Question

31 The beam deflective curve equation for sections 0 <I<L and L<I<L, The beam deflection dc at the mid-point of the beam where I = LL (12) (5) Total Marks: (25) Hints for Question 3 1) The bending moment M(x) for the beam AB may be simplified to two separate expressions for cach of the sections such that M (3) 90 24L L40 (51°c – 12x®L+8x"), for 06135 (-2+L), for ££<3<1 M2(x) 24

Answer 1

11 90 290 L 42 Lh Moment equation for Both the sections are giren as M(x) = {Sq. LX 29. (1-x) (9 24 15%La 24 190 1 2 ( x - - - os xah bast According to flexure Equation we can write El dy = M(1) da? For Cordition 1 From osasuh ( 2x ) 21 2 2 1 2 EA AM - SEMALAM PEC 202 2x² 24 5% La 290 24 59. LX² 290 / LX² 48 L 6 49X² L 49 x 9L 2 12L 290 Lx9 as ta 144 48 30 492 + (х 94 COL th 9 59.23 29.24 9.24 36

• Ely 59. LX? 29.1.X4 X5 144 L1 48 30 - 18 18 59012 (44 48 30 [Gori 944 IS L JtGx 290 / LXY A + q2s 30L 0 + Cox & C Car Berarclay cardition * co at 20 dre >> ogo at x=0 putting values in oq we gel. C200 and agair, a=0 59 LX² 290 / LX" FI(144) LGI 4.8 y. 9.XS SOLEI 30 Deftection Requation. Similanly Fog Mca) 2 qoll (x- El dy Bs. da? = M(x) 59.42 90L 24 de 59Lx? qoh px? 48 4. 59. LX² 90L 144 4 2 LX GAG © =) fly - Tuy 4 ( 3 12+CIX +62 D BRYAN

Appleging Boundary condition dx y = at putting in og o 0 = 0 +G ord 6 o Thus ge 59. Lx? Golc X LX By 6 14471 471 12 Thus deftcetion cut Sc=X :h yl 59.L(42% Yol L3 (4 1946) GET 8x6 5% LY qol L 13 1152 El 471 48 48 L 7:42 (5) w 5914 8c 1152EL 5919 1152 El dol