Question 7.6: At a point on the surface of a generator shaft the stresses ...
At a point on the surface of a generator shaft the stresses are \sigma_{x} = -50 MPa, \sigma_{y} = 10 MPa, and \tau_{xy} = -40 MPa, as shown in Fig. 7-21a.
Using Mohr’s circle, determine the following quantities: (a) the stresses acting on an element inclined at an angle θ = 45° , (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 axes for the normal and shear stresses are shown in Fig. 7-21b, with \sigma_{x_{1}} positive to the right and \tau_{x_{1}y_{1}} positive downward. The center C of the circle is located on the \sigma_{x_{1}} axis at the point where the stress equals the average normal stress (Eq. 7-31a):
\sigma_{aver}=\frac{\sigma_{x}+\sigma_{y}}{2}=\frac{-50 MPa+10 MPa}{2}=-20 MPaPoint A, representing the stresses on the x face of the element (θ = 0), has coordinates
\sigma_{x_{1}} = -50 MPa \tau_{x_{1}y_{1}} = -40 MPa
Similarly, the coordinates of point B, representing the stresses on the y face (θ = 90°), are
\sigma_{x_{1}} = 10 MPa \tau_{x_{1}y_{1}} = 40 MPa
The circle is now drawn through points A and B with center at C and radius R (from Eq. 7-31b) equal to
R=\sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right)^{2}+\tau^{2}_{xy}}=\sqrt{\left(\frac{-50 MPa-10 MPa}{2}\right)^{2}+\left(-40 MPa\right)^{2}}=50 MPa
(a) Stresses on an element inclined at θ = 45° . The stresses acting on a plane oriented at an angle θ = 45° are given by the coordinates of point D, which is at an angle 2θ = 90° from point A (Fig. 7-21b). To evaluate these coordinates, we need to know the angle between line CD and the negative \sigma_{x_{1}} axis (that is, angle DCP_{2}), which in turn requires that we know the angle between line CA and the negative \sigma_{x_{1}} axis (angle ACP_{2}). These angles are found from the geometry of the circle as follows:
\tan \overline{ACP_{2}}=\frac{40 MPa}{30 MPa}=\frac{4}{3} \overline{ACP_{2}}=53.13°\overline{DCP_{2}}=90^{\circ}-\overline{ACP_{2}}=90^{\circ}-53.13^{\circ}=36.87^{\circ}
Knowing these angles, we can obtain the coordinates of point D directly from the figure:
(Point D) \sigma_{x_{1}} = -20 MPa – (50 MPa)(cos 36.87°) = -60 MPa
\tau_{x_{1}y_{1}} = (50 MPa)(sin 36.87°) = 30 MPa
In an analogous manner, we can find the stresses represented by point D′ , which corresponds to a plane inclined at an angle θ = 135° (or 2θ = 270°):
(Point D′) \sigma_{x_{1}} = -20 MPa + (50 MPa)(cos 36.87°) = 20 MPa
\tau_{x_{1}y_{1}} = (-50 MPa)(sin 36.87°) = -30 MPa
These stresses are shown in Fig. 7-22a on a sketch of an element oriented at an angle θ = 45° (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 -40 MPa.
(b) Principal stresses. The principal stresses are represented by points P_{1} and P_{2} on Mohr’s circle. The algebraically larger principal stress (represented by point P_{1}) is
\sigma_{1} = -20 MPa + 50 MPa = 30 MPa
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} measured counterclockwise on the circle, that is,
\overline{ACP_{1}}=2\theta_{p_{1}} = 53.13° + 180° = 233.13° \theta_{p_{1}} = 116.6°
Thus, the plane of the algebraically larger principal stress is oriented at an angle \theta_{p_{1}} = 116.6° .
The algebraically smaller principal stress (point P_{2}) is obtained from the circle in a similar manner:
\sigma_{2} = -20 MPa – 50 MPa = -70 MPa
The angle 2\theta_{p_{2}} to point P_{2} on the circle is 53.13° ; thus, the second principal plane is defined by the angle \theta_{p_{2}} = 26.6° .
The principal stresses and principal planes are shown in Fig. 7-22b, and again we note that the sum of the normal stresses is equal to \sigma_{x} + \sigma_{y}, or -40 MPa.
(c) Maximum shear stresses. The maximum positive and negative shear stresses are represented by points S_{1} and S_{2} on Mohr’s circle (Fig. 7-21b). Their magnitudes, equal to the radius of the circle, are
\tau_{\max} = 50 MPa
The angle ACS_{1} from point A to point S_{1} is 90° + 53.13° = 143.13° , and therefore the angle 2\theta_{s_{1}} for point S_{1} is
2\theta_{s_{1}} = 143.13°
The corresponding angle \theta_{s_{1}} to the plane of the maximum positive shear stress is one-half that value, or \theta_{s_{1}} = 71.6° , as shown in Fig. 7-22c. The maximum negative shear stress (point S_{2} on the circle) has the same numerical value as the positive stress (50 MPa).
The normal stresses acting on the planes of maximum shear stress are equal to \sigma_{aver}, which is the coordinate of the center C of the circle (-20 MPa). These stresses are also shown in Fig. 7-22c. Note that the planes of maximum shear stress are oriented at 45° to the principal planes.
