Chapter 10 10-1 10-2 A = Sd m dim( A uscu ) = dim( S ) dim( d m ) = kpsi · in m dim( A SI ) = dim( S 1 ) dim ( d m 1 ) = MPa · mm m A SI = MPa kpsi · mm m in m A uscu = 6.894 757(25.40) m A uscu . = 6.895(25.4) m A uscu Ans. For music wire, from Table 10-4: A uscu = 201, m = 0.145; what is A SI ? A SI = 6.89(25.4) 0.145 (201) = 2214 MPa · mm m Ans. 10-3 Given: Music wire, d = 0.105 in, OD = 1.225 in, plain ground ends, N t = 12 coils. Table 10-1: N a = N t − 1 = 12 − 1 = 11 L s = dN t = 0.105(12) = 1.26 in Table 10-4: A = 201, m = 0.145 (a) Eq. (10-14): S ut = 201 (0.105) 0.145 = 278.7 kpsi Table 10-6: S sy = 0.45(278.7) = 125.4 kpsi D = 1.225 − 0.105 = 1.120 in C = D d = 1.120 0.105 = 10.67 Eq. (10-6): K B = 4(10.67) + 2 4(10.67) − 3 = 1.126 Eq. (10-3): F | S sy = π d 3 S sy 8 K B D = π (0.105) 3 (125.4)(10 3 ) 8(1.126)(1.120) = 45.2 lbf Eq. (10-9): k = d 4 G 8 D 3 N a = (0.105) 4 (11.75)(10 6 ) 8(1.120) 3 (11) = 11.55 lbf/in L 0 = F | S sy k + L s = 45.2 11.55 + 1.26 = 5.17 in Ans. 1 2 " 4" 1" 1 2 " 4" 1"
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FIRST PAGES
Chapter 10
10-1
10-2 A = Sdm
dim( Auscu) = dim(S) dim(dm) = kpsi · inm
dim( ASI) = dim(S1) dim(dm
1
) = MPa · mmm
ASI = MPa
kpsi· mmm
inm Auscu = 6.894 757(25.40)m Auscu.= 6.895(25.4)m Auscu Ans.
For music wire, from Table 10-4:
Auscu = 201, m = 0.145; what is ASI?
ASI = 6.89(25.4)0.145(201) = 2214 MPa · mmm Ans.
10-3 Given: Music wire, d = 0.105 in, OD = 1.225 in, plain ground ends, Nt = 12 coils.
10-17 Given: A232 (Cr-V steel), SQ&GRD ends, d = 4.3 mm, OD = 76.2 mm, L0 =228.6 mm, Nt = 8 turns.
Table 10-4: A = 2005 MPa · mmm , m = 0.168
Table 10-5: G = 77.2 GPa
D = OD − d = 76.2 − 4.3 = 71.9 mm
C = D/d = 71.9/4.3 = 16.72 (large)
KB = 4(16.72) + 2
4(16.72) − 3= 1.078
Na = Nt − 2 = 8 − 2 = 6 turns
Sut = 2005
(4.3)0.168= 1569 MPa
Table 10-6:
Ssy = 0.50(1569) = 784.5 MPa
k = d4G
8D3 Na= (4.3)4(77.2)
8(71.9)3(6)
[(10−3)4(109)
(10−3)3
]= 0.001 479(106)
= 1479 N/m or 1.479 N/mm
Ls = d Nt = 4.3(8) = 34.4 mm
Fs = kys
ys = L0 − Ls = 228.6 − 34.4 = 194.2 mm
τs = K B
[8(kys)D
πd3
]= 1.078
[8(1.479)(194.2)(71.9)
π(4.3)3
]= 713.0 MPa (1)
τs < Ssy , that is, 713.0 < 784.5; the spring is solid safe. With ns = 1.2
Eq. (1) becomes
y′s = (Ssy/n)(πd3)
8K Bk D= (784.5/1.2)(π)(4.3)3
8(1.078)(1.479)(71.9)= 178.1 mm
L ′0 = Ls + y′
s = 34.4 + 178.1 = 212.5 mm
Wind the spring to a free length of L ′0 = 212.5 mm. Ans.
10-18 For the wire diameter analyzed, G = 11.75 Mpsi per Table 10-5. Use squared and groundends. The following is a spread-sheet study using Fig. 10-3 for parts (a) and (b). For Na ,k = 20/2 = 10 lbf/in.
