1
Tension, Compression,
and Shear
Statics Review
Problem 1.2-1 Segments AB and BC of beam
136 N·m at joint B
3m
ABC are pin connected a small distance to the
right of joint B (see figure). Axial loads act at A
and at mid-span of AB. A concentrated moment
is applied at joint B.
440 N
220 N
A
6m
(a) Find reactions at supports A, B, and C.
(b) Find internal stress resultants N, V, and
M at x ⫽ 4.5 m.
x
B
C
3m
Pin
connection
Solution 1.2-1
(a) APPLY LAWS OF STATICS
⌺Fx ⫽ 0
Cx ⫽ NA – N 2 ⫽ 220 N
FBD of BC:
©MB ⫽ 0
Cy ⫽
1
(0) ⫽ 0
L1
Entire FBD:
©MA ⫽ 0
By ⫽
1
(⫺MB) ⫽ ⫺22.667 N
L2
©Fy ⫽ 0
Reactions are
Ay ⫽ ⫺By ⫽ 22.667 N
Ay ⫽ 22.7 N
By ⫽ ⫺22.7 N
Cx ⫽ 220 N
Cy ⫽ 0
(b) INTERNAL STRESS RESULTANTS N, V, AND M AT x ⫽ 4.5 m
Use FBD of segment from A to x ⫽ 4.5 m.
©Fx ⫽ 0
Nx ⫽ NA ⫺ N2 ⫽ 220 N
©Fy ⫽ 0
Vx ⫽ Ay ⫽ 22.7 N
©M ⫽ 0
Mx ⫽ Ay (4.5 m) ⫽ 102 N # m
1
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2
CHAPTER 1
Tension, Compression, and Shear
Problem 1.2-2 Segments AB and BCD of beam ABCD are pin connected at x ⫽ 4 m. The beam is supported by a sliding
support at A and roller supports at C and D (see figure). A triangularly distributed load with peak intensity of 80 N/m acts on
BC. A concentrated moment is applied at joint B.
(a) Find reactions at supports A, C, and D.
(b) Find internal stress resultants N, V, and M at x ⫽ 5 m.
(c) Repeat parts (a) and (b) for the case of the roller support at C replaced by a linear spring of stiffness ky ⫽ 200 kN/m.
80 N/m
200 N.m at joint D
A
C
4m
x
4m
B
D
C
ky
3m
Pin
connection
Part (c)
Solution 1.2-2
(a) APPLY LAWS OF STATICS
©Fx ⫽ 0
Ax ⫽ 0
FBD of AB:
©MB ⫽ 0
Entire FBD:
Reactions are
MA ⫽ 0
©MC ⫽ 0
Dy ⫽
©Fy ⫽ 0
Cy ⫽
MA ⫽ 0
Ax ⫽ 0
1
2
1
c200 N # m ⫺ (80 N/m) 4 m a b 4 m d ⫽ ⫺75.556 N
3m
2
3
1
180 N/m2 4 m ⫺ Dy ⫽ 235.556 N
2
Cy ⫽ 236 N
Dy ⫽ ⫺75.6 N
(b) INTERNAL STRESS RESULTANTS N, V, AND M AT x ⫽ 5 m
Use FBD of segment from A to x ⫽ 5 m; ordinate on triangular load at x ⫽ 5 m is
©Fx ⫽ 0
©Fy ⫽ 0
©M ⫽ 0
3
180 N/m2 ⫽ 60 N/m.
4
Nx ⫽ ⫺Ax ⫽ 0
⫺1
3(80 N/m + 60 N/m) 1 m4 ⫽ ⫺70 N
V ⫽ ⫺70 N
Upward
V⫽
2
1
2
1
1
M ⫽ ⫺MA ⫺ (80 N/m) 1 m a 1 mb ⫺ (60 N/m) 1 m a 1 mb ⫽ ⫺36.667 N # m
2
3
2
3
(break trapezoidal load into two triangular loads in moment expression)
M ⫽ ⫺36.7 N # m
CW
(c) REPLACE ROLLER SUPPORT AT C WITH SPRING SUPPORT
Structure remains statically determinate so all results above in (a) and (b) are unchanged.
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.2
Statics Review
3
Problem 1.2-3 Segments AB and BCD of beam ABCD are pin connected at x ⫽ 3.0 m. The beam is supported by a pin
support at A and roller supports at C and D; the roller at D is rotated by 30⬚ from the x axis (see figure). A trapezoidal distributed load on BC varies in intensity from 74 N/m at B to 37 N/m at C. A concentrated moment is applied at joint A and a
180 N inclined load is applied at mid-span of CD.
(a) Find reactions at supports A, C, and D.
(b) Find the resultant force in the pin connection at B.
(c) Repeat parts (a) and (b) if a rotational spring (kr ⫽ 68 N # m/rad) is added at A and the roller at C is removed.
200 N·m at joint A
74 N/m
180 N
37 N/m
A
A
kr
3m
Part (c)
4
3 1.5 m D
C 1.5 m
x
B
3m
Pin
connection
3m
30º
Remove roller at C in
part (c)
Solution 1.2-3
(a) STATICS
FBD of AB (cut through beam at pin): ©MB ⫽ 0
©MD ⫽ 0
Entire FBD:
Cy ⫽
Ay ⫽
1
200 N
(⫺MA) ⫽ ⫺
⫽ ⫺66.667 N
3m
3
L1 ⫽ 3 m
L1
1 4
1
1
2
c P (1.5 m) + q2L1 aL1 +
b + q1L1 aL1 + L1 b ⫺ MA ⫺ Ay A 3L1 B d ⫽ 464.333 N
3m 5
2
3
2
3
©Fy ⫽ 0
Dy ⫽
⫺Dy
4
1
⫽ 50.326 N
P + (q1 + q2) L1 ⫺ Ay ⫺ Cy ⫽ ⫺87.167 N so Dx ⫽
5
2
tan(60⬚)
©Fx ⫽ 0
Ax ⫽
3
P ⫺ Dx ⫽ 57.674 N
5
CHECK
Ax ⫽ 57.7 N
Ay ⫽ ⫺66.7 N
Dx ⫽ 50.3 N
Dy ⫽ ⫺87.2 N
Cy ⫽ 464.3 N
Ax + Dx ⫽ 108 N
Ay + Cy + Dy ⫽ 310.5 N
(b) USE FBD OF AB ONLY; MOMENT AT PIN IS ZERO
FBx ⫽ ⫺Ax
FBx ⫽ ⫺57.674 N
FBy ⫽ ⫺Ay
FBy ⫽ 66.667 N
3
⫽ 108 N
5
4
1
P + A q1 + q2 B L1 ⫽ 310.5 N
5
2
P
ResultantB ⫽ 2FBx2 + FBy2 ⫽ 88.2 N
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
4
CHAPTER 1
Tension, Compression, and Shear
(c) ADD ROTATIONAL SPRING AT A & REMOVE ROLLER AT C; APPLY EQUATIONS OF STATICAL EQUILIBRIUM
Use FBD of BCD:
©MB ⫽ 0
Dy ⫽
so
Dx ⫽
©MB ⫽ 0
SO REACTIONS ARE
tan(60⬚)
⫽ ⫺83.716 N
Ay ⫽
©Fx ⫽ 0
Use FBD of AB:
⫺Dy
1
4
(q1 + q2)L1 + P ⫺ Dy ⫽ 165.5 N
2
5
3
Ax ⫽ P ⫺ Dx ⫽ 191.716 N
5
©Fy ⫽ 0
Use entire FBD:
1 1
2
1
1
4
3
c q L a L b + q1L1 a L1 b + P a L1 b d ⫽ 145 N
2L1 2 2 1 3 1
2
3
5
2
MAz ⫽ MA + Ay L1 ⫽ 696.5 N # m
Ax ⫽ 191.7 N
Ay ⫽ 165.5 N
MAz ⫽ 697 N # m
Dx ⫽ ⫺83.72 N
Dy ⫽ 145 N
RESULTANT FORCE IN PIN CONNECTION AT B
ResultantB ⫽ 2FBx2 + FBy2 ⫽ 253 N
FBy ⫽ ⫺Ay
FBx ⫽ ⫺Ax
Problem 1.2-4 Consider the plane truss with
20 N
a pin support at joint 3 and a vertical roller
support at joint 5 (see figure).
4
5
(a) Find reactions at support joints 3 and 5.
(b) Find axial forces in truss members
11 and 13.
45 N
5
6
2.5 m
2
2
2m
9
13
1
1
8
10
8
60 N
6
12
11
7
7
2m
3
3
4
1m
Solution 1.2-4
(a) STATICS
©Fy ⫽ 0
R3y ⫽ 20 N ⫺ 45 N ⫽ ⫺25 N
©M3 ⫽ 0
R5x ⫽
©Fx ⫽ 0
R3x ⫽ ⫺R5x + 60 N ⫽ 40 N
1
(20 N * 2 m) ⫽ 20 N
2m
(b) MEMBER FORCES IN MEMBERS 11 and 13
Number of unknowns:
m ⫽ 13
Number of equations:
j⫽8
r⫽3
2j ⫽ 16
m ⫹ r ⫽ 16
so statically determinate
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SECTION 1.2
20 N
4
5
5
1
2
2.5 m
3
3
2m
2
2 m (4) Cut vertically through 4, 11, 12, and 1;
use left FBD; sum moments about joint 2:
1
F11 V ⫽
1R ⫺ F42 so F11 ⫽ 0
2.5 m 5x
9
13
8
(2) ⌺FV ⫽ 0 at joint 8 so F12 ⫽ 0
(3) ⌺FH ⫽ 0 at joint 5 so F4 ⫽ ⫺R5x ⫽ ⫺20 N
10
1
60 N
8
12
11
7
6
7
5
TRUSS ANALYSIS
(1) ⌺FV ⫽ 0 at joint 4 so F10 ⫽ 0
45 N
6
Statics Review
4
1m
(5) Sum vertical forces at joint 3; F9 ⫽ R3y
F9 ⫽ 25 N
Section cut for left FBD
F13V ⫽ 45 N – F9 ⫽ 20 N
(6) Sum vertical forces at joint 7:
F13 ⫽ 12 F13V ⫽ 28.3 N
Problem 1.2-5 A plane truss has a pin support at A and a roller support at E (see figure).
(a) Find reactions at all supports.
(b) Find the axial force in truss member FE.
A
3m
B
4.5 kN
9.0 kN
13.5 kN
3m
C
3m
D
3m
E
4.5 m
1m
F
G
Section cut for left FBD
Solution 1.2-5
(a) STATICS
P ⫽ 4.5 kN
©Fx ⫽ 0
Ax ⫽ 0
©MA ⫽ 0
Ey ⫽
©Fy ⫽ 0
L1 ⫽ 3 m
1
[3P (3 m) + 2P(6 m) + P(9 m)] ⫽ 22.5 kN
6m
Ay ⫽ 3P + 2P + P ⫺ Ey ⫽ 4.5 kN
(b) MEMBER FORCE IN MEMBER FE
Number of unknowns:
m ⫽ 11
Number of equations:
j⫽7
r⫽3
2 j ⫽ 14
m + r ⫽ 14
so statically determinate
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
6
CHAPTER 1
Tension, Compression, and Shear
TRUSS ANALYSIS
(1) Cut vertically through AB, GC, and GF; use left FBD; sum moments about C.
FGFx ( 4.5 m) ⫺ FGFy (6 m) ⫽ Ay(6 m) ⫽ 27 kN # m
FGFy ⫽ FGF
so
3m
2(0.5 m)2 + (3 m)2
0.5 m
2(0.5 m)2 + (3 m)2
Ay (6 m)
FGF ⫽
4.5 m
FGFx ⫽ FGF
FGFx ⫽ FGF
3m
2(0.5 m)2 + (3 m)2
3m
2(0.5 m)2 + (3 m)2
(2) Sum horizontal forces at joint F:
⫺ 6m
⫽ 7.821 kN
0.5 m
and
2(0.5 m)2 + (3 m)2
⫽ 7.714 kN
FFEx ⫽ FGFx ⫽ 7.714 kN
FFE ⫽
2(3 m)2 + (1 m)2
FFEx ⫽ 8.132 kN
3m
FFE ⫽ 8.13 kN
Problem 1.2-6 A plane truss has a pin support at F and a roller support at D (see figure).
(a) Find reactions at
both supports.
(b) Find the axial force
in truss member FE.
6 kN
9 kN
A
3m
B
3m
C
3 kN
3m
D
3m
E
4.5 m
F
1m
G
Section cut for left FBD
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.2
7
Statics Review
Solution 1.2-6
(a) STATICS
©Fx ⫽ 0
©MF ⫽ 0
©Fy ⫽ 0
Fx ⫽ 0
1
Dy ⫽
C 3 kN (6 m) + 6 kN (3 m) D ⫽ 6 kN
6m
Fy ⫽ 9 kN + 6 kN + 3 kN ⫺ Dy ⫽ 12 kN
(b) MEMBER FORCE IN MEMBER FE
Number of unknowns:
m ⫽ 11
r⫽3
m ⫹ r ⫽ 14
Number of equations:
j⫽7
2j ⫽ 14
so statically determinate
TRUSS ANALYSIS
(1) Cut vertically through AB, GD, and GF; use left FBD; sum moments about D to get FGF ⫽ 0.
(2) Sum horizontal forces at joint F:
FFEx ⫽ ⫺Fx ⫽ 0
so
FFE ⫽ 0
y
Problem 1.2-7 A space truss has three-dimensional pin supports at joints
Cy
O, B, and C. Load P is applied at joint A and acts toward point Q.
Coordinates of all joints are given in meters (see figure).
Cz
C(0, 4, 0)
Cx
(a) Find reaction force components Bx, Bz, and Oz.
(b) Find the axial force in truss member AC.
O(0, 0, 0)
Ox
Oz
Joint B
coordinates (m)
B(2, 0, 0)
(0, 0, 5) A
z
P
Oy
Bz
Bx
x
By
Q(4, −3, 5)
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
8
CHAPTER 1
Tension, Compression, and Shear
Solution 1.2-7
m⫽3
(a) FIND REACTIONS USING STATICS
r⫽9
m ⫹ r ⫽ 3j
rAQ ⫽
4
⫺3
P 0 Q
rOA ⫽
0
0
P5Q
eAQ ⫽
m ⫹ r ⫽ 12
j⫽4
3 j ⫽ 12
so truss is statically determinate
0.8
⫽ ⫺0.6
ƒ rAQ ƒ
P 0 Q
rAQ
PA ⫽ P eAQ ⫽
0.8 P
⫺0.6 P
P 0 Q
0
4
P0Q
rOC ⫽
rOB ⫽
2
0
P0Q
©M ⫽ 0
MO ⫽ rOA * PA + rOC *
Cx
Bx
4 Cz + 3.0 P
Cy + rOB * By ⫽ 4.0 P ⫺ 2 Bz
PC Q
PB Q P 2B ⫺ 4C Q
z
z
y
x
©Mx ⫽ 0 gives Cz ⫽
so
©My ⫽ 0 gives
⫺3
P
4
Bz ⫽ 2 P
©F ⫽ 0
RO ⫽ PA +
Ox
Bx
Cx
Bx + Cx + Ox + 0.8 P
Oy + By + Cy ⫽ By + Cy + Oy + ⫺0.6 P
≤
PO Q
PB Q
PC Q ±
5P
z
z
z
Oz +
4
METHOD OF JOINTS
Joint O
©Fx ⫽ 0
Ox ⫽ 0
Joint B
©Fy ⫽ 0
By ⫽ 0
Joint C
©Fx ⫽ 0
Cx ⫽ 0
©Fx ⫽ 0 gives Bx ⫽ ⫺0.8 P
For entire structure:
©Mz ⫽ 0 gives Oz ⫽
so
©Fy ⫽ 0
⫺5
P
4
Oy ⫽ 0
©Fy ⫽ 0 Cy ⫽ 0.6 P ⫺ By ⫽ Oy
Cy ⫽ 0.6 P
(b) FORCE IN MEMBER AC
©Fz ⫽ 0 at joint C:
FAC ⫽
3 241 ƒ P ƒ
242 + 52
ƒ Cz ƒ ⫽
5
20
FAC ⫽
3 241
P
20
3 241
⫽ 0.96
20
tension
Problem 1.2-8 A space truss is restrained at joints O, A, B, and C, as shown in
the figure. Load P is applied at joint A and load 2P acts downward at joint C.
y
(a) Find reaction force components Ax, By, and Bz in terms of load variable P.
(b) Find the axial force in truss member AB in terms of load variable P.
2P
C
Cx
0.6L
Ox
0.8 L
A
z
P
Ay
Ax
Oz
L
O
B
Oy
x
Bz By
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.2
9
Statics Review
Solution 1.2-8
(a) FIND REACTIONS USING STATICS
rOA ⫽
0
0
P 0.8 L Q
rOB ⫽
m⫽4
r⫽8
m + r ⫽ 12
j⫽4
m + r ⫽ 3j
so truss is statically determinate
L
0
P0Q
rOC ⫽
0
0.6 L
P 0 Q
FB ⫽
0
By
PB Q
z
⫺0.8 Ay L
* FC ⫽ 0.8 Ax L ⫺ Bz L
P B L ⫺ 0.6 C L Q
y
x
so
FA ⫽
Ax
Ay
PPQ
3 j ⫽ 12
Fc ⫽
Cx
⫺2 P
P 0 Q
FO ⫽
Ox
Oy
PO Q
z
©M ⫽ 0
Resultant moment at O:
MO ⫽ rOA * FA + rOB * FB + rOC
©Mx ⫽ 0 gives Ay ⫽ 0
©F ⫽ 0
Resultant force at O:
RO ⫽ FO + FA + FB + FC ⫽
METHOD OF JOINTS
Ax + Cx + Ox
Ay + By + Oy ⫺ 2 P
P
Q
Bz + Oz + P
©Fz ⫽ 0
Joint O:
Oz ⫽ 0
so from
©Fz ⫽ 0
Bz ⫽ ⫺P
Joint B:
©Fy ⫽ 0
By ⫽ 0
Joint C:
(b) FORCE IN MEMBER AB
©Fx ⫽ 0
Cx ⫽ 0
©Fz ⫽ 0 at joint B:
FAB ⫽
2(0.8 L)2 + L2
ƒ Bz ƒ
0.8 L
FAB ⫽ 1.601 P
©My ⫽ 0
and
ƒ Bz ƒ ⫽ ƒ P ƒ
Ax ⫽
Bz
0.8
⫽ ⫺1.25 P
2(0.8 L)2 + L2
⫽ 1.601
0.8 L
tension
Problem 1.2-9 A space truss is restrained at joints A, B, and C, as
shown in the figure. Load 2P is applied at in the ⫺x direction at
joint A, load 3P acts in the ⫹z direction at joint B and load P is
applied in the ⫹z direction at joint C. Coordinates of all joints are
given in terms of dimension variable L (see figure).
(a) Find reaction force components Ay and Az in terms of load
variable P.
(b) Find the axial force in truss member AB in terms of load
variable P.
y
By
3P(+z direction)
B(0, 4L, 0)
Bx
2L
C(0, 2L, 4L)
2L
3L
P
Cy
2P
Cx
4L
z
A(3L, 0, 0)
O(0, 0, 0)
x
Az
Ay
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
10
CHAPTER 1
Tension, Compression, and Shear
Solution 1.2-9
m⫽3
(a) FIND REACTIONS USING STATICS
m + r ⫽ 3j
rOA ⫽
3L
0
P0Q
rOB ⫽
0
4L
P0Q
rOC ⫽
m + r⫽9
r⫽6
j⫽3
3j ⫽ 9
so truss is statically determinate
0
2L
P 4L Q
FA ⫽
⫺2P
Ay
P A Q
z
FB ⫽
Bx
By
P 3P Q
so
©Mx ⫽ 0 gives Cy ⫽
FC ⫽
Cx
Cy
PPQ
©M ⫽ 0
Resultant moment at O:
MO ⫽ rOA * FA + rOB * FB + rOC
14 L P ⫺ 4 Cy L
* FC ⫽
4 Cx L ⫺ 3 Az L
P3A L ⫺ 4B L ⫺ 2C LQ
y
x
x
14
P
4
©F ⫽ 0
Resultant force at O:
RO ⫽ FA + FB + FC ⫽
Bx + Cx ⫺ 2 P
Ay + By + Cy
P A + 4P Q
z
©Fz ⫽ 0
so
gives
Az ⫽ ⫺4.0 P
METHOD OF JOINTS
Joint A:
©Fz ⫽ 0
FACz ⫽ ⫺Az ⫽ 4.0 P
FACy ⫽
so
©Fx ⫽ 0 FABx ⫽ ⫺2 P ⫺ FACx ⫽ ⫺3.0 P ⫺ 2 P
©Fy ⫽ 0 Ay ⫽ ⫺(FABy + FACy) ⫽
2
F
⫽ 2.0 P
4 ACz
so
8P
+ 4.0 P + ⫺2.0 P
3
FABy ⫽
FACx ⫽
3
F
⫽ 3.0 P
4 ACz
4
8P
F
⫽ ⫺4.0 P ⫺
3 ABx
3
Ay ⫽ 4.67 P
(b) FORCE IN MEMBER AB
FAB ⫽ 2FABx 2 + FABy 2
FAB ⫽ ⫺
C
FAB ⫽ ⫺8.33 P
52 + a
20 2
25 P
b P⫽⫺
3
3
25
⫽ 8.33
3
compression
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.2
Problem 1.2-10 A space truss is restrained at joints A, B, and C,
as shown in the figure. Load P acts in the ⫹z direction at joint B
and in the ⫺z direction at joint C. Coordinates of all joints are
given in terms of dimension variable L (see figure). Let P ⫽ 5 kN
and L ⫽ 2 m.
11
Statics Review
y
P(+z direction)
(0, 4L, 2L) B
Bx
(a) Find the reaction force components Az and Bx.
(b) Find the axial force in truss member AB.
4
1
z
P (–z direction)
2L
2L
C (0, 0, 4L)
Cy
3 2
4L
2
O(0, 0, 0)
3L
3
4
A(3L, 0, 0)
Ax
Az
Cx
x
Ay
Solution 1.2-10
m⫽3
(a) FIND REACTIONS USING STATICS
m + r ⫽ 3j
L⫽2m
rOA ⫽
3L
0
P0Q
m + r⫽9
r⫽6
j⫽3
3j ⫽ 9
so truss is statically determinate
P ⫽ 5 kN
rOB ⫽
0
4L
P 2L Q
rOC ⫽
0
0
P 4L Q
FA ⫽
Ax
Ay
PA Q
z
FB ⫽
Bx
0
PPQ
FC ⫽
Cx
Cy
P ⫺P Q
©F ⫽ 0
Resultant force at O:
RO ⫽ FA + FB + FC ⫽
Ax + Bx + Cx
Ay + Cy
P
Q
Az
so
©Fz ⫽ 0 gives
Az ⫽ 0
RESULTANT MOMENT AT A
rAC ⫽
MA ⫽ rAB * FB + rAC * FC ⫽
⫺3L
0
P 4L Q
eAC ⫽
120 kN ⫺ 24 Cy
12 Bx + 24 Cx
P
Q
⫺24 Bx ⫺ 18 Cy
rAC
ƒ rAC ƒ
⫽
⫺0.6
0
P 0.8 Q
rAB ⫽
⫺3L
4L
P 2L Q
MA eAC ⫽ ⫺19.2 Bx ⫺72.0 kN so Bx ⫽
⫺72
kN ⫽ ⫺3.75 kN
19.2
(b) FORCE IN MEMBER AB
Method of joints at B:
©Fx ⫽ 0
FABx ⫽ ⫺Bx
FAB ⫽
229
FAB x ⫽ 6.73 kN
3
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
12
CHAPTER 1
Tension, Compression, and Shear
Problem 1.2-11 A stepped shaft ABC consisting of two solid, circular segments is subjected to torques T1 and T2 acting in
opposite directions, as shown in the figure. The larger segment of the shaft has a diameter of d1 ⫽ 58 mm and a length
L1 ⫽ 0.75 m; the smaller segment has a diameter
T1
d2 ⫽ 44 mm and a length L2 ⫽ 0.5 m. The torques
T2
#
#
d1
are T1 ⫽ 2400 N m and T2 ⫽ 1130 N m.
d2
(a) Find reaction torque TA at support A.
(b) Find the internal torque T(x) at two locations:
x ⫽ L1/2 and x ⫽ L1 ⫹ L2/2. Show these internal
torques on properly drawn free-body diagrams
(FBDs).
x
C
B
A
L1
L2
Solution 1.2-11
(a) APPLY LAWS OF STATICS
©Mx ⫽ 0
TA ⫽ T1 ⫺ T2 ⫽ 1270 N # m
(b) INTERNAL STRESS RESULTANT T AT TWO LOCATIONS
Cut shaft at midpoint between A and B at x ⫽ L1/2
(use left FBD).
©Mx ⫽ 0
TAB ⫽ ⫺TA ⫽ ⫺1270 N # m
Cut shaft at midpoint between B and C at x ⫽ L1 ⫹ L2/2
(use right FBD).
©Mx ⫽ 0
TBC ⫽ T2 ⫽ 1130 N # m
Problem 1.2-12 A stepped shaft ABC consisting of two solid, circular segments is subjected to uniformly distributed torque
t1 acting over segment 1 and concentrated torque T2 applied at C, as shown in the figure. Segment 1 of the shaft has a diameter
of d1 ⫽ 57 mm and length of L1 ⫽ 0.75 m; segment 2
t1
has a diameter d2 ⫽ 44 mm and length L2 ⫽ 0.5 m.
T2
Torque intensity t1 ⫽ 3100 N # m/m and
d2
#
T2 ⫽ 1100 N m .
x
(a) Find reaction torque TA at support A.
(b) Find the internal torque T(x) at two locations:
x ⫽ L1/2 and x ⫽ L1 ⫹ L2/2. Show these internal
torques on properly drawn free-body diagrams
(FBDs).
(a) REACTION TORQUE AT A
Statics:
©Mx ⫽ 0
L1 ⫽ 0.75 m
d1
L1
Solution 1.2-12
L2 ⫽ 0.75 m
C
B
A
L2
t1 ⫽ 3100 N # m/m
TA ⫽ ⫺t1 L1 + T2 ⫽ ⫺1225 N # m
T2 ⫽ 1100 N # m
TA ⫽ ⫺1225 N # m
(b) INTERNAL TORSIONAL MOMENTS AT TWO LOCATIONS
L1
b ⫽ 62.5 N # m
2
Cut shaft between A and B
(use left FBD).
T1(x) ⫽ ⫺TA ⫺ t1 x
T1 a
Cut shaft between B and C
(use left FBD).
T2(x) ⫽ ⫺TA ⫺ t1 L1
T2 aL1 +
L2
b ⫽ ⫺1100 N # m
2
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.2
Problem 1.2-13 A plane frame is restrained at joints A and C, as shown in
the figure. Members AB and BC are pin connected at B. A triangularly
distributed lateral load with a peak intensity of 1300 N/m acts on AB.
A concentrated moment is applied at joint C.
13
Statics Review
680 N·m at joint C
Pin
connection
1300 N/m
B
(a) Find reactions at supports A and C.
(b) Find internal stress resultants N, V, and M at x ⫽ 1.0 m on column AB.
C
2.75 m
3.70 m
x
A
Solution 1.2-13
(a) STATICS
⫺1
N
a1300 b3.70 m ⫽ ⫺2405 N
2
m
©FH ⫽ 0
Ax ⫽
©FV ⫽ 0
Ay + Cy ⫽ 0
©MFBDBC ⫽ 0
©MA ⫽ 0
Cy ⫽
680 N # m
⫽ 247 N
2.75 m
MA ⫽ 680 N # m +
Ax ⫽ ⫺2405 N
Ay ⫽ ⫺247 N
Ay ⫽ ⫺Cy ⫽ ⫺247 N
N
2
1
a1300 b3.70 ma 3.70 mb ⫺ Cy 2.75 m ⫽ 5932 N # m
2
m
3
MA ⫽ 5932 N # m
Cy ⫽ 247 N
;
(b) INTERNAL STRESS RESULTANTS
N
Nx ⫽ ⫺Ay ⫽ 247 N
Vx ⫽ ⫺Ax ⫺
V
1 1.0
N
a
1300 b1.0 m ⫽ 2229 N
2 3.70
m
1 1.0
N
1
Mx ⫽ ⫺MA ⫺ Ax1.0 m ⫺ a
1300 b 1.0 ma 1.0 mb ⫽ ⫺3586 N # m
2 3.70
m
3
Nx ⫽ 247 N
Vx ⫽ 2229 N
Mx ⫽ ⫺3586 N # m
M
x
x = 1.0 m
A
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
14
CHAPTER 1
Tension, Compression, and Shear
Problem 1.2-14 A plane frame is restrained at joints A and D, as shown
in the figure. Members AB and BCD are pin connected at B. A triangularly
distributed lateral load with peak intensity of 80 N/m acts on CD.
An inclined concentrated force of 200 N acts at the mid-span of BC.
Pin
connection
C
B
1.5 m
(a) Find reactions at supports A and D.
(b) Find resultant forces in the pins at B and C.
80 N/m
1.5 m
4
200 N
3
4m
A
4m
D
Solution 1.2-14
(a) STATICS
©Fx ⫽ 0
Ax ⫽
1
3
(200 N) + (80 N/m) 4 m ⫽ 280 N
5
2
©MBRHFB ⫽ 0 Dy ⫽
1 4
1
1
c (200 N) (1.5 m) + (80 N/m) 4 m a 4 mb d
3m 5
2
3
⫽ 151.1 N 6 use right hand FBD (BCD only)
©Fy ⫽ 0
©MA ⫽ 0
Ay ⫽ ⫺Dy +
MA ⫽
4
(200 N) ⫽ 8.89 N
5
4
3
1
2
(200 N) (1.5 m) ⫺ (200 N) (4 m) ⫺ Dy 3 m ⫺ (80 N/m) 4 m a 4 mb ⫽ ⫺1120 N # m
5
5
2
3
(b) RESULTANT FORCE IN PIN AT B
FBy
LEFT HAND FBD (SEE FIGURE)
FBx ⫽ ⫺Ax ⫽ ⫺280 N
FBy ⫽ ⫺Ay ⫽ ⫺8.89 N
B
FBx
RIGHT HAND FBD
1
3
(200 N) + (80 N/m) 4 m ⫽ 280 N
5
2
4
FBy ⫽ (200 N) ⫺ Dy ⫽ 8.89 N
5
FBx ⫽
4m
ResultantB ⫽ 2FBx2 + FBy2 ⫽ 280 N
A
Left hand FBD
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.2
15
Statics Review
Problem 1.2-15 A 1900-N trap door (AB) is supported by a strut
B
(BC) which is pin connected to the door at B (see figure).
y
(a) Find reactions at supports A and C.
(b) Find internal stress resultants N, V, and M on the trap door at
0.5 m from A.
0.75 m
LAC ⫽
do
or
4
0.75 m
5
⫽
⫽ 0.671 m
2
15
t
Stru
p
1900 N
Tr
a
LBC
Pin or hinge
connection
2
4
3
1
(0.75 m) +
LBC ⫽ 0.75 m
5
15
A
1
3
C
x
Solution 1.2-15
(a) STATICS
©MA ⫽ 0
Cy ⫽
1
13
a1900 N
0.75 mb ⫽ 570 N
LAC
25
©Fx ⫽ 0
Ax ⫽ ⫺Cx ⫽ 285 N
©Fy ⫽ 0
Ay ⫽ 1900 N ⫺ Cy ⫽ 1330 N
;
Cx ⫽
⫺1
C ⫽ ⫺285 N
2 y
;
(resultant of Cx and Cy acts along line of strut)
;
(b) INTERNAL STRESS RESULTANTS N, V, M (SEE FIGURE)
Distributed weight of door in ⫺y direction:
w⫽
1900 N
N
⫽ 2533.333
0.75 m
m
Components of w along and perpendicular to door:
4
N
w ⫽ 2026.667
5
m
N
3
w ⫽ 1520
5
m
M
y
V
N
4
3
0.5 m
⫺ Ax(0.5 m) + Ay(0.5 m) ⫽ 95 N # m
2
5
5
Nx ⫽ ⫺222 N
0.75 m
Vx ⫽ ⫺190 N
Mx ⫽ 95 N # m
do
3
4
Ax + Ay ⫽ ⫺190 N
5
5
1900 N
p
Vx ⫽ ⫺wp(0.5 m) ⫺
or
3
4
A ⫺ Ay ⫽ ⫺221.667 N
5 x
5
Nx ⫽ wa(0.5 m) ⫺
Mx ⫽ ⫺wp(0.5 m)
wp ⫽
Tr
a
wa ⫽
4
A
3
x
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
16
CHAPTER 1
Tension, Compression, and Shear
Problem 1.2-16 A plane frame is constructed by using a pin
connection between segments ABC and CDE. The frame has pin
supports at A and E and has joint loads at B and D (see figure).
