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A beam of uniform section 10 metres long carries a uniformly distributed load of 10 kN per metre over the whole length and a concentrated load of 10 kN at the right end. If the beam is freely supported at the left end, then the position of the second support so that maximum bending moment for the beam shall be as small as possible is __________ m.
- a)2.30
- b)2.36
Correct answer is option ''. Can you explain this answer?
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Verified Answer
A beam of uniform section 10 metres long carries a uniformly distribu...
Let the supports be at A and B so that the overhang BC = a metre ∴ AB = (10 − a)metre
Taking moments about A,
Maximum positive B.M. will occur at a section in AB. Let this section be x metres from A.
∴ S. F. at this section = 0
∴ Va − 10x = 0
For the condition that the maximum bending moment shall be as small as possible, the maximum positive bending moment and the hogging moment over the support should be numerically equal.
Solving by trial and error, we get a = 2.33 m.
Maximum B.M. with the above value of a
= 5a(a + 2)
= 5 x 2.33(2.33 + 2)
= 50.44 kNm
Figure shows the beam showing the actual position of the supports.
SF and BM diagrams
78.23 kN
Reaction at A = Va = 110 − 78.23
= 31.77 kN.
Now the SF and BM diagram can be drawn easily.
Question_Type: 5
Most Upvoted Answer
A beam of uniform section 10 metres long carries a uniformly distribu...
Given data:
Length of the beam = 10 m
Uniformly distributed load = 10 kN/m
Concentrated load at the right end = 10 kN
To find: Position of the second support so that maximum bending moment for the beam shall be as small as possible.
Solution:
Let the distance of the second support from the left end be ‘x’ meters.
The reaction at the left support will be a vertical upward force R1 and a clockwise moment M1.
The reaction at the right support will be a vertical upward force R2 and a clockwise moment M2.
To find the value of ‘x’, we will use the concept of the maximum bending moment which occurs at a point where the shear force changes sign. At this point, the bending moment is maximum.
Calculating the reactions:
∑Fy = 0
R1 + R2 = 100 kN ——(1)
Taking moments about the right end:
∑M = 0
R1 × 10 + 10 × 10 = R2 × x ——(2)
Solving equations (1) and (2), we get:
R1 = 50 – 5x kN
R2 = 50 + 5x kN
Calculating the shear force and bending moment:
At a distance x from the left end:
Shear force, V = R1 = 50 – 5x kN (as there is no load to the left of x)
Bending moment, M = ∫Vdx = ∫(50 – 5x)dx = 50x – 2.5x^2 kNm
Differentiating M w.r.t. x:
dM/dx = 50 – 5x
Maximum bending moment occurs where dM/dx = 0
50 – 5x = 0
x = 10/5 = 2 m
However, this point is not valid as it lies beyond the length of the beam. Therefore, we need to check the bending moment at the ends of the beam and at x = 2.5 m.
At x = 0, M = 0
At x = 10, M = 0
At x = 2.5, M = 93.75 kNm
Therefore, the position of the second support so that maximum bending moment for the beam shall be as small as possible is 2.5 meters.
Hence, the correct answer is option (b) 2.36 m.
Length of the beam = 10 m
Uniformly distributed load = 10 kN/m
Concentrated load at the right end = 10 kN
To find: Position of the second support so that maximum bending moment for the beam shall be as small as possible.
Solution:
Let the distance of the second support from the left end be ‘x’ meters.
The reaction at the left support will be a vertical upward force R1 and a clockwise moment M1.
The reaction at the right support will be a vertical upward force R2 and a clockwise moment M2.
To find the value of ‘x’, we will use the concept of the maximum bending moment which occurs at a point where the shear force changes sign. At this point, the bending moment is maximum.
Calculating the reactions:
∑Fy = 0
R1 + R2 = 100 kN ——(1)
Taking moments about the right end:
∑M = 0
R1 × 10 + 10 × 10 = R2 × x ——(2)
Solving equations (1) and (2), we get:
R1 = 50 – 5x kN
R2 = 50 + 5x kN
Calculating the shear force and bending moment:
At a distance x from the left end:
Shear force, V = R1 = 50 – 5x kN (as there is no load to the left of x)
Bending moment, M = ∫Vdx = ∫(50 – 5x)dx = 50x – 2.5x^2 kNm
Differentiating M w.r.t. x:
dM/dx = 50 – 5x
Maximum bending moment occurs where dM/dx = 0
50 – 5x = 0
x = 10/5 = 2 m
However, this point is not valid as it lies beyond the length of the beam. Therefore, we need to check the bending moment at the ends of the beam and at x = 2.5 m.
At x = 0, M = 0
At x = 10, M = 0
At x = 2.5, M = 93.75 kNm
Therefore, the position of the second support so that maximum bending moment for the beam shall be as small as possible is 2.5 meters.
Hence, the correct answer is option (b) 2.36 m.
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A beam of uniform section 10 metres long carries a uniformly distributed load of 10 kN per metre over the whole length and a concentrated load of 10 kN at the right end. If the beam is freely supported at the left end, then the position of the second support so that maximum bending moment for the beam shall be as small as possible is __________ m.a) 2.30b) 2.36Correct answer is option ''. Can you explain this answer?
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