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Ftx = σxy ( x0 , y0 ) h x + h x y+ h , (8.19)
‚y ‚x 2
while
‚σxy x2
= σxy ( x0 , y0 ) h x +
Fbx h . (8.20)
‚x 2
139 8.3 Equilibrium

Force equilibrium in the x-direction now yields

’ Flx ’ Fbx + Frx + Ftx = 0. (8.21)

Use of the above results for the force components gives

‚σxx h 2
’ σxx (x0 , y0 ) h y ’ y ’ σxy (x0 , y0 ) h x
‚y 2
‚σxy h 2 ‚σxx ‚σxx h 2
’ x + σxx (x0 , y0 ) h y + h x y+ y
‚x 2 ‚x ‚y 2
‚σxy h 2
‚σxy
+ σxy (x0 , y0 ) h x + h x y+ x = 0, (8.22)
‚y ‚x 2

which implies that

‚σxy
‚σxx
+ = 0. (8.23)
‚x ‚y

This should hold for any position in space of the prism, hence for all values of x
and y. Performing a similar exercise in the y-direction yields the full set of partial
differential equations, known as the local equilibrium equations:

‚σxy
‚σxx
+ =0 (8.24)
‚x ‚y
‚σyx ‚σyy
+ = 0. (8.25)
‚x ‚y

Two different shear stresses are present: σxy and σyx . However, by using equilib-
rium of moment it can be proven that

σxy = σyx . (8.26)

This result is revealed by considering the sum of moments with respect to the
midpoint of the in¬nitesimal cube of Fig. 8.7. With respect to the midpoint, the
moments due to the normal forces, Flx , Frx , Fty and Fby are equal to zero,
while the shear forces generate a moment, hence enforcing the sum of moments
with respect to the midpoint to be equal to zero gives


x y
’ ( Fly + Fry ) + ( Ftx + Fbx ) = 0. (8.27)
2 2
Stress in three-dimensional continuous media
140

Using the above results for the force components it follows that
‚σyx ‚σyx
y2 y2
x
’ σyx ( x0 , y0 ) h y + + σyx ( x0 , y0 ) h y +
h h
‚y ‚y
2 2 2
‚σyx ‚σxy x2
y
+ hxy= σxy ( x0 , y0 ) h x + + σxy ( x0 , y0 ) h x
h
‚x ‚x
2 2
‚σxy ‚σxy
x2
+ + h x y = 0.
h (8.28)
‚y ‚x
2

x3 , y3 etc., reveals immediately that
Neglecting terms of order
σyx = σxy . (8.29)
Based on this result, the equilibrium equations Eqs. (8.24) and (8.25) may be
rewritten as
‚σxy
‚σxx
+ =0 (8.30)
‚x ‚y
‚σxy ‚σyy
+ = 0. (8.31)
‚x ‚y
Note, that strictly speaking the resulting forces on the faces of the prism as visu-
alized in Fig. 8.7 are not exactly located in the midpoints of the faces. This
should be accounted for in the equilibrium of moment. This would complicate
the derivations considerably, but it would lead to the same conclusions.
In the three-dimensional case a number of additional stress components is
present, see Fig. 8.8. In total there are six independent stress components: σxx ,
σxy , σxz , σyy , σyz and σzz . As due to moment equilibrium it can be derived that
σyx = σxy , σzx = σxz , σzy = σyz . (8.32)


σzz

σyz
σxz
σyz
σxz
σyy
σxy
σxy
ez
σxx
ey


ex

Figure 8.8
Stresses in three dimensions.
141 8.3 Equilibrium

The equilibrium equations Eqs. (8.30 and (8.31) can be generalized to three
dimensions as
‚σxy
‚σxx ‚σxz
+ + =0 (8.33)
‚x ‚y ‚z
‚σxy ‚σyy ‚σyz
+ + =0 (8.34)
‚x ‚y ‚z
‚σyz
‚σxz ‚σzz
+ + = 0. (8.35)
‚x ‚y ‚z
Some interpretation of the equilibrium equations is given by considering a number
of special cases.

