Question 1 |

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 1 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 2 |

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 2 Explanation:

(-PL) + (PL) + (-M_A) = 0

M_A = 0

For most economical,

Maximum cross-section is given where maximum bending moment occurs.

Question 3 |

A rigid weightless platform PQRS shown in the figure (not drawn to the scale) can slide
freely in the vertical direction. The platform is held in position by the weightless member
OJ and four weightless, frictionless rollers. Point O and J are pin connections. A block
of 90 kN rests on the platform as shown in the figure.

The magnitude of horizontal component of the reaction (in kN) at pin O, is

The magnitude of horizontal component of the reaction (in kN) at pin O, is

90 | |

120 | |

150 | |

180 |

Question 3 Explanation:

\begin{aligned} \Sigma y&=0\\ \Rightarrow \; & R_o \sin 36.87-90 =0\\ R_o&=\frac{90}{\sin 3687^{\circ}}=150kN\\ &\text{Horizontal reaction at O}\\ &=H_o\\ &=R_o \cos 36.87\\ &=150 \times \cos 36.87\\ &=120 kN \end{aligned}

Question 4 |

Consider the three prismatic beamwith the clamped supports P, Q and R as shown in the figures

Given that modulus of Elasticity, E is 2.5\times 10^{4}\, \, MPa; and the moment of inertia, I is 8\times 10^{8}\, \, mm^{4}, the correct Comparison of the magnitudes of the shear force S and the bending moment M developed at the supports is

Given that modulus of Elasticity, E is 2.5\times 10^{4}\, \, MPa; and the moment of inertia, I is 8\times 10^{8}\, \, mm^{4}, the correct Comparison of the magnitudes of the shear force S and the bending moment M developed at the supports is

S_{P} \lt S_{Q} \lt S_{R};\; M_{P}=M_{Q}=M_{R} | |

S_{P}= S_{Q} \gt S_{R};\; M_{P}=M_{Q} \gt M_{R} | |

S_{P} \lt S_{Q} \gt S_{R};\; M_{P}=M_{Q}=M_{R} | |

S_{P} \lt S_{Q} \lt S_{R};\; M_{P} \lt M_{Q} \lt M_{R} |

Question 4 Explanation:

\begin{array}{l} S_{P} \lt S_{Q} \gt S_{R} \\ M_{P}=M_{Q}=M_{R} \end{array}

Question 5 |

A simply supported beam is subjected to uniformly distributed load. Which one of the following statements is true?

Maximum or minimum shear force occurs where the curvature is zero. | |

Maximum or minimum bending moment occurs where the shear force is zero. | |

Maximum or minimum bending moment occurs where the curvature is zero. | |

Maximum bending moment and maximum shear force occur at the same section. |

Question 6 |

The portal frame shown in the figure is subjected to a uniformly distributed vertical load w (per unit length).

The bending moment in the beam at the joint 'Q' is

The bending moment in the beam at the joint 'Q' is

zero | |

\frac{wL^{2}}{24}(hogging) | |

\frac{wL^{2}}{12}(hogging) | |

\frac{wL^{2}}{8}(sagging) |

Question 6 Explanation:

As there is no horizontal force,

\begin{aligned} \text { Hence } \quad H_{P}=H_{S}&=0 \\ \Sigma M_{Q}=H_{P} \times \frac{L}{2}&=0 \\ \therefore \quad B M \text { at } Q&=0 \end{aligned}

Question 7 |

A rigid member ACB is shown in the figure. The member is supported at A and B by pinned and guided roller supports, respectively. A force P acts at C as shown. Let R_{Ah}\: and\: R_{Bh} be the horizontal reactions at supports A and B, respectively, and R_{Av} be the vertical reaction at support A. Self-weight of the member may be ignored.

Which one of the following sets gives the correct magnitudes of R_{Av},\: R_{Bh}\, and\, R_{Ah} ?

Which one of the following sets gives the correct magnitudes of R_{Av},\: R_{Bh}\, and\, R_{Ah} ?

R_{Av}=0;\: R_{Bh}=\frac{1}{3}P;\:and\, R_{Ah}=\frac{2}{3}P | |

R_{Av}=0;\: R_{Bh}=\frac{2}{3}P;\:and\, R_{Ah}=\frac{1}{3}P | |

R_{Av}=P;\: R_{Bh}=\frac{3}{8}P;\:and\, R_{Ah}=\frac{1.5}{8}P | |

R_{Av}=P;\: R_{Bh}=\frac{1.5}{8}P;\:and\, R_{Ah}=\frac{1.5}{8}P |

Question 7 Explanation:

\begin{aligned} \Sigma F_{y} &=0 \\ \Rightarrow\quad R_{a v} &=P \\ \Sigma F_{x} &=0 \\ \Rightarrow\quad R_{B h} &=R_{A h} \\ \Sigma M_{A} &=0 \\ R_{B h} \times 8-P \times 1.5 &=0 \\ \Rightarrow\quad R_{B h} &=\frac{1.5 P}{8} \\ \Rightarrow\quad R_{A h} &=\frac{1.5 P}{8} \\ R_{A V} &=P \\ R_{B h} &=\frac{1.5 P}{8} \\ R_{A h} &=\frac{1.5 P}{8} \end{aligned}

Question 8 |

A horizontal beam ABC is loaded as shown in the figure below. The distance of the point of contraflexure from end A (in m) is _________.

