HW 6-Sol - Homework #6 With = F F and E = , we can get = =...

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Homework #6 With A F = σ and ε = E , we can get EA F E = = . If GPa E Cu 110 = , then 3 39 . 1 ) ( 9 110 ) ( 3 1 . 19 3 2 . 15 44500 2 = × = = E Pa E m E E N EA F N F 62250 = , ) ( 5 67 . 5 2 3 5 . 8 2 2 m E E A = = π ) ( 1 . 1 GPa A F = = From the graph, we can approximately get 005 . 0 = corresponding to ) ( 1 . 1 GPa = ) ( 4 . 0 ) ( 80 005 . 0 0 mm mm l l = = = Δ (1) Take derivative with respect to r in order to obtain an expression for the force F () 1 2 d d d d d d + = + = = N n N r nB r A r B r r A r r E F (2) Set the derivative equal to zero and then solve for r ( 0 r = ). Therefore, we have ) 1 /( 1 0 n nB A r = (3) Take second derivative with respect to r , then we have ) 2 ( 3 1 2 ) 1 ( 2 d d d dF + + + + = = n N r B n n r A r nB r A r r (4) Substitute 0 r into the second derivative, and then we have ) 1 /( ) 2 ( ) 1 /( 2 ) 1 ( 2 d dF 0 n n n r nB A B n n nB A A r + + + = The expression where the modulus of elasticity is proportional
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(a) From the question, we can get strain ( ε ) and Young’s modulus ( E ). They are
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This note was uploaded on 12/10/2010 for the course MSEN 601 taught by Professor Zhang during the Spring '08 term at Texas A&M.

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HW 6-Sol - Homework #6 With = F F and E = , we can get = =...

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