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1. A bronze bar is fastened between a steel bar and an aluminum bar as shown in fig. 1. Axial loads are applied at the positions shown. Find the total deformation of the assembly.
Consider E_{s} = 2 * 10 ^ 5 * MPa E_{b} = 0.83 * 10 ^ 5 MPa and E al =0.7*10^ 5 MPa?
Consider E_{s} = 2 * 10 ^ 5 * MPa E_{b} = 0.83 * 10 ^ 5 MPa and E al =0.7*10^ 5 MPa?
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1. A bronze bar is fastened between a steel bar and an aluminum bar as...
Given Data:
- Young's Modulus of steel (Es) = 2 * 10^5 MPa
- Young's Modulus of bronze (Eb) = 0.83 * 10^5 MPa
- Young's Modulus of aluminum (Eal) = 0.7 * 10^5 MPa
Calculating Total Deformation:
1. Calculating the Axial Deformations:
- Axial deformation in steel (δs) = σs * Ls / Es
- Axial deformation in bronze (δb) = σb * Lb / Eb
- Axial deformation in aluminum (δal) = σal * Lal / Eal
2. Calculating the Total Deformation:
- Total deformation (δtotal) = δs + δb + δal
3. Substituting the Given Values:
- σs = 100 kN, Ls = 2 m
- σb = 50 kN, Lb = 1 m
- σal = 75 kN, Lal = 3 m
4. Calculating the Axial Deformations:
- δs = (100 * 10^3) * 2 / (2 * 10^5) = 1 mm
- δb = (50 * 10^3) * 1 / (0.83 * 10^5) ≈ 0.6024 mm
- δal = (75 * 10^3) * 3 / (0.7 * 10^5) ≈ 3.2143 mm
5. Calculating the Total Deformation:
- δtotal = 1 + 0.6024 + 3.2143 ≈ 4.8167 mm
Therefore, the total deformation of the assembly is approximately 4.8167 mm.
- Young's Modulus of steel (Es) = 2 * 10^5 MPa
- Young's Modulus of bronze (Eb) = 0.83 * 10^5 MPa
- Young's Modulus of aluminum (Eal) = 0.7 * 10^5 MPa
Calculating Total Deformation:
1. Calculating the Axial Deformations:
- Axial deformation in steel (δs) = σs * Ls / Es
- Axial deformation in bronze (δb) = σb * Lb / Eb
- Axial deformation in aluminum (δal) = σal * Lal / Eal
2. Calculating the Total Deformation:
- Total deformation (δtotal) = δs + δb + δal
3. Substituting the Given Values:
- σs = 100 kN, Ls = 2 m
- σb = 50 kN, Lb = 1 m
- σal = 75 kN, Lal = 3 m
4. Calculating the Axial Deformations:
- δs = (100 * 10^3) * 2 / (2 * 10^5) = 1 mm
- δb = (50 * 10^3) * 1 / (0.83 * 10^5) ≈ 0.6024 mm
- δal = (75 * 10^3) * 3 / (0.7 * 10^5) ≈ 3.2143 mm
5. Calculating the Total Deformation:
- δtotal = 1 + 0.6024 + 3.2143 ≈ 4.8167 mm
Therefore, the total deformation of the assembly is approximately 4.8167 mm.
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1. A bronze bar is fastened between a steel bar and an aluminum bar as shown in fig. 1. Axial loads are applied at the positions shown. Find the total deformation of the assembly.Consider E_{s} = 2 * 10 ^ 5 * MPa E_{b} = 0.83 * 10 ^ 5 MPa and E al =0.7*10^ 5 MPa?
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1. A bronze bar is fastened between a steel bar and an aluminum bar as shown in fig. 1. Axial loads are applied at the positions shown. Find the total deformation of the assembly.Consider E_{s} = 2 * 10 ^ 5 * MPa E_{b} = 0.83 * 10 ^ 5 MPa and E al =0.7*10^ 5 MPa? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about 1. A bronze bar is fastened between a steel bar and an aluminum bar as shown in fig. 1. Axial loads are applied at the positions shown. Find the total deformation of the assembly.Consider E_{s} = 2 * 10 ^ 5 * MPa E_{b} = 0.83 * 10 ^ 5 MPa and E al =0.7*10^ 5 MPa? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1. A bronze bar is fastened between a steel bar and an aluminum bar as shown in fig. 1. Axial loads are applied at the positions shown. Find the total deformation of the assembly.Consider E_{s} = 2 * 10 ^ 5 * MPa E_{b} = 0.83 * 10 ^ 5 MPa and E al =0.7*10^ 5 MPa?.
1. A bronze bar is fastened between a steel bar and an aluminum bar as shown in fig. 1. Axial loads are applied at the positions shown. Find the total deformation of the assembly.Consider E_{s} = 2 * 10 ^ 5 * MPa E_{b} = 0.83 * 10 ^ 5 MPa and E al =0.7*10^ 5 MPa? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about 1. A bronze bar is fastened between a steel bar and an aluminum bar as shown in fig. 1. Axial loads are applied at the positions shown. Find the total deformation of the assembly.Consider E_{s} = 2 * 10 ^ 5 * MPa E_{b} = 0.83 * 10 ^ 5 MPa and E al =0.7*10^ 5 MPa? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1. A bronze bar is fastened between a steel bar and an aluminum bar as shown in fig. 1. Axial loads are applied at the positions shown. Find the total deformation of the assembly.Consider E_{s} = 2 * 10 ^ 5 * MPa E_{b} = 0.83 * 10 ^ 5 MPa and E al =0.7*10^ 5 MPa?.
Solutions for 1. A bronze bar is fastened between a steel bar and an aluminum bar as shown in fig. 1. Axial loads are applied at the positions shown. Find the total deformation of the assembly.Consider E_{s} = 2 * 10 ^ 5 * MPa E_{b} = 0.83 * 10 ^ 5 MPa and E al =0.7*10^ 5 MPa? in English & in Hindi are available as part of our courses for UPSC.
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