instructor's solutions manual...(d) x1/x2: 1111 11 2222 22 1 22 11 12 1 22 22 12 1 11 11 2 22 2 1 2...
TRANSCRIPT
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Typewritten TextINSTRUCTOR'S SOLUTIONS 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
2
1.10 1.43 .0.85 0.95d
n A ns
______________________________________________________________________________ 1-10 (a) X1 + X2:
1 2 1 1 2 2
1 2 1 2
1 2
error .
x x X e X ee x x X Xe e Ans
(b) X1 X2:
1 2 1 1 2 2
1 2 1 2 1 2 .
x x X e X e
e x x X X e e Ans
(c) X1 X2:
1 2 1 1 2 2
1 2 1 2 1 2 2 1 1 2
1 21 2 2 1 1 2
1 2
.
x x X e X ee x x X X X e X e e e
e eX e X e X X AnsX X
Chapter 1 Solutions - Rev. B, Page 1/6
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(d) X1/X2:
1 1 1 1 1 1
2 2 2 2 2 2
1
2 2 1 1 1 2 1
2 2 2 2 1 2 1
1 1 1 1 2
2 2 2 1 2
11
11 1 then 1 1 11
Thus, .
x X e X e Xx X e X e X
e e e X e e e 22
eX X e X X X X
x X X e ee Ansx X X X X
X
______________________________________________________________________________ 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 ______________________________________________________________________________
1-12 3
3
25 1016 10000.799 in .
2.5d
S d An d
ns
Table A-17: d = 78
in Ans.
Factor of safety:
3
378
25 103.29 .
16 1000Sn A
ns
______________________________________________________________________________
1-13 Eq. (1-5): R =1
n
ii
R = 0.98(0.96)0.94 = 0.88
Overall reliability = 88 percent Ans. ______________________________________________________________________________
Chapter 1 Solutions - Rev. B, Page 2/6
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1-14 a = 1.500 0.001 in b = 2.000 0.003 in c = 3.000 0.004 in d = 6.520 0.010 in (a) d a b c w = 6.520 1.5 2 3 = 0.020 in = 0.001 + 0.003 + 0.004 +0.010 = 0.018 allt w t w = 0.020 0.018 in Ans. (b) From part (a), wmin = 0.002 in. Thus, must add 0.008 in to d . Therefore, d = 6.520 + 0.008 = 6.528 in Ans. ______________________________________________________________________________ 1-15 V = xyz, and x = a a, y = b b, z = c c, V abc
V a a b b c cabc bc a ac b ab c a b c b c a c a b a b c
The higher order terms in are negligible. Thus, V bc a ac b ab c
and, .V bc a ac b ab c a b c a b c AnsV abc a b c a b c
For the numerical values given, 31.500 1.875 3.000 8.4375 inV
30.002 0.003 0.004 0.00427 0.00427 8.4375 0.036 in1.500 1.875 3.000
V VV
V = 8.438 0.036 in3 Ans. ______________________________________________________________________________
Chapter 1 Solutions - Rev. B, Page 3/6
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1-16 wmax = 0.05 in, wmin = 0.004 in
0.05 0.004 0.027 in2
w =
Thus, w = 0.05 0.027 = 0.023 in, and then, w = 0.027 0.023 in.
0.027 0.042 1.51.569 in
a b ca
a
w =
tw = 0.023 = tallt a + 0.002 + 0.005 ta = 0.016 in Thus, a = 1.569 0.016 in Ans. ______________________________________________________________________________ 1-17 2 3.734 2 0.139 4.012 ino iD D d all 0.028 2 0.004 0.036 inoDt t Do = 4.012 0.036 in Ans. ______________________________________________________________________________ 1-18 From O-Rings, Inc. (oringsusa.com), Di = 9.19 0.13 mm, d = 2.62 0.08 mm 2 9.19 2 2.62 14.43 mmo iD D d all 0.13 2 0.08 0.29 mmoDt t Do = 14.43 0.29 mm Ans. ______________________________________________________________________________ 1-19 From O-Rings, Inc. (oringsusa.com), Di = 34.52 0.30 mm, d = 3.53 0.10 mm 2 34.52 2 3.53 41.58 mmo iD D d all 0.30 2 0.10 0.50 mmoDt t Do = 41.58 0.50 mm Ans. ______________________________________________________________________________
Chapter 1 Solutions - Rev. B, Page 4/6
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1-20 From O-Rings, Inc. (oringsusa.com), Di = 5.237 0.035 in, d = 0.103 0.003 in 2 5.237 2 0.103 5.443 ino iD D d all 0.035 2 0.003 0.041 inoDt t Do = 5.443 0.041 in Ans. ______________________________________________________________________________ 1-21 From O-Rings, Inc. (oringsusa.com), Di = 1.100 0.012 in, d = 0.210 0.005 in 2 1.100 2 0.210 1.520 ino iD D d all 0.012 2 0.005 0.022 inoDt t Do = 1.520 0.022 in Ans. ______________________________________________________________________________ 1-22 From Table A-2, (a) = 150/6.89 = 21.8 kpsi Ans. (b) F = 2 /4.45 = 0.449 kip = 449 lbf Ans. (c) M = 150/0.113 = 1330 lbf in = 1.33 kip in Ans. (d) A = 1500/ 25.42 = 2.33 in2 Ans. (e) I = 750/2.544 = 18.0 in4 Ans. (f) E = 145/6.89 = 21.0 Mpsi Ans. (g) v = 75/1.61 = 46.6 mi/h Ans. (h) V = 1000/946 = 1.06 qt Ans. ______________________________________________________________________________ 1-23 From Table A-2,
(a) l = 5(0.305) = 1.53 m Ans.
(b) = 90(6.89) = 620 MPa Ans.
(c) p = 25(6.89) = 172 kPa Ans.
Chapter 1 Solutions - Rev. B, Page 5/6
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Chapter 1 Solutions - Rev. B, Page 6/6
(d) Z =12(16.4) = 197 cm3 Ans. (e) w = 0.208(175) = 36.4 N/m Ans. (f) = 0.001 89(25.4) = 0.0480 mm Ans. (g) v = 1200(0.0051) = 6.12 m/s Ans. (h) = 0.002 15(1) = 0.002 15 mm/mm Ans. (i) V = 1830(25.43) = 30.0 (106) mm3 Ans.
______________________________________________________________________________ 1-24 (a) = M /Z = 1770/0.934 = 1895 psi = 1.90 kpsi Ans. (b) = F /A = 9440/23.8 = 397 psi Ans. (c) y =Fl3/3EI = 270(31.5)3/[3(30)106(0.154)] = 0.609 in Ans. (d) = Tl /GJ = 9740(9.85)/[11.3(106)( /32)1.004] = 8.648(102) rad = 4.95 Ans. ______________________________________________________________________________ 1-25 (a) =F / wt = 1000/[25(5)] = 8 MPa Ans. (b) I = bh3 /12 = 10(25)3/12 = 13.0(103) mm4 Ans. (c) I = d4/64 = (25.4)4/64 = 20.4(103) mm4 Ans. (d) =16T / d 3 = 16(25)103/[ (12.7)3] = 62.2 MPa Ans. ______________________________________________________________________________ 1-26 (a) =F /A = 2 700/[ (0.750)2/4] = 6110 psi = 6.11 kpsi Ans. (b) = 32Fa/ d 3 = 32(180)31.5/[ (1.25)3] = 29 570 psi = 29.6 kpsi Ans. (c) Z = (do4 di4)/(32 do) = (1.504 1.004)/[32(1.50)] = 0.266 in3 Ans. (d) k = (d 4G)/(8D 3 N) = 0.06254(11.3)106/[8(0.760)3 32] = 1.53 lbf/in Ans. ______________________________________________________________________________
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Chapter 2 2-1 From Tables A-20, A-21, A-22, and A-24c, (a) UNS G10200 HR: Sut = 380 (55) MPa (kpsi), Syt = 210 (30) Mpa (kpsi) Ans. (b) SAE 1050 CD: Sut = 690 (100) MPa (kpsi), Syt = 580 (84) Mpa (kpsi) Ans. (c) AISI 1141 Q&T at 540C (1000F): Sut = 896 (130) MPa (kpsi), Syt = 765 (111) Mpa (kpsi) Ans. (d) 2024-T4: Sut = 446 (64.8) MPa (kpsi), Syt = 296 (43.0) Mpa (kpsi) Ans. (e) Ti-6Al-4V annealed: Sut = 900 (130) MPa (kpsi), Syt = 830 (120) Mpa (kpsi) Ans. ______________________________________________________________________________ 2-2 (a) Maximize yield strength: Q&T at 425C (800F) Ans. (b)Maximize elongation: Q&T at 650C (1200F) Ans. ______________________________________________________________________________ 2-3 Conversion of kN/m3 to kg/ m3 multiply by 1(103) / 9.81 = 102 AISI 1018 CD steel: Tables A-20 and A-5
3370 1047.4 kN m/kg .
76.5 102yS Ans
2011-T6 aluminum: Tables A-22 and A-5
3169 1062.3 kN m/kg .
26.6 102yS Ans
Ti-6Al-4V titanium: Tables A-24c and A-5
3830 10187 kN m/kg .
43.4 102yS Ans
ASTM No. 40 cast iron: Tables A-24a and A-5.Does not have a yield strength. Using the ultimate strength in tension
342.5 6.89 1040.7 kN m/kg
70.6 102utS Ans
______________________________________________________________________________ 2-4 AISI 1018 CD steel: Table A-5
66
30.0 10106 10 in .
0.282E Ans
2011-T6 aluminum: Table A-5
66
10.4 10106 10 in .
0.098E Ans
Ti-6Al-6V titanium: Table A-5
Chapter 2 - Rev. D, Page 1/19
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66
16.5 10103 10 in .
0.160E Ans
No. 40 cast iron: Table A-5
66
14.5 1055.8 10 in .
0.260E Ans
______________________________________________________________________________ 2-5
22 (1 )2
E GG v E vG
From Table A-5
Steel:
30.0 2 11.50.304 .
2 11.5v A
ns
Aluminum: 10.4 2 3.90
0.333 .2 3.90
v A
ns
Beryllium copper:
18.0 2 7.00.286 .
2 7.0v A
ns
Gray cast iron:
14.5 2 6.00.208 .
2 6.0v A
ns
______________________________________________________________________________ 2-6 (a) A0 = (0.503)2/4, = Pi / A0 For data in elastic range, = l / l0 = l / 2
For data in plastic range, 0 00 0 0
1 1l l Al ll l l A
On the next two pages, the data and plots are presented. Figure (a) shows the linear part of the curve from data points 1-7. Figure (b) shows data points 1-12. Figure (c) shows the complete range. Note: The exact value of A0 is used without rounding off.
