GATE CE · Structural Analysis

Methods Of Structural Analysis – GATE CE Structural AnalysisPractice Questions & PYQs

Generate GATE-level questions on Methods Of Structural Analysis in Structural Analysis. Focus on core concepts, previous year patterns, and numerical problem-solving techniques.

32 questions · 20 PYQs · 0 AI practice · GATE CE 2027

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Q1GATE 2025MSQ

Consider the frame shown in the figure under the loading of $100 \mathrm{kN} . \mathrm{m}$ couples at the joints $B$ and $G$ . Considering only the effects of flexural deformations, which of the following statements is/are true:

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📖 Explanation

Step 1: Analyze the structure - It's a symmetric frame with members: AB, BC, CD, DE, EF, and FG. - Moments of $100 \, \text{kN} \cdot \text{m}$ are applied at joints B and G (clockwise and counterclockwise respectively). - All vertical columns (AB, DE, FG) have different stiffnesses as per EI values. - Assume fixed supports at A, D, and F. Step 2: Moment Distribution - focus on joints B, C, and D Let's denote member stiffness factors: - Stiffness of member BC from joint B is: $K_{BC} = \frac{4EI}{L} = \frac{4EI}{6} = \frac{2EI}{3}$ - Stiffness of member AB from joint B: $K_{BA} = \frac{2EI}{6} = \frac{EI}{3}$ Distribution Factor (DF) at Joint B: $DF_{BC} = \frac{K_{BC}}{K_{BA} + K_{BC}} = \frac{\frac{2EI}{3}}{\frac{EI}{3} + \frac{2EI}{3}} = \frac{2}{3}$ $DF_{BA} = \frac{1}{3}$ Similarly at Joint C (connecting BC, CD, and CE): All members have same length and stiffness $2EI$ , so: $K_{CB} = \frac{4EI}{6} = \frac{2EI}{3}$ $K_{CD} = \frac{2EI}{8} = \frac{EI}{4}$ $DF_{CB} = \frac{\frac{2EI}{3}}{\frac{2EI}{3} + \frac{EI}{4}} = \frac{8}{11}$ $DF_{CD} = \frac{3}{11}$ Step 3: Apply the moment at B and perform moment distribution - Moment at B = $100 \, \text{kN} \cdot \text{m}$ (clockwise), so fixed-end moment to BC is $+100$ - Apply moment distribution considering carry-over and distribution factors. After a few iterations (or by symmetry), we get: $M_{BC} = 50 \, \text{kN} \cdot \text{m}$ $M_{CB} = 50 \, \text{kN} \cdot \text{m}$ Because of symmetric load and structure, the moment gets equally distributed between BC and CD. Conclusion: - Moment at joint C is exactly $50 \, \text{kN} \cdot \text{m}$ , not more. - Thus, option D ("more than 50") is False . - Joint C has no rotation due to symmetry - option C is True . - CD has zero axial force and zero shear since there?s no direct force acting - options A and B are True .

Q2GATE 2025MCQ

In the rigid-jointed frame shown in the figure, the distribution factor of the member $A D$ is closest to

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📖 Explanation

To find the distribution factor of the member AD at joint A, we use the formula: $\text{Distribution Factor (DF)} = \frac{\text{Stiffness of Member}}{\text{Sum of Stiffnesses at the Joint}}$ We need the stiffness of all members connected at joint A: Members connected at A: AD, AC, and AF. All are fixed at far ends, so stiffness for each is: $K = \frac{4EI}{L}$ (for member with far end fixed) From the figure: For AD: Length = $\frac{L}{2}$ , Flexural Rigidity = $EI$ $\Rightarrow K_{AD} = \frac{4EI}{L/2} = \frac{8EI}{L}$ For AC: Length = $L$ , Flexural Rigidity = $2EI$ $\Rightarrow K_{AC} = \frac{4(2EI)}{L} = \frac{8EI}{L}$ For AF: Length = $L$ , Flexural Rigidity = $2EI$ $\Rightarrow K_{AF} = \frac{4(2EI)}{L} = \frac{8EI}{L}$ Total stiffness at A: $K_{\text{total}} = K_{AD} + K_{AC} + K_{AF} = \frac{8EI}{L} + \frac{8EI}{L} + \frac{8EI}{L} = \frac{24EI}{L}$ Distribution factor for AD: $DF_{AD} = \frac{K_{AD}}{K_{\text{total}}} = \frac{8EI/L}{24EI/L} = \frac{1}{3}$ Final Answer: $\boxed{\frac{1}{3}}$

