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A prismatic beam of length L is simply supported at its ends and subjected to a total UDL of W spread over its entire span. It is then propped at its Centre to neutralize the deflection. The net B.M. at its Centre will be
- a)WL
- b)WL/3
- c)WL/24
- d)WL/32
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
A prismatic beam of length L is simply supported at its ends and subj...
The propped reaction neutralize Downward deflection due to udl = upward Deflection due to support at C
∴ moment at C Mc = 
RA + RB + RC = W
∴ Rc =
RA + RB = W - 
Due to symmetry RA = RB = 
∴ Mc = 
= 
= 
= 
(Clockwise)
(Clockwise)Most Upvoted Answer
A prismatic beam of length L is simply supported at its ends and subj...
Given data:
Length of prismatic beam = L
Total UDL = W
Beam is propped at its Centre
To find:
Net B.M. at the Centre
Solution:
1. Calculation of reaction forces:
As the beam is simply supported, the reactions at both ends will be equal and half of the total load.
Reaction force at each end = W/2
2. Calculation of bending moment at the Centre:
Before the beam is propped at the Centre, the maximum bending moment occurs at the Centre and can be calculated as follows:
Maximum bending moment (M) = W*L^2/8
After the beam is propped at the Centre, the deflection is neutralized, and the bending moment at the Centre becomes zero.
Therefore, the net bending moment at the Centre due to the UDL can be calculated as:
Net bending moment = Maximum bending moment - Bending moment due to the prop
Bending moment due to the prop = (W/2)*(L/2)^2
Net bending moment = W*L^2/8 - (W/2)*(L/2)^2 = WL^2/32
3. Calculation of net B.M. at the Centre:
Net bending moment at the Centre = WL^2/32
Therefore, option (d) is the correct answer: WL/32.
Length of prismatic beam = L
Total UDL = W
Beam is propped at its Centre
To find:
Net B.M. at the Centre
Solution:
1. Calculation of reaction forces:
As the beam is simply supported, the reactions at both ends will be equal and half of the total load.
Reaction force at each end = W/2
2. Calculation of bending moment at the Centre:
Before the beam is propped at the Centre, the maximum bending moment occurs at the Centre and can be calculated as follows:
Maximum bending moment (M) = W*L^2/8
After the beam is propped at the Centre, the deflection is neutralized, and the bending moment at the Centre becomes zero.
Therefore, the net bending moment at the Centre due to the UDL can be calculated as:
Net bending moment = Maximum bending moment - Bending moment due to the prop
Bending moment due to the prop = (W/2)*(L/2)^2
Net bending moment = W*L^2/8 - (W/2)*(L/2)^2 = WL^2/32
3. Calculation of net B.M. at the Centre:
Net bending moment at the Centre = WL^2/32
Therefore, option (d) is the correct answer: WL/32.
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A prismatic beam of length L is simply supported at its ends and subjected to a total UDL of W spread over its entire span. It is then propped at its Centre to neutralize the deflection. The net B.M. at its Centre will bea) WLb) WL/3c) WL/24d) WL/32Correct answer is option 'D'. Can you explain this answer?
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A prismatic beam of length L is simply supported at its ends and subjected to a total UDL of W spread over its entire span. It is then propped at its Centre to neutralize the deflection. The net B.M. at its Centre will bea) WLb) WL/3c) WL/24d) WL/32Correct answer is option 'D'. Can you explain this answer? for Class 7 2024 is part of Class 7 preparation. The Question and answers have been prepared according to the Class 7 exam syllabus. Information about A prismatic beam of length L is simply supported at its ends and subjected to a total UDL of W spread over its entire span. It is then propped at its Centre to neutralize the deflection. The net B.M. at its Centre will bea) WLb) WL/3c) WL/24d) WL/32Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 7 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A prismatic beam of length L is simply supported at its ends and subjected to a total UDL of W spread over its entire span. It is then propped at its Centre to neutralize the deflection. The net B.M. at its Centre will bea) WLb) WL/3c) WL/24d) WL/32Correct answer is option 'D'. Can you explain this answer?.
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