Question

For the following Beam E=200GPa I=6000 cm Use the Slope-Deflection method to determine The reaction at support Bif this support settles 55 mm Remember Forces to the right and up are entered as positive, left and down are entered as negative. Counterclockwise moments are entered as positive, clockwise moments are entered as negative A B С 10 m 4 m 4 m

Answer 1

Hence; het RA, RB and Rc be Let MA Hence, at be the rotational reaction at A- MA- MAB = - 37, 365 KNM the support reactions A, B and C respectively - 1 35:13 37,365 PRBI ÎRA lom RBI X 10 = 0 EMA=0 -> -37,365 - 35.13 -1 RB, = -7.25 KN Efy=0 => 1 RA = 7.25 KN 8m 35.13 po RB2 Rc EMB=0 35:13 RCX8=0 & Rc = 4.39 KN { Fy=0 4.39 KN

Hence, Final support reactions - RA= 7.25KN MA=-37.365 KNm anticlock wise) RB = RBI RB2 -7.25 -4,39 = -11.64 KN Re= 4.39 KN '+' sign indicates upwards reaction indicates downwardy reaction. (-) sign

Guven data- E 200 x 10 3 N/mm2 I= 6000 cm" = 6000x104 mm4 = 6x107 mmy २ Nmm Hence, EI 2008 103 x 6X107 200 x103 x6x107x10-S KN ma 12000 KNMP EI- 0.055m SB = 55mm с A 1s = 0.055m tom 8m Unknown displacements & OB and oc Since there acting structure, is no external load on the Hence the fired end moments ine. are o. MEBA M FABI MFBC and MF CB

MAB=MEABT REI e Applying slope deflection equation 38] EI = 12000 KNma MABE QEI (08 - 3x 0.055 10 10 39.6 MAB= 0.2 EI OB MBA = 0+ QE1 (208- 3x0.059 10. MBA= 0.4 EI OB – 39.6 31-01053)) MBC = of ZEI 20B + Oc 8 IMBC= 0.5 EI OB +0.25 ETO + 61.875 o+2EI (2 Oct OB MCB = 31-0.055) 31-0,05)) MCB = 0.5 EI Oct 025 0.25 ETOB & 61.875

using equilibdium equations EMB=0 7 MBA +MBC =0 => 0.4 ETOR - 39.6 + 0.5 €100+ 0.25 E 1 Oc+61.875=0 - 22.275 → 0.9 E 1 OR +0.25 E IOC E Me=0 » MCB=0 - 61.875 => 0.25 E IOB +0.5 E IOC From and > E IOB = 11:177 EIO = -129.34 Hence, Putting the values of EI OB and FI oc to find the values of MAB, MBA, MBC and McB -37. 365 KNm ☆ MAB= 0.24 11.177 -39.6 MBA = 0.48 11.177 - 39.6= -35113 kNm MBC = 0.5% 11.177+ 0.25% (-129.34) +61.875 = 35.13 km MCB =0