5
INDETERMINACY
Contents
5 Indeterminacy ...................................................................................................... 1
5.1
Background .................................................................................................... 1
5.2
Kinematic indeterminacy ............................................................................... 1
5.2.1
Degrees of freedom ................................................................................ 1
5.2.2
Restraints................................................................................................ 2
5.2.3
Degree of kinematic indeterminacy ....................................................... 2
5.3
Statical indeterminacy.................................................................................... 3
5.3.1
Definitions.............................................................................................. 3
5.3.2
General ................................................................................................... 4
5.3.3
Degree of Statical Indeterminacy - Reductive Approach ..................... 5
5.3.4
Degree of Statical Indeterminacy - Deductive Approach ..................... 8
5.3.5
Compensation for negative internal redundancy ................................. 10
5.1
Background
For a statically determinate structure, the internal forces can be calculated using only
the principle of equilibrium. For a statically indeterminate structure the stiffness of the
parts affect the distribution of internal forces. In the past the flexibility method was a
common approach to achieving solutions for such systems. This involves establishing
the degree of statical indeterminacy - dsi - (Section 5.3.4) and solving a set of
simultaneous equations of order dsi. This method has been adapted for computer
analysis) but the stiffness method is much easier to formalise and is the standard
approach for finite element solutions.
5.2
Kinematic indeterminacy
5.2.1 Degrees of freedom
The concept of a degree of freedom is central to the stiffness method used in finite
element solutions.
In the context of an analysis model a degree of freedom (often denoted as a 'freedom')
can be viewed as the position and direction which are common to a displacement and
the corresponding force actions at a node in a model. A freedom is normally shown as
a line with an end arrow - a straight line for a translational freedom and a curved line
for a rotational freedom - Figure For example Figure 5-1(a) shows a freedom in the x
direction which can relate to either a direct force or to a translational displacement.
A node is a point at which degrees of freedom are defined as shown in Figure 5.1.
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Arrows represent directions of the freedoms
y
Node
x
Position
Direction
x
(a) Degree of freedom in the x direction
(b) Freedoms in a plane
y
y
z
x
x
z
(d) Freedoms in 3 dimensions
(c) Freedoms out of a plane
Figure 5-1 Degrees of freedom
Another view is that a force and a displacement are vectors which share the properties
of position and direction at a node. These shared properties are represented by the
degree of freedom. Typical sets of freedoms are shown in Figure 5-1
5.2.2 Restraints
A restraint is where the movement relative to a degree of freedom is inhibited at a
support. For example the plane frame of Figure 5-2 has two supports (i.e. nodes which
are restrained by foundations). The left hand support node is a fixed support having 3
restraints i.e. the three freedoms shown in Figure 5.2(b) are all restrained. The right
hand support is a pinned support having two restraints(i.e. of the translational freedoms
in the x and in the y directions) -Figure 5.2(b). At a roller support there is one restraint.
The total number of restraints for the model is denoted here as s
5.2.3 Degree of kinematic indeterminacy
Unlike statical indeterminacy, kinematic indeterminacy is not a fundamental property
of the structure. It depends on how you define the model.
The degree of kinematic indeterminacy - dki - is equal to the number of unrestrained
degrees of freedom used in the model. It is the number of unknowns to be solved in the
stiffness method.
For example, a plane frame model using beam elements has three degrees of freedom at
each node and when this is incorporated into an analysis model, the degree of kinematic
indeterminacy is:
dki = 3n - s
(1)
where n is the number of nodes in the model.
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2
3
1
4
(a) Structure showing nodes
(b) Restrained freedoms
Figure 5-2 Portal Frame
The degree of kinematic indeterminacy for the portal frame of Figure 5-2 with nodes
indicated as shown, is therefore:
s = 1x3 + 1x2 = 5, n = 4
dki = 3n -s = 3x4 -5 = 7
For a system in 3 dimensions the normal degree of kinematic indeterminacy would be:
dki = 6n - s
( 2)
It is not normally necessary for users of software to calculate dki.
