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Home›Structural Analysis›Influence Line Diagram And Rolling Loads

GATE CE · Structural Analysis

Influence Line Diagram And Rolling Loads Practice Questions

Generate GATE-level questions on Influence Line Diagram And Rolling Loads in Structural Analysis. Focus on core concepts, previous year patterns, and numerical problem-solving techniques.

17 questions · 17 PYQs · 0 AI practice · GATE CE 2027

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Q1GATE 2025MCQ

Consider the pin-jointed truss shown in the figure. Influence line is drawn for the axial force in the member G-L when a unit load travels on the bottom chord of the truss. Identify the CORRECT influence line from the following options: Note: Positive value corresponds to tension and negative value corresponds to compression in the member.

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📖 Explanation

To determine the correct influence line for axial force in member GL: Step 1: Member G-L is a diagonal member connecting the bottom chord joint G to the top chord joint L. It is part of a vertical and diagonal web system in a pin-jointed truss. Step 2: Apply a unit moving load on the bottom chord and analyze the influence of this load on the axial force in member G-L. When the unit load is directly under joint G, it induces the maximum positive influence (tension) in member G-L. When the unit load is under joint L, it causes a negative influence (compression). Between G and L, the influence is linear due to statically determinate nature of the truss and linear response of the system. Outside G and L (i.e., A-G and L-P), the influence is zero since the load does not affect the member significantly. Step 3: From the options, only Option (D) correctly shows: A positive peak (1.33) under G A negative value under L Zero elsewhere Final Answer: $\boxed{\text{Option (D)}}$

Q2GATE 2024NAT

The horizontal beam $P Q R S$ shown in the figure has a fixed support at point $P$ , an internal hinge at point $Q$ , and a pin support at point $R$ . A concentrated vertically downward load $(V)$ of $10 \mathrm{kN}$ can act at any point over the entire length of the beam. The maximum magnitude of the moment reaction (in $\mathrm{kN} . \mathrm{m}$ ) that can act at the support $\mathrm{P}$ due to $\mathrm{V}$ is ______ (in integer).

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📖 Explanation

Beam is determinate ILD for $\mathrm{BM}$ at ' $\mathrm{P}$ ' is given by Hence when point load is at ' $Q$ ', the point of $\max$ ordinate for ILD for BM at $\mathrm{P}, \mathrm{BM}$ at $\mathrm{P}$ will be max. $\Rightarrow \mathrm{M}_{\mathrm{P} \text { max }}=10 \times 15=150 \mathrm{kN}-\mathrm{m}$

Explanation Figure 2

Q3GATE 2023MCQ

Muller-Breslau principle is used in analysis of structures for

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📖 Explanation

Muller Breslass principle is used to draw influence line diagram for determinate and indeterminate structures. It states that influence line for any stress function may be obtained by removing the restraint offered by that function and introducing a directly related generalised unit displacement at that location in the direction of the stress function.

Q4GATE 2022NAT

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)

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📖 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}

Explanation Figure 2

Q5GATE 2021MCQ

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

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📖 Explanation

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

Explanation Figure 6

Q6GATE 2020MCQ

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

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📖 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

Explanation Figure 2

Q7GATE 2019NAT

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 _____

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📖 Explanation

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

Explanation Figure 2

Q8GATE 2017NAT

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 _____

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📖 Explanation

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

Explanation Figure 2

Explanation Figure 3

Q9GATE 2015NAT

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 ___________.

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📖 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}

Explanation Figure 2

Q10GATE 2014MCQ

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

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📖 Explanation

ILD for SF at X-X

Explanation Figure 2

Q11GATE 2013MCQ

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:

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📖 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 . $\Rightarrow \mathrm{SF}=\left[\left(\frac{1}{2} \times 0.25 \times 10\right)+\left(\frac{1}{2} \times 1 \times 20\right)\right] \times 4$ $\mathrm{SF}=45 \mathrm{kN}$

Explanation Figure 2

Q12GATE 2008MCQ

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

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📖 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.

Explanation Figure 2

Q13GATE 2007MCQ

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

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📖 Explanation

Note: Diagonal member subjected to reversal of stresses.

Explanation Figure 2

Q14GATE 2006MCQ

Consider the beam ABCD and the influence line as shown below. The influence line pertains to

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📖 Explanation

Currently no explanation available

Explanation Figure 2

Q15GATE 2003MCQ

A beam PQRS is 18 m long and is simply supported at points Q and R, 10 m apart. Overhangs PQ and RS are 3 m and 5 m respectively. A train of two point loads of 150 kN and 100kN, 5 m apart, crosses this beam from left to right with 100 kN load leading. The maximum hogging moment in the beam anywhere is

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📖 Explanation

Maximum hogging BM will be at support R because overhang is maximum on that side. ILD for hogging BM at R is, Maximum hogging moment in the beam occurs when 150 kN load is at S. BM at S $= -150 \times 5 = -750 kNm$

Explanation Figure 1

Q16GATE 2003MCQ

Muller Breslau's principle in structural analysis is used for

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📖 Explanation

Muller Breslau's principle is a principle on the basis of which it is possible to draw influence lines for various quantities pertaining to a structure. It states that: "The ordinates of the influence lines for any stress element (such as axial stress, moment or reaction) of any structure are proportional to those of deflection curve, which is obtained by removing the restraint corresponding to that element from the structure and introducing in its place a deformation into the primary structure which remains."

Q17GATE 2001MCQ

Identify, from the following, the correct value of the bending moment $M_A$ (in kNm units) at the fixed end A in the statically determinate beam shown below (with internal hinges at B and D), when a uniformly distributed load of 10 kN/m is placed on all spans. (Hint: Sketching the influence line for $M_A$ or applying the Principle of Virtual Displacements makes the solution easy).

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📖 Explanation

ILD for BM at A by using Muller Breslau principle will be

\begin{aligned} M_{A}&=\text { Net area of } 1 L D \times 10 \\ M_{A}&=\left[-\left(\frac{1}{2} \times 2 \times 4\right)+\left(\frac{1}{2} \times 2 \times 4\right)\right] \times 10=0 \mathrm{kNm} \end{aligned}

Explanation Figure 2

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