The position of a centroid of a plane area is an important geometric property. In order to obtain formulae for locating the centroid of a shape, we need to consider an arbitrary shape, such as the one pictured below:
If we take a very small piece of that shape (pictured as the square), its differential area will be dA and its coordinates will be x1 and y1. The area of the whole shape may be expressed as:
where dA is the differential area represented by coordinates x1 and y1. In other words, what the above expression is saying is that the area of the whole shape is simply the sum of all the individual differential areas.
The first or static moment of an area is defined by:
It is evident from Eq 2 that the first moment of an area is the sum of the products of the differential areas and their respective coordinates. However, if we are dealing with a simple shape such as a rectangle, or we are already given the location of centroid and area, we can avoid using integration.
If we were now to take Qx and Qy and divide them by the area of the whole shape, we would obtain yc and xc - the coordinates of the centroid! This is expressed as:
 |
---eq(1) |
Note that if an area is symmetric about an axis, its centroid must lie on that axis and therefore its first moment of area will be zero. Similarly, if the area is only symmetric about one axis, then only the remaining coordinates about the axes where the shape is not symmetrical must be calculated to locate the centroid.
Example 1
Find Qx and Qy of the following shape:
Solution
The differential area (dA) of the hatched section can be written as b × dy. For Qx, the range is
.
Similar process would be applied to obtain Qy, except that the differential area would now be vertical and thus have a width of dx:
For geometric shapes with irregular boundaries, the procedure for finding the centroid is by numerical evaluation of eq(1) above. First we divide the shape into a number of small, finite elements and then we take their sum. This is expressed as:
This is what all these confusing terms
mean: A is the sum of small (
) areas of
all terms from i=1 (first term) to i=n (nth or last term). Qx
and Qy are then just simply the products of those areas multiplied by
each individual areas' x and y (respectively) coordinates of its centroid. In other
words, we divide our irregular shape into a number of equal but very small finite
elements. Then we find Qx and Qy (static moments) of each
element by multiplying its area by x and y coordinates of its centroid.
The expression for the whole shape's x and y coordinates of the centroid then becomes:
The expressions are really not as bad as they look. All it is is just the sum of static moments of all finite elements divided by the sum of areas of all finite areas.
Example 2
| Locate the centroid C of the semisegment pictured below using both horizontal and vertical strips. Equation of the semisegment is |  |
Solution
Vertical strip: Differential area of the vertical strip, dA will be distance in x-direction (dx) multiplied by distance in y-direction (from y=0 to the curved part of semisegment, ie y=f(x)). Since the distance in the x-direction is the differential distance, then we take the range in the same direction, that is from 0 to any value, b, to obtain the area of the whole semisegment:
Now we find the static moment of each finite element (this will apply to all of them):
Note that we assumed that each strip was rectangular in shape (because the strip is very small), so the centroid of each strip was located at y/2 and x.
Finally, we just substitute the above three results into our expression for irregular shapes:
Horizontal strip: Differential area of the horizontal strip will be the distance in x-direction (from x=0 to x=f(y)) multiplied by the distance in y-direction (dy). The range of the differential distance, dy, will stretch from 0 to any value, h. With these 2 pieces of information, we can find the area of the semisegment. First, let's express the equation of the semisegment in terms of y:
Onto first moments:
Again, since we assumed that each strip was a rectangle, then naturally the centroid of each strip was located at x/2 and y.
And so finally,
It's obvious from the above example that it makes no difference which strip you choose, however, the work load in deriving the solution may be drastically reduced if the right strip is chosen (this you will be able to do in time after regular practice).
When we are dealing with an object made up of a number of smaller shapes, it is useful to treat each smaller shape individually and work out the centroid of the whole object by summing the results. This is simply expressed as:
Again, don't be put of by the terminology! The sigma sign with n on top and i=1 on bottom means sum of all shapes from the first one (i=1) to as many as there are (n). The x (or y) with a bar on top and i on bottom right is the x (or y) coordinate of the centroid of shape number i and A with an i on the bottom right is its area. So if there are 10 little shapes, we would need to multiply each individual centroidal x (or y) coordinate with the shape's individual area and add all the results together to get the location of the centroid for the whole object. Let's look at an example:
Example 3
Find the location of the centroid of the following shape:
Solution
Let's consider this object as being made up of two smaller shapes. Shape 1 we will consider to stretch from 0 to b in the y direction and 0 to r in the x direction. Shape two will therefore go from 0 to s in the y direction and r to c in the x direction.
Now you're ready for second moment of plane area >>>
