Material selection

In the first situation the beam experiences two forces: the weight of gravity ${\displaystyle w}$ and tension ${\displaystyle P}$. The material variables are density ${\displaystyle \rho }$ and strength ${\displaystyle \sigma }$. Assume that the length ${\displaystyle L}$ and tension ${\displaystyle P}$ are fixed, making them design variables. Lastly the cross sectional area ${\displaystyle A}$ is a free variable. The objective in this situation is to minimize the weight ${\displaystyle w}$ by choosing a material with the best combination of material variables ${\displaystyle \rho ,\sigma }$. Figure 1 illustrates this loading.

Figure 1. Beam under Tensile stress loading to minimize weight.

The stress in the beam is measured as ${\displaystyle P/A}$ whereas weight is described by ${\displaystyle w=\rho AL}$. Deriving a performance index requires that all free variables are removed, leaving only design variables and material variables. In this case that means that ${\displaystyle A}$ must be removed. The axial stress equation can be rearranged to give ${\displaystyle A=P/\sigma }$. Substituting this into the weight equation gives ${\displaystyle w=\rho (P/\sigma )L=\rho LP/\sigma }$. Next, the material variables and design variables are grouped separately, giving ${\displaystyle w=(\rho /\sigma )LP}$.

Since both ${\displaystyle L}$ and ${\displaystyle P}$ are fixed, and since the goal is to minimize ${\displaystyle w}$, then the ratio ${\displaystyle \rho /\sigma }$ should be minimized. By convention, however, the performance index is always a quantity which should be maximized. Therefore, the resulting equation is ${\displaystyle {\text{Performance index}}=P_{cr}=\sigma /\rho }$

Next, suppose that the material is also subjected to bending forces. The max tensile stress equation of bending is ${\displaystyle \sigma =(-My)/I}$, where ${\displaystyle M}$ is the bending moment, ${\displaystyle y}$ is the distance from the neutral axis, and ${\displaystyle I}$ is the moment of inertia. This is shown in Figure 2. Using the weight equation above and solving for the free variables, the solution arrived at is ${\displaystyle w={\sqrt {6MbL^{2}}}(\rho /{\sqrt {\sigma }})}$, where ${\displaystyle L}$ is the length and ${\displaystyle b}$ is the height of the beam. Assuming that ${\displaystyle b}$, ${\displaystyle L}$, and ${\displaystyle M}$ are fixed design variables, the performance index for bending becomes ${\displaystyle P_{CR}={\sqrt {\sigma }}/\rho }$.

Figure 2. Beam under bending stress. Trying to minimize weight

At this point two performance indices that have been derived: for tension ${\displaystyle \sigma /\rho }$ and for bending ${\displaystyle {\sqrt {\sigma }}/\rho }$. The first step is to create a log-log plot and add all known materials in the appropriate locations. However, the performance index equations must be modified before being plotted on the log-log graph.

For the tension performance equation ${\displaystyle P_{CR}=\sigma /\rho }$, the first step is to take the log of both sides. The resulting equation can be rearranged to give ${\displaystyle \log(\sigma )=\log(\rho )+\log(P_{CR})}$. Note that this follows the format of ${\displaystyle y=x+b}$, making it linear on a log-log graph. Similarly, the y-intercept is the log of ${\displaystyle P_{CR}}$. Thus, the fixed value of ${\displaystyle P_{CR}}$ for tension in Figure 3 is 0.1.

The bending performance equation ${\displaystyle P_{CR}={\sqrt {\sigma }}/\rho }$ can be treated similarly. Using the power property of logarithms it can be derived that ${\displaystyle \log(\sigma )=2\times (\log(\rho )+\log(P_{CR}))}$. The value for ${\displaystyle P_{CR}}$ for bending is ≈ 0.0316 in Figure 3. Finally, both lines are plotted on the Ashby chart.

Figure 3. Ashby chart with performance indices plotted for maximum result

First, the best bending materials can be found by examining which regions are higher on the graph than the ${\displaystyle {\sqrt {\sigma }}/\rho }$ bending line. In this case, some of the foams (blue) and technical ceramics (pink) are higher than the line. Therefore those would be the best bending materials. In contrast, materials which are far below the line (like metals in the bottom-right of the gray region) would be the worst materials.

Lastly, the ${\displaystyle \sigma /\rho }$ tension line can be used to "break the tie" between foams and technical ceramics. Since technical ceramics are the only material which is located higher than the tension line, then the best-performing tension materials are technical ceramics. Therefore, the overall best material is a technical ceramics in the top-left of the pink region such as boron carbide.

The performance index can then be plotted on the Ashby chart by converting the equation to a log scale. This is done by taking the log of both sides, and plotting it similar to a line with ${\displaystyle P_{cr}}$ being the y-axis intercept. This means that the higher the intercept, the higher the performance of the material. By moving the line up the Ashby chart, the performance index gets higher. Each materials the line passes through, has the performance index listed on the y-axis. So, moving to the top of the chart while still touching a region of material is where the highest performance will be.

As seen from figure 3 the two lines intercept near the top of the graph at Technical ceramics and Composites. This will give a performance index of 120 for tensile loading and 15 for bending. When taking into consideration the cost of the engineering ceramics, especially because the intercept is around the Boron carbide, this would not be the optimal case. A better case with lower performance index but more cost effective solutions is around the Engineering Composites near CFRP.