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Okay so good morning everyone. So last time we had done force balance and got an equation

for an elastic rod. Now we have to do moment balance, you know just as we do in beam theory.

So we are going to do moment balance. So let me just redraw the deformed rod.

So let me just redraw the deformed rod. And you guys can recall that on this cross section

which is at s equal to s 2, we have the internal traction acting which are you know any little

force that is there. So we have to do moment balance and we have to do moment balance.

So we have the internal traction acting which are you know any little on a little area here

the traction would be or the force would be P the first Pk E3 dA. And over here we will

have negative of it because the normal is pointing in the negative direction. So minus

P E3 dA. So that s equal to s 2 and then you have on the lateral surface, you have the

external traction which I guess we denoted by T external and you have S equal to s 2

and then you have on the lateral surface the external tractions which I guess we denoted

by T external. And you also have the body force of course acting everywhere within the volume. So you could denote that by you know B.

So we had done the force balance. So we want to do now the moment balance. So for moment

balance we have to choose a reference point about which we will compute the moment. So

let s take that to be the origin of the coordinate system itself. So we will do moment balance

about the origin of the coordinate system. So we will do moment balance about the origin

of the coordinate system.

So we have let s first get the moment of the external loads which are the body forces and

the traction on the lateral surface. So we are going to do moment about the origin. So

let s do first due to external loads. So for the body force part, the body force is acting

everywhere within the volume. So any point within the volume of the rod, since it is

a deformed rod, we could denote that with little x. And maybe let s take it cross section

wise. So here is your cross section and a point on the cross section we denote by little

x. So the body force that acts on this point will generate a moment of the force that is

x cross b. So this is the moment of the force that is x cross b. So this is the moment of

this point will generate a moment of x cross b dv. Isn t it? Where dv is the undeformed

volume of a small volume over here. And then we do its triple integration. And we can maybe

call it dv0 for the undeformed volume, v0. But now we want to write this little x in

terms of the center line position and the extra thing. So you see this is your center

line and then you have extra vector. So we could always write this as little r of s cross,

sorry plus little x minus little r of s cross product b dv0. Now what do you think, what

is this equal to? Little x minus little r of s. So that is the extra vector and what

would that be equal to? So if we have to use our constrained kinematics for a spatial cos

dot rod, then this thing is simply x alpha d alpha. Isn t it? x alpha d alpha of s. However,

as we said that we don t want to restrict to that constrained kinematics because it

makes our system stiff. So there is some extra, on top of this there is some extra displacement

u for warping. Isn t it? So let s just, so this was just to give a perspective. So let

s just forget about this and think that we have this thing. So this is anyway general,

little x denoting the deformed coordinate of any point and then little x minus small

r is that extra vector. I hope I am not confusing you. So this is same as for spatial cos dot

rod. This then becomes big x alpha big r e alpha plus your u.

Well last time we had to u also multiplied with r.

Okay you can write that, sure. You can write either way. So this then becomes a different

u of course. Okay since I did that so maybe I will restrict to what I used last time.

So this becomes big r of x alpha e alpha plus your little u. Okay? But I am not going to

use this really in this derivation. I will just keep it as it is here. There is no confusion

here, right? Little x denotes any point in the beam and then I am just taking this difference.

Okay? Alright so then let us see what does this become. So we have, we are going to split