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Chapter 1
D
B G F Facc
A
E
f f
1 1
C^ cr^
Impending motion to left
F cr
Consider force F at G , reactions at B and D. Extend lines of action for fully-developed fric- tion D E and B E to find the point of concurrency at E for impending motion to the left. The critical angle is θcr. Resolve force F into components Facc and F cr. Facc is related to mass and acceleration. Pin accelerates to left for any angle 0 < θ < θcr. When θ > θcr , no magnitude of F will move the pin.
D
B G F F acc
A
E E
f f
1 1
C
d
Impending motion to right
F cr cr
Consider force F ′^ at G , reactions at A and C. Extend lines of action for fully-developed fric- tion AE ′^ and C E ′^ to find the point of concurrency at E ′^ for impending motion to the left. The critical angle is θ cr′. Resolve force F ′^ into components F acc ′ and F cr′. F acc ′ is related to mass and acceleration. Pin accelerates to right for any angle 0 < θ′^ < θ cr′. When θ′^ > θ cr′ , no mag- nitude of F ′^ will move the pin. The intent of the question is to get the student to draw and understand the free body in order to recognize what it teaches. The graphic approach accomplishes this quickly. It is im- portant to point out that this understanding enables a mathematical model to be constructed, and that there are two of them. This is the simplest problem in mechanical engineering. Using it is a good way to begin a course. What is the role of pin diameter d? Yes, changing the sense of F changes the response.
Problems 1-1 through 1-4 are for student research.
1-
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(a)
Fy = − F − f N cos θ + N sin θ = 0 (1) ∑ Fx = f N sin θ + N cos θ −
T
r
F = N (sin θ − f cos θ) Ans. T = Nr ( f sin θ + cos θ)
Combining
T = Fr
1 + f tan θ tan θ − f
= KFr Ans. (2)
(b) If T → ∞ detent self-locking tan θ − f = 0 ∴ θcr = tan−^1 f Ans. (Friction is fully developed.)
Check: If F = 10 lbf, f = 0.20, θ = 45 ◦, r = 2 in
N =
− 0 .20 cos 45◦^ + sin 45◦^
= 17 .68 lbf
T
r
= 17 .28(0.20 sin 45◦^ + cos 45◦) = 15 lbf
f N = 0 .20(17.28) = 3 .54 lbf
θcr = tan−^1 f = tan−^1 (0.20) = 11. 31 ◦
11.31° < θ < 90°
(a) F = F 0 + k (0) = F 0 T 1 = F 0 r Ans.
(b) When teeth are about to clear F = F 0 + kx 2
From Prob. 1-
T 2 = Fr
f tan θ + 1 tan θ − f
T 2 = r
( F 0 + kx 2 )( f tan θ + 1) tan θ − f
Ans.
Given, F = 10 + 2. 5 x lbf, r = 2 in, h = 0 .2 in, θ = 60 ◦, f = 0 .25, x (^) i = 0, x (^) f = 0. 2
Fi = 10 lbf; F (^) f = 10 + 2 .5(0.2) = 10 .5 lbf Ans.
x
y
F
f N
N
Tr
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(c) % increase in speed
Modest change in optimal speed Ans.
1-10 This and the following problem may be the student’s first experience with a figure of merit.
- Formulate fom to reflect larger figure of merit for larger merit.
- Use a maximization optimization algorithm. When one gets into computer implementa- tion and answers are not known, minimizing instead of maximizing is the largest error one can make. ∑ FV = F 1 sin θ − W = 0 ∑ F (^) H = − F 1 cos θ − F 2 = 0 From which F 1 = W /sin θ F 2 = − W cos θ/sin θ fom = − S = −¢γ (volume) = −^. ¢γ( l 1 A 1 +^ l 2 A 2 )
A 1 =
F 1
S
W
S sin θ
, l 2 =
l 1 cos θ
A 2 =
F 2
S
∣ =^
W cos θ S sin θ
fom = −¢γ
l 2 cos θ
W
S sin θ
l 2 W cos θ S sin θ
−¢γ W l 2 S
1 + cos^2 θ cos θ sin θ
Set leading constant to unity
θ◦^ fom
0 −∞ 20 −5. 30 −4. 40 −3. 45 −3. 50 −2. 54.736 −2. 60 −2.
