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INSTRUCTOR'S SOLUTIONS MANUAL
TO ACCOMPANY
MECHANICS
OF MATERIALS
EIGHTH EDITION
JAMES M. GERE
BARRY J. GOODNO
Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States
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Contents
1. Tension, Compression, and Shear
1
2. Axially Loaded Members 117
3. Torsion
283
4. Shear Forces and Bending Moments 385
5. Stresses in Beams (Basic Topics) 435
6. Stresses in Beams (Advanced Topics) 557
7. Analysis of Stress and Strain 637
8. Applications of Plane Stress (Pressure Vessels, Beams, and Combined Loadings) 725
9. Deflections of Beams 793
10. Statically Indeterminate Beams 885
11. Columns
943
12. Review of Centroids and Moments of Inertia 1025
Answers to Problems 1057
Appendix A: FE Exam Review Problems 1083
v
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Answers to Problems
CHAPTER 1
1.2-1
1.2-2
1.2-3
1.2-4
1.2-5
1.2-6
1.2-7
1.2-8
1.2-9
1.2-10
1.2-11
1.2-12
1.2-13
1.2-14
1.2-15
1.2-16
(a) Ay 5 lb, By 5 lb, Cx 50 lb, Cy 0;
(b) N 50 lb, V 5 lb, M 75 ft-lb
(a) MA 0, Cy 236 N, Dy 75.6 N;
(b) N 0, V 70 N, M 36.7 Nm;
(c) MA 0, Cy 236 N, Dy 75.6 N;
N 0, V 70 N, M 36.7 Nm
(a) Ax 12.55 lb, Ay 15 lb, Cy 104.3 lb,
Dx 11.45 lb, Dy 19.83 lb;
(b) ResultantB 19.56 lb; (c) Ax 42.7 lb,
Ay 37.2 lb, MA 522 lb-ft, Dx 18.67 lb,
Dy 32.3 lb, ResultantB 56.6 lb
(a) R3x 40 N, R3y 25 N, R5x 20 N;
(b) F11 0, F13 28.3 N
(a) Ax 0, Ay 1.0 kip, Ey 5 kips;
(b) FFE 1.898 kips
(a) Fx 0, Fy 12.0 kN, Dy 6.0 kN; (b) FFE 0
(a) Bx 0.8 P, Bz 2.0 P, Oz 1.25 P;
(b) FAC 0.960 P
(a) Ax 1.25P, By 0, Bz P;
(b) FAB 1.601P
(a) Ay 4.67P, Az 4.0P; (b) FAB 8.33P
(a) Az 0, Bx 3.75 kN; (b) FAB 6.73 kN
(a) TA 11,000 lb-in.;
(b) T(L1/2) TA 11,000 lb-in.,
T (L1 L2/2) T2 10,000 lb-in.
(a) TA 1225 Nm; (b) T(L1/2) 62.5 Nm,
T (L1 L2/2) T2 1100 Nm
(a) Ax 540 lb, Ay 55.6 lb,
MA 4320 lb-ft, Cy 55.6 lb;
(b) N 55.6 lb, V 506 lb, M 2374 lb-ft
(a) Ax 280 N, Ay 8.89 N, MA 1120 Nm,
Dy 151.1 N; (b) ResultantB 280 N
(a) Ax 30 lb, Ay 140 lb, Cx 30 lb,
Cy 60 lb; (b) N 23.3 lb, V 20 lb,
M 33.3 lb-ft
(a) Ax 10.98 kN, Ay 29.0 kN,
Ex 8.05 kN, Ey 22 kN;
(b) ResultantC 23.4 kN
1.2-17
1.2-18
1.2-19
1.2-20
1.2-21
1.2-22
1.2-23
1.2-24
1.2-25
1.2-26
1.3-1
1.3-2
1.3-3
1.3-4
1.3-5
(a) Ay 1250 lb, Ex 0, Ey 1750 lb;
(b) N 1750 lb, V 500 lb, M 575 lb-ft
(a) Ax 320 N, Ay 240 N, Cy 192 N,
Ey 192 N; (b) N 312 N, V 57.9 N,
M 289 Nm; (c) ResultantC 400 N
(a) Ax 28.9 lb, Ay 50.0 lb,
Bx 65.0 lb; (b) Fcable 71.6 lb
(a) Ax 10 kN, Ay 2.17 kN,
Cy 9.83 kN, Ey 1.333 kN;
(b) ResultantD 12.68 kN
(a) Ox 48.3 lb, Oy 40 lb, Oz 12.94 lb,
MOx 331 lb-in., MOy 690 lb-in.,
MOz 338 lb-in.; (b) N 40 lb,
V 50 lb, T 690 lb-in., M 473 lb-in.
(a) Ay 120 N, Az 60 N,
MAx 70 Nm, MAy 142.5 Nm,
MAz 180 Nm, Dx 60 N,
Dy 120 N, Dz 30 N; (b) N 120 N,
V 41.3 N, T 142.5 Nm, M 180.7 Nm
(a) Ax 5.77 lb, Ay 47.3 lb,
Az 2.31 lb, MAz 200 lb-in.;
(b) TDC 3.81 lb, TEC 6.79 lb
Cx 120 N, Cy 160 N, Cz 506 N,
Dz 466 N, Hy 320 N, Hz 499 N
Ay 57.2 lb, Bx 44.2 lb (to the left),
By 112.4 lb, Cx 28.8 lb, Cy 5.88 lb
(a) HB 104.6 N, VB 516 N, VF 336 N;
(b) N 646 N, V 176.8 N, M 44.9 kNm
(a) sAB 1443 psi; (b) P2 1487.5 lbs;
(c) tBC 0.5 in.
(a) s 130.2 MPa; (b) 4.652 * 104
(a) RB lb (cantilever), 191.3 lb (V-brakes);
sC 144 psi (cantilever), 306 psi (V-brakes);
(b) scable 26,946 psi (both)
(a) s 3.101 * 104; (b) d 0.1526 mm;
(c) Pmax 89.5 kN
(a) sC 2.46 ksi; (b) xC 19.56 in.,
yC 19.56 in.
1057
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Answers to Problems
(a) st 132.7 MPa; (b) amax 34.4
(a) s1 34.4 ksi, s2 30.6 ksi;
(b) d1new 3.18 102 in.;
(c) s1 19.6 ksi, s2 18.78 ksi, s3 22.7 ksi
1.3-8
sC 5.21 MPa
1.3-9
(a) T 184 lb, s 10.8 ksi; (b) cable 5 104
1.3-10 (a) T 819 N, s 74.5 MPa;
(b) cable 4.923 104
1.3-6
1.3-7
5877
48,975
1.3-11 (a) T £ 4679 ≥ lb; (b) s £ 38,992 ≥ psi;
7159
59,658
4278
35,650
6461
53,842
(c) T §
¥ psi
¥ lb, s §
27,842
3341
35,650
4278
(a) sx gv2(L2 x2)/2g; (b) smax gv2L2/2g
(a) TAB 1620 lb, TBC 1536 lb,
TCD 1640 lb; (b) sAB 13,501 psi,
sBC 12,799 psi, sCD 13,667 psi
1.3-14 (a) TAQ TBQ 50.5 kN; (b) s 166 MPa
1.4-1
(a) Lmax 11,800 ft; (b) Lmax 13,500 ft
1.4-2
(a) Lmax 7143 m; (b) Lmax 8209 m
1.4-3
% elongation 6.5, 24.0, 39.0;
% reduction 8.1, 37.9, 74.9;
Brittle, ductile, ductile
1.4-4
11.9 103 m; 12.7 103 m; 6.1 103 m;
6.5 103 m; 23.9 103 m
1.4-5
s 52.3 ksi
1.4-6
spl ⬇ 47 MPa, Slope ⬇ 2.4 GPa,
s g ⬇ 53 MPa; Brittle
1.4-7
spl ⬇ 65,000 psi, Slope ⬇ 30 106 psi, s Y ⬇
69,000 psi, sU ⬇ 113,000 psi; Elongation 6%,
Reduction 31%
1.5-1
0.13 in. longer
1.5-2
4.0 mm longer
1.5-3
(a) dpset 1.596 in.; (b) sB 30 ksi
1.5-4
(a) dpset 4.28 mm; (b) sB 65.6 MPa
1.5-5
(b) 0.71 in.; (c) 0.58 in.; (d) 49 ksi
1.6-1
Pmax 157 k
1.6-2
P 27.4 kN (tension)
1.6-3
P 15.708 kips
1.6-4
(a) P 74.1 kN; (b) d # L 0.469 mm
1.6-5
1.6-6
1.6-7
1.6-8
1.7-1
1.7-2
1.7-3
1.7-4
1.7-5
1.7-6
1.3-12
1.3-13
shortening;
Af A
A
0.081%,
¢V1 V1f Vol 1 207 mm3;
1.7-7
1.7-8
1.7-9
1.7-10
1.7-11
1.7-12
1.7-13
1.7-14
1.7-15
d 1.56 104 in., P 2.154 kips
(a) E 104 GPa; (b) 0.34
(a) dBC inner 8 104 in.;
(b) brass 0.34; (c) tAB 2.732 104 in.,
dAB inner 1.366 104 in.
(a) L1 12.66 mm; L2 5.06 mm;
L3 3.8 mm; (b) Vol1 21,548 mm3;
Vol2 21,601 mm3; Vol3 21,610 mm3
sb 7.04 ksi, tave 10.756 ksi
sb 139.86 MPa; Pult 144.45 kN
(a) t 12.732 ksi; (b) sbf 20 ksi,
sbg 26.667 ksi
(a) Bx 252.8 N, Ax Bx,
Ay 1150.1 N; (b) Aresultant 1178 N;
(c) t 5.86 MPa, sbshoe 7.36 MPa
(a) tmax 2979 psi; (b) sbmax 936 psi
T1 13.176 kN, T2 10.772 kN,
t1ave 25.888 MPa, t2ave 21.166 MPa,
sb1 9.15 MPa, sb2 7.48 MPa
(a) Resultant 1097 lb; (b) sb 4999 psi;
(c) tnut 2793 psi, tpl 609 psi
G 2.5 MPa
(a) gaver 0.004; (b) V 89.6 k
(a) gaver 0.50; (b) d 4.92 mm
(a) sb 69.5 ksi, sbrg 39.1 ksi, tf 21 ksi;
(b) sb 60.4 ksi, sbrg 34 ksi, tf 18.3 ksi
taver 42.9 MPa
(a) Ax 0, Ay 170 lb, MA 4585 in.-lb;
(b) Bx 253.6 lb, By 160 lb, Bres 299.8 lb,
Cx Bx ; (c) tB 3054 psi, tC 1653 psi;
(d) sbB 4797 psi, sbC 3246 psi
For a bicycle wih L/R 1.8: (a) T 1440 N;
(b) taver 147 MPa
P
b
P
(a) t ; (b) d ln
2prh
2p hG
d
(a) t1 2.95 MPa , t4 0;
(b) sb1 1.985 MPa, sb4 0;
(c) sb4 41 MPa; (d) t 10.62 MPa;
(e) s3 75.1 MPa
1.7-17 (a) Ox 12.68 lb, Oy 1.294 lb,
Ores 12.74 lb; (b) tO 519 psi,
sbO 816 psi; (c) t 362 psi
1.7-18 (a) Fx 153.9 N, s 3.06 MPa;
(b) tave 1.96 MPa; (c) sb 1.924 MPa
1.7-19 (a) P 395 lb; (b) Cx 374 lb,
Cy 237 lb, Cres 443 lb; (c) t 18.04 ksi,
sbC 4.72 ksi
1.8-1
Pallow 3140 lb
1.7-16
(c) d3 65.4 mm
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Answers to Problems
1.8-2
1.8-3
1.8-4
1.8-5
1.8-6
1.8-7
1.8-8
1.8-9
1.8-10
1.8-11
1.8-12
1.8-13
1.8-14
1.8-15
1.8-16
1.9-1
1.9-2
1.9-3
1.9-4
1.9-5
1.9-6
1.9-7
1.9-8
1.9-9
1.9-10
1.9-11
1.9-12
1.9-13
1.9-14
1.9-15
Tmax 33.4 kN.m
Pallow 607 lb
(a) Pallow 8.74 kN; (b) Pallow 8.69 kN;
(c) Pallow 21.2 kN, Pallow 8.69 kN
(shear controls)
P 294 k
(a) F 1.171 kN; (b) Shear: Fa 2.86 kN
Wmax 5110 lb
(a) FA 22T, FB 2T, FC T;
(b) Shear at A: Wmax 66.5 kN
Pa 10.21 kips
Cult 5739 N: Pmax 445 N
Wmax 0.305 kips
Shear in rivets in CG & CD controls:
Pallow 45.8 kN
(a) Pa sa (0.587d 2); (b) Pa 21.6 kips
Pallow 96.5 kN
Pmax 11.98 psf
(a) Pallow sc (pd 2/4) 21 (R/L)2;
(b) Pallow 9.77 kN
(a) dmin 3.75 in; (b) dmin 4.01 in.
(a) dmin 164.6 mm; (b) dmin 170.9 mm
(a) dmin 0.704 in.; (b) dmin 0.711 in.
dmin 63.3 mm
dpin 1.029 in.
(b) Amin 435 mm2
dmin 0.372 in.
dmin 5.96 mm
n 11.6, or 12 bolts
(d2)min 131 mm
Ac 1.189 in.2
(a) tmin 18.8 mm, use t 20 mm;
(b) Dmin 297 mm
(a) sDF 10.38 ksi s allow,
sbF 378 psi sba; (b) new sBC 25 ksi, so
increase rod BC to 14-in. diameter; required
5
diameter of washer 1 in. 1.312 in.
16
(a) dm 24.7 mm; (b) Pmax 49.4 kN
u arccos 1/ 23 54.7
CHAPTER 2
4W
6W
; (b) d 5k
5k
2.2-1
(a) d 2.2-2
(a) d 12.5 mm; (b) n 5.8
da
Es
da
Es
30
; (b)
1.651;
ds
Ea
11
ds
C Ea
La
Ea
(c)
1.5
0.55;
Ls
Es
Es
(d) E 1 17,647 ksi (cast iron or copper
1.7
alloy) (see App. I)
2.2-3
(a)
2.2-4
2.2-5
2.2-6
h 13.4 mm
h L prmaxd 2/4k
(a) x 102.6 mm; (b) x 205 mm;
(c) Pmax 12.51 N; (d) uinit 1.325;
(e) P 20.4 N
26P
(a) d4 ;
3k
2.2-7
(b) d4 2.2-8
2.2-9
2.2-10
2.2-11
2.2-12
2.2-13
1059
104P
15
, ratio 3.75
45k
4
(a) dB 1.827 mm; (b) Pmax 390 kN;
(c) dBx 6.71 mm, Pmax 106.1 kN
Pmax 72.3 lb
(a) x 134.7 mm; (b) k1 0.204 N/mm;
(c) b 74.1 mm; (d) k3 0.638 N/mm
(a) tc,min 0.021 in.; (b) dr 0.031 in.;
(c) hmin 0.051 in.
dA 0.200 mm, dD 0.880 mm
L1
P
27
(a) dD (28f2 9f1); (b)
16
L2
16
d1
365L
(c)
1.225; (d) x d2
236
(a) u 35.1, d 44.6 mm, RA 25 N,
RC 25 N; (b) u 43.3, d 8.19 mm,
RA 31.5 N, RC 18.5 N,
MA 1.882 Nm
2.2-15 (a) u 35.1, d 1.782 in., RA 5 lb,
RC 5 lb; (b) u 43.3, d 0.327 in.,
RA 6.3 lb, RC 3.71 lb ,
MA 1.252 lb-ft
2.3-1
(a) d 0.0276 in.; (b) dB 1.074 in.
2.3-2
(a) d 0.675 mm; (b) Pmax 267 kN
2.3-3
(a) d 0.01125 in. (elongation); (b) So new value
of P3 is 1690 lb, an increase of 390 lb.
(c) AAB 0.78 in.2
7PL
2.3-4
(a) d ; (b) d 0.5 mm; (c) Lslot 244 mm
6Ebt
2.2-14
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Answers to Problems
7PL
; (b) d 0.021 in.; (c) Lslot 10 in.
6Ebt
2.3-5
(a) d 2.3-6
2.3-7
(a) dAC 3.72 mm; (b) P0 44.2 kN
(a) da 0.0589 in.; (b) db 0.0501 in.;
dc
dc
(c)
0.58,
0.681
da
db
2.3-8
2.3-9
Page 1060
2.4-8
2.4-9
2.4-10
(a) dmax 23.9 mm; (b) b 4.16 mm;
(c) x 183.3 mm
P y
PL
(a) d ; (b) s(y) a b ;
2EA
A L
(c) d y
P y
PL 2
a b, s(y) c a 2 b d
EA 3
A L
L
(a) d2-4 0.024 mm; (b) Pmax 8.15 kN;
(c) L2 9.16 mm
2.3-11 (a) R1 3P/2; (b) N1 3P/2 (tension),
N2 P/2 (tension); (c) x L/3;
(d) d2 2PL/3EA; (e) b 1/11
2
2
2.3-12 (a) dc W(L h )/2EAL;
(b) dB WL/2EA; (c) b 3;
WL
(d) d 359 mm (in sea water);
2EA
WL
d
412 mm (in air)
2EA
d (x) c
2.3-10
2.3-13
2.3-14
2.3-15
2.3-16
2.3-17
2.4-1
2.4-2
2.4-3
2.4-4
2.4-5
2.4-6
2.4-7
(a) RA 10.5 kN (to the left), RD 2.0 kN (to the
right); (b) FBC 15.0 kN (compression)
(b) sa 1610 psi (compression),
ss 9350 psi (tension)
(a) P 13.73 kN, R1 9.07 kN,
R2 4.66 kN, s2 7 MPa;
(b) dcap 190.9 mm, Axial Force Diagram:
N(x) R2 if x L2, N(x) R1
if x L2; Axial Displacement Diagram:
R2
(x) d if x L2,
d(x) c
EA2
2.4-11
2.4-12
2.4-13
2.4-14
(b) d 0.010 in.
d 2PH/3Eb2
d 2WL/pd 2E
(a) d 2.18 mm; (b) d 6.74 mm
(b) d 11.14 ft
(a) P 1330 lb; (b) Pallow 1300 lb
(a) P 104 kN; (b) Pmax 116 kN
(a) PB /P 3/11; (b) sB /sA 1/2; (c) Ratio 1
(a) If x L/2, RA (3PL)/(2(x 3L)),
RB P(2x 3L)/(2(x 3L)). If x L/2,
RA (P(x L))/(x 3L),
RB (2PL)/(x 3L). (b) If x L/2,
d PL(2x 3L)/[(x 3L)Epd 2]. If
x L/2, d 8PL(x L)/[3(x 3L)Epd 2].
(c) x 3L/10 or x 2L/3;
(d) RB rgpd 2L/8, RA 3 rgpd2L/32
(a) 41.7%; (b) sM 32.7 ksi, sO 51.4 ksi
(a) d 1.91 mm; (b) d 1.36 mm; (c) d 2.74 mm
(a) RA 2P/3, RE 5P/3;
LP
LP
5LP
(b) dB ;
,d ,d 6EA C 6EA D 6EA
5LP
(c) dmax (to the right), dA dE 0
6EA
2.4-15
(d) Pmax 12.37 kip
2.5-14
2.4-16
2.4-17
2.5-1
2.5-2
2.5-3
2.5-4
2.5-5
2.5-6
2.5-7
2.5-8
2.5-9
2.5-10
2.5-11
2.5-12
2.5-13
R2L 2
R1
+
(x L 2) d if x L2;
EA2
EA1
(c) q 1.552 kN/m
(a) P1 PE1/(E1 E2);
(b) e b(E2 E1)/[2(E2 E1)];
(c) s1/s2 E1/E2
(a) Pallow 1504 N; (b) Pallow 820 N;
(c) Pallow 703 N
d2 0.338 in., L2 48.0 in.
(a) Ax 41.2 kN, Ay 71.4 kN,
Bx 329 kN, By 256 kN;
(b) Pmax 233 kN
(a) sc 10,000 psi, sD 12,500 psi;
(b) dB 0.0198 in.
Pmax 1800 N
ss 3.22 ksi, sb 1.716 ksi, sc 1.93 ksi
s 11,700 psi
T 40.3C
T 185F
(a) T 24C, srod 57.6 MPa;
(b) Clevis: sbc 42.4 MPa, Washer:
sbw 74.1 MPa; (c) db 10.68 mm
(a) sc Ea ( TB)/4;
(b) sc Ea( TB)/[4(EA/kL 1)]
(a) N 51.8 kN, max. sc 26.4 MPa,
dC 0.314 mm; (b) N 31.2 kN,
max. sc 15.91 MPa, dC 0.546 mm
d 0.123 in.
T 34C
t 15.0 ksi
Pallow 39.5 kN
(a) TA 400 lb, TB 200 lb;
(b) TA 454 lb, TB 92 lb; (c) T 153F
(a) s 98 MPa; (b) T 35 C
(a) s 957 psi; (b) Fk 3006 lbs (C);
(c) s 2560 psi
s PL/6EA
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Answers to Problems
2.5-15
2.5-16
2.5-17
2.5-18
2.5-19
2.5-20
2.5-21
2.5-22
2.5-23
2.5-24
2.5-25
2.6-1
2.6-2
2.6-3
2.6-4
2.6-5
2.6-6
2.6-7
2.6-8
2.6-9
2.6-10
(a) P1 231 k, RA 55.2 k, RB 55.2 k;
(b) P2 145.1 k, RA 55.2 k, RB 55.2 k;
(c)For P1, tmax 13.39 ksi, for P2,
tmax 19.44 ksi; (d) T 65.8F, RA 0, RB 0;
(e) RA 55.2 k, RB 55.2 k
(a) RA [s a T(L1 L2)]/[(L1/EA1)
(L2/EA2) (L/k3)], RD RA;
(b) dB a T(L1) RA(L1/EA1),
dC a T(L1 L2) RA[(L1/EA1) L2/EA2)]
TB 660 lb, TC 780 lb
Pallow 1.8 MN
(a) sp 0.196 ksi, sr 3.42 ksi;
(b) sb 2.74 ksi, tc 0.285 ksi
sp 25.0 MPa
sp 2400 psi
(a) PB 25.4 kN, Ps PB;
(b) Sreqd 25.7 mm; (c) dfinal 0.35 mm
(a) Fx 0.174 k; (b) F1 0.174 k;
(c) Lf 12.01 in.; (d) T 141.9F
sa 500 MPa (tension), sc 10 MPa (compression)
(a) Fk 0.174 k; (b) Ft 0.174 k;
(c) Lf 11.99 in.; (d) T 141.6F
Pmax 42,600 lb
dmin 6.81 mm
Pmax 24,000 lb
(a) Tmax 46C; (b) T 9.93C
(a) tmax 10,800 psi; (b) Tmax 49.9F;
(c) T 75.9F
(a) sx 84 MPa; (b) tmax 42 MPa;
(c) On rotated x face: sx1 42 MPa,
tx1y1 42 MPa; On rotated y face:
sy1 42 MPa, (d) On rotated x face:
sx1 71.7 MPa, txlyl 29.7 MPa
On rotated y face: sy1 12.3 MPa
(a) smax 18,000 psi; (b) tmax 9000 psi
(a) Element A: sx 105 MPa (compression),
Element B: tmax 52.5 MPa; (b) u 33.1
sAC
(a) tmaxAC 1.859 ksi,
2
sAB
7.42 ksi,
tmaxAB 2
sBC
9.41 ksi;
tmaxDC 2
(b) Pmax 36.5 kip
(a) (1) sx 945 kPa; (2) su 807 kPa,
tu 334 kPa; (3) su 472 kPa,
tu 472 kPa, smax 945 kPa,
tmax 472 kPa; (b) smax 378 kPa,
tmax 189 kPa
2.6-11
2.6-12
2.6-13
2.6-14
2.6-15
2.6-16
2.6-17
2.6-18
2.6-19
2.7-1
2.7-2
2.7-3
2.7-4
2.7-5
2.7-6
2.7-7
2.7-8
2.7-9
1061
(a) tpq 11.54 psi; (b) spq 1700 psi, s(pq p/2) 784 psi; (c) Pmax 14,688 lb
(a) Tmax 31.3C; (b) spq 21.0 MPa
(compression), tpq 30 MPa (CCW);
(c) b 0.62
NAC 5.77 kips; dmin 1.08 in.
(a) su 0.57 MPa, tu 1.58 MPa;
(b) a 33.3; (c) a 26.6
(a) u 35.26, t0 7070 psi;
(b) smax 15,000 psi, tmax 7500 psi
su1 54.9 MPa, su2 18.3 MPa,
tu 31.7 MPa
smax 10,000 psi, tmax 5000 psi
(a) u 30.96; (b) Pmax 1.53 kN
(a) tu 348 psi, u 20.1;
(b) sx1 950 psi, sy1 127.6 psi;
(c) kmax 15,625 lb/in.; (d) Lmax 1.736 ft;
(e) Tmax 92.8F
(a) U 23P2L/12EA; (b) U 125 in.-lb
(a) U 5P2L/4pEd 2; (b) U 1.036 J
U 5040 in.-lb
(c) U P2L/2EA PQL/2EA Q2L/4EA
Aluminum: 171 psi, 1740 in.
(a) U P2L/EA; (b) dB 2PL/EA
(a) U1 0.0375 in.-lb; (b) U2 2.57 in.-lb;
(c) U3 2.22 in.-lb
(a) U 5kd 2; (b) d W/10k;
(c) F1 3W/10, F2 3W/20, F3 W/10
b2
P 2L
(a) U ln ;
2Et(b2 b1)
b1
(b) d b2
PL
ln
Et(b2 b1)
b1
(a) P1 270 kN; (b) d 1.321 mm;
(c) U 243 J
2.7-11 (a) x 2s, P 2(k1 k2)s;
(b) U1 (2k1 k2)s2
2.7-12 (a) U 6.55 J; (b) dC 168.8 mm
2.8-1
(a) dmax 0.0361 in.; (b) smax 22,600 psi;
(c) Impact factor 113
2.8-2
(a) dmax 6.33 mm; (b) smax 359 MPa;
(c) Impact factor 160
2.8-3
(a) dmax 0.0312 in.; (b) smax 26,000 psi;
(c) Impact factor 130
2.8-4
(a) dmax 215 mm; (b) Impact factor 3.9
2.8-5
(a) dmax 9.21 in.; (b) Impact factor 4.6
2.8-6
v 13.1 m/s
2.8-7
hmax 8.55 in.
2.8-8
Lmin 9.25 m
2.7-10
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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10:47 AM
Page 1062
Answers to Problems
Lmin 500 in.
Vmax 5.40 m/s
dmax 11.0 in.
L 25.5 m
(a) Impact factor 1 (1 2EA/W)1/ 2; (b) 10
smax 33.3 MPa
(a) smax ⬇ 6.2 ksi and 6.9 ksi;
(b) smax ⬇ 11.0 ksi and 9.0 ksi
2.10-2 (a) smax ⬇ 26 MPa and 29 MPa;
(b) smax ⬇ 25 MPa and 22 MPa
2.10-3 Pmax stbt/3
2.10-4 smax ⬇ 46 MPa
2.10-5 smax ⬇ 6100 psi
2.10-6 (a) No, it makes it weaker: P1 25.1 kN,
P2 ⬇ 14.4 kN; (b) d0 ⬇ 15.1 mm
2.10-7 dmax ⬇ 0.51 in.
2.11-2 (a) dC 1.67 mm; (b) dC 5.13 mm;
(c) dC 11.88 mm
2.11-3 (b) P 17.7 k
2.11-4 For P 30 kN: d 6.2 mm; for P 40 kN:
d 12.0 mm
2.11-5 For P 24 k: d 0.18 in.; for P 40 k:
d 0.68 in.
2.11-6 For P 3.2 kN: dB 4.85 mm; for
P 4.8 kN: dB 17.3 mm
2.12-1 PY PP 2sY A sin u
2.12-2 PP 201 kN
2.12-3 (a) PP 5sY A
2.12-4 PP 2sY A(1 sin a)
2.12-5 PP 47.9 k
2.12-6 PP 82.5 kN
2.12-7 PP 20.4 k
2.12-8 (a) PY sY A, dY 3sY L/2E;
(b) PP 4sY A/3, dP 3sY L/E
2.12-9 (a) PY sY A, dY sY L/E;
(b) PP 5sY A/4, dP 2sY L/E
2.12-10 (a) WY 28.8 kN, dY 125 mm;
(b) WP 48 kN, dP 225 mm
2.12-11 (a) PY 70.1 k, dY 0.01862 in.;
(b) PP 104.3 k, dP 0.0286 in.
2.8-9
2.8-10
2.8-11
2.8-12
2.8-13
2.8-14
2.10-1
CHAPTER 3
3.2-1
3.2-2
3.2-3
3.2-4
3.2-5
(a) dmax 0.413 in.; (b) Lmin 21.8 in.
(a) Lmin 162.9 mm; (b) dmax 68.8 mm
(a) g1 267 106 rad; (b) r2, min 2.2 in.
(a) g1 393 106 rad; (b) r2, max 50.9 mm
(a) g1 195 106 rad; (b) r2, max 2.57 in.
3.3-1
3.3-2
3.3-3
3.3-4
(a) tmax 8344 psi; (b) dmin 0.651 in.
(a) tmax 23.8 MPa; (b) Tmax 0.402 Nm;
(c) u 9.12/m
(a) tmax 18,300 psi; (b) f 3.32
(a) kT 2059 Nm; (b) tmax 27.9 MPa,
gmax 997 106 radians;
(c)
3.3-5
3.3-6
3.3-7
3.3-8
3.3-9
3.3-10
3.3-11
3.3-12
3.3-13
3.3-14
3.3-15
3.3-16
3.3-17
3.4-1
3.4-2
3.4-3
3.4-4
3.4-5
3.4-6
3.4-7
3.4-8
3.4-9
3.4-10
3.4-11
k T hollow
0.938,
k Tsolid
tmaxH
1.067;
tmaxS
(d) d2 32.5 mm
(a) Lmin 38.0 in.; (b) Lmin 40.9 in.
Tmax 6.03 Nm, f 2.20
(a) tmax 7965 psi; gmax 0.00255 radians,
G 3.13 106 psi; (b) Tmax 5096 lbin.
(a) Tmax 9164 Nm; (b) Tmax 7765 Nm;
tmax 4840 psi
(a) dmin 63.3 mm; (b) dmin 66 mm (4.2%
increase in diameter)
(a) t2 5170 psi; (b) t1 3880 psi;
(c) u 0.00898/in.
(a) t2 30.1 MPa; (b) t1 20.1 MPa;
(c) u 0.306/m
in # kip
(a) dmin 2.50 in.; (b) kT 2941
rad
(c) dmin 1.996 in.
(a) dmin 64.4 mm;
(b) kT 134.9 kNm/rad; (c) dmin 50 mm
(a) T1,max 4.60 in.-k; (b) T1,max 4.31 in.-k;
(c) Torque: 6.25%, Weight: 25%
(a) f 5.19; (b) d 88.4 mm;
(c) Ratio 0.524
(a) r2 1.399 in. (b) Pmax 1387 lb
(a) tmax tBC 7602 psi, fC 0.16;
(b) dBC 1.966 in., fC 0.177
(a) tbar 79.6 MPa, ttube 32.3 MPa;
(b) fA 9.43
(a) tmax tBC 4653 psi, fD 0.978;
(b) dAB 3.25 in., dBC 2.75 in., dCD 2.16 in.,
fD 1.303
Tallow 439 Nm
d1 0.818 in.
(a) d 77.5 mm; (b) d 71.5 mm
(a) d 1.78 in.; (b) d 1.83 in.
(b) dB /dA 1.45
Minimum dA 2.52 in.
Minimum dB 48.6 mm
(a) R1 3T/2; (b) T1 1.5T, T2 0.5T;
(c) x 7L/17; (d) f2 (12/17)(TL/GIp)
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
78572_ans_ptg01_hr_1057-1082.qxd
1/19/12
10:47 AM
Page 1063
Answers to Problems
f 3TL/2pGtd3A
(a) f 2.79; (b) f 2.21
T
19 TL
; (b) w3 –
3.4-14 (a) R1 2
8 pGtd3
3.4-12
3.4-13
3.4-15
fD `
4Fd L 1
`
pG t 01d 301
L2
L0
L A2
(d01L 2 d01x + d03x)3 (t 01L 2 t 01x + t 01x)
L3
dx ,
t 03d 303
fD 0.142
(a) tmax 16tL/pd 3; (b) f 16tL2/pGd 4
(a) tmax 8tAL/pd3; (b) f 16tAL2/3pGd 4
T0
3.4-18 (a) RA ;
6
3.4-16
3.4-17
3.7-2
3.7-3
3.7-4
3.7-5
3.7-6
3.7-7
3.7-8
3.7-9
3.7-10
3.8-1
3.8-2
3.8-3
3.8-4
3.8-5
3.8-6
(b) TAB(x) a
T0 x 2
L
2 T0 b 0 … x … ,
6
2
L
3.8-7
TBC(x) c a
x L 2 T0
b– d
L
3
3.8-8
T0L
L
… x … L; (c) fc ;
2
144GIp
8 T0
–
3p d 3AB
3.4-19 (a) L max 4.42 m; (b) f 170
3.8-9
(a) Tmax 875 Nm; (b) tmax 25.3 MPa
(a) smax 6280 psi; (b) T 74,000 lb-in.
(a) max 320 106; (b) smax 51.2 MPa;
(c) T 20.0 kNm
3.5-3
(a) d1 2.40 in.; (b) f 2.20;
(c) gmax 1600 106 rad
3.5-4
G 30.0 GPa
3.5-5
T 4200 lb-in.
3.5-6
(a) dmin 37.7 mm; (b) Tmax 431 Nm
3.5-7
(a) d1 0.6 in.; (b) d1max 0.661 in.
3.5-8
(a) d2 79.3 mm; (b) d2 80.5 mm
3.5-9
(a) tmax 5090 psi:
(b) gmax 432 106 rad
3.5-10 (a) tmax 23.9 MPa:
(b) gmax 884 106 rad
3.5-11 (a) T1allow 17.84 k-in., T2allow 13.48 k-in.;
(b) Lmid 18.54 in.; (c) d3new 2.58 in.;
(d) Tmax1 17.41 k-in., Tmax2 13.15 k-in.,
wmax1 1.487, wmax2 1.245
3.7-1
(a) tmax 4950 psi; (b) dmin 3.22 in.
3.8-11
(d) t max 3.4-20
3.5-1
3.5-2
3.8-10
3.8-12
3.8-13
3.8-14
3.8-15
3.8-16
1063
(a) tmax 50.0 MPa; (b) dmin 32.3 mm
(a) H 6560 hp: (b) Shear stress is halved.
(a) tmax 16.8 MPa; (b) Pmax 267 kW
dmin 4.28 in.
dmin 110 mm
Minimum d1 1.221d
Pmax 91.0 kW
d 2.75 in.
d 53.4 mm
9LT0
(a) fmax 3T0L/5GIP; (b) wmax 25GIp
(a) x L/4; (b) fmax T0L/8GIP
fmax 2btallow /Gd
Pallow 2710 N
(a) T0,max 3678 lb-in.;
(b) T0,max 3898 lb-in.
(a) T0,max 150 Nm;
(b) T0,max 140 Nm
(a) a/L dA/(dA dB):
(b) a/L d A4/1d A4 + d B42
Lt 0
Lt 0
, TB ,
6
3
13L2t 0
L
(b) fmax f a
b 27GIP
13
(a) x 30.12 in.; (b) fmax 1
(at x 30.12 in.)
(a) t1 32.7 MPa, t2 49.0 MPa;
(b) f 1.030; (c) kT 22.3 kNm
(a) t1 1790 psi, t2 2690 psi;
(b) f 0.354; (c) kT 809 k-in.
(a) Tmax 1.521 kNm; (b) d2 56.9 mm
(a) Tmax 9.13 k-in.; (b) d2 2.27 in.
(a) T1,allow 7.14 kNm;
(b) T2,allow 6.35 kNm; (c) T3,allow 7.41 kNm;
(d) Tmax 6.35 kNm;
(a) TA 15,292 in.-lb, TB 24,708 in.-lb;
(b) TA 8734 in.-lb, TB 31,266 in.-lb
(a) R1 0.77T, R2 0.23T;
(b) Tmax 2.79 kNm; (c) fmax 7.51;
(a) TA (d) Tmax 2.48 kNm (shear in flange plate bolts
b
,
controls); (e) R2 fT1 + fT2
R1 R2, with fT1 3.9-1
3.9-2
L1
L2
, fT2 ;
G1Ip1
G2Ip2
(f) bmax 29.1
(a) U 32.0 in.-lb; (b) f 0.775
(a) U 5.36 J; (b) f 1.53
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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1064
3.9-3
3.9-4
3.9-5
3.9-6
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10:47 AM
Page 1064
Answers to Problems
U 22.6 in.-lb
U 1.84 J
(c) U3 T 2L/2GIP TtL2/2GIP t2L3/6GIP
U 19T 20 L/32GIP
f T0LALB /[G(LBIPA LA IPB)]
U t 20L3/40GIP
T 2L1dA + dB2
3.9-9
(a) U ;
pGtd A2d B2
2TL(dA + dB)
(b) w pGtdA2dB2
2
b GIPAIPB
3.9-10 U 2L(IPA + IPB)
3.9-7
3.9-8
3.9-11
f
2n
15d
2C
2pImL
2pGJm
n
; t max G
15d C L
(a) tapprox 6310 psi; (b) texact 6830 psi
tmin pd/64
(a) t 1250 psi; (b) f 0.373
(a) t 9.17 MPa; (b) f 0.140
U1/U2 2
t 35.0 MPa, f 0.570
t 2390 psi, u 0.00480/in.
t T 13/9b2t, u 2T/9Gb3t
2
3.11-9 (a) f1/f2 1 1/4b
2
2
3.11-10 t 2T(1 b) /tL mb
3.11-11 tmin 0.140 in.
3.11-12 (a) t 6.66 mm; (b) t 7.02 mm
3.12-1 Tmax ⬇ 6200 lb-in.
3.12-2 Rmin ⬇ 4.0 mm
3.12-3 For D1 0.8 in.: tmax ⬇ 6400 psi
3.12-4 D2 ⬇ 115 mm; lower limit
3.12-5 D1 ⬇ 1.31 in.
3.11-1
3.11-2
3.11-3
3.11-4
3.11-5
3.11-6
3.11-7
3.11-8
CHAPTER 4
4.3-1
4.3-2
4.3-3
4.3-4
4.3-5
4.3-6
4.3-7
4.3-8
V 333 lb, M 50,667 lb-in.
V 0.938 kN, M 4.12 kNm
V 0, M 0
V 7.0 kN, M 9.5 kNm
(a) V 190 lb, M 16,580 ft-lb;
(b) q 370.4 lb/ft (upward)
(a) V 1.0 kN, M 7 kNm;
(b) P2 4 kN; (c) P1 8 kN
(acts to right)
b/L 1/2
M 108 Nm
N P sin u, V P cos u, M Pr sin u
V 6.04 kN, M 15.45 kNm
(a) P 1200 lb; (b) P 133.3 lb
V 4.17 kN, M 75 kNm
(a) VB 6000 lb, MB 9000 lb-ft;
(b) Vm 0, Mm 21,000 lb-ft
4.3-14 (a) N 21.6 kN (compression), V 7.2 kN,
M 50.4 kNm; (b) N 21.6 kN (compression),
V 5.4 kN, M 0 (at moment release)
2
3
4.3-15 Vmax 91wL a/30g, Mmax 229wL a/75g
4.5-1
Vmax P, Mmax Pa
4.5-2
Vmax M0/L, Mmax M0a/L
4.5-3
Vmax qL/2, Mmax 3qL2/8
4.5-4
Vmax P, Mmax PL/4
4.5-5
Vmax 2P/3, Mmax PL/9
4.5-6
Vmax 2M1/L, Mmax 7M1/3
P
4.5-7
(a) Vmax (on AB),
2
3L
3LP
(just right of B);
b Mmax RC a
4
8
P
(b) Nmax P (tension on AB), Vmax ,
5
P 3L
3LP
(just right of B)
Mmax –a b 5
4
20
4.3-9
4.3-10
4.3-11
4.3-12
4.3-13
4.5-8
4.5-9
4.5-10
4.5-11
4.5-12
4.5-13
4.5-14
4.5-15
4.5-16
4.5-17
4.5-18
4.5-19
4.5-20
4.5-21
(a) Vmax P, Mmax Pa;
(b) M 3Pa (CCW); Vmax 2P, Mmax 2Pa
Vmax qL/2, Mmax 5qL2/72
(a) Vmax q0L/2, Mmax q0L2/6;
4L2q0
2Lq0
, Mmax (at B)
(b) Vmax 3
15
RB 207 lb, RA 73.3 lb,
Vmax 207 lb, Mmax 2933 lb-in.
Vmax 1200 N, Mmax 960 N.m
Vmax 200 lb, Mmax 1600 lb-ft
Vmax 4.5 kN, Mmax 11.33 kNm
Vmax 1300 lb, Mmax 28,800 lb-in.
Vmax 15.34 kN, Mmax 9.80 kNm
The first case has the larger maximum
6
moment: a PLb
5
The third case has the larger maximum
6
moment: a PLb
5
Vmax 900 lb, Mmax 900 lb-ft
Vmax 10.0 kN, Mmax 16.0 kNm
Two cases have the same maximum moment: (PL).
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
78572_ans_ptg01_hr_1057-1082.qxd
1/19/12
10:47 AM
Page 1065
Answers to Problems
4.5-22
4.5-23
4.5-24
4.5-25
4.5-26
4.5-27
4.5-28
4.5-29
4.5-30
4.5-31
4.5-32
4.5-33
4.5-34
4.5-35
4.5-36
4.5-37
4.5-38
Vmax 33.0 kN, Mmax 61.2 kNm
(a) Vmax 4P/9, Mmax 8PL/3;
(b) Vmax 4P/5, Mmax 4.8PL
MAz PL(clockwise), Ax 0, Ay 0,
1
1
Cy P (upward), Dy P (upward),
12
6
Vmax P/12, Mmax PL
(a) RA 1.25 k, RB 13.75 k;
(b) P 8 k (upward)
Vmax 4.6 kN, Mmax 6.24 kNm
(a) RA 197.1 lb, RB 433 lb;
(b) a 4.624 ft; (c) a 3.143 ft
Vmax 2.8 kN, Mmax 1.450 kNm
a 0.5858L, Vmax 0.2929qL,
Mmax 0.02145qL2
Vmax 2.5 kN, Mmax 5.0 kNm
Lq0 ,
(a) Vmax RB 2
L2q0
Mmax MA ;
6
2Lq0 ,
(b) Vmax RB 3
4L2q0
Mmax MA 15
Mmax 10 kNm
Mmax Mpos 897.6 lb-ft (at x 9.6 ft);
Mneg 600 lb-ft (at x 20 ft)
Vmax w0L/3, Mmax woL2/12
w0
MA L2 (clockwise),
30
Ax 3w0L/10 (leftward), Ay 3w0L/20
(downward), Cy w0L/12 (upward),
Dy w0L/6 (upward), Vmax w0L/4,
Mmax w0L2/24 at B
(a) x 9.6 m, Vmax 28 kN; (b) x 4.0 m,
Mmax 78.4 kNm
(a) Ax 50.38 lb (right), Ay 210 lb (upward),
Bx 50.38 lb (left),
Nmax 214.8 lb, Vmax 47.5 lb,
Mmax 270 lb-ft; (b) Ax 0,
Ay 67.5 lb, Bx 0, By 142.5 lb,
Nmax 134.4 lb, Vmax 47.5 lb,
Mmax 270 lb-ft
(a) Ax q0L/2 (leftward), Ay 17q0L/18
(upward), Dx q0L/2 (leftward),
Dy 4q0L/9 (downward), MD 0,
Nmax q0L2, Vmax 17q0L/18,
1065
Mmax q0L2; (b) Bx q0 L/2 (rightward),
By q0L/2 5q0L/3 7q0L/6 (upward);
Dx q0L/2 (rightward), Dy 5q0L/3
(downward), MD 0, Nmax 5q0L/3
Vmax 5q0L/3, Mmax q0L2
4.5-39 (a) MA 0, RAx 0, RAy q0L/6
(upward), RCy q0L/3; Nmax q0L/6,
Vmax q0L/3, Mmax 0.06415q0L2;
(b)MA (16/15)q0L2, RAx 4q0L/3,
RAy q0L/6 (upward), RCy q0L/3;
Nmax q0L/6, Vmax 4q0L/3 (in
column), Vmax q0L/3 (in beam),
Mmax (16/15)q0L2 (in column),
Mmax 0.06415q0L2 (in beam)
4.5-40 MA 0, Ax 0, Ay 18.41 kN (downward),
MD 0, Dx 63.0 kN (leftward), Dy 62.1 kN
(upward), Nmax 62.1 kN, Vmax 63.0 kN,
Mmax 756 kNm
CHAPTER 5
5.4-1
5.4-2
5.4-3
5.4-4
5.4-5
5.4-6
5.5-1
5.5-2
5.5-3
5.5-4
5.5-5
5.5-6
5.5-7
5.5-8
5.5-9
5.5-10
5.5-11
5.5-12
5.5-13
(a) max 8.67 104; (b) Rmin 9.35 in.;
(c) dmax 0.24 in.
(a) Lmin 5.24 m; (b) dmax 4.38 mm
(a) max 5.98 103; (b) dmax 4.85 in.;
(c) Lmin 51 ft
1
(a) r 85 m, k 0.0118 , d 23.5 mm;
m
(b) hmax 136 mm; (c) d 75.3 mm
(a) 9.14 104; (b) tmax 0.241 in.;
(c) d 0.744 in.; (d) Lmax 37.1 in.
(a) 4.57 104; (b) Lmax 2 m
(a) smax 52.4 ksi; (b) 33.3%;
(c) Lnew 120 in.
(a) smax 250 MPa; (b) 19.98%; (c) 25%
(a) smax 38.2 ksi; (b) 10%; (c) 10%
(a) smax 8.63 MPa; (b) smax 6.49 MPa
smax 21.6 ksi
smax 203 MPa
smax 3420 psi
smax 101 MPa
smax 10.82 ksi
smax 7.0 MPa
(a) smax 432 psi; (b) s 0.58579L,
smin 153.7 psi; (c) s 0 or L, smax 896 psi
smax 2.10 MPa
(a) st 30.93M/d 3; (b) st 360M/(73bh2);
(c) st 85.24M/d 3
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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1/19/12
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Answers to Problems
smax 10.965M/d3
(a) smax 21.4 ksi; (b) L 20.9 ft;
(c) d 8.56 ft
5.5-16 (a) st 35.4 MPa, sc 61 MPa;
L
(b) dmax , st 37.1 MPa, sc 64.1 MPa
2
5.5-14
5.5-15
5.5-17
5.5-18
5.5-19
5.5-20
5.5-21
5.5-22
5.5-23
5.5-24
5.5-25
5.6-1
5.6-2
5.6-3
5.6-4
5.6-5
5.6-6
5.6-7
5.6-8
5.6-9
5.6-10
5.6-11
5.6-12
5.6-13
5.6-14
5.6-15
5.6-16
5.6-17
5.6-18
5.6-19
5.6-20
(a) st 4.34 ksi, sc 15.96 ksi;
(b) Pmax 214 lb; (c) 4.28 ft
(a) sc 1.456 MPa, st 1.514 MPa;
(b) sc 1.666 MPa (14%),
st 1.381 MPa (9%); (c) sc 0.728 MPa
( 50%), st 0.757 MPa ( 50%)
(a) st 7810 psi, sc 13,885 psi;
(b) a 12.73 ft
smax 3pL2a0/t
(a) st 20,360 psi, sc 13,188 psi;
(b) h 3.20 in.; (c) q 97.2 lb/ft,
P 675 lb
s 25.1 MPa, 17.8 MPa, 23.5 MPa
d 3 ft, smax 171 psi, d 6 ft,
smax 830 psi
(a) c1 91.7 mm, c2 108.3 mm,
Iz 7.969 107 mm4; (b) st 4659 kPa (top of
beam at C), sc 5506 kPa (bottom of beam at C)
(a) Fres 104.8 lb; (b) smax 36.0 ksi (compression at base); (c) smax 32.4 ksi (tension at base)
dmin 4.00 in.
(a) dmin 12.62 mm; (b) Pmax 39.8 N
(a) C 15 33.9; (b) S 8 18.4; (c) W 8 35
(a) W 360 39; (b) W 250 89
(a) S 10 35; (b) Pmax 3152 lb
(a) bmin 161.6 mm; (b) bmin 141.2 mm,
area(b)/area(a) 1.145
(a) 2 12; (b) wmax 137.3 lb/ft2
(a) smax 429 mm; (b) hmin 214 mm
(a) q0,allow 424 lb/ft; (b) q0,allow 268 lb/ft
hmin 30.6 mm
(a) Sreqd 15.37 in.3; (b) S 8 23
(a) dmin 37.6 mm; (b) dmin 45.2 mm,
area(b)/area(a) 0.635
(a) 4 12; (b) qmax 14.2 lb/ft
b 152 mm, h 202 mm
b 10.25 in.
t 13.61 mm
W1:W2:W3:W4 1:1.260:1.408:0.888
(a) qmax 6.61 kN/m; (b) qmax 9.37 kN/m
6.57%
(a) bmin 11.91 mm; (b) bmin 11.92 mm
5.6-21
5.6-22
5.6-23
5.7-1
5.7-2
5.7-3
5.7-4
5.7-5
5.7-6
5.7-7
5.7-8
5.8-2
5.8-3
5.8-4
5.8-5
5.8-6
5.8-7
5.8-8
5.8-9
5.8-10
5.8-11
5.8-12
5.9-1
5.9-2
5.9-3
5.9-4
5.10-1
5.10-2
5.10-3
5.10-4
(a) smax 49.2 in.; (b) d 12.65 in.
(a) b 1/9; (b) 5.35%
Increase when d/h 0.6861;
decrease when d/h 0.6861
(a) x L/4, smax 4PL/9hA3, smax/sB 2;
(b) x 0.209L, smax 0.394PL/hA3,
smax/sB 3.54
(a) x 4 m, smax 37.7 MPa,
smax/sB 9/8; (b) x 2 m,
smax 25.2 MPa, smax/sm 4/3
(a) x 8 in., smax 1250 psi,
smax/sB 1.042; (b) x 4.64 in.,
smax 1235 psi, smax/sm 1.215
(a) sA 210 MPa; (b) sB 221 MPa;
(c) x 0.625 m; (d) smax 231 MPa;
(e) smax 214 MPa
(a) 1 dB /dA 1.5;
(b) smax sB 32PL/p d 3B
hx hB x/L
by 2bB x/L
hx hB 2x/L
(a) tmax 731 kPa, smax 4.75 MPa;
(b) tmax 1462 kPa, smax 19.01 MPa
(a) Mmax 25.4 k-ft; (b) Mmax 4.95 k-ft
tmax 500 kPa
tmax 2400 psi
(a) L0 h(sallow /tallow);
(b) L0 (h/2)(sallow /tallow)
(a) Pmax 1.914 kip; (b) Pmax 2.05 kip
(a) Mmax 72.2 Nm; (b) Mmax 9.01 Nm
(a) 8 12-in. beam; (b) 8 12-in. beam
(a) P 38.0 kN; (b) P 35.6 kN
(a) w1 121 1b/ft2; (b) w2 324 lb/ft2;
(c) wallow 121 lb/ft2
(a) b 89.3 mm; (b) b 87.8 mm
dmin 5.70 in.
(a) W 28.6 kN; (b) W 38.7 kN
(a) d 10.52 in.; (b) d 2.56 in.
(a) q0,max 55.7 kN/m; (b) Lmax 2.51 m
(a) tmax 5795 psi; (b) tmin 4555 psi;
(c) taver 5714 psi; (d) Vweb 28.25 k
(a) tmax 28.43 MPa; (b) tmin 21.86 MPa;
(c) taver 27.41 MPa; (d) Vweb 119.7 kN
(a) tmax 4861 psi; (b) tmin 4202 psi;
(c) taver 4921 psi; (d) Vweb 9.432 k
(a) tmax 32.28 MPa; (b) tmin 21.45 MPa;
(c) taver 29.24 MPa; (d) Vweb 196.1 kN
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
78572_ans_ptg01_hr_1057-1082.qxd
1/19/12
10:47 AM
Page 1067
Answers to Problems
(a) tmax 2634 psi; (b) tmin 1993 psi;
(c) taver 2518 psi; (d) Vweb 20.19 k
5.10-6 (a) tmax 28.40 MPa; (b) tmin 19.35 MPa;
(c) taver 25.97 MPa; (d) Vweb 58.63 kN
5.10-7 qmax 1270 lb/ft
5.10-8 (a) qmax 184.7 kN/m; (b) qmax 247 kN/m
5.10-9 S 8 23
5.10-10 V 273 kN
5.10-11 tmax 1.42 ksi, tmin 1.03 ksi
5.10-12 tmax 19.7 MPa
5.10-13 tmax 2221 psi
5.11-1 Vmax 676 lb
5.11-2 Vmax 1.924 MN
5.11-3 F 1994 lb/in.
5.11-4 Vmax 10.7 kN
5.11-5 (a) smax 5.08 in.; (b) smax 4.63 in.
5.11-6 (a) sA 78.3 mm; (b) sB 97.9 mm
5.11-7 (a) smax 2.77 in.; (b) smax 1.85 in.
5.11-8 smax 92.3 mm
5.11-9 Vmax 18.30 k
5.11-10 smax 236 mm
5.11-11 (a) Case (1); (b) Case (3); (c) Case (1); (d) Case (3)
5.11-12 smax 180 mm
5.12-1 st 14,660 psi, sc 14,990 psi
5.12-2 st 5770 kPa, sc 6668 kPa
5.12-3 tmin 0.477 in.
5.12-4 st 11.83 MPa, sc 12.33 MPa,
tmin 12.38 mm
5.12-5 st 302 psi, sc 314 psi
5.12-6 Tmax 108.6 kN
5.12-7 a arctan [( d 22 + d 12)/14hd22]
5.10-5
5.12-8 (a) dmin 8.46 cm; (b) dmin 8.91 cm
5.12-9 Hmax 32.2 ft
5.12-10 W 33.3 kN
5.12-11 (a) st 87.6 psi, sc 99.6 psi;
(b) dmax 28.9 in.
5.12-12 (a) b p d/6; (b) b p d/3;
(c) Rectangular post
5.12-13 (a) st 1900 psi, sc 1100 psi;
(b) Both stresses increase in magnitude.
5.12-14 (a) st 8P/b , sc 4P/b ;
2
2
(b) st 9.11P/b , sc 6.36P/b2
5.12-15 (a) st 857 psi, sc 5711 psi;
(b) y0 4.62 in.; (c) st 453 psi,
sc 2951 psi, yo 6.33 in.
5.12-16 (a) st 3.27 MPa, sc 24.2 MPa;
(b) y0 76.2 mm; (c) st 1.587 MPa,
sc 20.3 MPa, y0 100.8 mm
2
1067
5.12-17 (a) st 15.48 ksi; (b) st 2.91 ksi
5.12-18 (a) y0 21.5 mm; (b) P 67.3 kN;
5.13-1
5.13-2
5.13-3
5.13-4
5.13-5
(c) y0 148.3 mm, P 149.6 kN
(a) d 0.50 in., smax 15,500 psi;
(b) R 0.10 in., smax ⬇ 49,000 psi
(a) d 16 mm, smax 81 MPa;
(b) R 4mm, smax ⬇ 200 MPa
bmin ⬇ 0.24 in.
bmin ⬇ 0.33 mm
(a) Rmin ⬇ 0.45 in.; (b) dmax 4.13 in.
CHAPTER 6
6.2-1
6.2-2
6.2-3
6.2-4
sface 1980 psi, score 531 psi
(a) Mmax 58.7 kNm;
(b) Mmax 90.9 kNm; (c) t 7.08 mm
(a) Mmax 172 k-in.; (b) Mmax 96 k-in.
pss1E B d 14 E sd 14 + E sd 242
(a) M allow,steel ,
32E s d2
M allow,brass psB1E B d 14 E sd 14 + E sd 242
;
32E s d1
(b) Mmax,brass 1235 Nm; (c) d1 33.3 mm
6.2-5
(a) sw 666 psi, ss 13,897 psi;
(b) qmax 665 lb/ft; (c) M0,max 486 lb-ft
6.2-6
(a) Mallow 768 Nm; (b) ssa 47.9 MPa,
Mmax 1051 Nm
6.2-7
(a) sface 3610 psi, score 4 psi;
(b) sface 3630 psi, score 0
6.2-8
(a) sface 14.1 MPa, score 0.214 MPa;
(b) sface 14.9 MPa, score 0
6.2-9
sa 4120 psi, sc 5230 psi
6.2-10 (a) sw 5.1 MPa (compression),
ss 37.6 MPa (tension); (b) ts 3.09 mm
6.2-11 (a) splywood 1131 psi, spine 969 psi;
(b) qmax 95.5 lb/ft
6.2-12 Q0.max 15.53 kN/m
6.3-1
(a) Mmax 442 k-in.; (b) Mmax 189 k-in.
6.3-2
tmin 15.0 mm
6.3-3
(a) qallow 454 lb/ft; (b) swood 277 psi,
ssteel 11,782 psi
6.3-4
(a) sB 60.3 MPa, sw 7.09 MPa;
(b) tB 25.1 mm, Mmax 80 kNm
6.3-5
sa 1860 psi, sP 72 psi
6.3-6
sa 12.14 MPa, sP 0.47 MPa
6.3-7
(a) qallow 264 lb/ft; (b) qallow 280 lb/ft
6.3-8
(a) ss 93.5 MPa; (b) hs 5.08 mm,
ha 114.92 mm
6.3-9
Mmax 81.1 k-in.
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
78572_ans_ptg01_hr_1057-1082.qxd
1068
6.3-10
6.3-11
6.3-12
6.3-13
6.3-14
6.3-15
6.4-1
6.4-2
6.4-3
6.4-4
6.4-5
6.4-6
6.4-7
6.4-8
6.4-9
6.4-10
6.4-11
6.4-12
6.4-13
6.5-1
6.5-2
6.5-3
6.5-4
6.5-5
6.5-6
6.5-7
6.5-8
6.5-9
6.5-10
6.5-11
6.5-12
1/19/12
10:47 AM
Page 1068
Answers to Problems
SA 50.6 mm3; Metal A
ss 13,400 psi (tension), sc 812 psi
(compression).
(a) sc 8.51 MPa, ss 118.3 MPa;
(b) Mmax Mc 172.9 kNm;
(c) As 2254 mm2, Mallow 167.8 kNm
(a) sc 649 psi, ss 15,246 psi;
(b) Mallow Ms 207 kip-ft
(a) Mmax Ms 10.59 kNm;
(b) As 1262 mm2, Mallow 15.79 kNm
Mallow MW 12.58 kip-ft
tan b h/b, so NA lies along other diagonal
b 51.8, smax 17.5 MPa
b 42.8, smax 1036 psi
b 78.9, sA sE 102 MPa,
sB sD 48 MPa
b 72.6, sA sE 14,554 psi,
sB sD 4953 psi
b 79.3, smax 8.87 MPa
b 78.8, smax 1660 psi
b 81.8, smax 69.4 MPa
b 72.9, smax 8600 psi
b 60.6, smax 20.8 MPa
(a) sA 45,420 sin a 3629 cos a (psi);
(b) tan b 37.54 tan a
b 79.0, smax 16.6 MPa
(a) b 76.2, smax 8469 psi;
(b) b 79.4, smax 8704 psi
b 83.1, st 5,060 psi,
sc 10,420 psi
b 83.4, st 10.5 MPa,
sc 23.1 MPa
b 75.6, st 3,080 psi, sc 3450 psi
b 75.8, st 31.7 MPa, sc 39.5 MPa
(a) b 28.7, st 4263 psi,
sc 4903 psi; (b) b 38.5,
st 5756 psi, sc 4868 psi
b 78.1, st 40.7 MPa,
sc 40.7 MPa
b 82.3, st 1397 psi, sc 1157 psi
b 2.93,st 6.56 MPa, sc 6.54 MPa
For u 0: st sc 2.546M/r3; for
u 45: st 4.535M/r3, sc 3.955M/r3;
for u 90: st 3.867M/r3,
sc 5.244M/r3;
b 78.9, st 131.1 MPa,
st 148.5 MPa
b 11.7, st 28.0 ksi, sc 24.2 ksi
b 56.5, st 31.0 MPa, sc 29.0 MPa
6.8-1
6.8-2
6.8-3
6.8-4
6.9-1
6.9-2
6.9-6
6.9-8
6.10-1
(a) tmax 3584 psi; (b) tB 430 psi
(a) tmax 29.7 MPa; (b) tB 4.65 MPa
(a) tmax 3448 psi; (b) tmax 3446 psi
(a) tmax 27.04 MPa; (b) tmax 27.02 MPa
e 1.027 in.
e 22.1 mm
63 p r
1.745r
(b) e 24p + 38
(a) e b 2h + 3b
a
b;
2 h + 3b
(b) e b 43h + 48b
a
b
2 23h + 48b
f 2(2b1 b2)/(3b1 b2)
(a) f 16t2(r 23 r 13)/3p (r 24 r 14);
(b) f 4/p
6.10-3 q 1000 lb/in.
6.10-4 (a) 56.7%: (b) M 12.3 kNm
6.10-5 f 1.12
6.10-6 f 1.15
3
6.10-7 Z 16.98 in. , f 1.14
6
3
6.10-8 Z 1.209 10 mm , f 1.11
6.10-9 MY 525 k-ft, MP 591 k-ft, f 1.13
6.10-10 MY 378 kNm, MP 427 kNm, f 1.13
6.10-11 MY 4320 k-in., MP 5450 k-in., f 1.26
6.10-12 MY 672 kNm, MP 878 kNm, f 1.31
6.10-13 MY 1619 k-in., MP 1951 k-in., f 1.21
6.10-14 MY 122 kNm, MP 147 kNm, f 1.20
6.10-15 (a) M 5977 k-in.; (b) 22.4%
6.10-16 (a) M 524 kNm; (b) 36%
6.10-17 (a) M 2551 k-in.; (b) 7.7%
3
3
6.10-18 Z 136 10 mm , f 1.79
6.10-19 MP 1120 k-in.
6.10-20 MP 295 kNm
6.10-2
CHAPTER 7
7.2-1
7.2-2
7.2-3
7.2-4
7.2-5
7.2-6
For u 55: sx1 4221 psi, sy1 4704 psi,
txlyl 3411 psi
For u 40: sx1 117.2 MPa,
sy1 62.8 MPa, txlyl 10.43 MPa
For u 30: sx1 3041 psi,
sy1 8959 psi, txlyl 12,725 psi
For u 52: sx1 136.6 MPa,
sy1 16.6 MPa, txlyl 84 MPa
For u 50: sx1 1243 psi,
sy1 6757 psi, txlyl 1240 psi
For u 40: sx1 5.5 MPa,
sy1 27 MPa, txlyl 28.1 MPa
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
78572_ans_ptg01_hr_1057-1082.qxd
1/19/12
10:47 AM
Page 1069
Answers to Problems
7.2-7
7.2-8
7.2-9
7.2-10
7.2-11
7.2-12
7.2-13
7.2-14
7.2-15
7.2-16
7.2-17
7.2-18
7.2-19
7.3-1
7.3-2
7.3-3
7.3-4
7.3-5
7.3-6
7.3-7
7.3-8
7.3-9
7.3-10
7.3-11
7.3-12
7.3-13
7.3-14
7.3-15
For u 38: sx1 13,359 psi,
sy1 3671 psi, txlyl 4960 psi
For u 40: sx1 66.5 MPa,
sy1 6.52 MPa, txlyl 14.52 MPa
Normal stress on seam, 187 psi tension. Shear
stress, 163 psi (clockwise)
Normal stress on seam, 1440 kPa tension. Shear
stress, 1030 kPa (clockwise)
sw 125 psi, tw 375 psi
sw 10.0 MPa, tw 5.0 MPa
u 51.3, sy1 500 psi, txlyl 1122 psi
u 36.6, sy1 9 MPa,
txlyl 14.83 MPa
For u 36: sxl 12,068 psi,
syl 4732 psi, txlyl 4171 psi
For u 50: sxl 51.4 MPa,
syl 14.4 MPa, txlyl 31.3 MPa
sy 3673 psi, txy 1405 psi
sy 77.7 MPa, txy 27.5 MPa
sb 4700 psi, tb 2655 psi, ul 48.04
s1 5868 psi, s2 982 psi, up1 8.94
s1 119.2 MPa, s2 60.8 MPa,
upl 29.52
s1 6333 psi, s2 1167 psi,
upl 23.68
s1 53.6 MPa, up1 14.2
s1 5771 psi, s2 18,029 psi,
s1 s2
6129 psi
tmax 2
upl 14.12
tmax 24.2 MPa, sx1 14.25 MPa,
syl 14.25 MPa, usl 60.53
tmax 6851 psi, usl 61.8
tmax 26.7 MPa, usl 19.08
(a) s1 180 psi, up1 20.56;
(b) tmax 730 psi, usl 65.56
(a) s1 25 MPa, s2 130 MPa;
(b) tmax 77.5 MPa, save 52.5 MPa
(a) s1 2693 psi, s2 732 psi;
(b) tmax 980 psi, save 1713 psi
(a) s1 2262 kPa, upl 13.70;
(b) tmax 1000 kPa, us l 58.7
(a) s1 14,764 psi, upl 7.90;
(b) tmax 6979 psi, usl 37.1
(a) s1 29.2 MPa, up1 17.98;
(b) tmax 66.4 MPa, usl 63.0
(a) s1 1228 psi, up1 24.7;
(b) tmax 5922 psi, usl 20.3
7.3-16
7.3-17
7.3-18
7.3-19
7.3-20
7.4-1
7.4-2
7.4-3
7.4-4
7.4-5
7.4-6
7.4-7
7.4-8
7.4-9
7.4-10
7.4-11
7.4-12
7.4-13
7.4-14
7.4-15
7.4-16
1069
(a) s1 76.3 MPa, upl 107.5;
(b) tmax 101.3 MPa, usl 62.5
3030 psi sy 9470 psi
18.5 MPa sy 85.5 MPa
(a) sy 3961 psi; (b) up1 38.93,
s1 6375 psi, up2 51.07, s2 2386 psi
(a) sy 23.3 MPa; (b) up1 65.6,
s1 41 MPa, up2 24.4, s2 62.7 MPa
(a) sx1 10,901 psi, sy1 3349 psi,
txlyl 6042 psi; (b) tmax 7125 psi,
save 7125 psi
(a) sx1 40.1 MPa, sy1 16.91 MPa,
txlyl 26 MPa; (b) tmax 28.5 MPa,
save 28.5 MPa
(a) sx1 5400 psi, sy1 1350 psi,
txlyl 2700 psi; (b) tmax 3375 psi,
saver 3375 psi
For u 25: (a) sx1 36.0 MPa,
txlyl 25.7 MPa; (b) tmax 33.5 MPa,
usl 45.0
For u 55: (a) sx1 882 psi, txlyl 3759 psi,
sy1 3618 psi, saver 2250 psi;
(b) tmax 4000 psi, ux1 45.0
For u 21.80: (a) sx1 17.1 MPa,
txlyl 29.7 MPa; (b) tmax 43.0 MPa,
0x1 45.0
For u 52: (a) sx1 2620 psi,
txlyl 653 psi; (b) s1 2700 psi,
up1 45.0
(a) sx1 60.8 MPa, sy1 128.8 MPa,
txlyl 46.7 MPa; (b) s1 139.6 MPa,
s2 71.6 MPa, tmax 105.6 MPa
For u 36.87: (a) sx1 3600 psi,
txlyl 1050 psi; (b) s1 3750 psi, up1 45.0
For u 40: sx1 27.5 MPa,
txlyl 5.36 MPa
For u 51: sx1 11,982 psi
txlyl 3569 psi
For u 33: sx1 61.7 MPa,
txlyl 51.7 MPa, sy1 171.3 MPa
For u 14: sx1 1509 psi,
txlyl 527 psi, sy1 891 psi
For u 35: sx1 46.4 MPa,
txlyl 9.81 MPa
For u 65: sx1 1846 psi,
txlyl 3897 psi
(a) s1 10,865 kPa, up1 115.2;
(b) tmax 4865 kPa, us1 70.2
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
78572_ans_ptg01_hr_1057-1082.qxd
1070
7.4-17
7.4-18
7.4-19
7.4-20
7.4-21
7.4-22
7.4-23
7.5-1
7.5-2
7.5-3
7.5-4
7.5-5
7.5-6
7.5-7
7.5-8
7.5-9
7.5-10
7.5-11
7.5-12
7.6-1
1/19/12
10:47 AM
Page 1070
Answers to Problems
(a) s1 2565 psi, up1 31.3;
(b) tmax 3265 psi, us1 13.70
(a) s1 18.2 MPa, up1 123.3;
(b) tmax 15.4 MPa, us1 78.3
(a) s1 6923 psi, up1 32.4;
(b) tmax 7952 psi, us1 102.6
(a) s1 40.0 MPa, up1 68.8;
(b) tmax 40.0 MPa, us1 23.8
(a) s1 7490 psi, up1 63.2;
(b) tmax 3415 psi, us1 18.20
(a) s1 3.43 MPa, up1 19.68:
(b) tmax 15.13 MPa, us1 64.7
(a) s1 7525 psi, up1 9.80;
(b) tmax 3875 psi, us1 35.2
sx 25,385 psi, sy 19,615 psi,
t 2.81 104 in.
sx 102.6 MPa, sy 11.21 MPa,
t 1.646 103 mm
(a) z v (x y)/(1 v);
(b) e (1 2v)( x y)/(1 v)
0.24, E 112.1 GPa
0.3, E 29,560 ksi
(a) gmax 5.85 104;
(b) t 1.32 103 mm;
(c) V 387 mm3
(a) gmax 1900 106;
(b) t 141 106 in. (decrease);
(c) V 0.0874 in.3 (increase)
(a) Vb 49.2 mm3, Ub 3.52 J;
(b) Va 71.5 mm3, Ua 4.82 J
V 0.0377 in.3, U 55.6 in.-lb
(a) V 2766 mm3, U 56 J;
(b) tmax 36.1 mm; (c) bmin 640 mm
(a) V 0.0385 in.3, U 574 lb-in.;
(b) tmax 0.673 in.; (c) bmin 10.26 in.
(a) ac xd 0.1296 mm (increase);
(b) bc yd 0.074 mm (decrease);
(c) t zt 2.86 103 mm
(decrease); (d) V eV0 430 mm3;
(e) U uV0 71.2 Nm; (f) tmax 22.0 mm;
(g) sxmax 63.9 MPa
s1 s3
(a) t max 8750 psi;
2
(b) a ax 7.73 103 in.,
b yb 3.75 103 in.,
c zc 1.3 103 in.;
7.6-2
7.6-3
(c) V eV0 0.0173 in.3;
(d) U uV0 964 in.-lb;
(e) sxmax 12,824 psi;
(f) sxmax 11,967 psi
s1 s3
8.5 MPa;
(a) t max 2
(b) a ax 0.0525 mm,
b yb 9.67 103 mm,
c zc 9.67 103 mm;
(c) V eV0 2.052 103 mm3;
(d) U uV0 56.2 Nm;
(e) sxmax 50 MPa;
(f) sxmax 65.1 MPa
(a) sx 4200 psi, sy 2100 psi,
sz 2100 psi;
s1 s3
(b) tmax 1050 psi;
2
(c) V eV0 0.0192 in.3;
(d) U uV0 35.3 in.-lb;
(e) sxmax 3864 psi
(f) x max 235 # (10 6)
7.6-4
(a) sx 82.6 MPa, sy 54.7 MPa,
sz 54.7 MPa;
s1 s3
(b) t max 13.92 MPa;
2
(c) V eV0 846 mm3;
(d) U uV0 29.9 Nm;
(e) sxmax 73 MPa
(f) xmax 741 # (10 6)
7.6-5
(a) KA1 1107 psi; (b) E 6139 ksi,
0.35
7.6-6
(a) K 4.95 GPa; (b) E 1.297 GPa, 0.40
7.6-7
(a) p F/[A(1 )];
(b) d FL(1 )(1 2 )/EA(1 )]
7.6-8
(a) p p0; (b) e p0(1 )(1 2 )/E;
(c) u p 02(1 2)/2E
7.6-9
(a) d 1.472 103 in., V 0.187 in.3,
U 332 in.-lb; (b) h 5282 ft
7.6-10 (a) p 700 MPa; (b) K 175 GPa;
(c) U 2470 J
6
7.6-11 0 276 10
, e 828 106,
u 4.97 psi
7.7-1
(a) d 1.878 103 in.;
(b) f a 1.425 104 (decrease, radians);
(c) a 1.425 104 (increase, radians)
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
78572_ans_ptg01_hr_1057-1082.qxd
1/19/12
10:47 AM
Page 1071
Answers to Problems
(a) d x1Ld 0.062 mm;
(b) f a 1.89 104 (decrease, radians);
(c) a 1.89 104 (increase, radians)
7.7-3
(a) d 0.00458 in. (increase);
(b) f 157 106 rad (decrease);
(c) g 314 106 rad (angle ced increases)
7.7-4
(a) d 0.168 mm (increase);
(b) f 317 106 rad (decrease);
(c) g 634 106 rad (angle ced increases)
7.7-5
x1 3.97 104, y1 3.03 104,
gxlyl 1.829 104
7.7-6
x1 9.53 105, y1 1.353 104,
gxlyl 3.86 104
7.7-7
1 554 106, up1 22.9,
gmax 488 106
7.7-8
1 172 106, up1 163.9,
gmax 674 106
7.7-9
For u 75: (a) x1 202 106,
gx1y1 569 106; (b) 1 568 106,
uP1 22.8; (c) gmax 587 106
6
7.7-10 For u 45: (a) x1 385 10 ,
6
gx1y1 690 10 ;
(b) 1 254 106, up1 65.7;
(c) gmax 1041 106
sx sy
7.7-11 tmaxxy 4076 psi,
2
7.7-2
gxymax 2
C
x y
a
b + a
2
4
6.83 * 10
,
x z
2
gxzmax 2
C
a
gyzmax 2
7.7-12
2
tmaxxy 2
C
a
gxymax 2
2
C
2
1.244 * 10
C
a
x z
2
C
a
gxy
2
,
2
2.15 * 104
7.7-18
7.7-19
7.7-20
7.7-21
7.7-22
7.7-23
7.7-24
7.7-25
7.7-26
7.7-27
7.7-28
b
2
8.2-1
8.2-2
8.2-3
8.2-4
2
b + gxz2
y z
7.7-17
For u 30: (a) x1 756 106,
gx1y1 868 106; (b) 1 426 106,
up1 99.8; (c) gmax 1342 106
For u 50: (a) x1 1469 106,
gx1y1 717 106; (b) 1 732 106,
up1 166.0; (c) gmax 911 106
1 551 106, up1 12.5,
gmax 662 106
1 332 106, up1 12.0,
gmax 515 106
(a) P 5154 1b, T 978 in.-1b;
(b) gmax 2.84 104, tmax 3304 psi
P 121.4 kN, a 56.7
P 9726 1b, a 75.2
x a, y (2b 2c a)/3,
gxy 2(b c)/ 13
For up1 30: 1 1550 106,
2 250 106, s1 10,000 psi,
s2 2000 psi
sx 91.6 MPa
x1 3.97 104, y1 3.03 104,
gx1y1 1.829 104
x1 9.53 105, y1 1.353 104,
gx1y1 3.86 104
1 554 106, up1 157.1,
gmax 488 106
1 172 106, up1 163.9,
gmax 674 106
For u 75: (a) x1 202 106,
gx1y1 569 106: (b) 1 568 106,
up1 22.8; (c) gmax 587 106
For u 45: (a) x1 385 106,
gx1y1 690 106; (b) 1 254 106,
up1 65.7; (c) gmax 1041 106
CHAPTER 8
1.459 * 103,
gyzmax 2
7.7-16
2
2
2
7.7-15
b + gyz2 + 2.13 * 104
b + a
3
gxzmax 2
b
33.7 MPa,
x y
a
2
7.7-14
2
b + gxz2 + 8.96 * 104,
y z
sx sy
gxy
7.7-13
1071
8.2-5
8.2-6
8.2-7
2
b + gyz2
8.2-8
(a) Use t 2.5 in. (b) pmax 381 psi
(a) Use t 98 mm. (b) pmax 3.34 MPa
(a) F 1073 1b, s 255 psi;
(b) db 0.286 in.; (c) r 7.35 in.
(a) smax 3.12 MPa, max 0.438;
(b) treqd 1.29 mm
(a) smax 425 psi, max 1.105;
(b) pmax 7.77 psi
(a) pmax 3.51 MPa; (b) pmax 2.93 MPa
(a) f 26.4 k/in.; (b) tmax 7543 psi;
(c) max 3.57 104
(a) f 5.5 MN/m; (b) tmax 57.3 MPa;
(c) max 3.87 104
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
78572_ans_ptg01_hr_1057-1082.qxd
1072
1/19/12
10:47 AM
Page 1072
Answers to Problems
(a) tmin 0.291 in.; (b) p 1904 psi
(a) tmin 7.17 mm; (b) p 19.25 MPa
D0 90 ft
(a) tmin 0.289 in.; (b) pmax 2286 psi
(a) h 22.2 m; (b) zero
n 2.25
(a) F 3ppr2; (b) treqd 10.91 mm
(a) p 55 psi; (b) r 9.18 104
(a) max 6.67 105; (b) r 2.83 104
tmin 0.113 in.
tmin 3.71 mm
(a) h 25 ft; (b) s1 L 125 psi
8.3-10 (a) sh 24.9 MPa; (b) sc 49.7 MPa;
(c) sw 24.9 MPa; (d) th 12.43 MPa;
(e) tc 24.9 MPa
8.3-11 (a) tmin 0.675 in.; (b) tmin 0.338 in.
8.3-12 (a) s1 93.3 MPa, s2 46.7 MPa;
(b) t1 23.2 MPa, t2 46.7 MPa;
(c) 1 3.97 104, 2 9.33 105;
(d) u 35, sx1 62.0 MPa,
8.2-9
8.2-10
8.2-11
8.3-1
8.3-2
8.3-3
8.3-4
8.3-5
8.3-6
8.3-7
8.3-8
8.3-9
sy1 78.0 MPa, tx 1 y1 21.9 MPa
8.3-13
8.4-1
(a) s1 7015 psi, s2 3508 psi;
(b) t1 1754 psi, t2 3508 psi;
(c) 1 1.988 104, 2 4.68 105;
(d) u 28, sx1 4281 psi, sy1 6242 psi,
tx1y1 1454 psi
h
M ad b
2
(a) sy 0, sx 5 ksi,
I
VQ
txy 1.111 ksi, s1 5.24 ksi,
Ib
up1 11.98, s2 0.236 ksi, up2 78.02;
tmax 2.74 # ksi
(b) sy 0,
h
Mad b
2
N
sx A
I
5.95 ksi, txy 8.4-2
VQ
1.11 ksi,
Ib
s1 6.15 ksi, up1 10.24,
s2 0.201 ksi, up2 79.76; tmax 3.18 ksi
(a) sy 0,
h
Mad b
2
sx 37.4 MPa,
I
txy VQ
7.49 MPa,
Ib
s1 38.9 MPa, up1 10.9,
s2 1.442 MPa, up2 79.1,
tmax 20.2 MPa;
(b) sy 0,
h
Ma d b
2
N
44.4 MPa,
sx A
I
txy VQ
7.49 MPa,
Ib
s1 45.7 MPa, up1 9.3,
s2 1.227 MPa, up2 80.7,
tmax 23.4 MPa
8.4-3
(a) s1 219 psi, s2 219 psi,
tmax 219 psi; (b) s1 49.6 psi,
s2 762 psi, tmax 406 psi;
(c) s1 0 psi, s2 2139 psi,
tmax 1069 psi
8.4-4
P 20 kN
8.4-5
P 2.91 k
8.4-6
(b) s1 4.5 MPa, s2 76.1 MPa,
tmax 40.3 MPa
8.4-7
(b) s1 14,100 psi, s2 220 psi,
tmax 7160 psi
8.4-8
(b) s1 8.27 MPa, s2 64.3 MPa,
tmax 36.3 MPa
8.4-9
(b) s1 159.8 psi, s2 3393 psi,
tmax 1777 psi
8.4-10 s1 17.86 MPa, s2 0.145 MPa,
tmax 9.00 MPa
s1
8.4-11
184
s2
s1
663
8.4-12
s2
8.5-1
8.5-2
8.5-3
8.5-4
8.5-5
8.5-6
8.5-7
8.5-8
8.5-9
tmin 0.125 in.
pmax 9.60 MPa
(a) smax s1 11.09 ksi, tmax 3.21 ksi;
(b) Tmax 178 k-ft; (c) tmin 0.519 in.
(a) Pmax 52.7 kN; (b) pmax 6 MPa
st 10,680 psi: No compressive stresses.
tmax 5340 psi
fmax 0.552 rad 31.6
st 3963 psi, sc 8791 psi, tmax 6377 psi
st 16.93 MPa, sc 41.4 MPa,
tmax 28.9 MPa
P 194.2 k
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
78572_ans_ptg01_hr_1057-1082.qxd
1/19/12
10:47 AM
Page 1073
Answers to Problems
8.5-14
8.5-15
(a) smax s1 35.8 MPa, tmax 18.05 MPa;
(b) Pmax 6.73 kN
(a) sx 0 psi, sy 6145 psi, txy 345 psi;
(b) s1 6164 psi, s2 19.30 psi, tmax 3092 psi
tA 76.0 MPa, tB 19.94 MPa, tC 23.7 MPa
s1 1094 psi (max. tensile stress at base of pole),
s2 7184 psi (max. compressive stress at base
of pole), tmax 3731 psi (max. shear stress at base
of pole)
dmin 48.4 mm
st 39,950 psi, sc 2226 psi, tmax 21,090 psi
8.5-16
(a) st 29.15
8.5-10
8.5-11
8.5-12
8.5-13
8.5-17
8.5-18
8.5-19
8.5-20
8.5-21
8.5-22
8.5-23
8.5-24
8.5-25
qR2
, sc 8.78
qR2
,
d3
d3
qR2
qR2
tmax 18.97 3 ; (b) st 14.04 3 ,
d
d
2
2
qR
qR
sc 2.41 3 , tmax 8.22 3
d
d
st 4320 psi, sc 1870 psi, tmax 3100 psi
Pure shear: tmax 0.804 MPa
(a) dmin 1.65 in.; (b) Pmax 19.25 lb
(a) s1 29.3 MPa, s2 175.9 MPa,
tmax 102.6 MPa; (b) s1 156.1 MPa,
s2 33 MPa, tmax 94.5 MPa
(a) s1 0 psi, s2 20,730 psi,
tmax 10,365 psi; (b) s1 988 psi,
s2 21,719 psi, tmax 11,354 psi
Maximum: st 18.35 MPa,
sC 18.35 MPa, tmax 9.42 MPa
Top of beam: s1 8591 psi, s2 0 psi,
tmax 4295 psi
(a) dAl 26.3 mm; (b) dTi 21.4 mm
d2
1FL2
2
sy 0, sx 1943 psi,
Ip
2
d2
T
2
547 psi, s1 2087 psi,
txy Ip
8.5-26
s2 143.2 psi, tmax 1115 psi
(a) s1 0, s2 sx 108.4 MPa,
sx
54.2 MPa;
tmax 2
(b) s1 0.703 MPa, s2 1.153 MPa,
tmax 0.928 MPa; (c) Pmax 348 N
8.5-27 sx 18.6 ksi, sy 0, txy 4.45 ksi,
s1 1.012 ksi, s2 19.62 ksi, tmax 10.31 ksi
1073
CHAPTER 9
9.2-1
9.2-2
9.2-3
9.2-4
9.3-1
9.3-2
9.3-3
9.3-4
9.3-5
9.3-6
9.3-7
q q0x/L; Triangular load, acting downward
(a) q q0 sin px/L, Sinusoidal load;
(b) RA RB q0L/p ; (c) Mmax qoL2/p 2
q q0(1 x/L); Triangular load, acting
downward
(a) q q0(L2 x2)/L2; Parabolic load,
acting downward; (b) RA 2q0L/3,
MA q0L2/A
dmax 0.182 in., u 0.199
h 96 mm
L 120 in. 10 ft
dmax 15.4 mm
d/L 1/400
Eg 80.0 GPa
dC
Let b a/L:
dmax
3 2311 + 8b 4b 22
The deflection at the midpoint is close to the
maximum deflection. The maximum difference
is only 2.6%.
2
2
9.3-11 v mx (3L x)/6EI, dB mL /3EI,
2
uB mL /2EI
q
9.3-12 v1x2 12x 4 12x 2L2 + 11L42,
48EI
qL4
48EI
See Table H-2, Case 9.
See Table H-1, Case 2.
q0L 3
L,
v1x2 1x 2Lx 22 for 0 x
24 EI
2
q0
(160L2x3 160L3x2 v1x2 960LEI
80Lx4 16x5 25L4x 3L5) for
L
7 q0L4 ,
1 q0L4
dC x L, dB 2
160 EI
64 EI
q0x
2 2
1200x L 240x 3L
v1x2 5760LEI
L
96x 4 53L42 for 0 … x … ,
2
q0L
140x 3 120Lx 2 + 83L2x 3L32
v1x2 5760EI
3q0L4
L
for
x L, dC 2
1280EI
dB 9.3-13
9.3-14
9.3-15
9.3-16
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
78572_ans_ptg01_hr_1057-1082.qxd
1074
9.3-17
1/19/12
10:47 AM
Page 1074
Answers to Problems
PL
14104x 2 + 3565L22
10,368EI
L
P
x
, v1x2 3
1152EI
v1x2 for 0
(648Lx2 192x3 64L2x 389L3)
L
L
P
for
… x … , v1x2 3
2
144EIL
(72L2x2 12Lx3 6x4 5L3x 49L4)
L
3565PL3 ,
3109PL3
… x … L , dA for
dC 2
10,368EI
10,368EI
9.4-3
9.4-4
9.4-5
9.4-6
9.4-7
9.4-8
9.4-9
9.4-10
v M0x(L x)2/2LEI,
dmax 2M0L2/27EI (downward)
q
12x 4 12x 2L2 + 11L42,
v1x2 48EI
qL3
uB 3EI
See Table H-1, Case 10.
v q0x2(45L4 40L3x 15L2x2 x4)/
360L2EI, dB 19q0L4/360EI, uB q0L3/15EI
v q0x(3L5 5L3x2 3Lx4 x5)/90L2EI,
dmax 61q0L4/5760EI
q0
v1x2 1x 5 5Lx 4 + 20L3x 2 16L52,
120EIL
2q0L4
dmax 15EI
qL2 2
1x L22 for 0 x L,
v1x2 16EI
q
120L3x + 27L2x 2 v1x2 48EI
3L ,
12Lx 3 + 2x 4 + 3L42 for L … x …
2
9qL4
7qL3
, uC dC 128EI
48EI
v1x2 q0L2
L
120x 2 + 19L22 for 0 … x … ,
480EI
2
v1x2 9.5-5
9.5-6
9.5-7
9.5-8
9.5-9
9.5-10
9.5-11
9.5-12
9.5-13
9.5-14
9.5-15
9.5-16
9.5-17
9.5-18
9.5-19
9.5-20
uB 7PL2/9EI, dB 5PL3/9EI
(a) d1 11PL3/144EI; (b) d2 25PL3/384EI;
(c) d1/d 2 88/75 1.173
(a) a/L 2/3; (b) a/L 1/2
(a) dC 6.25 mm (upward)
(b) dC 18.36 mm (downward)
y Px2(L x)2/3LEI
uB 7qL3/162EI, dB 23qL4/648EI
dC 0.0905 in., dB 0.293 in.
(a) M PL/2; (b) M 5PL/24,
uB PL2/12EI; (c) M PL/8,
dB PL3/24EI
M (19/180)q0L2
(a) dA PL2(10L 9a)/324EI (positive upward);
(b) Upward when a/L 10/9, downward when a/L
10/9
(a) dC PH2(L H)/3EI; (b) dmax PHL2/9 13EI
dC 3.5 mm
uB qoL3/10EI, dB 13q0L4/180EI
uA q(L3 6La2 4a3)/24EI,
dmax q(5L4 24L2a2 16a4)/384EI
(a) P/Q 9a/4L; (b) P/Q 8a(3L a)/9L2;
(c) P/qa 9a/8L for dB 0,
P/qa a(4L a)/3L2 for dD 0
d 19WL3/31,104EI
k 3.33 lb/in.
M1 7800 Nm, M2 4200 Nm
6Pb 3
d
EI
47Pb 3
dE 12EI
dC 0.120 in.
q 16cEI/7L4
dh Pcb2/2EI, dv Pc2(c 3b)/3EI
d PL2(2L 3a)/3EI
M
9.5-25 (a) HB 0, VB , VC VB ;
L
5ML ,
ML ,
ML ,
(b) uA u uC 6EI B 3EI
6EI
9.5-21
9.5-22
9.5-23
9.5-24
uD uC;
q0
(80Lx 4 16x 5 960EIL
120L2x 3 + 40L3x 2 25L4x + 41L52
L
for … x … L,
2
19q0L4 ,
13q0L3 ,
7q0L4
dA uB dC 480EI
192EI
240EI
9.5-1
9.5-2
9.5-3
9.5-4
(c) dA (7/24)ML2/EI (to the left),
dD (1/12)ML2/EI (to the right);
(d) L CD 9.5-26
114
L 1.871L
2
P,
2P
;
V 3 C
3
4 ML ,
(b) uA a
uB uA,
b
81 EI
(a) HB 0, VB uC a
5 ML ,
b
uD uC ;
81 EI
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
78572_ans_ptg01_hr_1057-1082.qxd
1/19/12
10:47 AM
Page 1075
Answers to Problems
2L2M
L
(to the right),
(c) dA uB a b 2
81EI
9.7-9
v
8PL3
2x
1
L
c
EIA 2L + x
9L
9
L
5L2M
dD uC a b (to the left);
2
162EI
(d) L CD (a) b/L 0.403; (b) dC 0.00287qL4/EI
a 22.5, 112.5, 67.5, or 157.5
uB 7qL3/162EI, dB 23qL4/648EI
dB 0.443 in., dC 0.137 in.
dB 11.8 mm, dC 4.10 mm
P 64 kN
uA M0L/6EI, uB M0L/3EI,
d M0L2/16EI
9.6-10 uA Pa(L a)(L 2a)/6LEI,
d1 Pa2(L 2a)2/6LEI, d2 0
2
9.6-11 uA M0L/6EI, uB 0, d M0L /27EI
(downward)
9.7-1
(a) dB PL3(1 7I1/I2)/24EI1;
(b) r (1 7I1/I2)/8
9.7-2
(a) dB qL4(1 15I1/I2)/128EI1;
(b) r (1 15I1/I2)/16
9.7-3
(a) dC 0.31 in. (upward);
(b) dC 0.75 in. (downward)
9.7-4
v qx(21L3 64Lx2 32x3)/768EI for
0 x L/4,
v q(13L4 256L3x 512Lx3
256x4)/12,288EI for L/4 x L/2,
uA 7qL3/256EI, dmax 31qL4/4096EI
9.7-5
uA 8PL2/243EI, dB 8PL3/729EI,
dmax 0.01363PL3/EI
9.7-6
v 2Px(19L2 27x2)/729EI for
0 x L/3,
v P(13L3 175L2x 243Lx2 81x3)/
1458EI for L/3 x L,
uA 38PL2/729EI, uC 34PL2/729EI,
dB 32PL3/2187EI
3x
PL3
L
1
v
c
+
9.7-7
EIA 21L + x2
8L
8
9.5-27
9.5-28
9.6-4
9.6-5
9.6-6
9.6-8
9.6-9
L + x
+ ln a
b d,
2L
PL3
18 ln 2 52
8EIA
4L12L + 3x2
PL3
2x
v
c7 d,
24EIA
L
1L + x22
dA 9.7-8
dA + ln a
2 15L
0.894L
5
PL3
24EIA
dA 9.7-10
2L + x
b d,
3L
8PL3
3
7
a ln b
EIA
2
18
v1x2 a
19,683PL3
81L
a
+ 2 ln b
2000EIA 81L + 40x
81
40x
6440x
3361
+
b a
b,
121
121L
14,641L
14,641
dA 19,683PL3
7,320,500EIA
a2820 + 14,641 lna
9.7-11
v1x2 dB 11
bb
9
19,683PL3
2000EIA
a
81L
40x
+ 2 ln a1 +
b
81L + 40x
81L
6440x
1b ,
14,641L
19,683PL3
7,320,500EIA
a2820 + 14,641 lna
9.7-12
1075
(a) v¿ 11
bb
9
qL3
8Lx 2
c1 d
16EIA
1L + x23
for 0 x L,
2
2
qL 19L + 14Lx + x 2x
v
c
2EIA
8L1L + x22
x
ln a1 + b d for 0 x L;
L
qL413 4 ln 22
qL3
, dC (b) uA 16EIA
8EIA
4
9.8-1
U 4bhLs 2max /45E
9.8-2
(a) and (b) U P2L3/96EI;
(c) d PL3/48EI
q 2L3
(a) and (b) U 15EI
9.8-3
9.8-4
9.8-5
(a) U 32EId 2/L3; (b) U p 4EId 2/4L3
(a) U P2a2(L a)/6EI;
(b) dC Pa2(L a)/3EI;
(c) U 241 in.-lb, dC 0.133 in.
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
78572_ans_ptg01_hr_1057-1082.qxd
1076
1/19/12
10:47 AM
Answers to Problems
L
117L4q 2 + 280qL2M 0
15,360EI
+ 2560M 202
9.8-6
U
9.8-7
9.9-11
9.9-12
9.10-1
dB 2PL3/3EI 8 12PL/EA
dD Pa2b2/3LEI
dC Pa2(L a)/3EI
dC L3(2P1 5P2)/48EI,
dB L3(5P1 16P2)/48EI
uA 7qL3/48EI
dC Pb2(b 3h)/3EI,
uC Pb(b 2h)/2EI
dC 31qL4/4096EI
uA MA(L 3a)/3EI,
dA MAa(2L 3a)/6EI
dC Pa2(L a)/3EI P(L a)2/kL2
dD 37qL4/6144EI (upward)
smax sst[1 (1 2h/dst)1/2]
9.10-2
smax 318WEh/AL
9.9-2
9.9-3
9.9-6
9.9-7
9.9-8
9.9-9
9.9-10
9.10-3
9.10-4
9.10-5
9.10-6
dmax 0.302 in., smax 21,700 psi
d 281 mm
W 14 53
h 360 mm
9.10-7
R 33EIIm2/L3
9.11-1
v a(T2 T1)(x)(L x)/2h
(pos. upward),
uA aL(T2 T1)/2h (clockwise),
dmax aL2(T2 T1)/8h (downward)
v a(T2 T1)(x2)/2h (upward),
uB aL(T2 T1)/h (counterclockwise),
dB aL2(T2 T1)/2h (upward)
a1T2 T121x 2 L22
9.11-3 v1x2 ,
2h
a1T2 T121L + a2
uC h
(counterclockwise),
a1T2 T1212La + a 22
(upward)
dC 2h
9.11-2
(a) dmax aT0L3
(downward),
6h
aT0L4
dmax (downward)
12h
(c) dmax CHAPTER 10
RA RB 3M0/2L, MA M0/2,
v M0x2(L x)/4LEI
2
10.3-2 RA RB qL/2, MA MB qL /12,
2
2
v qx (L x) /24EI
3
2
10.3-3 RA RB 3EIdB/L , MA 3EIdB/L ,
2
3
v dBx (3L x)/2L
10.3-1
10.3-4
uB qL3
61k RL EI2
,
k RqL5
1
dB qL4 +
8
121k RL EI2
10.3-5
RA V102 9
q L,
40 0
RB V1L2 MA 10.3-6
11
q L,
40 0
7
q0L2
120
7
q L,
60 0
13
RB V1L2 q L,
60 0
1
MA q L2,
30 0
q0
n
1x 6 + 7L3x 3 6q0L4x 22;
360L2EI
(a) RA V102 (b) RA V102 0.31q0L
a
2
6
p 2 4p + 8
p4
p
RB V1L2 0.327q0L
a6
aT0L3
(downward);
913h
aT0L41212 12
(b) dmax (downward)
48h
aT0L3
9.11-5 (a) dmax (downward);
6h
aT0L4
(b) dmax (downward);
12h
9.11-4
Page 1076
p 2 4p + 8
M A 2q0L
n
≥
b q0L,
b q0L,
p4
2
2 p 12p + 24 ,
p4
1
EI
q0 a
2L 4
px
p 2 4p + 8 x3
b sin a b 6q0L
p
2L
6
p4
2q0L2
p 2 12p + 24 x2
2L 3
+ q0 a b x
4
p
2
p
¥
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
78572_ans_ptg01_hr_1057-1082.qxd
1/19/12
10:47 AM
Page 1077
Answers to Problems
61Lq0
,
120
19Lq0
,
RB 120
(a) RA 10.3-7
n 1x2 (b) RA MA 11L2q0
,
120
q0x 2133L4 61L3x + 30L2x 2 2x 42
720EIL2
48Lq0
p4
,
L
2Lq0
48Lq0 ,
RB q1x2 dx RA p4
30
p
MA 2L2q01p 22
L
0
3
q1x2x dx RBL La
2Lq0
48Lq0
p4
p
p2
b,
16L4q0 24L2q0x 2 + 8Lq0x 3 16L4q0 cos A px
2L B
n 1x2 10.3-8
p 4EI
(a) RA V102 24
p4
RB V1L2 MA a
MB a
v
12
p4
12
p4
1
4
1
p2
1
p2
q0L,
24
p4
q0L,
bq0L2
(counterclockwise),
bq0L2
(counterclockwise),
[q0L4 cos a
px
b
L
p EI
4q0Lx 3 6q0L2x 2 + q0L4];
(b) RA RB q0L/p, MA MB 2q0L2/p3,
v q0L2(L2 sin px/L px2 pLx)/p 4EI
4814 p2
10.3-9 (a) RA V102 q0L,
p4
4814 p2
2
RB V1L2 a b q0L,
p4
p
M A q0 a
MB 2L
2
b +
p
321p 32
p4
1616 p2
p4
q0L2,
1
px
[16q0L4 cos a b
2L
p 4EI
3
814 p2q0Lx
v
q0L2,
1077
13
q L,
30 0
7
RB V1L2 q L,
30 0
1
(counterclockwise),
MA q L2
15 0
1
q L2 (counterclockwise),
MB 20 0
q0
[x 6 15L2x 4 + 26L3x 3
v
360L2EI
12L4x 2]
3
10.3-10 RA V102 q0L,
20
7
RB V1L2 q L,
20 0
1
q L2,
MA 30 0
1
1q0x 5 + 3q0Lx 3 2q0L2x 22
v
120LEI
10.3-11 RA RB 3M0/2L, MA MB M0 /4,
v M0x2(L 2x)/8LEI for 0 x L/2
9 M0,
10.3-12 RB 8 L
9 M0,
RA 8 L
1 M0,
MA 8 L
M0 2
1 9M 0 3
L
v
a
x x b a0 … x … b ,
EI 48L
16
2
1
v
EI
9M 0 3
9M 0 2
M 0L
M 0L2
a
x x +
x
b
48L
16
2
8
L
a … x … Lb
2
(b) RA V102 ;
RA Pb(3L2 b2)/2L3,
RB Pa2(3L a)/2L3, MA Pab(L b)/2L2
qL2
qL2
10.4-2 RA qL, M A , MB 3
6
1
17
1
10.4-3 RA qL, RB qL, M A qL2
8
8
8
10.4-4 (a) RA M0/3L, HA 4M0 /3L,
RB RA, RC HA;
(b) uA M0L/18EI, uB M0L/9EI,
uC uA; (c) LBC 2L
10.4-1
816 p2q0L2x 2 + 16q0L4];
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
78572_ans_ptg01_hr_1057-1082.qxd
1078
1/19/12
10:47 AM
Page 1078
Answers to Problems
4M 0 ,
2M 0 ,
4M 0 ,
HA RB 3L
3L
3L
2M 0
;
RC 3L
M 0L ,
5 M 0L ,
(b) uA uB 18 EI
18EI
M 0L
;
uC 36EI
(c) LAB 2.088L
M 0k R
M0
,
+
10.4-6 (a) RA L
213EI + Lk R2
RB RA,
LM 0k R
MB (CCW);
6EI + 2Lk R
LM 0
LM 0
+
(b) uA 4EI
413EI + Lk R2
10.4-5
10.4-11 RA RB q0L/4, MA MB 5q0L /96
10.4-12 RA qL/8, RB 33qL/16,
2
(a) RA RC 13qL/16
10.4-13 RA 1100 lb (downward), RB 2800 lb (upward),
MA 30,000 lb-in. (clockwise)
10.4-14 RB 6.44 kN
10.4-15 (a) The tension force in the tie rod For kR goes to zero:
uA LM 0
LM 0
LM 0
+
4EI
3EI
413EI2
For kR goes to infinity: uA M 0L
4EI
For kR goes to 6EI/L:
uA 10.4-7
LM 0
+
4EI
LM 0
4c3EI + L a
6EI
bd
L
3 M0,
2 L
HB 0, VB 0, VC 0, HD HA;
M 0L ,
(b) uA uD uA,
16EI
M 0L ,
uC uB;
uB 8EI
M0,
M0,
(c) HA HB 2
L
L
M0 ,
M 0 ,
HD HA,
VB VC L
L
M 0L ,
M 0L ,
uD uA, uB uA 24EI
12EI
(a) HA uC uB
tAB /tCD LAB /LCD
7
17
7
10.4-9 RA qL, RB qL, M A qL2
12
12
12
10.4-8
10.4-10 RA 2qL, M B 5LM 0
18EI
RD 604 lb; (b) RA 796 lb,
MA 1308 lb-ft 1.567 104 lb-in.
10.4-16 RA 31qL/48, RB 17qL/48,
MA 7qL2/48
10.4-17 (a) RA 23P/17, RD RE 20P/17,
MA 3PL/17; (b) Mmax PL/2
10.4-18 RA RD 2qL/5, RB RC 11qL/10
10.4-19 MB(q) (800 q) lb-in. for q
250 lb/in., MB(q)
(200 q 150,000) lb-in. for
q 250 lb/in.
3
10.4-20 RA RB 6M0ab/L ,
2
MA M0b(3a L)/L ,
MB M0a(3b L)/L2
10.4-21 s 509 psi
2
10.4-22 (MAB)max 121qL /2048 6.05 kNm;
2
(MCD)max 5qL /64 8.0 kNm
10.4-23 F 3160 lb, MAB 18,960 lb-ft,
MDE 7320 lb-ft
3
3
10.4-24 k 48EI(6 5 12)/7L 89.63EI/L
10.4-25 (a) VA VC 3P/32, HA P,
MA 13PL/32; (b) Mmax 13PL/32
35
29
10.4-26 HA P, HC P,
64
64
35
M max PL
128
10.4-27 RA RB 3000 lb, RC 0
2
2
10.4-28 (a) MA MB qb(3L b )/24L;
2
(b) b/L 1.0, MA qL /12;
(c) For a b L/3, (Mmax)pos 19qL2/648
4
10.4-29 (a) d2/d1 148 1.682;
(b) Mmax qL2 (3 2 12)/2 0.08579qL2;
(c) Point C is below points A and B by the amount
0.01307qL4/EI
2
10.4-30 Mmax 19q0L /256, smax 13.4 MPa,
4
smax 19q0L /7680EI 0.00891 mm
243E SE WIAHa1¢T2
10.5-1 S 4AL3E S + 243IHE W
7
qL2
12
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
78572_ans_ptg01_hr_1057-1082.qxd
1/19/12
10:47 AM
Page 1079
Answers to Problems
10.5-2
a1T2 T12L2
(a) RA RB 2h
a1T2 T12L2
2h
#a
3EI # k
b,
3EI + L3 # k
# a 3EI # k3 b ,
3EI + L # k
M A RB L
a1T2 T12L3
# a 3EI # k3 b ;
2h
3EI + L # k
3EIa1T2 T12
(b) RA RB (upward),
2hL
3EIa1T2 T12
RB (downward),
2hL
3EIa1T2 T12
M A RB L (counterclockwise)
2h
a1T2 T12L2
10.5-3 RA RB 2h
b (upward),
3EI + L3 # k
a1T2 T12L2
# a 3EI # k3 b (downward),
RB 2h
3EI + L # k
a1T2 T12L3
M A RB L 2h
#a
3EI # k
b (counterclockwise)
3EI + L3 # k
a1T1 T22L2
10.5-4 (a) RB h
#a
a
3
b (upward),
36EI + L # k
a(T1 T2)L2
3
RC RB 4
2h
b (upward);
36EI + L3 # k
6Ela (T1 T2)
(downward),
(b) RB Lh
a
9EI # k
3Ela (T1 T2)
(upward),
RA 2Lh
9Ela (T1 T2)
(upward)
RC 2Lh
a (T1 T2)L2
6EI # k
b
a
h
36EI + L3 # k
(downward),
a(T1 T2)L
1
RA RB 4
2h
b (upward),
36EI + L3 # k
a1T1 T22L2
3
RC RB 4
2k
a
3EI # k
a
9EI # k
36EI + L3 # k
b (upward)
(a) H p 2EAd 2/4L2, st p 2Ed 2/4L2;
(b) st 617, 154, and 69 psi
2 7
2 2
2
10.6-2 (a) l 17q L /40,320E I , sb qhL /16I;
2 6
2
(b) st 17q L /40,320EI ;
(c) l 0.01112 mm, sb 117.2 MPa,
st 0.741 MPa
10.6-1
CHAPTER 11
11.2-1
Pcr bR /L
11.2-2
(a) Pcr 11.2-3
Pcr 6bR/L
11.2-4
(a) Pcr (L a)(ba 2 + b R)
;
aL
(b) Pcr bL2 + 20b R
4L
6EI # k
3EI # k
RB 2
3EI # k
b (downward),
36EI + L3 # k
a1T1 T22L3
1
RA RB 4
2h
a
10.5-5
1079
ba 2 + 2b R
ba 2 + b R
; (b)Pcr L
L
11.3-1
11.3-2
11.3-3
11.3-4
11.3-5
3b R
L
3
Pcr bL
5
7
Pcr bL
4
(a) Pcr 453 k; (b) Pcr 152 k
(a) Pcr 2803 kN; (b) Pcr 953 kN
(a) Pcr 650 k; (b) Pcr 140 k
Mallow 1143 kNm
Qallow 23.8 k
11.3-6
(a) Qcr 11.3-7
(a) Qcr 11.2-5
11.2-6
11.2-7
Pcr p 2EI
2
; (b) Qcr L
2p 2EI
L2
2p 2EI
; (b) Mcr 9L2
3dp 2EI
L2
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
78572_ans_ptg01_hr_1057-1082.qxd
1080
1/19/12
10:47 AM
Page 1080
Answers to Problems
11.3-8
T p I/aAL
11.3-9 h/b 2
3
4
2
3
4
2
11.3-10 (a) Pcr 3p Er /4L ; (b) Pcr 11p Er /4L
11.3-11 P1 : P2 : P3 1.000 : 1.047 : 1.209
11.3-12 Pallow 604 kN
11.3-13 Fallow 54.4 k
11.3-14 Wmax 124 kN
11.3-15 tmin 0.165 in.
11.3-16 Pcr 497 kN
11.3-17 Wcr 51.9 k
11.3-18 u arctan 0.5 26.57
4
11.3-19 (a) qmax 142.4 lb/ft; (b) Ib,min 38.5 in. ;
2
2
(c) s 0.264 ft, 2.42 ft
11.3-20 Pcr 3.56 kN
11.3-21 Pcr 16.28 k
11.4-1 Pcr 235 k, 58.7 k, 480 k, 939 k
11.4-2 Pcr 62.2 kN, 15.6 kN, 127 kN, 249 kN
11.4-3 Pallow 253 k, 63.2 k, 517 k, 1011 k
11.4-4 Pallow 678 kN, 169.5 kN, 1387 kN, 2712 kN
11.4-5 Pcr 229 k
11.4-6 Tallow 18.1 kN
11.4-7 (a) Qcr 4575 lb; (b) Qcr 10,065 lb,
a 0 in.
11.4-8 Pcr 447 kN, 875 kN, 54.7 kN, 219 kN
2
2
11.4-9 Pcr 4p EI/L , vd(lcos 2px/L)/2
11.4-10 tmin 10.0 mm
2
11.4-11 (b) Pcr 413.89EI/L
11.5-1 d 0.112 in., Mmax 1710 lb-in.
11.5-2 d 8.87 mm, Mmax 2.03 kNm
11.5-3 For P 0.3Pcr:
M/Pe 1.162 (sin 1.721 x/L) cos 1.72lx/L
2
11.5-4 P 583.33{arccos [5/(5 d)]} , in which
P kN and d mm; P 884 kN when
d 10 mm
2
11.5-5 P 125.58{arccos [0.2/(0.2 d)]} , in which
P k and d in.; P 190 k when
d 0.4 in.
11.5-6 Pallow 49.9 kN
11.5-7 Lmax 150.5 in. 12.5 ft
11.5-8 Lmax 3.14 m
11.5-9 d e(sec kL 1), Mmax Pe sec kL
11.5-10 Lmax 2.21 m
11.5-11 Lmax 130.3 in. 10.9 ft
11.5-12 Tmax 8.29 kN
11.5-13 (a) q0 2230 lb/ft 186 lb/in.;
(b) Mmax 37.7 kin., ratio 0.47
11.6-1 (a) smax 17.3 ksi; (b) Lmax 46.2 in.
11.6-2 Pallow 37.2 kN
11.6-3 bmin 4.10 in.
11.6-4 (a) smax 38.8 MPa; (b) Lmax 5.03 m
11.6-5 (a) smax 9.65 ksi; (b) Pallow 3.59 k
11.6-6 d2 131 mm
11.6-7 (a) smax 10.9 ksi; (b) Pallow 160 k
11.6-8 (a) smax 104.5 MPa; (b) Lmax 3.66 m
11.6-9 (a) smax 9.60 ksi; (b) Pallow 53.6 k
11.6-10 (a) smax 47.6 MPa; (b) n 2.30
11.6-11 (a) smax 13.4 ksi; (b) n 2.61
11.6-12 (a) smax 120.4 MPa; (b) P2 387 kN
11.6-13 (a) smax 17.6 ksi; (b) n 1.89
11.6-14 (a) smax 115.2 MPa; (b) P2 193 kN
254
8
177
16
11.9-1 Pallow §
¥ k for L § ¥ ft
97
24
55
32
3019
3m
2193
6m
11.9-2 Pallow §
¥ kN for §
¥
1285
9m
723
12 m
10
338
20
240
11.9-3 Pallow §
¥ k for L § ¥ ft
30
135
40
76
11.9-4
11.9-5
11.9-6
W 250 67
W 12 87
W 360 122
60.7
6
42.4
12
11.9-7 Pallow §
¥ k for L § ¥ ft
23.3
18
13.1
24
1104
2.5
919
5.0
11.9-8 Pallow §
¥ kN for L §
¥m
678
7.5
441
10.0
96.9
6.0
73.9
9.0
11.9-9 Pallow §
¥ k for L §
¥ ft
50.6
12.0
32.6
15.0
2.6
229
2.8
207
11.9-10 Pallow §
¥ kN for L § ¥ m
3.0
185
3.2
164
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
78572_ans_ptg01_hr_1057-1082.qxd
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10:47 AM
Page 1081
Answers to Problems
11.9-11 Lmax 5.13 ft
11.9-12 Lmax 3.52 m
11.9-13 Lmax 13.9 ft
11.9-14 Pallow 5520 kN
11.9-15 (a) L max 21.25 ft; (b) L max 14.10 ft
11.9-16 (a) L max 6.44 m; (b) L max 4.68 m
11.9-17 d 4.88 in.
11.9-18 d 99.8 mm
11.9-19 d 5.25 in.
11.9-20 d 190 mm
96.5
0.6
84.1
0.8
11.9-24 Pallow §
¥ kN for L § ¥ m
61.0
1.0
42.4
1.2
11.9-25 (a) L max 25.2 in.; (b) dmin 2.11 in.
11.9-26 (a) L max 468 mm; (b) dmin 42.8 mm
11.9-27 (a) L max 14.9 in.; (b) dmin 1.12 in.
11.9-28 (a) L max 473 mm; (b) dmin 33.4 mm
5.0
11.9-29 Pallow £ 28.0 ≥ k for L £ 7.5 ≥ ft
193.8
10.0
1.5
11.9-30 Pallow £ 177.3 ≥ kN for L £ 2.0 ≥ m
153.5
2.5
22.8
6.0
11.9-31 Pallow £ 20.2 ≥ k for L £ 8.0 ≥ ft
16.7
10.0
310
2.5
11.9-32 Pallow £ 255 ≥ kN for L £ 3.5 ≥ m
190
12.5-1
12.5-2
12.5-3
x y 5a/12
y 1.10 in.
2c2 ab
y 13.94 in.
y 52.5 mm
x 0.99 in., y 1.99 in.
x 137 mm., y 132 mm
Ix 518 103 mm4
Ix 36.1 in.4, Iy 10.9 in.4
Ix Iy 194.6 106 mm4,
rx ry 80.1 mm
I1 1480 in.4, I2 186 in.4, r1 7.10 in.,
r2 2.52 in.
Ib 940 in.4
Ic 11a4/192
Ixc 7.23 in.4
12.5-4
I2 405 103 mm4
12.5-5
Ixc 6050 in.4, Iyc 475 in.4
12.5-6
Ixc 106 106 mm4
12.5-7
Ixc 17.40 in.4, Iyc 6.27 in.4
12.5-8
12.6-1
12.6-2
12.6-3
12.6-4
12.6-5
12.7-2
12.7-3
12.7-4
12.7-5
12.7-6
12.7-7
b 250 mm
IP bh(b2 12h2)/48
(IP)C r4(9a 2 8 sin2 a)/18a
IP 233 in.4
IP bh(b2 h2)/24
(IP)C r 4(176 84p 9p 2)/[72(4 p)]
Ixy r 4/24
b 2r
Ixy t2(2b2 t2)/4
I12 20.5 in.4
Ixy 24.3 106 mm4
Ixc yc 6.079 in.4
12.8-1
Ix1 Iy1 b4/12, Ix1y1 0
12.8-2
Ix1 12.3-4
2.0
18.1
14.8
3.0
11.9-23 Pallow §
¥ k for L § ¥ ft
8.4
4.0
5.4
5.0
19.9
CHAPTER 12
12.3-3
315
1.0
245
2.0
11.9-22 Pallow §
¥ kN for L § ¥ m
127
3.0
71
4.0
34.6
11.9-33 (a) L max 10.37 ft; (b) bmin 5.59 in.
11.9-34 (a) L max 3.45 m; (b) bmin 154.9 mm
11.9-35 (a) L max 10.25 ft; (b) bmin 3.47 in.
11.9-36 (a) L max 2.50 m; (b) bmin 134.8 mm
12.3-2
143.5
6.0
124.0
8.0
11.9-21 Pallow §
¥ k for L §
¥ ft
85.5
10.0
59.4
12.0
1081
12.3-5
12.3-6
12.3-7
12.3-8
12.4-6
12.4-7
12.4-8
12.4-9
b 3h3
4
4
, I bh(b + h ) ,
y
2
2
6(b + h ) 1 12(b 2 + h2)
b 2h2(h2 b 2)
Ix1y1 12(b 2 + h2)
4.5
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
78572_ans_ptg01_hr_1057-1082.qxd
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1/19/12
10:47 AM
Page 1082
Answers to Problems
12.8-3
Id 159 in.4
12.8-4
Ix1 12.44 10 mm ,
Iy1 9.68 106 mm4, Ix1y1 6.03 106 mm4
12.8-5
12.9-4
6
4
Ix1 13.50 in.4, Iy1 3.84 in.4,
12.9-5
12.9-6
Ix1y1 4.76 in.4
12.8-6
Ix1 8.75 106 mm4, Iy1 1.02 106 mm4,
Ix1y1 0.356 106 mm4
(a) c 2a 2 b 2/2; (b) a/b 15;
(c) 1 a/b 15
12.9-2 Shows that two different sets of principal axes exist
at each point.
4
12.9-3 up1 29.87, up2 60.13, I1 311.1 in. ,
4
I2 88.9 in.
12.9-1
12.9-7
12.9-8
12.9-9
up1 8.54, up2 81.46,
I1 17.24 106 mm4, I2 4.88 106 mm4
up1 37.73, up2 127.73, I1 15.45 in.4,
I2 1.89 in.4
up1 32.63, up2 122.63,
I1 8.76 106 mm4, I2 1.00 106 mm4
up1 16.85, up2 106.85, I1 0.2390b4,
I2 0.0387b4
up1 74.08, up2 15.92,
I1 8.29 106 mm4, I2 1.00 106 mm4
up1 75.73, up2 14.27, I1 20.07 in.4,
I2 2.12 in.4
APPENDIX A
A1.1
A1.2
A1.3
A1.4
A1.5
A1.6
A1.7
A1.8
A1.9
A1.10
A1.11
A1.12
A1.13
A1.14
A1.15
A1.16
A2.1
A2.2
A2.3
A2.4
A2.5
A2.6
A2.7
A2.8
A2.9
A2.10
A2.11
A2.12
A2.13
A2.14
C
C
D
A
B
A
A
D
A
C
D
D
D
A
B
C
D
B
A
A
D
A
B
C
A
D
D
C
A
C
A2.15
A2.16
A3.1
A3.2
A3.3
A3.4
A3.5
A3.6
A3.7
A3.8
A3.9
A3.10
A3.11
A3.12
A3.13
A3.14
A3.15
A4.1
A4.2
A4.3
A4.4
A4.5
A4.6
A4.7
A5.1
A5.2
A5.3
A5.4
A5.5
A5.6
D
C
D
A
C
A
D
D
B
B
C
B
D
B
B
D
B
D
C
D
A
A
C
B
A
C
D
D
A
B
A5.7
A5.8
A5.9
A5.10
A5.11
A5.12
A6.1
A6.2
A6.3
A6.4
A6.5
A7.1
A7.2
A7.3
A7.4
A7.5
A7.6
A7.7
A7.8
A8.1
A8.2
A8.3
A8.4
A8.5
A8.6
A8.7
A8.8
A8.9
A8.10
A8.11
B
C
C
A
B
B
B
C
B
B
D
C
C
D
A
B
A
C
D
A
C
D
B
C
A
D
D
A
D
D
A8.12
A8.13
A8.14
A8.15
A9.1
A9.2
A9.3
A9.4
A9.5
A9.6
A9.7
A10.1
A10.2
A10.3
A10.4
A10.5
A10.6
A10.7
A11.1
A11.2
A11.3
A11.4
A11.5
A11.6
A11.7
A11.8
A11.9
A11.10
A11.11
A
C
B
C
C
C
B
C
B
C
D
B
B
B
D
B
D
A
D
B
D
A
D
A
B
B
B
C
D
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 1083
Appendix A
FE Exam Review Problems
A-1.1: A plane truss has downward applied load P at joint 2 and another load P
applied leftward at joint 5. The force in member 3–5 is:
(A) 0
(B) P/2
(C) P
(D) 1.5 P
Solution
M1 0
V6 (3 L) P L P L 0
so
3
5
P
V6 0
Method of sections
L
Cut through members 3-5,
2-5 and 2-4; use right hand
FBD
1
2
L
L
P
M 2 0
6
4
L
5
P
F35 L P L 0
3
F35 P
L
1
H1
V1
2
L
P
6
4
L
5
L
V6
P
F35
F25
L
2
F24
4
6
L
V6
1083
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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12:15 PM
Page 1084
APPENDIX A FE Exam Review Problems
A-1.2: The force in member FE of the plane truss below is approximately:
(A) 1.5 kN
(B) 2.2 kN
(C) 3.9 kN
(D) 4.7 kN
Solution
3m
A
15 kN
3m
B
10 kN
3m
C
5 kN
D
3m
E
4.5 m
F
G
1m
Statics
MA 0
Ey (6 m) 15 kN (3 m) 10 kN (6 m) 5 kN (9 m) 0
E y 25 kN
Ax
A
Ay
4.5 m
G
3m
B
15 kN
3m
C
10 kN
3m
5 kN
D
3m
E
F
1m
Ey
Method of sections: cut through BC, BE and FE; use right-hand FBD;
sum moments about B
1
3
FFE (3 m) FFE (3 m) 10 kN (3 m) 5 kN (6 m) Ey (3 m) 0
110
110
Solving
FFE 5
110 kN
4
5 kN
10 kN
B
FFE 3.95 kN
3
·FFE
10
FFE
3m
C
D
3m
E
1m
1
·FFE
10
Ey
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 1085
1085
APPENDIX A FE Exam Review Problems
A-1.3: The moment reaction at A in the plane frame below is approximately:
(A) 1400 Nm
(B) 2280 Nm
(C) 3600 Nm
(D) 6400 Nm
Solution
900 N
1.2 m
1200 N/m
900 N
1.2 m
C
B
3m
4m
Pin
connection
Bx
B
C
3m
By
Cy
A
Statics: use FBD of member BC to find reaction C yy
MB 0
Cy Cy (3 m) 900 N (1.2 m) 0
900 N (1.2 m)
360 N
3m
Sum moments about A for entire structure
M A 0
1
N
2
MA Cy (3 m) 900 N (1.2 m) a1200 b 4 m a 4 mb 0
m
2
3
Solving for MA
M A 6400 Nm
900 N
1.2 m
1200 N/m
B
C
3m
Cy
4m
MA
A
Ax
Ay
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
78572_app_ptg01_hr_1083-1168.qxd
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1/24/12
12:15 PM
Page 1086
APPENDIX A FE Exam Review Problems
A-1.4: A hollow circular post ABC (see figure) supports a load P1 16 kN
acting at the top. A second load P2 is uniformly distributed around the cap plate
at B. The diameters and thicknesses of the upper and lower parts of the post are
dAB 30 mm, tAB 12 mm, dBC 60 mm, and tBC 9 mm, respectively. The
lower part of the post must have the same compressive stress as the upper part.
The required magnitude of the load P2 is approximately:
(A) 18 kN
(B) 22 kN
(C) 28 kN
(D) 46 kN
P1
Solution
P1 16 kN
dAB 30 mm
tAB 12 mm
dBC 60 mm
tBC 9 mm
p
AAB [dAB2 (dAB 2 tAB)2] 679 mm2
4
A
tAB
dAB
P2
B
p
ABC [dBC2 (dBC 2 tBC)2] 1442 mm2
4
Stress in AB:
tBC
P1
sAB 23.6 MPa
AAB
C
P1 P2
must equal sAB
ABC
Stress in BC:
sBC Solve for P2
P2 AB ABC P1 18.00 kN
Check:
sBC dBC
P1 P2
23.6 MPa
ABC
same as in AB
A-1.5: A circular aluminum tube of length L 650 mm is loaded in compression by forces P. The outside and inside diameters are 80 mm and 68 mm,
respectively. A strain gage on the outside of the bar records a normal strain
in the longitudinal direction of 400
106. The shortening of the bar is
approximately:
(A) 0.12 mm
(B) 0.26 mm
(C) 0.36 mm
(D) 0.52 mm
Solution
400 (106)
d
L 650 mm
L 0.260 mm
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 1087
APPENDIX A FE Exam Review Problems
1087
Strain gage
P
P
L
A-1.6: A steel plate weighing 27 kN is hoisted by a cable sling that has a clevis
at each end. The pins through the clevises are 22 mm in diameter. Each half of
the cable is at an angle of 35 to the vertical. The average shear stress in each pin
is approximately:
(A) 22 MPa
(B) 28 MPa
(C) 40 MPa
(D) 48 MPa
Solution
W 27 kN
dp 22 mm
35
Cross sectional area of each pin:
p
Ap d p2 380 mm2
4
P
Cable sling
Tensile force in cable:
W
a b
2
T
16.48 kN
cos(u)
Shear stress in each clevis
pin (double shear):
T
21.7 MPa
t
2 AP
35°
35°
Clevis
Steel plate
A-1.7: A steel wire hangs from a high-altitude balloon. The steel has unit weight
77kN/m3 and yield stress of 280 MPa. The required factor of safety against yield
is 2.0. The maximum permissible length of the wire is approximately:
(A) 1800 m
(B) 2200 m
(C) 2600 m
(D) 3000 m
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
78572_app_ptg01_hr_1083-1168.qxd
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1/24/12
12:15 PM
Page 1088
APPENDIX A FE Exam Review Problems
Solution
g 77
kN
m3
Y 280 MPa
sallow Allowable stress:
Weight of wire of length L:
sY
140.0 MPa
FSY
W Max. axial stress in wire of length L:
Lmax Max. length of wire:
FSY 2
AL
smax W
A
max L
sallow
1818 m
g
A-1.8: An aluminum bar (E 72 GPa, 0.33) of diameter 50 mm cannot
exceed a diameter of 50.1 mm when compressed by axial force P. The maximum
acceptable compressive load P is approximately:
(A) 190 kN
(B) 200 kN
(C) 470 kN
(D) 860 kN
Solution
E 72 GPa
Lateral strain:
Axial strain:
Axial stress:
dinit 50 mm
L a dfinal 50.1 mm
dfinal dinit
dinit
0.33
L 0.002
L
0.006
n
E a 436.4 MPa
below yield stress of 480 MPa
so Hooke’s Law applies
Max. acceptable compressive load:
p
Pmax s a dinit2b 857 kN
4
A-1.9: An aluminum bar (E 70 GPa, 0.33) of diameter 20 mm is stretched
by axial forces P, causing its diameter to decrease by 0.022 mm. The load P is
approximately:
(A) 73 kN
(B) 100 kN
(C) 140 kN
(D) 339 kN
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
78572_app_ptg01_hr_1083-1168.qxd
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Page 1089
APPENDIX A FE Exam Review Problems
1089
Solution
E 70 GPa
dinit 20 mm
L Lateral strain:
d
dinit
d 0.022 mm
0.33
d
P
P
L 0.001
L
3.333
v
103
Axial strain:
a Axial stress:
E a 233.3 MPa
below yield stress of 270 MPa
so Hooke’s Law applies
Max. acceptable load:
p
Pmax s a dinit 2b 73.3 kN
4
A-1.10: A polyethylene bar (E 1.4 GPa, 0.4) of diameter 80 mm is
inserted in a steel tube of inside diameter 80.2 mm and then compressed by axial
force P. The gap between steel tube and polyethylene bar will close when compressive load P is approximately:
(A) 18 kN
(B) 25 kN
(C) 44 kN
(D) 60 kN
Solution
E 1.4 GPa
Lateral strain:
d1 80 mm
L d1 0.2 mm
d1
d1
0.4
Steel
tube
L 0.003
d1 d2
Polyethylene
bar
Axial strain:
Axial stress:
a L
6.250
v
E a 8.8 MPa
103
well below ultimate stress of 28
MPa so Hooke’s Law applies
Max. acceptable compressive load:
p
Pmax s a d1 2 b 44.0 kN
4
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
78572_app_ptg01_hr_1083-1168.qxd
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12:15 PM
Page 1090
APPENDIX A FE Exam Review Problems
A-1.11: A pipe (E 110 GPa) carries a load P1 120 kN at A and a uniformly
distributed load P2 100 kN on the cap plate at B. Initial pipe diameters and
thicknesses are: dAB 38 mm, tAB 12 mm, dBC 70 mm, tBC 10 mm.
Under loads P1 and P2, wall thickness tBC increases by 0.0036 mm. Poisson’s
ratio v for the pipe material is approximately:
(A) 0.27
(B) 0.30
(C) 0.31
(D) 0.34
Solution
E 110 GPa
dAB 38 mm
tAB 12 mm
tBC 10 mm
P1 120 kN
P2 100 kN
ABC dBC 70 mm
p
[dBC2 (dBC 2 tBC)2] 1885 mm2
4
tBC dBC
C
Cap plate
tAB dAB
B
A
P1
P2
BC Axial strain of BC:
(P1 P2)
1.061
E ABC
103
BC E BC 116.7 MPa
Axial stress in BC:
(well below yield stress of 550 MPa so Hooke’s
Law applies)
Lateral strain of BC:
L tBC 0.0036 mm
tBC
3.600
tBC
Poisson’s ratio:
v
L
BC
104
0.34
confirms value for brass
given in properties table
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 1091
APPENDIX A FE Exam Review Problems
1091
A-1.12: A titanium bar (E 100 GPa, v 0.33) with square cross section
(b 75 mm) and length L 3.0 m is subjected to tensile load P 900 kN. The
increase in volume of the bar is approximately:
(A) 1400 mm3
(B) 3500 mm3
(C) 4800 mm3
(D) 9200 mm3
Solution
E 100 GPa
b 75 mm
L 3.0 m
b
P 900 kN
v 0.33
b
P
P
L
Initial volume of bar:
Vinit b2 L 1.6875000
Normal strain in bar:
L v
Lateral strain in bar:
Final length of bar:
P
1.60000
E b2
103
5.28000
104
Lf L L 3004.800 mm
Final lateral dimension of bar:
Final volume of bar:
107 mm3
bf b L b 74.96040 mm
Vfinal bf2 Lf 1.68841562
Increase in volume of bar:
107 mm3
V Vfinal Vinit 9156 mm3
V
0.000543
Vinit
A-1.13: An elastomeric bearing pad is subjected to a shear force V during a static
loading test. The pad has dimensions a 150 mm and b 225 mm, and thickness t 55 mm. The lateral displacement of the top plate with respect to the
bottom plate is 14 mm under a load V 16 kN. The shear modulus of elasticity
G of the elastomer is approximately:
(A) 1.0 MPa
(B) 1.5 MPa
(C) 1.7 MPa
(D) 1.9 MPa
Solution
V 16 kN
a 150 mm
b 225 mm
d 14 mm
t 55 mm
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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APPENDIX A FE Exam Review Problems
Ave. shear stress:
t
b
V
0.474 MPa
ab
a
V
Ave. shear strain:
d
g arctana b 0.249
t
t
Shear modulus of elastomer:
t
G 1.902 MPa
g
A-1.14: A bar of diameter d 18 mm and length L 0.75 m is loaded in tension by forces P. The bar has modulus E 45 GPa and allowable normal stress
of 180 MPa. The elongation of the bar must not exceed 2.7 mm. The allowable
value of forces P is approximately:
(A) 41 kN
(B) 46 kN
(C) 56 kN
(D) 63 kN
Solution
d 18 mm
da 2.7 mm
L 0.75 m
E 45 GPa
sa 180 MPa
d
P
P
L
(1) allowable value of P based on elongation
da
3.600 103
smax E a 162.0 MPa
L
p
elongation governs
Pa1 smax a d 2 b 41.2 kN
4
a (2) allowable load P based on tensile stress
p
Pa2 sa a d 2 b 45.8 kN
4
A-1.15: Two flanged shafts are connected by eight 18 mm bolts. The diameter of
the bolt circle is 240 mm. The allowable shear stress in the bolts is 90 MPa.
Ignore friction between the flange plates. The maximum value of torque T0 is
approximately:
(A) 19 kNm
(B) 22 kNm
(C) 29 kNm
(D) 37 kNm
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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APPENDIX A FE Exam Review Problems
1093
Solution
db 18 mm
d 240 mm
ta 90 MPa
Bolt shear area:
n8
T0
2
As p db
254.5 mm2
4
T0
Max. torque:
Tmax n (ta As)
d
22.0 kNm
2
A-1.16: A copper tube with wall thickness of 8 mm must carry an axial tensile
force of 175 kN. The allowable tensile stress is 90 MPa. The minimum required
outer diameter is approximately:
(A) 60 mm
(B) 72 mm
(C) 85 mm
(D) 93 mm
Solution
t 8 mm
P 175 kN
sa 90 MPa
d
P
P
Required area based on allowable stress:
Areqd P
1944 mm2
sa
Area of tube of thickness t but unknown outer diameter d:
A
p 2
[d (d 2 t)2]
4
A t(d t)
Solving for dmin:
P
sa
dmin t 85.4 mm
pt
so
dinner dmin 2 t 69.4 mm
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APPENDIX A FE Exam Review Problems
A-2.1: Two wires, one copper and the other steel, of equal length stretch the same
amount under an applied load P. The moduli of elasticity for each is: Es 210 GPa,
Ec 120 GPa. The ratio of the diameter of the copper wire to that of the steel wire
is approximately:
(A) 1.00
(B) 1.08
(C) 1.19
(D) 1.32
Solution
Es 210 GPa
Copper
wire
Ec 120 GPa
ds dc
Displacements are equal:
or
PL
PL
Es As Ec Ac
so
Es As Ec Ac
and
Ac Es
As Ec
Steel
wire
P
P
Express areas in terms of wire diameters then find ratio:
p dc2
4
Es
2
Ec
p ds
a
b
4
so
dc
Es
1.323
ds B Ec
A-2.2: A plane truss with span length L 4.5 m is constructed using cast iron pipes
(E 170 GPa) with cross sectional area of 4500 mm2. The displacement of joint B
cannot exceed 2.7 mm. The maximum value of loads P is approximately:
(A) 340 kN
(B) 460 kN
(C) 510 kN
(D) 600 kN
Solution
L 4.5 m
E 170 GPa
A 4500 mm2
dmax 2.7 mm
Statics: sum moments about A to find reaction at B
P
RB L
L
P
2
2
L
RB P
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Page 1095
1095
APPENDIX A FE Exam Review Problems
P
P
C
45°
A
45°
B
L
Method of Joints at B:
FAB P
(tension)
Force-displ. relation:
Pmax EA
dmax 459 kN
L
s
Check normal stress in bar AB:
Pmax
102.0 MPa
A
well below yield stress of
290 MPa in tension
A-2.3: A brass rod (E 110 GPa) with cross sectional area of 250 mm2 is loaded
by forces P1 15 kN, P2 10 kN, and P3 8 kN. Segment lengths of the bar
are a 2.0 m, b 0.75 m, and c 1.2 m. The change in length of the bar is
approximately:
(A) 0.9 mm
(B) 1.6 mm
(C) 2.1 mm
(D) 3.4 mm
Solution
E 110 GPa
a2m
A 250 mm2
A
c 1.2 m
P1 15 kN
P2
P1
b 0.75 m
C
B
a
b
D
P3
c
P2 10 kN
P3 8 kN
Segment forces (tension is positive):
NAB P1 P2 P3 17.00 kN
NBC P2 P3 2.00 kN
NCD P3 8.00 kN
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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APPENDIX A FE Exam Review Problems
Change in length:
dD 1
(NAB a NBC b NCD c) 0.942 mm
EA
dD
2.384
abc
104
positive so elongation
Check max. stress:
NAB
68.0 MPa
A
well below yield stress for brass so OK
A-2.4: A brass bar (E 110 MPa) of length L 2.5 m has diameter d1 18 mm
over one-half of its length and diameter d2 12 mm over the other half. Compare
this nonprismatic bar to a prismatic bar of the same volume of material with
constant diameter d and length L. The elongation of the prismatic bar under the
same load P 25 kN is approximately:
(A) 3 mm
(B) 4 mm
(C) 5 mm
(D) 6 mm
Solution
L 2.5 m
P 25 kN
d1 18 mm
d2 12 mm
E 110 GPa
d2
P
P
p
A1 d12 254.469 mm2
4
A2 d1
L/2
L/2
p 2
d2 113.097 mm2
4
Volume of nonprismatic bar:
Vol nonprismatic (A1 A2)
L
459458 mm3
2
Diameter of prismatic bar of same volume: d Aprismatic p 2
d 184 mm2
4
Volnonprismatic
15.30 mm
p
L
H
4
Vprismatic Aprismatic L 459458 mm3
Elongation of prismatic bar:
d
PL
3.09 mm
E Aprismatic
less than d for nonprismatic bar
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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APPENDIX A FE Exam Review Problems
1097
Elongation of nonprismatic bar shown in fig. above:
PL 1
1
a
b 3.63 mm
2 E A1
A2
A-2.5: A nonprismatic cantilever bar has an internal cylindrical hole of diameter
d/2 from 0 to x, so the net area of the cross section for Segment 1 is (3/4)A. Load
P is applied at x, and load -P/2 is applied at x L. Assume that E is constant. The
length of the hollow segment, x, required to obtain axial displacement
d PL/EA at the free end is:
(A) x L/5
(B) x L/4
(C) x L/3
(D) x 3L/5
Solution
Forces in Segments 1 & 2:
N1 3P
2
N2 P
2
Segment 1
Segment 2
3
—A
4
d
A
P
—
2
P
Displacement at free end:
d3 d
—
2
x
3
2
L–x
N1 x
N2 (L x)
EA
3
E a Ab
4
3P
P
x
(L x)
2
2
P (L 5 x)
d3 EA
2 AE
3
E a Ab
4
Set d3 equal to PL/EA and solve for x
P (L 5 x) P L
or
2AE
EA
P (L 5 x)
PL
P (3 L 5 x)
0 simplify S 0
2AE
EA
2AE
So x 3L/5
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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APPENDIX A FE Exam Review Problems
A-2.6: A nylon bar (E 2.1 GPa) with diameter 12 mm, length 4.5 m, and
weight 5.6 N hangs vertically under its own weight. The elongation of the bar at
its free end is approximately:
(A)
0.05 mm
(B)
0.07 mm
(C)
0.11 mm
(D)
0.17 mm
Solution
E 2.1 GPa
L 4.5 m
d 12 mm
A
2
pd
A
113.097 mm2
4
g 11
L
kN
m3
W L A 5.598 N
dB WL
2EA
B
dB or
(g L A) L
2 EA
2
so
dB gL
0.053 mm
2E
Check max. normal stress at top of bar smax W
0.050 MPa
A
ok - well below ult.
stress for nylon
A-2.7: A monel shell (Em 170 GPa, d3 12 mm, d2 8 mm) encloses a brass
core (Eb 96 GPa, d1 6 mm). Initially, both shell and core are of length 100 mm.
A load P is applied to both shell and core through a cap plate. The load P
required to compress both shell and core by 0.10 mm is approximately:
(A) 10.2 kN
(B) 13.4 kN
(C) 18.5 kN
(D) 21.0 kN
Solution
Em 170 GPa
Eb 96 GPa
d1 6 mm
d2 8 mm
d3 12 mm
L 100 mm
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APPENDIX A FE Exam Review Problems
1099
P
Monel shell
Brass core
L
d1
d2
d3
Am p 2
(d3 d22) 62.832 mm2
4
Ab p 2
d1 28.274 mm2
4
dm db
Compatibility:
Pm L
Pb L
Em Am Eb Ab
Pm Pm Pb P
Statics:
Em Am
Pb
Eb Ab
so
Pb P
Em Am
a1 b
Eb Ab
Set dB equal to 0.10 mm and solve for load P:
db Pb L
Eb Ab
and then
so
P
Pb Eb Ab
db
L
with
db 0.10 mm
Eb Ab
Em Am
db a1 b 13.40 kN
L
Eb Ab
A-2.8: A steel rod (Es 210 GPa, dr 12 mm, as 12 106 > C ) is held
stress free between rigid walls by a clevis and pin (dp 15 mm) assembly at each
end. If the allowable shear stress in the pin is 45 MPa and the allowable normal
stress in the rod is 70 MPa, the maximum permissible temperature drop T is
approximately:
(A) 14 C
(B) 20 C
(C) 28 C
(D) 40 C
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APPENDIX A FE Exam Review Problems
Solution
Es 210 GPa
dp 15 mm
dr 12 mm
p 2
Ar dr 113.097 mm2
4
p
Ap dp2 176.715 mm2
4
pin, dp
ΔT
rod, dr
s 12(10 ) > C
Clevis
6
ta 45 MPa
sa 70 MPa
Force in rod due to temperature drop T:
Fr Es Ar (s)T
and normal stress in rod:
sr Fr
Ar
So Tmax associated with normal stress in rod
Tmaxrod sa
27.8
Es as
degrees Celsius (decrease) Controls
Now check T based on shear stress in pin (in double shear):
Tmaxpin ta (2 Ap)
Es Ar as
tpin Fr
2 Ap
55.8
A-2.9: A threaded steel rod (Es 210 GPa, dr 15 mm, s 12 106 > C)
is held stress free between rigid walls by a nut and washer (dw 22 mm)
assembly at each end. If the allowable bearing stress between the washer and
wall is 55 MPa and the allowable normal stress in the rod is 90 MPa, the
maximum permissible temperature drop T is approximately:
(A) 25 C
(B) 30 C
(C) 38 C
(D) 46 C
Solution
Es 210 GPa
dr 15 mm
Ar p 2
dr 176.7 mm2
4
Aw p 2
(dw dr2) 203.4 mm2
4
s 12(106) > C
sba 55 MPa
dw 22 mm
sa 90 MPa
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APPENDIX A FE Exam Review Problems
rod, dr
washer, dw
ΔT
Force in rod due to temperature drop T:
and normal stress in rod:
sr Fr Es Ar ( s)T
1101
Fr
Ar
So Tmax associated with normal stress in rod
Tmaxrod sa
35.7
Es as
degrees Celsius (decrease)
Now check T based on bearing stress beneath washer:
sba (Aw)
25.1
Es Ar as
Tmaxwasher sb Fr
Aw
degrees Celsius (decrease)
Controls
A-2.10: A steel bolt (area 130 mm2, Es 210 GPa) is enclosed by a copper tube
(length 0.5 m, area 400 mm2, Ec 110 GPa) and the end nut is turned until
it is just snug. The pitch of the bolt threads is 1.25 mm. The bolt is now tightened
by a quarter turn of the nut. The resulting stress in the bolt is approximately:
(A) 56 MPa
(B) 62 MPa
(C) 74 MPa
(D) 81 MPa
Solution
Es 210 GPa
Ec 110 GPa
Ac 400 mm
As 130 mm
2
n 0.25
L 0.5 m
2
Copper tube
p 1.25 mm
Compatibility: shortening of tube
and elongation of bolt applied
displacement of n p
Steel bolt
Ps L
Pc L
np
Ec Ac
Es As
Statics:
Pc Ps
Solve for Ps
Ps L
Ps L
np
Ec Ac
Es As
or
Ps np
10.529 kN
1
1
La
b
Ec Ac Es As
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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APPENDIX A FE Exam Review Problems
Stress in steel bolt:
ss Ps
81.0 MPa
As
tension
Stress in copper tube:
sc Ps
26.3 MPa
Ac
compression
A-2.11: A steel bar of rectangular cross section (a 38 mm, b 50 mm)
carries a tensile load P. The allowable stresses in tension and shear are
100 MPa and 48 MPa respectively. The maximum permissible load Pmax is
approximately:
(A) 56 kN
(B) 62 kN
(C) 74 kN
(D) 91 kN
Solution
a 38 mm
b 50 mm
A a b 1900 mm2
b
sa 100 MPa
P
P
ta 48 MPa
a
Bar is in uniaxial tension so Tmax smax/2; since 2 ta sa, shear stress governs
Pmax ta A 91.2 kN
A-2.12: A brass wire (d 2.0 mm, E 110 GPa) is pretensioned to T 85 N.
The coefficient of thermal expansion for the wire is 19.5 106 > C. The temperature change at which the wire goes slack is approximately:
(A) 5.7 C
(B) 12.6 C
(C) 12.6 C
(D) 18.2 C
Solution
E 110 GPa
d 2.0 mm
b 19.5 (10 ) > C
6
p
A d 2 3.14 mm2
4
T 85 N
T
d
T
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APPENDIX A FE Exam Review Problems
1103
Normal tensile stress in wire due to pretension T and temperature increase T:
s
T
E ab T
A
Wire goes slack when normal stress goes to zero; solve for T
T
A
T 12.61
E ab
degrees Celsius (increase in temperature)
A-2.13: A copper bar (d 10 mm, E 110 GPa) is loaded by tensile load
P 11.5 kN. The maximum shear stress in the bar is approximately:
(A) 73 MPa
(B) 87 MPa
(C) 145 MPa
(D) 150 MPa
Solution
E 110 GPa
d 10 mm
p 2
A d 78.54 mm2
4
P 11.5 kN
d
P
P
Normal stress in bar:
p
s 146.4 MPa
A
For bar in uniaxial stress, max. shear stress is on a plane at 45 deg. to axis of
bar and equals 1/2 of normal stress:
s
tmax 73.2 MPa
2
A-2.14: A steel plane truss is loaded at B and C by forces P 200 kN. The cross
sectional area of each member is A 3970 mm2. Truss dimensions are H 3 m
and L 4 m. The maximum shear stress in bar AB is approximately:
(A) 27 MPa
(B) 33 MPa
(C) 50 MPa
(D) 69 MPa
Solution
P 200 kN
A 3970 mm2
H3m
L4m
Statics: sum moments about A to find vertical reaction at B
Bvert P H
150.000 kN
L
(downward)
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APPENDIX A FE Exam Review Problems
P
P
C
H
B
L
A
P
Method of Joints at B:
CBhoriz CBvert Bvert
L
CBvert 200.0 kN
H
AB P CBhoriz 400.0 kN (compression)
So bar force in AB is:
Max. normal stress in AB:
sAB AB
100.8 MPa
A
Max. shear stress is 1/2 of max. normal stress for bar in uniaxial stress and is
on plane at 45 deg. to axis of bar:
tmax sAB
50.4 MPa
2
A-2.15: A plane stress element on a bar in uniaxial stress has tensile stress of
s 78 MPa (see fig.). The maximum shear stress in the bar is approximately:
(A) 29 MPa
(B) 37 MPa
(C) 50 MPa
(D) 59 MPa
Solution
u 78 MPa
σθ /2
Plane stress transformation formulas for
uniaxial stress:
sx su
cos(u)2
and
⵩ on element face
at angle
τθ τθ
σθ
θ
su
2
sx sin(u)2
⵩ on element face
at angle 90
τθ τθ
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Page 1105
1105
APPENDIX A FE Exam Review Problems
Equate above formulas and solve for sx
tan(u)2 u atana
so
sx 1
2
1
b 35.264
12
su
117.0 MPa
cos(u)2
also u sx sin(u) cos(u) 55.154 MPa
Max. shear stress is 1/2 of max. normal stress for bar in uniaxial stress and is
on plane at 45 deg. to axis of bar:
tmax sx
58.5 MPa
2
A-2.16: A prismatic bar (diameter
d0 18 mm) is loaded by force P1.
A stepped bar (diameters d1 20 mm,
d2 25 mm, with radius R of fillets
2 mm) is loaded by force P2. The
allowable axial stress in the material is
75 MPa. The ratio P1/P2 of the maximum permissible loads that can be
applied to the bars, considering stress
concentration effects in the stepped
bar, is:
(A) 0.9
(B) 1.2
(C) 1.4
(D) 2.1
P1
P2
d0
d1
P1
d2
d1
P2
FIG. 2-66 Stress-concentration factor K for round bars with shoulder fillets.
The dashed line is for a full quater-circular fillet.
