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A cantilever beam of 1.5 m carries a uniformly distributed load of 24 kN/m over its full length and a force of 36 kN acts in upward direction at its free end. The magnitude of net deflection of the free end is X/El m. The value of X is __________. (Answer up to two decimal places)
Correct answer is '25.31'. Can you explain this answer?
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A cantilever beam of 1.5 m carries a uniformly distributed load of 24...
Calculation of Net Deflection of Cantilever Beam
Given:
Length of cantilever beam (L) = 1.5 m
Uniformly distributed load (w) = 24 kN/m
Force at free end (P) = 36 kN
To find: Net deflection of free end (δ)
Formula:
δ = (5PL^4)/(384EI) + (wL^4)/(185EI)
where,
P = Force at free end
L = Length of cantilever beam
w = Uniformly distributed load
E = Modulus of elasticity of beam material
I = Moment of inertia of beam cross-section
Calculation:
The moment of inertia of the cantilever beam can be calculated as follows:
I = (bd^3)/12
where,
b = width of the beam
d = depth of the beam
Assuming the beam to be rectangular in cross-section, we can take b = 0.1 m and d = 0.2 m.
I = (0.1 x 0.2^3)/12
I = 0.000133 m^4
Next, we can calculate the modulus of elasticity of the beam material. Assuming it to be made of steel, we can take E = 200 GPa (200 x 10^9 Pa).
Now, we can substitute the given values in the formula to get the net deflection of the free end.
δ = (5 x 36 x 1.5^4)/(384 x 200 x 10^9 x 0.000133) + (24 x 1.5^4)/(185 x 200 x 10^9 x 0.000133)
δ ≈ 0.02531 m
Therefore, the net deflection of the free end is 0.02531 m or 25.31 mm (rounded to two decimal places).
Given:
Length of cantilever beam (L) = 1.5 m
Uniformly distributed load (w) = 24 kN/m
Force at free end (P) = 36 kN
To find: Net deflection of free end (δ)
Formula:
δ = (5PL^4)/(384EI) + (wL^4)/(185EI)
where,
P = Force at free end
L = Length of cantilever beam
w = Uniformly distributed load
E = Modulus of elasticity of beam material
I = Moment of inertia of beam cross-section
Calculation:
The moment of inertia of the cantilever beam can be calculated as follows:
I = (bd^3)/12
where,
b = width of the beam
d = depth of the beam
Assuming the beam to be rectangular in cross-section, we can take b = 0.1 m and d = 0.2 m.
I = (0.1 x 0.2^3)/12
I = 0.000133 m^4
Next, we can calculate the modulus of elasticity of the beam material. Assuming it to be made of steel, we can take E = 200 GPa (200 x 10^9 Pa).
Now, we can substitute the given values in the formula to get the net deflection of the free end.
δ = (5 x 36 x 1.5^4)/(384 x 200 x 10^9 x 0.000133) + (24 x 1.5^4)/(185 x 200 x 10^9 x 0.000133)
δ ≈ 0.02531 m
Therefore, the net deflection of the free end is 0.02531 m or 25.31 mm (rounded to two decimal places).
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Community Answer
A cantilever beam of 1.5 m carries a uniformly distributed load of 24...
Given: w = 24 kN/m, L = 1.5 m and W = 36 kN = wL
Net deflection of the free end is
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A cantilever beam of 1.5 m carries a uniformly distributed load of 24 kN/m over its full length and a force of 36 kN acts in upward direction at its free end. The magnitude of net deflection of the free end is X/El m. The value of X is __________. (Answer up to two decimal places)Correct answer is '25.31'. Can you explain this answer?
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A cantilever beam of 1.5 m carries a uniformly distributed load of 24 kN/m over its full length and a force of 36 kN acts in upward direction at its free end. The magnitude of net deflection of the free end is X/El m. The value of X is __________. (Answer up to two decimal places)Correct answer is '25.31'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A cantilever beam of 1.5 m carries a uniformly distributed load of 24 kN/m over its full length and a force of 36 kN acts in upward direction at its free end. The magnitude of net deflection of the free end is X/El m. The value of X is __________. (Answer up to two decimal places)Correct answer is '25.31'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cantilever beam of 1.5 m carries a uniformly distributed load of 24 kN/m over its full length and a force of 36 kN acts in upward direction at its free end. The magnitude of net deflection of the free end is X/El m. The value of X is __________. (Answer up to two decimal places)Correct answer is '25.31'. Can you explain this answer?.
A cantilever beam of 1.5 m carries a uniformly distributed load of 24 kN/m over its full length and a force of 36 kN acts in upward direction at its free end. The magnitude of net deflection of the free end is X/El m. The value of X is __________. (Answer up to two decimal places)Correct answer is '25.31'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about A cantilever beam of 1.5 m carries a uniformly distributed load of 24 kN/m over its full length and a force of 36 kN acts in upward direction at its free end. The magnitude of net deflection of the free end is X/El m. The value of X is __________. (Answer up to two decimal places)Correct answer is '25.31'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A cantilever beam of 1.5 m carries a uniformly distributed load of 24 kN/m over its full length and a force of 36 kN acts in upward direction at its free end. The magnitude of net deflection of the free end is X/El m. The value of X is __________. (Answer up to two decimal places)Correct answer is '25.31'. Can you explain this answer?.
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