# Lecture 18 -Tutorial on Force Displacement Relationship and Geometric Compatibility -3

Welcome back to the course Mechanics of Solids.
So, basically we were discussing about, in the last lecture we were discussing about
one numerical problem and that problem was like this right, where A E B member was a
stiff member, that is a rigid member we are not considering it deformation there and that
member is connected by three steel rods that is D E, B D and B C. So, in the last lecture we found out the member
forces, forces in all members right B C, B D and D E, and we found that B C, B D is not
carrying any load. So, therefore, B D must be equal to B prime D and at the same time
B E must be equal to B prime E prime because that member itself is a stiff member we are
not having any deformation there and A E is equal to A prime E prime and based on that
we found out the horizontal movement of point A which was basically our one of the objectives
right for this problem. So, horizontal displacement of point A was found as 0.024 inch if you
recall ok. So, now today basically we are going to find
out the vertical movement of point A that is nothing, but delta AV. So, delta already
has been obtained. Now let us see how we can find out delta AV. So, let me draw it very
simple line diagram this is your A. So, this is your delta AV correct ok, now this is this
point was E prime. So, this point was A prime and this point was B prime. So, if you just
take out this configuration and if you blow it up, so it will look like this ok and this
E double prime is the point where if you draw a normal from point E to the line AB, so that
will be intersecting at E double prime. Similarly if you draw a normal from point B prime to
member A B that is intersecting at B double prime and this is nothing, but your delta
BV by definition ok. This angle is say theta this angle is also
theta and this is nothing, but the vertical movement of point E that is delta EV. I hope
you have understood because this problem the conceptualizing this problem is difficult,
once you conceptualize this problem then rest of the things is pretty simple ok the calculation
is pretty simple, only thing is that you have to you have to find out or you have to determine
how it will deform ok. Fine. So, now this point I am saying as point O
ok. So, what I can write, if I consider OE double prime is say a feet therefore, OB double
prime OE double prime if I say a feet then OB double prime should be 6 minus a feet because
the total length B E or B double prime E prime both are equal both I mean are equal and they
are nothing, but 6 feet ok. So, now if you look at this geometry I can
simply write just say what I am writing tan theta is equal to E prime E double prime by
OE double prime. Can I write that? Very much so ok. So this is nothing but delta EV E prime
E double prime is nothing, but delta EV as shown in the figure and OE double prime is
nothing, but your A into 12 that is nothing, but you are in inch actually ok, a feet basically.
So, OE double prime is nothing, but a feet. So, I am multiplying 12 because I want to
convert that thing any inch, fine. So, similarly tan theta can be written as if you look at
the figure tan theta can be written as delta BV plus delta EV divided by the total length
6 feet ok. So, if you consider one triangle like this.
So, delta BV plus delta EV divided by 6 feet, which can be written as delta BV plus delta
EV into 6 into 12 that is an inch I am expressing everything in inch ok. So, from this you know
delta BV right, you know delta EV right, in the last lecture we calculated that delta
BV and delta EV both we have calculated in the last lecture when you are talking about
or when we started this problem ok. So, from there if you put the value, so delta B V we
obtained as 0.024 and delta EV you obtained as 0.018 divided by 72 that gives me tan theta
ok. So, I can write here that gives me tan theta ok.
Now, from the geometry, from the geometry ok what we can say? That delta AV. If you look at this delta AV plus delta BV
by 18 feet 18 into 12 that is inch. So, what I am doing? I am considering a big triangle,
something like this a big triangle I am considering just simple geometry class 8 standard geometry
or 7 standard. So, that is nothing, but your tan theta ok. So, you know delta BV you know
tan theta from there. So, you can find out delta AV which is equal to 0.102 inch ok.
I hope you have understood this. So, delta AV is equal to 0.102 inch and in the previous
lecture we found delta AH. So, our objective is fulfilled. So, we have got the deformation
or the movement or point A in the vertical as well as horizontal direction ok.
So, this is all about your second chapter. Now we will be moving to the next chapter
that is concept of stress and strain ok. So, now, you should know what is stress first
we will talk about stress and then we will go to the strain part well. So, now, we will
start the new chapter that is concept of stress at a point plane stress case, transformations
of stresses at a point, principal stresses and Mohr’s circle ok. So, first we should
know what is stress ok, how how to define stress. Now if you look at a body, any any
body under the action of several external forces like here whatever is shown F 1 F 2
F 3 F 4 and so on up to Fn. So, there are n number of forces acting on this body. Now, we are considering one plane, we are
we are cutting this body in such a way that we can get this is the plane. So, this is
the plane ok, this plane is containing one points say point O ok. And now after cutting this body on this plane
we are getting this plane like this ok. So, now, this plane is discretized with very small
segments in number of segments or infinite number of segments and we are considering
one plane which is containing point O and that plane is defined by the direction normal
n and you know from your knowledge of physics that any plane can be defined by its direction
normal ok. So, this plane this small segment can be defined by this direction normal and
this on the small segments some delta F that force is acting.
So now, how this force has been developed? So, when you are applying some externally
applied force right every system will be generating some internal internal forces ok; it will
be developing some internal forces which will resist this external force. So, these are
