Material selection: Introduction

When it comes to structural design, a variety of options is available for each aspect of the design. A structural analysis might be elastic / plastic, static / time-history / pushover / pushdown, etc. The actions considered might include the self-weight of the structure, the usage (live) loads, wind actions, snow, earthquake, accidents, etc. Hence, it becomes obvious that an engineer needs to make decisions on each aspect, which will affect the final design.

One of the first in a list of such decisions is the selection of the material for each structural member (e.g. beams, columns, slabs, footings). In practice, there are many different materials available, such as timber, concrete, steel, aluminium, and masonry. However, the selection of the appropriate material is not that simple, as various different types of the aforementioned are available in the market, each with different properties. The engineer needs to consider these properties against specific design criteria and assess their fitness for each purpose, even before the preliminary design.

Lists of civil engineering materials

As explained before, in civil engineering works not only are different materials used for the same purpose (e.g. for load-bearing elements), but there are different types of each material available in the market. The following tables provide non-exhaustive lists of these materials for timber (Table 1.1), concrete (Table 1.2), steel (Table 1.3) and masonry (Table 1.4).

Note: The above tables are intentionally left blank. See Research Activity 1 for more details.

Research activities 

Material selection

As discussed in the introductory section, numerous materials are available for civil engineering works. To select the most appropriate material, an engineer needs to consider a number of parameters. Namely:

1. In similar projects, which materials are typically used? Previous experience is often used by engineers in the industry as a guidance towards the most preferable solution. While this is not restrictive and innovative solutions are realised in practice, this can often narrow down the range of the available options considered.
2. Are there any non-technical restrictions that prohibit the material from being used in the specific project? Such restrictions can be architectural requirements, environmental concerns, policy issues, local planning restrictions, etc.
3. Is it possible to find trained personnel to work on this material? Whenever it is not possible, or the use of specialised personnel would increase the cost of the construction significantly, this option can be disadvantaged over other available ones. The same restriction applies to the design of the structural members; if the designers lack the experience, sub-contracting the design might affect the overall budget for the project.
4. Is the material produced locally? If the material needs to be transported to the site from significant distance, or even to be imported, this can increase the total cost significantly and prohibit its use for the particular project.
5. Are the properties of the material adequate for the intended use? To consider this parameter, one needs to quantify not only the properties but also the expected performance of different materials in order to compare them on a fair basis. This is often achieved with the use of material indices.

For a preliminary selection of the most appropriate material, there are two methods typically used:

1. Material indices
2. Material selection charts

Material indices

The material indices method was developed 60 years ago (Niemann, 1960). It proposes that the selection of materials should be based on the comparison of volume, mass, and material cost of a given structural member. The material cost typically incorporates the aforementioned parameters (availability in the market, transportation cost, labour cost, etc.). The volume and mass are related to the performance of a structural member made of this material under given conditions. For civil engineering purposed, these are all the design cases considered.

Nowadays, the definition has evolved, in an effort to define parameters that can be independent of each other. Hence, the performance of a material (P) can be expressed as a function of the functional requirements (F) of the member, its geometry (G) and the material properties (M).

$P=f(f_1\left(F\right),f_2\left(G\right),f_3\left(M\right))$

where:

f₁(F) is a function that incorporates all functional requirements

f₂(G) is a function that incorporates all geometrical requirements

f₃(M) is a function that incorporates the related material properties

The above function is often found in literature as:

$P=f(f_1,\left(F\right)\times f_2\left(G\right)\times f_3\left(M\right))$

The reason is that, to simplify the procedure, all of these parameters are combined in a single performance function: a function that is relevant to the intended use of the element in design and the material index is the part of the function that incorporates the material properties, i.e. f₃(M).

To select the most appropriate material for a given scenario, one needs to select the material that would optimise the performance of the element. If the performance is measured as the strength or the stiffness of the element, then the performance needs to be maximised. On the contrary, if the performance is measured as the weight or the cost of the element, this would need to be minimised. In other words, the selection of the material is an optimization problem, in which the engineer considers the function of the element, selects the objective function, and defines the constraints. The main advantage in the use of Performance Indices is that this optimization problem is expressed by a single function which is related only to the properties of the material.

