Appendix B Solution of the Stress Cubic Equation

B.1. Principal Stresses

There are many methods in common usage for solving a cubic equation. A simple approach for dealing with Eq. (1.28) is to find one root, say σ1, by plotting it (σ as abscissa) or by trial and error. The cubic equation is then factored by dividing by (σp – σ1) to arrive at a quadratic equation. The remaining roots can be obtained by applying the familiar general solution of a quadratic equation. This process requires considerable time and algebraic work, however.

What follows is a practical approach for determining the roots of stress cubic equation (1.28):

(a)

Image

where

(B.1)

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According to the method, expressions that provide direct means for solving both two- and three-dimensional stress problems are [Refs. B.1 and B.2]

(B.2)

Image

Here the constants are given by

(B.3)

Image

and invariants I1, I2, and I3 are represented in terms of the given stress components by Eqs. (B.1).

The principal stresses found from Eqs. (B.2) are redesignated using numerical subscripts so that σ1 > σ2 > σ3. This procedure is well adapted to a pocket calculator or digital computer.

B.2. Direction Cosines

The values of the direction cosines of a principal stress are determined through the use of Eqs. (1.26) and (1.21), as already discussed in Sec. 1.13. That is, substitution of a principal stress, say σ1, into Eqs. (1.26) results in two independent equations in three unknown direction cosines. From these expressions, together with Image, we obtain l1, m1 and n1.

However, instead of solving one second-order and two linear equations simultaneously, the following simpler approach is preferred. Expressions (1.26) are expressed in matrix form as follows:

Image

The cofactors of the determinant of this matrix on the elements of the first row are

(B.4)

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Upon introduction of the notation

(B.5)

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the direction cosines are then expressed as

(B.6)

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It is clear that Eqs. (B.6) lead to Image

Application of Eqs. (B.2) and (B.6) to the sample problem described in Example 1.6 provides some algebraic exercise. Substitution of the given data into Eqs. (B.1) results in

I1 = −22.7,        I2 = −170.8125,        I3 = 2647.521

We then have

Image

Hence, Eqs. (B.2) give

                σa = 11.618 MPa,       σb = −25.316 MPa,       σc = –9.001 MPa

Reordering and redesignating these values,

                σ1 = 11.618 MPa,       σ2 = −9.001 MPa,       σ3 = −25.316 MPa

from which it follows that

Image

and

Image

Thus, Eqs. (B.6) yield

              l1 = 0.0266,       m1 = −0.8638,       n1 = −0.5031

As a check, Image. Repeating the same procedure for σ2 and σ3, we obtain the values of direction cosines given in Example 1.6.

A FORTRAN computer program is listed in Table B.1 to expedite the solution for the principal stresses and associated direction cosines. Input data and output values are also provided. The program was written and tested on a digital computer. Note that this listing may readily be extended to obtain the factors of safety according to the various theories of failure (Chapter 4).

Table B.1 FORTRAN Program for Principal Stresses

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