The examples thus far aimed at the design of components that did not bend too much—that is, they met a stiffness constraint. Elasticity can be used in another way: to design components that are strong but not stiff, arranging that they bend easily in a certain direction. Think, for instance, of a windscreen wiper blade. The frame to which the rubber squeegees are attached must adapt to the changing profile of the windscreen as it sweeps across it. It does so by flexing, maintaining an even pressure on the blades. This is a deliberately bendy structure.
Figure 5.17 shows two ways of making a spring-loaded plunger. The one on the left with a plunger and a spring involves sliding surfaces. The one on the right has none: it uses elastic bending to both locate and guide the plunger and to give the restoring force provided by the spring in the first design.
FIGURE 5.17 A sliding mechanism replaced by an elastic mechanism.
Exploiting elasticity in this way has many attractions. There is no friction, no wear, no need for lubrication, no precise clearances between moving parts. And in design with polymers there is another bonus; since there are no sliding surfaces it is often possible to mould the entire device as a single unit, reducing the part-count and doing away with the need for assembly. Reducing part-count is music to the ears of production engineers: it is cost-effective.
Figure 5.18 shows examples you will recognize: the ‘living hinge’ used on toothpaste tubes, moulded plastic boxes, clips and clothes pegs. Here the function of a rotational hinge is replaced by an elastic connecting strip, wide but thin. The width gives lateral registration; the thinness allows easy flexure. The traditional three-part hinge has been reduced to a single moulding.
FIGURE 5.18 Elastic or ‘natural’ hinges allowing flexure with no sliding parts.
There is much scope for imaginative design here. Figure 5.19 shows three section shapes, each allowing one or more degrees of elastic freedom, while retaining stiffness and strength in the other directions. By incorporating these into structures, parts can be allowed to move relative to the rest in controlled ways.
FIGURE 5.19 The flexural degrees of freedom of three alternative section shapes. (a) Thin plates are flexible about any axis in the plane of the plate, but are otherwise stiff. (b) Ribbed plates are flexible about one in-plane axis but not in others. (c) Cruciform beams are stiff in bending but can be twisted easily.
How do we choose materials for such elastic mechanisms? It involves a balance between stiffness and strength. Strength keeps appearing here—that means waiting until Chapter 7 for a full answer.
5.6 Summary and conclusions
Adequate stiffness is central to the design of structures that are deflection limited. The wing spar of an aircraft is an example: too much deflection compromises aerodynamic performance. Sports equipment is another: the feel of golf clubs, tennis rackets, skis and snowboards has much to do with stiffness. The sudden buckling of a drinking straw when you bend it is a stiffness-related problem—one that occurs, more disastrously, in larger structures. And when, as the turbine of an aircraft revs up, it passes through a speed at which vibration suddenly peaks, the cause is resonance linked to the stiffness of the turbine blades. Stiffness is influenced by the size and shape of the cross-section and the material of which it is made.
The property that matters here is the elastic modulus E.
In selecting materials, adequate stiffness is frequently a constraint. This chapter explained how to meet it for various modes of loading and for differing objectives: minimising mass, or volume or cost. Simple modeling delivers expressions for the objective: for the mass, or for the material cost. These expressions contain material properties, either singly or in combination. It is these that we call the material index. Material indices measure the excellence of a material in a given application. They are used to rank materials that meet the other constraints.
Stiffness is useful, but lack of stiffness can be useful too. The ability to bend or twist allows elastic mechanisms: single components that behave as if they had moving parts with bearings. Elastic mechanisms have limitations, but—where practical—they require no assembly, they have no maintenance requirements and they are cheap.
The chapter ended by illustrating how indices are plotted onto material property charts to find the best selection. The method is a general one that we apply in later chapters to strength thermal, electrical, magnetic and optical properties.
5.7 Further reading
1. Ashby MF. Materials Selection in Mechanical Design. 3rd ed. Oxford, UK: Butterworth-Heinemann; 2005; Chapter 4. ISBN 0-7506-6168-2.
