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