Baixe Soluções do Capítulo 10 do Livro Shigley's Mechanical Engineering Design, 10ª Edição e outras Exercícios em PDF para Engenharia Mecânica, somente na Docsity!
th
Chapter 10
10-1 From Eqs. (10-4) and (10-5)
W B
C C
K K
C C C
Plot 100( KW − KB )/ KW vs. C for 4 ≤ C ≤ 12 obtaining
We see the maximum and minimum occur at C = 4 and 12 respectively where
Maximum = 1.36 % Ans ., and Minimum = 0.743 % Ans.
10-2 A = Sd
m
dim( A uscu) = [dim ( S ) dim( d
m )]uscu = kpsi⋅in
m
dim( A SI) = [dim ( S ) dim( d
m )]SI = MPa⋅mm
m
SI uscu (^ )^ uscu (^ ) uscu
MPa mm 6.894 757 25.4 6.895 25.. kpsi in
m m m
m
A = ⋅ A = A A Ans
For music wire, from Table 10-4:
A uscu = 201 kpsi⋅in
m , m = 0.145; what is A SI?
A SI = 6.895(25.4)
(201) = 2215 MPa⋅mm
m Ans.
10-3 Given: Music wire, d = 2.5 mm, OD = 31 mm, plain ground ends, Nt = 14 coils.
1
4 6 8 10 12
100( KW-KB )/ KW
C
th
( a ) Table 10-1: Na = Nt − 1 = 14 − 1 = 13 coils
D = OD − d = 31− 2.5 = 28.5 mm
C = D/d = 28.5/2.5 = 11.
Table 10-5: d = 2.5/25.4 = 0.098 in ⇒ G = 81.0(
3 ) MPa
Eq. (10-9):
( )
4 4 3
(^3 )
1.314 N / mm (^8) a 8 28.5 13
d G k D N
= = = Ans.
(b) Table 10-1: Ls = d Nt = 2.5(14) = 35 mm
Table 10-4: m = 0.145, A = 2211 MPa⋅mm
m
Eq. (10-14):
1936 MPa
ut (^) m
A
S
d
Table 10-6: Ssy = 0.45(1936) = 871.2 MPa
Eq. (10-5):
4 2 4 11.4^2
B
C
K
C
+^ +
Eq. (10-7):
( )
3 3 2.5 871. 167.9 N 8 8 1.117 28.
sy s B
d S F K D
π^ π = = = Ans.
(c) (^) 0
35 162.8 mm.
s s
F
L L Ans k
( d ) ( )
(^0) cr
149.9 mm
L = =. Spring needs to be supported.^ Ans.
______________________________________________________________________________
10-4 Given: Design load, F 1 = 130 N.
Referring to Prob. 10-3 solution, C = 11.4, Na = 13 coils, Ssy = 871.2 MPa, Fs = 167.9 N,
L 0 = 162.8 mm and ( L 0 )cr = 149.9 mm_._
Eq. (10-18): 4 ≤ C ≤ 12 C = 11.4 O.K.
Eq. (10-19): 3 ≤ Na ≤ 15 Na = 13 O.K.
Eq. (10-17):
1
F s
F
ξ = − = − =
th
Eq. (10-14):
198.6 kpsi
ut (^) m
A
S
d
Table 10-6: Ssy = 0.50 Sut = 0.50(198.6) = 99.3 kpsi
sy s s
S
n Ans τ
______________________________________________________________________________
10-6 Given: Oil-tempered wire, d = 4 mm, C = 10, plain ends, L 0 = 80 mm, and at F = 50 N,
y = 15 mm.
( a ) k = F/y = 50/15 = 3.333 N/mm Ans.
( b ) D = Cd = 10(4) = 40 mm
OD = D + d = 40 + 4 = 44 mm Ans.
( c ) From Table 10-5, G = 77.2 GPa
Eq. (10-9):
4 4 3
3 3
11.6 coils 8 8 3.333 40
a
d G N kD
Table 10-1: Nt = Na = 11.6 coils Ans.
( d ) Table 10-1: Ls = d ( Nt + 1) = 4(11.6 + 1) = 50.4 mm Ans.
