5.98 kips
Q = 49.0 kips 
P = 5.98 kips ↓ 
Q = 49.0 kips ↓ 
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PROBLEM 4.126
The eccentric axial force P acts at point D, which must be located
25 mm below the top surface of the steel bar shown. For P = 60 kN,
determine (a) the depth d of the bar for which the tensile stress at
point A is maximum, (b) the corresponding stress at point A.
SOLUTION
1
bd 3
12
1
1
c= d
e= d −a
2
2
P Pec
σA = +
A
I

1
1 
P  1 12 2 d − a 2 d  P  4 6a 
σA =  +
=  − 2
b d
d3
 b d d 


A = bd
I=
(
(a)
)( )
Depth d for maximum σ A : Differentiate with respect to d.
dσ A P  4 12a 
= − 2 + 3  = 0
dd
b d
d 
(b)
σA =
60 × 103
40 × 10−3
4
(6)(25 × 10−3 ) 

6
−

 = 40 × 10 Pa
−3
(75 × 10−3 ) 2 
 75 × 10
d = 3a
d = 75 mm 
σ A = 40 MPa 
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PROBLEM 4.127
The couple M is applied to a beam of the cross section shown in a
plane forming an angle β with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
SOLUTION
1
(80)(100)3 = 6.6667 × 106 mm 4 = 6.6667 × 10−6 m 4
12
1
I y = (100)(80)3 = 4.2667 × 106 mm 4 = 4.2667 × 10−6 m 4
12
y A = − yB = − yD = 50 mm
Iz =
z A = z B = − z D = 40 mm
M y = −250 sin 30° = −125 N ⋅ m
M z = 250 cos 30° = 216.51 N ⋅ m
(a)
σA = −
M z yA M y zA
(216.51)(0.050) (−125)(0.040)
+
=−
+
Iz
Iy
6.6667 × 10−6
4.2667 × 10−6
= −2.80 × 106 Pa
(b)
σB = −
M z yB M y z B
(216.51)(−0.050) ( −125)(0.040)
+
=−
+
Iz
Iy
6.6667 × 10−6
4.2667 × 10−6
= 0.452 × 103 Pa
(c)
σD = −
σ A = −2.80 MPa 
σ B = 0.452 MPa 
M z yD M y z D
(216.51)(−0.050) (−125)(−0.040)
+
=−
+
Iz
Iy
6.6667 × 10−6
4.2667 × 10−6
= 2.80 × 106 Pa
σ D = 2.80 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.128
The couple M is applied to a beam of the cross section shown in a plane
forming an angle β with the vertical. Determine the stress at (a) point A,
(b) point B, (c) point D.
SOLUTION
1
(80)(32)3 = 218.45 × 103 mm 4 = 218.45 × 10−9 m 4
12
1
I y = (32)(80)3 = 1.36533 × 106 mm 4 = 1.36533 × 10−6 m 4
12
y A = yB = − yD = 16 mm
Iz =
z A = − z B = − z D = 40 mm
M y = 300 cos 30° = 259.81 N ⋅ m
(a)
σA = −
M z = 300sin 30° = 150 N ⋅ m
M z yA M y zA
(150)(16 × 10−3 ) (259.81)(40 × 10−3 )
+
=−
+
Iz
Iy
218.45 × 10−9
1.36533 × 10−6
= −3.37 × 106 Pa
(b)
σB = −
M z yB M y z B
(150)(16 × 10−3 ) (259.81)(−40 × 10−3 )
+
=−
+
Iz
Iy
218.45 × 10−9
1.36533 × 10−6
= −18.60 × 106 Pa
(c)
σD = −
σ A = −3.37 MPa 
σ B = −18.60 MPa 
M z yD M y z D
(150)(−16 × 10−3 ) (259.81)(−40 × 10−3 )
+
=−
+
Iz
Iy
218.45 × 10−9
1.36533 × 10−6
= 3.37 × 106 Pa
σ D = 3.37 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.129
The couple M is applied to a beam of the cross section shown in a
plane forming an angle β with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
SOLUTION
M z = −60 sin 40° = −38.567 kip ⋅ in
M y = 60 cos 40° = 45.963 kip ⋅ in
y A = yB = − yD = 3 in.
z A = − z B = − z D = 5 in.
(a)
σA = −
Iz =
1
π

(10)(6)3 − 2  (1)2  = 178.429 in 4
12
4


Iy =
1
π

(6)(10)3 − 2  (1)4 + π (1)2 (2.5) 2  = 459.16 in 4
12
4


M z yA M y zA
( −38.567)(3) (45.963)(5)
+
=−
+
178.429
459.16
Iz
Iy
= 1.149 ksi
(b)
σB = −
M z yB M y z B
(−38.567)(3) (45.963)(−5)
+
=−
+
178.429
459.16
Iz
Iy
= 0.1479
(c)
σD = −
σ A = 1.149 ksi 
σ B = 0.1479 ksi 
M z yD M y z D
(−38.567)(−3) (45.963)(−5)
+
=−
+
178.429
459.16
Iz
Iy
= −1.149 ksi
σ D = −1.149 ksi 
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PROBLEM 4.130
The couple M is applied to a beam of the cross section shown in a plane
forming an angle β with the vertical. Determine the stress at (a) point A,
(b) point B, (c) point D.
SOLUTION
Locate centroid.
A, in 2
z , in.
Az , in 3

16
−1
−16

8
2
16
Σ
24
0
The centroid lies at point C.
1
1
(2)(8)3 + (4)(2)3 = 88 in 4
12
12
1
1
I y = (8)(2)3 + (2)(4)3 = 64 in 4
3
3
y A = − yB = 1 in.,
yD = −4 in.
Iz =
z A = z B = −4 in.,
zD = 0
M z = 10 cos 20° = 9.3969 kip ⋅ in
M y = 10 sin 20° = 3.4202 kip ⋅ in
(a)
σA = −
M z yA M y zA
(9.3969)(1) (3.4202)(−4)
+
=−
+
88
64
Iz
Iy
(b)
σB = −
M z yB M y z B
(9.3969)( −1) (3.4202)(−4)
+
=−
+
88
64
Iz
Iy
σ B = −0.107 ksi 
(c)
σD = −
M z yD M y z D
(9.3969)(−4) (3.4202)(0)
+
=−
+
88
64
Iz
Iy
σ D = 0.427 ksi 
σ A = 0.321 ksi 
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PROBLEM 4.131
The couple M is applied to a beam of the cross section shown in a
plane forming an angle β with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
SOLUTION
M y = 25sin 15° = 6.4705 kN ⋅ m
M z = 25cos15° = 24.148 kN ⋅ m
1
1
(80)(90)3 + (80)(30)3 = 5.04 × 106 mm 4
12
12
−6
4
I y = 5.04 × 10 m
Iy =
1
1
1
I z = (90)(60)3 + (60)(20)3 + (30)(100)3 = 16.64 × 106 mm 4 = 16.64 × 10−6 m 4
3
3
3
Stress:
(a)
σA =
σ =
M yz
Iy
−
Mzy
Iz
(6.4705 kN ⋅ m)(0.045 m) (24.148 kN ⋅ m)(0.060 m)
−
5.04 × 10−6 m 4
16.64 × 10−6 m 4
= 57.772 MPa − 87.072 MPa
(b)
σB =
(6.4705 kN ⋅ m)( −0.045 m) (24.148 kN ⋅ m)(0.060 m)
−
5.04 × 10−6 m 4
16.64 × 10−6 m 4
= −57.772 MPa − 87.072 MPa
(c)
σD =
σ A = −29.3 MPa 
σ B = −144.8 MPa 
(6.4705 kN ⋅ m)(−0.015 m) (24.148 kN ⋅ m)( −0.100 m)
−
5.04 × 10−6 m 4
16.64 × 10−6 m 4
= −19.257 MPa + 145.12 MPa
σ D = −125.9 MPa 
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PROBLEM 4.132
The couple M is applied to a beam of the cross section shown in a
plane forming an angle β with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
SOLUTION
Flange:
1
(8)(0.5)3 + (8)(0.5)(4.75) 2
12
= 90.333 in 4
Iz =
Iy =
1
(0.5)(8)3 = 21.333 in 4
12
1
(0.3)(9)3 = 18.225 in 4
12
1
I y = (9)(0.3)3 = 0.02025 in 4
12
Iz =
Web:
I z = (2)(90.333) + 18.225 = 198.89 in 4
Total:
I y = (2)(21.333) + 0.02025 = 42.687 in 4
y A = yB = − yD = 5 in.
z A = − z B = − zC = 4 in.
M z = 250 cos 30° = 216.51 kip ⋅ in
M y = −250 sin 30° = −125 kip ⋅ in
(a)
σA = −
M z yA M y zA
(216.51)(5) (−125)(4)
+
=−
+
198.89
42.687
Iz
Iy
(b)
σB = −
M z yB M y z B
(216.51)(5) (−125)(−4)
+
=−
+
198.89
42.687
Iz
Iy
(c)
σD = −
M z yD M y z D
(216.51)( −5) ( −125)(−4)
+
=−
+
198.89
42.687
Iz
Iy
σ A = −17.16 ksi 
σ B = 6.27 ksi 
σ D = 17.16 ksi 
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PROBLEM 4.133
The couple M is applied to a beam of the cross section shown in a
plane forming an angle β with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
SOLUTION
1
1
(4.8)(2.4)3 − (4)(1.6)3 = 4.1643 in 4
12
12
1
1
I y = (2.4)(4.8)3 − (1.6)(4)3 = 13.5851 in 4
12
12
y A = yB = − yD = 1.2 in.
Iz =
z A = − z B = − z D = 2.4 in.
M z = 75sin15° = 19.4114 kip ⋅ in
M y = 75cos15° = 72.444 kip ⋅ in
(a)
σA = −
M z yA M y zA
(19.4114)(1.2) (72.444)(2.4)
+
=−
+
4.1643
13.5851
Iz
Iy
σ A = 7.20 ksi 
(b)
σB = −
M z yB M y z B
(19.4114)(1.2) (72.444)(−2.4)
+
=−
+
4.1643
13.5851
Iz
Iy
σ B = −18.39 ksi 
(c)
σD = −
M z yD M y z D
(19.4114)(−1.2) (72.444)(−2.4)
+
=−
+
4.1643
13.5851
Iz
Iy
σ D = −7.20 ksi 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.134
The couple M is applied to a beam of the cross section shown in a
plane forming an angle β with the vertical. Determine the stress at
(a) point A, (b) point B, (c) point D.
SOLUTION
π
2
8  4
 π  4r   π
I z = r −  r 2 
 = −
r
8
 2  3π   8 9π 
4
= (0.109757)(20) 4 = 17.5611 × 10−3 mm 4 = 17.5611 × 10−9 m 4
Iy =
π
8
r4 =
π (20)4
8
= 62.832 × 10−3 mm 4 = 62.832 × 10−9 m 4
4r
(4)(20)
=−
= −8.4883 mm
3π
3π
yB = 20 − 8.4883 = 11.5117 mm
y A = yD = −
z A = − z D = 20 mm
zB = 0
M z = 100 cos 30° = 86.603 N ⋅ m
M y = 100sin 30° = 50 N ⋅ m
(a)
σA = −
(86.603)(−8.4883 × 10−3 ) (50)(20 × 10−3 )
M z yA M y z A
+
=−
+
Iz
Iy
17.5611 × 10−9
62.832 × 10−9
= 57.8 × 106 Pa
(b)
σB = −
σ A = 57.8 MPa 
−3
(86.603)(11.5117 × 10 )
(50)(0)
M z yB M y z B
+
=−
+
−9
Iz
Iy
17.5611 × 10
62.832 × 10−9
= −56.8 × 106 Pa
(c)
σD = −
σ B = −56.8 MPa 
−3
3
(86.603)(−8.4883 × 10 ) (50)(−20 × 10 )
M z yD M y z D
+
=−
+
Iz
Iy
17.5611 × 10−9
62.832 × 10−9
= 25.9 × 106 Pa
σ D =25.9 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.135
The couple M acts in a vertical plane and is applied to a beam oriented
as shown. Determine (a) the angle that the neutral axis forms with the
horizontal, (b) the maximum tensile stress in the beam.
SOLUTION
For C 200 × 17.1 rolled steel shape,
I z = 0.538 × 106 mm 4 = 0.538 × 10−6 m 4
I y = 13.4 × 106 mm 4 = 13.4 × 10−6 m 4
zE = zD = − z A = − zB =
yD = yB = −14.4 mm
1
(203) = 101.5 mm
2
yE = y A = 57 − 14.4 = 42.6 mm
M z = (2.8 × 103 ) cos 10° = 2.7575 × 103 N ⋅ m
M y = (2.8 × 103 ) sin 10° = 486.21 N ⋅ m
(a)
Angle of neutral axis.
tan ϕ =
Iz
0.538
tan θ =
tan10° = 0.007079
13.4
Iy
ϕ = 0.4056°
α = 10° − 0.4056°
(b)
α = 9.59° 
Maximum tensile stress occurs at point D.
σD = −
M z yD M y z D
(2.7575 × 103 )(−14.4 × 10−3 ) (486.21)(0.1015)
+
=−
+
Iz
Iy
0.538 × 10−6
13.4 × 10−6
= 73.807 × 106 + 3.682 × 106 = 77.5 × 106 Pa
σ D = 77.5 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.136
The couple M acts in a vertical plane and is applied to a beam oriented as shown.
Determine (a) the angle that the neutral axis forms with the horizontal, (b) the
maximum tensile stress in the beam.
SOLUTION
For W 310 × 38.7 rolled steel shape,
I z = 85.1 × 106 mm 4 = 85.1 × 10−6 m 4
I y = 7.27 × 106 mm 4 = 7.27 × 10−6 m 4
1
y A = yB = − yD = − yE =   (310) = 155 mm
2
1
z A = z E = − z B = − z D =   (165) = 82.5 mm
2
M z = (16 × 103 ) cos 15° = 15.455 × 103 N ⋅ m
M y = (16 × 103 ) sin 15° = 4.1411 × 103 N ⋅ m
(a)
Angle of neutral axis.
tan ϕ =
Iz
85.1 × 10−6
tan θ =
tan15° = 3.1365
Iy
7.27 × 10−6
ϕ = 72.3°
α = 72.3° − 15°
(b)
α = 57.3° 
Maximum tensile stress occurs at point E.
σE = −
=−
M z yE M y z E
+
Iz
Iy
(15.455 × 103 )(−155 × 10−3 ) (4.1411 × 103 )(82.5 × 10−3 )
+
85.1 × 10−6
7.27 × 10−6
= 75.1 × 106 Pa
σ E = 75.1 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.137
The couple M acts in a vertical plane and is applied to a beam oriented
as shown. Determine (a) the angle that the neutral axis forms with the
horizontal, (b) the maximum tensile stress in the beam.
SOLUTION
I z′ = 21.4 in 4
I y′ = 6.74 in 4
z ′A = z B′ = 0.859 in.
y A = −4 in.
z D′ = −4 + 0.859 in. = −3.141 in.
yB′ = 4 in.
yD′ = −0.25 in.
M y′ = −15 sin 45° = −10.6066 kip ⋅ in
M z′ = 15 cos 45° = 10.6066 kip ⋅ in
(a)
Angle of neutral axis.
tan ϕ =
I z′
21.4
tan θ =
tan (−45°) = 3.1751
6.74
I y′
ϕ = −72.5°
α = 72.5° − 45°
α = 27.5° 



(b)
The maximum tensile stress occurs at point D.
σD = −
M z yD M y z D
(10.6066)(−0.25) (−10.6066)(−3.141)
+
=−
+
21.4
6.74
Iz
Iy
= 0.12391 + 4.9429
σ D = 5.07 ksi 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.138
The couple M acts in a vertical plane and is applied to a beam oriented
as shown. Determine (a) the angle that the neutral axis forms with the
horizontal, (b) the maximum tensile stress in the beam.
SOLUTION
I z′ = 176.9 × 103 mm 4 = 176.9 × 10−9 m 4
I y′ = 281 × 103 mm 4 = 281 × 10−9 m 4
y′E = −18.57 mm, z E = 25 mm
M z ′ = 400 cos 30° = 346.41 N ⋅ m
M y′ = 400sin 30° = 200 N ⋅ m
(a)
tan ϕ =
I z′
176.9 × 10−9
tan θ =
⋅ tan 30° = 0.36346
I y′
281 × 10−9
ϕ = 19.97°
α = 30° − 19.97°
(b)
α = 10.03° 
Maximum tensile stress occurs at point E.
σE = −
M y ′ z′E
(346.41)(−18.57 × 10−3 ) (200)(25 × 10−3 )
M z′ y′E
+
=−
+
I z′
I y′
176.9 × 10−9
281 × 10−9
= 36.36 × 106 + 17.79 × 106 = 54.2 × 106 Pa
σ E = 54.2 MPa 
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PROBLEM 4.139
The couple M acts in a vertical plane and is applied to a beam
oriented as shown. Determine (a) the angle that the neutral axis forms
with the horizontal, (b) the maximum tensile stress in the beam.
SOLUTION
Bending moments:
M y′ = −35sin15° = −9.059 kip ⋅ in
M z ′ = −35cos15° = 33.807 kip ⋅ in
Moments of inertia:
1
1
(2.4)(4)3 − (2 × 0.4)(2)3 = 12.267 in 4
12
12
1
1
(2)(2.4)3 +
(2)(1.6)3 = 2.987 in 4
=
12
12
I y′ =
I z′
(a)
Neutral axis:
tan φ =
I z′
2.987 in 4
tan θ =
tan(−15°) = −0.06525
I y′
12.267 in 4
φ = 3.73°
α = 15° − φ = 15° − 3.73° = 11.27°
α = 11.3°
(b)

Maximum tensile stress at D:
y′D = −1.2 in.
σD =
M y′ z D
I y′
z′D = −2 in.
−
(−9.059 kip ⋅ in)(−2 in.) (33.807 kip ⋅ in)(−1.2 in.)
M z ′ y′D
=
−
I z′
12.267 in 4
2.987 in 4
= 1.477 ksi + 13.582 ksi = 15.059 ksi
σ D = 15.06 ksi 
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PROBLEM 4.140
The couple M acts in a vertical plane and is applied to a beam
oriented as shown. Determine (a) the angle that the neutral
axis forms with the horizontal, (b) the maximum tensile stress
in the beam.
SOLUTION
I z′ = 53.6 × 103 mm 4 = 53.6 × 10−9 m 4
I y′ = 14.77 × 103 mm 4 = 14.77 × 10−9 m 4
M z′ = 120 sin 70° = 112.763 N ⋅ m
M y′ = 120 cos 70° = 41.042 N ⋅ m
(a)
Angle of neutral axis.
θ = 20°.
tan ϕ =
I z′
I y′
tan θ =
53.6 × 10−9
tan 20° = 1.32084
14.77 × 10−9
ϕ = 52.871°
α = 52.871° − 20°
(b)
α = 32.9° 
The maximum tensile stress occurs at point E.
yE′ = −16 mm = −0.016 m
z E = 10 mm = 0.010 m
M y ′ M y′ z E′
σ E = − z′ E +
I z′
I y′
=−
(112.763)(−0.016) (41.042)(0.010)
+
53.6 × 10−9
14.77 × 10−9
= 61.448 × 106 Pa
σ E = 61.4 MPa 
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PROBLEM 4.141
The couple M acts in a vertical plane and is applied to a beam oriented as
shown. Determine the stress at point A.
SOLUTION
1

I y = 2  (7.2)(2.4)3  = 66.355 in 4
3

1

I z = 2  (2.4)(7.2)3 + (2.4)(7.2)(1.2) 2  = 199.066 in 4
12

I yz = 2 {(2.4)(7.2)(1.2)(1.2)} = 49.766 in 4
Using Mohr’s circle determine the principal axes and principal moments of inertia.
Y : (66.355 in 4 , 49.766 in 4 )
Z : (199.066 in 4 , − 49.766 in 4 )
E : (132.710 in 4 , 0)
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PROBLEM 4.141 (Continued)
DY 49.766
=
DE 66.355
2θ m = 36.87° θ m = 18.435°
tan 2θ m =
2
2
R = DE + DY = 82.944 in 4
I u = 132.710 − 82.944 = 49.766 in 4
I v = 132.710 + 82.944 = 215.654 in 4
M u = 125sin18.435° = 39.529 kip ⋅ in
M v = 125cos18.435° = 118.585 kip ⋅ in
u A = 4.8 cos 18.435° + 2.4 sin 18.435° = 5.3126 in.
ν A = −4.8 sin 18.435° + 2.4 cos 18.435° = 0.7589 in.
M u
Mν
σA = − ν A + u A
Iν
=−
Iu
(118.585)(5.3126) (39.529)(0.7589)
+
215.654
49.766
= −2.32 ksi
σ A = −2.32 ksi 
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PROBLEM 4.142
The couple M acts in a vertical plane and is applied to a beam oriented
as shown. Determine the stress at point A.
SOLUTION
Using Mohr’s circle determine the principal axes and principal moments of inertia.
Y : (8.7, 8.3) in 4
Z : (24.5, − 8.3) in 4
E : (16.6, 0) in 4
EF = 7.9 in 4
FZ = 8.3 in 4
R = 7.92 + 8.32 = 11.46 in 4
FZ 8.3
=
= 1.0506
EF 7.9
I v = 16.6 + 11.46 = 28.06 in 4
tan 2θ m =
θ m = 23.2° I u = 16.6 − 11.46 = 5.14 in 4
M u = M sin θ m = (60) sin 23.2° = 23.64 kip ⋅ in
M v = M cos θ m = (60) cos 23.2° = 55.15 kip ⋅ in
u A = y A cos θ m + z A sin θ m = −3.92cos 23.2° − 1.08 sin 23.2° = −4.03 in.
v A = z A cos θ m − y A sin θ m = −1.08cos 23.2° + 3.92 sin 23.2° = 0.552 in.
σA = −
M vu A M u vA
(55.15)( −4.03) (23.64)(0.552)
+
+
=−
28.06
5.14
Iv
Iu
σ A = 10.46 ksi 
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PROBLEM 4.143
The couple M acts in a vertical plane and is applied to a beam oriented
as shown. Determine the stress at point A.
SOLUTION
Using Mohr’s circle determine the principal axes and principal moments of inertia.
Y : (1.894, 0.800) × 106 mm 4
Z : (0.614, 0.800) × 106 mm 4
E : (1.254, 0) × 106 mm 4
2
2
R = EF + FZ = 0.6402 + 0.8002 × 10−6 = 1.0245 × 106 mm 4
I v = (1.254 − 1.0245) × 106 mm 4 = 0.2295 × 106 mm 4 = 0.2295 × 10−6 m 4
I u = (1.254 + 1.0245) × 106 mm 4 = 2.2785 × 106 mm 4 = 2.2785 × 10−6 m 4
FZ 0.800 × 106
θ m = 25.67°
=
= 1.25
FE 0.640 × 106
M v = M cos θ m = (1.2 × 103 ) cos 25.67° = 1.0816 × 103 N ⋅ m
tan 2θ m =
M u = − M sin θ m = −(1.2 × 103 ) sin 25.67° = −0.5198 × 103 N ⋅ m
u A = y A cos θ m − z A sin θ m = 45 cos 25.67° − 45 sin 25.67° = 21.07 mm
v A = z A cos θ m + y A sin θ m = 45 cos 25.67° + 45 sin 25.67° = 60.05 mm
σA = −
M vu A M u vA
(1.0816 × 103 )(21.07 × 10−3 ) (−0.5198 × 103 )(60.05 × 10−3 )
=−
+
+
Iu
Iv
0.2295 × 10−6
2.2785 × 10−6
= 113.0 × 106 Pa
σ A = 113.0 MPa 
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PROBLEM 4.144
The tube shown has a uniform wall thickness of 12 mm. For the
loading given, determine (a) the stress at points A and B, (b) the point
where the neutral axis intersects line ABD.
SOLUTION
Add y- and z-axes as shown. Cross section is a 75 mm × 125 mm rectangle with a 51 mm × 101 mm rectangular
cutout.
1
1
(75)(125)3 − (51)(101)3 = 7.8283 × 106 mm 4 = 7.8283 × 10−6 m 4
12
12
1
1
I y = (125)(75)3 − (101)(51)3 = 3.2781 × 103 mm 4 = 3.2781 × 10−6 m 4
12
12
A = (75)(125) − (51)(101) = 4.224 × 103 mm 2 = 4.224 × 10−3 m 2
Iz =
Resultant force and bending couples:
P = 14 + 28 + 28 = 70 kN = 70 × 103 N
M z = −(62.5 mm)(14 kN) + (62.5 mm)(28kN) + (62.5 mm)(28 kN) = 2625 N ⋅ m
M y = −(37.5 mm)(14 kN) + (37.5 mm)(28 kN) + (37.5 mm)(28 kN) = −525 N ⋅ m
(a)
σA =
70 × 103
(2625)(−0.0625) (−525)(0.0375)
P M z yA M y zA
−
+
=
−
+
−3
A
Iz
Iy
4.224 × 10
7.8283 × 10−6
3.2781 × 10−6
= 31.524 × 106 Pa
σB =
σ A = 31.5 MPa 
70 × 103
(2625)(0.0625) (−525)(0.0375)
P M z yB M y z B
−
+
=
−
+
−3
A
Iz
Iy
4.224 × 10
7.8283 × 10−6
3.2781 × 10−6
= −10.39 × 106 Pa
(b)
σ B = −10.39 MPa 
Let point H be the point where the neutral axis intersects AB.
z H = 0.0375 m,
0=
yH = ?, σ H = 0
P M z yH M y z H
−
+
A
Iz
Iy
 P Mz H  7.8283 × 10−6  70 × 103
( −525)(0.0375) 
+
 +
=


−
3
A
2625
I y 
3.2781 × 10−6 
 4.224 × 10

= 0.03151 m = 31.51 mm
yH =
Iz
Mz
31.51 + 62.5 = 94.0 mm
Answer: 94.0 mm above point A. 
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PROBLEM 4.145
Solve Prob. 4.144, assuming that the 28-kN force at point E is
removed.
PROBLEM 4.144 The tube shown has a uniform wall thickness of
12 mm. For the loading given, determine (a) the stress at points A and
B, (b) the point where the neutral axis intersects line ABD.
SOLUTION
Add y- and z-axes as shown. Cross section is a 75 mm × 125 mm rectangle with a 51 mm × 101 mm rectangular
cutout.
1
1
(75)(125)3 − (51)(101)3 = 7.8283 × 106 mm 4 = 7.8283 × 10−6 m 4
12
12
1
1
3
I y = (125)(75) − (101)(51)3 = 3.2781 × 106 mm 4 = 3.2781 × 10−6 m 4
12
12
A = (75)(125) − (51)(101) = 4.224 × 103 mm 2 = 4.224 × 10−3 m 2
Iz =
Resultant force and bending couples:
P = 14 + 28 = 42 kN = 42 × 103 N
M z = −(62.5 mm)(14 kN) + (62.5 mm)(28 kN) = 875 N ⋅ m
M y = −(37.5 mm)(14 kN) + (37.5 mm)(28 kN) = 525 N ⋅ m
(a)
σA =
42 × 103
(875)(−0.0625) (525)(0.0375)
P M z yA M y zA
−
+
=
−
+
−3
A
Iz
Iy
4.224 × 10
7.8283 × 10−6
3.2781 × 10−6
= 22.935 × 106 Pa
σB =
σ A = 22.9 MPa 
P M z yB M y z B
42 × 103
(875)(0.0625) (525)(0.0375)
−
+
=
−
+
−3
A
Iz
Iy
4.224 × 10
7.8283 × 10−6
3.2781 × 10−6
= 8.9631 × 106 Pa
(b)
σ B = 8.96 MPa 
Let point K be the point where the neutral axis intersects BD.
z K = ?,
0=
yK = 0.0625 m, σ H = 0
P M z yH M y z H
−
+
A
Iz
Iy
I y  M z yH P  3.2781 × 10−6  (875)(0.0625)
42 × 103 
− =
−



−6
M y  Iz
A
525
4.224 × 10−3 
 7.8283 × 10
= −0.018465 m = −18.465 mm
zH =
37.5 + 18.465 = 56.0 mm
Answer: 56.0 mm to the right of point B. 
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PROBLEM 4.146
A rigid circular plate of 125-mm radius is attached to a solid
150 × 200-mm rectangular post, with the center of the plate directly above
the center of the post. If a 4-kN force P is applied at E with θ = 30°,
determine (a) the stress at point A, (b) the stress at point B, (c) the point
where the neutral axis intersects line ABD.
SOLUTION
P = 4 × 103 N (compression)
M x = − PR sin 30° = −(4 × 103 )(125 × 10−3 ) sin 30° = −250 N ⋅ m
M z = − PR cos 30° = −(4 × 103 )(125 × 10−3 ) cos 30° = −433 N ⋅ m
1
(200)(150)3 = 56.25 × 106 mm 4 = 56.25 × 10−6 m 4
12
1
(150)(200)3 = 100 × 106 mm 4 = 100 × 10−6 m 4
Iz =
12
− x A = xB = 100 mm z A = z B = 75 mm
Ix =
A = (200)(150) = 30 × 103 mm 2 = 30 × 10−3 m 2
(a)
σA = −
P M x zA M z xA
4 × 103
(−250)(75 × 10−3 ) (−433)(−100 × 10−3 )
−
+
=−
−
+
−3
A
Ix
Iz
30 × 10
56.25 × 10−6
100 × 10−6
σ A = 633 × 103 Pa = 633 kPa 
(b)
σB = −
P M x z B M z xB
4 × 103
(−250)(75 × 10−3 ) (−433)(100 × 10−3 )
−
+
=−
−
+
A
Ix
Iz
30 × 10−3
56.25 × 10−6
100 × 10−6
σ B = −233 × 103 Pa = −233 kPa 
(c)
Let G be the point on AB where the neutral axis intersects.
σG = 0
zG = 75 mm
σG = −
P M x zG M z xG
−
+
=0
A
Ix
Iz
xG =
xG = ?
I z  P M x Z G  100 × 10−6
 +
=
Mz  A
Ix 
−433
= 46.2 × 10−3 m = 46.2 mm
(−250)(75 × 10−3 ) 
 4 × 103
+


−3
56.25 × 10−6 
 30 × 10
Point G lies 146.2 mm from point A 
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PROBLEM 4.147
4.147 In Prob. 4.146, determine (a) the value of θ for which the stress at D
reaches it largest value, (b) the corresponding values of the stress, at A, B,
C, and D.
4.146 A rigid circular plate of 125-mm radius is attached to a solid
150 × 200-mm rectangular post, with the center of the plate directly above
the center of the post. If a 4-kN force P is applied at E with θ = 30°,
determine (a) the stress at point A, (b) the stress at point B, (c) the point
where the neutral axis intersects line ABD.
SOLUTION
P = 4 × 103 N
(a)
PR = (4 × 103 )(125 × 10−3 ) = 500 N ⋅ m
M x = − PR sin θ = −500sin θ
M x = − PR cos θ = −500cos θ
1
(200)(150)3 = 56.25 × 106 mm 4 = 56.25 × 10−6 m 4
2
1
I z = (150)(200)3 = 100 × 106 mm 4 = 100 × 10−6 m 4
2
xD = 100 mm
z D = −75 mm
Ix =
A = (200)(150) = 30 × 103 mm 2 = 30 × 10−3 m 2
σ =−
 1 Rz sin θ
P M xz M z x
Rx cos θ 
−
+
= −P  −
+

A
Ix
Iz
Ix
Iz 
A
For σ to be a maximum,
dσ
=0
dθ
with z = z D , x = xD

dσ D
Rz cos θ
Rx sin θ 
= − P 0 + D
+ D
=0
dθ
Ix
IZ


sin θ
I z
(100 × 10−6 )(−75 × 10−3 )
4
= tan θ = − z D = −
=
−6
−3
cos θ
I x xD
(56.25 × 10 )(100 × 10 ) 3
sin θ = 0.8,
(b)
σA = −
cos θ = 0.6,
θ = 53.1° 
P M x z A M z xA
4 × 103
(500)(0.8)(75 × 10−3 ) (500)(0.6)(−100 × 10−3 )
−
+
=−
+
−
A
Ix
Iz
30 × 10−3
56.25 × 10−6
100 × 10−6
= (−0.13333 + 0.53333 + 0.300) × 106 Pa = 0.700 × 106 Pa
σ A = 700 kPa 
σ B = (−0.13333 + 0.53333 − 0.300) × 106 Pa = 0.100 × 106 Pa
σ B = 100 kPa 
σ C = (−0.13333 + 0 + 0) × 106 Pa
σ D = (−0.13333 − 0.53333 − 0.300) × 106 Pa = − 0.967 × 106 Pa
σ C = −133.3kPa 
σ D = −967 kPa 
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PROBLEM 4.148
Knowing that P = 90 kips, determine the largest distance a for
which the maximum compressive stress dose not exceed 18 ksi.
SOLUTION
A = (5 in.)(6 in.) − 2(2 in.)(4 in.) = 14 in 2
1
1
(5 in.)(6 in.)3 − 2 (2 in.)(4 in.)3 = 68.67 in 4
12
12
1
1
I z = 2 (1 in.)(5 in.)3 +
(4 in.)(1 in.)3 = 21.17 in 4
12
12
Ix =
Force-couple system at C:
P= P
For P = 90 kips:
P = 90 kips M x = (90 kips)(2.5 in.) = 225 kip ⋅ in
Maximum compressive stress at B:
σB = −
−18 ksi = −
M x = P(2.5 in.)
M z = Pa
M z = (90 kips) a
σ B = −18 ksi
P M x (3 in.) M z (2.5 in.)
−
−
A
Ix
Iz
90 kips (225 kip ⋅ in)(3 in.) (90 kips) a (2.5 in.)
−
−
14 in 2
68.67 in 4
21.17 in 4
−18 = −6.429 − 9.830 − 10.628a
−1.741 = −10.628 a
a = 0.1638 in. 
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PROBLEM 4.149
Knowing that a = 1.25 in., determine the largest value of P that
can be applied without exceeding either of the following
allowable stresses:
σ ten = 10 ksi
σ comp = 18 ksi
SOLUTION
A = (5 in.)(6 in.) − (2)(2 in.)(4 in.) = 14 in 2
1
1
(5 in.)(6 in.)3 − 2 (2 in.)(4 in.)3 = 68.67 in 4
12
12
1
1
I z = 2 (1 in.)(5 in.)3 + (4 in.)(1 in.)3 = 21.17 in 4
12
12
Ix =
For a = 1.25 in.,
Force-couple system at C:
P = P M x = P(2.5 in.)
M y = Pa = (1.25in.)
Maximum compressive stress at B:
σB = −
−18 ksi = −
σ B = −18 ksi
P M x (3 in.) M z (2.5 in.)
−
−
A
Ix
Iz
P
P(2.5 in.)(3 in.) P(1.25in.)(2.5in.)
−
−
2
14 in
68.67 in 4
21.17 in 4
−18 = − 0.0714P − 0.1092P − 0.1476 P
−18 = 0.3282 P
P = 54.8 kips
Maximum tensile stress at D:
σD = −
σ D = +10 ksi
P M x (3 in.) M z (2.5 in.)
+
+
A
Ix
Iz
+10 ksi = − 0.0714P + 0.1092P + 0.1476P
10 = 0.1854 P
P = 53.9 kips
The smaller value of P is the largest allowable value.
P = 53.9 kips 
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PROBLEM 4.150
The Z section shown is subjected to a couple M 0 acting in a vertical
plane. Determine the largest permissible value of the moment M 0 of the
couple if the maximum stress is not to exceed 80 MPa. Given:
I max = 2.28 × 10−6 mm 4 ,
I min = 0.23 × 10−6 mm 4 , principal axes
25.7°
and 64.3°
.
SOLUTION
I v = I max = 2.28 × 106 mm 4 = 2.28 × 10−6 m 4
I u = I min = 0.23 × 106 mm 4 = 0.23 × 10−6 m 4
M v = M 0 cos 64.3°
M u = M 0 sin 64.3°
θ = 64.3°
I
tan ϕ = v tan θ
Iu
2.28 × 10−6
tan 64.3°
0.23 × 10−6
= 20.597
=
ϕ = 87.22°
Points A and B are farthest from the neutral axis.
uB = yB cos 64.3° + z B sin 64.3° = (−45) cos 64.3° + (−35) sin 64.3°
= −51.05 mm
vB = z B cos 64.3° − yB sin 64.3° = (−35) cos 64.3° − (−45) sin 64.3°
= +25.37 mm
σB = −
80 × 106 = −
M vu B M u v B
+
Iv
Iu
(M 0 cos 64.3°)(−51.05 × 10−3 ) (M 0 sin 64.3°)(25.37 × 10−3 )
+
0.23 × 10−6
2.28 × 10−6
= 109.1 × 103 M 0
M0 =
80 × 106
109.1 × 103
M 0 = 733 N ⋅ m 
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PROBLEM 4.151
Solve Prob. 4.150 assuming that the couple M 0 acts in a horizontal
plane.
PROBLEM 4.150 The Z section shown is subjected to a couple M 0
acting in a vertical plane. Determine the largest permissible value of the
moment M 0 of the couple if the maximum stress is not to exceed 80
MPa. Given:
I max = 2.28 × 10−6 mm 4 ,
and 64.3°
.
principal axes 25.7°
I min = 0.23 × 10−6 mm 4 ,
SOLUTION
I v = I min = 0.23 × 106 mm 4 = 0.23 × 106 m 4
I u = I max = 2.28 × 106 mm 4 = 2.28 × 106 m 4
M v = M 0 cos 64.3°
M u = M 0 sin 64.3°
θ = 64.3°
tan ϕ =
Iv
tan θ
Iu
0.23 × 10−6
tan 64.3°
2.28 × 10−6
= 0.20961
=
ϕ = 11.84°
Points D and E are farthest from the neutral axis.
uD = yD cos 25.7° − z D sin 25.7° = (−5) cos 25.7° − 45 sin 25.7°
= −24.02 mm
vD = z D cos 25.7° + yD sin 25.7° = 45cos 25.7° + (−5) sin 25.7°
= 38.38 mm
σD = −
M v u D M u vD
(M D cos 64.3°)(−24.02 × 10−3 )
+
=−
Iv
Iu
0.23 × 10−6
+
(M 0 sin 64.3°)(38.38 × 10−3 )
2.28 × 10−6
80 × 106 = 60.48 × 103 M 0
M 0 = 1.323 × 103 N ⋅ m
M 0 = 1.323 kN ⋅ m 
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PROBLEM 4.152
A beam having the cross section shown is subjected to a couple M 0 that
acts in a vertical plane. Determine the largest permissible value of the
moment M 0 of the couple if the maximum stress in the beam is
not |to exceed 12 ksi. Given: I y = I z = 11.3 in.4 , A = 4.75 in.2 ,
kmin = 0.983 in. (Hint: By reason of symmetry, the principal axes form
2
an angle of 45° with the coordinate axes. Use the relations I min = Akmin
and I min + I max = I y + I z )
SOLUTION
M u = M 0 sin 45° = 0.70711 M 0
M v = M 0 cos 45° = 0.7071 M 0
2
= (4.75)(0.983) 2 = 4.59 in 4
I min = Akmin
I max = I y + I z − I min = 11.3 + 11.3 − 4.59 = 18.01 in 4
u B = yB cos 45° + z B sin 45° = −3.57 cos 45° + 0.93 sin 45° = −1.866 in.
vB = z B cos 45° − yB sin 45° = 0.93 cos 45° − (−3.57) sin 45° = 3.182 in.
σB = −
 u
M v u B M u vB
v 
+
= −0.70711 M 0  − B + B 
Iv
Iu
I max 
 I min
 (−1.866) 3.182 
= 0.70711 M 0  −
+
= 0.4124 M 0
4.59
18.01 

M0 =
σB
0.4124
=
12
0.4124
M 0 = 29.1 kip ⋅ in 
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PROBLEM 4.153
Solve Prob. 4.152, assuming that the couple M 0 acts in a horizontal
plane.
PROBLEM 4.152 A beam having the cross section shown is subjected
to a couple M 0 that acts in a vertical plane. Determine the largest
permissible value of the moment M 0 of the couple if the maximum
stress in the beam is not to exceed 12 ksi. Given:
I y = I z = 11.3 in.4, A = 4.75 in.2, kmin = 0.983 in. (Hint: By reason of
symmetry, the principal axes form an angle of 45° with the coordinate
2
and I min + I max = I y + I z )
axes. Use the relations I min = Akmin
SOLUTION
M u = M 0 cos 45° = 0.70711 M 0
M v = − M 0 sin 45° = − 0.70711 M 0
2
= (4.75)(0.983) 2 = 4.59 in 4
I min = Akmin
I max = I y + I z − I min = 11.3 + 11.3 − 4.59 = 18.01 in 4
u D = yD cos 45° + z D sin 45° = −0.93 cos 45° + (−3.57 sin 45°) = −1.866 in.
vD = z D cos 45° − yD sin 45° = (−3.57) cos 45° − (0.93) sin 45° = 3.182 in.
σD = −
 u
M vu D M u v D
v 
+
= −0.70711 M 0  − D + D 
Iv
Iu
I max 
 I min
 (−1.866) 3.182 
= 0.70711 M 0  −
+
= 0.4124 M 0
4.59
18.01 

M0 =
σD
0.4124
=
12
0.4124
M 0 = 29.1 kip ⋅ in 
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PROBLEM 4.154
An extruded aluminum member having the cross section shown
is subjected to a couple acting in a vertical plane. Determine the
largest permissible value of the moment M 0 of the couple if the
maximum stress is not to exceed 12 ksi. Given:
I max = 0.957 in 4 , I min = 0.427 in 4 , principal axes 29.4° and
60.6° .
SOLUTION
I u = I max = 0.957 in 4
I v = I min = 0.427 in 4
M u = M 0 sin 29.4°, M v = M 0 cos 29.4°
θ = 29.4°
tan ϕ =
Iv
0.427
tan θ =
tan 29.4°
Iu
0.957
= 0.2514
ϕ = 14.11°
Point A is farthest from the neutral axis.
y A = −0.75 in., z A = −0.75 in.
u A = y A cos 29.4° + z A sin 29.4° = −1.0216 in.
v A = z A cos 29.4° − y A sin 29.4° = −0.2852 in.
σA = −
M vu A
MV
(M cos 29.4°)(−1.0216) ( M 0 sin 29.4°)(−0.2852)
+ u A =− 0
+
Iv
Iu
0.427
0.957
= 1.9381 M 0
M0 =
σA
1.9381
=
12
1.9381
M 0 = 6.19 kip ⋅ in 
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PROBLEM 4.155
A couple M 0 acting in a vertical plane is applied to a W12 × 16 rolled-steel beam,
whose web forms an angle θ with the vertical. Denoting by σ 0 the maximum stress
in the beam when θ = 0, determine the angle of inclination θ of the beam for which
the maximum stress is 2σ 0.
SOLUTION
For W 12 × 16 rolled steel section,
I z = 103 in 4
I y = 2.82 in 4
d = 11.99 in.
b f = 3.990 in.
yA = −
tan ϕ =
d
2
zA =
bf
2
Iz
103
tan θ =
tan θ = 36.52 tan θ
Iy
2.82
Point A is farthest from the neutral axis.
M y = M 0 sin θ
σA = −
For
θ = 0,
M z = M 0 cos θ

M 0b f
I zb f
M z yA M y zA
M d
M d
+
= 0 cos θ +
sin θ = 0 1 +
tan θ  cos θ

Iz
Iy
I yd
2I z
2I y
2 I z 

σ0 =
M 0d
2I z



σ A = σ 0 1 +
I 2b f
I yd
tan θ =

tan θ  cos θ = 2σ 0

I yd

I zb f
(103)(3.990)  2

− 1

(2.82)(11.99)  cos θ

 2

− 1
tan θ = 0.082273 
 cos θ

Assuming cos θ ≈ 1, we get
θ = 4.70° 
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PROBLEM 4.156
Show that, if a solid rectangular beam is bent by a couple applied
in a plane containing one diagonal of a rectangular cross section,
the neutral axis will lie along the other diagonal.
SOLUTION
b
h
M z = M cos θ , M z = M sin θ
tan θ =
Iz =
1 3
bh
12
Iy =
1 3
hb
12
1 3
bh
Iz
b
h
tan ϕ =
tan θ = 12
⋅ =
1
Iy
b
hb3 h
12
The neutral axis passes through corner A of the diagonal AD. 
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PROBLEM 4.157
A beam of unsymmetric cross section is subjected to a couple M 0 acting in the
horizontal plane xz. Show that the stress at point A, of coordinates y and z, is
σA =
zI z − yI yz
2
I y I z − I yz
My
where Iy, Iz, and Iyz denote the moments and product of inertia of the cross section
with respect to the coordinate axes, and My the moment of the couple.
SOLUTION
The stress σ A varies linearly with the coordinates y and z. Since the axial force is zero, the y- and z-axes are
centroidal axes:
σ A = C1 y + C2 z
where C1 and C2 are constants.
M z = −  yσ AdA = −C1 y 2dA − C2  yzdA
= − I zC1 − I yzC2 = 0
C1 = −
I yz
Iz
C2
M y =  zσ AdA = C1 yz dA + C2  z 2dA
= I yzC1 + I yC2
− I yz
(
I yz
Iz
C2 + I yC2
)
2
I z M y = I y I z − I yz
C2
C2 =
IzM y
2
I y I z − I yz
C1 = −
σA =
I yz M y
2
I y I z − I yz
I z z − I yz y
2
I y I z − I yz
My

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PROBLEM 4.158
A beam of unsymmetric cross section is subjected to a couple M 0 acting in the
vertical plane xy. Show that the stress at point A, of coordinates y and z, is
σA =
yI y − zI yz
2
I y I z − I yz
Mz
where I y , I z , and I yz denote the moments and product of inertia of the cross
section with respect to the coordinate axes, and M z the moment of the couple.
SOLUTION
The stress σ A varies linearly with the coordinates y and z. Since the axial force is zero, the y- and z-axes are
centroidal axes:
σ A = C1 y + C2 z
where C1 and C2 are constants.
M y =  zσ AdA = C1  yzdA + C2  z 2dA
= I yzC1 + I yC2 = 0
C2 = −
I yz
Iy
C1
M z = −  yσ Adz = −C1 y 2dA + C2  yzdA
I yz
C1
= − I Z C1 − I yz
Iy
(
)
2
I y M z = − I y I z − I yz
C1
C1 = −
C2 = +
σA = −
I yM z
2
I y I z − I yz
I yz M z
2
I y I z − I yz
I y y − I yz 2
2
I y I z − I yz
Mz

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PROBLEM 4.159
(a) Show that, if a vertical force P is applied at point A of the section
shown, the equation of the neutral axis BD is
 zA 
 xA 
 2  x +  2  z = −1
 rz 
 rx 
where rz and rx denote the radius of gyration of the cross section with
respect to the z axis and the x axis, respectively. (b) Further show that, if
a vertical force Q is applied at any point located on line BD, the stress at
point A will be zero.
SOLUTION
Definitions:
rx2 =
(a)
M x = Pz A
σE = −
=−
Ix
I
, rz2 = z
A
A
M z = − Px A
P M z xE M x z E
P Px A xE
Pz z
+
−
=− −
− A 2E
2
A
Iz
Ix
A
Arz
Arx
z  
P   xA 
1 +  2  xE +  A2  z E  = 0
A   rz 
 rx  
if E lies on neutral axis.
z 
x 
1 +  A2  x +  A2  z = 0,
 rz 
 rx 
(b)
M x = PzE
σA = −
 zA 
 xA 
 2  x +  2  z = −1
 rz 
 rx 

M z = − PxE
P M z xA M x z A
P Px x
Pz z
+
−
= − − E 2 A − E 2A
A
Iz
Iy
A
Arz
Arx
= 0 by equation from part (a).

