Shigley's Mechanical Engineering Design 8th Edition (Solution Manual) PDF

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Philadelphia University Mechanical Engineering Design 8th eng.ahmad jabali Chapter 1 Problems 1-1 through 1-4 are for student research. 1-5 (a) Point vehicles v x cars v 42.1v − v 2 Q= = = hour x 0.324 Seek stationary point maximum dQ 42.1 − 2v =0= ∴ v* = 21.05 mph dv 0.324 42.1(21.05) − 21.052 Q* =...

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Philadelphia University Mechanical Engineering Design 8th eng.ahmad jabali

Chapter 1 Problems 1-1 through 1-4 are for student research. 1-5 (a) Point vehicles v x

Q=

v 42.1v − v 2 cars = = hour x 0.324

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

42.1(21.05) − 21.052 = 1368 cars/h Ans. 0.324

(b)

v l 2

x

l 2


 000001 0.324 v l −1 = Q= + x +l v(42.1) − v 2 v Maximize Q with l = 10/5280 mi Q

v 22.18 22.19 22.20 22.21 22.22

1221.431 1221.433 1221.435 ← 1221.435 1221.434

% loss of throughput =

1368 − 1221 = 12% 1221

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

(c) % increase in speed

Ans.

Philadelphia University Mechanical Engineering Design 8th eng.ahmad jabali 2

1-6

Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design

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 = −$ = −¢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*.

Philadelphia University Mechanical Engineering Design 8th eng.ahmad jabali Chapter 1

1-7 (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 1-8 (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.

3

Philadelphia University Mechanical Engineering Design 8th eng.ahmad jabali 4

Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design

1-9 (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

1-10

1-11

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

(a) σ =

1-12

(c) y =

1200(800) 3 (10−3 ) 3 = 1.546(10−2 ) m = 15.5 mm 3(207)109 (64)103 (10−3 ) 4

(d) θ =

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

Philadelphia University Mechanical Engineering Design 8th eng.ahmad jabali Chapter 1

1-13 (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 (10−3 ) 4 (79.3)(109 ) = 286.8 N/m 8(19.2) 3 (10−3 ) 3 (32)

5

Philadelphia University Mechanical Engineering Design 8th eng.ahmad jabali

Chapter 2 2-1 From Table A-20 Sut = 470 MPa (68 kpsi), S y = 390 MPa (57 kpsi)

Ans.

2-2 From Table A-20 Sut = 620 MPa (90 kpsi), S y = 340 MPa (49.5 kpsi) Ans. 2-3 Comparison of yield strengths: Sut of G10 500 HR is Syt of SAE1020 CD is

620 = 1.32 times larger than SAE1020 CD 470 390 = 1.15 times larger than G10500 HR 340

Ans. Ans.

From Table A-20, the ductilities (reduction in areas) show, SAE1020 CD is

40 = 1.14 times larger than G10500 35

Ans.

The stiffness values of these materials are identical Ans.

Sut MPa (kpsi)

Sy MPa (kpsi)

Table A-20 Ductility R%

SAE1020 CD 470(68) UNS10500 HR 620(90)

390 (57) 340(495)

40 35

Table A-5 Stiffness GPa (Mpsi) 207(30) 207(30)

2-4 From Table A-21 1040 Q&T S¯ y = 593 (86) MPa (kpsi) at 205◦C (400◦F)

Ans.

2-5 From Table A-21 1040 Q&T

R = 65%

at 650◦C (1200◦F)

Ans.

2-6 Using Table A-5, the specific strengths are: Sy 39.5(103 ) = = 1.40(105 ) in UNS G10350 HR steel: W 0.282 2024 T4 aluminum:

Sy 43(103 ) = = 4.39(105 ) in W 0.098

Ti-6Al-4V titanium:

Sy 140(103 ) = = 8.75(105 ) in W 0.16

ASTM 30 gray cast iron has no yield strength.

Ans.

Ans. Ans. Ans.

Philadelphia University Mechanical Engineering Design 8th eng.ahmad jabali 7

Chapter 2

2-7 The specific moduli are: E 30(106 ) = = 1.06(108 ) in W 0.282

UNS G10350 HR steel:

Ans.

2024 T4 aluminum:

E 10.3(106 ) = = 1.05(108 ) in W 0.098

Ans.

Ti-6Al-4V titanium:

E 16.5(106 ) = = 1.03(108 ) in W 0.16

Ans.

Gray cast iron:

E 14.5(106 ) = = 5.58(107 ) in W 0.26

Ans.

