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Chapter 4
4-1
1
R
C
R
A
R
B
R
D
C
A
B
W
D
1
2
3
R
B
R
A
W
R
B
R
C
R
A
2
1
W
R
A
R
Bx
R
Bx
R
By
R
By
R
B
2
1
1
Scale of
corner magnified
W
A
B
(e)
(f)
(d)
W
A
R
A
R
B
B
1
2
W
A
R
A
R
B
B
11
2
(a)
(b)
(c)
shi20396_ch04.qxd 8/18/03 10:35 AM Page 50
Chapter 4 51
4-2
(a)
R
A
= 2 sin 60 = 1.732
kN Ans.
R
B
= 2 sin 30 = 1
kN Ans.
(b)
S = 0.6
m
α = tan
−1
0.6
0.4 + 0.6
= 30.96
◦
R
A
sin 135
=
800
sin 30.96
⇒ R
A
= 1100
N Ans.
R
O
sin 14.04
=
800
sin 30.96
⇒ R
O
= 377
N Ans.
(c)
R
O
=
1.2
tan 30
= 2.078
kN Ans.
R
A
=
1.2
sin 30
= 2.4
kN Ans.
(d) Step 1: Find
R
A
and
R
E
h =
4.5
tan 30
= 7.794 m
ۗ
+
M
A
= 0
9R
E
− 7.794(400 cos 30) − 4.5(400 sin 30) = 0
R
E
= 400
N Ans.
F
x
= 0 R
Ax
+ 400 cos 30 = 0 ⇒ R
Ax
=−346.4N
F
y
= 0 R
Ay
+ 400 − 400 sin 30 = 0 ⇒ R
Ay
=−200 N
R
A
=
346.4
2
+ 200
2
= 400 N
Ans.
D
C
h
B
y
E
x
A
4.5 m
9 m
400 N
3
4
2
30°
60°
R
Ay
R
A
R
Ax
R
E
1.2 kN
60°
R
A
R
O
60°90°
30°
1.2 kN
R
A
R
O
45Њ Ϫ 30.96Њ ϭ 14.04Њ
135°
30.96°
30.96°
800 N
R
A
R
O
O
0.4 m
45°
800 N
␣
0.6 m
A
s
R
A
R
O
B
60°
90°
30°
2 kN
R
A
R
B
2
1
2 kN
60°
30°
R
A
R
B
shi20396_ch04.qxd 8/18/03 10:35 AM Page 51
52 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
Step 2: Find components of
R
C
on link 4 and
R
D
ۗ
+
M
C
= 0
400(4.5) − (7.794 −1.9)R
D
= 0 ⇒ R
D
= 305.4N
Ans.
F
x
= 0 ⇒ (R
Cx
)
4
= 305.4N
F
y
= 0 ⇒ (R
Cy
)
4
=−400 N
Step 3: Find components of
R
C
on link 2
F
x
= 0
( R
Cx
)
2
+ 305.4 − 346.4 = 0 ⇒ (R
Cx
)
2
= 41 N
F
y
= 0
( R
Cy
)
2
= 200 N
4-3
(a)
ۗ
+
M
0
= 0
−18(60) + 14R
2
+ 8(30) − 4(40) = 0
R
2
= 71.43
lbf
F
y
= 0: R
1
− 40 + 30 + 71.43 − 60 = 0
R
1
=−1.43
lbf
M
1
=−1.43(4) =−5.72
lbf · in
M
2
=−5.72 − 41.43(4) =−171.44
lbf · in
M
3
=−171.44 − 11.43(6) =−240
lbf · in
M
4
=−240 + 60(4) = 0
checks!
4" 4" 6" 4"
Ϫ1.43
Ϫ41.43
Ϫ11.43
60
40 lbf 60 lbf
30 lbf
x
x
x
O
AB CD
y
R
1
R
2
M
1
M
2
M
3
M
4
O
V (lbf)
M
(lbf• in)
O
C
C
DB
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°
305.4 N
400 N
400 N
200 N
41 N
305.4 N
200 N
346.4 N
305.4 N
(R
Cx
)
2
(R
Cy
)
2
C
B
A
2
400 N
4
R
D
(R
Cx
)
4
(R
Cy
)
4
D
C
E
Ans.
shi20396_ch04.qxd 8/18/03 10:35 AM Page 52
Chapter 4 53
(b)
F
y
= 0
R
0
= 2 + 4(0.150) = 2.6
kN
M
0
= 0
M
0
= 2000(0.2) + 4000(0.150)(0.425)
= 655 N · m
M
1
=−655 + 2600(0.2) =−135
N · m
M
2
=−135 + 600(0.150) =−45
N · m
M
3
=−45 +
1
2
600(0.150) = 0
checks!
