Shi20396 ch04

Chapter 4
4-1
1
RC
RA
RB
RD
C
A B
W
D
1
3 2
RB
RA
W
RB
RC
RA
2
1
W
RA
RBx
RB
RBx RBy
RBy
2
1
1
Scale of
corner magnified
W
A
B
(e)
(f)
(d)
W
A
RA RB
B
1
2
W
2
A
RA RB
B
1 1
(a) (b)
(c)

Chapter 4 51
4-2
(a) RA = 2 sin 60 = 1.732 kN Ans.
RB = 2 sin 30 = 1 kN Ans.
RB
(b) S = 0.6 m
α = tan−1 0.6
0.4 + 0.6
= 30.96◦
RA
sin 135
= 800
sin 30.96
⇒ RA = 1100 N Ans.
RO
sin 14.04
= 800
sin 30.96
⇒ RO = 377 N Ans.
(c)
RO = 1.2
tan 30
= 2.078 kN Ans.
RA = 1.2
sin 30
= 2.4 kN Ans.
800 N
30°
RO RA
135°
30.96°
(d) Step 1: Find RA and RE
h = 4.5
tan 30
= 7.794m
+

MA = 0
9RE − 7.794(400 cos 30) − 4.5(400 sin 30) = 0
RE = 400 N Ans.
2 4

4.5 m
Fx = 0 RAx + 400 cos 30 = 0 ⇒ RAx = −346.4N

Fy = 0 RAy + 400 − 400 sin 30 = 0 ⇒ RAy = −200N
RA =

346.42 + 2002 = 400N Ans.
D
C
h
B
y
A E x
9 m
400 N
3
30°
60°
RAy
RAx
RA
RE
1.2 kN
60°
RA
RO
90° 60°
1.2 kN
45 30.96 14.04
30.96°
RA
RO
O
0.4 m
45°
800 N

0.6 m
A
s
RA
RO
B
60°
90°
30°
2 kN
RA
2
1
2 kN
30° 60°
RA
RB

52 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
Step 2: Find components of RC on link 4 and RD
+

MC = 0
400(4.5) − (7.794 − 1.9)RD = 0 ⇒ RD = 305.4N Ans.

Fx = 0 ⇒ (RCx)4 = 305.4N

Fy = 0 ⇒ (RCy)4 = −400N
Step 3: Find components of RC on link 2

Fx = 0
(RCx)2 + 305.4 − 346.4 = 0 ⇒ (RCx)2 = 41N

Fy = 0
(RCy)2 = 200N
4-3
(a)
+

M0 = 0
−18(60) + 14R2 + 8(30) − 4(40) = 0
R2 = 71.43 lbf

Fy = 0: R1 − 40 + 30 + 71.43 − 60 = 0
R1 = −1.43 lbf
M1 = −1.43(4) = −5.72 lbf · in
M2 = −5.72 − 41.43(4) = −171.44 lbf · in
M3 = −171.44 − 11.43(6) = −240 lbf · in
M4 = −240 + 60(4) = 0 checks!
200 N 400 N
305.4 N
40 lbf 60 lbf
4 4 6 4
V (lbf)
1.43
41.43
60
30 lbf
11.43
x
x
x
O
A B C D
y
R1 R2
M1
M2 M3
M4
O
M
(lbf • in)
O
C C
B D
A
B D
E
305.4 N
346.4 N
305.4 N
41 N
400 N
200 N
400 N
200 N
400 N
Pin C
30°
400 N
41 N
305.4 N
200 N
346.4 N
305.4 N
(RCx)2
(RCy)2
C
B
A
2
400 N
4
RD
(RCx)4
(RCy)4
D
C
E
Ans.

Chapter 4 53
(b)

Fy = 0
R0 = 2 + 4(0.150) = 2.6kN
M0 = 0
M0 = 2000(0.2) + 4000(0.150)(0.425)
= 655 N · m
M1 = −655 + 2600(0.2) = −135 N · m
M2 = −135 + 600(0.150) = −45 N · m
M3 = −45 + 1
2
600(0.150) = 0 checks!
(c)
B C

M0 = 0: 10R2 − 6(1000) = 0 ⇒ R2 = 600 lbf

Fy = 0: R1 − 1000 + 600 = 0 ⇒ R1 = 400 lbf
M1 = 400(6) = 2400 lbf · ft
M2 = 2400 − 600(4) = 0 checks!
O
M
(N•m)
y 1000 lbf
R1 R2
(d) +

MC = 0
−10R1 + 2(2000) + 8(1000) = 0
R1 = 1200 lbf

Fy = 0: 1200 − 1000 − 2000 + R2 = 0
R2 = 1800 lbf
M1 = 1200(2) = 2400 lbf · ft
M2 = 2400 + 200(6) = 3600 lbf · ft
M3 = 3600 − 1800(2) = 0 checks!
1000 lbf 2000 lbf
R1
O
1200
O
M1
M2
R2
M3
2 ft 6 ft 2 ft
A B C
y
M
1800
200
x
x
x
6 ft 4 ft
A
O
O
O
B
600
M1
M2
V (lbf)
400
M
(lbf •ft)
x
x
x
V (kN)
200 mm 150 mm 150 mm
2.6
655
0.6
M1
M2
M3
y 2 kN 4 kN/m
A
O
O
O
RO MO
x
x
x

(e) +

MB = 0
−7R1 + 3(400) − 3(800) = 0
R1 = −171.4 lbf

Fy = 0: −171.4 − 400 + R2 − 800 = 0
R2 = 1371.4 lbf
M1 = −171.4(4) = −685.7 lbf · ft
M2 = −685.7 − 571.4(3) = −2400 lbf · ft
M3 = −2400 + 800(3) = 0 checks!
V (lbf )
400 lbf 800 lbf
4 ft 3 ft 3 ft
(f) Break at A
R1 = VA = 1
2
40(8) = 160 lbf
+

MD = 0
12(160) − 10R2 + 320(5) = 0
R2 = 352 lbf

Fy = 0
−160 + 352 − 320 + R3 = 0
R3 = 128 lbf
M1 = 1
2
160(4) = 320 lbf · in
M2 = 320 − 1
2
160(4) = 0 checks! (hinge)
M3 = 0 − 160(2) = −320 lbf · in
M4 = −320 + 192(5) = 640 lbf · in
M5 = 640 − 128(5) = 0 checks!
A
R1 VA
40 lbf/in
O
V (lbf)
160
O
O
160
320 lbf
128
A
192
M
320 lbf
160 lbf 352 lbf 128 lbf
M1
M2
M3
M4
M5
x
x
x
8
5
2
5
40 lbf/in
160 lbf
A
y
B D
C
R2 R3
O
O
O
C
M
800
171.4
571.4
B
y
M1
M2
M3
R1 R2
x
x
x