Q3GATE 2025MCQ

The figure shows a propped cantilever with uniform flexural rigidity $EI$ (in $\text{N} \cdot \text{m}^2$ ) and subjected to a moment $M$ (in $\text{N} \cdot \text{m}$ ). Consider forces and displacements in the upward direction as positive. Find the upward reaction at the propped support B (in N) when this support settles by $(-\Delta )$ , given in metres.

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📖 Explanation

We use the principle of superposition: Reaction at B due to $M$ only: $\frac{3M}{2L}$ Reaction at B due to downward settlement $-\Delta$ : $+\frac{3EI \Delta}{L^3}$ Total reaction at B: $R_B = \frac{3M}{2L} - \frac{3EI \Delta}{L^3}$

Q4GATE 2023MSQ

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?

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📖 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.

Explanation Figure 2

Q5GATE 2022NAT

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)

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📖 Explanation

$\frac{M}{6}=10kNm\Rightarrow M=60kNm$

Q6GATE 2022MSQ

Which of the following statement(s) is/are correct?

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Q7GATE 2022NAT

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)

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📖 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}

Explanation Figure 2

Q8GATE 2021MCQ

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

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📖 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}

Explanation Figure 2

Q9GATE 2021NAT

A propped cantilever beam XY, with an internal hinge at the middle, is carrying a uniformly distributed load of 10 kN/m, as shown in the figure. The vertical reaction at support X ( in kN, in integer) is _____

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📖 Explanation

\begin{aligned} B M &=0 \text { at hinge } \\ R_{Y} \times 2-10 \times 2 \times 1 &=0 \\ R_{Y} &=10 \mathrm{kN} \\ R_{X}+R_{Y} &=10 \times 4 \\ R_{X}+10 &=40 \\ R_{X} &=30 \mathrm{kN} \end{aligned}

Explanation Figure 2

Q10GATE 2020MCQ

A prismatic linearly elastic bar of length, L, cross-sectional area A, and made up of a material with Young's modulus E, is subjected to axial tensile force as shown in the figures. When the bar is subjected to axial tensile force $P_1\; and\; P_2$ , the strain energies stored in the bar are $U_1\; and\; U_2$ , respectively. If U is the strain energy stored in the same bar when subjected to an axial tensile force ( $P_1 + P_2$ ), the correct relationship is

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📖 Explanation

$(P_1+P_2)^2 \gt P_1^{2+}P_2^2$ $U \gt U_1+U_2$

Explanation Figure 2

Q11GATE 2020MCQ

The planar structure RST shown in the figure is roller-supported at S and pin-supported at R. Members RS and ST have uniform flexural rigidity (EI) and S is a rigid joint. Consider only bending deformation and neglect effects of self-weight and axial stiffening When the structure is subjected to a concentrated horizontal load P at the end T, the magnitude of rotation at the support R, is

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📖 Explanation

$\theta _R=\frac{(PL/2)L}{6EI}$ $\theta _R=\frac{PL^2}{12EI}$

Explanation Figure 2

Q12GATE 2019MCQ

The rigid-jointed plane frame QRS shown in the figure is subjected to a load P at the joint R. Let the axial deformations in the frame be neglected. If the support S undergoes a settlement of $\Delta =\frac{PL^3}{\beta EI}$ , the vertical reaction at the support S will become zero when $\beta$ is equal to

