# Area and Centroid

#### Area of a Section

The definition of a section's area is common knowledge, in structural mechanics the area of a section is useful for determining both the axial stiffness of a section and also the axial stress that applies for a section under a given load. Calculating the area of standard shapes will not be talked about in detail here. The calculation of area of more complex sections is often undertaken by breaking the section down into it's constiuent shapes and then adding up each component.

$$A_{total} = \sum A_{1} + A_{2} + ... + A_{n}$$

#### Centroid of a Section

The centroid of a section is arithmetic average of the geometric layout of the section. The centroid is analogous to the centre of gravity of a section, if section geometry had a uniform thickness and density then the centre of gravity and the centroid of the section would be the same point. Similarly, if the section had a uniform thickness and density then the whole section could be balanced on a single point, if the point was positioned at the centroid

In the metric system, the centroid is a position on a section, given as a length from a known point and therefore has units of m, cm or mm.

The centroid is useful in structural mechanics for the following reasons:

• It is the point that load on whole section can be considered to act.
• When considering elastic Euler-Bernoulli beam bending theory, it is the point where the neutral axis will pass through. This is the axis that the section under pure bending will bend about, giving rise to no bending stress at the neutral axis, but the largest positive stress and negative stresses in the opposite fibres of the section. This is discussed in more detail in the article on "Euler-Bernoulli Beam bending equation".

The centroid of individual simple shapes can either be determined through inspection, symmetry or through standard solutions. For example it is clear by inspection that for a square and a circle the centroid will be at the centre of the shape. For shapes such as rectangles, it can be seen that through a line of symmetry, one part of the section will balance the other and therefore the centroid will be at the point where these lines of symmetry intersection.

However, the centroid for non-symmetric shapes such as triangles cannot be found by inspection or by symmetry. For these we must use a standard solution that states: The centroid of a triangle in which the three lines that connect one corner, to the mid-point of the opposite edge of the triangle, all intersect. For a right angled triangle, this point lies a third of the way along the base and a third of the way up the height of the perpendicular sides :

This method of establishing the centroid of a triangle was first recorded in a book by Heron of Alexandria, around the first century AD.
Amoung his other inventions Heron of Alexandria also demontrated a primative steam engine, the aeolipile.

Unfortunately, most Engineering sections do not consist of simple shapes that have pre-existing standard solutions. What is needed is a way to combine the known solutions for simple shapes together to approximate a more complex section. To do this we use the "Geometric Decomposition" method:

$$\overline{x} = \frac{\sum A_{i}\overline{x}_{i}}{\sum A_{i}}\text{ gives the offset in the x-direction}$$ $$\text{Where, } \overline{x} \text{ is the distance in x to the overall centroid } \overline{x}_{i} \text{ is the distance in x to the centroid of the sub shape } A_{i} \text{ is the area of the sub shape}$$ $$\overline{y} = \frac{\sum A_{i}\overline{y}_{i}}{\sum A_{i}}\text{ gives the offset in the y-direction}$$ $$\text{Where, } \overline{y} \text{ is the distance in y to the overall centroid } \overline{y}_{i} \text{ is the distance in y to the centroid of the sub shape } A_{i} \text{ is the area of the sub shape}$$

It is easy to see how the above equations work, as the total area is both on the top and bottom of the division, so dividing through leaves the weighted average centroid. An example calculation for determining both the area and centroid of an asymmetrical I-section is shown below:

Calculation of centroid for an I-section using "Geometric Decomposition"
$$\text{Using the values of } \sum A_{i}\overline{y}_{i} \text{ and } \sum A_{i} \text{ from the above table, we can find:}$$ $$\overline{y} = \frac{\sum A_{i}\overline{y}_{i}}{\sum A_{i}}= \frac{19,704,000}{64,800} = 304mm$$