A simply supported beam of length 6m carries a triangular load whose intensity varies uniformly from zero at the left end to 60 kN/m at the right end. It has one support at 1.5m from the left and the other support at the right end. Draw the S.F and B.M diagrams for the beam.

1. Total load on beam = area of triangular region

$ = \frac{1}{2}*60*(1.5+4.5) = 180KN$ which is acting at $\frac{1}{3}8(1.5+4.5) = 2m$ from B

Taking moment about B,

$V_a*4.5 = 180*2$

$V_a = 80KN$

$V_b = 180 - 80 = 100 KN $

$\sum F_y = 0 i.e V_a+V_b = 180$

Consider any section x distance x from the left end

w = load intensity at the section $X = \frac{60}{1.5+4.5}* x =10x KN/m$

S = external load acting b/w the left end C and section $X = \frac{-1}{2}*x*10X = -5x^2 KN$

OR

$S = -\int_0^x W dx = -\int_0^x 10x.dx = \frac{-10x^2}{2} = -5x^2KN$

M = moment of the above external load about x

$ = \int_0^x -S dx = \int_0^x 5x^2 dx = \frac{-5x^3}{3} KNm$

Shear force analysis:

At any section in CA distant x from C

$S_x = -5x^2 At x = 0, S = 0$

At x =1.5m, $S = -5*1.5^2 = -11.25KN$

At any section in AB, distant x from C,

$S = 80-5x^2 At x = 1.5m, S = 80-50*1.5^2 = 68.75KN$

At x = 6m, $S = 80-5*6^2 = -100KN$

Section of zero shear,

equating the general expression for shear to zero $80-5x^2=0$

x = 4m from C

Bending moment analysis:

At any section in CA distant x from C

$M_x = \frac{5x^3}{3} At x = 0, M = 0$

At x =1.5m, $M = \frac{-5}{3}*1.5^3 = -5.625 KNm$

At any section in AB, distant x from C

$M = 80(x-1.5)-\frac{5}{3}x^3$

At x = 1.5m, $M = \frac{5}{3}*1.5^3 = -5.625 KNm$

At x = 6m, $M = 80*(6-4.5)-\frac{5}{3}*6^3 = 0$

$M_{max} at x = 4m i.e M_{max} = 8(4-1.5)-\frac{5}{3}*4^3=\frac{280}{3} KNm = 93.33KNm$

Point of contraflexure : BM = 0 (equating general expression)

$80(x-1.5)-\frac{5x^3}{3} = 0$

On solving, x = 1.5826m from C