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σxy st


Figure 8.11
Stress vector s acting on an inclined plane with normal n, decomposed into a stress vector normal
and stress vector tangent to the plane.

Eq. (8.45) can also be written in column notation as
s = σ n. (8.49)

The purpose of introducing the stress tensor σ , de¬ned as the sum of four dyads,
is to compute the stress vector that acts on an in¬nitesimally small area that is
oriented in space as de¬ned by the normal n. For any given normal n this stress
vector is computed via s = σ · n.
At any point in a body and for any plane in that point this stress vector can be
computed. This stress vector itself may be decomposed into a stress vector normal
to the plane (normal stress) and a vector tangent to the plane (shear stress), see
Fig. 8.11. Hence let
s = σ · n, (8.50)
then the stress vector normal to the plane, sn follows from
sn = ( s · n) n
= ( ( σ · n) · n) n = σ · n · ( nn) . (8.51)
The stress vector tangent to the plane is easily obtained via
s = s n + st , (8.52)
st = s ’ sn
= σ · n ’ (( σ · n) · n) n
= σ · n ·( I ’ nn) . (8.53)

Example 8.3 If the stress state is speci¬ed as depicted in Fig. 8.12, then
σ = 10ex ex + 3( ex ey + ey ex ) .
145 8.4 Stress tensor


10 10
3 3


Figure 8.12
Stress components.

If the normal to the plane of interest equals
n = ex ,
then, the stress vector on this plane follows from
s=σ ·n
= ( 10ex ex + 3( ex ey + ey ex ) ) · ex
= 10ex + 3ey .
Clearly, as expected, the operation σ · ex extracts the stress components acting
on the right face of the rectangle shown in Fig. 8.12. The stress vector s may be
decomposed into a component normal to this face and a component tangent to
this face. Clearly, the normal component should be sn = 10ex , while the tangent
component should be st = 3ey . This also follows from
sn = ( ( σ · n) · n) n
= ( ( 10ex + 3ey ) · ex ) ex
= 10ex .

Generalization to three dimensions As noted before, there are six independent
stress components in the three-dimensional case. These may be stored in the three-
dimensional stress tensor using the sum of nine dyads:
σ = σxx ex ex + σyy ey ey + σzz ez ez
+ σxy ( ex ey + ey ex ) + σxz ( ex ez + ez ex ) + σyz ( ey ez + ez ey ) ,
and also in the symmetric stress matrix:
⎡ ¤
σxx σxy σxz
⎢ ⎥
σ = ⎣ σyx σyy σyz ¦ . (8.55)
σzx σzy σzz
Stress in three-dimensional continuous media

In this case the normal on a plane has three components: n = nx ex + ny ey + nz ez ,

s=σ ·n
= σxx ex ex + σyy ey ey + σzz ez ez
+ σxy ( ex ey + ey ex ) + σxz ( ex ez + ez ex )
+ σyz ( ey ez + ez ey ) · ( nx ex + ny ey + nz ez )
= σxx nx ex + σyy ny ey + σzz nz ez
+ σxy ( ny ex + nx ey ) + σxz ( nz ex + nx ez )
+ σyz ( nz ey + ny ez ) . (8.56)

Hence the components of the stress vector s = sx ex + sy ey + sz ez satisfy

sx = σxx nx + σxy ny + σxz nz
sy = σxy nx + σyy ny + σyz nz
sz = σxz nx + σyz ny + σzz nz . (8.57)

Application of Eq. (8.55) and using
⎡ ¤ ⎡ ¤
sx nx
⎢ ⎥ ⎢ ⎥
s = ⎣ sy ¦ , n = ⎣ ny ¦ , (8.58)

sz nz
gives an equivalent, but much shorter, expression for Eq. (8.57):

s = σ n. (8.59)

If all the shear stress components are zero, i.e.: σxy = σxz = σyz = 0, and all the
normal stresses are equal, i.e. σxx = σyy = σzz , this normal stress is called the
pressure p such that

p = ’σxx = ’σyy = ’σzz , (8.60)


σ = ’p ( ex ex + ey ey + ez ez ) = ’p I, (8.61)

with I the unit tensor. This is illustrated in Fig. 8.13.

