Question

PROBLEM 5.53 w=w0(1-1/2 Determine (a) the equations of the shear and bending-moment curves for the beam and loading shown, (b) the maximum absolute value of the bending moment in the beam. Answ: V(x) = -W. (x-x?/2L- L/3); M(x) = wo (x3/6L – x²/2 +Lx/3); Mmax = 0.0642wol?

Answer 1

W= W. IRVB RVA Rya - Vulical Reachion at bt. A at pointB RvB - ч upto any cuoss sechion. 2- distance from þoint A = total downwavd fove RA + R8 %3D » Arce of brangle. Total down ward fove

IXL x W So, RA + Re = WL Moment at bt. A MA = 0 RanX xXxw xL %3D RB WL %3D RA - WL 6. 3L - WL RA = 6. RA - WL for Shear fovu V:→ V (n). I(T) W. (L-x) %3D 6. WOL Wo 6. - Lu

W.xt A kwon 2L W. L W.L Wox? Won 6. 2. 2L (-1 +3) WOL Wout Wok 3 - W. 2L v(). -.(* -) Shear foru -Wo 2L for Bending moment:> (L-x) wL (L-x) (L-m) Wo -Wol 3L (u-b) = a-b°-3a' b + 3ab? %3D 3)

+ W,LX Wo 6. -W.L? 3. + Wolk 6. Wou' Wo X3 Lu 6K 6L + Woln Wox W. LX Wox? i. 2. 2. W.LK _Wo Lk Wox? Wok? 2. 6L 2. (1-3) (woLn) 6. Won' Wont - W. Lu Wor? Wont 62 2. Mx = W. tikending Momont Equahion

For Max Bending Momont ; Sheaw force So, Wo RL - (A) 느+t 2L "(i-) %3D *(- ) 2L Mulhply egy-(4) by 2L (may be solua wing Calculator) formulo > for quadrahic n- - bt b6²-4ac 2L + 422-4 XI x 2.L" २८

L# JVi L (1+0.577) |-577L, 0.42L Pulhing in moment Mx = W. w:( taking X = 0.423L (०.५२३)x (6.423) W./ Lxo.42L Mmax WoL? 6.141 + 0.이26 Mnax = woL" (0.0646) A bprox. Mmax Ans. So, Mmax = 0.0646 Wo L the quation for Shear lasly maximum Bending Momenh. by this, we Con clud, Can fovu, bunding moment L