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I need to calculate the moment of Inertia Iyz but I find 0 information online about how to get it or solve it in a real problem, only that is equal to

$$\bar{I}_{yz}=\int yz\text{ d}A$$

How can I solve this in the case of an antisymmetric beam section?

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  • $\begingroup$ I think this is called the Product moment of area, which is related to the second moment of area. Take a look at Wikipedia. $\endgroup$ – Andrew May 27 '18 at 7:54
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The products of inertia aren’t something that is encountered often. This is mainly because we can break up most cross section into small pieces and combine them with the parallel axis theorem. As such, we simply skip the integration step.

The product comes into play when a load is applied at a strange angle or you are dealing with a non-symmetrical cross section. In the case of an equal legged angle shape, for example, one can compute the product of inertia is 0 for each of the legs. However, when the legs are joined, the centroid of the shape moves. As a result, since the centroid moves in both x and y axis, the net change in the product inertia would be:

$$A*x*y$$

Where x would be the distance the centroid moved in the x direction and y would be the distance the centroid moved in the y direction.

This becomes especially relevant in the Mohr’s circle of inertia to determine the primary axis of inertia. You can rotate the reference frame by an angle $\theta$ to remove the product of inertia, where $\theta$ is found as follows:

$$\theta = \frac{1}{2} atan\left(\frac{-2I_{xy}}{I_x-I_y}\right)$$

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Second moment of inertia with respect to Z axis is in the form$$ \bar{I}_{yz}=\int yz^2\text{ d}A $$ By this we mean I-z but when the beam's base is considered Y. Alternatively we could seek I_xz which would have x as base and z as beam height.

Otherwise if we are looking for diagonal I_yz we know it is $ \space Iyz = \sqrt{[(I_z)^2 + (I_y)^2]}$

What you have shown is first moment of area, not second. Second moment of area is the sum of all small parts of a beam's surface, multiplied by their moment arm WRT the beam's neutral axis.

For example for a rectangular cross section beam with dimensions$$ X = 6cm,\space Y =12 cm\space and\space Z =180 cm $$ which could be a log of timber the

$$ \bar{I}_{yz}=\int yz^2\text{ d}A = 12 \times 180^3 /12 = 5832000 cm^4 $$

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  • $\begingroup$ Second moment of inertia always has length to the fourth - this would give length to the fifth. $\endgroup$ – Mark May 31 '18 at 12:10
  • $\begingroup$ where do you see power of five? This is width times height cubed divided by a constant, so it is forth power. The reason they don't use I_z is because the beam is not slender on that direction and planes of strain do not necessarily remains plain. $\endgroup$ – kamran May 31 '18 at 15:17
  • $\begingroup$ y = L^1, z^2 = L^2, dA = L^2, so integral is L^5. $\endgroup$ – Mark May 31 '18 at 15:53
  • $\begingroup$ dA is actually d(width) times d(height). Because the width is constant it gets out from integral, so dA is just dH. Therefore it is power of four. I will look for a freshman statics book and post the link, However one could google to see how prismatic section properties are calculated. $\endgroup$ – kamran May 31 '18 at 16:07
  • $\begingroup$ But the width would still be part of the formula. Just because you move it to the front doesn't change the formula for I. The dimensions still equate to L^5. $\endgroup$ – Mark May 31 '18 at 16:09

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