Mechanical engineering design shigley 7th edition solutions PDF

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Chapter 1 Problems 1-1 through 1-4 are for student research. 1-5 Impending motion to left E 1 1 f f A B G Fcr F ␪ D C ␪cr Facc Consider force F at G, reactions at B and D. Extend lines of action for fully-developed fric- tion D E and B E to find the point of concurrency at E for impending motion to ...

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Chapter 1 Problems 1-1 through 1-4 are for student research. 1-5

Impending motion to left E

1

1

f

f

A

B

D

G Fcr ␪cr

C

F

␪

Facc

Consider force F at G, reactions at B and D. Extend lines of action for fully-developed friction D E and B E to find the point of concurrency at E for impending motion to the left. The critical angle is θcr . Resolve force F into components Facc and Fcr . Facc is related to mass and acceleration. Pin accelerates to left for any angle 0 < θ < θcr . When θ > θcr , no magnitude of F will move the pin. Impending motion to right E⬘

1 f

⭈E

1 f B

A

G

d

⬘ Fcr D

C

⬘ ␪cr

F⬘ ␪⬘ F⬘acc

Consider force F ′ at G, reactions at A and C. Extend lines of action for fully-developed friction AE ′ and C E ′ to find the point of concurrency at E ′ for impending motion to the left. The ′ and F ′ . F ′ is related to mass critical angle is θcr′ . Resolve force F ′ into components Facc acc cr and acceleration. Pin accelerates to right for any angle 0 < θ ′ < θcr′ . When θ ′ > θcr′ , no magnitude of F ′ will move the pin. The intent of the question is to get the student to draw and understand the free body in order to recognize what it teaches. The graphic approach accomplishes this quickly. It is important to point out that this understanding enables a mathematical model to be constructed, and that there are two of them. This is the simplest problem in mechanical engineering. Using it is a good way to begin a course. What is the role of pin diameter d? Yes, changing the sense of F changes the response.

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1-6 (a)




y F

T r

␪ N

x




Fy = −F − f N cos θ + N sin θ = 0

(1)

T =0 r Ans.

Fx = f N sin θ + N cos θ −

fN

F = N (sin θ − f cos θ) T = Nr( f sin θ + cos θ)

Combining T = Fr

1 + f tan θ = KFr tan θ − f

(b) If T → ∞ detent self-locking tan θ − f = 0 (Friction is fully developed.) Check: If

F = 10 lbf, N=

∴ θcr = tan−1 f

f = 0.20, θ = 45◦ ,

(2)

Ans. Ans.

r = 2 in

10 = 17.68 lbf −0.20 cos 45◦ + sin 45◦

T = 17.28(0.20 sin 45◦ + cos 45◦ ) = 15 lbf r f N = 0.20(17.28) = 3.54 lbf θcr = tan−1 f = tan−1 (0.20) = 11.31◦ 11.31° < θ < 90° 1-7 (a) F = F0 + k(0) = F0 T1 = F0r Ans.

(b) When teeth are about to clear F = F0 + kx2

From Prob. 1-6 T2 = Fr T2 = r

f tan θ + 1 tan θ − f

( F0 + kx2 )( f tan θ + 1) tan θ − f

Ans.

1-8 Given, F = 10 + 2.5x lbf, r = 2 in, h = 0.2 in, θ = 60◦ , f = 0.25, xi = 0, x f = 0.2 Fi = 10 lbf; Ff = 10 + 2.5(0.2) = 10.5 lbf Ans.

3

Chapter 1

From Eq. (1) of Prob. 1-6 N=

F − f cos θ + sin θ 10 = 13.49 lbf Ni = −0.25 cos 60◦ + sin 60◦

Ans.

