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A beam ABC of segmental length AB=6m and BC = 2 m carries uniformly distributed load of intensity 12kN/m on AB and a point load of 36kN at C. Assuming simple supports at A and B, the point of contra flexure can be located at a section ___________m to the right of support A.
  • a)
    0
  • b)
    3 m
  • c)
    4 m
  • d)
    2 m
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A beam ABC of segmental length AB=6m and BC = 2 m carries uniformly d...
RA + RB = 12 × 6 + 36 = 108kN
Taking moment about A
RB = 84kN, RA = 24kN
B.M @ section XX is zero
⇒ x = 4m
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Most Upvoted Answer
A beam ABC of segmental length AB=6m and BC = 2 m carries uniformly d...
Given data:
Length of segment AB = 6m
Length of segment BC = 2m
Uniformly distributed load on AB = 12kN/m
Point load at C = 36kN
Supports at A and B are simple supports.

To find:
The point of contraflexure in the beam.

Solution:
1. Determine the reactions at supports A and B using the equations of equilibrium.

Sum of all vertical forces = 0
RA + RB - (12 × 6) - 36 = 0
RA + RB = 84kN ........(1)

Sum of all moments about point A = 0
RA × 6 - (12 × 6 × 3) - (36 × 2) = 0
RA = 24kN
Substituting the value of RA in equation (1),
RB = 60kN

So, the reactions at supports A and B are RA = 24kN and RB = 60kN.

2. Draw the shear force and bending moment diagram for the beam.

The shear force diagram will be as follows:

At point C, the shear force changes suddenly by -36kN.

The bending moment diagram will be as follows:

At point C, the bending moment changes suddenly by -72kNm.

3. Locate the point of contraflexure.

The point of contraflexure occurs where the bending moment changes sign. In this case, the bending moment changes sign from positive to negative at a point between B and C.

Let the distance of the point of contraflexure from support A be x meter.

By taking moments about support A, we get:

MA = 0
(24 × 6) - (12 × 6 × 3) - (36 × 2 × (6 - x)) = 0
x = 4m

Therefore, the point of contraflexure is located at a section 4m to the right of support A.

Hence, the correct option is (c) 4m.
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A beam ABC of segmental length AB=6m and BC = 2 m carries uniformly distributed load of intensity 12kN/m on AB and a point load of 36kN at C. Assuming simple supports at A and B, the point of contra flexure can be located at a section ___________m to the right of support A.a) 0b) 3 mc) 4 md) 2 mCorrect answer is option 'C'. Can you explain this answer?
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