( e ) Table 10-4: m = 0.187, A = 1855 MPa⋅mm
m
Eq. (10-14):
1431 MPa 4
ut (^) m
A
S
d
Table 10-6: Ssy = 0.50 Sut = 0.50(1431) = 715.5 MPa
ys = L 0 − Ls = 80 − 50.4 = 29.6 mm
Fs = k ys = 3.333(29.6) = 98.66 N
Eq. (10-5):
B
C
K
C
Eq. (10-7):
( )
3 3
1.135 178.2 MPa 4
s s B
F D
K
d
τ π π
th
sy s s
S
n Ans τ
______________________________________________________________________________
10-7 Static service spring with: HD steel wire, d = 0.080 in, OD = 0.880 in, Nt = 8 coils, plain
and ground ends.
Preliminaries
Table 10-5: A = 140 kpsi · in
m , m = 0_._ 190
Eq. (10-14):
226.2 kpsi
ut (^) m
A
S
d
Table 10-6: Ssy = 0_._ 45(226.2) = 101.8 kpsi
Then,
D = OD − d = 0.880 − 0.080 = 0.8 in
Eq. (10-1): C = D/d = 0.8/0.08 = 10
Eq. (10-5):
B
C
K
C
Table 10-1: Na = Nt − 1 = 8 − 1 = 7 coils
Ls = dNt = 0.08(8) = 0.64 in
Eq. (10-7) For solid-safe, ns = 1.2 :
( ) ( )
3 3 3 / 0.08^ 101.8 10^ / 1. 18.78 lbf 8 8(1.135)(0.8)
sy s s B
d S n F K D
π π ^ = = =
Eq. (10-9):
( )
4 4 6
3 3
16.43 lbf/in (^8) a 8 0.8 7
d G k D N
1.14 in
s s
F
y k
(a) L 0 = ys + Ls = 1.14 + 0.64 = 1.78 in Ans.
(b) Table 10-1:
0 1.^
0.223 in.
t^8
L
p Ans N
(c) From above: Fs = 18.78 lbf Ans.
(d) From above: k = 16.43 lbf/in Ans.
(e) Table 10-2 and Eq. (10-13): (^0) c r
( ) 4.21 in
D
L
α
Since L 0 < ( L 0 )cr, buckling is unlikely Ans.
10-8 Given: Design load, F 1 = 16.5 lbf.
Referring to Prob. 10-7 solution, C = 10, Na = 7 coils, Ssy = 101.8 kpsi, Fs = 18.78 lbf,
ys = 1.14 in, L 0 = 1.78 in, and ( L 0 )cr = 4.21 in.
Eq. (10-18): 4 ≤ C ≤ 12 C = 10 O.K.
th
Eq. (10-14):
412.7 kpsi
ut (^) m
A
S
d
Table 10-6: Ssy = 0.45 Sut = 0.45(412.7) = 185.7 kpsi
τ s > Ssy , that is, 325.1 > 185.7 kpsi, the spring is not solid-safe. Return to Eq. (1) with
Fs = kys and τ s = Ssy /ns , and solve for ys , giving
( ) (^ )^ (^ )
3 3 3 / 185.7 10^ /1.2^ 0. 0.149 in 8 8 1.340 3.358 0.
sy s s B
S n d y K kD
π ^ ^ π = = =
The free length should be wound to
L 0 = Ls + ys = 0.266 + 0.149 = 0.415 in Ans.
This only addresses the solid-safe criteria. There are additional problems.
10-10 Given: B159 phosphor-bronze, squared and ground. ends, d = 0.014 in, OD = 0.128 in,
L 0 = 0.50 in, Nt = 16 coils.
D = OD − d = 0.128 − 0.014 = 0.114 in
Eq. (10-1): C = D/d = 0.114/0.014 = 8.
