# Statics of Partial Dentures

Statics of Partial
Dentures
2 11
Basic Terms in Mechanics
This section outlines the basic terms in statics and dynamics that are used in dental tech-
nology. Statics is the study of the equilibrium of forces acting on a rigid body. In statics,
rigid bodies are structures that deform so little when loaded by forces that the points of
application of force undergo minimal displacement. Equilibrium refers to the state when
a rigid body is at rest or in uniform motion.
There are seven variables fundamental to physics and engineering: length, time, mass,
temperature, current intensity, amount of substance, and light intensity. Other variables
are derived from these, including velocity, work, and density. Quantities are either scalar
or vector. Scalar quantities are represented by a numeric value and a unit—for example,
time (t), temperature (T), mass (m), and electric charge (Q). Vector (or vectorial) quanti-
ties are represented by a numeric value, a unit, and a direction—for example, velocity
(v = m/s), electric  eld strength, and force. The character of a vector quantity is deter-
mined by the directional dependence of its action.
Forces are bound, aligned vectors that can be displaced along their line of action. They
can be depicted as arrows in a diagram. The of cial unit of force is the Newton (1 N =
1 kg × 1 m/s
2
), and force is characterized by the following:
• Magnitude, number, and unit (eg, 30 N)
Direction, represented by the vector arrow, in which the tip points the direction and the
length represents the measure of force
• Position of the line of action on which the force can be moved
In equilibrium, any force (action force or input force) causes a counterforce (reaction
force or output force) that is equal and in the opposite direction (Newton’s third law).
If, for example, a body rests on a solid support, its weight presses on that support; the
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212
Statics of Partial Dentures
action force (G) is absorbed by the support by an
equivalently sized reaction force (F), or else the
body is not in the resting state (action force =
reaction force).
Forces that act on a body are combined by geo-
metric addition to form a resultant force. This is
done with a parallelogram of forces for two forc-
es or with a polygon of forces for several forces.
Single forces (components) on a common line
of action can be added together or subtracted; the
sum or difference is the resultant (Fig 7-1). Single
forces on different lines of action are combined
to form a polygon of forces, and the resultant is
determined from the diagram.
In a parallelogram of forces, two forces with dif-
ferent directions but the same point of force ap-
plication are combined in the diagram to produce
a resultant. This resultant is obtained by drawing a
straight line from the origin to the intersection of
the parallel forces (Fig 7-2). If there are more than
two forces, a polygon of forces is constructed, and
the resultant is drawn from the origin of the rst
vector to the tip of the last vector (Fig 7-3).
Where there are different points of force appli-
cation, the forces are moved along their lines of
action to a common point of intersection to form
a polygon of forces (Fig 7-4). Now the line of ac-
tion and magnitude of the resultant are known
but not its point of force application.
Forces can also be resolved into two compo-
nents with the parallelogram of forces, provided
the direction or magnitude of the components is
known. If the direction is known, the lines of ac-
tion are placed through the point of force applica-
tion, and the parallelogram of forces is construct-
ed. If the magnitude of the components is known,
Fig 7-1 Forces that have a common line of action can be combined by simple
addition or subtraction. The resultant force lies on this same line of action.
Line of
action
F
1
F
2
F
3
(F
1
+ F
2
– F
3
)
= F
res
Fig 7-2 If two forces are not parallel but have
a common point of application, the resultant
force is determined by means of a parallelo-
gram of forces. A parallelogram is formed, the
diagonal is plotted, and the resultant is deter-
mined based on magnitude and direction.
F
1
F
2
F
res
α
F
2
F
1
F
3
F
2
F
1
F
3
(F
2
)
F
res
(F
3
)
Fig 7-3 The resultant of several nonparallel forces is determined by
means of a polygon of forces. The force vectors are moved parallel
to each other and strung together so that the resultant can be plot-
ted from the origin of the rst vector to the tip of the last vector.
F
1
F
2
F
res
F
1
F
2
(F
2
)
Fig 7-4 Force vectors that are not acting at one point can
be extended on their lines of action until they intersect.
