Question 7.3: An element in plane stress is subjected to stresses σx = 12,...
An element in plane stress is subjected to stresses \sigma_{x} = 12,300 psi, \sigma_{y} = -4,200 psi, and \tau_{xy} = -4,700 psi, as shown in Fig. 7-13a.
(a) Determine the principal stresses and show them on a sketch of a properly oriented element.
(b) Determine the maximum shear stresses and show them on a sketch of a properly oriented element. (Consider only the in-plane stresses.)
(a) Principal stresses. The principal angles \theta_{p} that locate the principal planes can be obtained from Eq. (7-11):
\tan2\theta_{p}=\frac{2\tau_{xy}}{\sigma_{x}-\sigma_{y}}=\frac{2(-4,700 psi)}{12,300 psi-(-4,200 psi)}=-0.5697Solving for the angles, we get the following two sets of values:
2\theta_{p} = 150.3° and \theta_{p} = 75.2°
2\theta_{p} = 330.3° and \theta_{p} = 165.2°
The principal stresses may be obtained by substituting the two values of 2\theta_{p} into the transformation equation for \sigma_{x_{1}} (Eq. 7-4a). As a preliminary calculation, we determine the following quantities:
\frac{\sigma_{x}+\sigma_{y}}{2}=\frac{12,300 psi-4,200 psi}{2}=4,050 psi\frac{\sigma_{x}-\sigma_{y}}{2}=\frac{12,300 psi+4,200 psi}{2}= 8,250 psi
Now we substitute the first value of 2\theta_{p} into Eq. (7-4a) and obtain
\sigma_{x_{1}}=\frac{\sigma_{x}+\sigma_{y}}{2}+\frac{\sigma_{x}-\sigma_{y}}{2}\cos2\theta+\tau_{xy}\sin2\theta (7-4a)
= 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 2\theta_{p} and obtain \sigma_{x_{1}} = 13,540 psi. Thus, the principal stresses and their corresponding principal angles are
\sigma_{1} = 13,540 psi and \theta_{p_{1}} = 165.2°
\sigma_{2} = -5,440 psi and \theta_{p_{2}} = 75.2°
Note that \theta_{p_{1}} and \theta_{p_{2}} differ by 90° and that \sigma_{1}+\sigma_{2}=\sigma_{x}+\sigma_{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):
=4,050 psi\pm\sqrt{(8,250 psi)^{2}+(-4,700 psi)^{2}}
\sigma_{1,2} = 4,050 psi ± 9,490 psi
Therefore,
\sigma_{1} = 13,540 psi \sigma_{2} = -5,440 psi
The angle \theta_{p_{1}} to the plane on which \sigma_{1} acts is obtained from Eqs. (7-18a) and (7-18b):
\cos2\theta_{p_{1}}=\frac{\sigma_{x}-\sigma_{y}}{2R}=\frac{8,250 psi}{9,490 psi}=0.869\sin2\theta_{p_{1}}=\frac{\tau_{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 \sigma_{1} and \sigma_{2}.
R=\sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right)^{2}+\tau^{2}_{xy}} (7-12)
The only angle between 0 and 360° having the specified sine and cosine is 2\theta_{p_{1}} = 330.3° ; hence, \theta_{p_{1}} = 165.2° . This angle is associated with the algebraically larger principal stress \sigma_{1} = 13,540 psi. The other angle is 90° larger or smaller than \theta_{p_{1}} ; hence, \theta_{p_{2}} = 75.2° . This angle corresponds to the smaller principal stress \sigma_{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):
=\sqrt{(8,250 psi)^{2}+(-4,700 psi)^{2}}=9,490 psi
The angle \theta_{s_{1}} to the plane having the maximum positive shear stress is calculated from Eq. (7-24):
\theta_{s_{1}}=\theta_{p_{1}} – 45° = 165.2° – 45° = 120.2°
It follows that the maximum negative shear stress acts on the plane for which \theta_{s_{2}} = 120.2° – 90° = 30.2° .
The normal stresses acting on the planes of maximum shear stresses are calculated from Eq. (7-27):
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 \theta_{s}, and then we can use the second transformation equation (Eq. 7-4b) to obtain the corresponding shear stresses.
\tan2\theta_{s}=-\frac{\sigma_{x}-\sigma_{y}}{2\tau_{xy}} (7-20)
\tau_{x_{1}y_{1}}=-\frac{\sigma_{x}-\sigma_{y}}{2}\sin2\theta+\tau_{xy}\cos2\theta (7-4b)
