Essentials Of Materials Science And Engineering, Si Edition
Essentials Of Materials Science And Engineering, Si Edition
4th Edition
ISBN: 9781337629157
Author: Donald R. Askeland, Wendelin J. Wright
Publisher: Cengage Learning
Question
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Chapter 4, Problem 4.42P
Interpretation Introduction

Interpretation:

The resolved shear stress acting on the (111) slip plane in the [1¯10] and [101¯] slip directions needs to be calculated. The slip system(s) that become active first needs to be identified.

Concept Introduction:

The resolved shear stress operating on a slip direction/plane is given by π=σCosλcosϕ

The expression for resolved shear stress is as follows:

  τr=σcosλcosφ

Here, σ is the applied stress, λ is the angle between slip direction and applied force, and φ is the angle between the normal to the slip plane and the applied force.

Expert Solution & Answer
Check Mark

Answer to Problem 4.42P

The resolved shear stress acting on the slip plane (111) in the [110] direction is zero.

Explanation of Solution

Considering the slip occurring on the (111) plane in a direction of [1¯10].

The angle between the applied stress direction [001] and normal to the (111) plane can be calculated as follows:

  φ=cos1[u1u2+v1v2+w1w2( u 1 2 + v 1 2 + w 1 2 )( u 2 2 + v 2 2 + w 2 2 )]

Here, u1,v1,w1 are the components of direction of applied stress and u2,v2,w2 are the components of slip plane.

Substitute 0 for u1,0 for v1,1 for w1,1for u2,1 for v2,and1 for w2.

  φ=cos1[ u 1 u 2+ v 1 v 2+ w 1 w 2 ( u 1 2 + v 1 2 + w 1 2 )( u 2 2 + v 2 2 + w 2 2 )]=cos1[0+0+1 ( 0 2 + 0 2 + 1 2 )( 1 2 + 1 2 + 1 2 )]=54.73o

Find the angle between the direction of the applied stress [001] and normal to the [1¯10] plane.

  λ=cos1[u1u2+v1v2+w1w2( u 1 2 + v 1 2 + w 1 2 )( u 2 2 + v 2 2 + w 2 2 )]..............(1)

Here, u1,v1,w1 are the components of direction of applied stress and u2,v2,w2 are the components of slip plane.

Substitute 0 for u1,0 for v1,1 for w1,1for u2,1 for v2,and0 for w2.

  λ=cos1[ u 1 u 2+ v 1 v 2+ w 1 w 2 ( u 1 2 + v 1 2 + w 1 2 )( u 2 2 + v 2 2 + w 2 2 )]=cos1[0 ( 0 2 + 0 2 + 1 2 )( ( 1 ) 2 + 1 2 + 0 2 )]=90o

Find the resolved shear stress acting on the (111) plane.

  τr=σcosλcosφ.......(2)

Here, σ is the applied stress, λ is the angle between slip direction and applied force, and φ is the angle between the normal to the slip plane and the applied force.

Substitute 5000psi for σ,90o for λand 54.73ofor φ.

  τr=σcosλcosφ=5000×cos90×cos54.73=0

Therefore, the resolved shear stress acting on the slip plane (111) in the [1¯10] direction is 0.

Find the Schmidt factor.

  cosλcosφ=cos90cos54.73=0

Now, considering the slip occurring on the (111) plane in a direction [101¯].

The angle between the applied stress [001] direction and the slip direction [101¯] can be calculated as follows:

  λ=cos1[u1u2+v1v2+w1w2( u 1 2 + v 1 2 + w 1 2 )( u 2 2 + v 2 2 + w 2 2 )]

Substitute 0 for u1,0 for v1,1 for w1,1for u2,0 for v2,and1 for w2 in equation (1).

  λ=cos1[ u 1 u 2+ v 1 v 2+ w 1 w 2 ( u 1 2 + v 1 2 + w 1 2 )( u 2 2 + v 2 2 + w 2 2 )]=cos1[0+01 ( 0 2 + 0 2 + 1 2 )( ( 1 ) 2 + 0 2 + ( 1 ) 2 )]=135o

Substitute 5000psi for σ, 135o for λand 54.73ofor φ in equation (2).

  τr=σcosλcosφ=5000×cos135×cos54.73=2041.5psi ]

Therefore, the resolved shear stress acting on the slip plane (111) in the [101¯] direction is 2041.5psi.

Find the Schmidt factor.

  cosλcosφ=cos135cos54.73=0.408

The Schmidt factor is more for (111) plane in a [101¯] direction. Therefore, the slip system (111),[101¯] will become active first.

Conclusion

The resolved shear stress acting on the slip plane (111) in the [110] direction is zero.

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