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Unformatted text preview: Beams 13 A steel (E=29.000 ksi) beam of rectangular crosssection is bent over a rigid mandrel
(R=20 in) as shown. If the maximum ﬂexural stress in the beam is not to exceed the yield
strength (q,.=36 ksi) of the steel detennine the maximum allowable thickness II of the bar. A M f M
R
Solution
2R0 ox=—Ey—>oy=—;hmu/2—>hm= ’ = p R+hmax/2 E—(Sy
ﬂ = 0.04972 in
29,000—36 Note that if R instead of R + h"1ka /2 is placed in the denominator above. the answer is 0.04965
in. which in incorrect. Beams 14 The wooden (E=1.6 ksi) beam/bar is subjected to the loads R N. and moment M as
shown. For P=2 kips and M=12 k—in. ﬁnd the minimum axial compressive load N such that there
will be no tensile stress OK at any location in the beam/bar. P=2 kips C rosssection:
12x12 inches Solution
P=2000 lb M=12000 lbin T8333 lb 1167.7 111T
36in 36in
—l— 833.3 lb v —ll67.7 lb
41998.8 lbin 29998.8lbin ..... .mlllllllllllllllllli""lllllllllillllllllll M _M=__— =145.83psi m“ _ I h“ /12 123
Then, Nmm = cggfmA = 145.83  122 = 20,9994 psi tension From bending: o Mmaxh/Z _ 41,998.8'6 Beams 15 The “legger” at the leﬁ end of
the trailer consists of a horizontal beam
and two legs. The horizontal beam can be
considered as a pinned beam as shown,
and the force from the trailer onto the
beam is 2001b/ft imposed over 1.0ft at the
middle of the beam. For the beam cross
section being 2.5” x 2.5”, calculate: (a) the maximum tensile bending stress <3x . . . . 1 ft m the entire beam, 1n ps1; (b) the s—9 maximum shear stress 1: in the entire c6
beam, in psi. 200 Ib/ﬂ cross'se on 2.5x2.5 in
5 ft Solution The maximum moment for this beam occurs at the middlepoint, and the maximum shear at the
two ends. The reactions at the two ends R, are such that R = %ZOO(lb / ft) 1(ft)= 100 lb. Thus Vm=100 1b. At the middle, the moment is
Mmx = 100(1b)  2.5(ft) 200(1b / ft) 0.5(ft) (ft): 225(b — ft) = 2,700(lb in)
_ 2.5253 I 12 = 3.25521(in4), c = ymax =1.25(in) .
a?“ = Mmc = 2,7001.25 =1036.8(psi)
I 3.25521
Beams: 8 A suppOIt beam 011 a ham is used for lifting hay into the hay left. A paiticularly heavy bale (125 lb) is being hoisted 11p. Find the shear and normal stress distribution in the beam
where it is ﬁxed to the ham wall. (E = 1.5 x 106 lb/inz) Solution: N01mal Stress 0' = g
m = F orcex distance: 125 lbx 72 in : 9000 in  lb
1 =ibh3 =ix21nx(10m)’ =166.7 m“
12 12 —
o_ 9000ian _54 ﬂ
166.7 in4 y y in2 0(5): 54x5 = 270 %
m 270 Ib/in2 (tension) 1
“1 11 n;
11 q—ﬁdn
. W
= 1’
—‘
w
*—
.ﬂ
ml 270 lb/in2 (compression) Shear stress QM = Ay = (5 inX2 inX2.5 in): 25 in3
1251bx25 in}
=166.7in" x2 in
QM = (2 inX2 inX4 in): 16 in3
125x16 in3 T,_ =—=6lb ‘nz
"—3 166.7 in4><2in /l r“ = 9.375 Ila/in2 Beams: 10. Beam ABCD is pinned at B, loadead by force P at D and supported through a spring
and bar CE at A and C, respectively. Is the beam statically determinate or not ?. If statically
indeterminate, what is the degree of indeterminancy ? Explain your answer. p
B C ...
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This note was uploaded on 04/28/2010 for the course CE 215 taught by Professor Frantziskonis during the Spring '08 term at Arizona.
 Spring '08
 FRANTZISKONIS

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