# Lecture 16 – Tutorial on Force Displacement Relationship and Geometric Compatibility

Welcome back to the course Mechanics of Solids.
So, in the last lecture we were talking about the thin walled pressure vessel that is and
we we found out the the force developed in that thin walled pressure vessel as well as
so we have found out the radial expansion due to the application of the internal all
round internal uniform pressure ok. And then we talked about the mechanism of the belt
around a wheel, so the when the wheel is not frictionless, so there we found out the the
the tangential force developed at one point of belt will not be same at the other points.
So, because of your friction, it will be that force will be increasing in the exponential
order ok. Now, we will be taking couple of numerical
problems. So, first problem it says a triangular frame supporting a load 20-kilo Newton is
shown in figure. I will show the figure later on. Determine the displacement at the point
D due to the 20-kilo Newton load carried by chain hoist ok. The problem says, so this is the figure. Ok.
So, the problem says a triangular frame supporting a load, this, triangular frame is this, so,
B C D ok is supporting a load of 20 kilo Newton as shown in the figure. So, determine the
displacement at the point D ok. So, this is displacement of point D due to the 20 kilo
Newton load carried by the chain hoist. So, this is the chain hoist because of that you
need to find out the displacement of D ok, where it is given the cross sectional area
of CD equal to 3200 millimeter square, cross sectional area of BD is equal to 491 millimeter
square. And the modulus of elasticity of both the members are made of same material, so
E is given as 205 into 10 to the power 6 kilo Newton per meter square. Ok.
So, before going to find out the geometry compatibility and all what is your first step
first step is to study the forces right, by exploiting or by satisfying the equilibrium
conditions ok. So, let us draw the free body diagram of this ok of this frame. So, the
free body diagram will look like this. This is point C, you
will be having the force F C from the support, this is point D, this is point B. So, similar
kind of problem already we have solved earlier. So, you will be having the reactions F B y,
F B x and this will be the resultant FB ok. Now, again for the sake of completeness, I
can ask you that why I have considered we can we can discuss this thing that why we
are considered F C in the horizontal direction and F B along the member BD. So, these things
already should be known to you already we have discussed this thing earlier when we
are discussing similar kind of problem ok when we are talking about the I mean when
we are discussing or when you are determining the support reactions anyway.
So, now basically what we are going to do we are going to find out the support reactions,
because without knowing the support reactions basically you cannot proceed so that is why
the free body diagram has been drawn. And then support from the support reactions basically
once you know the support reactions, then you can you can you can consider different
joints and you can find out the member force. So, once you know the member force then only
you will be able to find out the geometric deformation right. The second step is that
study of deformation, is not it? The first step is the study of forces. So, now we are studying the forces. So, deformation
will be coming later on when you know the forces are acting on different members ok.
So, now, considering the equilibrium, that is the we are considering moment equilibrium
that is summation of moment with respect to point C is 0. So, what we can write? F B x
into 3 ok minus 20 into 3 is 0. So, from that we can get F B x equal to 20 kilo Newton.
Fine. Similarly, we can exploit this condition summation
of F x where you can think of, this is my x y direction ok, summation of F x equal to
0. So, I can simply write F c minus F B x equal to 0. So, from which I can get F c equal
to F B x equal to 20 kilo Newton ok. Similarly, I can exploit this condition as well, summation
of F y equal to 0 from which I can get F B y what are the forces acting in y direction?
F B y and 20 kilo Newton ok. So, F B y will be simply 20 kilo Newton.
So, therefore, we can get F B equal to root over F B x square plus F B y square similar
thing we have already done for earlier in the lecture, so that will be giving me 28.3
kilo Newton ok. So, now, we will try to find out the forces in each member. So, what are
the members I mean do exist in the problem member BD and member CD. So, we need to find
out the forces carried by each member of the frame ok. Now, we we know the support reactions;
that means, F B and F C both are known to me. Now, if we consider each joint, so joint
C if we consider or if I consider joint B, I can find out the member force in CD and
BD. Ok. So, let us do that. So, we are considering.
So, force now we are talking about and also still we are studying the forces say we are
try to find out the member forces. So, this is my member CD, and we are considering the
member force in CD is compressive in nature. I mean I told you several times I am telling
