08A-Bending pbs.pdf

# 08A-Bending pbs.pdf - 661 The beam is subjected to a moment...

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Unformatted text preview: 6—61. The beam is subjected to a moment of 20 kN ‘ m. Determine the percentage of this moment that is resisted by the web D of the beam. ; = ELA = 12(25)(125) + 125(200)(25) + 237(75)(25) = 2A 25(125) + 200(25) + 75(25) 1 I = E(125)(253) + 125(25)(111 — 12)2 + %(25)(2003) + 200(25)(125 — 111)2 + 11—2(75)(253) + 75(25)(237 — 111)2 = 78.3 106 ( ) 25 mm/ . My Usmg ﬂexure formula 0' = — I 200 mm _ 20(10°)(111 — 25)_ 22 MP ”A _ 78.3(106) _ a 25 m 20(106)(225 — 111) 29 12 MP ! a = — = . a B 78.3(106) 75 m, FC = %(22)(86)(25) = 23.65 kN 86 mm FT = l(29.12)(114)(25) = 41.5 kN 2 114 mm M = 23.65(58) + 41.5(76) = 40.623 kip-in. = 4.52 kN -m . (I, 4.52 { % of moment carried by web = 2—0 X 100 = 22.6 % Ans FT . >"6—80. If the beam is subjected to an internal moment of M = 100 kNom, determine the bending stress developed at points A, B, and C. Sketch the bending stress distribution on the cross section. Section Properties: The neutral axis passes through centroid C of the cross section as shown in Fig. a.The location of C is 29A _ 0.015(0.03)(0.3) + 0.18(0.3)(0.03) 7 = = 0.0975 y 2A 0.03(0.3) + 0.3(003) m Thus, the moment of inertia of the cross section about the neutral axis is 1 1 1 = E(O.3)(0.033) + 0.3(0.03)(0.0975 — 0.015)2 + E (0.03)(0.33) + 0.03(0.3)(0.18 — 00975)2 = 0.1907(10’3) m4 Bending Stress: The distance from the neutral axis to points A, B, and C is y A = 0.33 — 00975 = 0.2325 m, y3 = 0.0975 in, and yc = 0.0975 — 0.03 = 00675 mi. _ MyA _ 100(103)(0.2325) = 122 MP C A . ”A I 0.1907(10’3) M ) "s M 100 103 0.0975 03 = M = ¥j = 51.1 MPa (T) Ans. I 0.1907(10 ) M 100 103 0.0675 ac = y” = M = 35.4 MPa (T) Ans. I 0.1907(10’3) Using these results, the bending stress distribution across the cross section is shown in Fig. I). «=53 ‘2- (a) Cb) 6—103. If the overhanging beam is made of wood having the allowable tensile and compressive stresses of (O'anow)r : 4MPa and (callow)c : 5 MPa, determine the 25 mm maximum concentrated force P that can applied at the free end. Support Reactions: Shown on the free—body diagram of the beam, Fig. a. Maximum Moment: The shear and moment diagrams are shown in Figs. b and c. As indicated on the moment diagram,the maximum moment is leaxl : P. Section Properties: The neutral axis passes through the centroid C of the cross section as shown in Fig. d.The location of C is given by am H" 7 2m 2[0.1(0.2)(0.025)] + 0.2125(0.025)(0.15) = |— 2. 7 2A 7 2(0.2)(0.025) + 0.025(015) 7 01306811] A? GOP (tug? l5}, The moment of inertia of the cross section about the neutral axis is I : 27 + Ad2 1 1 = 2 E(0.025)(0.23) + 0.025(0.2)(0.13068 i 0.1)2 + 5(0.15)(0.0253) + 0.15(0.025)(0.2125 , 0.13068)2 : 68.0457(10—6) m4 Absolute Maximum Bending Stress: The maximum tensile and compressive stresses occur at the top and bottom—most fibers of the cross section. For the top fiber, P(0.225 7 0.13068) 6804570076) M maxy (gallow)r : I ; 4(106) : P : 288579 N : 2.89 kN For the top fiber, P(0.13068) M c an“: max5106:— ( '1 ) ( ) 6804570076) I P : 2603.49 N : 2.60 kN (controls) Ans. (a ) Ans. P : 2.60 kN ...
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