GATE CE · Solid Mechanics

Theory Of Columns And Shear Centre – GATE CE Solid MechanicsPractice Questions & PYQs

Generate GATE-level questions on Theory Of Columns And Shear Centre in Solid Mechanics. Focus on core concepts, previous year patterns, and numerical problem-solving techniques.

15 questions · 15 PYQs · 0 AI practice · GATE CE 2027

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Q1GATE 2024MCQ

For a thin-walled section shown in the figure, points $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$ are located on the major bending axis $\mathrm{X}-\mathrm{X}$ of the section. Point $\mathrm{Q}$ is located on the web whereas point $S$ is located at the intersection of the web and the top flange of the section. Qualitatively, the shear center of the section lies at

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Q2GATE 2022MCQ

Consider the cross-section of a beam made up of thin uniform elements having thickness $t(t \lt \lt a)$ shown in the figure. The $(x,y)$ coordinates of the points along the center-line of the cross-section are given in the figure. The coordinates of the shear center of this cross-section are:

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Q3GATE 2017NAT

Consider two axially loaded columns, namely, 1 and 2, made of a linear elastic material with Young's modulus $2 \times 10^{5}$ MPa, square cross- section with side 10 mm, and length 1 m. For Column 1, one end is fixed and the other end is free. For Column 2, one end is fixed and the other end is pinned. Based on the Euler's theory, the ratio (up to one decimal place) of the buckling load of Column 2 to the buckling load of Column 1 is ________

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📖 Explanation

\begin{aligned} P_{c r 1}&=\frac{\pi^{2} E \cdot I}{(2 l)^{2}}=\frac{\pi^{2} E I}{4 l^{2}} \\ P_{c r 2}&=\frac{\pi^{2} E \cdot I}{\left(\frac{l}{\sqrt{2}}\right)^{2}}=\frac{2 \pi^{2} E I}{l^{2}}\\ \therefore \text{Required ratio }&=\frac{P_{C r 2}}{P_{c r 1}}=\frac{\frac{2 \pi^{2} E I}{l^{2}}}{\frac{\pi^{2} E I}{4 l^{2}}}=8 \end{aligned}

Explanation Figure 1

Q4GATE 2015MCQ

In a system,two connected rigid bars AC and BC are of identical length, L with pin supports at A and B. The bars are interconnected at C by a frictionless hinge. The rotation of the hinge is restrained by a rotational spring of stiffness, k. The system initially assumes a straight line configuration, ACB. Assuming both the bars as weightless, the rotation at supports, A and B, due to a transverse load, P applied at C is

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📖 Explanation

Deflection under load $P=\theta L$ Work done by force $P=\frac{1}{2} P(\theta L)$ Strain energy stored in spring

\begin{aligned} &=\frac{1}{2}[k(2 \theta)](2 \theta) \\ \frac{1}{2} P(\theta L) &=\frac{4 k \theta^{2}}{2} \\ \Rightarrow \quad \theta &=\frac{P L}{4 k} \end{aligned}

Explanation Figure 1

Q5GATE 2014MCQ

The possible location of shear centre of the channel section, shown below, is

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📖 Explanation

For no twisting,

\begin{aligned} V \times e &=H \times h\\ \Rightarrow \;\; e&=\frac{Hh}{V}\\ \end{aligned}

Hence, possible location of shear center is P.

Explanation Figure 2

Q6GATE 2013MCQ

Two steel columns P (length L and yield strength $f_{y}$ = 250 MPa) and Q (length 2L and yield strength $f_{y}$ = 500 MPa) have the same cross-sections and end-conditions. The ratio of buckling load of column P to that of column Q is:

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📖 Explanation

Buckling load,

\begin{aligned} P &=\frac{\pi^{2} E \mid}{\left(l_{e f f}\right)^{2}} \\ \therefore \quad P_{p} &=\frac{\pi^{2} E \mid}{(L)^{2}} ; P_{Q}=\frac{\pi^{2} E \mid}{(2 L)^{2}} \\ \therefore \quad \frac{P_{P}}{P_{a}} &=4 \end{aligned}

Q7GATE 2012MCQ

The ratio of the theoretical critical buckling load for a column with fixed ends to that of another column with the same dimensions and material, but with pinned ends, is equal to

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📖 Explanation

Eulars critical load,

\begin{aligned} P_{e}&=\frac{\pi^{2}EI}{I^{2}_{\text{eft}}}\\ \frac{P_{e1}}{P_{e2}}=\left[\frac{(I_{\text{elf}})_{2}}{(I_{\text{elf}})_{1}} \right]^{2}&=\left[\frac{I}{1/2}\right]^{2}=4 \end{aligned}