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PROBLEM 4.160
(a) Show that the stress at corner A of the prismatic member
shown in part a of the figure will be zero if the vertical force
P is applied at a point located on the line
x
z
+
=1
b /6 h /6
(b) Further show that, if no tensile stress is to occur in the
member, the force P must be applied at a point located
within the area bounded by the line found in part a and three
similar lines corresponding to the condition of zero stress at
B, C, and D, respectively. This area, shown in part b of the
figure, is known as the kern of the cross section.
SOLUTION
1 3
1 3
hb
Ix =
bh
12
12
h
b
zA = −
xA = −
2
2
Iz =
A = bh
Let P be the load point.
M z = − PxP
σA = −
σ A = 0,
1−
M x = Pz P
P M z xA M x z A
+
−
A
Iz
Ix
( )
=−
(− PxP ) − b2
( Pz P − 2h )
P
+
−
3
1 hb
1 bh3
bh
12
12
=−
P 
x
z 
1− P − P 

bh 
b/6 h/6 
x
z
−
=0
b/6 h/6
(a)
For
(b)
At point E:
z = 0 ∴ xE = b/6
At point F:
x = 0 ∴ z F = h /6
x
z
+
=1
b/6 h/6
If the line of action ( xP , z P ) lies within the portion marked TA , a tensile stress will occur at corner A.
By considering σ B = 0, σ C = 0, and σ D = 0, the other portions producing tensile stresses are identified.
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PROBLEM 4.161
For the machine component and loading shown, determine the
stress at point A when (a) h = 2 in., (b) h = 2.6 in.
SOLUTION
M = −4 kip ⋅ in
Rectangular cross section:
r =
(a)
A = bh r2 = 3 in. r1 = r2 − h
1
h
(r1 + r2 ), R =
, e=r −R
r
2
ln 2
r1
h = 2 in.
A = (0.75)(2) = 1.5 in 2
r1 = 3 − 2 = 1 in.
R=
2
= 1.8205 in.
ln 13
At point A:
σA =
r =
1
(3 + 1) = 2 in.
2
e = 2 − 1.8205 = 0.1795 in.
r = r1 = 1 in.
M (r − R) (−4)(1 − 1.8205)
=
= 12.19 ksi
Aer
(1.5)(0.1795)(1)
σ A = 12.19 ksi 
(b)
h = 2.6 in.
A = (0.75)(2.6) = 1.95 in 2
r =
r1 = 3 − 2.6 = 0.4 in.
R=
2.6
= 1.2904 in.
3
ln 0.4
At point A:
σA =
1
(3 + 0.4) = 1.7 in.
2
e = 1.7 − 1.2904 = 0.4906 in.
r = r1 = 0.4 in.
M (r − R )
(−4)(0.4 − 1.2904)
=
= 11.15 ksi
Aer
(1.95)(0.4096)(0.4)
σ A = 11.15 ksi 
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PROBLEM 4.162
For the machine component and loading shown, determine the
stress at points A and B when h = 2.5 in.
SOLUTION
M = −4 kip ⋅ in
Rectangular cross section:
h = 2.5 in. b = 0.75 in. A = 1.875 in.2
r2 = 3 in. r1 = r2 − h = 0.5 in.
1
1
(r1 + r2 ) = (0.5 + 3.0) = 1.75 in.
2
2
2.5
h
R = r = 3.0 = 1.3953 in.
2
ln 0.5
ln
r =
r1
e = r − R = 1.75 − 1.3953 = 0.3547 in.
At point A:
r = r1 = 0.5 in.
σA =
At point B:
M (r − R )
(−4 kip ⋅ in)(0.5 in. − 1.3953 in.)
=
Aer
(0.75 in.)(2.5 in.)(0.3547 in.)(0.5 in.)
σ A = 10.77 ksi 
r = r2 = 3 in.
σB =
M (r − R )
(−4 kip ⋅ in)(3 in. − 1.3953 in.)
=
Aer
(0.75 in. × 2.5 in.)(0.3547 in.)(3 in.)
σ B = −3.22 ksi 
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PROBLEM 4.163
The curved portion of the bar shown has an inner radius of 20 mm.
Knowing that the allowable stress in the bar is 150 MPa, determine the
largest permissible distance a from the line of action of the 3-kN force to
the vertical plane containing the center of curvature of the bar.
SOLUTION
Reduce the internal forces transmitted across section AB to a force-couple system at the centroid of the
section. The bending couple is
M = P (a + r )
For the rectangular section, the neutral axis for bending couple only lies at
R=
h
ln
r2
r1
Also e = r − R
The maximum compressive stress occurs at point A. It is given by
σA = −
P My A
P P (a + r ) y A
P
−
=− −
= −K
A Aer1
A
Aer1
A
with
y A = R − r1
Thus,
K =1+
Data:
(a + r )( R − r1)
er1
h = 25 mm, r1 = 20 mm, r2 = 45 mm, r = 32.5 mm
R=
25
= 30.8288 mm, e = 32.5 − 30.8288 = 1.6712 mm
45
ln 20
b = 25 mm,
A = bh = (25)(25) = 625 mm 2 = 625 × 10−6 m 2
R − r1 = 10.8288 mm
P = 3 × 103 N ⋅ m, σ A = −150 × 106 Pa
(−150 × 106 )(625 × 10−6 )
= 31.25
P
3 × 103
( K − 1)er1 (30.25)(1.6712)(20)
a+r =
=
= 93.37 mm
R − r1
10.8288
K =−
σ AA
=−
a = 93.37 − 32.5
a = 60.9 mm 
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PROBLEM 4.164
The curved portion of the bar shown has an inner radius of 20 mm.
Knowing that the line of action of the 3-kN force is located at a distance
a = 60 mm from the vertical plane containing the center of curvature
of the bar, determine the largest compressive stress in the bar.
SOLUTION
Reduce the internal forces transmitted across section AB to a force-couple system at the centroid of the
section. The bending couple is
M = P (a + r )
For the rectangular section, the neutral axis for bending couple only lies at
R=
h
ln
r2
r1
.
Also e = r − R
The maximum compressive stress occurs at point A. It is given by
σA = −
P My A
P P (a + r ) y A
P
−
=− −
= −K
A Aer1
A
Aer1
A
with
y A = R − r1
Thus,
K =1+
Data:
h = 25 mm, r1 = 20 mm, r2 = 45 mm, r = 32.5 mm
R=
(a + r )( R − r1)
er1
25
= 30.8288 mm, e = 32.5 − 30.8288 = 1.6712 mm
45
ln 20
b = 25 mm,
A = bh = (25)(25) = 625 mm 2 = 625 × 10−6 m 2
a = 60 mm, a + r = 92.5 mm, R − r1 = 10.8288 mm
(92.5)(10.8288)
K = 1+
= 30.968 P = 30 × 103 N
(1.6712)(20)
σA =
KP
(30.968)(3 × 103 )
=−
= −148.6 × 106 Pa
A
625 × 10−6
σ A = −148.6 MPa 
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PROBLEM 4.165
The curved bar shown has a cross section of 40 × 60 mm and an
inner radius r1 = 15 mm. For the loading shown, determine the
largest tensile and compressive stresses.
SOLUTION
h = 40 mm, r1 = 15 mm, r2 = 55mm
A = (60)(40) = 2400 mm 2 = 2400 × 10−6 m 2
R=
r =
h
40
=
= 30.786 mm
r2
55
ln
ln
r1
40
1
(r1 + r2 ) = 35 mm
2
e = r − R = 4.214 mm
At r = 15mm,
σ =−
σ =−
My
Aer
y = 30.786 − 15 = 15.756 mm
(120)(15.786 × 10−3 )
= −12.49 × 10−6 Pa
(2400 × 10−6 )(4.214 × 10−3 )(15 × 10−3 )
σ = −12.49MPa 
At r = 55mm,
σ =−
y = 30.786 − 55 = −24.214 mm
(120)(−24.214 × 10−3 )
= 5.22 × 106 Pa
(2400 × 10−6 )(4.214 × 10−3 )(55 × 10−3 )
σ = 5.22 MPa 
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PROBLEM 4.166
For the curved bar and loading shown, determine the percent
error introduced in the computation of the maximum stress by
assuming that the bar is straight. Consider the case when
(a) r1 = 20 mm, (b) r1 = 200 mm, (c) r1 = 2 m.
SOLUTION
h = 40 mm, A = (60)(40) = 2400 mm2 = 2400 × 10−6 m 2 , M = 120 N ⋅ m
1 3
1
bh =
(60)(40)3 = 0.32 × 106 mm 4 = 0.32 × 10 −6 mm 4 ,
12
12
I =
c=
1
h = 20 mm
2
Assuming that the bar is straight
σs = −
(a)
Mc
(120)(20 × 10−8 )
=−
= 7.5 × 106 Pa = 7.5 MPa
I
(0.32 × 10−6 )
r1 = 20 mm r2 = 60 mm
h
40
=
= 36.4096 mm
60
r2
ln
ln
r1
20
R=
r =
1
(r1 + r2 ) = 40 mm
2
σa =
r1 − R = −16.4096 mm
e = r − R = 3.5904 mm
M (r1 − R)
(120)(−16.4096 × 10−3 )
=
= −11.426 × 106 Pa = −11.426 MPa
Aer
(2400 × 10−6 )(3.5904 × 10−3 )(20 × 10−3 )
% error =
−11.426 − (−7.5)
× 100% = −34.4%
−11.426

For parts (b) and (c), we get the values in the table below:
(a)
(b)
(c)
r1, mm
r2 , mm
20
200
2000
60
240
2040
R, mm
36.4096
219.3926
2019.9340
r , mm
40
220
2020
e, mm
3.5904
0.6074
0.0660
σ , MPa
−11.426
−7.982
−7.546
% error
−34.4 %
6.0 % 
0.6 % 
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PROBLEM 4.167
The curved bar shown has a cross section of 30 × 30 mm. Knowing
that a = 60 mm, determine the stress at (a) point A, (b) point B.
SOLUTION
Reduce the internal forces transmitted across section AB to a force-couple system at the centroid of the
section. The bending couple is
M = P (a + r )
For the rectangular section, the neutral axis for bending couple only
lies at
R=
Also
h
.
r
ln 2
r1
e=r −R
The maximum compressive stress occurs at point A. It is given by
σA = −
P My A
P P (a + r ) y A
−
=− −
A Aer1
A
Aer1
= −K
Thus,
P
A
with
K =1+
y A = R − r1
(a + r )( R − r1)
er1
h = 30 mm, r1 = 20 mm, r2 = 50 mm, r = 35 mm
Data:
R=
30
= 32.7407 mm, e = 35 − 32.7407 = 2.2593 mm
50
ln
20
b = 30 mm,
A = bh = (30)(30) = 900 mm 2 = 900 × 10−6 m 2
a = 60 mm, a + r = 95 mm, R − r1 = 12.7407 mm
K =1+
(95)(12.7407)
= 27.786
(2.2593)(20)
P = 5 × 103 N
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PROBLEM 4.167 (Continued)
(a)
σA = −
KP
(27.786)(5 × 103 )
=−
= −154.4 × 106 Pa
A
900 × 10−6
σ A = −154.4 MPa 
(b)
At point B,
yB = r2 − R = 50 − 32.7407 = 17.2953 mm
P MyB
P
(a + r ) y B 
+
=  −1 +

A Aer2
A
er2

K ′P
(a + r ) y B
where K ′ =
=
−1
A
er2
σB = −
K′ =
(95)(17.2953)
− 1 = 13.545
(2.2593)(50)
σB =
(13.545)(5 × 103 )
= 75.2 × 106 Pa
−6
900 × 10
σ B = 75.2 MPa 
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PROBLEM 4.168
The curved bar shown has a cross section of 30 × 30 mm. Knowing
that the allowable compressive stress is 175 MPa, determine the largest
allowable distance a.
SOLUTION
Reduce the internal forces transmitted across section AB to a force-couple system at the centroid of the
section. The bending couple is
M = P (a + r )
For the rectangular section, the neutral axis for bending couple only lies at
R=
h
.
r
ln 2
r1
Also e = r − R
The maximum compressive stress occurs at point A . It is given by
σA = −
P My A
P P (a + r ) y A
−
=− −
A Aer1
A
Aer1
P
with y A = R − r1
A
(a + r )( R − r1)
Thus, K = 1 +
er1
= −K
h = 30 mm, r1 = 20 mm, r2 = 50 mm, r = 35 mm, R =
Data:
(1)
30
= 32.7407 mm
ln 50
20
e = 35 − 32.7407 = 2.2593 mm, b = 30 mm, R − r1 = 12.7407 mm, a = ?
σ A = −175 MPa = −175 × 106 Pa, P = 5 kN = 5 × 103 N
K =−
Solving (1) for
a+r =
Aσ A
(900 × 10−6 )(−175 × 106 )
=−
= 31.5
P
5 × 103
a + r,
a+r =
( K − 1)er1
R − r1
(30.5)(2.2593)(20)
= 108.17 mm
12.7407
a = 108.17 mm − 35 mm
a = 73.2 mm 
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PROBLEM 4.169
Steel links having the cross section shown are available with different
central angles β . Knowing that the allowable stress is 12 ksi, determine
the largest force P that can be applied to a link for which β = 90°.
SOLUTION
Reduce section force to a force-couple system at G; the centroid of the cross section AB.
β

a = r 1 − cos 
2

The bending couple is M = − Pa.
For the rectangular section, the neutral axis for bending couple only lies at
R=
h
ln
r2
r1
.
Also e = r − R
At point A the tensile stress is
K =1+
where
P=
Data:
σA =
P My A
P Pay A
P
ay 
P
−
=
+
= 1 + A  = K
A Aer1
A
Aer1
A
er1 
A
ay A
er1
and
y A = R − r1
Aσ A
K
r = 1.2 in., r1 = 0.8 in., r2 = 1.6 in., h = 0.8 in., b = 0.3 in.
0.8
A = (0.3)(0.8) = 0.24 in 2
R = 1.6 = 1.154156 in.
ln 0.8
e = 1.2 − 1.154156 = 0.045844 in.,
y A = 1.154156 − 0.8 = 0.35416 in.
a = 1.2(1 − cos 45°) = 0.35147 in.
K =1+
P=
(0.35147)(0.35416)
= 4.3940
(0.045844)(0.8)
(0.24)(12)
= 0.65544 kips
4.3940
P = 655 lb 
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PROBLEM 4.170
Solve Prob. 4.169, assuming that β = 60°.
PROBLEM 4.169 Steel links having the cross section shown are
available with different central angles β. Knowing that the allowable
stress is 12 ksi, determine the largest force P that can be applied to a
link for which β = 90°.
SOLUTION
Reduce section force to a force-couple system at G, the centroid of the cross section AB.
β

a = r 1 − cos 
2

The bending couple is M = − Pa.
For the rectangular section, the neutral axis for bending couple only lies at
R=
h
r2
r1
ln
.
Also e = r − R
At point A, the tensile stress is
σA =
K =1+
where
P=
Data:
P My A
P Pay A
P
ay 
P
−
=
+
= 1 + A  = K
A Aer1
A
Aer1
A
er1 
A
ay A
er1
and
y A = R − r1
Aσ A
K
r = 1.2 in., r1 = 0.8 in., r2 = 1.6 in., h = 0.8 in., b = 0.3 in.
0.8
A = (0.3)(0.8) = 0.24 in 2
R = 1.6 = 1.154156 in.
ln 0.8
e = 1.2 − 1.154156 = 0.045844 in.
y A = 1.154156 − 0.8 = 0.35416 in.
a = (1.2)(1 − cos 30°) = 0.160770 in.
K =1+
P=
(0.160770)(0.35416)
= 2.5525
(0.045844)(0.8)
(0.24)(12)
= 1.128 kips
2.5525
P = 1128 lb 
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PROBLEM 4.171
A machine component has a T-shaped cross section that is orientated
as shown. Knowing that M = 2500 N ⋅ m, determine the stress at
(a) point A, (b) point B.
SOLUTION
Properties of the cross section.
r , mm
Part
b, mm
R=
A
 bi hi
 Ai
=
=
1
ri +1
r
  dA  bi ln
 bi ln i +1
r
ri
ri
h, mm
40
90
110
1
20
2
60

2200
R=
= 77.852 mm,
28.2588
A, mm 2
bi ln
50
20
ri +1
, mm
ri
r =
 Ai ri
 Ai
ri , mm
1000
16.2186
65
1200
12.0402
100
2200
28.2588
185000
r =
= 84.091 mm, e = r − R = 6.239 mm
2200
Ai ri , mm3
65,000
120,000
185,000
Stresses.
(a)
Point A:
σA =
(b)
M (r − R )
(2.5 kN ⋅ m)(0.040 m − 0.0778517 m)
=
Aer
(2.2 × 10−3 m 2 )(6.2392 × 10−3 m)(0.040 m)
Point B:
σB =
r = rA = 40 mm
σ A = −172.4 MPa 
r = rB = 110 mm
M (r − R )
(2.5 kN ⋅ m)(0.110 m − 0.0778517 m)
=
Aer
(2.2 × 10−3 m 2 )(6.2392 × 10−3 m)(0.110 m)
σ B = 53.2 MPa 
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PROBLEM 4.172
Assuming that the couple shown is replaced by a vertical 10-kN force
attached at point D and acting downward, determine the stress at
(a) point A, (b) point B.
SOLUTION
Properties of the cross section. R =
r , mm
40
90
110
Part
b, mm
1
20
2
60

2200
R=
= 77.852 mm,
28.2588
A
 bi hi
 Ai
=
=
1
r
r
  dA  bi ln i +1
 bi ln i +1
r
ri
ri
h, mm
50
20
A, mm 2
ri +1
, mm
ri
16.2186
12.0402
28.2588
bi ln
1000
1200
2200
185000
r =
= 84.091 mm,
2200
r =
 Ai ri
 Ai
ri , mm
Ai ri , mm3
65
100
65,000
120,000
185,000
e = r − R = 6.239 mm
Force-couple system at the centroid E.
P = 10 kN
M = (10 kN)(100 mm + 84.0909 mm) = 1840.9 N ⋅ m
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PROBLEM 4.172 (Continued)
Stresses.
(a)
Point A:
σA = −
r = rA = 40 mm
P M ( r − R)
10 kN
(1840.9 N ⋅ m)(0.040 m − 0.0778517 m)
+
=−
+
−3
2
A
Aer
2.2 × 10 m
(2.2 × 10−3 m 2 )(6.2392 × 10−3 m)(0.040 m)
= −4.545 MPa − 126.913 MPa
σ A = −131.5 MPa 
(b)
Point B:
σB = −
r = rB = 110 mm
P M ( r − R)
10 kN
(1840.9 N ⋅ m)(0.110 m − 0.0778517 m)
+
=−
+
−3
2
A
Aer
2.2 × 10 m
(2.2 × 10−3 m 2 )(6.2392 × 10−3 m)(0.110 m)
= −4.545 MPa + 39.196 MPa
σ B = 34.7 MPa 
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PROBLEM 4.173
Three plates are welded together to form the curved beam shown. For
the given loading, determine the distance e between the neutral axis
and the centroid of the cross section.
SOLUTION
R=
r =
r
3
3.5
5.5
6
Part
ΣA
Σbi hi
ΣA
=
=
1
r
Σ  r dA Σb ln ri +1
Σbi ln i +1
i
ri
ri
ΣAri
ΣA
b
h
A
b ln
ri +1
ri
r
Ar

3
0.5
1.5
0.462452
3.25
4.875

0.5
2
1.0
0.225993
4.5
4.5

2
0.5
1.0
0.174023
5.75
5.75
3.5
0.862468
Σ
15.125
3.5
15.125
= 4.05812 in., r =
= 4.32143 in.
0.862468
3.5
e = r − R = 0.26331 in.
R=
e = 0.263 in. 
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PROBLEM 4.174
Three plates are welded together to form the curved beam
shown. For M = 8 kip ⋅ in., determine the stress at (a) point A,
(b) point B, (c) the centroid of the cross section.
SOLUTION
R=
r =
r
3
3.5
5.5
6
Part
b
ΣA
Σbi hi
ΣA
=
=
1
r
r
Σ  r dA Σb ln i +1
Σbi ln i +1
i
ri
ri
ΣAri
ΣA
h
A
b ln
r
Ar

3
0.5
1.5
0.462452
3.25
4.875

0.5
2
1.0
0.225993
4.5
4.5

2
0.5
1.0
0.174023
5.75
5.75
3.5
0.862468
Σ
3.5
= 4.05812 in.,
0.862468
e = r − R = 0.26331 in.
R=
(a)
ri +1
ri
15.125
15.125
= 4.32143 in.
3.5
M = −8 kip ⋅ in
r =
y A = R − r1 = 4.05812 − 3 = 1.05812 in.
σA = −
My A
(−8)(1.05812)
=−
Aer1
(3.5)(0.26331)(3)
(b)
yB = R − r2 = 4.05812 − 6 = −1.94188 in. 

σB = −
(c)
yC = R − r = −e
σC = −
MyB
(−8)(−1.94188)

=−
Aer2
(3.5)(0.26331)(6)
MyC
Me
M
−8
=−
=−
=−
Aer
Aer
Ar
(3.5)(4.32143)
σ A = 3.06 ksi 
σ B = −2.81 ksi  
σ C = 0.529 ksi 
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PROBLEM 4.175
The split ring shown has an inner radius r1 = 20 mm and a
circular cross section of diameter d = 32 mm. For the loading
shown, determine the stress at (a) point A, (b) point B.
SOLUTION
1
d = 16 mm, r1 = 20 mm, r2 = r1 + d = 52 mm
2
r = r1 + c = 36 mm
c=
1
1
r + r 2 − c 2  = 36 + 362 − 162 



 2 
2
= 34.1245 mm
R=
e = r − R = 1.8755 mm
A=
π
4
d2 =
π
4
(32)2 = 804.25 mm 2 = 804.25 × 10−6 m 2
P = 2.5 × 103 N
(a)
Point A:
M = Pr = (2.5 × 103 )(36 × 10−3 ) = 90 N ⋅ m
y A = R − r1 = 34.1245 − 20 = 14.125 mm
σA = −
P My A
2.5 × 103
(90)(14.1245 × 10−3 )
−
=−
−
A Aer1
804.25 × 10−6 (804.25 × 10−6 )(1.8755 × 10−3 )(20 × 10−3 )
= −45.2 × 106 Pa
(b)
Point B:
σ A = −45.2 MPa 
yB = R − r2 = 34.1245 − 52 = −17.8755 mm
σB = −
P MyB
2.5 × 103
(90)(−17.8755 × 10−3 )
−
=−
−
A Aer2
804.25 × 10−6 (804.25 × 10−6 )(1.8755 × 10−3 )(52 × 10−3 )
= 17.40 × 106 Pa
σ B = 17.40 MPa 
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without permission.
PROBLEM 4.176
The split ring shown has an inner radius r1 = 16 mm and a
circular cross section of diameter d = 32 mm. For the loading
shown, determine the stress at (a) point A, (b) point B.
SOLUTION
1
d = 16 mm, r1 = 16 mm, r2 = r1 + d = 48 mm
2
r = r1 + c = 32 mm
c=
1
1
r + r 2 − c 2  = 32 + 322 − 162 



 2 
2
= 29.8564 mm
R=
e = r − R = 2.1436 mm
A=
π
4
d2 =
π
4
(32)2 = 804.25 mm 2 = 804.25 × 10−6 m 2
P = 2.5 × 103 N
(a)
Point A:
M = Pr = (2.5 × 103 )(32 × 10−3 ) = 80 N ⋅ m
y A = R − r1 = 29.8564 − 16 = 13.8564 mm
σA = −
P My A
2.5 × 103
(80)(13.8564 × 10−3 )
−
=−
−
A Aer1
804.25 × 10−6 (804.25 × 10−6 )(2.1436 × 10−3 )(16 × 10−3 )
= −43.3 × 106 Pa
(b)
Point B:
σ A = −43.3 MPa 
yB = R − r2 = 29.8564 − 48 = −18.1436 mm
σB = −
P MyB
2.5 × 106
(80)(−18.1436 × 10−3 )
−
=−
−
A Aer2
804.25 × 10−6 (804.25 × 10−6 )(2.1436 × 10−3 )(48 × 10−3 )
= 14.43 × 106 Pa
σ B = 14.43 MPa 
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PROBLEM 4.177
The curved bar shown has a circular cross section of 32-mm
diameter. Determine the largest couple M that can be applied to
the bar about a horizontal axis if the maximum stress is not to
exceed 60 MPa.
SOLUTION
c = 16 mm
r = 12 + 16 = 28 mm
1
r + r 2 − c2 



2
1
=  28 + 282 − 162  = 25.4891 mm

2 
R=
e = r − R = 28 − 25.4891 = 2.5109 mm
σ max occurs at A, which lies at the inner radius.
It is given by σ max =
My A
from which
Aer1
M =
Aer1 σ max
Also,
A = π c 2 = π (16) 2 = 804.25 mm 2
Data:
y A = R − r1 = 25.4891 − 12 = 13.4891 mm
M =
yA
.
(804.25 × 10−6 )(2.5109 × 10−3 )(12 × 10−3 )(60 × 106 )
13.4891 × 10−3
M = 107.8 N ⋅ m 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.178
The bar shown has a circular cross section of 0.6 in. diameter.
Knowing that a = 1.2 in., determine the stress at (a) point A,
(b) point B.
SOLUTION
1
d = 0.3 in.
r = 0.5 + 0.3 = 0.8 in.
2
1
1
R =  r + r 2 − c 2  = 0.8 + 0.82 − 0.32 
 2 

2 
= 0.77081 in.
c=
e = r − R = 0.02919 in.
A = π c 2 = π (0.3)2 = 0.28274 in 2
P = 50 lb
M = − P(a + r ) = −50(1.2 + 0.8) = −100 lb ⋅ in
y A = R − r1 = 0.77081 − 0.5 = 0.27081 in.
yB = R − r2 = 0.77081 − 1.1 = −0.32919 in.
(a)
σA =
P My A
50
(−100)(0.27081)
−
=
−
= 6.74 × 103 psi
A Aer1
0.28274 (0.28274)(0.02919)(0.5)
σ A = 6.74 ksi 
(b)
σB =
P MyB
50
(−100)(−0.32919)
−
=
−
= −3.45 × 103 psi
A Aer2
0.28274 (0.28274)(0.02919)(1.1)
σ B = −3.45 ksi 
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PROBLEM 4.179
The bar shown has a circular cross section of 0.6 in. diameter.
Knowing that the allowable stress is 8 ksi, determine the largest
permissible distance a from the line of action of the 50-lb forces
to the plane containing the center of curvature of the bar.
SOLUTION
1
d = 0.3 in.,
r = 0.5 + 0.3 = 0.8 in.
2
1
1
R =  r + r 2 − c 2  = 0.8 + 0.82 − 0.32 





2
2 
= 0.77081 in.
e = r − R = 0.02919 in.
c=
A = π c 2 = π (0.3)2 = 0.28274 in 2
M = − P (a + r )
y A = R − r1 = 0.77081 − 0.5 = 0.27081 in.
σA =
=
P My A
P P( a + r ) y A
P
(a + r ) y A 
−
=
+
= 1 +

A Aer1
A
Aer1
A
er1

KP
A
where
K =1+
(a + r ) y A
er1
(8 × 103 )(0.28274)
= 45.238
P
50
( K − 1)er1
(44.238)(0.02919)(0.5)
a+r =
=
= 2.384 in.
yA
0.27081
K =
σ AA
=
a = 2.384 − 0.8
a = 1.584 in. 
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PROBLEM 4.180
Knowing that P = 10 kN , determine the stress at (a) point A, (b) point B.
SOLUTION
Locate the centroid D of the cross section.
r = 100 mm +
90 mm
= 130 mm
3
Force-couple system at D.
P = 10 kN
M = Pr = (10 kN)(130 mm) = 1300 N ⋅ m
Triangular cross section.
A=
1
1
bh = (90 mm)(80 mm)
2
2
= 3600 mm 2 = 3600 × 10−6 m 2
1
1
h
(90)
45 mm
2
2
R=
=
=
r2 r2
190 190
ln − 1
ln
− 1 0.355025
h r1
90 100
R = 126.752 mm
e = r − R = 130 mm − 126.752 mm = 3.248 mm
(a)
Point A:
σA = −
rA = 100 mm = 0.100 m
P M (rA − R)
10 kN
(1300 N ⋅ m)(0.100 m − 0.126752 m)
+
=−
+
−6 2
A
AerA
3600 × 10 m
(3600 × 10 −6 m 2 )(3248 × 10−3 m)(0.100 m)
= −2.778 MPa − 29.743 MPa
(b)
Point B:
σB = −
σ A = −32.5 MPa 
rB = 190 mm = 0.190 m
P M (rB − R)
10 kN
(1300 N ⋅ m)(0.190 m − 0.126752 m)
+
=−
+
−6 2
A
AerB
3600 × 10 m
(3600 × 10−6 m 2 )(3.248 × 10−3 m)(0.190 m)
= −2.778 MPa + 37.01 MPa
σ B = +34.2 MPa 
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PROBLEM 4.181
Knowing that M = 5 kip ⋅ in., determine the stress at (a) point A,
(b) point B.
SOLUTION
1
1
bh = (2.5)(3) = 3.75 in 2
2
2
r = 2 + 1 = 3.00000 in.
A=
b1 = 2.5 in., r1 = 2 in., b2 = 0, r2 = 5 in.
Use formula for trapezoid with
b2 = 0.
1 h 2 (b + b )
1
2
2
R=
r2
(b1r2 − b2r1) ln − h(b1 − b2 )
r1
=
(0.5)(3) 2 (2.5 + 0)
= 2.84548 in.
[(2.5)(5) − (0)(2)] ln 52 − (3)(2.5 − 0)
e = r − R = 0.15452 in.
(a)
y A = R − r1 = 0.84548 in.
σA = −
(b)
M = 5 kip ⋅ in
My A
(5) (0.84548)
=−
Aer1
(3.75)(0.15452) (2)
σ A = −3.65 ksi 
yB = R − r2 = −2.15452 in.
σB = −
MyB
(5)(−2.15452)
=−
Aer2
(3.75)(0.15452)(5)
σ B = 3.72 ksi 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.182
Knowing that M = 5 kip ⋅ in, determine the stress at (a) point A,
(b) point B.
SOLUTION
A = 1 (2.5)(3) = 3.75 in 2
2
r = 2 + 2 = 4.00000 in.
b1 = 0, r1 = 2 in.,
Use formula for trapezoid with
b2 = 2.5 in.,
r2 = 5 in.
b1 = 0.
1 h 2 (b + b )
1
2
2
R=
r2
(b1r2 − b2r1) ln − h (b1 − b2 )
r1
(0.5) (3) 2 (0 + 2.5)
= 3.85466 in.
[(0)(5) − (2.5)(2)] ln 5 − (3)(0 − 2.5)
2
e = r − R = 0.14534 in.
M = 5 kip ⋅ in
=
(a)
y A = R − r1 = 1.85466 in.
σA = −
(b)
My A
(5) (1.85466)
=−
Aer1
(3.75)(0.14534) (2)
σ A = −8.51 ksi 
yB = R − r2 = −1.14534 in.
σB = −
MyB
(5) (−1.14534)
=−
Aer2
(3.75)(0.14534) (5)
σ B = 2.10 ksi 
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PROBLEM 4.183
For the curved beam and loading shown, determine the stress at
(a) point A, (b) point B.
SOLUTION
Locate centroid.
A, mm 2
r , mm
Ar , mm3

600
45
27 × 103

300
55
16.5 × 103
Σ
900
r =
R=
=
43.5 × 103
43.5 × 103
= 48.333 mm
900
1
2
h 2 (b1 + b2 )
(b1r2 − b2r1) ln
r2
r1
− h(b2 − b1)
(0.5)(30) 2 (40 + 20)
= 46.8608 mm
− (30)(40 − 20)
[(40)(65) − (20)(35)]ln 65
35
e = r − R = 1.4725 mm
(a)
y A = R − r1 = 11.8608 mm
σA = −
(b)
M = −250 N ⋅ m
My A
(−250)(11.8608 × 10−3 )
=−
= 63.9 × 106 Pa
Aer1
(900 × 10−6 )(1.4725 × 10−3 )(35 × 10−3 )
σ A = 63.9 MPa 
yB = R − r2 = −18.1392 mm
σB = −
MyB
(−250)(−18.1392 × 10−3 )
=−
= −52.6 × 106 Pa
Aer2
(900 × 10−6 )(1.4725 × 10−3 )(65 × 10−3 )
σ B = −52.6 MPa 
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PROBLEM 4.184
For the crane hook shown, determine the largest tensile stress in
section a-a.
SOLUTION
Locate centroid.
A, mm 2
r , mm
Ar , mm3

1050
60
63 × 103

750
80
60 × 103
Σ
1800
r=
Force-couple system at centroid:
103 × 103
103 × 103
= 63.333 mm
1800
P = 15 × 103 N
M = − Pr = −(15 × 103 )(68.333 × 10−3 ) = −1.025 × 103 N ⋅ m
1
2
R=
=
h 2 (b1 + b2 )
(b1r2 − b2 r1 ) ln
r2
r1
− h(b1 − b2 )
(0.5)(60) 2 (35 + 25)
= 63.878 mm
[(35)(100) − (25)(40)]ln 100
− (60)(35 + 25)
40
e = r − R = 4.452 mm
Maximum tensile stress occurs at point A.
y A = R − r1 = 23.878 mm
σA =
=
P My A
−
A Aer1
15 × 103
−(1.025 × 103 )(23.878 × 10−3 )
−
1800 × 10−6 (1800 × 10 −6 )(4.452 × 10−3 )(40 × 10−3 )
= 84.7 × 106 Pa
σ A = 84.7 MPa 
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PROBLEM 4.185
Knowing that the machine component shown has a trapezoidal cross
section with a = 3.5 in. and b = 2.5 in., determine the stress at
(a) point A, (b) point B.
SOLUTION
Locate centroid.
A, in 2
r , in.
Ar , in 3

10.5
6
63

7.5
8
60
Σ
18
r=
R=
=
123
123
= 6.8333 in.
18
1 2
h (b1 + b2 )
2
(b1r2 − b2 r1 ) ln
r2
r1
− h(b1 − b2 )
(0.5)(6) 2 (3.5 + 2.5)
= 6.3878 in.
[(3.5)(10) − (2.5)(4)]ln 104 − (6)(3.5 − 2.5)
e = r − R = 0.4452 in.
(a)
(b)
y A = R − r1 = 6.3878 − 4 = 2.3878 in.
My
(80)(2.3878)
σA = − A = −
(18)(0.4452)(4)
Aer1
yB = R − r2 = 6.3878 − 10 = −3.6122 in.
My
(80)( −3.6122)
σB = − B = −
(18)(0.4452)(10)
Aer2
M = 80 kip ⋅ in
σ A = −5.96 ksi 
σ B = 3.61 ksi 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 4.186
Knowing that the machine component shown has a trapezoidal cross
section with a = 2.5 in. and b = 3.5 in., determine the stress at
(a) point A, (b) point B.
SOLUTION
Locate centroid.
A, in 2
r , in.
Ar , in 3

7.5
6
45

10.5
8
84
Σ
18
r=
R=
=
129
129
= 7.1667 in.
18
1 2
h (b1 + b2 )
2
(b1r2 − b2 r1 ) ln
r2
r1
− h(b1 − b2 )
(0.5)(6)2 (2.5 + 3.5)
= 6.7168 in.
[(2.5)(10) − (3.5)(4)]ln 104 − (6)(2.5 − 3.5)
e = r − R = 0.4499 in.
(a)
y A = R − r1 = 2.7168 in.
σA = −
(b)
M = 80 kip ⋅ in
My A
(80)(2.7168)
=−
(18)(0.4499)(4)
Aer1
σ A = −6.71 ksi 
yB = R − r2 = −3.2832 in.
σB = −
MyB
(80)( −3.2832)
=−
(18)(0.4499)(10)
Aer2
σ B = 3.24 ksi 
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PROBLEM 4.187
Show that if the cross section of a curved beam consists of two or more
rectangles, the radius R of the neutral surface can be expressed as
R=
A
 r   r b2  r b3 
ln  2   3   4  
 r1   r2   r3  


b1
where A is the total area of the cross section.
SOLUTION
R=
=
Note that for each rectangle,

ΣA
A
=
1
Σ dA Σ b ln ri +1
i
r
r

i
A
r 
Σ ln  i +1 
 ri 
bi
=
A
 r b1  r b2  r b3 
ln  2   3   4  
 r1   r2   r3  



ri + 1 dr
1
d A=
bi
ri
r
r
ri + 1 dr
ri +1
= bi
= bi ln
r2 r
ri


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PROBLEM 4.188
Using Eq. (4.66), derive the expression for R given in Fig. 4.73 for a circular cross section.
SOLUTION
Use polar coordinate β as shown. Let w be the width as a function of β
w = 2 c sin β
r = r − c cos β
dr = c sin β d β
dA = w dr = 2c 2 sin 2 β d β

dA
=
r

dA
=
r


=2
=2
2 c 2 sin β
dβ
r − c cos β
π
0
c 2 (1 − cos 2 β )
dβ
r − c cos β
π
0


r 2 − c 2 cos 2 β − (r 2 − c 2 )
dβ
r − c cos β
π
0
π
0
(r + c cos β ) d β − 2(r 2 − c 2 )
= 2r β
π
0
2
+ 2c sin β
2
− 2( r − c )

π
0
dr
r − c cos β
π
0
2
r 2 − c2
tan
−1
( )
r 2 − c 2 tan 1 β
2
r +c
π
0
π

= 2r (π − 0) + 2c(0 − 0) − 4 r 2 − c 2 ⋅  − 0 
2

= 2π r − 2π r 2 − c 2
A = π c2
R=

1
=
2
A
c2
r + r 2 − c2
1
π c2
=
=
×
dA 2π r − 2π r 2 − c 2 2 r − r 2 − c 2 r + r 2 − c 2
r
(
c 2 r + r 2 − c2
2
2
2
r − (r − c )
) = 1 c (r +
2
2
r 2 − c2
c
2
)=1 r+
2(
)
r 2 − c2 

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PROBLEM 4.189
Using Eq. (4.66), derive the expression for R given in Fig. 4.73 for a trapezoidal cross section.
SOLUTION
The section width w varies linearly with r.
w = c0 + c1r
w = b1 at r = r1 and w = b2 at r = r2
b1 = c0 + c1r1
b2 = c0 + c1r2
b1 − b2 = c1 (r1 − r2 ) = −c1h
b −b
c1 = − 1 2
h
r2 b1 − r1b2 = ( r2 − r1 )c0 = hc0
r b − rb
c0 = 2 1 1 2
h
r2 w
r2 c + c r
dA
0
1
dr =
dr
=
r1 r
r1
r
r
r2
r2
= c0 ln r + c1 r



r1
r1
r
= c0 ln 2 + c1 (r2 − r1 )
r1
r2b1 − r1b2 r2 b1 − b2
=
ln −
h
h
r1
h
r b − rb
r
= 2 1 1 2 ln 2 − (b1 − b2 )
h
r1
1
A = (b1 + b2 ) h
2
1 2
h (b1 + b2 )
A
2
R=
=
r
dA
(r2 b1 − r1b2 ) ln r2 − h(b1 − b2 )
r


1
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PROBLEM 4.190
Using Equation (4.66), derive the expression for R given in Fig. 4.73 for a triangular cross section.
SOLUTION
The section width w varies linearly with r.

w = c0 + c1r
w = b at r = r1 and w = 0 at r = r2
b = c0 + c1r1
0 = c0 + c1r2
b = c1 ( r1 − r2 ) = −c1h
br
b
c1 = −
and c0 = −c1r2 = 2
h
h
r
r
2 w
2 c + c r
dA
0
1
dr =
dr
=
r
r
r
r
1 r
1
r2
r2
= c0 ln r + c1 r


r1
r1
r
= c0 ln 2 + c1 (r2 − r1 )
r1
br
r b
= 2 ln 2 − h
h
r1 h
r

br
r
r
= 2 ln 2 − b = b  2 ln 2 − 1
h
r1
 h r1

1
A = bh
2
1
1
bh
h
A
2
R=
=
= r 2r
r
r
2
dA
ln r2 − 1
b h2 ln r2 − 1
h
r
1

(
1
)

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PROBLEM 4.191
For a curved bar of rectangular cross section subjected to a
bending couple M, show that the radial stress at the neutral
surface is
r1
M 
R
σr =
1 − − ln 
Ae 
R
r1 
and compute the value of σ r for the curved bar of Examples
4.10 and 4.11. (Hint: consider the free-body diagram of the
portion of the beam located above the neutral surface.)
SOLUTION
M (r − R) M MR
=
−
Aer
Ae Aer
For portion above the neutral axis, the resultant force is
σr =
At radial distance r,

H = σ r dA =

R
r1
σ r bdr
Mb R
MRb R dr
dr −
Ae r1
Ae r1 r
r1
Mb
MRb R MbR 
R
=
( R − r1 ) −
ln =
1 − − ln 
Ae
Ae
r1
Ae 
R
r1 

=


Fr = σ r cos β dA
Resultant of σ n :
=
θ /2
θ
− /2
σ r cos β (bRd β ) = σ r bR
θ /2
= σ r bR sin β
For equilibrium: Fr − 2 H sin
2σ r bR sin
θ
2
θ
2
−θ /2
θ /2
θ
− /2
cos β d β
= 2σ r bR sin θ
2
=0
−2
θ
MbR  r1
R
1 − − ln  sin = 0
Ae  R
r1 
2
σr =
M  r1
R
1 − − ln 
Ae 
R
r1 

Using results of Examples 4.10 and 4.11 as data,
M = 8 kip ⋅ in,
σr =
A = 3.75 in 2 , R = 5.9686 in., e = 0.0314 in., r1 = 5.25 in.
8
5.25
5.9686 

− ln
1−
(3.75)(0.0314) 
5.9686
5.25 
σ r = −0.54 ksi 
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PROBLEM 4.192
Two vertical forces are applied to a beam of the cross section shown.
Determine the maximum tensile and compressive stresses in portion BC
of the beam.
SOLUTION
A
y0
A y0

