Question

The wheel loads shown in the figure given below rolls from left to right over a simply supported beam of span 15 m. For a section 5 m from the left end, the maximum bending moment is _________ kNm.

• A. 2696

Answer 1

The correct option is A 2696
Let the loads cross the section 'D' and compare the average load on span AD and DE as shown in the table given below.

	Average load on AD	Average load on DB
(i) When all loads are on AD	$\left(\frac{250+200+350+140}{5}\right)=188$	= 0
(ii) When 140 kN crosses the section D	$\left(\frac{250+200+350}{5}\right)=160$	$>\frac{140}{10}=14$
(iii) When 350 kN crosses the section D	$\left(\frac{250+200}{5}\right)=90$	$>\frac{350+140}{10}=49$
(iv) When 200 kN crosses the section D	$\left(\frac{250}{5}\right)=50$	$<\frac{200+350+140}{10}=69$

i.e. when the 200 kN load crosses the section 'D' average load on AD becomes less than the average load on DB.
So keep 200 kN at D to get maximum BM at 'D'



The ordinages of ILD at various load positions are found by proportionality as shown below:

$\frac{3.33}{5}=\frac{c{c}^{\prime }}{4}$

$\Rightarrow c{c}^{\prime }=2.67$

$\frac{3.33}{10}=\frac{E{E}^{\prime }}{10-1.2}$

$\Rightarrow E{E}^{\prime }=2.93$

$\frac{3.33}{10}=\frac{F{F}^{\prime }}{(10-1.2-1.6)}$

$\Rightarrow F{F}^{\prime }=2.4$
$\therefore$ Maximum BM at D = 250 × 2.67 + 200 × 3.33 + 350 × 2.93 + 140 × 2.4
= 2695 kNm