10.5 13.49 = 14.17 lbf Ans. 10

Nf = From Eq. (2) of Prob. 1-6 K =

1 + f tan θ 1 + 0.25 tan 60◦ = = 0.967 Ans. tan θ − f tan 60◦ − 0.25

Ti = 0.967(10)(2) = 19.33 lbf · in

Tf = 0.967(10.5)(2) = 20.31 lbf · in

1-9 (a) Point vehicles v x

Q=

v 42.1v − v 2 cars = = hour x 0.324

Seek stationary point maximum 42.1 − 2v dQ =0= ∴ v* = 21.05 mph dv 0.324 Q* =

42.1(21.05) − 21.052 = 1367.6 cars/h Ans. 0.324

(b)

v l 2

v Q= = x +l

x



l 2

l 0.324 + v(42.1) − v 2 v

−1

Maximize Q with l = 10/5280 mi v 22.18 22.19 22.20 22.21 22.22 % loss of throughput

Q 1221.431 1221.433 1221.435 ← 1221.435 1221.434 1368 − 1221 = 12% 1221

Ans.

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22.2 − 21.05 = 5.5% 21.05 Modest change in optimal speed Ans.

(c) % increase in speed

1-10

This and the following problem may be the student’s first experience with a figure of merit. • Formulate fom to reflect larger figure of merit for larger merit. • Use a maximization optimization algorithm. When one gets into computer implementation and answers are not known, minimizing instead of maximizing is the largest error one can make.
FV = F1 sin θ − W = 0
FH = −F1 cos θ − F2 = 0

From which

F1 = W/sin θ F2 = −W cos θ/sin θ

fom = −S = −¢γ (volume) . = −¢γ(l1 A1 + l2 A2 ) F1 W l1 A1 = = , l2 = S S sin θ cos θ    F2  W cos θ A2 =   = S S sin θ   l2 W cos θ l2 W + fom = −¢γ cos θ S sin θ S sin θ   −¢γ W l2 1 + cos2 θ = S cos θ sin θ Set leading constant to unity θ◦

fom

0 20 30 40 45 50 54.736 60

−∞ −5.86 −4.04 −3.22 −3.00 −2.87 −2.828 −2.886

θ* = 54.736◦ fom* = −2.828

Ans.

Alternative:   d 1 + cos2 θ =0 dθ cos θ sin θ And solve resulting transcendental for θ*.

Check second derivative to see if a maximum, minimum, or point of inflection has been found. Or, evaluate fom on either side of θ*.

Chapter 1

1-11 (a) x1 + x2 = X 1 + e1 + X 2 + e2 error = e = (x1 + x2 ) − ( X 1 + X 2 ) = e1 + e2 Ans. (b) x1 − x2 = X 1 + e1 − ( X 2 + e2 ) e = (x1 − x2 ) − ( X 1 − X 2 ) = e1 − e2 Ans. (c) x1 x2 = ( X 1 + e1 )( X 2 + e2 ) e = x1 x2 − X 1 X 2 = X 1 e2 + X 2 e1 + e1 e2   e2 e1 . Ans. + = X 1 e2 + X 2 e1 = X 1 X 2 X1 X2   x1 X 1 + e1 X 1 1 + e1 / X 1 = = (d) x2 X 2 + e2 X 2 1 + e2 / X 2  −1    e2 e2 e1 e1 e2 . e2 . 1+ =1− and 1+ =1+ 1− − X2 X2 X1 X2 X1 X2   x1 X 1 . X 1 e1 e2 e= − − = Ans. x2 X2 X2 X1 X2 1-12 (a)

x1 =

√ 5 = 2.236 067 977 5

X 1 = 2.23 3-correct digits √ x2 = 6 = 2.449 487 742 78

X 2 = 2.44 3-correct digits √ √ x1 + x2 = 5 + 6 = 4.685 557 720 28 √ e1 = x1 − X 1 = 5 − 2.23 = 0.006 067 977 5 √ e2 = x2 − X 2 = 6 − 2.44 = 0.009 489 742 78 √ √ e = e1 + e2 = 5 − 2.23 + 6 − 2.44 = 0.015 557 720 28 Sum = x1 + x2 = X 1 + X 2 + e = 2.23 + 2.44 + 0.015 557 720 28 = 4.685 557 720 28 (Checks) Ans. (b) X 1 = 2.24, X 2 = 2.45 √ e1 = 5 − 2.24 = −0.003 932 022 50 √ e2 = 6 − 2.45 = −0.000 510 257 22 e = e1 + e2 = −0.004 442 279 72 Sum = X 1 + X 2 + e = 2.24 + 2.45 + (−0.004 442 279 72) = 4.685 557 720 28 Ans.

