Solutions completo elementos de maquinas de shigley 8th edition. 1. 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.431.433.435 ←.435.434% 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. Xl 2 l 2 v x v budysmch01.qxd 15:23 Page 1. 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(l1 A1 + l2 A2) 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. Budysmch01.qxd 15:23 Page 2. 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 − X1 X2 = X1e2 + X2e1 + e1e2. = X1e2 + X2e1 = X1 X2 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. Takeuchi tb 025 manual.
= 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.
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