Eq. (10-5):
4 2 4 8.143^2
B
C
K
C
+^ +
Table 10-1: Na = Nt − 2 = 16 − 2 = 14 coils
Table 10-5: G = 6 Mpsi
Eq. (10-9):
( )
4 4 6
(^3 )
1.389 lbf/in (^8) a 8 0.114 14
d G k D N
Table 10-1: Ls = dNt = 0.014(16) = 0.224 in
ys = L 0 − Ls = 0.50 − 0.224 = 0.276 in
Fs = kys = 1.389(0.276) = 0.3834 lbf
Eq. (10-7):
( )
( )
3 (^3 )
1.169 47.42 10 psi
s s B
F D
K
d
τ π π
Table 10-4: A = 145 kpsi⋅in
m , m = 0
Eq. (10-14): 0
145 kpsi
ut (^) m
A
S
d
Table 10-6: Ssy = 0.35 Sut = 0.35(135) = 47.25 kpsi
τ s > Ssy , that is, 47.42 > 47.25 kpsi, the spring is not solid-safe. Return to Eq. (1) with
Fs = kys and τ s = Ssy /ns , and solve for ys , giving
( ) (^ )^ (^ )
3 3 3 / 47.25 10^ /1.2^ 0. 0.229 in 8 8 1.169 1.389 0.
sy s s B
S n d y K kD
π ^ ^ π = = =
The free length should be wound to
th
L 0 = Ls + ys = 0.224 + 0.229 = 0.453 in Ans.
10-11 Given: A313 stainless steel, squared and ground ends, d = 0.050 in, OD = 0.250 in,
L 0 = 0.68 in, Nt = 11.2 coils.
D = OD − d = 0.250 − 0.050 = 0.200 in
Eq. (10-1): C = D/d = 0.200/0.050 = 4
Eq. (10-5):
4 2 4 4^2
B
C
K
C
+^ +
Table 10-1: Na = Nt − 2 = 11.2 − 2 = 9.2 coils
Table 10-5: G = 10 Mpsi
Eq. (10-9):
( )
4 4 6
(^3 )
106.1 lbf/in (^8) a 8 0.2 9.
d G k D N
Table 10-1: Ls = dNt = 0.050(11.2) = 0.56 in
ys = L 0 − Ls = 0.68 − 0.56 = 0.12 in
Fs = kys = 106.1(0.12) = 12.73 lbf
Eq. (10-7):
( )
( )
3 3 3
1.385 71.8 10 psi
s s B
F D
K
d
τ π π
Table 10-4: A = 169 kpsi⋅in
m , m = 0.
Eq. (10-14):
261.7 kpsi
ut (^) m
A
S
d
Table 10-6: Ssy = 0.35 Sut = 0.35(261.7) = 91.6 kpsi
sy s s
S
n τ
= = = Spring is solid-safe ( ns > 1.2) Ans.
______________________________________________________________________________
10-12 Given: A227 hard-drawn wire, squared and ground ends, d = 0.148 in, OD = 2.12 in,
L 0 = 2.5 in, Nt = 5.75 coils.
D = OD − d = 2.12 − 0.148 = 1.972 in
Eq. (10-1): C = D/d = 1.972/0.148 = 13.32 (high)
Eq. (10-5):
4 2 4 13.32^2
B
C
K
C
+^ +
Table 10-1: Na = Nt − 2 = 5.75 − 2 = 3.75 coils
Table 10-5: G = 11.4 Mpsi
Eq. (10-9):
( )
4 4 6
3 3
23.77 lbf/in (^8) a 8 1.972 3.
d G k D N
Table 10-1: Ls = dNt = 0.148(5.75) = 0.851 in
ys = L 0 − Ls = 2.5 − 0.851 = 1.649 in
Fs = kys = 23.77(1.649) = 39.20 lbf
th
L 0 = Ls + ys = 1.656 + 0.857 = 2.51 in Ans.
10-14 Given: A232 chrome-vanadium steel, squared and ground ends, d = 0.185 in, OD = 2.
in, L 0 = 7.5 in, Nt = 8 coils.
D = OD − d = 2.75 − 0.185 = 2.565 in
Eq. (10-1): C = D/d = 2.565/0.185 = 13.86 (high)
Eq. (10-5):
4 2 4 13.86^2
B
C
K
C
+^ +
Table 10-1: Na = Nt − 2 = 8 − 2 = 6 coils
Table 10-5: G = 11.2 Mpsi.