Check second derivative to see if a maximum, minimum, or point of inflection has been found. Or, evaluate fom on either side of θ*.
θ* = 54. 736 ◦^ Ans. fom* = − 2. 828 Alternative: d d θ
1 + cos^2 θ cos θ sin θ
And solve resulting tran- scendental for θ*.
Chapter 1 5
(a) x 1 + x 2 = X 1 + e 1 + X 2 + e 2 error = e = ( x 1 + x 2 ) − ( X 1 + X 2 ) = e 1 + e 2 Ans. (b) x 1 − x 2 = X 1 + e 1 − ( X 2 + e 2 ) e = ( x 1 − x 2 ) − ( X 1 − X 2 ) = e 1 − e 2 Ans. (c) x 1 x 2 = ( X 1 + e 1 )( X 2 + e 2 ) e = x 1 x 2 − X 1 X 2 = X 1 e 2 + X 2 e 1 + e 1 e 2 . = X 1 e 2 + X 2 e 1 = X 1 X 2
e 1 X 1
e 2 X 2
Ans.
(d)
x 1 x 2
X 1 + e 1 X 2 + e 2
X 1
X 2
1 + e 1 / X 1 1 + e 2 / X 2
e 2 X 2
e 2 X 2
and
e 1 X 1
e 2 X 2
e 1 X 1
e 2 X 2
e =
x 1 x 2
X 1
X 2
X 1
X 2
e 1 X 1
e 2 X 2
Ans.
(a) x 1 =
X 1 = 2. 23 3-correct digits x 2 =
X 2 = 2. 44 3-correct digits
x 1 + x 2 =
e 1 = x 1 − X 1 =
e 2 = x 2 − X 2 =
e = e 1 + e 2 =
Sum = x 1 + x 2 = X 1 + X 2 + e = 2. 23 + 2. 44 + 0 .015 557 720 28 = 4.685 557 720 28 (Checks) Ans.
(b) X 1 = 2 .24, X 2 = 2. 45
e 1 =
e 2 =
e = e 1 + e 2 = − 0 .004 442 279 72 Sum = X 1 + X 2 + e = 2. 24 + 2. 45 + (− 0 .004 442 279 72) = 4 .685 557 720 28 Ans.
Chapter 1 7
(a) τ =
120(10^3 )
(π/4)(20^2 )
= 382 MPa
(b) σ =
32(800)(800)(10−^3 )
π(32) 3 (10−^3 ) 3
= 198 .9(10^6 ) N/m^2 = 198 .9 MPa
(c) Z =
π 32(36)
(36^4 − 264 ) = 3334 mm^3
(d) k =
(1.6) 4 (79.3)(10−^3 ) 4 (10^9 )
8(19.2) 3 (32)(10−^3 ) 3
= 286 .8 N/m
(b) f /( N x ) = f /(69 · 10) = f / 690
Eq. (2-9) x ¯ =
= 122 .9 kcycles
Eq. (2-10) s (^) x =
[
] 1 / 2
= 30.3 kcycles Ans.