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FIRST PAGES
Chapter 10 273
(a) Spring over a Rod (b) Spring in a Hole
Source Parameter Values Source Parameter Values
d 0.075 0.08 0.085 d 0.075 0.08 0.085D 0.875 0.88 0.885 D 0.875 0.870 0.865ID 0.800 0.800 0.800 ID 0.800 0.790 0.780OD 0.950 0.960 0.970 OD 0.950 0.950 0.950
Eq. (10-2) C 11.667 11.000 10.412 Eq. (10-2) C 11.667 10.875 10.176Eq. (10-9) Na 6.937 8.828 11.061 Eq. (10-9) Na 6.937 9.136 11.846Table 10-1 Nt 8.937 10.828 13.061 Table 10-1 Nt 8.937 11.136 13.846Table 10-1 Ls 0.670 0.866 1.110 Table 10-1 Ls 0.670 0.891 1.1771.15y + Ls L0 2.970 3.166 3.410 1.15y + Ls L0 2.970 3.191 3.477Eq. (10-13) (L0)cr 4.603 4.629 4.655 Eq. (10-13) (L0)cr 4.603 4.576 4.550Table 10-4 A 201.000 201.000 201.000 Table 10-4 A 201.000 201.000 201.000Table 10-4 m 0.145 0.145 0.145 Table 10-4 m 0.145 0.145 0.145Eq. (10-14) Sut 292.626 289.900 287.363 Eq. (10-14) Sut 292.626 289.900 287.363Table 10-6 Ssy 131.681 130.455 129.313 Table 10-6 Ssy 131.681 130.455 129.313Eq. (10-6) K B 1.115 1.122 1.129 Eq. (10-6) K B 1.115 1.123 1.133Eq. (10-3) ns 0.973 1.155 1.357 Eq. (10-3) ns 0.973 1.167 1.384Eq. (10-22) fom −0.282 −0.391 −0.536 Eq. (10-22) fom −0.282 −0.398 −0.555
For ns ≥ 1.2, the optimal size is d = 0.085 in for both cases.
10-19 From the figure: L0 = 120 mm, OD = 50 mm, and d = 3.4 mm. Thus
D = OD − d = 50 − 3.4 = 46.6 mm
(a) By counting, Nt = 12.5 turns. Since the ends are squared along 1/4 turn on each end,
Na = 12.5 − 0.5 = 12 turns Ans.
p = 120/12 = 10 mm Ans.
The solid stack is 13 diameters across the top and 12 across the bottom.
Ls = 13(3.4) = 44.2 mm Ans.
(b) d = 3.4/25.4 = 0.1339 in and from Table 10-5, G = 78.6 GPa
k = d4G
8D3 Na= (3.4)4(78.6)(109)
8(46.6)3(12)(10−3) = 1080 N/m Ans.
(c) Fs = k(L0 − Ls) = 1080(120 − 44.2)(10−3) = 81.9 N Ans.
(d) C = D/d = 46.6/3.4 = 13.71
K B = 4(13.71) + 2
4(13.71) − 3= 1.096
τs = 8K B Fs D
πd3= 8(1.096)(81.9)(46.6)
π(3.4)3= 271 MPa Ans.
10-20 One approach is to select A227-47 HD steel for its low cost. Then, for y1 ≤ 3/8 atF1 = 10 lbf, k ≥10/ 0.375 = 26.67 lbf/in. Try d = 0.080 in #14 gauge
For a clearance of 0.05 in: ID = (7/16) + 0.05 = 0.4875 in; OD = 0.4875 + 0.16 =0.6475 in
D = 0.4875 + 0.080 = 0.5675 in
C = 0.5675/0.08 = 7.094
G = 11.5 Mpsi
Na = d4G
8k D3= (0.08)4(11.5)(106)
8(26.67)(0.5675)3= 12.0 turns
Nt = 12 + 2 = 14 turns, Ls = d Nt = 0.08(14) = 1.12 in O.K.
L0 = 1.875 in, ys = 1.875 − 1.12 = 0.755 in
Fs = kys = 26.67(0.755) = 20.14 lbf
K B = 4(7.094) + 2
4(7.094) − 3= 1.197
τs = K B
(8Fs D
πd3
)= 1.197
[8(20.14)(0.5675)
π(0.08)3
]= 68 046 psi
Table 10-4: A = 140 kpsi · inm , m = 0.190
Ssy = 0.45140
(0.080)0.190= 101.8 kpsi
n = 101.8
68.05= 1.50 > 1.2 O.K.
τ1 = F1
Fsτs = 10
20.14(68.05) = 33.79 kpsi,
n1 = 101.8
33.79= 3.01 > 1.5 O.K.
