Module Selection of Materials and Shapes
Lecture 3 Selection of Materials - II
Instructional objectives This is a continuation of the previous lecture. By the end of this lecture, the student will further learn (a) how to develop material index for structural application. (b) how to use the typical material indices for the selection of material for common engineering parts. Example 5: Selection of Material for a Cheap and Stiff Column Figure.3.1 Schematic presentation of a cylindrical column with a compressive load, F Figure 1 shows the schematic picture of a typical cylindrical column with uniform cross-section subjected to a compressive load, F. The task is to develop a suitable material index that can help in the selection of the material for the column that is cheap and sufficiently stiff to avoid a buckling failure. The above problem can be translated into functional requirement, objective, constraints to be considered, and the free variables that the designers are allowed to change. Function: Column to withstand a compressive load of F Objective: Minimise cost (C) where C = (ALρ)Cm, where ρ is the material density. Constraints: Free variable: L is the length of the column, A is the uniform circular cross-section of the column, and C m is the cost per unit mass of the material. (i) Length L is specified, (ii) Must not buckle under compressive load, F (i) Cross-sectional area, A, (ii) Material
Performance Equation: The Performance Equation can be developed considering the fact that the compressive load, F, must not exceed the critical buckling load (F CR ) for the given column considering its important dimension and material property. Usually, the length of typical columns is always a constraint. Thus, the Performance Equation can be given as nπ EI F FCR where FCR = (1) L where r is the radius of the cylindrical column, E is the young s modulus, I is the second moment π r A of area and can be given as I = =, and n depends on the typical end constraints of the 4 4π column. The Performance Equation can now be rewritten by substituting the expression of I in equation (1) and eliminating the term A thereafter from equation (1) as 1/ 1/ 4 F 1/ 3 C ρ C (L) m () nπ 1/ L E The material index, M, in this case becomes E C 1/ m ρ and a material with higher value of M would be a better candidate both in terms stiffness and cost in comparison to a material with lower value of M. Table.3.1 illustrates the evaluated values of the above material index for a range of common engineering materials. 1/ Table.3.1 Estimated values of ( E C ρ ) for common engineering materials Materials E (MPa) C m (Rs./kg) (app.) ρ (kg/m Mild Steel 05000 60.00 7870 9.59 x 10-4 Aluminium 68000 50.00 698 3.87 x 10-4 m 3 ) E 1/ C mρ Concrete 0000 10 010 70.35 x 10-4 Stainless Steel 193000 350.00 7860 1.59 x 10-4 The above example in addition to the four examples explained in the previous lecture show the general approach to develop Material Indices for the selection of suitable material for various
structural requirements. Once a typical material index is developed for an intended application, the same can be used next to choose a range of suitable materials either from a Material Property Chart in a graphical manner or analytically by computing and ranking the values of the material index for a range of engineering materials. The actual worked out examples on how to evaluate the material indices for common engineering materials are presented in subsequent chapter. General Approach to develop a Material Index The general approach to develop a Material Index for an intended application lies in the appropriate realization of the functional requirements, constraints, objective, and the free variables (unconstrained). It is easy to understand these aspects through a set of queries. For example, the function requirement can be realized by asking a question: what does the component intend to do? The constraints can be realized by asking the query: what specific constraints must be met e.g. stiffness and / or strength and / or dimensions? The constraints can often be specified as hard constrains that are non-negotiable. For example, a component must carry a certain load without elastic deflection or plastic deformation or failure. The constraints can also soft that are primarily relation to the aesthetic aspects or cost and hence, negotiable The objective primarily stands for what is to be minimized or maximized? The free variables refer to the features that the designers are free to change (e.g. dimensions, materials, etc.). Once the functional requirements, constraints, objective, and the free variables are identified for a typical application, all the constraints related to the task should be listed and if possible, the constraints can be presented in the form of a set of a single or multiple expressions. Next, the objective of the design must be expressed in terms of functional requirements, geometry and materials properties. This expression is referred to as the performance equation as shown in the examples 1 to 5. If the performance equation contains a free variable, we have to identify the constraint that limits the free variable. Next, we use this constrain to eliminate the free variable in the performance equation. Lastly, we should be able to select the combination of the material properties, referred to as the material index, which would maximize the performance. A set of examples are given below to show how a suitable material index can be developed for the selection of material for different applications.
Example 6: Selection of Material for legs of a typical Reading Table. Figure.3. Schematic picture of a Reading Table with four supporting legs Figure shows the schematic view of a typical table with four supporting legs. For simplicity, we can assume that the supporting lags confirm to a uniform circular cross-section and each leg must support a compressive load, F, without buckling distortion. Thus, the problem can be envisaged as to develop a suitable material index to aid to the selection of material for a slender and light supporting leg that will be able to support an applied design load without buckling distortion and will not break if struck accidentally. Thus, the nature of the problem is similar to the Example-5, as outlined above. The above problem can be translated into functional requirement, objective, constraints to be considered, and the free variables that the designers are allowed to change as follows. Function: Column to withstand a compressive load of F Objective: Minimise the mass (m) and Maximize the slenderness Constraints: (i) Length L is specified, (ii) Must not buckle under a compressive load, F, which is envisaged as the design load. (iii) Must not fracture if struck accidentally. Free variable: (i) Cross-sectional area, A, or Diameter of legs, (r) (ii) Material Performance Equation: Considering that the supporting leg to be a slender column of any material with density ρ and length L, the mass, to be minimized, can be given as m = π r Lρ (3)
The maximum buckling load that each leg can support without a buckling distortion can be given as 3 4 π EI π Er π r FCR = = where I = (4) L 4L 4 Substituting the free variable, r, from equation (4) to equation (3), we can write 1 1 4F ρ 4F 1 m (L ) m (L ) (5) π 1/ E π M1 where M 1 represents the material index. It is easy to understand from equation (5), that the mass, m, of a supporting table leg can be minimized as the material index, M 1, will be maximized for a given set of candidate materials. A comparison of equations (5) and (3) further shows that thinnest possible leg that will not buckle under a designed compressive load, F, can be given as 1 1 1/ 4 1/ 4 4F 4 1/ 1 4F 1 r (L ) r (L ) 3 E M (6) π π It is clear from equation (6), that the diameter (r) of the supporting table legs with uniform cross-section would be minimized, which will in turn enhance the slenderness, with the increase in the value of the material index, M. To meet the constraint that the table legs should not fracture if struck accidentally, a third material index, M 3, may be considered corresponding to the fracture toughness (K 1C ) of the material. Thus, the problem of selecting a suitable material for the table leg can be envisaged as an optimization problem where all the three material indices ( K ) ρ M 1 1/ E =, ( E) M = and M3 = 1C are required to be maximized within a set of candidate materials. Based on initial screening, an active set of candidate materials may be considered as wood, steel, aluminum, titanium, composite (e.g. CFRP), etc. The values of all the three material indices for the initially screened materials can be evaluated either from material handbook or material property charts and accordingly, a ranking of these materials based on the corresponding values of the material indices can be prepared. At this stage, the manufacturability and the cost evaluation of each
material should be undertaken. The final selection of material at the end of such an exercise would be a trade-off among greater values of the corresponding material indices, manufacturability aspects and cost evaluation. Exercise 1. Develop a suitable material index for the selection of material for Oars used for rowing.. Develop a suitable material index for the selection of material for Spatula used for cooking. References 1. G Dieter, Engineering Design - a materials and processing approach, McGraw Hill, NY, 000.. M F Ashby, Material Selection in Mechanical Design, Butterworth-Heinemann, 1999.