Essentials of Materials Science and Engineering, SI Edition
Essentials of Materials Science and Engineering, SI Edition
4th Edition
ISBN: 9781337672078
Author: ASKELAND, Donald R., WRIGHT, Wendelin J.
Publisher: Cengage Learning
Question
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Chapter 4, Problem 4.28P
Interpretation Introduction

Interpretation:

The interplanar spacing, the length of the Burgers vector and the ratio between the shear stresses required for slip for the two systems should be determined.

Concept introduction:

The formula used is:

  b=(12)(a0)(A˙)

Here, a0 and A˙ are lattice parameters.

Expert Solution & Answer
Check Mark

Answer to Problem 4.28P

The length of Burgers vector for slip system (111)/[11¯0] in FCC aluminum is 2.863A.

The inter planar spacing for slip system of plane (111) in FCC aluminum is 2.338A.

The length of Burgers vector for slip systems (110)/[11¯1] in FCC aluminum is 7.014A.

The inter planar spacing for slip system of plane (110) in FCC aluminum is 2.863A.

The ratio between the interplanar spacing and length of Burgers vector of the slip system

  of 111/[110] Is 0.8166

The ratio between the interplanar spacing and length of Burgers vector of the slip system

  of 110/[111] Is 0.408.

The ratio of shear stresses required for two slip systems is 0.44.

Explanation of Solution

Plane: (110) and direction [111]

The length of Burgers vector for slip system (111)/[11¯0] in FCC aluminum is calculated as:

  b=(12)(a0)(A˙)

Here, a0 and A˙ are lattice parameter.

The lattice parameter a0 and A˙ of aluminum is 2 and 4.04958 .

Substitute 2 for a0 and 4.04958 for A˙

  b=(12)(2)(4.04958)=2.863A˙

Hence, the length of Burgers vector for slip system (111)/[11¯0] in FCC aluminum is 2.863A.

The inter planar spacing for slip system of plane (111) in FCC aluminum is calculated as:

  dhkl=A˙h2+k2+l2

Here, Miller indices of adjacent planes are h,k, and l.

Substituting 1 for h, 1 for k, 1for l, and 4.04958 for A˙

  d111=4.04958A˙ 1 2 + 1 2 + 1 2 =2.338A˙

Hence, the inter planar spacing for slip system of plane (111) in FCC aluminum is 2.338A.

The length of Burgers vector for slip system (110)/[11¯1] in FCC aluminum is calculated as:

  b=(a0)(A˙)

Here, a0 and A˙ are lattice parameter.

The lattice parameter a0 and A˙ of aluminum as 3 and 4.04958.

Substituting 3 for a0and 4.04958for A˙.

  b=(3)(4.04958)=7.014A˙

Hence, the length of Burgers vector for slip systems (110)/[11¯1] in FCC aluminum is 7.014A.

The inter planar spacing for slip system of plane (110) in FCC aluminum is calculated as:

  dhkl=A˙h2+k2+l2

Here, Miller indices of adjacent planes are h,k and l.

Substituting 1 for h,1 for k,0for l,and4.04958 for A˙.

  d110=4.04958A˙ 1 2 + 1 2 + 0 2 =2.863A˙

Hence, the inter planar spacing for slip system of plane (110) in FCC aluminum is 2.863A˙.

The ratio between the inter planar spacing and length of Burgers vector for slip system of (111)/[11¯0] is calculated as:

  (d/b)(111)/[11¯0]=d111b

Substituting 2.338A˙ for d111 and 2.863A˙ for b

  (d/b)( 111)/[1 1 ¯0]=2.338A˙2.863A˙=0.8166

The ratio between the inter planar spacing and length of Burgers vector for slip system of (111)/[11¯0] is calculated as:

  (d/b)(110)/[11¯1]=d111b

Substitute 2.338A˙ for d111 and 2.863A˙ for b

  (d/b)( 110)/[1 1 ¯1]=2.863A˙7.014A˙=0.408

The ratio of shear stresses required for two slip system is calculated as:

  τ( 110)/[1 1 ¯1]τ( 110)/[1 1 ¯1]=exp[k( d/b)( 111)/[ 1 1 ¯ 0]]exp[k( d/b)( 110)/[ 1 1 ¯ 1]]

Substituting 2 for k,0.8166for (d/b)(111)/[11¯0], and 0.408 for (d/b)(110)/[11¯1]

  τ ( 111 )/[ 1 1 ¯ 0]τ ( 110 )/[ 1 1 ¯ 1]=exp[2( 0.8166)]exp[2( 0.408)]=0.44

Hence, the ratio of shear stresses required for two slip systems is 0.44.

Conclusion

The length of Burgers vector for slip system (111)/[11¯0] in FCC aluminum is 2.863A.

The inter planar spacing for slip system of plane (111) in FCC aluminum is 2.338A.

The length of Burgers vector for slip systems (110)/[11¯1] in FCC aluminum is 7.014A.

The inter planar spacing for slip system of plane (110) in FCC aluminum is 2.863A.

The ratio between the interplanar spacing and length of Burgers vector of the slip system

  of 111/[110] Is 0.8166

The ratio between the interplanar spacing and length of Burgers vector of the slip system

  of 110/[111] Is 0.408.

The ratio of shear stresses required for two slip systems is 0.44.

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Manual solution only, no Al used

Chapter 4 Solutions

Essentials of Materials Science and Engineering, SI Edition

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