10 kN B
3m
Pin connection
C
10 kN
(a) Find reactions at supports A and E.
(b) Find resultant force in the pin at C.
3m
D
3m
E
6m
90 kN·m
A
Solution 1.2-16
(a) STATICS
©MA ⫽ 0
10 kN (6 m) ⫺ 10 kN a
so
or
©MCRHFB
1
b (6 m) + 90 kN # m + Ey (6 m) ⫺ Ex 3 m ⫽ 6 Ey m ⫺ 3 Ex m + 150 kN # m ⫺ 30 12 kN # m
12
6 Ey m ⫺ 3 Ex m + 150 kN # m ⫺ 30 12 kN # m ⫽ 0
⫺(150 kN # m ⫺ 30 12 kN # m)
⫽ ⫺35.858 kN
3m
⫽ 0 6 right hand FBD (CDE) - see figure.
⫺Ex + 2 Ey ⫽
1Ex + Ey2 3 m ⫽ ⫺90 kN # m
Solving
FCy
a
Ex
⫺1 2
b⫽a
b
Ey
1 1
Ex + Ey ⫽
FCx
C
⫺90 kN # m
⫽ ⫺30 kN
3m
3m
D
⫺1
a
⫺35.858 kN
⫺8.05
b⫽a
b kN
⫺30 kN
⫺21.95
©Fy ⫽ 0
Ax ⫽ ⫺Ex + 10 kN ⫺ 10 kN a
Ay ⫽ ⫺Ey + 10 kN a
(b) RIGHT HAND FBD:
1
1
22
b ⫽ 10.98 kN
E
90 kN·m
Ex ⫽ ⫺8.05 kN
©Fx ⫽ 0
3m
Ey ⫽ ⫺22 kN
Ax ⫽ 10.98 kN
b ⫽ 29.07 kN
Ay ⫽ 29.1 kN
22
Cy ⫽ ⫺Ey ⫽ 22 kN
Cx ⫽ ⫺Ex ⫽ 8.05 kN
ResultantC ⫽ 2Cx2 + Cy2 ⫽ 23.4 kN
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.2
17
Statics Review
Problem 1.2-17 A plane frame with pin supports at A and E has a cable attached at C, which runs over a frictionless pulley
at F (see figure). The cable force is known to be 2.25 kN.
(a) Find reactions at supports A and E.
(b) Find internal stress resultants, N, V, and M at point H.
D
E
150 mm
150 mm
240 mm
C
Cable
F
H
360 mm
y
180 mm
B
150 mm
750 mm
G
2.25 kN
A
x
Solution 1.2-17
(a) STATICS
©Fx ⫽ 0
Ex ⫽ 0
©ME ⫽ 0
Ay ⫽
©Fy ⫽ 0
1
[⫺2250 N(750 mm)] ⫽ ⫺5625 N
300 mm
Ey ⫽ 2250 N ⫺ Ay ⫽ 7875 N
Ay ⫽ ⫺5625 N
Ex ⫽ 0
Ey ⫽ 7875 N
;
(b) USE UPPER (SEE FIGURE BELOW) OR LOWER FBD TO FIND STRESS RESULTANTS N, V, AND M AT H
D
©Fy ⫽ 0
150 mm
240 mm
180 mm
V
©Fx ⫽ 0
E
H
Cable
Nx ⫽ Ey ⫽ 7875 N
©MH ⫽ 0
2.25 kN
C
Vx ⫽ Ex + 2250 N ⫽ 2250 N
MH ⫽ ⫺(180 mm)(2250 N) ⫺ Ex(180 mm + 240 mm) + Ey(150 mm)
⫽ 776 N # m
Nx ⫽ 7875 N
Vx ⫽ 2250 N
Mx ⫽ 776 N # m
;
N
M
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
18
CHAPTER 1
Tension, Compression, and Shear
Problem 1.2-18 A plane frame with a
pin support at A and roller supports at C
and E has a cable attached at E, which
runs over frictionless pulleys at D and B
(see figure). The cable force is known to
be 400 N. There is a pin connection just
to the left of joint C.
(a) Find reactions at supports A, C,
and E.
(b) Find internal stress resultants N, V,
and M just to the right of joint C.
(c) Find resultant force in the pin
near C.
D
Cable is attached
at E and passes over
frictionless pulleys at B and D
4m
4
400 N
3
4
3
C
A
4m
B
E
3m
5m
Pin connection
just left of C
Solution 1.2-18
(a) STATICS
©Fx ⫽ 0
Ax ⫽
4
(400 N) ⫽ 320 N
5
Ax ⫽ 320 N
Use left hand FBD (cut through pin just left of C).
©MC ⫽ 0
Use entire FBD:
Ay ⫽
1
⫺3
4
cc
(400 N) ⫺ (400 N) d (3 m) d ⫽ ⫺240 N
7m
5
5
1
3
cAy (7 m) + a 400 Nb (3 m) d ⫽ ⫺192 N
5m
5
©MC ⫽ 0
Ey ⫽
©Fy ⫽ 0
Cy ⫽ ⫺Ay ⫺ Ey ⫺
3
(400 N) ⫽ 192 N
5
(b) N, V, AND M JUST RIGHT OF C; USE RIGHT HAND FBD
5
M
Fcable X
E
24 + 52
b ⫽ 312.348 N
©Fx ⫽ 0
©Fy ⫽ 0
V ⫽ ⫺FcableY ⫺ Ey ⫽ ⫺57.9 N
©MC ⫽ 0
5m
3
4
⫺ b 400 N ⫽ ⫺240 N
5
5
ResC ⫽ 2FCx2 + FCy2 ⫽ 400 N
5
2
4
F
⫽ 249.878 N
5 cableX
Nx ⫽ ⫺FcableX ⫽ ⫺312 N
(c) RESULTANT FORCE IN PIN JUST LEFT OF C; USE LEFT HAND FBD
FCx ⫽ ⫺Ax + a
Cy ⫽ 192 N
FcableX ⫽ 400 N a
N
V
Ey ⫽ ⫺192 N
FcableY ⫽
FcableY
4
Ay ⫽ ⫺240 N
M ⫽ A FcableY + Ey B (5 m) ⫽ 289 N # m
Ax ⫽ 320 N
FCy ⫽ ⫺Ay ⫺ a
3
4
+ b 400 N ⫽ ⫺320 N
5
5
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.2
Statics Review
Problem 1.2-19 A 650-N rigid bar AB, with frictionless rollers at
each end, is held in the position shown in the figure by a continuous
cable CAD. The cable is pinned at C and D and runs over a pulley at A.
19
C
(a) Find reactions at supports A and B.
(b) Find the force in the cable.
0.9 m
Ca
ble
B
id
y
0-N
rig
ar
b
D
65
0.6 m
30°
A
x
1.2 m
Solution 1.2-19
(a) STATICS
W ⫽ 650 N
©MA ⫽ 0
Bx (0.6 m + 1.2 m sin(30⬚)) + Wc
so
©FX ⫽ 0
©Fy ⫽ 0
A
a b⫽±
T
Bx ⫽ ⫺
650.0 N cos(30.0⬚)
2.0 sin(30.0⬚) + 1.0
650.0 N cos(30.0⬚)
1.2 m
(cos(30⬚)) d ⫽ 0 solve, Bx ⫽
2
2.0 sin(30.0⬚) + 1.0
⫽ ⫺281.458 N
⫺ A sin(30⬚) + Bx + T cos(30⬚) + T cos aarctan a
A cos(30⬚) + T sin(30⬚) + T sin a arctan a
cos(30⬚) + cosaarctan a
⫺sin(30⬚)
cos(30⬚)
A
250.09
a b⫽a
b N
T
310.412
sin(30⬚) + sin aarctan a
0.9 + 0.6 + 1.2 sin(30⬚)
1.2 cos(30⬚)
0.9 + 0.6 + 1.2 sin(30⬚)
1.2 cos(30⬚)
0.9 + 0.6 + 1.2 sin(30⬚)
1.2 cos(30⬚)
0.9 + 0.6 + 1.2 sin(30⬚)
1.2 cos(30⬚)
⫺1
bb
bb
≤
bb ⫽ 0
bb ⫽ W
a
⫺Bx
b
W
SUPPORT REACTIONS
Bx ⫽ ⫺281.5 N
A ⫽ 250.1 N
Ax ⫽ ⫺A sin(30⬚) ⫽ ⫺125 N
Ay ⫽ A cos(30⬚) ⫽ 216.6 N
2A2x + A2y ⫽ 250.09 N
(b) CABLE FORCE IS T (NEWTONS) FROM ABOVE SOLUTION
T ⫽ 310.4 N
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
20
CHAPTER 1
Tension, Compression, and Shear
Problem 1.2-20 A plane frame has a pin support at
A and roller supports at C and E (see figure). Frame
segments ABD and CDEF are joined just left of joint
D by a pin connection.
1.5 kN/m
Pin connection
just left of D
16 kN
(a) Find reactions at supports A, C, and E.
(b) Find the resultant force in the pin just
left of D.
B
6m
D
6m
4m
4m
A
C
3 kN/m
F
4m
E
Solution 1.2-20
(a) STATICS
RIGHT-HAND FBD
©Mpin ⫽ 0
ENTIRE FBD
©MA ⫽ 0 Cy ⫽
Cy ⫽ 9.83 kN
Ey ⫽
1
1 1
c (3 kN/m) 4 m a 4 mb d ⫽ 1.333 kN
6m 2
3
Ey ⫽ 1.333 kN
1
2
1
c⫺Ey 12 m + (16 kN) 4 m + (1.5 kN/m) 6 m (3 m) ⫺ (3 kN/m) 4 m a 4 mb d ⫽ 9.833 kN
6m
2
3
©Fy ⫽ 0
Ay ⫽ ⫺Cy ⫺ Ey + (1.5 kN/m) 6 m ⫽ ⫺2.167 kN
©Fx ⫽ 0
Ax ⫽ ⫺16 kN +
1
(3 kN/m) 4 m ⫽ ⫺10 kN
2
Ay ⫽ ⫺2.17 kN
Ax ⫽ ⫺10 kN
(b) RESULTANT FORCE IN PIN; USE EITHER RIGHT HAND OR LEFT HAND FBD (CUT
AND FDy) THEN SUM FORCES IN x AND y DIRECTIONS FOR EITHER FBD
LHFB:
FDx ⫽ ⫺16 kN ⫺ Ax ⫽ ⫺6 kN
FDy ⫽ ⫺Ay + (1.5 kN/m) 6 m ⫽ 11.167 kN
ResultantD ⫽ 3FDx2 + FDy2 ⫽ 12.68 kN
THROUGH PIN EXPOSING PIN FORCES
FDx
RHFB:
1
FDx ⫽ (3 kN/m) 4 m ⫽ 6 kN
2
FDy ⫽ ⫺Cy ⫺ Ey ⫽ ⫺11.167 kN
ResultantD ⫽ 12.68 kN
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.2
21
Statics Review
z
Problem 1.2-21 A special vehicle brake is clamped at O, (when the brake force P1 is
applied—see figure). Force P1 ⫽ 220 N and lies in a plane which is parallel to the xz
plane and is applied at C normal to line BC. Force P2 ⫽ 180 N and is applied at B in
the ⫺y direction.
x
178 mm
(a) Find reactions at support O.
(b) Find internal stress resultants N, V, T, and M at the midpoint of segment OA.
O
A
y
y'
150 mm
B
P2
15°
200 mm
P1
C
15°
x
Solution 1.2-21
(a) STATICS
P1 ⫽ 220 N
P2 ⫽ 180 N
©Fx ⫽ 0
Ox ⫽ ⫺P1 cos(15⬚) ⫽ ⫺212.5 N
©Fz ⫽ 0
Oz ⫽ P1 sin(15⬚) ⫽ 56.94 N
©Fy ⫽ 0
Oy ⫽ P2 ⫽ 180 N
©Mx ⫽ 0
MOx ⫽ P2 150 mm + P1 sin(15⬚)(178 mm) ⫽ 37.1 N # m
©My ⫽ 0
MOy ⫽ P1 sin(15⬚)(200 mm sin(15⬚)) + P1 cos(15⬚)(150 mm + 200 mm cos(15⬚))
©Mz ⫽ 0
MOz ⫽ ⫺P1 cos(15⬚)(178 mm) ⫽ ⫺37.8 N # m
MOy ⫽ 75.9 N # m
Ox ⫽ ⫺213 N
MOx ⫽ 37.1 N # m
Oy ⫽ 180 N
Oz ⫽ 56.9 N
MOy ⫽ 75.9 N # m
MOz ⫽ ⫺37.8 N # m
;
(b) INTERNAL STRESS RESULTANTS AT MIDPOINT OF OA
Nx ⫽ ⫺Oy ⫽ ⫺180 N
Vx ⫽ ⫺Ox ⫽ 212.5 N
Vz ⫽ ⫺Oz ⫽ ⫺56.9 N
Tx ⫽ ⫺MOy ⫽ ⫺75.9 N # m
Mx ⫽ ⫺MOx ⫽ ⫺37.1 N # m
Nx ⫽ ⫺180 N
Tx ⫽ ⫺75.9 N # m
Vres ⫽ 2V2x + V2z ⫽ 220 N
Mz ⫽ ⫺MOz ⫽ 37.8 N # m
Mres ⫽ 2M2x + M2z ⫽ 53 N # m
Vres ⫽ 220 N
Mres ⫽ 53 N # m
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
22
CHAPTER 1
Tension, Compression, and Shear
Problem 1.2-22 Space frame ABCD is clamped at A, except
it is free to translate in the x-direction. There is also a roller
support at D, which is normal to line CDE. A triangularly
distributed force with peak intensity q0 ⫽ 75 N/m acts along
AB in the positive z direction. Forces Px ⫽ 60 N and
Pz ⫽⫺45 N are applied at joint C and a concentrated
moment My ⫽ 120 N # m acts at the mid-span of member BC.
y
My
B(0, 2, 0)
C(1.5, 2, 0)
q0
Joint coordinates
in meters
Px
Pz
(a) Find reactions at supports A and D.
(b) Find internal stress resultants N, V, T, and M at the
mid-height of segment AB.
0.75 m
D
E(2.5, 0, −0.5)
A(0, 0, 0)
z
x
Solution 1.2-22
FORCES
Px ⫽ 60 N
My ⫽ 120 N # m
Pz ⫽ ⫺45 N
Px
60
FC ⫽ 0 ⫽
0
N
P P Q P ⫺45 Q
z
q0 ⫽ 75 N/m
0
RA ⫽ Ay
PA Q
z
VECTOR ALONG MEMBER CD
rEC ⫽
1.5 ⫺ 2.5
⫺1
2⫺0
⫽ 2
J 0 ⫺ (⫺0.5) K J 0.5 K
冷rEC 冷 ⫽ 2.291
eEC ⫽
rEC
冷rEC 冷
⫽
⫺0.436
0.873
P 0.218 Q
(a) STATICS (FORCE AND MOMENT EQUILIBRIUM)
©F ⫽ 0
0
0
Px
Dx
Ay + 0 + 0 + Dy ⫽ 0
PA Q
PR Q
PP Q
PD Q
z
z
z
T
where
Dx
Dy ⫽ D eEC
PD Q
z
resultant of triangular load:
RT ⫽
1
q (2 m) ⫽ 75 N
2 0
SOLVING ABOVE THREE EQUATIONS:
⫺Px
eEC1
©Fx ⫽ 0
Dx ⫽ ⫺Px
©Fy ⫽ 0
Dy ⫽ eEC2 D
Dy ⫽ 120 N
Ay ⫽ ⫺Dy ⫽ ⫺120 N
©Fz ⫽ 0
Dz ⫽ eEC3 D
Dz ⫽ 30 N
3Dx2 + Dy2 + Dz2 ⫽ 137.477 N
so
so
D⫽
Az ⫽ ⫺Dz ⫺ RT ⫺ Pz
D ⫽ 137.477 N
Dx ⫽ ⫺60 N
Az ⫽ ⫺60 N
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.2
23
Statics Review
©MA ⫽ 0
MAx
Px
0
0
MAy + rAE * D + rAC * 0 + My + rcg * 0 ⫽ 0
PM Q
PP Q
P0Q
PR Q
Az
z
T
rAE ⫽
2.5 ⫺ 0
0⫺0
m
P ⫺0.5 ⫺ 0 Q
1.5 ⫺ 0
rAC ⫽ 2 ⫺ 0 m
P 0⫺0 Q
D⫽
rAC
Dx
Dy
PD Q
z
D⫽
⫺60
120 N
P 30 Q
ƒ D ƒ ⫽ 137.477 N
rAE * D ⫽
0
Px
⫺90
* 0 ⫽ 67.5 J
P P Q P ⫺120 Q
z
2
rcg ⫽ ± (2 m) ≤
3
0
rcg *
MAx
Px
0
0
⫺70
MAy ⫽ ⫺ rAE * D + rAC * 0 + My + rcg * 0
⫽ ⫺142.5 N # m
PM Q
J
PP Q P 0 Q
P R Q K P ⫺180 Q
Az
z
T
60
⫺45 N # m
P 300 Q
0
100
0 ⫽
0 N#m
PR Q P 0 Q
T
MAx
⫺70
MAy ⫽ ⫺142.5 N # m
P M Q P ⫺80 Q
Az
(b) RESULTANTS AT MID-HEIGHT OF AB (SEE FBD IN FIGURE BELOW)
N ⫽ ⫺Ay ⫽ 120 N
Vx ⫽ ⫺Dx ⫺ Px ⫽ 0 N
Vz ⫽ ⫺Az ⫺
T ⫽ ⫺MAy ⫽ 142.5 N # m
Mx ⫽ ⫺MAx + Az (1 m) +
Mz ⫽ ⫺MAz ⫽ 180 N # m
1 q0
(2 m)/2 ⫽ 41.25 N V ⫽ Vz ⫽ 41.3 N
2 2
1 q0
1
1 m a 1 mb ⫽ 16.25 N # m
2 2
3
T
Mresultant ⫽ 3Mx2 + Mz2 ⫽ 180.732 N # m
Mresultant ⫽ 180.7 N # m
N
q0 /2
Mz
Vz
Vx
A(0, 0, 0)
z
Mx
x
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
24
CHAPTER 1
Tension, Compression, and Shear
Problem 1.2-23 Space frame ABC is clamped at A except
it is free to rotate at A about the x and y axes. Cables DC
and EC support the frame at C. Force Py ⫽ ⫺220 N is
applied at mid-span of AB and a concentrated moment
Mx ⫽ ⫺2.25 N # m acts at joint B.
y
D(0, 0.25, −0.5)
Joint coordinates
in meters
E(0, 0.2, 0.25)
Cable DC
(a) Find reactions at supports A.
(b) Find cable tension forces.
C(0.25, 0.1, −0.1)
Cable EC
A(0, 0, 0)
z
Py
Mx
x
B(0.25, 0 , 0)
Solution 1.2-23
POSITION AND UNIT VECTORS
rAB ⫽
0.25
0
P 0 Q
rAP ⫽
0.125
0
P 0 Q
rAC ⫽
0.25
0.1
P ⫺0.1 Q
APPLIED FORCE AND MOMENT
Py ⫽ ⫺220 N
Mx ⫽ ⫺2.25 N # m
rCD ⫽
rCE ⫽
0 ⫺ 0.25
⫺0.25
⫽ 0.15
0.25 ⫺ 0.1
J ⫺0.5 ⫺ (⫺0.1) K P ⫺0.4 Q
0 ⫺ 0.25
⫺0.25
0.2 ⫺ 0.1
0.1
⫽
J 0.25 ⫺ (⫺0.1) K P 0.35 Q
eCD ⫽
eCE ⫽
rCD
ƒ rCD ƒ
rCE
ƒ rCE ƒ
⫽
⫺0.505
0.303
P ⫺0.808 Q
⫽
⫺0.566
0.226
P 0.793 Q
STATICS–FORCE AND MOMENT EQUILIBRIUM
First sum moment about point A.
©MA ⫽ 0
MA ⫽
0
0
Mx
⫺0.050508 TD + 0.1019 TE ⫺ 2.25
≤
+ rAP * Py + 0 + rAC * A TD eCD + TE eCE B ⫽ ±
0
0.25254 TD + ⫺0.14153 TE
PM Q
P0Q P 0 Q
M
+
0.12627
T
+
0.11323
T
⫺
27.5
Az
Az
D
E
Solve moment equilibrium equations for moments about x and y axes to get cable tension forces.
⫺0.050508 0.1019 ⫺1 2.25
17.1
TD
b⫽a
b a
b⫽a
bN
6 Part (b)
TE
0.25254 ⫺0.14153
0
30.6
Next, solve moment equilibrium equation about z axis now that cable forces are known.
a
MAz ⫽ (0.12627 TD + 0.11323 TE ⫺ 27.5) ⫽ 21.9 N # m
;
6 Part (a)
Finally, use force equilibrium to find reaction forces at point A.
©F ⫽ 0
Ax
0
25.96
Ay ⫽ ⫺ Py ⫺ (TD eCD + TE eCE) ⫽ 207.88
PA Q
P0Q
P ⫺10.39 Q
z
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.2
Problem 1.2-24 A soccer goal is subjected
25
Statics Review
1.22 m
to gravity loads (in the ⫺z direction,
w ⫽ 73 N/m for DG, BG, and BC;
w ⫽ 29 N/m for all other members;
see figure) and a force F ⫽ 200 N applied
eccentrically at mid-height of member DG.
Find reactions at supports C, D, and H.
P
G
w = 73 N/m
z
Gravity
2.44 m
x
3
B
Q
4
2.44 m
F = 200 N
D
R
H
3.65 m
S
C
3.65 m
y
w = 29 N/m
Reaction force
Solution 1.2-24
FIND MEMBER LENGTHS
LRS ⫽ 3(2.44 m)2 + (2.44 m ⫺ 1.22 m)2 ⫽ 2.728 m LPQ ⫽ LRS
LQS ⫽ 2 (3.65 m) ⫽ 7.3 m
Assume that soccer goal is supported only at points C, H, and D (see reaction force components at each location
in figure).
STATICS–SUM MOMENT ABOUT EACH AXIS AND FORCES IN EACH AXIS DIRECTION F ⫽ 200 N
©Mx ⫽ 0
TO FIND REACTION COMPONENT
Hy:
Find moments about x for component Fy and also for distributed weight of each frame component.
MxGP ⫽
(1.22 m)2
129 N/m2
2
MxBR ⫽ MxGP
MxRS ⫽ LRS129 N/m2 a1.22 m +
Hz ⫽
©My ⫽ 0
(2.44 m)2
2
129 N/m2
MxCS ⫽ MxDQ
MxQS ⫽ LQS 129 N/m2 (2.44 m)
MxPQ ⫽ MxRS
1
4
2.44 m
c Fa
b + 2 MxGP + 2 MxDQ + 2 MxPQ + MxQS d ⫽ 498.818 N
2.44 m 5
2
TO FIND REACTION FORCE
MyPQ ⫽ LRS 129 N/m2 LQS
Hz ⫽ 499 N
Dz:
MyGD ⫽ 2.44 m 173 N/m2 LQS
Dz ⫽
1.22 m
b
2
MxDQ ⫽
MyGP ⫽ 1.22 m 129 N/m2 LQS
MyBG ⫽ LQS 173 N/m2
LQS
2
MyDQ ⫽ 2.44 m 129 N/m2 LQS
MyQS ⫽ LQS 129 N/m2
LQS
2
LQS
1
3
2.44 m
cMyGD + MyGP + MyDQ + MyPQ + MyBG + MyQS ⫺ Hz
⫺ Fa
b d ⫽ 466.208 N
LQS
2
5
2
Dz ⫽ 466 N
©Mz ⫽ 0 TO FIND REACTION FORCE Hy:
Hy ⫽
1
4
a F L b ⫽ 320 N
3.65 m 5 QS
Hy ⫽ 320 N
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
26
CHAPTER 1
Tension, Compression, and Shear
©Fx ⫽ 0 TO FIND REACTION FORCE Cx:
©Fy ⫽ 0 TO FIND REACTION FORCE Cy:
©Fz ⫽ 0 TO FIND REACTION FORCE Cz:
3
F ⫽ 120 N
5
4
Cy ⫽ ⫺Hy + F ⫽ ⫺160 N
5
Cx ⫽
Cy ⫽ ⫺160 N
Cz ⫽ ⫺Dz ⫺ Hz + 129 N/m2 12 1.22 m + 2 2.44 m + 2 LRS + LQS2
+ 173 N/m2 12 2.44 m + LQS2 ⫽ 506.318 N
Cz ⫽ 506 N
Geometry of Back rail
Problem 1.2-25 An elliptical exerciser
D
machine is composed of front and back rails. A
simplified plane-frame model of the back rail is
shown in the figure. Analyze the plane-frame
model to find reaction forces at supports A, B,
and C for the position and applied loads given
in the figure. Note that there are axial and
moment releases at the base of member 2, so
member 2 can lengthen and shorten as the
roller support at B moves along the 30° incline.
(These releases indicate that the internal axial
force N and moment M must be zero at this
location.)
400 mm
90 N
Cy
3
Member no.
C
Cx
a
y
780 N
860 mm
Axial
release
2
860 mm
400 mm a
1 250 mm
A
Moment
release
B
a
Joint no.
a = 11.537°
Bx
x
30°
By
B
Ay
Solution 1.2-25
a ⫽ arcsina
250
b ⫽ 11.444⬚
860 + 400
Analysis below pertains to this
position of exerciser only.
STATICS–UFBD (CUT AT AXIAL AND MOMENT RELEASES JUST ABOVE B)
Inclined vertical component of reaction at C ⫽ 0 (due to axial release)
Sum moments about moment release to get inclined normal reaction at C.
C⫽
90 N (860 mm + 400 mm)
⫽ 131.86 N
860 mm
Cx ⫽ C cos(a) ⫽ 129.2 N
Cy ⫽ C sin(a) ⫽ 26.2 N
3Cx2 + Cy2 ⫽ 131.86 N
STATICS–LFBD (CUT THROUGH AXIAL AND MOMENT RELEASES)
Sum moments to find reaction Ay:
Ay ⫽
780 N (400 mm)
⫽ 253 N
(860 mm + 400 mm) cos(a)
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.2
Statics Review
STATICS–SUM FORCES FOR ENTIRE FBD TO FIND REACTION AT B
Sum forces in x-direction:
Bx ⫽ Cx + 780 N (sin(a)) ⫺ 90 N (cos(a)) ⫽ 195.8 N
Sum forces in y-direction:
6 acts leftward
By ⫽ ⫺Ay ⫺ Cy + 780 N (cos(a)) + 90 N (sin(a)) ⫽ 503.5 N
Bx ⫽ 195.8 N
Resultant reaction force at B:
27
By ⫽ 504 N
B ⫽ 2B2x + B2y ⫽ 504 N
Problem 1.2-26 A mountain bike is moving along a flat path at constant velocity.
670 N
At some instant, the rider (weight ⫽ 670 N) applies pedal and hand forces, as
shown in the figure part a.
(a) Find reactions forces at the front and rear hubs. (Assume that the bike is pin
supported at the rear hub and roller supported at the front hub).
(b) Find internal stress resultants N, V, and M in the inclined seat post (see figure
part b).
V
15.3°
M
(b)
N
N
M
V
Solution 1.2-26
(a) REACTIONS: SUM MOMENTS ABOUT REAR HUB TO FIND VERTICAL REACTION AT FRONT HUB (Fig. part a)
©MB ⫽ 0
1
[670 (241) ⫺ 90 (cos(5⬚)) 254 + 200 cos(15⬚) 660 + 2 (45) cos(30⬚)1021 + 2 (45) sin(30⬚) 752]
1130
VF ⫽ 335.945 N
VF ⫽
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
28
CHAPTER 1
Tension, Compression, and Shear
Sum forces to get force components at rear hub.
©Fvert ⫽ 0
VB ⫽ 670 ⫺ 90 cos(5 ⬚) + 200 cos(15 ⬚) + 2 (45) cos(30 ⬚) ⫺ VF ⫽ 515.525 N
©Fhoriz ⫽ 0
HB ⫽ ⫺90 sin(5 ⬚) ⫺ 200 sin(15 ⬚) ⫺ 2 (45) sin(30 ⬚) ⫽ ⫺104.608 N
Figure part a
y
VF ⫽ 336 N
45 N at 30° to vertical
on each grip
670 N
HB ⫽ ⫺104.6 N
1021 mm
VB ⫽ 516 N
m
254
241 mm
m
15.3°
Origin at
B (0, 0, 0)
752 mm
90 N at 5°
x
HB
VF
VB
200 N at 15° to vertical
254 mm
660 mm
1130 mm
(b) STRESS RESULTANTS N, V, AND M IN SEAT POST (Fig. part b)
670 N
SEAT POST RESULTANTS
N ⫽ ⫺670 cos(15.3⬚) ⫽ ⫺646.253 N
N ⫽ ⫺646 N
V ⫽ 670 sin(15.3⬚) ⫽ 176.795 N
V ⫽ 176.8 N
M ⫽ 670 sin(15.3⬚) 254 ⫽ 44,905.916 N # mm
V
M ⫽ 44.9 N # m
15.3°
M
N
N
M
V
Figure part b (repeated)
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.3
Normal Stress and Strain
29
Normal Stress and Strain
P1
Problem 1.3-1 A hollow circular post ABC (see figure) supports a load
p1 ⫽ 7.5 KN acting at the top. A second load P2 is uniformly distributed
around the cap plate at B. The diameters and thicknesses of the upper and
lower parts of the post are dAB ⫽ 32 mm, tAB ⫽ 12 mm, dBC ⫽ 57 mm, and
tBC ⫽ 9 mm, respectively.
A
tAB
dAB
P2
(a) Calculate the normal stress ␴AB in the upper part of the post.
(b) If it is desired that the lower part of the post have the same
compressive stress as the upper part, what should be the magnitude
of the load P2?
(c) If P1 remains at 7.5 kN and P2 is now set at 10 kN, what new
thickness of BC will result in the same compressive stress in both
parts?
B
dBC
tBC
C
Solution 1.3-1
PART (a)
PART (b)
P1 ⫽ 75 kN
dAB ⫽ 32 mm
dBC ⫽ 57 mm
AAB ⫽
tAB ⫽ 12 mm
tBC ⫽ 9 mm
p [ dAB2 ⫺ (dAB ⫺ 2 tAB)2]
4
AAB ⫽ 7.54 ⫻ 10⫺4 m2
sAB ⫽ 9.95 MPa
P1
sAB ⫽
AAB
ABC ⫽
p[dBC2 ⫺ 1dBC ⫺ 2tBC22]
4
ABC ⫽ 1.357 ⫻ 10⫺3 m2
P2 ⫽ 6 kN
CHECK:
;
P1 + P2
⫽ ABC
sAB
dBC ⫺
tBC ⫽
B
2
(dBC ⫺ 2tBC)
⫽ dBC
2
4 P1 + P2
a
⫺
b
p
sAB
dBC ⫽ 2tBC ⫽
;
P1 + P2
⫽ 9.947 * 106 Pa
ABC
PART (c)
P2 ⫽ 10 kN
P2 ⫽ sAB ABC ⫺ P1
B
d2BC ⫺
tBC ⫽ 12.62 mm
d2BC ⫺
2
4 P1 + P2
b
a
p
s AB
;
4 P1 + P2
b
a
p
sAB
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
30
CHAPTER 1
Tension, Compression, and Shear
Problem 1.3-2 A force P of 70 N is applied by
a rider to the front hand brake of a bicycle (P is
the resultant of an evenly distributed pressure). As
the hand brake pivots at A, a tension T develops in
the 460-mm long brake cable (Ae ⫽ 1.075 mm2)
which elongates by d ⫽ 0.214 mm. Find normal
stress s and strain ␧ in the brake cable.
Brake cable, L = 460 mm
Hand brake pivot A
37.5 mm A
T
P (Resultant
of distributed
pressure)
50
mm
100
mm
Uniform hand
brake pressure
Solution 1.3-2
Ae ⫽ 1.075 mm2
P ⫽ 70 N
L ⫽ 460 mm
d ⫽ 0.214 mm
Statics: sum moments about A to get T ⫽ 2P
s⫽
T
Ae
␧⫽
d
L
E⫽
s ⫽ 103.2 MPa
␧ ⫽ 4.65 * 10⫺4
;
;
s
⫽ 1.4 * 105 MPa
␧
NOTE: (E for cables is approximately 140 GPa.)
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.3
31
Normal Stress and Strain
Problem 1.3-3 A bicycle rider would like to compare the effectiveness of cantilever hand brakes [see figure part (a)] versus
V brakes [figure part (b)].
