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Shigley’s MED, 10
th edition
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Notes to the Instructor.
1. For Problems 8-33 through 8
Prob. 8-33 which covers a change which will be
edition.
2. For Probs. 8-41 to 8-44. These problems, as well as many others in this chapter are best
implemented using a spreadsheet.
8-1 (a) Thread depth= 2_._ 5 mm
Width = 2_._ 5 mm Ans. dm = 25 - 1_._ 25 - 1_._ 25 = 22 dr = 25 - 5 = 20 mm
l = p = 5 mm Ans.
(b) Thread depth = 2.5 mm
Width at pitch line = 2.5 mm dm = 22_._ 5 mm
dr = 20 mm l = p = 5 mm Ans.
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8-2 From Table 8-1,
1.226 869
0.649 519
1.226 869 0.649 519
r
m
d d p
d d p
d p d p d d p
A t d p Ans
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8-3 From Eq. ( c ) of Sec. 8-
P R = F
Chap. 8 Solutions, Page
Chapter 8
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33 through 8-44 and 8-51 through 8-58. There is a statement preceding 33 which covers a change which will be corrected in the second printing of the 10
. These problems, as well as many others in this chapter are best
implemented using a spreadsheet.
5 mm Ans.
Ans. 25 = 22_._ 5 mm
Thread depth = 2.5 mm Ans.
Width at pitch line = 2.5 mm Ans.
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d d p
d d p
d p d p d d p
2 2 ( 0.938 194 ). 4 4
t
d A d p Ans
π π = = −
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tan
1 tan
f P F f
λ
λ
Chap. 8 Solutions, Page 1/
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There is a statement preceding corrected in the second printing of the 10
th
. These problems, as well as many others in this chapter are best
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d d 0.938 194 p
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Shigley’s MED, 10
th edition
R
P d Fd f T = =
0
R m
T Fl f f e Ans T Fd f f
Using f = 0_._ 08, form a table and plot the efficiency curve.
λ, deg. e
0 0 0 0.
20 0. 30 0.
40 0. 45 0.
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8-4 Given F = 5 kN, l = 5 mm, and
raise the load is found using Eqs. (
T R Ans π
The torque required to lower the load, from Eqs. (
T L Ans π
Since TL is positive, the thread is self
e Ans π
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8-5 Collar (thrust) bearings, at the bottom of the screws, must bear on the collars. The bottom
segment of the screws must be in compression. Where grips must be in tension. Both screws must be of the same
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Chap. 8 Solutions, Page
tan
2 2 1 tan
P d R m Fd (^) m f
f
λ
λ
0 / (2^ ) 1^ tan^ tan 1 tan.
R m^ / 2^ tan^ tan
T Fl f f e Ans T Fd f f
π λ λ λ λ λ
08, form a table and plot the efficiency curve.
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= 5 mm, and dm = d − p /2 = 25 − 5/2 = 22.5 mm, the torque required to
raise the load is found using Eqs. (8-1) and (8-6)
15.85 N m. 2 22.5 0.09 5 2
T Ans
π
π
The torque required to lower the load, from Eqs. (8-2) and (8-6) is
7.83 N m. 2 22.5 0.09 5 2
T Ans
π
π
is positive, the thread is self-locking. From Eq.(8-4) the efficiency is
e Ans π
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Collar (thrust) bearings, at the bottom of the screws, must bear on the collars. The bottom
s must be in compression. Whereas, tension specimens and the grips must be in tension. Both screws must be of the same-hand threads.
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Chap. 8 Solutions, Page 2/
e tan Ans. T Fd f f
π λ λ
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5/2 = 22.5 mm, the torque required to
T = + = 15.85 N m⋅ Ans.
T = + = 7.83 N m⋅ Ans.
- the efficiency is
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Collar (thrust) bearings, at the bottom of the screws, must bear on the collars. The bottom
tension specimens and their hand threads.
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th
( c ) The column has one end fixed and the other end pivoted. Base the decision on the
mean diameter column. Input: C = 1.2, D = 0.685 in, A = π(0.
