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Test: Shear Force & Bending Moment - 1

10 Questions MCQ Test Strength of Materials (SOM) | Test: Shear Force & Bending Moment - 1

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Attempt Test: Shear Force & Bending Moment - 1 | 10 questions in 30 minutes | Mock test for Civil Engineering (CE) preparation | Free important questions MCQ to study Strength of Materials (SOM) for Civil Engineering (CE) Exam | Download free PDF with solutions
QUESTION: 1

For the loaded beam shown in the figure, the correct shear force diagram is Solution:

The shear force in the span BC will be zero. The shear force in the span AB and CD will be of opposite sign.

QUESTION: 2

For the simply supported beam, shown in the figure below at what distance from the support A is the shear force zero? Solution: Let the point of zero shear force occur at section X-Xfrom left support A
∴ Shear force at X-X ∴ For zero shear force we have, ∴ QUESTION: 3

For the beam shown ih the given figure, the maximum positive bending moment is equal to the maximum negative bending moment. The value of L1 is Solution:

The BM can be found from the area of SFD.
The shear force diagram is Maximum negative, bending moment, Maximum positive, bending moment, For M1 = M2 QUESTION: 4

The maximum bending moment due to a moving load on a fixed ended beam occurs .

Solution:
QUESTION: 5

If a beam is subjected to a constant bending moment along its length then the shear force will

Solution:

If M is constant then,
S.F = 0, as
S.F = dM/dx

QUESTION: 6

If the shear force acting at every section of a beam is of the same magnitude and of the same direction, then it represents a

Solution: QUESTION: 7

The shear force diagram shown in the figure is that of a Solution:
QUESTION: 8

If the bending moment diagram for a simply suppoted beam is of the form as given in figure then the load acting on the beam is Solution:  Span AC: Span CB: QUESTION: 9

A beam is simply supported at its ends and is loaded by a couple at its mid-span as shown in figure. Shear force diagram for the beam is given by the figure Solution: QUESTION: 10

A horizontal beam carrying uniformly distributed load is supported with equal overhangs as shown in the figure beiow. The resultant bending moment at the mid-span shall be zero if a/b is Solution:

Since both the support reactions and loadings are symmetrically place about the center both the support will be of same magnitude and equal to ½ of total load acting on beam.
RA = RB = [w (2a+b)] / 2
= (w (2a+b))/2=w(a+b/2) From the fig. bending moment at center (C) i.e. mid-span will be equal to:
(RA) (b/2) – w (a+b/2)( ((a+b/2))/2 ) = BMc =0
w(a+b/2) (b/2) – w (a+b/2)( ((a+b/2))/2 ) =0
b−(a+b/2)=0
b = 2a
a/b  = 1/2 Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code

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