5

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1-13 (a) (b) (c) (d) (e) (f) (g) (h) (i)

σ = 20(6.89) = 137.8 MPa F = 350(4.45) = 1558 N = 1.558 kN M = 1200 lbf · in (0.113) = 135.6 N · m A = 2.4(645) = 1548 mm2 I = 17.4 in4 (2.54) 4 = 724.2 cm4 A = 3.6(1.610) 2 = 9.332 km2 E = 21(1000)(6.89) = 144.69(103 ) MPa = 144.7 GPa v = 45 mi/h (1.61) = 72.45 km/h V = 60 in3 (2.54) 3 = 983.2 cm3 = 0.983 liter

(a) (b) (c) (d) (e) (f) (g) (h) (i)

l = 1.5/0.305 = 4.918 ft = 59.02 in σ = 600/6.89 = 86.96 kpsi p = 160/6.89 = 23.22 psi Z = 1.84(105 )/(25.4) 3 = 11.23 in3 w = 38.1/175 = 0.218 lbf/in δ = 0.05/25.4 = 0.00197 in v = 6.12/0.0051 = 1200 ft/min ǫ = 0.0021 in/in V = 30/(0.254) 3 = 1831 in3

1-14

1-15

200 = 13.1 MPa 15.3 42(103 ) = 70(106 ) N/m2 = 70 MPa (b) σ = 6(10−2 ) 2

(a) σ =

1-16

(c) y =

1200(800) 3 (10−3 ) 3 = 1.546(10−2 ) m = 15.5 mm 3(207)(6.4)(109 )(10−2 ) 4

(d) θ =

1100(250)(10−3 ) = 9.043(10−2 ) rad = 5.18◦ 4 9 − 3 4 79.3(π/32)(25) (10 )(10 )

600 = 5 MPa 20(6) 1 (b) I = 8(24) 3 = 9216 mm4 12 π (c) I = 324 (10−1 ) 4 = 5.147 cm4 64 16(16) = 5.215(106 ) N/m2 = 5.215 MPa (d) τ = π(253 )(10−3 ) 3

(a) σ =

Chapter 1

1-17 (a) τ =

120(103 ) = 382 MPa (π/4)(202 )

32(800)(800)(10−3 ) = 198.9(106 ) N/m2 = 198.9 MPa π(32) 3 (10−3 ) 3 π (364 − 264 ) = 3334 mm3 (c) Z = 32(36) (b) σ =

(d) k =

(1.6) 4 (79.3)(10−3 ) 4 (109 ) = 286.8 N/m 8(19.2) 3 (32)(10−3 ) 3

7

Chapter 2 2-1 (a)

12 10 8 6 4 2 0

60

70

80

90 100 110 120 130 140 150 160 170 180 190 200 210

(b) f/(N
x) = f/(69 · 10) = f/690

Eq. (2-9) Eq. (2-10)

x

f

fx

f x2

f/(N
x)

60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210

2 1 3 5 8 12 6 10 8 5 2 3 2 1 0 1 69

120 70 240 450 800 1320 720 1300 1120 750 320 510 360 190 0 210 8480

7200 4900 19 200 40 500 80 000 145 200 86 400 169 000 156 800 112 500 51 200 86 700 64 800 36 100 0 44 100 1 104 600

0.0029 0.0015 0.0043 0.0072 0.0116 0.0174 0.0087 0.0145 0.0116 0.0174 0.0029 0.0043 0.0029 0.0015 0 0.0015

8480 = 122.9 kcycles 69 1/2
1 104 600 − 84802 /69 sx = 69 − 1 x¯ =

= 30.3 kcycles

Ans.

