ProblemSet4%20key

# ProblemSet4%20key - L = 0.50 m solve for delta L k= EA/L...

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Problem Set #4 Key 1. Design the thinnest tendon that could support a typical man’s body weight, i.e. 700 Newtons. First, figure out what cross sectional area it needs to be to avoid reaching a failure stress. Failure stress = 100Mpa = 100 x 10^6 N/m^2 = F/A = 700Newtons/A Solve for A = 700/100 x 10^6 = 7 x 10^-6 m^2 If the tendon was 1 meter long, and had the cross sectional area that you just calculated, would it fail due to too much strain? Failure strain = 8% = 0.08 = delta L/L; L = 1.0m, solve for delta L k= EA/L = {(1.5 x 10^9) x (7 x 10^-6 m^2)}/1.0 = 10,500N/m k = F/delta L; delta L = F/k = 700/10,500 = 0.067m delta L/L = 0.067m/1.0m = 6.7% strain. No, it won’t fail. What if the tendon was only 50cm long? Failure strain = 8% = 0.08 = delta L/L;

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Unformatted text preview: L = 0.50 m, solve for delta L k= EA/L = {(1.5 x 10^9) x (7 x 10^-6 m^2)}/0.50 = 21,000N/m k = F/delta L; delta L = F/k = 700/21,000 = 0.033m delta L/L = 0.033m/0.50m = 6.7% strain. No, it won’t fail. That problem should make sense to you. Imagine you are rock climbing. Do you decide to trust the rope based on its length or it’s thickness (cross sectional area)? Obviously, the thickness. The length does not determine its strength. QuickTime s and a TIFF (Uncompres ed) decompres or are needed to see this picture. QuickTime and a TIFF (Uncompressed) decompressor are needed to see this picture. QuickTime and a TIFF (Uncompressed) decompressor are needed to see this picture....
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ProblemSet4%20key - L = 0.50 m solve for delta L k= EA/L...

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