Internal Force for Various Systems

We have discussed the general procedure for obtaining internal force variations in a planer system. We can
apply that procedure for various types of structural system. Here, we discuss the significance of internal
forces (and internal force diagrams) for different structural system types.
Truss : A truss members carries only axial force (tension or compression) and no shear force or bending
moment. The axial force comes from loads applied only at the two ends of a member. Therefore, the axial
force remains constant along the length of a single truss member. So, we do not really need to plot
diagrams or express axial force as function of length ( x ) in case of a truss member.


Cable : A cable is similar to a truss member except for that it carries only axial tension. For further detail
on internal forces in cables, see chapter 3.


Axially Loaded Bar : Only axial force exists in these members (such as columns ). However, unlike a truss
member a bar may be acted upon by external forces along its length. Hence, it is important to study the
variation of axial force through diagrams/mathematical expressions.
Beam and Beam-Column : A beam carries shear force and bending moment and if it carries axial force as
well, then we call it a beam-column. It is for these structural members that internal force diagrams are most
important, because deformation and failure behaviour of these members can be directly linked to these
diagrams.


Frame : Frames are two/three-dimensional structural systems made of beams and columns. A frame
member, in general, carries internal forces similar to a beam-column. Therefore, it is equally important to
obtain internal force diagrams for these systems. Note that for a frame, we may need to specify sign
convention for each member individually, as these members may have different orientations.


Arch : Arches can be treated as curved beams (or beam-columns).
How to deal with a curved centroidal axis, and with orientations of axial and shear forces.

In the next few sections we will discuss specific cases of determining forces in different types of statically
determinate systems, such as trusses, beams, arches, etc.

Analysis of Statically Determinate Structures 3

Another alternative of studying internal force variations in a structural member is to express the internal
force as a mathematical function of the longitudinal dimension ( x ). Thus, the axial force, shear force and
bending moment at a section are expressed as P ( x ), V ( x ) and M ( x ), respectively, where x is the
distance measured along the primary dimension from one end of the member . For this course,
we will consider the left end of the member as origin unless otherwise specified. Note that equations
involving these internal forces change if the direction for positive x or its origin changes.

Considering the example of Figure 2.7 again, let us obtain these internal force functions for the whole

length. After obtaining the support reactions, we can investigate internal forces at different sections. Let us

first consider the portion x = 0 → 6 m . Since no force or moment is acting between these two points, the

internal force functions will be continuous in this section. We draw the free body diagram of the beam upto a

distance x from the left end of the beam (Figure a). Using equilibrium equations, we can find the

internal forces:

Similarly, we can find out the internal forces in the portions x = 6 m → 10m (Figure b) and x = 10 m

→ 12 m (Figure c). For x = 6 m → 10 m :

and for x = 10 m → 12 m :

If we look at these expressions carefully, we see that:

•We measure x always from the same origin and in the same direction. As noted earlier, it is not absolutely

necessary to follow this convention, but it is easier this way.

•The internal force expressions change at points where concentrated forces/moments (including support reactions) act. We will see later that these forces also change if a distributed force changes its distribution.

Using singularity functions , we can combine different expressions for different segments of the beam

together into a single expression, which we will discuss later.

•We need to obtain mathematical expressions of internal forces first in order to plot the force variation

diagrams. Although these expressions provide adequate information on variation of internal forces, a pictorial

representation is always very useful.

Analysis of Statically Determinate Structures 2

The general method of obtaining internal forces at certain cross-section of a system under a given loading
(and support) condition is by applying the concepts of equilibrium . To illustrate, let us consider
the beam-column AB in Figure below for which we have to find the internal forces at section a - a . As we have
learned earlier, equilibrium conditions are best studied through free body diagrams. We can find the
reactions at supports A and B using a free body diagram of the whole beam-column AB . We
solve the three equations for static equilibrium for this free body:

If a system is in static equilibrium condition, then every segment of it is also in equilibrium. So, we can

consider the equilibrium for each of AC or CB independently. Let us consider the equilibrium of part AC , and

draw its free body diagram . In addition to externally applied forces and the support reaction 

and , this free body is acted upon by forces P , V and M on the surface a - a . These are nothing but the

internal forces (axial force, shear force and bending moment, respectively) acting at the cross-section a - a

of AB . Note that these forces are drawn in their respective positive directions in order to avoid sign

confusion. Solving the three static equilibrium equations for AC we find these internal forces:

Thus we obtain the internal forces at section a - a . These could also be obtained by considering the

equilibrium of the part at the other side of section a - a , that is of part CB . Figure  shows the free

body diagram of CB . Again, the internal forces are drawn in their positive directions on surface a - a , which

is a negative x-surface for this free body. Solving the three equations we find the values for these internal

forces:

Note :- That these values match exactly with the values obtained previously by considering the equilibrium of

segment AC . This is true for any system because there is always a unique set of internal forces on an

internal surface for a given loading condition.