x f f x f x^2 f /( N x )
60 2 120 7200 0. 70 1 70 4900 0. 80 3 240 19 200 0. 90 5 450 40 500 0. 100 8 800 80 000 0. 110 12 1320 145 200 0. 120 6 720 86 400 0. 130 10 1300 169 000 0. 140 8 1120 156 800 0. 150 5 750 112 500 0. 160 2 320 51 200 0. 170 3 510 86 700 0. 180 2 360 64 800 0. 190 1 190 36 100 0. 200 0 0 0 0 210 1 210 44 100 0. 69 8480 1 104 600
Chapter 2
(a)
(^0 60 708090100110120130140150160170180) 190 200 210
2
4
6
8
10
12
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2-4 (a)
y f f y f y^2 y f /( N w) f ( y ) g ( y )
5.625 1 5.625 31.640 63 5.625 0.072 727 0.001 262 0.000 295 5.875 0 0 0 5.875 0 0.008 586 0.004 088 6.125 0 0 0 6.125 0 0.042 038 0.031 194 6.375 3 19.125 121.9219 6.375 0.218 182 0.148 106 0.140 262 6.625 3 19.875 131.6719 6.625 0.218 182 0.375 493 0.393 667 6.875 6 41.25 283.5938 6.875 0.436 364 0.685 057 0.725 002 7.125 14 99.75 710.7188 7.125 1.018 182 0.899 389 0.915 128 7.375 15 110.625 815.8594 7.375 1.090 909 0.849 697 0.822 462 7.625 10 76.25 581.4063 7.625 0.727 273 0.577 665 0.544 251 7.875 2 15.75 124.0313 7.875 0.145 455 0.282 608 0.273 138 8.125 1 8.125 66.015 63 8.125 0.072 727 0.099 492 0.106 72
55 396.375 2866.
For a normal distribution,
y ¯ = 396. 375 / 55 = 7 .207, s (^) y =
f ( y ) =
2 π
exp
[
x − 7. 207
- 4358
) 2 ]
For a lognormal distribution,
x ¯ = ln 7.206 818 − ln
1 + 0 .060 474^2 = 1 .9732, s (^) x = ln
1 + 0 .060 474^2 = 0. 0604
g ( y ) =
x (0.0604)(
2 π)
exp
[
ln x − 1. 9732
- 0604
) 2 ]
(b) Histogram
0
1
5.63 5.88 6.13 6.38 6.63 6. log N
7.13 7.38 7.63 7.88 8.
Data N LN
f
Chapter 2 11
2-5 Distribution is uniform in interval 0.5000 to 0.5008 in, range numbers are a = 0.5000, b = 0.5008 in.
(a) Eq. (2-22) μ x =
a + b 2
Eq. (2-23) σ x =
b − a 2
(b) PDF from Eq. (2-20)
f ( x ) =
1250 0. 5000 ≤ x ≤ 0 .5008 in 0 otherwise
(c) CDF from Eq. (2-21)
F ( x ) =
0 x < 0. 5000 ( x − 0 .5)/ 0. 0008 0. 5000 ≤ x ≤ 0. 5008 1 x > 0. 5008
If all smaller diameters are removed by inspection, a = 0.5002, b = 0.
μ x =
= 0 .5005 in
σ ˆ x =
= 0 .000 173 in
f ( x ) =
- 7 0. 5002 ≤ x ≤ 0. 5008 0 otherwise
F ( x ) =
0 x < 0. 5002
- 7 ( x − 0. 5002 ) 0. 5002 ≤ x ≤ 0. 5008 1 x > 0. 5008
2-6 Dimensions produced are due to tool dulling and wear. When parts are mixed, the distribution is uniform. From Eqs. (2-22) and (2-23),
a = μ x −
3 s = 0. 6241 −
3(0.000 581) = 0 .6231 in
b = μ x +
3 s = 0. 6241 +
3(0.000 581) = 0 .6251 in
We suspect the dimension was
in Ans.
Chapter 2 13
2-8 Cramer’s rule
a 1 =
y x^2 x y x^3
x x^2 x^2 x^3
y x^3 − x y x^2 x x^3 − ( x^2 ) 2
Ans.
a 2 =
x y x^2 x y
x x^2 x^2 x^3
x x y − y x^2 x x^3 − ( x^2 ) 2
Ans.
0.
0
0 0.2 0.4 0.6 0.8 1
Data Regression
x
y
x y x^2 x^3 xy
0 0.01 0 0 0 0.2 0.15 0.04 0.008 0. 0.4 0.25 0.16 0.064 0. 0.6 0.25 0.36 0.216 0. 0.8 0.17 0.64 0.512 0. 1.0 −0.01 1.00 1.000 −0. 3.0 0.82 2.20 1.800 0.
a 1 = 1 .040 714 a 2 = − 1 .046 43 Ans.