3.0
R
D2
=2
D1
P
D2
1.5
2.5
s
K = s max
1.2
K
D1
nom
1.1
s nom =
P
P
p D21/4
2.0
R=
1.5
0
D2 – D1
2
0.05
0.10
0.15
R
D1
0.20
0.25
0.30
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APPENDIX A FE Exam Review Problems
Solution
Prismatic bar P1 max sallow a
p d02
p (18 mm)2
b (75 MPa) c
d 19.1 kN
4
4
2 mm
d2 25 mm
R
0.100
1.250 so K 1.75
d1 20 mm
d1 20 mm
Stepped bar
from stress conc. Fig. 2-66
3.0
R
D2
=2
D1
P
D2
1.5
2.5
s
K = s max
1.2
K
D1
s nom =
nom
1.1
P
P
p D21/4
2.0
K = 1.75
R=
1.5
0
P2 max D2 – D1
2
0.05
0.10
0.15
R
D1
0.20
0.25
0.30
75 MPa p(20 mm)2
sallow p d12
a
ba
bc
d 13.5 kN
K
4
K
4
P1 max
19.1 kN
1.41
P2 max
13.5 kN
A-3.1: A brass rod of length L 0.75 m is twisted by torques T until the angle
of rotation between the ends of the rod is 3.5°. The allowable shear strain in
the copper is 0.0005 rad. The maximum permissible diameter of the rod is
approximately:
(A) 6.5 mm
(B) 8.6 mm
(C) 9.7 mm
(D) 12.3 mm
Solution
L 0.75 m
d
T
f 3.5°
ga 0.0005
L
Max. shear strain:
d
a fb
2
gmax so
L
T
dmax 2 ga L
12.28 mm
f
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APPENDIX A FE Exam Review Problems
1107
A-3.2: The angle of rotation between the ends of a nylon bar is 3.5°. The bar
diameter is 70 mm and the allowable shear strain is 0.014 rad. The minimum
permissible length of the bar is approximately:
(A) 0.15 m
(B) 0.27 m
(C) 0.40 m
(D) 0.55 m
Solution
d 70 mm
d
T
f 3.5
T
ga 0.014
L
Max. shear strain:
g
rf
L
so
L min df
0.15 m
2 ga
A-3.3: A brass bar twisted by torques T acting at the ends has the following
properties: L 2.1 m, d 38 mm, and G 41 GPa. The torsional stiffness of the
bar is approximately:
(A) 1200 Nm
(B) 2600 Nm
(C) 4000 Nm
(D) 4800 Nm
Solution
G 41 GPa
L 2.1 m
d
T
T
d 38 mm
L
Polar moment of inertia, Ip:
Ip p 4
d 2.047
32
105 mm4
Torsional stiffness, kT:
kT G Ip
L
3997 Nm
A-3.4: A brass pipe is twisted by torques T 800 Nm acting at the ends causing
an angle of twist of 3.5 degrees. The pipe has the following properties: L 2.1 m,
d1 38 mm, and d2 56 mm. The shear modulus of elasticity G of the pipe is
approximately:
(A) 36.1 GPa
(B) 37.3 GPa
(C) 38.7 GPa
(D) 40.6 GPa
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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APPENDIX A FE Exam Review Problems
Solution
T
T
d1
L
L 2.1 m d1 38 mm
d2
d2 56 mm
Polar moment of inertia: Ip f 3.5° T 800 N m
p 4
(d2 d14) 7.608
32
105 mm4
Solving torque-displacement relation for shear modulus G:
G
TL
36.1 GPa
f Ip
A-3.5: An aluminum bar of diameter d 52 mm is twisted by torques T1 at the
ends. The allowable shear stress is 65 MPa. The maximum permissible torque T1
is approximately:
(A) 1450 Nm
(B) 1675 Nm
(C) 1710 Nm
(D) 1800 Nm
Solution
d 52 mm
d
T1
T1
ta 65 MPa
Ip p 4
d 7.178
32
4
105 mm
From shear formula:
T1 max
ta Ip
d
a b
2
1795 Nm
A-3.6: A steel tube with diameters d2 86 mm and d1 52 mm is twisted by
torques at the ends. The diameter of a solid steel shaft that resists the same torque
at the same maximum shear stress is approximately:
(A) 56 mm
(B) 62 mm
(C) 75 mm
(D) 82 mm
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APPENDIX A FE Exam Review Problems
1109
Solution
d2 86 mm
IPpipe d1 52 mm
p 4
(d2 d14 ) 4.652
32
106 mm4
Shear formula for hollow pipe:
d2
Ta b
2
tmax IPpipe
d
d1
d2
Shear formula for solid shaft:
d
Ta b
2
16 T
tmax p 4
p
d3
d
32
1
Equate and solve for d of solid shaft:
da
dD
16 T
p
3
d2
Ta b
2
IPpipe
T
1
32 IPpipe 3
b 82.0 mm
p d2
A-3.7: A stepped steel shaft with diameters d1 56 mm and d2 52 mm is
twisted by torques T1 3.5 kNm and T2 1.5 kNm acting in opposite directions. The maximum shear stress is approximately:
(A) 54 MPa
(B) 58 MPa
(C) 62 MPa
(D) 79 MPa
Solution
d1 56 mm
d2 52 mm
T1 3.5 kNm T2 1.5 kNm
T1
d1
d2
B
A
L1
T2
C
L2
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APPENDIX A FE Exam Review Problems
Polar moments of inertia:
p 4
d1 9.655
32
p
Ip2 d24 7.178
32
Ip1 105 mm4
105 mm4
Shear formula - max. shear stresses in segments 1 & 2:
(T1 T2)
tmax1 d1
2
Ip1
d2
T2 a b
2
tmax2 54.3 MPa
Ip2
58.0 MPa
A-3.8: A stepped steel shaft (G 75 GPa) with diameters d1 36 mm and
d2 32 mm is twisted by torques T at each end. Segment lengths are L1 0.9 m
and L2 0.75 m. If the allowable shear stress is 28 MPa and maximum allowable twist is 1.8 degrees, the maximum permissible torque is approximately:
(A) 142 Nm
(B) 180 Nm
(C) 185 Nm
(D) 257 Nm
Solution
d1 36 mm
d1
d2
T
d2 32 mm
G 75 GPa
A
ta 28 MPa
C
B
L1
T
L2
L1 0.9 m L2 0.75 m
fa 1.8
Polar moments of inertia:
p 4
d1 1.649
32
p
Ip2 d24 1.029
32
Ip1 105 mm4
105 mm4
Max torque based on allowable shear stress - use shear formula:
T
tmax1 d1
2
Ip1
Tmax1 ta a
d2
Ta b
2
tmax2 Ip2
2 Ip1
2 Ip2
b 257 Nm Tmax2 ta a
b 180 Nm controls
d1
d2
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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APPENDIX A FE Exam Review Problems
1111
Max. torque based on max. rotation & torque-displacement relation:
f
L2
T L1
a
b
G Ip1
Ip2
Tmax G fa
185 Nm
L1
L2
a
b
Ip1
Ip2
A-3.9: A gear shaft transmits torques TA 975 Nm, TB 1500 Nm, TC 650 Nm and TD 825 Nm. If the allowable shear stress is 50 MPa, the
required shaft diameter is approximately:
(A) 38 mm
(B) 44 mm
(C) 46 mm
(D) 48 mm
Solution
ta 50 MPa
TA
TA 975 Nm
TB
TB 1500 Nm
TC
TC 650 Nm
A
TD 825 Nm
TD
B
C
Find torque in each
segment of shaft:
TAB TA 975.0 Nm
TBC TA TB 525.0 Nm
D
TCD TD 825.0 Nm
Shear formula:
d
Ta b
2
16 T
t
p 4
p d3
d
32
Set t to tallowable and T to torque in each segment; solve for required diameter d
(largest controls)
1
Segment AB:
16 |TAB| 3
da
b 46.3 mm
p ta
1
16 |TBC| 3
b 37.7 mm
Segment BC: d a
p ta
1
Segment CD:
16 |TCD| 3
b 43.8 mm
da
p ta
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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APPENDIX A FE Exam Review Problems
A-3.10: A hollow aluminum shaft (G 27 GPa, d2 96 mm, d1 52 mm)
has an angle of twist per unit length of 1.8°/m due to torques T. The resulting
maximum tensile stress in the shaft is approximately:
(A) 38 MPa
(B) 41 MPa
(C) 49 MPa
(D) 58 MPa
Solution
G 27 GPa
d2
T
T
d2 96 mm
d1 52 mm
L
u 1.8 >m
Max. shear strain due to
twist per unit length:
d1 d2
d2
gmax a b u 1.508
2
103
radians
tmax Ggmax 40.7 MPa
Max. shear stress:
Max. tensile stress on plane at 45 degrees & equal to max. shear stress:
smax tmax 40.7 MPa
A-3.11: Torques T 5.7 kNm are applied to a hollow aluminum shaft (G 27 GPa, d1 52 mm). The allowable shear stress is 45 MPa and the allowable
normal strain is 8.0
104. The required outside diameter d2 of the shaft is
approximately:
(A) 38 mm
(B) 56 mm
(C) 87 mm
(D) 91 mm
Solution
T 5.7 kNm
G 27 GPa
ta1 45 MPa
a 8.0(10
d1 52 mm
4
)
d1 d2
Allowable shear strain based on allowable normal strain for pure shear
ga 2 a 1.600
103
so resulting allow. shear stress is:
ta2 Gga 43.2 MPa
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APPENDIX A FE Exam Review Problems
So allowable shear stress based on normal strain governs
ta ta2
Use torsion formula to relate required d2 to allowable shear stress:
tmax d2
Ta b
2
p 4
(d2 d14)
32
16 T
4
4
and rearrange equation to get d 2 d 1 p t d2
a
Solve resulting 4th order equation numerically, or use a calculator and trial & error
d1 52 mm
T 5700000 Nmm
f(d2) d2 4 a
16 T
b d d1 4
p ta 2
ta 43.2 MPa
gives
d2 91 mm
A-3.12: A motor drives a shaft with diameter d 46 mm at f 5.25 Hz
and delivers P 25 kW of power. The maximum shear stress in the shaft is
approximately:
(A) 32 MPa
(B) 40 MPa
(C) 83 MPa
(D) 91 MPa
Solution
f 5.25 Hz
d 46 mm
P 25 kW
p 4
Ip d 4.396
32
5
4
10 mm
Power in terms of torque T:
P 2pf T
f
d
Solve for torque T:
T
P
757.9 Nm
2pf
P
Max. shear stress using torsion formula:
d
Ta b
2
tmax 39.7 MPa
Ip
A-3.13: A motor drives a shaft at f 10 Hz and delivers P 35 kW of power.
The allowable shear stress in the shaft is 45 MPa. The minimum diameter of the
shaft is approximately:
(A) 35 mm
(B) 40 mm
(C) 47 mm
(D) 61 mm
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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APPENDIX A FE Exam Review Problems
Solution
f 10 Hz
f
P 35 kW
d
ta 45 MPa
P
Power in terms of torque T:
P 2pf T
Solve for torque T:
T
P 557.0 Nm
2pf
Shear formula:
d
Ta b
2
t
p 4
d
32
t
or
16 T
pd
3
1
Solve for diameter d:
16 T 3
da
b 39.8 mm
p ta
A-3.14: A drive shaft running at 2500 rpm has outer diameter 60 mm and inner
diameter 40 mm. The allowable shear stress in the shaft is 35 MPa. The maximum
power that can be transmitted is approximately:
(A) 220 kW
(B) 240 kW
(C) 288 kW
(D) 312 kW
Solution
n 2500 rpm
ta 35 (106)
N
m2
d
d2 0.060 m
n
d1
d1 0.040 m
Ip d2
p 4
(d2 d14) 1.021
32
106 m4
Shear formula:
d2
Ta b
2
t
Ip
or
Tmax 2 ta Ip
d2
1191.2 Nm
Power in terms of torque T:
P 2p f T 2p(n/60) T
2pn
Pmax 60 Tmax 3.119
105 W
Pmax 312 kW
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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APPENDIX A FE Exam Review Problems
1115
A-3.15: A prismatic shaft (diameter d0 19 mm) is loaded by torque T1.
A stepped shaft (diameters d1 20 mm, d2 25 mm, radius R of fillets
2 mm) is loaded by torque T2. The allowable shear stress in the material is 42
MPa. The ratio T1/T2 of the maximum permissible torques that can be applied to
the shafts, considering stress concentration effects in the stepped shaft is:
(A) 0.9
(B) 1.2
(C) 1.4
(D) 2.1
T1
d0
T1
D2
R
D1
T2
T2
FIG. 3-59 Stress-concentration factor K for a stepped shaft in torsion. (The
dashed line is for a full quarter-circular fillet.)
2.00
R
T
1.2
K
D2
1.1
tmax = Ktnom
1.5
1.50
D1
T
16T
tnom = ——
p D13
D2
—–
=
D1 2
D2 = D1 + 2R
1.00
0
0.10
0.20
R–
—
D1
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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APPENDIX A FE Exam Review Problems
Solution
Prismatic shaft
T1max p
tallow IP
p d03
b 42 MPa c (19 mm)3 d 56.6 Nm
tallow a
d0
16
16
2
Stepped shaft
d2 25 mm
R
2 mm
1.250
0.100 so from graph (see Fig. 3-59)
d1 20 mm
d1 20 mm
K 1.35
T2 max tallow p d13
a
b
K
16
42 MPa p
c (20 mm)3 d 48.9 Nm
1.35
16
T1 max 56.6
1.16
T2 max 48.9
A-4.1: A simply supported beam with proportional loading (P 4.1 kN) has
span length L 5 m. Load P is 1.2 m from support A and load 2P is 1.5 m from
support B. The bending moment just left of load 2P is approximately:
(A) 5.7 kNm
(B) 6.2 kNm
(C) 9.1 kNm
(D) 10.1 kNm
Solution
a 1.2 m
b 2.3 m
L a b c 5.00 m
P
c 1.5 m
2P
A
B
P 4.1 kN
a
b
L
c
Statics to find reaction force at B:
RB 1
[P a 2 P (a b)] 6.724 kN
L
Moment just left of load 2P:
M RB c 10.1 kNm
compression on top of beam
A-4.2: A simply-supported beam is loaded as shown in the figure. The bending
moment at point C is approximately:
(A) 5.7 kNm
(B) 6.1 kNm
(C) 6.8 kNm
(D) 9.7 kNm
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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APPENDIX A FE Exam Review Problems
1117
Solution
7.5 kN
1.8 kN/m
C
A
1.0 m
B
0.5 m
1.0 m
3.0 m
5.0 m
Statics to find reaction force at A:
RA 1
kN (3 m 0.5 m)2
cc1.8
d 7.5 kN (3 m 1 m)d 7.125 kN
m
5m
2
Moment at point C, 2 m from A:
M RA (2 m) 7.5 kN (1.0m) 6.75 kNm
compression
on top of beam
A-4.3: A cantilever beam is loaded as shown in the figure. The bending moment
at 0.5 m from the support is approximately:
(A) 12.7 kNm
(B) 14.2 kNm
(C) 16.1 kNm
(D) 18.5 kNm
Solution
4.5 kN
1.8 kN/m
A
1.0 m
B
1.0 m
3.0 m
Cut beam at 0.5 m from support; use statics and right-hand FBD to find
internal moment at that point
M 0.5 m (4.5 kN) a0.5 m 1.0 m 18.5 kNm
3.0 m
kN
(3.0 m)
b 1.8
m
2
(tension on top of beam)
A-4.4: An L-shaped beam is loaded as shown in the figure. The bending moment
at the midpoint of span AB is approximately:
(A) 6.8 kNm
(B) 10.1 kNm
(C) 12.3 kNm
(D) 15.5 kNm
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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APPENDIX A FE Exam Review Problems
Solution
4.5 kN
9 kN
1.0 m
A
B
5.0 m
C
1.0 m
Use statics to find reaction at B; sum moments about A
RB 1
[9 kN (6 m) 4.5 kN (1. m)] 9.90 kN
5m
Cut beam at midpoint of AB; use right hand FBD, sum moments
M RB a
5m
5m
b 9 kN a
1 mb 6.75 kNm
2
2
tension on
top of beam
A-4.5: A T-shaped simple beam has a cable with force P anchored at B and
passing over a pulley at E as shown in the figure. The bending moment just left
of C is 1.25 kNm. The cable force P is approximately:
(A) 2.7 kN
(B) 3.9 kN
(C) 4.5 kN
(D) 6.2 kN
Solution
MC 1.25 kNm
Sum moments about D to
find vertical reaction at A:
VA 1
[P (4 m)]
7m
VA 4
P
7
E
P
Cable
4m
A
B
C
D
(downward)
Now cut beam & cable just
left of CE & use left FBD;
2m
show VA downward & show
vertical cable force component
of (4/5)P upward at B; sum moments
at C to get MC and equate to given
numerical value of MC to find P:
3m
2m
4
MC P (3) VA (2 3)
5
4
4
16 P
MC P (3) a Pb (2 3) 5
7
35
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APPENDIX A FE Exam Review Problems
1119
Solve for P:
P
35
(1.25) 2.73 kN
16
A-4.6: A simple beam (L 9 m) with attached bracket BDE has force P 5 kN
applied downward at E. The bending moment just right of B is approximately:
(A) 6 kNm
(B) 10 kNm
(C) 19 kNm
(D) 22 kNm
Solution
Sum moments about A to find
reaction at C:
RC B
A
C
1
L
L
P
cP a b d L
6
3
2
D
E
P
Cut through beam just
right of B, then use FBD of BC
to find moment at B:
L
—
6
L
—
3
L
—
2
L
L
5LP
L
MB RC a b 2
3
12
Substitute numbers for L and P:
L9m
MB P 5 kN
5LP
18.8 kN m
12
A-4.7: A simple beam AB with an overhang BC is loaded as shown in the figure.
The bending moment at the midspan of AB is approximately:
(A) 8 kNm
(B) 12 kNm
(C) 17 kNm
(D) 21 kNm
Solution
4.5 kN · m
15 kN/m
A
C
B
1.6 m
1.6 m
1.6 m
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APPENDIX A FE Exam Review Problems
Sum moments about B to
get reaction at A:
RA 1
1.6
b 4.5d 19.40625 kN
c15 s1.6) a1.6 3.2
2
Cut beam at midspan, use left FBD & sum moments to find moment at
midspan:
Mmspan RA s1.6) 15 s1.6) a
1.6
b 11.85 kNm
2
A-5.1: A copper wire (d 1.5 mm) is bent around a tube of radius R 0.6 m.
The maximum normal strain in the wire is approximately:
(A) 1.25 103
(B) 1.55 103
(C) 1.76 103
(D) 1.92 103
Solution
max d
2
d
d
R
2
d
2 aR d 1.5 mm
max d
b
2
R
R 0.6 m
d
d
2 aR b
2
1.248
103
A-5.2: A simply supported wood beam (L 5 m) with rectangular cross section
(b 200 mm, h 280 mm) carries uniform load q 6.5 kN/m which includes
the weight of the beam. The maximum flexural stress is approximately:
(A) 8.7 MPa
(B) 10.1 MPa
(C) 11.4 MPa
(D) 14.3 MPa
Solution
L5m
q 9.5
b 200 mm
h 280 mm
kN
m
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APPENDIX A FE Exam Review Problems
Section modulus:
q
2
S
bh
2.613
6
106 m3
A
h
B
Max. moment at midspan:
Mmax q L2
29.7 kNm
8
b
L
Max. flexural stress at midspan:
smax M max
11.4 MPa
S
A-5.3: A cast iron pipe (L 12 m, weight density 72 kN/m3, d2 100 mm,
d1 75 mm) is lifted by a hoist. The lift points are 6 m apart. The maximum
bending stress in the pipe is approximately:
(A) 28 MPa
(B) 33 MPa
(C) 47 MPa
(D) 59 MPa
Solution
d1
d2
s
L
L 12 m
s4m
d2 100 mm
d1 75 mm
gCI 72
kN
m3
Pipe cross sectional properties:
A
p
p 2
sd2 d12) 3436 mm2 I sd24 d14) 3.356
4
64
Uniformly distributed weight of pipe, q:
Vertical force at each lift point:
F
q gCI A 0.247
106 mm4
kN
m
qL
1.484 kN
2
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APPENDIX A FE Exam Review Problems
Max. moment is either at lift points (M1) or at midspan (M2):
Ls Ls
ba
b 3.958 kNm
2
2
M1 q a
M2 F
s
L L
q a b 1.484 kNm
2
2 4
controls, tension on top
tension on top
d2
u M1 u a b
2
Max. bending stress at lift point: smax 59.0 MPa
I
A-5.4: A beam with an overhang is loaded by a uniform load of 3 kN/m over its
entire length. Moment of inertia Iz 3.36 106 mm4 and distances to top and
bottom of the beam cross section are 20 mm and 66.4 mm, respectively. It is
known that reactions at A and B are 4.5 kN and 13.5 kN, respectively. The
maximum bending stress in the beam is approximately:
(A) 36 MPa
(B) 67 MPa
(C) 102 MPa
(D) 119 MPa
Solution
3 kN/m
A
y
C
B
20 mm
z
4m
C
2m
RA 4.5 kN
Iz 3.36 (106) mm4
q3
66.4 mm
kN
m
Location of max. positive moment in AB (cut beam at location of zero shear &
use left FBD):
x max
RA
1.5 m
q
M pos RA x max 3
kN x max 2
3.375 kNm
m 2
compression on top of
beam
Compressive stress on top of beam at xmax:
sc1 M pos (20 mm)
Iz
20.1 MPa
Tensile stress at bottom of beam at xmax:
st1 M pos (66.4 mm)
Iz
66.696 MPa
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APPENDIX A FE Exam Review Problems
1123
Max. negative moment at B (use FBD of BC to find moment; compression on
bottom of beam):
M neg a3
sc2 st2 kN (2 m)2
6.000 kNm
b
m
2
M neg (66.4 mm)
Iz
M neg (20 mm)
Iz
118.6 MPa
35.7 MPa
A-5.5: A steel hanger with solid cross section has horizontal force P 5.5 kN
applied at free end D. Dimension variable b 175 mm and allowable normal
stress is 150 MPa. Neglect self weight of the hanger. The required diameter of
the hanger is approximately:
(A) 5 cm
(B) 7 cm
(C) 10 cm
(D) 13 cm
Solution
P 5.5 kN
b 175 mm
a 150 MPa
Reactions at support:
6b
A
B
NA P
(leftward)
2b
D
MA P(2 b) 1.9 kNm
(tension on bottom)
C
P
2b
Max. normal stress at bottom of cross section at A:
d
(2 P b) a b
2
P
smax 4
2
pd
pd
b
a
b
a
64
4
smax 4 P (16 b d)
pd3
Set smax sa and solve for required diameter d:
(sa)d3 (4 P)d 64Pb 0
solve numerically or by trial &
error to find
dreqd 5.11 cm
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APPENDIX A FE Exam Review Problems
A-5.6: A cantilever wood pole carries force P 300 N applied at its free end, as
well as its own weight (weight density 6 kN/m3). The length of the pole is
L 0.75 m and the allowable bending stress is 14 MPa. The required diameter
of the pole is approximately:
(A) 4.2 cm
(B) 5.5 cm
(C) 6.1 cm
(D) 8.5 cm
Solution
P 300 N
L 0.75 m
sa 14 MPa
gw 6
kN
m3
Uniformly distributed weight of pole:
w gw a
p d2
b
4
A
B
d
Max. moment at support:
L
Mmax P L w L
2
P
L
Section modulus of pole cross section:
Set Mmax equal to sa
P L cgw a
S
I
d
a b
2
p d4
64
p d3
S
32
d
a b
2
S and solve for required min. diameter d:
pd
L
p d3
b d L sa a
b0
4
2
32
2
Or
a
p sa 3
p gw L2 2
bd a
b d P L 0 solve numerically or by trial
32
8
& error to find
dreqd 5.50 cm
Since wood pole is light, try simpler solution which ignores self weight:
PL sa S
Or
p sa 3
b d PL
32
a
dreqd cP L a
1
32 3
b d 5.47 cm
p sa
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APPENDIX A FE Exam Review Problems
1125
A-5.7: A simply supported steel beam of length L 1.5 m and rectangular cross
section (h 75 mm, b 20 mm) carries a uniform load of q 48 kN/m, which
includes its own weight. The maximum transverse shear stress on the cross section at 0.25 m from the left support is approximately:
(A) 20 MPa
(B) 24 MPa
(C) 30 MPa
(D) 36 MPa
Solution
kN
m
L 1.5 m
q 48
h 75 mm
b 20 mm
q
Cross section properties:
A bh 1500 mm2
h
h h
Q ab b 14062 mm3
2 4
b h3
I
7.031
12
L
5
b
4
10 mm
Support reactions:
R
qL
36.0 kN
2
Transverse shear force at 0.25 m from left support:
V0.25 R q (0.25 m) 24.0 kN
Max. shear stress at NA at 0.25 m from left support:
V0.25 Q
24.0 MPa
Ib
3 V0.25
tmax 24.0 MPa
2A
tmax Or more simply . . .
A-5.8: A simply supported laminated beam of length L 0.5 m and square cross
section weighs 4.8 N. Three strips are glued together to form the beam, with the
allowable shear stress in the glued joint equal to 0.3 MPa. Considering also the
weight of the beam, the maximum load P that can be applied at L/3 from the left
support is approximately:
(A) 240 N
(B) 360 N
(C) 434 N
(D) 510 N
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APPENDIX A FE Exam Review Problems
Solution
q P at L/3
12 mm
12 mm 36 mm
12 mm
L
36 mm
L 0.5 m
W 4.8 N
h 36 mm
b 36 mm
q
N
W
9.60
m
L
ta 0.3 MPa
Cross section properties:
h h h
Qjoint ab b a b 5184 mm3
3 2 6
A bh 1296 mm2
I
b h3
1.400
12
105 mm4
Max. shear force at left support:
Vmax qL
2
Pa b
2
3
Shear stress on glued joint at left support; set t ta then solve for Pmax:
t
Vmax Qjoint
ta Ib
Or
t
4 qL
2
c
P a bd
3bh 2
3
Vmax a
b h2
b
9
b h3
a bb
12
Or
ta 4 Vmax
3 bh
so for ta 0.3 MPa
qL
3 3 b h ta
Pmax a
b 434 N
2
4
2
A-5.9: An aluminum cantilever beam of length L 0.65 m carries a distributed
load, which includes its own weight, of intensity q/2 at A and q at B. The beam
cross section has width 50 mm and height 170 mm. Allowable bending stress is
95 MPa and allowable shear stress is 12 MPa. The permissible value of load
intensity q is approximately:
(A) 110 kN/m
(B) 122 kN/m
(C) 130 kN/m
(D) 139 kN/m
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APPENDIX A FE Exam Review Problems
1127
Solution
L 0.65 m
b 50 mm
sa 95 MPa
h 170 mm
ta 12 MPa
Cross section properties:
q
—
2
q
B
A
L
A bh 8500 mm2
b h3
2.047 107 mm4
12
Reaction force and moment at A:
I
1 q
RA a qb L
2 2
5
MA q L2
12
S
3
RA q L
4
b h2
2.408
6
105 mm3
q L 1 q 2L
MA L L
2 2 22 3
Compare max. permissible values of q based on shear and moment allowable
stresses; smaller value controls
tmax 3 RA
2 A
3
qL
3 4
ta ±
≤
2
A
kN
8 ta A
139
m
9 L
5
q L2
MA
12
sa smax S
S
12 sa S
kN
qmax2 130.0
m
5 L2
So, since ta 12 MPa
qmax1 So, since sa 95 MPa
A-5.10: An aluminum light pole weighs 4300 N and supports an arm of weight
700 N, with arm center of gravity at 1.2 m left of the centroidal axis of the pole.
A wind force of 1500 N acts to the right at 7.5 m above the base. The pole cross
section at the base has outside diameter 235 mm and thickness 20 mm. The
maximum compressive stress at the base is approximately:
(A) 16 MPa
(B) 18 MPa
(C) 21 MPa
(D) 24 MPa
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APPENDIX A FE Exam Review Problems
Solution
W2 = 700 N
H 7.5 m
B 1.2 m
W1 4300 N
W2 700 N
1.2 m
P1 1500 N
d2 235 mm
P1 = 1500 N
W1 = 4300 N
t 20 mm
d1 d2 2t 195 mm
7.5 m
20 mm
z
y
Pole cross sectional properties at base:
x
p
A (d2 2 d1 2) 13509 mm2
4
p
I
(d2 4 d1 4) 7.873
64
y
235 mm
x
107 mm4
Compressive (downward) force at base of pole:
N W1 W2 5.0 kN
Bending moment at base of pole:
M W2 B P1 H 10.410 kNm
results in compression at right
Compressive stress at right side at base of pole:
N
sc A
d2
|M| a b
2
15.9 MPa
I
A-5.11: Two thin cables, each having diameter d t/6 and carrying tensile loads
P, are bolted to the top of a rectangular steel block with cross section dimensions
b t. The ratio of the maximum tensile to compressive stress in the block due
to loads P is:
(A) 1.5
(B) 1.8
(C) 2.0
(D) 2.5
Solution
Cross section properties of block:
A bt
bt3
I
12
t
d
6
b
P
P
t
Tensile stress at top of block:
P
st A
Pa
d
t
t
ba b
2
2 2
9P
I
2bt
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APPENDIX A FE Exam Review Problems
1129
Compressive stress at bottom of block:
P
sc A
d
t
t
Pa b a b
2
2 2
5P
I
2bt
Ratio of max. tensile to compressive stress in block:
ratio `
st
9
`
sc
5
9
1.8
5
A-5.12: A rectangular beam with semicircular notches has dimensions h 160 mm and h 1 140 mm. The maximum allowable bending stress
in the plastic beam is s max 6.5 MPa, and the bending moment is
M 185 Nm. The minimum permissible width of the beam is:
(A) 12 mm
(B) 20 mm
(C) 28 mm
(D) 32 mm
2R
M
M
h
h1
3.0
2R
h
— = 1.2
h1
K
M
M
h
h1
2.5
h = h1 + 2R
1.1
2.0
s
K = s max s nom= 6M2
nom
bh 1
b = thickness
1.05
1.5
0
0.05
0.10
0.15
0.20
0.25
0.30
R
—
h1
FIG. 5-50 Stress-concentration factor K for a notched beam of rectangular
cross section in pure bending (h height of beam; b thickness of beam,
perpendicular to the plane of the figure). The dashed line is for semicircular
notches (h h1 2R)
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APPENDIX A FE Exam Review Problems
Solution
R
1
1
(h h1) (160 mm 140 mm) 10.000 mm
2
2
10
R
0.071
h 1 140
h
160
1.143
h 1 140
From Fig 5-50:
K 2.25
sallow 6 M
2 so
K
b h1
b min 6MK
6 (185 Nm) (2.25)
19.6 mm
sallow h21
6.5 MPa C (140 mm)2D
3.0
2R
h
— = 1.2
h1
K
M
M
h
h1
2.5
h = h1 + 2R
K = 2.25
1.1
2.0
s
K = s max s nom= 6M2
nom
bh 1
b = thickness
1.05
1.5
0
0.05 0.071 0.10
0.15
0.20
0.25
0.30
R
—
h1
A-6.1: A composite beam is made up of a 200 mm 300 mm core (Ec 14 GPa)
and an exterior cover sheet (300 mm
12 mm, Ee 100 GPa) on each side.
Allowable stresses in core and exterior sheets are 9.5 MPa and 140 MPa, respectively. The ratio of the maximum permissible bending moment about the z-axis to
that about the y-axis is most nearly:
(A) 0.5
(B) 0.7
(C) 1.2
y
(D) 1.5
b 200 mm
t 12 mm
h 300 mm
Ec 14 GPa
z
Ee 100 GPa
sac 9.5 MPa
sae 140 MPa
C
300 mm
Solution
200 mm
12 mm
12 mm
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APPENDIX A FE Exam Review Problems
1131
Composite beam is symmetric about both axes so each NA is an axis of
symmetry
Moments of inertia of cross section about z and y axes:
b h3
h b3
4.500 108 mm4
2.000
Icy 12
12
2 t h3
Iez 5.400 107 mm4
12
2 h t3
b
t 2
Iey 2 (t h) a b 8.099 107 mm4
12
2
2
Icz 108 mm4
Bending about z axis based on allowable stress in each material (lesser value
controls)
Mmax_cz sac
Mmax_ez sae
aEc Icz Ee Iez b
h
Ec
2
aEc Icz Ee Iez b
h
Ee
2
52.9 kNm
109.2 kNm
Bending about y axis based on allowable stress in each material (lesser value
controls)
Mmax_cy sac
Mmax_ey sae
ratioz_to_y (Ec Icy Ee Iey)
b
Ec
2
(Ec Icy Ee Iey)
b
a tbEe
2
Mmax_cz
0.72
Mmax_cy
74.0 kNm
136.2kNm
allowable stress in the core, not
exterior cover sheet, controls
moments about both axes
A-6.2: A composite beam is made up of a 90 mm 160 mm wood beam (Ew 11 GPa) and a steel bottom cover plate (90 mm
8 mm, Es 190 GPa).
Allowable stresses in wood and steel are 6.5 MPa and 110 MPa, respectively.
The allowable bending moment about the z-axis of the composite beam is most
nearly:
(A) 2.9 kN?m
(B) 3.5 kN?m
(C) 4.3 kN?m
(D) 9.9 kN?m
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APPENDIX A FE Exam Review Problems
Solution
b 90 mm
t 8 mm
y
h 160 mm
Ew 11 GPa
Es 190 GPa
160 mm
saw 6.5 MPa
z
sas 110 MPa
O
8
mm
Aw bh 14400 mm2
As bt 720 mm2
90 mm
Locate NA (distance h2 above base) by summing 1st moments of EA about
base of beam; then find h1 dist. from NA to top of beam:
Es As
h2 t
h
Ew Aw at b
2
2
49.07 mm
Es As Ew Aw
h1 h t h2 118.93 mm
Moments of inertia of wood and steel about NA:
Is b t3
t 2
As ah2 b 1.467
12
2
106 mm4
Iw b h3
h 2
Aw ah1 b 5.254
12
2
107 mm4
Allowable moment about z axis based on allowable stress in each material
(lesser value controls)
Mmax_w saw
Mmax_s sas
(Ew Iw Es Is)
4.26 kNm
h1 Ew
(Ew Iw Es Is)
10.11 kNm
h2 Es
A-6.3: A steel pipe (d3 104 mm, d2 96 mm) has a plastic liner with inner
diameter d1 82 mm. The modulus of elasticity of the steel is 75 times that of
the modulus of the plastic. Allowable stresses in steel and plastic are 40 MPa and
550 kPa, respectively. The allowable bending moment for the composite pipe is
approximately:
(A) 1100 N?m
(B) 1230 N?m
(C) 1370 N?m
(D) 1460 N?m
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APPENDIX A FE Exam Review Problems
1133
Solution
d3 104 mm
d2 96 mm
d1 82 mm
sas 40 MPa
y
sap 550 kPa
Cross section properties:
z
p
(d3 2 d2 2) 1256.6 mm2
4
p
Ap (d2 2 d1 2) 1957.2 mm2
4
p
Is (d3 4 d2 4) 1.573 106 mm4
64
p
Ip (d2 4 d1 4) 1.950 106 mm4
64
C
As d1
d2 d3
Due to symmetry, NA of composite beam is the z axis
Allowable moment about z axis based on allowable stress in each material
(lesser value controls)
Mmax_s sas
(Ep Ip Es Is)
Modular ratio:
Mmax_p sap
d3
a b Es
2
n
(Ep Ip Es Is)
d2
a b Ep
2
Es
n 75
Ep
Divide through by Ep in moment expressions above
Mmax_s sas
(Ip nIs)
Mmax_ p sap
d3
a bn
2
1230 Nm
(Ip nIs)
d2
a b
2
1374 Nm
A-6.4: A bimetallic beam of aluminum (Ea 70 GPa) and copper (Ec 110 GPa)
strips has width b 25 mm; each strip has thickness t 1.5 mm. A bending
moment of 1.75 N?m is applied about the z axis. The ratio of the maximum stress in
the aluminum to that in the copper is approximately:
(A) 0.6
(B) 0.8
(C) 1.0
(D) 1.5
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APPENDIX A FE Exam Review Problems
Solution
b 25 mm t 1.5 mm Aa b t 37.5 mm2 Ac Aa 37.5 mm2
M 1.75 Nm
Ea 70 GPa
y
Ec 110 GPa
t
Equate 1st moments of EA
about bottom of beam to locate NA
(distance h2 above base); then find
h1 dist. from NA to top of beam:
Ec Ac
h2 A
z
O
C
b
t
t
t
Ea Aa at b
2
2
1.333 mm
Ec Ac Ea Aa
h1 2t h2 1.667 mm
h1 h2 3.000 mm
2 t 3.000 mm
Moments of inertia of aluminum and copper strips about NA:
Ic bt 3
t 2
Ac ah2 b 19.792 mm4
12
2
Ia bt3
t 2
Aa ah1 b 38.542 mm4
12
2
Bending stresses in aluminum and copper:
sa Mh1 Ea
41.9 MPa
Ea Ia Ec Ic
sc Mh2 Ec
52.6 MPa
Ea Ia Ec Ic
sa
0.795
sc
Ratio of the stress in the aluminum to that of the copper:
A-6.5: A composite beam of aluminum (Ea 72 GPa) and steel (Es 190 GPa)
has width b 25 mm and heights ha 42 mm, hs 68 mm. A bending moment
is applied about the z axis resulting in a maximum stress in the aluminum of
55 MPa. The maximum stress in the steel is approximately:
(A) 86 MPa
(B) 90 MPa
(C) 94 MPa
(D) 98 MPa
y
Aluminum
Solution
b 25 mm
ha 42 mm
Ea 72 GPa
Es 190 GPa
ha
hs 68 mm
sa 55 MPa
Steel
z
O hs
Aa bha 1050.0 mm2
As bhs 1700.0 mm2
b
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APPENDIX A FE Exam Review Problems
1135
Locate NA (distance h2 above base) by summing 1st moments of EA about
base of beam; then find h1 dist. from NA to top of beam:
Es As
h2 hs
ha
Ea Aa ahs b
2
2
44.43 mm
Ea Aa Es As
h1 ha hs h2 65.57 mm
h1 h2 110.00 mm
Moments of inertia of aluminum and steel parts about NA:
Is b hs3
hs 2
As ah2 b 8.401
12
2
105 mm4
Ia b ha3
ha 2
Aa ah1 b 2.240
12
2
106 mm4
Set max. bending stress in aluminum to given value then solve for moment M:
M
sa (Ea Ia Es Is)
3.738 kNm
h1Ea
Use M to find max. bending stress in steel: ss M h2 Es
98.4 MPa
Ea Ia Es Is
A-7.1: A rectangular plate (a 120 mm, b 160 mm) is subjected to compressive stress sx 4.5 MPa and tensile stress sy 15 MPa. The ratio of the
normal stress acting perpendicular to the weld to the shear stress acting along the
weld is approximately:
(A) 0.27
(B) 0.54
(C) 0.85
(D) 1.22
Solution
a 120 mm
a
u arctana b 36.87
b
sx 4.5 MPa
σy
b 160 mm
sy 15 MPa
ld
We
a
b
σx
txy 0
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APPENDIX A FE Exam Review Problems
Plane stress transformation: normal and shear stresses on y-face of element
rotated through angle u (perpendicular to & along weld seam):
su sx sy
2
tu `
sx sy
2
sx sy
2
cosc2 au sinc2 au p
p
b d txy sinc2 au b d 7.98 MPa
2
2
p
p
b d txy cosc2 au b d 9.36 MPa
2
2
su
` 0.85
tu
A-7.2: A rectangular plate in plane stress is subjected to normal stresses sx and
sy and shear stress txy. Stress sx is known to be 15 MPa but sy and txy are
unknown. However, the normal stress is known to be 33 MPa at counterclockwise angles of 35° and 75° from the x axis. Based on this, the normal stress sy
on the element below is approximately:
(A) 14 MPa
(B) 21 MPa
(C) 26 MPa
(D) 43 MPa
Solution
sx 15 MPa
s35 33 MPa
s75 s35
Plane stress transformations for 35 & 75 :
su sx sy
2
sx sy
2
cos(2 u) txy sin(2 u)
y
σy
τxy
σx
O
x
For u 35 :
sx sy
2
Or
u35 35
sx sy
2
cos[2 (u35)] txy sin[2 (u35)] s35
sy 2.8563txy 69.713
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APPENDIX A FE Exam Review Problems
And for u 75 :
sx sy
2
1137
u75 75
sx sy
2
cos[2 (u75)] txy sin[2 (u75)] s75
sy 0.5359 txy 34.292
Or
Solving above two equations for sy and txy gives:
sy
1 2.8563 1 69.713
26.1
bc
d a
ba
b MPa
txy
1 0.5359
34.292
15.3
a
so y 26.1 MPa
A-7.3: A rectangular plate in plane stress is subjected to normal stresses sx 35 MPa, sy 26 MPa, and shear stress txy 14 MPa. The ratio of the magnitudes of the principal stresses (s1/s2) is approximately:
(A) 0.8
(B) 1.5
(C) 2.1
(D) 2.9
Solution
y
sx 35 MPa
sy 26 MPa
txy 14 MPa
σy
Principal angles:
uP1 2txy
1
arctana
b 36.091
s
2
x sy
uP2 uP1 τxy
O
σx
p
126.091
2
x
Plane stress transformations:
s1 s2 sx sy
2
sx sy
2
sx sy
2
sx sy
2
cos(2uP1) txy sin(2uP1) 45.21 MPa
cos(2uP2) txy sin(2uP2) 15.79 MPa
Ratio of principal stresses:
s1
2.86
s2
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APPENDIX A FE Exam Review Problems
A-7.4: A drive shaft resists torsional shear stress of 45 MPa and axial compressive stress of 100 MPa. The ratio of the magnitudes of the principal stresses
(s1/s2) is approximately:
(A) 0.15
(B) 0.55
(C) 1.2
(D) 1.9
Solution
sx 100 MPa
sy 0
txy 45 MPa
Principal angles:
2txy
1
arctana
b 20.994
s
2
x sy
p
uP2 uP1 110.994
2
uP1 100 MPa
45 MPa
Plane stress transformations:
suP1 sx sy
2
actually s2
suP2 sx sy
2
sx sy
2
sx sy
2
cos(2uP1) txy sin(2uP1) 117.27 MPa
cos(2uP2) txy sin(2uP2) 17.27 MPa
this is s1
So
s1 max(suP1, suP2) 17.268 MPa s2 min(suP1, suP2) 117.268 MPa
Ratio of principal stresses:
`
s1
` 0.15
s2
A-7.5: A drive shaft resists torsional shear stress of 45 MPa and axial compressive stress of 100 MPa. The maximum shear stress is approximately:
(A) 42 MPa
(B) 67 MPa
(C) 71 MPa
(D) 93 MPa
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APPENDIX A FE Exam Review Problems
1139
Solution
sx 100 MPa
sy 0
txy 45 MPa
Max. shear stress:
tmax sx sy 2
b txy2 67.3 MPa
B
2
a
100 MPa
45 MPa
A-7.6: A drive shaft resists torsional shear stress of txy 40 MPa and axial compressive stress sx 70 MPa. One principal normal stress is known to be 38 MPa
(tensile). The stress sy is approximately:
(A) 23 MPa
(B) 35 MPa
(C) 62 MPa
(D) 75 MPa
Solution
sx 70 MPa
txy 40 MPa
sy is unknown
sprin 38 MPa
y
σy
τxy
Stresses sx and sy must be smaller than
the given principal stress so:
σx
O
x
s1 sprin
Substitute into stress transformation equation
and solve for sy:
sx sy
2
sx sy 2
626 MPa
b txy2 s1 solve, sy B
2
27
23.2 MPa
a
A-7.7: A cantilever beam with rectangular cross section (b 95 mm, h 300 mm)
supports load P 160 kN at its free end. The ratio of the magnitudes of the principal
stresses (s1/s2) at point A (at distance c 0.8 m from the free end and distance
d 200 mm up from the bottom) is approximately:
(A) 5
(B) 12
(C) 18
(D) 25
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APPENDIX A FE Exam Review Problems
Solution
P 160 kN
b 95 mm
c 0.8 m
d 200 mm
h 300 mm
d
0.667
h
P
Cross section properties:
A bh 28500 mm2
I
b h3
2.138
12
108 mm4
QA [b (h d)] c
A
h
c
b
h (h d)
d 9.500
2
2
d
105 mm3
Moment, shear force and normal and shear stresses at A:
MA Pc 1.280
VA QA
tA 7.485 MPa
Ib
h
MA ad b
2
29.942 MPa
sA I
sx sA
Plane stress state at A:
VA P
105 kNmm
txy tA
sy 0
Principal stresses:
uP s1 s2 2 txy
1
b 13.283
arctana
sx sy
2
sx sy
2
sx sy
2
sx sy 2
b txy2 31.709 MPa
B
2
sx sy 2
b txy2 1.767 MPa
B
2
a
a
Ratio of principal stresses (s1 / s2):
s1
` s ` 17.9
2
A-7.8: A simply supported beam (L 4.5 m) with rectangular cross section
(b 95 mm, h 280 mm) supports uniform load q 25 kN/m. The ratio of
the magnitudes of the principal stresses (s1/s2) at a point a 1.0 m from the
left support and distance d 100 mm up from the bottom of the beam is
approximately:
(A) 9
(B) 17
(C) 31
(D) 41
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APPENDIX A FE Exam Review Problems
1141
Solution
kN
m
L 4.5 m
b 95 mm
h 280 mm
a 1.0 m
d 100 mm
q 25
q
h
b
a
L
Cross section properties:
A bh 26600 mm2
I
b h3
1.738
12
Q [b (h d)] c
108 mm4
h (h d)
d 8.550
2
2
105 mm3
Moment, shear force and normal and shear stresses at distance a from left
support:
Va q a2
qL
qL
q a 31.250 kN Ma a
4.375
2
2
2
Va Q
t
1.618 MPa
Ib
Plane stress state: sx s
104 kNmm
h
Ma ad b
2
s
10.070 MPa
I
txy t
sy 0
Principal stresses:
uP s1 s2 2 txy
1
arctana
b 8.909
s
2
x sy
sx sy
2
sx sy
2
sx sy 2
b txy2 10.324 MPa
B
2
sx sy 2
b txy2 0.254 MPa
B
2
a
a
Ratio of principal stresses (s1 / s2):
`
s1
` 40.7
s2
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APPENDIX A FE Exam Review Problems
A-8.1: A thin wall spherical tank of diameter 1.5 m and wall thickness 65 mm
has internal pressure of 20 MPa. The maximum shear stress in the wall of the
tank is approximately:
(A) 58 MPa
(B) 67 MPa
(C) 115 MPa
(D) 127 MPa
Solution
d 1.5 m
t 65 mm
p 20 MPa
Weld
t
0.087
d
a b
2
Thin wall tank since:
Biaxial stress:
d
pa b
2
s
2t
115.4 MPa
Max. shear stress at 45 deg. rotation is 1/2 of s
tmax s
57.7 MPa
2
A-8.2: A thin wall spherical tank of diameter 0.75 m has internal pressure
of 20 MPa. The yield stress in tension is 920 MPa, the yield stress in shear is
475 MPa, and the factor of safety is 2.5. The modulus of elasticity is
210 GPa, Poisson’s ratio is 0.28, and maximum normal strain is 1220
106. The minimum permissible thickness of the tank is approximately:
(A) 8.6 mm
(B) 9.9 mm
(C) 10.5 mm
(D) 11.1 mm
Solution
d 0.75 m
sY 920 MPa
0.28
p 20 MPa
E 210 GPa
tY 475 MPa
FSY 2.5
6
a 1220(10
)
Weld
Thickness based on tensile stress:
d
pa b
2
10.190 mm
t1 sY
2a
b
FSY
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APPENDIX A FE Exam Review Problems
1143
Thickness based on shear stress:
d
pa b
2
t2 9.868 mm
tY
4a
b
FSY
Thickness based on normal strain:
d
pa b
2
t3 (1 n)
2 aE
t3 10.54 mm
largest value controls
A-8.3: A thin wall cylindrical tank of diameter 200 mm has internal pressure of
11 MPa. The yield stress in tension is 250 MPa, the yield stress in shear is
140 MPa, and the factor of safety is 2.5. The minimum permissible thickness of
the tank is approximately:
(A) 8.2 mm
(B) 9.1 mm
(C) 9.8 mm
(D) 11.0 mm
Solution
d 200 mm
sY 250 MPa
p 11 MPa
tY 140 MPa
FSY 2.5
Wall thickness based on tensile stress:
d
pa b
2
11.00 mm
t1 sY
FSY
larger value governs
t1
0.110
d
a b
2
Wall thickness based on shear stress:
d
pa b
2
9.821 mm
t2 tY
2a
b
FSY
t2
0.098
d
a b
2
A-8.4: A thin wall cylindrical tank of diameter 2.0 m and wall thickness 18 mm
is open at the top. The height h of water (weight density 9.81 kN/m3) in the
tank at which the circumferential stress reaches 10 MPa in the tank wall is
approximately:
(A) 14 m
(B) 18 m
(C) 20 m
(D) 24 m
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APPENDIX A FE Exam Review Problems
Solution
d2m
t 18 mm
Pressure at height h:
sa 10 MPa
gw 9.81
kN
m3
d
ph gw h
d
ph a b
2
Circumferential stress: sc t
d
(gw h) a b
2
sc t
h
Set sc equal to sa and solve for h:
h
sa t
18.3 m
d
(gw) a b
2
A-8.5: The pressure relief valve is opened on a thin wall cylindrical tank, with
radius to wall thickness ratio of 128, thereby decreasing the longitudinal strain
by 150 106. Assume E 73 GPa and v 0.33. The original internal pressure in the tank was approximately:
(A) 370 kPa
(B) 450 kPa
(C) 500 kPa
(D) 590 kPa
Solution
rt r
t
rt 128
L 148 (10
E 73 GPa
6
)
n 0.33
strain gage
Longitudinal strain:
Set
to
p
p r
a b (1 2)
2E t
L and solve for pressure p:
2E L
497 kPa
rt (1 2n)
A-8.6: A cylindrical tank is assembled by welding steel sections circumferentially. Tank diameter is 1.5 m, thickness is 20 mm, and internal pressure is
2.0 MPa. The maximum stress in the heads of the tank is approximately:
(A) 38 MPa
(B) 45 MPa
(C) 50 MPa
(D) 59 MPa
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APPENDIX A FE Exam Review Problems
1145
Solution
d 1.5 m
t 20 mm
p 2.0 MPa
d
pa b
2
sh 37.5 MPa
2t
Welded seams
A-8.7: A cylindrical tank is assembled by welding steel sections circumferentially.
Tank diameter is 1.5 m, thickness is 20 mm, and internal pressure is 2.0 MPa. The
maximum tensile stress in the cylindrical part of the tank is approximately:
(A) 45 MPa
(B) 57 MPa
(C) 62 MPa
(D) 75 MPa
Solution
d 1.5 m
t 20 mm
p 2.0 MPa
d
pa b
2
sc 75.0 MPa
t
Welded seams
A-8.8: A cylindrical tank is assembled by welding steel sections circumferentially. Tank diameter is 1.5 m, thickness is 20 mm, and internal pressure is 2.0 MPa.
The maximum tensile stress perpendicular to the welds is approximately:
(A) 22 MPa
(B) 29 MPa
(C) 33 MPa
(D) 37 MPa
Solution
d 1.5 m
t 20 mm
d
pa b
2
sw 37.5 MPa
2t
p 2.0 MPa
Welded seams
A-8.9: A cylindrical tank is assembled by welding steel sections circumferentially. Tank diameter is 1.5 m, thickness is 20 mm, and internal pressure is
2.0 MPa. The maximum shear stress in the heads is approximately:
(A) 19 MPa
(B) 23 MPa
(C) 33 MPa
(D) 35 MPa
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APPENDIX A FE Exam Review Problems
Solution
d 1.5 m
t 20 mm
p 2.0 MPa
d
pa b
2
th 18.8 MPa
4t
Welded seams
A-8.10: A cylindrical tank is assembled by welding steel sections circumferentially.
Tank diameter is 1.5 m, thickness is 20 mm, and internal pressure is 2.0 MPa. The
maximum shear stress in the cylindrical part of the tank is approximately:
(A) 17 MPa
(B) 26 MPa
(C) 34 MPa
(D) 38 MPa
Solution
d 1.5 m
t 20 mm
p 2.0 MPa
d
pa b
2
tmax 37.5 MPa
2t
Welded seams
A-8.11: A cylindrical tank is assembled by welding steel sections in a helical
pattern with angle a 50 degrees. Tank diameter is 1.6 m, thickness is 20 mm,
and internal pressure is 2.75 MPa. Modulus E 210 GPa and Poisson’s ratio
n 0.28. The circumferential strain in the wall of the tank is approximately:
(A) 1.9 104
(B) 3.2 104
(C) 3.9 104
(D) 4.5 104
Solution
d 1.6 m
E 210 GPa
t 20 mm
p 2.75 MPa
a 50
n 0.28
Circumferential stress:
d
pa b
2
sc 110.000 MPa
t
Helical weld
α
Circumferential strain:
c sc
(2 n) 4.50
2E
104
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APPENDIX A FE Exam Review Problems
1147
A-8.12: A cylindrical tank is assembled by welding steel sections in a helical
pattern with angle a 50 degrees. Tank diameter is 1.6 m, thickness is
20 mm, and internal pressure is 2.75 MPa. Modulus E 210 GPa and
Poisson’s ratio n 0.28. The longitudinal strain in the the wall of the tank is
approximately:
(A) 1.2 104
(B) 2.4 104
(C) 3.1 104
(D) 4.3 104
Solution
d 1.6 m
t 20 mm
E 210 GPa
n 0.28
p 2.75 MPa
a 50
Longitudinal stress:
Helical weld
d
pa b
2
sL 55.000 MPa
2t
α
Longitudinal strain:
L sL
(1 2n) 1.15
E
104
A-8.13: A cylindrical tank is assembled by welding steel sections in a helical
pattern with angle a 50 degrees. Tank diameter is 1.6 m, thickness is
20 mm, and internal pressure is 2.75 MPa. Modulus E 210 GPa and
Poisson’s ratio n 0.28. The normal stress acting perpendicular to the weld
is approximately:
(A) 39 MPa
(B) 48 MPa
(C) 78 MPa
(D) 84 MPa
Solution
d 1.6 m
t 20 mm
p 2.75 MPa
E 210 GPa
0.28 a 50
Helical weld
α
Longitudinal stress:
d
pa b
2
55.000 MPa So sx sL
sL 2t
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APPENDIX A FE Exam Review Problems
Circumferential stress:
d
pa b
2
sc 110.000 MPa So sy sc
t
u 90 a 40.000
Angle perpendicular to the weld:
Normal stress perpendicular to the weld:
s40 sx sy
2
sx sy
2
cos (2 u) 77.7 MPa
A-8.14: A segment of a drive shaft (d2 200 mm, d1 160 mm) is subjected
to a torque T 30 kN?m. The allowable shear stress in the shaft is 45 MPa. The
maximum permissible compressive load P is approximately:
(A) 200 kN
(B) 286 kN
(C) 328 kN
(D) 442 kN
Solution
d2 200 mm
d1 160 mm
ta 45 MPa
P
T 30 kNm
Cross section properties:
T
p 2
(d2 d12) 11310 mm2
4
p 4
Ip (d2 d14) 9.274 107 mm4
32
A
T
Normal and in-plane shear stresses:
sx 0
sy P
A
txy d2
Ta b
2
IP
32.349 MPa
P
Maximum in-plane shear stress: set max allow then solve for sy
t max sx sy 2
b txy2
B
2
a
Finally solve for P sy A:
So sy #4 (ta txy)2 25.303 MPa
Pmax y A 286 kN
A-8.15: A thin walled cylindrical tank, under internal pressure p, is compressed
by a force F 75 kN. Cylinder diameter is d 90 mm and wall thickness
t 5.5 mm. Allowable normal stress is 110 MPa and allowable shear stress is
60 MPa. The maximum allowable internal pressure pmax is approximately:
(A) 5 MPa
(B) 10 MPa
(C) 13 MPa
(D) 17 MPa
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APPENDIX A FE Exam Review Problems
1149
Solution
d 90 mm
t 5.5 mm
F 75 kN
A 2p
sa 110 MPa
d
t 1555 mm2
2
d
pmax a b
2
sc t
F
F
Circumferential normal stress:
and setting sc sa and solving for pmax:
2t
pmaxc sa a b 13.4 MPa
d
controls
Longitudinal normal stress:
d
pmax a b
2
F
sL 2t
A
Or
sL pmax d F
4t
A
So set sL sa and solve for pmax:
F 4t
pmaxL asa b
38.7 MPa
A d
Check also in-plane & out-of-plane shear stresses: all are below allowable
shear stress so circumferential normal stress controls as noted above.
A-9.1: An aluminum beam (E 72 GPa) with a square cross section and
span length L 2.5 m is subjected to uniform load q 1.5 kN/m. The allowable bending stress is 60 MPa. The maximum deflection of the beam is
approximately:
(A) 10 mm
(B) 16 mm
(C) 22 mm
(D) 26 mm
Solution
E 72 (103)MPa
q 1.5
sa 60 MPa
q = 1.5 kN/m
N
mm
L 2500 mm
Max. moment and deflection at L/2:
Mmax q L2
8
dmax L = 2.5 m
5 q L4
384 E I
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APPENDIX A FE Exam Review Problems
Moment of inertia and section modulus for square cross section (height width b)
I
b4
12
S
I
b3
6
b
a b
2
Flexure formula
qL2
8
smax 3
b
a b
6
Mmax
S
smax so
b3 3 qL2
4 smax
Max. deflection formula
dmax dmax 5q L4
so
b4
384 E a b
12
5q L4
3 qL
ca
smax
4
384 E ≥
2
solve for dmax if smax sa
1
b 3d
12
22.2 mm
4
¥
A-9.2: An aluminum cantilever beam (E 72 GPa) with a square cross section
and span length L 2.5 m is subjected to uniform load q 1.5 kN/m. The allowable bending stress is 55 MPa. The maximum deflection of the beam is approximately:
(A) 10 mm
(B) 20 mm
(C) 30 mm
(D) 40 mm
Solution
E 72(103) MPa
q 1.5
sa 55 MPa
q
N
mm
L 2500 mm
L
Max. moment at support & max. deflection at L:
Mmax q L2
q L4
dmax 2
8EI
Moment of inertia and section modulus for square cross section (height width b)
I
b4
12
S
b3
I
6
b
a b
2
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APPENDIX A FE Exam Review Problems
1151
Flexure formula
q L2
2
smax 3
b
a b
6
Mmax
S
smax so
b3 3
q L2
smax
Max. deflection formula
dmax dmax q L4
b4
8E a b
12
so solve for dmax if smax sa
q L4
q L2 1 4
b 3d
ca3
smax
8E ≥
¥
12
29.9 mm
A-9.3: A steel beam (E 210 GPa) with I 119 106 mm4 and span length
L 3.5 m is subjected to uniform load q 9.5 kN/m. The maximum deflection
of the beam is approximately:
(A) 10 mm
(B) 13 mm
(C) 17 mm
(D) 19 mm
Solution
E 210(103) MPa
q 9.5
I 119(106) mm4
strong axis I for W310 52
N
mm
L 3500 mm
y
MA
q
A
L
x
B
k = 48EI/L3
RB = kδB
Max. deflection at A by superposition of SS beam mid-span deflection & RB/k:
dmax 5 q (2 L)4
(q L)
13.07 mm
384 E I
48 E I
a 3 b
L
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APPENDIX A FE Exam Review Problems
A-9.4: A steel bracket ABC (EI 4.2 106 N?m2) with span length L 4.5 m
and height H 2 m is subjected to load P 15 kN at C. The maximum rotation of joint B is approximately:
(A) 0.1
(B) 0.3
(C) 0.6
(D) 0.9
Solution
E 210 GPa
EI 4.20
I 20
6
106 mm4
strong axis I for W200
10 Nm
C
P 15 kN
L 4.5 m
22.5
2
P
H2m
H
B
A
L
Max. rotation at B: apply statically-equivalent moment P H at B on SS beam
uBmax (P H) L
0.614
3EI
uBmax 0.011 rad
A-9.5: A steel bracket ABC (EI 4.2 106 N?m2) with span length L 4.5 m
and height H 2 m is subjected to load P 15 kN at C. The maximum horizontal displacement of joint C is approximately:
(A) 22 mm
(B) 31 mm
(C) 38 mm
(D) 40 mm
Solution
E 210 GPa
EI 4.20
I 20
106 mm4
strong axis I for W200
106 Nm2
C
P 15 kN
L 4.5 m
22.5
P
H
H2m
B
A
L
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APPENDIX A FE Exam Review Problems
Max. rotation at B: apply statically-equivalent moment P
uBmax (P H) L
0.614
3EI
H at B on SS beam
uBmax 0.011 rad
Horizontal deflection of vertical cantilever BC:
Finally, superpose uB
1153
dBC P H3
9.524 mm
3EI
H and dBC
dC uBmax H dBC 31.0 mm
A-9.6: A nonprismatic cantilever beam of one material is subjected to load P at
its free end. Moment of inertia I2 2 I1. The ratio r of the deflection dB to the
deflection d1 at the free end of a prismatic cantilever with moment of inertia I1
carrying the same load is approximately:
(A) 0.25
(B) 0.40
(C) 0.56
(D) 0.78
Solution
Max. deflection of prismatic cantilever
(constant I1)
A
P L3
d1 3 E I1
I2
L
—
2
P
C
I1
B
L
—
2
Rotation at C due to both load P & moment
PL/2 at C for nonprismatic beam:
L 2
L L
Pa b
aP b
2
2 2 3 L2 P
uC 2 E I2
E I2
8 E I2
Deflection at C due to both load P & moment
PL/2 at C for nonprismatic beam:
L 3
L L 2
Pa b
aP b a b
2
2 2
5 L3 P
dCl 3 E I2
2 E I2
48 E I2
Total deflection at B:
L 3
Pa b
2
L
dB dCl uC 2
3 E I1
L 3
P
b
a
2
5 L3 P
3 L2 P L
L3 P (7 I1 I2)
dB a
b 48 E I2
8 E I2 2
3 E I1
24 E I1 I2
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APPENDIX A FE Exam Review Problems
Ratio dB/d1:
L3 P (7 I1 I2)
24 E I1 I2
7 I1 1
r
8 I2 8
P L3
3 E I1
1
7 1
so r a b 0.563
8 2
8
A-9.7: A steel bracket ABCD (EI 4.2 106 N?m2), with span length L 4.5 m
and dimension a 2 m, is subjected to load P 10 kN at D. The maximum
deflection at B is approximately:
(A) 10 mm
(B) 14 mm
(C) 19 mm
(D) 24 mm
Solution
E 210 GPa
EI 4.20
I 20
106 mm4
strong axis I for W200 22.5
106 Nm2
L
A
P 10 kN
B
D
L 4.5 m a 2 m
a
h 206 mm
C
P
Statically-equivalent loads at end of cantilever AB:
• downward load P
• CCW moment P a
Downward deflection at B by superposition:
dB P L3 (P a) L2
24.1 mm
3EI
2EI
dB
0.005
L
A-10.1: Propped cantilever beam AB has moment M1 applied at joint B. Framework
ABC has moment M2 applied at C. Both structures have constant flexural rigidity EI.
If the ratio of the applied moments M 1/M 2 3/2, the ratio of the reactive moments
M A1/M A2 at clamped support A is:
(A) 1
(B) 3/2
(C) 2
(D) 5/2
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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APPENDIX A FE Exam Review Problems
Solution
M1
L/2
B
A
C
M2
y
y
A
B
x
x
MA2
MA1
L
L
From Prob. 10.3-1:
MA1 M1
2
M2
2
Statically-equivalent moment at B is M2
so
MA2 MA1 M1
MA2 M2
so
MA1 3
MA2 2
Ratio of reactive moments is
A-10.2: Propped cantilever beam AB has moment M1 applied at joint B. Framework
ABC has moment M2 applied at C. Both structures have constant flexural rigidity EI.
If the ratio of the applied moments M 1/M 2 3/2, the ratio of the joint rotations at
B, uB1/uB2, is:
(A) 1
(B) 3/2
(C) 2
(D) 5/2
Solution
M2
y
y
M1
L/2
B
A
A
x
MA1
θB1
L
C
B
MA2
x
θB2
L
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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APPENDIX A FE Exam Review Problems
From Prob 10.3-1:
uB1 M1 L
4 EI
Statically-equivalent moment at B is M 2
Ratio of joint rotations is
u B1
M1
u B2
M2
uB2 so
M2 L
4 EI
uB1
3
u B2
2
so
A-10.3: Structure 1 with member BC of length L/2 has force P1 applied at joint C.
Structure 2 with member BC of length L has force P2 applied at C. Both structures
have constant flexural rigidity EI. If the ratio of the applied forces P1/P2 5/2, the
ratio of the joint B rotations uB1/uB2 is:
(A) 1
(B) 5/4
(C) 3/2
(D) 2
Solution
P2
C
L
P1
C
y
y
L/2
A
B
θB1
MA1
B
A
x
x
θB2
MA2
L
L
Statically-equivalent moment at B is M1 P1
From Prob. 10.3-1:
u B1 aP1
L
bL
2
L2 P1
4 EI
8 EI
Statically-equivalent moment at B is M2 P2
AP 2 LB L L2 P 2
4 EI
4 EI
u B1
P1
Ratio of joint rotations is
u B2 2 P2
so
L/2
L
u B2 so
u B1 5
u B2 4
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APPENDIX A FE Exam Review Problems
A-10.4: Structure 1 with member BC of length L/2 has force P1 applied at
joint C. Structure 2 with member BC of length L has force P2 applied at C. Both
structures have constant flexural rigidity EI. The required ratio of the applied
forces P1/P2 so that joint B rotations uB1 and uB2 are equal is approximately:
(A) 1
(B) 5/4
(C) 3/2
(D) 2
Solution
P2
C
L
P1
C
y
y
L/2
A
B
x
θB1
MA1
B
A
x
θB2
MA2
L
L
Statically-equivalent moment at B is M1 P1
From Prob. 10.3-1:
uB1 aP1
L
bL
2
L2 P 1
4 EI
8 EI
Statically-equivalent moment at B is M 2 P2
so
uB2 L/2
L
(P2 L) L L2 P 2
4 EI
4 EI
Ratio of joint rotations is
uB1
P1
uB2 2 P2
and
uB1
1 so
uB2
P1
2
P2
A-10.5: Structure 1 with member BC of length L/2 has force P1 applied at
joint C. Structure 2 with member BC of length L has force P2 applied at C. Both
structures have constant flexural rigidity EI. If the ratio of the applied forces
P1/P2 5/2, the ratio of the joint B reactions RB1/RB2 is:
(A) 1
(B) 5/4
(C) 3/2
(D) 2
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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APPENDIX A FE Exam Review Problems
Solution
P2
C
L
P1
C
y
y
L/2
A
B
B
A
x
x
RB1
RB2
L
L
Statically-equivalent moment at B is M 1 P1
From Prob. 10.3-1:
RB1 3 M1
2 L
L
aP1 b
2
3
3P1
or RB1 2
L
4
Statically-equivalent moment at B is M 2 P2
so
RB2 L/2
L
3 AP2 LB
3 P2
2
L
2
Ratio of reactions at B is
RB1
P1
RB2 2 P 2
so
5
RB1 1 5
a b
RB2 2 2
4
A-10.6: Structure 1 with member BC of length L/2 has force P1 applied at
joint C. Structure 2 with member BC of length L has force P2 applied at C. Both
structures have constant flexural rigidity EI. The required ratio of the applied
forces P1/P2 so that joint B reactions RB1 and RB2 are equal is:
(A) 1
(B) 5/4
(C) 3/2
(D) 2
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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APPENDIX A FE Exam Review Problems
Solution
P2
C
L
P1
C
y
y
L/2
A
B
B
A
x
x
RB2
RB1
L
L
Statically-equivalent moment at B is M 1 P1
From Prob. 10.3-1: RB1 3 M1
2 L
RB1 or
Statically-equivalent moment at B is M 2 P2
so
RB2 3 AP2 LB
3 P2
2
L
2
Ratio of reactions at B is
RB1
P1
RB2 2 P2
L/2
and
3
2
aP1
L
b
2
3 P1
L
4
L
RB1
1 so
RB2
P1
2
P2
A-10.7: Structure 1 with member BC of length L/2 has force P1 applied at
joint C. Structure 2 with member BC of length L has force P2 applied at C. Both
structures have constant flexural rigidity EI. If the ratio of the applied forces
P1/P2 5/2, the ratio of the joint C lateral deflections dC1/dC2 is approximately:
(A) 1/2
(B) 4/5
(C) 3/2
(D) 2
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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APPENDIX A FE Exam Review Problems
Solution
P1
δC1
C
y
L/2
A
B
x
L
L 3
L
P1 a b
P1 L
2
2 L 5 L3 P 1
d C1 3 EI
4 EI 2
48 EI
d C2 P 2 L3 P 2 L L
7 L3 P 2
L
3 EI
4 EI
12 EI
dC1
5P1
dC2 28 P 2
5 5 25
28 2 56
25
0.446
56
A-11.1: Beam ACB has a sliding support at A and is supported at C by a pinned end
steel column with square cross section (E 200 GPa, b 40 mm) and height
L 3.75 m. The column must resist a load Q at B with a factor of safety 2.0 with
respect to the critical load. The maximum permissible value of Q is approximately:
(A) 10.5 kN
(B) 11.8 kN
(C) 13.2 kN
(D) 15.0 kN
Solution
E 200 GPa
b 40 mm
I
b4
2.133
12
n 2.0
L 3.75 m
105 mm4
Statics: sum vertical forces to
find reaction at D:
RD Q
So force in pin-pin column is Q
p2 E I
Qcr 2 29.9 kN
Pcr Qcr
L
A
C
B
d
2d
Q
L
D
Allowable value of Q:
Qcr
Qallow 15.0 kN
n
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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APPENDIX A FE Exam Review Problems
1161
A-11.2: Beam ACB has a pin support at A and is supported at C by a steel column
with square cross section (E 190 GPa, b 42 mm) and height L 5.25 m.
The column is pinned at C and fixed at D. The column must resist a load Q at
B with a factor of safety 2.0 with respect to the critical load. The maximum permissible value of Q is approximately:
(A) 3.0 kN
(B) 6.0 kN
(C) 9.4 kN
(D) 10.1 kN
Solution
E 190 GPa
n 2.0
b 42 mm
L 5.25 m
A
d
Effective length of pinned-fixed column:
Le 0.699 L 3.670 m
b4
I
2.593
12
2d
Q
L
105 mm4
D
Statics: use FBD of ACB and sum moments
about A to find force in column as a multiple of Q:
FCD B
C
Q (3 d)
3Q
d
So force in pin-fixed column is 3Q
Pcr 3 Qcr Pcr p2 E I
36.1 kN
Le2
So
Qcr Pcr
12.0 kN
3
Allowable value of Q:
Qallow Qcr
6.0 kN
n
A-11.3: A steel pipe column (E 190 GPa, a 14 1026 per degree Celsius,
d2 82 mm, d1 70 mm) of length L 4.25 m is subjected to a temperature
increase T. The column is pinned at the top and fixed at the bottom. The temperature increase at which the column will buckle is approximately:
(A) 36 °C
(B) 42 °C
(C) 54 °C
(D) 58 °C
Solution
E 190 GPa
d2 82 mm
L 4.25 m
d1 70 mm
a [14 (106)]/ C
A
p 2
(d2 d12) 1432.57 mm2
4
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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APPENDIX A FE Exam Review Problems
Effective length of pinned-fixed column:
Le 0.699L 3.0 m
B
p
(d24 d14) 1.04076
I
64
106 mm4
ΔT
L
Axial compressive load in bar: P EA a(T)
Equate to Euler buckling load and solve for T:
p2 E I
L2
T e
EAa
A
Or
T p2 I
58.0 C
a A L2e
A-11.4: A steel pipe (E 190 GPa, a 14 106/ C, d2 82 mm, d1 70 mm)
of length L 4.25 m hangs from a rigid surface and is subjected
to a temperature increase T 50 °C. The column is fixed at the top and has a
small gap at the bottom. To avoid buckling, the minimum clearance at the bottom
should be approximately:
(A) 2.55 mm
(B) 3.24 mm
(C) 4.17 mm
(D) 5.23 mm
Solution
E 190 GPa
L 4250 mm
a [14(106)] / °C
d2 82 mm
A
T 50 °C
d1 70 mm
p 2
(d2 d12) 1433 mm2
4
p
I (d24 d14) 1.041
64
ΔT
6
L
4
10 mm
Effective length of fixed-roller support column:
gap
frictionless
surface
Le 2.0L 8500.0 mm
Column elongation due to temperature increase:
d1 aTL 2.975 mm
Euler buckling load for fixed-roller column:
Pcr p2 E I
27.013 kN
L2e
Column shortening under load of P Pcr:
d2 Pcr L
0.422 mm
EA
Mimimum required gap size to avoid buckling: gap d1 d2 2.55 mm
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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APPENDIX A FE Exam Review Problems
1163
A-11.5: A pinned-end copper strut (E 110 GPa) with length L 1.6 m is constructed of circular tubing with outside diameter d 38 mm. The strut must
resist an axial load P 14 kN with a factor of safety 2.0 with respect to the critical load. The required thickness t of the tube is:
(A) 2.75 mm
(B) 3.15 mm
(C) 3.89 mm
(D) 4.33 mm
Solution
E 110 GPa
Pcr nP
L 1.6 m
d 38 mm
n 2.0
P 14 kN
Pcr 28.0 kN
t
Solve for required moment of inertia I in
terms of Pcr then find tube thickness
Pcr p2 E I
L2
I
Pcr L2
p2 E
d
I 66025 mm
4
Moment of inertia
I
Solve numerically for min. thickness t:
p 4
[d (d 2 t)4]
64
tmin 4.33 mm
d 4 (d 2 t)4 I
64
p
d 2tmin 29.3 mm inner diameter
A-11.6: A plane truss composed of two steel pipes (E 210 GPa, d 100 mm,
wall thickness 6.5 mm) is subjected to vertical load W at joint B. Joints A and
C are L 7 m apart. The critical value of load W for buckling in the plane of the
truss is nearly:
(A) 138 kN
(B) 146 kN
(C) 153 kN
(D) 164 kN
Solution
E 210 GPa
L7m
d 100 mm
t 6.5 mm
Moment of inertia
p 4
[d (d 2 t)4] 2.097
I
64
Member lengths:
LBA L cos(40 ) 5.362 m
LBC L cos(50 ) 4.500 m
B
d
106 mm4
W
40°
50°
A
C
L
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Page 1164
APPENDIX A FE Exam Review Problems
Statics at joint B to find member forces FBA and FBC:
• Sum horizontal forces at joint B:
FBA cos(40 ) FBC cos(50 )
cos(50 )
cos(40 )
FBA FBC
cos(50 )
0.839
cos(40 )
• Sum vertical forces at joint B:
a
where
W FBA sin(40 ) FBC sin(50 )
cos(50 )
W FBC
sin(40 ) FBC sin(50 )
cos(40 )
1
FBC W b where b cos(50 )
a
sin(40 ) sin(50 )b
cos(40 )
0.766
So member forces in terms of W are:
FBC Wb and FBA FBC a
with ab 0.643
or
FBA W(ab)
Euler buckling loads in BA & BC:
p2 E I
151.118 kN
FBA_cr LBA2
b
so WBA_cr FBA_cr 138 kN
a
FBC_cr p2 E I
214.630 kN
LBC2
lower value controls
so
WBC_cr b FBC_cr 164 kN
A-11.7: A beam is pin-connected to the tops of two identical pipe columns, each
of height h, in a frame. The frame is restrained against sidesway at the top of
column 1. Only buckling of columns 1 and 2 in the plane of the frame is of
interest here. The ratio (a/L) defining the placement of load Qcr, which causes
both columns to buckle simultaneously, is approximately:
(A) 0.25
(B) 0.33
(C) 0.67
(D) 0.75
Qcr
Solution
a
Draw FBD of beam only; use
statics to show that Qcr causes
forces P1 and P2 in columns 1 & 2
respectively:
La
b Qcr
L
P1 a
P2 EI
h
L-a
EI
1
h
2
a
Qcr
L
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APPENDIX A FE Exam Review Problems
1165
Buckling loads for columns 1 & 2:
Pcr1 p2 EI
La
b Qcr
a
L
(0.699 h)2
Pcr2 p2 EI
a
a b Qcr
L
h2
Solve above expressions for Qcr, then solve for required a/L so that columns
buckle at the same time:
p2 EI
L
p2 EI L
a
b
a b
(0.699 h)2 L a
h2 a
Or
L
p2 EI L
p2 EI
b 2 a b0
2 a
a
(0.699 h) L a
h
Or
L
L
0
a
0.6992 (L a)
So
a
0.6992
0.328
L (1 0.6992)
Or
a
0.6992
La
A-11.8: A steel pipe column (E 210 GPa) with length L 4.25 m is
constructed of circular tubing with outside diameter d2 90 mm and inner
diameter d1 64 mm. The pipe column is fixed at the base and pinned at the
top and may buckle in any direction. The Euler buckling load of the column
is most nearly:
(A) 303 kN
(B) 560 kN
(C) 690 kN
(D) 720 kN
Solution
E 210 GPa
L 4.25 mm
d2 90 mm
d1 64 mm
d1
Moment of inertia
I
p
(d24 d14)
64
I 2.397
d2
106 mm4
Effective length of column for fixed-pinned case:
Le 0.699 L 2.971 m
Pcr p2 E I
563 kN
L2e
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APPENDIX A FE Exam Review Problems
A-11.9: An aluminum tube (E 72 GPa) AB of circular cross section has a
pinned support at the base and is pin-connected at the top to a horizontal beam
supporting a load Q 600 kN. The outside diameter of the tube is 200 mm and
the desired factor of safety with respect to Euler buckling is 3.0. The required
thickness t of the tube is most nearly:
(A) 8 mm
(B) 10 mm
(C) 12 mm
(D) 14 mm
Solution
E 72 GPa
SMc 0
L 2.5 m
n 3.0
2.5 Q
1.5
P 1000 kN
P
Find required I based on critical
buckling load
Critical load
Q 600 kN
d 200 mm
Q = 600 kN
C
B
Pcr Pn
Pcr 3000 kN
Pcr 1.5 m
1.0 m
p2 E I
L2
2.5 m
Pcr L2
I 2
p E
d 200 mm
I 26.386
6
4
10 mm
Moment of inertia
p 4
I
[d (d 2 t)4]
64
d
t min 4
B
d4 I
2
64
p
A
tmin 9.73 mm
A-11.10: Two pipe columns are required to have the same Euler buckling load
Pcr. Column 1 has flexural rigidity E I and height L1; column 2 has flexural
rigidity (4/3)E I and height L2. The ratio (L2/L1) at which both columns will
buckle under the same load is approximately:
(A) 0.55
(B) 0.72
(C) 0.81
(D) 1.10
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APPENDIX A FE Exam Review Problems
1167
Solution
Equate Euler buckling load expressions
for the two columns considering their
different properties, base fixity conditions
and lengths:
2
p2 a Eb (2 I)
3
p EI
(0.699 L1)2
L22
2
Pcr
Pcr
E
I
2E/3
2I
L1
L2
Simplify then solve for L2 /L1:
2
L2
4
a
b 0.699 L1
3
L2
4
0.6992 0.807
L1 B 3
A-11.11: Two pipe columns are required to have the same Euler buckling load
Pcr. Column 1 has flexural rigidity E I1 and height L; column 2 has flexural
rigidity (2/3)E I2 and height L. The ratio (I2/I1) at which both columns will
buckle under the same load is approximately:
(A) 0.8
(B) 1.0
(C) 2.2
(D) 3.1
Solution
Equate Euler buckling load expressions
for the two columns considering their
different properties, base fixity conditions
and lengths:
2
p2 a Eb (I2)
3
p E I1
(0.699 L)2
L2
2
Pcr
Pcr
E
I1
2E/3
I2
L
L
Simplify then solve for I2/I1:
L2
E
I2
I1 (0.699 L)2 2
E
3
3
I2
2
3.07
I1 (0.699)2
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© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 1
1
Tension, Compression,
and Shear
Statics Review
Problem 1.2-1 Segments AB and BC of beam
100 ft-lb at joint B
10 ft
ABC are pin connected a small distance to the
right of joint B (see figure). Axial loads act at A
and at mid-span of AB. A concentrated moment
is applied at joint B.
100 lb
50 lb
A
20 ft
(a) Find reactions at supports A, B, and C.
(b) Find internal stress resultants N, V, and
M at x 15 ft.
x
B
C
10 ft
Pin
connection
Solution 1.2-1
(a) APPLY LAWS OF STATICS
Fx 0
Cx 100 lb – 50 lb 50 lb
FBD of BC
©MB 0
Cy 1
(0) 0
10 ft
Entire FBD
©MA 0
By 1
(100 lb-ft) 5 lb
20 ft
©Fy 0
Ay By 5 lb-ft
Ay 5 lb
By 5 lb
Reactions are
Cx 50 lb
Cy 0
(b) INTERNAL STRESS RESULTANTS N, V, AND M AT x 15 ft
Use FBD of segment from A to x 15 ft
©Fx 0
N 100 lb 50 lb 50 lb
©Fy 0
V Ay 5 lb
©M 0
M Ay 15 ft 75 lb-ft
1
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Page 2
CHAPTER 1 Tension, Compression, and Shear
Problem 1.2-2 Segments AB and BCD of beam ABCD are pin connected at x 4 m. The beam is supported by a sliding
support at A and roller supports at C and D (see figure). A triangularly distributed load with peak intensity of 80 N/m acts on
BC. A concentrated moment is applied at joint B.
(a) Find reactions at supports A, C, and D.
(b) Find internal stress resultants N, V, and M at x 5 m.
(c) Repeat parts (a) and (b) for the case of the roller support at C replaced by a linear spring of stiffness ky 200 kN/m.
80 N/m
200 N.m at joint D
A
C
4m
4m
B
C
ky
3m
Pin
connection
x
D
Part (c)
Solution 1.2-2
(a) APPLY LAWS OF STATICS
©Fx 0
Ax 0
FBD of AB
©MB 0
MA 0
Entire FBD
©MC 0
Dy 1
1
2
c200 N # m (80 N/m) 4 m a b 4 m d 75.556 N
3m
2
3
©Fy 0
Cy 1
180 N/m2 4 m Dy 235.556 N
2
MA 0
Ax 0
Reactions are
Cy 236 N
Dy 75.6 N
(b) INTERNAL STRESS RESULTANTS N, V, AND M AT x 5 m
Use FBD of segment from A to x 5 m; ordinate on triangular load at x 5 m is
©Fx 0
©Fy 0
©M 0
3
180 N/m2 60 N/m.
4
Nx Ax 0
1
V
3(80 N/m + 60 N/m) 1 m4 70 N
V 70 N
Upward
2
1
2
1
1
M MA (80 N/m) 1 m a 1 mb (60 N/m) 1 m a 1 mb 36.667 N # m
2
3
2
3
(break trapezoidal load into two triangular loads in moment expression)
M 36.7 N # m
CW
(c) REPLACE ROLLER SUPPORT AT C WITH SPRING SUPPORT
Structure remains statically determinate so all results above in (a) and (b) are unchanged.
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 1.2 Statics Review
3
Problem 1.2-3 Segments AB and BCD of beam ABCD are pin connected at x 10 ft. The beam is supported by a pin
support at A and roller supports at C and D; the roller at D is rotated by 30 from the x axis (see figure). A trapezoidal distributed load on BC varies in intensity from 5 lb/ft at B to 2.5 lb/ft at C. A concentrated moment is applied at joint A and a
40 lb inclined load is applied at mid-span of CD.
(a) Find reactions at supports A, C, and D.
(b) Find the resultant force in the pin connection at B.
(c) Repeat parts (a) and (b) if a rotational spring (kr 50 ft-lb/rad) is added at A and the roller at C is removed.
150 lb-ft at joint A
5 lb/ft
40 lb
2.5 lb/ft
A
A
kr
10 ft
Part (c)
4
3 5 ft D
C 5 ft
B
x
10 ft
Pin
connection
10 ft
30º
Remove roller at C in
part (c)
Solution 1.2-3
(a) STATICS
FBD of AB (cut through beam at pin): ©MB 0
©MD 0
Entire FBD:
Cy Ay 1
(150 lb-ft) 15 lb
10 ft
1 4
1
10 ft
1
2
c 40 lb (5 ft) + (2.5 lb/ft)10 fta 10 ft +
b + (5 lb/ft)10 ft a 10 ft + 10 ftb
10 ft 5
2
3
2
3
150 lb-ft Ay 30 ft d 104.333 lb
©Fy 0
Dy Dy
1
4
11.451 lb
40 lb + (5 lb/ft + 2.5 lb/ft) 10 ft Ay Cy 19.833 lb so Dx 5
2
tan(60)
©Fx 0
Ax 3
40 lb Dx 12.549 lb
5
Ax 12.55 lb, Ay 15 lb, Cy 104.3 lb, Dx 11.45 lb, Dy 19.83 lb
(b) USE FBD OF AB ONLY; MOMENT AT PIN IS ZERO
FBx Ax
FBx 12.55 lb
FBy Ay
FBy 15 lb
ResultantB 2FBx2 + FBy2 19.56 lb
(c) ADD ROTATIONAL SPRING AT A AND REMOVE ROLLER AT C; APPLY EQUATIONS OF STATICAL EQUILIBRIUM
Use FBD of BCD ©MB 0
Dy so
Use entire FBD
1
4
1 1
2
1
c (2.5 lb/ft) 10 fta 10 ft b + (5 lb/ft) 10 fta 10 ft b + 40 lb (15 ft) d 32.333 lb
20 ft 2
3
2
3
5
Dy
18.668 lb
Dx tan(60)
©Fy 0
©Fx 0
1
4
(5 lb/ft + 2.5 lb/ft) 10 ft + (40 lb) Dy 37.167 lb
2
5
3
Ax (40 lb) Dx 42.668 lb
5
Ay © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 1 Tension, Compression, and Shear
Use FBD of AB
©MB 0
MA 150 lb-ft + Ay 10 ft 521 .667 lb-ft
SO REACTIONS ARE
Ax 42.7 lb
Ay 37.2 lb
MA 522 lb-ft
Dx 18.67 lb
Dy 32.3 lb
RESULTANT FORCE IN PIN CONNECTION AT B
FBx Ax
FBy Ay
ResultantB 2FBx2 + FBy2 56.6 lb
Problem 1.2-4 Consider the plane truss with
20 N
a pin support at joint 3 and a vertical roller
support at joint 5 (see figure).
4
5
(a) Find reactions at support joints 3 and 5.
(b) Find axial forces in truss members
11 and 13.
45 N
5
6
10
1
2
2.5 m
1
2
2m
9
13
8
60 N
8
12
11
7
6
7
3
3
2m
4
1m
Solution 1.2-4
(a) STATICS
©Fy 0
R3y 20 N 45 N 25 N
©M3 0
R5x ©Fx 0
R3x R5x + 60 N 40 N
1
(20 N * 2 m) 20 N
2m
(b) MEMBER FORCES IN MEMBERS 11 and 13
Number of unknowns:
m 13
r3
m r 16
Number of equations:
j8
2 j 16
So statically determinate
20 N
4
5
5
6
2.5 m
8
2 m (4) Cut vertically through 4, 11, 12, and 1;
use left FBD; sum moments about joint 2
1
1R F42 so F11 0
F11 V 2.5 m 5x
9
2
2
(2) FV 0 at joint 8 so F12 0
(3) FH 0 at joint 5 so F4 R5x 20 N
13
8
1
6
10
1
60 N
7
12
11
7
TRUSS ANALYSIS
(1) FV 0 at joint 4 so F10 0
45 N
2m
3
3
1m
4
(5) Sum vertical forces at joint 3; F9 R3y
F9 25 N
Section cut for left FBD
(6) Sum vertical forces at joint 7
F13V 45 N – F9 20 N
F13 12 F13V 28.3 N
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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5
SECTION 1.2 Statics Review
Problem 1.2-5 A plane truss has a pin support at A and a roller support at E (see figure).
(a) Find reactions at all supports.
(b) Find the axial force in truss member FE.
A
B
10 ft
1 kips
2 kips
3 kips
C
10 ft
D
10 ft
10 ft
E
15 ft
3 ft
F
G
Section cut for left FBD
Solution 1.2-5
(a) STATICS
©Fx 0
Ax 0
©MA 0
Ey 1
(3 k * 10 ft + 2 k * 20 ft + 1 k * 30 ft) 5 k
20 ft
Ay 3 k + 2 k + 1 k Ey 1 k
©Fy 0
(b) MEMBER FORCE IN MEMBER FE
Number of unknowns:
m 11
r3
m + r 14
Number of equations:
j7
2 j 14
So statically determinate
TRUSS ANALYSIS
(1) Cut vertically through AB, GC, and GF; use left FBD; sum moments about C
FGFx (15 ft) FGFy (20 ft) Ay (20 ft) 20 ft-k
so
Ay (20 ft)
FGF 15 ft
10
222 + 102
20 ft
(2) Sum horizontal forces at joint F
2
FGFx FGF
1.854 k
10
22 + 10
2
and
2
FGFy FGF
FGFx FGF
2
22 + 102
10
22 + 102
2
2
1.818 k
222 + 102
FFEx FGFx 1.818 k
2102 + 32
FFEx 1.898 k
10
FFE 1.898 k
FFE © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 6
CHAPTER 1 Tension, Compression, and Shear
Problem 1.2-6 A plane truss has a pin support at F and a roller support at D (see figure).
(a) Find reactions at
both supports.
(b) Find the axial force
in truss member FE.
6 kN
9 kN
A
3m
B
3 kN
C
3m
D
3m
3m
E
4.5 m
1m
F
G
Section cut for left FBD
Solution 1.2-6
(a) STATICS
©Fx 0
©MF 0
©Fy 0
Fx 0
1
Dy C 3 kN (6 m) + 6 kN (3 m) D 6 kN
6m
Fy 9 kN + 6 kN + 3 kN Dy 12 kN
(b) MEMBER FORCE IN MEMBER FE
Number of unknowns:
m 11
r3
m r 14
Number of equations:
j7
2 j 14
So statically determinate
TRUSS ANALYSIS
(1) Cut vertically through AB, GD, and GF; use left FBD; sum moments about D to get FGF 0
(2) Sum horizontal forces at joint F
FFEx Fx 0
so
FFE 0
y
Problem 1.2-7 A space truss has three-dimensional pin supports at joints
Cy
O, B, and C. Load P is applied at joint A and acts toward point Q.
Coordinates of all joints are given in feet (see figure).
Cz
C(0, 4, 0)
Cx
(a) Find reaction force components Bx, Bz, and Oz.
(b) Find the axial force in truss member AC.
O(0, 0, 0)
Ox
Oz
Joint B
coordinates (ft)
B(2, 0, 0)
(0, 0, 5) A
z
P
Oy
Bz
Bx
x
By
Q(4, −3, 5)
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 7
7
SECTION 1.2 Statics Review
Solution 1.2-7
m3
(a) FIND REACTIONS USING STATICS
r9
m r 3j
rAQ 4
3
P 0 Q
rOA ©M 0
0
0
P5Q
MO rOA * PA + rOC *
eAQ m r 12
j4
3 j 12
So truss is statically determinate
0.8
0.6
ƒ rAQ ƒ
P 0 Q
rAQ
PA P eAQ 0.8 P
0.6 P
P 0 Q
Cx
Bx
4 Cz + 3.0 P
Cy + rOB * By 4.0 P 2 Bz
PC Q
PB Q P 2B 4C Q
z
z
y
x
0
4
P0Q
rOC ©Mx 0 gives Cz so
©My 0 gives
2
0
P0Q
rOB 3
P
4
Bz 2 P
©F 0
RO PA +
Ox
Bx
Cx
Bx + Cx + Ox + 0.8 P
Oy + By + Cy By + Cy + Oy + 0.6 P
≤
PO Q
PB Q
PC Q ±
5P
z
z
z
Oz +
4
METHOD OF JOINTS
Joint O
©Fx 0
Ox 0
Joint B
©Fy 0
By 0
Joint C
©Fx 0
Cx 0
©Fx 0 gives Bx 0.8 P
For entire structure
©Mz 0 gives Oz so
©Fy 0
5
P
4
Oy 0
©Fy 0 Cy 0.6 P By Oy
Cy 0.6 P
3 241
P
20
3 241
0.96
20
(b) FORCE IN MEMBER AC
©Fz 0 at joint C
FAC 3 241 ƒ P ƒ
242 + 52
ƒ CZ ƒ 5
20
FAC tension
Problem 1.2-8 A space truss is restrained at joints O, A, B, and C, as shown in
y
the figure. Load P is applied at joint A and load 2P acts downward at joint C.
(a) Find reaction force components Ax, By, and Bz in terms of load variable P.
(b) Find the axial force in truss member AB in terms of load variable P.
2P
C
Cx
0.6L
0.8 L
A
z
P
Ay
Ax
Ox
Oz
O
L
B
Oy
x
Bz By
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 8
CHAPTER 1 Tension, Compression, and Shear
Solution 1.2-8
(a) FIND REACTIONS USING STATICS
rOA 0
0
P 0.8 L Q
rOB ©M 0
L
0
P0Q
m4
r8
m + r 12
j4
m + r 3j
so truss is statically determinate
rOC 0
0.6 L
P 0 Q
FA Ax
Ay
PPQ
FB 0
By
PB Q
z
3 j 12
Fc Cx
2 P
P 0 Q
FO Ox
Oy
PO Q
z
Resultant moment at O
0.8 Ay L
MO rOA * FA + rOB * FB + rOC * FC 0.8 Ax L Bz L
P B L 0.6 C L Q
y
x
©Mx 0 gives Ay 0
so
©F 0
Resultant force at O
RO FO + FA + FB + FC METHOD OF JOINTS
Ax + Cx + Ox
Ay + By + Oy 2 P
P
Q
Bz + Oz + P
©Fz 0
Joint O
Oz 0
so from
©Fz 0
Bz P
Joint B
©Fy 0
By 0
Joint C
(b) FORCE IN MEMBER AB
©Fx 0
Cx 0
©Fz 0 at joint B
FAB 2(0.8 L)2 + L2
ƒ Bz ƒ
0.8 L
FAB 1.601 P
©My 0
and
ƒ Bz ƒ ƒ P ƒ
Ax Bz
0.8
1.25 P
2(0.8 L)2 + L2
1.601
0.8 L
tension
Problem 1.2-9 A space truss is restrained at joints A, B, and C, as
y
shown in the figure. Load 2P is applied at in the x direction at
joint A, load 3P acts in the z direction at joint B and load P is
applied in the z direction at joint C. Coordinates of all joints are
given in terms of dimension variable L (see figure).
(a) Find reaction force components Ay and Az in terms of load
variable P.
(b) Find the axial force in truss member AB in terms of load
variable P.
By
3P(+z-direction)
B(0, 4L, 0)
Bx
2L
C(0, 2L, 4L)
2L
3L
P
Cy
Cx
4L
z
2P
A(3L, 0, 0)
Az
Ay
O(0, 0, 0)
x
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 1.2 Statics Review
9
Solution 1.2-9
m3
(a) FIND REACTIONS USING STATICS
r6
m + r 3j
rOA ©M 0
3L
0
P 0 Q
rOB 0
4L
P 0 Q
rOC m + r9
j3
3j 9
So truss is statically determinate
0
2L
P4 LQ
FA 2 P
Ay
P A Q
z
FB Bx
By
P3PQ
FC Cx
Cy
PPQ
Resultant moment at O
14 L P 4 Cy L
MO rOA * FA + rOB * FB + rOC * FC 4 Cx L 3 Az L
P3A L 4B L 2C LQ
y
x
x
so
©Mx 0 gives Cy 14
P
4
©F 0
Resultant force at O
RO FA + FB + FC METHOD OF JOINTS
Joint A
©Fz 0
Bx + Cx 2 P
Ay + By + Cy
P A + 4P Q
z
FACz Az 4.0 P
so
so
©Fz 0
FACy 2
F
2.0 P
4 ACz
©Fx 0 FABx 2 P FACx 3.0 P 2 P
©Fy 0 Ay (FABy + FACy) so
8P
+ 4.0 P + 2.0 P
3
gives
FABy Az 4.0 P
FACx 3
F
3.0 P
4 ACz
4
8P
F
4.0 P 3 ABx
3
Ay 4.67 P
(b) FORCE IN MEMBER AB
FAB 2FABx 2 + FABy 2
FAB C
FAB 8.33 P
52 + a
25 P
20 2
b P
3
3
25
8.33
3
compression
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 10
CHAPTER 1 Tension, Compression, and Shear
Problem 1.2-10 A space truss is restrained at joints A, B, and C,
y
as shown in the figure. Load P acts in the z direction at joint B
and in the z direction at joint C. Coordinates of all joints are
given in terms of dimension variable L (see figure). Let P 5 kN
and L 2 m.
P(z direction)
(0, 4L, 2L) B
Bx
(a) Find the reaction force components Az and Bx.
(b) Find the axial force in truss member AB.
4
1
z
P (–z direction)
2L
2L
C (0, 0, 4L)
3 2
4L
2
4
3L
3
A(3L, 0, 0)
Ax
Az
Cx
Cy
O(0, 0, 0)
x
Ay
Solution 1.2-10
(a) FIND REACTIONS USING STATICS
r6
m3
m + r 3j
L2m
rOA 3L
0
P0Q
m + r9
j3
3j 9
so truss is statically determinate
P 5 kN
rOB ©F 0
0
4L
P2LQ
rOC 0
0
P4LQ
FA RO FA + FB + FC Resultant force at O
RESULTANT MOMENT AT A
rAC MA rAB * FB + rAC * FC 3 L
0
P 4L Q
eAC 120 kN 24 Cy
12 Bx + 24 Cx
P
Q
24 Bx 18 Cy
Ax
Ay
PA Q
z
Ax + Bx + Cx
Ay + Cy
P
Q
Az
rAC
ƒ rAC ƒ
0.6
0
P 0.8 Q
FB Bx
0
PPQ
so
rAB FC Cx
Cy
P P Q
©Fz 0 gives
3 L
4L
P 2L Q
MA eAC 19.2 Bx 72.0 kN so Bx Az 0
72
kN 3.75 kN
19.2
(b) FORCE IN MEMBER AB
Method of joints at B
©Fx 0
FABx BX
FAB 229
FABx 6.73 kN
3
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SECTION 1.2 Statics Review
11
Problem 1.2-11 A stepped shaft ABC consisting of two solid, circular segments is subjected to torques T1 and T2 acting in
opposite directions, as shown in the figure. The larger segment of the shaft has a diameter of d1 2.25 in. and a length of
L1 30 in.; the smaller segment has a diameter
T1
d2 1.75 in. and a length L2 20 in. The torques are
T2
d1
T1 21,000 lb-in. and T2 10,000 lb-in.
d2
(a) Find reaction torque TA at support A.
(b) Find the internal torque T(x) at two locations:
x L1/2 and x L1 L2/2. Show these internal
torques on properly drawn free-body diagrams
(FBDs).
x
C
B
A
L1
L2
Solution 1.2-11
(a) APPLY LAWS OF STATICS
©Mx 0
L1 30 in.
L2 20 in.
T1 21000 lb-in.
T2 10000 lb-in.
TA T1 T2 11,000 lb-in.
(b) INTERNAL STRESS RESULTANT T AT TWO LOCATIONS
Cut shaft at midpoint between A and B at x L1/2
(use left FBD)
©Mx 0
TAB TA 11,000 lb-in.
Cut shaft at midpoint between B and C at x L1 L2/2
(use right FBD)
©Mx 0
TBC T2 10,000 lb-in.
Problem 1.2-12 A stepped shaft ABC consisting of two solid, circular segments is subjected to uniformly distributed torque
t1 acting over segment 1 and concentrated torque T2 applied at C, as shown in the figure. Segment 1 of the shaft has a diameter
of d1 57 mm and length of L1 0.75 m; segment 2
t1
has a diameter d2 44 mm and length L2 0.5 m.
T2
Torque intensity t1 3100 N # m/m and
d2
#
T2 1100 N m .
x
(a) Find reaction torque TA at support A.
(b) Find the internal torque T(x) at two locations:
x L1/2 and x L1 L2/2. Show these internal
torques on properly drawn free-body diagrams
(FBDs).
C
B
A
d1
L1
L2
Solution 1.2-12
(a) REACTION TORQUE AT A
Statics
©Mx 0
L1 0.75 m
L2 0.75 m
t1 3100 N # m/m
TA t1 L1 + T2 1225 N # m
T2 1100 N # m
TA 1225 N # m
(b) INTERNAL TORSIONAL MOMENTS AT TWO LOCATIONS
L1
b 62.5 N # m
2
Cut shaft between A and B
(use left FBD)
T1(x) TA t1 x
T1 a
Cut shaft between B and C
(use left FBD)
T2(x) TA t1 L1
T2 a L1 +
L2
b 1100 N # m
2
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 12
CHAPTER 1 Tension, Compression, and Shear
Problem 1.2-13 A plane frame is restrained at joints A and C, as shown
in the figure. Members AB and BC are pin connected at B. A triangularly
distributed lateral load with peak intensity of 90 lb/ft acts on AB.
A concentrated moment is applied at joint C.
500 lb-ft at joint C
Pin
connection
90 lb/ft
(a) Find reactions at supports A and C.
(b) Find internal stress resultants N, V, and M at x 3 ft on column AB.
B
C
9 ft
12 ft
x
A
Solution 1.2-13
(a) STATICS
1
(90 lb/ft) 12 ft 540 lb
2
©FH 0
Ax ©FV 0
Ay + Cy 0
©MFBDBC 0
©MA 0
Cy 500 lb-ft
55.6 lb
9 ft
MA 500 lb-ft +
Ay Cy 55.6 lb
1
2
(90 lb/ft) 12 ft a 12 ft b Cy 9 ft 4320 lb-ft
2
3
(b) INTERNAL STRESS RESULTANTS
N Ay 55.6 lb
V Ax 1 3
a 90 lb/ftb 3 ft 506 lb
2 12
1 3
1
M MA Ax 3 ft a 90 lb/ftb 3 ft a 3 ft b 2734 lb-ft
2 12
3
N
V
M
x
x = 3 ft
A
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
78572_ch01_ptg01_hr_001-116.qxd
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Page 13
13
SECTION 1.2 Statics Review
Problem 1.2-14 A plane frame is restrained at joints A and D, as shown
in the figure. Members AB and BCD are pin connected at B. A triangularly
distributed lateral load with peak intensity of 80 N/m acts on CD.
An inclined concentrated force of 200 N acts at the mid-span of BC.
Pin
connection
C
B
1.5 m
(a) Find reactions at supports A and D.
(b) Find resultant forces in the pins at B and C.
80 N/m
1.5 m
4
200 N
3
4m
4m
A
D
Solution 1.2-14
(a) STATICS
©Fx 0
Ax 1
3
(200 N) + (80 N/m) 4 m 280 N
5
2
©MBRHFB 0 Dy 1 4
1
1
c (200 N) (1.5 m) + (80 N/m) 4 m a 4 mb d
3m 5
2
3
151.1 N 6 use right hand FBD (BCD only)
©Fy 0
Ay Dy +
©MA 0
MA 4
(200 N) 8.89 N
5
4
3
1
2
(200 N) (1.5 m) (200 N) (4 m) Dy 3 m (80 N/m) 4 m a 4 mb 1120 N # m
5
5
2
3
(b) RESULTANT FORCE IN PIN AT B
FBy
LEFT HAND FBD (SEE FIGURE)
FBx Ax 280 N
FBy Ay 8.89 N
B
FBx
RIGHT HAND FBD
1
3
(200 N) + (80 N/m) 4 m 280 N
5
2
4
FBy (200 N) Dy 8.89 N
5
FBx 4m
ResultantB 2FBx2 + FBy2 280 N
A
Left hand FBD
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 14
CHAPTER 1 Tension, Compression, and Shear
Problem 1.2-15 A 200 lb trap door (AB) is supported by a strut (BC)
which is pin connected to the door at B (see figure).
B
y
(a) Find reactions at supports A and C.
(b) Find internal stress resultants N, V, and M on the trap door at
20 in. from A.
30 in.
do
or
t
Stru
4
30 in.
5
26.833 in.
LBC 2
15
LAC Pin or hinge
connection
Tr
ap
200 lb
2
4
3
1
(30 in.) +
LBC 30 in.
5
15
1
3
A
C
x
Solution 1.2-15
(a) STATICS
1
1 3
c200 lb a b a b 30 in. d 60 lb
LAC
2 5
©MA 0
Cy ©Fx 0
Ax Cx 30 lb
©Fy 0
Ay 200 lb Cy 140 lb
Cx 1
Cy 30 lb
2
(resultant of Cx and Cy acts along line of strut)
(b) INTERNAL STRESS RESULTANTS N, V, M (SEE FIGURE)
w
Distributed weight of door in y direction
200 lb
6.667 lb/in.
30 in.
Components of w along and perpendicular to door
wa 4
w 5.333 lb/in.
5
wp 3
w 4 lb/in.
5
M
y
V
N
M wp (20 in.)
N 23.3 lb
do
20 in.
20 in.
3
4
Ax (20 in.) + Ay (20 in.) 33.333 lb-ft
2
5
5
V 20 lb
200 lb
p
4
3
A + Ay 20 lb
5 x
5
Tr
a
V wp (20 in.) or
3
4
N wa (20 in.) Ax Ay 23.333 lb
5
5
M 33.3 lb-ft
4
A
3
x
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 15
SECTION 1.2 Statics Review
Problem 1.2-16 A plane frame is constructed by using a pin
connection between segments ABC and CDE. The frame has pin
supports at A and E and has joint loads at B and D (see figure).
10 kN B
15
3m
Pin connection
C
10 kN
(a) Find reactions at supports A and E.
(b) Find resultant force in the pin at C.
3m
D
3m
E
6m
90 kN·m
A
Solution 1.2-16
(a) STATICS
©MA 0
10 kN (6 m) 10 kN a
so
1
b (6 m) + 90 kN # m + Ey (6 m) Ex 3 m 6 Ey m 3 Ex m + 150 kN # m 30 12 kN # m
12
FCy
6 Ey m 3 Ex m + 150 kN # m 30 12 kN # m 0
(150 kN # m 30 12 kN # m)
35.858 kN
3m
©MCRHFB 0 6 right hand FBD (CDE) - see figure.
or
Ex + 2 Ey 1Ex + Ey2 3 m 90 kN # m
Ex + Ey FCx
C
90 kN # m
30 kN
3m
3m
D
Solving
a
Ex
1 2
b a
b
Ey
1 1
1
a
3m
E
35.858 kN
8.05
b a
b kN
30 kN
21.95
90 kN·m
Ex 8.05 kN
©Fx 0
Ax Ex + 10 kN 10 kN a
©Fy 0
Ay Ey + 10 kN a
(b) RIGHT HAND FBD
1
22
b 10.98 kN
Ey 22 kN
Ax 10.98 kN
Ay 29.1 kN
b 29.07 kN
22
Cx Ex 8.05 kN
Cy Ey 22 kN
1
ResultantC 2Cx2 + Cy2 23.4 kN
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 16
CHAPTER 1 Tension, Compression, and Shear
Problem 1.2-17 A plane frame with pin supports at A and E has a cable attached at C, which runs over a frictionless pulley
at F (see figure). The cable force is known to be 500 lb.
(a) Find reactions at supports A and E.
(b) Find internal stress resultants, N, V, and M at point H.
D
E
0.5 ft
0.5 ft
0.8 ft
C
Cable
F
H
1.2 ft
y
0.6 ft
0.5 ft
B
2.5 ft
G
500 lb
A
x
Solution 1.2-17
(a) STATICS
©Fx 0
Ex 0
©ME 0
Ay ©Fy 0
Ey 500 lb Ay 1750 lb
1
(500 lb * 2.5 ft) 1250 lb
1 ft
(b) USE UPPER (SEE FIGURE BELOW) OR LOWER FBD TO FIND STRESS RESULTANTS N, V, AND M AT H
D
©Fx 0
E
©Fy 0
0.5 ft
N Ey 1750 lb
©MH 0
0.8 ft
500 lb
C
V Ex + 500 lb 500 lb
M 0.6 ft (500 lb) Ex 1.4 ft + Ey 0.5 ft 575 lb-ft
Cable
0.6 ft
H
V
N
M
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 17
SECTION 1.2 Statics Review
Problem 1.2-18 A plane frame with a
pin support at A and roller supports at C
and E has a cable attached at E, which
runs over frictionless pulleys at D and B
(see figure). The cable force is known to
be 400 N. There is a pin connection just
to the left of joint C.
(a) Find reactions at supports A, C,
and E.
(b) Find internal stress resultants N, V,
and M just to the right of joint C.
(c) Find resultant force in the pin
near C.
17
D
Cable is attached at E and
passes over
frictionless pulleys at B and D
4m
4
400 N
3
4
3
C
A
4m
B
E
3m
5m
Pin connection
just left of C
Solution 1.2-18
(a) STATICS
©Fx 0
Ax 4
(400 N) 320 N
5
Ax 320 N
Use left hand FBD (cut through pin just left of C)
©MC 0
Use entire FBD
Ay 1
3
4
cc
(400 N) (400 N) d (3 m) d 240 N
7m
5
5
1
3
cAy (7 m) + a 400 N b (3 m) d 192 N
5m
5
©MC 0
Ey ©Fy 0
Cy Ay Ey 3
(400 N) 192 N
5
(b) N, V, AND M JUST RIGHT OF C; USE RIGHT HAND FBD
5
M
Fcable X
5
24 + 52
2
b 312.348 N
©Fx 0
4
F
249.878 N
5 cableX
Nx FcableX 312 N
©Fy 0
V FcableY Ey 57.9 N
E
©MC 0
M A FcableY + Ey B (5 m) 289 N # m
5m
(c) RESULTANT FORCE IN PIN JUST LEFT OF C; USE LEFT HAND FBD
FCx Ax + a
Cy 192 N
FcableX 400 N a
N
V
Ey 192 N
FcableY FcableY
4
Ay 240 N
4
3
b 400 N 240 N
5
5
Ax 320 N
FCy Ay a
3
4
+ b 400 N 320 N
5
5
ResC 2FCx2 + FCy2 400 N
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.2-19 A 150-lb rigid bar AB, with frictionless rollers at each
C
end, is held in the position shown in the figure by a continuous cable
CAD. The cable is pinned at C and D and runs over a pulley at A.
(a) Find reactions at supports A and B.
(b) Find the force in the cable.
3 ft
Ca
ble
B
y
id
ig
br
0-l
15
ar
b
D
2 ft
30°
A
x
4 ft
Solution 1.2-19
(a) STATICS
W 150 lb
Bx (4) + W a
©MA 0
2 13
7513
b 0 solve, Bx 2
2
Bx so
75 13
64.952
2
©Fx 0
A sin(30) + Bx + T cos(30) + T cos a arctan a
©Fy 0
A cos(30) + T sin(30) + T sin a arctan a
A
a b ±
T
sin(30)
cos(30)
cos(30) + cos a arctan a
7
7
bb 0
2 13
7
bb W
2 13
bb
1
223
≤
7
sin(30) + sin aarctan a
bb
223
a
Bx
b
W
A
57.713
a b a
b lb
T
71.634
SUPPORT REACTIONS
Bx 65
A 57.7
Units lbs
Ax A sin(30) 28.9 lb
Ay A cos(30) 50 lb
3A2x + A2y 57.713
(b) CABLE FORCE IS T (LBS) FROM ABOVE SOLUTION
T 71.6 lb
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 19
19
SECTION 1.2 Statics Review
Problem 1.2-20 A plane frame has a pin support at
A and roller supports at C and E (see figure). Frame
segments ABD and CDEF are joined just left of joint
D by a pin connection.
1.5 kN/m
Pin connection
just left of D
16 kN
(a) Find reactions at supports A, C, and E.
(b) Find the resultant force in the pin just
left of D.
B
6m
D
4m
6m
4m
A
C
3 kN/m
F
4m
E
Solution 1.2-20
(a) STATICS
RIGHT-HAND FBD
©Mpin 0
Ey 1 1
1
c (3 kN/m) 4 m a 4 mb d 1.333 kN
6m 2
3
Ey 1.333 kN
ENTIRE FBD
©MA 0 Cy 1
1
2
cEy 12 m + (16 kN) 4 m + (1.5 kN/m) 6 m (3 m) (3 kN/m) 4 m a 4 m b d 9.833 kN
6m
2
3
Cy 9.83 kN
©Fy 0
Ay Cy Ey + (1.5 kN/m) 6 m 2.167 kN
©Fx 0
Ax 16 kN +
1
(3 kN/m) 4 m 10 kN
2
Ay 2.17 kN
Ax 10 kN
(b) RESULTANT FORCE IN PIN; USE EITHER RIGHT HAND OR LEFT HAND FBD (CUT THROUGH PIN EXPOSING PIN FORCES FDx
AND FDy) THEN SUM FORCES IN x AND y DIRECTIONS FOR EITHER FBD
LHFB:
FDx 16 kN Ax 6 kN
FDy Ay + (1.5 kN/m) 6 m 11.167 kN
ResultantD 3FDx2 + FDy2 12.68 kN
RHFB:
1
FDx (3 kN/m) 4 m 6 kN
2
FDy Cy Ey 11.167 kN
ResultantD 12.68 kN
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 20
CHAPTER 1 Tension, Compression, and Shear
z
Problem 1.2-21 A special vehicle brake is clamped at O, (when the brake force P1 is
applied–see figure). Force P1 50 lb and lies in a plane which is parallel to the xz
plane and is applied at C normal to line BC. Force P2 40 lb and is applied at B in the
y direction.
x
7 in.
(a) Find reactions at support O.
(b) Find internal stress resultants N, V, T, and M at the midpoint of segment OA.
O
A
y
y′
6 in.
B
P2
15°
8 in.
P1
C
15°
x′
Solution 1.2-21
(a) STATICS
P1 50 lb
P2 40 lb
©Fx 0
Ox P1 cos(15) 48.3 lb
©Fy 0
Oy P2 40 lb
©Fz 0
Oz P1 sin(15) 12.94 lb
©Mx 0
MOx P2 6 in. + P1 sin(15) (7 in.) 331 lb-in.
©My 0
MOy P1 sin(15) (8 in. sin(15)) + P1 cos(15) (6 in. + 8 in. cos(15))
MOy 690 lb-in.
©Mz 0
MOz P1 cos(15) (7 in.) 338 lb-in.
(b) INTERNAL STRESS RESULTANTS AT MIDPOINT OF OA
N Oy 40 lb
Vx Ox 48.3 lb
Vz Oz 12.94 lb
V 3V 2x + V 2z 50 lb
T MOy 690 lb-in.
Mx MOx 330.59 lb-in.
Mz MOz 338.07 lb-in.
M 3M 2x + M 2z 473 lb-in.
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 21
SECTION 1.2 Statics Review
Problem 1.2-22 Space frame ABCD is clamped at A, except
it is free to translate in the x-direction. There is also a roller
support at D, which is normal to line CDE. A triangularly
distributed force with peak intensity q0 75 N/m acts along
AB in the positive z direction. Forces Px 60 N and
Pz 45 N are applied at joint C and a concentrated
moment My 120 N # m acts at the mid-span of member BC.
y
21
My
B(0, 2, 0)
C(1.5, 2, 0)
q0
Joint coordinates
in meters
Px
Pz
(a) Find reactions at supports A and D.
(b) Find internal stress resultants N, V, T, and M at the
mid-height of segment AB.
0.75 m
D
E(2.5, 0, −0.5)
A(0, 0, 0)
z
x
Solution 1.2-22
FORCES
Px 60 N
Pz 45 N
Px
60
FC 0 0
N
P P Q P 45 Q
z
VECTOR ALONG MEMBER CD
rEC My 120 N # m
q0 75 N/m
0
RA Ay
PA Q
z
1.5 2.5
1
20
2
J 0 (0.5) K J 0.5 K
冷rEC 冷 2.291
eEC rEC
冷rEC 冷
(a) STATICS (FORCE AND MOMENT EQUILIBRIUM)
0
0
Px
Dx
Ay + 0 + 0 + Dy 0
PA Q
PR Q
PP Q
PD Q
z
z
z
T
©F 0
where
0.436
0.873
P 0.218 Q
resultant of triangular load:
RT 1
q (2 m) 75 N
2 0
Dx
Dy D eEC
PD Q
z
SOLVING ABOVE THREE EQUATIONS:
Dx Px so
©Fy 0
Dy eEC2 D
Dy 120 N
Ay Dy 120 N
©Fz 0
Dz eEC3 D
Dz 30 N
3Dx2 + Dy2 + Dz2 137.477 N
so
D
Px
eEC1
©Fx 0
Az Dz RT Pz
D 137.477 N
Dx 60 N
Az 60 N
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 1 Tension, Compression, and Shear
©MA 0
MAx
Px
0
0
MAy + rAE * D + rAC * 0 + My + rcg * 0 0
PM Q
PP Q
P0Q
PR Q
Az
z
T
rAE 2.5 0
00
m
P 0.5 0 Q
1.5 0
rAC 2 0 m
P 00 Q
D
Dx
Dy
PD Q
z
D
60
120 N
P 30 Q
Px
90
rAC * 0 67.5 J
P P Q P 120 Q
z
ƒ D ƒ 137.477 N
rAE * D 0
2
rcg ± (2 m) ≤
3
0
rcg *
MAx
Px
0
0
70
MAy rAE * D + rAC * 0 + My + rcg * 0
142.5 N # m
PM Q
J
PP Q P 0 Q
P R Q K P 180 Q
Az
z
T
(b) RESULTANTS AT MID-HEIGHT OF AB (SEE FBD IN FIGURE BELOW)
N Ay 120 N
Vx Dx Px 0 N
Vz Az T MAy 142.5 N # m
Mx MAx + Az (1 m) +
60
45 N # m
P 300 Q
0
100
0 0 N#m
PR Q P 0 Q
T
MAx
70
MAy 142.5 N # m
P M Q P 80 Q
Az
1 q0
(2 m)/2 41.25 N V Vz 41.3 N
2 2
1 q0
1
1 m a 1 mb 16.25 N # m
2 2
3
Mz MAz 180 N # m
T
Mresultant 3Mx2 + Mz2 180.732 N # m
Mresultant 180.7 N # m
N
q0 /2
Mz
Vz
Vx
A(0, 0, 0)
z
Mx
x
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 23
23
SECTION 1.2 Statics Review
Problem 1.2-23 Space frame ABC is clamped at A except
it is free to rotate at A about the x and y axes. Cables DC and
EC support the frame at C. Forces Py 50 lb is applied at
mid-span of AB and a concentrated moment Mx 20 in-lb
acts at joint B.
y
D(0, 10, −20)
Joint coordinates
in inches
E(0, 8, 10)
Cable DC
(a) Find reactions at supports A.
(b) Find cable tension forces.
C(10, 4, −4)
Cable EC
A(0, 0, 0)
z
Py
Mx
x
B(10, 0, 0)
Solution 1.2-23
POSITION AND UNIT VECTORS
rAB 10
0
P0Q
rAP 5
0
P0Q
rAC 10
4
P 4 Q
APPLIED FORCE AND MOMENT
Py 50 lb
rCD rCE Mx 20 lb-in.
0 10
10
10 4
6
J 20 (4) K P 16 Q
0 10
10
84
4
J 10 (4) K P 14 Q
eCD eCE rCD
ƒ rCD ƒ
rCE
ƒ rCE ƒ
0.505
0.303
P 0.808 Q
0.566
0.226
P 0.793 Q
STATICS FORCE AND MOMENT EQUILIBRIUM
First sum moment about point A
©MA 0
0
0
Mx
2.0203 TD + 4.0762 TE 20.0
≤
MA 0
+ rAP * Py + 0 + rAC * A TD eCD + TE eCE B ±
10.102 TD + 5.6614 TE
PM Q
P0Q P 0 Q
MAz + 5.0508 TD + 4.5291 TE 250.0
Az
Solve moment equilibrium equations for moments about x and y axes to get cable tension forces
T
2.0203 4.0762 1 20
3.81
a Db a
b a b a
b lb
TE
10.102 5.6614
0
6.79
(b)
Next, solve moment equilibrium equation about z axis now that cable forces are known
MAz (5.0508 TD + 4.5291 TE 250.0) 200 lb-in.
(a)
Finally, use force equilibrium to find reaction forces at point A
©F 0
Ax
0
5.77
Ay Py (TD eCD + TE eCE) 47.31 lb
PA Q
P0Q
P 2.31 Q
z
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 24
CHAPTER 1 Tension, Compression, and Shear
Problem 1.2-24 A soccer goal is subjected
to gravity loads (in the z direction,
w 73 N/m for DG, BG, and BC;
1.22 m
w 29 N/m for all other members;
see figure) and a force F 200 N applied
eccentrically at mid-height of member DG.
Find reactions at supports C, D, and H.
P
G
w = 73 N/m
z
Gravity
2.44 m
x
3
B
Q
4
2.44 m
F = 200 N
D
R
H
3.65 m
S
C
3.65 m
y
w = 29 N/m
Reaction force
Solution 1.2-24
FIND MEMBER LENGTHS
LQS 2 (3.65 m) 7.3 m
LRS 3(2.44 m)2 + (2.44 m 1.22 m)2 2.728 m LPQ LRS
Assume that soccer goal is supported only at points C, H, and D (see reaction force components at each location
in figure)
STATICS
SUM MOMENT ABOUT EACH AXIS AND FORCES IN EACH AXIS DIRECTION F 200 N
©Mx 0 TO FIND REACTION COMPONENT Hy:
Find moments about x due to for component Fy and also for distributed weight of each frame component
MxGP (1.22 m)2
129 N/m2
2
MxBR MxGP
MxRS LRS129 N/m2 a1.22 m +
Hz 1.22 m
b
2
MxDQ (2.44 m)2
2
129 N/m2
MxCS MxDQ
MxQS LQS 129 N/m2 (2.44 m)
MxPQ MxRS
1
4
2.44 m
c Fa
b + 2 MxGP + 2 MxDQ + 2 MxPQ + MxQS d 498.818 N
2.44 m 5
2
Hz 499 N
©My 0 TO FIND REACTION FORCE Dz:
MyGD 2.44 m 173 N/m2 LQS
MyPQ LRS 129 N/m2 LQS
Dz MyGP 1.22 m 129 N/m2 LQS
MyBG LQS 173 N/m2
LQS
2
MyDQ 2.44 m 129 N/m2 LQS
MyQS LQS 129 N/m2
LQS
2
LQS
1
3
2.44 m
cMyGD + MyGP + MyDQ + MyPQ + MyBG + MyQS Hz
Fa
b d 466.208 N
LQS
2
5
2
Dz 466 N
©Mz 0 TO FIND REACTION FORCE Hy:
Hy 1
4
a F LQS b 320 N
3.65 m 5
Hy 320 N
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 25
25
SECTION 1.2 Statics Review
©Fx 0 TO FIND REACTION FORCE Cx:
©Fy 0 TO FIND REACTION FORCE Cy:
©Fz 0 TO FIND REACTION FORCE Cz:
3
F 120 N
5
4
Cy Hy + F 160 N
5
Cx Cy 160 N
Cz Dz Hz + 129 N/m2 12 1.22 m + 2 2.44 m + 2 LRS + LQS2
+ 173 N/m2 12 2.44 m + LQS2 506.318 N
Cz 506 N
Geometry of Back rail
Problem 1.2-25 An elliptical exerciser
D
machine (see figure part a) is composed of
front and back rails. A simplified plane frame
model of the back rail is shown in figure part b.
Analyze the plane frame model to find reaction
forces at supports A, B, and C for the position
and applied loads given in figure part b. Note
that there are axial and moment releases at the
base of member 2 so that member 2 can
lengthen and shorten as the roller support at B
moves along the 30 incline. (These releases
indicate that the internal axial force N and
moment M must be zero at this location.)
20 lb
16 in.
Member no.
y
C
Cx
175 lb
2
34 in.
16 in.
1
10 in.
α
34 in.
Axial
release
Moment
release
B
α
A
Joint no.
Cy
3
α = 11.537°
x
Bx
30°
By
B
Ay
Solution 1.2-25
a arcsina
STATICS
10
b 11.537
50
Analysis pertains to this
position of exerciser only
UFBD (CUT AT AXIAL AND MOMENT RELEASES JUST ABOVE B)
Inclined vertical component of reaction at C 0 (due to axial release)
Sum moments about moment release to get inclined normal reaction at C
C
20 lb (34 in. + 16 in.)
29.412 lb
34 in.
Cx C cos(a) 28.8 lb
Cy C sin(a) 5.88 lb
STATICS
3Cx2 + Cy2 29.412 lb
LFBD (CUT THROUGH AXIAL AND MOMENT RELEASES)
Sum moments to find reaction Ay
Ay 175 lb (16 in.)
57.2 lb
(34 in. + 16 in.) cos(a)
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 1 Tension, Compression, and Shear
STATICS
SUM FORCES FOR ENTIRE FBD TO FIND REACTION AT B
Bx Cx + 175 lb (sin(a)) 20 lb (cos(a)) 44.2 lb
Sum forces in x-direction:
Sum forces in y-direction:
By Ay Cy + 175 lb (cos(a)) + 20 lb (sin(a)) 112.4 lb
Bx 44.2 lb
Resultant reaction force at B:
6 acts leftward
By 112.4 lb
B 3Bx2 + By2 120.8 lb
Problem 1.2-26 A mountain bike is moving along a flat path at constant velocity.
At some instant, the rider (weight 670 N) applies pedal and hand forces, as shown
in the figure part a.
(a) Find reactions forces at the front and rear hubs. (Assume that the bike is pin
supported at the rear hub and roller supported at the front hub).
(b) Find internal stress resultants N, V, and M in the inclined seat post (see figure
part b).
V
15.3°
M
N
N
M
V
Solution 1.2-26
(a) REACTIONS: SUM MOMENTS ABOUT REAR HUB TO FIND VERTICAL REACTION AT FRONT HUB (FIG. 1)
©MB 0
1
[670 (241) 90 (cos(5)) 254 + 200 cos(15) 660 + 2 (45) cos(30) 1021 + 2 (45) sin(30) 752]
1130
VF 335.945 N
VF © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 1.2 Statics Review
27
Sum forces to get force components at rear hub
©Fvert 0
VB 670 90 cos(5 ) + 200 cos(15 ) + 2 (45) cos(30 ) VF 515.525 N
©Fhoriz 0
HB 90 sin(5 ) 200 sin(15 ) 2 (45) sin(30 ) 104.608 N
y
VF 336 N
45 N at 30° to vertical
on each grip
670 N
HB 104.6 N
1021 mm
VB 516 N
254
241 mm
mm
15.3°
Origin at
B (0, 0, 0)
752 mm
90 N at 5°
x
HB
VF
VB
200 N at 15° to vertical
254 mm
660 mm
1130 mm
(b) STRESS RESULTANTS N, V, AND M IN SEAT POST (Fig. 2)
SEAT POST RESULTANTS (FIG. 2)
N 670 cos(15.3) 646.253 N
N 646 N
V 670 sin(15.3) 176.795 N
V 176.8 N
M 670 sin(15.3) 254 44,905.916 N # mm
M 44.9 N # m
V
15.3°
M
N
N
M
V
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 1 Tension, Compression, and Shear
Normal Stress and Strain
P1
Problem 1.3-1 A hollow circular post ABC (see figure) supports a load
P1 1700 lb acting at the top. A second load P2 is uniformly distributed
around the cap plate at B. The diameters and thicknesses of the upper and
lower parts of the post are dAB 1.25 in., tAB 0.5 in., dBC 2.25 in., and
tBC 0.375 in., respectively.
A
tAB
dAB
P2
(a) Calculate the normal stress sAB in the upper part of the post.
(b) If it is desired that the lower part of the post have the same
compressive stress as the upper part, what should be the magnitude
of the load P2?
(c) If P1 remains at 1700 lb and P2 is now set at 2260 lb, what new
thickness of BC will result in the same compressive stress in both
parts?
B
dBC
tBC
C
Solution 1.3-1
PART (a)
PART (b)
P1 1700 lb
dAB 1.25 in.
dBC 2.25 in.
AAB tAB 0.5 in.
tBC 0.375 in.
p [ dAB2 (dAB 2 tAB)2]
4
AAB 1.178 in.2
sAB 1443 psi
P1
sAB AAB
p[dBC2 1dBC 2tBC22]
4
ABC 2.209 in.2
P2 sABABC P1
P2 1488 lbs
CHECK:
;
P1 + P2
1443 psi
ABC
;
PART (c)
P2 2260
ABC P1 + P2
ABC
sAB
dBC 2tBC P1 + P2
2.744
sAB
dBC 4 P 1 + P2
(dBC 2tBC)2 dBC 2 a
b
p
sAB
tBC A
dBC2 2
A
tBC 0.499 in.
dBC 4 P1 + P2
b
a
p
sAB
4 P1 + P2
b
a
p
sAB
2
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 1.3 Normal Stress and Strain
Problem 1.3-2 A force P of 70 N is applied by
a rider to the front hand brake of a bicycle (P is
the resultant of an evenly distributed pressure). As
the hand brake pivots at A, a tension T develops in
the 460-mm long brake cable (Ae 1.075 mm2)
which elongates by d 0.214 mm. Find normal
stress s and strain in the brake cable.
Brake cable, L = 460 mm
29
Hand brake pivot A
37.5 mm A
T
P (Resultant
of distributed
pressure)
50
mm
100
mm
Uniform hand
brake pressure
Solution 1.3-2
P 70 N
Ae 1.075 mm2
L 460 mm
d 0.214 mm
Statics: sum moments about A to get T 2P
s
T
Ae
s 103.2 MPa
d
L
4.65 * 104
E
s
1.4 * 105 MPa
;
;
NOTE: (E for cables is approximately 140 GPa)
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.3-3 A bicycle rider would like to compare the effectiveness of cantilever hand brakes [see figure part (a)] versus
V-brakes [figure part (b)].
(a) Calculate the braking force RB at the wheel rims for each of the bicycle brake systems shown. Assume that all forces act in
the plane of the figure and that cable tension T 45 lbs. Also, what is the average compressive normal stress sc on the
brake pad (A 0.625 in.2)?
(b) For each braking system, what is the stress in the brake cable
T
(assume effective cross-sectional area of 0.00167 in.2)?
(HINT: Because of symmetry, you only need to use the right
half of each figure in your analysis.)
T
D
4 in.
TDC = TDE
45°
T
4 in.
TDC
TDE
TDE
E
C
D
T
TDCv
C
5 in.
4.25 in.
TDCh
B
E
2 in.
RB
A
G
Pivot points
anchored to frame
RB
1 in.
B
F
HA
1 in.
F
1 in.
HA
Pivot points
anchored to frame
A
VA
VA
(b) V-brakes
(a) Cantilever brakes
Solution 1.3-3
T 45 lbs
Apad 0.625 in.2
Acable 0.00167 in.2
(a) CANTILEVER BRAKES—BRAKING FORCE
RB and PAD PRESSURE
STATICS
SUM FORCES AT D TO GET TDCv T / 2
a MA 0
RB(1) TDCh(3) TDCv(1)s
TDCh TDCv
TDCh T / 2
RB 2T
RB 90 lbs
;
so RB 2T versus 4.25T for V-brakes (next)
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 1.3 Normal Stress and Strain
spad RB
Apad
scable spad 144 psi
T
4.25
2.125
2
;
scable 26,946 psi
Acable
31
;
(same for
V-brakes (below))
(b) V-BRAKES—BRAKING FORCE RB AND PAD PRESSURE
a MA 0
RB 4.25T
RB 191.3 lbs
;
RB
Apad
spad 306 psi
;
spad Problem 1.3-4 A circular aluminum tube of length L 420 mm is loaded in compression by forces P (see figure). The
hollow segment of length L/3 has outside and inside diameters of 60 mm and 35 mm, respectively. The solid segment of
length 2L/3 has diameter of 60 mm. A strain gage is placed on the outside of the hollow segment of the bar to measure
normal strains in the longitudinal direction.
(a) If the measured strain in the hollow segment is h 470 * 106, what is the strain s in the solid part? (Hint: The
strain in the solid segment is equal to that in the hollow
L/3
segment multiplied by the ratio of the area of the hollow to
Strain gage
that of the solid segment).
P
P
(b) What is the overall shortening d of the bar?
(c) If the compressive stress in the bar cannot exceed 48 MPa,
L = 420 mm
what is the maximum permissible value of load P?
Solution 1.3-4
L 420 mm
d2 60 mm
d1 35 mm
h 470 11062
sa 48 MPa
PART (a)
As p 2
d 2.827 * 103m2
4 2
h Ah
3.101 * 104
As h
Ah p
a d 2 d1 2 b 1.865 * 103m2
4 2
PART (b)
d h
L
2L
+ s a b 0.1526 mm
3
3
h
L
0.066 mm
3
s a
2L
b 0.087 mm
3
PART (c)
Pmaxh sa Ah 89.535 kN
Pmaxs sa As 135.717 kN
6 lesser value controls
Pmax Pmaxh 89.5 kN
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.3-5 The cross section of a concrete corner column that is loaded
y
uniformly in compression is shown in the figure. A circular pipe chase cut-out
of 10 in. in diameter runs the height of the column (see figure).
24 in.
(a) Determine the average compressive stress sc in the concrete if the load is
equal to 3500 kips.
(b) Determine the coordinates xc and yc of the point where the resultant load
must act in order to produce uniform normal stress in the column.
20 in.
Cirular pipe
chase cutout
20 in.
16 in.
5 in.
8 in.
x
8 in.5 in.
Solution 1.3-5
P 3500 kips
1
p
A (24 + 20) (20 + 16 + 8) a 82 b 202 102
2
4
A 1425.46 in.2
(a) AVERAGE COMPRESSIVE STRESS
sc P
A
sc 2.46 ksi
(b) CENTROID
(24 + 20)2
xc (24 + 20)
1
8
A 202 B (24 + 10) 82 a b
2
2
3
A
p
a 102 b (8 + 5)
4
A
xc 19.56 in.
(24 + 20)2
yc (24 + 20)
1
8
A 202 B (24 + 10) 82 a b
2
2
3
A
p
a 102 b (8 + 5)
4
A
yc 19.56 in.
ˆ xc and yc are the same as expected due to symmetry about a diagonal
Problem 1.3-6 A car weighing 130 kN when fully loaded is
pulled slowly up a steep inclined track by a steel cable (see figure).
The cable has an effective cross-sectional area of 490 mm2, and the
angle of the incline is 30.
(a) Calculate the tensile stress st in the cable.
(b) If the allowable stress in the cable is 150 MPa, what is the
maximum acceptable angle of the incline for a fully loaded
car?
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SECTION 1.3 Normal Stress and Strain
33
Solution 1.3-6
W 130 kN
a 30
A 490 mm2
sa 150 MPa
PART (a)
st W sin(a)
132.7 MPa
A
PART (b)
amax arcsina
sa A
b 34.4
W
Problem 1.3-7 Two steel wires support a
moveable overhead camera weighing W 28 lb
(see figure part a) used for close-up viewing of
field action at sporting events. At some instant,
wire 1 is at an angle a 22 to the horizontal
and wire 2 is at an angle b 40. Wires 1 and 2
have diameters of 30 and 35 mils, respectively.
(Wire diameters are often expressed in mils;
one mil equals 0.001 in.)
(a) Determine the tensile stresses s1 and s2 in
the two wires.
(b) If the stresses in wires 1 and 2 must be
the same, what is the required diameter
of wire 1?
(c) Now, to stabilize the camera for windy
outdoor conditions, a third wire is added
(see figure part b). Assume the three wires
meet at a common point (coordinates (0, 0,
0) above the camera at the instant shown
in figure part b). Wire 1 is attached to a
support at coordinates (75 ft, 48 ft, 70 ft).
Wire 2 is supported at (70 ft, 55 ft,
80 ft). Wire 3 is supported at (10 ft,
85 ft, 75 ft). Assume that all three wires
have a diameter of 30 mils. Find the tensile
stresses in wires 1 to 3.
T2
T1
b
a
W
(a)
Plan view of camera suspension system
All coordinates in feet
y
(−70, 55, 80)
(75, 48, 70)
Wire 2
Wire 1
Camera
x
Wire 3
(−10, −85, 75)
(b)
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.3-7
d1 30 A 103 B in.
d2 35 A 103 B in.
A1 A2 W 28 lb
a 22
p 2
d1 7.069 * 104 in.2
4
p 2
d 9.621 * 104 in.2
4 2
b 40
(a) FIND NORMAL STRESS IN WIRES
T2 W
29.403 lb
cos(b)
sin(a) + sin(b)
cos(a)
T1 T2
cos(b)
24.293 lb
cos(a)
s2 T2
30.6 ksi
A2
s1 T1
34.4 ksi
A1
(b) FIND NEW d1 S.T. NORMAL STRESSES IN WIRES IS THE SAME
A1 new T1
7.949 * 104 in.2
s2
s1new T1
30.6 ksi
p
d1new2
4
d1 new 4
A
3.18 * 102 in.
A p 1 new
or
31.8 mils
(c) Now, to stabilize the camera for windy outdoor conditions, a third wire is added (see figure b); assume the 3 wires
meet at a common point (coordinates (0, 0, 0) above the camera at the instant shown in figure b); wire 1 is attached
to a support at coordinates (75', 48', 70'); wire 2 is supported at (70', 55', 80'); and wire 3 is supported at
(10', 85', 75'); assume that all three wires have diameter of 30 mils. Find tensile stresses in wires 1 to 3.
d 30 A 103 B in.
A
p 2
d 7.069 * 104 in.2
4
75
Position vectors from camera r1 48 ft
P 70 Q
to each support
L1 ƒ r1 ƒ 113.265
L2 ƒ r2 ƒ 119.687
Unit vectors along wires 1 to 3
T1 F1 e1
T2 F2 e2
r2 e1 r1
ƒ r1 ƒ
T3 F3 e3
70
55 ft
P 80 Q
10
85 ft
P 75 Q
L3 ƒ r3 ƒ 113.798
0.662
0.424
P 0.618 Q
i
r3 1
0
P0Q
e2 j
r2
ƒ r2 ƒ
0
1
P0Q
0.585
0.46
P 0.668 Q
k
0
0
P1Q
0
W 28 0 lb
P1Q
e3 r3
ƒ r3 ƒ
0.088
0.747
P 0.659 Q
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 1.3 Normal Stress and Strain
Equilibrium of forces
819
T
e1
FT
1
829
T
35
T1 + T2 + T3 W
e2
839
T
e3
13.854
W 13.277 lb
P 16.028 Q
s1 0.585
0.46
0.668
0.662
T 0.424
P 0.618
F1
19.6 ksi
A
0.088
0.747
0.659 Q
s2 s1 19.6 ksi
F2
18.78 ksi
A
s3 s2 18.78 ksi
F3
22.7 ksi
A
s3 22.7 ksi
Problem 1.3-8 A long retaining wall is braced by wood shores
set at an angle of 30° and supported by concrete thrust blocks, as
shown in the first part of the figure. The shores are evenly spaced,
3 m apart.
For analysis purposes, the wall and shores are idealized as
shown in the second part of the figure. Note that the base of the
wall and both ends of the shores are assumed to be pinned. The
pressure of the soil against the wall is assumed to be triangularly distributed, and the resultant force acting on a 3-meter
length of the wall is F 190 kN.
If each shore has a 150 mm 150 mm square cross section,
what is the compressive stress sc in the shores?
Solution 1.3-8 Retaining wall braced by wood shores
F 190 kN
A area of one shore
A (150 mm)(150 mm)
22,500 mm2
0.0225 m2
FREE-BODY DIAGRAM OF WALL AND SHORE
SUMMATION OF MOMENTS ABOUT POINT A
MA 0 哵哴
F(1.5 m) CV (4.0 m) CH (0.5 m) 0
or
(190 kN)(1.5 m) C(sin 30°)(4.0 m)
C(cos 30°)(0.5 m) 0
⬖ C 117.14 kN
C compressive force in wood shore
CH horizontal component of C
CV vertical component of C
CH C cos 30°
CV C sin 30°
COMPRESSIVE STRESS IN THE SHORES
sc C
117.14 kN
A
0.0225 m2
5.21 MPa
;
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.3-9 A pickup truck tailgate supports a crate
(WC 150 lb), as shown in the figure. The tailgate weighs
WT 60 lb and is supported by two cables (only one is
shown in the figure). Each cable has an effective crosssectional area Ae 0.017 in2.
(a) Find the tensile force T and normal stress s in each
cable.
(b) If each cable elongates d 0.01 in. due to the weight
of both the crate and the tailgate, what is the average
strain in the cable?
WC = 150 lb
dc = 18 in.
Ca
ble
H = 12 in.
Crate
Tail gate
Truck
dT = 14 in.
WT = 60 lb
L = 16 in.
Solution 1.3-9
(a) T 2 Tv 2 + T h2 T 184.4 lb
Wc 150 lb
Ae 0.017 in.2
WT 60
d 0.01
;
scable T
Ae
scable 10.8 ksi
;
(b) cable d
Lc
cable 5
104
;
dc 18
dT 14
H 12
L 16
L c 2 L2 + H2
a Mhinge 0
Lc 20
2TvL Wcdc WT dT
Tv Wc dc + WT dT
2L
Th L
T
H v
Tv 110.625 lb
T h 147.5
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SECTION 1.3 Normal Stress and Strain
37
Problem 1.3-10 Solve the preceding problem if the
mass of the tail gate is MT 27 kg and that of the
crate is MC 68 kg. Use dimensions H 305 mm,
L 406 mm, dC 460 mm, and dT 350 mm. The
cable cross-sectional area is Ae 11.0 mm2.
(a) Find the tensile force T and normal stress s in
each cable.
(b) If each cable elongates d 0.25 mm due to the
weight of both the crate and the tailgate, what
is the average strain in the cable?
MC = 68 kg
dc = 460 mm
Ca
H = 305 mm
ble
Crate
Tail gate
Truck
dT = 350 mm
MT = 27 kg
L = 406 mm
Solution 1.3-10
(a) T 2 T2v + T2h
Mc 68
g 9.81 m/s2
MT 27 kg
Wc Mcg
WT 264.87
;
scable T
Ae
scable 74.5 MPa
(b) cable d
Lc
cable 4.92
WT MTg
Wc 667.08
T 819 N
104
;
;
N kg # m/s2
Ae 11.0 mm2
d 0.25
dc 460
dT 350
H 305
L 406
L c 2 L2 + H2
a Mhinge 0
Tv Lc 507.8 mm
2TvL Wc dc WT dT
Wc dc + WT dT
2L
Th L
T
H v
Tv 492.071 N
Th 655.019 N
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.3-11 An L-shaped reinforced concrete slab
12 ft * 12 ft (but with a 6 ft * 6 ft cutout) and thickness
t 9.0 in. is lifted by three cables attached at O, B and D, as
shown in the figure. The cables are combined at point Q,
which is 7.0 ft above the top of the slab and directly above the
center of mass at C. Each cable has an effective crosssectional area of Ae 0.12 in.2.
(a) Find the tensile force T1(i 1, 2, 3) in each cable due
to the weight W of the concrete slab (ignore weight of
cables).
(b) Find the average stress si in each cable. (see Table I-1
in Appendix I for the weight density of reinforced
concrete.)
(c) Add cable AQ so that OQA is one continuous cable, with
each segment having force T1, which is connected to
cables BQ and DQ at point Q. Repeat parts (a) and (b).
(Hint: There are now three force equilibrium equations
and one constraint equation, T1 T4)
F
Coordinates of D in feet
Q
(5, 5, 7)
T3
1
T1
D (5, 12, 0)
1
A (0, 12, 0)
T2
7
5
5
z
O
(0, 0, 0)
C (5, 5, 0) 5 7
y
7
x
6 ft
6 ft
W
6 ft
B (12, 0, 0)
Concrete slab g = 150 lb/ft3
Thickness t, c, g at (5 ft, 5 ft, 0)
Solution 1.3-11
CABLE LENGTHS (FT)
L1 252 + 52 + 72 L1 9.95
L2 252 + 72 + 72
L2 11.091
L3 272 + 72
L3 9.899
(a) SOLUTION FOR CABLE FORCES USING STATICS (THREE EQUATIONS, THREE UNKNOWNS); UNITS lb, ft
rOQ eOQ 5
5
P7Q
rOQ
ƒ rOQ ƒ
rBQ 0.503
0.503
P 0.704 Q
W 150 1122 622
STATICS
7
5
P 7 Q
rDQ eBQ rBQ
ƒ rBQ ƒ
9
12,150 lbs
12
0
7
P 7 Q
0.631
0.451
P 0.631 Q
©F 0 T1 eOQ + T2 eBQ + T3 eDQ eDQ rDQ
ƒ rDQ ƒ
0
0.707
P 0.707 Q
0
0.50252 T1 0.63117 T2
0 0.50252 T1 + 0.45083 T2 0.70711 T3
P W Q P 0.70353 T + 0.63117 T + 0.70711 T 12,150 Q
1
2
3
or in matrix form; solve simultaneous equations to get cable tension forces
T1
eOQ1, 1
T2 eOQ2, 1
PT Q Pe
3
OQ3, 1
eBQ1, 1
eBQ2, 1
eBQ3, 1
eDQ1, 1 1 0
5877
eDQ2, 1
0 4679 lb
P 7159 Q
eDQ3, 1 Q P W Q
(b) AVERAGE NORMAL STRESS IN EACH CABLE
i1Á3
si Ti
Ae
s
48975
38992 psi
P 59658 Q
T
5877
4679 lb
P 7159 Q
Ae 0.12 in.2
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SECTION 1.3 Normal Stress and Strain
39
(c) ADD CONTINUOUS CABLE OQA
5
rOQ 5
P7Q
eOQ STATICS
rOQ
ƒ rOQ ƒ
5
rAQ 7
P 7 Q
0.503
0.503
P 0.704 Q
7
rBQ 5
P 7 Q
eAQ rAQ
ƒ rAQ ƒ
0
rDQ 7
P 7 Q
0.451
0631
P 0.631 Q
eBQ 0.451
eAQ 0.631
ƒ rAQ ƒ P 0.631 Q
rAQ
rBQ
ƒ rBQ ƒ
0.631
0.451
P 0.631 Q
Solve simultaneous equations to get cable tension forces
T1
eOQ1, 1
T2
eOQ2, 1
± ≤ ±
T3
eOQ3, 1
T4
1
eBQ1, 1
eBQ2, 1
eBQ3, 1
0
Normal stresses in cables
i1Á4
Ti
si Ae
eDQ1, 1
eDQ2, 1
eDQ3, 1
0
eAQ1, 1 1 0
4278
4278
eAQ2, 1
0
6461
6461
≤
lbs
T
lb
W
3341
eAQ3, 1
P Q P
Q
P 3341 Q
0
4278
4278
1
<for case of T1 ⴝ T4
35650
53842
≤ psi
s ±
27842
35650
Problem 1.3-12
A round bar ACB of length 2L (see figure) rotates about an axis
through the midpoint C with constant angular speed v (radians per second). The
material of the bar has weight density g.
(a) Derive a formula for the tensile stress sx in the bar as a function of the
distance x from the midpoint C.
(b) What is the maximum tensile stress smax?
Solution 1.3-12
Rotating Bar
Consider an element of mass dM at distance j from the
midpoint C. The variable j ranges from x to L.
g
dM g A dj
dF Inertia force (centrifugal force) of element of mass dM
g
dF (dM)(j 2) g A 2jdj
v angular speed (rad/s)
A cross-sectional area
g weight density
g
mass density
g
gA 2 2
(L x2)
2g
LD
Lx g
(a) TENSILE STRESS IN BAR AT DISTANCE x
We wish to find the axial force Fx in the bar at
Section D, distance x from the midpoint C.
The force Fx equals the inertia force of the part
of the rotating bar from D to B.
(b) MAXIMUM TENSILE STRESS
Fx B
sx dF Lg
A 2jdj g 2 2
Fx
(L x2)
A
2g
x 0 smax g 2L2
2g
;
;
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.3-13 Two gondolas on a ski lift are locked in
A
the position shown in the figure while repairs are being made
elsewhere. The distance between support towers is L 100 ft.
The length of each cable segment under gondola weights
WB 450 lb and WC 650 lb are DAB 12 ft, DBC 70 ft,
and DCD 20 ft. The cable sag at B is B 3.9 ft and that at
C( C) is 7.1 ft. The effective cross-sectional area of the cable
is Ae 0.12 in2.
D
u1
DB
u2
B
DC
u3
C
WB
(a) Find the tension force in each cable segment; neglect
the mass of the cable.
(b) Find the average stress (s ) in each cable segment.