(b) From Fig. (a) the slope of the line from a linear regression is E = 30.5 Mpsi Ans. From Fig. (b) the equation for the dotted offset line is found to be = 30.5(106) 61 000 (1) The equation for the line between data points 8 and 9 is = 7.60(105) + 42 900 (2)
Chapter 2 - Rev. D, Page 2/19
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Solving Eqs. (1) and (2) simultaneously yields = 45.6 kpsi which is the 0.2 percent offset yield strength. Thus, Sy = 45.6 kpsi Ans.
The ultimate strength from Figure (c) is Su = 85.6 kpsi Ans. The reduction in area is given by Eq. (2-12) is
00
0.1987 0.1077100 100 45.8 % .0.1987
fA AR AnsA
Data Point Pi l, Ai
1 0 0 0 0 2 1000 0.0004 0.00020 5032 3 2000 0.0006 0.00030 10065 4 3000 0.001 0.00050 15097 5 4000 0.0013 0.00065 20130 6 7000 0.0023 0.00115 35227 7 8400 0.0028 0.00140 42272 8 8800 0.0036 0.00180 44285 9 9200 0.0089 0.00445 46298
10 8800 0.1984 0.00158 44285 11 9200 0.1978 0.00461 46298 12 9100 0.1963 0.01229 45795 13 13200 0.1924 0.03281 66428 14 15200 0.1875 0.05980 76492 15 17000 0.1563 0.27136 85551 16 16400 0.1307 0.52037 82531 17 14800 0.1077 0.84506 74479
(a) Linear range
Chapter 2 - Rev. D, Page 3/19
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(b) Offset yield
(c) Complete range (c) The material is ductile since there is a large amount of deformation beyond yield. (d) The closest material to the values of Sy, Sut, and R is SAE 1045 HR with Sy = 45 kpsi,
Sut = 82 kpsi, and R = 40 %. Ans. ______________________________________________________________________________ 2-7 To plot true vs., the following equations are applied to the data.
truePA
Eq. (2-4)
Chapter 2 - Rev. D, Page 4/19
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00
ln for 0 0.0028 in
ln for 0.0028 in
l llA lA
where 2
20
(0.503) 0.1987 in4
A
The results are summarized in the table below and plotted on the next page. The last 5 points of data are used to plot log vs log
The curve fit gives m = 0.2306
log 0 = 5.1852 0 = 153.2 kpsi Ans. For 20% cold work, Eq. (2-14) and Eq. (2-17) give,
A = A0 (1 – W) = 0.1987 (1 – 0.2) = 0.1590 in2
0
0.23060
0.1987ln ln 0.22310.1590
Eq. (2-18): 153.2(0.2231) 108.4 kpsi .
Eq. (2-19), with 85.6 from Prob. 2-6, 85.6 107 kpsi .
1 1 0.2
my
u
uu
AA
S AS
SS AnsW
ns
P L A true log log true 0 0 0.198 713 0 0
1000 0.0004 0.198 713 0.000 2 5032.388 -3.699 01 3.701 7742000 0.0006 0.198 713 0.000 3 10 064.78 -3.522 94 4.002 8043000 0.001 0.198 713 0.000 5 15 097.17 -3.301 14 4.178 8954000 0.0013 0.198 713 0.000 65 20 129.55 -3.187 23 4.303 8347000 0.0023 0.198 713 0.001 149 35 226.72 -2.939 55 4.546 8728400 0.0028 0.198 713 0.001 399 42 272.06 -2.854 18 4.626 0538800 0.0036 0.198 4 0.001 575 44 354.84 -2.802 61 4.646 9419200 0.0089 0.197 8 0.004 604 46 511.63 -2.336 85 4.667 5629100 0.196 3 0.012 216 46 357.62 -1.913 05 4.666 121
13200 0.192 4 0.032 284 68 607.07 -1.491 01 4.836 36915200 0.187 5 0.058 082 81 066.67 -1.235 96 4.908 84217000 0.156 3 0.240 083 108 765.20 -0.619 64 5.036 49 16400 0.130 7 0.418 956 125 478.20 -0.377 83 5.098 56814800 0.107 7 0.612 511 137 418.80 -0.212 89 5.138 046
Chapter 2 - Rev. D, Page 5/19
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______________________________________________________________________________ 2-8 Tangent modulus at = 0 is
63
5000 0 25 10 psi0.2 10 0
E
Ans.
At = 20 kpsi
Chapter 2 - Rev. D, Page 6/19
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36
20 3
26 19 1014.0 10 psi
1.5 1 10E
Ans.
(10-3) (kpsi)
0 0 0.20 5 0.44 10 0.80 16 1.0 19 1.5 26 2.0 32 2.8 40 3.4 46 4.0 49 5.0 54
______________________________________________________________________________ 2-9 W = 0.20, (a) Before cold working: Annealed AISI 1018 steel. Table A-22, Sy = 32 kpsi, Su = 49.5
kpsi, 0 = 90.0 kpsi, m = 0.25, f = 1.05 After cold working: Eq. (2-16), u = m = 0.25
Eq. (2-14), 0 1 1 1.251 1 0.20i
AA W
Eq. (2-17), 0ln ln1.25 0.223i ui
AA
Eq. (2-18), S 93% increase Ans. 0.250 90 0.223 61.8 kpsi .my i Ans
Eq. (2-19), 49.5 61.9 kpsi .1 1 0.20
uu
SS AW
ns 25% increase Ans.
(b) Before: 49.5 1.5532
u
y
SS
After: 61.9 1.0061.8
u
y
SS
Ans.
Lost most of its ductility ______________________________________________________________________________ 2-10 W = 0.20, (a) Before cold working: AISI 1212 HR steel. Table A-22, Sy = 28 kpsi, Su = 61.5 kpsi,
0 = 110 kpsi, m = 0.24, f = 0.85 After cold working: Eq. (2-16), u = m = 0.24
Chapter 2 - Rev. D, Page 7/19
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Eq. (2-14), 0 1 1 1.251 1 0.20i
AA W
Eq. (2-17), 0ln ln1.25 0.223i ui
AA
Eq. (2-18), 174% increase Ans. 0.240 110 0.223 76.7 kpsi .my iS A ns
Eq. (2-19), 61.5 76.9 kpsi .1 1 0.20
uu
SS AW
ns 25% increase Ans.
(b) Before: 61.5 2.2028
u
y
SS
After: 76.9 1.0076.7
u
y
SS
Ans.
Lost most of its ductility ______________________________________________________________________________ 2-11 W = 0.20, (a) Before cold working: 2024-T4 aluminum alloy. Table A-22, Sy = 43.0 kpsi, Su =
64.8 kpsi, 0 = 100 kpsi, m = 0.15, f = 0.18 After cold working: Eq. (2-16), u = m = 0.15
Eq. (2-14), 0 1 1 1.251 1 0.20i
AA W
Eq. (2-17), 0ln ln1.25 0.223ii
AA f
Material fractures. Ans.
______________________________________________________________________________ 2-12 For HB = 275, Eq. (2-21), Su = 3.4(275) = 935 MPa Ans. ______________________________________________________________________________ 2-13 Gray cast iron, HB = 200. Eq. (2-22), Su = 0.23(200) 12.5 = 33.5 kpsi Ans. From Table A-24, this is probably ASTM No. 30 Gray cast iron Ans. ______________________________________________________________________________ 2-14 Eq. (2-21), 0.5HB = 100 HB = 200 Ans. ______________________________________________________________________________
Chapter 2 - Rev. D, Page 8/19
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2-15 For the data given, converting HB to Su using Eq. (2-21)
HB Su (kpsi) Su2 (kpsi)230 115 13225 232 116 13456 232 116 13456 234 117 13689 235 117.5 13806.25235 117.5 13806.25235 117.5 13806.25236 118 13924 236 118 13924 239 119.5 14280.25
Su = 1172 Su2 = 137373
1172 117.2 117 kpsi .10
uu
SS A
N ns
Eq. (20-8),
10
2 22
1 137373 10 117.2 1.27 kpsi .1 9u
u ui
S
S NSs A
N
ns
______________________________________________________________________________ 2-16 For the data given, converting HB to Su using Eq. (2-22)
HB Su (kpsi) Su2 (kpsi)230 40.4 1632.16 232 40.86 1669.54 232 40.86 1669.54 234 41.32 1707.342235 41.55 1726.403235 41.55 1726.403235 41.55 1726.403236 41.78 1745.568236 41.78 1745.568239 42.47 1803.701
Su = 414.12 Su2 =17152.63
Chapter 2 - Rev. D, Page 9/19
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414.12 41.4 kpsi .10
uu
SS A
N ns
Eq. (20-8),
10
2 22
1 17152.63 10 41.4 1.20 .1 9u
u ui
S
S NSs A
N
ns
______________________________________________________________________________
2-17 (a) 2
345.5 34.5 in lbf / in .2(30)R
u A ns
(b)
P L A A0 / A – 1 = P/A0 0 0 0 0
1000 0.0004 0.0002 5 032.39 2000 0.0006 0.0003 10 064.78 3000 0.0010 0.0005 15 097.17 4000 0.0013 0.000 65 20 129.55 7000 0.0023 0.001 15 35 226.72 8400 0.0028 0.0014 42 272.06 8800 0.0036 0.0018 44 285.02 9200 0.0089 0.004 45 46 297.97 9100 0.1963 0.012 291 0.012 291 45 794.73
13200 0.1924 0.032 811 0.032 811 66 427.53 15200 0.1875 0.059 802 0.059 802 76 492.30 17000 0.1563 0.271 355 0.271 355 85 550.60 16400 0.1307 0.520 373 0.520 373 82 531.17 14800 0.1077 0.845 059 0.845 059 74 479.35
From the figures on the next page,
5
1
3 3
1 (43 000)(0.001 5) 45 000(0.004 45 0.001 5)2
1 45 000 76 500 (0.059 8 0.004 45)2
81 000 0.4 0.059 8 80 000 0.845 0.4
66.7 10 in lbf/in .
