Question 20.3: Determine the maximum positive and negative values of shear ...
Determine the maximum positive and negative values of shear force and the maximum value of bending moment at the section \mathrm{K} in the simply supported beam AB shown in Fig. 20.6(a) when it is crossed by the system of loads shown in Fig. 20.6(b).
The influence lines for the shear force and bending moment at \mathrm{K} are constructed using either of the methods described in Sections 20.1 and 20.2 as shown in Fig. 20.6(c) and (d).
Maximum positive shear force at K
It is clear from inspection that S_{\mathrm{K}} will be a maximum with the 5 kN load just to the left of K, in which case the 3 kN load is off the beam and the ordinate under the 4 kN load in the S_{\mathrm{K}} influence line is, from similar triangles, 0.1. Then
S_{\mathrm{K}}(\operatorname*{max})=5\times0.3+4\times0.1=1.9\,\mathrm{kN}
Maximum negative shear force at K
There are two possible load positions which could give the maximum negative value of shear force at K; neither can be eliminated by inspection. First we shall place the 3 kN load just to the right of K. The ordinates under the 4 and 5 kN loads are calculated from similar triangles and are −0.5 and −0.3, respectively. Then
S_{\mathrm{K}}=3\times(-0.7)+4\times(-0.5)+5\times(-0.3)=-5.6\,\mathrm{KN}
Now with the 4 kN load just to the right of K, the ordinates under the 3 and 5 kN loads are 0.1 and −0.5, respectively. Then
S_{\mathrm{K}}=3\times(0.1)+4\times(-0.7)+5\times(-0.5)=-5.0\,\mathrm{KN}
Therefore the maximum negative value of SK is −5.6 kN and occurs with the 3 kN load immediately to the right of K.
Maximum bending moment at K
We position the loads in accordance with the criterion of Eq. (20.14). The load per unit length of the complete beam is (3 + 4 + 5)/20 = 0.6 kN/m. Therefore if we position the 4 kN load at K and allocate 0.6 kN of the load to AK the load per unit length on AK is (3 + 0.6)/6 = 0.6 kN/m and the load per unit length on KB is (3.4 + 5)/14 = 0.6 kN/m. The maximum bending moment at K therefore occurs with the 4 kN load at K; in this example the critical load position could have been deduced by inspection.
With the loads in this position the ordinates under the 3 and 5 kN loads in the M_{\mathrm{K}} influence line are 1.4 and 3.0, respectively. Then
M_{\mathrm{K}}(\mathrm{max})=3\times1.4+4\times4.2+5\times3.0=36.0\,\mathrm{kNm}
{\frac{1}{L}}\sum_{j=1}^{n}W_{j}={\frac{1}{a}}\sum_{j=1}^{n}W_{j,\mathrm{L}}={\frac{1}{L-a}}\sum_{j=1}^{n}W_{j,\mathrm{R}} (20.14)