18
5
90

18
1
18
Σ
36
Y0 =
108
108
= 3 in.
36
Neutral axis lies 3 in. above the base.
1
1
b1h13 + A1d12 = (3)(6)3 + (18)(2) 2 = 126 in 4
12
12
1
1
I 2 = b2 h23 + A2 d 22 = (9)(2)3 + (18)(2) 2 = 78 in 4
12
12
I = I1 + I 2 = 126 + 78 = 204 in 4
I1 =
ytop = 5 in. ybot = −3 in.
M − Pa = 0
M = Pa = (15)(40) = 600 kip ⋅ in
σ top = −
σ bot = −
M ytop
I
=−
(600)(5)
204
M ybot
(600)(−3)
=−
I
204
σ top = −14.71 ksi (compression) 
σ bot = 8.82 ksi (tension) 
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PROBLEM 4.193
Straight rods of 6-mm diameter and 30-m length are stored by coiling the
rods inside a drum of 1.25-m inside diameter. Assuming that the yield
strength is not exceeded, determine (a) the maximum stress in a coiled rod,
(b) the corresponding bending moment in the rod. Use E = 200 GPa.
SOLUTION
D = inside diameter of the drum.
Let
1
d,
2
ρ = radius of curvature of centerline of rods when bent.
d = diameter of rod,
ρ=
I=
(a)
σ max =
(b)
M=
Ec
EI
ρ
ρ
=
=
c=
1
1
1
1
D − d = (1.25) − (6 × 10−3 ) = 0.622 m
2
2
2
2
π
4
c4 =
π
4
(0.003) 4 = 63.617 × 10−12 m 4
(200 × 109 )(0.003)
= 965 × 106 Pa
0.622
(200 × 109 )(63.617 × 10−12 )
= 20.5 N ⋅ m
0.622
σ = 965 MPa 
M = 20.5 N ⋅ m 
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PROBLEM 4.194
Knowing that for the beam shown the allowable stress is 12 ksi in
tension and 16 ksi in compression, determine the largest couple M that
can be applied.
SOLUTION
 = rectangle
 = semi-circular cutout
A1 = (2.4)(1.2) = 2.88 in 2
A2 =
π
2
(0.75) 2 = 0.8836 in 2
A = 2.88 − 0.8836 = 1.9964 in 2
y1 = 0.6 in.
4r
(4)(0.75)
=
= 0.3183 in.
3π
3π
ΣA y
(2.88)(0.6) − (0.8836)(0.3183)
Y =
=
= 0.7247 in.
ΣA
1.9964
y2 =
Neutral axis lies 0.7247 in. above the bottom.
Moment of inertia about the base:
1
π
1
π
I b = bh3 − r 4 = (2.4)(1.2)3 − (0.75)4 = 1.25815 in 4
3
8
3
8
Centroidal moment of inertia:
I = I b − AY 2 = 1.25815 − (1.9964)(0.7247) 2 = 0.2097 in 4
ytop = 1.2 − 0.7247 = 0.4753 in.,
ybot = −0.7247 in.
|σ | =
My
I
M=
σI
y
Top: (tension side)
M=
(12)(0.2097)
= 5.29 kip ⋅ in
0.4753
Bottom: (compression)
M=
(16)(0.2097)
= 4.63 kip ⋅ in
0.7247
Choose the smaller value.
M = 4.63 kip ⋅ in 
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PROBLEM 4.195
In order to increase corrosion resistance, a 2-mm-thick cladding of
aluminum has been added to a steel bar as shown. The modulus of
elasticity is 200 GPa for steel and 70 GPa for aluminum. For a
bending moment of 300 N ⋅ m, determine (a) the maximum stress
in the steel, (b) the maximum stress in the aluminum, (c) the radius
of curvature of the bar.
SOLUTION
Use aluminum as the reference material.
n = 1 in aluminum.
n=
Es
200
=
= 2.857 in steel.
Ea
70
Cross section geometry:
Steel: As = (46 mm)(26 mm) = 1196 mm 2
Is =
1
(46 mm)(26 mm)3 = 67,375 mm 4
12
Aluminum: Aa = (50 mm)(30 mm) − 1196 mm 2 = 304 mm 2
Ia =
1
(50 mm)(30 mm3 ) − 67,375mm 4 = 45,125mm 4
12
Transformed section.
I = na I a + ns I s = (1)(45,125) + (2.857)(67,375) = 237, 615 mm 4 = 237.615 × 10−9 m 4
Bending moment.
(a)
ys = 13mm = 0.013m
ns Mys
(2.857)(300)(0.013)
=
= 46.9 × 106 Pa
I
237.615 × 10−9
Maximum stress in aluminum:
σa =
(c)
ns = 2.857
Maximum stress in steel:
σs =
(b)
M = 300 N ⋅ m
na = 1,
ya = 15 mm = 0.015 m
na Mya
(1)(300)(0.015)
=
= 18.94 × 106 Pa
I
237.615 × 10−9
Radius of curvative: ρ =
EI
M
ρ =
σ s = 46.9 MPa 
(70 × 109 )(237.615 × 10−9 )
300
σ a = 18.94 MPa 
ρ = 55.4 m 
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PROBLEM 4.196
A single vertical force P is applied to a short steel post as shown.
Gages located at A, B, and C indicate the following strains:
∈A = −500μ
∈B = −1000μ
∈C = −200μ
Knowing that E = 29 × 106 psi, determine (a) the magnitude of
P, (b) the line of action of P, (c) the corresponding strain at the
hidden edge of the post, where x = −2.5 in. and z = −1.5 in.
SOLUTION
1
(5)(3)3 = 11.25 in 4
12
M x = Pz
M z = − Px
Ix =
Iz =
1
(3)(5)3 = 31.25 in 4
12
A = (5)(3) = 15 in 2
x A = −2.5 in., xB = 2.5 in., xC = 2.5 in., xD = −2.5 in.
z A = 1.5 in., z B = 1.5 in., zC = −1.5 in., z D = −1.5 in.
σ A = Eε A = (29 × 106 )(−500 × 10−6 ) = −14,500 psi = −14.5 ksi
σ B = Eε B = (29 × 106 )(−1000 × 10−6 ) = −29,000 psi = −29 ksi
σ C = Eε C = (29 × 106 )(−200 × 10−6 ) = −5800 psi = −5.8 ksi
σA = −
P M x z A M z xA
−
+
= −0.06667 P − 0.13333 M x − 0.08 M z
A
Ix
Iz
(1)
σB = −
P M x z B M z xB
−
+
= −0.06667 P − 0.13333 M x + 0.08 M z
A
Ix
Iz
(2)
σC = −
P M x zC
M x
−
+ z C = −0.06667 P + 0.13333 M x + 0.08 M z
A
Ix
Iz
(3)
Substituting the values for σ A , σ B , and σ C into (1), (2), and (3) and solving the simultaneous equations
gives
M x = 87 kip ⋅ in, M z = −90.625 kip ⋅ in,
(a) P = 152.25 kips 

x=−

z =
Mz
−90.625
=−

P
152.25
Mx
87
=

P
152.25
σD = −

(c)
(b) x = 0.595 in. 
z = 0.571 in. 
P M x z D M z xD
−
+
= −0.06667 P + 0.13333 M x − 0.08 M z
A
Ix
Iz
= −(0.06667)(152.25) + (0.13333)(87) − (0.08)(−90.625) = 8.70 ksi 
Strain at hidden edge:
ε =
σD
E
=
8.70 × 103
29 × 106
 
ε = 300 u 
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PROBLEM 4.197
For the split ring shown, determine the stress at (a) point A,
(b) point B.
SOLUTION
r1 =
1
40 = 20 mm,
2
r2 =
A = (14)(25) = 350 mm 2
1
(90) = 45 mm h = r2 − r1 = 25 mm
2
25
h
R = r = 45 = 30.8288 mm
2
ln 20
ln r
1
1
r = (r1 + r2 ) = 32.5 mm
2
e = r − R = 1.6712 mm
Reduce the internal forces transmitted across section AB to a force-couple system at the centroid of the cross
section. The bending couple is
M = Pa = P r = (2500)(32.5 × 10−3 ) = 81.25 N ⋅ m
(a)
Point A:
rA = 20 mm
y A = 30.8288 − 20 = 10.8288 mm
P My A
2500
(81.25)(10.8288 × 10−3 )
−
=−
−
A AeR
350 × 10−6 (350 × 10−6 )(1.6712 × 10−3 )(20 × 10−3 )
6
= −82.4 × 10 Pa
σA = −
(b)
Point B:
rB = 45 mm
σ A = −82.4 MPa 
yB = 30.8288 − 45 = −14.1712 mm
2500
(81.25)( −14.1712 × 10−3 )
P MyB
−
=−
−
A AerB
350 × 10−6 (350 × 10−6 )(1.6712 × 10−3 )(45 × 10−3 )
= 36.6 × 106 Pa
σB = −
σ B = 36.6 MPa 
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PROBLEM 4.198
A couple M of moment 8 kN ⋅ m acting in a vertical plane is
applied to a W200 × 19.3 rolled-steel beam as shown. Determine
(a) the angle that the neutral axis forms with the horizontal plane,
(b) the maximum stress in the beam.
SOLUTION
For W200 × 19.3 rolled steel sector,
I z′ = 16.6 × 106 mm 4 = 16.6 × 10−6 m 4
I y′ = 1.15 × 106 mm 4 = 1.15 × 10−6 m 4
203
= 101.5 mm
2
102
= zE =
= 51 mm
2
y A = yB = − yD = − yE =
z A = − zB = − zD
M z = (8 × 103 ) cos 5° = 7.9696 × 103 N ⋅ m
M y = −(8 × 103 )sin 5° = −0.6972 × 103 N ⋅ m
(a)
tan ϕ =
Iz
16.6 × 10−6
tan θ =
tan(−5°) = −1.2629
Iy
1.15 × 10−6
ϕ = −51.6°
α = 51.6° − 5°
(b)
α = 46.6° 
Maximum tensile stress occurs at point D.
σD = −
(7.9696 × 103 )(−101.5 × 10−3 ) (−0.6972 × 103 )(−51 × 10−3 )
M z yD M y z D
+
=−
+
Iz
Iy
16.6 × 10−6
1.15 × 10−6
= 79.6 × 106 Pa
σ D = 79.6 MPa 
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PROBLEM 4.199
Determine the maximum stress in each of the two machine
elements shown.
SOLUTION
For each case, M = (400)(2.5) = 1000 lb ⋅ in
At the minimum section,
1
(0.5)(1.5)3 = 0.140625 in 4
12
c = 0.75 in.
I =
(a)
D/d = 3/1.5 = 2
r/d = 0.3/1.5 = 0.2
From Fig 4.32,
σ max =
(b)
KMc
(1.75)(1000)(0.75)
=
= 9.33 × 103 psi
I
0.140625
D/d = 3/1.5 = 2
From Fig. 4.31,
σ max =
K = 1.75
σ max = 9.33 ksi 
r/d = 0.3/1.5 = 0.2
K = 1.50
KMc
(1.50)(1000)(0.75)
=
= 8.00 × 103 psi
I
0.140625
σ max = 8.00 ksi 
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PROBLEM 4.200
The shape shown was formed by bending a thin steel plate. Assuming that the
thickness t is small compared to the length a of a side of the shape, determine the
stress (a) at A, (b) at B, (c) at C.
SOLUTION
Moment of inertia about centroid:
 a 
1
2 2t 

12
 2
1
= ta3
12
(
I=
Area:
)
3
 a 
a
A = 2 2t 
 = 2at , c =
2 2
 2
(
)
(a)
P
P Pec
P
σA = −
=
−
A
I
2at
(b)
P
P Pec
P
σB = +
=
+
2at
A
I
(c)
σC = σ A
( )( )
a
2 2
1
ta
12
a
2 2
3
( )( )
a
2
1
ta 3
12
a
2
σA = −
P

2at
σB = −
2P

at
σC = −
P

2at
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PROBLEM 4.201
Three 120 × 10-mm steel plates have been welded together to form the
beam shown. Assuming that the steel is elastoplastic with
E = 200 GPa and σ Y = 300 MPa, determine (a) the bending moment
for which the plastic zones at the top and bottom of the beam are
40 mm thick, (b) the corresponding radius of curvature of the beam.
SOLUTION
A1 = (120)(10) = 1200 mm 2
R1 = σ Y A1 = (300 × 106 )(1200 × 10− 6 ) = 360 × 103 N
A2 = (30)(10) = 300 mm 2
R2 = σ Y A2 = (300 × 106 )(300 × 10−6 ) = 90 × 103 N
A3 = (30)(10) = 300 mm 2
1
1
R3 = σ Y A2 = (300 × 106 )(300 × 10−6 ) = 45 × 103 N
2
2
y1 = 65 mm = 65 × 10−3 m y2 = 45 mm = 45 × 10−3 m
(a)
M = 2( R1 y1 + R2 y2 + R3 y3 ) = 2{(360)(65) + (90)(45) + (45)(20)}
= 56.7 × 103 N ⋅ m
(b)
y3 = 20 mm = 20 × 10−3 m
yY
ρ
=
σY
E
ρ=
EyY
σY
M = 56.7 kN ⋅ m 
=
(200 × 109 )(30 × 10−3 )
300 × 106
ρ = 20 m 
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PROBLEM 4.202
A short column is made by nailing four 1 × 4-in. planks to a 4 × 4-in.
timber. Determine the largest compressive stress created in the column by a
16-kip load applied as shown in the center of the top section of the timber if
(a) the column is as described, (b) plank 1 is removed, (c) planks 1 and 2 are
removed, (d) planks 1, 2, and 3 are removed, (e) all planks are removed.
SOLUTION
(a)
M =0
Centric loading:
σ =−
P
A
A = (4)(4) + (4)(1)(4) = 32 in 2
σ =−
(b)
Eccentric loading:
M = Pe
σ =−
16 × 103
32
σ = −500 psi 
P Pec
−
A
I
A = (4) (4) + (3)(1)(3) = 28 in 2
e= y
ΣAy (1)(4)(2.5)
y=
=
= 0.35714 in.
28
A
I = Σ( I + Ad 2 ) =
+
1
(4)(1)3 + (4)(1)(2.14286) 2 = 53.762 in 4
12
σ =−
(c)
Centric loading:
1
(6)(4)3 + (6) (4)(0.35714) 2
12
16 × 103 (16 × 103 ) (0.35714) (2.35714)
−
28
53.762
M =0
σ =−
σ = −822 psi 
P
A
A = (6) (4) = 24 in 2
σ =−
16 × 103
24
σ = −667 psi 
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PROBLEM 4.202 (Continued)
(d)
Eccentric loading:
M = Pe
σ =−
P Pec
−
A
I
A = (4) (1) = (1)(4)(1) = 20 in 2
x = 2.5 − 2 = 0.5 in.
1
I = (4)(5)3 = 41.667 in 4
12
σ =−
(e)
Centric loading:
M =0
16 × 103 (16 × 103 ) (0.5) (2.5)
−
20
41.667
σ =−
e=x
σ = −1280 psi 
P
A
A = (4)(4) = 16 in 2
σ =−
16 × 103
16
σ = −1000 psi 
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PROBLEM 4.203
Two thin strips of the same material and same cross
section are bent by couples of the same magnitude
and glued together. After the two surfaces of
contact have been securely bonded, the couples are
removed. Denoting by σ1 the maximum stress and
by ρ1 the radius of curvature of each strip while
the couples were applied, determine (a) the final
stresses at points A, B, C, and D, (b) the final radius
of curvature.
SOLUTION
Let b = width and t = thickness of one strip.
Loading one strip, M = M1
I1 =
1 3
1
bt , c = t
12
2
M 1c σ M1
=
I
bt 2
1
M
12 M1
= 1 =
ρ1 EI1
Et 3
σ1 =
After M1 is applied to each of the strips, the stresses are those given in the sketch above. They are
σ A = −σ1, σ B = σ1, σ C = −σ1, σ D = σ 1
The total bending couple is 2M1.
After the strips are glued together, this couple is removed.
M ′ = 2 M 1, I ′ =
1
2
b(2t )3 = bt 3
12
3
c=t
The stresses removed are
σ′ = −
M ′y
2M y
3M1 y
= − 2 13 = −
I
bt
bt 2
3
σ ′A = −
3M1
1
3M 1 1
= − σ1, σ ′B = σ C′ = 0, σ ′D =
= σ1
2
2
2
bt
bt 2
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PROBLEM 4.203 (Continued)
(a)
Final stresses:
1
2
σ A = −σ1 − (− σ 1)
1
2
σ A = − σ1 
σ B = σ1 
σ C = −σ1 
1
2
σ D = σ1 − σ1
1
ρ′
(b)

=
σD =
1
σ1 
2
M′
2M
3M 1 1 1
= 2 13 =
=
EI ′
4 ρ′
E 3 bt
Et 3
Final radius:
1
ρ
=
1
ρ1
−
1
1
1 1
3 1
=
−
=
ρ ′ ρ1 4 ρ1 4 ρ1
ρ =
4
ρ1 
3
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CHAPTER 5
PROBLEM 5.1
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the equations of the shear and bending-moment
curves.
SOLUTION
Reactions:

From A to B:
ΣM C = 0: LA − bP = 0
A=
Pb
L
ΣM A = 0: LC − aP = 0
C =
Pa
L
0< x<a
ΣFy = 0:
Pb
−V = 0
L

Pb

L
V =
ΣM J = 0: M −
Pb
x=0
L
M =
From B to C:
Pbx

L
a< x< L
ΣFy = 0: V +
Pa
=0
L
V =−
ΣM K = 0: − M +
Pa
( L − x) = 0
L
M =
At section B:
Pa

L
Pa( L − x)

L
M =
Pab

L2
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PROBLEM 5.2
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the equations of the shear and bending-moment
curves.
SOLUTION
Reactions:
ΣM B = 0: − AL + wL ⋅
L
=0
2
A=
wL
2
ΣM A = 0:
L
=0
2
B=
wL
2
BL − wL ⋅
Free body diagram for determining reactions.
Over whole beam,
0< x< L
Place section at x.
Replace distributed load by equivalent concentrated load.
ΣFy = 0:
wL
− wx − V = 0
2
L

V = w − x  
2

ΣM J = 0: −
M =
wL
x
x + wx + M = 0
2
2
w
( Lx − x 2 )
2
M =
Maximum bending moment occurs at x =
w
x( L − x) 
2
L
.
2
M max =
wL2

8
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PROBLEM 5.3
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the equations of the shear and
bending-moment curves.
SOLUTION
From A to B (0 < x < a) :
 Fy = 0 : − wx − V = 0
V = −wx 
 M J = 0 : (wx)
x
+M =0
2
M =−
wx 2

2
From B to C (a < x < L) :
 Fy = 0 : − wa − V = 0
a

 M J = 0 : (wa)  x −  + M = 0
2

V = − wa 
a

M = −wa  x −  
2

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PROBLEM 5.4
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the equations of the shear and bending-moment
curves.
SOLUTION
ΣFy = 0:
ΣM J = 0:

−
1 w0 x
⋅ x −V = 0
2 L
V =−
w0 x 2

2L
M =−
w0 x3

6L
1 w0 x
x
⋅x⋅ +M = 0
2 L
3
At x = L,
V =−
w0 L
2
M =−
w0 L2
6
|V |max =
|M |max =
w0 L

2
w0 L2

6
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PROBLEM 5.5
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the equations of the shear and bending-moment
curves.
SOLUTION
Reactions:
A = D = wa
From A to B:
0< x<a
 Fy = 0 :
wa − wx − V = 0
V = w(a − x) 
MJ = 0 :
− wax + (wx)
x
+M =0
2

x2 
M = w  ax −
 
2 

From B to C:
a< x< L−a
 Fy = 0 :
wa − wa − V = 0
V =0 
MJ = 0 :
From C to D:
 Fy = 0:
a

− wax + wa  x −  + M = 0
2

M =
1 2
wa 
2
L−a< x< L
V − w( L − x) + wa = 0
V = w( L − x − a) 
 M J = 0:
L − x
− M − w( L − x) 
 + wa( L − x) = 0
 2 
M = wa[( L − x) −
1
( L − x) 2 ] 
2
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PROBLEM 5.6
For the beam and loading shown, (a) draw the shear and bending-moment
diagrams, (b) determine the equations of the shear and bending-moment
curves.
SOLUTION
Calculate reactions after replacing distributed load by an equivalent
concentrated load.


Reactions are

A= D=


From A to B:

0< x<a
ΣFy = 0:

1
w ( L − 2a)
2
1
w ( L − 2a) − V = 0
2
V =
1
w ( L − 2a) 
2
1
ΣM = 0: − w ( L − 2a) + M = 0
2
M =
From B to C:
a< x< L−a
b=


ΣFy = 0:

x−a
2
Place section cut at x. Replace distributed load by equivalent concentrated
load.


1
w ( L − 2a ) x 
2
1
w ( L − 2a) − w ( x − a) − V = 0
2
L

V = w − x  
2

1
x − a
M J = 0: − w ( L − 2a) x + w ( x − a) 
+M =0
2
 2 


M =


1
w [( L − 2a) x − ( x − a) 2 ] 
2
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
PROBLEM 5.6 (Continued)



From C to D:

L−a< x< L
ΣFy = 0: V +

1
w ( L − 2a ) = 0
2

V =−


ΣM J = 0: −M +




1
w ( L − 2a)( L − x) = 0
2
M =

At x =
L
,
2
w
( L − 2a ) 
2
1
w ( L − 2a)( L − x) 
2
 L2 a 2 
M max = w 
−
 
2 
 8

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PROBLEM 5.7
Draw the shear and bending-moment diagrams for the beam and
loading shown, and determine the maximum absolute value (a) of the
shear, (b) of the bending moment.
SOLUTION
Reactions:
 M C = 0 : (300)(4) − (240)(3) − (360)(7) + 12 B = 0
 Fy = 0 :
− 300 + C − 240 − 360 + 170 = 0
B = 170 lb ↑
C = 730 lb ↑
From A to C:
 Fy = 0 :
 M1 = 0 :
− 300 − V = 0
V = −300 lb
(300)( x) + M = 0
M = −300 x
From C to D:
 Fy = 0 :
 M2 = 0 :
− 300 + 730 − V = 0
V = +430 lb
(300) x − (730)( x − 4) + M = 0
M = −2920 + 430 x
From D to E:
 Fy = 0 : V − 360 + 170 = 0
 M3 = 0 :
V = +190 lb
(170)(16 − x) − (360)(11 − x) − M = 0
M = −1240 + 190 x
From E to B:
 Fy = 0 : V + 170 = 0
 M4 = 0 :
V = −170lb
(170)(16 − x) − M = 0
M = 2720 − 170 x

(a)
(b)
V
M
max
max
= 430 lb 
= 1200 lb ⋅ in 

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PROBLEM 5.8
Draw the shear and bending-moment diagrams for the beam and
loading shown, and determine the maximum absolute value (a) of the
shear, (b) of the bending moment.
SOLUTION
V = 200N, M = 0
At B,
At E + ,
 Fy = 0 :
V − 200 = 0 V = 200N
ME = 0 :
− M − (0.225)(200) = 0
M = −45 N ⋅ m
At D + ,
 Fy = 0 :
V + 500 − 200 = 0
(a) V = −300 N 
MD = 0 :
− M + (0.3)(500) − (0.525)(200) = 0
M = 45 N ⋅ m

At C + ,
 Fy = 0 :
 MC = 0 :
V − 200 + 500 − 200 = 0 V = −100N
− M − (0.225)(200) + (0.525)(500) − (0.75)(200) = 0
(b) M = 67.5 N ⋅ m 
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PROBLEM 5.8 (Continued)
At A,
 Fy = 0 :
MA = 0 :
V − 200 − 200 + 500 − 200 = 0
V = 100 N
− M − (0.3)(200) − (0.525)(200) + (0.825)(500) − (1.05)(200) = 0
M = 37.5 N ⋅ m
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PROBLEM 5.9
Draw the shear and bending-moment diagrams for the beam and loading
shown, and determine the maximum absolute value (a) of the shear,
(b) of the bending moment.
SOLUTION
Reactions:
 M C = 0 : − 2 A + (1)(24) − (1)(40) = 0
A = −8kN = 8kN ↓
MA = 0 :
2C − (1)(24) − (3)(40) = 0
C = 72 kN = 72 kN ↑
A to C.
0 < x < 2m
 Fy = 0 : − 8 − 12 x − V = 0 V = (−8 − 12 x) kN
MJ = 0 :
x
− 8x − (12 x)   − M = 0
2
M = (−8 x − 6 x 2 ) kN ⋅ m
C to B.
2m < x < 3m
 Fy = 0 : V − 40 = 0
V = 40 kN
MK = 0 :
− M − (3 − x)(40) = 0
M = (40 x − 120) kN ⋅ m
From the diagrams,
(a) V
(b)
M
max
max
= 40.0 kN 
= 40.0 kN ⋅ m 
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PROBLEM 5.10
Draw the shear and bending-moment diagrams for the beam and loading
shown, and determine the maximum absolute value (a) of the shear, (b)
of the bending moment.
SOLUTION
A to C:
0 < x < 4 ft
 Fy = 0 :
− V − 2x = 0
V = −2 x kips
 x
 M J = 0 : M + (2 x)   = 0
2
M = − x kip ⋅ ft
V = −8 kips
At C,
M = −16 kip ⋅ ft
At D − ,
 Fy = 0: − 8 − V = 0
 M D = 0:
V = −8 kips
(6)(8) − M = 0
M = −48 kip ⋅ ft
At B − ,
 Fy = 0 :
− 8 − 15 − V = 0
V = −23 kips
MB = 0 :
− (10)(8) − (4)(15) − M = 0
M = −140 kip ⋅ ft
From the diagrams:

(a) V

(b)
M
max
max
= 23.0 kips 
= 140.0 kip ⋅ ft 
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PROBLEM 5.11
Draw the shear and bending-moment diagrams for the beam and
loading shown, and determine the maximum absolute value (a) of
the shear, (b) of the bending moment.
SOLUTION
Reactions:
 M A = 0 : 3FEF − (8)(60) − (24)(60) = 0
FEF = 640 kips
 Fx = 0 : Ax − 640 = 0
Ax = 640 kips →
 Fy = 0 : Ay − 60 − 60 = 0
Ay = 120 kips ↑
(0 < x < 8 in.)
From A to C:
 Fy = 0 :
120 − V = 0
V = 120 kips
 M J = 0 : M − 120 x = 0
M = 120 x kip ⋅ in
(8 in. < x < 16 in.)
From C to D:
 FY = 0 : 120 − 60 − V = 0 V = 60 kips
MJ = 0 :
M − 120 x + 60( x − 8) = 0
M = (60 x + 480) kips ⋅ in
(16 in. < x < 24 in.)
From D to B:
 Fy = 0 : V − 60 = 0 V = 60 kips
 M J = 0 : − M − 60(24 − x) = 0
M = (60 x − 1440) kip ⋅ in
(a) V
(b)
M
max
max
= 120.0 kips 
= 1440 kip ⋅ in = 120.0 kip ⋅ ft 
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PROBLEM 5.12
Draw the shear and bending-moment diagrams for the beam and loading
shown, and determine the maximum absolute value (a) of the shear, (b) of
the bending moment.
SOLUTION
M B = 0: − 0.750 RA + (0.550) (75) + (0.300) (75) = 0
Reaction at A:
RA = 85 N ↑
RB = 65 N ↑
Also,
A to C :
V = 85 N
C to D :
V = 10 N
D to B :
V = −65 N
At A and B,
M =0
Just to the left of C,
Σ M C = 0: − (0.25) (85) + M = 0
M = 21.25 N ⋅ m
Just to the right of C,
Σ M C = 0:
− (0.25) (85) + (0.050)(75) + M = 0
M = 17.50 N ⋅ m
Just to the left of D,
Σ M D = 0:
−(0.50) (85) + (0.300)(75) + M = 0
M = 20 N ⋅ m
Just to the right of D,
Σ M D = 0:
−M + (0.25) (65) = 0
M = 16.25 kN
(a)
(b)
|V |max = 85.0 N 
|M |max = 21.25 N ⋅ m 

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PROBLEM 5.13
Assuming that the reaction of the ground is uniformly distributed, draw
the shear and bending-moment diagrams for the beam AB and determine
the maximum absolute value (a) of the shear, (b) of the bending
moment.
SOLUTION
Over the whole beam,
ΣFy = 0: 12w − (3)(2) − 24 − (3)(2) = 0
A to C:
w = 3 kips/ft
(0 ≤ x < 3 ft)
ΣFy = 0: 3x − 2 x − V = 0
+ΣM J = 0: − (3x)
At C,
V = ( x) kips
x
x
+ (2 x) + M = 0
2
2
M = (0.5x 2 ) kip ⋅ ft
x = 3 ft
V = 3 kips, M = 4.5 kip ⋅ ft
C to D:
(3 ft ≤ x < 6 ft)
ΣFy = 0: 3x − (2)(3) − V = 0
V = (3x − 6) kips
3
x

ΣMK = 0: −(3x)   + (2)(3)  x −  + M = 0
2
2

M = (1.5 x 2 − 6 x + 9) kip ⋅ ft
At D −,
x = 6 ft
V = 12 kips,
D to B:
M = 27 kip ⋅ ft
Use symmetry to evaluate.
(a)
|V |max = 12.00 kips 
(b)
|M |max = 27.0 kip ⋅ ft 
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PROBLEM 5.14
Assuming that the reaction of the ground is uniformly distributed, draw
the shear and bending-moment diagrams for the beam AB and determine
the maximum absolute value (a) of the shear, (b) of the bending
moment.
SOLUTION
Over the whole beam,
ΣFy = 0: 1.5w − 1.5 − 1.5 = 0
A to C:
w = 2 kN/m
0 ≤ x < 0.3 m
ΣFy = 0: 2 x − V = 0
V = (2 x) kN
 x
ΣM J = 0: −(2 x)   + M = 0
 2
At C −,
M = ( x 2 ) kN ⋅ m
x = 0.3 m
V = 0.6 kN, M = 0.090 kN ⋅ m
= 90 N ⋅ m
C to D:
0.3 m < x < 1.2 m
ΣFy = 0: 2 x − 1.5 − V = 0
V = (2 x − 1.5) kN
x
ΣM J = 0: − (2 x)   + (1.5)( x − 0.3) + M = 0
2
M = ( x 2 − 1.5x + 0.45) kN ⋅ m
At the center of the beam: x = 0.75 m
V =0
M = −0.1125 kN ⋅ m
= −112.5 N ⋅ m
At C +,
x = 0.3 m,
V = −0.9 kN
(a) Maximum |V | = 0.9 kN = 900 N 
(b)
Maximum |M | = 112.5 N ⋅ m 
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PROBLEM 5.15
For the beam and loading shown, determine the maximum
normal stress due to bending on a transverse section at C.
SOLUTION
Reaction at A:
M B = 0:
− 4.5 A + (3.0)(3) + (1.5)(3) + (1.8)(4.5)(2.25) = 0
A = 7.05 kN ↑
Use AC as free body.
ΣM C = 0: M C − (7.05)(1.5) + (1.8)(1.5)(0.75) = 0
M C = 8.55 kN ⋅ m = 8.55 × 103 N ⋅ m
I =
1 3
1
(80)(300)3 = 180 × 106 mm 4
bh =
12
12
= 180 × 10−6 m 4
c=
1
(300) = 150 mm = 0.150 m
2
σ =
Mc (8.55 × 103 )(0.150)
=
= 7.125 × 106 Pa
−6
I
180 × 10
σ = 7.13 MPa 
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PROBLEM 5.16
For the beam and loading shown, determine the maximum normal
stress due to bending on a transverse section at C.
SOLUTION
Use CB as free body.
6
M C = 0: −M − (200)(6)   = 0
2
M = −3600 lb ⋅ ft
= −43.2 × 103 lb ⋅ in
For rectangular section,
I =
1 3
1
bh =
(4)(8)3 = 170.667 in 3
12
12
c=
1
h = 4 in.
2
σ =
|M |c (43.2 × 103 )(4)
=
= 1.0125 × 103 psi
170.667
I
σ = 1.013 ksi 
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PROBLEM 5.17
For the beam and loading shown, determine the maximum
normal stress due to bending on a transverse section at C.
SOLUTION
Use portion CB as free body.
 MC = 0 :
− M + (3)(2.1)(1.05) + (8)(2.1) = 0
M = 23.415 kN ⋅ m = 23.415 × 103 N ⋅ m
For W310 × 60 : S = 844 × 103 mm3
= 844 × 10−6 m3
Normal stress: σ =
M
23.415 × 103
=
= 27.7 × 106 Pa
S
844 × 10−6
σ = 27.7 MPa 

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PROBLEM 5.18
For the beam and loading shown, determine the maximum normal
stress due to bending on section a-a.
SOLUTION
Reactions:
A= B
By symmetry,
 Fy = 0 :
A = B = 80 kN
Using left half of beam as free body,
MJ = 0 :
−(80)(2) + (30)(1.2) + (50)(0.4) + M = 0
M = 104 kN ⋅ m = 104 × 103 N ⋅ m
For
W310 × 52 : S = 747 × 103 mm3
= 747 × 10−6 m3
Normal stress:
σ =
M
104 × 103
=
= 139.2 × 106 Pa
−6
S
747 × 10
σ = 139.2 MPa 

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PROBLEM 5.19
For the beam and loading shown, determine the maximum normal
stress due to bending on a transverse section at C.
SOLUTION
Use entire beam as free body.
MB = 0 :
−90 A + (75)(5) + (60)(5) + (45)(2) + (30)(2) + (15)(2) = 0
A = 9.5 kips
Use portion AC as free body.
 M C = 0 : M − (15)(9.5) = 0
M = 142.5 kip ⋅ in
For S 8 × 18.4, S = 14.4 in 3
Normal stress:
σ =
M
142.5
=
S
14.4
σ = 9.90 ksi 
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PROBLEM 5.20
For the beam and loading shown, determine the maximum
normal stress due to bending on a transverse section at C.
SOLUTION
Use entire beam as free body.
MB = 0 :
−4.8 A + (3.6)(216) + (1.6)(150) + (0.8)(150) = 0
A = 237 kN
Use portion AC as free body.
 MC = 0 :
M − (2.4)(237) + (1.2)(216) = 0
M = 309.6 kN ⋅ m
For W460 × 113, S = 2390 × 106 mm3
Normal stress:
σ =
M
309.6 × 103 N ⋅ m
=
S
2390 × 10−6 m3
= 129.5 × 106 Pa
σ = 129.5 MPa 
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PROBLEM 5.21
Draw the shear and bending-moment diagrams for the beam
and loading shown and determine the maximum normal stress
due to bending.
SOLUTION
MB = 0 :
(11)(25) − 10C + (8)(25) + (2)(25) = 0
C = 52.5 kips
 MC = 0 :
(1)(25) − (2)(25) − (8)(25) + 10B = 0 B = 22.5 kips
Shear:
A to C − :
V = −25 kips
C+ to D−:
V = 27.5 kips
D+ to E−:
V = 2.5 kips
E+ to B:
V = −22.5 kips
Bending moments:
At C,
 M C = 0 : (1)(25) + M = 0
M = −25 kip ⋅ ft
At D,
 M D = 0 : (3)(25) − (2)(52.5) + M = 0
M = 30 kip ⋅ ft
At E,
ME = 0 :
− M + (2)(22.5) = 0 M = 45 kip ⋅ ft
max M = 45 kip ⋅ ft = 540 kip ⋅ in
For S12 × 35 rolled steel section:
Normal stress:
σ =
S = 38.1 in 3
M
540
=
= 14.17 ksi
S
38.1
σ = 14.17 ksi 
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PROBLEM 5.22
Draw the shear and bending-moment diagrams for the beam
and loading shown and determine the maximum normal stress
due to bending.
SOLUTION
Reactions:
 M D = 0 : 4 A − 64 − (24)(2)(1) = 0
A = 28 kN
 Fy = 0 : − 28 + D − (24)(2) = 0 D = 76 kN
A to C:
0 < x < 2m
 Fy = 0 : − V − 28 = 0
V = −28 kN
 M J = 0 : M + 28x = 0
M = (−28 x) kN ⋅ m
C to D:
2m < x < 4m
 Fy = 0 : − V − 28 = 0
V = −28 kN
MJ = 0 :
M + 28x − 64 = 0
M = (−28x + 64) kN ⋅ m
D to B:
4m < x < 6m
 Fy = 0 :
V − 24(6 − x) = 0
V = (−24 x + 144) kN
MJ = 0 :
6 − x
− M − 24(6 − x) 
=0
 2 
M = −12(6 − x) 2 kN ⋅ m
max M = 56 kN ⋅ m = 56 × 103 N ⋅ m
S = 482 × 103 mm3
For S250 × 52 section,
Normal Stress: σ =
M
S
=
56 × 103 N ⋅ m
482 × 10 −6 m 3
= 116.2 × 106 Pa
σ = 116.2 MPa 
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PROBLEM 5.23
Draw the shear and bending-moment diagrams for the beam
and loading shown and determine the maximum normal
stress due to bending.
SOLUTION
Statics: Consider portion AB and BE separately.
Portion BE:
ME = 0 :
(96)(3.6) + (48)(3.3) − C (3) + (160)(1.5) = 0
C = 248kN ↑
E = 56 kN ↑
MA = MB = ME = 0
At midpoint of AB:
 Fy = 0 : V = 0
 M = 0 : M = (96)(1.2) − (96)(0.6) = 57.6 kN ⋅ m
Just to the left of C:
 Fy = 0 : V = −96 − 48 = −144 kN
 M C = 0 : M = −(96)(0.6) − (48)(0.3) = −72 kN
Just to the left of D:
 Fy = 0 : V = 160 − 56 = +104 kN
MD = 0 :
M = (56)(1.5) = +84 kN ⋅ m


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PROBLEM 5.23 (Continued)

From the diagram:

M
max
= 84 kN ⋅ m = 84 × 103 N ⋅ m 
For W310 × 60 rolled steel shape,
S x = 844 × 103 mm3
= 844 × 10−6 m3
Stress: σ m =
σm =
M
max
S
84 × 103
= 99.5 × 106 Pa
844 × 10−6
σ m = 99.5 MPa 
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PROBLEM 5.24
Draw the shear and bending-moment diagrams for the
beam and loading shown and determine the maximum
normal stress due to bending.
SOLUTION
Reaction at A:
 M B = 0 : − 4.8 A + 40 + (25)(3.2)(1.6) = 0
A = 35 kN
A to C:
0 < x < 1.6 m
 Fy = 0 : 35 − V = 0 V = 35 kN
 M J = 0 : M + 40 − 35x = 0
M = (30 x − 40) kN ⋅ m
C to B:
1.6 m < x < 4.8m
 Fy = 0 : 35 − 25( x − 1.6) − V = 0
V = (−25 x + 75) kN
 M K = 0 : M + 40 − 35x
 x − 1.6 
+ (25)( x − 1.6) 
=0
 2 
M = (−12.5x 2 + 75x − 72) kN ⋅ m
Normal stress:
σ =
For W200 × 31.3, S = 298 × 103 mm3
M
40.5 × 103 N ⋅ m
=
= 135.9 × 106 Pa
S
298 × 10−6 m3
σ = 135.9 MPa 
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PROBLEM 5.25
Draw the shear and bending-moment diagrams for the beam
and loading shown, and determine the maximum normal stress
due to bending.
SOLUTION
Reaction at C:
Σ M B = 0: (18)(5) − 13C +(5)(10) = 0
C = 10.769 kips
Reaction at B:
M C = 0: (5)(5) − (8)(10) + 13 B = 0
B = 4.231 kips
Shear diagram:
A to C −:
V = −5 kips
C + to D −:
V = −5 + 10.769 = 5.769 kips
D + to B :
V = 5.769 − 10 = −4.231 kips
At A and B,
M =0
At C,
M C = 0: (5) (5) + M C = 0
M C = −25 kip ⋅ ft
At D,
Σ M D = 0: −M D + (5) (4.231)
M D = 21.155 kip ⋅ ft
V
max
= 5.77 kips 
|M |max = 25 kip ⋅ ft = 300 kip ⋅ in 
|M |max occurs at C.
For W14 × 22 rolled steel section,
S = 29.0 in 3
Normal stress:
σ =
M
300
=
S
29.0
σ = 10.34 ksi 
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PROBLEM 5.26
Knowing that W = 12 kN , draw the shear and bending-moment
diagrams for beam AB and determine the maximum normal
stress due to bending.
SOLUTION
By symmetry, A = B
Σ Fy = 0: A − 8 + 12 − 8 + B = 0
A = B = 2 kN
Shear:
A to C −:
V = 2 kN


C + to D −:
V = −6 kN 


D + to E −:
V = 6 kN 


E + to B :
V = −2 kN 

Bending moment:
Σ M C = 0: M C − (1)(2) = 0
At C,
M C = 2 kN ⋅ m 
At D,
+Σ M D = 0: M D − (2)(2) + (8) (1) = 0
M D − 4 kN ⋅ m 
By symmetry,
M = 2 kN ⋅ m at E.
M E = 2 kN ⋅ m 
max|M | = 4 kN ⋅ m occurs at E.
For W310 × 23.8,
Normal stress:
S x = 280 × 103 mm3 = 280 × 10−6 m3
σ max =
|M |max
4 × 103
=
Sx
280 × 10−6
= 14.29 × 106 Pa
σ max = 14.29 MPa 
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PROBLEM 5.27
Determine (a) the magnitude of the counterweight W for which
the maximum absolute value of the bending moment in the beam
is as small as possible, (b) the corresponding maximum normal
stress due to bending. (Hint: Draw the bending-moment diagram
and equate the absolute values of the largest positive and negative
bending moments obtained.)
SOLUTION
By symmetry,
A=B
Σ Fy = 0: A − 8 + W − 8 + B = 0
A = B = 8 − 0.5W
Bending moment at C:
Σ M C = 0: −(8 − 0.5W )(1) + M C = 0
M C = (8 − 0.5W ) kN ⋅ m
Bending moment at D:
Σ M D = 0: − (8 − 0.5W )(2) + (8) (1) + M D = 0
M D = (8 − W ) kN ⋅ m
−M D = M C
Equate:
W − 8 = 8 − 0.5W
W = 10.67 kN 
(a)
W = 10.6667 kN
M C = −2.6667 kN ⋅ m
M D = 2.6667 kN ⋅ m = 2.6667.103 N ⋅ m
|M |max = 2.6667 kN ⋅ m
For W310 × 23.8 rolled steel shape,
S x = 280 × 103 mm3 = 280 × 10−6 m3
(b)
σ max =
|M |max
2.6667 × 103
=
= 9.52 × 106 Pa
Sx
280 × 10−6
σ max = 9.52 MPa 
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PROBLEM 5.28
Determine (a) the distance a for which the absolute value
of the bending moment in the beam is as small as possible,
(b) the corresponding maximum normal stress due to bending.
(See hint of Prob. 5.27.)
SOLUTION
For W14 × 68,
S x = 103 in 3
Let b = (18 − a) ft
Segment BC:
By symmetry,
VB = C
Σ Fy = 0: VB + C − 4b = 0
VB = 2b
 x
Σ M J = 0: −VB x + (4 x)   − M = 0
2
M = VB x − 2 x 2 = 2bx − 2 x 2 lb ⋅ ft
dM
1
= 2b − xm = 0
xm = b
dx
2
1
1
M max = b 2 − b 2 = b 2
2
2
Segment AB:
( a − x)
2
−VB (a − x) − M = 0
Σ M K = 0: − 4 (a − x)
M = −2(a − x) 2 + 2b (a − x)
|M max | occurs at x = 0.
|M max | = −2a 2 − 2ab = −2a 2 − 2a(18 − a) = 36a
(a)
Equate the two values of |M max |:
36a =
1 2 1
1
b = (18 − a)2 = 162 − 18a + a 2
2
2
2
1 2
a − 54a + 162 = 0
2
a = 54 ± (54) 2 − (4)
a = 54 ± 50.9118 = 3.0883 ft
(b)
( 12 ) (162)
a = 3.09 ft 
|M |max = 36 a = 111.179 kip ⋅ ft = 1334.15 kip ⋅ in
σ =
|M |max 1334.15
=
= 12.95 kips/in 2
103
Sx
σ m = 12.95 ksi 
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PROBLEM 5.29
Determine (a) the distance a for which the absolute value of the
bending moment in the beam is as small as possible, (b) the
corresponding maximum normal stress due to bending. (See hint of
Prob. 5.27.)
SOLUTION
 M C = 0 : 0.8a − (1.5)(1.2) − (2.7)(1.2) + (3.6) B = 0
B = 1.4 − 0.22222 a ↑
 M B = 0 : (0.8)(3.6 + a) − 3.6C + (2.1)(1.2) + (0.9)(1.2) = 0
Bending moment at C:
C = 1.8 + 0.22222a ↑
 M C = 0 : M C + (0.8)(a) = 0
M C = −0.8a
Bending moment at D:
MD = 0 :
M D + (0.8)(a + 1.5) − 1.5C = 0
M D = 1.5 − 0.46667a
Bending moment at E:
 M E = 0 : − M E + 0.9 B = 0
M E = 1.26 − 0.2a
Note that M D < 1.008 kip ⋅ ft
For rolled steel section S3 × 5.7 :
Normal stress:
σ =
Assume − M C = M E :
0.8a = 1.26 − 0.2a
M C = −1.008 kip ⋅ ft
M E = 1.008 kip ⋅ ft
a = 1.26 ft 
M D = 0.912 kip ⋅ ft
max M = 1.008 kip ⋅ ft = 12.096 kip ⋅ in
S = 1.67 in 3
M
12.096
=
S
1.67
σ = 7.24 ksi 
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PROBLEM 5.30
Knowing that P = Q = 480 N, determine (a) the distance a
for which the absolute value of the bending moment in the
beam is as small as possible, (b) the corresponding maximum
normal stress due to bending. (See hint of Prob. 5.27.)
SOLUTION
P = 480 N
Q = 480 N
Σ M D = 0: − Aa + 480(a − 0.5)
Reaction at A:
− 480(1 − a) = 0
720 

A =  960 −
N
a 

Bending moment at C:
Σ M C = 0: − 0.5 A + M C = 0
360 

M C = 0.5 A =  480 −
N⋅m
a 

Bending moment at D:
Σ M D = 0: − M D − 480 (1 − a) = 0
M D = −480 (1 − a) N ⋅ m
(a)
Equate:
−M D = M C
480 (1 − a) = 480 −
360
a
a = 0.86603 m
A = 128.62 N
(b)
For rectangular section, S =
S =
M C = 64.31 N ⋅ m
a = 866 mm 
M D = −64.31 N ⋅ m
1 2
bh
6
1
(12)(13)2 = 648 mm3 = 648 × 10−9 m3
6
σ max =
|M |max
64.31
=
= 99.2 × 106 Pa
S
6.48 × 10−9
σ max = 99.2 MPa 
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PROBLEM 5.31
Solve Prob. 5.30, assuming that P = 480 N and Q = 320 N.
PROBLEM 5.30 Knowing that P = Q = 480 N, determine
(a) the distance a for which the absolute value of the bending
moment in the beam is as small as possible, (b) the
corresponding maximum normal stress due to bending. (See
hint of Prob. 5.27.)
SOLUTION
P = 480 N
Reaction at A:
Q = 320 N
Σ M D = 0: Aa + 480(a − 0.5) − 320(1 − a) = 0
560 