Data Regression x y y
0 0.01 0 0.2 0.15 0.166 286 0.4 0.25 0.248 857 0.6 0.25 0.247 714 0.8 0.17 0.162 857 1.0 −0.01 −0.005 71
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0
20
40
60
80
100
120
140
0 100 200 S^ u
Se
300 400
Data Regression
Data Regression Su S ′ e S e ′ S u^2 Su S e ′
0 20.356 75 60 30 39.080 78 3 600 1 800 64 48 40.329 05 4 096 3 072 65 29.5 40.641 12 4 225 1 917. 82 45 45.946 26 6 724 3 690 101 51 51.875 54 10 201 5 151 119 50 57.492 75 14 161 5 950 120 48 57.804 81 14 400 5 760 130 67 60.925 48 16 900 8 710 134 60 62.173 75 17 956 8 040 145 64 65.606 49 21 025 9 280 180 84 76.528 84 32 400 15 120 195 78 81.209 85 38 025 15 210 205 96 84.330 52 42 025 19 680 207 87 84.954 66 42 849 18 009 210 87 85.890 86 44 100 18 270 213 75 86.827 06 45 369 15 975 225 99 90.571 87 50 625 22 275 225 87 90.571 87 50 625 19 575 227 116 91.196 51 529 26 332 230 105 92.132 2 52 900 24 150 238 109 94.628 74 56 644 25 942 242 106 95.877 01 58 564 25 652 265 105 103.054 6 70 225 27 825 280 96 107.735 6 78 400 26 880 295 99 112.416 6 87 025 29 205 325 114 121.778 6 105 625 37 050 325 117 121.778 6 105 625 38 025 355 122 131.140 6 126 025 43 310 5462 2274.5 1 251 868 501 855.
m = 0.312 067 b = 20.356 75 Ans.
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Data Regression x y y x^2 y^2 xy x − ¯ x ( x − ¯ x ) 2
0.2 7.1 7.931 803 0.04 50.41 1.42 −0.633 333 0.401 111 111 0.4 10.3 9.884 918 0.16 106.09 4.12 −0.433 333 0.187 777 778 0.6 12.1 11.838 032 0.36 146.41 7.26 −0.233 333 0.054 444 444 0.8 13.8 13.791 147 0.64 190.44 11.04 −0.033 333 0.001 111 111 1 16.2 15.744 262 1.00 262.44 16.20 0.166 666 0.027 777 778 2 25.2 25.509 836 4.00 635.04 50.40 1.166 666 1.361 111 111 5 84.7 6.2 1390.83 90.44 0 2.033 333 333
m ˆ = k =
b^ ˆ = Fi = 84.^7 −^9 .7656(5) 6
(a) x ¯^ =^
; y ¯ =
Eq. (2-37)
s (^) yx =
Eq. (2-36)
s (^) b ˆ = 0. 556
= 0 .3964 lbf
Fi = (5.9787, 0.3964) lbf (^) Ans.
F
0 x
5
10
15
20
25
30
0 0.5 1 1.5 2 2.
Data Regression
Chapter 2 17
(b) Eq. (2-35)
s (^) m ˆ =
= 0 .3899 lbf/in
k = (9.7656, 0.3899) lbf/in Ans.
2-12 The expression = δ/ l is of the form x / y. Now δ = ( 0. 0015 , 0 .000 092) in, unspecified distribution; l = (2.000, 0.0081) in, unspecified distribution;
C (^) x = 0 .000 092/ 0. 0015 = 0. 0613
C (^) y = 0. 0081 / 2. 000 = 0 .000 75
From Table 2-6, ¯ = 0. 0015 / 2. 000 = 0 .000 75
σ ˆ = 0 .000 75
[
0. 06132 + 0 .004 05^2
1 + 0 .004 05^2
] 1 / 2
= 4. 607 ( 10 −^5 ) = 0 .000 046
We can predict ¯ and σˆ but not the distribution of .
