Structural Analysis I

Prof Amit Shaw

Department of Civil Engineering
Indian Institute of Technology Kharagpur
Lecture 33
Analysis of Statically Intermediate Structures

Hello everyone! Welcome to the second part of our of our of our course I am calling it second
part because what you have what we learnt in last 6 week is we several methods to to analyse
that say to analyse statically determinate structures right. But many structures are not
determinate because the number of constant in those structures are more than the number
equations statically equilibrium equations available. So whatever concept we have learned
those concept cannot be directly applied to the structures which are not determinate which are
indeterminate structures.

(Refer Slide Time: 01:03)

What will be what we're going to do this week and next 4 5 week is we will try to understand
what is statically indeterminate structure when it becomes indeterminate and if it becomes
indeterminate then what are the what are the methods available to analyse them to find out
the internal forces and and displacements. Ok! So since this is the first lecture what we will
be doing is we will understand the will understand the underlying philosophy of all these
methods and also give you some of the basic concepts that we learnt in previous classes. And
those concepts will be useful for subsequent classes as well.
(Refer Slide Time: 01:47)

Ok! So this week we're going to start analysis of statically indeterminate structures and today
what we will doing is we will just understand what is indeterminacy and what are the possible
way out we have when a structure becomes indeterminate ok.

(Refer Slide Time: 02:06)

Ok! You see you I believe that you are familiar with this sliide because we discussed this
slide in one of our earlier earlier lectures. What it is? We have an arch here which is subjected
to uniformly distributed load and the support conditions at A and B are we have hinge at a
and b. And this is this is an indeterminate structure because the number of reactions support
reaction are 4- 2 here and 2 here. So essentially we have total 4 numbers of 4 equations, 4
reactions and number of equations available is 3.

Now if we apply equilibrium equation sation of f x is equal to zero sation f o is equal to the
sation M a is equal to zero, Moment about a a and finally what we have is we can determine
the you see if the structure indeterminate it doesn't mean that we cannot find out support
reactions or internal forces in any number we can probly partially partially some of the
unknown (())(03:09) we can determine but if the structure is indeterminate then just we are
applying the equilibrium conditions we cannot determine all the unknowns ok.

(Refer Slide Time: 03:19)

For instance in this case we can determine what is the value of what is the support reaction in
vertical direction A y is equal to ql by 2 and B y is equal to ql by 2 that is obvious from the
symmetry but what we have is we cannot determine whatever information available here we
cannot determine what is the value of horizontal reaction A x and B x atmost what
information we have that A x has to be is equal to minus B x means A x plus B x has to be is
equal to 0, right?

So if AX and if whatever may be reaction whatever may be the value of AX and BX if they
satisfy this condition then they satisfy the equilibrium condition equilibrium equation and this
system is called for any 4 system which is we satisfy this, this is statically admissible but then
again we mention that it is statically admissible but it is not kinematically admissible but that
time we did not discuss what is kinematic admissibility.
Now we are going to discuss that so this is an indeterminate structure where the solutions are
called partial solutions are ql EY is equal to ql by 2 B y is equal to ql by 2 and another
condition is horizontal reactions at A and B they are equal and opposite. That is if any force
system satisfy these condition means they are statically admissible.

(Refer Slide Time: 04:45)

Now suppose now these let us find out an equivalent system this is our actual system right?
Let us find out an equivalent system Now here equivalent system is only the support are
replaced by there characteristic reactions. So we know the value of A y and value of B y is ql
by 2 it is ql by 2 ql by 2 and value of only thing another thing we know that it is value of A x
and B x they are equal and oppsosite but they can have any values but as per static
admissibility is concerned but they should satisfy these conditions.

So let us horizontal reaction at A is equal to H corresponding horizontal reaction at B is equal

to minus H ok and this is statically admissible for any H. Reason you apply the equilibrium
condition here draw the free body it is the free body diagram you apply the equilibrium
condition here you can see that these four systems satisfies the equilibrium condition sation
of f x sation of f o and sation of moment ok.

