Solution Set 1

Solution Set 1 - 1-1For hot rolled C11000 copper alloy E =...

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Unformatted text preview: 1-1For hot rolled C11000 copper alloy: E = 115 GPa = 115,000 MPa, v= 0.33 ( )2422oom1084.2m019.4d4A-×=π=π=, MPa69.156mN1069.156m1084.2N500,44AF2624o=×=×==σ-(a) 001363.MPa000,115MPa69.156E==σ=εor 0.1363% (b)σT= σ(1 + ε) = (156.69 MPa)(1 + 0.001363) = 156.90 MPa (c)εT= ln(1 + ε) = ln(1 + 0.001363) = 0.001362 (d)Δl= εlo= (0.001363)(1000 mm) = 1.36 mm (e)εtranverse= -vεlongitudinal= -v(0.33)(0.001363) = -0.000450 = 1919ddddiooi-=-→di= 18.99 mm Change in cross sectional area = Ai– Ao= ( )2o2i2o2idd4d4d4-=-πππ( )2222mm30.mm1999.184-=-=π(area reduction) 1-2θ= 90o-30o= 60osince θis the angle between the applied force and the normal to the plane σ’ = σcos2θ= (156.69 MPa)(cos260o) = 39.17MPa τ’ = σsinθcosθ= (156.69 MPa)(sin60o)(cos60o) = 67.85 MPa 1-3(a) ( )2422oom1013.1m012.4d4A-×=π=π=σ= F/Ao→F = σAo= (345 ×106N/m2)(1.13 ×10-4m2) = 38,985 N (b) 00335....
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Solution Set 1 - 1-1For hot rolled C11000 copper alloy E =...

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