9

Chapter 2

2-2 Data represents a 7-class histogram with N = 197. x

f

fx

f x2

174 182 190 198 206 214 220

6 9 44 67 53 12 6 197

1044 1638 8360 13 266 10 918 2568 1320 39 114

181 656 298 116 1 588 400 2 626 688 2 249 108 549 552 290 400 7 789 900

39 114 = 198.55 kpsi Ans. 197 1/2
7 783 900 − 39 1142 /197 sx = 197 − 1 x¯ =

= 9.55 kpsi Ans. 2-3 Form a table: x

f

fx

f x2

64 68 72 76 80 84 88 92

2 6 6 9 19 10 4 2 58

128 408 432 684 1520 840 352 184 4548

8192 27 744 31 104 51 984 121 600 70 560 30 976 16 928 359 088

4548 = 78.4 kpsi 58 1/2
359 088 − 45482 /58 sx = = 6.57 kpsi 58 − 1 x¯ =

From Eq. (2-14)

   1 1 x − 78.4 2 f (x) = √ exp − 2 6.57 6.57 2π

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2-4 (a) y

f

fy

f y2

y

f/(Nw)

f (y)

g(y)

5.625 5.875 6.125 6.375 6.625 6.875 7.125 7.375 7.625 7.875 8.125

1 0 0 3 3 6 14 15 10 2 1 55

5.625 0 0 19.125 19.875 41.25 99.75 110.625 76.25 15.75 8.125 396.375

31.640 63 0 0 121.9219 131.6719 283.5938 710.7188 815.8594 581.4063 124.0313 66.015 63 2866.859

5.625 5.875 6.125 6.375 6.625 6.875 7.125 7.375 7.625 7.875 8.125

0.072 727 0 0 0.218 182 0.218 182 0.436 364 1.018 182 1.090 909 0.727 273 0.145 455 0.072 727

0.001 262 0.008 586 0.042 038 0.148 106 0.375 493 0.685 057 0.899 389 0.849 697 0.577 665 0.282 608 0.099 492

0.000 295 0.004 088 0.031 194 0.140 262 0.393 667 0.725 002 0.915 128 0.822 462 0.544 251 0.273 138 0.106 72

For a normal distribution, 2866.859 − (396.3752 /55) y¯ = 396.375/55 = 7.207, sy = 55 − 1    1 x − 7.207 2 1 f ( y) = √ exp − 2 0.4358 0.4358 2π 

1/2

= 0.4358

For a lognormal distribution, √ √ x¯ = ln 7.206 818 − ln 1 + 0.060 4742 = 1.9732, sx = ln 1 + 0.060 4742 = 0.0604    1 ln x − 1.9732 2 1 exp − g( y) = √ 2 0.0604 x(0.0604)( 2π) (b) Histogram f 1.2

Data N LN

1 0.8 0.6 0.4 0.2 0 5.63

5.88

6.13

6.38

6.63

6.88 7.13 log N

7.38

7.63

7.88

8.13

Chapter 2

11

2-5 Distribution is uniform in interval 0.5000 to 0.5008 in, range numbers are a = 0.5000, b = 0.5008 in. (a) Eq. (2-22)

µx =

a+b 0.5000 + 0.5008 = = 0.5004 2 2

Eq. (2-23)

σx =

0.5008 − 0.5000 b−a = 0.000 231 √ = √ 2 3 2 3

(b) PDF from Eq. (2-20) f (x) =



1250 0.5000 ≤ x ≤ 0.5008 in 0 otherwise

(c) CDF from Eq. (2-21)  x < 0.5000 0 F(x) = (x − 0.5)/0.0008 0.5000 ≤ x ≤ 0.5008  1 x > 0.5008

If all smaller diameters are removed by inspection, a = 0.5002, b = 0.5008 µx =

0.5002 + 0.5008 = 0.5005 in 2

0.5008 − 0.5002 = 0.000 173 in √ 2 3  1666.7 0.5002 ≤ x ≤ 0.5008 f (x) = 0 otherwise σˆ x =

 x < 0.5002 0 F(x) = 1666.7(x − 0.5002) 0.5002 ≤ x ≤ 0.5008  1 x > 0.5008

2-6 Dimensions produced are due to tool dulling and wear. When parts are mixed, the distribution is uniform. From Eqs. (2-22) and (2-23), √ √ a = µx − 3s = 0.6241 − 3(0.000 581) = 0.6231 in √ √ b = µx + 3s = 0.6241 + 3(0.000 581) = 0.6251 in We suspect the dimension was

0.623 in 0.625

Ans.