(a) Calculate the braking force RB at the wheel rims for each of the bicycle brake systems shown. Assume that all forces
act in the plane of the figure and that cable tension T ⫽ 200 N. Also, what is the average compressive normal stress
␴c on the brake pad (A ⫽ 4 cm2)?
(b) For each braking system, what is the stress in the brake cable
(assume effective cross-sectional area of 1.077 mm2)?
T
(HINT: Because of symmetry, you only need to use the right half
of each figure in your analysis.)
T
D
100 mm
TDC = TDE
C
D
45°
TDE
100 mm
TDC
TDE
C
TDCh
90°
125 mm
E
TDCv
106 mm
50 mm
B
F
RB
B
E
25 mm
25
mm
25 mm
A
G
F
Pivot points
anchored to frame
(a) Cantilever brakes
Pivot points
anchored to frame
A
(b) V brakes
Solution 1.3-3
T ⫽ 200 N
Apad ⫽ 4 cm2
Acable ⫽ 1.077 mm2
(a) CANTILEVER BRAKES—BRAKING FORCE
RB AND PAD PRESSURE
©MA ⫽ 0
RB(25) ⫽ TDCh(75) ⫹ TDCv(25)
TDCh ⫽ T / 2
TDCh ⫽ TDCv
4
RB ⫽ T
2
RB ⫽ 400 N
;
⬍ 2T here vs 4.25T for V brakes
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
32
CHAPTER 1
spad ⫽
Tension, Compression, and Shear
RB
Apad
spad ⫽ 1.0 MPa
;
V-BRAKES—BRAKING FORCE RB AND PAD PRESSURE
©MA ⫽ 0
spad ⫽
RB ⫽
RB
Apad
106
T
25
RB ⫽ 848 N
spad ⫽ 2.12 MPa
;
⬍ much more effective
;
(b) STRESS IN BRAKE CABLES
scable ⫽
T
Acable
scable ⫽ 185.7 MPa
← ⬍ same for both brake systems
Problem 1.3-4 A circular aluminum tube of length L ⫽ 420 mm is loaded in compression by forces P (see figure). The
hollow segment of length L/3 has outside and inside diameters of 60 mm and 35 mm, respectively. The solid segment of
length 2L/3 has diameter of 60 mm. A strain gage is placed on the outside of the hollow segment of the bar to measure
normal strains in the longitudinal direction.
(a) If the measured strain in the hollow segment is ␧h ⫽ 470 * 10⫺6, what is the strain ␧s in the solid part? (Hint: The
strain in the solid segment is equal to that in the hollow
L/3
segment multiplied by the ratio of the area of the hollow to
Strain gage
that of the solid segment).
P
P
(b) What is the overall shortening d of the bar?
(c) If the compressive stress in the bar cannot exceed 48 MPa,
L = 420 mm
what is the maximum permissible value of load P?
Solution 1.3-4
L ⫽ 420 mm
d2 ⫽ 60 mm
d1 ⫽ 35 mm
PART (a)
As ⫽
p 2
d ⫽ 2.827 * 10⫺3m2
4 2
␧h ⫽
Ah
␧ ⫽ 3.101 * 10⫺4
As h
Ah ⫽
␧h ⫽ 470 110⫺62
sa ⫽ 48 MPa
p
ad 2 ⫺ d1 2 b ⫽ 1.865 * 10⫺3 m2
4 2
PART (b)
d ⫽ ␧h
L
2L
+ ␧s a b ⫽ 0.1526 mm
3
3
␧h
L
⫽ 0.066 mm
3
PART (c)
Pmaxh ⫽ sa Ah ⫽ 89.535 kN
Pmaxs ⫽ sa As ⫽ 135.717 kN
␧s a
2L
b ⫽ 0.087 mm
3
6 lesser value controls
Pmax ⫽ Pmaxh ⫽ 89.5 kN
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.3
Normal Stress and Strain
Problem 1.3-5 The cross section of a concrete corner column that is
y
600 mm
loaded uniformly in compression is shown in the figure. A circular pipe
chase cut-out of 250 mm in diameter runs the height of the column
(see figure).
(a) Determine the average compressive stress sc in the concrete if the
load is equal to 14.5 MN.
(b) Determine the coordinates xc and yc of the point where the resultant
load must act in order to produce uniform normal stress in the column.
33
500 mm
Circular pipe
chase cutout
500 mm
400 mm
125
mm
200 mm
x
200 125
mm mm
Solution 1.3-5
P ⫽ 14.5 MN
p
1
A ⫽ (600 mm + 500 mm)(500 mm + 400 mm + 200 mm) ⫺ c (200 mm)2 d ⫺ (50 mm)2 ⫺ (250 mm)2
2
4
A ⫽ 11384.12615 cm2
(a) AVERAGE COMPRESSIVE STRESS
sc⫽
P
A
s c ⫽ 12.74 MPa
p
(250 mm)2 d(200 mm + 125 mm)
4
pipe ⫽
(b) CENTROID
A
(600 mm + 500 mm)
200 mm
1
2
⫺ C (500 mm)2 D (600 mm + 250 mm) ⫺ (200 mm)2 a
b
(600 mm + 500 mm)
2
2
3
⫺ pipe
xc ⫽
A
c
xc ⫽ 382.7376 mm
xc ⫽ 383 mm
(600 mm + 500 mm)2
yc ⫽
yc ⫽ 382.74 mm
(600 mm + 500 mm)
2
1
200 mm
⫺ [(500 mm)2](600 mm + 250 mm) ⫺ (200 mm)2 a
b
2
3
⫺ pipe
A
yc ⫽ 383 mm
Note that xc and yc are the same as expected due to symmetry about a diagonal.
Problem 1.3-6 A car weighing 130 kN when fully loaded is
pulled slowly up a steep inclined track by a steel cable (see
figure). The cable has an effective cross-sectional area of
490 mm2, and the angle ␣ of the incline is 30⬚.
Cable
(a) Calculate the tensile stress st in the cable.
(b) If the allowable stress in the cable is 150 MPa, what is the
maximum acceptable angle of the incline for a fully
loaded car?
a
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
34
CHAPTER 1
Tension, Compression, and Shear
Solution 1.3-6
W ⫽ 130 kN
a ⫽ 30⬚
A ⫽ 490 mm2
sa ⫽ 150 MPa
PART (a)
st ⫽
W sin(a)
⫽ 132.7 MPa
A
PART (b)
amax ⫽ arcsin a
sa A
b ⫽ 34.4⬚
W
Problem 1.3-7 Two steel wires support a
moveable overhead camera weighing
W ⫽ 125 N (see figure part a) used for
close-up viewing of field action at sporting
events. At some instant, wire 1 is at an angle
a ⫽ 22⬚ to the horizontal and wire 2 is at an
angle b ⫽ 40⬚. Wires 1 and 2 have diameters
of 0.75 mm and 0.90 mm, respectively.
(a) Determine the tensile stresses s1 and
s2 in the two wires.
(b) If the stresses in wires 1 and 2 must
be the same, what is the required
diameter of wire 1?
(c) Now, to stabilize the camera for windy
outdoor conditions, a third wire is
added (see figure part b). Assume the
three wires meet at a common point
(coordinates (0, 0, 0) above the camera
at the instant shown in figure part b).
Wire 1 is attached to a support at
coordinates (25 m, 16 m, 23 m). Wire
2 is supported at (⫺23 m, 18 m, 27 m).
Wire 3 is supported at (⫺3 m,
⫺28 m, 25 m). Assume that all three
wires have a diameter of 30 mils. Find
the tensile stresses in wires 1 to 3.
T2
T1
b
a
W
(a)
Plan view of camera suspension system
y
(−23 m, 18 m, 27 m)
(25 m, 16 m, 23 m)
Wire 2
Wire 1
Camera
x
Wire 3
(−3 m, −28 m, 25 m)
(b)
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.3
Normal Stress and Strain
35
Solution 1.3-7
NUMERICAL DATA
d1 ⫽ 0.75 mm
a ⫽ 22⬚
d2 ⫽ 0.90 mm
W ⫽ 125 N
A1 ⫽
p 2
d ⫽ 0.44179 mm2
4 1
A2 ⫽
p 2
d ⫽ 0.63617 mm2
4 2
b ⫽ 40⬚
(a) FIND NORMAL STRESS IN WIRES
T2 ⫽
W
⫽ 131.263 N
cos(b)
sin(a) + sin(b)
cos(a)
T1 ⫽ T2
s2 ⫽
cos(b)
⫽ 108.45 N
cos(a)
s1 ⫽
T2
⫽ 206 MPa
A2
T1
⫽ 245 MPa
A1
(b) FIND NEW d1 SO THAT NORMAL STRESSES IN WIRES ARE THE SAME
A1 new ⫽
T1
⫽ 0.526 mm2
s2
s1new ⫽
d1 new ⫽
4
A1 new ⫽ 0.818 mm
Ap
T1
⫽ 206.3 MPa
p
d1new2
4
d1new ⫽ 0.818 mm
p 2
d ⫽ 0.50265 mm2
4
25
Position vectors from camera r1 ⫽ £ 16 ≥
to each support:
23
(c) d ⫽ 0.8 mm
A⫽
L1 ⫽ ƒ r1 ƒ ⫽ 37.55
L2 ⫽ ƒ r2 ƒ ⫽ 39.774
Unit vectors along wires 1 to 3:
T1 ⫽ F1 e1
T2 ⫽ F2 e2
Equilibrium of forces:
T819 ⫽ e1
ORIGIN ⫽ 1
0
W ⫽ 125£ 0 ≥N
1
L3 ⫽ ƒ r3 ƒ ⫽ 37.656
0.666
e1 ⫽
⫽ £ 0.426 ≥
ƒ r1 ƒ
0.613
r1
T3 ⫽ F3 e3
⫺3
r3 ⫽ £ ⫺28 ≥
25
i⫽
1
0
P0Q
⫺0.578
e2 ⫽
⫽ £ 0.453 ≥
ƒ r2 ƒ
0.679
r2
j⫽
0
1
P0Q
k⫽
⫺0.08
e3 ⫽
⫽ £ ⫺0.744 ≥
ƒ r3 ƒ
0.664
r3
0
0
P1Q
T1 + T2 + T3 ⫽ W
T829 ⫽ e2
F ⫽ T ⫺1 W ⫽
⫺23
r2 ⫽ £ 18 ≥
27
60.55
59.91 N
P 71.161 Q
T839 ⫽ e3
s1 ⫽
T⫽
0.666
0.426
P 0.613
F1
⫽ 120.46 MPa
A
s1 ⫽ 120.5 MPa
⫺0.578
0.453
0.679
⫺0.08
⫺0.744
0.664 Q
s2 ⫽
F2
⫽ 119.19 MPa
A
s2 ⫽ 119.2 MPa
s3 ⫽
F3
⫽ 141.6 MPa
A
s3 ⫽ 141.6 MPa
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
36
CHAPTER 1
Tension, Compression, and Shear
Problem 1.3-8 A long retaining wall is braced by wood shores
Soil
set at an angle of 30° and supported by concrete thrust blocks,
as shown in the first part of the figure. The shores are evenly
spaced, 3 m apart.
For analysis purposes, the wall and shores are idealized
as shown in the second part of the figure. Note that the base
of the wall and both ends of the shores are assumed to be
pinned. The pressure of the soil against the wall is assumed
to be triangularly distributed, and the resultant force acting
on a 3-meter length of the wall is F ⫽ 190 kN.
If each shore has a 150 mm ⫻ 150 mm square cross
section, what is the compressive stress sc in the shores?
Retaining
wall
Concrete
Shore thrust
block
30⬚
B
F
30⬚
1.5 m
A
C
0.5 m
4.0 m
Solution 1.3-8 Retaining wall braced by wood shores
F ⫽ 190 kN
A ⫽ area of one shore
A ⫽ (150 mm)(150 mm)
⫽ 22,500 mm2
⫽ 0.0225 m2
FREE-BODY DIAGRAM OF WALL AND SHORE
SUMMATION OF MOMENTS ABOUT POINT A
⌺MA ⫽ 0 哵哴
⫺F(1.5 m) ⫹ CV (4.0 m) ⫹ CH (0.5 m) ⫽ 0
or
⫺(190 kN)(1.5 m) ⫹ C(sin 30°)(4.0 m)
⫹ C(cos 30°)(0.5 m) ⫽ 0
⬖
C ⫽ compressive force in wood shore
CH ⫽ horizontal component of C
CV ⫽ vertical component of C
CH ⫽ C cos 30°
CV ⫽ C sin 30°
C ⫽ 117.14 kN
COMPRESSIVE STRESS IN THE SHORES
sc ⫽
C
117.14 kN
⫽
A
0.0225 m2
⫽ 5.21 MPa
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.3
37
Normal Stress and Strain
Problem 1.3-9 A pickup truck tailgate supports a crate
h = 275 mm
(a) Find the tensile force T and normal stress ␴ in each
cable.
(b) If each cable elongates ␦ ⫽ 0.42 mm due to the
weight of both the crate and the tailgate, what is the
average strain in the cable?
WC = 900 N
F = 450 N
dc = 450 mm
Ca
ble
H = 300 mm
(WC ⫽ 900 N), as shown in the figure. The tailgate weighs
WT ⫽ 270 N and is supported by two cables (only one is
shown in the figure). Each cable has an effective crosssectional area Ae ⫽ 11 mm2.
Crate
Tail gate
Truck
dT = 350 mm
WT = 270 N
L = 400 mm
Solution 1.3-9
Fh ⫽ 450 N
T ⫽ 2Tv2 + Th2
h ⫽ 275 mm
Wc ⫽ 900 N
Ac ⫽ 11 mm2
WT ⫽ 270 N
d ⫽ 0.42 mm
T ⫽ 1.298 kN
;
;
(a) scable ⫽
T
Ac
scable ⫽ 118.0 MPa
(b) ␧cable ⫽
d
Lc
␧cable ⫽ 8.4 * 10⫺4
;
dc ⫽ 450 mm
dT ⫽ 350 mm
H ⫽ 300 mm
L ⫽ 400 mm
Le ⫽ 2L2 + H2
©Mhinge ⫽ 0
Lc ⫽ 0.5 m
2 Tv L ⫽ Wc d c + WT d T + Fh h
Tv ⫽
Wc dc + WT dT + Fhh
2L
Th ⫽
L
T
H v
Tv ⫽ 779.063 N
Th ⫽ 1.039 * 103 N
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
38
CHAPTER 1
Tension, Compression, and Shear
Problem 1.3-10 Solve the preceding problem if the
mass of the tail gate is MT ⫽ 27 kg and that of the
crate is Mc ⫽ 68 kg. Use dimensions H ⫽ 305 mm,
L ⫽ 406 mm, dc ⫽ 460 mm, and dT ⫽ 350 mm. The
cable cross-sectional area is Ae ⫽ 11.0 mm2.
(a) Find the tensile force T and normal stress s in
each cable.
(b) If each cable elongates d ⫽ 0.25 mm due to the
weight of both the crate and the tailgate, what
is the average strain in the cable?
Mc = 68 kg
dc = 460 mm
Ca
H = 305 mm
ble
Crate
Tail gate
Truck
dT = 350 mm
MT = 27 kg
L = 406 mm
Solution 1.3-10
(a) T ⫽ 2 T v2 + T 2h
Mc ⫽ 68 kg
g ⫽ 9.81 m/s2
MT ⫽ 27 kg
Wc ⫽ Mcg
scable ⫽
WT ⫽ MTg
Wc ⫽ 667.08
WT ⫽ 264.87
(b) ␧cable ⫽
T
Ae
d
Lc
T ⫽ 819 N
;
scable ⫽ 74.5 MPa
␧cable ⫽ 4.92 ⫻ 10⫺4
;
;
N ⫽ kg # m/s2
Ae ⫽ 11.0 mm2
d ⫽ 0.25 mm
dc ⫽ 460 mm
dT ⫽ 350 mm
H ⫽ 305 mm
L ⫽ 406 mm
L c ⫽ 2 L2 + H2
a Mhinge ⫽ 0
Tv ⫽
Lc ⫽ 507.8 mm
2TvL ⫽ Wc dc ⫹ WT d T
Wc dc + WT dT
2L
Th ⫽
L
T
H v
Tv ⫽ 492.071 N
Th ⫽ 655.019 N
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.3
Normal Stress and Strain
39
Problem 1.3-11 An L-shaped reinforced concrete slab
3.6 m * 3.6 m (but with a 1.8 m * 1.8 m cut-out) and thickness t ⫽ 230 mm is lifted by three cables attached at O, B
and D, as shown in the figure. The cables are combined at
point Q, which is 2.1 m above the top of the slab and directly
above the center of mass at C. Each cable has an effective
cross-sectional area of Ae ⫽ 77 mm2.
(a) Find the tensile force T1(i ⫽ 1, 2, 3) in each cable due
to the weight W of the concrete slab (ignore weight of
cables).
(b) Find the average stress si in each cable. (see Table H-1
in Appendix H for the weight density of reinforced
concrete.)
(c) Add cable AQ so that OQA is one continuous cable, with
each segment having force T1, which is connected to
cables BQ and DQ at point Q. Repeat parts (a) and (b).
(Hint: There are now three force equilibrium equations
and one constraint equation, T1 ⫽ T4.)
F
Coordinates of D in m
Q
(1.5, 1.5, 2.1)
T1
T3
1
D (1.5, 3.6, 0)
A 1
(0, 3.6, 0)
T2
7
5
5
z
C (1.5, 1.5, 0) 5 7
y
7
O
(0, 0, 0)
x
1.8 m
1.8 m
W
1.8 m
B (3.6, 0, 0)
kN
m3
Thickness t, c.g at (1.5 m, 1.5 m, 0)
Concrete slab g = 24
Solution 1.3-11
CABLE LENGTHS (M)
L1 ⫽ 21.52 + 1.52 + 2.12
L1 ⫽ 2.985
L3 ⫽ 2(1.5 ⫺ 3.6) + 2.1
2
(a)
2
2
eOQ ⫽
0.503
⫽ £ 0.503 ≥
ƒ rOQ ƒ
0.704
rOQ
W ⫽ 24
STATICS
rBQ ⫽ £
kN
m3
L2 ⫽ 3.327
L3 ⫽ 2.97
SOLUTION FOR CABLE FORCES USING STATICS
1.5
rOQ ⫽ £ 1.5 ≥
2.1
L2 ⫽ 2(1.52 ⫺ 3.6)2 + 1.52 + 2.12
1.5 ⫺ 3.6
1.5 ≥
2.1
eBQ ⫽
(3 EQUATIONS, 3 UNKNOWNS)
0
rDQ ⫽ £ 1.5 ⫺ 3.6 ≥
2.1
⫺0.631
⫽ £ 0.451 ≥
ƒ rBQ ƒ
0.631
rBQ
C (3.6 m)2 ⫺ (1.8 m)2 D (0.230 m) ⫽ 53654.4 N
eDQ ⫽
0
⫽ £ ⫺0.707 ≥
ƒ rDQ ƒ
0.707
rDQ
Ae ⫽ 77 mm2
0
0.50252 T1 + ⫺0.63117T2
0.50252 T1 + 0.45083 T2 + ⫺0.70711T3
©F ⫽ 0 T1 eOQ + T2 eBQ + T3 eDQ ⫺ 0 ⫽
PWQ P⫺53654.0 N + 0.70353 T + 0.63117 T + ⫺0.70711T Q
1
2
3
or in matrix form, solve simultaneous equations to get cable tension forces.
T1
eOQ1, 1
T2 ⫽ eOQ2, 1
PT Q Pe
3
OQ3, 1
eBQ1, 1
eBQ2, 1
eBQ3, 1
eDQ1, 1 ⫺1 0
25,951
eDQ2, 1
0 ⫽ 20,662 N
P 31,616 Q
eDQ3, 1 Q P W Q
T⫽
25,951
20,662 N
P 31,616 Q
(b) AVERAGE NORMAL STRESS IN EACH CABLE
i ⫽ 1 .. 3
si ⫽
Ti
Ae
s⫽
337
268 MPa
P 411 Q
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
40
CHAPTER 1
Tension, Compression, and Shear
(c) ADD CONTINUOUS CABLE OQA
rOQ
1.5
⫽ £ 1.5 ≥
2.1
eOQ ⫽
1.5
rAQ ⫽ £ 1.5 ⫺ 3.6 ≥
2.1
0.503
⫽ £ 0.503 ≥
ƒ rOQ ƒ
0.704
rOQ
eAQ ⫽
⫺2.1
rBQ ⫽ £ 1.5 ≥
2.1
0.451
⫽ £ ⫺0.631 ≥
ƒ rAQ ƒ
0.631
rAQ
0
rDQ ⫽ £ ⫺2.1 ≥
2.1
eBQ ⫽
0.451
eAQ ⫽
⫽ £ ⫺0.631 ≥
ƒ rAQ ƒ
0.631
rAQ
⫺0.631
⫽ £ 0.451 ≥
ƒ rBQ ƒ
0.631
rBQ
STATICS–Solve simultaneous equations to get cable tension forces.
T1
eOQ1, 1
T2
eOQ2, 1
± ≤ ⫽ ±
T3
eOQ3, 1
T4
1
eBQ1, 1
eBQ2, 1
eBQ3, 1
0
eDQ1, 1
eDQ2, 1
eDQ3, 1
0
Normal stresses in cables:
i ⫽ 1 .. 4
Ti
si ⫽
Ae
eAQ1, 1 ⫺1 0
18,890
18,890
eAQ2, 1
0
28,532
28,532
≤
⫽
N
T⫽
N
W
14,755
eAQ3, 1
P Q P
Q
P 14,755 Q
0
18,890
18,890
⫺1
<for case of T1 ⴝ T4
245
371
≤ MPa
s⫽ ±
192
245
Problem 1.3-12
A round bar ACB of length 2L (see figure) rotates about an
axis through the midpoint C with constant angular speed v (radians per
second). The material of the bar has weight density g.
v
A
C
(a) Derive a formula for the tensile stress sx in the bar as a function of the
distance x from the midpoint C.
(b) What is the maximum tensile stress smax?
Solution 1.3-12
L
B
x
L
Rotating Bar
Consider an element of mass dM at distance j from the
midpoint C. The variable j ranges from x to L.
g
dM ⫽ g A dj
dF ⫽ Inertia force (centrifugal force) of element of mass dM
v ⫽ angular speed (rad/s)
A ⫽ cross-sectional area
g ⫽ weight density
g
⫽ mass density
g
We wish to find the axial force Fx in the bar at
Section D, distance x from the midpoint C.
The force Fx equals the inertia force of the part
of the rotating bar from D to B.
g
dF ⫽ (dM)(j␻2) ⫽ g
A␻2jdj
gA␻2 2
(L ⫺ x2)
2g
LD
Lx g
(a) TENSILE STRESS IN BAR AT DISTANCE x
B
Fx ⫽
sx ⫽
dF ⫽
Lg
A␻2jdj ⫽
g␻2 2
Fx
⫽
(L ⫺ x2)
A
2g
;
(b) MAXIMUM TENSILE STRESS
x ⫽ 0 smax ⫽
g␻2L2
2g
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.3
Problem 1.3-13 Two gondolas on a ski lift are locked in the
position shown in the figure while repairs are being made elsewhere. The distance between support towers is L ⫽ 30.5 m.
The length of each cable segment under gondola weights
WB ⫽ 2000 N and WC ⫽ 2900 N are DAB ⫽ 3.7 m,
DBC ⫽ 21.4 m, and DCD ⫽ 6.1 m. The cable sag at B is
⌬B ⫽ 1.3 m and that at C(⌬C) is 2.3 m. The effective
cross-sectional area of the cable is Ae ⫽ 77 mm2.
41
Normal Stress and Strain
A
D
u1
DB
u2
B
DC
u3
C
WB
(a) Find the tension force in each cable segment; neglect
the mass of the cable.
(b) Find the average stress (␴) in each cable segment.
WC
Support
tower
L = 30.5 m
Solution 1.3-13
DATA
CONSTRAINT EQUATIONS
WB ⫽ 2000 N
DAB cos(u1) + DBC cos(u2) + DCD cos(u3) ⫽ L
Wc ⫽ 2900 N
DAB sin(u1) + DBC sin(u2) ⫽ DCD sin(u3)
¢ B ⫽ 1.3 m
(a) SOLVE SIMULTANEOUS EQUATIONS NUMERICALLY FOR
TENSION FORCE (NEWTONS) IN EACH CABLE SEGMENT
¢ C ⫽ 2.3 m
L ⫽ 30.5 m
TAB ⫽ 6757 N TBC ⫽ 6358 N TCD ⫽ 6859 N
DAB ⫽ 3.7
CHECK EQUILIBRIUM AT B AND C
DBC ⫽ 21.4
TAB sin(u1) ⫺ TBC sin(u2) ⫽ 2000
DCD ⫽ 6.1
TBC sin(u2) + TCD sin(u3) ⫽ 2900
(b) COMPUTE STRESSES IN CABLE SEGMENTS
DAB ⫹ DBC ⫹ DCD ⫽ 31.2
Ae ⫽
77
3 2
(10 )
m2
sAB ⫽
COMPUTE INITIAL VALUES OF THETA ANGLES (RADIANS)
u1 ⫽ arcsin a
u2 ⫽ arcsin a
u3 ⫽ arcsin a
GIVEN
¢B
b
DAB
¢ C⫺¢ B
b
DBC
¢C
b
DCD
u1 ⫽ 0.359
TAB
Ae
sBC ⫽
TBC
Ae
sAB ⫽ 87.8 MPa
sBC ⫽ 82.6 MPa
sCD ⫽ 89.1 MPa
;
sCD ⫽
TCD
Ae
u2 ⫽ 0.047
u3 ⫽ 0.387
⫺TAB cos(u1) + TBC cos(u2) ⫽ 0
TAB sin(u1) ⫺ TBC sin(u2) ⫽ WB
⫺TBC cos(u2) + TCD cos(u3) ⫽ 0
TBC sin(u2) + TCD sin(u3) ⫽ WC
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
42
CHAPTER 1
Tension, Compression, and Shear
z
Problem 1.3-14 A crane boom of mass 450 kg with its
D
center of mass at C is stabilized by two cables AQ and BQ
(Ae ⫽ 304 mm2 for each cable) as shown in the figure. A
load P ⫽ 20 kN is supported at point D. The crane boom lies
in the y–z plane.
y
bo
om
C
1
B
2
an
e
2
2m
Cr
(a) Find the tension forces in each cable: TAQ and
TBQ (kN). Neglect the mass of the cables, but include
the mass of the boom in addition to load P.
(b) Find the average stress (s) in each cable.
P
Q
2m
55°
TBQ
O
5m
TAQ
5m
x
5m
A
3m
Solution 1.3-14
Data
Mboom ⫽ 450 kg
g ⫽ 9.81 m/s2
2TAQz (3000) ⫽ Wboom(5000) + P(9000)
TAQz ⫽
Wboom(5000) + P(9000)
2(3000)
P ⫽ 20 kN
TAQ ⫽
222 + 22 + 12
TAQz
2
Ae ⫽ 304 mm2
TAQ ⫽ 50.5 kN ⫽ TBQ
Wboom ⫽ Mboom g
Wboom ⫽ 4415 N
(a) Symmetry: TAQ ⫽ TBQ
a Mx ⫽ 0
(b) s ⫽
TAQ
Ae
;
s ⫽ 166.2 MPa
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.4
Mechanical Properties and Stress-Strain Diagrams
43
Mechanical Properties and Stress-Strain Diagrams
Problem 1.4-1 Imagine that a long steel wire hangs vertically from a high-altitude balloon.
(a) What is the greatest length (meters) it can have without yielding if the steel yields at 260 MPa?
(b) If the same wire hangs from a ship at sea, what is the greatest length? (Obtain the weight densities of steel and sea
water from Table H-1, Appendix H.)
Solution 1.4-1
W ⫽ total weight of steel wire
gS ⫽ weight density of steel
gS ⫽ 77 kN/m3
gw ⫽ weight density of sea water
gw ⫽ 10 kN/m3
A ⫽ cross-sectional area of wire
smax ⫽ 260 MPa
(b) WIRE HANGING IN SEA WATER
F ⫽ tensile force at top of wire
F ⫽ A gS ⫺ gW B AL sy ⫽
Lmax ⫽
sy
gS ⫺ gW
⫽ 3881 m
(a) WIRE HANGING IN AIR
F
⫽ A gS ⫺ gW B L
A
⬍hanging from ship at sea
W ⫽ gS AL
sy ⫽
Lmax
W
⫽ gS L
A
sy
⫽ 3377 m
⫽
gS
⬍hanging from balloon
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44
CHAPTER 1
Tension, Compression, and Shear
Problem 1.4-2 Steel riser pipe hangs from a drill rig located offshore
in deep water (see figure).
(a) What is the greatest length (meters) it can have without
breaking if the pipe is suspended in air and the ultimate strength
(or breaking strength) is 550 MPa?
(b) If the same riser pipe hangs from a drill rig at sea, what is the
greatest length? (Obtain the weight densities of steel and sea
water from Appendix H, Table H-1. Neglect the effect of buoyant
foam casings on the pipe.)
2
Riser
3
BOP
Drill pipe
Solution 1.4-2
(a) PIPE SUSPENDED IN AIR
sU ⫽ 550 MPa
gs ⫽ 77 kN/m3
W ⫽ gs AL
Lmax ⫽
sU
⫽ 7143 m
gs
(b) PIPE SUSPENDED IN SEA WATER
gw ⫽ 10 kN/m3
Force at top of pipe:
Stress at top of pipe:
smax ⫽
F
A
F ⫽ A gs ⫺ gw B AL
smax ⫽ A gs ⫺ gw B L
Set max stress equal to ultimate and then solve for
Lmax:
Lmax ⫽
sU
A gs ⫺ gw B
⫽ 8209 m
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SECTION 1.4
45
Mechanical Properties and Stress-Strain Diagrams
Problem 1.4-3 Three different materials, designated A, B, and C,
are tested in tension using test specimens having diameters of
12 mm and gage lengths of 50 mm (see figure). At failure, the
distances between the gage marks are found to be 54.5 mm,
63.2 mm, and 69.4 mm, respectively. Also, at the failure cross
sections the diameters are found to be 11.46, 9.48, and 6.06 mm,
respectively.
Determine the percent elongation and percent reduction in
area of each specimen, and then, using your own judgment,
classify each material as brittle or ductile.
Gage
length
P
P
Solution 1.4-3 Tensile tests of three materials
where Lf is in millimeters.
⫽
Percent reduction in area
Percent elongation ⫽
Lf ⫺ L0
L0
L0 ⫽ 50 mm
Percent elongation ⫽ a
Af
df
Lf
50
(100) ⫽ a
Lf
L0
⫺ 1b(100)
A0
⫽ a1 ⫺
⫺ 1b 100
d0 ⫽ initial diameter
A0 ⫺ Af
Af
A0
(100)
b (100)
df ⫽ final diameter
(Eq. 1)
2
% Elongation
(Eq. 1)
% Reduction
(Eq. 2)
Brittle or
Ductile?
Material
Lf
(mm)
df
(mm)
Percent reduction in area
A
54.5
11.46
9.0%
8.8%
Brittle
df
B
63.2
9.48
26.4%
37.6%
Ductile
C
69.4
6.06
38.8%
74.5%
Ductile
A0
⫽a
d0
b
⫽ c1 ⫺ a
d0 ⫽ 12 mm
12
2
b d(100)
(Eq. 2)
where df is in millimeters.
Problem 1.4-4 The strength-to-weight ratio of a structural material is defined as its load-carrying capacity divided by its
weight. For materials in tension, we may use a characteristic tensile stress (as obtained from a stress-strain curve) as a measure of
strength. For instance, either the yield stress or the ultimate stress could be used, depending upon the particular application. Thus,
the strength-to-weight ratio RS/W for a material in tension is defined as
RS/W ⫽
s
g
in which s is the characteristic stress and g is the weight density. Note that the ratio has units of length.
Using the ultimate stress sU as the strength parameter, calculate the strength-to-weight ratio (in units of meters) for
each of the following materials: aluminum alloy 6061-T6, Douglas fir (in bending), nylon, structural steel ASTM-A572, and
a titanium alloy. (Obtain the material properties from Appendix H, Tables H-1 and H-3. When a range of values is given in a
table, use the average value.)
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46
CHAPTER 1
Tension, Compression, and Shear
Solution 1.4-4 Strength-to-weight ratio
The ultimate stress sU for each material is obtained
from Appendix H, Tables H-3, and the weight density g is obtained from Table H-1.
sU ( MPa)
g( kN/m3)
g
(kN/m3)
RS/W
(m)
310
26.0
11.9 ⫻ 103
Douglas fir
65
5.1
12.7 ⫻ 103
Nylon
60
9.8
6.1 ⫻ 103
Structural steel
ASTM-A572
500
77.0
6.5 ⫻ 103
Titanium alloy
1050
44.0
23.9 ⫻ 103
Aluminum alloy
6061-T6
The strength-to-weight ratio (meters) is
RS/W ⫽
sU
(MPa)
(103)
Values of sU, g, and RS/W are listed in the table.
