Question 7.5: An element in plane stress at the surface of a large machine...
An element in plane stress at the surface of a large machine is subjected to stresses \sigma_{x} = 15,000 psi, \sigma_{y} = 5,000 psi, and \tau_{xy} = 4,000 psi, as shown in Fig.7-19a.
Using Mohr’s circle, determine the following quantities: (a) the stresses acting on an element inclined at an angle θ = 40° , (b) the principal stresses, and (c) the maximum shear stresses. (Consider only the in-plane stresses, and show all results on sketches of properly oriented elements.)
Construction of Mohr’s circle. The first step in the solution is to set up the axes for Mohr’s circle, with \sigma_{x_{1}} positive to the right and \tau_{x_{1}y_{1}} positive downward (Fig. 7-19b). The center C of the circle is located on the \sigma_{x_{1}} axis at the point where \sigma_{x_{1}} equals the average normal stress (Eq. 7-31a):
\sigma_{aver}=\frac{\sigma_{x}+\sigma_{y}}{2}=\frac{15,000 psi+5,000 psi}{2}=10,000 psiPoint A, representing the stresses on the x face of the element (θ = 0), has coordinates
\sigma_{x_{1}}=15,000 psi \tau_{x_{1}y_{1}}=4,000 psiSimilarly, the coordinates of point B, representing the stresses on the y face (θ = 90°) are
\sigma_{x_{1}}=5,000 psi \tau_{x_{1}y_{1}}=-4,000 psiThe circle is now drawn through points A and B with center at C. The radius of the circle, from Eq. (7-31b), is
R=\sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right)^{2}+\tau^{2}_{xy}}=\sqrt{\left(\frac{15,000 psi-5,000 psi}{2}\right)^{2}+\left(4,000 psi\right)^{2}}=6,403 psi
(a) Stresses on an element inclined at θ = 40° . The stresses acting on a plane oriented at an angle θ = 40° are given by the coordinates of point D, which is at an angle 2θ = 80° from point A (Fig. 7-19b). To evaluate these coordinates, we need to know the angle between line CD and the \sigma_{x_{1}} axis (that is, angle DCP_{1}), which in turn requires that we know the angle between line CA and the \sigma_{x_{1}} axis (angle ACP_{1}). These angles are found from the geometry of the circle, as follows:
\tan \overline{ACP_{1}}=\frac{4,000 psi}{5,000 psi}=0.8 \overline{ACP_{1}}=38.66°\overline{DCP_{1}}=80^{\circ}-\overline{ACP_{1}}=80^{\circ}-38.66^{\circ}=41.34^{\circ}
Knowing these angles, we can determine the coordinates of point D directly from the figure:
(Point D) \sigma_{x_{1}} = 10,000 psi + (6,403 psi)(cos 41.34°) = 14,810 psi
\tau_{x_{1}y_{1}} = -(6,403 psi)(sin 41.34°) = -4,230 psi
In an analogous manner, we can find the stresses represented by point D′, which corresponds to a plane inclined at an angle θ = 130° (or 2θ = 260°):
(Point D′) \sigma_{x_{1}} = 10,000 psi – (6,403 psi)(cos 41.34°) = 5,190 psi
\tau_{x_{1}y_{1}} = (6,403 psi)(sin 41.34°) = 4,230 psi
These stresses are shown in Fig. 7-20a on a sketch of an element oriented at an angle θ = 40° (all stresses are shown in their true directions). Also, note that the sum of the normal stresses is equal to \sigma_{x} + \sigma_{y}, or 20,000 psi.
(b) Principal stresses. The principal stresses are represented by points P_{1} and P_{2} on Mohr’s circle (Fig. 7-19b). The algebraically larger principal stress (point P_{1}) is
\sigma_{1} = 10,000 psi + 6,400 psi = 16,400 psi
as seen by inspection of the circle. The angle 2\theta_{p_{1}} to point P_{1} from point A is the angle ACP_{1} on the circle, that is,
\overline{ACP_{1}}=2\theta_{p_{1}} = 38.66° \theta_{p_{1}} = 19.3°
Thus, the plane of the algebraically larger principal stress is oriented at an angle \theta_{p_{1}} = 19.3° , as shown in Fig. 7-20b.
The algebraically smaller principal stress (represented by point P_{2}) is obtained from the circle in a similar manner:
\sigma_{2} = 10,000 psi – 6,400 psi = 3,600 psi
The angle 2\theta_{p_{2}} to point P_{2} on the circle is 38.66° + 180° = 218.66° ; thus, the second principal plane is defined by the angle \theta_{p_{2}} = 109.3° . The principal stresses and principal planes are shown in Fig. 7-20b, and again we note that the sum of the normal stresses is equal to 20,000 psi.
(c) Maximum shear stresses. The maximum shear stresses are represented by points S_{1} and S_{2} on Mohr’s circle; therefore, the maximum in-plane shear stress (equal to the radius of the circle) is
The angle ACS_{1} from point A to point S_{1} is 90° – 38.66° = 51.34° , and therefore the angle 2\theta_{s_{1}} for point S_{1} is
2\theta_{s_{1}} = -51.34°
This angle is negative because it is measured clockwise on the circle. The corresponding angle \theta_{s_{1}} to the plane of the maximum positive shear stress is one-half that value, or \theta_{s_{1}} = -25.7° , as shown in Figs. 7-19b and 7-20c. The maximum negative shear stress (point S_{2} on the circle) has the same numerical value as the maximum positive stress (6,400 psi).
The normal stresses acting on the planes of maximum shear stress are equal to \sigma_{aver}, which is the abscissa of the center C of the circle (10,000 psi). These stresses are also shown in Fig. 7-20c. Note that the planes of maximum shear stress are oriented at 45° to the principal planes.
