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467_Mechanics Homework Mechanics of Materials Solution

467_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: mm :14 PROBLEM 4 207 4- 207 through 4. 209 Using Eq“ (4. 66), derive the expression for R given in Fig. 4.79 SOLUTION for “.207 A circular cross section Use. Papas" coo¢c££na+e f3 45 shown waH. w = ZQSEnfi V‘ = 17" — egos/3 09%:- .-c sinfiafl dA = w air = {fist-nae a3 51f IZC(‘-CGSi/3)J/g= F~Cca~sfi Fz-c‘ J. F+c Y‘ 2w(1T-o)+2c(o- 07- 41/1?‘ H-c 4% 2.11? - 21r-JF‘—c.‘ ...
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