5.3
Statical indeterminacy
5.3.1
Definitions
Constraint A constraint (in this context) is where the relative movement between
adjacent parts of a structure are prevented. For example, in Figure 5-3(a) the ends of
the two members which meet at the pinned connection are constrained to move together
horizontally and vertically (but not rotationally). In Figure 5-3(b) the gusset plate and
extra bolts additionally constrain the ends of the members to rotate by the same amount
making the connection ‘rigid’.
single bolt
Bolt 1
Bolt 2
Bolt 3
Gusset plate
(a) Pin connection
(b) Rigid connection
Figure 5-3 Pinned and rigid connections
Release A release is the opposite of a constraint where connections between members
are removed so as to allow free movement between adjacent parts. Thus, taking away
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either bolt 2 or bolt 3 in Figure 5-3( b) would release a member end and convert a rigid
connection to a pin connection. Note that in the situation of Figure 5-3(b) it is not
necessary to release both member ends to create a pin connection. To make a fully
rigid connection into a fully pinned connection one needs to apply 'mc-1' rotational
releases at the joint where 'mc’ is the number of members meeting at the connection.
Statically Determinate Structure A statically determinate structure has just enough
constraints and restraints in combination to allow it to support a static load (sometimes
called a ‘simply stiff’ structure). If one release is made to a statically determinate
system it becomes a mechanism which cannot support a static load. For example if the
diagonal member of the frame of Figure 5.4(a) is removed the system would not have
any stiffness and could not therefore support a static load.
Statically Indeterminate structure A statically indeterminate structure (also described
as a redundant structure) has at least one more constraint than the statically determinate
form.
Degree of statical indeterminacy The degree of statical indeterminacy (or the degree
of redundancy) is the number of releases which need to be applied to make the system
statically determinate. It denote here as - dsi
5.3.2 General
For a structure with a high degree of indeterminacy a knowledge of a precise value for
the degree of indeterminacy may not be needed but it can be important to understand
the features of the structure which contribute to its redundancy.
Formerly the degree of statical indeterminacy was needed for the flexibility method of
solution but this is no longer used in practice
There are positive and negative issues associated with indeterminacy. On the positive
side, having redundancy means that if one part of the structure fails, the load may
redistribute to other parts. That is, there are alternative load paths. This is a very
important safety feature in relation to the robustness of the system.
On the negative side, redundant structures can have ‘lack of fit’ problems. Consider
the pin-jointed frame of Figure Figure 5-4(a) which is statically
determinate(internally - see Section 5.3.4). The shape of the frame is defined by the
5 members. An additional diagonal - Figure 5-4(b) - makes the system singly
redundant. The five members of Figure 5.4(a) can be fitted together without difficulty
but to add the extra member of Figure 5-4(b), the holes for the bolts must line up
accurately. Special care is needed in the fabrication of the parts of statically
indeterminate structures to avoid difficulties in erecting them..
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(a) Statically
determinate frame
(b) Single redundancy
Figure 5-4 Extra member in a frame
In what follows two methods of establishing the degree of statical indeterminacy are
described. In the Reductive Method (Section 5.3.3) releases are applied to the system
until a statically determinate subset is defined. The Deductive Method (Section 5.3.4)
uses formulae. It is recommended that the former is the basic approach since it
promotes understanding of behaviour with the Deductive Method used for checking
purposes.
5.3.3 Degree of Statical Indeterminacy - Reductive Approach
A useful way of assessing the degree of indeterminacy is to make releases to the system
until one has reached a basic determinate system. This is a system which is known to
be determinate. This condition is also know as being simply stiff. The number of
releases to be applied is then the degree of statical indeterminacy.
