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# How to Calculate the Centroid of a Beam Section

The centroid or center of mass of beam sections is useful for beam analysis when the moment of inertia is required for calculations such as shear/bending stress and deflection. Beam sections are usually made up of one or more shapes. So to find the centroid of an entire beam section area, it first needs to be split into appropriate segments. After this, the area and centroid of each individual segment needs to be considered to find the centroid of the entire section.

Consider the I-beam section shown below. To calculate the vertical centroid (in the y-direction) it can be split into 3 segments as illustrated:

Now we simply need to use the formula for calculating the vertical (y) centroid of a multi-segment shape:

$\bar{y}=\frac{\sum{y}_{i}{A}_{i}}{\sum{A}_{i}}\textup{&space;where:}\\\\&space;{A}_{i}&space;=&space;\textup{The&space;individual&space;segment's&space;area}\\&space;{y}_{i}&space;=&space;\textup{The&space;individual&space;segment's&space;centroid&space;distance&space;from&space;a&space;reference&space;line&space;or&space;datum}$

We will take the datum or reference line from the bottom fo the beam section. Now let's find Ai and yi for each segment of the I-beam section shown above so that the vertical or y centroid can be found.

$\\&space;\textup{Segment&space;1:}\\&space;{A}_{1}&space;=&space;250\times38&space;=&space;9500&space;{\textup{&space;mm}}^{2}\\&space;{y}_{1}&space;=&space;38+300+\tfrac{38}{2}&space;=&space;357&space;\textup{&space;mm}\\\\&space;\textup{Segment&space;2:}\\&space;{A}_{2}&space;=&space;300\times25&space;=&space;7500&space;{\textup{&space;mm}}^{2}\\&space;{y}_{2}&space;=&space;38+\tfrac{300}{2}&space;=&space;188&space;\textup{&space;mm}\\\\&space;\textup{Segment&space;3:}\\&space;{A}_{3}&space;=&space;38\times150&space;=&space;5700&space;{\textup{&space;mm}}^{2}\\&space;{y}_{3}&space;=&space;\tfrac{38}{2}&space;=19&space;\textup{&space;mm}\\\\&space;\therefore&space;\bar{y}=\frac{\sum{y}_{i}{A}_{i}}{\sum{A}_{i}}=\frac{{y}_{1}{A}_{1}&space;+&space;{y}_{2}{A}_{2}&space;+&space;{y}_{3}{A}_{3}}{{A}_{1}+{A}_{2}+{A}_{3}}=\frac{(357\times9500)+(188\times7500)+(19\times5700)}{9500+7500+5700}\\\\&space;\bar{y}=216.29\textup{&space;mm}$

Of course you don't need to do all these calculations manually because you can use our fantastic Free Moment of Inertia Calculator to find the vertical (y) and horizontal (x) centroids of beam sections.

Visit the next step: How to Calculate the Moment of Inertia of a Beam Section.

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• Safi

Hi dear! I have problem to find d value. Is there anyone to help me?

• Alex Martin

When applying this equation for the y-centroid it works perfectly, but I can’t seem to get this to work for the x-centroid. Is there anything different when calculating the x-centroid (other than values)?

• Hi Alex. It should all be the same. I guess be careful where your datum (reference) line is. Share a pic of your section/shape and I can put the working down so you can see if/where you’re going wrong