Question 1 |

Consider a simply supported beam PQ as shown in the figure. A truck having
100 kN on the front axle and 200 kN on the rear axle, moves from left to right.
The spacing between the axles is 3 m. The maximum bending moment at point
R is ______ kNm. (in integer)

124 | |

180 | |

147 | |

582 |

Question 1 Explanation:

\frac{ab}{L}=\frac{1 \times 4}{5}=0.8m

To get maximum BN at R

\begin{aligned} BM_{max}&=200 \times \frac{ab}{L}+100 \times y\\ \frac{ab/l}{b}&=\frac{y}{4-3}\Rightarrow \frac{0.8}{4}=\frac{y}{1}=0.2m\\ BM_{max}&=200 \times 0.8+100 \times 0.2=180kNm \end{aligned}

Question 2 |

A propped cantilever beam EF is subjected to a unit moving load as shown in the figure (not to scale). The sign convention for positive shear force at the left and right sides of any section is also shown.

The CORRECT qualitative nature of the influence line diagram for shear force at G is

The CORRECT qualitative nature of the influence line diagram for shear force at G is

Question 2 Explanation:

As per Muller Breslau principle ILD for stress function (shear -V_{G}) will be a combination of curves (3^{\circ} curves).

Question 3 |

Distributed load(s) of 50 kN/m may occupy any position(s) (either continuously or in
patches) on the girder PQRST as shown in the figure

The maximum negative (hogging) bending moment (in kNm) that occurs at point R, is

The maximum negative (hogging) bending moment (in kNm) that occurs at point R, is

22.5 | |

56.25 | |

93.75 | |

150 |

Question 3 Explanation:

ILD for BM at R:

To get maximum hogging BM at R, keep UDL over PQ and ST.

Max. -ve BM at R =50\left [ -\frac{1}{2} \times 1.5 \times 0.6 \right ] +50 \left [ -\frac{1}{2} \times 1.5 \times 0.9 \right ]

=56.25 kNm

Question 4 |

A long uniformly distributed load of 10 kN/m and a concentrated load of 60 kN are moving together on the beam ABCD shown in the figure. The relative positions of the two loads are not fixed. The maximum shear force (in kN, round off to the nearest integer) caused at the internal hinge B due to the two loads is _____

50 | |

70 | |

90 | |

120 |

Question 4 Explanation:

ILD for V_B

Maximum shear V_B=-\left [ \left ( \frac{1}{2}\times 2 \times 1 \times 10 \right )+(60 \times 1) \right ]=-70kN

Maximum shear V_B=-\left [ \left ( \frac{1}{2}\times 2 \times 1 \times 10 \right )+(60 \times 1) \right ]=-70kN

Question 5 |

Consider the beam ABCD shown in the figure.

For a moving concentrated load of 50 kN on the beam, the magnitude of the maximum bending moment (in kN-m) obtained at the support C will be equal to _____

For a moving concentrated load of 50 kN on the beam, the magnitude of the maximum bending moment (in kN-m) obtained at the support C will be equal to _____

200 | |

155 | |

250 | |

450 |

Question 5 Explanation:

\text { B.M. at } C=50 \times 4=200 \mathrm{kNm}

ILD for BM at C

ILD for BM at C

Question 6 |

A simply supported beam AB of span, L=24 m is subjected to two wheel loads acting at a distance, d=5m apart as shown in the figure below. Each wheel transmits a load, P=3 kN and may occupy any position along the beam. If the beam is an I-section having section modulus, S=16.2 cm^{3}, the maximum bending stress (in GPa) due to the wheel loads is ___________.

1.78 | |

1.24 | |

2.21 | |

0.81 |

Question 6 Explanation:

C.G. of system= 2.5 m from any load
For maximum bending moment, system of load
should be,

Maximum BM will occur below load at C.

Ordinate of ILD at C

\begin{aligned} &=\frac{(12-1.25) \times(12+1.25)}{24} \\ &=5.935 \end{aligned}

Ordinate of LLD at D

\begin{aligned} &=\frac{5.935 \times 8.25}{13.25}=3.695 \\ \text { Maximum } \mathrm{BM} &=5.935 \times 3+3.695 \times 3 \\ &=28.89 \mathrm{kN}-\mathrm{m} \end{aligned}

Maximum bending stress,

\begin{aligned} \frac{M}{Z}&=\frac{28.89 \times 10^{3}}{16.2 \times 10^{-6} \times 10^{9}}\\ &=1.783 \mathrm{GPa} \end{aligned}

Maximum BM will occur below load at C.

