Basic Equations: Differential Form - Mechanics of Structures: Variational and Computational Methods - page 33

The eigenvalues (principal stresses) of T are the three real roots of the cubic equation
given by Eq. (1.82)

3 2

| T − λ I | =− λ + I 1 λ − I 2 λ + I 3 = 0 (1.87)

where

I 1 = σ + σ + σ

x y z

2
xy

2
yz

2
xz

I 2 = σ σ − τ + σ σ − τ + σ σ − τ

x y y z x z

yz xz xy

I 3 = σ σ σ − σ τ − σ τ − σ τ + 2 τ τ τ

yz
Since the principal stresses must be independent of the choice of the coordinate system,
the coefficients of this cubic polynomial are uniquely determined. This means that the
quantities I 1 ,I 2 ,I 3 , called stress invariants , have the same values regardless of the choice of
axes x, y, z in which the state of stress is given. If, in particular, these axes are chosen to be
a set of principal axes, then the shear stresses are all zero and the invariants are expressed
in terms of the principal stresses

x y z x y z xy xz

σ 1 , σ 2 , σ 3 as

I 1 = σ + σ + σ

1 2 3

I 2 = σ σ + σ σ + σ σ (1.88)

1 2 2 3 1 3

= σ σ σ

I 3

1 2 3

1.7.2 Extreme Shear Stresses
The extreme values assumed by the shear stresses as cutting plane orientation varies will
be determined in this section. The calculations are simplified if the x, y, z axes are chosen
to be a set of principal axes at the point in question. We relabel these axes 1 , 2 , 3, and write
the stress vectors on coordinate planes as

σ 1 = σ

1 e 1

σ 2 = σ

2 e 2

3 e 3
These are the stress vectors of Eq. (1.77) expressed along the principal axes for which the
corresponding shear stresses are zero. As in Eq. (1.79), the stress vector on a plane whose
unit normal is the vector a is written as

σ 3 = σ

3 e 3
The stress vector σ a can be decomposed into two orthogonal components σ a

σ a = a 1 σ

1 e 1

+ a 2 σ

2 e 2

+ a 3 σ

·

a and a shear

τ.

component of magnitude Then, this shear component is given by

− a 1 σ

3 2

τ 2

2

a 1 σ

2
1

a 2 σ

2
2

a 3 σ

2
3

a 2 σ a 3 σ

= σ a · σ a − ( σ a · ) = + + + +

a

1 2

A more convenient, equivalent expression for calculating extreme values is

2 2 a 1

a 2 + (σ

2 a 2

a 3 + (σ

2 a 3

a 1 τ = (σ − σ ) − σ ) − σ ) (1.89)

1 2 2 3 3 1

For τ to attain an extreme value the function

− λ a 1 + 1

2

a 2 + a 3 − F ( a 1 ,a 2 ,a 3 ) = τ

in which λ is a Lagrange multiplier that must assume its extremum. The conditions

∂ F ∂ F ∂ F ∂ F

∂λ = a 1 =

0

a 2 =

0

a 3 =

0 0

∂ ∂ ∂