Question

Question No. 3 (25%) Determine the reactions at A, B and D, where A is hinge, B and D are rollers. Use the compatibility method of analysis. Draw the shear and bending moment diagrams FREE BODY DIAGRAMS, SECTIONS AND ALL CALCULATION STEPS MUST BE SHOWN 250 KN 25 kN/m -В с 12 m 6 m 6 m 1.57 E = constant

Answer 1

250 15KN 25KN/m A Its D III/. с. K om om 12m (I) (21) E - Constant the beam is end of one The relative rotation of zero with respect to the other and of beam · DELKE * MB CAB 8) •MB LBB" ته i O B + a+(.'+ [Xeo") ms] = Me [dbo] (@o'+ 09")+ Mo [cee't do where Os+og" = Og

See - Hee + Yaa slopes ont can be determined from the slope equations of a simply supported beam with ude. is, op' = W.LS 25 x 123 24E I 24 EI = 1800 EL for beam BD, with point load at center, Slope, Ob" = PLA ੨50xਕ੨ 16 ET(1.51) 16 EI 2250 1500 EI LBB ML 3 EI 1x12 3 EI 4 EI L 1 ML 1x (12) 3 x Ex1.51 3 ET 2.67 ET

substitute the values of op', os", age', deel in equation 1 Mereka ) -- 1500 1800 EL + + MB [4 EI Me [4 2.67 - 3300 + EI EI EI MB -3300 EI x EI 6.67 = - 494. 75 kNm (-ve -ve, sign means MB Moment at B=-494.75 km acts in opposite direction] Bean BD (Free body diagram of BD -494.75 250KN 1166. C 23 KN om '83.77 KN Consider mo ment at B = 0 : , Mozo

Taking moment, Rox12 + 494.75 E 250x6 Rpx 12 + 494.75 250X6 Ro = 83.77 KN Reaction at D = 83.77 KN RBIRD = 250 KN RB = 250-83.77 = 166.23 KN -494,75 250 KN B с 166.23 KN 83.77KN Free body diagsam of AB consider the moment at B=0, Mg=0

25 KNIN 494.75 Hotely B A K iam Ro X12 – 25x12? 495 494.75 = 0 2 Ro 494.75 + 1800 12 = 191.22 KN RA+RB = 25x12 = 300 KN RA= 2 300 –191.22 = 108-78 KN Зоок 300 kN (UDL as point load) 191.2a kn 108.78KN la felmo

SFD 166.23 KM 166.23 KN 108.78 KN + + E B А K 4.35m 83.77KN 83.77KN 191.22KN >K om 1am BMD BM 0 A BMB - 494.75 km 4,35.m from BM at poidpoiot of Beam AB RAXC-WXoca 3 = 108.78 x4.35-25x4.35*4.38 BME = 473.19 – 236.153 = 236.66 KNM

BMC 83.77 X6 = 502.62 KNM BM b =0 502.69 KNMD 236.66 KNM c D B A 4.35 -494.75 kNm 12m OM om