12

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2-7 F(x) = 0.555x − 33 mm (a) Since F(x) is linear, the distribution is uniform at x = a F(a) = 0 = 0.555(a) − 33 ∴ a = 59.46 mm. Therefore, at x = b F(b) = 1 = 0.555b − 33 ∴ b = 61.26 mm. Therefore,  x < 59.46 mm 0 F(x) = 0.555x − 33 59.46 ≤ x ≤ 61.26 mm  1 x > 61.26 mm

The PDF is d F/dx , thus the range numbers are:  0.555 59.46 ≤ x ≤ 61.26 mm f (x) = 0 otherwise

Ans.

From the range numbers, µx =

59.46 + 61.26 = 60.36 mm Ans. 2 1

61.26 − 59.46 = 0.520 mm Ans. σˆ x = √ 2 3 (b) σ is an uncorrelated quotient F¯ = 3600 lbf, A¯ = 0.112 in2 C F = 300/3600 = 0.083 33,

C A = 0.001/0.112 = 0.008 929

From Table 2-6, for σ µF 3600 = 32 143 psi Ans. = µA 0.112 1/2
(0.083332 + 0.0089292 ) σˆ σ = 32 143 = 2694 psi Ans. (1 + 0.0089292 ) σ¯ =

Cσ = 2694/32 143 = 0.0838 Ans. Since F and A are lognormal, division is closed and σ is lognormal too. σ = LN(32 143, 2694) psi

Ans.

13

Chapter 2

2-8 Cramer’s rule    y x 2    x y x 3  yx 3 − x yx 2  = a1 =  Ans.  x x 2  xx 3 − (x 2 ) 2   x 2 x 3    y   x  2  xx y − yx 2 x x y  = a2 = Ans.  x x 2  xx 3 − (x 2 ) 2   x 2 x 3 

x

y

x2

x3

xy

0 0.2 0.4 0.6 0.8 1.0 3.0

0.01 0.15 0.25 0.25 0.17 −0.01 0.82

0 0.04 0.16 0.36 0.64 1.00 2.20

0 0.008 0.064 0.216 0.512 1.000 1.800

0 0.030 0.100 0.150 0.136 −0.010 0.406

a1 = 1.040 714

x 0 0.2 0.4 0.6 0.8 1.0

y 0.3

a2 = −1.046 43

Ans.

Data y

Regression y

0.01 0.15 0.25 0.25 0.17 −0.01

0 0.166 286 0.248 857 0.247 714 0.162 857 −0.005 71

Data Regression

0.25 0.2 0.15 0.1 0.05 0 ⫺0.05 0

0.2

0.4

0.6

0.8

1

x

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2-9 Su 0 60 64 65 82 101 119 120 130 134 145 180 195 205 207 210 213 225 225 227 230 238 242 265 280 295 325 325 355 5462

Data Se′ 30 48 29.5 45 51 50 48 67 60 64 84 78 96 87 87 75 99 87 116 105 109 106 105 96 99 114 117 122 2274.5

m = 0.312 067 Se⬘ 140

Regression Se′ 20.356 75 39.080 78 40.329 05 40.641 12 45.946 26 51.875 54 57.492 75 57.804 81 60.925 48 62.173 75 65.606 49 76.528 84 81.209 85 84.330 52 84.954 66 85.890 86 86.827 06 90.571 87 90.571 87 91.196 92.132 2 94.628 74 95.877 01 103.054 6 107.735 6 112.416 6 121.778 6 121.778 6 131.140 6

Su2

Su Se′

3 600 4 096 4 225 6 724 10 201 14 161 14 400 16 900 17 956 21 025 32 400 38 025 42 025 42 849 44 100 45 369 50 625 50 625 51 529 52 900 56 644 58 564 70 225 78 400 87 025 105 625 105 625 126 025 1 251 868

1 800 3 072 1 917.5 3 690 5 151 5 950 5 760 8 710 8 040 9 280 15 120 15 210 19 680 18 009 18 270 15 975 22 275 19 575 26 332 24 150 25 942 25 652 27 825 26 880 29 205 37 050 38 025 43 310 501 855.5

b = 20.356 75

Ans.

