Diseño en ingenieria mecanica-shigley solucionario octava edicion

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Chapter 1
Problems 1-1 through 1-4 are for student research. 1-5
Impending motion to left E

1 f A

1 f B G Fcr F
cr

D

C

Facc

Consider force F at G, reactions at B and D. Extend lines of action for fully-developed friction D E and B E to find the point of concurrency at E for impending motion to the left. The critical angle is θcr . Resolve force F into components Facc and Fcr .Facc is related to mass and acceleration. Pin accelerates to left for any angle 0 < θ < θcr . When θ > θcr , no magnitude of F will move the pin.
Impending motion to right E E

1 f A d D

1 f B G Fcr C
cr

F Facc

Consider force F at G, reactions at A and C. Extend lines of action for fully-developed friction AE and C E to find the point of concurrency at E for impending motion to theleft. The critical angle is θcr . Resolve force F into components Facc and Fcr . Facc is related to mass and acceleration. Pin accelerates to right for any angle 0 < θ < θcr . When θ > θcr , no magnitude of F will move the pin. The intent of the question is to get the student to draw and understand the free body in order to recognize what it teaches. The graphic approach accomplishes this quickly.It is important to point out that this understanding enables a mathematical model to be constructed, and that there are two of them. This is the simplest problem in mechanical engineering. Using it is a good way to begin a course. What is the role of pin diameter d? Yes, changing the sense of F changes the response.

2

Solutions Manual • Instructor’s Solution Manual to Accompany MechanicalEngineering Design

1-6 (a)
y F

Fy = −F − f N cos θ + N sin θ = 0 Fx = f N sin θ + N cos θ −
T r x

(1)

N

F = N (sin θ − f cos θ) T = Nr( f sin θ + cos θ) Combining T = Fr

T =0 r Ans.

fN

1 + f tan θ = KFr tan θ − f

Ans. Ans.

(2)

(b) If T → ∞ detent self-locking tan θ − f = 0 (Friction is fully developed.) Check: If F = 10 lbf, N= f = 0.20, θ = 45◦ ,

∴ θcr = tan−1f r = 2 in

10 = 17.68 lbf −0.20 cos 45◦ + sin 45◦

T = 17.28(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 ofthroughput Q 1221.431 1221.433 1221.435 ← 1221.435 1221.434 1368 − 1221 = 12% 1221 Ans.

4

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 toreflect 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...
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