Question

The steel beam has the cross-sectional area shown in (Figure 1) Part A Determine the largest intensity of the distributed load to that the beam can support so that the maximum bending stress in the beam does not exceed 160 MPa Express your answer to three significant figures and include the appropriate units. Figure 1 of 1 Value Units Submit Request Answer Provide Feedback

Answer 1

- area / WL = Total load 02 12 w x6 = 36 Due to syste symmetry (RA = Rp) RA+ R₂ = Wol Y 6 WoX 6 Wo.3 RA = RB = WOL 4 wolle 4 2. SF at ped x = Wol a Wox X X X 4 L co L cq x² 4 SF At no SFA = lot = 300 مر SF At x = 3 8P (= Wit - wo (+) W X6 4 6 اليا ساوالا SF g = wolle اولا ۔ at SF x = 6 maximum S At a where Bending moment こ Bm at a - Wol. L wol a 2 Woll Woll 24 3 w l - well 8 2 24 2 3W. W6 x 36 اندرا WX (6) c +2 max Bm at c 2 12

200 mm 200 A = 200 x6 = ܩܢܟ 2 1800 hm A= 300 X 6 = 6t 306 1200 mm 200 X 6- 타 6 g 3 mm F ge 6 + 15o = 156 miny 9² 6+ 300+ 3 = 309 mm O : 100x 3 + 1800X isot 1200X 309 A. I + ₂ Y₂ + A₂ Zz A3 AtA 1200 + 1800 t Roo 150mm Y I, - 200X6 3600 mm 3600 t I' 1200 X (153) = 28.0944x 10 mm I + Ay y = 7 -3 = 156-3 = 153 my 3 6 x 300 13.5x100mm 12 3 200XG 3000mmy 2 3 2- 3600 + 1200 X (153) = 28.0944x100mmy I. Iz + Azy y=J_3= 156-3= 153ren I= I't I t I 28.0 944X10 + 135x10° +28.0944x lor I 69.6888 × 100 mm y

alg I Max Bending moment at c = 3 W. Norm = 3000 W Noming E I = 6.96888 × 100 mm o こ 160 mpa= 160 N/mm² so 153 mm, 3W xlooo 160 69.6393X10 153 لا = 24 292.4 N/m 24 KN/m 그 24. 2924 No
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