The centroid of a cross section occurs at the location where there is a juncture in the shape between positive and negative stresses due to bending that occurs with respect to an axis in the cross section. It is the location where above and below the centroid there is an equivalent amount of cross sectional area distributed. This distribution of area with respect to the centroid is what determines the moment of inertia for a section which is its level of resistance to bending. Determination of the location of a centroid is necessary when calculating the moment of inertia for a section that is not symmetric and therefore requies the use of the parallel axis theorem.

We are concerned with determining the centroid of a section about one or both of its axes as it relates to bending. When a simple wooden beam supports a floor system, we see that it undergoes bending about its x-axis (the x-axis conventionally is assumed to be the axis for which a section has its greatest resistance to bending, I_{X}) to resist loads acting due to gravity. The wooden beam does not undergo bending about its y-axis (weak axis bending) simply because there are no loads for it to resist. Nonetheless, we can still determine the location of the centroid for each axis.

For a symmetric cross section, the centroid with respect to each axis is quite simply at the very center of the section. For example, let's consider a wooden beam of actual dimensions 2" (horizontal) x 8" (vertical). The best way to get started is to draw the x-axis along the bottom horizontal surface of the section and likewise the y-axis along the left vertical surface of the section. Since the section is symmetric with respect to the x and y axes individually, we can move the x-axis up vertically 4 inches and the y-axis to the right 1 inch, these respectively become the centroids of the section with respect to the x and y axis.

If we are considering an unsymmetric section, determining the locations of the centroids takes a little more work but is needed to calculate the moment of inertia with respect to each axis. Let's imagine that the same 2" x 8" wooden beam is found to be inadequate, to strengthen it up we add an actual dimension 2" x 4" to the bottom of it. This 2" x 4" is positioned where 2" is in the vertical and 4" is horizontal. The 2" x 4" is placed symmetrically with respect to the y-axis and so this vertical centroid remains in the center of the section. The centroid with respect to the x-axis of course has now changed and shifted downward. We can determine it new location by employing the following, simple formulation.

y = [(A_{1})(d_{1})+ (A_{2})( d_{2})] / (A_{1} + A_{2})

A_{1 }and A_{2}, respectively are the areas of the cross section broken up into its individual components, and d_{1 }and d_{2} are the distances from the centroid of each individual area to the x-axis line that is drawn along the bottom base of the 2' x 8" and 2" x 4" composite section.

So using this formulation we will calculate the new location of the centroid with respect to the x-axis.

A_{1} = (2in.)(8in.) = 16 in.^{2 }

A_{2} = (2in.)(4in.) = 8 in.^{2 }

d_{1} = 4in. + 2in. = 6in. Thisis the location of the centroid of the 2" x 8" section to the x-axis line drawn along the base of the 4" surface.

d_{2} = 1in. Thisis the location of the centroid of the 2" x 4" section to the x-axis line drawn along the base of the 4" surface.

y = [(16 in.^{2})( 6in.)+ (8 in.^{2})( 1in.)] / (16 in.^{2 }+ 8 in.^{2})

y = [96 in.^{3} + 8 in.^{3}] / (24 in.^{2})

y = (104 in.^{3}) / (24 in.^{2})

y = 4.33 in.

So this means the location of the centroid is 4.33 in. above the x-axis line drawn at the base. When we consider the total depth of the composite section is 10 inches, it seems reasonable that we have a distance less than 5 inches since we have more cross sectional area concentrated at the base of the section. This same formulation could also be applied to a composite section consisting of more than 2 individual areas. With the piece of information we can now calculate the moment of inertia for this composite section using the parallel axis theorem.

RESOURCE:

Gere, James M. (2004). Mechanics of Materials (6^{th} Ed.). Belmont, CA: Brooks/Cole â€“

Thomson Learning.

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