# Advanced Mechanics of Materials and Applied Elasticity: Analysis of Stress

This chapter is from the book

## 1.11 Mohr's Circle for Two-Dimensional Stress

A graphical technique, predicated on Eq. (1.18), permits the rapid transformation of stress from one plane to another and leads also to the determination of the maximum normal and shear stresses. In this approach, Eqs. (1.18) are depicted by a stress circle, called Mohr's circle.* In the Mohr representation, the normal stresses obey the sign convention of Section 1.5. However, for the purposes only of constructing and reading values of stress from Mohr's circle, the sign convention for shear stress is as follows: If the shearing stresses on opposite faces of an element would produce shearing forces that result in a clockwise couple, as shown in Fig. 1.15c, these stresses are regarded as positive. Accordingly, the shearing stresses on the y faces of the element in Fig. 1.15a are taken as positive (as before), but those on the x faces are now negative.

Given s x , s y , and t xy with algebraic sign in accordance with the foregoing sign convention, the procedure for obtaining Mohr's circle (Fig. 1.15b) is as follows:

1. Establish a rectangular coordinate system, indicating +t and +s. Both stress scales must be identical.
2. Locate the center C of the circle on the horizontal axis a distance from the origin.
3. Locate point A by coordinates s x and –t xy . These stresses may correspond to any face of an element such as in Fig. 1.15a. It is usual to specify the stresses on the positive x face, however.
4. Draw a circle with center at C and of radius equal to CA.
5. Draw line AB through C.

The angles on the circle are measured in the same direction as q is measured in Fig. 1.15a. An angle of 2q on the circle corresponds to an angle of q on the element. The state of stress associated with the original x and y planes corresponds to points A and B on the circle, respectively. Points lying on diameters other than AB, such as A' and B', define states of stress with respect to any other set of x' and y' planes rotated relative to the original set through an angle q.

It is clear that points A 1 and B 1 on the circle locate the principal stresses and provide their magnitudes as defined by Eqs. (1.19) and (1.20), while D and E represent the maximum shearing stresses, defined by Eqs. (1.21) and (1.22). The radius of the circle is

Equation a

where

Thus, the radius equals the magnitude of the maximum shearing stress. Mohr's circle shows that the planes of maximum shear are always located at 45° from planes of principal stress, as already indicated in Fig. 1.14. The use of Mohr's circle is illustrated in the first two of the following examples.

#### Example 1.3. Principal Stresses in a Member

At a point in the structural member, the stresses are represented as in Fig. 1.16a. Employ Mohr's circle to determine (a) the magnitude and orientation of the principal stresses and (b) the magnitude and orientation of the maximum shearing stresses and associated normal stresses. In each case, show the results on a properly oriented element; represent the stress tensor in matrix form.

Solution

Mohr's circle, constructed in accordance with the procedure outlined, is shown in Fig. 1.16b. The center of the circle is at (40 + 80)/2 = 60 MPa on the s axis.

1. The principal stresses are represented by points A 1 and B 1 Hence, the maximum and minimum principal stresses, referring to the circle, are

or

 s 1 = 96.05 MPa and s 2 = 23.95 MPa

The planes on which the principal stresses act are given by

Hence

Mohr's circle clearly indicates that locates the s 1 plane. The results may readily be checked by substituting the two values of q p into Eq. (1.18a). The state of principal stress is shown in Fig. 1.16c.

2. The maximum shearing stresses are given by points D and E. Thus,

It is seen that (s 1s 2)/2 yields the same result. The planes on which these stresses act are represented by

As Mohr's circle indicates, the positive maximum shearing stress acts on a plane whose normal x' makes an angle with the normal to the original plane (x plane). Thus, +t max on two opposite x' faces of the element will be directed so that a clockwise couple results. The normal stresses acting on maximum shear planes are represented by OC, s' = 60 MPa on each face. The state of maximum shearing stress is shown in Fig. 1.16d. The direction of the t max's may also be readily predicted by recalling that they act toward the shear diagonal. We note that, according to the general sign convention (Sec. 1.5), the shearing stress acting on the x' plane in Fig. 1.16d is negative. As a check, if and the given initial data are substituted into Eq. (1.18b), we obtain t x'y' = –36.05 MPa, as already found.

We may now describe the state of stress at the point in the following matrix forms:

These three representations, associated with the q = 0°, q = 28.15°, and q = 73.15° planes passing through the point, are equivalent.

Note that if we assume s z = 0 in this example, a much higher shearing stress is obtained in the planes bisecting the x' and z planes (Problem 1.56). Thus, three-dimensional analysis, Section 1.15, should be considered for determining the true maximum shearing stress at a point.

#### Example 1.4. Stresses in a Frame

The stresses acting on an element of a loaded frame are shown in Fig. 1.17a. Apply Mohr's circle to determine the normal and shear stresses acting on a plane defined by q = 30°.

Solution

Mohr's circle of Fig. 1.17b describes the state of stress given in Fig. 1.17a. Points A1 and B 1 represent the stress components on the x and y faces, respectively. The radius of the circle is (14 + 28)/2 = 21. Corresponding to the 30° plane within the element, it is necessary to rotate through 60° counterclockwise on the circle to locate point A'. A 240° counterclockwise rotation locates point B'. Referring to the circle,

 s x' = 7 + 21 cos 60° = 17.5 MPa s y' = –3.5 MPa and t x'y' = ±21 sin 60° = ±18.19 MPa

Figure 1.17c indicates the orientation of the stresses. The results can be checked by applying Eq. (1.18), using the initial data.

#### Example 1.5. Cylindrical Vessel Under Combined Loads

A thin-walled cylindrical pressure vessel of 250-mm diameter and 5-mm wall thickness is rigidly attached to a wall, forming a cantilever (Fig. 1.18a). Determine the maximum shearing stresses and the associated normal stresses at point A of the cylindrical wall. The following loads are applied: internal pressure p = 1.2 MPa, torque T = 3 kN · m, and direct force P = 20 kN. Show the results on a properly oriented element.

Solution

The internal force resultants on a transverse section through point A are found from the equilibrium conditions of the free-body diagram of Fig. 1.18b. They are V = 20 kN, M = 8 kN · m, and T = 3 kN · m. In Fig. 1.18c, the combined axial, tangential, and shearing stresses are shown acting on a small element at point A. These stresses are (Tables 1.1 and C.1)

We thus have s x = 47.6 MPa, s y = 30 MPa, and t xy = –6.112 MPa. Note that for element A, Q = 0; hence, the direct shearing stress t d = t xz = VQ/Ib = 0.

The maximum shearing stresses are from Eq. (1.22):

Equation (1.23) yields

To locate the maximum shear planes, we use Eq. (1.21):

Applying Eq. (1.18b) with the given data and 2q s = 55.2°, t x'y' = –10.71 MPa. Hence, , and the stresses are shown in their proper directions in Fig. 1.18d.

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