Question

Beam PQRS has internal hinges in spans PQ and RS as shown.The beam may be subjected to a moving distributed vertical load of maximum intensity 4 kN/m of any length anywhere on the beam.The maximum absolute value of the shear force (in kN) that can occur due to this loading just to the right of support Q shall be.

• A. 55
• B. 30
• C. 40
• D. 45

Answer 1

The correct option is C 40

When a cut is made just to the right of Q and displacements are given such that A'B' is parallel to B'C'.As B' is very close to Q,displacement of B' to the left will be zero and that to the right will be, 1

Hence Slope of B'C' =$\frac{1}{2}$
$\Rightarrow Slopeof{B}^{\prime }{A}^{\prime }=\frac{1}{20}$
$\Rightarrow Ordinateat{A}^{\prime }=\frac{1}{20}\times 5=0.25$
$Ordinateat{D}^{\prime }=\frac{1}{20}\times 5=0.25$
if udl is loading span PR,we get max.SF just to the right of Q
$\Rightarrow SF=(\frac{1}{2}\times 0.25\times 10+\frac{1}{2}\times 20\times 1)\times 4$

SF=45kN
Alternative solution

$R_s=0$
$R_P\times 5-1(5-x)=0$
$R_p=\frac{5-x}{5}$
$R_R+R_Q+R_P=1$
$R_Q+R_R=1-\left(\frac{5-x}{5}\right)=\frac{5-5+x}{5}=\frac{x}{5}$
$R_Q+R_R=\frac{x}{5}$
$R_P\times 10-1(10-x)=R_R\times 20$
$\Rightarrow 20\times R_R=\frac{5-x}{5}\times 10-(10-x)$
$\text{/}10-2x-\text{/}10+x=R_R\times 20$
$\frac{-x}{20}=R_R$
SF to the right of Q, for x $\le 5$=$-R_R=\frac{x}{20}$

$R_Q+R_R=1$
$1\times (10-x)+R_R\times 20=0$

$R_R=\frac{x-10}{20}$

$\Rightarrow$ SF to right of Q=-$R_R=\frac{10-x}{20}$


$R_Q=\frac{30-x}{20}$
SF to right of
$Q=\frac{(30-x)}{20}$

$R_Q\times 20+1\times (x-30)=0$


$R_Q=\frac{30-x}{20}=\frac{-(x-30)}{20}$
SF to right of
$Q=\frac{-(x-30)}{20}$




SF when PR is loaded,
=$(\frac{1}{2}\times 0.25\times 10+\frac{1}{2}\times 1\times 20)\times 4$
$=(1.25+10)\times 4=40+5=45$