Question

A two hinged parabolic arch of span 40 m and rise 5 m carries a uniformly distributed load of 5 kN/m on the left half of the span and also a concentrated load of 40 kN at the crown. Determine the maximum positive bending moment (in kN-m upto two decimal places)

• A. 100.84

Answer 1

The correct option is A 100.84

$\sum M_B=0$
$V_a\times 40-5\times 20\times 30-40\times 20=0$
$\Rightarrow V_a=95kNand{V}_b=45kN$

$Also,H=\frac{w{l}^2}{16h}+\frac{25Pl}{128h}=\frac{5\times {40}^2}{16\times 5}+\frac{25}{128}\times \frac{40\times 40}{5}$

$\Rightarrow H=162.5kN$

Maximum positive bending moment:

$M=95x-\frac{5x^2}{2}-\frac{162.5}{80}x\times (40-x)$

$\left[\begin{array}{c}\because y_x=\frac{4\times 5\times x}{{40}^2}(40-x)\\ \Rightarrow y_x=\frac{x}{80}(40-x)\end{array}\right]$

$=95x-2.5x^2-81.25x+2.03125x^2$

$\Rightarrow M=13.75x-0.46875x^2$

$\text{For M to be maximum,}\frac{dM}{dx}=0$

$\frac{dM}{dx}=13.75-0.46875\times 2x=0$

$\Rightarrow x=14.667m$

Maximum positive BM:

$M_{max}=13.75\times 14.667-0.46875\times {14.667}^2$
$=100.84kN-m$