2 )/4 = 0.369 in
2 , Sy = 41 kpsi, E = 30(
6 ) psi, L = 6 in, k = D/ 4 =0_._ 171 25 in, L/k = 35.04. From Eq. (4-45),
( ) ( )
1/2 1/ 2 2 6
1
y^ 41 000
l CE
k S
^ π^ ^ π =^ ^ =^ ^ = ^ ^
From Eq. (4-46), the limiting clamping force for buckling is
( )
( )
( )
2
clamp c r
3 2 3 3 6
0.369 41 10 35.04 14.6 10 lbf 2 1.2 30 10
y y
S (^) l F P A S k CE
Ans
π
π
^
( d ) This is a subject for class discussion.
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8-8 T = 8(3.5) = 28 lbf ⋅ in
0.6667 in 4 12
d m = − =
l =
= 0.1667 in, α =
0 29
2
0 , sec 14.
0 = 1.
From Eqs. (8-5) and (8-6)
total
0.6667 0.1667^ 0.15^ 0.6667^ 1.033^ 0.15 1
F^ F
T F
π
π
182 lbf.
F = = Ans
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8-9 dm = 1.5 − 0.25/2 = 1.375 in, l = 2(0.25) = 0.5 in
From Eq. (8-1) and Eq. (8-6),
th
( ) ( )
3 3 2.2 10 (1.375) (^) 0.5 (0.10)(1.375) 2.2 10 (0.15)(2.25)
2 (1.375) 0.10(0.5) 2
330 371 701 lbf · in
T R
π
π
Since n = V/l = 2 / 0.5 = 4 rev/s = 240 rev/min
so the power is
2.67 hp. 63 025 63 025
Tn H = = = Ans
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8-10 dm = 40 − 4 = 36 mm, l = p = 8 mm
From Eqs. (8-1) and (8-6)
(3.831 4.5) 8.33 N · m ( in kN)
2 2 (1) 2 rad/s
477 N · m 2 477 57.3 kN.
F F
T
F F F
n
H T
H T
F Ans
π
π
ω π π π
ω
ω π
Fl e Ans π T π
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8-11 ( a ) Table A-31, nut height H = 12.8 mm. L ≥ l + H = 2(15) + 12.8 = 42.8 mm. Rounding
up,
L = 45 mm Ans.
( b ) From Eq. (8-14), LT = 2 d + 6 = 2(14) +6 = 34 mm
From Table 8-7, ld = L − LT = 45 − 34 = 11 mm, lt = l − ld = 2(15) − 11 = 19 mm,
Ad = π (
2 ) / 4 = 153.9 mm
2
. From Table 8-1, At = 115 mm
2
. From Eq. (8-17)
874.6 MN/m. 153.9 19 115 11
d t b d t t d
A A E
k Ans A l A l
( c ) From Eq. (8-22), with l = 2(15) = 30 mm
th
From Table 8-7, ld = L − LT = 40 − 34 = 6 mm, lt = l − ld = 22 − 6 = 16 mm
Ad = π (
2 ) / 4 = 153.9 mm
2
. From Table 8-1, At = 115 mm
2
. From Eq. (8-17)
1 162.2 MN/m. 153.9 16 115 6
d t b d t t d
A A E
k Ans A l A l
( c ) From Eq. (8-22), with l = 22 mm
0.5774 0.5774^ 207 14
3 624.4 MN/m. 0.5774 0.5 (^) 0.5774 22 0.5 14 2 ln (^5) 2 ln 5 0.5774 2.5 (^) 0.5774 22 2.5 14
m
Ed k Ans l d
l d
π^ π = = = + (^) + (^) + (^) +
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8-14 ( a ) From Table A-31, the nut height is H = 7/16 in. L ≥ l + H = 2 + 1 + 7/16 = 3 7/16 in.
Rounding up, L = 3.5 in Ans.
