Question

#1) (65p.) Draw the Shear Force (V) and Bending Moment (M) diagrams of statically indeterminate beam shown in figure using “Force Method”. The (roller) support at “B” settles 35 mm. The moment of inertia is given by (1) for regions "AB", "BC" and "CD"; however it is equal to (21) for the region “DE”. (“B” is the roller and "E" is the fixed type of support). [The flexural rigidity: EI=40000 kNm-] 60 KN 10 kN/m I. B (21) X 1.5 m 1 m X X X

Answer 1

oll GORN at 10KN/m mmm B C 21 A T I I 3m Im 1.5m Am method, Hence taking we will apply castigliano's RB lie reaction at B downwards and hence- 35X10"3 m 1. IM.dk 35 X10-ºm -45 SM . ET 17 2 RB du ƏRB 3. Hence, Occa RB GORN 10KN/m my al С 1 D F Il I 0 2 (3 <im 3m 1.5m Am

Limit of x 0 to 1.5m For section ♡ ♡ AB x from A M = 0 (1) aM =0 a RB section ® 2, limit of a o to im For BC, from B M= - RBox (I) IM ƏRB 3 For, CD, section x from o to im x from a (I) RB (x+1) - 60x as M= JM ƏRB -(x+1) x limit → o to 3m for DE, section wy , x from D (1 ऐति - RB (x+2) - 60 (x+1)- 10x/ = -(x+2) ƏM ƏRB 35 710-3 Hence, => 2 (mam, dx ARB 0+ (RBS) {-x) dx R$(x+1) – 60x}{+(x+) } dx EI EI 3 +({-Relx+ 2) - 60(x+1) -box/) } { -(x+2}} ox - 35x10 -3 2 EI

RB, a + RB. (x+1) 3 + 50 + / RB (x+2) 3 BEI BEI EI O ЗЕТ 0 ס + 950.625 35x10-3 EI 8 RB 6 EI + 950.625 I Ro + 125 RB 6EI 8 RB Ro 3/E I ЗЕТ 3 EI 35810-3 t 50 E E IS 40000 kNme Couven) 133 RB + 10000.625= 35X10-3 40 40000 Ý 6 | RB 18.017 KN For shear force diagram 18.017 KN I GOKNI nh 11 10KN/m mmmal mod A 00 с im 1.5 m Figure - 1

x from a SF00 x from B SF@@ - - 18.017 KN Hence, SFA=0 - 18.017 KN SFB left SF Bright SF e left = =-18.017 KN x from a - 18.017-60= -78.017 KN -78.017 KN SF 6) = sferight = SED left SF9 - 78.017RN x from - 18.017-60- 10x SFD - 78.017 KN x=0 sight SFE= - 108.017 KN at x = 3m, Hence, с - 18.017 KN - 78.017KN - 10.8.017 KN diagram Shear force

for Bending moment diagram- Taking sections from figure (1 M00 x from B - 18.017 a at x=0, MB=0 at x = im Mc = -18.017 KNm M@ ®) = -18,017 (2+1) – 60 x x from C at 2l=0, Me= - 18.017 KNM at o im, MD= -96,034 kNm Ma = - 18.017 (x+2) - 60(x+1) - 100% Gefrons) at x=0, Mo = -96•034 KNm at x= 3m, ME= - 3751085 KNm Bending moment diagram - D B A El 18.017 kNm Ĉ -96.034 KNM - 375.085 kNm