Titanium has a high strength-to-weight ratio, which is why
it is used in space vehicles and high-performance airplanes.
Aluminum is higher than steel, which makes it desirable for
commercial aircraft. Some woods are also higher than steel,
and nylon is about the same as steel.
Problem 1.4-5 A symmetrical framework consisting of three pinconnected bars is loaded by a force P (see figure). The angle between
the inclined bars and the horizontal is ␣ ⫽ 52°. The axial strain in the
middle bar is measured as 0.027.
Determine the tensile stress in the outer bars if they are constructed
of a copper alloy having the following stress-strain relationship:
s⫽
124,020␧
1 + 300␧
A
B
C
a
0 … ␧ … 0.03 1s ⫽ MPa2
D
P
Solution 1.4-5
B
DATA
␧BD ⫽ 0.027
a ⫽ 52 ⬚
LBD ⫽ 1
Strain in CE:
L2 ⫽
LBD
sin1a2
a
6 assume unit length to facilitate
numerical calculations below
L
L3 ⫺ L2
␧CE ⫽
L2
C
L2
L3
LBC ⫽
LBD
tan1a2
D
e BDL
E
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SECTION 1.4
Mechanical Properties and Stress-Strain Diagrams
47
Increased length of CE (see figure):
L3 ⫽ 3LBC 2 + 1LBD + ␧BD LBD22 ⫽
1
B tan152 ⬚22
+ 1.054729
Compute strain in CE then substitute strain value into stress-strain relationship to find tensile stress in outer bars:
␧CE ⫽
L3 ⫺ L2
⫽ 0.017
L2
s⫽
124,020 ␧CE
1 + 300 ␧CE
s ⫽ 345 MPa
Problem 1.4-6 A specimen of a methacrylate plastic is tested in tension at room temperature (see figure), producing the stress-strain data
listed in the accompanying table.
Plot the stress-strain curve and determine the proportional limit,
modulus of elasticity (i.e., the slope of the initial part of the stress-strain
curve), and yield stress at 0.2% offset. Is the material ductile or brittle?
STRESS-STRAIN DATA FOR PROBLEM 1.4-6
P
P
Stress (MPa)
Strain
8.0
17.5
25.6
31.1
39.8
44.0
48.2
53.9
58.1
62.0
62.1
0.0032
0.0073
0.0111
0.0129
0.0163
0.0184
0.0209
0.0260
0.0331
0.0429
Fracture
Solution 1.4-6 Tensile test of a plastic
Using the stress-strain data given in the problem
statement, plot the stress-strain curve:
60
sPL ⫽ proportional limit
sPL ⬇ 47 MPa
Modulus of elasticity (slope)
⬇ 2.4 GPa
;
;
sY ⫽ yield stress at 0.2% offset
Stress
(MPa)
sY ⬇ 53 MPa
sY
sPL
;
Material is brittle, because the strain after the proportional
limit is exceeded is relatively small.
;
40
Slope ≈
40 MPa
= 2.4 GPa
0.017
20
0.2% offset
0
0.01
0.02
0.03
Strain
0.04
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48
CHAPTER 1
Tension, Compression, and Shear
Problem 1.4-7 The data shown in the table below were obtained from a tensile
test of high-strength steel. The test specimen had a diameter of 13 mm and a
gage length of 50 mm (see figure for Prob. 1.4-3). At fracture, the elongation
between the gage marks was 3.0 mm and the minimum diameter was 10.7 mm.
Plot the conventional stress-strain curve for the steel and determine the
proportional limit, modulus of elasticity (i.e., the slope of the initial part of the
stress-strain curve), yield stress at 0.1% offset, ultimate stress, percent
elongation in 50 mm, and percent reduction in area.
TENSILE-TEST DATA FOR PROBLEM 1.4-7
Load (kN)
Elongation (mm)
5
10
30
50
60
64.5
67.0
68.0
69.0
70.0
72.0
76.0
84.0
92.0
100.0
112.0
113.0
0.005
0.015
0.048
0.084
0.099
0.109
0.119
0.137
0.160
0.229
0.259
0.330
0.584
0.853
1.288
2.814
Fracture
Solution 1.4-7 Tensile test of high-strength steel
d ⫽ 13 mm
A0 ⫽
(MPa)
L 0 ⫽ 50 mm
1130
pd 2
⫽ 132.7 mm2
4
753.6
NORMAL STRESS AND STRAIN
s⫽
P
A0
␧⫽
376.8
d
L0
Proportional limit L 486 MPa
0
;
Modulus of elasticity (slope) L 224 GPa
Yield stress at 0.1% offset L 520 MPa
0.002 0.004 0.006
;
YP
PL
527.5
Ultimate stress (maximum stress)
L 852 MPa
(MPa)
;
;
slope = 376.8 MPa/0.00168
= 224 GPa
452.1
0.2% offset
Percent elongation
⫽ ec
Lf ⫺ L0
⫽ 6%
L0
;
d f(100) ⫽ c
3
d(100)
50
376.8
0
0.002 0.004
Percent reduction in area
⫽ ec
⫽ ec
A0 ⫺ A1
d f(100)
A0
133 mm2 ⫺ 1p4 2(10.7 mm)2
⫽ 32%
133 mm2
;
d f(100)
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SECTION 1.5
49
Elasticity, Plasticity, and Creep
Elasticity, Plasticity, and Creep
s (MPa)
420
Problem 1.5-1 A bar made of structural steel having the stressstrain diagram shown in the figure has a length of 1.5 m. The yield
stress of the steel is 290 MPa and the slope of the initial linear part
of the stress-strain curve (modulus of elasticity) is 207 GPa. The
bar is loaded axially until it elongates 7.6 mm, and then the load is
removed.
How does the final length of the bar compare with its original
length? (Hint: Use the concepts illustrated in Fig. 1-18b.)
280
140
0
0
0.002
0.004
0.006
ε
Solution 1.5-1 Steel bar in tension
ELASTIC RECOVERY ␧E ⫽ ␧I ⫺ ␧R
From s⫺␧ diagram,
␧E ⫽
sYP
290 MPa
⫽
⫽ 0.00140
Slope
207 GPa
RESIDUAL STRAIN
εE
εR
ε
ε1
␧R ⫽ ␧1 ⫺ ␧E ⫽ 0.0057 ⫺ 0.00140
⫽ 0.00367
L ⫽ 1.5 m ⫽ 1500 mm
PERMANENT SET
Yield stress sY ⫽ 290 MPa
␧RL ⫽ (0.00367)(1500 mm)
Slope ⫽ 207 GPa
d ⫽ 7.6 mm
␧1 ⫽
7.6 mm
d
⫽
⫽ 0.00507
L
1500 mm
Problem 1.5-2 A bar of length 2.0 m is made of a structural
steel having the stress-strain diagram shown in the figure. The
yield stress of the steel is 250 MPa and the slope of the initial
linear part of the stress-strain curve (modulus of elasticity) is
200 GPa. The bar is loaded axially until it elongates 6.5 mm, and
then the load is removed.
How does the final length of the bar compare with its original length? (Hint: Use the concepts illustrated in Fig. 1-36b.)
⫽ 5.5 mm
Final length of bar is 5.5 mm greater than the original
length.
;
s (MPa)
300
200
100
0
0
0.002
0.004
0.006
ε
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50
CHAPTER 1
Tension, Compression, and Shear
Solution 1.5-2 Steel bar in tension
L ⫽ 2.0 m ⫽ 2000 mm
s
Yield stress sY ⫽ 250 MPa
B
A
sY
Slope ⫽ 200 GPa
d ⫽ 6.5 mm
ELASTIC RECOVERY ␧E
␧E ⫽
sB
250 MPa
⫽
⫽ 0.00125
Slope
200 GPa
RESIDUAL STRAIN ␧R
␧R ⫽ ␧B ⫺ ␧E ⫽ 0.00325 ⫺ 0.00125
εE
⫽ 0.00200
0
εR
εB
Permanent set ⫽ ␧RL ⫽ (0.00200)(2000 mm)
ε
⫽ 4.0 mm
STRESS AND STRAIN AT POINT B
Final length of bar is 4.0 mm greater than its original
length.
;
sB ⫽ sY ⫽ 250 MPa
␧B ⫽
d
6.5 mm
⫽
⫽ 0.00325
L
2000 mm
Problem 1.5-3 An aluminum bar has length L ⫽ 1.8 m and diameter d ⫽ 34 mm. The stress-strain curve for the aluminum
is shown in Fig. 1-31 of Section 1.4. The initial straight-line part of the curve has a slope (modulus of elasticity) of 73 GPa.
The bar is loaded by tensile forces P ⫽ 200 kN and then unloaded.
(a) What is the permanent set of the bar?
(b) If the bar is reloaded, what is the proportional limit? (Hint: Use the concepts illustrated in Figs. 1-36b and 1-37.)
Solution 1.5-3
DATA
L ⫽ 1.8 m
d ⫽ 34 mm
s (MPa)
280
E ⫽ 73 GPa
P ⫽ 200 kN
210
NORMAL STRESS IN BAR
140
P
sB ⫽
⫽ 220 MPa
p 2
d
4
from curve, say that ␧B ⫽ 0.03
70
0
ELASTIC RECOVERY: unloading parallel to initial straight
line
␧E ⫽
sB
⫽ 3.018 * 10⫺3
E
0
0.05 0.10 0.15 0.20 0.25
e
Typical stress-strain diagram for an
aluminum alloy
RESIDUAL STRAIN
␧R ⫽ ␧B ⫺ ␧E ⫽ 0.027
(a) PERMANENT SET
dpset ⫽ ␧RL ⫽ 48.568 mm
dpset ⫽ 48.6 mm
(b) PROPORTIONAL LIMIT WHEN RELOADED IS sB ⫽ 220 MPa
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SECTION 1.5
51
Elasticity, Plasticity, and Creep
Problem 1.5-4 A circular bar of magnesium alloy is 750 mm long. The stress-strain diagram for the material is shown in
the figure. The bar is loaded in tension to an elongation of 6.0 mm, and then the load is removed.
(a) What is the permanent set of the bar?
(b) If the bar is reloaded, what is the proportional limit? (Hint: Use the concepts illustrated in Figs. 1-36b and 1-37.)
Solution 1.5-4
70
L ⫽ 750 mm
NUMERICAL DATA
63
d ⫽ 6 mm
56
a␧
d
Et 1␧2 ⫽
d␧ 1 + b␧
Et 102 ⫽ 41,000 MPa
49
sSI (e) 42
35
(or 41 GPa 7 magnesium alloy)
28
d
␧B ⫽ ⫽ 8 * 10⫺3
L
␧E ⫽ 0.0023
21
sB 65.6 MPa 6 from curve
(see figure)
14
7
6 elastic recovery (see figure)
␧R ⫽ ␧B ⫺ ␧E ⫽ 5.7 * 10
⫺3
0
0
6 residual strain
e
(a) PERMANENT SET
dpset ⫽ ␧R L ⫽ 4.275
0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01
70
63
dpset ⫽ 4.28 mm
56
(b) PROPORTIONAL LIMIT WHEN RELOADED
49
sB ⫽ 65.6 MPa
sSI (e)
42
35
28
21
14
7
0
0
0.002
0.004
0.006
0.008
0.01
e
Residual strain
= 0.0057
Elastic recovery
= 0.008 – 0.0057 = 0.0023
Problem 1.5-5 A wire of length L ⫽ 2.5 m and diameter d ⫽ 1.6 mm is stretched by tensile forces P ⫽ 600 N. The wire is
made of a copper alloy having a stress-strain relationship that may be described mathematically by the following equation:
124,020␧
1 + 300␧
in which ␧ is nondimensional and ␴ has units of MPa.
s⫽
(a)
(b)
(c)
(d)
0 … ␧ … 0.03 1s ⫽ MPa2
Construct a stress-strain diagram for the material.
Determine the elongation of the wire due to the forces P.
If the forces are removed, what is the permanent set of the bar?
If the forces are applied again, what is the proportional limit?
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52
CHAPTER 1
Tension, Compression, and Shear
Solution 1.5-5 Wire stretched by forces P
L ⫽ 2.5 m ⫽ 2500 mm
d ⫽ 1.6 mm
␧⫽
P ⫽ 660 N
s
124,020 ⫺ 300s
0 … s … 372 MPa (s ⫽ MPa)
(Eq. 2)
COPPER ALLOY
124,020␧
s⫽
1 + 300␧
0 … ␧ … 0.03 (s ⫽ MPa) (Eq. 1)
This equation may also be used when plotting the stressstrain diagram.
(b) ELONGATION d OF THE WIRE
(a) STRESS-STRAIN DIAGRAM
INITIAL SLOPE OF STRESS-STRAIN CURVE
600 N
P
⫽
⫽ 328 MPa
p
A
(1.6 mm)2
4
From Eq. (2) or from the stress-strain diagram:
s⫽
300
␧ ⫽ 0.0128
σ
(MPa)
d ⫽ ␧L ⫽ (0.0128)(2500 mm) ⫽ 32 mm
;
STRESS AND STRAIN AT POINT B
200
sB ⫽ 328 MPa
␧B ⫽ 0.0128
(Diagram not to scale)
σ
100
B
σB
0
0.001
0.003
0.005
ε
0.007
0.009
εE
Take the derivative of s with respect to ␧:
(1 + 300␧)(124,020) ⫺ (124,020␧)(300)
ds
⫽
d␧
(1 + 300␧)2
⫽
At ␧ ⫽ 0,
⬖
124,020
(1 + 300␧)2
ds
⫽ 124 GPa
d␧
Initial slope ⫽ 124 GPa
ALTERNATIVE FORM OF THE STRESS-STRAIN
RELATIONSHIP
Solve Eq. (1) for ␧ in terms of s:
0 εR
ε
εB
ELASTIC RECOVERY ␧E
␧E ⫽
sB
328 MPa
⫽
⫽ 0.00265
Slope
124 GPa
RESIDUAL STRAIN ␧R
␧R ⫽ ␧B ⫺ ␧E ⫽ 0.0128 ⫺ 0.0027 ⫽ 0.0120
(c) PERMANENT SET ⫽ ␧RL ⫽ (0.0120)(2500 mm)
⫽ 30 mm
;
(d) PROPORTIONAL LIMIT WHEN RELOADED ⫽ sB
sB ⫽ 328 MPa
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SECTION 1.6
53
Linear Elasticity, Hooke’s Law, and Poisson’s Ratio
Linear Elasticity, Hooke’s Law, and Poisson’s Ratio
When solving the problems for Section 1.6, assume that the material behaves
linearly elastically.
d
P
Problem 1.6-1 A high-strength steel bar used in a large crane has diameter
P
d ⫽ 50 mm (see figure). The steel has modulus of elasticity E ⫽ 200 GPa and
Poisson’s ratio ␯ ⫽ 0.3. Because of clearance requirements, the diameter of
the bar is limited to 50.025 mm when it is compressed by axial forces.
What is the largest compressive load Pmax that is permitted?
Solution 1.6-1 Steel bar in compression
STEEL BAR
E ⫽ 200 GPa
AXIAL STRESS
max. ⌬d ⫽ 0.025 mm
d ⫽ 50 mm
v ⫽ 0.3
¢d
d
E¢d
vd
(compressive stress)
MAXIMUM PERMISSIBLE COMPRESSIVE LOAD
LATERAL STRAIN
␧¿ ⫽
s ⫽ Ee ⫽
(increase in diameter)
P ⫽ sA ⫽
EA¢d
nd
SUBSTITUTE NUMERICAL VALUES:
AXIAL STRAIN
¢d
␧¿
␧ ⫽⫺ ⫽ ⫺
v
vd
(decrease in length)
Assume Hooke’s law is valid for the material.
P⫽
(200 GPa)(p/4)(50 mm)(0.025 mm)
(0.3)(50 mm)
⫽ 654 kN
;
Problem 1.6-2 A round bar of 10 mm diameter is made of aluminum
alloy 7075-T6 (see figure). When the bar is stretched by axial forces P,
its diameter decreases by 0.016 mm.
Find the magnitude of the load P. (Obtain the material properties
from Appendix H.)
Solution 1.6-2
d ⫽ 10 mm
d = 10 mm
P
7075-T6
Aluminum bar in tension
⌬d ⫽ 0.016 mm
(Decrease in diameter)
7075-T6
AXIAL STRESS
s ⫽ E␧ ⫽ (72 GPa)(0.004848)
⫽ 349.1 MPa (Tension)
From Table H-2: E ⫽ 72 GPa v ⫽ 0.33
Because s ⬍ sY, Hooke’s law is valid.
From Table H-3: Yield stress sY ⫽ 480 MPa
LOAD P (TENSILE FORCE)
LATERAL STRAIN
p
P ⫽ sA ⫽ (349.1 MPa) a b(10 mm)2
4
⫽ 27.4 kN
;
␧¿ ⫽
P
⫺0.016 mm
¢d
⫽
⫽ ⫺0.0016
d
10 mm
AXIAL STRAIN
␧¿
0.0016
⫽
v
0.33
⫽ 0.004848 (Elongation)
␧⫽⫺
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54
CHAPTER 1
Tension, Compression, and Shear
Steel
tube
Problem 1.6-3 A polyethylene bar having diameter d1 ⫽ 70 mm is placed inside
a steel tube having inner diameter d2 ⫽ 70.2 mm (see figure). The polyethylene bar
is then compressed by an axial force P.
At what value of the force P will the space between the polyethylene bar and
the steel tube be closed? (For polyethylene, assume E ⫽ 1.4 GPa and ␯ ⫽ 0.4.)
d1 d2
Polyethylene
bar
Solution 1.6-3
NORMAL STRAIN
NUMERICAL DATA
d1 ⫽ 70 mm
d2 ⫽ 70.2 mm
E ⫽ 1.4 GPa
␯ ⫽ 0.4
¢d1 ⫽ d2 ⫺ d1
p 2
p 2
A1 ⫽
d
d
A2 ⫽
4 1
4 2
⫺3 2
A1 ⫽ 3.848 * 10 m
s1 ⫽ E␧1
␧¿ ⫽
␧1 ⫽ ⫺ 6.25 * 10⫺3
s1 ⫽ ⫺ 8.75 * 106 Pa
COMPRESSION FORCE
P ⫽ EA1␧1
P ⫽ ⫺33.7 kN
LATERAL STRAIN
¢d1
d1
⫺␧¿
␯
AXIAL STRESS
A2 ⫽ 3.87 * 10⫺3 m2
␧¿ ⫽
␧1 ⫽
0.01
4
;
␧œ ⫽ 2.5 * 10⫺3
Strain gage
Problem 1.6-4 A circular aluminum tube of length L ⫽ 600 mm
is loaded in compression by forces P (see figure). The outside and
inside diameters are d2 ⫽ 75 mm and d1= 63 mm, respectively. A
strain gage is placed on the outside of the tube to measure normal
strains in the longitudinal direction. Assume that E ⫽ 73 GPa and
Poisson’s ration v ⫽ 0.33.
(a) If the compressive stress in the tube is 57 MPa, what is the
load P?
(b) If the measured strain is ␧ ⫽ 781 ⫻ 10⫺6, what is the shortening
d of the tube? What is the percent change in its cross-sectional
area? What is the volume change of the tube?
(c) If the tube has a constant outer diameter of d2 ⫽ 75 mm
along is entire length L but now has increased inner
diameter d3 with a normal stress of 70 MPa over the middle
third, while the rest of the tube remains at normal stress of
57 MPa, what is the diameter d3?
P
d2
d1
P
L
(a)
P
d3
L/3
d1
d2
L/3
P
L/3
(b)
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.6
55
Linear Elasticity, Hooke’s Law, and Poisson’s Ratio
Solution 1.6-4
(a) GIVEN STRESS, FIND FORCE P IN BAR FIGURE (A)
(b) GIVEN STRAIN, FIND CHANGE IN LENGTH IN BAR FIGURE (A) AND ALSO
VOLUME CHANGE
s ⫽ 57 MPa
E ⫽ 73 GPa
v ⫽ 0.33
L ⫽ 600 mm
d2 ⫽ 75 mm
d1 ⫽ 63 mm
p
A ⫽ A d22 ⫺ d12 B ⫽ 1301 mm2
4
P ⫽ s A ⫽ 74.1 kN
d2 ⫺ d1
⫽ 6 mm
2
d ⫽ ␧L ⫽ ⫺0.469 mm shortening
Vol1 ⫽ L (A) ⫽ 7.804 * 105 mm3
␧ ⫽ ⫺781 A 10⫺6 B
t⫽
␧lat ⫽ ⫺v␧ ⫽ 2.577 * 10⫺4
¢d2 ⫽ ␧lat d2 ⫽ 0.019 mm
¢t ⫽ ␧latt ⫽ 1.546 * 10⫺3 mm
¢d1 ⫽ ␧lat d1 ⫽ 0.016 mm
p
c A d2 + ¢d2 B 2 ⫺ A d1 + ¢d1 B 2 d
4
Af ⫽
Af ⫽ 1301.29 mm2
Af ⫺ A
A
⫽
1301.29 ⫺ 1300.62
⫽ 0.052%
1300.62
V1f ⫽ (L + d) A Af B ⫽ 7.802 * 105 mm3
¢V1 ⫽ V1f ⫺ Vol 1 ⫽ ⫺207.482 mm3
¢V1 ⫽ ⫺207 mm3 change
d2 ⫽ 75 mm ALONG ITS ENTIRE LENGTH L BUT NOW HAS INCREASED INNER
d3 OVER THE MIDDLE THIRD WITH NORMAL STRESS OF 70 MPa, WHILE THE REST OF THE BAR REMAINS AT NORMAL
STRESS OF 57 MPa, WHAT IS THE DIAMETER d3?
(c) IF
THE TUBE HAS CONSTANT OUTER DIAMETER OF
DIAMETER
sM3 ⫽ 70 MPa P ⫽ 74.135 kN AM3 ⫽
d22 ⫺ d32 ⫽
4
A
p M3
SO
d3 ⫽
A
P
⫽ 1059.076 mm2
sM3
d22 ⫺
4
A ⫽ 65.4 mm
p M3
d2 ⫽ 75 mm
tM3 ⫽
d1 ⫽ 63 mm
d2 ⫺ d3
⫽ 4.802 mm
2
d3 ⫽ 65.4 mm
Problem 1.6-5 A bar of monel metal as in the figure (length
L ⫽ 230 mm, diameter d ⫽ 6 mm) is loaded axially by a tensile
force P. If the bar elongates by 0.5 mm, what is the decrease in
diameter d? What is the magnitude of the load P? Use the data
in Table H-2, Appendix H.
d
P
P
L
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56
CHAPTER 1
Tension, Compression, and Shear
Solution 1.6-5
NUMERICAL DATA
DECREASE IN DIAMETER
E ⫽ 170 GPa
⌬d ⫽ ␧¿ d
n ⫽ 0.32
¢d ⫽ ⫺4.17 * 10⫺3 mm
L ⫽ 230 mm
INITIAL CROSS SECTIONAL AREA
d ⫽ 0.5 mm
Ai ⫽
d ⫽ 6 mm
p 2
d Ai ⫽ 2.827 * 10⫺5 m2
4
MAGNITUDE OF LOAD P
NORMAL STRAIN
d
␧⫽
L
;
P ⫽ EAi␧
⫺3
␧ ⫽ 2.174 ⫻ 10
P ⫽ 10.45 kN
;
LATERAL STRAIN
␧¿ ⫽ ⫺n␧
␧¿ ⫽ ⫺6.957 ⫻ 10⫺4
Problem 1.6-6 A tensile test is performed on a brass specimen
10 mm in diameter using a gage length of 50 mm (see figure).
When the tensile load P reaches a value of 20 kN, the distance
between the gage marks has increased by 0.122 mm.
10 mm
50 mm
P
P
(a) What is the modulus of elasticity E of the brass?
(b) If the diameter decreases by 0.00830 mm, what is Poisson’s ratio?
Solution 1.6-6 Brass specimen in tension
d ⫽ 10 mm Gage length L ⫽ 50 mm
P ⫽ 20 kN d ⫽ 0.122 mm
⌬d ⫽ 0.00830 mm
AXIAL STRESS
P
20 k
⫽ 254.6 MPa
s⫽ ⫽
p
A
(10 mm)2
4
Assume s is below the proportional limit so that
Hooke’s law is valid.
(a) MODULUS OF ELASTICITY
E⫽
s
254.6 MPa
⫽
⫽ 104 GPa
␧
0.002440
;
(b) POISSON’S RATIO
␧⬘ ⫽ v␧
⌬d ⫽ ␧⬘d ⫽ v␧d
v⫽
¢d
0.00830 mm
⫽
⫽ 0.34
␧d
(0.002440)(10 mm)
;
AXIAL STRAIN
␧⫽
0.122 mm
d
⫽
⫽ 0.002440
L
50 mm
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SECTION 1.6
57
Linear Elasticity, Hooke’s Law, and Poisson’s Ratio
Problem 1.6-7 A hollow, brass circular pipe ABC (see figure) supports a load
P1
P1 ⫽ 118 kN acting at the top. A second load P2 ⫽ 98 kN is uniformly distributed
around the cap plate at B. The diameters and thicknesses of the upper and lower
parts of the pipe are dAB ⫽ 31 mm, tAB ⫽ 12 mm, dBC ⫽ 57 mm, and tBC ⫽ 9 mm,
respectively. The modulus of elasticity is 96 GPa. When both loads are fully
applied, the wall thickness of pipe BC increases by 5 ⫻ 10⫺3 mm.
A
dAB
(a) Find the increase in the inner diameter of pipe segment BC.
(b) Find Poisson’s ratio for the brass.
(c) Find the increase in the wall thickness of pipe segment AB and the increase
in the inner diameter of AB.
tAB
P2
B
Cap plate
dBC
tBC
C
Solution 1.6-7
NUMERICAL DATA
P1 ⫽ 118 kN
P2 ⫽ 98 kN
dAB ⫽ 31 mm
tAB ⫽ 12 mm
dBC ⫽ 57 mm
tBC ⫽ 9 mm
E ⫽ 96 GPa
⌬tBC ⫽ 5(10⫺3) mm
(c) INCREASE IN THE WALL THICKNESS OF PIPE SEGMENT AB
AND THE INCREASE IN THE INNER DIAMETER OF AB
AAB ⫽
␧AB ⫽
p
cd 2 ⫺ 1dAB ⫺ 2tAB22 d
4 AB
⫺P1
EAAB
␧AB ⫽ ⫺1.716 ⫻ 10⫺3
␧pAB ⫽ ⫺nbrass␧AB
⌬tAB ⫽ ␧pABtAB
(a) INCREASE IN THE INNER DIAMETER OF PIPE
SEGMENT BC
¢tBC
␧pBC ⫽ 5.556 ⫻ 10⫺4
␧pBC ⫽
tBC
␧pAB ⫽ 5.75 ⫻ 10⫺4
¢tAB ⫽ 6.90 * 10⫺3 mm
;
⌬dABinner ⫽ ␧pAB(dAB ⫺ 2tAB)
¢dABinner ⫽ 4.02 * 10⫺3 mm
⌬dBCinner ⫽ ␧pBC(dBC ⫺ 2tBC)
¢ dBCinner ⫽ 0.022 mm
;
(b) POISSON’S RATIO FOR THE BRASS
ABC ⫽
p
c dBC 2 ⫺ 1dBC ⫺ 2tBC22 d
4
ABC ⫽ 1.357 ⫻ 10⫺3m2
␧BC ⫽
⫺1P1 + P22
␯ brass ⫽
1EABC2
⫺␧pBC
␧BC
␧BC ⫽ ⫺1.658 ⫻ 10⫺3
␯brass ⫽ 0.34
(agrees with App. H (Table H-2))
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58
CHAPTER 1
Tension, Compression, and Shear
Problem 1.6-8 Three round, copper alloy bars having the
Bar 1
same length L but different shapes are shown in the figure.
The first bar has a diameter d over its entire length, the
second has a diameter d over one-fifth of its length, and
the third has a diameter d over one-fifteenth of its length.
Elsewhere, the second and third bars have diameter 2d. All
three bars are subjected to the same axial load P.
Use the following numerical data: P ⫽ 1400 kN,
L ⫽ 5 m, d ⫽ 80 mm, E ⫽ 110 GPa, and ␯ ⫽ 0.33.
Bar 2
d
Bar 3
2d
L
2d
d
L
5
d
L
15
(a) Find the change in length of each bar.
(b) Find the change in volume of each bar.
P
P
P
Solution 1.6-8
P ⫽ 1400 kN
L⫽5m
d ⫽ 80 mm
E ⫽ 110 GPa
Ad ⫽
␯ ⫽ 0.33
p 2
d ⫽ 5026.5 mm2
4
(a) FIND CHANGE IN LENGTH OF EACH BAR
A2d ⫽
p
12d22 ⫽ 20,106.2 mm2
4
⬍ Appendix H, Table H-3: copper alloys can
have yield stress in range 55–760 MPa, so
assume this is below proportional limit so that
¢L1 ⫽ ␧1L ⫽ 12.66 mm
Lf1 ⫽ L + ¢L1 ⫽ 5012.66 mm Hooke’s Law applies
␧2a
P
P
BAR #2
⫽ 2.532 * 10⫺3 ␧2b ⫽
⫽ 6.33 * 10⫺4
⫽ 6.33 * 10⫺4
␧2a ⫽
E Ad
E A2d
4
BAR #1
␧1 ⫽
P
⫽ 2.532 * 10⫺3
E Ad
¢L2a ⫽ ␧2a
L
⫽ 2.532 mm
5
¢L2 ⫽ ¢L2a + ¢L2b ⫽ 5.06 mm
s1 ⫽ E ␧1 ⫽ 279 MPa
¢L2b ⫽ ␧2b a
4L
b ⫽ 2.532 mm
5
Lf 2 ⫽ L + ¢L2 ⫽ 5005.06 mm
¢L2
⫽ 0.4
¢L1
BAR #3
¢L2a ⫽ ␧2a
L
⫽ 0.844 mm
15
¢L3 ⫽ ¢L2a + ¢L2b ⫽ 3.8 mm
¢L2b ⫽ ␧2b a
14 L
b ⫽ 2.954 mm
15
Lf 3 ⫽ L + ¢L3 ⫽ 5003.08 mm
¢L3
⫽ 0.3
¢L1
(b) FIND CHANGE IN VOLUME OF EACH BAR
Use lateral strain (␧p) in each segment to find change in diameter ⌬d, then find change in cross sectional area, then
volume.
BAR #1
p
␧p1 ⫽ ⫺v␧1 ⫽ ⫺8.356 * 10⫺4 ¢d1 ⫽ ␧p1 d ⫽ ⫺0.067 mm A1 ⫽ 1d + ¢d122 ⫽ 5018.152 mm2
4
¢Vol
1
¢Vol1 ⫽ A1 Lf1 ⫺ Ad L ⫽ 21,548 mm3
⫽ 8.574 * 10⫺4
Ad L
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SECTION 1.7
Shear Stress and Strain
59
BAR #2
␧p2a ⫽ ␧p1 ␧p2b ⫽ ⫺v ␧2b ⫽ ⫺2.089 * 10⫺4
¢d2b ⫽ ␧p2b 12d2 ⫽ ⫺0.33 mm A2a ⫽ A1
␧p1
4
A2b ⫽
⫽ ⫺2.089 * 10⫺4
p
12d + ¢d2b22 ⫽ 20,097.794 mm2
4
L
4L
⫽ 2.532 mm ¢L2b ⫽ ␧2b a
b ⫽ 2.532 mm
5
5
4L
4L
L
L
+ ¢L2b b d ⫺ cA2d a
b + Ad a b d
¢Vol2 ⫽ cA1 a + ¢L2a b + A2b a
5
5
5
5
¢L2a ⫽ ␧2a
¢Vol2
⫽ 1.002
¢Vol1
⫽ 21,601 mm3
BAR #3
L
14L
⫽ 0.844 mm
¢L2b ⫽ ␧2b a
b ⫽ 2.954 mm
15
15
14L
14L
L
L
+ ¢L2a b + A2b a
+ ¢L2b b d ⫺ cA2d a
b + Ad a b d
¢Vol3 ⫽ cA1 a
15
15
15
15
¢L2a ⫽ ␧2a
¢Vol3
⫽ 1.003
¢Vol2
⫽ 21,610 mm3
¢Vol1 ⫽ 21,548 mm3
¢Vol2 ⫽ 21,601 mm3
¢Vol3 ⫽ 21,610 mm3
Shear Stress and Strain
Problem 1.7-1 An angle bracket having thickness t ⫽ 19 mm is attached to the flange of a column by two 16 mm
diameter bolts (see figure). A uniformly distributed load from a floor joist acts on the top face of the bracket with a pressure
p ⫽ 1.9 MPa. The top face of the bracket has length L ⫽ 200 mm and width b ⫽ 75 mm.