this thing that you need not to think about whether it should be compressive or tensile;
right now I am considering compressive that does not mean that I know the answer that
is why I am writing ok. So, you can take tension, if it is really tension then it will become
positive; if it is not really tension then it will becoming negative.
Now, for example, if I consider compression ok and if I get positive value of F C D then
it is really compressive; if it is not really compressive then it will becoming negative.
So, this thing I mean several times I am saying you. So, I mean you need not to think much
about these things the direction of the force. Well, and I am considering BD say tensile
ok. So, I am considering BD as, BD is under tension that is F B D is tensile in nature
and F C D is compressive, compression force ok. So, now, you need to find out the force
F C D and F B D; F C D and F B D. Do you know the magnitude, numerical value
of these forces? Yes, you know. How do you know? If you consider the individual joint,
so suppose I am considering joint C this is my joint C, this is F C ok this is F C and
this will be F C D. So, F C D is nothing but F C, agreed? So, therefore, whatever direction
we consider that is compressive force F C D is compression in nature then that comes
perfectly fine ok. So, F C D is equal to F C. So, I can write F C D equal to, F C D was
F C was 20 kilo Newton. So, this is 20 kilo Newton: and what about F B D? So, you consider
joint B, so this your joint B ok. So, from joint at joint B you are having F B. So, this
is the vertical direction, this is the horizontal direction ok and this is your F BD. Right?
So, from this from the equilibrium of joint B, you can simply say F B D is nothing but
equal to F B. So, F B already you have calculated. So, F B D will be F B is nothing but 28.3
kilo Newton. So, my study on the forces is over. So, I have got the information about
the support reactions; and from the support reactions, I have got the information about
the member forces right. So, the study of forces is all over. Now, we will be studying
the deformation. Now, if you look at this member CD and I am
calling I am I am I am saying that member CD as well as member BD both are deformable
bodies right, I am not talking about by rigid body or something like that, they are deformable
bodies. Now, under this action of force F C D, the member CD will be compressed. Now,
what will be the amount of compression that we can calculate by using your Hooke’s law
P L by A E. Similarly, if you consider member BD under the action of this tensile force
F B D that member BD will be elongated how much will be the elongation that also we can
calculate by using the simple Hooke’s law. So, let us calculate those things ok and let
us get the numerical values. So, your delta BD So, delta BD is the expansion
or the elongation of the member BD because that BD member is under tension ok so that
is given by F B D length of BD member cross sectional area of BD member and E BD ok. So,
you put the numerical values here F B D is 28.3 into L B D you can calculate 4.242 into
10 to the power 3 by area is given 491 into 10 to the power minus 6 into E is given 205
into 10 to the power 6. Ok fine. So, from this you can get the value will be coming
as, if you calculate the value it will be coming as 1.19 millimeter. Similarly you can
calculate F S delta CD which will be compressive which will be the the shortening of the member
CD ok. So, that is F CD L CD A CD E CD. So, if you put all the values you will be getting
0.0915 millimeter this is you can write extension and this is compression ok.
Now, whatever values you are getting, what it will indicate that means if you do not
have the frame, suppose you have individual member BD and CD under the forces F BD and
F CD respectively then member BD will be extended or will elongated by this much amount, and
member CD will be compressed by this much amount. if I do not consider any assembly,
but when they are actually connected or when they are present in an in an assembly then
basically they I mean this these values these values have to be satisfied, but at the same
time they cannot be, they cannot behave individually. What I mean to say the joint suppose if if
I consider the elongation of BD, the BD will be elongated like that and CD will be compressed
like that, then where is the point D. So, now CD after compression D point will be coming
here, and after elongation of BD, D point will be coming here that is not acceptable
because they are acting in an assembly. So, the D point will be the common point even
after deformation; before deformation D point was the common point where CD member and BD
member both members are meeting; even after deformation D, D point will be the common
point for both the members, but at the same time they have to experience this much of
deformation. Now, how to do that? Ok so, let us draw the
geometric compatibility figure ok and then I will explain. Let me draw it. Ok. Now, let
us understand this figure first. As I told you that if BD and CD are not in the assembly
ok, so this BD is free to elongate up to BD 1 agreed? If what I what I said if BD and
CD are not making this frame or they are free to expand or they are free to deform then
