Question

The maximum bending moment due to an isolated load in a three-hinged parabolic arch (span l, rise h) having one of its hinges at the crown, occurs on either side of the crown at a distance

• A. $L/\left(2\sqrt{3}\right)$
• B. $L/4$
• C. $h/4$
• D. $L/\left(3\sqrt{2}\right)$

Answer 1

The correct option is A $L/\left(2\sqrt{3}\right)$

Given :



Let us assume as isolated load W acting at distance x from end A.

$\text{Since there is no horizontal external force so horizontal reactions at end A and B will be}{H}_A=H_B=H.$

Taking moment about (A), we have

$wx-V_B\times 2=0$

$\Rightarrow V_B=\frac{Wx}{L}$

$\because TakingbendingmomentatC:$

$Wehave(BM{)}_C=0$

$\Rightarrow V_B\times \frac{L}{2}-H\times h=0$

$\Rightarrow H=\frac{V_{B}L}{2h}=\frac{Wx}{2h}$

$Now,(BM{)}_D=V_B(L-x)-H\times y$

$Since,y=\frac{4hx}{L^2}(L-x)\left(Equationofparabolicarch\right)$

$(BM{)}_D=\frac{Wx}{L}(L-x)-H\times \frac{4hx}{L^2}(L-x)$

$\Rightarrow (BM{)}_D=\frac{Wx(L-x)}{L}-\frac{Wx}{2h}\times \frac{4hx}{L^2}(L-x)$

$\Rightarrow (BM{)}_D=\frac{Wx(L-x)}{L}-\frac{2W{x}^2}{L^2}(L-x)$

$\therefore Forabsolutemaximumbendingmoment$
$\frac{d(BM{)}_D}{dx}=0$

$\Rightarrow \frac{d\left(\frac{PWx(L-x)}{2}\right)}{dx}-\frac{d\left(\frac{2W{x}^2}{L^2}\right(L-x\left)\right)}{dx}=0$

$\Rightarrow \frac{W}{L}(L-2x)-\frac{2W}{L^2}(2Lx-3x^2)=0$

$\Rightarrow (L-2x)-\frac{2}{L}(2Lx-3x^2)=0$

$\Rightarrow 6x^2-6xL+L^2=0$

$\Rightarrow x=\left(\frac{6\pm 2\sqrt{3}}{12}\right)L$

x = 0.78867L, 0.2113l

So, distance of maximum moments for given load condition either side of crown.

$(\frac{L}{2}-0.2113L)or(0.7886L-\frac{L}{2})$

= 0.2886L or 0.2886L from crown

$\frac{L}{2\sqrt{3}}=0.2886L$

So, option (c) is correct.