Chapter 4 55
4-4
(a) q = R1x−1 − 40x − 4−1 + 30x − 8−1 + R2x − 14−1 − 60x − 18−1
V = R1 − 40x − 40 + 30x − 80 + R2x − 140 − 60x − 180 (1)
M = R1x − 40x − 41 + 30x − 81 + R2x − 141 − 60x − 181 (2)
for x = 18+ V = 0 and M = 0 Eqs. (1) and (2) give
0 = R1 − 40 + 30 + R2 − 60 ⇒ R1 + R2 = 70 (3)
0 = R1(18) − 40(14) + 30(10) + 4R2 ⇒ 9R1 + 2R2 = 130 (4)
Solve (3) and (4) simultaneously to get R1 = −1.43 lbf, R2 = 71.43 lbf. Ans.
From Eqs. (1) and (2), at x = 0+, V = R1 = −1.43 lbf, M = 0
x = 4+: V = −1.43 − 40 = −41.43, M = −1.43x
x = 8+: V = −1.43 − 40 + 30 = −11.43
M = −1.43(8) − 40(8 − 4)1 = −171.44
x = 14+: V = −1.43 − 40 + 30 + 71.43 = 60
M = −1.43(14) − 40(14 − 4) + 30(14 − 8) = −240.
x = 18+: V = 0, M = 0 See curves of V and M in Prob. 4-3 solution.
(b) q = R0x−1 − M0x−2 − 2000x − 0.2−1 − 4000x − 0.350 + 4000x − 0.50
V = R0 − M0x−1 − 2000x − 0.20 − 4000x − 0.351 + 4000x − 0.51 (1)
M = R0x − M0 − 2000x − 0.21 − 2000x − 0.352 + 2000x − 0.52 (2)
at x = 0.5+m, V = M = 0, Eqs. (1) and (2) give
R0 − 2000 − 4000(0.5 − 0.35) = 0 ⇒ R1 = 2600N = 2.6 kN Ans.
R0(0.5) − M0 − 2000(0.5 − 0.2) − 2000(0.5 − 0.35)2 = 0
with R0 = 2600 N, M0 = 655 N · m Ans.
With R0 and M0, Eqs. (1) and (2) give the same V and M curves as Prob. 4-3 (note for
V, M0x−1 has no physical meaning).
(c) q = R1x−1 − 1000x − 6−1 + R2x − 10−1
V = R1 − 1000x − 60 + R2x − 100 (1)
M = R1x − 1000x − 61 + R2x − 101 (2)
at x = 10+ ft, V = M = 0, Eqs. (1) and (2) give
R1 − 1000 + R2 = 0 ⇒ R1 + R2 = 1000
10R1 − 1000(10 − 6) = 0 ⇒ R1 = 400 lbf , R2 = 1000 − 400 = 600 lbf
0 ≤ x ≤ 6: V = 400 lbf, M = 400x
6 ≤ x ≤ 10: V = 400 − 1000(x − 6)0 = 600 lbf
M = 400x − 1000(x − 6) = 6000 − 600x
See curves of Prob. 4-3 solution.

(d) q = R1x−1 − 1000x − 2−1 − 2000x − 8−1 + R2x − 10−1
V = R1 − 1000x − 20 − 2000x − 80 + R2x − 100 (1)
M = R1x − 1000x − 21 − 2000x − 81 + R2x − 101 (2)
At x = 10+, V = M = 0 from Eqs. (1) and (2)
R1 − 1000 − 2000 + R2 = 0 ⇒ R1 + R2 = 3000
10R1 − 1000(10 − 2) − 2000(10 − 8) = 0 ⇒ R1 = 1200 lbf ,
R2 = 3000 − 1200 = 1800 lbf
0 ≤ x ≤ 2: V = 1200 lbf, M = 1200x lbf · ft
2 ≤ x ≤ 8: V = 1200 − 1000 = 200 lbf
M = 1200x − 1000(x − 2) = 200x + 2000 lbf · ft
8 ≤ x ≤ 10: V = 1200 − 1000 − 2000 = −1800 lbf
M = 1200x − 1000(x − 2) − 2000(x − 8) = −1800x + 18 000 lbf · ft
Plots are the same as in Prob. 4-3.
(e) q = R1x−1 − 400x − 4−1 + R2x − 7−1 − 800x − 10−1
V = R1 − 400x − 40 + R2x − 70 − 800x − 100 (1)
M = R1x − 400x − 41 + R2x − 71 − 800x − 101 (2)
at x = 10+, V = M = 0
R1 − 400 + R2 − 800 = 0 ⇒ R1 + R2 = 1200 (3)
10R1 − 400(6) + R2(3) = 0 ⇒ 10R1 + 3R2 = 2400 (4)
Solve Eqs. (3) and (4) simultaneously: R1 = −171.4 lbf, R2 = 1371.4 lbf
0 ≤ x ≤ 4: V = −171.4 lbf, M = −171.4x lbf · ft
4 ≤ x ≤ 7: V = −171.4 − 400 = −571.4 lbf
M=−171.4x − 400(x − 4) lbf · ft =−571.4x + 1600
7 ≤ x ≤ 10: V = −171.4 − 400 + 1371.4 = 800 lbf
M=−171.4x − 400(x − 4) + 1371.4(x − 7) = 800x − 8000 lbf · ft
Plots are the same as in Prob. 4-3.
(f ) q = R1x−1 − 40x0 + 40x − 80 + R2x − 10−1 − 320x − 15−1 + R3x − 20
V = R1 − 40x + 40x − 81 + R2x − 100 − 320x − 150 + R3x − 200 (1)
M = R1x − 20x2 + 20x − 82 + R2x − 101 − 320x − 151 + R3x − 201 (2)
M = 0 at x = 8 in ∴8R1 − 20(8)2 = 0 ⇒ R1 = 160 lbf
at x = 20+, V and M = 0
160 − 40(20) + 40(12) + R2 − 320 + R3 = 0 ⇒ R2 + R3 = 480
160(20) − 20(20)2 + 20(12)2 + 10R2 − 320(5) = 0 ⇒ R2 = 352 lbf
R3 = 480 − 352 = 128 lbf
0 ≤ x ≤ 8: V = 160 − 40x lbf, M = 160x − 20x2 lbf · in
8 ≤ x ≤ 10: V = 160 − 40x + 40(x − 8) = −160 lbf ,
M = 160x − 20x2 + 20(x − 8)2 = 1280 − 160x lbf · in

Chapter 4 57
10 ≤ x ≤ 15: V = 160 − 40x + 40(x − 8) + 352 = 192 lbf
M = 160x − 20x2 + 20(x − 8) + 352(x − 10) = 192x − 2240
15 ≤ x ≤ 20: V = 160 − 40x + 40(x − 8) + 352 − 320 = −128 lbf
M = 160x − 20x2 − 20(x − 8) + 352(x − 10) − 320(x − 15)
=−128x + 2560
Plots of V and M are the same as in Prob. 4-3.
4-5 Solution depends upon the beam selected.
4-6
(a) Moment at center, xc = (l − 2a)/2
Mc = w
2

l
2
(l − 2a) −

l
2
2

= wl
2

l
4
− a

At reaction, |Mr| = wa2/2
a = 2.25, l = 10 in, w = 100 lbf/in
Mc = 100(10)
2

10
4

= 125 lbf · in
− 2.25
Mr = 100(2.252)
2
= 253.1 lbf · in Ans.
(b) Minimum occurs when Mc = |Mr |
wl
2

l
4
− a

= wa2
2
⇒ a2 + al − 0.25l2 = 0
Taking the positive root
a = 1
2

−l +

l2 + 4(0.25l2)

= l
2
√

= 0.2071l Ans.
2 − 1
for l = 10 in and w = 100 lbf, Mmin = (100/2)[(0.2071)(10)]2 = 214.5 lbf · in
4-7 For the ith wire from bottom, from summing forces vertically
(a)
Ti = (i + 1)W
Ti
xi
a
From summing moments about point a,

Ma = W(l − xi ) − iWxi = 0
Giving,
xi = l
i + 1
W iW

So
W = l
1 + 1
= l
2
x = l
2 + 1
= l
3
y = l
3 + 1
= l
4
z = l
4 + 1
= l
5
(b) With straight rigid wires, the mobile is not stable. Any perturbation can lead to all wires
becoming collinear. Consider a wire of length l bent at its string support:

