Question 7.1: An element in plane stress is subjected to stresses σx = 16,......

To determine the stresses acting on an inclined element, we will use the transformation equations (Eqs. 7-4a and 7-4b). From the given numerical data, we obtain the following values for substitution into those equations:

\frac{σ_x+σ_y}{2} = 11,000 psi          \frac{σ_x\ -\ σ_y}{2} = 5,000 psi                τ_{xy} = 4,000 psi

sin 2θ = sin 90° = 1        cos 2θ = cos 90° = 0

Substituting these values into Eqs. (7-4a) and (7-4b), we get

σ_{x_1}=\frac{σ_x+σ_y}{2}+\frac{σ_x\ -\ σ_y}{2}cos\ 2θ+τ_{xy}\ sin\ 2θ

= 11,000 psi + (5,000 psi)(0) + (4,000 psi)(1) = 15,000 psi

= -(5,000 psi)(1) + (4,000 psi)(0) = -5,000 psi

In addition, the stress σ_{y_1} may be obtained from Eq. (7-5):

σ_{y_1}=\frac{σ_x+σ_y}{2}+\frac{σ_x\ -\ σ_y}{2}cos\ 2θ\ -\ τ_{xy}\ sin\ 2θ

= 11,000 psi – (5,000 psi)(0) – (4,000 psi)(1) = 7,000 psi

From these results we can readily obtain the stresses acting on all sides of an element oriented at θ = 45°, as shown in Fig. 7-7b. The arrows show the true directions in which the stresses act. Note especially the directions of the shear stresses, all of which have the same magnitude. Also, observe that the sum of the normal stresses remains constant and equal to 22,000 psi (see Eq. 7-6).
The stresses shown in Fig. 7-7b represent the same intrinsic state of stress as do the stresses shown in Fig. 7-7a. However, the stresses have different values because the elements on which they act have different orientations.

σ_{x_1}+σ_{y_1}=σ_{x}+σ_{y}                    (7-6)