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be written as

s = sx ( x) ex + sy ( x) ey + sz ( x) ez . (8.5)

Let this stress vector act on a plane z = 0, that spans the range 0 < x < L and that
has a width h in the y-direction. The resulting force vector on the plane considered
due to this stress vector is denoted by F = Fx ex + Fy ey + Fz ez . If sx , sy and sz are
constant, the net force is simply computed by multiplication of the stress vector
components with the surface area hL in this case:

F = sx hLex + sy hLey + sy hLez . (8.6)

For non-constant stress vector components (e.g. as visualized in Fig. 8.3),
the force components in the x-, y- and z-direction due to the stress vector s are
obtained via integration of these components over the domain in the x-direction
and multiplication with the width h of the plane (which is allowed because the
stress components are constant in the y-direction):
Stress in three-dimensional continuous media
134

z


s


y


x




Figure 8.2
The stress vector s working on a cut section of a body.


sx (x)




y
h

L x

Figure 8.3
A stress (sx ) distribution where sx is a function of x only.


L
Fx = h sx ( x) dx, (8.7)
0
L
Fy = h sy ( x) dx, (8.8)
0
L
Fz = h sz ( x) dx, (8.9)
0
respectively. In conclusion, the force vector is found by integration of the stress
vector over the plane on which it acts.



8.3 Equilibrium

In Chapter 6 the equilibrium equation for the one-dimensional case has been
formulated by demanding the balance of forces of an isolated thin slice of a con-
tinuous body. In analogy with this, we consider a continuous, arbitrary material
135 8.3 Equilibrium

”Ft

y0 + ”y

”Fl ”Fr
”y
y0
”x ey ”Fb
ey
ex
ex
x0 + ”x
x0
ez
(a) An in¬nitesimal volume (b) Free body diagram of the in¬nites-
element in a continuous body imal volume element with the lower
left corner located at x = x0 and y = y0

Figure 8.4
Free body diagram of an in¬nitesimal volume element of a continuous body.




volume, in particular a cross section in the xy-plane as depicted in Fig. 8.4(a).
In the direction perpendicular to the drawing, hence in the ez direction, the body
has a thickness h. It is assumed that all stress components are constant across
the thickness. The free body diagram of an in¬nitesimal volume element, which
in this case is a rectangular prism having dimension x — y — h, the cross
section of which is also shown in Fig. 8.4(a), is examined. The lower left cor-
ner of the rectangle as depicted in Fig. 8.4(b) is located at x = x0 and y = y0 .
Forces have been introduced on all faces of the prism in the xz- or yz-plane, see
Fig. 8.4(b). These forces are a consequence of the interaction of the prism with its
surroundings.
The force vectors F i , i = l, b, t, r, represent forces that are acting on the left,
bottom, top and right face of the prism. These forces are in¬nitesimally small
because they act on an in¬nitesimally small surface element (the faces of the
prism) that experiences only a small part of the total force that is exerted on the
body. Moreover, it is assumed that the in¬nitesimal volume element is suf¬ciently
small, such that the individual force vectors can be transformed to stress vectors
in the usual way, see Fig. 8.5.
Each of the stress vectors may be additively decomposed into a component
acting in the ex -direction and a component acting in the ey -direction. These com-
ponents are sketched in Fig. 8.6. The double subscript notation is interpreted as
follows: the second subscript indicates the direction of the normal to the plane or
face on which the stress component acts. The ¬rst subscript relates to the direction
of the stress itself. There is also a sign convention. When both the outer normal
and the stress component are oriented in positive direction relative to the coordi-
nate axes, the stress is positive also, which is required because of Newton™s law of
Stress in three-dimensional continuous media
136

st

y0 + ”y

sl sr
y0

sb
ey

ex
x0 + ”x
x0

Figure 8.5
Free body diagram of the in¬nitesimal volume element.




