Question

A portal frame ABCD is shown in figure. If the support 'D' sinks by an amount of 15 mm and EI = 48000 kN-$m^2$, the moment developed at joint 'D' (in kN-m, upto two decimal places) is____.

• A. 196.32

Answer 1

The correct option is A 196.32
Given EI = 48000 kN-$m^2$

$\because$ No external load is applied on the frame, the fixed end moments of all members will be zero.

$\therefore$ Slope deflection equations are ;

$M_{BA}=\frac{3EI}{3}\left[{\theta }_B\right]$

$M_{BC}=\frac{2EI}{2}[2{\theta }_B+{\theta }_c-\frac{3\times 15}{2000}]$

$M_{CB}=\frac{2EI}{2}[2{\theta }_c+{\theta }_B-\frac{3\times 15}{2000}]$

$M_{CD}=\frac{2EI}{2}\left[2{\theta }_C\right]$

Under equilibrium

$M_{BA}+M_{BC}=0$

$48000{\theta }_B+96000{\theta }_B+48000{\theta }_C-1080=0$

$\Rightarrow 144000{\theta }_B+{48000}_C=1080...\left(1\right)$

$M_{CB}+M_{CD}=0$

$96000{\theta }_C+48000{\theta }_B-1080+96000{\theta }_c=0$

$\Rightarrow 48000{\theta }_B+192000{\theta }_C=1080..\left(2\right)$

From (1) and (2) , we get

${\theta }_B=0.00613rad{\theta }_C=0.00409rad$

$M_{DC}=\frac{2EI}{2}\left[{\theta }_C\right]=48000\times 0.00409$

$M_{DC}=196.32kN-m$