Question

Vertical reaction developed at B in the frame given below due to the applied load of $50kN$ (with $75000m{m}^2$ cross-sectional area and $1.5625\times {10}^{9}m{m}^4$ moment of inertia for both members ) is

• A. 5.884 kN
• B. 47.058 kN
• C. 2.942 kN
• D. None of these

Answer 1

The correct option is C 2.942 kN

Deflection at A in Beam = compression in column AC.

Let us draw FBD of structure


$\frac{(50-R)L^3}{3EI}=\frac{R.L}{AE}$

$\frac{(50-R)L^2}{3I}=\frac{R}{A}$

$R=\frac{(50-R)\times {1000}^2\times 75000}{1.5625\times {10}^9\times 3}$

$R=(50-R)\times 16$

$17R=800$

$R=47.058kN$

$\therefore V_B=50-47.058=2.942kN$