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Test: Bending Moment - Civil Engineering (CE) MCQ


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10 Questions MCQ Test RCC & Prestressed Concrete - Test: Bending Moment

Test: Bending Moment for Civil Engineering (CE) 2024 is part of RCC & Prestressed Concrete preparation. The Test: Bending Moment questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: Bending Moment MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Bending Moment below.
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Test: Bending Moment - Question 1

The point of contra flexure is a point where

Detailed Solution for Test: Bending Moment - Question 1


The points of contraflexure (or inflection) are points of zero bending moment, i.e. where the beam changes its curvature from hogging to sagging.
In a bending beam, a point of contraflexure is a location where the bending moment is zero (changes its sign). In a bending moment diagram, it is the point at which the bending moment curve intersects with the zero lines. 
At the point of contra flexure, the bending moment is zero.

Test: Bending Moment - Question 2

The shapes of the bending moment diagram for a uniform cantilever beam carrying a uniformly distributed load over its length is

Detailed Solution for Test: Bending Moment - Question 2


is the equation of parabola.

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Test: Bending Moment - Question 3

In a simply supported beam of span (L + 2a) with equal overhang (a) carries a uniformly distributed load over the whole length. Bending moment changes sign if -

Detailed Solution for Test: Bending Moment - Question 3



L > 2a

Test: Bending Moment - Question 4

A concentrated load P acts on a simply supported beam of span L at a distance L/3 from the left support. The bending moment at the point of application of the load is given by

Detailed Solution for Test: Bending Moment - Question 4

For vertical equilibrium

Taking moment about R1 , we have


The bending moment at the point of application of load =Area under OABC

Test: Bending Moment - Question 5

At what height from the base of a pillar must the end of a rope of length l be fixed so that a man standing on the ground and pulling it at the other end may have the greatest tendency to overturn the pillar

Detailed Solution for Test: Bending Moment - Question 5


The pillar will have the greatest tendency of overturning when the Moment of the force F induced in the rope will have a maximum moment about the base of the pillar A.
Hence, To find the greatest overturning tendency, we have to maximise MA.
Calculation:
Given,
Length of the rope = l, 
let y = height from the ground at which rope is fixed to the pillar.
Taking moment about A we get,
MA = F × cos θ × y
From ΔABC, y = l × sin θ 
MA = Fl × cos θ × sin θ  = Fl/2 x sin2θ
For the maximum value of MA, sin 2θ = 90° , θ = 45 ° 
y = l × sin θ y = l × sin 45° = l/√ 2

Test: Bending Moment - Question 6

A cantilever beam carriers the anti symmetric load as shown, where w is the peak intensity of the distributed load. Qualitatively, the correct bending moment diagram for the beam is

Detailed Solution for Test: Bending Moment - Question 6

Here make hit and trial method
Consider first option (A) of moment diagram.



At right end B, slope must be zero as there is no shear force at B so option A is wrong.
Now, consider option (B)
Due to symmetric load intensity at left side too the shear force equal to zero so slope at left side must be zero. So option (B) wrong.


Now consider option (C)
Here at both ends, slope is zero means shear force is zero and also when we move from right to left, the rate of increase of shear force decreases due triangular shape of load intensity and at middle slope should be maximum and there after decreases so in option (C) all these criterion fulfills. Here (C) is the correct option.


As far as (D) option is concerned its middle part slope and right most part slope is strictly not agreeing with the load shown.

Test: Bending Moment - Question 7

A mobile phone has a small motor with an eccentric mass used for vibrator mode. The location of the eccentric mass on motor with respect to center of gravity (CG) of the mobile and the rest of the dimensions of the mobile phone are shown. The mobile is kept on a flat horizontal surface.

Given in addition that the eccentric mass = 2 grams, eccentricity = 2.19 mm, mass of the mobile = 90 grams, g = 9.81 m/s2. Uniform speed of the motor in RPM for which the mobile will get just lifted off the ground at the end Q is approximately.

Detailed Solution for Test: Bending Moment - Question 7
  • When a rotating body is involved then there comes a force which is known as Centrifugal force (Fc)
    It is calculated by using formula
    Fc = mr(ω)2
  • Now in the Question, it is mentioned that end Q is just lifted off the ground. So as it is just lifted off the ground there will be no reaction force from that point.

∴ FBD of mobile

M = mass of mobile
m = eccentric mass
r = eccentricity
Hence, Taking a moment about point P = O
∴ Mg × 0.06 - Fc × (0.09) = 0
∴ 90 × 10-3 × 9.81 × 0.06 = mr(ω)2 × (0.09)
90 × 10-3 × 9.81 × 0.06 = 2 × 10-3 × 2.19 × 10-3 (ω)2 × 0.09
∴ ω2 = 134380.8964
∴ ω = 366.58
∴2πN/60 = 366.58
∴ N = 3500 rpm

Test: Bending Moment - Question 8

A beam is made up of two identical bars AB and BC by hinging them together at B. the end A is built in (cantilevered) and the end C is simply supported. With the load P acting as shown, the bending moment at A is

Detailed Solution for Test: Bending Moment - Question 8

Test: Bending Moment - Question 9

A uniformly loaded propped cantilever beam and its free body diagram are shown below. The reaction are

Detailed Solution for Test: Bending Moment - Question 9

 



Moment about (1),

Moment about (2),

From (i), (ii), (iii), we get 

Test: Bending Moment - Question 10

In a simply-supported beam loaded as shown below, the maximum bending moment is Nm is

Detailed Solution for Test: Bending Moment - Question 10

 


Taking moment about A,



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