Solution Manual for Shigleys Mechanical Engineering Design 11th Edition Budynas / All Chapters 1 - 20 / Full Complete 2023

Chaptera31

Problemsa31-1a3througha31-6a3area3fora3studenta3research.a3Noa3standarda3solutionsa3area3provided.

1-7 Froma3Fig.a31-
2,a3costa3ofa3grindinga3toa3a30.0005a3ina3isa3270%.a3Costa3ofa3turninga3toa3a30.003a3in
a3isa360%.
Relativea3costa3ofa3grindinga3vs.a3turninga 3 =a3270/60a3=a 3 4.5a3times Ans.

1-8 CAa3 =a 3 CB,

10a3+a30.8a3Pa3=a360a3+a30.8a3Pa3a30.005a3Pa32

Pa32a3=a350/0.005  Pa3=a3100a3partsa 3 Ans.


1-9 Max.a3loada3=a31.10a3
Pa3Min.a3areaa3=a3(0.9
5)2Aa3Min.a3strengtha3
=a30.85a3S
Toa3offseta3thea3absolutea3uncertainties,a3thea3designa3factor,a3froma3Eq.a3(1-1)a3shoulda3be

1.10
nda 3  a31.4 Ans.
0.850.95 3
2




1-10 (a)a 3 X1a3 +a
3X2: x1a 3 a3x2a 3 a 3 X1a 3 a3e1a 3 a3Xa32a 3 a3e2
error a3ea3a3a3x1a3a3x2a3a3a3a3X1a3a3Xa32a3
a3e1a3 a3e2 Ans.
(b) X1a 3 a3 
X2: x1a3a3x2a3a3X1a3a3e1a3a3a3Xa32a3a

3e2a3 Ans.
ea3a3a3x1a3a3x2a3a3a3a3X1a3a3Xa32a
(c) X1a 3 X2:
3a3a3e1a3a3e2



x1x2a3a3a3X1a3a3e1a3a3Xa32a3a3e2a3

ea3a3 x1a3x2a 3 a3X1a3Xa32a 3 a 3 X1e2a 3 a3 Xa32e1a 3 a3e1e2a 3
a3Xa3e  3 Xe X a3 Xa a 3 e2a Ans.
e1  3 
a 3

1a32 2a 3 1 1a 3 a 3 2a3 a3  
X Xa 3 a 3
Chaptera31a3Solutionsa3-
a3Rev.a3B,a3Pagea31/6

, 1 2a 3 




Chaptera31a3Solutionsa3-
a3Rev.a3B,a3Pagea32/6

, (d) X1/X2:
x1a 3 X1a 3 a3e1a 3 X1a 3 
X1a 3aa331 a3e1 a
a3 
x3 Xa 3 a3e Xa 3 a 3 1a3e Xa 3
2 2 2 2a 3  2 2a 3 
1
 ea 3 a 3  e a31a3ea 3 a 3 Xa ea 3  ea 3 a 3  e e
3a3 
a 3 a 3 2a 3 a3 a31a3a3a then
1a3  3 2a 3 a3 1a3a3
1
a 3 1a 3 a3 1 a3a 3 a 3 2a 3 a3 a 3 1a 3 a3 a3
a 3 2a 3 a3 a31a3  a3 a31 a3
 Xa32a 3  a31 a3 Xa32a 3   X1a 3 Xa32 X1 Xa32
Xa32 e2  a 3 
x1a 3
Thus, ea3 X a3 a X1a 3  e Ans.
3
1a 3
a 3 e1 a 3 2
a 3
x X X a3 X Xa 3 
2 2 2a 3  1 2a 3 


1-11 (a) x1a 3 = 7a 3 =a32.645a3751a3311a31
X1a 3 =a32.64 (3a3correcta3digits)
x2a 3 = 8a 3 =a32.828a3427a3124a37
X2a 3 =a32.82 (3a3correcta3digits)
x1a3 +a3x2a3 =a35.474a3178a3435a38
e1a3 =a3x1a3 a 3 X1a3 =a30.005a3751a3311a31
e2a3 =a3x2a3 a 3 X2a3 =a30.008a3427a3124a37
ea3=a3e1a3+a3e2a3=a30.014a3178a3
435a38a3Suma3=a3x1a3+a3x2a3=a3X
1a3+a3X2a3+a3e
=a32.64a3+a32.82a3+a30.014a3178a3435a38a3=a35.474a3178a3435a38 Checks
(b) X1a3 =a32.65,a 3 X2a3 =a32.83a 3 (3a3digita3significanta3numbers)
e1a3 =a3x1a3 a 3 X1a3 =a3a30.004a3248a3688a39
e2a3 =a3x2a3 a 3 X2a3 =a3a30.001a3572a3875a33
ea3=a3e1a3+a3e2a3=a3a30.005a382
1a3564a32a3Suma3=a3x1a3+a3x2a3=
a3X1a3+a3X2a3+a3e
=a32.65a3+2.83a3a30.001a3572a3875a33a3=a35.474a3178a3435a38 Checks


S 
25 103a3 
16a31000

1-12 a 3     da3 a30.799 Ans.
a 3 in
nd 2.5
da
3
3
Tablea3A-17: da3=a8 3 7a 3 in Ans.
S 25 103a3
Factora3ofa3safet y: na3a3 a3 a33.29
Chaptera31a3Solutionsa3-
a3Rev.a3B,a3Pagea33/6

, Ans.
 16a31000
a3 a37a3
3

8


n

1-13a 3 a 3 Eq.a3(1-5): Ra3=a3a3Ria 3 =a30.98(0.96)0.94a3=a30.88
i1

Overalla3reliabilitya3=a388a3percent Ans.




Chaptera31a3Solutionsa3-
a3Rev.a3B,a3Pagea34/6