Now let us what is the solution of these problems ok solution of these problem is if we solve
it I do we have not yet discussed how to solve indeterminate structure but if we can solve it
probably as we understand some of the methods in one of the class we will probably address
this problem and try and solve what is the value of it. But till now take it for granted that if
we solve this problem for the given value given value means if we take q is equal to 1.
(Refer Slide Time: 06:26)

And then h reaction is h is equal to minus 2.5 ok? And the deflected shape of this is shown
here. So this is the solution of the problem how you arrive at this solution for the time being
you take this solution for granted we will come to this point as we proceed Now what point I
want to make it this is as I said just into previous slide the four systems are the is at any value
of H the equilibrium conditions are satisfied. Now let us give some arbitrary value of H and
then say what happens, ok.
(Refer Slide Time: 07:05)

Now you see these are some solutions that we have obtained by giving some arbitrary value
of H, but we satisfy that equilibrium condition, ok. Now here H is taken as 3 kilo Newton in
this case H is taken as minus 3, H is minus 0.5 and H is 1. In all these cases equilibrium
equations are satisfied, ok in all these cases. But just by looking at the (())(07:36) we can
easily say they cannot be the solution of this problem.

One simple reason they cannot be solution of this problem because if you see this problem it
was in support so there should not be any deflection here and there should not be any
deflection both the deflection in both the directions are restrained and not allowed but if you
see in all these cases the beam the support they deflects like this they change their position
like this

So these solutions are not the solution, now it is for some value of H you can have you can
put any value of H and they satisfy the equilibrium conditions but the deformation you get
did not satisfy these support condition that we have in real structure, right. And we can have
such infinite so what is the point we can make here we can have infinitely many possibilities,
possibilities of four system which is statically admissible which satisfy the equilibrium
condition, but among all these infinite four system only one system one force system will be
the solution of this problem.

And that solution will satisfy these boundary conditions these conditions ok. Now this is
called kinematically kinematic admissibility means all these problems if we look at this slide
all these result solutions they the force system that we have here they are statically admissible
they satisfy the equilibrium conditions but they are not kinematically admissible solution
because they do not satisfy the kinematics that structure with given boundary condition and
given load must follow ok.

So then what but at least the information that we have in this slide we cannot arrived at this
solution we can keep on trying substituting different values of either by doing for different
values of H we will get different kinds of solution but we shall not kinematically admissible.
Now so you don’t need to find out we have study equilibrium equations five they give some
information.

But based on that information we cannot find out the solution uniquely we can put some
value of H and get the solution where those that we can get the different shape and the force
system. But that will not be the solution of the real structure, the structure what we have in
this case.

(Refer Slide Time: 10:32)

So in addition to the equilibrium equation we need some more condition in order to find out
the solution uniquely, ok. So among all these infinite possibilities only one case will be the
solution of the system and in order to pin point that in order to uniquely identify that
equilibrium conditions along are not enough we need some additional condition which will
help us to find out the solution uniquely, ok.
Now entire all the process that we have all the methods we have to solve statically
indeterminate structure those methods are different because how these additional conditions
are formed that is different in different methods ok.

(Refer Slide Time: 11:25)

Now we have broadly two kinds of methods one is force methods one is displacement
methods ok. Now these methods are different as I say it in a sense that the additional
conditions we have equilibrium equations but we need some additional conditions in order to
find out solutions uniquely and those additional conditions how those conditions are
formulated what conditions are taken in analysis based on that we have broadly two methods
one is force method one is displacement method.

We are not going to tell you right now we are not going to discuss what is force method and
what is displacement method as we proceed we will understand those but there are two
methods available one is force method one is displacement method. Well in this week we will
introduce this method force method and displacement method. And rest of the weeks we will
apply those methods in different structural problems ok.

Again we will consider three idealizations as as before and those idealizations are being plain
frame and plane process ok. But before we do so before we actually formally introduce these
methods force method and displacement method some of the concept that we need
throughout the journey let us review them, let us understand them.
(Refer Slide Time: 12:51)

Ok! First is degree of static indeterminacy ok we know what is static indeterminacy?

Statically indeterminate structure and what is the degree of that indeterminacy? Degree of
indeterminacy is the total number of unknowns that unknowns could be external or internal
what is external and what is internal unknown we are coming to that point shortly.