WC
Support
tower
L = 100 ft
Solution 1.3-13
WB 450
A
Wc 650 lb
D
u1
DB
u2
B
¢ B 3.9 ft
DC
u3
C
¢ C 7.1 ft
L 100 ft
WB
DAB 12 ft
WC
Support
tower
DBC 70 ft
DCD 20 ft
L = 100 ft
DAB DBC DCD 102 ft
CONSTRAINT EQUATIONS
Ae 0.12 in.2
DAB cos(u1) + DBC cos (u2) + DCD cos(u3) L
COMPUTE INITIAL VALUES OF THETA ANGLES (RADIANS)
DAB sin(u1) + DBC sin (u2) DCD sin(u3)
u1 arcsin a
SOLVE SIMULTANEOUS EQUATIONS NUMERICALLY FOR TENSION
¢B
b
DAB
u1 0.331
FORCE IN EACH CABLE SEGMENT
¢ C¢ B
b
u2 arcsin a
DBC
u2 0.046
¢C
b
u3 arcsin a
DCD
u3 0.363
TAB 1620 lb
TCB 1536 lb TCD 1640 lb
;
CHECK EQUILIBRIUM AT B AND C
(a) STATICS AT B AND C
TAB cos(u1) + TBC cos (u2) 0
TAB sin(u1) TBC sin(u2) WB
TAB sin(u1) TBC sin(u2) 450
TBC sin(u2) + TCD sin(u3) 650
(b) COMPUTE STRESSES IN CABLE SEGMENTS
sAB TAB
Ae
sBC TBC
Ae
TBC cos(u2) + TCD cos (u3) 0
sAB 13.5 ksi
sBC 12.8 ksi
TBC sin(u2) + TCD sin (u3) WC
sCD 13.67 ksi
;
sCD TCD
Ae
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41
SECTION 1.3 Normal Stress and Strain
z
Problem 1.3-14 A crane boom of mass 450 kg with its
center of mass at C is stabilized by two cables AQ and BQ
(Ae 304 mm2 for each cable) as shown in the figure. A
load P 20 kN is supported at point D. The crane boom lies
in the y–z plane.
P
Q
y
oo
m
C
1
B
2
eb
2
Cr
an
(a) Find the tension forces in each cable: TAQ and
TBQ (kN). Neglect the mass of the cables, but include
the mass of the boom in addition to load P.
(b) Find the average stress (s) in each cable.
D
2m
2m
55°
TBQ
O
5m
TAQ
5m
x
5m
A
3m
Solution 1.3-14
Data
Mboom 450 kg
g 9.81 m/s2
Wboom Mboom g
2TAQZ (3000) Wboom(5000) + P(9000)
TAQZ Wboom(5000) + P(9000)
2(3000)
P 20 kN
TAQ 222 + 22 + 12
TAQz
2
Ae 304 mm2
TAQ 50.5 kN TBQ
Wboom 4415 N
(a) Symmetry: TAQ TBQ
(b) s TAQ
Ae
;
s 166.2 MPa
;
a Mx 0
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CHAPTER 1 Tension, Compression, and Shear
Mechanical Properties of Materials
Problem 1.4-1 Imagine that a long steel wire hangs vertically from a high-altitude balloon.
(a) What is the greatest length (feet) it can have without yielding if the steel yields at 40 ksi?
(b) If the same wire hangs from a ship at sea, what is the greatest length? (Obtain the weight densities of steel and sea
water from Appendix I, Table I-1.)
Solution 1.4-1
Hanging wire of length L
W total weight of steel wire
gS weight density of steel
490 lb/ft3
gw weight density of sea water
63.8 lb/ft3
A cross-sectional area of wire
smax 40 ksi (yield strength)
F tensile force at top of wire
F A gS gW B AL smax Lmax (a) WIRE HANGING IN AIR
W gS AL
W
gS L
A
Lmax smax 40,000 psi
A 144 in.2/ft2 B
gS
490 lb/ft3
F
A gS gW B L
A
smax
gSgW
40,000 psi
(490 63.8) lb/ft3
13,500 ft
smax 11,800 ft
(b) WIRE HANGING IN SEA WATER
A 144 in.2/ft2 B
;
;
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SECTION 1.4 Mechanical Properties of Materials
43
Problem 1.4-2 Steel riser pipe hangs from a drill rig located offshore
in deep water (see figure).
(a) What is the greatest length (meters) it can have without
breaking if the pipe is suspended in air and the ultimate strength
(or breaking strength) is 550 MPa?
(b) If the same riser pipe hangs from a drill rig at sea, what is the
greatest length? (Obtain the weight densities of steel and sea
water from Appendix I, Table I-1. Neglect the effect of buoyant
foam casings on the pipe).
2
Riser
3
BOP
Drill pipe
Solution 1.4-2
(a) PIPE SUSPENDED IN AIR
(b) PIPE SUSPENDED IN SEA WATER
sU 550 MPa
gw 10 kN/m3
gs 77 kN/m3
Force at top of pipe:
W gs AL
Stress at top of pipe:
Lmax sU
7143 m
gs
smax F A gs gw B A L
smax A gs gw B L
F
A
Set max stress equal to ultimate and then solve for
Lmax
Lmax sU
A gs gw B
8209 m
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.4-3 Three different materials, designated A, B, and C,
are tested in tension using test specimens having diameters of 0.505
in. and gage lengths of 2.0 in. (see figure). At failure, the distances
between the gage marks are found to be 2.13, 2.48, and 2.78 in.,
respectively. Also, at the failure cross sections the diameters are
found to be 0.484, 0.398, and 0.253 in., respectively.
Determine the percent elongation and percent reduction in area
of each specimen, and then, using your own judgment, classify each
material as brittle or ductile.
Solution 1.4-3 Tensile tests of three materials
where L1 is in inches.
A0 A1
(100)
A0
A1
a1 b(100)
A0
Percent reduction in area
Percent elongation L1
L1 L0
(100) a 1b 100
L0
L0
L0 2.0 in.
Percent elongation a
L1
1b (100)
2.0
A1
d1 2
a b d0 0.505 in.
A0
d0
Percent reduction in area
c1 a
d1 2
b d(100)
0.505
(Eq. 2)
d0 initial diameter
d1 final diameter
(Eq. 1)
Material
L1
(in.)
d1
(in.)
% Elongation
(Eq. 1)
% Reduction
(Eq. 2)
Brittle or
Ductile?
A
2.13
0.484
6.5%
8.1%
Brittle
B
2.48
0.398
24.0%
37.9%
Ductile
C
2.78
0.253
39.0%
74.9%
Ductile
where d1 is in inches.
Problem 1.4-4 The strength-to-weight ratio of a structural material is defined as its load-carrying capacity divided by its
weight. For materials in tension, we may use a characteristic tensile stress (as obtained from a stress-strain curve) as a measure of
strength. For instance, either the yield stress or the ultimate stress could be used, depending upon the particular application. Thus,
the strength-to-weight ratio RS/W for a material in tension is defined as
RS/W s
g
in which s is the characteristic stress and g is the weight density. Note that the ratio has units of length.
Using the ultimate stress sU as the strength parameter, calculate the strength-to-weight ratio (in units of meters) for
each of the following materials: aluminum alloy 6061-T6, Douglas fir (in bending), nylon, structural steel ASTM-A572, and
a titanium alloy. (Obtain the material properties from Appendix I, Tables I-1 and I-3. When a range of values is given in a
table, use the average value.)
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45
SECTION 1.4 Mechanical Properties of Materials
Solution 1.4-4 Strength-to-weight ratio
The ultimate stress sU for each material is obtained
from Appendix I, Tables I-3, and the weight density
g is obtained from Table I-1.
The strength-to-weight ratio (meters) is
RS/W sU ( MPa)
g( kN/m3)
(103)
Values of sU, g, and RS/W are listed in the table.
sU
(MPa)
g
(kN/m3)
Aluminum alloy
6061-T6
310
26.0
11.9
103
Douglas fir
65
5.1
12.7
103
Nylon
60
9.8
6.1
103
Structural steel
ASTM-A572
500
77.0
6.5
103
Titanium alloy
1050
44.0
23.9
103
RS/W
(m)
Titanium has a high strength-to-weight ratio, which is why
it is used in space vehicles and high-performance airplanes.
Aluminum is higher than steel, which makes it desirable for
commercial aircraft. Some woods are also higher than steel,
and nylon is about the same as steel.
Problem 1.4-5 A symmetrical framework consisting of three pinconnected bars is loaded by a force P (see figure). The angle between
the inclined bars and the horizontal is a 52. The axial strain in the
middle bar is measured as 0.036.
Determine the tensile stress in the outer bars if they are constructed
of a copper alloy having the following stress-strain relationship:
s
18,000
1 + 300
A
B
C
a
0 … … 0.03 (s ksi)
D
P
Solution 1.4-5
DATA
BD 0.036
a 52 LBD 1
6 assume unit length to facilitate
numerical calculations below
Strain in CE
CE L3 L2
L2
L2 LBD
sin1a2
LBC LBD
tan1a2
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CHAPTER 1 Tension, Compression, and Shear
Increased length of CE (see figure)
L3 3LBC 2 + 1LBD + BD LBD22 1
B tan152 22
+ 1.073296
Compute strain in CE then substitute strain value into stress-strain relationship to find tensile stress in outer bars:
CE L3 L2
0.023
L2
s
18000 CE
1 + 300 CE
s 52.3 ksi
Problem 1.4-6 A specimen of a methacrylate plastic is tested in tension at room temperature (see figure), producing the stress-strain data
listed in the accompanying table.
Plot the stress-strain curve and determine the proportional limit,
modulus of elasticity (i.e., the slope of the initial part of the stress-strain
curve), and yield stress at 0.2% offset. Is the material ductile or brittle?
STRESS-STRAIN DATA FOR PROBLEM 1.4-6
Stress (MPa)
Strain
8.0
17.5
25.6
31.1
39.8
44.0
48.2
53.9
58.1
62.0
62.1
0.0032
0.0073
0.0111
0.0129
0.0163
0.0184
0.0209
0.0260
0.0331
0.0429
Fracture
Solution 1.4-6 Tensile test of a plastic
Using the stress-strain data given in the problem
statement, plot the stress-strain curve:
sPL proportional limit
sPL ⬇ 47 MPa
;
Modulus of elasticity (slope) ⬇ 2.4 GPa
;
sY yield stress at 0.2% offset
sY ⬇ 53 MPa
;
Material is brittle, because the strain after the proportional
;
limit is exceeded is relatively small.
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SECTION 1.4 Mechanical Properties of Materials
47
Problem 1.4-7 The data shown in the accompanying table were
obtained from a tensile test of high-strength steel. The test specimen
had a diameter of 0.505 in. and a gage length of 2.00 in. (see figure for
Prob. 1.4-3). At fracture, the elongation between the gage marks was 0.12 in.
and the minimum diameter was 0.42 in.
Plot the conventional stress-strain curve for the steel and determine the
proportional limit, modulus of elasticity (i.e., the slope of the initial part of
the stress-strain curve), yield stress at 0.1% offset, ultimate stress, percent
elongation in 2.00 in., and percent reduction in area.
TENSILE-TEST DATA FOR PROBLEM 1.4-7
Load (lb)
Elongation (in.)
1,000
2,000
6,000
10,000
12,000
12,900
13,400
13,600
13,800
14,000
14,400
15,200
16,800
18,400
20,000
22,400
22,600
0.0002
0.0006
0.0019
0.0033
0.0039
0.0043
0.0047
0.0054
0.0063
0.0090
0.0102
0.0130
0.0230
0.0336
0.0507
0.1108
Fracture
Solution 1.4-7 Tensile test of high-strength steel
d0 0.505 in.
A0 pd02
4
L0 2.00 in.
0.200 in.2
CONVENTIONAL STRESS AND STRAIN
s
P
A0
d
L0
Load P
(lb)
Elongation d
(in.)
Stress s
(psi)
1,000
2,000
6,000
10,000
12,000
12,900
13,400
13,600
13,800
14,000
14,400
15,200
16,800
18,400
20,000
22,400
22,600
0.0002
0.0006
0.0019
0.0033
0.0039
0.0043
0.0047
0.0054
0.0063
0.0090
0.0102
0.0130
0.0230
0.0336
0.0507
0.1108
Fracture
5,000
10,000
30,000
50,000
60,000
64,500
67,000
68,000
69,000
70,000
72,000
76,000
84,000
92,000
100,000
112,000
113,000
Strain 0.00010
0.00030
0.00100
0.00165
0.00195
0.00215
0.00235
0.00270
0.00315
0.00450
0.00510
0.00650
0.01150
0.01680
0.02535
0.05540
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CHAPTER 1 Tension, Compression, and Shear
STRESS-STRAIN DIAGRAM
RESULTS
Proportional limit ⬇ 65,000 psi
;
150,000
Stress
(psi)
Yield stress at 0.1% offset ⬇ 69,000 psi
100,000
Ultimate stress (maximum stress)
Modulus of elasticity (slope) ⬇ 30
⬇ 113,000 psi
106 psi
;
;
;
Percent elongation in 2.00 in.
50,000
0
0.0200
0.0400
Strain
0.0600
L1 L0
(100)
L0
0.12 in.
(100) 6%
2.00 in.
Percent reduction in area
ENLARGEMENT OF PART OF THE STRESS-STRAIN CURVE
Stress
(psi)
0.200 in.2 p4 (0.42 in.) 2
0.200 in.2
(100)
;
sYP ≈ 69,000 psi
(0.1% offset)
sPL ≈ 65,000 psi
sPL
60,000
A0 A1
(100)
A0
31%
sYP
70,000
;
0.1% pffset
50,000 psi
0.00165
≈ 30 × 106 psi
Slope ≈
50,000
0
0.0020
0.0040
Strain
Elasticity, Plasticity, and Creep
Problem 1.5-1 A bar made of structural steel having the stressstrain diagram shown in the figure has a length of 48 in. The yield
stress of the steel is 42 ksi and the slope of the initial linear part of
the stress-strain curve (modulus of elasticity) is 30 103 ksi. The
bar is loaded axially until it elongates 0.20 in., and then the load is
removed.
How does the final length of the bar compare with its original
length? (Hint: Use the concepts illustrated in Fig. 1-36b.)
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SECTION 1.5 Elasticity, Plasticity, and Creep
49
Solution 1.5-1 Steel bar in tension
ELASTIC RECOVERY E
E sB
42 ksi
0.00140
Slope
30 * 103 ksi
RESIDUAL STRAIN R
R B E 0.00417 0.00140
0.00277
PERMANENT SET
L 48 in.
Yield stress sY 42 ksi
Slope 30
103 ksi
d 0.20 in.
RL (0.00277)(48 in.)
0.13 in.
Final length of bar is 0.13 in. greater than its original
;
length.
STRESS AND STRAIN AT POINT B
sB sY 42 ksi
B d
0.20 in.
0.00417
L
48 in.
Problem 1.5-2 A bar of length 2.0 m is made of a structural steel
having the stress-strain diagram shown in the figure. The yield stress
of the steel is 250 MPa and the slope of the initial linear part of the
stress-strain curve (modulus of elasticity) is 200 GPa. The bar is
loaded axially until it elongates 6.5 mm, and then the load is
removed.
How does the final length of the bar compare with its original
length? (Hint: Use the concepts illustrated in Fig. 1-36b.)
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.5-2 Steel bar in tension
L 2.0 m 2000 mm
Yield stress sY 250 MPa
Slope 200 GPa
d 6.5 mm
ELASTIC RECOVERY E
E sB
250 MPa
0.00125
Slope
200 GPa
RESIDUAL STRAIN R
R B E 0.00325 0.00125
0.00200
Permanent set RL (0.00200)(2000 mm)
4.0 mm
STRESS AND STRAIN AT POINT B
sB sY 250 MPa
B d
6.5 mm
0.00325
L
2000 mm
Final length of bar is 4.0 mm greater than its original
;
length.
Problem 1.5-3 An aluminum bar has length L 6 ft and diameter d 1.375 in.
The stress-strain curve for the aluminum is shown in Fig. 1-31 of Section 1.4.
The initial straight-line part of the curve has a slope (modulus of elasticity) of 10.6 * 106 psi. The bar is loaded by tensile
forces P 44.6 k and then unloaded.
(a) What is the permanent set of the bar?
(b) If the bar is reloaded, what is the proportional limit? (Hint: Use the concepts illustrated in Figs. 1-36b and 1-37.)
Solution 1.5-3
DATA
P 44.6 kip
d 1.375 in.
L 6 ft
E 10.6 11062 psi
NORMAL STRESS IN BAR
P
sB 30036 psi
p 2
d
4
from curve, say that B 0.025
ELASTIC RECOVERY unloading parallel to initial straight line
E sB
2.834 * 103
E
s (ksi)
40
30
20
10
0
0
0.05 0.10 0.15 0.20 0.25
eR
e
FIG 1-31 Typical stress-strain diagram
for an aluminum alloy
RESIDUAL STRAIN
R B E 0.022
(a) PERMANENT SET
R L 1.596 in.
(b) PROPORTIONAL LIMIT WHEN RELOADED IS sB 30 ksi
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51
SECTION 1.5 Elasticity, Plasticity, and Creep
Problem 1.5-4 A circular bar of magnesium alloy is 750 mm long. The stress-strain diagram for the material is shown in
the figure. The bar is loaded in tension to an elongation of 6.0 mm, and then the load is removed.
(a) What is the permanent set of the bar?
(b) If the bar is reloaded, what is the proportional limit? (Hint: Use the concepts illustrated in Figs. 1-36b and 1-37.)
Solution 1.5-4
numerical data
70
L 750 mm
d
a
Et 12 d 1 + b
d 6 mm
Et 102 41000 MPa
63
56
49
sSI (e) 42
35
(or 41 GPa 7 magnesium alloy)
28
d
B 8 * 103
L
E 0.0023
21
sB 65.6 MPa 6 from curve
(see figure)
14
7
6 elastic recovery (see figure)
3
R B E 5.7 * 10
0
0
6 residual strain
e
(a) PERMANENT SET
dpset R L 4.275
0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01
70
dpset 4.28 mm
63
56
(b) PROPORTIONAL LIMIT WHEN RELOADED
49
sB 65.6 MPa
sSI (e)
42
35
28
21
14
7
0
0
0.002
0.004
0.006
0.008
0.01
e
Residual strain
= 0.0057
Elastic recovery
= 0.008 – 0.0057 = 0.0023
Problem 1.5-5 A wire of length L 4 ft and diameter d 0.125 in. is stretched by tensile forces P 600 lb. The wire is
made of a copper alloy having a stress-strain relationship that may be described mathematically by the following equation:
s
18,000P
0 … P … 0.03 (s ksi)
1 + 300P
in which P is nondimensional and s has units of kips per square inch (ksi).
(a) Construct a stress-strain diagram for the material.
(b) Determine the elongation of the wire due to the forces P.
(c) If the forces are removed, what is the permanent set of the bar?
(d) If the forces are applied again, what is the proportional limit?
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.5-5 Wire stretched by forces P
L 4 ft 48 in.
d 0.125 in.
ALTERNATIVE FORM OF THE STRESS-STRAIN RELATIONSHIP
P 600 lb
Solve Eq. (1) for in terms of s:
COPPER ALLOY
18,000
0 … … 0.03 (s ksi) (Eq. 1)
s
1 + 300
(a) STRESS-STRAIN DIAGRAM (From Eq. 1)
s
0 … s … 54 ksi (s ksi)
18,000300s
(Eq. 2)
This equation may also be used when plotting the stressstrain diagram.
(b) ELONGATION d OF THE WIRE
P
600 lb
48,900 psi 48.9 ksi
p
A
(0.125 in.)2
4
From Eq. (2) or from the stress-strain diagram:
s
0.0147
d L (0.0147)(48 in.) 0.71 in.
;
STRESS AND STRAIN AT POINT B (see diagram)
sB 48.9 ksi
B 0.0147
ELASTIC RECOVERY E
E sB
48.9 ksi
0.00272
Slope
18,000 ksi
INITIAL SLOPE OF STRESS-STRAIN CURVE
RESIDUAL STRAIN R
Take the derivative of s with respect to :
R B E 0.0147 0.0027 0.0120
(1 + 300)(18,000) (18,000)(300)s
ds
d
(1 + 300)2
(c) Permanent set RL (0.0120)(48 in.)
At 0,
0.58 in.
;
(d) Proportional limit when reloaded sB
18,000
(1 + 300)2
sB 49 ksi
;
ds
18,00 ksi
d
⬖ Initial slope 18,000 ksi
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SECTION 1.6 Linear Elasticity, Hooke’s Law, and Poisson’s Ratio
53
Linear Elasticity, Hooke’s Law, and Poisson’s Ratio
When solving the problems for Section 1.5, assume that the material behaves
linearly elastically.
Problem 1.6-1 A high-strength steel bar used in a large crane has diameter
d 2.00 in. (see figure). The steel has modulus of elasticity E 29 106 psi
and Poisson’s ratio v 0.29. Because of clearance requirements, the diameter
of the bar is limited to 2.001 in. when it is compressed by axial forces.
What is the largest compressive load Pmax that is permitted?
Solution 1.6-1 Steel bar in compression
STEEL BAR
d 2.00 in.
E 29
AXIAL STRESS
Maximum d 0.001 in.
6
10 psi
v 0.29
LATERAL STRAIN
0.001 in.
¢d
0.0005
¿ d
2.00 in.
AXIAL STRAIN
s E (29
0.0005
¿
0.001724
v
0.29
106 psi)(0.001724)
50.00 ksi (compression)
Assume that the yield stress for the high-strength steel is
greater than 50 ksi. Therefore, Hooke’s law is valid.
MAXIMUM COMPRESSIVE LOAD
p
Pmax sA (50.00 ksi)a b(2.00 in.)2
4
157 k
;
(shortening)
Problem 1.6-2 A round bar of 10 mm diameter is made of aluminum
alloy 7075-T6 (see figure). When the bar is stretched by axial forces P, its
diameter decreases by 0.016 mm.
Find the magnitude of the load P. (Obtain the material properties
from Appendix I.)
Solution 1.6-2
Aluminum bar in tension
d 10 mm
d 0.016 mm
AXIAL STRESS
s E (72 GPa)(0.004848)
(Decrease in diameter)
349.1 MPa (Tension)
7075-T6
From Table I-2: E 72 GPa
v 0.33
From Table I-3: Yield stress sY 480 MPa
LATERAL STRAIN
¿ 0.016 mm
¢d
0.0016
d
10 mm
Because s
sY, Hooke’s law is valid.
LOAD P (TENSILE FORCE)
p
P sA (349.1 MPa)a b(10 mm)2
4
27.4 kN
;
AXIAL STRAIN
¿
0.0016
v
0.33
0.004848 (Elongation)
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.6-3 A polyethylene bar having diameter d1 4.0 in. is placed inside a
Steel
tube
steel tube having inner diameter d2 4.01 in. (see figure). The polyethylene bar is
then compressed by an axial force P.
At what value of the force P will the space between the nylon bar and the steel
tube be closed? (For nylon, assume E 400 ksi and v 0.4.)
d1 d2
Polyethylene
bar
Solution 1.6-3
NORMAL STRAIN
NUMERICAL DATA
d1 4 in.
d2 4.01 in.
v 0.4
p
A1 d1 2
4
d1 0.01 in.
p
A2 d2 2
4
E 200 ksi
A1 12.566 in.2
103
1 6.25
v
AXIAL STRESS
s1 1.25 ksi
COMPRESSION FORCE
LATERAL STRAIN
0.01
p 4
p
s1 E 1
A2 12.629 in.2
¢d1
p d1
1 P EA11
p 2.5
3
10
P 15.71 kips
Strain gage
Problem 1.6-4 A circular aluminum tube of length L 600 mm
is loaded in compression by forces P (see figure). The outside and
inside diameters are d2 75 mm and d1= 63 mm, respectively. A
strain gage is placed on the outside of the tube to measure normal
strains in the longitudinal direction. Assume that E 73 GPa and
Poisson’s ration v 0.33.
(a) If the compressive stress in the tube is 57 MPa, what is the
load P?
(b) If the measured strain is 781 106, what is the shortening
d of the tube? What is the percent change in its cross-sectional
area? What is the volume change of the tube?
(c) If the tube has a constant outer diameter of d2 75 mm
along is entire length L but now has increased inner
diameter d3 with a normal stress of 70 MPa over the middle
third, while the rest of the tube remains at normal stress of
57 MPa, what is the diameter d3?
;
P
d2
d1
d2
d1
P
L
(a)
P
d3
L/3
L/3
P
L/3
(b)
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SECTION 1.6 Linear Elasticity, Hooke’s Law, and Poisson’s Ratio
55
Solution 1.6-4
(a) GIVEN STRESS, FIND FORCE P IN BAR FIGURE (A)
(b) GIVEN STRAIN, FIND CHANGE IN LENGTH IN BAR FIGURE (A) AND ALSO
VOLUME CHANGE
s 57 MPa
E 73 GPa
v 0.33
L 600 mm
d2 75 mm
d1 63 mm
p
A A d22 d12 B 1301 mm2
4
P s A 74.1 kN
781 A 106 B
d2 d1
6 mm
2
d L 0.469 mm shortening
Vol1 L (A) 7.804 * 105 mm3
t
lat v 2.577 * 104
¢d2 lat d2 0.019 mm
¢t lat t 1.546 * 103 mm
¢d1 lat d1 0.016 mm
p
c A d2 + ¢d2 B 2 A d1 + ¢d1 B 2 d
4
Af Af 1301.29 mm2
Af A
A
1301.29 1300.62
0.052%
1300.62
V1f (L + d) A Af B 7.802 * 105 mm3
¢V1 V1f Vol 1 207.482 mm3
¢V1 207 mm3 change
(c) IF THE TUBE HAS CONSTANT OUTER DIAMETER OF d2 75 mm ALONG ITS ENTIRE LENGTH L BUT NOW HAS INCREASED INNER
DIAMETER d3 OVER THE MIDDLE THIRD WITH NORMAL STRESS OF 70 MPa, WHILE THE REST OF THE BAR REMAINS AT NORMAL
STRESS OF 57 MPa, WHAT IS THE DIAMETER d3?
sM3 70 MPa P 74.135 kN AM3 d22 d32 4
A
p M3
SO
d3 A
P
1059.076 mm2
sM3
d22 4
A 65.4 mm
p M3
d2 75 mm
tM3 d2 d3
2
d1 63 mm
4.802 mm
d3 65.4 mm
Problem 1.6-5 A bar of monel metal as in the figure (length L 9 in., diameter d 0.225 in.) is loaded axially by a tensile
force P. If the bar elongates by 0.0195 in., what is the decrease in diameter d? What is the magnitude of the load P? Use the
data in Table I-2, Appendix I.
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.6-5
NUMERICAL DATA
DECREASE IN DIAMETER
E 25,000 ksi
d pd
n 0.32
¢d 1.56 * 104 in.
L 9 in.
INITIAL CROSS SECTIONAL AREA
d 0.0195 in.
Ai d 0.225 in.
2.167
p 2
d Ai 0.04 in.2
4
MAGNITUDE OF LOAD P
NORMAL STRAIN
d
L
;
P EAi
3
10
P 2.15 kips
;
LATERAL STRAIN
p n
p 6.933
104
Problem 1.6-6 A tensile test is peformed on a brass specimen
10 mm in diameter using a gage length of 50 mm (see figure).
When the tensile load P reaches a value of 20 kN, the distance
between the gage marks has increased by 0.122 mm.
(a) What is the modulus of elasticity E of the brass?
(b) If the diameter decreases by 0.00830 mm, what is Poisson’s ratio?
Solution 1.6-6 Brass specimen in tension
d 10 mm Gage length L 50 mm
P 20 kN d 0.122 mm
d 0.00830 mm
AXIAL STRESS
P
20 k
s 254.6 MPa
p
A
(10 mm)2
4
Assume s is below the proportional limit so that
Hooke’s law is valid.
(a) MODULUS OF ELASTICITY
E
s
254.6 MPa
104 GPa
0.002440
;
(b) POISSON’S RATIO
v
d d vd
v
¢d
0.00830 mm
0.34
d
(0.002440)(10 mm)
;
AXIAL STRAIN
0.122 mm
d
0.002440
L
50 mm
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57
SECTION 1.6 Linear Elasticity, Hooke’s Law, and Poisson’s Ratio
Problem 1.6-7 A hollow, brass circular pipe ABC (see figure) supports a load
P1 26.5 kips acting at the top. A second load P2 22.0 kips is uniformly
distributed around the cap plate at B. The diameters and thicknesses of the upper
and lower parts of the pipe are dAB 1.25 in., tAB 0.5 in., dBC 2.25 in., and
tAB 0.375 in., respectively. The modulus of elasticity is 14,000 ksi. When both
loads are fully applied, the wall thickness of pipe BC increases by 200 106 in.
P1
A
dAB
(a) Find the increase in the inner diameter of pipe segment BC.
(b) Find Poisson’s ratio for the brass.
(c) Find the increase in the wall thickness of pipe segment AB and the increase
in the inner diameter of AB.
tAB
P2
B
Cap plate
dBC
tBC
C
Solution 1.6-7
NUMERICAL DATA
P1 26.5 k
P2 22 k
dAB 1.25 in.
tAB 0.5 in.
dBC 2.25 in.
tBC 0.375 in.
E 14000 ksi
tBC 200 106
(c) INCREASE IN THE WALL THICKNESS OF PIPE SEGMENT AB
AND THE INCREASE IN THE INNER DIAMETER OF AB
AAB p
cd 2 1dAB 2tAB22 d
4 AB
AB P1
EAAB
AB 1.607
pAB nbrassAB
tAB pABtAB
(a) INCREASE IN THE INNER DIAMETER OF PIPE
SEGMENT BC
¢tBC
pBC 5.333
pBC tBC
4
103
pAB 5.464
104
¢tAB 2.73 * 104 in.
;
dABinner pAB(dAB 2tAB)
¢dABinner 1.366 * 104 in.
10
dBCinner pBC(dBC 2tBC)
¢ dBCinner 8 * 104 in.
;
(b) POISSON’S RATIO FOR THE BRASS
ABC p
c dBC 2 1dBC 2tBC22 d
4
ABC 2.209 in.2
BC 1P1 + P22
brass 1EABC2
pBC
BC
BC 1.568
103
brass 0.34
(agrees with App. I (Table I-2))
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.6-8 Three round, copper alloy bars having the
Bar 1
same length L but different shapes are shown in the figure.
The first bar has a diameter d over is entire length, the
second has a diameter d over one-fifth of is length, and
the third has a diameter d over one-fifteenth of its length.
Elsewhere, the second and third bars have diameter 2d. All
three bars are subjected to the same axial load P.
Use the following numerical date: P 1400 kN,
L 5 m, d 80 mm, E 110 GPa, and 0.33.
Bar 2
d
Bar 3
2d
L
2d
d
L
5
d
L
15
(a) Find the change in length of each bar.
(b) Find the change in volume of each bar.
P
P
P
Solution 1.6-8
P 1400 kN
L5m
d 80 mm
E 110 GPa
Ad 0.33
p 2
d 5026.5 mm2
4
A2d p
12 d22 20,106.2 mm2
4
(a) FIND CHANGE IN LENGTH OF EACH BAR
Appendix I, Table I-3: copper alloys can
have yield stress in range 55–760 MPa so
assume this is below proportional limit so that
¢L1 1 L 12.66 mm
Lf1 L + ¢L1 5012.66 mm Hooke’s Law applies
2a
P
P
2a 2.532 * 103 2b 6.33 * 104
6.33 * 104
BAR #2
E Ad
E A2d
4
BAR #1
1 P
2.532 * 103
E Ad
¢L2a 2a
L
2.532 mm
5
¢L2 ¢L2a + ¢L2b 5.06 mm
s1 E 1 279 MPa
¢L2b 2b a
4L
b 2.532 mm
5
Lf 2 L + ¢L2 5005.06 mm
¢L2
0.4
¢L1
BAR #3
¢L2a 2a
L
0.844 mm
15
¢L3 ¢L2a + ¢L2b 3.8 mm
¢L2b 2b a
14 L
b 2.954 mm
15
Lf 3 L + ¢L3 5003.08 mm
¢L3
0.3
¢L1
(b) FIND CHANGE IN VOLUME OF EACH BAR
Use lateral strain (p) in each segment to find change in diameter d, then find change in cross sectional area, then
volume
BAR #1
p
p1 v 1 8.356 * 104 ¢d1 p1 d 0.067 mm A1 1d + ¢d122 5018.152 mm2
4
¢Vol
1
¢Vol1 A1 Lf1 Ad L 21548 mm3
8.574 * 104
Ad L
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SECTION 1.7 Shear Stress and Strain
59
BAR #2
p2a p1 p2b v 2b 2.089 * 104
¢d2b p2b 12 d2 0.33 mm A2a A1
p1
4
A2b 2.089 * 104
p
12 d + ¢d2b22 20097.794 mm2
4
L
4L
2.532 mm ¢L2b 2b a
b 2.532 mm
5
5
L
4L
4L
L
¢Vol2 cA1 a + ¢L2a b + A2b a
+ ¢L2b b d cA2d a
b + Ad a b d
5
5
5
5
¢L2a 2a
¢Vol2
1.002
¢Vol1
21601 mm3
BAR #3
L
14 L
0.844 mm
¢L2b 2b a
b 2.954 mm
15
15
L
14 L
14 L
L
+ ¢L2a b + A2b a
+ ¢L2b b d cA2d a
b + Ad a b d
¢Vol3 cA1 a
15
15
15
15
¢L2a 2a
21610 mm3
¢Vol3
1.003
¢Vol2
¢Vol1 21548 mm3
¢Vol2 21601 mm3
¢Vol3 21610 mm3
Shear Stress and Strain
Problem 1.7-1 An angle bracket having thickness t 0.75 in. is attached to the flange of a column by two 5/8-inch diameter
bolts (see figure). A uniformly distributed load from a floor joist acts on the top face of the bracket with a pressure
p 275 psi. The top face of the bracket has length L 8 in. and width b 3.0 in.
Determine the average bearing pressure sb between the angle bracket and the bolts and the average shear stress taver in the
bolts. (Disregard friction between the bracket and the column.)
Distributed pressure on angle bracket
P
b
Floor slab
L
Floor joist
Angle bracket
Angle bracket
t
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.7-1
NUMERICAL DATA
t 0.75 in.
L 8 in.
b 3. in.
p
275
ksi
1000
Ab dt
d
5
in.
8
Ab 0.469 in.2
BEARING STRESS
sb F
2Ab
sb 7.04 ksi
;
BEARING FORCE
F pbL
SHEAR STRESS
F 6.6 k
SHEAR AND BEARING AREAS
AS p 2
d
4
tave F
2AS
tave 10.76 ksi
;
AS 0.307 in.2
Roof structure
Problem 1.7-2
Truss members supporting a roof are connected to a 26-mm-thick gusset plate by a 22-mm diameter
pin as shown in the figure and photo. The two end plates on
the truss members are each 14 mm thick.
Truss
member
(a) If the load P 80 kN, what is the largest bearing
stress acting on the pin?
(b) If the ultimate shear stress for the pin is 190 MPa, what
force Pult is required to cause the pin to fail in shear?
P
End
plates
(Disregard friction between the plates.)
P
Pin
t = 14 mm
Gusset
plate
26 mm
Solution 1.7-2
NUMERICAL DATA
(b) ULTIMATE FORCE IN SHEAR
tep 14 mm
Cross sectional area of pin
tgp 26 mm
Ap P 80 kN
dp 22 mm
4
Ap 380.133 mm2
tult 190 MPa
(a) BEARING STRESS ON PIN
sb p d2p
Pult 2tultAp
Pult 144.4 kN
;
P
gusset plate is thinner than
dptgp
(2 tep) so gusset plate controls
sb 5 139.9 MPa
;
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SECTION 1.7 Shear Stress and Strain
61
Problem 1.7-3 The upper deck of a football stadium is supported by braces each of which transfers a load P 160 kips to
the base of a column [see figure part (a)]. A cap plate at the bottom of the brace distributes the load P to four flange plates
(tf 1 in.) through a pin (dp 2 in.) to two gusset plates (tg 1.5 in.) [see figure parts (b) and (c)].
Determine the following quantities.
(a) The average shear stress taver in the pin.
(b) The average bearing stress between the flange plates and the pin (sbf), and also between the gusset
plates and the pin (sbg).
(Disregard friction between the plates.)
Cap plate
Flange plate
(tf = 1 in.)
Pin (dp = 2 in.)
Gusset plate
(tg = 1.5 in.)
(b) Detail at bottom of brace
P
P = 160 k
Cap plate
(a) Stadium brace
Pin (dp = 2 in.)
P
Flange plate
(tf = 1 in.)
Gusset plate
(tg = 1.5 in.)
P/2
P/2
(c) Section through bottom of brace
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.7-3
(b) BEARING STRESS ON PIN FROM FLANGE PLATE
NUMERICAL DATA
P 160 kips
dp 2 in.
tg 1.5 in.
P
4
sbf dp tf
tf 1 in.
s bf 20 ksi
;
(a) SHEAR STRESS ON PIN
t
V
a
p d2p
4
t
b
t 12.73 ksi
BEARING STRESS ON PIN FROM GUSSET PLATE
P
4
a
p d2p
4
P
2
sbg dp tg
b
sbg 26.7 ksi
;
;
Problem 1.7-4 The inclined ladder AB supports a house painter (85 kg) at C and the self weight (q 40 N/m) of the
ladder itself. Each ladder rail (tr 4 mm) is supported by a shoe (ts 5mm) which is attached to the ladder rail by a bolt of
diameter dp 8 mm.
(a) Find support reactions at A and B
(b) Find the resultant force in the shoe bolt at A.
(c) Find maximum average shear (t) and bearing (sb) stresses in
the shoe bolt at A.
B
C
H = 7.5 m
N/
m
Typical rung
Shoe bolt at A
q=
Ladder rail (tr = 4 mm)
36
tr
Shoe bolt (dp = 8 mm)
Ladder shoe (ts = 5 mm)
A
ts
Ay
2
a = 1.8 m
Ay
2
Section at base
b = 0.7 m
Assume no slip at A.
Solution 1.7-4
NUMERICAL DATA
tr 4 mm ts 5 mm dp 8 mm
a 1.8 m
P 85 19.812 P 833.85 N
b 0.7 m H 7.5 m q 40 N/m
(a) SUPPORT REACTIONS
L 31a + b22 + H2
L 7.906 m
LAC a
L LAC 5.692
a + b
LCB a
L LCB 2.214
a + b
LAC + LCB 7.906
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63
SECTION 1.7 Shear Stress and Strain
SUM MOMENTS ABOUT A
Bx Pa + qL a
a + b
b
2
H
Ay P + q L
Bx 252.829 N (left) and Ax Bx (Ax acts to right)
Ay 1150.078 N
(b) RESULTANT FORCE IN SHOE BOLT AT A
Aresultant 1177.54 N
Bx 252.8 N
Ax Bx
Ax Bx
Ay 1150.1 N
Aresultant 3A2x + A2y
Aresultant 1178 N
(c) MAXIMUM SHEAR AND BEARING STRESSES IN SHOE BOLT AT A
dp 8 mm
Shear area:
p
As d2p
4
ts 5 mm
As 50.256 mm2
tr 4 mm
Shear stress:
Aresultant
2
t
2 As
t 5.86 MPa
Aresultant
Bearing area:
Ab 2 dp ts
Ab 80 mm2
Bearing stress:
Problem 1.7-5 The force in the brake cable of the V-brake
system shown in the figure is T 45 lb. The pivot pin at A has
diameter dp 0.25 in. and length Lp 5/8 in.
Use dimensions show in the figure. Neglect the weight of the
brake system.
sbshore 2
Ab
sbshoe 7.36 MPa
T
(a) Find the average shear stress taver in the pivot pin where it
is anchored to the bicycle frame at B.
(b) Find the average bearing stress sb,aver in the pivot pin over
segment AB.
Lower end of front brake cable
D
T
T
3.25 in.
Brake pads
C
HC
1.0 in.
HB
HE
B
A
HF
VF
Pivot pins
anchored to
frame (dP)
VB
LP
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.7-5
(a) FIND THE AVE SHEAR STRESS tave IN THE PIVOT PIN WHERE
IT IS ANCHORED TO THE BICYCLE FRAME AT B:
NUMERICAL DATA
L
dp 0.25 in.
BC 1 in.
5
in.
8
CD 3.25 in.
AS T 45 lb
EQUILIBRIUM - FIND HORIZONTAL FORCES
AT B AND C [VERTICAL REACTION VB 0]
T(BC + CD)
HC BC
a MB 0
HC 191.25 lb
a FH 0
HB T HC
HB 146.25 lb
tave pdp 2
4
ƒ HB ƒ
AS
As 0.049 in.2
tave 2979 psi
(b) FIND THE AVE BEARING STRESS sb,ave IN THE PIVOT PIN
OVER SEGMENT AB.
Ab dpL
sb,ave Ab 0.156 in.2
ƒ HB ƒ
Ab
sb,ave 936 psi
Problem 1.7-6 A steel plate of dimensions 2.5
1.5 0.08 m
and weighing 23.1kN is hoisted by steel cables with lengths
L1 3.2 m and L2 3.9 m that are each attached to the plate by
a clevis and pin (see figure). The pins through the clevises are
18 mm in diameter and are located 2.0 m apart. The orientation
angles are measured to be u 94.4° and a 54.9°.
For these conditions, first determine the cable forces T1 and
T2, then find the average shear stress taver in both pin 1 and pin 2,
and then the average bearing stress sb between the steel plate
and each pin. Ignore the mass of the cables.
;
;
P
a=
Clevis
L1
and pin 1
0.6
m
b1
u
b2
L2
a
2.0
Clevis
and pin 2
m
Center of mass
of plate
b=
1.0
m
Steel plate
(2.5 × 1.5 × 0.08 m)
Solution 1.7-6
SOLUTION APPROACH
NUMERICAL DATA
L1 3.2 m
u 94.4a
L2 3.9 m
a 54.9a
p
b rad
180
p
b rad
180
STEP (1) d 2a2 + b2
d 1.166 m
a
STEP (2) u1 arctan a b
b
u1
180
30.964
p
STEP (3)-Law of cosines
a 0.6 m
b1m
W 77.0(2.5
1.5
H 2d2 + L21 2dL1cos(u + u1)
W 23.1 kN
0.08)
H 3.99 m
(77 wt density of steel, kN/m )
3
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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65
SECTION 1.7 Shear Stress and Strain
STEP (4) b 1 arccosa
b1
L22 + H2 d2
b
2L1H
L22 + H 2 d 2
b
2L2H
180
b2
16.95
p
T1 13.18 kN
180
p
180.039
STATICS
T1sin( b 1) T2sin( b 2)
;
checks
SHEAR & BEARING STRESSES
dp 18 mm
AS STEP (6)
Check (b 1 + b 2 + u + a)
sin(b 2)
b
sin(b 1)
T1cos( b 1) T2cos( b 2) 23.1
180
13.789
p
STEP (5) b 2 arccosa
T1 T2 a
p 2
dp
4
T1
2
t1ave AS
T2
2
t2ave AS
t 80 mm
Ab tdp
t1ave 25.9 MPa
;
t2ave 21.2 MPa
;
sin(b 2)
T1 T2 a
b
sin(b 1)
sb1 T1
Ab
s b1 9.15 MPa
;
T1cos( b 1) T2cos( b 2) W
sb2 T2
Ab
sb2 7.48 MPa
;
T2 W
sin(b 2)
cos(b 1)
+ cos(b 2)
sin(b 1)
T2 10.77 kN
;
Problem 1.7-7 A special-purpose eye bolt of shank diameter d 0.50 in. passes
y
through a hole in a steel plate of thickness tp 0.75 in. (see figure) and is secured by
a nut with thickness t 0.25 in. The hexagonal nut bears directly against the steel
plate. The radius of the circumscribed circle for the hexagon is r 0.40 in. (which
means that each side of the hexagon has length 0.40 in.). The tensile forces in three
cables attached to the eye bolt are T1 800 lb., T2 550 lb., and T3 1241 lb.
(a) Find the resultant force acting on the eye bolt.
(b) Determine the average bearing stress sb between the hexagonal nut on the
eye bolt and the plate.
(c) Determine the average shear stress taver in the nut and also in the steel
plate.
T1
tp
T2
d
30
2r
x
Cables
Nut
t
30
Eye bolt
T3
Steel plate
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.7-7
(c) AVERAGE SHEAR THROUGH NUT
CABLE FORCES
T1 800 lb
T2 550 lb
T3 1241 lb
(a) RESULTANT
P T2
23
+ T30.5
2
P 1097 lb
;
d 0.5 in.
t 0.25 in.
Asn pdt
Asn 0
tnut 2793 psi
Aspl 6rtp
Ab 0.2194 in.2 hexagon (Case 25, Appendix E)
P
sb Ab
s b 4999 psi
;
tpl P
Aspl
P
Asn
;
tp 0.75
SHEAR THROUGH PLATE
(b) AVERAGE BEARING STRESS
tnut r 0.40
Aspl 2
tpl 609 psi
;
Problem 1.7-8 An elastomeric bearing pad consisting of two steel
plates bonded to a chloroprene elastomer (an artificial rubber) is
subjected to a shear force V during a static loading test (see figure).
The pad has dimensions a 125 mm and b 240 mm, and the elastomer has thickness t 50 mm. When the force V equals 12 kN, the top
plate is found to have displaced laterally 8.0 mm with respect to the
bottom plate.
What is the shear modulus of elasticity G of the chloroprene?
Solution 1.7-8
NUMERICAL DATA
V 12 kN
a 125 mm
b 240 mm
t 50 mm
d 8 mm
AVERAGE SHEAR STRESS
tave V
ab
tave 0.4 MPa
AVERAGE SHEAR STRAIN
gave SHEAR MODULUS G
d
t
gave 0.16
G
tave
gave
G 2.5 MPa
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 1.7 Shear Stress and Strain
67
Problem 1.7-9 A joint between two concrete slabs A and B is filled
with a flexible epoxy that bonds securely to the concrete (see figure).
The height of the joint is h 4.0 in., its length is L 40 in., and its
thickness is t 0.5 in. Under the action of shear forces V, the slabs
displace vertically through the distance d 0.002 in. relative to
each other.
(a) What is the average shear strain gaver in the epoxy?
(b) What is the magnitude of the forces V if the shear modulus
of elasticity G for the epoxy is 140 ksi?
Solution 1.7-9 Epoxy joint between concrete slabs
(a) AVERAGE SHEAR STRAIN
gaver d
0.004
t
;
(b) SHEAR FORCES V
Average shear stress: taver Ggaver
h 4.0 in.
t 0.5 in.
L 40 in.
d 0.002 in.
G 140 ksi
V taver(hL) Ggaver(hL)
(140 ksi)(0.004)(4.0 in.)(40 in.)
89.6 k
;
Problem 1.7-10 A flexible connection consisting of rubber pads
(thickness t 9 mm) bonded to steel plates is shown in the figure.
The pads are 160 mm long and 80 mm wide.
(a) Find the average shear strain gaver in the rubber if the
force P 16 kN and the shear modulus for the rubber
is G 1250 kPa.
(b) Find the relative horizontal displacement d between the
interior plate and the outer plates.
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.7-10 Rubber pads bonded to steel plates
(a) SHEAR STRESS AND STRAIN IN THE RUBBER PADS
taver gaver Rubber pads: t 9 mm
Length L 160 mm
P/2
8 kN
625 kPa
bL
(80 mm)(160 mm)
taver
625 kPa
0.50
G
1250 kPa
;
(b) HORIZONTAL DISPLACEMENT
d t * tan1gave2 4.92 mm
Width b 80 mm
G 1250 kPa
P 16 kN
Problem 1.7-11 Steel riser pipe hangs from a drill rig located offshore in deep water (see figure). Separate segments are
joined using bolted flange plates (see figure and photo). Assume that there are six bolts at each pipe segment connection.
Assume that the total length of riser pipe is L 5000 ft; outer and inner diameters are d2 16 in., d1 15 in.; flange plate
thickness is tf 1.75 in.; and bolt and washer diameters are db 1.125 in., dw 1.875 in.
(a) If the entire length of the riser pipe were suspended in air, find the average normal stress s in each bolt, the average
bearing stress sb beneath each washer, and the average shear stress t through the flange plate at each bolt location
for the topmost bolted connection.
(b) If the same riser pipe hangs from a drill rig at sea, what are the normal, bearing, and shear stresses in the connection?
(Obtain the weight densities of steel and sea water from Table I-1, Appendix I. Neglect the effect of buoyant foam
casings on the riser pipe).
Flange plate (tf), typical
bolt (db), and washer (dw)
d2
tf
r
Flange plate on riser pipe
L
d1
db t
dw
d2
x
f
r
d2
60°
Riser pipe (d2, d1 L)
(a)
y
Flange plate on riser pipe–
plan view (n = 6 bolts shown)
(b)
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 1.7 Shear Stress and Strain
69
Photo Courtesy of
Transocean
Solution 1.7-11
(a) PIPE SUSPENDED IN AIR
L 5000 ft
gs 490 lb/ft3
d2 16 in. d1 15 in.
t
gw 63.8 lb/ft3
d2 d1
0.5 in. tf 1.75 in.
2
Apipe p 2
1d2 d122 24.347 in.2
4
Wpipe gs Apipe L 414.243 k
n6
sb db 1.125 in.
Wpipe
n Ab
69.5 ksi
(b) PIPE SUSPENDED IN SEA WATER
sb dw 1.875 in.
sbrg Wpipe
n Aw
Ab 39.1 ksi
p 2
db 0.994 in.2
4
tf Wpipe
n dw tf
Aw p 2
adw d2b b 1.8 in.2
4
21 ksi
Winwater (gs gw) Apipe L 360.307 kip
Winwater
60.4 ksi
n Ab
sbrg Winwater
34 ksi
n Aw
tf Winwater
18.3 ksi
n dw tf
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.7-12 The clamp shown in the figure is used to
support a load hanging from the lower flange of a steel beam.
The clamp consists of two arms (A and B) joined by a pin at C.
The pin has diameter d 12 mm. Because arm B straddles arm
A, the pin is in double shear.
Line 1 in the figure defines the line of action of the
resultant horizontal force H acting between the lower flange of
the beam and arm B. The vertical distance from this line to the
pin is h 250 mm. Line 2 defines the line of action of the
resultant vertical force V acting between the flange and arm B.
The horizontal distance from this line to the centerline of the
beam is c 100 mm. The force conditions between arm A and
the lower flange are symmetrical with those given for arm B.
Determine the average shear stress in the pin at C when the
load P 18 kN.
Solution 1.7-12
Clamp supporting a load P
FREE-BODY DIAGRAM OF CLAMP
h 250 mm
c 100 mm
P 18 kN
From vertical equilibrium:
V
P
9 kN
2
d diameter of pin at C 12 mm
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SECTION 1.7 Shear Stress and Strain
FREE-BODY DIAGRAMS OF ARMS A AND B
71
SHEAR FORCE F IN PIN
F
H 2
P 2
b
+
a
b
a
A 4
2
4.847 kN
AVERAGE SHEAR STRESS IN THE PIN
©MC 0 哵哴
taver F
F
42.9 MPa
Apin
pa2
VC Hh 0
H
;
4
VC
Pc
3.6 kN
h
2h
FREE-BODY DIAGRAM OF PIN
Problem 1.7-13 A hitch-mounted bicycle rack is designed to carry up to four 30-lb. bikes mounted on and strapped to two arms
GH [see bike loads in the figure part (a)]. The rack is attached to the vehicle at A and is assumed to be like a cantilever beam
ABCDGH [figure part (b)]. The weight of fixed segment AB is W1 10 lb, centered 9 in. from A [see the figure part (b)] and the
rest of the rack weighs W2 40 lb, centered 19 in. from A. Segment ABCDG is a steel tube, 2 2 in., of thickness t 1/8 in.
Segment BCDGH pivots about a bolt at B of diameter dB 0.25 in. to allow access to the rear of the vehicle without removing the
hitch rack. When in use, the rack is secured in an upright position by a pin at C (diameter of pin dp 5/16 in.) [see photo and
figure part (c)]. The overturning effect of the bikes on the rack is resisted by a force couple Fh at BC.
(a) Find the support reactions at A for the fully loaded rack.
(b) Find forces in the bolt at B and the pin at C.
(c) Find average shear stresses taver in both the bolt at B and the pin at C.
(d) Find average bearing stresses sb in the bolt at B and the pin at C.
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CHAPTER 1 Tension, Compression, and Shear
Bike loads
y
Four bike loads
19 in.
G
27 in.
G
Release pins
at C and G
5
(dp = — in.)
16
3 @ 4 in.
W2
H
a
2 in.
C
Fixed
support
at A
2 in.
1
( — in.)
8
MA A
6 in.
D
C
y
D
A
H
F
2.125 in.
F
a
F
B
Bolt at B
1
(dB = — in.)
4
h = 7 in.
W1
x
B
Ax A
9 in.
h = 7 in.
F
8 in.
(a)
(b)
Pin at C
C
Pin at C
2.125 in.
D
Bolt at B
2 2
tube
1/8 in.
(c) Section a–a
Solution 1.7-13
Ay 170 lb
NUMERICAL DATA
t
1
in.
8
h 7 in.
P 30 lb
L1 17 2.125 6
b 2 in.
W1 10 lb
W2 40 lb
5
in.
dp 16
dB 0.25 in.
(a) REACTIONS AT A
Ax 0
L1 25 in.
(dist from A to first bike)
MA W1(9) W2(19) P(4L1 4 8 12)
MA 4585 in.-lb
(b) FORCES IN BOLT AT B AND PIN AT C
Fy 0
;
Ay W1 W2 4P
;
;
By W2 4P
By 160 lb
;
MB 0
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SECTION 1.7 Shear Stress and Strain
Right hand FBD
AsC 2
[W2(19 17) + P(6 + 2.125)
+ P(8.125 + 4) + P(8.125 + 8)
tC + P(8.125 + 12)]
Bx h
Bx 254 lb
;
Bres 2Bx 2 + By 2
pdB2
4
Bres
tB AsB
AsB 0.098 in.2
tB 3054 psi
Bx
AsC
AsC 0.153 in.2
tC 1653 psi
;
t 0.125 in.
;
(c) AVERAGE SHEAR STRESSES tave IN BOTH THE BOLT
AT B AND THE PIN AT C
AsB 2
4
(d) BEARING STRESSES sB IN THE BOLT AT B AND THE PIN AT C
Cx Bx
Bres 300 lb
pdp2
73
;
AbB 2tdB
AbB 0.063 in.2
Bres
AbB
sbB 4797 psi
AbC 2tdp
AbC 0.078 in.2
Cx
AbC
sbC 3246 psi
sbB sbC ;
;
Problem 1.7-14 A bicycle chain consists of a series of small
links, each 12 mm long between the centers of the pins (see
figure). You might wish to examine a bicycle chain and observe its
construction. Note particularly the pins, which we will assume to
have a diameter of 2.5 mm.
In order to solve this problem, you must now make two measurements on a bicycle (see figure): (1) the length L of the crank
arm from main axle to pedal axle, and (2) the radius R of the
sprocket (the toothed wheel, sometimes called the chainring).
(a) Using your measured dimensions, calculate the tensile force
T in the chain due to a force F 800 N applied to one of
the pedals.
(b) Calculate the average shear stress taver in the pins.
Solution 1.7-14
Bicycle chain
F force applied to pedal 800 N
R radius of sprocket
L length of crank arm
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CHAPTER 1 Tension, Compression, and Shear
(b) SHEAR STRESS IN PINS
MEASUREMENTS (FOR AUTHOR’S BICYCLE)
(1) L 162 mm
(2) R 90 mm
taver (a) TENSILE FORCE T IN CHAIN
©M axle 0
FL TR
T
FL
R
Substitute numerical values:
T
(800 N)(162 mm)
1440 N
90 mm
2T
T/2
T
2
Apin
pd 2
pd
2
(4)
2FL
pd2R
Substitute numerical values:
2(800 N)(162 mm)
147 MPa
taver p(2.5 mm)2(90 mm)
;
;
Problem 1.7-15 A shock mount constructed as shown in the
figure is used to support a delicate instrument. The mount consists
of an outer steel tube with inside diameter b, a central steel bar of
diameter d that supports the load P, and a hollow rubber cylinder
(height h) bonded to the tube and bar.
(a) Obtain a formula for the shear t in the rubber at a radial
distance r from the center of the shock mount.
(b) Obtain a formula for the downward displacement d of the
central bar due to the load P, assuming that G is the shear
modulus of elasticity of the rubber and that the steel tube
and bar are rigid.
Solution 1.7-15
Shock mount
(a) SHEAR STRESS t AT RADIAL DISTANCE r
As shear area at distance r 2prh
t
P
P
As
2prh
;
(b) DOWNWARD DISPLACEMENT d
g shear strain at distance r
g
t
P
G
2prhG
dd downward displacement for element dr
dd gdr d
r radial distance from center of shock mount to
element of thickness dr
L
dd Pdr
2prhG
b/2
Pdr
Ld/2 2prhG
d
b/2
P
dr
P
b/2
[ln r]d/2
2phG Ld/2 r
2phG
d
P
b
ln
2phG
d
;
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75
SECTION 1.7 Shear Stress and Strain
Problem 1.7-16
A removable sign post on a hurricane evacuation route has a square base plate with four slots (or cut
outs) at bolts 1 through 4 (see drawing and photo) for ease of installation and removal. The upper portion of the post
has a separate base plate which is bolted to an anchored base (see photo). Each of the four bolts has diameter db and a
washer of diameter dw. The bolts are arranged in a rectangular pattern (b h). Consider wind force Wy applied in the
y-direction at the center of pressure of the sign structure at a height z L above the base. Neglect the weight of the
sign and post, and also neglect friction between the upper and lower base plates. Assume that the lower base plate and
short anchored post are rigid.
(a) Find average shear stress t, (MPa) at bolt #1 due to the wind force Wy;
repeat for bolt #4.
(b) Find average bearing stress sb, (MPa) between the bolt and the base plate
(thickness t) at bolt #1; repeat for bolt #4.
(c) Find average bearing stress sb, (MPa) between base plate and washer at bolt #4
due to the wind force Wy (assume initial bolt pretension is zero)
(d) Find average shear stress t, (MPa) through the base plate at bolt #4 due to the
wind force Wy.
(e) Find an expression for normal stress s in bolt #3 due to the wind force Wy.
z
C.P
Wy
See Prob. 1.8-15 for additional discussion of wind on a sign, and the resulting forces
acting on a conventional base plate.
NUMERICAL DATA
L 2.75 m
b 96 mm
Wy 667 N
t 14 mm
H 150 mm
db 12 mm
x
y
h 108 mm
dw 22 mm
L
h
(a)
Bolt and washer
(db, dw)
y
2
Wy
Square base plate
(H × H) of
thickness t
1
y
h
3
Slot in base plate
(db = slot width)
h
x
x
Base of position
anchored in
ground
4
b
Upper removable
sign post
with base plate
Plan view
of upper
base plate
Slotted upper
and lower
base plates
(b)
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.7-16
(a) FIND AVERAGE SHEAR STRESS (t, MPa) AT BOLT #1 DUE TO THE WIND FORCE Wy; REPEAT FOR BOLT #4
wy
Ab p 2
d 113.097 mm2
4 b
t1 2
2.95 MPa
Ab
t1 2.95 MPa
wy
t4 0
2
333.5 N
(only bolts 1 and 2 resist wind force shear in y-direction)
(b) FIND AVERAGE BEARING STRESS (sb, MPa) BETWEEN THE BOLT AND THE BASE PLATE (THICKNESS t) AT BOLT #1; REPEAT
FOR BOLT #4
wy
Abrg db t 168 mm2
sb1 2
1.985 MPa
Abrg
sb1 1.985 MPa
sb4 0
(only bolts 1 and 2 resist wind force bearing in y-direction)
(c) FIND AVERAGE BEARING STRESS (sb, MPa) BETWEEN BASE PLATE AND WASHER AT BOLT #4 DUE TO THE WIND FORCE Wy
(ASSUME INITIAL BOLT PRETENSION IS ZERO)
Assume wind force creates overturning moment about x axis OTMx
OTMx Wy L 1834.25 N # m
OTM is resisted by force couples pairs at bolts 1 to 4 and 2 to 3; so force in bolt 4 is
F4 OTMx
8491.898 N
2h
Bearing area is donut shaped area of washer in contact with the plate minus approximate rectangular cutout for slot
Abrg dw db
p 2
b 207.035 mm2
A d db2B db a
4 w
2
sb4 F4
41 MPa
Abrg
sb4 41 MPa
(d) FIND AVERAGE SHEAR STRESS (t, MPa) THROUGH THE BASE PLATE AT BOLT #4 DUE TO THE WIND FORCE Wy;
Use force F4 above; shear stress is on cylindrical surface at perimeter of washer; must deduct approximate
rectangular area due to slot
F4
Ash A p dw db B t 799.611 mm2
t
10.62 MPa
t 10.62 MPa
Ash
(e) FIND AN EXPRESSION FOR NORMAL STRESS (s) IN BOLT #3 DUE TO THE WIND FORCE Wy.
Force in bolt #3 due to OTMx is same as that in bolt #4
s3 F4
75.1 MPa
Ab
s3 75.1 MPa
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SECTION 1.7 Shear Stress and Strain
77
Problem 1.7-17 A spray nozzle for a garden hose requires a force F 5 lb. to open the spring-loaded spray chamber AB.
The nozzle hand grip pivots about a pin through a flange at O. Each of the two flanges has thickness t 1/16 in., and the pin
has diameter dp 1/8 in. [see figure part (a)]. The spray nozzle is attached to the garden hose with a quick release fitting at B
[see figure part (b)]. Three brass balls (diameter db 3/16 in.) hold the spray head in place under water pressure force fp 30 lb
at C [see figure part (c)]. Use dimensions given in figure part (a).
(a) Find the force in the pin at O due to applied force F.
(b) Find average shear stress taver and bearing stress sb in the pin at O.
Pin
Flange
t
dp
Pin at O
A
F
Top view at O
B
O
a = 0.75 in.
Spray
nozzle Flange
F
b = 1.5 in.
F
F
15
c = 1.75 in.
F
Sprayer
hand grip
Water pressure
force on nozzle, fp
C
(b)
C
Quick
release
fittings
Garden hose
(c)
(a)
Three brass retaining
balls at 120,
3
diameter db = — in.
16
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.7-17
Ox 12.68 lb
NUMERICAL DATA
F 5 lb
t
fp 30 lb
1
in.
16
dN a 0.75 in.
dp 5
in.
8
b 1.5 in.
1
in.
8
db 3
in.
16
u 15
p
rad
180
c 1.75 in.
(a) FIND THE FORCE IN THE PIN AT O DUE TO APPLIED
FORCE F
Mo 0
FAB [ F cos (u)( b a)] + F sin (u)(c)
a
FAB 7.849 lb
a FH 0
Ox FAB F cos (u)
Oy 1.294 lb
Ores 2O2x + O2y
Ores 12.74 lb
;
(b) FIND AVERAGE SHEAR STRESS tave AND BEARING STRESS
sb IN THE PIN AT O
As 2
pdp2
tO Ores
As
tO 519 psi
;
sbO Ores
Ab
sbO 816 psi
;
4
Ab 2tdp
(c) FIND THE AVERAGE SHEAR STRESS tave IN THE BRASS
RETAINING BALLS AT B DUE TO WATER PRESSURE FORCE Fp
As 3
pdb2
4
tave fp
tave 362 psi
As
;
Oy F sin (u)
y
Problem 1.7-18
A single steel strut AB with diameter
ds 8 mm. supports the vehicle engine hood of mass 20 kg
which pivots about hinges at C and D [see figures (a) and (b)].
The strut is bent into a loop at its end and then attached to a
bolt at A with diameter db 10 mm. Strut AB lies in a vertical
plane.
(a) Find the strut force Fs and average normal stress s in the
strut.
(b) Find the average shear stress taver in the bolt at A.
(c) Find the average bearing stress sb on the bolt at A.
h = 660 mm
W hc = 490 mm
C
B
45
C
x
A
30
D
(a)
W
Fs
C
z
Hinge
Strut
ds = 8 mm
D
H = 1041 mm
h = 660 mm
b = 254 mm c = 506 mm
y
a = 760 mm
d = 150 mm
B
C
Hood
A
(b)
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 1.7 Shear Stress and Strain
79
Solution 1.7-18
NUMERICAL DATA
ds 8 mm
db 10 mm
m 20 kg
cd
Fsz 2H + (c d)2
2
a 760 mm
b 254 mm
c 506 mm
d 150 mm
H
h 660 mm
hc 490 mm
2H + (c d)2
H h atan a30
p
p
b + tan a45
bb
180
180
2H + (c d)2
where
2
cd
2
H 1041 mm
a + b + c
760 mm
2
M
VECTOR rAB
Fsy 145.664
0
rAB H
Pc dQ
eAB rAB
冷rAB冷
eAB 0
W W
P 0 Q
M
D
hc
rDC hc
Pb + cQ
MD rDB
0
0.946
P 0.324 Q
ƒ eAB ƒ 1
0
W 196.2
P 0 Q
490
rDC 490
P 760 Q
Fs eAB W
rDC
(ignore force at hinge C since it will vanish with
moment about line DC)
Fsx 0
Fsy H
2H 2 + (c d)2
0.324
Fsy lineDC 0
0
rAB 1.041 * 103
P
Q
356
UNIT VECTOR eAB
0.946
(a) FIND THE STRUT FORCE FS AND AVERAGE NORMAL STRESS
s IN THE STRUT
W 196.2 N
W m (9.81m/s2)
Fs
Fs
Fsy
Fs 冷W 冷hc
h
Fs 153.9 N
H
;
2H2 + (c d)2
Astrut s
p 2
d
4 s
Fs
Astrut
Astrut 50.265 mm2
s 3.06 MPa
;
(b) FIND THE AVERAGE SHEAR STRESS tave IN THE
BOLT AT A
db 10 mm
As p 2
d
4 b
As 78.54 mm2
tave Fs
As
tave 1.96 MPa
;
(c) FIND THE BEARING STRESS sb ON THE BOLT AT A
Ab dsdb
sb Fs
Ab
Ab 80 mm2
sb 1.924 MPa
;
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.7-19
The top portion of a pole saw used to trim
small branches from trees is shown in the figure part (a).
The cutting blade BCD [see figure parts (a) and (c)] applies
a force P at point D. Ignore the effect of the weak return
spring attached to the cutting blade below B. Use properties
and dimensions given in the figure.
B
Rope, tension = T
a
T
y
Weak return spring
2T
x
(a) Find the force P on the cutting blade at D if the tension force in the rope is T 25 lb (see free body
diagram in part (b)].
(b) Find force in the pin at C.
(c) Find average shear stress tave and bearing stress sb in
the support pin at C [see Section a–a through cutting
blade in figure part (c)].
C
Cutting
blade
Collar
Saw blade
D
a
P
(a) Top part of pole saw
B
T
B
2T
BC
= 6 in.
50
20
Cy
DC =
70
C
D
.
1 in
P
Cx
x
20
20
Cutting blade
3
(tb = — in.)
32
6 in. C
Collar
3
(tc = — in.)
8
1 in.
D
70
(b) Free-body diagram
Pin at C
1
(dp = — in.)
8
(c) Section a–a
Solution 1.7-19
NUMERICAL PROPERTIES
dp 1
in.
8
T 25 lb
tb 3
in.
32
SOLVE EQUATION FOR P
tc 3
in.
8
dBC 6 in.
dCD 1 in.
[T(6 sin (70)) + 2T cos (20)
P
6 sin (70)) 2T sin (20)(6 cos (70))]
cos (20)
P 395 lbs
;
(b) SOLVE FOR FORCES ON PIN AT C
(a) FIND THE CUTTING FORCE P ON THE CUTTING BLADE
AT D IF THE TENSION FORCE IN THE ROPE IS T 25 lb:
Mc 0
MC T(6 sin(70))
2T cos (20)(6 sin (70))
2T sin (20)(6 cos (70))
P cos (20)(1)
Fx 0
Cx T 2T cos (20) P cos (40)
Cx 374 lbs
Fy 0
;
Cy 2T sin (20) P sin (40)
Cy 237 lbs
;
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SECTION 1.8 Allowable Stresses and Allowable Loads
RESULTANT AT C
81
BEARING STRESSES ON PIN ON EACH SIDE OF COLLAR
Cres 2Cx2 + Cy2
Cres 443 lbs
;
(c) FIND MAXIMUM SHEAR AND BEARING STRESSES IN THE
SUPPORT PIN AT C (SEE SECTION A–A THROUGH SAW).
Cres
2
sbC dp tc
sbC 4.72 ksi
;
BEARING STRESS ON PIN AT CUTTING BLADE
SHEAR STRESS—PIN IN DOUBLE SHEAR
p
As d2p
4
tave As 0.012 in.2
Cres
2As
sbcb Cres
dp tb
s bcb 37.8 ksi
;
tave 18.04 ksi
Allowable Stresses and Allowable Loads
Problem 1.8-1 A bar of solid circular cross section is loaded in
tension by forces P (see figure). The bar has length L 16.0 in. and
diameter d 0.50 in. The material is a magnesium alloy having
modulus of elasticity E 6.4 106 psi. The allowable stress in
tension is sallow 17,000 psi, and the elongation of the bar must not
exceed 0.04 in.
What is the allowable value of the forces P?
Solution 1.8-1
Magnesium bar in tension
p
Pmax smaxA (16.000 psi)a b (0.50 in.)2
4
3140 lb
L 16.0 in.
E 6.4
d 0.50 in.
106 psi
sallow 17,000 psi
dmax 0.04 in.
MAXIMUM LOAD BASED UPON ELONGATION
max dmax
0.04 in.
0.00250
L
16 in.
MAXIMUM LOAD BASED UPON TENSILE STRESS
p
Pmax sallowA (17,000 psi)a b (0.50 in.)2
4
3340 Ib
ALLOWABLE LOAD
Elongation governs.
Pallow 3140 lb
;
smax EPmax (6.4 * 106 psi)(0.00250)
16,000 psi
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.8-2 A torque T0 is transmitted between two flanged
T0
d
shafts by means of ten 20-mm bolts (see figure and photo). The
diameter of the bolt circle is d 250 mm.
If the allowable shear stress in the bolts is 90 MPa, what is the
maximum permissible torque? (Disregard friction between the
flanges.)
T0
T0
Solution 1.8-2 Shafts with flanges
NUMERICAL DATA
MAXIMUM PERMISSIBLE TORQUE
r 10
^ bolts
d
Tmax ta As a r b
2
d 250 mm
^ flange
As p r2
Tmax 3.338 * 107 N # mm
As 314.159 m2
Tmax 33.4 kN # m
;
ta 85 MPa
Problem 1.8-3 A tie-down on the deck of a sailboat consists of a bent bar bolted at both ends, as shown in the figure. The
diameter dB of the bar is 1/4 in. , the diameter dW of the washers is 7/8 in. , and
the thickness t of the fiberglass deck is 3/8 in.
If the allowable shear stress in the fiberglass is 300 psi, and the allowable
bearing pressure between the washer and the fiberglass is 550 psi, what is the
allowable load Pallow on the tie-down?
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83
SECTION 1.8 Allowable Stresses and Allowable Loads
Solution 1.8-3 Bolts through fiberglass
dB 1
in.
4
P1
309.3 lb
2
dW 7
in.
8
P1 619 lb
t
3
in.
8
ALLOWABLE LOAD BASED UPON SHEAR STRESS IN
FIBERGLASS
tallow 300 psi
Shear area As pdWt
ALLOWABLE LOAD BASED UPON BEARING PRESSURE
sb 550 psi
Bearing area Ab 2
2
P2
p
1
7
sb Ab (550 psi) a b c a in. b a in. b d
2
4
8
4
303.7 lb
P1
t allow As t allow (pdWt)
2
7
3
(300 psi)(p)a in. b a in.b
8
8
P2 607 lb
ALLOWABLE LOAD
Bearing pressure governs.
Pallow 607 lb
;
Problem 1.8-4
Two steel tubes are joined at B by four pins
(dp 11 mm), as shown in the cross section a–a in the
figure. The outer diameters of the tubes are dAB 41 mm and
dBC 28 mm. The wall thicknesses are tAB 6.5 mm and
tBC 7.5 mm. The yield stress in tension for the steel is
sY 200 MPa and the ultimate stress in tension is
sU 340 MPa. The corresponding yield and ultimate values
in shear for the pin are 80 MPa and 140 MPa, respectively.
Finally, the yield and ultimate values in bearing between the
pins and the tubes are 260 MPa and 450 MPa, respectively.
Assume that the factors of safety with respect to yield stress
and ultimate stress are 3.5 and 4.5, respectively.
p 2
(d d2B)
4 W
a
Pin
tAB
dAB
A
tBC
B
dBC
C
P
a
(a) Calculate the allowable tensile force Pallow considering
tension in the tubes.
(b Recompute Pallow for shear in the pins.
(c) Finally, recompute Pallow for bearing between the pins
and the tubes. Which is the controlling value of P?
tAB
dp
tBC
dAB
dBC
Section a–a
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.8-4
Yield and ultimate stresses (all in MPa)
TUBES:
sY 200
su 340
PIN (SHEAR):
tY 8
PIN (BEARING): sbY 260
TUBES AND PIN DIMENSIONS (MM)
FSy 3.5
dAB 41
tAB 6.5
tu 140 FSu 4.5
dBC dAB 2 tAB
dBC 28
sbu 450
tBC 7.5
dp 11
(a) PALLOW CONSIDERING TENSION IN THE TUBES
p
C d 2 1dAB 2 tAB22 D 4 dp tAB
4 AB
p
AnetBC C dBC 2 1dBC 2 tBC22 D 4 dp tBC
4
sy
A
PaT1 8743.993 N
PaT1 FSy netBC
AnetAB PaT 2 su
A
FSu netBC
tY
FSy
Pas2 (4 As)
tu
FSu
AnetBC 153.02
6 controls
6 use smaller
Pallow 8.74 kN
PaT 2 11,561.501 N
(b) Pallow CONSIDERING SHEAR IN THE PINS
PaS1 A 4 As B
AnetAB 418.502 mm2
As p 2
d As 95.033 mm2
4 p
PaS1 8688.748 N
Pallow 8.69 kN
< controls
Pas2 11,826.351 N
(c) Pallow CONSIDERING BEARING IN THE PINS
AbAB 4 dp tAB
AbAB 286 mm2
smaller controls
AbBC 4 dp tBC
AbBC 330
sby
b
Pab1 21,245.714 N
Pab1 AbAB a
FSy
Pab2 AbAB a
sbu
b
FSu
Pab2 28,600 N
< controls
Pallow 21.2 kN
Overall, shear controls (Part (b))
Problem 1.8-5 A steel pad supporting heavy machinery rests on
four short, hollow, cast iron piers (see figure). The ultimate
strength of the cast iron in compression is 50 ksi. The outer diameter of the piers is d 4.5 in. and the wall thickness is t 0.40 in.
Using a factor of safety of 3.5 with respect to the ultimate
strength, determine the total load P that may be supported by
the pad.
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85
SECTION 1.8 Allowable Stresses and Allowable Loads
Solution 1.8-5 Cast iron piers in compression
Four piers
A
sU 50 ksi
p 2
p
(d d02) [(4.5 in.)2 (3.7 in.)2]
4
4
5.152 in.2
n 3.5
sU
50 ksi
s allow 14.29 ksi
n
3.5
d 4.5 in.
P1 allowable load on one pier
sallow A (14.29 ksi)(5.152 in.2)
73.62 k
Total load P 4P1 294 k
t 0.4 in.
;
d0 d 2t 3.7 in.
Problem 1.8-6 The rear hatch of a van [BDCF in figure part (a)] is supported by two hinges at B1 and B2 and by two struts
A1B1 and A2B2 (diameter ds 10 mm) as shown in figure part (b). The struts are supported at A1 and A2 by pins, each with
diameter dp 9 mm and passing through an eyelet of thickness t 8 mm at the end of the strut [figure part (b)]. If a closing
force P 50 N is applied at G and the mass of the hatch Mh 43 kg is concentrated at C:
(a) What is the force F in each strut? [Use the free-body diagram of one half of the hatch in the figure part (c)]
(b) What is the maximum permissible force in the strut, Fallow, if the allowable stresses are as follows: compressive stress in
the strut, 70 MPa; shear stress in the pin, 45 MPa; and bearing stress between the pin and the end of the strut, 110 MPa.
127 mm
B1
505 mm
505 mm
F
B2
C
G
Mh
D
Bottom
part of
strut
P
F
A1
B
710 mm
Mh
—
2
g
By
ds = 10 mm
A2
75 mm
Bx
G
C
D
P
2
10
460 mm
Eyelet
Pin support
A
F
t = 8 mm
(b)
(a)
(c)
Solution 1.8-6
(a) FORCE F IN EACH STRUT FROM STATICS (SUM
MOMENTS ABOUT B)
NUMERICAL DATA
Mh 43 kg
sa 70 MPa
ta 45 MPa
sba 110 MPa
ds 10 mm
dp 9 mm
P 50 N
g 9.81 m/s2
FV F cos1102 FH F sin1102
t 8 mm
g MB 0
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CHAPTER 1 Tension, Compression, and Shear
FV (127) + FH (75) Mh
g (127 + 505)
2
P
+ [127 + 2(505)]
2
F (127cos(10) + 75sin(10))
Mh
P
g (127 + 505) +
+ 2(505)]
2
2
Mh
P
g (127 + 505) + [127 + 2(505)]
2
2
F
(127 cos(10) + 75 sin(10))
F 1.171 kN
(b) MAXIMUM PERMISSIBLE FORCE F IN EACH STRUT
Fmax IS SMALLEST OF THE FOLLOWING
p 2
d
Fa1 5.50 kN
4 s
p
Fa2 ta dp 2
4
Fa2
;
2.445
Fa2 2.86 kN
F
Fa1 sa
Fa3 sba dp t
Fa3 7.92 kN
;
Problem 1.8-7 A lifeboat hangs from two ship’s davits, as shown in the
figure. A pin of diameter d 0.80 in. passes through each davit and supports two pulleys, one on each side of the davit.
Cables attached to the lifeboat pass over the pulleys and wind around
winches that raise and lower the lifeboat. The lower parts of the cables are
vertical and the upper parts make an angle a 15° with the horizontal. The
allowable tensile force in each cable is 1800 lb, and the allowable shear
stress in the pins is 4000 psi.
If the lifeboat weighs 1500 lb, what is the maximum weight that should
be carried in the lifeboat?
Solution 1.8-7 Lifeboat supported by four cables
FREE-BODY DIAGRAM OF ONE PULLEY
Pin diameter d 0.80 in.
T tensile force in one cable
Tallow 1800 lb
tallow 4000 psi
W weight of lifeboat
1500 lb
©Fhoriz 0
RH T cos 15 0.9659T
©Fvert 0
RV T T sin 15 0.7412T
V shear force in pin
V 2(RH)2 + (Rv)2 1.2175T
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87
SECTION 1.8 Allowable Stresses and Allowable Loads
ALLOWABLE TENSILE FORCE IN ONE CABLE BASED
MAXIMUM WEIGHT
UPON SHEAR IN THE PINS
Shear in the pins governs.
p
Vallow tallow A pin (4000 psi)a b(0.80 in.)2
4
2011 lb
Vallow
1652 lb
T1 V 1.2175T
1.2175
Total tensile force in four cables
ALLOWABLE FORCE IN ONE CABLE BASED UPON
6608 lb 1500 lb
Tmax T1 1652 lb
4Tmax 6608 lb
Wmax 4Tmax W
TENSION IN THE CABLE
5110 lb
T2 Tallow 1800 lb
;
Problem 1.8-8 A cable and pulley system in figure part (a) supports a cage of mass 300 kg at B. Assume that this includes the
mass of the cables as well. The thickness of each the three steel pulleys is t 40 mm. The pin diameters are dpA 25 mm,
dpB 30 mm and dpC 22 mm [see figure, parts (a) and part (b)].
(a) Find expressions for the resultant forces acting on the pulleys at A, B, and C in terms of cable tension T.
(b) What is the maximum weight W that can be added to the cage at B based on the following allowable stresses? Shear
stress in the pins is 50 MPa; bearing stress between the pin and the pulley is 110 MPa.
a
C
dpA = 25 mm
L1
A
Cable
Cable
Pulley
t
a
dpB
L2
tB
Pin
dpC = 22 mm
dp
Support
bracket
B
dpB = 30 mm
Cage
W
(a)
Section a–a: pulley support
detail at A and C
Cage at B
Section a–a: pulley
support detail at B
(b)
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.8-8
NUMERICAL DATA
g 9.81 m/s2
M 300 kg
ta 50 MPa
sba 110 MPa
tA 40 mm
tB 40 mm
tC 50
dpA 25 mm
dpB 30
dpC 22 mm
(a) RESULTANT FORCES F ACTING ON PULLEYS A, B, AND C
FA 22 T
FC T
FB 2T
T
Mg
Wmax
+
2
2
Wmax 2T M g
(b) MAXIMUM LOAD W THAT CAN BE ADDED AT B DUE TO
ta AND sba IN PINS AT A, B, AND C
22
a sba Ab b M g
a sba tA dpA b M g
22
Wmax2 152.6 kN (bearing at A)
Wmax2 2
PULLEY AT B
Wmax3 2T taAs
2
1ta As2 M g
2
p
Wmax3 cta a 2 dpB2 b d M g
4
Wmax4 (shear at B)
2
(s A ) M g
2 ba b
Wmax4 129.1 kN
FA
As
22T ta As
FA ta As
Mg
Wmax
ta As
+
2
2
22
2
22
2
22
ataAs b M g
ata2
p
d A2 b M g
4 p
(bearing at B)
T ta As
PULLEY AT C
DOUBLE SHEAR
Wmax1 2
Wmax4 sba tB dpB M g
PULLEY AT A
Wmax1 Wmax2 Wmax3 67.7 kN
From statics at B
tA OR check bearing stress
Wmax5 21t a As2 M g
Wmax5 c2ta a 2
p 2
d b d Mg
4 pC
Wmax5 7.3 * 104
Wmax5 73.1 kN
(shear at C)
Wmax6 2sbatC dp C M g
Wmax6 239.1 kN
(bearing at C)
Wmax1
22.6
Mg
Wmax1 66.5 kN
;
(shear at A controls)
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89
SECTION 1.8 Allowable Stresses and Allowable Loads
Problem 1.8-9 A ship’s spar is attached at the base of a mast by a pin connection
Mast
(see figure). The spar is a steel tube of outer diameter d2 3.5 in. and inner diameter
d1 2.8 in. The steel pin has diameter d 1 in., and the two plates connecting the spar
to the pin have thickness t 0.5 in. The allowable stresses are as follows: compressive
stress in the spar, 10 ksi; shear stress in the pin, 6.5 ksi; and bearing stress between the
pin and the connecting plates, 16 ksi.
Determine the allowable compressive force Pallow in the spar.
P
Pin
Spar
Connecting
plate
Solution 1.8-9
COMPRESSIVE STRESS IN SPAR
p
Pa1 sa 1d 2 2 d1 22
4
Pa1 34.636 k
SHEAR STRESS IN PIN
Pa2 ta a 2
Pa2 10.21 kips
NUMERICAL DATA
d2 3.5 in.
dp 1 in.
sa 10 ksi
d1 2.8 in.
controls
;
^double shear
t 0.5 in.
ta 6.5 ksi
p 2
d b
4 p
sba 16 ksi
BEARING STRESS BETWEEN PIN AND CONECTING PLATES
Pa3 sba(2dpt)
Problem 1.8-10 What is the maximum possible value of the
clamping force C in the jaws of the pliers shown in the figure if
the ultimate shear stress in the 5-mm diameter pin is 340 MPa?
What is the maximum permissible value of the applied load
P if a factor of safety of 3.0 with respect to failure of the pin is
to be maintained?
Pa3 16 k
P
y
15
mm
90
38
Ry
50
x
Rx 140
b=
10
mm
90
P
C
Pin
m
C
50 mm
125 m
a
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.8-10
NUMERICAL DATA
FS 3
ta tu
ta FS
tu 340 MPa
Pmax 2Rx 2 + Ry 2
pin at C in single shear
As
Rx C cos (40)
Ry P C sin (40)
a 50 cos (40) 125
a 163.302 mm
b 38 mm
Rx 2
2
a
a
c cos (40) d + c1 + sin(40) d
C b
b
Pmax 445 N
here
;
a
4.297
b
P(a)
cos(40)
b
Ry Pc1 +
a
Cult PmaxFS a b
b
a
sin(40) d
b
2
2
a
a
P c cos1402 d + c1 + sin(40) d ta A s
C b
b
As ta As
a/b mechanical advantage
FIND MAXIMUM CLAMPING FORCE
P(a)
C
b
a Mpin 0
STATICS
ta 113.333 MPa
Find Pmax
d 5 mm
ta tu
FS
Pult PmaxFS
Cult 5739 N
;
Pult 1335
Cult
4.297
Pult
p 2
d
4
Problem 1.8-11 A metal bar AB of weight W is suspended by a system of steel
2.0 ft
2.0 ft
7.0 ft
wires arranged as shown in the figure. The diameter of the wires is 5/64 in., and the
yield stress of the steel is 65 ksi.
Determine the maximum permissible weight Wmax for a factor of safety of
1.9 with respect to yielding.
5.0 ft
5.0 ft
W
A
B
Solution 1.8-11
NUMERICAL DATA
d
5
in.
64
sY
sa FSy
sY 65 ksi
sa 34.211 ksi
FORCES IN WIRES AC, EC, BD, AND FD
FSy 1.9
a FV 0
at A, B, E, or F
322 + 52 W
*
5
2
Wmax = 0.539 sa A
FW 322 + 52
0.539
10
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SECTION 1.8 Allowable Stresses and Allowable Loads
Wmax 0.539a
sY
p
b a d2b
FSy
4
Wmax 0.305 kips
FCD 2a
;
FCD 2c
CHECK ALSO FORCE IN WIRE CD
a FH 0
FCD at C or D
2
22 + 52
2
2
322 + 52
2
W
5
91
Fw b
a
322 + 52 W
* bd
5
2
less than FAC so AC controls
Problem 1.8-12 A plane truss is subjected to loads 2P and P at joints B and C, respectively, as shown in the figure
part (a). The truss bars are made of two L102 76 6.4 steel angles [see Table F-5(b): cross sectional area of the two
angles, A 2180 mm2, figure part (b)] having an ultimate stress in tension equal to 390 MPa. The angles are connected
to an 12 mm-thick gusset plate at C [figure part (c)] with 16-mm diameter rivets; assume each rivet transfers an equal
share of the member force to the gusset plate. The ultimate stresses in shear and bearing for the rivet steel are 190 MPa
and 550 MPa, respectively.
Determine the allowable load Pallow if a safety factor of 2.5 is desired with respect to the ultimate load that can be
carried. (Consider tension in the bars, shear in the rivets, bearing between the rivets and the bars, and also bearing between
the rivets and the gusset plate. Disregard friction between the plates and the weight of the truss itself.)
F
FCF
G
a
a
A
B
a
FCG
Truss bars
C
a
a
D
Gusset
plate
Rivet
C
P
2P
a
FBC
FCD
(a)
P
(c)
Gusset plate
6.4 mm
12 mm
Rivet
(b) Section a–a
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.8-12
PCG 45.8 kN ; < so shear in rivets in CG and CD
controls Pallow here
NUMERICAL DATA
A 2180 mm
2
tg 12 mm
dr 16 mm
tang 6.4 mm
su 390 MPa
tu 190 MPa
sbu 550 MPa
FS 2.5
sa su
FS
ta tu
FS
sba 4
FCD P
3
22
0.471
3
sbu
FS
22
P
FCF 3
4
FCG P
3
Pallow FOR TENSION ON NET SECTION IN TRUSS BARS
Anet A 2drtang
PCD 45.8 kN
;
Next, Pallow for bearing of rivets on truss bars
rivet bears on each angle in two angle
Ab 2drtang
pairs
MEMBER FORCES FROM TRUSS ANALYSIS
5
FBC P
3
3
PCD 2 a b(ta As)
4
Anet 1975 mm2
Anet
0.906
A
Fmax
sba Ab
N
3
PBC 3 a b(sba Ab)
5
PBC 81.101 kN
PCF 2 a
PCF 191.156 kN
3
22
b (sba Ab)
3
PCG 2 a b(sba Ab)
4
PCG 67.584 kN
3
PCD 2 a b(sba Ab)
4
PCD 67.584 kN
Finally, Pallow for bearing of rivets on gusset plate
Fallow saAnet
allowable force in a member
so BC controls since it has the largest
member force for this loading
3
3
Pallow (sa Anet)
Pallow FBCmax
5
5
Pallow 184.879 kN
Ab drtg
(bearing area for each rivert on gusset plate)
tg 12 mm
2tang 12.8 mm
so gusset will control over angles
Next, Pallow for shear in rivets (all are in double shear)
3
PBC 3a b(sba Ab)
5
PBC 76.032 kN
p
As 2 dr 2
4
PCF 2 a
PCF 179.209 kN
Fmax
ta As
N
for one rivet in DOUBLE shear
N number of rivets in a particular
member (see drawing of connection
detail)
3
PBC 3a b(ta As)
5
PCF 2a
3
22
b (ta As)
PBC 55.0 kN
3
22
b (sba Ab)
3
PCG 2a b(sba Ab)
4
PCG 63.36 kN
3
PCD 2a b(sba Ab)
4
PCD 63.36 kN
So, shear in rivets controls: Pallow 45.8 kN
;
PCF 129.7 kN
3
PCG 2a b(ta As)
4
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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93
SECTION 1.8 Allowable Stresses and Allowable Loads
Problem 1.8-13
A solid bar of circular cross section (diameter d)
has a hole of diameter d/5 drilled laterally through the center of the
bar (see figure). The allowable average tensile stress on the net
cross section of the bar is sallow.
d
d/5
P
d/5
P
d
(a) Obtain a formula for the allowable load Pallow that the bar
can carry in tension.
(b) Calculate the value of Pallow if the bar is made of brass with
diameter d 1.75 in. and sallow 12 ksi.
(Hint: Use the formulas of Case 15 Appendix E.)
Solution 1.8-13
NUMERICAL DATA
d 1.75 in.
2
1
1
26b d
Pa sa c d 2 a arccosa b 2
5
25
sa 12 ksi
1
2
arccosa b 26
5
25
(a) FORMULA FOR PALLOW IN TENSION
From Case 15, Appendix E:
A 2r 2 aa ab
r
a
a acos a b
r
a
2
b
r
d
2
r 0.875 in.
180
78.463
p
a
d
10
a 0.175 in.
2
Pa sa10.587 d22
0.587
p
0.785
4
;
0.587
0.748
0.785
(b) EVALUATE NUMERICAL RESULT
b 2r2 a2
d 1.75 in.
sa 12 ksi
Pa 21.6 k
;
d 2
d 2
b ca b a b d
C 2
10
b
6
a d2 b
B 25
d
b 26
5
Pa sa A
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.8-14 A solid steel bar of diameter d1 60 mm has a
hole of diameter d2 32 mm drilled through it (see figure). A steel
pin of diameter d2 passes through the hole and is attached to supports.
Determine the maximum permissible tensile load Pallow in the bar
if the yield stress for shear in the pin is tY 120 MPa, the yield stress
for tension in the bar is sY 250 MPa and a factor of safety of 2.0 with
respect to yielding is required. (Hint: Use the formulas of Case 15,
Appendix E.)
d2
d1
d1
P
Solution 1.8-14
SHEAR AREA (DOUBLE SHEAR)
NUMERICAL DATA
d1 60 mm
p
As 2a d 2 2 b
4
d2 32 mm
tY 120 MPa
sY 250 MPa
As 1608 mm2
FSy 2
NET AREA IN TENSION (FROM APPENDIX E)
ALLOWABLE STRESSES
Anet 2a
ta tY
FSy
ta 60 MPa
sa sY
FSy
sa 125 MPa
From Case 15, Appendix E:
A 2r aa 2
a
d2
2
ab
r2
b
E arccosa
r
d1
2
d2
d 2/2
arc cos
a arc cos
d1/2
d1
b 2r 2 a2
d1 2
b
2
d2
b d1
d2
d1 2
d2 2
B a b a b R
2 C 2
2
d1 2
a b
2
U
Anet 1003 mm2
Pallow in tension: smaller of values based on either shear
or tension allowable stress x appropriate area
Pa1 ta As Pa1 96.5 kN 6 shear governs
Pa2 saAnet
;
Pa2 125.4 kN
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SECTION 1.8 Allowable Stresses and Allowable Loads
Problem 1.8-15 A sign of weight W is supported at its base by
four bolts anchored in a concrete footing. Wind pressure p acts
normal to the surface of the sign; the resultant of the uniform
wind pressure is force F at the center of pressure. The wind force
is assumed to create equal shear forces F/4 in the y-direction at
each bolt [see figure parts (a) and (c)]. The overturning effect of
the wind force also causes an uplift force R at bolts A and C and
a downward force (R) at bolts B and D [see figure part (b)].
The resulting effects of the wind, and the associated ultimate
stresses for each stress condition, are: normal stress in each bolt
(su 60 ksi); shear through the base plate (tu 17 ksi); horizontal shear and bearing on each bolt (thu 25 ksi and
sbu 75 ksi); and bearing on the bottom washer at B (or D)
(sbw 50 ksi).
Find the maximum wind pressure pmax (psf) that can be
carried by the bolted support system for the sign if a safety
factor of 2.5 is desired with respect to the ultimate wind load
that can be carried.
Use the following numerical data: bolt db 3⁄4 in.;
washer dw 1.5 in.; base plate tbp 1 in.; base plate
dimensions h 14 in. and b 12 in.; W 500 lb; H 17 ft;
sign dimensions (Lv 10 ft. Lh 12 ft.); pipe column
diameter d 6 in., and pipe column thickness t 3/8 in.
Sign (Lv
Resultant
of wind
pressure
Lh)
Lh
2
C.P.
F
W
Pipe
column
Lv
z b
2
D
H
95
y Overturning
moment
B about x-axis
FH
C
A
F
at each
4
bolt
h
x
W at each
4
bolt
(a)
W
Pipe column
db
dw
FH
— = Rh
2
One half of over turning
moment about x-axis acts
on each bolt pair
z
Base
plate (tbp)
B
A
y
Footing
F/4
Tension
h
R
Compression
R
(b)
z
2 in
.
R
F
4
W
4
x
A
F
4 R W
4
y
B
in.
C
h
4
=1
FH
—
2
b=
1
D
FH
2
R
W
4
(c)
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.8-15
(1) COMPUTE pmax BASED ON NORMAL STRESS IN EACH BOLT
(GREATER AT B AND D)
Numerical Data
su 60 ksi
tu 17 ksi
thu 25 ksi
sbu 75 ksi
sbw 50 ksi
FSu 2.5
db 3
in.
4
h 14 in.
dw 1.5 in.
tbp 1 in.
b 12 in.
W 0.500 kips
3
in.
8
H 204 in.
d 6 in.
H 17(12)
Lv 10(12)
Lh 12(12)
Lv 120 in.
Lh 144 in.
su
FSu
ta 6.8
sba sa 24
tha sbu
FSu
ta thu
FSu
sba 30
tu
FSu
sbwa FORCES F AND R IN TERMS OF pmax
R pmax
p
W
sa a db 2 b 4
4
pmax1 Lv Lh H
2h
pmax1 11.98 psf
FH
2h
;
controls
W
4
t
p dw tbp
Rmax ta(p dw tbp) sbw
FSu
pmax2 R
p
W
Rmax sa a db 2 b 4
4
p 2
d
4 b
R +
tha 10
sbwa 20
F pmaxLvLh
s
W
4
(2) COMPUTE pmax BASED ON SHEAR THROUGH BASE PLATE
(GREATER AT B AND D)
ALLOWABLE STRESSES (ksi)
sa t
R +
W
4
ta1p dw tbp2 W
4
Lv Lh H
2h
pmax2 36.5 psf
Lv Lh H
2h
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97
SECTION 1.8 Allowable Stresses and Allowable Loads
(3) COMPUTE pmax BASED ON HORIZONTAL SHEAR
ON EACH BOLT
th F
4
p
Fmax 4tha a db 2 b
4
p
a db 2 b
4
pmax3 tha(p db 2)
Lv Lh
pmax3 147.3 psf
(4) COMPUTE pmax BASED ON HORIZONTAL BEARING ON
EACH BOLT
F
4
sb (tbp db)
pmax4 Fmax 4sba(tbpdb)
(5) COMPUTE pmax BASED ON BEARING UNDER THE TOP WASHER
AT A (OR C) AND THE BOTTOM WASHER AT B (OR D)
R +
sbw W
4
p
A d 2 db2 B
4 w
Rmax sbwa c
p
W
A dw2 db2 B d 4
4
p
W
A dw2 db2 B d 4
4
pmax5 Lv Lh H
2h
pmax5 30.2 psf
sbwa c
So, normal/stress in bolts controls; pmax 11.98 psf
4sba(tbpdb)
Lv Lh
pmax4 750 psf
Problem 1.8-16 The piston in an engine is attached to a
connecting rod AB, which in turn is connected to a crank arm
BC (see figure). The piston slides without friction in a cylinder
and is subjected to a force P (assumed to be constant) while
moving to the right in the figure. The connecting rod, which has
diameter d and length L, is attached at both ends by pins. The
crank arm rotates about the axle at C with the pin at B moving
in a circle of radius R. The axle at C, which is supported by
bearings, exerts a resisting moment M against the crank arm.
Cylinder
P
Piston
Connecting rod
A
M
d
C
B
L
R
(a) Obtain a formula for the maximum permissible force
Pallow based upon an allowable compressive stress c in
the connecting rod.
(b) Calculate the force Pallow for the following data:
sc 160 MPa, d 9.00 mm, and R 0.28L.
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.8-16
The maximun allowable force P occurs when cos has its
smallest value, which means that has its largest value.
LARGEST VALUE OF d diameter of rod AB
FREE-BODY DIAGRAM OF PISTON
The largest value of occurs when point B is the farthest
distance from line AC. The farthest distance is the radius R
of the crank arm.
P applied force (constant)
C compressive force in connecting rod
RP resultant of reaction forces between cylinder
and piston (no friction)
Therefore,
—
BC R
—
Also, AC 2L2R2
cos a R 2
2L2R2
A 1 a b
L
L
(a) MAXIMUM ALLOWABLE FORCE P
a Fhoriz 0 Pallow c Ac cos P C cos 0
sc a
: ;
P C cos MAXIMUM COMPRESSIVE FORCE C IN CONNECTING ROD
Cmax cAc
in which Ac area of connecting rod
pd2
4
b
R 2
1a b
A
L
;
(b) SUBSTITUTE NUMERICAL VALUES
c 160 MPa
R 0.28L
d 9.00 mm
R/L 0.28
Pallow 9.77 kN
;
pd2
Ac 4
MAXIMUM ALLOWABLE FORCE P
P Cmax cos c Ac cos © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 1.9 Design for Axial Loads and Direct Shear
99
Design for Axial Loads and Direct Shear
Problem 1.9-1 An aluminum tube is required
to transmit an axial tensile force P 33 k
[see figure part (a)]. The thickness of the wall
of the tube is to be 0.25 in.
d
P
(a) What is the minimum required outer
diameter dmin if the allowable tensile
stress is 12,000 psi?
(b) Repeat part (a) if the tube will have a
hole of diameter d/10 at mid-length
[see figure parts (b) and (c)].
P
(a)
d
Hole of diameter d/10
d/10
P
P
d
(b)
(c)
Solution 1.9-1
NUMERICAL DATA
P 33 kips
(b) MINIMUM DIAMETER OF TUBE (WITH HOLES)
t 0.25 in.
sa 12 ksi
(a) MINIMUM DIAMETER OF TUBE (NO HOLES)
A1 p 2
C d 1d2t22 D
4
A2 P
sa
A2 2.75 in.2
P
+ t
psat
p 2
d
C d 1d2t22 D 2a bt d
4
10
A1 d apt t
b pt2
5
Equating A1 and A2 and solving for d:
Equating A1 and A2 and solving for d:
d
A1 c
d 3.75 in.
;
P
+ p t2
sa
d
t
pt 5
d 4.01 in.
;
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.9-2
A copper alloy pipe having yield stress
sY 290 MPa is to carry an axial tensile load P 1500 kN
[see figure part (a)]. A factor of safety of 1.8 against yielding
is to be used.
d
t =—
8
P
(a) If the thickness t of the pipe is to be one-eighth of its
outer diameter, what is the minimum required outer
diameter dmin?
(b) Repeat part (a) if the tube has a hole of diameter d/10
drilled through the entire tube as shown in the figure
[part (b)].
d
(a)
P
Hole of diameter d/10
d
t =—
8
d
(b)
Solution 1.9-2
NUMERICAL DATA
Equate A1 and A2 and solve for d:
sY 290 MPa
P
7
s
Y
d 64p P
FSy Q
2
P 1500 kN
FSy 1.8
(a) MINIMUM DIAMETER (NO HOLES)
A1 p 2
d 2
cd a d b d
4
4
A1 7
p d2
64
A2 P
sY
FSy
7
64p
dmin
R
P
sY
P FSy Q
dmin 164.6 mm
A2 9.31
;
103 mm2
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SECTION 1.9 Design for Axial Loads and Direct Shear
(b) MINIMUM DIAMETER (WITH HOLES)
Redefine A1—subtract area for two holes—then
equate to A2
d
p
d 2
d
A1 c cd2a d b d2a b a b d
4
4
10 8
A1 d2 a
P
sy
P
Q
FSy
7
1
p
b
64
40
7
1 2
pd 2 d
64
40
A1 d 2 a
7
1
p
b
64
40
7
1
p
0.319
64
40
dmin Problem 1.9-3 A horizontal beam AB with cross-sectional dimensions
(b 0.75 in.) (h 8.0 in.) is supported by an inclined strut CD and
carries a load P 2700 lb at joint B [see figure part (a)]. The strut,
which consists of two bars each of thickness 5b/8, is connected to the
beam by a bolt passing through the three bars meeting at joint C
[see figure part (b)].
(a) If the allowable shear stress in the bolt is 13,000 psi, what is the
minimum required diameter dmin of the bolt at C?
(b) If the allowable bearing stress in the bolt is 19,000 psi, what is
the minimum required diameter dmin of the bolt at C?
101
P
P
sy
FSy
Q
dmin 170.9 mm
7
1
a p
b
64
40
c
4 ft
;
5 ft
B
C
A
3 ft
P
D
(a)
b
Beam AB (b
h
—
2
h)
Bolt (dmin)
h
—
2
5b
—
8
Strut CD
(b)
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.9-3
NUMERICAL DATA
P 2.7 k
ta 13 ksi
(b) dmin BASED ON ALLOWABLE BEARING AT JT C
b 0.75 in.
sba 19 ksi
h 8 in.
Bearing from beam ACB
(a) dmin BASED ON ALLOWABLE SHEAR—DOUBLE SHEAR
dmin IN STRUT
ta FDC
As
As 2 a
FDC 15 P/4
bd
dmin 0.711 in.
;
15
P
4
Bearing from strut DC sb 5
2 bd
8
15
P
4
p 2
d b
4
15
P
4
dmin p
ta a b
a
2
15 P/4
b sba
sb sb 3
dmin 0.704 in.
P
bd
(lower than ACB)
;
Problem 1.9-4 Lateral bracing for an elevated pedestrian walkway is shown in the figure part (a). The thickness of the clevis
plate tc 16 mm and the thickness of the gusset plate tg 20 mm [see figure part (b)]. The maximum force in the diagonal
bracing is expected to be F 190 kN.
If the allowable shear stress in the pin is 90 MPa and the allowable bearing stress between the pin and both the clevis and
gusset plates is 150 MPa, what is the minimum required diameter dmin of the pin?
Clevis
Gusset plate
Gusset plate
tc
Pin
tg
Cl
ev
is
dmin
Diagonal brace
F
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103
SECTION 1.9 Design for Axial Loads and Direct Shear
Solution 1.9-4
NUMERICAL DATA
(2) dmin BASED ON ALLOW BEARING IN GUSSET AND CLEVIS
F 190 kN
ta 90 MPa
tg 20 mm
tc 16 mm
PLATES
sba 150 MPa
Bearing on gusset plate
sb (1) dmin BASED ON ALLOW SHEAR—DOUBLE SHEAR
IN PIN
p
As 2 a d2 b
4
F
t
As
dmin F
p
t a b
Q a 2
F
Ab
dmin 63.3 mm
6 controls
Bearing on clevis
Ab d(2tc)
dmin dmin 36.7 mm
dmin Ab tgd
F
2tcsba
;
dmin 39.6 mm
Problem 1.9-5 A plane truss has joint loads P, 2P, and 3P at
3P
joints D, C, and B, respectively (see figure) where load variable
P 5200 lb. All members have two end plates (see figure for
Prob. 1.7-2) which are pin connected to gusset plates (see also
figure for Prob. 1.8-12). Each end plate has thickness
tp 0.625 in., and all gusset plates have thickness tg 1.125 in.
If the allowable shear stress in each pin is 12,000 psi and the
allowable bearing stress in each pin is 18,000 psi, what is the
minimum required diameter dmin of the pins used at either end of
member BE?
A
9 ft
F
tgsba
B
P
2P
9 ft
C
9 ft
D
9 ft
13.5 ft
E
F
3 ft
G
Typical gusset plate
Solution 1.9-5
P 5200 lb
tp 5
in.
8
FBE 3.83858 P 19,960.616 lb
tg 1.125 in.
from plane truss analysis
(see Probs. 1.2-4 to 1.2-6)
tp 0.625 in.
2 tp 1.25 in.
ta 12 ksi
sba 18 ksi
PIN DIAMETER BASED ON ALLOWABLE SHEAR STRESS (PINS IN DOUBLE SHEAR)
FBE
2
dp1 1.029 in.
p
ta
a4
< controls
dpin 1.029 in.
PIN DIAMETER BASED ON BEARING BETWEEN PIN AND EACH OF TWO END PLATES
dp2 2tp is greater than tg so gusset will
control
FBE
0.887 in.
2 tp sba
PIN DIAMETER BASED ON BEARING BETWEEN PIN AND GUSSET PLATE
dp3 FBE
0.986 in.
tg sba
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.9-6 A suspender on a suspension bridge consists of a cable that
passes over the main cable (see figure) and supports the bridge deck, which is
far below. The suspender is held in position by a metal tie that is prevented from
sliding downward by clamps around the suspender cable.
Let P represent the load in each part of the suspender cable, and let u
represent the angle of the suspender cable just above the tie. Finally, let sallow
represent the allowable tensile stress in the metal tie.
(a) Obtain a formula for the minimum required cross-sectional area of the tie.
(b) Calculate the minimum area if P 130 kN, u 75°, and sallow 80 MPa.
Solution 1.9-6 Suspender tie on a suspension bridge
F tensile force in cable
above tie
P tensile force in cable
below tie
sallow allowable tensile
stress in the tie
(a) MINIMUM REQUIRED AREA OF TIE
Amin T
sallow
Pcotu
sallow
;
(b) SUBSTITUTE NUMERICAL VALUES:
P 130 kN
u 75°
sallow 80 MPa
Amin 435 mm2
;
FREE-BODY DIAGRAM OF HALF THE TIE
Note: Include a small amount of the cable in the free-body
diagram
T tensile force in the tie
FORCE TRIANGLE
cotu T
P
T P cot u
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SECTION 1.9 Design for Axial Loads and Direct Shear
105
Problem 1.9-7 A square steel tube of length L 20 ft and
width b2 10.0 in. is hoisted by a crane (see figure). The
tube hangs from a pin of diameter d that is held by the cables
at points A and B. The cross section is a hollow square with
inner dimension b1 8.5 in. and outer dimension b2 10.0
in. The allowable shear stress in the pin is 8,700 psi, and the
allowable bearing stress between the pin and the tube is
13,000 psi.
Determine the minimum diameter of the pin in order to
support the weight of the tube. (Note: Disregard the rounded
corners of the tube when calculating its weight.)
Solution 1.9-7 Tube hoisted by a crane
T tensile force in cable
W gs AL
W weight of steel tube
(490 lb/ft3)(27.75 in.2)a
d diameter of pin
1,889 lb
1 ft2
b (20 ft)
144 in.
b1 inner dimension of tube
8.5 in.
b2 outer dimension of tube
10.0 in.
L length of tube 20 ft
tallow 8,700 psi
sb 13,000 psi
DIAMETER OF PIN BASED UPON SHEAR
Double shear. 2tallow Apin W
2(8,700 psi)a
p d2
b 1889 lb
4
d2 0.1382 in.2
d1 0.372 in.
DIAMETER OF PIN BASED UPON BEARING
WEIGHT OF TUBE
sb(b2 b1)d W
gs weight density of steel
(13,000 psi)(10.0 in. 8.5 in.) d 1,889 lb
490 lb/ft
3
A area of tube
b22 b21 (10.0 in.)2 (8.5 in.)2
27.75 in.
d2 0.097 in.
MINIMUM DIAMETER OF PIN
Shear governs.
dmin 0.372 in.
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.9-8 A cable and pulley system at D is used to
bring a 230-kg pole (ACB) to a vertical position as shown in
the figure part (a). The cable has tensile force T and is attached
at C. The length L of the pole is 6.0 m, the outer diameter is
d 140 mm, and the wall thickness t 12 mm. The pole
pivots about a pin at A in figure part (b). The allowable shear
stress in the pin is 60 MPa and the allowable bearing stress
is 90 MPa.
Find the minimum diameter of the pin at A in order to support
the weight of the pole in the position shown in the figure part (a).
B
Pole
C
1.0 m
Cable
30
Pulley
5.0 m
a
T
A
D
4.0 m
a
(a)
d
ACB
Pin support
plates
A
Pin
Ay
2
Ay
2
(b)
Solution 1.9-8
ALLOWABLE SHEAR AND BEARING STRESSES
STATICS TO FIND CABLE FORCE T
ta 60 MPa
a MA 0
sba 90 MPa
FIND INCLINATION OF AND FORCE IN CABLE, T
let a angle between pole and cable at C; use law
of cosines
DC A
52 + 42 2(5)(4)cos a120 p b
180
DC 7.81 m
a 26.33
u 33.67
a arccosc
52 + DC 2 42
d
2DC(5)
u 60 a
< ange between cable and horizontal
at D
W 2.256 103 N
W 230 kg(9.81 m/s2)
W(3 sin(30º)) TX(5 cos(30º)) Ty(5 sin(30º)) 0
substitute for Tx and Ty in terms of T and solve for T:
T
3
W
2
523
5
sin(u)
cos(u)
2
2
T 1.53
103 N
Ty T sin(u )
Tx T cos(u)
Tx 1.27
103 N
Ty 846.11 N
(1) dmin BASED ON ALLOWABLE SHEAR–DOUBLE SHEAR AT A
Ax Tx
Ay Ty W
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SECTION 1.9 Design for Axial Loads and Direct Shear
CHECK SHEAR DUE TO RESULTANT FORCE ON PIN AT A
RA 2 A2x + A2y
RA 3.35
(2) dmin BASED ON ALLOWABLE BEARING ON PIN
dpole 140 mm
Lpole 6000 mm
3
10 N
RA
2
dmin p
t a b
a a 4
dmin 5.96 mm 6controls
107
tpole 12 mm
MEMBER AB BEARING ON PIN
sb ;
RA
Ab
dmin Ab 2tpoled
RA
2tpole sba
dmin 1.55 mm
Problem 1.9-9 A pressurized circular cylinder has a sealed cover plate fastened
with steel bolts (see figure). The pressure p of the gas in the cylinder is 290 psi,
the inside diameter D of the cylinder is 10.0 in., and the diameter dB of the bolts
is 0.50 in.
If the allowable tensile stress in the bolts is 10,000 psi, find the number n of
bolts needed to fasten the cover.
Solution 1.9-9 Pressurized cylinder
P
ppD2
F
n
4n
Ab area of one bolt p 2
d
4 b
P sallow Ab
sallow p 290 psi
D 10.0 in.
sallow 10,000 psi
db 0.50 in.
n number of bolts
F total force acting on the cover plate from the
internal pressure
F pa
n
pD2
b
4
NUMBER OF BOLTS
ppD2
pD2
P
Ab
(4n)(p4 )d 2b
nd 2b
pD2
d2bsallow
SUBSTITUTE NUMERICAL VALUES:
n
(290 psi)(10 in.)2
(0.5 in.)2(10,000 psi)
Use 12 bolts
11.6
;
P tensile force in one bolt
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.9-10 A tubular post of outer diameter d2 is guyed by two cables
fitted with turnbuckles (see figure). The cables are tightened by rotating the
turnbuckles, thus producing tension in the cables and compression in the post.
Both cables are tightened to a tensile force of 110 kN. Also, the angle between the
cables and the ground is 60°, and the allowable compressive stress in the post is
sc 35 MPa.
If the wall thickness of the post is 15 mm, what is the minimum permissible
value of the outer diameter d2?
Solution 1.9-10
Tubular post with guy cables
d2 outer diameter
AREA OF POST
d1 inner diameter
A
t wall thickness
15 mm
T tensile force in a cable
110 kN
sallow 35 MPa
P compressive force in post
2T cos 30°
REQUIRED AREA OF POST
A
P
s allow
2Tcos 30
s allow
p
p 2
(d2 d21) [d22(d22t)2 ]
4
4
pt (d2 t)
EQUATE AREAS AND SOLVE FOR d2:
2T cos 30
pt (d2 t)
sallow
d2 2T cos 30
+ t
ptsallow
;
SUBSTITUTE NUMERICAL VALUES:
(d2)min 131 mm
;
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SECTION 1.9 Design for Axial Loads and Direct Shear
109
Problem 1.9-11 A large precast concrete panel for a warehouse is being raised to a vertical position using two sets of
cables at two lift lines as shown in the figure part (a). Cable 1 has length L1 22 ft and distances along the panel (see figure
part (b)) are a L1/2 and b L1/4. The cables are attached at lift points B and D and the panel is rotated about its base
at A. However, as a worst case, assume that the panel is momentarily lifted off the ground and its total weight must be
supported by the cables. Assuming the cable lift forces F at each lift line are about equal, use the simplified model of
one half of the panel in figure part (b) to perform your analysis for the lift position shown. The total weight of the panel
is W 85 kips. The orientation of the panel is defined by the following angles: g 20° and u 10°.
Find the required cross-sectional area AC of the cable if its breaking stress is 91 ksi and a factor of safety of 4 with respect
to failure is desired.
F
H
F
F
T2
b2
T1
B
a
b1
W
D
b
—
2
B
g
u
y
a
C
W
—
2
D
b
g
A
b
A
(a)
x
(b)
Solution 1.9-11
GEOMETRY
1
a L1
2
L1 22 ft
u 10º
1
b L1
4
a 2.5b 24.75 ft
g 20º
Using law of cosines
L2 2(a + b)2 + L21 2(a + b)L1 cos(u )
b arccosc
L21 + L22 (a + b)2
d
2L1L2
b 26.484
b1 p (u p g)
b2 b b1
b 1 10
b 2 16.484
SOLUTION APPROACH: FIND T THEN AC T/(sU/FS)
L2 6.425 ft
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CHAPTER 1 Tension, Compression, and Shear
STATICS at point H
g Fx 0
H
So
T2 T1
g FY 0
H
and
T1 27.042 k
T1sin(b1) T2sin(b2)
T2 T1
sin(b 1)
sin(b 2)
T2 16.549 k
sin(b 1)
sin(b 2)
COMPUTE REQUIRED CROSS-SECTIONAL AREA
T1cos(b 1) + T2 cos(b 2) F
su 91 ksi
F W/2,
W 85 k
sin(b 1)
cos(b 2)b F
So
T1 acos(b 1) +
sin(b 2)
W
2
T1 sin(b 1)
acos(b 1) +
cos(b 2)b
sin(b 2)
Ac T1
su
FS
FS 4
su
22.75 ksi
FS
Ac 1.189 in.2
;
Problem 1.9-12 A steel column of hollow circular cross section
is supported on a circular steel base plate and a concrete pedestal
(see figure). The column has outside diameter d 250 mm and
supports a load P 750 kN.
(a) If the allowable stress in the column is 55 MPa, what is the
minimum required thickness t? Based upon your result, select
a thickness for the column. (Select a thickness that is an even
integer, such as 10, 12, 14, . . . , in units of millimeters.)
(b) If the allowable bearing stress on the concrete pedestal is
11.5 MPa, what is the minimum required diameter D of the
base plate if it is designed for the allowable load Pallow that
the column with the selected thickness can support?
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SECTION 1.9 Design for Axial Loads and Direct Shear
111
Solution 1.9-12 Hollow circular column
SUBSTITUTE NUMERICAL VALUES IN EQ. (1):
t2 250 t +
(750 * 103 N)
p(55 N/mm2)
0
(Note: In this eq., t has units of mm.)
t2 250t 4,340.6 0
Solve the quadratic eq. for t:
d 250 mm
(a) THICKNESS t OF THE COLUMN
pt ptd +
2
t2 td +
P
sallow
p
(4t)(d t) pt(d t)
4
D2 P
sallow
Pallow
pD2
4
sb
4s allowt(d t)
sb
4(55 MPa)(20 mm)(230 mm)
11.5 MPa
D2 88,000 mm2
0
P
0
psallow
Area of base plate sallowpt(d t)
pD2
4
sb
p
pd2
A
(d 2t)2
4
4
pt(d t) Pallow sallow A
A pt(d t) Pallow sallow pt(d t)
sb 11.5 MPa (allowable pressure on concrete)
;
;
where A is the area of the column with t 20 mm.
D diameter of base plate
s allow
Use t 20 mm
For the column,
t thickness of column
A
tmin 18.8 mm
(b) DIAMETER D OF THE BASE PLATE
P 750 kN
sallow 55 MPa (compression in column)
P
t 18.77 mm
Dmin 297 mm
D 296.6 mm
;
(Eq. 1)
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CHAPTER 1 Tension, Compression, and Shear
Problem 1.9-13 An elevated jogging track is supported at
intervals by a wood beam AB (L 7.5 ft) which is pinned at A
and supported by steel rod BC and a steel washer at B. Both
the rod (dBC 3/16 in.) and the washer (dB 1.0 in.) were
designed using a rod tension force of TBC 425 lb. The rod
was sized using a factor of safety of 3 against reaching the
ultimate stress su 60 ksi. An allowable bearing stress
sba 565 psi was used to size the washer at B.
Now, a small platform HF is to be suspended below a section
of the elevated track to support some mechanical and electrical
equipment. The equipment load is uniform load q 50 lb/ft and
concentrated load WE 175 lb at mid-span of beam HF. The
plan is to drill a hole through beam AB at D and install the same
rod (dBC) and washer (dB) at both D and F to support beam HF.
(a) Use su and sba to check the proposed design for rod DF
and washer dF; are they acceptable?
(b) Also re-check the normal tensile stress in rod BC and
bearing stress at B; if either is inadequate under the
additional load from platform HF, redesign them to
meet the original design criteria.
Original
structure
C
Steel rod,
3
dBC = — in.
16
TBC =
425 lb.
L
—
25
L = 7.5 ft
A
Wood beam supporting track
D
B
Washer,
dB = 1.0 in.
3
New steel rod, dDF = — in.
16
WE = 175 lb
q = 50 lb/ft
H
New beam to support equipment
L
—
2
Hy
L
—
2
Hx
L
—
25
F
Washer, dF
(same at D
above)
Solution 1.9-13
NUMERICAL DATA
sDF 10.38 ksi
L 7.5(12)
L 90 in.
su 60 ksi
FSu 3
50
12
q 4.167 lb/in.
q
dBC 3
in.
16
TBC 425 lb
sba 0.565 ksi
WE 175 lb
dB 1.0 in.
(a) FIND FORCE IN ROD DF AND FORCE ON WASHER AT F
MH 0
L
L
WE qL
2
2
TDF L
aL
b
25
TDF 286.458 lb
NORMAL STRESS IN ROD DF:
TDF
s DF p 2
d
4 BC
sa su
FSu
OK—less than sa; rod is
;
acceptable
sa 20 ksi
BEARING STRESS ON WASHER AT F:
sbF TDF
p 2
2
1d d BC2
4 B
sbF 378 psi
OK—less than sba; washer is
;
acceptable
(b) FIND NEW FORCE IN ROD BC—SUM MOMENT ABOUT A FOR
UPPER FBD—THEN CHECK NORMAL STRESS IN BC and
BEARING STRESS AT B
MA 0
TBC2 TBCL + TDF a L L
b
25
L
TBC2 700 lb
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113
SECTION 1.9 Design for Axial Loads and Direct Shear
REVISED NORMAL STRESS IN ROD BC:
s BC2 Original
structure
TBC2
C
Steel rod,
3
dBC = — in.
16
p
a dBC 2 b
4
sBC2 25.352 ksi
exceeds sa = 20 ksi
A
Wood beam supporting track
dBCreqd dBCreqd
^say 4/16 1/4 in.
L
—
25
L = 7.5 ft
SO RE-DESIGN ROD BC:
TBC2
p
sa
Q4
dBCreqd 0.211 in.
TBC =
425 lb.
D
B
Washer,
dB = 1.0 in.
3
New steel rod, dDF = — in.
16
WE = 175 lb
q = 50 lb/ft
16 3.38
1
dBC2 in.
4
H
L
—
25
New beam to support equipment
F
RE-CHECK BEARING STRESS IN WASHER AT B:
s bB2 TBC2
p
c 1dB2 dBC 22 d
4
L
—
2
sbB2 924 psi
^ exceeds
sba = 565 psi
Hy
Washer, dF
(same at D
above)
L
—
2
Hx
SO RE-DESIGN WASHER AT B:
TBC2
+ dBC 2
dBreqd 1.281 in.
p
sba
Q4
use 1 5/16 in washer at B: 1 + 5/16 1.312 in.
dBreqd ;
Problem 1.9-14 A flat bar of width b 60 mm and thickness t 10 mm
is loaded in tension by a force P (see figure). The bar is attached to a
support by a pin of diameter d that passes through a hole of the same size
in the bar. The allowable tensile stress on the net cross section of the bar is
sT 140 MPa, the allowable shear stress in the pin is tS 80 MPa, and
the allowable bearing stress between the pin and the bar is sB 200 MPa.
(a) Determine the pin diameter dm for which the load P will be a
maximum.
(b) Determine the corresponding value Pmax of the load.
d
P
b
t
P
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CHAPTER 1 Tension, Compression, and Shear
Solution 1.9-14
Bar with a pin connection
SHEAR IN THE PIN
PS 2tS Apin 2tS a
pd 2
b
4
p
1
2(80 MPa)a b(d 2)a
b
4
1000
0.040 pd 2 0.12566d 2
(Eq. 2)
BEARING BETWEEN PIN AND BAR
PB sB td
b 60 mm
(200 MPa)(10 mm)(d )a
t 10 mm
2.0 d
d diameter of hole and pin
1
b
1000
(Eq. 3)
GRAPH OF EQS. (1), (2), AND (3)
sT 140 MPa
tS 80 MPa
sB 200 MPa
UNITS USED IN THE FOLLOWING CALCULATIONS:
P is in kN
s and t are in N/mm2 (same as MPa)
b, t, and d are in mm
TENSION IN THE BAR
PT sT (Net area) st(t)(b d )
(140 MPa)(10 mm) (60 mm d) a
1.40 (60 d)
1
b
1000
(Eq. 1)
(a) PIN DIAMETER dm
PT PB or 1.40(60 d) 2.0 d
84.0
mm 24.7 mm
Solving, dm 3.4
;
(b) LOAD Pmax
Substitute dm into Eq. (1) or Eq. (3):
Pmax 49.4 kN
;
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SECTION 1.9 Design for Axial Loads and Direct Shear
Problem 1.9-15 Two bars AC and BC of the same material support a vertical load P
(see figure). The length L of the horizontal bar is fixed, but the angle u can be varied by
moving support A vertically and changing the length of bar AC to correspond with the
new position of support A. The allowable stresses in the bars are the same in tension and
compression.
We observe that when the angle u is reduced, bar AC becomes shorter but the crosssectional areas of both bars increase (because the axial forces are larger). The opposite
effects occur if the angle u is increased. Thus, we see that the weight of the structure
(which is proportional to the volume) depends upon the angle u.
Determine the angle u so that the structure has minimum weight without exceeding
the allowable stresses in the bars. (Note: The weights of the bars are very small compared to the force P and may be disregarded.)
115
A
θ
B
C
L
P
Solution 1.9-15 Two bars supporting a load P
LENGTHS OF BARS
Joint C
T
LAC L
cos u
LBC L
WEIGHT OF TRUSS
θ
C
g weight density of material
C
P
T tensile force in bar AC
C compressive force in bar BC
P
sin u
a Fvert 0
T
a Fhoriz 0
P
C
tan u
AREAS OF BARS
AAC T
P
sallow
sallow sin u
ABC C
P
sallow
sallow tan u
W g(AAC LAC + ABC LBC)
gPL
1
1
a
+
b
sallow sin u cos u
tan u
gPL 1 + cos2u
a
b
sallow sin u cos u
Eq. (1)
g, P, L, and sallow are constants
W varies only with u
Let k gPL
(k has unis of force)
sallow
W
1 + cos2u
(Nondimensional)
k
sin u cos u
Eq. (2)
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CHAPTER 1 Tension, Compression, and Shear
GRAPH OF EQ. (2):
(sin u cos u)(2)(cos u) (sin u)
(1 + cos2u)(sin2u + cos2 u)
df
du
sin2u cos2u
sin2u cos2 u + sin2u cos2 u cos4u
sin2 u cos2 u
SET THE NUMERATOR 0 AND SOLVE FOR u:
sin2u cos2u sin2u cos2u cos4u 0
Replace sin2u by 1 cos2u:
(1 cos2u)(cos2u) 1 cos2u cos2u cos4u 0
Combine terms to simplify the equation:
ANGLE u THAT MAKES WA MINIMUM
Use Eq. (2)
Let f 2
1 + cos u
sin u cos u
1 3 cos2u 0
u 54.7°
cos u 1
23
;
df
0
du
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2
Axially Loaded
Members
Changes in Lengths of Axially Loaded Members
Problem 2.2-1 The L-shaped arm ABCD shown in the figure lies
in a vertical plane and pivots about a horizontal pin at A. The arm
has constant cross-sectional area and total weight W.
A vertical spring of stiffness k supports the arm at point B.
(a) Obtain a formula for the elongation of the spring due to the
weight of the arm.
(b) Repeat part (a) if the pin support at A is moved to D.
k
A
B
C
b
b
—
2
b
(a)
D
Solution 2.2-1
(a) SUM MOMENTS ABOUT A
©MA 0
b
2b
2
Wb +
W (2 b) k d b
5
5
b
b
2
2
k
A
b
2
2b
Wb +
W (2 b)
5
5
b
b
2
2
6W
d
kb
5k
(b) ©M D 0
kbd 2b
4Wb
Wb 5
5
b
2
B
C
b
b
—
2
b
D
(b)
so
2b
Wb
5
b
2
4W
d
kb
5k
117
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CHAPTER 2 Axially Loaded Members
Problem 2.2-2 A steel cable with nominal diameter
25 mm (see Table 2-1) is used in a construction yard
to lift a bridge section weighing 38 kN, as shown in
the figure. The cable has an effective modulus of
elasticity E 140 GPa.
(a) If the cable is 14 m long, how much will it
stretch when the load is picked up?
(b) If the cable is rated for a maximum load of
70 kN, what is the factor of safety with respect
to failure of the cable?
Solution 2.2-2 Bridge section lifted by a cable
A 304 mm2 (from
Table 2-1)
W 38 kN
E 140 GPa
L 14 m
(b) FACTOR OF SAFETY
PULT 406 kN (from Table 2-1)
Pmax 70 kN
n
PULT
406 kN
5.8
Pmax
70 kN
;
(a) STRETCH OF CABLE
d
(38 kN)(14 m)
WL
EA
(140 GPa)(304 mm2)
12.5 mm
;
Problem 2.2-3 A steel wire and an aluminum alloy wire have equal lengths
Aluminum alloy
wire
and support equal loads P (see figure). The moduli of elasticity for the steel and
aluminum alloy are Es 30,000 ksi and Ea 11,000 ksi, respectively.
(a) If the wires have the same diameters, what is the ratio of the
elongation of the aluminum alloy wire to the elongation of the
steel wire?
(b) If the wires stretch the same amount, what is the ratio of the
diameter of the aluminum alloy wire to the diameter of the steel wire?
(c) If the wires have the same diameters and same load P, what is the ratio
of the initial length of the aluminum alloy wire to that of the steel wire
if the aluminum alloy wire stretches 1.5 times that of the steel wire?
(d) If the wires have the same diameters, same initial length, and same
load P, what is the material of the upper wire if it elongates 1.7 times
that of the steel wire?
Steel
wire
P
P
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SECTION 2.2 Changes in Lengths of Axially Loaded Members
119
Solution 2.2-3
PL
Ea A
da
(a)
ds
a
PL
b
Es A
:
Es
Ea
Es 30,000 ksi Ea 11,000 ksi
Es
2.727
Ea
(b) da ds
so
30
2.727
11
PL
PL
Ea Aa
Es As
so
Aa
Es
As
Ea
and
Es
da
1.651
ds
C Ea
(c) SAME DIAMETER, SAME LOAD, FIND RATIO OF LENGTH OF ALUMINUM TO STEEL WIRE IF ELONGATION OF ALUMINUM IS 1.5 TIMES
THAT OF STEEL WIRE
P La
Ea A
da
ds
P Ls
a
b
Es A
P La
Ea A
1.5
P Ls
a
b
Es A
La
Ea
1.5
0.55
Ls
Es
(d) SAME DIAMETER, SAME LENGTH, SAME LOAD—BUT WIRE 1 ELONGATION 1.7 TIMES THE STEEL WIRE WHAT IS WIRE
1 MATERIAL?
d1
ds
PL
E1 A
a
PL
b
Es A
PL
E1 A
a
PL
b
Es A
1.7
E1 Es
17,647 ksi
1.7
cast iron or copper alloy (see App. I)
Problem 2.2-4 By what distance h does the cage shown in the figure
move downward when the weight W is placed inside it?
Consider only the effects of the stretching of the cable, which
has axial rigidity EA 10,700 kN. The pulley at A has diameter
dA 300 mm and the pulley at B has diameter dB 150 mm. Also,
the distance L1 4.6 m, the distance L2 10.5 m, and the weight
W 22 kN. (Note: When calculating the length of the cable, include
the parts of the cable that go around the pulleys at A and B.)
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CHAPTER 2 Axially Loaded Members
Solution 2.2-4 Cage supported by a cable
dA 300 mm
dB 150 mm
LENGTH OF CABLE
L L 1 + 2L 2 +
1
1
1pdA2 + (pdB)
4
2
L1 4.6 m
4,600 mm + 21,000 mm + 236 mm + 236 mm
L2 10.5 m
26,072 mm
EA 10,700 kN
W 22 kN
ELONGATION OF CABLE
d
(11 kN)(26,072 mm)
TL
26.8 mm
EA
(10,700 kN)
LOWERING OF THE CAGE
h distance the cage moves downward
TENSILE FORCE IN CABLE
W
T
11 kN
2
h
1
d 13.4 mm
2
;
Problem 2.2-5 A safety valve on the top of a tank containing steam
under pressure p has a discharge hole of diameter d (see figure).
The valve is designed to release the steam when the pressure reaches the
value pmax.
If the natural length of the spring is L and its stiffness is k, what
should be the dimension h of the valve? (Express your result as a
formula for h.)
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SECTION 2.2 Changes in Lengths of Axially Loaded Members
Solution 2.2-5
121
Safety valve
pmax pressure when valve opens
L natural length of spring (L h)
k stiffness of spring
FORCE IN COMPRESSED SPRING
F k(L h) (From Eq. 2-1a)
PRESSURE FORCE ON SPRING
P pmax a
pd 2
b
4
EQUATE FORCES AND SOLVE FOR h:
h height of valve (compressed length of the spring)
d diameter of discharge hole
p pressure in tank
F P k1L h2 hL
ppmax d 2
4k
ppmaxd 2
4
;
Problem 2.2-6 The device shown in the figure consists of a prismatic rigid pointer ABC supported by a uniform translational spring of stiffness k 950 N/m. The spring is positioned at distance b 165 mm from the pinned end A of the
pointer. The device is adjusted so that when there is no load P, the pointer reads zero on the angular scale.
(a) If the load P 11 N, at what distance x should the load be placed so that the pointer will read u 2.5° on the scale
(see figure part a)?
(b) Repeat part (a) if a rotational spring kr kb2 is added at A (see figure part b).
(c) Let x 7b/8. What is Pmax (N) if u cannot exceed 2°? Include spring kr in your analysis.
(d) Now, if the weight of the pointer ABC is known to be Wp 3 N and the weight of the spring is Ws 2.75 N, what initial angular position (i.e., u in degrees) of the pointer will result in a zero reading on the angular scale once the pointer
is released from rest? Assume P kr 0.
(e) If the pointer is rotated to a vertical position (see figure part c), find the required load P, applied at mid-height of the
pointer, that will result in a pointer reading of u 2.5° on the scale. Consider the weight of the pointer Wp in your
analysis.
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CHAPTER 2 Axially Loaded Members
Solution 2.2-6
NUMERICAL DATA
k 950 N/m
Wp 3 N
Ws 2.75 N
b 165 mm P 11 N u 2.5
(a) If the load P 11 N, at what distance x should the load be
placed so that the pointer will read u 2.5° on the scale
(see Fig. a)?
Sum moments about A, then solve for x:
x
k u b2
102.6 mm
P
umax 2
P
x
A
B
C
θ
0
k
x 102.6 mm
b
b/2
(a)
(b) Repeat (a) if a rotational spring kr kb2 is added at A (see
Fig. b).
P
x
A
B
C
0
kr k b 25864 N # mm
2
Sum moments about A, then solve for x:
k u b2 + kr u
x
x
205 mm
1.244
P
b
kr
x 205 mm
(c) Now if x 7b/8, what is Pmax (N) if u cannot exceed 2 ?
Sum moments about A, then solve for P:
Pmax k
b
x
7
b 144.375 mm
8
k umax b2 + kr umax
12.51 N
7
b
8
b/2
(b)
Pmax 12.51 N
(d) Now, if the weight of the pointer ABC is known to be Wp 3 N and the weight of the spring is Ws 2.75 N,
what initial angular position (i.e., u in degrees) of the pointer will result in a zero reading on the angular scale
once the pointer is released from rest? Assume P kr 0.
Deflection at spring due to Wp:
dBp Wp a
3
bb
4
kb
Deflection at B due to self weight of spring:
dB dBp + dBk 3.816 mm
OR
dBk 2.368 mm
uinit arctan a
uinit dB
b 1.325
b
Ws
1.447 mm
2k
dB
1.325
b
uinit 1.325
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123
SECTION 2.2 Changes in Lengths of Axially Loaded Members
(e) If the pointer is rotated to a vertical position (figure part c), find the required load P,
applied at mid-height of the pointer that will result in a pointer reading of u 2.5°
on the scale. Consider the weight of the pointer, Wp, in your analysis.
k 950 N/m
0
b 165 mm Wp 3 N
C
k r k b 25.864 N # m u 2.5 2
b/2
Sum moments about A to get P:
P
k
5
3b
u
b d 20.388 N
ckr + k a b2 b Wp a
4
4
3b
b
a
4
B
P 20.4 N
P
Wp
b
3b/4
A
kr
(c)
Problem 2.2-7 Two rigid bars are connected to each other by two linearly elastic springs. Before loads are applied, the
lengths of the springs are such that the bars are parallel and the springs are without stress.
(a) Derive a formula for the displacement d4 at point 4 when the load P is applied at joint 3 and moment PL is applied at
joint 1, as shown in the figure part a. (Assume that the bars rotate through very small angles under the action of the
load P.)
(b) Repeat part (a) if a rotational spring, kr kL2, is now added at joint 6. What is the ratio of the deflection d4 in the figure part a to that in the figure part b?
P
PL
1
2
Rigid bar
2L/3
L/3
3
1
L
4
5
δ4
(a)
2
Rigid bar
2L/3
2k
k
P
PL
Rigid bar
6
L/3
3
2k
k
L
4
5
δ4
Rigid bar
kr = kL2
6
(b)
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CHAPTER 2 Axially Loaded Members
Solution 2.2-7
(a) Derive a formula for the displacement d4 at point 4 when the load P is applied at joint 3 and moment PL is applied
at joint 1, as shown.
Cut horizontally through both springs to create upper and lower FBD’s. Sum moments about joint 1 for upper FBD and
also sum moments about joint 6 for lower FBD to get two equations of equilibrium; assume both springs are in tension.
d2 Note that
2
d
3 3
and
d5 3
d
4 4
Force in left spring:
k a d4 Force in right spring:
2k a
2
d b
3 3
3
d d3 b
4 4
Summing moments about joint 1 (upper FBD) and about joint 6 (lower FBD) then dividing through by k gives
22
9
±
26
9
13
2 P
6
d3
≤a b k
17
d4
P
Q
0
6
22
9
d3
a b ±
26
d4
9
13 1
17 P
2 P
6
2k
≤
≤
±
k
26 P
17
P 0 Q
6
3k
17
8.5
2
26
8.667
3
d4 26 P
3k
^ deltas are positive downward
(b) Repeat part (a) if a rotational spring kr kL2 is now added at joint 6. What is the ratio of the deflection d4 in part (a)
to that in (b)?
Upper FBD—sum moments about joint 1:
k ad4 2
2L
3
d b
+ 2 k a d4 d3 b L 2 P L OR
3 3 3
4
a
22 L k
13 L k
b d3 +
d4 2 P L
9
6
Lower FBD—sum moments about joint 6:
k ad4 2
4L
3
d3 b
+ 2 k a d4 d3 b L kr u6 0
3
3
4
ck ad4 d4
2
4L
3
26 L k
43 L k
d b
+ 2 k a d4 d3 b L d + (k L2)
0 OR a
b d3 +
d4 0
3 3 3
4
4
9
12
P LQ
3
Divide matrix equilibrium equations through by k to get the following displacement equations:
22
9
±
26
9
13
2 P
6
d3
≤a b k
43
d4
P
Q
0
6
22
d3
9
a b ±
d4
26
9
13 1
43 P
2 P
6
15 k
≤
≤
±
k
43
104 P
P 0 Q
12
45 k
43
2.867
15
104
2.311
45
d4 104 P
45 k
^ deltas are positive downward
Ratio of the deflection d4 in part (a) to that in (b):
26
15
3
104
4
45
Ratio 15
3.75
4
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SECTION 2.2 Changes in Lengths of Axially Loaded Members
125
Problem 2.2-8 The three-bar truss ABC shown in figure part a has a span L 3 m and is constructed of steel pipes having
cross-sectional area A 3900 mm2 and modulus of elasticity E 200 GPa. Identical loads P act both vertically and horizontally at joint C, as shown.
(a) If P 475 kN, what is the horizontal displacement of joint B?
(b) What is the maximum permissible load value Pmax if the displacement of joint B is limited to 1.5 mm?
(c) Repeat parts (a) and (b) if the plane truss is replaced by a space truss (see figure part b).
Solution 2.2-8
P
NUMERICAL DATA
A 3900 mm
2
P 475 kN
P
C
E 200 GPa
L 3000 mm
dBmax 1.5 mm
(a) FIND HORIZONTAL DISPLACEMENT OF JOINT B
STATICS TO FIND SUPPORT REACTIONS AND THEN MEMBER FORCES:
g MA 0
By 45°
45°
A
1
L
a2 P b
L
2
L
(a)
By P
g FH 0
Ax P
g FV 0
Ay P By
METHOD OF JOINTS:
B
Ay 0
ACV AY ACV 0
Force in AC 0
AB AX
Force in AB is P (tension) so elongation of AB is the horizontal displacement of joint B.
dB FAB L
EA
dB PL
EA
dB 1.82692 mm
dB 1.827 mm
(b) FIND Pmax IF DISPLACEMENT OF JOINT B dBmax 1.5 mm
Pmax EA
d
Pmax 390 kN
L Bmax
(c) REPEAT PARTS (a) AND (b) IF THE PLANE TRUSS IS REPLACED BY
A SPACE TRUSS (SEE FIGURE PART b).
FIND MISSING DIMENSIONS a AND c:
y
P
P 475 kN L 3 m
Cz
P
L
C
L/2
L/2
a
B
x
L/2
A
c
z
Bz
Ax
Az
By
Ay
(b)
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CHAPTER 2 Axially Loaded Members
a
C
L2 2 a
L 2
b 2.12132 m
2
c 2L2 + a2 3.67423 m
c
a
0.707
L
C
L2 + a
a
L
2.12132 m
12
L 2
b 3.67423 m
12
cL
3
3.67423 m
C2
(1) SUM MOMENTS ABOUT A LINE THRU A WHICH IS PARALLEL TO THE y-AXIS
Bz P
L
671.751 kN
a
(2) SUM MOMENTS ABOUT THE z-AXIS
By Pa
L
b
2
a
335.876 kN
Ay P By 139.124 kN
SO
(3) SUM MOMENTS ABOUT THE x-AXIS
Ay L P
Cz L
2
L
2
196.751 kN
Ax P 475 kN
(4) SUM FORCES IN THE x- AND z-DIRECTIONS
(5) USE METHOD OF JOINTS TO FIND MEMBER FORCES
Sum forces in x-direction at joint A:
Sum forces in y-direction at joint A:
a
F + Ax 0
c AB
L
2
12
Sum forces in y-direction at joint B:
L
2
FAB FAC + Ay 0
L
2
F + By 0
L BC
Az Cz Bz 868.503 kN
c
A 823 kN
a x
FAC 12 1Ay2 196.8 kN
FBC 2 By 672 kN
(6) FIND DISPLACEMENT ALONG x-AXIS AT JOINT B
Find change in length of member AB then find its projection along x axis:
dAB FAB c
L
3.875 mm b arctan a b 54.736
EA
a
dBx dAB
6.713 mm dBx 6.71 mm
cos(b)
(7) FIND Pmax FOR SPACE TRUSS IF Bx MUST BE LIMITED TO 1.5 mm
Displacements are linearly related to the loads for this linear elastic small displacement problem, so reduce load
variable P from 475 kN to
1.5
475 106.145 kN
6.71254
Repeat space truss analysis using vector operations
Pmax 106.1 kN
a = 2.121 m
L=3m
P = 475 kN
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127
SECTION 2.2 Changes in Lengths of Axially Loaded Members
POSITION AND UNIT VECTORS:
a
rAB 0
P L Q
0
L
rAC • 2 μ
L
2
0.577
rAB
eAB 0
冷rAB 冷 P 0.816 Q
eAC rAC
冷rAC 冷
0
0.707
P 0.707 Q
FIND MOMENT AT A:
M A rAB * RB + rAC * RC
0
2.P
3.0 m RBy + 1.5 m RCz 712.5 kN # m
M A rAB * RBy + rAC * P 2.1213 m RBZ 1425.0 kN # m
P RB Q
P RC Q P
Q
2.1213 m RBy 1425.0 kN # m
z
z
FIND MOMENTS ABOUT LINES OR AXES:
MA eAB 1.732 m RBy + 1.7321 m RBy + 0.86603 m RCz + 752.15 kN # m
RCz 244.12
338.262
0.72169
MA eAC 1.5 m RBy + 1.5 m RBz
So
Cz 196.751 kN
RBy RBz
0
462.5
MA 1 2.1213 m RBz + 1425.0 kN # m So RBz 261.625 Bz 671.75 kN
1.7678
P0Q
0
MA 1 2.1213 m RBy + 1425.0 kN # m So RBy RBz 261.625 By 335.876 kN
P0Q
gFy 0
Ay P By 139.124 kN
Reactions obtained using vector operations agree with those based on scalar operations.
Problem 2.2-9 An aluminum wire having a diameter d 1/10 in.
and length L 12 ft is subjected to a tensile load P (see figure).
The aluminum has modulus of elasticity E 10,600 ksi
If the maximum permissible elongation of the wire is 1/8 in. and the
allowable stress in tension is 10 ksi, what is the allowable load Pmax?
P
d
L
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CHAPTER 2 Axially Loaded Members
Solution 2.2-9
1
in.
10
1
d a in.
8
d
A
pd2
4
L 12(12) in. E 10,600 (103) psi
s a 10 * (103) psi
A 7.854 103 in.2
EA 8.325 104 lb
Maximum load based on elongation:
EA
Pmax1 d Pmax1 72.3 lb ; controls
L a
Maximum load based on stress:
Pmax2 sa A Pmax2 78.5 lb
Problem 2.2-10 A uniform bar AB of weight W 25 N is supported by
two springs, as shown in the figure. The spring on the left has stiffness
k1 300 N/m and natural length L1 250 mm. The corresponding quantities for the spring on the right are k2 400 N/m and L2 200 mm. The
distance between the springs is L 350 mm, and the spring on the right
is suspended from a support that is distance h 80 mm below the point
of support for the spring on the left. Neglect the weight of the springs.
(a) At what distance x from the left-hand spring (figure part a)
should a load P 18 N be placed in order to bring the bar to a
horizontal position?
(b) If P is now removed, what new value of k1 is required so that the
bar (figure part a) will hang in a horizontal position under
weight W?
(c) If P is removed and k1 300 N/m, what distance b should spring
k1 be moved to the right so that the bar (figure part a) will hang in
a horizontal position under weight W?
(d) If the spring on the left is now replaced by two springs in series
(k1 300N/m, k3) with overall natural length L1 250 mm (see
figure part b), what value of k3 is required so that the bar will hang
in a horizontal position under weight W?
New position of
k1 for part (c) only
k1
L1
h
b
k2
L2
W
A
B
P
x
Load P for
part (a) only
L
(a)
k3
L1
—
2
k1
L1
—
2
h
k2
L2
W
A
B
L
(b)
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 2.2 Changes in Lengths of Axially Loaded Members
129
Solution 2.2-10
NUMERICAL DATA
W 25 N
k1 0.300 N/mm
k2 0.400 N/mm
L 350 mm
L1 250 mm
L2 200 mm
h 80 mm
P 18 N
(a) LOCATION OF LOAD P TO BRING BAR TO HORIZONTAL
POSITION
Use statics to get forces in both springs:
a MA 0
a FV 0
F2 L
1
a W + Px b
L
2
F2 W
x
+ P
2
L
F1 W + P F2
F1 W
x
+ Pa1 b
2
L
Use constraint equation to define horizontal
position, then solve for location x:
L1 +
F1
F2
L2 + h +
k1
k2
Substitute expressions for F1 and F2 above into constraint equilibrium and solve for x:
2L1 L k1 k2 k2WL 2k2 P L + 2L2 L k1 k2 + 2 h L k1 k2 + k1W L
2P1k1 + k22
x 134.7 mm ;
x
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CHAPTER 2 Axially Loaded Members
(b) NEXT REMOVE P AND FIND NEW VALUE OF SPRING
CONSTANT k1 SO THAT BAR IS HORIZONTAL
UNDER WEIGHT W
Now, F1 W
2
F2 W
since P 0
2
Same constraint equation as above but now P 0:
PART (C)—CONTINUED
STATICS
wa
a Mk1 0
F2 L
bb
2
Lb
a FV 0
W
W
a b
2
2
L1 +
1L2 + h2 0
k1
k2
F1 W F2
Solve for k1:
F1 W Wk2
[2k2[L1 (L2 + h)]] W
k1 k1 0.204 N/mm
F1 ;
(c) USE k1 0.300 N/mm BUT RELOCATE
SPRING k1 (x b) SO THAT BAR ENDS UP
IN HORIZONTAL POSITION UNDER WEIGHT W
Wa
L
bb
2
Lb
WL
2( L b)
Constraint equation—substitute above expressions
for F1 and F2 and solve for b:
F1
F2
( L2 + h) 0
k1
k2
Use the following data:
L1 +
L /2 – b
F1
b
F2
k1 0.300 N/mm
k2 0.4 N/mm L1 250 mm
L2 200 mm
L 350 mm
b 74.1 mm
;
L /2
L /2
W
L–b
FBD
b
2L1k1k2L + WLk2 2L2k1k2L 2hk1k2L Wk1L
(2L1k1k2) 2L2k1k2 2hk1k2 2Wk1
(d) REPLACE SPRING k1 WITH SPRINGS IN SERIES:
k1 0.3 N/mm, L1/2, AND k3, L1/2. FIND k3
SO THAT BAR HANGS IN HORIZONTAL POSITION
STATICS
k3 F1 W
2
F2 W
2
Wk1k2
2L1k1k2 Wk2 + 2L2k1k2 + 2hk1k2 + Wk1
NOTE—equivalent spring constant for series springs:
k 1k 3
ke k1 + k3
New constraint equation; solve for k3:
L1 +
F1
F1
F2
+
( L2 + h) 0
k1
k3
k2
L1 +
W/2
W/2
W/2
+
( L2 + h) 0
k1
k3
k2
k3 0.638 N/mm
k e 0.204 N/mm
;
;
checks—same as (b) above
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 2.2 Changes in Lengths of Axially Loaded Members
Problem 2.2-11 A hollow, circular, cast-iron pipe (Ec 12,000 ksi)
supports a brass rod (Eb 14,000 ksi) and weight W 2 kips, as
shown. The outside diameter of the pipe is dc 6 in.
