SS Beams with a Point Load Anywhere on Span:

AB = l,                        AC = a,           CB = b,           a + b = l

∑ y = 0              R_A  + R_B  = W

Taking moments around A,

W × AC = RB  × AB    or Wa = R_B l

∴  R_B  =   (Wa /l )(↑)

From Equation (3) & (4)

R_A  =   W_b/  l (↑)

Figure

Let a section X-X at a distance x from A,

M = R_A x - W [ x - a]

Note down that if x < a, M = R_A x, that means second term is not applicable.

 M = (Wb/l) x - W [ x - a]

                  or,

The governing equation for deflection is :

EI(d ²y /dx²)  = M= (W b /l )x-  [ x - a]

Integrating the Eq. (9.17), we can get

EI(d y /dx)  = (W b /l )(x2/2) - (W/2) [ x - a]²  + C₁

EIy = (W b / 2l )(x³/3) - (W/6) [ x - a]³  + C₁x+C₂

The constants C₁ & C₂ may be found through the boundary conditions. The boundary conditions are following:

at A, x = 0, y = 0              --------------- (1)

at B, x = l, y = 0      ----------------- (2)

By applying BC (1) to Eq. (19) and noting that W (x - a) is not applicable if x < a, as is needed for BC (1) or BC at A.

0 =  Wb /6l  (0) + C₁ (0) + C₂

∴          C₂ = 0

By applying BC (2) to the Eq. (19), C₂ = 0.

0 = (Wb/6) l ²  - (W/6) b³ + C₁ l

C₁ = Wb/6l  (b²  - l ² )