T ii
u A
Ans
Chapter 2 - Rev. D, Page 10/19
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Chapter 2 - Rev. D, Page 11/19
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2-18, 2-19 These problems are for student research. No standard solutions are provided. ______________________________________________________________________________ 2-20 Appropriate tables: Young’s modulus and Density (Table A-5)1020 HR and CD (Table A-
20), 1040 and 4140 (Table A-21), Aluminum (Table A-24), Titanium (Table A-24c) Appropriate equations:
For diameter, 2
4/ 4 y y
F F S d FA d S
Weight/length = A, Cost/length = $/in = ($/lbf) Weight/length, Deflection/length = /L = F/(AE) With F = 100 kips = 100(103) lbf,
Material Young's Modulus Density
Yield Strength Cost/lbf Diameter
Weight/ length
Cost/ length
Deflection/ length
units Mpsi lbf/in^3 kpsi $/lbf in lbf/in $/in in/in 1020 HR 30 0.282 30 $0.27 2.060 0.9400 $0.25 1.000E‐031020 CD 30 0.282 57 $0.30 1.495 0.4947 $0.15 1.900E‐031040 30 0.282 80 $0.35 1.262 0.3525 $0.12 2.667E‐034140 30 0.282 165 $0.80 0.878 0.1709 $0.14 5.500E‐03Al 10.4 0.098 50 $1.10 1.596 0.1960 $0.22 4.808E‐03Ti 16.5 0.16 120 $7.00 1.030 0.1333 $0.93 7.273E‐03
The selected materials with minimum values are shaded in the table above. Ans. ______________________________________________________________________________ 2-21 First, try to find the broad category of material (such as in Table A-5). Visual, magnetic,
and scratch tests are fast and inexpensive, so should all be done. Results from these three would favor steel, cast iron, or maybe a less common ferrous material. The expectation would likely be hot-rolled steel. If it is desired to confirm this, either a weight or bending test could be done to check density or modulus of elasticity. The weight test is faster. From the measured weight of 7.95 lbf, the unit weight is determined to be
3 327.95 lbf 0.281 lbf/in 0.28 lbf/in
[ (1 in) / 4](36 in)WAl
w
which agrees well with the unit weight of 0.282 lbf/in3 reported in Table A-5 for carbon steel. Nickel steel and stainless steel have similar unit weights, but surface finish and darker coloring do not favor their selection. To select a likely specification from Table
Chapter 2 - Rev. D, Page 12/19
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A-20, perform a Brinell hardness test, then use Eq. (2-21) to estimate an ultimate strength of . Assuming the material is hot-rolled due to the rough surface finish, appropriate choices from Table A-20 would be one of the higher carbon steels, such as hot-rolled AISI 1050, 1060, or 1080. Ans.
0.5 0.5(200) 100 kpsiu BS H
______________________________________________________________________________ 2-22 First, try to find the broad category of material (such as in Table A-5). Visual, magnetic,
and scratch tests are fast and inexpensive, so should all be done. Results from these three favor a softer, non-ferrous material like aluminum. If it is desired to confirm this, either a weight or bending test could be done to check density or modulus of elasticity. The weight test is faster. From the measured weight of 2.90 lbf, the unit weight is determined to be
3 322.9 lbf 0.103 lbf/in 0.10 lbf/in
[ (1 in) / 4](36 in)WAl
w
which agrees reasonably well with the unit weight of 0.098 lbf/in3 reported in Table A-5 for aluminum. No other materials come close to this unit weight, so the material is likely aluminum. Ans.
______________________________________________________________________________ 2-23 First, try to find the broad category of material (such as in Table A-5). Visual, magnetic,
and scratch tests are fast and inexpensive, so should all be done. Results from these three favor a softer, non-ferrous copper-based material such as copper, brass, or bronze. To further distinguish the material, either a weight or bending test could be done to check density or modulus of elasticity. The weight test is faster. From the measured weight of 9 lbf, the unit weight is determined to be
3 329.0 lbf 0.318 lbf/in 0.32 lbf/in
[ (1 in) / 4](36 in)WAl
w
which agrees reasonably well with the unit weight of 0.322 lbf/in3 reported in Table A-5 for copper. Brass is not far off (0.309 lbf/in3), so the deflection test could be used to gain additional insight. From the measured deflection and utilizing the deflection equation for an end-loaded cantilever beam from Table A-9, Young’s modulus is determined to be
33
4
100 2417.7 Mpsi
3 3 (1) 64 (17 / 32)FlEIy
which agrees better with the modulus for copper (17.2 Mpsi) than with brass (15.4 Mpsi). The conclusion is that the material is likely copper. Ans.
______________________________________________________________________________ 2-24 and 2-25 These problems are for student research. No standard solutions are provided. ______________________________________________________________________________
Chapter 2 - Rev. D, Page 13/19
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2-26 For strength, = F/A = S A = F/S For mass, m = Al = (F/S) l Thus, f 3(M ) = /S , and maximize S/ ( = 1) In Fig. (2-19), draw lines parallel to S/
From the list of materials given, both aluminum alloy and high carbon heat treated
steel are good candidates, having greater potential than tungsten carbide or polycarbonate. The higher strength aluminum alloys have a slightly greater potential. Other factors, such as cost or availability, may dictate which to choose. Ans.
______________________________________________________________________________ 2-27 For stiffness, k = AE/l A = kl/E For mass, m = Al = (kl/E) l =kl2 /E Thus, f 3(M) = /E , and maximize E/ ( = 1) In Fig. (2-16), draw lines parallel to E/
Chapter 2 - Rev. D, Page 14/19
-
From the list of materials given, tungsten carbide (WC) is best, closely followed by
aluminum alloys, and then followed by high carbon heat-treated steel. They are close enough that other factors, like cost or availability, would likely dictate the best choice. Polycarbonate polymer is clearly not a good choice compared to the other candidate materials. Ans.
______________________________________________________________________________ 2-28 For strength, = Fl/Z = S (1) where Fl is the bending moment and Z is the section modulus [see Eq. (3-26b), p. 90 ].
The section modulus is strictly a function of the dimensions of the cross section and has the units in3 (ips) or m3 (SI). Thus, for a given cross section, Z =C (A)3/2, where C is a
number. For example, for a circular cross section, C = 14 . Then, for strength, Eq. (1) is
2/3
3/2
Fl FlS ACA CS
(2)
Chapter 2 - Rev. D, Page 15/19
-
For mass, 2/3 2/3
5/32/3
Fl Fm Al l lCS C S
Thus, f 3(M) = /S 2/3, and maximize S 2/3/ ( = 2/3) In Fig. (2-19), draw lines parallel to S 2/3/
From the list of materials given, a higher strength aluminum alloy has the greatest
potential, followed closely by high carbon heat-treated steel. Tungsten carbide is clearly not a good choice compared to the other candidate materials. .Ans.
______________________________________________________________________________ 2-29 Eq. (2-26), p. 65, applies to a circular cross section. However, for any cross section shape
it can be shown that I = CA 2, where C is a constant. For example, consider a rectangular section of height h and width b, where for a given scaled shape, h = cb, where c is a
Chapter 2 - Rev. D, Page 16/19
-
constant. The moment of inertia is I = bh 3/12, and the area is A = bh. Then I = h(bh2)/12 = cb (bh2)/12 = (c/12)(bh)2 = CA 2, where C = c/12 (a constant).
Thus, Eq. (2-27) becomes
1/23
3klACE
and Eq. (2-29) becomes
1/2
5/21/23
km Al lC E
Thus, minimize 3 1/2f M E
, or maximize1/2EM
. From Fig. (2-16)
From the list of materials given, aluminum alloys are clearly the best followed by steels
and tungsten carbide. Polycarbonate polymer is not a good choice compared to the other candidate materials. Ans.
______________________________________________________________________________ 2-30 For stiffness, k = AE/l A = kl/E For mass, m = Al = (kl/E) l =kl2 /E
Chapter 2 - Rev. D, Page 17/19
-
So, f 3(M) = /E, and maximize E/ . Thus, = 1. Ans. ______________________________________________________________________________ 2-31 For strength, = F/A = S A = F/S For mass, m = Al = (F/S) l So, f 3(M ) = /S, and maximize S/ . Thus, = 1. Ans. ______________________________________________________________________________ 2-32 Eq. (2-26), p. 65, applies to a circular cross section. However, for any cross section shape
it can be shown that I = CA 2, where C is a constant. For example, consider a rectangular section of height h and width b, where for a given scaled shape, h = cb, where c is a constant. The moment of inertia is I = bh 3/12, and the area is A = bh. Then I = h(bh2)/12 = cb (bh2)/12 = (c/12)(bh)2 = CA 2, where C = c/12.
Thus, Eq. (2-27) becomes
1/23
3klACE
and Eq. (2-29) becomes
1/2
5/21/23
km Al lC E
So, minimize 3 1/2f M E
, or maximize1/2EM
. Thus, = 1/2. Ans.
______________________________________________________________________________ 2-33 For strength, = Fl/Z = S (1) where Fl is the bending moment and Z is the section modulus [see Eq. (3-26b), p. 90 ].
The section modulus is strictly a function of the dimensions of the cross section and has the units in3 (ips) or m3 (SI). Thus, for a given cross section, Z =C (A)3/2, where C is a
number. For example, for a circular cross section, C = 14 . Then, for strength, Eq. (1) is
2/3
3/2
Fl FlS ACA CS
(2)
For mass, 2/3 2/3
5/32/3
Fl Fm Al l lCS C S
So, f 3(M) = /S 2/3, and maximize S 2/3/. Thus, = 2/3. Ans. ______________________________________________________________________________ 2-34 For stiffness, k=AE/l, or, A = kl/E.
Chapter 2 - Rev. D, Page 18/19
-
Chapter 2 - Rev. D, Page 19/19
Thus, m = Al = (kl/E )l = kl 2 /E. Then, M = E / and = 1. From Fig. 2-16, lines parallel to E / for ductile materials include steel, titanium,
molybdenum, aluminum alloys, and composites. For strength, S = F/A, or, A = F/S. Thus, m = Al = F/Sl = Fl /S. Then, M = S/ and = 1. From Fig. 2-19, lines parallel to S/ give for ductile materials, steel, aluminum alloys,
nickel alloys, titanium, and composites. Common to both stiffness and strength are steel, titanium, aluminum alloys, and
composites. Ans.
-
Chapter 3
3-1
0oM
18 6(100) 0BR
33.3 lbf .BR Ans
0yF
100 0o BR R
66.7 lbf .oR Ans
33.3 lbf .C BR R A ns ______________________________________________________________________________ 3-2 Body AB:
0xF Ax BxR R
0yF Ay ByR R
0BM (10) (10) 0Ay AxR R
Ax AyR R Body OAC:
0OM (10) 100(30) 0AyR
300 lbf .AyR Ans
0xF 300 lbf .Ox AxR R A ns
0yF 100 0Oy AyR R
200 lbf .OyR Ans ______________________________________________________________________________
Chapter 3 - Rev. A, Page 1/100
-
3-3
0.8 1.39 kN .
tan 30OR Ans
0.8 1.6 kN .
sin 30AR Ans
______________________________________________________________________________ 3-4 Step 1: Find RA & RE
4.5 7.794 mtan 30
0
9 7.794(400cos30 )
4.5(400sin 30 ) 0400 N .