Eq. (10-9):
( )
4 4 6
(^3 )
16.20 lbf/in (^8) a 8 2.565 6
d G k D N
Table 10-1: Ls = dNt = 0.185(8) = 1.48 in
ys = L 0 − Ls = 7.5 − 1.48 = 6.02 in
Fs = kys = 16.20(6.02) = 97.5 lbf
Eq. (10-7):
( )
( )
3 (^3 )
1.095 110.1 10 psi
s s B
F D
K
d
τ π π
Table 10-4: A = 169 kpsi⋅in
m , m = 0.
Eq. (10-14):
224.4 kpsi
ut (^) m
A
S
d
Table 10-6: Ssy = 0.50 Sut = 0.50(224.4) = 112.2 kpsi
sy s s
S
n τ
= = = Spring is not solid-safe ( ns < 1.2)
Return to Eq. (1) with Fs = kys and τ s = Ssy /ns , and solve for ys , giving
( ) (^ )^ (^ )
3 3 3 / 112.2 10^ /1.2^ 0. 5.109 in 8 8 1.095 16.20 2.
sy s s B
S n d y K kD
π ^ ^ π = = =
The free length should be wound to
L 0 = Ls + ys = 1.48 + 5.109 = 6.59 in Ans.
______________________________________________________________________________
10-15 Given: A313 stainless steel, squared and ground ends, d = 0.25 mm, OD = 0.95 mm,
L 0 = 12.1 mm, Nt = 38 coils.
D = OD − d = 0.95 − 0.25 = 0.7 mm
Eq. (10-1): C = D/d = 0.7/0.25 = 2.8 (low)
Eq. (10-5):
4 2 4 2.8^2
B
C
K
C
+^ +
th
Table 10-1: Na = Nt − 2 = 38 − 2 = 36 coils (high)
Table 10-5: G = 69.0(
3 ) MPa.
Eq. (10-9):
( )
4 4 3
(^3 )
2.728 N/mm (^8) a 8 0.7 36
d G k D N
Table 10-1: Ls = dNt = 0.25(38) = 9.5 mm
ys = L 0 − Ls = 12.1 − 9.5 = 2.6 mm
Fs = kys = 2.728(2.6) = 7.093 N
Eq. (10-7):
( )
(^3 )
1.610 1303 MPa
s s B
F D
K
d
τ π π
Table 10-4 (dia. less than table): A = 1867 MPa⋅mm
m , m = 0.
Eq. (10-14):
2286 MPa
ut (^) m
A
S
d
Table 10-6: Ssy = 0.35 Sut = 0.35(2286) = 734 MPa
τ s > Ssy , that is, 1303 > 734 MPa, the spring is not solid-safe. Return to Eq. (1) with
Fs = kys and τ s = Ssy /ns , and solve for ys , giving
( ) (^ )^ ( )
3 3 / 734 / 1.2 0. 1.22 mm 8 8 1.610 2.728 0.
sy s s B
S n d y K kD
π π = = =
The free length should be wound to
L 0 = Ls + ys = 9.5 + 1.22 = 10.72 mm Ans.
This only addresses the solid-safe criteria. There are additional problems.
10-16 Given: A228 music wire, squared and ground ends, d = 1.2 mm, OD = 6.5 mm, L 0 = 15.
mm, Nt = 10.2 coils.
D = OD − d = 6.5 − 1.2 = 5.3 mm
Eq. (10-1): C = D/d = 5.3/1.2 = 4.
Eq. (10-5):
4 2 4 4.417^2
B
C
K
C
+^ +
Table (10-1): Na = Nt − 2 = 10.2 − 2 = 8.2 coils
Table 10-5 ( d = 1.2/25.4 = 0.0472 in): G = 81.7(
3 ) MPa.
Eq. (10-9):
( )
4 4 3
(^3 )
17.35 N/mm (^8) a 8 5.3 8.
d G k D N
Table 10-1: Ls = dNt = 1.2(10.2) = 12.24 mm
ys = L 0 − Ls = 15.7 − 12.24 = 3.46 mm
Fs = kys = 17.35(3.46) = 60.03 N
th
L 0 = Ls + ys = 19.25 + 48.96 = 68.2 mm Ans.
10-18 Given: B159 phosphor-bronze, squared and ground ends, d = 3.8 mm, OD = 31.4 mm,
L 0 = 71.4 mm, Nt = 12.8 coils.