A =  800 −
N
a 

Bending moment at C:
Σ M C = 0: − 0.5 A + M C = 0
280 

M C = 0.5 A =  400 −
 N⋅m
a 

Bending moment at D:
Σ M D = 0: −M D − 320(1 − a) = 0
M D = (−320 + 320 a) N ⋅ m
(a)
−M D = M C
Equate:
320 a 2 + 80a − 280 = 0
320 − 320 a = 400 −
280
a
a = 0.81873 m, − 1.06873 m
a = 819 mm 
Reject negative root.
A = 116.014 N
(b)
M C = 58.007 N ⋅ m
M D = −58.006 N ⋅ m
1 2
bh
6
1
S = (12)(18) 2 = 648 mm3 = 648 × 10−9 m3
6
For rectangular section, S =
σ max =
|M |max
58.0065
=
= 89.5 × 106 Pa
S
648 × 10−9
σ max = 89.5 MPa 
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PROBLEM 5.32
A solid steel bar has a square cross section of side b and is
supported as shown. Knowing that for steel ρ = 7860 kg / m3 ,
determine the dimension b for which the maximum normal stress
due to bending is (a) 10 MPa, (b) 50 MPa.
SOLUTION
Weight density: γ = ρ g
Let L = total length of beam.
W = AL ρ g = b 2 L ρ g
Reactions at C and D:
C = D=
W
2
Bending moment at C:
 L  W 
Σ M C = 0:    + M = 0
 6  3 
WL
M =−
18
Bending moment at center of beam:
 L  W   L  W 
Σ M E = 0:    −    + M = 0
 4  2   6  2 
max|M | =
S =
For a square section,
Normal stress:
Solve for b:
Data:
σ =
M =−
WL
24
WL b 2 L2 ρ g
=
18
18
1 3
b
6
|M | b 2 L2 ρ g /18 L2 ρ g
=
=
S
3b
b3 /6
b=
L = 3.6 m ρ = 7860 kg/m3
L2 ρ g
3σ
g = 9.81 m/s 2
(a) σ = 10 × 106 Pa
(b) σ = 50 × 106 Pa
(a)
b=
(3.6) 2 (7860) (9.81)
= 33.3 × 10−3 m
6
(3) (10 × 10 )
b = 33.3 mm 
(b)
b=
(3.6) 2 (7860) (9.81)
= 6.66 × 10−3 m
(3) (50 × 106 )
b = 6.66 mm 
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PROBLEM 5.33
A solid steel rod of diameter d is supported as shown. Knowing that for
steel γ = 490 lb/ft 3 , determine the smallest diameter d that can be
used if the normal stress due to bending is not to exceed 4 ksi.
SOLUTION
Let W = total weight.
W = ALγ =
π
d 2 Lγ
4
Reaction at A:
A=
1
W
2
Bending moment at center of beam:
 W  L   W  L 
Σ M C = 0: −    +    + M = 0
 2  2   2  4 
π 2 2
WL
=
M =
d Lγ
8
32
(
For circular cross section, c = 1 d
2
)
I =
π
4
c4,
S =
I
π
π 3
= c3 =
d
c
4
32
Normal stress:
σ =
Solving for d,
Data:
d =
M
=
S
d 2 L2γ
π
32
π
32
d3
=
L2γ
d
L2γ
σ
L = 10 ft = (12)(10) = 120 in.
γ = 490 lb/ft 3 =
490
= 0.28356 lb/in 3
123
σ = 4 ksi = 4000 lb/in 2
d =
(120) 2 (0.28356)
4000
d = 1.021 in. 
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PROBLEM 5.34
Using the method of Sec. 5.3, solve Prob. 5.1a.
PROBLEM 5.1 For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the equations of the shear and
bending-moment curves.
SOLUTION
At A+,
ΣM C = 0: LA − bP = 0
A=
Pb
L
ΣM A = 0: LC − aP = 0
C =
Pa
L
V = A=
A to B − :
Pb
L
M =0
0< x<a
x
0 wdx = 0
w=0
V − VA = 0
a Pb
a
M B − M A = 0 Vdx =  0
At B +,
V = A−P =
B + to C:
Pb

L
V =
L
dx =
Pba
L
MB =
Pba

L
Pb
Pa
−P=−
L
L
a< x< L
x
a wdx = 0
w=0
VC − VB = 0
V =−
Pa

L
Pa
Pab
( L − a) = −
L
L
Pab
Pba Pab
= MB −
=
−
=0
L
L
L
L
M C − M B = a Vdx = −
MC
|M | max =
Pab

L
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PROBLEM 5.35
Using the method of Sec. 5.3, solve Prob. 5.2a.
PROBLEM 5.2 For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the equations of the shear and
bending-moment curves.
SOLUTION
Σ M B = 0: −AL + wL ⋅
Σ M A = 0: BL − wL ⋅
L
=0
2
L
=0
2
A=
wL
2
B=
wL
2
dV
= −w
dx
x
V − VA = −0 wdx = − wx
V = VA − wx = A − wx
V =
wL
− wx 
2
dM
=V
dx
x
x  wL

M − M A = 0 Vd x = 0 
− wx  dx
2


=
wLx wx 2
−
2
2
M = MA +
Maximum M occurs at x =
V =
wLx wx 2
−
2
2
M =
w
( Lx − x 2 ) 
2
1
, where
2
dM
=0
dx
|M |max =
wL2

8
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PROBLEM 5.36
Using the method of Sec. 5.3, solve Prob. 5.3a.
PROBLEM 5.3 For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the equations of the shear and
bending-moment curves.
SOLUTION
Over AB:
VA = 0 M A = 0
x
V = − 0 wdx = −wx
dM
= V = −wx
dx
x
Vdx
0
M =
At B:
x=a
Over BC:
w=0
dV
=0
dx
wx 2
=−
2
x
M =−
wx 2

2
MB = −
wa 2

2
0
VB = −wa
V = constant = VB
V = − wa 
dM
= V = − wa
dx
x
x
M − M B = a Vdx = − wax a = −wa( x − a)
M = − wa( x − a) −
wa 2
2
a

M = −wa  x −  
2

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PROBLEM 5.37
Using the method of Sec. 5.3, solve Prob. 5.4a.
PROBLEM 5.4 For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the equations of the shear and
bending-moment curves.
SOLUTION
x
L
w = w0
VA = 0,
MA = 0
dV
Wx
= −w = − 0
dx
L
x
V − VA = −0
w0 x
w x2
=− 0
L
2L





V =−
w0 x 2

2L
M =−
w0 x3

6L
dM
w x2
=V = − 0
2L
dx
x
x
M − M A = 0 V dx = −0


w0 x 2
dx
2L

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PROBLEM 5.38
Using the method of Sec. 5.3, Solve Prob. 5.5a.
PROBLEM 5.5 For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the equations of the shear and
bending-moment curves.
SOLUTION
Reactions:
A = D = wa
A to B:
0< x<a
w=w
VA = A = wa,
MA = 0
x
V − VA = − 0 w dx = −wx
V = w(a − x) 
dM
= V = wa − wx
dx
M − MA =
x
x
0 Vdx = 0 (wa − wx)dx
M = wax −
VB = 0
B to C:
MB =
1 2
wx 
2
1 2
wa
2
a< x< L−a
V =0 
dM
=V =0
dx
M − MB =
M = MB
x
a V dx = 0
M =
1 2
wa 
2






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PROBLEM 5.38 (Continued)
C to D:
x
V − VC = −L − a w dx = −w[ x − ( L − a)]
V = −w[ x − ( L − a)] 
M − MC =
x
x
L − a V dx = L − a −[wx − (L − a)]dx
 x2

= − w  − ( L − a) x 
2

x
L−a
 x2

( L − a) 2
= − w  − ( L − a) x −
+ ( L − a) 2 
2
2

 x2
( L − a)2 
= − w  − ( L − a) x +

2
2

M =
 x2
1 2
( L − a) 2 
wa − w  − ( L − a) x +

2
2
2


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PROBLEM 5.39
Using the method of Sec. 5.3, solve Prob. 5.6a.
PROBLEM 5.6 For the beam and loading shown, (a) draw the shear and
bending-moment diagrams, (b) determine the equations of the shear and
bending-moment curves.
SOLUTION
Reactions.
A= D =
1
w( L − 2a)
2
At A.
VA = A =
1
w( L − 2a), M A = 0
2
A to B.
0< x<a












VB = VA =
1
w( L − 2a)
2
MB =

B to C.
a
VB − VA = −
MB − MA =


w=0
0

a
0
w dx = 0
V dx =
0
1
w( L − 2a)dx
2
1
w( L − 2a)a
2
a< x< L−a
V − VB = −
V =

a

x
a
w=w
w dx = − w( x − a)
1
1
w( L − 2a) − w( x − a) = w( L − 2 x)
2
2
dM
1
= V = w( L − 2 x)
dx
2
M − MB =
=
M =
=
1
x
x
2
a V dx = 2 w ( Lx − x ) a
1
w( Lx − x 2 − La + a 2 )
2
1
1
w( L − 2a)a + w( Lx − x 2 − La + a 2 )
2
2
1
w( Lx − x 2 − a 2 )
2
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PROBLEM 5.39 (Continued)
1
1
VC = − w( L − 2a) M C = ( L − 2a)a
2
2
At C.
x= L−a
C to D.
1
V = VC = − w( L − 2a)
2
MD = 0
At x =
L
,
2
 L2 a 2 
M max = w 
−
 
2 
 8

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PROBLEM 5.40
Using the method of Sec. 5.3 solve Prob. 5.7.
PROBLEM 5.7 Draw the shear and bending-moment diagrams for the
beam and loading shown, and determine the maximum absolute value (a)
of the shear, (b) of the bending moment.
SOLUTION
Reaction at C:
 M B = 0 : (16)(300) − 12C + (9)(240) + (5)(360) = 0
C = 730 lb ↑
Shear diagram:
A to C:
V = −300 lb
C to D:
V = −300 + 730 = 430 lb
D to E:
V = 430 – 240 = 190 lb
E to B:
V = 190 – 360 = −170 lb
Areas of shear diagram:
A to C:
AAC = (−300)(4) = −1200 lb ⋅ in
C to D:
ACD = (430)(3) = 1290 lb ⋅ in
D to E:
ADE = (190)(4) = 760lb ⋅ in
E to B:
AEB = (−170)(5) = −850lb ⋅ in
Bending moments:
MA = 0
M C = 0 − 1200 = −1200 lb ⋅ in
M D = −1200 + 1290 = 90 lb ⋅ in
M E = 90 + 760 = 850 lb ⋅ in
M B = 850 – 850 = 0
(a)
Maximum V = 430 lb 
(b)
Maximum M = 1200 lb ⋅ in 
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PROBLEM 5.41
Using the method of Sec. 5.3, Solve Prob. 5.8
PROBLEM 5.8 Draw the shear and bending-moment diagram for the
beam and loading shown, and determine the maximum absolute value
(a) of the shear, (b) of the bending moment.
SOLUTION
 M A = 0 : − M A − (0.3)(200) − (0.525)(200)
+ (0.825)(500) − (1.05)(200) = 0
M A = 37.5 N ⋅ m
 Fy = 0 : VA − 200 − 200 + 500 − 200 = 0
VA = 100 N
Shear:
A to C:
V = 100 N
C to D:
V = 100 − 200 = −100 N
D to E:
V = −100 − 200 = −300 N
E to B:
V = −300 + 500 = 200 N
Areas under shear diagram:
A to C:
 Vdx = (100)(0.3) = 30 N ⋅ m
C to D:
 Vdx = (−100)(0.225) = −22.5 N ⋅ m
D to E:
 Vdx = (−300)(0.3) = −90 N ⋅ m
E to B:
 Vdx = (200)(0.225) = 45 N ⋅ m
Bending moments:
M A = 37.5 N ⋅ m
MC = M A +
C
A V dx = 37.5 + 30 = 67.5 N ⋅ m
M D = MC +
D
C V dx = 67.5 − 22.5 = 45 N ⋅ m
ME = MD +
E
D V dx = 45 − 90 = −45 N ⋅ m
MB = ME +
D
E V dx = −45 + 45 = 0
(a) Maximum V = 300 N 
(b) Maximum M = 67.5 N ⋅ m 
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PROBLEM 5.42
Using the method of Sec. 5.3, Solve Prob. 5.9
PROBLEM 5.9 Draw the shear and bending-moment diagrams for the
beam and loading shown, and determine the maximum absolute value
(a) of the shear, (b) of the bending moment.
SOLUTION
Reactions:
 M C = 0 : 2 A + (12)(2)(1) − (40)(1) = 0
A = 8kN ↓
 M A = 0 : 2C − (12)(2)(1) − (40)(3) = 0
C = 72 kN ↑
Shear diagram: VA = −8 kN
A to C:
0 < x < 2 m w = 12kN/m
VC − VA = −

2
0

2
wdx = − 12dx = −24 kN
0
VC = −24 − 8 = −32 kN
C to B:
VB = −32 + 72 = 40 kN
Areas of shear diagram:
1
( −8 − 32)(2) = −40 kN ⋅ m
2
A to C:
 Vdx =
C to B:
 Vdx = (1)(40) = 40 kN ⋅ m
Bending moments:
MA = 0
M C = M A +  Vdx = 0 − 40 = −40 kN ⋅ m
M B = M C +  Vdx = −40 + 40 = 0
(a) Maximum V = 40.0 kN 
(b) Maximum M = 40.0 kN ⋅ m 
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PROBLEM 5.43
Using the method of Sec. 5.3, solve Prob. 5.10
PROBLEM 5.10 Draw the shear and bending-moment diagrams for the
beam and loading shown, and determine the maximum absolute value
(a) of the shear, (b) of the bending moment.
SOLUTION
Shear:
VA = 0
B
VB = VA − A wdx = 0 − (4)(2) = −8 kips
C to D: V = −8 kips
D to B: V = −8 − 15 = −23 kips
Areas under shear diagram:
A to C:
1
 Vdx =   (4)(−8) = −16 kip ⋅ ft
2
C to D:
 Vdx = (4)(−8) = −32 kip ⋅ ft
D to B:
 Vdx = (4)(−23) = −92 kip ⋅ ft
Bending moments:
MA = 0
M C = M A +  Vdx = 0 − 16 = −16 kip ⋅ ft
M D = M C +  Vdx = −16 − 32 = −48 kip ⋅ ft
M B = M D +  Vdx = −48 − 92 = −140 kip ⋅ ft
(a) Maximum V = 23 kips 
(b) Maximum M = 140 kip ⋅ ft 

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PROBLEM 5.44
Draw the shear and bending-moment diagrams for the beam and loading
shown, and determine the maximum absolute value (a) of the shear, (b) of
the bending moment.
SOLUTION
Reaction at A:
ΣM B = 0: −3.0 A + (1.5)(3.0)(3.5) + (1.5)(3) = 0
A = 6.75 kN ↑
B = 6.75 kN ↑
Reaction at B:
Beam ACB and loading: (See sketch.)
Areas of load diagram:
A to C:
(2.4)(3.5) = 8.4 kN
C to B:
(0.6)(3.5) = 2.1 kN
Shear diagram:
VA = 6.75 kN
VC − = 6.75 − 8.4 = −1.65 kN
VC + = −1.65 − 3 = −4.65 kN
VB = −4.65 − 2.1 = −6.75 kN
Over A to C,
V = 6.75 − 3.5x
At G,
V = 6.75 − 3.5xG = 0 xG = 1.9286 m
Areas of shear diagram:
A to G:
1
(1.9286)(6.75) = 6.5089 kN ⋅ m
2
G to C:
1
(0.4714)(−1.65) = −0.3889 kN ⋅ m
2
C to B:
1
(0.6)(−4.65 − 6.75) = −3.42 kN ⋅ m
2
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PROBLEM 5.44 (Continued)
Bending moments:
MA = 0
M G = 0 + 6.5089 = 6.5089 kN ⋅ m
M C − = 6.5089 − 0.3889 = 6.12 kN ⋅ m
M C + = 6.12 − 2.7 = 3.42 kN ⋅ m
M B = 3.42 − 3.42 = 0
(a)
(b)
|V |max = 6.75 kN 
|M |max = 6.51 kN ⋅ m 
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PROBLEM 5.45
Draw the shear and bending-moment diagrams for the beam and loading
shown, and determine the maximum absolute value (a) of the shear, (b) of
the bending moment.
SOLUTION
 M B = 0:
− 3 A + (1)(4) + (0.5)(4) = 0
A = 2 kN ↑
 M A = 0: 3B − (2)(4) − (2.5)(4) = 0
B = 6 kN ↑
Shear diagram:
A to C:
V = 2 kN
C to D:
V = 2 − 4 = −2 kN
D to B:
V = −2 − 4 = −6 kN
Areas of shear diagram:
 Vdx = (1)(2) = 2 kN ⋅ m
 Vdx = (1)(−2) = −2 kN ⋅ m
 Vdx = (1)(−6) = −6 kN ⋅ m
A to C:
C to D:
D to E:
Bending moments:
MA = 0
M C − = 0 + 2 = 2 kN ⋅ m
M C + = 2 + 4 = 6 kN ⋅ m
M D − = 6 − 2 = 4 kN ⋅ m
M D + = 4 + 2 = 6 kN ⋅ m
MB = 6 − 6 = 0
(a)
(b)
V
M
max
max
= 6.00 kN 
= 6.00 kN ⋅ m 
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PROBLEM 5.46
Using the method of Sec. 5.3, solve Prob. 5.15
PROBLEM 5.15 For the beam and loading shown, determine the
maximum normal stress due to bending on a transverse section at C.
SOLUTION
By symmetry, A = B.
ΣFy = 0: A + B − 3 − 3 − (4.5)(1.8) = 0
A = B = 7.05 kN
VA = 7.05 kN
Shear diagram:
A to C − :
w = 1.8 kN/m
−
V = 7.05 − (1.8)(1.5) = 4.35 kN
At C ,
At C +,
+
V = 4.35 − 3 = 1.35 kN
−
w = 1.8 kN/m
C to D :
−
At D ,
V = 1.35 − (1.5)(1.8) = −1.35 kN
At D +,
V = −1.35 − 3 = −4.35 kN
D + to B:
w = 1.8 kN
At B,
V = −4.35 − (1.5)(1.8) = −7.05 kN
Draw the shear diagram:
V = 0 at point E, the midpoint of CD.
Areas of the shear diagram:
A to C:
1
(7.05 + 4.35)(1.5) = 8.55 kN ⋅ m
2
C to E:
1
(1.35)(0.75) = 0.50625 kN ⋅ m
2
E to D:
1
(−1.35)(0.75) = −0.50625 kN ⋅ m
2
D to B:
1
(−4.35 − 7.05)(1.5) = −8.55 kN ⋅ m
2
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PROBLEM 5.46 (Continued)
For a rectangular section,
Bending moments:
MA = 0
M C = 0 + 8.55 = 8.55 kN ⋅ m
M E = 8.55 + 0.50625 = 9.05625 kN ⋅ m
M D = 9.05625 − 0.50625 = 8.55 kN ⋅ m
M B = 8.55 − 8.55 = 0
3
M C = 8.55 × 10 N ⋅ m
1
1
S = bh 2 =   (80)(300) 2
6
6
= 1.2 × 106 mm3 = 1.2 × 10−3 m3
Maximum normal stress at C:
M C 8.55 × 103
=
S
1.2 × 10−3
= 7.125 × 106 Pa
σ=
σ = 7.13 MPa 

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PROBLEM 5.47
Using the method of Sec. 5.3, solve Prob. 5.16.
PROBLEM 5.16 For the beam and loading shown, determine the
maximum normal stress due to bending on a transverse section at C.
SOLUTION
ΣM C = 0: −8 A + (4)(2000) − (3)(6)(200) = 0
A = 550 lb ↑
ΣM A = 0: 8C − (4)(2000) − (11)(6)(200) = 0
C = 2650 lb ↑
Check:
ΣFy = 550 − 2000 + 2650 − (6)(200) = 0
Shear diagram:
V = 550 lb
A to D:
−
V = 550 − 2000 = −1450 lb
D to C :
+
V = −1450 + 2650 = 1200 lb
At C ,
+
w = 200 lb/ft
C to B:

At B,
14
8
200 dx = 1200 lb
V = 0 as expected.
Areas of shear diagram:
A to D:
AAD = (550)(4) = 2200 lb ⋅ ft
D to C:
ADC = (−1450)(4) = −5800 lb ⋅ ft
C to B:
ACB =
Bending moments:
MA = 0
1
(1200)(6) = 3600 lb ⋅ ft
2
M D = 0 + 2200 = 2200 lb ⋅ ft
M C = 2200 + (−5800) = −3600 lb ⋅ ft
M B = −3600 + 3600 = 0 as expected
|M |max = 3600 lb ⋅ ft = 43200 lb ⋅ in
For a rectangular section,
1
1
S = bh 2 =   (4)(8) 2 = 42.667 in 3
6
6
σ max =
|M |max 43200
=
= 1012.5 psi
S
42.667
σ max = 1013 psi 
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PROBLEM 5.48
Using the method of Sec. 5.3, solve Prob. 5.18.
PROBLEM 5.18 For the beam and loading shown, determine
the maximum normal stress due to bending on section a-a.
SOLUTION
Reactions:
By symmetry, A = B.
 Fy = 0 : A = B = 80 kN ↑
Shear diagram:
A to C:
V = 80 kN
C to D:
V = 80 − 30 = 50 kN
D to E:
V = 50 − 50 = 0
Areas of shear diagram:
A to C:
 Vdx = (80)(0.8) = 64 kN ⋅ m
C to D:
 Vdx = (50)(0.8) = 40 kN ⋅ m
D to E:
 Vdx = 0
Bending moments:
MA = 0
M C = 0 + 64 = 64 kN ⋅ m
M D = 64 + 40 = 104 kN ⋅ m
M E = 104 + 0 = 104 kN ⋅ m
M
max
= 104 kN ⋅ m = 104 × 103 N ⋅ m
For W310 × 52, S = 747 × 103 mm3 = 747 × 10−6 m3
Normal stress:
σ =
M
104 × 103
=
= 139.2 × 106 Pa
S
747 × 10−6
σ = 139.2 MPa 

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PROBLEM 5.49
Using the method of Sec. 5.3, solve Prob.5.19.
PROBLEM 5.19 For the beam and loading shown, determine the
maximum normal stress due to bending on a transverse section at C.
SOLUTION
Use entire beam as free body.
MB = 0 :
−90 A + ( 75 )( 5) + ( 60 )( 5 ) + ( 45 )( 2 ) + ( 30 )( 2 )
+ (15)( 2 ) = 0
A = 9.5 kips ↑
Shear A to C:
V = 9.5 kips
 Vdx = (15)(9.5)
Area under shear curve A to C:
= 142.5 kip ⋅ in
MA = 0
M C = 0 + 142.5 = 142.5 kip ⋅ in
For S8 × 18.4, S = 14.4in 3
Normal stress:
σ =
M
142.5
=
S
14.4
σ = 9.90 ksi 

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PROBLEM 5.50
For the beam and loading shown, determine the equations of the shear
and bending-moment curves, and the maximum absolute value of the
bending moment in the beam, knowing that (a) k = 1, (b) k = 0.5.
SOLUTION
w0 x kw0 ( L − x)
wx
−
= (1 + k ) 0 − kw.
L
L
L
w0 x
= − w = kw0 − (1 + k )
L
2
wx
= kw0 x − (1 + k ) 0 + C1
2L
= 0 at x = 0 C1 = 0
w=
dV
dx
V
V
w x2
dM
= V = kw0 x − (1 + k ) 0
dx
2L
kw0 x 2
w x3
− (1 + k ) 0 + C2
2
6L
M = 0 at x = 0 C2 = 0
M=
M=
(a)
w0 x 2

L
w x 2 w x3
M= 0 − 0

2
3L
k = 1.
V = w0 x −
x = L.
Maximum M occurs at
(b)
kw0 x 2 (1 + k ) w0 x3
−
2
6L
k=
M
1
.
2
V = 0 at

At
2
x = L,
3
At
x = L,
M=
x=
w0 ( 23 L )
4
2
−
max
=
w0 L2

6
V=
w0 x 3w0 x 2
−

2
4L
M=
w0 x 2 w0 x3
−

4
4L
2
L
3
w0 ( 23 L )
4L
3
=
w0 L2
= 0.03704 w0 L2
27
M =0
|M |max =
w0 L2

27
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PROBLEM 5.51
Determine (a) the equations of the shear and bending-moment curves for
the beam and loading shown, (b) the maximum absolute value of the
bending moment in the beam.
SOLUTION
dV
x
= − w = − w0
dx
L
2
1
x
dM
+ C1 =
V = − w0
2
L
dx
3
1
x
M = − w0
+ C1 x + C2
6
L
(a)
M = 0 at
x=0
C2 = 0
M = 0 at
x=L
1
0 = − w0 L2 + C1 L
6
1
x2 1
V = − w0
+ w0 L2
2
L 6
M max occurs when
1
w0 L
6
V=
1
x3 1
M = − w0
+ w0 Lx
L 6
6
(b)
C1 =
1
w0 ( L2 − 3x 2 )/L 
6
M=
1
w0 ( Lx − x3 /L) 
6
dM
= V = 0. L2 − 3 xm2 = 0
dx
xm =
L
3
M max =
1  L2
L2 
w0 
−

6  3 3 3 
M max = 0.0642w0 L2 
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PROBLEM 5.52
Determine (a) the equations of the shear and bending-moment curves for
the beam and loading shown, (b) the maximum absolute value of the
bending moment in the beam.
SOLUTION
dV
πx
= − w = − w0 sin
dx
L
w0 L
dM
πx
cos
+ C1 =
V=
L
dx
π
2
wL
πx
+ C1 x + C2
M = 0 2 sin
L
π
M = 0 at x = 0
C2 = 0
M = 0 at x = L
0 = 0 + C1 L + 0
C1 = 0
V=
(a)
M=
dM
= V = 0 at
dx
(b)
M max =
w0 L2
π2
sin
x=
π
2
w0 L
π
w0 L2
π
2
cos
πx
sin
πx
L
L


L
2
M max =
w0 L2
π2

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PROBLEM 5.53
Determine (a) the equations of the shear and bending-moment curves for
the beam and loading shown, (b) the maximum absolute value of the
bending moment in the beam.
SOLUTION
πx
dV
= − w = − w0 cos
2L
dx
2Lw0
πx
dM
sin
+ C1 =
V =−
π
2L
dx
M =
4 L2 w0
πx
V = 0 at
+ C1x + C2
2L
x = 0. Hence, C1 = 0.
M = 0 at
x = 0. Hence, C2 = −
π
2
cos
4L2 w0
π2
.
(a) V = −(2 Lw0 / π ) sin(π x / 2L) 
M = −(4 L2 w0 /π 2 )[1 − cos(π x/2L)] 
(b)
M
max
= 4w0 L2/π 2 

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PROBLEM 5.54
Draw the shear and bending-moment diagrams for the beam and
loading shown and determine the maximum normal stress due to
bending.
SOLUTION
M D = 0: −12 A + (9)(6)(2) − (2)(6) = 0
A = 8 kips
M A = 0: − (3)(6) + 12 D − (14)(6) = 0
D = 10 kips
VA = 8 kips
Shear:
VC = 8 − (6)(2) = −4 kips
C to D :
D to B :
V = −4 kips
V = −4 + 10 = 6 kips
Locate point E where V = 0.
e 6−e
12e = 48
=
8
4
e = 4 ft
6 − e = 2 ft
Areas of the shear diagram:
1
A to E :
 Vdx =  2  (4)(8) = 16 kip ⋅ ft
E to C :
 Vdx =  2  (2)(−4) = −4 kip ⋅ ft
 Vdx = (6)(−4) = −24 kip ⋅ ft
 Vdx = (2)(6) = 12 kip ⋅ ft
C to D :
D to B :
1
Bending moments:
MA = 0

ME
MC
MD
MB
= 0 + 16 = 16 kip ⋅ ft 
= 16 − 4 = 12 kip ⋅ ft
= 12 − 24 = −12 kip ⋅ ft
= −12 + 12 = 0
Maximum |M | = 16 kip ⋅ ft = 192 kip ⋅ in
For W8 × 31 rolled steel section,
Normal stress:
S = 27.5 in 3
|M | 192
σ=
=
S
27.5
σ = 6.98 ksi 
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PROBLEM 5.55
Draw the shear and bending-moment diagrams for the beam and
loading shown and determine the maximum normal stress due to
bending.
SOLUTION
 M C = 0 : (2)(1) − (3)(4)(2) + 4B = 0
B = 5.5 kN
 M B = 0 : (5)(2) + (3)(4)(2) − 4C = 0
C = 8.5 kN
Shear:
A to C:
V = −2 kN
C+ :
V = −2 + 8.5 = 6.5 kN
B:
V = 6.5 − (3)(4) = −5.5 kN
Locate point D where V = 0.
d
4−d
=
6.5
5.5
d = 2.1667 m
12d = 26
4 − d = 3.8333 m
Areas of the shear diagram:
A to C:
 Vdx = (−2.0)(1) = −2.0 kN ⋅ m
C to D:
 Vdx =
1
(2.16667)(6.5) = 7.0417 kN ⋅ m
2
D to B:
 Vdx =
1
(3.83333)(−5.5) = −5.0417 kN ⋅ m
2
Bending moments:
MA = 0
M C = 0 − 2.0 = −2.0 kN ⋅ m
M D = −2.0 + 7.0417 = 5.0417 kN ⋅ m
M B = 5.0417 − 5.0417 = 0
Maximum M = 5.0417 kN ⋅ m = 5.0417 × 103 N ⋅ m
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PROBLEM 5.55 (Continued)
For pipe:
I =
Normal stress:
1
1
1
1
do = (160) = 80 mm, ci = di = (140) = 70 mm
2
2
2
2
co =
π
(c
4
4
o
)
− ci4 =
π
4
4
6
4
(80) − (70)  = 13.3125 × 10 mm
4
S =
I
13.3125 × 106
=
= 166.406 × 103 mm3 = 166.406 × 10−6 m3
co
80
σ =
M
5.0417 × 103
=
= 30.3 × 106 Pa
S
166.406 × 10−6
σ = 30.3 MPa 

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PROBLEM 5.56
Draw the shear and bending-moment diagrams for the beam
and loading shown, and determine the maximum normal
stress due to bending.
SOLUTION
 M A = 0 : (1600 lb)(1.5 ft) + (80 lb/ft ) (9 ft)  (7.5 ft) − 12B = 0
B = +650 lb
B = 650 lb ↑
 Fy = 0 : A − 1600 lb − ( 80 lb/ft ) (9 ft)  + 650 lb = 0
A = +1670 lb
x
9−x
=
70 lb 650 lb
x = 0.875 ft
A = 1670 lb ↑
2641 lb ⋅ ft = M max
c=
1
(11.5 in.) = 5.75 in.
2
I =
1
(1.5 in.)(11.5 in.)3 = 190.1 in 4
12

M max = 2641 lb ⋅ ft = 31, 690 lb ⋅ in
σm =
M max c
(31, 690 lb ⋅ in)(5.75 in.)
=
I
190.1 in 4
σ m = 959 psi 
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PROBLEM 5.57
Draw the shear and bending-moment diagrams for the beam
and loading shown and determine the maximum normal stress
due to bending.
SOLUTION
w=0
MD = 0 :
− 4 RA + (2)(250) − (2)(150) = 0
RA = 50 kN ↑
MA = 0 :
4RD − (2)(250) − (6)(150) = 0
RD = 350 kN ↑
Shear:
VA = 50 kN
A to C:
V = 50 kN
C to D:
V = 50 − 250 = −200 kN
D to B:
V = −200 + 350 = 150 kN
Areas of shear diagram:
A to C:
 Vdx = (50)(2) = 100 kN ⋅ m
C to D:
 Vdx = (−200)(2) = −400 kN ⋅ m
D to B:
 Vdx = (150)(2) = 300 kN ⋅ m
Bending moments: M A = 0
M C = M A +  Vdx = 0 + 100 = 100 kN ⋅ m
M D = M C +  Vdx = 100 − 400 = −300 kN ⋅ m
M B = M D +  Vdx = −300 + 300 = 0
Maximum M = 300 kN ⋅ m = 300 × 103 N ⋅ m
For W410 × 114 rolled steel section,
σm =
M
max
Sx
=
S x = 2200 × 103 mm3 = 2200 × 10−6 m3
300 × 103
= 136.4 × 106 Pa
2200 × 10−6
σ m = 136.4 MPa 
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PROBLEM 5.58
Draw the shear and bending-moment diagrams for the
beam and loading shown and determine the maximum
normal stress due to bending.
SOLUTION
Reaction:
 M B = 0 : − 4 A + 60 + (80)(1.6)(2) − 12 = 0
A = 76 kN ↑
Shear:
VA = 76 kN
V = 76 kN 
A to C :
VD = 76 − (80)(1.6) = −52 kN
V = −52 kN
D to C :
Locate point where V = 0:
V ( x) = −80 x + 76 = 0
x = 0.95 m
Areas of shear diagram:
A to C:  Vdx = (1.2)(76) = 91.2 kN ⋅ m
C to E:  Vdx =
1
(0.95)(76) = 36.1 kN ⋅ m
2
E to D:  Vdx =
1
(0.65)(−52) = −16.9 kN ⋅ m
2
D to B:  Vdx = (1.2)(−52) = −62.4 kN ⋅ m
Bending moments:
M A = −60 kN ⋅ m
M C = −60 + 91.2 = 31.2 kN ⋅ m
M E = 31.2 + 36.1 = 67.3 kN ⋅ m

M D = 67.3 − 16.9 = 50.4 kN ⋅ m
M B = 50.4 − 62.4 = −12 kN ⋅ m
For W250 × 80, S = 983 × 103 mm3
Normal stress:
σ max =
M
67.3 × 103 N ⋅ m
=
= 68.5 × 106 Pa
−6 3
S
983 × 10 m
σ m = 68.5 MPa 
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PROBLEM 5.59
Draw the shear and bending-moment diagrams for the beam and
loading shown, and determine the maximum normal stress due to
bending.
SOLUTION
ΣM B = 0: −20 A + (6)(28)(800) = 0
A = 6.72 × 103 lb
ΣM A = 0:
20 B − (14)(28)(800) = 0
B = 15.68 × 103 lb
VA = 6.72 × 103 lb
Shear:
B −:
VB − = 6.72 × 103 − (20)(800) = −9.28 × 103 lb
B +:
VB + = −9.28 × 103 + (15.68 × 103 ) = 6.4 × 103 lb
VC = 6.4 × 103 − (8)(800) = 0
C:
Locate point D where
V = 0.
d
20 − d
=
6.72
9.28
d = 8.4 in.
16d = 134.4
20 − d = 11.6 in.
Areas of the shear diagram:
1
A to D:
 Vdx =  2  (8.4)(6.72 × 10 ) = 28.224 × 10 lb ⋅ in
D to B:
 Vdx =  2  (11.6)(−9.28 × 10 ) = −53.824 × 10 lb ⋅ in
B to C:
 Vdx =  2  (8)(6.4 × 10 ) = 25.6 × 10 lb ⋅ in
3
1
1
3
3
3
3
3
Bending moments:
MA = 0
M D = 0 + 28.224 × 103 = 28.224 × 103 lb ⋅ in
M B = 28.224 × 103 − 53.824 × 103 = − 25.6 × 103 lb ⋅ in
M C = −25.6 × 103 + 25.6 × 103 = 0
Maximum |M | = 28.224 × 103 lb ⋅ in
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PROBLEM 5.59 (Continued)
Locate centroid of cross section. See table below.
For each triangle,
Y =
7.5
= 1.3333 in. from bottom.
5.625
I =
1 3
bh
36
Moment of inertia:
I = ΣI + ΣAd 2
= 1.25 + 2.8125 = 4.0625 in 4
σ=
Normal stress:
Mc (28.224 × 103 )(1.6667)
=
= 11.58 × 103 psi
I
4.0625
σ = 11.58 ksi 
Part
A, in 2
y , in.
Ay , in 3
d, in.
Ad 2 in 4
I in 4

1.875
2
3.75
0.6667
0.8333
0.9375

3.75
1
3.75
0.3333
0.4167
1.875
Σ
5.625
1.25
2.8125
7.5
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PROBLEM 5.60
Beam AB, of length L and square cross section of side a, is supported
by a pivot at C and loaded as shown. (a) Check that the beam is in
equilibrium. (b) Show that the maximum stress due to bending occurs
at C and is equal to w0 L2 /(1.5a)3.
SOLUTION
(a)
Replace distributed load by equivalent concentrated load at the centroid of the area of the load diagram.
For the triangular distribution, the centroid lies at x =
(a)
 Fy = 0 : RD − W = 0
RD =
1
w0 L
2
2L
.
3
W =
1
w0 L
2
 M C = 0 : 0 = 0 equilibrium

V = 0, M = 0, at x = 0
0< x<
2L
,
3
dV
wx
= −w = − 0
dx
L
dM
w x2
w x2
= V = − 0 + C1 = − 0
dx
2L
2L
M =−
V =−
Just to the left of C,
w0 x3
w x3
+ C2 = − 0
6L
6L
w0 (2 L / 3)2
2
= − w0 L
2L
9
2
5
V = − w0 L + RD =
w0 L
9
18
Just to the right of C,
Note sign change. Maximum M occurs at C.
Maximum M =
σm =
M
max
I
w0 (2L / 3)3
4
= − w0 L2
6L
81
4
w0 L2
81
For square cross section,
(b)
MC = −
c
I =
1 4
a
12
c=
1
a
2
3
=
4 w0 L2 6
8 w0 L2  2  w0 L2
=
= 
3
3
81 a
27 a3
3 a
σm =
w0 L2

(1.5a)3
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PROBLEM 5.61
Knowing that beam AB is in equilibrium under the loading
shown, draw the shear and bending-moment diagrams and
determine the maximum normal stress due to bending.
SOLUTION
 Fy = 0: (1)(w0 ) − (0.4)(400) = 0
w0 = 160 kN/m
VA = 0
Shear diagram:
VC = 0 + (0.3)(160) = 48kN
VD = 48 − (0.3)(400) + (0.3)(160) = −48kN
VB = −48 + (0.3)(160) = 0
Locate point E where V = 0.
By symmetry, E is the midpoint of CD.
Areas of shear diagram:
A to C:
1
(0.3)(48) = 7.2 kN ⋅ m
2
C to E:
1
(0.2)(48) = 4.8 kN ⋅ m
2
E to D:
1
(0.2)(−48) = −4.8 kN ⋅ m
2
D to B:
1
(0.3)(−48) = −7.2 kN ⋅ m
2
Bending moments:
MA = 0
M C = 0 + 7.2 = 7.2 kN
M E = 7.2 + 4.8 = 12.0 kN
M D = 12.0 − 4.8 = 7.2 kN
M B = 7.2 − 7.2 = 0
M
max
= 12.0 kN ⋅ m = 12.0 × 103 N ⋅ m
For W200 × 22.5 rolled steel shape, S x = 193 × 103 mm3 = 193 × 10−6 m3
Normal stress:
σ =
M
12.0 × 103
=
= 62.2 × 106 Pa
−6
S
193 × 10
σ = 62.2 MPa 

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PROBLEM 5.62
The beam AB supports a uniformly distributed load of 480 lb/ft and two
concentrated loads P and Q. The normal stress due to bending on the
bottom edge of the lower flange is +14.85 ksi at D and +10.65 ksi at E.
(a) Draw the shear and bending-moment diagrams for the beam. (b)
Determine the maximum normal stress due to bending that occurs in the
beam.
SOLUTION
(a) For W8 × 31 rolled steel section, S = 27.5 in 3
M = Sσ
At D,
M D = (27.5)(14.85) = 408.375 kip ⋅ in
At E,
M E = (27.5)(10.65) = 292.875 kip ⋅ in
M D = 34.03 kip ⋅ ft
M E = 24.41 kip ⋅ ft
Use free body DE:
 M E = 0 : − 34.03 + 24.41 + (1.5)(3)(0.48) − 3VD = 0
VD = −2.487 kips
M D = 0 : − 34.03 + 24.41 − (1.5)(3)(0.48) − 3VE = 0
VE = −3.927 kips
Use free body ACD:
 M A = 0 : − 1.5P − (1.25)(2.5)(0.48) + (2.5)(2.487) + 34.03 = 0
P = 25.83 kips ↓
 Fy = 0 : A − (2.5)(0.48) + 2.487 − 25.83 = 0
A = 24.54 kips ↑
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PROBLEM 5.62 (Continued)
Use free body EFB:
 M B = 0 : 1.5Q + (1.25)(2.5)(0.48) + (2.5)(3.927) − 24.41 = 0
Q = 8.728 kips
 Fy = 0 : B − 3.927 − (2.5)(0.48) − 8.7 = 0
B = 13.855 kips
Areas of load diagram:
A to C:
(1.5)(0.48) = 0.72 kip ⋅ ft
C to F:
(5)(0.48) = 2.4 kip ⋅ ft
F to B:
(1.5)(0.48) = 0.72 kip ⋅ ft
VA = 24.54 kips
Shear diagram:
VC− = 24.54 − 0.72 = 23.82 kips
VC+ = 23.82 − 25.83 = −2.01 kips
VF− = −2.01 − 2.4 = 4.41 kips
VF+ = −4.41 − 8.728 = −13.14 kips
VB = −13.14 − 0.72 = −13.86 kips
Areas of shear diagram:
A to C:
1
(1.5)(24.52 + 23.82) = 36.23 kip ⋅ ft
2
C to F:
1
(5)(−2.01 − 4.41) = −16.05 kip ⋅ ft
2
F to B:
1
(1.5)(−13.14 − 13.86) = 20.25 kip ⋅ ft
2
Bending moments:
MA = 0
M C = 0 + 36.26 = 36.26 kip ⋅ ft
M F = 36.26 − 16.05 = 20.21 kip ⋅ ft
M B = 20.21 − 20.25 ≈ 0
Maximum M occurs at C: M
(b) Maximum stress:
max
= 36.26 kip ⋅ ft = 435.1 kip ⋅ in
σ =
M
max
S
=
435.1
27.5
σ = 15.82 ksi 
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PROBLEM 5.63*
Beam AB supports a uniformly distributed load of 2 kN/m and two
concentrated loads P and Q. It has been experimentally determined that
the normal stress due to bending in bottom edge of the beam is
−56.9 MPa at A and −29.9 MPa at C. Draw the shear and bendingmoment diagrams for the beam and determine the magnitudes of the
loads P and Q.
SOLUTION

1
(18)(36)3 = 69.984 × 103 mm 4
12
1
c = d = 18 mm
2
I
S = = 3.888 × 103 mm3 = 3.888 × 10−6 m3
c
I=



At A,
M A = Sσ A = (3.888 × 10−6 )(−56.9) = −221.25 N ⋅ m
At C,
M C = Sσ C = (3.888 × 10−6 )( −29.9) = −116.25 N ⋅ m
ΣM A = 0: 221.23 − (0.1)(400) − 0.2 P − 0.325Q = 0
0.2 P + 0.325Q = 181.25

(1)
ΣM C = 0: 116.25 − (0.05)(200) − 0.1P − 0.225Q = 0

0.1P + 0.225Q = 106.25
Solving (1) and (2) simultaneously,
(2)
P = 500 N 
Q = 250 N 
RA − 400 − 500 − 250 = 0 RA = 1150 N ⋅ m
Reaction force at A:
VA = 1150 N
VD = 250
M A = −221.25 N ⋅ m
M C = −116.25 N ⋅ m
M D = −31.25 N ⋅ m
|V |max = 1150 N 
|M |max = 221 N ⋅ m 

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PROBLEM 5.64*
The beam AB supports two concentrated loads P and Q. The normal
stress due to bending on the bottom edge of the beam is +55MPa at
D and +37.5 MPa at F. (a) Draw the shear and bending-moment
diagrams for the beam. (b) Determine the maximum normal stress
due to bending that occurs in the beam.
SOLUTION
At D,
1
(24)(60)3 = 432 × 103 mm 4 c = 30 mm
12
I
S = = 14.4 × 103 mm3 = 14.4 × 10−6 m3 M = Sσ
c
M D = (14.4 × 10−6 )(55 × 106 ) = 792 N ⋅ m
At F,
M F = (14.4 × 10−6 )(37.5 × 106 ) = 540 N ⋅ m
(a)
I=
Using free body FB,
ΣM F = 0: −540 + 0.3B = 0
B=
540
= 1800 N
0.3
ΣM D = 0: −792 − 3Q + (0.8)(1800) = 0
Using free body DEFB,
Q = 2160 N
Using entire beam,
ΣM A = 0: − 0.2 P − (0.7)(2160) + (1.2)(1800) = 0
P = 3240 N
ΣFy = 0: A − 3240 − 2160 + 1800 = 0
A = 3600 N
Shear diagram and its areas:
A to C − :
+
−
C to E :
+
E to B:
V = 3600 N
AAC = (0.2)(3600) = 720 N ⋅ m
V = 3600 − 3240 = 360 N
ACE = (0.5)(360) = 180 N ⋅ m
V = 360 − 2160 = −1800 N
AEB = (0.5)(−1800) = −900 N ⋅ m
Bending moments:
MA = 0
M C = 0 + 720 = 720 N ⋅ m
M E = 720 + 180 = 900 N ⋅ m
|M |max = 900 N ⋅ m
M B = 900 − 900 = 0
(b)
Normal stress.
σ max =
|M |max
900
=
= 62.5 × 106 Pa
S
14.4 × 10−6
σ max = 62.5 MPa 
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PROBLEM 5.65
For the beam and loading shown, design the cross section of the
beam, knowing that the grade of timber used has an allowable normal
stress of 12 MPa.
SOLUTION
Reactions:
ΣM D = 0: − 2.4 A + (1.6)(1.8) + (0.8)(3.6) = 0
A = 2.4 kN
ΣM A = 0: − (0.8)(1.8) − (1.6)(3.6) + 2.4 D = 0
D = 3 kN
Construct shear and bending moment diagrams:
|M |max = 2.4 kN ⋅ m = 2.4 × 103 N ⋅ m
σ all = 12 MPa
= 12 × 106 Pa
Smin =
|M |max
σ all
=
2.4 × 103
12 × 106
= 200 × 10−6 m3
= 200 × 103 mm3
1
1
S = bh 2 = (40)h 2
6
6
= 200 × 103
(6)(200 × 103 )
40
= 30 × 103 mm 2
h2 =
h = 173.2 mm 
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PROBLEM 5.66
For the beam and loading shown, design the cross section of the beam,
knowing that the grade of timber used has an allowable normal stress
of 12 MPa.
SOLUTION
 Fy = 0 : A − (1.2)(18) = 0
Reactions:
A = 21.6 kN ↑
 M A = 0 : − M A − (1.8)(1.2)(18) = 0
M A = −38.88 kN ⋅ m
VA = VB = 21.6 kN
Shear diagram:
VC = 21.6 − (1.2)(18) = 0
Areas of shear diagram:
A to B : (1.2)(21.6) = 25.92 kN ⋅ m
B to C :
1
(1.2)(21.6) = 12.96 kN ⋅ m
2
Bending moments: M A = −38.88 kN ⋅ m
M B = −38.88 + 25.92 = −12.96 kN ⋅ m
M C = −12.96 + 12.96 = 0
M
max
= 38.88 kN ⋅ m = 38.8 × 103 N ⋅ m
σ max =
S =
M
max
σ max
=
6S
=
b
max
S
38.8 × 103 N ⋅ m
= 3240 × 10−6 m3 = 3240 × 103 mm3
6
12 × 10 Pa
For a rectangular section, S =
h=
M
1 2
bh
6
6(3240 × 103 )
= 394 mm
125
h = 394 mm 
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PROBLEM 5.67
For the beam and loading shown, design the cross section of the
beam, knowing that the grade of timber used has an allowable
normal stress of 1750 psi.
SOLUTION
By symmetry, A = D.
Reactions:
1
1
(3)(1.5) − (6)(1.5) − (3)(1.5) − D = 0
2
2
A = D = 6.75 kips ↑
 Fy = 0 : A −
Shear diagram:
VB = 6.75 −
VA = 6.75 kips
1
(3)(1.5) = 4.5 kips
2
VC = 4.5 − (6)(1.5) = −4.5 kips
VD = −4.5 −
1
(3)(1.5) = −6.75 kips
2
Locate point E where V = 0 :
By symmetry, E is the midpoint of BC.
Areas of the shear diagram:
A to B : (3)(4.5) +
2
(3)(2.25) = 18 kip ⋅ ft
3
B to E :
1
(3)(4.5) = 6.75 kip ⋅ ft
2
E to C :
1
(3)(−4.5) = −6.75 kip ⋅ ft
2
C to D : By antisymmetry, − 18 kip ⋅ ft
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PROBLEM 5.67 (Continued)
Bending moments: M A = 0
M B = 0 + 18 = 18 kip ⋅ ft
M E = 18 + 6.75 = 24.75 kip ⋅ ft
M C = 24.75 − 6.75 = 18 kip ⋅ ft
M D = 18 − 18 = 0
σ max =
M
max
S
S =
M
max
σ max
For a rectangular section,
=
(24.75 kip ⋅ ft)(12 in/ft)
= 169.714 in 3
1.750 ksi
S =
1 2
bh
6
h=
6S
=
b
6(169.714)
= 14.27 in.
5
h = 14.27 in. 
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PROBLEM 5.68
For the beam and loading shown, design the cross section of the beam,
knowing that the grade of timber used has an allowable normal stress of 1750
psi.
SOLUTION
Equivalent concentrated load:
1
P =   (6)(1.2) = 3.6 kips
2
Bending moment at A:
M A = (2)(3.6) = 7.2 kip ⋅ ft = 86.4 kip ⋅ in
Smin =
M
max
σ all
=
86.4
= 49.37 in 3
1.75
For a square section,
a =
amin =
3
3
S =
1 3
a
6
6S
(6)(49.37)
amin = 6.67 in. 

