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).
12 | |
16 | |
18 | |
24 |
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}
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).
197 | |
125.2 | |
362.1 | |
148.2 |
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?
Support reactions are zero | |
Shear force is zero everywhere | |
Bending moment is zero everywhere
| |
Deflection is zero everywhere |
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
The magnitude of the horizontal component of reaction (in kN) at S, is
5 | |
10 | |
15 | |
20 |
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
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 | |
B | |
C | |
D |
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.
if I have mark the question option that would be better for future or say revision