(c)
M
0
= 0: 10R
2
− 6(1000) = 0 ⇒ R
2
= 600
lbf
F
y
= 0: R
1
− 1000 + 600 = 0 ⇒ R
1
= 400
lbf
M
1
= 400(6) = 2400
lbf · ft
M
2
= 2400 − 600(4) = 0
checks!
(d)
ۗ
+
M
C
= 0
−10R
1
+ 2(2000) + 8(1000) = 0
R
1
= 1200
lbf
F
y
= 0: 1200 − 1000 − 2000 + R
2
= 0
R
2
=
1800 lbf
M
1
= 1200(2) = 2400
lbf · ft
M
2
= 2400 + 200(6) = 3600
lbf · ft
M
3
= 3600 − 1800(2) = 0
checks!
2000 lbf
1000 lbf
R
1
O
O
M
1
M
2
M
3
R
2
6 ft 2 ft2 ft
AB
C
y
M
1200
Ϫ1800
200
x
x
x
6 ft 4 ft
A
O
O
O
B
Ϫ600
M
1
M
2
V (lbf)
1000 lbf
y
R
1
R
2
400
M
(lbf
•ft)
x
x
x
V (kN)
150 mm200 mm 150 mm
2.6
Ϫ655
M
(N• m)
0.6
M
1
M
2
M
3
2 kN
4 kN/m
y
A
O
O
O
O
BC
R
O
M
O
x
x
x
shi20396_ch04.qxd 8/18/03 10:35 AM Page 53
54 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
(e) ۗ
+
M
B
= 0
−7R
1
+ 3(400) − 3(800) = 0
R
1
=−171.4
lbf
F
y
= 0: −171.4 − 400 + R
2
− 800 = 0
R
2
= 1371.4
lbf
M
1
=−171.4(4) =−685.7
lbf · ft
M
2
=−685.7 − 571.4(3) =−2400
lbf · ft
M
3
=−2400 + 800(3) = 0
checks!
(f) Break at A
R
1
= V
A
=
1
2
40(8) = 160
lbf
ۗ
+
M
D
= 0
12(160) − 10R
2
+ 320(5) = 0
R
2
= 352
lbf
F
y
= 0
−160 + 352 − 320 + R
3
= 0
R
3
= 128
lbf
M
1
=
1
2
160(4) = 320
lbf · in
M
2
= 320 −
1
2
160(4) = 0
checks! (hinge)
M
3
= 0 − 160(2) =−320
lbf · in
M
4
=−320 + 192(5) = 640
lbf · in
M
5
= 640 − 128(5) = 0
checks!
40 lbf/in
V (lbf)
O
O
160
Ϫ160
Ϫ128
192
M
320 lbf
160 lbf 352 lbf 128 lbf
M
1
M
2
M
3
M
4
M
5
x
x
x
8"
5"
2"
5"
40 lbf/in
160 lbf
O
A
y
BD
C
A
320 lbf
R
2
R
3
R
1
V
A
A
O
O
O
C
M
V (lbf )
800
Ϫ171.4
Ϫ571.4
3 ft 3 ft4 ft
800 lbf400 lbf
B
y
M
1
M
2
M
3
R
1
R
2
x
x
x
shi20396_ch04.qxd 8/18/03 10:35 AM Page 54
Chapter 4 55
4-4
(a)
q = R
1
x
−1
− 40x − 4
−1
+ 30x − 8
−1
+ R
2
x − 14
−1
− 60x − 18
−1
V = R
1
− 40x − 4
0
+ 30x − 8
0
+ R
2
x − 14
0
− 60x − 18
0
(1)
M = R
1
x − 40x − 4
1
+ 30x − 8
1
+ R
2
x − 14
1
− 60x − 18
1
(2)
for
x = 18
+
V = 0
and
M = 0
Eqs. (1) and (2) give
0 = R
1
− 40 + 30 + R
2
− 60 ⇒ R
1
+ R
2
= 70
(3)
0 = R
1
(18) − 40(14) + 30(10) + 4R
2
⇒ 9R
1
+ 2R
2
= 130
(4)
Solve (3) and (4) simultaneously to get
R
1
=−1.43
lbf,
R
2
= 71.43
lbf. Ans.