Ma = 0

Ma = iWl
i + 1
cos α − ilW
i + 1
cos β = 0
iWl
i + 1
(cos α − cos β) = 0
W iW
Moment vanishes when α = β for any wire. Consider a ccw rotation angle β, which
makes α →α + β and β →α − β
Ma = iWl
i + 1
[cos(α + β) − cos(α − β)]
= 2iWl
i + 1
sin α sin β
.=
2iWlβ
i + 1
sin α
There exists a correcting moment of opposite sense to arbitrary rotation β. An equation
for an upward bend can be found by changing the sign of W. The moment will no longer
be correcting. A curved, convex-upward bend of wire will produce stable equilibrium
too, but the equation would change somewhat.
4-8
(a)
C = 12 + 6
2
= 9
CD = 12 − 6
2
= 3
R =

32 + 42 = 5
σ1 = 5 + 9 = 14
σ2 = 9 − 5 = 4
2s
(12, 4cw)
1
C
R
D
2
1

2
2p
(6, 4ccw)
y
x
cw
ccw
il
i 1
Ti

l
i 1

Chapter 4 59
φp = 1
2
tan−1

4
3

= 26.6◦ cw
τ1 = R = 5, φs = 45◦ − 26.6◦ = 18.4◦ ccw
(b)
C = 9 + 16
2
= 12.5
CD = 16 − 9
2
= 3.5
R =

52 + 3.52 = 6.10
σ1 = 6.1 + 12.5 = 18.6
φp = 1
2
tan−1 5
3.5
= 27.5◦ ccw
σ2 = 12.5 − 6.1 = 6.4
y
2 1
x
1
τ1 = R = 6.10, φs = 45◦ − 27.5◦ = 17.5◦ cw
(c)
C = 24 + 10
2
= 17
CD = 24 − 10
2
= 7
R =

72 + 62 = 9.22
σ1 = 17 + 9.22 = 26.22
σ2 = 17 − 9.22 = 7.78
6.10
2 1
2s
(24, 6cw)
C
R
D
1
2
2p
(10, 6ccw)
y
x

cw
ccw
x
12.5
12.5
17.5
6.4
18.6
27.5
2s
(16, 5ccw)
C
R
D
2
2p
(9, 5cw)
x

cw
ccw
3
5
3
3
3
18.4
x
x
4
14
26.6

φp = 1
2

90 + tan−1 7
6

= 69.7◦ ccw
x
τ1 = R = 9.22, φs = 69.7◦ − 45◦ = 24.7◦ ccw
(d)
C = 9 + 19
2
= 14
CD = 19 − 9
2
= 5
R =

52 + 82 = 9.434
σ1 = 14 + 9.43 = 23.43
σ2 = 14 − 9.43 = 4.57
2 1
φp = 1
2

90 + tan−1 5
8

= 61.0◦ cw
2s
1
τ1 = R = 9.434, φs = 61◦ − 45◦ = 16◦ cw
x
14
14
9.434
16
x
23.43
4.57
61
(9, 8cw)
C
R
D
2
2p
(19, 8ccw)
y
x

cw
ccw
x
17
17
9.22
24.7
26.22
7.78
69.7

Chapter 4 61
4-9
(a)
C = 12 − 4
2
= 4
CD = 12 + 4
2
= 8
R =

82 + 72 = 10.63
σ1 = 4 + 10.63 = 14.63
σ2 = 4 − 10.63 = −6.63
φp = 1
2

90 + tan−1 8
7

= 69.4◦ ccw
x
1
τ1 = R = 10.63, φs = 69.4◦ − 45◦ = 24.4◦ ccw
(b)
C = 6 − 5
2
= 0.5
CD = 6 + 5
2
= 5.5
R =

5.52 + 82 = 9.71
σ1 = 0.5 + 9.71 = 10.21
σ2 = 0.5 − 9.71 = −9.21
φp = 1
2
tan−1 8
5.5
= 27.75◦ ccw
x
10.21
9.21
2s
27.75
(5, 8cw)
1
C
R
D
2
1

2
2p
(6, 8ccw)
y
x
cw
ccw
x
4
4
10.63
24.4
14.63
6.63
69.4
2s
(12, 7cw)
C
R
D
2
1

2
2p
(4, 7ccw)
y
x
cw
ccw

τ1 = R = 9.71, φs = 45◦ − 27.75◦ = 17.25◦ cw
(c)
C =
−8 + 7
2
= −0.5
CD = 8 + 7
2
= 7.5
R =

7.52 + 62 = 9.60
σ1 = 9.60 − 0.5 = 9.10
σ2 = −0.5 − 9.6 = −10.1
9.71
2 1
φp = 1
2

90 + tan−1 7.5
6

= 70.67◦ cw
x
1
τ1 = R = 9.60, φs = 70.67◦ − 45◦ = 25.67◦ cw
(d)
C = 9 − 6
2
= 1.5
CD = 9 + 6
2
= 7.5
R =

7.52 + 32 = 8.078
σ1 = 1.5 + 8.078 = 9.58
σ2 = 1.5 − 8.078 = −6.58
2s
(9, 3cw)
1
2 1
R C D
2
2p
(6, 3ccw)
y
x

cw
ccw
x
0.5
0.5
9.60
25.67
10.1
9.1
70.67
2s
(8, 6cw)
C
R
D
2
2p
(7, 6ccw)
x
y

cw
ccw
x
0.5
0.5
17.25

Chapter 4 63
φp = 1
2
tan−1 3
7.5
= 10.9◦ cw
τ1 = R = 8.078, φs = 45◦ − 10.9◦ = 34.1◦ ccw
4-10
(a)
C = 20 − 10
2
= 5
CD = 20 + 10
2
= 15
R =

152 + 82 = 17
σ1 = 5 + 17 = 22
σ2 = 5 − 17 = −12
2s (20, 8cw)
2 1
φp = 1
2
tan−1 8
15
= 14.04◦ cw
1
τ1 = R = 17, φs = 45◦ − 14.04◦ = 30.96◦ ccw
5
17
5
30.96
x
12
22
14.04
x
C
R
D
2
2p
(10, 8ccw)
y
x

cw
ccw
x
1.5
8.08
1.5
34.1
x
6.58
9.58
10.9

(b)
C = 30 − 10
2
= 10
CD = 30 + 10
2
= 20
R =

202 + 102 = 22.36
σ1 = 10 + 22.36 = 32.36
σ2 = 10 − 22.36 = −12.36
φp = 1
2
tan−1 10
20
= 13.28◦ ccw
2s
1
τ1 = R = 22.36, φs = 45◦ − 13.28◦ = 31.72◦ cw
(c)
C =
−10 + 18
2
= 4
CD = 10 + 18
2
= 14
R =

142 + 92 = 16.64
σ1 = 4 + 16.64 = 20.64
σ2 = 4 − 16.64 = −12.64
10
22.36
10
2 1
φp = 1
2

90 + tan−1 14
9

= 73.63◦ cw
1
τ1 = R = 16.64, φs = 73.63◦ − 45◦ = 28.63◦ cw
4
x
16.64
4
28.63
12.64
20.64
73.63
x
(10, 9cw) 2s
C
R
D
2
2p
(18, 9ccw)
y
x

cw
ccw
31.72
x
12.36
32.36
x
13.28
(10, 10cw)
C
R
D
2
1

2 2p
(30, 10ccw)
y
x
cw
ccw

Chapter 4 65
(d)
C =
−12 + 22
2
= 5
CD = 12 + 22
2
= 17
R =

172 + 122 = 20.81
σ1 = 5 + 20.81 = 25.81
σ2 = 5 − 20.81 = −15.81
2 1
φp = 1
2

90 + tan−1 17
12

= 72.39◦ cw
1
τ1 = R = 20.81, φs = 72.39◦ − 45◦ = 27.39◦ cw
4-11
(a)
(b)
C = 0 + 10
2
= 5
CD = 10 − 0
2
= 5
R =

