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

Chapter 7 solutions solution manual shigleys mechanical en
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gineering design g




Mechanical engineering (The University of Lahore)
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, Chapter 7 g




7-1 (a)gDE-Gerber,gEq.g(7-10):

= 4(2.2)(70) +g3(1.8)(45) =g338.4gNggm
2 2 2g 2
Ag= a ƒsg a



( = 4(2.2)(55) +g3(1.8)(35) =g265.5gNggm
2 2 2g 2
Bg= 4g K g M +g3g K ƒsTg m
2 g 1/g2  1/3

g 8(2)(338.4)g 
  g2(265.5)g(210) 106g   
 g
( )
dg =g g(210) 106g 1+g1+gg 338.4g(700) 106g g 

( )   ( )
g  
g g
  
  
dg=g25.85g(10−3)gmg=g25.85gmm Ans.

(b) DE-elliptic,gEq.g(7-12)gcangbegshowngtogbe
1/3
g16n 1/3  16(2) 
A2g B2  =g (338.4) (265.5)
2 2
dg =g 
g  +g    
Sge2 Sgy2 
2 2

 
6 6
 
dg=g25.77g(10−3)gmg=g25.77gmm Ans.

(c) DE-Soderberg,gEq.g(7-14)gcangbegshowngtogbe

1/3
   1/3
16(2) 
g g g 338.4g g g g g 265.5g g

dg =gg  gSe +g Syg  =gg g  gg210g(106g)+g 560g(106g)g
16ng gAg gBg
g 
  
− 3
dg=g27.70g(10 )gmg=g27.70gmm Ans.

(d) DE-Goodman:gEq.g(7-8)gcangbegshowngtogbe
  Bg 
1/3
 1/3
 16(2)  338.4 265.5  
16ng Ag
dg =g  ggSe + Sutg  =gg g  g210g(106g) + 700g(106g)g
g


g 
  
− 3
dg=g27.27g(10 )gmg=g27.27gmm Ans.

Criterion dg(mm) ComparedgtogDE-Gerber
DE-Gerber 25.85
DE-Elliptic 25.77 0.31%gLower Lessgconservative
DE-Soderberg 27.70 7.2%gHigher Moregconservative
DE-Goodman 27.27 5.5%gHigher Moregconservative


7-2 Thisgproblemghasgtogbegdonegbygsuccessivegtrials,gsincegSegisgagfunctiongofgshaftgsize.gTheg
materialgisgSAEg2340gforgwhichgSutg=g175gkpsi,gSyg=g160gkpsi,gandgHBg≥g370.


Chapter 7g-gRev.gA,gPageg1/45

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,Eq.g(6-19),gp.g287: kgag =g2.70(175)−0.265g =g0.69

Trialg#1:gChoosegdrg =g0.75gin

Eq.g(6-20),gp.g288: kgbg =g0.879(0.75)−0.107g =g0.91
Eq.g(6-8),gp.282: Sgeg=g0.5Sg utg =g0.5g(175)g=g87.5gkpsi
Eq.g(6-18),gp.g287: Seg =g0.69g(0.91)(87.5)g=g54.9gkpsi

drg =gdg−g2rg=g0.75Dg−g2Dg/g20g=g0.65D
d 0.75g
Dg=g r =g =g1.15gin
0.65 0.65
Dg 1.15g
rg = g = g =g0.058gin
20 20

Fig.gA-15-14:
dg =gdrg +g2rg=g0.75g+g2(0.058)g=g0.808gin
dg 0.808g
=g =g1.08
dr 0.75
rg 0.058g
=g =g0.077
dr 0.75
Ktg =g1.9
Fig.g6-20,gp.g295: rg=g0.058gin,gqg=g0.90
Eq.g(6-32),gp.g295: Kƒg =g1g+g0.90g(1.9g–
g1)g=g1.81gFig.gA-15-15: Ktsg=g1.5
Fig.g6-21,gp.g296: rg=g0.058gin,gqsg =g0.92
Eq.g(6-32),gp.g295: Kƒsg =g1g+g0.92g(1.5g–g1)g=g1.46

WegselectgthegDE-ASMEgEllipticgfailuregcriteria,gEq.g(7-12),gwithgdgasgdr,gand
Mmg =gTag =g0,
1/3
16(2.5)g g1.81(600)g 2 g1.46(400)g2g g1/g2g
drg =g g4g g 
  + 3 
g54.9g(10g3)g g160g(10g )g
3g g

  
     g 

drg =g0.799gin

Trialg#2:gChoosegdrg =g0.799gin.

kgbg =g0.879(0.799)−0.107g =g0.90
Seg =g0.69g(0.90)(0.5)(175)g=g54.3gkpsi
d 0.799g
Dg=g r =g =g1.23gin
0.65 0.65
rg=gDg/g20g=g1.23/20g=g0.062gin


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, Figs.gA-15-14gandgA-15-15:
dg =gdrg +g2rg =g0.799g+g2(0.062)g=g0.923gin
dg 0.923g
=g =g1.16
dr 0.799
rg 0.062g
=g =g0.078
dr 0.799

Withgthesegratiosgonlygslightlygdifferentgfromgthegpreviousgiteration,gwegaregatgtheglimitgofgreadabi
litygofgthegfigures.gWegwillgkeepgthegsamegvaluesgasgbefore.

Ktg =g1.9, Ktsg =g1.5, qg=g0.90, qsg =g0.92
gKgƒ =g1.81, Kgƒsg =g1.46
UsinggEq.g(7-12)gproducesgdrg =g0.802gin.gFurthergiterationgproducesgnogchange.g With
drg =g0.802gin,
0.802g
Dg=g =g1.23gin
0.65
dg=g0.75(1.23)g=g0.92gin

Aglookgatgagbearinggcataloggfindsgthatgthegnextgavailablegboregdiametergisg0.9375gin.g Ingn
ominalgsizes,gwegselectgdg=g0.94gin,gDg=g1.25gin,grg=g0.0625gin Ans.


7-3 Fgcosg20(dg/g2)g=gTA,g Fg=g2gTAg /g(gdgcosg20)g=g2(340)g/g(0.150gcosg20)g=g4824gN.
ThegmaximumgbendinggmomentgwillgbegatgpointgC,gwithgMCg =g4824(0.100)g=g482.4gN·m.gD
uegtogthegrotation,gthegbendinggisgcompletelygreversed,gwhilegthegtorsiongisgconstant.
Thus,gMag =g482.4gN·m,gTmg =g340gN·m,gMmg =gTag =g0.

Forgsharpgfilletgradiigatgthegshoulders,gfromgTableg7-1,gKtg =g2.7,gandgKtsg =g2.2.g Examining
Figs.g6-20gandg6-21g(pp.g295gandg296grespectively)gwith Sutg =g560gMPa,gconservatively
estimategqg=g0.8gandg qsg =g0.9.gThesegestimatesgcangbegcheckedgoncegagspecificgfilletgradiusgi
sgdetermined.

Eq.g(6-32): Kgƒg =g1g+g0.8(2.7g−1)g=g2.4
Kgƒsg =g1g+g0.9(2.2g−1)g=g2.1

(a) Wegwillgchoosegtogincludegfatiguegstressgconcentrationgfactorsgevengforgthegstaticga
nalysisgtogavoidglocalizedgyielding.
g32Kg Mg 2 g16Kg T 2g1/g2
Eq.g(7-15):   =    +g3
ƒ
 
g g a ƒsg m
max

d g 
3
 d g g 
3




Chapter 7g-gRev.gA,gPageg3/45


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