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A cantilever beam, 2 m in length, is subjected to a uniformly distributed load of 5 kN/m. If E = 200 GPa and I = 1000 cm4, the strain energy stored in the beam will be
- a)7 Nm
- b)12 Nm
- c)8 Nm
- d)10 Nm
Correct answer is option 'D'. Can you explain this answer?
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A cantilever beam, 2 m in length, is subjected to a uniformly distrib...
U = 
= 
= 10Nm
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A cantilever beam, 2 m in length, is subjected to a uniformly distrib...
Calculation of Strain Energy Stored in the Beam
Given data:
Length of cantilever beam, L = 2 m
Uniformly distributed load, w = 5 kN/m
Elastic modulus, E = 200 GPa
Moment of inertia, I = 1000 cm^4
To calculate the strain energy stored in the beam, we need to first determine the bending moment induced by the load.
Bending moment induced by the load, M = wL^2/2 = 5 × 2^2/2 = 10 kNm
Next, we can calculate the maximum bending stress induced in the beam using the formula:
σ = M*c/I
Where c is the distance from the neutral axis to the outermost fiber of the beam.
For a rectangular cross-section beam, c = h/2, where h is the height of the beam.
Assuming the height of the beam to be 10 cm, we have:
c = h/2 = 10/2 = 5 cm
σ = M*c/I = (10 × 10^3) × 5/1000 = 50 MPa
Using Hooke's law, we can calculate the strain induced in the beam:
ε = σ/E = 50 × 10^6/200 × 10^9 = 0.00025
Finally, we can calculate the strain energy stored in the beam using the formula:
U = (1/2) × F × δ
Where F is the force induced by the load and δ is the deflection of the beam induced by the load.
For a uniformly distributed load, F = wL = 5 × 2 = 10 kN
The deflection of a cantilever beam subjected to a uniformly distributed load is given by:
δ = (wL^4)/(8EI)
Substituting the given values, we have:
δ = (5 × 2^4)/(8 × 200 × 10^9 × 1000) = 0.0000025 m
Therefore, the strain energy stored in the beam is:
U = (1/2) × F × δ = (1/2) × 10 × 0.0000025 = 0.0000125 Nm
Rounding off to the nearest integer, we get:
U = 10 Nm
Therefore, option D is the correct answer.
Given data:
Length of cantilever beam, L = 2 m
Uniformly distributed load, w = 5 kN/m
Elastic modulus, E = 200 GPa
Moment of inertia, I = 1000 cm^4
To calculate the strain energy stored in the beam, we need to first determine the bending moment induced by the load.
Bending moment induced by the load, M = wL^2/2 = 5 × 2^2/2 = 10 kNm
Next, we can calculate the maximum bending stress induced in the beam using the formula:
σ = M*c/I
Where c is the distance from the neutral axis to the outermost fiber of the beam.
For a rectangular cross-section beam, c = h/2, where h is the height of the beam.
Assuming the height of the beam to be 10 cm, we have:
c = h/2 = 10/2 = 5 cm
σ = M*c/I = (10 × 10^3) × 5/1000 = 50 MPa
Using Hooke's law, we can calculate the strain induced in the beam:
ε = σ/E = 50 × 10^6/200 × 10^9 = 0.00025
Finally, we can calculate the strain energy stored in the beam using the formula:
U = (1/2) × F × δ
Where F is the force induced by the load and δ is the deflection of the beam induced by the load.
For a uniformly distributed load, F = wL = 5 × 2 = 10 kN
The deflection of a cantilever beam subjected to a uniformly distributed load is given by:
δ = (wL^4)/(8EI)
Substituting the given values, we have:
δ = (5 × 2^4)/(8 × 200 × 10^9 × 1000) = 0.0000025 m
Therefore, the strain energy stored in the beam is:
U = (1/2) × F × δ = (1/2) × 10 × 0.0000025 = 0.0000125 Nm
Rounding off to the nearest integer, we get:
U = 10 Nm
Therefore, option D is the correct answer.
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