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A strut in steel beam comprised two equal angles of ISA 150 x 150 x 10 mm connected back to the back gusset plate. The cross section of the angle is A = 2921 mm2 and moment of inertia is I = 6335000 mm4 . The distance of Cg from the surface (Cx = Cy ) is 40.8mm Find the minimum radius of gyration of the strut.
- a)93.2 mm
- b)62.7 mm
- c)46.6 mm
- d)29.8 mm
Correct answer is option 'C'. Can you explain this answer?
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A strut in steel beam comprised two equal angles of ISA 150 x 150 x 1...
Radius of gyration or gyradius of a body about an axis of rotation is defined as the radial distance of a point from the axis of rotation at which, if the whole mass of the body is assumed to be concentrated, its moment of inertia about the given axis would be the same as with its actual distribution of mass.
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Total cross section area A=2×2921 =5842mm2
Total Moment of Inertia 1=2×6335000mm4
=12670000mm4
Min Radius of gyration rmin= √(I/A) = 46.6mm
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A strut in steel beam comprised two equal angles of ISA 150 x 150 x 1...
Given Data:
- Strut in steel beam comprised of two equal angles of ISA 150 x 150 x 10 mm
- Cross section of the angle: A = 2921 mm2
- Moment of inertia: I = 6335000 mm4
- Distance of Cg from the surface (Cx = Cy): 40.8 mm
To find: Minimum radius of gyration of the strut.
Formula:
The radius of gyration (r) is given by the formula: r = √(I/A)
Calculation:
1. Substitute the given values into the formula to calculate the radius of gyration.
r = √(6335000/2921)
r ≈ 46.6 mm
Therefore, the minimum radius of gyration of the strut is 46.6 mm.
The correct answer is option 'C' (46.6 mm).
Explanation:
- The radius of gyration is a property of the cross-sectional shape of a structural member. It represents the distance from the axis of bending at which the entire area of the member can be considered to be concentrated.
- The formula for calculating the radius of gyration is r = √(I/A), where I is the moment of inertia of the cross-sectional shape and A is the area of the cross-sectional shape.
- In this case, the given angle section has a cross-sectional area of 2921 mm2 and a moment of inertia of 6335000 mm4.
- By substituting these values into the formula, we can calculate the radius of gyration as approximately 46.6 mm.
- The distance of Cg from the surface (Cx = Cy) is not directly used in the calculation of the radius of gyration. It is mentioned in the question, but it does not affect the calculation.
- Therefore, the correct answer is option 'C' (46.6 mm).
- Strut in steel beam comprised of two equal angles of ISA 150 x 150 x 10 mm
- Cross section of the angle: A = 2921 mm2
- Moment of inertia: I = 6335000 mm4
- Distance of Cg from the surface (Cx = Cy): 40.8 mm
To find: Minimum radius of gyration of the strut.
Formula:
The radius of gyration (r) is given by the formula: r = √(I/A)
Calculation:
1. Substitute the given values into the formula to calculate the radius of gyration.
r = √(6335000/2921)
r ≈ 46.6 mm
Therefore, the minimum radius of gyration of the strut is 46.6 mm.
The correct answer is option 'C' (46.6 mm).
Explanation:
- The radius of gyration is a property of the cross-sectional shape of a structural member. It represents the distance from the axis of bending at which the entire area of the member can be considered to be concentrated.
- The formula for calculating the radius of gyration is r = √(I/A), where I is the moment of inertia of the cross-sectional shape and A is the area of the cross-sectional shape.
- In this case, the given angle section has a cross-sectional area of 2921 mm2 and a moment of inertia of 6335000 mm4.
- By substituting these values into the formula, we can calculate the radius of gyration as approximately 46.6 mm.
- The distance of Cg from the surface (Cx = Cy) is not directly used in the calculation of the radius of gyration. It is mentioned in the question, but it does not affect the calculation.
- Therefore, the correct answer is option 'C' (46.6 mm).
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A strut in steel beam comprised two equal angles of ISA 150 x 150 x 10 mm connected back to the back gusset plate. The cross section of the angle is A = 2921 mm2 and moment of inertia is I = 6335000 mm4 . The distance of Cg from the surface (Cx = Cy ) is 40.8mm Find the minimum radius of gyration of the strut.a)93.2 mmb)62.7 mmc)46.6 mmd)29.8 mmCorrect answer is option 'C'. Can you explain this answer?
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A strut in steel beam comprised two equal angles of ISA 150 x 150 x 10 mm connected back to the back gusset plate. The cross section of the angle is A = 2921 mm2 and moment of inertia is I = 6335000 mm4 . The distance of Cg from the surface (Cx = Cy ) is 40.8mm Find the minimum radius of gyration of the strut.a)93.2 mmb)62.7 mmc)46.6 mmd)29.8 mmCorrect answer is option 'C'. 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 strut in steel beam comprised two equal angles of ISA 150 x 150 x 10 mm connected back to the back gusset plate. The cross section of the angle is A = 2921 mm2 and moment of inertia is I = 6335000 mm4 . The distance of Cg from the surface (Cx = Cy ) is 40.8mm Find the minimum radius of gyration of the strut.a)93.2 mmb)62.7 mmc)46.6 mmd)29.8 mmCorrect answer is option 'C'. 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 strut in steel beam comprised two equal angles of ISA 150 x 150 x 10 mm connected back to the back gusset plate. The cross section of the angle is A = 2921 mm2 and moment of inertia is I = 6335000 mm4 . The distance of Cg from the surface (Cx = Cy ) is 40.8mm Find the minimum radius of gyration of the strut.a)93.2 mmb)62.7 mmc)46.6 mmd)29.8 mmCorrect answer is option 'C'. Can you explain this answer?.
A strut in steel beam comprised two equal angles of ISA 150 x 150 x 10 mm connected back to the back gusset plate. The cross section of the angle is A = 2921 mm2 and moment of inertia is I = 6335000 mm4 . The distance of Cg from the surface (Cx = Cy ) is 40.8mm Find the minimum radius of gyration of the strut.a)93.2 mmb)62.7 mmc)46.6 mmd)29.8 mmCorrect answer is option 'C'. 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 strut in steel beam comprised two equal angles of ISA 150 x 150 x 10 mm connected back to the back gusset plate. The cross section of the angle is A = 2921 mm2 and moment of inertia is I = 6335000 mm4 . The distance of Cg from the surface (Cx = Cy ) is 40.8mm Find the minimum radius of gyration of the strut.a)93.2 mmb)62.7 mmc)46.6 mmd)29.8 mmCorrect answer is option 'C'. 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 strut in steel beam comprised two equal angles of ISA 150 x 150 x 10 mm connected back to the back gusset plate. The cross section of the angle is A = 2921 mm2 and moment of inertia is I = 6335000 mm4 . The distance of Cg from the surface (Cx = Cy ) is 40.8mm Find the minimum radius of gyration of the strut.a)93.2 mmb)62.7 mmc)46.6 mmd)29.8 mmCorrect answer is option 'C'. Can you explain this answer?.
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ample number of questions to practice A strut in steel beam comprised two equal angles of ISA 150 x 150 x 10 mm connected back to the back gusset plate. The cross section of the angle is A = 2921 mm2 and moment of inertia is I = 6335000 mm4 . The distance of Cg from the surface (Cx = Cy ) is 40.8mm Find the minimum radius of gyration of the strut.a)93.2 mmb)62.7 mmc)46.6 mmd)29.8 mmCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice Railways tests.
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