Problem 2-6 - E S el el E 30.0 10 6 psi . = E 207 GPa = 2....

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MACHINE DESIGN - An 2-6-1 PROBLEM 2-6 Statement: A metal has a strength of 60 kpsi (414 MPa) at its elastic limit and the strain at that point is 0.002. What is the modulus of elasticity? What is the strain energy at the elastic limit? Assume that the test speimen is 0.505-in dia and has a 2-in gage length. Can you define the type of metal based on the given data? Units: ksi 10 3 psi . MPa 10 6 Pa . GPa 10 9 Pa . kN 10 3 newton . Given: Elastic limit: Strength S el 60 ksi . Strain ε el 0.002 S el 414 MPa = Test specimen Diameter d o 0.505 in . Length L o 2.00 in . Solution: See Mathcad file P0206. 1. The modulus of elasticity is the slope of the stress-strain curve, which is a straight line, in the elastic region. Since one end of this line is at the origin, the slope (modulus of elasticity) is
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Unformatted text preview: E S el el E 30.0 10 6 psi . = E 207 GPa = 2. The strain energy per unit volume at the elastic limit is the area under the stress-strain curve up to the elastic limit. Since the curve is a straight line up to this limit, the area is one-half the base times the height, or U' el 1 2 S el . el . U' el 60 lbf in . in 3 = U' el 414 kN m . m 3 = The total strain energy in the specimen is the strain energy per unit volume times the volume, U el U' el π d o 2 . 4 . L o . U el 24.04 in lbf . = 3. Based on the modulus of elasticity and using Table C-1, the material is steel . P
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This note was uploaded on 04/26/2010 for the course MAE 3242 taught by Professor N/a during the Spring '10 term at University of Texas-Tyler.

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