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PROBLEM 5.69
For the beam and loading shown, design the cross section of the beam,
knowing that the grade of timber used has an allowable normal stress of
12 MPa.
SOLUTION
By symmetry,
B=C
ΣFy = 0: B + C + 2.5 + 2.5 − (3)(6) = 0
B = C = 6.5 kN
Shear:
V = 2.5 kN
VB+ = 2.5 + 6.5 = 9 kN
VC − = 9 − (3)(6) = −9 kN
A to B:
V = −9 + 6.5 = −2.5 kN
C to D:
Areas of the shear diagram:
 Vdx = (0.6)(2.5) = 1.5 kN ⋅ m
1
 Vdx =  2  (1.5)(9) = 6.75 kN ⋅ m
A to B:
B to E:
 Vdx = −6.75 kN ⋅ m
 Vdx = −1.5 kN ⋅ m
E to C:
C to D:
Bending moments:
MA
MB
ME
MC
MD
=0
= 0 + 1.5 = 1.5 kN ⋅ m
= 1.5 + 6.75 = 8.25 kN ⋅ m
= 8.25 − 6.75 = 1.5 kN ⋅ m
= 1.5 − 1.5 = 0
Maximum |M | = 8.25 kN ⋅ m = 8.25 × 103 N ⋅ m
σ all = 12 MPa = 12 × 106 Pa
Smin =
For a rectangular section,
|M |max
σ all
=
8.25 × 103
= 687.5 × 10−6 m3 = 687.5 × 103 mm3
6
12 × 10
1
S = bh 2
6
1

687.5 × 103 =   (100) h 2
6
(6)(687.5 × 103 )
h2 =
= 41.25 × 103 mm 2
100
h = 203 mm 
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PROBLEM 5.70
For the beam and loading shown, design the cross section of the
beam, knowing that the grade of timber used has an allowable
normal stress of 12 MPa.
SOLUTION
Shear:
M B = 0: − 2.4 A + (0.6)(3.6)(3) = 0
A = 2.7 kN
M A = 0: −(1.8)(3.6)(3) + 2.4 B = 0
B = 8.1 kN
VA = 2.7 kN
VB − = 2.7 − (2.4)(3) = −4.5 kN
VB+ = −4.5 + 8.1 = 3.6 kN
VC = 3.6 − (1.2)(3) = 0
d
2.4 − d
=
2.7
4.5
d = 0.9 m
Locate point D where V = 0.
7.2 d = 6.48
2.4 − d = 1.5 m
Areas of the shear diagram:
1
A to D:
 Vdx =  2  (0.9)(2.7) = 1.215 kN ⋅ m
D to B:
 Vdx =  2  (1.5)(−4.5) = −3.375 kN ⋅ m
B to C:
 Vdx =  2  (1.2)(3.6) = 2.16 kN ⋅ m
Bending moments:
1
1
MA = 0
M D = 0 + 1.215 = 1.215 kN ⋅ m
M B = 1.215 − 3.375 = −2.16 kN ⋅ m
M C = −2.16 + 2.16 = 0
Maximum |M | = 2.16 kN ⋅ m = 2.16 × 103 N ⋅ m
Smin =
For rectangular section,
|M |
σ all
=
σ all = 12 MPa = 12 × 106 Pa
2.16 × 103
= 180 × 10−6 m3 = 180 × 103 mm3
6
12 × 10
1
1
S = bh 2 = b(150)2 = 180 × 103
6
6
b=
(6)(180 × 103 )
1502
b = 48.0 mm 
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PROBLEM 5.71
Knowing that the allowable stress for the steel used is 24 ksi, select the
most economical wide-flange beam to support the loading shown.
SOLUTION
ΣM C = 0: (17)(62) − 12 B + (5)(62) = 0
B = 113.667 kips ↑
ΣM B = 0: (5)(62) + 12C + (17)(62) = 0
C = −113.667 kips or C = 113.667 kips ↓
Shear diagram:
A to B − :
V = −62 kips
B+ to C − :
V = −62 + 113.667 = 51.667 kips
C+ to D:
V = 51.667 − 113.667 = −62 kips.
Areas of shear diagram:
(5)(−62) = −310 kip ⋅ ft
A to B:
B to C:
(12)(51.667) = 620 kip ⋅ ft
(5)(−62) = −310 kip ⋅ ft
C to D:
Bending moments:
MA = 0
M B = 0 − 310 = −310 kip ⋅ ft
M C = −310 + 620 = 310 kip ⋅ ft
M D = 310 − 310 = 0
|M |max = 310 kip ⋅ ft = 3.72 × 103 kip ⋅ in
Smin =
Required Smin :
Shape
S (in 3 )
W27 × 84
213
W21 × 101
227
W18 × 106
204
W14 × 232
232
|M |max
σ all
=
3.72 × 103
= 155 in 3
24
Use W27 × 84 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 5.72
Knowing that the allowable stress for the steel used is 24 ksi, select the
most economical wide-flange beam to support the loading shown.
SOLUTION
ΣM C = 0: − 24 A + (12)(24)(2.75) + (15)(24) = 0
A = 48 kips
ΣM A = 0: 24 C − (12)(24)(2.75) − (9)(24) = 0
C = 42 kips
VA = 48
Shear:
VB− = 48 − (9)(2.75) = 23.25 kips
VB+ = 23.25 − 24 = −0.75 kips
VC = −0.75 − (15)(2.75) = −42 kips
Areas of the shear diagram:
1
A to B:
 Vdx =  2  (9)(48 + 23.25) = 320.6 kip ⋅ ft
B to C:
 Vdx =  2 (15)(−0.75 − 42) = −320.6 kip ⋅ ft
1
Bending moments:
MA = 0
M B = 0 + 320.6 = 320.6 kip ⋅ ft
M C = 320.6 − 320.6 = 0
Maximum |M | = 320.6 kip ⋅ ft = 3848 kip ⋅ in
σ all = 24 ksi
Smin =
Shape
W30 × 99
W27 × 84
W24 × 104
W21 × 101
W18 × 106
S , (in 3)
269
213 ←
258
227
204
|M |
σ all
=
3848
= 160.3 in 3
24
Lightest wide flange beam: W27 × 84 @ 84 lb/ft 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 5.73
Knowing that the allowable stress for the steel used is 160 MPa, select the
most economical wide-flange beam to support the loading shown.
SOLUTION
18 − 6 

w = 6 +
x  = (6 + 2 x) kN/m
6


dV
= −w = −6 − 2 x
dx
V = −6 x − x 2 + C1
V = 0 at
x = 0, C1 = 0
dM
= V = −6 x − x 2
dx
1
M = −3x 2 − x3 + C2
3
M = 0 at x = 0, C2 = 0
M = −3x 2 −
M
M
max
max
1 3
x
3
occurs at x = 6 m.
1
= −(3)(6)2 −   (6)3 = 80 kN ⋅ m = 180 × 103 N ⋅ m
3
σ all = 160 MPa = 160 × 106 Pa
Smin =
Shape
W530 × 66
M
σ all
=
180 × 103
= 1.125 × 10−3 m3 = 1125 × 103 mm3
160 × 106
S, ( 103 mm3 )
W460 × 74
1340 ←
1460
W410 × 85
1510
W360 × 79
1270
W310 × 107
1600
W250 × 101
1240
Lightest acceptable wide flange beam: W530 × 66 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 5.74
Knowing that the allowable stress for the steel used is 160 MPa, select the most
economical wide-flange beam to support the loading shown.
SOLUTION
Shape
Section modulus
σ all = 160 Mpa
Smin =
M max
σ all
=
286 kN ⋅ m
= 1787 × 10−6 m3
160 MPa
= 1787 × 103 mm3
S, ( 103 mm3 )
W610 × 101
2520
W530 × 92
2080 ←
W460 × 113
2390
W410 × 114
2200
W360 × 122
2020
W310 × 143
2150

Use W530 × 92 
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PROBLEM 5.75
Knowing that the allowable stress for the steel used is 160 MPa, select the
most economical S-shape beam to support the loading shown.
SOLUTION
Reaction:
 M D = 0 : − 10 A + (7.5)(60) + (5)(40) = 0
A = 65 kN ↑
Shear diagram:
A to B:
V = 65 kN
B to C:
V = 65 − 60 = 5 kN
C to D:
V = 5 − 40 = −35 kN
Areas of shear diagram:
A to B: (2.5)(65) = 162.5 kN ⋅ m
B to C: (2.5)(5) = 12.5 kN ⋅ m
C to D: (5)(−35) = −175 kN ⋅ m
Bending moments: M A = 0
M B = 0 + 162.5 = 162.5 kN ⋅ m
M C = 162.5 + 12.5 = 175 kN ⋅ m
M D = 175 − 175 = 0
M
max
= 175 kN ⋅ m = 175 × 103 N ⋅ m
σ all = 160 MPa = 160 × 106 Pa
Shape
Sx, ( 103 mm3 )
S610 × 119
2870
S510 × 98.2
1950
S460 × 81.4
1460 ←
Smin =
M
σ all
=
175 × 103
= 1093.75 × 10−6 m3
160 × 106
= 1093.75 × 103 mm3
Lightest S-section:
S460 × 81.4 
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PROBLEM 5.76
Knowing that the allowable stress for the steel used is 160 MPa, select the
most economical S-shape beam to support the loading shown.
SOLUTION
Reactions:
By symmetry,
B = C.
 Fy = 0 : − 70 + B − (9)(45) + C − 70 = 0
B = C = 272.5 kN ↑
VA = −70 kN
Shear:
VB − = −70 + 0 = −70 kN
VB + = −70 + 272.5 = 202.5 kN
VC − = 202.5 − (9)(45) = −202.5 kN
VC + = −202.5 + 272.5 = 70 kN
VD = 70 kN
Draw shear diagram. Locate point E where V = 0 .
E is the midpoint of BC.
Areas of the shear diagram:
A to B:
 Vdx = (3)(−70) = −210 kN ⋅ m
B to E:
 Vdx = 2 (4.5)(202.5) = 455.625 kN ⋅ m
E to C:
 Vdx = 2 (4.5)(−202.5) = −455.625 kN ⋅ m
C to D:
 Vdx = (3)(70) = 210 kN ⋅ m
1
1

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PROBLEM 5.76 (Continued)
Bending moments: M A = 0
M B = 0 − 210 = −210.5 kN ⋅ m
M E = −210 + 455.625 = 245.625 kN
M C = 245.625 − 455.625 = −210 kN
M D = −210 + 210 = 0
M
max
S =
S( 103 mm3 )
S610 × 119
2870
S510 × 98.2
1950 ←
S460 × 104
1690
= 245.625kN ⋅ m = 245.625 × 103 N ⋅ m
σ all = 160 MPa = 160 × 106 Pa
σ =
Shape
Lightest S-shape S510 × 98.2 
M
S
M
σ
=
245.625 × 103
= 1.5352 × 10−3 m3
6
160 × 10
= 1535.2 × 103 mm3
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PROBLEM 5.77
Knowing that the allowable stress for the steel used is 24 ksi, select the
most economical S-shape beam to support the loading shown.
SOLUTION
 M E = 0 : (12)(48) − 10 B + (8)(48) + (2)(48) = 0
 M B = 0 : (2)(48) − (2)(48) − (8)(48) + 10 E = 0
Shear:
Areas:
B = 105.6 kips ↑
E = 38.4 kips ↑
A to B:
V = −48 kips
B to C:
V = −48 + 105.6 = 57.6 kips
C to D:
V = 57.6 − 48 = 9.6 kips
D to E:
V = 9.6 − 48 = −38.4 kips
A to B:
(2)(−48) = −96 kip ⋅ ft
B to C:
(2)(57.6) = 115.2 kip ⋅ ft
C to D:
(6)(9.6) = 57.6 kip ⋅ ft
D to E:
(2)(−38.4) = 76.8 kip ⋅ ft
Bending moments: M A = 0
M B = 0 − 96 = −96 kip ⋅ ft
M C = −96 + 115.2 = 19.2 kip ⋅ ft
M D = 19.2 + 57.2 = 76.8 kip ⋅ ft
M E = 76.8 − 76.8 = 0
Maximum M = 96 kip ⋅ ft = 1152 kip ⋅ in
σ all = 24 ksi
Smin =
Shape
S15 × 42.9
S12 × 50
M
σ all
=
1152
= 48 in 3
24
S (in 3 )
59.4
Lightest S-shaped beam: S15 × 42.9 
50.6

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PROBLEM 5.78
Knowing that the allowable stress for the steel used is 24 ksi, select the
most economical S-shape beam to support the loading shown.
SOLUTION
 M C = 0 : − 12 A + (9)(6)(3) − (3)(18) = 0
A = 9 kips
 M A = 0 : 12C − (3)(6)(3) − (15)(18) = 0
C = 27 kips
Shear:
VA = 9 kips
B to C:
V = 9 − (6)(3) = −9 kips
C to D:
V = −9 + 27 = 18 kips
Areas:
A to E:
(0.5)(3)(9) = 13.5 kip ⋅ ft
E to B:
(0.5)(3)(−9) = −13.5 kip ⋅ ft
B to C:
(6)(−9) = −54 kip ⋅ ft
C to D:
(3)(18) = 54 kip ⋅ ft
Bending moments: M A = 0
M E = 0 + 13.5 = 13.5 kip ⋅ ft
M B = 13.5 − 13.5 = 0
M C = 0 + 54 = 54 kip ⋅ ft
M D = 54 − 54 = 0
Maximum M = 54 kip ⋅ ft = 648 kip ⋅ in
Smin =
Shape

σ all = 24 ksi
648
= 27 in 3
24
S (in 3 )
S12 × 31.8
36.2
S10 × 35
29.4
Lightest S-shaped beam: S12 × 31.8 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 5.79
Two L102 × 76 rolled-steel angles are bolted together and used to
support the loading shown. Knowing that the allowable normal
stress for the steel used is 140 MPa, determine the maximum angle
of thickness that can be used.
SOLUTION
By symmetry, A = C
Reactions:
Σ Fy = 0: A − (2)(4.5) − 9 + C = 0
A = C = 9 kN ↑
VA = 9 kN
Shear:
VB − = 9 − (1)(4.5) = 4.5 kN
VB + = 4.5 − 9 = −4.5 kN
VC = −4.5 − (1)(4.5) = −9 kN
Areas of shear diagram:
1
A to B:
 Vdx = 2 (1)(9 + 4.5) = 6.75 kN ⋅ m
B to C:
 Vdx = 2 (1)(−9 − 4.5) = −6.75 kN ⋅ m
1
MA = 0
Bending moments:
M B = 0 + 6.75 = 6.75 kN ⋅ m
M C = 6.75 − 6.75 = 0
Maximum |M | = 6.75 kN ⋅ m = 6.75 × 103 N ⋅ m
σ all = 140 MPa = 140 × 106 Pa
For the section of two angles, Smin =
|M |
σ all
=
6.75 × 103
= 48.21 × 10−6 m3
6
140 × 10
= 48.21× 103 mm3
Smin =
For each angle,
Shape
S (103 mm3 )
L102 × 76 × 12.7
31.1
L102 × 76 × 9.5
24.0
L102 × 76 × 6.4
16.6
←
1
(48.21) = 24.105 × 103 mm3
2
Lightest angle is L102 × 76 × 12.7 
tmin = 12.7 mm 

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PROBLEM 5.80
Two rolled-steel channels are to be welded back to back and used to
support the loading shown. Knowing that the allowable normal stress
for the steel used is 30 ksi, determine the most economical channels that
can be used.
SOLUTION
Σ M D = 0: −12 A + 9(20) + (6)(2.25)(3) = 0
Reaction:
A = 18.375 kips ↑
Shear diagram:
A to B:
V = 18.375 kips
B to C:
V = 18.375 − 20 = −1.625 kips
VD = −1.625 − (6)(2.25) = −15.125 kips
Areas of shear diagram:
A to B:
(3)(18.375) = 55.125 kip ⋅ ft
B to C:
(3)(−1.625) = −4.875 kip ⋅ ft
C to D:
0.5(6) (−1.625 − 15.125) = −50.25 kip ⋅ ft
Bending moments: M A = 0
M B = 0 + 55.125 = 55.125 kip ⋅ ft
M C = 55.125 − 4.875 = 50.25 kip ⋅ ft
Shape
S (in)3
M D = 50.25 − 50.25 = 0
C10 × 15.3
13.5
|M |max = 55.125 kip ⋅ ft = 661.5 kip ⋅ in
C9 × 15
11.3 ←
σ all = 30 ksi
C8 × 18.7
11.0
For double channel, Smin =
For single channel,
|M |
σ all
=
661.5
= 22.05 in 3
30
Smin = 0.5(22.05) = 11.025 in 3
Lightest channel section: C9 × 15 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 5.81
Three steel plates are welded together to form the beam
shown. Knowing that the allowable normal stress for the
steel used is 22 ksi, determine the minimum flange
width b that can be used.
SOLUTION
Σ M E = 0: − 42 A + (37.5)(8) + (23.5)(32) − (9.5)(32) = 0
Reactions:
A = 32.2857 kips ↑
Σ M A = 0: 42 E − (4.5)(8) − (18.5) (32) − (32.5)(32) = 0
E = 39.7143 kips ↑
Shear:
A to B :
32.2857 kips
B to C :
32.2857 − 8 = 24.2857 kips
C to D :
24.2857 − 32 = −7.7143 kips
D to E :
−7.7143 − 32 = −39.7143 kips
Areas:
(4.5)(32.2857) = 145.286 kip ⋅ ft
A to B :
B to C :
(14)(24.2857) = 340 kip ⋅ ft
C to D :
(14) (−7.7143) = −108 kip ⋅ ft
(9.5) (−39.7143) = −377.286 kip ⋅ ft
D to E :
MA = 0
Bending moments:
M B = 0 + 145.286 = 145.286 kip ⋅ ft
M C = 145.286 + 340 = 485.29 kip ⋅ ft
M D = 485.29 − 108 = 377.29 kip ⋅ ft
M E = 377.29 − 377.286 = 0
Maximum |M | = 485.29 kip ⋅ ft = 5.2834 × 103 kip ⋅ in
Smin =
|M |
σ all
=
σ all = 22 ksi
5.2834 × 103
= 264.70 in 3
22
1 3
1

(19)3 + 2  (b) (1)3 + (b)(1)(10) 2  = 428.69 + 200.17b


12  4 
12

c = 9.5 + 1 = 10.5 in.
I=
Smin =
I
= 40.828 + 19.063b = 264.70
c
b = 11.74 in. 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 5.82
A steel pipe of 100-mm diameter is to support the loading shown.
Knowing that the stock of pipes available has thicknesses varying
from 6 mm to 24 mm in 3-mm increments, and that the allowable
normal stress for the steel used is 150 MPa, determine the minimum
wall thickness t that can be used.
SOLUTION
 M A = 0 : − M A − (1)(1.5) − (1.5)(1.5) − (2)(1.5) = 0
M
max
M A = −6.75 kN ⋅ m
= M A = 6.75 kN ⋅ m
Smin =
Smin =
I m in =
M
max
σ all
=
6.75 × 103 N ⋅ m
= 45 × 10−6 m3 = 45 × 103 mm3
6
150 × 10 Pa
I min
c2
π
4
I min = c2 Smin = (50)(45 × 103 ) = 2.25 × 106 mm 4
(c
4
2
4
c1max
= c24 −
4
− c1max
4
π
)
I min = (50) 4 −
4
π
(2.25 × 106 ) = 3.3852 × 106 mm 4
c1max = 42.894 mm
tmin = c2 − c1max = 50 − 42.894 = 7.106 mm
t = 9 mm 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 5.83
Assuming the upward reaction of the ground to be uniformly distributed and
knowing that the allowable normal stress for the steel used is 24 ksi, select
the most economical wide-flange beam to support the loading shown.
SOLUTION
q =
Distributed reaction:
Shear:
400
= 33.333 kip/ft
12
VA = 0
VB− = 0 + (4)(33.333) = 133.33 kips
VB+ = 133.33 − 200 = −66.67 kips
VC − = −66.67 + 4(33.333) = 66.67 kips
VC + = 66.67 − 200 = −133.33 kips
VD = −133.33 + (4)(33.333) = 0 kips
Areas:
A to B:
1
 2  (4) (133.33) = 266.67 kip ⋅ ft
 
B to E:
1
 2  (2) (−66.67) = −66.67 kip ⋅ ft
 
E to C:
1
 2  (2) (66.67) = 66.67 kip ⋅ ft
 
C to D:
1
 2  (4) (−133.33) = −266.67 kip ⋅ ft
 
Bending moments:
MA = 0
M B = 0 + 266.67 = 266.67 kip ⋅ ft
M E = 266.67 − 66.67 = 200 kip ⋅ ft
M C = 200 + 66.67 = 266.67 kip ⋅ ft
M D = 266.67 − 266.67 = 0
Maximum |M | = 266.67 kip ⋅ ft = 3200 kip ⋅ in.
σ all = 24 ksi
Smin =
|M |
σ all
=
3200
= 133.3 in 3
24
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 5.83 (Continued)
Shape
S (in 3 )
W 27 × 84
213
W 24 × 68
154
W 21 × 101
227
W18 × 76
146
W16 × 77
134
W14 × 145
232
←
Lightest W-shaped section: W 24 × 68 

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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without permission.
PROBLEM 5.84
Assuming the upward reaction of the ground to be uniformly distributed and
knowing that the allowable normal stress for the steel used is 170 MPa,
select the most economical wide-flange beam to support the loading shown.
SOLUTION
Downward distributed load:
w=
2
= 2 MN/m
1.0
Upward distributed reaction:
q =
2
= 0.8 MN/m
2.5
Net distributed load over BC:
1.2 MN/m
VA = 0
Shear:
VB = 0 + (0.75)(0.8) = 0.6 MN
VC = 0.6 − (1.0)(1.2) = −0.6 MN
VD = −0.6 + (0.75)(0.8) = 0
Areas:
A to B:
1
 2  (0.75) (0.6) = 0.225 MN ⋅ m
 
B to E:
1
 2  (0.5) (0.6) = 0.150 MN ⋅ m
 
E to C:
1
 2  (0.5) (−0.6) = −0.150 MN ⋅ m
 
C to D:
1
 2  (0.75) (−0.6) = −0.225 MN ⋅ m
 
Bending moments:
MA = 0
M B = 0 + 0.225 = 0.225 MN ⋅ m
M E = 0.225 + 0.150 = 0.375 MN ⋅ m
M C = 0.375 − 0.150 = 0.225 MN ⋅ m
M D = 0.225 − 0.225 = 0
Maximum |M | = 0.375 MN ⋅ m = 375 × 103 N ⋅ m
σ all = 170 MPa = 170 × 106 Pa
Smin =
|M |
σ all
=
375 × 103
= 2.206 × 10−3 m3 = 2206 × 103 mm3
170 × 106
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PROBLEM 5.84 (Continued)
Shape
S (103 mm3 )
W 690 × 125
3510
W 610 × 101
2530
W 530 × 150
3720
W 460 × 113
2400
←
Lightest wide flange section: W 610 × 101 

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PROBLEM 5.85
Determine the largest permissible value of P for the beam
and loading shown, knowing that the allowable normal stress
is +6 ksi in tension and −18 ksi in compression.
SOLUTION
Y =
 yA (1)(0.5)(2) + (2.25)(4)(0.5)
=
A
(0.5 × 2) + (4 × 0.5)
= 1.83333 in.
 1

I x′ =   bh3 + Ad 2 
 12

1

=  (0.5)(2)3 + (0.5)(2)(1.83333 − 1) 2 
12

1

+  (4)(0.5)3 + (4)(0.5)(2.25 − 1.83333)2 
12

= 1.41667 in 4
Σ M C = 0: − 30 A + (20) P − (6) P = 0
A = 0.46667 P ↑
Σ Fy = 0:
0.46667 P − 2 P + C = 0
C = 1.53333 P ↑
σ all = +6 ksi
σ all = −18 ksi
At section B:
For σ b = 18 ksi (compression):
σb = M B
cb
I x′
18 = (4.6667 P)
0.66667
1.41667
P = 8.20 kips
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PROBLEM 5.85 (Continued)
For σ a = 6 ksi (tension),
σa = MB
ca
I x′
6 = (4.6667 P)
At section C:
1.83333
1.41667
P = 0.994 kips
For σ b = 6 ksi (tension),
σb = MC
cb
I x′
6 = (6 P)
0.66667
1.41667
P = 2.13 kips
For σ a = 18 ksi (Compression),
σ a = MC
Choose smallest value of P:
ca
I x′
18 = (6P)
1.83333
1.41667
P = 2.32 kips
P = 0.994 kips 
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PROBLEM 5.86
Determine the largest permissible value of P for the beam and
loading shown, knowing that the allowable normal stress is
+6 ksi in tension and −18 ksi in compression.
SOLUTION
Y =
 yA (1.5)(0.5)(2) + (0.25)(4)(0.5)
=
A
(0.5)(2) + (4)(0.5)
= 0.66667 in.
 1

I =   bh3 + Ad 2 
 12

1
=
(0.5)(2)3 + (0.5)(2)(1.5 − 0.66667) 2
12
1
(4)(0.5)3 + (4)(0.5)(0.66667 − 0.25) 2
+
12
= 1.41667 in 4
σ all = +6 ksi, − 18 ksi
For σ a = 18 ksi (compression),
σ a = M max
18 = (8P)
ca
I x′
1.83333
1.41667
P = 1.739 kips
For σ b = 6 ksi (tension),
σ b = M max
6 = (8P)
cb
I x′
0.66667
1.41667
P = 1.594 kips
Choose smallest value of P: P = 1.594 kips 

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PROBLEM 5.87
Determine the largest permissible distributed load w for the
beam shown, knowing that the allowable normal stress is
+80 MPa in tension and −130 MPa in compression.
SOLUTION
By symmetry, B = C
Reactions.
 Fy = 0 : B + C − 0.9w = 0
B = C = 0.45w ↑
VA = 0
Shear:
VB − = 0 − 0.2w = −0.2w
VB + = −0.2w + 0.45w = 0.25w
VC − = 0.25w − 0.5w = −0.25w
VC + = −0.25w + 0.45w = 0.2w
VD = 0.2w − 0.2w = 0
Areas:
A to B.
1
(0.2)(−0.2 w) = −0.02w
2
B to E
1
(0.25)(0.25w) = 0.03125w
2
Bending moments:
MA = 0
M B = 0 − 0.02 w = −0.02 w
M E = −0.02w + 0.03125w = 0.01125w
Centroid and moment of inertia:
Part
A, mm 2
y , mm
Ay (103 mm3 )
d , mm.
Ad 2 (103 mm 4 ) I (103 mm 4 )

1200
70
84
20
480
40

1200
30
36
20
480
360
Σ
2400
960
400
Y =
120 × 103
= 50 mm
2400
I =
 Ad 2 +  I = 1360 × 103 mm 4
120
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PROBLEM 5.87 (Continued)
Top:
I /y = (1360 × 103 )/30 = 45.333 × 103 mm3 = 45.333 × 10−6 m3
Bottom:
I /y = (1360 × 103 ) / (−50) = −27.2 × 103 mm3 = −27.2 × 10−6 m3
Bending moment limits ( M = −σ I / y ) and load limits w.
Tension at B and C:
−0.02 w = −(80 × 106 ) (45.333 × 10−6 )
w = 181.3 × 103 N/m
Compression at B and C:
−0.02 w = −(−130 × 106 ) (27.2 × 10−6 )
w = 176.8 × 103 N/m
Tension at E:
0.01125 w = −(80 × 106 ) (27.2 × 10−6 )
Compression at E:
0.01125 w = −(−130 × 10) (45.333 × 10−6 ) w = 523.8 × 103 N/m
The smallest allowable load controls:
w = 193.4 × 103 N/m
w = 176.8 × 103 N/m
w = 176.8 kN/m 
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PROBLEM 5.88
Solve Prob. 5.87, assuming that the cross section of the beam is
reversed, with the flange of the beam resting on the supports at
B and C.
PROBLEM 5.87 Determine the largest permissible distributed
load w for the beam shown, knowing that the allowable normal
stress is +80 MPa in tension and −130 MPa in compression.
SOLUTION
By symmetry, B = C
Reactions:
Σ Fy = 0: B + C − 0.9 w = 0
B = C = 0.45 w ↑
VA = 0
Shear:
VB − = 0 − 0.2 w = −0.2 w
VB + = −0.2 w + 0.45 w = 0.25 w
VC − = 0.25 w − 0.5 w = −0.25 w
VC + = −0.25 w + 0.45 w = 0.2 w
VD = 0.2 w − 0.2 w = 0
Areas:
1
(0.2)(−0.2 w) = −0.02 w
2
1
(0.25) (0.25 w) = 0.03125 w
2
A to B:
B to E:
MA = 0
Bending moments:
M B = 0 − 0.02 w = −0.02 w
M E = −0.02 w + 0.03125 w = 0.01125 w
Centroid and moment of inertia:
Part
A, mm 2
y , mm
Ay , (103 mm3 )
d , mm
Ad 2 (103 mm 4 )
I , (103 mm 4 )

1200
50
60
20
480
360

1200
10
12
20
480
40
Σ
2400
960
400
72
Y =
72 × 103
= 30 mm
2400
I = Σ Ad 2 + ΣI = 1360 × 103 mm3
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PROBLEM 5.88 (Continued)
Top:
I /y = (1360 × 103 ) / (50) = 27.2 × 103 mm3 = 27.2 × 10−6 m3
Bottom:
I /y = (1360 × 103 ) / (−30) = −45.333 × 108 mm3 = −45.333 × 10−6 m3
Bending moment limits ( M = −σ I / y ) and load limits w.
Tension at B and C:
−0.02 w = −(80 × 106 ) (27.2 × 10−6 )
w = 108.8 × 103 N/m
Compression at B and C:
−0.02 w = −( −130 × 106 ) (−45.333 × 10−6 )
w = 294.7 × 103 N/m
Tension at E:
0.01125 w = −(80 × 106 ) (−45.333 × 10−6 )
w = 322.4 × 103 N/m
Compression at E:
0.01125 w = −( −130 × 106 ) (27.2 × 10−6 )
w = 314.3 × 103 N/m
The smallest allowable load controls:
w = 108.8 × 103 N/m
w = 108.8 kN/m 
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PROBLEM 5.89
A 54-kip is load is to be supported at the center of the 16-ft span shown.
Knowing that the allowable normal stress for the steel used is 24 ksi,
determine (a) the smallest allowable length l of beam CD if the
W12 × 50 beam AB is not to be overstressed, (b) the most economical
W shape that can be used for beam CD. Neglect the weight of both
beams.
SOLUTION
(a)
d = 8ft −
l
l = 16 ft − 2d
2
(1)
Beam AB (Portion AC):
For W12 × 50,
S x = 64.2 in 3 σ all = 24 ksi
M all = σ all S x = (24)(64.2) = 1540.8 kip ⋅ in = 128.4 kip ⋅ ft
M C = 27d = 128.4 kip ⋅ ft
Using (1),
d = 4.7556 ft
l = 16 − 2d = 16 − 2(4.7556) = 6.4888 ft
l = 6.49 ft  
(b)
l = 6.4888 ft σ all = 24 ksi
Beam CD:
M
(87.599 × 12) kip ⋅ in
Smin = max =
24 ksi
σ all
= 43.800 in 3
Shape
W18 × 35
W16 × 31
W14 × 38
W12 × 35
W10 × 45
S (in 3 )
57.6
47.2
←
54.6
45.6
49.1
W16 × 31. 
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PROBLEM 5.90
A uniformly distributed load of 66 kN/m is to be supported over
the 6-m span shown. Knowing that the allowable normal stress for
the steel used is 140 MPa, determine (a) the smallest allowable
length l of beam CD if the W460 × 74 beam AB is not to be
overstressed, (b) the most economical W shape that can be used
for beam CD. Neglect the weight of both beams.
SOLUTION
For W460 × 74,
S = 1460 × 103 mm3 = 1460 × 10−6 m3
σ all = 140 MPa = 140 × 106 Pa
M all = Sσ all = (1460 × 10−6 )(140 × 106 )
= 204.4 × 103 N ⋅ m = 204.4 kN ⋅ m
A = B,
Reactions: By symmetry,
C=D
+↑ ΣFy = 0: A + B − (6)(66) = 0
A = B = 198 kN = 198 × 103 N
+ΣFy = 0: C + D − 66l = 0
C = D = (33l ) kN
(1)
Shear and bending moment in beam AB:
0 < x < a,
V = 198 − 66 x kN
M = 198 x − 33x 2 kN ⋅ m
At C, x = a.
M = M max
M = 198a − 33a 2 kN ⋅ m
Set M = M all .
198a − 33a 2 = 204.4
33a 2 − 198a + 204.4 = 0
a = 4.6751 m ,
(a)
By geometry,
From (1),
1.32487 m
l = 6 − 2a = 3.35 m
l = 3.35 m 
C = D = 110.56 kN
Draw shear and bending moment diagrams for beam CD. V = 0 at point E, the midpoint of CD.
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PROBLEM 5.90 (Continued)
1 
1
 Vdx = 2 (110.560)  2 l  = 92.602 kN ⋅ m
Area from A to E:
M E = 92.602 kN ⋅ m = 92.602 × 103 N ⋅ m
Smin =
ME
σ all
=
92.602 × 103
= 661.44 × 10−6 m3
140 × 106
= 661.44 × 103 mm3
Shape
S (103 mm3 )
W410 × 46.1
774
W360 × 44
693
W310 × 52
748
W250 × 58
693
W200 × 71
709
←
(b) Use W360 × 44. 

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PROBLEM 5.91
Each of the three rolled-steel beams
shown (numbered 1, 2, and 3) is to
carry a 64-kip load uniformly
distributed over the beam. Each of
these beams has a 12-ft span and is to
be supported by the two 24-ft rolledsteel girders AC and BD. Knowing
that the allowable normal stress for
the steel used is 24 ksi, select (a) the
most economical S shape for the
three beams, (b) the most economical
W shape for the two girders.
SOLUTION
For beams 1, 2, and 3
1
Maximum M =   (6)(32) = 96 kip ⋅ ft = 1152 kip ⋅ in
2
Smin =



















M
σ all
1152
=
= 48 in 3
24
Shape
S15 × 42.9
S (in 3 )
59.4
S12 × 50
50.6
(a) Use S15 × 42.9. 
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PROBLEM 5.91 (Continued)
For beams AC and BC
Areas under shear digram:
(4)(48) = 192 kip ⋅ ft
(8)(16) = 128 kip ⋅ ft
Maximum M = 192 + 128 = 320 kip ⋅ ft = 3840 kip ⋅ in
Smin =
M
σ all
Shape
W30 × 99
W27 × 84
W24 × 104
W21 × 101
W18 × 106
=
3840
= 160 in 3
24
S (in 3 )
269
213
258
227
204
(b) Use W27 × 84. 

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PROBLEM 5.92
Beams AB, BC, and CD have the cross section
shown and are pin-connected at B and C. Knowing
that the allowable normal stress is +110 MPa in
tension and −150 MPa in compression, determine (a)
the largest permissible value of w if beam BC is not
to be overstressed, (b) the corresponding maximum
distance a for which the cantilever beams AB and
CD are not overstressed.
SOLUTION
M B = MC = 0
1
VB = −VC =   (7.2) w = 3.6w
2
Area B to E of shear diagram:
1
 2  (3.6) (3.6 w) = 6.48 w
 
M E = 0 + 6.48 w = 6.48 w
Centroid and moment of inertia:
Part
A (mm 2 )
y (mm)
Ay (mm3 )
d (mm)
Ad 2 (mm 4 )
I (mm 4 )

2500
156.25
390625
34.82
3.031 × 106
0.0326 × 106

1875
140625
46.43
4.042 × 106
3.516 × 106
Σ
4375
7.073 × 106
3.548 × 106
75
531250
531250
= 121.43 mm
4375
I = ΣAd 2 + ΣI = 10.621 × 106 mm 4
Y =
Location
Top
Bottom
y (mm)
41.07
−121.43
I / y (103 mm3 )
← also (10−6 m3 )
258.6
−87.47
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PROBLEM 5.92 (Continued)
M = −σ I / y
Bending moment limits:
Tension at E:
− (110 × 106 ) (−87.47 × 10−6 ) = 9.622 × 103 N ⋅ m
Compression at E:
− (−150 × 10−6 )(258.6 × 10−6 ) = 38.8 × 103 N ⋅ m
Tension at A and D:
− (110 × 106 ) (258.6 × 10−6 ) = −28.45 × 103 N ⋅ m
Compression at A and D: − (−150 × 106 )(−87.47 × 10−6 ) = −13.121 × 103 N ⋅ m
(a)
Allowable load w:
Shear at A:
6.48 w = 9.622 × 103
w = 1.485 × 103 N/m
w = 1.485 kN/m 
VA = (a + 3.6) w
Area A to B of shear diagram:
1
1
a (VA + VB ) = a(a + 7.2) w
2
2
1
Bending moment at A (also D): M A = − a(a + 7.2) w
2
1
− a (a + 7.2)(4.485 × 103 ) = −13.121 × 103
2
(b)
Distance a:
1 2
a + 3.6a − 8.837 = 0
2
a = 1.935 m 
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PROBLEM 5.93
Beams AB, BC, and CD have the cross section shown
and are pin-connected at B and C. Knowing that the
allowable normal stress is +110 MPa in tension and
−150 MPa in compression, determine (a) the largest
permissible value of P if beam BC is not to be
overstressed, (b) the corresponding maximum
distance a for which the cantilever beams AB and CD
are not overstressed.
SOLUTION
M B = MC = 0
VB = −VC = P
Area B to E of shear diagram: 2.4 P
M E = 0 + 2.4 P = 2.4 P = M F
Centroid and moment of inertia:
Part
A (mm 2 )
y (mm)
Ay (mm3 )
d (mm)
Ad 2 (mm 4 )
I (mm 4 )

2500
156.25
390625
34.82
3.031 × 106
0.0326 × 106

1875
140625
46.43
4.042 × 106
3.516 × 106
Σ
4375
7.073 × 106
3.548 × 106
75
531250
531250
= 121.43 mm
4375
I = ΣAd 2 + ΣI = 10.621 × 106 mm 4
Y =
Location
Top
Bottom
y (mm)
I / y (103 mm3 )
41.07
258.6
−121.43
← also (10−6 m3 )
−87.47
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PROBLEM 5.93 (Continued)
M = −σ I / y
Bending moment limits:
− (110 × 106 ) ( −87.47 × 10−6 ) = 9.622 × 103 N ⋅ m
Tension at E and F:
Compression at E and F:
−(110 × 106 ) (258.6 × 10−6 ) = −28.45 × 103 N ⋅ m
Tension at A and D:
Compression at A and D:
(a)
Allowable load P:
−( −150 × 106 )( −87.47 × 10−6 ) = −13.121 × 103 N ⋅ m
2.4 P = 9.622 × 103
Shear at A:
(b)
−(−150 × 106 )(258.6 × 10−6 ) = 38.8 × 103 N ⋅ m
P = 4.01 × 103 N
VA = P
Area A to B of shear diagram:
aVA = aP
Bending moment at A:
M A = −aP = −4.01 × 103 a
Distance a:
P = 4.01 kN 
−4.01 × 103 a = −13.121 × 103
a = 3.27 m 

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PROBLEM 5.94∗
A bridge of length L = 48 ft is to be built on a secondary road whose
access to trucks is limited to two-axle vehicles of medium weight. It will
consist of a concrete slab and of simply supported steel beams with an
ultimate strength of σ U = 60 ksi. The combined weight of the slab and
beams can be approximated by a uniformly distributed load
w = 0.75 kips/ft on each beam. For the purpose of the design, it is
assumed that a truck with axles located at a distance a = 14 ft from each
other will be driven across the bridge and that the resulting concentrated
loads P1 and P2 exerted on each beam could be as large as 24 kips and
6 kips, respectively. Determine the most economical wide-flange shape
for the beams, using LRFD with the load factors γ D = 1.25, γ L = 1.75
and the resistance factor φ = 0.9. [Hint: It can be shown that the
maximum value of |M L | occurs under the larger load when that load is
located to the left of the center of the beam at a distance equal
to aP2 /2( P1 + P2 ) .]
SOLUTION
L = 48 ft a = 14 ft
P1 = 24 kips
P2 = 6 kips W = 0.75 kip/ft
Dead load:
1
RA = RB =   (48)(0.75) = 18 kips
2
Area A to E of shear diagram:
1
 2  (8)(18) = 216 kip ⋅ ft
 
M max = 216 kip ⋅ ft = 2592 kip ⋅ in at point E.
Live load:
u=
aP2
(14)(6)
=
= 1.4 ft
2( P1 + P2 ) (2)(30)
L
− u = 24 − 1.4 = 22.6 ft
2
x + a = 22.6 + 14 = 36.6 ft
L − x − a = 48 − 36.6 = 11.4 ft
x=
ΣM B = 0: −48 RA + (25.4)(24) + (11.4)(6) = 0
RA = 14.125 kips
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PROBLEM 5.94* (Continued)
Shear:
A to C:
V = 14.125 kips
C to D:
V = 14.125 − 24 = −9.875 kips
D to B:
V = −15.875 kips
Area:
(22.6)(14.125) = 319.225 kip ⋅ ft
A to C:
M C = 319.225 kip ⋅ ft = 3831 kip ⋅ in
Bending moment:
Design:
γ D M D + γ L M L = ϕ M U = ϕσ U Smin
Smin =
=
γ DM D + γ LM L
ϕσ U
(1.25)(2592) + (1.75)(3831)
(0.9)(60)
= 184.2 in 3

