Question

Problem Statement: Two designs for a diving board are shown in the figures below. A diver umping at the end of the board exerts a force of 1500N. Assume in design (a) the strap provides a simple support, while in design (b) the rigid clamp block provides a cantilever support. Assume the board is 60 cm wide. 1500 N Strap provides 2.0-- 3.0 simple support Diving board Pipe roller (a) 1500 N Rigid clamp block 2.0 m- -3.0 m Diving board Pipe roller Part I: Analytical Work in the Report (Due date - April 15, 2020) 1. For each of the proposed designs, compute the reaction forces and derive the analytical equations for the shearing force and bending moment along the length of the beam 2. For each of the proposed designs, derive the analytical equations for the elastic curve for vertical deflection, v(x), and slope, v(x), in terms of distance between two supports, a, and total beam length, L, and El. 3. For each of the proposed designs, show the equation to compute the maximum normal bending stress in the diving board.

Answer 1

@) 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