Complete Solutions for Shigley's Mechanical Engineering Design (11th Edition) - Budynas & Nisbett

All 20 Chapters Covered
j j j




SOLUTION MANUAL
j

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Chapter 1 j




Problemsj1-1jthroughj1-6jarejforjstudentjresearch.jNojstandardjsolutionsjarejprovided.

1-7 FromjFig.j1-2,jcostjofjgrindingjtojj0.0005jinjisj270%.jCostjofjturningjtojj0.003jinjisj60%.
Relativejcostjofjgrindingjvs.jturningj =j270/60j=j 4.5jtimes Ans.


1-8 CAj=j CB,

10j+j0.8jPj=j60j+j0.8jPjj0.005jPj2

Pj2j=j50/0.005  Pj=j100jpartsj Ans.


1-9 Max.jloadj=j1.10jPjMin
.jareaj=j(0.95)2AjMin.jst
rengthj=j0.85jS
Tojoffsetjthejabsolutejuncertainties,jthejdesignjfactor,jfromjEq.j(1-1)jshouldjbe

1.10
nj  j j1.43
j

Ans.
0.850.95
2
d




1-10 (a)j X1j+jX2:
x1j jx2j j X1j je1j jXj2j je2
error jejjxj 1jjx2j
j jX
j 1jjXj2j



 je1jje2 Ans.
(b) X1jjX2:
x1jjx2jjX1jje1jjjXj2jje2j

ejjjx1jjx2jjjjX1jjXj2jjje1jje2 Ans.
(c) X1jX2:
x1x2jjX
j 1jje1jX
j j2jje2j


ejj x1x2j jX1jXj2j j X1e2j jXj2e1j je1e2
j e ej j 
j Xj e jXj e j Xj X j j 1j j jj j 2j j Ans.
1j 2 2j 1 1j j 2  j 
 X1 Xj2j 




Shigley’sj MED,j10thjedition Chapterj 1jSolutions,jPagej1/12

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




(d) X1/X2:
x Xj je Xj j1jej j Xj j 
j 1jj
jj j 1
jj j 1jjj
1j j 1 1j j

x2 Xj2j je2 Xj2j j1je2 Xj2j 
1
 ej j  e j1 jej j Xj   ej  ej j  e e
 1 j
j j 2j j
 j1jjj j
2
then j
1 1j j

j 
j 1 j
j j 1j j
 1 j
j j 2j j
j 1
j  j
j j 1j j
 j
j j 2j


 Xj2j  Xj2 j1je2 Xj2j   X1j  Xj2j  X1 Xj2
x X X j e e 
Thus, ejjj 1jj jj j 1jj jj j 1jjjj j 1j j jj j 2j j Ans.
x2 Xj2 X 2j j 1
j X X 2j 
j


 





1-11 (a) x1j= 7 =j2.645j751j311j1
X1j=j2.64 (3jcorrectjdigits)
x2j= 8 =j2.828j427j124j7
X2j=j2.82 (3jcorrectjdigits)
x1j+jx2j=j5.474j178j435j8
e1j=jx1jj X1j=j0.005j751j311j1
e2j=jx2jj X2j=j0.008j427j124j7
ej=je1j+je2j=j0.014j178j435j8jSumj
=jx1j+jx2j=jX1j+jX2j+je
=j2.64j+j2.82j+j0.014j178j435j8j=j5.474j178j435j8 Checks
(b) X1j=j2.65,j X2j=j2.83j j (3jdigitjsignificantjnumbers)
e1j=jx1jj X1j=jj0.004j248j688j9
e2j=jx2jj X2j=jj0.001j572j875j3
ej=je1j+je2j=jj0.005j821j564j2jSu
mj=jx1j+jx2j=jX1j+jX2j+je
=j2.65j+2.83jj0.001j572j875j3j=j5.474j178j435j8 Checks


S 32j1000  
25 103j
1-12 j     dj j1.006j in Ans.
nd d
j j
3
2.5
1
TablejA-17: dj=j 1j j4in Ans.

Factorjofjsafety: nj 
S

   4.79
25 103j
j Ans.
 32j1000
 1.25
3
j




Shigley’sj MED,j10thjedition Chapterj 1jSolutions,jPagej1/12

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1-13 (a)
x f fjx fjx2
60 2 120 7200
70 1 70 4900
80 3 240 19200
90 5 450 40500
100 8 800 80000
110 12 1320 145200
120 6 720 86400
130 10 1300 169000
140 8 1120 156800
150 5 750 112500
160 2 320 51200
170 3 510 86700
180 2 360 64800
190 1 190 36100
200 0 0 0
210 1 210 44100
 69 8480 1j104j600



k 8 480j
Eq.j(1-6) x  j 1j
 fjixj ji j
Nj i1 69
j122
.9j kcycles


Eq.j(1-7)

fj xj j j Nj x
1/j2
 1104j 600j j69(122.9) j
2
sjxj  j 30.3j kcycles Ans.
j ij1
69j j1
Nj j1  

xj j  x j x 115j j122.9
(b) Eq.j(1-5) z115j  ˆ x j 115js j
30.3 j 0.2607
x x



InterpolatingjfromjTablej (A-10)

0.2600 0.3974
0.2607 x  xj =j 0.3971
0.2700 0.3936

N(0.2607)j =j 69j(0.3971)j =j27.4j j27 Ans.

Fromjthejdata,jthejnumberjofjinstancesjlessjthanj115jkcyclesjis


Shigley’sj MED,j10thjedition Chapterj 1jSolutions,jPagej1/12