52 + 42 = 6.40
σ1 = 5 + 6.40 = 11.40
σ2 = 0, σ3 = 5 − 6.40 = −1.40
14
2
1/3 7
4 y x 10
1/3
τ1/3 = R = 6.40, τ1/2 = 11.40
2
= 5.70, τ2/3 = 1.40
2
= 0.70
2 1

3
D
x
y

C
R
(0, 4cw)
(10, 4ccw)
2/3
1/2
x 1

3 y

2 0
2/3 2
1/2 5
5
20.81
5
27.39
x
15.81
25.81
72.39
x
(12, 12cw) 2s
C
R
D
2
2p
(22, 12ccw)
y
x

cw
ccw

(c)
C =
−2 − 8
2
= −5
CD = 8 − 2
2
= 3
R =

32 + 42 = 5
σ1 = −5 + 5 = 0, σ2 = 0
σ3 = −5 − 5 = −10
τ1/3 = 10
2
= 5, τ1/2 = 0, τ2/3 = 5
(d)
C = 10 − 30
2
= −10
CD = 10 + 30
2
= 20
R =

202 + 102 = 22.36
σ1 = −10 + 22.36 = 12.36
σ2 = 0
σ3 = −10 − 22.36 = −32.36
2/3
τ1/3 = 22.36, τ1/2 = 12.36
2
1
= 6.18, τ2/3 = 32.36
2
= 16.18
4-12
(a)
C =
−80 − 30
2
= −55
CD = 80 − 30
2
= 25
R =

252 + 202 = 32.02
σ1 = 0
σ2 = −55 + 32.02 = −22.98 = −23.0
σ3 = −55 − 32.0 = −87.0
x
τ1/2 = 23
2
C
= 11.5, τ2/3 = 32.0, τ1/3 = 87
2
= 43.5
1
(80, 20cw)
(30, 20ccw)
D
R
2/3
1/2
1/3
3 2
y

(30, 10cw)
(10, 10ccw)
C D
R
1/2
1/3
3 2

y
x
12

3
(2, 4cw)
Point is a circle
2 circles
C
D
y
x
(8, 4ccw)

Chapter 4 67
(b)
C = 30 − 60
2
= −15
CD = 60 + 30
2
= 45
R =

452 + 302 = 54.1
σ1 = −15 + 54.1 = 39.1
σ2 = 0
σ3 = −15 − 54.1 = −69.1
τ1/3 = 39.1 + 69.1
2
R
= 54.1, τ1/2 = 39.1
2
= 19.6, τ2/3 = 69.1
2
= 34.6
(c)
C = 40 + 0
2
= 20
CD = 40 − 0
2
= 20
R =

202 + 202 = 28.3
σ1 = 20 + 28.3 = 48.3
σ2 = 20 − 28.3 = −8.3
σ3 = σz = −30
C
y
τ1/3 = 48.3 + 30
2
= 39.1, τ1/2 = 28.3, τ2/3 = 30 − 8.3
2
= 10.9
(d)
C = 50
2
= 25
CD = 50
2
= 25
R =

252 + 302 = 39.1
σ1 = 25 + 39.1 = 64.1
σ2 = 25 − 39.1 = −14.1
σ3 = σz = −20
1/3
C D
τ1/3 = 64.1 + 20
2
(50, 30cw)
= 42.1, τ1/2 = 39.1, τ2/3 = 20 − 14.1
2
= 2.95
1
(0, 30ccw)
2/3
1/2
3 2
x
y

1
(0, 20cw)
(40, 20ccw)
D
R
2/3
1/2
1/3
3 2
x

1

(60, 30ccw)
(30, 30cw)
C D
2/3
1/2
1/3
3 2
x
y

4-13
σ = F
A
= 2000
(π/4)(0.52)
= 10 190 psi = 10.19 kpsi Ans.
δ = FL
AE
= σ
L
E
= 10 190
72
30(106)
= 0.024 46 in Ans.
1 = δ
L
= 0.024 46
72
= 340(10−6) = 340μ Ans.
From Table A-5, ν = 0.292
2 = −ν1 = −0.292(340) = −99.3μ Ans.

d = 2d = −99.3(10−6)(0.5) = −49.6(10−6) in Ans.
4-14 From Table A-5, E = 71.7 GPa
δ = σ
L
E
= 135(106)
3
71.7(109)
= 5.65(10−3) m = 5.65 mm Ans.
4-15 From Table 4-2, biaxial case. From Table A-5, E = 207 GPa and ν = 0.292
σx = E(x + νy)
1 − ν2
= 207(109)[0.0021 + 0.292(−0.000 67)]
1 − 0.2922 (10−6) = 431MPa Ans.
σy = 207(109)[−0.000 67 + 0.292(0.0021)]
1 − 0.2922 (10−6) = −12.9MPa Ans.
4-16 The engineer has assumed the stress to be uniform. That is,

Ft = −F cos θ + τ A = 0 ⇒ τ = F
A
cos θ
When failure occurs in shear
Ssu = F
A
cos θ
The uniform stress assumption is common practice but is not exact. If interested in the
details, see p. 570 of 6th edition.
4-17 From Eq. (4-15)
σ3 − (−2 + 6 − 4)σ2 + [−2(6) + (−2)(−4) + 6(−4) − 32 − 22 − (−5)2]σ
−[−2(6)(−4) + 2(3)(2)(−5) − (−2)(2)2 − 6(−5)2 − (−4)(3)2] = 0
σ3 − 66σ + 118 = 0
Roots are: 7.012, 1.89, −8.903 kpsi Ans.

t

F

Chapter 4 69
τ1/2 = 7.012 − 1.89
2
= 2.56 kpsi
τ2/3 = 8.903 + 1.89
2
= 5.40 kpsi
τmax = τ1/3 = 8.903 + 7.012
2
= 7.96 kpsi Ans.
Note: For Probs. 4-17 to 4-19, one can also find the eigenvalues of the matrix
[σ] =

σx τxy τzx
τxy σy τyz
τzx τyz σz

for the principal stresses
4-18 From Eq. (4-15)
σ3 − (10 + 0 + 10)σ2 +

10(0) + 10(10) + 0(10) − 202 −
√
2

−10

2 − 02
σ
−

10(0)(10) + 2(20)
√
2

−10

(0) − 10
√
2

−10

2 − 0(0)2 − 10(20)2
= 0
σ3 − 20σ2 − 500σ + 6000 = 0
Roots are: 30, 10, −20 MPa Ans.
τ1/2 = 30 − 10
2
= 10MPa
τ2/3 = 10 + 20
2
= 15MPa
τmax = τ1/3 = 30 + 20
2
= 25 MPa Ans.
4-19 From Eq. (4-15)
2/3
1/2
1/3
σ3 − (1 + 4 + 4)σ2 + [1(4) + 1(4) + 4(4) − 22 − (−4)2 − (−2)2]σ
−[1(4)(4) + 2(2)(−4)(−2) − 1(−4)2 − 4(−2)2 − 4(2)2] = 0
σ3 − 9σ2 = 0
20 10 30
(MPa)
(MPa)
1.89 7.012
8.903
2/3
1/2
1/3
(kpsi)
(kpsi)

Roots are: 9, 0, 0 kpsi
(kpsi)
τ2/3 = 0, τ1/2 = τ1/3 = τmax = 9
2
= 4.5 kpsi Ans.
4-20
(a) R1 = c
l
F Mmax = R1a = ac
l
F
σ = 6M
bh2
= 6
bh2
ac
l
F ⇒ F = σbh2l
6ac
Ans.
(b)
Fm
F
= (σm/σ )(bm/b) (hm/h)2 (lm/l)
(am/a) (cm/c)
= 1(s)(s)2(s)
(s)(s)
= s2 Ans.
For equal stress, the model load varies by the square of the scale factor.
4-21
R1 = wl
2
, Mmax|x=l/2 = w
2
l
2

l − l
2

= wl2
8
σ = 6M
bh2
= 6
bh2
wl2
8
= 3Wl
4bh2
⇒ W = 4
3
σbh2
l
Ans.
Wm
W
= (σm/σ)(bm/b) (hm/h)2
lm/l
= 1(s)(s)2
s
= s2 Ans.
wmlm
wl
= s2 ⇒ wm
w
= s2
s
= s Ans.
For equal stress, the model load w varies linearily with the scale factor.
4-22
(a) Can solve by iteration or derive equations for the general case.
Find maximum moment under wheel W3

