The microstructures and properties of polymers are manipulated by cross-linking and by adjusting the molecular weight and degree of crystallinity. Figures 8.8 and 8.9 showed that the strengths of polymers span around a factor of 5, whereas fracture toughness spans a factor of 20. More dramatic changes are possible by blending, by adding fillers and by reinforcement with chopped or continuous fibres to form composites. Figure 8.19 shows how these influence the modulus E and fracture toughness K1c of polypropylene (PP). Blending or co-polymerization with elastomers such as EPR or EDPM (‘impact modifiers’) reduces the modulus but increases the fracture toughness K1c and toughness Gc. Filling with cheap powdered glass, talc or calcium carbonate more than doubles the modulus, but at the expense of some loss of toughness.
FIGURE 8.19 The fracture toughness and modulus of polypropylene, showing the effect of fillers, impact modifiers and fibres.
Reinforcement with glass or carbon fibres most effectively increases both modulus and fracture toughness K1c, and in the case of glass, the toughness Gc as well. How is it that a relatively brittle polymer (K1c ≈ 3 MPa.m1/2) mixed with even more brittle fibres (glass K1c ≈ 0.8 MPa.m1/2) can give a composite K1c as high as 10 MPa.m1/2? The answer is illustrated in Figure 8.20. The fine fibres contain only tiny flaws and consequently have high strengths. When a crack grows in the matrix, the fibres remain intact and bridge the crack. This promotes multiple cracking—each contributing its own energy and thereby raising the overall dissipation. When the fibres do break, the breaks are statistically distributed, leaving ligaments of fibre buried in the matrix. Fibre pull-out, as the cracks open up, dissipates more energy by friction. Composites are processed to control adhesion between the fibres and the matrix to maximise the toughening by these mechanisms.
FIGURE 8.20 Toughening by fibres. The pull-out force opposes the opening of the crack.
8.8 Summary and conclusions
Toughness is the resistance of a material to the propagation of a crack. Tough materials are forgiving: they tolerate the presence of cracks, they absorb impact without shattering and, if overloaded, they yield rather than fracture. The dominance of steel as a structural material derives from its unbeatable combination of low cost, high stiffness and strength and high toughness.
Toughness is properly measured by loading a pre-cracked sample with one of a number of standard geometries, measuring the load at which the crack first propagates. From this is calculated the fracture toughness, K1c, as the value of the stress intensity, K1, at which the crack advances. Values of K1c above about 15 MPa.m1/2 are desirable for damage- tolerant design—design immune to the presence of small cracks. The charts of Figures 8.8 and 8.9 show that most metals and fibre-reinforced composites meet this criterion, but polymers and ceramics fall below it. We shall see in Chapter 10 that some designs require a high value not of K1c but of the toughness , also shown in the chart of Figure 8.8. Polymers do well by this criterion and it is in applications that require it that they become a good choice. Ceramics are
poor by this criterion too, making design with them much more difficult—at least in tension; in compression, they have considerable strength.
High toughness means that crack advance absorbs energy. It does so in the plastic zone that forms at the crack tip; the larger this becomes, the higher is the fracture toughness. As the material within the plastic zone deforms, voids nucleate at inclusions and link up, advancing the crack; the cleaner the material, the fewer are the inclusions, contributing further to high toughness. The higher the yield strength, the smaller is the zone until, as in ceramics with their very high yield strengths, the zone becomes vanishingly small, the stresses approach the ideal strength and the material fails by brittle cleavage fracture.
This is the background we need for design. That comes in Chapter 10, after we have examined fracture under cyclic loads.
8.9 Further reading
1. Broek D. Elementary Engineering Fracture Mechanics. 3rd ed. Boston, USA: Martinus Nijhoff; 1981; ISBN 90-247- 2580-1.
(A standard, well-documented introduction to the intricacies of fracture mechanics.)
2. Ewalds HL, Wanhill RJH. Fracture Mechanics. London, UK: Edward Arnold; 1984; ISBN 0-7131-3515-8.
(An introduction to fracture mechanics and testing for both static and cyclic loading.)
