practice_final_sol

# practice_final_sol - Course Code Practice Final Materials...

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Course Code: ____________ 1/8 pages total P.M. Anderson (614) 292-0176 (614) 292-1537• e-mail: [email protected] Practice Final Materials Science and Engineering 765 Introduction to Mechanical Behavior of Materials Spring Quarter, 2005 SOLUTIONS Instructions: Please Read Before Beginning! You have1 hour and 48 minutes to answer all 6 problems on this exam. You may use a calculator and an 8.5" by 11" double-sided sheet of notes in your own handwriting. Write all answers on the question sheet in the space provided. Continue on the reverse side of the sheet if more space is needed. Good Luck! 1. Interaction Between Dislocations Determine the glide and climb forces exerted on an edge dislocation that lies along the z- direction, and is located at a position (x, y) from a parallel screw dislocation. (Note: if you did not write down the stress state produced by dislocations, then you may borrow a textbook from me). We use the diagram below and note that the glide and climb forces on the edge dislocation are given by: F glide edge L = F x edge L = σ xy screw (x,y)b edge F climb edge L = F y edge L = −σ xx screw edge Since the screw dislocation does not produce components σ xy and σ xx , the glide and climb forces on the edge dislocation are both zero.

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2/8 pages total 2. Crystal Slip Systems (a) List three slip systems that are in the family of <110> slip directions and {111} slip planes. We can take several possibilities here. For the sake of an example, we take: n (1) as {1 1 1}/sqrt(3), s (1) as {1 0 -1}/sqrt(2) n (2) as {1 1 1}/sqrt(3), s (2) as {1 -1 0}/sqrt(2) n (3) as {1 -1 1}/sqrt(3), s (3) as {1 1 0}/sqrt(2) (b) What direct plastic strain in the 1-direction would be produced if a single crystal of volume 1mm 3 deformed by having each of the systems you mention above slip together, so that the slipped area for each of the systems is 1mm 2 . Assume that the Burgers vector magnitude is b = 2.5 10 -10 m We realize that: ε ij P = ε ij P(1) + ε ij P(2) + ε ij P(3) where ε ij P( α ) = A s ( α ) b ( α ) 2V n i s j + n j s i ( ) ( α ) If we insert the values mentioned above and set i = 1 and j = 1, then: ε ij P = 2.5 × 10 7
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practice_final_sol - Course Code Practice Final Materials...

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