Question

The simply supported beam of length L is subjected to uniformly distributed load of w and a vertical point load P at its middle, as shown in Figure Q3. Both young's modulus and second moment of area of this structure are given as E and I. Please provide your answers in terms of letters w, P,L,1, E. Self-weight of the beam is neglected. P W L/2 L/2 Figure Q3 (a) Determine the reactions, bending moment equation along the beam and draw the corresponding bending moment diagram. [10] (b) Determine the deflection of the beam at the midspan of the beam. [15]

Answer 1

Given that P al mbong Rax=0 42 1/2 Ray = Pour Rg² RIWL 2 PAWL) Pla SFD C А eln (Ptwy woh tt) BAD » from the eavinbrium Conditions EMA 20 (RB) (L) = (RB) (L) = 0 {/2) + wee Rex P+ wy 2

Efyco Ray+ RB = WLAP 2) RAYE Wet? Efnzo Ranzo

Section between A&C : - Consider a section x-x between A&C ata distance x from A. Sfa sw ends Rayzo K اد RAY= Mr = Watp) (x)- SFn = (Watt) - wx olig Wv SFA - (Why) ; Mazo SFC = wht power to = P/2 Me= (WL+P) (L) - WL? ab n=o - SFA aknath -> 21 wr² + PL 4 Section between c2B sta Mx monty beton = WLKP) 0) - PC4-412) - WMO (with) Helena Spor= (Witp)- wn-P at n=4/2-) SPG = =P/2's meo (WLAP) (42) - - (2) k- DA M = WLR + PL at n=L - SFB- (WL+P) г\%= о

deflection of beam doing Mx = (WL+P) (2) - PG-7) (with ww from the defection eavation. EI day dre My -> EJ day - (White) (-o) -PC2-413) - daz W(2-0)2 EI dy dr (WLxP) (mo). P(x-4) 2 w (n-o)3 G toi 3) EJ (u) = (WLt) (n-o). P(x-423 W (n-oj4 24 6 fciat ca Boundary. Condition at n= oy=o. for a singularity function S-a)n = Soo-ajn C220 Ya rea at net y zo (WL+P) (L) 3 - P(13/6) 2 +6,64)

& PLWLY az - WL3 dale 4 W(n-o) 24. E5 (9)= (wzO) (XO)? PG1=452) 3 -(ws + Ple) Cad at n = 1/2 - y = Iman Ymana to WID))-0 16 EI 24 9 сч PES 32 لا را Iman z z (9_4 WLY EI PL3 96 384 32 I mana / EI -PLS - 5wer 384 n 18 defection at the midpoint.