# Structural Analysis

 Question 1
The linearly elastic planar structure shown in the figure is acted upon by two vertical concentrated forces. The horizontal beams UV and WX are connected with the help of the vertical linear spring with spring constant k = 20 kN/m. The fixed supports are provided at U and X. It is given that flexural rigidity $EI = 10^5 kN-m^2, P = 100 kN,$ and $a = 5 m$. Force Q is applied at the center of beam WX such that the force in the spring VW becomes zero.

The magnitude of force Q (in kN) is ________. (round off to the nearest integer)
 A 640 B 458 C 251 D 325
GATE CE 2022 SET-2      Methods of Structural Analysis
Question 1 Explanation:
If force in spring is zero, there will be no deformation in spring i.e., deflection of point V will be equal to deflection of point W

\begin{aligned} \delta _v&=\delta _w\\ \frac{P(2a)^3}{3EI}&=\frac{Q(a)^3}{3(2EI)}+\frac{Q(a)^2}{2(2EI)} +a\\ \frac{8}{3}P&=\left ( \frac{1}{6}+\frac{1}{4} \right )Q\\ Q&=\frac{32}{5}P=640kN \end{aligned}
 Question 2
Which of the following statement(s) is/are correct?
 A If a linearly elastic structure is subjected to a set of loads, the partial derivative of the total strain energy with respect to the deflection at any point is equal to the load applied at that point. B If a linearly elastic structure is subjected to a set of loads, the partial derivative of the total strain energy with respect to the load at any point is equal to the deflection at that point. C If a structure is acted upon by two force system Pa and Pb , in equilibrium separately, the external virtual work done by a system of forces Pb during the deformations caused by another system of forces Pa is equal to the external virtual work done by the Pa system during the deformation caused by the Pb system. D The shear force in a conjugate beam loaded by the M/EI diagram of the real beam is equal to the corresponding deflection of the real beam.
GATE CE 2022 SET-2      Methods of Structural Analysis
 Question 3
Consider a beam PQ fixed at P, hinged at Q, and subjected to a load F as shown in figure (not drawn to scale). The static and kinematic degrees of indeterminacy, respectively, are

 A 2 and 1 B 2 and 0 C 1 and 2 D 2 and 2
GATE CE 2022 SET-2      Determinacy and Indeterminacy
Question 3 Explanation:

Static indeterminacy, SI=r-3=(3+2)-3=2
Kinematic indeterminacy=0+1=1
 Question 4
Consider a simply supported beam PQ as shown in the figure. A truck having 100 kN on the front axle and 200 kN on the rear axle, moves from left to right. The spacing between the axles is 3 m. The maximum bending moment at point R is ______ kNm. (in integer)

 A 124 B 180 C 147 D 582
GATE CE 2022 SET-1      Influence Line Diagram and Rolling Loads
Question 4 Explanation:

$\frac{ab}{L}=\frac{1 \times 4}{5}=0.8m$
To get maximum BN at R
\begin{aligned} BM_{max}&=200 \times \frac{ab}{L}+100 \times y\\ \frac{ab/l}{b}&=\frac{y}{4-3}\Rightarrow \frac{0.8}{4}=\frac{y}{1}=0.2m\\ BM_{max}&=200 \times 0.8+100 \times 0.2=180kNm \end{aligned}
 Question 5
Consider the linearly elastic plane frame shown in the figure. Members HF, FK and FG are welded together at joint F. Joints K, G and H are fixed supports. A counter-clockwise moment M is applied at joint F. Consider flexural rigidity EI=$10^5kN-m^2$ for each member and neglect axial deformations.

If the magnitude (absolute value) of the support moment at H is 10 kN-m, the magnitude (absolute value) of the applied moment M (in kN-m) to maintain static equilibrium is ___________. (round off to the nearest integer)
 A 60 B 25 C 85 D 45
GATE CE 2022 SET-1      Methods of Structural Analysis
Question 5 Explanation:
$\frac{M}{6}=10kNm\Rightarrow M=60kNm$
 Question 6
The plane truss shown in the figure is subjected to an external force $P$. It is given that $P = 70 kN, a = 2 m,$ and $b = 3 m$.

The magnitude (absolute value) of force (in kN) in member $EF$ is _______. (round off to the nearest integer)
 A 12 B 30 C 87 D 92
GATE CE 2022 SET-1      Trusses
Question 6 Explanation:

\begin{aligned} \Sigma M_J&=0\\ V_A \times 8-H_A \times 1-70 \times 4&=0\\ 8V_A-H_A&=280 \;\;\;(i) \end{aligned}
To find $H_A$ (Cut the truss by 1-1)
Consider left hand side
\begin{aligned} \Sigma M_E&=0\\ V_A \times 4-H_A \times 4&=0\\ V_A&=H_A \;\;\;(ii)\\ \text{using (i) and (ii)}&\\ 8V_A-V_A&=280\\ 7V_A&=280\\ V_A&=40kN\\ H_A&=40kN\\ H_J&=40kN\\ V_J&=70-40=30kN\\ \end{aligned}
To find force in member EF (Cit the truss by 2-2)
Consider right hand side
Force in member EF
$F_{EF}=V_J=30kN$
 Question 7
A semi-circular bar of radius R m, in a vertical plane, is fixed at the end G, as shown in the figure. A horizontal load of magnitude P kN is applied at the end H. The magnitude of the axial force, shear force, and bending moment at point Q for $\theta =45 ^{\circ}$, respectively, are

