Question

Problem 2: i. For the cantilever beam (free at A and fixed at B) determine the expression for moment as a function of x, that is M(x). Take A as origin. [10 Points 4 kN/m T - 2 m 27 - 2 m The expression for moment as a function of x, that is M(x), is shown below for a cantilever beam (fixed at A and free at B), considering A as origin. Taking EI constant, determine the slope and deflection at point B. (10 Points 4 kN/m M(x) = -8 + 8x - 2x2

Answer 1

Soda x = 22 For portion AC se<2 → center of gravity 30 M a 2 Mex) Mars = (4.3.22)Ix Mex) = x (cubic cunde) For portion cB [2Ex=4] M(x) ) T -2 2 AC Mcx) = (2x2x4) (2-2 +}) M(x) = +2 -16 (dinear) 1 [2= x=4] < z a klon 2 m - Using M(x) = -8 + 8x - 242 double integration method · Er der = Meny

ant 2 2 EI e - Mesc) = -8 + 8x-272 - © Integrating the above eq". EI de - - 8x + 83?- nte, euro st ET y = - 3x2 + - 3 +6,+62 - > As the support is fixed. The values obtained from boundary conditions goe at man dy-0 By substituting n=o de la = o in eq ② »c,=0 at B e di = 80 = 6-8x2 +16-19) 10 B=-lo se k slope at point B 0. n=2m 1 EI (Negative makes direction sign shows that an angle in with AB I the the tangent at B anti clock wise

At B x=2 -4 (2) +4x2 EI NIM By substituting xao, .yo in eq@ ¢=0 Portal = [4633* +972-32 ] (Ypres) -- deglection at point B (Negative sign shows the deflection is downwards )