Question 7.4: At a point on the surface of a pressurized cylinder, the mat......

Construction of Mohr’s circle. We begin by setting up the axes for the normal and shear stresses, with σ_{x_1} positive to the right and τ_{x_1y_1} positive downward, as shown in Fig. 7-17b. Then we place the center C of the circle 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{90\ MPa+20\ MPa}{2} = 55 MPa

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

σ_{x_1} = 90 MPa         τ_{x_1y_1} = 0

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

σ_{x_1} = 20 MPa         τ_{x_1y_1} = 0

Now we draw the circle through points A and B with center at C and radius R (see Eq. 7-31b) equal to

R=\sqrt{\left(\frac{σ_x\ -\ σ_y}{2}\right)+τ_{xy}^2}=\sqrt{\left(\frac{90\ MPa\ -\ 20\ MPa}{2}\right)+0} = 35 MPa

Stresses on an element inclined at θ = 30°. The stresses acting on a plane oriented at an angle θ = 30° are given by the coordinates of point D, which is at an angle 2θ = 60° from point A (Fig. 7-17b). By inspection of the circle, we see that the coordinates of point D are

(Point D)        σ_{x_1}=σ_{\text{aver}}+R\ cos\ 60°

= 55 MPa + (35 MPa)(cos 60°) = 72.5 MPa

τ_{x_1y_1} = -R sin 60° = -(35 MPa)(sin 60°) = -30.3 MPa

In a similar manner, we can find the stresses represented by point D′, which corresponds to an angle θ = 120° (or 2θ = 240°):

(Point D′)        σ_{x_1}=σ_{\text{aver}}+R\ cos\ 60°

= 55 MPa – (35 MPa)(cos 60°) = 37.5 MPa

τ_{x_1y_1} = -R sin 60° = (35 MPa)(sin 60°) = 30.3 MPa

These results are shown in Fig. 7-18 on a sketch of an element oriented at an angle θ = 30°, with all stresses shown in their true directions. Note that the sum of the normal stresses on the inclined element is equal to σ_x + σ_y, or 110 MPa.