From Eqs. (1) and (2), at
x = 0
+
,
V = R
1
=−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 = R
0
x
−1
− M
0
x
−2
− 2000x − 0.2
−1
− 4000x − 0.35
0
+ 4000x − 0.5
0
V = R
0
− M
0
x
−1
− 2000x − 0.2
0
− 4000x − 0.35
1
+ 4000x − 0.5
1
(1)
M = R
0
x − M
0
− 2000x − 0.2
1
− 2000x − 0.35
2
+ 2000x − 0.5
2
(2)
at
x = 0.5
+
m,
V = M = 0
, Eqs. (1) and (2) give
R
0
− 2000 − 4000(0.5 − 0.35) = 0 ⇒ R
1
= 2600 N = 2.6
kN Ans.
R
0
(0.5) − M
0
− 2000(0.5 − 0.2) − 2000(0.5 − 0.35)
2
= 0
with
R
0
= 2600
N,
M
0
= 655
N · m Ans.
With R
0
and M
0
, Eqs. (1) and (2) give the same V and M curves as Prob. 4-3 (note for
V,
M
0
x
−1
has no physical meaning).
(c)
q = R
1
x
−1
− 1000x − 6
−1
+ R
2
x − 10
−1
V = R
1
− 1000x − 6
0
+ R
2
x − 10
0
(1)
M = R
1
x − 1000x − 6
1
+ R
2
x − 10
1
(2)
at
x = 10
+
ft,
V = M = 0
, Eqs. (1) and (2) give
R
1
− 1000 + R
2
= 0 ⇒ R
1
+ R
2
= 1000
10R
1
− 1000(10 − 6) = 0 ⇒ R
1
= 400 lbf
,
R
2
= 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.
shi20396_ch04.qxd 8/18/03 10:35 AM Page 55
56 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
(d)
q = R
1
x
−1
− 1000x − 2
−1
− 2000x − 8
−1
+ R
2
x − 10
−1
V = R
1
− 1000x − 2
0
− 2000x − 8
0
+ R
2
x − 10
0
(1)
M = R
1
x − 1000x − 2
1
− 2000x − 8
1
+ R
2
x − 10
1
(2)
At
x = 10
+
,
V = M = 0
from Eqs. (1) and (2)
R
1
− 1000 − 2000 + R
2
= 0 ⇒ R
1
+ R
2
= 3000
10R
1
− 1000(10 − 2) − 2000(10 − 8) = 0 ⇒ R
1
= 1200 lbf
,
R
2
= 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 = R
1
x
−1
− 400x − 4
−1
+ R
2
x − 7
−1
− 800x − 10
−1
V = R
1
− 400x − 4
0
+ R
2
x − 7
0
− 800x − 10
0
(1)
M = R
1
x − 400x − 4
1
+ R
2
x − 7
1
− 800x − 10
1
(2)
at
x = 10
+
,
V = M = 0
R
1
− 400 + R
2
− 800 = 0 ⇒ R
1
+ R
2
= 1200
(3)
10R
1
− 400(6) + R
2
(3) = 0 ⇒ 10R
1
+ 3R
2
= 2400
(4)
Solve Eqs. (3) and (4) simultaneously:
R
1
=−171.4
lbf,
R
2
= 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 = R
1
x
−1
− 40x
0
+ 40x − 8
0
+ R
2
x − 10
−1
− 320x − 15
−1
+ R
3
x − 20
V = R
1
− 40x + 40x − 8
1
+ R
2
x − 10
0
− 320x − 15
0
+ R
3
x − 20
0
(1)
M = R
1
x − 20x
2
+ 20x − 8
2
+ R
2
x − 10
1
− 320x − 15
1
+ R
3
x − 20
1
(2)
M = 0 at x = 8 in ∴
8R
1
− 20(8)
2
= 0 ⇒ R
1
= 160 lbf
at
x = 20
+
, V and M = 0
160 − 40(20) + 40(12) + R
2
− 320 + R
3
= 0 ⇒ R
2
+ R
3
= 480
160(20) − 20(20)
2
+ 20(12)
2
+ 10R
2
− 320(5) = 0 ⇒ R
2
= 352
lbf
R
3
= 480 − 352 = 128 lbf