There is much latitude for reducing the amount of material. Iterate on y1 using a spreadsheet. The final results are: y1 = 0.32 in, k = 31.25 lbf/in, Na = 10.3 turns, Nt =12.3 turns, Ls = 0.985 in, L0 = 1.820 in, ys = 0.835 in, Fs = 26.1 lbf, K B = 1.197,τs = 88 190 kpsi, ns = 1.15, and n1 = 3.01.
ID = 0.4875 in, OD = 0.6475 in, d = 0.080 in
Try other sizes and/or materials.
10-21 A stock spring catalog may have over two hundred pages of compression springs with upto 80 springs per page listed.
• Students should be aware that such catalogs exist.• Many springs are selected from catalogs rather than designed.• The wire size you want may not be listed.• Catalogs may also be available on disk or the web through search routines. For exam-
ple, disks are available from Century Spring at
1 − (800) − 237 − 5225www.centuryspring.com
• It is better to familiarize yourself with vendor resources rather than invent them yourself.
• Sample catalog pages can be given to students for study.
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Chapter 10 275
10-22 For a coil radius given by:
R = R1 + R2 − R1
2π Nθ
The torsion of a section is T = P R where d L = R dθ
δp = ∂U
∂ P= 1
G J
∫T
∂T
∂ Pd L = 1
G J
∫ 2π N
0P R3 dθ
= P
G J
∫ 2π N
0
(R1 + R2 − R1
2π Nθ
)3
dθ
= P
G J
(1
4
)(2π N
R2 − R1
) [(R1 + R2 − R1
2π Nθ
)4]∣∣∣∣∣
2π N
0
= π P N
2G J (R2 − R1)
(R4
2 − R41
) = π P N
2G J(R1 + R2)
(R2
1 + R22
)J = π
32d4 ∴ δp = 16P N
Gd4(R1 + R2)
(R2
1 + R22
)k = P
δp= d4G
16N (R1 + R2)(R2
1 + R22
) Ans.
10-23 For a food service machinery application select A313 Stainless wire.
G = 10(106) psi
Note that for 0.013 ≤ d ≤ 0.10 in A = 169, m = 0.146
The shaded areas depict conditions outside the recommended design conditions. Thus,one spring is satisfactory–A313, as wound, unpeened, squared and ground,
d = 0.0915 in, OD = 0.879 + 0.092 = 0.971 in, Nt = 15.59 turns
10-24 The steps are the same as in Prob. 10-23 except that the Gerber-Zimmerli criterion isreplaced with Goodman-Zimmerli:
Sse = Ssa
1 − (Ssm/Ssu)The problem then proceeds as in Prob. 10-23. The results for the wire sizes are shownbelow (see solution to Prob. 10-23 for additional details).
Iteration of d for the first triald1 d2 d3 d4 d1 d2 d3 d4
Without checking all of the design conditions, it is obvious that none of the wire sizessatisfy ns ≥ 1.2. Also, the Gerber line is closer to the yield line than the Goodman. Settingn f = 1.5 for Goodman makes it impossible to reach the yield line (ns < 1) . The tablebelow uses n f = 2.
Iteration of d for the second triald1 d2 d3 d4 d1 d2 d3 d4
The satisfactory spring has design specifications of: A313, as wound, unpeened, squaredand ground, d = 0.0915 in, OD = 0.811 + 0.092 = 0.903 in, Nt = 19.3 turns.
10-25 This is the same as Prob. 10-23 since Sse = Ssa = 35 kpsi. Therefore, design the springusing: A313, as wound, un-peened, squared and ground, d = 0.915 in, OD = 0.971 in,Nt = 15.59 turns.
10-26 For the Gerber fatigue-failure criterion, Ssu = 0.67Sut ,
Sse = Ssa
1 − (Ssm/Ssu)2, Ssa = r2S2
su
2Sse
−1 +
√1 +
(2Sse
r Ssu
)2
The equation for Ssa is the basic difference. The last 2 columns of diameters of Ex. 10-5are presented below with additional calculations.
d = 0.105 d = 0.112 d = 0.105 d = 0.112
Sut 278.691 276.096 Na 8.915 6.190Ssu 186.723 184.984 Ls 1.146 0.917Sse 38.325 38.394 L0 3.446 3.217Ssy 125.411 124.243 (L0)cr 6.630 8.160Ssa 34.658 34.652 K B 1.111 1.095α 23.105 23.101 τa 23.105 23.101β 1.732 1.523 n f 1.500 1.500C 12.004 13.851 τs 70.855 70.844D 1.260 1.551 ns 1.770 1.754ID 1.155 1.439 fn 105.433 106.922OD 1.365 1.663 fom −0.973 −1.022
There are only slight changes in the results.
