## Moment of Inertia of a Trapezoid

This tool calculates the moment of inertia I (second moment of area) of a trapezoid. A trapezoid has at least one pair of parallel sides, called bases. Enter the shape dimensions 'b', 'a' and 'h' below.The calculated results will have the same units as your input. Please use consistent units for any input.

 b = a = h = The results are: Ix = Ix0 = Axis offset (from centroid or base): d = or... d0 = Ix' = Enter a rotation angle: φ = deg rad Enter length b1 b1 = Iu = Iu' = Complete results: Ixx = Iyy = Ixy = Principal II = Principal III = Principal θ (°) = Iuu = Ivv = Iuv = Ix0x0 = Iy0y0 = Ix0y0 = Iu0u0 = Iv0v0 = Iu0v0 = Area A = Centroid xc = Centroid yc =

## Definitions

The moments of inertia of a trapezoid can be found, if the total area is divided into three, smaller ones, A, B, C, as shown in figure below. The final area, may be considered as the additive combination of A+B+C. Therefore, the moment of inertia Ix0 of the trapezoid, relative to axis x0, is determined like this:

$\begin{split} I_{x0} & = I_{x0}^{A} + I_{x0}^{B} + I_{x0}^{C}\Rightarrow \\ I_{x0} & = \frac{b_1 h^3}{12} + \frac{a h^3}{3} + \frac{(b-a-b_1)h^3}{12} \Rightarrow \\ I_{x0} & = \frac{h^3}{12} (3a+b) \end{split}$

The moment of inertia relative to centroidal axis x-x, can be found by application of the Parallel Axes Theorem (see below). The distance of the parallel axes x,x0 is essentially the centroid coordinate yc, relative the bottom base (see figure above). It can be calculated using the first moments of area, of the three sub-areas A,B,C. Assuming, that the triangular areas A and C, have bases aligned with the trapezoid base 'b', the following expression is found:

$\begin{split} y_{c} & = \frac{1}{A}\left( \frac{b_1 h}{2}\frac{h}{3} + {a h}\frac{h}{2} + \frac{(b-a-b_1) h}{2}\frac{h}{3}\right) \Rightarrow \\ y_{c} & = \frac{h}{3}\frac{2a+b}{a+b} \end{split}$

where A is the area of the trapezoid, equal to $$(a+b)h/2$$ .

Making similar considerations, the moment of inertia, relative to axes y or y0 (perpendicular to the bases) as well as the product of inertia, can be found. However, the formulation of simple equations is not possible for these quantities.

### Parallel Axes Theorem

The moment of inertia of any shape, in respect to an arbitrary, non centroidal axis, can be found if its moment of inertia in respect to a centroidal axis, parallel to the first one, is known. The so-calledParallel Axes Theorem is given by the following equation:

$I' = I + A d^2$

where I' is the moment of inertia in respect to an arbitrary axis, I the moment of inertia in respect to a centroidal axis, parallel to the first one, d the distance between the two parallel axes and A the area of the shape (=bh/2 in case of a triangle).

For the product of inertia Ixy, the parallel axes theorem takes a similar form:

$I_{xy'} = I_{xy} + A d_{x}d_{y}$

where Ixy is the product of inertia, relative to centroidal axes x,y, and Ixy' is the product of inertia, relative to axes that are parallel to centroidal x,y ones, having offsets from them $$d_{x}$$ and $$d_{y}$$ respectively.

### Rotated axes

For the transformation of the moments of inertia from one system of axes x,y to another one u,v, rotated by an angle φ, the following equations are used:

$\begin{split} I_u & = \frac{I_x+I_y}{2} + \frac{I_x-I_y}{2} \cos{2\varphi} -I_{xy} \sin{2\varphi} \\ I_v & = \frac{I_x+I_y}{2} - \frac{I_x-I_y}{2} \cos{2\varphi} +I_{xy} \sin{2\varphi} \\ I_{uv} & = \frac{I_x-I_y}{2} \sin{2\varphi} +I_{xy} \cos{2\varphi} \end{split}$

where Ix, Iy the moments of inertia about the initial axes and Ixy the product of inertia. Iu, Iv and Iuv are the respective quantities for the rotated axes u,v. The product of inertia for a trapezoid is generally nonzero, unless symmetry exists, in any of the relevant axes.

### Principal axes

In principal axes, that are rotated by an angle θ relative to original centroidal ones x,y, the product of inertia becomes zero. Because of this, any symmetry axis of the shape, is also a principal axis. The moments of inertia about principal axes, $$I_I, I_{II}$$ are called principal moments of inertia, and are the maximum and minimum ones, for any angle of rotation of the coordinate system. If Ix, Iy and Ixy are known for the arbitrary centroidal coordinate system x,y, then the principal moments of inertia and the rotation angle θ of the principal axes can be found, through the next expressions:

$\begin{split} I_{I,II} & = \frac{I_x+I_y}{2} \pm \sqrt{\left(\frac{I_x-I_y}{2}\right)^2 + I_{xy}^2} \\ \tan 2\theta & = -\frac{2I_{xy}}{I_x-I_y} \end{split}$

### Dimensions

The dimensions of moment of inertia (second moment of area) are $$[Length]^4$$.

### Mass moment of inertia

In Physics the term moment of inertia has a different meaning. It is related with the mass distribution of an object (or multiple objects) about an axis. This is different from the definition usually given in Engineering disciplines (also in this page) as a property of the area of a shape, commonly a cross-section, about the axis. The termsecond moment of area seems more accurate in this regard.

### Applications

The moment of inertia (second moment or area) is used in beam theory to describe the rigidity of a beam against flexure (see beam bending theory). The bending moment M applied to a cross-section is related with its moment of inertia with the following equation:

$M = E\times I \times \kappa$

where E is the Young's modulus, a property of the material, and κ the curvature of the beam due to the applied load. Beam curvature κ describes the extent of flexure in the beam and can be expressed in terms of beam deflection w(x) along longitudinal beam axis x, as: $$\kappa = \frac{d^2 w(x)}{dx^2}$$. Therefore, it can be seen from the former equation, that when a certain bending moment M is applied to a beam cross-section, the developed curvature is reversely proportional to the moment of inertia I. Integrating curvatures over beam length, the deflection, at some point along x-axis, should also be reversely proportional to I.