STUDY NOTES & REFERENCE GUIDE
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SOLUTION MANUAL
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STUDY NOTES & REFERENCE GUIDE
Chapter 1
Problems 1-1 through 1-6 are for student research. No standard solutions are provided.
1-7 From Fig. 1-2, cost of grinding to 0.0005 in is 270%. Cost of turning to 0.003 in is
60%.
Relative cost of grinding vs. turning = 270/60 = 4.5 times Ans.
1-8 CA = CB,
10 + 0.8 P = 60 + 0.8 P − 0.005 P 2
P 2 = 50/0.005 P = 100 parts Ans.
·
1-9 Max. load = 1.10 P
Min. area = (0.95)2A
Min. strength = 0.85 S
To offset the absolute uncertainties, the design factor, from Eq. (1-1) should be
1.10
n = = 1.43 Ans.
0.85(0.95)
2
d
1-10 (a) X1 + X2:
x1 + x2 = X1 + e1 + X 2 + e2
error = e = ( x1 + x2 ) − ( X1 + X 2 )
= e1 + e2 Ans.
(b) X1 − X2:
x1 − x2 = X1 + e1 − ( X 2 + e2 )
e = ( x1 − x2 ) − ( X1 − X2 ) = e1 − Ans.
(c) X1 X2: e2
x1x2 = ( X1 + e1 )( X 2 + e2 )
e = x1x2 − X1 X 2 = X1e2 + X 2e1 + e1e2
e e
Xe +X =X 1
+ 2 Ans.
e X
1 2 2 1 1 2
X1 X2
Shigley’s MED, 10th edition Chapter 1 Solutions,
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(d) X1/X2:
x X +e X 1+ e X ···
1
= 1 1
= 1
1 1
x2 X 2 + e2 X 2 1+ X 2
e2
−1
e 1+ e X
e e e e
e
·
1−
2
1+
2 then 1 1
1+ 1−
1 2
1+
1
− 2
X2 X2 1+ e2 X 2 X1 X2 X1 X 2
···
x X e e
X
Thus, e = 1 − 1 1 1 − 2 Ans.
x2 X 2 X 2 X 2
X1
1-11 (a) x1 = 7 = 2.645 751 311 1
X1 = 2.64 (3 correct digits)
x2 = 8 = 2.828 427 124 7
X2 = 2.82 (3 correct digits)
x1 + x2 = 5.474 178 435 8
e1 = x1 − X1 = 0.005 751 311 1
e2 = x2 − X2 = 0.008 427 124 7
e = e1 + e2 = 0.014 178 435 8
Sum = x1 + x2 = X1 + X2 + e
= 2.64 + 2.82 + 0.014 178 435 8 = 5.474 178 435 8 Checks
(b) X1 = 2.65, X2 = 2.83 (3 digit significant numbers)
e1 = x1 − X1 = − 0.004 248 688 9
e2 = x2 − X2 = − 0.001 572 875 3
e = e1 + e2 = − 0.005 821 564 2
Sum = x1 + x2 = X1 + X2 + e
= 2.65 +2.83 − 0.001 572 875 3 = 5.474 178 435 8 Checks
S 32 (
25 103 )
(1000)
1-12 = = d = 1.006 Ans.
in
nd d 2.5
3
Table A-17: d = 1 14 in Ans.
Factor of safety:
S
n= =
25 103
=
( ) Ans.
4.79
32 (1000)
(1.25)
3
Shigley’s MED, 10th edition Chapter 1 Solutions,
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1-13 (a)
x f fx f x2
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 1 104 600
k 8 480
Eq. (1-6) x = 1 fi xi = = 122
.9 kcycles
N i=1 69
Eq. (1-7)
1/ 2
fx − N
= 1104 600 − 69(122.9)
2
sx = = 30.3 kcycles Ans.
i =1 x 69 − 1
N −1
x− x115 − 115 − 122.9
(b) Eq. (1-5) z115 = = = −0.2607
x x 30.3
= s
ˆ
x x
Interpolating from Table (A-10)
0.2600 0.3974
0.2607 x x = 0.3971
0.2700 0.3936
N(−0.2607) = 69 (0.3971) = 27.4 27 Ans.
