Question

1. For the simply supported beam given below, if the deflection at 14 in. from the left end is 0.03 in., calculate (i) maximum deflection (in.), (ii) reaction force at A, RA (lb), and (iii) reaction force at B, RB (lb). E = 30 x 10 psi, and I = 325 in*. Web A FB 3ft Ro

Answer 1

Salution L=3ff = 36in Wat web (in) MXZ B Hn RA RB It aff To find Reaction forces EM B20 1 x x x al IRA - WL RANY K= . 2 6 RA = (0L t E fyzo RA + RB 1 WL 2 RB BuL-WL =WlLWL 2 6 z llell 63 Ro. I EL 3

Maz + ND M L x2 (0х4- 12 24L 1 wer5 - W L x3 + (12+2 1C1 z 7 ul 13 measuring from , A By double integration Eidy -welz wer3 dx2 6 EL Integrating eq to get slope & deflection equation Erdy + da EIJE + Boundary condition are at y zo From eq ETO) = 0 +62 (zo EI(0) = + 3Go 136 120 L rzo KIL -uel (L)334 vie > C1 (4) 36 120L

$o deflection equation become cus EI Y = -ulx] to + 7 col3 x x 3C w x5 120L 360 given y = 0.03 in at x= 14 in E=304106 Psi ,I= 325 int 0:03 2 1 -lee x 362(14)3, 40(1435 + (30 X106x325) +3* 360 36 120136 ce ( 36)3x 14 = 29350.69 462 lblin I To find maximum deflection (dy = WeLx²7 cu L²=0 du 241 12 360 From con TV The solution of which gives bc= 0.5191 Ymeix = 0.00652 we (1.1% 4 L-0006152129350694(36) 304606 9325 EI imax z 0.032966 in Are

Reaction at Á 293500 69 462 x 36 RA= CUL 6 RA= 176 104-16.77 lb Ang Ľ Reaction at B Ro B = WL 3. = 29 350.69462K 36 3 Rei 35 2208.3354 lb