( b ) From Eq. (8-13), LT = 2 d + 1/4 = 2(0.5) + 0.25 = 1.25 in
From Table 8-7, ld = L − LT = 3.5 − 1.25 = 2.25 in, lt = l − ld = 3 − 2.25 = 0.75 in
Ad = π (0.
2 )/4 = 0.1963 in
2
. From Table 8-2, At = 0.1419 in
2
. From Eq. (8-17)
1.79 Mlbf/in. 0.1963 0.75 0.1419 2.
d t b d t t d
A A E
k Ans A l A l
th
( c )
Top steel frustum: t = 1.5 in, d = 0.5 in, D = 0.75 in, E = 30 Mpsi. From Eq. (8-20)
1
22.65 Mlbf/in 1.155 1.5 0.75 0.5 0.75 0. ln 1.155 1.5 0.75 0.5 0.75 0.
k
π = = +^ −^ +
+ + −
Lower steel frustum: t = 0.5 in, d = 0.5 in, D = 0.75 + 2(1) tan 30° = 1.905 in, E = 30
Mpsi. Eq. (8-20) ⇒ k 2 = 210.7 Mlbf/in
Cast iron: t = 1 in, d = 0.5 in, D = 0.75 in, E = 14.5 Mpsi (Table 8-8). Eq. (8-20) ⇒ k 3 = 12.27 Mlbf/in
From Eq. (8-18)
km = (1/ k 1 + 1/ k 2 +1/ k 3 )
− 1 = (1/22.65 + 1/210.7 + 1/12.27)
− 1 = 7.67 Mlbf/in Ans.
8-15 ( a ) From Table A-32, the washer thickness is 0.095 in. Thus, l = 2 + 1 + 2(0.095) = 3.
in. From Table A-31, the nut height is H = 7/16 in. L ≥ l + H = 3.19 + 7/16 = 3.63 in.
Rounding up, L = 3.75 in Ans.
( b ) From Eq. (8-13), LT = 2 d + 1/4 = 2(0.5) + 0.25 = 1.25 in
From Table 8-7, ld = L − LT = 3.75 − 1.25 = 2.5 in, lt = l − ld = 3.19 − 2.5 = 0.69 in
Ad = π (0.
2 )/4 = 0.1963 in
2
. From Table 8-2, At = 0.1419 in
2
. From Eq. (8-17)
th
Ad = π (0.
2 )/4 = 0.1963 in
2
. From Table 8-2, At = 0.1419 in
2
. From Eq. (8-17)
2.321 Mlbf/in. 0.1963 0.75 0.1419 1.
d t b d t t d
A A E
k Ans A l A l
( c )
Top steel frustum: t = 1.125 in, d = 0.5 in, D = 0.75 in, E = 30 Mpsi. From Eq. (8-20)
1
24.48 Mlbf/in 1.155 1.125 0.75 0.5 0.75 0. ln 1.155 1.125 0.75 0.5 0.75 0.
k
π = = + − +
+ + −
Lower steel frustum: t = 0.875 in, d = 0.5 in, D = 0.75 + 2(0.25) tan 30° = 1.039 in, E =
30 Mpsi. Eq. (8-20) ⇒ k 2 = 49.36 Mlbf/in
Cast iron: t = 0.25 in, d = 0.5 in, D = 0.75 in, E = 14.5 Mpsi (Table 8-8). Eq. (8-20) ⇒
k 3 = 23.49 Mlbf/in
From Eq. (8-18)
km = (1/ k 1 + 1/ k 2 +1/ k 3 )
− 1 = (1/24.48 + 1/49.36 + 1/23.49)
− 1 = 9.645 Mlbf/in Ans.
8-17 a ) Grip, l = 2(2 + 0.095) = 4.19 in. L ≥ 4.19 + 7/16 = 4.628 in.
Rounding up, L = 4.75 in Ans.
( b ) From Eq. (8-13), LT = 2 d + 1/4 = 2(0.5) + 0.25 = 1.25 in
th
From Table 8-7, ld = L − LT = 4.75 − 1.25 = 3.5 in, lt = l − ld = 4.19 − 3.5 = 0.69 in
Ad = π (0.