Analysis of Statically Determinate Structures

We will restrict our discussions to primarily one-dimensional members (in reality these are three-

dimensional structural members, but the other two dimensions are relatively much smaller). When the

loading on such a member is on a plane same as the member itself, we call it a two-dimensional (planar)

case (see Figure 2.3a for example). In such cases, the internal forces also lie on the same plane. The

internal forces on any cross-section can be expressed with two orthogonal force components and one

moment in the plane of loading ( , , M in Figure 2.3b). We can align x -axis along the centroidal axis

of the member and we can also align one of the forces, let's say , along this centroidal axis (along the

primary dimension). Then this internal force will be known as the axial force (Figure 2.3c). In general, we

consider an internal surface perpendicular to the centroidal axis (transverse cross-section, also called the yz-

plane or x-plane ). Then the other force component acts tangentially to this surface and it is known as

the shear force . The internal moment, which is acting on the transverse cross-section, is known as the

bending moment .

Axial force, shear force and bending moment on a cross-section of a two-dimensional

(planar) system

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 For a three-dimensional case, that is when the loading is not restricted to one plane, we have three
orthogonal force components and three orthogonal moment components on an internal surface (Figures 2.4
a & b).
We align the centroidal axis of the member along, say, x -axis, and consider an internal surface
perpendicular to it (Figures 2.4c). Then is the axial force and and are the two shear forces .
Moments and are the two bending moments . The moment is known as torsion . This set of
forces is the most generic case of internal forces for such structural members.

Axial force, two shear forces and two bending moments for three-dimensional

systems

Note that these internal forces are defined according to their orientation respective to the structural

member. The axial force acts along the centroidal axis of the member. The shear forces act in a plane which

is perpendicular to this centroidal axis and the bending moments act along directions perpendicular to this

axis as well. The torsion acts along the centroidal axis.

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The sign convention for internal forces depends on the internal surface on which these forces are being

considered. So, we need to define the internal surface first. Let us assume that the centroidal axis of a

longitudinal structural member is aligned along the x -axis, and we consider the internal forces on an x-

plane (or x-surface ). If this cross-section is facing positive x -direction, then it is called a positive x -

surface, and vice-versa. Figures 2.5a & b show the positive internal forces on positive and negative x -

surfaces, respectively.

Direction of positive internal forces on a positive x -surface (a) and a negative x -

surface (b)

This sign convention is followed throughout this course and relations involving these forces (and other

parameters, such as stresses and deformations) are derived based on this sign convention. It will not be

illogical to adopt any other sign convention for internal forces. However, in that case one will have to develop the equations involving these forces independently following the new sign convention.

Let us restrict our discussion to planer loading (that is, two-dimensional) cases with no torsion. For an

internal segment of a member, we can show the internal forces both on the positive and the negative

internal surfaces. Positive directions for each internal force (an axial force, a shear force and a bending

moment) are shown individually in Figure 2.6. This is an easy and standard way of defining sign conventions

for two-dimensional cases, and students are encouraged to define (for each specific case) their adopted sign

convention similarly.

Notations that we follow for these internal forces are: P for axial force, V for shear force, and M for bending

moment. However, please note that in a three-dimensional case, we need suffixes to distinguish between

the two shear forces and also between the two bending moments. General notation for torsion is T .

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Internal forces in a system

Internal forces (or moments) are generated within a solid body (or structural system) when it is acted upon
by external forces (including support reactions and other contact forces as well). To illustrate how internal
forces are generated or why they exist, let us consider a three-dimensional solid body (Figure 2.1a),
supported at points A and B. , and are external loads applied on the system. To study the
equilibrium of the whole body, we draw its free body diagram (Figure 2.1b). The supports are replaced by
reactions and in the free body diagram. We consider an internal surface by taking an arbitrary cut
through the system (Figure 2.1c). For equilibrium of the part at the right of the section, there has to be
forces acting on the internal surface which balance the external loads and (Figure 2.1d). and
are internal forces acting on the surface of the cut.

