Problem 2-9 - E S el el E 16.7 10 6 psi . = E 115 GPa = 2....

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
MACHINE DESIGN - An 2-9-1 PROBLEM 2-9 Statement: A metal has a strength of 100 kpsi (689 MPa) at its elastic limit and the strain at that point is 0.006. 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 100 ksi . Strain ε el 0.006 S el 689 MPa = Test specimen Diameter d o 0.505 in . Length L o 2.00 in . Solution: See Mathcad file P0209. 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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: E S el el E 16.7 10 6 psi . = E 115 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 300 lbf in . in 3 = U' el 2 10 3 . 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 d o 2 . 4 . L o . U el 120.18 in lbf . = 3. Based on the modulus of elasticity and using Table C-1, the material is titanium . P 0209.mcd...
View Full Document

Ask a homework question - tutors are online