Mechanical Engineering Assignment Help

Mechanical Engineering Assignment Help

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Mechanical Engineering Yield Modeling Assignment Question With Solution 1. A steel bar of length 1 meter and diameter 25 mm is subjected to a tensile load of 20 kN. Determine the yield strength of the material if the extension produced in the bar is 0.25 mm.

Solution: Given: Length of steel bar, L = 1 m Diameter of steel bar, d = 25 mm = 0.025 m Tensile load, P = 20 kN = 20000 N Extension produced, deltaL = 0.25 mm = 0.00025 m The cross-sectional area of the steel bar can be calculated as: A = (pi/4) * d^2 = (3.14/4) * 0.025^2 = 0.00049 m^2 The stress in the steel bar can be calculated as: sigma = P/A = 20000/0.00049 = 40816.33 Pa https://www.mechanicalengineeringassignmenthelp.com/

The strain in the steel bar can be calculated as: epsilon = deltaL/L = 0.00025/1 = 0.00025 The yield strength can be calculated as: Yield strength = sigma/epsilon = 40816.33/0.00025 = 163265.3 Pa Therefore, the yield strength of the material is 163265.3 Pa.

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2. A cylindrical steel specimen of diameter 25 mm and length 100 mm is subjected to a tensile load of 30 kN. Determine the yield strength of the material if the elongation produced in the specimen is 0.3 mm.

Solution: Given: Diameter of steel specimen, d = 25 mm = 0.025 m Length of steel specimen, L = 100 mm = 0.1 m Tensile load, P = 30 kN = 30000 N Elongation produced, deltaL = 0.3 mm = 0.0003 m The cross-sectional area of the steel specimen can be calculated as: A = (pi/4) * d^2 = (3.14/4) * 0.025^2 = 0.00049 m^2 The stress in the steel specimen can be calculated as: sigma = P/A = 30000/0.00049 = 61224.49 Pa The strain in the steel specimen can be calculated as: https://www.mechanicalengineeringassignmenthelp.com/

epsilon = deltaL/L = 0.0003/0.1 = 0.003 The yield strength can be calculated as: Yield strength = sigma/epsilon = 61224.49/0.003 = 20408163.27 Pa Therefore, the yield strength of the material is 20408163.27 Pa.

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3. A steel rod of length 500 mm and diameter 12 mm is subjected to a compressive load of 10 kN. Determine the yield strength of the material if the reduction in diameter produced in the rod is 0.2 mm.

Solution: Given: Length of steel rod, L = 500 mm = 0.5 m Diameter of steel rod, d = 12 mm = 0.012 m Compressive load, P = 10 kN = 10000 N Reduction in diameter produced, deltaD = 0.2 mm = 0.0002 m The cross-sectional area of the steel rod can be calculated as: A = (pi/4) * d^2 = (3.14/4) * 0.012^2 = 0.000113 m^2 The stress in the steel rod can be calculated as: sigma = P/A = 10000/0.000113 = 88495.58 Pa The strain in the steel rod can be calculated as: https://www.mechanicalengineeringassignmenthelp.com/

epsilon = deltaD/(d/2) = 0.0002/(0.012/2) = 0.0333 The yield strength can be calculated as: Yield strength = sigma/epsilon = 88495.58/0.0333 = 2658604.8 Pa Therefore, the yield strength of the material is 2658604.8 Pa.

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