Shape
S (in 3 )
W 30 × 99
269
W 27 × 84
213
W 24 × 104
258
W 21 × 101
227
W18 × 106
204
←
Use W 27 × 84. 
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PROBLEM 5.95
Assuming that the front and rear axle loads remain in the same ratio as
for the truck of Prob. 5.94, determine how much heavier a truck could
safely cross the bridge designed in that problem.
PROBLEM 5.94∗ A bridge of length L = 48 ft is to be built on a
secondary road whose access to trucks is limited to two-axle vehicles of
medium weight. It will consist of a concrete slab and of simply
supported steel beams with an ultimate strength of σ U = 60 ksi. The
combined weight of the slab and beams can be approximated by a
uniformly distributed load w = 0.75 kips/ft on each beam. For the
purpose of the design, it is assumed that a truck with axles located at a
distance a = 14 ft from each other will be driven across the bridge and
that the resulting concentrated loads P1 and P2 exerted on each beam
could be as large as 24 kips and 6 kips, respectively. Determine the
most economical wide-flange shape for the beams, using LRFD with
the load factors γ D = 1.25, γ L = 1.75 and the resistance factor φ = 0.9.
[Hint: It can be shown that the maximum value of |M L | occurs under
the larger load when that load is located to the left of the center of the
beam at a distance equal to aP2 /2( P1 + P2 ) .]
SOLUTION
L = 48 ft
a = 14 ft
P1 = 24 kips
P2 = 6 kips W = 0.75 kip/ft
See solution to Prob. 5.94 for calculation of the following:
M D = 2592 kip ⋅ in
M L = 3831 kip ⋅ in
For rolled steel section W27 × 84, S = 213 in 3
Allowable live load moment M L* :
γ D M D + γ L M L* = ϕ M U = ϕσ U S
ϕσ U S − γ D M D
M L* =
γL
(0.9)(60)(213) − (1.25)(2592)
1.75
= 4721 kip ⋅ in
=
Ratio:
M L* 4721
=
= 1.232 = 1 + 0.232
M L 3831
Increase 23.2%. 
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PROBLEM 5.96
A roof structure consists of plywood and roofing material
supported by several timber beams of length L = 16 m. The
dead load carried by each beam, including the estimated weight
of the beam, can be represented by a uniformly distributed load
wD = 350 N/m. The live load consists of a snow load,
represented by a uniformly distributed load wL = 600 N/m, and
a 6-kN concentrated load P applied at the midpoint C of each
beam. Knowing that the ultimate strength for the timber used is
σ U = 50 MPa and that the width of the beam is b = 75 mm,
determine the minimum allowable depth h of the beams, using
LRFD with the load factors γ D = 1.2, γ L = 1.6 and the
resistance factor φ = 0.9.
SOLUTION
L = 16 m, wD = 350 N/m = 0.35 kN/m
wL = 600 N/m = 0.6 kN/m, P = 6 kN
Dead load:
1
RA =   (16)(0.35) = 2.8 kN
2
Area A to C of shear diagram:
1
 2  (8)(2.8) = 11.2 kN ⋅ m
 
Bending moment at C:
11.2 kN ⋅ m = 11.2 × 103 N ⋅ m
Live load:
1
RA = [(16)(0.6) + 6] = 7.8 kN
2
Shear at C −:
V = 7.8 − (8)(0.6) = 3 kN
Area A to C of shear diagram:
1
 2  (8)(7.8 + 3) = 43.2 kN ⋅ m
 
Bending moment at C:
43.2 kN ⋅ m = 43.2 × 103 N ⋅ m
Design:
γ D M D + γ L M L = ϕ M U = ϕσ U S
S=
γ D M D + γ L M L (1.2)(11.2 × 103 ) + (1.6)(43.2 × 103 )
=
ϕσ U
(0.9)(50 × 106 )
= 1.8347 × 10−3 m3 = 1.8347 × 106 mm3
For a rectangular section,
1
S = bh 2
6
h=
6S
(6)(1.8347 × 106 )
=
75
b
h = 383 mm 
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PROBLEM 5.97∗
Solve Prob. 5.96, assuming that the 6-kN concentrated load P
applied to each beam is replaced by 3-kN concentrated loads P1
and P2 applied at a distance of 4 m from each end of the beams.
PROBLEM 5.96∗ A roof structure consists of plywood and
roofing material supported by several timber beams of length
L = 16 m. The dead load carried by each beam, including the
estimated weight of the beam, can be represented by a uniformly
distributed load wD = 350 N/m. The live load consists of a snow
load, represented by a uniformly distributed load wL = 600 N/m,
and a 6-kN concentrated load P applied at the midpoint C of each
beam. Knowing that the ultimate strength for the timber used is
σ U = 50 MPa and that the width of the beam is b = 75 mm,
determine the minimum allowable depth h of the beams, using
LRFD with the load factors γ D = 1.2, γ L = 1.6 and the resistance
factor φ = 0.9.
SOLUTION
L = 16 m,
a = 4 m,
wD = 350 N/m = 0.35 kN/m
wL = 600 N/m = 0.6 kN/m, P = 3 kN
Dead load:
1
RA =   (16)(0.35) = 2.8 kN
2
Area A to C of shear diagram:
1
 2  (8)(2.8) = 11.2 kN ⋅ m
 
Bending moment at C:
11.2 kN ⋅ m = 11.2 × 103 N ⋅ m
1
RA = [(16)(0.6) + 3 + 3] = 7.8 kN
2
Live load:
Shear at D −:
7.8 − (4)(0.6) = 5.4 kN
Shear at D +:
5.4 − 3 = 2.4 kN
Area
A
to
D:
1
 2  (4)(7.8 + 5.4) = 26.4 kN ⋅ m
 
Area D to C:
1
 2  (4)(2.4) = 4.8 kN ⋅ m
 
Bending moment at C:
26.4 + 4.8 = 31.2 kN ⋅ m
= 31.2 × 103 N ⋅ m
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PROBLEM 5.97* (Continued)
γ D M D + γ L M L = ϕ M U = ϕσ U S
Design:
S=
γ D M D + γ L M L (1.2)(11.2 × 103 ) + (1.6)(31.2 × 103 )
=
ϕσ U
(0.9)(50 × 106 )
= 1.408 × 10−3 m3 = 1.408 × 106 mm3
For a rectangular section,
1
S = bh 2
6
h=
6S
(6)(1.408 × 106 )
=
75
b
h = 336 mm 
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PROBLEM 5.98
(a) Using singularity functions, write the equations defining the shear
and bending moment for the beam and loading shown. (b) Use the
equation obtained for M to determine the bending moment at point C
and check your answer by drawing the free-body diagram of the entire
beam.
SOLUTION
w = w0 −
=−
(a)
V = − w0 x +
w0 x w0
 x − a1
+
a
a
dV
dx
w0 x 2 w0
dM
 x − a 2 =
−
2a
2a
dx

M =−
At point C,
(b)
MC = −
Check:
w0 x 2 w0 x3 w0
+
−
 x − a 3 
2
6a
6a
x = 2a
w0 (2a )2 w0 (2a)3 w0 a 3
+
−
2
6a
6a
5
M C = − w0 a 2 
6
 5  1

ΣM C = 0:  a  w0 a  + M C = 0
 3  2

5
M C = − w0 a 2
6

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PROBLEM 5.99
(a) Using singularity functions, write the equations defining the shear and
bending moment for the beam and loading shown. (b) Use the equation
obtained for M to determine the bending moment at point C and check
your answer by drawing the free-body diagram of the entire beam.
SOLUTION
w = w0 − w0  x − a 0
=−
(a)
V = −w0 x + w0  x − a1 =
dV
dx
dM
dx

1
1
M = − w0 x 2 + w0  x − a 2 
2
2
At point C,
(b)
x = 2a
1
1
M C = − w0 (2a) 2 + w0 a 2
2
2
Check:
3
M C = − w0 a 2 
2
 3a 
ΣM C = 0:   ( w0 a) + M C = 0
 2 
3
M C = − w0 a 2
2

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PROBLEM 5.100
(a) Using singularity functions, write the equations defining the shear
and bending moment for the beam and loading shown. (b) Use the
equation obtained for M to determine the bending moment at point C
and check your answer by drawing the free-body diagram of the entire
beam.
SOLUTION
w0 x
w
− w0  x − a 0 − 0  x − a1
a
a
dV
=−
dx
w=
(a)
V =−
w0 x 2
w
dM
+ w0  x − a1 + 0  x − a 2 =
2a
2a
dx

M =−
At point C,
(b)
MC = −
Check:
w0 x3 w0
w
+
 x − a 2 + 0  x − a 3 
6a
2
6a
x = 2a
w0 (2a)3 w0 a 2 w0 a 3
+
+
6a
2
6a
2
M C = − w0 a 2 
3
 4a  1

ΣM C = 0:   w0 a  + M C = 0
 3  2

2
M C = − w0 a 2
3

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PROBLEM 5.101
(a) Using singularity functions, write the equations defining the shear
and bending moment for the beam and loading shown. (b) Use the
equation obtained for M to determine the bending moment at point E and
check your answer by drawing the free-body diagram of the portion of
the beam to the right of E.
SOLUTION
a
ΣM C = 0: −2aA −   + (3aw0 ) = 0
2
3
A = − w0 a
4
 5a 
ΣM A = 0: 2aC −   + (3aw0 ) = 0
 2 
C=
w = w0  x − a 0 = −

15
w0 a
4
dV

dx

(a)
V = − w0  x − a1 −
3
15
dM
w0 a + w0 a x − 2a 0 =
4
4
dx

1
3
15
M = − w0  x − a 2 − w0 ax + w0 a  x − 2a1 + 0 
2
4
4
At point E ,
(b)
x = 3a
1
3
15
M E = − w0 (2a )2 − w0 a(3a) + w0 a(a )
2
4
4
1
M E = − w0 a 2 
2
Check:
ΣM E = 0: −M E −
a
( w0 a) = 0
2
1
M E = − w0 a 2 
2
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PROBLEM 5.102
(a) Using singularity functions, write the equations defining the shear
and bending moment for the beam and loading shown. (b) Use the
equation obtained for M to determine the bending moment at point E and
check your answer by drawing the free-body diagram of the portion of
the beam to the right of E.
SOLUTION
ΣM D = 0: − 4aA + 3aP + 2aP = 0
(a)
A = 1.25P
V = 1.25 P − P x − a 0 − P x − 2a 0

M = 1.25Px − P x − a1 − P x − 2a1 
(b)
At point E ,
x = 3a
M E = 1.25 P(3a) − P(2a) − P(a ) = 0.750 Pa 
Reaction:
ΣFy = 0: A − P − P + D = 0
D = 0.750P ↑
ΣM E = 0: −M E + 0.750 Pa = 0
M E = 0.750 Pa


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PROBLEM 5.103
(a) Using singularity functions, write the equations defining the shear and
bending moment for the beam and loading shown. (b) Use the equation
obtained for M to determine the bending moment at point E and check your
answer by drawing the free-body diagram of the portion of the beam to the
right of E.
SOLUTION
0
w = w0 + w0 x − a − w0 x − 3a
(a)

V = 3w0 a − wdx
= 3w0a − w0 x − w0 x − a

M = Vdx = 3w0 ax − w0 x 2 /2 − w0 x − a
2
2
At point E,
1
+ w0 x − 3a
1

/2
+ w0 x − 3a /2
(b)
0

x = 3a
M E = 3w0 a (3a ) − w0 (3a ) 2 /2 − w0 (2a )2 /2
M E = 5w0 a 2 / 2 
 M E = 0 : 3w0 a(a) − ( w0 a)( a2 ) − M E = 0
M E = 5w0 a 2 /2 (checks)
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PROBLEM 5.104
(a) Using singularity functions, write the equations for the shear and
bending moment for beam ABC under the loading shown. (b) Use the
equation obtained for M to determine the bending moment just to the
right of point B.
SOLUTION
(a)
V = − P  x − a 0
dM
= − P  x − a 0
dx

M = − P x − a1 − Pa x − a 0 
Just to the right of B, x = a1.
(b)
M = − 0 − Pa
M = − Pa 


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PROBLEM 5.105
(a) Using singularity functions, write the equations for the shear and
bending moment for beam ABC under the loading shown. (b) Use the
equation obtained for M to determine the bending moment just to the
right of point D.
SOLUTION
(a)
V = −P − P x −
2L
3
0
=
dM
dx

M = − Px +
Just to the right of D, x =
(b)
PL
2L
−P x−
3
3
1
−
PL
2L
x−
3
3
0

2L
.
3
PL
 2 L  PL
M D+ = − P 
+
− P(0) −

3
 3  3
M D+ = −
4 PL

3
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PROBLEM 5.106
(a) Using singularity functions, write the equations for the shear and
bending moment for the beam and loading shown. (b) Determine the
maximum value of the bending moment in the beam.
SOLUTION
w = 1.5 kN/m
By statics,
C = D = 3 kN ↑
(a)
V = −1.5 x + 3 x − 0.8 0 + 3 x − 3.2 0 kN

M = −0.75 x 2 + 3 x − 0.81 + 3 x − 3.21 kN ⋅ m 
Locate point E where V = 0.
Assume xC < xE < xD
0 = −1.5 xE + 3( xE − 0.8) + 0
xE = 2.0 m
2
M C = −(0.75)(0.8) + 0 + 0 = −0.480 kN ⋅ m
M E = −(0.75)(2.0) 2 + (3)(1.2) + 0 = 0.600 kN ⋅ m
M D = −(0.75)(3.2) 2 + (3)(2.4) + 0 = −0.480 kN ⋅ m
(b)
|M |max = 0.600 kN ⋅ m
M
max
= 600 N ⋅ m 
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PROBLEM 5.107
(a) Using singularity functions, write the equations for the shear and
bending moment for the beam and loading shown. (b) Determine the
maximum value of the bending moment in the beam.
SOLUTION
ΣM E = 0: − 4.5 RA + (3.0)(48) + (1.5)(60) − (0.9)(60) = 0
RA = 40 kN
(a)
(b)
V = 40 − 48 x − 1.5 0 − 60 x − 3.0 0 + 60 x − 3.6 0 kN

M = 40 x − 48 x − 1.51 − 60 x − 3.01 + 60 x − 3.61 kN ⋅ m

Pt.
x(m)
M (kN ⋅ m)
A
0
0
B
1.5
(40)(1.5) = 60 kN ⋅ m
C
3.0
(40)(3.0) − (48)(1.5) = 48 kN ⋅ m
D
3.6
(40)(3.6) − (48)(2.1) − (60)(0.6) = 7.2 kN ⋅ m
E
4.5
(40)(4.5) − (48)(3.0) − (60)(1.5) + (60)(0.9) = 0
M max = 60 kN ⋅ m 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 5.108
(a) Using singularity functions, write the equations for the shear and
bending moment for the beam and loading shown. (b) Determine the
maximum value of the bending moment in the beam.
SOLUTION
ΣM B = 0: − 14 A + (12.5)(3)(3) + (7)(8) + (1.5)(3)(3) = 0
A = 13 kips ↑
w = 3 − 3 x − 3 0 + 3 x − 11 0 = −
(a)
dV
dx
V = 13 − 3x + 3 x − 31 − 8 x − 7 0 − 3 x − 111 kips

M = 13x − 1.5 x 2 + 1.5 x − 3 2 − 8 x − 71 − 1.5 x − 11 2 kip ⋅ ft

VC = 13 − (3)(3) = 4 kips
VD − = 13 − (3)(7) + (3)(4) = 4 kips
VD + = 13 − (3)(7) + (3)(4) − 8 = −4 kips
VE = 13 − (3)(11) + (3)(8) − 8 = −4 kips
VB = 13 − (3)(14) + (3)(11) − 8 − (3)(3) = −13 kips
(b)
Note that V changes sign at D.
|M |max = M D = (13)(7) − (1.5)(7) 2 + (1.5)(4) 2 − 0 − 0
|M |max = 41.5 kip ⋅ ft 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 5.109
(a) Using singularity functions, write the equations for the shear and
bending moment for the beam and loading shown. (b) Determine the
maximum value of the bending moment in the beam.
SOLUTION
ΣM B = 0: (15)(3) − 12C + (8)(6)C + (4)(6) = 0
C = 9.75 kips ↑
(a)
(b)
V = −3 + 9.75 x − 3 0 − 6 x − 7 0 − 6 x − 11 0 kips

M = −3x + 9.75 x − 31 − 6 x − 71 − 6 x − 111 kip ⋅ ft

M (kip ⋅ ft)
Pt.
x(ft)
A
0
C
3
−(3)(3) = −9
D
7
−(3)(7) + (9.75)(4) = 18
E
11
−(3)(11) + (9.75)(8) − (6)(4) = 21
B
15
−(3)(15) + (9.75)(12) − (6)(8) − (6)(4) = 0
0
← maximum
|M |max = 21.0 kip ⋅ ft 
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PROBLEM 5.110
(a) Using singularity functions, write the equations for the
shear and bending moment for the beam and loading
shown. (b) Determine the maximum stress due to
bending.
SOLUTION
ΣM D = 0:
(1.2)(50) − 0.9 B + (0.5)(125) − (0.2)(50) = 0
B = 125 kN ↑
ΣM B = 0:
(0.3)(50) − (0.4)(125) + 0.9 D − (1.1)(50) = 0
D = 100 kN ↑
(a)
V = −50 + 125 x − 0.3 0 − 125 x − 0.7 0 + 100 x − 1.2 0 kN

M = −50 x + 125 x − 0.31 − 125 x − 0.71 + 100 x − 1.21 kN ⋅ m

M (kN ⋅ m)
Point
x(m)
B
0.3
C
0.7
−(50)(0.7) + (125)(0.4) − 0 + 0 = 15 kN ⋅ m
D
1.2
−(50)(1.2) + (125)(0.9) − (125)(0.5) + 0 = −10 kN ⋅ m
E
1.4
−(50)(1.4) + (125)(1.1) − (125)(0.7) + (100)(0.2) = 0 (checks)
−(50)(0.3) + 0 − 0 + 0 = −15 kN ⋅ m
Maximum M = 15 kN ⋅ m = 15 × 103 N ⋅ m
For S150 × 18.6 rolled steel section, S = 120 × 103 mm3 = 120 × 10−6 m3
(b)
Normal stress:
σ=
M
S
=
15 × 103
= 125 × 106 Pa
120 × 10−6
σ = 125.0 MPa 
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PROBLEM 5.111
(a) Using singularity functions, write the equations for
the shear and bending moment for the beam and
loading shown. (b) Determine the maximum normal
stress due to bending.
SOLUTION
ΣM E = 0: − 3RA + (2.25)(24) − (1.5)(24) − (0.75)(24) + (0.75)(24) = 0
RA = 30 kips
ΣM A = 0: − (0.75)(24) − (1.5)(24) − (2.25)(24) + 3RE − (3.75)(24) = 0
RE = 66 kips
(a)
V = 30 − 24 x − 0.75 0 − 24 x − 1.5 0 − 24 x − 2.25 0 + 66 x − 3 0 kN

M = 30 x − 24 x − 0.751 − 24 x − 1.51 − 24 x − 2.251 + 66 x − 31 kN ⋅ m

Point
x(m)
M (kN ⋅ m)
B
0.75
(30)(0.75) = 22.5 kN ⋅ m
C
1.5
(30)(1.5) − (24)(0.75) = 27 kN ⋅ m
D
2.25
(30)(2.25) − (24)(1.5) − (24)(0.75) = 13.5 kN ⋅ m
E
3.0
(30)(3.0) − (24)(2.25) − (24)(1.5) − (24)(0.75) = −18 kN ⋅ m
F
3.75
(30)(3.75) − (24)(3.0) − (24)(2.25) − (24)(1.5) + (66)(0.75) = 0 
Maximum |M | = 27 kN ⋅ m = 27 × 303 N ⋅ m
For rolled steel section W250 × 28.4, S = 308 × 103 mm3 = 308 × 10−6 m3
(b)
Normal stress:
σ=
|M |
27 × 103
=
= 87.7 × 106 Pa
S
308 × 10−6
σ = 87.7 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 5.112
(a) Using singularity functions, find the
magnitude and location of the maximum
bending moment for the beam and loading
shown. (b) Determine the maximum normal
stress due to bending.
SOLUTION
M c = 0 : 18 − 3.6 A + (1.2)(2.4)(40) − 27 = 0
A = 29.5 kN ↑
1
V = 29.5 − 40 x − 1.2 kN
Point D.
V =0
29.5 − 40( xD − 1.2) = 0
xD = 1.9375 m
2
M = −18 + 29.5 x − 20 x − 1.2 kN ⋅ m
M A = −18 kN ⋅ m
M D = −18 + (29.5)(1.9375) − (20)(0.7375) 2 = 28.278 kN ⋅ m
M E = −18 + (29.5)(3.6) − (20)(2.4) 2 = −27 kN ⋅ m
(a)
M = 28.278 kN ⋅ m at x = 1.9375 m 
Maximum
For S310 × 52 rolled steel section,
S = 624 × 103 mm3
= 624 × 10−6 m3
(b)
Normal stress:
σ =
M
28.278 × 103
=
= 45.3 × 106 Pa
−6
S
624 × 10
σ = 45.3 MPa 
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PROBLEM 5.113
(a) Using singularity functions, find the magnitude and
location of the maximum bending moment for the beam and
loading shown. (b) Determine the maximum normal stress
due to bending.
SOLUTION
M D = 0 : (6)(10) − 5RB + (2)(4)(80) = 0
RB = 140 kN
w = 80 x − 2
0
kN/m = −dV/dx
V = −10 + 140 x − 1
0
A to B:
V = −10 kN
B to C:
V = −10 + 140 = 130 kN
( x = 6)
D:
1
− 80 x − 2 kN
V = −10 + 140 − 80(4) = −190 kN
V changes sign at B and at point E ( x = xE ) between C and D.
V = 0 = −10 + 140 xE − 1
0
= −10 + 140 − 80( xE − 2)
− 80 xE − 2
1
xE = 3.625 m
1
2
M = −10 x + 140 x − 1 − 40 x − 2 kN ⋅ m
At pt. B,
x =1
At pt. E,
x = 3.625
M B = −(10)(1) = −10 kN ⋅ m

M E = −(10)(3.625) + (140)(2.625) − (40)(1.625) 2 = 225.6 kN ⋅ m
(a)
M
For W530 × 150,
S = 3720 × 103 mm3 = 3720 × 10−6 m3
(b)
σ =
Normal stress:
max
= 225.6 kN ⋅ m at x = 3.625 m 
M
225.6 × 103
=
= 60.6 × 106 Pa
S
3720 × 10−6
σ = 60.6 MPa 
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PROBLEM 5.114
A beam is being designed to be supported and loaded as shown.
(a) Using singularity functions, find the magnitude and location of the
maximum bending moment in the beam. (b) Knowing that the
allowable normal stress for the steel to be used is 24 ksi, find the most
economical wide-flange shape that can be used.
SOLUTION
ΣM D = 0: −(16)(12) − 12 B + (8)(24) − (4)(12) = 0
B = −4 kips B = 4 kips ↓
ΣM B = 0: −(4)(12) − (4)(24) + 12 D − (16)(12) = 0
D = 28 kips ↑
ΣFy = 12 − 4 − 24 + 28 − 12 = 0 
Check:
V = 12 − 4 x − 4 0 − 24 x − 8 0 + 28 x − 16 0
M = 12 x − 4 x − 41 − 24 x − 81 + 28 x − 161
At A,
x = 0,
M =0
At B,
x = 4 ft,
M = (12)(4) = 48 kip ⋅ ft
At C,
x = 8 ft,
M = (12)(8) − (4)(4) = 80 kip ⋅ ft
At D,
x = 16 ft,
M = (12)(16) − (4)(12) − (24)(8) = −48 kip ⋅ ft
At E,
x = 20 ft,
M = (12)(20) − (4)(16) − (24)(12) + (28)(4) = 0 (checks)
|M |max = 80 kip ⋅ ft at C. 
(a)
(b)
σ=
Shape
W18 × 35
W16 × 31
W14 × 30
W12 × 35
W10 × 39
W 8 × 48
σ all = 24 ksi
|M |max = 960 kip ⋅ in
|M |
S
S x (in 3 )
57.6
47.2
42.0 ←
45.6
42.1
43.5
Smin =
|M |max
σ
=
960
= 40 in 3
24
Lightest wide flange shape: W14 × 30 
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PROBLEM 5.115
A beam is being designed to be supported and loaded as shown. (a) Using
singularity functions, find the magnitude and location of the maximum
bending moment in the beam. (b) Knowing that the allowable normal
stress for the steel to be used is 24 ksi, find the most economical wideflange shape that can be used.
SOLUTION
ΣM C = 0: − 15RA + (7.5)(15)(3) + (12)(22.5) = 0
RA = 40.5 kips ↑
w = 3 kips/ft = −
dV
dx
V = 40.5 − 3x − 22.5 x − 3 0 kips
M = 40.5 x − 1.5 x 2 − 22.5 x − 31 kip ⋅ ft
(a)
Location of point D where V = 0.
Assume 3 ft < xD < 12 ft.
0 = 40.5 − 3xD − 22.5
At point D, ( x = 6 ft).
xD = 6 ft
M = (40.5)(6) − (1.5)(6) 2 − (22.5)(3)
= 121.5 kip ⋅ ft = 1458 kip ⋅ in
|M |max = 121.5 kip ⋅ ft at
Maximum |M|:
Smin =
(b)
Shape
W 21 × 44
W18 × 50
W16 × 40
W14 × 43
W12 × 50
W10 × 68
S (in 3 )
81.6
88.9
64.7 ←
62.6
64.2
75.7
M
σ all
=
x = 6.00 ft 
1458
= 60.75 in 3
24
Wide-flange shape: W16 × 40 
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PROBLEM 5.116
A timber beam is being designed with supports and loads as
shown. (a) Using singularity functions, find the magnitude and
location of the maximum bending moment in the beam.
(b) Knowing that the available stock consists of beams with an
allowable stress of 12 MPa and a rectangular cross section of
30-mm width and depth h varying from 80 mm to 160 mm in
10-mm increments, determine the most economical cross section
that can be used.
SOLUTION
480 N/m = 0.48 kN/m
ΣM C = 0:
1
− 4 RA + (3)   (1.5)(0.48) + (1.25)(2.5)(0.48) = 0
2
RA = 0.645 kN ↑
dV
0.48
0.48
x−
 x − 1.51 = 0.32 x − 0.32 x − 1.51 kN/m = −
dx
1.5
1.5
2
2
V = 0.645 − 0.16 x + 0.16 x − 1.5 kN
w=
M = 0.645 x − 0.05333x3 + 0.05333 x − 1.5 3 kN ⋅ m
(a)
Locate point D where V = 0.
Assume 1.5 m < xD < 4 m.
0 = 0.645 − 0.16 xD2 + 0.16( xD − 1.5)2
= 0.645 − 0.16 xD2 + 0.16 xD2 − 0.48 xD + 0.36
xD = 2.09375 m 
At point D,
M D = (0.645)(2.09375) − (0.05333)(2.09375)3 + (0.05333)(0.59375)3
M D = 0.87211 kN ⋅ m 
Smin =
MD
σ all
=
1
S = bh 2
6
For a rectangular cross section,
hmin =
(b)
0.87211 × 103
= 72.6758 × 10−6 m3 = 72.6758 × 103 mm3
12 × 106
At next larger 10-mm increment,
h=
6S
b
(6)(72.6758 × 103 )
= 120.56 mm
30
h = 130 mm 
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PROBLEM 5.117
A timber beam is being designed with supports and loads as
shown. (a) Using singularity functions, find the magnitude and
location of the maximum bending moment in the beam.
(b) Knowing that the available stock consists of beams with an
allowable stress of 12 MPa and a rectangular cross section of
30-mm width and depth h varying from 80 mm to 160 mm in
10-mm increments, determine the most economical cross section
that can be used.
SOLUTION
500 N/m = 0.5 kN/m
1
ΣM C = 0: − 4 RA + (3.2)(1.6)(0.5) + (1.6)   (2.4)(0.5) = 0
2
RA = 0.880 kN ↑
0.5
dV
 x − 1.61 = 0.5 − 0.20833 x − 1.61 kN/m = −
2.4
dx
2
V = 0.880 − 0.5 x + 0.104167 x − 1.6 kN
VA = 0.880 kN
w = 0.5 −
VB = 0.880 − (0.5)(1.6) = 0.080 kN

 Sign change
VC = 0.880 − (0.5)(4) + (0.104167)(2.4) = −0.520 kN 
2
Locate point D (between B and C) where V = 0.
0 = 0.880 − 0.5 xD + 0.104167 ( xD − 1.6) 2
0.104167 xD2 − 0.83333 xD + 1.14667 = 0
xD =
0.83333 ± (0.83333)2 − (4)(0.104167)(1.14667)
= 6.2342 , 1.7658 m
(2)(0.104167)

M = 0.880 x − 0.25 x 2 + 0.347222 x − 1.6 3 kN ⋅ m
M D = (0.880)(1.7658) − (0.25)(1.7658) 2 + (0.34722)(0.1658)3 = 0.776 kN ⋅ m
M max = 0.776 kN ⋅ m at
(a)
Smin =
M max
σ all
=
0.776 × 103
= 64.66 × 10−6 m3 = 64.66 × 103 mm3
6
12 × 10
1
For a rectangular cross section, S = bh 2
6
(b)
x = 1.7658 m 
At next higher 10-mm increment,
h=
6S
b
hmin =
(6)(64.66 × 103 )
= 113.7 mm
30
h = 120 mm 
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PROBLEM 5.118
Using a computer and step functions, calculate the shear and bending
moment for the beam and loading shown. Use the specified increment ΔL,
starting at point A and ending at the right-hand support.
SOLUTION
3
3
x − 3 x − 4.5 0 −
 x − 4.51
4.5
4.5
2
2
dV
= x − 3 x − 4.5 0 −  x − 4.51 = −
3
3
dx
1 2
1
V = − x + 3 x − 4.51 +  x − 4.5 2 − 4 x − 6 0
3
3
1 3 3
1
M = − x +  x − 4.5 2 +  x − 4.5 3 − 4 x − 61
9
2
9
w=
x
V
M
ft
kips
kip ⋅ ft
0.0
0.00
0.00
0.5
−0.08
−0.01
1.0
−0.33
−0.11
1.5
−0.75
−0.38
2.0
−1.33
−0.89
2.5
−2.08
−1.74
3.0
−3.00
−3.00
3.5
−4.08
−4.76
4.0
−5.33
−7.11
4.5
−6.75
−10.13
5.0
−6.75
−13.50
5.5
−6.75
−16.88
6.0
−10.75
−20.25
6.5
−10.75
−25.63
7.0
−10.75
−31.00
7.5
−10.75
−36.38
8.0
−10.75
−41.75
8.5
−10.75
−47.13
9.0
−10.75
−52.50
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PROBLEM 5.119
Using a computer and step functions, calculate the shear and bending
moment for the beam and loading shown. Use the specified increment
ΔL, starting at point A and ending at the right-hand support.
SOLUTION
1
ΣM C = 0: −12 RA + (6)(12)(1.8) + (10)   (6)(1.8) = 0
2
RA = 15.3 kips
1.8
1.8
w = 3.6 −
x+
 x − 61
6
6
= 3.6 − 0.3x + 0.3 x − 61
V = 15.3 − 3.6 x + 0.15 x 2 − 0.15 x − 6 2 kips 
x
V
M
ft
kips
kip ⋅ ft
0.0
0.5
15.30
13.54
0.0
7.2
1.0
11.85
13.6
1.5
10.24
19.1
2.0
8.70
2.5
M = 15.3x − 1.8 x 2 + 0.05 x3 − 0.05 x − 6 3 kip ⋅ ft 
x
V
M
ft
kips
kip ⋅ ft
9.5
−7.20
23.6
23.8
10.0
−8.10
19.8
7.24
27.8
10.5
−9.00
15.5
3.0
5.85
31.1
11.0
4.54
33.6
−9.90
10.8
3.5
4.0
3.30
35.6
11.5
−10.80
5.6
4.5
2.14
37.0
12.0
−11.70
0.0
5.0
1.05
37.8
5.5
0.04
38.0
6.0
−0.90
37.8
6.5
−1.80
37.1
7.0
−2.70
36.0
7.5
−3.60
34.4
8.0
−4.50
32.4
8.5
−5.40
29.9
9.0
−6.30
27.0
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PROBLEM 5.120
Using a computer and step functions, calculate the shear and bending
moment for the beam and loading shown. Use the specified increment ΔL,
starting at point A and ending at the right-hand support.
SOLUTION
1
ΣM D = 0: −6 RA + (4)(120) + (1)   (3)(36) = 0
2
RA = 89 kN
36
w =  x − 31 = 12 x − 31
3
V = 89 − 120 x − 2 0 − 6 x − 3 2 kN 
M = 89 x − 120 x − 21 − 2 x − 3 3 kN ⋅ m 
x
V
M
m
kN
kN ⋅ m
0.0
89.0
0.0
x
V
M
0.3
89.0
22.3
m
kN
0.5
89.0
44.5
kN ⋅ m
0.8
89.0
66.8
5.0
−55.0
69.0
1.0
89.0
89.0
5.3
−61.4
54.5
1.3
89.0
111.3
5.5
−68.5
38.3
1.5
89.0
133.5
5.8
−76.4
20.2
1.8
89.0
155.8
6.0
−85.0
−0.0
2.0
−31.0
178.0
2.3
−31.0
170.3
2.5
−31.0
162.5
2.8
−31.0
154.8
3.0
−31.0
147.0
3.3
−31.4
139.2
3.5
−32.5
131.3
3.8
−34.4
122.9
4.0
−37.0
114.0
4.3
−40.4
104.3
4.5
−44.5
93.8
4.8
−49.4
82.0
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PROBLEM 5.121
Using a computer and step functions, calculate the shear and bending
moment for the beam and loading shown. Use the specified increment
ΔL, starting at point A and ending at the right-hand support.
SOLUTION
ΣM C = 0: (5.2)(12) − 4 B + (2)(4)(16) = 0
B = 47.6 kN ↑
ΣM B = 0: (1.2)(12) − (2)(4)(16) + 4C = 0
C = 28.4 kN ↑
w = 16 x − 1.2 0 = −
dV
dx
V = −16 x − 1.21 − 12 + 47.6 x − 1.2 0 
M = −8 x − 1.2 2 − 12 x + 47.6 x − 1.21 

x
V
M
m
kN
kN ⋅ m
0.0
−12.0
0.00
0.4
−12.0
−4.80
0.8
−12.0
−9.60
1.2
35.6
−14.40
1.6
29.2
−1.44
2.0
22.8
8.96
2.4
16.4
16.80
2.8
10.0
22.08
3.2
3.6
24.80
3.6
−2.8
24.96
4.0
−9.2
22.56
4.4
−15.6
17.60
4.8
−22.0
10.08
5.2
−28.4
−0.00
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 5.122
For the beam and loading shown, and using a computer and
step functions, (a) tabulate the shear, bending moment, and
maximum normal stress in sections of the beam from x = 0
to x = L, using the increments ΔL indicated, (b) using
smaller increments if necessary, determine with a 2%
accuracy the maximum normal stress in the beam. Place the
origin of the x-axis at end A of the beam.
SOLUTION
ΣM D = 0:
−5 RA + (4.0)(2.0)(3) + (1.5)(3)(5) + (1.5)(3) = 0
RA = 10.2 kN
w = 3 + 2 x − 2 0 kN/m = −
dV
dx
V = 10.2 − 3x − 2 x − 21 − 3 x − 3.5 0 kN
(a)

M = 10.2 x − 1.5x 2 −  x − 2 2 − 3 x − 3.51 kN ⋅ m
(b)

For rolled steel section W200 × 22.5,
S = 193 × 103 mm3 = 193 × 10−6 m3
σ max =
M max 16.164 × 103
=
= 83.8 × 106 Pa
S
193 × 10−6
σ = 83.8 MPa 
x
V
M
σ
m
kN
kN ⋅ m
MPa
0.00
10.20
0.00
0.0
0.25
9.45
2.46
12.7
0.50
8.70
4.73
24.5
0.75
7.95
6.81
35.3
1.00
7.20
8.70
45.1
1.25
6.45
10.41
53.9
1.50
5.70
11.93
61.8
1.75
4.95
13.26
68.7
2.00
4.20
14.40
74.6
2.25
2.95
15.29
79.2
2.50
1.70
15.88
82.3
2.75
0.45
16.14
83.6
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PROBLEM 5.122 (Continued)
x
V
M
σ
m
kN
kN ⋅ m
MPa
3.00
−0.80
16.10
83.4
3.25
−2.05
15.74
81.6
3.50
−6.30
15.08
78.1
3.75
−7.55
13.34
69.1
4.00
−8.80
11.30
58.5
4.25
−10.05
8.94
46.3
4.50
−11.30
6.28
32.5
4.75
−12.55
3.29
17.1
5.00
−13.80
0.00
0.0
2.83
2.84
0.05
0.00
16.164
16.164
83.8
83.8
2.85
−0.05
16.164
83.8
←
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 5.123
For the beam and loading shown, and using a computer and
step functions, (a) tabulate the shear, bending moment, and
maximum normal stress in sections of the beam from x = 0
to x = L, using the increments ΔL indicated, (b) using
smaller increments if necessary, determine with a 2%
accuracy the maximum normal stress in the beam. Place the
origin of the x-axis at end A of the beam.
SOLUTION
ΣM D = 0: −4 RB + (6)(5) + (2.5)(3)(20) = 0
w = 20 x − 2 0 − 20 x − 5 0 kN/m = −
dV
dx
V = −5 + 45 x − 2 0 − 20 x − 21 + 20 x − 51 kN

M = −5 x + 45 x − 21 − 10 x − 2 2 + 10 x − 5 2 kN ⋅ m

(a)
(b)
RB = 45 kN
Maximum |M | = 30 kN ⋅ m at
x
V
M
stress
m
kN
kN ⋅ m
MPa
0.00
−5
0.00
0.0
0.50
−5
−2.50
−3.3
1.00
−5
−5.00
−6.7
1.50
−5
−7.50
−10.0
2.00
40
−10.00
−13.3
2.50
30
7.50
10.0
3.00
20
20.00
26.7
3.50
10
27.50
36.7
4.00
0
30.00
40.0
4.50
−10
27.50
36.7
5.00
−20
20.00
26.7
5.50
−20
10.00
13.3
6.00
−20
0.00
0.0
←
x = 4.0 m
1
1
For rectangular cross section, S = bh 2 =   (50)(300)2 = 750 × 103 mm3 = 750 × 10−6 m3
6
6
σ max =
M max
30 × 103
=
= 40 × 106 Pa
−6
S
750 × 10
σ max = 40.0 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 5.124
For the beam and loading shown, and using a computer and
step functions, (a) tabulate the shear, bending moment, and
maximum normal stress in sections of the beam from x = 0 to
x = L, using the increments ΔL indicated, (b) using smaller
increments if necessary, determine with a 2% accuracy the
maximum normal stress in the beam. Place the origin of the x
axis at end A of the beam.
SOLUTION
300 lb = 0.3 kips
ΣM D = 0: −5 RA + (4.25)(1.5)(2) + (2.5)(2)(1.2) + (1.5)(0.3) = 0
RA = 3.84 kips
w = 2 − 0.8 x − 1.5 0 − 1.2 x − 3.5 0 kip/ft
V = 3.84 − 2 x + 0.8 x − 1.51 + 1.2 x − 3.51 − 0.3 x − 3.5 0 kips

M = 3.84 x − x 2 + 0.4 x − 1.5 2 + 0.6 x − 3.5 2 − 0.3 x − 3.51 kip ⋅ ft

x
V
M
stress
ft
kips
kip ⋅ ft
ksi
0.00
3.84
0.00
0.000
0.25
3.34
0.90
0.224
0.50
2.84
1.67
0.417
0.75
2.34
2.32
0.579
1.00
1.84
2.84
0.710
1.25
1.34
3.24
0.809
1.50
0.84
3.51
0.877
1.75
0.54
3.68
0.921
2.00
0.24
3.78
0.945
2.25
−0.06
3.80
0.951
2.50
−0.36
3.75
0.937
2.75
−0.66
3.62
0.906
3.00
−0.96
3.42
0.855
3.25
−1.26
3.14
0.786
3.50
−1.86
2.79
0.697
3.75
−1.86
2.32
0.581
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 5.124 (Continued)
x
V
M
stress
ft
kips
kip ⋅ ft
ksi
4.00
−1.86
1.86
0.465
4.25
−1.86
1.39
0.349
4.50
−1.86
0.93
0.232
4.75
−1.86
0.46
0.116
5.00
−1.86
−0.00
−.000
2.10
0.12
3.80
0.949
2.20
0.00
3.80
0.951 ←
2.30
−0.12
3.80
0.949
Maximum |M | = 3.804 kip ⋅ ft = 45.648 kip ⋅ in at
x = 2.20 ft
Rectangular section:
2 in. × 12 in.
1
1
S = bh 2 =   (2)(12) 2
6
6
= 48 in 3
σ=
M 45.648
=
S
48
σ = 0.951 ksi 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 5.125
For the beam and loading shown, and using a computer and
step functions, (a) tabulate the shear, bending moment, and
maximum normal stress in sections of the beam from x = 0
to x = L, using the increments ΔL indicated, (b) using
smaller increments if necessary, determine with a 2%
accuracy the maximum normal stress in the beam. Place the
origin of the x axis at end A of the beam.
SOLUTION
ΣM D = 0: − 12.5 RB + (12.5)(5.0)(4.8) + (5)(10)(3.2) = 0
RB = 36.8 kips
w = 4.8 − 1.6 x − 5 0 kips/ft
V = −4.8 x + 36.8 x − 2.5 0 + 1.6 x − 51 kips

M = −2.4 x 2 + 36.8 x − 2.51 + 0.8 x − 5 2 kip ⋅ ft

x
V
M
stress
ft
kips
kip ⋅ ft
ksi
0.00
0.00
0.00
0.00
1.25
−6.0
−3.75
−1.17
2.50
24.8
−15.00
−4.66
3.75
18.8
12.25
3.81
5.00
12.8
32.00
9.95
6.25
8.8
45.50
14.15
7.50
4.8
54.00
16.79
8.75
0.8
57.50
17.88
10.00
−3.2
56.00
17.41
11.25
−7.2
49.50
15.39
12.50
−11.2
38.00
11.81
13.75
−15.2
21.50
6.68
15.00
−19.2
0.00
0.00
8.90
0.32
57.58
17.90
9.00
−0.00
57.60
17.91 ←
9.10
−0.32
57.58
17.90
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 5.125 (Continued)
Maximum |M |= 57.6 kip ⋅ ft = 691.2 kip ⋅ in at
x = 9.0 ft
For rolled steel section W12 × 30,
S = 38.6 in 3
Maximum normal stress:
σ=
M 691.2
=
S
38.6
σ = 17.91 ksi 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 5.126
The beam AB, consisting of an aluminum plate of uniform thickness b and
length L, is to support the load shown. (a) Knowing that the beam is to be
of constant strength, express h in terms of x, L, and h0 for portion AC of the
beam. (b) Determine the maximum allowable load if L = 800 mm,
h0 = 200 mm, b = 25 mm, and σ all = 72 MPa.
SOLUTION
RA = RB =
ΣM J = 0: −
M=
S=
For a rectangular cross section,
Equating,
(a)
At x =
L
,
2
For x >
(b)
P
↑
2
P
x+M =0
2
L

0 < x < 2 


Px
2
M
σ all
=
Px
2σ all
1
S = bh 2
6
1 2
Px
bh =
6
2σ all
h=
h = h0 =
3PL
2σ all b
3Px
σ allb
h = h0
2x
L
, 0< x< 
L
2
L
, replace x by L − x.
2
Solving for P,
P=
2σ all bh02 (2)(72 × 106 )(0.025)(0.200) 2
=
= 60 × 103 N
3L
(3)(0.8)
P = 60 kN 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 5.127
The beam AB, consisting of an aluminum plate of uniform thickness b and
length L, is to support the load shown. (a) Knowing that the beam is to be
of constant strength, express h in terms of x, L, and h0 for portion AC of the
beam. (b) Determine the maximum allowable load if L = 800 mm,
h0 = 200 mm, b = 25 mm, and σ all = 72 MPa.
SOLUTION
A = M 0 /L ↓
B = M 0 /L ↑
M0
L
M 0x
M =−
L
ΣM J = 0:
For x >
L
,
2
S =
For x >
M 0 ( L − x)
L
M =
M
σ all
=
M 0x
σ all L
x+M =0
L

0 < x< 
2

L

 2 < x< L


L

0 < x 2 


for
L
, replace x by L − x.
2
S =
For a rectangular cross section,
1 2
bh
6
1 2 M0 x
bh =
σ all L
6
Equating,
L
,
2
(a)
At x =
(b)
Solving for M 0 , M 0 =
h = h0 =
σ allbh02
3
=
h=
6M 0 x
σ allbL
3M 0
σ allb
h = h0 2 x/L 
(72 × 106 )(0.025)(0.200) 2
= 24 × 103 N ⋅ m
3
M 0 = 24 kN ⋅ m 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 5.128
The beam AB, consisting of a cast-iron plate of uniform thickness b and
length L, is to support the load shown. (a) Knowing that the beam is to
be of constant strength, express h in terms of x, L, and h0. (b) Determine
the maximum allowable load if L = 36 in., h0 = 12 in., b = 1.25 in., and
σ all = 24 ksi.
SOLUTION
V = −P
M = − Px
|M | = Px
|M |
P
S=
x
=
σ all
σ all
1
S = bh 2
6
For a rectangular cross section,
1 2 Px
bh =
6
σ all
Equating,
1/2
 6 Px 
h=

 σ all b 
(1)
1/2
 6PL 
h = h0 = 

 σ all b 
At x = L,
(a)
Divide Eq. (1) by Eq. (2) and solve for h.
(b)
Solving for P,
P=
σ all bh02
6L
(2)
h = h0 ( x/L)1/2 
=
(24)(1.25)(12)2
(6)(36)
P = 20.0 kips 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 5.129
The beam AB, consisting of a cast-iron plate of uniform thickness b and
length L, is to support the load shown. (a) Knowing that the beam is to
be of constant strength, express h in terms of x, L, and h0. (b) Determine
the maximum allowable load if L = 36 in., h0 = 12 in., b = 1.25 in., and
σ all = 24 ksi.
SOLUTION
ΣFy = 0: RA + RB − wL = 0
RA = RB =
wL
2
ΣM J = 0:
wL
x
x − wx + M = 0
2
2
S=
For a rectangular cross section,
Equating,
M=
|M |
σ all
=
w
x( L − x)
2
wx( L − x)
2σ all
1
S = bh 2
6
1/2
 3wx( L − x) 
h=

