Question 7.5: An element in plane stress at the surface of a large machine......

Construction of Mohr’s circle. The first step in the solution is to set up the axes for Mohr’s circle, with σ_{x_1} positive to the right and τ_{x_1y_1} positive downward (Fig. 7-19b). The center C of the circle is located on the σ_{x_1} axis at the point where σ_{x_1} equals the average normal stress (Eq. 7-31a):

σ_{\text{aver}}=\frac{σ_x+σ_y}{2}=\frac{15,000\ psi+5,000\ psi}{2} = 10,000 psi

Point A, representing the stresses on the x face of the element (θ = 0), has coordinates

σ_{x_1} = 15,000 psi        τ_{x_1y_1} = 4,000 psi

Similarly, the coordinates of point B, representing the stresses on the y face (θ = 90°). are

σ_{x_1} = 5,000 psi        τ_{x_1y_1} = -4,000 psi

The 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{σ_x\ -\ σ_y}{2}\right)^2+τ_{xy}^2}\\ \quad =\sqrt{\left(\frac{15,000\ psi\ -\ 5,000\ psi}{2}\right)^2+(4,000\ psi)^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 σ_{x_1} axis (that is, angle DCP_1), which in turn requires that we know the angle between line CA and the σ_{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°\ -\ \overline{ACP_1} = 80° – 38.66° = 41.34°

Knowing these angles, we can determine the coordinates of point D directly from the figure:

(Point D)        σ_{x_1} = 10,000 psi + (6,403 psi)(cos 41.34°) = 14,810 psi

τ_{x_1y_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′)        σ_{x_1} = 10,000 psi – (6,403 psi)(cos 41.34°) = 5,190 psi

τ_{x_1y_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 σ_x+σ_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

σ_{1} = 10,000 psi + 6,400 psi = 16,400 psi

as seen by inspection of the circle. The angle 2θ_{p_1} to point P_1 from point A is the angle ACP_1 on the circle, that is.

\overline{ACP_1}=2θ_{p_1} = 38.66°             θ_{p_1} = 19.3°

Thus, the plane of the algebraically larger principal stress is oriented at an angle θ_{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:

σ_2 = 10,000 psi – 6,400 psi = 3,600 psi

The angle 2θ_{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 θ_{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

τ_{max} = 6,400 psi

The angle ACS_1 from point A to point S_1 is 90° – 38.66° = 51.34°, and therefore the angle 2θ_{s_1} for point S_1 is

2θ_{s_1} = -51.34°

This angle is negative because it is measured clockwise on the circle. The corresponding angle θ_{s_1} to the plane of the maximum positive shear stress is one-half that value, or θ_{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 σ_{\text{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.