# wss1 - Worksheet #1 solutions Aug. 27, 2007 (a) To...

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Worksheet #1 solutions Aug. 27, 2007 (a) To determine the stress, we need the force and cross-sectional area. The force is not too accurate, with a standard deviation of 15% of the mean. The cross-sectional area is related to the square of the diameter, which is pretty accurate. Note that assuming a natural tolerance, the standard deviation in the diameter is 1/3 the tolerance, thus, the standard deviation is 6 . 667 × 10 - 4 , or 0.13% of the mean. Even if we look at the variability in the cross-sectional area, we see that the standard deviation is only about 0.27%, which is small compared to the deviation in the force. Thus, the deviation in the cross-sectional area is relatively negligible. Neglecting the variability in the deviation in the cross-sectional area then the mean and standard deviation of the amplitude of the applied stress is: ˆ μ σ = 10 , 000 lb π (0 . 25 in ) 2 = 50 , 930 psi ˆ σ σ = 1 , 500 lb π (0 . 25 in ) 2 = 7 , 639 psi (b) 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14

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## This note was uploaded on 04/20/2008 for the course MANE EMD taught by Professor All during the Spring '08 term at Rensselaer Polytechnic Institute.

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wss1 - Worksheet #1 solutions Aug. 27, 2007 (a) To...

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