Question

The magnitude of base shear at support A for the given frame is

• A. 4.85 kN
• B. 7.88 kN
• C. 3.64 kN
• D. 1.82 kN

Answer 1

The correct option is C 3.64 kN

$D{F}_{CA}=\frac{\frac{4}{15}}{\frac{4}{15}+\frac{4}{15}+\frac{3}{15}}=\frac{4}{11}$

$\left[\text{Far end hinge}\right(k)=\frac{3EI}{l}\text{far end fixed}\left(k\right)=\frac{4EI}{l}]$

So,
$M_{CA}=\frac{4}{11}\times 100=36.36kN-m$

$\therefore M_{AC}=\frac{M_{CA}}{2}\left[\text{carry over moment}\right]$

$=\frac{36.36}{2}$

$M_{AC}=18.18kN-m$


$H_A=\text{Base shear at A}=\frac{M_{AC}+M_{CA}}{l_{CA}}=\frac{36.36+18.18}{15}$
$=3.64kN$