Determine the average bearing pressure ␴b between the angle bracket and the bolts and the average shear stress ␶aver in
the bolts. (Disregard friction between the bracket and the column.)
p
Distributed pressure on angle bracket
b
Floor slab
L
Floor joist
Angle bracket
Angle bracket
t
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60
CHAPTER 1
Tension, Compression, and Shear
Solution 1.7-1
NUMERICAL DATA
BEARING STRESS
t ⫽ 19 mm
L ⫽ 200 mm
sb ⫽
b ⫽ 75 mm
p ⫽ 1.9 MPa
d ⫽ 16 mm
sb ⫽ 46.9 MPa
;
SHEAR STRESS
BEARING FORCE
F ⫽ 2.85 * 104 N
F ⫽ pbL
F
2Ab
SHEAR AND BEARING AREAS
p 2
AS ⫽
d
AS ⫽ 2.011 * 10⫺4 m2
4
tave ⫽
F
2AS
tave ⫽ 70.9 MPa
;
Ab ⫽ 3.04 * 10⫺4 m2
Ab ⫽ dt
Roof structure
Problem 1.7-2
Truss members supporting a roof are connected to a 26-mm-thick gusset plate by a 22-mm diameter
pin as shown in the figure and photo. The two end plates on
the truss members are each 14 mm thick.
Truss
member
(a) If the load P ⫽ 80 kN, what is the largest bearing
stress acting on the pin?
(b) If the ultimate shear stress for the pin is 190 MPa, what
force Pult is required to cause the pin to fail in shear?
P
End
plates
(Disregard friction between the plates.)
P
Pin
t = 14 mm
Gusset
plate
26 mm
Solution 1.7-2
NUMERICAL DATA
(b) ULTIMATE FORCE IN SHEAR
tep ⫽ 14 mm
Cross-sectional area of pin:
tgp ⫽ 26 mm
Ap ⫽
P ⫽ 80 kN
dp ⫽ 22 mm
4
Ap ⫽ 380.133 mm2
tult ⫽ 190 MPa
(a) BEARING STRESS ON PIN
sb ⫽
p d2p
Pult ⫽ 2tultAp
Pult ⫽ 144.4 kN
;
P
gusset plate is thinner than
dptgp
(2tep) so gusset plate controls
sb ⫽ 139.9 MPa
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.7
Shear Stress and Strain
61
Problem 1.7-3 The upper deck of a football stadium is supported by braces, each of which transfers a load P ⫽ 700 kN to
the base of a column (see figure part a). A cap plate at the bottom of the brace distributes the load P to four flange plates
(tf ⫽ 25 mm) through a pin (dp ⫽ 50 mm) to two gusset plates (tg ⫽ 38 mm) (see figure parts b and c).
Determine the following quantities.
(a) The average shear stress ␶aver in the pin.
(b) The average bearing stress between the flange plates and the pin (␴bf), and also between the gusset plates
and the pin (␴bg).
(Disregard friction between the plates.)
Cap plate
Flange plate
(tf = 25 mm)
Pin (dp = 50 mm)
Gusset plate
(tg = 38 mm)
(b) Detail at bottom of brace
(© Barry Goodno)
P
P = 700 kN
Cap plate
(a) Stadium brace
(© Barry Goodno)
Pin (dp = 50 mm)
P
Flange plate
(tf = 25 mm)
Gusset plate
(tg = 38 mm)
P/2
P/2
(c) Section through bottom of brace
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62
CHAPTER 1
Tension, Compression, and Shear
Solution 1.7-3
(b) BEARING STRESS ON PIN FROM FLANGE PLATE
NUMERICAL DATA
P ⫽ 700 kN
dp ⫽ 50 mm
tg ⫽ 38 mm
tf ⫽ 25 mm
P
4
sbf ⫽
dptf
(a) SHEAR STRESS ON PIN
V
t⫽
a
pd2p
4
b
t ⫽ 89.1 MPa
;
a
;
BEARING STRESS ON PIN FROM GUSSET PLATE
P
4
t⫽
sbf ⫽ 140 MPa
pd2p
4
P
2
sbg ⫽
dptg
b
sbg ⫽ 184.2 MPa
;
Problem 1.7-4 The inclined ladder AB supports a house painter (85 kg) at C and the self-weight (q ⫽ 40 N/m) of the
ladder itself. Each ladder rail (tr ⫽ 4 mm) is supported by a shoe (ts ⫽ 5mm) that is attached to the ladder rail by a bolt of
diameter dp ⫽ 8 mm.
(a) Find support reactions at A and B.
(b) Find the resultant force in the shoe bolt at A.
(c) Find maximum average shear (t) and bearing (sb) stresses in
the shoe bolt at A.
B
C
H = 7.5 m
N/
m
Typical rung
Shoe bolt at A
q=
Ladder rail (tr = 4 mm)
36
tr
Shoe bolt (dp = 8 mm)
Ladder shoe (ts = 5 mm)
A
ts
Ay
2
a = 1.8 m
Ay
2
Section at base
b = 0.7 m
Assume no slip at A.
Solution 1.7-4
NUMERICAL DATA
tr ⫽ 4 mm ts ⫽ 5 mm dp ⫽ 8 mm
a ⫽ 1.8 m
P ⫽ 85 19.812 P ⫽ 833.85 N
b ⫽ 0.7 m H ⫽ 7.5 m q ⫽ 40 N/m
(a) SUPPORT REACTIONS
L ⫽ 31a + b22 + H2
L ⫽ 7.906 m
LAC ⫽
a
L LAC ⫽ 5.692
a + b
LCB ⫽
a
L LCB ⫽ 2.214
a + b
LAC + LCB ⫽ 7.906
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.7
Shear Stress and Strain
63
SUM MOMENTS ABOUT A
Bx ⫽
Pa + qL a
a + b
b
2
⫺H
Ay ⫽ P + q L
Bx ⫽ ⫺252.829 N (left) and Ax ⫽ ⫺Bx (Ax acts to right)
Ay ⫽ 1150.078 N
(b) RESULTANT FORCE IN SHOE BOLT AT A
Aresultant ⫽ 1177.54 N
Bx ⫽ ⫺252.8 N
Ax ⫽ ⫺Bx
Ax ⫽ ⫺Bx
Ay ⫽ 1150.1 N
Aresultant ⫽ 3A2x + A2y
Aresultant ⫽ 1178 N
(c) MAXIMUM SHEAR AND BEARING STRESSES IN SHOE BOLT AT A
dp ⫽ 8 mm
Shear area:
p
As ⫽ d2p
4
As ⫽ 50.256 mm2
ts ⫽ 5 mm
tr ⫽ 4 mm
Aresultant
2
t⫽
2 As
Shear stress:
t ⫽ 5.86 MPa
Aresultant
Bearing area:
Ab ⫽ 2 dp ts
Ab ⫽ 80 mm2
Bearing stress:
sbshore ⫽
2
Ab
sbshoe ⫽ 7.36 MPa
Problem 1.7-5
The force in the brake cable of the V-brake
system shown in the figure is T ⫽ 200 N. The pivot pin at A has
diameter dp ⫽ 6 mm and length Lp ⫽ 16 mm.
Use dimensions shown in the figure. Neglect the weight of the
brake system.
(a) Find the average shear stress ␶aver in the pivot pin where it is
anchored to the bicycle frame at B.
(b) Find the average bearing stress sb,aver in the pivot pin over
segment AB.
T
Lower end of front brake cable
D
81 mm
Brake pads
C
25 mm
B
A
Pivot pins
anchored to
frame (dP)
LP
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64
CHAPTER 1
Tension, Compression, and Shear
Solution 1.7-5
NUMERICAL DATA
L ⫽ 16 mm
dp ⫽ 6 mm
BC ⫽ 25 mm
CD ⫽ 81 mm
T ⫽ 200 N
As ⫽
EQUILIBRIUM - FIND HORIZONTAL FORCES AT
B AND C [VERTICAL REACTION VB ⫽ 0]
a MB ⫽ 0
HC ⫽
T(BC + CD)
BC
HB ⫽ T ⫺ HC
pdp2
As ⫽ 2.827 * 10⫺5 m2
4
t max ⫽
ƒ HB ƒ
tmax ⫽ 22.9 MPa
AS
;
(b) FIND THE MAXIMUM BEARING STRESS sb,max IN THE PIVOT
PIN OVER SEGMENT AB:
a FH ⫽ 0
HC ⫽ 848 N
(a) FIND THE MAXIMUM SHEAR STRESS tmax IN THE PIVOT PIN
WHERE IT IS ANCHORED TO THE BICYCLE FRAME AT B:
Ab ⫽ 9.6 * 10⫺5m2
Ab ⫽ dpL
Hn ⫽ ⫺648 N
sb,max ⫽
ƒ HB ƒ
Ab
sb,max ⫽ 6.75 MPa
Problem 1.7-6 A steel plate of dimensions 2.5 ⫻ 1.5 ⫻ 0.08 m
and weighing 23.1 kN is hoisted by steel cables with lengths
L1 ⫽ 3.2 m and L2 ⫽ 3.9 m that are each attached to the plate by
a clevis and pin (see figure). The pins through the clevises are
18 mm in diameter and are located 2.0 m apart. The orientation
angles are measured to be u ⫽ 94.4° and a ⫽ 54.9°.
For these conditions, first determine the cable forces T1 and
T2, then find the average shear stress taver in both pin 1 and pin 2,
and then the average bearing stress sb between the steel plate
and each pin. Ignore the mass of the cables.
;
P
a=
Clevis
L1
and pin 1
0.6
m
b1
u
b2
L2
a
2.0
Clevis
and pin 2
m
Center of mass
of plate
b=
1.0
m
Steel plate
(2.5 × 1.5 × 0.08 m)
Solution 1.7-6
NUMERICAL DATA
L1 ⫽ 3.2 m
u ⫽ 94.4 a
a ⫽ 0.6 m
L2 ⫽ 3.9 m
p
b rad
180
SOLUTION APPROACH
a ⫽ 54.9 a
p
b rad
180
STEP (1) d ⫽ 2a2 + b2
a
STEP (2) u1 ⫽ arctan a b
b
d ⫽ 1.166 m
u1
180
⫽ 30.964⬚
p
STEP (3)-Law of cosines
H ⫽ 2d2 + L21 ⫺ 2dL1cos(u + u1)
b⫽1m
W ⫽ 77.0(2.5 ⫻ 1.5 ⫻ 0.08)
W ⫽ 23.1 kN
H ⫽ 3.99 m
3
(77 ⫽ wt density of steel, kN/m )
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.7
STEP (4) b 1 ⫽ arccosa
b1
L22 + H2 ⫺ d2
b
2L1H
180
⫽ 13.789⬚
p
STEP (5) b 2 ⫽ arccosa
L22 + H 2 ⫺ d 2
b
2L2H
180
⫽ 16.95⬚
b2
p
180
p
⫽ 180.039⬚
STATICS
T1sin( b 1) ⫽ T2sin( b 2)
sin(b 2)
T1 ⫽ T2 a
b
sin(b 1)
T1cos( b 1) ⫹ T2cos( b 2) ⫽ W
T2 ⫽
sin(b 2)
b
sin(b 1)
T1 ⫽ 13.18 kN
;
T1cos( b 1) ⫹ T2cos( b 2) ⫽ 23.1 ⬍ checks
SHEAR AND BEARING STRESSES
dp ⫽ 18 mm
AS ⫽
STEP (6)
Check (b 1 + b 2 + u + a)
T1 ⫽ T2 a
65
Shear Stress and Strain
p 2
d
4 p
T1
2
t1ave ⫽
AS
T2
2
t2ave ⫽
AS
sb1 ⫽
T1
Ab
sb2 ⫽
T2
Ab
t ⫽ 80 mm
Ab ⫽ tdp
t1ave ⫽ 25.9 MPa
;
t2ave ⫽ 21.2 MPa
;
s b1 ⫽ 9.15 MPa
sb2 ⫽ 7.48 MPa
;
;
W
sin(b 2)
cos(b 1)
+ cos(b 2)
sin(b 1)
T2 ⫽ 10.77 kN
;
Problem 1.7-7 A special-purpose eye bolt of shank diameter d ⫽ 12 mm passes
y
through a hole in a steel plate of thickness tp ⫽ 19 mm (see figure) and is secured
by a nut with thickness t ⫽ 6 mm. The hexagonal nut bears directly against the
steel plate. The radius of the circumscribed circle for the hexagon is r ⫽ 10 mm
(which means that each side of the hexagon has length 10 mm). The tensile
forces in three cables attached to the eye bolt are T1 ⫽ 3560 N, T2 ⫽ 2448 N, and
T3 ⫽ 5524 N.
(a) Find the resultant force acting on the eye bolt.
(b) Determine the average bearing stress ␴b between the hexagonal nut on the
eye bolt and the plate.
(c) Determine the average shear stress ␶aver in the nut and also in the steel
plate.
T1
tp
T2
d
30⬚
2r
x
Cables
Nut
t
30⬚
Eye bolt
T3
Steel plate
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
66
CHAPTER 1
Tension, Compression, and Shear
Solution 1.7-7
(c)
CABLE FORCES
T1 ⫽ 3560 N
T2 ⫽ 2448 N
T3 ⫽ 5524 N
d ⫽ 0.012 m
(a) RESULTANT
P ⫽ T2 a
⬍ x-dir.
13
b + T3(0.5)
2
n⫽6
t ⫽ 6 mm
Asn ⫽ 226 mm2
Asn ⫽ pdt
P ⫽ 4882 N
;
tnut ⫽ 21.6 MPa
b ⫽ 10 mm
b⫽
360
n
Aspl ⫽ 6 rtp
tpl ⫽
2
b
nb
cota b
Ah ⫽
4
2
P
Aspl
tnut ⫽
P
Asn
;
tp ⫽ 19 mm
SHEAR THROUGH PLATE
(b) AVE. BEARING STRESS
d ⫽ 12 mm
SHEAR THROUGH NUT
Aspl ⫽ 1140 mm
tpl ⫽ 4.28 MPa
r ⫽ 10 mm
2
;
⬍ hexagon (Case 25, App. D)
Ab ⫽ Ah ⫺
sb ⫽
P
Ab
p 2
d
4
Ab ⫽ 146.71 mm2
;
sb ⫽ 33.3 MPa
b
Problem 1.7-8 An elastomeric bearing pad consisting of two steel
plates bonded to a chloroprene elastomer (an artificial rubber) is
subjected to a shear force V during a static loading test (see figure).
The pad has dimensions a ⫽ 125 mm and b ⫽ 240 mm, and the elastomer has thickness t ⫽ 50 mm. When the force V equals 12 kN, the
top plate is found to have displaced laterally 8.0 mm with respect to
the bottom plate.
What is the shear modulus of elasticity G of the chloroprene?
a
V
t
Solution 1.7-8
d = 8.0 mm
NUMERICAL DATA
V ⫽ 12 kN
b ⫽ 240 mm
V
a ⫽ 125 mm
t ⫽ 50 mm
d ⫽ 8 mm
g
t = 50 mm
AVERAGE SHEAR STRESS
tave ⫽
V
ab
b = 250 mm
tave ⫽ 0.4 MPa
AVERAGE SHEAR STRAIN
SHEAR MODULUS G
gave ⫽
d
t
gave ⫽ 0.16
G⫽
tave
gave
G ⫽ 2.5 MPa
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.7
Problem 1.7-9 A joint between two concrete slabs A and B
67
Shear Stress and Strain
A
is filled with a flexible epoxy that bonds securely to the
concrete (see figure). The height of the joint is h ⫽ 100 mm,
its length is L ⫽ 1.0 m, and its thickness is t ⫽ 12 mm. Under
the action of shear forces V, the slabs displace vertically
through the distance d ⫽ 0.048 mm relative to each other.
B
L
h
(a) What is the average shear strain ␥aver in the epoxy?
(b) What is the magnitude of the forces V if the shear
modulus of elasticity G for the epoxy is 960 MPa?
t
d
A
h
B
V
V
t
Solution 1.7-9 Epoxy joint between concrete slabs
(a) AVERAGE SHEAR STRAIN
d
A
B
V
V
gaver ⫽
h
d
⫽ 0.004
t
;
(b) SHEAR FORCES V
t
Average shear stress: taver ⫽ Ggaver ⫽ 3.84 MPa
h ⫽ 100 mm
t ⫽ 12 mm
L ⫽ 1.0 m
d ⫽ 0.048 mm
V ⫽ taver(hL) ⫽ 384 kN
;
G ⫽ 960 MPa
Problem 1.7-10 A flexible connection consisting of rubber
pads (thickness t ⫽ 9 mm) bonded to steel plates is shown in the
figure. The pads are 160 mm long and 80 mm wide.
(a) Find the average shear strain gaver in the rubber if the
force P ⫽ 16 kN and the shear modulus for the rubber
is G ⫽ 1250 kPa.
(b) Find the relative horizontal displacement d between the
interior plate and the outer plates.
P
—
2
160 mm
Rubber pad
X
P
P
—
2
Rubber pad
X
80 mm
t = 9 mm
t = 9 mm
Section X-X
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68
CHAPTER 1
Tension, Compression, and Shear
Solution 1.7-10
Rubber pads bonded to steel plates
P
—
2
Thickness t
P
(a) SHEAR STRESS AND STRAIN IN THE RUBBER PADS
taver ⫽
Rubber pad
P
—
2
gaver ⫽
Rubber pads: t ⫽ 9 mm
Length L ⫽ 160 mm
8 kN
P/2
⫽
⫽ 625 kPa
bL
(80 mm)(160 mm)
taver
625 kPa
⫽
⫽ 0.50
G
1250 kPa
;
(b) HORIZONTAL DISPLACEMENT
d ⫽ t * tan1gave2 ⫽ 4.92 mm
Width b ⫽ 80 mm
G ⫽ 1250 kPa
P ⫽ 16 kN
Problem 1.7-11 Steel riser pipe hangs from a drill rig located offshore in deep water (see figure). Separate segments are
joined using bolted flange plages (see figure part b and photo). Assume that there are six bolts at each pipe segment
connection. Assume that the total length of riser pipe is L ⫽ 1500 m; outer and inner diameters are d2 ⫽ 405 mm and
d1 ⫽ 380 mm; flange plate thickness tf ⫽ 44 mm; and bolt and washer diameters are db ⫽ 28 mm, and dw ⫽ 47 mm.
(a) If the entire length of the riser pipe were suspended in air, find the average normal stress ␴ in each bolt, the average
bearing stress ␴b beneath each washer, and the average shear stress ␶ through the flange plate at each bolt location for
the topmost bolted connection.
(b) If the same riser pipe hangs from a drill rig at sea, what are the normal, bearing, and shear stresses in the connection?
(Obtain the weight densities of steel and sea water from Table H-1, Appendix H. Neglect the effect of buoyant foam
casings on the riser pipe.)
Flange plate (tf), typical
bolt (db), and washer (dw)
d2
tf
r
Flange plate on riser pipe
L
d1
db t
f
dw
d2
x
r
d2
60°
Riser pipe (d2, d1 L)
(a)
y
Flange plate on riser pipe–
plan view (n = 6 bolts shown)
(b)
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.7
Shear Stress and Strain
69
Photo Courtesy of
Transocean
(c)
Solution 1.7-11
(a) PIPE SUSPENDED IN AIR
kN
L ⫽ 1500 m
gs ⫽ 77
db ⫽ 28 mm
dw ⫽ 47 mm
Apipe ⫽
m
p
1d 2 ⫺ d122 ⫽ 154.134 cm2
4 2
Wpipe
n Ab
⫽ 482 MPa
(b) PIPE SUSPENDED IN SEA WATER
sb ⫽
Winwater
⫽ 419 MPa
n Ab
m3
d2 ⫽ 405 mm
d1 ⫽ 380 mm
tf ⫽ 44 mm
n⫽6
Wpipe ⫽ gs Apipe L ⫽ 1780.252 kN
sb ⫽
kN
gw ⫽ 10
3
sbrg ⫽
t⫽
Ab ⫽
Wpipe
n Aw
d2 ⫺ d1
⫽ 12.5 mm
2
p 2
d ⫽ 6.158 cm2
4 b
⫽ 265 MPa
tf ⫽
Aw ⫽
Wpipe
n dw tf
p
1d 2 ⫺ db22 ⫽ 11.2 cm2
4 w
⫽ 143.5 MPa
Winwater ⫽ 1gS ⫺ gW2 Apipe L ⫽ 1549.051 kN
sbrg ⫽
Winwater
⫽ 231 MPa
n Aw
tf ⫽
Winwater
⫽ 124.8 MPa
n dw tf
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70
CHAPTER 1
Tension, Compression, and Shear
Problem 1.7-12 The clamp shown in the figure is used to
support a load hanging from the lower flange of a steel beam.
The clamp consists of two arms (A and B) joined by a pin at C.
The pin has diameter d ⫽ 12 mm. Because arm B straddles arm
A, the pin is in double shear.
Line 1 in the figure defines the line of action of the
resultant horizontal force H acting between the lower flange of
the beam and arm B. The vertical distance from this line to the
pin is h ⫽ 250 mm. Line 2 defines the line of action of the
resultant vertical force V acting between the flange and arm B.
The horizontal distance from this line to the centerline of the
beam is c ⫽ 100 mm. The force conditions between arm A and
the lower flange are symmetrical with those given for arm B.
Determine the average shear stress in the pin at C when the
load P ⫽ 18 kN.
c Line 2
Arm A
Arm B
Line 1
P
h
Arm A
C
P
Solution 1.7-12
Clamp supporting a load P
FREE-BODY DIAGRAM OF CLAMP
h ⫽ 250 mm
c ⫽ 100 mm
P ⫽ 18 kN
From vertical equilibrium:
V⫽
P
⫽ 9 kN
2
d ⫽ diameter of pin at C ⫽ 12 mm
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.7
FREE-BODY DIAGRAMS OF ARMS A AND B
Shear Stress and Strain
71
SHEAR FORCE F IN PIN
F⫽
P 2
H 2
a
b
+
a
b
A 4
2
⫽ 4.847 kN
AVERAGE SHEAR STRESS IN THE PIN
©MC ⫽ 0 哵哴
VC ⫺ Hh ⫽ 0
H⫽
taver ⫽
F
F
⫽
⫽ 42.9 MPa
Apin
pa2
4
;
VC
Pc
⫽
⫽ 3.6 kN
h
2h
FREE-BODY DIAGRAM OF PIN
Problem 1.7-13 A hitch-mounted bicycle rack is designed to carry up to four 135 N bikes mounted on and strapped to two
arms GH (see bike loads in the figure part a). The rack is attached to the vehicle at A and is assumed to be like a cantilever
beam ABCDGH (figure part b). The weight of fixed segment AB is W1 ⫽ 45 N, centered 225 mm from A (see the figure
part b) and the rest of the rack weighs W2 ⫽ 180 N, centered 480 mm from A. Segment ABCDG is a steel tube, 50 ⫻ 50 mm,
of thickness t ⫽ 3 mm. Segment BCDGH pivots about a bolt at B of diameter dB ⫽ 6 mm to allow access to the rear of the
vehicle without removing the hitch rack. When in use, the rack is secured in an upright position by a pin at C (diameter of
pin dp ⫽ 8 mm) (see photo and figure part c). The overturning effect of the bikes on the rack is resisted by a force couple
F # h at BC.
(a)
(b)
(c)
(d)
Find the support reactions at A for the fully loaded rack.
Find forces in the bolt at B and the pin at C.
Find average shear stresses ␶aver in both the bolt at B and the pin at C.
Find average bearing stresses ␴b in the bolt at B and the pin at C.
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72
CHAPTER 1
Tension, Compression, and Shear
Bike loads
y
4 bike loads
480 mm
G
685 mm
G
Release pins
at C & G
(dp = 8 mm)
H
3 @ 100 mm
W2
H
150 mm
a
50 mm ⫻ 50 mm ⫻ 3 mm
C
Fixed
support
at A
D
C
D
A
F
54 mm
F
F
B
Bolt at B
(dB = 6 mm)
a
W1
B
h = 175 mm
h = 175 mm
x
F
230 mm 200 mm
(a)
(b)
Pin at C
C
Pin at C
54 mm
D
Bolt at B
50 ⫻ 50 ⫻ 3 mm
tube
(c) Section a–a
Solution 1.7-13
L1 ⫽ 430 mm + 54 mm + 150 mm
NUMERICAL DATA
t ⫽ 3 mm
(dist. from A to 1st bike)
b ⫽ 50 mm
h ⫽ 175 mm
W1 ⫽ 45 N
W2 ⫽ 180 N
P ⫽ 135 N
dB ⫽ 6 mm
dp ⫽ 8 mm
(a) REACTIONS AT A
Ax ⫽ 0
MA ⫽ W1(230 mm) + W2(480 mm) + P(4L1 + 100 mm
MA ⫽ 520 N # m
+ 200 mm + 300 mm)
(b) FORCES IN BOLT AT B AND PIN AT C
;
Ay ⫽ W1 + W2 + 4P
Ay ⫽ 765 N
L1 ⫽ 634 mm
;
;
©Fy ⫽ 0
By ⫽ W2 + 4P
By ⫽ 720 N
;
©MB ⫽ 0
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.7
d ⫽ 54 + 150 mm
AsC ⫽ 2
W2[(480 ⫺ 230) mm] + P[(d) mm]
+ P[(d + 100)] + P[(d + 200) mm]
⫹P[(d + 300) mm]
Bx ⫽
h
Bx ⫽ 1.349 kN
;
Bres ⫽ 2B2x + B2y
Cx ⫽ ⫺Bx
tC ⫽
AsB ⫽ 2
tB ⫽
4
Bres
AsB
Bres ⫽ 1.530 kN
Bx
AsC
tC ⫽ 13.4 MPa
AbB ⫽ 36 mm2
AbB ⫽ 2 tdB
AsB ⫽ 57 mm2
Bres
AbB
sbB ⫽ 42.5 MPa
;
sbC ⫽
;
AbC ⫽ 48 mm2
AbC ⫽ 2 tdp
Cx
28.1
sbC ⫽ 28.1 MPa
Problem 1.7-14 A bicycle chain consists of a series of small
;
Links
links, each 12 mm long between the centers of the pins (see
figure). You might wish to examine a bicycle chain and observe
its construction. Note particularly the pins, which we will
assume to have a diameter of 2.5 mm.
In order to solve this problem, you must now make two
measurements on a bicycle (see figure): (1) the length L of the
crank arm from main axle to pedal axle, and (2) the radius R
of the sprocket (the toothed wheel, sometimes called the
chainring).
Pin
12 mm
2.5 mm
T
F
Sprocket
(a) Using your measured dimensions, calculate the tensile
force T in the chain due to a force F ⫽ 800 N applied to
one of the pedals.
(b) Calculate the average shear stress taver in the pins.
R
Chain
L
Solution 1.7-14
;
t ⫽ 3 mm
sbB ⫽
tB ⫽ 27.1 MPa
AsC ⫽ 101 mm2
4
(d) BEARING STRESSES ␴B IN THE BOLT AT B AND THE
PIN AT C
(c) AVERAGE SHEAR STRESSES ␶ave IN BOTH THE BOLT
AT B AND THE PIN AT C
2
pdB
pdp2
73
Shear Stress and Strain
Bicycle chain
Pin
T
—
2
T
—
2
T
—
2
T
—
2
T
F
Sprocket
R
12 mm
2.5 mm
F ⫽ force applied to pedal ⫽ 800 N
L
Chain
R ⫽ radius of sprocket
L ⫽ length of crank arm
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74
CHAPTER 1
Tension, Compression, and Shear
MEASUREMENTS (FOR AUTHOR’S BICYCLE)
(1) L ⫽ 162 mm
(b) SHEAR STRESS IN PINS
(2) R ⫽ 90 mm
taver ⫽
(a) TENSILE FORCE T IN CHAIN
©M axle ⫽ 0
FL ⫽ TR
T⫽
FL
R
⫽
Substitute numerical values:
T⫽
(800 N)(162 mm)
⫽ 1440 N
90 mm
T/2
T
2T
⫽
⫽
2
Apin
pd
pd 2
2
(4)
2FL
pd2R
Substitute numerical values:
2(800 N)(162 mm)
taver ⫽
⫽ 147 MPa
p(2.5 mm)2(90 mm)
;
;
Problem 1.7-15 A shock mount constructed as shown in the
figure is used to support a delicate instrument. The mount consists
of an outer steel tube with inside diameter b, a central steel bar of
diameter d that supports the load P, and a hollow rubber cylinder
(height h) bonded to the tube and bar.
(a) Obtain a formula for the shear t in the rubber at a radial
distance r from the center of the shock mount.
(b) Obtain a formula for the downward displacement d of the
central bar due to the load P, assuming that G is the shear
modulus of elasticity of the rubber and that the steel tube
and bar are rigid.
Solution 1.7-15
Shock mount
(a) SHEAR STRESS t AT RADIAL DISTANCE r
As ⫽ shear area at distance r ⫽ 2prh
t⫽
P
P
⫽
As
2prh
;
(b) DOWNWARD DISPLACEMENT d
g ⫽ shear strain at distance r
g⫽
P
t
⫽
G
2prhG
dd ⫽ downward displacement for element dr
dd ⫽ gdr ⫽
r ⫽ radial distance from center of shock mount to
element of thickness dr
Pdr
2prhG
b/2
d⫽
L
d⫽
b/2
P
dr
P
b/2
⫽
[ln r]d/2
2phG Ld/2 r
2phG
d⫽
b
P
ln
2phG
d
dd ⫽
Ld/2
Pdr
2prhG
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.7
75
Shear Stress and Strain
Problem 1.7-16
A removable sign post on a hurricane evacuation route has a square base plate with four slots (or
cutouts) at bolts 1 through 4 (see drawing and photo) for ease of installation and removal. The upper portion of the
post has a separate base plate that is bolted to an anchored base (see photo). Each of the four bolts has diameter db and
a washer of diameter dw. The bolts are arranged in a rectangular pattern (b ⫻ h). Consider wind force Wy applied in
the y-direction at the center of pressure of the sign structure at a height z ⫽ L above the base. Neglect the weight of the
sign and post, and also neglect friction between the upper and lower base plates. Assume that the lower base plate and
short anchored post are rigid.
(a) Find average shear stress t (MPa) at bolt #1 due to the wind force Wy; repeat
for bolt #4.
(b) Find average bearing stress sb (MPa) between the bolt and the base plate
(thickness t) at bolt #1; repeat for bolt #4.
(c) Find average bearing stress sb (MPa) between base plate and washer at bolt #4
due to the wind force Wy (assume initial bolt pretension is zero).
(d) Find average shear stress t (MPa) through the base plate at bolt #4 due to the
wind force Wy.
(e) Find an expression for normal stress s in bolt #3 due to the wind force Wy.
z
C.P.
Wy
See Prob. 1.8-15 for additional discussion of wind on a sign, and the resulting forces
acting on a conventional base plate.
NUMERICAL DATA
L ⫽ 2.75 m
b ⫽ 96 mm
Wy ⫽ 667 N
t ⫽ 14 mm
H ⫽ 150 mm
db ⫽ 12 mm
x
y
h ⫽ 108 mm
dw ⫽ 22 mm
L
h
(a)
Bolt and washer
(db, dw)
y
2
Wy
Square base plate
(H × H) of
thickness t
1
y
h
3
Slot in base plate
(db = slot width)
h
x
x
Base of position
anchored in
ground
4
b
Upper removable
sign post
with base plate
Plan view
of upper
base plate
Slotted upper
and lower
base plates
(b)
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76
CHAPTER 1
Tension, Compression, and Shear
Solution 1.7-16
(a) FIND AVERAGE SHEAR STRESS (t, MPa) AT BOLT #1 DUE TO THE WIND FORCE Wy; REPEAT FOR BOLT #4
wy
Ab ⫽
p 2
d ⫽ 113.097 mm2
4 b
t1 ⫽
2
⫽ 2.95 MPa
Ab
t1 ⫽ 2.95 MPa
wy
t4 ⫽ 0
2
⫽ 333.5 N
(only bolts 1 and 2 resist wind force shear in ⫹y-direction)
(b) FIND AVERAGE BEARING STRESS (sb, MPa) BETWEEN THE BOLT AND THE BASE PLATE (THICKNESS t) AT BOLT #1; REPEAT
FOR BOLT #4
wy
Abrg ⫽ db t ⫽ 168 mm2
sb1 ⫽
2
⫽ 1.985 MPa
Abrg
sb1 ⫽ 1.985 MPa
sb4 ⫽ 0
(only bolts 1 and 2 resist wind force bearing in ⫹y-direction)
(c) FIND AVERAGE BEARING STRESS (sb, MPa) BETWEEN BASE PLATE AND WASHER AT BOLT #4 DUE TO THE WIND FORCE Wy
(ASSUME INITIAL BOLT PRETENSION IS ZERO)
Assume wind force creates overturning moment about x axis ⫽ OTMx
OTMx ⫽ Wy L ⫽ 1834.25 N # m
OTM is resisted by force couple pairs at bolts 1 to 4 and 2 to 3; so force in bolt 4 is
F4 ⫽
OTMx
⫽ 8491.898 N
2h
Bearing area is donut-shaped area of washer in contact with the plate minus approximate rectangular cutout for slot.