0 ≤ x ≤ 8: V = 160 − 40x lbf, M = 160x −20x
2
lbf · in
8 ≤ x ≤ 10: V = 160 −40x + 40(x − 8) =−160 lbf
,
M = 160x − 20x
2
+ 20(x −8)
2
= 1280 − 160x lbf · in
shi20396_ch04.qxd 8/18/03 10:35 AM Page 56
Chapter 4 57
10 ≤ x ≤ 15: V = 160 −40x + 40(x −8) + 352 = 192 lbf
M = 160x − 20x
2
+ 20(x −8) + 352(x − 10) = 192x − 2240
15 ≤ x ≤ 20: V = 160 − 40x + 40(x −8) + 352 − 320 =−128 lbf
M = 160x − 20x
2
− 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,
x
c
= (l − 2a)/2
M
c
=
w
2
l
2
(l − 2a) −
l
2
2
=
wl
2
l
4
− a
At reaction,
|M
r
|=wa
2
/2
a = 2.25, l = 10 in, w = 100
lbf/in
M
c
=
100(10)
2
10
4
− 2.25
= 125
lbf · in
M
r
=
100(2.25
2
)
2
= 253.1
lbf · in Ans.
(b) Minimum occurs when
M
c
=|M
r
|
wl
2
l
4
− a
=
wa
2
2
⇒ a
2
+ al − 0.25l
2
= 0
Taking the positive root
a =
1
2
−l +
l
2
+ 4(0.25l
2
)
=
l
2
√
2 − 1
= 0.2071l
Ans.
for l = 10 in and
w
= 100 lbf,
M
min
= (100/2)[(0.2071)(10)]
2
= 214.5
lbf · in
4-7 For the ith wire from bottom, from summing forces vertically
(a)
T
i
= (i + 1)W
From summing moments about point a,
M
a
= W(l − x
i
) − iWx
i
= 0
Giving,
x
i
=
l
i + 1
WiW
T
i
x
i
a
shi20396_ch04.qxd 8/18/03 10:35 AM Page 57
58 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
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:
M
a
= 0
M
a
=
iWl
i + 1
cos α −
ilW
i + 1
cos β = 0
iWl
i + 1
(cos α −cos β) = 0
Moment vanishes when
α = β
for any wire. Consider a ccw rotation angle
β
, which
makes
α → α +β
and
β → α − β
M
a
=
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 =
3
2
+ 4
2
= 5
σ
1
= 5 + 9 = 14
σ
2
= 9 − 5 = 4
2
s
(12, 4
cw
)
C
R
D
2
1
1
2
2
p
(6, 4
ccw
)
y
x
cw
ccw
W
i
W
il
i ϩ 1
T
i

␣
l
i ϩ 1
shi20396_ch04.qxd 8/18/03 10:35 AM Page 58
Chapter 4 59
φ
p
=
1
2
tan
−1
4
3
= 26.6
◦
cw
τ
1
= R = 5, φ
s
= 45
◦
− 26.6
◦
= 18.4
◦
ccw
(b)
C =
9 + 16
2
= 12.5
CD =
16 − 9
2
= 3.5
R =
5
2
+ 3.5
2
= 6.10
σ
1
= 6.1 + 12.5 = 18.6
φ
p
=
1
2
tan
−1
5
3.5
= 27.5
◦
ccw
σ
2
= 12.5 − 6.1 = 6.4
τ
1
= R = 6.10
,
φ
s
= 45
◦
− 27.5
◦
= 17.5
◦
cw
(c)
C =
24 + 10
2
= 17
CD =
24 − 10
2
= 7
R =
7
2
+ 6
2
= 9.22
σ
1
= 17 + 9.22 = 26.22
σ
2
= 17 − 9.22 = 7.78
2
s
(24, 6
cw
)
C
R
D
2
1
1
2
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y
x
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12.5
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6.10
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x
