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Let’s go back to this line diagram. This
is a nice line diagram, I can use this for
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analysis. Next thing I have to do is to do
analysis of this structure. Let’s just for
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now have a force vertically acting here P.
Before analysis I should tell you what are
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all the dimensions. Let us finish that. Whenever
we give examples, we give nice simple examples
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and ask you for difficult problems but we
want to take representative examples. So here
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without loss of that representation, let’s
say I have equal lengths of B E and A B and
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let’s say the height is such that this length
is also L. What will this be? The height is
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root of 3 by 2 L. Find the required reactions
from the supports? Where are the supports
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here? This is one support, this is another
support. Support meaning supports this structure
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to a particular fixed frame.
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Find the required reactions from the supports,
find the internal forces in members. It could
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be in all members so here I could have it
as all or a set or a single member. Depending
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on my design requirement of having to build
this, I would ask these questions and given
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these answers, I will be able to go back and
say the structure is well built. It can take
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the load P that is acting on it. For example
if this is a piece of chalk and I wish to
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know whether this piece of chalk will hold
attention.
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Now you notice I am going to increase, increase,
increase, increase, increase, increase, increase,
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increase and then at a particular point of
increased force, it breaks. If I wish to find
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out the force that is acting on each of these
members is such that they are well within
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the breaking load then it is all right to
have such a structure and safe at it. If I
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am going to use this as a piece of chalk,
please remember I am applying a load at the
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center like this. Therefore there will be
a moment also coming into each other. We will
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come to that particular type of structure
at a later stage. If this is clear we will
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move on. For now let’s assume that we have
to find out all the reactions and all the
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member forces. How do I go about doing?
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If I have to find out the reactions from the
supports, one of the important things that
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I have to know is I have to remove these supports
and draw a free body that is devoid of the
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supports. Let’s do that exercise as a first
step. Let me redraw now. I have removed at
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E the support, I have removed the support
at A. I am not bothered about the others so
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I am just going to just take it as a particular
body. At point E or at joint E, if you notice
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carefully this is a pin joint or hinged support
which will introduce two reactions E along
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y direction and E along x direction. At A
going back to this, we have a roller support
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which means the reaction from the fixed frame
will be only in the vertical direction, I
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will have Ay. This is something that I have
done when we dealt with simple rigid bodies
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that’s basically what I have done here.
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Now I will insert the external force on this.
This is the external force. I don’t bother
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about anything else inside. There are no other
forces acting and this is a free body. I can
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always apply the equations of equilibrium.
In this particular case it is stationary or
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it is a static equilibrium. I can employ the
three equations sigma, let’s say this is
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the x direction, this is y direction. I can
employ these three equations. This moment
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can be found out with respect to any of them.
Let me just take it as moment about E. These
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are the three equations or equivalent equations
that I need to use. I will get three sets
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of equations. How many unknowns we have found
out? This is P, this is a force that is applied.
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These are the three unknowns Ay, Ey and Ex
to be found out. Three equations I can as
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well solve for it. Is it clear? This is something
I have already done.
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So let’s just jump to the answers sigma
Fx equals zero immediately tells us this is
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equal to zero. Instead of taking vertical
equilibrium and moment equilibrium, I will
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take moment about this particular point which
will involve Ay and P. This distance is L
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plus 1 by 2 which is 3 by 2 L and this distance
is L and therefore if I take moment about
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this Ay gives a positive sense, P gives a
negative sense. This implies Ay times 2L minus
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p times 3 by 2 L and this is equal to 0, I
can cut off L. Net result will be Ay equal
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to P times 3 by 4 so 3 by 4 P. Since Ay plus
Ey should be equal to P, Ey will automatically
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be 1 by 4 P.