Abrg ⫽
dw ⫺ db
p 2
b ⫽ 207.035 mm2
A d ⫺ db2B ⫺ db a
4 w
2
sb4 ⫽
F4
⫽ 41 MPa
Abrg
sb4 ⫽ 41 MPa
(d) FIND AVERAGE SHEAR STRESS (t, MPa) THROUGH THE BASE PLATE AT BOLT #4 DUE TO THE WIND FORCE Wy
Use force F4 above; shear stress is on cylindrical surface at perimeter of washer; must deduct approximate
rectangular area due to slot.
F4
Ash ⫽ A p dw ⫺ db B t ⫽ 799.611 mm2
t⫽
⫽ 10.62 MPa
t ⫽ 10.62 MPa
Ash
(e) FIND AN EXPRESSION FOR NORMAL STRESS (s) IN BOLT #3 DUE TO THE WIND FORCE Wy
Force in bolt #3 due to OTMx is same as that in bolt #4.
s3 ⫽
F4
⫽ 75.1 MPa
Ab
s3 ⫽ 75.1 MPa
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.7
Shear Stress and Strain
77
Problem 1.7-17 A spray nozzle for a garden hose requires a force F ⫽ 22 N to open the spring-loaded spray chamber AB.
The nozzle hand grip pivots about a pin through a flange at O. Each of the two flanges has thickness t ⫽ 1.5 mm, and the pin
has diameter dp ⫽ 3 mm (see figure part a). The spray nozzle is attached to the garden hose with a quick release fitting at B
(see figure part b). Three brass balls (diameter db ⫽ 4.5 mm) hold the spray head in place under water pressure force
fp ⫽ 135 N at C (see figure part c). Use dimensions given in figure part a.
(a) Find the force in the pin at O due to applied force F.
(b) Find average shear stress ␶aver and bearing stress ␴b in the pin at O.
(c) Find the average shear stress ␶aver in the brass retaining balls at C due to water pressure force fp.
Pin
Flange
t
dp
Pin at O
A
F
Top view at O
B
O
a = 19 mm
Spray
nozzle Flange
F
b = 38 mm
F
F
15⬚
c = 44 mm
F
Sprayer
hand grip
Water pressure
force on nozzle, fp
C
(b)
C
Quick
release
fittings
Garden hose
(a)
(c)
3 brass retaining
balls at 120⬚,
diameter db = 4.5 mm
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
78
CHAPTER 1
Tension, Compression, and Shear
Solution 1.7-17
Ox ⫽ 55.7 N
NUMERICAL DATA
F ⫽ 22 N
t ⫽ 1.5 mm
db ⫽ 4.5 mm
u ⫽ 15⬚
fp ⫽ 135 N
a ⫽ 19 mm
dp ⫽ 3 mm
dN ⫽ 16 mm
b ⫽ 38 mm
Ores ⫽ 2O2x + O2y
FAB ⫽
As ⫽ 2
[Fcos(u)(b ⫺ a) + Fsin(u)(c)]
a
Ox ⫽ FAB + Fcos(u)
;
4
tO ⫽
Ores
As
tO ⫽ 3.96 MPa
sbO ⫽
Ores
Ab
sbO ⫽ 6.22 MPa
;
;
(c) FIND THE AVERAGE SHEAR STRESS ␶ave IN THE BRASS
RETAINING BALLS AT B DUE TO WATER PRESSURE FORCE Fp
FAB ⫽ 34.437 N
a FH ⫽ 0
pd2p
Ab ⫽ 2tdp
© Mo ⫽ 0
Ores ⫽ 56.0 N
(b) FIND AVERAGE SHEAR STRESS ␶ave AND BEARING STRESS
␴b IN THE PIN AT O
c ⫽ 44 mm
(a) FIND THE FORCE IN THE PIN AT O DUE TO APPLIED
FORCE F
Oy ⫽ 5.69 N
As ⫽ 3
pd2b
4
fp ⫽ 135 N
t⫽
fp
As
t ⫽ 2.83 MPa
Oy ⫽ Fsin(u)
y
Problem 1.7-18
A single steel strut AB with diameter
ds ⫽ 8 mm supports the vehicle engine hood of mass 20 kg,
which pivots about hinges at C and D [see figures (a) and (b)].
The strut is bent into a loop at its end and then attached to a
bolt at A with diameter db ⫽ 10 mm. Strut AB lies in a vertical
plane.
(a) Find the strut force Fs and average normal stress s in the
strut.
(b) Find the average shear stress taver in the bolt at A.
(c) Find the average bearing stress sb on the bolt at A.
h = 660 mm
W hc = 490 mm
C
B
45⬚
C
x
A
30⬚
D
(a)
C
Hinge
W
Fs
D
z
Strut
ds = 8 mm
H = 1041 mm
h = 660 mm
b = 254 mm c = 506 mm
y
a = 760 mm
d = 150 mm
B
C
Hood
A
(b)
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.7
Shear Stress and Strain
79
Solution 1.7-18
NUMERICAL DATA
ds ⫽ 8 mm
db ⫽ 10 mm
m ⫽ 20 kg
c⫺d
Fsz ⫽
2H + (c ⫺ d)2
2
a ⫽ 760 mm
b ⫽ 254 mm
c ⫽ 506 mm
d ⫽ 150 mm
H
h ⫽ 660 mm
hc ⫽ 490 mm
2H + (c ⫺ d)2
H ⫽ h atan a30
H ⫽ 1041 mm
where
2
p
p
b + tan a 45
bb
180
180
W ⫽ m (9.81 m/s2)
c⫺d
2H + (c ⫺ d)2
2
(a) FIND THE STRUT
s IN THE STRUT
W ⫽ 196.2 N
a + b + c
⫽ 760 mm
2
⌺M
VECTOR rAB
Fsy ⫽ 145.664
0
rAB ⫽
H
Pc ⫺ dQ
lineDC
0
rAB ⫽ 1.041 * 103
P
Q
356
UNIT VECTOR eAB
eAB ⫽
rAB
冷rAB冷
0
W ⫽ ⫺W
P 0 Q
0
0.946
P 0.324 Q
0
W ⫽ ⫺196.2
P 0 Q
490
rDC ⫽ 490
P 760 Q
hc
rDC ⫽
hc
Pb + cQ
⌺M
D
ƒ eAB ƒ ⫽ 1
MD ⫽ rDB ⫻ Fs eAB ⫹ W ⫻ rDC
(ignore force at hinge C since it will vanish with
moment about line DC)
Fsx ⫽ 0
Fsy ⫽
H
2H 2 + (c ⫺ d)2
Fs
⫽0
⫽ 0.946
⫽ 0.324
FORCE
FS
Fsy ⫽
Fsy
Fs ⫽
AND AVERAGE NORMAL STRESS
冷W 冷hc
h
Fs ⫽ 153.9 N
H
;
2H2 + (c ⫺ d)2
Astrut ⫽
eAB ⫽
Fs
s⫽
p 2
d
4 s
Fs
Astrut
Astrut ⫽ 50.265 mm2
s ⫽ 3.06 MPa
;
(b) FIND THE AVERAGE SHEAR STRESS tave IN THE
BOLT AT A
db ⫽ 10 mm
As ⫽
p 2
d
4 b
As ⫽ 78.54 mm2
tave ⫽
Fs
As
tave ⫽ 1.96 MPa
;
(c) FIND THE BEARING STRESS sb ON THE BOLT AT A
Ab ⫽ ds db
sb ⫽
Fs
Ab
Ab ⫽ 80 mm2
sb ⫽ 1.924 MPa
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
80
CHAPTER 1
Tension, Compression, and Shear
Problem 1.7-19
The top portion of a pole saw used to
trim small branches from trees is shown in the figure
part a. The cutting blade BCD (see figure parts a and c)
applies a force P at point D. Ignore the effect of the
weak return spring attached to the cutting blade below
B. Use properties and dimensions given in the figure.
B
Rope, tension = T
a
T
y
Weak return spring
2T
x
(a) Find the force P on the cutting blade at D if the
tension force in the rope is T ⫽ 110 N (see
free-body diagram in figure part b).
(b) Find force in the pin at C.
(c) Find average shear stress ␶aver and bearing stress
␴b in the support pin at C (see section a–a
through cutting blade in figure part c).
C
Cutting
blade
Collar
Saw blade
D
a
P
(a) Top part of pole saw
B
T
⫻
B
2T
BC
= 150 mm
50⬚
20⬚
Cy
5
2
DC =
70⬚
Cx
C
m
m D
x
20⬚
20⬚
P
Cutting blade
(tb = 2.4 mm)
Collar
150 mm C
(tc = 9 mm)
⫻
25 mm
⫻
D
70⬚
(b) Free-body diagram
Pin at C
(dp = 3 mm)
(c) Section a–a
Solution 1.7-19
NUMERICAL PROPERTIES
dp ⫽ 3 mm
T ⫽ 110 N
tb ⫽ 2.4 mm
dBC ⫽ 150 N
SOLVE ABOVE EQUATION FOR P
tc ⫽ 9 mm
dCD ⫽ 25 mm
(a) FIND THE CUTTING FORCE P ON THE CUTTING BLADE
AT D IF THE TENSION FORCE IN THE ROPE IS
T ⫽ 110 N
⌺MC ⫽ 0
MC ⫽ T(dBC sin(70⬚))
⫹ 2T cos(20⬚)(dBC sin(70⬚))
⫺ 2T sin(20⬚)(dBC cos(70⬚))
⫺ P cos(20⬚)(dCD)
P ⫽ T(dBC sin(70⬚)) ⫹ 2T cos(20⬚)
(dBC sin(70⬚)) ⫺ 2T sin(20⬚)
(dBC cos(70⬚))
cos(20⬚)dCD
⫽ 1736 N
;
(b) FIND FORCE IN THE PIN AT C
P ⫽ 1736 N
SOLVE FOR FORCES ON PIN AT C
©Fx ⫽ 0
Cx ⫽ T + 2T cos(20⬚) + P cos(40⬚)
Cx ⫽ 1647 N
©Fy ⫽ 0
;
Cy ⫽ 2T sin(20⬚) ⫺ P sin(40⬚)
Cy ⫽ ⫺1041 N
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.8
Allowable Stresses and Allowable Loads
81
Allowable Stresses and Allowable Loads
Problem 1.8-1 A bar of solid circular cross section is loaded in
tension by forces P (see figure). The bar has length L ⫽ 380 mm
and diameter d ⫽ 6 mm. The material is a magnesium alloy having
modulus of elasticity E ⫽ 42.7 GPa. The allowable stress in tension
is ␴allow ⫽ 89.6 GPa, and the elongation of the bar must not exceed
0.08 mm.
What is the allowable value of the forces P?
Solution 1.8-1
Magnesium bar in tension
MAXIMUM STRESS BASED UPON ELONGATION
smax ⫽ E␧max ⫽ (44,850 MPa)(0.00200)
⫽ 89.7 MPa
MAXIMUM STRESS BASED UPON ELONGATION
L ⫽ 380 mm
d ⫽ 6 mm
E ⫽ 42.7 GPa
sallow ⫽ 89.6 MPa
␧max ⫽
sallow ⫽ 89.6 MPa
ELONGATION GOVERNS
d max ⫽ 0.8 mm
dmax
0.8 mm
⫽
⫽ 0.002
L
380 mm
Problem 1.8-2 A torque T0 is transmitted between two flanged
p
Pallow ⫽ smaxA ⫽ (89.7 MPa)a b(6 mm)2
4
⫽ 2.53 kN
;
T0
d
shafts by means of ten 20-mm bolts (see figure and photo). The
diameter of the bolt circle is d ⫽ 250 mm.
If the allowable shear stress in the bolts is 90 MPa, what is the
maximum permissible torque? (Disregard friction between the
flanges.)
T0
T0
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82
CHAPTER 1
Tension, Compression, and Shear
Solution 1.8-2 Shafts with flanges
NUMERICAL DATA
MAXIMUM PERMISSIBLE TORQUE
r ⫽ 10
^ bolts
d
Tmax ⫽ ta As ar b
2
d ⫽ 250 mm
^ flange
Tmax ⫽ 3.338 * 107 N # mm
As ⫽ pr 2
Tmax ⫽ 33.4 kN # m
As ⫽ 314.159 m2
;
ta ⫽ 85 MPa
Problem 1.8-3 A tie-down on the deck of a sailboat consists of a bent bar bolted
at both ends, as shown in the figure. The diameter dB of the bar is 6 mm, the
diameter dW of the washers is 22 mm, and the thickness t of the fiberglass deck
is 10 mm.
If the allowable shear stress in the fiberglass is 2.1 MPa, and the allowable
bearing pressure between the washer and the fiberglass is 3.8 MPa, what is the
allowable load Pallow on the tie-down?
P
dB
dB
t
dW
dW
Solution 1.8-3 Bolts through fiberglass
dB ⫽ 6 mm
P1 ⫽ 2(1451 N)
dW ⫽ 22 mm
P1 ⫽ 2902 N
t ⫽ 10 mm
ALLOWABLE LOAD BASED UPON BEARING PRESSURE
sb ⫽ 3.8 MPa
Bearing area Ab ⫽
ALLOWABLE LOAD BASED UPON SHEAR STRESS
p 2
1d ⫺ dB22
4 W
tallow ⫽ 2.1 MPa
P2
p
⫽ sbAb ⫽ (3.8 MPa)a b [(22 mm)2 ⫺ (6 mm)2]
2
4
Shear area As ⫽ pdWt
P2 ⫽ 2(1337 N)
P1
⫽ tallow As ⫽ tallow(pdWt)
2
P2 ⫽ 2674 N
IN FIBERGLASS
⫽ (2.1 MPa)(p)(22 mm)(10 mm)
⫽ 1451 N
⫽ 1337 N
ALLOWABLE LOAD
Bearing pressure governs.
Pallow ⫽ 2.67 kN
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.8
Problem 1.8-4 Two steel tubes are joined at B by four pins
(dp ⫽ 11 mm), as shown in the cross section a–a in the
figure. The outer diameters of the tubes are dAB ⫽ 41 mm and
dBC ⫽ 28 mm. The wall thicknesses are tAB ⫽ 6.5 mm and
tBC ⫽ 7.5 mm. The yield stress in tension for the steel is
sY ⫽ 200 MPa and the ultimate stress in tension is
sU ⫽ 340 MPa. The corresponding yield and ultimate values
in shear for the pin are 80 MPa and 140 MPa, respectively.
Finally, the yield and ultimate values in bearing between the
pins and the tubes are 260 MPa and 450 MPa, respectively.
Assume that the factors of safety with respect to yield stress
and ultimate stress are 3.5 and 4.5, respectively.
Allowable Stresses and Allowable Loads
83
a
Pin
tAB
dAB
A
tBC
B
dBC
C
P
a
(a) Calculate the allowable tensile force Pallow considering
tension in the tubes.
(b Recompute Pallow for shear in the pins.
(c) Finally, recompute Pallow for bearing between the pins
and the tubes. Which is the controlling value of P?
dp
tAB
tBC
dAB
dBC
Section a–a
Solution 1.8-4
Yield and ultimate stresses (all in MPa)
TUBES AND PIN DIMENSIONS (MM)
TUBES:
sY ⫽ 200
su ⫽ 340
FSy ⫽ 3.5
dAB ⫽ 41
PIN (SHEAR):
tY ⫽ 8
tu ⫽ 140
FSu ⫽ 4.5
dBC ⫽ dAB ⫺ 2tAB
PIN (BEARING): sbY ⫽ 260
sbu ⫽ 450
tBC ⫽ 7.5
tAB ⫽ 6.5
dBC ⫽ 28
dp ⫽ 11
(a) Pallow CONSIDERING TENSION IN THE TUBES
p
C d 2 ⫺ 1dAB ⫺ 2 tAB22 D ⫺ 4 dp tAB
4 AB
p
AnetBC ⫽ C dBC 2 ⫺ 1dBC ⫺ 2 tBC22 D ⫺ 4 dp tBC
4
sy
AnetBC
PaT1 ⫽ 8743.993 N
PaT1 ⫽
FSy
AnetAB ⫽
PaT 2 ⫽
su
A
FSu netBC
AnetAB ⫽ 418.502 mm2
AnetBC ⫽ 153.02
6 controls
6 use smaller
Pallow ⫽ 8.74 kN
PaT 2 ⫽ 11,561.501 N
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
84
CHAPTER 1
Tension, Compression, and Shear
(b) Pallow CONSIDERING SHEAR IN THE PINS
PaS1 ⫽ A 4 As B
tY
p 2
d As ⫽ 95.033 mm2
4 p
PaS1 ⫽ 8688.748 N
FSy
Pas2 ⫽ (4 As)
As ⫽
tu
FSu
Pallow ⫽ 8.69 kN
< controls
Pas2 ⫽ 11,826.351 N
(c) Pallow CONSIDERING BEARING IN THE PINS
AbAB ⫽ 286 mm2
AbAB ⫽ 4dp tAB
⬍ smaller controls
AbBC ⫽ 4dp tBC
AbBC ⫽ 330
sby
Pab1 ⫽ AbAB a
b
Pab1 ⫽ 21,245.714 N
FSy
Pab2 ⫽ AbAB a
sbu
b
FSu
Pab2 ⫽ 28,600 N
< controls
Pallow ⫽ 21.2 kN
Overall, shear controls (Part (b))
Problem 1.8-5 A steel pad supporting heavy machinery rests on
four short, hollow, cast iron piers (see figure). The ultimate
strength of the cast iron in compression is 344.5 MPa. The outer
diameter of the piers is d ⫽ 114 mm and the wall thickness is
t = 10 mm.
Using a factor of safety of 4.0 with respect to the ultimate
strength, determine the total load P that may be supported by
the pad.
t
d
Solution 1.8-5 Cast iron piers in compression
sa ⫽
344.5
MPa
4
sa ⫽ 86.125 MPa
Atotal ⫽ 4
p
[1142 ⫺ [114 ⫺ 2(10)]2]
4
Atotal ⫽ 0.013 m2
Ptotal ⫽ sa Atotal
Ptotal ⫽ 1.126 MN
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.8
85
Allowable Stresses and Allowable Loads
Problem 1.8-6 The rear hatch of a van [BDCF in figure part (a)] is supported by two hinges at B1 and B2 and by two struts
A1B1 and A2B2 (diameter ds ⫽ 10 mm) as shown in figure part (b). The struts are supported at A1 and A2 by pins, each with
diameter dp ⫽ 9 mm and passing through an eyelet of thickness t ⫽ 8 mm at the end of the strut [figure part (b)]. If a closing
force P ⫽ 50 N is applied at G and the mass of the hatch Mh ⫽ 43 kg is concentrated at C:
(a) What is the force F in each strut? [Use the free-body diagram of one half of the hatch in the figure part (c).]
(b) What is the maximum permissible force in the strut, Fallow, if the allowable stresses are as follows: compressive stress in
the strut, 70 MPa; shear stress in the pin, 45 MPa; and bearing stress between the pin and the end of the strut, 110 MPa.
127 mm
B1
505 mm
505 mm
F
B2
C
Mh
D
Bottom
part of
strut
G
P
F
75 mm
Bx
A1
Mh
—⫻g
2
710 mm
By
ds = 10 mm
A2
B
G
C
D
P
2
10⬚
460 mm
Eyelet
Pin support
A
F
t = 8 mm
(b)
(a)
(c)
Solution 1.8-6
NUMERICAL DATA
F (127cos(10⬚) + 75sin(10⬚))
Mh ⫽ 43 kg
Mh
P
+ 2(505)]
g (127 + 505) +
2
2
Mh
P
g (127 + 505) + [127 + 2(505)]
2
2
F⫽
(127 cos(10⬚) + 75 sin(10⬚))
sa ⫽ 70 MPa
ta ⫽ 45 MPa
sba ⫽ 110 MPa
ds ⫽ 10 mm
dp ⫽ 9 mm
P ⫽ 50 N
g ⫽ 9.81 m/s2
⫽
t ⫽ 8 mm
F ⫽ 1.171 kN
(a) FORCE F IN EACH STRUT FROM STATICS (SUM
MOMENTS ABOUT B)
FV ⫽ F cos110⬚2 FH ⫽ F sin110⬚2
g MB ⫽ 0
FV (127) + FH (75) ⫽
;
(b) MAXIMUM PERMISSIBLE FORCE F IN EACH STRUT;
Fmax IS SMALLEST OF THE FOLLOWING
p 2
d
Fa1 ⫽ 5.50 kN
4 s
p
⫽ ta dp 2
4
Fa2
⫽ 2.86 kN
;
⫽ 2.445
F
Fa1 ⫽ sa
Mh
g (127 + 505)
2
P
+ [127 + 2(505)]
2
Fa2
Fa2
Fa3 ⫽ sba dp t
Fa3 ⫽ 7.92 kN
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
86
CHAPTER 1
Tension, Compression, and Shear
Problem 1.8-7 A lifeboat hangs from two ship’s davits, as shown in
the figure. A pin of diameter d ⫽ 20 mm passes through each davit and
supports two pulleys, one on each side of the davit.
Cables attached to the lifeboat pass over the pulleys and wind
around winches that raise and lower the lifeboat. The lower parts of the
cables are vertical and the upper parts make an angle ␣ ⫽ 15° with
the horizontal. The allowable tensile force in each cable is 8 kN, and the
allowable shear stress in the pins is 27.5 MPa.
If the lifeboat weighs 6.7 kN, what is the maximum weight that
should be carried in the lifeboat?
T T
Davit
a = 15⬚
Pulley
Pin
Cable
Solution 1.8-7 Lifeboat supported by four cables
FREE-BODY DIAGRAM OF ONE PULLEY
ALLOWABLE LOAD BASED UPON ALLOWABLE TENSION
IN CABLES
Pin diameter d ⫽ 20 mm
T ⫽ tensile force in one cable
Tallow ⫽ 8 kN
Tallow ⫽ 27.5 MPa
W ⫽ weight of lifeboat
⫽ 6.7 kN
©Fhoriz ⫽ 0
ALLOWABLE FORCE IN CABLE BASED UPON SHEAR IN
©Fvert ⫽ 0
RH ⫽ T cos 15⬚ ⫽ 0.9659T
RV ⫽ T ⫺ T sin 15⬚ ⫽ 0.7412T
THE PINS
V ⫽ shear force in pin
p
V ⫽ tallow Apin ⫽ (27.5 MPa)a b(20 mm)2
4
V ⫽ 2(RH)2 + (RV)2 ⫽ 1.2175T
⫽ 8.6 kN
V ⫽ 1.2175T
T1 ⫽
V
⫽ 7.1 kN
1.2175
TOTAL ALLOWABLE LOAD ⫽ 4T ⫽ 28.3 kN
TOTAL ALLOWABLE LOAD
⫽ 4Tallow ⫽ 4(8 kN) ⫽ 32 kN
Shear in the pins governs.
Total allowable load ⫽ 28.3 kN
Maximum load that should be carried ⫽ 28.3 ⫺ 6.7
⫽ 21.6 kN
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.8
87
Allowable Stresses and Allowable Loads
Problem 1.8-8 A cable and pulley system in figure part (a) supports a cage of mass 300 kg at B. Assume that this includes the
mass of the cables as well. The thickness of each of the three steel pulleys is t ⫽ 40 mm. The pin diameters are dpA ⫽ 25 mm,
dpB ⫽ 30 mm and dpC ⫽ 22 mm [see figure parts (a) and part (b)].
(a) Find expressions for the resultant forces acting on the pulleys at A, B, and C in terms of cable tension T.
(b) What is the maximum weight W that can be added to the cage at B based on the following allowable stresses? Shear
stress in the pins is 50 MPa; bearing stress between the pin and the pulley is 110 MPa.
a
dpA = 25 mm
L1
C
A
Cable
Cable
Pulley
t
a
dpB
L2
tB
Pin
dpC = 22 mm
dp
Support
bracket
B
dpB = 30 mm
Cage
W
Section a–a: pulley support
detail at A and C
Cage at B
Section a–a: pulley
support detail at B
(a)
(b)
Solution 1.8-8
NUMERICAL DATA
M ⫽ 300 kg
g ⫽ 9.81 m/s2
ta ⫽ 50 MPa
sba ⫽ 110 MPa
tA ⫽ 40 mm
tC ⫽ 50
dpB ⫽ 30
tB ⫽ 40 mm
dpA ⫽ 25 mm
dpC ⫽ 22 mm
(a) RESULTANT FORCES F ACTING ON PULLEYS A, B, AND C
FA ⫽ 22 T
FC ⫽ T
(b) MAXIMUM LOAD W THAT CAN BE ADDED AT B DUE TO
ta AND sba IN PINS AT A, B, AND C
PULLEY AT A
tA ⫽
FA
As
DOUBLE SHEAR
22T ⫽ ta As
FA ⫽ ta As
FB ⫽ 2T
Mg
Wmax
ta As
+
⫽
2
2
22
Mg
Wmax
+
2
2
Wmax1 ⫽
T⫽
Wmax ⫽ 2T ⫺ Mg
From statics at B
Wmax1 ⫽
2
22
2
22
ataAs b ⫺ Mg
ata2
p
d A2 b ⫺ Mg
4 p
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88
CHAPTER 1
Tension, Compression, and Shear
Wmax1
⫽ 22.6
Mg
Wmax4 ⫽
Wmax1 ⫽ 66.5 kN
;
(shear at A controls)
OR check bearing stress
Wmax2 ⫽
22
2
asba Ab b ⫺ Mg
asba tA dpA b ⫺ Mg
22
⫽ 152.6 kN (bearing at A)
Wmax2 ⫽
Wmax2
2
PULLEY AT B
Wmax3 ⫽
2T ⫽ taAs
2
1t A 2 ⫺ Mg
2 a s
2
(s A ) ⫺ Mg
2 ba b
Wmax4 ⫽ sba tB dpB ⫺ Mg
Wmax4 ⫽ 129.1 kN
(bearing at B)
T ⫽ ta As
PULLEY AT C
Wmax5 ⫽ 21t a As2 ⫺ Mg
Wmax5 ⫽ c2ta a 2
p 2
d b d ⫺ Mg
4 pC
Wmax5 ⫽ 7.3 * 104
Wmax5 ⫽ 73.1 kN
(shear at C)
Wmax6 ⫽ 2sbatC dp C ⫺ Mg
Wmax6 ⫽ 239.1 kN
(bearing at C)
p
Wmax3 ⫽ cta a 2 dpB2 b d ⫺ Mg
4
Wmax3 ⫽ 67.7 kN
(shear at B)
Problem 1.8-9 A ship’s spar is attached at the base of a mast by a pin connection
(see figure). The spar is a steel tube of outer diameter d2 ⫽ 80 mm and inner diameter
d1 ⫽ 70 mm. The steel pin has diameter d ⫽ 25 mm, and the two plates connecting the
spar to the pin have thickness t ⫽ 12 mm. The allowable stresses are as follows: compressive stress in the spar, 75 MPa; shear stress in the pin, 50 MPa; and bearing stress
between the pin and the connecting plates, 120 MPa.
Determine the allowable compressive force Pallow in the spar.
Mast
Pin
P
Spar
Connecting
plate
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SECTION 1.8
89
Allowable Stresses and Allowable Loads
Solution 1.8-9
ALLOWABLE LOAD P BASED UPON SHEAR IN THE PIN
(DOUBLE SHEAR)
tallow ⫽ 50 MPa
As ⫽ 2(pd2/4) ⫽ (p /2)(25 mm)2 ⫽ 981.7 mm2
P2 ⫽ tallowAs ⫽ (50 MPa)(981.7 mm2) ⫽ 49.1 kN
ALLOWABLE LOAD P BASED UPON BEARING
Spar: d2 ⫽ 80 mm
d1 ⫽ 70 mm
Pin: d ⫽ 25 mm
Plates: t ⫽ 12 mm
ALLOWABLE LOAD P BASED UPON COMPRESSION IN THE
sb ⫽ 120 MPa
Ab ⫽ 2dt ⫽ 2(25 mm)(12 mm) ⫽ 600 mm2
P3 ⫽ sbAb ⫽ (120 MPa)(600 mm2) ⫽ 72.0 kN
ALLOWABLE COMPRESSIVE LOAD IN THE SPAR
SPAR
Shear in the pin governs.
sc ⫽ 75 MPa
Pallow ⫽ 49.1 kN
;
p
p
Ac ⫽ a b(d22 ⫺ d21) ⫽ a b [(80 mm)2 ⫺ (70 mm)2]
4
4
⫽ 1178.1 mm2
P1 ⫽ scAc ⫽ (75 MPa)(1174.1 mm2) ⫽ 88.4 kN
Problem 1.8-10 What is the maximum possible value of the
clamping force C in the jaws of the pliers shown in the figure if
the ultimate shear stress in the 5-mm-diameter pin is 340 MPa?
What is the maximum permissible value of the applied load
P if a factor of safety of 3.0 with respect to failure of the pin is
to be maintained?
P
y
15
mm
90⬚
38
Ry
50⬚
x
Rx 140⬚
b=
10⬚
mm
90⬚
P
C
Pin
m
C
50 mm
125 m
a
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
90
CHAPTER 1
Tension, Compression, and Shear
Solution 1.8-10
NUMERICAL DATA
FS ⫽ 3
ta ⫽
tu
ta ⫽
FS
tu ⫽ 340 MPa
tu
FS
ta ⫽ 113.333 MPa
Find Pmax:
d ⫽ 5 mm
Pmax ⫽
ta ⫽
2Rx 2 + Ry 2
As
⬍ pin at C in single shear
Rx ⫽ ⫺C cos (40⬚)
Ry ⫽ P ⫹ C sin (40⬚)
a ⫽ 50 cos (40⬚) ⫹ 125
a ⫽ 163.302 mm
b ⫽ 38 mm
a Mpin ⫽ 0
STATICS
Rx ⫽ ⫺
P(a)
cos(40⬚)
b
2
2
a
a
c⫺ cos (40⬚) d + c1 + sin(40⬚) d
C b
b
Pmax ⫽ 445 N
here
;
a
⫽ 4.297 ⬍ a/b ⫽ mechanical advantage
b
FIND MAXIMUM CLAMPING FORCE
P(a)
C⫽
b
Ry ⫽ Pc1 +
a
sin(40⬚) d
b
2
2
a
a
P c⫺ cos140⬚2 d + c1 + sin(40⬚) d ⫽ ta A s
C b
b
As ⫽
ta As
a
Cult ⫽ PmaxFS a b
b
Pult ⫽ PmaxFS
Cult ⫽ 5739 N
;
Pult ⫽ 1335
Cult
⫽ 4.297
Pult
p 2
d
4
Problem 1.8-11 A metal bar AB of weight W is suspended by a system of
steel wires arranged as shown in the figure. The diameter of the wires is 2 mm,
and the yield stress of the steel is 45 MPa.
Determine the maximum permissible weight Wmax for a factor of safety of
1.9 with respect to yielding.
0.6 m
0.6 m
2.1 m
1.5 m
1.5 m
W
A
B
Solution 1.8-11
NUMERICAL DATA
d ⫽ 2 mm
sa ⫽
sY
FSy
sY ⫽ 450 MPa
sa ⫽ 236.842 MPa
FORCES IN WIRES AC, EC, BD, FD
FSy ⫽ 1.9
h1 ⫽ 0.6 m h2 ⫽ 2.1 m v1 ⫽ 1.5 m
a Fv ⫽ 0
at A, B, E, or F
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.8
Fw ⫽
2h21 + v21 W
v1
2
Wmax ⫽
2 v1
2h21
+
v21
Allowable Stresses and Allowable Loads
91
CHECK ALSO FORCE IN WIRE CD
2h21 + v21
⫽ 0.539
2v1
©FH ⫽ 0
sY p 2
d
FSy 4
at C or D
2
Fw b
22 + 52
2
222 + 52 W
FCD ⫽ 2c
a
*
bd
5
2
222 + 52
2
less than FAC so AC controls
FCD ⫽ W
5
FCD ⫽ 2a
Wmax ⫽ 1.382 kN
2
Problem 1.8-12 A plane truss is subjected to loads 2P and P at joints B and C, respectively, as shown in the figure
part (a). The truss bars are made of two L102 ⫻ 76 ⫻ 6.4 steel angles [see Table F-5(b): cross-sectional area of the two
angles, A ⫽ 2180 mm2, figure part (b)] having an ultimate stress in tension equal to 390 MPa. The angles are connected
to an 12 mm-thick gusset plate at C [figure part (c)] with 16-mm diameter rivets; assume each rivet transfers an equal
share of the member force to the gusset plate. The ultimate stresses in shear and bearing for the rivet steel are 190 MPa
and 550 MPa, respectively.