6.4
18.6
27.5Њ
2
s
(16, 5
ccw
)
C
R
D
2
1
1
2
2
p
(9, 5
cw
)
y
x
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3
5
3
3
3
18.4Њ
x
x
4
14
26.6Њ
shi20396_ch04.qxd 8/18/03 10:35 AM Page 59
[...]... θ(deg) 0 0.125 0.225 0.325 0 .42 5 0 .47 5 0.902 5 0.889 087 0.859 043 0.811 831 0. 747 45 0 0.708 822 3.8 3.585 398 3 .41 3 717 3. 242 035 3.070 3 54 2.9 84 513 0 0.10 0.20 0.30 0 .40 0 .45 1037.9 1022.5 987.9 933.6 859.6 815.1 0 0.10 0.20 0.30 0 .40 0 .45 4. 825 4. 621 4. 553 4. 576 4. 707 4. 825 1200 1000 800 T (lbf • in) shi20396_ ch 04. qxd 600 40 0 200 0 0 0.1 0.2 0.3 ri (in) 0 .4 0.5 shi20396_ ch 04. qxd 82 8/18/03 10:36 AM... 4. 57 y φp = 4. 57 1 5 90 + tan−1 = 61.0◦ cw 2 8 x 61Њ 23 .43 τ1 = R = 9 .43 4, φs = 61◦ − 45 ◦ = 16◦ cw 14 x 16Њ 14 9 .43 4 9 + 19 = 14 2 shi20396_ ch 04. qxd 8/18/03 10:35 AM Page 61 61 Chapter 4 4-9 (a) 1 cw C= y (12, 7cw) R 2 D 1 2p σ1 = 4 + 10.63 = 14. 63 σ2 = 4 − 10.63 = −6.63 2s x 2 ccw 12 + 4 =8 2 R = 82 + 72 = 10.63 CD = C ( 4, 7ccw) 12 − 4 =4 2 φp = 14. 63 69 .4 1 8 90 + tan−1 = 69 .4 ccw 2 7... = 45 ◦ − 13.28◦ = 31.72◦ cw 10 22.36 x 31.72Њ 10 (c) 1 cw 2s (Ϫ10, 9cw) 2 2p D 1 C R (18, 9ccw) y ccw −10 + 18 =4 2 10 + 18 CD = = 14 2 R = 142 + 92 = 16. 64 C= x σ1 = 4 + 16. 64 = 20. 64 σ2 = 4 − 16. 64 = −12. 64 2 φp = 12. 64 1 14 90 + tan−1 = 73.63◦ cw 2 9 x 73.63Њ 20. 64 τ1 = R = 16. 64, φs = 73.63◦ − 45 ◦ = 28.63◦ cw 4 x 28.63Њ 4 16. 64 shi20396_ ch 04. qxd 8/18/03 10:35 AM Page 65 65 Chapter 4. .. 8/18/03 10:36 AM Page 85 85 Chapter 4 (τmax ) rd = Round: 16 T 16T 3. 545 T = = 3 3/2 π d π(4A/π) ( A) 3/2 (τmax ) sq 4. 8 = 1.3 54 = (τmax ) rd 3. 545 Square stress is 1.3 54 times the round stress 4- 45 s= √ A, d= Ans 4A/π Square: Eq (4- 44) with b = c, β = 0. 141 θsq = Tl Tl = 4G 0. 141 c 0. 141 ( A) 4/ 2 G Round: θrd = Tl Tl 6.2832T l = = JG ( A) 4/ 2 G (π/32) (4A/π) 4/ 2 G θsq 1/0. 141 = 1.129 = θrd 6.2832 Square has... ji = 0 WT = 1 04. 4, W1 = W2 = W3 = W4 = 1 04.4 = 26.1 kip 4 (1200 − 238) 2 M1 = (1 04. 4) = 20 128 kip · in 4( 1200) 47 6 = 238 in, Wheel 1: d1 = 2 Wheel 2: i−1 d2 = 238 − 84 = 1 54 in (1200 − 1 54) 2 M2 = (1 04. 4) − 26.1( 84) = 21 605 kip · in = Mmax 4( 1200) Check if all of the wheels are on the rail 84" 77" 84" 315" xmax 600" 600" (b) xmax = 600 − 77 = 523 in (c) See above sketch (d) inner axles 4- 23 (a) D c1... psi 2 (2)(1) Ans shi20396_ ch 04. qxd 8/18/03 10:36 AM Page 77 77 Chapter 4 4-26 Mmax = wl 2 8 ⇒ σmax = wl 2 c 8I ⇒ w= 8σ I cl 2 (a) l = 12(12) = 144 in, I = (1/12)(1.5)(9.5) 3 = 107.2 in4 w= 8(1200)(107.2) = 10 .4 lbf/in 4. 