3. Hertzberg RW. Deformation and Fracture of Engineering Materials. 3rd ed. New York, USA: Wiley; 1989; ISBN 0- 471-63589-8.
(A readable and detailed coverage of deformation, fracture and fatigue.)
4. Kinloch AJ, Young RJ. Fracture Behavior of Polymers. London, UK: Elsevier Applied Science; 1983; ISBN 0-85334- 186-9.
(An introduction both to fracture mechanics as it is applied to polymeric systems and to the fracture behaviour of different classes of polymers and composites.)
5. Knott JF. The Mechanics of Fracture. London, UK: Butterworths; 1973; ISBN 0-408-70529-9.
(One of the first texts to present a systematic development of the mechanics of fracture; dated, but still a good introduction.)
6. Tada H, Paris G, Irwin GR. The Stress Analysis of Cracks Handbook. 3rd ed. 2000; ISBN 1-86058-304-0.
(Here we have another ‘Yellow Pages’, like Roark for stress analysis of uncracked bodies—this time of stress intensity factors for a great range of geometries and modes of loading.)
8.10 Exercises
Exercise
E8.1 What is meant by stress intensity factor? How does this differ from the stress concentration factor?
Exercise E8.2
What is meant by toughness? How does it differ from strength, and from fracture toughness?
Exercise
E8.3 Why does a plastic zone form at the tip of a crack when a cracked body is loaded in tension?
Exercise E8.4
Why is there a transition from ductile to brittle behaviour at a transition crack length, ccrit ? Exercise
E8.5 Compare the critical crack length ccrit of a Titanium alloy with σy = 550 MPa and K1c = 40 MPam1/2 with that of Silicon Carbide, with σy = 490 MPa and K1c = 4 MPam1/2. Comment on the ductility of the two materials.
Exercise
E8.6 A tensile sample of width 10 mm contains an internal crack of length 0.3 mm. When loaded in tension the crack suddenly propagates when the stress reaches 450 MPa. What is the fracture toughness of the material of the sample? If the material has a modulus of 200 GPa, what is its toughness ? (Assume the geometric factor Y = 1).
Exercise
E8.7 The aircraft designer in Example 7.5 changes the design so that the window has a corner radius of 20 mm instead of 5 mm.
(a) What is the new stress concentration factor at the corner of the window, and what is the maximum stress?
(b) Now suppose there is a small sharp defect near the corner of the window, such that the effective length of the crack in the fuselage is the size of the window itself. If the fracture toughness of the fuselage material is 25 MPa.m1/2 and the window is of length 160 mm, what stress would cause fast fracture? Is the design safe?
Exercise
E8.8 Two long wooden beams of square cross-section t × t = 0.1 m are butt-jointed using an epoxy adhesive. The adhesive was stirred before application, trapping air bubbles that, under pressure in forming the joint, deformed to flat, penny-shaped cracks of diameter 2 mm (see figure below). For these cracks, assume a geometry factor Y = 1 in the expression for stress intensity factor. The fracture toughness of the epoxy K1c = 1.3 MPa.m1/2.
(a) If the beams are loaded in axial tension, what is the load Ft that leads to fast fracture?
(b) The beams are now loaded in three-point bending, with a central load Fb and a total span L = 2 m.
Show that the maximum tensile stress at the surface of the beam is given by . Assuming that there is a bubble close to the beam surface, find the maximum load that the beam can support without fast fracture.
Exercise
E8.9 Use the chart of Figure 8.8 to establish whether:
1. CFRP has a higher fracture toughness than aluminum alloys.
2. Polypropylene (PP) has a higher toughness than aluminum alloys.
3. Polycarbonate (PC) has a higher fracture toughness than glass.
Exercise
E8.10 Find epoxy, soda glass and GFRP (epoxy reinforced with glass fibres) on the chart of Figure 8.8 and read off an approximate mean value for the toughness for each. Explain how it is that the toughness of the GFRP is so much larger than that of either of its components.