But we have total number of unknowns and we have certain equilibrium equations available
some equilibrium equations available independent equations available then the static
indeterminacy will be the total number of unknowns minus total number of independent
equilibrium equations available ok. So as I said we need some additional conditions
additional equations, how many equations are required, the number of equations required
equal to the in the static indeterminacy of that problem.

There is another kind of indeterminacy if remember in one class we mentioned that is called
kinematic indeterminacy will also come to that point, what is kinematic indeterminacy but for
the time being what we need is static indeterminacy will discuss kinematic indeterminacy as
we proceed, Ok. So this is degree of static indeterminacy ok.
(Refer Slide Time: 14:10)

Now let us see some example for instance the first example it is prop cantilever beam it is
very obvious we have considered this problem many times in earlier classes, you see how
many equations are available number of equations available is 3, ok. And equilibrium
equations available is 3. So equations number of equations is available is 3. One is sation of f
x is equal to 0 sation of f y and sation of moment is equal to 0.

And how many reactions we have? We have one is vertical reaction b and then we have A y
horizontal here we have horizontal reaction and then a moment. So total unknowns are 4, so
static indeterminacy which is sometime you know data is n s is equal to 1. So in this problem
the degree of static indeterminacy is 4 minus 3 is equal to 1.

Let us see for this case again number of equations available is 3 and how many unknowns we
have? We have here 2 unknowns horizontal and vertical because it is hinge support horizontal
and vertical and then we have here 3 because it is fixed support so total 3 plus 2 plus 2 total 7
unknown is 7.

So in this case n s becomes 7 minus 3 4. And similarly in this case again equations available
is 3 and unknown available is, it is roller support so only vertical reaction in this case vertical
reaction and horizontal reaction so total 5, total 1, 2, 3, 4, 5 total 5 and then n s become 5
minus 2,ok. In this case now equation equilibrium equation available is 3 but you see there is
a hinge here, ok.
And hinge will provide you one more condition and what is that condition at big point
moment is 0. So total equations available in this case is 4 3 equilibrium conditions sation of f
x sation of y and sation of moments is 0 from any point and additional equation is at the
hinge point moment is 0. So equation becomes 4 and number of unknown becomes 5 in this
case n s becomes 1, ok. So for a beam we can find out for a continuous beam or any statically
what is the degree of static indeterminacy, ok.

Now let us see some more example for frame. Now see all these indeterminacy were external
indeterminacy because these structures are indeterminate because of the external support
system. For instance if in this problem if we remove this support then it becomes the
cantilever beam and it is statically indeterminate structure in this problem if we remove this
support and remove this again it becomes simply supported beam statically determinate.

In this case if we remove one support say one support if we remove keeping one support as it
is then it becomes statically determinate. So in all these problems indeterminacy created due
to the external due to the reactions, reactions from the support. So these kind of
indeterminacy is called external indeterminacy.

(Refer Slide Time: 18:07)

Now there could be another kind of indeterminacy, what is internal indeterminacy? Now for
consider this problem this is a frame, two story frame now you see if we see the support
condition this is how many how many reactions we have?
We have one reaction here and one reaction and then another reaction here and then one
reaction this is roller this is hinge. What should be 3 reactions and we can apply 3 equilibrium
conditions to find out these reactions so these reactions can be determined.

But so externally this structure is determinate as long as our intention is to determine the
support reactions. But now in addition to support reaction we need to find out internal forces
in different members as well. If we try to do that in this case let us see suppose we have
already completed the support reaction, let us draw the free body diagram of AB and then this
will be the free body diagram of AB at point A this is A and this is B.

So at point A we have support reaction A y and A x and point B this is frame so we have three
forces horizontal vertical shear force action force and the moment ok. Now here we can apply
three equilibrium condition and determine with G F M so you can determine that fine. Now
let us draw the free body diagram of joint of this part join B so this is B this is member BE
this is member BC and this is member BE.

Now you see here we have 3 unknown here we have 3 unknown here we have 3 unknown, so
total 9 unknowns we have. You see number of equations available in this case are 3 so we
cannot determine all these unknowns. And if you take any other beam any other joint or any
other beam you can see that the determination of member 4 is not possible in this case just by
applying the equilibrium conditions whereas as far as support reactions are concerned we can
determine the support reactions.

