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The centroid (center) of a beam.


Several posts ago I talked about finding the center of mass of an object. That procedure when applied to a flat surface (a plane) is used to locate the point called the centroid.  The the plane is a flat plate made completely from a uniform material, the centroid is where a pin could be stuck and the plate would be balanced.

For simple symmetric objects this is simply the center of the object.   This definition will work for the majority of objects we will work with such as circular rods, tubes, and pipes as well as rectangular objects such as lumber including popsicle sticks.  If our plane surface is the cross section of a pipe, or stick of lumber and cross section area is uniform throughout the length of the material we can imagine the centroid forming a line down the length of the object.   This line will become our neutral axis as we start thinking about bending stresses.

For circles the centroid is the center of the circle.  On triangles the centroid can be located by bisecting each of the angles and finding the point where these bisectors intersect as shown in the first picture.  The point where the diagonals cross on a rectangle locates the centroid of it.  On more complex objects the centroid is found by summing all the areas of component parts and locating their centroid in relation to a reference point. This is done in both the horizontal and vertical direction.   For each of those directions the centroid of the total is the sum of each of the component centroids X the area of each component part divided by the total area.   An example is probably the best way to explain that.

A T shaped area.

Imagine we have a beam with the cross sectional area in the shape of a T as shown in the picture to the right.  We  look at the object and divide it into component parts as shown with the dotted line  Each component is a rectangle with an area of  2 X 6 or 12.

The center of each part will 3 in one direction and 1 in the other direction from the corner of the object and since we have located our reference point at the point located by the blue center symbol. The following will be the centers of each part in reference to the reference point.

Horizontal (X):  Bx = 1 + 2 = 3  and Ax = 3.   Our total center will be located at:

3 X 12 + 3 X 12) / (12 + 12)  = *(36+36)/24 = 72/24 = 3 in the horizontal direction.   (Looking at symmetry would have been a whole easier).’

Vertical (Y): By = 3 and Ay = 1+ 6 = 7 .  Our total center in the vertical direction is located at:

(3 X 12 + 7 X12) / (12 + 12) =( 36+84)/24 = 120/24 = 5

So the centroid lies in the center of the T leg, 1 inch below the top cross.

A slightly more interesting one is shown to the right of this.  We will use exactly the same procedure, expect in the vertical direction we can see based on symmetry that the centroid will be 2 + 3 or 5 above the bottom of the shape.

The horizontal will be:
(3X12+3X12+1X12)/(12X3)
or
(3+3+1)/3 = 7/3 or 2.3333

This centroid is not even on the object, but as you look at the object this answer makes a lot of sense.

This is the general procedure and is good enough for anything most of us will ever do.  However, if we were to use a computer program to find the center in each direction, it would make lots and lots of very small squares or rectangles (stirips) and do exactly the same procedure but at a very fine detail.   This would enable it to easily determine the centoid of very unusual shapes.

The other procedure is to do this same operation using calculus.  However, since most of us have forgotten most of the calculus we once knew that is not really a viable possibility.

If we had a shape that included triangles, semi circles, or quarter circles we can look up the centroid and area in tables and add and subtract the components just as I did earlier.    An example of a table can be found in this Wikipedia article:  List of Centroids. Math and physics books often also have similar lists.

As I constructed the main columns in my popsicle tower I made each one so the centroid was in the center of the beam so the neutral axis would run the length of the column straight down the middle.   In the sections that consisted of 3 sticks this was in the middle of a stick.   In the sections that only consisted of 2 sticks it was in the space between the sticks.    In the next structural post on this blog you will see why that was done.

Again the emphasis is not on the math.  The math is there to show why, but the main emphasis is developing the intuition and understanding.   We do not know the actual tensile and compressive strength of our popsicle sticks so the math will not get us very far in calculating the strength of the assembly.  But our understanding will save us in the effort.

As always if you enjoyed this post, please consider signing up using one of the subscription methods.   Also, if you have not yet done so you may want to check out the Resources Page.  I have added several CAD files there of drawings I am using to create the diagrams for these posts.

Thank you,

Gary


 

 

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