It is important to know the internal forces acting at different sections of a system. The material, of which

the body is made, should be strong enough to carry these forces. Otherwise the system fails (by crushing,

breaking, etc.) under the loading condition.

The general procedure of obtaining internal forces includes these following steps:

Obtain the system configuration (dimension and support conditions) and external loadings applied on it.

1. Draw a free body diagram of the whole system.

2. Find the support reactions by using equations of (static) equilibrium.

3. Take a cut through the body where internal forces have to be obtained.

4. Consider equilibrium of the part of the system at any one side of the cut by drawing a free body

diagram of that part.

5. Obtain the unknown internal forces acting on the cut surface by solving these equilibrium equations.

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One method of classifying structural systems is on the basis of their load
transfer mechanisms. To elaborate, a system (or a structural member) is identified based on the
predominant types of internal forces carried by it. Thus we have: bars , cables , beams , columns , arches ,
etc. Below is a list of such members along with the predominant internal forces that they carry. Cable : A
cable or wire can carry axial tension only. Internal forces in cables are not discussed in this chapter
because cables are very different from all other systems due to their flexible geometry. Internal forces and
geometry of cable systems are discussed in detail in another post.

1. Cable : A cable or wire can carry axial tension only. Internal forces in cables are not discussed in this
chapter because cables are very different from all other systems due to their flexible geometry.Internal
forces and geometry of cable systems are discussed in detail in another post.

2.Bar : A bar carries only axial forces – tension and compression both. That is why it is also known as
axially loaded bar .

 3.Beam : A beam's primary function is to transfer lateral loads applied externally on the beam. These loads
produce bending moments and shear forces on beam a cross-section.

4.Column : The predominant internal force in a column is axial compression.

5.Beam-Column : A beam-column, as the name suggests, carries all kinds of internal forces that are
produced in a beam or a column, which include: bending moment, shear force and axial force.

6.Arch: An arch is a curved member which carries primarily axial compression under lateral loads applied
externally.
There is no difference in the shapes of a beam, a column, a beam-column or a bar. All are straight
longitudinal members (one dimension is much larger than the other two) and we will not be able to
distinguish one from the other unless we know the load transfer mechanism. Figure  illustrates this
issue.

All the members discussed above are primarily one-dimensional geometrically. Two-dimensional members

are also categorized similarly, such as: plates , shells ( thin & thick ), slabs , etc. We also have specific

names for systems formed by combination of members, such as a truss or a frame . A frame is a

combination of beams and columns, whereas all the members in a truss are axially loaded bars.

Kinematic Indeterminacy of a structure

Indeterminacy is the minimum number of independent displacement quantities that you need to
define to be able to get the complete displaced geometry of the structure.
I will look at two different types of structures. One set of structures which are made up of only
actually loaded members if you remember those are trusses. I am going to start off with trusses
because even trusses and beams and frames which are essentially structures made up of members
which can deform flexurally, the treatment is different. Let us look at a truss and again I go back
to the simple truss that I had last time. We want to find out what the Kinematic Indeterminacy of
this truss structure is.
Before that I need to define what a Kinematically determinate remember, it is very easy for you
to know how to get the statically determinate structure. You know when you look at it what a
statically determinate structure looks like. How do you see a Kinematically indeterminate
structure? Let us take this same structure; I want to make this Kinematically indeterminate. How
will I make it Kinematically indeterminate? This is a Kinematically determinate structure. Why
is this Kinematically determinate? Let us look at this structure. Understand what Kinematic
Indeterminacy is.
It is the minimum number of displacement quantities that you need to define, the displaced shape
of a structure. In this structure, how will it displace under loads? How will it deform? Can this
point go anywhere? This point cannot go anywhere because it is restrained both in this direction
as well as this direction; it cannot move anywhere on the plane that means this point cannot go
anywhere. Can this point go anywhere? Neither can this point go anywhere. Can this point go
anywhere?
Again, it cannot, simply because every point has a hinge which restraints the displacements in
the plane of the structure. So what would be the displaced shape of the structure if it was
subjected to loads? Let us put some loads on the structure. These are the possible loads on a truss
structure because a truss structure can only be loaded at the joints. I put all possible loads that
this structure can be subjected to. What will be the displaced shape under these loads?
Remember, all these loads would get directly transferred to the supports and the structure will
not be subjected to any stress. Therefore there is going to be zero strain and since this going to be
zero strain there is going to be no displacement. What is a Kinematically Determinate structure?
It is a structure that does not displace. Is that interesting? Currently, it is not interesting. Later on,
we will see that we need to define a Kinematically Determinate structure before we can use what
is known as the displacement method of analyzing statically indeterminate structures.
Once we have a Kinematically determinate structure, it is very easy to look at this structure and
see what is going to be the Kinematic Indeterminacy or the Degrees of Freedom of this structure.
Let us look at it. Can this point go anywhere? No. Again there is a hinge which is restraining it
from moving anywhere. Can this point go anywhere? What is the restraint? The only restraint is
it cannot move vertically. But is there any restrain for moving it horizontally? No. So, this is a
possible displacement. This node can displace in this direction and therefore this horizontal
displacement is a degree of freedom of the structure. Let us look at this point now. Can this point
go anywhere? Note that this point can go anywhere on this plane and in a plane how many
displacements do you require to define? If you define two orthogonal displacement quantities,
they completely define the displacement of this point anywhere in space.