So this kind of indeterminacy is called internal indeterminacy. This indeterminacy is due to

the fact that the number of members provided in this structure is more than required for the
stability of the structure. So this is statically indeterminate structure but it is internal
indeterminacy, ok.
(Refer Slide Time: 20:32)

Now see the next problem now for a frame there is a rule that external indeterminacy can be
computed at 3 into C , ok, where c is the number of close loop you have in the structure. For
instance in the structure let us go back to the previous structure, ok! Let us see firstly for this
problem this is one loop complete loop this is second loop. This cannot be taken as loop
because these are not close this is open loop.

So external indeterminacy for these become 3 into 2 this is 6 and what is the internal
indeterminacy we have? Internal indeterminacy here we have two reactions then two
reactions these are hinge two reactions so total 6 reactions equations available is 3 so it is
internal indeterminacy is 3. So total indeterminacy, total n s is external plus internal means 9,
ok.So you can verify this for different frame as well.
(Refer Slide Time: 22:09)

So let us see for next another problem. For this again apply the same concept we have this is
one close loop this is not close so external indeterminacy is 3 into 1 3 and what is internal
again internal indeterminacy in this case internal becomes 3. So total n s is 6. So this is a total
indeterminacy of the structure.

Now if you want to find out this only the support reactions we need only if you want to
completely analyze the structure we need in addition to equilibrium equation we need atleast
we need 6 additional conditions, ok. Now these are 4 frame if you take any book of structure
analysis there are different kind of problems given and their static internal and external
indeterminacy are given you can do some exercise for that, ok.
(Refer Slide Time: 23:15)

For trus also you can have external or internal indeterminacy for instance if you remember
for trus the condition was if we have number of member is m and support reactions are and
joint number of joint is j then total number of equations total number of unknowns are m plus
r because m member forces and r reactions and total number of equations available is 2 into j
because per joint you have two equations sation of f x and sation of f y is equal to 0 and for a
statically determinate structure this has to be equal to 1.

And if it less than 2 j then the structure become unstable and if it is greater than 2 j then
structure is statically indeterminate but having said that please note when we discuss about
class we said that this is there are some examples where we can show that as per this
condition structure is indeterminate and stable.

And sometimes stable and determinate but still for some loading conditions structure
becomes unstable so this you cannot take just as for granted this is a genuine this is necessary
condition. But whether the stability of the structure needs to be assessed must be assessed by
visual inspection as well, ok.
(Refer Slide Time: 24:38)

Now in this case let us see for this problem let us consider for second one, second one you
see number of joints in both the cases number of joints is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 number of
joint is 10 and for second case number of member is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,11, 12, 13, 14,
15, 16, 17 and number of support reactions are 2 here and 2 here because both are hinge
support total 4.

Now n plus this is 21 and into 2 this becomes 20 so this is greater than 20 so number of
unknowns are more than number of equations so this is statically indeterminate structure. And
what is n s? N s is 1 21 minus 21. Now you see in this case what causes the indeterminacy it
is the support condition if we remove the hinge support instead provide a roller support then
the structure becomes determinate.

So this is the structure is determinate but indeterminacy is external indeterminacy. Now take
the first one again number of joints are 10 and number of members in this case we have one
additional member this one, so number of members are 18 and number of supports are 2 here
an 1 here so number of reactions are 3 so again this becomes 21 and this into 2 become 20.

So number of unknowns are more than number of equations available it is again

indeterminate structure but what causes this indeterminacy the additional member this
additional member causes the indeterminacy if we remove this member the structure becomes
indeterminate. So this is an indeterminate structure but this kind of indeterminacy is internal
indeterminacy, ok.
(Refer Slide Time: 26:45)

Now some more concept one is principle of super position. What is Principle of Super
position? Principle of Super position is you see in one of the classes probably we discussed
what is Principle of Super position since in this all the problems that we are doing here it is
linear problem means force and displacement relations are linear, ok space change relations
are linear.