The resultant can then be determined from a polygon of
forces.
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213
Basic Terms in Mechanics
one arc of a circle is drawn at the origin of the
force being resolved and another arc at the tip;
the magnitude of the force forms the radius of the
arc. The point of intersection of the arcs provides
the corners of the parallelogram (Fig 7-5).
A force couple comprises two equal and paral-
lel but oppositely directed forces. A force couple
exerts a static moment; if rotation takes place, it
is known as the turning moment (torque). Torque
is produced by a force that acts on a rigid body
when the line of action runs at some distance
from the fulcrum (Figs 7-6 and 7-7). Torque is the
product of force and the perpendicular distance
of its line of action from the fulcrum:
Moment (M) = Force (F) × Distance of action (X)
F
F
1
F
2
Draw the arcs of a circle
F
1
F
2
F
(F
1
)
F
Fig 7-5 A given force can be resolved if the magnitudes of the single forces
are known. To do this, a polygon of forces is constructed by drawing arcs of
circles at the origin and at the tip of the force being resolved, along with the
radius of the components. The intersection of the arcs establishes the apices
of the components. If the angle between the given force and the unknown
forces is known, the forces are drawn in the corresponding angle without the
magnitude; they are moved parallel to the polygon of forces until they enclose
the given force as a diagonal.
L
F
M
L
M
F
Fig 7-6 Torque is an axial vector. Such vectors
can be classied as free vectors, which are not
bound to any lines of action but can be moved
in parallel. Counterclockwise moments are
identied with a negative sign, and clockwise
moments are identied with a positive sign. If
a force acting on a tooth is not in the direction
of the tooth axis or runs at a distance from the
axis, a torque acts on the tooth. The torques
countering the acting forces have to be applied
by the periodontium.
F F
F
L
L
F
S
F
H
F
S
F
H
F
S
L
H
L
H
L
S
M
M
L
S
M
Fig 7-7 The periodon-
tium is the least protect-
ed against eccentric
forces (not acting axial-
ly). In multiple-rooted
teeth, eccentric forces
can be absorbed to a
certain degree. In the
case of tilted teeth, forc-
es encountered have di-
sastrous effects when
they produce torques
that tip the tooth.
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214
Statics of Partial Dentures
Newtons Laws of Motion
The forces acting on a rigid body can produce
either a progressive motion (translation) or a
circular motion (rotation). Newton’s rst law of
motion states that a rigid body is in equilibrium
if the resultant of all the forces and the sum of
all the torques equal zero. Without any additional
external force being exerted, the body continues
in a state of rest or in uniform motion in a straight
line. This property of a body can also be referred
to as inertia. From this, it may naturally be de-
duced that any change in the state of a motion or
rest is based on the action of forces. Furthermore,
the degree of force can actually be determined by
measuring the strength of the change in the state
of motion.
According to Newton’s second law of motion,
the acting force and the acceleration achieved be-
have in proportion to each other based on the fol-
lowing equation:
Force (F) = Mass (m) × Acceleration (a)
Here a new term is introduced: mass. Based on
what was outlined earlier, mass is bound to have
something to do with inertia. The mass of a body
depends on the speed at which it moves. Thus,
if a body moves very quickly, inertia will be very
great, or a great deal of force needs to be applied
to brake the motion of the body or make it move
even faster.
The unit of force is dened on the basis of this
equation: F = m × a. According to this, 1 N is the
force required to accelerate a mass of 1 kg by 1 m
per second squared:
1 N = 1 kg × 1 m/s
2
Newton’s third law of motion states that forces
always occur in pairs in the nature of an action
and reaction force. If a body exerts a force on an-
other body, it is reacted to with an equal and op-
posite force. This is also known as Newton’s law
of reaction. These interactive forces include:
Gravitational forces or forces of attraction be-
tween two bodies
Attraction and repulsion forces between electri-
cally charged bodies or magnets
• Intermolecular forces
Forces between the nucleons in the nucleus of
an atom
Volume is the spatial extent of the mass. Bodies
that have the same volume but are made of dif-
ferent materials therefore have a different mass.