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This is pretty clear. We have accomplished
the first task as finding out what are the
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reactions. The next task that we have in hand
is to find out the internal forces in let
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us say for now all members. As a first step
in finding out internal forces, we know very
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well that we have to cut any one of those
members for which we want to find out the
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internal force. Supposing I want to find out
the internal force in B E, I need to cut that
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particular number. Let’s seek as a first
step to solve for internal forces. We need
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to cut and let’s cut and see if we can find
out how to solve for internal forces. If I
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cut a single member, please remember it is
a part of a larger system. I cannot just draw
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free bodies and that’s not going to give
me anything that is going to be used. For
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example if I had cut this like this, let’s
say this is the cut.
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It is not going to give me much of it. I will
have equal and opposite forces acting on this
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like this. Since they are equal and opposite
in this structure, I will not be able to solve
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for this internal force. This is an important
thing to understand. If I can extract the
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free body out of this entire body then it
is possible to write down separate equations
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from which we can solve for internal forces.
Given this, how do we go about picking out
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or removing certain free bodies from which
we can get some results. If you look at it,
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these are the pins and if I draw a particular
domain that engulfs each of these joints
and then look at this particular zone alone.
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I can pick out each one of these as rigid
bodies and draw the forces acting on that
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free body. Let’s take E for example.
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We have cut here this is E, we have B over
here, we have D over here that these two members
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have been cut. There is an external force
acting on it equal to p by 4. Since we have
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cut here, we will be exposing an internal
force in each of these members that act along
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the axis of these members. I will tell you
in a moment as to what direction force I have
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to insert over here. For now I will just put
the internal forces to be like this. Since
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this is an internal force for the member B
E, I am going to call this as FBE or simply
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B E. Some people follow FBE, some people follow
just B E. We are going to use just BE as the
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force and use that notation all through and
therefore this is D E. There are two forces
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B E and D E which are yet to be found out
that are acting on this particular free body.
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Is this clear? Apart from the external force
that is already known, these are the two forces
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that act on it. If we examine this free body
and ask the question how many equations can
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I write. Since these three forces are coincident
at E, you can see that from E, B E is emanating,
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D E is emanating. The force E P by 4 is emerging
which means these three forces are coincident
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forces and therefore I cannot write any moment
equation separately. I will only have two
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equations for vertical and horizontal equilibrium.
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If I take x as horizontal and y as vertical
this way, I can now write the total force
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in y direction is equal to zero, the total
force in x direction equal to zero. This lets
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say if this angle is theta which can be found
out from geometry here. In this particular
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case we already know that this angle is equal
to 60 degrees. Now which one would I choose,
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horizontal equilibrium and then vertical equilibrium
or other way it depends on your convenience.
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Supposing I take the horizontal equilibrium,
what will I have? I will have B E acting along
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the positive direction. I am going to write
this as positive and a component of D E along
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x direction which is again positive notion
plus D E cos 60 degrees is equal to, there
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is no external force acting on it, is equal
to 0. This is one equation that I can get.
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The other equation concerns vertical equilibrium.
Upward is positive which means P by 4 plus
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DE sin sixty equals zero for static equilibrium.
From these two equations, I can solve for
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B E and D E. Now which one will I use first
to solve? Obviously B E and D E take part
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in this equation whereas in this equation
only D E takes part and therefore I will solve
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this first and then go for this. That is one
of the reasons why I started with sigma Fy
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equal to 0 and then said sigma Fx equal to
0.
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It’s a matter of choice as to which order
you will go, so that you can solve in a simple
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way. If I had started from this, I will only
get relationship between these two and then
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I have to use this equation in order to find
out what each one is. Going back to this,
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we looked at this free body and from this
free body it was possible to find out what
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is the force acting on this particular member
B E and the member D E. Similarly I can do
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it for each one of these free bodies. Free
bodies of each of the joints I am just going
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to write… What I have essentially done is
I have engulfed so that I have each of these
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joints separated. I can have free bodies associated
with each of the joints. So I am just going
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to call those as free body of the joints are
used to solve for member forces.
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Now what’s the guarantee that this particular
free body is a stable body in or in other
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words, it’s a stable and stationary in this
particular set of forces that are acting here.
