**1 Answer**

written 6.1 years ago by |

1) Whenever a body is under tensile load or compressive load it experiences a direct stress. When the body under goes plastic deformation, the crystal structure gets deformed.

2) Due to this deformation layers begins to slip over each other and this slip occurs along slip plane.

3) As slipping action takes place the body experiences a shear stress along with tensile stress and the component of this shear stress along the slip plane is known as CRITICALLY RESOLVED SHEAR STRESS.

4) Consider a cylindrical body is under the tensile load, F is the applied force and A is the cross section area of it as shown in the image section area of it as shown in the image.

5) As is any slip area, N is the normal direction to slip plane and D is direction along slip plane, Ө be angle of normal with geometric axis of cylinder and Ф be angle between direction along slip plane and geometric axis.

6) Resolving force F along the plane FD = F cos Ф,

Area of slip plane be As = A/cosθ ,

The tensile stress on cylinder is σ = F/A and

CRSS be T_CRSS.

7) So now in vertical plane Ө = 90 degrees, so τ_CRSS = 0, and in horizontal plane Ф = 90 degrees so their also τ_CRSS = 0. Therefore, no shear stress occurs in vertical and horizontal plane, but it occurs in between both.

8) The maximum value of Ө = Ф = 45 degrees so the CRSS becomes now,

This is the maximum critically resolved shear stress for slip deformation phenomenon.