Page 255 - Mechanical Behavior of Materials

256 Chapter 6 Review of Complex and Principal States of Stress and Strain

6.4.2 Principal Normal Stresses from Equilibrium of Forces

We are now ready to apply equilibrium of forces to the cube portion for the special case σ = σ i ,
τ = 0, where σ i is any one of the principal normal stresses σ 1 , σ 2 , σ 3 . First, note that the components
of σ i in the x, y, z directions are l i σ i , m i σ i , n i σ i , respectively, where subscripts i have been added
to indicate direction cosines for the particular principal normal stress σ i . Multiplying stresses by
areas to obtain forces and summing in the x-direction of Fig. 6.14(a), we obtain

l i σ i A PQR − σ x A yz − τ xy A zx − τ zx A xy = 0 (a)
(6.24)
−l i σ i A PQR + σ x l i A PQR + τ xy m i A PQR + τ zx n i A PQR = 0 (b)

where (b) is obtained from (a) by invoking Eq. 6.23 and multiplying through by −1. Dividing (b)
by A PQR , we obtain the first of the following three equations:

(σ x − σ i )l i + τ xy m i + τ zx n i = 0
τ xy l i + (σ y − σ i )m i + τ yz n i = 0 (6.25)

τ zx l i + τ yz m i + (σ z − σ i )n i = 0

The second two are similarly obtained by summing forces in the other two directions.
Equation 6.25 represents a homogeneous, linear system in variables l i , m i , n i , which has a
nontrivial solution only if the following determinant relationship is satisfied:

(σ x − σ)
τ xy τ zx

τ xy (σ y − σ) τ yz = 0 (6.26)

τ zx τ yz (σ z − σ)
Expanding this determinant gives a cubic equation:

2
2
2
3
2
σ − σ (σ x + σ y + σ z ) + σ(σ x σ y + σ y σ z + σ z σ x − τ xy − τ yz − τ )
zx
(6.27)
2
2
2
−(σ x σ y σ z + 2τ xy τ yz τ zx − σ x τ yz − σ y τ zx − σ z τ ) = 0
xy
This cubic always has three real roots, which are the principal normal stresses, σ 1 , σ 2 , σ 3 .
An alternative means of expressing this relationship is
2
3
σ − σ I 1 + σ I 2 − I 3 = 0 (6.28)