the external forces right, these are the external forces now due to the application of these
external forces when you are cutting the body with this plane ok and on this plane if you
consider different or infinite number of segments small small segments and all those segments
some internal forces will be acting and these internal forces will try to balance the external
forces right. So, one such force, one such internal force is say delta x which is acting
on a small segment which is containing point O and this small segment is defined by the
direction normal n, I hope you have understood this ok. So, if that is there then I can define the
stress vector T n, I can define the stress vector T n like this. So, T n can be given
by limit delta A tends to 0, delta F that is the force over delta A here that is the
limiting condition, that is the limiting condition of the force. Now what is delta A? Delta A
is the area, cross sectional area of this small segment which is containing point O.
So, on that small segment ok on that small segment, this is a small segment whose area
is say delta A on that small segment your delta F is acting right, the limiting condition
when delta A tends to 0, try to understand when delta A tends to 0; that means, when
delta A will tends will tend to 0 basically that will be becoming a point kind of thing.
So, when delta A tends to 0 then the limiting condition of the force is known as stress
ok. So, now few things you need to remember. So, T n is a vector as I told you earlier
and that it acts on a plane passing through the point O whose normal is n, direction normal
ok and the stress vector does not act in general ok in the direction of n. It is not necessary for the T n is acting
along the direction of n. So, n is the direction normal; that means any plane. So, suppose
this is my plane ok this is this is the plane how to define this plane? You draw a normal
direction normal from that plane and by that normal this plane will be defined. So, if
the direction normally is n. So, I can call that plane as n plane right.
So, what does it say? So, T n is a vector which is quite obvious and that it acts on
a plane passing through the point O as I showed you whose normal is n that also I have shown
you right in the figure if you come back to this figure. So, this is the direction normal.
So, this is the point O, this is the direction normal n ok and T n does not T n so delta
F if you see delta F, it is not necessary that delta F will be always acting along the
n direction or the direction normal right. So, F could be any any direction it is not
necessary, in generally it is not acting along the n direction ok, along the direction normal
fine. So, now what I can write then, then the stress
vector in terms of its component that is x y z if you if you have the mutually perpendicular
axis coordinate system then in terms of components what I can write this stress vector is equal
to T x that is the x component ok multiplied by the unit vector along x direction, similarly
T n y unit vector along y direction plus T n z into unit vector along z direction. So,
where T x, T y, T z they are the x component, y component and z component of the stress
vector T n that I can have because I have any stress vector that I can decompose or
the resolved in three mutually perpendicular coordinate system xyz. So, they can be written
like that. So, now we should understand, now one thing
is very clear from this discussion that we should know the plane. So, without knowing
the plane you cannot define the stress. So, when you are going to define the stress basically
you need to define the plane also on which the stress is acting. So, the definition of
plane is very very important, right. So, now, let us talk about that. Now we will define
the positive and negative faces, positive and negative faces ok. So, now if you come back to this figure, they
are xyz coordinate system and we have drawn one parallelepiped in this xyz coordinate
system and this is the parallel. Now, we are going to define a face which is
positive. So, if I define a face which is positive; that means, the face is positive
when its outward directed normal is normal vector is towards the positive coordinate
direction right. So, what does it mean? So, let us let us write down that statement and
then we will we will basically explain a face because you need to define a plane before
understanding the stress ok. So, a face will be positive when its outward directed normal
vector points in the direction of the positive coordinate
axis. So, what is the definition? The definition
says a face will be defined as positive when its outward directed normal vector points
in the direction of the positive coordinate axis. Now if you look at this figure. So,
if I consider one two three four plane or one two three four face this is the this is
my face 1 2 3 4 face ok. So, this face will be defined as positive as per the definition
because the outward directed normal vector of this face is towards the direction of positive
coordinate axis that is nothing, but positive x axis. Similarly 5 6 7 8 if you consider
this face, this is the face what should I say? This face should be positive or negative
this face will be negative because the outward directed normal vector from this face is towards
the negative direction of coordinate axis that is negative x direction. So, that face
will be defined as negative face. Similarly you will be having three positive
faces and three negative faces. So, this is my positive face this is my positive y face,
this is my positive x face, this is my positive z face, positive z face whereas this will
be my negative x face this will be my negative y face and this back side face will be my
negative z face ok. I hope you have understood that. So, a face will be defined as positive
when its outward directed normal vector points in the direction of the positive coordinate
axis. Similarly a face will be defined as negative when its outward directed normal
vectors points in the direction of the negative coordinator axis. So, then that that face
will be defined as negative ok I hope you have understood.
Well, so I will stop here today. So, in the next class we will continue with the further
discussion on concept of stress. Thank you very much.

## 1 thought on “Lecture 18 -Tutorial on Force Displacement Relationship and Geometric Compatibility -3”

1. Anuj Yadav says:

Mind blowing explanation