Material Selection - Example

Example

A common method for choosing an appropriate material is an “Ashby chart.” By plotting a performance index for a specific case of loading on the Ashby chart, a material with maximum performance can be selected. The performance index takes into consideration the dimensional constraints, material constraints, and free variable constraints of a specific application. The following example will show the how to come up with the performance index and how to plot and interpret the Ashby chart.

This example will take into consideration a beam that will undergo two different loads with the goal of minimizing weight. The first load is a beam in tension. Figure 1 illustrates this loading.

The parameters for the beam can be organized into categories. These categories are material variables, which include density, modulus, and yield stress, free variables which are variables that can change during the loading cycle, for example applied force. The final category is design variables which usually are a limit of how thick the beam can be, how much it can deflect, or any other limiting factor for the specific application. For this loading cycle, the stress in the beam is measured as =P/A, where P is the load and A is the cross sectional area. The weight is measure as w=ρAL, where ρ is the density, and L is the length. By looking at the equation, we see that for a fixed length of L, the material variables are σ and ρ. There is one free variable, A, and a variable that needs to be minimized, w.

In order to find the performance index, an equation for w in terms of fixed and material variables needs to be found. This means that the variable A has to somehow be replaced. By rearranging the axial stress equation, A can be represented as A=P/σ. Substituting this into the weight equation,w=ρ P/σ L, gives an equation for weight that has only fixed and material variables.

The next step is to separate the material variables form all other variables and constants. The equation becomes w=(ρ/σ)LP. Since the goal is to minimize weight, the material variables have to be minimized. This means that (ρ/σ) has to be minimized, or the inverse equation, (σ/ρ) has to be maximized. We call the equation that needs to be maximized our performance index. P_cr=(σ/ρ). It is important to note that the performance index is always an equation that needs to be maximized, so inverting an equation that needs to be minimized is necessary.

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 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.

The next loading cycle will have a different performance index with a different equation. For example, if you also want to maximize this beam for bending, using the max tensile stress equation of bending σ=(-My)/I, where M is the bending moment, y is the distance from the neutral axis, and I is the moment of inertia. This is shown in Figure 2. Using the weight equation above and solving for the free variables, you arrive at w=(√(6MbL^2 ))*(ρ/√σ), where L is the length and b is the height of the beam. This turns the material performance index into P_CR=√σ/ρ

By plotting the two performance indices on the shame Ashby chart, the maximum performance index of both loading types together will be at the intercept of the two lines. This is shown in figure 3

As seen from figure 3 the two lines intercept near the top of the graph at engineering ceramics. 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 “diamond” area, 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 CRFP.