Principal stresses

We now wish to find the orientation, 6, of the plane on which aN is either a maximum or minimum. Take the derivative of Equation (9.1) with respect to 6 and set the result equal to 0, thus:

This equation may be satisfied by either of two values of 26, 180° apart. Thus, there are two solutions for 6 that are 90° apart. One is the plane of maximum aN and the other is the plane of minimum aN. We call the stresses acting in these directions the principal stresses. This is an important concept to understand, and we will return to it frequently.

Figure 9.2. Illustration of relation among (oxx - Oyy), oXy, and 20 in Equation (9.3b).

sxx syy sxx syy la xy

Figure 9.2. Illustration of relation among (oxx - Oyy), oXy, and 20 in Equation (9.3b).

The magnitude of the principal stresses is obtained by substituting for 20 in Equation (9.1). Equation (9.3b) and the diagram in Figure 9.2 are used to get expressions for cos 20 and sin 20. The result is:

2&xy

a\ is «Nmax and «2 is «Nmin. Thus, («1 + «2) = («xx + «yy).

Comparing Equations (9.2) and (9.3a), it will be seen that (daN/d0) = 2aS. Thus when (9aN/90) = 0, 2aS = 0. This is another important principle. Shear stresses vanish on planes on which the normal stresses are a maximum or minimum.

The orientations and magnitudes of the maximum shear stresses can be obtained in a similar manner. This is left as an exercise for the reader.

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