(a) If the allowable compressive stress in the pipe is 5000 psi and
the allowable shortening of the pipe is 0.02 in., what is the
minimum required wall thickness tc,min? (Include the weights
of the rod and steel cap in your calculations.)
(b) What is the elongation of the brass rod dr due to both load
W and its own weight?
(c) What is the minimum required clearance h?
131
Nut & washer
3
dw = — in.
4
(
)
Steel cap
(ts = 1 in.)
Cast iron pipe
(dc = 6 in., tc)
Lr = 3.5 ft
Lc = 4 ft
(
Brass rod
1
dr = — in.
2
)
W
h
Solution 2.2-11
(a) MINIMUM REQUIRED WALL THICKNESS OF CAST IRON
PIPE, tcmin
The figure shows a section cut through the pipe, cap,
and rod.
First check allowable stress then allowable
shortening:
NUMERICAL DATA
Ec 12000 ksi
W2k
Eb 14,000 ksi
dc 6 in.
sa 5 ksi
g b 3.009 * 104 k/in.3
ts 1 in.
Wcap 8.018 103 k
da 0.02 in.
Unit weights (see Table I-1):
Lc 48 in.
p
Wcap g s a d 2c t s b
4
1
dr in.
2
Lr 42 in.
4
gs 2.836 * 10
3
k/in.
p
Wrod gb a d2r Lr b
4
Wrod 2.482 103 k
Wt W
Wcap
Wt
Amin sa
A pipe Wrod
Wt 2.01 k
Amin 0.402 in.2
p 2
[d (dc 2 tc)2]
4 c
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 2 Axially Loaded Members
Apipe ptc(dc tc)
t c( d c t c) Let a tc2 dctc
tc Wt
ps a
tc2 dctc
b
Wt
ps a
a 0.128:
tc dc 2 d2c 4b
2
tc 0.021 in.
d c 2 dc2 4a
2
WtLc
pEc da
b 0.142
a0
; minimum based on da and sa
controls
tc 0.021 in.
^ minimum based
on sa
Now check allowable shortening requirement:
WtLc
dpipe EcAmin
b0
WtLc
Amin Ec da
Amin 0.447 in.2 larger than value based on
(b) ELONGATION OF ROD DUE TO SELF WEIGHT AND
ALSO WEIGHT W
dr aW +
Wrod
bLr
2
p
E b a dr 2 b
4
dr 0.031 in.
;
(c) MINIMUM CLEARANCE h
hmin da
dr
hmin 0.051 in.
;
sa above
ptc(dc tc) Wt Lc
Ec da
Problem 2.2-12 The horizontal rigid beam ABCD is supported
by vertical bars BE and CF and is loaded by vertical forces
P1 400 kN and P2 360 kN acting at points A and D,
respectively (see figure). Bars BE and CF are made of steel
(E 200 GPa) and have cross-sectional areas ABE 11,100 mm2
and ACF 9,280 mm2. The distances between various points on
the bars are shown in the figure.
Determine the vertical displacements dA and dD of points A
and D, respectively.
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 2.2 Changes in Lengths of Axially Loaded Members
133
Solution 2.2-12 Rigid beam supported by vertical bars
SHORTENING OF BAR BE
dBE (296 kN)(3.0 m)
FBEL BE
EABE
(200 GPa)(11,100 mm2)
0.400 mm
SHORTENING OF BAR CF
dCF (464 kN)(2.4 m)
FCFL CF
EACF
(200 GPa)(9,280 mm2)
0.600 mm
DISPLACEMENT DIAGRAM
ABE 11,100 mm2
ACF 9,280 mm2
E 200 GPa
LBE 3.0 m
LCF 2.4 m
P1 400 kN; P2 360 kN
dBE dA dCF dBE or dA 2dBE dCF
dA 2(0.400 mm) 0.600 m
0.200 mm ;
(Downward)
2.1
(d dBE)
1.5 CF
12
7
d D dCF dBE
5
5
12
7
(0.600 mm) (0.400 mm)
5
5
0.880 mm ;
(Downward)
dD dCF or
MB 0 哵哴
(400 kN)(1.5 m)
FCF(1.5 m) (360 kN)(3.6 m) 0
FCF 464 kN
MC 0
(400 kN)(3.0 m) FBE(1.5 m) (360 kN)(2.1 m) 0
FBE 296 kN
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 2 Axially Loaded Members
Problem 2.2-13 Two pipe columns (AB, FC) are pin-connected to a rigid beam (BCD) as shown in the figure. Each pipe
column has modulus E, but heights (L1 or L2) and outer diameters (d1 or d2) are different for each column. Assume the inner
diameter of each column is 3⁄4 of outer diameter. Uniformly distributed downward load q 2P/L is applied over a distance
of 3L/4 along BC, and concentrated load P/4 is applied downward at D.
(a) Derive a formula for the displacement dD at
point D in terms of P and column flexibilities
f1 and f2.
(b) If d1 (9/8) d2, find the L1/L2 ratio so that beam
BCD displaces downward to a horizontal position under the load system in (a).
(c) If L1 2 L2, find the d1/d2 ratio so that beam
BCD displaces downward to a horizontal position under the load system in (a).
(d) If d1 (9/8) d2 and L1/L2 1.5, at what horizontal distance x from B should load P/4 be placed so
that beam BCD displaces downward to a horizontal position under the load system in part (a)?
B
q = 2P/L
Rigid beam
3L/4
L/4
P/4
Pin
C
Pin
D
3L/4
x
1
2
L 1, E
L 2, E
d2
d1
F
A
Solution 2.2-13
(a) DISPLACEMENT dD
Use FBD of beam BCD
gMB 0 RC g FV 0 RB a 2
Downward displacements at B and C:
Geometry: dD dB + (dC dB) P
1
P
3
3
P
3
c a 2 b a L b a L b + a L + Lb d P 6compression force
L
L
4
8
4
4
in column CF
P
3
P
3P
b a Lb +
RC L
4
4
4
dB RB f1 L +
L
3
L
4
Q
(b) DISPLACEMENT TO HORIZONTAL POSITION, SO dC dB
L1
E A1
4
L2
3
E A2
or
L1
4 A1
a b
L2
3 A2
L1
4 9 2
27
a b L2
3 8
16
3 P f1
4
dC RC f2 P f2
7 P f2
9 P f1
4
16
and
compression force in column BA
dD 3 P f1
P f2 or
4
p 2
d
L1
4 d1 2
4
4 1
≤ ±
L2
3 p 2
3 d2 2
d2
4
7 P f2
9 P f1
P
128 f2 9 f12
4
16
16
f1
4
f2
3
L1
4 d1 2
a b with
L2
3 d2
d1
9
d2
8
L1
27
L2
16
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SECTION 2.2 Changes in Lengths of Axially Loaded Members
135
(c) IF L1 2 L2, FIND THE d1/d2 RATIO SO THAT BEAM BCD DISPLACES DOWNWARD TO A HORIZONTAL POSITION
L1
2 and dC dB from part (b).
L2
a
d1 2 3 L1
b a b
d2
4 L2
so
d1
3
(2) 1.225
d2 A 4
(d) IF d1 = (9/8) d2 AND L1/L2 = 1.5, AT WHAT HORIZONTAL DISTANCE X FROM B SHOULD LOAD P/4 AT D BE PLACED?
Given
d1
9
d2
8
L1
1.5 or
L2
and
f1
L1 A2
a b
f2
L2 A1
f1
L1 d2 2 3 8 2 32
a b a b f2
L2 d1
2 9
27
Recompute column forces RB and RC but now with load P/4 positioned at distance x from B.
Use FBD of beam BCD:
gMB 0
gFV 0
9LP
Px
+
P
16
4
1
3
3
P
RC c a 2 b a L b a L b + (x) d L
L
4
8
4
L
9LP
Px
+
P
P
16
4
3
7P
RC RB a2 b a Lb +
L
4
4
4
L
Horizontal displaced position under load q and load P/4 so
dC dB or RC f2 RB f1.
9LP
Px
9LP
Px
+
+
L 19 f2 19 f12
9 L f2 19 L f1
16
4
7P
16
4
f2 P
f1 solve, x Q
P
Q
L
4
L
4 f1 + 4f2
4 1f1 + f22
x
L 19 f2 19 f12
4 1f1 + f22
f1
9
f2
x L≥
¥
f1
4a
+ 1b
f2
19
or
32
9
27
365 L
x L≥
¥ 236
32
4a
+ 1b
27
19
Now substitute f1/f2 ratio from above:
365
1.547
236
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CHAPTER 2 Axially Loaded Members
Problem 2.2-14 A framework ABC consists of two rigid bars AB and
B
BC, each having a length b (see the first part of the figure part a). The
bars have pin connections at A,B, and C and are joined by a spring of
stiffness k. The spring is attached at the midpoints of the bars. The
framework has a pin support at A and a roller support a C, and the bars
are at an angle a to the horizontal.
When a vertical load P is applied at joint B (see the second part of
the figure part a) the roller support C moves to the right, the spring is
stretched, and the angle of the bars decreases from a to the angle u.
b
—
2
k
C
(a) Initial position of structure
P
B
B
A
A
C
k
u
kr
(a) - cont’d: displaced position of structure
a
A
P
u
b
—
2
a
(a) Determine the angle u and the increase d in the distance between
points A and C. Also find reactions at A and C. (Use the following
data: b 200 mm, k 3.2 kN/m, a 45, and P 50 N.)
(b) Repeat part (a) if a translational spring k1 k/2 is added at C and
a rotational spring kr kb2/2 is added at A (see figure part b).
u
b
—
2
b
—
2
u
C
k1
(b) Displaced structure
Solution 2.2-14
Apply the laws of statics to the structure in its displaced position; also use FBD’s of the left and right bars alone
(referred to as LHFB and RHFB below).
OVERALL FBD:
LHFB:
gFH 0
HA k 1 d 0
gFV 0
RA + RC P
gM A 0
k r (a u) P
gM B 0
HA h + k
RA RHFB:
g M B 0 k
so
HA k 1 d
L2
+ RC L 2 0
2
RC L2
1
cP
k r (a u) d
L2
2
L2
d h
a b RA a b + k r (a u) 0
2 2
2
2
d h
ck d h + k a b + k r (a u) d
L2 1
2 2
L2
d h
a b k 1 d h + RC
0
2 2
2
RC 2
d h
ck a b + k 1 d h d
L2 2 2
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SECTION 2.2 Changes in Lengths of Axially Loaded Members
137
Equate the two expressions for RC then substitute expressions for L2, kr, k1, h and d
L2
2
d h
1
cP
k r (a u) d ck a b + k 1 d h d
L2
2
L2 2 2
OR
2 b 1cos1u2 cos1a22 b sin1u2
L2
2
1
cP
kr (a u) d c
ck
+ k1 [2 b 1cos1u2 cos1a22] 1b sin1u22 d d 0
L2
2
L2
2
2
(a) SUBSTITUTE NUMERICAL VALUES, THEN SOLVE NUMERICALLY FOR ANGLE u AND DISTANCE INCREASE d
b 200 mm k 3.2 kN/m
a 45
P 50 N k1 0 kr 0
L2 2 b cos1u2 L1 2 b cos1a2 d L2 L 1 d 2 b 1cos1u2 cos1a22 h b sin1u2
2 b 1cos1u2 cos1a22 b sin1u2
L2
1
1
kr 1a u2 d c
+ k1 [2 b 1cos1u2 cos1a22] 1b sin1u22 d d 0
cP
ck
L2
2
L2
2
2
Solving above equation numerically gives
COMPUTE REACTIONS
u 35.1 d 44.6 mm
RC 2 d h
ck a b + k1 d h d 25 N
L2 2 2
RA 2
d h
ck d h + k a b + kr 1a u2 d 25 N
L2 1
2 2
RA + RC 50 N
RC L2
1
cP
kr 1a u2 d 25 N
LC
2
MA kr 1a u2 0
RA 25 N
6 check
RC 25 N
(b) SUBSTITUTE NUMERICAL VALUES, THEN SOLVE NUMERICALLY FOR ANGLE u AND DISTANCE INCREASE d
b 200 mm
L2 2 b cos1u2
k 3.2 kN/m
a 45
L1 2 b cos1a2
P 50 N
d L2 L 1
k1 k
2
kr k 2
b
2
d 2 b 1cos1u2 cos1a22
h b sin1u2
L2
2 2 b 1cos1u2 cos1a22 b sin1u2
1
cP
kr 1a u2 d c ck
+ k1 [2 b 1cos1u2 cos1a22] 1b sin1u22 d d 0
L2
2
L2
2
2
Solving above equation numerically gives
COMPUTE REACTIONS
u 43.3 d 8.19 mm
RC L2
2
d h
1
ck a b + k1 d h d 18.5 N R2 cP
kr 1a u2 d 18.5 N
L2
2 2
L2
2
RA 2
d h
ck 1 d h + k a b + k r 1a u2 d 31.5 N M A k r 1a u2 1.882 N # m
L2
2 2
RA + RC 50 N
6 check
RA 31.5 N
RC 18.5 N
M A 1.882 N # m
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CHAPTER 2 Axially Loaded Members
Problem 2.2-15 Solve the preceding problem for the following data:
b 8.0 in., k 16 1b/in., a 45, and P 10 1b.
Solution 2.2-15
Apply the laws of statics to the structure in its displaced position; also use FBD’s of the left and right bars alone
(referred to as LHFB and RHFB below)
OVERALL FBD
LHFB
g FH 0
HA k 1 d 0
gFV 0
RA + RC P
g MA 0
kr (a u) P
gMB 0
L2
d h
HA h + k a b RA
+ kr (a u) 0
2 2
2
RA so
HA k 1 d
L2
+ RC L2 0
2
RC L2
1
cP
kr (a u) d
L2
2
2
d h
ck d h + k a b + kr (a u) d
L2 1
2 2
L2
d h
2
d h
k a b k1 d h + RC 0
RC ck a b + k1 d h d
2 2
2
L2 2 2
Equate the two expressions above for RC then substitute expressions for L2, kr, k1, h, and d
g MB 0
RHFB
L2
2
d h
1
cP
kr (a u) d ck a b + k1 d h d
L2
2
L2
2 2
OR
2 b 1cos1u2 cos1a22 b sin1u2
L2
1
2
cP
kr (a u) d c
ck
+ k1 [2 b 1cos1u2 cos1a22] 1b sin1u22 d d 0
L2
2
L2
2
2
(a) SUBSTITUTE NUMERICAL VALUES, THEN SOLVE NUMERICALLY FOR ANGLE u AND DISTANCE INCREASE d
b 8 in.
k 16 lb/in.
L2 2 b cos1u2
a 45
L1 2 b cos1a2
P 101b
k1 0
kr 0
d 2 b 1cos1u2 cos1a22
d L2 L 1
h b sin1u2
L2
2 2 b 1cos1u2 cos1a22 b sin1u2
1
cP
kr 1a u2 d c ck
+ k1 [2 b 1cos1u2 cos1a22] 1b sin1u22 d d 0
L2
2
L2
2
2
Solving above equation numerically gives
u 35.1 d 1.782 in.
COMPUTE REACTIONS
RC 2 d h
ck a b + k1 d h d 5 lb
L2 2 2
RA 2
d h
ck d h + k a b + k1 1a u2 d 5 lb
L2 1
2 2
RA + RC 10 lb
6 check
RC L2
1
cP
kr 1a u2 d 5 lb
LC
2
MA kr1a u2 0
RA 5 lb RC 5 lb
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139
SECTION 2.3 Changes in Lengths under Nonuniform Conditions
(b) SUBSTITUTE NUMERICAL VALUES, THEN SOLVE NUMERICALLY FOR ANGLE u AND DISTANCE INCREASE d
b 8 in. k 16 lb/in.
a 45 P 101b k1 k
2
kr k 2
b
2
L2 2 b cos1u2 L1 2 b cos1a2 d L2 L 1 d 2 b 1cos1u2 cos1a22 h b sin1u2
2 b 1cos1u2 cos1a22 b sin1u2
L2
2
1
cP
kr 1a u2 d c ck
+ k1 [2 b 1cos1u2 cos1a22] 1b sin1u22 d d 0
L2
2
L2
2
2
Solving above equation numerically gives
COMPUTE REACTIONS
u 43.3 d 0.327 in.
RC 2
d h
ck a b + k1 d h d 3.71 lb
L2 2 2
RA d h
2
ck 1 d h + k a b + k r 1a u2 d 6.3 lb
L2
2 2
RA + RC 10.01 lb
RC L2
1
cP
kr 1a u2 d 3.71 lb
L2
2
RA 6.3 lb
6 check
M A k r1a u2 1.252 ft # lb
RC 3.71 lb
M A 1.252 lb # ft
Changes in Lengths under Nonuniform Conditions
Problem 2.3-1
(a) Calculate the elongation of a copper bar of solid circular cross section with tapered ends when it is stretched 3.0 k
by axial loads of magnitude 3.0 k (see figure).
(The length of the end segments is 20 in. and the
length of the prismatic middle segment is 50 in. Also,
the diameters at cross sections A, B, C, and D are 0.5,
1.0, 1.0, and 0.5 in., respectively, and the modulus of
elasticity is 18,000 ksi. (Hint: Use the result of Example 2-4.)
A
B
C
D
20 in.
50 in.
20 in.
3.0 k
(b) If the total elongation of the bar cannot exceed 0.025 in., what are the required diameters at B and C? Assume that
diameters at A and D remain at 0.5 in.
Solution 2.3-1
NUMERICAL DATA
P3k
L1 20 in.
L2 50 in.
dA 0.5 in.
dB 1 in.
E 18000 ksi
(a) TOTAL ELONGATION
d1 4 P L1
0.00849 in.
p E dA dB
d 2 d1 + d2 0.0276 in.
d2 P L2
0.01061 in.
p
E dB 2
4
d 0.0276 in.
(b) FIND NEW DIAMETERS AT B AND C IF TOTAL ELONGATION CANNOT EXCEED 0.025 in.
2a
4 P L1
b +
p E dA dB
P L2
0.025 in.
p 2
E dB
4
Solving for dB:
dB 1.074 in.
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CHAPTER 2 Axially Loaded Members
Problem 2.3-2 A long, rectangular copper bar under a tensile load P
hangs from a pin that is supported by two steel posts (see figure). The
copper bar has a length of 2.0 m, a cross-sectional area of 4800 mm2,
and a modulus of elasticity Ec 120 GPa. Each steel post has a height
of 0.5 m, a cross-sectional area of 4500 mm2, and a modulus of elasticity
Es 200 GPa.
Steel
post
(a) Determine the downward displacement d of the lower end of the
copper bar due to a load P 180 kN.
(b) What is the maximum permissible load Pmax if the displacement
d is limited to 1.0 mm?
Copper
bar
P
Solution 2.3-2 Copper bar with a tensile load
(a) DOWNWARD DISPLACEMENT d (P 180 kN)
dc (180 kN)(2.0 m)
PL c
E c Ac
(120 GPa)(4800 mm2)
0.625 mm
ds (P/2)L s
(90 kN)(0.5 m)
E sAs
(200 GPa)(4500 mm2)
0.050 mm
d dc + ds 0.625 mm + 0.050 mm
Lc 2.0 m
Ac 4800 mm2
Ec 120 GPa
Ls 0.5 m
As 4500 mm2
Es 200 GPa
0.675 mm
;
(b) MAXIMUM LOAD Pmax (dmax 1.0 mm)
dmax
Pmax
P
d
Pmax Pa
Pmax (180 kN)a
dmax
b
d
1.0 mm
b 267 kN
0.675 mm
;
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SECTION 2.3 Changes in Lengths under Nonuniform Conditions
Problem 2.3-3 An aluminum bar AD (see figure) has a cross-
sectional area of 0.40 in.2 and is loaded by forces P1 1700 lb,
P2 1200 lb, and P3 1300 lb. The lengths of the segments of the
bar are a 60 in., b 24 in., and c 36 in.
P1
A
a
P2
C
B
b
141
D
P3
c
(a) Assuming that the modulus of elasticity E 10.4 10 psi,
calculate the change in length of the bar. Does the bar elongate or shorten?
(b) By what amount P should the load P3 be increased so that the bar does not change in length when the three loads are
applied?
(c) If P3 remains at 1300 lb, what revised cross-sectional area for segment AB will result in no change of length when all
three loads are applied?
6
Solution 2.3-3
NUMERICAL DATA
A 0.40 in.2
P1 1700 lb
P2 1200 lb
P3 1300 lb
E 10.4 110 2 psi
6
a 60 in.
b 24 in.
c 36 in.
(a) TOTAL ELONGATION
d
1
C 1P1 + P2 P32 a + 1P2 P32 b + 1P32 c D 0.01125 in. d 0.01125 in. (elongation)
EA
(b) INCREASE P3 SO THAT BAR DOES NOT CHANGE LENGTH
1
C 1P1 + P2 P32 a + 1P2 P32 b + 1P32 c D 0 solve, P3 1690 lb
EA
So new value of P3 is 1690 lb,
an increase of 390 lb.
(c) NOW CHANGE CROSS-SECTIONAL AREA OF AB SO THAT BAR DOES NOT CHANGE LENGTH
P3 1300 lb
1
a
b
c
c1P1 + P2 P32
+ 1P2 P32 + 1P32 d 0
E
AAB
A
A
Solving for AAB:
AAB 0.78 in.2
AAB
1.951
A
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CHAPTER 2 Axially Loaded Members
Problem 2.3-4 A rectangular bar of length L has a slot in the
middle half of its length (see figure). The bar has width b,
thickness t, and modulus of elasticity E. The slot has width b/4.
b
—
4
P
b
t
(a) Obtain a formula for the elongation d of the bar due
to the axial loads P.
L
P
—
(b) Calculate the elongation of the bar if the material is
4
L
—
high-strength steel, the axial stress in the middle
2
L
region is 160 MPa, the length is 750 mm, and the
—
4
modulus of elasticity is 210 GPa.
(c) If the total elongation of the bar is limited to
dmax 0.475 mm, what is the maximum length of the slotted region? Assume that the axial stress in the middle region
remains at 160 MPa.
Solution 2.3-4
L
L
2
P
4
2
7LP
≤ (a) d ±
+
E
bt
3
6 Ebt
bt
4
(b) NUMERICAL DATA
so smid d
P
3
bt
4
7LP
6 Ebt
(c) dmax P
E
or
and
d
7PL
6 Ebt
L 750 mm
smid 160 MPa
3
P
smid
bt
4
7L 3
a s b 0.5 mm
6 E 4 mid
L Lslot
Lslot
+
bt
3
P
bt Q
4
or dmax a
Lslot E 210 GPa
d
or dmax a
Lslot
3
1
smid b a b a L +
b
4
E
3
d 0.5 mm
P
1
4
L b
b a b a L Lslot +
bt
E
3 slot
Solving for Lslot with dmax 0.475 mm
4 E dmax 3 L smid
244 mm Lslot 244 mm
smid
Lslot
0.325
L
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143
SECTION 2.3 Changes in Lengths under Nonuniform Conditions
Problem 2.3-5 Solve the preceding problem if the axial stress
in the middle region is 24,000 psi, the length is 30 in., and the
modulus of elasticity is 30 * 106 psi. In part (c), assume that
dmax 0.02 in.
b
—
4
P
t
b
L
—
4
P
L
—
2
L
—
4
Solution 2.3-5
L
L
P
4
2
7LP
±
≤ d
+
E
bt
3
6 Ebt
bt
4
2
(a)
(b) E 30,000 ksi
smid So
d
7 LP
6 Ebt
(c) dmax or
L 30 in.
P
3
bt
4
or
and
d
smid 24 ksi
3
P
smid
bt
4
7L 3
a s b 0.021 in.
6 E 4 mid
Lslot
P L Lslot
+
E
bt
3
bt Q
P
4
or
dmax a
Lslot
3
1
dmax a smid b a b a L +
b
4
E
3
Lslot 4 E dmax 3 L smid
10 in.
smid
d 0.021 in.
P
1
4
b a b a L Lslot + Lslot b
bt
E
3
Solving for Lslot with dmax 0.02 in.:
Lslot 10 in.
Lslot
0.333
L
Problem 2.3-6 A two-story building has steel columns AB in the first floor
and BC in the second floor, as shown in the figure. The roof load P1 equals
400 kN and the second-floor load P2 equals 720 kN. Each column has length
L 3.75 m. The cross-sectional areas of the first- and second-floor columns
are 11,000 mm2 and 3,900 mm2, respectively.
(a) Assuming that E 206 GPa, determine the total shortening dAC
of the two columns due to the combined action of the loads P1 and P2.
(b) How much additional load P0 can be placed at the top of the column
(point C) if the total shortening dAC is not to exceed 4.0 mm?
P1 = 400 kN
C
L = 3.75 m
P2 = 720 kN
B
L = 3.75 m
A
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CHAPTER 2 Axially Loaded Members
Solution 2.3-6
Steel columns in a building
(b) ADDITIONAL LOAD P0 AT POINT C
(dAC)max 4.0 mm
d0 additional shortening of the two columns
due to the load P0
d0 (dAC)max dAC 4.0 mm 3.7206 mm
0.2794 mm
Also, d0 P0L
P0L
P0L 1
1
a
+
+
b
EAAB
EABC
E AAB
ABC
Solve for P0:
(a) SHORTENING dAC OF THE TWO COLUMNS
P0 NiL i
NABL
NBCL
dAC g
+
E iAi
EAAB
EABC
SUBSTITUTE NUMERICAL VALUES:
(1120 kN)(3.75 m)
E 206 109 N/m2
(206 GPa)(11,000 mm2)
L 3.75 m
(400 kN)(3.75 m)
+
Ed0 AAB ABC
a
b
L AAB + ABC
d0 0.2794 103 m
AAB 11,000 106 m2
ABC 3,900 106 m2
2
(206 GPa)(3,900 mm )
1.8535 mm + 1.8671 mm 3.7206 mm
dAC 3.72 mm
;
P0 44,200 N 44.2 kN
;
Problem 2.3-7 A steel bar 8.0 ft long has a circular cross section of diameter d1 0.75 in. over one-half of its length and
diameter d2 0.5 in. over the other half (see figure part a). The modulus of elasticity E 30 * 106 psi.
(a) How much will the bar elongate under a tensile load P 5000 lb?
(b) If the same volume of material is made into a bar of constant diameter d and length 8.0 ft, what will be the elongation
under the same load P?
(c) If the uniform axial centroidal load q 1250 lb/ft is applied to the left over segment 1 (see figure part b), find the ratio
of the total elongation of the bar to that in parts (a) and (b).
d1 = 0.75 in.
q = 1250 lb/ft
d1 = 0.75 in.
d2 = 0.50 in.
d2 = 0.50 in.
P = 5000 lb
P
P = 5000 lb
4.0 ft
4.0 ft
(a)
4.0 ft
4.0 ft
(b)
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SECTION 2.3 Changes in Lengths under Nonuniform Conditions
145
Solution 2.3-7
NUMERICAL DATA E 30 11062 psi
(a) da P 5000 lb
1
1
PL
+
0.0589 in.
p
p
2
E P d1
d2 2 Q
4
4
P 12 L2
EA
0.0501 in.
d1 0.75 in.
d2 0.5 in.
da 0.0589 in.
p
p
(b) Va a d1 2 + d2 2 b L 30.631 in.3 d 4
4
db L 4 ft
Va
p
12 L2
R4
0.637 in. A p 2
d 0.31907 in.2
4
db 0.0501 in.
(c) q 1250 lb/ft L 4 ft
dc q L2
p
2 E a d21 b
4
+
PL
0.0341 in.
EA
dc
0.58
da
dc
0.681
db
Problem 2.3-8 A bar ABC of length L consists of two
parts of equal lengths but different diameters. Segment AB
has diameter d1 100 mm, and segment BC has diameter
d2 60 mm. Both segments have length L/2 0.6 m.
A longitudinal hole of diameter d is drilled through segment
AB for one-half of its length (distance L/4 0.3 m). The bar
is made of plastic having modulus of elasticity E 4.0 GPa.
Compressive loads P 110 kN act at the ends of the bar.
B
d2
C
d1
P
L
—
4
(a) If the shortening of the bar is limited to 8.0 mm, what
is the maximum allowable diameter dmax of the hole?
(See figure part a.)
(b) Now, if dmax is instead set at d2/2, at what distance b
from end C should load P be applied to limit the bar
shortening to 8.0 mm? (See figure part b.)
(c) Finally, if loads P are applied at the ends and
dmax d2/2, what is the permissible length x of the
hole if shortening is to be limited to 8.0 mm? (See
figure part c.)
dmax
A
P
L
—
4
L
—
2
(a)
d
dmax = —2
2
A
B
P
d2
C
d1
P
L
—
4
L
—
4
L
—
2
b
(b)
d
dmax = —2
2
A
B
d2
C
d1
P
x
P
L
—
2
L
—
x
2
(c)
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CHAPTER 2 Axially Loaded Members
Solution 2.3-8
b c
NUMERICAL DATA
d1 100 mm
d2 60 mm
L 1200 mm
E 4.0 GPa
b 4.16 mm
P 110 kN
OF THE HOLE IF SHORTENING IS TO BE LIMITED TO
(a) FIND dmax IF SHORTENING IS LIMITED TO da
p
A1 d1 2
4
da 8.0 mm?
p
A2 d2 2
4
L
4
P x
+
d J
E A0
L
L
4
2
+
¥
+
A1
A2
P
≥
E p
1d 2 dmax 22
4 1
x
Eda p d1 d2 2PLd2 2PLd1
2
2
2
a
L
a b
2
L
xb
2
+
A1
A2
Set d da and solve for x:
Set d to da, and solve for dmax:
dmax d1
;
(c) FINALLY IF LOADS P ARE APPLIED AT THE ENDS AND
dmax d2 2, WHAT IS THE PERMISSIBLE LENGTH x
da 8.0 mm
d
Eda
L
L
L
+
bdd
A2 c
a
2
P
4A0
4A1
2
c A0 A1a
K
Ed a
L
1
b d A0 L
P
2 A2
2
A1 A0
x 183.3 mm
;
9 Edapd1 2d2 2 PLd2 2 2PLd1 2
dmax 23.9 mm
;
(b) NOW, IF dmax IS INSTEAD SET AT d2 2, AT WHAT DISTANCE
b FROM END C SHOULD LOAD P BE APPLIED TO LIMIT THE
BAR SHORTENING TO da 8.0 mm?
d2 2
p 2
c d1 a b d
4
2
p
p
A1 d1 2
A2 d2 2
4
4
A0 d
L
P L
J
+
+
E 4A0
4A1
a
L
bb
2
A2
K
No axial force in segment at end of length b; set d da
and solve for b:
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147
SECTION 2.3 Changes in Lengths under Nonuniform Conditions
Problem 2.3-9 A wood pile, driven into the earth, supports a
P
P
load P entirely by friction along its sides (see figure part a).
The friction force f per unit length of pile is assumed to be
uniformly distributed over the surface of the pile. The pile has
length L, cross-sectional area A, and modulus of elasticity E.
(a) Derive a formula for the shortening d of the pile in terms
of P, L, E, and A.
(b) Draw a diagram showing how the compressive stress sc
varies throughout the length of the pile.
(c) Repeat parts (a) and (b) if skin friction f varies linearly
with depth (see figure part b).
f( y) = f0 (1 – y/L)
Skin
friction f
uniform
L
Skin friction f
varies linearly
with depth
f
L
f
y
f0
f
(a)
(b)
Solution 2.3-9
AFD LINEAR
L
d
(a) N(y) f y
(b) s (y) N(y)
A
(f y)
L2 f
dy 2AE
L0 E A
s (y) fy
A
s (L) d
PL
2EA
fL
P
A
A
s (0) 0
So linear variation, zero at bottom, P/A at top (i.e., at ground surface)
N(L) f
s (y) P y
a b
A L
P
A
1
1
0.8
0.8
0.6
0.6
Py
σc =
AL
σ ( y)
0.4
Compressive stress
in pile
0.2
0
σ ( y)
0.4
0.2
0
0
0.5
y
f(y) is constant
and AFD is linear
1
0
0
0.5
y
1
f(y) is linear and
AFD quadratic
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CHAPTER 2 Axially Loaded Members
(c) N(y) f(y) y
y
N(y) d
a
L0
f0 L
b
2
3
EA
2
f0 a1 bd f0 y (y 2)
2
1
f L
2 0
d
L
P
N(L) PL 2
a b
EA 3
s (y) f0
2
N(0) 0
y
P y
c a2 b d
A L
L
s (0) 0
s (L) f0
P/A
2
Problem 2.3-10 Consider the copper tubes joined below using a “sweated” joint. Use the properties and dimensions
given.
(a) Find the total elongation of segment 2-3-4 (d2-4) for an applied tensile force of P 5 kN. Use Ec 120 GPa.
(b) If the yield strength in shear of the tin-lead solder is ty 30 MPa and the tensile yield strength of the copper is
sy 200 MPa, what is the maximum load Pmax that can be applied to the joint if the desired factor of safety in
shear is FSt 2 and in tension is FSs 1.7?
(c) Find the value of L2 at which tube and solder capacities are equal.
Sweated
joint
P
Segment number
Solder joints
1
2
3
4
5
L2
L3
L4
d0 = 18.9 mm
t = 1.25 mm
P
d0 = 22.2 mm
t = 1.65 mm
L3 = 40 mm
L2 = L4 = 18 mm
Tin-lead solder in space
between copper tubes;
assume thickness of
solder equal zero
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SECTION 2.3 Changes in Lengths under Nonuniform Conditions
149
Solution 2.3-10
NUMERICAL DATA
(b) MAXIMUM LOAD Pmax THAT CAN BE APPLIED TO THE
P 5 kN
Ec 120 GPa
L2 18 mm
L4 L2
JOINT
First check normal stress:
p
A1 [ d2o5 1 do5 2 t522]
4
L3 40 mm
do3 22.2 mm
t3 1.65 mm
do5 18.9 mm
t5 1.25 mm
tY 30 MPa
sY 200 MPa
FSt 2
FSs 1.7
tY
ta FSt
ta 15 MPa
sa sY
FSs
sa 117.6 MPa
A1 69.311 mm2 smallest cross-sectional area
controls normal stress
; smaller than
Pmaxs sa A1 Pmaxs 8.15 kN
Pmax based on shear below so normal stress controls
Next check shear stress in solder joint:
Ash pdo5L2
Pmaxt taAsh
d24 Pmaxt 16.03 kN
(c) FIND THE VALUE OF L2 AT WHICH TUBE AND SOLDER
CAPACITIES ARE EQUAL
(a) ELONGATION OF SEGMENT 2-3-4
p
A2 [d2o3 (d o5 2 t5)2]
4
p 2
A3 [d o3 1d o3 2t322]
4
A2 175.835 mm2
Ash 1.069 103 mm2
Set Pmax based on shear strength equal to Pmax based
on tensile strength and solve for L2:
L2 A3 106.524 mm2
saA1
ta1pd o52
L2 9.16 mm
;
L3
P L2 + L4
a
+
b
Ec
A2
A3
d24 0.024 mm
;
Segment 1
Segment 2
Problem 2.3-11 The nonprismatic cantilever circular bar shown
has an internal cylindrical hole of diameter d/2 from 0 to x, so
the net area of the cross section for Segment 1 is (3/4)A. Load
P is applied at x, and load P/2 is applied at x L. Assume that
E is constant.
(a) Find reaction force R1.
(b) Find internal axial forces Ni in segments 1 and 2.
(c) Find x required to obtain axial displacement at joint 3 of
d3 PL/EA.
(d) In (c), what is the displacement at joint 2, d2?
(e) If P acts at x 2L/3 and P/2 at joint 3 is replaced by bP,
find b so that d3 PL/EA.
(f) Draw the axial force (AFD: N(x), 0 x L) and axial
displacement (ADD: d(x), 0 x L) diagrams using
results from (b) through (d) above.
3
—A
4
d
R1
A
P
—
2
P
d
—
2
x
3
2
L–x
3P
—
2
P
—
2
0
AFD 0
δ3
δ2
ADD 0
0
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CHAPTER 2 Axially Loaded Members
Solution 2.3-11
R1 P (a) STATICS a FH 0
R1 3
P
2
P
2
;
(b) DRAW FBD’S CUTTING THROUGH SEGMENT 1 AND
AGAIN THROUGH SEGMENT 2
3P
P
N1 6 tension N2 6 tension
2
2
(c) FIND x REQUIRED TO OBTAIN AXIAL DISPLACEMENT AT
JOINT 3 OF d3 PL/EA
Add axial deformations of segments 1 and 2, then
set to d3; solve for x:
N2( L x)
N1x
PL
+
3
EA
EA
E A
4
3P
P
x
( L x)
PL
2
2
+
EA
EA
3
E A
4
3
L
L
x
x
;
2
2
3
(e) IF x 2L/3 AND P/2 AT JOINT 3 IS REPLACED BY bP,
FIND b SO THAT d3 PL/EA
2L
3
substitute in axial deformation expression above
and solve for b
N1 (1
b)P N2 bP
x
2L
3
2L
b
3
[(1 + b)P]
bPaL +
3
E A
4
EA
PL
EA
1 8 + 11b
PL
PL
9
EA
EA
(8
11b) 9
1
;
11
b 0.091
b
(f) Draw AFD, ADD—see plots for x L
3
(d) WHAT IS THE DISPLACEMENT AT JOINT 2, d2?
N1x
d2 3
E A
4
d2 d2 a
3P L
b
2 3
3
E A
4
2 PL
3 EA
Problem 2.3-12 A prismatic bar AB of length L, cross-sectional area A, modulus of elasticity E, and
A
weight W hangs vertically under its own weight (see figure).
(a) Derive a formula for the downward displacement C of point C, located at distance h from the
lower end of the bar.
(b) What is the elongation B of the entire bar?
(c) What is the ratio b of the elongation of the upper half of the bar to the elongation of the lower half
of the bar?
(d) If bar AB is a riser pipe hanging from a drill rig at sea, what is the total elongation of the pipe? Let
L 1500 m, A 0.0157 m2, E 210 GPa. See Appendix I for weight densities of steel and sea
water. (See Problems 1.4-2 and 1.7-11 for additional figures).
C
L
h
B
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SECTION 2.3 Changes in Lengths under Nonuniform Conditions
151
Solution 2.3-12 Prismatic bar hanging vertically
W Weight of bar
A
(a) DOWNWARD DISPLACEMENT dC
Consider an element at distance y from the lower end.
dy
L
C
y
(b) ELONGATION OF BAR (h 0)
dB Elongation of upper half of bar a h B
Wy
L
dd L
dupper N(y) dy
Wydy
EA
EAL
L
dC 1h dd 1h
dC ;
(c) RATIO OF ELONGATIONS
h
N(y) WL
2EA
3WL
8EA
Elongation of lower half of bar:
Wydy
W
(L2 h2)
EAL
2EAL
W
(L2 h2)
2EAL
L
b:
2
dlower dB dupper b
;
dupper
dlower
3/8
3
1/8
3WL
WL
WL
2EA
8EA
8EA
;
(d) NUMERICAL DATA
gs 77 kN/m3
In sea water:
gw 10 kN/m3
L 1500 m
A 0.0157 m2
E 210 GPa
W (gs gw) A L 1577.85 kN
d
WL
359 mm
2EA
d
2.393 * 104
L
W (gs) A L 1813.35 kN
d
WL
412 mm
2EA
d
2.75 * 104
L
In air:
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CHAPTER 2 Axially Loaded Members
Problem 2.3-13 A flat bar of rectangular cross section,
b2
length L, and constant thickness t is subjected to tension by
forces P (see figure). The width of the bar varies linearly
from b1 at the smaller end to b2 at the larger end. Assume
that the angle of taper is small.
t
(a) Derive the following formula for the elongation of
the bar:
d
b2
PL
ln
Et(b2 b1)
b1
P
b1
L
P
(b) Calculate the elongation, assuming L 5 ft, t 1.0 in., P 25 k, b1 4.0 in., b2 6.0 in., and
E 30 106 psi.
Solution 2.3-13 Tapered bar (rectangular cross section)
t thickness (constant)
L0 + L
x
b b1 a b b2 b1 a
b
L0
L0
x
A(x) bt b1 t a b
L0
From Eq. (1):
(Eq. 1)
Solve Eq. (3) for L0: L 0 La
PL 0 dx
Pdx
EA(x)
Eb1 tx
L0 L
d
LL0
d
b1
b
b2 b1
(Eq. 4)
b2
PL
ln
Et (b2 b1) b1
(Eq. 5)
(b) SUBSTITUTE NUMERICAL VALUES:
PL0 L0 L dx
dd Eb1 t LL0
x
L0 L
PL 0
L0 + L
PL 0
ln x `
ln
Eb1 t
Eb
t
L0
L0
1
(Eq. 3)
Substitute Eqs. (3) and (4) into Eq. (2):
(a) ELONGATION OF THE BAR
dd L0 + L
b2
L0
b1
(Eq. 2)
L 5 ft 60 in.
t 10 in.
P 25 k
b1 4.0 in.
b2 6.0 in.
E 30 106 psi
From Eq. (5): d 0.010 in.
;
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SECTION 2.3 Changes in Lengths under Nonuniform Conditions
Problem 2.3-14 A post AB supporting equipment in a laboratory
is tapered uniformly throughout its height H (see figure). The cross
sections of the post are square, with dimensions b b at the top
and 1.5b 1.5b at the base.
Derive a formula for the shortening d of the post due to the
compressive load P acting at the top. (Assume that the angle of
taper is small and disregard the weight of the post itself.)
153
P
A
A
b
b
H
B
1.5b
B
1.5b
Solution 2.3-14
Tapered post
Ay cross-sectional area at distance y
1by22 b2
H2
(H + 0.5y)2
SHORTENING OF ELEMENT dy
dd Pdy
EAy
Pdy
Ea
b
2
H2
b1H + 0.5y22
SHORTENING OF ENTIRE POST
d
L
dd PH 2
Eb L0 (H + 0.5y)2
From Appendix D:
b width at A
2
1.5b width at B
by width at distance y
y
b + (1.5b b)
H
b
1H + 0.5y2
H
c
2
PH
Eb
PH 2
Eb 2
2PH
c
3Eb 2
dy
2
Square cross sections:
d
H
dx
L (a + bx)
2
1
b(a + bx)
H
1
d
(0.5)(H + 0.5y) 0
1
1
+
d
(0.5)(1.5H )
0.5H
;
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CHAPTER 2 Axially Loaded Members
Problem 2.3-15 A long, slender bar in the shape of a right circular cone
d
with length L and base diameter d hangs vertically under the action of its
own weight (see figure). The weight of the cone is W and the modulus of
elasticity of the material is E.
Derive a formula for the increase d in the length of the bar due to
its own weight. (Assume that the angle of taper of the cone is small.)
L
Solution 2.3-15
Conical bar hanging vertically
ELEMENT OF BAR
W weight of cone
TERMINOLOGY
ELONGATION OF ELEMENT dy
Ny dy
Wy dy
4W
y dy
dd E Ay
E ABL
pd 2 EL
Ny axial force acting on element dy
ELONGATION OF CONICAL BAR
Ay cross-sectional area at element dy
AB cross-sectional area at base of cone
pd 2
4
1
ABL
3
d
L
dd 4W
pd EL L0
2
L
y dy 2WL
;
pd 2 E
V volume of cone
Vy volume of cone below element dy
1
Ay y Wy weight of cone below element dy
3
Ay yW
Vy
Ny Wy
(W ) V
AB L
x
P
Problem 2.3-16 A uniformly tapered plastic tube AB of circular
dA
B
A
P
L
cross section and length L is shown in the figure. The average diameters at the ends are dA and dB 2dA. Assume E is constant. Find the
elongation d of the tube when it is subjected to loads P acting at the
ends. Use the following numerial data: dA 35 mm, L 300 mm,
E 2.1 GPa, P 25 kN. Consider two cases as follows:
(a) A hole of constant diameter dA is drilled from B toward A to
form a hollow section of length x L/2 (see figure part a).
dA
dB
(a)
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155
SECTION 2.3 Changes in Lengths under Nonuniform Conditions
(b) A hole of variable diameter d(x) is drilled from
B toward A to form a hollow section of length x L/2
and constant thickness t (see figure part b). (Assume
that t dA/20.)
x
B
A
P
P
dA
L
d(x)
t constant
dB
(b)
Solution 2.3-16
(a) ELONGATION d FOR CASE OF CONSTANT DIAMETER HOLE
d( ) dA a1 +
L
b
d
P
1
a
d b
E L A( )
d
P
E
d
d
J
p
d( )2
solid portion of length L x
4
p
hollow portion of length x
A( ) (d( )2 dA 2)
4
A( ) d
Lx
L
P
4
4
d d
d
+
c
2
2
E L0
pd( )
LLx p1d( ) d A 22
L
1
Lx
2
p
c cdA a1 + b d d
4
L
L0
d +
L
P
L2
+
4
4
+
2
2
E J (2 x)pdA
J J pdA
1
2
p
c
c
d
b
d
c
a
1
+
dA 2 d d
A
LLx 4
L
1
L
2
p
c c c dA a 1 + b d dA 2 d d
LLx 4
L
d
d
K
KKK
ln(Lx) + ln(3Lx)
ln(3)
L
L2
P
a4
2L
+ 2L
bd
c4
2 +
2
2
E (2 x)pdA
pdA
pdA
pdA 2
if x L/2
ln(3)
P 4 L
±
d
2L 2
E 3 pd2A
pdA
2L
1
5
ln a L b + ln a Lb
2
2
p d2A
≤
Substitute numerical data:
d 2.18 mm
;
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CHAPTER 2 Axially Loaded Members
(b) ELONGATION d FOR CASE OF VARIABLE DIAMETER HOLE BUT CONSTANT WALL THICKNESS t dA/20 OVER SEGMENT x
p
A( ) d( )2 solid portion of length L x
d( ) dA a1 + b
L
4
A( ) d
P
1
a
d b
E L A( )
d
Lx
P
≥
E L0
d
d
4
pcdA a1 + L b d
dA 2
p
cd( )2 a d( ) 2 b d
4
20
Lx
L
P
4
d
+
≥
E L0
pd( )2
LLx
L
d +
L2
P
L
c4
+ 4
2
E (2L + x)pdA
pdA 2
4
pc d( )2 a d( ) 2
4
LLx
20L
hollow portion of length x
pc c dA a 1 +
2
L
b d cdA a 1 +
L
b 2
dA 2
d d
20
dA 2
b d
20
d ¥
d ¥
ln(3) + ln(13) + 2ln( dA) + ln( L)
20L
pdA 2
2ln( dA) + ln (39L 20x)
pdA 2
d
if x L/2
d
2ln( dA) + ln(29L)
ln(3) + ln(13) + 2ln( dA) + ln( L)
P 4 L
20L
+ 20L
b
a
E 3 pdA 2
pdA 2
pdA 2
Substitute numerical data:
d 6.74 mm
;
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157
SECTION 2.3 Changes in Lengths under Nonuniform Conditions
Problem 2.3-17 The main cables of a suspension bridge
[see part (a) of the figure] follow a curve that is nearly parabolic
because the primary load on the cables is the weight of the
bridge deck, which is uniform in intensity along the horizontal.
Therefore, let us represent the central region AOB of one of the
main cables [see part (b) of the figure] as a parabolic cable supported at points A and B and carrying a uniform load of intensity
q along the horizontal. The span of the cable is L, the sag is h,
the axial rigidity is EA, and the origin of coordinates is at
midspan.
(a)
y
L
—
2
A
(a) Derive the following formula for the elongation of cable
AOB shown in part (b) of the figure:
L
—
2
B
h
qL3
16h2
)
(1 +
d
8hEA
3L2
O
q
(b) Calculate the elongation d of the central span of one of
the main cables of the Golden Gate Bridge, for which the
dimensions and properties are L 4200 ft, h 470 ft,
q 12,700 lb/ft, and E 28,800,000 psi. The cable
consists of 27,572 parallel wires of diameter 0.196 in.
x
(b)
Hint: Determine the tensile force T at any point in the cable from a free-body diagram of part of the cable; then determine
the elongation of an element of the cable of length ds; finally, integrate along the curve of the cable to obtain an equation for
the elongation d.
Solution 2.3-17 Cable of a suspension bridge
dy
8hx
2
dx
L
FREE-BODY DIAGRAM OF HALF OF CABLE
MB 0 哵哴
Hh +
H
qL L
a b 0
2 4
qL2
8h
Fhorizontal 0
HB H qL2
8h
(Eq. 1)
Fvertical 0
VB qL
2
(Eq. 2)
Equation of parabolic curve:
y
4hx 2
L2
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CHAPTER 2 Axially Loaded Members
FREE-BODY DIAGRAM OF SEGMENT DB OF CABLE
dd Tds
EA
ds 2(dx)2 + (dy)2 dx 1 + a
A
dx 1 + a
A
dx 1 +
8hx
L2
b
dy 2
b
dx
2
64h2x2
A
(Eq. 6)
L4
(a) ELONGATION d OF CABLE AOB
d
©F horiz 0
TH HB
©F vert 0 VB Tv q a
Tv VB q a
qL2
8h
(Eq. 3)
L
xb 0
2
Substitute for T from Eq. (5) and for ds from
Eq. (6):
qL2
1
64h 2x 2
a1 +
b dx
EA L 8h
L4
For both halves of cable:
(Eq. 4)
qL2 2
b + (qx)2
A 8h
a
qL2
64h2x2
1 +
8h A
L4
ELONGATION dd OF AN ELEMENT OF LENGTH ds
(Eq. 5)
L/2
qL2
64h2x 2
a1 +
b dx
8h
L4
d
2
EA L0
d
qL3
16h2
a1 +
b
8hEA
3L4
TENSILE FORCE T IN CABLE
T 2T 2H + T2v T ds
L EA
d
qL
qL
L
xb + qx
2
2
2
qx
L
dd ;
(Eq. 7)
(b) GOLDEN GATE BRIDGE CABLE
L 4200 ft
h 470 ft
q 12,700 lb/ft E 28,800,000 psi
27,572 wires of diameter d 0.196 in.
p
A (27,572)a b(0.196 in.)2 831.90 in.2
4
Substitute into Eq. (7):
d 133.7 in 11.14 ft
;
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159
SECTION 2.3 Changes in Lengths under Nonuniform Conditions
Problem 2.3-18 A bar ABC revolves in a horizontal plane about a
A
vertical axis at the midpoint C (see figure). The bar, which has length
2L and cross-sectional area A, revolves at constant angular speed v.
Each half of the bar (AC and BC) has weight W1 and supports a
weight W2 at its end.
Derive the following formula for the elongation of one-half of the
bar (that is, the elongation of either AC or BC):
W2
v
C
W1
B
W1
L
W2
L
L22
(W + 3W2)
3gEA 1
in which E is the modulus of elasticity of the material of the bar
and g is the acceleration of gravity.
d
Solution 2.3-18 Rotating bar
Centrifugal force produced by weight W2
a
W2
b (L2)
g
AXIAL FORCE F(x)
F(x) v angular speed
A cross-sectional area
E modulus of elasticity
g acceleration of gravity
F(x) axial force in bar at distance x from point C
Consider an element of length dx at distance x from
point C.
To find the force F(x) acting on this element, we must
find the inertia force of the part of the bar from distance
x to distance L, plus the inertia force of the weight W2.
Since the inertia force varies with distance from point C,
we now must consider an element of length dj at distance j, where j varies from x to L.
Mass of element d d W1
a
b
L g
L W 2
1
Lx
gL
d +
W2L2
g
W12 2
W2L2
(L x 2) +
2gL
g
ELONGATION OF BAR BC
L
d
F(x) dx
L0 EA
L
L
W12 2
W2L2dx
(L x 2)dx +
gEA
L0
L0 2gL
W2L2dx
W1L2
2
2
L dx x dx d +
dx
c
2gLEA L0
gEA L0
L0
W2L22
W1L22
+
3gEA
gEA
L22
+ (W1 + 3W2)
3gEA
L
L
L
;
Acceleration of element jv2
Centrifugal force produced by element
(mass)( acceleration) W12
d
gL
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CHAPTER 2 Axially Loaded Members
Statically Indeterminate Structures
Problem 2.4-1 The assembly shown in the figure consists of a brass
core (diameter d1 0.25 in.) surrounded by a steel shell (inner diameter
d2 0.28 in., outer diameter d3 0.35 in.). A load P compresses the
core and shell, which have length L 4.0 in. The moduli of elasticity of
the brass and steel are Eb 15 106 psi and Es 30 106 psi,
respectively.
(a) What load P will compress the assembly by 0.003 in.?
(b) If the allowable stress in the steel is 22 ksi and the allowable stress
in the brass is 16 ksi, what is the allowable compressive load
Pallow? (Suggestion: Use the equations derived in Example 2-6.)
Solution 2.4-1 Cylindrical assembly in compression
Substitute numerical values:
E s As + E b Ab (30 * 106 psi)(0.03464 in.2)
+ (15 * 106 psi)(0.04909 in.2)
1.776 * 106 lb
0.003 in.
b
4.0 in.
P (1.776 * 106 lb)a
1330 lb
;
(b) ALLOWABLE LOAD
d1 0.25 in.
Eb 15 106 psi
d2 0.28 in.
Es 30 106 psi
d3 0.35 in.
As L 4.0 in.
p 2
(d 3 d 22) 0.03464 in.2
4
p
Ab d 21 0.04909 in.2
4
(a) DECREASE IN LENGTH (d 0.003 in.)
Use Eq. (2-18) of Example 2-6.
d
PL
E s As + E b Ab
or
d
P (E s As + E s Ab)a b
L
ss 22 ksi sb 16 ksi
Use Eqs. (2-17a and b) of Example 2-6.
For steel:
ss PE s
E s As + E b Ab
Ps (E s As + E b Ab)
Ps (1.776 * 106 lb)a
22 ksi
30 * 106 psi
ss
Es
b 1300 lb
For brass:
sb PE b
E s As + E b Ab
Ps (E s As + E b Ab)
Ps (1.776 * 106 lb)a
Steel governs.
16 ksi
15 * 106 psi
Pallow 1300 lb
sb
Eb
b 1890 lb
;
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161
SECTION 2.4 Statically Indeterminate Structures
Problem 2.4-2 A cylindrical assembly consisting of a brass core and
an aluminum collar is compressed by a load P (see figure). The length
of the aluminum collar and brass core is 350 mm, the diameter of the
core is 25 mm, and the outside diameter of the collar is 40 mm. Also, the
moduli of elasticity of the aluminum and brass are 72 GPa and 100 GPa,
respectively.
(a) If the length of the assembly decreases by 0.1% when the load
P is applied, what is the magnitude of the load?
(b) What is the maximum permissible load Pmax if the allowable
stresses in the aluminum and brass are 80 MPa and 120 MPa,
respectively? (Suggestion: Use the equations derived in
Example 2-6.)
Solution 2.4-2 Cylindrical assembly in compression
d
PL
E a Aa + E b Ab
or
d
P (E a Aa + E b Ab)a b
L
Substitute numerical values:
E a Aa + E b Ab (72 GPa)(765.8 mm2)
(100 GPa)(490.9 mm2)
55.135 MN + 49.090 MN
104.23 MN
P (104.23 MN)a
104.2 kN
A aluminum
;
(b) ALLOWABLE LOAD
B brass
sa 80 MPa
L 350 mm
sb 120 MPa
Use Eqs. (2-17a and b) of Example 2-6.
da 40 mm
For aluminum:
db 25 mm
sa p
Aa (d 2a d 2b)
4
Eb 100 GPa
PE a
E a Aa + E b Ab
Pa (104.23 MN)a
765.8 mm2
Ea 72 GPa
0.350 mm
b
350 mm
Ab 490.9 mm2
(a) DECREASE IN LENGTH
(d 0.1% of L 0.350 mm)
Use Eq. (2-18) of Example 2-6.
p 2
d
4 b
Pa (E a Aa + E b Ab) a
sa
b
Ea
80 MPa
b 115.8 kN
72 GPa
For brass:
sb PE b
E a Aa + E b Ab
Pb (104.23 MN)a
Pb (E a Aa + E b Ab) a
sb
b
Eb
120 MPa
b 125.1 kN
100 GPa
Aluminum governs. Pmax 116 kN
;
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CHAPTER 2 Axially Loaded Members
Problem 2.4-3 Three prismatic bars, two of material A and one of material B,
transmit a tensile load P (see figure). The two outer bars (material A) are identical.
The cross-sectional area of the middle bar (material B) is 50% larger than the
cross-sectional area of one of the outer bars. Also, the modulus of elasticity of
material A is twice that of material B.
(a) What fraction of the load P is transmitted by the middle bar?
(b) What is the ratio of the stress in the middle bar to the stress in the outer bars?
(c) What is the ratio of the strain in the middle bar to the strain in the outer bars?
Solution 2.4-3 Prismatic bars in tension
FREE-BODY DIAGRAM OF END PLATE
STRESSES:
EQUATION OF EQUILIBRIUM
Fhoriz 0
PB P 0
PA
PA
EA P
AA
E A AA + E B AB
sB PB
EB P
AB
E A AA + E B AB
(a) LOAD IN MIDDLE BAR
(2)
PB
E B AB
1
E A AA
P
E A AA + E B AB
+ 1
E B AB
Given:
FORCE-DISPLACEMENT RELATIONS
AA total area of both outer bars
‹
dA PA L
E A Ak
dB PB L
E B AB
(7)
(1)
EQUATION OF COMPATIBILITY
dA dB
sA (3)
Substitute into Eq. (2):
AA
EA
4
1 + 1
2
EB
AB
1.5
3
PB
P
1
3
1
8
11
E A AA
+ 1
a ba b + 1
3
E B AB
;
(b) RATIO OF STRESSES
PA L
PB L
E A AA
E B AB
(4)
sB
EB
1
sA
EA
2
;
SOLUTION OF THE EQUATIONS
(c) RATIO OF STRAINS
Solve simultaneously Eqs. (1) and (4):
E A AAP
PA E A AA + E B AB
E B AB P
PB E A AA + E B AB
All bars have the same strain
(5)
Ratio 1
;
Substitute into Eq. (3):
d dA dB PL
E A AA + E B AB
(6)
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SECTION 2.4 Statically Indeterminate Structures
Problem 2.4-4 A circular bar ACB of diameter d having a cylindrical hole of
(a) Obtain formulas for the reactions RA and RB at supports A and B, respectively, due to the load P (see figure part a).
(b) Obtain a formula for the displacement d at the point of load application
(see figure part a).
(c) For what value of x is RB (6/5) RA? (See figure part a.)
(d) Repeat part (a) if the bar is now rotated to a vertical position, load P is
removed, and the bar is hanging under its own weight (assume mass
density r). (See figure part b.) Assume that x L/2.
P, d
L
—
2
length x and diameter d/2 from A to C is held between rigid supports at A and
B. A load P acts at L/2 from ends A and B. Assume E is constant.
163
d
—
2
d
B
C
A
x
L–x
(a)
B
L–x
C
d
—
2
x
d
A
(b)
Solution 2.4-4
(a) REACTIONS AT A AND B DUE TO LOAD P AT L/2
AAC p 2
d 2
cd a b d
4
2
ACB p 2
d
4
AAC 3 2
pd
16
Select RB as the redundant; use superposition and a compatibility equation at B:
if x
L/2
d B1a Px
+
EA AC
Pa
L
xb
2
L
x
P
x
2
≤
dB1a ±
+
p 2
E 3
2
d
pd
4
16
EACB
2 2x + 3L
dB1a P
3
Epd2
P
if x L/2
dB1b L
2
EA AC
P
d B1b L
2
3
Ea pd 2 b
16
dB1b 8 PL
3 Epd2
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CHAPTER 2 Axially Loaded Members
The following expression for d B2 is good for all x:
dB2 RB x
Lx
+
b
a
E AAC
ACB
dB2 RB 16 x
Lx
a
+ 4
b
E 3 pd 2
pd 2
RB
x
Lx
+
p 2 Q
E P 3
d
pd2
4
16
dB2 Solve for RB and RA assuming that x … L/2:
Compatibility:
dB1a
dB2 0
RBa 2 2x + 3L
a P
b
3
pd 2
a
16 x
Lx
+ 4
b
3 pd 2
pd 2
RBa 1 2x + 3L
P
2
x + 3L
;
^ check—if x 0, RB P/2
Statics:
RAa P RBa
RAa P 1 2x + 3L
P
2
x + 3L
RAa 3
L
P
2
x + 3L
;
^ check—if x 0, RAa P/2
Solve for RB and RA assuming that x Ú L/2:
Compatibility:
dB1b
dB2 0
RBb 8 PL
3 pd 2
16 x
Lx
a
+ 4
b
2
3 pd
pd 2
RBb 2PL
x + 3L
;
^ check—if x L, RB P/2
Statics:
RAb P RBb
RAb P a
2PL
b
x + 3L
RAb P
x + L
x + 3L
;
(b) FIND d AT POINT OF LOAD APPLICATION; AXIAL FORCE FOR SEGMENT 0 TO L/2 RA AND d ELONGATION OF THIS SEGMENT
Assume that x … L/2:
L
x
RAa x
2
da +
Q
P
E
AAC
ACB
da PL
a
da 3
L
P
b
2
x + 3L
E
L
x
2
±
≤
+
p 2
3
d
pd2
4
16
x
2x + 3L
(x + 3L)Epd2
For x L/2,
da 8
P
L
7 Epd2
;
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SECTION 2.4 Statically Indeterminate Structures
165
ASSUME THAT x Ú L/2:
db 1RAb2
L
2
EAAC
db aP
x + L L
b
x + 3L 2
3
E a pd 2 b
16
db 8
x + L
L
Pa
b
3
x + 3L Epd 2
for x L/2
db 8
L
P
7 Epd 2
;
same as da above (OK)
(c) FOR WHAT VALUE OF x is RB (6/5) RA?
Guess that x L/2 here and use RBa expression above to find x:
1 2x + 3L
6 3
L
P
a
P
b 0
2
x + 3L
5 2
x + 3L
1 10x 3L
P
0
10
x + 3L
x
3L
10
;
Now try RBb (6/5)RAb, assuming that x L/2
2PL
6
x + L
aP
b 0
x + 3L
5
x + 3L
2 2L + 3x
P
0
5
x + 3L
x
2
L
3
;
So, there are two solutions for x.
(d) FIND REACTIONS IF THE BAR IS NOW ROTATED TO A VERTICAL POSITION, LOAD P IS REMOVED, AND THE BAR IS HANGING
UNDER ITS OWN WEIGHT (ASSUME MASS DENSITY ρ). ASSUME THAT x L/2.
AAC 3
pd2
16
ACB p 2
d
4
Select RB as the redundant; use superposition and a compatibility equation at B
from (a) above. compatibility: dB1
dB2 dB1 RB x
Lx
a
+
b
E AAC
ACB
L
2
NAC
d
EA
L0
AC
L N
For x L/2, dB2 CB
L EACB
L
2
dB2 0
RB 14 L
a
b
E 3 pd 2
d
L
Where axial forces in bar due to self weight are WAC rgAAC
2
(assume z is measured upward from A):
NAC crgACB
L
L
+ rgAAC a b d
2
2
AAC 3
pd2
16
WCB rgACB
ACB L
2
p 2
d
4
NCB [r gACB(L z)]
NAC 1
3
1
rgp d2 L rgp d2 a L b
8
16
2
1
NCB c rgp d2( L ) d
4
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CHAPTER 2 Axially Loaded Members
dB1 L
2
1
3
1
rgpd 2L rgpd 2 a L b
8
16
2
L0
dB1 a
3
E a pd 2 b
16
11
L2
1
L2
rg
+
rg
b
24
E
8
E
Compatibility: dB1
RB a
a
Statics:
RA RB 14 L
b
3 Epd2
RA c c rga
1
rgpd2 (L ) d
4
LL2
Ea p4 d2 b
7
L2
rg
12
E
7
0.583
12
d
dB2 0
7
L2
rg
b
12
E
RA (WAC
dB1 d +
L c
1
rgpd 2L
8
;
WCB) RB
3
L
p
L
1
pd2 b + rga d2 b d rgpd2L d
16
2
4
2
8
3
rgp d 2 L
32
;
Problem 2.4-5 Three steel cables jointly support a load of 12 k (see figure). The diameter of
the middle cable is 3⁄4 in. and the diameter of each outer cable is 1⁄2 in. The tensions in the
cables are adjusted so that each cable carries one-third of the load (i.e., 4 k). Later, the load is
increased by 9 k to a total load of 21 k.
(a) What percent of the total load is now carried by the middle cable?
(b) What are the stresses sM and sO in the middle and outer cables, respectively?
(NOTE: See Table 2-1 in Section 2.2 for properties of cables.)
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SECTION 2.4 Statically Indeterminate Structures
167
Solution 2.4-5 Three cables in tension
SUBSTITUTE INTO COMPATIBILITY EQUATION:
PML
POL
EAM
EAO
PM
PO
AM
AO
(5)
SOLVE SIMULTANEOUSLY EQS. (1) AND (5):
PM P2 a
AM
0.268 in.2
b (9 k) a
b
AM + 2AO
0.506 in.2
4.767 k
AREAS OF CABLES (from Table 2-1)
Po P2 a
Middle cable: AM 0.268 in.2
Outer cables: AO 0.119 in.2
Ao
0.119 in.2
b (9 k) a
b
AM + 2AO
0.506 in.2
2.117 k
(for each cable)
FORCES IN CABLES
FIRST LOADING
P1 12 k aEach cable carries
P1
or 4 k.b
3
Middle cable: Force 4 k
4.767 k 8.767 k
Outer cables: Force 4 k
2.117 k 6.117 k
(for each cable)
SECOND LOADING
(a) PERCENT OF TOTAL LOAD CARRIED BY MIDDLE CABLE
Percent 8.767 k
(100%) 41.7%
21 k
;
(b) STRESSES IN CABLES (s P/A)
Middle cable: sM EQUATION OF EQUILIBRIUM
Fvert 0
2PO
PM P2 0
32.7 ksi
;
51.4 ksi
;
(1)
EQUATION OF COMPATIBILITY
dM dO
8.767 k
0.268 in.2
Outer cables: sO (2)
6.117 k
0.119 in.2
FORCE-DISPLACEMENT RELATIONS
dM PML
Po L
dO EAM
EAo
(3, 4)
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Page 168
CHAPTER 2 Axially Loaded Members
Problem 2.4-6 A plastic rod AB of length L 0.5 m has a
diameter d1 30 mm (see figure). A plastic sleeve CD of length
c 0.3 m and outer diameter d2 45 mm is securely bonded to the
rod so that no slippage can occur between the rod and the sleeve.
The rod is made of an acrylic with modulus of elasticity E1 3.1
GPa and the sleeve is made of a polyamide with E2 2.5 GPa.
(a) Calculate the elongation d of the rod when it is pulled by
axial forces P 12 kN.
(b) If the sleeve is extended for the full length of the rod, what is
the elongation?
(c) If the sleeve is removed, what is the elongation?
Solution 2.4-6 Plastic rod with sleeve
P 12 kN
d1 30 mm
b 100 mm
L 500 mm
d2 45 mm
c 300 mm
Part CD: dCD 0.81815 mm
Rod: E1 3.1 GPa
Sleeve: E2 2.5 GPa
Rod: A1 pd 21
706.86 mm2
4
Sleeve: A2 E1A1
p 2
(d d 12) 883.57 mm2
4 2
E2A2 4.400 MN
(a) ELONGATION OF ROD
Part AC: dAC Pb
0.5476 mm
E 1A1
Pc
E 1A1 + E 2A2
(From Eq. 2-18 of Example 2-6)
d 2dAC
dCD 1.91 mm
;
(b) SLEEVE AT FULL LENGTH
L
500 mm
d dCD a b (0.81815 mm)a
b
c
300 mm
1.36 mm
;
(c) SLEEVE REMOVED
PL
d
2.74 mm
E 1A1
;
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SECTION 2.4 Statically Indeterminate Structures
Problem 2.4-7 A tube structure is acted on by
x
loads at B and D, as shown in the figure. The
tubes are joined using two flange plates at C,
which are bolted together using six 0.5 in. diameter bolts.
3P at x = 3L/4
Flange
plate
169
Bolt
2P at x = L/4
EA/2
EA
(a) Derive formulas for the reactions RA and
RE at the ends of the bar.
L/4
L/4
L/4
L/4
Flange plates at C joined
(b) Determine the axial displacement B, c,
by six bolts
A
B
C
D
E
and D at points B, C, and D, respectively.
(c) Draw an axial-displacement diagram
(ADD) in which the abscissa is the distance x from support A to any point on the bar and the ordinate is the horizontal
displacement d at that point.
(d) Find the maximum value of the load variable P if allowable normal stress in the bolts is 14 ksi.
Solution 2.4-7
NUMERICAL DATA
n6
db 0.5 in.
sa 14 ksi
Ab p 2
d 0.196 in.2
4 b
(a) FORMULAS FOR REACTIONS F
2a
Segment ABC flexibility: f1 Segment CDE flexibility: f2 L
b
4
EA
L
2a b
4
1
EA
2
L
2 EA
L
EA
Loads at points B and D:
PB 2 P
PD 3 P
(1) Select RE as the redundant; find axial displacement d1 displacement at E due to loads PB and PD:
d1 1PB + PD2
L
4
EA
L
L
PD
5LP
4
4
+
+
EA
1
2 EA
EA
2
PD
(2) Next apply redundant RE and find axial displacement d2 displacement at E due to redundant RE:
d2 RE 1f1 + f22 3LRE
2 EA
(3) Use compatibility equation to find redundant RE then use statics to find RA:
5P
3
RE 2P
3
RA d1 + d2 0 solve, RE RA RE PB PD 5
P
3
2P
3
RA 2P
3
RE 5P
3
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CHAPTER 2 Axially Loaded Members
(b) DETERMINE THE AXIAL DISPLACEMENTS dB, dC, AND dD AT POINTS B, C, AND D, RESPECTIVELY.
dB
a
2 P L
ba b
3
4
EA
LP
6 EA
leftward
dc dB +
a2P 2P L
ba b
3
4
EA
LP
6 EA
to the right
dD a
5P L
ba b
3
4
EA
2
5LP
6 EA
to the right
(c) DRAW AN AXIAL-DISPLACEMENT DIAGRAM (ADD) IN WHICH THE ABSCISSA IS THE DISTANCE X FROM SUPPORT A TO ANY
POINT ON THE BAR AND THE ORDINATE IS THE HORIZONTAL DISPLACEMENT d AT THAT POINT.
AFD for use below in Part (d)
Axial Force Diagram (AFD)
2
Axial force (times P )
AFD is composed of 4 constant segments, so
ADD is linear with zero displacements at
supports A and E.
1
N(x)
0
RA
−1
RE
−2
dmax dD
dmax 5LP
6 EA
to the right
Boundary conditions at supports:
dA dE 0
0.25
0.5
0.75
x
Distance x (times L)
1
Axial Displacement Diagram (ADD)
1
Axial displ (times L /EA)
Plot displacements dB, dC, and dD from part (b)
above, then connect points using straight lines
showing linear variation of axial displacement
Between points
0
5
6
0.5
δ (x)
0
−
−0.5
0
0.25
0.5
0.75
x
Distance (times L)
1
6
1
(d) MAXIMUM PERMISSIBLE VALUE OF LOAD VARIABLE P BASED ON ALLOWABLE NORMAL STRESS IN FLANGE BOLTS
FROM AFD, FORCE AT L/2:
Fmax 4
P
3
Pmax 3
Fmax 12.37 k
4
and
Fmax n sa Ab 16.493 k
Pmax 12.37 k
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171
SECTION 2.4 Statically Indeterminate Structures
Problem 2.4-8 The fixed-end bar ABCD consists of three prismatic
segments, as shown in the figure. The end segments have cross-sectional
area A1 840 mm2 and length L1 200 mm. The middle segment has
cross-sectional area A2 1260 mm2 and length L2 250 mm. Loads PB
and PC are equal to 25.5 kN and 17.0 kN, respectively.
(a) Determine the reactions RA and RD at the fixed supports.
(b) Determine the compressive axial force FBC in the middle
segment of the bar.
Solution 2.4-8 Bar with three segments
PB 25.5 kN
PC 17.0 kN
L1 200 mm
L2 250 mm
A1 840 mm
A2 1260 mm2
2
m meter
SOLUTION OF EQUATIONS
Substitute Eqs. (3), (4), and (5) into Eq. (2):
FREE-BODY DIAGRAM
RA
RA
1
1
a 238.095 b +
a 198.413 b
E
m
E
m
EQUATION OF EQUILIBRIUM
Simplify and substitute PB 25.5 kN:
Fhoriz 0 : ;
PB
RD
PB
1
1
a 198.413 b +
a 238.095 b 0
E
m
E
m
RA a 436.508
RD PC RA 0 or
RA RD PB PC 8.5 kN
(Eq. 1)
1
1
b + RD a 238.095 b
m
m
5,059.53 kN/m
(Eq. 6)
EQUATION OF COMPATIBILITY
(a) REACTIONS RA AND RD
dAD elongation of entire bar
dAD dAB
dBC
dCD 0
(Eq. 2)
From (1): RD RA 8.5 kN
FORCE-DISPLACEMENT RELATIONS
dAB RAL 1
RA
1
a238.05 b
EA1
E
m
dBC (RA PB)L 2
EA2
RA
PB
1
1
a198.413 b a 198.413 b
E
m
E
m
RDL 1
RD
1
dCD a238.095 b
EA1
E
m
Solve simultaneously Eqs. (1) and (6).
Substitute into (6) and solve for RA:
(Eq. 3)
RA a 674.603
1
b 7083.34 kN/m
m
RA 10.5 kN
;
RD RA 8.5 kN 2.0 kN
(Eq. 4)
;
(b) COMPRESSIVE AXIAL FORCE FBC
FBC PB RA PC RD 15.0 kN
;
(Eq. 5)
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CHAPTER 2 Axially Loaded Members
Problem 2.4-9 The aluminum and steel pipes shown in the figure are fastened to rigid
supports at ends A and B and to a rigid plate C at their junction. The aluminum pipe is
twice as long as the steel pipe. Two equal and symmetrically placed loads P act on the
plate at C.
(a) Obtain formulas for the axial stresses sa and ss in the aluminum and steel pipes,
respectively.
(b) Calculate the stresses for the following data: P 12 k, cross-sectional area of
aluminum pipe Aa 8.92 in.2, cross-sectional area of steel pipe As 1.03 in.2,
modulus of elasticity of aluminum Ea 10 106 psi, and modulus of elasticity
of steel Es 29 106 psi.
Solution 2.4-9 Pipes with intermediate loads
SOLUTION OF EQUATIONS
Substitute Eqs. (3) and (4) into Eq. (2):
RB(2L)
RAL
0
E s As
E aAa
(Eq. 5)
Solve simultaneously Eqs. (1) and (5):
RA 4E s As P
2E aAaP
RB E aAa + 2E sAs
E aAa + 2E sAs
(Eqs. 6, 7)
(a) AXIAL STRESSES
RB
2E aP
Aa
E aAa + 2E sAs
;
(compression)
(Eq. 8)
Aluminum: sa RA
4EsP
Steel: ss As
EaAa + 2Es As
Pipe 1 is steel.
Pipe 2 is aluminum.
EQUATION OF EQUILIBRIUM
Fvert 0
RA
RB 2P
dCB 0
(b) NUMERICAL RESULTS
P 12 k
(Eq. 2)
Aa 8.92 in.2
Ea 10 106 psi
As 1.03 in.2
Es 29 106 psi
(A positive value of d means elongation.)
Substitute into Eqs. (8) and (9):
FORCE-DISPLACEMENT RELATIONS
sa 1,610 psi (compression)
dAC RB(2L)
RAL
d E sAs BC
E aAa
(Eq. 9)
(tension)
(Eq. 1)
EQUATION OF COMPATIBILITY
dAB dAC
;
ss 9,350 psi (tension)
;
;
(Eqs. 3, 4))
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173
SECTION 2.4 Statically Indeterminate Structures
Problem 2.4-10 A hollow circular pipe (see figure) supports a load P which is uniformly distributed around a cap plate at the top of the lower pipe. The inner and outer
diameters of the upper and lower parts of the pipe are d1 50 mm, d2 60 mm,
d3 57 mm, and d4 64 mm, respectively. Pipe lengths are L1 2m and L2 3 m.
Neglect the self-weight of the pipes. Assume that cap plate thickness is small compared to L1 and L2. Let E 110 MPa.
d1
d2
(a) If the tensile stress in the upper part is s1 10.5 MPa, what is load P? Also,
what are reactions R1 at the upper support and R2 at the lower support. What is
the stress s2 MPa in the lower part?
(b) Find displacement (mm) at the cap plate. Plot the Axial Force Diagram,
AFD [N(x)] and Axial Displacement Diagram, ADD [(x)].
(c) Add the uniformly distributed load q along the centroidal axis of pipe segment 2.
Find q(kN/m) so that R2 0. Assume that load P from part (a) is also applied.
L1
P
q
Cap plate
(Part (c) only)
d3
x
d4
L2
Solution 2.4-10
(a) STRESSES AND REACTIONS: SELECT R1 AS REDUNDANT AND DO SUPERPOSITION ANALYSIS (HERE q 0; DEFLECTION
POSITION UPWARD)
d1 50 mm
d2 60 mm
d3 57 mm
d4 64 mm
E 110 MPa
L1 2 m
SEGMENT FLEXIBILITIES
f1 L1
0.02105 mm/N
E A1
f2 d1b R1 1f1 + f22
Solve for P:
P R1 a
L2
0.041 mm/N
E A2
p
1d 2 d3 22 665.232 mm2
4 4
f1
0.513
f2
s1 10.5 MPa
R1 s1 A1 9.07 kN
Compatibility: d1a + d1b 0
R2 P R1 4.66 kN
P 13.73 kN
(b) DISPLACEMENT AT CAP PLATE
dc R1 f1 190.909 mm
AFD and ADD:
A2 f1 + f2
b 13.73 kN
f2
Finally, use statics to find R2:
dcap dc 0.191 m
p
1d 2 d1 22 863.938 mm2
4 2
L2 3 m
TENSILE stress (s1) is known in upper segment so R1 s1 * A1
d1a P f2
A1 s2 R2
7 MPa
A2
R1 9.07 kN
6 downward OR dc 1R22 f2 190.909 mm
compressive since R2 is
positive (upward)
R2 4.66 kN
s2 7 MPa
downward (neg. x-direction)
dcap 190.9 mm
R 1 9.071
R2 4.657
L1 2
A1 863.938
A2 665.232
E 110
L2 3
NOTE: x is measured up from lower support.
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CHAPTER 2 Axially Loaded Members
Axial Displacement Diagram (ADD)
Axial Force Diagram (AFD)
10
0
–5 ×10–5
5
δ (x)
N (x)
0
−5
–1 ×10–4
–1.5 ×10–4
0
1
2
3
4
–2 ×10–4
5
1
0
2
3
4
5
x
x
(c) UNIFORM LOAD Q ON SEGMENT 2 SUCH THAT R2 0
P 13.728 kN
Equilibrium:
R1 s1 A1 9.071 kN
L2 3 m
R1 + R2 P q L2 6 set R2 0, solve for req’d q
q
P R1
1.552 kN/m
L2
q 1.552 kN/m
Problem 2.4-11 A bimetallic bar (or composite bar) of square cross
section with dimensions 2b 2b is constructed of two different metals
having moduli of elasticity E1 and E2 (see figure). The two parts of the
bar have the same cross-sectional dimensions. The bar is compressed by
forces P acting through rigid end plates. The line of action of the loads
has an eccentricity e of such magnitude that each part of the bar is
stressed uniformly in compression.
(a) Determine the axial forces P1 and P2 in the two parts of the bar.
(b) Determine the eccentricity e of the loads.
(c) Determine the ratio s1/s2 of the stresses in the two parts of the bar.
E2
P
b
b
e
e
E1
b
b
2b
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SECTION 2.4 Statically Indeterminate Structures
175
Solution 2.4-11 Bimetallic bar in compression
FREE-BODY DIAGRAM
(a) AXIAL FORCES
(Plate at right-hand end)
Solve simultaneously Eqs. (1) and (3):
P1 PE 1
PE 2
P E1 + E2 2 E1 + E2
;
(b ECCENTRICITY OF LOAD P
Substitute P1 and P2 into Eq. (2) and solve for e:
e EQUATIONS OF EQUILIBRIUM
F 0 P1
P2 P
(Eq. 1)
M 0 哵哴
b
b
Pe + P1 a b P2 a b 0
2
2
(Eq. 2)
b(E 2E 1)
2(E 2 E 1)
;
(c) RATIO OF STRESSES
s1 P1
P2
s2 A
A
s1
P1
E1
s2
P2
E2
;
EQUATION OF COMPATIBILITY
d2 d1
P2L
P1L
E 2A
E 1A
or
P2
P1
E2
E1
(Eq. 3)
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CHAPTER 2 Axially Loaded Members
Problem 2.4-12 A rigid bar of weight W 800 N hangs from three equally
spaced vertical wires (length L 150 mm, spacing a 50 mm): two of steel
and one of aluminum. The wires also support a load P acting on the bar. The
diameter of the steel wires is ds 2 mm, and the diameter of the aluminum wire
is da 4 mm. Assume Es 210 GPa and Ea 70 GPa.
a
L
S
a
A
S
Rigid bar
of weight W
(a) What load Pallow can be supported at the midpoint of the bar (x a) if the
allowable stress in the steel wires is 220 MPa and in the aluminum wire is
80 MPa? (See figure part a.)
(b) What is Pallow if the load is positioned at x a/2? (See figure part a.)
(c) Repeat (b) above if the second and third wires are switched as shown in figure
part b.
x
P
(a)
a
L
S
a
S
A
Rigid bar
of weight W
x
P
(b)
Solution 2.4-12
Numerical data:
W 800 N
L 150 mm
a 50 mm
dS 2 mm
dA 4 mm
ES 210 GPa
EA 70 GPa
sSa 220 MPa
AA p 2
dA
4
AA 13 mm2
sAa 80 MPa
AS p 2
dS
4
AS 3 mm2
(a) Pallow AT CENTER OF BAR
One-degree statically indeterminate - use reaction (RA) at top of aluminum bar as the redundant
compatibility: d1 d2 0
d1 P +W
L
b
a
2
ESAS
Statics:
2RS
RA P
W
downward displacement due to elongation of each steel wire under P
aluminum wire is cut at top
W if
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SECTION 2.4 Statically Indeterminate Structures
d2 RA a
177
L
L
+
b
2E SAS
EAAA
upward displ. due to shortening of steel wires and elongation of aluminum
wire under redundant RA
Enforce compatibility and then solve for RA:
d1 d2
RA so
P +W
L
a
b
2
ESAS
L
L
+
2ESAS
EAAA
RA ( P + W)
EAAA
EAAA + 2ESAS
and
s Aa RA
AA
Now use statics to find RS:
P + W (P + W)
P + W RA
RS 2
RS E AAA
E AAA + 2E S AS
2
RS (P + W)
and
E S AS
E AAA + 2E S AS
RS
s Sa AS
Compute stresses and apply allowable stress values:
s Aa ( P + W)
EA
E AAA + 2E SAS
s Sa ( P + W)
ES
E AAA + 2E SAS
Solve for allowable load P:
PAa s Aa a
E AAA + 2E SAS
b W
EA
PAa 1713 N
PSa sSa a
E AAA + 2E SAS
b W
ES
PSa 1504 N
(lower value of P controls)
; Pallow is controlled by steel wires
(b) Pallow IF LOAD P AT x a/2
Again, cut aluminum wire at top, then compute elongations of left and right steel wires:
d 1L a
d1 W
L
3P
+
ba
b
4
2
E SAS
d 1L + d 1R
2
d1 d 1R a
W
L
P
+ ba
b
4
2
E SAS
L
P +W
a
b where d1 displacement at x a
2
E SAS
Use d2 from part (a):
d2 RA a
L
L
+
b
2E S AS
E AAA
So equating d1 and d2, solve for RA: RA ( P + W)
E AAA
E AAA + 2E S AS
^ same as in part (a)
RSL RSL RA
W
3P
+
4
2
2
3P
W
+
4
2
stress in left steel wire exceeds that in right steel wire
(P + W)
E AAA
E AAA + 2E S AS
2
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CHAPTER 2 Axially Loaded Members
RSL PE A AA + 6PE S AS + 4WE S AS
4E A AA + 8E S AS
s Sa PE AAA + 6PE S AS + 4WE S AS 1
a b
4E AAA + 8E S AS
AS
Solve for Pallow based on allowable stresses in steel and aluminum:
sSa(4ASE A AA + 8E S AS2) (4WE S AS)
E A AA + 6E S AS
PSa PSa 820 N
;
PAa 1713 N
same as in part(a)
steel controls
(c) Pallow IF WIRES ARE SWITCHED AS SHOWN AND x a/2
Select RA as the redundant; statics on the two released structures:
(1) Cut aluminum wire—apply P and W, compute forces in left and right steel wires, then compute displacements
at each steel wire:
RSL d1L P
2
RSR L
P
a
b
2 E SAS
P
+W
2
d1R a
L
P
+ Wb a
b
2
E SAS
By geometry, d at aluminum wire location at far right is
d1 a
P
L
b
+ 2W b a
2
E SAS
(2) Next apply redundant RA at right wire, compute wire force and displacement at aluminum wire:
RSL RA
RSR 2RA
d2 RA a
5L
L
+
b
E S AS
E A AA
(3) Compatibility equate d1, d2 and solve for RA, then Pallow for aluminum wire:
RA a
P
L
+ 2Wb a
b
2
E S AS
5L
L
+
E S AS
E A AA
RA E A AAP + 4E A AAW
10E A AA + 2E S AS
s Aa RA
AA
sAa E AP + 4E AW
10E AAA + 2E S AS
PAa sAa(10E A AA + 2E S AS) 4E AW
EA
PAa 1713 N
(4) Statics or superposition—find forces in steel wires, then Pallow for steel wires:
RSL P
+ RA
2
RSL E A AAP + 4E A AAW
P
+
2
10E A AA + 2E S AS
RSL 6E A AAP + PE S AS + 4E A AAW
10E A AA + 2E S AS
larger than RSR, so use in allowable stress
calculations
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179
SECTION 2.4 Statically Indeterminate Structures
RSR P
+ W 2RA
2
RSR E AAAP + 4E AAAW
P
+ W
2
5E AAA + E S AS
RSR sSa RSL
AS
3E AAAP + PE SAS + 2E AAAW + 2WE S AS
10E AAA + 2E SAS
PSa sSaAS a
10E AAA + 2E S AS
4E A AAW
b
6E AAA + E S AS
6E AAA + E S AS
2
PSa 10sSaASE AAA + 2sSaA S E S 4E A AAW
6E A AA + E S AS
PSa 703 N
^ steel controls
;
Problem 2.4-13 A horizontal rigid bar of weight W 7200 lb is supported by
three slender circular rods that are equally spaced (see figure). The two outer
rods are made of aluminum (E1 10 106 psi) with diameter d1 0.4 in. and
length L1 40 in. The inner rod is magnesium (E2 6.5 106 psi) with diameter d2 and length L2. The allowable stresses in the aluminum and magnesium
are 24,000 psi and 13,000 psi, respectively.
If it is desired to have all three rods loaded to their maximum allowable
values, what should be the diameter d2 and length L2 of the middle rod?
Solution 2.4-13 Bar supported by three rods
BAR 1 ALUMINUM
E1 10 10 psi
6
FREE-BODY DIAGRAM OF RIGID BAR
EQUATION OF EQUILIBRIUM
Fvert 0
d1 0.4 in.
2F1
L1 40 in.
F2 W 0
(Eq. 1)
FULLY STRESSED RODS
s1 24,000 psi
F1 s1A1
F2 s2A2
BAR 2 MAGNESIUM
E2 6.5 106 psi
d2 ?
L2 ?
s2 13,000 psi
A1 pd 21
4
A2 pd 22
4
Substitute into Eq. (1):
2s1 a
pd 21
4
b + s2 a
pd 22
4
b W
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CHAPTER 2 Axially Loaded Members
Diameter d1 is known; solve for d2:
4W
d 22 ps2
2
2s1d 1
;
s2
d2 (Eq. 2)
4(7200 lb)
2(24,000 psi)(0.4 in.)2
p(13,000 psi)
13,000 psi
Length L1 is known; solve for L2:
0.70518 in.2 0.59077 in.2 0.11441 in.2
d2 0.338 in.
(Eq. 5)
Substitute (4) and (5) into Eq. (3):
L1
L2
s1 a b s2 a b
E1
E2
SUBSTITUTE NUMERICAL VALUES:
d 22 F2L 2
L2
s2 a b
E 2A2
E2
;
L2 L1 a
s1E 2
b
s2E 1
;
(Eq. 6)
SUBSTITUTE NUMERICAL VALUES:
EQUATION OF COMPATIBILITY
d1 d2
L 2 (40 in.) a
(Eq. 3)
FORCE-DISPLACEMENT RELATIONS
F1L 1
L1
s1 a b
d1 E 1A1
E1
24,000 psi 6.5 * 106 psi
ba
b
13,000 psi
10 * 106 psi
48.0 in.
(Eq. 4)
Problem 2.4-14 Three-bar truss ABC (see figure) is constructed of steel
y
pipes having a cross-sectional area A 3500 mm2 and a modulus of
elasticity E 210 GPa. Member BC is of length L 2.5 m, and the
angle between members AC and AB is known to be 60. Member AC
length is b 0.71L. Loads P 185 kN and 2P 370 kN act vertically
and horizontally at joint C, as shown. Joints A and B are pinned supports.
(Use the law of sines and law of cosines to find missing dimensions and
angles in the figure.)
(a) Find the support reactions at joints A and B. Use horizontal reaction Bx as the redundant.
(b) What is the maximum permissible value of load variable P if the
allowable normal stress in each truss member is 150 MPa?
P
C
2P
uC
L
b
uA = 60°
A
uB
B
c
x
Solution 2.4-14
NUMERICAL DATA
L 2.5 m
b 0.71
L 1.775 m
E 210 GPa
A 3500 mm2
P 185 kN
uA 60
sa 150 MPa
FIND MISSING DIMENSIONS AND ANGLES IN PLANE TRUSS FIGURE
xc b cos 1uA2 0.8875 m
b
L
sin(uB)
sin(uA)
so
yc b sin1uA2 1.5372 m
uB a sin a
b sin(uA)
b 37.94306
L
uC 180 (uA + uB) 82.05694
c
L
sin(uC) 2.85906 m
sin(uA)
or
c 3b2 + L2 2 b L cos (uC) 2.85906 m
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181
SECTION 2.4 Statically Indeterminate Structures
(a) SELECT Bx AS THE REDUNDANT; PERFORM SUPERPOSITION ANALYSIS TO FIND Bx THEN USE STATICS TO FIND REMAINING
REACTIONS. FINALLY USE METHOD OF JOINTS TO FIND MEMBER FORCES (SEE EXAMPLE 1-1)
dBx1 displacement in x-direction in released structure acted upon by loads P and 2P at joint C:
dBx1 1.2789911 mm
this displacement equals force in AB divided by flexibility of AB
dBX2 Bx
dBx2 displacement in x-direction in released structure acted upon by redundant Bx:
COMPATIBILITY EQUATION:
STATICS:
dBX1 + dBX2 0
BX so
©FX 0
AX BX 2 P 41.2 kN
©M A 0
By ©Fy 0
Ay P By 71.361 kN
c
EA
E A
dBX1 328.8 kN
c
1
[2 P (b sin(uA)) + P (b cos(uA))] 256.361 kN
c
REACTIONS:
Ax 41.2 kN
Ay 71.4 kN
Bx 329 kN
By 256 kN
(b) FIND MAXIMUM PERMISSIBLE VALUE OF LOAD VARIABLE P IF ALLOWABLE NORMAL STRESS IS 150 MPA
(1) Use reactions and Method of Joints to find member forces in each member for above loading.
Results: FAB 0
FBC 416.929 kN
FAC = 82.40 kN
(2) Compute member stresses:
sAB 0
sBC 416.93 kN
119.123 MPa
A
sAC 82.4 kN
23.543 MPa
A
(3) Maximum stress occurs in member BC. For linear analysis, the stress is proportional to the load so
Pmax sa
P 233 kN
sBC
So when downward load P 233 kN is applied at C and
horizontal load 2P 466 kN is applied to the right at C,
the stress in BC is 150 MPa
Problem 2.4-15 A rigid bar AB of length L 66 in. is hinged to a
support at A and supported by two vertical wires attached at points C
and D (see figure). Both wires have the same cross-sectional area
(A 0.0272 in.2) and are made of the same material (modulus E 30
106 psi). The wire at C has length h 18 in. and the wire at D has
length twice that amount. The horizontal distances are c 20 in. and
d 50 in.
(a) Determine the tensile stresses sC and sD in the wires due to the
load P 340 lb acting at end B of the bar.
(b) Find the downward displacement dB at end B of the bar.
2h
h
A
C
D
B
c
d
P
L
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CHAPTER 2 Axially Loaded Members
Solution 2.4-15 Bar supported by two wires
EQUATION OF COMPATIBILITY
dc
dD
c
d
FORCE-DISPLACEMENT RELATIONS
dC TCh
EA
dD TD(2h)
EA
(Eq. 2)
(Eqs. 3, 4)
SOLUTION OF EQUATIONS
Substitute (3) and (4) into Eq. (2):
TD(2h)
TCh
TC
2TD
or
cEA
dEA
c
d
TENSILE FORCES IN THE WIRES
h 18 in.
(Eq. 5)
Solve simultaneously Eqs. (1) and (5):
2h 36 in.
TC c 20 in.
d 50 in.
2cPL
2c2 + d 2
TD dPL
2c2 + d 2
(Eqs. 6, 7)
TENSILE STRESSES IN THE WIRES
TC
2cPL
sC A
A(2c2 + d 2)
L 66 in.
E 30 106 psi
A 0.0272 in.2
TD
dPL
A
A(2c2 + d 2)
P 340 lb
sD FREE-BODY DIAGRAM
DISPLACEMENT AT END OF BAR
(Eq. 8)
(Eq. 9)
2hTD L
2hPL2
L
dB dD a b a b d
EA d
EA(2c2 + d 2)
(Eq. 10)
SUBSTITUTE NUMERICAL VALUES
2c2
d2 2(20 in.)2
(a) sC 2cPL
2
2
A(2c + d )
(50 in.)2 3300 in.2
10,000 psi
DISPLACEMENT DIAGRAM
sD 2
A(2c + d )
12,500 psi
(b) dB EQUATION OF EQUILIBRIUM
MA 0 哵 哴 TC (c)
TD(d) PL
(Eq. 1)
(0.0272 in.2)(3300 in.2)
;
dPL
2
2(20 in.)(340 lb)(66 in.)
(50 in.)(340 lb)(66 in.)
(0.0272 in.2)(3300 in.2)
;
2hPL2
EA(2c2 + d 2)
2(18 in.)(340 lb)(66 in.)2
(30 * 106 psi)(0.0272 in.2)(3300 in.2)
0.0198 in.
;
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183
SECTION 2.4 Statically Indeterminate Structures
Problem 2.4-16 A rigid bar ABCD is pinned at point B and sup-
a = 250 mm
ported by springs at A and D (see figure). The springs at A and D
have stiffnesses k1 10 kN/m and k2 25 kN/m, respectively, and
the dimensions a, b, and c are 250 mm, 500 mm, and 200 mm,
respectively. A load P acts at point C.
If the angle of rotation of the bar due to the action of the load P
is limited to 3°, what is the maximum permissible load Pmax?
A
b = 500 mm
B
C
D
P
c = 200 mm
k 2 = 25 kN/m
k1 = 10 kN/m
Solution 2.4-16 Rigid bar supported by springs
a
A
EQUATION OF EQUILIBRIUM
b
B
C
MB 0
D
FD(b) 0
EQUATION OF COMPATIBILITY
dA
dD
a
b
FORCE-DISPLACEMENT RELATIONS
FA
FD
dD dA k1
k2
k2
k1
FA(a) P(c)
P
c
NUMERICAL DATA
a 250 mm
b 500 mm
SOLUTION OF EQUATIONS
c 200 mm
Substitute (3) and (4) into Eq. (2):
FA
FD
ak 1
bk 2
k1 10 kN/m
k2 25 kN/m
FREE-BODY DIAGRAM AND DISPLACEMENT DIAGRAM
b
FA
RB
P
(Eq. 2)
(Eqs. 3, 4)
(Eq. 5)
SOLVE SIMULTANEOUSLY EQS. (1) AND (5):
ack 1P
bck 2P
FA 2
FD 2
2
a k1 + b k2
a k 1 + b 2k 2
ANGLE OF ROTATION
FD
dD
bcP
cP
dD 2
2
u
2
k2
b
a k1 + b k2
a k 1 + b 2k 2
p
umax 3 rad
60
a
(Eq. 1)
FD
c
MAXIMUM LOAD
u
P (a 2k 1 + b 2k 2)
c
Pmax umax 2
(a k 1 + b 2k 2)
c
;
SUBSTITUTE NUMERICAL VALUES:
δA
Pmax A
B
C
D
p/60 rad
[(250 mm)2(10 kN/m)
200 mm
+ (500 mm)2(25 kN/m)]
1800 N
θ
;
δD
δC
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CHAPTER 2 Axially Loaded Members
Problem 2.4-17 A trimetallic bar is uniformly compressed by an
axial force P 9 kips applied through a rigid end plate (see figure).
The bar consists of a circular steel core surrounded by brass and
copper tubes. The steel core has diameter 1.25 in., the brass tube has
outer diameter 1.75 in., and the copper tube has outer diameter 2.25 in.
The corresponding moduli of elasticity are Es 30,000 ksi, Eb 16,000 ksi, and Ec 18,000 ksi.
Calculate the compressive stresses ss, sb, and sc in the steel, brass,
and copper, respectively, due to the force P.
P=9k
Copper tube
Brass tube
Steel core
1.25
in.
1.75
in.
2.25
in.
Solution 2.4-17
Numerical properties (kips, inches):
dc 2.25 in.
db 1.75 in.
Ec 18,000 ksi
ds 1.25 in.
Eb 16,000 ksi
Es 30000 ksi
As p 2
d
4 s
Ab p 2
1d d s22
4 b
Ac p
1d 2 d b22
4 c
P9k
EQUATION OF EQUILIBRIUM
Fvert 0
Ps
Pb
Pc P
(Eq. 1)
EQUATIONS OF COMPATIBILITY
ds db
dc ds
(Eqs. 2)
FORCE-DISPLACEMENT RELATIONS
ds PsL
PbL
PcL
db dc E sAs
E bAb
E cAc
(Eqs. 3, 4, 5)
SOLUTION OF EQUATIONS
Substitute (3), (4), and (5) into Eqs. (2):
Pb Ps
E bAb
E cAc
Pc Ps
E sAs
E sAs
(Eqs. 6, 7)
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SECTION 2.4 Statically Indeterminate Structures
185
SOLVE SIMULTANEOUSLY EQS. (1), (6), AND (7):
Ps P
E s As
3.95 k
E s As + E b Ab + E c Ac
Pb P
E bAb
2.02 k
E s As + E b Ab + E c Ac
Pc P
E c Ac
E s As + E b Ab + E c Ac
Ps
Pb
Pc 9
3.03 k
statics check
COMPRESSIVE STRESSES
Let EA EsAs
EbAb
EcAc
Ps
PE s
As
©EA
ss 3.22 ksi
;
sb Pb
PE b
Ab
©EA
sb 1.716 ksi
;
sc Pc
PE c
Ac
©EA
sc 1.93 ksi
;
ss © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 2 Axially Loaded Members
Thermal Effects
Problem 2.5-1 The rails of a railroad track are welded together at their ends (to form continuous rails and thus eliminate
the clacking sound of the wheels) when the temperature is 60°F.
What compressive stress s is produced in the rails when they are heated by the sun to 120°F if the coefficient of thermal
expansion a 6.5 106/°F and the modulus of elasticity E 30 106 psi?
Solution 2.5-1 Expansion of railroad rails
The rails are prevented from expanding because of
their great length and lack of expansion joints.
Therefore, each rail is in the same condition as a bar
with fixed ends (see Example 2-7).
The compressive stress in the rails may be calculated
as follows.
T 120°F 60°F 60°F
s Ea(T)
(30 106 psi)(6.5 106/°F)(60°F)
s 11,700 psi
;
Problem 2.5-2 An aluminum pipe has a length of 60 m at a temperature of 10°C. An adjacent steel pipe at the same temperature is 5 mm longer than the aluminum pipe.
At what temperature (degrees Celsius) will the aluminum pipe be 15 mm longer than the steel pipe? (Assume that the coefficients of thermal expansion of aluminum and steel are aa 23 106/°C and as 12 106/°C, respectively.)
Solution 2.5-2
Aluminum and steel pipes
INITIAL CONDITIONS
or, aa(T )La
La 60 m
T0 10°C
Ls 60.005 m
T0 10°C
aa 23 106/°C
as 12 106/°C
La L
as(T)Ls
Solve for T:
¢T ¢L + (L s L a)
aaL a asL s
;
FINAL CONDITIONS
Substitute numerical values:
Aluminum pipe is longer than the steel pipe by the
amount L 15 mm.
aaLa asLs 659.9 106 m/°C
T increase in temperature
¢T da aa(T )La
ds as(T )Ls
Ls
15 mm + 5 mm
659.9 * 106 m/C 30.31C
T T0 + ¢T 10C + 30.31C
40.3C
;
From the figure above:
da
La L
ds
Ls
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SECTION 2.5 Thermal Effects
187
Problem 2.5-3 A rigid bar of weight W 750 lb hangs from three equally spaced wires, two of steel and one of aluminum
(see figure). The diameter of the wires is 1/8 in . Before they were loaded, all three wires had the same length.
What temperature increase T in all three wires will result in the entire load being carried by the steel wires? (Assume
Es 30 106 psi, as 6.5 106/°F, and aa 12 106/°F.)
Solution 2.5-3 Bar supported by three wires
d1 increase in length of a steel wire due to temperature increase T
as (T)L
d2 increase in length of a steel wire due to load
W/2
WL
2E sAs
d3 increase in length of aluminum wire due to temperature increase T
aa(T)L
S steel
A aluminum
W 750 lb
For no load in the aluminum wire:
d1
d2 d3
d
1
in.
8
as(¢T)L +
As pd 2
0.012272 in.2
4
or
Es 30 106 psi
EsAs 368,155 lb
as 6.5 106/°F
aa 12 106/°F
L Initial length of wires
¢T WL
aa(¢T)L
2E sAs
W
2E sAs(aa as)
;
Substitute numerical values:
¢T 750 lb
(2)(368,155 lb)(5.5 * 106/F )
;
185F
NOTE: If the temperature increase is larger than T,
the aluminum wire would be in compression, which is
not possible. Therefore, the steel wires continue to
carry all of the load. If the temperature increase is less
than T, the aluminum wire will be in tension and
carry part of the load.
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CHAPTER 2 Axially Loaded Members
Problem 2.5-4 A steel rod of 15-mm diameter is held
snugly (but without any initial stresses) between rigid walls
by the arrangement shown in figure part a. (For the steel
rod, use a 12 * 106/C and E 200 GPa.)
Washer,
dw = 20 mm
12-mm diameter bolt
ΔT
(a) Calculate the temperature drop ¢T (degrees
Celsius) at which the average shear stress in the
12-mm diameter bolt becomes 45 MPa. Also, what
is the normal stress in the rod?
(b) What are the average bearing stresses in the bolt and
clevis at A and between the washer (dw 20 mm)
and wall (t 18 mm) at B?
(c) If the connection to the wall at B is changed to an
end plate with two bolts (see figure part b), what is
the required diameter db of each bolt if the temperature drop is ¢T 38C and the allowable bolt stress
is 90 MPa?
B
A
18 mm
15 mm
Clevis,
t = 10 mm
(a)
Bolt and washer
(db, dw)
12-mm diameter bolt
ΔT
A
15 mm
Clevis,
t = 10 mm
Mounting
plate (t)
(b)
Solution 2.5-4
NUMERICAL PROPERTIES
dr 15 mm
db 12 mm
tb 45 MPa
dw 20 mm
a 12 110 2
(a) TEMPERATURE DROP RESULTING IN BOLT SHEAR STRESS
p
Rod force P (E a ¢T) dr 2
4
tb 2
p db
c(E a ¢T)
2
p 2
d d
4 r
tc 10 mm
6
E 200 GPa
a¢T
s E a ¢T
twall 18 mm
and bolt in double shear with shear stress
tb P
2
t
As
t
P
p
2 db 2
4
E a ¢T dr 2
a b
2
db
tb 45 MPa
¢T srod db 2
2 tb
a b
E (1000) a dr
P 1000
p 2
d
4 r
¢T 24C
P (E a ¢T)
p 2
d
4 r
P 10 kN
srod 57.6 MPa
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189
SECTION 2.5 Thermal Effects
(b) BEARING STRESSES
BOLT AND CLEVIS
WASHER AT WALL
P
2
sbc db tc
sbw sbc 42.4 MPa
P
p
1dw 2 dr 22
4
sbw 74.1 MPa
(c) If the connection to the wall at B is changed to an end plate with two bolts (see Fig. b), what is the required diameter db of each bolt if temperature drop ¢T 38C and the allowable bolt stress is 90 MPa?
Find force in rod due to temperature drop.
¢T 38C
P (E a ¢T)
p 2
d
4 r
P 200 GPa
p
(15 mm)2 C 12 A 106 B D (38) 16116 N
4
P 16.12 kN
Each bolt carries one half of the force P:
16 12 kN
2
db 10.68 mm)
p
(90 MPa)
a4
Problem 2.5-5 A bar AB of length L is held between rigid supports
and heated nonuniformly in such a manner that the temperature increase
T at distance x from end A is given by the expression T TBx3/L3,
where TB is the increase in temperature at end B of the bar
(see figure part a).
db 10.68 mm
ΔTB
ΔT
0
A
(a) Derive a formula for the compressive stress sc in the bar. (Assume that
the material has modulus of elasticity E and coefficient of thermal
expansion a).
(b) Now modify the formula in (a) if the rigid support at A is replaced by
an elastic support at A having a spring constant k (see figure part b).
Assume that only bar AB is subject to the temperature increase.
B
x
L
(a)
ΔTB
ΔT
0
k
A
B
x
L
(b)
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CHAPTER 2 Axially Loaded Members
Solution 2.5-5
(a) ONE DEGREE STATICALLY INDETERMINATE—USE
SUPERPOSITION SELECT REACTION RB AS THE
REDUNDANT; FOLLOW PROCEDURE
Bar with nonuniform temperature change.
COMPRESSIVE FORCE P REQUIRED TO SHORTEN THE BAR BY
THE AMOUNT d
EAd
1
EAa(¢TB)
L
4
P
COMPRESSIVE STRESS IN THE BAR
sc x3
L
3
dB2 RB a
b
REMOVE THE SUPPORT AT THE END B OF THE BAR:
1
L
+
b
k
EA
Compatibility: solve for RB:
a a¢TB
RB Consider an element dx at a distance x from end A.
dd Elongation of element dx
dd a(¢T)dx a(¢TB)a
x3
L3
b dx
dd elongation of bar
L
;
(b) ONE DEGREE STATICALLY INDETERMINATE—USE
SUPERPOSITION.
Select reaction RB as the redundant then compute
bar elongations due to T and due to RB
L
dB1 a¢TB
due to temperature from above
4
At distance x:
¢T ¢TB a
Ea(¢TB)
P
A
4
L
1
dd a(¢TB) a 3 b dx a(¢TB)L
d
4
L
L0
L0
dB2 0
L
b
4
L
1
+
b
k
EA
RB a¢TB
EA
EA
J 4a
+ 1b K
kL
So compressive stress in bar is
sc x3
a
dB1
RB
A
E a1¢TB2
sc 4a
;
EA
+ 1b
kL
NOTE: sc in part (b) is the same as in part (a) if spring
constant k goes to infinity.
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SECTION 2.5 Thermal Effects
Problem 2.5-6 A plastic bar ACB having two different solid circular
cross sections is held between rigid supports as shown in the figure.
The diameters in the left- and right-hand parts are 50 mm and
75 mm, respectively. The corresponding lengths are 225 mm and
300 mm. Also, the modulus of elasticity E is 6.0 GPa, and the coefficient of thermal expansion a is 100 106/°C. The bar is subjected
to a uniform temperature increase of 30°C.
(a) Calculate the following quantities: (1) the compressive force
N in the bar; (2) the maximum compressive stress sc; and
(3) the displacement dC of point C.
(b) Repeat (a) if the rigid support at A is replaced by an elastic
support having spring constant k 50 MN/m (see figure
part b; assume that only the bar ACB is subject to the
temperature increase).
75 mm
50 mm C
A
225 mm
191
B
300 mm
(a)
75 mm
50 mm C
A
k
225 mm
B
300 mm
(b)
Solution
NUMERICAL DATA
d1 50 mm
d2 75 mm
L1 225 mm
L2 300 mm
dC a ¢T( L 1) RB
6
E 6.0 GPa
a 100 10 /°C
T 30°C
k 50 MN/m
(a) COMPRESSIVE FORCE N, MAXIMUM COMPRESSIVE STRESS
AND DISPLACEMENT OF PT. C
p
p
A1 d1 2 A2 d2 2
4
4
One-degree statically indeterminate—use RB as
redundant
dB1 aT(L1
L2)
L1
L2
dB2 RB a
+
b
E A1
E A2
Compatibility: dB1 dB2, solve for RB
RB a¢T( L 1 + L 2)
L1
L2
+
E A1
E A2
N RB
dC 0.314 mm
left
L1
E A1
; () sign means joint C moves
(b) COMPRESSIVE FORCE N, MAXIMUM COMPRESSIVE
STRESS AND DISPLACEMENT OF PART C FOR ELASTIC
SUPPORT CASE
Use RB as redundant as in part (a):
dB1 aT(L1
dB2 RB a
L2)
L1
L2
1
+
+ b
E A1
E A2
k
Now add effect of elastic support; equate dB1 and dB2
then solve for RB:
RB a¢T1L 1 + L 22
L1
L2
1
+
+
E A1
EA2
k
N 31.2 kN
N RB
;
N 51.8 kN ;
Maximum compressive stress in AC since
it has the smaller area (A1 A2):
N
scmax 15.91 MPa
A1
Superposition:
N
scmax 26.4 MPa
A1
Displacement dC of point C superposition of
displacements in two released structures at C:
dC a¢T( L 1) RBa
scmax scmax dC 0.546 mm
moves left
;
L1
1
+ b
E A1
k
; () sign means joint C
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CHAPTER 2 Axially Loaded Members
Problem 2.5-7 A circular steel rod AB (diameter d1 1.0 in., length L1 3.0 ft) has a bronze sleeve (outer diameter d2 1.25 in., length L2 1.0 ft)
shrunk onto it so that the two parts are securely bonded (see figure).
Calculate the total elongation d of the steel bar due to a temperature rise
T 500°F. (Material properties are as follows: for steel, Es 30 106 psi and
as 6.5 106/°F; for bronze, Eb 15 106 psi, and ab 11 106/°F.)
Solution 2.5-7 Steel rod with bronze sleeve
SUBSTITUTE NUMERICAL VALUES
as 6.5 106/°F
ab 11 106/°F
Es 30 106 psi
Eb 15 106 psi
d1 1.0 in.
L1 36 in.
L2 12 in.
ELONGATION OF THE TWO OUTER PARTS OF THE BAR
d1 as(T)(L1 L2)
6
(6.5 10 /°F)(500°F)(36 in. 12 in.)
0.07800 in.
ELONGATION OF THE MIDDLE PART OF THE BAR
The steel rod and bronze sleeve lengthen the same
amount, so they are in the same condition as the bolt
and sleeve of Example 2-8. Thus, we can calculate the
elongation from Eq. (2-23):
d2 As p 2
d 0.78540 in.2
4 1
d2 1.25 in.
Ab p
(d 2 d1 2) 0.44179 in.2
4 2
T 500°F
L2 12.0 in.
d2 0.04493 in.
TOTAL ELONGATION
d d1
d2 0.123 in.
;
(as E s As + ab E b Ab)(¢T)L 2
E s As + E b Ab
Problem 2.5-8 A brass sleeve S is fitted over a steel bolt B (see
figure), and the nut is tightened until it is just snug. The bolt has a
diameter dB 25 mm, and the sleeve has inside and outside diameters
d1 26 mm and d2 36 mm, respectively.
Calculate the temperature rise T that is required to produce a compressive stress of 25 MPa in the sleeve. (Use material properties as
follows: for the sleeve, aS 21 106/°C and ES 100 GPa; for
the bolt, aB 10 106/°C and EB 200 GPa.)
(Suggestion: Use the results of Example 2-8.)
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SECTION 2.5 Thermal Effects
193
Solution 2.5-8 Brass sleeve fitted over a Steel bolt
sS
E S AS
b
a1 +
E S(aS aB)
E B AB
¢T ;
SUBSTITUTE NUMERICAL VALUES:
sS 25 MPa
d2 36 mm
d1 26 mm
Subscript S means “sleeve”.
ES 100 GPa
EB 200 GPa
Subscript B means “bolt”.
aS 21 106/°C
Use the results of Example 2-8.
sS compressive force in sleeve
EQUATION (2-22a):
(aS aB)(¢T)E S E B AB
(Compression)
E S AS + E B AB
SOLVE FOR T:
sS ¢T sS(E S AS + E B AB)
(aS aB)E S E B AB
dB 25 mm
aB 10 106/°C
AS p 2
p
(d 2 d 21) (620 mm2)
4
4
AB E S AS
p
p
1.496
(dB)2 (625 mm2) 1 +
4
4
E B AB
¢T 25 MPa (1.496)
(100 GPa)(11 * 106/C)
T 34°C
;
(Increase in temperature)
or
Problem 2.5-9 Rectangular bars of copper and aluminum are held by
pins at their ends, as shown in the figure. Thin spacers provide a separation
between the bars. The copper bars have cross-sectional dimensions
0.5 in. 2.0 in., and the aluminum bar has dimensions 1.0 in. 2.0 in.
Determine the shear stress in the 7/16 in. diameter pins if the temperature is raised by 100°F. (For copper, Ec 18,000 ksi and ac 9.5 106/°F; for aluminum, Ea 10,000 ksi, and aa 13 106/°F.)
Suggestion: Use the results of Example 2-8.
Solution 2.5-9 Rectangular bars held by pins
Diameter of pin: dP Area of pin: AP 7
in. 0.4375 in.
16
p 2
d 0.15033 in.2
4 P
Area of two copper bars: Ac 2.0 in.2
Area of aluminum bar: Aa 2.0 in.2
T 100°F
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CHAPTER 2 Axially Loaded Members
Copper: Ec 18,000 ksi
ac 9.5 106/°F
Aluminum: Ea 10,000 ksi
SUBSTITUTE NUMERICAL VALUES:
Pa Pc (3.5 * 106/F)(100F)(18,000 ksi)(2 in.2)
1 + a
aa 13 106/°F
Use the results of Example 2-8.
Find the forces Pa and Pc in the aluminum bar and
copper bar, respectively, from Eq. (2-21).
18 2.0
ba
b
10 2.0
4,500 lb
FREE-BODY DIAGRAM OF PIN AT THE LEFT END
Replace the subscript “S” in that equation by “a” (for
aluminum) and replace the subscript “B” by “c” (for
copper), because a for aluminum is larger than a for
copper.
Pa Pc (aa ac)(¢T)E a Aa E c Ac
E a Aa + E c Ac
Note that Pa is the compressive force in the aluminum
bar and Pc is the combined tensile force in the two
copper bars.
Pa Pc (aa ac)(¢T)E c Ac
E c Ac
1 +
E a Aa
V shear force in pin
Pc/2
2,250 lb
t average shear stress on cross section of pin
t
2,250 lb
V
AP
0.15033 in.2
t 15.0 ksi
;
Problem 2.5-10 A rigid bar ABCD is pinned at end A and supported by
two cables at points B and C (see figure). The cable at B has nominal
diameter dB 12 mm and the cable at C has nominal diameter
dC 20 mm. A load P acts at end D of the bar.
What is the allowable load P if the temperature rises by 60°C and each
cable is required to have a factor of safety of at least 5 against its ultimate
load?
(Note: The cables have effective modulus of elasticity E 140 GPa and
coefficient of thermal expansion a 12 106/°C. Other properties of
the cables can be found in Table 2-1, Section 2.2.)
Solution 2.5-10 Rigid bar supported by two cables
FREE-BODY DIAGRAM OF BAR ABCD
From Table 2-1:
AB 76.7 mm2 E 140 GPa
T 60°C
AC 173 mm2
6
a 12 10 /°C
EQUATION OF EQUILIBRIUM
MA 0 哵 哴 TB(2b)
or 2TB 4TC 5P
TC(4b) P(5b) 0
(Eq. 1)
TB force in cable B TC force in cable C
dB 12 mm
dC 20 mm
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195
SECTION 2.5 Thermal Effects
DISPLACEMENT DIAGRAM
COMPATIBILITY:
dC 2dB
(Eq. 2)
FORCE-DISPLACEMENT AND TEMPERATURE-DISPLACEMENT
SOLVE SIMULTANEOUSLY EQS. (1) AND (6):
TB 0.2494 P 3,480
TC 1.1253 P 1,740
in which P has units of newtons.
(Eq. 7)
(Eq. 8)
SOLVE EQS. (7) AND (8) FOR THE LOAD P:
PB 4.0096 TB 13,953
PC 0.8887 TC 1,546
(Eq. 3)
(Eq. 4)
Factor of safety 5
(TB)allow 20,400 N
TCL
2TBL
+ a(¢T)L + 2a(¢T)L
EAC
EAB
(Eq. 9)
(Eq. 10)
(TC)ULT 231,000 N
(TC)allow 46,200 N
From Eq. (9): PB (4.0096)(20,400 N)
95,700 N
SUBSTITUTE EQS. (3) AND (4) INTO EQ. (2):
13,953 N
From Eq. (10): PC (0.8887)(46,200 N) 1546 N
39,500 N
Cable C governs.
or
2TBAC TCAB Ea(T)AB AC
(Eq. 6)
ALLOWABLE LOADS
From Table 2-1:
(TB)ULT 102,000 N
RELATIONS
TBL
+ a(¢T)L
dB EAB
TCL
+ a(¢T)L
dC EAC
SUBSTITUTE NUMERICAL VALUES INTO EQ. (5):
TB(346) TC(76.7) 1,338,000
in which TB and TC have units of newtons.
(Eq. 5)
Pallow 39.5 kN
;
Problem 2.5-11 A rigid triangular frame is pivoted at C and held by two identical
horizontal wires at points A and B (see figure). Each wire has axial rigidity EA 120 k
and coefficient of thermal expansion a 12.5 106/°F.
(a) If a vertical load P 500 lb acts at point D, what are the tensile forces TA and
TB in the wires at A and B, respectively?
(b) If, while the load P is acting, both wires have their temperatures raised by
180°F, what are the forces TA and TB?
(c) What further increase in temperature will cause the wire at B to become slack?
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CHAPTER 2 Axially Loaded Members
Solution 2.5-11 Triangular frame held by two wires
FREE-BODY DIAGRAM OF FRAME
(b) LOAD P AND TEMPERATURE INCREASE T
Force-displacement and temperature-displacement
relations:
TAL
+ a(¢T)L
(Eq. 8)
dA EA
TBL
+ a(¢T)L
EA
Substitute (8) and (9) into Eq. (2):
TAL
2TBL
+ a(¢T)L + 2a(¢T)L
EA
EA
dB (Eq. 9)
TA 2TB EAa(T)
EQUATION OF EQUILIBRIUM
or
MC 0 哵哴
Solve simultaneously Eqs. (1) and (10):
P(2b) TA(2b) TB(b) 0 or 2TA
TB 2P (Eq. 1)
DISPLACEMENT DIAGRAM
(Eq. 10)
1
TA [4P + EAa(¢T)]
5
2
TB [P EAa(¢T)]
5
(Eq. 11)
(Eq. 12)
Substitute numerical values:
P 500 lb
EA 120,000 lb
T 180°F
a 12.5 106/°F
1
TA (2000 lb + 270 lb) 454 lb
5
EQUATION OF COMPATIBILITY
dA 2dB
(Eq. 2)
(a) LOAD P ONLY
Force-displacement relations:
;
(c) WIRE B BECOMES SLACK
TAL
TBL
dB dA EA
EA
(L length of wires at A and B.)
Substitute (3) and (4) into Eq. (2):
(Eq. 3, 4)
Set TB 0 in Eq. (12):
P EAa(T)
or
¢T 2TBL
TAL
EA
EA
or TA 2TB
Solve simultaneously Eqs. (1) and (5):
4P
2P
TB 5
5
Numerical values:
P 500 lb
⬖TA 400 lb TB 200 lb
2
TB (500 lb 270 lb) 92 lb
5
;
TA (Eq. 5)
500 lb
P
EAa
(120,000 lb)(12.5 * 106/F)
333.3°F
Further increase in temperature:
(Eqs. 6, 7)
T 333.3°F 180°F
153°F
;
;
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197
SECTION 2.5 Thermal Effects
Misfits and Prestrains
Problem 2.5-12 A steel wire AB is stretched between rigid supports (see
figure). The initial prestress in the wire is 42 MPa when the temperature is 20°C.
(a) What is the stress s in the wire when the temperature drops to 0°C?
(b) At what temperature T will the stress in the wire become zero?
(Assume a 14 106/°C and E 200 GPa.)
Solution 2.5-12 Steel wire with initial prestress
s2 Ea(T)
s s1
s2 s1
Ea(T)
Initial prestress: s1 42 MPa
42 MPa
(200 GPa)(14 106/°C)(20°C)
Initial temperature: T1 20°C
42 MPa
56 MPa 98 MPa
E 200 GPa
a 14 106/°C
(a) STRESS s WHEN TEMPERATURE DROPS TO 0°C
T2 0°C
T 20°C
;
(b) TEMPERATURE WHEN STRESS EQUALS ZERO
s s1
s2 0
¢T s1
Ea
Ea(T) 0
s1
NOTE: Positive T means a decrease in temperature
and an increase in the stress in the wire.
(Negative means increase in temp.)
Negative T means an increase in temperature and a
decrease in the stress.
¢T Stress s equals the initial stress s1 plus the additional
stress s2 due to the temperature drop.
42 MPa
15C
(200 GPa)(14 * 106/C
T 20°C 15°C 35°C
;
0.008 in.
Problem 2.5-13 A copper bar AB of length 25 in. and diameter 2 in. is placed in position at
A
room temperature with a gap of 0.008 in. between end A and a rigid restraint (see figure). The
bar is supported at end B by an elastic spring with spring constant k 1.2 106 lb/in.
(a) Calculate the axial compressive stress sc in the bar if the temperature
rises 50°F. (For copper, use a 9.6 106/°F and E 16 106 psi.)
(b) What is the force in the spring? (Neglect gravity effects.)
(c) Repeat (a) if k : .
25 in.
d = 2 in.
B
k
C
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CHAPTER 2 Axially Loaded Members
Solution 2.5-13
Numerical data:
Compressive stress in bar:
RA
s
s 957 psi
A
L 25 in. d 2 in. d 0.008 in.
k 1.2 (106) lb/in. E 16 (106) psi
a 9.6 (106)/°F T 50°F
p
A d 2 A 3.14159 in.2
4
(a) ONE-DEGREE STATICALLY INDETERMINATE IF GAP CLOSES
aTL 0.012 in.
exceeds gap
Select RA as redundant and do superposition
analysis:
dA1 L
1
+ b
dA2 RAa
EA
k
Compatibility:
RA dA1
d¢
L
1
+
EA
k
dA2 d
dA2 d dA1
(b) FORCE IN SPRING Fk RC
STATICS
RA RC 0
RC RA
RC 3006 lb
(c) FIND COMPRESSIVE STRESS IN BAR IF k GOES TO INFINITY
FROM EXPRESSION FOR RA ABOVE, 1/k GOES TO ZERO
RA d ¢
L
EA
s 2560 psi
RA 8042 lb
s
RA
A
;
RA 3006 lb
2L
—
3
Problem 2.5-14 A bar AB having length L and axial rigidity EA is fixed
at end A (see figure). At the other end a small gap of dimension s exists
between the end of the bar and a rigid surface. A load P acts on the bar at
point C, which is two-thirds of the length from the fixed end.
If the support reactions produced by the load P are to be equal in
magnitude, what should be the size s of the gap?
Solution 2.5-14
;
A
s
L
—
3
C
B
P
Bar with a gap (load P )
FORCE-DISPLACEMENT RELATIONS
d1 P A 2L
3 B
EA
L length of bar
s size of gap
EA axial rigidity
d2 RBL
EA
Reactions must be equal; find s.
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199
SECTION 2.5 Thermal Effects
COMPATIBILITY EQUATION
Reactions must be equal.
d1 d2 s
⬖ RA RB P 2RB
or
RBL
2PL
s
3EA
EA
(Eq. 1)
RA reaction at end A (to the left)
or s PL
6EA
;
NOTE: The gap closes when the load reaches the
value P/4. When the load reaches the value P, equal
to 6EAs/L, the reactions are equal (RA RB P/2).
When the load is between P/4 and P, RA is greater than
RB. If the load exceeds P, RB is greater than RA.
RB reaction at end B (to the left)
P RA
P
2
Substitute for RB in Eq. (1):
PL
2PL
s
3EA
2EA
EQUILIBRIUM EQUATION
RB RB
Problem 2.5-15 Pipe 2 has been inserted snugly into Pipe 1,
but the holes for a connecting pin do not line up: there is a gap s.
The user decides to apply either force P1 to Pipe 1 or force P2 to
Pipe 2, whichever is smaller. Determine the following using the
numerical properties in the box.
Pipe 1 (steel)
Pipe 2 (brass)
Gap s
L1
RA
L
P1 2
(a) If only P1 is applied, find P1 (kips) required to close gap s;
if a pin is then inserted and P1 removed, what are reaction
P2
P1 at L1
forces RA and RB for this load case?
(b) If only P2 is applied, find P2 (kips) required to close gap s;
L
P2 at —2
if a pin is inserted and P2 removed, what are reaction
2
forces RA and RB for this load case?
Numerical properties
(c) What is the maximum shear stress in the pipes, for the
E1 = 30,000 ksi, E2 = 14,000 ksi
loads in parts (a) and (b)?
a1 = 6.5 10–6/°F, a2 = 11 10–6/°F
(d) If a temperature increase T is to be applied to the entire
Gap s = 0.05 in.
structure to close gap s (instead of applying forces P1 and
L1 = 56 in., d1 = 6 in., t1 = 0.5 in., A1 = 8.64 in.2
P2), find the T required to close the gap. If a pin is inserted
L2 = 36 in., d2 = 5 in., t2 = 0.25 in., A2 = 3.73 in.2
after the gap has closed, what are reaction forces RA and RB
for this case?
(e) Finally, if the structure (with pin inserted) then cools to the
original ambient temperature, what are reaction forces RA and RB?
RB
Solution 2.5-15
(a) FIND REACTIONS AT A AND B FOR APPLIED FORCE P1
First compute P1, required to close gap:
P1 E 1A1
s
L1
P1 231.4 k
;
Statically indeterminate analysis with RB as the
redundant:
dB1 s
L1
L2
+
b
d B2 RB a
E 1A1
E 2 A2
Compatibility: dB1
dB2 0
RB s
L1
L2
a
+
b
E 1A1
E 2A2
RA RB
RB 55.2 k
;
;
(b) FIND REACTIONS AT A AND B FOR APPLIED FORCE P2
E 2A2
;
s P2 145.1 k
L2
2
Analysis after removing P2 is same as in part (a), so
reaction forces are the same
P2 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 2 Axially Loaded Members
(c) MAXIMUM SHEAR STRESS IN PIPE 1 OR 2 WHEN EITHER
P1
P1 OR P2
A1
;
IS APPLIED t maxa tmaxa 13.39 ksi
2
P2
A2
t maxb ;
tmaxb 19.44 ksi
2
(d) REQUIRED ¢T AND REACTIONS AT A AND B
s
¢T reqd Treqd 65.8°F
a1L 1 + a2L 2
If pin is inserted but temperature remains at T
above ambient temperature, reactions are zero.
(e) IF TEMPERATURE RETURNS TO ORIGINAL AMBIENT TEMPERATURE, FIND REACTIONS AT A AND B
statically indeterminate analysis with RB as the
redundant Compatibility: dB1 dB2 0
Analysis is the same as in parts (a) and (b) above
since gap s is the same, so reactions are the same.
;
Problem 2.5-16 A nonprismatic bar ABC made up of
a, T
RA
segments AB (length L1, cross-sectional area A1) and BC
(length L2, cross-sectional area A2) is fixed at end A and free at
A
L1, EA1 B
end C (see figure). The modulus of elasticity of the bar is E.
A small gap of dimension s exists between the end of the bar and
an elastic spring of length L3 and spring constant k3. If bar ABC only
(not the spring) is subjected to temperature increase T determine the following.
s
D
RD
L3, k3
L2, EA2 C
(a) Write an expression for reaction forces RA and RD if the elongation of ABC exceeds gap length s.
(b) Find expressions for the displacements of points B and C if the elongation of ABC exceeds gap length s.
Solution 2.5-16
With gap s closed due to T, structure is one-degree
statically-indeterminate; select internal force (Q) at
juncture of bar and spring as the redundant. Use superposition of two released structures in the solution.
Compatibility: drel1
drel2 s
drel2 s aT(L1
L2)
drel1 relative displacement between end of bar at C
and end of spring due to T
drel1 aT(L1 L2)
drel1 is greater than gap length s
L1
L2
1
+
+
E A1
E A2
k3
E A1A2 k 3
Q
L 1A2 k 3 + L 2A1k 3 + EA1A2
drel2 relative displacement between ends of bar and
spring due to pair of forces Q, one on end of
bar at C and the other on end of spring
Q
L1
L2
drel2 Q a
+
b +
E A1
E A2
k3
L1
L2
1
+
+
b
drel2 Q a
EA1
E A2
k3
Q
drel2 s drel1
s a¢T1 L 1 + L 22
[ s a ¢T1 L 1 + L 22]
(a) REACTIONS AT A AND D
Statics: RA Q RD Q
RA s + a¢T1 L 1 + L 22
L1
L2
1
+
+
E A1
E A2
k3
RD RA
;
;
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SECTION 2.5 Thermal Effects
(b) DISPLACEMENTS AT B AND C
Use superposition of displacements in the two
released structures:
dB a¢T1 L 12 RA a
L1
b
E A1
dB a ¢T 1 L 12 [ s + a ¢T1 L 1 + L 22]
L1
L2
1
+
+
E A1
EA2
k3
;
dC a¢T1 L 1 + L 22 RAa
L1
L2
+
b
E A1
E A2
;
dC a ¢T1 L1 + L22 [ s + a ¢T1 L 1 + L 22]
a
201
L1
L2
1
+
+
E A1
EA2
k3
L1
b
EA1
a
L1
L2
+
b
EA1
EA2
Problem 2.5–17 Wires B and C are attached to a support at the left-hand
end and to a pin-supported rigid bar at the right-hand end (see figure).
Each wire has cross-sectional area A 0.03 in.2 and modulus of elasticity
E 30 106 psi. When the bar is in a vertical position, the length of each
wire is L 80 in. However, before being attached to the bar, the length of
wire B was 79.98 in. and of wire C was 79.95 in.
Find the tensile forces TB and TC in the wires under the action of a force
P 700 lb acting at the upper end of the bar.
700 lb
B
b
C
b
b
80 in.
Solution 2.5–17 Wires B and C attached to a bar
EQUILIBRIUM EQUATION
Mpin 0
TC(b)
2TB
;
TB(2b) P(3b)
TC 3P
(Eq. 1)
P 700 lb
A 0.03 in.2
E 30 106 psi
LB 79.98 in.
LC 79.95 in.
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CHAPTER 2 Axially Loaded Members
DISPLACEMENT DIAGRAM
Combine Eqs. (3) and (5):
SB 80 in. LB 0.02 in.
TCL
SC + d
EA
SC 80 in. LC 0.05 in.
(Eq. 7)
Eliminate between Eqs. (6) and (7):
TB 2TC EASB
2EASC
L
L
(Eq. 8)
Solve simultaneously Eqs. (1) and (8):
Elongation of wires:
dB SB
2d
(Eq. 2)
dC SC
d
(Eq. 3)
FORCE-DISPLACEMENT RELATIONS
TB L
TC L
dB dC EA
EA
EASB
2EASC
6P
+
5
5L
5L
;
TC 2EASB
4EASC
3P
+
5
5L
5L
;
SUBSTITUTE NUMERICAL VALUES:
EA
2250 lb/in.
5L
TB 840 lb
(Eqs. 4, 5)
SOLUTION OF EQUATIONS
45 lb 225 lb 660 lb
TC 420 lb 90 lb
;
;
450 lb 780 lb
(Both forces are positive, which means tension, as
required for wires.)
Combine Eqs. (2) and (4):
TBL
SB + 2d
EA
TB (Eq. 6)
P
Problem 2.5-18 A rigid steel plate is supported by three posts of
high-strength concrete each having an effective cross-sectional area
A 40,000 mm2 and length L 2 m (see figure). Before the load
P is applied, the middle post is shorter than the others by an amount
s 1.0 mm.
Determine the maximum allowable load Pallow if the allowable
compressive stress in the concrete is sallow 20 MPa.
(Use E 30 GPa for concrete.)
S
s
C
C
C
L
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SECTION 2.5 Thermal Effects
203
Solution 2.5-18 Plate supported by three posts
P
EQUILIBRIUM EQUATION
Steel plate
P2 P
2P1
P
(Eq. 1)
COMPATIBILITY EQUATION
s
1 shortening of outer posts
2 shortening of inner post
C
C
C
L
P1
P2
P1
1 2
s
(Eq. 2)
FORCE-DISPLACEMENT RELATIONS
d1 s size of gap 1.0 mm
L length of posts 2.0 m
A 40,000 mm2
allow 20 MPa
E 30 GPa
C concrete post
DOES THE GAP CLOSE?
Stress in the two outer posts when the gap is just
closed:
s
1.0 mm
s E Ea b (30 GPa) a
b
L
2.0 m
15 MPa
Since this stress is less than the allowable stress, the
allowable force P will close the gap.
P1 L
P2 L
d2 EA
EA
(Eqs. 3, 4)
SOLUTION OF EQUATIONS
Substitute (3) and (4) into Eq. (2):
P1L
P2L
EAs
+ s or P1 P2 EA
EA
L
(Eq. 5)
Solve simultaneously Eqs. (1) and (5):
P 3P1 EAs
L
By inspection, we know that P1 is larger than P2.
Therefore, P1 will control and will be equal to allow A.
Pallow 3sallow A EAs
L
2400 kN 600 kN 1800 kN
1.8 MN
;
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CHAPTER 2 Axially Loaded Members
Problem 2.5-19 A capped cast-iron pipe is compressed by a brass rod,
Nut & washer
3
dw = — in.
4
as shown. The nut is turned until it is just snug, then add an additional
quarter turn to pre-compress the CI pipe. The pitch of the threads of the
bolt is p 52 mils (a mil is one-thousandth of an inch). Use the numerical
properties provided.
(
)
Steel cap
(tc = 1 in.)
(a) What stresses sp and sr will be produced in the cast-iron pipe and
brass rod, respectively, by the additional quarter turn of the nut?
(b) Find the bearing stress sb beneath the washer and the shear stress
tc in the steel cap.
Cast iron pipe
(do = 6 in.,
di = 5.625 in.)
Lci = 4 ft
Brass rod
1
dr = — in.
2
)
(
Modulus of elasticity, E:
Steel (30,000 ksi)
Brass (14,000 ksi)
Cast iron (12,000 ksi)
Solution 2.5-19
The figure shows a section through the pipe, cap and rod
Arod 0.196 in.2
NUMERICAL PROPERTIES
Compatibility equation:
Lci 48 in.
Es 30000 ksi
Eb 14,000 ksi
1
Ec 12,000 ksi tc 1 in. p 52 (10 ) in. n 4
3
1
dr in. do 6 in. di 5.625 in.
dw in.
4
2
3
(a) FORCES AND STRESSES IN PIPE AND ROD
One degree statically indeterminate—cut rod at cap
and use force in rod (Q) as the redundant:
drel1 relative displacement between cut ends of
rod due to 1/4 turn of nut
Ends of rod move apart, not
drel1 np
together, so this is ().
drel2 relative displacement between cut ends of
rod due pair of forces Q
L + 2t c
L ci
+
b
E bArod
E cApipe
p
p
Apipe (do2 di2)
Arod dr2
4
4
Q
Apipe 3.424 in.2
drel1
drel2 0
np
L ci + 2t c
L ci
+
E bArod
E cApipe
Q 0.672 k
Frod Q
Statics:
Fpipe Q
Stresses:
sc sb Fpipe
Apipe
Frod
Arod
sc 0.196 ksi
sb 3.42 ksi
;
;
(b) BEARING AND SHEAR STRESSES IN STEEL CAP
sb Frod
p
(d 2 dr 2)
4 w
tc Frod
pdwt c
d rel2 Qa
sb 2.74 ksi
tc 0.285 ksi
;
;
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SECTION 2.5 Thermal Effects
205
Problem 2.5-20 A plastic cylinder is held snugly between a rigid plate
and a foundation by two steel bolts (see figure).
Determine the compressive stress sp in the plastic when the nuts on the
steel bolts are tightened by one complete turn.
Data for the assembly are as follows: length L 200 mm, pitch of the bolt
threads p 1.0 mm, modulus of elasticity for steel Es 200 GPa, modulus of
elasticity for the plastic Ep 7.5 GPa, cross-sectional area of one bolt As 36.0 mm2, and cross-sectional area of the plastic cylinder Ap 960 mm2.
Steel
bolt
L
Solution 2.5-20 Plastic cylinder and two steel bolts
COMPATIBILITY EQUATION
L 200 mm
P 1.0 mm
ds elongation of steel bolt
Es 200 GPa
dp shortening of plastic cylinder
As 36.0 mm2 (for one bolt)
ds
Ep 7.5 GPa
dp np
(Eq. 2)
FORCE-DISPLACEMENT RELATIONS
Ap 960 mm2
n 1 (See Eq. 2-24)
EQUILIBRIUM EQUATION
ds PpL
PsL
dp E sAs
E p Ap
(Eq. 3, Eq. 4)
SOLUTION OF EQUATIONS
Substitute (3) and (4) into Eq. (2):
Pp L
PsL
+
np
E s As
E p Ap
(Eq. 5)
Solve simultaneously Eqs. (1) and (5):
Ps tensile force in one steel bolt
Pp compressive force in plastic cylinder
Pp 2Ps
Pp (Eq. 1)
2npE s AsE p Ap
L(E p Ap + 2E s As)
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CHAPTER 2 Axially Loaded Members
STRESS IN THE PLASTIC CYLINDER
sp Pp
Ap
2EsAs 21.6 106 N
D EpAp
2np E s As E p
sp L(E p Ap + 2E s As)
2np N
2(1)(1.0 mm) N
a b a b
L D
200 mm
D
25.0 MPa
SUBSTITUTE NUMERICAL VALUES:
;
N Es As Ep 54.0 10 N /m
15
2
2
Problem 2.5-21 Solve the preceding problem if the data for the assembly are
as follows: length L 10 in., pitch of the bolt threads p 0.058 in., modulus
of elasticity for steel Es 30 106 psi, modulus of elasticity for the plastic
Ep 500 ksi, cross-sectional area of one bolt As 0.06 in.2, and crosssectional area of the plastic cylinder Ap 1.5 in.2
Steel
bolt
L
Solution 2.5-21 Plastic cylinder and two steel bolts
COMPATIBILITY EQUATION
L 10 in.
ds elongation of steel bolt
p 0.058 in.
Es 30 106 psi
dp shortening of plastic cylinder
ds
dp np
(Eq. 2)
As 0.06 in.2 (for one bolt)
Ep 500 ksi
Ap 1.5 in.2
n 1 (see Eq. 2-24)
FORCE-DISPLACEMENT RELATIONS
EQUILIBRIUM EQUATION
Ps tensile force in one steel bolt
ds Pp compressive force in plastic cylinder
Pp 2Ps
(Eq. 1)
Pp L
Ps L
dp E s As
E p Ap
(Eq. 3, Eq. 4)
SOLUTION OF EQUATIONS
Substitute (3) and (4) into Eq. (2):
Pp L
Ps L
+
np
E s As
E p Ap
(Eq. 5)
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207
SECTION 2.5 Thermal Effects
SUBSTITUTE NUMERICAL VALUES:
Solve simultaneously Eqs. (1) and (5):
Pp 2 np E s As E p Ap
N Es As Ep 900 109 lb2/in.2
L(E p Ap + 2E s As)
D Ep Ap
STRESS IN THE PLASTIC CYLINDER
sp Pp
Ap
sp 2 np E s As E p
;
L(E p Ap + 2E s As)
2Es As 4350 103 lb
2np N
2(1)(0.058in.) N
a b a b
L D
10 in.
D
2400 psi
;
d = np
Problem 2.5-22 Consider the sleeve made from two copper tubes joined by tin-lead
L1 = 40 mm,
d1 = 25 mm,
t1 = 4 mm
(a) Find the forces in the sleeve and bolt, Ps and PB, due to both the prestress
in the bolt and the temperature increase. For copper, use Ec 120 GPa and
ac 17 106/°C; for steel, use Es 200 GPa and as 12 106/°C.
The pitch of the bolt threads is p 1.0 mm. Assume s 26 mm and bolt
diameter db 5 mm.
(b) Find the required length of the solder joint, s, if shear stress in the sweated
joint cannot exceed the allowable shear stress taj 18.5 MPa.
(c) What is the final elongation of the entire assemblage due to both temperature change T and the initial prestress in the bolt?
Brass
cap
T
S
T
L2 = 50 mm,
d2 = 17 mm,
t2 = 3 mm
solder over distance s. The sleeve has brass caps at both ends, which are held in
place by a steel bolt and washer with the nut turned just snug at the outset. Then, two
“loadings” are applied: n 1/2 turn applied to the nut; at the same time the internal
temperature is raised by T 30°C.
Copper
sleeve
Steel
bolt
Solution 2.5-22
p 2
d
4 b
The figure shows a section through the sleeve, cap, and
bolt.
Ab NUMERICAL PROPERTIES
Ab 19.635 mm2
n
1
2
p 1.0 mm
A2 T 30°C
ac 17 (106)/°C
Ec 120 GPa
6
Es 200 GPa
as 12 (10 )/°C
taj 18.5 MPa
s 26 mm
L1 40 mm
t1 4 mm
d1 25 mm
d1 2t1 17 mm
db 5 mm
L2 50 mm
t2 3 mm
A1 p
[d 2 1 d1 2 t 122]
4 1
A1 263.894 mm2
p
[ d 2 2 1 d2 2t 222]
4
A2 131.947 mm2
(a) FORCES IN SLEEVE AND BOLT
One-degree statically indeterminate—cut bolt and
use force in bolt (PB) as redundant (see sketches):
dB1 np
asT(L1
L2 s)
d2 17 mm
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CHAPTER 2 Axially Loaded Members
dB2 PB c
L1 + L2 s
L1 s
L2 s
s
+
+
+
d
E s Ab
E cA1
E c A2
E c( A1 + A2)
Compatibility:
PB dB1
dB2 0
[ np + a s ¢T( L 1 + L 2 s)]
L1 + L2 s
L1 s
L2 s
s
c
+
+
+
d
E sAb
E c A1
E c A2
E c ( A1 + A2)
PB 25.4 kN
;
Ps PB
;
Sketches illustrating superposition procedure for statically-indeterminate analysis
δ = np
Cap
ΔT
L1
=
S
ΔT
L2
1° SI superposition analysis using
internal force in bolt as the
redundant
Actual
indeterminate
structure
under load(s)
Sleeve
Two released structures (see below) under:
(1)load(s); (2) redundant applied as a load
Bolt
δ = np
ΔT
+
Ps
S
S
ΔT
Cut
bolt
δB1
δB1
relative
displacement
across cut bolt, δB1
due to both δ and
ΔT (positive if pieces
move together)
Relative
displacement
across cut bolt, δB2
due to Pb (positive
if pieces move
together)
PB
δB2
PB
δB2
Apply redundant
internal force Ps &
find relative
displacement
across cut bolt,
δB2
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SECTION 2.5 Thermal Effects
(b) REQUIRED LENGTH OF SOLDER JOINT≈
P
t
As pd2s
As
PB
sreqd pd2taj
sreqd 25.7 mm
ds Ps c
209
L1 s
L2 s
s
+
+
d
E c A1
E c A2
E c (A1 + A2)
ds 0.064 mm
df db
ds
df 0.35 mm
;
(c) FINAL ELONGATION
df net of elongation of bolt (db) and shortening of
sleeve (ds)
db PB a
L1 + L2 s
b
E s Ab
db 0.413 mm
Problem 2.5-23 A polyethylene tube (length L) has a cap which when installed
compresses a spring (with undeformed length L1 L) by amount d (L1 L).
Ignore deformations of the cap and base. Use the force at the base of the spring as
the redundant. Use numerical properties in the boxes given.
(a) What is the resulting force in the spring, Fk?
(b) What is the resulting force in the tube, Ft?
(c) What is the final length of the tube, Lf?
(d) What temperature change T inside the tube will result in zero force in
the spring?
d = L1 – L
Cap (assume rigid)
Tube
(d0, t, L, at, Et)
Spring (k, L1 > L)
Modulus of elasticity
Polyethylene tube (Et = 100 ksi)
Coefficients of thermal expansion
at = 80 10–6/°F, ak = 6.5 10–6/°F
Properties and dimensions
1
d0 = 6 in. t = — in.
8
L1 = 12.125 in. > L = 12 in. k = 1.5 k /in.
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CHAPTER 2 Axially Loaded Members
Solution 2.5-23
Solve for redundant Q:
The figure shows a section through the tube, cap, and
spring.
Q
Properties and dimensions:
Fk 0.174 k
do 6 in.
t
1
in.
8
At 2.307 in.2
L1 12.125 in. L 12 in.
k 1.5 k/in.
(b) Ft force in tube Q
Lf L
note that Q result below is for
zero temperature (until part(d))
Lf L Qft
f
1
k
ft L
E tAt
d2 relative displacement across cut spring due to
redundant Q(f ft)
d1 relative displacement across cut spring due to
precompression and T d akTL1 atTL
Compatibility: d1
dc2
at(T)L
i.e., add displacements
for the two released structures
to initial tube length L
Lf 12.01 in.
;
(d) SET Q 0 TO FIND T REQUIRED TO REDUCE SPRING
(a) FORCE IN SPRING FK REDUNDANT Q
Flexibilities:
dc1
at 80 (106)/F
;
NOTE: If tube is rigid, Fk k 0.1875 k
d 0.125 in.
ak 6.5(106)/F
T 0
spring (Fk) and also
tensile force in tube
(c) FINAL LENGTH OF TUBE
Spring is 1/8 in. longer than tube
d L1 L
; compressive force in
Et 100 ksi
p
[ d o2 ( do 2 t)2]
4
At d + ¢T (a kL 1 + a tL)
Fk
f + ft
FORCE TO ZERO
¢T reqd d
(ak L 1 + at L)
Treqd 141.9F
Since at ak, a temp. increase is req’d to expand
tube so that spring force goes to zero.
d2 0
Steel wires
Problem 2.5-24 Prestressed concrete beams are sometimes
manufactured in the following manner. High-strength steel wires
are stretched by a jacking mechanism that applies a force Q, as
represented schematically in part (a) of the figure. Concrete is then
poured around the wires to form a beam, as shown in part (b).
After the concrete sets properly, the jacks are released and the
force Q is removed [see part (c) of the figure]. Thus, the beam is left
in a prestressed condition, with the wires in tension and the concrete
in compression.
Let us assume that the prestressing force Q produces in the steel
wires an initial stress s0 620 MPa. If the moduli of elasticity of the
steel and concrete are in the ratio 12:1 and the cross-sectional areas
are in the ratio 1:50, what are the final stresses ss and sc in the two
materials?