(A more advanced text that develops the ideas presented here, including a much fuller discussion of shape factors and an expanded catalog of simple solutions to standard problems.)
2. Gere JM. Mechanics of Materials. 6th ed. Toronto, Canada: Thompson Publishing; 2006; ISBN 0-534-41793-0.
(An intermediate level text on statics of structures by one of the fathers of the field; his books with Timoshenko introduced an entire generation to the subject.)
3. Hosford WF. Mechanical Behavior of Materials. Cambridge, UK: Cambridge University Press; 2005; ISBN 0-521- 84670-6.
(A text that nicely links stress–strain behavior to the micromechanics of materials.)
4. Jenkins CHM, Khanna SK. Mechanics of Materials. Boston, MA: USA: Elsevier Academic; 2006; ISBN 0-12-383852-
5.
(A simple introduction to mechanics, emphasising design.)
5. Riley WF, Sturges LD, Morris DH. Statics and Mechanics of Materials. 2nd ed. Hoboken, NJ, USA: McGraw-Hill;
2003; ISBN 0-471-43446-9.
(An intermediate level text on the stress, strain and the relationships between them for many modes of loading. No discussion of micromechanics—response of materials to stress at the microscopic level.)
6. Vable M. Mechanics of Materials. Oxford, UK: Oxford University Press; 2002; ISBN 0-19-513337-4.
(An introduction to stress–strain relations, but without discussion of the micromechanics of materials.)
7. Young WC. Roark’s Formulas for Stress and Strain. 6th ed. New York, USA: McGraw-Hill; 1989; ISBN 0-07-100373- 8.
(This is the ‘Yellow Pages’ of formulae for elastic problems—if the solution is not here, it doesn’t exist.)
5.8 Exercises
Exercise
E5.1 Distinguish between tension, torsion, bending and buckling.
Exercise
E5.2 Polymer ropes and lines for use on water are often designed to float, to aid in their retrieval and to avoid applying a downwards load to an object or person attached to them in the water. Excessive stretch is undesirable, so a lower limit of 0.5 GPa is also imposed on Young’s modulus. Identify suitable polymers, using Figure 5.14.
Exercise E5.3
A beam is to be made by gluing and screwing together four wooden planks of length 2 m, thickness 20 mm and width 120 mm.
(a) Determine the second moment of area I for the three configurations shown in cross-section in the figure below.
(a) Calculate the mid-span deflection δ of each of the beams when they are simply supported at their ends and loaded at mid- span with a force F of 100N.
(c) Determine the stiffness F/δ for each of the beam configurations.
Exercise E5.4
What is meant by a material index?
Exercise
E5.5 Your task is to design a lightweight tie of length L with a circular cross-section of radius R. It has to carry an axial force F, without stretching by more than δ. You will need to choose the material (with Young’s modulus E and density ρ) and the corresponding cross-section radius to suit your choice of material, that is, R is a ‘free variable’.
(a) Show that the extension of the tie is given by δ = FL/AE, where A = πR2.
(b) Rearrange this equation to find an expression for the radius R of the tie that will carry the load without excessive deflection.
(c) If the length of the tie is L = 0.3m and the extension is not to exceed 0.1mm for a load of F = 100N, what value of R is needed if the tie is made of: (i) PEEK; (ii) Butyl Rubber; (iii) Titanium; (iv) Copper? Use the material properties in the table below.
(d) Write an expression for the mass m of the tie and determine the mass of each of the ties in (c). Which one would you choose?
(e) Substitute the expression for radius R from (b) into the expression for mass m from (d) to show that the material index to be minimised is ρ/ E. Determine the value of the material index for the four ties.
(f) Comment on the relationship between the mass of each beam and its material index ρ/E.
(g) Examine Figure 5.14 to see how these three materials compare in terms of the reciprocal material index E/ρ.
