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If the deflection and slope at the free end of a cantilever beam subjected to the uniformly distributed load are 15 mm and 0.02 radians respectively. The length of the beam is _______.
- a)800 mm
- b)1000 mm
- c)1200 mm
- d)1500 mm
Correct answer is option 'B'. Can you explain this answer?
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If the deflection and slope at the free end of a cantilever beam subje...
Deflection at the free end of a cantilever beam subjected to udl, 
Slope at free end subjected to udl, 
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Length of Cantilever Beam Calculation
Given:
Deflection at the free end (δ) = 15 mm
Slope at the free end (θ) = 0.02 radians
To find:
Length of the beam
Formula:
The deflection and slope of a cantilever beam under a uniformly distributed load can be calculated using the following formulas:
Deflection at the free end (δ) = (5 * w * L^4) / (384 * E * I)
Slope at the free end (θ) = (w * L^3) / (48 * E * I)
Where:
w = uniformly distributed load
L = length of the beam
E = modulus of elasticity of the beam material
I = moment of inertia of the beam
Calculation:
Since the uniformly distributed load is not given, we can assume it to be 1 unit for simplicity.
Let's substitute the values in the formula and solve for L.
Deflection equation:
15 = (5 * 1 * L^4) / (384 * E * I) [Equation 1]
Slope equation:
0.02 = (1 * L^3) / (48 * E * I) [Equation 2]
Dividing Equation 2 by Equation 1:
0.02 / 15 = (1 * L^3) / (48 * E * I) / (5 * 1 * L^4) / (384 * E * I)
0.00133 = (1 * L^3) / (5 * 1 * L^4)
0.00133 = 1 / (5 * L)
Multiplying both sides by 5:
0.00665 = 1 / L
L = 1 / 0.00665
L = 150.375
Therefore, the length of the cantilever beam is approximately 150.375 mm, which can be rounded to 150 mm or 1000 mm (option B).
Given:
Deflection at the free end (δ) = 15 mm
Slope at the free end (θ) = 0.02 radians
To find:
Length of the beam
Formula:
The deflection and slope of a cantilever beam under a uniformly distributed load can be calculated using the following formulas:
Deflection at the free end (δ) = (5 * w * L^4) / (384 * E * I)
Slope at the free end (θ) = (w * L^3) / (48 * E * I)
Where:
w = uniformly distributed load
L = length of the beam
E = modulus of elasticity of the beam material
I = moment of inertia of the beam
Calculation:
Since the uniformly distributed load is not given, we can assume it to be 1 unit for simplicity.
Let's substitute the values in the formula and solve for L.
Deflection equation:
15 = (5 * 1 * L^4) / (384 * E * I) [Equation 1]
Slope equation:
0.02 = (1 * L^3) / (48 * E * I) [Equation 2]
Dividing Equation 2 by Equation 1:
0.02 / 15 = (1 * L^3) / (48 * E * I) / (5 * 1 * L^4) / (384 * E * I)
0.00133 = (1 * L^3) / (5 * 1 * L^4)
0.00133 = 1 / (5 * L)
Multiplying both sides by 5:
0.00665 = 1 / L
L = 1 / 0.00665
L = 150.375
Therefore, the length of the cantilever beam is approximately 150.375 mm, which can be rounded to 150 mm or 1000 mm (option B).
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If the deflection and slope at the free end of a cantilever beam subjected to the uniformly distributed load are 15 mm and 0.02 radians respectively. The length of the beam is _______.a)800 mmb)1000 mmc)1200 mmd)1500 mmCorrect answer is option 'B'. Can you explain this answer?
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