 σ all b 
1 2 wx( L − x)
bh =
6
2σ all
1/2
(a)
L
At x = ,
2
2
 3wL 
h = h0 = 

 4σ all b 
(b)
Solving for w,
w=
4σ all bh02
3L2
=
(4)(24)(1.25)(12) 2
(3)(36) 2
1/2
x
x 
h = h0  1 −  
L 
L

w = 4.44 kip/in 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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without permission.
PROBLEM 5.130
The beam AB, consisting of a cast-iron plate of uniform thickness b and
length L, is to support the distributed load w(x) shown. (a) Knowing
that the beam is to be of constant strength, express h in terms of x, L,
and h0 . (b) Determine the smallest value of h0 if L = 750 mm,
b = 30 mm, w0 = 300 kN/m, and σ all = 200 MPa.
SOLUTION
wx
dV
= −w = − 0
dx
L
2
wx
dM
V =− 0 =
2L
dx
3
wx
M =− 0
6L
w x3
|M |
S=
= 0
σ all 6 Lσ all
For a rectangular cross section,
At x = L,
w0 x3
6L
1
S = bh 2
6
w x3
1 2
bh = 0
6
6 Lσ all
Equating,
|M |=
h = h0 =
h=
w0 x3
σ allbL
w0 L2
σ allb
x
h = h0  
L
(a)
Data:
3/2

L = 750 mm = 0.75 m, b = 30 mm = 0.030 m
w0 = 300 kN/m = 300 × 103 N/m, σ all = 200 MPa = 200 × 106 Pa
(b)
h0 =
(300 × 103 )(0.75) 2
= 167.7 × 10−3 m
6
(200 × 10 )(0.030)
h0 = 167.7 mm 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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PROBLEM 5.131
The beam AB, consisting of a cast-iron plate of uniform thickness b and
length L, is to support the distributed load w(x) shown. (a) Knowing
that the beam is to be of constant strength, express h in terms of x, L, and
h0 . (b) Determine the smallest value of h0 if L = 750 mm, b = 30 mm,
w0 = 300 kN/m, and σ all = 200 MPa.
SOLUTION
πx
dV
= − w = − w0 sin
2L
dx
2w L
πx
+ C1
V = 0 cos
2L
π
V = 0 at
x = 0 → C1 =
2 w0 L
π
2w L 
dM
πx
= V = − 0 1 − cos

dx
π 
2L 
2 w0 L 
2w L 
2L
2L
πx
πx
sin
|M | = 0  x −
sin
x−


2L 
L 
π 
π
π 
π
|M | 2 w0 L 
2L
πx
=
sin
S=
x−
2 L 
σ all πσ all 
π
M =−
1
S = bh 2
6
For a rectangular cross section,
1 2 2 w0 L 
2L
πx
sin
bh =
x−
6
2 L 
πσ all 
π
Equating,
1/2
12w0 L 
π x 
2L
sin
h=
x−

π
2L 
 πσ all b 
At x = L,
2
12 w0 L
h = h0 = 
 πσ all b
1/ 2
2  

1 −  
 π  
= 1.178
w0 L2
σ allb
1/2
 x 2
π x   2 
h = h0  − sin
 1 − 
2 L   π  
 L π
(a)
Data:
1/2
πx
x 2
h = 1.659 h0  − sin
2 L 
L π

L = 750 mm = 0.75 m, b = 30 mm = 0.030 m
w0 = 300 kN/m = 300 × 103 N/m, σ all = 200 MPa = 200 × 106 Pa
(b)
h0 = 1.178
(300 × 103 )(0.75) 2
= 197.6 × 10−3 m
6
(200 × 10 )(0.030)
h0 = 197.6 mm 
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PROBLEM 5.132
A preliminary design on the use of a simply supported prismatic timber
beam indicated that a beam with a rectangular cross section 50 mm wide
and 200 mm deep would be required to safely support the load shown in
part a of the figure. It was then decided to replace that beam with a
built-up beam obtained by gluing together, as shown in part b of the
figure, four pieces of the same timber as the original beam and of
50 × 50-mm cross section. Determine the length l of the two outer
pieces of timber that will yield the same factor of safety as the original
design.
SOLUTION
P
2
RA = RB =
0< x<
1
2
P
x+M =0
2
M x
or M = max
1.2
ΣM J = 0:
M=
−
Px
2
Bending moment diagram is two straight lines.
At C,
SC =
1 2
bhC
6
M C = M max
Let D be the point where the thickness changes.
At D,
SD =
1 2
bhD
6
MD =
M max xD
1.2
2
S D hD2  100 mm 
1 M D xD
= 2 =
=
 = =
4 M C 1.2
SC hC  200 mm 
l
= 1.2 − xD = 0.9
2
xD = 0.3 m
l = 1.800 m 
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PROBLEM 5.133
A preliminary design on the use of a simply supported prismatic timber
beam indicated that a beam with a rectangular cross section 50 mm
wide and 200 mm deep would be required to safely support the load
shown in part a of the figure. It was then decided to replace that beam
with a built-up beam obtained by gluing together, as shown in part b of
the figure, four pieces of the same timber as the original beam and of
50 × 50-mm cross section. Determine the length l of the two outer
pieces of timber that will yield the same factor of safety as the original
design.
SOLUTION
RA = RB =
0.8 N
= 0.4 w
2
Shear:
A to C:
V = 0.4 w
D to B:
V = −0.4 w
Areas:
A to C:
(0.8)(0.4) w = 0.32 w
C to E:
1
 2  (0.4)(0.4) w = 0.08 w
 
Bending moments:
M C = 0.40 w
At C,
M = 0.40 wx
A to C:
At C,
SC =
1 2
bhC
6
M C = M max = 0.40 w
Let F be the point were the thickness changes.
At F,
SF =
1 2
bhF
6
M F = 0.40 wxF
2
S F hF2  100 mm 
1 M F 0.40 wxF
= 2 =
=
 = =
4 MC
0.40 w
SC hC  200 mm 
xF = 0.25 m
l
= 1.2 − xF = 0.95 m
2
l = 1.900 m 
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PROBLEM 5.134
A preliminary design on the use of a cantilever prismatic timber
beam indicated that a beam with a rectangular cross section 2 in.
wide and 10 in. deep would be required to safely support the load
shown in part a of the figure. It was then decided to replace that
beam with a built-up beam obtained by gluing together, as shown in
part b of the figure, five pieces of the same timber
as the original beam and of 2 × 2-in. cross section. Determine the
respective lengths l1 and l2 of the two inner and outer pieces of
timber that will yield the same factor of safety as the original
design.
SOLUTION
ΣM J = 0: Px + M = 0
M = − Px
|M | = Px
At B,
|M |B = M max
At C,
|M |C = M max xC /6.25
At D,
|M |D = M max xD /6.25
1
1
25
S B = bh 2 = ⋅ b(5b) 2 = b3
6
6
6
A to C:
SC =
1
1
⋅ b(b) 2 = b3
6
6
C to D:
SD =
1
9
b (3b)2 = b3
6
6
|M |C
x
S
1
= C = C =
|M |B 6.25 S B 25
xC =
(1)(6.25)
= 0.25 ft
25
l1 = 6.25 − 0.25
| M |D
x
S
9
= D = D =
|M |B 6.25 S B 25
xD =
l1 = 6.00 ft 
(9)(6.25)
= 2.25 ft
25
l2 = 6.25 − 2.25
l2 = 4.00 ft 
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PROBLEM 5.135
A preliminary design on the use of a cantilever prismatic timber beam
indicated that a beam with a rectangular cross section 2 in. wide and
10 in. deep would be required to safely support the load shown in part a
of the figure. It was then decided to replace that beam with a built-up
beam obtained by gluing together, as shown in part b of the figure, five
pieces of the same timber as the original beam and of 2 × 2-in. cross
section. Determine the respective lengths l1 and l2 of the two inner and
outer pieces of timber that will yield the same factor of safety as the
original design.
SOLUTION
ΣM J = 0: wx
M =−
wx 2
2
x
+M =0
2
wx 2
|M |=
2
At B,
|M |B = |M |max
At C,
|M |C = |M |max ( xC /6.25) 2
At D,
|M |D = |M |max ( xD /6.25) 2
At B,
SB =
1 2 1
25
bh = b(5b) 2 = b3
6
6
6
A to C:
SC =
1 2 1
1
bh = b(b) 2 = b3
6
6
6
C to D:
SD =
1 2 1
9
bh = b (3b)2 = b3
6
6
6
2
|M |C  xC 
S
1
=
= C =

S B 25
|M |B  6.25 
xC =
6.25
25
= 1.25 ft
l1 = 6.25 − 1.25 ft
2
| M | D  xD 
S
9
=
= D =

|M |B  6.25 
S B 25
xD =
6.25 9
25
l1 = 5.00 ft 
= 3.75 ft
l2 = 6.25 − 3.75 ft
l2 = 2.50 ft 
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PROBLEM 5.136
A machine element of cast aluminum and in the shape of a solid of
revolution of variable diameter d is being designed to support the load
shown. Knowing that the machine element is to be of constant strength,
express d in terms of x, L, and d 0 .
SOLUTION
RA = RB =
ΣM J = 0: −
For a solid circular cross section,
Equating,
L
At x = ,
2
c=
wL
x
x + wx + M = 0
2
2
w
x( L − x)
2
M=
S=
wL
2
|M |
σ all
d
2
=
wx( L − x)
2σ all
I=
π
4
c3
S=
I π d3
=
c
32
1/ 3
16 wx ( L − x) 
d =

πσ all


π d3
wx( L − x)
=
32
2σ all
1/ 3
2
 4wL 
d = d0 = 

 πσ all 
1/3
 x
x 
d = d 0 4  1 −  
L 
 L

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PROBLEM 5.137
A machine element of cast aluminum and in the shape of a solid of
revolution of variable diameter d is being designed to support the load
shown. Knowing that the machine element is to be of constant strength,
express d in terms of x, L, and d 0 .
SOLUTION
Draw shear and bending moment diagrams.
0≤ x≤
L
,
2
Px
2
P( L − x)
M=
2
M=
L
≤ x ≤ L,
2
For a solid circular section, c =
I=
1
d
2
π
4
c4 =
π
64
d4
S=
For constant strength design, σ = constant.
For
For
S=
π
L
,
2
32
L
≤ x ≤ L,
2
32
0≤ x≤
π
Dividing Eq. (1b) by Eq. (2),
L
,
2
L
≤ x ≤ L,
2
σ
Px
2
(1a)
d3 =
P( L − x)
2
(1b)
d 03 =
PL
4
32
0≤ x≤
M
d3 =
π
At point C,
Dividing Eq. (1a) by Eq. (2),
I π 3
=
d
c 32
d 3 2x
=
L
d 03
d 3 2( L − x)
=
L
d 03
(2)
d = d 0 (2 x/L)1/3 
d = d 0 [2( L − x)/L]1/3 
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PROBLEM 5.138
A cantilever beam AB consisting of a steel plate of uniform depth h and variable
width b is to support the distributed load w along its centerline AB. (a) Knowing
that the beam is to be of constant strength, express b in terms of x, L, and b0 .
(b)
Determine
the
maximum
allowable
value
of
w
if
L = 15 in., b0 = 8 in., h = 0.75 in., and σ all = 24 ksi.
SOLUTION
L−x
=0
2
w ( L − x)2
|M | =
2
ΣM J = 0: − M − w ( L − x)
w ( L − x) 2
2
|M | w ( L − x ) 2
S=
=
σ all
2σ all
M =−
For a rectangular cross section,
1
S = bh 2
6
1 2 w( L − x) 2
bh =
6
2σ all
b=
3w( L − x) 2
σ all h 2
(a)
At
3wL2
x = 0, b = b0 =
σ all h 2
(b)
Solving for w,
w=
σ allb0 h 2
3L2
=
(24)(8)(0.75)2
= 0.160 kip/in
(3)(15) 2
2
x

b = b0 1 −  
L

w = 160.0 lb/in 
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PROBLEM 5.139
A cantilever beam AB consisting of a steel plate of uniform depth h and variable
width b is to support the concentrated load P at point A. (a) Knowing that the beam
is to be of constant strength, express b in terms of x, L, and b0 . (b) Determine the
smallest allowable value of h if L = 300 mm, b0 = 375 mm, P = 14.4 kN, and
σ all = 160 MPa.
SOLUTION
ΣM J = 0: −M − P( L − x) = 0
M = − P( L − x)
|M | = P ( L − x )
|M | P ( L − x )
S=
=
σ all
For a rectangular cross section,
At x = 0,
b = b0 =
h=
Solving for h,
Data:
1
S = bh 2
6
1 2 P( L − x)
bh =
σ all
6
Equating,
(a)
σ all
b=
6 P( L − x)
σ all h 2
6PL
σ all h 2
x

b = b0 1 −  
L


6PL
σ allb0
L = 300 mm = 0.300 m, b0 = 375 mm = 0.375 m
P = 14.4 kN = 14.4 × 103 N ⋅ m, σ all = 160 MPa = 160 × 106 Pa
(b)
h=
(6)(14.4 × 103 )(0.300)
= 20.8 × 10−3 m
6
(160 × 10 )(0.375)
h = 20.8 mm 
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PROBLEM 5.140
Assuming that the length and width of the cover plates used with the beam of Sample Prob. 5.12 are,
respectively, l = 4 m and b = 285 mm, and recalling that the thickness of each plate is 16 mm, determine the
maximum normal stress on a transverse section (a) through the center of the beam, (b) just to the left of D.
SOLUTION
A = B = 250 kN ↑
ΣM J = 0: −250 x + M = 0
M = 250 x kN ⋅ m
At center of beam,
x = 4 m M C = (250)(4) = 1000 kN ⋅ m
At D,
x=
At center of beam,
I = I beam + 2I plate
1
1
(8 − l ) = (8 − 4) = 2 m
2
2
M 0 = 500 kN ⋅ m
2


1
 678 16 
= 1190 × 106 + 2 (285)(16) 
+  +
(285)(16)3 
2
12
 2


= 2288 × 106 mm 4
c=
678
+ 16 = 355 mm
2
S =
I
= 6445 × 103 mm3
c
= 6445 × 10−6 m3
(a)
Normal stress:
At D,
(b)
Normal stress:
σ =
M
1000 × 103
=
= 155.2 × 106 Pa
−6
S
6445 × 10
σ = 155.2 MPa 
S = 3490 × 103 mm3 = 3510 × 10−6 m3
σ =
500 × 103
M
=
= 143.3 × 106 Pa
S
3490 × 10−6
σ = 143.3 MPa 
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PROBLEM 5.141
Knowing that σ all = 150 MPa, determine the largest concentrated
load P that can be applied at end E of the beam shown.
SOLUTION
ΣM C = 0: −4.8 A − 2.2P = 0
A = −0.45833P
A = 0.45833 P ↓
ΣM A = 0: 4.8D − 7.0P = 0
D = 1.45833P ↑
Shear:
A to C:
V = −0.45833P
C to E:
V = P
Bending moments:
MA = 0
M C = 0 + (4.8)(−0.45833P) = −2.2P
M E = −2.2 P + 2.2P = 0
 4.8 − 2.25 
MB = 
 ( −2.2 P ) = −1.16875P
48


 2.2 − 1.25 
MD = 
 (−2.2P) = −0.95P
2.2


M D | < |M B |
<figure>
For W410 × 85,
Allowable value of P based on strength at B. σ =
150 × 106 =
1.16875P
1510 × 10−6
S = 1510 × 103 mm3 = 1510 × 10−6 m3
|M B |
S
P = 193.8 × 103 N
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PROBLEM 5.141 (Continued)
Section properties over portion BCD:
1
d = 208.5 mm, I x = 316 × 106 mm 4
2
W410 × 85: d = 417 mm,
Plate:
A = (18)(220) = 3960 mm 2
I =
1
d = 208.5 +   (18) = 217.5 mm
2
1
(220)(18)3 = 106.92 × 103 mm 4
12
Ad 2 = 187.333 × 106 mm 4
I x = I + Ad 2 = 187.440 × 106 mm 4
For section,
I = 316 × 106 + (2)(187.440 × 106 ) = 690.88 × 106 mm 4
c = 208.5 + 18 = 226.5 mm
S =
I
690.88 × 106
=
= 3050.2 × 103 mm3 = 3050.2 × 10−6 m3
c
226.5
Allowable load based on strength at C:
150 × 106 =
σ =
|M C |
S
2.2P
3050.2 × 10−6
The smaller allowable load controls.
P = 208.0 × 103 N
P = 193.8 × 103 N
P = 193.8 kN 
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PROBLEM 5.142
Two cover plates, each 58 in. thick, are welded to a W 30 × 99
beam as shown. Knowing that l = 9 ft and b = 12 in., determine
the maximum normal stress on a transverse section (a) through the
center of the beam, (b) just to the left of D.
SOLUTION
A = B = 240 kips ↑
ΣM J = 0: −240 x + 30 x
x
+M =0
2
M = 240 x − 15x 2 kip ⋅ ft
x = 8 ft
At center of beam,
M C = 960 kip ⋅ ft = 11,520 kip ⋅ in
x=
At point D,
1
(16 − 9) = 3.5 ft
2
M D = 656.25 kip ⋅ ft = 7875 kip ⋅ in
At center of beam,
I = I beam + 2I plate
2


1
 29.7 0.625 
3
4
+
+
I = 3990 + 2 (12)(0.625) 
(12)(0.625)
 = 7439 in

2
2
12




c=
(a)
(b)
29.7
+ 0.625 = 15.475 in.
2
Mc (11,520)(15.475)
=
I
7439
Normal stress:
σ =
At point D,
S = 269 in 3
Normal stress:
σ =
M
7875
=
S
269
σ = 24.0 ksi 
σ = 29.3 ksi 
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PROBLEM 5.143
Two cover plates, each 58 in. thick, are welded to a W30 × 99 beam
as shown. Knowing that σ all = 22 ksi for both the beam and the
plates, determine the required value of (a) the length of the plates,
(b) the width of the plates.
SOLUTION
RA = RB = 240 kips ↑
ΣM J = 0: −240 x + 30 x
x
+M =0
2
M = 240 x − 15x 2 kip ⋅ ft
For W30 × 99 rolled steel section, S = 269 in 3
Allowable bending moment:
M all = σ all S = (22)(269) = 5918 kip ⋅ in = 493.167 kip ⋅ ft
To locate points D and E, set M = M all .
240 x − 15x 2 = 493.167
x=
15x 2 − 240 x + 493.167 = 0
240 ± (240) 2 − (4)(15)(493.167)
= 2.42 ft, 13.58 ft
(2)(15)
l = 11.16 ft 
l = xE − xD = 13.58 − 2.42
(a)
M = 960 kip ⋅ ft = 11520 kip ⋅ in
Center of beam:
S =
M
σ all
=
11520
= 523.64 in 3
22
Required moment of inertia:
But
c=
29.7
+ 0.625 = 15.475 in.
2
I = Sc = 8103.3 in 4
I = I beam + 2I plate
2


1
 29.7 0.625 
+
8103.3 = 3990 + 2 (b)(0.625) 
(b)(0.625)3 
 +
2 
12
 2


= 3990 + 287.42b
(b)
b = 14.31 in. 
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PROBLEM 5.144
Two cover plates, each 7.5 mm thick, are welded to
a W460 × 74 beam as shown. Knowing that l = 5 m and
b = 200 mm, determine the maximum normal stress on
a transverse section (a) through the center of the beam,
(b) just to the left of D.
SOLUTION
RA = RB = 160 kN ↑
x
+M =0
2
M = 160 x − 20 x 2 kN ⋅ m
ΣM J = 0: −160 x + (40 x)
At center of beam,
x = 4m
At D,
x=
At center of beam,
M C = 320 kN ⋅ m
1
(8 − l ) = 1.5 m
2
M D = 195 kN ⋅ m
I = I beam + 2 I plate
2


1
 457 7.5 
= 333 × 106 + 2 (200)(7.5) 
+
+ (200)(7.5)3 

2  12
 2


= 494.8 × 106 mm 4
457
+ 7.5 = 236 mm
2
I
S = = 2097 × 103 mm3 = 2097 × 10−6 m3
c
c=
(a)
Normal stress:
σ=
M
320 × 103
=
= 152.6 × 106 Pa
−6
S
2097 × 10
σ = 152.6 MPa 
(b)
At D,
S = 1460 × 103 mm3 = 1460 × 10−6 m3
Normal stress:
σ=
M
195 × 103
=
= 133.6 × 106 Pa
S 1460 × 10−6
σ = 133.6 MPa 
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PROBLEM 5.145
Two cover plates, each 7.5 mm thick, are welded to a
W460 × 74 beam as shown. Knowing that σ all = 150 MPa
for both the beam and the plates, determine the required
value of (a) the length of the plates, (b) the width of the
plates.
SOLUTION
RA = RB = 160 kN ↑
x
ΣM J = 0: −160 x + (40 x)   + M = 0
2
M = 160 x − 20 x 2 kN ⋅ m
For W460 × 74 rolled steel beam,
S = 1460 × 103 mm3 = 1460 × 10−6 m3
Allowable bending moment:
M all = σ all S = (150 × 106 )(1460 × 10−6 )
= 219 × 103 N ⋅ m = 219 kN ⋅ m
M = M all
To locate points D and E, set
160 x − 20 x 2 = 219
x=
(a)
20 x 2 − 160 x + 219 = 0
160 + 1602 − (4)(20)(219)
(2)(20)
xD = 1.753 ft
At center of beam,
x = 1.753 m and x = 6.247 m
xE = 6.247 ft
l = xE − xD = 4.49 m 
M = 320 kN ⋅ m = 320 × 103 N ⋅ m
S=
M
σ all
=
c=
457
+ 7.5 = 236 mm 4
2
320 × 103
= 2133 × 10−6 m3 = 2133 × 103 mm3
6
150 × 10
Required moment of inertia:
I = Sc = 503.4 × 106 mm 4
But
I = I beam + 2 I plate
2


1
 457 7.5 
+
+ (b)(7.5)3 
503.4 × 10 = 333 × 10 + 2 (b)(7.5) 

2  12
 2


6
(b)
6
= 333 × 106 + 809.2 × 103 b
b = 211 mm 
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PROBLEM 5.146
Two cover plates, each 12 in. thick, are welded to a
W 27 × 84 beam as shown. Knowing that l = 10 ft and
b = 10.5 in., determine the maximum normal stress on a
transverse section (a) through the center of the beam, (b) just
to the left of D.
SOLUTION
RA = RB = 80 kips ↑
ΣM J = 0: −80 x + M = 0
M = 80 x kip ⋅ ft
At C,
x = 9 ft M C = 720 kip ⋅ ft = 8640 kip ⋅ in
At D,
x = 9 − 5 = 4 ft
M D = (80)(4) = 320 kip ⋅ ft = 3840 kip ⋅ in
At center of beam,
I = I beam + 2 I plate
2


1
 26.71 0.500 
+
+ (10.5)(0.500)3 
I = 2850 + 2 (10.5)(0.500) 

2  12
 2


= 4794 in 3
26.71
+ 0.500 = 13.855 in.
c=
2
(a)
(b)
Mc (8640)(13.855)
=
I
4794
Normal stress:
σ=
At point D,
S = 213 in 3
Normal stress:
σ=
M 3840
=
S
213
σ = 25.0 ksi 
σ = 18.03 ksi 
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PROBLEM 5.147
Two cover plates, each 12 in. thick, are welded to a
W27 × 84 beam as shown. Knowing that σ all = 24 ksi
for both the beam and the plates, determine the required
value of (a) the length of the plates, (b) the width of the
plates.
SOLUTION
RA = RB = 80 kips ↑
ΣM J = 0: −80 x + M = 0
M = 80 x kip ⋅ ft
At D,
S = 213 in 3
Allowable bending moment:
M all = σ all S = (24)(213) = 5112 kip ⋅ in
= 426 kip ⋅ ft
Set M D = M all .
80 xD = 426
xD = 5.325 ft
l = 18 − 2 xD
(a)
At center of beam,
M = (80)(9) = 720 kip ⋅ ft = 8640 kip ⋅ in
M
S=
c=
σ all
=
8640
= 360 in 3
24
26.7
+ 0.500 = 13.85 in.
2
Required moment of inertia:
But
l = 7.35 ft 
I = Sc = 4986 in 4
I = I beam + 2 I plate
2


1
 26.7 0.500 
+
+ (b)(0.500)3 
4986 = 2850 + 2 (b)(0.500) 

2  12
 2


= 2850 + 184.981b
(b)
b = 11.55 in. 
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PROBLEM 5.148
For the tapered beam shown, determine (a) the transverse
section in which the maximum normal stress occurs, (b) the
largest distributed load w that can be applied, knowing
that σ all = 140 MPa.
SOLUTION
1
wL ↑
L = 1.2 m
2
1
x
Σ M J = 0: − wL + wx + M = 0
2
2
w
M = ( Lx − x 2 )
2
w
= x( L − x)
2
RA = RB =
h = a + kx
For the tapered beam,
a = 120 mm
300 − 120
k=
= 300 mm/m
0.6
For rectangular cross section,
1
1
S = bh 2 = b (a + kx) 2
6
6
Bending stress:
σ=
M 3w Lx − x 2
=
S
b (a + kx)2
To find location of maximum bending stress, set
dσ
= 0.
dx
dσ 3w d  Lx − x 2  3w  (a + kx) 2 ( L − 2 x) − ( Lx − x 2 )2(a + kx)k 
=

=


dx
b dx  (a + kx) 2  b 
(a + kx)3

3w  (a + kx)( L − 2 x) − 2k ( Lx − x 2 ) 
=


b 
(a + kx)3

3w  aL + kLx − 2ax − 2kx2 − 2kLx + 2kx2 
=


b 
( a + kx)3

=
3w  aL − (2a + kL) x 

=0
b  ( a + kx)3

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PROBLEM 5.148 (Continued)
(a)
xm =
aL
(120) (1.2)
=
2a + kL (2) (120) + (300) (1.2)
xm = 0.24 m 
hm = a + kxm = 120 + (300)(0.24) = 192 mm
1
1
Sm = bhm2 = (20) (192) 2 = 122.88 × 103 mm3 = 122.88 × 10−6 m3
6
6
Allowable value of M m: M m = Smσ all = (122.88 × 10−6 ) (140 × 106 )
= 17.2032 × 103 N ⋅ m
(b)
Allowable value of w:
w=
2M m
(2) (17.2032 × 103 )
=
xm ( L − xm )
(0.24) (0.96)
=149.3 × 103 N/m
w = 149.3 kN/m 
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PROBLEM 5.149
For the tapered beam shown, knowing that w = 160 kN/m,
determine (a) the transverse section in which the maximum
normal stress occurs, (b) the corresponding value of the
normal stress.
SOLUTION
1
wL ↑
2
1
x
Σ M J = 0: − wLx + wx + M = 0
2
2
w
M = ( Lx − x 2 )
2
w
= x( L − x)
2
RA = RB =
where w = 160 kN/m and L = 1.2 m.
h = a + kx
For the tapered beam,
a = 120 mm
300 − 120
k=
= 300 mm/m
0.6
1
1
For a rectangular cross section, S = bh 2 = b(a + kx) 2
6
6
σ=
Bending stress:
M 3w Lx − x 2
=
S
b (a + kx)2
dσ
To find location of maximum bending stress, set dx = 0.
dσ 3w d  Lx − x 2  3w  (a + kx) 2 ( L − 2 x) − ( Lx − x 2 )2(a + kx)k 
=

=


dx
b dx  (a + kx) 2  b 
(a + kx) 4

=
3w  (a + kx)( L − 2 x) − 2k ( Lx − x 2 ) 


b 
(a + kx)3

=
3w  aL + kLx − 2ax − 2kx2 − 2kLx + 2kx2

b 
( a + kx)3
=
3w  aL − 2ax + kLx 

=0
b  (a + k x)3 



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PROBLEM 5.149 (Continued)
(a)
xm =
aL
(120) (1.2)
=
2a + kL (2) (120) + (300) (1.2)
xm = 0.240 m 
hm = a + k xm = 120 + (300)(0.24) = 192 mm
1
1
Sm = bhm2 = (20) (192)2 = 122.88 × 103 mm3 = 122.88 × 10−6 m3
6
6
160 × 103
w
(0.24) (0.96) = 18.432 × 103 N ⋅ m
M m = xm ( L − xm ) =
2
2
(b)
Maximum bending stress:
σm =
M m 18.432 × 103
=
= 150 × 106 Pa
−6
S m 122.88 × 10
σ m = 150.0 MPa 
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PROBLEM 5.150
For the tapered beam shown, determine (a) the transverse section
in which the maximum normal stress occurs, (b) the largest
distributed load w that can be applied, knowing that
σ all = 24 ksi.
SOLUTION
RA = RB =
1
wL ↑ L = 60 in.
2
1
x
ΣM J = 0: − wLx + wx + M = 0
2
2
w
M = (Lx − x 2 )
2
w
= x( L − x)
2
For the tapered beam,
h = a + kx
a = 4 in. k =
8−4 2
=
in./in.
30
15
For a rectangular cross section,
1
1
S = bh 2 = b(a + kx) 2
6
6
Bending stress:
σ=
M 3w Lx − x 2
=
⋅
S
b (a + kx)2
To find location of maximum bending stress, set
dσ
= 0.
dx
dσ 3w d  Lx − x 2 
=


dx
b dx  ( a + kx) 2 
=
3w  (a + kx) 2 ( L − 2 x) − ( Lx − x 2 )2(a + kx)k 


b 
(a + kx) 4

=
3w  (a + kx)( L − 2 x) − 2k ( Lx − x 2 ) 


b 
(a + kx)3

=
3w  aL + kLx − 2ax − 2kx 2 − 2kLx + 2kx 2 


b 
(a + kx)3

=
3w  aL − (2a + kL) x 

=0
b  ( a + kx)3

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PROBLEM 5.150 (Continued)
xm =
(a)
aL
(4)(60)
=
2a + kL (2)(4) + ( 152 ) (6.0)
xm = 15 in. 
 2
hm = a + kxm = 4 +   (15) = 6.00 in.
 15 
1
 1  3 
Sm = bhm3 =    (6.00) 2 = 4.50 in 3
6
 6  4 
Allowable value of M m : M m = Sm σ all = (4.50)(24) = 180.0 kip ⋅ in
(b)
Allowable value of w:
w=
2M m
(2)(108.0)
=
= 0.320 kip/in
xm ( L − xm )
(15)(45)
w = 320 lb/in 
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PROBLEM 5.151
For the tapered beam shown, determine (a) the transverse
section in which the maximum normal stress occurs, (b) the
largest concentrated load P that can be applied, knowing
that σ all = 24 ksi.
SOLUTION
P
↑
2
Px
Σ M J = 0: −
+M =0
2
Px 
L
M=
0 < x < 2 
2


RA = RB =
For a tapered beam,
h = a + kx
For a rectangular cross section,
1
1
S = bh 2 = b (a + kx)2
6
6
Bending stress:
σ=
To find location of maximum bending stress, set
M
3 Px
=
S b ( a + kx) 2
dσ
= 0.
dx
 3P (a + kx)2 − x − 2(a + kx)k
dσ 3P d 
x
=

=
dx
b dx  (a + kx) 2  b
(a + kx)4
3P a − kx
a
=
=0
xm =
b (a + kx)3
k
a = 4 in.,
Data:
(a)
xm =
4
= 30 in.
0.13333
k=
8−4
= 0.13333 in/in
30
xm = 30.0 in. 
hm = a + kxm = 8 in.
1
 1  3 
Sm = bhm2 =    (8) 2 = 8 in 3
6
 6  4 
=
σ
=
(24) (8) = 192 kip ⋅ in
Mm
all S m
(b)
P=
2M m (2) (192)
=
= 12.8 kips
xm
30
P = 12.80 kips 
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PROBLEM 5.152
Draw the shear and bending-moment diagrams for the beam and
loading shown, and determine the maximum absolute value (a) of
the shear, (b) of the bending moment.
SOLUTION
ΣM G = 0: − 16C + (36)(400) + (12)(1600)
− (12)(400) = 0
C = 1800 lb
ΣFx = 0: − C + Gx = 0
Gx = 1800 lb
ΣFy = 0: −400 − 1600 + G y − 400 = 0
G y = 2400 lb
A to E:
V = −400 lb
E to F:
V = −2000 lb
F to B:
V = 400 lb
At A and B,
M =0
At D −,
ΣM D = 0: (12)(400) + M = 0
At D +,
ΣM D = 0: (12)(400) − (8)(1800) + M = 0
M = 9600 lb ⋅ in
At E,
ΣM E = 0: (24)(400) − (8)(1800) + M = 0
M = 4800 lb ⋅ in
At F,
ΣM F = 0: −M − (8)(1800) − (12)(400) = 0
M = −19200 lb ⋅ in
At F ,+
ΣM F = 0: −M − (12)(400) = 0
−
(a)
(b)
M = −4800 lb ⋅ in
M = −4800 lb ⋅ in
Maximum |V | = 2000 lb 
Maximum |M | = 19200 lb ⋅ in 
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PROBLEM 5.153
Draw the shear and bending-moment diagrams for the beam
and loading shown, and determine the maximum normal
stress due to bending.
SOLUTION

Free body EFGH. Note that M E = 0 due to hinge.

Σ M E = 0: 0.6 H − (0.2) (40) − (0.40)(300) = 0

H = 213.33 N

Σ Fy = 0: VE − 40 − 300 + 213.33 = 0

VE = 126 ⋅ 67 N


Shear:

E to F :
V = 126.67 N ⋅ m

F to G :
V = 86.67 N ⋅ m
G to H :
V = −213.33 N ⋅ m
Bending moment at F:
Σ M F = 0: M F − (0.2) (126.67) = 0
M F = 25.33 N ⋅ m
Bending moment at G:
Σ M G = 0: −M G + (0.2) (213.33) = 0
M G = 42.67 N ⋅ m
Free body ABCDE.
Σ M B = 0: 0.6 A + (0.4) (300) + (0.2)(300)
− (0.2)(126.63) = 0
A = 257.78 N
Σ M A = 0: −(0.2)(300) − (0.4)(300) − (0.8)(126.67) + 0.6 D = 0
D = 468.89 N
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PROBLEM 5.153 (Continued)
Bending moment at B.
Σ M B = 0: − (0.2)(257.78) + M B = 0
M B = 51.56 N ⋅ m
Bending moment at C.
Σ M C = 0: − (0.4)(257.78) + (0.2)(300)
+ MC = 0
M C = 43.11 N ⋅ m
max M = 51.56 N ⋅ m
S =
1 2 1
bh = (20)(30) 2
6
6
= 3 × 103 mm3 = 3 × 10−6 m3
Normal stress.
σ =
51.56
= 17.19 × 106 Pa
3 × 10−6
σ = 17.19 MPa 
Bending moment at D.
Σ M D = 0: − M D − (0.2)(213.33) = 0
M D = −25.33 N ⋅ m
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PROBLEM 5.154
Determine (a) the distance a for which the maximum absolute
value of the bending moment in the beam is as small as
possible, (b) the corresponding maximum normal stress due to
bending. (See hint of Prob. 5.27.)
SOLUTION
Reaction at B:
Σ M C = 0: 5a − (8)(10) + 13RB = 0
RB =
1
(80 − 5 a)
18
Bending moment at D:
Σ M D = 0: −M D + 5RB = 0
M D = 5RB =
5
(80 − 5 a)
13
Bending moment at C:
M C = 0 5a + M C = 0
M C = −5a
Equate:
−M C = M D
5a =
5
(80 − 5a)
13
a = 4.4444 ft
Then
(a) a = 4.44 ft 
− M C = M D = (5)(4.4444) = 22.222 kip ⋅ ft
|M |max = 22.222 kip ⋅ ft = 266.67 kip ⋅ in
For W14 × 22 rolled steel section, S = 29.0 in 3
Normal stress:
σ =
M
266.67
=
= 9.20 ksi
S
29.0
(b) 9.20 ksi 
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PROBLEM 5.155
Determine (a) the equations of the shear and bending-moment curves
for the beam and loading shown, (b) the maximum absolute value of
the bending moment in the beam.
SOLUTION

dV
x2 
= − w = − w0 1 + 2 
dx
L 


x3 
V = − w0  x + 2  + C1
3L 

V = 0 at x = L.
(a)
1 
4

0 = − w0  L + L  + C1 C1 = w0 L
3
3


4
dM
1 x3 
= V = w0  L − x −

dx
3 L2 
3
4
1
1 x4
M = w0  Lx − x 2 −
2
12 L2
3
M = 0 at


 + C2

1
1 
4
x = L. w0  L2 − L2 − L2  + C2 = 0
2
12 
3
3
C2 = − w0 L2
4
4
1
1 x4 3 2 
M = w0  Lx − x 2 −
− L 
2
12 L2 4 
3
(b)
|M | max occurs at x = 0.

|M | max =
3
w0 L2 
4
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PROBLEM 5.156
Draw the shear and bending-moment diagrams for the beam
and loading shown, and determine the maximum normal stress
due to bending.
SOLUTION
ΣM B = 0: −2.5 A + (1.75)(1.5)(16) = 0
A = 16.8 kN
ΣM A = 0: −(0.75) + (1.5)(16) + 2.5B = 0
B = 7.2 kN
Shear diagram:
VA = 16.8 kN
VC = 16.8 − (1.5)(16) = −7.2 kN
VB = −7.2 kN
Locate point D where V = 0.
d
1.5 − d
=
24d = 25.2
16.8
7.2
d = 1.05 m 1.5 − d = 0.45 m
Areas of the shear diagram:
1
A to D:
 Vdx =  2  (1.05)(16.8) = 8.82 kN ⋅ m
D to C:
 Vdx =  2  (0.45)(−7.2) = −1.62 kN ⋅ m
C to B:
 Vdx = (1)(−7.2) = −7.2 kN ⋅ m
1
Bending moments:
MA = 0
M D = 0 + 8.82 = 8.82 kN ⋅ m
M C = 8.82 − 1.62 = 7.2 kN ⋅ m
M B = 7.2 − 7.2 = 0
Maximum |M | = 8.82 kN ⋅ m = 8.82 × 103 N ⋅ m
For S150 × 18.6 rolled steel section, S = 120 × 103 mm3 = 120 × 10−6 m3
Normal stress:
σ=
|M | 8.82 × 103
=
= 73.5 × 106 Pa
S
120 × 10−6
σ = 73.5 MPa 
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PROBLEM 5.157
Knowing that beam AB is in equilibrium under the loading shown,
draw the shear and bending-moment diagrams and determine the
maximum normal stress due to bending.
SOLUTION
0 < x < 1.2 ft
A to C:
x 

w = 50 1 −
 = 50 − 41.667 x
 1.2 
dV
= − w = 41.667 x − 50
dx
V = VA +

x
0
(41.667 x − 50)dx
= 0 + 20.833x 2 − 50 x =
M = MA +
=0+

x
0
dM
dx
x
 V dx
0
(20.833x 2 − 50 x)dx
= 6.944 x3 − 25 x 2
x = 1.2 ft,
At
V = −30 lb
M = −24 lb ⋅ in
C to B: Use symmetry conditions.
Maximum |M | = 24 lb ⋅ ft = 288 lb ⋅ in
Cross section:
c=
I=
Normal stress:
σ=
d 1
=
(0.75) = 0.375 in.
2  2 
π
π 
c 4 =   (0.375) = 15.532 × 10−3 in 4
4
4
|M | c (2.88)(0.375)
=
= 6.95 × 103 psi
−3
I
15.532 × 10
σ = 6.95 ksi 
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PROBLEM 5.158
For the beam and loading shown, design the cross section of the
beam, knowing that the grade of timber used has an allowable
normal stress of 1750 psi.
SOLUTION
B = E = 2.8 kips
For equilibrium,
Shear diagram:
A to B −:
+
V = −4.8 kips
−
B to C :
V = −4.8 + 2.8 = −2 kips
C + to D −:
V = −2 + 2 = 0
+
−
V = 0 + 2 = 2 kips
D to E :
+
V = 2 + 2.8 = 4.8 kips
E to F:
Areas of shear diagram:
(2)(−4.8) = −9.6 kip ⋅ ft
A to B:
B to C:
(2)(−2) = −4 kip ⋅ ft
C to D:
(3)(0) = 0
D to E:
(2)(2) = 4 kip ⋅ ft
E to F:
(2)(4.8) = 9.6 kip ⋅ ft
Bending moments:
MA = 0
M B = 0 − 9.6 = −9.6 kip ⋅ ft
M C = −9.6 − 4 = −13.6 kip ⋅ ft
M D = −13.6 + 0 = −13.6 kip ⋅ ft
M E = −13.6 + 4 = −9.6 kip ⋅ ft
M F = −9.6 + 9.6 = 0
|M |max = 13.6 kip ⋅ ft = 162.3 kip ⋅ in = 162.3 × 103 lb ⋅ in
Required value for S:
S=
For a rectangular section, I =
|M |max
σ all
=
162.3 × 103
= 93.257 in 3
1750
1 3
1
bh , c = h
12
2
Equating the two expressions for S,
S=
I bh 2 (b)(9.5)2
=
=
= 15.0417b
6
6
c
15.0417 b = 93.257
b = 6.20 in. 
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PROBLEM 5.159
Knowing that the allowable stress for the steel used is 160 MPa, select the
most economical wide-flange beam to support the loading shown.
SOLUTION
ΣM D = 0: −3.2 B + (24)(3.2)(50) = 0
B = 120 kN
ΣM B = 0: 3.2 D − (0.8)(3.2)(50) = 0
D = 40 kN
VA = 0
Shear:
VB− = 0 − (0.8)(50) = −40 kN
VB+ = −40 + 120 = 80 kN
VC = 80 − (2.4)(50) = −40 kN
VD = −40 + 0 = −40 kN
Locate point E where V = 0.
e 2.4 − e
=
80
40
e = 1.6 m
Areas:
120e = 192
2.4 − e = 0.8 m
1
A to B :
 Vdx =  2  (0.8)(−40) = −16 kN ⋅ m
B to E :
 Vdx =  2 (1.6)(80) = 64 kN ⋅ m
E to C :
 Vdx =  2  (0.8)(−40) = −16 kN ⋅ m
C to D :
 Vdx = (0.8)(−40) = −32 kN ⋅ m
Bending moments:
1
1
MA = 0
M B = 0 − 16 = −16 kN ⋅ m
M E = −16 + 64 = 48 kN ⋅ m
M C = 48 − 16 = 32 kN ⋅ m
M D = 32 − 32 = 0
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PROBLEM 5.159 (Continued)
Maximum |M | = 48 kN ⋅ m = 48 × 103 N ⋅ m
σ all = 160 MPa = 160 × 106 Pa
Smin =
Shape
S (103 mm3 )
W 310 × 32.7 415
W 250 × 28.4 308 ←
W 200 × 35.9 342
|M |
σ all
=
48 × 103
= 300 × 10−6 m3 = 300 × 103 mm3
160 × 106
Lightest wide flange beam:
W 250 × 28.4 @ 28.4 kg/m 
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PROBLEM 5.160
Determine the largest permissible value of P for the beam and
loading shown, knowing that the allowable normal stress is
+8 ksi in tension and −18 ksi in compression.
SOLUTION
B = D = 1.5 P ↑
Reactions:
Shear diagram:
A to B −:
V = −P
B + to C −:
V = − P + 1.5 P = 0.5 P
C + to D −:
V = 0.5 P − P = −0.5 P
+
V = −0.5 P + 1.5 P = P
D to E :
Areas:
A to B :
(10)(− P ) = −10 P
B to C :
(60) (0.5 P) = 30 P
C to D :
(60) ( −0.5 P) = −30 P
D to E :
(10) ( P) = 10 P
MA = 0
M B = 0 − 10 P = −10 P
M C = −10 P + 30 P = 20 P
M D = 20 P − 30 P = −10 P
M E = −10 P + 10 P = 0
Bending moments:
Largest positive bending moment: 20 P
Largest negative bending moment: −10 P
Centroid and moment of inertia:
Part
A, in 2
y0 , in.
Ay0 , in 3
d , in.
Ad 2, in 4
I , in 4

5
3.5
17.5
1.75
15.3125
10.417

7
0.5
3.5
1.25
10.9375
0.583
Σ
12
Y =
21
21
= 1.75 in.
12
Top: y = 4.25 in.
26.25
11.000
I = Σ Ad 2 + ΣI = 37.25 in 4
Bottom: y = −1.75 in.
σ =−
My
I
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PROBLEM 5.160 (Continued)
Top, tension:
Top, comp.:
Bottom. tension:
Bottom. comp.:
8=−
−18 = −
8=−
−18 = −
( −10 P)(4.25)
37.25
P = 7.01 kips
(20 P) (4.25)
37.25
P = 7.89 kips
(20 P)(−1.75)
37.25
P = 8.51 kips
(−10 P)(−1.75)
37.25
P = 38.3 kips
Smallest value of P is the allowable value.
P = 7.01 kips 
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PROBLEM 5.161
(a) Using singularity functions, find the magnitude and
location of the maximum bending moment for the beam and
loading shown. (b) Determine the maximum normal stress
due to bending.
SOLUTION
ΣM B = 0: − 4.5 A + (2.25)(4.5)(40) + (2.7)(60) + (0.9)(60) = 0
A = 138 kN ↑
ΣM A = 0: − (2.25)(4.5)(40) − (1.8)(60) − (3.6)(60) + 4.5 B = 0
B = 162 kN ↑
w = 40 kN/m =
dV
dx
V = −40 x + 138 − 60 x − 1.8 0 − 60 x − 3.6 0 =
dM
dx
M = −20 x 2 − 138 x − 60 x − 1.81 − 60 x − 3.61
VC+ = −(40)(1.8) + 138 − 60 = 6 kN
VD− = −(40)(3.6) + 138 − 60 = −66 kN
Locate point E where V = 0. It lies between C and D.
VE = − 40 xE + 138 − 60 + 0 = 0
xE = 1.95 m
M E = −(20)(1.95)2 + (138)(1.95) − (60)(1.95 − 1.8) = 184 kN ⋅ m
|M |max = 184 kN ⋅ m = 184 × 103 N ⋅ m at
(a)
x = 1.950 m 
For W 530 × 66 rolled steel section, S = 1340 × 103 mm3 = 1340 × 10−6 m3
(b)
Normal stress:
σ=
|M |max
184 × 103
=
= 137.3 × 106 Pa
S
1340 × 10−6
σ = 137.3 MPa 
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PROBLEM 5.162
The beam AB, consisting of an aluminum plate of uniform thickness b
and length L, is to support the load shown. (a) Knowing that the beam is
to be of constant strength, express h in terms of x, L, and h0 for portion
AC of the beam. (b) Determine the maximum allowable load if
L = 800 mm, h0 = 200 mm, b = 25 mm, and σ all = 72 MPa.
SOLUTION
A= B
By symmetry,
ΣFy = 0: A −
For 0 ≤ x ≤
L
,
2
At x = 0,
w=
2w0 x
L
V=
1
w0 L :
4
1
w0 L + B = 0
2
2w x
dV
= −w = − 0
dx
L
C1 =
M = 0:
M=
(a)
L
At x = ,
2
1
w0 L ↑
4
V = C1 −
w0 x 2
L
1
w0 L
4
w x2
dM
1
= V = w0 L − 0
dx
L
4
At x = 0,
A=B=
M = C2 +
1
1 w0 x3
w0 Lx −
4
3 L
C2 = 0
1 w0
(3L2 − 4 x3 )
2 L
3
1 w0  2  L 
L  1
M = MC =
3L   − 4    = w0 L2
12 L 
2
 2   12
M
For constant strength,
S=
For a rectangular section,
1
S = bh 2
6
σ all
,
S0 =
M0
σ all
MC
=
S
M
1
=
= 3 (3L2 x − 4 x3 )
S0 M 0 L
σ all
1
S0 = bh02
6
S  h 
= 
S0  h0 
2
h = h0
(b)
Data:
L = 800 mm
h0 = 200 mm
b = 25 mm
3L2 x − 4 x3

