What is the shear stress on a plane 42^o from the horizontal (plane a-a)?

Stress Element Rotated 42^o

The stress state at other orientations can be determined using the stress rotation equations. This problems asks for the shear stress on a plane 42^o from the horizontal. The basic parameters are

     σ_x = -10 ksi
     σ_y = -20 ksi
     τ_xy = 30 ksi
     θ = 42^o

Notice, the two normal stress are negative since the arrows in the original problem diagram are pointing in the negative direction.

Using the shear stress rotation equation gives,

     τ_x´y´ = - [(σ_x -σ_y) sin 2θ] / 2 + τ_xy cos 2θ
           = -[(-10 -(-20)) sin84] / 2 + 30 cos84
           = -4.973 + 3.136 ksi

     τ_x'y' = -1.837 ksi