WT = W at centroid of W’s
RA = l − x3 − d3
l
WT
RA
W1
W2 W3 . . . WT . . .
A B
d3
Wn
RB
a23
a13
x3
l
2/3
(kpsi)
1/2 1/3
O0 9

Chapter 4 71
Under wheel 3
M3 = RAx3 − W1a13 − W2a23 = (l − x3 − d3)
l
WT x3 − W1a13 − W2a23
For maximum,
dM3
dx3
= 0 = (l − d3 − 2x3)
WT
l
⇒ x3 = l − d3
2
substitute into M, ⇒ M3 = (l − d3)2
4l
WT − W1a13 − W2a23
This means the midpoint of d3 intersects the midpoint of the beam
For wheel i xi = l − di
2
, Mi = (l − di )2
4l
WT −
i−1
j=1
Wjaji
Note for wheel 1:

Wjaji = 0
WT = 104.4, W1 = W2 = W3 = W4 = 104.4
4
= 26.1 kip
Wheel 1: d1 = 476
2
= 238 in, M1 = (1200 − 238)2
4(1200)
(104.4) = 20 128 kip · in
Wheel 2: d2 = 238 − 84 = 154 in
M2 = (1200 − 154)2
4(1200)
(104.4) − 26.1(84) = 21 605 kip · in = Mmax
Check if all of the wheels are on the rail
(b) xmax = 600 − 77 = 523 in
(c) See above sketch.
(d) inner axles
4-23
(a)
84 77 84
315
xmax
a a
C
Ga
c1 0.833
0.75
1 1
Gb
b
1 1
1
2 4

3
8

1
4
1
4

D
0.167
Aa = Ab = 0.25(1.5) = 0.375 in2
A = 3(0.375) = 1.125 in2
0.083
0.5
1.5
y c2 0.667
B
A
600 600

¯y = 2(0.375)(0.75) + 0.375(0.5)
1.125
= 0.667 in
Ia = 0.25(1.5)3
12
= 0.0703 in4
Ib = 1.5(0.25)3
12
= 0.001 95 in4
I1 = 2[0.0703 + 0.375(0.083)2] + [0.001 95 + 0.375(0.167)2] = 0.158 in4 Ans.
σA = 10 000(0.667)
0.158
= 42(10)3 psi Ans.
σB = 10 000(0.667 − 0.375)
0.158
= 18.5(10)3 psi Ans.
σC = 10 000(0.167 − 0.125)
0.158
= 2.7(10)3 psi Ans.
σD = −10 000(0.833)
0.158
= −52.7(10)3 psi Ans.
(b)
D
C
1 1
B
A
y
Here we treat the hole as a negative area.
Aa = 1.732 in2
Ab = 1.134

0.982
2

= 0.557 in2
A = 1.732 − 0.557 = 1.175 in2
¯y = 1.732(0.577) − 0.557(0.577)
1.175
a b
Ga 0.327 Gb
= 0.577 in Ans.
Ia = bh3
36
= 2(1.732)3
36
= 0.289 in4
Ib = 1.134(0.982)3
36
= 0.0298 in4
I1 = Ia − Ib = 0.289 − 0.0298 = 0.259 in4 Ans.
A
0.25
c1 1.155
c2 0.577
2
1.732
0.577
0.982
0.577
1.134

Chapter 4 73
because the centroids are coincident.
σA = 10 000(0.577)
0.259
= 22.3(10)3 psi Ans.
σB = 10 000(0.327)
0.259
= 12.6(10)3 psi Ans.
σC = −10 000(0.982 − 0.327)
0.259
= −25.3(10)3 psi Ans.
σD = −10 000(1.155)
0.259
= −44.6(10)3 psi Ans.
(c) Use two negative areas.
D
C
1 1
B
Gb Gc
Aa = 1 in2, Ab = 9 in2, Ac = 16 in2, A = 16 − 9 − 1 = 6 in2;
¯ya = 0.25 in, ¯yb = 2.0 in, ¯yc = 2 in
¯y = 16(2) − 9(2) − 1(0.25)
6
= 2.292 in Ans.
c1 = 4 − 2.292 = 1.708 in
Ia = 2(0.5)3
12
= 0.020 83 in4
Ib = 3(3)3
12
= 6.75 in4
Ic = 4(4)3
12
= 21.333 in4
I1 = [21.333 + 16(0.292)2] − [6.75 + 9(0.292)2]
−[0.020 83 + 1(2.292 − 0.25)2]
= 10.99 in4 Ans.
σA = 10 000(2.292)
10.99
= 2086 psi Ans.
σB = 10 000(2.292 − 0.5)
10.99
= 1631 psi Ans.
σC = −10 000(1.708 − 0.5)
10.99
= −1099 psi Ans.
σD = −10 000(1.708)
10.99
= −1554 psi Ans.
c
a
b
A
Ga
c1 1.708
c2 2.292
2
1.5
0.25

(d) Use a as a negative area.
B
1 1
Aa = 6.928 in2, Ab = 16 in2, A = 9.072 in2;
¯ya = 1.155 in, ¯yb = 2 in
¯y = 2(16) − 1.155(6.928)
9.072
= 2.645 in Ans.
c1 = 4 − 2.645 = 1.355 in
Ia = bh3
36
= 4(3.464)3
36
= 4.618 in4
Ib = 4(4)3
12
= 21.33 in4
I1 = [21.33 + 16(0.645)2] − [4.618 + 6.928(1.490)2]
= 7.99 in4 Ans.
σA = 10 000(2.645)
7.99
= 3310 psi Ans.
σB = −10 000(3.464 − 2.645)
7.99
= −1025 psi Ans.
σC = −10 000(1.355)
7.99
= −1696 psi Ans.
(e) Aa = 6(1.25) = 7.5 in2
Ab = 3(1.5) = 4.5 in2
A = Ac + Ab = 12 in2
¯y = 3.625(7.5) + 1.5(4.5)
12
= 2.828 in Ans.
I = 1
12
C
(6)(1.25)3 + 7.5(3.625 − 2.828)2 + 1
12
(1.5)(3)3 + 4.5(2.828 − 1.5)2
= 17.05 in4 Ans.
σA = 10 000(2.828)
17.05
= 1659 psi Ans.
σB = −10 000(3 − 2.828)
17.05
= −101 psi Ans.
σC = −10 000(1.422)
17.05
= −834 psi Ans.
a
b
A
B
c1 1.422
c2 2.828
3.464
Ga
b
a
C
A
c1 1.355
c2 2.645
1.490
1.155

Chapter 4 75
(f) Let a = total area
A = 1.5(3) − 1(1.25) = 3.25 in2
I = Ia − 2Ib = 1
12
(1.5)(3)3 − 1
12
(1.25)(1)3
= 3.271 in4 Ans.
σA = 10 000(1.5)
3.271
= 4586 psi, σD = −4586 psi
Ans.
σB = 10 000(0.5)
3.271
= 1529 psi, σC = −1529 psi
4-24
(a) The moment is maximum and constant between A and B
M = −50(20) = −1000 lbf · in , I = 1
12
(0.5)(2)3 = 0.3333 in4
ρ =

