@) simply supported stoop 1.) =) consider the FBD am 3m AXE r a 15OON e zm - → Efy z o - Ay+By z 1500 EMAzo - cih - 1500x5 + Byx220 1. By a 3750 N put By in eis we get 1. Ay=-2250N so By 23750 N( upwards) Ayz 2250 N (downwards) =) Shearing force along the beam o Shear force at e (se) (COB) Scz - 1500N Shear fore at B (SD) (BMA) SB2-1500+ 3750 582 2250N • Shear force at A (SA) SAZ Ay z - 2250 N
>>Equation for bending moment along the length com be given by -1500x x +3750(x-3) = Meat 66-320, ij 153] we can calculate bending moment at diffrent points by using this equation 9 Bending moment at e (Me) 220, X<3 Me .z - 1500X0 z o → at B CMB) [x23] , MB 2 -1500832 -4500 N-m at A (MA) (x=5,273) MAZ-1500x5+ 3750X2 MA zo 1 bending com be equation of given as - MZ ET I Here, Mye -1500X +3750(x-3) son •f7d942 - 1500304 3750 (2-3) [(x-7)zo ij x53]
* Integrate winnt be! ET DY 2 - 750x461/+ 1875 (43) - (11) Izak dot. I' EI Y2 - 2503344X+62/+625(x-3)? -ci 3) The boundary conditions are i Yzo, at x 25 so, from cili), we get Oz -26250 +6,5+62 - Civ) ii) yz o at xaz so from cilis, we get 02-6750 +673 +C2 -CV) by solving civ) and eV) we get ei= 9750 - C22 - 22500 so, the equation of deflection VCX) . along the beam is ye va)=(2504 +97501-2500 E7 ] +625 (2-3)') {(2-3) =0 if LE3}
EI - The equation of slope along the beam NO) e dyzet (-750L2+9750/ +1875 (2-3)) - {(2-320 ,if Ls3} 3) The maximum normal bending stress overy at point of maximum bendling moment • from calculatim , we can observe maximum bending moment Mmar a : 4500 at point B • Bending stress is given by oba Momave to AV here yz thickness (t) so, обе M max x(t) I b) by considering the rigid damp -> consider the FBD 1 SOON rie X LAY 9 BY Emaye -Ay+By 2 1500N EMAZ MA = 150005 - By X 2 MA= 7500-2 By EMezo - МА - Ayx5 +СуX 3
N by moment ared method MAZ - MB z 2250 N-m so, Byz 2625 N and Aya 1125N 2 =) shearing force along the beam at c (se) 1. Sez-1500N at B (SB) SBz-1500 + 2625 SB2 1125N at A (SA) SAZ - 1125N - z) Bending moment along the bean ; Myz -1500 XX + 1125 (2-3) [62-3) 20, if «53) 2) Bending moment atc (820) 1 Mezo at B (X2 3) - Mez -4500 N-m at A Bez 5) MA 2 - 95008 5 of 2625 X 2 MAZ-2250 Nom 2) Equation of bending can be given as Mxz Ez dy
dx - conditions Here , Mula -1500W +2825(x-3) So, EZ DP -1500X+2625 (-3) integrate ww.g.& ET DYZ -750x?+4/+1312.5 -91 integrate w.ot-x ET Y Z -250x3+4x+62 4937.5 (x-3)* > The boundary conditions in % 20, at 225 1. so from (b) 02-277504 95+ lz -(e) ii) y 2o, at x = 3 so from (6) Oz - 67504301tle -(d) on solving ce) and (d) G2 10500, C22-17250 so the equation of deflection vese) yz Verde -250£?+105004-17250 /+937.5 (2-3) 3 {(L-3) €0, YLI 3} equation of slope vere) dy z vex)2 -75022710500 H1312:5(2-3)? CL-3)20 , ig 253) 2) Maximum bending stress oba Mmare (14) 2 - 4500 (²) M Oh I I