Now, if you look at this structure, a simple truss has three nodes: one node completely restrained
and cannot go anywhere, this node vertically restraint and can move horizontally, this node is
free to displace in any direction in the plane. We can define two displacement quantities which
will define the motion of this point anywhere in space. Now how many Degrees of Freedom or
what is the Kinematic Indeterminacy of this structure? 1, 2, 3 the Kinematic Indeterminacy (K.I)
of this structure is equal to 3. Therefore if you have a truss, any joint in the truss can move in the
plane and what you need to look at is what the restraints are. Let me redefine this problem. I am
not going to do an algorithmic way of getting the Kinematic Indeterminacy. How many nodes
does this structure have? It has three nodes. Now for the movement of a node or a joint in a plane
is given by two independent displacement quantities. In other words, if I know those two
displacements quantities I know where the joint is in the space. Let us look at it this way. How
many joints? 3. How many Degrees of Freedom per joint? 2. Therefore if this plane truss were to
have no supports what will be the Degrees of Freedom? It would be 2 into the number of joints
so let it be 6. The Degrees of Freedom 2 per joint multiplied by 3 is equal to 6. 6 Degrees of
Freedom since this truss has three joints. This is the unrestrained Degrees of Freedom.
In other words, the truss did not have any support, then the number of degree of freedom would
be 6. However, you do have two supports in the structure. What are the supports? At one joint,
you have a hinge which has two restraints; it stops it from moving vertically and it also stops it
from moving horizontally. It does not allow this joint to go anywhere so it gives two restraints to
this joint. What about this joint? This is a kind of roller joint. How many restraints does it give?
It gives exactly 1 it restrains the vertical motion of this joint. 1 meaning 2, plus 1. The total
number of restraints in this is 3. What is the restrained structure? How many Degrees of Freedom
does it have? 3.
Kinematic Indeterminacy by finding out displacements and also the algorithmic way: Find out
the number of joints in a truss, multiply that by 2. That gives you the unrestrained Degrees of
Freedom. You find out the number of restraints that the body has and then subtract those
restraints and you have the Degrees of Freedom or the Kinematic Indeterminacy of the truss. Let
us look at one another problem. Let us look at the slightly more complicated truss.
....................................................................................