A
E
E
h
MR
R Ans
2 2
0 400cos30 0 346.4 N
0 400 400sin 30 0
200 N
346.4 200 400 N .
x Ax
Ax
y Ay
Ay
A
F RR
F RR
R Ans
Step 2: Find components of RC on link 4 and RD
44
0
400(4.5) 7.794 1.9 0 305.4 N .
0 305.4 N
0 ( ) 400 N
C
D
D
x Cx
y Cy
M
RR Ans
F R
F R
Chapter 3 - Rev. A, Page 2/100
-
Step 3: Find components of RC on link 2
2
2
2
0
305.4 346.4 0
41 N
0
200 N
x
Cx
Cx
y
Cy
F
R
R
F
R
____________________________________________________________________________________________________________________
_
Chapter 3 - Rev. A, Page 3/100
-
3-5
0CM 11500 300(5) 1200(9) 0R 1 8.2 kN .R Ans
0yF 28.2 9 5 0R 2 5.8 kN .R Ans
1 8.2(300) 2460 N m .M Ans
2 2460 0.8(900) 1740 N m .M Ans
3 1740 5.8(300) 0 checks!M _____________________________________________________________________________ 3-6
0yF 0 500 40(6) 740 lbf .R Ans
0 0M 0 500(8) 40(6)(17) 8080 lbf in .M Ans 1 8080 740(8) 2160 lbf in .M Ans
2 2160 240(6) 720 lbf in .M Ans
3
1720 (240)(6) 0 checks!2
M
______________________________________________________________________________
Chapter 3 - Rev. A, Page 4/100
-
3-7
0BM
12.2 1(2) 1(4) 0R 1 0.91 kN .R Ans
0yF
20.91 2 4 0R
2 6.91 kN .R Ans
1 0.91(1.2) 1.09 kN m .M Ans
2 1.09 2.91(1) 4 kN m .M Ans 3 4 4(1) 0 checks!M ______________________________________________________________________________ 3-8 Break at the hinge at B Beam OB: From symmetry, 1 200 lbf .BR V Ans Beam BD: 0DM 2200(12) (10) 40(10)(5) 0R 2 440 lbf .R Ans 0yF 3200 440 40(10) 0R 3 160 lbf .R Ans
Chapter 3 - Rev. A, Page 5/100
-
1 200(4) 800 lbf in .M Ans
2 800 200(4) 0 checks at hingeM
3 800 200(6) 400 lbf in .M Ans
4
1400 (240)(6) 320 lbf in .2
M Ans
5
1320 (160)(4) 0 checks!2
M
______________________________________________________________________________ 3-9
1 1 11 2
0 0 01 2
1 1 11 2
9 300 5 1200 1500
9 300 5 1200 1500 (1)
9 300 5 1200 1500 (2)
q R x x x R x
V R x x R x
M R x x x R x
1
At x = 1500+ V = M = 0. Applying Eqs. (1) and (2),
1 2 1 29 5 0 14R R R R
1 11500 9(1500 300) 5(1500 1200) 0 8.2 kN .R R A
2 14 8.2 5.8 kN .ns
R Ans 0 300 : 8.2 kN, 8.2 N m300 1200 : 8.2 9 0.8 kN 8.2 9( 300) 0.8 2700 N m1200 1500 : 8.2 9 5 5.8 kN 8.2 9( 300
x V M xx V
M x x xx V
M x x
) 5( 1200) 5.8 8700 N mx x
Plots of V and M are the same as in Prob. 3-5. ______________________________________________________________________________
Chapter 3 - Rev. A, Page 6/100
-
3-10
1 2 1 0 00 0
1 0 1 10 0
1 2 20 0
500 8 40 14 40 20
500 8 40 14 40 20 (1)
500 8 20 14 20 20 (2)
at 20 in, 0, Eqs. (1) and (2) give
q R x M x x x x
V R M x x x x
M R x M x x x
x V MR
0 02
0 0 0
500 40 20 14 0 740 lbf .
(20) 500(20 8) 20(20 14) 0 8080 lbf in .
R Ans
R M M
Ans
0 8 : 740 lbf, 740 8080 lbf in8 14 : 740 500 240 lbf 740 8080 500( 8) 240 4080 lbf in14 20 : 740 500 40( 14) 40 800 lbf 740 8080
x V M xx V
M x x xx V x x
M x
2 2500( 8) 20( 14) 20 800 8000 lbf inx x x x
Plots of V and M are the same as in Prob. 3-6. ______________________________________________________________________________ 3-11
1 1 1 11 2
0 0 01 2
1 1 11 2
2 1.2 2.2 4 3.2
2 1.2 2.2 4 3.2 (1)
2 1.2 2.2 4 3.2 (2)
q R x x R x x
V R x R x x
M R x x R x x
at x = 3.2+, V = M = 0. Applying Eqs. (1) and (2),
Solving Eqs. (3) and (4) simultaneously,
1 2 1 2
1 2 1 2
2 4 0 6 (3)3.2 2(2) (1) 0 3.2 4 (4)R R R R
R R R R
R1 = -0.91 kN, R2 = 6.91 kN Ans. 0 1.2 : 0.91 kN, 0.91 kN m1.2 2.2 : 0.91 2 2.91 kN 0.91 2( 1.2) 2.91 2.4 kN m2.2 3.2 : 0.91 2 6.91 4 kN 0.91 2(
x V M xx V
M x x xx V
M x x
1.2) 6.91( 2.2) 4 12.8 kN mx x
Plots of V and M are the same as in Prob. 3-7. ______________________________________________________________________________
Chapter 3 - Rev. A, Page 7/100
-
3-12
1 1 1 0 0 11 2 3
0 0 1 1 01 2 3
1 1 2 2 11 2 3
1
400 4 10 40 10 40 20 20
400 4 10 40 10 40 20 20 (1)
400 4 10 20 10 20 20 20 (2)0 at 8 in 8 400(
q R x x R x x x R x
V R x R x x x R x
M R x x R x x x R xM x R
18 4) 0 200 lbf .R Ans
at x = 20+, V =M = 0. Applying Eqs. (1) and (2), 2 3 2 3
22 2
200 400 40(10) 0 600
200(20) 400(16) (10) 20(10) 0 440 lbf .
R R R RR R A
3 600 440 160 lbf .ns
R Ans
0 4 : 200 lbf, 200 lbf in4 10 : 200 400 200 lbf, 200 400( 4) 200 1600 lbf in10 20 : 200 400 440 40( 10) 640 40 lbf
200 400( 4)
x V M xx V
M x x xx V x x
M x x
2 2440( 10) 20 10 20 640x x x
4800 lbf inx Plots of V and M are the same as in Prob. 3-8.
______________________________________________________________________________ 3-13 Solution depends upon the beam selected. ______________________________________________________________________________ 3-14
(a) Moment at center,
2
22
22 2 2 2 4
c
c
l ax
l l lM l a a
w wl
At reaction, 2 2rM a w
a = 2.25, l = 10 in, w = 100 lbf/in
2
100(10) 10 2.25 125 lbf in2 4
100 2.25253 lbf in .
2
c
r
M
M Ans
(b) Optimal occurs when c rM M
Chapter 3 - Rev. A, Page 8/100
-
22 20.25 0
2 4 2l l aa a al l
w w
Taking the positive root
2 21 4 0.25 2 1 0.207 .2 2
la l l l l A ns
for l = 10 in, w = 100 lbf, a = 0.207(10) = 2.07 in 2min 100 2 2.07 214 lbf inM
______________________________________________________________________________ 3-15
(a) 20 10 5 kpsi
2C
20 10 15 kpsi2
CD
2 215 8 17 kpsiR
1 5 17 22 kpsi
2 5 17 12 kpsi
11 8tan 14.04 cw
2 15p
1 17 kpsi
45 14.04 30.96 ccws
R
(b) 9 16 12.5 kpsi
2C
16 9 3.5 kpsi2
CD
2 25 3.5 6.10 kpsiR
1 12.5 6.1 18.6 kpsi 2 12.5 6.1 6.4 kpsi
11 5tan 27.5 ccw2 3.5p
1 6.10 kpsi
45 27.5 17.5 cws
R
Chapter 3 - Rev. A, Page 9/100
-
(c)
2 2
1
2
24 10 17 kpsi2
24 10 7 kpsi2
7 6 9.22 kpsi17 9.22 26.22 kpsi17 9.22 7.78 kpsi
C
CD
R
11 790 tan 69.7 ccw2 6p
1 9.22 kpsi
69.7 45 24.7 ccws
R
(d)
2 2
1
2
12 22 5 kpsi2
12 22 17 kpsi2
17 12 20.81 kpsi5 20.81 25.81 kpsi5 20.81 15.81 kpsi
C
CD
R
11 1790 tan 72.39 cw2 12p
Chapter 3 - Rev. A, Page 10/100
-
1 20.81 kpsi
72.39 45 27.39 cws
R
______________________________________________________________________________
Chapter 3 - Rev. A, Page 11/100
-
3-16
(a)
2 2
1
2
8 7 0.5 MPa2
8 7 7.5 MPa2
7.5 6 9.60 MPa9.60 0.5 9.10 MPa
0.5 9.6 10.1 Mpa
C
CD
R
11 7.590 tan 70.67 cw
2 6p
1 9.60 MPa
70.67 45 25.67 cws
R
(b)
2 2
1
2
9 6 1.5 MPa29 6 7.5 MPa
27.5 3 8.078 MPa
1.5 8.078 9.58 MPa1.5 8.078 6.58 MPa
C
CD
R
11 3tan 10.9 cw2 7.5p
1 8.078 MPa
45 10.9 34.1 ccws
R
Chapter 3 - Rev. A, Page 12/100
-
(c)
2 2
1
2
12 4 4 MPa2
12 4 8 MPa2
8 7 10.63 MPa4 10.63 14.63 MPa4 10.63 6.63 MPa
C
CD
R
11 890 tan 69.4 ccw2 7p
1 10.63 MPa
69.4 45 24.4 ccws
R
(d)
2 2
1
2
6 5 0.5 MPa26 5 5.5 MPa
25.5 8 9.71 MPa
0.5 9.71 10.21 MPa0.5 9.71 9.21 MPa
C
CD
R
11 8tan 27.75 ccw
2 5.5p
1 9.71 MPa
45 27.75 17.25 cws
R
______________________________________________________________________________
Chapter 3 - Rev. A, Page 13/100
-
3-17
(a)
2 2
1
2
12 6 9 kpsi2
12 6 3 kpsi2
3 4 5 kpsi5 9 14 kpsi9 5 4 kpsi
C
CD
R
11 4tan 26.6 ccw2 3p
1 5 kpsi
45 26.6 18.4 ccws
R
(b)
2 2
1
2
30 10 10 kpsi2
30 10 20 kpsi2
20 10 22.36 kpsi10 22.36 32.36 kpsi10 22.36 12.36 kpsi
C
CD
R
11 10tan 13.28 ccw2 20p
1 22.36 kpsi
45 13.28 31.72 cws
R
Chapter 3 - Rev. A, Page 14/100
-
(c)
2 2
1
2
10 18 4 kpsi2
10 18 14 kpsi2
14 9 16.64 kpsi4 16.64 20.64 kpsi4 16.64 12.64 kpsi
C
CD
R
11 1490 tan 73.63 cw2 9p
1 16.64 kpsi
73.63 45 28.63 cws
R
(d)
2 2
1
2
9 19 14 kpsi2
19 9 5 kpsi2
5 8 9.434 kpsi14 9.43 23.43 kpsi14 9.43 4.57 kpsi
C
CD
R
11 590 tan 61.0 cw2 8p
1 9.34 kpsi
61 45 16 cws
R
______________________________________________________________________________
Chapter 3 - Rev. A, Page 15/100
-
3-18
(a)
2 2
1
2
3
80 30 55 MPa2
80 30 25 MPa2
25 20 32.02 MPa0 MPa
55 32.02 22.98 23.0 MPa55 32.0 87.0 MPa
C
CD
R
1 2 2 3 1 323 8711.5 MPa, 32.0 MPa, 43.5 MPa2 2
(b)
2 2
1
2
3
30 60 15 MPa2
60 30 45 MPa2
45 30 54.1 MPa15 54.1 39.1 MPa
0 MPa15 54.1 69.1 MPa
C
CD
R
1 3
1 2
2 3
39.1 69.1 54.1 MPa2
39.1 19.6 MPa2
69.1 34.6 MPa2
Chapter 3 - Rev. A, Page 16/100
-
(c)
2 2
1
2
3
40 0 20 MPa240 0 20 MPa
220 20 28.3 MPa
20 28.3 48.3 MPa20 28.3 8.3 MPa
30 MPaz
C
CD
R
1 3 1 2 2 348.3 30 30 8.339.1 MPa, 28.3 MPa, 10.9 MPa
2 2
(d)
2 2
1
2
3
50 25 MPa250 25 MPa2
25 30 39.1 MPa25 39.1 64.1 MPa25 39.1 14.1 MPa
20 MPaz
C
CD
R
1 3 1 2 2 364.1 20 20 14.142.1 MPa, 39.1 MPa, 2.95 MPa
2 2
______________________________________________________________________________ 3-19
(a) Since there are no shear stresses on the stress element, the stress element already represents principal stresses.
