Shigleys mechanical engineering design 11th edition solutions

SKU: Solutions Manual Shigley's Mechanical Engineering Design Budynas, 11th Edition

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Solutions Manual Shigley's Mechanical Engineering Design Budynas, 11th Edition

Chapter 1

Problems 1-1 through 1-6 are for student research. No standard solutions are provided.

1-7 From Fig. 1-2, cost of grinding to ± 0.0005 in is 270%. Cost of turning to ± 0.003 in is 60%.

Relative cost of grinding vs. turning = 270/60 = 4.5 times Ans.

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1-8 CA = CB,

10 + 0.8 P = 60 + 0.8 P - 0.005 P 2

P 2 = 50/0.005 Þ P = 100 parts Ans.

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1-9 Max. load = 1.10 P

Min. area = (0.95)2A

Min. strength = 0.85 S

To offset the absolute uncertainties, the design factor, from Eq. (1-1) should be

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1-10 (a) X1 + X2:

(b) X1 - X2:

(c) X1 X2:

(d) X1/X2:

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1-11 (a) x1 = = 2.645 751 311 1

X1 = 2.64 (3 correct digits)

x2 = = 2.828 427 124 7

X2 = 2.82 (3 correct digits)

x1 + x2 = 5.474 178 435 8

e1 = x1 - X1 = 0.005 751 311 1

e2 = x2 - X2 = 0.008 427 124 7

e = e1 + e2 = 0.014 178 435 8

Sum = x1 + x2 = X1 + X2 + e

= 2.64 + 2.82 + 0.014 178 435 8 = 5.474 178 435 8 Checks

(b) X1 = 2.65, X2 = 2.83 (3 digit significant numbers)

e1 = x1 - X1 = - 0.004 248 688 9

e2 = x2 - X2 = - 0.001 572 875 3

e = e1 + e2 = - 0.005 821 564 2

Sum = x1 + x2 = X1 + X2 + e

= 2.65 +2.83 - 0.001 572 875 3 = 5.474 178 435 8 Checks

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1-12

Table A-17: d = in Ans.

Factor of safety:

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1-13 (a)

Eq. (1-6)

Eq. (1-7)

(b) Eq. (1-5)

Interpolating from Table (A-10)

0.2600 0.3974

0.2607 x x = 0.3971

0.2700 0.3936

NF(-0.2607) = 69 (0.3971) = 27.4 » 27 Ans.

From the data, the number of instances less than 115 kcycles is

2 + 1 + 3 + 5 + 8 + 12 = 31 (the data is not perfectly normal)

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1-14

Eq. (1-6)

Eq. (1-7)

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1-15

Eq. (1-5)

Thus, x10 = 122.9 + 30.3 z10 = L10

From Table A-10, for 10 percent failure, z10 = -1.282. Thus,

L10 = 122.9 + 30.3(-1.282) = 84.1 kcycles Ans.

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1-16

Eq. (1-6)

Eq. (1-7)

Note, for accuracy in the calculation given above, needs to be of more significant figures than the rounded value.

For a normal distribution, from Eq. (1-5), and a yield strength exceeded by 99 percent

(R = 0.99, pf = 0.01),

Solving for the yield strength gives

x0.01 = 98.26 + 4.30 z0.01

From Table A-10, z0.01 = - 2.326. Thus

x0.01 = 98.26 + 4.30(- 2.326) = 88.3 kpsi Ans.

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1-17 Eq. (1-9): R = = 0.98(0.96)0.94 = 0.88

Overall reliability = 88 percent Ans.

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1-18 Obtain the coefficients of variance for strength and stress

For R = 0.99, from Table A-10, z = - 2.326.

Eq. (1-12):

From the given equation for stress,

Solving for d gives

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1-19 Obtain the coefficients of variance for stress and strength

(a)

Eq. (1-11):

Interpolating Table A-10,

1.61 0.0537

1.6127 F F = 0.0534

1.62 0.0526

R = 1 - 0.0534 = 0.9466 Ans.

(b)

3.6 0.000159

3.605 F F = 0.00015645

3.7 0.000108

R = 1 - 0.00015645 = 0.9998 Ans.

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1-20

From footnote 9 of text,

Eq. (1-11):

From Table A-10, F(- 1.635) = 0.05105

R = 1 - 0.05105 = 0.94895 = 94.9 percent Ans.

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1-21 a = 1.500 ± 0.001 in

b = 2.000 ± 0.003 in

c = 3.000 ± 0.004 in

d = 6.520 ± 0.010 in

(a) = 6.520 - 1.5 - 2 - 3 = 0.020 in

= 0.001 + 0.003 + 0.004 +0.010 = 0.018

w = 0.020 ± 0.018 in Ans.

