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

Q= Seek stationary point maximum

cars v 42.1v − v 2 = = hour x 0.324

dQ 42.1 − 2v =0= ∴ v* = 21.05 mph dv 0.324 Q* = (b) 42.1(21.05) − 21.052 = 1368 cars/h Ans. 0.324 v l 2 x l 2

v = Q= x +l Maximize Q with l = 10/5280 mi v 22.18 22.19 22.20 22.21 22.22

0.324 l + 2 v(42.1) − v v

−1

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

% loss of throughput = (c) % increase in speed

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

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

1-6 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 ) W F1 l1 = , l2 = A1 = S S sin θ cos θ A2 = W cos θ F2 = S S sin θ l2 W cos θ l2 W + cos θ S sin θ S sin θ 1 + cos2 θ cos θ sin θ

fom = −¢γ = Set leading constant to unity θ◦ 0 20 30 40 45 50 54.736 60

−¢γ W l2 S

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

θ* = 54.736◦ fom* = −2.828 Alternative:

Ans.

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

3

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 (d) x1 X 1 + e1 X1 = = x2 X 2 + e2 X2 1+ e= 1-8 (a) e2 X2
−1

1 + e1 / X 1 1 + e2 / X 2 and 1+ e1 X1 1− e2 X2 e1 e2 . =1+ − X1 X2

e2 . =1− X2

x1 X1 . X1 − = x2 X2 X2 x1 =

e2 e1 − X1 X2

Ans.

√ 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.

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

1-9 (a) (b) (c) (d) (e) (f) (g) (h) (i) 1-10 (a) (b) (c) (d) (e) (f) (g) (h) (i) 1-11 (a) σ = 200 = 13.1 MPa 15.3 42(103 ) = 70(106 ) N/m2 = 70 MPa (b) σ = 6(10−2 ) 2 1200(800) 3 (10−3 ) 3 = 1.546(10−2 ) m = 15.5 mm 3(207)109 (64)103 (10−3 ) 4 1100(250)(10−3 ) = 9.043(10−2 ) rad = 5.18◦ 9 )(π/32)(25) 4 (10−3 ) 4 79.3(10 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 = 0.0021 in/in V = 30/(0.254) 3 = 1831 in3 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

(c) y = (d) θ = 1-12

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

5

1-13 (a) τ = (b) σ = 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) (d) k = (1.6) 4 (10−3 ) 4 (79.3)(109 ) = 286.8 N/m 8(19.2) 3 (10−3 ) 3 (32)