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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.

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Answer:

The answer is 0.4490Mpa

Explanation:

Given :

P in direction of 001 , P = 1.1MPa

slip   plane  =  111, slip plane normal direction of 111,

slip direction = 101

[tex]T_{R}[/tex] = δcosФcosλ

= (001 * 101) = 1 = [tex]\sqrt{2}[/tex]cosλ

(111 * 001) 1 [tex]\sqrt{3}[/tex]cosФ

= 1.1 MPa * [tex]\frac{1}{\sqrt{2} }[/tex] * [tex]\frac{1}{\sqrt{3} }[/tex]

1.1 Mpa * 1/1.4142 * 1/1.7320

1.1 Mpa * 0.7072 * 0.5773

= 0.4490 MPa

therefore the resolved shear stress = 0.4490MPa

Cricetus

The critical resolved shear stress will be "0.4490 MPa".

Stress and Strain

According to the question,

Tensile stress in direction of 001, P = 1.1 MPa

Slip plane = 111

Slip direction = 101

We know the relation,

→ [tex]T_R[/tex] = δCosΦ Cosλ

By substituting the values,

       = 001 × 101

       = 1

hence,

The critical resolved stress be:

= (111 × 001) × [tex]\frac{1}{\sqrt{2} }[/tex] × [tex]\frac{1}{\sqrt{3} }[/tex]

By putting the values,

= 1.1 × [tex]\frac{1}{1.4142 }[/tex] × [tex]\frac{1}{1.7320}[/tex]

= 1.1× 0.7072 × 0.5773

= 0.4490 MPa

Thus the above approach is correct.  

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