2G(1 + ν) = E

2-8

⇒ ν=

E − 2G 2G

From Table A-5 Steel: ν = Aluminum: ν = Beryllium copper: ν = Gray cast iron: ν =

30 − 2(11.5) = 0.304 2(11.5)

Ans.

10.4 − 2(3.90) = 0.333 2(3.90) 18 − 2(7) = 0.286 2(7) 14.5 − 2(6) = 0.208 2(6)

Ans.

Ans. Ans.

2-9 E U 80 70

Stress P兾A0 kpsi

60 50 Y

40

Su ⫽ 85.5 kpsi Ans. Sy ⫽ 45.5 kpsi Ans.

30

E ⫽ 90兾0.003 ⫽ 30 000 kpsi Ans.

20

R⫽

A0 ⫺ AF 0.1987 ⫺ 0.1077 ⫽ (100) ⫽ 45.8% Ans. A0 0.1987

10

⑀⫽

A0 l ⫺ l0 ⌬l l ⫽ ⫺ 1⫽ ⫺1 ⫽ A l0 l0 l0

0

0

0.002 0.1

0.004 0.2

0.006 0.3

0.008 0.4 Strain, ⑀

0.010 0.5

0.012 0.6

0.014 0.7

0.016 (Lower curve) 0.8 (Upper curve)

Philadelphia University Mechanical Engineering Design 8th eng.ahmad jabali 8

Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design

2-10

To plot σtrue vs. ε, the following equations are applied to the data. A0 = Eq. (2-4)

π(0.503) 2 = 0.1987 in2 4

ε = ln

l l0

ε = ln

A0 A

σtrue =

for 0 ≤ L ≤ 0.0028 in for L > 0.0028 in

P A

The results are summarized in the table below and plotted on the next page. The last 5 points of data are used to plot log σ vs log ε m = 0.2306

The curve fit gives

Ans.

log σ0 = 5.1852 ⇒ σ0 = 153.2 kpsi For 20% cold work, Eq. (2-10) and Eq. (2-13) give, A = A0 (1 − W ) = 0.1987(1 − 0.2) = 0.1590 in2 ε = ln

A0 0.1987 = ln = 0.2231 A 0.1590

Eq. (2-14): S y′ = σ0 εm = 153.2(0.2231) 0.2306 = 108.4 kpsi

Ans.

Eq. (2-15), with Su = 85.5 kpsi from Prob. 2-9, Su′ = P 0 1 000 2 000 3 000 4 000 7 000 8 400 8 800 9 200 9 100 13 200 15 200 17 000 16 400 14 800

Su 85.5 = = 106.9 kpsi 1−W 1 − 0.2

Ans.

L

A

ε

σtrue

log ε

log σtrue

0 0.0004 0.0006 0.0010 0.0013 0.0023 0.0028 0.0036 0.0089

0.198 713 0.198 713 0.198 713 0.198 713 0.198 713 0.198 713 0.198 713 0.198 4 0.197 8 0.196 3 0.192 4 0.187 5 0.156 3 0.130 7 0.107 7

0 0.000 2 0.000 3 0.000 5 0.000 65 0.001 149 0.001 399 0.001 575 0.004 604 0.012 216 0.032 284 0.058 082 0.240 083 0.418 956 0.612 511

0 5032.388 10 064.78 15 097.17 20 129.55 35 226.72 42 272.06 44 354.84 46 511.63 46 357.62 68 607.07 81 066.67 108 765.2 125 478.2 137 418.8

−3.699 01 −3.522 94 −3.301 14 −3.187 23 −2.939 55 −2.854 18 −2.802 61 −2.336 85 −1.913 05 −1.491 01 −1.235 96 −0.619 64 −0.377 83 −0.212 89

3.701 774 4.002 804 4.178 895 4.303 834 4.546 872 4.626 053 4.646 941 4.667 562 4.666 121 4.836 369 4.908 842 5.036 49 5.098 568 5.138 046

Philadelphia University Mechanical Engineering Design 8th eng.ahmad jabali 9

Chapter 2

160000 140000

␴true (psi)

120000 100000 80000 60000 40000 20000 0

0

0.1

0.2

0.3

0.4 ␧true

0.5

0.6

0.7

5.2

y ⫽ 0.2306x ⫹ 5.1852 5

log ␴

5.1

4.9

⫺1.6

⫺1.2

⫺1

⫺0.8 log ␧

⫺0.6

⫺0.4

⫺0.2

0

4.8

Tangent modulus at σ = 0 is E0 =

σ . 5000 − 0 = 25(106 ) psi = ε 0.2(10−3 ) − 0

At σ = 20 kpsi . (26 − 19)(103 ) = 14.0(106 ) psi E 20 = − 3 (1.5 − 1)(10 ) ε(10−3 ) 0 0.20 0.44 0.80 1.0 1.5 2.0 2.8 3.4 4.0 5.0

0 5 10 16 19 26 32 40 46 49 54

Ans.

