Principal stresses/Invariants

1
Principal stresses/Invariants
2

• 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.

3
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

4
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)
5

• 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)
6
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)
7

• 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)
8

• 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)
9

• 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

10

• 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)
11

• 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.

12

• 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.

13

• 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

14

• 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

15

• 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)
16

• 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.

17
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.

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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

19
Examples

• (2) Repeat example 1, where all the stresses are
  the same except that ?33 8 instead of zero.

20
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

21
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

22
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)