The answer to the question is sound
Explanation:
Given that,
Angle by the normal to the slip α= 60°
Angle by the slip direction with the tensile axis β= 35°
Shear stress = 6.2 MPa
Applied stress = 12 MPa
We need to calculate the shear stress applied at the slip plane
Using formula of shear stress

Put the value into the formula


Since, the shear stress applied at the slip plane is less than the critical resolved shear stress
So, The crystal will not yield.
Now, We need to calculate the applied stress necessary for the crystal to yield
Using formula of stress

Put the value into the formula


Hence, This is the required solution.
Answer:
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Answer:
I think the answer must be A.
That is the answer to the question
I hope this helps you.
Thank you for your question