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📖 Explanation

Adopting moment distribution method: Thus, fnal vertical reaction at S (both due $f_0$ sinking and P)

\begin{aligned} R_S&=P-V \\ &= P-\frac{7.5EI}{L^3} \times \frac{PL^3}{\beta EI}\\ R_S&=\frac{P-7.5T}{\beta}\;\;\text{when}\; \beta =\text{ reaction at S = 0} \end{aligned}

Explanation Figure 2

Q13GATE 2019NAT

A portal frame shown in figure has a hinge support at joint P and a roller support at joint R. A point load of 50 kN is acting at joint R in the horizontal direction. The flexural rigidity, EI, of each member is 10 $kNm^2$ . Under the applied load, the horizontal displacement (in mm, round off to 1 decimal place) of joint R would be _______

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📖 Explanation

\begin{aligned} \Sigma M_p&=0\;\;[\text{Take clockwise as positive}]\\ R_R \times 5-50 \times 10&=0\\ R_R&=-100kN(\downarrow )\\ R_P&=100kN(\uparrow )\\ EI&=10^6kN-m^2\\ \end{aligned}

Using unit load method: The horizontal displacement at 'R' $\delta _{HR}=\int_{0}^{L}\frac{M_xm_x}{EI}dx$ where $M_x=$ BM at X-X due to real loads $m_x=$ BM at X-X due to vertical unit load applied where we want to find the deflection.

\begin{aligned} \Sigma M_R &=0\;\;[\text{Take clockwise as positive}]\\ R_P \times 5-1 \times 10&=0\\ R_P&=2kN(\uparrow )\end{aligned}

Sign conventions: Sagging +ve Hogging -ve $\delta _{HR}=\int_{0}^{5}\frac{200x^2}{10^6}dx+\int_{0}^{10}\frac{50x^2}{10^6}dx=\frac{1}{40}=0.025m=25mm$

Explanation Figure 2

Explanation Figure 3

Explanation Figure 4

Q14GATE 2018MCQ

The figure shows a simply supported beam PQ of uniform flexural rigidity EI carrying two moments M and 2M. The slope at P will be

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📖 Explanation

\begin{array}{c} \text { Now } \quad R_{1}+R_{2}=\frac{1}{2} \times \frac{L}{3} \times \frac{M}{E I}=\frac{M L}{6 E I} \\ \Sigma M_{O}=0 \\ R_{1} L-\frac{1}{2} \times \frac{L}{3} \times \frac{M}{E I}\left(\frac{2 L}{3}+\frac{L}{9}\right)-\frac{1}{2} \times \frac{L}{3} \\ \times \frac{M}{E I}\left(\frac{L}{3}+\frac{L}{9}\right)+\frac{1}{2} \times \frac{L}{3} \times \frac{M}{E I} \times \frac{2 L}{9}=0\\ \quad R_{1}=\frac{7 M L}{54 E I}+\frac{4 M L}{54 E I}-\frac{M L}{27 E I}=\frac{M L}{6 E I} \\ \therefore \text{Slope at }P, \theta_{P}=R_{1}=\frac{M L}{6 E I} \end{array}

Explanation Figure 2

Q15GATE 2018MCQ

A vertical load of 10 kN acts on a hinge located at a distance of L/4 from the roller support Q of a beam of length L (see figure). The vertical reaction at support Q is

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📖 Explanation

Bending moment about hinge point A=0 (consider the right hand side of A )

\begin{aligned} R_{0} \times \frac{L}{4} &=0 \\ R_{0} &=0 \mathrm{kN} \end{aligned}

Explanation Figure 2

Q16GATE 2018NAT

A prismatic beam P-Q-R of flexural rigidity $EI=1\times 10^{4}kNm^{2}$ is subjected to a moment of 180 kNm at Q as shown in the figure. The rotation at Q (in rad, up to two decimal places) is ______

