Solution Manual for Fundamentals of Aerodynamics 7th Edition by John D. Anderson PDF | Complete Step-by-Step Solutions and Detailed Explanations | Covers Airfoil and Wing Theory, Inviscid and Viscous Flow, Compressible Flow, Boundary Layers, Shock Waves,

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SOLUTION MANUAL

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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 Ans.
1.43
2
d 0.85 0.95


1-10 (a) X1 + X2:
x1 x2 X1 e1 X2 e2
error e x1 x2 X1 X2
e1 e2 Ans.
(b) X1
X2:
x1 x2 X1 e1 X2 e2
Ans.
(c) X1 X2: e x1 x2 X1 X2
e1 e2

x1x2 X1 e1 X2 e2


e x1x2 X1 X 2 X1e2 X 2e1 e1e2
e e
Xe X X 1 2 Ans.
e X

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1 2 2 1 1 2
 X1 X2




Shigley’s MED, 10th edition Chapter 1 Solutions, Page 1/12

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(d) X1 /X2 :
x X e X 1 e X
1 1 1 1 1 1

x2 X2 e2 X2 1 X2
e2
1
e e 1 e X e e e e

1 2
2
1 then 1 1
1 1
1 2
1 1 2


X2 X2 1 e2 X 2 X1 X2 X1 X2
x X X e e


Thus, e2 1 1 1 1 Ans.
x2 X2 X2 X2
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.
S 25 103
Factor of safety: n Ans.
4.79
32 1000

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3
1.25




Shigley’s MED, 10th edition Chapter 1 Solutions, Page 1/12

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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
fi xi   122
Eq. (1-6) x  1
N

i 1 69
.9 kcycles


Eq. (1-7)

fx  Nx
1/ 2
sx 1104 600 69(122.9)2 30.3 Ans.
kcycles i 1
69 1
N 1

x x115 115
(b) Eq. (1-5) z115 0.2607
x 122.9
x s 30.3
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.

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From the data, the number of instances less than 115 kcycles is




Shigley’s MED, 10th edition Chapter 1 Solutions, Page 1/12

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2 + 1 + 3 + 5 + 8 + 12 = 31 (the data is not perfectly normal)



1-14
x f fx f x2
174 6 1044 181656
182 9 1638 298116
190 44 8360 1588400
198 67 13266 2626668
206 53 10918 2249108
214 12 2568 549552
222 6 1332 295704
 197 39126 7789204


1
fx
k
x   39 126  198.61 kpsi
Eq. (1-6) i i
N i 1 197

k

Eq. (1-7) skpsi
x
fx
i1
i i
2
 N x2 7 789 204 197(198.61)2
1 2
9.68 Ans.
197 1
N 1


1-15 L 122.9 kcycles sL 30.3 kcycles
and

Eq. (1-5) x x x10 L x10 122.9
z10
ˆ sL 30.3

Thus, x10 = 122.9 + 30.3 z10 = L10

From Table A-10, for 10 percent failure, z10 = 1.282. Thus,

L10 = 122.9 + 30.3( 1.282) = 84.1 kcycles Ans.

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1-16
x f fx fx2
93 19 1767 164331
95 25 2375 225625
97 38 3686 357542
99 17 1683 166617
101 12 1212 122412
103 10 1030 106090
105 5 525 55125
107 4 428 45796
109 4 436 47524
111 2 222 24642
136 13364 1315704

k

Eq. (1-6) x fi 13 364 / 98.26471 = 98.26 kpsi
1 136
xi
N i 1




k 1 315 704 136(98.26471)2
Eq. (1-7)  fi xi2  N x 2 1 2 4.30 kpsi
sx i1

N 1 136 1

Note, for accuracy in the calculation given above, x needs to be of more significant
figures than the rounded value.

For a normal distribution, from Eq. (1-5), and a yield strength exceeded by 99 percent
(R = 0.99, pf = 0.01),
x x0.01 x0.01 98.26
z0.01 4.30
x x
ˆ s
x x
Solving for the yield strength gives

x0.01 = 98.26 + 4.30 z0.01

From Table A-10, z0.01 = 2.326. Thus


x0.01 = 98.26 + 4.30( 2.326) = 88.3 kpsi Ans.
n

1-17 Eq. (1-9): R= Ri = 0.98(0.96)0.94 = 0.88

Shigley’s MED, 10th edition Chapter 1 Solutions, Page 1/12

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i 1
Overall reliability = 88 percent Ans.