σyy (x, y0 + ”y)
σxy (x, y0 + ”y)
y0 + ”y σyx (x0 + ”x, y)
σxx (x0, y ) σxx (x0 + ”x, y)

y0
σyx (x0, y )
σxy (x, y0)

σyy (x, y0)
ey

ex
x0 + ”x
x0

Figure 8.6
Stresses acting on the surfaces of an in¬nitesimal volume element.




action and reaction. When both the outer normal and the stress component are ori-
ented in a negative direction relative to the coordinate axes, the stress is positive.
When the normal points in a positive direction while the stress points in a negative
direction (or vice versa), the stress is negative.
Clearly, all stress components are a function of the position in space, hence
of the x- and y-coordinate (but not of the z-coordinate because all quantities are
assumed constant in the z-direction):

σxx = σxx (x, y)
σyy = σyy (x, y)
σxy = σxy (x, y)
σyx = σyx (x, y) . (8.10)
137 8.3 Equilibrium

Using the notation from Fig. 8.6 it holds that

sl = ’σxx (x0 , y) ex ’ σyx ( x0 , y) ey
sb = ’σxy (x, y0 ) ex ’ σyy ( x, y0 ) ey
st = σxy (x, y0 + y) ex + σyy ( x, y0 + (8.11)
y) ey
sr = σxx (x0 + x, y) ex + σyx ( x0 + x, y) ey .

Notice that the stress components of the vector sl that act on the left face are
taken at a constant value of x (x = x0 ), but that these stress components may be a
function of y. Likewise, the stress components on the right face are also taken at a
constant x-coordinate (x = x0 + x) and are also a function of y, while the stress
components on the bottom and top faces are both a function of x but are taken at
y = y0 and y = y0 + y, respectively.
Force equilibrium of the in¬nitesimal volume element depicted in Fig. 8.6 is
established next. The force components acting on the faces of the prism have been
identi¬ed in Fig. 8.7.
The net force in the negative x-direction acting on the left face is denoted by
Flx and is obtained by integration of the stress σxx at constant value of x (i.e.
x = x0 ), from y0 to y0 + y:
y0 + y
Flx = h σxx ( x0 , y) dy, (8.12)
y0




”Fty

”Ftx
y0 + ”y ”Fry
”Flx ”Frx

”Fly
y0
”Fbx


ey ”Fby

ex
x0 + ”x
x0

Figure 8.7
Forces acting on surfaces of in¬nitesimal volume element.
Stress in three-dimensional continuous media
138

where h is the thickness of the cube. The stress ¬eld σxx ( x0 , y) may be approxi-
mated by
‚σxx
σxx ( x0 , y) ≈ σxx ( x0 , y0 ) + ( y ’ y0 ) , (8.13)
‚y x=x0 , y=y0

which is allowed if y is suf¬ciently small. Notice that the partial derivative of
σxx in the y-direction is taken at a ¬xed value of x and y, i.e. at x = x0 and
y = y0 (for readability reasons, in the following the addition |x=x0 ,y=y0 to the
partial derivatives is omitted from the equations). Consequently, the force Flx
can be written as
y0 + y ‚σxx
Flx = h σxx ( x0 , y0 ) + ( y ’ y0 ) dy. (8.14)
‚y
y0

To compute this integral it must be realized that σxx ( x0 , y0 ), ‚σxx /‚y and y0 denote
quantities that are not a function of the integration parameter y. Bearing this in
mind, it is straightforward to show that
‚σxx y2
Flx = σxx ( x0 , y0 ) h y+ h . (8.15)
‚y 2
The stress on the right face of the cube, σxx ( x0 + x, y) may be integrated to give
the force in the (positive) x-direction on this face:
y0 + y
Frx = h σxx ( x0 + x, y) dy. (8.16)
y0

Clearly, the stress component σxx on the right face, hence at constant x0 + x, may
be approximated by
‚σxx ‚σxx
σxx ( x0 + x, y) ≈ σxx ( x0 , y0 ) + x+ ( y ’ y0 ) . (8.17)
‚x ‚y
Therefore the force Frx can be computed as
‚σxx ‚σxx y2
Frx = σxx ( x0 , y0 ) h y + h x y+ h . (8.18)
‚x ‚y 2
A similar exercise can be performed with respect to the forces in the x-direction
on the top and bottom faces, giving
‚σxy ‚σxy x2

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