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We will postpone that and look at it separately
in a clipping. Such a method where I used
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the free body of the joints or free body associated
with joints in finding out the member forces,
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I am going to introduce here. Member internal
forces or simply member forces is called method
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of joints. Now in this particular case, I
can draw free body of A B C D and E so that
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I can solve for all the member forces. But
I am not clear which one to start with. Shall
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I start with B, shall I start with D, C. That’s
a question that will arise. What do you think,
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what criteria will you adopt in order to choose?
This is one of the problems that many students
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will face. The best thing that you can do
is examine.
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If you look at this particular joint E, I
have B E and D E as unknown member forces
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acting. Are there any other unknowns? No,
and for every joint how many equations can
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I write? If I go back to this, I can write
two equations for every joint and therefore
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if I am able to generate two equations for
two unknowns, I can directly solve for those
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unknowns. Going back to this and therefore
the first attempt that i will make is to find
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out, if there are joints with only two member
forces. What are they? E A are the two joints
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which have only two member forces. From these
what will I find out? I will be able to find
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out B E, I will be able to find out D E, I
will be able to find out A C and I will be
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able to find out A B.
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Now if I ask the question, what next? Then
I will go to any one of these and ask the
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question, how many are still unknown here.
Since this is already known, there are two
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of them that are unknown here. Since this
is already known, there are two of them that
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are unknown. Since these two are already known,
these two are unknown and therefore they actually
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have equal importance with each other but
if I take this particular joint, there are
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no external forces and therefore this will
be a simpler one to do compared to C or B.
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The next choice of taking equilibrium will
be D and naturally the other one would be
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P or B. The choice may be P, if I am taking
vertical equilibrium where P will directly
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come in. Here I have to take the components.
Mind you if I do D, I would have already found
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out these two. The only one that is left is
here and if I take C, I would have found out
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this. The equilibrium of B is a redundant
one. That’s how I will solve for all the
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member forces. Is this clear? Just to recap,
I draw a line diagram of this particular truss
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member, find out the reaction forces so that
I can draw a free body diagram of this truss.
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Assuming that I already know that this entire
structural system forms a rigid system. I
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can now start to examine and find out the
internal forces by drawing the free body of
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each of these domains that encircle as single
joint or in other words free bodies associated
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with each of these joints. After doing that
I will pick the appropriate order of these
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joints for writing equilibrium equations to
solve for the internal forces in these members.
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Mind you this is a simpler problem because
over the entire length of the members, the
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axial force is the same and therefore it’s
a single value that I need to find out. This
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is a simpler structure that I can think of.
In a more complex structure, I may need to
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find out the internal forces over the entire
length of the member. That is a simplicity
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that this system, called the truss system
offers. Thank you.
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Now one of the most important concept that
we have to understand here is about sign convention.
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I skipped it when I was solving for the problem.
Let’s go back to that and ask the question,
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how do I make a sense of positive, negative
and that becomes a big confusion here. I will
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just tell you in a moment, how it becomes
a confusion.
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We started with this problem. I have just
redrawn in a bigger way here. These are the
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free bodies that I have taken for just an
understanding and to just indicate the forces
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over here. Assume we have cut like this. We
have cut each of these members and therefore
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there is an exposure of forces. Let me just
draw this small one here also. This will have
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equal and opposite forces. Now at this stage
I will have a doubt which direction should
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I take it as positive. This direction or this
direction? Is this positive or is this positive
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for this particular internal force? Now this
force, this force, this force, this force,
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all these forces indicate internal force in
this particular member and therefore this
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confusion is a very common confusion. In order
to understand this, the first thing that you
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have to remember is the entire member B E
is having a single internal force. Let’s
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say B E in this particular case. This internal
force B E is taken as positive
if B E is in tension.