Young’s Modulus E (GPa) Density ρ (kg/m3) Radius R (mm) Mass m (kg) Material index ρ/E
PEEK 3.8 1300
Butyl Rubber 0.0015 2400
Titanium alloy 110 4600
Copper alloy 120 8900
Exercise
E5.6 You are asked to design a lightweight cantilever beam of length L with a square cross-section of width b and depth b. The beam is to be built-in to a wall at one end and loaded at the free end with a point load F. The tip deflection is δ. When designing the beam, you are free to specify the value of b to suit your choice of material, that is, b is a ‘free variable’.
(a) Find an expression for the size of the beam b that can carry F with deflection δ.
(b) If the length of the cantilever is L = 1 m and the deflection δ is not to exceed 5 mm for a load of F = 1000N, what value of b is needed if the beam is made of: (i) Steel; (ii) Aluminium; (iii) CFRP? Use the material properties in the table below.
(c) Write an expression for the mass m of the beam and determine the mass of each of the beams in (b). Which one would you choose?
(d) Substitute the expression for beam dimension b from (a) into the expression for mass m from (c) to show that the material index to be minimised is ρ/E1/2. Determine the value of the material index for the three beams.
(e) Comment on the relationship between the mass of each beam and its material index ρ/E1/2.
(f) Examine Figure 5.14 to see how these three materials compare in terms of the reciprocal material index E1/2/ρ.
Young’s Modulus E (GPa) Density ρ (kg/m3) Section size b (mm) Mass m (kg) Material index ρ/ E1/2
Steel 210 7800
Aluminium 70 2700
CFRP 100 1500
Exercise
E5.7 The objective in selecting a material for a panel of given in-plane dimensions for the casing of a portable computer is that of minimising the panel thickness while meeting a constraint on bending stiffness, . What is the appropriate material index?
Exercise
E5.8 A sailing enthusiast is seeking materials for lightweight panels to use in a sea-going yacht. The panels are of rectangular cross- section and will be loaded in bending, as shown in the figure. The span L and width b of the panels are fixed, but the depth d may vary (up to a maximum specified value). The required stiffness is specified as a maximum allowable deflection δ under a given central load W.
The central deflection for the simply supported span shown is given by:
The designer is interested in two scenarios: (i) minimum mass; (ii) minimum material cost.
(a) Show that the stiffness and geometric constraints lead to the relationship , where E is Young’s modulus. Explain why the design specification leads to a minimum allowable value for E, and find an expression for this minimum value.
(b) Write down an expression for the first objective to be minimised (mass) and use the stiffness constraint to eliminate the free variable (thickness). Hence define the material performance index to maximise for a minimum mass design.
(c) On a Young’s modulus–density property chart (Figure 5.14), how are materials with the best values of the performance index identified? Use this method to identify a short-list of candidate materials. Exclude ceramics and glasses—why is this?
(d) How does the short-list of materials in (c) change if the limit on depth leads to a requirement for a minimum required Young’s modulus of 5 GPa?
(e) Modify your material performance index to describe the alternative objective of minimum material cost. Use the material property chart in Figure 5.16 to identify a short-list of materials, noting any that may be excluded because they have excessive thickness.
Exercise
E5.9 Derive the material index for a torsion bar with a solid circular section. The length and the stiffness are specified, and the torsion bar is to be as light as possible. Follow the steps used in the text for the beam, but replace the bending stiffness
with the torsional stiffness (equation
5.8), using the expression for given in
Figure 5.2. Exercise
E5.10 The body of a precision instrument can be modelled as a beam with length L, second moment of area I, and mass per unit length mo. It is made of a material with density ρ and Young’s modulus E. For the instrument to have the highest possible accuracy, it is desired to maximise the first natural frequency of vibration, which is given by equation (5.11), with the value of C2 from Figure 5.6:
If the beam has a square cross-section b × b, where the value of b is free to be selected by the designer, and the stiffness S is fixed by the design, show that the best material is one with a high value of the material index E1/2/ρ.
Exercise
E5.11 A material is required for a cheap column with a solid circular cross-section that must support a load without buckling. It is to have a height . Write down an equation for the material cost of the column in terms of its dimensions, the price per kg of the material, , and the material density . The cross-section area is a free variable: eliminate it by using the constraint that the buckling load must not be less than (equation
5.9). Hence read off the index for finding the cheapest tie. Plot the index on a copy of the appropriate chart and identify three possible candidates.