D = OD − d = 31.4 − 3.8 = 27.6 mm
Eq. (10-1): C = D/d = 27.6/3.8 = 7.
Eq. (10-5):
4 2 4 7.263^2
B
C
K
C
+^ +
Table 10-1: Na = Nt − 2 = 12.8 − 2 = 10.8 coils
Table 10-5: G = 41.4(
3 ) MPa.
Eq. (10-9):
( )
4 4 3
(^3 )
4.752 N/mm (^8) a 8 27.6 10.
d G k D N
Table 10-1: Ls = dNt = 3.8(12.8) = 48.64 mm
ys = L 0 − Ls = 71.4 − 48.64 = 22.76 mm
Fs = kys = 4.752(22.76) = 108.2 N
Eq. (10-7):
( )
(^3 )
1.192 165.2 MPa
s s B
F D
K
d
τ π π
Table 10-4 ( d = 3.8/25.4 = 0.150 in): A = 932 MPa⋅mm
m , m = 0.
Eq. (10-14):
855.7 MPa
ut (^) m
A
S
d
Table 10-6: Ssy = 0.35 Sut = 0.35(855.7) = 299.5 MPa
sy s s
S
n τ
= = = Spring is solid-safe ( ns > 1.2) Ans.
______________________________________________________________________________
10-19 Given: A232 chrome-vanadium steel, squared and ground ends, d = 4.5 mm, OD = 69.
mm, L 0 = 215.6 mm, Nt = 8.2 coils.
D = OD − d = 69.2 − 4.5 = 64.7 mm
Eq. (10-1): C = D/d = 64.7/4.5 = 14.38 (high)
Eq. (10-5):
4 2 4 14.38^2
B
C
K
C
+^ +
Table 10-1: Na = Nt − 2 = 8.2 − 2 = 6.2 coils
Table 10-5: G = 77.2(
3 ) MPa.
Eq. (10-9):
( )
4 4 3
3 3
2.357 N/mm (^8) a 8 64.7 6.
d G k D N
Table 10-1: Ls = dNt = 4.5(8.2) = 36.9 mm
th
ys = L 0 − Ls = 215.6 − 36.9 = 178.7 mm
Fs = kys = 2.357(178.7) = 421.2 N
Eq. (10-7):
( )
(^3 )
1.092 832 MPa
s s B
F D
K
d
τ π π
Table 10-4: A = 2005 MPa⋅mm
m , m = 0.
Eq. (10-14):
1557 MPa
ut (^) m
A
S
d
Table 10-6: Ssy = 0.50 Sut = 0.50(1557) = 779 MPa
τ s > Ssy , that is, 832 > 779 MPa, the spring is not solid-safe. Return to Eq. (1) with
Fs = kys and τ s = Ssy /ns , and solve for ys , giving
( ) ( ) ( )
3 3 / 779 / 1.2 4. 139.5 mm 8 8 1.092 2.357 64.
sy s s B
S n d y K kD
π π = = =
The free length should be wound to
L 0 = Ls + ys = 36.9 + 139.5 = 176.4 mm Ans.
This only addresses the solid-safe criteria. There are additional problems.
10-20 Given: A227 HD steel.
From the figure: L 0 = 4.75 in, OD = 2 in, and d = 0.135 in. Thus
D = OD − d = 2 − 0.135 = 1.865 in
(a) By counting, Nt = 12_._ 5 coils. Since the ends are squared along 1 / 4 turn on each end,
12.5 0.5 12 turns.
4.75 / 12 0.396 in.
N (^) a Ans
p Ans
The solid stack is 13 wire diameters
Ls = 13(0.135) = 1.755 in Ans.
(b) From Table 10-5, G = 11.4 Mpsi
( )
( )
4 4 6
3 3
6.08 lbf/in. (^8) a 8 1.865 (12)
d G k Ans D N
(c) Fs = k ( L 0 - Ls ) = 6.08(4.75 − 1.755) = 18.2 lbf Ans.
(d) C = D/d = 1.865/0.135 = 13_._ 81
th
OD = D + d = 0.950 in. (3)
and,
d C
Again, for ns ≥ 1.2, the optimal size is = 0.085 in.
