Objectives:

When you finish this module (especially the appropriate textbook sections and problems), you will be able to:
· define the principal planes of stress at a point on a structural element
· use Mohr's circle to find maximum normal and shearing stresses
· use yield criteria for ductile materials to predict safety
· apply the analysis of plane stress to thin-walled pressure vessels

Reference: §7.1 - 7.9 of Mechanics of Materials, 3rd Edition, Beer, Johnston, & DeWolf, McGraw-Hill

Definitions and Concepts: Suppose a structural member is under stress. At a point on or inside the material, the normal and shearing stresses can be computed by the methods presented in earlier chapters. For example, the normal stress may be a sum of the effects of a tensile force and a bending moment. The previous chapter (module) explained how to compute the horizontal shearing stress; you already knew how to obtain the vertical shearing stress from your Statics course and from the first chapter of the textbook. And section 1.11, page 23, explained that the shearing stress reaches a maximum at 45^o when the forces are purely normal forces. But it's more complicated when the forces are not purely normal; now we consider what the stresses are if a small cubic element in the material is rotated. Section 7.2, starting on page 425, develops the equations for the normal and shearing stress for any such planar angle of rotation:

It is not too difficult to show (please see this explanation) that the equations above are the parametric equations of a circle (see equations 7.9 and 7.10 on page 428), with radius and position given as:

Please look at Figures 7.9 and 7.10 on page 428. (your author, as usual, uses the incorrect abbreviation "ave" for average) is the position of the center of the circle from the origin. The axes are not x and y, but .

When these are plotted (page 428, figures 7.9 & 7.10), the circle's center C is from the origin. Since the radius R can be calculated from the formula, point A may be found by adding  and R; point B is found by subtracting R from . Note that these points represent the maximum and minimum values of the normal stress, and a zero value for the shear. At some rotation angleon the circle shown by point M (representing an actual rotation angle of of the element), different values of the normal and shearing stresses apply; your authors use primes to distinguish these from the other values. At point D, the shearing stress is a maximum and the two normal stresses are equal. You should not continue unless you understand these statements!

It's probably best to look at an example before considering any more theory. In Example 7.01 on page 431, two normal stresses and two shearing stresses are given (in practice, you would determine these by the methods of the previous chapters). Carefully note the signs! The 10 MPa is a compression! The two shearing stresses (on adjacent faces of the element) have to be equal and are positive in the directions shown; if you don't know why, return to the earlier discussions of shearing stresses on page 26.

Let's examine the values that are given to see what can be deduced from them. Since the normal stresses are not equal (50 MPa and -10 MPa), you know that point M (refer back to Figure 7.9) is not coincident with point D and the shear (40 MPa) is not a maximum value. Furthermore, the shear is not zero so M is not coincident with A, and the 50 MPa and -10 MPa are not principal values (maximum and minimum values).

Since M is not coincident with either D or A, your authors first find the angle through which the element has to be rotated (to obtain the principal stresses, which are the values on the normal stress axis through the circle - points A and B on Figure 7.9). There are two possible results: 26.6^o and 116.6^o. Next, the text computes the maximum & minimum normal stresses from the formula, getting 70 MPa & -30 MPa. Skipping over the check, in part c we see that the text computes the maximum shearing stress (again using the formula). Remember (from the circle)that the maximum and minimum normal stresses are equal when the shearing stress is a maximum? This value of the normal stress is , the coordinate on the horizontal (or normal stress) axis of the circle's center.

Mohr's Circle: This is a geometric method for doing the same analysis as above. Look at Figure 7.17 on page 437. To assure that the geometric method gives the same results as above, we have to follow certain standards when drawing the circle. From Figure 7.17a, we have positive (i.e. tensile) normal stresses and a positive shearing stress. Suppose we want to find the angle (for theta-sub-principal) through which the element must be rotated (counterclockwise) so that the shearing stress is zero and the normal stresses are at their maximum and minimum values (called the principal values). We set up a circle where A and B are on the axis and are the maximum and minimum values, respectively (we don't know the values yet, of course). We place X and Y on the circle at some angle which is twice the desired angle (in accordance with the equations) and with point Y having the correct sign for the shear (positive in this case, since our given shear value is positive). (Or, you may use your text's method with point X - see the footnote - "...when the shearing stress on a given face tends to rotate the element clockwise, the point on Mohr's circle corresponding to that face is located above the axis."  For counterclockwise rotations, the point is below the axis.)  Then follow these steps:

1. Compute the projection of XC on the horizontal 
2. Find the radius R = XC (or YC) using 
3. Calculate the angle using
4. Calculate the position of the center using 
5. Compute from
6. is, of course, merely R itself!