Q
Q
(a)
Concrete
Q
Q
(b)
(c)
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SECTION 2.5 Thermal Effects
211
Solution 2.5-24 Prestressed concrete beam
L length
s0 initial stress in wires
Q
620 MPa
As
As total area of steel wires
Ac area of concrete
50 As
Es 12 Ec
Ps final tensile force in steel wires
Pc final compressive force in concrete
EQUILIBRIUM EQUATION
Ps Pc
COMPATIBILITY EQUATION AND
FORCE-DISPLACEMENT RELATIONS
STRESSES
(Eq. 1)
d1 initial elongation of steel wires
ss sc QL
s0L
E sAs
Es
d2 final elongation of steel wires
Ps
As
s0
E sAs
1 +
E cAc
Pc
s0
Ac
Es
Ac
+
As
Ec
s0 620 MPa
d3 shortening of concrete
Pc L
E c Ac
or
s0L
PsL
PcL
Es
E sAs
E cAc
;
SUBSTITUTE NUMERICAL VALUES:
PsL
E sAs
d1 d2 d3
;
(Eq. 2, Eq. 3)
Es
As
1
12
Ec
Ac
50
ss 620 MPa
500 MPa (Tension)
12
1 +
50
sc 620 MPa
10 MPa (Compression)
50 + 12
;
;
Solve simultaneously Eqs. (1) and (3):
Ps Pc s0 As
Es As
1 +
Ec Ac
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CHAPTER 2 Axially Loaded Members
Problem 2.5-25 A polyethylene tube (length L) has a cap which is held in place
by a spring (with undeformed length L1 L). After installing the cap, the spring
is post-tensioned by turning an adjustment screw by amount d. Ignore deformations of the cap and base. Use the force at the base of the spring as the redundant.
Use numerical properties in the boxes below.
(a) What is the resulting force in the spring, Fk?
(b) What is the resulting force in the tube, Ft?
(c) What is the final length of the tube, Lf?
(d) What temperature change T inside the tube will result in zero force in the
spring?
Cap (assume rigid)
Tube
(d0, t, L, at, Et)
Spring (k, L1 < L)
d = L – L1
Adjustment
screw
Modulus of elasticity
Polyethylene tube (Et = 100 ksi)
Coefficients of thermal expansion
at = 80 10–6/°F, ak = 6.5 10–6/°F
Properties and dimensions
1
d0 = 6 in. t = — in.
8
L = 12 in. L1 = 11.875 in. k = 1.5 k/in.
Solution 2.5-25
The figure shows a section through the tube, cap, and
spring.
Pretension and temperature:
Spring is 1/8 in. shorter than tube.
Properties and dimensions:
d 0.125 in.
T 0
d L L1
Note that Q result below is for zero temperature (until
part (d)).
t
do 6 in.
1
in.
8
Et 100 ksi
L 12 in. L1 11.875 in.
k 1.5 k/in.
ak 6.5(106) at 80 (106)
At p 2
[ d 1 do 2t22]
4 o
At 2.307 in.
2
Flexibilities:
f
1
k
ft L
E tAt
(a) FORCE IN SPRING (Fk) REDUNDANT (Q)
Follow solution procedure outlined in Prob. 2.5-23
solution:
Q
d + ¢T 1a k L 1 + a t L2
f + ft
Fk
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SECTION 2.5 Thermal Effects
Fk 0.174 k
; also the compressive force in the
tube
at(T)L Lf 11.99 in.
(d) SET Q 0 TO FIND T REQUIRED TO REDUCE SPRING
FORCE TO ZERO
;
(b) FORCE IN TUBE Ft Q 0.174 k
(c) FINAL LENGTH OF TUBE AND SPRING Lf L dc1
dc2
Lf L Qft
213
;
¢ Treqd d
1ak L 1 + at L2
Treqd 141.6F
Since at ak, a temperature drop is required to
shrink tube so that spring force goes to zero.
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CHAPTER 2 Axially Loaded Members
Stresses on Inclined Sections
Problem 2.6-1 A steel bar of rectangular cross section
(1.5 in. 2.0 in.) carries a tensile load P (see figure). The allowable stresses in tension and shear are 14,500 psi and 7,100 psi,
respectively. Determine the maximum permissible load Pmax.
2.0 in.
P
P
1.5 in.
Solution 2.6-1
MAXIMUM LOAD—TENSION
2.0 in.
P
P
Pmax1 sa A
Pmax1 43500 lbs
MAXIMUM LOAD—SHEAR
Pmax2 2ta A
1.5 in.
Because tallow is less than one-half of sallow, the shear
stress governs.
NUMERICAL DATA
A 3 in.2
Pmax2 42,600 lbs
sa 14500 psi
ta 7100 psi
Problem 2.6-2 A circular steel rod of diameter d is subjected to a tensile
force P 3.5 kN (see figure). The allowable stresses in tension and shear
are 118 MPa and 48 MPa, respectively. What is the minimum permissible
diameter dmin of the rod?
d
P
P = 3.5 kN
Solution 2.6-2
P
d
P = 3.5 kN
Pmax 2t a a
dmin P 3.5 kN
sa 118 MPa
ta 48 MPa
Find Pmax then rod diameter.
since ta is less than 1/2 of sa, shear governs.
NUMERICAL DATA
p
d 2b
4 min
2
A pt a
P
dmin 6.81 mm
;
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215
SECTION 2.6 Stresses on Inclined Sections
Problem 2.6-3 A standard brick (dimensions 8 in. 4 in. 2.5 in.) is compressed
P
lengthwise by a force P, as shown in the figure. If the ultimate shear stress for brick is
1200 psi and the ultimate compressive stress is 3600 psi, what force Pmax is required to break
the brick?
8 in.
4 in.
2.5 in.
Solution 2.6-3 Standard brick in compression
Maximum shear stress:
P
t max 8 in.
4 in.
2.5 in.
sx
P
2
2A
sult 3600 psi
tult 1200 psi
Because tult is less than one-half of sult, the shear stress
governs.
t max A 2.5 in. 4.0 in. 10.0 in.2
Maximum normal stress:
sx P
2A
or P max 2Atult
P max 2(10.0 in.2)(1200 psi) 24,000 lb
;
P
A
Problem 2.6-4 A brass wire of diameter d 2.42 mm is stretched tightly
between rigid supports so that the tensile force is T 98 N (see figure). The
coefficient of thermal expansion for the wire is 19.5 * 106/ C and the
modulus of elasticity is E 110 GPa.
T
d
T
(a) What is the maximum permissible temperature drop T if the allowable
shear stress in the wire is 60 MPa?
(b) At what temperature changes does the wire go slack?
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CHAPTER 2 Axially Loaded Members
Solution 2.6-4 Brass wire in tension
d
T
ta 60 MPa
T
A
p 2
d
4
T
2 ta
A
¢Tmax Ea
NUMERICAL DATA
d 2.42 mm
T 98 N
a 19.5 (106)/°C E 110 GPa
¢Tmax 46C (drop)
(a) ¢Tmax (DROP IN TEMPERATURE)
(b) ¢T AT WHICH WIRE GOES SLACK
ta Increase ¢T until s 0:
s
tmax 2
T
s (E a ¢T)
A
T
E aA
¢T 9.93C (increase)
¢T T
E a ¢T
2A
2
Problem 2.6-5 A brass wire of diameter d 1/16 in. is stretched between
rigid supports with an initial tension T of 37 lb (see figure). Assume that the
coefficient of thermal expansion is 10.6 106/°F and the modulus of
elasticity is 15 106 psi.)
d
T
T
(a) If the temperature is lowered by 60°F, what is the maximum
shear stress tmax in the wire?
(b) If the allowable shear stress is 10,000 psi, what is the
maximum permissible temperature drop?
(c) At what temperature change T does the wire go slack?
Solution 2.6-5
d
T
T
(b) ¢Tmax FOR ALLOWABLE SHEAR STRESS
ta 10000 psi
NUMERICAL DATA
d
1
in.
16
T 37 lb
a 10.6 (106)/F
E 15 (106) psi
T 60F
p 2
A d
4
(a) tmax (DUE TO DROP IN TEMPERATURE)
sx
tmax 2
T
(E a ¢T)
A
tmax 2
tmax 10,800 psi
T
2ta
A
¢Tmax Ea
Tmax 49.9F
;
(c) T AT WHICH WIRE GOES SLACK
Increase T until s 0:
¢T T
E aA
T 75.9F (increase)
;
;
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SECTION 2.6 Stresses on Inclined Sections
Problem 2.6-6 A steel bar with diameter d 12 mm is subjected to
a tensile load P 9.5 kN (see figure).
d = 12 mm
P
217
P = 9.5 kN
(a) What is the maximum normal stress smax in the bar?
(b) What is the maximum shear stress tmax?
(c) Draw a stress element oriented at 45 to the axis of the bar and
show all stresses acting on the faces of this element.
(d) Repeat part (c) for a stress element oriented at 22.5 to the axis of the bar.
Solution 2.6-6
(a) d 12 mm
sx P 9.5 kN
A
p 2
d 1.131 * 104 m2
4
P
84 MPa
A
(b) tmax sx
42 MPa
2
On plane stress element rotated 45
(C) ROTATED STRESS ELEMENT (45) HAS NORMAL TENSILE STRESS sx/2 ON ALL FACES, Tmax (CW) ON
Tmax (CCW) ON y-FACE
txy1y1 tmax
sx1 sx
2
sy1 sx1
On rotated x-face:
sx1 42 MPa
On rotated y-face:
sy1 42 MPa
(d) u 22.5
tx1y1 42 MPa
CCW ROTATION OF ELEMENT
su sx cos(u)2 71.7 MPa
Eq. 2-31b
x-FACE, AND
tu on rotated x face
sx
sin(2 u) 29.7 MPa
2
sy sx cos a u +
p 2
b 12.3 MPa
2
on rotated y face
CW on rotated x-face
On rotated x-face:
sx1 71.7 MPa
On rotated y-face:
sy1 12.3 MPa
tx1y1 29.7 MPa
Problem 2.6-7 During a tension test of a mild-steel specimen
(see figure), the extensometer shows an elongation of 0.00120 in.
with a gage length of 2 in. Assume that the steel is stressed
below the proportional limit and that the modulus of elasticity
E 30 106 psi.
2 in.
T
T
(a) What is the maximum normal stress smax in
the specimen?
(b) What is the maximum shear stress tmax?
(c) Draw a stress element oriented at an angle of 45° to the
axis of the bar and show all stresses acting on the
faces of this element.
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CHAPTER 2 Axially Loaded Members
Solution 2.6-7
Tension test
(b) MAXIMUM SHEAR STRESS
The maximum shear stress is on a 45° plane and
equals sx/2.
tmax Elongation: d 0.00120 in.
(2 in. gage length)
sx
9,000 psi
2
;
(c) STRESS ELEMENT AT u 45°
d
0.00120 in.
Strain: 0.00060
L
2 in.
Hooke’s law: sx E (30 106 psi)(0.00060)
18,000 psi
(a) MAXIMUM NORMAL STRESS
sx is the maximum normal stress.
smax 18,000 psi
NOTE: All stresses have units of psi.
;
Problem 2.6-8 A copper bar with a rectangular cross section is held
45°
without stress between rigid supports (see figure). Subsequently, the
temperature of the bar is raised 50C.
A
B
(a) Determine the stresses on all faces of the elements A and B,
and show these stresses on sketches of the elements. (Assume
a 17.5 * 106/C and E 120 GPa.)
(b) If the shear stress at B is known to be 48 MPa at some inclination u, find angle u and show the stresses on a sketch of a
properly oriented element.
Solution 2.6-8
(a) a 17.5 A 106 B
¢T 50
sx E a ¢T 105 MPa
E 120 GPa
sx
tmax 52.5 MPa
2
at u 45
Element A: sx 105 MPa (compression);
Element B: tmax 52.5 MPa
(compression)
(b) tu 48 MPa
Eq. 2-31b
tu sx
sin(2 u)
2
so
u
2 tu
1
asin a
b 33.1
2
sx
su ux cos(u)2 73.8 MPa
sy sx cos a u +
2
CCW rotation of element
u 33.1
on rotated x face
p
b 31.2 MPa
2
on rotated y face
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219
SECTION 2.6 Stresses on Inclined Sections
Problem 2.6-9 The plane truss below is assembled from steel
y
P
C10 * 20 shapes (see Table 3(a) in Appendix F). Assume that
L 10 ft and b 0.71 L.
C
(a) If load variable P 49 k, what is the maximum shear stress
tmax in each truss member?
(b) What is the maximum permissible value of load variable P if
the allowable normal stress is 14 ksi and the allowable shear
stress is 7.5 ksi?
2P
uC
L
b
uA = 60°
A
uB
c
B
x
Solution 2.6-9
NUMERICAL DATA
L 10 ft
b 0.71 L
P 49 k
sa 14 ksi
ta 7.5 ksi
A 5.87 in.2
(a) FOR LINEAR ANALYSIS, MEMBER FORCES ARE PROPORTIONAL TO LOADING
P
FAC FROM EXAMPLE 1-1:
15.59 21.826 k
35
(solution for P 35 k)
P
(78.9)
FBC 35
FAB FBC 110.46 k
Normal stresses in each member:
sAC FAC
3.718 ksi
A
From Eq. 2-33:
sBC FBC
18.818 ksi
A
tmaxAC sAC
1.859 ksi
2
tmaxBC sBC
9.41 ksi
2
tmaxAB P
62.2 87.08 k
35
sAB FAB
14.835 ksi
A
sAB
7.42 ksi
2
(b) sa 6 2 * Ta so normal stress will control; lowest value governs here
MEMBER AC:
Pmaxs P
(sa A) 184.496 k
FAC
Pmaxt P
(2 ta A) 197.675 k
FAC
MEMBER AB:
Pmaxs P
(s A) 46.243 k
FAB a
Pmaxt P
(2 ta A) 49.546 k
FAB
MEMBER BC:
Pmaxs `
P
` 1sa A2 36.5 k
FBC
Pmaxt `
P
` 12 ta A2 39.059 k
FBC
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CHAPTER 2 Axially Loaded Members
Problem 2.6-10 A plastic bar of diameter d 32 mm is
compressed in a testing device by a force P 190 N applied as shown in the figure.
(a) Determine the normal and shear stresses acting on all
faces of stress elements oriented at (1) an angle u 0°,
(2) an angle u 22.5°, and (3) an angle u 45°. In each
case, show the stresses on a sketch of a properly oriented
element. What are smax and tmax?
(b) Find smax and tmax in the plastic bar if a re-centering
spring of stiffness k is inserted into the testing device, as
shown in the figure. The spring stiffness is 1/6 of the axial
stiffness of the plastic bar.
P = 190 N
100 mm
300 mm
200 mm
Re-centering
spring
(Part (b) only)
Plastic bar
u
d = 32 mm
k
Solution
NUMERICAL DATA
p
A d2
4
A 804.25 mm2
d 32 mm
P 190 N
a 100 mm
(a) STATICS—FIND COMPRESSIVE FORCE F AND STRESSES
IN PLASTIC BAR
F
P( a + b)
a
sx F
A
sx 945 kPa
or
(1) u 0
smax 945 kPa
sx
472 kPa
2
sx 945 kPa
(2) u 22.50
On
x-face:
uu +
p
2
su sxcos(u)2
su 138.39 kPa
(3) u 45
On
x-face:
su sxcos(u)2
su 472 kPa
From (1), (2), and (3) below:
tmax 472 kPa
y-face:
tu sxsin(u) cos(u)
tu 334.1 kPa
F 760 N
sx 0.945 MPa
;
tu sxsin(u) cos(u)
tu 334 kPa ;
On
b 300 mm
smax sx
su sxcos(u)2
su 807 kPa
;
;
tu sxsin(u) cos(u)
tu 472 kPa ;
On
y-face:
su sxcos(u)2
uu +
p
2
su 472.49 kPa
tu sxsin(u) cos(u)
tu 472.49 kPa
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SECTION 2.6 Stresses on Inclined Sections
(b) ADD SPRING—FIND MAXIMUM NORMAL AND SHEAR
100 mm
200 mm
δ/3
P
100 mm
6k
k
2kδ
kδ
F
δ
d
4P
b
5 k
8
P
5
F 304 N
Normal and shear stresses in plastic bar:
sx a Mpin 0
P (400) [2kd (100)
F (2k)a
Force in plastic bar:
STRESSES IN PLASTIC BAR
221
tmax kd (300)]
F
A
sx 0.38
sx
2
smax 378 kPa
;
tmax 189 kPa
;
4P
5 k
Problem 2.6-11 A plastic bar of rectangular cross section (b 1.5 in.
and h 3 in.) fits snugly between rigid supports at room temperature (68°F)
but with no initial stress (see figure). When the temperature of the bar is
raised to 160°F, the compressive stress on an inclined plane pq at midspan
becomes 1700 psi.
L
—
2
L
—
2
L
—
4
p
6
(a) What is the shear stress on plane pq? (Assume a 60 10 /°F
and E 450 103 psi.)
(b) Draw a stress element oriented to plane pq and show the stresses
Load P for part (c) only
acting on all faces of this element.
(c) If the allowable normal stress is 3400 psi and the allowable shear
stress is 1650 psi, what is the maximum load P (in x direction)
which can be added at the quarter point (in addition to
thermal effects above) without exceeding allowable stress values in the bar?
b
u
P
h
q
Solution 2.6-11
NUMERICAL DATA
b 1.5 in.
h 3 in.
A bh
T (160 68)F
T 92F
A 4.5 in.2
spq 1700 psi
(a) SHEAR STRESS ON PLANE PQ
Statically indeterminate analysis gives,
for reaction at right support:
R
A
A sx
b
s pq
sx
u 34.2°
tpq sxsin(u)cos(u)
E 450 (103) psi
sx u acos a
s pq
cos1u22 Now with u, can find shear stress on plane pq:
a 60 (106)/F
R EAaT
Using su sxcos(u)2:
R 11178 lb
tpq 1154 psi
spq sxcos(u)
2
Stresses at u
p/2 (y-face):
s y sx cos a u +
;
spq 1700 psi
p 2
b
2
sy 784 psi
sx 2484 psi
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CHAPTER 2 Axially Loaded Members
(b) STRESS ELEMENT FOR PLANE PQ
784
Set tmax ta and solve for Pmax1:
psi
si
4p
115
si θ = 34.2°
0p
170
)
t (b
Par
3 Pmax1
Ea¢T
+
2
8A
tmax 1650 psi
check
tmax sa 3400 psi
(c) MAXIMUM LOAD AT QUARTER POINT
ta 1650 psi
2ta 3300
less than sa,
so shear controls
Statically indeterminate analysis for P at L/4 gives
for reactions:
RR2 P
4
E a¢T
3P
+
2
8A
4A
Pmax1 12ta + Ea¢ T2
3
Pmax1 34,704 lb
ta RL2 3
P
4
s x Ea ¢ T +
sx 3300 psi
3Pmax1
4A
less than sa
Stresses in bar (L/4 to L):
sx
P
tmax 4A
2
Set tmax ta and solve for Pmax2:
s x E a¢ T Pmax2 4A(2ta
EaT)
Pmax2 14,688 lb
;
shear in segment (L/4
to L) controls
(tension for 0 to L/4 and compression for rest of bar)
From part (a) (for temperature increase T):
RR1 EAaT
RL1 EAaT
Stresses in bar (0 to L/4):
s x Ea¢ T +
3P
4A
tmax tmax Pmax2
E a¢ T
2
8A
s x Ea¢ T tmax 1650 psi
Pmax2
4A
sx 3300 psi
sx
2
Problem 2.6-12 A copper bar of rectangular cross section (b 18 mm
L
—
2
and h 40 mm) is held snugly (but without any initial stress) between
rigid supports (see figure). The allowable stresses on the inclined plane pq
bL
at midspan, for which 55°, are specified as 60 MPa in compression
p
and 30 MPa in shear.
P
(a) What is the maximum permissible temperature rise T if the
allowable stresses on plane pq are not to be exceeded? (Assume A
a 17 106/°C and E 120 GPa.)
Load for part (c) only
(b) If the temperature increases by the maximum permissible amount,
what are the stresses on plane pq?
(c) If the temperature rise T 28°C, how far to the right of end A
(distance bL, expressed as a fraction of length L) can load P 15 kN
be applied without exceeding allowable stress values in the bar? Assume
that sa 75 MPa and ta 35 MPa.
L
—
2
b
u
B
h
q
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SECTION 2.6 Stresses on Inclined Sections
223
Solution 2.6-12
(c) ADD LOAD P IN
x-DIRECTION TO TEMPERATURE
CHANGE AND FIND LOCATION OF LOAD
T 28C
P 15 kN from one-degree statically indeterminate
analysis, reactions RA and RB due to load P:
NUMERICAL DATA
u 55a
p
b rad
180
RA (1 b)P
RB bP
Now add normal stresses due to P to thermal
stresses due to T (tension in segment 0 to bL,
compression in segment bL to L).
b 18 mm
h 40 mm
A bh
A 720 mm2
spqa 60 MPa
tpqa 30 Mpa
E 120 GPa
a 17 (106)/C
T 20C
P 15 kN
Stresses in bar (0 to bL):
(a) FIND Tmax BASED ON ALLOWABLE NORMAL AND
SHEAR STRESS VALUES ON PLANE pq
s x
Ea
spq sxcos(u)2
tpq sxsin(u)cos(u)
Set each equal to corresponding allowable and
solve for sx:
spqa
sx1 sx1 182.38 MPa
cos1u22
¢ Tmax sx EaTmax
sx2 tpqa
sx2 63.85 MPa
sin1u2cos1u2
Lesser value controls, so allowable shear stress governs.
¢Tmax sx2
Ea
Tmax 31.3C
;
(b) STRESSES ON PLANE PQ FOR MAxIMUM TEMPERATURE
sx EaTmax
spq sxcos(u)2
sx 63.85 MPa
spq 21.0 MPa
tpq sxsin(u)cos(u)
tpq 30 MPa
RA
sx
tmax A
2
Shear controls so set tmax ta and solve for b:
sx Ea¢ T +
2ta E a¢ T +
b1
A
[2 t a + Ea¢ T]
P
b 5.1
Impossible so evaluate segment (bL to L):
Stresses in bar (bL to L):
RB
sx
tmax A
2
set tmax ta and solve for Pmax2
s x E a¢ T 2t a Ea¢ T b
;
(1 b)P
A
bP
A
A
[2 t a + E a ¢T]
P
b 0.62
;
;
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CHAPTER 2 Axially Loaded Members
Problem 2.6-13 A circular brass bar of diameter d is member AC
in truss ABC which has load P 5000 lb applied at joint C. Bar AC
is composed of two segments brazed together on a plane pq making
an angle a 36° with the axis of the bar (see figure). The allowable
stresses in the brass are 13,500 psi in tension and 6500 psi in shear.
On the brazed joint, the allowable stresses are 6000 psi in tension
and 3000 psi in shear. What is the tensile force NAC in bar AC?
What is the minimum required diameter dmin of bar AC?
NAC
A
a
p
q
θ = 60°
B
d
C
P
NAC
Solution 2.6-13
(2) Check tension and shear on brazed joint:
NUMERICAL DATA
P5k
a 36°
sa 13.5 ksi
ta 6.5 ksi
u
p
a
2
u 54
NAC
A
sx NAC
p 2
d
4
Tensile force NAC using Method of Joints at C:
P
sin(60)
Set equal to sja and solve for sx, then dreqd:
sja
sx 17.37 ksi
sx cos(u)2
(tension)
dreqd 4 NAC
dreqd 0.65 in.
;
A p sx
Minimum required diameter of bar AC:
Shear on brazed joint:
NAC 5.77 k
(1) Check tension and shear in bars; ta sa/2 so shear
sx
controls tmax :
2
NAC
A
sx 2ta = 13 ksi
NAC
2ta
Areqd 0.44 in.2
Areqd dmin 4 NAC
A p sX
su sxcos(u)2
tja 3.0 ksi
2ta dreqd Tension on brazed joint:
sja 6.0 ksi
NAC sx 4
Ap
Areqd
tu sxsin(u)cos(u)
sx `
dreqd tja
(sin(u) cos(u))
4 NAC
A p sX
`
sx 6.31 ksi
dreqd 1.08 in.
;
dmin 0.75 in.
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SECTION 2.6 Stresses on Inclined Sections
Problem 2.6-14 Two boards are joined by gluing along a scarf
joint, as shown in the figure. For purposes of cutting and gluing,
the angle a between the plane of the joint and the faces of the
boards must be between 10° and 40°. Under a tensile load P,
the normal stress in the boards is 4.9 MPa.
225
P
P
a
(a) What are the normal and shear stresses acting on the glued joint if a 20°?
(b) If the allowable shear stress on the joint is 2.25 MPa, what is the
largest permissible value of the angle a?
(c) For what angle a will the shear stress on the glued joint be
numerically equal to twice the normal stress on the joint?
Solution 2.6-14 Two boards joined by a scarf joint
Therefore: sin 2u 2(0.4592) 0.9184
Solving: 2u 66.69°
u 33.34°
10°
a
or
a 90° u
40°
or
113.31°
56.66°
⬖a 56.66°
or
33.34°
Since a must be between 10° and 40°, we select
Due to load P: sx 4.9 MPa
a 33.3°
(a) STRESSES ON JOINT WHEN a 20°
;
NOTE: If a is between 10° and 33.3°,
| tu | 2.25 MPa.
If a is between 33.3° and 40°,
| tu | 2.25 MPa.
(c) WHAT IS a if tu 2su?
u 90° a 70°
su sx cos2u (4.9 MPa)(cos 70°)2
0.57 MPa
;
tu sx sin u cos u
(4.9 MPa)(sin 70°)(cos 70°)
1.58 MPa
| tu | sx sin u cos u
`
| su | sx cos2u
t0
` 2
s0
sx sin u cos u 2sxcos2u
;
(b) LARGEST ANGLE a IF tallow 2.25 MPa
tallow sx sin u cos u
The shear stress on the joint has a negative sign. Its
numerical value cannot exceed tallow 2.25 MPa.
Therefore,
2.25 MPa (4.9 MPa)(sin u)(cos u) or sin u cos
u 0.4592
From trigonometry: sin u cos u Numerical values only:
1
sin 2u
2
sin u 2 cos u or tan u 2
u 63.43°
a 90° u
a 26.6°
;
NOTE: For a 26.6° and u 63.4°, we find
su 0.98 MPa and tu 1.96 MPa.
Thus, `
t0
` 2 as required.
s0
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CHAPTER 2 Axially Loaded Members
Problem 2.6-15 Acting on the sides of a stress element cut from a bar in
5000 psi
uniaxial stress are tensile stresses of 10,000 psi and 5,000 psi, as shown
in the figure.
su = 10,000 psi
tu tu
u
(a) Determine the angle u and the shear stress tu and show all stresses on a
sketch of the element.
(b) Determine the maximum normal stress smax and the maximum shear
stress tmax in the material.
10,000 psi
tu tu
5000 psi
Solution 2.6-15 Bar in uniaxial stress
1
1
tanu u 35.26
2
12
From Eq. (1) or (2):
tan2u ;
sx 15,000 psi
tu sx sin u cos u
(15,000 psi)(sin 35.26°)(cos 35.26°)
7,070 psi
;
Minus sign means that tu acts clockwise on the plane
for which u 35.26°.
(a) ANGLE u AND SHEAR STRESS tu
su sx cos2u
su 10,000 psi
sx s0
2
cos u
10,000 psi
cos2u
PLANE AT ANGLE u
su
(1)
90°
90° sx[cos(u
90°)]2 sx[sin u]2
sx sin2u
su
NOTE: All stresses have units of psi.
90° 5,000 psi
sx s 0 90
sin2u
(b) MAXIMUM NORMAL AND SHEAR STRESSES
5,000 psi
sin2u
Equate (1) and (2):
10,000 psi
2
cos u
smax sx 15,000 psi
(2)
tmax sx
7,500 psi
2
;
;
5,000 psi
sin2u
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227
SECTION 2.6 Stresses on Inclined Sections
Problem 2.6-16 A prismatic bar is subjected to an axial force that produces a
tensile stress su 65 MPa and a shear stress tu 23 MPa on a certain inclined plane
(see figure). Determine the stresses acting on all faces of a stress element oriented at
30° and show the stresses on a sketch of the element.
65 MPa
u
23 MPa
Solution 2.6-16
(4754 + 65s x)
s 2x
sx 4754
65
sx 73.1 MPa
Find u and sx for stress state shown in figure.
cos (u) su sxcos(u)2
sin (u) so
su
A
u acos P
su 65 MPa
A sx Q
su
u 19.5
Pa
3M
18.
A sx
1
0
a
7
31.
su
MP
a
9
54.
sx
MP
θ = 30°
tu sxsin(u) cos(u)
tu
sx
tu
2
b A
1
su
su
su
sx A sx
sx
a
65 2
23 2 65
b a b
sx
sx
sx
a
65 2
65
23 2
b a b + a b 0
sx
sx
sx
sx
a
su
a
sx
b
Now find su and tu for u 30°:
su1 sxcos(u)2
su1 54.9 MPa
tu sxsin(u) cos(u)
su2 s x cos a u +
p 2
b
2
;
tu 31.7 MPa
su2 18.3 MPa
;
;
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CHAPTER 2 Axially Loaded Members
Problem 2.6-17 The normal stress on plane pq of a prismatic bar in
p
tension (see figure) is found to be 7500 psi. On plane rs, which makes
an angle b 30° with plane pq, the stress is found to be 2500 psi.
Determine the maximum normal stress smax and maximum shear
stress tmax in the bar.
r b
P
P
s
q
Solution 2.6-17
Bar in tension
SUBSTITUTE NUMERICAL VALUES INTO EQ. (2):
cosu1
)
cos(u1 + 30
7500 psi
A 2500 psi
23 1.7321
Solve by iteration or a computer program:
u1 30°
Eq. (2-31a):
MAXIMUM NORMAL STRESS (FROM EQ. 1)
su sxcos2u
b 30°
smax sx PLANE pq: s1 sxcos2u1
PLANE rs: s2 sxcos2(u1
s1 7500 psi
b)
s1
cos2u1
s2
cos2(u1 + b)
2
cos u1
10,000 psi
s2 2500 psi
7500 psi
cos2 30
;
MAXIMUM SHEAR STRESS
Equate sx from s1 and s2:
sx s1
(Eq. 1)
tmax sx
5,000 psi
2
;
or
cos2u1
2
cos (u1 + b)
cosu1
s1
s1
s2 cos(u1 + b) A s2
(Eq. 2)
Problem 2.6-18 A tension member is to be constructed of two
pieces of plastic glued along plane pq (see figure). For purposes of
cutting and gluing, the angle u must be between 25° and 45°.
The allowable stresses on the glued joint in tension and shear are
5.0 MPa and 3.0 MPa, respectively.
P
u
p
P
q
(a) Determine the angle u so that the bar will carry the largest
load P. (Assume that the strength of the glued joint
controls the design.)
(b) Determine the maximum allowable load Pmax if the
cross-sectional area of the bar is 225 mm2.
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SECTION 2.6 Stresses on Inclined Sections
Solution 2.6-18
229
Bar in tension with glued joint
(a) DETERMINE ANGLE Q FOR LARGEST LOAD
Point A gives the largest value of sx and hence the
largest load. To determine the angle u corresponding to point A, we equate Eqs. (1) and (2).
25° u 45°
5.0 MPa
A 225 mm2
cos2u
On glued joint: sallow 5.0 MPa
tan u tallow 3.0 MPa
ALLOWABLE STRESS sx IN TENSION
su sxcos2u
sx su
2
cos u
3.0 MPa
sin u cos u
3.0
u 30.96
5.0
;
(b) DETERMINE THE MAXIMUM LOAD
5.0 MPa
2
cos u
(1)
tu sxsin u cos u
Since the direction of tu is immaterial, we can write:
tu | sxsin u cos u
From Eq. (1) or Eq. (2):
sx 5.0 MPa
2
cos u
3.0 MPa
6.80 MPa
sin u cos u
Pmax sxA (6.80 MPa)(225 mm2)
1.53 kN
;
or
sx |tu|
sin u cosu
3.0 MPa
sin u cosu
(2)
GRAPH OF EQS. (1) AND (2)
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CHAPTER 2 Axially Loaded Members
Problem 2.6-19 Plastic bar AB of rectangular cross section
(b 0.75 in. and h 1.5 in.) and length L 2 ft. is fixed at
A and has a spring support (k 18 k/in.) at C (see figure).
Initially, the bar and spring have no stress. When the temperature of the bar is raised by 100F, the compressive stress on an
inclined plane pq at Lu 1.5 ft becomes 950 psi. Assume the
spring is massless and is unaffected by the temperature
change. Let a 55 * 106/F and E 400 ksi.
L = 2 ft p
A
B
u
Lu = 1.5 ft
C
k
b
h
q
(a) What is the shear stress tu on plane pq? What is angle u?
(b) Draw a stress element oriented to plane pq, and show the stresses acting on all faces of this element.
(c) If the allowable normal stress is ;1000 psi and the allowable shear stress is ;560 psi, what is the maximum permissible value of spring constant k if allowable stress values in the bar are not to be exceeded?
(d) What is the maximum permissible length L of the bar if allowable stress values in the bar are not to be exceeded?
(Assume k 18 k/in.)
(e) What is the maximum permissible temperature increase ( ¢T) in the bar if allowable stress values in the bar are not to
be exceeded? (Assume L 2 ft and k 18 k/in.)
Solution 2.6-19
NUMERICAL DATA
a 55 11062
E 400 ksi
su 950 psi
sa 1000 psi
L 2 ft
¢T 100
ta 560 psi
k 18 k/in.
L u 1.5 ft
A bh
b 0.75 in.
f
h 1.5 in.
1
5.556 * 105 in./lb
k
(a) FIND u AND Tu
R2 redundant
R2 su
b 0.351
u acos a
A sx
sx cos(u)2 950 psi
u 0.351
a ¢T L
1.212 * 103 lb
L
a
b + f
EA
cos(2 u) 0.763
or
u 20.124
sx 1077.551 psi
or
R2
1077.551 psi
A
su
A sx
0.939
u 20.124
sx
(1 + cos(2 u)) 950 psi
2
tu sx sin(u) cos(u) 348.1 psi
tu 348 psi
sx sy sx cos a u +
p 2
b 127.551 psi
2
2 u 0.702
sx
tu sin(2 u) 348.1 psi
2
u 20.1
(b) FIND sx1 AND sy1
sx1 sx cos(u)2
sy1 sx cos a u +
sx1 950 psi
sy1 127.6 psi
p 2
b
2
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SECTION 2.6 Stresses on Inclined Sections
231
(c) GIVEN L 2 ft, FIND k max
sa A
k max1 or
a ¢T L sa A a
k max2 L
b
EA
15625 lb/in. 6 controls (based on sallow)
2 ta A
L
a ¢T L 2 ta A a
b
EA
19444.444 lb/in.
based on allowable shear stress
k max 15625 lb/in.
(d) GIVEN ALLOWABLE NORMAL AND SHEAR STRESSES, FIND Lmax
k 18000 lb/in.
sx or
R2
A
sa A L max2 a ¢T L
L
a
b + f
EA
L max1 sa A (f)
sa
a a ¢T +
b
E
1.736 ft 6 controls (based on sallow)
2 ta A (f)
2.16 ft 6 based on Tallow
2 ta
b
a a ¢T +
E
L max 1.736 ft
(e) FIND Tmax GIVEN L, k, AND ALLOWABLE STRESSES
k 18000 lb/in.
L 2 ft
sa 1000 psi
ta 560 psi
¢Tmax1 ¢Tmax2 a
L
+ f b sa A
EA
a L
a
92.803F
L
+ f b 2 ta A
EA
a L
6 based on sallow
¢T 100
103.939F 6 based on Tallow
¢Tmax 92.8F
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CHAPTER 2 Axially Loaded Members
Strain Energy
When solving the problems for Section 2.7, assume that the
material behaves linearly elastically.
Problem 2.7-1 A prismatic bar AD of length L, cross-sectional
area A, and modulus of elasticity E is subjected to loads 5P, 3P,
and P acting at points B, C, and D, respectively (see figure).
Segments AB, BC, and CD have lengths L/6, L/2, and L/3,
respectively.
5P
A
3P
B
L
—
6
P
C
D
L
—
2
L
—
3
(a) Obtain a formula for the strain energy U of the bar.
(b) Calculate the strain energy if P 6 k, L 52 in.,
A 2.76 in.2, and the material is aluminum with
E 10.4 106 psi.
Solution 2.7-1 Bar with three loads
(a) STRAIN ENERGY OF THE BAR (EQ. 2-40)
P6k
L 52 in.
U g
E 10.4 106 psi
A 2.76 in.2
L
L
L
1
c(3P)2 a b + (2P)2 a b + (P)2 a b d
2EA
6
2
3
23P 2L
P 2L 23
a b 2EA 6
12EA
INTERNAL AXIAL FORCES
NAB 3P
NBC 2P
NCD P
LENGTHS
L AB L
6
L BC L
2
L CD L
3
N 2i L i
2E iAi
;
(b) SUBSTITUTE NUMERICAL VALUES:
U
23(6 k)2(52 in.)
12(10.4 * 106 psi)(2.76 in.2)
125 in.-lb
;
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Page 233
SECTION 2.7 Strain Energy
Problem 2.7-2 A bar of circular cross section having two different
2d
diameters d and 2d is shown in the figure. The length of each segment
of the bar is L/2 and the modulus of elasticity of the material is E.
d
P
(a) Obtain a formula for the strain energy U of the bar due to the load P.
(b) Calculate the strain energy if the load P 27 kN, the length L 600 mm, the diameter d 40 mm, and the material is brass with E 105 GPa.
L
—
2
233
P
L
—
2
Solution 2.7-2 Bar with two segments
2d
d
P
L
—
2
L
—
2
(b) SUBSTITUTE NUMERICAL VALUES:
(a) STRAIN ENERGY OF THE BAR
Add the strain energies of the two segments of the
bar (see Eq. 2-42).
P 2(L/2)
N 2i L i
1
cp
2
i1 2 E iAi
2E
4 (2d)
2
U g
1
5P 2L
P 2L 1
a 2 + 2b pE 4d
d
4pEd 2
P
p 2 d
4 (d )
1
P 27 kN
L 600 mm
d 40 mm
E 105 GPa
U
5(27 kN 2)(600 mm)
4p(105 GPa)(40 mm)2
;
1.036 N # m 1.036 J
Problem 2.7-3 A three-story steel column in a building supports roof
;
P1
and floor loads as shown in the figure. The story height H is 10.5 ft, the cross-sectional area
A of the column is 15.5 in.2, and the modulus of elasticity E of the steel is 30 106 psi.
Calculate the strain energy U of the column assuming P1 40 k and P2 P3 60 k.
P2
P3
H
H
H
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CHAPTER 2 Axially Loaded Members
Solution 2.7-3 Three-story column
Upper segment: N1 P1
P1
P2
H
Middle segment: N2 (P1
P2)
Lower segment: N3 (P1
P2
STRAIN ENERGY
U g
P3
P3)
H
H
N 2i L i
2E iAi
H
[P 2 + (P1 + P2)2 + (P1 + P2 + P3)2]
2EA 1
H
[Q]
2EA
[Q] (40 k)2 + (100 k)2 + (160 k)2 37,200 k 2
H 10.5 ft
E 30 106 psi
A 15.5 in.2
P1 40 k
2EA 2(30 * 106 psi)(15.5 in.2) 930 * 106 lb
P2 P3 60 k
To find the strain energy of the column, add the strain
energies of the three segments (see Eq. 2-42).
U
(10.5 ft)(12 in./ft)
6
[37,200 k 2]
930 * 10 lb
5040 in.-lb
;
Problem 2.7-4 The bar ABC shown in the figure is loaded by a
force P acting at end C and by a force Q acting at the midpoint B. The
bar has constant axial rigidity EA.
(a) Determine the strain energy U1 of the bar when the force P acts
alone (Q 0).
(b) Determine the strain energy U2 when the force Q acts alone
(P 0).
(c) Determine the strain energy U3 when the forces P and Q act
simultaneously upon the bar.
Q
A
L
—
2
B
P
L
—
2
C
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235
SECTION 2.7 Strain Energy
Solution 2.7-4 Bar with two loads
Q
A
B
L
—
2
(c) FORCES P AND Q ACT SIMULTANEOUSLY
P
L
—
2
C
(a) FORCE P ACTS ALONE (Q 0)
U1 P 2L
2EA
Segment BC: UBC P 2(L/2)
P 2L
2EA
4EA
Segment AB: UAB (P + Q)2(L/2)
2EA
;
PQL
Q 2L
P 2L
+
+
4EA
2EA
4EA
U3 UBC + UAB PQL
Q 2L
P 2L
+
+
2EA
2EA
4EA
(b) FORCE Q ACTS ALONE (P 0)
U2 Q 2(L/2)
Q 2L
2EA
4EA
;
;
(Note that U3 is not equal to U1 U2. In this case,
U3 U1 U2. However, if Q is reversed in direction,
U3 U1 U2. Thus, U3 may be larger or smaller than
U1 U2.)
Problem 2.7-5 Determine the strain energy per unit volume (units of psi) and the strain energy per unit weight (units of in.)
that can be stored in each of the materials listed in the accompanying table, assuming that the material is stressed to the proportional limit.
DATA FOR PROBLEM 2.7-5
Material
Weight
density
(lb/in.3)
Modulus of
elasticity
(ksi)
Proportional
limit
(psi)
Mild steel
Tool steel
Aluminum
Rubber (soft)
0.284
0.284
0.0984
0.0405
30,000
30,000
10,500
0.300
36,000
75,000
60,000
300
Solution 2.7-5 Strain-energy density
STRAIN ENERGY PER UNIT VOLUME
DATA:
Material
Weight
density
(lb/in.3)
Modulus of
elasticity
(ksi)
Proportional
limit
(psi)
Mild steel
Tool steel
Aluminum
Rubber (soft)
0.284
0.284
0.0984
0.0405
30,000
30,000
10,500
0.300
36,000
75,000
60,000
300
U
P 2L
2EA
Volume V AL
Stress s u
P
A
s2PL
U
V
2E
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CHAPTER 2 Axially Loaded Members
At the proportional limit:
At the proportional limit:
u uR modulus of resistance
uW uR s2PL
2E
s2PL
2gE
(Eq. 2)
(Eq. 1)
RESULTS
STRAIN ENERGY PER UNIT WEIGHT
U
2
P L
2EA
Mild steel
Tool steel
Aluminum
Rubber (soft)
Weight W gAL
g weight density
uW uR (psi)
uw (in.)
22
94
171
150
76
330
1740
3700
U
s2
W
2gE
P
B
Problem 2.7-6 The truss ABC shown in the figure is subjected to a horizontal
load P at joint B. The two bars are identical with cross-sectional area A and modulus
of elasticity E.
(a) Determine the strain energy U of the truss if the angle b 60°.
(b) Determine the horizontal displacement dB of joint B by equating the strain
energy of the truss to the work done by the load.
b
b
A
C
L
Solution 2.7-6 Truss subjected to a load P
P
B
b
b
A
C
L
b 60°
Fvert 0
↓
LAB LBC L
FAB sin b
FBC sin b 0
sin b 13/2
FAB FBC
cos b 1/2
Fhoriz 0 : ←
FREE-BODY DIAGRAM OF JOINT B
FAB cos b FBC cos b
FAB FBC ↓
(Eq. 1)
P0
P
P
P
2 cos b
2(1/2)
(Eq. 2)
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SECTION 2.7 Strain Energy
(b) HORIZONTAL DISPLACEMENT OF JOINT B (EQ. 2-44)
Axial forces: NAB P (tension)
NBC P (compression)
dB (a) STRAIN ENERGY OF TRUSS (EQ. 2-42)
U g
(NBC)2L
(NAB)2L
N 2i L i
P 2L
+
2E iAi
2EA
2EA
EA
237
2 P 2L
2PL
2U
a
b P
P EA
EA
;
;
Problem 2.7-7 The truss ABC shown in the figure supports a
A
horizontal load P1 300 lb and a vertical load P2 900 lb. Both
bars have cross-sectional area A 2.4 in.2 and are made of steel with
E 30 106 psi.
(a) Determine the strain energy U1 of the truss when the load P1
acts alone (P2 0).
(b) Determine the strain energy U2 when the load P2 acts alone
(P1 0).
(c) Determine the strain energy U3 when both loads act
simultaneously.
30°
C
B P1 = 300 lb
P2 = 900 lb
60 in.
Solution 2.7-7 Truss with two loads
LAB LBC
120
in. 69.282 in.
cos 30
13
2EA 2(30 106 psi)(2.4 in.2) 144 106 lb
FORCES FAB AND FBC IN THE BARS
From equilibrium of joint B:
FAB 2P2 1800 lb
FBC P1 P2 13 300 lb 1558.8 lb
P1 300 lb
P2 900 lb
A 2.4 in.2
Force
P1 alone
P2 alone
P1 and P2
FAB
FBC
0
300 lb
1800 lb
1558.8 lb
1800 lb
1258.8 lb
E 30 106 psi
LBC 60 in.
(a) LOAD P1 ACTS ALONE
b 30°
U1 sin b sin 30 1
2
cos b cos 30 13
2
(FBC)2L BC
(300 lb)2(60 in.)
2EA
144 * 106 lb
0.0375 in.-lb
;
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CHAPTER 2 Axially Loaded Members
(b) LOAD P2 ACTS ALONE
U2 1
c(F )2L + (FBC)2L BC d
2EA AB AB
+ (1258.8 lb)2(60 in.) d
1
c(1800 lb)2(69.282 in.)
2EA
+ (1558.8 lb)2(60 in.) d
370.265 * 106 lb 2-in.
144 * 106 lb
2.57 in.-lb ;
1
c(1800 lb)2(69.282 in.)
2EA
319.548 * 106 lb 2-in.
144 * 106 lb
2.22 in.- lb
;
NOTE: The strain energy U3 is not equal to U1
U2.
(c) LOADS P1 AND P2 ACT SIMULTANEOUSLY
U3 1
c(FAB)2L AB + (FBC)2L BC d
2EA
Problem 2.7-8 The statically indeterminate structure shown
in the figure consists of a horizontal rigid bar AB supported by
five equally spaced springs. Springs 1, 2, and 3 have stiffnesses
3k, 1.5k, and k, respectively. When unstressed, the lower ends of
all five springs lie along a horizontal line. Bar AB, which has
weight W, causes the springs to elongate by an amount d.
(a) Obtain a formula for the total strain energy U of the
springs in terms of the downward displacement
d of the bar.
(b) Obtain a formula for the displacement d by equating the
strain energy of the springs to the work done by the
weight W.
(c) Determine the forces F1, F2, and F3 in the springs.
(d) Evaluate the strain energy U, the displacement d, and the
forces in the springs if W 600 N and k 7.5 N/mm.
1
3k
k
1.5k
2
3
1.5k
2
A
1
3k
B
W
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239
SECTION 2.7 Strain Energy
Solution 2.7-8 Rigid bar supported by springs
(c) FORCES IN THE SPRINGS
F1 3kd F3 kd ;
W 600 N k 7.5 N/mm 7500 N/mm
k2 1.5k
k3 k
U 5kd2 5ka
d downward displacement of rigid bar
kd2
Eq. (2-40b)
2
(a) STRAIN ENERGY U OF ALL SPRINGS
For a spring: U 2
2
3kd
1.5kd
kd
b + 2a
b +
2
2
2
2
5kd2
Work done by the weight W equals
;
d
W
8.0 mm
10k
;
F1 3W
180 N
10
;
F2 3W
90 N
20
;
F3 W
60 N
10
;
;
Wd
2
Strain energy of the springs equals 5kd2
Wd
5kd2
2
and d W
10k
;
NOTE: W 2F1
Problem 2.7-9 A slightly tapered bar AB of rectangular cross
section and length L is acted upon by a force P (see figure). The
width of the bar varies uniformly from b2 at end A to b1 at end B.
The thickness t is constant.
(a) Determine the strain energy U of the bar.
(b) Determine the elongation d of the bar by equating the
strain energy to the work done by the force P.
W 2
W2
b 10k
20k
2.4 N # m 2.4 J
(b) DISPLACEMENT d
...
W
10
;
(d) NUMERICAL VALUES
k1 3k
U 2a
3W
3W
F2 1.5kd 10
20
F3 600 N (Check)
2F2
A
B
b2
b1
P
L
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CHAPTER 2 Axially Loaded Members
Solution 2.7-9 Tapered bar of rectangular cross section
Apply this integration formula to Eq. (1):
U
(b2 b1)x L
P2
1
ln cb2 c
dd
2Et (b2 b1)11 2
L
0
L
b(x) b2 (b2 b1)x
L
U
A(x) tb(x)
t cb2 (b2 b1)x
d
L
d
[N(x)]2dx
( Eq. 2-43)
L 2EA(x)
L
;
b2
PL
2U
ln
P
Et(b2 b1) b1
;
NOTE: This result agrees with the formula derived in
Prob. 2.3-13.
L
P2
P 2dx
dx
2Et L0 b2 (b2 b1)Lx
L0 2Etb(x)
From Appendix C:
b2
P 2L
ln
2Et(b2 b1) b1
(b) ELONGATION OF THE BAR (EQ. 2-44)
(a) STRAIN ENERGY OF THE BAR
U
P
L
L
c
ln b1 ln b2 d
2Et (b2 b1)
(b2 b1)
2
(1)
1
dx
ln (a + bx)
a
+
bx
b
L
Problem 2.7-10 A compressive load P is transmitted through a rigid plate to three
magnesium-alloy bars that are identical except that initially the middle bar is slightly
shorter than the other bars (see figure). The dimensions and properties of the assembly
are as follows: length L 1.0 m, cross-sectional area of each bar A 3000 mm2,
modulus of elasticity E 45 GPa, and the gap s 1.0 mm.
(a) Calculate the load P1 required to close the gap.
(b) Calculate the downward displacement d of the rigid plate when P 400 kN.
(c) Calculate the total strain energy U of the three bars when P 400 kN.
(d) Explain why the strain energy U is not equal to Pd/2.
(Hint: Draw a load-displacement diagram.)
P
s
L
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SECTION 2.7 Strain Energy
241
Solution 2.7-10 Three bars in compression
(c) STRAIN ENERGY U FOR P 400 kN
U g
EAd2
2L
Outer bars:
d 1.321 mm
Middle bar:
d 1.321 mm s
0.321 mm
U
s 1.0 mm
L 1.0 m
EA
[2(1.321 mm)2 + (0.321 mm)2]
2L
1
(135 * 106 N/m)(3.593 mm2)
2
For each bar:
A 3000 mm2
243 N # m 243 J
E 45 GPa
(d) LOAD-DISPLACEMENT DIAGRAM
EA
135 * 106 N/m
L
U 243 J 243 N # m
(a) LOAD P1 REQUIRED TO CLOSE THE GAP
EAd
PL
In general, d and P EA
L
For two bars, we obtain:
P1 2 a
;
Pd
1
(400 kN)(1.321 mm) 264 N # m
2
2
Pd
because the
2
load-displacement relation is not linear.
The strain energy U is not equal to
EAs
b 2(135 * 106 N/m)(1.0 mm)
L
P1 270 kN
;
(b) DISPLACEMENT d FOR P 400 kN
Since P P1, all three bars are compressed.
The force P equals P1 plus the additional force
required to compress all three bars by the amount
d s.
P P1 + 3 a
EA
b(d s)
L
or 400 kN 270 kN
(d 0.001 m)
U area under line OAB.
3(135 106 N/m)
Solving, we get d 1.321 mm
Pd
area under a straight line from O to B, which is
2
larger than U.
;
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CHAPTER 2 Axially Loaded Members
Problem 2.7-11 A block B is pushed against three springs by a force P
(see figure). The middle spring has stiffness k1 and the outer springs each
have stiffness k2. Initially, the springs are unstressed and the middle spring
is longer than the outer springs (the difference in length is denoted s).
s
k2
P
(a) Draw a force-displacement diagram with the force P as ordinate
and the displacement x of the block as abscissa.
(b) From the diagram, determine the strain energy U1 of the springs
when x 2s.
(c) Explain why the strain energy U1 is not equal to Pd/2, where d 2s.
k1
B
k2
x
Solution 2.7-11 Block pushed against three springs
s
k2
P
k1
B
k2
x
Force P0 required to close the gap:
P0 k1s
(a) FORCE-DISPLACEMENT DIAGRAM
(1)
FORCE-DISPLACEMENT RELATION BEFORE GAP IS CLOSED
P k1x
(0
x
s)(0
P
P0)
(2)
FORCE-DISPLACEMENT RELATION AFTER GAP IS CLOSED
All three springs are compressed. Total stiffness equals
k1 2k2. Additional displacement equals x s. Force
P equals P0 plus the force required to compress all three
springs by the amount x s.
P P0
(k1
2k2)(x s)
k1s
(k1
2k2)x k1s 2k2s
P (k1
2k2)x 2k2s
(x s); (P P0)
(b) STRAIN ENERGY U1 WHEN x 2s
(3)
P1 force P when x 2s
Substitute x 2s into Eq. (3):
P1 2(k1
k2)s
U1 Area below force-displacement curve
(4)
1
1
1
P0s + P0s + (P1 P0)s P0s + P1s
2
2
2
k 1s 2 + (k 1 + k 2)s 2
U1 (2k1
k2)s2
;
(5)
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243
SECTION 2.7 Strain Energy
(c) STRAIN ENERGY U1 IS NOT EQUAL TO
Pd
2
Pd
1
P1(2 s) P1s 2(k 1 + k 2)s 2
2
2
(This quantity is greater than U1.)
For d 2s:
Pd
area under a straight line from O to B, which
2
is larger than U1.
Pd
Thus,
is not equal to the strain energy because
2
the force-displacement relation is not linear.
U1 area under line OAB.
Problem 2.7-12 A bungee cord that behaves linearly
elastically has an unstressed length L0 760 mm and a
stiffness k 140 N/m.The cord is attached to two pegs, distance b 380 mm apart, and pulled at its midpoint by a
force P 80 N (see figure).
b
A
B
(a) How much strain energy U is stored in the cord?
(b) What is the displacement dC of the point where the
load is applied?
(c) Compare the strain energy U with the quantity
PdC/2.
(Note: The elongation of the cord is not small compared
to its original length.)
Solution 2.7-12
C
P
Bungee cord subjected to a load P.
DIMENSIONS BEFORE THE LOAD P IS APPLIED
From triangle ACD:
1
d 2L20 b2 329.09 mm
2
DIMENSIONS AFTER THE LOAD P IS APPLIED
L0 760 mm
(1)
L0
380 mm
2
b 380 mm
Let x distance CD
k 140 N/m
Let L1 stretched length of bungee cord
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CHAPTER 2 Axially Loaded Members
From triangle ACD:
L1
b
a b + x2
2
A 2
(2)
L1 2b + 4x
2
L1 L0 +
or
2
2
(3)
L0 a1 P 1 2
2
2
b + 4x2 1b + 4x
4kx
P 1 2
b b + 4x2
4kx
(7)
This equation can be solved for x.
EQUILIBRIUM AT POINT C
SUBSTITUTE NUMERICAL VALUES INTO EQ. (7):
Let F tensile force in bungee cord
760 mm c1 * 1(380 mm)2 + 4x2
(8)
142.857 1
b 144,400 + 4x2
x
(9)
760 a1 L 1/2
F
P L1 1
F a ba ba b
P/2
x
2
2
x
(80 N)(1000 mm/m)
d
4(140 N/m)x
Units: x is in millimeters
Solve for x (Use trial-and-error or a computer program):
x 497.88 mm
P
b 2
1 + a b
2A
2x
(4)
kd2
2
From Eq. (5):
U
ELONGATION OF BUNGEE CORD
Let d elongation of the entire bungee cord
d
F
P
b2
1 + 2
k
2k A
4x
(5)
Final length of bungee cord original length
P
b2
L1 L0 + d L0 +
1 + 2
2k A
4x
SOLUTION OF EQUATIONS
Combine Eqs. (6) and (3):
L1 L0 +
2
P
b
1 + 2 1b2 + 4x2
2k A
4x
(a) STRAIN ENERGY U OF THE BUNGEE CORD
d
(6)
d
k 140 N/m
P 80 N
b2
P
1 + 2 305.81 mm
2k A
4x
1
U (140 N/m)(305.81 mm)2 6.55 N # m
2
U 6.55 J
;
(b) DISPLACEMENT dC OF POINT C
dC x d 497.88 mm 329.09 mm
168.8 mm
;
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SECTION 2.7 Strain Energy
245
(c) COMPARISON OF STRAIN ENERGY U WITH THE
QUANTITY PdC/2
U 6.55 J
PdC
1
(80 N)(168.8 mm) 6.75 J
2
2
The two quantities are not the same. The work done by
the load P is not equal to PdC/2 because the loaddisplacement relation (see below) is non-linear when
the displacements are large. (The work done by the
load P is equal to the strain energy because the bungee
cord behaves elastically and there are no energy
losses.)
U area OAB under the curve OA.
PdC
area of triangle OAB, which is greater than U.
2
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CHAPTER 2 Axially Loaded Members
Impact Loading
The problems for Section 2.8 are to be solved on the basis of the assumptions
and idealizations described in the text. In particular, assume that the material
behaves linearly elastically and no energy is lost during the impact.
Collar
Problem 2.8-1 A sliding collar of weight W 150 lb falls from a height
h 2.0 in. onto a flange at the bottom of a slender vertical rod (see figure).
The rod has length L 4.0 ft, cross-sectional area A 0.75 in.2, and modulus
of elasticity E 30 106 psi.
Calculate the following quantities: (a) the maximum downward
displacement of the flange, (b) the maximum tensile stress in the rod,
and (c) the impact factor.
L
Rod
h
Flange
Probs. 2.8-1, 2.8-2, 2.8-3
Solution 2.8-1
Collar falling onto a flange
(a) DOWNWARD DISPLACEMENT OF FLANGE
dst WL
0.00032 in.
EA
Eq. (2-55):
dmax dst c1 + a1 +
0.0361 in.
2h 1/2
b d
dst
;
(b) MAXIMUM TENSILE STRESS (EQ. 2-57)
smax Edmax
22,600 psi
L
;
(c) IMPACT FACTOR (EQ. 2-63)
Impact factor W 150 lb
h 2.0 in.
L 4.0 ft 48 in.
E 30 10 psi
6
dmax
0.0361 in.
dst
0.00032 in.
113
;
A 0.75 in.2
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SECTION 2.8 Impact Loading
247
Problem 2.8-2 Solve the preceding problem if the collar has mass
M 80 kg, the height h 0.5 m, the length L 3.0 m, the cross-sectional
area A 350 mm2, and the modulus of elasticity E 170 GPa.
Solution 2.8-2
Collar falling onto a flange
(a) DOWNWARD DISPLACEMENT OF FLANGE
dst WL
0.03957 mm
EA
Eq. (2-53): dmax dst c1 + a 1 +
6.33 mm
2h 1/2
b d
dst
;
(b) MAXIMUM TENSILE STRESS (EQ. 2-57)
smax Edmax
359 MPa
L
;
(c) IMPACT FACTOR (EQ. 2-63)
M 80 kg
Impact factor W Mg (80 kg)(9.81 m/s2)
784.8 N
h 0.5 m
L 3.0 m
E 170 GPa
A 350 mm2
dmax
6.33 mm
dst
0.03957 mm
160
;
Problem 2.8-3 Solve Problem 2.8-1 if the collar has weight W 50 lb,
the height h 2.0 in., the length L 3.0 ft, the cross-sectional area
A 0.25 in.2, and the modulus of elasticity E 30,000 ksi.
Solution 2.8-3
Collar falling onto a flange
W 50 lb
h 2.0 in.
L 3.0 ft 36 in.
E 30,000 psi
A 0.25 in.2
(a) DOWNWARD DISPLACEMENT OF FLANGE
dst WL
0.00024 in.
EA
2h 1/2
b d
dst
;
Eq. (2-55): dmax dst c1 + a 1 +
0.0312 in.
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CHAPTER 2 Axially Loaded Members
(b) MAXIMUM TENSILE STRESS (EQ. 2-57)
smax Edmax
26,000 psi
L
(c) IMPACT FACTOR (EQ. 2-63)
Impact factor ;
dmax
0.0312 in.
dst
0.00024 in.
130 ;
Problem 2.8-4 A block weighing W 5.0 N drops inside a cylinder
from a height h 200 mm onto a spring having stiffness k 90 N/m
(see figure).
Block
(a) Determine the maximum shortening of the spring due to the
impact and (b) determine the impact factor.
Cylinder
h
k
Prob. 2.8-4 and 2.8-5
Solution 2.8-4
Block dropping onto a spring
W 5.0 N
h 200 mm
k 90 N/m
(a) MAXIMUM SHORTENING OF THE SPRING
dst W
5.0 N
55.56 mm
k
90 N/m
Eq. (2-55): dmax dst c1 + a1 +
215 mm
(b) IMPACT FACTOR (EQ. 2-63)
Impact factor dmax
215 mm
dst
55.56 mm
3.9 ;
2h 1/2
b d
dst
;
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SECTION 2.8 Impact Loading
249
Problem 2.8-5 Solve the preceding problem if the block weighs
W 1.0 lb, h 12 in., and k 0.5 lb/in.
Solution 2.8-5
Block dropping onto a spring
(a) MAXIMUM SHORTENING OF THE SPRING
dst W
1.0 lb
2.0 in.
k
0.5 lb/in.
Eq. (2-55): dmax dst c1 + a 1 +
9.21 in.
2h 1/2
b d
dst
;
(b) IMPACT FACTOR (EQ. 2-63)
dmax
9.21 in.
dst
2.0 in.
4.6 ;
Impact factor W 1.0 lb
h 12 in.
k 0.5 lb/in.
Problem 2.8-6 A small rubber ball (weight W 450 mN) is attached by a rubber cord to
a wood paddle (see figure). The natural length of the cord is L0 200 mm, its crosssectional area is A 1.6 mm2, and its modulus of elasticity is E 2.0 MPa. After being
struck by the paddle, the ball stretches the cord to a total length L1 900 mm.
What was the velocity v of the ball when it left the paddle? (Assume linearly elastic
behavior of the rubber cord, and disregard the potential energy due to any change in
elevation of the ball.)
Solution 2.8-6
Rubber ball attached to a paddle
WHEN THE RUBBER CORD IS FULLY STRETCHED:
U
EAd2
EA
(L L 0)2
2L 0
2L 0 1
CONSERVATION OF ENERGY
KE U
v2 g 9.81 m/s2
E 2.0 MPa
2
A 1.6 mm
L0 200 mm
L1 900 mm
W 450 mN
WHEN THE BALL LEAVES THE PADDLE
KE Wv 2
2g
Wv 2
EA
(L 1 L 0)2
2g
2L 0
gEA
(L L 0)2
WL 0 1
gEA
A WL0
v (L1 L0)
;
SUBSTITUTE NUMERICAL VALUES:
(9.81 m/s2) (2.0 MPa) (1.6 mm2)
A
(450 mN) (200 mm)
13.1 m/s ;
v (700 mm)
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CHAPTER 2 Axially Loaded Members
Problem 2.8-7 A weight W 4500 lb falls from a height h onto
a vertical wood pole having length L 15 ft, diameter d 12 in.,
and modulus of elasticity E 1.6 106 psi (see figure).
If the allowable stress in the wood under an impact load is 2500 psi,
what is the maximum permissible height h?
W = 4,500 lb
h
d = 12 in.
L = 15 ft
Solution 2.8-7
Weight falling on a wood pole
E 1.6 106 psi
sallow 2500 psi ( smax)
Find hmax
STATIC STRESS
sst 4500 lb
W
39.79 psi
A
113.10 in.2
MAXIMUM HEIGHT hmax
Eq. (2-61): smax sst c1 + a1 +
2hE 1/2
b d
Lsst
or
smax
2hE 1/2
1 a1 +
b
sst
Lsst
Square both sides and solve for h:
h h max W 4500 lb
d 12 in.
L 15 ft 180 in.
A
pd 2
113.10 in.2
4
Lsmax smax
a
2b
2E
sst
;
SUBSTITUTE NUMERICAL VALUES:
h max (180 in.) (2500 psi) 2500 psi
2b
a
2(1.6 * 106 psi) 39.79 psi
8.55 in.
;
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SECTION 2.8 Impact Loading
251
Problem 2.8-8 A cable with a restrainer at the bottom hangs
vertically from its upper end (see figure). The cable has an
effective cross-sectional area A 40 mm2 and an effective
modulus of elasticity E 130 GPa. A slider of mass M 35 kg
drops from a height h 1.0 m onto the restrainer.
If the allowable stress in the cable under an impact load is
500 MPa, what is the minimum permissible length L of the cable?
Cable
Slider
L
h
Restrainer
Probs. 2.8-8, 2.8-2, 2.8-9
Solution 2.8-8
Slider on a cable
STATIC STRESS
sst W
343.4 N
8.585 MPa
A
40 mm2
MINIMUM LENGTH Lmin
Eq. (2-61): smax sst c1 + a1 +
2hE 1/2
b d
Lsst
or
smax
2hE 1/2
1 a1 +
b
sst
Lsst
Square both sides and solve for L:
L L min 2Ehsst
smax(smax 2sst)
;
SUBSTITUTE NUMERICAL VALUES:
W Mg (35 kg)(9.81 m/s2) 343.4 N
A 40 mm2
h 1.0 m
E 130 GPa
sallow smax 500 MPa
L min 2(130 GPa) (1.0 m) (8.585 MPa)
(500 MPa) [500 MPa 2(8.585 MPa)]
9.25 mm
;
Find minimum length Lmin.
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CHAPTER 2 Axially Loaded Members
Problem 2.8-9 Solve the preceding problem if the slider has
weight W 100 lb, h 45 in., A 0.080 in.2, E 21 106 psi,
and the allowable stress is 70 ksi.
Cable
Slider
L
h
Restrainer
Solution 2.8-9
Slider on a cable
STATIC STRESS
sst 100 lb
W
1250 psi
A
0.080 in.2
MINIMUM LENGTH Lmin
Eq. (2-61): smax sst c1 + a1 +
2hE 1/2
b d
Lsst
or
smax
2hE 1/2
1 a1 +
b
sst
Lsst
Square both sides and solve for L:
L L min 2Ehsst
smax(smax 2sst)
;
SUBSTITUTE NUMERICAL VALUES:
L min W 100 lb
A 0.080 in.2
h 45 in
E 21 106 psi
sallow smax 70 ksi
2(21 * 106 psi) (45 in.) (1250 psi)
(70,000 psi) [70,000 psi 2(1250 psi)]
500 in.
;
Find minimum length Lmin.
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253
SECTION 2.8 Impact Loading
Problem 2.8-10 A bumping post at the end of a track in a
railway yard has a spring constant k 8.0 MN/m (see figure).
The maximum possible displacement d of the end of the
striking plate is 450 mm.
What is the maximum velocity nmax that a railway car of
weight W 545 kN can have without damaging the bumping
post when it strikes it?
v
k
d
Solution 2.8-10 Bumping post for a railway car
STRAIN ENERGY WHEN SPRING IS COMPRESSED TO THE
MAXIMUM ALLOWABLE AMOUNT
v
U
k
CONSERVATION OF ENERGY
d
k 8.0 MN/m
KE U
k
A W/g
KINETIC ENERGY BEFORE IMPACT
Wv
Mv
2
2g
8.0 MN/m
vmax (450 mm)
A (545 kN)/(9.81 m/s2)
Find nmax.
KE ;
SUBSTITUTE NUMERICAL VALUES:
d dmax 450 mm
2
Wv 2
kd 2 2
kd 2
v 2g
2
W/g
v vmax d
W 545 kN
d maximum displacement of spring
kd2max
kd 2
2
2
5400 mm/s 5.4 m/s
;
2
Problem 2.8-11 A bumper for a mine car is constructed with
a spring of stiffness k 1120 lb/in. (see figure). If a car weighing
3450 lb is traveling at velocity n 7 mph when it strikes the
spring, what is the maximum shortening of the spring?
v
k
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CHAPTER 2 Axially Loaded Members
Solution 2.8-11 Bumper for a mine car
v
k
k 1120 lb/in.
W 3450 lb
n 7 mph 123.2 in./sec
g 32.2 ft/sec2 386.4 in./sec2
Find the shortening dmax of the spring.
KINETIC ENERGY JUST BEFORE IMPACT
Mv 2
Wv 2
KE 2
2g
Conservation of energy
KE U
Solve for dmax: dmax Wv2
A gk
;
SUBSTITUTE NUMERICAL VALUES:
dmax STRAIN ENERGY WHEN SPRING IS FULLY COMPRESSED
U
kd2max
Wv 2
2g
2
kd2max
2
(3450 lb) (123.2 in./sec)2
A (386.4 in./sec2) (1120 lb/in.)
11.0 in.
;
Problem 2.8-12 A bungee jumper having a mass of 55 kg leaps
from a bridge, braking her fall with a long elastic shock cord having
axial rigidity EA 2.3 kN (see figure).
If the jumpoff point is 60 m above the water, and if it is desired to
maintain a clearance of 10 m between the jumper and the water, what
length L of cord should be used?
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SECTION 2.8 Impact Loading
Solution 2.8-12
255
Bungee jumper
SOLVE QUADRATIC EQUATION FOR dmax:
dmax WL
WL 2
WL 1/2
+ ca
b + 2L a
bd
EA
EA
EA
WL
2EA 1/2
c1 + a1 +
b d
EA
W
VERTICAL HEIGHT
h C + L + dmax
hCL +
W Mg (55 kg)(9.81 m/s2)
539.55 N
SOLVE FOR L:
L
EA 2.3 kN
Height: h 60 m
2EA 1/2
WL
c1 + a 1 +
b d
EA
W
hC
W
2EA 1/2
1 +
c1 + a 1 +
b d
EA
W
;
SUBSTITUTE NUMERICAL VALUES:
Clearance: C 10 m
Find length L of the bungee cord.
W
539.55 N
0.234587
EA
2.3 kN
P.E. Potential energy of the jumper at the top of
bridge (with respect to lowest position)
Numerator h C 60 m 10 m 50 m
W(L
dmax)
U strain energy of cord at lowest position
EAd2max
2L
or
W(L + dmax) d2max * c1 + a 1 +
1.9586
50 m
25.5 m
L
1.9586
CONSERVATION OF ENERGY
P.E. U
Denominator 1 + (0.234587)
1/2
2
b d
0.234587
;
EAd2max
2L
2WL
2WL2
dmax 0
EA
EA
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CHAPTER 2 Axially Loaded Members
Problem 2.8-13 A weight W rests on top of a wall and is attached to one
end of a very flexible cord having cross-sectional area A and modulus of
elasticity E (see figure). The other end of the cord is attached securely
to the wall. The weight is then pushed off the wall and falls freely the full
length of the cord.
W
W
(a) Derive a formula for the impact factor.
(b) Evaluate the impact factor if the weight, when hanging statically,
elongates the band by 2.5% of its original length.
Solution 2.8-13
Weight falling off a wall
CONSERVATION OF ENERGY
P.E. U
or
W(L + dmax) d2max EAd2max
2L
2WL
2WL2
dmax 0
EA
EA
SOLVE QUADRATIC EQUATION FOR dmax:
W Weight
dmax WL 2
WL
WL 1/2
+ ca
b + 2L a
bd
EA
EA
EA
Properties of elastic cord:
E modulus of elasticity
STATIC ELONGATION
A cross-sectional area
dst L original length
dmax elongation of elastic cord
WL
EA
IMPACT FACTOR
P.E. potential energy of weight before fall (with
respect to lowest position)
dmax
2EA 1/2
1 + c1 +
d
dst
W
P.E. W(L
NUMERICAL VALUES
dmax)
Let U strain energy of cord at lowest position.
EAd2max
U
2L
;
dst (2.5%)(L) 0.025L
dst WL
EA
W
0.025
EA
EA
40
W
Impact factor 1 + [1 + 2(40)]1/2 10
;
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257
SECTION 2.8 Impact Loading
Problem 2.8-14 A rigid bar AB having mass M 1.0 kg and
length L 0.5 m is hinged at end A and supported at end B by a nylon
cord BC (see figure). The cord has cross-sectional area A 30 mm2,
length b 0.25 m, and modulus of elasticity E 2.1 GPa.
If the bar is raised to its maximum height and then released, what is
the maximum stress in the cord?
C
b
A
B
W
L
Solution 2.8-14
Falling bar AB
GEOMETRY OF BAR AB AND CORD BC
RIGID BAR:
W Mg (1.0 kg)(9.81 m/s2)
9.81 N
L 0.5 m
NYLON CORD:
A 30 mm2
CD CB b
AD AB L
h height of center of gravity of raised bar AD
dmax elongation of cord
From triangle ABC:sin u cos u b
2b + L2
L
2
E 2.1 GPa
2b2 + L2
2h
2h
From line AD: sin 2 u AD
L
Find maximum stress smax in cord BC.
From Appendix D: sin 2 u 2 sin u cos u
b 0.25 m
‹
L
2bL
b
2h
ba
b 2
2a
2
2
2
2
L
b + L2
2b + L
2b + L
and h bL2
b 2 + L2
(Eq. 1)
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CHAPTER 2 Axially Loaded Members
CONSERVATION OF ENERGY
P.E. potential energy of raised bar AD
Substitute from Eq. (1) into Eq. (3):
s2max dmax
b
W ah +
2
dmax
EAd2max
b
2
2b
smaxb
For the cord: dmax E
Substitute into Eq. (2) and rearrange:
s2max W
2WhE
s
0
A max
bA
(Eq. 4)
SOLVE FOR smax:
EAd2max
U strain energy of stretched cord 2b
P.E. U W a h +
W
2WL2E
0
smax A
A(b 2 + L2)
(Eq. 2)
smax W
8L2EA
c1 + 1 +
d
2A
A
W(b2 + L2)
;
SUBSTITUTE NUMERICAL VALUES:
smax 33.3 MPa
;
(Eq. 3)
Stress Concentrations
The problems for Section 2.10 are to be solved by considering
the stress-concentration factors and assuming linearly elastic behavior.
P
Problem 2.10-1 The flat bars shown in parts (a) and (b) of the figure are
P
d
b
subjected to tensile forces P 3.0 k. Each bar has thickness t 0.25 in.
(a) For the bar with a circular hole, determine the maximum stresses for
hole diameters d 1 in. and d 2 in. if the width b 6.0 in.
(b) For the stepped bar with shoulder fillets, determine the maximum
stresses for fillet radii R 0.25 in. and R 0.5 in. if the bar
widths are b 4.0 in. and c 2.5 in.
(a)
R
P
c
b
P
(b)
Probs. 2.10-1 and 2.10-2
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SECTION 2.10 Stress Concentrations
Solution 2.10-1
P 3.0 k
259
Flat bars in tension
(b) STEPPED BAR WITH SHOULDER FILLETS
t 0.25 in.
(a) BAR WITH CIRCULAR HOLE (b 6 in.)
s nom Obtain K from Fig. 2-63
FOR d 1 in.:
c b d 5 in.
3.0 k
P
2.40 ksi
s nom ct
(5 in.) (0.25 in.)
1
K L 2.60
d/b 6
smax ksnom ⬇ 6.2 ksi
b 4.0 in.
c 2.5 in.; Obtain k from Fig. 2-65
P
3.0 k
4.80 ksi
ct
(2.5 in.) (0.25 in.)
FOR R 0.25 in.: R/c 0.1
b/c 1.60
k ⬇ 2.30 smax Ksnom ⬇ 11.0 ksi
FOR R 0.5 in.: R/c 0.2
K ⬇ 1.87
;
b/c 1.60
smax Ksnom ⬇ 9.0 ksi
;
;
FOR d 2 in.: c b d 4 in.
s nom d/b P
3.0 k
3.00 ksi
ct
(4 in.) (0.25 in.)
1
K L 2.31
3
smax Ksnom ⬇ 6.9 ksi
;
Problem 2.10-2 The flat bars shown in parts (a) and (b) of the figure
are subjected to tensile forces P 2.5 kN. Each bar has thickness
t 5.0 mm.
P
(a) For the bar with a circular hole, determine the maximum stresses for
hole diameters d 12 mm and d 20 mm if the width b 60 mm.
(b) For the stepped bar with shoulder fillets, determine the maximum
stresses for fillet radii R 6 mm and R 10 mm if the bar widths are
b 60 mm and c 40 mm.
P
d
b
(a)
R
P
c
b
P
(b)
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CHAPTER 2 Axially Loaded Members
Solution 2.10-2
Flat bars in tension
P 2.5 kN
t 5.0 mm
(b) STEPPED BAR WITH SHOULDER FILLETS
(a) BAR WITH CIRCULAR HOLE (b 60 mm)
Obtain K from Fig. 2-63
b 60 mm
FOR d 12 mm: c b d 48 mm
s nom s nom d/b P
2.5 kN
10.42 MPa
ct
(48 mm) (5 mm)
smax Ksnom ⬇ 26 MPa
Obtain K from Fig. 2-65
P
2.5 kN
12.50 MPa
ct
(40 mm) (5 mm)
FOR R 6 mm: R/c 0.15
K ⬇ 2.00
1
K L 2.51
5
c 40 mm;
smax Ksnom ⬇ 25 MPa
FOR R 10 mm: R/c 0.25
;
b/c 1.5
K ⬇ 1.75
;
b/c 1.5
smax Ksnom ⬇ 22 MPa
;
FOR d 20 mm: c b d 40 mm
s nom d/b 1
3
2.5 kN
P
12.50 MPa
ct
(40 mm) (5 mm)
K L 2.31
smax Ksnom ⬇ 29 MPa
;
Problem 2.10-3 A flat bar of width b and thickness t has a hole
of diameter d drilled through it (see figure). The hole may have
any diameter that will fit within the bar.
What is the maximum permissible tensile load Pmax if the allowable
tensile stress in the material is st?
P
b
d
P
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261
SECTION 2.10 Stress Concentrations
Solution 2.10-3
P
Flat bar in tension
b
d
P
t thickness
st allowable tensile stress
Find Pmax
Find K from Fig. 2-63
Pmax s nom ct smax
st
ct (b d)t
K
K
st
d
bt a1 b
K
b
Because st, b, and t are constants, we write:
P *
Pmax
1
stbt
K
a1 d
b
K
P*
0
0.1
0.2
0.3
0.4
3.00
2.73
2.50
2.35
2.24
0.333
0.330
0.320
0.298
0.268
We observe that Pmax decreases as d/b increases.
Therefore, the maximum load occurs when the hole
becomes very small.
d
a :0
b
Pmax and K : 3b
stbt
3
;
d
b
b
Problem 2.10-4 A round brass bar of diameter d1 20 mm has
upset ends of diameter d2 26 mm (see figure). The lengths of
the segments of the bar are L1 0.3 m and L2 0.1 m.
Quarter-circular fillets are used at the shoulders of the bar, and
the modulus of elasticity of the brass is E 100 GPa.
If the bar lengthens by 0.12 mm under a tensile load P, what is
the maximum stress smax in the bar?
P
d2 = 26 mm
L2
d1 = 20 mm
L1
L2
Probs. 2.10-4 and 2.10-5
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CHAPTER 2 Axially Loaded Members
Solution 2.10-4 Round brass bar with upset ends
P
d2 = 26 mm
d1 = 20 mm
P
Use Fig. 2-66 for the stress-concentration factor:
s nom L1
L2
L2
E 100 GPa
d 0.12 mm
dEA2
P
A1
2L 2A1 + L 1A2
dE
2L 2 a
d1 2
b + L1
d2
L2 0.1 m
SUBSTITUTE NUMERICAL VALUES:
L1 0.3 m
s nom R radius of fillets 26 mm 20 mm
3 mm
2
PL 1
PL 2
b +
d 2a
EA2
EA1
Solve for P:
P
dEA1A2
2L 2A1 + L 1A2
(0.12 mm) (100 GPa)
2(0.1 m) a
20 2
b + 0.3 m
26
metal having the following properties: d1 1.0 in., d2 1.4 in.,
L1 20.0 in., L2 5.0 in., and E 25 106 psi. Also, the bar
lengthens by 0.0040 in. when the tensile load is applied.
Solution 2.10-5
28.68 MPa
3 mm
R
0.15
D1
20 mm
Use the dashed curve in Fig. 2-66. K ⬇ 1.6
smax Ksnom ⬇ (1.6) (28.68 MPa)
⬇ 46 MPa
Problem 2.10-5 Solve the preceding problem for a bar of monel
dE
A1
2L 2 a b + L 1
A2
d2
P
;
d2
d1
L1
L2
L2
Round bar with upset ends
d 2a
PL 1
PL 2
b +
EA2
EA1
Solve for P: P E 25 106 psi
d 0.0040 in.
Use Fig. 2-66 for the stress-concentration factor.
s nom L1 20 in.
L2 5 in.
1.4 in. 1.0 in.
R radius of fillets R 2
0.2 in.
dEA1A2
2L 2A1 + L 1A2
dEA2
P
A1
2L 2A1 + L 1A2
dE
A1
2L 2 a b + L 1
A2
dE
2L 2 a
d1 2
b + L1
d2
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SECTION 2.10 Stress Concentrations
SUBSTITUTE NUMERICAL VALUES:
s nom (0.0040 in.)(25 * 106 psi)
1.0 2
b + 20 in.
2(5 in.)a
1.4
0.2 in.
R
0.2
D1
1.0 in.
263
Use the dashed curve in Fig. 2-66. K ⬇ 1.53
3,984 psi
smax Ksnom ⬇ (1.53)(3984 psi)
⬇ 6100 psi
Problem 2.10-6 A prismatic bar of diameter d0 20 mm is being compared
;
P1
with a stepped bar of the same diameter (d1 20 mm) that is enlarged
in the middle region to a diameter d2 25 mm (see figure).
The radius of the fillets in the stepped bar is 2.0 mm.
(a) Does enlarging the bar in the middle region make it stronger than the
prismatic bar? Demonstrate your answer by determining the maximum
permissible load P1 for the prismatic bar and the maximum permissible
load P2 for the enlarged bar, assuming that the allowable stress for the
material is 80 MPa.
(b) What should be the diameter d0 of the prismatic bar if it is to have the same
maximum permissible load as does the stepped bar?
P2
d0
d1
P1
d2
d1
P2
Solution 2.10-6 Prismatic bar and stepped bar
Fillet radius: R 2 mm
Allowable stress: st 80 MPa
(a) COMPARISON OF BARS
Prismatic bar: P1 stA0 st a
pd 20
b
4
p
(80 MPa)a b(20mm)2 25.1 kN
4
;
Stepped bar: See Fig. 2-66 for the stress-concentration
factor.
d0 20 mm
d1 20 mm
d2 25 mm
R 2.0 mm
D1 20 mm
D2 25 mm
R/D1 0.10
D2/D1 1.25
K ⬇ 1.75
s nom P2
P2
smax
s nom p 2
A1
K
d
4 1
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CHAPTER 2 Axially Loaded Members
P2 s nom A1 a
s max
st
A1 A1
K
K
80 MPa p
b a b(20 mm)2
1.75
4
ALLOWABLE LOAD
P1 P2 st a
st pd 21
pd 20
b a
b
4
K 4
d 20 d1
20 mm
L 15.1 mm
11.75
;
d0 ;
L 14.4 kN
(b) DIAMETER OF PRISMATIC BAR FOR THE SAME
1K
L
d 21
K
Enlarging the bar makes it weaker, not stronger. The
ratio of loads is P1/P2 K 1.75
Problem 2.10-7 A stepped bar with a hole (see figure) has widths
b 2.4 in. and c 1.6 in. The fillets have radii equal to 0.2 in.
What is the diameter dmax of the largest hole that can be drilled
through the bar without reducing the load-carrying capacity?
Solution 2.10-7
Stepped bar with a hole
b 2.4 in.
BASED UPON HOLE (Use Fig. 2-63)
c 1.6 in.
Fillet radius: R 0.2 in.
b 2.4 in.
c1 b d
Find dmax
Pmax s nom c1t smax
(b d)t
K
d
1
a1 bbtsmax
K
b
BASED UPON FILLETS (Use Fig. 2-65)
b 2.4 in.
c 1.6 in.
R 0.2 in.
R/c 0.125
b/c 1.5
K ⬇ 2.10
Pmax s nomct smax
smax c
ct a b(bt)
K
K
b
L 0.317 bt smax
d diameter of the hole (in.)
d(in.)
d/b
K
Pmax/btsmax
0.3
0.4
0.5
0.6
0.7
0.125
0.167
0.208
0.250
0.292
2.66
2.57
2.49
2.41
2.37
0.329
0.324
0.318
0.311
0.299
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SECTION 2.11 Nonlinear Behavior (Changes in Lengths of Bars)
265
Nonlinear Behavior (Changes in Lengths of Bars)
A
Problem 2.11-1 A bar AB of length L and weight density g hangs vertically
under its own weight (see figure). The stress-strain relation for the material is
given by the Ramberg-Osgood equation (Eq. 2-73):
P
s0a s m
s
+
a b
E
E s0
L
Derive the following formula
d
gL2
gL m
s0aL
+
a b
2E
(m + 1)E s0
B
for the elongation of the bar.
Solution 2.11-1
Bar hanging under its own weight
STRAIN AT DISTANCE x
Let A cross-sectional area
Let N axial force at distance x
N gAx
N
s gx
A
s0a s m gx
s0 gx m
s
+
a b +
a b
E
E s0
E
aE s0
ELONGATION OF BAR
L
L
L
gx
gx m
s0a
d
dx a b dx
dx +
E L0 s0
L0
L0 E
gL2
gL m
s0aL
+
a b
2E
(m + 1)E s0
Q.E.D.
;
A
B
P1 C
Problem 2.11-2 A prismatic bar of length L 1.8 m and cross-sectional
area A 480 mm is loaded by forces P1 30 kN and P2 60 kN
(see figure). The bar is constructed of magnesium alloy having a stress-strain
curve described by the following Ramberg-Osgood equation:
P2
2
P
2L
—
3
L
—
3
s
1
s 10
+
a
b
(s MPa)
45,000
618 170
in which s has units of megapascals.
(a) Calculate the displacement dC of the end of the bar when the load
P1 acts alone.
(b) Calculate the displacement when the load P2 acts alone.
(c) Calculate the displacement when both loads act simultaneously.
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CHAPTER 2 Axially Loaded Members
Solution 2.11-2
Axially loaded bar
(c) BOTH P1 AND P2 ARE ACTING
AB:s L 1.8 m
A 480 mm2
P1 30 kN
P2 60 kN
0.008477
dAB a
Ramberg–Osgood equation:
1
s
s
+
a
b
45,000
618 170
10
BC:s (s MPa)
Find displacement at end of bar.
P1
30 kN
62.5 MPa
A
480 mm2
0.001389
dc a
2L
b 1.67 mm
3
2L
b 10.17 mm
3
P2
60 kN
125 MPa
A
480 mm2
0.002853
L
dBC a b 1.71 mm
3
(a) P1 ACTS ALONE
AB: s P1 + P2
90 kN
187.5 MPa
A
480 mm2
;
dC dAB + dBC 11.88 mm
;
(Note that the displacement when both loads act
simultaneously is not equal to the sum of the displacements when the loads act separately.)
(b) P2 ACTS ALONE
P2
60 kN
125 MPa
A
480 mm2
0.002853
dc L 5.13 mm ;
ABC:s Problem 2.11-3 A circular bar of length L 32 in. and diameter
d 0.75 in. is subjected to tension by forces P (see figure).
The wire is made of a copper alloy having the following hyperbolic
stress-strain relationship:
s
18,000P
0 … P … 0.03 (s ksi)
1 + 300P
d
P
P
L
(a) Draw a stress-strain diagram for the material.
(b) If the elongation of the wire is limited to 0.25 in. and
the maximum stress is limited to 40 ksi, what is the allowable load P?
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SECTION 2.11 Nonlinear Behavior (Changes in Lengths of Bars)
267
Solution 2.11-3 Copper bar in tension
(b) ALLOWABLE LOAD P
d
P
P
Maximum elongation dmax 0.25 in.
Maximum stress smax 40 ksi
L
Based upon elongation:
L 32 in.
A
d 0.75 in.
pd 2
0.4418 in.2
4
max dmax
0.25 in.
0.007813
L
32 in.
smax 18,000max
42.06 ksi
1 + 300max
(a) STRESS-STRAIN DIAGRAM
s
18,000
0 … … 0.03 (s ksi)
1 + 300
BASED UPON STRESS:
smax 40 ksi
Stress governs. P smax A (40 ksi)(0.4418 in.2)
17.7 k
Problem 2.11-4 A prismatic bar in tension has length L 2.0 m
and cross-sectional area A 249 mm . The material of the bar has the stressstrain curve shown in the figure.
Determine the elongation d of the bar for each of the following axial
loads: P 10 kN, 20 kN, 30 kN, 40 kN, and 45 kN. From these results,
plot a diagram of load P versus elongation d (load-displacement diagram).
;
200
2
s (MPa)
100
0
0
0.005
e
0.010
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CHAPTER 2 Axially Loaded Members
Solution 2.11-4
Bar in tension
L 2.0 m
A 249 mm2
STRESS-STRAIN DIAGRAM
(See the problem statement for the diagram)
LOAD-DISPLACEMENT DIAGRAM
P
(kN)
s P/A
(MPa)
(from diagram)
d L
(mm)
10
20
30
40
45
40
80
120
161
181
0.0009
0.0018
0.0031
0.0060
0.0081
1.8
3.6
6.2
12.0
16.2
NOTE: The load-displacement curve has the same
shape as the stress-strain curve.
Problem 2.11-5 An aluminum bar subjected to tensile forces P has length
L 150 in. and cross-sectional area A 2.0 in. The stress-strain behavior of the
aluminum may be represented approximately by the bilinear stress-strain diagram
shown in the figure.
Calculate the elongation d of the bar for each of the following axial loads:
P 8 k, 16 k, 24 k, 32 k, and 40 k. From these results, plot a diagram of load
P versus elongation d (load-displacement diagram).
s
2
12,000
psi
E2 = 2.4 × 106 psi
E1 = 10 × 106 psi
0
e
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SECTION 2.11 Nonlinear Behavior (Changes in Lengths of Bars)
Solution 2.11-5
269
Aluminum bar in tension
LOAD-DISPLACEMENT DIAGRAM
L 150 in.
A 2.0 in.2
STRESS-STRAIN DIAGRAM
P (k)
s P/A
(psi)
(from Eq.
1 or Eq. 2)
d L
(in.)
8
16
24
32
40
4,000
8,000
12,000
16,000
20,000
0.00040
0.00080
0.00120
0.00287
0.00453
0.060
0.120
0.180
0.430
0.680
E1 10 106 psi
E2 2.4 106 psi
s1 12,000 psi
1 12,000 psi
s1
E1
10 * 106 psi
0.0012
For 0 s s1:
s
s
(s psi)
E1
10 * 106psi
Eq. (1)
For s s1:
1 +
s
s 12,000
s s1
0.0012 +
E2
2.4 * 106
2.4 * 106
0.0038 (s psi)
Eq. (2)
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CHAPTER 2 Axially Loaded Members
Problem 2.11-6 A rigid bar AB, pinned at end A, is supported by a
wire CD and loaded by a force P at end B (see figure). The wire is
made of high-strength steel having modulus of elasticity E 210 GPa
and yield stress sY 820 MPa. The length of the wire is L 1.0 m
and its diameter is d 3 mm. The stress-strain diagram for the steel is
defined by the modified power law, as follows:
C
L
A
D
B
s EP 0 … s … sY
s sY a
EP n
b s Ú sY
sY
P
2b
(a) Assuming n 0.2, calculate the displacement dB at the end of
the bar due to the load P. Take values of P from 2.4 kN to
5.6 kN in increments of 0.8 kN.
b
(b) Plot a load-displacement diagram showing P versus dB.
Solution 2.11-6 Rigid bar supported by a wire
sY s 1/n
a b
E sY
3P
Axial force in wire: F 2
3P
F
Stress in wire: s A
2A
PROCEDURE: Assume a value of P
Calculate s from Eq. (6)
Calculate from Eq. (4) or (5)
Calculate dB from Eq. (3)
From Eq. (2): C
L
A
D
B
P
2b
b
Wire: E 210 GPa
sY 820 MPa
L 1.0 m
d 3 mm
A
pd 2
7.0686 mm2
4
STRESS-STRAIN DIAGRAM
s E
(0
s sY a
E n
b
sY
s
sY)
(1)
(5)
(6)
P
(kN)
s (MPa)
Eq. (6)
Eq. (4)
or (5)
dB (mm)
Eq. (3)
2.4
3.2
4.0
4.8
5.6
509.3
679.1
848.8
1018.6
1188.4
0.002425
0.003234
0.004640
0.01155
0.02497
3.64
4.85
6.96
17.3
37.5
For s sY 820 MPa:
0.0039048
P 3.864 kN
dB 5.86 mm
(b) LOAD-DISPLACEMENT DIAGRAM
(s sY)
(n 0.2)
(2)
(a) DISPLACEMENT dB AT END OF BAR
3
3
d elongation of wire dB d L
2
2
Obtain from stress-strain equations:
(3)
From Eq. (1): sE
(0 … s … sY)
(4)
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271
SECTION 2.12 Elastoplastic Analysis
Elastoplastic Analysis
The problems for Section 2.12 are to be solved assuming that the material is
elastoplastic with yield stress sY, yield strain Y, and modulus of elasticity
E in the linearly elastic region (see Fig. 2-72).
A
u
u
C
Problem 2.12-1 Two identical bars AB and BC support a vertical load
P (see figure). The bars are made of steel having a stress-strain curve that
may be idealized as elastoplastic with yield stress sY. Each bar has
cross-sectional area A.
Determine the yield load PY and the plastic load PP.
B
P
Solution 2.12-1 Two bars supporting a load P
JOINT B
Fvert 0
Structure is statically determinate. The yield load PY
and the plastic lead PP occur at the same time, namely,
when both bars reach the yield stress.
(2sYA) sin u P
PY PP 2sYA sin u
Problem 2.12-2 A stepped bar ACB with circular cross sections
is held between rigid supports and loaded by an axial force P at
midlength (see figure). The diameters for the two parts of the bar are
d1 20 mm and d2 25 mm, and the material is elastoplastic with
yield stress sY 250 MPa.
Determine the plastic load PP.
A
d1
;
C
L
—
2
d2
P
B
L
—
2
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CHAPTER 2 Axially Loaded Members
Solution 2.12-2 Bar between rigid supports
FAC sYA1
FCB sYA2
P FAC
FCB
PP sYA1
sYA2 sY(A1
A2)
;
SUBSTITUTE NUMERICAL VALUES:
d1 20 mm
d2 25 mm
sY 250 MPa
DETERMINE THE PLASTIC LOAD PP:
At the plastic load, all parts of the bar are stressed to the
yield stress.
p
PP (250 MPa)a b(d 21 + d 22)
4
p
(250 MPa)a b[(20 mm)2 + (25 mm)2]
4
201 kN
Point C:
;
Problem 2.12-3 A horizontal rigid bar AB supporting a load P is hung
from five symmetrically placed wires, each of cross-sectional area A
(see figure). The wires are fastened to a curved surface of radius R.
R
(a) Determine the plastic load PP if the material of the wires is
elastoplastic with yield stress sY.
(b) How is PP changed if bar AB is flexible instead of rigid?
(c) How is PP changed if the radius R is increased?
A
B
P
Solution 2.12-3 Rigid bar supported by five wires
(b) BAR AB IS FLEXIBLE
At the plastic load, each wire is stressed to the yield
stress, so the plastic load is not changed. ;
(a) PLASTIC LOAD PP
At the plastic load, each wire is stressed to the yield
stress. ⬖ PP 5sYA ;
(c) RADIUS R IS INCREASED
Again, the forces in the wires are not changed, so the
plastic load is not changed. ;
F sYA
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SECTION 2.12 Elastoplastic Analysis
273
Problem 2.12-4 A load P acts on a horizontal beam that is supported by four
rods arranged in the symmetrical pattern shown in the figure. Each rod has
cross-sectional area A and the material is elastoplastic with yield stress sY.
Determine the plastic load PP.
a
a
P
Solution 2.12-4 Beam supported by four rods
a
a
P
At the plastic load, all four rods are stressed to the yield
stress.
F sYA
Sum forces in the vertical direction and solve for the
load:
PP 2F
2F sin a
PP 2sYA (1
21 in.
Problem 2.12-5 The symmetric truss ABCDE shown in the figure
is constructed of four bars and supports a load P at joint E. Each of
the two outer bars has a cross-sectional area of 0.307 in.2, and each of
the two inner bars has an area of 0.601 in.2 The material is elastoplastic with yield stress sY 36 ksi.
Determine the plastic load PP.
sin a)
A
;
54 in.
21 in.
C
B
D
36 in.
E
P
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CHAPTER 2 Axially Loaded Members
Solution 2.12-5 Truss with four bars
PLASTIC LOAD PP
At the plastic load, all bars are stressed to the yield
stress.
FAE sYAAE
PP FBE sYABE
6
8
sY AAE + sY ABE
5
5
;
SUBSTITUTE NUMERICAL VALUES:
AAE 0.307 in.2 ABE 0.601 in.2
LAE 60 in.
JOINT E
LBE 45 in.
sY 36 ksi
6
8
PP (36 ksi) (0.307 in.2) + (36 ksi) (0.601 in.2)
5
5
Equilibrium:
3
4
2FAE a b + 2FBE a b P
5
5
or
6
8
P FAE + FBE
5
5
13.26 k + 34.62 k 47.9 k
Problem 2.12-6 Five bars, each having a diameter of 10 mm, support a
b
b
;
b
b
load P as shown in the figure. Determine the plastic load PP if the material is
elastoplastic with yield stress sY 250 MPa.
2b
P
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SECTION 2.12 Elastoplastic Analysis
275
Solution 2.12-6 Truss consisting of five bars
At the plastic load, all five bars
are stressed to the yield stress
F sYA
Sum forces in the vertical direction and solve for the load:
PP 2Fa
d 10 mm
pd 2
78.54 mm2
A
4
sY 250 MPa
1
2
b + 2Fa
b + F
12
15
sYA
(5 12 + 415 + 5)
5
4.2031sYA
;
Substitute numerical values:
PP (4.2031)(250 MPa)(78.54 mm2)
82.5 kN
Problem 2.12-7 A circular steel rod AB of diameter d 0.60 in.
;
B
A
is stretched tightly between two supports so that initially the
tensile stress in the rod is 10 ksi (see figure). An axial force P is
then applied to the rod at an intermediate location C.
d
(a) Determine the plastic load PP if the material is elastoplastic
with yield stress sY 36 ksi.
(b) How is PP changed if the initial tensile stress is doubled
to 20 ksi?
A
P
B
C
Solution 2.12-7 Bar held between rigid supports
POINT C:
sYA
sYA
P
— C ¡ —
d 0.6 in.
sY 36 ksi
Initial tensile stress 10 ksi
(a) PLASTIC LOAD PP
The presence of the initial tensile stress does not
affect the plastic load. Both parts of the bar must
yield in order to reach the plastic load.
p
PP 2sYA (2) (36 ksi)a b(0.60 in.)2
4
20.4 k
;
(B) INITIAL TENSILE STRESS IS DOUBLED
PP is not changed.
;
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CHAPTER 2 Axially Loaded Members
Problem 2.12-8 A rigid bar ACB is supported on a fulcrum
at C and loaded by a force P at end B (see figure). Three
identical wires made of an elastoplastic material
(yield stress sY and modulus of elasticity E) resist the load P.
Each wire has cross-sectional area A and length L.
(a) Determine the yield load PY and the corresponding
yield displacement dY at point B.
(b) Determine the plastic load PP and the corresponding
displacement dP at point B when the load just reaches
the value PP.
(c) Draw a load-displacement diagram with the load
P as ordinate and the displacement dB of point B
as abscissa.
L
A
C
B
P
L
a
a
a
a
Solution 2.12-8 Rigid bar supported by wires
(b) PLASTIC LOAD PP
(a) YIELD LOAD PY
Yielding occurs when the most highly stressed wire
reaches the yield stress sY
At the plastic load, all wires reach the yield stress.
MC 0
PP 4sYA
3
;
At point A:
dA (sYA)a
sYL
L
b EA
E
At point B:
dB 3dA dP MC 0
PY sYA
At point A:
3sYL
E
;
(c) LOAD-DISPLACEMENT DIAGRAM
;
sYA
sYL
L
dA a
ba
b 2
EA
2E
4
PP PY
3
dP 2dY
At point B:
dB 3dA dY 3sYL
2E
;
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277
SECTION 2.12 Elastoplastic Analysis
Problem 2.12-9 The structure shown in the figure consists of a
horizontal rigid bar ABCD supported by two steel wires, one of length L
and the other of length 3L/4. Both wires have cross-sectional area A and
are made of elastoplastic material with yield stress sY and modulus of
elasticity E. A vertical load P acts at end D of the bar.
L
A
(a) Determine the yield load PY and the corresponding yield
displacement dY at point D.
(b) Determine the plastic load PP and the corresponding displacement
dP at point D when the load just reaches the value PP.
(c) Draw a load-displacement diagram with the load P as ordinate
and the displacement dD of point D as abscissa.
3L
4
B
C
D
P
2b
b
FB(2b)
2FB
FC(3b) P(4b)
3FC 4P
b
Solution 2.12-9 Rigid bar supported by two wires
FREE-BODY DIAGRAM
A cross-sectional area
EQUILIBRIUM:
sY yield stress
MA 0 哵哴
E modulus of elasticity
DISPLACEMENT DIAGRAM
(3)
FORCE-DISPLACEMENT RELATIONS
dB FBL
dC EA
3
FC a Lb
4
EA
(4, 5)
Substitute into Eq. (1):
COMPATIBILITY:
3
dC dB
2
(1)
dD 2dB
(2)
3FCL
3FBL
4EA
2EA
FC 2FB
(6)
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CHAPTER 2 Axially Loaded Members
STRESSES
From Eq. (3):
FB
FC
sC sC 2sB
(7)
A
A
Wire C has the larger stress. Therefore, it will yield first.
2(sYA)
sB (a) YIELD LOAD
5
P PP sYA
4
;
From Eq. (4):
sC sY
sC
sY
sB 2
2
FC sY A
FB (From Eq. 7)
1
s A
2 Y
From Eq. (3):
1
2a sYA b + 3(sYA) 4P
2
P PY sYA
3(sYA) 4P
FBL
sY L
EA
E
From Eq. (2):
dB dD dP 2dB 2sYL
E
;
(c) LOAD-DISPLACEMENT DIAGRAM
5
PP PY
4
;
From Eq. (4):
dP 2dY
FB L
sY L
dB EA
2E
From Eq. (2):
dD dY 2dB sY L
E
;
(b) PLASTIC LOAD
At the plastic load, both wires yield.
sB sY sC
FB FC sY A
Problem 2.12-10 Two cables, each having a length L of approximately 40 m, support a l
oaded container of weight W (see figure). The cables, which have effective cross-sectional area
A 48.0 mm2 and effective modulus of elasticity E 160 GPa, are identical except that one
cable is longer than the other when they are hanging separately and unloaded. The difference
in lengths is d 100 mm. The cables are made of steel having an elastoplastic stress-strain
diagram with sY 500 MPa. Assume that the weight W is initially zero and is slowly increased
by the addition of material to the container.
L
(a) Determine the weight WY that first produces yielding of the shorter cable. Also, determine
the corresponding elongation dY of the shorter cable.
(b) Determine the weight WP that produces yielding of both cables. Also, determine the
elongation dP of the shorter cable when the weight W just reaches the value WP.
(c) Construct a load-displacement diagram showing the weight W as ordinate and the
elongation d of the shorter cable as abscissa. (Hint: The load displacement diagram is
not a single straight line in the region 0 W WY.)
W
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SECTION 2.12 Elastoplastic Analysis
Solution 2.12-10
279
Two cables supporting a load
L 40 m
A 48.0 mm2
(b) PLASTIC LOAD WP
E 160 GPa
F1 sYA
d difference in length 100 mm
WP 2sYA 48 kN
sY 500 MPa
d2P elongation of cable 2
INITIAL STRETCHING OF CABLE 1
Initially, cable 1 supports all of the load.
Let W1 load required to stretch cable 1
to the same length as cable 2
W1 EA
d 19.2 kN
L
F2 a
F2 sYA
;
sYL
L
b 0.125 mm 125 mm
EA
E
d1P d2P
d 225 mm
dP d1P 225 mm
;
(c) LOAD-DISPLACEMENT DIAGRAM
d1 100 mm (elongation of cable 1)
s1 W1
Ed
400 MPa (s1 6 sY ‹ 7 OK)
A
L
(a) YIELD LOAD WY
Cable 1 yields first. F1 sYA 24 kN
d1Y total elongation of cable 1
d1Y total elongation of cable 1
d1Y F1L
sY L
0.125 m 125 mm
EA
E
dY d1Y 125 mm
;
d2Y elongation of cable 2
d1Y d 25 mm
EA
F2 d2Y 4.8 kN
L
WY F1 + F2 24 kN + 4.8 kN
28.8 kN
dY
WY
1.5
1.25
W1
d1
dP
WP
1.667
1.8
WY
dY
0 W W1: slope 192,000 N/m
W1 W WY: slope 384,000 N/m
WY W WP: slope 192,000 N/m
;
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CHAPTER 2 Axially Loaded Members
Problem 2.12-11 A hollow circular tube T of length L 15 in.
is uniformly compressed by a force P acting through a rigid plate
(see figure). The outside and inside diameters of the tube are 3.0 and
2.75 in., repectively. A concentric solid circular bar B of 1.5 in.
diameter is mounted inside the tube. When no load is present, there is
a clearance c 0.010 in. between the bar B and the rigid plate. Both
bar and tube are made of steel having an elastoplastic stress-strain
diagram with E 29 103 ksi and sY 36 ksi.
(a) Determine the yield load PY and the corresponding shortening
dY of the tube.
(b) Determine the plastic load PP and the corresponding shortening
dP of the tube.
(c) Construct a load-displacement diagram showing the load P as
ordinate and the shortening d of the tube as abscissa. (Hint: The
load-displacement diagram is not a single straight line in the
region 0 P PY.)
Solution 2.12-11
L 15 in.
c 0.010 in.
P
c
T
T
B
T
L
B
Tube and bar supporting a load
TUBE:
d2 3.0 in.
E 29 103 ksi
d1 2.75 in.
sY 36 ksi
AT p 2
(d d 21) 1.1290 in.2
4 2
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SECTION 2.12 Elastoplastic Analysis
BAR
281
(b) PLASTIC LOAD PP
d 1.5 in.
AB FT sYAT
PP FT
2
pd
1.7671 in.2
4
FB sY(AT
104,300 lb
Initially, the tube supports all of the load.
Let P1 load required to close the clearance
FB a
dTP dBP
EAT
c 21,827 lb
L
Let d1 shortening of tube
AB)
;
dBP shortening of bar
INITIAL SHORTENING OF TUBE T
P1 P1
19,330 psi
s1 AT
FB sYAB
d1 c 0.010 in.
sYL
L
b 0.018621 in.
EAB
E
c 0.028621 in.
dP dTP 0.02862 in.
;
(c) LOAD-DISPLACEMENT DIAGRAM
(s1 sY ⬖ OK)
(a) YIELD LOAD PY
Because the tube and bar are made of the same
material, and because the strain in the tube is larger
than the strain in the bar, the tube will yield first.
FT sYAT 40,644 lb
d TY shortening of tube at the yield stress
s TY FTL
sYL
0.018621 in.
EAT
E
dY dTY 0.018621 in.
;
dBY shortening of bar
dTY c 0.008621 in.
dY
PY
3.21
1.86
P1
d1
EAB
d 29,453 lb
L BY
dP
PP
1.49
1.54
PY
dY
FB PY FT
FB 40,644 lb
70,097 lb
PY 70,100 lb
29,453 lb
0 P P1: slope 2180 k/in.
P1 P PY: slope 5600 k/in.
;
PY P PP: slope 3420 k/in.
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3
Torsion
Torsional Deformations
Problem 3.2-1 A copper rod of length L 18.0 in. is to be
twisted by torques T (see figure) until the angle of rotation between
the ends of the rod is 3.0.
d
T
T
(a) If the allowable shear strain in the copper is 0.0006 rad, what
is the maximum permissible diameter of the rod?
(b) If the rod diameter is 0.5 in., what is the minimum permissible length of the rod?
L
Probs. 3.2-1 and 3.2-2
Solution 3.2-1
(a) L 18 in. f 3
ga 0.0006 rad
From Eq. (3-5):
gmax rf
df
L
2L
Solve for d: dmax 2 L ga
f
0.413 in.
dmax 0.413 in.
(b) d 0.5 in.
Now solve for L:
L min df
21.817 in.
2 ga
L min 21.8 in.
Problem 3.2-2 A plastic bar of diameter d 56 mm is to be twisted by torques T (see figure) until the angle of rotation
between the ends of the bar is 4.0.
(a) If the allowable shear strain in the plastic is 0.012 rad, what is
the minimum permissible length of the bar?
(b) If the length of the bar is 200 mm, what is the maximum permissible diameter of the bar?
d
T
T
L
283
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CHAPTER 3 Torsion
Solution 3.2-2
(a) d 56 mm ga 0.012 rad f 4
Solution based on Eq. (3-5):
Lmin df
df
Lmin 162.897 mm
2ga
2 ga
L min 162.9 mm
(b) L 200 mm
dmax Now solve for d:
2 L ga
f
68.755 mm
Problem 3.2-3 A circular aluminum tube subjected to pure
dmax 68.8 mm
T
T
torsion by torques T (see figure) has an outer radius r2 equal to
1.5 times the inner radius r1.
L
(a) If the maximum shear strain in the tube is measured as
400 106 rad, what is the shear strain g1 at the inner
surface?
(b) If the maximum allowable rate of twist is 0.125 degrees
per foot and the maximum shear strain is to be kept at
400 106 rad by adjusting the torque T, what is the
minimum required outer radius (r2)min?
r2
r1
Probs. 3.2-3, 3.2-4, and 3.2-5
Solution 3.2-3
NUMERICAL DATA
(b) MINIMUM REQUIRED OUTER RADIUS
gmax
u
r2 1.5r1
gmax 400 (106) rad
r2min u 0.125a
p
1
ba b
180 12
r2min 2.2 in.
r2min gmax
u
;
u 1.818 104 rad /m
(a) SHEAR STRAIN AT INNER SURFACE AT RADIUS r1
g1 r1
gmax
r2
g1 g1 267 106 rad
1
gmax
1.5
;
Problem 3.2-4 A circular steel tube of length L 1.0 m is loaded in torsion by torques T (see figure).
(a) If the inner radius of the tube is r1 45 mm and the measured angle of twist between the ends is 0.5°, what is the
shear strain g1 (in radians) at the inner surface?
(b) If the maximum allowable shear strain is 0.0004 rad and the angle of twist is to be kept at 0.45° by adjusting the torque T,
what is the maximum permissible outer radius (r2)max?
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SECTION 3.2 Torsional Deformations
285
Solution 3.2-4
(b) MAXIMUM PERMISSIBLE OUTER RADIUS
NUMERICAL DATA
L 1000 mm
f 0.45 a
r1 45 mm
f 0.5a
p
b rad
180
(a) SHEAR STRAIN AT INNER SURFACE
f
g1 r1
g1 393 106 rad
L
p
b rad
180
gmax 0.0004 rad
r2max 50.9 mm
gmax r2
f
L
r2max gmax
L
f
;
;
Problem 3.2-5 Solve the preceding problem if the length L 56 in., the inner radius r1 1.25 in., the angle of twist is
0.5°, and the allowable shear strain is 0.0004 rad.
Solution 3.2-5
NUMERICAL DATA
L 56 in.
f 0.5 a
(b) MAXIMUM PERMISSIBLE OUTER RADIUS (r2)max
r1 1.25 in.
f 0.5 a
p
b rad
180
f
gmax r2
L
p
b rad
180
ga 0.0004 rad
(a) SHEAR STRAIN g1 (IN RADIANS) AT THE INNER
SURFACE
g1 r1
f
L
g1 195 106 rad
;
ga 0.0004 rad
L
r2max ga
f
r2max 2.57 in.
;
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CHAPTER 3 Torsion
Circular Bars and Tubes
Problem 3.3-1 A prospector uses a hand-powered winch (see
P
figure) to raise a bucket of ore in his mine shaft. The axle of the
winch is a steel rod of diameter d 0.625 in. Also, the distance
from the center of the axle to the center of the lifting rope is
b 4.0 in.
(a) If the weight of the loaded bucket is W 100 lb, what
is the maximum shear stress in the axle due to torsion?
(b) If the maximum bucket load is 125 lb, and the allowable
shear stress in the axle is 9250 psi, what is the minimum
permissible axle diameter?
d
W
b
W
Solution 3.3-1
(a) MAXIMUM SHEAR STRESS IN THE AXLE
W 100 lb
b 4 in.
tmax 16 T
p d3
d
5
in. 0.625 in.
8
8344.303 lb/in.2
T W b 33.333 ft-lb
tmax 8344 psi
(b) IF THE MAXIMUM BUCKET LOAD IS 125 LB, AND THE ALLOWABLE SHEAR STRESS IN THE AXLE IS 9250 PSI, WHAT IS THE
MINIMUM PERMISSIBLE AXLE DIAMETER?
W 125 lb
T Wb
dmin a
ta 9250 psi
1
3
16 T
b 0.651 in.
p ta
dmin 0.651 in.
Problem 3.3-2 When drilling a hole in a table leg, a furniture maker uses
a hand-operated drill (see figure) with a bit of diameter d 4.0 mm.
(a) If the resisting torque supplied by the table leg is equal to 0.3 N # m,
what is the maximum shear stress in the drill bit?
(b) If the allowable shear stress in the drill bit is 32 MPa, what is the
maximum resisting torque before the drill binds up?
(c) If the shear modulus of elasticity of the steel is G 75 GPa, what is
the rate of twist of the drill bit (degrees per meter)?
d
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287
SECTION 3.3 Circular Bars and Tubes
Solution 3.3-2
(a) MAXIMUM SHEAR STRESS IN DRILL BIT
From Eq. (3-14):
tmax tmax (c) RATE OF TWIST
From Eq. (3-16):
16T
u
pd3
16(0.3 N # m)
u
p(4.0 mm)3
tmax 23.8 MPa
;
(b) MAXIMUM RESISTING TORQUE BEFORE THE DRILL
T
GIp
0.3 N # m
p
(75 GPa) a b(4.0 mm)4
32
u 0.1592 rad/m 9.12/m
;
BINDS UP
ta 32 MPa d 4 mm
Tmax ta p d 3
0.402 N # m
16
Tmax 0.402 N # m
Problem 3.3-3 While removing a wheel to change a tire,
a driver applies forces P 25 lb at the ends of two of the arms
of a lug wrench (see figure). The wrench is made of steel with
shear modulus of elasticity G 11.4 106 psi. Each arm of
the wrench is 9.0 in. long and has a solid circular cross section
of diameter d 0.5 in.
(a) Determine the maximum shear stress in the arm
that is turning the lug nut (arm A).
(b) Determine the angle of twist (in degrees)
of this same arm.
P
9.0
in.
A
9.0
in.
d = 0.5 in.
P = 25 lb
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CHAPTER 3 Torsion
Solution 3.3-3
Lug wrench
(a) MAXIMUM SHEAR STRESS
P 25 lb
From Eq. (3-14):
(16)(450 lb - in.)
16T
tmax pd3
p(0.5 in.)3
L 9.0 in.
d 0.5 in.
G 11.4 106 psi
T torque acting on
arm A
tmax 18,300 psi
;
(b) ANGLE OF TWIST
From Eq. (3-17):
(450 lb-in.)(9.0 in.)
TL
f
GIP
p
b (0.5 in.)4
(11.4 * 106 psi)a
32
T P(2L) 2(25 lb)
(9.0 in.)
450 lb-in.
f 0.05790 rad 3.32
Problem 3.3-4 An aluminum bar of solid circular cross section is twisted by torques T acting at the ends (see figure). The
dimensions and shear modulus of elasticity are as follows:
L 1.4 m, d 32 mm, and G 28 GPa.
;
d
T
T
L
(a) Determine the torsional stiffness of the bar.
(b) If the angle of twist of the bar is 5, what is the maximum shear stress? What is the maximum shear strain
(in radians)?
(c) If a hole of diameter d/2 is drilled longitudinally through
the bar, what is the ratio of the torsional stiffnesses of the
hollow and solid bars? What is the ratio of their maximum
shear stresses if both are acted on by the same torque?
(d) If the hole diameter remains at d/2, what new outside diameter
d2 will result in equal stiffnesses of hollow and solid bars?
Solution 3.3-4
(a) TORSIONAL STIFFNESS OF SOLID BAR
d
G 28 GPa L 1.4 m
2
G IpS
p 4
d 1.029 * 107 m4 kTsolid kTsolid 2059 N # m
IpS 32
L
d 32 mm d2 d d1 (b) MAXIMUM SHEAR STRESS AND STRAIN OF SOLID BAR
k Tsolid f
f 5
gmax Tsolid k Tsolid f 179.671 N # m tmaxS tmax
G
gmax 997 * 106 rad
Ips
d
2
27.9 MPa
;
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SECTION 3.3 Circular Bars and Tubes
289
(c) STIFFNESS AND SHEAR STRESS RATIOS FOR SOLID AND HOLLOW BARS
IpH G IpH
p
1930.195 N # m
A d24 d14 B 9.651 * 108 m4 k T hollow 32
L
tmaxH A Tsolid B
d2
2
IpH
29.787 MPa
tmaxS
0.938
tmaxH
k T hollow
0.938
k T solid
tmaxH
1.067
tmaxS
(d) IF THE HOLE DIAMETER REMAINS AT d/2, WHAT NEW OUTSIDE DIAMETER d2 WILL RESULT IN EQUAL STIFFNESSES OF
HOLLOW AND SOLID BARS?
Equate torsional stiffnesses of solid and hollow bars; solve for new value of d2 for hollow bar.
Must have
d2 4 d1 4 d 4,
d2 32.5 mm
so solving for d2 gives
Problem 3.3-5 A high-strength steel drill rod used for boring
d = 0.5 in.
T
a hole in the earth has a diameter of 0.5 in. (see figure). The
allowable shear stress in the steel is 40 ksi and the shear modulus of elasticity is 11,600 ksi.
T
L
(a) What is the minimum required length of the rod so that
one end of the rod can be twisted 30 with respect to the
other end without exceeding the allowable stress?
(b) If the shear strain in part (a) is limited to 3.2 * 103, what
is the minimum required length of the drill rod?
Solution 3.3-5 Steel drill rod
d = 0.5 in.
T
T
L
T
(a) MINIMUM REQUIRED LENGTH
d 0.5 in.
p
brad 0.52360 rad
180
tallow 40 ksi
16T
Gpd 4f
Gdf
b 32L
2L
(11,600 ksi)(0.5 in.)(0.52360 rad)
2(40 ksi)
Lmin 38.0 in.
MINIMUM LENGTH
From Eq. (3-14): tmax ba
3
16
pd
Gdf
Lmin 2tallow
TL
32TL
GIP
Gpd4
Gpd 4f
, substitute T into Eq. (1):
32L
tmax a
G 11,600 psi
f 30 (30 ) a
From Eq. (3-17): f ;
(1)
pd3
(b) IF THE SHEAR STRAIN IN PART (a) IS LIMITED TO 3.2 * 103, WHAT IS THE MINIMUM REQUIRED LENGTH OF THE DRILL ROD?
G 11600 ksi f 30
gact df
3.445 * 103
2L
ta 40 ksi L 38 in. ga 3.2 A 103 B
L min d 0.5 in.
df
40.9 in.
2 ga
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CHAPTER 3 Torsion
Problem 3.3-6 The steel shaft of a socket wrench has a
diameter of 8.0 mm. and a length of 200 mm (see figure).
If the allowable stress in shear is 60 MPa, what is the
maximum permissible torque Tmax that may be exerted with
the wrench?
Through what angle f (in degrees) will the shaft twist
under the action of the maximum torque? (Assume G 78 GPa
and disregard any bending of the shaft.)
d = 8.0 mm
T
L = 200 mm
Solution 3.3-6 Socket wrench
ANGLE OF TWIST
From Eq. (3-17): f d 8.0 mm
L 200 mm
tallow 60 MPa
G 78 GPa
MAXIMUM PERMISSIBLE TORQUE
16T
From Eq. (3-14): tmax pd3
pd 3tmax
Tmax 16
Tmax p(8.0 mm)3(60 MPa)
16
Tmax 6.03 N # m
;
Problem 3.3-7 A circular tube of aluminum is subjected to torsion by torques T applied at the ends (see figure). The bar is 24
in. long, and the inside and outside diameters are 1.25 in. and
1.75 in., respectively. It is determined by measurement that the
angle of twist is 4 when the torque is 6200 lb-in.
(a) Calculate the maximum shear stress tmax in the tube, the
shear modulus of elasticity G, and the maximum shear
strain gmax (in radians).
(b) If the maximum shear strain in the tube is limited to
2.5 * 103 and the inside diameter is increased to
1.375 in., what is the maximum permissible torque?
TmaxL
GIP
From Eq. (3-14): Tmax f a
f
f
pd3t max
L
ba
b
16
GIP
pd3tmaxL(32)
4
16G(pd )
pd3tmax
16
IP pd4
32
2tmaxL
Gd
2(60 MPa)(200 mm)
0.03846 rad
(78 GPa)(8.0 mm)
f 10.03846 rad2a
180
/radb 2.20
p
T
;
T
24 in.
1.25 in.
1.75 in.
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SECTION 3.3 Circular Bars and Tubes
291
Solution 3.3-7
(a) MAXIMUM SHEAR STRESS, SHEARING MODULUS, AND
MAXIMUM SHEAR STRAIN
NUMERICAL DATA
L 24 in. r2 f 4a
MAXIMUM SHEAR STRAIN
1.75
1.25
in. r1 in.
2
2
p
b rad T 6200 lb-in.
180
MAXIMUM SHEAR STRESS
Ip p 4
A r r14 B
2 2
tmax gmax 0.00255 rad
tmax r2
f
L
;
SHEAR MODULUS OF ELASTICITY G
G
Tr2
Ip
tmax
gmax
G 3.129 * 106 psi
or G Ip 0.681 in.4
Tr2
tmax 7965 psi
Ip
gmax TL
fIp
G 3.13 * 106 psi
;
;
(b) IF THE MAXIMUM SHEAR STRAIN IN THE TUBE IS LIMITED TO 2.5 * 103 AND THE INSIDE DIAMETER IS INCREASED TO
1.375 in., WHAT IS THE MAXIMUM PERMISSIBLE TORQUE?
G 3.13 A 106 B psi d2 1.75 in. d1 1.375 in. Ip ga 2.5 A 103 B
Tmax 2 G Ip
d2
p
a d 4 d1 4 b 0.56985 in.4
32 2
ga 5096 lb-in.
Problem 3.3-8 A propeller shaft for a small yacht is made of a
solid steel bar 104 mm in diameter. The allowable stress in shear
is 48 MPa, and the allowable rate of twist is 2.0 in
3.5 meters.
(a) Assuming that the shear modulus of elasticity is
G 80 GPa, determine the maximum torque Tmax that
can be applied to the shaft.
(b) Repeat (a) if the shaft is now hollow with inner diameter
of 5d/8. Compare Tmax values to corresponding values
from part (a).
d
T
T
L
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CHAPTER 3 Torsion
Solution 3.3-8
(a) MAXIMUM TORQUE
NUMERICAL DATA
FIND MAXIMUM TORQUE BASED ON ALLOWABLE RATE OF
TWIST
d 104 mm ta 48 MPa u 2a
u
Ip p
b
180
3.5
f
L
Tmax GIpf
Tmax GIpu
L
Tmax 9164 N # m
¿
governs
rad/m G 80 GPa
;
FIND MAXIMUM TORQUE BASED ON ALLOWABLE SHEAR
STRESS
p 4
d Ip 1.149 * 107 mm4
32
Tmax ta Ip
d
2
Tmax 10,602 N # m
(b) REPEAT PART (a) IF THE SHAFT IS NOW HOLLOW WITH INNER DIAMETER OF 5d/8. COMPARE Tmax VALUES TO
CORRESPONDING VALUES FROM PART (a).
d2 104 mm d1 u
5
d 65 mm G 80 GPa ta 48 MPa
8 2
2
9.973 * 103 rad/m
3.5 m
Ip 9.733
p
0.847
A d24 d14 B 9.733 * 106 mm4
32
11.49
Tmax BASED ON ALLOWABLE RATE OF TWIST:
T1b G Ip u 7765.282 N # m
Tmax BASED ON ALLOWABLE SHEAR STRESS:
T2b For hollow shaft:
ta Ip
d2
2
T1b
0.847
9164 N # m
6 controls
T2b
0.847
10602 N # m
8983.919 N # m
Tmax T1b 7765 N # m
Problem 3.3-9 Three identical circular disks A, B, and C are welded
P3
to the ends of three identical solid circular bars (see figure). The bars
lie in a common plane and the disks lie in planes perpendicular to the
axes of the bars. The bars are welded at their intersection D to form a
rigid connection. Each bar has diameter d1 0.5 in. and each disk has
diameter d2 3.0 in.
Forces P1, P2, and P3 act on disks A, B, and C, respectively, thus
subjecting the bars to torsion. If P1 28 lb, what is the maximum shear
stress tmax in any of the three bars?
C
135∞
P1
P3
d1
A
D
135∞
P1
90∞
d2
P2
P2
B
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SECTION 3.3 Circular Bars and Tubes
293
Solution 3.3-9 Three circular bars
THE THREE TORQUES MUST BE IN EQUILIBRIUM
T3 is the largest torque:
T3 T1 12 P1d2 12
MAXIMUM SHEAR STRESS (Eq. 3-14)
16T3
16P1d2 12
16T
tmax pd3
pd31
pd31
d1 diameter of bars
0.5 in.
tmax d2 diameter of disks
16(28 lb)(3.0 in.) 12
p(0.5 in.)3
4840 psi
;
3.0 in.
P1 28 lb
T1 P1d2
T2 P2d2
T3 P3d2
Problem 3.3-10 The steel axle of a large winch on an ocean
T
liner is subjected to a torque of 1.65 kN m (see figure).
(a) What is the minimum required diameter dmin if the allowable shear stress is
48 MPa and the allowable rate of twist is 0.75/m? (Assume that the shear
modulus of elasticity is 80 GPa.)
(b) Repeat part (a) if the shaft is now hollow with an inner diameter of 5d/8. Compare dmin
values to corresponding values from (a).
d
T
Solution 3.3-10
(a) MINIMUM REQUIRED DIAMETER
NUMERICAL DATA
T 1.65 kN # m ta 48 MPa G 80 GPa
ua 0.75 a
p
b rad/m
180
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CHAPTER 3 Torsion
MINIMUM REQUIRED DIAMETER OR SHAFT BASED ON
MINIMUM REQUIRED DIAMETER OR SHAFT BASED ON
ALLOWABLE RATE OF TWIST
ALLOWABLE SHEAR STRESS
u
T
T
Ip GIp
Gu
32T
d pGua
4
p 4
T
d 32
Gu
t
Td
p
2a d 4 b
32
t
1
32T 4
dmin a
b
pGua
dmin 0.063 m
Td
2Ip
1
16T 3
d dmin 0.056 m
dmin c
pta
dmin 63.3 mm ;
¿
governs
dmin 55.9 mm
(b) REPEAT PART (a) IF THE SHAFT IS NOW HOLLOW WITH INNER DIAMETER OF 5d/8. COMPARE dmin VALUES TO CORRESPONDING
VALUES FROM PART (a).
u
0.75
0.013 rad/m
1m
Ip p
5 4
3471 p d 4
T
cd 4 a db d 32
8
131,072
Gu
1
dmin BASED ON ALLOWABLE RATE OF TWIST:
131,072 T
4
dmin 1 a
b 66 mm
3471 p G u
dmin 1
1.042
63.3 mm
6 controls
1
dmin BASED ON ALLOWABLE SHEAR STRESS:
dmin2
T
3
dmin2 59.116 mm
1.058
3471
p
55.9
mm
P2t
Q
a
dmin 66 mm
131,072
Problem 3.3-11 A hollow steel shaft used in a construction auger has
outer diameter d2 6.0 in. and inner diameter d1 4.5 in. (see figure).
The steel has shear modulus of elasticity G 11.0 106 psi.
For an applied torque of 150 k-in., determine the following quantities:
(a) Shear stress t2 at the outer surface of the shaft.
(b) Shear stress t1 at the inner surface.
(c) Rate of twist u (degrees per unit of length).
d2
Also, draw a diagram showing how the shear stresses vary in magnitude
along a radial line in the cross section.
d1
d2
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295
SECTION 3.3 Circular Bars and Tubes
Solution 3.3-11 Construction auger
d2 6.0 in.
r2 3.0 in.
d1 4.5 in.
r1 2.25 in.
(c) RATE OF TWIST
u
G 11 106 psi
u 157 * 106 rad/ in. 0.00898/ in.
T 150 k-in.
IP (150 k-in.)
T
GIP
(11 * 106 psi)(86.98 in.)4
p 4
(d d14) 86.98 in.4
32 2
;
(d) SHEAR STRESS DIAGRAM
(a) SHEAR STRESS AT OUTER SURFACE
t2 (150 k-in.)(3.0 in.)
Tr2
IP
86.98 in.4
5170 psi
;
(b) SHEAR STRESS AT INNER SURFACE
t1 Tr1
r1
t 3880 psi
IP
r2 2
;
Problem 3.3-12 Solve the preceding problem if the shaft has outer diameter d2 150 mm and inner diameter d1 100 mm.
Also, the steel has shear modulus of elasticity G 75 GPa and the applied torque is 16 kN # m.
Solution 3.3-12 Construction auger
d2 150 mm
r2 75 mm
d1 100 mm
r1 50 mm
G 75 GPa
T 16 kN # m
p 4
IP (d d14) 39.88 * 106 mm4
32 2
(a) SHEAR STRESS AT OUTER SURFACE
(16 kN # m)(75 mm)
Tr2
t2 IP
39.88 * 106 mm4
30.1 MPa
(b) SHEAR STRESS AT INNER SURFACE
t1 Tr1
r1
t 2 20.1 MPa
IP
r2
;
(c) RATE OF TWIST
T
16 kN # m
u
GIP
(75 GPa)(39.88 * 106 mm4)
u 0.005349 rad/m 0.306°/m
;
(d) SHEAR STRESS DIAGRAM
;
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CHAPTER 3 Torsion
Problem 3.3-13 A vertical pole of solid circular cross section is twisted by horizontal forces P 1100 lb acting at the
ends of a rigid horizontal arm AB (see figure part a). The distance from the outside of the pole to the line of action of each
force is c 5.0 in. (see figure part b) and the pole height L 14 in.
(a) If the allowable shear stress in the pole is 4500 psi, what is the minimum required diameter dmin of the pole?
(b) Find the torsional stiffness of the pole (kip-in./rad). Assume that G 10,800 ksi.
(c) If two translational springs, each with stiffness k 33 kips/in., are added at 2c/5 from A and B (see figure part c),
repeat part (a) to find dmin. [Hint: Consider the pole and pair of springs as “springs in parallel.”]
c
P
c
d
c
P
c
d
c
P
c
B
A
A
B
A
B
P
P
P
(b)
d
k
k
L = 14 in.
3c/5
3c/5
(c)
(a)
Solution 3.3-13
Vertical pole
NUMERICAL DATA
P 1100 lb
c 5 in.
d
2
16 T
T
(a) t (b) k T Ip
t
G Ip
L
pd
Ip 3
L 14 in.
ta 4500 psi
T P (2 c + d) P (2 c + d) a
p 4
d 3.813 in.4
32
kT G Ip
L
G 10800 ksi
16
p d3
b ta
Solving gives d 2.5 in.
2941 k-in./rad
(c) k 33 k/in.
Total torsional stiffness torsional stiffness of pole plus effect of parallel springs when a unit rotation is applied:
kTtotal G Ip
L
+ 2k a
3
d 2
c + b 4132.428 k-in./rad
5
2
f
[P (2 c + d)]
3.326 * 103
kTtotal
For pole:
tpole Gfd
3202.942 psi
2L
fa Tmax L
G Ip
fa 2 ta L
4.673 * 103
Gd
and Tmax ta Ip
d
2
ta a
pd 3
b
16
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297
SECTION 3.3 Circular Bars and Tubes
Equate expressions for f and fa, then solve for dmin:
[P (2 c + d)]
4
Ga
pd
b
32
L
2 ta L
Gd
P (2 c + d)
OR
2k a
3
d 2
+ 2k a c + b
5
2
Solving:
2
d
3c
pGd
+ b +
5
2
32 L
4
2 L ta
0
Gd
dmin 1.996 in.
Problem 3.3-14 A vertical pole of solid circular cross section is twisted by horizontal forces P 5 kN acting at the ends
of a rigid horizontal arm AB (see figure part a). The distance from the outside of the pole to the line of action of each force is
c 125 mm (see figure part b) and the pole height L 350 mm.
(a) If the allowable shear stress in the pole is 30 MPa, what is the minimum required diameter dmin of the pole?
(b) What is the torsional stiffness of the pole (kN # m/rad)?
(c) If two translational springs each with stiffness k 2550 kN/m are added at 2c/5 from A and B (see figure part c),
repeat part (a) to find dmin. (Hint: Consider the pole and pair of springs as “springs in parallel.”)
P
c
c
d
c
c
P
d
c
P
c
B
A
A
B
A
B
P
P
P
(b)
d
k
k
L = 14 in.
3c/5
3c/5
(c)
(a)
Solution 3.3-14
Vertical pole
NUMERICAL DATA
P 5 kN c 125 mm
T
(a) t (b) k T d
2
Ip
G Ip
L
t
16 T
pd
Ip 3
L 350 mm
ta 30 MPa
G 28 GPa
T P (2 c + d) P (2 c + d) a
16
pd 3
b ta
Solving gives:
G Ip
p 4
d 4.052 in.4 k T 134.9 kN # m/rad
32
L
d 64.4 mm
k T 134.9 kN # m/rad
(c) TOTAL TORSIONAL STIFFNESS
k 2550 kN/m
k Ttotal G Ip
3
d 2
+ 2 k a c + b 193.533 kN # m/rad
L
5
2
f
[P (2 c + d)]
8.122 * 103
k Ttotal
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CHAPTER 3 Torsion
FOR POLE
Gfd
Tmax L
20.917 MPa fa 2L
G Ip
tpole fa and Tmax ta Ip
d
2
ta a
pd 3
b
16
2 ta L
0.012
Gd
Equate expressions for f and fa, then solve for dmin:
[P (2 c + d)]
Ga
4
pd
b
32
L
3
d
+ 2k a c + b
5
2
2 ta L
Gd
or
2
Solving:
P (2 c + d)
2k a
2
3c
d
pGd
+ b +
5
2
32 L
4
2 L ta
0
Gd
dmin 50 mm
Problem 3.3-15 A solid brass bar of diameter d 1.25 in. is
d
T1
subjected to torques T1, as shown in part (a) of the figure. The
allowable shear stress in the brass is 12 ksi.
(a) What is the maximum permissible value of the torques T1?
(b) If a hole of diameter 0.625 in. is drilled longitudinally through
the bar, as shown in part (b) of the figure, what is the maximum
permissible value of the torques T2?
(c) What is the percent decrease in torque and the percent decrease
in weight due to the hole?
T1
(a)
d
T2
T2
(b)
Solution 3.3-15
(b) MAXIMUM PERMISSIBILE VALUE OF TORQUE T2 —
HOLLOW BAR
(a) MAXIMUM PERMISSIBILE VALUE OF TORQUE T1—SOLID
BAR
T1max taIp
p
ta d 4
32
T1max d
2
d
2
1
T1max tapd 3
16
1
T1max (12)p (1.25)3
16
T1max 4.60 in.-k
;
d1
d2
d2 1.25 in.
d1 0.625 in.
ta 12 ksi
p
1 d 24 d142
32
T2max d2
2
d24d12
1
T2max tap
16
d2
ta
T2max 4.31 in.-k
;
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299
SECTION 3.3 Circular Bars and Tubes
(c) PERCENT DECREASE IN TORQUE AND PERCENT
DECREASE IN WEIGHT DUE TO HOLE IN PART (b)
Percent decrease in torque:
T1max T2max
(100) 6.25%
T1max
Percent decrease in weight (weight is proportional to
x-sec area):
A1 ;
p 2
d
4 2
A2 p 2
1d d122
4 2
A1 A2
(100) 25 %
A1
;
Problem 3.3-16 A hollow aluminum tube used in a roof structure has an outside
diameter d2 104 mm and an inside diameter d1 82 mm (see figure). The tube is
2.75 m long, and the aluminum has shear modulus G 28 GPa.
(a) If the tube is twisted in pure torsion by torques acting at the ends, what is the
angle of twist (in degrees) when the maximum shear stress is 48 MPa?
(b) What diameter d is required for a solid shaft (see figure) to resist the same torque
with the same maximum stress?
(c) What is the ratio of the weight of the hollow tube to the weight of the solid shaft?
d
d1
d2
Solution 3.3-16
Set tmax expression equal to
NUMERICAL DATA
d2 104 mm
Td22
d1 82 mm
p
a d 4 d14 b
32 2
L 2.75 103 mm
d3 I p (p/32)(d2 4 d14 )
f (tmax)
;
(c) RATIO OF WEIGHTS OF HOLLOW AND SOLID SHAFTS
WEIGHT IS PROPORTIONAL TO CROSS-SECTIONAL AREA
Td2 2L
b
2Ip Gd2
p 2
A d d12 B
4 2
Ah
p
As d reqd2
0.524
4
As
Ah 2L
Gd2
f 0.091 rad
f 5.19
d24 d14
d2
1
(a) FIND ANGLE OF TWIST tmax 48 MPa
f a
pa d24 d14 b
d24 d14 3
dreqd a
b dreqd88.4 mm
d2
Ip 7.046 106 mm4
TL
GIp
32Td2
Then solve for d:
G 28 GPa
f
;
So the weight of the tube is 52% of the solid shaft,
but they resist the same torque.
;
(b) REPLACE HOLLOW SHAFT WITH SOLID SHAFT—FIND
DIAMETER
d
2
p
T
tmax 32d
tmax 16T
d3p
4
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CHAPTER 3 Torsion
Problem 3.3-17 A circular tube of inner radius r1 and outer radius r2 is subjected to a torque produced by forces P 900 lb
(see figure part a). The forces have their lines of action at a distance b 5.5 in. from the outside of the tube.
(a) If the allowable shear stress in the tube is 6300 psi and the inner radius r1 1.2 in., what is the minimum permissible
outer radius r2?
(b) If a torsional spring of stiffness kR 450 kip-in./rad is added at the end of the tube (see figure part b), what is the maximum value of forces P if the allowable shear stress is not to be exceeded? Assume that the tube has length L 18 in.,
outer radius r2 1.45 in., and shear modulus G 10,800 ksi. (Hint: Consider the tube and torsional spring as “springs
in parallel.”)
P
P
kR
P
P
(b)
P
r2
r1
P
b
b
2r2
(a)
Solution 3.3-17 Circular tube in torsion
NUMERICAL DATA
P 900 lb b 5.5 in. ta 6300 psi r1 1.2 in. L 18 in.
(a) USING THE TORSION FORMULA; SET SHEAR STRESS EQUAL TO ALLOWABLE VALUE THEN SOLVE FOR RADIUS r2
P 12 b + 2 r22 r2
T r2
t
or
ta
Solving: r2 1.399 in.
p
Ip
A r 24 r 14 B
2
p
(b) LET r2 1.45 in. Ip A r 2 4 r 1 4 B 3.687 in.4 G 10800 ksi k R 450 kip-in./rad
2
Allowable twist of shaft fa based on allowable shear stress Ta:
ta Ip
Tmax L
ta L
fa where Tmax So fa 7.24138 * 103
G Ip
r2
G r2
Twist of shaft and torsional spring due to torque P(2b 2r2):
f
P (2 b + 2 r2)
4.7 * 103
G IP
a
+ kR b
L
Equate f and fa and solve for Pmax:
Pmax (2 b + 2 r2)
ta L
G r2
G IP
+ kRb
a
L
Solving for pmax:
Pmax 1387 lb
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SECTION 3.4 Nonuniform Torsion
301
Nonuniform Torsion
T1
Problem 3.4-1 A stepped shaft ABC consisting of two solid
d1
circular segments is subjected to torques T1 and T2 acting in
opposite directions, as shown in the figure. The larger segment
of the shaft has a diameter of d1 2.25 in. and length of L1 30 in.;
the smaller segment has a diameter of d2 1.75 in. and length
L2 20 in. The material is steel with shear modulus
G 11 106 psi, and the torques are T1 20,000 lb-in.
and T2 8,000 lb-in.
d2
B
A
L1
T2
C
L2
(a) Calculate the maximum shear stress tmax in the shaft, and the
angle of twist fc (in degrees) at end C.
(b) If the maximum shear stress in BC must be the same as that in
AB, what is the required diameter of segment BC? What is the
resulting twist at end C?
Solution 3.4-1
Stepped shaft
(a) d2 1.75 in. d1 2.25 in.
L1 30 in. L2 20 in.
G 11(106) psi
T1 20,000 in.-lb T2 8000 in.-lb
IpAB tAB p 4
p
d 2.516 in.4 IpBC d 4 0.921 in.4
32 1
32 2
1T1 T22 a
d1
b
2
IpAB
fC fAB + fBC
5365 psi
fAB tBC 1T2 T12 L 1
G IpAB
T2 a
d2
b
2
IpBC
7602 psi
0.745 fBC T2 L 2
0.905
G IpBC
fC fAB + fBC 0.16
(b) tBC T2 a
Check:
16
p d 23
b
So T2 a
tBC T2 a
fBC 16
p d 23
16
p d 2reqd3
b tAB Solving for d2:
b 5365 psi
T2 L 2
0.569
G IpBC
1
16 3
d2 reqd aT2
b 1.966 in.
p tAB
6 same as tAB above
IpBC p
4
d
1.465 in.4
32 2 reqd
fC fAB + fBC 0.177
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CHAPTER 3 Torsion
Problem 3.4-2 A circular tube of outer diameter d3 70 mm and
inner diameter d2 60 mm is welded at the right-hand end to a fixed
plate and at the left-hand end to a rigid end plate (see figure). A solid
circular bar of diameter d1 40 mm is inside of, and concentric with,
the tube. The bar passes through a hole in the fixed plate and is
welded to the rigid end plate.
The bar is 1.0 m long and the tube is half as long as the bar. A
torque T 1000 N # m acts at end A of the bar. Also, both the bar
and tube are made of an aluminum alloy with shear modulus of
elasticity G 27 GPa.
Tube
Fixed
plate
End
plate
Bar
T
A
(a) Determine the maximum shear stresses in both the bar
and tube.
(b) Determine the angle of twist (in degrees) at end A of the bar.
Tube
Bar
d1
d2
d3
Solution 3.4-2
Bar and tube
TORQUE
T 1000 N # m
(a) MAXIMUM SHEAR STRESSES
Bar: t bar 16T
79.6 MPa
;
pd31
T(d3/2)
32.3 MPa
Tube: t tube (Ip) tube
;
TUBE
d3 70 mm
d2 60 mm
Ltube 0.5 m
G 27 GPa
(Ip) tube (b) ANGLE OF TWIST AT END A
Bar: fbar p 4
(d3 d24)
32
Tube: ftube 1.0848 * 106 mm4
fA 9.43°
(Ip) bar pd14
32
Lbar 1.0 m
TL tube
0.0171 rad
G(Ip) tube
fA fbar ftube 0.1474 0.0171 0.1645 rad
BAR
d1 40 mm
TL bar
0.1474 rad
G(Ip) bar
;
G 27 GPa
251.3 * 103 mm4
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SECTION 3.4 Nonuniform Torsion
Problem 3.4-3 A stepped shaft ABCD consisting of solid circular
12.5 k-in.
segments is subjected to three torques, as shown in the figure. The
torques have magnitudes 12.5 k-in., 9.8 k-in., and 9.2 k-in. The
length of each segment is 25 in. and the diameters of the segments
are 3.5 in., 2.75 in., and 2.5 in. The material is steel with shear
modulus of elasticity G 11.6 103 ksi.
3.5 in.
9.8 k-in.
9.2 k-in.
2.75 in.
2.5 in.
25 in.
(a) Calculate the maximum shear stress tmax in the shaft and the
angle of twist fD (in degrees) at end D.
(b) If each segment must have the same shear stress, find the
required diameter of each segment in part (a) so that all three
segments have shear stress tmax from part (a). What is the
resulting angle of twist at D?
D
C
B
A
303
25 in.
25 in.
Solution 3.4-3
(a)
TB 12.5 k-in. TC 9.8 k-in. TD 9.2 k-in. L 25 in.
G 11.6 (103) ksi dAB 3.5 in. dBC 2.75 in. dCD 2.5 in.
IpAB tAB tCD p
p
p
d 4 I
d 4 I
d 4
32 AB pBC 32 BC pCD 32 CD
1TB + TC + TD2 a
dAB
b
2
IpAB
TD a
dCD
b
2
IpCD
3742 psi tBC 2999 psi or tCD TD a
fD fAB + fBC + fCD fD 1TC + TD2 a
dBC
b
2
IpBC
16
p d CD3
4653 psi
max. shear stress
b 2999 psi
1TC + TD2
TD
L 1TB + TC + TD2
c
+
+
d 0.978 fD 0.017 rad
G
IpAB
IpBC
IpCD
(b)
1
3
tmax tBC 4653 psi
so
16
dABreqd c A TB + TC + TD B a
b d 3.25 in.
p tmax
1
dBC 2.75 in.
IpAB p
4
d
32 ABreqd
fD 3
16
dCDreqd cTD a
b d 2.16 in.
p tmax
IpCD p
4
d
32 CDreqd
(TC + TD)
TD
L 1TB + TC + TD2
c
+
+
d 1.303 fD 0.023 rad
G
IpAB
IpBC
IpCD
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CHAPTER 3 Torsion
Problem 3.4-4 A solid circular bar ABC consists of two
segments, as shown in the figure. One segment has diameter
d1 56 mm and length L1 1.45 m; the other segment has
diameter d2 48 mm and length L2 1.2 m.
What is the allowable torque Tallow if the shear stress is not
to exceed 30 MPa and the angle of twist between the ends of the
bar is not to exceed 1.25°? (Assume G 80 GPa.)
d1
d2
T
A
C
B
L1
T
L2
Solution 3.4-4
Tallow based on angle of twist:
NUMERICAL DATA
d1 56 mm
d2 48 mm
L1 1450 mm
ta 30 MPa
L2 1200 mm
f a 1.25 a
G 80 GPa
fmax p
b rad
180
Tallow Allowable torque:
16T
d32p
Tallow tapd23
16
Ja
L1
p 4
d1 b
32
+
L2
a
p 4
d2 b K
32
Gf a
L1
p
a d1 4 b
32
Tallow based on shear stress
tmax T
G
+
L2
p
a d2 4 b
32
Tallow 459 N # m
;
T1 =
T2 =
1000 lb-in. 500 lb-in.
T3 =
T4 =
800 lb-in. 500 lb-in.
governs
Tallow 651.441 N # m
Problem 3.4-5 A hollow tube ABCDE constructed of
monel metal is subjected to five torques acting in the directions
shown in the figure. The magnitudes of the torques are
T1 1000 lb-in., T2 T4 500 lb-in., and T3 T5 800 lb-in.
The tube has an outside diameter d2 1.0 in. The allowable
shear stress is 12,000 psi and the allowable rate of twist is
2.0°/ft.
Determine the maximum permissible inside diameter d1
of the tube.
A
B
C
D
d2 = 1.0 in.
T5 =
800 lb-in.
E
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305
SECTION 3.4 Nonuniform Torsion
Solution 3.4-5
Hollow tube of monel metal
REQUIRED POLAR MOMENT OF INERTIA BASED UPON
ALLOWABLE SHEAR STRESS
tmax d2 1.0 in.
tallow 12,000 psi
uallow 2°/ft 0.16667°/in.
