Structural Analysis

Question 1
In the frame shown in the figure (not to scale), all four members (A B, B C, C D, and A D) have the same length and same constant flexural rigidity. All the joints A, B, C, and D are rigid joints. The midpoints of A B, B C, C D, and A D, are denoted by E, F, G, and H, respectively. The frame is in unstable equilibrium under the shown forces of magnitude P acting at E and G. Which of the following statements is/are TRUE?

Shear forces at \mathrm{H} and \mathrm{F} are zero
Horizontal displacement at \mathrm{H} and \mathrm{F} are zero
Vertical displacement at \mathrm{H} and \mathrm{F} are zero
Slopes at E, F, G, and \mathrm{H} are zero
GATE CE 2023 SET-2      Methods of Structural Analysis
Question 1 Explanation: 

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.
Question 2
Muller-Breslau principle is used in analysis of structures for
drawing an influence line diagram for any force response in the structure
writing the virtual work expression to get the equilibrium equation
superposing the load effects to get the total force response in the structure
relating the deflection between two points in a member with the curvature diagram inbetween
GATE CE 2023 SET-2      Influence Line Diagram and Rolling Loads
Question 2 Explanation: 
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.

Question 3
An idealised bridge truss is shown in the figure. The force in Member U_{2} L_{3} is _____ kN (round off to one decimal place).

GATE CE 2023 SET-1      Trusses
Question 3 Explanation: 

\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.
Question 4
Consider the pin-jointed truss shown in the figure (not to scale). All members have the same axial rigidity, \mathrm{AE}. Members \mathrm{QR}, \mathrm{RS}, and \mathrm{ST} have the same length L. Angles OBT, RCT, SDT are all 90^{\circ}. Angles BQT, CRT, DST are all 30^{\circ}. The joint T carries a vertical load P. The vertical deflection of joint T is k \frac{P L}{A E}. What is the value of k ?

GATE CE 2023 SET-1      Trusses
Question 4 Explanation: 

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
Question 5
Consider the following three structures:

Structure I: Beam with hinge support at A, roller at C, guided roller at E, and internal hinges at B and D.

Structure II: Pin-jointed truss, with hinge support at A, and rollers at B and D.

Structure III: Pin-jointed truss, with hinge support at A and roller at C.

Which of the following statements is/are TRUE?
Structure I is unstable
Structure II is unstable
Structure III is unstable
All three structures are stable
GATE CE 2023 SET-1      Determinacy and Indeterminacy
Question 5 Explanation: 
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

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.

There are 5 questions to complete.

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