 # 1- Easy illustration of the theory of pure bending.

### Brief content of the video.

We will talk about the third assumption the beam is initially straight and all the longitudinal filaments bend into a circular arc with a common center of curvature. This is a part of the video that has a subtitle and a closed caption in English.

You can click on any picture to enlarge then press the small arrow at the right to review all the other images as a slide show.

### Lecture objectives.

The Theory of pure bending assumptions and the relation is our first item. The second item is why we estimate the product of inertia. and the third Item we are going to talk about the Mohr circle of inertia and how can we use it to get the maximum and minimum values of the moment of inertia.

### Assumptions to follow for the theory of pure bending.

First, we have some assumptions when there is a beam acted upon by pure bending, it is going to be deformed.
But first, let us examine the beam before deformation. We have two axes X and Y and we consider the section of the beam as a rectangular section at the first stage before the deformation.

We have a vertical section perpendicular to the neutral axis and in our assumption that the plain section perpendicular to the neutral axis before the deformation remains perpendicular to the neutral axis after deformation.

After deformation, the beam will take the form of a curved shape, and perpendicular to that it will not remain any more vertical but the section will be it will remain perpendicular to the neutral axis after deformation.
The second assumption is that deformation due to the bending is small.

The third assumption the beam is initially straight and all the longitudinal filaments bend into a circular arc with a common center of curvature.

This means that all these perpendicular sections in this section and if we assume there is another section that would be a plain section perpendicular to the neutral axis. And the third portion there will be another section perpendicular to the neutral axis.

All these sections if you are going to extend they will meet at a common point too. And the line constituted will be a form of a circular arc.

The fourth assumption, of the theory of pure bending, is the deformation in the vertical direction i.e the transverse strain, epsilon εyy. It may be neglected in driving the expression of the longitudinal strain when we have a bending moment, there will be an extension in the lower chord accompanied by compression in the upper chord.

The compression is always accompanied by compression or transverse strain. This transverse strain in our case will be represented by the εyy =dh/h , it will be negative. The longitudinal deformation will be= the extension of length/ the original length.

An extension of length will be replaced by dl/ l, which will be positive in εyy. We consider if it has a small value and it will be deleted.

The fifth assumption, the radius of curvature is larger as compared to the dimension of the cross-section.

The sixth assumption- the value of the young’s modulus of elasticity in both tension and compression will be the same.

What is pure bending? the pure bending means that the beam is subjected to moments on both sides and they are both equals, the shear value will be zero.

Why is the shear value will be zero?  because this bending moment on the right-hand side will produce a couple. It is anti-clockwise so the couple will be upward here the shear here will be accompanied by a downward shear force on the right-hand side, while this moment will produce a resisting couple anti-clockwise.

so the shear force will be downward and the other shear force will be upward so downward and upward opposite to each other and both, suppose will cancel each other.

That is why the shear force will be equal to zero, while for the bending moment diagram. If we want to draw this is sagging. It will be positive and value will go to the right support and that will be closed here. So this, from left to right.

It will be subjected to bending of equal value positive bending, for the left-hand side moment and the right-hand side moment they will cause the beam to bend in a form of Arc and the enclosed angle would be α  as we can see in the diagram.

The second case when we have a pure bending in case if you have a simply supported beam, hinged, simply supported beam acted upon by two shear forces a P and P are of equal values and they are acting on the third point of the beam and L/3 and L/3 and, in that regard.

We have shear forces diagram we have P positive shear force coming up and go down until coming to B and from the right-hand side, we have negative shear that will continue until point d and closing.

While for the bending moment diagram,  we have (Pl/3) this is the positive bending moment that we have, so for this, the intermediate section will be having a bending moment while shear force between C and d =0.

### Stress formula for the case of pure bending.

Assumption number #3 is that beam is initially straight. If we can see that is a rectangular section and this is the x-axis and y-axis before the deformation.
If we take, a section with apart distance by (dx) and there is an element small element apart by distance y
before deformation.

Let us have a look after the deformation and find that the beam will bend in this form and the neutral axis which was horizontal becomes curved as that and we can see here the x-axis and y and perpendicular axis 3. If we assume that the arc radius will be R and that is an enclosed arc angle will be considered as the dα.

We are going to equate and we are going to find the strain value before deformation.
There was no strain, after deformation.

This segment with length the (dx) its length it will become the Rdα. For the element at Y distance up, its length originally was dx while after deformation the length will become (R-Y) multiplied by (dα).

So the length for this red line=(R-y*dα), was originally the length it was dx.
If we are going to compare it’s becoming in the neutral axis.

There is no change so we can take the dx as =(Rdα), the deformation is tension or compression.
For this case, we are above the neutral axis. So the deformation will become negative.

We have (R-y)*dα-Rdα,+Rdα and -Rdα will cancel each other and we are left with (-y*dα). And since it a strain is deformation over the origin length the deformation is (-ydα) and the original length was R*dα so we left with(-y/R) and since due to Young’s modulus of elasticity = stress/strain, we have our strain = (-y)/R.

This is a strain, so the f= (e *E )the strain * young ‘s modulus of elasticity. It means that f=(-y/R)*(E) .

And since we have the assumption of 4 that Young’s modulus of elasticity is the same for the case of tension and compression will continue into the next topic.

This is the pdf file used in the illustration of this post.

This is a link to a useful external resource. By Jus.
This is a link to the next post, 2-Derive the expression for the first moment of the area at the CG.

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