E I
M

= 1.6(106)(0.3333)
1000
= 533.3 in
(x, y) = (30, −533.3) in Ans.
(b) The moment is maximum and constant between A and B
M = 50(5) = 250 lbf · in, I = 0.3333 in4
ρ = 1.6(106)(0.3333)
250
= 2133 in Ans.
(x, y) = (20, 2133) in Ans.
4-25
(a)
I = 1
12
(0.75)(1.5)3 = 0.2109 in4
A = 0.75(1.5) = 1.125 in
Mmax is at A. At the bottom of the section,
σmax = Mc
I
= 4000(0.75)
0.2109
= 14 225 psi Ans.
Due to V, τmax constant is between A and B
at y = 0
τmax = 3
2
V
A
= 3
2
667
1.125
= 889 psi Ans.
1000 lbf
O B
333 lbf 667 lbf
4000
x
A
333
O
667
O
12 6
V (lbf)
M
(lbf •in)
x
C
B
A
b
a
b
D
c 1.5
c 1.5
1.5

(b)
I = 1
12
(1)(2)3 = 0.6667 in4
Mmax is at A at the top of the beam
σmax = 8000(1)
0.6667
= 12 000 psi Ans.
|Vmax| = 1000 lbf from O to B at y = 0
τmax = 3
2
V
A
= 3
2
1000
(2)(1)
= 750 psi Ans.
(c)
I = 1
12
(0.75)(2)3 = 0.5 in4
M1 = −1
2
600(5) = −1500 lbf · in = M3
M2 = −1500 + 1
2
(900)(7.5) = 1875 lbf · in
Mmax is at span center. At the bottom of the
beam,
σmax = 1875(1)
0.5
= 3750 psi Ans.
At A and B at y = 0
τmax = 3
2
900
(0.75)(2)
= 900 psi Ans.
(d)
I = 1
12
(1)(2)3 = 0.6667 in4
M1 = −600
2
(6) = −1800 lbf · in
M2 = −1800 + 1
2
750(7.5) = 1013 lbf · in
At A, top of beam
σmax = 1800(1)
0.6667
= 2700 psi Ans.
At A, y = 0
τmax = 3
2
750
(2)(1)
= 563 psi Ans.
O B
100 lbf/in
V (lbf)
M
(lbf •in)
900
O A B
1350 lbf 450 lbf
V (lbf)
O
M
(lbf •in)
O
6 12
7.5
750
M1
450
M2
600
x
x
x
120 lbf/in
1500 lbf 1500 lbf
O
A B C
O
O
5 15 5
900
M1
M2
x
x
M3
600
600
x
x
1000 lbf 1000 lbf
1000
1000
8000
2000 lbf
A
O
O
8 8
V (lbf)
M
(lbf •in)
x
x

Chapter 4 77
4-26
Mmax = wl2
8
⇒ σmax = wl2c
8I
⇒ w = 8σ I
cl2
(a) l = 12(12) = 144 in, I = (1/12)(1.5)(9.5)3 = 107.2 in4
w = 8(1200)(107.2)
4.75(1442)
= 10.4 lbf/in Ans.
(b) l = 48 in, I = (π/64)(24 − 1.254) = 0.6656 in4
w = 8(12)(103)(0.6656)
1(48)2
= 27.7 lbf/in Ans.
(c) l = 48 in, I .=
(1/12)(2)(33) − (1/12)(1.625)(2.6253) = 2.051 in4
w = 8(12)(103)(2.051)
1.5(48)2
= 57.0 lbf/in Ans.
(d) l = 72 in; Table A-6, I = 2(1.24) = 2.48 in4
cmax = 2.158
w = 8(12)(103)(2.48)
2.158(72)2
= 21.3 lbf/in Ans.
(e) l = 72 in; Table A-7, I = 3.85 in4
w = 8(12)(103)(3.85)
2(722)
= 35.6 lbf/in Ans.
(f) l = 72 in, I = (1/12)(1)(43) = 5.333 in4
w = 8(12)(103)(5.333)
(2)(72)2
= 49.4 lbf/in Ans.
4-27 (a) Model (c)
I = π
64
(0.54) = 3.068(10−3) in4
A = π
4
(0.52) = 0.1963 in2
σ = Mc
I
= 218.75(0.25)
3.068(10−3)
= 17 825 psi = 17.8 kpsi Ans.
τmax = 4
3
V
A
= 4
3
500
0.1963
= 3400 psi Ans.
500 lbf 500 lbf
1.25 in
0.4375
500 lbf 500 lbf
500
500
V (lbf)
O
M
(lbf ¥in)
O
Mmax 500(0.4375)
218.75 lbf ¥in
2
0.842
2.158

(b) Model (d)
Mmax = 500(0.25) + 1
2
(500)(0.375)
= 218.75 lbf · in
Vmax = 500 lbf
Same M and V
∴ σ = 17.8 kpsi Ans.
τmax = 3400 psi Ans.
0.25
1333 lbf/in
V (lbf)
M Mmax
4-28 If support RB is between F1 and F2 at position x = l, maximum moments occur at x = 3 and l.

MB = RAl − 2000(l − 3) + 1100(7.75 − l) = 0
RA = 3100 − 14 525/l
Mx=3 = 3RA = 9300 − 43 575/l
MB = RAl − 2000(l − 3) = 1100 l − 8525
To minimize the moments, equate Mx=3 to −MB giving
9300 − 43 575/l = −1100l + 8525
Multiply by l and simplify to
l2 + 0.7046l − 39.61 = 0
The positive solution for l is 5.95 in and the magnitude of the moment is
M = 9300 − 43 575/5.95 = 1976 lbf · in
Placing the bearing to the right of F2, the bending moment would be minimized by placing
it as close as possible to F2. If the bearing is near point B as in the original figure, then we
need to equate the reaction forces. From statics, RB = 14 525/l, and RA = 3100 − RB.
For RA = RB, then RA = RB = 1550 lbf, and l = 14 575/1550 = 9.37 in.
4-29
l p2
p1
q = −Fx−1 + p1x − l0 − p1 + p2
a
a
b
x − l1 + terms for x l + a
V = −F + p1x − l1 − p1 + p2
2a
M = −Fx + p1
2
x − l2 − p1 + p2
6a
F
1.25
500 lbf 500 lbf
500
500
O
O

Chapter 4 79
At x = (l + a)+, V = M = 0, terms for x l + a = 0
−F + p1a − p1 + p2
2a
a2 = 0 ⇒ p1 − p2 = 2F
a
(1)
−F(l + a) + p1a2
2
− p1 + p2
6a
a3 = 0 ⇒ 2p1 − p2 = 6F(l + a)
a2 (2)
From (1) and (2) p1 = 2F
a2 (3l + 2a), p2 = 2F
a2 (3l + a) (3)
From similar triangles
b
p2
= a
p1 + p2
⇒ b = ap2
p1 + p2
(4)
Mmax occurs where V = 0
xmax = l + a − 2b
l a 2b
p2
p2
p2
b b
Mmax = −F(l + a − 2b) + p1
2
(a − 2b)2 − p1 + p2
6a
(a − 2b)3
= −Fl − F(a − 2b) + p1
2
(a − 2b)2 − p1 + p2
6a
(a − 2b)3
F
p1
Normally Mmax = −Fl
The fractional increase in the magnitude is

= F(a − 2b) − ( p1/2)(a − 2b)2 − [( p1 + p2)/6a](a − 2b)3
Fl
(5)
For example, consider F = 1500 lbf, a = 1.2 in, l = 1.5 in
(3) p1 = 2(1500)
1.22 [3(1.5) + 2(1.2)] = 14 375 lbf/in
p2 = 2(1500)
1.22 [3(1.5) + 1.2] = 11 875 lbf/in
(4) b = 1.2(11 875)/(14 375 + 11 875) = 0.5429 in
Substituting into (5) yields

= 0.036 89 or 3.7% higher than −Fl
4-30 Computer program; no solution given here.