Data Regression

120 100 80 60 40 20 0

0

100

200

300

400

Su

15

Chapter 2

2-10

E

=



y − a0 − a2 x 2

2

 ∂E y − a0 − a2 x 2 = 0 = −2 ∂a0 y = na0 + a2 x2 x2 = 0 ⇒ y − na0 − a2

 ∂E =2 y − a0 − a2 x 2 (2x) = 0 ⇒ x3 x + a2 x y = a0 ∂a2

Cramer’s rule

  y  x y a0 =   n  x   n x a2 =   n  x

 x 2  x 3  x 3 y − x 2 x y  = x 2  nx 3 − xx 2 3 x  y  nx y − xy x y  2  = x  nx 3 − xx 2 x 3

x

Data y

Regression y

20 40 60 80 200

19 17 13 7 56

19.2 16.8 12.8 7.2

a0 = a2 =

x2 400 1600 3600 6400 12 000

x3

xy

8 000 380 64 000 680 216 000 780 512 000 560 800 000 2400

800 000(56) − 12 000(2400) = 20 4(800 000) − 200(12 000)

4(2400) − 200(56) = −0.002 4(800 000) − 200(12 000)

y 25

Data Regression

20

15

10

5

0

0

20

40

60

80

100

x

Ans.

16

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2-11

x

Data y

Regression y

0.2 0.4 0.6 0.8 1 2 5

7.1 10.3 12.1 13.8 16.2 25.2 84.7

7.931 803 9.884 918 11.838 032 13.791 147 15.744 262 25.509 836

x2

y2

xy

x − x¯

(x − x) ¯ 2

0.04 0.16 0.36 0.64 1.00 4.00 6.2

50.41 106.09 146.41 190.44 262.44 635.04 1390.83

1.42 4.12 7.26 11.04 16.20 50.40 90.44

−0.633 333 −0.433 333 −0.233 333 −0.033 333 0.166 666 1.166 666 0

0.401 111 111 0.187 777 778 0.054 444 444 0.001 111 111 0.027 777 778 1.361 111 111 2.033 333 333

mˆ = k =

6(90.44) − 5(84.7) = 9.7656 6(6.2) − (5) 2

84.7 − 9.7656(5) = 5.9787 bˆ = Fi = 6 F 30

Data Regression

25 20 15 10 5 0

0

0.5

5 x¯ = ; 6

(a)

1

y¯ =

1.5

2

2.5

x

84.7 = 14.117 6

Eq. (2-37) s yx = Eq. (2-36)



1390.83 − 5.9787(84.7) − 9.7656(90.44) 6−2

= 0.556 

sbˆ = 0.556

1 (5/6) 2 + = 0.3964 lbf 6 2.0333

Fi = (5.9787, 0.3964) lbf

Ans.

17

Chapter 2

(b) Eq. (2-35) 0.556 = 0.3899 lbf/in smˆ = √ 2.0333 k = (9.7656, 0.3899) lbf/in 2-12

Ans.

The expression ǫ = δ/l is of the form x/y. Now δ = (0.0015, 0.000 092) in, unspecified distribution; l = (2.000, 0.0081) in, unspecified distribution; C x = 0.000 092/0.0015 = 0.0613

C y = 0.0081/2.000 = 0.000 75

From Table 2-6, ǫ¯ = 0.0015/2.000 = 0.000 75 1/2
0.06132 + 0.004 052 σˆ ǫ = 0.000 75 1 + 0.004 052 = 4.607(10−5 ) = 0.000 046

We can predict ǫ¯ and σˆ ǫ but not the distribution of ǫ. 2-13

σ = ǫE ǫ = (0.0005, 0.000 034) distribution unspecified; E = (29.5, 0.885) Mpsi, distribution unspecified; C x = 0.000 034/0.0005...