Young's Modulus Cross Sectional Area
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Young's Modulus Cross Sectional Area

[From: ] [author: ] [Date: 12-08-02] [Hit: ]
5 x 10^-3) ---> A= (80)*(5)/(20x10^10)(1.5x10^-3)= 1.......
Young's Modulus is defined as the ratio of the applied stress (F/A) to the resulting strain (Delta Length/ Original Length) A stell rod ha a Young's Modulus value of 20 X 10^10 Newton Meter^2 If a 80 Newton force is applied longitudinally to a 5 meter length of this steel rod and the rod stretches by 1.5 X 10^-3 meters then calculate the cross sectional area of the rod.

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An equation that relates all these variables is d = F*L / E*A

Rearrange and:

A = F*L / d*E

A = 80N * 5m / (1.5E-3m*20E10Nm^-2) = 1.33333333 × 10^-6 m^2

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20 x10^10 = (80/A)*(5/1.5 x 10^-3) ---> A= (80)*(5)/(20x10^10)(1.5x10^-3)= 1.333x10^-6
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keywords: Young,039,Cross,Sectional,Area,Modulus,Young's Modulus Cross Sectional Area
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