Ordinate of ILD at C

\begin{aligned} &=\frac{(12-1.25) \times(12+1.25)}{24} \\ &=5.935 \end{aligned}

Ordinate of LLD at D

\begin{aligned} &=\frac{5.935 \times 8.25}{13.25}=3.695 \\ \text { Maximum } \mathrm{BM} &=5.935 \times 3+3.695 \times 3 \\ &=28.89 \mathrm{kN}-\mathrm{m} \end{aligned}

Maximum bending stress,

\begin{aligned} \frac{M}{Z}&=\frac{28.89 \times 10^{3}}{16.2 \times 10^{-6} \times 10^{9}}\\ &=1.783 \mathrm{GPa} \end{aligned}

Question 7 |

In a beam of length L, four possible influence line diagrams for shear force at a section located at a distance of \frac{L}{4} from the left end support (marked as P, Q, R and S) are shown below. The correct influence line diagram is

P | |

Q | |

R | |

S |

Question 7 Explanation:

ILD for SF at X-X

Question 8 |

Beam PQRS has internal hinges in spans PQ and RS as shown. The beammay be subjected to a moving distributed vertical load of maximum intensity 4 kN/m of any length anywhere on the beam. The maximum absoulate value of the shear force that can occur due to this loading just to the right of support Q shall be:

30 | |

40 | |

45 | |

55 |

Question 8 Explanation:

Drawing the ILD of shear force just to right of Q
by using Muller Breslau's principle. A cut is
made just to the right of 0, since cut is very
close to support Q, therefore displacement of
left portion is almost zero and that to the right
portion will be 1.

When unit load-is at T:

Vertical reaction at P and Q equal to 0

\Sigma M_{0}=0

\begin{aligned} \Rightarrow \quad 1 \times 4&=R_{R} \times 20 \\ \therefore \quad R_{A}&=0.25 \mathrm{kN} \\ V_{O \text { (Righit) }}&=0.25 \mathrm{kN}\\ \end{aligned}

Now, if moving distributed load is present over span P R then we get maximum shear force just to the right of Q .

\Rightarrow \mathrm{SF}=\left[\left(\frac{1}{2} \times 0.25 \times 10\right)+\left(\frac{1}{2} \times 1 \times 20\right)\right] \times 4

\mathrm{SF}=45 \mathrm{kN}

When unit load-is at T:

Vertical reaction at P and Q equal to 0

\Sigma M_{0}=0

\begin{aligned} \Rightarrow \quad 1 \times 4&=R_{R} \times 20 \\ \therefore \quad R_{A}&=0.25 \mathrm{kN} \\ V_{O \text { (Righit) }}&=0.25 \mathrm{kN}\\ \end{aligned}

Now, if moving distributed load is present over span P R then we get maximum shear force just to the right of Q .

\Rightarrow \mathrm{SF}=\left[\left(\frac{1}{2} \times 0.25 \times 10\right)+\left(\frac{1}{2} \times 1 \times 20\right)\right] \times 4

\mathrm{SF}=45 \mathrm{kN}

Question 9 |

The span(s) to be loaded uniformly for maximum positive (upward) reaction at support P, as shown in the figure below, is (are)

PQ Only | |

PQ and QR | |

QR and RS | |

PQ and RS |

Question 9 Explanation:

With the help of Muller Breslau principle, we can 1 unit
draw the ILD for reaction at P

The reaction is positive between P and Q and R and S respectively. Hence?the spans PQ and RS should be loaded uniformly for maximum positive reaction at P.

The reaction is positive between P and Q and R and S respectively. Hence?the spans PQ and RS should be loaded uniformly for maximum positive reaction at P.

Question 10 |

The influence line diagram (ILD) shown is for the member

PS | |

RS | |

PQ | |

QS |

Question 10 Explanation:

Note: Diagonal member subjected to reversal of stresses.

There are 10 questions to complete.