Question

This is only One Question. Q25,26,27,28 and 29 are its sub parts. No need to calculate Maximum bending Moment. Just do the five parts mentioned below.

For the simply supported beam shown in Figure 6a, calculate the maximum Bending Moment in the beam produced by the loads acting on this beam. 5% 15KN D * ut Figure 6a Assume that the beam in Figure 6a has the cross-section shown in Figure 5b, and the maximum allowable tensile and compressive stress in the material used to manufacture the beam is 30 N/mm2. Calculate the maximum Moment of Resistance for this beam. 225 toru Figure 6b

Answer 1

5 KN 45 KN * A 1 4m um 4m 3 TUB 2 om confused it is is in point wad or 45 IN as the writing is is not deal I have tupon 45kN at point C. &- find Reactions & 3 Mp 20 YAX 12 = 5x +45 X 4 YA = 18:34 www Ng = 50-18:34 In MB 31.67 IN Scross-section is $OX 225 mm. welchilwg м. T: y 틍 R. I of the section 50 x 2253 12 -44.460 8106 mm 4. I Moment of Resistance of section My 38 - 2 X 30 12.65 1. 22.5 2 y My F 12.65 km

926). Max B-M due to wed induced in the boam :- I BMmax BMCs YBX 4 = 31.64x4 ium 5 126.68 renom. FB Mimar > B.M. Hence the Rocickince section is neet adequate enough to resist com bear such loading 00 Beam Rocierince gy the given sechos 50 X 205 mm 2 M D Me 225x50x30 6 MR 2.812 in BMmax (126.66 167-6) > BMR of the section section is neet adequate enough to bear the wad rence the B. Mmax = 126-68 section as 50 X 225 & Max tensile = Max comp. stress - max 2 [ ? selon Mo-dules Bole? Snov € go 2) 225 2 126.66* 106 50 X 2252 max & rc = = 300.27 N/mm2

orato. A e for N-A2 y = 0) (30024) oc INA M. My D. ТА 629 aerob ebees diagam is drawn the cross section. T (300.27)