Centroid Equations of Various Beam Sections

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The Fundamentals of the Centroid

It is important to note that on a cross section, whose area is uniform throughout, the centroid can be found by taking the summation of moments with respect to an axis arbitrarily set, but is usually set at the top or bottom fiber. You can visit this page for a more in-depth discussion of the topic.

Fundamentally, the centroid can be obtained by taking the summation of moments over the summation of the area. Which is expressed in this way.

[math] \bar{x}=\frac{1}{A}\int xf\left ( x \right )dx [math]

In the equation above, f(x) is the function and x is the moment arm. To better illustrate this, we will derive the y centroid of an arbitrary triangle with its base coincident to the x-axis. In this situation, the shape of the triangle, whether be it equilateral, isosceles or scalene is irrelevant as everything is relative to only the x-axis. Note the the shape is irrelevant if the triangle's base is coincident or parallel with respect to the axis. This will not be the case when solving for the x centroid. Instead, you can imagine it as getting the centroid of two right triangles with respect to the y-axis. For the sake of convenience, let us imagine an isosceles triangle similar to the refernce table below. Finding the relationship of b and h will yield the following relationship.

[math] \frac{-y}{x}=\frac{-h}{b} [math]

Note that the slope is negative as we are imagining the triangle to be upright. If we imagine the triangle to be inverted, the slope would be positive. Regardless, the relationship remains the same. As x = f(y), the relationship above can rewritten as the following.

[math] x=f\left ( y \right )=\frac{b}{h}y [math]

We can now solve for centroid. Adjusting the first equation above, we get the following.

[math] \bar{y}=\frac{1}{A}\int yf\left ( y \right )dy [math]

Plugging additional values and substituting the relationship above will yield the following equation.

[math] \bar{y}=\frac{2}{bh}\int_{0}^{h} \frac{b}{h}y^{2}dy [math]

Simplifying,

[math] \bar{y}=\frac{2}{h^{2}}\left [ \frac{y^{3}}{3} \right ]_{0}^{h} [math] [math] \bar{y}=\frac{2}{h^{2}}\left [ \frac{h^{3}}{3}-0 \right ] [math] [math] \bar{y}=\frac{2}{3}h [math]

Note that this solution is taken from the top. The centroid taken from the bottom must then be equal to 1/3 of h.

Centroids of common shapes and beam sections

Below is a list of a variety of beam section shapes and the distance to the centroids of the section. The equations show how to find the centroid of a particular section from the base or the furthest left point of the section. For SkyCiv Student and Structural subscriptions, this reference can also be downloaded as a PDF Reference to take with you wherever you go. Centroids of a beam section are extremely important as they locate the Neutral Axis and are one of the earliest steps required when analysing a beam section. SkyCiv's Free Moment of Inertia Calculator is a valuable resource to check the below equations for centroids have been applied correctly.

SkyCiv also offers a comprehensive Summary of Sections Table which contains all the equations and formula relating to beam sections (moment of inertia, area etc...).

The equation for various centroids are listed below:


REFERENCE Cy
(Distance from bottom)
Cx
(Distance from furthest left point)

Centroid of Rectangle or Rectangular Sections

rectangle beam section [math] \dfrac{h}{2} [math] [math] \dfrac{b}{2} [math]

Centroid of a Hollow Rectangular Section

hollow rectangle beam section [math] \dfrac{b}{2} [math] [math] \dfrac{h}{2} [math]

Centroid of a Circle or Circular Section

circle beam section for centroid [math] \dfrac{D}{2} [math] [math] \dfrac{D}{2} [math]

Centroid Equation of a Hollow Circular Section

hollow circle section for centroid [math] \dfrac{D}{2} [math] [math] \dfrac{D}{2} [math]

Centroid of an Isosceles Triangle

triangular section for centroid [math] \dfrac{h}{3} [math] [math] \dfrac{b}{2} [math]

Centroid of an I-Section

I Beam [math] \frac{TFw\times TFt\times \left ( BFt + Wh + \frac{TFt}{2} \right )}{TFw\times TFt + Wt\times Wh + BFw\times BFt} + [math] [math] \frac{Wt\times Wh\times \left ( BFt + \frac{Wh}{2} \right )}{TFw\times TFt + Wt\times Wh + BFw\times BFt} + [math] [math] \frac{BFw\times BFt\times \left ( \frac{BFt}{2} \right )}{TFw\times TFt + Wt\times Wh + BFw\times BFt} [math] [math] TFw > BFw, \frac{TFw}{2}[math] [math] BFw > TFw, \frac{BFw}{2} [math]

Centroid of a T-Section

T Beam [math] \frac{Wt\times Wh\times \left ( \frac{Wh}{2} \right )}{TFw\times TFt + Wt\times Wh } + [math] [math] \frac{TFw\times TFt\times \left ( Wh + \frac{TFt}{2} \right ) }{TFw\times TFt + Wt\times Wh } [math] [math] \frac{TFw}{2} [math]

Centroid of an C-Section

Channel Beam [math] \frac{TFw\times TFt\times \left ( h - \frac{TFt}{2} \right )}{TFw\times TFt + Wt\times Wh + BFw\times BFt} + [math] [math] \frac{Wt\times h\times \left ( \frac{h}{2} \right )}{TFw\times TFt + Wt\times Wh + BFw\times BFt} + [math] [math] \frac{BFw\times BFt\times \left ( \frac{BFt}{2} \right )}{TFw\times TFt + Wt\times Wh + BFw\times BFt} [math] [math] \frac{TFt\times TFw\times \left ( Wt + \frac{TFw}{2} \right )}{TFt\times TFw + h\times Wt + BFt\times BFw} + [math] [math] \frac{h\times Wt\times \left ( \frac{Wt}{2} \right )}{TFt\times TFw + h\times Wt + BFt\times BFw} + [math] [math] \frac{BFt\times BFw\times \left ( Wt + \frac{BFw}{2} \right )}{TFt\times TFw + h\times Wt + BFt\times BFw} [math]

Centroid of an Angles

Angle Beam [math] \frac{LFt\times LFh\times \left ( BFt + \frac{LFh}{2} \right ) }{LFt\times LFh + BFw\times BFt} + [math] [math] \frac{BFw\times BFt\times \left ( \frac{BFt}{2} \right )}{LFt\times LFh + BFw\times BFt} [math] [math] \frac{LFh\times LFt\times \left ( \frac{LFt}{2} \right )}{LFh\times LFt + BFt\times BFw} + [math] [math] \frac{BFt\times BFw\times \left ( \frac{BFw}{2} \right )}{LFh\times LFt + BFt\times BFw} [math]


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