A portal frame ABC is shown in figure. The horizontal movement of joint 'B' (in mm, upto two significant places) is____. $[T a k e E I = 1780 k N - m^{2}]$

1.11

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Solution

The correct option is A 1.11

Member FEM (kN-m)

AB 4.5 kN-m

BA -4.5 kN-m

BC -20 kN-m

$M_{A B} = 4.5 + \frac{2 E (2 I)}{3} [θ_{B} - δ]$

$M_{B C} = - 4.5 + \frac{2 E (2 I)}{3} [2 θ_{B} - \frac{3 δ}{3}]$

$M_{A B} = - 20 + \frac{2 E I}{2} [2 θ_{B}]$

$U n d e r e q u i l i b r i u m, M_{B A} + M_{B C} = 0$

$\Rightarrow - 4.5 + 4746.67 θ_{B} - 2373.34 δ - 20 + 3560 θ_{B} = 0$

$\Rightarrow 8306.67 θ_{B} - 2373.34 δ = 24.5... (1)$

Shear equation,

$H_{A} = 12 K N$

$\frac{M_{A B} + M_{B A} + 12 \times 1.5}{3} = H_{A}$

$\Rightarrow 4 \times 1780 \times θ_{B} - \frac{8 \times 1780 \times δ}{3} + 18 = 36$

$\Rightarrow 7120 θ_{B} - 4746.67 δ = 18 . . . (2)$

From (1) and (2), we get

$θ_{B} = 0.0033 r a d, δ = 1.106 m m$

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