Explanation Figure 1

Q8GATE 2012MCQ

The sketch shows a column with a pin at the base and rollers at the top. It is subjected to an axial force P and a moment M at mid-height. The reaction(s) at R is/are

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📖 Explanation

\begin{aligned} \Sigma F_y &=0 \\ \Rightarrow \;\; V_R-P&=0\\ \Rightarrow \;\;V_R&=P\\ \Sigma M_Q&=0\\ \Rightarrow \;\; H_Rh-M&=0\\ \Rightarrow \;\;H_R&=\frac{M}{h} \end{aligned}

Explanation Figure 2

Q9GATE 2010MCQ

The effective length of a column of length L fixed against rotation and translation at one end and free at the other end is

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Q10GATE 2009MCQ

The point within the cross sectional plane of a beam through which the resultant of the external loading on the beam has to pass through to ensure pure bending without twisting of the cross-section of the beam is called

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Q11GATE 2008MCQ

A rigid bar GH of length L is supported by a hinge and a spring of stiffness K as shown in the figure below. The buckling load, $P_{Cr}$ , for the bar will be

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📖 Explanation

Let the deflection in the spring be $\delta$ and force in the spring be F. Taking moments about G, we get

\begin{aligned} P_{c r} \times \delta&=F \times L\\ \Rightarrow \quad P_{c r}&=\frac{K \delta \times L}{\delta} &\left[\because F=K\delta \right]\\ \Rightarrow \quad P_{c r}&=K L \end{aligned}

Explanation Figure 2

Q12GATE 2008MCQ

Cross-section of a column consisting to two steel strips, each of thickness t and width b is shown in the figure below. The critical loads of the column with perfect bond and without bond between the strips are $P$ and $P_0$ respectively. The ratio $P/P_0$ is

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📖 Explanation

We know that critical load for a column is proportional to moment of inertia irrespective of end conditions of the column i.e. $P_{c r} \propto 1$ When the steel strips are perfectly bonded, then $f_{p b}=\frac{b \times (2 t)^{3}}{12}=\frac{8 b t^{3}}{12}$ When the steel strips are not bonded, then

\begin{aligned} I_{wb}&=2 \times \frac{b t^{3}}{12}=\frac{2 b t^{3}}{12} \\ \therefore \quad \frac{P}{P_{0}}&=\frac{8 b t^{3} / 12}{2 b t^{3} / 12}=4 \end{aligned}

Q13GATE 2007MCQ

A steel column, pinned at both end, has a buckling load of 200 kN. If the column is restrained against lateral movement at its mid-height, it buckling load will be

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📖 Explanation

When both ends are hinged, the buckling load is given by, $P_{c r}=\frac{\pi^{2} E I}{L^{2}} \Rightarrow 200=\frac{\pi^{2} E l}{L^{2}}$ When the lateral movement at the mid-height is not available, then buckling load

\begin{aligned} &=\frac{\pi^{2} E \mid}{\left(L_{1}\right)^{2}} \quad \text { where } L_{1}=\frac{L}{2} \\ &=\frac{4 \pi^{2} E \mid}{L^{2}}=4 \times 200=800 \mathrm{kN} \end{aligned}

Q14GATE 2006MCQ

The buckling load $P=P_{cr}$ for the column AB shown in figure, as $K_T$ approaches infinity, becomes $\alpha \frac{\pi ^{2}EI}{L^{2}}$ , where $\alpha$ is equal to

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Q15GATE 2003MCQ

A long structural column (length= L) with both ends hinged is acted upon by an axial compressive load, P. The differential equation governing the bending of column is given by : $EI\frac{d^{2}y}{dx^{2}}=-Py$ Where y is the structural lateral deflection and EI is the flexural rigidity. The first critical load on column responsible for its buckling is given by

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📖 Explanation

Differential equation governing the bending of column

\begin{aligned} \frac{E I d^{2} y}{d x^{2}}&=-P y\\ \Rightarrow \frac{d^{2} y}{d x^{2}}+\frac{P}{E I} y&=0 \end{aligned}

Solution of differential equation

\begin{aligned} y&=A\cos\left(x \sqrt{\frac{P}{E I}}\right)+B \sin \left(x \sqrt{\frac{P}{E I}}\right)\\ \text{when }x=0 . y&=0 \\ \Rightarrow \quad 0&=B \sin \left(L \sqrt{\frac{P}{E I}}\right) \\ \therefore \quad L \sqrt{\frac{P}{E I}}&=x\text{ (First value)}\\ \Rightarrow P&=\frac{\pi^{2} E I}{L^{2}} \end{aligned}

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