2 )/4 = 0.1963 in
2
. From Table 8-2, At = 0.1419 in
2
. From Eq. (8-17)
1.322 Mlbf/in. 0.1963 0.69 0.1419 3.
d t b d t t d
A A E
k Ans A l A l
( c )
Upper and lower halves are the same. For the upper half,
Steel frustum: t = 0.095 in, d = 0.531 in, D = 0.75 in, and E = 30 Mpsi. From Eq. (8-20)
1
89.20 Mlbf/in 1.155 0.095 0.75 0.531 0.75 0. ln 1.155 0.095 0.75 0.531 0.75 0.
k
π = = + − +
+ + −
Aluminum: t = 2 in, d = 0.5 in, D =0.75 + 2(0.095) tan 30° = 0.860 in, and E = 10.
Mpsi. Eq. (8-20) ⇒ k 2 = 9.24 Mlbf/in
For the top half, km ′ = (1/ k 1 + 1/ k 2 )
− 1 = (1/89.20 + 1/9.24)
− 1 = 8.373 Mlbf/in
Since the bottom half is the same, the overall stiffness is given by
km = (1/ km ′^ + 1/ km ′^ )
− 1 = km ′^ /2 = 8.373/2 = 4.19 Mlbf/in Ans
8-18 ( a ) Grip, l = 2(2 + 0.095) = 4.19 in. L ≥ 4.19 + 7/16 = 4.628 in.
Rounding up, L = 4.75 in Ans.
( b ) From Eq. (8-13), LT = 2 d + 1/4 = 2(0.5) + 0.25 = 1.25 in
th
( b ) From Eq. (8-14), LT = 2 d + 6 = 2(10) + 6 = 26 mm, ld = L − LT = 60 − 26 =
34 mm, lt = l − l = 50 − 34 = 16 mm. Ad = π (
2 ) / 4 = 78.54 mm
2
. From Table 8-1,
At = 58 mm
2
. From Eq. (8-17)
292.1 MN/m. 78.54 16 58.0 34
d t b d t t d
A A E
k Ans A l A l
( c )
Upper and lower frustums are the same. For the upper half,
Aluminum: t = 10 mm, d = 10 mm, D = 15 mm, and from Table 8-8, E = 71 GPa. From Eq. (8-20)
1
1576 MN/m 1.155 10 15 10 15 10 ln 1.155 10 15 10 15 10
k
π = = +^ −^ +
+ + −
Steel: t = 15 mm, d = 10 mm, D = 15 + 2(10) tan 30° = 26.55 mm, and E = 207 GPa. From Eq. (8-20)
2
11 440 MN/m 1.155 15 26.55 10 26.55 10 ln 1.155 15 26.55 10 26.55 10
k
π = = + − +
+ + −
For the top half, km ′ = (1/ k 1 + 1/ k 2 )
− 1 = (1/1576 + 1/11 440)
− 1 = 1385 MN/m
Since the bottom half is the same, the overall stiffness is given by
km = (1/ km ′ + 1/ km ′ )
− 1 = km ′ /2 = 1385/2 = 692.5 MN/m Ans.
th
8-20 ( a ) From Table A-31, the nut height is H = 8.4 mm. L > l + H = 60 + 8.4 = 68.4 mm.
Rounding up, L = 70 mm Ans.
( b ) From Eq. (8-14), LT = 2 d + 6 = 2(10) + 6 = 26 mm, ld = L − LT = 70 − 26 =
44 mm, lt = l − ld = 60 − 44 = 16 mm. Ad = π (
2 ) / 4 = 78.54 mm
2
. From Table 8-1,
At = 58 mm
2
. From Eq. (8-17)
247.6 MN/m. 78.54 16 58.0 44
d t b d t t d
A A E
k Ans A l A l
( c )
Upper aluminum frustum: t = 10 mm, d = 10 mm, D = 15 mm, and E = 71 GPa. From Eq.
