Elementos De Maquinas

7.28(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 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. f tan θ + 1 tan θ − f Ans.

( F0 + kx2 )( f tan θ + 1) tan θ − f

Chapter 1

3

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◦ 10.5 13.49 = 14.17 lbf Ans. 10

Ans.

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= Seek stationary point maximum

v 42.1v − v 2 cars = = hour x 0.324

42.1 − 2v dQ =0= ∴ v* = 21.05 mph dv 0.324 Q* = (b) 42.1(21.05) − 21.052 = 1367.6 cars/h Ans. 0.324

v l 2 x l 2

Q=

v = x +l

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

(c) % increase in speed

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

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 ) 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

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