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(a) 2 bay portal frame
(b) Releases to create
a 'tree cantilever'
(d) releases to create a simply
supported beam
(c) Continuous deck bridge
(e) 2 storey frame
(f) Releases to create
a 'tree cantilever'
(g) Releases to create
a singly braced system
(i) Diagonal removed to create a
statically determinate system
(h) Pin-jointed truss
(j) Truss with moment connections
(k) Truss with moment releases
Figure 5-5 Structures reduced to statically determinate systems
Reductive Process:
1. Define an analysis model of the system with one element per member - define the
restraints, constraints and releases.
2. Apply releases to the analysis model to reduce it to a statically determinate form.
The number of releases is the degree of statical indeterminacy - dsi
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Applying releases
When applying releases to a structure the following rules may be used:
 In a pin connected structure (plane or 3D) taking one member away provides one
release.
 In a structure with rigid joints (no pin connections) taking away a member causes:
*
3 releases in a plane structure
*
6 releases in a space frame
 Where ‘m’ members connect at a joint only 'm-1' member end releases should be
applied. Releasing ‘m’ members causes a local mechanism.
 A local mechanism is also created if the torsional freedoms at both ends of a beam
element are released. If this is done, then the element would be able to rotate about
its centroidal axis. Only one torsional release on a member in a rigid jointed model
can be applied
Examples
1. Error! Reference source not found. (a) shows a 2 bay portal frame model with fixed
supports. Taking away two of the supports Error! Reference source not
found.(b)) reduces it to a 'tree cantilever' which is statically determinate. Six
restraints need to be released to do this; therefore the degree of statical
indeterminacy - dsi = 6
2. Error! Reference source not found.(c) shows a continuous deck bridge model with
two column supports. If the restraints at the column supports are removed (Error!
Reference source not found.(d)) the result is a statically determinate simply
supported beam. With 6 released restraints ksi = 6.
3. Figure 20.5(e) shows 2 bay, 2 storey vierendeel frame with fixed supports. The
statically determinate form of Error! Reference source not found.(f) is achieved
by removing 2 members and releasing two supports.
The number of releases is therefore:
Remove 2 moment connected members: 6
Remove 6 support releases
6
Total = dsi =
12
Error! Reference source not found.(g) shows a different release system for the
frame of Error! Reference source not found.(e). The fixed support and
continuity of the left hand column is not altered but moment releases are applied
at all other joints of the frame and at the other two supports.
Note that only m-1 releases are made at each joint. The total number of moment
releases is 12 confirming the previous result.
4. Error! Reference source not found.(h) shows a pin-jointed, simply supported truss
with a cross braced central panel. To make this system statically determinate one
the cross braces can be removed (Error! Reference source not found.(i)) to give
dsi = 1
5. Error! Reference source not found.(j) shows the same frame as Error! Reference
source not found.(h) but with moment connections at all joints but maintaining
the simply supports. To make this frame statically determinate a cross braces is
again removed and pins are inserted at all joints. The releases to achieve this are:
Moment releases at joints
12
Total = dsi =
12
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5.3.4 Degree of Statical Indeterminacy - Deductive Approach
For this method, formulae are used to establish the degree of statical indeterminacy.
Internal Redundancy
For internal redundancy the structure is considered to be separated from its supports
(equivalent to looking at a free body diagram of the whole structure) The degree of
internal redundancy - i - is the number of releases that one needs to make to this
system to make it simply stiff.
Triangulation
An important issue in internal redundancy is triangulation. If the system is formed by
building out from a triangle with other triangles (all connections pinned) then it is
statically determinate provided the system is singly connected. Error! Reference
source not found. shows a singly connected triangulation where each triangle has been
constructed by creating a new node which is then connected to two existing nodes.
This pattern would be broken by adding a member in the position of the dotted line in
the diagram. With a member in this position the frame would not be singly connected
and would not be statically determinate.
Figure 5-6 Triangulated frame
The following rules can be used to calculate the degree of internal redundancy of a
structure:
1. Pin jointed triangulated plane frame
A pin jointed frame has all connections between members are pin connected. The
internal redundancy of such a frame is given by:
i = 2m - n + 3
(3)
where m is the number of members of the system and n is the number of nodes of
the system.
