# Hw6 - Beams 13 A steel (E=29.000 ksi) beam of rectangular...

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Unformatted text preview: Beams 13 A steel (E=29.000 ksi) beam of rectangular cross-section 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 ross-section: 12x12 inches Solution P=2000 lb M=12000 lb-in T8333 lb 1167.7 111T 36in 36in |—l—| 833.3 lb v —ll67.7 lb 41998.8 lb-in 29998.8lb-in ..... .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 middle-point, 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) _ 2.5-253 I 12 = 3.25521(in4), c = ymax =1.25(in) . a?“ = Mmc = 2,700-1.25 =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—ﬁ-dn -. 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.

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Hw6 - Beams 13 A steel (E=29.000 ksi) beam of rectangular...

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