Solutions Manual for Shigley's Mechanical Engineering Design 11th Edition By Richard Budynas

Chapter 1 g




Problemsg1-1gthroughg1-4garegforgstudentgresearch.

1-5 Impendinggmotiongtogleft
E




1 1

f f
A B
G
Fcr F θ
D C θcrg
Facc



ConsidergforcegFgatgG,greactionsgatgBgandgD.gExtendglinesgofgactiongforgfully-developedgfric-
gtiongDE g andgBE g togfindgthegpointgofgconcurrencygatgEgforgimpendinggmotiongtogthegleft.gThegcr

iticalganglegisgθcr.gResolvegforcegFgintogcomponentsgFaccgandgFcr.gFaccgisgrelatedgtogmassgandgacc
eleration.gPingacceleratesgtogleftgforganygangleg0g<gθg <gθcr.gWhengθg >gθcr,g nogmagnitudegofgFg
willgmovegthegpin.

Impendinggmotiongtogright




1 1

f f
A B
G
d
θ'
D C θc'r




ConsidergforcegFgrg atgG,greactionsgatgAgandgC.gExtendglinesgofgactiongforgfully-developedgfric-
gtiongAE
r g andgCE r g togfindgthegpointgofgconcurrencygatgEgrgforgimpendinggmotiongtogthegleft.gTheg
criticalganglegisgθcrr.gResolvegforcegFgrgintogcomponentsgFargccg andgFcrr.gFargccg isgrelatedgtogmassg
andgacceleration.gPingacceleratesgtogrightgforganygangleg0g<gθgrg <gθcrr.gWhengθgrg >gθcrr,g nogmag-
nitudegofg Fgrg willgmovegthegpin.
Thegintentgofgthegquestiongisgtoggetgthegstudentgtogdrawgandgunderstandgthegfreegbodygingord
ergtogrecognizegwhatgitgteaches.gTheggraphicgapproachgaccomplishesgthisgquickly.gItgisgim-
gportantgtogpointgoutgthatgthisgunderstandinggenablesgagmathematicalgmodelgtogbegconstructed,ga

ndgthatgtheregaregtwogofgthem.
Thisgisgthegsimplestgproblemgingmechanicalgengineering.gUsinggitgisgaggoodgwaygtogbegingagc
ourse.
Whatgisgthegrolegofgpingdiametergd?
Yes,gchanginggthegsensegofgFgchangesgthegresponse.

,2 SolutionsgManualg •g Instructor’sgSolutiongManualgtogAccompanygMechanicalgEngineeringgDesign


1-6
Σg
(a) y Fyg =g−Fg−g fgNgcosgθg +gNgsingθg =g0 (1)
F Σg T
Fx =g fgNg singθg +g Ng cosgθg −g =0
r
θ
T
r x Fg =g Ng(singθg −g fg cosgθg) Ans.
T = Nr g(gfg singθ + cosgθg)
N
fNg
Combining
1g+g fg tangθ
T = Frg =g KFr Ans. (2)
tangθg−g f
(b)g g IfgT → g∞ detentgself-locking tangθ − f = 0 ∴gθcr = tan−1g f Ans.
(Frictiongisgfullygdeveloped.)

Check: If Fg=g 10glbf, fg =g 0.20, θg =g 45◦, rg =g 2gin
10
N= =g17.68glbf
−0.20gcosg45◦g +g sing45◦
T
=g17.28(0.20gsing45◦g+gcosg45◦)g=g15glbf
rg
fgNg=g0.20(17.28)g=g3.54glbf

θcrg=gtan−1g fg =gtan−1(0.20)g=g11.31◦

11.31°g<gθg <g90°

1-7
(a) Fg=gF0g+gk(0)g=gF0g
T1g=gF0r Ans.
(b) Whengteethgaregaboutgtogclear
Fg =g F0g+gkx2
FromgProb.g1-6
T
2g= gfg tang
Frg
θg+g1gtang
θg−g f
Ans.
(gF0g+gkx2)(gfg tangθg +g1)
T2g =grg
tangθg−g f

1-8
Given,gFg =g 10g+g2.5xglbf,grg =g 2gin,ghg =g0.2gin,g θg =g60◦,g fg =g 0.25,g xig =g 0,gxgfg =g0.2
Fig =g10glbf;gFfg =g10g+g2.5(0.2)g=g10.5glbf Ans.

