# Shear Force and Bending Moment

 Question 1
An idealized frame supports a load as shown in the figure. The horizontal component of the force transferred from the horizontal member PQ to the vertical member $\mathrm{RS}$ at $\mathrm{P}$ is $\mathrm{N}$ (round off to one decimal place). A 12 B 16 C 18 D 24
GATE CE 2023 SET-2   Solid Mechanics
Question 1 Explanation:
As member UT is a link member, it will carry axial load only (assuming as R).
FBD of $\mathrm{PQ}^{\prime}$ $\theta=\tan ^{-1} \left(\frac{1}{2}\right)$
$\theta=39.805 ^{\circ}$
$\sum M_{P}=0$
$\Rightarrow-\mathrm{R} \sin \theta * 1.2+10 * 1.8=0$
$\Rightarrow R \sin \left(39.805^{\circ}\right)=\frac{10 * 1.8}{1.2}$
$\Rightarrow \quad \mathrm{R}=23.43 \mathrm{~N}$

Now,
\begin{aligned} & \sum F_{H}=0 \\ & \Rightarrow \mathrm{H}_{\mathrm{P}}-\mathrm{R} \cos \left(39.805^{\circ}\right)=0 \\ & \Rightarrow \quad \mathrm{H}_{\mathrm{P}}=23.43 * \cos \left(39.805^{\circ}\right) \\ & =17.988 \mathrm{~N} \end{aligned}
 Question 2
A beam is subjected to a system of coplanar forces as shown in the figure. The magnitude of vertical reaction at Support $\mathrm{P}$ is $\mathrm{N}$ (round off to one decimal place). A 197 B 125.2 C 362.1 D 148.2
GATE CE 2023 SET-2   Solid Mechanics
Question 2 Explanation: Taking components of inclined $500 \mathrm{~N}$ load along horizontal and vertical direction. Now, using equilibrium equations.
$\Rightarrow \quad \mathrm{H}_{\mathrm{P}}=350 \mathrm{~N}$
$\left.\sum \mathrm{M}_{\mathrm{Q}}\right)=0$ $\Rightarrow R_{P} * 6-\left(500 \sin 60^{\circ}\right) * 4+200 * 2.5+100 * 0.5=0$
$\Rightarrow \quad \mathrm{R}_{\mathrm{P}}=197 \mathrm{~N}$

 Question 3
Consider the beam shown in the figure (not to scale), on a hinge support at end $A$ and a roller support at end $B$. The beam has a constant flexural rigidity, and is subjected to the external moments of magnitude $M$ at one-third spans, as shown in the figure. Which of the following statements is/ are TRUE? A Support reactions are zero B Shear force is zero everywhere C Bending moment is zero everywhere D Deflection is zero everywhere
GATE CE 2023 SET-1   Solid Mechanics
Question 3 Explanation: Using equilibrium equations
$\left.\sum \mathrm{M}_{\mathrm{A}}\right)=0$
$\Rightarrow \quad-\mathrm{M}+\mathrm{M}-\mathrm{R}_{\mathrm{B}} \times(3 \mathrm{~L})=0$
$\Rightarrow \mathrm{R}_{\mathrm{B}}=0$
$\sum F_{V}=0 \Rightarrow R_{A}+R_{B}=0$
$\Rightarrow \mathrm{R}_{\mathrm{A}}=0$

SFD $\Rightarrow \quad$ No shear force throughout the span.

BMD As there is $B M$ in span $C D$, which leads to curvature in CD. i.e. deflection is not zero everywhere.
 Question 4
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
 A 5 B 10 C 15 D 20
GATE CE 2020 SET-2   Solid Mechanics
Question 4 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.
 Question 5
A weightless cantilever beam of span L is loaded as shown in the figure. For the entire span of the beam, the material properties are identical and the cross-section is rectangular with constant width. From the flexure-critical perspective, the most economical longitudinal profile of the beam to carry the given loads amongst the options given below, is A A B B C C D D
GATE CE 2020 SET-2   Solid Mechanics
Question 5 Explanation: $(-PL) + (PL) + (-M_A) = 0$
$M_A = 0$ For most economical,
Maximum cross-section is given where maximum bending moment occurs. There are 5 questions to complete.

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