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📖 Explanation

\begin{aligned} K_{Q} &=K_{Q P}+K_{Q R} \\ &=\frac{4 E I}{5}+\frac{4 E I}{4}=1.8 \mathrm{EI} \\ \theta_{Q} &=\frac{M_{Q}}{K_{Q}}=\frac{180}{1.8 \times 10^{4}}=0.01 \mathrm{rad} \end{aligned}

Alternate solution

\begin{aligned} &\begin{aligned} M_{Q P} &=M_{f Q P}+\frac{2 E I}{l}\left[2 \theta_{Q}+\theta_{P}\right] \\ &=\frac{4 \times 10^{4}}{5} \times \theta_{Q}=8000 \theta_{Q} \\ M_{Q R}=& M_{f Q R}+\frac{2 E I}{l}\left[2 \theta_{Q}+\theta_{R}\right] \\ & \quad\left\{M_{f Q R}=0, \theta_{R}=0\right\} \\ &=\frac{2 \times 10^{4}}{4}\left[2 \theta_{Q}\right]=10000 Q \\ \Sigma M_{Q} &=0 \end{aligned} \\ &\Rightarrow \quad M_{Q P}+M_{Q R}+180=0 \\ & \Rightarrow \quad 18000 \cdot \theta_{Q}=-180 \\ & \Rightarrow \quad \theta_{Q}=0.01 \;\text{radian (anticlockwise)} \end{aligned}

Explanation Figure 2

Q17GATE 2017NAT

The value of M in the beam ABC shown in the figure is such that the joint B does not rotate. The value of support reaction (in kN) at B should be equal to ______

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📖 Explanation

As the slope $\theta_{B}$ is an anticlockwise sense, hence the moment at joint B must be applied in clockwise sense $=40 \mathrm{kNm}$ to make $\theta_{B}=0$

Explanation Figure 2

Explanation Figure 3

Q18GATE 2017NAT

Consider the portal frame shown in the figure and assume the modulus of elasticity $E=2.5\times 10^{4}$ MPa and the moment of inertial, $I=8\times 10^{8}\; mm^{4}$ for all the members of the frame. The rotations (in degrees, up to decimal place) at the rigid joint Q would be ___

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📖 Explanation

\begin{aligned} \text { Moment at } Q &=\frac{1650 \times 2^{2}}{2}-2000 \times 2 \\ &=-700 \mathrm{kNm}(\mathrm{ACW}) \\ &=2.5 \times 10^{4} \times 8 \times 10^{8} \\ &=20 \times 10^{12} \mathrm{Nmm}^{2} \\ &=20000 \mathrm{kNm}^{2} \\ \theta_{Q} &=\frac{M l}{4 E I}=\frac{350 \times 4}{4 \times 20000} \\ &=0.0175 \mathrm{rad} \\ &=\frac{0.0175 \times 180^{\circ}}{\pi} \\ &=1.0027^{\circ} \end{aligned}

Explanation Figure 2

Q19GATE 2015NAT

For the beam shown below, the value of the support moment M is __________ kN-m.

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📖 Explanation

By symmetry we can consider two parts of beam separately with half the load Moment at joint $B=10 \mathrm{kNm}$ Carry over moment at joint A due to 10 kNm moment at propped end $=\frac{10}{2}=5 \mathrm{kN}-\mathrm{m}$

Explanation Figure 2

Q20GATE 2014MCQ

Considering the symmetry of a rigid frame as shown below, the magnitude of the bending moment (in kNm) at P (preferably using the moment distributing method) is

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📖 Explanation

Distribution Factor Fixed End Moment

\begin{array}{l} \overline{M_{A B}}=\overline{M_{B A}}=\overline{M_{P E}}=\overline{M_{E P}}=\overline{M_{\infty }}=\overline{M_{D C}}=0 \\ \overline{M_{B P}}=-\frac{24 \times (8)^{2}}{12}=-128 \mathrm{kNm} \\ \overline{M_{P B}}=+128 \mathrm{kNm} \\ \overline{M_{P C}}=-128 \mathrm{kNm} \end{array}

Distribution Table

Explanation Figure 2

Explanation Figure 3

Explanation Figure 4

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