Tmax r
Ip
IP Tmax(d2/2)
0.05417 in.4
tallow
REQUIRED POLAR MOMENT OF INERTIA BASED UPON
ALLOWABLE ANGLE OF TWIST
0.002909 rad/in.
From Table I-2, Appendix I: G 9500 ksi
TORQUES
u
Tmax
GIP
IP Tmax
0.04704 in.4
Gu allow
SHEAR STRESS GOVERNS
Required IP 0.05417 in.4
IP T1 1000 lb-in.
T2 500 lb-in.
T4 500 lb-in.
T5 800 lb-in.
T3 800 lb-in.
INTERNAL TORQUES
p 4
(d 2 d 41)
32
d 41 d 43 32(0.05417 in.4)
32IP
(1.0 in.)4 p
p
0.4482 in.4
TAB T1 1000 lb-in.
d1 0.818 in.
TBC T1 T2 500 lb-in.
(Maximum permissible inside diameter)
;
TCD T1 T2 T3 1300 lb-in.
TDE T1 T2 T3 T4 800 lb-in.
Largest torque (absolute value only):
Tmax 1300 lb-in.
80 mm
Problem 3.4-6 A shaft of solid circular cross section consisting of two segments is shown in the first part of the figure. The left-hand segment has a
diameter of 80 mm and length of 1.2 m; the right-hand segment has a diameter
of 60 mm and length of 0.9 m.
Shown in the second part of the figure is a hollow shaft made of the same
material and having the same length. The thickness t of the hollow shaft is d/10,
where d is the outer diameter. Both shafts are subjected to the same torque.
(a) If the hollow shaft is to have the same torsional stiffness as the solid
shaft, what should be its outer diameter d?
(b) If torque T is applied at either end of both shafts, and the hollow shaft
is to have the same maximum shear stress as the solid shaft, what should
be its outer diameter d?
1.2 m
60 mm
0.9 m
d
d
t=—
10
2.1 m
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CHAPTER 3 Torsion
Solution 3.4-6
Solid and hollow shafts
(a) L1 1.2 m
L2 0.9 m
d1 80 mm
d2 60 mm
L 2.1 m
HOLLOW SHAFT
SOLID SHAFT CONSISTING OF TWO SEGMENTS
d0 inner diameter 0.8d
TLi
f1 g
GIPi
T(1.2 m)
T(0.9 m)
p
Ga b(80 mm)4
32
+
32T
(29,297 m3 + 69,444 m3)
pG
32T
(98,741 m3)
pG
TL
GIp
f2 p
G a b (60 mm)4
32
T(2.1 m)
p
G a b[d 4 (0.8d)4]
32
32T
2.1 m
32T 3.5569 m
a
b a
b
pG 0.5904 d 4
pG
d4
UNITS: d meters
TORSIONAL STIFFNESS
kT T
f
Torque T is the same for both shafts.
For equal stiffnesses, f1 f2.
98,741 m3 d4 3.5569 m
d4
3.5569
36.023 * 106 m4
98,741
d 0.0775 m 77.5 mm
;
(b) FIRST, FIND EXPRESSIONS FOR MAXIMUM SHEAR STRESS IN SEGMENTS 1 AND 2 OF THE STEPPED SHAFT
tmax1 T a
16
p d 13
b
T
3
32000 p mm
tmax2 T a
16
p d 23
b T
3
13500 p mm
Next, find an expression for max. shear stress in the hollow shaft:
larger shear stress
occurs on surface of
segment 2 of stepped
solid shaft
tmaxHollow d
Ta b
2
8d 4
p
cd 4 a b d
32
10
Equate tmax for each segment of stepped shaft to tmax for the hollow
shaft and then solve for required outer diameter d of the hollow shaft.
The two solutions are as follows.
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SECTION 3.4 Nonuniform Torsion
(1) If tmax1 tmaxHollow,
d 95.4 mm
(2) If tmax2 tmaxHollow,
d 71.5 mm
307
Use d 71.5 mm (and t d/10 7.15 mm)
Check:
Let
T 1.0 kN # m
tmaxHollow tmax1 T a
d 71.5 mm
d
Ta b
2
p
8d 4
cd 4 a b d
32
10
16
p d13
23.6 MPa
b 9.95 MPa
tmax2 T a
16
p d23
Problem 3.4-7 Four gears are attached to a circular shaft and
transmit the torques shown in the figure. The allowable shear
stress in the shaft is 10,000 psi.
(a) What is the required diameter d of the shaft if it has a
solid cross section?
(b) What is the required outside diameter d if the shaft is
hollow with an inside diameter of 1.0 in.?
b 23.6 MPa
8,000 lb-in.
19,000 lb-in.
4,000 lb-in.
A
7,000 lb-in.
B
C
D
Solution 3.4-7 Shaft with four gears
(b) HOLLOW SHAFT
Inside diameter d0 1.0 in.
tallow 10,000 psi
TBC 11,000 lb-in.
TAB 8000 lb-in.
TCD 7000 lb-in.
Tr
tallow Ip
10,000 psi (a) SOLID SHAFT
tmax tmax 16T
pd3
d
Tmax a b
2
Ip
d
(11,000 lb-in.) a b
2
a
p
b[d 4 (1.0 in.)4]
32
UNITS: d inches
16(11,000 lb-in.)
16Tmax
5.602 in.3
d3 pt allow
p(10,000 psi)
10,000 Required d 1.78 in.
or
;
56,023 d
d4 1
d 4 5.6023 d 1 0
Solving: d 1.832
Required: d 1.83 in.
;
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CHAPTER 3 Torsion
Problem 3.4-8 A tapered bar AB of solid circular cross section is
T
twisted by torques T (see figure). The diameter of the bar varies
linearly from dA at the left-hand end to dB at the right-hand end.
B
A
T
(a) Confirm that the angle of twist of the tapered bar is
L
1
1
32TL
3b
a
f
3pG(dB dA) d3A
dB
dA
(b) For what ratio dB/dA will the angle of twist of the tapered bar be
one-half the angle of twist of a prismatic bar of diameter dA?
(The prismatic bar is made of the same material, has the
same length, and is subjected to the same torque as the
tapered bar.)
Solution 3.4-8
f
d(x) dA a 1 L
L0
x
x
b + dB
L
L
T
32 T L
1
1
a
3b
3 p G (dB dA) d3A
dB
Now set ftapered dB
dA
(b + b + 1)
IPA fprismatic
2
2
3 b3
d(0) dA d(L) dB
p
x
x 4
G
cdA a1 b + dB d
32
L
L
where b or
Problems 3.4-8, 3.4-9 and 3.4-10
Tapered bar AB
(a) LINEAR VARIATION
(b) ftapered dB
1
2
dx 1
32TL
1
a
3b
3pG(dB dA) d3A
dB
or ftapered b2 + b + 1
TL
a
b
3 G IPA
b3
p 4
TL
dA and fprismatic 32
G IPA
or
L T (b 2 + b + 1)
3 G IPA b
Solve numerically:
3
LT
2 G IPA
b 1.446
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SECTION 3.4 Nonuniform Torsion
309
Problem 3.4-9 A tapered bar AB of solid circular cross section is twisted
by torques T 36,000 lb-in. (see figure). The diameter of the bar varies
linearly from dA at the left-hand end to dB at the right-hand end. The bar
has length L 4.0 ft and is made of an aluminum alloy having shear
modulus of elasticity G 3.9 106 psi. The allowable shear stress in
the bar is 15,000 psi and the allowable angle of twist is 3.0°.
If the diameter at end B is 1.5 times the diameter at end A, what is
the minimum required diameter dA at end A? [Hint: Use the angle of
twist expression from Prob. 3.4-8(a)].
Solution 3.4-9
Tapered bar
MINIMUM DIAMETER BASED UPON ALLOWABLE ANGLE OF
dB 1.5 dA
TWIST-MODIFYING THE ANGLE OF TWIST EXPRESSION FROM
T 36,000 lb-in.
PROB. 3.4-8a:
L 4.0 ft 48 in.
b dB/dA 1.5
G 3.9 10 psi
6
f
tallow 15,000 psi
fallow 3.0°
0.0523599 rad
MINIMUM DIAMETER BASED UPON ALLOWABLE SHEAR
tmax pd3A
d3A (36,000 lb-in.)(48 in.)
p
(3.9 * 10 psi) a bdA4
32
(0.469136)
6
STRESS
16T
b2 + b + 1
TL
TL
a
b (0.469136)
3
G(IP)A
G(I
3b
P)A
16(36,000 lb-in.)
16 T
ptallow
p(15,000 psi)
12.2231 in.3
dA 2.30 in.
dA4 2.11728 in.4
dA4
2.11728 in.4
2.11728 in.4
0.0523599 rad
fallow
40.4370 in.4
dA 2.52 in.
ANGLE OF TWIST GOVERNS
Minimum: dA 2.52 in.
;
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CHAPTER 3 Torsion
Problem 3.4-10 The bar shown in the figure is tapered linearly from
end A to end B and has a solid circular cross section. The diameter at the
smaller end of the bar is dA 25 mm and the length is L 300 mm.
The bar is made of steel with shear modulus of elasticity G 82 GPa.
If the torque T 180 N # m and the allowable angle of twist is 0.3°,
what is the minimum allowable diameter dB at the larger end of the bar?
[Hint: Use the angle of twist expression from Prob. 3.4-8(a)].
Solution 3.4-10
Tapered bar
p
(0.3)a 180 rad/ b
dA 25 mm
L 300 mm
G 82 GPa
T 180 N # m
(180 N # m)(0.3 m)
(82 GPa)a
fallow 0.3°
p
b(25 mm)4
32
a
b2 + b + 1
3b 3
b
b2 + b + 1
Find dB.
0.304915 DIAMETER BASED UPON ALLOWABLE ANGLE OF TWIST-
0.914745b3 b2 1 0
3b 3
MODIFYING THE ANGLE OF TWIST EXPRESSION FROM
PROB. 3.4-8a:
b
SOLVE NUMERICALLY:
dB
dA
b2 + b + 1
TL
f
a
b
G(IP)A
3b 3
b 1.94452
Minimum: dB bdA 48.6 mm
;
p 4
(IP)A d
32 A
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Page 311
311
SECTION 3.4 Nonuniform Torsion
Problem 3.4-11 The nonprismatic cantilever circular bar
shown has an internal cylindrical hole from 0 to x, so the net
polar moment of inertia of the cross section for segment 1 is
(7/8)Ip. Torque T is applied at x and torque T/2 is applied
at x L. Assume that G is constant.
(a) Find reaction moment R1.
(b) Find internal torsional moments Ti in segments 1 and 2.
(c) Find x required to obtain twist at joint 3 of w3 TL/GIp
(d) What is the rotation at joint 2, w2?
(e) Draw the torsional moment (TMD: T(x), 0 x L) and
displacement (TDD: w(x), 0 x L) diagrams.
Segment 1
Segment 2
7
—Ip
8
Ip
x
R1
T
1
2
T
—
2
3
x
L–x
T1
T2
TMD 0
φ2
TDD 0
0
φ3
0
Solution 3.4-11
(a) REACTION TORQUE R1
T
3
a Mx 0 R1 a T + 2 b R1 2 T ;
L
1
17
x + L
14
2
x
14 L
a b
17 2
(b) INTERNAL MOMENTS IN SEGMENTS 1 AND 2
T1 R1
T1 1.5 T
T2 T
2
(c) FIND X REQUIRED TO OBTAIN TRWIST AT JOINT 3
f3 a
TiLi
GIPi
TL
GIP
T 1x
TL
GIP
L
7
Ga IP b
8
+
3
a Tbx
2
7
Ga IP b
8
3
a bx
2
7
a b
8
+
T2(L x)
GIP
T
a b(L x)
2
+
GIP
7
L
17
x
;
(d) ROTATION AT JOINT 2 FOR X VALUE IN PART (C)
f2 f2 T1x
7
G a Ip b
8
12TL
17GIP
f2 3
7
a T b a Lb
2
17
7
Ga Ip b
8
;
(e) TMD AND TDD—SEE PLOTS ABOVE
TMD is constant—T1 for 0 to x and T2 for x to L;
hence TDD is linear—zero at joint 1, f2 at joint 2
and f3 at joint 3
1
(L x)
2
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Page 312
CHAPTER 3 Torsion
Problem 3.4-12 A uniformly tapered tube AB of hollow circular cross
section is shown in the figure. The tube has constant wall thickness t and
length L. The average diameters at the ends are dA and dB 2dA. The polar
moment of inertia may be represented by the approximate formula IP L pd3t/4
(see Eq. 3-21).
Derive a formula for the angle of twist f of the tube when it is subjected
to torques T acting at the ends.
B
A
T
T
L
t
t
dA
dB = 2dA
Solution 3.4-12
Tapered tube
t thickness (constant)
dA, dB average diameters at the ends
dB 2dA
Ip pd3t
(approximate formula)
4
ANGLE OF TWIST
Take the origin of coordinates at point O.
d(x) x
x
(dB) dA
2L
L
For element of length dx:
df Tdx
GIP(x)
3
p[d(x)] t
ptd 3A 3
x
Ip(x) 3
4
4L
2L
f
LL
df Tdx
ptd3A
G a 3 b x3
4L
4TL3
pGtd3A LL
2L
dx
3
x
4TL3dx
pGtdA3 x3
3TL
2pGtd3A
;
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313
SECTION 3.4 Nonuniform Torsion
Problem 3.4-13 A uniformly tapered aluminum-alloy tube AB
of circular cross section and length L is shown in the figure. The
outside diameters at the ends are dA and dB 2dA. A hollow
section of length L/2 and constant thickness t dA/10 is cast
into the tube and extends from B halfway toward A.
(a) Find the angle of twist w of the tube when it is subjected
to torques T acting at the ends. Use numerical values as
follows: dA 2.5 in., L 48 in., G 3.9 106 psi, and
T 40,000 in.-lb.
(b) Repeat part (a) if the hollow section has constant diameter dA. (See figure part b.)
A
T
t constant
dB – 2t
L
—
2
B
T
dA
L
dB
(a)
L
—
2
T
dA
A
dA
B
T
L
dB
(b)
Solution 3.4-13
PART (a)—CONSTANT THICKNESS
Use x as integration variable measured from B toward A.
FROM B TO CENTERLINE
Outer and inner diameters as function of x.
0 … x …
L
2
d0(x) 2dA d0(x) dB a
xdA
L
di (x) (dB 2t) di (x) dB dA
bx
L
[(2dA 2t)(dA 2t)]
x
L
1 9L + 5x
d
5 A
L
SOLID FROM CENTERLINE TO A
L
… x … L
2
d0(x) 2dA x dA
L
L
L
T 32
1
1
2
f a b
dx
+
dx
L
4
4
p
G
P L0 d0 d i
L2 d04 Q
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CHAPTER 3 Torsion
L
T 32
2
f a b≥
G p
L0
f 32
L
1
xdA 4
1
9L + 5x
a2dA b a
d
b
L
5 A
L
dx +
4
L2
L
1
xdA 4
a 2dA b
L
dx ¥
T 125 3ln(2) + 2ln(7) ln(197)
125 2ln(19) + ln(181)
19
a
L
L
+
Lb
4
4
Gp
2
2
dA
dA
81dA 4
16TL
Simplifying: f 81GpdA4
a38 + 10125 lna
Use numerical properties: L 48 in.
fa 0.049 rad
fa 2.79
71117
bb
70952
or
G 3.9 106 psi
fa 3.868
TL
GdA4
dA 2.5 in.
t
dA
10
T 40000 in.-lb
;
PART (b)—CONSTANT HOLE DIAMETER
L
2
d0( x) dB a
d B dA
bx
L
L
… x … L
2
d0(x) 2dA xdA
L
0 … x …
L
L
T 1
± L
fb 32
Gp 4
xdA
L
di (x) dA
L
2
T 32
1
1
f a b
dx +
dx
4
4
L
G p P L0 d0 di
L2 d0 4 Q
2
T 32
f a b
p
G
J L0
d0(x) 2 dA L
1
1
dx +
dx
4
1
xdA
xdA 4 K
L2
4
a2dA b dA
a 2dA b
L
L
3
ln(5) + 2 arctan a b
2
dA
Simplifying: fb 3.057
4
1 ln(3) + 2 arctan(2)
19
L
+
L≤
4
4
dA
81dA 4
TL
Gd A4
Use numerical properties given above:
fb 0.039 rad
fa
fb
1.265
fb 2.21
;
So tube (a) is more flexible than tube (b).
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Page 315
SECTION 3.4 Nonuniform Torsion
Problem 3.4-14 For the thin nonprismatic steel pipe of constant
thickness t and variable diameter d shown with applied torques at
joints 2 and 3, determine the following.
(a) Find reaction moment R1.
(b) Find an expression for twist rotation w3 at joint 3. Assume
that G is constant.
(c) Draw the torsional moment diagram (TMD: T(x),
0 x L).
2d
t
d
t
d
T, f3
T/2
R1
2
1
315
L
—
2
x
3
L
—
2
T
T
—
2
0
TMD
Solution 3.4-14
(a) REACTION TORQUE R1
L
f3 Statics:
T0
T
R1 + T0
2
R1 T
2
2
2T
Gpt L0
;
(b) ROTATION AT JOINT 3
p
3
Ga d12(x) t b
4
L
+
L
2
LL2
T
dx
L
Gpd3t LL2
dx
L
2
2T
f3 Gpt L0
1
x 3
c2d a 1 b d
L
dx
2TL
+
T
2
L
2
L
0
0 … x …
L
… x … L
2
d23(x) d
f3 x
b
L
x 3
c2da 1 b d
L
4T
+
d12( x) 2da 1 1
dx
dx
p
Ga d23(x)3t b
4
Use IP expression for thin walled tubes.
f3 f3 Gpd3t
2TL
3TL
8Gp d3t
19TL
8Gpd3t
+
Gpd3t
;
(c) TMD
TMD is piecewise constant: T(x) T/2 for
segments 1 and 2 and T(x) T for segments 2
and 3 (see plot above).
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 3 Torsion
Problem 3.4-15 A mountain-bike rider going uphill applies
Handlebar extension
d01, t01
torque T Fd (F 15 lb, d 4 in.) to the end of the handlebars
ABCD (by pulling on the handlebar extenders DE). Consider
the right half of the handlebar assembly only (assume the
bars are fixed at the fork at A). Segments AB and CD are
prismatic with lengths L1 2 in. and L3 8.5 in.,
and with outer diameters and thicknesses d01 1.25 in.,
t01 0.125 in., and d03 0.87 in., t03 0.115 in., respectively
as shown. Segment BC of length L2 1.2 in., however,
is tapered, and outer diameter and thickness vary linearly
between dimensions at B and C.
B
A
E
d03, t03
T = Fd
D
C
L3
L1 L2
d
Consider torsion effects only. Assume G 4000 ksi is
constant.
Derive an integral expression for the angle of twist wD
of half of the handlebar tube when it is subjected to torque
T Fd acting at the end. Evaluate wD for the given numerical
values.
45∞
Handlebar
extension
F
D
Handlebar
Solution 3.4-15
ASSUME THIN WALLED TUBES
Segments AB and CD:
p
p
IP1 d01 3t01
IP3 d03 3t03
4
4
Segment BC:
0 x L2
d02(x) d01 a1 d02(x) d01L2 d01x + d03x
L2
t02(x) t01 a1 t02(x) fD x
x
b + d03 a b
L2
L2
x
x
b + t03 a b
L2
L2
fD L2
L2 4
L1
4Fd
c 3 dx
Gp d01 t01 L0 (d01L2 d01x + d03x)3
* (t01L2 t01x + t03x)
+
L3
d03 3t03
d
;
NUMERICAL DATA
L1 2 in.
L2 1.2 in.
t03 0.115 in.
t01 0.125 in.
F 15 lb
d03 0.87 in.
G 4 (106) psi
f D 0.142
L3 8.5 in.
d01 1.25 in.
d 4 in.
;
t01L2 t01x + t03x
L2
L2
L3
1
Fd L1
+
dx +
G P IP1
IP3 Q
L0 p d (x)3t (x)
02
02
4
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Page 317
SECTION 3.4 Nonuniform Torsion
317
Problem 3.4-16 A prismatic bar AB of length L and solid circular cross
section (diameter d) is loaded by a distributed torque of constant intensity
t per unit distance (see figure).
t
A
(a) Determine the maximum shear stress tmax in the bar.
(b) Determine the angle of twist f between the ends of the bar.
B
L
Solution 3.4-16 Bar with distributed torque
(a) MAXIMUM SHEAR STRESS
Tmax tL
tmax 16Tmax
3
pd
16tL
pd3
;
(b) ANGLE OF TWIST
T(x) tx
df t intensity of distributed torque
d diameter
IP T(x)dx
32 tx dx
GIP
pGd 4
L
f
pd 4
32
L0
df L
32t
pGd L0
4
x dx 16tL2
pGd 4
;
G shear modulus of elasticity
Problem 3.4-17 A prismatic bar AB of solid circular cross section
(diameter d) is loaded by a distributed torque (see figure). The intensity
of the torque, that is, the torque per unit distance, is denoted t(x) and
varies linearly from a maximum value tA at end A to zero at end B. Also,
the length of the bar is L and the shear modulus of elasticity of the
material is G.
(a) Determine the maximum shear stress tmax in the bar.
(b) Determine the angle of twist f between the ends of the bar.
t(x)
A
L
B
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CHAPTER 3 Torsion
Solution 3.4-17 Bar with linearly varying torque
(a) Maximum shear stress
16Tmax
tmax pd
3
16TA
pd
3
8tAL
pd3
;
(b) ANGLE OF TWIST
T(x) torque at distance x from end B
t(x)x
tAx2
pd4
IP 2
2L
32
T(x) df T(x) dx
16tAx2 dx
GIP
pGLd4
L
f
L0
df L
16tA
pGLd L0
4
x2 dx 16tA L2
3pGd4
;
t(x) intensity of distributed torque
tA maximum intensity of torque
d diameter
G shear modulus
TA maximum torque
12 tAL
Problem 3.4-18 A nonprismatic bar ABC of solid circular
cross section is loaded by distributed torques (see figure).
The intensity of the torques, that is, the torque per unit
distance, is denoted t(x) and varies linearly from zero at A to a
maximum value T0/L at B. Segment BC has linearly distributed
torque of intensity t(x) T0/3L of opposite sign to that applied
along AB. Also, the polar moment of inertia of AB is twice that
of BC, and the shear modulus of elasticity of the material is G.
(a) Find reaction torque RA.
(b) Find internal torsional moments T(x) in segments AB
and BC.
(c) Find rotation fC.
(d) Find the maximum shear stress tmax and its location
along the bar.
(e) Draw the torsional moment diagram (TMD: T(x),
0 x L).
T
—0
L
A
T0
—
6
Fc
IP
2Ip
RA
C
B
L
—
2
T0
—
3L
L
—
2
2°
2°
0
TMD
–T0
—
12
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Page 319
SECTION 3.4 Nonuniform Torsion
319
Solution 3.4-18
(a) TORQUE REACTION RA
(d) MAXIMUM SHEAR STRESS ALONG BAR
T0
STATICS:
RA +
1 T0 L
1 T0
L
a ba b a ba b 0
2 L
2
2 3L
2
RA +
1
T 0
6 0
RA T0
6
p
d 4
32 AB
p
For BC: IP d 4
32 BC
For AB: 2IP 1
1 4
dBC a b dAB
2
;
(b) INTERNAL TORSIONAL MOMENTS IN AB AND BC
T0
T0 x
x
TAB (x) a b
6
L
P Q L 2
2
TAB (x) a
TBC (x) T0
x2
2 T0 b
6
L
0 … x …
(L x) T0 (L x)
a b
L
3L
2
2
TBC (x) c a
x L 2 T0
b
d
L
3
L
… x … L
2
;
L
2
L
TAB(x)
TBC(x)
dx
dx +
L
GIP
L0 G(2IP)
L2
2
T0
x T0
2
6
3L
fC dx
G(2IP)
L0
L
2
L c a
+
LL2
x L 2 T0
b
d
L
3
GIP
fC T0L
T0L
48GIP
72GIP
fC T0L
144GIP
L
2
;
tmax 8T0
3pdAB3
; controls
Just to right of B, T T0/12
T0 dBC
a
b
12 2
tmax p
d 4
32 BC
T0 0.841dAB
a
b
12
2
tmax p
(0.841dAB)4
32
(c) ROTATION AT C
fC At A, T T0/6
T0 dAB
6 2
tmax p
d 4
32 AB
tmax 2.243T0
pdAB 3
(e) TMD two second-degree curves: from T0/6 at A,
to T0/12 at B, to zero at C (with zero slopes at A
and C, since slope on TMD is proportional to ordinate on torsional loading)—see plot of T(x) above.
dx
;
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Page 320
CHAPTER 3 Torsion
Problem 3.4-19 A magnesium-alloy wire of diameter d 4 mm and
length L rotates inside a flexible tube in order to open or close
a switch from a remote location (see figure). A torque T is applied
manually (either clockwise or counterclockwise) at end B, thus
twisting the wire inside the tube. At the other end A, the rotation of
the wire operates a handle that opens or closes the switch.
A torque T0 0.2 N # m is required to operate the switch.
The torsional stiffness of the tube, combined with friction between
the tube and the wire, induces a distributed torque of constant
intensity t 0.04 N # m/m (torque per unit distance) acting along the
entire length of the wire.
T0 = torque
Flexible tube
B
d
A
T
t
(a) If the allowable shear stress in the wire is tallow 30 MPa,
what is the longest permissible length Lmax of the wire?
(b) If the wire has length L 4.0 m and the shear modulus of
elasticity for the wire is G 15 GPa, what is the angle of twist f
(in degrees) between the ends of the wire?
Solution 3.4-19 Wire inside a flexible tube
(b) ANGLE OF TWIST f
d 4 mm
T0 0.2 N # m
t 0.04 N # m/m
L 4 m G 15 GPa
f1 angle of twist due to distributed torque t
(a) MAXIMUM LENGTH Lmax
tallow 30 MPa
Equilibrium: T tL T0
pd3tmax
16T
T
From Eq. (3-14): tmax 16
pd3
tL + T0 1
(pd3tmax 16T0)
16t
Lmax 1
(pd3tallow 16T0)
16t
pGd 4
(from Problem 3.4-16)
f2 angle of twist due to torque T0
pd 3tmax
16
L
16tL2
32 T0 L
T0 L
(from Eq. 3 -17)
GIP
pGd 4
f total angle of twist
f1 f2
f
(tL + 2T0)
;
Substitute numerical values:
f 2.971 rad 170°
;
;
Substitute numerical values: Lmax 4.42 m
16L
pGd 4
;
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Page 321
321
SECTION 3.4 Nonuniform Torsion
Problem 3.4-20 Two tubes (AB, BC) of the
Diameter d1
same material are connected by three pins (pin
diameter dp) just left of B as shown in the
figure. Properties and dimensions for each tube
are given in the figure. Torque 2T is applied at
x 2L/5 and uniformly distributed torque
intensity t0 3T/L is applied on tube BC.
[Hint: See Example 3-5 for torsional moment
and displacement diagrams]
A
d2
t0 = 3T/L
2T
d3
d2
x
B
3L/5
2L/5
C
2L/5
dp
(a) Find the maximum value of load variable T(N # m) based on allowable shear (ta) and bearing (sba) stresses in the three
pins which connect the two tubes at B. Use the following numerical properties: L 1.5 m, E 74 GPa, y 0.33,
dp 18 mm, ta 45 MPa, sba 90 MPa, d1 85 mm, d2 73 mm, and d3 60 mm.
(b) What is the maximum shear stress in the tubes for the applied torque in part(a)?
Solution 3.4-20
See Example 3-5 for determination of internal torques and TMD
(a) FIND THE MAXIMUM VALUE OF LOAD VARIABLE T (N # m)
Based on allowable shear (ta) and bearing (sba) stresses in the three pins which connect the two tubes at B; use the
following numerical properties: L 1.5 m, E 74 GPa, y 0.33, dp 18 mm, ta 45 MPa, sba 90 MPa, d1 85 mm, d2 73 mm, d3 60 mm
From Example 3-5, Tmax 6T/5
E 74 GPa
0.33
G
E
27.82 GPa
2 (1 + )
dp 18 mm d1 85 mm d2 73 mm d3 60 mm ta 45 MPa sba 90 MPa
t1 d1 d2
d2 d3
6 mm t2 6.5 mm
2
2
BEARING STRESSES ON PIN
Convert Tmax above to force-couple at distribution (d1 t1) or (d2 t2) [see figure]; set sb to sba and
solve for Tmax.
F1
F1
F1
F1 a
6T
b
5
d1
t1
3a b
2
2
T1max (sba dp t 1)
F1
sb1 dp t1
SOLUTION: Convert maximum torque into
bearing forces on pin: either F1 or F2; do same
for shear on interface of two tubes.
F2 6T
5
d2
t2
3a
b
2
2
sb2 F2
dp t2
3
5
3
5
(d t 1) 960 N # m T2max (sba dp t 2) (d2 t 2) 875 N # m
2 1
6
2
6
lowest controls;
here for bearing
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CHAPTER 3 Torsion
SHEAR STRESSES ON PIN
Convert Tmax to force-couple at distribution d2; set t to ta and solve for Tmax:
6T
5
Ft d2
3a b
2
t
Ft
p 2
d
4 p
T3max cta a
p 2
3 5
d b d d2 1045 N # m
4 p
2 6
(b) WHAT IS THE MAXIMUM SHEAR STRESS IN THE TUBES FOR THE APPLIED TORQUE IN PART (a)?
Tmax Ip1 6
T
1050.4 N # m
5 2max
p
p
1d 4 d242 2.337 * 106m4 Ip2 1d 4 d342 1.516 * 106m4
32 1
32 2
tmaxAB Tmax a
Ip1
d1
b
2
19.1 MPa tmaxBC Tmax a
Ip2
d2
b
2
25.3 MPa
Ip1
Ip2
1.542
both are less than 45 MPa
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Page 323
SECTION 3.5 Pure Shear
323
Pure Shear
Problem 3.5-1 A hollow aluminum shaft (see figure) has outside
diameter d2 4.0 in. and inside diameter d1 2.0 in. When twisted
by torques T, the shaft has an angle of twist per unit distance equal
to 0.54°/ft. The shear modulus of elasticity of the aluminum is
G 4.0 106 psi.
d2
T
T
L
(a) Determine the maximum tensile stress smax in the shaft.
(b) Determine the magnitude of the applied torques T.
d1 d2
Probs. 3.5-1, 3.5-2, and 3.5-3
Solution 3.5-1
d2 4.0 in.
Hollow aluminum shaft
d1 2.0 in.
u 0.54°/ft
(a) MAXIMUM TENSILE STRESS
G 4.0 106 psi
smax occurs on a 45° plane and is equal to tmax.
MAXIMUM SHEAR STRESS
smax tmax 6280 psi
tmax Gru (from Eq. 3-9a)
r d2 /2 2.0 in.
u (0.54°/ft)a
1 ft
p rad
b
ba
12 in.
180
785.40 106 rad/in.
tmax (4.0 106 psi)(2.0 in.)(785.40 106 rad/in.)
6283.2 psi
;
(b) APPLIED TORQUE
Use the torsion formula tmax T
Tr
IP
tmaxIP
p
IP [(4.0 in.)4 (2.0 in.)4]
r
32
23.562 in.4
T
(6283.2 psi) (23.562 in.4)
2.0 in.
74,000 lb-in.
;
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CHAPTER 3 Torsion
Problem 3.5-2 A hollow steel bar (G 80 GPa) is twisted by torques T (see figure). The twisting of the bar produces
a maximum shear strain gmax 640 106 rad. The bar has outside and inside diameters of 150 mm and 120 mm,
respectively.
(a) Determine the maximum tensile strain in the bar.
(b) Determine the maximum tensile stress in the bar.
(c) What is the magnitude of the applied torques T ?
Solution 3.5-2 Hollow steel bar
G 80 GPa
gmax 640 106 rad
d2 150 mm
d1 120 mm
IP tmax Ggmax (80 GPa)(640 106)
51.2 MPa
p 4
(d d 41)
32 2
smax tmax 51.2 MPa
p
[(150 mm)4 (120 mm)4]
32
Torsion formula: tmax (a) MAXIMUM TENSILE STRAIN
gmax
320 * 106
2
;
(c) APPLIED TORQUES
29.343 * 106 mm4
max (b) MAXIMUM TENSILE STRESS
T
;
Td2
Tr
IP
2IP
2(29.343 * 106 mm4)(51.2 MPa)
2IPtmax
d2
150 mm
20,030 N # m
20.0 kN # m
;
Problem 3.5-3 A tubular bar with outside diameter d2 4.0 in. is twisted by torques T 70.0 k-in. (see figure). Under the
action of these torques, the maximum tensile stress in the bar is found to be 6400 psi.
(a) Determine the inside diameter d1 of the bar.
(b) If the bar has length L 48.0 in. and is made of aluminum with shear modulus G 4.0 106 psi, what is the angle of
twist f (in degrees) between the ends of the bar?
(c) Determine the maximum shear strain gmax (in radians)?
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Page 325
SECTION 3.5 Pure Shear
Solution 3.5-3
d2 4.0 in.
Tubular bar
T 70.0 k-in. 70,000 lb-in.
smax 6400 psi
f
tmax smax 6400 psi
TL
GIp
From torsion formula, T (a) INSIDE DIAMETER d1
Td2
Tr
IP
2IP
‹ f
(70.0 k-in.)(4.0 in.)
Td2
2tmax
2(6400 psi)
Torsion formula: tmax IP 21.875 in.4
Also, Ip 325
2(48 in.)(6400 psi)
(4.0 * 106 psi)(4.0 in.)
0.03840 rad
;
(c) MAXIMUM SHEAR STRAIN
gmax Equate formulas:
p
[256 in.4 d14] 21.875 in.4
32
Solve for d1: d1 2.40 in.
2IPtmax
2Ltmax
L
a
b d2
GIP
Gd2
f 2.20
p 4
p
(d d 14) [(4.0 in.)4 d14]
32 2
32
2IP tmax
d2
6400 psi
tmax
G
4.0 * 106 psi
1600 * 106 rad
;
;
(b) ANGLE OF TWIST f
L 48 in. G 4.0 106 psi
Problem 3.5-4 A solid circular bar of diameter d 50 mm
(see figure) is twisted in a testing machine until the applied
torque reaches the value T 500 N # m. At this value of torque,
a strain gage oriented at 45° to the axis of the bar gives a reading
P 339 106.
What is the shear modulus G of the material?
d = 50 mm
Strain gage
T = 500 N·m
T
45°
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CHAPTER 3 Torsion
Solution 3.5-4 Bar in a testing machine
Strain gage at 45°:
SHEAR STRESS (FROM Eq. 3-14)
6
max 339 10
tmax d 50 mm
16T
pd
3
16(500 N # m)
p(0.050 m)3
20.372 MPa
SHEAR MODULUS
T 500 N # m
G
SHEAR STRAIN (FROM Eq. 3-32)
gmax 2 max 678 106
tmax
20.372 MPa
30.0 GPa
gmax
678 * 106
;
Problem 3.5-5 A steel tube (G 11.5 106 psi) has an outer diameter d2 2.0 in. and an inner diameter d1 1.5 in.
When twisted by a torque T, the tube develops a maximum normal strain of 170 106.
What is the magnitude of the applied torque T ?
Solution 3.5-5
Steel tube
G 11.5 106 psi
d2 2.0 in.
d1 1.5 in.
6
max 170 10
IP p 2
p
1d 2 d142 [(2.0 in.)4 (1.5 in.)4]
32
32
1.07379 in.
Equate expressions:
Td2
Ggmax
2IP
SOLVE FOR TORQUE
4
T
SHEAR STRAIN (FROM Eq. 3-32)
2GIPgmax
d2
2(11.5 * 106 psi)(1.07379 in.4)(340 * 106)
2.0 in.
gmax 2 max 340 106
SHEAR STRESS (FROM TORSION FORMULA)
4200 lb-in.
;
Td2
Tr
tmax IP
2IP
Also, tmax Ggmax
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Page 327
SECTION 3.5 Pure Shear
327
Problem 3.5-6 A solid circular bar of steel (G 78 GPa) transmits a torque T 360 N # m. The allowable stresses in tension,
compression, and shear are 90 MPa, 70 MPa, and 40 MPa, respectively. Also, the allowable tensile strain is 220 106.
(a) Determine the minimum required diameter d of the bar.
(b) If the bar diameter d 40 mm, what is Tmax?
Solution 3.5-6 Solid circular bar of steel
6
a 220 110 2
(a) G 78 GPa saT 90 MPa saC 70 MPa ta1 40 MPa
ga 2 a ta2 G ga 34.32 MPa
tmax T a
16
pd
3
b
Ta
16
p d3
b ta2
T 360 N # m
6 controls
Solving for d:
d 37.7 mm
(b) At 45, smax plus or minus t for pure shear (Fig. 3-29b), so lowest allowable shear stress governs
d 40 mm Tmax a
16
pd
b ta2 Tmax ta2
3
pd 3
431 N # m
16
Problem 3.5-7 The normal strain in the 45 direction on the
surface of a circular tube (see figure) is 880 106 when the
torque T 750 lb-in. The tube is made of copper alloy with
G 6.2 106 psi and n 0.35.
Strain gage
T = 750 lb-in.
d 2 = 0.8 in.
45°
(a) If the outside diameter d2 of the tube is 0.8 in., what is
the inside diameter d1?
(b) If the allowable normal stress in the tube is 14 ksi, what
is the maximum permissible inside diameter d1?
Solution 3.5-7 Circular tube with strain gage
(a)
880 11062 T 750 lb-in. G 6.2 11062 psi d2 0.8 in.
3
g 2 1.76 * 10
Ta
d2
b
2
p
1d 4 d142
32 2
(b) sa 14 ksi
t
0.35
t G g 10.912 ksi
solving for d1
pure shear, so
d1 0.6 in.
ta sa: ta 14 ksi
1
d2
Ta b
2
p
1d 4 d1max 42
32 2
ta
Solving:
d1max J d2 4 d1max 0.661 in.
d2
32
Ta b
p
2
ta
4
K 0.661 in.
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CHAPTER 3 Torsion
Problem 3.5-8 An aluminum tube has inside diameter d1 50 mm, shear modulus of elasticity G 27 GPa, n 0.33, and
torque T 4.0 kN # m. The allowable shear stress in the aluminum is 50 MPa and the allowable normal strain is 900 106.
(a) Determine the required outside diameter d2.
(b) Re-compute the required outside diameter d2 if allowable normal stress is 62 MPa and allowable shear strain is
1.7 103.
Solution 3.5-8
Aluminum tube
(a) d1 50 mm G 27 GPa
0.33 T 4 kN # m ta 50 MPa
ga 2 a 1.8 * 103 ta2 G ga 48.6 MPa
Ta
d2
b
2
p
1d24 d142
32
ta2
Solving for d2:
6
a 900 110 2
6 controls
d2 79.3 mm
(b) sa 62 MPa Pure shear, so ta sa 62 MPa However, ta G ga 45.9 MPa
Ta
d2
b
2
p
1d 4 d142
32 2
ta
Solving for d2:
d2 80.5 mm
Problem 3.5-9 A solid steel bar (G 11.8 106 psi) of
diameter d 2.0 in. is subjected to torques T 8.0 k-in.
acting in the directions shown in the figure.
(a) Determine the maximum shear, tensile, and compressive
stresses in the bar and show these stresses on sketches of
properly oriented stress elements.
(b) Determine the corresponding maximum strains (shear, tensile, and compressive) in the bar and show these
strains on sketches of the deformed elements.
T
d = 2.0 in.
T = 8.0 k-in.
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SECTION 3.5 Pure Shear
329
Solution 3.5-9 Solid steel bar
T 8.0 k-in.
(b) MAXIMUM STRAINS
G 11.8 10 psi
6
gmax (a) MAXIMUM STRESSES
tmax 16T
pd
3
432 * 106 rad
16(8000 lb-in.)
5093 psi
st 5090 psi
3
p(2.0 in.)
max ;
sc 5090 psi
gmax
5093 psi
G
11.8 * 106 psi
Problem 3.5-10 A solid aluminum bar (G 27 GPa) of
diameter d 40 mm is subjected to torques T 300 N # m
acting in the directions shown in the figure.
gmax
216 * 106
2
6
t 216 10
;
;
d = 40 mm
6
c 216 10
;
T = 300 N·m
T
(a) Determine the maximum shear, tensile, and compressive
stresses in the bar and show these stresses on sketches of
properly oriented stress elements.
(b) Determine the corresponding maximum strains (shear,
tensile, and compressive) in the bar and show these
strains on sketches of the deformed elements.
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CHAPTER 3 Torsion
Solution 3.5-10
Solid aluminum bar
(b) MAXIMUM STRAINS
(a) MAXIMUM STRESSES
tmax 16T
pd
3
16(300 N # m)
p(0.040 m)
23.87 MPa
st 23.9 MPa
gmax 3
884 * 106 rad
;
sc 23.9 MPa
tmax
23.87 MPa
G
27 GPa
;
max gmax
442 * 106
2
6
t 442 10
Problem 3.5-11 Two circular aluminum pipes of equal length
L 24 in. are loaded by torsional moments T (see figure). Pipe 1 has Pipe 1
outside and inside diameters d2 3 in. and d1 2.5 in., respectively.
Pipe 2 has a constant outer diameter of d2 along its entire length L
and an inner diameter of d1 but has an increased inner diameter of
d3 2.65 in. over the middle third.
Pipe 2
Assume that E 10,400 ksi, n 0.33, and allowable shear
stress ta 6500 psi.
;
6
c 442 10
T
d2
d1
;
T
L
(a)
T
L/3
d3
d2 d1
L/3
(b)
L/3
T
(a) Find the maximum acceptable torques that can be applied
to Pipe 1; repeat for Pipe 2.
(b) If the maximum twist f of Pipe 2 cannot exceed 5/4 of that of Pipe 1, what is the maximum acceptable length of the
middle segment? Assume both pipes have total length L and the same applied torque T.
(c) Find the new value of inner diameter d3 of Pipe 2 if the maximum torque carried by Pipe 2 is to be 7/8 of that
for Pipe 1.
(d) If the maximum normal strain in each pipe is known to be max 811 * 106, what is the applied torque on each
pipe? Also, what is the maximum twist of each pipe? Use original properties and dimensions.
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SECTION 3.5 Pure Shear
Solution 3.5-11
331
Numerical data
L 24 in. d2 3 in. d1 2.5 in. d3 2.65 in. E 10400 ksi
ta 6500 psi
G
0.33
E
3910 ksi
2 (1 + )
(a) FIND Tmax1 AND Tmax2
Ip1 p
p
1d 24 d 142 4.1172 in.4 Ip2 1d 24 d 342 3.111 in.4
32
32
Ip1
Ip2
1.324
Based on torsion formula:
Tmax1 Ip1 ta a
2
2
b 17.841 k-in. Tmax2 Ip2 ta a b 13.479 k-in.
d2
d2
Tmax1
1.324
Tmax2
T1allow 17.84 k-in. T2allow 13.48 k-in.
(b) MAX. LENGTH OF MIDDLE SEGMENT OF PIPE 2 BASED ON ALLOWABLE TWIST
Using the torque-displacement relation:
L Lmid
Lmid
5 L
+
Ip1
Ip2
4 Ip1
Solving for Lmid:
L mid 18.54 in.
Lmid
0.772
L
(c) REQUIRED NEW DIAMETER d3 SO THAT Tmax2 7/8 Tmax1
Ip2 7
7
I
so d24 d34 1d24 d142
8 p1
8
1
d 24
7
4
Solving for d3: d3new a
+ d 14 b 2.57996 in. d3new 2.58 in.
8
8
(d) GIVEN MAXIMUM STRAIN IN EACH PIPE, FIND APPLIED TORQUE IN EACH; ALSO MAXIMUM TWIST
6
max 811 110 2
gmax 2 max 1.622 * 103 for pure shear only
G gmax 6.342 ksi
Using the expressions above for Tmax1 and Tmax2:
Tmax1 G gmax IP1 a
2
2
b 17.407 k-in. Tmax2 G gmax Ip2 a b 13.153 kip in.
d2
d2
Tmax1
1.323
Tmax2
Tmax1 17.41 k-in.
Tmax2 13.15 kip in.
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CHAPTER 3 Torsion
max1 Tmax1 L
0.026 rad or
G Ip1
max1 gmax a
2
b L 0.026
d2
2L
L
Tmax2
3
3
+
0.022
max2 G P Ip1
Ip2 Q
max1
1.195
gout2 max2 1.245
only middle third of Pipe 2 is at max. strain so it
twists less; outer third of Pipe 2 on either end has
lower strain (see below)
max2
Tmax2 a
max1 1.487
d2
b
2
G Ip1
1.226 * 103
gout2
0.756
gmax
Transmission of Power
Problem 3.7-1 A generator shaft in a small hydroelectric plant turns
120 rpm
at 120 rpm and delivers 50 hp (see figure).
d
(a) If the diameter of the shaft is d 3.0 in., what is the maximum
shear stress tmax in the shaft?
(b) If the shear stress is limited to 4000 psi, what is the minimum
permissible diameter dmin of the shaft?
50 hp
Solution 3.7-1 Generator shaft
n 120 rpm
TORQUE
H
H 50 hp
d diameter
2pnT
H hp n rpm T 1b-ft
33,000
33,000 H
(33,000)(50 hp)
T
2pn
2p(120 rpm)
2188 1b-ft 26,260 1b-in.
(a) MAXIMUM SHEAR STRESS tmax
d 3.0 in.
tmax 16T
3
pd
16(26,260 1b-in.)
tmax 4950 psi
p (3.0 in.)3
;
(b) MINIMUM DIAMETER dmin
tallow 4000 psi
d3 16(26,260 1b-in.)
16T
33.44 in.3
ptallow
p (4000 psi)
dmin 3.22 in.
;
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SECTION 3.7 Transmission of Power
333
Problem 3.7-2 A motor drives a shaft at 12 Hz and delivers 20 kW
of power (see figure).
12 Hz
d
(a) If the shaft has a diameter of 30 mm, what is the maximum shear
stress tmax in the shaft?
(b) If the maximum allowable shear stress is 40 MPa, what is the
minimum permissible diameter dmin of the shaft?
20 kW
Solution 3.7-2 Motor-driven shaft
f 12 Hz
P 20 kW 20,000 N # m/s
tmax TORQUE
P 2pfT P W
T
20,000 W
P
265.3 N # m
2pf
2p(12 Hz)
16(265.3 N # m)
pd
p(0.030 m)3
3
50.0 MPa
f Hz s1
T Newton meters
16T
;
(b) MINIMUM DIAMETER dmin
tallow 40 MPa
d3 (a) MAXIMUM SHEAR STRESS tmax
16(265.3 N # m)
16T
pt allow
p(40 MPa)
33.78 106 m3
d 30 mm
dmin 0.0323 m 32.3 mm
Problem 3.7-3 The propeller shaft of a large ship has outside
;
100 rpm
18 in.
diameter 18 in. and inside diameter 12 in., as shown in the figure.
The shaft is rated for a maximum shear stress of 4500 psi.
(a) If the shaft is turning at 100 rpm, what is the maximum
horsepower that can be transmitted without exceeding the
allowable stress?
12 in.
18 in.
(b) If the rotational speed of the shaft is doubled but the power
requirements remain unchanged, what happens to the shear
stress in the shaft?
Solution 3.7-3 Hollow propeller shaft
d2 18 in. d1 12 in. tallow 4500 psi
p 4
(d d24) 8270.2 in.4
IP 32 2
TORQUE
T(d2/2)
2t allowIP
tmax T
IP
d2
T
2(4500 psi)(8270.2 in.4)
18 in.
(a) HORSEPOWER
n 100 rpm
n rpm
H
H
2pnT
33,000
T lb-ft H hp
2p(100 rpm)(344,590 lb-ft)
33,000
6560 hp
;
4.1351 * 106 1b-in.
344,590 1b-ft.
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CHAPTER 3 Torsion
(b) ROTATIONAL SPEED IS DOUBLED
H
2pnT
33,000
If n is doubled but H remains the same, then T is
halved.
If T is halved, so is the maximum shear stress.
Shear stress is halved
;
Problem 3.7-4 The drive shaft for a truck (outer diameter
60 mm and inner diameter 40 mm) is running at 2500 rpm
(see figure).
2500 rpm
60 mm
(a) If the shaft transmits 150 kW, what is the maximum shear
stress in the shaft?
(b) If the allowable shear stress is 30 MPa, what is the
maximum power that can be transmitted?
40 mm
60 mm
Solution 3.7-4 Drive shaft for a truck
d2 60 mm
IP d1 40 mm
n 2500 rpm
p 4
(d d14) 1.0210 * 106 m4
32 2
P 150 kW 150,000 W
T torque (newton meters)
(572.96 N # m)(0.060 m)
Td2
2 IP
2(1.0210 * 106 m4)
16.835 MPa
tmax 16.8 MPa
(a) MAXIMUM SHEAR STRESS tmax
P power (watts)
tmax n rpm
P
2pnT
60P
T
60
2pn
T
60(150,000 W)
572.96 N # m
2p(2500 rpm)
;
(b) MAXIMUM POWER Pmax
tallow 30 MPa
Pmax P
tallow
30 MPa
(150 kW) a
b
tmax
16.835 MPa
267 kW
;
Problem 3.7-5 A hollow circular shaft for use in a pumping station is being designed with an inside diameter equal to
0.75 times the outside diameter. The shaft must transmit 400 hp at 400 rpm without exceeding the allowable shear stress
of 6000 psi.
Determine the minimum required outside diameter d.
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SECTION 3.7 Transmission of Power
Solution 3.7-5
Hollow shaft
d outside diameter
H hp
d0 inside diameter
n 400 rpm
tallow 6000 psi
IP T lb-ft
5252.1 lb-ft 63,025 lb-in.
MINIMUM OUTSIDE DIAMETER
p 4
[d (0.75 d)4] 0.067112 d 4
32
TORQUE
H
n rpm
(33,000)(400 hp)
33,000 H
T
2pn
2p(400 rpm)
0.75 d
H 400 hp
335
tmax Td
Td
Td
I 2IP P 2tmax
2t allow
0.067112 d 4 2pnT
33,000
(63,025 lb-in.)(d)
2(6000 psi)
d3 78.259 in.3
dmin 4.28 in.
;
Problem 3.7-6 A tubular shaft being designed for use on a construction site must transmit 120 kW at 1.75 Hz. The inside
diameter of the shaft is to be one-half of the outside diameter.
If the allowable shear stress in the shaft is 45 MPa, what is the minimum required outside diameter d?
Solution 3.7-6
Tubular shaft
d outside diameter
T
d0 inside diameter
MINIMUM OUTSIDE DIAMETER
0.5 d
P 120 kW 120,000 W
f 1.75 Hz
tallow 45 MPa
IP p 4
[d (0.5 d)4] 0.092039 d 4
32
tmax Td
Td
Td
I 2IP P 2tmax
2t allow
0.092039 d 4 (10,913.5 N # m)(d)
2(45 MPa)
d3 0.0013175 m3
TORQUE
P 2pfT P watts
120,000 W
P
10,913.5 N # m
2pf
2p(1.75 Hz)
f Hz
dmin 110 mm
d 0.1096 m
;
T in newton meters
Problem 3.7-7 A propeller shaft of solid circular cross section
and diameter d is spliced by a collar of the same material
(see figure). The collar is securely bonded to both parts
of the shaft.
What should be the minimum outer diameter d1 of the collar in
order that the splice can transmit the same power as the solid shaft?
d1
d
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CHAPTER 3 Torsion
Solution 3.7-7 Splice in a propeller shaft
EQUATE TORQUES
SOLID SHAFT
tmax 16 T1
pd3
T1 pd3tmax
16
For the same power, the torques must be the same.
For the same material, both parts can be stressed to the
same maximum stress.
HOLLOW COLLAR
IP T2 T2(d1/2)
T2r
p 4
(d d 4) tmax 32 1
IP
IP
2tmaxIP
2tmax p
a
b(d14 d 4)
d1
d1 32
ptmax 4
(d1 d 4)
16 d1
‹ T1 T2
or a
pd3tmax
ptmax 4
(d d 4)
16
16d1 1
d1 4
d1
b 10
d
d
(Eq. 1)
MINIMUM OUTER DIAMETER
Solve Eq. (1) numerically:
Min. d1 1.221 d
;
Problem 3.7-8 What is the maximum power that can be delivered by a hollow propeller shaft (outside diameter 50 mm,
inside diameter 40 mm, and shear modulus of elasticity 80 GPa) turning at 600 rpm if the allowable shear stress is 100 MPa
and the allowable rate of twist is 3.0°/m?
Solution 3.7-8 Hollow propeller shaft
d2 50 mm
d1 40 mm
G 80 GPa
n 600 rpm
tallow 100 MPa
IP uallow 3.0°/m
p 4
(d d41) 362.3 * 109 m4
32 2
BASED UPON ALLOWABLE SHEAR STRESS
tmax T1(d2/2)
2t allowIP
T1 IP
d2
2(100 MPa)(362.3 * 109 m4)
T1 0.050 m
1449 N # m
BASED UPON ALLOWABLE RATE OF TWIST
T2
T2 GIPuallow
u
GIP
9 4
m )(3.0/m)
T (80 GPa) (362.3 * 10
2
* a
p
rad /degreeb
180
T2 1517 N # m
SHEAR STRESS GOVERNS
Tallow T1 1449 N # m
MAXIMUM POWER
2p(600 rpm)(1449 N # m)
2pnT
P
60
60
P 91,047 W
Pmax 91.0 kW
;
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SECTION 3.7 Transmission of Power
337
Problem 3.7-9 A motor delivers 275 hp at 1000 rpm to the end of a shaft (see figure). The gears at B and C take out 125
and 150 hp, respectively.
Determine the required diameter d of the shaft if the allowable shear stress is 7500 psi and the angle of twist between the
motor and gear C is limited to 1.5°. (Assume G 11.5 106 psi, L1 6 ft, and L2 4 ft.)
Motor
C
A
d
B
L2
L1
PROBS. 3.7-9 and 3.7-10
Solution 3.7-9 Motor-driven shaft
FREE-BODY DIAGRAM
L1 6 ft
L2 4 ft
TA 17,332 lb-in.
d diameter
TC 9454 lb-in.
n 1000 rpm
d diameter
tallow 7500 psi
(fAC)allow 1.5° 0.02618 rad
G 11.5 106 psi
TORQUES ACTING ON THE SHAFT
H
2pnT
33,000
T
33,000 H
2pn
TB 7878 lb-in.
INTERNAL TORQUES
TAB 17,332 lb-in.
TBC 9454 lb-in.
H hp n rpm T lb-ft
DIAMETER BASED UPON ALLOWABLE SHEAR STRESS
The larger torque occurs in segment AB
At point A: TA 33,000(275 hp)
2p(1000 rpm)
1444 lb-ft
17,332 lb-in.
At point B: TB At point C: TC 125
TA 7878 lb-in.
275
150
TA 9454 lb-in.
275
tmax 16TAB
pd
d3 3
16TAB
pt allow
16(17,332 lb-in.)
11.77 in.3
p(7500 psi)
d 2.27 in.
DIAMETER BASED UPON ALLOWABLE ANGLE OF TWIST
IP pd4
32
f
TL
32TL
GIP
pGd 4
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CHAPTER 3 Torsion
Segment AB:
fAB fAB fBC 32TAB LAB
pGd 4
32(17,330 lb in.)(6 ft)(12 in./ft)
p(11.5 * 106 psi)d 4
1.1052
d
d4
(fAC)allow 0.02618 rad
0.02618 1.5070
d4
d 2.75 in.
32 TBCLBC
pGd
1.5070
and
d 2.75 in.
Angle of twist governs
Segment BC:
fBC d4
From A to C: fAC fAB + fBC 4
0.4018
;
4
32(9450 lb-in.)(4 ft)(12 in./ft)
p(11.5 * 106 psi)d 4
Problem 3.7-10 The shaft ABC shown in the figure is driven by a motor that delivers 300 kW at a rotational speed of
32 Hz. The gears at B and C take out 120 and 180 kW, respectively. The lengths of the two parts of the shaft are L1 1.5 m
and L2 0.9 m.
Determine the required diameter d of the shaft if the allowable shear stress is 50 MPa, the allowable angle of twist
between points A and C is 4.0°, and G 75 GPa.
Solution 3.7-10 Motor-driven shaft
L1 1.5 m
L2 0.9 m
At point B: TB 120
T 596.8 N # m
300 A
At point C: TC 180
T 895.3 N # m
300 A
FREE-BODY DIAGRAM
d diameter
f 32 Hz
tallow 50 MPa
G 75 GPa
(fAC)allow 4° 0.06981 rad
TA 1492 N # m
TORQUES ACTING ON THE SHAFT
TB 596.8 N # m
P 2pfT P watts
TC 895.3 N # m
f Hz
T in newton meters
T
INTERNAL TORQUES
P
2pf
At point A: TA d diameter
TAB 1492 N # m
300,000 W
1492 N # m
2p(32 Hz)
TBC 895.3 N # m
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339
SECTION 3.8 Statically Indeterminate Torsional Members
DIAMETER BASED UPON ALLOWABLE SHEAR STRESS
Segment BC:
The larger torque occurs in segment AB
tmax fBC 16(1492 N # m)
16 TAB
d pt allow
p(50 MPa)
16 TAB
pd 3
d 0.0001520 m
3
fBC d 0.0534 m 53.4 mm
DIAMETER BASED UPON ALLOWABLE ANGLE OF TWIST
TL
32TL
f
GIP
pGd 4
fAB pGd 4
32(895.3 N # m)(0.9 m)
p(75 GPa)d 4
0.1094 * 106
d4
0.4133 * 106
d4
(fAC)allow 0.06981 rad
0.06981 32 TABL AB
4
From A to C: fAC fAB + fBC Segment AB:
fAB pGd
3
3
pd 4
IP 32
32 TBCLBC
32(1492 N # m)(1.5 m)
0.1094 * 106
d4
and d 0.04933 m
p(75 GPa)d 4
49.3 mm
0.3039 * 106
SHEAR STRESS GOVERNS
d4
d 53.4 mm
;
Statically Indeterminate Torsional Members
Problem 3.8-1 A solid circular bar ABCD with fixed supports is acted
A
upon by torques T0 and 2T0 at the locations shown in the figure.
(a) Obtain a formula for the maximum angle of twist fmax of the
bar. [Hint: Use Eqs. (3-50a and b) of Example 3-9 to obtain the
reactive torques.]
(b) What is fmax if the applied torque T0 at B is reversed in direction?
3L
—
10
T0
2T0
B
C
3L
—
10
D
4L
—
10
L
Solution 3.8-1 Circular bar with fixed ends
(a) SELECT TD AS THE REDUNDANT; USE SUPERPOSITION OF RELEASED STRUCTURES TO FIND TWIST ANGLES DUE TO APPLIED
TORQUES AT B AND C (i.e., f1) AND ALSO FOR APPLIED REDUNDANT TD APPLIED AT D (i.e., f2)
f1 3 T0 a
12 T02 a
3L
b
10
3L
b
10
+
GIp
GIp
Compatibility
f1 + f2 0
so TD TORSIONAL MOMENT DIAGRAM (TMD)
f2 TD a
3 L T0
2 GIp
GIp
L
a
L
b
GIp
3 L T0
3 T0
b 2 GIp
2
TA 3 T0
2
and TA 3 T0 TD 3 T0
2
TD TA
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CHAPTER 3 Torsion
Torsional Moment Diagram (TMD)
3
2
Times T
TMD(x) 1
0
0
–1
–2
0
0.1
0.2
0.3
0.4
0.5
x
0.6
0.7
0.8
0.9
1
Times L
TORSIONAL DISPLACEMENT DIAGRAM (TDD)
Torsional Displacement Diagram (TDD, twist angle)
1
Times TL / GIp
0.5
TDD(x)
0
0
–0.5
–3
5
–1
0
0.1
0.2
0.3
0.4
0.5
x
Times L
0.6
0.7
0.8
0.9
1
4L
3 L T0
10
;
GIp
5 GIp
TD
So maximum twist is at x 3L/5 and is equal to
fmax
TA1 T0 + 2 T0 T0
(B) REPEAT SUPERPOSITION PROCEDURE USING REDUNDANT TD BUT REVERSE SIGN OF TORQUE APPLIED AT B
f1 T0 a
12 T02a
3L
b
10
3L
b
10
+
GIp
GIp
compatibility
f1 + f2 0
TA T0 TD T0
10
9 L T0
L
f2 TD a
b
10 GIp
GIp
GIp 9 L T0
9 T0
TD a
b L
10 GIp
10
TORSIONAL MOMENT DIAGRAM (TMD)
TA T0
10
TD 9
T
10 0
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SECTION 3.8 Statically Indeterminate Torsional Members
341
Torsional Moment Diagram (TMD)
1
Times T
0.4
TMD(x) –0.2
0
–0.8
–1.4
–2
0
0.1
0.2
0.3
0.4
0.5
x
Times L
0.6
0.7
0.8
0.9
1
TORSIONAL DISPLACEMENT DIAGRAM (TDD)
Torsional Displacement Diagram (TDD, twist angle)
Times TL /G Ip
0.1
0
– 0.025
TDD(x)
– 0.15
– 0.275
–9
25
– 0.4
0
0.1
0.2
0.3
0.4
0.5
x
Times L
0.6
0.7
0.8
4L
9 L T0
10
fmax GIp
25
0.9
1
TD
So maximum twist is at x 3L/5 and is equal to
;
9
0.36
25
Problem 3.8-2 A solid circular bar ABCD with fixed supports at
ends A and D is acted upon by two equal and oppositely directed
torques T0, as shown in the figure. The torques are applied at points
B and C, each of which is located at distance x from one end of the
bar. (The distance x may vary from zero to L/2.)
(a) For what distance x will the angle of twist at points B and
C be a maximum?
(b) What is the corresponding angle of twist fmax?
[Hint: Use Eqs. (3-50a and b) of Example 3-9 to obtain the
reactive torques.]
TA
A
T0
T0
B
C
x
D
TD
x
L
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CHAPTER 3 Torsion
Solution 3.8-2 Circular bar with fixed ends
(a) ANGLE OF TWIST AT SECTIONS B AND C
φB
φ max
0
From Eqs. (3-50a and b):
fB fAB L/4
T0 LB
L
dfB
T0
(L 4x)
dx
GIPL
TB T0 LA
L
dfB
0; L 4x 0
dx
or x L
4
x
TAx
T0
(L 2x)(x)
GIP
GIPL
TA APPLY THE ABOVE FORMULAS TO THE GIVEN BAR:
L /2
;
(b) MAXIMUM ANGLE OF TWIST
fmax (fB)max (fB)x L4 TA T0L
8GIP
;
T0(L x)
T0x
T0
(L 2x) TD TA
L
L
L
Problem 3.8-3 A solid circular shaft AB of diameter d is fixed against rotation at
both ends (see figure). A circular disk is attached to the shaft at the location shown.
What is the largest permissible angle of rotation fmax of the disk if the allowable
shear stress in the shaft is tallow? (Assume that a b. Also, use Eqs. 3-50a and b of
Example 3-9 to obtain the reactive torques.)
Disk
A
d
B
a
Solution 3.8-3
b
Shaft fixed at both ends
Assume that a torque T0 acts at the disk.
The reactive torques can be obtained from Eqs. (3-50a
and b):
T0b
T0a
TB L
L
Since a b, the larger torque (and hence the larger
stress) is in the right hand segment.
TA Lab
ab
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SECTION 3.8 Statically Indeterminate Torsional Members
tmax TB(d/2)
T0 ad
IP
2LIP
2LIPtmax
T0 ad
2L IPt allow
(T0)max ad
343
ANGLE OF ROTATION OF THE DISK (FROM Eq. 3-53)
f
T0 ab
GLIP
(T0)maxab
2bt allow
GLIp
Gd
fmax ;
Problem 3.8-4 A hollow steel shaft ACB of outside diameter 50 mm
and inside diameter 40 mm is held against rotation at ends A and B
(see figure). Horizontal forces P are applied at the ends of a vertical
arm that is welded to the shaft at point C.
Determine the allowable value of the forces P if the maximum
permissible shear stress in the shaft is 45 MPa. (Hint: Use Eqs. 3-50a
and b of Example 3-9 to obtain the reactive torques.)
200 mm
A
P
200 mm
C
B
P
600 mm
400 mm
Solution 3.8-4
Hollow shaft with fixed ends
GENERAL FORMULAS:
T0 P(400 mm)
LB 400 mm
LA 600 mm
L LA LB 1000 mm
d2 50 mm
d1 40 mm
tallow 45 MPa
APPLY THE ABOVE FORMULAS TO THE GIVEN SHAFT
TA P(0.4 m)(400 mm)
T0 LB
0.16 P
L
1000 mm
TB P(0.4 m)(600 mm)
T0 LA
0.24 P
L
1000 mm
UNITS: P in Newtons T in Newton meters
From Eqs. (3-50a and b):
TA T0 LB
L
T0 LA
L
The larger torque, and hence the larger shear
stress, occurs in part CB of the shaft.
TB © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 3 Torsion
Tmax TB 0.24 P
Substitute numerical values into (Eq. 1):
SHEAR STRESS IN PART CB
tmax Tmax(d/2)
2tmaxIP
Tmax IP
d
0.24P (Eq. 1)
652.07 N # m
UNITS: Newtons and meters
P
tmax 45 10 N/m
6
Ip 2(45 * 106 N/m2)(362.26 * 109 m4)
0.05 m
2
652.07 N # m
2717 N
0.24 m
Pallow 2720 N
p 4 4
(d d ) 362.26 * 109m4
32 2 1
;
d d2 0.05 mm
Problem 3.8-5 A stepped shaft ACB having solid circular cross sections with
two different diameters is held against rotation at the ends (see figure).
1.50 in.
0.75 in.
C
A
(a) If the allowable shear stress in the shaft is 6000 psi, what is the
maximum torque (T0)max that may be applied at section C? [Hint: Use
Eqs. (3-49a and b) of Example 3-9 to obtain the reactive torques.]
(b) Find (T0)max if the maximum angle of twist is limited to 0.55. Let
G 10,600 ksi.
B
T0
6.0 in.
15.0 in.
Solution 3.8-5
(a) NUMERICAL DATA
G 10600 ksi dA 3
3
in. dB in. L A 6 in. L B 15 in. ta 6 ksi
4
2
p
p 4
1d 42 IpB d
32 A
32 B
Using Eqs 3-49a, b
IpA TA a
LB IpA
LB IpA + LA IpB
b T0 :
5 T0
37
TB a
LA IpB
LB IpA + LA IpB
b T0 :
32 T0
37
Now use torsion formula to find shear stresses in each segment; set each shear equal to the allowable value then
solve for the max. permissible value of torque T0.
tA TA a
16
p d3A
b
tB TB a
16
p d3B
b
From segment 1:
T0max1 ta a
p d 3A
16
5
37
From segment 2:
b
3678 lb-in.
controls
T0max2 ta a
p d3B
b
16
32
37
4597 lb-in.
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SECTION 3.8 Statically Indeterminate Torsional Members
345
(b) THE MAX. TWIST ANGLE IS AT C; BOTH SEGMENTS TWIST THE SAME AMOUNT. USE THE TORQUE-DISPL. RELATION, SUBSITITUTE FOR TA AND TB IN TERMS OF T0
fallow 0.55
fAB TA LA
GIpA
T0max fBC G IpA
5
LA a b
37
TBLB
GIpB
fallow 3898 lb-in.
or
T0max G IpB
32
LB a b
37
fallow 3898 lb-in.
20 mm
Problem 3.8-6 A stepped shaft ACB having solid circular cross
sections with two different diameters is held against rotation at the
ends (see figure).
B
C
A
(a) If the allowable shear stress in the shaft is 43 MPa, what is the
maximum torque (T0)max that may be applied at section C? [Hint:
Use Eqs. (3-49a and b) of Example 3-9 to obtain the reactive
torques.]
(b) Find (T0)max if the maximum angle of twist is limited to 1.85. Let
G 28 GPa.
25 mm
T0
225 mm
450 mm
Solution 3.8-6
NUMERICAL DATA
G 28 GPa
dA 20 mm
LA 225 mm
LB 450 mm
IpA dB 25 mm
ta 43 MPa
p
p 4
1d 42 IpB d
32 A
32 B
(a) Using Eqs 3-49a, b
TA a
L B IpA
L BIPA + L AIpB
bT0 512 T0
1137
TB a
L A IpB
L B IpA + L A IpB
b T0 625 T0
1137
Now use torsion formula to find shear stresses in each segment; set each shear equal to the allowable value then
solve for the maximum permissible value of torque T0.
tA TA a
16
p d3A
b
tB TB a
16
p d3B
From segment 1:
T0max1 ta a
p d 3A
b
16
512
1137
b
From segment 2:
150 N # m
6 controls T0max2 ta a
p d 3B
b
16
625
1137
240 N # m
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CHAPTER 3 Torsion
(b) The maximum twist angle is at C; both segments twist the same amount. Use the torque-displacement relation,
subsititute for TA and TB in terms of T0.
fallow 1.85
TA LA
TB LB
fBC G IpA
G IpB
G IpA
G IpB
fallow 140 N # m or T0max fallow 140 N # m
T0max 512
625
LA a
b
LB a
b
1137
1137
fAB Problem 3.8-7 A stepped shaft ACB is held against rotation at ends
dA
A and B and subjected to a torque T0 acting at section C (see figure).
The two segments of the shaft (AC and CB) have diameters dA and dB,
respectively, and polar moments of inertia IPA and IPB, respectively.
The shaft has length L and segment AC has length a.
dB
IPA
A
C
IPB
B
T0
a
(a) For what ratio a/L will the maximum shear stresses be the same
in both segments of the shaft?
(b) For what ratio a/L will the internal torques be the same in
both segments of the shaft? (Hint: Use Eqs. 3-49a and b of
Example 3-9 to obtain the reactive torques.)
L
Solution 3.8-7 Stepped shaft
SEGMENT AC: dA, IPA LA a
or (L a)dA adB
SEGMENT CB: dB, IPB LB L a
REACTIVE TORQUES (from Eqs. 3-49a and b)
Solve for a/L:
TA T0 a
LBIPA
LAIPB
b; TB T0 a
b
LBIPA + LAIPB
LBIPA + LAIPB
TAdA
TBdB
IPA
IPB
;
(b) EQUAL TORQUES
TA TB or LBIPA LAIPB
or (L a)IPA aIPB
(a) EQUAL SHEAR STRESSES
TA(dA/2)
TB(dB/2)
tCB tAC IPA
IPB
tAC tCB or
dA
a
L
dA + dB
Solve for a/L:
(Eq. 1)
or
IPA
a
L
IPA + IPB
dA4
a
4
L
dA + dB4
;
Substitute TA and TB into Eq. (1):
LBIPAdA
LAIPBdB
IPA
IPB
or
LBdA LAdB
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SECTION 3.8 Statically Indeterminate Torsional Members
Problem 3.8-8 A circular bar AB of length L is fixed against
347
t0
rotation at the ends and loaded by a distributed torque t(x) that
varies linearly in intensity from zero at end A to t0 at end B (see
figure).
t(x)
(a) Obtain formulas for the fixed-end torques TA and TB.
(b) Find an expression for the angle of twist f (x). What is
fmax and where does it occur along the bar?
A
B
x
L
Solution 3.8-8
(a) USE SUPERPOSITION WITH TB AS THE REDUNDANT. TWIST AT B DUE TO LOADING T(X) IS fB1, AND TWIST AT B DUE TO
REDUNDANT TB IS fB2
L
fB1 3
c
t 0 L
1
x
+ a t0 b x d
2
2
L
G Ip
0
Compatibility
dx fB1 + fB2 0 TB L2 t 0
3 G Ip
G Ip
L
a
fB2 TB L
G Ip
L2 t 0
L t0
b 3 G Ip
3
t0 L
L t0
Statics TA TB 2
6
(b) USE INTEGRAL EXPRESSION FOR TORQUE-DISPLACEMENT RELATION TO FIND f(X)
f(x) x
t 0 x 1L2 x 22
t 0 L
1
1
c
c
+
t 0 a b dd d G Ip L0
6
2
L
6 G Ip L
;
Differentiate expression for f(x) and set equal to zero to find location xm at which fmax occurs.
t 0 1L2 x 22
t0 x 2
d
f(x) dx
3 G Ip L
6 G Ip L
fmax f a
13 L2 t 0
L
b 27 G Ip
13
t 0 1L2 x 22
t0 x 2
L
0 Solving: x m 3 G Ip L
6 G Ip L
13
13
0.064
27
1
0.577
13
Plot torsional moment (TMD) and torsional displacement (TDD) diagrams showing variation of T(x) and f(x),
respectively, over the length of the bar from A to B
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CHAPTER 3 Torsion
TMD
Times t0 x L
0.4
T(x)
0
.
L
t0.L
3
3
0.2
0
–t0.L
–0.2
0
0.2
0.4
0.6
x
times L
6
0.8
1
TDD
Times t0 L2/GIp
0
L
3
–0.02
φ ( x)
φ
–0.04
L
.
3
–0.06
–0.08
–0.1
0
0.1
0.2
0.3
0.4
0.5
x
times L
0.6
0.7
Problem 3.8-9 A circular bar AB with ends fixed against rotation has
a hole extending for half of its length (see figure). The outer diameter of
the bar is d2 3.0 in. and the diameter of the hole is d1 2.4 in. The
total length of the bar is L 50 in.
(a) At what distance x from the left-hand end of the bar should a
torque T0 be applied so that the reactive torques at the supports
will be equal?
(b) Based on the solution for x in part (a), what is fmax, and where
does it occur? Assume that T0 87.4 kip-in. and G 10,600 ksi.
A
0.8
0.9
1
25 in.
25 in.
3.0 in.
T0
B
x
2.4
in.
3.0
in.
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SECTION 3.8 Statically Indeterminate Torsional Members
349
Solution 3.8-9
(a) SUPERPOSITION WITH TB AS THE REDUNDANT
fB1 T0
TB 25 in.
C
p (3 in.)4
Gc
d
32
+
32 (x 25 in.)
800
x 25 in.
d
+
T0 c
3
p
S
81
p
G
in.
p
G(81
in.4 33.1776 in.4)
4
4
Gc
C (3 in.) (2.4 in.) D d
32
T0
2
fB2 TB
25 in.
25 in.
+
p
S
p (3 in.)4
4
4
G
c
(3
in.)
(2.4
in.)
d
C
D
d
Gc
32
32
fB1 + fB2 0 Solve: x 30.12 in.
T0 c
800
800 in.
3
+
81 p G in.
C
(b) x 30.12 in. G 10600 ksi T0 87.4 k-in. TA T0
2
p G 181 in.4 33.1776 in.42
d
2
TB TB L 50 in.
fmax occurs at x, point of application of torque T0 (TMD has zero ordinate so TDD has horiz. tangent)
fmax TA
25 in.
C
p (3 in.)4
Gc
d
32
+
x 25 in.
1
p
4
4 S
Gc
C (3 in.) (2.4 in.) D d
32
CW when viewed from B toward A
or
fmax TB
C
Lx
1
p
4
4 S
Gc
C (3 in.) (2.4 in.) D d
32
fmax 1 at x 30.12 in.
Problem 3.8-10 A solid steel bar of diameter d1 25.0 mm is
enclosed by a steel tube of outer diameter d3 37.5 mm and inner
diameter d2 30.0 mm (see figure). Both bar and tube are held
rigidly by a support at end A and joined securely to a rigid plate
at end B. The composite bar, which has a length L 550 mm,
is twisted by a torque T 400 N m acting on the end plate.
(a) Determine the maximum shear stresses t1 and t2 in the bar and
tube, respectively.
(b) Determine the angle of rotation f (in degrees) of the end plate,
assuming that the shear modulus of the steel is G 80 GPa.
(c) Determine the torsional stiffness kT of the composite bar.
(Hint: Use Eqs. 3-48a and b to find the torques in the bar and tube.)
Tube
A
B
T
Bar
End
plate
L
d1
d2
d3
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CHAPTER 3 Torsion
Solution 3.8-10
Bar enclosed in a tube
TORQUES IN THE BAR (1) AND TUBE (2)
FROM EQS. (3-44A AND B)
Bar: T1 T a
IP1
b 100.2783 N # m
IP1 + IP2
Tube: T2 T a
IP2
b 299.7217 N # m
IP1 + IP2
(a) MAXIMUM SHEAR STRESSES
Bar: t1 T1(d1/2)
32.7 MPa
IP1
;
T2(d3/2)
49.0 MPa
IP2
;
Tube: t2 d1 25.0 mm
d2 30.0 mm
d3 37.5 mm
(b) ANGLE OF ROTATION OF END PLATE
T 2L
T1L
0.017977 rad
GIP1
GIP2
G 80 GPa
f
POLAR MOMENTS OF INERTIA
f 1.03°
p 4
Bar: IP1 d 38.3495 * 109 m4
32 1
Tube: IP2 ;
(c) TORSIONAL STIFFNESS
p
1d 4 d242 114.6229 * 109 m4
32 3
kT T
22.3 kN # m
f
Problem 3.8-11 A solid steel bar of diameter d1 1.50 in. is enclosed
by a steel tube of outer diameter d3 2.25 in. and inner diameter
d2 1.75 in. (see figure). Both bar and tube are held rigidly by a support
at end A and joined securely to a rigid plate at end B. The composite bar,
which has length L 30.0 in., is twisted by a torque T 5000 lb-in.
acting on the end plate.
(a) Determine the maximum shear stresses t1 and t2 in the bar and
tube, respectively.
(b) Determine the angle of rotation f (in degrees) of the end plate,
assuming that the shear modulus of the steel is G 11.6 106 psi.
(c) Determine the torsional stiffness kT of the composite bar. (Hint: Use
Eqs. 3-48a and b to find the torques in the bar and tube.)
;
Tube
A
B
T
Bar
End
plate
L
d1
d2
d3
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SECTION 3.8 Statically Indeterminate Torsional Members
Solution 3.8-11
351
Bar enclosed in a tube
TORQUES IN THE BAR (1) AND TUBE (2)
FROM EQS. (3-48a AND b)
Bar: T1 T a
IPI
b 1187.68 lb-in.
IPI + IPI
Tube: T2 T a
IP2
b 3812.32 lb-in.
IP1 + IP2
(a) MAXIMUM SHEAR STRESSES
Bar: t1 T1(d1/2)
1790 psi
IP1
;
T2(d3/2)
2690 psi
IP2
;
Tube: t2 (b) ANGLE OF ROTATION OF END PLATE
d1 1.50 in.
d2 1.75 in.
d3 2.25 in.
f
G 11.6 106 psi
f 0.354°
POLAR MOMENTS OF INERTIA
Bar: IP1 T 2L
T1L
0.006180015 rad
GIP1
GIP2
p 4
d 0.497010 in.4
32 1
;
(c) TORSIONAL STIFFNESS
kT p
Tube: IP2 1d34 d242 1.595340 in.4
32
T
809 k- in.
f
Problem 3.8-12 The composite shaft shown in the figure is manufactured by
shrink-fitting a steel sleeve over a brass core so that the two parts act as a single
solid bar in torsion. The outer diameters of the two parts are d1 40 mm for
the brass core and d2 50 mm for the steel sleeve. The shear moduli of elasticity are Gb 36 GPa for the brass and Gs 80 GPa for the steel.
T
;
Steel sleeve
Brass core
T
(a) Assuming that the allowable shear stresses in the brass and steel are tb
48 MPa and ts 80 MPa, respectively, determine the maximum permissible torque Tmax that may be applied to the shaft. [Hint: Use Eqs.
(3-48a and b) to find the torques.]
(b) If the applied torque T 2500 kN # m, find the required diameter d2 so
that allowable shear stress ts is reached in the steel.
d1 d2
Probs. 3.8-12 and 3.8-13
Solution 3.8-12
NUMERICAL DATA
(a) d1 40 mm
IpB d2 50 mm
GB 36 GPa
GS 80 GPa
tb 48 MPa
ts 80 MPa
p 4
p
d1 IpS 1d 4 d142
32
32 2
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CHAPTER 3 Torsion
FROM EQ. (3-48a, b)
GB IpB
TB T a
b
GB IpB + GS IpS
And
2 tb IpB
TBmax TSmax and TS T a
GS IpS
GB IpB + GS IpS
b
2 ts IpS
d1
d2
Solving for Tmax for either brass or steel
TmaxB TmaxS (b) Let
2 tb IpB
d1
2 ts IpS
d2
a
a
GB IpB + GS IpS
GB IpB
GB IpB + GS IpS
GS IpS
T 2500 N m
b 2535.265 N # m
b 1521.159 N # m
controls so
Tmax 1521 N # m
so TmaxS T
Expand IpS and IpB in above expression for TmaxS then for required diameter d2
2 ts c
p
A d 4 d14 B d
32 2
d2
≥
GB IpB + GS c
GS c
p
A d 4 d14 B d
32 2
p
A d 4 d14 B d
32 2
¥ T
Solving: d2 56.868 mm
d2 56.9 mm
Check
IpS Recompute:
TS T a
p
A d 4 d 14 B 7.778 * 107 m4
32 2
GS IpS
GB IpB + GS IpS
b
ts TS a
d2
b
2
IpS
80 MPa
OK, at specified allowable value
Find associated max. shear stress in brass core:
TB (T) a
GB IpB
GB IpB + GS IpS
b
TB a
d1
b
2
IpB
25.257 MPa
OK, less than allowable value of 48 MPa
Problem 3.8-13 The composite shaft shown in the figure is manufactured
by shrink-fitting a steel sleeve over a brass core so that the two parts act as
a single solid bar in torsion. The outer diameters of the two parts are
d1 1.6 in. for the brass core and d2 2.0 in. for the steel sleeve.
The shear moduli of elasticity are Gb 5400 ksi for the brass and
Gs 12,000 ksi for the steel.
(a) Assuming that the allowable shear stresses in the brass and
steel are tb 4500 psi and ts 7500 psi, respectively,
determine the maximum permissible torque Tmax that may
be applied to the shaft. [Hint: Use Eqs. (3-48a and b) to find the torques.]
(b) If the applied torque T 15 kip-in., find the required diameter
d2 so that allowable shear stress ts is reached in the steel.
T
Steel sleeve
Brass core
T
d1 d2
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SECTION 3.8 Statically Indeterminate Torsional Members
353
Solution 3.8-13
NUMERICAL DATA
(a) d1 1.6 in. d2 2 in.
GB 5400 ksi
GS 12000 ksi
tb 4500 psi
ts 7500 psi
p 4
p
d1 IpS A d 4 d14 B
32
32 2
FROM Eqs. (3-48a, and b)
GB IpB
GS IpS
TB T a
b and TS T a
b
GB IpB + GS IpS
GB IpB + GS IpS
And
2 tb IpB
2 ts IpS
TBmax TSmax d1
d2
Solving for Tmax for either brass or steel:
2 tb IpB GB IpB + GS IpS
TmaxB a
b 15.212 k-in.
d1
GB IpB
IpB TmaxS (b) Let
2 ts IpS
d2
a
GB IpB + GS IpS
GS IpS
b 9.127 k-in.
controls, so
Tmax 9.13 k-in.
T 15 k-in. so TmaxS T
Expand IpS and IpB in above expression for TmaxS then for required diameter d2:
2 ts c
p
A d24 d14 B d
32
d2
≥
GB IpB + GS c
GS c
p
A d24 d14 B d
32
p
A d24 d14 B d
32
¥ T
Solving:
d2 2.275 in.
d2 2.27 in.
Check
Recomputed:
TS T a
IpS p
A d 4 d14 B 9.574 * 105 ft4
32 2
GS IpS
GB IpB + GS IpS
b
ts TS a
d2
b
2
IpS
7500 psi
OK, at specified allowable value
Find associated max. shear stress in brass core:
TB (T) a
GB IpB
GB IpB + GS IpS
b
TB a
d1
b
2
IpB
2374 psi
OK, less than allowable value of 4500 psi
Problem 3.8-14 A steel shaft (Gs 80 GPa) of total length L 3.0 m is encased for one-third of its length by a brass
sleeve (Gb 40 GPa) that is securely bonded to the steel (see figure). The outer diameters of the shaft and sleeve are
d1 70 mm and d2 90 mm. respectively.
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CHAPTER 3 Torsion
(a) Determine the allowable torque T1 that may be applied
to the ends of the shaft if the angle of twist between the
ends is limited to 8.0°.
(b) Determine the allowable torque T2 if the shear stress in
the brass is limited to tb 70 MPa.
(c) Determine the allowable torque T3 if the shear stress
in the steel is limited to ts 110 MPa.
(d) What is the maximum allowable torque Tmax if all three
of the preceding conditions must be satisfied?
Brass
sleeve
Steel
shaft
d2 = 90 mm
d1 = 70 mm
T
T
A
B
1.0 m
L
= 2.0 m
2
d1
C
L
= 2.0 m
2
d1
Brass
sleeve
d2
d1
Steel
shaft
d2
Solution 3.8-14
(a) ALLOWABLE TORQUE T1 BASED ON TWIST AT ENDS OF 8 DEGREES
first find torques in steel (Ts) & brass (Tb) in segment in which they are joined 1 degree statically indeterminate;
use Ts as the internal redundant; see Eq. 3-48a and b
Ts T1 a
Gs IPs
Gb IPb
b statics 7 Tb T1 Ts Tb T1 a
b
Gs IPs + Gb IPb
Gs IPs + Gb IPb
Now find twist of three segments:
L
L
L
T1
Ts
T1
4
4
2
for middle term, brass sleeve and steel shaft twist the same so could use
+
+
f
Gb IPb
Gs IPs
Gs IPs
Tb(L/4)/(GbIPb) instead
T1
fa T1 a
L
4
Gs IPs
L
b
Gs IPs + Gb IPb 4
+
Gb IPb
T1
L
2
+
Gs IPs
Gs IPs
fa T1
L
1
1
2
a
+
+
b
4 Gb IPb
Gs IPs + Gb IPb
Gs IPs
fa T1
L Gs2 IPs2 + 4 Gb IPb Gs IPs + 2 Gb2 IPb2
c
d
4
Gb IPb1Gs IPs + Gb IPb2 GsIPs
T1allow L
L
T1
4
2
fa
+
+
Gb IPb Gs IPs + Gb IPb
GsIPs
T1
L
4
T1
4 fa
Gb IPb (Gs IPs + Gb IPb) Gs IPs
c 2 2
d
L
Gs IPs + 4 Gb IPb Gs IPs + 2 Gb2 IPb2
NUMERICAL VALUES
fa 8 1
p
2 rad Gs 80 GPa Gb 40 GPa L 4.0 m
180
d1 70 mm d2 90 mm
4
4
d1 4
p
p 1d2 d1 2
a
b IPs 2.357 * 106 m4 IPb IPb 4.084 * 106 m4
32 1000
32
110324
Gb IPb (Gs IPs + Gb IPb) Gs Ips
4 fa
c 2 2
d T1allow (109) 7135.197 N # m
T1allow L
Gs IPs + 4 Gb IPb Gs IPs + 2 Gb2 IPb2
Ips T1allow 7.14 kN # m
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SECTION 3.8 Statically Indeterminate Torsional Members
355
(b) ALLOWABLE TORQUE T1 BASED ON ALLOWABLE SHEAR STRESS IN BRASS, tb tb 70 MPa
FIRST CHECK HOLLOW SEGMENT 1 (BRASS SLEEVE ONLY)
T2
t
d2
2
T2allow IPb
2 tb IPb
d2
T2allow A 109 B 6.353 * 103 N # m
T2allow 6.35 kN # m
controls over T2 below
ALSO CHECK SEGMENT 2 WITH BRASS SLEEVE OVER STEEL SHAFT
Tb
t
d2
2
IPb
where from statically indeterminate analysis above
T2allow 2 tb 1Gs Ips + Gb IPb2
d2 Gb
Tb T2 a
Gb IPb
b
Gs Ips + Gb IPb
T2allow A 109 B 1.369 * 104 N # m
T2allow 13.69 kN # m
(c) ALLOWABLE TORQUE T1 BASED ON ALLOWABLE SHEAR STRESS IN STEEL, ts
FIRST CHECK SEGMENT 2 WITH BRASS SLEEVE OVER STEEL SHAFT
Ts
t
d1
2
IPs
T3allow where from statically indeterminate analysis above
2 ts1Gs Ips + Gb IPb2
d 1 Gs
ts 110 MPa
Ts T3 a
Gs IPs
b
Gs IPs + Gb IPb
T3allow A 109 B 1.383 * 104 N # m
T3allow 13.83 kN # m
ALSO CHECK SEGMENT 3 WITH STEEL SHAFT ALONE
T3
t
d1
2
IPs
T3allow 2 ts IPs
d1
T3allow A 109 B 7.408 * 103 N # m
T3allow 7.41 kN # m
6 controls over T3 above
(d) TMAX IF ALL PRECEDING CONDITIONS MUST BE CONSIDERED
From part (b):
Tmax 6.35 kN # m max. shear stress in holow brass sleeve in segement 1 controls overall
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CHAPTER 3 Torsion
Problem 3.8-15 A uniformly tapered aluminum-alloy tube AB of circular cross section and length L is fixed against rota-
tion at A and B, as shown in the figure. The outside diameters at the ends are dA and dB 2dA. A hollow section of length
L/2 and constant thickness t dA/10 is cast into the tube and extends from B halfway toward A. Torque To is applied at L/2.
(a) Find the reactive torques at the supports, TA and TB. Use numerical values as follows: dA 2.5 in., L 48 in.,
G 3.9 106 psi, T0 40,000 in.-lb.
(b) Repeat (a) if the hollow section has constant diameter dA.
Solution 3.8-15
Solution approach-superposition: select TB as the redundant (1° SI )
L
—
2
TA1
A
B
f1(same results for parts a and b)
T0
f1 L
+
TA2
L
—
2
A
81GpdA
4
f1 2.389
T0L
GdA 4
See solution to Prob. 3.4-13 for
results for w2 for parts a and b
B
f2
TB
L
608T0 L
f2a 3.868
f2a 3.057
T0 L
Gd A 4
T0L
Gd A 4
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357
SECTION 3.8 Statically Indeterminate Torsional Members
(a) REACTIVE TORQUES, TA AND TB, FOR CASE OF
(b) REACTIVE TORQUES, TA AND TB, FOR CASE OF
CONSTANT THICKNESS OF HOLLOW SECTION OF TUBE
compatibility equation:
TB redundant
T0 40000 in.-lb
f1 f2 0
TB a
TB 24708 in.-lb
TA T0 TB TA 15292 in.-lb
608T0 L
81Gpd A4
TB 2.45560
Gd A4
b
ba
TB a
4
3.86804L
81G pd A
608T0 L
T0
TB 1.94056
p
CONSTANT DIAMETER OF HOLE
T0
p
ba
Gd A4
b
3.05676L
TB 31266 in.-lb
TA T0 TB TA 8734 in.-lb
;
;
;
TA TB 40,000 in.-1b (check)
;
TA TB 40,000 in.-lb (check)
Problem 3.8-16 Two pipes (L1 2.5 m and L2 1.5 m) are
joined at B by flange plates (thickness tf 14 mm) with five bolts
(dbf 13 mm) arranged in a circular pattern (see figure). Also, each
pipe segment is attached to a wall (at A and C, see figure) using a
base plate (tb 15 mm) and four bolts (dbb 16 mm). All bolts are
tightened until just snug. Assume E1 110 GPa, E2 73 GPa, 1
0.33, and 2 0.25. Neglect the self-weight of the pipes, and
assume the pipes are in a stress-free state initially. The cross-sectional areas of the pipes are A1 1500 mm2 and A2 (3/5)A1. The
outer diameter of Pipe 1 is 60 mm. The outer diameter of Pipe 2 is
equal to the inner diameter of Pipe 1. The bold radius r 64 mm for
both base and flange plates.
Flange plate 2
Base plate
(4 bolts)
A
x
T
E1
dbf
dbb
L1
B
E2
C
L2
(a) If torque T is applied at x L1, find an expression for reactive torques R1 and R2 in terms of T.
Bolt hole in flange
(b) Find the maximum load variable T (i.e., Tmax) if allowable torplate 2
Bolt group radius
sional stress in the two pipes is tallow 65 MPa.
Bolt hole in flange
for both base
(c) Draw torisonal moment (TMD) and torisonal displacement
plate 1
plates
and
(TDD) diagrams. Label all key ordinates. What is fmax?
flange
plates
r
β
(d) Find Tmax if allowable shear and bearing stresses in the
base plate and flange bolts cannot be exceeded. Assume
allowable stresses in shear and bearing for all bolts are
Cross section at
tallow 45 MPa and sallow 90 MPa.
flange plate
(e) Remove torque T at x L1. Now assume the flange-plate bolt
holes are misaligned by some angle b (see figure). Find the
expressions for reactive torques R1 and R2 if the pipes are twisted to align the flange-plate bolt holes, bolts are then
inserted, and the pipes released.
(f) What is the maximum permissible misalignment angle b max if allowable stresses in shear and bearing for all bolts
(from part (d) above) are not to be exceeded?
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CHAPTER 3 Torsion
Solution 3.8-16
NUMERICAL DATA AND CROSS SECTIONAL AND MATERIAL PROPERTIES
A1 1500 mm2 A2 di1 4
Cp
3
A 900 mm2 do1 60 mm
5 1
A1 + do12 41.111 mm
do2 di1 41.111 mm di2 p
1d 2 di122 1500 mm2
4 o1
4
d o22 a b A2 23.329 mm
p
C
p
1d o22 d i222 900 mm2
4
t1 do1 di1
do2 di2
9.444 mm t 2 2
2
t 2 8.891 mm L 1 2.5 m L 2 1.5 m
Ip1 p
1d 4 d i142 9.919 * 107 m4 r 64 mm
32 o1
Ip2 p
1d 4 d i242 2.514 * 107 m4 t f 14 mm dbf 13 mm t b 15 mm dbb 16 mm
32 o2
E 1 110 Gpa E 2 73 GPa v1 0.33 v2 0.25
Ip1
Ip2
3.946
G1 E1
E2
41.4 GPa G2 29.2 GPa
2 (1 + v1)
2 (1 + v2)
(a) IF TORQUE T IS APPLIED AT x L1, FIND AN EXPRESSION FOR REACTIVE TORQUES R1 AND R2 IN TERMS OF APPLIED T
FROM ONE-DEGREE STATICALLY-INDETERMINATE ANALYSIS WITH REACTION AT RIGHT SUPPORT R2 AS THE REDUNDANT:
fa T fT1 fT1 R2 L1
G1 Ip1
fb R2(fT1 + fT2) fT2 L2
G2 Ip2
using fT segment torsional flexibility
fT1
T fT1
0.23
0.23 T R1 T R2 0.77 T
fT1 + fT2
1fT1 + fT22
R1 0.77 T
R2 0.23 T
fT2
1fT1 + fT22
0.77
(b) FIND MAX. LOAD VARIABLE TMAX IF ALLOWABLE TORSIONAL STRESS IN THE TWO PIPES IS
tallow 65 MPa tallow 65 MPa
Tmax1 tallow Ip1
2.149 kN # m Tmax2 tallow Ip2
0.795 kN # m
do1
do2
2
2
Now, must relate these Tmax values above back to TMD to find max. APPLIED (load variable)T (i.e., see reaction
moment expressions above)
Tmax1
Tmax2
ON SEGMENT 1: Tmax ON SEGMENT 2: Tmax 2.791 kN # m
3.456 kN # m
0.77 ^ controls
0.23
Tmax 2.79 kN # m
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SECTION 3.8 Statically Indeterminate Torsional Members
359
(c) DRAW TORSIONAL MOMENT (TMD) AND TORSIONAL DISPLACEMENT (TDD) DIAGRAMS; LABEL ALL KEY ORDINATES;
WHAT IS f MAX?
fmax
0.77 T
TMD
(Part c)
0
0
TDD
(Part c)
or
−0.23 T
0.77 T fT1
or 0.23 T fT2
fT1 fT2
T
7.51
fT1 + fT2 max
fmax 0
L 1 L 2 Tmax
7.51
G1Ip1 L 2 + G2 + Ip2 L 1
0
(d) FIND TMAX IF ALLOWABLE SHEAR AND BEARING STRESSES IN THE BASE PLATE AND FLANGE PLATE BOLTS CANNOT BE EXCEEDED;
ASSUME ALLOWABLE STRESSES IN SHEAR AND BEARING FOR ALL BOLTS ARE tallow 45 MPa and sballow 90 MPa
tallow 45 MPa sballow 90 MPa tf 14 mm tb 15 mm dbf 13 mm dbb 16 mm
p
p
⵩
Abb d bb2 201.062 mm2 Abf d bf 2 132.732 mm2 ⵩diameter
diameter of
4
4
of flange
base plate
plate bolts
bolts
Bolt radius in both base and flange plates: r 64 mm
Number of base plate bolts: nb 4 Number of flange plate bolts: nf 5
Replace torque by n forces Fb in the bolts at distributed r (radius) from center of shaft (shaft axis).
TORQUE REACTION R1 CONTROLS (0.77 T ): AT Base plate
R1 0.77 T
fT2
0.77
fT1 + fT2
FIRST, CHECK SHEAR STRESS IN BASE AND FLANGE PLATE BOLTS; MAXIMUM FORCE IN EACH BOLT Tallow x AREA OF BOLT
nb r 1tallow Abb2
fT2
1Tmax2 nb r 1tallow Abb2 Tmax1 3.007 kN # m
fT2
fT1 + fT2
fT1 + fT2
n f r 1tallow Abf2
2.481 kN # m shear in flange plate
Tmax2 fT2
bolts controls over that in
fT1 + fT2
base plate bolts
THEN, CHECK BEARING stress IN BASE AND FLANGE PLATE BOLTS; MAXIMUM FORCE IN EACH BOLT sballow Abrg-bolt
AbrgBasePl dbb tb 240 mm
⵩
2
bearing stress area between base plate and each bolt
Tmax3 n b r 1sballow Abrg BaseP12
fT2
fT1 + fT2
7.179 kN # m
AbrgFlangePl dbf tf 182 mm2
bearing stress area between flange plate and each bolt
⵩
Tmax4 n f r 1sballow AbrgFlangeP12
fT2
fT1 + fT2
Both Tmax3 and Tmax4 exceed Tmax1 and Tmax2 above (based on bolt shear) so
6.805 kN # m
Tmax Tmax2 2.48 kN # m
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CHAPTER 3 Torsion
(e) REMOVE TORQUE T AT x L1; NOW ASSUME THE FLANGE PLATE BOLT HOLES ARE MISALIGNED BY SOME ANGLE b (SEE
FIGURE); FIND EXPRESSIONS FOR REACTIVE TORQUES R1 AND R2 IF THE PIPES ARE TWISTED TO ALIGN THE FLANGE PLATE
BOLT HOLES, BOLTS ARE THEN INSERTED AND THE PIPES RELEASED
Use superposition analysis with R2 as the redundant. Twist due to misfit f1 b and twist due to redundant R2 is f2 where
f2 R2 (fT1 + fT2)
and the compatibility equation is f1 + f2 0
b
then statics gives R1 R2
Solving for R2 gives R2 fT1 + fT2
where
fT1 L1
G1 IP1
and
fT2 L2
G2 IP2
(f) WHAT IS THE MAXIMUM PERMISSIBLE MISALIGNMENT ANGLE bMAX IF ALLOWABLE STRESSES IN SHEAR AND BEARING FOR
ALL BOLTS (FROM PART (d) ABOVE) ARE NOT TO BE EXCEEDED?
Maximum reaction torque n r (allow stress bolt or bearing area), so bmax (fT1 fT2) [n r (allow stress * bolt or bearing area)].
CHECK SHEAR IN BOTH BASE PLATE AND FLANGE PLATE BOLTS
b max1 (fT1 + fT2) [n b r (tallow Abb)] 0.615
b max2 (fT1 + fT2) [n f r (tallow Abf)] 0.507
b max2
1.911 kN # m
fT1 + fT2
shear in flange plage bolts controls; associated reactive torques
at each end:
b max2
7.264/m
L1 + L2
CHECK BEARING STRESS IN BOTH BASE PLATE AND FLANGE PLATE BOLTS
b max3 (fT1 + fT2) [nb r (sballow AbrgBaseP1)] 1.467
b max4 (fT1 + fT2) [nf r (sballow AbrgFlangeP1)] 1.391
b max b max2 29.1
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SECTION 3.9 Strain Energy in Torsion
361
Strain Energy in Torsion
Problem 3.9-1 A solid circular bar of steel (G 11.4 106 psi)
with length L 30 in. and diameter d 1.75 in. is subjected to
pure torsion by torques T acting at the ends (see figure).
d
T
T
(a) Calculate the amount of strain energy U stored in the bar
when the maximum shear stress is 4500 psi.
(b) From the strain energy, calculate the angle of twist f
(in degrees).
Solution 3.9-1
L
Steel bar
pd2Lt2max
16G
(Eq. 2)
Substitute numerical values:
U 32.0 in.-lb
G 11.4 10 psi
;
6
(b) ANGLE OF TWIST
L 30 in.
d 1.75 in.
U
tmax 4500 psi
tmax IP 16 T
pd3
Substitute for T and U from Eqs. (1) and (2):
pd3tmax
T
16
pd 4
32
Tf
2U
f
2
T
f
(Eq. 1)
2Ltmax
Gd
(Eq. 3)
Substitute numerical values:
f 0.013534 rad 0.775°
;
(a) STRAIN ENERGY
U
pd3tmax 2 L
32
T2L
a
b a
ba
b
2GIP
16
2G pd 4
Problem 3.9-2 A solid circular bar of copper (G 45 GPa) with length L 0.75 m and diameter d 40 mm is subjected
to pure torsion by torques T acting at the ends (see figure).
(a) Calculate the amount of strain energy U stored in the bar when the maximum shear stress is 32 MPa.
(b) From the strain energy, calculate the angle of twist f (in degrees)
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CHAPTER 3 Torsion
Solution 3.9-2
Copper bar
(a) STRAIN ENERGY
U
pd3tmax 2 L
32
T2L
a
b a
ba
b
2GIP
16
2G pd 4
L 0.75 m
pd2Lt2max
16G
Substitute numerical values:
d 40 mm
U 5.36 J
tmax 32 MPa
(b) ANGLE OF TWIST
G 45 GPa
pd3tmax
16T
tmax T
3
16
pd
IP ;
Tf
2U
f
2
T
Substitute for T and U from Eqs. (1) and (2):
2Ltmax
f
(Eq. 3)
Gd
Substitute numerical values:
U
pd 4
32
(Eq. 1)
f 0.026667 rad 1.53°
Problem 3.9-3 A stepped shaft of solid circular cross sections
(see figure) has length L 45 in., diameter d2 1.2 in., and
diameter d1 1.0 in. The material is brass with G 5.6 106 psi.
Determine the strain energy U of the shaft if the angle of
twist is 3.0°.
d2
d1
T
L
—
2
Stepped shaft
8T 2L 1
1
a
+ 4b
pG d24
d1
(Eq. 1)
Tf
2
(Eq. 2)
Also, U Equate U from Eqs. (1) and (2) and solve for T:
d1 1.0 in.
T
d2 1.2 in.
L 45 in.
pGd14 d24f
16L(d14 + d24)
Tf
pGf2 d 14 d 24
a
b
2
32L d 14 + d 24
G 5.6 106 psi (brass)
U
f 3.0° 0.0523599 rad
SUBSTITUTE NUMERICAL VALUES:
STRAIN ENERGY
U a
2
;
T
L
—
2
Solution 3.9-3
(Eq. 2)
U 22.6 in.-lb
2
f in radians
;
2
16 T (L/2)
16 T (L/2)
TL
+
4
2GIP
pGd2
pGd14
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SECTION 3.9 Strain Energy in Torsion
363
Problem 3.9-4 A stepped shaft of solid circular cross sections (see figure) has length L 0.80 m, diameter d2 40 mm,
and diameter d1 30 mm. The material is steel with G 80 GPa.
Determine the strain energy U of the shaft if the angle of twist is 1.0°.
Solution 3.9-4
Stepped shaft
Also, U Tf
2
(Eq. 2)
Equate U from Eqs. (1) and (2) and solve for T:
T
d1 30 mm
d2 40 mm
L 0.80 m
G 80 GPa (steel)
U
pG d14 d24f
16L(d14 + d24)
Tf
pGf2 d 14 d 24
b
a
2
32L d 14 + d 24
f in radians
f 1.0° 0.0174533 rad
SUBSTITUTE NUMERICAL VALUES:
STRAIN ENERGY
U a
2
U 1.84 J
2
;
2
16T (L/2)
16T (L/2)
TL
+
4
2GIP
pGd2
pGd14
1
8T2L 1
a 4 + 4b
pG d2
d1
(Eq. 1)
Problem 3.9-5 A cantilever bar of circular cross section and length L is
fixed at one end and free at the other (see figure). The bar is loaded by a
torque T at the free end and by a distributed torque of constant intensity
t per unit distance along the length of the bar.
(a) What is the strain energy U1 of the bar when the load T acts alone?
(b) What is the strain energy U2 when the load t acts alone?
(c) What is the strain energy U3 when both loads act simultaneously?
t
L
T
Solution 3.9-5 Cantilever bar with distributed torque
G shear modulus
IP polar moment of inertia
T torque acting at free end
t torque per unit distance
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CHAPTER 3 Torsion
(a) LOAD T ACTS ALONE (Eq. 3-55a)
U1 T2L
2GIP
(c) BOTH LOADS ACT SIMULTANEOUSLY
;
(b) LOAD t ACTS ALONE
From Eq. (3-64) of Example 3-11:
At distance x from the free end:
t2L3
U2 6GIP
T(x) T + tx
;
L
U3 L
[T(x)]2
1
dx (T + tx)2dx
2GIP L0
L0 2GIP
T2L
TtL2
t2L3
+
+
2GIP
2GIP
6GIP
;
NOTE: U3 is not the sum of U1 and U2.
Problem 3.9-6 Obtain a formula for the strain energy U of the statically
2T0
indeterminate circular bar shown in the figure. The bar has fixed supports
at ends A and B and is loaded by torques 2T0 and T0 at points C and D,
respectively.
Hint: Use Eqs. 3-50a and b of Example 3-9, Section 3.8, to obtain the
reactive torques.
T0
A
B
C
L
—
4
D
L
—
2
L
—
4
Solution 3.9-6 Statically indeterminate bar
REACTIVE TORQUES
STRAIN ENERGY (from Eq. 3-57)
From Eq. (3-50a):
U a
(2T0)a
TA 3L
b
4
L
T0 a b
4
+
L
L
TB 3T0 TA Ti2Li
i1 2GiIPi
n
7T0
4
5T0
4
L
L
L
1
2
2
2
cTAC
a b + TCD
a b + TDB
a bd
2GIp
4
2
4
7T0 2 L
1
c a
b a b
2GIP
4
4
+ a
INTERNAL TORQUES
TAC 7T0
4
TCD T0
4
TDB 5T0
4
U
5T0 2 L
T0 2 L
b a b + a
b a bd
4
2
4
4
19T02L
32GIP
;
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Page 365
SECTION 3.9 Strain Energy in Torsion
365
Problem 3.9-7 A statically indeterminate stepped shaft ACB is fixed at
ends A and B and loaded by a torque T0 at point C (see figure). The two
segments of the bar are made of the same material, have lengths LA and LB,
and have polar moments of inertia IPA and IPB.
Determine the angle of rotation f of the cross section at C by using
strain energy.
[Hint: Use Eq. 3-55b to determine the strain energy U in terms of the
angle f. Then equate the strain energy to the work done by the torque T0.
Compare your result with Eq. (3-52) of Example 3-9, Section 3.8.]
A
IPA
T0
C
IPB
LA
B
LB
Solution 3.9-7 Statically indeterminate bar
WORK DONE BY THE TORQUE T0
W
T0f
2
EQUATE U AND W AND SOLVE FOR f
T0f
Gf2 IPA
IPB
a
+
b 2
LA
LB
2
f
STRAIN ENERGY (FROM Eq. 3-55b)
;
(This result agrees with Eq. (3-52) of Example 3-9,
Section 3.8.)
GIPif2i
GIPAf2
GIPBf2
+
2LA
2LB
i1 2Li
n
U a
T0LALB
G(LBIPA + LAIPB)
Gf2 IPA
IPB
+
b
a
2
LA
LB
Problem 3.9-8 Derive a formula for the strain energy U of the cantilever bar shown in the figure.
The bar has circular cross sections and length L. It is subjected to a distributed torque of intensity t per unit distance. The
intensity varies linearly from t 0 at the free end to a maximum value t t0 at the support.
t0
t
L
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Page 366
CHAPTER 3 Torsion
Solution 3.9-8 Cantilever bar with distributed torque
x distance from right-hand end of the bar
ELEMENT dj
STRAIN ENERGY OF ELEMENT dx
Consider a differential element dj at distance j from the
right-hand end.
dU [T(x)]2dx
t0 2
1
a b x4dx
2GIP
2GIP 2L
t20
8L2GIP
x4 dx
STRAIN ENERGY OF ENTIRE BAR
L
U
L0
dT external torque acting on this element
dT t()d
t 0 a bd
L
U
t20
dU t20L3
40GIP
L
x4 dx
8L2GIP L0
t20
L5
2
a b
8L GIP 5
;
ELEMENT dx AT DISTANCE x
T(x) internal torque acting on this element
T(x) total torque from x 0 to x x
x
T(x) L0
dT x
L0
t0 a
bd
L
t0x2
2L
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Page 367
367
SECTION 3.9 Strain Energy in Torsion
Problem 3.9-9 A thin-walled hollow tube AB of conical shape has
constant thickness t and average diameters dA and dB at the ends
(see figure).
B
A
T
T
(a) Determine the strain energy U of the tube when it is subjected
to pure torsion by torques T.
(b) Determine the angle of twist f of the tube.
L
t
Note: Use the approximate formula IP ⬇ pd3t/4 for a thin circular
ring; see Case 22 of Appendix E.
t
dB
dA
Solution 3.9-9
Thin-walled, hollow tube
Therefore,
L
dx
3
d
L0
B dA
cdA + a
bx d
L
t thickness
dA average diameter at end A
dB average diameter at end B
d(x) average diameter at distance x from end A
d(x) dA + a
dB dA
bx
L
pd3t
4
U
p[d(x)] t
dB dA
pt
cdA + a
bx d
4
4
L
3
IP(x) 3
T2dx
L0 2GIP(x)
2T2
pGt L0
L
L (a + bx)
3
dx
3
dB dA
cdA + a
bx d
L
2(dB dA)(dB)
+
2(dB dA)(dA)2
L(dA + dB)
2dA2 dB2
2T2 L(dA + dB)
T2L dA + dB
a 2 2 b
2
2
pGt 2dAdB
pGt
dA dB
WU
From Appendix D:
dx
L
2
Work of the torque T: W L
L
;
(b) ANGLE OF TWIST
(a) STRAIN ENERGY (FROM Eq. 3-58)
U
2 †
2(dB dA)
dB dA
cdA + a
bx d
L
L
0
Substitute this expression for the integral into the equation
for U (Eq. 1):
POLAR MOMENT OF INERTIA
IP L
1
(Eq. 1)
Tf
2
Tf
T2L(dA + dB)
2
pGt d2Ad2B
Solve for f:
f
2TL(dA + dB)
pGt d2A d2B
;
1
2b(a + bx)2
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Page 368
CHAPTER 3 Torsion
Problem 3.9-10 A hollow circular tube A fits over the end of
IPA
a solid circular bar B, as shown in the figure. The far ends of both
bars are fixed. Initially, a hole through bar B makes an angle b with
a line through two holes in tube A. Then bar B is twisted until the
holes are aligned, and a pin is placed through the holes.
When bar B is released and the system returns to equilibrium,
what is the total strain energy U of the two bars? (Let IPA and IPB
represent the polar moments of inertia of bars A and B, respectively.
The length L and shear modulus of elasticity G are the same for
both bars.)
IPB
Tube A
Bar B
L
L
b
Tube A
Bar B
Solution 3.9-10 Circular tube and bar
TUBE A
COMPATIBILITY
fA fB b
FORCE-DISPLACEMENT RELATIONS
fA TL
TL
fB GIPA
GIPB
T torque acting on the tube
Substitute into the equation of compatibility and solve
for T:
fA angle of twist
T
BAR B
STRAIN ENERGY
bG
IPAIPB
a
b
L IPA + IPB
U g
T 2L
T 2L
T 2L
+
2GIP
2GIPA
2GIPB
T2L 1
1
a
+
b
2G IPA
IPB
Substitute for T and simplify:
U
b 2G IPA IPB
a
b
2L IPA + IPB
;
T torque acting on the bar
fB angle of twist
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Page 369
SECTION 3.9 Strain Energy in Torsion
369
Problem 3.9-11 A heavy flywheel rotating at n revolutions per minute is
rigidly attached to the end of a shaft of diameter d (see figure). If the
bearing at A suddenly freezes, what will be the maximum angle of twist f
of the shaft? What is the corresponding maximum shear stress in the shaft?
(Let L length of the shaft, G shear modulus of elasticity, and
Im mass moment of inertia of the flywheel about the axis of the shaft.
Also, disregard friction in the bearings at B and C and disregard the mass
of the shaft.)
A
d
n (rpm)
B
C
Hint: Equate the kinetic energy of the rotating flywheel to the strain
energy of the shaft.
Solution 3.9-11 Rotating flywheel
p 4
d
32
IP d diameter of shaft
U
pGd 4f2
64L
UNITS:
d diameter
n rpm
1
I v2
2 m
2pn
v
60
n rpm
2pn 2
1
b
K.E. Im a
2
60
p2n2Im
1800
Im (force)(length)(second)2
v radians per second
K.E. (length)(force)
STRAIN ENERGY OF SHAFT (FROM Eq. 3-55b)
GIPf
2L
L length
U (length)(force)
EQUATE KINETIC ENERGY AND STRAIN ENERGY
2
pGd 4f2
p2n2Im
1800
64 L
K.E. U
Solve for f:
f
2n
2A
15d
2pImL
G
;
MAXIMUM SHEAR STRESS
t
UNITS:
U
IP (length)4
f radians
KINETIC ENERGY OF FLYWHEEL
K.E. G (force)/(length)2
T(d/2)
TL
f
IP
GIP
Eliminate T:
t
tmax tmax Gdf
2L
Gd2n
2L15d
2A
2pImL
G
2pGIm
n
15d A L
;
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Page 370
CHAPTER 3 Torsion
Thin-Walled Tubes
Problem 3.11-1 A hollow circular tube having an inside diameter of 10.0 in.
and a wall thickness of 1.0 in. (see figure) is subjected to a torque T 1200 k-in.
Determine the maximum shear stress in the tube using (a) the approximate
theory of thin-walled tubes, and (b) the exact torsion theory. Does the approximate
theory give conservative or nonconservative results?
10.0 in.
1.0 in.
Solution 3.11-1 Hollow circular tube
APPROXIMATE THEORY (Eq. 3-83)
t1 T
2pr2 t
1200 k-in.
2p(5.5 in.)2(1.0 in.)
tapprox 6310 psi
6314 psi
;
EXACT THEORY (Eq. 3-13)
T 1200 k-in.
t 1.0 in.
t2 T(d2/2)
IP
r radius to median line
r 5.5 in.
d2 outside diameter 12.0 in.
d1 inside diameter 10.0 in.
Td2
2a
p
b 1d24 d142
32
16(1200k-in.)(12.0 in.)
p[(12.0 in.)4 (10.0 in.)4]
6831 psi
t exact 6830 psi
;
Because the approximate theory gives stresses that are
too low, it is nonconservative. Therefore, the approximate theory should only be used for very thin tubes.
Problem 3.11-2 A solid circular bar having diameter d is to be
replaced by a rectangular tube having cross-sectional dimensions
d 2d to the median line of the cross section (see figure).
Determine the required thickness tmin of the tube so that the
maximum shear stress in the tube will not exceed the maximum
shear stress in the solid bar.
t
t
d
d
2d
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Page 371
SECTION 3.11 Thin-Walled Tubes
Solution 3.11-2
371
Bar and tube
SOLID BAR
tmax 16T
pd3
(Eq. 3-14)
Am (d)(2d) 2d2
(Eq. 3-84)
T
T
2tAm
4td2
(Eq. 3-81)
tmax EQUATE THE MAXIMUM SHEAR STRESSES AND SOLVE FOR t
RECTANGULAR TUBE
16T
3
pd
T
4td2
tmin pd
64
;
If t tmin, the shear stress in the tube is less than the
shear stress in the bar.
Problem 3.11-3 A thin-walled aluminum tube of rectangular
cross section (see figure) has a centerline dimensions b 6.0 in.
and h 4.0 in. The wall thickness t is constant and equal to 0.25 in.
t
h
(a) Determine the shear stress in the tube due to a torque
T 15 k-in.
(b) Determine the angle of twist (in degrees) if the length L of
the tube is 50 in. and the shear modulus G is 4.0 106 psi.
b
Probs. 3.11-3 and 3.11-4
Solution 3.11-3
Thin-walled tube
Eq. (3-84): Am bh 24.0 in.2
J
Eq. (3-71) with t1 t2 t:
2b2h2t
b + h
J 28.8 in.4
(a) SHEAR STRESS (Eq. 3-81)
t
b 6.0 in.
h 4.0 in.
t 0.25 in.
T 15 k-in.
T
1250 psi
2tAm
;
(b) ANGLE OF TWIST (Eq. 3-17)
f
TL
0.0065104 rad
GJ
0.373
;
L 50 in.
G 4.0 106 psi
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Page 372
CHAPTER 3 Torsion
Problem 3.11-4 A thin-walled steel tube of rectangular cross section (see figure) has centerline dimensions b 150 mm
and h 100 mm. The wall thickness t is constant and equal to 6.0 mm.
(a) Determine the shear stress in the tube due to a torque T 1650 N # m.
(b) Determine the angle of twist (in degrees) if the length L of the tube is 1.2 m and the shear modulus G is 75 GPa.
Solution 3.11-4
Thin-walled tube
b 150 mm
(a) SHEAR STRESS (Eq. 3-81)
h 100 mm
t
t 6.0 mm
T 1650 N # m
f
G 75 GPa
TL
0.002444 rad
GJ
0.140
Eq. (3-84): Am bh 0.015 m2
J
;
(b) ANGLE OF TWIST (Eq. 3-17)
L 1.2 m
Eq. (3-94) with t1 t2 t:
T
9.17 MPa
2tAm
;
2b2h2t
b + h
J 10.8 106 m4
Tube (1)
Problem 3.11-5 A thin-walled circular tube and a solid circular bar of
Bar (2)
the same material (see figure) are subjected to torsion. The tube and bar
have the same cross-sectional area and the same length.
What is the ratio of the strain energy U1 in the tube to the strain
energy U2 in the solid bar if the maximum shear stresses are the same in
both cases? (For the tube, use the approximate theory for thin-walled bars.)
Solution 3.11-5 THIN-WALLED TUBE (1)
Am pr2
tmax J 2pr3t A 2prt
SOLID BAR (2)
T
T
2tAm
2pr 2t
A pr22
T 2pr 2ttmax
12pr2ttmax22L
T 2L
U1 2GJ
2G(2pr3t)
prtt2maxL
G
A
But rt 2p
‹ U1 At2max L
2G
tmax U2 IP Tr2
2T
3
IP
pr2
p 4
r
2 2
T
pr23tmax
2
2
(pr23tmax)2L
L
pr22tmax
T2L
2GIP
4G
p
8G a r24 b
2
But pr22 A
RATIO
U1
2
U2
‹ U2 2
L
Atmax
4G
;
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Page 373
SECTION 3.11 Thin-Walled Tubes
Problem 3.11-6 Calculate the shear stress t and the angle of twist f (in
373
t = 8 mm
degrees) for a steel tube (G 76 GPa) having the cross section shown
in the figure. The tube has length L 1.5 m and is subjected to a torque
T 10 kN # m.
r = 50 mm
r = 50 mm
b = 100 mm
Solution 3.11-6
Steel tube
SHEAR STRESS
G 76 GPa
t
10 kN # m
T
2tAm
2(8 mm)(17,850 mm2)
L 1.5 m
T 10 kN # m
Am pr2 2br
Am p (50 mm)2 2(100 mm)(50 mm)
17,850 mm2
35.0 MPa
;
ANGLE OF TWIST
f
(10 kN # m)(1.5 m)
TL
GJ
(76 GPa)(19.83 * 106 mm4)
0.00995 rad
0.570
;
Lm 2b 2pr
2(100 mm) 2p(50 mm)
514.2 mm
J
4(8 mm)(17,850 mm2)2
4tA2m
Lm
514.2 mm
19.83 * 106 mm4
Problem 3.11-7 A thin-walled steel tube having an elliptical cross
t
section with constant thickness t (see figure) is subjected to a torque
T 18 k-in.
Determine the shear stress t and the rate of twist u (in degrees
per inch) if G 12 106 psi, t 0.2 in., a 3 in., and b 2 in.
(Note: See Appendix E, Case 16, for the properties of an ellipse.)
2b
2a
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Page 374
CHAPTER 3 Torsion
Solution 3.11-7
Elliptical tube
FROM APPENDIX E, CASE 16:
Am pab p(3.0 in.)(2.0 in.) 18.850 in.2
Lm L p[1.5(a + b) 1ab]
p[1.5(5.0 in.) 26.0 in.2] 15.867 in.
J
17.92 in.4
T 18 k-in.
SHEAR STRESS
G 12 106 psi
t
t constant
t 0.2 in.
4(0.2 in.)(18.850 in.2)2
4tA2m
Lm
15.867 in.
a 3.0 in.
b 2.0 in.
18 k-in.
T
2tAm
2(0.2 in.)(18.850 in.2)
2390 psi
;
ANGLE OF TWIST PER UNIT LENGTH (RATE OF TWIST)
u
f
T
18 k-in.
L
GJ
(12 * 106 psi)(17.92 in.)4
u 83.73 * 106 rad/in. 0.0048/in.
Problem 3.11-8 A torque T is applied to a thin-walled tube having
a cross section in the shape of a regular hexagon with constant wall
thickness t and side length b (see figure).
Obtain formulas for the shear stress t and the rate of twist u.
;
t
b
Solution 3.11-8
Regular hexagon
b Length of side
t Thickness
Lm 6b
FROM APPENDIX E, CASE 25:
b 60°
Am n6
2
b
nb
6b2
cot cot 30
4
2
4
313b2
2
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Page 375
SECTION 3.11 Thin-Walled Tubes
SHEAR STRESS
t
u
T
T13
2tAm
9b2t
;
2T
T
2T
3
GJ
G(9b t)
9Gb3t
375
;
(radians per unit length)
ANGLE OF TWIST PER UNIT LENGTH (RATE OF TWIST)
4A2mt
J
Lm d
s
L0
4A2mt
9b3t
Lm
2
t
Problem 3.11-9 Compare the angle of twist f1 for a thin-walled circular tube
(see figure) calculated from the approximate theory for thin-walled bars with the
angle of twist f2 calculated from the exact theory of torsion for circular bars.
t
r
(a) Express the ratio f1/f2 in terms of the nondimensional ratio b r/t.
(b) Calculate the ratio of angles of twist for b 5, 10, and 20. What conclusion
about the accuracy of the approximate theory do you draw from these results?
Solution 3.11-9
C
Thin-walled tube
(a) RATIO
f1
f2
4r 2 + t 2
Let b APPROXIMATE THEORY
TL
f1 GJ
J 2pr t
3
(b)
f1 TL
GIP
2pGr3t
From Eq. (3-17): Ip TL
2TL
f2 GIP
pGrt(4r 2 + t 2)
r
t
f1
f2
t2
4r 2
1 +
1
;
4b 2
b
f1/f2
5
10
20
1.0100
1.0025
1.0006
TL
EXACT THEORY
f2 4r 2
1 +
prt
(4r 2 + t 2)
2
As the tube becomes thinner and b becomes larger, the
ratio f1/f2 approaches unity. Thus, the thinner the tube,
the more accurate the approximate theory becomes.
Problem 3.11-10 A thin-walled rectangular tube has uniform thickness t
and dimensions a b to the median line of the cross section (see figure).
How does the shear stress in the tube vary with the ratio b a/b if
the total length Lm of the median line of the cross section and the torque T
remain constant?
From your results, show that the shear stress is smallest when the
tube is square (b 1).
t
b
a
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CHAPTER 3 Torsion
Solution 3.11-10 Rectangular tube
T, t, and Lm are constants.
Let k 2T
constant t k
tL2m
(1 + b)2
b
t thickness (constant)
a, b dimensions of the tube
b
a
b
t
a b
4
k min
Lm 2(a b) constant
T constant
T
2tAm
ALTERNATE SOLUTION
Am ab bb2
t
Lm 2b(1 b) constant
b
Am 8T
tL2m
From the graph, we see that t is minimum when b 1
and the tube is square.
SHEAR STRESS
t
tmin Lm
2(1 + b)
Am b c
2T b(2)(1 + b) (1 + b)2(1)
dt
2c
d 0
db
tLm
b2
2
Lm
d
2(1 + b)
or 2b (1 b) (1 b)2 0
bL2m
4(1 + b)2
T(4)(1 + b)2
2T(1 + b)2
T
t
2tAm
2tbL2m
tL2m b
2T (1 + b)2
c
d
b
tL2m
;
b 1
Thus, the tube is square and t is either a minimum or a
maximum. From the graph, we see that t is a minimum.
Problem 3.11-11 A tubular aluminum bar (G 4 106 psi) of square
cross section (see figure) with outer dimensions 2 in. 2 in. must resist a
torque T 3000 lb-in.
Calculate the minimum required wall thickness tmin if the allowable
shear stress is 4500 psi and the allowable rate of twist is 0.01 rad/ft.
t
2 in.
2 in.
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377
SECTION 3.11 Thin-Walled Tubes
Solution 3.11-11 Square aluminum tube
THICKNESS t BASED UPON SHEAR STRESS
t
T
2tAm
tAm UNITS: t in.
T
2t
t(b t)2 b in.
t(2.0 in. t)2 T
2t
T lb-in. t psi
3000 lb-in.
1
in.3
2(4500 psi)
3
3t(2 t)2 1 0
Outer dimensions:
2.0 in. 2.0 in.
G 4 106 psi
T 3000 lb-in.
Solve for t: t 0.0915 in.
THICKNESS t BASED UPON RATE OF TWIST
u
tallow 4500 psi
0.01
u allow 0.01 rad/ft rad/in.
12
T
T
GJ
Gt(b t)3
UNITS: t in.
G psi
6
Centerline dimension b t
10t(2 t)3 9 0
Am (b t)2
Lm 4(b t)
Solve for t:
4t(b t)4
t(b t)3
4(b t)
t 0.140 in.
Lm
u rad/in.
(4 * 10 psi)(0.01/12 rad/in.)
9
10
2.0 in.
J
T
Gu
3000 lb-in.
t(2.0 in. t)3 Let b outer dimension
4tA2m
t(b t)3 ANGLE OF TWIST GOVERNS
tmin 0.140 in.
Problem 3.11-12 A thin tubular shaft of circular cross section
(see figure) with inside diameter 100 mm is subjected to a torque
of 5000 N # m.
If the allowable shear stress is 42 MPa, determine the required
wall thickness t by using (a) the approximate theory for a thin-walled
tube, and (b) the exact torsion theory for a circular bar.
;
100 mm
t
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CHAPTER 3 Torsion
Solution 3.11-12
Thin tube
(b) EXACT THEORY
T 5,000 N # m
d1 inner diameter 100 mm
tallow 42 MPa
t is in millimeters.
Ip p 4
p
(r r41) [(50 + t)4 (50)4]
2 2
2
42 MPa (5,000 N # m)(50 + t)
p
[(50 + t)4 (50)4]
2
t
2
r1 inner radius
5 * 106
mm3
21p
Solve for t:
50 mm
t 7.02 mm
r2 Outer radius
50 mm t Am pr2
(a) APPROXIMATE THEORY
t
Tr2
Ip
(50 + t)4 (50)4
(5000 N # m)(2)
50 + t
(p)(42 MPa)
r average radius
50 mm +
t
;
The approximate result is 5% less than the exact
result. Thus, the approximate theory is nonconservative and should only be used for thin-walled tubes.
T
T
T
2
2tAm
2t(pr )
2pr 2 t
5,000 N # m
42 MPa 2pa50 +
t 2
b t
2
or
ta 50 +
5,000 N # m
t 2
5 * 106
b mm3
2
2p(42 MPa)
84p
Solve for t:
t 6.66 mm
;
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379
SECTION 3.11 Thin-Walled Tubes
Problem 3.11-13 A long, thin-walled tapered tube AB of circular cross
section (see figure) is subjected to a torque T. The tube has length L and
constant wall thickness t. The diameter to the median lines of the cross
sections at the ends A and B are dA and dB, respectively.
Derive the following formula for the angle of twist of the tube:
f
T
2TL dA + dB
a 2 2 b
pGt
dAdB
T
L
t
t
Hint: If the angle of taper is small, we may obtain approximate
results by applying the formulas for a thin-walled prismatic tube to a
differential element of the tapered tube and then integrating along the
axis of the tube.
Solution 3.11-13
B
A
dB
dA
Thin-walled tapered tube
For entire tube:
f
4T
pGT L0
L
dx
3
dB dA
cdA + a
bx d
L
From table of integrals (see Appendix D):
1
t thickness
dx
3
(a + bx)
1
2b(a + bx)2
dA average diameter at end A
dB average diameter at end B
T torque
d(x) average diameter at distance x from end A.
d(x) dA + a
J 2pr 3t dB dA
bx
L
3
pd t
4
3
dB dA
pt
pt
J(x) [d(x)]3 cdA + a
bx d
4
4
L
L
4T
f
pGt J
f
2
dB dA
dB dA
# xb K0
b adA +
L
L
1
2a
4T
L
L
c
+
d
2
pGt 2(dB dA)dB
2(dB dA)d2A
2TL dA + dB
a 2 2 b
pGt
dAdB
;
For element of length dx:
df Tdx
GJ(x)
4Tdx
3
dB dA
GptcdA + a
bx d
L
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CHAPTER 3 Torsion
Stress Concentrations in Torsion
D2
The problems for Section 3.12 are to be solved by considering the
stress-concentration factors.
Problem 3.12-1 A stepped shaft consisting of solid circular
R
D1
T
T
segments having diameters D1 2.0 in. and D2 2.4 in. (see figure)
is subjected to torques T. The radius of the fillet is R 0.1 in.
If the allowable shear stress at the stress concentration is 6000 psi,
what is the maximum permissible torque Tmax?
Probs. 3.12-1 through 3.12-5
Solution 3.12-1 Stepped shaft in torsion
USE FIG. 3-59 FOR THE STRESS-CONCENTRATION FACTOR
D2
2.4 in.
1.2
D1
2.0 in.
R
0.1 in.
0.05
D1
2.0 in.
K ⬇ 1.52
D1 2.0 in.
D2 2.4 in.
R 0.1 in.
tallow 6000 psi
tmax Kt nom K a
16 Tmax
pD31
b
pD31tmax
16K
p(2.0 in.)3(6000 psi)
6200 lb-in.
16(1.52)
Tmax Tmax ⬇ 6200 lb-in.
;
Problem 3.12-2 A stepped shaft with diameters D1 40 mm and D2 60 mm is loaded by torques T 1100 N # m
(see figure).
If the allowable shear stress at the stress concentration is 120 MPa, what is the smallest radius Rmin that may be used for
the fillet?
Solution 3.12-2 Stepped shaft in torsion
USE FIG. 3-59 FOR THE STRESS-CONCENTRATION FACTOR
tmax Kt nom Ka
16T
pD31
b
p(40 mm)3(120 MPa)
pD 31tmax
1.37
16 T
16(1100 N # m)
D2
60 mm
1.5
D1
40 mm
K
D1 40 mm
D2 60 mm
T 1100 N # m
From Fig. (3-59) with
tallow 120 MPa
we get
D2
1.5 and K 1.37,
D1
R
L 0.10
D1
Rmin ⬇ 0.10(40 mm) 4.0 mm
;
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381
SECTION 3.12 Stress Concentrations in Torsion
Problem 3.12-3 A full quarter-circular fillet is used at the shoulder of a stepped shaft having diameter D2 1.0 in.
(see figure). A torque T 500 lb-in. acts on the shaft.
Determine the shear stress tmax at the stress concentration for values as follows: D1 5 0.7, 0.8, and 0.9 in. Plot a graph
showing tmax versus D1.
Solution 3.12-3 Stepped shaft in torsion
D1 (in.)
D2/D1
R(in.)
R/D1
K
tmax(psi)
0.7
0.8
0.9
1.43
1.25
1.11
0.15
0.10
0.05
0.214
0.125
0.056
1.20
1.29
1.41
8900
6400
4900
D2 1.0 in.
T 500 lb-in.
D1 0.7, 0.8, and 0.9 in.
Full quarter-circular fillet (D2 D1 2R):
R
D2 D1
D1
0.5 in. 2
2
USE FIG. 3-59 FOR THE STRESS-CONCENTRATION FACTOR
tmax Kt nom K a
K
16 T
pD31
16(500 lb-in.)
pD31
b
2546
K
D31
Note: tmax gets smaller as D1 gets larger, even though K
is increasing.
Problem 3.12-4 The stepped shaft shown in the figure is required to transmit 600 kW of power at 400 rpm.
The shaft has a full quarter-circular fillet, and the smaller diameter D1 100 mm.
If the allowable shear stress at the stress concentration is 100 MPa, at what diameter D2 will this stress be reached?
Is this diameter an upper or a lower limit on the value of D2?
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Page 382
CHAPTER 3 Torsion
Solution 3.12-4 Stepped shaft in torsion
P 600 kW
D1 100 mm
Use the dashed line for a full quarter-circular fillet.
n 400 rpm
tallow 100 MPa
R
L 0.075
D1
R ⬇ 0.075 D1 0.075 (100 mm)
Full quarter-circular fillet
POWER P 7.5 mm
2pnT
( Eq. 3-42 of Section 3.7)
60
P in watts, n in rpm T in Newton meters.
60(600 * 103 W)
60P
14,320 N # m
T
2pn
2p(400 rpm)
D2 D1 2R 100 mm 2(7.5 mm) 115 mm
D2 ⬇ 115 mm
;
This value of D2 is a lower limit
;
(If D2 is less than 115 mm, R/D1 is smaller, K is larger,
and tmax is larger, which means that the allowable stress
is exceeded.)
USE FIG. 3-59 FOR THE STRESS-CONCENTRATION FACTOR
tmax Kt nom K a
K
16T
pD31
b
tmax(pD31)
16T
(100 MPa)(p)(100 mm)3
1.37
16(14,320 N # m)
Problem 3.12-5 A stepped shaft (see figure) has diameter D2 1.5 in. and a full quarter-circular fillet. The allowable shear
stress is 15,000 psi and the load T 4800 lb-in.
What is the smallest permissible diameter D1?
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SECTION 3.12 Stress Concentrations in Torsion
383
Solution 3.12-5 Stepped shaft in torsion
D2 1.5 in.
Use trial-and-error. Select trial values of D1
tallow 15,000 psi
D1 (in.)
R (in.)
R/D1
K
tmax(psi)
T 4800 lb-in.
1.30
1.35
1.40
0.100
0.075
0.050
0.077
0.056
0.036
1.38
1.41
1.46
15,400
14,000
13,000
Full quarter-circular fillet D2 D1 2R
R
D1
D2 D1
0.75 in. 2
2
USE FIG. 3-59 FOR THE STRESS-CONCENTRATION FACTOR
tmax Ktnom K a
16T
pD 31
K 16(4800 lb-in.)
3c
d
p
D1
24,450
b
K
D31
From the graph, minimum D1 ⬇ 1.31 in.
;
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Page 384
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Page 385
4
Shear Forces and
Bending Moments
Shear Forces and Bending Moments
800 lb
1600 lb
Problem 4.3-1 Calculate the shear force V and bending moment M at a cross
section just to the left of the 1600-1b load acting on the simple beam AB shown
in the figure.
A
B
30 in.
50 in.
120 in.
40 in.
Solution 4.3-1
3800
1267 lb
3
gFVERT 0: V 1600 lb 1267 lb
3400
g MB 0: RA 1133 lb
3
g MD 0: M 11267 lb2(40 in.)
FREE-BODY DIAGRAM OF SEGMENT DB
gMA 0: RB 333 lb
;
152000 #
lb in 50667 lb-in.
3
;
1600 lb
D
B
40 in.
RB
385
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CHAPTER 4 Shear Forces and Bending Moments
Problem 4.3-2 Determine the shear force V and bending moment M at the
6.0 kN
midpoint C of the simple beam AB shown in the figure.
2.0 kN/m
C
A
B
0.5 m
1.0 m
2.0 m
4.0 m
1.0 m
Solution 4.3-2
FREE-BODY DIAGRAM OF SEGMENT AC
g MA 0:
RB 3.9375 kN
g MB 0:
RA 5.0625 kN
g FVERT 0:
V RA 6 0.938 kN
g MC 0:
M RA # 2 m 6 kN # 1 m
4.12 kN # m
;
Problem 4.3-3 Determine the shear force V and bending moment M at the
Pb
P
midpoint of the beam with overhangs (see figure). Note that one load acts
downward and the other upward. Also clockwise moments Pb are applied
at each support.
;
Pb
b
L
P
b
Solution 4.3-3
Pb
Pb
Pb
FREE-BODY DIAGRAM (C IS THE MIDPOINT)
1
(2Pb (b + L)P Pb)
L
P (upward)
gMB 0: RA g MA 0:
g FVERT 0:
V RA P 0
g MC 0: M Pa b +
RB P (downward)
+ RA
;
L
b
2
L
+ Pb 0
2
;
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Page 387
SECTION 4.3 Shear Forces and Bending Moments
Problem 4.3-4 Calculate the shear force V and bending moment M at a cross
4.0 kN
section located 0.5 m from the fixed support of the cantilever beam AB shown
in the figure.
387
1.5 kN/m
A
B
1.0 m
1.0 m
2.0 m
Solution 4.3-4 Cantilever beam
4.0 kN
g FVERT 0:
1.5 kN/m
A
V 4.0 kN (1.5 kN/m)(2.0 m)
B
4.0 kN 3.0 kN 7.0 kN
1.0 m
1.0 m
2.0 m
g MD 0:
FREE-BODY DIAGRAM OF SEGMENT DB
;
M (4.0 kN)(0.5 m)
(1.5 kN/m)(2.0 m)(2.5 m)
2.0 kN # m 7.5 kN # m
Point D is 0.5 m from support A.
9.5 kN # m
Problem 4.3-5 Consider the beam with an overhang shown
18 ft
in the figure.
q = 300 lb/ft
400 lb/ft
(a) Determine the shear force V and bending moment M
at a cross section located 18 ft from the left-hand end A.
(b) Find the required magnitude of load intensity q, acting
on the right half of member BC that will result in a
zero shear force on the cross section 18 ft from A.
;
B
C
A
10 ft
10 ft
6 ft
6 ft
Solution 4.3-5
(a) V AND M AT X 18 FT
gM B 0
RA 1
10 ft
6 ft
c400 lb/ft 110 ft2 a 10 ft +
b + q 16 ft2 a 6 ft +
b d 3810 lb
20 ft
2
2
V18 RA 400 lb/ft 110 ft2 190 lb
M 18 RA 118 ft2 400 lb/ft 110 ft2 118 ft 5 ft2 16,580 lb-ft
400 lb/ft
M
RA
10 ft
8 ft
V
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CHAPTER 4 Shear Forces and Bending Moments
(b) REQUIRED q SO THAT SHEAR AT x 18 FT IS ZERO
RA 54 q ft 2 + 60,000 lb ft
1
10 ft
6 ft
b + q 16 ft2 a 6 ft +
bd c400 lb/ft 110 ft2 a10 ft +
20 ft
2
2
20 ft
V18 RA 400 lb/ft 110 ft2 V18 0 solve, q 54 q ft 2 + 60,000 lb # ft
4000 lb
20 ft
10,000 lb
370.4 lb/ft
27 ft
q is upward
Problem 4.3-6 The beam ABC shown in the figure is simply supported at A and B and has an overhang from B to C. The
loads consist of a horizontal force P1 4.0 kN acting at the end of a vertical arm and a vertical force P2 8.0 kN acting at
the end of the overhang.
(a) Determine the shear force V and bending moment
M at a cross section located 3.0 m from the lefthand support. (Note: Disregard the widths of the
beam and vertical arm and use centerline dimensions when making calculations.)
(b) Find the value of load P2 that results in V 0 at
a cross section located 2.0 m from the left-hand
support.
(c) If P2 8 kN, find the value of load P1 that results
in M 0 at a cross section located 2.0 m from the
left-hand support.
P1 = 4.0 kN
P2 = 8.0 kN
1.0 m
A
B
4.0 m
C
1.0 m
Solution 4.3-6
P1 4 P2 8
units: kN, m
(a) RHFB
RB 1
1P1 1 + P2 52 9
4
V3 RB + P2 1 kN M 3 RB 1 P2 2 7 kN # m
(b)
5 P2
1
14 1 + P2 52 1
4
4
P2
V2 RB + P2 1 V2 0
4
RB gives
P2 4 kN
(c) LHFB
P2 8
P1
1
1P1 1 + P2 52 10 4
4
P1
M 2 RA 2 P1 1 4
2
RB RA P2 RB P1
2
4
Let M 2 0 so P1 8 kN (acting to the right)
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Page 389
389
SECTION 4.3 Shear Forces and Bending Moments
Problem 4.3-7 The beam ABCD shown in the figure has overhangs at each
q
end and carries a uniform load of intensity q.
For what ratio b/L will the bending moment at the midpoint of the beam
be zero?
A
D
B
b
C
L
b
Solution 4.3-7 Beam with overhangs
FREE-BODY DIAGRAM OF LEFT-HAND HALF OF BEAM:
Point E is at the midpoint of the beam.
From symmetry and equilibrium of vertical forces:
L
RB RC q ab + b
2
gME 0 哵 哴
L
1
L 2
RB a b + qa b ab + b 0
2
2
2
L L
1
L 2
b a b + qa b ab + b 0
2
2
2
2
qa b +
Solve for b/L:
b
1
L
2
;
Problem 4.3-8 At full draw, an archer applies a pull of 130 N to the bowstring
of the bow shown in the figure. Determine the bending moment at the midpoint
of the bow.
70°
1400 mm
350 mm
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CHAPTER 4 Shear Forces and Bending Moments
Solution 4.3-8 Archer’s bow
FREE-BODY DIAGRAM OF SEGMENT BC
g MC 0
T(cos b)a
MTa
P 130 N
b 70
H 1400 mm
哵哴
H
b + T(sin b)(b) M 0
2
H
cos b + b sin b b
2
P H
a + b tan b b
2 2
SUBSTITUTE NUMERICAL VALUES:
1.4 m
b 350 mm
M
0.35 m
FREE-BODY DIAGRAM OF POINT A
130 N 1.4 m
c
+ (0.35 m)(tan 70) d
2
2
M 108 N # m
;
T tensile force in the bowstring
g FHORIZ 0: 2T cos b P 0
T
P
2 cos b
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Page 391
391
SECTION 4.3 Shear Forces and Bending Moments
Problem 4.3-9 A curved bar ABC is subjected to loads in the form of
two equal and opposite forces P, as shown in the figure. The axis of the
bar forms a semicircle of radius r.
Determine the axial force N, shear force V, and bending moment M
acting at a cross section defined by the angle u.
M
B
V
r
u
P
N
P
u
P
A
O
C
g FN 0
Q b
N P sin u 0
g FV 0
R a
g MO 0
哵哴
A
Solution 4.3-9 Curved bar
N P sin u
;
V P cos u 0
V P cos u
;
M Nr 0
M Nr Pr sin u
;
Problem 4.3-10 Under cruising conditions the distributed
load acting on the wing of a small airplane has the idealized
variation shown in the figure.
Calculate the shear force V and bending moment M at the
inboard end of the wing.
1600 N/m
2.6 m
Solution 4.3-10
900 N/m
2.6 m
1.0 m
Airplane wing
(Minus means the shear force acts opposite to the direction shown in the figure.)
LOADING (IN THREE PARTS)
SHEAR FORCE
g FVERT 0
V +
+
c T
1
(700 N/m)(2.6 m) + (900 N/m)(5.2 m)
2
1
1900 N/m2(1.0 m) 0
2
V 6040 N 6.04 kN
;
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CHAPTER 4 Shear Forces and Bending Moments
BENDING MOMENT
M 788.67 N # m 12,168 N # m 2490 N # m
g MA 0 哵哴
15,450 N # m
15.45 kN # m
1
2.6 m
M + (700 N/m) (2.6 m) a
b
2
3
+ (900 N/m) (5.2 m) (2.6 m)
1
1.0 m
+ (900 N/m) (1.0 m) a 5.2 m +
b 0
2
3
;
Problem 4.3-11 A beam ABCD with a vertical arm CE is supported as a simple beam at A and D (see figure a). A cable
passes over a small pulley that is attached to the arm at E. One end of the cable is attached to the beam at point B.
(a) What is the force P in the cable if the bending moment in the beam just to the left of point C is equal numerically to
640 lb-ft? (Note: Disregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.)
(b) Repeat (a) if a roller support is added at C and a shear release is inserted just left of C (see figure part b).
E
P
Cable
A
B
E
Cable
8 ft
C
P
D
A
8 ft
B
C
D
Shear release
6 ft
6 ft
6 ft
6 ft
(a)
6 ft
6 ft
(b)
Solution 4.3-11
(a) LHFB
RA 1
4P
18 P2 18
9
M C RA12 +
4
8P
P6 5
15
8 P
640 Solve: P 1200 lb
15
(b) LHFB
VC RA +
MC RA 12 +
4
P0
5
4
P 6 640
5
so
RA 4
P
5
4
400
4
P 12 + P 6 640 Solve: P lb
5
5
3
400
133.333
3
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393
SECTION 4.3 Shear Forces and Bending Moments
Problem 4.3-12 A simply supported beam AB supports a trapezoidally
distributed load (see figure). The intensity of the load varies linearly from
50 kN/m at support A to 25 kN/m at support B.
Calculate the shear force V and bending moment M at the midpoint of
the beam.
50 kN/m
25 kN/m
A
B
4m
Solution 4.3-12
FREE-BODY DIAGRAM OF SECTION CB
Point C is at the midpoint of the beam.
gMB 0: RA (4m) + (25 kN/m) (4m) (2m)
1
2
(25 kN/m)(4 m)a b a4 m b 0
2
3
RA 83.33 kN
gFVERT 0: RA + RB
1
(50 kN/m + 25 kN/m)(4 m) 0
2
RB 66.67 kN
gFVERT 0: V (25 kN/m)(2 m)
(12.5 kN/m)(2 m)
V 4.17 kN
1
+ RB 0
2
;
g MC 0: M (25 kN/m)(2 m)(1 m)
(12.5 kN/m)(2 m)
1
1
a2 m b
2
3
+ RB (2 m) 0
M 75 kN # m
;
Problem 4.3-13 Beam ABCD represents a reinforced-concrete foundation beam
that supports a uniform load of intensity q1 3500 lb/ft (see figure). Assume that
the soil pressure on the underside of the beam is uniformly distributed with
intensity q2.
(a) Find the shear force VB and bending moment MB at point B.
(b) Find the shear force Vm and bending moment Mm at the midpoint
of the beam.
q1 = 3500 lb/ft
B
C
A
D
3.0 ft
q2
8.0 ft
3.0 ft
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CHAPTER 4 Shear Forces and Bending Moments
Solution 4.3-13
Foundation beam
(b) V AND M AT MIDPOINT E
g FVERT 0:
‹ q2 q2(14 ft) q1(8 ft)
8
q 2000 lb/ft
14 1
g FVERT 0:
(a) V AND M AT POINT B
Vm 0
g MB 0:
;
g ME 0:
a FVERT 0:
VB 6000 lb
Vm (2000 lb/ft)(7 ft)
(3500 lb/ft)(4 ft)
Mm (2000 lb/ft)(7 ft)(3.5 ft)
(3500 lb/ft)(4 ft)(2 ft)
;
Mm 21,000 lb-ft
MB 9000 lb-ft
;
;
Problem 4.3-14 The simply-supported beam ABCD is loaded by a weight W 27 kN through the arrangement shown in
the figure part a. The cable passes over a small frictionless pulley at B and is attached at E to the end of the vertical arm.
