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A steel column of rectangular section (15 mm x10 mm) and length 1.5 m is simply supported at both ends. Assuming modulus of elasticity, E = 200GPa for steel, the critical axial load (in kN) is _______ (correct to two decimal places)
Correct answer is '(1.097)'. Can you explain this answer?
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A steel column of rectangular section (15 mm x10 mm) and length 1.5 m ...
For Simply Supported column at both ends,
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A steel column of rectangular section (15 mm x10 mm) and length 1.5 m ...
- **Given Data**
- Section of the column: 15 mm x 10 mm
- Length of the column: 1.5 m
- Modulus of elasticity, E = 200 GPa
- **Calculating the Critical Axial Load**
The critical axial load for a column can be calculated using the Euler's Formula:
\[ P_{cr} = \dfrac{\pi^2 \cdot E \cdot I}{(K \cdot L)^2} \]
where:
- \( P_{cr} \) = Critical axial load
- \( E \) = Modulus of elasticity (200 GPa = 200 x \(10^9\) N/m²)
- \( I \) = Moment of inertia of the column section
- \( K \) = Effective length factor (for simply supported both ends, K = 1)
- \( L \) = Length of the column
- **Calculating Moment of Inertia**
Moment of inertia for a rectangle is given by the formula:
\[ I = \dfrac{b \cdot h^3}{12} \]
Substituting the values, we get:
\[ I = \dfrac{10 \times 15^3}{12} = 5625 \, mm^4 = 5.625 \times 10^{-6} \, m^4 \]
- **Calculating Critical Axial Load**
Substituting the values into the Euler's Formula:
\[ P_{cr} = \dfrac{\pi^2 \times 200 \times 10^9 \times 5.625 \times 10^{-6}}{(1.5)^2} \]
\[ P_{cr} = 1.097 \, kN \]
Therefore, the critical axial load for the steel column of given dimensions and conditions is 1.097 kN.
- Section of the column: 15 mm x 10 mm
- Length of the column: 1.5 m
- Modulus of elasticity, E = 200 GPa
- **Calculating the Critical Axial Load**
The critical axial load for a column can be calculated using the Euler's Formula:
\[ P_{cr} = \dfrac{\pi^2 \cdot E \cdot I}{(K \cdot L)^2} \]
where:
- \( P_{cr} \) = Critical axial load
- \( E \) = Modulus of elasticity (200 GPa = 200 x \(10^9\) N/m²)
- \( I \) = Moment of inertia of the column section
- \( K \) = Effective length factor (for simply supported both ends, K = 1)
- \( L \) = Length of the column
- **Calculating Moment of Inertia**
Moment of inertia for a rectangle is given by the formula:
\[ I = \dfrac{b \cdot h^3}{12} \]
Substituting the values, we get:
\[ I = \dfrac{10 \times 15^3}{12} = 5625 \, mm^4 = 5.625 \times 10^{-6} \, m^4 \]
- **Calculating Critical Axial Load**
Substituting the values into the Euler's Formula:
\[ P_{cr} = \dfrac{\pi^2 \times 200 \times 10^9 \times 5.625 \times 10^{-6}}{(1.5)^2} \]
\[ P_{cr} = 1.097 \, kN \]
Therefore, the critical axial load for the steel column of given dimensions and conditions is 1.097 kN.
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A steel column of rectangular section (15 mm x10 mm) and length 1.5 m is simply supported at both ends. Assuming modulus of elasticity, E = 200GPa for steel, the critical axial load (in kN) is _______ (correct to two decimal places)Correct answer is '(1.097)'. Can you explain this answer?
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A steel column of rectangular section (15 mm x10 mm) and length 1.5 m is simply supported at both ends. Assuming modulus of elasticity, E = 200GPa for steel, the critical axial load (in kN) is _______ (correct to two decimal places)Correct answer is '(1.097)'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A steel column of rectangular section (15 mm x10 mm) and length 1.5 m is simply supported at both ends. Assuming modulus of elasticity, E = 200GPa for steel, the critical axial load (in kN) is _______ (correct to two decimal places)Correct answer is '(1.097)'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A steel column of rectangular section (15 mm x10 mm) and length 1.5 m is simply supported at both ends. Assuming modulus of elasticity, E = 200GPa for steel, the critical axial load (in kN) is _______ (correct to two decimal places)Correct answer is '(1.097)'. Can you explain this answer?.
A steel column of rectangular section (15 mm x10 mm) and length 1.5 m is simply supported at both ends. Assuming modulus of elasticity, E = 200GPa for steel, the critical axial load (in kN) is _______ (correct to two decimal places)Correct answer is '(1.097)'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about A steel column of rectangular section (15 mm x10 mm) and length 1.5 m is simply supported at both ends. Assuming modulus of elasticity, E = 200GPa for steel, the critical axial load (in kN) is _______ (correct to two decimal places)Correct answer is '(1.097)'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A steel column of rectangular section (15 mm x10 mm) and length 1.5 m is simply supported at both ends. Assuming modulus of elasticity, E = 200GPa for steel, the critical axial load (in kN) is _______ (correct to two decimal places)Correct answer is '(1.097)'. Can you explain this answer?.
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A steel column of rectangular section (15 mm x10 mm) and length 1.5 m is simply supported at both ends. Assuming modulus of elasticity, E = 200GPa for steel, the critical axial load (in kN) is _______ (correct to two decimal places)Correct answer is '(1.097)'. Can you explain this answer?, a detailed solution for A steel column of rectangular section (15 mm x10 mm) and length 1.5 m is simply supported at both ends. Assuming modulus of elasticity, E = 200GPa for steel, the critical axial load (in kN) is _______ (correct to two decimal places)Correct answer is '(1.097)'. Can you explain this answer? has been provided alongside types of A steel column of rectangular section (15 mm x10 mm) and length 1.5 m is simply supported at both ends. Assuming modulus of elasticity, E = 200GPa for steel, the critical axial load (in kN) is _______ (correct to two decimal places)Correct answer is '(1.097)'. Can you explain this answer? theory, EduRev gives you an
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