Determine the allowable load Pallow if a safety factor of 2.5 is desired with respect to the ultimate load that can be
carried. (Consider tension in the bars, shear in the rivets, bearing between the rivets and the bars, and also bearing between
the rivets and the gusset plate. Disregard friction between the plates and the weight of the truss itself.)
F
FCF
G
a
a
A
B
a
FCG
Truss bars
C
a
a
D
Gusset
plate
Rivet
C
P
2P
a
FBC
FCD
(a)
P
(c)
Gusset plate
6.4 mm
12 mm
Rivet
(b) Section a–a
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92
CHAPTER 1
Tension, Compression, and Shear
Solution 1.8-12
PCG ⫽ 45.8 kN ; < so shear in rivets in CG and CD
controls Pallow here
NUMERICAL DATA
A ⫽ 2180 mm
2
tg ⫽ 12 mm
dr ⫽ 16 mm
su ⫽ 390 MPa
tu ⫽ 190 MPa
sbu ⫽ 550 MPa
sa ⫽
su
FS
tang ⫽ 6.4 mm
tu
FS
sba ⫽
sbu
FS
MEMBER FORCES FROM TRUSS ANALYSIS
5
FBC ⫽ P
3
4
FCD ⫽ P
3
22
⫽ 0.471
3
22
FCF ⫽
P
3
Pallow FOR TENSION ON NET SECTION IN TRUSS BARS
Anet ⫽ 1975 mm2
Anet
⫽ 0.906
A
⬍ allowable force in a member
so BC controls since it has the largest
member force for this loading
3
3
Pallow ⫽ FBCmax
Pallow ⫽ (sa Anet)
5
5
Pallow ⫽ 184.879 kN
Next, Pallow for shear in rivets (all are in double shear)
Fmax
⫽ ta As
N
⬍ for one rivet in DOUBLE shear
N ⫽ number of rivets in a particular
member (see drawing of connection
detail)
3
PBC ⫽ 3a b (ta As)
5
PCF ⫽ 2a
3
22
3
PBC ⫽ 3a b(sba Ab)
5
3
22
b(sba Ab)
3
PCG ⫽ 2a b(sba Ab)
4
3
PCD ⫽ 2a b(sba Ab)
4
PBC ⫽ 81.101 kN
PCF ⫽ 191.156 kN
PCG ⫽ 67.584 kN
PCD ⫽ 67.584 kN
Finally, Pallow for bearing of rivets on gusset plate
Fallow ⫽ saAnet
p
As ⫽ 2 dr 2
4
;
Fmax
⫽ sba Ab
N
PCF ⫽ 2a
4
FCG ⫽ P
3
Anet ⫽ A ⫺ 2drtang
PCD ⫽ 45.8 kN
Next, Pallow for bearing of rivets on truss bars
Ab ⫽ 2drtang ⬍ rivet bears on each angle in two angle
pairs.
FS ⫽ 2.5
ta ⫽
3
PCD ⫽ 2 a b(ta As)
4
b (ta As)
PBC ⫽ 55.0 kN
Ab ⫽ drtg
(bearing area for each rivet on gusset plate)
tg ⫽ 12 mm ⬍ 2tang ⫽ 12.8 mm
so gusset will control over angles
3
PBC ⫽ 3a b(sba Ab)
5
PCF ⫽ 2 a
3
22
b(sba Ab)
3
PCG ⫽ 2a b(sba Ab)
4
3
PCD ⫽ 2a b(sba Ab)
4
PBC ⫽ 76.032 kN
PCF ⫽ 179.209 kN
PCG ⫽ 63.36 kN
PCD ⫽ 63.36 kN
So, shear in rivets controls: Pallow ⫽ 45.8 kN
;
PCF ⫽ 129.7 kN
3
PCG ⫽ 2 a b(ta As)
4
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SECTION 1.8
Problem 1.8-13
A solid bar of circular cross section
(diameter d) has a hole of diameter d/5 drilled laterally through
the center of the bar (see figure). The allowable average tensile
stress on the net cross section of the bar is ␴allow.
93
Allowable Stresses and Allowable Loads
d
d/5
P
d/5
P
d
(a) Obtain a formula for the allowable load Pallow that the bar
can carry in tension.
(b) Calculate the value of Pallow if the bar is made of brass
with diameter d ⫽ 45 mm and ␴allow ⫽ 83 MPa.
(Hint: Use the formulas of Case 15, Appendix D.)
Solution 1.8-13
NUMERICAL DATA
d ⫽ 45 mm
Pa ⫽ sa c
sa ⫽ 83 MPa
1
2
arccosa b ⫺
16
5
25
(a) FORMULA FOR Pallow IN TENSION
From Case 15, Appendix D:
A ⫽ 2r2 aa ⫺
ab
r2
b
r⫽
1 2
1
2
d aarccosa b ⫺
16b d
2
5
25
2
d
2
a
a ⫽ arccosa b r ⫽ 0.023 m
r
a ⫽ 78.463⬚
a⫽
d
10
a ⫽ 4.5 * 10⫺3 m
Pa ⫽ sa(0.587d 2)
⫽ 0.587
;
(b) EVALUATE NUMERICAL RESULT
d ⫽ 0.045 m
sa ⫽ 83 MPa
Pa ⫽ 98.7 kN
;
b ⫽ 2r2 ⫺ a2
b⫽
b⫽
d 2
d 2
ca b ⫺ a b d
B 2
10
6
a d2 b
B 25
b⫽
d
16
5
Pa ⫽ sa A
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94
CHAPTER 1
Tension, Compression, and Shear
Problem 1.8-14 A solid steel bar of diameter d1 ⫽ 60 mm has a
hole of diameter d2 ⫽ 32 mm drilled through it (see figure). A steel
pin of diameter d2 passes through the hole and is attached to supports.
Determine the maximum permissible tensile load Pallow in the bar
if the yield stress for shear in the pin is tY ⫽ 120 MPa, the yield stress
for tension in the bar is sY ⫽ 250 MPa and a factor of safety of 2.0 with
respect to yielding is required. (Hint: Use the formulas of Case 15,
Appendix D.)
d2
d1
d1
P
Solution 1.8-14
SHEAR AREA (DOUBLE SHEAR)
NUMERICAL DATA
d1 ⫽ 60 mm
p
As ⫽ 2a d 2 2 b
4
d2 ⫽ 32 mm
tY ⫽ 120 MPa
sY ⫽ 250 MPa
As ⫽ 1608 mm2
FSy ⫽ 2
NET AREA IN TENSION (FROM APPENDIX D)
ALLOWABLE STRESSES
Anet ⫽ 2a
ta ⫽
tY
FSy
ta ⫽ 60 MPa
sa ⫽
sY
FSy
sa ⫽ 125 MPa
From Case 15, Appendix D:
A ⫽ 2r aa ⫺
2
a⫽
d2
2
ab
r2
b
E arccosa
r⫽
d1
2
d2
d 2/2
⫽ arccos
a ⫽ arccos
d1/2
d1
b ⫽ 2r 2 ⫺ a2
d1 2
b
2
d2
b ⫺
d1
d2
d1 2
d2 2
B a b ⫺ a b R
2 C 2
2
d1 2
a b
2
U
Anet ⫽ 1003 mm2
Pallow in tension: smaller of values based on either shear
or tension allowable stress * appropriate area
Pa1 ⫽ ta As Pa1 ⫽ 96.5 kN 6 shear governs
Pa2 ⫽ saAnet
;
Pa2 ⫽ 125.4 kN
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.8
Allowable Stresses and Allowable Loads
Problem 1.8-15 A sign of weight W is supported at its base
by four bolts anchored in a concrete footing. Wind pressure p
acts normal to the surface of the sign; the resultant of the
uniform wind pressure is force F at the center of pressure. The
wind force is assumed to create equal shear forces F/4 in the
y direction at each bolt (see figure parts a and c). The overturning effect of the wind force also causes an uplift force R at
bolts A and C and a downward force (–R) at bolts B and D
(see figure part b). The resulting effects of the wind, and the
associated ultimate stresses for each stress condition, are:
normal stress in each bolt (␴u ⫽ 410 MPa); shear through the
base plate (␶u ⫽ 115 MPa); horizontal shear and bearing on
each bolt (thu ⫽ 170 MPa and sbu ⫽ 520 MPa); and bearing on
the bottom washer at B (or D) (sbw ⫽ 340 MPa).
Find the maximum wind pressure pmax (Pa) that can be
carried by the bolted support system for the sign if a safety
factor of 2.5 is desired with respect to the ultimate wind load
that can be carried.
Use the following numerical data: bolt db ⫽ 19 mm; washer
dw ⫽ 38 mm; base plate tbp ⫽ 25 mm; base plate dimensions
h ⫽ 350 mm and b ⫽ 300 mm; W ⫽ 2.25 kN; H ⫽ 5.2 m; sign
dimensions (Lv ⫽ 3 m * Lh ⫽ 3.7 m); pipe column diameter
d ⫽ 150 mm, and pipe column thickness t ⫽ 10 mm.
95
Sign (Lv ⫻ Lh)
Resultant
of wind
pressure
Lh
2
C.P.
F
W
Pipe
column
z b
2
D
H
Lv
y Overturning
moment
B about x-axis
FH
C
A
F
at each
4
bolt
h
W
at each
4
bolt
x
(a)
W
Pipe column
db
dw
FH
— = Rh
2
One half of overturning
moment about x-axis acts
on each bolt pair
z
B
A
y
Footing
F/4
Tension
h
R
Base
plate (tbp)
R
Compression
(b)
z
mm
R⫹
F
4
W
4
x
A
F
4 R⫺ W
4
y
B
mm
FH
—
2
b=
300
C
h=
0
35
D
FH
2
R⫺
W
4
(c)
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96
CHAPTER 1
Tension, Compression, and Shear
Solution 1.8-15
(1) COMPUTE pmax BASED ON NORMAL STRESS IN EACH BOLT
(GREATER AT B AND D)
Numerical Data
su ⫽ 410 MPa
tu ⫽ 115 MPa
sbu ⫽ 520 MPa
thu ⫽ 170 MPa
sbw ⫽ 340 MPa
FSu ⫽ 2.5
db ⫽ 19 mm
dw ⫽ 38 mm
tbp ⫽ 25 mm
h ⫽ 350 mm
b ⫽ 300 mm
d ⫽ 150 mm
t ⫽ 10 mm
H ⫽ 5.2 mm
Lh ⫽ 3.7 m
sba ⫽
sbu
FSu
tu
ta ⫽
FSu
sbwa ⫽
thu
tha ⫽
FSu
sbw
FSu
FORCES F AND R IN TERMS OF pmax
F ⫽ pmaxLvLh
R ⫽ pmax
Lv Lh H
2h
R⫽
W
4
W
p
Rmax ⫽ sa a db 2 b ⫺
4
4
p 2
d
4 b
p
W
sa a db 2 b ⫺
4
4
⫽
Lv Lh H
2h
pmax1 ⫽ 557 Pa
ALLOWABLE STRESSES (MPa)
su
sa ⫽
FSu
s⫽
pmax1
W ⫽ 2.25 kN
Lv ⫽ 3 m
R +
controls
;
(2) COMPUTE pmax BASED ON SHEAR THROUGH BASE PLATE
(GREATER AT B AND D)
W
4
t⫽
p dw tbp
R +
Rmax ⫽ ta(p dw tbp) ⫺
FH
2h
pmax2 ⫽
W
4
ta1p dw tbp2 ⫺
W
4
Lv Lh H
2h
pmax2 ⫽ 1.658 ⫻ 103 Pa
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SECTION 1.8
(3) COMPUTE pmax BASED ON HORIZONTAL SHEAR
ON EACH BOLT
th ⫽
F
4
p
a db 2 b
4
pmax3 ⫽
Fmax
p
⫽ 4tha a db 2 b
4
tha(p db 2)
Lv Lh
pmax3 ⫽ 6.948 ⫻ 103 Pa
(4) COMPUTE pmax BASED ON HORIZONTAL BEARING ON
EACH BOLT
F
4
sb ⫽
(tbp db)
pmax4 ⫽
97
Allowable Stresses and Allowable Loads
(5) COMPUTE pmax BASED ON BEARING UNDER THE TOP WASHER
AT A (OR C) AND THE BOTTOM WASHER AT B (OR D)
R +
sbw ⫽
W
4
p
A d 2 ⫺ db 2 B
4 w
Rmax ⫽ sbwa c
W
p
A dw2 ⫺ db2 B d ⫺
4
4
p
W
A dw2 ⫺ db2 B d ⫺
4
4
pmax5 ⫽
Lv Lh H
2h
pmax5 ⫽ 1.396 * 103 Pa
sbwa c
Fmax ⫽ 4sba(tbpdb)
4sba(tbpdb)
Lv Lh
pmax4 ⫽ 3.56 ⫻ 104 Pa
Problem 1.8-16 The piston in an engine is attached to a
connecting rod AB, which in turn is connected to a crank arm
BC (see figure). The piston slides without friction in a cylinder
and is subjected to a force P (assumed to be constant) while
moving to the right in the figure. The connecting rod, which has
diameter d and length L, is attached at both ends by pins. The
crank arm rotates about the axle at C with the pin at B moving
in a circle of radius R. The axle at C, which is supported by
bearings, exerts a resisting moment M against the crank arm.
Cylinder
P
Piston
Connecting rod
A
M
d
C
B
L
R
(a) Obtain a formula for the maximum permissible force
Pallow based upon an allowable compressive stress ␴c in
the connecting rod.
(b) Calculate the force Pallow for the following data:
sc ⫽ 160 MPa, d ⫽ 9.00 mm, and R ⫽ 0.28L.
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98
CHAPTER 1
Tension, Compression, and Shear
Solution 1.8-16
The maximun allowable force P occurs when cos ␣ has its
smallest value, which means that ␣ has its largest value.
LARGEST VALUE OF ␣
d ⫽ diameter of rod AB
FREE-BODY DIAGRAM OF PISTON
The largest value of ␣ occurs when point B is the farthest
distance from line AC. The farthest distance is the radius R
of the crank arm.
P ⫽ applied force (constant)
C ⫽ compressive force in connecting rod
RP ⫽ resultant of reaction forces between cylinder
and piston (no friction)
:
a Fhoriz ⫽ 0 ⫹
;
⫺
P ⫺ C cos ␣ ⫽ 0
P ⫽ C cos ␣
Therefore,
—
BC ⫽ R
—
Also, AC ⫽ 2L2⫺R2
cos a ⫽
R 2
2L2⫺R2
⫽ A 1⫺a b
L
L
(a) MAXIMUM ALLOWABLE FORCE P
Pallow ⫽ ␴c Ac cos ␣
⫽ sc a
pd2
4
b
R 2
1⫺a b
A
L
;
(b) SUBSTITUTE NUMERICAL VALUES
MAXIMUM COMPRESSIVE FORCE C IN CONNECTING ROD
␴c ⫽ 160 MPa
Cmax ⫽ ␴cAc
R ⫽ 0.28L
in which Ac ⫽ area of connecting rod
Pallow ⫽ 9.77 kN
d ⫽ 9.00 mm
R/L ⫽ 0.28
;
pd2
Ac ⫽
4
MAXIMUM ALLOWABLE FORCE P
P ⫽ Cmax cos ␣
⫽ ␴c Ac cos ␣
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SECTION 1.9
Design for Axial Loads and Direct Shear
99
Design for Axial Loads and Direct Shear
Problem 1.9-1 An aluminum tube is
d
required to transmit an axial tensile force
P ⫽ 148 kN (see figure part a). The thickness
of the wall of the tube is to be 6 mm.
P
(a) What is the minimum required outer
diameter dmin if the allowable tensile
stress is 84 MPa?
(b) Repeat part (a) if the tube will have a
hole of diameter d/10 at mid-length
(see figure parts b and c).
P
(a)
d
Hole of diameter d/10
d/10
P
P
d
(b)
(c)
Solution 1.9-1
(b) MIN. DIAMETER OF TUBE (WITH HOLES)
NUMERICAL DATA
P ⫽ 148 kN
t ⫽ 6 mm
sa ⫽ 84 MPa
A1 ⫽ c
(a) MIN. DIAMETER OF TUBE (NO HOLES)
A1 ⫽
p 2
C d ⫺1d⫺2t22 D
4
A2 ⫽
P
sa
A1 ⫽ d apt ⫺
A2 ⫽ 1.762 ⫻ 10⫺3m2
P
+ t
psat
d ⫽ 99.5 mm
t
b ⫺ pt2
5
Equating A1 and A2 and solving for d:
Equating A1 and A2 and solving for d:
d⫽
d
p 2
C d ⫺1d⫺2t22 D ⫺2a bt d
4
10
;
P
+ p t2
sa
d⫽
t
pt ⫺
5
d ⫽ 106.2 mm
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
100
CHAPTER 1
Tension, Compression, and Shear
Problem 1.9-2 A copper alloy pipe having yield stress
sY ⫽ 290 MPa is to carry an axial tensile load P ⫽ 1500 kN
[see figure part (a)]. A factor of safety of 1.8 against yielding
is to be used.
d
t =—
8
P
(a) If the thickness t of the pipe is to be one-eighth of its
outer diameter, what is the minimum required outer
diameter dmin?
(b) Repeat part (a) if the tube has a hole of diameter d/10
drilled through the entire tube as shown in the figure
[part (b)].
d
(a)
P
Hole of diameter d/10
d
t =—
8
d
(b)
Solution 1.9-2
NUMERICAL DATA
Equate A1 and A2 and solve for d:
sY ⫽ 290 MPa
FSy ⫽ 1.8
P
7
s
Y
d ⫽
64p P
FSy Q
(a) MINIMUM DIAMETER (NO HOLES)
dmin ⫽
2
P ⫽ 1500 kN
A1 ⫽
A1 ⫽
A2 ⫽
d 2
p 2
cd ⫺ad ⫺ b d
4
4
7
p d2
64
P
sY
FSy
7
64p
R
P
sY
P FSy Q
dmin ⫽ 164.6 mm
;
A2 ⫽ 9.31 ⫻ 103 mm2
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SECTION 1.9
(b) MINIMUM DIAMETER (WITH HOLES)
A1 ⫽
d2 ⫽
a
1 2
7
pd 2 ⫺
d
64
40
A1 ⫽ d 2 a
7
1
p⫺
b
64
40
7
1
p⫺
⫽ 0.319
64
40
7
1
p⫺
b
64
40
dmin ⫽
P
sY
P
Q
FSy
(a) If the allowable shear stress in the bolt is 90 MPa, what is
the minimum required diameter dmin of the bolt at C?
(b) If the allowable bearing stress in the bolt is 130 MPa, what is
the minimum required diameter dmin of the bolt at C?
dmin ⫽ 170.9 mm
7
1
a p⫺
b
64
40
c
Problem 1.9-3 A horizontal beam AB with cross-sectional
dimensions (b ⫽ 19 mm) ⫻ (h ⫽ 200 mm) is supported by an
inclined strut CD and carries a load P ⫽ 12 kN at joint B (see figure
part a). The strut, which consists of two bars each of thickness 5b/8,
is connected to the beam by a bolt passing through the three bars
meeting at joint C (see figure part b).
101
P
sY
P
Q
FSy
Redefine A1—subtract area for two holes—then
equate to A2:
d
d 2
d
p
A1 ⫽ c cd2⫺ad ⫺ b d⫺2a b a b d
4
4
10 8
Design for Axial Loads and Direct Shear
1.2 m
;
1.5 m
B
C
A
0.9 m
P
D
(a)
b
Beam AB (b ⫻ h)
h
—
2
Bolt (dmin)
h
—
2
5b
—
8
Strut CD
(b)
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102
CHAPTER 1
Tension, Compression, and Shear
Solution 1.9-3
NUMERICAL DATA
P ⫽ 12 kN
ta ⫽ 90 MPa
(b) dmin BASED ON ALLOWABLE BEARING AT JOINT C
b ⫽ 19 mm
h ⫽ 200 mm
sba ⫽ 130 MPa
Bearing from beam ACB:
(a) dmin BASED ON ALLOWABLE SHEAR—DOUBLE SHEAR
IN STRUT
ta ⫽
FDC
As
FDC ⫽
dmin
p 2
d b
4
sb ⫽ 3
15
P
4
⫽
p
t a b
a a 2
15P/4
bsba
15P/4
bd
dmin ⫽ 18.22 mm
;
15
P
4
Bearing from strut DC: sb ⫽
5
2a bbd
8
15
P
4
ta ⫽ 90 MPa
As ⫽ 2a
dmin ⫽
sb ⫽
dmin ⫽ 17.84 mm
P
bd
(lower than ACB)
;
Problem 1.9-4 Lateral bracing for an elevated pedestrian walkway is shown in the figure part (a). The thickness of the clevis
plate tc ⫽ 16 mm and the thickness of the gusset plate tg ⫽ 20 mm [see figure part (b)]. The maximum force in the diagonal
bracing is expected to be F ⫽ 190 kN.
If the allowable shear stress in the pin is 90 MPa and the allowable bearing stress between the pin and both the clevis and
gusset plates is 150 MPa, what is the minimum required diameter dmin of the pin?
Clevis
Gusset plate
Gusset plate
tc
Pin
tg
Cl
ev
is
dmin
Diagonal brace
F
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SECTION 1.9
103
Design for Axial Loads and Direct Shear
Solution 1.9-4
(2) dmin BASED ON ALLOW BEARING IN GUSSET AND CLEVIS
NUMERICAL DATA
F ⫽ 190 kN
ta ⫽ 90 MPa
tg ⫽ 20 mm
tc ⫽ 16 mm
sba ⫽ 150 MPa
PLATES
Bearing on gusset plate:
(1) dmin BASED ON ALLOW SHEAR—DOUBLE SHEAR
IN PIN
sb ⫽
F
Ab
dmin ⫽
F
p
t a b
Q a 2
dmin ⫽
Problem 1.9-5 A plane truss has joint loads P, 2P, and 3P at
joints D, C, and B, respectively (see figure) where load variable
P ⫽ 23 kN. All members have two end plates (see figure for
Prob. 1.7-2) which are pin-connected to gusset plates (see
also figure for Prob. 1.8-12). Each end plate has thickness
tp ⫽ 16 mm, and all gusset plates have thickness tg ⫽ 28 mm.
If the allowable shear stress in each pin is 83 MPa and the
allowable bearing stress in each pin is 124 MPa, what is the
minimum required diameter dmin of the pins used at either end
of member BE?
F
2tcsba
dmin ⫽ 39.6 mm
3m
3m
B
P
2P
3P
A
;
Ab ⫽ d(2tc)
Bearing on clevis:
dmin ⫽ 36.7 mm
F
tgsba
6 controls
dmin ⫽ 63.3 mm
p
As ⫽ 2a d2 b
4
F
t⫽
As
dmin ⫽
Ab ⫽ tgd
C
3m
D
3m
4.5 m
E
F
1m
G
Typical gusset plate
Solution 1.9-5
P ⫽ 23 kN
tp ⫽ 16 mm
FBE ⫽ 3.83858P ⫽ 88.287 kN ⬍ from plane truss analysis
(see Probs. 1.2-4 to 1.2-6)
tg ⫽ 28 mm
tp ⫽ 16 mm
2tp ⫽ 32 mm
ta ⫽ 83 MPa
sba ⫽ 124 MPa
PIN DIAMETER BASED ON ALLOWABLE SHEAR STRESS (PINS IN DOUBLE SHEAR)
dp1
FBE
2
⫽
⫽ 26.023 mm
p
t
a4 a
< controls
dmin ⫽ dp1 ⫽ 26 mm
PIN DIAMETER BASED ON BEARING BETWEEN PIN AND EACH OF TWO END PLATES
dp2 ⫽
⬍ 2tp is greater than tg so gusset will
control
FBE
⫽ 22.25 mm
2 tp sba
PIN DIAMETER BASED ON BEARING BETWEEN PIN AND GUSSET PLATE
dp3 ⫽
FBE
⫽ 25.428 mm
tg sba
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104
CHAPTER 1
Tension, Compression, and Shear
Problem 1.9-6 A suspender on a suspension bridge consists of a cable that
passes over the main cable (see figure) and supports the bridge deck, which is
far below. The suspender is held in position by a metal tie that is prevented from
sliding downward by clamps around the suspender cable.
Let P represent the load in each part of the suspender cable, and let u
represent the angle of the suspender cable just above the tie. Finally, let sallow
represent the allowable tensile stress in the metal tie.
Main
cable
Suspender
Collar
(a) Obtain a formula for the minimum required cross-sectional area of the tie.
(b) Calculate the minimum area if P ⫽ 130 kN, u ⫽ 75°, and sallow ⫽ 80 MPa.
u
u
Tie
Clamp
P
Solution 1.9-6
P
Suspender tie on a suspension bridge
F ⫽ tensile force in cable
above tie
P ⫽ tensile force in cable
below tie
sallow ⫽ allowable tensile
stress in the tie
(a) MINIMUM REQUIRED AREA OF TIE
Amin ⫽
Pcotu
T
⫽
sallow
sallow
;
(b) SUBSTITUTE NUMERICAL VALUES:
P ⫽ 130 kN
u ⫽ 75°
sallow ⫽ 80 MPa
Amin ⫽ 435 mm2
;
FREE-BODY DIAGRAM OF HALF THE TIE
Note: Include a small amount of the cable in the free-body
diagram.
T ⫽ tensile force in the tie
FORCE TRIANGLE
cotu ⫽
T
P
T ⫽ Pcot u
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SECTION 1.9
Problem 1.9-7 A square steel tube of length L ⫽ 6 m and width
b2 ⫽ 250 mm is hoisted by a crane (see figure). The tube hangs
from a pin of diameter d that is held by the cables at points A and
B. The cross section is a hollow square with inner dimension
b1 ⫽ 210 mm and outer dimension b2 ⫽ 250 mm. The allowable
shear stress in the pin is 60 MPa, and the allowable bearing stress
between the pin and the tube is 90 MPa.
Determine the minimum diameter of the pin in order to support the weight of the tube. (Note: Disregard the rounded corners
of the tube when calculating its weight.)
105
Design for Axial Loads and Direct Shear
d
A
B
Square
tube
Square
tube
Pin
L
d
b2
A
B
b1
b2
Solution 1.9-7 Tube hoisted by a crane
T ⫽ tensile force in cable
W ⫽ weight of steel tube
d ⫽ diameter of pin
b1 ⫽ inner dimension of tube
⫽ 210 mm
W ⫽ gsAL
⫽ (77.0 kN/m3)(18,400 mm2)(6.0 m)
⫽ 8.501 kN
DIAMETER OF PIN BASED UPON SHEAR
Double shear: 2tallow Apin ⫽ W
b2 ⫽ outer dimension of tube
⫽ 250 mm
L ⫽ length of tube ⫽ 6.0 m
2(60 MPa)a
pd 2
b ⫽ 8.501 kN
4
d2 ⫽ 90.20 mm2
d1 ⫽ 9.497 mm
tallow ⫽ 60 MPa
sb ⫽ 90 MPa
WEIGHT OF TUBE
gs ⫽ weight density of steel
⫽ 77.0 kN/m3
A ⫽ area of tube
⫽ b22 ⫺ b21 ⫽ 18,400 mm2
DIAMETER OF PIN BASED UPON BEARING
sb(b2 ⫺ b1)d ⫽ W
(90 MPa)(40 mm)d ⫽ 8.501 kN
d2 ⫽ 2.361 mm
MINIMUM DIAMETER OF PIN
Shear governs.
dmin ⫽ 9.50 mm
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106
CHAPTER 1
Tension, Compression, and Shear
Problem 1.9-8 A cable and pulley system at D is used to
bring a 230-kg pole (ACB) to a vertical position as shown in
the figure part (a). The cable has tensile force T and is attached
at C. The length L of the pole is 6.0 m, the outer diameter is
d ⫽ 140 mm, and the wall thickness t ⫽ 12 mm. The pole
pivots about a pin at A in figure part (b). The allowable shear
stress in the pin is 60 MPa and the allowable bearing stress
is 90 MPa.
Find the minimum diameter of the pin at A in order to support
the weight of the pole in the position shown in the figure part (a).
B
1.0 m
Pole
C
Cable
30⬚
Pulley
5.0 m
a
T
A
D
4.0 m
a
(a)
d
ACB
A
Pin support
plates
Ay
2
Pin
(b)
Ay
2
Solution 1.9-8
ALLOWABLE SHEAR AND BEARING STRESSES
ta ⫽ 60 MPa
sba ⫽ 90 MPa
FIND INCLINATION OF AND FORCE IN CABLE, T
Let a ⫽ angle between pole and cable at C; use law
of cosines.
DC ⫽
A
52 + 42 ⫺ 2(5)(4)cos a120 p b
180
DC ⫽ 7.81 m
a ⫽ 26.33⬚
u ⫽ 33.67⬚
a ⫽ arccosc
u ⫽ 60 ⫺ a
52 + DC 2 ⫺ 42
d
2DC(5)
< angle between cable and horizontal
at D
W ⫽ 2.256 ⫻ 103 N
W ⫽ 230 kg(9.81 m/s2)
STATICS TO FIND CABLE FORCE T
a MA ⫽ 0
W(3 sin(30º)) ⫺ TX(5 cos(30º)) ⫹
Ty(5 sin(30º)) ⫽ 0
Substitute for Tx and Ty in terms of T and solve for T:
3
W
2
T⫽
523
⫺5
sin(u)⫹
cos(u)
2
2
T ⫽ 1.53 ⫻ 103 N
Ty ⫽ T sin(u )
Tx ⫽ T cos(u )
Tx ⫽ 1.27 ⫻ 103 N
Ty ⫽ 846.11 N
(1) dmin BASED ON ALLOWABLE SHEAR–DOUBLE SHEAR AT A
Ax ⫽ ⫺Tx
Ay ⫽ Ty ⫹ W
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SECTION 1.9
CHECK SHEAR DUE TO RESULTANT FORCE ON PIN AT A
RA ⫽ 2
dmin
A2x
+
A2y
dpole ⫽ 140 mm
Lpole ⫽ 6000 mm
RA ⫽ 3.35 ⫻ 10 N
MEMBER
sb ⫽
dmin ⫽ 5.96 mm 6controls
107
(2) dmin BASED ON ALLOWABLE BEARING ON PIN
3
RA
2
⫽
p
t a b
a a 4
Design for Axial Loads and Direct Shear
;
AB BEARING ON PIN
RA
Ab
dmin ⫽
tpole ⫽ 12 mm
Ab ⫽ 2tpoled
RA
2tpole sba
dmin ⫽ 1.55 mm
Problem 1.9-9 A pressurized circular cylinder has a sealed cover plate fastened
Cover plate
with steel bolts (see figure). The pressure p of the gas in the cylinder is 1900 kPa,
the inside diameter D of the cylinder is 250 mm, and the diameter dB of the bolts
is 12 mm.
If the allowable tensile stress in the bolts is 70 MPa, find the number n of
bolts needed to fasten the cover.
Steel bolt
p
Cylinder
D
Solution 1.9-9 Pressurized cylinder
Cover plate
P⫽
Steel bolt
Ab ⫽ area of one bolt ⫽
p
sallow ⫽
D
sallow ⫽ 70 kPa
D ⫽ 250 mm
p 2
d
4 b
P ⫽ sallow Ab
Cylinder
p ⫽ 1900 kPa
ppD2
F
⫽
n
4n
n⫽
db ⫽ 12 mm
ppD2
pD2
P
⫽
⫽
Ab
(4n)(p4 )d 2b
nd 2b
pD2
d2bsallow
n ⫽ number of bolts
F ⫽ total force acting on the cover plate from the
internal pressure
2
pD
b
F ⫽ pa
4
NUMBER OF BOLTS
SUBSTITUTE NUMERICAL VALUES:
n⫽
(1900 kPa)(250 mm)2
(12 mm)2(70 MPa)
Use 12 bolts.
⫽ 11.8
;
P ⫽ tensile force in one bolt
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108
CHAPTER 1
Tension, Compression, and Shear
Problem 1.9-10 A tubular post of outer diameter d2 is guyed by two cables
fitted with turnbuckles (see figure). The cables are tightened by rotating the
turnbuckles, thus producing tension in the cables and compression in the post.