(8-20)
1
1576 MN/m 1.155 2.095 15 10 15 10 ln 1.155 2.095 15 10 15 10
k
π = = + − +
+ + −
Lower aluminum frustum: t = 20 mm, d = 10 mm, D = 15 mm, and E = 71 GPa. Eq.
(8-20) ⇒ k 2 = 1 201 MN/m
Top steel frustum: t = 20 mm, d = 10 mm, D = 15 + 2(10) tan 30° = 26.55 mm, and E =
207 GPa. Eq. (8-20) ⇒ k 3 = 9 781 MN/m
th
Lower steel frustum: t = 17.5 mm, d = 10 mm, D = 15 + 2(5) tan 30° = 20.77 mm, and E
= 207 GPa. Eq. (8-20) ⇒ k 4 = 6 806 MN/m
From Eq. (8-18)
km = (1/ k 1 + 1/ k 2 +1/ k 3 +1/ k 4 )
− 1 = (1/1 576 + 1/2 300 + 1/12 759 +1/6 806)
− 1
= 772.4 MN/m Ans.
8-22 Equation ( f ), p. 428:
b
b m
k C k k
Eq. (8-17):
d t b d t t d
A A E
k A l A l
Eq. (8-22):
2 ln 5 0.5774 40 2.
m
d k d
d
π
(^) + + See Table 8-7 for other terms used.
Using a spreadsheet, with coarse-pitch bolts (units are mm, mm
2 , MN/m):
d At Ad H L > L LT
10 58 78.53982 8.4 48.4 50 26
12 84.3 113.0973 10.8 50.8 55 30
14 115 153.938 12.8 52.8 55 34
16 157 201.0619 14.8 54.8 55 38
20 245 314.1593 18 58 60 46
24 353 452.3893 21.5 61.5 65 54
30 561 706.8583 25.6 65.6 70 66
d l ld lt kb km C
10 40 24 16 356.0129 1751.566 0.
12 40 25 15 518.8172 2235.192 0.
14 40 21 19 686.2578 2761.721 0.
16 40 17 23 895.9182 3330.796 0.
20 40 14 26 1373.719 4595.515 0.
24 40 11 29 1944.24 6027.684 0.
30 40 4 36 2964.343 8487.533 0.
Use a M14 × 2 bolt, with length 55 mm. Ans.
8-23 Equation ( f ), p. 428:
b
b m
k C k k
th
Eq. (8-17): (^) b d^ t
d t t d
A A E
k A l A l
For upper frustum, Eq. (8-20), with D = 1.5 d and t = 1.5 in:
1
ln ln 5 1.155 1.5 2.5 0.5 1.733^ 2.
d d k d d d
d d^ d
π π = = (^) + (^) + ^ + ^ + ^
Lower steel frustum, with D = 1.5 d + 2(1) tan 30° = 1.5 d + 1.155, and t = 0.5 in:
2
ln 1.733 2.5 0.5 1.
d k d d
d d
π
(^) + + +^ +
For cast iron frustum, let E = 14. 5 Mpsi, and D = 1.5 d , and t = 1 in:
3
ln 5 1.155 2.
d k d
d
π
+ +
Overall, km = (1/ k 1 +1/ k 2 +1/ k 3 )
− 1
See Table 8-7 for other terms used.
Using a spreadsheet, with coarse-pitch bolts (units are in, in
2 , Mlbf/in):
th
Top frustum, Eq. (8-20), with E = 10.3Mpsi, D = 1.5 d , and t = l /2:
1
ln 5 1.155 / 2 2.
d k l d
l d
π
(^) + +
Middle frustum, with E = 10.3 Mpsi, D = 1.5 d + 2( l − 0.5) tan 30°, and t = 0.5 − l /
{ (^ )^ (^ )^ (^ ) }
{ (^ )^ (^ )^ (^ ) }
(^2 0 )
0 0
1.155 0.5 0.5 0.5 2 0.5 tan 30 2.5 2 0.5 tan 30 ln 1.155 0.5 0.5 2.5 2 0.5 tan 30 0.5 2 0.5 tan 30
d k l d l d l
l d l d l
π
(^) − + + − (^) + −
(^) − + + − (^) + −
Lower frustum, with E = 30Mpsi, D = 1.5 d , t = l − 0.