2-13 σ = E = (0.0005, 0.000 034) distribution unspecified; E = (29.5, 0.885) Mpsi, distribution unspecified; C (^) x = 0 .000 034/ 0. 0005 = 0. 068 , C (^) y = 0. 0885 / 29. 5 = 0. 030 σ is of the form x , y
Table 2-
σ ¯ = ¯ E ¯ = 0. 0005 ( 29. 5 ) 106 = 14 750 psi
σ ˆσ = 14 750(0. 0682 + 0. 0302 + 0. 0682 + 0. 0302 ) 1 /^2
= 1096 .7 psi
C σ = 1096. 7 /14 750 = 0 .074 35
δ =
Fl AE
F = (14.7, 1.3) kip, A = (0.226, 0.003) in^2 , l = (1.5, 0.004) in, E = (29.5, 0.885) Mpsi dis- tributions unspecified.
C (^) F = 1. 3 / 14. 7 = 0 .0884 ; C (^) A = 0. 003 / 0. 226 = 0 .0133 ; C (^) l = 0. 004 / 1. 5 = 0 .00267 ;
C (^) E = 0. 885 / 29. 5 = 0. 03
Mean of δ:
δ =
Fl AE
= Fl
A
E
Chapter 2 19
Fraction discarded is α + β. The area under the PDF was unity. Having discarded α + β fraction, the ordinates to the truncated PDF are multiplied by a.
a =
1 − (α + β) New PDF,^ g ( x ) , is given by
g ( x ) =
f ( x )/[1 − (α + β)] x 1 ≤ x ≤ x 2 0 otherwise
More formal proof: g ( x ) has the property
∫ (^) x 2
x 1
g ( x ) dx = a
∫ (^) x 2
x 1
f ( x ) dx
1 = a
[∫ ∞
−∞
f ( x ) dx −
∫ (^) x 1
0
f ( x ) dx −
x 2
f ( x ) dx
]
1 = a { 1 − F ( x 1 ) − (^) [1 − F ( x 2 )]}
a =
F ( x 2 ) − F ( x 1 )
(1 − β) − α
1 − (α + β)
(a) d = U [0.748, 0.751]
μ d =
= 0 .7495 in
σ ˆ d =
= 0 .000 866 in
f ( x ) =
b − a
= 333 .3 in−^1
F ( x ) =
x − 0. 748
- 751 − 0. 748
= 333 .3( x − 0 .748)
x 1
f ( x )
x 2^ x
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(b) F ( x 1 ) = F (0.748) = 0
F ( x 2 ) = (0. 750 − 0 .748)333. 3 = 0. 6667
If g ( x ) is truncated, PDF becomes
g ( x ) =
f ( x ) F ( x 2 ) − F ( x 1 )
= 500 in−^1
μ x =
a ′^ + b ′ 2
= 0 .749 in
σ ˆ x =
b ′^ − a ′ 2
= 0 .000 577 in
2-18 From Table A-10, 8.1% corresponds to z 1 = −^1 .4 and 5.5% corresponds to z 2 = +^1 .6.
k 1 = μ + z 1 σˆ
k 2 = μ + z 2 σˆ From which
μ =
z 2 k 1 − z 1 k 2 z 2 − z 1
σ ˆ =
k 2 − k 1 z 2 − z 1
The original density function is
f ( k ) =
2 π
exp
[
k − 9. 933
- 6667
) 2 ]
Ans.
2-19 From Prob. 2-1, μ^ =^122 .9 kcycles and σˆ^ =^30 .3 kcycles.
z 10 =
x 10 − μ σ ˆ
x 10 − 122. 9
- 3
x 10 = 122. 9 + 30. 3 z 10
From Table A-10, for 10 percent failure, z 10 = −^1.^282
x 10 = 122. 9 + 30. 3 (− 1. 282 )
= 84 .1 kcycles Ans.
g ( x ) 500
x
f ( x ) 333.
0.749 0.750 0.
