Question 13.20: Determine the fixed-end moments and the fixed-end reactions ...

The resultant bending moment diagram is shown in Fig. 13.28(b) where the line AB represents the datum from which values of bending moment are measured. Again the net area of the resultant bending moment diagram is zero since the change in slope between the ends of the beam is zero. Hence

\frac{1}{2} (M_A + M_B)L = \frac{1}{2} L\frac{Wab}{L}

which gives

M_A + M_B = \frac{Wab}{L}                                           (i)

We require a further equation to solve for M_A and M_B . This we obtain using Eq. 13.10 and taking the origin for x at A; hence we have

x_B \left(\frac{dv}{dx} \right) _B – x_A\left(\frac{dv}{dx} \right) _A – (v_B – v_A) = \int_{A}^{B}{} \frac{M}{EI} x \ dx                                                         (ii)

In Eq. (ii) (dv/dx)_B = (dv/dx)_A = 0 and v_B = v_A = 0 so that

0 = \int_{A}^{B}{\frac{M}{EI}x } \ dx                                           (iii)

and the moment of the area of the M/EI diagram between A and B about A is zero. Since EI is constant for the beam, we need only consider the bending moment diagram. Therefore from Fig. 13.28(b)

M_AL\frac{L}{2} +(M_B – M_A)\frac{L}{2} \frac{2}{3} L = \frac{1}{2} a\frac{Wab}{L} \frac{2a}{3} + \frac{1}{2} b \frac{Wab}{L} \left(a+\frac{1}{3} b \right)

Simplifying, we obtain

M_A + 2M_B = \frac{Wab}{L^2} (2a+b)                                             (iv)

Solving Eqs (i) and (iv) simultaneously we obtain

M_A = \frac{Wab^2}{L^2}           M_B = \frac{Wa^2b}{L^2}                                     (v)

We can now use statics to obtain R_A and R_B ; hence, taking moments about B

R_AL − M_A + M_B − Wb = 0

Substituting for M_A and M_B from Eq. (v) we have

R_AL = \frac{Wab^2}{L^2} – \frac{Wa^2b}{L^2} +Wb

whence

R_A = \frac{Wb^2}{L^3} (3a+b)

Similarly

R_B = \frac{Wa^2}{L^3} (a+3b)