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.

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

= δcosФcosλ

= (001 * 101) = 1 = cosλ

(111 * 001) 1 cosФ

= 1.1 MPa * *

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

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