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If you remember we started with this particular
notion. I showed this particular member, asked
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my friend venketesh rao to pull it. When you
pull it, it is in tension which means I have
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a positive internal force and when I push
it, I have a negative internal force. Now
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examine these to get an idea of what kind
of force I have drawn. Let me just encircle
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them to get you an idea. Let’s take this
member B E. I will just cut both sides, expose
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the force. Remember this is equal and opposite,
this is equal and opposite. What kind of internal
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force is acting here? It is a tension force.
How about here? This is member D E so this
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is D E, again I have indicated forces in such
a way that this particular member is being
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pulled both sides.
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This is also under tension. If you notice
all the other members, I have drawn forces
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in such a way that each of these members are
assumed to be in tension or in other words,
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this is a positive value, if they are in tension.
I will have a positive value here, if this
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is what it is. In the result of the forces,
supposing this value that I get is a negative
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value, it means that it is pushing this member
into compression. What I am going to do is
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I am going to start with a value which is
positive all through. So I will take the positive
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notion for each one of these. If I draw this,
I know that this member is in tension. Equal
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and opposite is what you have here, so I will
indicate it to make sure that you understand.
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Look at how the direction is here. The same
force as a direction which is opposite are
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the other free body. Why? because equal and
opposite have to be exposed.
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Similarly in each one of these. So that I
get an understanding of how to draw the positive
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values of forces for each of the free body.
As you can see in this particular case, I
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don’t have to waste any time in order to
draw this particular notion, having drawn
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each of these members separately as tensile
forces on these members. To repeat, if I snap
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out a part of this particular member B E and
assume that it is under tension, I will have
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these two forces. Since I have removed this
part, the other part of it should have equal
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and opposite reaction and therefore these
two conventions. If this is clear, I will
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tell you a simple way of drawing the correct
positive directions of each of these internal
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forces. For that I am just going to erase
these individual members that I have drawn.
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The question is how do I make sure that I
draw forces that indicate positive values
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and if I get a negative value for this, it
means that the member is in compression.
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Let me just write down the member forces here.
This is B E, this is D E and so on. Note that
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if I choose this particular joint E, each
of these forces seem to be pulling this particular
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joint away from this particular joint. If
this is the body, they are all away from the
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free body that I have drawn. If I take this,
they are all emanating from this particular
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joint. Emanating like this, emanating like
this, emanating like this, emanating like
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this, emanating like this.
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Similarly point D emanating like this B C,
emanating like this A B, emanating like this
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B E and emanating like this B D. Very simple.
When I draw these free bodies of every joint,
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I will take the directions which are emanating
from that particular joint and all these are
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treated as positive internal forces. If I
get a negative value for any one of these,
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it means that particular member that I have
for example if I get a negative value for
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B E it means that member is in compression,
as simple as that. This is one of the most
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simplest one but yet people make lots of mistakes
in inserting this particular sign. Thank you.
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In this particular problem let’s examine,
I have a force P acting on this. We started
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with this particular joint E and found out
what is B E and D E. B E happens to be…
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can you tell me what it is? D E is, we get
minus P by 2 root 3. We happen to get from
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equilibrium at E in the vertical equilibrium,
we get D E is equal to… Let me just insert
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support reaction here. We happen to get D
E equals minus P by 2 root 3. What is this
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equal to? This is equal to minus, so what
should happen is this direction happens to
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be like this. Mind you this direction also
should be this way. This is an important thing
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to note. All you have to do is if I get a
negative value, if I retain the negative value,
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I can have a direction like this and insert
a negative or if I change the direction, it
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is equal to P by 2 root 3.
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Other way of looking at it is I can retain
the direction that I have which is like this
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and like this and insert the value minus P
by 2 root 3, same here minus P by 2 root 3.
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One of the common mistakes done by the people
is since this direction is minus, this should
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be plus. This is an internal force, if this
I got as minus, this also should be equal
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to minus. When I solve at D, this particular
direction is taken as it is and the magnitude
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is taken as minus P by 2 root 3. This is important
to note. If this sign convention unless done
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properly, you will get into results that are
not correct.
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