This is the truss; we have to find out the Kinematic Indeterminacy. How would I go about it? We
are going to use only the algorithmic way of getting it, and then I will show you which are those
displacements. How many joints? 4. What is the number of Degrees of Freedom? Note that
Kinematic Indeterminacy and Degrees of Freedom are used interchangeably. What are the
Degrees of Freedom per joint for a planar truss? 2 Degrees of Freedom. What is the total number
of unrestrained Degrees of Freedom? 2 into 4 is equal to 8. We need to define what the restraints
are? Restraints here are one hinge support, two restraints - one roller support and one restraint.
So a total of 3. The total actual Degrees of Freedom is 8 minus 3 is 5. The Kinematic
Indeterminacy is 5, for this structure. Once I have defined the kinematic Degrees of Freedom,
you need to understand what is Kinematic Indeterminacy, it is the minimum number of
displacement quantities, that you need to define, to be able to get the displaced geometry of the
structure. I need to find out and define those displacement Degrees of Freedom. Note that I
should have only 5 independent displacement Degrees of Freedom. Let us look at this. Are there
any Degrees of Freedom at this node? No. Both of them are restrained. Let us go to joint 2;
restraint none. What are the Degrees of Freedom? 1 2. Let us go to 3. Any restraint? No.
What are the Degrees of Freedom 1 2 the vertical displacement and the horizontal displacement.
What about 4? It is restrained in this direction and is not restrained in this direction, so how
many do I have? I have 1 2 3 4 and 5. 5 Degrees of Freedom; 5 Kinematic Indeterminacy. It
should be fairly easy for you now, to get the Kinematic Indeterminacy of a truss type structure.
Planar truss - 2 Degrees of Freedom, space truss - 3 Degrees of Freedom because you need 3
displacement quantities to define the position of a point in space. 3 independent: x displacement,
y displacement and z displacement. What will be the unrestrained Degrees of Freedom for a
space truss? 3 into the number of joints; that will be the unrestrained and then you find out the
number of restraints that you have which are the support conditions or the supports. Subtract the
restraints and you got the Degrees of Freedom for a truss and that defines the Kinematic
Indeterminacy for a truss structure. Let us move on to that for a beam.

This is a structure that you are all familiar with; a simply supported beam. What is the Static
Indeterminacy of the simply supported beam? 0. You know that it is a statically determinate
structure but now we are not looking at Static Indeterminacy we are looking at Kinematic
Indeterminacy. Again there are 2 joints 2 nodes in the structure and you can ask why we do not
use the same method as a truss. For every node to move in the plane, there will be two Degrees
of Freedom. 2 nodes, 2 Degrees of Freedom and three restraints. This is kinematical so if you use
the truss model 2 into 2 so 2 nodes 2 Degrees of Freedom per node that is 4 and there are 3
restraints; 2 here and 1 here one degree of freedom. What is the one degree of freedom this one
(Refer Slide Time: 20:43). Is that correct? Is this single kinematically indeterminate structure
single degree of freedom? No.
Understand that members in a beam and frame behaved differently from a truss; a truss is only
an actually loaded member. All we need to know is where the two joints at the end of a truss
member is and that gives us elongation of the truss and from that we can find out the force in the
truss. Can we do that for a beam? No, because the behaviour of a beam is in the way it is loaded,
it goes like this. The entire concept of displacement, Degrees of Freedom for a beam and a frame
would be different from a truss and that is why we need to look at it differently. How do we look
at it differently? An important point is that a joint that connects beam members or frame
members in addition to the two Degrees of Freedom which define its movement also has another
additional degree of freedom. What does that do?
That degree of freedom actually defines the movement of the members, relative to each other.
What is that? That degree of freedom is the rotation degree of freedom. Whenever you look at a
beam or a frame every joint when it is unrestrained has 3 Degrees of Freedom. So a joint has 3
Degrees of Freedom. What does this degree of freedom do? If I fix this and this, this disappears
because this cannot go up or down.

Remember, that if I have a Kinematically determinate structure I can define the displaced
geometry of the structure completely. If I restrain this and I restrain this, both the displacements
are restrained which means no displacement. But is that true? No. Under this load it displaces in
this manner. Understand that I required additional displacement quantities to define the displaced
shape of the structure and what are those? Look at this rotation and this rotation; these two
rotations gives me the displaced shape of the structure. That is that additional degree of freedom
that I need to define. Whenever you have a beam or a frame you have 3 Degrees of Freedom per
joint. Let us come back to this structure; the simply supported beam. How many Degrees of
Freedom does it have? How many joints? 2. How many unrestrained Degrees of Freedom? 3 per
joint, so a total of 6. Note that this hinge support still restraints only 2 which is this displacement
and this displacement. It has 2 restraints and this one. What was the total number of restraints? 3.
What are the Degrees of Freedom? 3. This simply supported beam has 3 Degrees of Freedom or
its Kinematic Indeterminacy is equal to 3.
What is Kinematic Indeterminacy? Number of unknown displacement quantities. Let us find
those out for this structure just like we did in the truss. Note that it restraints this but allows this,
so this is my first degree of freedom. This is a pin roller; it allows free rotation so this is another
degree of freedom. This joint restraints both the motions but allows it to rotate and that is my
third degree of freedom. In other words, this simply supported beam, if I knew the horizontal
displacement of this point, the rotation at this point and the rotation at this point, I would be able
to give you how this structure has displaced. If I knew the displacement quantities I could draw
the displaced geometry of the structure. This, in essence, defines the Kinematic Indeterminacy
for beams and frames.
I want to introduce one more joint that you are all familiar with; another statically determinate
structure; this is a cantilever beam. What is the Kinematic Indeterminacy of this cantilever
beam? How many joints? 1 and 2 so 2 joints. How many unrestrained Degrees of Freedom? 6.
You need to find out the number of restraints. How many restraints? Let us look at this joint, this
is a fixed support; a fixed support is one which restrains the point from displacing in any
direction and it also restrains the rotation of that point. How will this displace? This will displace
something like this. At this point, the slope would be 0, the displacements both vertically and
horizontally would be 0, so completely restrained. At this point, it is not restrained; it does not
have any support. What does this restrain? It restrains vertical, horizontal and rotation, so there
are three restraints. So the Kinematic Indeterminacy is 3. What are those 3? It is 1, 2, 3. So, for a
beam you should be able to get the Kinematic Indeterminacy or the number of Degrees of
Freedom for the structure. Let us move on to Kinematic Indeterminacy for a frame.