1
2
3
10 kpsi0 kpsi
4 kpsi
x
y
1 3
1 2
2 3
10 ( 4) 7 kpsi2
10 5 kpsi20 ( 4) 2 kpsi
2
Chapter 3 - Rev. A, Page 17/100
-
(b)
2 2
1
2 3
0 10 5 kpsi2
10 0 5 kpsi2
5 4 6.40 kpsi5 6.40 11.40 kpsi0 kpsi, 5 6.40 1.40 kpsi
C
CD
R
1 3 1 2 311.40 1.406.40 kpsi, 5.70 kpsi, 0.70 kpsi
2 2R
(c)
2 2
1 2
3
2 8 5 kpsi2
8 2 3 kpsi2
3 4 5 kpsi5 5 0 kpsi, 0 kpsi5 5 10 kpsi
C
CD
R
1 3 1 2 2 310 5 kpsi, 0 kpsi, 5 kpsi2
(d)
2 2
1
2
3
10 30 10 kpsi2
10 30 20 kpsi2
20 10 22.36 kpsi10 22.36 12.36 kpsi
0 kpsi10 22.36 32.36 kpsi
C
CD
R
1 3 1 2 2 312.36 32.3622.36 kpsi, 6.18 kpsi, 16.18 kpsi
2 2
______________________________________________________________________________
Chapter 3 - Rev. A, Page 18/100
-
3-20 From Eq. (3-15),
3 2 2 2
2 2 2
3
( 6 18 12) 6(18) ( 6)( 12) 18( 12) 9 6 ( 15)
6(18)( 12) 2(9)(6)( 15) ( 6)(6) 18( 15) ( 12)(9) 0
594 3186 0
2
Roots are: 21.04, 5.67, –26.71 kpsi Ans.
1 2
2 3
max 1 3
21.04 5.67 7.69 kpsi2
5.67 26.71 16.19 kpsi221.04 26.71 23.88 kpsi .
2Ans
_____________________________________________________________________________ 3-21 From Eq. (3-15)
23 2 2
2 2 2
3 2
(20 0 20) 20(0) 20(20) 0(20) 40 20 2 0
20(0)(20) 2(40) 20 2 (0) 20 20 2 0(0) 20(40) 0
40 2 000 48 000 0
2
Roots are: 60, 20, –40 kpsi Ans.
1 2
2 3
max 1 3
60 20 20 kpsi2
20 40 30 kpsi2
60 40 50 kpsi .2
Ans
_____________________________________________________________________________
Chapter 3 - Rev. A, Page 19/100
-
3-22
From Eq. (3-15)
2 23 2 2
2 2 2
3 2
(10 40 40) 10(40) 10(40) 40(40) 20 40 20
10(40)(40) 2(20)( 40)( 20) 10( 40) 40( 20) 40(20) 0
90 0
Roots are: 90, 0, 0 MPa Ans.
2 3
1 2 1 3 max
0
90 45 MPa .2
Ans
_____________________________________________________________________________ 3-23
2
6
61
15000 33 950 psi 34.0 kpsi .4 0.75
6033 950 0.0679 in .30 10
0.0679 1130 10 1130 .60
F AnsA
FL L AnsAE E
AnsL
From Table A-5, v = 0.292
2 1
6 62
0.292(1130) 330 .
330 10 (0.75) 248 10 in .
v A
d d An
ns
s
_____________________________________________________________________________ 3-24
2
6
61
3000 6790 psi 6.79 kpsi .4 0.75
606790 0.0392 in .10.4 10
0.0392 653 10 653 .60
F AnsA
FL L AnsAE E
AnsL
From Table A-5, v = 0.333
2 1
6 62
0.333(653) 217 .
217 10 (0.75) 163 10 in .
v Ans
d d Ans
Chapter 3 - Rev. A, Page 20/100
-
_____________________________________________________________________________ 3-25
20.0001 0.0001d d
d d
From Table A-5, v = 0.326, E = 119 GPa
621
6 91
26
0.0001 306.7 100.326
and , so
= 306.7 10 (119) 10 36.5 MPa
0.0336.5 10 25 800 N 25.8 kN .
4
vFL FAE AE EL
F A An
s
Sy = 70 MPa > , so elastic deformation assumption is valid. _____________________________________________________________________________ 3-26
68(12)20 000 0.185 in .
10.4 10FL L AnsAE E
_____________________________________________________________________________ 3-27
6
9
3140 10 0.00586 m 5.86 mm .71.7 10
FL L AnsAE E
_____________________________________________________________________________ 3-28
610(12)15 000 0.173 in .
10.4 10FL L AnsAE E
_____________________________________________________________________________ 3-29 With 0,z solve the first two equations of Eq. (3-19) simulatenously. Place E on the
left-hand side of both equations, and using Cramer’s rule,
2 2
11 1 1
1
x
y xx yx
E vE EE vE
v v vv
yv
Likewise,
Chapter 3 - Rev. A, Page 21/100
-
21
y xy
Ev
From Table A-5, E = 207 GPa and ν = 0.292. Thus,
96
2 2
96
2
207 10 0.0019 0.292 0.000 7210 382 MPa .
1 1 0.292207 10 0.000 72 0.292 0.0019
10 37.4 MPa .1 0.292
x yx
y
E vAns
v
Ans
_____________________________________________________________________________ 3-30 With 0,z solve the first two equations of Eq. (3-19) simulatenously. Place E on the
left-hand side of both equations, and using Cramer’s rule,
2 2
11 1 1
1
x
y xx yx
E vE EE vE
v v vv
yv
Likewise,
21
y xy
Ev
From Table A-5, E = 71.7 GPa and ν = 0.333. Thus,
96
2 2
96
2
71.7 10 0.0019 0.333 0.000 7210 134 MPa .
1 1 0.33371.7 10 0.000 72 0.333 0.0019
10 7.04 MPa .1 0.333
x yx
y
E vAns
v
Ans
_____________________________________________________________________________ 3-31
(a) 1 max 1 c acR F M R a Fl l
2
2 2
6 66
M ac bh lF Fbh bh l ac
Ans.
(b)
2 21 21( )( ) ( ) .
( )( )m m m mm
m m
b b h h l lF s s s s AnsF a a c c s s
3-32
For equal stress, the model load varies by the square of the scale factor.
_____________________________________________________________________________
Chapter 3 - Rev. A, Page 22/100
-
2
1 max /2,
2 2 2 2x ll l lR M l
w w 8l
ww(a)
2 2
2 2 2
6 6 3 4 .8 4 3
M l Wl bhW Abh bh bh l
w ns
(b) 2 2
2( / )( / )( / ) 1( )( ) ./
m m m m
m
W b b h h s s s AnW l l s
s
22 .m m ml ss s
l s
w ww w
Ans
For equal stress, the model load w varies linearly with the scale factor. _____ _____________
-33 (a) Can solve by iteration or derive
_ __________________________________________________________ 3
equations for the general case. Find maximum moment under wheel 3W .
W W at centroid of W’s T3 3d
A Tl xR W
l
Under wheel 3,
3 3
3 3 1 13 2 23 3 1 13 2 23A T
l x dM R x W a W a W x W a W a
l
For maximum, 3 33 3 33
0 22
TdM l dWl d x xdx l
Substitute into 2
33 1 14 T
l d3 2 23M M W W al
W a
intersects the midpoint of the beam.