Shigley’s MED, 10th edition Chapter 1 Solutions,
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COURSEROOMSITY | Page 4 of 1092
All 20 Chapters Covered ·
SOLUTION MANUAL
COURSEROOMSITY | Page 1 of 1092
, www.konkur.in
STUDY NOTES & REFERENCE GUIDE
Chapter 1
Problems 1-1 through 1-6 are for student research. No standard solutions are provided.
1-7 From Fig. 1-2, cost of grinding to 0.0005 in is 270%. Cost of turning to 0.003 in is
60%.
Relative cost of grinding vs. turning = 270/60 = 4.5 times Ans.
1-8 CA = CB,
10 + 0.8 P = 60 + 0.8 P − 0.005 P 2
P 2 = 50/0.005 P = 100 parts Ans.
·
1-9 Max. load = 1.10 P
Min. area = (0.95)2A
Min. strength = 0.85 S
To offset the absolute uncertainties, the design factor, from Eq. (1-1) should be
1.10
n = = 1.43 Ans.
0.85(0.95)
2
d
1-10 (a) X1 + X2:
x1 + x2 = X1 + e1 + X 2 + e2
error = e = ( x1 + x2 ) − ( X1 + X 2 )
= e1 + e2 Ans.
(b) X1 − X2:
x1 − x2 = X1 + e1 − ( X 2 + e2 )
e = ( x1 − x2 ) − ( X1 − X2 ) = e1 − Ans.
(c) X1 X2: e2
x1x2 = ( X1 + e1 )( X 2 + e2 )
e = x1x2 − X1 X 2 = X1e2 + X 2e1 + e1e2
e e
Xe +X =X 1
+ 2 Ans.
e X
1 2 2 1 1 2
X1 X2
Shigley’s MED, 10th edition Chapter 1 Solutions,
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(d) X1/X2:
x X +e X 1+ e X ···
1
= 1 1
= 1
1 1
x2 X 2 + e2 X 2 1+ X 2
e2
−1
e 1+ e X
e e e e
e
·
1−
2
1+
2 then 1 1
1+ 1−
1 2
1+
1
− 2
X2 X2 1+ e2 X 2 X1 X2 X1 X 2
···
x X e e
X
Thus, e = 1 − 1 1 1 − 2 Ans.
x2 X 2 X 2 X 2
X1
1-11 (a) x1 = 7 = 2.645 751 311 1
X1 = 2.64 (3 correct digits)
x2 = 8 = 2.828 427 124 7
X2 = 2.82 (3 correct digits)
x1 + x2 = 5.474 178 435 8
e1 = x1 − X1 = 0.005 751 311 1
e2 = x2 − X2 = 0.008 427 124 7
e = e1 + e2 = 0.014 178 435 8
Sum = x1 + x2 = X1 + X2 + e
= 2.64 + 2.82 + 0.014 178 435 8 = 5.474 178 435 8 Checks
(b) X1 = 2.65, X2 = 2.83 (3 digit significant numbers)
e1 = x1 − X1 = − 0.004 248 688 9
e2 = x2 − X2 = − 0.001 572 875 3
e = e1 + e2 = − 0.005 821 564 2
Sum = x1 + x2 = X1 + X2 + e
= 2.65 +2.83 − 0.001 572 875 3 = 5.474 178 435 8 Checks
S 32 (
25 103 )
(1000)
1-12 = = d = 1.006 Ans.
in
nd d 2.5
3
Table A-17: d = 1 14 in Ans.
Factor of safety:
S
n= =
25 103
=
( ) Ans.
4.79
32 (1000)
(1.25)
3
Shigley’s MED, 10th edition Chapter 1 Solutions,
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STUDY NOTES & REFERENCE GUIDE
1-13 (a)
x f fx f x2
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 1 104 600
k 8 480
Eq. (1-6) x = 1 fi xi = = 122
.9 kcycles
N i=1 69
Eq. (1-7)
1/ 2
fx − N
= 1104 600 − 69(122.9)
2
sx = = 30.3 kcycles Ans.
i =1 x 69 − 1
N −1
x− x115 − 115 − 122.9
(b) Eq. (1-5) z115 = = = −0.2607
x x 30.3
= s
ˆ
x x
Interpolating from Table (A-10)
0.2600 0.3974
0.2607 x x = 0.3971
0.2700 0.3936
N(−0.2607) = 69 (0.3971) = 27.4 27 Ans.
Shigley’s MED, 10th edition Chapter 1 Solutions,
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