## The First moment of area and product of inertia at the CG.

### The video for The First moment of area

### The expression of the first moment of area at the Cg.

We will continue talking about pure bending and the first moment of area. This is a beam but in isometric form.

We have set the x-axis as longitudinal and y as perpendicular to that X and the Z as pointing to the reader we draw it in the isometric form, in case we have a pure bending as we can see in the red arrow, this is rotating about Z. Please refer to the previous post for directions.

So we will call it MZ, and rotate around Z, if we have a look at the transverse section we will see the Y in the vertical direction, and that Z -axis will go to the left.

Since we have pure bending and this pure bending is not accompanied by compression or tension.

For the whole section, we don’t have any tension or compression force, which means that the sum of the forces will be equal to zero for that strip dA in which you have a stress =(-y*E/R), we will make the integration.

For the whole section and we equate it to zero since this stress =(-y*E/R)=0. And E over R has a value, so it cannot be equal to zero.

will examine small strip dA apart from the Z-axis by y distance according to the expression, which we have obtained that the strain is (-y/R) at distance, y (-) sign will indicate that when y has a positive value we have a compression from the neutral axis to the uppermost part of the section, while for the lower part, it will become positive when we multiply the strain by young modulus for Elasticity E we will give the stress value (+).

So the only thing which would be considered equal to zero is the integration of ∫y*dA.

=0 and this expression we are familiar with, which is the first moment of area, it is the first moment of area about any section =0, which means that this section is passing by the C G.

### The expression of the product of inertia at the Cg.

From that relation, we get it is that the z-axis is passing by the centroid. We have a moment and the moment value is the moment around Z this moment value will be =∫f *dA*y.

We have the strip the dA is apart by y and the (f*dA) is the giving the force (F*y) will give the moment we are going to make integration of these forces around Z.

We know that the stress= (-y*E/R), so we replace the value of f, in M z equation ∫f*dA*y we are left with ∫(-E/R) ∫(Y^2*dA).

We are familiar with the expression of ∫ dA*y^2, it is the moment of inertia but this time it is a moment of inertia around Z, we can write Mz. =-(E/R)*Iz. Again the (-) sign will indicate this in the upper portion in the y it is compression and since (f=-E/R*y), we will multiply this left side by y.

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And this right-hand side by y and looking to this (f=-E*y/R), we consider (f=-E*y/R). We have (y*Mz)= f*Iz.

We get a new expression for the stress due to pure bending that is stress f=(-)Mz*y/Iz and again the negative sign indicates compression for the stress above the neutral axis. The stress will become tension and since we don’t have a bending moment around y we can write that ∫dA*f*z=0, which is the product of inertia expression.

How to express local and global access using SAP 2000?

How to express local and global access using SAP 2000 for vertical elements?

This is the link to CE 160 files.

This is the link for an external resource. Bending Equation Derivation

This is the link for the next post, Estimation of Max and minimum moment of Inertia for a section.