Note: The term “Performance Index” is often used in the literature instead of “Material Index”.

How to derive a performance index for a given problem

To derive an appropriate Material Index, one needs to define a Performance Function (P) and extract the Material Function (i.e. f₃(M)) from it. The Material Function is the Material Index for the specific scenario. The whole procedure to derive a material index can be summarised in the following steps:

1. Identify the primary function of the member considered. In structural design, this is the one that causes the critical load/deformation in the element. This will correlate the material properties to the performance of the member.
2. Define an appropriate objective function for the problem. The objective function is the one that best represents the intended outcome. E.g. if the goal is to minimise the cost, the objective function would be the total mass of the material, while the material properties can be modified to consider for the cost of the material in the market.
3. Identify all free variables. Free variables are variables that can be expressed as a function of the independent variables and known variables of the problem.
4. Identify the constraints. The constraints here are not necessarily constraints of an optimization problem, but they can be e.g. design requirements.
5. Develop formulas for the constraints. These formulas might come from literature (e.g. the bending moment resistance of a beam must be larger than the maximum bending moment in the element), the availability of options (e.g. the section height should be between 200mm and 800mm), or the characteristic of the particular problem (e.g. the beam’s deflection should be less than its span over 500, because of the equipment it carries).
6. Identify all independent variables. While in many problems it is possible to define independent variables, in some engineering problems they can be correlated. If this is the case and a function that describes their correlation cannot be implemented in the performance function, this might be depicted in the constraints.
7. Eliminate the independent variables in the objective function using the constraints. Solve the constraints for the free variables and substitute in the performance function.
8. Group the variables into 3 groups (e. Function (F), Geometry (G) and Material (M)). The part of the performance function that contains the F-variables is f₁(F), the part of that contains the G-variables is f₂(G), and the part which contains the M-variables is f₂(M), or the Material Index for this scenario.
9. Substitute in the Material Index the properties of the available materials, aiming to optimise it.

Example 1.1

Derive an appropriate Material Index for a cantilever beam with a point load at its mid-span, for structural design. The beam carries brittle elements.

1. Identify the primary function of the member considered.

   This is a beam in bending. Hence, the critical force (or moment) is the bending moment in the section.

2. Define an appropriate objective function for the problem.

   In structural design, an engineer tries to minimise the cost while at the same time the desirable performance is achieved. Hence, as the minimisation of the cost is the goal, the total mass of the element needs to be minimised

   $m=A\times L\times \rho$

   where:

   A          is the cross-sectional area of the beam

   L           is the length of the beam

   ρ          is the density of the material

3. Identify all free variables.

   The free variable for this problem is the area of the section (A).

    

   For a square cross-section: A = b².

4. Identify the constraints.

   It is mentioned that the beam carries brittle elements, so its maximum deflection should not exceed specific limits (e.g. L/400). So, its stiffness should be larger than a minimum limit.

5. Develop formulas for the constraints.

   As a function the element’s stiffness can be expressed as:

   $S=\frac{F}{\delta }=\frac{C_{1}EI}{L^3}$

   The second moment of area for a square section is:

   $S=\frac{b^4}{\mathrm{12}}=\frac{A^2}{\mathrm{12}}$

6. Identify all independent variables.

   In the above expressions, the independent variables are: ρ, E, L, F, δ and C₁. The deflection δ will be substituted with the maximum deflection δ_max.

7. Eliminate the independent variables in the objective function using the constraints.

   Substituting the second moment of area in the constraint on the element’s stiffness, we get:

   $S=\frac{F}{{\delta }_{\max }}\le \frac{C_{1}EI}{L^3}\to \frac{F}{{\delta }_{\max }}\le \frac{C_{1}E\left(\raisebox{1ex}{$A^2$}\!\left/ \!\raisebox{-1ex}{$12$}\right.\right)}{L^3}\to \frac{F}{{\delta }_{\max }}\le \frac{C_{1}E{A}^2}{L^3}$

   Solving the above for A, we get:

