a steel rod of 20mm in diameter of 1.2m long is heated through 120°C and at the same time subjected to a pull "p". if the total elongation of the rod is 3mm, what should be the magnitude of "p"
take young modulus of steel = 200
αs = 12x10^-6/°C
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We have a steel rod that is 20 millimeters in diameter and 1.2 meters long. The rod is heated through 120 degrees Celsius and simultaneously subjected to a pull force P. The total elongation is 3 millimeters. We need to find the magnitude of force P using the given material properties.
The first step is to calculate the cross-sectional area of the steel rod. Using the given diameter of 20 millimeters, we apply the formula A equals pi times the radius squared. The radius is half the diameter, which is 10 millimeters. Therefore, the area equals pi times 100 square millimeters, which is approximately 314.16 square millimeters.
The second step is to calculate the thermal elongation due to the temperature increase. We use the thermal expansion formula: delta L thermal equals length times alpha s times delta T. Substituting our values: 1200 millimeters times 12 times 10 to the minus 6 per degree Celsius times 120 degrees Celsius. This gives us 1.728 millimeters of thermal elongation.
The third step is to calculate the mechanical elongation caused by the applied force. Since the total elongation is the sum of thermal and mechanical elongations, we subtract the thermal elongation from the total. Three millimeters minus 1.728 millimeters equals 1.272 millimeters. This is the elongation purely due to the mechanical force P.
The final step is to calculate the force P using the mechanical elongation formula. We rearrange the formula to solve for P: P equals delta L mechanical times A times E divided by L. Substituting our values: 1.272 millimeters times 314.16 square millimeters times 200,000 megapascals divided by 1200 millimeters. This gives us approximately 66,586 Newtons. Therefore, the magnitude of the required pull force P is 66,586 Newtons.