(a) Calculate the axial force N, shear force V, and bending moment M at section C, which is just to the left of the vertical
arm. (Note: Disregard the widths of the beam and vertical arm and use centerline dimensions when making calculations.)
(b) Repeat (a) if a roller support is added at C and a moment release is inserted just left of C (see figure part b).
E
E
Cable
A
B
2.0 m
C
2.0 m
W = 27 kN
D
2.0 m
Cable
Moment release
1.5 m
A
1.5 m
B
2.0 m
C
2.0 m
D
2.0 m
W = 27 kN
(a)
(b)
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SECTION 4.3 Shear Forces and Bending Moments
395
Solution 4.3-14
(a) SUM MOMENT ABOUT A TO FIND RC, THEN USE RHFB TO FIND N, V, AND M AT C
RDy 1
W
12 W2 6
3
M C RDy 122 +
W 27 kN
VC RDy +
3
W
5
Simplify :
4W
15
NC 4
W (compression)
5
4
W 11.52 1.8667 W
5
NC 4
W 21.6 kN
5
VC 4
W 7.2 kN
15
M C 50.4 kN # m
(b)
RHFB:
g M Mrel 0
RDy 1 4
a W 1.5b 16.2 kN
2 5
RDy 162 + W 122
Entire FBD: g M A 0
RCy RHFB: NC 21.6 kN
VC 1RCy + RDy2 +
4
37.8 kN
3
W 5.4 kN
5
M C 0 (at moment release)
Problem 4.3-15 The centrifuge shown in the figure rotates in a horizontal plane
y
(the xy plane) on a smooth surface about the z axis (which is vertical) with an
angular acceleration a. Each of the two arms has weight w per unit length and
supports a weight W 2.0 wL at its end.
Derive formulas for the maximum shear force and maximum bending moment
in the arms, assuming b L/9 and c L/10.
c
L
b
W
x
W
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CHAPTER 4 Shear Forces and Bending Moments
Solution 4.3-15 Rotating centrifuge
SUBSTITUTE NUMERICAL DATA:
Tangential acceleration ra
Inertial force Mr a W 2.0 wL b W
ra
g
Maximum V and M occur at x b.
Vmax W
(L + b + c)a +
g
Lb
wa
x dx
g
91wL2a
30g
Mmax 229wL3a
75g
;
;
Wa
(L + b + c)
g
+
Mmax Lb
Vmax L
L
c
9
10
wLa
(L + 2b)
2g
;
Wa
(L + b + c)(L + c)
g
Lb
+
wa
x(x b)dx
g
Lb
Wa
(L + b + c)(L + c)
g
+
wL2a
(2L + 3b)
6g
;
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397
SECTION 4.5 Shear-Force and Bending-Moment Diagrams
Shear-Force and Bending-Moment Diagrams
When solving the problems for Section 4.5, draw the shear-force
and bending-moment diagrams approximately to scale and label all
critical ordinates, including the maximum and minimum values.
Problems 4.5-1 through 4.5-10 are symbolic problems and
Problems 4.5-11 through 4.5-24 are mostly numerical problems. The
remaining problems (4.5-25 through 4.5-40) involve specialized topics,
such as optimization, beams with hinges, and moving loads.
P
a
P
a
A
B
Problem 4.5-1 Draw the shear-force and bending-moment diagrams for
a simple beam AB supporting two equal concentrated loads P (see figure).
L
Solution 4.5-1
Simple beam
Problem 4.5-2 A simple beam AB is subjected to a counterclockwise
couple of moment M0 acting at distance a from the left-hand support
(see figure).
Draw the shear-force and bending-moment diagrams for this beam.
M0
A
B
a
L
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CHAPTER 4 Shear Forces and Bending Moments
Solution 4.5-2
Simple beam
Problem 4.5-3 Draw the shear-force and bending-moment diagrams for
q
a cantilever beam AB carrying a uniform load of intensity q over one-half of
its length (see figure).
A
B
L
—
2
L
—
2
Solution 4.5-3 Cantilever beam
Moment varies linearly from A to L/2 on moment diagram.
3qL2
MA = —
8
q
A
B
qL
RA = —
2
L
—
2
L
—
2
qL
2
—
V
0
M
0
3qL2
–—
8
qL2
–—
8
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399
SECTION 4.5 Shear-Force and Bending-Moment Diagrams
Problem 4.5-4 The cantilever beam AB shown in the figure
PL
M1 = —–
4
P
is subjected to a concentrated load P at the midpoint and
a counterclockwise couple of moment M1 PL/4 at
the free end.
Draw the shear-force and bending-moment diagrams for this
beam.
A
B
L
—
2
L
—
2
Solution 4.5-4 Cantilever beam
RA P
Problem 4.5-5 The simple beam AB shown in the figure is subjected
to a concentrated load P and a clockwise couple M1 PL/3 acting at the
third points.
Draw the shear-force and bending-moment diagrams for this beam.
MA PL
4
P
PL
M1 = —–
3
A
B
L
—
3
L
—
3
L
—
3
Solution 4.5-5
PL
M1 = —–
3
P
A
B
L
—
3
P
RA= —–
3
L
—
3
L
—
3
2P
RB= —–
3
P/3
V
0
Vmax = –2P/3
PL/9
Mmax = 2PL/9
M 0
–PL/9
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CHAPTER 4 Shear Forces and Bending Moments
Problem 4.5-6 A simple beam AB subjected to couples M1 and 3M1
M1
acting at the third points is shown in the figure.
Draw the shear-force and bending-moment diagrams for this beam.
3M1
A
B
L
—
3
L
—
3
L
—
3
Solution 4.5-6
M1
3M1
A
B
L
—
3
L
—
3
L
—
3
RA
V
RB
2M 1
L
0
7M1
3
5M 1
3
M
0
2M 1
3
2M 1
3
Problem 4.5-7 A simply supported beam ABC is loaded by a vertical
load P acting at the end of a bracket BDE (see figure).
B
A
C
(a) Draw the shear-force and bending-moment diagrams for beam ABC.
(b) Now assume that load P at E is directed to the right. Vertical dimension BD is L/5. Draw axial-force, shear-force and bending-moment
diagrams for ABC.
D
E
P
L
—
4
L
—
4
L
—
2
L
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401
SECTION 4.5 Shear-Force and Bending-Moment Diagrams
Solution 4.5-7
(a) REPLACE DOWNWARD LOAD P AT E WITH STATICALLY-EQUIVALENT LOAD P AND CW MOMENT PL /4 AT B
a MA 0
Vmax P
2
1
L
PL
P
aP +
b L
4
4
2
RC M max RC a
on AB
a Fy 0
RA P RC P
2
3L
3LP
b (just right of B)
4
8
(b) REMOVE DOWNWARD LOAD P; REPLACE HORIZONTAL LOAD P AT E WITH STATICALLY-EQUIVALENT x-LOAD P AT B, AND
CCW MOMENT PL/5 AT B
1 P L
P
a
(downward)
b L
5
5
g MA 0
RCy gFy 0
RAy RCy gFx 0
RAx P
Nmax P (tension on AB)
P
5
(upward)
(leftward)
Vmax P
5
M max 3LP
P 3 L
a b 5
4
20
P
Problem 4.5-8 A beam ABC is simply supported at A and B
and has an overhang BC (see figure). The beam is loaded by
two forces P and a clockwise couple of moment Pa at D that
act through the arrangement shown.
(just right of B)
P
Pa
D
A
(a) Draw the shear-force and bending-moment diagrams for
beam ABC.
(b) If moment Pa at D is replaced by moment M, find an
expression for M in terms of variables P and a so that
the reaction at B goes to zero. Plot the associated shearforce and bending-moment diagrams for beam ABC.
C
B
a
a
a
a
Solution 4.5-8
(a)
UFBD:
LFBD:
1
1Pa + Pa2 P
2a
g MO 0
VC gFV 0
VO 2 P P P
DOWNWARD LOAD P AT O AND ALSO AT C
1
[P a + P 13 a2] 2 P
2a
gM A 0
RB gFV 0
RA RB + 2 P 0
Vmax P,
Mmax Pa
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CHAPTER 4 Shear Forces and Bending Moments
(b)
UFBD:
VC 1
M + Pa
1Pa + M2 2a
2a
VO 2 P VC
LFBD:
RB sum moments about left end of upper beam to find downward
load VC applied on lower beam
sum vertical forces for upper beam to find downward load VO applied downward on lower beam by upper beam
1
[V a + VC 13 a2] 2a O
3M
3 Pa
M + Pa
+
+ a a2 P b
2
2
2a
RB 0 solve, M 3 P a
2a
6 CCW
find reaction at B
M 3 Pa
RB 0
RA RB + VO + VC 2 P
VC 1
1P a + M2 P
2a
VO 2 P VC 3 P
Problem 4.5-9 Beam ABCD is simply supported at B and C and has
overhangs at each end (see figure). The span length is L and each overhang
has length L/3. A uniform load of intensity q acts along the entire length of
the beam.
Draw the shear-force and bending-moment diagrams for this beam.
q
A
D
B
L
3
C
L
L
3
Solution 4.5-9 Beam with overhangs
x1 L
15
0.3727 L
6
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SECTION 4.5 Shear-Force and Bending-Moment Diagrams
403
Problem 4.5-10 Draw the shear-force and bending-moment diagrams for a cantilever beam AB acted upon by two
different load cases:
(a) A distributed load with linear variation and maximum intensity q0 (see figure part a).
(b) A distributed load with parabolic variation and maximum intensity q0 (see figure part b).
q(x) = x . q0
L
q0
q(x) =
⎯x
q0
√ L . q0
A
B
x
A
L
(a)
B
x
L
(b)
Solution 4.5-10
(a) SHEAR DIAGRAM IS QUADRATIC; MOMENT DIAGRAM IS CUBIC
L
x
L q0
q0 x 3
x
q0 dx at B
M1x2 q0 1x 2d 2
6. L
L0 L
L0 L
2
2
L q0
L q0
M max M1L2 M max at B
6
6
Vmax (b) SHEAR DIAGRAM IS CUBIC; MOMENT DIAGRAM IS QUARTIC
L
V max Vmax 2 L q0
x
q0 dx at B
A
L
3
L0
2 L q0
3
M max M1L2 x
M1x2 4 L2 q0
15
q0 1x 2 d A
L
L0
M max x
AL
15
4 q0x 2
4 L2 q0
at B
15
Problem 4.5-11 The simple beam AB supports a triangular load
of maximum intensity q0 10 lb/in. acting over one-half of the
span and a concentrated load P 80 lb acting at midspan
(see figure).
Draw the shear-force and bending-moment diagrams for this
beam.
q0 = 10 lb/in.
P = 80 lb
A
B
L =
—
40 in.
2
L =
—
40 in.
2
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CHAPTER 4 Shear Forces and Bending Moments
Solution 4.5-11
Simple beam
q0 = 10 lb/in.
MA 0: RB (80 in.) (80 lb)(40 in.)
P = 80 lb
1
2
(10 lb/in. )140 in.2(40 + 40 in.) 0
2
3
A
RB 206.7 lb
L =
—
40 in.
2
1
g FVERT 0: RA + RB80 lba10 lb/in. b(40 in.) 0
2
RA 73.3 lb
B
L =
—
40 in.
2
RA
RB
73.3 lb
V
0
–6.67 lb
–207 lb
2933 lb-in.
M 0
Problem 4.5-12 The beam AB shown in the figure supports a uniform load
of intensity 3000 N/m acting over half the length of the beam. The beam rests on
a foundation that produces a uniformly distributed load over the entire length.
Draw the shear-force and bending-moment diagrams for this beam.
3000 N/m
A
B
0.8 m
1.6 m
0.8 m
Solution 4.5-12 Beam with distributed loads
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SECTION 4.5 Shear-Force and Bending-Moment Diagrams
Problem 4.5-13 A cantilever beam AB supports a couple and a concentrated
load, as shown in the figure.
Draw the shear-force and bending-moment diagrams for this beam.
405
200 lb
400 lb-ft
A
B
5 ft
5 ft
Solution 4.5-13 Cantilever beam
Problem 4.5-14 The cantilever beam AB shown in the figure is
2.0 kN/m
subjected to a triangular load acting throughout one-half of its length
and a concentrated load acting at the free end.
Draw the shear-force and bending-moment diagrams for this beam.
2.5 kN
B
A
2m
2m
Solution 4.5-14
4.5 kN
V
2.5 kN
2.5 kN
0
M 0
0
–5 kN • m
–11.33 kN • m
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CHAPTER 4 Shear Forces and Bending Moments
Problem 4.5-15 The uniformly loaded beam ABC has simple supports
25 lb/in.
at A and B and an overhang BC (see figure).
A
Draw the shear-force and bending-moment diagrams for this beam.
C
B
72 in.
Solution 4.5-15
Beam with an overhang
3 kN • m
12 kN/m
Problem 4.5-16 A beam ABC with an overhang at one end supports
a uniform load of intensity 12 kN/m and a concentrated moment of
magnitude 3 kN # m at C (see figure).
Draw the shear-force and bending-moment diagrams for this beam.
A
C
B
1.6 m
Solution 4.5-16
48 in.
1.6 m
1.6 m
Beam with an overhang
3 kN • m
12 kN/m
A
C
B
1.6 m
1.6 m
RA
1.6 m
RB
15.34 kN
V
0 kN
0
–3.86 kN
max 9.80 kN • m 9.18 kN • m
3 kN • m
M 0
1.28 m
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407
SECTION 4.5 Shear-Force and Bending-Moment Diagrams
Problem 4.5-17 Consider the two beams below; they are loaded the same but have different support conditions. Which
beam has the larger maximum moment?
First, find support reactions, then plot axial force (N), shear (V) and moment (M) diagrams for both beams. Label all
critical N, V, and M values and also the distance to points where N, V, and/or M is zero.
PL
L
—
2
A
B
L
—
2
L
—
4
C
P
4 L
—
4
3
D
PL
Ay
Ax
Cy
(a)
PL
A
L
—
2
B
L
—
2
L
—
4
P
4 L
—
4 D
3
PL
Cy
Dy
Dx
(b)
Solution 4.5-17
BEAM (a):
g M A 0: Cy 0
N 0
1 4 5
a P Lb P (upward)
L 5 4
g FV 0: Ay 4
P
P Cy (downward)
5
5
g FH 0: Ax 3
P (right)
5
–3P/5(compression)
4P/5
0
V 0
–P/5
0
M 0
–PL /10
–PL /5
–11PL /10
–6PL /5
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CHAPTER 4 Shear Forces and Bending Moments
BEAM (b):
3P/5
g MD 0: Cy 2
2 4 1
a P Lb P (upward)
L 5 4
5
g FV 0: Dy 4
2
P Cy P (upward)
5
5
g FH 0: Dx N
0
V
0
2P/5
3
P (right)
5
–2P/5
⬖ The first case has the larger maximum moment
6
a PL b
5
PL /10
M
0
0
;
–PL
Problem 4.5-18 The three beams below are loaded the same and have the same support conditions. However, one has a
moment release just to the left of C, the second has a shear release just to the right of C, and the third has an axial release
just to the left of C. Which beam has the largest maximum moment?
First, find support reactions, then plot axial force (N), shear (V) and moment (M) diagrams for all three beams. Label all
critical N, V, and M values and also the distance to points where N, V, and/or M is zero.
PL at C
A
L
—
2
B
L
—
2
L
—
4
C
P
4 L
—
4 D
3
PL at B
Moment
release
Ax
Ay
Cy
Dy
(a)
PL at C
A
L
—
2
B
L
—
2
L
—
4
C
PL at B
Ax
Ay
P
3
Shear
release
Cy
4 L
—
4 D
Dv
(b)
PL at C
A
L
—
2
B
PL at B
Ay
Ax
L
—
2
Axial force
release
C
L P 4 L
—
—
4
4
3
Cx
Cy
(c)
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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409
SECTION 4.5 Shear-Force and Bending-Moment Diagrams
Solution 4.5-18
BEAM (a): MOMENT RELEASE
N
0
0
Ay P (upward)
Cy Dy –3P/5 (compression)
13
P (downward)
5
12
P (upward)
5
P
V
0
–8P/5
3
Ax P (right)
5
PL /2
M
–12P/5
PL
3PL /5
0
–PL /2
BEAM (b): SHEAR RELEASE
N
0
0
1
Ay P (upward)
5
1
Cy P (downward)
5
Dy 4
P (upward)
5
Ax 3
P (right)
5
–3P/5 (compression)
P/5
V
0
– 4P/5
M
PL /10
0
–9PL /10
BEAM (c): AXIAL RELEASE
N
–3P/5 (compression)
4P/5
V
0
M
0
Ax 0
Cx 3
P (right)
5
⬖ The third case has the largest maximum moment
6
a PL b
5
– 4PL /5
0
1
Ay P (downward)
5
Cy P (upward)
PL /5
–P/5
–PL /10
–11PL /10
–PL /5
– 6PL /5
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 4 Shear Forces and Bending Moments
Problem 4.5-19 A beam ABCD shown in the figure is simply
supported at A and B and has an overhang from B to C. The loads
consist of a horizonatal force P1 400 lb acting at the end of the
vertical arm and a vertical force P2 900 lb acting at the end of
the overhang.
Draw the shear-force and bending-moment diagrams for this
beam. (Note: Disregard the widths of the beam and vertical arm
and use centerline dimensions when making calculations.)
Solution 4.5-19 Beam with vertical arm
Problem 4.5-20 A simple beam AB is loaded by two segments
of uniform load and two horizontal forces acting at the ends of a vertical arm
(see figure).
Draw the shear-force and bending-moment diagrams for this beam.
Solution 4.5-20
Simple beam
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411
SECTION 4.5 Shear-Force and Bending-Moment Diagrams
Problem 4.5-21 The two beams below are loaded the same and have the same support conditions. However, the location of
internal axial, shear and moment releases is different for each beam (see figures). Which beam has the larger maximum
moment?
First, find support reactions, then plot axial force (N), shear (V) and moment (M) diagrams for both beams. Label all
critical N, V, and M values and also the distance to points where N, V, and/or M is zero.
MAz
PL
A
L
—
2
B
L
—
2
L
—
4
C
P
4 L
—
4 D
3
Ax
PL
Axial force
release
Ay
Shear
release
Moment
release
Cy
Dy
Dx
(a)
MAz
PL
A
L
—
2
B
L
—
2
L
—
4
C
P
3
4 L
—
4 D
Ax
PL
Ay
Shear
release
Axial force
release
Moment
release
Cy
Dy
Dx
(b)
Solution 4.5-21
Support reactions for both beams:
MAz 0, Ax 0, Ay 0
Cy 2
2
P ( upward), Dy P ( upward)
5
5
Dx 3
P ( rightward)
5
3P/5(tension)
N
0
2P/ 5
V 0
–2P/ 5
⬖ These two cases have the same maximum
moment (PL) ;
(Both beams have the same N, V, and M diagrams)
–PL /10
M 0
–PL
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CHAPTER 4 Shear Forces and Bending Moments
Problem 4.5-22 The beam ABCD shown in the figure has
10.6 kN/m
overhangs that extend in both directions for a distance of 4.2 m
from the supports at B and C, which are 1.2 m apart.
Draw the shear-force and bending-moment diagrams for this
overhanging beam.
5.1 kN/m
5.1 kN/m
A
D
B
C
4.2 m
4.2 m
1.2 m
Solution 4.5-22 Beam with overhangs
Problem 4.5-23 A beam ABCD with a vertical arm CE is supported as a simple beam at A and D (see figure). A cable
passes over a small pulley that is attached to the arm at E. One end of the cable is attached to the beam at point B. The
tensile force in the cable is 1800 lb.
(a) Draw the shear-force and bending-moment diagrams for beam ABCD. (Note: Disregard the widths of the beam and
vertical arm and use centerline dimensions when making calculations.)
(b) Repeat part (a) if a roller support is added at C and a shear release is inserted just left of C (see figure part b).
P
E
Cable
A
B
Cable
8 ft
C
D
P
E
A
8 ft
B
C
D
Shear release
6 ft
6 ft
(a)
6 ft
6 ft
6 ft
6 ft
(b)
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SECTION 4.5 Shear-Force and Bending-Moment Diagrams
413
Solution 4.5-23
(a) NO SHEAR RELEASE
1
4P
RA 18 P2 18
9
RD RA 4P
9
VBC RA + 0.8 P 0.35556 P
SHEAR DIAGRAM
0.356 P
0.444 P
0.444 P
MOMENT DIAGRAM
2.67 PL
2.67 PL
(b) ADD SUPPORT AT C AND SHEAR RELEASE JUST LEFT OF C
4
4
P
LHFB: VC RA + P 0 so RA 5
5
1
4P
1RA 12 + P 82 6
15
16 P
g Fy 0 RC 1RA + RD2 15
gM C 0 RD Entire FBD:
SHEAR DIAGRAM
0.267 P
0.8 P
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 414
CHAPTER 4 Shear Forces and Bending Moments
MOMENT DIAGRAM
1.6 PL
4.8 PL
MAz
Problem 4.5-24 Beams ABC and CD are supported
at A, C, and D, and are joined by a hinge (or moment release)
just to the left of C and a shear release just to the right of C.
The support at A is a sliding support (hence reaction
Ay 0 for the loading shown below). Find all support
reactions then plot shear (V) and moment (M) diagrams.
Label all critical V and M values and also the distance to
points where either V and/or M is zero.
q0 = P/L
A
L
—
2
L
—
B 2
C
Ax
L
—
2
PL
Ay
Moment
release
Sliding
support
Cy
Dy
Solution 4.5-24
MAz PL (clockwise), Ax 0, Ay 0
;
1
1
P (upward), Dy P (upward)
12
6
;
Cy Vmax P
Mmax PL
6
P/12
V 0
0.289L
–P/6
PL
;
0.016PL
M 0
x = 13.5 ft
Problem 4.5-25 The simple beam AB shown in the figure supports a
P=5k
concentrated load and a segment of uniform load.
(a) Draw the shear-force and bending-moment diagrams for this beam.
(b) Find the value of P that will result in zero shear at x 13.5 ft. Draw
shear-force and bending-moment diagrams for this case.
2.0 k/ft
C
A
5 ft
B
10 ft
x
20 ft
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Page 415
SECTION 4.5 Shear-Force and Bending-Moment Diagrams
415
Solution 4.5-25
(a)
1
15 * 5 + 2 * 10 * 152 13.75 k RA 5 RB + 2 * 10 1.25 k
20
V13a RB + 2 6.5 0.75
RB (b) 6.5 FT FROM B, USE RHFB
1
P
15 P + 2 * 10 * 152 15 RB 20
4
P8
RB V13b RB + 2 * 6.5 P
2.0 20 6.5 13.5
4
So apply P 8 k upward.
1
15 P + 2 * 10 * 152 13
20
RA RB P + 2 1102 1
Problem 4.5-26 The cantilever beam shown in the figure supports a concentrated
3 kN
load and a segment of uniform load.
Draw the shear-force and bending-moment diagrams for this cantilever beam.
1.0 kN/m
A
0.8 m
B
0.8 m
1.6 m
Solution 4.5-26 Cantilever beam
Problem 4.5-27 The simple beam ACB shown in the figure is subjected
to a triangular load of maximum intensity q0 180 lb/ft at a 6.0 ft, and
a concentrated moment M 300 lb-ft at A.
(a) Draw the shear-force and bending-moment diagrams for this beam.
(b) Find the value of distance a that results in the maximum moment
occurring at L/2. Draw the shear-force and bending-moment diagrams
for this case.
(c) Find the value of distance a for which Mmax is the largest possible
value.
q0 = 180 lb/ft
M = 300 lb-ft
A
B
C
a = 6.0 ft
L = 7.0 ft
a 6 L 7 M 0 300 q 180 (Units in feet and pounds)
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CHAPTER 4 Shear Forces and Bending Moments
Solution 4.5-27
(a) FIND REACTIONS THEN USE TO FORM EXPRESSIONS FOR V (x) AND M(x); M Mmax AT LOCATION OF ZERO V
RB 1
1
2
1
L a
cM 0 + q a a b a + q 1L a2 a a +
b d 432.85 lb
L
2
3
2
3
V RA 1
x
qa bx
2
a
M max M 0 + RA x If V 0: x m 2 RA a
3.625 ft
B q
RA 1
q L RB 197.143 lb
2
x xm
1
x
x
q a b x 776.469 lb-ft
2
a
3
180 lb/ft
300 lb-ft
A
B
C
6.0 ft
7.0 ft
197.1
0 lb
V 0
(lb)
3.625 ft
Max 776
–343
–433
403
300
M
0
(lb-ft)
(b) FIND VALUE OF DISTANCE A FOR WHICH SHEAR V IS ZERO (i.e., LOCATION OF Mmax) AT X L/2 3.5 ft
V 0 when
1
q a RA—substitute expression for RA, then solve numerically for a.
2
1
1
1
2
1
La
bdda
2 c q L cM 0 + q a a b a + q 1L a2 a a +
2
L
2
3
2
3
S
q
7.9471
b
4.624328
Second solution for a is valid:
3.5
Solve: a a
RB xm a 4.624328 ft
1
1
2
1
L a
cM 0 + q a a b a + q 1L a2 a a +
b d 391.58 lb
L
2
3
2
3
B
2 RA a
3.5 ft
q
M max M 0 + RA x m RA 1
q L RB 238.413 lb
2
xm
xm
1
q a b xm
856.29 lb-ft
2
a
3
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SECTION 4.5 Shear-Force and Bending-Moment Diagrams
417
(c) FIND VALUE OF a (i.e., LOCATION WHERE Q(x) HAS PEAK INTENSITY) FOR WHICH Mmax IS LARGEST POSSIBLE
(1) Write expressions for reactions and maximum moment in terms of variable a:
RB 1
1
2
1
L a
1770
b d 30 a +
cM 0 + q a a b a + q 1L a2 a a +
L
2
3
2
3
7
RA 1
2640
q L RB 30 a
2
7
xm 2 RA a
121 288 a 7 a 2
B q
21
x xm
1
x
x
1760 21848 a 147 a 2
20 a 21848 a 147 a 2
qa bx + 300
2
a
3
147
21
M max M 0 + RA x (2) Differentiate expressions for Mmax with respect to a. Set expression equal to zero and solve for a:
10 a 1294 a 18482
880 1294 a 18482
d
20 21848 a 147 a2
Mmax :
2
da
21
21 21848 a 147 a
147 21848 a 147 a2
22
3.143
7
Substitute a into reaction and moment expressions to get larget possible Mmax:
Solve numerically:
RB 30 a +
a 1770
1
347.143 lb RA q L RB 282.857 lb
7
2
M maxx M 0 + RA x xm 2 RAa
3.143 ft
B q
1
x
x
q a b x 892.653 lb-ft
2
a
3
Problem 4.5-28 A beam with simple supports is subjected to a trapezoidally
3.0 kN/m
distributed load (see figure). The intensity of the load varies from 1.0 kN/m
at support A to 3.0 kN/m at support B.
Draw the shear-force and bending-moment diagrams for this beam.
1.0 kN/m
A
B
2.4 m
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Page 418
CHAPTER 4 Shear Forces and Bending Moments
Solution 4.5-28
V 2.0 x Set V 0:
Simple beam
x2
2.4
(x meters; V kN)
x1 1.2980 m
Problem 4.5-29 A beam of length L is being designed to support a uniform load
q
of intensity q (see figure). If the supports of the beam are placed at the ends,
creating a simple beam, the maximum bending moment in the beam is ql 2/8.
However, if the supports of the beam are moved symmetrically toward the middle
of the beam (as pictured), the maximum bending moment is reduced.
Determine the distance a between the supports so that the maximum bending
moment in the beam has the smallest possible numerical value.
Draw the shear-force and bending-moment diagrams for this condition.
A
B
a
L
Solution 4.5-29 Beam with overhangs
Solve for a:
a (2 22) L 0.5858L
M1 M2 q
(L a)2
8
qL2
(3 222) 0.02145qL2
8
;
;
The maximum bending moment is smallest when
M1 M2 (numerically).
M1 q(L a)2
8
qL2
qL
a
(2a L)
M2 RA a b 2
8
8
M1 M2
(L a)2 L(2a L)
x1 0.3536 a
0.2071 L
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419
SECTION 4.5 Shear-Force and Bending-Moment Diagrams
Problem 4.5-30 The compound beam ABCDE shown in the figure
consists of two beams (AD and DE) joined by a hinged connection at D.
The hinge can transmit a shear force but not a bending moment. The loads
on the beam consist of a 4-kN force at the end of a bracket attached at
point B and a 2-kN force at the midpoint of beam DE.
Draw the shear-force and bending-moment diagrams for this
compound beam.
4 kN
1m
B
C
1m
A
E
2m
Solution 4.5-30
2 kN
D
2m
2m
2m
Compound beam
Problem 4.5-31 Draw the shear-force and bending-moment diagrams for beam AB, with a sliding support at A and an
elastic support with spring constant k at B acted upon by two different load cases.
(a) A distributed load with linear variation and maximum intensity q0 (see figure part a).
(b) A distributed load with parabolic variation and maximum intensity q0 (see figure part b).
q(x) =
y
(x)
A
q
Linear
⎯
q0
√ Lx q0
q0
B x
B x
L
L
k
k
(a)
(b)
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CHAPTER 4 Shear Forces and Bending Moments
Solution 4.5-31
(a) SHEAR DIAGRAM IS QUADRATIC; MOMENT DIAGRAM IS CUBIC
RB L2 q0
1
1
2
q0 L M A RB L + q0 L a L b 2
2
3
6
Vmax RB L q0
2
M max M A L2 q0
6
(b) SHEAR DIAGRAM IS CUBIC; MOMENT DIAGRAM IS QUARTIC
L
RB 2L q0
x
q0 dx 3
L0 A L
Vmax RB 2 L q0
3
M A RB L + RB
M max M A 4L2 q0
3L
5
15
4 L2 q0
15
Problem 4.5-32 The shear-force diagram for a simple beam is
shown in the figure.
Determine the loading on the beam and draw the bendingmoment diagram, assuming that no couples act as loads on the
beam.
5 kN
5 kN
V
0
0
1m
1m
–10 kN
1m
–10 kN
1m
Solution 4.5-32
BEAM WITH LOADING AND REACTION FORCES (kN, meters)
5.000 kN/m
5.000 kN/m
15 kN
R = 10 kN
R = 5 kN
5.000 kN/m
5.000 kN/m
10 kN• m
MOMENT DIAGRAM (kN # m)
5 kN • m
2.5 kN • m
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421
SECTION 4.5 Shear-Force and Bending-Moment Diagrams
Problem 4.5-33 The shear-force diagram for a beam is
4 ft
16 ft
shown in the figure. Assuming that no couples act as loads
on the beam, determine the forces acting on the beam and
draw the bending-moment diagram.
180 lb
150 lb
V
150 lb
150 lb
0
0
4 ft
30 lb
–300 lb
Solution 4.5-33
BEAM WITH LOADING AND REACTION FORCES (lbs, ft)
150 lb
R = 450 lb
R = 150 lb
30 lb/ft
150 lb
30 lb/ft
897.6 lb-ft
MOMENT DIAGRAM (lb-ft)
360 lb-ft
60 lb-ft
Problem 4.5-34 The compound beam below has an internal
moment release just to the left of B and a shear release just to
the right of C. Reactions have been computed at A, C, and D
and are shown in the figure.
First, confirm the reaction expressions using statics, then
plot shear (V) and moment (M) diagrams. Label all critical
V and M values and also the distance to points where either
V and/or M is zero.
w0 L2
MA = ––––
12
w0
w0
A
B
L
L
Ax = 0
—
—
2
2 Moment
release
w0 L
w0 L
Ay = ––––
Cy = ––––
6
3
C
D
L
—
2
Shear
release
–w0 L
Dy = ––––
4
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CHAPTER 4 Shear Forces and Bending Moments
Solution 4.5-34
FREE-BODY DIAGRAM
w0 L2
––––
12
w0
0
w0 L
––––
6
w0 L
––––
6
V
w0 L
––––
6
w0 L
––––
6
w0
w0 L2
––––
24
w0 L2
––––
24
w0 L
––––
3
w0 L
––––
4
–w0 L
––––
4
0
–w0 L
––––
3
L
6
––––
w0 L2
––––
72
M 0
–w0 L2
––––
12
L
––––
3
–w0 L2
––––
24
Problem 4.5-35 The compound beam below has an shear
release just to the left of C and a moment release just to the
right of C. A plot of the moment diagram is provided below
for applied load P at B and triangular distributed loads w(x) on
segments BC and CD.
First, solve for reactions using statics, then plot axial
force (N) and shear (V) diagrams. Confirm that the moment
diagram is that shown below. Label all critical N, V, and M values
and also the distance to points where N, V, and/or M is zero.
w0
A
B
L
—
2
w0 L2
––––
30
M
w0
w0 L
P = ––––
2
4
3
C
L
—
2
Shear
release
D
L
—
2
Moment
release
2w0 L2
–––––
125
–w0 L2
–––––
24
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SECTION 4.5 Shear-Force and Bending-Moment Diagrams
423
Solution 4.5-35
Solve for reactions using statics.
M A w0 L2/30 (clockwise),
Ax 3
w0L (left)
10
Ay 3
w L (downward)
20 0
;
;
Cy w0
L (upward)
12
;
Dy w0
L (upward)
6
;
;
Vmax w0 L/4, M max w0 L2/24 at B
FREE-BODY DIAGRAM
–w0 L2/30
3w0 L /10
3w0 L/20
w0 L2/24 w0 L2/24
w0 L/2
w0 L/4
w0
w0
w0 L/4
w0 L/12
w0 L /6
3w0L/10 (tension)
N 0
w0L/4
w0L/12
V
0
0.289L
–3w0L/20
w0L2/30
–w0L/6
2w0L2/125
M 0
–w0L2/24
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CHAPTER 4 Shear Forces and Bending Moments
Problem 4.5-36 A simple beam AB supports two connected wheel loads
P and 2P that are distance d apart (see figure). The wheels may be placed
at any distance x from the left-hand support of the beam.
P
x
(a) Determine the distance x that will produce the maximum shear force
in the beam, and also determine the maximum shear force Vmax.
(b) Determine the distance x that will produce the maximum bending
moment in the beam, and also draw the corresponding bendingmoment diagram. (Assume P 10 kN, d 2.4 m, and L 12 m.)
Solution 4.5-36
2P
d
A
B
L
Moving loads on a beam
P = 10 kN
d = 2.4 m
L = 12 m
Reaction at support B:
2P
P
P
x +
(x + d) (2d + 3x)
L
L
L
Bending moment at D:
RB MD RB (L x d)
(a) MAXIMUM SHEAR FORCE
By inspection, the maximum shear force occurs at support B when the larger load is placed close to, but not
directly over, that support.
P
(2d + 3x) (L x d)
L
P
[3x2 + (3L 5d)x + 2d(L d)]
L
Eq.(1)
dMD
P
(6x + 3L 5d) 0
dx
L
x
Solve for x:
L
5d
a3 b 4.0 m
6
L
;
Substitute x into Eq (1):
P
L 2
5d 2
c3a b a 3 b + (3L 5d)
L
6
L
Mmax x L d 9.6 m
;
d
Vmax RB P a3 b 28 kN
L
5d
L
b + 2d(L d) d
* a b a3 6
L
;
(b) MAXIMUM BENDING MOMENT
By inspection, the maximum bending moment occurs at
point D, under the larger load 2P.
Note:
PL
d 2
a3 b 78.4 kN # m
12
L
RA P
d
a 3 + b 16 kN
2
L
RB P
d
a 3 b 14 kN
2
L
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 425
425
SECTION 4.5 Shear-Force and Bending-Moment Diagrams
Problem 4.5-37 The inclined beam represents a ladder with
the following applied loads: the weight (W) of the house painter
and the distributed weight (w) of the ladder itself.
6f
t
B
W = 150 lb
(a) Find support reactions at A and B, then plot axial force (N ),
shear (V), and moment (M) diagrams. Label all
critical N,V, and M values and also the distance to points
where any critical ordinates are zero. Plot N,V, and M
diagrams normal to the inclined ladder.
(b) Repeat part (a) for the case of the ladder suspended from
a pin at B and traveling on a roller support perpendicular
to the floor at A.
θ
Bx
w
18
=2
.5
ft
lb/
ft
θ
θ
θ
A
Ax
θ
Ay
θ
8 ft
Solution 4.5-37
(a) LADDER WITH PIN AT BASE
8
1
2 12
, sin u 18 + 6
3
3
Solution procedure:
(1) Use statics to find reaction forces at A and B.
cos u g FV 0: Ay 150 + 2.5118 + 62 210 lb
Ay 210 lb 1upward2
;
Bx 50.38 lb 1left2
;
g M A 0: Bx (24 sin u) + 150 * 6 + 2.5 * 24 * 4 0
g FH 0; Ax 50.38 lb 1right2
;
(2) Use u to find forces at ends A and B which are along and perpendicular to member AB (see free-body diagram);
also resolve forces W and w into components along and perpendicular to member AB.
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 426
CHAPTER 4 Shear Forces and Bending Moments
(3) Starting at end A, plot N, V, and M diagrams (see plots).
lb
1.4
14
θ=
B
sin
–30.98 lb
W
270 lb·ft
7.5 lb
–172.4 lb
–42.5 lb
b/f
t
lb/
wc
os
θ=
57
2.3
θ=
in
ws
Ax cos θ
+ Ay sin θ = 214.8 lb
Bx sin θ = 47.5 lb
0.8
33
l
ft
W cos θ = 50 lb
–47.5 lb
–16.79 lb
Bx cos θ = –16.79 lb
N
22.5 lb
V
–214.8 lb
M
A
Ay cos θ – Ax sin θ = 22.5 lb
(b) LADDER WITH PIN AT TOP
Use statics to find reactions at A and B.
Ax 0, Ay 67.5 lb, Bx 0, By 142.5 lb
V DIAGRAM
N DIAGRAM
134.4 lb
–47.5 lb
22.5 lb
–63.6 lb
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 427
427
SECTION 4.5 Shear-Force and Bending-Moment Diagrams
M DIAGRAM
270 ft-lb
Problem 4.5-38 Beam ABC is supported by
MD
a tie rod CD as shown (see Prob. 10.4-15).
Two configurations are possible: pin support
at A and downward triangular load on AB, or
pin at B and upward load on AB. Which has
the larger maximum moment?
First, find all support reactions, then plot
axial force (N), shear (V) and moment (M)
diagrams for ABC only and label all critical
N, V, and M values. Label the distance to
points where any critical ordinates are zero.
Dy
Dx
D
Moment
releases
q0 at B
y
L
—
4
r q(x)
Linea
Ax
A
L
L
—
4P=q L
0
B
x
C
L
—
2
PL
Ay
(a)
Solution 4.5-38
FREE-BODY DIAGRAM—BEAM (a)
4q0L/9
q0L/2
7q0L2/9
7q0L2/9
q0L/2
q0L/2
q0L/2
q0L/2
4q0L/9
q0L
q0L2
17q0L/18
4q0L/9
q0L/2
4q0L/9
4q0L/9
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
78572_ch04_ptg01_hr_385-434.qxd
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Page 428
CHAPTER 4 Shear Forces and Bending Moments
Use statics to find reactions at A and D for beam (a).
1
Ax q0L (left)
2
1
Dx q0L (left)
2
;
17
Ay q L (upward)
18 0
4
Dy q0L (downward)
9
;
MD 0
–4q0L/9
q0L/2(tension)
0
4q0L/9
0
q0L2
q0L2/4
0
7q0L2/9
M
q0L/2
17q0L/18
V
C 0
B
–q0L/2
A
;
;
D
N
;
(compression)
428
1/6/12
0
0
MD
Dy
Dx
D
Moment
releases
q0 at B
y
r q(x)
Linea
A
L
—
4 P=q L
0
L
—
4
B
L
L
—
2
By
x
C
PL
Bx
(b)
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Page 429
SECTION 4.5 Shear-Force and Bending-Moment Diagrams
429
FREE-BODY DIAGRAM—BEAM (b)
5q0L/3
q0L/2
q0L2/6
q0L2/6
q0L2
q0L/2 q0L/2
q0L/2
5q0L/3
q0L/2
q0L
5q0L/3
5q0L/3
Use statics to find reactions at B and D for beam (b).
Bx 1
q0L (right)
2
;
1
5
7
By q0L + q0L q0L (upward)
2
3
6
Dx 1
q L (right)
2 0
;
5
Dy q0L (downward)
3
MD 0
;
;
;
0
B
A
C
0
q0L/2
q0L/2
(compression)
5q0L/3
q0L/2
0
0
–q0L2/4
V
–q0L/2
N
–5q0L/3
(compression)
D
q0L2
q0L2/6
M 0
0
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 430
CHAPTER 4 Shear Forces and Bending Moments
Problem 4.5-39
The plane frame below consists of column AB and beam BC which carries a triangular distributed
load (see figure part a). Support A is fixed, and there is a roller support at C. Column AB has a moment release just
below joint B.
(a) Find support reactions at A and C, then plot axial force (N), shear-force (V), and bending-moment (M) diagrams
for both members. Label all critical N, V, and M values and also the distance to points where any critical ordinates
are zero.
(b) Repeat part (a) if a parabolic lateral load acting to the right is now added on column AB (figure part b).
q0
q(x) =
x
B
C
B
L
x q
L 0
q0
C
L
Moment
release
RCy
√
Moment
release
⎯
q(y) = 1– y q0
2L
2L
2L
y
A
q0
A
RAx
RAy
(b)
MA
(a)
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Page 431
SECTION 4.5 Shear-Force and Bending-Moment Diagrams
431
Solution 4.5-39
(a) USE STATICS TO FIND REACTIONS AT A AND C
MA 0
RAy RCy ;
q0
L (upward)
6
;
B
B
RAx 0
;
B
0
;
B
N
–w0L/6 (compression)
q0
L (upward)
3
C
0
w0L/6
0
0
V
0
0.5774L
–w0L/3
0.06415w0L2
A
A
0
0
A
0
N
V
M
M 0
(b) REACTIONS AT A AND C
(1) Sum moments about moment release of FBD of BC to find reaction RCy, then sum vertical forces to get RAy.
RCy L q0
1 1
2
c q0 L a Lb d L 2
3
3
RAy L q0
1
q0 L RCy 2
6
(2) Sum forces in x-direction for entire FBD to get RAx.
2L
y
4 L q0
b q0 dy A
2L
3
L0
(3) Sum moments about A for entire FBD to get reaction moment MA.
RAx 2L
MA L0
a
1
y
16 L2 q0
1
2
a 1
b q0 y dy + q0 L a Lb RCy L A
2L
2
3
15
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 432
CHAPTER 4 Shear Forces and Bending Moments
Problem 4.5-40 The plane frame shown below is part of an
elevated freeway system. Supports at A and D are fixed but
there are moment releases at the base of both columns (AB and
DE), as well near in column BC and at the end of beam BE.
Find all support reactions, then plot axial force (N), shear (V)
and moment (M) diagrams for all beam and column members.
Label all critical N, V, and M values and also the distance to
points where any critical ordinates are zero.
750 N/m
C
F
45 kN
Moment
release
7m
1500 N/m
E
B
18 kN
7m
19 m
A
D
Dx
Ax
MA
Ay
MD
Dy
Solution 4.5-40
Solution procedure:
(4) g MB 0 for AB: Ax 0
(1) MA MD 0 due to moment releases
(2) g MA 0: Dy 61,164 N 61.2 kN
(5) g FH 0: Dx 63 kN
(6) Draw separate FBD’s of each member (see below)
to find N, V, and M for each member; plot diagrams
(see below).
(3) g Fy 0: Ay 18,414 N 18.41 kN
756 kN•m
756 kN•m
FREE-BODY DIAGRAM
750 N/m
C C
441kN·m
32.7 kN
1500 N/m
B
32.7 kN
B
18.41 kN
B
A
18.41 kN
14.25 kN
E
14.25 kN
46.9 kN
F
F
46.9 kN
441kN·m
32.7 kN
E
45 kN
45 kN
46.9 kN
61.2 kN
E 63 kN
D
63 kN
61.2 kN
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Page 433
SECTION 4.5 Shear-Force and Bending-Moment Diagrams
433
0
F
–46.9 kN
32.7 kN
C
B
E
A
D
–61.2 kN
18.41 kN
0
AXIAL FORCE DIAGRAM
() COMPRESSION
F
C
–32.7 kN
–46.9 kN
45 kN
0
14.25 kN
E
–14.25 kN
B
63 kN
0
A
D
SHEAR FORCE DIAGRAM
756 kN•m F
C
0
0
756 kN•m
67.7 kN•m
B
0
E
D
A
BENDING MOMENT DIAGRAM
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78572_ch04_ptg01_hr_385-434.qxd
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Page 434
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
78572_ch05_ptg01_hr_435-556.qxd
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4:25 AM
Page 435
5
Stresses in Beams
(Basic Topics)
d
Longitudinal Strains in Beams
Problem 5.4-1 A steel wire of diameter d 1/16 in. is bent around a cylindrical drum of radius
R 36 in. (see figure).
R
(a) Determine the maximum normal strain max.
(b) What is the minimum acceptable radius of the drum if the maximum normal strain must
remain below yield? Assume E 30,000 ksi and sY 100 ksi.
(c) If R 36 in., what is the maximum acceptable diameter of the wire if the maximum normal
strain must remain below yield?
Solution 5.4-1
Steel wire
(a) sY 100 ksi
E 30,000 ksi
R 36 in.
(b)
d
1
in.
16
1
1
1.093 * 103
r
mm
d
2
Rmin +
CHECK
d
2
Y
sY
3.333 * 103
E
d
r R + 915.194 mm
2
Y max y
r
d
2
max 8.673 * 104
r
solving for Rmin 9.34375 in.
d
2
d
Rmin +
2
Rmin 9.35 in.
3.331 * 103
435
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Page 436
CHAPTER 5 Stresses in Beams (Basic Topics)
(c) R 36 in.
dmax
2
Y
dmax
R +
2
solving for dmax gives
dmax 0.24 in.
CHECK
dmax
2
3.322 * 103
dmax
R +
2
Problem 5.4-2 A copper wire having diameter d 4 mm is bent into a circle and held
d = diameter
with the ends just touching (see figure).
(a) If the maximum permissible strain in the copper is max 0.0024, what is the
shortest length L of wire that can be used?
(b) If L 5.5 m, what is the maximum acceptable diameter of the wire if the maximum
normal strain must remain below yield? Assume E 120 GPa and sY 300 MPa.
Solution 5.4-2
L = length
Copper wire
d
2
r
d
2
pd
L
L
2p
L
(a) max 0.0024 r d 4 mm
2p
Lmin pd
5.24 m
max
(b) E 120 GPa
dmax sY 300 MPa
Y sY
2.5 * 103
E
L 5.5 m
Y L
4.38 mm
p
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78572_ch05_ptg01_hr_435-556.qxd
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Page 437
SECTION 5.4 Longitudinal Strains in Beams
437
Problem 5.4-3 A 4.75-in. outside diameter polyethylene pipe
designed to carry chemical wastes is placed in a trench and bent
around a quarter-circular 90 bend (see figure). The bent section of the
pipe is 52 ft long.
(a) Determine the maximum compressive strain max in the pipe.
(b) If the normal strain cannot exceed 6.1 * 103, what is the maximum diameter of the pipe?
(c) If d 4.75 in., what is the minimum acceptable length of the
bent section of the pipe?
Solution 5.4-3
(a) L 52 ft
Polyethylene pipe
r
max (b) a 6.1 A 103 B
(c) d 4.75 in.
90°
L
397.251 in.
p
2
d 4.75 in.
d
2
pd
4L
2L
a b
p
pd
5.98 * 103
4L
dmax Lmin a 14 L2
p
4.85 in.
pd
51 ft
4 a
Problem 5.4-4 A cantilever beam AB is loaded by a couple M0 at its free end (see figure).
The length of the beam is L 2.0 m, and the longitudinal normal strain at the top surface
is 0.0010. The distance from the top surface of the beam to the neutral surface is
c 85 mm.
(a) Calculate the radius of curvature r, the curvature k, and the vertical deflection d at
the end of the beam.
(b) If allowable strain a 0.0008, what is the maximum acceptable depth of the
beam? (Assume that the curvature is unchanged from Part (a)).
(c) If allowable strain a 0.0008, c 85 mm, and L 4 m, what is deflection d?
d
A
B
L
M0
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4:25 AM
Page 438
CHAPTER 5 Stresses in Beams (Basic Topics)
Solution 5.4-4
Let b 40 mm.
(a) 0.0010
c 85 mm
L2m
r
c
85 m
u
u
L
0.02353 d L tan a b
r
2
h 2c 170 mm
so M0a E a
1
1
0.01176
r
m
E 210 GPa
bh3
b 40.46 kN # m
12
d
0.012
L
d 23.5 mm
or d r 11 cos1u22 23.5 mm
(b) b 0.0008 cmax rb 68 mm hmax 2cmax 136 mm
so M0b E a
L
84.996
d
ASSUMES THAT CURVATURE IS SAME AS IN (a)
bh max3
b 20.716 kN # m
12
ALTERNATE SOLUTION: assume that APPLIED MOMENT M0 is SAME AS IN (A) (minus sign cancels in final result)
a a
M0
bc
EIa a
so
b a
M0
bc
EIb b
cb b 12 cb23
cb a
12
c
b 12 ca2
3
12
d
M0 EIa
ca a
EIb
M0 b
or
b
c3b b
b ca 2
a
ca a
hb 2 cb hb 2 185 mm2
0.0010
C 0.0008
cb b 12cb23
Ib b
c
Ia a a
so c2b cb M0a (c) L 4 m
12
85 mm
10.00102 40.46 kN # m M0b a 0.0008
u
d L tan a b
2
d
0.019
L
d
r
b12ca2
12
d
190.1 mm where cb 185 mm2
a
b
bc
a a
c
106.25 m
a
u
0.0010
C 0.0008
95.033 mm
hb
95.033 mm
2
Now in alternate solution:
b 1170 mm23
c
3
a 2
a
ca or cb ca
b
A b
or
Ec
12
Ec
b 1hb23
12
hb
2
d
10.00082 40.46 kN # m
L
0.038
r
d 75.3 mm or d r 11 cos 1u22 75.285 mm
L
53.131
d
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78572_ch05_ptg01_hr_435-556.qxd
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Page 439
SECTION 5.4 Longitudinal Strains in Beams
Problem 5.4-5 A thin strip of steel with a length of L 19 in. and thickness of
M0
439
M0
t 0.275 in. is bent by couples M0 (see figure). The deflection at the midpoint
of the strip (measured from a line joining its end points) is found to be 0.30 in.
d
t
(a) Determine the longitudinal normal strain at the top surface of the strip.
(b) If allowable strain a 0.0008, what is the maximum acceptable thickness
of the strip?
(c) If allowable strain a 0.0008, t 0.275 in., and L 32 in., what is
deflection d?
(d) If allowable strain a 0.0008, t 0.275 in., and the deflection cannot
exceed 1.0 in., what is the maximum permissible length of the strip?
L
—
2
L
—
2
Solution 5.4-5
(a) L 19 in.
t 0.275 in.
d r 11 cos 1u22
d 0.30 in.
L
2
sin 1u2 r
t
2
0.1375 in.
r
r
or approx. u L
2r
so d r a 1 cos a
L
bb
2r
SOLVE NUMERICALLY FOR r:
r a1 cos a
L
bb d
2r
t
2
9.144 * 104
r
compressive
r 150.367 in.
(b) a 0.0008 tmax 2 r a 0.241 in.
1.143
a
t
2
(c) r 171.875 in.
a
d r a 1 cos a
L 32 in.
d
0.023
L
(d) a 0.0008 t 0.275 in. dmax 1.0 in. r L
b b 0.744 in.
2r
L
43
d
t
171.875 in.
2 a
SOLVE NUMERICALLY FOR L:
r a1 cos a
L
b b dmax
2r
Lmax 37.1 in.
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Page 440
CHAPTER 5 Stresses in Beams (Basic Topics)
Problem 5.4-6 A bar of rectangular cross section is loaded and
h
P
supported as shown in the figure. The distance between supports is
L 1.75 m, and the height of the bar is h 140 mm. The deflection
at the midpoint is measured as 2.5 mm.
(a) What is the maximum normal strain at the top and bottom
of the bar?
(b) If allowable strain a 0.0006 and the deflection cannot exceed
4.3 mm, what is the maximum permissible length of the bar?
d
P
a
L
—
2
L
—
2
a
Solution 5.4-6
(a) L 1.75 m
h 140 mm
d r 11 cos 1u22
d 2.5 mm
L
2
sin 1u2 r
h
2
h
r
2r
or approx. u L
2r
so d r a 1 cos a
L
bb
2r
SOLVE NUMERICALLY FOR r:
L
r a1 cos a b b d
2r
(b) a 0.0006 r h
2
r 153.125 m 4.57 * 104
r
h
116.667 m dmax 4.3 mm
2 a
SOLVE NUMERICALLY FOR L:
r a1 cos a
L
bb dmax
2r
Lmax 2 m
Normal Stresses in Beams
A thin strip of hard copper (E 16,000 ksi) having length
L 90 in. and thickness t 3/32 in. is bent into a circle and held with the
ends just touching (see figure).
3
t = — in.
32
Problem 5.5-1
(a) Calculate the maximum bending stress smax in the strip.
(b) By what percent does the stress increase or decrease if the thickness of
the strip is increased by 1/32 in.?
(c) Find the new length of the strip so that the stress in part (b) (t 1/8 in.
and L 90 in.) is equal to that in part (a) (t 3/32 in. and L 90 in.).
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78572_ch05_ptg01_hr_435-556.qxd
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4:25 AM
Page 441
441
SECTION 5.5 Normal Stresses in Beams
Solution 5.5-1
E 16,000 ksi
r
L 90 in.
t
3
in.
32
L
14.324 in.
2p
t
2
smax E ± ≤ 52.4 ksi
r
(a)
E
(b) s maxnew ±
r
t +
1
in.
32
≤ 69.813 ksi
2
s maxnew smax
33.3% same as
smax
t +
(c) E ±
1
in.
32
2
Lnew
2p
≤ smax
solving
at +
1
in.b t
32
t
33.3%
Lnew 120 in.
Lnew L
33.3%
L
Problem 5.5-2 A steel wire (E 200 GPa) of diameter d 1.25 mm
is bent around a pulley of radius R0 500 mm (see figure).
(a) What is the maximum stress smax in the wire?
(b) By what percent does the stress increase or decrease if the radius of the pulley is
increased by 25%?
(c) By what percent does the stress increase or decrease if the diameter of the wire
increased by 25% while the pulley radius remains at R0 500 mm?
R0
d
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CHAPTER 5 Stresses in Beams (Basic Topics)
Solution 5.5-2
E 200 GPa
(a) smax E ±
d 1.25 mm
d
2
R0 +
(b) s maxnew E ±
d
2
R0 500 mm
1.25 d
2
≤ 312.012 MPa
s maxnew E ±
1.25 d
R0 +
2
≤ 250 MPa
d
2
1.25 R0 +
d
2
1.25d 1.563 mm
(c)
s maxnew smax
25%
smax
≤ 199.8 MPa
s maxnew smax
19.98%
smax
A thin, high-strength steel rule (E 30 106 psi)
having thickness t 0.175 in. and length L 48 in. is bent by couples
M0 into a circular arc subtending a central angle a 40° (see figure).
L = length
Problem 5.5-3
t
(a) What is the maximum bending stress smax in the rule?
(b) By what percent does the stress increase or decrease if the central angle is
increased by 10%?
(c) What percent increase or decrease in rule thickness will result in the maximum
stress reaching the allowable value of 42 ksi?
M0
M0
a
Solution 5.5-3
E 30,000 ksi
t 0.175 in. L 48 in.
a 40
(a) r L
68.755 in.
a
t
2
smax E ± ≤ 38.2 ksi
r
(c) sa 42 ksi t 0.175 in. a 40
L 48 in.
t
2
≤ sa tnew 0.193 in.
E±
L
a
tnew t
10.01%
t
t
2
≤ 41.997 ksi
(b) s maxnew E ±
L
1.1 a
s maxnew smax
10%
smax
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443
SECTION 5.5 Normal Stresses in Beams
q
Problem 5.5-4
A simply supported wood beam AB with span length
L 4 m carries a uniform load of intensity q 5.8 kN/m (see figure).
(a) Calculate the maximum bending stress smax due to the load q if the
beam has a rectangular cross section with width b 140 mm and
height h 240 mm.
(b) Repeat part (a) but use the trapezoidal distributed load shown in the
figure part b.
A
h
B
b
L
(a)
q
—
2
q
A
B
L
(b)
Solution 5.5-4
(a) MAX. BENDING STRESS DUE TO UNIFORM LOAD q
2
qL
8
Mmax bh3
12
S
h
2
smax S
S
RA c
I
h
2
find x location of zero shear
qL2
8
smax 1
a bh2 b
6
RA L4m
6 L 1184 L22
2(3)
x
1
a 1 + 184b
L
6
xmax 0.52753 L
qL2
8
Mmax 11.6 kN # m
Mmax smax 8.63 MPa
b 140 mm
q
1 x q
x a
bx 0
2
2 L2
3x 2 + 6Lx 4L2 0
x
3 L2
smax q 2
4 bh
kN
q 5.8
m
h 240 mm
1 q1
1 q
a bL + a
b Ld
2 2
3 22
uniform load (q/2) & triang. load (q/2)
1
RA qL
3
1 2
bh
6
Mmax
S
(b) MAX. BENDING STRESS DUE TO TRAPEZOIDAL LOAD q
Mmax RAxmax q xmax2
1 xmax q xmax2
a
b
2 2
2 L 2
3
Mmax 9.40376 * 102 qL2
;
Mmax 8.727 kN # m
smax Mmax
S
smax 6.493 * 103
N
m2
smax 6.49 MPa
;
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CHAPTER 5 Stresses in Beams (Basic Topics)
Problem 5.5-5 Each girder of the lift bridge (see figure) is
180 ft long and simply supported at the ends. The design load
for each girder is a uniform load of intensity 1.6 k/ft. The girders
are fabricated by welding three steel plates so as to form an
I-shaped cross section (see figure) having section modulus
S 3600 in.3.
What is the maximum bending stress smax in a girder due
to the uniform load?
Solution 5.5-5
Bridge girder
L 180 ft
q 1.6 k/ft
S 3600 in.
3
Mmax qL2
8
smax qL2
Mmax
S
8S
smax (1.6 k/ft)(180 ft)2(12 in./ft)
8(3600 in.3)
Problem 5.5-6 A freight-car axle AB is loaded approximately as shown in
the figure, with the forces P representing the car loads (transmitted to the axle
through the axle boxes) and the forces R representing the rail loads (transmitted
to the axle through the wheels). The diameter of the axle is d 82 mm, the
distance between centers of the rails is L, and the distance between the forces
P and R is b 200 mm.
Calculate the maximum bending stress smax in the axle if P 47 kN.
21.6 ksi
P
;
P
B
A
d
d
R
b
R
L
b
Solution 5.5-6
NUMERICAL DATA
d 82 mm
b 220 mm
P 50 kN
I
pd 4
I 2.219 * 106 m4
64
Mmax Pb
MAX. BENDING STRESS
smax Md
2I
smax 203 MPa
;
Mmax 11 kN # m
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445
SECTION 5.5 Normal Stresses in Beams
Problem 5.5-7 A seesaw weighing 3 lb/ft of length is occupied by two
children, each weighing 90 lb (see figure). The center of gravity of each
child is 8 ft from the fulcrum. The board is 19 ft long, 8 in. wide, and
1.5 in. thick.
What is the maximum bending stress in the board?
Solution 5.5-7
Seesaw
b 8 in.
h 1.5 in.
q 3 lb/ft
P 90 lb
d 8.0 ft
L 9.5 ft
2
Mmax Pd +
qL
720 lb-ft + 135.4 lb-ft
2
855.4 lb-ft 10,264 lb-in.
S
2
bh
3.0 in.3
6
smax 10,264 lb-in.
M
3420 psi
S
3.0 in.3
Problem 5.5-8 During construction of a highway bridge,
the main girders are cantilevered outward from one pier toward
the next (see figure). Each girder has a cantilever length of
48 m and an I-shaped cross section with dimensions shown
in the figure. The load on each girder (during construction)
is assumed to be 9.5 kN/m, which includes the weight of
the girder.
Determine the maximum bending stress in a girder due to
this load.
;
52 mm
2600 mm
28 mm
620 mm
Solution 5.5-8
NUMERICAL DATA
tf 52 mm
h 2600 mm
L 48 m
I
tw 28 mm
bf 620 mm
q 9.5
kN
m
L
Mmax qL a b
2
Mmax h
smax 2I
smax 101 MPa
Mmax 1.094 * 104 kN # m
;
1
1
(b ) h3 (b tw) [ h 2 (tf)]3
12 f
12 f
I 1.41 * 1011 mm4
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CHAPTER 5 Stresses in Beams (Basic Topics)
Problem 5.5-9 The horizontal beam ABC of an oil-well pump
has the cross section shown in the figure. If the vertical pumping force acting at
end C is 9 k and if the distance from the line of action of that force to point B is
16 ft, what is the maximum bending stress in the beam due to the pumping
force?
Horizontal beam transfers loads as part of oil well pump
C
B
A
0.875 in.
22 in.
0.625 in.
8.0 in.
Solution 5.5-9
NUMERICAL DATA
FC 9 k
MAX. BENDING STRESS AT B
BC 16 ft
Mmax FC (BC)
Mmax 144 k-ft
1
1
I
(8)(22)3 (8 0.625)
12
12
* [22 2 (0.875)]3
smax Mmax (12) a
22
b
2
I
smax 9.53 ksi
;
I 1.995 * 103 in.4
Problem 5.5-10 A railroad tie (or sleeper) is subjected to two rail
loads, each of magnitude P 175 kN, acting as shown in the figure.
The reaction q of the ballast is assumed to be uniformly distributed
over the length of the tie, which has cross-sectional dimensions
b 300 mm and h 250 mm.
Calculate the maximum bending stress smax in the tie due to
the loads P, assuming the distance L 1500 mm and the overhang
length a 500 mm.
P
a
P
L
a
b
h
q
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447
SECTION 5.5 Normal Stresses in Beams
Solution 5.5-10 Railroad tie (or sleeper)
DATA
P 175 kN
Substitute numerical values:
a 500 mm
M1 17,500 N # m
2P
bh2
S
3.125 * 103 m3
L + 2a
6
Mmax 21,875 N # m
L 1500 mm
q
h 250 mm
b 300 mm
M2 21,875 N # m
MAXIMUM BENDING STRESS
BENDING-MOMENT DIAGRAM
smax 21,875 N # m
Mmax
7.0 MPa
5
3.125 * 103 m3
;
(Tension on top; compression on bottom)
M1 qa2
Pa2
2
L + 2a
M2 2
q L
PL
a + ab 2 2
2
2
L
PL
P
a + ab L + 2a 2
2
P
(2a L)
4
Problem 5.5-11 A fiberglass pipe is lifted by a sling, as shown in the
figure. The outer diameter of the pipe is 6.0 in., its thickness is 0.25 in., and
its weight density is 0.053 lb/in.3 The length of the pipe is L 36 ft and the
distance between lifting points is s 11 ft.
(a) Determine the maximum bending stress in the pipe due to its own
weight.
(b) Find the spacing s between lift points which will minimize the
bending stress. What is the minimum bending stress?
(c) What spacing s will lead to maximum bending stress? What is that
stress?
s
L
Solution 5.5-11
NUMERICAL DATA AND CROSS-SECTIONAL PROPERTIES OF PIPE
L 36 ft d2 6 in. t 0.25 in. d1 d2 2 t 5.5 in. A g 0.053
lb
3
in.
q g A 0.239
lb
in.
s 11 ft a p
1d 2 d1 22 4.516 in.2
4 2
Ls
p
12.5 ft I 1d 4 d1 42 18.699 in.4
2
64 2
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CHAPTER 5 Stresses in Beams (Basic Topics)
BENDING-MOMENT DIAGRAM
M1 qa2
2,692.7 lb-in.
2
M2 qL L
a sb 2,171.4 lb-in.
4 2
Mmax 2,692.7 lb-in.
(a) MAXIMUM BENDING STRESS
smax smax Mmaxc
1
c
d2
3.0 in.
2
12,692.7 lb- in.213.0 in.2
18.699 in.4
432 psi
;
(Tension on top)
(b) IF M1
M2 0, THE BENDING STRESS IS MINIMIZED
M11s2 q
a
Ls 2
b
2
M21s2 2
q L L
a sb
4
2
Let M1 1s2 + M2 1s2 0 solving s 0.58579 L
M1 10.58579 L2 0.021446 L2 q M2 10.58579 L2 0.021447 L2 q s 21.1 ft
MINIMUM BENDING STRESS
c
d2
2
s min1 M11s2 c
I
153.692 psi s min2 Compare to solution for stresses in Part (a):
s max1 M1 111 ft2 c
I
432 psi s max2 M21s21c2
I
M2111 ft21c2
I
153.701 psi smin s min1 153.7 psi
348 psi
(c) EITHER M1,max (S 0) OR M2,max (S L) WILL LEAD TO MAXIMUM BENDING STRESS
(1) Support at L/2 so 1/2 of beam is a cantilever with max. moment
and s max1 M1102 c
I
895.795 psi s max2 M21021c2
I
q L2
8
so
q L2 c
895.795 psi
8 I
895.795 psi
(2) OR simply supported beam (s L) under uniform load, so max. moment is once again
and
s max1 M11L2 (c)
I
^ moment is zero
at pin support
0 psi
so maximum bending stress is
s max2 M21L2 (c)
q L2
8
895.795 psi
I
^ moment is qL2/8 at L/2
smax 896 psi
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SECTION 5.5 Normal Stresses in Beams
Problem 5.5-12 A small dam of height h 2.0 m is constructed of
vertical wood beams AB of thickness t 120 mm, as shown in the figure.
Consider the beams to be simply supported at the top and bottom.
Determine the maximum bending stress smax in the beams, assuming
that the weight density of water is g 9.81 kN/m3
449
A
h
t
B
Solution 5.5-12 Vertical wood beam
MAXIMUM BENDING MOMENT
RA q0 L
6
q0 x 3
6L
q0 Lx
q0 x 3
6
6L
q0L
q0x 2
dM
L
0 x
dx
6
2L
13
M RAx h 2.0 m
t 120 mm
g 9.81 kN/ m3 (water)
Let b = width of beam perpendicular to the plane
of the figure
Substitute x L/13 into the equation for M:
Let q0 = maximum intensity of distributed load
Mmax q0 gbh S bt 2
6
q0L
q0
q0 L2
L
L3
a
b a
b 6
6L 313
13
9 13
For the vertical wood beam: L h; Mmax q0 h 2
9 13
MAXIMUM BENDING STRESS
smax 2q0 h 2
2gh 3
Mmax
S
313 bt 2
313 t 2
Substitute numerical values:
smax 2.10 MPa
;
NOTE: For b 1.0 m, we obtain q0 19,620 N/m,
S 0.0024 m3,
Mmax 5,034.5 N # m, and smax Mmax/S 2.10 MPa
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CHAPTER 5 Stresses in Beams (Basic Topics)
y
Problem 5.5-13
Determine the maximum tensile stress st
(due to pure bending about a horizontal axis through C
by positive bending moments M ) for beams having cross
sections as follows (see figure).
x
xc
C
(a) A semicircle of diameter d
(b) An isosceles trapezoid with bases b1 b and b2 4b/3,
and altitude h
(c) A circular sector with p/3 and r d/2
x
xc
b1
C
h
y
a
xc
C
a
r
d
b2
O
(a)
(b)
(c)
x
Solution 5.5-13
MAX. TENSILE STRESS DUE TO POSITIVE BENDING MOMENT
IS ON BOTTOM OF BEAM CROSS-SECTION
r4
(a + sin (a) cos(a))
4
Ix (a) SEMICIRCLE
ybar From Appendix E, Case 10:
(9p 2 64)r 4
(9p 2 64)d 4
72p
1152p
Ic c
;
A d2 a
c
2a
p
b
12
d
b
2
3
From Appendix E, Case 8:
c
h3(b21 + 4b1b2 + b22)
36(b1 + b2)
73bh
756
3
Mc
360M
Ic
73bh2
Ix a
d 4
b
2
4
A 0.2618 d 2
p
sin a b
3
±
≤
p
3
a
d 2 p
b a b
2
3
c 0.276 d
p
p
p
+ sin a b cos a b b
3
3
3
Ix 0.02313 d 4
h(2b1 + b2)
10h
3(b1 + b2)
21
st A a
For a p/3, r d/2:
(b) ISOSCELES TRAPEZOID
IC c ybar
d1
2d
4r
3p
3p
Mc
768M
M
st 30.93 3
2
3
Ic
(9p 64)d
d
2r sin (a)
a
b
3
a
;
(c) CIRCULAR SECTOR WITH a p/3, r d/2
IC Ix A y2bar
IC cd 4
(4p 3 13)
p
d 13 2
d 2 a b c a
bd d
768
12
2 p
IC 3.234 * 103 d 4
max. tensile stress st From Appendix E, Case 13:
Mc
IC
st 85.24
M
d3
;
A r 2 (a)
Problem 5.5-14 Determine the maximum bending stress smax
(due to pure bending by a moment M) for a beam having a cross
section in the form of a circular core (see figure). The circle has
diameter d and the angle b 60. (Hint: Use the formulas given
in Appendix E, Cases 9 and 15.)
C
b
b
d
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SECTION 5.5 Normal Stresses in Beams
Solution 5.5-14
Circular core
From Appendix E, Cases 9 and 15:
C
b
Iy b
r
d
r4
ab
pr 4
2ab 3
aa 2 +
b
4
2
r
r4
p
d
a b
2
2
b radians a radians a r sin b b r cos b
Iy pd 4
d4 p
a b sin b cos b + 2 sin b cos3 b b
64
32 2
pd 4
d4 p
a b (sin b cos b)(1 2 cos2 b) b
64
32 2
4
451
4
d p
1
pd
a b a sin 2b b (cos 2b)b
64
32 2
2
d4 p
1
pd 4
a b + sin 4b b
64
32 2
4
d4
(4 b sin4 b)
128
MAXIMUM BENDING STRESS
smax smax c r sin b 64M sin b
d
sin b
2
;
d (4b sin 4b)
3
For b 60 p/3 rad:
576M
M
smax 10.96 3
(8p 13 + 9)d3
d
Problem 5.5-15 A simple beam AB of span length L 24 ft is subjected
to two wheel loads acting at distance d 5 ft apart (see figure).
Each wheel transmits a load P 3.0 k, and the carriage may
occupy any position on the beam.
(a) Determine the maximum bending stress max due to the
wheel loads if the beam is an I-beam having section modulus S 16.2 in.3
(b) If d 5 ft, find the required span length L to reduce the
maximum stress in part (a) to 18 ksi.
(c) If L 24 ft, find the required wheel spacing s to reduce the
maximum stress in part (a) to 18 ksi.
Mc
Iy
P
d
A
;
P
B
C
L
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CHAPTER 5 Stresses in Beams (Basic Topics)
Solution 5.5-15 Wheel loads on a beam
NUMERICAL DATA
P
x
L 24 ft d 5 ft P 3 kip
x
L
d
10.75 ft
2
4
x +
d
13.25 ft S 16.2 in.3
2
P
d
A
RA
B
L
MAXIMUM BENDING MOMENT
P
P
P
L x + (L x d) (2L d 2x)
L
L
L
P
M RA x (2L x dx 2x2)
L
dM
P
L
d
(2L d 4x) 0 x dx
L
2
4
RA (a) SUBSTITUTE x INTO THE EQUATION FOR M:
Mmax P
d 2
aL b
2L
2
MAXIMUM BENDING STRESS
Mmax
P
d 2
aL b
S
2LS
2
Substitute numerical values:
smax smax 3k
21288 in.2 116.2 in.32
21.4 ksi
;
1288 in. 30 in.22
;
(b) MAX. MOMENT AND STRESS EXPRESSIONS FROM ABOVE
Mmax P
d 2
P
d 2
aL b smax 1L, d2 aL b
2L
2
2LS
2
Set smax expression equal to 18 ksi, then solve for L; only first solution below is valid.
Pd + 36S ksi + 622 2S ksi (Pd + 18S ksi)
d 2
P
20.901
2P
aL b 18 ksi solving for L ≥
¥ a
b ft Lreqd 20.9 ft
2L S
2
0.299
Pd + 36S ksi 622 2S ksi (Pd + 18S ksi)
2P
(c) IF L 24 FT, FIND REQUIRED WHEEL SPACING d TO REDUCE THE MAXIMUM STRESS IN (A) TO 18 KSI
Set smax expression equal to 18 ksi, then solve for d:
d 2
LS
P
aL b 18 ksi Solving gives d 2 L 12
8.564 ft so dreqd 8.56 ft
2LS
2
B P
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453
SECTION 5.5 Normal Stresses in Beams
Problem 5.5-16
Determine the maximum tensile stress st and
maximum compressive stress sc due to the load P acting on the
simple beam AB (see figure).
(a) Data are as follows: P 6.2 kN, L 3.2 m, d 1.25 m,
b 80 mm, t 25 mm, h 120 mm, and h1 90 mm.
(b) Find the value of d for which tensile and compressive
stresses will be largest. What are these stresses?
t
P
d
A
B
h1
h
L
b
Solution 5.5-16
NUMERICAL DATA AND CROSS SECTION PROPERTIES
P 6.2 kN L 3.2 m d 1.25 m b 80 mm
Sum moments about base to find NA:
b
c2 Ib
1h h122
2
b 1h h12 + h1 t
1h h123
12
+ h1 t ah h1 +
h1
b
2
+ b 1h h12 cc2 a
t 25 mm
h 120 mm
h1 90 mm
44.032 mm c1 h c2 75.968 mm
h h1 2
th1 3
h1 2
bd +
+ th1 a c1 b 5.879 * 106m4
2
12
2
(a) MAX. TENSILE AND COMPRESSIVE STRESSES
Mmax Pd
1L d2 4.72 kN # m
L
sc Mmax c1
61 MPa
I
st Mmax c2
35.4 MPa
I
(b) FIND EXPRESSION FOR MOMENT AS A FUNCTION OF DISTANCE d, THEN TAKE FIRST DERIVATION TO FIND VALUE OF
DISTANCE d AT WHICH MOMENT IS MAXIMUM
M1d2 Pd
1L d2
L
Pdmax
d
L
M1d2 0 solving for d 0.5 L so dmax and Mmax 1L dmax2 4.96 kN # m
dd
2
L
Tensile and compressive stresses are:
st Mmax c2
37.1 MPa
I
sc Mmax c1
64.1 MPa
I
Problem 5.5-17 A cantilever beam AB, loaded by a uniform load and a
concentrated load (see figure), is constructed of a channel section.
(a) Find the maximum tensile stress st and maximum compressive stress sc
if the cross section has the dimensions indicated and the moment of
inertia about the z axis (the neutral axis) is I 3.36 in.4 (Note: The uniform load represents the weight of the beam.)
(b) Find the maximum value of the concentrated load if the maximum tensile
stress cannot exceed 4 ksi and the maximum compressive stress is limited
to 14.5 ksi.
(c) How far from A can load P 250 lb be positioned if the maximum tensile stress cannot exceed 4 ksi and the maximum compressive stress is
limited to 14.5 ksi?
250 lb
22.5 lb/ft
B
A
5.0 ft
3.0 ft
y
z
C
0.617 in.
2.269 in.
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 5 Stresses in Beams (Basic Topics)
Solution 5.5-17
NUMERICAL DATA
c1 0.617 in. c2 2.269 in. I 3.36 in.4 a 5 ft b 3 ft q 22.5
(a) Mmax q (L)2
+ Pa
2
sc Mmax c2
15.96 ksi
I
st lb
ft
L a + b P 250 lb
Mmax c1
4.34 ksi
I
(b) FIND AN EXPRESSION FOR THE MAX. TENSILE STRESS IN TERMS OF LOAD P, EQUATE IT TO THE ALLOWABLE VALUE, THEN
SOLVE FOR Pmax
c
q 1L22
2
+ Pa d c1
4 ksi solving for P I
Repeat for max. compressive stress:
c
q 1L22
2
+ Pa dc2
I
4.0 I ksi 32.0 c1 ft2 q
219 lb
ac1
14.5 ksi solving for P 0.5 (29.0 I ksi 64.0 c2 ft2 q
214 lb compression on bottom
ac2
controls Pmax 214 lb
(c) FIND AN EXPRESSION FOR THE MAX. TENSILE STRESS IN TERMS OF DISTANCE a TO LOAD P FROM SUPPORT A, EQUATE IT
TO THE ALLOWABLE VALUE, THEN SOLVE FOR amax
c
q 1L22
2
+ Pa d c1
4 ksi solving for a I
Repeat for max. compressive stress:
c
q 1L22
2
+ Pa d c2
I
4.0 I ksi 32.0 c1 ft2 q
4.38 ft
Pc1
14.5 ksi solving for a 0.5 (29.0 I ksi 64.0 c2 ft2 q
4.28 ft
Pc2
compression on bottom
controls a 4.28 ft
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SECTION 5.5 Normal Stresses in Beams
q
Problem 5.5-18 A cantilever beam AB of isosceles trapezoidal
cross section has length L 0.8 m, dimensions b1 80 mm,
b2 90 mm, and height h 110 mm (see figure). The beam is
made of brass weighing 85 kN/m3.
455
b1
C
h
L
(a) Determine the maximum tensile stress st and maximum
compressive stress sc due to the beam’s own weight.
(b) If the width b1 is doubled, what happens to the stresses?
(c) If the height h is doubled, what happens to the stresses?
b2
Solution 5.5-18
NUMERICAL DATA
MAX. TENSILE STRESS AT SUPPORT (TOP)
g 85
L 0.8 m
b1 80 mm
kN
st 3
m
b 2 90 mm
(a) MAX. STRESSES DUE TO BEAM’S OWN WEIGHT
q L2
2
q gA
A
1
(b1 + b 2) h
2
A 9.35 * 103 mm2
q 7.9475 * 102
Ih
3
h (2b1 b2)
3 (b1 b2)
1 b21 4 b1 b2
sc Mmax ybar
I
ybar 53.922 mm
;
1
(b1 + b2) h A 1.375 * 104 mm2
2
q 1.169 * 103
N
m
qL2
2
Mmax 374 N # m
Mmax 36 (b1 b2)
I 9.417 * 106 mm4
sc 1.456 MPa
b1 160 mm
q gA
b222
;
(b) DOUBLE b1 & RECOMPUTE STRESSES
A
N
m
Mmax 254.32 N # m
ybar st 1.514 MPa
MAX. COMPRESSIVE STRESS AT SUPPORT (BOTTOM)
h 110 mm
Mmax Mmax (h ybar)
I
ybar h (2 b1 + b2)
3 (b1 + b2)
ybar 60.133 mm
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CHAPTER 5 Stresses in Beams (Basic Topics)
I h3
1b21 + 4 b1 b2 + b222
Mmax 36 (b1 + b2)
I 1.35 * 107 mm4
ybar MAX. TENSILE STRESS AT SUPPORT (TOP)
st Mmax (h ybar)
I
st 1.381 MPa
I h3
;
qL2
2
Mmax 508.64 N # m
h(2b1 + b2)
3(b1 + b2)
ybar 107.843 mm
(b12 + 4b1b2 + b22)
361b1 + b22
I 7.534 * 107 mm4
MAX. COMPRESSIVE STRESS AT SUPPORT (BOTTOM)
sc Mmax ybar
2
MAX. TENSILE STRESS AT SUPPORT (TOP)
sc 1.666 MPa
;
st Mmax (h ybar)
I
st 0.757 MPa
;
(c) DOUBLE h & RECOMPUTE STRESSES
MAX. COMPRESSIVE STRESS AT SUPPORT (BOTTOM)
b1 80 mm h 220 mm
A
1
(b + b2) h
2 1
q gA
A 1.87 * 104 mm2
q 1.589 * 103
sc Mmax ybar
I
sc 0.728 MPa
;
N
m
Problem 5.5-19 A beam ABC with an overhang from B to C supports a
uniform load of 200 lb/ft throughout its length (see figure). The beam is a
channel section with dimensions as shown in the figure. The moment of
inertia about the z axis (the neutral axis) equals 8.13 in.4
q = 200 lb/ft
A
B
(a) Calculate the maximum tensile stress st and maximum compressive
stress sc due to the uniform load.
(b) Find required span length a that results in the ratio of larger to smaller
compressive stress being equal to the ratio of larger to smaller tensile
stress for the beam. Assume that the total length L a b 18 ft
remains unchanged.
a = 12 ft
C
b = 6 ft
y
0.787 in.
z
C
2.613 in.
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 5.5 Normal Stresses in Beams
457
Solution 5.5-19
NUMERICAL DATA
q 200
lb
ft
I 8.13 in.4
a 12 ft b 6 ft L a + b
c1 0.787 in. c2 2.613 in.
(a) STATICS:
q L2
2
RB a
MB qb2
2
Find reactions and moment at B and max. moment on span AB.
RA qL RB
xmax M maxAB RA xmax q
MmaxAB aqL qL RA
q
xmax qL2
2a
q
9 ft
2
x2max
2
2
qL
L2
b aL b q
2a
2a
L2 2
b
2a
aL 2
2025 ft-lb MmaxAB L2q 1L 2 a22
8 a2
COMPUTE MAX. TENSILE AND COMPRESSIVE STRESSES on span AB and at B:
MmaxAB c1
2352 psi 6 comp.
I
stAB MB c1
4182 psi 6 tens.
I
scB scAB stB (b) THE STRESS RATIOS FROM PART (A) ARE:
scB
5.903
scAB
MmaxAB c2
7810 psi 6 tens.
I
MB c2
13,885 psi 6 comp.
I
stAB
1.868 in Part (b), we want these ratios to be
stB
the same for tension and compression
Write general expression for ratio of compressive stresses, then repeat for tensile stresses.
MB c2
scB
I
MmaxAB c1
scAB
I
q b2
c2
2
I
scB
2
scAB
L q 1L 2 a22
8 a2
I
c1
L2q 1L 2a22
MmaxAB c2
stAB
I
MB c1
stB
I
stAB
stB
8 a2
I
qb2
c1
2
I
4 a2b2c2
L2c1 1L 2 a22
c2
L2c2 1L 2 a22
4 a2b2c2
compression
tension
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CHAPTER 5 Stresses in Beams (Basic Topics)
Equate above expressions and then solve for a:
4 a2 1L a22 c2
L2 c2 1L 2a22
L2 c1 1L 2a22
4 a2 1L a22 c1
solving a L
12.728 ft
12
a 12.73 ft
CONFIRM THAT STRESS RATIOS ARE EQUAL IF a 12.73 ft
a
L
12.728 ft b L a 5.272 ft L a + b 18 ft
12
scAB MmaxAB c1
3229 psi
I
6 comp.
stB MBc1
3229 psi
I
stAB MmaxAB c2
10,720 psi
I
6 tens.
scB MBc2
10,720 psi
I
So ratios are:
scB
3.32
scAB
L2 q 1L 2a22
MmaxAB 8a
2
MB qb2
2
6 tens.
6 comp.
stAB
3.32
stB
A
Problem 5.5-20
A frame ABC travels horizontally with an acceleration
a0 (see figure). Obtain a formula for the maximum stress smax in the
vertical arm AB, which had length L, thickness t, and mass density r.
t
a0 = acceleration
L
B
C
Solution 5.5-20 Accelerating frame
L length of vertical arm
t thickness of vertical arm
r mass density
a0 acceleration
Let b width of arm perpendicular to the plane of the figure
Let q inertia force per unit distance along vertical arm
TYPICAL UNITS FOR USE
VERTICAL ARM
t meters (m)
IN THE PRECEDING EQUATION
SI units: r kg/m3 N # s2/m4
L meters (m)
a0 m/s2
smax N/m2 (pascals)
q rbta0
USCS units: r slug/ft3 lb-s2/ft4
qL2
rbta0L2
Mmax 2
2
3rL2a0
Mmax
bt2
S
smax 6
S
t
L ft a0 ft/s2 t ft
smax lb/ft2 (Divide by 144 to obtain psi.)
;
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459
SECTION 5.5 Normal Stresses in Beams
Problem 5.5-21 A beam of T-section is supported and loaded
as shown in the figure. The cross section has width b 2 1/2 in.,
height h 3 in., and thickness t 3/8 in.
3
t = —8 in.
P = 750 lb
q = 110 lb/ft
L1 = 3 ft
3
t = —8 in.
(a) Determine the maximum tensile and compressive
stresses in the beam.
(b) If the allowable stresses in tension and compression are
18 ksi and 12 ksi, respectively, what is the required
depth h of the beam? Assume that thickness t remains at
3/8 in. and that flange width b 2.5 in.
(c) Find the new values of loads P and q so that allowable
tension (18 ksi) and compression (12 ksi) stresses are
reached simultaneously for the beam. Use the beam
cross section in part (a) (see figure) and assume that L1, L2
and L3 are unchanged.
L2 = 8 ft
L3 = 5 ft
h=
3 in.
b = 2 —1 in.
2
Solution 5.5-21
NUMERICAL DATA
P 750 lb q 110
lb
ft
L1 3 ft L2 8 ft L3 5 ft t 3
in. b 2.5 in. h 3 in.
8
LOCATE NEUTRAL AXIS AND COMPUTE MOMENT OF INERTIA I
bt
c2 I
t
ht
+ t 1h t2 a
+ tb
2
2
bt + 1h t2 t
0.956 in. c1 h c2 2.044 in.
1
t 2
1
ht 2
bt3 + bt ac2 b +
t 1h t23 + t (h + t) cc1 a
b d 1.657 in.4
12
2
12
2
STATICS
RB L3
1
b d 1.003 * 103 lb RA P + q L3 RB 296.875 lb
cPL1 + qL3 aL2 +
L2
2
MOMENT AT LOAD P
MP RA L1 890.625 ft-lb
STRESSES AT P AND AT B
MOMENT AT SUPPORT B
MB scP MP c1
MP c2
13,188 psi stP 6166 psi
I
I
scB MB c2
9520 psi
I
stB qL23
1375 ft-lb
2
MB c1
20,360 psi
I
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CHAPTER 5 Stresses in Beams (Basic Topics)
(b) FOR COMPRESSION, USE TOP OF BEAM AT P
c1
MP a b 12 ksi where c1 h I
bt
t
ht
+ t (h t) a
+ tb
2
2
bt + (h t) t
and
I 1
bt 3 + bt ≥
12
bt
t
ht
+ t (h t) a
+ tb
2
2
bt + (h t) t
2
t
1
¥ +
t (h t)3 + t (h t)
2
12
≥ ≥h MP a
c1
b 12 ksi
I
bt
t
ht
+ t (h t) a
+ tb
2
2
bt + (h t) t
solve this expression numerically for depth h: h 3.1486 in.
2
¥ a
ht
b¥
2
compression controls
h 3.15 in.
Substitute value of h to confirm compressive stress value
bt
c2 I t
ht
+ t (h t) a
+ tb
2
2
bt + (h t) t
1 3
bt + bt ≥
12
bt
1.015 in. c1 h c2 2.133 in.
t
ht
+ t (h t) a
+ tb
2
2
bt + (h t) t
2
t
1
¥ +
t (h t)3 + t (h t)
2
12
≥ ≥h MP 890.625 ft-lb
bt
t
ht
+ t (h t) a
+ tb
2
2
bt + (h t) t
2
¥ a
ht
b¥
2
MP c1
12,000 psi
I
FOR TENSION, USE TOP OF BEAM AT B
MB
c1
18 ksi
I
solving for h
h 3.196 in. so value here based on tension (h 3.20 in.) controls
h 3.20 in.
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SECTION 5.5 Normal Stresses in Beams
(c) MUST HAVE RATIO MB/MP 18/12 1.5
461
IF ALLOWABLE TENSION AND COMPRESSION STRESSES ARE TO BE REACHED
SIMULTANEOUSLY
MB c1
saT
1.0 MB
I
18
3
MP c1
saC
MP
12
2
I
Use expression above for MB and MP.
qL 3 2
2
L3
1
cP + qL3 cPL1 + qL3 a L2 +
b d d L1
L2
2
88
57
88
1.544
57
we want ratio to be 1.50.
Divide through by q in above expression to get the following where a P/q, then solve for a:
1L L222
2
L L2
1
ca + (L L2) caL1 + (L L2) a L2 +
b d d L1
L2
2
3
2
solving numerically gives
a 6.944 ft
Now find q so that tension at top at B is 18 ksi, then use that q to find P so that compression at top at P is 12 ksi:
qL 3 2 c1
18 ksi
2 I
solving for q gives
q 97.249
q 97.2
lb
ft
lb
ft
so P aq 675.297 lb
P 675 lb
P
6.944 ft
q
Final check on stresses at B and P:
qL 3 2 c1
18 ksi
2 I
cP + qL3 L3
c1
1
b d d L1 12 ksi
cPL1 + qL3 a L3 +
L2
2
I
OK
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CHAPTER 5 Stresses in Beams (Basic Topics)
Problem 5.5-22 A cantilever beam AB with a rectangular
cross section has a longitudinal hole drilled throughout its length
(see figure). The beam supports a load P 600 N. The cross
section is 25 mm wide and 50 mm high, and the hole has a
diameter of 10 mm.
Find the bending stresses at the top of the beam, at the top
of the hole, and at the bottom of the beam.
10 mm
50 mm
A
B
12.5 mm
37.5 mm
P = 600 N
L = 0.4 m
25 mm
Solution 5.5-22 Rectangular beam with a hole
MOMENT OF INERTIA ABOUT THE NEUTRAL AXIS
(THE z AXIS)
All dimensions in millimeters.
Rectangle:
Iz Ic + Ad2
1
(25)(50)3 + (25)(50)(25 24.162)2
12
MAXIMUM BENDING MOMENT
M PL (600 N)(0.4 m) 240 N # m
260,420 + 878 261,300 mm4
PROPERTIES OF THE CROSS SECTION
Hole:
A1 area of rectangle
Iz Ic + Ad2 (25 mm)(50 mm) 1250 mm2
A2 area of hole
p
(10 mm)2 78.54 mm2
4
A area of cross section
A1 A2 1171.5 mm
Using line BB as reference axis:
©Aiyi A1(25 mm) A2(37.5 mm) 28,305 mm3
Aiyi
28,305 mm3
y a
24.162 mm
A
1171.5 mm2
Distances to the centroid C:
c2 y 24.162 mm
c1 50 mm c2 25.838 mm
p
(10)4 + (78.54)(37.5 24.162)2
64
490.87 + 13,972 14,460 mm4
Cross-section:
I 261,300 14,460 246,800 mm4
STRESS AT THE TOP OF THE BEAM
(240 N # m)(25.838 mm)
Mc1
s1 I
246,800 mm4
25.1 MPa
(tension)
;
STRESS AT THE TOP OF THE HOLE
My
s2 y c1 7.5 mm 18.338 mm
I
(240 N # m)(18.338 mm)
s2 17.8 MPa
246,800 mm4
(tension)
;
STRESS AT THE BOTTOM OF THE BEAM
(240 N # m)(24.162 mm)
Mc2
I
246,800 mm4
23.5 MPa
;
(compression)
s3 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 5.5 Normal Stresses in Beams
Problem 5.5-23 A small dam of height h 6 ft is constructed of
463
Steel beam
vertical wood beams AB, as shown in the figure. The wood beams,
which have thickness t 2.5 in., are simply supported by horizontal
steel beams at A and B.
Construct a graph showing the maximum bending stress smax in
the wood beams versus the depth d of the water above the lower
support at B. Plot the stress smax (psi) as the ordinate and the
depth d (ft) as the abscissa. (Note: The weight density g of water
equals 62.4 lb/ft3.)
A Wood beam
t
t
Wood beam
Steel beam
h
d
B
Side view
Solution 5.5-23
Vertical wood beam in a dam
h 6 ft
t 2.5 in.
g 62.4 lb/ft3
Let b width of beam
(perpendicular to the
figure)
Let q0 intensity of
load at depth d
q0 gbd
ANALYSIS OF BEAM
L h 6 ft
q0d2
RA 6L
q0d
d
RB a3 b
6
L
MAXIMUM BENDING STRESS
1
Section modulus: S bt2
6
Mmax
6 q0d2
d
2d d
bd
2c
a1 +
S
6
L
3L A 3L
bt
q0 g bd
smax smax gd3
t
2
a1 d
2d d
b
+
L
3L A 3L
;
SUBSTITUTE NUMERICAL VALUES:
d depth of water (ft) (Max. d h 6 ft)
L h 6 ft g 62.4 lb/ft3 t 2.5 in.
smax psi
smax d
A 3L
x0 d
q0d2
d
a1 b
Mc RA(L d) 6
L
Mmax Top view
(62.4)d3
2
(2.5)
a1 d
d d
+
b
6
9 A 18
0.1849d3(54 9d + d12d )
d(ft)
0
1
2
3
4
5
6
;
smax(psi)
0
9
59
171
347
573
830
q0d2
d
2d d
a1 +
b
6
L
3L A 3L
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CHAPTER 5 Stresses in Beams (Basic Topics)
Problem 5.5-24 Consider the compound beam
MA = 600 N.m
with segments AB and BCD joined by a pin
connection (moment release) just right of B
(see figure part a). The beam cross section is a
double-Tee made up from three 50 mm 150 mm
wood members (actual dimensions, see figure
part b).
q1 = 920 N/m
A
C 1.5 m
3m
B
x
(a) Find the centroid C of the double-T cross
section (c1, c2), then compute the moment
of inertia Iz (mm4).
(b) Find the maximum tensile normal stress st
and maximum compressive normal stress sc
(kPa) for the loading shown. (Ignore the
weight of the beam.)
P = 1730 N
q2 = 460 N/m
3m
Pin
connection
(a)
1.5 m D
3m
y
z
c1
Each piece is
a 50 mm 150 mm
wood plank
(actual
dimensions)
C
c2
(b)
Solution 5.5-24
(a) CROSS-SECTIONAL PROPERTIES
b 50 mm h 150 mm Ax bh 7.5 * 103 m2
CENTROID
c2 h
b
2 Ax a b + Ax ah + b
2
2
3Ax
108.333 mm
c1 h + b c2 91.667 mm
c1 91.7 mm c2 108.3 mm
Iz 2 a
1
h 2
1
b 2
b b (h)3 + 2 Ax ac2 b +
(h) (b)3 + Ax a c1 b
12
2
12
2
Iz 7.969 * 107 mm4
SECTION MODULUS for top and also for bottom of beam
Stop Iz
c1
8.693 * 105 mm3 Sbot Iz
c2
7.356 * 105 mm3
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SECTION 5.5 Normal Stresses in Beams
STATICS:
465
find reactions, then shear and MOMENT DIAGRAMS for this beam. (See Prob. 1.2-3 for similar problem.)
600 N•m
4050 N•m
MA 600 N # m MC 4050 N # m max. moments at A (compression on top) and C (compression on bottom)
STRESSES AT A
sAtop MA
690 kPa
Stop
compression
sAbot MA
816 kPa
Sbot
tension
STRESSES AT C:
maximum tension and compression occur at C.
sCtop MC
4659 kPa
Stop
tension
sCbot MC
5506 kPa
Sbot
compression
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CHAPTER 5 Stresses in Beams (Basic Topics)
Problem 5.5-25 A steel post (E 30 106 psi) having thickness t 1/8 in. and
height L 72 in. supports a stop sign (see
figure: s 12.5 in.). The height of the post L
is measured from the base to the centroid of
the sign. The stop sign is subjected to wind
pressure p 20 lb/ft2 normal to its surface.
Assume that the post is fixed at its base.
s
L
(a) What is the resultant load on the
sign? [See Appendix E, Case 25, for
properties of an octagon, n 8].
(b) What is the maximum bending stress
smax in the post?
(c) Repeat part (b) if the circular cutouts
are eliminated over the height of the
post.
y
5/8 in.
Section A–A
z
Circular cut-out, d = 0.375 in.
Post, t = 0.125 in.
c1
1.5 in.
C
c2
Stop sign
0.5 in. 1.0 in.
1.0 in. 0.5 in.
Wind load
Numerical properties of post
A = 0.578 in.2, c1 = 0.769 in., c2 = 0.731 in.,
Iy = 0.44867 in.4, Iz = 0.16101 in.4
A
A
Elevation
view of post
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 467
SECTION 5.5 Normal Stresses in Beams
467
Solution 5.5-25
(a) RESULTANT LOAD F ON SIGN
A
b
n s2
cot a b
4
2
p 20 psf s 12.5 in. n 8 b A 754.442 in.2 A A
144
360 p
a
b
n 180
b 0.785 rad
A 5.239 ft2
F p A F 104.784 lb Fres 104.8 lb
(b) MAX. BENDING STRESS IN POST
L 72 in. Iz 0.16101 in.4 c1 0.769 in. c2 0.731 in.
Mmax FL Mmax 628.7 ft-lb
sc Mmax c1
Iz
sc 36.0 ksi
st Mmax c2
Iz
st 34.2 ksi
max. bending stress at base of post
(c) REPEAT (b) IF CUTOUT HOLES ARE ELIMINATED FROM POST
Re-compute cross sectional properties (see below):
p 20
sc lb
ft2
c1 0.716 in. c2 0.784 in. Iz 0.18269 in.4
A 754.442 in.2 F pA L 6 ft Mmax FL 628.702 ft-lb
Mmax c1
Mmax c2
29.6 ksi st 32.4 ksi
Iz
Iz
tension at base
Re-compute cross-sectional properties with holes eliminated:
A t (2 a1 + 2b + 2a2 d) A 0.625 in.2 b cos(b 0) 0.375
CENTROID
c1 t
1.5
t
t (2 a2 d) + 2bt a
b + 2a1 a 1.5 b
2
2
2
A
c2 1.5 c1
c1 0.716 in.
c2 0.784 in.
MOMENTS OF INERTIA WRT Y & Z AXES
Iy a1 2
1
1
tb3
13 2
t C (2 a2)3 d3 D + a 2ta31 b + 2 (a1t) a 1.5 b + 2
cos(b 0)2 + 2bt a a2 +
b
12
12
2
12
28
Iy 0.44922 in.4
Iz 1 3
t 2
t3
t 2 2 t b3
1.5 2
t (2a2 d) + t (2 a2 d) ac1 b + 2a1
+ t (2 a1) ac2 b +
sin(b 0)2 + 2 bt ac1 b
12
2
12
2
12
2
Iz 0.18269 in.4
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 5 Stresses in Beams (Basic Topics)
Design of Beams
P
Problem 5.6-1 The cross section of a narrow-gage railway
bridge is shown in part a of the figure. The bridge is
constructed with longitudinal steel girders that support the
wood cross ties. The girders are restrained against lateral
buckling by diagonal bracing, as indicated by the dashed lines.
The spacing of the girders is s1 50 in. and the spacing
of the rails is s2 30 in. The load transmitted by each rail to a
single tie is P 1500 lb. The cross section of a tie, shown in
part b of the figure, has width b 5.0 in. and depth d.
Determine the minimum value of d based upon an
allowable bending stress of 1125 psi in the wood tie.
(Disregard the weight of the tie itself.)
P
s2
Steel rail
Wood
tie
d
b
Steel
girder
(b)
s1
(a)
Solution 5.6-1 Railway cross tie
Mmax S
P(s1 s2)
15,000 lb-in.
2
1
5d 2
bd 2
(5.0 in.)(d 2) 6
6
6
d inches
Mmax s allow S 15,000 (1125) a
s1 50 in.
b 5.0 in.
d depth of tie
Solving, d 2 16.0 in.
s2 30 in.
P 1500 lb
sallow 1125 psi
5d 2
b
6
dmin 4.0 in.
NOTE: Symbolic solution: d 2 ;
3P(s1 s2)
bsallow
Problem 5.6-2 A fiberglass bracket ABCD of solid circular cross section has the shape and dimensions shown in the figure.
A vertical load P 40 N acts at the free end D.
(a) Determine the minimum permissible diameter dmin of the bracket if the allowable bending stress in the material is
30 MPa and b 37 mm. (Note: Disregard the weight of the bracket itself.)
(b) If d 10 mm, b 37 mm, and sallow 30 MPa, what is the maximum value of load P if vertical load P at D is
replaced with horizontal loads P at B and D (see figure part b)?
6b
6b
A
B
A
B
2b
D
P
2b
C
D
C
P
2b
P
(a)
2b
(b)
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Page 469
SECTION 5.6 Design of Beams
469
Solution 5.6-2
(a) NORMAL STRESS AT A DUE TO FLEXURE (COMPRESSION AT TOP)
sa (4Pb) a
a
dmin
b
2
4
p dmin
b
64
128 Pb
p dmin 3
Solve above equation for required diameter d:
1
128 Pb 3
dmin a
b
psa
SUBSTITUTE NUMERICAL VALUES THEN SOLVE FOR d
P 40 N b 37 mm sa 30 MPa
dmin c
1
128 (40 N) (37 mm) 3
d dmin 12.62 mm
p (30 MPa)
(b) REACTIVE MOMENT AT A IS P(2b); HORIZONTAL AND VERTICAL REACTIONS AT A ARE BOTH ZERO
MA P (2b) so sallow solving for P Pmax d
(2Pb) a b
2
a
4
pd
b
64
sallow pd3
64b
and
94,720 N # mm
pd 3
Pmax (30 MPa) p (10 mm)3
39.8 N
64 (37 mm)
Problem 5.6-3 A cantilever beam AB is loaded by a uniform load q and a
q
P
concentrated load P as shown in the figure.
(a) Select the most economical steel C shape from Table F-3(a) in
A
B
Appendix F; use q 20 lb/ft and P 300 lb (assume allowable
normal stress is sd 18 ksi).
6 ft
4 ft
(b) Select the most economical steel S shape from Table F-2(a) in Appendix F;
use q 45 lb/ft and P 2000 lb (assume allowable normal stress is
sd 20 ksi).
(c) Select the most economical steel W shape from Table F-1(a) in Appendix F; use q 45 lb/ft and P 2000 lb (assume
allowable normal stress is sa 20 ksi). However, assume that the design requires that the W shape must be used in
weak axis bending, i.e., it must bend about the 2-2 (or y) axis of the cross section.
Note: For parts (a), (b), and (c), revise your initial beam selection as needed to include the distributed weight of the
beam in addition to uniform load q.
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Page 470
CHAPTER 5 Stresses in Beams (Basic Topics)
Solution 5.6-3
(a) SELECT THE MOST ECONOMICAL STEEL C SHAPE FROM TABLE F-3(a) IN APPENDIX F
L 10 ft
q 20
lb
P 300 lb
ft
Mmax1 (q)
3
L2
+ P a Lb 2 8 kip-ft
2
5
does not yet include beam self-weight
sas 18 ksi
Sreqd1 TRIAL #1
lb
ft
w 30
Mmax1
sas
Sreqd1 1.867 in.3
Mmax2 (q + w)
use 2.2 axis in Table F-3(a); try C12 : 30, S22 2.05 in.3
Mmax2
L2
3
2.867 in.3
+ P a L b 4.3 kip-ft Sreqd2 2
5
sas
<C12 : 30 will not work
Try C15 * 33.9 (S22 3.09 in.3)
TRIAL #2:
w 33.9
Sreqd3 3
lb
L2
3 Mmax3 (q + w)
+ P a L b 4.495 kip- ft
ft
2
5
Mmax3
2.997 in.3 <C15 : 33.9 will work
sas
(b) SELECT THE MOST ECONOMICAL STEEL S SHAPE FROM TABLE F-2(a) IN APP. F:
q 45
lb
ft
P 2000 lb Mmax1 (q)
L2
3
+ P a Lb 14.25 kip-ft does not yet include beam self-weight
2
5
sas 20 ksi
TRIAL #1
Sreqd1 Mmax1
sas
Sreqd1 8.55 in.3
use 1–1 axis in Table F-2(a); try S6 : 17.2,
S11 8.74 in.3
w 17.3
lb
ft
Mmax2 (q + w)
L2
Mmax2
3
+ P a Lb 15.115 kip-ft Sreqd2 9.069 in.3
2
5
sas
<C6 : 17.2 will not work
TRIAL #2: TRY S8 * 18.4 (S11 14.4 in.3)
w 18.4
lb
ft
Mmax3 (q + w)
Mmax3
L2
3
+ P a L b 15.17 kip-ft Sreqd3 9.102 in.3
2
5
sas
<S8 : 18.4 will work
(c) SELECT THE MOST ECONOMICAL STEEL W SHAPE FROM TABLE F-1(a) IN APP. F:
From (b), assume that required S will be approx. 10 in.3 so from Table F-1(a), select W shape from S22 column.
TRY W8 * 35: S22 10.6 in.3 w 35
Mmax (q + w)
lb
ft
Mmax
L2
3
+ P a Lb 16 kip-ft S22reqd 9.6 in.3 <W8 : 35 will work
2
5
sas
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 471
SECTION 5.6 Design of Beams
Problem 5.6-4 A simple beam of length L 5 m carries a uniform load of
intensity q 5.8 kN/m and a concentrated load 22.5 kN (see figure).
(a) Assuming sallow 110 MPa, calculate the required section modulus S.
Then select the most economical wide-flange beam (W shape) from
Table F-1(b) in Appendix F, and recalculate S, taking into account the
weight of the beam. Select a new beam if necessary.
(b) Repeat part (a), but now assume that the design requires that the W shape
must be used in weak axis bending, (i.e., it must bend about the 2–2 (or y)
axis of the cross section).
471
P = 22.5 kN 1.5 m
q = 5.8 kN/m
L=5m
Solution 5.6-4
NUMERICAL DATA
q 5.8
L5m
kN
m
P 22.5 kN
b 1.5 m
aLb
a 3.5 m
sallow 110 MPa
qL
Pb
+
RA 21.25 kN
2
L
RA + RB 51.5
qL + P 51.5
(a) STATICS
RA LOCATE POINT OF ZERO SHEAR
Mmax RA a qa2
2
xm RA
q
RB qL
Pa
+
2
L
RB 30.25 kN
xm 3.664 m greater than dist. a to load P so zero shear is at load
point
Mmax 38.85 kN # m
FIND REQUIRED SECTION MODULUS
Sreqd Mmax 110322
sallow
Sreqd
103
¿
353.182 103 mm3
select W360 : 39 (Sact 578 * 103 mm3)
RECOMPUTE MAX. MOMENT WITH BEAM MASS INCLUDED & THEN CHECK ALLOWABLE STRESS
w (39) 9.81
RA aq +
w
bL
1000
+
2
Mmax RA a smax kg m
N
6 N/m w 382.59
2
m s
m
aq +
Mmax11062
Sact
Pb
L
RA 22.206 kN
w
b a2
1000
2
smax 68.952
Sact 57811032 mm3
xm RA
w
q +
1000
xm 3.592
^ greater than a so max.
moment at load pt.
Mmax 39.854 kN # m
OK, less than 110 MPa
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CHAPTER 5 Stresses in Beams (Basic Topics)
(b) REPEAT (a) BUT ASSUME WEAK AXIS BENDING
Sreqd 35311032 mm3
Mmax 11062
Select W250 * 89 Sact 37711032 mm3 Mmax 41.142 kN # m
¿
¿
S22
Sact
109.13 MPa
OK, less than 110 MPa
Problem 5.6-5 A simple beam AB is loaded as shown in the figure.
(a) Calculate the required section modulus S if sallow 18,000 psi, L 32 ft,
P 2900 lb, and q 450 lb/ft.
Then select a suitable I-beam (S shape) from Table F-2(a), Appendix F,
and recalculate S taking into account the weight of the beam. Select a
new beam size if necessary.
(b) What is the maximum load P that can be applied to your final beam selection in part (a)?
P
q
q
B
A
L
—
4
L
—
4
L
—
4
L
—
4
Solution 5.6-5
(a) RA qL
P
+
4
2
Mmax RA
L L
L
L
q a + b 2
4 4
8
sa 18 ksi L 32 ft P 2900 lb q 450
La
Sreqd Mmax
sa
Sreqd Lq
P
+
b
2
4
2
qL
P
wL
+
+
4
2
2
Sreqd2 so Mmax RA
Mmax
28.053 in.3
sa
3L2q
32
sa
Recompute Mmax including beam weight
RA La
Lq
P
+
b
2
4
2
3 L2 q
32
lb
ft
25.067 in.3
w 35
6 try S10 : 35 (Sact 29.4 in.3)
lb
ft
L
L L
L
LL
q a + b w
42,080 ft-lb
2
4 4
8
2 4
6 less than Sact for S10 * 35 so S10 : 35 will work
qL
P
wL
+
+
so
4
2
2
P Lq
Lw
La
+
b
2 4
2
3L2q
L
L L
L
LL
L2w
Mmax RA q a + b w
2
4 4
8
2 4
2
8
32
(b) Sact 29.4 in.3
La
Mmax sa Sact 44,100 ft-lb
Lq
P
Lw
+
+
b
2
4
2
3 L2q
L2 w
sa Sact
8
32
solving for P L2q + 4L2 w 32Sact sa
3152 lb
8L
Pmax 3152 lb
2
Pmax RA L2q + 4L2 w 32 Sact sa
8L
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Page 473
473
SECTION 5.6 Design of Beams
Problem 5.6-6 A pontoon bridge (see figure) is constructed of
two longitudinal wood beams, known as balks, that span between
adjacent pontoons and support the transverse floor beams, which
are called chesses. For purposes of design, assume that a uniform
floor load of 7.5 kPa acts over the chesses. (This load includes an
allowance for the weights of the chesses and balks.) Also, assume
that the chesses are 2.5 m long and that the balks are simply supported with a span of 3.0 m. The allowable bending stress in the
wood is 15 MPa.
Chess
Pontoon
Balk
(a) If the balks have a square cross section, what is their minimum required width bmin?
(b) Repeat part (a) if the balk width is 1.5 b and the balk depth
is b; compare the cross-sectional areas of the two designs.
Solution 5.6-6
(a) Lc 2.5 m Lb 3 m w 7.5 kPa sa 15 MPa
q
wLc
kN
9.375
2
m
Mmax qLb 2
8
and Mmax sa S Mmax sa a
b3
b
6
1
6 qLb 2 3
bmin a
b 161.6 mm
8 sa
(b) S 1.5 b 1b22
bmin a
6
Aa bmin 2 2.611 * 104 mm2
: 0.25 b3 Mmax sa a
b3
b
4
1
4 q Lb 2 3
b 141.2 mm
8 sa
Ab 1.5 bmin 2 2.989 * 104 mm2
Ab
1.145
Aa
Problem 5.6-7 A floor system in a small building consists of wood planks
supported by 2-in. (nominal width) joists spaced at distance s and measured
from center to center (see figure). The span length L of each joist is 12 ft, the
spacing s of the joists is 16 in., and the allowable bending stress in the wood
is 1250 psi. The uniform floor load is 120 lb/ft2, which includes an allowance
for the weight of the floor system itself.
(a) Calculate the required section modulus S for the joists, and then select a
suitable joist size (surfaced lumber) from Appendix G, assuming that
each joist may be represented as a simple beam carrying a uniform load.
(b) What is the maximum floor load that can be applied to your final beam
selection in part (a)?
Planks
s
s
L
Joists
s
Probs. 5.6-7 and 5.6-8
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Page 474
CHAPTER 5 Stresses in Beams (Basic Topics)
Solution 5.6-7
Floor joists
(a) sa 1250 psi L 12 ft w 120
lb
2
ft
s 16 in. q ws 160
lb
ft
2
Mmax qL
Mmax
2880 lb-ft Sreqd 27.648 in.3
8
sa
6 use 2 : 12, Sact 31.64 in.3
Sact 31.64 in.3
(b) Mallow sa Sact 3295.833 lb-ft
qmax Mallow a
8
2
L
b 183.102
lb
ft
wmax qmax
lb
137.3 2
s
ft
Problem 5.6-8 The wood joists supporting a plank floor (see figure) are 38 mm 220 mm in cross section (actual dimensions) and have a span length L 4.0 m. The floor load is 5.0 kPa, which includes the weight of the joists and the floor.
(a) Calculate the maximum permissible spacing s of the joists if the allowable bending stress is 14 MPa. (Assume that
each joist may be represented as a simple beam carrying a uniform load.)
(b) If spacing s 406 mm, what is the required depth h of the joist? Assume all other variables remain unchanged.
Solution 5.6-8 Spacing of floor joists
(a) sa 14 MPa
L4m
Mmax Sact qL
8
2
bh
6
(b) sact 406 mm
w 5 kPa
1ws2 L
q ws
2
2
sa Sact smax 8
so
smax b 38 mm
2
bh
a
b
2
6
wL
8sa
h 220 mm
smax 4 bh2 sa
3 wL2
4 bh2sa
429 mm
3 wL2
hmin 3 wL2sact
214 mm
C 4bsa
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Page 475
SECTION 5.6 Design of Beams
Problem 5.6-9 A beam ABC with an overhang from B to C is constructed of
a C 10 30 channel section with flanges facing upward (see figure). The beam
supports its own weight (30 lb/ft) plus a triangular load of maximum intensity
q0 acting on the overhang. The allowable stresses in tension and compression
are 18 ksi and 12 ksi, respectively.
(a) Determine the allowable triangular load intensity q0,allow if the distance
L equals 4 ft.
(b) What is the allowable triangular load intensity q0,allow if the beam is
rotated 180 about its longitudinal centroidal axis so that the
flanges are downward?
475
q0
A
C
B
L
3.03 in.
L
2.356 in.
0.674 in.
C
10.0 in.
Solution 5.6-9
NUMERICAL VALUES
C10 : 30 I22 3.93 in.4
c2 0.674 in. c1 3.03 in. c2 2.356 in. w 30
lb
ft
sat 18 ksi sac 12 ksi L 4 ft
L2 q0
L
1
2
L2 w
(a) Mmax MB MB wL a b + q0 L a L b :
+
2
2
3
3
2
sBtop MB c1
I22
q0allowtop I22
wL2
cs
a
b
d
at
c1
2
L2
q0allowtop I22
lb
wL2
d 424
cs
a
b
at
2
c
2
ft
L
1
3
sBbot tension (allowable 20 ksi)
q0allowbot 3
MB c2
I22
compression (allowable 11 ksi)
I22
wL2
cs
a
b
d
ac
c2
2
L2
3
q0allowbot I22
lb
wL2
d 1048
cs
a
b
ac
2
c
2
ft
L
2
3
(b) If flanges point downward, we must switch c1 and c2 in above formulas:
q0allowtop I22
wL2
lb
cs
a
b
d 1595
at
2
c
2
ft
L
2
3
q0allowbot Problem 5.6-10 A so-called “trapeze bar” in a hospital
room provides a means for patients to exercise while in bed
(see figure). The bar is 2.1 m long and has a cross section in
the shape of a regular octagon. The design load is 1.2 kN applied
at the midpoint of the bar, and the allowable bending stress
is 200 MPa.
Determine the minimum height h of the bar. (Assume
that the ends of the bar are simply supported and that the
weight of the bar is negligible.)
I22
wL2
lb
cs
a
b
d 268
ac
2
c
2
ft
L
1
3
C
h
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Page 476
CHAPTER 5 Stresses in Beams (Basic Topics)
Solution 5.6-10 Trapeze bar (regular octagon)
P 1.2 kN L 2.1 m
sallow 200 MPa
b 0.41421h
‹ Ic 1.85948(0.41421h)4 0.054738h4
Determine minimum height h.
SECTION MODULUS
MAXIMUM BENDING MOMENT
S
Mmax (1.2 kN)(2.1 m)
PL
630 N # m
4
4
PROPERTIES OF THE CROSS SECTION
Use Appendix E, Case 25, with n 8.
b length of one side
b
360
360
45
n
8
tan
b
b
(from triangle)
2
h
cot
b
h
2
b
Ic
0.054738h4
0.109476h3
h/2
h/2
MINIMUM HEIGHT h
M
M
S
S
s
630 N # m
3.15 * 106 m3
0.109476h3 200 MPa
s
h3 28.7735 * 106 m3 h 0.030643 m
‹ hmin 30.6 mm
;
ALTERNATIVE SOLUTION (n 8)
M
PL
4
b 45 tan
b
b
121 cot 12 1
2
2
b (12 1)h h (12 + 1)b
b
45
For b 45: tan
0.41421
h
2
45
h
cot
2.41421
b
2
Ic a
S a
11 + 812 4
412 5 4
bb a
bh
12
12
3PL
4 12 5 3 3
bh h 6
2(4 12 5)sallow
;
Substitute numerical values:
MOMENT OF INERTIA
h3 28.7735 * 106 m3 hmin 30.643 mm
;
4
Ic b
b
nb
acot b a3 cot2
1b
192
2
2
Ic 8b4
(2.41421)[3(2.41421)2 1] 1.85948b4
192
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Page 477
477
SECTION 5.6 Design of Beams
Problem 5.6-11 A two-axle carriage that is part of an overhead traveling
crane in a testing laboratory moves slowly across a simple beam AB
(see figure). The load transmitted to the beam from the front axle is 2200 lb
and from the rear axle is 3800 lb. The weight of the beam itself may be
disregarded.
3800 lb
2200 lb
5 ft
A
B
(a) Determine the minimum required section modulus S for the beam
if the allowable bending stress is 17.0 ksi, the length of the beam is
18 ft, and the wheelbase of the carriage is 5 ft.
(b) Select the most economical I-beam (S shape) from Table F-2(a), Appendix F.
18 ft
Solution 5.6-11
NUMERICAL DATA
xm P1 2200 lb
L 18 ft
P2 3800 lb
L (xm + d)
L xm
b + P1 c
d
L
L
RA 2694 lb
(a) FIND REACTION RA, THEN AN EXPRESSION FOR MOMENT
UNDER LARGER LOAD P2; LET X DIST. FROM A
TO LOAD P2
Mmax xm cP2 a
Sreqd M2 RA x
L (x + d )
Lx
b + P1 c
dd
L
L
L(xm d)
Lxm
b P1 c
dd
L
L
Mmax 21,780 ft-lb
L (x + d )
Lx
b + P1 c
d
L
L
M2 x cP2 a
xm 8.083 ft
RA P2 a
d 5 ft
sa 17 ksi
RA P2 a
(P1 + P2) L P1d
2 (P1 + P2)
Mmax
sa
Sreqd 15.37 in.3
;
(b) SELECT MOST ECONOMICAL S SHAPE FROM
TABLE F-2(a)
xP2LP2x2 xP1LP1x2xP1d
L
Take derivative of MA & set to zero to find max.
bending moment at x x m
select S 8 * 23
M2 Sact 16.2 in.3
;
d xP2LP2x2 xP1LP1x2xP1d
a
b
dx
L
P2L 2P2x + P1L 2P1x P1d
L
P2L 2P2x + P1L 2P1x P1d 0
Problem 5.6-12 A cantilever beam AB of circular cross
section and length L 750 mm supports a load P 800 N
acting at the free end (see figure). The beam is made of
steel with an allowable bending stress of 120 MPa.
(a) Determine the required diameter dmin (figure part a) of the
beam, considering the effect of the beam’s own weight.
(b) Repeat part (a) if the beam is hollow with wall thickness
t d/8; (figure part b) compare the cross-sectional areas
of the two designs.
t=
A
d
8
B
d
P
L
d
(a)
(b)
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 5 Stresses in Beams (Basic Topics)
Solution 5.6-12 Cantilever beam
NUMERICAL DATA
L 750 mm P 800 N
sa 120 MPa gs 77
kN
m3
d 50 mm
L
p d2
p d2
w gs a
b Mmax PL + gs L2 a
b
2
4
8
Equate Mmax expressions and then solve for d:
(a) Mmax PL + wL
S
p d3
32
Aa p gs L2d 2
p sa d 3
PL 0
32
8
solving
and Mmax sa S
dmin 37.6 mm
p dmin 2
1.108 * 103 m2
4
(b) w gs ≥
d 2
p cd2 ad 2 b d
8
4
Mmax PL + gs L2 ≥
¥ 2
7p d gs
64
d 2
p cd2 a d 2 b d
8
4
S
¥ d 3
p cd 3 a d 2 b d
8
32
37p d 3
2048
7p gsL2d2
+ PL and Mmax sa S
64
Equate Mmax expressions and then solve for d:
sa a
7pgsL2d2
37 p d 3
b a
+ PL b 0
2048
64
dmin 45.2 mm
Ab 2
p
3
cdmin 2 a dmin b d 7.031 * 104 m2
4
4
Ab
0.635
Aa
Problem 5.6-13
A propped cantilever beam ABC (see figure) has a shear release just right of the mid-span.
(a) Select the most economical wood beam from the table in Appendix G; assume q 55 lb/ft, L 16 ft,
saw 1750 psi, taw 375 psi; include the self-weight of the beam in your design.
(b) If a C1025 steel beam is now used for beam ABC, what is the maximum permissible value of load variable q?
Assume sas 16 ksi and L 10 ft. Include the self-weight of the beam in your analysis.
q
P = qL
C
A
B
Shear release
L
L/2
L/2
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SECTION 5.6 Design of Beams
479
Solution 5.6-13
First, plot shear (V ) and moment (M) diagrams; label all critical ordinates.
qL
1
2
1
2
3
−qL
Shear diagram
Moment diagram
1
qL 2/2
2
1
3
2
(a) SELECT THE MOST ECONOMICAL WOOD BEAM FROM THE TABLE IN APP. G;
q 55
lb
ft
Mmax qL2
7040 lb-ft
2
w 6.8
lb
ft
TRIAL #2:
w 9.6
L 16 ft saw 1750 psi taw 375 psi
Mmax2 Sreqd1 Mmax
48.274 in.3
saw
6 TRY 3 : 12 (S11 52.73 in.3, w 6.8 lb/ft)
qL2
Mmax2
wL2
54.243 in.3 6 3 : 12 will not work
+
7910.4 lb-ft Sreqd2 2
2
saw
4 : 12 (S11 73.83 in.3, w 9.6 lb/ft)
lb
ft
Mmax2 qL2
wL2
+
8268.8 lb-ft
2
2
(b) S22 1.47 in.3 sas 16 ksi L 10 ft w 25
Mmax qL2
wL2
+
2
2
qmax wL2
lb
b 14.2
as
S
as
22
2
2
ft
L
Sreqd2 Mmax2
56.7 in.3
saw
6 4 : 12 will work
lb
ft
and Mmax sas S22
2
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CHAPTER 5 Stresses in Beams (Basic Topics)
Problem 5.6-14 A small balcony constructed of wood is
supported by three identical cantilever beams (see figure).
Each beam has length L1 2.1 m, width b, and height h 4b/3.
The dimensions of the balcony floor are L1 * L2,
with L2 2.5 m. The design load is 5.5 kPa acting over the
entire floor area. (This load accounts for all loads except
the weights of the cantilever beams, which have a weight
density g 5.5 kN/m3.) The allowable bending stress in the
cantilevers is 15 MPa.
Assuming that the middle cantilever supports 50% of the
load and each outer cantilever supports 25% of the load,
determine the required dimensions b and h.
Solution 5.6-14
4b
h= —
3
L2
b
L1
Compound beam
MAXIMUM BENDING MOMENT
(q q0)L12
1
(6875 N/m 7333b2)(2.1 m)2
Mmax 2
2
15,159 + 16,170b2 (N # m)
bh2
8b3
6
27
Mmax sallow S
L1 2.1 m L2 2.5 m Floor dimensions: L1 * L2
Design load w 5.5 kPa
g 5.5 kN/m3 (weight density of wood beam)
sallow 15 MPa
S
MIDDLE BEAM SUPPORTS 50% OF THE LOAD.
Rearrange the equation:
‹ q wa
L2
2.5 m
b (5.5 kPA)a
b 6875 N/m
2
2
WEIGHT OF BEAM
q0 gbh 4gb2
4
(5.5 kN/m2) b2
3
3
7333b2 (N/m)
(b meters)
15,159 + 16,170b2 (15 * 106 N/m2) a
8b3
b
27
(120 * 106)b3 436,590b2 409,300 0
SOLVE NUMERICALLY FOR DIMENSION b
4b
h
0.2023 m
b 0.1517 m
3
REQUIRED DIMENSIONS
b 152 mm
h 202 mm
;
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481
SECTION 5.6 Design of Beams
Problem 5.6-15 A beam having a cross section in the form of an
unsymmetric wide-flange shape (see figure) is subjected to a negative
bending moment acting about the z axis.
Determine the width b of the top flange in order that the stresses at
the top and bottom of the beam will be in the ratio 4:3, respectively.
y
b
1.5 in.
1.25 in.
z
12 in.
C
1.5 in.
16 in.
Solution 5.6-15 Unsymmetric wide-flange beam
AREAS OF THE CROSS SECTION (in.2)
A1 1.5b A2 (12)(1.25) 15 in.2
A3 (16)(1.5) 24 in.2
A A1 + A2 + A3 39 + 1.5b (in.2)
FIRST MOMENT OF THE CROSS-SECTIONAL
AREA ABOUT THE LOWER EDGE B-B
QBB gyi Ai (14.25)(1.5b) + (7.5)(15) + (0.75)(24)
Stresses at top and bottom are in the ratio 4:3.
Find b (inches)
h height of beam 15 in.
LOCATE CENTROID
stop
c1
4
sbottom
c2
3
4
60
8.57143 in.
c1 h 7
7
3
45
c2 h 6.42857 in.
7
7
130.5 + 21.375b (in.3)
DISTANCE c2 FROM LINE B-B TO THE CENTROID C
c2 QBB
130.5 + 21.375b
45
in.
A
39 + 1.5b
7
SOLVE FOR b
(39 + 1.5b)(45) (130.5 + 21.375b)(7)
82.125b 841.5 b 10.25 in.
;
y
Problem 5.6-16 A beam having a cross section in the form of a channel
(see figure) is subjected to a bending moment acting about the z axis.
Calculate the thickness t of the channel in order that the bending stresses
at the top and bottom of the beam will be in the ratio 7:3, respectively.
t
z
t
C
t
55 mm
152 mm
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Page 482
CHAPTER 5 Stresses in Beams (Basic Topics)
Solution 5.6-16
ratio of top to bottom stresses c1/c2 7/3
NUMERICAL DATA
h 152 mm
b 55 mm
1 11,385 186 t + t 2
2
131 + t
Take 1st moments to find distances c1 & c2.
1st moments about base.
c2 J
t
b
(h 2t)(t) + 2bt a b
2
2
2bt + t (h 2t)
c1 b c2
c2 2.55t + t (152 2t)
111,385 186 t + t 22
76 t + t 2 3025
K
7/3
7a 76 t t 23025b d 0
t
55
(152 2t)(t) + 2.55t a b
2
2
t2 109 t + 1298 0
2.55t + t(152 2t)
1 11,385 186 t + t
2
131 + t
2.55 t + t (152 2 t)
c3 ca 11,385186 t t2 b d
t
55
(152 2t)(t) + 2.55t a b
2
2
c1 55 c1 55
t
(152 2t)(t) + 2.55 t a b
2
2
t
2
Problem 5.6-17 Determine the ratios of the weights
of four beams that have the same length, are made
of the same material, are subjected to the same
maximum bending moment, and have the same
maximum bending stress if their cross sections
are (1) a rectangle with height equal to twice the
width, (2) a square, (3) a circle, and (4) a hollow
pipe with outer diameter d and wall thickness
t d/8 (see figures).
109 110924(1298)
2
t 13.61 mm
;
t=
h = 2b
d
8
a
b
a
d
d
Part 1
Part 2
Part 3
Part 4
Solution 5.6-17 Ratio of weights of three beams
Since M and s are the same, the section moduli must be the same.
(1) RECTANGLE: S 2b3
bh2
6
3
A1 2b2 2 a
(2) SQUARE: S b a
3S 1/3
b
2
3S 2/3
b 2.6207S2/3
2
a3
a (6S)1/3
6
A2 a2 (6S)2/3 3.3019 S2/3
(3) CIRCLE: S A3 pd3
32S 1/3
d a
b
p
32
pd2
p 32S 2/3
a
b 3.6905S2/3
p
4
4
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Page 483
483
SECTION 5.6 Design of Beams
(4) HOLLOW PIPE:
S
p
d 3
37p d 3
cd 3 a d 2 b d 32
8
2048
1
A4 2
p 2
3
7pd
cd a db d 4
4
64
2
A4 2048
7p c a
Sb 3 d
37p
64
2
2
7p1d32 3
64
2
7
2048 3
pa
b 2.327
64
37p
W1 : W2 : W3 : W4 1 : 1.260 : 1.408 : 0.888
denom. 2.6207
3.3019
1.26
denom.
3.6905
1.408
denom.
2.327
0.888
denom.
A horizontal shelf AD of length L 1215 mm, width
b 305 mm, and thickness t 22 mm is supported by brackets at B and C
(see part a of the figure). The brackets are adjustable and may be placed in
any desired positions between the ends of the shelf. A uniform load of
intensity q, which includes the weight of the shelf itself, acts on the shelf
(see part b of the figure).
t
Problem 5.6-18
(a) Determine the maximum permissible value of the load q if the
allowable bending stress in the shelf is sallow 8.5 MPa and
the position of the supports is adjusted for maximum load carrying
capacity.
(b) The bookshelf owner decides to reinforce the shelf with a bottom
wood plate b/2 t/2 along its entire length (see figure part c). Find the
new maximum permissible value of the load q if the allowable bending
stress in the shelf remains at sallow 8.5 MPa.
A
B
D
C
b
L
(a)
q
A
D
B
C
L
(b)
Solution 5.6-18
(a) FOR MAXIMUM LOAD-CARRYING CAPACITY, PLACE THE SUPPORTS SO THAT M1 |M2|.
Let x length of overhang.
M1 qL
1L 4x2
8
ƒ M2 ƒ qx 2
2
qL
qx 2
(L 4x) 8
2
L
Solve for x: x 112 12
2
NUMERICAL VALUES L 1215 mm b 305 mm t 22 mm sallow 8.5 MPa
‹
x
L
A 12 1 B
2
Solve for q
qmax 4bt 2sallow
3L2 A 3 2 12 B
6.61
kN
m
S
bt 2
2.46 * 104 mm3
6
Ia S
t
2.706 * 105 mm4
2
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Page 484
CHAPTER 5 Stresses in Beams (Basic Topics)
b
2t
(b) a b 0.305 m t 0.022 m
b t
A bt +
22
a 6.932
b t t
t
t
+ bt a + b
2 2 4
2
2
c2 A
3
13t
c1 t c2 2
20
I
b t 3
a b
2 2
+
12
Stop b×t
C1
C2
αt × t/2
17t
20
(c)
c1 + c2 33 mm
bt
t 2
bt 3
t 2 193 bt 3
ac2 b +
+ bt a c1 b 22
4
12
2
960
I
193 bt 2
c1
624
Sbot I
193 bt 2
c2
816
Stop 4.566 * 104 mm3 Sbot 3.492 * 104 mm3
% increase in smaller section modulus:
Stop S
S
85.6 %
Sbot S
41.912 %
S
qL2
13 2 122 and Mmax sallow Stop
8
Check if Stop controls:
Mmax qmax 8 sallow
13 2 122 L
2
1Stop2 12.26
kN
m
Check if Sbot controls:
qmax 8 sallow
13 2 122 L2
1Sbot2 9.37
kN
m
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Page 485
SECTION 5.6 Design of Beams
Problem 5.6-19 A steel plate (called a cover plate) having cross sectional
dimensions 6.0 in. * 0.5 in. is welded along the full length of the bottom flange
of a W 12 * 50 wide-flange beam (see figure, which shows the beam cross
section).
What is the percent increase in the smaller section modulus (as compared to
the wide-flange beam alone)?
485
W 12 50
6.0 0.5 in. cover plate
Solution 5.6-19
FIND I ABOUT HORIZ. CENTROIDAL AXIS
NUMERICAL PROPERTIES FOR W 12 * 50
(FROM TABLE F-1(a))
A 14.6 in.2
d 12.2 in.
c1 c2
d
2
c1 I 391 in.4
Ih I + A a c1 + (6)(0.5) a c2 c1 d
0.5
+ (6)(0.5)ad +
b
2
2
A + (6)(0.5)
c2 (d + 0.5) c1
0.5 2
b
2
Ih 491.411 in.4
S 64.2 in.3
FIND SMALLER SECTION MODULUS
Ih
Stop Stop 68.419 in.3
c1
% increase in smaller section modulus
Stop S
;
(100) 6.57%
S
FIND CENTROID OF BEAM WITH COVER PLATE (TAKE
1ST MOMENTS ABOUT TOP TO FIND c1 7 c2)
A
d 2
1
b +
(6)(0.5)3
2
12
c1 7.182 in.
c2 5.518 in.
Problem 5.6-20 A steel beam ABC is simply supported
at A and B and has an overhang BC of length L 150 mm
(see figure). The beam supports a uniform load of intensity
q 4.0 kN/m over its entire span AB and 1.5q over BC.
The cross section of the beam is rectangular with width b
and height 2b. The allowable bending stress in the steel is
sallow 60 MPa, and its weight density is 77.0 kN/m3.
1.5 q
q
C
A
2b
B
2L
L
b
(a) Disregarding the weight of the beam, calculate the required width b of the rectangular cross section.
(b) Taking into account the weight of the beam, calculate the required width b.
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Page 486
CHAPTER 5 Stresses in Beams (Basic Topics)
Solution 5.6-20
NUMERICAL DATA
L 150 mm
(b) NOW MODIFY; INCLUDE BEAM WEIGHT
kN
m
kN
g 77 3
m
w gA
q4
sa 60 MPa
and
and
L2
2
2
Mmax s a a b3 b
3
Mmax (1.5q + w)
(a) IGNORE BEAM SELF-WEIGHT; FIND bmin
Mmax1 1.5q
w g12b22
Equate Mmax1 to Mmax2 & solve for bmin
L2
2
at B
Mmax2 s a S
2
3
a sa b b3 1gL22 b2 qL2 0
3
4
2
S b3
3
Insert numerical values, then solve for b
Equate Mmax1 to Mmax2 & solve for bmin
bmin 11.92 mm
1
9 qL2 3
bmin a
b
8 sa
bmin 11.91 mm
;
;
Problem 5.6-21 A retaining wall 6 ft high is constructed
of horizontal wood planks 2.5 in. thick (actual dimension)
that are supported by vertical wood piles of 12-in. diameter
(actual dimension), as shown in the figure. The lateral earth
pressure is p1 125 lb/ft2 at the top of the wall and
p2 425 lb/ft2 at the bottom.
(a) Assuming that the allowable stress in the wood is
1175 psi, calculate the maximum permissible spacing
s of the piles.
(b) Find the required diameter of the wood piles so that
piles and planks (t 2.5 in.) reach the allowable
stress at the same time.
(Hint: Observe that the spacing of the piles may be governed
by the load-carrying capacity of either the planks or the piles.
Consider the piles to act as cantilever beams subjected to a
trapezoidal distribution of load, and consider the planks to
act as simple beams between the piles. To be on the safe side,
assume that the pressure on the bottom plank is uniform and
equal to the maximum pressure.)
2.5 in.
p1 = 125 lb/ft2
12-in.
diam.
12-in.
diam.
s
6 ft
2.5 in.
Top view
p2 = 425 lb/ft2
Side view
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Page 487
487
SECTION 5.6 Design of Beams
Solution 5.6-21
Retaining wall
(a) sa 1175 psi t 2.5 in. p2 425
splank lb
2
ft
p1 125
lb
ft2
h 6 ft d 12 in.
4 sat2
3psad 3
49.2 in.
57.599 in. spole C 3p2
16 h2 12 p1 + p22
6 controls
smax spole
smax 49.2 in.
(b) EQUAL SPACING IF t ⴝ 2.14 in. OR d ⴝ 12.65 in.; HERE t IS SET AT 2.5 in., SO FIND d
3p sad3
16h212p1 + p22
4sat2
C 3p2
solving for d
d 12.65 in.
Problem 5.6-22 A beam of square cross section (a length of each side) is bent in
the plane of a diagonal (see figure). By removing a small amount of material at the top
and bottom corners, as shown by the shaded triangles in the figure, we can increase the
section modulus and obtain a stronger beam, even though the area of the cross section
is reduced.
(a) Determine the ratio b defining the areas that should be removed in order to obtain
the strongest cross section in bending.
(b) By what percent is the section modulus increased when the areas are removed?
y
ba
a
z
C
a
ba
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Page 488
CHAPTER 5 Stresses in Beams (Basic Topics)
Solution 5.6-22 Beam of square cross section with corners
removed
RATIO OF SECTION MODULI
S
(1 + 3b)(1 b)2
S0
Eq. (1)
GRAPH OF EQ. (1)
a length of each side
ba amount removed
Beam is bent about the z axis.
ENTIRE CROSS SECTION (AREA 0)
I0 a4
12
c0 I0
a
a3 12
S0 c0
12
12
(a) VALUE OF b FOR A MAXIMUM VALUE OF S/S0
d S
a b 0
db S0
SQUARE mnpq (AREA 1)
I1 (1 b)4a4
12
Take the derivative and solve this equation for b .
b
PARALLELOGRAM mm, n, n (AREA 2)
1
I2 (base)(height)3
3
1
9
;
(b) MAXIMUM VALUE OF S/S0
(1 b)a 3
ba4
1
I2 (ba12)c
d (1 b)3
3
6
12
Substitute b 1/9 into Eq. (1). (S/S0)max 1.0535
The section modulus is increased by 5.35% when
;
the triangular areas are removed.
REDUCED CROSS SECTION (AREA qmm, n, p, pq)
a4
I I1 + 2I2 (1 + 3b)(1 b)3
12
c
(1 b)a
12
S
I
12a3
(1 + 3b)(1 b)2
c
12
b
—
9
Problem 5.6-23
The cross section of a rectangular beam having
width b and height h is shown in part a of the figure. For reasons
unknown to the beam designer, it is planned to add structural projections
of width b/9 and height d to the top and bottom of the beam (see part
b of the figure).
For what values of d is the bending-moment capacity of the beam
increased? For what values is it decreased?
d
h
b
(a)
h
d
b
—
9
(b)
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 489
SECTION 5.6 Design of Beams
489
Solution 5.6-23 Beam with projections
Graph of
(1) ORIGINAL BEAM
I1 bh3
12
c1 h
2
S1 S2
d
versus
S1
h
d
h
S2
S1
0
0.25
0.50
0.75
1.00
1.000
0.8426
0.8889
1.0500
1.2963
I1
bh2
c1
6
(2) BEAM WITH PROJECTIONS
I2 1 8b 3
1 b
a bh +
a b (h + 2d)3
12 9
12 9
b
[8h3 + (h + 2d)3]
108
1
h
c2 + d (h + 2d)
2
2
S2 b[8h3 + (h + 2d)3]
I2
c2
54(h + 2d)
RATIO OF SECTION MODULI
b [8h3 + (h + 2d)3]
S2
S1
9(h + 2d)(bh2)
8 + a1 +
2d
b
9a 1 +
h
EQUAL SECTION MODULI
Set
S2
d
1 and solve numerically for .
S1
h
d
0.6861 and
h
d
0
h
2d 3
b
h
Moment capacity is increased when
d
7 0.6861
;
h
Moment capacity is decreased when
d
6 0.6861
;
h
NOTES:
S2
2d 3
2d
1 when a1 +
b 9a 1 +
b + 80
S1
h
h
or
d
0.6861 and 0
h
3 1
S2
d
14
is minimum when 0.2937
S1
h
2
a
S2
b 0.8399
S1 min
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 490
CHAPTER 5 Stresses in Beams (Basic Topics)
Nonprismatic Beams
Problem 5.7-1 A tapered cantilever beam AB of length
L has square cross sections and supports a concentrated load
P at the free end (see figure part a). The width and height of
the beam vary linearly from hA at the free end to hB at the
fixed end.
Determine the distance x from the free end A to the cross
section of maximum bending stress if hB 3hA.
(a) What is the magnitude smax of the maximum bending
stress? What is the ratio of the maximum stress to the
largest stress B at the support?
(b) Repeat part (a) if load P is now applied as a uniform load
of intensity q P/L over the entire beam, A is restrained
by a roller support and B is a sliding support (see figure
part b).
q = P/L
B
hA
A
B
A
hB
x
P
Sliding
support
x
L
L
(a)
(b)
Solution 5.7-1
(a) FIND MAX. BENDING STRESS FOR TAPERED
sB s(L)
CANTILEVER
h(x) hA a1 +
2x
b
L
M(x)
s(x) S(x)
s(x) s(x) sB S(x) h(x)
6
4PL
6(P)(x)
chA a 1 +
2x
bd
L
3
6PxL3
smax
9hA 3
sB
2PL
smax
2
sB
9hA 3
a Fv 0
6PxL3
d
c 3
d 0
dx hA (L + 2x)3
M(x) c c RA x L3
L3
36Px
d 0
hA3 (L 2x)3
hA3 (L 2x)4
hA3(L + 2x)4
0
L
smax s a b
4
smax 4PL
9hA 3
so x smax ;
(b) REPEAT (A), BUT NOW FOR DISTRIBUTED UNIFORM
LOAD OF P/L OVER ENTIRE BEAM
hA3(L + 2x)3
L + 4x
9hA 3
3
d
s(x) 0 then solve for xmax
dx
c6P
2PL
L
4
L
6P L3
4
L 3
hA3 a L + 2 b
4
M(x) Px M(x)
s(x) S(x)
RA P
P
x
xa b d d
L
2
1 2P
x
2 L
Px s(x) s(x) 3xP (2L + x)
1 2P
x
2 L
c hA a 1 +
2x 3
bd
L
6
L2
hA3 (L + 2x)3
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 5.7 Nonprismatic Beams
xmax 0.20871L
d
s(x) 0 then solve for xmax
dx
smax s (0.20871L)
PL
smax 0.394 3
;
hA
d
L2
c3xP (2L x) 3
d 0
dx
hA (L 2x)3
c3P (2L + x)
3xP
L2
sB s(L) so
hA3(L + 2x)3
L2
sB hA3 (L + 2x)3
2
+ 18xP (2L + x)
491
L
hA3 (L + 2x)4
d 0
smax
sB
PL
9hA3
a 0.39385
PL
hA3
b
PL
9hA3
smax
3.54
sB
Simplifying:
;
L2 5xL + x 2 0 so
xmax
5 152 4
L
2
Problem 5.7-2 A tall signboard is supported by two vertical beams consisting of thin-walled, tapered circular tubes
(see figure part a). For purposes of this analysis, each beam may be represented as a cantilever AB of length L 8.0 m
subjected to a lateral load P 2.4 kN at the free end. The tubes have constant thickness t 10.0 mm and average diameters
dA 90 mm and dB 270 mm at ends A and B, respectively.
Because the thickness is small compared to the diameters, the moment of inertia at any cross section may be obtained
from the formula I pd 3t/8 (see Case 22, Appendix E), and therefore, the section modulus may be obtained from the formula
S pd 2t/4.
(a) At what distance x from the free end does the maximum bending stress occur? What is the magnitude smax of the
maximum bending stress? What is the ratio of the maximum stress to the largest stress sB at the support?
(b) Repeat part (a) if concentrated load P is applied upward at A and downward uniform load q(x) 2P/L is applied over
the entire beam as shown. What is the ratio of the maximum stress to the stress at the location of maximum moment?
2P
q(x) = —
L
P = 2.4 kN
Wind
load
B
A
t
B
A
x
P
d
L = 8.0 m
t = 10.0 mm
x
L = 8.0 m
(b)
dA = 90 mm
dB = 270 mm
(a)
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Page 492
CHAPTER 5 Stresses in Beams (Basic Topics)
Solution 5.7-2
(a) FIND MAX. BENDING STRESS FOR TAPERED CANTILEVER
d(x) dA a1 +
2x
b
L
2
S(x) pd(x) t
4
P 2.4 kN
dA 90 mm
M(x)
s(x) S(x)
4P
s(x) pt
4P
xL2
s(x) c 2
d
pt dA (L + 2x)2
J c dA a 1 + L b d K
2x
xL2
d 4P
c
c 2
dd 0
dx pt dA (L + 2x)2
xL2
P
d 0
2
pt dA (L + 2x)3
so xmax L + 2x
ptdA2 (L + 2x)3
L
4m
2
smax 4P
≥
pt
d 0
(b) REPEAT (A), BUT NOW ADD DISTRIBUTED LOAD
L 2
L
2
L 2
b
2
PL
M(x) Px a
L + x
b
L
s(x) Px a
L + x
b
L
pt
2x 2
cdA a 1 +
bd
4
L
L
ptdA2 (L 2x)2
tension on top, compression on bottom of beam
¥
c4P (L + x)
4Px
L
ptdA2 (L + 2x)2
L
ptdA2 (L + 2x)2
16Px (L x)
2ptdA2
4 P L
9 pt dA2
P x
x b
L 2
M(x) aPx 2
L
d
c4Px (L x)
d 0
dx
ptdA2 (L 2x)2
Stress at support sB s(L)
sB 2
d
s(x) 0 then solve for xmax
dx
;
dA2 aL + 2
(2400)(8)
2p(0.010)(0.090)2
;
smax 37.7 MPa
s(x) 4Px (L x)
L
smax s a b
2
smax ;
M(x)
s(x) S(x)
L2
P
pt dA2 (L + 2x)2
c4PL2
4 P L
b
9 pt dA2
smax x
d
s(x) 0 then solve for xmax
dx
or
a
Evaluate using numerical data:
dB 270 mm
16
2ptdA2
smax
9
sB
8
L 8 m t 10 mm
c4
PL
smax
sB
OR simplifying
L
4
xmax 2 m
L
ptdA2 (L
c4PL2
2x)3
d 0
L + 4x
ptdA2 (L + 2x)3
d 0
so xmax ;
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SECTION 5.7 Nonprismatic Beams
Stress at support:
L
smax s a b
4
sB s(L)
L
L
smax 4P aL
b
4
4
J
smax L
ptdA2 aL 2
PL
sB 4PL (L + L)
L 2K
b
4
L
ptdA2 (L + 2L2)
sB 0 so no ratio of smax/sB is possible.
MAX. MOMENT AT L/2, SO COMPARE
3ptdA2
Stress at location of max. moment.
Evaluate using numerical data:
L
L
L
b
sa b 4P aL
2
2
2
L8m
P 2.4 kN
d A 90 mm
t 10 mm
dB 270 mm
smax 493
L
ptdA2 aL 2
L 2
b
2
1
L
L
sa b P
2
4 ptdA2
(2400)(8)
3p(0.010)(0.090)2
smax 25.2 MPa
;
PL
smax/s(L/2) 3ptdA2
L
1
a P
b
4 ptdA2
4
3
;
Problem 5.7-3 A tapered cantilever beam AB having rectangular cross sections is subjected to a concentrated
load P 50 lb and a couple M0 800 lb-in. acting at the free end (see figure part a). The width b of the beam is
constant and equal to 1.0 in., but the height varies linearly from hA 2.0 in. at the loaded end to hB 3.0 in.
at the support.
(a) At what distance x from the free end does the maximum bending stress smax occur? What is the magnitude
smax of the maximum bending stress? What is the ratio of the maximum stress to the largest stress sB at
the support?
(b) Repeat (a) if, in addition to P and M0, a triangular distributed load with peak intensity q0 3P/L acts upward over
the entire beam as shown in the figure part b. What is the ratio of the maximum stress to the stress at the location of
maximum moment?
P = 50 lb
P = 50 lb
A M0 = 800 lb-in.
hA =
2.0 in.
B
hB =
3.0 in.
x
b = 1.0 in.
3P
q0 = —
L
A M0 = 800 lb-in.
B
x
L = 20 in.
(a)
b = 1.0 in.
L = 20 in.
(b)
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Page 494
CHAPTER 5 Stresses in Beams (Basic Topics)
Solution 5.7-3
(a) FIND MAX. BENDING STRESS FOR TAPERED CANTILEVER
FIG. (A)
x
b
h(x) hA a1 +
2L
NUMERICAL DATA
b c hA a 1 +
2
x
bd
2L
6
L2
bhA2 (2L + x)2
d
s(x) 0 then solve for xmax
dx
4
M0 PL M0 800 in.-lb
5
bh(x)3
I(x)
I(x) S(x) h(x)
12
2
bh(x)2
S(x) 6
d
L2
c24 1Px + M02
d 0
dx
bhA2 (2L + x)2
24P
L2
bhA2 (2L + x)2
2124Px 24M02
2
x
b chA a1 +
bd
2L
L2
so x 0
x)3
2 2PL + Px + 2M0
bhA2 (2L
OR simplifying c24L
6
M(x) Px + M0
s(x) Px + M0
s(x) 24 1Px + M02
P 50 lb L 20 in.
hA 2 in. hB 3 in. b 1 in.