2. Pin jointed triangulated space frames The degree of internal redundancy is given
by:
i = m - 3n + 6
(4)
3. Rigid jointed plane frames
i = 3(n - m + 1) - r
(5)
where r is the number of releases in the system (normally moment releases).
4. Rigid jointed space frames.
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i
= 6(m - n +1) - r
(6)
External Redundancy
In order to be fixed in space, a plane a structure needs a minimum of 3 restraints (i.e. 3
reaction components) which must be non-collinear (i.e. not all in one line) and nonconcurrent (i.e. they must not all meet at one point). A body in three dimensions needs
a minimum of 6 restraints.
The external redundancy - e - of a plane frame is therefore:
e =s-3
where s is the number of restraints i.e constraints at supports (s for supports).
(7)
For a structure in three dimensions the degree of external redundancy is given by:
e = s - 6
(8)
Overall Redundancy
Overall redundancy is the degree of statical indeterminacy of the overall system - dsi
which is calculated using:
dsi = e + i
(9)
Examples
1. Figure 5-5(a): m = 5, n = 6 using Equation(4) gives:
i = 3(m-n+1) - r = 3*(5-6+1) - 0 = 0
s = 3x 3 = 9
e = s - 3 = 9 -3 = 6
hence dsi = i + e = 0 + 6 = 6
2. Figure 5-5(c): m = 5, n = 6 using (4) gives:
i = 3(m-n+1) - r = 3*(5-6+1) - 0 = 0
s = 2x3 + 1x2 +1x1 = 9
e = s - 3 = 9 -3 = 6
hence dsi = i + e = 0 + 6 = 6
3. Figure 5-5(e): m = 10, n = 9 using (4) gives:
i = 3(m -n + 1) - r = 3*(10 - 9 + 1) - 0 = 6
s = 3x3 = 9
e = s - 3 = 9 -3 = 6
hence dsi = i + e = 6 + 6 = 12
4. Figure 5-5(h): m = 10, n = 6 using (3) gives:
i = m - 2n + 3 = 10 -2*6+3 = 1
s = 1x2 + 1x1 = 3
e = s - 3 = 3 -3 = 0
hence dsi = i + e = 1 +0 = 1
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5. Figure 5-5(j): m =9, n = 6 using (4) gives:
i = 3(m - n +1) = 3*(9 - 6 + 1) = 12
s = 1x2 + 1x1 = 3
e = s - 3 = 3 -3 =0
hence dsi = i + e = 12 + 0 = 12
These values of dsi are the same a obtained by the reductive method.
Limitation to deductive method
The deductive method for assessing redundancy is not foolproof. There are situations
where the relationships will not give the correct answer. For example the model of
Figure 5-7 is a pin jointed frame for which the deductive method would give dsi = 0
but which is clearly a mechanism. The correct value is -1. However the addition of
only one member to brace the right hand panel will give dsi = +1
Figure 5-7 Pin-jointed frame
5.3.5 Compensation for negative internal redundancy
It is possible to have negative internal redundancy if this can be compensated by
external redundancy.
Pin connection
Figure 5-8 Continuous beam with pin connection in a span
Consider, for example, the beam system of Figure 5-8. Separated from the supports,
the two parts of the beam are articulated and therefore form a mechanism. The internal
redundancy using Equation (4) is:
m = 3 n = 4, r = 1
i = 3(m - n = 1) -r = 3*(3-4+1) - 1 = -1
s = 1x2 + 2x1 = 4
The external redundancy is e = s - 3 = 4 - 3 = 1 therefore:
dsi = e + i = 1 - 1 = 0
i.e. the system is statically determinate.
Therefore an external redundancy can (but may not necessarily) compensate for the lack
of internal determinacy.
Internal redundancy cannot compensate for external negative redundancy.
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