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Section 3.1 Solid Mechanics Part I Kelly 31
3.1 Surface and Contact Stress
The concept of the force is fundamental to mechanics and many important problems can be cast in terms of forces only, for example the problems considered in Chapter 2. However, more sophisticated problems require that the action of forces be described in terms of
stress
, that is, force divided by area. For example, if one hangs an object from a rope, it is not the weight of the object which determines whether the rope will break, but the weight divided by the cross-sectional area of the rope, a fact noted by Galileo in 1638.
3.1.1 Stress Distributions
As an introduction to the idea of stress, consider the situation shown in Fig. 3.1.1a: a block of mass
m
and cross sectional area
A
sits on a bench. Following the methodology of Chapter 2, an analysis of a free-body of the block shows that a force equal to the weight
mg
acts upward on the block, Fig. 3.1.1b. Allowing for more detail now, this force will actually be distributed over the surface of the block, as indicated in Fig. 3.1.1c. Defining the stress to be force divided by area, the stress acting on the block is
Amg
(3.1.1) The unit of stress is the Pascal (Pa): 1Pa is equivalent to a force of 1 Newton acting over an area of 1 metre squared. Typical units used in engineering applications are the kilopascal, kPa (Pa10
3
), the megapascal, MPa (Pa10
6
) and the gigapascal, GPa (Pa10
9
).
Figure 3.1.1: a block resting on a bench; (a) weight of the block, (b) reaction of the bench on the block, (c) stress distribution acting on the block
The stress distribution of Fig. 3.1.1c acts on the block. By Newton’s third law, an equal and opposite stress distribution is exerted by the block on the bench; one says that the weight force of the block is
transmitted
to the underlying bench. The stress distribution of Fig. 3.1.1 is
uniform
, i.e. constant everywhere over the surface. In more complex and interesting situations in which materials contact, one is more likely to obtain a
non-uniform
distribution of stress. For example, consider the case of a metal ball being pushed into a similarly stiff object by a force
F
, as
mg
(a)(c)
mg
(b)
Section 3.1 Solid Mechanics Part I Kelly 32
illustrated in Fig. 3.1.2.
1
Again, an equal force
F
acts on the underside of the ball, Fig. 3.1.2b. As with the block, the force will actually be distributed over a
contact region
. It will be shown in Part II that the ball (and the large object) will deform and a circular contact region will arise where the ball and object meet
2
, and that the stress is largest at the centre of the contact surface, dying away to zero at the edges of contact, Fig. 3.1.2c (
21
in Fig. 3.1.2c). In this case, with stress
not constant, one can only write, Fig. 3.1.2d,
A A
F dF dA
(3.1.2) The stress varies from point to point over the surface but the sum (or integral) of the stresses (times areas) equals the total force applied to the ball.
Figure 3.1.2: a ball being forced into a large object, (a) force applied to ball, (b) reaction of object on ball, (c) a non-uniform stress distribution over the contacting surface, (d) the stress acting on a small (infinitesimal) area
A given stress distribution gives rise to a resultant force, which is obtained by integration, Eqn. 3.1.2. It will also give rise to a resultant moment. This is examined in the following example.
Example
Consider the surface shown in Fig. 3.1.3, of length 2m and depth 2m (into the page). The stress over the surface is given by
x
kPa, with
x
measured in m from the left-hand side of the surface. The force acting on an element of length
dx
at position
x
is (see Fig. 3.1.3b)
kPam2m
dF dA x dx
1
the weight of the ball is neglected here
2
the radius of which depends on the force applied and the materials in contact
1
2
F
(a)(c)
F
(b)
F
contact region
(d)
Small region
dA
Section 3.1 Solid Mechanics Part I Kelly 33
The resultant force is then, from Eqn. 3.1.2
kN4mkPa2
220
xdxdF F
A
The moment of the stress distribution is given by
A A
dAldM M
0
(3.1.3) where
l
is the length of the moment-arm from the chosen axis. Taking the axis to be at 0
x
, the moment-arm is
xl
, Fig. 3.1.3b, and
mkN316mkPa2
3200
dx x xdM M
A x
Taking moments about the right-hand end, 2
x
, one has
mkN38mkPa22
3202
dx x xdM M
A x
Figure 3.1.3: a non-uniform stress acting over a surface; (a) the stress distribution, (b) stress acting on an element of size d
x
■
3.1.2 Equivalent Forces and Moments
Sometimes it is useful to replace a stress distribution
with an
equivalent force
F
, i.e. a force equal to the resultant force of the distribution and one which also give the same moment about any axis as the distribution. Formulae for equivalent forces are derived in what follows for triangular and arbitrary linear stress distributions.
x
m2
)(
x
dx
)(
x
x
(a)(b)
Section 3.1 Solid Mechanics Part I Kelly 34
Triangular Stress Distribution
Consider the triangular stress distribution shown in Fig. 3.1.4. The stress at the end is
0
, the length of the distribution is
L
and the thickness “into the page” is
t
. The equivalent force is, from Eqn. 3.1.2,
Lt dx L xt F
L
000
21
(3.1.4) which is just the average stress times area. The point of action of this force should be such that the moment of the force is equivalent to the moment of the stress distribution. Taking moments about the left hand end, for the distribution one has, from 3.1.3,
t Ldx x xt M
L
200o
31)(
Placing the force at position
c
x x
, Fig. 3.1.4, the moment of the force is
c
x Lt M
2/
0o
. Equating these expressions leads to the position at which the equivalent force acts:
L x
c
32
. (3.1.5)
Figure 3.1.4: triangular stress distribution and equivalent force
Note that the moment about
any
axis is now the same for both the stress distribution and the equivalent force.
■
Arbitrary Linear Stress Distribution
Consider the linear stress distribution shown in Fig. 3.1.5. The stress at the ends are
1
and
2
and this time the equivalent force is
2/)/)((
210121
Lt dx L xt F
L
(3.1.6)
0
L
o
equivalent force
)(
x
c
x x

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