Density denes the ratio of the mass of a body
to its volume:
Density =
Mass
Volume
or ρ =
m
v
where ρ stands for density, m stands for mass,
and v stands for volume.
The term density is initially hard to grasp be-
cause, based on the experience of our senses,
wood seems just as dense as metal: Both are
solid and both are opaque; wood is simply lighter
than metal. This property should be referred to as
density, namely, how much mass of a substance
is accommodated in a specic spatial volume.
The dimension of density is g/cm
3
(kg/dm
3
or kg/
m
3
). The atoms are most densely packed together
in metal because this is the only way the metallic
bond can function; thus, metal is very dense and
heavy, certainly heavier than wood. In the casting
technique, the amount of metal required can be
calculated from the mass of the wax pattern, the
density of the wax, and the density of the metal.
The density of solid and liquid bodies depends
on temperature; that is, density decreases with
rising temperature. In gaseous bodies, density
is also dependent on pressure, which is why the
term condensing is used when gaseous bodies
are compressed.
Principles of Mechanical
Systems
When partial and complete dentures are being
constructed, the aim is always to achieve a com-
promise that combines the esthetic concerns with
secure seating of the denture under functional
conditions. Secure seating is determined not only
by the possibilities of anchorage to the residual
dentition or mucosa but also by the forms of
dentures on the dental arches. For this purpose,
it is necessary to work on the basic terms in stat-
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215
Principles of Mechanical Systems
ics that can be applied to a rigid body that rests
on a base and is supported by what are known
as bearings. These bearings are meant to absorb
all the forces acting on the rigid body. Depending
on the type of bearing, different support reactions
occur; bearings can be classied as ranging from
single-value to six-value bearings (Fig 7-8).
A single-value bearing can only absorb com-
pressive forces from one direction in space; this
is known as a oating or movable bearing. A
oating bearing does not offer stability because
a body whose bearing can only absorb compres-
sive forces threatens to tip when external forces
are applied. A xed bearing absorbs forces but no
moments from three directions in space; this is a
three-value bearing. A xed clamp is a six-value
bearing that can absorb all forces and moments.
Support reactions are determined by means of
the equilibrium conditions that apply to the static
system. There are two types of static systems:
statically determinate and statically indeterminate
systems. The notion of statically determinate or
indeterminate systems becomes clear when one
looks at a three-legged and a four-legged table.
A three-legged table will always stand stably; a
four-legged table will wobble unless one of the ta-
ble legs is variable in length and adjusted to allow
for the base on which it stands. If a four-legged
table does not wobble, even without a variable
fourth leg, this is because the table is adapting
to the support by distorting slightly. Similarly, a
statically indeterminate system wobbles and will
only reach a state of equilibrium if it deforms. In
a statically indeterminate system, the equilibrium
conditions are insufcient to calculate support re-
actions; here the deformation conditions provide
the equations that are lacking.
In a statically determinate system, the support
reactions can be ascertained from its equilibrium
conditions alone. Adding the term equilibrium
further broadens our insight into static systems.
There are three static states in relation to the
equilibrium position (Fig 7-9):
1. Stable equilibrium exists when a body seeks to
ternal forces.
2. Unstable equilibrium exists when a body tries
to leave its original position.
3. Neutral equilibrium exists when any displace-
ment brings the body into a new equilibrium
position.
Floating bearing
Fixed bearing
Sliding bearing
Fixed joint
Clamp
F
x
F
z
F
z
F
z
F
x
F
y
F
z
F
x
F
y
M
z
M
x
M
y
2
1
3
6
Fig 7-8 This chart shows different types of
bearings, the structure and symbols for the
bearings, and the corresponding reaction
sizes and bearing values. Note that only a
xed clamp is able to absorb all the acting
forces and moments.
Structure Symbol Reaction size Value
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216
Statics of Partial Dentures
The static states become clear from the follow-
ing system: A bar rests on two bearings without
being xed and protrudes beyond the bearing
points on one side. A load can be applied at three
different points on the bar (Fig 7-10):
1. The load lies between the two bearing points
so that all the compressive forces are placed on
the bearings, and the bar rests in stable equi-
librium.
