Question

The rigid-joined plane frame QRS shown in the figure is subjected to a load P at the joint R. Let the axial deformation in the frame be neglected. If the support S undergoes a settlement of $\Delta =P{L}^3/\beta EI$, the vertical reaction at the support S will become zero when $\beta$ is equal to

• A. 3.0
• B. 7.5
• C. 0.1
• D. 48.0

Answer 1

The correct option is B 7.5


Using slope deflection method,

$M_{QR}=\frac{2EI}{l}({\theta }_R-\frac{3\Delta }{l})$
$M_{RQ}=\frac{2EI}{l}(2{\theta }_R-\frac{3\Delta }{l})$
$M_{RS}=\frac{2EI}{l}\left(2{\theta }_R\right)$

If reaction at S is equal to zero

$M_{RQ}+M_{QR}+Pl=0$

$\frac{6EI{\theta }_R}{l}-\frac{12EI\Delta }{l^2}+Pl=0$

$\frac{6EI{\theta }_R}{l}-\frac{12EI}{l^2}\times \frac{P{l}^3}{\beta EI}+Pl=0$

$\frac{6EI{\theta }_R}{l}-\frac{12Pl}{\beta }+Pl=0\text{...(i)}$

From equilibrium of joint

$M_{RQ}+M_{RS}=0$
$\frac{8EI{\theta }_R}{l}-\frac{6EI\Delta }{l^2}=0$

$\frac{6EI{\theta }_R}{l}=\frac{6}{8}(\frac{6EI}{l^2}\times \frac{P{l}^3}{\beta EI})$

$\frac{6EI{\theta }_R}{l}=\frac{36Pl}{8\beta }\text{...(ii)}$

From (i) & (ii)

$\frac{36Pl}{8\beta }-\frac{96Pl}{8\beta }+Pl=0$
$-\frac{60Pl}{8\beta }+Pl=0$
$\Rightarrow 8\beta =60$
$\beta =\frac{60}{8}=7.5$