GATE CE · Solid Mechanics

Shear Force And Bending Moment Practice Questions

Generate GATE-level questions on Shear Force And Bending Moment in Solid Mechanics. Focus on core concepts, previous year patterns, and numerical problem-solving techniques.

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

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Q1GATE 2025MSQ

Consider the beam $A C D E B$ given in the figure. Which of the following statements is/ are correct:

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📖 Explanation

Statement 1: Bending moment is zero between the points A and C This is incorrect. Between A and C, there is a fixed support at A and a point load at C. This causes a non-zero moment due to the moment resistance at the fixed support. $\text{Bending moment} \ne 0 \text{ between A and C}$ Statement 2: There is a sudden jump in shear force at the point D This is correct. At point D, there is a point load of 4 kN acting vertically downward. A point load causes a sudden change in shear force. $\Delta V = -4 , \text{kN at D}$ Statement 3: There is a sudden jump in bending moment at the point E This is incorrect. Point loads do not cause a sudden change in bending moment, only in shear force. Bending moment is continuous at point loads (like at E) Statement 4: Bending moment is zero somewhere between the points D and E This is correct. Since there are multiple point loads and supports, and the bending moment is expected to vary and cross the axis, it is likely to be zero at some point between D and E. There exists a point between D and E where M = 0

Q5GATE 2020MCQ

Joints I, J, K, L, Q and M of the frame shown in the figure (not drawn to the scale) are pins. Continuous members IQ and IJ are connected through a pin at N. Continuous members JM and KQ are connected through a pin at P. The frame has hinge supports at joints R and S. The loads acting at joints I, J and K are along the negative Y direction and the loads acting at joints I, M are along the positive X direction. The magnitude of the horizontal component of reaction (in kN) at S, is

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📖 Explanation

Remove hinge at support S and replace it with roller support as shown in the figure. Step One : Find coordinates of all the points where forces are acting

\begin{aligned} y_I&=\sqrt{2} \sin \theta \\ y_J&=\sqrt{2} \sin \theta \\ y_K&=\sqrt{2} \sin \theta \\ x_L&=\sqrt{2} \cos \theta \\ x_M&=5\sqrt{2} \cos \theta \\ x_S&=6\sqrt{2} \cos \theta \end{aligned}

Step Two: Find virtual displacements of all the points

\begin{aligned} \delta y_I&=\sqrt{2} \cos \theta d\theta \\ \delta y_J&=\sqrt{2} \cos \theta d\theta \\ \delta y_K&=\sqrt{2} \cos \theta d\theta \\ \delta x_L&=-\sqrt{2} \sin \theta d\theta \\ \delta x_M&=-5\sqrt{2} \sin \theta d\theta \\ \delta x_S&=-6\sqrt{2} \sin \theta d\theta \\ \end{aligned}

Step Three : Use principle of virtual work to find unknown horizontal force $H_s$ $\Rightarrow \; \delta U=0$ $=[-10 \times \sqrt{2} \cos \theta d\theta ] \times 3 + [-10 \times -\sqrt{2} \cos \theta d\theta ] +[-10 \times -5\sqrt{2} \sin \theta d\theta ] - [-H_S \times -6\sqrt{2} \sin \theta d\theta ]$ $H_S=\frac{30\sqrt{2} \cos \theta +10\sqrt{2} \sin \theta +50\sqrt{2} \sin \theta}{6\sqrt{2} \sin \theta}$ Substituting, $\theta =45^{\circ},\; H_S=\frac{90}{6}=15kN$ Note : Sign conventions If a force acts along positive x or positive y-axis, take it as positive. If a force acts along negative x or negative y-axis, take it as negative.