Question

A cantilever beam of length L and constant EI carries a concentrated load P at a distance of L/4 from the free end. Use any method that you chose to develop an expression for the deflection of the free end. VEI = ?

Loading Loading Function wwx) Shear V /w(xdx Moment M-Vax 1/4 M = M-a) V-M, M-M V-a) N-P) 03 V- M- (3 shop

Answer 1

Solution. We will solve this preoblem using virtual work method. P. L/4 B с A RA Ify=0; RA-P=0 RA=.P (*) Moment about A; 26 + MA - P(2-2/4)=0 3PL (6) MA= At firest we will find the sectional moments as the sections are shown above, Take section - Consider left portion of it Cosas 34) (x is taken from A) (AB portion) consider 24 31L 3PL 4. + Ma PR=0 Ma A Ma=Px-3PL e P Take section ② -②; consider. Right portion of it (osa<4) (x is taken from B) CBC porection) Consider 24 Ma C - My =o Mx=0 2 (4/4-x)

we will 1 Now put on unit load at free end to construct the virtual structure, Ify=0; "A-1=0 MA=I(+) Moment about A; ? MA-(1+2)=0 MA=L(Q) MA A 3 A ท A 10 We will again find the sectional moment in the virtual strincture. Section -(0; consider left portion of it Cosas 34) (x is taken from A) (AB poration) Consider ? L+mu Exx) = gone M = R-L L A 17924 section ② - ③; Consider right portion of it (osnah) (x is taken from B) (Be portion) Consider ? 11 -mx - 1x (44 wn) =0 ma=n-4 (44-) ma 4

Now using virtual work principle to evaluate the deflection at free end. lu 344 Mama Mama dat da Ef EI 344 so, As o 344 As PRL- 3 PLZ 4 + 그 EI E I J (PR-304) (x-2) de 3922) dx & [PLx2] - [3 92 ] + 34] 344 ✓ (Px² Pal Es [pixel 344 As EI EI 6 344 6 344 3 9 P23 9 PL 27 PL² + 9PL² 32 EI 128 EL 16 ET 64 EI A 128 EI 27 PL3 (t since the sign is positive: its indicating the deflection will occur along applied unit load cie downwards) This is the expression for the deflection of the free end. (Answer)