Although this approach used less iterations than in Prob. 10-21, this was due to the initial
values picked and not the approach.
10-23 One approach is to select A227 HD steel for its low cost. Try L 0 = 48 mm, then for
y = 48 − 37.5 = 10.5 mm when F = 45 N. The spring rate is k = F/y = 45/10.5 = 4.
N/mm.
For a clearance of 1.25 mm with screw, ID = 10 + 1.25 = 11.25 mm. Starting with
d = 2 mm,
D = ID + d = 11.25 + 2 = 13.25 mm
C = D/d = 13.25/2 = 6.625 (acceptable)
Table 10-5 ( d = 2/25.4 = 0.0787 in): G = 79.3 GPa
Eq. (10-9):
4 4 3
3 3
15.9 coils 8 8(4.286)13.
a
d G N kD
Assume squared and closed.
(a) Spring over a rod (b) Spring in a Hole
Source Parameter Values Source Parameter Values
C 10.000 10.5 C 10.
Eq. (2) d 0.089 0.084 Eq. (4) d 0.
Table A-28 d 0.090 0.085 Table A-28 d 0.
Eq. (1) D 0.890 0.885 Eq. (3) D 0.
Eq. (10-1) C 9.889 10.412 Eq. (10-1) C 10.
Eq. (10-9) Na 13.669 11.061 Eq. (10-9) Na 11.
Table 10-1 Nt 15.669 13.061 Table 10-1 Nt 13.
Table 10-1 Ls 1.410 1.110 Table 10-1 Ls 1.
1.15 y + Ls L 0 3.710 3.410 1.15 y + Ls L 0 3.
Eq. (10-13) ( L 0 )cr 4.681 4.655 Eq. (10-13) ( L 0 )cr 4.
Table 10-4 A 201.000 201.000 Table 10-4 A 201.
Table 10-4 m 0.145 0.145 Table 10-4 m 0.
Eq. (10-14) Sut 284.991 287.363 Eq. (10-14) Sut 287.
Table 10-6 Ssy 128.246 129.313 Table 10-6 Ssy 129.
Eq. (10-5) KB 1.135 1.128 Eq. (10-5) KB 1.
Eq. (10-7) τ s 81.167 95.223 Eq. (10-7) τ s 93.
ns = Ssy/ τ s ns 1.580 1.358 ns = Ssy/ τ s ns 1.
Eq. (10-22) fom -0.725 -0.536 Eq. (10-22) fom -0.
th
Table 10-1: Nt = Na + 2 = 15.9 + 2 = 17.9 coils
Ls = dNt = 2(17.9) =35.8 mm
ys = L 0 − Ls = 48 − 35.8 = 12.2 mm
Fs = kys = 4.286(12.2) = 52.29 N
Eq. (10-5):
4 2 4 6.625^2
B
C
K
C
+^ +
Eq. (10-7):
( )
(^3 )
1.213 267.5 MPa 2
s s B
F D
K
d
τ π π
Table 10-4: A = 1783 MPa · mm
m , m = 0_._ 190
Eq. (10-14):
1563 MPa 2
ut m
A
S
d
Table 10-6: Ssy = 0.45 Sut = 0.45(1563) = 703.3 MPa
sy s s
S
n O K τ
No other diameters in the given range work. So specify
A227-47 HD steel, d = 2 mm, D = 13.25 mm, ID = 11.25 mm, OD = 15.25 mm, squared
and closed, Nt = 17.9 coils, Na = 15.9 coils, k = 4.286 N/mm, Ls = 35.8 mm, and L 0 = 48
mm. Ans.
10-24 Select A227 HD steel for its low cost. Try L 0 = 48 mm, then for y = 48 − 37.5 = 10.5 mm
when F = 45 N. The spring rate is k = F/y = 45/10.5 = 4.286 N/mm.
For a clearance of 1.25 mm with screw, ID = 10 + 1.25 = 11.25 mm.
D − d = 11.25 (1)
and, D =Cd (2)
Starting with C = 8, gives D = 8 d. Substitute into Eq. (1) resulting in d = 1.607 mm.