Both cables are tightened to a tensile force of 110 kN. Also, the angle between
the cables and the ground is 60°, and the allowable compressive stress in the
post is sc ⫽ 35 MPa.
If the wall thickness of the post is 15 mm, what is the minimum permissible
value of the outer diameter d2?
Cable
Turnbuckle
d2
60⬚
Solution 1.9-10
Post
60⬚
Tubular post with guy cables
d2 ⫽ outer diameter
AREA OF POST
d1 ⫽ inner diameter
A⫽
t ⫽ wall thickness
⫽ 15 mm
T ⫽ tensile force in a cable
⫽ 110 kN
sallow ⫽ 35 MPa
P ⫽ compressive force in post
⫽ 2Tcos 30°
REQUIRED AREA OF POST
A⫽
P
s allow
⫽
2Tcos 30⬚
s allow
p
p 2
(d2 ⫺ d12) ⫽ [d 22⫺(d2⫺2t)2 ]
4
4
⫽ pt (d2 ⫺ t)
EQUATE AREAS AND SOLVE FOR d2:
2Tcos 30⬚
⫽ pt (d2 ⫺ t)
sallow
d2 ⫽
2Tcos 30⬚
+ t
ptsallow
;
SUBSTITUTE NUMERICAL VALUES:
(d2)min ⫽ 131 mm
;
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SECTION 1.9
Design for Axial Loads and Direct Shear
109
Problem 1.9-11 A large precast concrete panel for a warehouse is being raised to a vertical position using two sets of
cables at two lift lines as shown in the figure part a. Cable 1 has length L1 ⫽ 6.7 m and distances along the panel (see figure
part b) are a ⫽ L1/2 and b ⫽ L1/4. The cables are attached at lift points B and D and the panel is rotated about its base at A.
However, as a worst case, assume that the panel is momentarily lifted off the ground and its total weight must be supported
by the cables. Assuming the cable lift forces F at each lift line are about equal, use the simplified model of one half of the
panel in figure part b to perform your analysis for the lift position shown. The total weight of the panel is W ⫽ 378 kN. The
orientation of the panel is defined by the following angles: ␥ =20° and u ⫽ 10°.
Find the required cross-sectional area AC of the cable if its breaking stress is 630 MPa and a factor of safety of 4 with
respect to failure is desired.
F
H
F
F
T2
b2
T1
B
a
b1
W
D
b
—
2
B
g
u
y
a
C
W
—
2
D
b
g
A
b
A
(a)
x
(b)
Solution 1.9-11
GEOMETRY
1
a ⫽ L1
2
L1 ⫽ 6.7 m
u ⫽ 10º
1
b ⫽ L1
4
a ⫹ 2.5b ⫽ 7.537 m
g ⫽ 20º
Using law of cosines:
L2 ⫽ 2(a + b)2 + L21 ⫺ 2(a + b)L1 cos(u )
b ⫽ arccosc
L21 + L22 ⫺ (a + b)2
d
2L1L2
b ⫽ 26.484⬚
b1 ⫽ p ⫺ (u ⫹ p ⫺ g)
b2 ⫽ b ⫺ b1
b 1 ⫽ 10⬚
b 2 ⫽ 16.484⬚
SOLUTION APPROACH: FIND T THEN AC ⫽ T/(sU /FS)
L2 ⫽ 1.957 m
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110
CHAPTER 1
Tension, Compression, and Shear
STATICS AT POINT H
g Fx ⫽ 0
H
So
T2 ⫽ T1
g Fy ⫽ 0
H
and
T1 ⫽ 120.257 kN
T1sin(b1) ⫽ T2sin(b2)
T2 ⫽ T1
sin(b 1)
sin(b 2)
T2 ⫽ 73.595 kN
sin(b 1)
sin(b 2)
COMPUTE REQUIRED CROSS-SECTIONAL AREA
T1cos(b 1) + T2 cos(b 2) ⫽ F
su ⫽ 630 MPa
F ⫽ W/2,
W ⫽ 378 kN
sin(b 1)
So
cos(b 2)b ⫽ F
T1 acos(b 1) +
sin(b 2)
W
2
T1 ⫽
sin(b 1)
acos(b 1) +
cos(b 2)b
sin(b 2)
Ac ⫽
T1
su
FS
FS ⫽ 4
Ac ⫽ 764 mm2
su
⫽ 157.5 MPa
FS
;
⬍ see Table 2-1: use nominal diam. 44-mm cable with
area ⫽ 948 mm2
Problem 1.9-12 A steel column of hollow circular cross
section is supported on a circular steel base plate and a concrete
pedestal (see figure). The column has outside diameter
d ⫽ 250 mm and supports a load P ⫽ 750 kN.
(a) If the allowable stress in the column is 55 MPa, what is
the minimum required thickness t? Based upon your
result, select a thickness for the column. (Select a thickness that is an even integer, such as 10, 12, 14, . . . , in
units of millimeters.)
(b) If the allowable bearing stress on the concrete pedestal
is 11.5 MPa, what is the minimum required diameter D
of the base plate if it is designed for the allowable load
Pallow that the column with the selected thickness can
support?
d
P
Column
P
Base plate
t
D
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SECTION 1.9
Solution 1.9-12
Design for Axial Loads and Direct Shear
111
Hollow circular column
SUBSTITUTE NUMERICAL VALUES IN EQ. (1):
t2 ⫺ 250t +
(750 * 103 N)
p(55 N/mm2)
⫽0
(Note: In this eq., t has units of mm.)
t2 ⫺ 250t ⫹ 4340.6 ⫽ 0
Solve the quadratic eq. for t:
d ⫽ 250 mm
(a) THICKNESS t OF THE COLUMN
2
pt ⫺ ptd +
t2 ⫺ td +
P
sallow
p
(4t)(d ⫺ t) ⫽ pt(d ⫺ t)
4
P
D2 ⫽
⫽
sallow
Pallow
pD2
⫽
4
sb
4s allowt(d ⫺ t)
sb
4(55 MPa)(20 mm)(230 mm)
11.5 MPa
D2 ⫽ 88,000 mm2
⫽0
P
⫽0
psallow
Area of base plate ⫽
sallowpt(d ⫺ t)
pD2
⫽
4
sb
pd2
p
A⫽
⫺ (d ⫺ 2t)2
4
4
pt(d ⫺ t) ⫽
Pallow ⫽ sallow A
A ⫽ pt(d ⫺ t) Pallow ⫽ sallow pt(d ⫺ t)
sb ⫽ 11.5 MPa (allowable pressure on concrete)
⫽
;
;
where A is the area of the column with t ⫽ 20 mm.
D ⫽ diameter of base plate
s allow
Use t ⫽ 20 mm
For the column,
t ⫽ thickness of column
A⫽
tmin ⫽ 18.8 mm
(b) DIAMETER D OF THE BASE PLATE
P ⫽ 750 kN
sallow ⫽ 55 MPa (compression in column)
P
t ⫽ 18.77 mm
Dmin ⫽ 297 mm
D ⫽ 296.6 mm
;
(Eq. 1)
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112
CHAPTER 1
Tension, Compression, and Shear
Problem 1.9-13 An elevated jogging track is supported at
intervals by a wood beam AB (L ⫽ 2.3 m) which is pinned at A
and supported by steel rod BC and a steel washer at B. Both the
rod (dBC ⫽ 5 mm) and the washer (dB ⫽ 25 mm) were
designed using a rod tension force of TBC ⫽ 1890 N. The rod
was sized using a factor of safety of 3 against reaching the
ultimate stress ␴u ⫽ 410 MPa. An allowable bearing stress
␴ba ⫽ 3.9 MPa was used to size the washer at B.
Now, a small platform HF is to be suspended below a
section of the elevated track to support some mechanical and
electrical equipment. The equipment load is uniform load
q ⫽ 730 N/m and concentrated load WE ⫽ 780 N at mid-span
of beam HF. The plan is to drill a hole through beam AB at D
and install the same rod (dBC) and washer (dB) at both D and F
to support beam HF.
(a) Use ␴u and ␴ba to check the proposed design for rod DF
and washer dF; are they acceptable?
(b) Also re-check the normal tensile stress in rod BC and
bearing stress at B; if either is inadequate under the additional load from platform HF, redesign them to meet the
original design criteria.
Original
structure
C
Steel rod,
TBC =
1890 N
dBC = 5 mm
L
—
25
L = 2.3 m
Wood beam supporting track
A
D
B
Washer
dB = 25 mm
New steel rod, dDF = 5 mm
q = 730 N/m
L
—
25
WE = 780 N
New beam to support equipment
H
L
—
2
F
Washer, dF
(same at D
above)
L
—
2
Solution 1.9-13
REVISED NORMAL STRESS IN ROD BC:
NUMERICAL DATA
L ⫽ 2.3 m
TBC ⫽ 1890 N
su ⫽ 410 MPa
q ⫽ 730
N
m
dBC ⫽ 5 mm
sa ⫽
FSu ⫽ 3
sba ⫽ 3.9 MPa
WE ⫽ 780 N
dB ⫽ 25 mm
su
FSu
sa ⫽ 136.7 MPa
TBC2
sBC2 ⫽
a
sBC2 ⫽ 158.9 MPa
dBCreqd ⫽
NORMAL STRESS IN ROD DF:
TDF
sDF ⫽
p 2
d
4 BC
sDF ⫽ 65.2 MPa
TBC2
p
sa
R4
dBCreqd ⫽ 5.391 mm
WE
©MH ⫽ 0
exceeds ␴a
SO RE-DESIGN ROD BC:
(a) FIND FORCE IN ROD DF AND FORCE ON WASHER AT F
L
L
+ qL
2
2
TDF ⫽
L
aL ⫺
b
25
p 2
d b
4 BC
dB ⫽ 25 mm
⬍ increased rod diam.
RE-CHECK BEARING STRESS IN WASHER AT B:
sbB2 ⫽
TBC2
p
c 1d2B ⫺ d2BC22 d
4
sbB2 ⫽ 6.743 MPa
^ exceeds ␴ba
SO RE-DESIGN WASHER AT B:
OK—less than ␴a; rod is
acceptable ;
dBreqd ⫽
TBC2
+ 1dBC222
p
sba
R4
dBreqd ⫽ 32.472 mm
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.9
SO USE 34-mm WASHER AT B:
Original
structure
dB ⫽ 34 mm
C
Steel rod,
dBC2 ⫽ 6 mm
c
p 2
1d ⫺ d2BC22 d
4 B
⫽ 3.55 MPa 6 less than
sba ⫽ 3.9 MPa
BEARING STRESS ON WASHER AT F:
sbF ⫽
TBC =
1890 N
dBC = 5 mm
TBC2
sbB2 ⫽
L
—
25
L = 2.3 m
Wood beam supporting track
A
D
B
Washer
dB = 25 mm
TDF
p 2
1d ⫺ d2BC2
4 B
New steel rod, dDF = 5 mm
q = 730 N/m
OK—less than ␴ba; washer is
acceptable ;
sbF ⫽ 2.72 MPa
(b) FIND NEW FORCE IN ROD BC—SUM MOMENT ABOUT A
FOR UPPER FBD—THEN CHECK NORMAL STRESS IN BC
AND BEARING STRESS AT B
H
L
—
25
WE = 780 N
New beam to support equipment
L
—
2
TBC2 ⫽
L
F
Washer, dF
(same at D
above)
L
—
2
©MA ⫽ 0
TBCL + TDF aL ⫺
113
Design for Axial Loads and Direct Shear
L
b
25
TBC2 ⫽ 3.119 kN
Problem 1.9-14 A flat bar of width b ⫽ 60 mm and thickness t ⫽ 10 mm
is loaded in tension by a force P (see figure). The bar is attached to a
support by a pin of diameter d that passes through a hole of the same size
in the bar. The allowable tensile stress on the net cross section of the bar is
sT ⫽ 140 MPa, the allowable shear stress in the pin is tS ⫽ 80 MPa, and
the allowable bearing stress between the pin and the bar is sB ⫽ 200 MPa.
(a) Determine the pin diameter dm for which the load P will be a
maximum.
(b) Determine the corresponding value Pmax of the load.
d
P
b
t
P
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114
CHAPTER 1
Tension, Compression, and Shear
Solution 1.9-14
Bar with a pin connection
SHEAR IN THE PIN
PS ⫽ 2tS Apin ⫽ 2tS a
pd 2
b
4
1
p
b
⫽ 2(80 MPa)a b(d 2)a
4
1000
⫽ 0.040pd 2 ⫽ 0.12566d 2
(Eq. 2)
BEARING BETWEEN PIN AND BAR
PB ⫽ sBtd
b ⫽ 60 mm
⫽ (200 MPa)(10 mm)(d )a
t ⫽ 10 mm
⫽ 2.0d
1
b
1000
(Eq. 3)
d ⫽ diameter of hole and pin
GRAPH OF EQS. (1), (2), AND (3)
sT ⫽ 140 MPa
tS ⫽ 80 MPa
sB ⫽ 200 MPa
UNITS USED IN THE FOLLOWING CALCULATIONS:
P is in kN
s and t are in N/mm2 (same as MPa)
b, t, and d are in mm
TENSION IN THE BAR
PT ⫽ sT (Net area) ⫽ st(t)(b ⫺ d )
⫽ (140 MPa)(10 mm)(60 mm ⫺ d) a
⫽ 1.40(60 ⫺ d)
1
b
1000
(Eq. 1)
(a) PIN DIAMETER dm
PT ⫽ PB or 1.40(60 ⫺ d) ⫽ 2.0d
84.0
Solving, dm ⫽
mm ⫽ 24.7 mm
3.4
;
(b) LOAD Pmax
Substitute dm into Eq. (1) or Eq. (3):
Pmax ⫽ 49.4 kN
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
SECTION 1.9
Design for Axial Loads and Direct Shear
Problem 1.9-15 Two bars AC and BC of the same material support a vertical load P
115
A
(see figure). The length L of the horizontal bar is fixed, but the angle u can be varied by
moving support A vertically and changing the length of bar AC to correspond with the
new position of support A. The allowable stresses in the bars are the same in tension and
compression.
We observe that when the angle u is reduced, bar AC becomes shorter but the crosssectional areas of both bars increase (because the axial forces are larger). The opposite
effects occur if the angle u is increased. Thus, we see that the weight of the structure
(which is proportional to the volume) depends upon the angle u.
Determine the angle u so that the structure has minimum weight without exceeding
the allowable stresses in the bars. (Note: The weights of the bars are very small compared to the force P and may be disregarded.)
θ
B
C
L
P
Solution 1.9-15
Two bars supporting a load P
LENGTHS OF BARS
Joint C
T
LAC ⫽
L
cos u
LBC ⫽ L
WEIGHT OF TRUSS
θ
C
g ⫽ weight density of material
C
P
T ⫽ tensile force in bar AC
W ⫽ g(AAC LAC + ABC LBC)
⫽
⫽
gPL
1
1
a
+
b
sallow sin u cos u
tan u
gPL 1 + cos2u
b
a
sallow sin u cos u
C ⫽ compressive force in bar BC
g, P, L, and sallow are constants.
a Fvert ⫽ 0
T⫽
W varies only with u
a Fhoriz ⫽ 0
P
C⫽
tan u
P
sin u
AREAS OF BARS
AAC ⫽
P
T
⫽
sallow
sallow sin u
ABC ⫽
C
P
⫽
sallow
sallow tan u
Let k ⫽
Eq. (1)
gPL
(k has units of force)
sallow
W
1 + cos2u
⫽
k
sin u cos u
(Nondimensional)
Eq. (2)
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116
CHAPTER 1
Tension, Compression, and Shear
GRAPH OF EQ. (2):
(sin u cos u)(2)(cos u) (⫺sin u)
df
⫺ (1 + cos2u)(⫺sin2u + cos2u)
⫽
du
sin2u cos2u
⫽
⫺sin2u cos2u + sin2u ⫺ cos2u ⫺ cos4u
sin2u cos2u
SET THE NUMERATOR ⫽ 0 AND SOLVE FOR u:
⫺sin2u cos2u ⫹ sin2u ⫺ cos2u ⫺ cos4u ⫽ 0
Replace sin2u by 1 ⫺ cos2u:
⫺(1 ⫺ cos2u)(cos2u) ⫹ 1 ⫺ cos2u ⫺ cos2u ⫺ cos4u ⫽ 0
Combine terms to simplify the equation:
ANGLE u THAT MAKES WA MINIMUM
1 ⫺ 3 cos2u ⫽ 0
Use Eq. (2):
Let f ⫽
1 + cos2u
sin u cos u
u ⫽ 54.7°
cos u ⫽
1
23
;
df
⫽0
du
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Review Problems
117
R-1.1: A plane truss has downward applied load P at joint 2 and another load P applied leftward at joint 5. The force in
member 3-5 is:
(A)
(B)
(C)
(D)
0
⫺P/2
⫺P
⫹1.5 P
Solution
©M1 ⫽ 0
3
5
P
V6(3L) + PL ⫺ PL ⫽ 0
so
V6 ⫽ 0
L
L
METHOD OF SECTIONS
Cut through members 3-5, 2-5 and 2-4; use right hand FBD.
1
2
L
©M2 ⫽ 0
P
6
4
L
L
F35L + PL ⫽ 0
F35 ⫽ ⫺P
3
;
5
P
L
1
H1
V1
L
2
L
P
5
4
6
L
V6
P
F35
F25
L
2
F24
4
6
L
V6
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118
CHAPTER 1
Tension, Compression, and Shear
R-1.2: The force in member FE of the plane truss below is approximately:
(A)
(B)
(C)
(D)
⫺1.5 kN
⫺2.2 kN
3.9 kN
4.7 kN
Solution
3m
A
15 kN
3m
B
5 kN
10 kN
3m
C
D
3m
E
4.5 m
F
G
1m
STATICS
©MA ⫽ 0
Ey(6 m) ⫺ 15 kN(3 m) ⫺ 10 kN (6 m) ⫺ 5 kN(9 m) ⫽ 0
Ey ⫽ 25 kN
Ax
Ay
A
3m
B
15 kN
3m
10 kN
3m
5 kN
D
3m
E
4.5 m
G
C
F
1m
Ey
METHOD OF SECTIONS: cut through BC, BE, and FE; use right-hand FBD; sum moments about B.
1
⫺3
FFE(3 m) ⫺
FFE(3 m) ⫺ 10 kN(3 m) ⫺ 5 kN(6 m) + Ey(3 m) ⫽ 0
110
110
SOLVING
FFE ⫽ 3.95 kN
5 kN
10 kN
B
5
FFE ⫽ 110 kN
4
3
FFE
10
;
FFE
3m
C
D
3m
E
1m
1
⫺
FFE
10
Ey
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Review Problems
119
R-1.3: The moment reaction at A in the plane frame below is approximately:
(A)
(B)
(C)
(D)
⫹1400 N # m
⫺2280 N # m
⫺3600 N # m
⫹6400 N # m
Solution
900 N
900 N
1200 N/m
1.2 m
B
4m
1.2 m
C
Bx
3m
B
By
Pin
connection
C
3m
Cy
A
STATICS: use FBD of member BC to find reaction Cy.
©MB ⫽ 0
Cy(3 m) ⫺ 900 N(1.2 m) ⫽ 0
900 N(1.2 m)
⫽ 360 N
Cy ⫽
3m
900 N
1.2 m
1200 N/m
B
C
3m
Cy
SUM MOMENTS ABOUT A FOR ENTIRE STRUCTURE
4m
©MA ⫽ 0
MA + Cy(3 m) ⫺ 900 N(1.2 m) ⫺
Solving for MA:
MA ⫽ 6400 N # m
;
1
N
2
a 1200 b4 ma 4 mb ⫽ 0
2
m
3
MA
A
Ax
Ay
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120
CHAPTER 1
Tension, Compression, and Shear
A hollow circular post ABC (see figure) supports a load P1 ⫽ 16 kN acting at the top. A second load P2 is
uniformly distributed around the cap plate at B. The diameters and thicknesses of the upper and lower parts of the post are
dAB ⫽ 30 mm, tAB ⫽ 12 mm, dBC ⫽ 60 mm, and tBC ⫽ 9 mm, respectively. The lower part of the post must have the same
compressive stress as the upper part. The required magnitude of the load P2 is approximately:
R-1.4:
(A)
(B)
(C)
(D)
18 kN
22 kN
28 kN
46 kN
Solution
P1 ⫽ 16 kN
dAB ⫽ 30 mm
tAB ⫽ 12 mm
dBC ⫽ 60 mm
tBC ⫽ 9 mm
P1
A
tAB
dAB
p
AAB ⫽ [dAB2 ⫺ (dAB ⫺ 2tAB)2] ⫽ 679 mm2
4
ABC
P2
p
⫽ [dBC2 ⫺ (dBC ⫺ 2tBC)2] ⫽ 1442 mm2
4
B
Stress in AB:
P1
sAB ⫽
⫽ 23.6 MPa
AAB
Stress in BC:
sBC ⫽
Solve for P2:
P2 ⫽ sABABC ⫺ P1 ⫽ 18.00 kN
Check:
sBC ⫽
P1 + P2
ABC
dBC
tBC
⬍ must equal ␴AB
P1 + P2
⫽ 23.6 MPa
ABC
;
C
⬍ same as in AB
R-1.5: A circular aluminum tube of length L ⫽ 650 mm is loaded in compression by forces P. The outside and inside
diameters are 80 mm and 68 mm, respectively. A strain gage on the outside of the bar records a normal strain in the longitudinal direction of 400 * 10⫺6. The shortening of the bar is approximately:
(A) 0.12 mm
(B) 0.26 mm
(C) 0.36 mm
(D) 0.52 mm
Solution
␧ ⫽ 400 (10⫺6)
Strain gage
L ⫽ 650 mm
d ⫽ ␧L ⫽ 0.260 mm
;
P
P
L
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Review Problems
121
R-1.6: A steel plate weighing 27 kN is hoisted by a cable sling that has a clevis at each end. The pins through the clevises
are 22 mm in diameter. Each half of the cable is at an angle of 35° to the vertical. The average shear stress in each pin is
approximately:
(A)
(B)
(C)
(D)
22 MPa
28 MPa
40 MPa
48 MPa
P
Solution
W ⫽ 27 kN
dp ⫽ 22 mm
u ⫽ 35⬚
Cross-sectional area of each pin:
Ap ⫽
Cable sling
p 2
d ⫽ 380 mm2
4 p
35°
35°
Tensile force in cable:
W
a b
2
T⫽
cos (u)
Clevis
⫽ 16.48 kN
Shear stress in each clevis pin (double shear):
t⫽
T
⫽ 21.7 MPa
2Ap
Steel plate
;
R-1.7: A steel wire hangs from a high-altitude balloon. The steel has unit weight 77 kN/m3 and yield stress of 280 MPa.
The required factor of safety against yield is 2.0. The maximum permissible length of the wire is approximately:
(A)
(B)
(C)
(D)
1800 m
2200 m
2600 m
3000 m
Solution
g ⫽ 77
kN
m3
sY ⫽ 280 MPa
Allowable stress:
sallow ⫽
Weight of wire of length L:
sY
⫽ 140.0 MPa
FSY
W ⫽ gAL
Max. axial stress in wire of length L:
Max. length of wire:
FSY ⫽ 2
Lmax ⫽
smax ⫽
W
A
sallow
⫽ 1818 m
g
smax ⫽ gL
;
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122
CHAPTER 1
Tension, Compression, and Shear
R-1.8: An aluminum bar (E ⫽ 72 GPa, ␯ ⫽ 0.33) of diameter 50 mm cannot exceed a diameter of 50.1 mm when
compressed by axial force P. The maximum acceptable compressive load P is approximately:
(A)
(B)
(C)
(D)
190 kN
200 kN
470 kN
860 kN
Solution
E ⫽ 72 GPa
Lateral strain:
dinit ⫽ 50 mm
␧L ⫽
dfinal ⫽ 50.1 mm
dfinal ⫺ dinit
dinit
␯ ⫽ 0.33
␧L ⫽ 0.002
⫺␧L
⫽ ⫺0.006
␯
Axial strain:
␧a ⫽
Axial stress:
s ⫽ E␧a ⫽ ⫺436.4 MPa
⬍ below yield stress of 480 MPa
so Hooke’s Law applies
Max. acceptable compressive load:
p
Pmax ⫽ s a dinit2b ⫽ 857 kN
4
;
R-1.9: An aluminum bar (E ⫽ 70 GPa, ␯ ⫽ 0.33) of diameter 20 mm is stretched by axial forces P, causing its diameter
to decrease by 0.022 mm. The load P is approximately:
(A)
(B)
(C)
(D)
73 kN
100 kN
140 kN
339 kN
Solution
E ⫽ 70 GPa
Lateral strain:
Axial strain:
Axial stress:
dinit ⫽ 20 mm
␧L ⫽
¢d ⫽ ⫺0.022 mm
¢d
dinit
␯ ⫽ 0.33
P
d
P
␧L ⫽ ⫺0.001
⫺␧L
␧a ⫽
⫽ 3.333 * 10⫺3
␯
s ⫽ E␧a ⫽ 233.3 MPa
⬍ below yield stress of 270 MPa
so Hooke’s Law applies
Max. acceptable load:
p
Pmax ⫽ s a dinit2b ⫽ 73.3 kN
4
;
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Review Problems
123
R-1.10: A polyethylene bar (E ⫽ 1.4 GPa, ␯ ⫽ 0.4) of diameter 80 mm is inserted in a steel tube of inside diameter
80.2 mm and then compressed by axial force P. The gap between steel tube and polyethylene bar will close when
compressive load P is approximately:
(A)
(B)
(C)
(D)
18 kN
25 kN
44 kN
60 kN
Solution
E ⫽ 1.4 GPa
d1 ⫽ 80 mm
␧L ⫽
Lateral strain:
Axial strain:
Axial stress:
¢d1 ⫽ 0.2 mm
Steel
tube
␯ ⫽ 0.4
¢d1
d1
d1 d2
␧L ⫽ 0.003
⫺␧L
⫽ 6.250 * 10⫺3
␧a ⫽
␯
Polyethylene
bar
⬍ well below ultimate stress of 28 MPa
so Hooke’s Law applies
s ⫽ E␧a ⫽ ⫺8.8 MPa
Max. acceptable compressive load:
p
Pmax ⫽ s a d12b ⫽ 44.0 kN
4
;
R-1.11: A pipe (E ⫽ 110 GPa) carries a load P1 ⫽ 120 kN at A and a uniformly distributed load P2 ⫽ 100 kN on the cap
plate at B. Initial pipe diameters and thicknesses are: dAB ⫽ 38 mm, tAB ⫽ 12 mm, dBC ⫽ 70 mm, tBC ⫽ 10 mm. Under
loads P1 and P2, wall thickness tBC increases by 0.0036 mm. Poisson’s ratio ␯ for the pipe material is approximately:
(A)
(B)
(C)
(D)
0.27
0.30
0.31
0.34
Solution
E ⫽ 110 GPa
dAB ⫽ 38 mm
tBC ⫽ 10 mm
P1 ⫽ 120 kN
ABC ⫽
tAB ⫽ 12 mm
dBC ⫽ 70 mm
P2 ⫽ 100 kN
p
[d 2 ⫺ (dBC ⫺ 2tBC)2] ⫽ 1885 mm2
4 BC
tBC dBC
C
Cap plate
tAB dAB
B
A
P1
P2
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124
CHAPTER 1
Tension, Compression, and Shear
⫺(P1 + P2)
⫽ ⫺1.061 * 10⫺3
EABC
Axial strain of BC:
␧BC ⫽
Axial stress in BC:
s BC ⫽ E␧BC ⫽ ⫺116.7 MPa
(well below yield stress of 550 MPa so Hooke’s Law applies)
Lateral strain of BC:
¢tBC ⫽ 0.0036 mm
¢tBC
⫽ 3.600 * 10⫺4
tBC
⫺␧L
v⫽
⫽ 0.34
Poisson’s ratio:
␧BC
␧L ⫽
;
⬍ confirms value for brass given in properties table
R-1.12: A titanium bar (E ⫽ 100 GPa, v ⫽ 0.33) with square cross section (b ⫽ 75 mm) and length L ⫽ 3.0 m is
subjected to tensile load P ⫽ 900 kN. The increase in volume of the bar is approximately:
(A)
(B)
(C)
(D)
1400 mm3
3500 mm3
4800 mm3
9200 mm3
Solution
E ⫽ 100 GPa
b ⫽ 75 mm
L ⫽ 3.0 m
P ⫽ 900 kN
b
v ⫽ 0.33
b
P
P
L
Initial volume of bar:
Vinit ⫽ b2L ⫽ 1.6875000 * 107 mm3
Normal strain in bar:
␧⫽
Lateral strain in bar:
␧L ⫽ ⫺v␧ ⫽ ⫺5.28000 * 10⫺4
Final length of bar:
Eb2
⫽ 1.60000 * 10⫺3
Lf ⫽ L + ␧L ⫽ 3004.800 mm
Final lateral dimension of bar:
Final volume of bar:
P
bf ⫽ b + ␧Lb ⫽ 74.96040 mm
Vfinal ⫽ b2f Lf ⫽ 1.68841562 * 107 mm3
Increase in volume of bar:
¢V ⫽ Vfinal ⫺ Vinit ⫽ 9156 mm3
;
¢V
⫽ 0.000543
Vinit
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Review Problems
125
R-1.13: An elastomeric bearing pad is subjected to a shear force V during a static loading test. The pad has dimensions
a ⫽ 150 mm and b ⫽ 225 mm, and thickness t ⫽ 55 mm. The lateral displacement of the top plate with respect to the
bottom plate is 14 mm under a load V ⫽ 16 kN. The shear modulus of elasticity G of the elastomer is approximately:
(A)
(B)
(C)
(D)
1.0 MPa
1.5 MPa
1.7 MPa
1.9 MPa
Solution
V ⫽ 16 kN
a ⫽ 150 mm
d ⫽ 14 mm
t ⫽ 55 mm
b ⫽ 225 mm
b
a
V
V
⫽ 0.474 MPa
ab
Ave. shear stress:
t⫽
Ave. shear strain:
g ⫽ arctan a
Shear modulus of elastomer:
d
b ⫽ 0.249
t
G⫽
t
t
⫽ 1.902 MPa
g
;
R-1.14: A bar of diameter d ⫽ 18 mm and length L ⫽ 0.75 m is loaded in tension by forces P. The bar has modulus
E ⫽ 45 GPa and allowable normal stress of 180 MPa. The elongation of the bar must not exceed 2.7 mm. The allowable
value of forces P is approximately:
(A)
(B)
(C)
(D)
41 kN
46 kN
56 kN
63 kN
Solution
d ⫽ 18 mm
L ⫽ 0.75 m
E ⫽ 45 GPa
sa ⫽ 180 MPa
da ⫽ 2.7 mm
d
(1) Allowable value of P based on elongation:
␧a ⫽
da
⫽ 3.600 * 10⫺3
L
P
P
L
smax ⫽ E␧a ⫽ 162.0 MPa
p
Pa1 ⫽ smax a d 2b ⫽ 41.2 kN
4
;
⬍ elongation governs
(2) Allowable load P based on tensile stress:
p
Pa2 ⫽ sa a d 2b ⫽ 45.8 kN
4
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126
CHAPTER 1
Tension, Compression, and Shear
R-1.15: Two flanged shafts are connected by eight 18-mm bolts. The diameter of the bolt circle is 240 mm. The allowable
shear stress in the bolts is 90 MPa. Ignore friction between the flange plates. The maximum value of torque T0 is
approximately:
(A)
(B)
(C)
(D)
19 kN # m
22 kN # m
29 kN # m
37 kN # m
Solution
db ⫽ 18 mm
d ⫽ 240 mm
ta ⫽ 90 MPa
n⫽8
T0
Bolt shear area:
As ⫽
pdb2
⫽ 254.5 mm2
4
T0
Max. torque:
d
Tmax ⫽ n(taAs) ⫽ 22.0 kN # m
2
;
R-1.16: A copper tube with wall thickness of 8 mm must carry an axial tensile force of 175 kN. The allowable tensile
stress is 90 MPa. The minimum required outer diameter is approximately:
(A)
(B)
(C)
(D)
60 mm
72 mm
85 mm
93 mm
Solution
t ⫽ 8 mm
P ⫽ 175 kN
P
Required area based on allowable stress:
Areqd ⫽
d
sa ⫽ 90 MPa
P
P
⫽ 1944 mm2
sa
Area of tube of thickness t but unknown outer diameter d:
A⫽
p 2
[d ⫺ (d ⫺ 2t)2]
4
A ⫽ pt(d ⫺ t)
Solving for dmin:
dmin
P
sa
⫽
+ t ⫽ 85.4 mm
pt
;
so
dinner ⫽ dmin ⫺ 2t ⫽ 69.4 mm
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.