Example 8.1 If all the individual partial derivatives appearing in Eqs. (8.30) and (8.31) are zero,
that is if
‚σxy ‚σyy ‚σxy
‚σxx
= 0, = 0, = 0, = 0,
‚x ‚y ‚y ‚x
the stresses on the faces of the cube are shown in Fig. 8.9. Clearly the forces on
opposing faces are in equilibrium, as demanded by the equilibrium equations if
the individual partial derivatives are zero.


Example 8.2 In the absence of shear stresses (σxy = 0), or if the shear stresses are constant
(σxy = c) it follows that
‚σxy
‚σxy
= = 0.
‚x ‚y
Hence the equilibrium equations reduce to:
‚σyy
‚σxx
= 0, = 0.
‚x ‚y


σyy
σxy

σxx σxx

σxy
σxy
σxy
σyy

Figure 8.9
Stresses on the faces.
Stress in three-dimensional continuous media
142

This means that σxx may only be a function of y : σxx = σxx ( y). Likewise, σyy may
only be a function of x.



8.4 Stress tensor

Suppose that, in a two-dimensional con¬guration, all three stress components (σxx ,
σxy and σyy ) are known. How can the resulting stress vector acting on an arbitrary
cross section of a body be computed based on these stress components? To answer
this question, consider an arbitrary, but in¬nitesimally small prism, having a trian-
gular cross section as depicted in Fig. 8.10(a). Two faces of the prism are parallel
to ex and ey , respectively, while the third face, having length l, is oriented at
some angle ± with respect to ex . The orientation in space of this face is fully
characterized by the unit outward normal vector n, related to ± by
n = nx ex + ny ey = sin( ±) ex + cos( ±) ey . (8.36)
The face parallel to ex has length hx = ny l, while the face parallel to ey has
length hy = nx l. This follows immediately from
hy
nx = sin( ±) = ’ hy = nx l, (8.37)
l
while
hx
ny = cos( ±) =
’ hx = ny l. (8.38)
l
The stresses acting on the left and bottom face of the triangular prism are
depicted in Fig. 8.10(b). On the inclined face a stress vector s is introduced. Force
equilibrium in the x-direction yields


syey
s
nyey
n
σxx
sxex
”l
σxy
nxex
hy = nx ”l σxy
±

σyy
hx = ny ”l
(a) (b)

Figure 8.10
Stress vector s acting on an inclined plane with normal n.
143 8.4 Stress tensor

sx ( h l) ex = σxx ( h l nx ) ex + σxy ( h l ny ) ex , (8.39)
where h denotes the thickness of the prism in the z-direction. Dividing by the area
h l yields
sx = σxx nx + σxy ny . (8.40)
A similar exercise in the y-direction gives
sy ( h l) ey = σyy ( h l ny ) ey + σxy ( h l nx ) ey , (8.41)
hence dividing by h l yields
sy = σyy ny + σxy nx . (8.42)
So, the stress vector s is directly related to the stress components σxx , σyy and σxy
via the normal n to the in¬nitesimal surface element at which s acts. This can also
be written in a compact form by introducing the so-called stress tensor σ . Let this
tensor be de¬ned according to
σ = σxx ex ex + σyy ey ey + σxy ( ex ey + ey ex ) . (8.43)
In the two-dimensional case, the stress tensor σ is the sum of four dyads. The
components of the stress tensor σ can be assembled in the stress matrix σ
according to

σxx σxy
σ= , (8.44)
σyx σyy
where σyx equals σxy , see Eq. (8.29). This stress tensor σ has been constructed
such that the stress vector s (with components ∼, acting on an in¬nitesimal surface
s
element with outward unit normal n (with components n) may be computed via


s = σ · n. (8.45)
This follows immediately from
σ · n = ( σxx ex ex + σyy ey ey + σxy ( ex ey + ey ex ) ) ·n
= σxx ex ex · n + σyy ey ey · n + σxy ( ex ey · n + ey ex · n)
= σxx nx ex + σyy ny ey + σxy ( ny ex + nx ey )
= ( σxx nx + σxy ny ) ex + ( σyy ny + σxy nx ) ey . (8.46)
Hence, it follows immediately that, with s = sx ex + sy ey :
sx = σxx nx + σxy ny (8.47)
and
sy = σxy nx + σyy ny . (8.48)
Stress in three-dimensional continuous media
144

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