   $\frac{f}{{\delta }_{\max }}\le \frac{C_{1}E{A}^2}{12L^3}\to \frac{12F{L}^3}{{\delta }_{\text{max}}}\le C_{1}E{A}^2\to C_{1}E{A}^2\ge \frac{12F{L}^3}{{\delta }_{\text{max}}}\to A^2\ge \frac{12F{L}^3}{{\delta }_{\text{max}}}\to$

   $A\ge \sqrt{\frac{12F{L}^3}{{\delta }_{\text{max}}{C}_{1}E}}$

    

   The total mass of the element is:

   $m=A\times L\times \rho \to m\ge \sqrt{\frac{12F{L}^3}{{\delta }_{\text{max}}{C}_{1}E}}\times L\times \rho$

8. Group the variables into 3 groups (e. Function (F), Geometry (G) and Material (M)).

• The independent variables that are material properties are its Young’s modulus (E) and its density (ρ).
• The only independent variable that is related to its geometry is its length (L).
• The remaining independent variables, e. C₁, F and δ_max, are related to the element’s function.

Rearranging the above formula, it is:

$m\ge \sqrt{\frac{12F{L}^3}{{\delta }_{\text{max}}{C}_{1}E}}\times L\times \rho \to m\ge \sqrt{\frac{12f}{{\delta }_{\text{max}}{C}_1}}\times L^{2.5}\times \frac{\rho }{\sqrt{E}}$

The above function can be rewritten in the form of

$P=f(f_1,\left(F\right)\times f_2\left(G\right)\times f_3\left(M\right))$

as follows:

Τhe material index for this scenario is ρ/Ε^1/2.

Hence, one would need to minimise that in order to minimise the total material mass.

Note: In literature, the above be found as E^1/2/ρ, where one needs to maximise this quantity.

Example 1.2

Before you start: Finish Discussion Activity 1.1 before you read this example.

Derive an appropriate Material Index for a cantilever beam with a point load at its mid-span, for structural design. The beam carries brittle elements. The section of the beam is a solid rectangular one.

1. Identify the primary function of the member considered.

   This is a beam in bending. Hence, the critical force (or moment) is the bending moment in the section.

2. Define an appropriate objective function for the problem.

   In structural design, an engineer tries to minimise the cost while at the same time the desirable performance is achieved. Hence, as the minimisation of the cost is the goal, the total mass of the element needs to be minimised

   The total mass of the beam can be expressed as:

   $m=A\times L\times \rho$

   where:

   A          is the cross-sectional area of the beam

   L           is the length of the beam

   ρ          is the density of the material

3. Identify all free variables.

   The free variable for this problem is the area of the section (A).

    

   For a square cross-section: A = b².

4. Identify the constraints.

   It is mentioned that the beam carries brittle elements, so its maximum deflection should not exceed specific limits (e.g. L/400). So, its stiffness should be larger than a minimum limit.

5. Develop formulas for the constraints.

   As a function the element’s stiffness can be expressed as:

   $S=\frac{F}{\delta }=\frac{C_{1}EI}{L^3}$

   The second moment of area for a square section is:

   $I=\frac{b{h}^3}{12}=\frac{\beta A^2}{12}$

6. Identify all independent variables.

   In the above expressions, the independent variables are: ρ, E, L, F, δ and C₁. The deflection δ will be substituted with the maximum deflection δ_max.

7. Eliminate the independent variables in the objective function using the constraints.

   Substituting the second moment of area in the constraint on the element’s stiffness, we get:

   $\frac{F}{{\delta }_{\text{max}}}\le \frac{C_{1}EI}{L^3}\to \frac{F}{{\delta }_{\text{max}}}\le \frac{C_{1}E\left(\beta \raisebox{1ex}{$A^{12}$}\!\left/ \!\raisebox{-1ex}{$12$}\right.\right)}{L^3}\to \frac{F}{{\delta }_{\text{max}}}\le \frac{C_{1}E\beta A^{12}}{{12L}^3}$

   Solving the above for A, we get:

   $\frac{F}{{\delta }_{\text{max}}}\le \frac{C_{1}E\beta A^2}{12L^3}\to \frac{12F{L}^3}{{\delta }_{\text{max}}}\le C_{1}E\beta A^2\to C_{1}E\beta A^2\ge \frac{12F{L}^3}{{\delta }_{\text{max}}}\to A^2\ge \frac{12F{L}^3}{{\delta }_{\text{max}}{C}_{1}E\beta }\to$

