Structural Analysis

Due to symmetry horizontal displacement of \mathrm{H} \& F= zero
\Rightarrow option (B) correct
Also due to symmetry slopes at E, F, G, H =0 \Rightarrow option (D) is correct
Since A D and B C are subjected to pure bending Hence shear force at \mathrm{H} \& \mathrm{F} are zero
\Rightarrow option (A) correct
Vertical displacement at \mathrm{H} \& \mathrm{~F} \neq 0
\Rightarrow option (C) is incorrect.

Muller Breslass principle is used to draw influence line diagram for determinate and indeterminate structures. It states that influence line for any stress function may be obtained by removing the restraint offered by that function and introducing a directly related generalised unit displacement at that location in the direction of the stress function.

\sum F_{y}=0, \quad 50-20-20-F_{U_{2} L_{3}} \sin 45^{\circ}=0
\therefore \quad F_{U_{2} L_{3}}=\frac{10}{\sin 45^{\circ}}=14.14 \mathrm{kN}
\therefore Rounding off to one decimal.

Considering joint T,

\sum F_{y}=0, \quad F_{Q T} \sin 60^{\circ}=P
\therefore \quad \mathrm{F}_{\mathrm{QT}}=\frac{2 \mathrm{P}}{\sqrt{3}}
\sum \mathrm{F}_{\mathrm{X}}=0, \mathrm{~F}_{\mathrm{BT}}=-\mathrm{F}_{\mathrm{QT}} \cos 60^{\circ}=-\frac{P}{\sqrt{3}}
\therefore \quad Strain energy (U)=\left(\frac{P^{2} L}{2 A E}\right)_{Q T}+\left(\frac{P^{2} L}{2 A E}\right)_{B T}
=\frac{\left(\frac{2 \mathrm{P}}{\sqrt{3}}\right)^{2}(3 \mathrm{~L})}{2 \mathrm{AE}}+\frac{\left(-\frac{\mathrm{P}}{\sqrt{3}}\right)^{2}\left(\frac{3 \mathrm{~L}}{2}\right)}{2 \mathrm{AE}}
=\frac{2 P^{2} \times L}{A E}+\frac{P^{2} \times L}{4 A E}
\frac{8 P^{2} L+P^{2} L}{4 A E}=\frac{9 P^{2} L}{4 A E}
\therefore \quad Vertical deflection of joint T
=\frac{\partial \mathrm{U}}{\partial \mathrm{P}}=\frac{18 \mathrm{PL}}{4 \mathrm{AE}}=4.5 \frac{\mathrm{PL}}{\mathrm{AE}}
\therefore \quad \mathrm{K}=4.5

Unstability in beam can be checked

(i) If support reactions are not enough.
(ii) If reactions are concurrent.
(iii) If reactions are parallel.
(iv) If there is mechanism.

where as for truss also along with the above given points the triangular panels are generally stable.
But, a virtual inspection must be conducted to understand the stability

(I)
It is unstable as it shows mechanisms.
Also to understand if we apply a vertical force at slider side, the deflected shape will be as follows.

(II) The frame is internally stable but all the reactions are concurrent and meeting at hinge A, and the frame can rotate about A. Hence, it is unstable.

(III) The frame is unstable because it cannot resist shear in DE and AB since DE and AB are slender members.