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The deformation of a bar under its own weight as compared to that when subjected to a direct axial load equal to its own weight will be
- a)the same
- b)one fourth
- c)half
- d)double
Correct answer is option 'C'. Can you explain this answer?
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Deformation of a bar under its own weight:
When a bar is subjected to its own weight, it experiences a distributed load along its length. The deformation of the bar can be calculated using the formula:
δ = (wL^3)/(3EI)
where δ is the deflection of the bar, w is the weight per unit length, L is the length of the bar, E is the modulus of elasticity, and I is the moment of inertia of the cross-section.
Deformation of a bar under direct axial load equal to its own weight:
When a bar is subjected to a direct axial load equal to its own weight, the load is concentrated at one end of the bar. The deformation of the bar can be calculated using the formula:
δ = (wL)/(AE)
where δ is the deflection of the bar, w is the weight per unit length, L is the length of the bar, A is the cross-sectional area, and E is the modulus of elasticity.
Comparison of deformation:
The ratio of the deformation of the bar under its own weight to the deformation of the bar under direct axial load equal to its own weight can be calculated as follows:
δ1/δ2 = ((wL^3)/(3EI))/((wL)/(AE))
δ1/δ2 = (AL^2)/(3EI)
Since A, L, and E are constant for a given bar, the ratio of the deformation is proportional to L^2/I. As the moment of inertia of the cross-section is greater than the square of the length of the bar, the ratio of the deformation is less than 1. Hence, the deformation of the bar under its own weight is half that of the deformation of the bar under direct axial load equal to its own weight.
Therefore, the correct answer is option C, half.
When a bar is subjected to its own weight, it experiences a distributed load along its length. The deformation of the bar can be calculated using the formula:
δ = (wL^3)/(3EI)
where δ is the deflection of the bar, w is the weight per unit length, L is the length of the bar, E is the modulus of elasticity, and I is the moment of inertia of the cross-section.
Deformation of a bar under direct axial load equal to its own weight:
When a bar is subjected to a direct axial load equal to its own weight, the load is concentrated at one end of the bar. The deformation of the bar can be calculated using the formula:
δ = (wL)/(AE)
where δ is the deflection of the bar, w is the weight per unit length, L is the length of the bar, A is the cross-sectional area, and E is the modulus of elasticity.
Comparison of deformation:
The ratio of the deformation of the bar under its own weight to the deformation of the bar under direct axial load equal to its own weight can be calculated as follows:
δ1/δ2 = ((wL^3)/(3EI))/((wL)/(AE))
δ1/δ2 = (AL^2)/(3EI)
Since A, L, and E are constant for a given bar, the ratio of the deformation is proportional to L^2/I. As the moment of inertia of the cross-section is greater than the square of the length of the bar, the ratio of the deformation is less than 1. Hence, the deformation of the bar under its own weight is half that of the deformation of the bar under direct axial load equal to its own weight.
Therefore, the correct answer is option C, half.
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The deformation of a bar under its own weight as compared to that when...
1/2 is THE answer 1st case fix beam with udl deflection is pl^4/384EI 2nd case fix beam with point load deflection is pl^4/192EI
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