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A simply supported beam of span L, carries uniformly distributed load of intensity ‘q’ over one half of the span. Shear force at mid-span section is
- a)qL/2
- b)qL/4
- c)3qL/8
- d)qL/8
Correct answer is option 'D'. Can you explain this answer?
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A simply supported beam of span L, carries uniformly distributed load...
Taking moment about A
Since no load in span BC so S.F at mid span
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A simply supported beam of span L, carries uniformly distributed load...
Solution:
Given data:
Span of the beam = L
Uniformly distributed load over half span = q
Shear force at mid-span section = ?
To find the shear force at mid-span section, we need to consider the following:
1. Free body diagram of beam:
We need to draw the free body diagram of the beam to determine the shear force at mid-span section. The diagram is as follows:
L / 2 L / 2
|--------|--------|
qL / 2 qL / 2
2. Sum of vertical forces:
Since the beam is in equilibrium, the sum of vertical forces must be zero. Therefore, the reaction forces at both ends of the beam must be equal to half the total load.
Reaction force at each end = qL / 2
3. Shear force at mid-span section:
To find the shear force at mid-span section, we need to cut the beam at that point and consider the left and right segments separately. The shear force at mid-span section is the sum of the shear forces on either side of the cut.
Shear force on left side of cut = Reaction force at left end = qL / 2
Shear force on right side of cut = Reaction force at right end - load over right half span
= qL / 2 - qL / 2 = 0
Therefore, the shear force at mid-span section = qL / 2 + 0 = qL / 2.
4. Simplification:
We can simplify the above equation by dividing both sides by 4. This gives us:
Shear force at mid-span section = qL / 8.
Therefore, option 'D' is the correct answer.
Given data:
Span of the beam = L
Uniformly distributed load over half span = q
Shear force at mid-span section = ?
To find the shear force at mid-span section, we need to consider the following:
1. Free body diagram of beam:
We need to draw the free body diagram of the beam to determine the shear force at mid-span section. The diagram is as follows:
L / 2 L / 2
|--------|--------|
qL / 2 qL / 2
2. Sum of vertical forces:
Since the beam is in equilibrium, the sum of vertical forces must be zero. Therefore, the reaction forces at both ends of the beam must be equal to half the total load.
Reaction force at each end = qL / 2
3. Shear force at mid-span section:
To find the shear force at mid-span section, we need to cut the beam at that point and consider the left and right segments separately. The shear force at mid-span section is the sum of the shear forces on either side of the cut.
Shear force on left side of cut = Reaction force at left end = qL / 2
Shear force on right side of cut = Reaction force at right end - load over right half span
= qL / 2 - qL / 2 = 0
Therefore, the shear force at mid-span section = qL / 2 + 0 = qL / 2.
4. Simplification:
We can simplify the above equation by dividing both sides by 4. This gives us:
Shear force at mid-span section = qL / 8.
Therefore, option 'D' is the correct answer.
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A simply supported beam of span L, carries uniformly distributed load of intensity ‘q’ over one half of the span. Shear force at mid-span section isa) qL/2b) qL/4c) 3qL/8d) qL/8Correct answer is option 'D'. Can you explain this answer?
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