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Unformatted text preview: Materials Science and Engineering 765
mechanical behavior of materials
Spring 2005 Homework 3 Solutions (updated May10)
Due: On Tuesday, May 3, 2005 at 5pm. Please show all work necessary to arrive at your answers. Slide under my office door (345 Fontana). 1. Construction of a Burgers circuit Using the SF/RH (note the corrected typo in green) rule discussed in class, determine the Burgers vector and line sense of the dislocation in the figure below (Figure 4.7 in McClintock and Argon). As part of your answer, submit a copy of the figure showing your circuit. b We choose the dislocation line direction to point out of the plane of the paper and then construct a righthanded circuit as shown above. The circuit reveals a Burgers vector b pointing from left to right. A square or other circuit reveals the same b. The only requirement is that the circuit would be closed in a perfect crystal. 2. Independent slip systems In class, we derived an expression for the average plastic strain produced when a dislocation with slip plane normal n and Burgers vector b slips an area As of a crystal with volume V. A fcc crystal with {111} type slip planes and <110> slip directions has 12 distinct slip systems (4 {111} slip planes and 3 <110> slip directions in each plane). A set of slip systems is said to be independent when the plastic strain produced by any one slip system in the set cannot be reproduced by any combination of slip on the others in the set. Determine whether the following set of slip planes/slip directions is independent: (111)/ [ 1 10] , (1 11) /[110], (11 1) /[011] We use the formula for the average plastic strain induced by motion of a dislocation with slip plane normal (note sqrt(6) factors were inserted into the eqs below): 1/3 Anderson G4:Users:peterand:Central files:Courses:MSE 765:2005:hw3sol.doc bA s si n j + s j n i 2V The first slip system has n1 = n2= n3 = 1/ 3 and s1 = 1/ 2 , s2 = 1/ 2 , and s3 = 0. Inserting these into the formula gives bA (1) 2 0 1 p(1) s ij = 0 2 1 2V 6 1 1 0 Similarly, the plastic strains produced by operation of slip systems 2 and 3 are bA (2) 2 0 1 bA (3) 0 1 1 p(2) p(3) s s ij = 0 2 1 and ij = 1 2 0 2V 6 1 0 2 2V 6 1 1 0 It is clear that these three slip systems are independent since no linear combination of systems (2) and (3) can produce all of the proper strain components from slip system (1). For example, selecting A (2) = A (3) =1 and adding the plastic strain components s s from these systems will give the following strain state b 2 1 2 p(2) p(3) 1 0 1 ij + ij = (3) A A s =1 2V 6 2 s(2) =1 1 2 p(1) p(1) p(1) Although this combination reproduces 11 and 23 = 32 with A (1) = 1, it fails to s reproduce the other components properly. Therefore, these slip systems are independent. 3. Interaction between an edge and screw dislocation Consider two parallel dislocations. The character of one pure edge and the other is is pure screw. Does one dislocation exert a force on the other? Explain why or why not using the known elastic fields for dislocations. The dislocations do not exert any force on each other, since the PeachKoehler force exerted by one dislocation on the other is zero. Consider a "positive" edge dislocation with along the zdirection and b along the xdirection. A glide component of the PeachKoehler force is exerted on the dislocation provided a nonzero value of xy is present, and a climb component is exerted provided a nonzero value of xx is present. This can be confirmed by using the PeachKoehler formula Fk =  ijk i b l jl L Therefore, the positive edge dislocation has a glide force F1 =  3 j1 3 b1 21 = 21b L and a climb force F2 =  3 j 2 3 b1 11 =  11b L The stress field solution for a screw dislocation provides neither of these (only xz = zx and yz = zy are produced). Therefore, a parallel screw dislocation does not exert a force on an edge dislocation. The same conclusion is obtained by considering whether an edge dislocation exerts any force on a screw dislocation.
p ij = ( ) ( ) ( ) ( ) ( ) ( ) 2/3
Anderson G4:Users:peterand:Central files:Courses:MSE 765:2005:hw3sol.doc 4. HallPetch relation for yield strength as a function of grain size (Note: you do not need to hand this problem in for hmk 3.) In class, we derived a relation for the force per unit length acting on an obstacle such appl as a grain boundary, in terms of the uniform applied shear stress nb on the active
s slip plane, the critical local shear stress nb needed to operate a source in the center p of the grain, the Peierls stress nb needed to glide dislocations on the slip plane, grain size D, and isotropic elastic properties and : 2 Fobs appl p appl p p 2 (1 )D s = nb  nb  nb  nb nb  nb L s a. In the limit of sufficiently large grain size, (Fobs /L)(/D) << ( nb ) 2 . Solve for the yield strength, which is taken as the critical applied stress to push the pileup past the obstacle in this limit. Based on your answer, discuss what controls yield strength for polycrystals with such a large grain size. s b. In the limit of sufficiently small grain size, (Fobs /L)(/D) >> ( nb ) 2 . Solve for the yield strength (i.e., critical applied stress to push the pileup past the obstacle) in this limit. Based on your answer, what dependence does yield strength have on grain size? c. For case (b), what material/physical parameter(s) might be manipulated to make yield strength very dependent on grain size? ( ) ( )( ) 3/3
Anderson G4:Users:peterand:Central files:Courses:MSE 765:2005:hw3sol.doc ...
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This note was uploaded on 07/17/2008 for the course MSE 765 taught by Professor Anderson during the Spring '05 term at Ohio State.
 Spring '05
 Anderson
 Materials Science And Engineering

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