Note I just said that, the Kinematic Indeterminacy for both a beam and a frame are computed in 

exactly the same fashion. I can use the same algorithm that I developed for the beam. Look at 

this; how many nodes 1, 2, 3, 4 there are four joints in this structure. Following my procedure,

each joint unrestrained has three Degrees of Freedom. The number of unrestrained Degrees of 

Freedom equal to 3 into 4 is equal to 12. Unrestrained Degrees of Freedom is 12. Now, we need 

to find out what are the restraints in the structure. How many supports? Two supports, 1 here and 

1 here. What are they? fixed supports. How many restraints does the fixed support give? 3. How 

many fixed support? 2 What are the restraints 6 is equal to 2 into 3. Two supports with 3 

restrains each. How many degrees of freedom? 6. Kinematic Indeterminacy, remember, is a same 

as Degrees of Freedom 6. I should be able to draw them 1, 2, 3, 4, 5, 6. The vertical, horizontal 

and rotation of this joint and the vertical, horizontal and rotation of this joint; 3 per node; total 

six Degrees of Freedom. If I know these 6 displacement quantities I can draw the displaced 

shape of this frame. 

Now, I want to bring in the concept of constraints. Typically what we do, when we consider 

beams and frames we tend to neglect, the axial deformations because by and large in a beam and 

a frame, the main way that the loads are transferred is through flexural deformations. In other 

words, the structure bends to take the loads. While flexural deformations are very large the axial 

deformation are not zero, but are very small. We tend to neglect the axial deformations. What 

happens then?

Let me go back to a simply supported beam; I neglect the axial deformation of this member,
therefore I am saying that it is axially rigid. Note, in general, for a statically determinate beam,
the forces that you normally find out are the sheer force bending movement not the axial force.
Implicitly, without knowing, you are actually neglecting axial deformation; you never explicitly
mention that. When you are analyzing a frame or a beam the member is axially rigid. What
happens when you have a member which is axially rigid? What happens here is that you are
constraining the structure. It has no effect on the Static Indeterminacy of the structure.
Remember, that it has a tremendous effect on the Kinematic Indeterminacy or the number of
Degrees of Freedom that you get from a structure. You are constraining a structure into one
actually rigid. When I say that this member is actually rigid, this cannot go anywhere. It can
move horizontally. Forget about the rotational Degrees of Freedom; I am looking only at the
axial deformation. It can go this way so this goes this way. What effectively happens to this
member when this goes this way? This joint cannot go anywhere, this joint can move
horizontally. The member is either axially shortened or if the displacement is in this direction it
is elongated. In other words, this degree of freedom actually corresponds to the axial deformation
of the body. If this member is axially rigid, since this point is not going anywhere; it cannot
deform axially. That means this displacement has to be zero.
By making the structure axially rigid I have eliminated a degree of freedom. If I say that this
beam is axially rigid how many Degrees of Freedom does it have? It has 2 rotations. It is a 2
degree of freedom structure. How do I put it into my algorithm? I truly believe that unless you
have an algorithm way of computing Static Indeterminacy and Kinematic Indeterminacy, you are
always going to flounder because you would not know how many indeterminacies there are in
the structure both static and kinematic. Let us go back to algorithm way. How do I include this
effect in the algorithmic way? What we define is two joints unrestrained Degrees of Freedom: 2into 3 is equal to 6, restraints -3. Now, what have I done by making it axially rigid? I am
constraining the structure.
So, a new thing: constraints. Constraints, also reduce the number of Degrees of Freedom in the
structure. In this case how many constraints? It is axially rigid; whatever I have done I have
made this into 0. How many constraints do you think axial rigidity in a member brings in? 1.
Because a single degree of freedom defines the axial deformation of a member and when I make
it axially rigid that is the additional constraint that I am giving; that this body cannot deform
axially. I am making one constraint per member. How many members are there in this structure?
There is 1. How many Degrees of Freedom? 2. I have already shown you the Degrees of
Freedom in the structure. How about the algorithm for getting the Kinematic Indeterminacy of a
beam or a frame? Start off by finding out how many joints there are in the structure. Multiply the
number of joints by 3 that gives you the total number of unrestrained Degrees of Freedom. Then
find out the restraints. What are the restraints that are provided always by supports? You go to
every support and find out what kind of restraint it gives add up all the restraints from all the
supports. Those are the total number of restraints those are to be subtracted from the unrestrained
Degrees of Freedom. In addition to that, you need to look at the constraints. If, a member is
axially rigid how many constraints? Single; it is 1 per member. So, you subtract the total number
of constraints and you got your Degrees of Freedom or the Kinematic Indeterminacy of the
structure.

