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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 L_{1} 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 M_{1} = M_{2}

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

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