Fig 7-9 The three static states: (a) In the stable equilibrium position, the body returns to its initial position when it is displaced by
external forces. (b) In the neutral equilibrium position, any displacement of the body produces a new equilibrium position. (c) In the
unstable equilibrium position, the body tries to leave its original position.
a b c
Fig 7-10 The static states, when transferred to a real case, demonstrate the special features of a static system: (a) A bar is placed
on two bearings and loaded in the middle; both bearings must absorb equal forces, namely ½F. (b) If the bar is not loaded in the
middle but over bearing A, bearing A has to absorb the entire force. While a stable state existed in the rst case, a neutral state now
exists because bearing B remains fully unloaded; this support reaction cannot be calculated at all. (c) If the same bar is now loaded
outside the bearings, an unstable state exists whereby bearing B is unable to compensate for the torque that arises. If bearing B
is a oating bearing, the bar will be levered off.
a b c
F
A
F
B
F
BA BA
F
A
F
F
A
F
M
BA
–F
B
Fig 7-11 A lever is a rigid body that can turn around an axis. In a class 2 lever, the fulcrum lies at one end and forces can act on the
lever at different distances from the fulcrum. In a class 1 lever, the fulcrum lies between the forces applied. For both types, the law
of levers applies: Force × Force arm = Load × Load arm (F
1
× L
1
= F
2
× L
2
). In this condition, the lever is in equilibrium (ie, it does
not move). Torques of equal size are acting on the lever in opposite directions, so the lever stays in the resting state.
L
1
L
2
F
1
F
2
L
1
L
2
F
1
F
2
L
1
L
2
F
1
F
2
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217
Forces Acting on the Residual Dentition
2. The load is located outside the bearing points
so that the compressive force can no longer be
absorbed by the bearing and the bar will tip; the
bar is in unstable equilibrium.
3. The load lies on one bearing point so that all
the compressive forces rest on that one bear-
ing, and the bar is in neutral equilibrium.
Unstable equilibrium can be converted into a
stable state if the bar is xed in a bearing. That
bearing now absorbs the forces that arise from
bearing. Two terms now emerge that are taken
from the description of the rst law of levers. This
law expresses special equilibrium conditions per-
taining to a lever.
A lever is a rigid body that can be turned around
an axis (Fig 7-11). In a class 2 lever, the fulcrum
lies at one end, and in a class 1 lever, the fulcrum
lies in the middle. When equilibrium exists, the
following law of levers applies:
Forces Acting on the
Residual Dentition
Centric occlusion is dened as the hinge position
(maximal intercuspation) in which the teeth are
loaded axially (ie, centrically to their periodonti-
um). Eccentric loads during dynamic occlusion
are compensated for in a closed dental arch by
the approximal contacts, tissue coupling, and
anatomical double interlocking. In a partially
edentulous dentition, this functional interaction is
disrupted; the essential approximal support is in-
terrupted. Horizontally acting forces lead to tip-
ping and twisting of the remaining teeth.
On a solitary tooth, a horizontal action of force
occurs on the sloping surfaces of the cusps. If a
vertical force acts on a sloping surface, this load
is split into vertically and horizontally acting com-
ponents. The horizontally acting force component
will tip the tooth. The more sloping the surface, the
larger the horizontal force component becomes
in relation to the vertical component. Where tooth
surfaces slope by about 45 degrees, the vertical
and horizontal actions of force are equal. With
more pronounced tipping, the horizontal as well
as the vertical action becomes smaller in relation
to the perpendicular force being exerted. The geo-
metric and mathematic illustrations in Figs 7-12
to 7-14 show the relationship between the force
cally acting forces.
The mathematic depiction of the actions of
force demonstrates the basic relationships. How-
ever, this view needs to be qualied. In normal
punctiform occlusal contact in a healthy denti-
tion, the slopes of the cusps will ensure that all of
the teeth are loaded centrically to their periodon-
tal tissues.