, Chapterg1 3

FromgEq.g(1)gofgProb.g1-6
F
Ng =g
−gfg cosgθg +gsingθ
10
Nig= =g13.49glbf Ans.
−0.25gcosg60◦g +g sing60◦
10.5
N f g =g 13.49g=g14.17glbf Ans.
10
FromgEq.g(2)gofgProb.g1-6
K 1g+g0.25gtang60◦
= 1g+g fg ta = = 0.967 Ans.
ngθgtangθg tang60◦g −g0.25
−g f
Tig =g 0.967(10)(2)g =g 19.33g lbfg ·g in
Tfg =g 0.967(10.5)(2)g =g 20.31g lbfg ·g in

1-9
(a) Pointgvehicles
v

x

cars v 42.1vg−gv2
Qg=g =g =
hourg xg 0.324
Seekgstationarygpointgmaximum
dQ g 42.1g−g2v ∴gv*g=g21.05gmph
=g0g=g
dvg Q 0.324
42.1(21.05) — 21.052
*g= =g 1367.6gcars/h Ans.
0.324
(b) v

l l
x
2 2
µ ¶
g vg 0.324 lg −1
Qg=g = +g
xg+glg v(42.1)g−gv2g v

MaximizegQgwithglg=g 10/5280gmi

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

1368g−g1221
%glossgofgthroughputg = 12% Ans.
1221

, 4 SolutionsgManualg •g Instructor’sgSolutiongManualgtogAccompanygMechanicalgEngineeringgDesign

22.2g−g21.05g g
(c) %gincreasegingspeedg 5.5%
=
21.05
Modestgchangegingoptimalgspeed Ans.

1-10 Thisgandgthegfollowinggproblemgmaygbegthegstudent’sgfirstgexperiencegwithgagfiguregofgmerit.
• Formulategfomgtogreflectglargergfiguregofgmeritgforglargergmerit.
• Usegagmaximizationgoptimizationgalgorithm.gWhengoneggetsgintogcomputergimplementa-
gtiongandganswersgaregnotgknown,gminimizingginsteadgofgmaximizinggisgtheglargestgerrorgo

negcangmake. Σg
FVg =g F1gsingθg −g Wg =g0
Σg
FHg =g−F1gcosgθg −g F2g=g0
Fromgwhich
F1g=gW/singθ
F2g=g−Wgcosgθ/singθ
fomg= −Sg=g−¢γg (volume)
.
=g−¢γg(l1gA1g+gl2gA2)
F1g W l1
A1g=g = , l2g=
S Sgsingθ cosgθ
¯gF2g¯g Wgcosgθ
A2g =g ¯gSg ¯g=g Sgsingθ
µ l2 ¶
W l2Wgcosgθg
fomg=g −¢γ +
cosgθg Sgsingθg Sgsingθ
µg 2 ¶
−¢γgWl 2 g 1g+gcos gθg
=
S cosgθgsingθ
Setgleadinggconstantgtogunity

θg◦ fom
θg*g=g 54.736◦ Ans.
0 −∞ fom*g=g−2.828
20 −5.86 Alternative:
30 −4.04
40 −3.22 d µ 1g+ cos2gθ ¶ =g 0
45 −3.00 dθ cos g gθg singθ

50 −2.87
54.736 −2.828 Andgsolvegresultinggtran-
gscendentalgforgθg*.
60 −2.886


Checkg secondg derivativeg tog seeg ifg ag maximum,g minimum,g org pointg ofg inflectiong hasg beengf
ound.gOr,gevaluategfomgongeithergsidegofgθg*.