Question 13.21: The fixed beam shown in Fig. 13.29(a) carries a uniformly di...

From symmetry, M_A = M_B and R_A = R_B . Again the net area of the bending moment diagram must be zero since the change of slope between the ends of the beam is zero (Eq. (13.7)). Hence

\left(\frac{dv}{dx} \right) _B – \left(\frac{dv}{dx} \right) _A = \int_{A}^{B}{\frac{M}{EI} } \ dx                                                    (13.7)

M_AL = \frac{2}{3} \frac{wL^2}{8} L

so that

M_A = M_B = \frac{wL^2}{12}

From statics

R_A = R_B = \frac{wL}{2}