Principal stresses/Invariants

- In many real situations, some of the components

of the stress tensor (Eqn. 4-1) are zero. - E.g., Tensile test
- For most stress states there is one set of

coordinate axes (1, 2, 3) along which the shear

stresses vanish. The normal stresses, ?1, ?2,

and ?3 along these axes are principal stresses.

Stress Acting in a General Direction

Figure 4.3 Force p ds acting on surface element

ds.

- p ds force acting on a surface element ds
- Area ds is defined by the unit vector (normal to

it) ? that passes through R - In order to know how p ds changes with the

orientation of the area, we have to consider a

system of orthogonal axes and the summation of

forces

Figure 4.4 Same situation as Fig. 4-3 referred to

a system of orthogonal axes.

- The summation of the forces along the axes are

(4-17)

- Where lm1, lm 2, and lm3 are the direction

cosines between the normal to the oblique plane

and the x1, x2, x3 axes. - In indicial notation, Eqn. 4-17 can be written

as - This equation defines ?ij as a tensor, because it

relates to vectors p and ? according to the

relationship for tensor

(4-18)

Determination of Principal Stresses

- The shear stresses acting on the faces of a cube

referred to its principal axes are zero. - This means that the total stress is equal to the

normal stress. - The total stress is
- The normal stress is
- If pi and ?N coincide then

(4-18)

(4-19)

(4-20)

- Applying equations 4-18 and 4-20 to p1, we have
- OR
- Similarly, for p2 and p3,

(4-21)

(4-22)

(4-23)

(4-24)

- The solution of the system of Eqs. 4-22, 4-23 and

4-24 is given by the following determinant - Solution of the determinant results in a cubic

equation in ?,

(4-25)

(4-26)

- The three roots of Eq. 4-26 are the principal

stresses ?1, ?2 and ?3, of which ?1gt ?2 gt ?3 - To determine the direction of the principal

stresses with respect to the original x1, x2 and

x3 axes, - we substitute ?1, ?2 and ?3 successively back

into Eqs. (4-22), (4-23) and (4-24). - solve the resulting equations simultaneously for

the direction cosines - and use

- Notes
- The convention (notation) in your text book is

different. It uses -

respectively - There are three combinations of stress components

in Eq. 4-26 that make up the coefficient of the

cubic equation, and these are

(4-27)

(4-28)

(4-29)

- Notes (cont)
- The coefficients I1, I2 and I3 are independent of

the coordinate system, and are therefore called

invariants. - This means that the principal stresses for a

given stress state are unique. - Example
- The first invariant I1 states that the sum of

the normal stresses for any orientation in the

coordinate system is equal to the sum of the

normal stresses for any other orientation.

- Notes (cont)
- For any stress state that includes all shear

components as in Eq. 4-1, a determination of the

three principal stresses can be made only by

finding the three roots. - The invariants is important in the development of

the criteria that predict the onset of yielding.

- The invariants of the stress tensor may be

determined readily from the matrix of its

components. Since s12s21, etc., the stress

tensor is a symmetric tensor. - The first invariant is the trace of the matrix,

i.e. , the sum of the main diagonal terms. - I1 s11 s22 s33

- The second invariant is the sum of the principal

minors. - Thus taking each of the principal (main diagonal)

terms in order and suppressing that row and

column we have

- Finally, the third invariant is the

determinant of the entire - matrix of the components of the stress

tensor. - The cubic equation can be expressed in terms of

the stress - invariants.

(4-30)

- Since the principal normal stresses are roots of

an equation involving the stress invariants as

coefficients, their values are also invariant,

that is, not dependent on the choice of the

original coordinate system. - It is common practice to assign the subscripts

1, 2, and 3 in order to the maximum,

intermediate, and minimum values.

Examples

- (1) Consider a stress state where
- ?11 10, ?22 5, ?12 3 (all in ksi)
- and
- ?33 ?31 ?32 0
- Find the principal stresses.

Solution

- Using Eqs. 4-27 to 4-29, we obtain I1 15, I2

41 and I3 0 - Substitute values into Eq. 4-30, and we have
- The roots of this quadratic give the two

principal stresses in the x-y plane. They are

Examples

- (2) Repeat example 1, where all the stresses are

the same except that ?33 8 instead of zero.

Solution

- Using Eqs. 4-27 to 4-29, we obtain I1 23,
- I2 161 and I3 328
- Substitute values into Eq. 4-30, and we have
- The three roots are

Principal Shear Stresses

- Principal shear stresses, ?1, ?2 and ?3 are

define in analogy with the principal stresses. - In order to understand how to derive the values,

students are advised to see pages 29 and 30 of

the text. - The principal shear stresses occur along the

direction that bisect any two of the three

principal axes

Principal Shear Stresses (cont.)

- The numeric values of the Principal shear

stresses are - Since ?1 gt ?2 gt ?3, ?2 is the maximum shear

stress - i.e.
- In materials that fail by shear (as most metals

do) the orientation of the maximum shear is very

important.

(4-31)

(4-32)