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Assignment_3_sol(3).doc

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Assignment #3 ENGR 2220 Structure & Properties of Materials To get part marks, show all your work 1. Consider a single crystal of silver oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a (111) plane and in a [1 01] direction, and is initiated at an applied tensile stress of 1.1 MPa (160 psi), compute the critical resolved shear stress.  Solution This problem asks that we compute the critical resolved shear stress for silver. In order to do this, w
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  ENGR 2220 Assignment #3Structure & Properties of Materials To get part marks, show all your work  1. Consider a single crystal of silver oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a ( 111 )  plane and in a [ 1 01 ]  direction, and is initiated at an applied tensile stress of 1.1 MPa (160 psi, co!pute thecritical resolved shear stress. SolutionThis problem asks that e compute the critical resol!e shear stress for sil!er. n or er to o this$ e must emplo%'uation .$ but first it is necessar% to sol!e for the angles λ  an φ  hich are shon in the sketch belo.The angle λ  is the angle beteen the tensile a*is+i.e.$ along the ,--1 irection+an the slip irection+i.e.$ [ 1 -1 ] .The angle λ  ma% be etermine using 'uation ./ as   λ=  cos − 1  u 1 u 0 +  v 1 v 0 +  w 1 w 0 u 10 +  v 10 +  w 10 () u 00 +  v 00 +  w 00 ()  here (for ,--1) u 1   - $   v 1   -$ 1   1$ an (for [ 1 -1 ] ) u 0   21$ v 0   -$ 0   1. Therefore$ λ  is e'ual to  λ= cos − 1 (0)( − 1) + (0)(0) + (1)(1)(0) 2 + (0) 2 + (1) 2 [] ( − 1) 2 + (0) 2 + (1) 2 []      =  cos − 1 10       = 3 ° 1 of   ENGR 2220 Assignment #3Structure & Properties of Materials 4urthermore$ φ  is the angle beteen the tensile a*is+the ,--1 irection+an the normal to the slip plane+i.e.$ the (111) plane5 for this case this normal is along a ,111 irection. Therefore$ again using 'uation ./   φ=  cos − 1  ( - )( 1 ) + (-)(1 ) +  ( 1 )( 1 )( - ) 0 +  ( - ) 0 +  ( 1 ) 0 []  ( 1 ) 0 +  ( 1 ) 0 +  ( 1 ) 0 []     = cos − 1 13       = 54.7 ° An $ finall%$ using 'uation .$ the critical resol!e shear stress is e'ual to   τ crss  σ  y (cos φ  cos λ )   = (1.1 MPa)cos(54.7 ° )cos(45 ° ) [] = (1.1MPa)13        12        =0.45 MPa (65.1 psi) 0 of   ENGR 2220 Assignment #3Structure & Properties of Materials 0. Consider a single crystal of so!e hypothetical !etal that has the #CC crystal structure and is oriented such that a tensile stress is applied along a .,  121  direction. If slip occurs on a ( 111 )  plane and in a [ 1 01 ]  direction, co!pute the stressat hich the crystal yields if its critical resolved shear stress is $.0 MPa. SolutionThis problem asks for us to etermine the tensile stress at hich a 466 metal %iel s hen the stress is applie along a .,  121   irection such that slip occurs on a (111) plane an in a [ 1 -1 ]   irection5 the critical resol!e shear stress for this metal is 3.0 MPa. To sol!e this problem e use 'uation .5 hoe!er it is first necessar% to etermine the!alues of φ  an λ . These eterminations are possible using 'uation ./. 7o$ λ  is the angle beteen .,  121  an   [ 1 -1 ]   irections. Therefore$ relati!e to 'uation ./ let us take u 1   21$ v 1   1$ an 1   0$ as ell as u 0   21$ v 0   -$an 0   1. This lea s to   λ=  cos − 1  u 1 u 0 +  v 1 v 0 +  w 1 w 0 u 10 +  v 10 +  w 10 () u 00 +  v 00 +  w 00 ()  [][] ++−++− ++−− =  − 000000 1 )1()-()1()0()1()1( )1)(0()-)(1()1)(1( cos       = 103cos λ   7o for the etermination of φ $ the normal to the (111) slip plane is the ,111 irection. Again using 'uation ./$ heree no take u 1   21$ v 1   -$ 1   0 (for .,  121 )$ an u 0   1$ v 0   1$ 0   1 (for ,111). Thus$ [][] ++++− ++−=  − 000000 1 )1()1()1()0()1()1( )1)(0()1)(1()1)(1( cos φ        = 180cos φ  t is no possible to compute the %iel stress (using 'uation .) as  MPa MPa crss  y  0.10180103-.coscos =            == λ φ τ  σ   3 of   ENGR 2220 Assignment #3Structure & Properties of Materials 3. % o previously undefor!ed cylindrical speci!ens of an alloy are to &e strain hardened &y reducing their cross'sectional areas ( hile !aintaining their circular cross sections. #or one speci!en, the initial and defor!ed radii are 1 !! and )!!, respectively. %he second speci!en, ith an initial radius of 1* !!, !ust have the sa!e defor!ed hardness as the first  speci!en+ co!pute the second speci!ens radius after defor!ation. Solutionn or er for these to c%lin rical specimens to ha!e the same eforme har ness$ the% must be eforme to thesame percent col ork. 4or the first specimen   %CW=  - 0 −  - d   - 0 ×  100= π r  02 −π r  d  2 π r  02 ×  100 96:31-- )18();()18(  000 ×ππ−π !!!!!! 4or the secon specimen$ the eforme ra ius is compute using the abo!e e'uation an sol!ing for r  d   as   r  d   r  - 1 −  % 6:1-- mm-./ 1--931mm)(10  C.   −  of 
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