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 classied as free vectors, which are not

bound to any lines of action but can be moved

in parallel. Counterclockwise moments are

identied with a negative sign, and clockwise

moments are identied 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

Newton’s 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 dened 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 denes 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 specic 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

functional loading and the static relationships 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 classied 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 insufcient 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

return to its initial position if displaced by ex-

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

Radial 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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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

the torque of load and load arm up to the rst

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:

Force × Force arm = Load × Load arm

Forces Acting on the

Residual Dentition

Centric occlusion is dened 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

loading the tooth and the horizontally and verti-

cally acting forces.

The mathematic depiction of the actions of

force demonstrates the basic relationships. How-

ever, this view needs to be qualied. 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-

placed. As a result, the loading conditions be-

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 articial occlusal

eld of the denture so that no effect arises from

transverse thrusts. Therefore, the articial 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

controllable loading on abutment teeth. Statically

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

dened 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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