How would I find the second moment of area of a parallelogram, as shown?

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  • $\begingroup$ You can splits it in two triangle and a central rectangle, now you know all the shapes has the known second moment of area, you need to evaluate it for the rectangle and one of the triangles, the second triangle has the same second are of moment, then apply the Steiner theorem (parallel axis theorem). I find it way easier than evaluate all these integrals. it's up to you. $\endgroup$ – Sam Farjamirad Jul 1 '18 at 20:54
  • $\begingroup$ @SamFarjamirad - The goal was to show how to use integrals to solve for products of inertia, something that apparently causes a lot of confusion. However, other answers to this question would be interesting. $\endgroup$ – Mark Jul 3 '18 at 21:46

While this could be resolved using rectangle and triangle formulas, a generalized approach could be applied that would work for any shape using integration.

The area would be:

$$A = \int^c_0\int^{x\:=\:\frac{b}{c}y+a}_{x\:=\:\frac{b}{c}y}1\:dx\:dy = ac$$

From the origin, we evaluate the first moment of the areas:

$$Q_x = \int^c_0\int^{x\:=\:\frac{b}{c}y+a}_{x\:=\:\frac{b}{c}y}x\:dx\:dy = \frac{ac(a+b)}{2}$$ $$Q_y = \int^c_0\int^{x\:=\:\frac{b}{c}y+a}_{x\:=\:\frac{b}{c}y}y\:dx\:dy = \frac{ac^2}{2}$$

From here, the centroid of the parallelogram could be evaluated, as $\overline{x} = \frac{Q_x}{A}$, and $\overline{y} = \frac{Q_y}{A}$:

$$\overline{x} = \frac{Q_x}{A} = \frac{a+b}{2}$$ $$\overline{y} = \frac{Q_y}{A} = \frac{c}{2}$$

We can then evaluate the integrals. We need to change our origin, and set ourselves new limits of integration:

$$c-\overline{y} = \frac{c}{2}$$ $$0-\overline{y} = -\frac{c}{2}$$ $$\frac{b}{c}y+a-\overline{x} = \frac{b}{c}y+\frac{a-b}{2}$$ $$\frac{b}{c}y-\overline{x} = \frac{b}{c}y-\frac{a+b}{2}$$

As such, we can evaluate $I_{xx}$, $I_{xy}$, and $I_{yy}$:

$$I_{xx} = \int^\frac{c}{2}_{-\frac{c}{2}}\int^{x\:=\:\frac{b}{c}y+\frac{a-b}{2}}_{x\:=\:\frac{b}{c}y-\frac{a+b}{2}}x^2\:dx\:dy = \frac{ac(a^2+4b^2)}{12}$$ $$I_{xy} = \int^\frac{c}{2}_{-\frac{c}{2}}\int^{x\:=\:\frac{b}{c}y+\frac{a-b}{2}}_{x\:=\:\frac{b}{c}y-\frac{a+b}{2}}xy\:dx\:dy = \frac{abc^2}{12}$$ $$I_{yy} = \int^\frac{c}{2}_{-\frac{c}{2}}\int^{x\:=\:\frac{b}{c}y+\frac{a-b}{2}}_{x\:=\:\frac{b}{c}y-\frac{a+b}{2}}y^2\:dx\:dy = \frac{ac^3}{12}$$

Use of the perpendicular axis theorem would give the polar moment:

$$I_{zz} = I_{xx}+I_{yy} = \frac{ac(a^2+4b^2+c^2)}{12}$$

Because the product of inertia is not zero, the Mohr's circle of Inertia will be needed to generate the Inertia-Matrix. The principal moments of Inertia are:

$$I_{\overline{xx}} = \frac{I_{xx}+I_{yy}}{2} \pm \sqrt{\left(\frac{I_{xx}-I_{yy}}{2}\right)^2 + I_{xy}^2}$$

This yields a principal moment of:

$$I_{\overline{xx}} = \frac{ac}{24}\left[a^2+4b^2+c^2 \pm \sqrt{(a-c)^2(a+c)^2+4b^2(2a^2+4b^2-c^2)}\right]$$

This principle moment would be found by rotating the reference frame by an angle $\theta$ of:

$$\theta = \frac{1}{2} atan\left(\frac{-2I_{xy}}{I_{xx}-I_{yy}}\right) = \frac{1}{2} atan\left(\frac{2bc}{c^2-4b^2-a^2}\right)$$

Note that this can be brought into tensor format:

$$\begin{bmatrix} I_{xx} & I_{xy} & I_{xz} \\ I_{xy} & I_{yy} & I_{yz} \\ I_{xz} & I_{yz} & I_{zz} \end{bmatrix}$$

Which is useful in some tensor-based applications.


The best way is to create a rectangle by cutting the triangular portion from the left to the right side where it is missing and calculate from the original zero coordinate which is the base width of the triangular portion.

The logic is this the centroid of both triangles is calculated from $$[\frac {bc}{2}\times \frac {2}{3}b]_{left} + [\frac {bc}{2} \times \frac {1}{3}b]_{right}$$ By this statement we realise that the 2 triangular portions are equivalent to a single rectangle added to the rectangular portion of the p'gram but critically we cannot change the origin point to begin at the origin+b or the answer will be wrong.


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