Strength of materials/Appendix A

Centroids of Plane Areas
A centroid is the center of a given area. In physical terms, if the area was assigned a uniform density, the centroid would be the center of mass. In order to fully explore the idea of centroids, a firm understanding of calculus will be required, especially integration. We start by taking an area and dividing it up into differential areas. The total area is &int; dA. We find the first moment of area by the following equation: $$Q_x = \int y dA \;, \; Q_y = \int x dA$$ We define the centroid as the first moments divided by the total area: $$\bar{x}=\frac{Q_y}{A}=\frac{\int x dA}{\int dA} \;, \; \bar{y}=\frac{Q_x}{A}=\frac{\int y dA}{\int dA}$$ Using differential areas, we can find the centroid of any complicated area we have an equation for: In this example, we know the equation of the top line, so finding the centroid involves finding the area and then the first moment of area: $$A=\int dA=\int\limits_0^b y \, dx = \int\limits_0^b \left [ a+c \left ( \frac{x}{b} \right )^2 \right ] \, dx=\frac{b(3a+c)}{3}$$ $$Q_y=\int x\, dA=\int\limits_0^b x (y \, dx) = \int\limits_0^b x \, \left [ a+c \left ( \frac{x}{b} \right )^2 \right ] \, dx = \frac{b^2(2a+c)}{4}$$ $$Q_x=\int\limits_0^b \frac{y}{2} \, dA=\frac{1}{2} \int\limits_0^b y^2 \, dx = \frac{1}{2} \int\limits_0^b \left [ a+c \left ( \frac{x}{b} \right )^2 \right ]^2 \, dx = \frac{b(15a^2+10ac+3c^2)}{30}$$ $$\bar{x}=\frac{Q_y}{A}=\frac{3b(2a+c)}{4(3a+c)}$$ $$\bar{y}=\frac{Q_x}{A}=\frac{15a^2+10ac+3c^2}{10(3a+c)}$$ Equations can be built for all kinds of functions using similar techniques. For simple shapes like triangles and rectangles, the centroids and areas can be determined easily. Knowing this can greatly simplify finding centroids for more complicated areas.

Centroids of Composite Areas
Most common shapes for beams are composites of simple shapes, like triangles and rectangles. Since the centroids of those simple areas can be easily found, the centroid of a composite of those shapes can be found with little more effort. The equations for the centroid of a composite of shapes is: $$ \begin{align} A & =\sum A_i \\ Q_x & = \sum \bar{y_i}A_i \\ Q_y & = \sum \bar{x_i}A_i \\ \bar{x} & = \frac{Q_y}{A} = \frac{\sum \bar{x_i}A_i}{\sum A_i} \\ \bar{y} & = \frac{Q_x}{A} = \frac{\sum \bar{y_i}A_i}{\sum A_i} \\ \end{align}$$ These equations can easily be derived. The ith terms are the individual simple shapes. The centroids of the shapes are denoted with a bar over the variable.

Moments of Inertia of Plane Areas
The moments of inertia are also known as second moments of area. This can be seen in the equation: $$I_x=\int y^2 dA$$ $$I_y=\int x^2 dA$$ The physical meaning of the moment of inertia is related to its name. If the region of area was to be spun around the axis we are finding the oment of inertia about, this would be how "hard" it would be or how much torque is required if the area is assigned a density. The same types of techniques used to find the centroid can be used to find the moment of inertia. Moments of inertia can be found around any set of axes desired. In the next section, we will see how to more easily find moments of inertia at different axes. In addition to the above techniques, moments of inertia of composite shapes can be added together: $$I_x = \sum_{j=1}^n I_{x_j}$$ $$I_y = \sum_{j=1}^n I_{y_j}$$

The Parallel Axis Theorem
The parallel axis theorem allows us to "shift" the axes we want to find the moments of inertia about. Usually, we know the moments of inertia of a certain shape about the centroidal axes (axes that pass through the centroid), but not around a specific set of axes. For example, an I-beam consists of 3 rectangular pieces joined together. The moment of inertia about centroidal axes of each individual piece can be found, but that of the total I-beam can be solved for using piecewise functions and the integral method, but we can use the parallel axis theorem to find it very easy. In the figure above, the centroid C of the ellipse is already known. The moment of inertia of the ellipse is easy to find around the axes that run through the centroid, but what about the axes that run through the origin? The parallel axis theorem can be derived from this case: $$ \begin{align} I_x & = \int y^2 dA \\ & = \int ( \bar{y} + y_c)^2 dA \\ & = \int \bar{y}^2 dA + 2 \int y_c \, \bar{y} \, dA + \int y_c^2 \, dA \\ & = \bar{y}^2A+2\bar{y} \underbrace{\int y_c \, dA}_0 + I_{x_c} \\ & = \bar{y}^2A + I_{x_c} \end{align} $$ $$ \begin{align} I_y & = \int x^2 dA \\ & = \int ( \bar{x} + x_c)^2 dA \\ & = \int \bar{x}^2 dA + 2 \int x_c \, \bar{x} \, dA + \int x_c^2 \, dA \\ & = \bar{x}^2A+2\bar{x} \underbrace{\int x_c \, dA}_0 + I_{y_c} \\ & = \bar{x}^2A + I_{y_c} \end{align} $$ The c subscript denotes the y value in relation to the centroid. This theorem is really powerful because the moment of inertia about any set of axes can be found by finding the moment of inertia about the centroidal axes and adding the distance-area term to it. This works especially well when the general shape of the area can be decomposed into simpler shapes for which the moment of inertia is calculated for.

Polar Moments of Inertia
A polar moment of inertia is defined as the moment of inertia if you try to rotate a plane area around the axis coming out of the plane. It is defined as: $$ \begin{align} I_p & =\int \rho^2 dA \\ & = \int (x^2+y^2) dA \\ & = I_y + I_x \\ & = (\bar{I}_y+A\bar{x}^2)+(\bar{I}_x+A\bar{y}^2) \\ & = \bar{I}_y + \bar{I}_x + A(x^2+y^2) \\ & = \bar{I}_p + Ad^2 \end{align} $$ where d is the distance from the origin of the axes to the centroid of the area. The parallel axis theorem applies here as well.