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A beam with a rectangular section of 120 mm × 60 mm, designed to be placed vertically is placed horizontally by mistake. If the maximum stress is to be limited, the reduction in load carrying capacity would be
- a)1/4
- b)1/3
- c)1/2
- d)1/6
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
A beam with a rectangular section of 120 mm × 60 mm, designed to ...
-A beam with a rectangular section of 120 mm � 60 mm, designed to be placed vertically is placed horizontally by mistake. If the maximum stress is to be limited, the reduction in load carrying capacity would be 1/2.
When the beam is placed horizontally
b=120mm. d=60mm
= 120*60^2/6 = 72000mm^3
When the beam is placed vertically
b=60mm, d=120mm
= 60*120^2/6 = 144000mm^3
The ratio of section modulus is Z1/Z2 = 72000/144000 = 1/2
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A beam with a rectangular section of 120 mm × 60 mm, designed to ...
Given:
Rectangular section of beam: 120 mm x 60 mm (designed to be placed vertically)
Orientation: Placed horizontally by mistake
To find: Reduction in load carrying capacity to limit maximum stress
Solution:
When the rectangular section beam is placed horizontally, the width of the beam becomes the depth and the depth becomes the width. Therefore, the new dimensions of the beam are 60 mm x 120 mm.
The load carrying capacity of a beam is directly proportional to the moment of inertia (I) of the beam's cross-section. The moment of inertia of a rectangular section beam is given by the formula:
I = (bh^3)/12
where b is the width and h is the depth of the beam.
For the original beam (designed to be placed vertically), b = 60 mm and h = 120 mm. Therefore, the moment of inertia of the original beam is:
I1 = (60 x 120^3)/12 = 34,560,000 mm^4
For the beam placed horizontally by mistake, b = 120 mm and h = 60 mm. Therefore, the moment of inertia of the beam placed horizontally is:
I2 = (120 x 60^3)/12 = 5,184,000 mm^4
To limit the maximum stress in the beam, the load carrying capacity must be reduced by the same ratio as the moment of inertia. Therefore, the reduction in load carrying capacity is:
I2/I1 = 5,184,000/34,560,000 = 1/6
Therefore, the reduction in load carrying capacity to limit maximum stress is 1/2, or option C.
Rectangular section of beam: 120 mm x 60 mm (designed to be placed vertically)
Orientation: Placed horizontally by mistake
To find: Reduction in load carrying capacity to limit maximum stress
Solution:
When the rectangular section beam is placed horizontally, the width of the beam becomes the depth and the depth becomes the width. Therefore, the new dimensions of the beam are 60 mm x 120 mm.
The load carrying capacity of a beam is directly proportional to the moment of inertia (I) of the beam's cross-section. The moment of inertia of a rectangular section beam is given by the formula:
I = (bh^3)/12
where b is the width and h is the depth of the beam.
For the original beam (designed to be placed vertically), b = 60 mm and h = 120 mm. Therefore, the moment of inertia of the original beam is:
I1 = (60 x 120^3)/12 = 34,560,000 mm^4
For the beam placed horizontally by mistake, b = 120 mm and h = 60 mm. Therefore, the moment of inertia of the beam placed horizontally is:
I2 = (120 x 60^3)/12 = 5,184,000 mm^4
To limit the maximum stress in the beam, the load carrying capacity must be reduced by the same ratio as the moment of inertia. Therefore, the reduction in load carrying capacity is:
I2/I1 = 5,184,000/34,560,000 = 1/6
Therefore, the reduction in load carrying capacity to limit maximum stress is 1/2, or option C.
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