Solution Manual for Shigley-s Mechanical Engineering Design in SI Units 11th Edition By Budynas

All20 ChaptersCovered
k k k




SOLUTIONMANUAL
k

, Chapter1 k




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




E




1 1

f f
A B
G
Fcr F θ
D C k cr k k



Facc



Consider force F at G, reactions at B and D. Extend lines of action for fully-developed fric- tion
k k k k k k k k k k k k k k k k k




DE and BE to find the point of concurrency at E for impending motion to the left. The critical
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angle is θcr. Resolve force F into components Facc and Fcr. Facc is related to mass and acceleration.
k k k k k k k k k k k k k k k k k k




Pin accelerates to left for any angle 0 < θ < θcr. When θ > θcr, no magnitude of F will move the
k k k k k k k k k k k k k k k k k k k k k k k




pin.
k




Impending motion to right k k k




∙



1 1

f f
A B
G
d
cr θ'
D C cr
acc



Consider force F ′ at G, reactions at A and C. Extend lines of action for fully-developed fric- tion
k k k k k k k k k k k k k k k k k k




AE ′ and CE ′ to find the point of concurrency at E ′ for impending motion to the left. The critical
k
k
k
k
k k k k k k k k k k k k k k k k




angle is θc′r. Resolve force F ′ into components Fa′cc and Fc′r. Fa′cc is related to mass and
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k
k k
k
k k k k




acceleration. Pin accelerates to right for any angle 0 < θ′ < θc′r. When θ′ > θc′r, no mag-
k k k k k k k k k k k k k k k k k k k k k




nitude of F′ will move the pin.
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The intent of the question is to get the student to draw and understand the free body in order to
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recognize what it teaches. The graphic approach accomplishes this quickly. It is im- portant to
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point out that this understanding enables a mathematical model to be constructed, and that there
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are two of them.
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This is the simplest problem in mechanical engineering. Using it is a good way to begin a
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course.
k




What is the role of pin diameter d? k k k k k k k




Yes, changing the sense of F changes the response.
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,2 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
k k k k k k k k k k




1-6
Σ k




(a) y Fy = −F − f N cosθ + N sinθ = 0 k k k k k k k k k k k k k (1)
F Σ T
Fx = f N sinθ + N cosθ − =0
r
k k k k k k k k k k k




T
θ r x F = N(sinθ − f cosθ) Ans. k k k k k k k k k




T = Nr ( f sinθ + cosθ) k k k k k k k k k k


N
fN k




Combining
1 + f tanθ
T = Fr = KFr Ans. (2)
k k k k




tanθ − f
k k



k k k




(b) If T → ∞ detent self-locking
k k k k k tanθ − f = 0 k ∴ θcr = tan−1 f k k Ans.
(Friction is fully developed.) k k k




Check: If F = 10 lbf,
k k k θ = 45◦, r = 2in f = 0.20,
k k k k k k k




10
N= = 17.68lbf
−0.20cos45◦ + sin45◦
k k k k

k k
k k k k




T
= 17.28(0.20sin45◦ + cos45◦) = 15lbf
r
k k k k k k k k k


k




fN = 0.20(17.28) = 3.54lbf
k k k k k k




θcr = tan−1 f = tan−1(0.20) = 11.31◦
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11.31° < θ <90° k k k k




1-7
(a) F = F0 + k(0) = F0 T1
k k k k k k k




= F0r k Ans.
k




(b) When teeth are about to clear k k k k k




F = F0 + kx2k k k k




From Prob. 1-6 k k




f tanθ + 1
T2 = Fr
k k k k k




tan θ − f
k



k k k k




(F0 + kx2)( f tanθ + 1)
T2 = r
k k k k k k k



Ans.
k




tanθ − f
k



k k k




1-8
Given, F = 10 + 2.5x lbf, r = 2in, h = 0.2in, θ = 60◦, f = 0.25, xi = 0, xf = 0.2
k k k k k k k k k k k k k k k k k k k k k k k k k k k




Fi = 10lbf;Ff = 10+ 2.5(0.2) = 10.5lbf
k k k k k k k k k k k Ans.

, Chapter 1 k 3

From Eq. (1) of Prob. 1-6
F
k k k k k




N=
− f cosθ + sinθ
k



k k k




10
Ni = = 13.49lbf Ans.
−0.25cos60◦ + sin60◦
k k k
k
k k
k k k k




10.5
Nf = 13.49 = 14.17 lbf Ans.
10
k k k k k k


k




From Eq. (2) of Prob. 1-6
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1 + f tanθ 1 + 0.25 tan60◦
K= = = 0.967 Ans.
k k k k k k k k




tanθ − f tan60◦ − 0.25
k k k k
k
k




Ti = 0.967(10)(2) = 19.33 lbf · in
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Tf = 0.967(10.5)(2) = 20.31 lbf · in
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1-9
(a) Point vehicles k




v

x

cars v 42.1v − v2 k k




Q= = =
hour x 0.324
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k k




Seek stationary point maximum
k k k




dQ 42.1 − 2v
= 0= ∴ v* = 21.05mph
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k k k k k k k




dv 0.324
42.1(21.05) — 21.052
Q* = = 1367.6cars/h k
Ans.
0.324
k k




(b) v

l x l
2 2

−1
k



k v k k
l 0.324 k k




+
k




Q= = v(42.1) − v2 v
k




x +l
k k
k
k k
k k k




Maximize Q with l = 10/5280 mi k k k k k k




v Q
22.18 1221.431
22.19 1221.433
22.20 1221.435 ←
22.21 1221.435
22.22 1221.434

1368 − 1221
% loss of throughput = 12% Ans.
k k



k k k k




1221