For wheel i,
This means the midpoint of 3d
2 1il dl d1
,2 4
iiT j ji
ji ix M W W al
Note for wheel 1:
0j jiW a
1 2 3 4104.4104.4, 26.1 kips
4TW W W W W
Wheel 1: 2
1 1476 (1200 238)238 in, (104.4) 20128 kip in
2 4(1200)d M
Wheel 2: 238 84 154 ind 2
Chapter 3 - Rev. A, Page 23/100
-
2
2 max(1200 154) (104.4) 26.1(84) 21605 kip in .
4(1200)M M A ns
Check if all of the wheels are on the rail.
(b) max 600 77 523 in .x Ans (c) See above sketch. (d) Inner axles
_____________________________________________________________________________ 3-34
(a) Let a = total area of entire envelope Let b = area of side notch
2
3 3
6 4
2 40(3)(25) 25 34 2150 mm1 12 40 75 34 25
12 121.36 10 mm .
a b
A a b
I I I
I Ans
Dimensions in mm. (b)
2
2
2
0.375(1.875) 0.703 125 in
0.375(1.75) 0.656 25 in
2(0.703125) 0.656 25 2.0625 in
a
b
AAA
34
34
2 21
2(0.703 125)(0.9375) 0.656 25(0.6875) 0.858 in .2.0625
0.375(1.875) 0.206 in12
1.75(0.375) 0.007 69 in12
2 0.206 0.703 125(0.0795) 0.00769 0.656 25(0.1705) 0.448 in .
a
b
y A
I
I
4
ns
I Ans
(c) Use two negative areas.
2 2
2
625 mm , 5625 mm , 10 000 mm
10 000 5625 625 3750 mm ;a b cA A A
A
2
Chapter 3 - Rev. A, Page 24/100
-
13
4
36 4
36 4
6.25 mm, 50 mm, 50 mm10 000(50) 5625(50) 625(6.25) 57.29 mm .
3750100 57.29 42.71 mm .
50(12.5) 8138 mm12
75(75) 2.637 10 mm12
100(100) 8.333 10 in12
a b c
a
b
c
y y y
y Ans
c Ans
I
I
I
2 26 2 61
6 41
8.333 10 10000(7.29) 2.637 10 5625 7.29 8138 625 57.29 6.25
4.29 10 in .
I
I Ans
(d)
2
2
2
4 0.875 3.5 in
2.5 0.875 2.1875 in
5.6875 in2.9375 3.5 1.25(2.1875)
2.288 in .5.6875
a
b
a b
A
A
A A A
y Ans
3 2 3
4
1 1(4) 0.875 3.5 2.9375 2.288 0.875 2.5 2.1875 2.288 1.2512 125.20 in .
I
I Ans
2
_____________________________________________________________________________ 3-35
3 52
1 (20)(40) 1.067 10 mm1220(40) 800 mm
I
A
4
Mmax is at A. At the bottom of the section,
max 5450 000(20) 84.3 MPa .1.067 10
Mc AnsI
Due to V, max is between A and B at y = 0.
max3 3 3000 5.63 MPa .2 2 800
V AnsA
_____________________________________________________________________________
Chapter 3 - Rev. A, Page 25/100
-
3-36 3 41 (1)(2) 0.6667 in
12I
21(2) 2 inA
0oM 8 100(8)(12) 0AR 1200 lbfAR 1200 100(8) 400 lbfoR
is at A. At the top of the beam, maxM
max3200(0.5) 2400 psi .
0.6667Mc AnsI
Due to V, max is at A, at y = 0.
max3 3 800 600 psi .2 2 2
V AnsA
_____________________________________________________________________________ 3-37
3 41 (0.75)(2) 0.5 in12
I 2(0.75)(2) 1.5 inA
0AM 15 1000(20) 0BR 1333.3 lbfBR 3000 1333.3 1000 2666.7 lbfAR
is at B. At the top of the beam, maxM
max5000(1) 10000 psi .
0.5Mc AnsI
Due to V, max is between B and C at y = 0.
max3 3 1000 1000 psi .2 2 1.5
V AnsA
_____________________________________________________________________________
Chapter 3 - Rev. A, Page 26/100
-
3-38
4 4
3 4(50) 306.796 10 mm64 64dI
2 22(50) 1963 mm
4 4dA
0BM 6(300)(150) 200 0AR
1350 kNAR 6(300) 1350 450 kNBR
maxM is at A. At the top, maxMcI
Due to V, max is at A, at y = 0.
2max
4 4 750 0.509 kN/mm 509 MPa .3 3 1963
V AnsA
_____________________________________________________________________________ 3-39
2 2max
max max 2
8 8 8
Il l cMI cl
w w w
(a) 448 in; Table A-8, 0.537 inl I
3
2
8 12 10 0.53722.38 lbf/in .
1 48Ans w
(b) 3 360 in, 1 12 2 3 1 12 1.625 2.625 2.051 inl I 4
3
2
8 12 10 2.05136.5 lbf/in .
1.5 60Ans w
(c) 460 in; Table A-6, 2 0.703 1.406 inl I y = 0.717 in, cmax = 1.783 in
3
2
8 12 10 1.40621.0 lbf/in .
1.783 60Ans w
(d)
460 in, Table A-7, 2.07 inl I
3
2
8 12 10 2.0736.8 lbf/in .
1.5 60Ans w
_____________________________________________________________________________
Chapter 3 - Rev. A, Page 27/100
-
3-40
4 3 4 2 20.5 3.068 10 in , 0.5 0.1963 in64 4I A
Model
(c) 3
max
500(0.5) 500(0.75 / 2) 218.75 lbf in2 2
218.75(0.25)3.068 10
17 825 psi 17.8 kpsi .4 4 500 3400 psi 3.4 kpsi .3 3 0.1963
M
McI
AnsV AnsA
Model (d)
3
500(0.625) 312.5 lbf in312.5(0.25)3.068 10
25 464 psi 25.5 kpsi .
MMcI
Ans
max4 4 500 3400 psi 3.4 kpsi .3 3 0.1963
V AnsA
Model
(e) 3
max
500(0.4375) 218.75 lbf in218.75(0.25)3.068 10
17 825 psi 17.8 kpsi .4 4 500 3400 psi 3.4 kpsi .3 3 0.1963
MMcI
AnsV AnsA
_____________________________________________________________________________ 3-41
Chapter 3 - Rev. A, Page 28/100
-
4 4 212 1018 mm , 12 113.1 mm64 4I A2
Model (c)
2
2max
2000(6) 2000(9) 15 000 N mm2 2
15 000(6)1018
88.4 N/mm 88.4 MPa .4 4 2000 23.6 N/mm 23.6 MPa .3 3 113.1
M
McI
AnsV AnsA
Model (d)
2
2000(12) 24 000 N mm24 000(6)
1018141.5 N/mm 141.5 MPa .
MMcI
Ans
2max
4 4 2000 23.6 N/mm 23.6 MPa .3 3 113.1
V AnsA
Model (e)
2
2000(7.5) 15000 N mm15000(6)
101888.4 N/mm 88.4 MPa .
MMcI
Ans
2max
4 4 2000 23.6 N/mm 23.6 MPa .3 3 113.1
V AnsA
_____________________________________________________________________________
4 3/ 2 32
/ 64M dMc M
I d d
3-42 (a)
Chapter 3 - Rev. A, Page 29/100
-
3 332 32(218.75) 0.420 in .
(30 000)Md A
ns
(b)
2 / 4V VA d
4 4( 500) 0.206 in .
(15000)Vd Ans
(c)
2
4 43 3 / 4
V VA d
4 4 4 4(500) 0.238 in .3 3 (15000)
Vd A
ns
______________ __________________ ______________________________
_____________ _
_
3-43
1 0 11 21
1 21 21
2 31 1 2
terms for
terms for 2
terms for 2 6
p pq F x p x l x l x l aa
p pV F p x l x l x l aa
p p pM Fx x l x l x l aa
terms for x > l + a = 0 At x ( ) , 0,l a V M
21 21 1 2
2 Fp p
231 1 2
1 2 2
0 (1)2
6 ( )( ) 0 2 (2)2 6
p pF p a aa a
p a p p F l aF l a a p pa a
From (1) and (2) 1 22 22 2(3 2 ), (3 ) (3)F Fp l a p l aa a
From similar triang les 22 1 2 1 2
(4)apb a bp p p p p
Chapter 3 - Rev. A, Page 30/100
-
Mmax occurs where V = 0
max 2x l a b
2 31 1 2max
2 31 1 2
( 2 ) ( 2 ) ( 2 )2 6
( 2 ) ( 2 ) ( 2 )2 6
p p pM F l a b a b a ba
p p pFl F a b a b a ba
Normally Mmax = Fl
The fractional increase in the magnitude is
2 31 22 ( 2 ) 6 ( 2 ) (5)a b p p a a b
For example, consider F = 1500 lbf, a = 1.2 in, l = 1.5 in
(3)
1( 2 )F a b p Fl
1 22(1500) 3 1.5 2(1.2) 14 375 lbf/in
1.2p
2 2
2(1500) 3 1.5 1.2 11 875 lbf/in1.2
p
(4) b = 1.2(11 875)/(14 375 + 11 875) = 0.5429 in Substituting into (5) yields
_____________________________________________________________________________
-44
= 0.036 89 or 3.7% higher than -Fl
3
Chapter 3 - Rev. A, Page 31/100
-
1
2
300(30)R 401800 6900 lbf2 30
300(30) 101800 3900 lbf2 30
3900 13 in300
R
a
MB = 1800(10) = 18 000 lbfin
x = 27 in = (1/2)3900(13) = 25 350 lbfin
MB = 1800(10) = 18 000 lbfin
x = 27 in = (1/2)3900(13) = 25 350 lbfin MM
3 41
3 42
0.5(3) 2.5(3) 1.5 in6
1 (3)(1 ) 0.25 in 121 (1)(3 ) 2.25 in
12
y
I
I
Applying the parallel-axis theorem,
(a)
20.25 3(1.5 0.5) 2.25 3zI 2 4 (2.5 1.5) 8.5 in
18000( 1.5)At 10 in, 1.5 in, 3176 psi8.5
18000(2.5)At 10 in, 2.5 in, 5294 psi8.5
25350( 1.5)At 27 in, 1.5 in, 4474 psi8.5
At 27 in, 2.5 in,
x
x
x
x
x y
x y
x y
x y
25350(2.5) 7456 psi
8.5
Max tension 5294 psi .Max compression 7456 psi .
AnsAns
aximum shear stress due to V is at B, at the neutral axis.
(b) The m max 5100 lbfV
3
max
1.25(2.5)(1) 3.125 in5100(3.125) 1875 psi .