(b) From part (a), wmin = 0.002 in. Thus, must add 0.008 in to . Therefore,

= 6.520 + 0.008 = 6.528 in Ans.

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1-22 V = xyz, and x = a ± D a, y = b ± D b, z = c ± D c,

The higher order terms in D are negligible. Thus,

and,

For the numerical values given,

V = 8.4375 ± 0.0360 in3 Ans.

This answer yields in, whereas, exact is in

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1-23

wmax = 0.05 in, wmin = 0.004 in

Thus, D w = 0.05 - 0.027 = 0.023 in, and then, w = 0.027 ± 0.023 in.

tw = Þ 0.023 = ta + 0.002 + 0.005 Þ ta = 0.016 in

Thus, a = 1.569 ± 0.016 in Ans.

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1-24

Do = 4.012 ± 0.036 in Ans.

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1-25 From O-Rings, Inc. (oringsusa.com), Di = 9.19 ± 0.13 mm, d = 2.62 ± 0.08 mm

Do = 14.43 ± 0.29 mm Ans.

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1-26 From O-Rings, Inc. (oringsusa.com), Di = 34.52 ± 0.30 mm, d = 3.53 ± 0.10 mm

Do = 41.58 ± 0.50 mm Ans.

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1-27 From O-Rings, Inc. (oringsusa.com), Di = 5.237 ± 0.035 in, d = 0.103 ± 0.003 in

Do = 5.443 ± 0.041 in Ans.

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1-28 From O-Rings, Inc. (oringsusa.com), Di = 1.100 ± 0.012 in, d = 0.210 ± 0.005 in

Do = 1.520 ± 0.022 in Ans.

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1-29 From Table A-2,

(a) s = 150/6.89 = 21.8 kpsi Ans.

(b) F = 2 /4.45 = 0.449 kip = 449 lbf Ans.

(c) M = 150/0.113 = 1330 lbf × in = 1.33 kip × in Ans.

(d) A = 1500/ 25.42 = 2.33 in2 Ans.

(e) I = 750/2.544 = 18.0 in4 Ans.

(f) E = 145/6.89 = 21.0 Mpsi Ans.

(g) v = 75/1.61 = 46.6 mi/h Ans.

(h) V = 1000/946 = 1.06 qt Ans.

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1-30 From Table A-2,

(a) l = 5(0.305) = 1.53 m Ans.

(b) s = 90(6.89) = 620 MPa Ans.

(c) p = 25(6.89) = 172 kPa Ans.

(d) Z =12(16.4) = 197 cm3 Ans.

(e) w = 0.208(175) = 36.4 N/m Ans.

(f) d = 0.001 89(25.4) = 0.048 0 mm Ans.

(g) v = 1 200(0.0051) = 6.12 m/s Ans.

(h) ò = 0.002 15(1) = 0.002 15 mm/mm Ans.

(i) V = 1830(25.43) = 30.0 (106) mm3 Ans.

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1-31

(a) s = M /Z = 1770/0.934 = 1895 psi = 1.90 kpsi Ans.

(b) s = F /A = 9440/23.8 = 397 psi Ans.

(c) y =Fl3/3EI = 270(31.5)3/[3(30)106(0.154)] = 0.609 in Ans.

(d) q = Tl /GJ = 9 740(9.85)/[11.3(106)(p /32)1.004] = 8.648(10-2) rad = 4.95° Ans.

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1-32

(a) s =F / wt = 1000/[25(5)] = 8 MPa Ans.

(b) I = bh3 /12 = 10(25)3/12 = 13.0(103) mm4 Ans.

(c) I =p d4/64 = p (25.4)4/64 = 20.4(103) mm4 Ans.

(d) t =16T /p d 3 = 16(25)103/[p (12.7)3] = 62.2 MPa Ans.

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1-33

(a) t =F /A = 2 700/[p(0.750)2/4] = 6110 psi = 6.11 kpsi Ans.

(b) s = 32Fa/p d 3 = 32(180)31.5/[p (1.25)3] = 29 570 psi = 29.6 kpsi Ans.

(c) Z =p (do4 - di4)/(32 do) = p (1.504 - 1.004)/[32(1.50)] = 0.266 in3 Ans.

(d) k = (d 4G)/(8D 3 N) = 0.062 54(11.3)106/[8(0.760)3 32] = 1.53 lbf/in Ans.

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