So what we can do suppose this is th actual structure that we need to solve we need to find
out the we need to analyze find out the support reaction, suppose in this case the
displacement at point C we need to find out for this force system. What we can do is we can
break this structure into several small small parts.

First things suppose this is the structure and this structure same structure only consider load P
1 and second case consider the same structure but only consider the distributed load and say
third case same structure but only consider the P 3 Case. Now suppose in this case it is delta 1
delta 2 and this delta 3 so delta 1 is the displacement at the same point causes by P 1 delta 2 is
the displacement at the same point causes by Q and delta 3 is the displacement at the dame
point causes by P 3.

Then we can say that the actual this structure is this. So total delta, delta for this structure can
be obtained as delta 1 plus delta 2 plus delta 3. This is called Principle of Super position
means we have we obtain the solution for small small part and then super impose for those
solution to get the actual solution.
It is a very very effective way because again in advanced structure analysis or in different
method of analysis we will see that this is the underlined philosophy of most of the method
that you have a structure needs to be analyzed or you have a problem that needs to be
analyzed divide the problem into small small sub problems and then assemble them together.

And how you assemble them that depends on whether the problem is linear or non linear but
in this case since we are the linearity is one of the important assumption so we can just add
them to get the final deflection at this. Whether that is true for bending moment, shear force
support reactions in internal forces ok so this method this is called principle or super position
we will be using frequently this principle concept of Principle Super position.

(Refer Slide Time: 29:15)

And other important concept is compatibility what compatibility say just to demonstrate take
suppose this is a simply supported beam subjected to not simply supported beam it is a
continuous beam indeterminate beam subjected to some loading system and this blue line that
you can see that is the deflected shape of this beam.

Now if we draw a slope at b then suppose this slope is theta BA and this slope is theata BC,
ok. What compatibility says that this theta BA has to be equal to a theta BA this is one
condition and the second condition is at B, since it is supported at the hill supported at B the
deflection of B should be equal to 0. So delta B should be equal to 0 and this angle and this
angle this slope obtained from this segment and slope obtained from this segment at the same
point has to be same. Ok.
Just to one more example suppose this is another example and these blue lines that you can
see that is the deflected shape ok. And at this joint B we have three members member AB BC
and BD. Now suppose theta BC is the slope at B but obtain from segment BC similarly BA is
from BA and theta BD is from BD what compatibility condition says that because this is this
is a rigid joint and initially it was 90 degree so it remains 90 degree there will be no rotation
at this joint.

So these angles are 90 degree these angles are all 90 degree so it says that this angle this
angle and this angle should be same, ok. And we have used compatibility conditions earlier
also but probably without referring to the term compatibility for instance if you remember we
solved this problem right and we have a simply supported beam wich is subjected to a
concentrated load P, ok. And this is A this is B and this is C.

And if we want to solve it this is P if we want to solve it by direct integration method then the
direct integration equations that we had was d 2 y d x 2 is equal to minus M by E I, right? So
but what is the bending moment diagram for these beam? The bending moment diagram for
this beam is like this, like this ok. So bending moment in part A B and part B C they are
different. So we need to apply these we have to integrate this equation over B over AB and
over BC separately. Therefore each case we will get two constants because the second order
equation.

So for AB we get four constants C 1 and C 2 and for BC also we get two constants C 3 and C
4 what are conditions we have to determine this constant boundary conditions we have that A
is equal to deflection is equal to 0 at C is equal to deflection B is equal to 0 but that will give
us two constants but other two constants we need two equations but what are those two
conditions if you remember we use that slope at B is equal to 0 and deflection and slope is
continuous at B and deflection is continuous at B.

Means slope obtained from AB at B should be equal to slope obtained from BC at B similarly
deflection at B obtained from AB and deflection from deflection from B obtained at BC
should be same. This was compatibility equation ok and this figure and this figure exactly
tells you that so we are used compatibility equation even in even while solving statically
determinate structure ok.

So compatibility equation is another thing that will be used very frequently while studying
indeterminate structure ok. So what we do is stop here today next class as I said there are two
kinds of method one is broadly one force method one is displacement method what we do
next lecture we just introduce the force method.

Ok Thank you see you in the next class.