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 σ_x = -50 MPa, σ_y = 10 MPa, and τ_{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 σ_{x_1} positive to the right and τ_{x_1y_1} positive downward. The center C of the circle is located on the σ_{x_1} axis at the point where the stress equals the average normal stress (Eq. 7-31a):
σ_{\text{aver}}=\frac{σ_x+σ_y}{2}=\frac{-50\ MPa+10\ MPa}{2} = -20 MPa
Point A, representing the stresses on the x face of the element (θ = 0), has coordinates
σ_{x_1} = -50 MPa τ_{x_1y_1} = -40 MPa
Similarly, the coordinates of point B, representing the stresses on the y face (θ = 90°), are
σ_{x_1} = 10 MPa τ_{x_1y_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{σ_x\ -\ σ_y}{2}\right)^2+τ_{xy}^2}\\ \quad =\sqrt{\left(\frac{-50\ MPa\ -\ 10\ MPa}{2}\right)^2+(-40\ MPa)^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 σ_{x_1} axis (that is, angle DCP_2), which in turn requires that we know the angle between line CA and the negative σ_{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°\ -\ \overline{ACP_2} = 90° – 53.13° = 36.87°
Knowing these angles, we can obtain the coordinates of point D directly from the figure:
(Point D) σ_{x_1} = -20 MPa – (50 MPa)(cos 36.87°) = -60 psi
τ_{x_1y_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′) σ_{x_1} = -20 MPa + (50 MPa)(cos 36.87°) = 20 MPa
τ_{x_1y_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 σ_x+σ_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
σ_1 = -20 MPa + 50 MPa = 30 MPa
as seen by inspection of the circle. The angle 2θ_{p_1} to point P_1 from point A is the angle ACP_1 measured counterclockwise on the circle, that is,
\overline{ACP_1}=2θ_{p_1} = 53.13° + 180° = 233.13° θ_{p_1} = 116.6°
Thus, the plane of the algebraically larger principal stress is oriented at an angle θ_{p_1} = 116.6°.
The algebraically smaller principal stress (point P_2) is obtained from the circle in a similar manner:
σ_2 = -20 MPa – 50 MPa = -70 MPa
The angle 2θ_{p_2} to point P_2 on the circle is 53.13°; thus, the second principal plane is defined by the angle θ_{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 σ_x+σ_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
τ_{max} = 50 MPa
The angle ACS_1 from point A to point S_1 is 90° + 53.13° = 143.13°, and therefore the angle 2θ_{s_1} for point S_1 is
2θ_{s_1} = 143.13°
The corresponding angle θ_{s_1} to the plane of the maximum positive shear stress is one-half that value, or 2θ_{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 σ_{\text{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.