1.25 | |

0.25 | |

0.5 | |

0.75 |

Question 8 Explanation:

Reaction at B

\begin{aligned} \Delta_{\mathrm{B}}&=0 \text { (Compatibility condition) }\\ \frac{10 \times(0.75)^{3}}{3 E I}&+\frac{2.5 \times 0.75^{2}}{2 E I}-\frac{R_{B} \times 0.75^{3}}{3 E I}=0 \\ \frac{9 R_{B}}{64 E I}&=\frac{135}{64 E I}\\ \therefore \quad \mathrm{R}_{\mathrm{B}}&=15 \mathrm{kN} \end{aligned}

BM at a distance x from free end

Reaction at B

\begin{aligned} & & \mathrm{BM}_{x} &=10 \times x-15 \times(x-0.25)=0 \\ \Rightarrow & & 10 x &=15 x-3.75 \\ \Rightarrow & & 5 x &=3.75 \\ \therefore & & x &=0.75 \mathrm{m} \end{aligned}

\therefore From end A, distance is 0.25m

\begin{aligned} \Delta_{\mathrm{B}}&=0 \text { (Compatibility condition) }\\ \frac{10 \times(0.75)^{3}}{3 E I}&+\frac{2.5 \times 0.75^{2}}{2 E I}-\frac{R_{B} \times 0.75^{3}}{3 E I}=0 \\ \frac{9 R_{B}}{64 E I}&=\frac{135}{64 E I}\\ \therefore \quad \mathrm{R}_{\mathrm{B}}&=15 \mathrm{kN} \end{aligned}

BM at a distance x from free end

Reaction at B

\begin{aligned} & & \mathrm{BM}_{x} &=10 \times x-15 \times(x-0.25)=0 \\ \Rightarrow & & 10 x &=15 x-3.75 \\ \Rightarrow & & 5 x &=3.75 \\ \therefore & & x &=0.75 \mathrm{m} \end{aligned}

\therefore From end A, distance is 0.25m

Question 9 |

The following statements are related to bending of beams:

I The slope of the bending moment diagram is equal to the shear force.

II The slope of the shear force diagram is equal to the load intensity.

III The slope of the curvature is equal to the flexural rotation.

IV The second derivative of the deflection is equal to the curvature.

The only FALSE statement is

I The slope of the bending moment diagram is equal to the shear force.

II The slope of the shear force diagram is equal to the load intensity.

III The slope of the curvature is equal to the flexural rotation.

IV The second derivative of the deflection is equal to the curvature.

The only FALSE statement is

I | |

II | |

III | |

IV |

Question 9 Explanation:

We know that

\begin{aligned} \frac{d S}{d x} &=W ; \frac{d M}{d X}=S_{x} \\ E 1 \cdot \frac{d^{2} y}{d x^{2}} &=M \\ \therefore\quad \frac{d^{2} y}{d x^{2}} &=\frac{M}{E I} \\ \text{Also}\quad \frac{M}{I} &=\frac{f}{y}=\frac{E}{R} \quad \therefore \frac{M}{E I}=\frac{1}{R} \\ \therefore\quad \frac{d^{2} y}{d x^{2}} &=\frac{1}{R} \end{aligned}

\begin{aligned} \frac{d S}{d x} &=W ; \frac{d M}{d X}=S_{x} \\ E 1 \cdot \frac{d^{2} y}{d x^{2}} &=M \\ \therefore\quad \frac{d^{2} y}{d x^{2}} &=\frac{M}{E I} \\ \text{Also}\quad \frac{M}{I} &=\frac{f}{y}=\frac{E}{R} \quad \therefore \frac{M}{E I}=\frac{1}{R} \\ \therefore\quad \frac{d^{2} y}{d x^{2}} &=\frac{1}{R} \end{aligned}

Question 10 |

For the cantilever bracket, PQRS, loaded as shown in the below figure (
PQ = RS = L, and, QR = 2L), which of the following statements is FALSE?

The portion RS has a constant twisting moment with a value of 2WL | |

The portion QR has a varying twisting moment with a maximum value of WL. | |

The portion PQ has a varying bending moment with a maximum value of WL. | |

The portion PQ has no twisting moment. |

Question 10 Explanation:

Bending moment at a distance x from P, M_{x} = Wx

Member PQ has varying BM with a maximum value of WL at Q and it has no twisting moment.

Now moment WL will act as a twisting moment for the member QR. Portion QR has varying BM with a maximum value of 2WL at R.

Now moment 2WL will act as a twisting moment for portion RS which is constant in nature.

There are 10 questions to complete.