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A short hollow circular steel column having wall thickness 8 mm and outer diameter being 80 mm must carry a compressive load of 100 kN. Calculate the deformation of the column if the length of the column is 2 meters. (Take E = 2.1 × 105 N/mm2)
- a)0.256 mm
- b)0.324 mm
- c)0.432 mm
- d)0.526 mm
Correct answer is option 'D'. Can you explain this answer?
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A short hollow circular steel column having wall thickness 8 mm and o...
Deformation of an axially loaded bar is given by
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δl=P×L/A×E
Where
δl = deformation of a bar
L = Length of a bar
A = cross-sectional area of bar
E = Modulus of Elasticity of section
Calculation:
Given: L = 2000 mm, E = 2.1 × 105 N/mm2 and P = 105 N
Most Upvoted Answer
A short hollow circular steel column having wall thickness 8 mm and o...
To calculate the deformation of the column, we can use the formula for axial deformation of a column under compressive load:
δ = (P * L) / (A * E)
Where:
δ = deformation of the column
P = compressive load
L = length of the column
A = cross-sectional area of the column
E = Young's modulus of the column material
1. Find the cross-sectional area of the column:
The outer diameter of the column is given as 80 mm, and the wall thickness is 8 mm. So, the inner diameter of the column can be calculated as:
Inner diameter = Outer diameter - 2 * Wall thickness
Inner diameter = 80 mm - 2 * 8 mm = 64 mm
The cross-sectional area of the column can be calculated using the formula for the area of a hollow cylinder:
A = π * (Outer radius^2 - Inner radius^2)
A = π * ((80/2)^2 - (64/2)^2)
A = π * (40^2 - 32^2) mm^2
2. Convert the units:
The given Young's modulus, E, is in N/mm^2. To make it consistent with other units, we need to convert it to N/m^2. 1 N/mm^2 = 1 MPa = 1 x 10^6 N/m^2. Therefore, E = 2.1 x 10^5 N/mm^2 = 2.1 x 10^11 N/m^2.
3. Calculate the deformation:
Plugging the values into the formula, we get:
δ = (100 kN * 2 m) / ((π * (40^2 - 32^2) mm^2) * (2.1 x 10^11 N/m^2))
Simplifying the expression, we find:
δ = (200,000 N * 2 m) / ((π * (40^2 - 32^2) mm^2) * (2.1 x 10^11 N/m^2))
δ = (400,000 N*m) / ((π * (1600 - 1024) mm^2) * (2.1 x 10^11 N/m^2))
δ = (400,000 N*m) / ((π * 576 mm^2) * (2.1 x 10^11 N/m^2))
δ ≈ 0.526 mm
Therefore, the deformation of the column is approximately 0.526 mm. Thus, the correct answer is option 'D'.
δ = (P * L) / (A * E)
Where:
δ = deformation of the column
P = compressive load
L = length of the column
A = cross-sectional area of the column
E = Young's modulus of the column material
1. Find the cross-sectional area of the column:
The outer diameter of the column is given as 80 mm, and the wall thickness is 8 mm. So, the inner diameter of the column can be calculated as:
Inner diameter = Outer diameter - 2 * Wall thickness
Inner diameter = 80 mm - 2 * 8 mm = 64 mm
The cross-sectional area of the column can be calculated using the formula for the area of a hollow cylinder:
A = π * (Outer radius^2 - Inner radius^2)
A = π * ((80/2)^2 - (64/2)^2)
A = π * (40^2 - 32^2) mm^2
2. Convert the units:
The given Young's modulus, E, is in N/mm^2. To make it consistent with other units, we need to convert it to N/m^2. 1 N/mm^2 = 1 MPa = 1 x 10^6 N/m^2. Therefore, E = 2.1 x 10^5 N/mm^2 = 2.1 x 10^11 N/m^2.
3. Calculate the deformation:
Plugging the values into the formula, we get:
δ = (100 kN * 2 m) / ((π * (40^2 - 32^2) mm^2) * (2.1 x 10^11 N/m^2))
Simplifying the expression, we find:
δ = (200,000 N * 2 m) / ((π * (40^2 - 32^2) mm^2) * (2.1 x 10^11 N/m^2))
δ = (400,000 N*m) / ((π * (1600 - 1024) mm^2) * (2.1 x 10^11 N/m^2))
δ = (400,000 N*m) / ((π * 576 mm^2) * (2.1 x 10^11 N/m^2))
δ ≈ 0.526 mm
Therefore, the deformation of the column is approximately 0.526 mm. Thus, the correct answer is option 'D'.
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A short hollow circular steel column having wall thickness 8 mm and outer diameter being 80 mm must carry a compressive load of 100 kN. Calculate the deformation of the column if the length of the column is 2 meters. (Take E = 2.1 × 105 N/mm2)a)0.256 mmb)0.324 mmc)0.432 mmd)0.526 mmCorrect answer is option 'D'. Can you explain this answer?
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A short hollow circular steel column having wall thickness 8 mm and outer diameter being 80 mm must carry a compressive load of 100 kN. Calculate the deformation of the column if the length of the column is 2 meters. (Take E = 2.1 × 105 N/mm2)a)0.256 mmb)0.324 mmc)0.432 mmd)0.526 mmCorrect answer is option 'D'. Can you explain this answer? for Railways 2024 is part of Railways preparation. The Question and answers have been prepared according to the Railways exam syllabus. Information about A short hollow circular steel column having wall thickness 8 mm and outer diameter being 80 mm must carry a compressive load of 100 kN. Calculate the deformation of the column if the length of the column is 2 meters. (Take E = 2.1 × 105 N/mm2)a)0.256 mmb)0.324 mmc)0.432 mmd)0.526 mmCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Railways 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A short hollow circular steel column having wall thickness 8 mm and outer diameter being 80 mm must carry a compressive load of 100 kN. Calculate the deformation of the column if the length of the column is 2 meters. (Take E = 2.1 × 105 N/mm2)a)0.256 mmb)0.324 mmc)0.432 mmd)0.526 mmCorrect answer is option 'D'. Can you explain this answer?.
A short hollow circular steel column having wall thickness 8 mm and outer diameter being 80 mm must carry a compressive load of 100 kN. Calculate the deformation of the column if the length of the column is 2 meters. (Take E = 2.1 × 105 N/mm2)a)0.256 mmb)0.324 mmc)0.432 mmd)0.526 mmCorrect answer is option 'D'. Can you explain this answer? for Railways 2024 is part of Railways preparation. The Question and answers have been prepared according to the Railways exam syllabus. Information about A short hollow circular steel column having wall thickness 8 mm and outer diameter being 80 mm must carry a compressive load of 100 kN. Calculate the deformation of the column if the length of the column is 2 meters. (Take E = 2.1 × 105 N/mm2)a)0.256 mmb)0.324 mmc)0.432 mmd)0.526 mmCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Railways 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A short hollow circular steel column having wall thickness 8 mm and outer diameter being 80 mm must carry a compressive load of 100 kN. Calculate the deformation of the column if the length of the column is 2 meters. (Take E = 2.1 × 105 N/mm2)a)0.256 mmb)0.324 mmc)0.432 mmd)0.526 mmCorrect answer is option 'D'. Can you explain this answer?.
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