   $A\ge \sqrt{\frac{12F{L}^3}{{\delta }_{\text{max}}{C}_{1}E\beta }}$

    

   The total mass of the element is:

   $m=A\times L\times \rho \to m\ge \sqrt{\frac{12F{L}^3}{{\delta }_{\text{max}}{C}_{1}E\beta }}\times L\times \rho$

8. Group the variables into 3 groups (e. Function (F), Geometry (G) and Material (M)).

• The independent variables that are material properties are its Young’s modulus (E) and its density (ρ).
• The only independent variable that is related to its geometry is its length (L).
• The remaining independent variables, e. C₁, F and δ_max, are related to the element’s function.

Rearranging the above formula, it is:

$m\ge \sqrt{\frac{12F{L}^3}{{\delta }_{\text{max}}{C}_{1}E\beta }}\times L\times \rho \to m\ge \sqrt{\frac{12F}{{\delta }_{\text{max}}{C}_1}}\times \frac{L^{2.5}}{{\beta }^{0.5}}\times \frac{\rho }{\sqrt{E}}$

The above function can be rewritten in the form of

$P=f(f_1,\left(F\right)\times f_2\left(G\right)\times f_3\left(M\right))$

as follows:

Τhe material index for this scenario is ρ/Ε^1/2.

Hence, one would need to minimise that in order to minimise the total material mass.

Note: It can be seen here that the Material Index does not change because of the additional geometrical constraint. The reason is that this constraint is not the primary constraint of the problem.

Example 1.3

Derive an appropriate Material Index for structural member in tension, for structural design. The element is used as a tie.

1. Identify the primary function of the member considered.

   This is an element in pure tension (e.g. a cable). Hence, the critical force (or moment) is the tensile axial force in the section.

2. Define an appropriate objective function for the problem.

   In structural design, an engineer tries to minimise the cost while at the same time the desirable performance is achieved. Hence, as the minimisation of the cost is the goal, the total mass of the element needs to be minimised

    

   The total mass of the beam can be expressed as:

   $m=A\times L\times \rho$

   where:

   A          is the cross-sectional area of the beam

   L           is the length of the beam

   ρ          is the density of the material

    

3. Identify all free variables.

   The free variable for this problem is the area of the section (A).

4. Identify the constraints.

   It is mentioned that the element is used as a tie, i.e. it keeps other structural elements together. Hence, its maximum deformation should not exceed specific limits, otherwise, it will fail its purpose. So, its axial stiffness should be larger than a minimum limit.

5. Develop formulas for the constraints.

   As a function the element’s axial stiffness can be expressed as: 

   $K=\frac{F}{\delta }=\frac{EA}{L}$

6. Identify all independent variables.

   In the above expressions, the independent variables are: ρ, E, L, δ_max, and F.

7. Eliminate the independent variables in the objective function using the constraints.

   Solving the above for A, we get:

   $K=\frac{F}{{\delta }_{\text{max}}}\le \frac{EA}{L}\to \frac{FL}{{\delta }_{\text{max}}}\le EA\to \frac{FL}{E{\delta }_{\text{max}}}\le A\to$

   $A\ge \frac{FL}{E{\delta }_{\text{max}}}$

    The total mass of the element is:

   $m=A\times L\times \rho \to m\ge \frac{FL}{E{\delta }_{\text{max}}}\times L\times \rho$

8. Group the variables into 3 groups (e. Function (F), Geometry (G) and Material (M)).
• The independent variables that are material properties are its Young’s modulus (E) and its density (ρ).
• The only independent variable that is related to its geometry is its length (L).
• The remaining independent variables, e. F and δ_max, are related to the element’s function.

Rearranging the above formula, it is:

$m\ge \frac{FL}{E{\delta }_{\text{max}}}\times L\times \rho \to m\ge \frac{F}{{\delta }_{\text{max}}}\times L^2\times \frac{\rho }{E}$

The above function can be rewritten in the form of

$P=f(f_1,\left(F\right)\times f_2\left(G\right)\times f_3\left(M\right))$

 as follows:

Τhe material index for this scenario is ρ/Ε.