In a partially edentulous dentition with soli-
tary teeth, the horizontal force action cannot be
absorbed; the tooth is tipped and becomes dis-
come pathologic. A tooth that is already tilted will
be extremely stressed in its tipping by a vertically
acting force because the vertical force compo-
nent no longer loads the tooth exactly axially. The
pressure on the tilt becomes even greater.
Forces acting at a distance from the central
axis of the tooth act like lever forces, where the
distance from the central axis equates to the le-
ver arm. The vertical force component can work
with a lever arm that roughly corresponds to two-
thirds of the tooth length. If a tooth is already dis-
placed, horizontal and vertical force components
with different lever arms will act in the direction
of tipping.
When constructing partial dentures, it is impor-
tant to ensure that the remaining teeth are not
exposed to any eccentric action of force by the
retainers. Tipping of an abutment tooth by occlu-
sal rests must be prevented. In particular, a tooth
that is already tipped should not be engaged in its
sloping position.
Sagittal forces should run so that an abutment
tooth can brace itself via existing approximal con-
tacts. Therefore, in a shortened dental arch, sagit-
tal thrusts can be compensated for mesially if the
dental arch is closed mesially.
Transverse forces in a vestibular direction can
be prevented by contouring the articial occlusal
eld of the denture so that no effect arises from
transverse thrusts. Therefore, the articial teeth on
the partial denture must be placed as far lingually
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218
Statics of Partial Dentures
as possible so that the masticatory forces run
roughly through the middle of the alveolar ridge
of the edentulous segments of the dental arch.
Vertical masticatory forces should be absorbed
by occlusal rests in a supported denture. The
splinting parts of the retainers must absorb the
horizontally acting forces. The support points cre-
ated by the occlusal rests are assumed to be rigid.
The rigid anchorage of free-end saddles to the
residual dentition is the method of choice for
achieving statically determinate systems and
indeterminate systems are outside the area of
statics analysis. If a free-end saddle has mixed
support, the mechanical system is no longer at
rest and can no longer be calculated by statics
methods. For this purpose, kinetics analysis crite-
ria should be used, and because this is the sphere
of anatomy and biology, the approaches of bioki-
netics should be adopted. Kinetics is the study of
movements and deformations in response to
forces and stresses (Fig 7-15).
Fig 7-12 When calculating the force components, the size of normal force is rst
established, which is perpendicular to the sloping surface: F
N
= F
S
× cos α, where F
N
is the normal force, F
S
is the force impacting on the tooth, and α is the angle of inclina-
tion of the tooth surface. Normal force is now broken down into horizontally and verti-
cally acting forces. The angle of inclination of the tooth surface occurs again between
the vertical force component and normal force, so that it can be used for calculation.
α
α
F
S
F
N
F
S
F
N
F
V
F
H
α
α
F
S
F
N
F
V
F
H
Fig 7-13 The horizontal force is then
dened as follows: F
H
= F
S
× sin α = F
S
× cos α × sin α. The vertical component
F
V
is then calculated: F
V
= F
N
× cos α =
F
S
× cos
2
α. If the tooth surface slopes
by 45 degrees, the vertical component
is as large as the horizontal component.
F
S
F
V
F
H
F
N
Fig 7-14 If the ratio of vertical to horizontal force component
is to be calculated, the result is
The more the angle of the slope of the tooth surface in-
creases beyond 45 degrees, the larger the horizontal force
component on the tooth becomes in relation to the vertical
force. Having said that, the horizontal force action becomes
smaller in relation to the vertical masticatory force striking
the tooth surface (F
S
); that is, the horizontal force compo-
nent is largest when the cusp inclination is 45 degrees.
Mathematic deduction of the function yields a maximum at
α = 45 degrees; it thus becomes clear that the tipping force
for a tooth will not exceed half of the vertically striking force
if the tooth stands perpendicular.
F
H
=
F
N
· sin α
F
V
F
N
· cos α
= tan α, where F
H
= F
V
× tan α (tan 45° = 1).
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