L3
σ all = 72 MPa
1
1
S0 = bh02 = (25)(200)2 = 166.667 × 103 mm3 = 166.667 × 10−6 m3
6
6
M C = σ all S0 = (72 × 106 )(166.667 × 10−6 ) = 12 × 103 N ⋅ m
w0 =
12 M C
L2
=
(12)(12 × 103 )
= 225 × 103 N/m
2
(0.800)
w0 = 225 kN/m 
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PROBLEM 5.163
A transverse force P is applied as shown at end A of the conical taper AB.
Denoting by d 0 the diameter of the taper at A, show that the maximum
normal stress occurs at point H, which is contained in a transverse section
of diameter d = 1.5d0 .
SOLUTION
V = −P =
dM
dx
Let
d = d0 + k x
For a solid circular section,
I=
π
4
c4 =
π
64
M = − Px
d3
d
I π 3 π
S= =
d = ( d 0 + k x )3
c 32
2
32
dS 3π
3π 2
d k
=
(d 0 + k x) 2 k =
dx 32
32
c=
Stress:
At H,
σ=
|M | Px
=
S
S
dσ
1 
dS 
= 2  PS − PxH
=0
dx S 
dx 
3π 2
dS π 3
S − xH
d − xH
d k
=
dx 32
32
1
1
1
k xH = d = (d 0 + k H xH ) k xH = d0
3
3
2
d = d0 +
1
3
d0 = d0
2
2
d = 1.5d0 
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CHAPTER 6
PROBLEM 6.1
Three boards, each of 1.5 × 3.5-in. rectangular cross section, are nailed
together to form a beam that is subjected to a vertical shear of 250 lb.
Knowing that the spacing between each pair of nails is 2.5 in., determine
the shearing force in each nail.
SOLUTION
I =
1 3
1
(3.5)(4.5)3 = 26.578 in 4
bh =
12
12
A = (3.5)(1.5) = 5.25 in 2
y1 = 1.5 in.
Q = Ay1 = 7.875 in 3
q=
VQ (250)(7.875)
=
= 74.074 lb/in
I
26.578
qs = 2 Fnail
Fnail =
qs
(74.074)(2.5)
=
2
2
Fnail = 92.6 lb 
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PROBLEM 6.2
Three boards, each 2 in. thick, are nailed together to form a beam
that is subjected to a vertical shear. Knowing that the allowable
shearing force in each nail is 150 lb, determine the allowable shear
if the spacing s between the nails is 3 in.
SOLUTION
1 3
bh + Ad 2
12
1
(6)(2)3 + (6)(2)(3) 2 = 112 in 4
=
12
1 3
1
I2 =
bh =
(2)(4)3 = 10.667 in 4
12
12
I1 =
I 3 = I1 = 112 in 4
I = I1 + I 2 + I 3 = 234.667 in 4
Q = A1 y1 = (6)(2)(3) = 36 in 3
qs = Fnail
(1)
VQ
I
(2)
q=
Dividing Eq. (2) by Eq. (1),
1
VQ
=
s
Fnail I
V =
Fnail I
(150)(234.667)
=
Qs
(36)(3)
V = 326 lb 
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PROBLEM 6.3
Three boards are nailed together to form a beam shown, which is
subjected to a vertical shear. Knowing that the spacing between the
nails is s = 75 mm and that the allowable shearing force in each nail is
400 N, determine the allowable shear when w = 120 mm.
SOLUTION
Part
A (mm 2 )
d (mm)
7200
Middle Plank
Bottom Plank
Top plank
Ad 2 (106 mm 4 )
I (106 mm 4 )
60
25.92
2.16
12,000
0
0
3.60
7200
60
25.92
2.16
51.84
7.92
Σ
I = Ad 2 +  I = 59.76 × 106 mm 4 = 59.76 × 10−6 m 4
Q = (7200)(60) = 432 × 103 mm3 = 432 × 10−6 m3
VQ
I
Fnail
q=
s
q=
V=
Fnail = qs
V =
Iq IFnail
=
Q
Qs
(59.76 × 10−6 )(400)
(432 × 10−6 )(75 × 10−3 )
V = 738 N 
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PROBLEM 6.4
Solve Prob. 6.3, assuming that the width of the top and bottom boards
is changed to w = 100 mm .
PROBLEM 6.3 Three boards are nailed together to form a beam
shown, which is subjected to a vertical shear. Knowing that the spacing
between the nails is s = 75 mm and that the allowable shearing force
in each nail is 400 N, determine the allowable shear when
w = 120 mm.
SOLUTION
A (mm 2 )
d (mm)
6000
Middle Plank
Bottom Plank
Part
Top plank
Ad 2 (106 mm 4 )
I (106 mm 4 )
60
21.6
1.80
12,000
0
0
3.60
6000
60
21.6
1.80
43.2
7.20
Σ
I = Ad 2 +  I = 50.4 × 106 mm 4 = 50.4 × 10−6 m 4
Q = (6000)(60) = 360 × 103 mm3 = 360 × 10−6 m3
VQ
I
Fnail
q=
s
q=
V=
Fnail = qs
V =
Iq IFnail
=
Q
Qs
(50.4 × 10−6 )(400)
(360 × 10−6 )(75 × 10−3 )
V = 747 N 
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PROBLEM 6.5
The American Standard rolled-steel beam shown has been reinforced by
attaching to it two 16 × 200-mm plates, using 18-mm-diameter bolts
spaced longitudinally every 120 mm. Knowing that the average allowable
shearing stress in the bolts is 90 MPa, determine the largest permissible
vertical shearing force.
SOLUTION
Calculate moment of inertia:
A (mm 2 )
3200
6650
3200
Part
Top plate
S310 × 52
Bot. plate
Σ
*d
=
d (mm)
*
160.5
0
*
160.5
Ad 2 (106 mm 4 )
82.43
82.43
164.86
I (106 mm 4 )
0.07
95.3
0.07
95.44
305 16
+
= 160.5 mm
2
2
I = ΣAd 2 + ΣI = 260.3 × 106 mm 4 = 260.3 × 10−6 m 4
Q = Aplate d plate = (3200)(160.5) = 513.6 × 103 mm3 = 513.6 × 10−6 m3
Abolt =
π
4
2
d bolt
=
π
4
(18 × 10−3 ) 2 = 254.47 × 10−6 m 2
Fbolt = τ all Abolt = (90 × 106 )(254.47 × 10−6 ) = 22.90 × 103 N
qs = 2 Fbolt
q=
VQ
I
q =
V =
2Fbolt
(2)(22.90 × 103 )
=
= 381.7 × 103 N/m
−3
s
120 × 10
Iq
(260.3 × 10−6 )(381.7 × 103 )
=
= 193.5 × 103 N
Q
513.6 × 10−6
V = 193.5 kN 
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PROBLEM 6.6
Solve Prob. 6.5, assuming that the reinforcing plates are only 12 mm thick.
PROBLEM 6.5 The American Standard rolled-steel beam shown has been
reinforced by attaching to it two 16 × 200-mm plates, using 18-mm-diameter
bolts spaced longitudinally every 120 mm. Knowing that the average
allowable shearing stress in the bolts is 90 MPa, determine the largest
permissible vertical shearing force.
SOLUTION
Calculate moment of inertia:
A(mm 2 )
2400
6650
2400
Part
Top plate
S310 × 52
Bot. plate
Σ
*
d =
d (mm)
*158.5
0
*158.5
Ad 2 (106 mm 4 )
60.29
0
60.29
120.58
I (106 mm 4 )
0.03
95.3
0.03
95.36
305 12
+
= 158.5 mm
2
2
I = ΣAd 2 + ΣI = 215.94 × 106 mm 4 = 215.94 × 10−6 m 4
Q = Aplate d plate = (200)(12)(158.5) = 380.4 × 103 mm3 = 380.4 × 10−6 m3
Abolt =
π
4
2
d bolt
=
π
4
(18 × 10−3 ) 2 = 254.47 × 10−6 m 2
Fbolt = τ all Abolt = (90 × 106 )(254.47 × 10−6 ) = 22.902 × 103 N
qs = 2 Fbolt
q=
VQ
I
q =
V =
2Fbolt
(2)(22.903 × 103 )
=
= 381.7 × 103 N/m
s
120 × 10−3
Iq
(215.94 × 10−6 )(381.7 × 103 )
=
= 217 × 103 N
Q
380.4 × 10−6
V = 217 kN 
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PROBLEM 6.7
A columm is fabricated by connecting the rolled-steel members shown by
bolts of 34 -in. diameter spaced longitudinally every 5 in. Determine the
average shearing stress in the bolts caused by a shearing force of 30 kips
parallel to the y axis.
SOLUTION
Geometry:
d 
f =   + (t w )C
 2 s
10.0
=
+ 0.303 = 5.303 in.
2
x = 0.534 in.
y1 = f − x = 5.303 − 0.534 = 4.769 in.
Determine moment of inertia.
Part
C8 × 13.7
S 10 × 25.4
C8 × 13.7
Σ
d (in.)
4.769
0
4.769
A(in 2 )
4.04
7.4
4.04
Ad 2 (in 4 )
91.88
0
91.88
183.76
I (in 4 )
1.52
123
1.52
126.04
I = ΣAd 2 + ΣI = 183.76 + 126.04 = 309.8 in 4
Q = A y1 = (4.04)(4.769) = 19.267 in 3
VQ (30)(19.267)
=
= 1.8658 kip/in
309.8
I
1
1
= qs =   (1.8658)(5) = 4.664 kips
2
2
q=
Fbolt
Abolt =
τ bolt =
π
4
2
=
d bolt
π 3
2
2
  = 0.4418 in
4 4
4.664
Fbolt
=
= 10.56 ksi
Abolt 0.4418
τ bolt = 10.56 ksi 
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PROBLEM 6.8
The composite beam shown is fabricated by connecting two W6 × 20 rolled-steel
members, using bolts of 58 -in. diameter spaced longitudinally every 6 in. Knowing
that the average allowable shearing stress in the bolts is 10.5 ksi, determine the largest
allowable vertical shear in the beam.
SOLUTION
W6 × 20: A = 5.87 in 2 , d = 6.20 in., I x = 41.4 in 4
y =
Composite:
1
d = 3.1 in.
2
I = 2[41.4 + (5.87)(3.1)2 ]
= 195.621 in 4
Q = A y = (5.87)(3.1) = 18.197 in 3
Bolts:
d =
Abolt =
5
in., τ all = 10.5 ksi, s = 6 in.
8
π 5
2
2
  = 0.30680 in
4 8
Fbolt = τ all Abolt = (10.5)(0.30680) = 3.2214 kips
Shear:
q=
2Fbolt
(2)(3.2214)
=
= 1.07380 kips/in
s
6
q =
VQ
I
V =
Iq
(195.621)(1.0780)
=
Q
18.197
V = 11.54 kips 
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PROBLEM 6.9
For the beam and loading shown,
consider section n-n and determine
(a) the largest shearing stress in that
section, (b) the shearing stress at
point a.
SOLUTION
By symmetry, RA = RB .
 Fy = 0:
RA + RB − 15 − 20 − 15 = 0
RA = RB = 25 k ips
V = 30 kips at n-n.
From shear diagram,
Determine moment of inertia.
Part
A(in 2 )
Top Flng
6
Web
3.30
Bot. Flng
6
d (in.)
Ad 2 (in 4 )
I (in 4 )
4.7
132.54
0.18
0
0
21.30
4.7
132.54
0.18
265.08
21.66
Σ
I =  Ad 2 +  I = 286.74 in 4
(a)
Part
A(in 2 )

6

1.65
Σ
y (in.)
Ay (in 3 )
4.7
28.2
2.2
3.63
31.83
Q =  Ay
= 31.83 in 3
t = 0.375 in.
τ max =
VQmax
(25)(31.83)
=
It
(286.74)(0.375)
τ max = 7.40 ksi 
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PROBLEM 6.9 (Continued)
(b)
A(in 2 )
Part

6

0.15
Σ
y (in.)
Ay (in 3 )
4.7
28.2
4.2
0.63
28.83
Q =  Ay = 28.83 in 3
t = 0.375 in.
τ =
VQ
(23)(28.83)
=
It
(286.74)(0.375)
τ = 6.70 ksi 
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PROBLEM 6.10
For the beam and loading
shown, consider section
n-n and determine (a) the
largest shearing stress in
that section, (b) the
shearing stress at point a.
SOLUTION
At section n-n, V = 10 kN.
I = I1 + 4I 2
=
1
 1

b1h13 + 4  b2h23 + A2d 22 
12
 12

=
 1 

1
(100)(150)3 + 4   (50)(12)3 + (50)(12)(69) 2 
12
 12 

= 28.125 × 106 + 4 0.0072 × 106 + 2.8566 × 106 
= 39.58 × 106 mm 4 = 39.58 × 10−6 m 4
(a)
Q = A1 y1 + 2 A2 y2
= (100)(75)(37.5) + (2)(50)(12)(69)
= 364.05 × 103 mm3 = 364.05 × 10−6 m3
t = 100 mm = 0.100 m
τ max =
(b)
VQ (10 × 103 )(364.05 × 10−6 )
=
= 920 × 103 Pa
−6
It
(39.58 × 10 )(0.100)
τ max = 920 kPa 
Q = A1 y1 + 2 A2 y2
= (100)(40)(55) + (2)(50)(12)(69)
= 302.8 × 103 mm3 = 302.8 × 10−6 m3
t = 100 mm = 0.100 m
τa =
VQ
(10 × 103 )(302.8 × 10−6 )
=
= 765 × 103 Pa
It
(39.58 × 10−6 )(0.100)
τ a = 765 kPa 
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PROBLEM 6.11
For the beam and loading shown, consider section
n-n and determine (a) the largest shearing stress in
that section, (b) the shearing stress at point a.
SOLUTION
V = 80 kN
At section n-n,
Consider cross section as composed of rectangles of types , , and .
1
(12)(80)3 + (12)(80)(90)2 = 8.288 × 106 mm 4
12
1
(180)(16)3 + (180)(16)(42)2 = 5.14176 × 106 mm 4
I2 =
12
1
(16)(68)3 = 419.24 × 103 mm 4
I3 =
12
I1 =
I = 4 I1 + 2 I 2 + 2I 3 = 44.274 × 106 mm 4
= 44.274 × 10−6 m 4
(a)
Calculate Q at neutral axis.
Q1 = (12)(80)(90) = 86.4 × 103 mm 4
Q2 = (180)(16)(42) = 120.96 × 103 mm 4
Q3 = (16)(34)(17) = 9.248 × 103 mm 4
Q = 2Q1 + Q2 + 2Q3 = 312.256 × 103 mm3 = 312.256 × 10−6 m3
τ =
(b)
At point a,
VQ
(80 × 103 )(312.256 × 10−6 )
=
= 17.63 × 106 Pa
It
(44.274 × 10−6 )(2 × 16 × 10−3 )
τ = 17.63 MPa 
Q = Q1 = 86.4 × 103 mm 4 = 86.4 × 10−6 m 4
τ =
VQ
(80 × 103 )(86.4 × 10−6 )
=
= 13.01 × 106 Pa
−6
−3
It
(44.274 × 10 )(12 × 10 )
τ = 13.01 MPa 
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PROBLEM 6.12
For the beam and loading shown, consider
section n-n and determine (a) the largest
shearing stress in that section, (b) the shearing
stress at point a.
SOLUTION
By symmetry, RA = RB .
ΣFy = 0: RA + RB − 10 − 10 = 0
RA = RB = 10 kips
From the shear diagram,
V = 10 kips at n-n.
1
1
b2h23 − b1h13
12
12
1
1
=
(4)(4)3 − (3)(3)3 = 14.583 in 4
12
12
I =
(a)
1
1
Q = A1 y1 + A2 y2 = (3)   (1.75) + (2)   (2)(1) = 4.625 in 3
2
2
t =
τ max =
(b)
1 1
+ = 1 in.
2 2
VQ (10)(4.625)
=
It
(14.583)(1)
τ max = 3.17 ksi 
1
Q = Ay = (4)   (1.75) = 3.5 in 3
2
1 1
t = + = 1 in.
2 2
τ =
VQ
(10)(3.5)
=
It
(14.583)(1)
τ a = 2.40 ksi 
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PROBLEM 6.13
For a beam having the cross section shown, determine the largest
allowable vertical shear if the shearing stress is not to exceed
60 MPa.
SOLUTION
Calculate moment of inertia.
I =
1
1

(50 mm)(120 mm)3 − 2  (30 mm) 4 + (30 mm × 30 mm)(35 mm) 2 
12
12

I = 7.2 × 106 mm 4 − 2[1.170 × 106 mm 4 ] = 4.86 × 106 mm 4
= 4.86 × 10−6 m 4
Assume that τ m occurs at point a.
t = 2(10 mm) = 0.02 m
Q = (10 mm × 50 mm)(55mm) + 2[(10 mm × 30 mm)(35 mm)]
= 48.5 × 103 mm3 = 48.5 × 10−6 m3
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PROBLEM 6.13 (Continued)
For τ all = 60 MPa,
τ m = τ all =
60 × 106 Pa =
Check τ at neutral axis:
VQ
It
V (48.5 × 10−6 m3 )
(4.86 × 10−6 m 4 )(0.02 m)
V = 120.3 kN 
t = 50 mm = 0.05 m
Q = (50 × 60)(30) − (30 × 30)(35) = 58.5 × 103 mm3
τ =
VQ (120.3 kN)(58.5 × 10−6 m3 )
=
= 29.0 MPa < 60 MPa
It
(4.86 × 10−6 m 4 )(0.05m)
OK
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PROBLEM 6.14
For a beam having the cross section shown, determine the largest
allowable vertical shear if the shearing stress is not to exceed 60
MPa.
SOLUTION
Calculate moment of inertia.
1
 1
I = 2  (10 mm)(120 mm)3  +
(30 mm)(40 mm)3
12
 12
= 2[1.440 × 106 mm 4 ] + 0.160 × 106 mm 4
= 3.04 × 106 mm 4
I = 3.04 × 10−6 m 4
Assume that τ m occurs at point a.
t = 10 mm = 0.01m
Q = (10 mm × 40 mm)(40 mm)
= 16 × 103 mm3 Q = 16 × 10−6 m3
For
τ all = 60 MPa,
60 × 106 Pa =
τ m = τ all =
V (16 × 10−6 m3 )
(3.04 × 10−6 m 4 )(0.01m)
VQ
It
V = 114.0 kN 
Check τ at neutral axis:
t = 50 mm = 0.05 m
Q = 2[(10 × 60)(30)] + (30 × 20)(10) = 42 × 103 m3 = 42 × 10−6 m3
τ NA =
VQ (114.0 kN)(42 × 10−6 m3 )
=
= 31.5 MPa < 60 MPa
It
(3.04 × 10−6 m 4 )(0.05 m)
OK
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PROBLEM 6.15
For the beam and loading shown, determine the minimum required
depth h, knowing that for the grade of timber used, σ all = 1750 psi
and τ all = 130 psi.
SOLUTION
(750 lb/ft)(16 ft) = 12 × 103 lb
Total load:
Reaction at A:
RA = 6 × 103 lb ↑
Vmax = 6 × 103 lb
1
(8 ft)(6 × 103 ) = 24 × 103 lb ⋅ ft
2
M max =
= 288 × 103 lb ⋅ in
Bending: S =
1 2
bh for rectangular section.
6
S =
M max
σ all
h=
Shear: I =
=
6S
=
b
288 × 103
= 164.57 in 3
1750
(6)(164.57)
= 14.05 in.
5
1 3
bh for rectangular section.
12
1
bh
2
1
y = h
4
A=
 1  1  1
Q = Ay = (b)  h  h  = bh 2
 2  4  8
VQ 3Vmax
τ max =
=
2bh
Ib
h=
3Vmax
(3)(6 × 103 )
=
= 13.85 in.
2bτ max
(2)(5)(130)
The larger value of h is the minimum required depth.
h = 14.05 in. 
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PROBLEM 6.16
For the beam and loading shown, determine the minimum
required width b, knowing that for the grade of timber
used, σ all = 12 MPa and τ all = 825 kPa.
SOLUTION
M D = 0: − 3RA + (2)(2.4) + (1)(4.8) = 0
RA = 3.2 kN
Draw shear and bending moment diagrams.
V
max
= 4.0 kN
Bending: S =
M
max
σ all
M
=
max
= 4.0 kN ⋅ m
4.0 × 103
12 × 106
= 333.33 × 10−6 m3 = 333.33 × 103 mm3
For a rectangular cross section,
1 3
bh
I 12
1
= bh 2
S = =
1
c
6
h
2
6S
(6)(333.33 × 103 )
=
= 88.9 mm
h2
1502
1
1
A=
bh, y = h
2
4
1
1
Q = Ay = bh 2 , I =
bh3
8
12
VQ
3V
=
τ=
It
2bh
b=
bh =
3 V 3 4.0 × 103
=
2 τ 2 825 × 103
= 7.2727 × 10−3 m 2 = 7.2727 × 103 mm 2
b=
The required value for b is the larger one.
bh 7.2727 × 103
=
= 48.5 mm
h
150
b = 88.9 mm 
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PROBLEM 6.17
A timber beam AB of length L and rectangular cross section carries a
uniformly distributed load w and is supported as shown. (a) Show that
the ratio τ m /σ m of the maximum values of the shearing and normal
stresses in the beam is equal to 2h/L, where h and L are, respectively,
the depth and the length of the beam. (b) Determine the depth h and
the width b of the beam, knowing that L = 5 m, w = 8 kN/m,
τ m = 1.08 MPa, and σ m = 12 MPa.
SOLUTION
RA = RB =
From shear diagram,
For rectangular section,
wL
4
(1)
A = bh
(2)
|V |m =
τm =
From bending moment diagram,
wL
2
3 Vm
3wL
=
2 A
8bh
(3)
|M |m =
wL2
32
(4)
S =
1 2
bh
6
(5)
|M |m
3wL2
=
S
16 bh2
(6)
For a rectangular cross section,
σm =
(a)
Dividing Eq. (3) by Eq. (6),
(b)
Solving for h:
h=
Lτ m
(5)(1.08 × 106 )
=
= 225 × 10−3 m
6
2σ m
(2)(12 × 10 )
τm
2h

=
L
σm
h = 225 mm 
Solving Eq. (3) for b:
b=
3wL
(3)(8 × 103 )(5)
=
8hτ m
(8)(225 × 10−3 )(1.08 × 106 )
= 61.7 × 10−3 m
b = 61.7 mm 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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PROBLEM 6.18
A timber beam AB of length L and rectangular cross section carries a
single concentrated load P at its midpoint C. (a) Show that the ratio
τ m /σ m of the maximum values of the shearing and normal stresses in
the beam is equal to 2h/L, where h and L are, respectively, the depth and
the length of the beam. (b) Determine the depth h and the width b of the
beam, knowing that L = 2 m, P = 40 kN, τ m = 960 kPa, and
σ m = 12 MPa.
SOLUTION
RA = RB = P/2 ↑
Reactions:
P
2
(1)
Vmax = RA =
(2)
A = bh for rectangular section.
(3)
τm =
(4)
M max =
(5)
S =
(6)
σm =
3 Vmax
3P
=
for rectangular section.
2 A
4bh
PL
4
1 2
bh for rectangular section.
6
M max
3PL
=
S
2bh 2
τm
h

=
σ m 2L
(a)
(b)
Solving for h,
h=
(2)(2)(960 × 103 )
= 320 × 10−3 m
6
12 × 10
h = 320 mm 
3P
(3)(40 × 103 )
=
= 97.7 × 10−3 m
4hτ m
(4)(320 × 10−3 )(960 × 103 )
b = 97.7 mm 
2Lτ m
σm
=
Solving Eq. (3) for b,
b=
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PROBLEM 6.19
For the wide-flange beam with the loading shown, determine the
largest load P that can be applied, knowing that the maximum
normal stress is 24 ksi and the largest shearing stress using the
approximation τ m = V/Aweb is 14.5 ksi.
SOLUTION
 M C = 0: − 15RA + qP = 0
RA = 0.6 P
Draw shear and bending moment diagrams.
V
= 0.6 P
max
M
max
= 0.6 PLAB
LAB = 6 ft = 72 in.
S = 258 in 3
For W 24 × 104 ,
Bending.
M
S =
P=
max
σ all
σ all S
0.6LAB
=
=
0.6 PLAB
σ all
(24)(258)
= 143.3 kips
(0.6)(72)
Aweb = dtw
Shear.
= (24.1)(0.500)
= 12.05 in 2
τ =
P=
V
max
Aweb
τ Aweb
0.6
=
0.6 P
Aweb
=
(14.5)(12.05)
= 291 kips
0.6
The smaller value of P is the allowable value.
P = 143.3 kips 
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PROBLEM 6.20
For the wide-flange beam with the loading shown, determine the
largest load P that can be applied, knowing that the maximum
normal stress is 160 MPa and the largest shearing stress using the
approximation τ m = V/Aweb is 100 MPa.
SOLUTION
 M E = 0: − 3.6RA + 3.0P + 2.4P + 1.8P = 0
RA = 2 P ↑
Draw shear and bending moment diagrams.
M B = 2PLAB , M C = M D = 3PLAB
V
max
= 2P
M
max
= 3PLAB
Bending. For W 360 × 122, S = 2020 × 103 mm3
= 2020 × 10−6 m3
M
max
σ all
P=
Shear.
3PLAB
=
σ all
σ all S
=
3LAB
=S
(160 × 106 )(2020 × 103 )
= 179.6 × 103 N
(3)(0.6)
Aweb = dt w = (363)(13.0)
= 4.719 × 103 mm 2 = 4.719 × 10−3 m 2
τ =
P=
The smaller value of P is the allowable one.
V
max
Aweb
τ Aweb
2
=
2P
Aweb
=
(100 × 106 )(4.719 × 10−3 )
= 236 × 103 N
2
P = 179.6 kN 
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PROBLEM 6.21
For the beam and loading
shown, consider section n-n
and determine the shearing
stress at (a) point a, (b) point
b.
SOLUTION
|V |max = 90 kN
Draw the shear diagram.
Part
A(mm 2 )
y (mm)
A y (103 mm3 )
Ad 2 (106 mm 4 )
I (106 mm 4 )

3200
90
288
25
2.000
0.1067

1600
40
64
−25
1.000
0.8533

1600
40
64
−25
1.000
0.8533
Σ
6400
4.000
1.8133
d(mm)
416
ΣAy
416 × 103
=
= 65 mm
ΣA
6400
Y =
I = ΣAd 2 + ΣI = (4.000 + 1.8133) × 106 mm 4
= 5.8133 × 106 mm 4 = 5.8133 × 10−6 m 4
(a)
A = (80)(20) = 1600 mm 2
y = 25 mm
Qa = Ay = 40 × 103 mm3 = 40 × 10−6 m3
τa =
VQa
(90 × 103 )(40 × 10−6 )
=
= 31.0 × 106 Pa
−6
−3
It
(5.8133 × 10 )(20 × 10 )
τ a = 31.0 MPa 
(b)
A = (30)(20) = 600 mm 2
y = 65 − 15 = 50 mm
Qb = Ay = 30 × 10 mm = 30 × 10−6 m3
3
τb =
3
VQb
(90 × 103 )(30 × 10−6 )
=
= 23.2 × 106 Pa
It
(5.8133 × 10−6 )(20 × 10−3 )
τ b = 23.2 MPa 
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PROBLEM 6.22
For the beam and loading shown, consider
section n-n and determine the shearing stress
at (a) point a, (b) point b.
SOLUTION
RA = RB = 12 kips
Draw shear diagram.
V = 12 kips
Determine section properties.
Part
A(in 2 )
y (in.)
Ay (in 3 )
d(in.)
Ad 2 (in 4 )
I (in 4 )

4
4
16
2
16
5.333

8
1
8
−1
8
2.667
Σ
12
24
8.000
24
Y =
ΣAy
24
=
= 2 in.
ΣA
12
I = ΣAd 2 + ΣI = 32 in 4
(a)
A = 1 in 2
y = 3.5 in. Qa = Ay = 3.5 in 3
t = 1 in.
τa =
(b)
VQa
(12)(3.5)
=
It
(32)(1)
A = 2 in 2
τ a = 1.3125 ksi 
y = 3 in. Qb = Ay = 6 in 3
t = 1 in.
τb =
VQb
(12)(6)
=
It
(32)(1)
τ b = 2.25 ksi 
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PROBLEM 6.23
For the beam and loading
shown,
determine
the
largest shearing stress in
section n-n.
SOLUTION
|V |max = 90 kN
Draw the shear diagram.
Part
A (mm 2 )
x′
Ay (103 mm3 ) d(mm) Ad 2 (106 mm 4 ) I (106 mm 4 )

3200
90
288
25
2.000
0.1067

1600
40
64
−25
1.000
0.8533

1600
40
64
−25
1.000
0.8533
Σ
6400
4.000
1.8133
416
ΣAy
416 × 103
=
= 65 mm
ΣA
6400
Y =
I = ΣAd 2 + ΣI = (4.000 + 1.8133) × 106 mm 4
= 5.8133 × 106 mm 4 = 5.8133 × 10−6 m 4
Part
A(mm 2 )

3200

300
7.5
2.25

300
7.5
2.25
y (mm)
25
Σ
Ay (103 mm3 )
80
84.5
Q = ΣAy = 84.5 × 103 mm3 = 84.5 × 10−6 m3
t = (2)(20) = 40 mm = 40 × 10−3 m
τ max =
VQ
(90 × 103 )(84.5 × 10−6 )
=
It
(5.8133 × 10−6 )(40 × 10−3 )
= 32.7 × 106 Pa
τ m = 32.7 MPa 
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PROBLEM 6.24
For the beam and loading shown, determine
the largest shearing stress in section n-n.
SOLUTION
RA = RB = 12 kips
Draw shear diagram.
V = 12 kips
Determine section properties.
Part
A(in 2 )
y (in.)
Ay (in 3 )
d(in.)
Ad 2 (in 4 )
I (in 4 )

4
4
16
2
16
5.333

8
1
8
−1
8
2.667
Σ
12
24
8.000
24
Y =
ΣAy
24
=
= 2 in.
ΣA
12
I = ΣAd 2 + ΣI = 32 in 4
Q = A1 y1 = (4)(2) = 8 in 3
t = 1 in.
τ max =
VQ (12)(8)
=
It
(32)(1)
τ max = 3.00 ksi 
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PROBLEM 6.25
A beam having the cross section shown is subjected to a vertical shear V. Determine (a) the
horizontal line along which the shearing stress is maximum, (b) the constant k in the following
expression for the maximum shearing stress
τ max = k
V
A
where A is the cross-sectional area of the beam.
SOLUTION
I =
For semicircle,
As =
π
4
π
2
c4
c2
Q = As y =
(a)
τ max occurs at center where
(b)
τ max =
V ⋅ 2 c3
VQ
4V
4V
= π 43
=
=
2
3A
It
c ⋅ 2c 3π c
4
and
A = π c2
y =
π
2
c2 ⋅
4c
3π
4c
2
= c3
3π
3
t = 2c. 
k =
4
= 1.333 
3
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PROBLEM 6.26
A beam having the cross section shown is subjected to a vertical shear V. Determine
(a) the horizontal line along which the shearing stress is maximum, (b) the constant k in
the following expression for the maximum shearing stress
τ max = k
V
A
where A is the cross-sectional area of the beam.
SOLUTION
A=
1
bh
2
I =
1 3
bh
36
For a cut at location y,
A( y) =
1  by 
by 2
=
y
 
2 h 
2h
y ( y) =
2
2
h− y
3
3
Q( y) = Ay =
t ( y) =
by 2
(h − y )
3
by
h
2
by
VQ V 3 (h − y) 12 Vy(h − y) 12V
τ ( y) =
=
=
=
(hy − y 2 )
3
3
by
3
1
It
bh
bh
( 36 bh ) h
(a)
To find location of maximum of τ , set
dτ
= 0.
dy
dτ
12V
( h − 2 ym ) = 0
=
dy
bh3
(b)
τm
2
12V
12V  1 2  1   3V
3V
2
(hym − ym ) =
=
=
 h −  h  =
3
3
2A
bh
bh  2
 2   bh
ym =
1
h, i.e., at mid-height 
2
k =
3
= 1.500 
2
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PROBLEM 6.27
A beam having the cross section shown is subjected to a vertical shear V. Determine (a) the
horizontal line along which the shearing stress is maximum, (b) the constant k in the
following expression for the maximum shearing stress
τ max = k
V
A
where A is the cross-sectional area of the beam.
SOLUTION
A = 2π rmtm
For a thin-walled circular section,
J = Arm2 = 2π rm3tm ,
For a semicircular arc,
y =
I =
1
J = π rm3tm
2
2rm
π
As = π rmtm
Q = As y = π rmtm
(a)
t = 2tm
(b)
τ max =
2rm
π
= 2rm2tm
at neutral axis where maximum occurs. 
VQ
V (2rm2tm )
V
2V
=
=
=
3
It
A
(π rmtm )(2tm ) π rmtm
k = 2.00 
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PROBLEM 6.28
A beam having the cross section shown is subjected to a vertical shear V. Determine (a) the
horizontal line along which the shearing stress is maximum, (b) the constant k in the
following expression for the maximum shearing stress
τ max = k
V
A
where A is the cross-sectional area of the beam.
SOLUTION
1 
A = 2  bh  = bh
2 
 1
 1
I = 2  bh3  = bh3
 12
 6
For a cut at location y, where y ≤ h,
A( y) =
1  by 
by 2
 y =
2 h 
2h
y ( y) = h −
2
y
3
Q( y) = Ay =
t ( y) =
τ ( y) =
(a)
by 2 by 3
−
2
3h
by
h
2
VQ
6
h by 2 by 3
V   y
 y 
=V 3 ⋅
⋅
−
=
3   − 2   
3h
It
bh   h 
bh by 2
 h  
To find location of maximum of τ , set
dτ
= 0.
dy
dτ
V
y
= 2 [3 − 4 m ] = 0
dy
h
bh
(b)
τ ( ym ) =
2
V  3
V
3  9 V
= 1.125
3   − 2    =
bh   4 
A
 4   8 bh
ym
3
1
= , i.e., ± h from neutral axis. 
4
h
4
k = 1.125 
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PROBLEM 6.29
The built-up beam shown is made by gluing together five planks. Knowing that
in the glued joints the average allowable shearing stress is 350 kPa, determine
the largest permissible vertical shear in the beam.
SOLUTION
1
(240 mm)(160 mm)3
12
1
− (200 mm)(80 mm)3
12
I=
= 73.4 × 106 mm 4
I = 73.4 × 10−6 m 4
t = 40 mm = 0.04 m
Q = (100 mm × 40 mm)(60 mm)
= 240 × 103 mm3
Q = 240 × 10−6 m3
For
τ m = 350 kPa,
τm =
VQ
:
Tt
350 × 10 3 Pa =
V (240 × 10−6 m3 )
(73.4 × 10−6 m 4 )0.04 m
V = 4.28 kN 
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PROBLEM 6.30
For the beam of Prob. 6.29, determine the largest permissible horizontal shear.
PROBLEM 6.29 The built-up beam shown is made by gluing together five
planks. Knowing that in the glued joints the average allowable shearing stress is
350 kPa, determine the largest permissible vertical shear in the beam.
SOLUTION
I =2
1
1
(40)(240)3 + (80)(40)3
12
12
I = 92.6 × 10+6 mm 4 = 92.6 × 10−6 m 4
For
Q = (40 × 100)70
= 280 × 103 mm3
Q = (280 × 10−6 ) m3
τ = 350 kPa,
τ =
VQ
;
It
350 × 103 Pa =
V (280 × 10−6 m3 )
(92.6 × 10−6 m 4 )(0.04 m)
V = 4630 N
V = 4.63 kN 
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PROBLEM 6.31
Several wooden planks are glued together to form the box beam shown. Knowing
that the beam is subjected to a vertical shear of 3 kN, determine the average
shearing stress in the glued joint (a) at A, (b) at B.
SOLUTION
IA =
1 3
1
(60)(20)3 + (60)(20)(50) 2
bh + Ad 2 =
12
12
= 3.04 × 106 mm 4
1 3
1
(60)(20)3 = 0.04 × 106 mm4
bh =
12
12
1 3
1
(20)(120)3 = 2.88 × 106 mm 4
IC =
bh =
12
12
IB =
I = 2I A + I B + 2 I C = 11.88 × 106 mm 4 = 11.88 × 10−6 m 4
QA = Ay = (60)(20)(50) = 60 × 103 mm3 = 60 × 10−6 m3
t = 20 mm + 20 mm = 40 mm = 40 × 10−3 m
(a)
τA =
VQA
(3 × 103 )(60 × 10−6 )
=
= 379 × 103 Pa
It
(11.88 × 10−6 )(40 × 10−3 )
QB = 0
(b)
τB =
VQB
=0
It
τ A = 379 kPa 
τB = 0 
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PROBLEM 6.32
The built-up timber beam is subjected to a 1500-lb vertical shear. Knowing
that the longitudinal spacing of the nails is s = 2.5 in. and that each nail is
3.5 in. long, determine the shearing force in each nail.
SOLUTION
I1 =
1
(2)(4)3 + (2)(4)(3) 2
12
= 82.6667 in 4
1
(2)(6)3 = 36 in 4
12
I = 2 I1 + 2I 2
I2 =
= 237.333 in 4
Q = A1 y1 = (2)(4)(3) = 24 in 3
q=
VQ (1500)(24)
=
= 151.685 lb/in
237.333
I
2 Fnail = qs
Fnail =
1
1
qs =   (151.685)(2.5)
2
2
Fnail = 189.6 lb 
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PROBLEM 6.33
The built-up wooden beam shown is subjected to a vertical shear of
8 kN. Knowing that the nails are spaced longitudinally every 60 mm
at A and every 25 mm at B, determine the shearing force in the nails
(a) at A, (b) at B. (Given: I x = 1.504 × 109 mm 4.)
SOLUTION
I x = 1.504 × 109 mm 4 = 1504 × 10−6 m 4
s A = 60 mm = 0.060 m
sB = 25 mm = 0.025 m
(a)
QA = Q1 = A1 y1 = (50)(100)(150) = 750 × 103 mm3
= 750 × 10−6 m3
FA = q A s A
=
(8 × 103 )(750 × 10−6 )(0.060)
VQ1s A
=
I
1504 × 10−6
FA = 239 N 
(b)
Q2 = A2 y2 = (300)(50)(175) = 2625 × 103 mm3
QB = 2Q1 + Q2 = 4125 × 103 mm3
= 4125 × 10−6 m3
FB = qB sB =
VQB sB
(8 × 103 )(4125 × 10−6 )(0.025)
=
I
1504 × 10−6
FB = 549 N 
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PROBLEM 6.34
Knowing that a vertical shear V of 50 kips is exerted on W14 × 82 rolled-steel beam,
determine the shearing stress (a) at point a, (b) at the centroid C.
SOLUTION
For W14 × 82, d = 14.3 in., b f = 10.1 in., t f = 0.855 in., tw = 0.510 in., I = 881 in 4
(a)
Aa = (4.15)(0.855) = 3.5482 in 2
d t f 14.3 0.855
−
=
−
= 6.7225 in.
2
2
2
2
ya =
Qa = Aa ya = 23.853 in 3
ta = t f = 0.855 in.
τa =
(b)
VQa (50)(23.853)
=
Ita
(881)(0.855)
τ a = 1.583 ksi 
A1 = b f t f (10.1)(0.855) = 8.6355 in 2
d tf
− = 6.7225 in.
2
2
d

A2 = tw  − t f  = (0.510)(6.295) = 3.2105 in 2
2


y1 =
y2 =
1d
 1
 − t f  = ( 6.295 ) = 3.1475 in.
2 2
 2
QC = A1 y1 + A2 y2 (8.6355)(6.7225) + (3.2105)(3.1475)
= 68.157 in 3
tC = t w = 0.510 in.
τC =
VQC (50)(68.157)
=
ItC
(881)(0.510)
τ C = 7.59 ksi 
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PROBLEM 6.35
An extruded aluminum beam has the cross section shown. Knowing that the
vertical shear in the beam is 150 kN, determine the shearing stress at
(a) point a, (b) point b.
SOLUTION
I=
1
1
(80)(80)3 − (56)(68)3 = 1.9460 × 106 mm 4
12
12
= 1.946 × 10−6 m 4
(a)
Qa = A1 y1 + 2 A2 y2
= (56)(6)(37) + (2)(12)(40)(20) = 31.632 × 103 mm3
= 31.632 × 10−6 m3
ta = (2)(12) = 24 mm = 0.024 m
τa =
VQa (150 × 103 )(31.632 × 10−6 )
=
= 101.6 × 106 Pa
−6
Ita
(1.946 × 10 )(0.024)
τ a = 101.6 MPa 
(b)
Qb = A1 y1 = (56)(6)(37) = 12.432 × 103 mm3
= 12.432 × 10−6 m3
tb = (2)(6) = 12 mm = 0.012 m
τb =
VQb
(150 × 103 )(12.432 × 10−6 )
=
= 79.9 × 106 Pa
−6
Itb
(1.946 × 10 )(0.012)
τ b = 79.9 MPa 

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PROBLEM 6.36
Knowing that a given vertical shear V causes a maximum shearing
stress of 75 MPa in the hat-shaped extrusion shown, determine the
corresponding shearing stress (a) at point a, (b) at point b.
SOLUTION
Neutral axis lies 30 mm above bottom.
τc =
VQc
It
τ a Qatc
=
τ c Qcta
τa =
VQa
Ita
τb =
VQb
Itb
τ b Qbtc
=
τ c Qctb
Qc = (6)(30)(15) + (14)(4)(28) = 4260 mm3
tc = 6 mm
Qa = (14)(4)(28) = 1568 mm3
ta = 4 mm
Qb = (14)(4)(28) = 1568 mm3
tb = 4 mm
τ c = 75 MPa
(a)
τa =
Qa tc
1568 6
⋅ τc =
⋅ ⋅ 75
Qc ta
4260 4
(b)
τb =
Qb tc
1568 6
⋅ τc =
⋅ ⋅ 75
Qc tb
4260 4
τ a = 41.4 MPa 
τ b = 41.4 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 6.37
Knowing that a given vertical shear V causes a maximum shearing stress of 75
MPa in an extruded beam having the cross section shown, determine the shearing
stress at the three points indicated.
SOLUTION
τ =
VQ
It
τ is proportional to Q/t.
Qc = (30)(10)(75)
Point c:
= 22.5 × 103 mm3
tc = 10 mm
Qc /tc = 2250 mm 2
Qb = Qc + (20)(50)(55)
Point b:
= 77.5 × 103 mm3
tb = 20 mm
Qb /tb = 3875 mm 2
Qa = 2Qb + (120)(30)(15)
Point a:
= 209 × 103 mm3
ta = 120 mm
Qa /ta = 1741.67 mm 2
τ m = τ b = 75 MPa
(Q/t ) m occurs at b.
τa
Qa /ta
τa
1741.67 mm
2
=
=
τb
Qb /tb
=
τc
Qc /tc
75 MPa
τa
=
2
3875 mm
2250 mm 2
τ a = 33.7 MPa 
τ b = 75.0 MPa 
τ c = 43.5 MPa 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 6.38
An extruded beam has the cross section shown and a uniform wall thickness
of 0.20 in. Knowing that a given vertical shear V causes a maximum shearing
stress τ = 9 ksi, determine the shearing stress at the four points indicated.
SOLUTION
Qa = (0.2)(0.5)(0.5 − 0.25) = 0.125 in 3
Qb = (0.2)(0.5)(0.3 + 0.25) = 0.055 in 3
Qc = Qa + Qb + (1.4)(0.2)(0.9) = 0.432 in 3
Qd = 2Qa + 2Qb + (3.0)(0.2)(0.9) = 0.900 in.3
Qm = Qd + (0.2)(0.8)(0.4) = 0.964 in 3
τ =
VQ
It
Since V, I, and t are constant, τ is proportional to Q.
τa
0.125
=
τb
0.055
=
τc
0.432
=
τd
0.900
=
τm
0.964
=
9
0.964
τ a = 1.167 ksi; τ b = 0.513 ksi; τ c = 4.03 ksi; τ d = 8.40 ksi 
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it
without permission.
PROBLEM 6.39
Solve Prob. 6.38, assuming that the beam is subjected to a horizontal shear V.
PROBLEM 6.38 An extruded beam has the cros