Question 13.22: The fixed beam of Fig. 13.30 carries a uniformly distributed...

Consider a small element δx of the distributed load. We can use the results of Ex. 13.20 to write down the fixed-end moments produced by this elemental load since it may be regarded, in the limit as δx → 0, as a concentrated load. Therefore from Eq. (v) of Ex. 13.20 we have

$δM_A = w δx \frac{x(L-x)^2}{L^2}$

The total moment at A, $M_A$, due to all such elemental loads is then

$M_A = \int_{a}^{b}{\frac{w}{L^2} } x(L-x)^2 \ dx$

which gives

Similarly

$M_B = \frac{wb^3}{L^2} \left(\frac{L}{3} – \frac{b}{4} \right)$                                                     (ii)

If the load covers the complete span, a = 0, b = L and Eqs (i) and (ii) reduce to

$M_A = M_B = \frac{wL^2}{12}$        (as in Ex. 13.21.)