Learning topic
Second and Polar Moments of Area
Definitions of Ix, Iy, Ixy and Jp, geometric meaning, units, and standard formulas for rectangular, circular, and annular sections.
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A second moment of area characterizes how cross-sectional area is distributed relative to a selected axis. It is a geometric property and must not be confused with a mass moment of inertia.
The second moments of area about the x and y axes are:
$$I_x=\int_A y^2\,dA,\qquad I_y=\int_A x^2\,dA.$$
The product moment of area is:
$$I_{xy}=\int_A xy\,dA.$$
The polar second moment of area about the intersection of orthogonal axes is:
$$J_p=\int_A(x^2+y^2)dA=I_x+I_y.$$
Second and polar moments of area have dimensions of length to the fourth power, for example mm⁴.
Geometric meaning
Area elements located farther from an axis contribute much more strongly because the distance enters quadratically. This is why sections that place material far from the centroidal axis can achieve high bending stiffness with relatively modest area.
Standard shapes
| Shape | Area A | Centroidal second moment | Section modulus |
|---|---|---|---|
| Rectangle b × h, x-axis parallel to side b | $bh$ | $I_x=bh^3/12$ | $W_x=bh^2/6$ |
| Rectangle b × h, y-axis parallel to side h | $bh$ | $I_y=hb^3/12$ | $W_y=hb^2/6$ |
| Circle of diameter d | $\pi d^2/4$ | $I_x=I_y=\pi d^4/64$ | $W_x=W_y=\pi d^3/32$ |
| Circle of diameter d, polar property | $\pi d^2/4$ | $J_p=\pi d^4/32$ | $W_p=\pi d^3/16$ |
| Annulus D, d | $\pi(D^2-d^2)/4$ | $I_x=I_y=\pi(D^4-d^4)/64$ | $W_x=2I_x/D$ |
Polar moment
For two mutually perpendicular axes $x$ and $y$ passing through the same point, $J_p=I_x+I_y$. For a circular shaft, this property enters directly into the classical torsion formulas.
Units
If geometric dimensions are given in millimetres, second moments of area are expressed in mm⁴. Because characteristic dimensions enter to the fourth power, relatively small dimensional changes can strongly affect $I$.
Example
For a rectangle with $b=40\ \text{mm}$ and $h=80\ \text{mm}$, about its centroidal $x$-axis parallel to $b$:
$$I_x=\frac{40\cdot80^3}{12}\approx1.707\cdot10^6\ \text{mm}^4.$$