8.5 Principal stresses and principal stress directions

Assume, that in a certain point of the material volume the stress state is known by
speci¬cation of the tensor σ . One might ask, whether it is possible to chose the
147 8.5 Principal stresses and principal stress directions





Figure 8.13

orientation of a surface element in such a direction, that only a normal stress acts
on the surface and no shear stress. This means that we are trying to determine a
vector n for which the following equation holds:

σ · n = » n, (8.62)

with » to be interpreted as the normal stress. By shifting the right-hand side to the
left, this equation can also be written as

( σ ’ »I) · n = 0 and in components as ( σ ’ »I) n = 0, (8.63)
∼ ∼

with I the unit tensor (I the unit matrix) and 0 the zero vector (0 the column

with zeros). We can recognize an eigenvalue problem. The equation only has
non-trivial solutions for n (so solutions with n = 0) if

det( σ ’ »I) = 0 and also det( σ ’ »I) = 0. (8.64)

This third-order algebraic equation for » has, because the tensor σ (with matrix
representation σ ) is symmetric, three real solutions (sometimes coinciding), the
eigenvalues of the stress tensor (stress matrix). For each eigenvalue it is possi-
ble to determine a normalized eigenvector. If the three eigenvalues are different,
the three eigenvectors are unique and mutually perpendicular. If two (or three)
eigenvalues are the same, it is still possible to determine a set of three, associated,
mutually perpendicular eigenvectors of unit length, however they are no longer
The three eigenvalues (solutions for ») are called the principal stresses and
denoted as σ1 , σ2 and σ3 . The principal stresses are arranged in such a way that
σ1 ¤ σ2 ¤ σ3 .
Stress in three-dimensional continuous media

The associated, normalized, mutually perpendicular eigenvectors are the prin-
cipal stress directions and speci¬ed by n1 , n2 and n3 . It will be clear that

σ · ni = σi ni for i = 1, 2, 3 (8.65)

and also

ni · nj = 1 for: i = j
= 0 for: i = j. (8.66)

Based on the above it is obvious, that to every arbitrary stress cube, as depicted
in Fig. 8.8, another cube can be attributed, which is differently oriented in space,
upon which only normal stresses (the principal stresses) and no shear stresses
are acting. Such a cube is called a principal stress cube, see Fig. 8.14. Positive
principal stresses indicate extension, negative stresses indicate compression. The
principal stress cube makes it easier to interpret a stress state and to identify the
way a material is loaded. In the following section this will be discussed in more
detail. As observed earlier, the stress state in a certain point is determined com-
pletely by the stress tensor σ ; in other words, by all stress components that act
upon a cube, of which the orientation coincides with the xyz-coordinate system.
Because, actually, the choice of the coordinate system is arbitrary, it can also be
stated that the stress state is completely determined when the principal stresses
and principal stress directions are known. How the principal stresses σi (with
i = 1, 2, 3) and the principal stress directions ni can be determined when the stress
tensor σ is known is discussed above. The inverse procedure to reconstruct the


σ3 n3





Figure 8.14
The principal stress cube.
149 8.6 Mohr™s circles for the stress state

stress tensor σ , when the principal stresses and stress directions are known, will
not be outlined here.

8.6 Mohr™s circles for the stress state

When the stress tensor σ is given, the stress vector s on an arbitrary oriented plane,
de¬ned by the unit outward normal n can be determined by
s = σ · n. (8.67)
Then, the normal stress sn and shear stress st can be determined. The normal
stress sn is the inner product of the stress vector s with the unit outward normal n:
sn = s · n and can be either positive or negative (extension or compression). The
shear stress st is the magnitude of the component of s tangent to that surface, see
Section 8.4. In this way, it is possible to add to each n a combination ( sn , st ) that
can be regarded as a mapping in a graph with sn along the horizontal axis and st
along the vertical axis, see Fig. 8.15. It can be proven that all possible combina-
tions ( sn , st ) are located in the shaded area between the drawn circles (the three
Mohr™s circles). If the principal stresses σ1 , σ2 and σ3 are known, Mohr™s circles
(the centroid is located on the sn -axis) can be drawn at once! For this it is not nec-
essary to determine the principal stress directions. In the sn st -coordinate system
the combinations ( σ1 , 0), ( σ2 , 0) and ( σ3 , 0) constitute the image points associ-
ated with the faces of the principal stress cube in Fig. 8.14. Based on Fig. 8.15, it
can immediately be concluded that
( σn )max = σ3 , ( σn )min = σ1
( σs )max = ( σ3 ’ σ1 ) /2. (8.68)


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