4-31
R1 = c
l
F
M = c
l
Fx 0 ≤ x ≤ a
a c
σ = 6M
bh2
= 6(c/l)Fx
bh2
⇒ h =

6cFx
blσmax
0 ≤ x ≤ a Ans.
4-32
R1 = b
l
F
M = b
l
Fx
σmax = 32M
πd3
= 32
πd3
b
l
Fx
d =

32
π
bFx
lσmax

1/3
0 ≤ x ≤ a Ans.
4-33
F
l
Square: Am = (b − t)2
Tsq = 2Amtτall = 2(b − t)2tτall
Round: Am = π(b − t)2/4
Trd = 2π(b − t)2tτall/4
Ratio of torques
Tsq
Trd
= 2(b − t)2tτall
π(b − t)2tτall/2
= 4
π
= 1.27
Twist per unit length
square:
θsq = 2Gθ1t
tτall

L
A

m
= C

L
A

m
= C
4(b − t)
(b − t)2
Round:
θrd = C

L
A

m
= C
π(b − t)
π(b − t)2/4
= C
4(b − t)
(b − t)2
Ratio equals 1, twists are the same.
t
b
t
b
R1
F
a b
l
R2
x
y
R1 R2

Chapter 4 81
Note the weight ratio is
Wsq
Wrd
= ρl(b − t)2
ρlπ(b − t)(t)
= b − t
πt
thin-walled assumes b ≥ 20t
= 19
π
= 6.04 with b = 20
= 2.86 with b = 10t
4-34 l = 40 in, τall = 11 500 psi, G = 11.5(106) psi, t = 0.050 in
rm = ri + t/2 = ri + 0.025 for ri 0
= 0 for ri = 0
Am = (1 − 0.05)2 − 4

r2m
− π
4
r2m

= 0.952 − (4 − π)r2m
Lm = 4(1 − 0.05 − 2rm + 2πrm/4) = 4[0.95 − (2 − π/2)rm]
Eq. (4-45): T = 2Amtτ = 2(0.05)(11 500)Am = 1150Am
Eq. (4-46):
θ(deg) = θ1 l
180
π
= T Lml
4GA2
mt
180
π
= T Lm(40)
4(11.5)(106)A2
m(0.05)
180
π
= 9.9645(10−4)
T Lm
A2
m
Equations can then be put into a spreadsheet resulting in:
ri rm Am Lm ri T(lbf · in) ri θ(deg)
0 0 0.902 5 3.8 0 1037.9 0 4.825
0.10 0.125 0.889 087 3.585 398 0.10 1022.5 0.10 4.621
0.20 0.225 0.859 043 3.413 717 0.20 987.9 0.20 4.553
0.30 0.325 0.811 831 3.242 035 0.30 933.6 0.30 4.576
0.40 0.425 0.747 450 3.070 354 0.40 859.6 0.40 4.707
0.45 0.475 0.708 822 2.984 513 0.45 815.1 0.45 4.825
ri (in)
T (lbf •in)
1200
1000
800
600
400
200
0
0 0.1 0.2 0.3 0.4 0.5

0 0.1 0.2 0.3 0.4 0.5
ri (in)
(deg)
4.85
4.80
4.75
4.70
4.65
4.60
4.55
4.50
Torque carrying capacity reduces with ri . However, this is based on an assumption of uni-form
stresses which is not the case for small ri . Also note that weight also goes down with
an increase in ri .
4-35 From Eq. (4-47) where θ1 is the same for each leg.
T1 = 1
3
1, T2 = 1
Gθ1L1c3
3
Gθ1L2c3
2
T = T1 + T2 = 1
3
Gθ1

L1c3
1
+ L2c3
2

= 1
3
Gθ1

Li c3
i Ans.
τ1 = Gθ1c1, τ2 = Gθ1c2
τmax = Gθ1cmax Ans.
4-36
(a) τmax = Gθ1cmax
Gθ1 = τmax
cmax
= 11 500
3/32
= 1.227(105) psi · rad
T1/16 = 1
3
Gθ1(Lc3)1/16 = 1
3
(1.227)(105)(0.5)(1/16)3 = 4.99 lbf · in Ans.
T3/32 = 1
3
(1.227)(105)(0.5)(3/32)3 = 16.85 lbf · in Ans.
τ1/16 = 1.227(105)1/16 = 7669 psi, τ3/32 = 1.227(105)3/32 = 11 500 psi Ans.
(b) θ1 = 1.227(105)
11.5(106)
= 1.0667(10−2) rad/in = 0.611◦/in Ans.
4-37 Separate strips: For each 1/16 in thick strip,
T = Lc2τ
3
= (1)(1/16)2(11 500)
3
= 14.97 lbf · in
∴ Tmax = 2(14.97) = 29.95 lbf · in Ans.

Chapter 4 83
For each strip,
θ = 3Tl
Lc3G
= 3(14.97)(12)
(1)(1/16)3(11.5)(106)
= 0.192 rad Ans.
kt = T/θ = 29.95/0.192 = 156.0 lbf · in Ans.
Solid strip: From Example 4-12,
Tmax = 59.90 lbf · in Ans.
θ = 0.0960 rad Ans.
kt = 624 lbf · in Ans.
4-38 τall = 8000 psi, 50 hp
(a) n = 2000 rpm
Eq. (4-40) T = 63 025H
n
= 63 025(50)
2000
= 1575.6 lbf · in
τmax = 16T
πd3
⇒ d =

16T
πτmax
1/3
=

16(1575.6)
π(8000)

1/3
= 1.00 in Ans.
(b) n = 200 rpm ∴ T = 15 756 lbf · in
d =

16(15 756)
π(8000)

1/3
= 2.157 in Ans.
4-39 τall = 110 MPa, θ = 30◦, d = 15 mm, l = ?
τ = 16T
πd3
⇒ T = π
16
τd3
θ = Tl
JG

180
π

l = π
180
JGθ
T
= π
180

π
32
d4Gθ
(π/16) τd3

= π
360
dGθ
τ
= π
360
(0.015)(79.3)(109)(30)
110(106)
= 2.83 m Ans.
4-40 d = 70 mm, replaced by 70 mm hollow with t = 6 mm
(a) Tsolid = π
16
τ (703) Thollow = π
32
τ
(704 − 584)
35
%
T = (π/16)(703) − (π/32) [(704 − 584)/35]
(π/16)(703)
(100) = 47.1% Ans.

(b) Wsolid = kd2 = k(702) , Whollow = k(702 − 582)
%
W = k(702) − k(702 − 582)
k(702)
(100) = 68.7% Ans.
4-41 T = 5400 N · m, τall = 150 MPa
(a) τ = Tc
J
⇒ 150(106) = 5400(d/2)
(π/32)[d4 − (0.75d)4]
= 4.023(104)
d3
d =

4.023(104)
150(106)
1/3
= 6.45(10−2)m = 64.5 mm
From Table A-17, the next preferred size is d = 80 mm; ID = 60mm Ans.
(b) J = π
32
(0.084 − 0.064) = 2.749(10−6) mm4
τi = 5400(0.030)
2.749(10−6)
= 58.9(106) Pa = 58.9 MPa Ans.
4-42
(a) T = 63 025H
n
= 63 025(1)
5
= 12 605 lbf · in
τ = 16T
πd3C
⇒ dC =

16T
πτ
1/3
=

16(12 605)
π(14 000)

1/3
= 1.66 in Ans.
From Table A-17, select 13/4 in
τstart = 16(2)(12 605)
π(1.753)
= 23.96(103) psi = 23.96 kpsi
(b) design activity
4-43 ω = 2πn/60 = 2π(8)/60 = 0.8378 rad/s
T = H
ω
= 1000
0.8378
= 1194 N·m
dC =

