Question 20.6: The set of loads shown in Fig. 20.14(b) crosses the simply s...

The first step is to find the position of the centre of gravity of the set of loads. Thus, taking moments about the load W_{5} we have

(9+15+15+8+8) \bar{x}=15 \times 2+15 \times 4.3+8 \times 7.0+8 \times 9.3

whence

\bar{x}=4.09 \mathrm{~m}

Therefore the centre of gravity of the loads is 0.21 \mathrm{~m} to the left of the load W_{3}.

By inspection of Fig. 20.14(b) we see that it is probable that the maximum bending moment will occur under the load W_{3}. We therefore position W_{3} and the centre of gravity of the set of loads at equal distances either side of the mid-span of the beam as shown in Fig. 20.14(a). We now check to determine whether this position of the loads satisfies the load per unit length condition. The load per unit length on \mathrm{AB}=55 / 20=2.75  \mathrm{kN} / \mathrm{m}. Therefore the total load required on \mathrm{AK}=2.75 \times 10.105= 27.79  \mathrm{kN}. This is satisfied by W_{5}, W_{4} and part (3.79  \mathrm{kN}) of W_{3}.

Having found the load position, the bending moment at \mathrm{K} is most easily found by direct calculation. Thus taking moments about \mathrm{B} we have

R_{\mathrm{A}} \times 20-55 \times 10.105=0

which gives

R_{\mathrm{A}}=27.8  \mathrm{kN}

so that

M_{\mathrm{K}}=27.8 \times 10.105-9 \times 4.3-15 \times 2.3=207.7  \mathrm{kN} \mathrm{m}

It is possible that in some load systems there may be more than one load position which satisfies both criteria for maximum bending moment but the corresponding bending moments have different values. Generally the absolute maximum bending moment will occur under one of the loads between which the centre of gravity of the system lies. If the larger of these two loads is closer to the centre of gravity than the other, then this load will be the critical load; if not then both cases must be analysed.