Consider a metal single crystal oriented such that the normal to the slip plane and the slip directions are at angles of 43.1° and 47.9°, respectively, with the tensile axis. if the critical resolved shear stress is 20.7 mpa (3,000 psi), will an applied stress of 45 mpa (6,500 psi)cause the single crystal to yield? if not, what stress will be necessary?

resolve shear stress = 22 MPa

Explanation:

Given data

slip plane α = 43.1°

slip directions β = 47.9°

shear stress = 20.7 MPa (3,000 psi)

applied stress =45 mPa (6,500 psi)

to find out

what stress will be necessary

solution

we know that

resolve shear stress = aplied stress × cosα ×  cosβ

resolve shear stress = 45 × cos(43.1) ×  cos(47.9)

resolve shear stress = 22 MPa

we can say that here single cristal will be yield

because resolve shear stress is bigger than critical shear stress

f=ma newton's second law 7x8=56n

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