S(x) s(x) 21PL M02
bhA2 (2L + x)3
M(x)
S(x)
P
xmax 8 in.
;
agrees with plot at left
2000
Evaluate max. stress & stress at B using
numerical data:
smax s(8)
1500
1000
0
10
x (in.)
20
1260
(b) FIND MAX. BENDING STRESS FOR TAPERED
CANTILEVER, FIG. (B)
h(x) hA a 1 +
x
b
2L
4
PL
5
P
q0 3
L
M0 800 in.-lb
M0 1240
σ (x)
(psi)
1220
1200
I(x) 0
;
sB s(20)
sB 1200 psi
smax
1.042
;
sB
M(x)
(in.-lb)
500
smax 1250 psi
d 0
10
x (in.)
bh(x)3
12
S(x) I(x)
h(x)
2
S(x) bh(x)2
6
20
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Page 495
SECTION 5.7 Nonprismatic Beams
S(x) b chA a1 +
d
s(x) 0
dx
then solve for xmax
d
s(x) c124PL 12x2 q02
dx
6
M(x) Px + M0 +
s(x) 2
x
bd
2L
495
1 x
x
a q bx
2 L 0
3
L
bhA2 (2L + x)2
M(x)
S(x)
4x3q0) *
1500
2 (24PxL + 24M0L
L
bhA2 (2L + x)3
d 0
Simplifying:
12PL2 + 6PxL + 6x2 q0L
+ x3 q0 + 12M0 L 0
M(x)
1000
(in.-lb)
Solve for xmax:
xmax 4.642 in.
;
Max. stress & stress at B:
500
0
10
x (in.)
20
smax s (xmax)
smax 1235 psi
1400
sB s (20)
1200
MOMENT
1000
d
M(x) 0
dx
;
sB 867 psi
FIND MAX. MOMENT AND STRESS AT LOCATION OF MAX.
σ (x)
(psi)
xm 800
s(x) 0
10
x (in.)
q0 x3
Px + M0 6L
b chA a1 +
20
P(2L)
A q0
q0x3
d
aPx + M0 b 0
dx
6L
xm 16.33 in.
sm s(xm)
sm 1017 psi
smax
1.215
sm
;
2
x
bd
2L
6
s(x) 416PxL 6 M0 L + x3q02
L
*
bhA2 (2L + x)2
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Page 496
CHAPTER 5 Stresses in Beams (Basic Topics)
Problem 5.7-4 The spokes in a large flywheel are modeled as beams fixed at one end and loaded by a force P and a
couple M0 at the other (see figure). The cross sections of the spokes are elliptical with major and minor axes (height and width,
respectively) having the lengths shown in the figure part a. The cross-sectional dimensions vary linearly from end A to end B.
Considering only the effects of bending due to the loads P and M0, determine the following quantities.
(a) The largest bending stress sA at end A.
(b) The largest bending stress sB at end B.
(c) The distance x to the cross section of maximum bending stress.
(d) The magnitude smax of the maximum bending stress.
(e) Repeat part (d) if uniform load q(x) 10P/3L is added to loadings P and M0, as shown in the figure part b.
P = 12 kN
M0 = 10 kN.m
10P
q(x) = —
3L
B
A
P
x
M0
L = 1.25 m
A
B
x
L = 1.25 m
hA = 90 mm
hB = 120 mm
(b)
bA = 60 mm
bB = 80 mm
(a)
Solution 5.7-4
(a-d) FIND MAX. BENDING STRESS FOR TAPERED
30
CANTILEVER
NUMERICAL DATA
L 1.25 m bA 60 mm hA 90 mm
bB 80 mm hB 120 mm
M(x)
20
(kN•m)
P 12 kN M0 10 kN # m
h(x) hA a1
I(x) x
b
3L
pb(x)h(x)3
64
b(x) bA a 1 +
S(x) p b(x)h(x)2
S(x) 32
S(x) pbAhA2 a1 +
32
x 3
b
3L
I(x)
h(x)
2
x
b
3L
10
0
0.5
x (m)
1
0
0.5
x (m)
1
240
230
σ (x)
220
(MPa)
210
200
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SECTION 5.7 Nonprismatic Beams
M(x) Px + M0
s(x) s(x) M(x)
S(x)
I(x) p b(x) h(x)3
64
S(x) p b(x) h(x)2
32
Px + M0
p bA hA2 a1 +
x 3
b
3L
32
s(x) 864 a
Px + M0
L3
p bA hA
(3L + x)3
d
s (x) 0
dx
ba
2
b
S(x) s(x) p bAhA2 (3L + x)3
Px + M0
L3
0
2592
p bAhA2 (3L + x)4
M(x)
S(x)
10
5
3PL + 2Px + 3M0
p bAhA2 (3L + x)4
d 0
0
3(PL M0)
2P
xmax 0.625 m
;
so xmax σ (x)
(MPa)
Evaluate using numerical data:
smax 231 MPa
;
100
sA s(0)
sB s(L)
smax
1.045
sB
sA 210 MPa
sB 221 MPa
;
;
0
(e) FIND MAX. BENDING STRESS INCLUDING
UNIFORM LOAD
bA 60 mm
bB 80 mm
hB 120 mm
P 12 kN
M0 10 kN # m
x
b
h(x) hA a1 +
3L
x
b
3L
0.5
x (m)
1
0
0.5
x (m)
1
200
smax s(xmax)
L 1.25 m
0
300
agrees with plot above
b(x) bA a1 +
10 P x2
3 L 2
M(x)
(kN•m)
OR simplifying:
c864L3
32
15
L3
P
I(x)
h(x)
2
x 3
b
3L
p bA hA2 a 1 +
M(x) P x + M0 then solve for xmax
Px + M0
d
L3
c864
d 0
dx
p bAhA2 (3L + x)3
864
S(x) 497
hA 90 mm
P x + M0 s(x) ≥
10 P x2
3 L 2
p bAhA2 a 1 +
32
x 3
b
3L
¥
s(x) 288 13P xL 3 M0 L
L2
+ 5Px22
p bA hA2 (3L + x)3
d
s(x) 0 then solve for xmax
dx
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Page 498
CHAPTER 5 Stresses in Beams (Basic Topics)
OR
d
c288 13PxL 3 M0 L + 5Px22
dx
L2
*
pbAhA2 (3L + x)3
9PL2 36PxL + 5Px2 9M0L 0
d 0
Solving for x max: xmax 0.105 m
L2
d
s (x) c(864 PL 2880 Px)
dx
pbAhA2 (3 L x)3
31864PxL 864 M0 L1440Px22
L2
*
pbAhA2 (3L + x)4
d 0
Solution agrees with plot above; evaluate using
numerical data.
smax s(xmax)
sA s(0)
sB s(L)
smax 214 MPa
;
sA 210 MPa
;
sB 0 MPa
;
OR simplifying:
2
(288 L )
Problem 5.7-5
c9PL2 36PxL + 5Px2 9M0L d
cpbAhA2 (3L + x)4 d
0
Refer to the tapered cantilever beam of solid circular cross section shown in Fig. 5-24 of Example 5-9.
(a) Considering only the bending stresses due to the load P, determine the range of values of the ratio dB/dA for which
the maximum normal stress occurs at the support.
(b) What is the maximum stress for this range of values?
Solution 5.7-5 Tapered cantilever beam
FROM EQ. (5-33), EXAMPLE 5-9
s1 32Px
Eq. (1)
x 3
pcdA + (dB dA)a b d
L
After simplification:
FIND THE VALUE OF x THAT MAKES s1 A MAXIMUM
Let s1 u
v
ds1
dx
du
dv
va b ua b
dx
dx
v2
x 3
N p cdA + (dB dA)a b d [32P]
L
x 2 1
[32Px][p][3]cdA + (dB dA)a b d c (dB dA) d
L
L
N
D
x 2
x
N 32pPcdA + (dB dA)a b d cdA 2(dB dA) d
L
L
x 6
D p 2 cdA + (dB dA) d
L
ds1
N
dx
D
x
32PcdA 2(dB dA) d
L
x 4
pcdA + (dB dA) a b d
L
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Page 499
499
SECTION 5.7 Fully Stressed Beams
ds1
x
0 dA 2(dB dA)a b 0
dx
L
dA
x
L
2(dB dA)
‹
1
2a
dB
1b
dA
(a) GRAPH OF x/L VERSUS dB/dA (EQ. 2)
Maximum bending stress occurs at the support when
1 …
Eq. (2)
dB
… 1.5
dA
;
(b) MAXIMUM STRESS (AT SUPPORT B)
Substitute x/L 1 into Eq. (1):
smax 32PL
;
pdB3
Fully Stressed Beams
q
Problems 5.7-6 to 5.7-8 pertain to fully stressed beams of rectangular
cross section. Consider only the bending stresses obtained from the
flexure formula and disregard the weights of the beams.
B
Problem 5.7-6
A cantilever beam AB having rectangular cross sections
with constant width b and varying height hx is subjected to a uniform load
of intensity q (see figure).
How should the height hx vary as a function of x (measured from the
free end of the beam) in order to have a fully stressed beam? (Express hx
in terms of the height hB at the fixed end of the beam.)
A
hx
hB
x
L
hx
hB
b
b
Solution 5.7-6 Fully stressed beam with constant width and varying height
hx height at distance x
hB height at end B
b width (constant)
AT DISTANCE x: M sallow 3qx 2
M
S
bhx2
qx 2
2
AT THE FIXED END (x L):
hB L
S
bhx2
6
3q
A bsallow
Therefore,
hB x
hx
x
hx hB
L
L
;
3q
A bsallow
hx x
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 5 Stresses in Beams (Basic Topics)
Problem 5.7-7 A simple beam ABC having rectangular cross sections
with constant height h and varying width bx supports a concentrated load
P acting at the midpoint (see figure).
How should the width bx vary as a function of x in order to have a
fully stressed beam? (Express bx in terms of the width bB at the midpoint
of the beam.)
P
A
h
B
C
x
L
—
2
L
—
2
h
h
bx
bB
Solution 5.7-7 Fully stressed beam with constant height and varying width
h height of beam (constant)
L
bx width at distance x from end A a 0 … x … b
2
bB width at midpoint B (x L/2)
Px
1
AT DISTANCE x M S bx h2
2
6
3Px
M
3Px
sallow bx S
bxh2
sallow h2
AT MIDPOINT B (x L/2)
bB 3PL
2sallowh2
Therefore,
bx
2bB x
2x
and bx bb
L
L
;
NOTE: The equation is valid for 0 … x …
L
and the
2
beam is symmetrical about the midpoint.
q
Problem 5.7-8
A cantilever beam AB having rectangular cross sections
with varying width bx and varying height hx is subjected to a uniform
load of intensity q (see figure). If the width varies linearly with x
according to the equation bx bB x/L, how should the height hx vary as
a function of x in order to have a fully stressed beam? (Express hx in
terms of the height hB at the fixed end of the beam.)
B
hB
hx
A
x
L
hx
hB
bx
bB
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SECTION 5.8 Shear Stresses in Rectangular Beams
Solution 5.7-8
501
Fully stressed beam with varying width and varying height
hx height at distance x
hB height at end B
bx width at distance x
bB width at end B
3qLx
B bB sallow
hx AT THE FIXED END (x L)
x
bx bB a b
L
3qL2
C bB sallow
hx
x
x
Therefore,
hx hB
hB A L
AL
hB AT DISTANCE x
qx 2
M
2
bxh2x
bB x
S
(hx)2
6
6L
3qLx
M
sallow S
bB h2x
;
Shear Stresses in Rectangular Beams
Problem 5.8-1
The shear stresses t in a rectangular beam are given by
Eq. (5-43):
t
V h2
a y21 b
2I 4
in which V is the shear force, I is the moment of inertia of the cross-sectional
area, h is the height of the beam, and y1 is the distance from the neutral axis to
the point where the shear stress is being determined (Fig. 5-30).
By integrating over the cross-sectional area, show that the resultant
of the shear stresses is equal to the shear force V.
Solution 5.8-1 Resultant of the shear stresses
V shear force acting on the cross section
R resultant of shear stresses t
h/2
R
h/2
tbdy1 2
Lh/2
12V
h/2
(b)
a
L0
V h2
a y21 b bdy1
2I 4
2
h
y21 b dy1
4
bh
L0
12V 2h3
b V
3 a
24
h
I
bh3
12
t
V h2
a y21 b
2I 4
3
‹ R V Q.E.D.
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 5 Stresses in Beams (Basic Topics)
Problem 5.8-2
Calculate the maximum shear stress tmax
and the maximum bending stress smax in a wood beam
(see figure) carrying a uniform load of 22.5 kN/m (which
includes the weight of the beam) if the length is 1.95 m and the
cross section is rectangular with width 150 mm and height
300 mm, and the beam is either (a) simply supported as in the figure
part a, or (b) has a sliding support at right as in the figure
part b.
22.5 kN/m
300 mm
150 mm
1.95 m
(a)
22.5 kN/m
1.95 m
(b)
Solution 5.8-2
q 22.5 kN/m
h 300 mm
b 150 mm
smax L 1.95 m
M
S
smax 4.75 MPa
;
(b) MAXIMUM SHEAR STRESS
(a) MAXIMUM SHEAR STRESS
V qL
qL
A bh
V
2
tmax 3V
2A
tmax 731 kPa
tmax ;
;
qL2
2
M
qL2
bh2
S
8
6
smax Problem 5.8-3 Two wood beams, each of rectangular cross
section (3.0 in. 4.0 in., actual dimensions) are glued together
to form a solid beam of dimensions 6.0 in. 4.0 in. (see figure).
The beam is simply supported with a span of 8 ft.
(a) What is the maximum moment Mmax that may be applied
at the left support if the allowable shear stress in the
glued joint is 200 psi? (Include the effects of the beam’s
own weight, assuming that the wood weighs 35 lb/ft3.)
(b) Repeat part (a) if Mmax is based on allowable bending stress
of 2500 psi.
tmax 1462 kPa
MAXIMUM BENDING STRESS
MAXIMUM BENDING STRESS
M
3V
2A
M
S
smax 19.01 MPa
;
4.0 in.
M
6.0 in.
8 ft
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 5.8 Shear Stresses in Rectangular Beams
503
Solution 5.8-3
NUMERICAL DATA
L 8 ft b 4 in. h 6 in. tallow 200 psi sa 2500 psi
A bh g 35
lb
3
ft
q gA
weight of beam per unit distance
q 5.833
lb
ft
(a) MAXIMUM LOAD Mmax BASED ON SHEAR ALONE
V
qL
M
+
L
2
Mmax tmax qL
3V
3 M
a +
b
2A
2A L
2
qL2
2AL
tallow 3
2
Mmax 25.4 k-ft
(b) MAXIMUM LOAD Mmax BASED ON BENDING STRESS
Mmax sa S M qL2
2AL
tmax 3
2
based on allowable shear
Mm sa S S qL2
4.95 kip-ft Mmax 4.95 kip-ft
8
Problem 5.8-4 A cantilever beam of length L 2 m supports a
load P 8.0 kN (see figure). The beam is made of wood with
cross-sectional dimensions 120 mm * 200 mm.
Calculate the shear stresses due to the load P at points located
25 mm, 50 mm, 75 mm, and 100 mm from the top surface of the
beam. From these results, plot a graph showing the distribution of
shear stresses from top to bottom of the beam.
bh2
6
Mm M +
qL2
8
bending controls for this beam
P = 8.0 kN
200 mm
L=2m
120 mm
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CHAPTER 5 Stresses in Beams (Basic Topics)
Solution 5.8-4 Shear stresses in a cantilever beam
y1
(mm)
t
(MPa)
0
100
0
25
75
0.219
219
50
50
0.375
375
75
25
0.469
469
100 (N.A.)
0
0.500
500
Distance from the
top surface (mm)
Eq. (5-43): t V h2
a y21 b
2I 4
V P 8.0 kN 8,000 N
t
(kPa)
0
GRAPH OF SHEAR STRESS t
bh3
I
80 * 106 mm4
12
h 200 mm (y1 mm)
t
(200)2
y21 d
4
2(80 * 106)
8,000
c
(t N/mm2 MPa)
t 50 * 106(10,000 y21) (y1 mm; t MPa)
A steel beam of length L 16 in. and crosssectional dimensions b 0.6 in. and h 2 in. (see figure) supports a
uniform load of intensity q 240 lb/in., which includes the weight
of the beam.
Calculate the shear stresses in the beam (at the cross
section of maximum shear force) at points located 1/4 in., 1/2 in.,
3/4 in., and 1 in. from the top surface of the beam. From these
calculations, plot a graph showing the distribution of shear stresses
from top to bottom of the beam.
Problem 5.8-5
q = 240 lb/in.
h = 2 in.
L = 16 in.
b = 0.6 in.
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SECTION 5.8 Shear Stresses in Rectangular Beams
505
Solution 5.8-5 Shear stresses in a simple beam
y1
(in.)
t
(psi)
0
1.00
0
0.25
0.75
1050
0.50
0.50
1800
0.75
0.25
2250
0
2400
Distance from the
top surface (in.)
V h2
Eq. (5-43): t a y21 b
2I 4
V
1.00 (N.A.)
qL
bh3
1920 lb I 0.4 in.4
2
12
GRAPH OF SHEAR STRESS t
UNITS: POUNDS AND INCHES
t
1920 (2)2
c
y21 (2400)(1 y21) d
2(0.4) 4
(t psi; y1 in.)
Problem 5.8-6 A beam of rectangular cross section (width b and height h) supports a uniformly distributed load along its
entire length L. The allowable stresses in bending and shear are sallow and tallow, respectively.
(a) If the beam is simply supported, what is the span length L0 below which the shear stress governs the allowable load
and above which the bending stress governs?
(b) If the beam is supported as a cantilever, what is the length L0 below which the shear stress governs the allowable load
and above which the bending stress governs?
Solution 5.8-6
Beam of rectangular cross section
b width
L length
h height
Uniform load
SHEAR
q intensity of load
sallow and tallow
ALLOWABLE STRESSES
(a) SIMPLE BEAM
Vmax qL
A bh
2
tmax 3qL
3V
2A
4bh
qallow 4tallowbh
3L
BENDING
qL2
bh2
Mmax S
8
6
Equate (1) and (2) and solve for L0:
3qL3
Mmax
smax S
4bh2
qallow 4sallow bh
3L2
(2)
L 0 ha
2
sallow
b
tallow
;
(1)
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CHAPTER 5 Stresses in Beams (Basic Topics)
SHEAR
(b) CANTILEVER BEAM
Vmax qL A bh
BENDING
2
2
Mmax qL
2
bh
6
tmax 3qL
3V
2A
2bh
smax 3qL2
Mmax
S
bh2
qallow 2tallow bh
3L
S
sallowbh2
qallow Equate (3) and (4) and solve for L0:
(3)
2
3L
(4)
h sallow
L0 a
b
2 tallow
;
NOTE: If the actual length is less than L 0, the shear
stress governs the design. If the length is greater than
L0, the bending stress governs.
Problem 5.8-7
A laminated wood beam on simple
supports (figure part a) is built up by gluing together four
2 in. 4 in. boards (actual dimensions) to form a
solid beam 4 in. 8 in. in cross section, as shown
in figure part b. The allowable shear stress in the glued
joints is 62 psi, the allowable shear stress in the
wood is 175 psi, and the allowable bending stress
in the wood is 1650 psi.
P
L/3
2 in.
3 in.
2 in.
2 in.
2 in.
3 in.
2 in.
L = 12 ft
4 in.
(a)
4 in.
(c)
(a) If the beam is 12 ft long, what is the allowable
(b)
load P acting at the one-third point along the
beam, as shown? (Include the effects of the beam’s
own weight, assuming that the wood weighs 35 lb/ft3.)
(b) Repeat part (a) if the beam is assembled by gluing together two 3 in. 4 in. boards and a 2 in. 4 in. board
(see figure part c).
Solution 5.8-7
(a) L 12 ft b 4 in. h (2 in.) 4 tag 62 psi taw 175 psi saw 1650 psi
gw 35
lb
ft3
A bh 32 in.2
Pmax BASED ON ALLOWABLE SHEAR IN GLUE AT NA (CONTROLS OVER SHEAR IN ADJACENT WOOD)
2
L
Vmax P + gw A
3
2
and tmax 3Vmax
2A
so Vmax 2A
t
3 ag
Pmax 3AL gw
3 2A
L
a
tag gw A b A tag 2 3
2
4
Mmax 2 L
LL
L 1L
P + gw A
gw A
a
b
3 3
2 3
3 23
Pmax 9
LL
L 1L
csaw S cgw A
gw A
a
b d d 2.15 kip so Pmax based on shear in glued joint at NA
2L
2 3
3 23
controls
3ALgw
1.914 kip
4
Pmax BASED ON ALLOWABLE BENDING STRESS IN WOOD (AT TOP OR BOTTOM SURFACE); MAX. MOMENT IS AT LOCATION
OF LOAD P
Pmax A tag and Mmax saw S S b 1h22
6
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SECTION 5.8 Shear Stresses in Rectangular Beams
(b) Pmax BASED ON ALLOWABLE SHEAR IN GLUE AT 1 IN. ABOVE NA b 4 in. h (3 in.)2
Q 13 in. # 4 in.2 a1 in. +
Vmax tag
Pmax Ib
Q
2 in. 8 in. A bh
Vmax Q
3
bh3
in. b 30 in.3 I 170.667 in.4 tmax 2
12
Ib
and Vmax 2
L
P + gw A
3
2
3
Ib
L
gw A b 2.05 kip
atag
2
Q
2
controls
Pmax BASED ON ALLOWABLE SHEAR IN WOOD AT NA
Pmax 507
tmax 3 Vmax
2 A
3 2 taw A
L
a
gw A b 5.53 kip
2
3
2
Pmax BASED ON ALLOWABLE BENDING STRESS IN WOOD (AT TOP OR BOTTOM SURFACE); MAX. MOMENT IS AT LOCATION
OF LOAD P
Same as in PART (A):
Pmax 2.15 kip
Problem 5.8-8 A laminated plastic beam of square cross
section is built up by gluing together three strips, each
10 mm 30 mm in cross section (see figure). The beam has
a total weight of 3.6 N and is simply supported with span
length L 360 mm.
Considering the weight of the beam (q) calculate the
maximum permissible CCW moment M that may be placed
at the right support.
M
q
10 mm
10 mm 30 mm
10 mm
L
30 mm
(a) If the allowable shear stress in the glued joints is
0.3 MPa.
(b) If the allowable bending stress in the plastic is 8 MPa.
Solution 5.8-8
(a) FIND M BASED ON ALLOWABLE SHEAR STRESS IN GLUED
JOINT
ta 0.3 MPa
b 30 mm
h 30 mm
W 3.6 N
L 360 mm
q
ta Q
bh2
9
Q
2 3
Ib
b h
Q
4
Ib
3bh
N
m
beam distributed weight
MAX. SHEAR AT LEFT SUPPORT
Vm bh h
3 3
12
W
L
q 10
Q
bh2
9
qL
M
Ib
+
and Vm t a a b
2
L
Q
3
Vm Q
bh
I
Ib
12
Ib M L cta a
qL
Ib
b d
Q
2
M L cta a
qL
3bh
b d
4
2
Mmax 72.2 N # m
;
2 3
bh
12
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CHAPTER 5 Stresses in Beams (Basic Topics)
(b) FIND M BASED ON ALLOWABLE BENDING STRESS AT
h/2 FROM NA AT LOCATION (xm) OF MAX. BENDING
MOMENT, Mm
qL
qx2
M
M(x) a
+ bx
2
L
2
Mm a
d
M(x) 0
dx
qL
M
L
M
+ b a +
b
2
L
2
qL
qa
2
Use to find location of zero shear where
max. moment occurs:
Simplifying:
qx2
M
d qL
ca
+ bx
d
dx
2
L
2
Mm xm L
M 2
+
b
2
qL
2
2
1 1qL + 2M2
8q
L2
bh2
b
6
M
1
qL +
qx 0
2
L
Also Mm sa S
L
M
+
2
qL
Equating both Mm expressions & solving for M where
sa 8 MPa
MAX. MOMENT Mm
Mm a
qL
qxm
M
+ b xm 2
L
2
2
M
A
sa a
Mm sa a
bh2
b a8 qL2 b qL2
6
2
Mmax 9.01 N # m
Problem 5.8-9
A wood beam AB on simple supports with span length
equal to 10 ft is subjected to a uniform load of intensity 125 lb/ft acting
along the entire length of the beam, a concentrated load of magnitude
7500 lb acting at a point 3 ft from the right-hand support, and a moment at
A of 18,500 ft-lb (see figure). The allowable stresses in bending and shear,
respectively, are 2250 psi and 160 psi.
;
7500 lb
18,500 ft-lb
125 lb/ft
3 ft
A
(a) From the table in Appendix G, select the lightest beam that will
support the loads (disregard the weight of the beam).
(b) Taking into account the weight of the beam (weight density 35 lb/ft3),
verify that the selected beam is satisfactory, or if it is not, select a new beam.
B
10 ft
Solution 5.8-9
(a) q 125
L 10 ft
lb
P 7500 lb M 18,500 ft-lb
ft
d 3 ft
sallow 2250 psi
t allow 160 psi
RB 7.725 * 103 lb
Vmax RB
Vmax 7.725 * 103 lb
qd2
Mmax RB d 2
qL
d
M
+ P RA 2
L
L
Mmax 2.261 * 104 lb-ft
RA 1.025 * 103 lb
tmax RB qL
Ld
M
+ P
+
2
L
L
3Vmax
3V
Areq 2A
2tallow
Areq 72.422 in.2
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509
SECTION 5.8 Shear Stresses in Rectangular Beams
smax M
S
Sreq Mmax
sallow
Sreq 120.6 in.3
From Appendix G: Select 8 * 12-in. beam (nominal
;
dimensions).
S 165.3 in.3
A 86.25 in.2
Vmax RB
Areq Areq 73.405 in.2 < A
8 * 12 beam is still satisfactory for shear.
g 35
ft3
1b
ft
qtotal q + qbeam q total 145.964
(b) REPEAT (A) CONSIDERING THE WEIGHT OF THE BEAM
lb
3Vmax
2 tallow
Mmax RB d qbeam g A
qd2
2
Mmax 2.293 * 104 1b-ft
lb
q beam 20.964
ft
Sreq RB 7.725 * 103 lb +
qbeam L
2
Mmax
sallow
Sreq 122.3 in.3 < S
8 * 12 beam is still satisfactory for moment.
;
Use 8 * 12-in. beam
RB 7.83 * 103 lb
Problem 5.8-10 A simply supported wood beam of rectangular
cross section and span length 1.2 m carries a concentrated load
P at midspan in addition to its own weight (see figure). The cross
section has width 140 mm and height 240 mm. The weight density
of the wood is 5.4 kN/m3.
Calculate the maximum permissible value of the load P if
(a) the allowable bending stress is 8.5 MPa, and (b) the allowable
shear stress is 0.8 MPa.
P
240 mm
0.6 m
0.6 m
140 mm
Solution 5.8-10 Simply supported wood beam
(a) ALLOWABLE P BASED UPON BENDING STRESS
P
240 mm
0.6 m
b 140 mm
0.6 m
h 240 mm
A bh 33,600 mm2
S
bh2
1344 * 103 mm3
6
140 mm
sallow 8.5 MPa s Mmax
S
Mmax qL2
P(1.2 m)
PL
+
4
8
4
+
(181.44 N/m)(1.2 m)2
8
0.3P + 32.66 N # m
(P newtons; M N # m)
g 5.4 kN/m
Mmax Ssallow (1344 * 103 mm3)(8.5 MPa)
L 1.2 m q gbh 181.44 N/m
11,424 N # m
3
Equate values of Mmax and solve for P:
0.3P + 32.66 11,424 P 37,970 N
or P 38.0 kN
;
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CHAPTER 5 Stresses in Beams (Basic Topics)
(b) ALLOWABLE LOAD P BASED UPON SHEAR STRESS
tallow 0.8 MPa t V
3V
2A
qL
(181.44 N/m)(1.2 m)
P
P
+
+
2
2
2
2
P
+ 108.86 (N)
2
2At
2
V
(33,600 mm2)(0.8 MPa) 17,920 N
3
3
Equate values of V and solve for P:
P
+ 108.86 17,920 P 35,622 N
2
or P 35.6 kN
;
NOTE: The shear stress governs and
Pallow 35.6 kN
Problem 5.8-11 A square wood platform, 8 ft * 8 ft in area,
rests on masonry walls (see figure). The deck of the platform is
constructed of 2-in. nominal thickness tongue-and-groove planks
(actual thickness 1.5 in.; see Appendix G) supported on two
8-ft long beams. The beams have 4 in. * 6 in. nominal
dimensions (actual dimensions 3.5 in. * 5.5 in.).
The planks are designed to support a uniformly distributed
load w (lb/ft2) acting over the entire top surface of the platform.
The allowable bending stress for the planks is 2400 psi and
the allowable shear stress is 100 psi. When analyzing the
planks, disregard their weights and assume that their reactions
are uniformly distributed over the top surfaces of the supporting
beams.
8 ft
8 ft
(a) Determine the allowable platform load w1 (lb/ft2) based
upon the bending stress in the planks.
(b) Determine the allowable platform load w2 (lb/ft2) based
upon the shear stress in the planks.
(c) Which of the preceding values becomes the allowable
load wallow on the platform?
(Hints: Use care in constructing the loading diagram for the
planks, noting especially that the reactions are distributed loads
instead of concentrated loads. Also, note that the maximum shear
forces occur at the inside faces of the supporting beams.)
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 511
SECTION 5.8 Shear Stresses in Rectangular Beams
Solution 5.8-11
511
Wood platform with a plank deck
Load on one plank:
q c
w (lb/ft2)
144 in.2/ ft2
d(b in.) Reaction R qa
wb
(lb/in.)
144
96 in.
wb
wb
b a
b (48) 2
144
3
(R lb; w lb/ft2; b in.)
Mmax occurs at midspan.
Mmax Ra
q(48 in.)2
3.5 in.
89 in.
+
b 2
2
3
wb
wb
89
(46.25) (1152) wb
3
144
12
(M lb-in.; w lb/ft2; b in.)
Platform: 8 ft * 8 ft
t thickness of planks
Allowable bending moment:
1.5 in.
Mallow s allow S (2400 psi)(0.375b)
w uniform load on the deck (lb/ft )
900b (lb-in.)
2
sallow 2400 psi
Equate Mmax and Mallow and solve for w:
tallow 100 psi
89
wb 900b w1 121 lb/ft2
12
Find wallow (lb/ft2).
(a) ALLOWABLE LOAD BASED UPON BENDING STRESS IN THE
;
(b) ALLOWABLE LOAD BASED UPON SHEAR STRESS IN THE
PLANKS
PLANKS
Let b width of one plank (in.)
See the free-body diagram in part (a).
A 1.5b (in.2)
S
b
(1.5 in.)2
6
Vmax occurs at the inside face of the support.
Vmax qa
89 in.
b 44.5q
2
(44.5)a
0.375b (in.3)
Free-body diagram of one plank supported on the
beams:
89 wb
wb
b 144
288
(V lb; w lb/ft2; b in.)
Allowable shear force:
t
2Atallow
3V
Vallow 2A
3
2(1.5b)(100 psi)
100b (lb)
3
Equate Vmax and Vallow and solve for w:
89wb
100b w2 324 lb/ft2
288
;
(c) ALLOWABLE LOAD
Bending stress governs. wallow 121 lb/ft2
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 512
CHAPTER 5 Stresses in Beams (Basic Topics)
Problem 5.8-12 A wood beam ABC with simple supports
at A and B and an overhang BC has height h 300 mm
(see figure). The length of the main span of the beam is
L 3.6 m and the length of the overhang is L/3 1.2 m.
The beam supports a concentrated load 3P 18 kN at the
midpoint of the main span and a moment PL/2 10.8 kN . m
at the free end of the overhang. The wood has weight
density g 5.5 kN/m3.
3P
L
—
2
PL
M = –––
2
A
h=
300 mm
C
B
L
—
3
L
b
(a) Determine the required width b of the beam
based upon an allowable bending stress of 8.2 MPa.
(b) Determine the required width based upon an allowable
shear stress of 0.7 MPa.
Solution 5.8-12
Numerical data:
h 300 mm
L 3.6 m
A bh
g 5.5
s
P 6 kN
kN
M
PL
2
qbeam g A
m3
Reactions, max. shear and moment equations:
RA M
4
3P
4
+ qbeam L P + qbeam L
2
L
9
9
RB M
5
3P
5
+
+ qbeam L 2P + qbeam L
2
L
9
9
Vmax RB 2P +
MD RA
5
q
L
9 beam
PL
7
L
L2
qbeam
+
q
L2
2
2
2
72 beam
(a) REQUIRED WIDTH b BASED UPON BENDING STRESS
sallow 8.2 MPa
Mmax MD b
PL
7
+
q
L2
2
72 beam
6Mmax
Mmax
S
bh2
PL
2
a
sallow h2
7
gh L2 +
b
72
6
b 89.3 mm
(b) REQUIRED WIDTH b BASED UPON SHEAR STRESS
tallow 0.7 MPa
Vmax 2P +
t
b
5
q
L
9 beam
3Vmax
3Vmax
2A
2bh
3
5
3P
5
a2P + qbeam L b + gL
2bh
9
bh
6
3P
ha t allow 5
gLb
6
Shear stress governs.
b 87.8 mm
; (governs)
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 513
513
SECTION 5.9 Shear Stresses in Circular Beams
Shear Stresses in Circular Beams
Problem 5.9-1 A wood pole of solid circular cross section (d diameter)
q0 = 20 lb/in.
is subjected to a triangular distributed horizontal force of peak intensity
q0 20 lb/in (see figure). The length of the pole is L 6 ft, and the allowable
stresses in the wood are 1900 psi in bending and 120 psi in shear.
Determine the minimum required diameter of the pole based upon
(a) the allowable bending stress, and (b) the allowable shear stress.
d
d
L
Solution 5.9-1
q 20
1b
in.
3
L 6 ft
s allow 1900 psi
dmin 5.701 in.
t allow 120 psi
Vmax qL
2
Vmax 720 lb
qL 2L
Mmax 2 3
(b) BASED UPON SHEAR STRESS
t
Mmax 2.88 * 103 lb-ft
(a) BASED UPON BENDING STRESS
s
32 Mmax
A p sallow
dmin 32 M
M
S
pd3
4V
16V
3A
3pd2
dmin 16Vmax
A 3p tallow
dmin 3.192 in.
Bending stress governs. dmin 5.70 in.
;
Problem 5.9-2 A simple log bridge in a remote area consists of two
parallel logs with planks across them (see figure). The logs
are Douglas fir with average diameter 300 mm. A truck moves
slowly across the bridge, which spans 2.5 m. Assume that the weight
of the truck is equally distributed between the two logs.
Because the wheelbase of the truck is greater than 2.5 m, only
one set of wheels is on the bridge at a time. Thus, the wheel load on
one log is equivalent to a concentrated load W acting at any position
along the span. In addition, the weight of one log and the planks it
supports is equivalent to a uniform load of 850 N/m acting on the log.
Determine the maximum permissible wheel load W based upon
(a) an allowable bending stress of 7.0 MPa, and (b) an allowable
shear stress of 0.75 MPa.
x
W
850 N/m
300 mm
2.5 m
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Page 514
CHAPTER 5 Stresses in Beams (Basic Topics)
Solution 5.9-2
Log bridge
Diameter d 300 mm
sallow 7.0 MPa
tallow 0.75 MPa
Find allowable load W.
(b) BASED UPON SHEAR STRESS
Maximum shear force occurs when wheel is adjacent to support (x 0).
Vmax W +
(a) BASED UPON BENDING STRESS
Maximum moment occurs when wheel is at midspan
(x L/2).
Mmax qL
WL
+
4
8
A
0.625W + 664.1 (N # m) (W newtons)
S
W + 1062.5 N (W newtons)
2
W
1
(2.5 m) + (850 N/m)(2.5 m)2
4
8
pd3
2.651 * 103m3
32
qL
1
W + (850 N/m)(2.5 m)
2
2
pd2
0.070686 m2
4
tmax 4Vmax
3A
Vmax 3Atallow
3
(0.070686 m2)(0.75 MPa)
4
4
39,760 N
Mmax S # sallow (2.651 * 103 m3)(7.0 MPa)
‹ W + 1062.5 N 39,760 N
18,560 N # m
W 38,700 N 38.7 kN
;
‹ 0.625W + 664.1 18,560
W 28,600 N 28.6 kN
;
b
Problem 5.9-3 A sign for an automobile service station is supported by
two aluminum poles of hollow circular cross section, as shown in the
figure. The poles are being designed to resist a wind pressure of 75 lb/ft2
against the full area of the sign. The dimensions of the poles and sign
are h1 20 ft, h2 5 ft, and b 10 ft. To prevent buckling of the walls of
the poles, the thickness t is specified as one-tenth the outside diameter d.
(a) Determine the minimum required diameter of the poles based
upon an allowable bending stress of 7500 psi in the aluminum.
(b) Determine the minimum required diameter based upon an
allowable shear stress of 2000 psi.
h2
d
t=—
10
Wind
load
d
h1
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Page 515
SECTION 5.9 Shear Stresses in Circular Beams
Solution 5.9-3
Wind load on a sign
b width of sign
b 10 ft
p 75 lb/ft2
sallow 7500 psi
tallow 2000 psi
(b) REQUIRED DIAMETER BASED UPON SHEAR STRESS
Vmax W 1875 lb
d diameter W wind force on one pole
t
d
10
b
W ph2 a b 1875 lb
2
(a) REQUIRED DIAMETER BASED UPON BENDING STRESS
Mmax W ah1 +
h2
b 506,250 lb-in.
2
4
p
(d 4 d24) d2 d d1 d 2t d
64 2
5
I
t
4V r22 + r2r1 + r12
a
b
3A
r2 2 + r1 2
r1 d
d
d
2d
t 2
2
10
5
r2 d
2
r22 + r2r1 + r21
r2 2 + r1 2
2d 2
d 2d
d 2
a b + a ba b + a b
2
2
5
5
61
2
2
41
d
2d
a b + a b
2
5
A
p 2
p
4d 2
9pd2
(d2 d21) cd2 a b d 4
4
5
100
369pd 4
(in.4)
40,000
t
4V 61
100
V
a ba
b 7.0160 2
2
3 41 9pd
d
d
c
(d inches)
2
d2 4d 4
pd 4 369
p
a
b
I cd 4 a b d 64
5
64 625
515
s
M(d/2)
17.253 M
Mc
4
I
369pd /40,000
d3
d3 (17.253)(506,250 lb-in.)
17.253 Mmax
sallow
7500 psi
1164.6 in.3 d 10.52 in.
7.0160Vmax
tallow
(7.0160)(1875 lb)
6.5775 in.2
2000 psi
d 2.56 in.
;
(Bending stress governs.)
;
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Page 516
CHAPTER 5 Stresses in Beams (Basic Topics)
Problem 5.9-4 A steel pipe is subjected to a quadratic distributed load over its height,
q(x) =
with the peak intensity q0 at the base (see figure). Assume the following pipe properties and q0[1–(x/L)2]
dimensions: height L, outside diameter d 200 mm, wall thickness t 10 mm. Allowable
stresses for flexure and shear are sa 125 MPa and ta 30 MPa.
(a) If L 2.6 m, find q0,max (kN/m), assuming that allowable flexure and
shear stresses in the pipe are not to be exceeded.
(b) If q0 60 kN/m, find the maximum height Lmax (m) of the pipe if the
allowable flexure and shear stresses in the pipe are not to be exceeded.
L
x
q0
Solution 5.9-4
(a) SUM MOMENTS ABOUT A TO GET RB
L
RB 1
a
q(x) x dx b
L L0
RB L q0
1
x 2
L
c 0 q0 c1 a b d x dx d :
L L
L
4
L
RA RB +
L0
RB q0 L
4
L
q(x) dx RA RB +
5L q0
x 2
q0 c1 a b d dx :
L
12
L0
this is Vmax
Find x so that shear 0 to find Mmax:
x
V(x) RA 2
q0 c1 a b d d :
L
L0
q0 x3
5L q0
q0 x +
0
12
3L2
5L q0
q0 x3
5L q0
q0 x3
V(x)
x
+
q0 x +
q
0
12
12
3L2
3L2
Mmax at x 0.446298 L
FIND EXPRESSION FOR M(X)
x
M(x) RA x 2
q0 x4
q0 x4
5L q0 x
q0 c1 a b d (x ) d :
+
2
L
2
12
12
L
L0
Mmax M(0.446298L) 0.089673L q0
M(x) q0 x4
2
12 L
q0 x2
5L q0 x
+
2
12
this is Mmax
2
2
FIND q0,max BASED ON ALLOWABLE SHEAR STRESS
ta 30 MPa A 5.969 103 mm2
q1 ta a
tmax 4 Vmax r1 + r1 r2 + r2 2
a
b
3 A
r1 2 + r2 2
L 2.6 m
r1 2 + r2 2
3A
12
kN
b 82.801
ba ba 2
4
5L r1 + r1r2 + r2 2
m
FIND q0,max BASED ON ALLOWABLE FLEXURAL STRESS
smax Mmax
Sz
q2 sa Sz
2
0.089673 L
55.7
kN
m
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Page 517
SECTION 5.10 Shear Stresses in Beams with Flanges
(b) q0 60
517
kN
m
FIND Lmax BASED ON ALLOWABLE SHEAR STRESS
tmax 4 5L q0 r1 2 + r1r2 + r2 2
a
b
ba
3A
12
r21 + r22
so L1 ta a
r1 2 + r2 2
3A
12
b 2
3.588 m
ba
4
5q0 r1 + r1r2 + r2 2
FIND Lmax BASED ON ALLOWABLE FLEXURAL STRESS
smax 0.089673L2q0
Sz
so
L2 sa Sz
C 0.089673 q0
2.51 m
Shear Stresses in Beams with Flanges
Problem 5.10-1 through 5.10-6 A wide-flange beam (see figure) having
y
the cross section described below is subjected to a shear force V. Using the
dimensions of the cross section, calculate the moment of inertia and then
determine the following quantites:
(a) The maximum shear stress tmax in the web.
(b) The minimum shear stress tmin in the web.
(c) The average shear stress taver (obtained by dividing the shear
force by the area of the web) and the ratio tmax/taver.
(d) The shear force Vweb carried in the web and the ratio Vweb /V.
z
O h1
h
t
b
(Note: Disregard the fillets at the junctions of the web and flanges and
determine all quantities, including the moment of inertia, by considering
the cross section to consist of three rectangles.)
Probs 5.10-1 through 5.10-6
Problem 5.10-1 Dimensions of cross section: b 6 in., t 0.5 in.,
h 12 in., h1 10.5 in., and V 30 k.
Solution 5.10-1 Wide-flange beam
(b) MINIMUM SHEAR STRESS IN THE WEB (Eq. 5-53b)
b 6.0 in.
tmin t 0.5 in.
h 12.0 in.
taver V 30 k
V
5714 psi
th1
tmax
1.014
taver
MOMENT OF INERTIA (Eq. 5-52)
1
(bh3 bh31 + th31) 333.4 in.4
12
;
;
(d) SHEAR FORCE IN THE WEB (Eq. 5-54)
(a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-53a)
V
tmax (bh2 bh21 + th21) 5795 psi
8It
;
(c) AVERAGE SHEAR STRESS IN THE WEB (Eq. 5-55)
h1 10.5 in.
I
Vb 2
(h h12) 4555 psi
8It
;
Vweb th1
(2tmax + tmin) 28.25 k
3
Vweb
0.942
V
;
;
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Page 518
CHAPTER 5 Stresses in Beams (Basic Topics)
Problem 5.10-2 Dimensions of cross section: b 180 mm, t 12 mm,
h 420 mm, h1 380 mm, and V 125 kN.
Solution 5.10-2 Wide-flange beam
b 180 mm
(b) MINIMUM SHEAR STRESS IN THE WEB (Eq. 5-53b)
t 12 mm
tmin h 420 mm
taver V 125 kN
V
27.41 MPa
th1
tmax
1.037
taver
MOMENT OF INERTIA (Eq. 5-52)
1
(bh3 bh31 + th31) 343.1 * 106 mm4
12
;
;
(d) SHEAR FORCE IN THE WEB (Eq. 5-54)
Vweb (a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-53a)
tmax ;
(c) AVERAGE SHEAR STRESS IN THE WEB (Eq. 5-55)
h1 380 mm
I
Vb 2
(h h21) 21.86 MPa
8It
V
(bh2 bh21 + th21) 28.43 MPa
8It
;
th1
(2tmax + tmin) 119.7 kN
3
Vweb
0.957
V
;
;
Problem 5.10-3 Wide-flange shape, W 8 * 28 (see Table F-1(a),
Appendix F); V 10 k.
Solution 5.10-3 Wide-flange beam
(b) MINIMUM SHEAR STRESS IN THE WEB (Eq. 5-53b)
W 8 * 28
b 6.535 in.
tmin t 0.285 in.
h 8.06 in.
taver V 10 k
1
(bh3 bh31 + th31) 96.36 in.4
12
;
;
(d) SHEAR FORCE IN THE WEB (EQ. 5-54)
(a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-53a)
V
(bh2 bh21 + th21) 4861 psi
8It
V
4921 psi
th1
tmax
0.988
taver
MOMENT OF INERTIA (Eq. 5-52)
tmax ;
(c) AVERAGE SHEAR STRESS IN THE WEB (Eq. 5-55)
h1 7.13 in.
I
Vb 2
(h h21) 4202 psi
8It
;
Vweb th1
(2tmax + tmin) 9.432 k
3
Vweb
0.943
V
;
;
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Page 519
519
SECTION 5.10 Shear Stresses in Beams with Flanges
Problem 5.10-4 Dimensions of cross section: b 220 mm, t 12 mm,
h 600 mm, h1 570 mm, and V 200 kN.
Solution 5.10-4 Wide-flange beam
b 220 mm
(c) AVERAGE SHEAR STRESS IN THE WEB (EQ. 5-55)
t 12 mm
taver h 600 mm
V
29.24 MPa
th1
;
tmax
1.104
taver
h1 570 mm
V 200 kN
(d) SHEAR FORCE IN THE WEB (Eq. 5-54)
MOMENT OF INERTIA (Eq. 5-52)
Vweb 1
I
(bh3 bh31 + th31) 750.0 * 106 mm4
12
(a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-53a)
tmax V
(bh2 bh21 + th21) 32.28 MPa
8It
th1
(2tmax + tmin) 196.1 kN
3
Vweb
0.981
V
;
;
;
(b) MINIMUM SHEAR STRESS IN THE WEB (EQ. 5-53b)
tmin Vb 2
(h h21) 21.45 MPa
8It
;
Problem 5.10-5 Wide-flange shape, W 18 * 71
(see Table F-1(a), Appendix F); V 21 k.
Solution 5.10-5 Wide-flange beam
(b) MINIMUM SHEAR STRESS IN THE WEB (EQ. 5-53b)
W 18 * 71
b 7.635 in.
tmin t 0.495 in.
h 18.47 in.
taver V 21 k
1
(bh3 bh31 + th31) 1162 in.4
12
;
;
(d) SHEAR FORCE IN THE WEB (EQ. 5-54)
(a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-53a)
V
(bh2 bh21 + th21) 2634 psi
8It
V
2518 psi
th1
tmax
1.046
taver
MOMENT OF INERTIA (Eq. 5-52)
tmax ;
(c) AVERAGE SHEAR STRESS IN THE WEB (EQ. 5-55)
h1 16.85 in.
I
Vb 2
(h h21) 1993 psi
8It
;
Vweb th1
(2tmax + tmin) 20.19 k
3
Vweb
0.961
V
;
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 520
CHAPTER 5 Stresses in Beams (Basic Topics)
Problem 5.10-6 Dimensions of cross section: b 120 mm, t 7 mm,
h 350 mm, h1 330 mm, and V 60 kN.
Solution 5.10-6 Wide-flange beam
(b) MINIMUM SHEAR STRESS IN THE WEB (Eq. 5-53b)
b 120 mm
tmin t 7 mm
h 350 mm
taver V 60 kN
V
25.97 MPa
th1
tmax
1.093
taver
MOMENT OF INERTIA (Eq. 5-52)
1
(bh3bh31 + th31) 90.34 * 106 mm4
12
V
(bh2 bh21 + th21) 28.40 MPa
8It
;
;
(d) SHEAR FORCE IN THE WEB (Eq. 5-54)
Vweb (a) MAXIMUM SHEAR STRESS IN THE WEB (Eq. 5-53a)
tmax ;
(c) AVERAGE SHEAR STRESS IN THE WEB (Eq. 5-55)
h1 330 mm
I
Vb 2
(h h21) 19.35 MPa
8It
th1
(2tmax + tmin) 58.63 kN
3
Vweb
0.977
V
;
Problem 5.10-7 A cantilever beam AB of length L 6.5 ft supports a
trapezoidal distributed load of peak intensity q, and minimum intensity
q/2, that includes the weight of the beam (see figure). The beam is a
steel W 12 14 wide-flange shape (see Table F-1(a), Appendix F).
Calculate the maximum permissible load q based upon
(a) an allowable bending stress sallow 18 ksi and (b) an allowable
shear stress tallow 7.5 ksi. (Note: Obtain the moment of inertia
and section modulus of the beam from Table F-1(a).)
;
;
q
—
2
q
B
A
W 12 14
L = 6.5 ft
Solution 5.10-7
b 3.97 in.
I 88.6 in.4
t 0.2 in.
Vmax tf 0.225 in.
S 14.9 in.
3
h 11.9 in.
h1 h 2 tf
a
q
+ qb L
2
2
Vmax Mmax 1q 2
1 q 2L
L +
L
22
22 3
Mmax 5
qL2
12
3
qL
4
h1 11.45 in.
L 6.5 ft
s allow 18 ksi
t allow 7.5 ksi
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 521
SECTION 5.10 Shear Stresses in Beams with Flanges
3qL
1bh2 bh21 + th212
32It
tallow32It
q
2
3L1bh bh21 + th212
(a) MAXIMUM LOAD BASED UPON BENDING STRESS
5 2
qL
12
M
s
S
S
q
12Ssallow
5L2
q 1270 lb/ft
Vmax
1bh2 bh21 + th212
8It
Problem 5.10-8 A bridge girder AB on a simple span of
length L 14 m supports a distributed load of maximum
intensity q at mid-span and minimum intensity q/2 at supports
A and B that includes the weight of the girder (see figure).
The girder is constructed of three plates welded to form the
cross section shown.
Determine the maximum permissible load q based upon
(a) an allowable bending stress sallow 110 MPa, and
(b) an allowable shear stress tallow 50 MPa.
lb
ft
q 3210
(b) MAXIMUM LOAD UPON SHEAR STRESS
tmax 521
Shear stress governs.
q 1270 lb/ft
q
q
—
2
q
—
2
A
B
L = 14 m
;
450 mm
32 mm
16 mm
1800 mm
32 mm
450 mm
Solution 5.10-8
L 14 m
(a) MAXIMUM LOAD BASED UPON BENDING STRESS
h 1864 mm h1 1800 mm
b 450 mm
I
tf 32 mm tw 16 mm
1
1bh3 bh31 + tw h312
12
I 3.194 * 10 mm
10
S
2I
h
RA RB 4
S 3.427 * 107 mm3
qL
qL
3
+
qL
22
42
8
sallow 110 MPa
qLL
qLL
3
L
qL 8
2
22 4
24 6
Mmax 5
qL2
48
5
qL2
Mmax
48
s
S
S
qmax sallow S
5 2
L
48
qmax 184.7
kN
m
;
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 522
CHAPTER 5 Stresses in Beams (Basic Topics)
(b) MAXIMUM LOAD BASED UPON SHEAR STRESS
tallow 50 MPa
Vmax RA tmax 3
qL
8
qmax 3 qL
1bh2 bh21 + th212
64It
64 tallow Itw
3L1bh2bh12 tw h122
qmax 247 kN/m
Vmax
1bh2 bh21 + th212
8It
;
‹ Bending stress governs: qmax 184.7 kN/m
Problem 5.10-9 A simple beam with an overhang supports a uniform
load of intensity q 1200 lb/ft and a concentrated load P 3000 lb
at 8 ft to the right of A and also at C (see figure). The uniform load
includes an allowance for the weight of the beam. The allowable stresses
in bending and shear are 18 ksi and 11 ksi, respectively.
Select from Table F-2 (a), Appendix F, the lightest I-beam (S shape) A
that will support the given loads.
(Hint: Select a beam based upon the bending stress and then
calculate the maximum shear stress. If the beam is overstressed in
shear, select a heavier beam and repeat.)
Solution 5.10-9
Beam with an overhang
sallow 18 ksi
t allow 11 ksi
lb
ft
P 3000 lb
q 1200
L 12 ft
Sum moments about A & solve for RB:
P = 3000 lb
8 ft
;
P = 3000 lb
q = 1200 lb/ft
C
B
12 ft
4 ft
Find moment at D (at Load P between A and B):
MD RA (8 ft) q
(8 ft)2
2
MD 1.28 * 104 lb-ft
Mmax | MB|
Mmax 2.16 * 104 lb-ft
2
RB 4
1
q a Lb
+ P(8 ft + 16 ft)
3
2
12 ft
RB 1.88 * 104 lb
Required section modulus:
S
Mmax
S 14.4 in.3
sallow
Lightest beam is S 8 * 23 (from Table F-2(a)).
Sum forces in vertical direction:
I 64.7 in.4
S 16.2 in.3
RA q(16 ft) + 2P RB
b 4.17 in.
t 0.441 in.
t f 0.425 in.
h 8 in.
RA 6.4 * 103 lb
Vmax RB (P + q4 ft)
Vmax 1.1 * 104 lb at B
MB P (4 ft) q
(4 ft)2
2
MB 2.16 * 104 lb-ft
h1 h 2 tf h1 7.15 in.
Check max. shear stress.
tmax Vmax
1bh2 bh21 + th212
8It
tmax 3674 6 11,000 psi, so ok for shear
Select S 8 * 23 beam.
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 523
SECTION 5.10 Shear Stresses in Beams with Flanges
Problem 5.10-10 A hollow steel box beam has the rectangular
cross section shown in the figure. Determine the maximum allowable
shear force V that may act on the beam if the allowable shear
stress is 36 Mpa.
523
20
mm
450
10 mm mm
10 mm
20
mm
200 mm
Solution 5.10-10 Rectangular box beam
tallow 36 MPa
Find Vallow:
t
Vallow I
VQ
It
tallowIt
Q
1
1
(200)(450)3 (180)(410)3
12
12
484.9* 106mm4
Q (200) a
450 450
410 410
ba
b (180) a
ba
b
2
4
2
4
1.280 * 106 mm3
tallow It
Q
Vallow (36 MPa)(484.9 * 106 mm4)(20 mm)
273 kN
1.280 * 106 mm3
;
t 2(10 mm) 20 mm
1.0 in.
Problem 5.10-11 A hollow aluminum box beam has the square cross section
shown in the figure. Calculate the maximum and minimum shear stresses tmax
and tmin in the webs of the beam due to a shear force V 28 k.
1.0 in.
12 in.
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 524
CHAPTER 5 Stresses in Beams (Basic Topics)
Solution 5.10-11
Square box beam
Q a
1
(b3 b31) 91.0 in.3
8
V 28 k 28,000 lb
t1 1.0 in.
b 12 in.
b21 b1
b2 b
ba b a ba b
2
4
2
4
tmax VQ
(28,000 lb)(91.0 in.3)
1424 psi
It
(894.67 in.4)(2.0 in.)
1.42 ksi
b1 10 in.
;
MINIMUM SHEAR STRESS IN THE WEB (AT LEVEL A–A)
VQ
t
t 2t1 2.0 in.
It
bt1
b t1
Q Ay (bt1)a b a b (bt1)
2 2
2
MOMENT OF INERTIA
t1 b b1
2
MAXIMUM SHEAR STRESS IN THE WEB (AT NEUTRAL AXIS)
Q
(12 in.)
[(12 in.)2 (10 in.)2] 66.0 in.3
8
b
b2
Q A1y1A2y2 A1 ba b 2
2
tmin 1 4
I
(b b41) 894.67 in.4
12
A2 b1 a
b1
b21
b
2
2
b
Q (b2 b21)
8
VQ
(28,000 lb)(66.0 in.3)
1033 psi
It
(894.67 in4)(2.0 in.)
1.03 ksi
;
b1
b
1 b
1 b1
y1 a b y a b
2 2
4 2 2 2
4
y
Problem 5.10-12 The T-beam shown in the figure has cross-sectional dimensions
as follows: b 210 mm, t 16 mm, h 300 mm, and h1 280 mm. The beam
is subjected to a shear force V 68 kN.
Determine the maximum shear stress tmax in the web of the beam.
t
h1
z
h
C
c
b
Probs 5.10-12 and 5.10-13
Solution 5.10-12
h 300 mm
h1 280 mm
b 210 mm
t 16 mm
t f h h1
V 68 kN
t f 20 mm
LOCATION OF NEUTRAL AXIS
c
b1h h12 a
h h1
h1
b + th1 ah b
2
2
b1h h12 + th1
c 87.419 mm
c1 c
c1 87.419 mm
c2 h c c2 212.581 mm
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 525
SECTION 5.10 Shear Stresses in Beams with Flanges
MOMENT OF INERTIA ABOUT THE z AXIS
Iflange 2.531 * 107 mm4
1 3
1
tc + t1c1 tf 23
3 2
3
I Iweb + Iflange
Iweb Iweb 5.287 * 107 mm4
Iflange tf 2
1
btf3 + btf ac1 b
12
2
525
I 7.818 * 107 mm4
FIRST MOMENT OF AREA ABOVE THE z AXIS
c2
Q tc2
2
VQ
tmax tmax 19.7 MPa
;
It
Problem 5.10-13 Calculate the maximum shear stress tmax in the web
of the T-beam shown in the figure if b 10 in., t 0.5 in., h 7 in.,
h1 6.2 in., and the shear force V 5300 lb.
Solution 5.10-13
T-beam
h 7 in.
h1 6.2 in.
b 10 in.
t 0.5 in.
tf h h1
tf 0.8 in.
MOMENT OF INERTIA ABOUT THE z AXIS
Iweb 1 3
1
tc + t1c1 tf23
3 2
3
V 5300 lb
Iweb 29.656 in.4
LOCATION OF NEUTRAL AXIS
Iflange c
b1h h12 a
c 1.377 in.
c1 c
h h1
h1
b + th1 a h b
2
2
b 1h h12 + th1
c1 1.377 in.
c2 h c
c2 5.623 in.
tf 2
1
btf3 + btf ac1 b
12
2
Iflange 8.07 in.4
I Iweb + Iflange
I 37.726 in.4
FIRST MOMENT OF AREA ABOVE THE z AXIS
c2
Q tc2
2
VQ
tmax tmax 2221 psi
;
It
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 526
CHAPTER 5 Stresses in Beams (Basic Topics)
Built-Up Beams
Problem 5.11-1 A prefabricated wood I-beam serving as a floor joist
y
has the cross section shown in the figure. The allowable load in shear
for the glued joints between the web and the flanges is 65 lb/in. in the
longitudinal direction.
Determine the maximum allowable shear force Vmax for the beam.
0.75 in.
z
0.625 in.
8 in.
O
0.75 in.
5 in.
Solution 5.11-1
Wood I-beam
All dimensions in inches.
Find Vmax based upon shear in the glued joints.
Allowable load in shear for the glued joints is 65 lb/in.
‹ fallow 65 lb/in.
f
fallow I
VQ
Vmax I
Q
I
(b t)h31
bh3
12
12
1
1
(5)(9.5)3 (4.375)(8)3 170.57 in.4
12
12
;
Q Qflange Af df
(5)(0.75)(4.375) 16.406 in.3
Vmax fallowI
Q
(65 lb/in.)(170.57 in.4)
16.406 in.3
676 lb
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 527
SECTION 5.11 Built-Up Beams
Problem 5.11-2 A welded steel girder having the cross section shown in the figure
527
y
is fabricated of two 300 mm * 25 mm flange plates and a 800 mm * 16 mm
web plate. The plates are joined by four fillet welds that run continuously for the
length of the girder. Each weld has an allowable load in shear of 920 kN/m.
Calculate the maximum allowable shear force Vmax for the girder.
25 mm
z
16 mm
O
800 mm
25 mm
300 mm
Solution 5.11-2
h 850 mm
h1 800 mm
b 300 mm
t 16 mm
Qflange 3.094 * 106 mm3
f allow 920
tf 25 mm
I
(b t)h13
bh3
12
12
f 2 fallow
(2 welds, one either side of web)
f
I 3.236 * 109 mm4
Qflange Af df Qflange btf a
kN
m
h tf
2
b
VQ
I
Vmax Vmax 1.924 MN
fI
Qflange
;
y
Problem 5.11-3 A welded steel girder having the cross section shown in the figure
is fabricated of two 20 in. * 1 in. flange plates and a 60 in. * 5/16 in. web plate.
The plates are joined by four longitudinal fillet welds that run continuously throughout
the length of the girder.
If the girder is subjected to a shear force of 280 kips, what force F (per inch of
length of weld) must be resisted by each weld?
1 in.
z
O
60 in.
5
— in.
16
1 in.
20 in.
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Page 528
CHAPTER 5 Stresses in Beams (Basic Topics)
Solution 5.11-3
h 62 in.
h1 60 in.
b 20 in.
t
5
in.
16
Qflange btf a
I
bh
12
V 280 k
(b t)h13
12
F
I 4.284 * 10 in.
4
2
b
Qflange 610 in.3
tf 1 in.
3
h tf
4
f 2F VQflange
F 1994 * 103 lb-in.
21
F 1994 lb/in.
Qflange Af df
VQ
I
;
Problem 5.11-4 A box beam of wood is constructed of two
y
25 mm
260 mm * 50 mm boards and two 260 mm * 25 mm boards (see figure).
The boards are nailed at a longitudinal spacing s 100 mm.
If each nail has an allowable shear force F 1200 N, what is the
maximum allowable shear force Vmax?
z
O
50
mm
50
mm
260 mm
260 mm
Solution 5.11-4
Wood box beam
All dimensions in millimeters.
b 260 b1 260 2(50) 160
h 310 h1 260
s nail spacing 100 mm
F allowable shear force for one nail 1200 N
f shear flow between one
flange and both webs
fallow 25 mm
2(1200 N)
2F
24 kN/ m
s
100 mm
f
VQ
fallow I
Vmax I
Q
I
1
(bh3 b1h31) 411.125 * 106 mm4
12
Q Qflange Af df (260)(25)(142.5)
926.25 * 103 mm4
Vmax fallowI
(24 kN/m)(411.25 * 106 mm4)
.
Q
926.25 * 103 mm3
10.7 kN
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 529
SECTION 5.11 Built-Up Beams
Problem 5.11-5 A box beam is constructed of four wood
boards as shown in the figure part a. The webs are 8 in. 1 in.
and the flanges are 6 in. 1 in. boards (actual dimensions),
joined by screws for which the allowable load in shear is
F 250 lb per screw.
(a) Calculate the maximum permissible longitudinal
spacing Smax of the screws if the shear
force V is 1200 lb.
(b) Repeat part (a) if the flanges are attached to the webs
using a horizontal arrangement of screws as shown
in the figure part b.
Solution 5.11-5
Wood box beam
V 1200 lb
F 250 lb
y
1 in. Web
Web
8 in.
O
1 in.
1 in.
1 in.
(a)
(b) Horizontal screws:
h 8 in.
h1 6 in.
t 1 in.
b 6 in.
t 1 in.
3
(b 2t) h13
I
f
I 329.333 in.4
Qa 27 in.3
f
I 233.333 in.4
Qb 21 in.3
VQ
2F
I
s
smax 2FI
VQa
smax 5.08 in.
(b 2t) h13
bh3
12
12
Qb (b 2t) t (3.5 in.)
VQ
2F
I
S
smax 1 in.
(b)
b 8 in.
Qa bt (4.5 in.)
1 in.
6 in.
h1 8 in.
12
8 in.
1 in.
h 10 in.
bh
12
1 in.
6 in.
(a) Vertical screws:
I
Flange
Flange
z
529
2FI
VQb
smax 4.63 in.
;
;
y
Problem 5.11-6 Two wood box beams (beams A and B)
have the same outside dimensions (200 mm * 360 mm)
and the same thickness (t 20 mm) throughout, as shown
in the figure. Both beams are formed by nailing, with each
nail having an allowable shear load of 250 N. The beams are
designed for a shear force V 3.2 kN.
y
A
z
(a) What is the maximum longitudinal spacing sA for the
t=
nails in beam A?
20 mm
(b) What is the maximum longitudinal spacing sB for the
nails in beam B?
(c) Which beam is more efficient in resisting the shear force?
B
O
360
mm
z
O
360
mm
t=
20 mm
200 mm
200 mm
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 530
CHAPTER 5 Stresses in Beams (Basic Topics)
Solution 5.11-6 Two wood box beams
Cross-sectional dimensions are the same.
(a) BEAM A
All dimensions in millimeters.
Q Af df (bt)a
b 200 b1 200 2(20) 160
680 * 103 mm3
h 360 h1 360 2(20) 320
t 20
sA F allowable load per nail 250 N
V shear force 3.2 kN
I
1
(bh3 b1h31) 340.69 * 106 mm4
12
;
(b) BEAM B
Q Af df (b 2t)(t)a
f shear flow between one flange and both webs
‹ smax (2)(250 N)(340.7 * 106 mm4)
2FI
VQ
(3.2 kN)(680 * 103 mm3)
78.3 mm
s longitudinal spacing of the nails
VQ
2F
f
s
I
ht
1
b (200)(20)a b (340)
2
2
ht
b
2
1
(160)(20) (340)
2
2FI
VQ
544 * 103 mm3
sB (2)(250 N)(340.7 * 106 mm4)
2FI
VQ
(3.2 kN)(544 * 103 mm3)
97.9 mm
;
(c) BEAM B IS MORE EFFICIENT because the shear flow on
the contact surfaces is smaller and therefore fewer
;
nails are needed.
3 in.
—
16
Problem 5.11-7 A hollow wood beam with plywood webs has the
cross-sectional dimensions shown in the figure. The plywood is attached
to the flanges by means of small nails. Each nail has an allowable load
in shear of 30 lb.
Find the maximum allowable spacing s of the nails at cross sections
where the shear force V is equal to (a) 200 lb, and (b) 300 lb.
3 in.
—
16
3 in.
y
z
3
in.
4
8 in.
O
3
in.
4
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 531
531
SECTION 5.11 Built-Up Beams
Solution 5.11-7 Wood beam with plywood webs
(a) V 200 lb
All dimensions in inches.
b 3.375 b1 3.0
smax h 8.0 h1 6.5
2.77 in.
F allowable shear force for one nail 30 lb
s longitudinal spacing of the nails
f shear flow between one flange and both webs
f
VQ
2F
I
s
I
1
(bh3 b1h31) 75.3438 in.4
12
‹ smax 2FI
VQ
2(30 lb)(75.344 in.4)
2FI
VQ
(200 lb)(8.1563 in.3)
;
(b) V 300 lb
By proportion,
smax (2.77 in.)a
200
b 1.85 in.
300
;
Q Qflange Af df (3.0)(0.75)(3.625) 8.1563 in.3
Problem 5.11-8 A beam of T cross section is formed by nailing together two boards
y
having the dimensions shown in the figure.
If the total shear force V acting on the cross section is 1500 N and each nail may
carry 760 N in shear, what is the maximum allowable nail spacing s?
240 mm
60 mm
z
C
200 mm
60 mm
Solution 5.11-8
V 1500 N
F allow 760 N
h1 200 mm
b 240 mm
t 60 mm
h 260 mm
MOMENT OF INERTIA ABOUT THE NEUTRAL AXIS
I
A bt + h1t A 2.64 * 104 mm2
LOCATION OF NEUTRAL AXIS (z AXIS)
c2 bt ah1
h1
t
b + th1
2
2
1 3
1
tc2 + t1h1c223
3
3
+
1 3
t 2
bt + bt ac1 b
12
2
I 1.549 * 108 mm4
FIRST MOMENT OF AREA OF FLANGE
c2 170.909 mm
t
Q bt ac1 b
2
c1 h c2
Q 8.509 * 105 mm3
A
c1 89.091 mm
MAXIMUM ALLOWABLE SPACING OF NAILS
f
VQ
F
I
s
smax F allowI
VQ
smax 92.3 mm
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 532
CHAPTER 5 Stresses in Beams (Basic Topics)
Problem 5.11-9 The T-beam shown in the figure is fabricated by welding
together two steel plates. If the allowable load for each weld is 1.8 k/in. in the
longitudinal direction, what is the maximum allowable shear force V?
y
0.6 in.
5.5 in.
z
C
0.5 in.
4.5 in.
Solution 5.11-9
F allow 1.8
T-beam (welded)
MOMENT OF INERTIA ABOUT THE NEUTRAL AXIS
k
in.
h1 5.5 in.
b 4.5 in.
t1 0.6 in.
t2 0.5 in.
I
+
h 6 in.
A bt2 + h1t1 A 5.55 in.2
LOCATION OF NEUTRAL AXIS (z AXIS)
c2 t2
h1
bt2 + t1h1 a
+ t2 b
2
2
c2 2.034 in.
c1 3.966 in.
A
c1 h c 2
1
1
t c 3 + t1 1 c2 t223
3 1 1
3
t2 2
1
bt23 + bt2 ac2 b
12
2
I 20.406 in.4
FIRST MOMENT OF AREA OF FLANGE
Q bt2 ac2 t2
b
2
Q 4.014 in.3
MAXIMUM ALLOWABLE SHEAR FORCE
f
VQ
2F
I
Vmax 2Fallow I
Q
Vmax 18.30 k
;
y
Problem 5.11-10 A steel beam is built up from a W 410 * 85 wide flange beam
and two 180 mm * 9 mm cover plates (see figure). The allowable load in shear
on each bolt is 9.8 kN. What is the required bolt spacing s in the longitudinal
direction if the shear force V = 110 kN (Note: Obtain the dimensions and moment
of inertia of the W shape from Table F-1(b).)
z
180 mm 9 mm
cover plates
W 410 85
O
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Page 533
SECTION 5.11 Built-Up Beams
533
Solution 5.11-10
V 110 kN
F allow 9.8 kN
+ Acp a c W 410 * 85
I 4.57 * 108 mm4
Aw 10,800 mm2 hw 417 mm
Iw 310 * 106 mm4
First moment of area of one flange:
Acp (180)(9)(2) mm for two plates
2
Q 180 mm (9 mm)ac h hw + (9 mm)(2)
Maximum allowable spacing of nails:
LOCATION OF NEUTRAL AXIS (z AXIS)
h
2
9 mm
b
2
Q 3.451 * 105 mm3
A Aw + Acp A 1.404 * 104 mm2
c
9 mm 2
b
2
c 217.5 mm
Moment of inertia about the neutral axis:
f
VQ
2F
I
s
smax 2Fallow I
VQ
180 mm (9 mm)3
(2)
I Iw +
12
smax 236 mm
;
Problem 5.11-11 The three beams shown have approximately the same cross-sectional area. Beam 1 is a W 14 82 with
flange plates; Beam 2 consists of a web plate with four angles; and Beam 3 is constructed of 2 C shapes with flange plates.
(a) Which design has the largest moment capacity?
(b) Which has the largest shear capacity?
(c) Which is the most economical in bending?
(d) Which is the most economical in shear?
Assume allowable stress values are: sa 18 ksi and ta 11 ksi. The most economical beam is that having the largest capacityto-weight ratio. Neglect fabrication costs in answering (c) and (d) above. (Note: Obtain the dimensions and properties of all rolled
shapes from tables in Appendix F.)
8 0.52
4 0.375
Four angles
1
66—
2
W 14 82
Beam 1
8 0.52
14 0.675
C 15 50
4 0.375
Beam 2
Beam 3
Solution 5.11-11 Built-up steel beam
Beam 1: properties and dimensions for W14 * 82 with
flange plates
Aw 24 in.2
b1 8 in.
hw 14.3 in.
t1 0.52 in.
Iw 88l in.4
h1 hw + 2t1
bf 1 10.1 in.
tf1 0.855 in.
tw 0.51 in.
A1 Aw + 2b1t1
1
A1 32.32 in.2
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 534
CHAPTER 5 Stresses in Beams (Basic Topics)
I1 Iw +
t1 2
b1 * t31
hw
2 + b1t1 a
+ b 2
12
2
2
Q1 b1t1 a
I1 1.338 * 103 in.4
Beam 2: properties and dimensions for L6 * 6 * 1/2
angles with web plate
Aa 5.77 in.2
ca 1.67 in.
Ia 19.9 in.4
b2 14 in.
t2 0.675 in.
h2 b2
A2 4Aa + b2t2
I2 4Ia + Aa a
+ tw1
tf1
h1
t1
hw
b + bf1 tf1 a
b
2
2
2
2
Q2 2Aa a
h2
ca b + t2
2
Q3 b3 t3 a
hc 15 in.
Ic 404 in.4
t3 0.375 in.
h3 hc + 2t3
tf 0.65 in.
3
3
A3 2Ac + 2b3 t3
b2 2
b
2
2
Q2 78.046 in.
A2 32.53 in.2
tf3
h3
t3
hc
b + 2bf3 tf3 a b
2
2
2
2
+ 2tw3
Beam 3: properties and dimensions for C15 * 50 with
flange plates
bf 3.72 in.
a
3
I2 889.627 in.4
b3 4 in.
Q1 98.983 in.3
2
ha 6 in.
2
b2
t2 b32
ca b 4 +
2
12
Ac 14.7 in.2
2
hw
tf1 b
2
a
tw 0.716 in.
3
A3 32.4 in.2
a
2
hc
tf3 b
2
2
Q3 79.826 in.3
I1tw1
6.894 in.2
Q1
I2 t2
7.694 in.2
Q2
b3t33
hc
t3 2
I3 Ic 2 +
2 + b3 t3 a + b 2
12
2
2
I32tw3
17.676 in.2
Q3
I3 985.328 in.4
Itw
Case (3) with maximum
has the largest shear
Q
capacity. ;
(a) Beam with largest moment capacity; largest section
modulus controls.
Mmax sallow S
S1 2I1
h1
S1 174.449 in.3
S2 2I2
h2
S2 127.09 in.3
S3 2I3
h3
S3 125.121 in.3
largest value
Case (1) with maximum S has the largest moment
capacity. ;
(b) BEAM WITH LARGEST SHEAR CAPACITY: LARGEST Itw /Q
(c) MOST ECONOMICAL BEAM IN BENDING HAS LARGEST
BENDING CAPACITY-TO-WEIGHT RATIO
S3
3.862 in.
A3
6
S2
3.907 in.
A2
6
S1
5.398 in.
A1
;
Case (1) is the most economical in bending.
(d) MOST ECONOMICAL BEAM IN SHEAR HAS LARGEST
SHEAR CAPACITY-TO-WEIGHT RATIO
I1tw1
0.213
Q1A1
RATIO CONTROLS
tallow Itw
Vmax Q
largest value
6
6
I2 t2
0.237
Q2 A2
I3 tw3
0.273
Q3 A3
Case (3) is the most economical in shear.
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 535
SECTION 5.12 Beams with Axial Loads
Problem 5.11-12 Two W 310 * 74 steel wide-flange beams are bolted together to form
a built-up beam as shown in the figure. What is the maximum permissible bolt spacing s
if the shear force V 80 kN and the allowable load in shear on each bolt is F 13.5 kN
[Note: Obtain the dimensions and properties of the W shapes from Table F-1(b).]
535
W 310 74
W 310 74
Solution 5.11-12
V 80 kN
W 310 * 74
F allow 13.5 kN
Aw 9420 mm2
hw 310 mm
Iw 163 * 106 mm4
Location of neutral axis (z axis):
c hw
c 310 mm
FIRST MOMENT OF AREA OF FLANGE
Q Aw
hw 2
b d (2)
2
Q 1.46 * 106 mm3
MAXIMUM ALLOWABLE SPACING OF BOLTS
f
MOMENT OF INERTIA ABOUT THE NEUTRAL AXIS
I c Iw + Aw a
hw
2
VQ
2F
I
s
smax 2Fallow I
VQ
smax 180 mm
;
I 7.786 * 108 mm4
Beams with Axial Loads
When solving the problems for Section 5.12, assume that the bending
moments are not affected by the presence of lateral deflections.
P = 25 lb
Problem 5.12-1 While drilling a hole with a brace and bit, you exert
a downward force P 25 lb on the handle of the brace (see figure).
The diameter of the crank arm is d 7/16 in. and its lateral offset
is b 4-7/8 in.
Determine the maximum tensile and compressive stresses st and
sc, respectively, in the crank.
7
in.
d= —
16
7 in.
b = 4—
8
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 536
CHAPTER 5 Stresses in Beams (Basic Topics)
Solution 5.12-1
Brace and bit
P 25 lb (compression)
M Pb (25 lb)(4-7/8 in.)
121.9 lb-in.
MAXIMUM STRESSES
st P
121.9 lb-in.
M
25 lb
+
+
A
S
0.1503 in.2
0.008221 in.3
166 psi + 14,828 psi 14,660 psi
d diameter
d 7/16 in.
sc A
pd2
0.1503 in.2
4
S
pd3
0.008221 in.3
32
;
P
M
166 psi 14,828 psi
A
S
14,990 psi
;
Problem 5.12-2 An aluminum pole for a street light weights 4600 N
and supports an arm that weights 660 N (see figure). The center of gravity
of the arm is 1.2 m from the axis of the pole. A wind force of 300 N also acts
in the (y) direction at 9 m above the base. The outside diameter of the
pole (at its base) is 225 mm, and its thickness is 18 mm.
Determine the maximum tensile and compressive stresses st and sc,
respectively, in the pole (at its base) due to the weights and the wind force.
W2 = 660 N
1.2 m
P1 = 300 N
W1 = 4600 N
9m
18 mm
z
y
x
y
225 mm
x
Solution 5.12-2
W1 4600 N
b 1.2 m
Mx W2 b + P1h
W2 660 N
h9m
Mx 3.492 * 103 N # m
P1 300 N
d1 225 mm t 18 mm
MAXIMUM STRESS
d2 d1 2 t
A
p 2
1d d222
4 1
I
A 1.171 * 104 mm2
p
1d 4 d2 42
64 1
I 6.317 * 107 mm4
st a
Pz
Mx d1
+
b
A
I 2
st 5.77 * 103 kPa
5770 kPa
;
AT BASE OF POLE
Pz Mx d1
sc a b
A
I 2
Pz W1 + W2
sc 6.668 * 103
Pz 5.26 * 10 N
3
V y P1
(Axial force )
V y 300 N
(Moment)
6668 kPa
;
;
(Shear force)
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Page 537
537
SECTION 5.12 Beams with Axial Loads
Problem 5.12-3 A curved bar ABC having a circular axis (radius
h
B
r 12 in.) is loaded by forces P 400 lb (see figure). The cross
section of the bar is rectangular with height h and thickness t.
If the allowable tensile stress in the bar is 12,000 psi and the
height h 1.25 in., what is the minimum required thickness tmin?
C
A
P
P
45°
45°
r
h
t
Solution 5.12-3 Curved bar
TENSILE STRESS
st r radius of curved bar
e r r cos 45
r a1 M Pe P
r
c1 + 3(2 12) d
ht
h
MINIMUM THICKNESS
1
b
12
tmin Pr
(2 12)
2
P
r
c1 + 3(2 12) d
hsallow
h
SUBSTITUE NUMERICAL VALUES:
P 400 lb s allow 12,000 psi
CROSS SECTION
h height t thickness A ht S 3Pr(2 12)
P
M
P
+
+
A
S
ht
th2
1 2
th
6
r 12 in. h 1.25 in.
tmin 0.477 in.
;
B
Problem 5.12-4 A rigid frame ABC is formed by welding two
steel pipes at B (see figure). Each pipe has cross-sectional area
A 11.31 * 103 mm2, moment of inertia I 46.37 * 106 mm4,
and outside diameter d 200 mm.
Find the maximum tensile and compressive stresses st and sc,
respectively, in the frame due to the load P 8.0 kN if L H 1.4 m.
d
d
P
H
A
C
d
L
L
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Page 538
CHAPTER 5 Stresses in Beams (Basic Topics)
Solution 5.12-4
Rigid frame
P
sin a
2
AXIAL FORCE: N RA sin a PL
2
BENDING MOMENT: M RAL TENSILE STRESS
N
Mc
P sin a
PLd
+
+
A
I
2A
4I
st Load P at midpoint B:
P
REACTIONS: RA RC 2
BAR AB:
P 8.0 kN L H 1.4 m a 45
sin a 1/12 d 200 mm
A 11.31 * 103 mm2 I 46.37 * 106 mm4
H
tan a L
sin a SUBSTITUTE NUMERICAL VALUES
st H
1H2 + L2
2(11.31 * 103 mm2)
(8.0 kN)(1.4 m)(200 mm)
+
4(46.37 * 106 mm4)
d diameter
c d/2
(8.0 kN)(1/12)
0.250 MPa + 12.08 MPa
11.83 MPa (tension)
sc ;
N
Mc
0.250 MPa 12.08 MPa
A
I
12.33 MPa (compression)
Problem 5.12-5 A palm tree weighing 1000 lb is inclined
at an angle of 60 (see figure). The weight of the tree may be
resolved into two resultant forces, a force P1 900 lb acting at
a point 12 ft from the base and a force P2 100 lb acting at
the top of the tree, which is 30 ft long. The diameter at the base
of the tree is 14 in.
Calculate the maximum tensile and compressive stresses
st and sc, respectively, at the base of the tree due to its weight.
;
P2 = 100 lb
30 ft
12 ft
P1 = 900 lb
60°
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78572_ch05_ptg01_hr_435-556.qxd
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Page 539
539
SECTION 5.12 Beams with Axial Loads
Solution 5.12-5
Palm tree
M P1L1 cos 60 P2 L2 cos 60
[(900 lb)(144 in.) + (100 lb)(360 in.)] cos 60
82,800 lb-in.
N (P1 + P2) sin 60 (1000 lb) sin 60 866 lb
FREE-BODY DIAGRAM
MAXIMUM TENSILE STRESS
P1 900 lb
st P2 100 lb
L1 12 ft 144 in.
L2 30 ft 360 in.
d 14 in.
A
pd2
153.94 in.2
4
S
pd3
269.39 in.3
32
82,800 lb-in.
N
M
866 lb
+
+
2
A
S
153.94 in.
269.39 in.3
5.6 psi + 307.4 psi 302 psi
;
MAXIMUM COMPRESSIVE STRESS
sc 5.6 psi 307.4 psi 313 psi
Problem 5.12-6 A vertical pole of aluminum is fixed at the base and pulled
at the top by a cable having a tensile force T (see figure). The cable is attached
at the outer edge of a stiffened cover plate on top of the pole and makes an
angle a 20° at the point of attachment. The pole has length L 2.5 m and a
hollow circular cross section with outer diameter d2 280 mm and inner
diameter d1 220 mm. The circular cover plate has diameter 1.5d2.
Determine the allowable tensile force Tallow in the cable if the allowable
compressive stress in the aluminum pole is 90 MPa.
;
1.5 d2
a
T
L
d2
d1
d2
Solution 5.12-6
sallow 90 MPa
d1 220 mm
d2 280 mm
t
d2 d1
2
A
p
1d 2 d1 22
4 2
a 20
A 2.356 * 104 mm2
L 2.5 m
I
PN T cos (a)
(Axial force)
V T sin (a)
(Shear force)
M VL + PN a
1.5d2
b
2
(Moment)
p
1d 4 d1 42
64 2
I 1.867 * 108 mm4
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Page 540
CHAPTER 5 Stresses in Beams (Basic Topics)
ALLOWABLE TENSILE FORCE
sc T cos (a)
PN
M d2
A
I 2
A
T sin (a)L T cos (a) a
1.5d2
b
2
d2
I
sallow
Tallow cos (a)
+
A
sin (a)L + cos (a)a
Tallow 108.6 kN
I
1.5d2
b
2
d2
2
;
2
Problem 5.12-7 Because of foundation settlement,
a circular tower is leaning at an angle a to the vertical
(see figure). The structural core of the tower is a circular
cylinder of height h, outer diameter d2, and inner diameter d1.
For simplicity in the analysis, assume that the weight
of the tower is uniformly distributed along the height.
Obtain a formula for the maximum permissible angle
a if there is to be no tensile stress in the tower.
h
d1
d2
a
Solution 5.12-7
Leaning tower
CROSS SECTION
W weight of tower
a angle of tilt
A
p 2
(d d12)
4 2
I
p 4
(d d14)
64 2
p 2
(d d12)(d22 + d12)
64 2
d22 + d12
I
A
16
c
d2
2
AT THE BASE OF THE TOWER
h
N W cos a M Wa b sin a
2
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Page 541
SECTION 5.12 Beams with Axial Loads
TENSILE STRESS (EQUAL TO ZERO)
st ‹
Mc
W cos a
N
+
A
I
A
hd2 sin a
cos a
A
4I
MAXIMUM ANGLE a
d22 + d12
a arctan
4hd2
d2
W h
+ a sin a b a b 0
I 2
2
Problem 5.12-8 A steel bracket of solid circular cross section is subjected
to two loads, each of which is P 4.5 kN at D (see figure). Let the dimension
variable b 240 mm.
tan a 541
d22 + d12
4I
hd2A
4hd2
;
6b
A
(a) Find the minimum permissible diameter dmin of the bracket if the
allowable normal stress is 110 MPa.
(b) Repeat part (a), including the weight of the bracket. The weight density
of steel is 77.0 kN/m3.
B
2b
D
C
P
2b
P
Solution 5.12-8
(a) P 4.5 kN b 240 mm
gs 77
sa 110 MPa
kN
m3
Reactions at support A:
NA P
(compression)
MA P (2b 4b) 6.5 kN # m
(compression on bottom)
Max. normal stress at top of cross section at A (compression): N/A ⴙ My/I
smax d
16Pb2 a b
2
P
p d2
a
b
4
+
pd 4
a
b
64
4p
pd2
192Pb
+
pd3
smax 4P (48b + d)
pd3
Set smax sa and solve for required diameter d:
(psa)d3 (4P)d 192Pb 0
solve numerically or by trail & error to find dreqd = 8.46 cm
(b) DISTRIBUTED WEIGHT OF BRACKET w gA ; AFFECTS MOMENT M AT A , NOT AXIAL FORCE N
MA P(2b)
P(4b)
OR
MA 2Pb
so
smax w(6b)(3b)
w(2b)(6b)
w(2b)(5b) : 40wb2
w gs a
pd2
b
4
6Pb
compression on bottom
d
16Pb + 40wb22a b
2
2
40wb
P
pd 2
a
b
4
or
psa d3 (320p gs b2d2
8.91 8.46
5.3 %
8.46
+
a
192Pb
pd 4
b
64
4Pd) 0
solve numerically or by trial & error to find
dreqd 8.91 cm
increase in diameter
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Page 542
CHAPTER 5 Stresses in Beams (Basic Topics)
Problem 5.12-9 A cylindrical brick chimney of height H weighs
w 825 lb/ft of height (see figure). The inner and outer diameters are d1 3 ft
and d2 4 ft, respectively. The wind pressure against the side of the chimney is
p = 10 lb/ft2 of projected area.
Determine the maximum height H if there is to be no tension in the
brickwork.
p
w
H
d1
d2
Solution 5.12-9
Brick Chimney
I
d2
H
q
w
p 2
p 4
(d d14) (d d12)(d22 d12)
64 2
64 2
I
1 2
(d2 + d12)
A
16
c
d2
2
AT BASE OF CHIMNEY
N W wH
V
M
Md2
N
+
0
A
2I
p wind pressure
pd2 H 2
d22 + d12
2wH
8d2
q intensity of load pd2
SOLVE FOR H
d2 outer diameter
1
H
b pd2 H2
2
2
TENSILE STRESS (EQUAL TO ZERO)
s1 N
M qH a
H
2I
M
N
Ad2
or
w(d22 + d12)
d1 inner diameter
SUBSTITUTE NUMERICAL VALUES
W total weight of chimney wH
w 825 lb/ft
d2 4 ft
CROSS SECTION
q 10 lb/ft
Hmax 32.2 ft
A
2
;
4pd22
d1 3 ft
;
p 2
(d2 d12)
4
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Page 543
543
SECTION 5.12 Beams with Axial Loads
A flying buttress transmits a load P 25 kN, acting
at an angle of 60 to the horizontal, to the top of a vertical buttress AB
(see figure). The vertical buttress has height h 5.0 m and rectangular
cross section of thickness t 1.5 m and width b 1.0 m (perpendicular
to the plane of the figure). The stone used in the construction weighs
g 26 kN/m3.
What is the required weight W of the pedestal and statue above the
vertical buttress (that is, above section A) to avoid any tensile stresses
in the vertical buttress?
Problem 5.12-10
Flying
buttress
P
W
60°
A
A
—t
2
h
t
B
Solution 5.12-10
h
t
B
Flying buttress
FREE-BODY DIAGRAM OF VERTICAL BUTTRESS
CROSS SECTION
A bt (1.0 m)(1.5 m) 1.5 m2
1
1
S bt2 (1.0 m)(1.5 m)2 0.375 m3
6
6
AT THE BASE
N W + WB + P sin 60
W + 195 kN + (25 kN) sin 60
W + 216.651 kN
M (P cos 60) h (25 kN) (cos 60) (5.0 m)
62.5 kN # m
TENSILE STRESS (EQUAL TO ZERO)
P 25 kN
st h 5.0 m
t 1.5 m
b width of buttress perpendicular to the figure
M
N
+
A
S
62.5 kN # m
W + 216.651 kN
1.5 m2
+
0.375 m3
0
or W 216.651 kN + 250 kN 0
b 1.0 m
g 26 kN/m
3
W 33.3 kN
;
WB weight of vertical buttress
bthg
195 kN
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4:26 AM
Page 544
CHAPTER 5 Stresses in Beams (Basic Topics)
Problem 5.12-11
A plain concrete wall (i.e., a wall with no steel
reinforcement) rests on a secure foundation and serves as a small
dam on a creek (see figure). The height of the wall is h 6.0 ft
and the thickness of the wall is t 1.0 ft.
t
(a) Determine the maximum tensile and compressive stresses st and
sc, respectively, at the base of the wall when the water level reaches the
top (d h). Assume plain concrete has weight density gc 145 lb/ft3.
(b) Determine the maximum permissible depth dmax of the water if
there is to be no tension in the concrete.
Solution 5.12-11
h
d
Concrete wall
h height of wall
t thickness of wall
b width of wall (perpendicular to the figure)
STRESSES AT THE BASE OF THE WALL
(d DEPTH OF WATER)
st d 3gw
W
M
+
hgc +
A
S
t2
Eq. (1)
sc d 3gw
W
M
hgc 2
A
S
t
Eq. (2)
gc width density of concrete
gw weight density of water
d depth of water
W weight of wall
(a) STRESSES AT THE BASE WHEN d h
W bhtgc
h 6.0 ft 72 in. d 72 in.
F resultant force for the water pressure
t 1.0 ft 12 in.
MAXIMUM WATER PRESSURE = gw d
gc 145 lb/ft3 F
1
1
(d)(gw d)(b) bd2gw
2
2
d
1
M F a b bd3gw
3
6
145
lb/in.3
1728
gw 62.4 lb/ft3 62.4
lb/in.3
1728
1
A bt S bt2
6
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78572_ch05_ptg01_hr_435-556.qxd
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Page 545
SECTION 5.12 Eccentric Axial Loads
Substitute numerical values into Eqs. (1) and (2):
st 6.042 psi + 93.600 psi 87.6 psi
d3 (72 in.)(12 in.)2 a
;
sc 6.042 psi 93.600 psi 99.6 psi
dmax 28.9 in.
;
545
145
b 24,092 in.3
62.4
;
(b) MAXIMUM DEPTH FOR NO TENSION
Set st = 0 in Eq. (1):
hgc +
d3gw
2
t
0 d3 ht2 a
gc
b
gw
Eccentric Axial Loads
Problem 5.12-12 A circular post, a rectangular post, and a post of cruciform
cross section are each compressed by loads that produce a resultant force P acting
at the edge of the cross section (see figure). The diameter of the circular post and the
depths of the rectangular and cruciform posts are the same.
(a) For what width b of the rectangular post will the maximum tensile
stresses be the same in the circular and rectangular posts?
(b) Repeat part (a) for the post with cruciform cross section.
(c) Under the conditions described in parts (a) and (b), which post has the
largest compressive stress?
P
P
P
b
4 — = b
4
x
b
d
d
d
Load P here
d
4 — = d
4
Solution 5.12-12
(a) EQUAL MAXIMUM TENSILE STRESSES
COMPRESSION sc CIRCULAR POST
A
p 2
d
4
S
p 3
d
32
M
Pd
2
Tension
st P
M
A
S
4P
pd 2
16P
pd 2
20P
pd 2
RECTANGULAR POST
P
M
4P
16P
12P
+
2 +
2
A
S
pd
pd
pd2
A bd
TENSION
st S
bd2
6
M
Pd
2
M
P
3P
2P
P
+
+
A
S
bd
bd
bd
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 5 Stresses in Beams (Basic Topics)
COMPRESSION s c P
M
P
3P
4P
A
S
bd
bd
bd
Equate tensile stress expressions; solve for b:
12P
pd2
2P
bd
6
1
pd
b
b
pd
6
;
(b) CRUCIFORM CROSS SECTION
A cbd a
3
bd
b d 1 2
3
S c
+ a b
d bd 2
2 12
2 2 12 d
32
M
Pd
16P
3bd
3
2 a bd2 b
32
TENSION
st COMPRESSION
12 P
pd2
2P
3bd
1
3
pd
b
b
pd
3
;
(c) THE LARGEST COMPRESSIVE STRESS
Substitute expressions for b above & compare
compressive stresses:
CIRCULAR POST
bd
bd
22
3
Equate compressive stresses & solve for b:
16 P
12 P
4P
+
3bd
3bd
3bd
M
P
A
S
20 P
pd2
RECTANGULAR POST
sc M
P
+
A
S
sc sc 4P
24 P
2
pd
pd
a bd
6
CRUCIFORM POST
20 P
20 P
2
sc pd
pd
3
d
3
Rectangular post has the largest compressive
stress. ;
4P
16 P
20 P
3bd
3bd
3bd
Problem 5.12-13
Two cables, each carrying a tensile force
P 1200 lb, are bolted to a block of steel (see figure). The
block has thickness t 1 in. and width b 3 in.
b
P
(a) If the diameter d of the cable is 0.25 in., what are the maximum tensile
and compressive stresses st and sc, respectively, in the block?
(b) If the diameter of the cable is increased (without changing
the force P), what happens to the maximum tensile and compressive stresses?
P
t
Solution 5.12-13 Steel block loaded by cables
P 1200 lb d 0.25 in.
t 1.0 in. e t
d
+ 0.625 in.
2
2
b width of block
3.0 in.
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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547
SECTION 5.12 Eccentric Axial Loads
MAXIMUM COMPRESSIVE STRESS (AT BOTTOM OF BLOCK)
CROSS SECTION OF BLOCK
A bt 30 in.2
I
1 3
bt 0.25 in.4
12
t
y 0.5 in.
2
sc (a) MAXIMUM TENSILE STRESS (AT TOP OF BLOCK)
y
t
0.5 in.
2
Pey
P
st +
A
I
3 in.
(1200 lb)(0.625 in.)(0.5 in.)
1200 lb
3 in.
2
+
0.25 in.4
400 psi 1500 psi 1100 psi
;
(1200 lb)(0.625 in.)(0.5 in.)
1200 lb
2
Pey
P
+
A
I
+
(b) IF d IS INCREASED, the eccentricity e increases and
both stresses increase in magnitude.
0.25 in.4
400 psi + 1500 psi 1900 psi
;
Problem 5.12-14
A bar AB supports a load P acting at the centroid
of the end cross section (see figure). In the middle region of the bar
the cross-sectional area is reduced by removing one-half of the bar.
b
—
2
A
(a) If the end cross sections of the bar are square with sides of
length b, what are the maximum tensile and compressive stresses st
and sc, respectively, at cross section mn within the reduced region?
(b) If the end cross sections are circular with diameter b, what
are the maximum stresses st and sc?
b
b
b
b
—
2
m
(a)
b
—
2
n
B
P
b
(b)
Solution 5.12-14 Bar with reduced cross section
(a) SQUARE BAR
(b) CIRCULAR BAR
Cross section mn is a rectangle.
2
b
b
A (b) a b 2
2
b
M Pa b
4
c
I
Cross section mn is a semicircle
3
4
1
b
b
(b)a b 12
2
96
b
4
From Appendix E, Case 10:
b 4
I 0.1098a b 0.006860b4
2
STRESSES
st P
Mc
2P
6P
8P
+
2 + 2 2
A
I
b
b
b
sc P
Mc
2P
6P
4P
2 2 2
A
I
b
b
b
1 pb2
pb2
A a
b 0.3927b2
2 4
8
;
M Pa
2b
b 0.2122Pb
3p
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 5 Stresses in Beams (Basic Topics)
FOR TENSION:
ct 2.546
2
+ 6.564
2
2
8.903
b
P Mcc
sc A
I
2b
4r
0.2122b
3p
3p
FOR COMPRESSION:
cc r ct P
b
2b
0.2878b
2
3p
P
0.3927b
2.546
STRESSES
(0.2122Pb)(0.2122b)
Mct
P
P
st +
+
2
A
I
0.3927b
0.006860b4
Problem 5.12-15 A short column constructed
of a W 12 * 35 wide-flange shape is subjected to a resultant
compressive load P 12 k having its line of action at the
midpoint of one flange (see figure).
(a) Determine the maximum tensile and compressive
stresses st and sc, respectively, in the column.
(b) Locate the neutral axis under this loading condition.
(c) Recompute maximum tensile and compressive stresses
if a C 10 15.3 is attached to one flange, as shown.
P
b
P
P
9.11 2
b
b
;
2
(0.2122Pb)(0.2878b)
P = 25 k
0.006860b4
P
2
b
6.36
P
b2
;
y
C 10 15.3
(Part (c) only)
z
C
2
W 12 35
1
1
2
Solution 5.12-15
Column of wide-flange shape
PROPERTIES OF EACH SHAPE:
sc W 12 * 35
C 10 * 15.3
Aw 10.3 in.3
Ac 4.48 in.2
hw 12.5 in.
twc 0.24 in.
tf 0.52 in.
xp 0.634 in.
Iw 285 in.4
Ic 2.27 in.4 (2–2 axis)
hw
2
y0 P 25 k
st ew hw
2 2
Pew
P
+
c
Aw
Iw w
;
Iw
Aw ew
y0 4.62 in.
;
(C) COMBINED COLUMN, W 12 * 35 with C 10 * 15.3
h hw + twc
h 12.74 in.
cw 6.25 in.
tf
sc 5711 psi
(b) NEUTRAL AXIS (W SHAPE ALONE)
(a) THE MAXIMUM TENSILE AND COMPRESSIVE STRESSES
LOCATION OF CENTROID FOR W 12 35 ALONE
cw Pew
P
c
Aw
Iw w
A Aw + Ac
A 14.78 in.2
ew 5.99 in.
st 857 psi
;
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Page 549
549
SECTION 5.12 Eccentric Axial Loads
LOCATION OF CENTROID OF COMBINED SHAPE
c
Aw a
hw
b + Ac (h xp)
2
A
I Iw + Aw ac c 8.025 in.
hw 2
b
2
+ Ic + Ac (h xp c)2
I 394.334 in.
4
st P
Pe
+
c
A
I
sc P
Pe
(h c)
A
I
y0 I
Ae
Problem 5.12-16
A short column of wide-flange shape is
subjected to a compressive load that produces a resultant force
P 55 kN acting at the midpoint of one flange (see figure).
(a) Determine the maximum tensile and compressive
stresses st and sc, respectively, in the column.
(b) Locate the neutral axis under this loading condition.
(c) Recompute maximum tensile and compressive stresses
if a 120 mm 10 mm cover plate is added to one flange
as shown.
e hw P 25 k
tf
2
e 4.215 in.
c
st 453 psi
;
sc 2951 psi
;
y0 6.33 in. (from centroid)
;
P = 55 kN
z
y
Cover plate
(120 mm 10 mm)
(Part (c) only)
y
P
C
8 mm
z
200
mm
C
12 mm
160
mm
Solution 5.12-16
P 55 kN
(a) MAXIMUM TENSILE AND COMPRESSIVE STRESSES FOR W
SHAPE ALONE
PROPERTIES AND DIMENSIONS FOR W SHAPE
b 160 mm
d 200 mm
tf 12 mm
tw 8 mm
Aw bd (b tw)(d 2 tf)
Aw 5.248 * 103 mm2
(b tw)(d 2 tf)3
bd3
Iw 12
12
Iw 3.761 * 107 mm4
e
tf
d
2
2
P
Pe d
+
Aw
Iw 2
st 3.27 MPa
P
Pe d
Aw
Iw 2
sc 24.2 MPa
st sc e 94 mm
;
;
(b) NEUTRAL AXIS (W SHAPE ALONE)
y0 Iw
Aw e
y0 76.2 mm
;
(c) COMBINED COLUMN: W SHAPE & COVER PLATE
bp 120 mm tp 10 mm
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CHAPTER 5 Stresses in Beams (Basic Topics)
h d tp
I 4.839 * 107 mm4
tf
e 74.459 mm
ed c
2
h 210 mm
A Aw + bp tp
A 6.448 * 103 mm2
CENTROID OF COMPOSITE SECTION
tp
d
Aw + bp tp ad + b
2
2
c
A
st sc c 119.541 mm
+
12
d
b
2
2
y0 + bp tp a d +
tp
2
cb
P
Pe
(h c)
Aw
Iw
(a) Determine the maximum tensile stress st in the
angle section.
(b) Recompute the maximum tensile stress if two angles
are used and P is applied as shown in the figure part b.
Solution 5.12-17
AL 3.75 in.
I
Ae
y0 100.8 mm (from centroid)
1
L44—
2
C
1
3
1
P 2
3
P
(a)
(b)
A 2AL
t 0.5 in.
t 0.5 in.
c 1.18 in.
c 1.18 in.
t
b 12
2
IL 5.52 in.4 (2–2 axis)
e 1.315 in.
e ac P 12.5 k
c1 1.699 in.
Mc1
P
+
AL
I3
t
b
2
e 0.93 in.
P 12.5 k
I3 2.258 in.
4
M 16.44 k-in.
MAXIMUM TENSILE STRESS OCCURS AT CORNER
st 1
2L44—
2
C
(b) TWO ANGLES: L 4 * 4 * 1/2
rmin 0.776 in.
2
M Pe
;
Angle section in tension
(a) ONE ANGLE: L 4 * 4 * 1/2
I3 AL rmin2
sc 20.3 MPa
2
1
L 4 4 2 inch angle section (see Table F-4(a) in
Appendix F) is subjected to a tensile load P 12.5 kips
that acts through the point where the mid-lines of the legs
intersect (see figure part a).
c1 c 12
;
2
Problem 5.12-17 A tension member constructed of an
e ac st 1.587 MPa
NEUTRAL AXIS
I Iw + Aw ac bp t3p
P
Pe
+
c
A
I
st 15.48 ksi
;
I 2 IL
I 11.04 in.4
M Pe
M 11.625 k-in.
MAXIMUM TENSILE STRESS OCCURS AT THE LOWER EDGE
st P
Mc
+
A
I
st 2.91 ksi
;
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Page 551
SECTION 5.12 Eccentric Axial Loads
Two L 76 76 6.4 angles
Problem 5.12-18
A short length of a C 200 * 17.1 channel
is subjected to an axial compressive force P that has its line
of action through the midpoint of the web of the channel
(see figure part a).
(a) Determine the equation of the neutral axis under this
loading condition.
(b) If the allowable stresses in tension and compression
are 76 MPa and 52 MPa respectively, find the
maximum permissible load Pmax.
(c) Repeat parts (a) and (b) if two L 76 * 76 * 6.4 angles are
added to the channel as shown in the figure part b.
551
y
y
C 200 17.1
P
z
P
z
C
C
C 200 17.1
(a)
(b)
See Table F-3(b) in Appendix F for channel properties and Table F-4(b)
for angle properties.
Solution 5.12-18
P
tw = 5.59 mm
bf = 57.4 mm
sc
1
e
c1
Ac
Ic
Pmax 67.3 kN
;
(c) COMBINED COLUMN WITH 2 ANGLES
C 200 * 17.1
Ac 2170 mm2 dc 203 mm c1 14.5 mm
L 76 * 76 * 6.4
Ic 0.545 * 106 mm4 (z-axis)
AL 929 mm2
c2 bf c1 c2 42.9 mm
cL 21.2 mm
IL 0.512 * 106 mm4
ALLOWABLE STRESSES
COMPOSITE SECTION
st 76 MPa s c 52 MPa
A Ac + 2AL
A 4.028 * 103 mm2
ECCENTRICITY OF THE LOAD
h bf + 76 mm
h 133.4 mm
e c1 tw
2
e 11.705 mm
(a) LOCATION OF THE NEUTRAL AXIS (CHANNEL ALONE)
Ic
y0 Ac # e
y0 21.5 mm
;
P
Pe
+
c
A
I 2
P 165.025 kN
P
Pe
sc c
A
I 1
c 59.367 mm
I Ic + Ac 1bf c1 c22
+ 2 IL + 2AL 1bf + cL c22
I 2.845 * 106 mm4
(b) FIND PMAX
st CENTROID OF COMPOSITE SECTION
Ac 1bf c12 + 2AL 1bf + cL2
c
A
P
st
1
e
+ c2
Ac
Ic
e bf tw
c
2
e 4.762 mm
bf 57.4 mm
LOCATION OF THE NEUTRAL AXIS
y0 I
Ae
y0 148.3 mm
;
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Page 552
CHAPTER 5 Stresses in Beams (Basic Topics)
y0 148.3 mm 7 h 133.4 mm
;
P
Thus, this composite section has no tensile stress.
sc P
Pe
+
c
A
I
sc
1
e
+ c
A
I
Pmax 149.6 kN
;
Stress Concentrations
The problems for Section 5.13 are to be solved considering the
stress-concentration factors.
M
M
h
Problem 5.13-1 The beams shown in the figure are subjected
to bending moments M 2100 lb-in. Each beam has a rectangular
cross section with height h 1.5 in. and width b 0.375 in.
(perpendicular to the plane of the figure).
d
(a)
(a) For the beam with a hole at midheight, determine the
maximum stresses for hole diameters d 0.25, 0.50, 0.75,
and 1.00 in.
(b) For the beam with two identical notches (inside height
h1 1.25 in.), determine the maximum stresses for notch radii
R 0.05, 0.10, 0.15, and 0.20 in.
2R
M
M
h
Probs. 5.13-1 through 5.13-4
h1
(b)
Solution 5.13-1
M 2100 lb-in. h 1.5 in. b 0.375 in.
(b) BEAM WITH NOTCHES
(a) BEAM WITH A HOLE
h1 1.25 in.
d
1
6Mh
Eq. (5-62): sc …
h
2
b(h3 d3)
50,400
3.375 d3
Eq. (5-63):
(1)
1
12Md
d
Eq. (5-61): sB Ú
h
2
b(h3 d3)
d (in.)
d
h
0.25
0.50
0.75
1.00
0.1667
0.3333
0.5000
0.6667
67,200 d
3.375 d3
(2)
sc
sB
Eq. (1) (psi) Eq. (2) (psi)
15,000
15,500
17,100
—
—
—
17,100
28,300
h
1.5 in.
1.2
h1
1.25 in.
smax
(psi)
snom 6M
bh21
21,500 psi
R (in.)
R
h1
K
(Fig. 5-50)
smax Ks nom
smax (psi)
0.05
0.10
0.15
0.20
0.04
0.08
0.12
0.16
3.0
2.3
2.1
1.9
65,000
49,000
45,000
41,000
NOTE: The larger the notch radius, the smaller the stress.
15,000
15,500
17,100
28,300
NOTE: The larger the hole, the larger the stress.
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SECTION 5.13 Stress Concentrations
553
Problem 5.13-2 The beams shown in the figure are subjected to bending
moments M 250 N # m. Each beam has a rectangular cross section with height
h 44 mm and width b 10 mm (perpendicular to the plane of the figure).
(a) For the beam with a hole at midheight, determine the maximum stresses
for hole diameters d 10, 16, 22, and 28 mm.
(b) For the beam with two identical notches (inside height h1 40 mm),
determine the maximum stresses for notch radii R 2, 4, 6, and 8 mm.
Solution 5.13-2
M 250 N # m h 44 mm b 10 mm
(b) BEAM WITH NOTCHES
(a) BEAM WITH A HOLE
h1 40 mm
1
d
Eq. (5-62):
…
h
2
sc 6Mh
b(h3 d3)
6.6 * 106
85,180 d3
Eq. (5-63): snom MPa
d (mm)
10
16
22
28
300 * 103d
MPa
b(h3 d3)
85,180 d3
12Md
d
h
0.227
0.364
0.500
0.636
sB
sc
Eq. (2)
Eq. (1) (MPa) (MPa)
78
81
89
—
—
—
89
133
6M
bh21
93.8 MPa
(1)
d
1
Eq. (5-61):
Ú
h
2
sB 44 mm
h
1.1
h1
40 mm
(2)
smax
(MPa)
R
(mm)
R
h1
K
(Fig. 5-50)
smax Ks nom
smax (MPa)
2
4
6
8
0.05
0.10
0.15
0.20
2.6
2.1
1.8
1.7
240
200
170
160
NOTE: The larger the notch radius, the smaller the stress.
78
81
89
133
NOTE: The larger the hole, the larger the stress.
Problem 5.13-3 A rectangular beam with semicircular notches, as shown
in part b of the figure, has dimensions h 0.88 in. and h1 0.80 in. The
maximum allowable bending stress in the metal beam is smax 60 ksi, and
the bending moment is M 600 lb-in.
Determine the minimum permissible width bmin of the beam.
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CHAPTER 5 Stresses in Beams (Basic Topics)
Solution 5.13-3 Beam with semicircular notches
h 0.88 in.
h1 0.80 in.
smax 60 ksi M 600 lb-in.
smax Ksnom Ka
1
h h1 + 2R R (h h1) 0.04 in.
2
60 ksi 2.57 c
0.04 in.
R
0.05
h1
0.80 in.
Solve for b:
6M
bh21
b
6(600 lb-in.)
b(0.80 in.)2
d
;
bmin L 0.24 in.
From Fig. 5-50: K L 2.57
Problem 5.13-4 A rectangular beam with semicircular notches,
as shown in part b of the figure, has dimension h 120 mm and
h1 100 mm. The maximum allowable bending stress in the plastic
beam is smax 6 MPa, and the bending moment is M 150 N # m.
Determine the minimum permissible width bmin of the beam.
Solution 5.13-4 Beam with semicircular notches
h 120 mm
h1 100 mm
smax 6 MPa M 150 N # m
smax Ksnom Ka
1
h h1 + 2R R (h h1) 10 mm
2
6 MPa 2.20c
R
10 mm
0.10
h1
100 mm
Solve for b:
From Fig.5-50: K L 2.20
Problem 5.13-5 A rectangular beam with notches and a
hole (see figure) has dimensions h 5.5 in., h1 5 in., and width
b 1.6 in. The beam is subjected to a bending moment
M 130 k-in., and the maximum allowable bending stress
in the material (steel) is smax 42,000 psi.
6M
bh21
b
6(150 N # m)
b(100 mm)2
bmin L 33 mm
d
;
2R
M
M
h1
h
d
(a) What is the smallest radius Rmin that should be
used in the notches?
(b) What is the diameter dmax of the largest hole that should
be drilled at the midheight of the beam?
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Page 555
SECTION 5.13 Stress Concentrations
555
Solution 5.13-5 Beam with notches and a hole
h 5.5 in. h1 5 in. b 1.6 in.
M 130 k-in. smax 42,000 psi
(b) LARGEST HOLE DIAMETER
Assume
(a) MINIMUM NOTCH RADIUS
sB 5.5 in.
h
1.1
h1
5 in.
snom 1
d
7
and use Eq. (5-61).
h
2
12Md
b(h3 d3)
12(130 k-in.)d
6M
42,000 psi bh1
d3 + 23.21d 166.4 0
19,500 psi
2
42,000 psi
smax
K
2.15
snom
19,500 psi
From Fig. 5-50, with K 2.15 and
(1.6 in.)[(5.5 in.)3 d3]
or
Solve numerically:
h
1.1, we get
h1
dmax 4.13 in.
;
R
L 0.090
h1
‹ Rmin L 0.090h1 0.45 in.
;
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Page 556
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Page 557
6
Stresses in Beams
(Advanced Topics)
Composite Beams
When solving the problems for Section 6.2, assume that the
component parts of the beams are securely bonded by
adhesives or connected by fasteners. Also, be sure to use the
general theory for composite beams described in Sect. 6.2.
y
Problem 6.2-1 A composite beam consisting of fiberglass
faces and a core of particle board has the cross section
shown in the figure. The width of the beam is 2.0 in., the
thickness of the faces is 0.10 in., and the thickness of the
core is 0.50 in. The beam is subjected to a bending moment
of 250 lb-in. acting about the z axis.
Find the maximum bending stresses sface and score
in the faces and the core, respectively, if their respective
moduli of elasticity are 4 106 psi and 1.5 106 psi.
0.10 in.
z
0.50 in.
C
0.10 in.
2.0 in.
Solution 6.2-1 Composite beam
b 2 in.
h 0.7 in.
hc 0.5 in.
M 250 lb-in.
E1 4 10 psi
6
E2 1.5 106 psi
I1 b 3
(h h3c) 0.03633 in.4
12
I2 bh3c
0.02083 in.4
12
From Eq. (6-6b): score ;
E1I1 + E2I2 176,600 lb-in.
2
From Eq. (6-6a): sface ;
M(h/2)E1
E1I1 + E2I2
; 1980 psi
M(hc / 2)E2
E1I1 + E2I2
;531 psi
;
;
557
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 6 Stresses in Beams (Advanced Topics)
y
Problem 6.2-2 A wood beam with cross-sectional dimensions
200 mm 300 mm is reinforced on its sides by steel plates 12 mm
thick (see figure). The moduli of elasticity for the steel and wood
are Es 190 GPa and Ew 11 GPa, respectively. Also, the corresponding allowable stresses are ss 110 MPa and sw 7.5 MPa.
z
12 mm
C
200 mm
(a) Calculate the maximum permissible bending moment
z
Mmax when the beam is bent about the z axis.
(b) Repeat part (a) if the beam is now bent about its y axis.
(C) Find the required thickness of the steel plates on the beam
bent about the y axis so that Mmax is the same for both
12 mm
beam orientations.
300 mm
558
1/18/12
200 mm
300 mm
12 mm
12 mm
y
C
(a)
(b)
MAXIMUM MOMENT BASED UPON THE STEEL
Solution 6.2-2
Mmax_s sallow_s ≥
y
1
Ew Iw + Es Is
h
a b Es
2
¥
Mmax_s 58.7 kN # m
2
Mmax min (Mmax_w, Mmax_s)
z
300 mm
C
Mmax 58.7 kN # m
STEEL GOVERNS.
;
(b) BENT ABOUT THE Y AXIS
Iw 200 mm
12 mm
12 mm
b3 h
12
Is 2c
(a) BENT ABOUT THE Z AXIS
b 200 mm
t 12 mm
Ew 11 GPa
Es 190 GPa
sallow_w 7.5 MPa
Iw Is bh3
12
3
2th
12
h 300 mm
sallow_s 110 MPa
Iw 4.50 * 108 mm4
Is 5.40 * 107 mm4
EwIw EsIs 1.52 107 Nm2
Mmax_w sallow_w ≥
Mmax_w 69.1 kN # m
h
a b Ew
2
b + t 2
t3h
+ th a
b d
12
2
Is 8.10 * 107 mm4
Ew Iw + Es Is 1.76 * 107 N # m2
MAXIMUM MOMENT BASED UPON THE WOOD
Mmax_w sallow_w
J
Ew Iw + Es Is
b
a bEw
2
Mmax_w 119.9 kN # m
K
MAXIMUM MOMENT BASED UPON THE STEEL
MAXIMUM MOMENT BASED UPON THE WOOD
Ew Iw + Es Is
Iw 2.00 * 108 mm4
¥
Mmax_s sallow_s ≥
Ew Iw + Es Is
a
b
+ tb Es
2
Mmax_s 90.9 kN # m
¥
;
Mmax min (Mmax_w, Mmax_s)
STEEL GOVERNS.
Mmax 90.9 kN # m
;
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SECTION 6.2 Composite Beams
559
(c) MmaxX 58.7053 kN # m
Beam bent about y-axis (Fig. b):
Mmax Y sas
Ew Iw + Es Is
b
J Es a + t b K
2
Iw hb3
2 * 108 mm4
12
58.696 kN # m
sas
Find (t) 7.062 mm
Is 2
h t3
b
t 2
+ 2h t a + b
12
2
2
Ew Iw + Es Is
b
J Es a + tb K
2
Mmax X
t 7.08 mm
b
Ew a b
2
K
74.019 kN # m
y
Problem 6.2-3 A hollow box beam is constructed with webs
of Douglas-fir plywood and flanges of pine, as shown in the
figure in a cross-sectional view. The plywood is 1 in. thick
and 12 in. wide; the flanges are 2 in. 4 in. (nominal size).
The modulus of elasticity for the plywood is 1,800,000 psi
and for the pine is 1,400,000 psi.
(a) If the allowable stresses are 2000 psi for the plywood
and 1750 psi for the pine, find the allowable bending
moment Mmax when the beam is bent about the z axis.
(b) Repeat part (a) if the beam is now bent about its y axis.
z
1.5 in.
1.5 in.
z
C
1.5 in.
1 in.
3.5 in.
J
Ew Iw + Es Is
12 in.
saw
C
y
1 in.
1.5 in.
1 in.
12 in.
3.5 in.
1 in.
(b)
(a)
Solution 6.2-3
b 3.5 in.
(a) BENT ABOUT THE Z AXIS
t
t
1
h1
2
h
t 1 in.
h 12 in.
h1 9 in.
E1 1.4 * 106 psi
E2 1.8 * 106 psi
sallow_1 1750 psi
sallow_2 2000 psi
I1 b(h3 h31)
12
I2 2th3
12
I1 291 in.4
I2 288 in.4
E1I1 + E2I 2 9.26 * 108 lb # in.2
1
2
b
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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CHAPTER 6 Stresses in Beams (Advanced Topics)
MAXIMUM MOMENT BASED UPON THE WOOD
Mmax_1 sallow_1
J
E1 I1 + E2 I2
h
a b E1
2
Mmax_1 193 k-in.
;
MAXIMUM MOMENT BASED UPON THE WOOD
K
Mmax_1 sallow_1
J
E1 I1 + E2 I2
h
a b E2
2
Mmax_2 172 k-in.
;
K
MAXIMUM MOMENT BASED UPON THE PLYWOOD
K
Mmax_2 sallow_2
E1 I1 + E2 I2
Ja
Mmax_2 96 k-in.
Mmax min (Mmax_1, Mmax_2)
PLYWOOD GOVERNS.
b
a b E1
2
Mmax_1 170 k-in.
MAXIMUM MOMENT BASED UPON THE PLYWOOD
Mmax_2 sallow_2
J
E1 I1 + E2 I2
b
+ tb E2 K
2
Mmax min (Mmax_1, Mmax_1)
Mmax 172 k-in.
;
PLYWOOD GOVERNS.
Mmax 96 k- in.
;
(b) BENT ABOUT THE Y AXIS
I1 b3 (h h1)
12
I2 2 c
I1 11 in.4
b + t 2
t3h
+ th a
b d
12
2
I2 123 in.4
E1 I1 + E2 I2 2.37 * 108 lb # in.2
Problem 6.2-4 A round steel tube of outside diameter d2 and a brass core of diameter d1 are bonded to form a composite
beam, as shown in the figure.
(a) Derive formulas for the allowable bending moment M that can be carried by the beam based upon an allowable stress
ss in the steel and an allowable stress sB in the brass. (Assume that the moduli of elasticity for the steel and brass are
Es and EB, respectively.)
(b) If d2 50 mm, d1 40 mm, Es 210 GPa, EB 110 GPa, ss 150 MPa, sB 100 MPa, what is the maximum
bending moment M?
(c) What new value of brass diameter d1 will result in a balanced design? (Note that a balanced design is that in which
steel and brass reach allowable stress values at the same time).
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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SECTION 6.2 Composite Beams
561
Solution 6.2-4
(a) DERIVATION OF FORMULAS
p
Is Ad24 d14 B
64
MallowSteel IB ss1Es Is + EB IB2
Es a
MallowBrass sB
d2
b
2
1Es Is + EB IB2
Es a
d1
b
2
p
Ad14 B
64
pss A EB d1 4 Es d1 4 + Es d2 4 B
psB A EB d1 4 Es d1 4 + Es d2 4 B
32 Es d2
32 Es d1
(b) MAXIMUM BENDING MOMENT
d2 50 mm
d1 40 mm
Es 210 GPa
MmaxS pss 1EB d1 4 Es d1 4 + Es d2 42
MmaxB psB 1EB d1 4 Es d1 4 + Es d2 42
32 Es d2
32 Es d1
EB 110 GPa
ss 150 MPa
sB 100 MPa
1482 N # m
1235 N # m
(c) BALANCED DESIGN: equate allowable moments, then solve for d1.
Is p
p
1d 42
1d2 4 d1 42 IB 64
64 1
Mas ss 1Es Is + EB IB2
1481.738 N # m
1Es Is + EB IB2
1234.781 N # m
MaB sB
sB
S
Es a
d2
b
2
d1
Es a b
2
1Es Is + EB IB2
d1
Es a b
2
ss
B
d1
d2
1Es Is + EB IB2
d2
Es a b
2
or d1 sB
1d 2 33.333 mm
ss 2
d1 33.3 mm
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Problem 6.2-5 A beam with a guided support and 10-ft span
supports a distributed load of intensity q 660 lb/ft over its first
half (see figure part a) and a moment M0 300 ft-lb at joint B.
The beam consists of a wood member (nominal dimensions
6 in. 12 in., actual dimensions 5.5 in. 11.5 in. in cross
section, as shown in the figure part b) that is reinforced by
0.25-in.-thick steel plates on top and bottom. The moduli
of elasticity for the steel and wood are Es 30 106 psi
and Ew 1.5 106 psi, respectively.
y
0.25 in.
q
11.5 in.
M0
z
A
5 ft
C
C
B
5 ft
0.25 in.
(a) Calculate the maximum bending stresses ss in the steel
(a)
plates and sw in the wood member due to the applied loads.
(b) If the allowable bending stress in the steel plates is
sas 14,000 psi and that in the wood is saw 900 psi,
find qmax. (Assume that the moment at B, M0, remains
at 300 ft-lb.)
(c) If q 660 lb/ft and allowable stress values in (b) apply, what is M0,max at B?
5.5 in.
(b)
Solution 6.2-5
q 660 lb/in.
M0 300 lb # ft
L 10 ft
(b) MAXIMUM UNIFORM DISTRIBUTED LOAD
MAXIMUM MOMENT BASED UPON WOOD
(a) MAXIMUM BENDING STRESSES
sallow_w 900 psi
L 3L
Mmax q a b a b + M0
2
4
Mmax 25,050 lb-ft
b 5.5 in.
Wood (1):
From sallow_w h1 11.5 in.
b 3
1h h312
12
sallow_s 14,000 psi
t 0.25 in.
Es 30 * 106 psi
From sallow_s I2 94.93 in.4
sw ss h1
b Ew
2
Ew I1 + Es I2
h
Mmax a b Es
2
Ew I1 + Es I2
h
Mallow_s a b Es
2
Ew I1 + Es I2
Mallow_s 25,236 lb-ft
Ew I1 + Es I 2 3.894 * 109 lb-in.2
Mmax a
Ew I1 + Es I2
MAXIMUM MOMENT BASED UPON STEEL PLATE
I1 697.07 in.4
b 5.5 in.
h 12 in.
Plate (2):
I2 bh31
12
h1
b Ew
2
Mallow_w 33,857 lb-ft
Ew 1.5 * 106 psi
I1 Mallow_w a
MAXIMUM ALLOWABLE MOMENT
sw 666 psi
;
Mallow min (Mallow_s, Mallow_w)
STEEL PLATES GOVERN
ss 13,897 psi
;
Mallow 25,236 lb-ft
;
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SECTION 6.2 Composite Beams
(c) MAXIMUM APPLIED MOMENT
MAXIMUM UNIFORM DISTRIBUTED LOAD
L 3L
From Mallow qmax a b a b + M0
2
4
qmax 665 lb/ft
563
L 3L
From Mallow q a b a b + Mo_max
2
4
M0_max 486 lb-ft
;
;
y
Problem 6.2-6 A plastic-lined steel pipe has the cross-sectional shape shown
in the figure. The steel pipe has outer diameter d3 100 mm and inner diameter
d2 94 mm. The plastic liner has inner diameter d1 82 mm. The modulus
of elasticity of the steel is 75 times the modulus of the plastic.
(a) Determine the allowable bending moment Mallow if the allowable stress in
the steel is 35 MPa and in the plastic is 600 kPa.
(b) If pipe and liner diameters remain unchanged, what new value of allowable
stress for the steel pipe will result in the steel pipe and plastic liner
reaching their allowable stress values under the same maximum moment
(i.e., a balanced design)? What is the new maximum moment?
z
C
d1
d2 d3
Solution 6.2-6
d3 100 mm
n 75
d2 94 mm
d1 82 mm
Es/Ep
ssa 35 MPa spa 600 kPa
(a) FIND ALLOWABLE MOMENT BASED ON ALLOWABLE STRESSES IN STEEL AND PLASTIC
CROSS-SECTIONAL PROPERTIES
Is p
ad 4 d2 4 b 1.076 * 106 m4
64 3
Ip p
a d 4 d1 4 b 1.613 * 106 m4
64 2
MAXIMUM MOMENT BASED ON ALLOWABLE STRESS IN STEEL
Msmax ssa 1Es Is + Ep Ip2
d3
Es
2
ssa 1nIs + Ip2
d3
na b
2
768.428 N # m
6 steel governs.
Mallow 768 N # m
MAXIMUM MOMENT BASED ON ALLOWABLE STRESS IN PLASTIC
spa 1Es Is + Ep Ip2
spa 1nIs + Ip2
Mpmax 1051.042 N # m
d2
d2
Ep
a b
2
2
(b) BALANCED DESIGN - must equate expressions for Mmax, then solve for required ssa :
spa
d3
ssa
ssa nspa a b 47.87 MPa
ssa 47.9 MPa
nd3
d2
d2
ssa 1nIs + Ip2
spa 1nIs + Ip2
Msmax Mpmax 1052 N # m
1051 N # m
d3
d2
na b
a b
2
2
¿
same as above
increased due to increased ssa
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CHAPTER 6 Stresses in Beams (Advanced Topics)
y
Problem 6.2-7 The cross section of a sandwich beam consisting of aluminum
alloy faces and a foam core is shown in the figure. The width b of the beam
is 8.0 in., the thickness t of the faces is 0.25 in., and the height hc of the core
is 5.5 in. (total height h 6.0 in.). The moduli of elasticity are 10.5 106 psi for
the aluminum faces and 12,000 psi for the foam core. A bending moment
M 40 k-in. acts about the z axis.
Determine the maximum stresses in the faces and the core using (a) the
general theory for composite beams, and (b) the approximate theory for sandwich
beams.
t
z
Probs. 6.2-7 and 6.2-8
Solution 6.2-7
C
hc
b
t
h
Sandwich beam
I2 bh3c
110.92 in.4
12
M 40 k-in. E1I1 + E2I2 348.7 * 106 lb-in.2
(a) GENERAL THEORY (EQS. 6-7a AND b)
sface s1 M(h/2)E1
3610 psi
E1I1 + E2I2
score s2 M(hc / 2)E2
4 psi
E1I1 + E2I2
;
;
(1) ALUMINUM FACES:
b 8.0 in.
t 0.25 in.
h 6.0 in.
E1 10.5 * 106 psi
I1 I1 b 3
(h h3c ) 33.08 in.4
12
b 3
(h h3c ) 33.08 in.4
12
sface Mh
3630 psi
2I1
score 0
(2) Foam core:
b 8.0 in.
(b) APPROXIMATE THEORY (EQS. 6-9 AND 6-10)
hc 5.5 in.
;
;
E2 12,000 psi
Problem 6.2-8 The cross section of a sandwich beam consisting of fiberglass faces and a lightweight plastic core is shown in
the figure. The width b of the beam is 50 mm, the thickness t of the faces is 4 mm, and the height hc of the core is 92 mm (total
height h 100 mm). The moduli of elasticity are 75 GPa for the fiberglass and 1.2 GPa for the plastic. A bending moment
M 275 N # m acts about the z axis.
Determine the maximum stresses in the faces and the core using (a) the general theory for composite beams, and (b) the
approximate theory for sandwich beams.
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565
SECTION 6.2 Composite Beams
Solution 6.2-8
Sandwich beam
(a) GENERAL THEORY (EQS. 6-7a AND b)
sface s1 M(h/2)E1
14.1 MPa
E1I1 + E2I2
;
score s2 M(hc / 2)E2
0.21 MPa
E1I1 + E2I2
;
(b) APPROXIMATE THEORY (EQS. 6-9 AND 6-10)
I1 (1) Fiber glass faces:
b 50 mm
t 4 mm
h 100 mm
sface E1 75 GPa
I1 b 3
(h h3c ) 0.9221 * 106 m4
12
Mh
14.9 MPa
2I1
score 0
b 3
(h h3c ) 0.9221 * 106 m4
12
;
;
(2) Plastic core:
hc 92 mm
b 50 mm
E2 1.2 GPa
bh3c
I2 3.245 * 106 m4
12
M 275 N # m
E1I1 + E2I2 73,050 N # m2
Problem 6.2-9 A bimetallic beam used in a temperature-control switch
consists of strips of aluminum and copper bonded together as shown in the
figure, which is a cross-sectional view. The width of the beam is 1.0 in., and
each strip has a thickness of 1/16 in.
Under the action of a bending moment M 12 lb-in. acting about the
z axis, what are the maximum stresses sa and sc in the aluminum and copper,
respectively? (Assume Ea 10.5 106 psi and Ec 16.8 106 psi.)
y
1
— in.
16
A
z
O
1.0 in.
C
1
— in.
16
Solution 6.2-9 Bimetallic beam
NEUTRAL AXIS (EQ. 6-3)
CROSS SECTION
L1
y dA y1A1 (h1 t/2)(bt)
(h1 1/32)(1)(1/16) in.3
(1) Aluminum: E1 Ea 10.5 10 psi
6
(2) Copper:
E2 Ec 16.8 10 psi
L2
y dA y2A2 (h1 t t/2)(bt)
(h1 3/32)(1)(1/16) in.3
6
M 12 lb-in.
Eq. (6-4): E1
L1
y dA + E2
L2
y dA 0
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CHAPTER 6 Stresses in Beams (Advanced Topics)
(10.5 106)(h1 1/32)(1/16)
(16.8 106)(h1 3/32)(1/16) 0
MAXIMUM STRESSES (EQS. 6-7a AND b)
sa s1 Mh1E1
4120 psi
E1I1 + E2I2
;
sc s2 Mh2E2
5230 psi
E1I1 + E2I2
;
Solve for h1: h1 0.06971 in.
h2 2(1/16 in.) h1 0.05529 in.
MOMENTS OF INERTIA (FROM PARALLEL-AXIS THEOREM)
I1 bt3
+ bt(h1 t/2)2 0.0001128 in.4
12
I2 bt3
+ bt(h2 t/2)2 0.00005647 in.4
12
E1I1 + E2I2 2133 lb-in.2
Problem 6.2-10 A simply supported composite beam
y
3 m long carries a uniformly distributed load of intensity
q 3.0 kN/m (see figure). The beam is constructed of a wood
member, 100 mm wide by 150 mm deep, reinforced on its
lower side by a steel plate 8 mm thick and 100 mm wide.
q = 3.0 kN/m
150 mm
(a) Find the maximum bending stresses sw and ss in the
wood and steel, respectively, due to the uniform load if
the moduli of elasticity are Ew 10 GPa for the wood
and Es 210 GPa for the steel.
(b) Find the required thickness of the steel plate so that the
steel plate and wood reach their allowable stress values,
sas 100 MPa and saw 8.5 MPa, simultaneously
under the maximum moment.
z
O
8 mm
3m
100 mm
Solution 6.2-10 Simply supported composite beam
(a) BEAM: L 3 m
2
Mmax q 3.0 kN/m
qL
3375 N # m
8
CROSS SECTION
b 100 mm
h 150 mm
t 8 mm
(1) Wood: E1 Ew 10 GPa
(2) Steel: E2 Es 210 GPa
NEUTRAL AXIS
L1
y dA y1A1 (h1 h/2)(bh)
(h1 75)(100)(150) mm3
L2
y dA y2A2 (h + t/2 h1)(bt)
(154 h1)(100)(18) mm3
Eq. (6-4): E1
L1
y dA + E2
L2
y dA 0
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SECTION 6.2 Composite Beams
567
MAXIMUM STRESSES (EQS. 6-7a AND b)
(10 GPa)(h1 75)(100)(150)(109)
(210 GPa)(h1 154)(100)(8)(109) 0
sw s1 Solve for h1: h1 116.74 mm
Mh1E1
E1I1 + E2I2
5.1 MPa (Compression)
h2 h t h1 41.26 mm
MOMENTS OF INERTIA (FROM PARALLEL-AXIS THEOREM)
I1 bh3
+ bh(h1 h/2)2 54.26 * 106 mm4
12
I2 bt2
+ bt(h2 t/2)2 1.115 * 106 mm4
12
;
Mh2E2
ss s2 E1I1 + E2I2
37.6 MPa (Tension)
;
E1I1 + E2I2 776,750 N # m2
(b) as 100 MPa
b 100 mm
aw 8.5 MPa
Ew 10 GPa
Es 210 GPa
h 150 mm
M 3375 N # m
ss M h2 Es
Es Is + Ew Iw
sw M h1 Ew
Es Is + Ew Iw
Solving for M and equating expressions gives
saw
sas
h1 Ew
h2 Es
so
h1
saw Es
h2
sas Ew
From Eg. 6-3, Eq. g
ts 2 cnbs a1 n
saw
saw
saw
b d + ts c2 h cnbs b a n
b d d + bh2 a 1 n
b 0
sas
sas
sas
Solving for ts:
(g)
ts 3.09 mm
Problem 6.2-11 A simply supported wooden I-beam with a 12-ft span supports
a distributed load of intensity q 90 lb/ft over its length (see figure part a). The
beam is constructed with a web of Douglas-fir plywood and flanges of pine glued
2 in.
to the web as shown in the figure part b. The plywood is 3/8 in. thick; the flanges are
2 in. 2 in. (actual size). The modulus of
z
q
8 in.
elasticity for the plywood is 1,600,000 psi
and for the pine is 1,200,000 psi.
(a) Calculate the maximum bending
stresses in the pine flanges and in
the plywood web.
(b) What is qmax if allowable stresses
are 1600 psi in the flanges and
1200 psi in the web?
A
12 ft
(a)
B
y 2 in. 2 in. pine flange
1
— in.
2
C
3
— in. plywood
8 (Douglas fir)
2 in.
2 in.
(b)
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Solution 6.2-11
q 90 lb/ft
L 12 ft
(b) MAXIMUM UNIFORM DISTRIBUTED LOAD
MAXIMUM MOMENT BASED UPON PLYWOOD
(a) MAXIMUM BENDING STRESSES
sallow_plywood 1200 psi
qL2
Mmax 8
Mmax 1620 lb-ft
Plywood (1):
t
3
in.
8
From sallow_plywood h1 7 in.
th13
12
I1 10.72 in.4
h2 2 in.
Pine (2): b 2 in.
I2 2 c
1
in.
2
a
h1
h2 a 2
+ (b t) (h2 a) a b d
2
2
h1
Mmax a b Eplywood
2
Eplywood I1 + Epine I2
splywood 1131 psi
spine Mmax a
3
Mallow_pine a
h1
+ a b Epine
2
Eplywood I1 + Epine I2
Mallow_pine 2675 lb-ft
MAXIMUM ALLOWABLE MOMENT
Mallow min (Mallow_plywood, Mallow_pine)
PLYWOOD GOVERNS.
;
Eplywood I1 EpineI2 96.287 106 lb/in.2
splywood sallow_pine 1600 psi
From sallow_pine (b t)(h2 a)
h1 + a
ba
+ ba a
b +
12
2
12
2
I2 65.95 in.4
Eplywood I1 + Epine I2
MAXIMUM MOMENT BASED UPON PINE
Epine 1.2 * 106 psi
3
h1
b Eplywood
2
Mallow_plywood 1719 lb-ft
Eplywood 1.6 * 106 psi
I1 Mallow_plywood a
Mallow 1719 lb-ft
;
MAXIMUM UNIFORM DISTRIBUTED LOAD
From Mallow qmax L2
8
qmax 95.5 lb/ft
;
;
h1
+ a b Epine
2
Eplywood I1 + Epine I2
spine 969 psi
;
© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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569
SECTION 6.2 Composite Beams
Problem 6.2-12 A simply supported composite beam with a
6-mm 80-mm
steel plate
3.6-m span supports a triangularly distributed load of peak
intensity q0 at midspan (see figure part a). The beam is constructed
of two wood joists, each 50 mm 280 mm, fastened to two steel
plates, one of dimensions 6 mm 80 mm and the lower plate of
dimensions 6 mm 120 mm (see figure part b). The modulus of
elasticity for the wood is 11 GPa and for the steel is 210 GPa.
If the allowable stresses are 7 MPa for the wood and 120 MPa
for the steel, find the allowable peak load intensity q0,max when the
beam is bent about the z axis. Neglect the weight of the beam.
y
50-mm 280-mm
wood joist
C
280 mm
z
6-mm 120-mm
steel plate
q0
A
1.8 m
1.8 m
B
(a)
(b)
Steel (2): I2 t2 b31
b1 2
+ t2 b1 a h h1 b
12
2
Solution 6.2-12
L 3.6 m
DETERMINE NEUTRAL AXIS
t1 50 mm
WOOD (1):
h 280 mm
Ew 11 GPa
L
y1 dA y1 A1 a
Steel (2):
I2 10.47 106 mm4
h
h1b (2t1 h)
2
t2 6 mm
b1 80 mm
b2 120 mm
Es 210 GPa
b1
y2 dA y2 A 2 ah h1 b (t2 b1)
2
L
b2
a h1 b (t 2 b 2)
2
From E1
Ew a
L
y1 dA + E2
L
y2 dA 0
b1
h
h1 b (2t1 h) + Es c a h h1 b
2
2
(t2 b1) ah1 b2
b (t2 b 2) d 0
2
h1 136.4 mm
Ew I1 Es I2 4.22 1012 N # mm2
MAXIMUM MOMENT BASED UPON WOOD
sallow_w 7 MPa
From sallow_w Mallow_w (h h1)Ew
Ew I1 + Es I2
Mallow_w 18.68 kN # m
MAXIMUM MOMENT BASED UPON STEEL
sallow_s 120 MPa
From sallow_s Mallow_s (h h1)Es
Ew I1 + Es I2
Mallow_s 16.78 kN # m
MAXIMUM ALLOWABLE MOMENT
Mallow min (Mallow_w, Mallow_s)
STEEL GOVERNS.
Mallow 16.78 kN # m
;
MAXIMUM UNIFORM DISTRIBUTED LOAD
MOMENT OF INERTIA
Wood (1): I1 2 c
t2 b23
b2 2
+ t2 b2 a h1 b
12
2
+
3
2
t1 h
h
+ (t1 h)a h1 b d
12
2
I1 183.30 * 106 mm4
From Mallow q0max L2
12
q0max 15.53 kN/m
;
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Transformed-Section Method
When solving the problems for Section 6.3, assume that
the component parts of the beams are securely bonded by
adhesives or connected by fasteners. Also, be sure to use
the transformed-section method in the solutions.
y
Problem 6.3-1 A wood beam 8 in. wide and 12 in. deep
3.5 in.
(nominal dimensions) is reinforced on top and bottom by
0.25-in.-thick steel plates (see figure part a).
C
z
11.25 in.
0.25 in
11.5 in.
(a) Find the allowable bending moment Mmax about the
z
z axis if the allowable stress in the wood is 1100 psi
and in the steel is 15,000 psi. (Assume that the ratio
of the moduli of elasticity of steel and wood is 20.)
(b) Compare the moment capacity of the beam in part a with
that shown in the figure part b which has two 4-in. 12-in.
joists (nominal dimensions) attached to a 1/4-in. 11.0-in.
steel plate.
1
— -in. 11.0-in.
4
steel plate
y
C
4-in. 12-in.
joists
0.25 in
7.5 in.
(a)
(b)
Solution 6.3-1
Mmax min (M1, M2)
(a) FIND Mmax:
(1) Wood beam:
b 7.5 in.
h1 11.5 in.
STEEL GOVERNS
Mmax 422 k-in.
;
sallow_w 1100 psi
(2) Steel plates:
b 7.5 in.
h2 12 in.
t 0.25 in.
sallow_s 15,000 psi
TRANSFORMED SECTION (WOOD)
(1) Wood beam:
b 3.5 in.
h1 11.25 in.
(2) Steel plates:
h2 11 in.
t 0.25 in.
WIDTH OF STEEL PLATES
bT nt bT 5 in.
n 20
WIDTH OF STEEL PLATES
bT nb
(b) COMPARE MOMENT CAPACITIES
bT 150 in.
t 3 bT
h2 t 2
+ 2c
+ t bT a
b d
IT 12
12
2
bh31
IT 2
bh13 bT h23
+
12
12
MAXIMUM MOMENT BASED UPON THE WOOD (1)
M1 IT 3540 in.
4
MAXIMUM MOMENT BASED UPON THE WOOD (1)
M1 sallow_w IT
h1
2
M1 677 k-in.
MAXIMUM MOMENT BASED UPON THE STEEL (2)
M2 sallow_s I T
h2n
2
M2 442 k-in.
IT 1385 in.4
sallow_w IT
h1
2
M1 271 k-in.
MAXIMUM MOMENT BASED UPON THE STEEL (2)
M2 sallow_s IT
h2 n
2
M2 189 k-in.
Mmax min (M1, M2)
STEEL GOVERNS.
Mmax 189 k-in.
;
THE MOMENT CAPACITY OF THE BEAM IN (a) IS 2.3
TIMES MORE THAN THE BEAM IN (b)
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SECTION 6.3 Transformed-Section Method
Problem 6.3-2 A simple beam of span length 3.2 m carries a uniform load of intensity
y
48 kN/m. The cross section of the beam is a hollow box with wood flanges and steel side
plates, as shown in the figure. The wood flanges are 75 mm by 100 mm in cross section,
and the steel plates are 300 mm deep.
What is the required thickness t of the steel plates if the allowable stresses are
120 MPa for the steel and 6.5 MPa for the wood? (Assume that the moduli of elasticity
for the steel and wood are 210 GPa and 10 GPa, respectively, and disregard the weight
of the beam.)
75 mm
z
300 mm
C
75 mm
100 mm
t
Solution 6.3-2
Box beam
Width of steel plates
2
Mmax qL
61.44 kN # m
8
SIMPLE BEAM:
t
L 3.2 m
(1) Wood flanges: b 100 mm
nt 21t
q 48 kN/m
All dimensions in millimeters.
h 300 mm
IT h1 150 mm
(s1)allow 6.5 MPa
1
1
(100 + 42t)(300)3 (100)(150)3
12
12
196.9 * 106 mm4 + 94.5t * 106 mm4
Ew 10 GPa
(2) Steel plates:
t thickness
(s2)allow 120 MPa
Es 210 GPa
TRANSFORMED SECTION (WOOD)
h 300 mm
REQUIRED THICKNESS BASED UPON THE WOOD (1)
(EQ. 6-16)
s1 M(h/2)
IT
(IT)1 Mmax(h/2)
(s1)allow
1.418 * 109 mm4
Equate IT and (IT)1 and solve for t : t1 12.92 mm
REQUIRED THICKNESS BASED UPON THE STEEL (2) (EQ. 6-18b)
s2 M(h/2)n
IT
(IT)2 Mmax(h/2)n
(s2)allow
1.612 * 109 mm4
Equate IT and (IT)2 and solve for t: t2 14.97 mm
STEEL GOVERNS.
tmin 15.0 mm
;
Wood flanges are not changed.
n
Es
21
Ew
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CHAPTER 6 Stresses in Beams (Advanced Topics)
Problem 6.3-3 A simple beam that is 18 ft long supports a
uniform load of intensity q. The beam is constructed of two
C 8 11.5 sections (channel sections or C shapes) on either side
of a 4 8 (actual dimensions) wood beam (see the cross section
shown in the figure part a). The modulus of elasticity of the steel
(Es 30,000 ksi) is 20 times that of the wood (Ew).
y
C 8 11.5
z
C
z
y
(a) If the allowable stresses in the steel and wood are 12,000
C
psi and 900 psi, respectively, what is the allowable load
Wood
beam
Wood
beam
C
8
11.5
qallow? (Note: Disregard the weight of the beam, and see
Table F-3a, Appendix F for the dimensions and properties
(a)
(b)
of the C-shape beam.)
(b) If the beam is rotated 90° to bend about its y axis (see figure part b), and uniform load q 250 lb/ft is applied, find
the maximum stresses ss and sw in the steel and wood, respectively. Include the weight of the beam. (Assume weight
densities of 35 lb/ft3 and 490 lb/ft3 for the wood and steel, respectively.)
Solution 6.3-3
(b) BENT ABOUT THE Y AXIS (INCLUDING THE WEIGHT OF
THE BEAM) q 250 lb/ft
L 18 ft
(a) BENT ABOUT THE Z AXIS
b 4 in.
(1) Wood beam:
Iz 32.5 in.4
Iy 1.31 in.
4
c 0.572 in.
sallow_s 12,000 psi
qs 11.5 lb/ft
(2) Steel channels:
As 3.37 in.2
qtotal q + qw + 2qs
TRANSFORMED SECTION (WOOD)
n 20
TRANSFORMED SECTION (WOOD)
IT bh
+ 2 Iz n
12
IT 1471 in.4
IT sallow_w IT
h/2
M1 331 k-in.
M2 hn/ 2
sw_max M2 221 k-in.
Mmax min (M1, M2)
STEEL GOVERNS.
b
Mmax a b
2
IT
sw_max 277 psi
;
MAXIMUM MOMENT BASED UPON THE STEEL (2)
Mmax 221 k-in.
ss_max ALLOWABLE LOAD ON a 18-FT-LONG SIMPLE BEAM
qallow L2
From Mmax 8
b3h
b 2
+ 2n cIy + As a c + b d
12
2
MAXIMUM STRESS IN THE WOOD (1)
MAXIMUM MOMENT BASED UPON THE STEEL (2)
sallow_s IT
Mmax 11.4 k-ft
IT 987 in.4
MAXIMUM MOMENT BASED UPON THE WOOD (1)
M1 bs 2.26 in.
qtotal 281 lb/ft
qtotal L2
Mmax 8
3
qw lbhrw
qw 7.778 lb/ft
sallow_w 900 psi
(2) Steel channel: h 8.0 in.
rw 35 lb/ft
(1) Wood beam:
h 8 in.
qallow 454 lb/ft.
;
nMmax a
b
+ bsb
2
IT
ss_max 11,782 psi
;
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SECTION 6.3 Transformed-Section Method
Problem 6.3-4 The composite beam shown in the figure is simply
573
y
supported and carries a total uniform load of 40 kN/m on a span
length of 4.0 m. The beam is built of a southern pine wood member
having cross-sectional dimensions 150 mm * 250 mm and two
brass plates of cross-sectional dimensions 30 mm * 150 mm.
40 kN/m
30 mm
z
(a) Determine the maximum stresses B and w in the brass and
wood, respectively, if the moduli of elasticity are EB 96 GPa
and Ew 14 GPa. (Disrega