BD due to the action of F B D that is the member force BD will be elongated up to BD
1. So, this from D to D 1 distance is nothing but the elongation delta BD whatever we have
calculated fine. Similarly if CD is not in in in the assembly if CD is free to deform
then this CD will be compressed ok under the action of F C D, and it will take the position
of D 2 C D 2. So, D will be shifted to D 2. So, now, CD 2 is the compressed member if
we are allowing free deformation of the members right. So, D, D 2 is nothing but your delta
CD agreed. So, what and for example I mean for for the sake of completeness, we must
think that this problem we are defining with small deformation problem. If we consider
this is the same problem with large deformation problem the formulation and analysis will
be completely different, but so far as I told you in this particular course we are not talking
about or we are not concerned for that large deformation problem, we will be only restricting
ourselves to the small deformation problem. So, if it is the small deformation problem,
so and the figures of course it is little bit exaggerated figure to understand in a
better way ok. So, this D D 2. D D 2 is your delta CD, and D D1 is your delta BD. So, basically
BD will be taking the point D 1 and CD will be taking the point D 2 if they are free to
expand or compress, but that is not happening because D point is the common point. So, I
cannot have some different point D 1 and D 2. So, they should be in the same point.
So, what is happening then? D if you consider D will be taking the arc kind of thing is.
So, D point this is the common point for BD and CD this point will be becoming D prime
point. Where this BD is getting expanded and CD is getting compressed ok. Though these
angles and all those things are looking not looking like the same angle but for the small
deformation problem and as I told you this is a very exaggerated figure, but those angles
will be remaining same, anyway. Now how to get this D prime point, now what
we are doing here? From this point D 1, we are drawing one normal to line BD or the BD
1 rather we are drawing we are dropping one normal to this that line is coming like this
ok. And from this CD from point D 2, we are dropping another perpendicular line. So, these
two perpendicular lines are meeting with a common point D prime ok. So, frankly speaking
that should happen through the I mean I mean circular arc, but nevertheless if we consider
like this dropping the normal that will be also fine I mean that will not create any
significant difference because we are dealing with the small deformation problem.
So, what we did we did we draw we draw a normal line or the perpendicular line from D 1 and
another one perpendicular line from D 2, which will be meeting a common point D prime ok.
Now, this D prime is the final point after deformation before deformation it was at D
now after deformation it is becoming D prime. So, what is the displacement of point D. So,
D is coming to D prime. So, this much is your horizontal displacement or horizontal shift
of point D and D is coming to D prime. So, therefore, this is your vertical displacement
of point D. Fine, ok. So, if it is so then from the figure it is
very clear that is delta DH that means, the horizontal displacement or the horizontal
shift of point D is equal to delta CD. You just look at this figure is equal to delta
CD is nothing but 0.0915 milimeter ok; and what about your delta dv; that means, the
vertical displacement of point D that will be, if you look at figure D 2 F plus F D prime.
So, D 2 F D 2 F plus F D prime. It can be further written as D 2 F is nothing but D
2 it has D G, you look at here D G this dash line or the dotted line plus F D prime. Now,
this DG can be written in terms of delta BD in this fashion. So, delta BD by cos 45 degree
plus. Now, what is F D prime? What is F D prime? So, F D prime is nothing but your FG
because this angle is 45 degree right. So, if this angle is 45 degree, so FD prime is
nothing but equal to FG. Now, what is FG? FG is nothing but equal to delta CD. So, I
can simply write delta CD. So, by putting the values here you know delta
BD, you know delta CD the numerical values by putting the values here you will be getting
1.77 millimeter. So, what did you get, you got the horizontal displacement or the horizontal
shift of point D is this much and the vertical shift of point D is this much ok. So, this
is the way you solve the geometric, I mean basically this is this figure is coming from
your geometric fit or the geometric compatibility because D point cannot be different. So, you
cannot have D 1, D 2 two different points after deformation it will be the same common
point. So, D has to take the position of D prime.
Well, I will stop here today. In the next class, we will take another numerical problem
and then we will move to the next topic. Thank you very much.

## 3 thoughts on “Lecture 16 – Tutorial on Force Displacement Relationship and Geometric Compatibility”

1. Sathya B S says:

Excellent explanation.. Thank you

2. Keiana Simpson says:

How did you get the lengths for Lbd and Lcd??

3. Raj Badkar says:

Sir there is cos 45 or sin45 please explain i got answer 1.77 from sin45