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Design an axially loaded tied column pinned at both ends with an unsupported length of 3.5m for carrying a characteristic load of 1500 KN. Concrete grade M20, Steel grade Fe415. 1. Design load , Pu= 1.5 x 1500 KN = 2250 KN 2. Effective length, Le = 3.5 m (pinned - pinned)?
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Design an axially loaded tied column pinned at both ends with an unsup...
Design of an Axially Loaded Tied Column Pinned at Both Ends
Given Data:
- Characteristic load = 1500 KN
- Concrete grade = M20
- Steel grade = Fe415
- Design load = 1.5 x 1500 KN = 2250 KN
- Effective length = 3.5 m (pinned-pinned)
Calculations:
1. Effective Length Factor (K)
- For a pinned-pinned column, K = 1.0 (as per IS 456:2000)
2. Slenderness Ratio (l/r)
- unsupported length = 3.5m
- Assuming the column has a square cross-section, the radius of gyration (r) can be calculated as:
- r = (b/2) * √(0.833)
- where b is the side of the square cross-section
- 0.833 is the shape factor for a square cross-section
- Assuming b = 300 mm, r = (300/2) * √(0.833) = 97.43 mm
- Slenderness ratio (l/r) = 3500/97.43 = 35.91
3. Buckling Load (Pcr)
- Buckling load can be calculated using the formula:
- Pcr = (π^2 * E * I) / (K * Le)^2
- where E is the modulus of elasticity of steel, I is the moment of inertia of the cross-section, and Le is the effective length of the column
- Moment of inertia of a square cross-section can be calculated as:
- I = (b^4)/12
- Assuming E = 200 GPa, I = (300^4)/12 = 1.35 x 10^8 mm^4
- Pcr = (π^2 * 200 * 10^3 * 1.35 x 10^8) / (1.0 * 3500)^2 = 9589 KN
4. Design Load (Pu)
- Pu = 2250 KN (given)
5. Factor of Safety (FOS)
- FOS can be calculated as:
- FOS = Pcr / Pu
- FOS = 9589 / 2250 = 4.27
6. Cross-Sectional Area of the Column (A)
- Cross-sectional area of the column can be calculated using the formula:
- Pu = σc * A + σs * As
- where σc and σs are the stresses in concrete and steel respectively, and As is the area of steel
- Assuming a percentage of steel of 1%, As = 0.01 * A
- Taking the permissible stresses as:
- σc = 0.67 * fck = 0.67 * 20 = 13.4 N/mm^2 (as per IS 456:2000)
- σs = fy = 415 N/mm^2 (as per IS 456:2000)
- Solving the above equation, we get:
- A = (Pu - σs *
Given Data:
- Characteristic load = 1500 KN
- Concrete grade = M20
- Steel grade = Fe415
- Design load = 1.5 x 1500 KN = 2250 KN
- Effective length = 3.5 m (pinned-pinned)
Calculations:
1. Effective Length Factor (K)
- For a pinned-pinned column, K = 1.0 (as per IS 456:2000)
2. Slenderness Ratio (l/r)
- unsupported length = 3.5m
- Assuming the column has a square cross-section, the radius of gyration (r) can be calculated as:
- r = (b/2) * √(0.833)
- where b is the side of the square cross-section
- 0.833 is the shape factor for a square cross-section
- Assuming b = 300 mm, r = (300/2) * √(0.833) = 97.43 mm
- Slenderness ratio (l/r) = 3500/97.43 = 35.91
3. Buckling Load (Pcr)
- Buckling load can be calculated using the formula:
- Pcr = (π^2 * E * I) / (K * Le)^2
- where E is the modulus of elasticity of steel, I is the moment of inertia of the cross-section, and Le is the effective length of the column
- Moment of inertia of a square cross-section can be calculated as:
- I = (b^4)/12
- Assuming E = 200 GPa, I = (300^4)/12 = 1.35 x 10^8 mm^4
- Pcr = (π^2 * 200 * 10^3 * 1.35 x 10^8) / (1.0 * 3500)^2 = 9589 KN
4. Design Load (Pu)
- Pu = 2250 KN (given)
5. Factor of Safety (FOS)
- FOS can be calculated as:
- FOS = Pcr / Pu
- FOS = 9589 / 2250 = 4.27
6. Cross-Sectional Area of the Column (A)
- Cross-sectional area of the column can be calculated using the formula:
- Pu = σc * A + σs * As
- where σc and σs are the stresses in concrete and steel respectively, and As is the area of steel
- Assuming a percentage of steel of 1%, As = 0.01 * A
- Taking the permissible stresses as:
- σc = 0.67 * fck = 0.67 * 20 = 13.4 N/mm^2 (as per IS 456:2000)
- σs = fy = 415 N/mm^2 (as per IS 456:2000)
- Solving the above equation, we get:
- A = (Pu - σs *
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Design an axially loaded tied column pinned at both ends with an unsupported length of 3.5m for carrying a characteristic load of 1500 KN. Concrete grade M20, Steel grade Fe415. 1. Design load , Pu= 1.5 x 1500 KN = 2250 KN 2. Effective length, Le = 3.5 m (pinned - pinned)?
