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.

Mind blowing explanation