202 Lab Report 2 Questions

202 Lab Report 2 Questions - 3 Area of Steel =.05842 in2 a...

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3. Area of Steel = .05842 in 2 Area of Aluminum = .0632 in 2 a. Ours: Steel: Slope = 1.429*10 6 lb/in At 1000 lb: displacement = 1000 lb / 1.429*10 6 lb/in = 6.998*10 -4 in Stress at 1000 lb: 1000lb /.05842 in 2 = 17.12 ksi Strain at 1000 lb: displacement / initial length = 6.998*10 -4 in / 7 in = 9.997 * 10 -5 Young’s modulus: 17.12 ksi / 9.997*10 -5 = 171.23 Mpsi Aluminum: Slope = 6.67*10 5 lb/in At 1000 lb: displacement = 1000 lb / 6.67*10 5 lb/in = 1.499*10 -3 in Stress at 1000 lb: 1000lb /.0632 in 2 = 15.82 ksi Strain at 1000 lb: displacement / initial length = 1.499*10 -3 in / 7 in = 2.141 * 10 -4 Young’s modulus: 15.82 ksi / 2.141*10 -5 = 73.90 Mpsi Average: Steel: Slope = 1.392*10 6 lb/in At 1000 lb: displacement = 1000 lb / 1.392*10 6 lb/in = 7.163*10 -4 in Stress at 1000 lb: 1000lb /.05842 in 2 = 17.12 ksi Strain at 1000 lb: displacement / initial length = 7.163*10 -4 in / 7 in = 1.023 * 10 -4 Young’s modulus: 17.12 ksi / 1.023*10 -4 = 167.30 Mpsi Aluminum: Slope = 6.003*10 5 lb/in At 1000 lb: displacement = 1000 lb / 6.003*10 5 lb/in = 1.666*10 -3 in Stress at 1000 lb: 1000lb /.0632 in 2 = 15.82 ksi Strain at 1000 lb: displacement / initial length = 1.666*10 -3 in / 7 in = 2.380 * 10 -4 Young’s modulus: 15.82 ksi / 2.380*10 -4 = 66.48 Mpsi b. Our Max Stress of Steel: (5719 lb. / .05842 in 2 ) = 97.89 ksi Our Max Stress of Aluminum: (2819.2 lb. / .0632 in
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This note was uploaded on 07/14/2008 for the course ENGRD 2020 taught by Professor Zehnder during the Fall '06 term at Cornell.

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202 Lab Report 2 Questions - 3 Area of Steel =.05842 in2 a...

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