Exercise E5.12
The following table shows a selection of candidate materials for the lightweight, stiffness-limited design of a beam. The shape is initially constrained to be square, but the cross-sectional area can vary. Evaluate the relevant index for this problem, E1/2/ρ, for each material. Which will be the lightest? This chapter showed that if the shape is allowed to vary, the mass can be improved by a factor of Φ−1/2, which is equivalent to increasing the material index by Φ1/2. Evaluate the modified index (ΦE)1/2/ρ, using the maximum shape factor for each material (as listed in Table 5.3). Does the ranking of the materials change?
Young’s Modulus E (GPa) Density ρ (kg/m3)Material index without shape E1/2/ρ
Material index with shape (ΦE)1/2/ρ
Steel 210 7800
Al alloy 70 2700
GFRP 25 1900
Wood 10 650
Exercise E5.13
Devise an elastic mechanism that, when compressed, shears in a direction at right angles to the axis of compression.
Exercise
E5.14 Universal joints usually have sliding bearings. Devise a universal joint that could be moulded as a single elastic unit, using a polymer.
5.9 Exploring design with CES
Use Level 2, Materials, throughout the following exercises.
Exercise
E5.15 Use a ‘Limit’ stage to find materials with modulus > 180 GPa and price < 3 $/kg.
Exercise E5.16
Use a ‘Limit’ stage to find materials with modulus > 2 GPa, density < 1000 kg/m3 and price < 3 $/kg.
Exercise E5.17
Make a bar chart of modulus, . Add a tree stage to limit the selection to polymers alone. Which three polymers have the highest modulus?
Exercise E5.18
Repeat the material selection in Exercise E5.2 using CES. Examine the environmental resistance of the candidate materials and comment on any possible weaknesses for this application. Use the information for material applications to see if the materials identified are used in practice.
Exercise
E5.19 (a) A component is made from brass (a copper alloy). Identify three alternative classes of alloy that offer higher Young’s modulus for this component.
(b) Find materials with E > 40 GPa and ρ < 2 Mg/m3 and identify the cheapest.
(c) Find metals and composites that are both stiffer and lighter than (i) steels; (ii) Ti alloys; (iii) Al alloys.
(d) Compare the specific stiffness, E/ρ, of steels, Ti alloys, Al alloys, Mg alloys, GFRP, CFRP and wood (parallel to the grain).
(e) Comment on the usefulness of the approaches in (c) and (d) for seeking improved performance in lightweight, stiffness- limited design.
Exercise
E5.20 Make a chart showing modulus E and density ρ. Apply a selection line of slope 1, corresponding to the index E/ρ positioning the line such that six materials are left above it. Which are they and to which families do they belong?
Exercise
E5.21 A material is required for a tensile tie to link the front and back walls of a barn to stabilise both. It must meet a constraint on stiffness and be as cheap as possible. To be safe the material of the tie must have a fracture toughness K1c > 18 MPa.m1/2 (defined in Chapter 8). The relevant index is
Construct a chart of E plotted against Cmρ. Add the constraint of adequate fracture toughness, meaning K1c > 18 MPa.m1/2, using a ‘Limit’ stage. Then plot an appropriate selection line on the chart and report the three materials that are the best choices for the tie.
5.10 Exploring the science with CES Elements
Exercise
E5.22 There is nothing that we can do to change the modulus or the density of the building blocks of all materials: the elements. We have to live with the ones we have. Make a chart of modulus E plotted against the atomic number An to explore the modulus across the periodic table. (Use a linear scale for An. To do so, change the default log scale to linear by double-clicking on the axis name to reveal the axis-choice dialog box and choose ‘Linear’.) Which element has the highest modulus? Which has the lowest?
Exercise
E5.23 Repeat Exercise E5.15, exploring instead the density ρ. Which solid element has the lowest density? Which has the highest?
1Granta Design Ltd, Cambridge, UK (www.grantadesign.com).