The stiffness coefficient indicates

Stiffness(k) is the force required to produce unit deformation. Thus denotes force required in direction i due to unit deformation (displacement) in direction j.

A. force at i due to a unit deformation at j

B. deformation at j due to a unit force at i

C. deformation at i due to a unit force at j

D. force at j due to a unit deformation at i

GATE CE · Structural Analysis

Matrix Method Of Structural Analysis – GATE CE Structural AnalysisPractice Questions & PYQs

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

6 questions · 6 PYQs · 0 AI practice · GATE CE 2027

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Q1GATE 2021NAT

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 _______

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📖 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}

Q2GATE 2021NAT

Employ stiffness matrix approach for the simply supported beam as shown in the figure to calculate unknown displacements/rotations. Take length, L=8 m; modulus of elasticity, $E=3 \times 10^{4} \mathrm{~N} / \mathrm{mm}^{2}$ ; moment of inertia, $I=225 \times 10^{6} \mathrm{mm}^{4}$ . The mid-span deflection of the beam (in mm,round off to integer) under P=100 kN in downward direction will be ___________

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📖 Explanation

By stiffness matrix method Step-1 : Generation of stiffness matrix Column-1

\begin{aligned} \mathrm{k}_{11}&=\frac{3 E(2 I)}{(L / 2)^{3}}+\frac{3 E(I)}{(L / 2)^{3}}=\frac{72 \mathrm{E} I}{L^{3}}\\ k_{21}&=-\frac{3 E(2 I)}{(L / 2)^{2}}+\frac{3 E(I)}{(L / 2)^{2}}=-\frac{12 E I}{L^{2}} \end{aligned}

Column-2

\begin{aligned} k_{22}&=\frac{3 E(2 I)}{(L / 2)}+\frac{3 E(I)}{(L / 2)}=18 \frac{E I}{L} \\ \text { Stiffness matrix }[k]&=\left[\begin{array}{cc} 72 \frac{E I}{L^{3}} & -\frac{12 E I}{L^{2}} \\ -\frac{12 E I}{L^{2}} & \frac{E I}{L} \end{array}\right]\$ \end{aligned}

Step-2: Calculation of unknown Nodal displacements $\left(\Delta_{\mathrm{B}}, \theta_{\mathrm{B}}\right)$

\begin{aligned} \text{Using} \qquad \qquad \qquad \qquad [P]&=[k][\Delta]\\ \Rightarrow \qquad \qquad \qquad \qquad \qquad \quad & \quad\left[\begin{array}{l} P \\ O \end{array}\right]=\left[\begin{array}{cc} \frac{72 E I}{L^{2}} & -\frac{12 E I}{L^{2}} \\ -\frac{12 E I}{L^{2}} & \frac{8 E I}{L} \end{array}\right]\$\left[\begin{array}{l} \Delta_{B} \\ \theta_{B} \end{array}\right]\\ \text{On solving} \qquad \qquad \qquad \quad\\\Delta_{\mathrm{B}}&=\frac{P L^{3}}{64 E I}(\downarrow) \\ \therefore \qquad \qquad \qquad \qquad \qquad \theta_{\mathrm{B}}&=\frac{P L^{2}}{96 E I}(\mathrm{CW}) \\ \Delta_{\mathrm{B}}&=\frac{\left(100 \times 10^{3}\right) \times (8000)^{3}}{64 \times 3 \times 10^{4} \times 225 \times 10^{6}}\\ &=118.519 \mathrm{~mm} \simeq 119 \mathrm{~mm} \end{aligned}

Q3GATE 2015MCQ

For the beam shown below, the stiffness coefficient $K_{22}$ can be written as

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📖 Explanation

By giving unit displacement in $2^{\text {nd }}$ direction without giving displacement in any other direction, the force developed R_{6}, is, $K_{22}=\frac{12 E I}{L^{3}}$

Q4GATE 2007MCQ

The stiffness coefficient $k_{ij}$ indicates

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📖 Explanation

Stiffness(k) is the force required to produce unit deformation. Thus $k_{ij}$ denotes force required in direction i due to unit deformation (displacement) in direction j.

Q5GATE 2005MCQ

For a linear elastic frame, if stiffness matrix is doubled with respect to the existing stiffness matrix, the deflection of the resulting frame will be

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📖 Explanation

Stiffness matrix and deflection are related as, $P=K \delta$ Thus, when stiffness matrix is doubled, then deflection will reduce to half of the existing value.

Q6GATE 2003MCQ

The stiffness K of a beam deflecting in a symmetric mode, as shown in the figure, is

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📖 Explanation

The beam will deflect in symmetric mode when a constant moment M is applied at both ends. Slope at either ends will be given as $\frac{M L}{2 E I}$ . We know that stiffness is the moment required to produce unit slope.

\begin{aligned} \therefore \quad \theta&=\frac{M L}{2 E I} \\ \Rightarrow \quad M&=\frac{2 E I}{L} \theta &[\because \theta=1]\\ \Rightarrow \quad M&=\frac{2 E I}{L}=K \end{aligned}

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Q2GATE 2021NAT

🚀 Solve in Practice Mode

📖 Explanation

By stiffness matrix method Step-1 : Generation of stiffness matrix Column-1

\begin{aligned} \mathrm{k}_{11}&=\frac{3 E(2 I)}{(L / 2)^{3}}+\frac{3 E(I)}{(L / 2)^{3}}=\frac{72 \mathrm{E} I}{L^{3}}\\ k_{21}&=-\frac{3 E(2 I)}{(L / 2)^{2}}+\frac{3 E(I)}{(L / 2)^{2}}=-\frac{12 E I}{L^{2}} \end{aligned}

Column-2

\begin{aligned} k_{22}&=\frac{3 E(2 I)}{(L / 2)}+\frac{3 E(I)}{(L / 2)}=18 \frac{E I}{L} \\ \text { Stiffness matrix }[k]&=\left[\begin{array}{cc} 72 \frac{E I}{L^{3}} & -\frac{12 E I}{L^{2}} \\ -\frac{12 E I}{L^{2}} & \frac{E I}{L} \end{array}\right]\$ \end{aligned}

Step-2: Calculation of unknown Nodal displacements $\left(\Delta_{\mathrm{B}}, \theta_{\mathrm{B}}\right)$

\begin{aligned} \text{Using} \qquad \qquad \qquad \qquad [P]&=[k][\Delta]\\ \Rightarrow \qquad \qquad \qquad \qquad \qquad \quad & \quad\left[\begin{array}{l} P \\ O \end{array}\right]=\left[\begin{array}{cc} \frac{72 E I}{L^{2}} & -\frac{12 E I}{L^{2}} \\ -\frac{12 E I}{L^{2}} & \frac{8 E I}{L} \end{array}\right]\$\left[\begin{array}{l} \Delta_{B} \\ \theta_{B} \end{array}\right]\\ \text{On solving} \qquad \qquad \qquad \quad\\\Delta_{\mathrm{B}}&=\frac{P L^{3}}{64 E I}(\downarrow) \\ \therefore \qquad \qquad \qquad \qquad \qquad \theta_{\mathrm{B}}&=\frac{P L^{2}}{96 E I}(\mathrm{CW}) \\ \Delta_{\mathrm{B}}&=\frac{\left(100 \times 10^{3}\right) \times (8000)^{3}}{64 \times 3 \times 10^{4} \times 225 \times 10^{6}}\\ &=118.519 \mathrm{~mm} \simeq 119 \mathrm{~mm} \end{aligned}