List of area moments of inertia

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The following is a list of area moments of inertia. The area moment of inertia or second moment of area has a unit of dimension length4, and should not be confused with the mass moment of inertia. If the piece is thin, however, the mass moment of inertia equals the area density times the area moment of inertia. Each is with respect to a horizontal axis through the centroid of the given shape, unless otherwise specified.

Area moments of inertia[edit]

Description Figure Area moment of inertia Comment Reference
a filled circular area of radius r Area moment of inertia of a circle.svg I_x = \frac{\pi}{4} r^4

I_y = \frac{\pi}{4} r^4

I_z = \frac{\pi}{2} r^4
[1]
an annulus of inner radius r1 and outer radius r2 Area moment of inertia of a circular area.svg I_x = \frac{\pi}{4} \left({r_2}^4-{r_1}^4\right)

I_y = \frac{\pi}{4} \left({r_2}^4-{r_1}^4\right)

I_z = \frac{\pi}{2} \left({r_2}^4-{r_1}^4\right)
For thin tubes,  r \equiv r_1 \approx r_2 and r_2 \equiv r_1+t.

We can say that \left(r_2^4-r_1^4\right) = \left(\left(r_1+t\right)^4-r_1^4\right) = \left(4r_1^3t+6r_1^2t^2+4r_1t^3+t^4\right) and because r_1>>t this bracket can be simplified to   \left(4r_1^3t+6r_1^2t^2+4r_1t^3+t^4\right) \approx  4r_1^3t. Ultimately, for a thin tube, I_x = I_y = \pi {r}^3{t}.

a filled circular sector of angle θ in radians and radius r with respect to an axis through the centroid of the sector and the center of the circle Area moment of inertia of a circular sector.svg I_0 = \left( \theta -\sin \theta \right) \frac{r^{4}}{8} This formula is valid for only for 0 ≤ \theta\pi
a filled semicircle with radius r with respect to a horizontal line passing through the centroid of the area Area moment of inertia of a semicircle 2.svg I_0 = \left(\frac{\pi}{8} - \frac{8}{9\pi}\right)r^4 \approx 0.1098r^4 [2]
a filled semicircle as above but with respect to an axis collinear with the base Area moment of inertia of a semicircle.svg I = \frac{\pi r^4}{8} This is a consequence of the parallel axis theorem and the fact that the distance between these two axes is \frac{4r}{3\pi} [2]
a filled semicircle as above but with respect to a vertical axis through the centroid
Area moment of inertia of a semicircle 3.svg
I_0 = \frac{\pi r^4}{8} [2]
a filled quarter circle with radius r entirely in the 1st quadrant of the Cartesian coordinate system Area moment of inertia of a quartercircle.svg I = \frac{\pi r^4}{16} [3]
a filled quarter circle as above but with respect to a horizontal or vertical axis through the centroid Area moment of inertia of a quartercircle 2.svg I_0 = \left(\frac{\pi}{16}-\frac{4}{9\pi}\right)r^4 \approx 0.0549r^4 This is a consequence of the parallel axis theorem and the fact that the distance between these two axes is \frac{4r}{3\pi} [3]
a filled ellipse whose radius along the x-axis is a and whose radius along the y-axis is b Area moment of inertia of an ellipsis.svg I_x = \frac{\pi}{4} ab^3

I_y = \frac{\pi}{4} a^3b
a filled rectangular area with a base width of b and height h Area moment of inertia of a rectangle.svg I_x = \frac{bh^3}{12}

I_y = \frac{b^3h}{12}
[4]
a filled rectangular area as above but with respect to an axis collinear with the base Area moment of inertia of a rectangle 2.svg I = \frac{bh^3}{3} This is a result from the parallel axis theorem [4]
a filled rectangular area as above but with respect to an axis collinear, where r is the perpendicular distance from the centroid of the rectangle to the axis of interest I = \frac{bh^3}{12}+bhr^2 This is a result from the parallel axis theorem [4]
a filled triangular area with a base width of b and height h with respect to an axis through the centroid Area moment of inertia of a triangle.svg I_0 = \frac{bh^3}{36} [5]
a filled triangular area as above but with respect to an axis collinear with the base Area moment of inertia of a triangle 2.svg I = \frac{bh^3}{12} This is a consequence of the parallel axis theorem [5]
a filled regular hexagon with a side length of a Area moment of inertia of a regular hexagon.svg I_0 = \frac{5\sqrt{3}}{16}a^4 The result is valid for both a horizontal and a vertical axis through the centroid, and therefore is also valid for an axis with arbitrary direction that passes through the origin.
An equal legged angle Second Moment of Area Angle.jpg I_x = I_y = \frac{t(5L^2-5Lt+t^2)(L^2-Lt+t^2)}{12(2L-t)}

I_(xy) = \frac{L^2t(L-t)^2}{4(t-2L)}

I_a = \frac{t(2L^4-4L^3t+8L^2t^2-6Lt^3+t^4)}{12(2L-t)}

I_b = \frac{t(2L-t)(2L^2-2Lt+t^2)}{12}
I_(xy) is the often unused product of inertia, used to define inertia with a rotated axis
Any plane region with a known area moment of inertia for a parallel axis. (Main Article parallel axis theorem) Parallel Axes Compact.png I_z = I_x + Ar^2 This can be used to determine the second moment of area of a rigid body about any axis, given the body's moment of inertia about a parallel axis through the object's center of mass and the perpendicular distance (r) between the axes.

See also[edit]

References[edit]

  1. ^ "Circle". eFunda. Retrieved 2006-12-30. 
  2. ^ a b c "Circular Half". eFunda. Retrieved 2006-12-30. 
  3. ^ a b "Quarter Circle". eFunda. Retrieved 2006-12-30. 
  4. ^ a b c "Rectangular area". eFunda. Retrieved 2006-12-30. 
  5. ^ a b "Triangular area". eFunda. Retrieved 2006-12-30. 

External links[edit]