8.5(1)V
Q y AVQ AnsIb
(c) There are three potentially critical locations for the maximum shear stress, all at x = 27 in: (i) at the top where the bending stress is maximum, (ii) at the neutral axis where
Chapter 3 - Rev. A, Page 32/100
-
the transverse shear is maximum, or (iii) in the web just above the flange where bending stress and shear stress are in their largest combination. For (i):
The maximum bending stress was previously found to be 7456 psi, and the shear stress is zero. From Mohr’s circle,
maxmax7456 3728 psi
2 2
For (ii): The bending stress is zero, and the transverse shear stress was found previously to be 1875 psi. Thus, max = 1875 psi.
For (iii): The bending stress at y = – 0.5 in is
18000( 0.5) 1059 psi
8.5x
The transverse shear stress is
3(1)(3)(1) 3.0 in5100(3.0) 1800 psi
8.5(1)
Q y AVQIb
From Mohr’s circle,
22
max1059 1800 1876 psi 2
The critical location is at x = 27 in, at the top surface, where max = 3728 psi. Ans. _____________________________________________________________________________ 3-45 (a) L = 10 in. Element A:
34(1000)(10)(0.5) 10 101.9 kpsi( / 64)(1)AMyI
, 0A AVQ Q 0Ib
2 22 2
max101.9 (0) 50.9 kpsi .
2 2A
A Ans
Element B:
, 0 0B BMy yI
32 3 34 0.54 4 1/12 in3 2 6 6
r r rQ y A
Chapter 3 - Rev. A, Page 33/100
-
34(1000)(1/12) 10 1.698 kpsi( / 64)(1) (1)BVQIb
2
2max
0 1.698 1.698 kpsi .2
Ans
Element C:
34(1000)(10)(0.25) 10 50.93 kpsi( / 64)(1)CMyI
2 2
1 1 1
3/2 3/2 3/22 2 2 2 2 21
1
3/22 21
(2 ) 2
2 23 3
23
r r r
y y y
r
y
Q ydA y x dy y r y dy
r y r r r y
r y
For C, y1 = r /2 =0.25 in
3/22 22 0.5 0.25 0.054133Q in3
2 2 2 212 2 2 0.5 0.25 0.866 inb x r y
34(1000)(0.05413) 10 1.273 kpsi( / 64)(1) (0.866)CVQIb
2
2max
50.93 (1.273) 25.50 kpsi .2
Ans
(b) Neglecting transverse shear stress: Element A: Since the transverse shear stress at point A is zero, there is no change.
max 50.9 kpsi .Ans
% error 0% .Ans
Element B: Since the only stress at point B is transverse shear stress, neglecting the transverse shear stress ignores the entire stress.
2
max0 0 psi .2
Ans
1.698 0% error *(100) 100% .1.698
Ans
Chapter 3 - Rev. A, Page 34/100
-
Element C: 2
max50.93 25.47 kpsi .
2Ans
25.50 25.47% error *(100) 0.12% .25.50
Ans
(c) Repeating the process with different beam lengths produces the results in the table.
Bending stress, kpsi)
Transverse shear stress, kpsi)
Max shear stress,
max kpsi)
Max shear stress,
neglecting max kpsi)
% error
L = 10 in A 102 0 50.9 50.9 0 B 0 1.70 1.70 0 100 C 50.9 1.27 25.50 25.47 0.12 L = 4 in A 40.7 0 20.4 20.4 0 B 0 1.70 1.70 0 100 C 20.4 1.27 10.26 10.19 0.77 L = 1 in A 10.2 0 5.09 5.09 0 B 0 1.70 1.70 0 100 C 5.09 1.27 2.85 2.55 10.6 L = 0.1in A 1.02 0 0.509 0.509 0 B 0 1.70 1.70 0 100 C 0.509 1.27 1.30 0.255 80.4
Discussion:
The transverse shear stress is only significant in determining the critical stress element as the length of the cantilever beam becomes smaller. As this length decreases, bending stress reduces greatly and transverse shear stress stays the same. This causes the critical element location to go from being at point A, on the surface, to point B, in the center. The maximum shear stress is on the outer surface at point A for all cases except L = 0.1 in, where it is at point B at the center. When the critical stress element is at point A, there is no error from neglecting transverse shear stress, since it is zero at that location. Neglecting the transverse shear stress has extreme significance at the stress element at the center at point B, but that location is probably only of practical significance for very short beam lengths.
_____________________________________________________________________________
Chapter 3 - Rev. A, Page 35/100
-
3-46
1
0
cR FlcM Fx x al
2 2
max
66
6 0 .
c l FxMbh bh
Fcxh xlb
a Ans
_____________________________________________________________________________ 3-47
From Problem 3-46, 1 , 0cR F V x al
maxmax
3 3 ( / ) 3 .2 2 2
V c l F Fch Abh bh lb
ns
From Problem 3-46, max
6( ) Fcxh x . lb
Sub in x = e and equate to h above.
max max
max2max
3 62
3 .8
Fc Fcelb lb
Fce Alb
ns
_____________________________________________________________________________ 3-48 (a)
x-z plane
20 1.5(0.5) 2(1.5)sin(30 )(2.25) (3)O zM R
2 1.375 kN .zR Ans
10 1.5 2(1.5)sin(30 ) 1.375z zF R
1 1.625 kN .zR Ans x-y plane
20 2(1.5)cos(30 )(2.25) (3)O yM R
2 1.949 kN .yR Ans
10 2(1.5) cos(30 ) 1.949y yF R
1 0.6491 kN .yR Ans
Chapter 3 - Rev. A, Page 36/100
-
(b)
(c) The transverse shear and bending moments for most points of interest can readily be taken straight from the diagrams. For 1.5 < x < 3, the bending moment equations are parabolic, and are obtained by integrating the linear expressions for shear. For convenience, use a coordinate shift of x = x – 1.5. Then, for 0 < x < 1.5,
2
2
0.125
0.1252
At 0, 0.9375 0.5 0.125 0.9375
z
y z
y y
V x
xM V dx x C
x M C M x x
2
2
1.949 0.6491 1.732 0.64911.1251.732 0.6491
2At 0, 0.9737 0.8662 0.125 0.9375
y
z
z z
V x x
M x x C
x M C M x x
By programming these bending moment equations, we can find My, Mz, and their vector combination at any point along the beam. The maximum combined bending moment is found to be at x = 1.79 m, where M = 1.433 kN·m. The table below shows values at key locations on the shear and bending moment diagrams.
x (m) Vz (kN) Vy (kN) V (kN) My
(kNm) Mz
(kNm) M
(kNm) 0 –1.625 0.6491 1.750 0 0 0
0.5 –1.625 0.6491 1.750 –0.8125 0.3246 0.8749 1.5 –0.1250 0.6491 0.6610 0.9375 0.9737 1.352
1.625 0 0.4327 0.4327 –0.9453 1.041 1.406 1.875 0.2500 0 0.2500 –0.9141 1.095 1.427
3 1.375 –1.949 2.385 0 0 0
Chapter 3 - Rev. A, Page 37/100
-
(d) The bending stress is obtained from Eq. (3-27),
y Az Ax
z y
M zM yI I
The maximum tensile bending stress will be at point A in the cross section of Prob. 3-34 (a), where distances from the neutral axes for both bending moments will be maximum. At A, for Mz, yA = –37.5 mm, and for My, zA = –20 mm.
3 36 4 640(75) 34(25) 1.36(10 ) mm 1.36(10 ) m
12 12zI 4
3 35 4 725(40) 25(6)2 2.67(10 ) mm 2.67(10 ) m
12 12yI
4
It is apparent the maximum bending moment, and thus the maximum stress, will be in the parabolic section of the bending moment diagrams. Programming Eq. (3-27) with the bending moment equations previously derived, the maximum tensile bending stress is found at x = 1.77 m, where My = – 0.9408 kN·m, Mz = 1.075 kN·m, and x = 100.1 MPa. Ans.
_____________________________________________________________________________ 3-49 (a) x-z plane
3 6000 (1000)(4) (10)5 2O Oy
M M
1842.6 lbf in .OyM Ans 3 60 (1000)5 2z Oz
F R 00
175.7 lbf .OzR Ans x-y plane
4 6000 (1000)(4) (10)5 2O Oz
M M
7442.5 lbf in .OzM Ans 4 60 (1000)5 2y Oy
F R 00
1224.3 lbf .OyR Ans
Chapter 3 - Rev. A, Page 38/100
-
(b)
( (c)
1/22 2( ) ( ) ( )y zV x V x V x
1/22 2( ) ( ) ( )y zM x M x M x
x (m) Vz (kN) Vy (kN) V (kN) My (kNm) Mz (kNm) M (kNm)
0 –175.7 1224.3 1237 –1842.6 –7442.6 7667 4 –175.7 1224.3 1237 –2545.4 –2545.4 3600
10 424.3 424.3 600 0 0 0 (d) The maximum tensile bending stress will be at the outer corner of the cross section in
the positive y, negative z quadrant, where y = 1.5 in and z = –1 in. 3 3
42(3) (1.625)(2.625) 2.051 in12 12z
I
3 343(2) (2.625)(1.625) 1.601 in
12 12yI
At x = 0, using Eq. (3-27), yz
xz y
M zM yI I
( 7442.6)(1.5) ( 1842.6)( 1) 6594 psi2.051 1.601x
Check at x = 4 in, ( 2545.4)(1.5) ( 2545.4)( 1) 2706 psi
2.051 1.601x
The critical location is at x = 0, where x = 6594 psi. Ans. _____________________________________________________________________________
Chapter 3 - Rev. A, Page 39/100
-
3-50 The area within the wall median line, Am, is
Square: 2( )mA b t . From Eq. (3-45) 2
sq all all2 2( )mT A t b t t
Round: 2( ) /mA b t 4
2rd all2 ( ) / 4T b t t
Ratio of Torques
2sq all
2rd all
2( ) 4 1.27( ) / 2
T b t tT b t t
Twist per unit length from Eq. (3-46) is
all all1 2 2
24 4 2
m m m m
m m m
TL A t L L LCGA t GA t G A A
m
m
Square:
sq 24( )( )
b tCb t
Round:
rd 2 2( ) 4(
( ) / 4 ( )b t b tC C
b t b t)
Ratio equals 1. Twists are the same.
_____________________________________________________________________________ 3-51
(a) The area enclosed by the section median line is Am = (1 0.0625)2 = 0.8789 in2 and the length of the section median line is Lm = 4(1 0.0625) = 3.75 in. From Eq. (3-45),
2 2(0.8789)(0.0625)(12 000) 1318 lbf in .mT A t Ans
From Eq. (3-46),
1 2 6 2
(1318)(3.75) 360.0801 rad 4.59 .
