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

Chaptera11

Problemsa11-1a1througha11-6a1area1fora1studenta1research.a1Noa1standarda1solutionsa1area1provided.

1-7 Froma1Fig.a11-
2,a1costa1ofa1grindinga1toa1a10.0005a1ina1isa1270%.a1Costa1ofa1turninga1toa1a10.003a1in
a1isa160%.
Relativea1costa1ofa1grindinga1vs.a1turninga 1 =a1270/60a1=a 1 4.5a1times Ans.

1-8 CAa1 =a 1 CB,

10a1+a10.8a1Pa1=a160a1+a10.8a1Pa1a10.005a1Pa12

Pa12a1=a150/0.005  Pa1=a1100a1partsa 1 Ans.


1-9 Max.a1loada1=a11.10a1
Pa1Min.a1areaa1=a1(0.9
5)2Aa1Min.a1strengtha1
=a10.85a1S
Toa1offseta1thea1absolutea1uncertainties,a1thea1designa1factor,a1froma1Eq.a1(1-1)a1shoulda1be

1.10
nda 1  a11.4 Ans.
0.850.95 3
2




1-10 (a)a 1 X1a1 +a
1X2: x1a 1 a1x2a 1 a 1 X1a 1 a1e1a 1 a1Xa12a 1 a1e2
error a1ea1a1a1x1a1a1x2a1a1a1a1X1a1a1Xa12a1
a1e1a1 a1e2 Ans.
(b) X1a 1 a1 
X2: x1a1a1x2a1a1X1a1a1e1a1a1a1Xa12a1a

1e2a1 Ans.
ea1a1a1x1a1a1x2a1a1a1a1X1a1a1Xa12a
(c) X1a 1 X2:
1a1a1e1a1a1e2



x1x2a1a1a1X1a1a1e1a1a1Xa12a1a1e2a1

ea1a1 x1a1x2a 1 a1X1a1Xa12a 1 a 1 X1e2a 1 a1 Xa12e1a 1 a1e1e2a 1
a1Xa1e  1 Xe X a1 Xa a 1 e2a Ans.
e1  1 
a 1

1a12 2a 1 1 1a 1 a 1 2a1 a1  
X Xa 1 a 1
Chaptera11a1Solutionsa1-
a1Rev.a1B,a1Pagea11/6

, 1 2a 1 




Chaptera11a1Solutionsa1-
a1Rev.a1B,a1Pagea12/6

, (d) X1/X2:
x1a 1 X1a 1 a1e1a 1 X1a 1 
X1a 1aa111 a1e1 a
a1 
x1 Xa 1 a1e Xa 1 a 1 1a1e Xa 1
2 2 2 2a 1  2 2a 1 
1
 ea 1 a 1  e a11a1ea 1 a 1 Xa ea 1  ea 1 a 1  e e
1a1 
a 1 a 1 2a 1 a1 a11a1a1a then
1a1  1 2a 1 a1 1a1a1
1
a 1 1a 1 a1 1 a1a 1 a 1 2a 1 a1 a 1 1a 1 a1 a1
a 1 2a 1 a1 a11a1  a1 a11 a1
 Xa12a 1  a11 a1 Xa12a 1   X1a 1 Xa12 X1 Xa12
Xa12 e2  a 1 
x1a 1
Thus, ea1 X a1 a X1a 1  e Ans.
1
1a 1
a 1 e1 a 1 2
a 1
x X X a1 X Xa 1 
2 2 2a 1  1 2a 1 


1-11 (a) x1a 1 = 7a 1 =a12.645a1751a1311a11
X1a 1 =a12.64 (3a1correcta1digits)
x2a 1 = 8a 1 =a12.828a1427a1124a17
X2a 1 =a12.82 (3a1correcta1digits)
x1a1 +a1x2a1 =a15.474a1178a1435a18
e1a1 =a1x1a1 a 1 X1a1 =a10.005a1751a1311a11
e2a1 =a1x2a1 a 1 X2a1 =a10.008a1427a1124a17
ea1=a1e1a1+a1e2a1=a10.014a1178a1
435a18a1Suma1=a1x1a1+a1x2a1=a1X
1a1+a1X2a1+a1e
=a12.64a1+a12.82a1+a10.014a1178a1435a18a1=a15.474a1178a1435a18 Checks
(b) X1a1 =a12.65,a 1 X2a1 =a12.83a 1 (3a1digita1significanta1numbers)
e1a1 =a1x1a1 a 1 X1a1 =a1a10.004a1248a1688a19
e2a1 =a1x2a1 a 1 X2a1 =a1a10.001a1572a1875a13
ea1=a1e1a1+a1e2a1=a1a10.005a182
1a1564a12a1Suma1=a1x1a1+a1x2a1=
a1X1a1+a1X2a1+a1e
=a12.65a1+2.83a1a10.001a1572a1875a13a1=a15.474a1178a1435a18 Checks


S 
25 103a1 
16a11000

1-12 a 1     da1 a10.799 Ans.
a 1 in
nd 2.5
da
3
1
Tablea1A-17: da1=a8 1 7a 1 in Ans.
S 25 103a1
Factora1ofa1safet y: na1a1 a1 a13.29
Chaptera11a1Solutionsa1-
a1Rev.a1B,a1Pagea13/6

, Ans.
 16a11000
a1 a17a1
3

8


n

1-13a 1 a 1 Eq.a1(1-5): Ra1=a1a1Ria 1 =a10.98(0.96)0.94a1=a10.88
i1

Overalla1reliabilitya1=a188a1percent Ans.




Chaptera11a1Solutionsa1-
a1Rev.a1B,a1Pagea14/6