 A $\frac{P}{\sqrt{2}}kN,\frac{P}{\sqrt{2}}kN \text{ and } \frac{PR}{\sqrt{2}}kNm$ B $\frac{P}{\sqrt{2}}kN,\frac{P}{\sqrt{2}}kN \text{ and } 0 \; kNm$ C $0 \; kN,\frac{P}{\sqrt{2}}kN \text{ and } \frac{PR}{\sqrt{2}}kNm$ D $\frac{P}{\sqrt{2}}kN,0 \; kN \text{ and } \frac{PR}{\sqrt{2}}kNm$
GATE CE 2022 SET-1      Arches
Question 7 Explanation:

\begin{aligned} (F_Q)&=P\sin \theta =\frac{P}{\sqrt{2}}&(\; at\; \theta =45^{\circ})\\ (S_Q)&=P\cos \theta =\frac{P}{\sqrt{2}}&(\; at\; \theta =45^{\circ})\\ M_Q&=PR \sin \theta =\frac{PR}{\sqrt{2}}&( \; at\; \theta =45^{\circ}) \end{aligned}
 Question 8
A perfectly flexible and inextensible cable is shown in the figure (not to scale). The external loads at F and G are acting vertically.

The magnitude of tension in the cable segment FG (in kN, round off to two decimal places) is _______
 A 4.68 B 3.78 C 5.64 D 8.25
GATE CE 2021 SET-2      Arches
Question 8 Explanation:

\begin{aligned} \Sigma M_F=0\\ \left ( 10.667+\frac{H}{6} \right )2-H \times 3=0\\ H=8\; kN\\ V_E=10.667+\frac{8}{6}\\ V_E=12\; kN\\ V_H=10\; kN \end{aligned}

Consider left side of section (1)-(1)

\begin{aligned} \Sigma H=0\\ T \cos \theta =H\\ T \cos \theta =8\\ T \sin \theta =2\\ \therefore \;\; T^2 \cos ^2\theta + T^2 \sin ^2 \theta =8^2+2^2\\ T=8.246\; kN \end{aligned}

Tension in segment GF is 8.246 kN.
 Question 9
A prismatic fixed-fixed beam, modelled with a total lumped-mass of 10 kg as a single degree of freedom (SDOF) system is shown in the figure.

If the flexural stiffness of the beam is $4 \pi^{2} \mathrm{kN} / \mathrm{m}$, its natural frequency of vibration (in Hz, in integer) in the flexural mode will be _______
 A 8 B 16 C 10 D 18
GATE CE 2021 SET-2      Matrix Method of Structural Analysis
Question 9 Explanation:
Given Data:
\begin{aligned} m&=10\; kg\\ K&=4 \pi^2 \times 10^3\\ &\text{Natural frequency,}\\ f_n&=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}\\ &=\frac{1}{2 \pi} \sqrt{\frac{4 \pi ^2 \times 10^3}{10}}\\ &=\frac{2 \pi \times 10}{2 \pi}=10 \end{aligned}
 Question 10
A frame EFG is shown in the figure. All members are prismatic and have equal flexural rigidity. The member FG carries a uniformly distributed load w per unit length. Axial deformation of any member is neglected.

Considering the joint F being rigid, the support reaction at G is
 A 0.375 wL B 0.453 wL C 0.482 wL D 0.500 wL
GATE CE 2021 SET-2      Methods of Structural Analysis
Question 10 Explanation:
Compatibility condition

$\begin{array}{l} \qquad \qquad \qquad \qquad \qquad \frac{\partial U}{\partial R}=0 \\ \Rightarrow \qquad \qquad \frac{\partial}{\partial R}\left[\int \frac{M^{2}}{2 E \mid} d x\right]=0 \\ \Rightarrow \frac{\partial}{\partial R}\left[\int_{0}^{L} \frac{\left(R x-\frac{w x^{2}}{2}\right)^{2}}{2 E I} d x+\int_{0}^{2 L} \frac{\left(R L-\frac{w L^{2}}{2}\right)^{2}}{2 E I} d x\right]=0 \\ \Rightarrow \int_{0}^{L} \frac{2\left(R x-\frac{w x^{2}}{2}\right)}{2 E I}(x) d x+\int_{0}^{2 L} \frac{\left(R L-\frac{w L^{2}}{2}\right)}{E I}(L) d x=0 \\ \Rightarrow \frac{R L^{3}}{3}-\frac{W}{2} \times \frac{L^{4}}{4}+R L^{2}(2 L)-\frac{W^{3}}{2}(2 L)=0 \\ \Rightarrow \frac{R L^{3}}{3}-\frac{w^{4}}{8}+2 R L^{3}-w L^{4}=0 \\ \Rightarrow 2 R L^{3}+\frac{R L^{3}}{3}=\frac{w^{4}}{8}+w^{4}=\frac{9}{8} w^{4} \\ \Rightarrow \frac{7 R L^{3}}{3}=\frac{9}{8} w L^{4} \\ \Rightarrow R=\frac{27}{56} w L=0.482 w L \end{array}$
Alternative
\begin{aligned} \frac{w L^{4}}{8 E I}+\frac{w L^{2}(2 L)}{2 E I} \times L &=\frac{R L^{3}}{3 E I}+\frac{R L \times 2 L \times L}{E I} \\ \frac{w^{4}}{8 E I}+\frac{w^{4}}{E I} &=\frac{R L^{3}}{3 E I}+\frac{2 R L^{3}}{E I} \\ \frac{9 W^{4}}{8 E I} &=\frac{7 R L^{3}}{3 E I} \\ R &=\frac{27 w L}{56}=0.482 \mathrm{wL} \end{aligned}
There are 10 questions to complete.

### 1 thought on “Structural Analysis”

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