Question 20.4: A load of length 2 m and intensity 2 kN/m crosses the simply...

The shear force and bending moment influence lines for the quarter span point \mathrm{K} are constructed in the same way as before and are shown in Fig. 20.9(b) and (c).

Maximum shear force at K

The maximum positive shear force at K occurs with the head of the load at K. In this position the ordinate under the tail of the load is 0.05. Hence

S_{\mathrm{K}}(\operatorname*{max}+\mathrm{ve})=2\times\frac{1}{2}(0.05+0.25)\times2=0.6\,\mathrm{kN}

The maximum negative shear force at K occurs with the tail of the load at K. With the load in this position the ordinate under the head of the load is −0.55. Thus

S_{\mathrm{K}}(\operatorname*{max}-\mathrm{ve})=-2\times\frac{1}{2}(0.75+0.55)\times2=-2.6\,\mathrm{kN}

Maximum bending moment at K

We position the load so that K divides the load in the same ratio that it divides the span. Therefore 0.5 m of the load is to the left of K and 1.5 m to the right of K.

The ordinate in the M_{\mathrm{K}} influence line under the tail of the load is then 1.5 as is the ordinate under the head of the load. The maximum value of M_{\mathrm{K}} is thus given by

M_{\mathrm{K}}(\operatorname*{max})=2\left[\frac{1}{2}(1.5+1.875)\times0.5+\frac{1}{2}(1.875+1.5)\times1.5\right]

which gives

M_{\mathrm{K}}(\mathrm{max})=6.75\,\mathrm{kN}\,\mathrm{m}