60

σ (kpsi)

50 40 ␴ (kpsi)

2-11

⫺1.4

(Sy)0.001 ⫽ ˙ 35 kpsi Ans. 30 20 10 0

0

1

2

3 ␧ (10⫺3)

4

5

Philadelphia University Mechanical Engineering Design 8th eng.ahmad jabali 10

2-12

Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design

Since |εo | = |εi |






R+h




R R + N
ln
=
ln
=
−ln

R+ N

R+ N

R
R+N R+h = R+N R

( R + N ) 2 = R( R + h) N 2 + 2R N − Rh = 0     h 1/2 N = R −1 ± 1 + R

From which, The roots are:

The + sign being significant,    h 1/2 −1 1+ N=R R

Ans.

Substitute for N in εo = ln

Gives



  ε0 = ln  

R+h R+N

R+h   h 1/2 −R R+ R 1+ R



   h 1/2   = ln 1 +  R

Ans.

These constitute a useful pair of equations in cold-forming situations, allowing the surface strains to be found so that cold-working strength enhancement can be estimated. 2-13

From Table A-22 AISI 1212

S y = 28.0 kpsi, σ f = 106 kpsi, Sut = 61.5 kpsi σ0 = 110 kpsi,

ε f = 0.85

εu = m = 0.24

From Eq. (2-12) Eq. (2-10)

A0 1 1 = = = 1.25 Ai′ 1−W 1 − 0.2

Eq. (2-13)

εi = ln 1.25 = 0.2231 ⇒ εi < εu

Eq. (2-14)

S y′ = σ0 εim = 110(0.2231) 0.24 = 76.7 kpsi

Eq. (2-15) 2-14

m = 0.24,

Su′ =

Su 61.5 = = 76.9 kpsi 1−W 1 − 0.2

Ans.

Ans.

For H B = 250, Eq. (2-17)

Su = 0.495 (250) = 124 kpsi = 3.41 (250) = 853 MPa

Ans.

Philadelphia University Mechanical Engineering Design 8th eng.ahmad jabali 11

Chapter 2

2-15

For the data given, H B = 2530



H B2 = 640 226

640 226 − (2530) 2 /10 2530 = 3.887 = 253 σˆ H B = H¯ B = 9 10

Eq. (2-17)

2-16

From Prob. 2-15,

S¯u = 0.495(253) = 125.2 kpsi

Ans.

σ¯ su = 0.495(3.887) = 1.92 kpsi

Ans.

H¯ B = 253

and σˆ HB = 3.887

Eq. (2-18) S¯u = 0.23(253) − 12.5 = 45.7 kpsi σˆ su = 0.23(3.887) = 0.894 kpsi 2-17

. 45.52 uR = = 34.5 in · lbf/in3 2(30)

(a)

Ans.

Ans.

Ans.

(b) P

L

0 1 000 2 000 3 000 4 000 7 000 8 400 8 800 9 200 9 100 13 200 15 200 17 000 16 400 14 800

0 0.0004 0.0006 0.0010 0.0013 0.0023 0.0028 0.0036 0.0089

A

0.1963 0.1924 0.1875 0.1563 0.1307 0.1077

A0 /A − 1

ε

0.012 291 0.032 811 0.059 802 0.271 355 0.520 373 0.845059

0 0.0002 0.0003 0.0005 0.000 65 0.001 15 0.0014 0.0018 0.004 45 0.012 291 0.032 811 0.059 802 0.271 355 0.520 373 0.845 059

σ = P/A0 0 5 032.39 10 064.78 15 097.17 20 129.55 35 226.72 42 272.06 44 285.02 46 297.97 45 794.73 66 427.53 76 492.30 85 550.60 82 531.17 74 479.35

Philadelphia University Mechanical Engineering Design 8th eng.ahmad jabali 12

Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design

␴ 90000 80000 70000 60000 50000 40000 30000 20000 10000 0

␧ 0

0.2

0.4 All data points

0.6

0.8

␴ 50000 45000 40000 35000 30000 25000 20000 A1

15000

A2

10000 5000 0

␧ 0

0.001

0.002 0.003 First 9 data points

0.004

0.005

␴ 90000 80000 70000 60000 50000 40000

A4

A5

30000 20000 A3 10000 0

␧ 0

0.2

0.4 0.6 Last 6 data points

0.8

5 1 . Ai = (43 000)(0.001 5) + 45 000(0.004 45 − 0.001 5) uT = 2 i=1 1 + (45 000 + 76 500)(0.059 8 − 0.004 45) 2 + 81 000(0.4 − 0.059 8) + 80 000(0.845 − 0.4)

. = 66.7(103 )in · lbf/in3

Ans.

