Question 7.3: An element in plane stress is subjected to stresses σx = 12,......

(a) Principal stresses. The principal angles θ_p that locate the principal planes can be obtained from Eq. (7- 11):

tan\ 2θ_p=\frac{2τ_{xy}}{σ_x\ -\ σ_y}=\frac{2(-4,700\ psi)}{12,300\ psi\ -\ (-4,200\ psi)} = -0.5697

Solving for the angles, we get the following two sets of values:

2θ_p = 150.3°         and        θ_p = 75.2°

2θ_p = 330.3°         and        θ_p = 165.2°

The principal stresses may be obtained by substituting the two values of 2θ_p into the transformation equation for σ_{x_1} (Eq. 7-4a). As a preliminary calculation, we determine the following quantities:

σ_{x_1}=\frac{σ_x+σ_y}{2}+\frac{σ_x\ -\ σ_y}{2}cos\ 2θ+τ_{xy}sin\ 2θ                   (7-4a)

\frac{σ_x+σ_y}{2}=\frac{12,300\ psi\ -\ 4,200\ psi}{2} = 4,050 psi

\frac{σ_x\ -\ σ_y}{2}=\frac{12,300\ psi\ +\ 4,200\ psi}{2} = 8,250 psi

Now we substitute the first value of 2θ_p into Eq. (7-4a) and obtain

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

= 4,050 psi + (8,250 psi)(cos 150.3°) – (4,700 psi)(sin 150.3°)
= – 5,440 psi

In a similar manner, we substitute the second value of 2θ_p and obtain σ_{x_1} = 13,540 psi. Thus, the principal stresses and their corresponding principal angles are

σ_1 = 13,540 psi           and          θ_{p_1} = 165.2°

σ_2 = -5,440 psi           and          θ_{p_2} = 75.2°

Note that θ_{p_1} and θ_{p_2} differ by 90° and that σ_1+σ_2=σ_x+σ_y. The principal stresses are shown on a properly oriented element in Fig. 7- 13b. Of course, no shear stresses act on the principal planes.
Alternative solution for the principal stresses. The principal stresses may also be calculated directly from Eq. (7-17):

σ_{1,2}=\frac{σ_x+σ_y}{2}±\sqrt{\left(\frac{σ_x\ -\ σ_y}{2}\right)^2+τ_{xy}^2}\\ =4,050\ psi±\sqrt{(8,250\ psi)^2 + (-4,700\ psi)^2}\\ σ_{1,2}=4,050\ psi\ ±\ 9,490\ psi

Therefore

σ_1 = 13,540 psi        σ_2 = -5,440 psi

The angle θ_{p_1} to the plane on which σ_{1} acts is obtained from Eqs. (7-18a) and (7-18b):

cos\ 2θ_{p_1}=\frac{σ_x\ -\ σ_y}{2R}=\frac{8,250\ psi}{9,490\ psi} = 0.869

sin\ 2θ_{p_1}=\frac{τ_{xy}}{R}=\frac{-4,700\ psi}{9,490\ psi} = -0.495

in which R is given by Eq. (7-12) and is equal to the square-root term in the preceding calculation for the principal stresses σ_1 and σ_2. The only angle between 0 and 360° having the specified sine and cosine is 2θ_{p_1} = 330.3°; hence, 2θ_{p_1} = 165.2°. This angle is associated with the algebraically larger principal stress σ_1 = 13.540 psi. The other angle is 90° larger or smaller than 2θ_{p_1}; hence, 2θ_{p_2} = 75.2°. This angle corresponds to the smaller principal stress σ_2 = -5,440 psi. Note that these results for the principal stresses and principal angles agree with those found previously.
(b) Maximum shear stresses. The maximum in-plane shear stresses are given by Eq. (7-25):

R=\sqrt{\left(\frac{σ_x\ -\ σ_y}{2}\right)^2+τ_{xy}^2}              (7-12)

τ_{max}=\sqrt{\left(\frac{σ_x\ -\ σ_y}{2}\right)^2+τ_{xy}^2}\\ =\sqrt{(8,250\ psi)^2+(-4,700\ psi)^2} = 9,490 psi

The angle θ_{s_1} to the plane having the maximum positive shear stress is calculated from Eq. (7-24):

θ_{s_1}=θ_{p_1}\ -\ 45° = 165.2° – 45° = 120.2°

It follows that the maximum negative shear stress acts on the plane for which θ_{s_2} = 120.2° – 90° = 30.2°.
The normal stresses acting on the planes of maximum shear stresses are calculated from Eq. (7-27):

θ_{\text{aver}}=\frac{σ_x\ +\ σ_y}{2} = 4,050 psi

Finally, the maximum shear stresses and associated normal stresses are shown on the stress element of Fig. 7- 13c.
As an alternative approach to finding the maximum shear stresses, we can use Eq. (7-20) to determine the two values of the angles θ_{s}. and then we can use the second transformation equation (Eq. 7-4b) to obtain the corresponding shear stresses.

tan\ 2θ_s=-\frac{σ_x\ -\ σ_y}{2τ_{xy}}                       (7-20)

τ_{x_1y_1}=-\frac{σ_x\ -\ σ_y}{2}sin\ 2θ+τ_{xy}cos\ 2θ            (7-4b)