75( 144 2 ) Ans (b) l = 48 in, I = (π/ 64) ( 24 − 1.2 54 ) = 0.6656 in4 w= 8(12)(103 )(0.6656) = 27.7 lbf/in 1 (48 ) 2 Ans 8(12)(103 )(2.051) = 57.0 lbf/in 1.5 (48 ) 2 Ans l = 48 in, I = (1/12)(2)(33... A-17, select 1 3 /4 in τstart = 16(2)(12 605) = 23.96(103 ) psi = 23.96 kpsi 3) π(1.75 (b) design activity 4- 43 ω = 2πn/60 = 2π(8)/60 = 0.8378 rad/s T = dC = 1000 H = = 11 94 N · m ω 0.8378 16T πτ 1/3 = From Table A-17, select 45 mm 4- 44 s= √ A, d = 16(11 94) π(75)(106 ) 1/3 = 4. 328(10−2 ) m = 43 .3 mm Ans 4A/π Square: Eq (4- 43) with b = c 4. 8T c3 4. 8T (τmax ) sq = ( A) 3/2 τmax = shi20396_ ch 04. qxd 8/18/03... ⇒ 150(106 ) = 4. 023(1 04 ) 150(106 ) 1/3 4. 023(1 04 ) 540 0(d/2) = (π/32)[d 4 − (0.75d) 4 ] d3 = 6 .45 (10−2 ) m = 64. 5 mm From Table A-17, the next preferred size is d = 80 mm; ID = 60 mm π (0.0 84 − 0.0 64 ) = 2. 749 (10−6 ) mm4 32 540 0(0.030) = 58.9(106 ) Pa = 58.9 MPa τi = 2. 749 (10−6 ) Ans J= (b) Ans 4- 42 63 025(1) 63 025H = = 12 605 lbf · in n 5 16T 16T 1/3 16(12 605) τ= ⇒ dC = = 3 πτ π( 14 000) πdC T =... MPa 30 − 10 = 10 MPa τ1/2 = 2 Ans 10 + 20 = 15 MPa τ2/3 = 2 30 + 20 = 25 MPa τmax = τ1/3 = 2 4- 19 (MPa) 1/3 2/3 1/2 Ϫ20 10 Ans From Eq (4- 15) σ 3 − (1 + 4 + 4) σ 2 + [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 σ 3 − 9σ 2 = 0 30 (MPa) shi20396_ ch 04. qxd 70 8/18/03 10:36 AM Page 70 Solutions Manual • Instructor’s Solution Manual to Accompany... 2 Am = (1 − 0.05) 2 − 4 rm − rm = 0.952 − (4 − π)rm 4 L m = 4( 1 − 0.05 − 2rm + 2πrm /4) = 4[ 0.95 − (2 − π/2)rm ] T = 2Am tτ = 2(0.05)(11 500) Am = 1150Am Eq (4- 45): Eq (4- 46): θ(deg) = θ1 l T L m l 180 T L m (40 ) 180 180 = = π 4G A2 t π 4( 11.5)(106 ) A2 (0.05) π m m = 9.9 645 (10 4 ) T Lm A2 m Equations can then be put into a spreadsheet resulting in: ri 0 0.10 0.20 0.30 0 .40 0 .45 rm Am Lm ri T(lbf · . R
1
=−1 .43
lbf,
M = 0
x = 4
+
: V =−1 .43 − 40 = 41 .43 , M =−1 .43 x
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 . in)
O
C
C
DB
A
B
D
E
305 .4 N
346 .4 N
305 .4 N
41 N
40 0 N
200 N
40 0 N
200 N
40 0 N
Pin C
30°
305 .4 N
40 0 N
40 0 N
200 N
41 N
305 .4 N
200 N
346 .4 N
305 .4 N
(R
Cx
)
2
(R
Cy
)
2
C
B
A
2
40 0