Hence, one would need to minimise that in order to minimise the total material mass.

Note: In literature, the above be found as E/ρ, where one needs to maximise this quantity.

Example 1.4

Derive an appropriate Material Index for structural member currently in tension, for structural design. The element is part of a truss.

1. Identify the primary function of the member considered.

   This is an element in tension. However, it is mentioned that it is part of a truss, hence it could be under compression under different loading. Hence, the critical force (or moment) is the axial force in the section.

2. Define an appropriate objective function for the problem.

   In structural design, an engineer tries to minimise the cost while at the same time the desirable performance is achieved. Hence, as the minimisation of the cost is the goal, the total mass of the element needs to be minimised

    

   The total mass of the beam can be expressed as:

   $m=A\times L\times \rho$

   where:

   A          is the cross-sectional area of the beam

   L           is the length of the beam

   ρ          is the density of the material

3. Identify all free variables.

   The free variable for this problem is the area of the section (A).

4. Identify the constraints.

   It is mentioned that the element is part of a truss, so it is a load-bearing element. Hence, its axial resistance should not exceed the maximum internal force, otherwise, it will fail its purpose.

5. Develop formulas for the constraints.

   As a function the element’s axial resistance can be expressed as:

   $N_R={\sigma }_{\text{max}}A$

   The maximum internal force should not exceed its resistance, so:

   $F\le N_R\to F\le {\sigma }_{\text{max}}A$

6. Identify all independent variables.

   In the above expressions, the independent variables are: ρ, L, σ_max, and F.

7. Eliminate the independent variables in the objective function using the constraints.

   Solving the above for A, we get:

   $F\le {\sigma }_{\text{max}}A\to A\ge \frac{F}{{\sigma }_{\text{max}}}$

   The total mass of the element is:

   $m=A\times L\times \rho \to m\ge \frac{F}{{\sigma }_{\text{max}}}\times L\times \rho$

8. Group the variables into 3 groups (e. Function (F), Geometry (G) and Material (M)).

• The independent variables that are material properties are its strength (σ_max) and its density (ρ).
• The only independent variable that is related to its geometry is its length (L).
• The remaining independent variables are related to the element’s function,e. only F in this case.

Rearranging the above formula, it is:

$m\ge \frac{F}{{\sigma }_{\text{max}}}\times L\times \rho \to m\ge F\times L\times \frac{\rho }{{\sigma }_{\text{max}}}$

The above function can be rewritten in the form of

$P=f(f_1,\left(F\right)\times f_2\left(G\right)\times f_3\left(M\right))$

 as follows:

Τhe material index for this scenario is ρ/σ_max

Hence, one would need to minimise that in order to minimise the total material mass.

Note: In literature, the above be found as σ_max/ρ, where one needs to maximise this quantity.

Material Indices for different purposes and targets

Tables 1.5 to 1.9 contain the performance indices for different scenarios.

Note: All tables above were prepared after Ashby (1999). Please refer to this book for more Material Indices and further discussion on material selection.

How to use the tables

The following Figure illustrates the procedure one needs to follow to use the Material Index tables provided (or similar ones from the literature). The tables are prepared so that the selection of the appropriate Material Index follows the procedure to determine one manually.

For example:

Material selection charts

Material Selection Charts are charts used to define the most appropriate material when multiple options are available. These charts are used in order to eliminate materials whose properties do not meet the design criteria and, so, limit the search space. Material Selection Charts show the correlation between two properties of different materials and illustrate the available range for each type. They can be used as the primary means, or in conjunction with Material Indices. An example of a Material Selection Chart is given in Figure 1.1.

Swarde at English Wikipedia / Public domain

Figure 1.1. Material selection chart: Young’s modulus (E) versus density (ρ).

References

Ashby, M.F. Materials selection in mechanical design, 2^nd Edn. Butterworth-Heinemann, Oxford.

Niemann, G. Maschinenelemente: Band, I.; Grundlagen, Verbindungen, Lager, Wellen und Zubehör, 4th ed.; Springer: Berlin/Heidelberg, Germany, 1960.