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Q. 2 For the beam section shown in Figure Q. 2 (a), determine: (25 marks) (a) The depth to the neutral axis of bending. (3 ma
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lokalen 7 -Boom ho llom । 3m -M yoomm RA RB Solution PAT RB= b10) RA TRB cyo kn EmA =0 -RAM + १०१५ -२० BRB= 104442 १० Roa १ RFA + Чә RA+ 26.67 = 40 RA= You 26.67 RAS 13.33 kn 1 Shear force calculation - SPA= 13.33 kn SFB = 13.31 -10xB+ RD 13.33 30 +ec 13.33 ใจ x= 1.39m B Mmar 11 [2133m from A} RAXX- loxxxx 슬 13.33 X1+33 lo X6134 133 2 BM mara 17. 73 8.84 BMmax - 8.89 kNmwhere Al= (доо хто мл 91 = lo | - shem А, . Чоо- 9, ( \о Чэ. 쁜 9, - Blomm | — oo Xok S Чоо Хъ2-15 Zoo X (ə Чоо+ % 6% 1 ооо G6(6) Second Moment of area about neukel Asus INA = It Allý -9,3² + Iz + A2 (9- y) 3 300 x 103 12 + 3008 lo (110.6-5] + 8x4003Maximum compressive bending Stress (oc) occ INA Stop ma 8.89 kNm ma 8.69 X 100 Nomm INA 10.764 107 mmy 9 topa luig & mim =) 8Maximum Tensile bending stress (at) - ot INA Усе bottom =) 8.4944 101 784107 N] of 299.2 -) oti 8.89 xlor x 299.2 lor78x107 2Va 13.33 X los n A = - Area below Neutral Axis A - 299.2 x 8 mm2 y = (1 Come of below neutral axis 299.20 ý=149.6mm IN A = 16

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