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So we told that this beam is such that its center line does not really coincide with its cross sectional normal.

So we allow for that. So for example, if this is the center line tangent, then the cross section could be like this.

So, they work with 90 degree tangent. So similarly over here, we put something like this. And the cross section could be like this.

And if you want to construct your full beam.

So, and then we said that, see initially the cross section was vertical. So it's normal. This is the cross section normal.

So that's the cross section normal.

So initially the normal was horizontal. And then it has rotated. And since it has rotated in a negative sense, so the angle of rotation is therefore negative theta.

And if I look at the center line tangent, then it has also undergone rotation. And that's the horizontal line.

And this angle, assuming it is very small, then this becomes dy by dx. Otherwise, it's 10 equals to dy by dx.

Okay, and so when we look at these two lines, maybe I can make them thicker.

So this is the cross section.

And we see that this, the shear we found out was gamma equal to dy by dx minus theta. Right, so that implies dy by dx was equal to gamma plus theta.

So you see that the slope, the rotation of the slope has got contribution from two terms.

The first term, sorry, this theta, the second term is the rotation due to cross section's rotation, which we also call due to bending.

We will see why we call this due to bending. Whereas the gamma, that's also rotation due to shear.

Okay, so this is already, and then we wrote gamma in terms of shear force, and it turned out to be dy by dx is equal to shear force V of x

divided by the correction factor kappa k, sorry, correction factor k, gA plus theta.

So that's the first equation of Philos-Senkoubian theory.

So we have two unknowns, y and theta, and this is the first equation.

And then we need one more equation, and I just wrote down the other equation would be EI d theta by dx is equal to M of x, the bending moment.

And I have posed the question, why is this d theta by dx, why not d two by dx two? Because we all have this in our mind that EI times curvature is bending moment, right?

And curvature somehow appears to have a, it is d two by dx two.

So we'll see why we are saying this is d theta by dx and why not d two by dx two.

So let's just look at that. So I'll work out a very simple situation.

So think of pure bending of a beam. So here is a pure beam, and it undergoes pure bending.

So there will be a center somewhere.

So that's the center of the rotation. And this line is your neutral line, isn't it?

So we call the radius of the neutral line as capital R.

And let us suppose this angle is theta.

Okay, and then we know that neutral line does not undergo any stretching or compression. That implies that R theta is equal to L.

So theta by L is equal to R.

One by R, okay. And then if you want to know what's the strain, normal strain in different fibers of the beam, what do we do?

We simply let us, for example, set up our coordinate system. That's x, and let's call it y.

So if we are at a distance y, ever, then the fiber, what's the length of that fiber?

That's simply R minus y times theta. And because it isn't bent in the perfect circle, so if I go distance y, ever, the radius is R minus y, so that length would be R minus y times this theta.

So before bending, that length was R times theta, same as the length of the neutral line. So what's the strain?

That's going to be the form length minus the original length divided by original length. So that's something we have all seen. This becomes minus y over R.

So that's sigma xx, which is minus E times y over R. And then you compute the bending moment, right?

And the bending moment simply turns out to be, you compute over the, integrate over the cross-section, which is y times sigma xx dA. And you get it to be EI one over R.

Right? Something we have learned from our strength of materials course.

So here, what you see here is that this one by R, which is curvature, is really the curvature of the center line, or the neutral line.

So everything makes perfect sense as we have been doing. The bending moment is EI times the curvature, where curvature is that of the center line. Isn't it?

Now let's do something else. Let's compress our beam also.

Do we need a minus sign there in the last equation?

So J is coming out for...

Is the sigma xx a minus? And then we have... Well as you... So bending moment, the epsilon xx as you go up is negative. So stress sigma xx is also negative, compressing, pointing in that direction.

But the moment is positive. The moment due to that is positive.

Do we need a minus sign from the three?

Yeah, right. Okay. Yes.