3
ln 5 1.155 0.5 2.
d k l d
l d
π
(^) − + ^ −^ +
See Table 8-7 for other terms used. Using a spreadsheet, with coarse-pitch bolts (units are in, in
2 , Mlbf/in)
Size d At Ad L > L LT l ld
1 0.073 0.00263 0.004185 0.6095 0.75 0.396 0.5365 0.
2 0.086 0.0037 0.005809 0.629 0.75 0.422 0.543 0.
3 0.099 0.00487 0.007698 0.6485 0.75 0.448 0.5495 0.
4 0.112 0.00604 0.009852 0.668 0.75 0.474 0.556 0.
5 0.125 0.00796 0.012272 0.6875 0.75 0.5 0.5625 0.
6 0.138 0.00909 0.014957 0.707 0.75 0.526 0.569 0.
8 0.164 0.014 0.021124 0.746 0.75 0.578 0.582 0.
10 0.19 0.0175 0.028353 0.785 1 0.63 0.595 0.
Size d lt kb k 1 k 2 k 3 km C
1 0.073 0.1825 0.194841 1.084468 1.954599 7.09432 0.635049 0.
2 0.086 0.215 0.261839 1.321595 2.449694 8.357692 0.778497 0.
3 0.099 0.2475 0.333134 1.570439 2.993366 9.621064 0.930427 0.
4 0.112 0.28 0.403377 1.830494 3.587564 10.88444 1.090613 0.
5 0.125 0.3125 0.503097 2.101297 4.234381 12.14781 1.258846 0.
6 0.138 0.345 0.566787 2.382414 4.936066 13.41118 1.434931 0.
8 0.164 0.41 0.801537 2.974009 6.513824 15.93792 1.809923 0.
10 0.19 0.225 1.15799 3.602349 8.342138 18.46467 2.214214 0.
Use a 2−56 UNC × 0.75 in long bolt. Ans.
th
8-25 For half of joint, Eq. (8-20): t = 20 mm, d = 14 mm, D = 21 mm, and E = 207 GPa
1
5523 MN/m 1.155 20 21 14 21 14 ln 1.155 20 21 14 21 14
k
π = = + − +
km = (1/ k 1 + 1/ k 1 )
− 1 = k 1 /2 = 5523/2 = 2762 MN/m Ans.
From Eq. (8-22) with l = 40 mm
2762 MN/m. 0.5774 40 0.5 14 2 ln 5 0.5774 40 2.5 14
k (^) m Ans
π = = (^) + +
which agrees with the earlier calculation.
For Eq. (8-23), from Table 8-8, A = 0.787 15, B = 0.628 73
km = 207(14)(0.78 715) exp [0.628 73(14)/40] = 2843 MN/m Ans.
This is 2.9% higher than the earlier calculations.
8-26 ( a ) Grip, l = 10 in. Nut height, H = 41/64 in (Table A-31).
L ≥ l + H = 10 + 41/64 = 10.641 in. Let L = 10.75 in.
Table 8-7, LT = 2 d + 0.5 = 2(0.75) + 0.5 = 2 in, ld = L − LT = 10.75 − 2 = 8.75 in,
lt = l − ld = 10 − 8.75 = 1.25 in
Ad = π(0.
2 )/4 = 0.4418 in
2 , At = 0.373 in
2 (Table 8-2) Eq. (8-17),
1.296 Mlbf/in. 0.4418 1.25 0.373 8.
d t b d t t d
A A E
k Ans A l A l
Eq. (4-4), p. 163,
( )( )
2 2 / 4 1.125 0.75 30 1.657 Mlbf/in. 10
m m m
A E
k Ans l
π − = = =
Eq. ( f ), p. 428, C = kb /( kb + km ) = 1.296/(1.296 + 1.657) = 0.439 Ans.