10-27 As in Prob. 10-26, the basic change is Ssa.
For Goodman, Sse = Ssa
1 − (Ssm/Ssu)Recalculate Ssa with
Ssa = r SseSsu
r Ssu + Sse
Calculations for the last 2 diameters of Ex. 10-5 are given below.
d = 0.105 d = 0.112 d = 0.105 d = 0.112
Sut 278.691 276.096 Na 9.153 6.353Ssu 186.723 184.984 Ls 1.171 0.936Sse 49.614 49.810 L0 3.471 3.236Ssy 125.411 124.243 (L0)cr 6.572 8.090Ssa 34.386 34.380 K B 1.112 1.096α 22.924 22.920 τa 22.924 22.920β 1.732 1.523 n f 1.500 1.500C 11.899 13.732 τs 70.301 70.289D 1.249 1.538 ns 1.784 1.768ID 1.144 1.426 fn 104.509 106.000OD 1.354 1.650 fom −0.986 −1.034
There are only slight differences in the results.
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Chapter 10 279
10-28 Use: E = 28.6 Mpsi, G = 11.5 Mpsi, A = 140 kpsi · inm , m = 0.190, rel cost = 1.
Try d = 0.067 in, Sut = 140
(0.067)0.190= 234.0 kpsi
Table 10-6: Ssy = 0.45Sut = 105.3 kpsi
Table 10-7: Sy = 0.75Sut = 175.5 kpsi
Eq. (10-34) with D/d = C and C1 = C
σA = Fmax
πd2[(K ) A(16C) + 4] = Sy
ny
4C2 − C − 1
4C(C − 1)(16C) + 4 = πd2Sy
ny Fmax
4C2 − C − 1 = (C − 1)
(πd2Sy
4ny Fmax− 1
)
C2 − 1
4
(1 + πd2Sy
4ny Fmax− 1
)C + 1
4
(πd2Sy
4ny Fmax− 2
)= 0
C = 1
2
πd2Sy
16ny Fmax±
√(πd2Sy
16ny Fmax
)2
− πd2Sy
4ny Fmax+ 2
= 1
2
{π(0.0672)(175.5)(103)
16(1.5)(18)
+√[
π(0.067)2(175.5)(103)
16(1.5)(18)
]2
− π(0.067)2(175.5)(103)
4(1.5)(18)+ 2
= 4.590
D = Cd = 0.3075 in
Fi = πd3τi
8D= πd3
8D
[33 500
exp(0.105C)± 1000
(4 − C − 3
6.5
)]
Use the lowest Fi in the preferred range. This results in the best fom.
Fi = π(0.067)3
8(0.3075)
{33 500
exp[0.105(4.590)]− 1000
(4 − 4.590 − 3
6.5
)}= 6.505 lbf
For simplicity, we will round up to the next integer or half integer; therefore, use Fi = 7 lbf
This means (2.5 − 2.417)(360◦) or 29.9◦ from closed. Treating the hand force as in themiddle of the grip
r = 1 + 3.5
2= 2.75 in
F = My
r= 57.2
2.75= 20.8 lbf Ans.
10-33 The spring material and condition are unknown. Given d = 0.081 in and OD = 0.500,
(a) D = 0.500 − 0.081 = 0.419 inUsing E = 28.6 Mpsi for an estimate
k′ = d4 E
10.8DN= (0.081)4(28.6)(106)
10.8(0.419)(11)= 24.7 lbf · in/turn
for each spring. The moment corresponding to a force of 8 lbf
Fr = (8/2)(3.3125) = 13.25 lbf · in/spring
The fraction windup turn is
n = Fr
k′ = 13.25
24.7= 0.536 turns
The arm swings through an arc of slightly less than 180◦ , say 165◦ . This uses up165/360 or 0.458 turns. So n = 0.536 − 0.458 = 0.078 turns are left (or0.078(360◦) = 28.1◦ ). The original configuration of the spring was
Ans.
(b)C = 0.419
0.081= 5.17
Ki = 4(5.17)2 − 5.17 − 1
4(5.17)(5.17 − 1)= 1.168
σ = Ki32M
πd3
= 1.168
[32(13.25)
π(0.081)3
]= 296 623 psi Ans.
To achieve this stress level, the spring had to have set removed.