Stress Element


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


Stress Element Rotated 42o


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

     σx = -10 ksi
     σy = -20 ksi
     τxy = 30 ksi
     θ = 42o

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,

     τx´y´ = - [(σxy) sin 2θ] / 2 + τxy cos 2θ
           = -[(-10 -(-20)) sin84] / 2 + 30 cos84
           = -4.973 + 3.136 ksi

     τx'y' = -1.837 ksi

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