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## Principal%20stresses/Invariants

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### ... 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 ... – PowerPoint PPT presentation

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Title: Principal%20stresses/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.

18
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)