Selecting the nearest diameter in the given range, d = 1.6 mm. From this point, the
calculations are shown in the third column of the spreadsheet output shown. We see that
for d = 1.6 mm, the spring is not solid safe. Iterating on C we find that C = 6.5 provides
acceptable results with the specifications
A227-47 HD steel, d = 2 mm, D = 13.25 mm, ID = 11.25 mm, OD = 15.25 mm, squared
and closed, Nt = 17.9 coils, Na = 15.9 coils, k = 4.286 N/mm, Ls = 35.8 mm, and L 0 = 48
mm. Ans.
th
Table 10-5: G = 11.5 Mpsi
Eq. (10-9):
4 4 6
(^3 )
3.424 lbf/in. (^8) a 8 0.667 28
d G k Ans D N
( d ) Table 10-4: A = 140 kpsi⋅in
m , m = 0.
Eq. (10-14):
234.2 kpsi
ut (^) m
A
S
d
Table 10-6: Ssy = 0.45 Sut = 0.45 (234.2) = 105.4 kpsi
Fs = kys = 3.424(3) = 10.27 lbf
Eq. (10-5):
4 2 4 10^2
B
C
K
C
+^ +
Eq. (10-7):
(^3 )
3
66.72 10 psi 66.72 kpsi
s s B
F D
K
d
τ π π
sy s s
S
n Ans τ
( e ) τ a = τ m = 0.5 τ s = 0.5(66.72) = 33.36 kpsi, r = τ a / τ m = 1. Using the Gerber fatigue
failure criterion with Zimmerli data,
Eq. (10-30): Ssu = 0.67 Sut = 0.67(234.2) = 156.9 kpsi
The Gerber ordinate intercept for the Zimmerli data is
2 2
39.9 kpsi 1 / 1 55 / 156.
sa se sm su
S
S
S S
Table 6-7, p. 315,
2 2 2
2 2 2
1 1 37.61 kpsi 2 39.9 1 156.
su se sa se su
r S S S S rS
= ^ − + +
^
sa f a
S
n Ans τ
______________________________________________________________________________
10-27 Given: OD ≤ 0.9 in, C = 8, L 0 = 3 in, Ls = 1 in, ys = 3 − 1 = 2 in, sq. ends, unpeened,
music wire.
( a ) Try OD = D + d = 0.9 in, C = D/d = 8 ⇒ D = 8 d ⇒ 9 d = 0.9 ⇒ d = 0.1 Ans.
th
D = 8(0.1) = 0.8 in
( b ) Table 10-1: Ls = d ( Nt + 1) ⇒ Nt = Ls / d − 1 = 1/0.1− 1 = 9 coils Ans.
Table 10-1: Na = Nt − 2 = 9 − 2 = 7 coils
( c ) Table 10-5: G = 11.75 Mpsi
Eq. (10-9):
( )
4 4 6
(^3 )
40.98 lbf/in. (^8) a 8 0.8 7
d G k Ans D N
( d ) Fs = kys = 40.98(2) = 81.96 lbf
Eq. (10-5):
4 2 4 8^2
B
C
K
C
+^ +
Eq. (10-7):
( )
( )
3 (^3 )
1.172 195.7 10 psi 195.7 kpsi
s s B
F D
K
d
τ π π
Table 10-4: A = 201 kpsi⋅in
m , m = 0.
Eq. (10-14):
280.7 kpsi
ut (^) m
A
S
d
Table 10-6: Ssy = 0.45 Sut = 0.45(280.7) = 126.3 kpsi
sy s s
S
n Ans τ
( e ) τ a = τ m = τ s /2 = 195.7/2 = 97.85 kpsi. Using the Gerber fatigue failure criterion with
Zimmerli data,
Eq. (10-30): Ssu = 0.67 Sut = 0.67(280.7) = 188.1 kpsi
The Gerber ordinate intercept for the Zimmerli data is
2 2
38.3 kpsi 1 / 1 55 /188.
sa se sm su
S
S
S S
Table 6-7, p. 315,
( )
2 2 2
2 2 2
1 1 36.83 kpsi 2 38.3 1 188.
su se sa se su
r S S S S rS
= ^ − + +
^
