Question 13.10: Determine the deflected shape of the beam shown in Fig. 13.1...

In this problem an external moment M_0 is applied to the beam at B. The support reactions are found in the normal way and are

R_A = – \frac{M_0}{L}   (downwards)        R_C = \frac{M_0}{L}   (upwards)

The bending moment at any section X between B and C is then given by

M = R_Ax + M_0                                                  (i)

Equation (i) is valid only for the region BC and clearly does not contain a singularity function which would cause M_0 to vanish for x ≤ b. We overcome this difficulty by writing

M = R_Ax + M_0[x − b]^0           (Note: [x − b]^0 = 1 )                             (ii)

Equation (ii) has the same value as Eq. (i) but is now applicable to all sections of the beam since [x − b]^0 disappears when x ≤ b. Substituting for M from Eq. (ii) in Eq. (13.3) we obtain

\frac{d^2v}{dx^2} =\frac{M}{EI}                                       (13.3)

EI\frac{d^2v}{dx^2} = R_Ax + M_0[x-b]^0                                                   (iii)

Integration of Eq. (iii) yields

EI\frac{dv}{dx} = R_A\frac{x^2}{2} + M_0[x-b] + C_1                                              (iv)

and

EIv = R_A\frac{x^3}{6} + \frac{M_0}{2} [x-b]^2 +C_1x + C_2                                                  (v)

where C_1 and C_2 are arbitrary constants. The boundary conditions are v = 0 when x = 0 and x = L. From the first of these we have C_2 = 0 while the second gives

0=-\frac{M_0}{L} \frac{L^3}{6} +\frac{M_0}{2} [L-b]^2 + C_1L

from which

C_1 = – \frac{M_0}{6L} (2L^2-6Lb+3b^2)

The equation of the deflection curve of the beam is then