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Design an axially loaded tied column pinned at both ends with an unsupported length of 3.5m for carrying a characteristic load of 1500 KN. Concrete grade M20, Steel grade Fe415. 1. Design load , Pu= 1.5 x 1500 KN = 2250 KN 2. Effective length, Le = 3.5 m (pinned - pinned)? 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 Design an axially loaded tied column pinned at both ends with an unsupported length of 3.5m for carrying a characteristic load of 1500 KN. Concrete grade M20, Steel grade Fe415. 1. Design load , Pu= 1.5 x 1500 KN = 2250 KN 2. Effective length, Le = 3.5 m (pinned - pinned)? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Design an axially loaded tied column pinned at both ends with an unsupported length of 3.5m for carrying a characteristic load of 1500 KN. Concrete grade M20, Steel grade Fe415. 1. Design load , Pu= 1.5 x 1500 KN = 2250 KN 2. Effective length, Le = 3.5 m (pinned - pinned)?.
Design an axially loaded tied column pinned at both ends with an unsupported length of 3.5m for carrying a characteristic load of 1500 KN. Concrete grade M20, Steel grade Fe415. 1. Design load , Pu= 1.5 x 1500 KN = 2250 KN 2. Effective length, Le = 3.5 m (pinned - pinned)? 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 Design an axially loaded tied column pinned at both ends with an unsupported length of 3.5m for carrying a characteristic load of 1500 KN. Concrete grade M20, Steel grade Fe415. 1. Design load , Pu= 1.5 x 1500 KN = 2250 KN 2. Effective length, Le = 3.5 m (pinned - pinned)? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Design an axially loaded tied column pinned at both ends with an unsupported length of 3.5m for carrying a characteristic load of 1500 KN. Concrete grade M20, Steel grade Fe415. 1. Design load , Pu= 1.5 x 1500 KN = 2250 KN 2. Effective length, Le = 3.5 m (pinned - pinned)?.
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Design an axially loaded tied column pinned at both ends with an unsupported length of 3.5m for carrying a characteristic load of 1500 KN. Concrete grade M20, Steel grade Fe415. 1. Design load , Pu= 1.5 x 1500 KN = 2250 KN 2. Effective length, Le = 3.5 m (pinned - pinned)?, a detailed solution for Design an axially loaded tied column pinned at both ends with an unsupported length of 3.5m for carrying a characteristic load of 1500 KN. Concrete grade M20, Steel grade Fe415. 1. Design load , Pu= 1.5 x 1500 KN = 2250 KN 2. Effective length, Le = 3.5 m (pinned - pinned)? has been provided alongside types of Design an axially loaded tied column pinned at both ends with an unsupported length of 3.5m for carrying a characteristic load of 1500 KN. Concrete grade M20, Steel grade Fe415. 1. Design load , Pu= 1.5 x 1500 KN = 2250 KN 2. Effective length, Le = 3.5 m (pinned - pinned)? theory, EduRev gives you an
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