Question 7.4: At a point on the surface of a pressurized cylinder, the mat...
At a point on the surface of a pressurized cylinder, the material is subjected to biaxial stresses \sigma_{x} = 90 MPa and \sigma_{y} = 20 MPa, as shown on the stress element of Fig. 7-17a.
Using Mohr’s circle, determine the stresses acting on an element inclined at an angle θ = 30° . (Consider only the in-plane stresses, and show the results on a sketch of a properly oriented element.)
Construction of Mohr’s circle. We begin by setting up the axes for the normal and shear stresses, with \sigma_{x_{1}} positive to the right and \tau_{x_{1}y_{1}} positive downward, as shown in Fig. 7-17b. Then we place the center C of the circle 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{90_\ MPa+20_\ MPa}{2}=55_\ MPaPoint A, representing the stresses on the x face of the element (θ = 0), has coordinates
\sigma_{x_{1}}=90_\ MPa \tau_{x_{1}y_{1}}=0
Similarly, the coordinates of point B, representing the stresses on the y face (θ = 90°), are
\sigma_{x_{1}}=20_\ MPa \tau_{x_{1}y_{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{\sigma_{x}-\sigma_{y}}{2}\right)^{2}+\tau^{2}_{xy}}=\sqrt{\left(\frac{90_\ MPa-20_\ MPa}{2}\right)^{2}+0 }=35_\ MPaStresses 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) \sigma_{x_{1}}=\sigma_{aver}+R\cos 60^{\circ}
= 55 MPa + (35 MPa)(cos 60°) = 72.5 MPa
\tau_{x_{1}y_{1}}=-R\sin 60^{\circ}=-(35_\ MPa)(\sin 60^{\circ})=-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′) \sigma_{x_{1}}=\sigma_{aver}-R\cos 60^{\circ}
= 55 MPa – (35 MPa)(cos 60°) = 37.5 MPa
\tau_{x_{1}y_{1}}=R\sin 60^{\circ}=(35_\ MPa)(\sin 60^{\circ})=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 \sigma_{x}+\sigma_{y}, or 110 MPa.