Three-Dimensional Analysis of Stress: Until now, we have considered only two-dimensional rotations of the cubical element. Are there rotations into the third dimension that would result in even larger normal and shearing stresses? It turns out that there are, and since design has to be based on the largest values, the engineer must be able to compute them.

Consider Figure 7.28 on page 448. Suppose we have already done an appropriate rotation about the c-axis to obtain the principal values; let the circle representing this be the one shown as BA in Figure 7.29. Suppose we now have as the maximum value, represented by OA, and as the minimum value (in this case also positive or tensile), represented by OB. is R (half of BA); as we shall see, there are larger shearing stress values possible.

Now with the cubical element having been rotated about the c-axis by the appropriate amount so that there is a maximum normal stress from this rotation, we have new axes x and y in place of a and b. is now zero, but there may be shearing stresses on the other faces of the cube that are not zero.  was unaffected by this rotation. If we now rotate the cube about the y-axis by an appropriate amount so that becomes a (a minimum value), this would plot as point C in Figure 7.29.  will be unaffected by this new rotation; that's still point B on the first circle. Taken together, these two circles can be used to define a third (outside) circle representing a rotation about the third (x) axis. Then it's easy to pick off the largest possible shearing stress.

Example 7.03 on page 450 uses a first rotation about the z-axis to obtain the principal values of the normal stress, 8 and 1.5 ksi. These are plotted in Figure 7.37; OB = 1.5 ksi. Considering a second rotation about the b-axis (Figure 7.36), your text says: "Since the faces of the element which are perpendicular to the z-axis are free of stress...". I would be more specific by saying: "Since the faces of the element which are perpendicular to the z-axis are free of shearing stress, these faces are principal planes (remember that zero shear means we have max & min normal stress). Thus we can plot circle OB. Finally, we can plot circle OA and find geometrically.

You should read section 7.6 carefully, concentrating especially on Figures 7.30 and 7.32.

Yield Criteria for Ductile Materials under Plane Stress: As your text states, simple tension (Figure 7.38 page 451) is easy; , perhaps with the inclusion of an appropriate safety factor. A test can be made on the material using one of the standard testing machines. But as soon as the force is applied differently, such as a beam in pure bending (Figure 7.39a), a state of plane stress exists and the Mohr's circle method should be used to determine the maximum values. Since the maximum normal stress is not oriented along the beam's axis (Figure 7.39b), it is not possible to make predictions directly from experimental tests; i.e., cannot be applied experimentally at the angle shown.

The Maximum-Shearing Stress Criterion is based on the fact that, in ductile materials, failure occurs along oblique surfaces (we have seen, for example, that the shear is a maximum at 45^o for simple axial forces). Remembering that this section is restricted to plane stress, Figures 7.29 and 7.32 apply. So if is less than the specimen is safe (see Figure 7.30; half the diameter is the maximum shear). Likewise, if the values of the normal stress are all positive (Figure 7.32), the shearing stress must be less than . Putting this another way,

When these are plotted, we get Tresca's hexagon (Figure 7.40, page 452). The specimen is safe if a point represented by coordinates falls within the hexagon.

The Maximum Distortion Energy Criterion depends on the determination of the distortion energy in the specimen. In section 11.6, we shall discuss how to compute the distortion energy per unit volume (equation 7.25 page 452). In this case, the specimen is safe provided the stress coordinates fall within the ellipse shown in Figure 7.41 (page 452).

You may want to study Example 2 and Example 3.

Thin-Walled Pressure Vessels:

Cylindrical Pressure Vessels - I'll use subscripts of h (hoop) and L (longitudinal) in place of the textbook's "1" and "2". Please refer to Figure 7.51 (page 462) to see the directions of

In the following formulas, r is the tank's inside radius, t is the wall thickness, and p is the gage pressure; i.e., the amount that the inside pressure exceeds atmospheric pressure outside the tank:

A Mohr's circle analysis shows that the maximum shearing stress equals (numerically) the longitudinal stress:

Spherical Pressure Vessels: Both perpendicular normal stresses are equal, and the maximum shearing stress is half of either:

Here's an example that will help you understand the use of these formulas.

¹From the 2nd Edition of Beer & Johnston's Mechanics of Materials, here's a jingle that may help:  "In the kitchen, the clock is above, and the counter is below."