16T
πτ
1/3
=

16(1194)
π(75)(106)

1/3
= 4.328(10−2) m = 43.3 mm
From Table A-17, select 45 mm Ans.
4-44 s =
√
A, d =

4A/π
Square: Eq. (4-43) with b = c
τmax = 4.8T
c3
(τmax)sq = 4.8T
(A)3/2

Chapter 4 85
Round: (τmax)rd = 16
π
T
d3
= 16T
π(4A/π)3/2
= 3.545T
(A)3/2
(τmax)sq
(τmax)rd
= 4.8
3.545
= 1.354
Square stress is 1.354 times the round stress Ans.
4-45 s =
√
A, d =

4A/π
Square: Eq. (4-44) with b = c, β = 0.141
θsq = Tl
0.141c4G
= Tl
0.141(A)4/2G
Round:
θrd = Tl
JG
= Tl
(π/32) (4A/π)4/2 G
= 6.2832Tl
(A)4/2G
θsq
θrd
= 1/0.141
6.2832
= 1.129
Square has greater θ by a factor of 1.13 Ans.
4-46 Text Eq. (4-43) gives
τmax = T
αbc2
= T
bc2
· 1
α
From in-text table, p. 139, α is a function of b/c. Arrange equation in the form
b2cτmax
T
= 1
α
= y = a0 + a1
1
b/c
= a0 + a1x
To plot 1/α vs 1/(b/c) , first form a table.
x y
b/c α 1/(b/c) 1/α
1 0.208 1 4.807 692
1.5 0.231 0.666 667 4.329 004
1.75 0.239 0.571 429 4.184 100
2 0.246 0.5 4.065 041
2.5 0.258 0.4 3.875 969
3 0.267 0.333 333 3.745 318
4 0.282 0.25 3.546 099
6 0.299 0.166 667 3.344 482
8 0.307 0.125 3.257 329
10 0.313 0.1 3.194 888
∞ 0.333 0 3.003 003

5
4.5
4
3.5
y 1.867x 3.061
Plot is a gentle convex-upward curve. Roark uses a polynomial, which in our notation is
τmax = 3T
8(b/2)(c/2)2

1 + 0.6095
1
b/c

+ · · ·
τmax
.=
T
bc2

3 + 1.8285
1
b/c

Linear regression on table data
y = 3.06 + 1.87x
1
= 3.06 + 1.87
α
1
b/c
τmax = T
bc2

3.06 + 1.87
1
b/c

Eq. (4-43) τmax = T
bc2

3 + 1.8
b/c

4-47
Ft = 1000
2.5
= 400 lbf
Fn = 400 tan 20 = 145.6 lbf
Torque at C TC = 400(5) = 2000 lbf · in
P = 2000
3
= 666.7 lbf
10
145.6 lbf
C
1000 lbf•in
2.5R
Ft
Fn
Gear F
y
Shaft ABCD
z
A
RAy
RAz
3
B
666.7 lbf
2000 lbf•in
D x
C
400 lbf 5
RDy
RDz
2000 lbf•in
1(bc)
1
3
0 0.2 0.4 0.6 0.8 1

Chapter 4 87

(MA)z = 0 ⇒ 18RDy − 145.6(13) − 666.7(3) = 0 ⇒ RDy = 216.3 lbf

(MA)y = 0 ⇒ −18RDz + 400(13) = 0 ⇒ RDz = 288.9 lbf

Fy = 0 ⇒ RAy + 216.3 − 666.7 − 145.6 = 0 ⇒ RAy = 596.0 lbf

Fz = 0 ⇒ RAz + 288.9 − 400 = 0 ⇒ RAz = 111.1 lbf

5962 + 111.12 = 1819 lbf · in
MB = 3

216.32 + 288.92 = 1805 lbf · in
MC = 5
∴ Maximum stresses occur at B. Ans.
σB = 32MB
πd3
= 32(1819)
π(1.253)
= 9486 psi
τB = 16TB
πd3
= 16(2000)
π(1.253)
= 5215 psi
σmax = σB
2
+
σB
2
2
+ τ2B
= 9486
2
+

9486
2
2
+ 52152 = 11 792 psi Ans.
τmax =
σB
2
2
+ τ2B
= 7049 psi Ans.
4-48
(a) At θ = 90◦, σr = τrθ = 0, σθ = −σ Ans.
θ = 0◦, σr = τrθ = 0, σθ = 3σ Ans.
(b)
r σθ/σ
5 3.000
6 2.071
7 1.646
8 1.424
9 1.297
10 1.219
11 1.167
12 1.132
13 1.107
14 1.088
15 1.074
16 1.063
17 1.054
18 1.048
19 1.042
20 1.037
r (mm)

3
2.5
2
1.5
1
0.5
0
0 5 10 15 20

4-49
D/d = 1.5
1
= 1.5
r/d = 1/8
1
= 0.125
Fig. A-15-8: Kts
.=
1.39
Fig. A-15-9: Kt
.=
1.60
σA = Kt
Mc
I
= 32KtM
πd3
= 32(1.6)(200)(14)
π(13)
= 45 630 psi
τA = Kts
Tc
J
= 16KtsT
πd3
= 16(1.39)(200)(15)
π(13)
= 21 240 psi
σmax = σA
2
+
σA
2
2
+ τ2A
= 45.63
2
+

45.63
2
2
+ 21.242
= 54.0 kpsi Ans.
τmax =

45.63
2
2
+ 21.242 = 31.2 kpsi Ans.
4-50 As shown in Fig. 4-34, the maximum stresses occur at the inside fiber where r = ri . There-fore,
from Eq. (4-51)
σt,max =
r 2
i pi
r 2
o
− r 2
i

1 + r 2
o
r 2
i

= pi

r 2
o
+ r 2
i
r 2
o
− r 2
i

Ans.
σr,max =
r 2
i pi
r 2
o
− r 2
i

1 − r 2
o
r 2
i

= −pi Ans.
4-51 If pi = 0, Eq. (4-50) becomes
σt =
−por 2
o
− r 2
i r 2
o po/r 2
r 2
o
− r 2
i
= − por 2
o
r 2
o
− r 2
i

1 +
r 2
i
r 2

The maximum tangential stress occurs at r = ri . So
2
σt,= − 2por o
max r 2
o
− r 2
i
Ans.

Chapter 4 89
For σr , we have
σr =
−por 2
o
+ r 2
i r 2
o po/r 2
r 2
o
− r 2
i
= por 2
o
r 2
o
− r 2
i

r 2
i
r 2

− 1
So σr = 0 at r = ri . Thus at r = ro
por 2
σr,= o
max r 2
o
− r 2
i

r 2
i
− r 2
o
r 2
o

= −po Ans.
4-52
F = pA = πr 2
av p
σ1 = σ2 = F
Awall
= πr 2
av p
2πravt
= prav
2t
Ans.
4-53 σt σl σr
τmax = (σt − σr )/2 at r = ri where σl is intermediate in value. From Prob. 4-50
τmax = 1
2
(σt, max − σr, max)
τmax = pi
2

r 2
o
+ r 2
i
r 2
o
− r 2
i

+ 1
Now solve for pi using ro = 3 in, ri = 2.75 in, and τmax = 4000 psi. This gives pi =
639 psi Ans.
4-54 Given ro = 120mm, ri = 110mm and referring to the solution of Prob. 4-53,
τmax = 2.4 MPa
2

(120)2 + (110)2
(120)2 − (110)2

+ 1
= 15.0MPa Ans.
rav
p
t
F

Shi20396 ch04

More Related Content

What's hot

Viewers also liked

Similar to Shi20396 ch04

More from Paralafakyou Mens

Shi20396 ch04