4020-F2011-HW6Solutions - MSE 4020 HW6-SOLUTION 10/05/2011...

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1 MSE 4020 HW6-SOLUTION 10/05/2011 Problem 1 (a) Totally, there are 8 contact areas(see sketch above), which can be separate into two groups: Group 1: Areas 1~6; Group 2 : Areas 7~8; Focus on only one are in each group (say, area 3 and 7) For area 3 , it has a relative displacement, 2 ! where δ is the vertical displacement of the punch (see the sketch on the top right) and the contact area is l (see the sketch on the bottom right). Thus, work done by Group 1 areas is: 6 ! 2 " ! l ! k For area 7 , it has a relative displacement, 2 ! and the contact area is 1 2 3 4 5 6 7 8 δ δ 2 δ 2 l l 2 l Displacement Contact area
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2 ! l Thus, work done by Group 2 areas is: 2 ! 2 " ! 2 ! l ! k = 4 2 ! " ! l ! k Thus, total work is 6 ! 2 " ! l ! k + 4 ! 2 " ! l ! k = 10 2 " ! l ! k which should be equal to external work done by the punch. Note that the cross-sectional area of the punch is 2 ! l Thus, ! "#" 2 " l = 10 2 #" l " k $ ! = 10 " k = 5 ! yield (b) The flow field is sketched in big yellow arrows (c) Now we have totally 10 contact areas(see sketch below), which can be separate into two groups: Group 1: Areas 1~8; Group 2 : Areas 9~10; Similarly, focus on only one are in each group (say, area 3 and 9) For area 3 , it has a relative displacement, 2 ! 1
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This note was uploaded on 12/05/2011 for the course MSE 4020 at Cornell University (Engineering School).

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4020-F2011-HW6Solutions - MSE 4020 HW6-SOLUTION 10/05/2011...

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