Static Indeterminacy of Structures

If the number of independent static equilibrium equations

is not sufficient for solving for all
the external and internal forces (support reactions and member forces, respectively) in a system, then the system
is said to be statically indeterminate . A statically determinate system, as against an indeterminate one, is that for
which one can obtain all the support reactions and internal member forces using only the static equilibrium
equations.

The equilibrium equations are described as the necessary and sufficient conditions to maintain the

equilibrium of a body. However, these equations are not always able to provide all the information needed to obtain

the unknown support reactions and internal forces. The number of external supports and internal members in a

system may be more than the number that is required to maintain its equilibrium configuration. Such systems are

known as indeterminate systems and one has to use compatibility conditions and constitutive relations in addition to

equations of equilibrium to solve for the unknown forces in that system.

For an indeterminate system, some support(s) or internal member(s) can be removed without disturbing its

equilibrium. These additional supports and members are known as redundants . A determinate system has the exact

number of supports and internal members that it needs to maintain the equilibrium and no redundants. If a system

has less than required number of supports and internal members to maintain equilibrium, then it is considered

unstable. 

An indeterminate system is often described with the number of redundants it posses and this number is known as its

degree of static indeterminacy . Thus, mathematically:

Degree of static indeterminacy = Total number of unknown (external and internal)

forces

- Number of independent equations of equilibrium (1.21)

It is very important to know exactly the number of unknown forces and the number of independent equilibrium

equations. Let us investigate the determinacy/indeterminacy of a few two-dimensional pin-jointed truss systems.

Let m be the number of members in the truss system and n be the number of pin (hinge) joints connecting these

members. Therefore, there will be m number of unknown internal forces (each is a two-force member) and 2 n

numbers of independent joint equilibrium equations ( and for each joint, based on its free

body diagram). If the support reactions involve r unknowns, then:

Total number of unknown forces = m + r

Total number of independent equilibrium equations = 2 n

So, degree of static indeterminacy = ( m + r ) - 2 n. 

Sometimes, these two different types of redundancy are treated differently; as internal indeterminacy and

external indeterminacy . Note that a structure can be indeterminate either externally or internally or both externally

and internally.

We can group external and internal forces (and equations) separately, which will help us understand easily the cases

of external and internal indeterminacy. There are r numbers of external unknown forces, which are the support

reactions components. We can treat 3 system equilibrium equations as external equations. This will lead us to:

Degree of external static indeterminacy = r - 3.

The number of internal unknown forces is m and we are left with (2 n -3) equilibrium equations. The 3 system

equilibrium equations used earlier were not independent of joint equilibrium equations, so we are left with (2 n - 3)

equations instead of 2 n numbers of equations. So:

Degree of internal static indeterminacy = m - (2 n - 3).

Please note that the above equations are valid only for two-dimensional pin-jointed truss systems. For example, for

three-dimensional ( “space” ) pin-jointed truss systems, the degree of static indeterminacy is given by ( m + r - 3 n

). Similarly, the expression will be different for systems with rigid (fixed) joints, frame members, etc.

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