Problem 2-19

# Problem 2-19 - of elasticity divided by the weight density...

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MACHINE DESIGN - An 2-19-1 PROBLEM 2-19 Statement: Calculate the specific strength and specific stiffness of the following materials and pick one for use in an aircraft wing spar. Units: ksi 10 3 psi . Given: Material Code Ultimate Strength Young's Modulus Weight Density Steel st 0 Sut st 80 ksi . E st 30 10 6 . psi . γ st 0.28 lbf in 3 . Aluminum al 1 Sut al 60 ksi . E al 10.4 10 6 . psi . al 0.10 lbf in 3 . Titanium ti 2 Sut ti 90 ksi . E ti 16.5 10 6 . psi . ti 0.16 lbf in 3 . Index i 01 , 2 .. Solution: See Mathcad file P0219. 1. Specific strength is the ultimate tensile strength divided by the weight density and specific stiffness is the modulus
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Unformatted text preview: of elasticity divided by the weight density. The text does not give a symbol to these quantities. Specific strength Sut i i 1 in . 2.857·10 5 6·10 5 5.625·10 5 = Specific stiffness E i i 1 in . 1.071·10 8 1.04·10 8 1.031·10 8 = Steel Aluminum Titanium 2. Based on the results above, all three materials have the same specific stiffness but the aluminum has the largest specific strength. Aluminum for the aircraft wing spar is recommended. P 0219.mcd...
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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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