Philadelphia University Mechanical Engineering Design 8th eng.ahmad jabali Chapter 2

2-18

13

m = Alρ For stiffness, k = AE/l, or, A = kl/E. Thus, m = kl 2 ρ/E, and, M = E/ρ. Therefore, β = 1 From Fig. 2-16, ductile materials include Steel, Titanium, Molybdenum, Aluminum, and Composites. For strength, S = F/A, or, A = F/S. Thus, m = Fl ρ/S, and, M = S/ρ. From Fig. 2-19, lines parallel to S/ρ give for ductile materials, Steel, Nickel, Titanium, and composites. Common to both stiffness and strength are Steel, Titanium, Aluminum, and Composites. Ans.

Philadelphia University Mechanical Engineering Design 8th eng.ahmad jabali

Chapter 3 3-1 W

W 2

2 B

A RA

A

RB

B RA

1

RB 1

1

(a)

(b) 1

1 RD

RC

D

2

3

C

A

RA RB

B

W

(c)

W 1

RC RB

RA RB

2 RA W

(d) (e) A

2 W RBx B RBx 1

RB

RBy

RBy

Scale of corner magnified

(f)

RA

1

Philadelphia University Mechanical Engineering Design 8th eng.ahmad jabali 15

Chapter 3

3-2 (a)

2 kN

R A = 2 sin 60 = 1.732 kN

RB

60°

Ans.

90°

R B = 2 sin 30 = 1 kN

2 2 kN

30°

60° RB

Ans.

RA

30°

RA

1

(b)

RA

0.4 m

S = 0.6 m

B

A

α = tan−1

45° 800 N

0.6 m

RO

0.6 = 30.96◦ 0.4 + 0.6

␣

O s

RA 800 = sin 135 sin 30.96 RO 800 = sin 14.04 sin 30.96

RO RA 135° 30.96°

800 N

45⬚ ⫺ 30.96⬚ ⫽ 14.04⬚

⇒

R A = 1100 N

⇒

R O = 377 N

Ans. Ans.

30.96°

(c)

1.2 = 2.078 kN Ans. tan 30 1.2 = 2.4 kN Ans. RA = sin 30

1.2 kN

RO = 30° RA

RO

60°

90°

60° 1.2 kN

RA RO

(d) Step 1: Find R A and R E

4.5 = 7.794 m tan 30
MA = 0 哷+

h=

4.5 m

C

30° y

400 N

4

2

9R E − 7.794(400 cos 30) − 4.5(400 sin 30) = 0 R E = 400 N Ans.

h B RAx RA

3

D 60°

A

x

E 9m

RAy







RE

Fx = 0

R Ax + 400 cos 30 = 0

Fy = 0

R Ay + 400 − 400 sin 30 = 0 ⇒ R Ay = −200 N  R A = 346.42 + 2002 = 400 N Ans.

⇒

R Ax = −346.4 N

Philadelphia University Mechanical Engineering Design 8th eng.ahmad jabali 16

Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design

Step 2: Find components of RC on link 4 and R D
(RCy)4 MC = 0 哷+ (RCx)4

C

400(4.5) − (7.794 − 1.9) R D = 0 ⇒
Fx = 0 ⇒ ( RC x ) 4 = 305.4 N
Fy = 0 ⇒ ( RC y ) 4 = −400 N

4 RD

D

R D = 305.4 N

Ans.

E 400 N

Step 3: Find components of RC on link 2
(RCy)2 Fx = 0 (RCx)2

C

( RC x ) 2 + 305.4 − 346.4 = 0
Fy = 0

2

B

( RC x ) 2 = 41 N

⇒

( RC y ) 2 = 200 N

305.4 N

346.4 N A 200 N 400 N

200 N 41 N

200 N C

305.4 N

400 N C

41 N

305.4 N

30° Pin C

400 N

305.4 N

B

D

B

346.4 N

305.4 N

D

E

A 200 N

400 N

400...