MECH 420 Problems Problem Chap 1, Fall 2012

Chapter 1 Problems 1-1 through 1-4 are for student research. 1-5 (a) Point vehicles Q = cars hour = v x = 42.1v − v2 0.324 Seek stationary point maximum dQ dv = 0 = 42.1 − 2v 0.324 ∴ v* = 21.05 mph Q* = 42.1(21.05) − 21.052 0.324 = 1368 cars/h Ans. (b) Q = v x + l = � 0.324 v(42.1) − v2 + l v �−1 Maximize Q with l = 10/5280 mi v Q 22.18 1221.431 22.19 1221.433 22.20 1221.435 ← 22.21 1221.435 22.22 1221.434 % loss of throughput = 1368 − 1221 1221 = 12% Ans. (c) % increase in speed 22.2 − 21.05 21.05 = 5.5% Modest change in optimal speed Ans. x l 2 l 2 v x v 2 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 implementa- tion 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(l1A1 + l2A2) A1 = F1 S = W S sin θ , l2 = l1 cos θ A2 = ���� F2 S ���� = W cos θ S sin θ fom = −¢γ � l2 cos θ W S sin θ + l2W cos θ S sin θ � = −¢γ Wl2 S �1 + cos2 θ cos θ sin θ � Set leading constant to unity θ◦ fom 0 −∞ 20 −5.86 30 −4.04 40 −3.22 45 −3.00 50 −2.87 54.736 −2.828 60 −2.886 Check second derivative to see if a maximum, minimum, or point of inflection has been found. Or, evaluate fom on either side of θ*. θ* = 54.736◦ Ans. fom* = −2.828 Alternative: d dθ �1 + cos2 θ cos θ sin θ � = 0 And solve resulting tran- scendental for θ*. Chapter 1 3 1-7 (a) x1 + x2 = X1 + e1 + X2 + e2 error = e = (x1 + x2) − (X1 + X2) = e1 + e2 Ans. (b) x1 − x2 = X1 + e1 − (X2 + e2) e = (x1 − x2) − (X1 − X2) = e1 − e2 Ans. (c) x1x2 = (X1 + e1)(X2 + e2) e = x1x2 − X1X2 = X1e2 + X2e1 + e1e2 .= X1e2 + X2e1 = X1X2 � e1 X1 + e2 X2 � Ans. (d) x1 x2 = X1 + e1 X2 + e2 = X1 X2 �1 + e1/X1 1 + e2/X2 � � 1 + e2 X2 �−1 .= 1 − e2 X2 and � 1 + e1 X1 � � 1 − e2 X2 � .= 1 + e1 X1 − e2 X2 e = x1 x2 − X1 X2 .= X1 X2 � e1 X1 − e2 X2 � Ans. 1-8 (a) x1 = √ 5 = 2.236 067 977 5 X1 = 2.23 3-correct digits x2 = √ 6 = 2.449 487 742 78 X2 = 2.44 3-correct digits x1 + x2 = √ 5 + √ 6 = 4.685 557 720 28 e1 = x1 − X1 = √ 5 − 2.23 = 0.006 067 977 5 e2 = x2 − X2 = √ 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 = X1 + X2 + e = 2.23 + 2.44 + 0.015 557 720 28 = 4.685 557 720 28 (Checks) Ans. (b) X1 = 2.24, X2 = 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 = X1 + X2 + e = 2.24 + 2.45 + (−0.004 442 279 72) = 4.685 557 720 28 Ans. 4 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 1-9 (a) σ = 20(6.89) = 137.8 MPa (b) F = 350(4.45) = 1558 N = 1.558 kN (c) M = 1200 lbf · in (0.113) = 135.6 N · m (d) A = 2.4(645) = 1548 mm2 (e) I = 17.4 in4 (2.54)4 = 724.2 cm4 (f) A = 3.6(1.610)2 = 9.332 km2 (g) E = 21(1000)(6.89) = 144.69(103) MPa = 144.7 GPa (h) v = 45 mi/h (1.61) = 72.45 km/h (i) V = 60 in3 (2.54)3 = 983.2 cm3 = 0.983 liter 1-10 (a) l = 1.5/0.305 = 4.918 ft = 59.02 in (b) σ = 600/6.89 = 86.96 kpsi (c) p = 160/6.89 = 23.22 psi (d) Z = 1.84(105)/(25.4)3 = 11.23 in3 (e) w = 38.1/175 = 0.218 lbf/in (f) δ = 0.05/25.4 = 0.00197 in (g) v = 6.12/0.0051 = 1200 ft/min (h) ϵ = 0.0021 in/in (i) V = 30/(0.254)3 = 1831 in3 1-11 (a) σ = 200 15.3 = 13.1 MPa (b) σ = 42(103) 6(10−2)2 = 70(106) N/m2 = 70 MPa (c) y = 1200(800)3(10−3)3 3(207)109(64)103(10−3)4 = 1.546(10−2) m = 15.5 mm (d) θ = 1100(250)(10−3) 79.3(109)(π/32)(25)4(10−3)4 = 9.043(10−2) rad = 5.18◦ 1-12 (a) σ = 600 20(6) = 5 MPa (b) I = 1 128(24)3 = 9216 mm4 (c) I = π 64324(10−1)4 = 5.147 cm4 (d) τ = 16(16) π(253)(10−3)3 = 5.215(106) N/m2 = 5.215 MPa Chapter 1 5 1-13 (a) τ = 120(103) (π/4)(202) = 382 MPa (b) σ = 32(800)(800)(10−3) π(32)3(10−3)3 = 198.9(106) N/m2 = 198.9 MPa (c) Z = π 32(36)(364 − 264) = 3334 mm3 (d) k = (1.6)4 (10−3)4(79.3)(109) 8(19.2)3(10−3)3(32) = 286.8 N/m