4 4 11.5 10 (0.8789) 0.0625m
m
TL ll AGA t
ns
(b) The radius at the median line is rm = 0.125 + (0.5)(0.0625) = 0.15625 in. The area enclosed by the section median line is Am = (1 0.0625)2 – 4(0.15625)2 + 4(π /4)(0.15625)2 = 0.8579 in2. The length of the section median line is Lm = 4[1 – 0.0625 – 2(0.15625)] + 2π(0.15625) = 3.482 in.
Chapter 3 - Rev. A, Page 40/100
-
From Eq. (3-45), 2 2(0.8579)(0.0625)(12 000) 1287 lbf in .mT A t Ans
From Eq. (3-46),
1 2 6 2
(1287)(3.482) 360.0762 rad 4.37 .
4 4 11.5 10 (0.8579) 0.0625m
m
TL ll AGA t
ns
_____________________________________________________________________________
3-52
31
1 33
3i i
ii i
T GTGL c
i
L c
331
1 2 31
.3 i iiGT T T T L c Ans
From Eq. (3-47), G1c G and 1 are constant, therefore the largest shear stress occurs when c is a maximum.
max 1 max .G c Ans _____________________________________________________________________________
3-53 (b) Solve part (b) first since the twist is needed for part (a).
max allow 12 6.89 82.7 MPa
6max
1 9max
82.7 100.348 rad/m .
79.3 10 (0.003)Ans
Gc
(a) 9 331 1 1
1
0.348(79.3) 10 (0.020)(0.002 )1.47 N m .
3 3GL cT A ns
9 332 2 2
2
9 333 3 3
3
1 2 3
0.348(79.3) 10 (0.030)(0.003 )7.45 N m .
3 30.348(79.3) 10 (0)(0 )
0 .3 3
1.47 7.45 0 8.92 N m .
GL cT A
GL cT A
T T T T Ans
ns
ns
_____________________________________________________________________________
Chapter 3 - Rev. A, Page 41/100
-
3-54 (b) Solve part (b) first since the twist is needed for part (a).
3max
1 6max
12000 8.35 10 rad/in .11.5 10 (0.125)
AnsGc
(a)
3 6 331 1 1
1
3 6 332 2 2
2
3 6 333 3 3
3
1 2 3
8.35 10 11.5 10 0.75 0.06255.86 lbf in .
3 38.35 10 11.5 10 1 0.125
62.52 lbf in .3 3
8.35 10 11.5 10 0.625 0.06254.88 lbf in .
3 35.86 62.52 4
GL cT A
GL cT A
GL cT A
T T T T
.88 73.3 lbf in .Ans
ns
ns
ns
_____________________________________________________________________________
3-55 (b) Solve part (b) first since the twist is needed for part (a).
max allow 12 6.89 82.7 MPa
6max
1 9max
82.7 100.348 rad/m .
79.3 10 (0.003)Ans
Gc
(a) 9 331 1 1
1
0.348(79.3) 10 (0.020)(0.002 )1.47 N m .
3 3GL cT A ns
9 332 2 2
2
9 333 3 3
3
1 2 3
0.348(79.3) 10 (0.030)(0.003 )7.45 N m .
3 30.348(79.3) 10 (0.025)(0.002 )
1.84 N m .3 3
1.47 7.45 1.84 10.8 N m .
GL cT A
GL cT A
T T T T Ans
ns
ns
_____________________________________________________________________________
3-56 (a) From Eq. (3-40), with two 2-mm strips,
6 22max
max
80 10 0.030 0.0023.08 N m
3 1.8 / ( / ) 3 1.8 / 0.030 / 0.0022(3.08) 6.16 N m .
bcTb c
T Ans
Chapter 3 - Rev. A, Page 42/100
-
From the table on p. 102, with b/c = 30/2 = 15, and has a value between 0.313 and 0.333. From Eq. (3-40),
1 0.3213 1.8 / (30 / 2)
From Eq. (3-41),
3 3 93.08(0.3) 0.151 rad .
0.321 0.030 0.002 79.3 10
6.16 40.8 N m .0.151t
Tl Ansbc G
Tk Ans
From Eq. (3-40), with a single 4-mm strip,
6 22max
max
80 10 0.030 0.00411.9 N m .
3 1.8 / ( / ) 3 1.8 / 0.030 / 0.004bcT A
b c
ns
Interpolating from the table on p. 102, with b/c = 30/4 = 7.5,
7.5 6 (0.307 0.299) 0.299 0.3058 6
From Eq. (3-41)
3 3 911.9(0.3) 0.0769 rad .
0.305 0.030 0.004 79.3 10
11.9 155 N m .0.0769t
Tl Ansbc G
Tk Ans
(b) From Eq. (3-47), with two 2-mm strips,
2 62
max
0.030 0.002 80 103.20 N m
3 32(3.20) 6.40 N m .
LcT
T Ans
3 3 93 3(3.20)(0.3) 0.151 rad .
0.030 0.002 79.3 10
6.40 0.151 42.4 N m .t
Tl AnsLc G
k T Ans
From Eq. (3-47), with a single 4-mm strip,
2 62max
0.030 0.004 80 1012.8 N m .
3 3LcT A ns
Chapter 3 - Rev. A, Page 43/100
-
3 3 93 3(12.8)(0.3) 0.0757 rad .
0.030 0.004 79.3 10
12.8 0.0757 169 N m .t
Tl AnsLc G
k T Ans
The results for the spring constants when using Eq. (3-47) are slightly larger than when using Eq. (3-40) and Eq. (3-41) because the strips are not infinitesimally thin (i.e. b/c does not equal infinity). The spring constants when considering one solid strip are significantly larger (almost four times larger) than when considering two thin strips because two thin strips would be able to slip along the center plane.
_____________________________________________________________________________ 3-57
(a) Obtain the torque from the given power and speed using Eq. (3-44).
(40000)9.55 9.55 152.8 N m2500
HTn
max 316Tr T
J d
1 31 3
6max
16 152.816 0.0223 m 22.3 mm .70 10
Td A
ns
(b) (40000)9.55 9.55 1528 N m250
HTn
1 3
616(1528) 0.0481 m 48.1 mm .
70 10d A
ns
_____________________________________________________________________________ 3-58
(a) Obtain the torque from the given power and speed using Eq. (3-42). 63025 63025(50) 1261 lbf in
2500HT
n
max 316Tr T
J d
1 31 3
max
16 126116 0.685 in .(20000)
Td A
ns
(b) 63025 63025(50) 12610 lbf in
250HT
n
1 316(12610) 1.48 in .
(20000)d A
ns
_____________________________________________________________________________
Chapter 3 - Rev. A, Page 44/100
-
3-59 6 33max
max 3
50 10 0.0316 265 N m16 16
dT Td
Eq. (3-44), 3265(2000) 55.5 10 W 55.5 kW .9.55 9.55
TnH A ns _____________________________________________________________________________
3-60
3 6 33
4 94
16 110 10 0.020 173 N m16 16
0.020 79.3 10 15180
32 32(173)1.89 m .
T T dd
Tl d GlJG T
l Ans
_____________________________________________________________________________ 3-61
3 33
4 4 6
16 30 000 0.75 2485 lbf in16 16
32 32(2485)(24) 0.167 rad 9.57 .0.75 11.5 10
T T dd
Tl Tl AnsJG d G
_____________________________________________________________________________ 3-62
(a) 4 4
max max max maxsolid hollow
( ) 16 16
o o
o o
J d J d dT Tr d r d
4i
44solid hollow
4 4solid
36% (100%) (100%) (100%) 65.6% .
40i
o
T T dT AT d
ns
(b) 2 2solid hollow, o oW kd W k d d 2i
22
solid hollow2 2
solid
36% (100%) (100%) (100%) 81.0% .
40i
o
W W dW AW d
ns
_____________________________________________________________________________ 3-63
(a) 444 maxmax max max
solid hollow 16 16
d xdJ d JT Tr d r d
44solid hollow
4soli
( )% (100%) (100%) (100%) .d
T T xdT xT d
Ans
Chapter 3 - Rev. A, Page 45/100
-
(b) 22 2solid hollow W kd W k d xd 2 2solid hollow
2solid
% (100%) (100%) (100%) .xdW WW x
W d
Ans
Plot %T and %W versus x.
The value of greatest difference in percent reduction of weight and torque is 25% and occurs at 2 2x .
_____________________________________________________________________________ 3-64
(a)
46344
2.8149 104200 2 120 10
32 0.70
dTcJ dd d
1 34
26
2.8149 106.17 10 m 61.7 mm
120(10 )d
d From Table A-17, the next preferred size is d = 80 mm. Ans.
i = 0.7d = 56 mm. The next preferred size smaller is di = 50 mm Ans.
(b)
4 4 44
4200 2 4200 0.050 230.8 MPa .
32 0.080 0.05032i
i
dTc AnsJ d d
_____________________________________________________________________________
Chapter 3 - Rev. A, Page 46/100
-
3-65
(1500)9.55 9.55 1433 N m10
HTn
1 31 3
3 6
16 143316 16 = 0.045 m 45 mm 80 10
CC
T Tdd
From Table A-17, select 50 mm. Ans.
(a)
6start 3
16 2 1433117 10 Pa 117 MPa .
0.050Ans
(b) Design activity _____________________________________________________________________________
3-66
1 31 3
3
63 025 63 025(1) 7880 lbf in8
16 788016 16 = 1.39 in 15 000CC
HTn
T Tdd
From Table A-17, select 1.40 in. Ans. _____________________________________________________________________________
3-67 For a square cross section with side length b, and a circular section with diameter d, 2 2
square circular 4 2A A b d b d
From Eq. (3-40) with b = c,
3
max 2 3 3square
1.8 1.8 23 3 (4.8) 6.896/ 1
T T Tbc b c b d d
3
T
For the circular cross section,
max 3 3circular16 5.093T T
d d
3max square
max circular3
6.8961.354
5.093
TdTd
The shear stress in the square cross section is 35.4% greater. Ans. (b) For the square cross section, from the table on p. 102, β = 0.141. From Eq. (3-41),
Chapter 3 - Rev. A, Page 47/100
-
square 43 4 11.50
0.1412
Tl Tl Tl Tlbc G b G d G
d G
4
For the circular cross section,
4410.19
32rdTl Tl TlGJ d GG d
4
4
11.501.129
10.19
sq
rd
Tld GTl
d G
The angle of twist in the square cross section is 12.9% greater. Ans. _____________________________________________________________________________
3-68 (a)
1 2
2 1 2 2
2 2
1
0.150 (500 75)(4) 5 1700 0.15 5