Exercise
E8.11 Use the chart of
Figure 8.8 to compare the fracture toughness, , of the two composites GFRP and CFRP. Do the same for their toughness, . What do the values suggest about applications they might best fill?
Exercise
E8.12 Use the chart of Figure 8.9 to find:
1. The range of transition crack sizes for stainless steel.
2. The range of transition crack sizes for polycarbonate (PC).
3. The range of transition crack sizes for silicon nitride (Si3N4).
Exercise
E8.13 The cylindrical ceramic samples from Example 8.4 (50 mm long and 5 mm radius) were found to have a survival probability of 50%
when the applied stress was 120 MPa. The Weibull modulus of the ceramic is m = 10. Find the reference stress σo for this volume of material. What stress can be carried with a probability of failure lower than 1%?
Exercise
E8.14 The strengths of a square ceramic beam loaded in pure bending and in tension were compared in this chapter, equation (8.18).
Plot how the strength ratio varies with the value of the Weibull modulus, m, and explain the shape of the curve.
Exercise
E8.15 Square section silicon nitride beams were loaded in pure bending. When the maximum bending stress was below 500 MPa, 50% of the beams had broken. Use equation (8.18) to find the tensile stress applied to the same samples that gives the same
probability of failure. The Weibull modulus for silicon nitride is m = 10.
8.11 Exploring design with CES
Use the Level 2 database unless otherwise stated.
Exercise E8.16
Use the ‘Advanced’ facility to make a bar chart showing for an internal crack of length mm plotted on the y-axis (i.e. using the CES parameter ‘Fracture toughness’ for K1c).
Which materials have the highest values? Add an axis of density, . Use the new chart to find the two materials with highest values of .
Exercise
E8.17 Suppose that the resolution limit of the non-destructive testing (NDT) facility available to you is 1 mm, meaning that it can detect cracks of this length or larger. You are asked to explore which materials will tolerate cracks equal to or smaller than this without brittle fracture. Make a bar chart with for an internal crack of length mm plotted on the y-axis, as in the previous exercise. Add yield strength on the x-axis. The material will fracture in tension if , and it will yield, despite being cracked, if . Plot a line of slope 1 along which , either on a printout of the chart or in CES (using the line selection tool). All the materials above the line will yield, all those below will fracture. Do age- hardening aluminum alloys lie above the line? Does CFRP?
Exercise E8.18
Find data for PVCs in the Level 3 database and make a plot like that of Figure 8.19 showing how fillers, blending and fibres influence modulus and toughness.
8.12 Exploring the science with CES Elements
Exercise
E8.19 Explore the origins of surface energy, γ J/m2. The toughness, Gc, cannot be less than 2γ because two new surfaces are created when a material is fractured. What determines γ and how big is it? The text explained how bonds are broken and atoms separated when a new surface is created. It was shown in Section 8.4 that
where Hc is the cohesive energy in J/m3 and ro is the atomic radius in m. To convert Hc in kJ/mol (as it is in the database) into these units, multiply it by 106/molar volume, and to convert ro from nm into m multiply by 10−9. All this can be done using the
‘Advanced’ facility in the axis-choice dialog box.
Make a chart with γ calculated in this way on the x-axis and the measured value, ‘Surface energy, solid’ on the y-axis, and see how good the agreement is.
Exercise E8.20
If you pull on an atomic bond, it breaks completely at a strain of about 0.1. If the atom spacing is ao, then breaking it requires a displacement δ = 0.1ao and doing so creates two new surfaces, each of area . If the bond stiffness is S, then the work done to stretch the bonds to breaking point, and hence to create the new surfaces, per unit area, is
(using equation (4.20) for S), where ro is the atomic radius. Use the CES Elements database to explore whether real surface energies can be explained in this way. (Watch out for the units.)
Exercise
E8.21 Observe the general magnitudes of surface energies γ: they are about 1.5 J/m2. Thus, the minimum value for Gc should be about 2γ or 3 J/m2. Return to the CES Edu Level 3 database and find the material with the lowest value of Gc, which you can calculate as
. Is it comparable with 2 γ
? Limit the selection to metals and alloys, polymers, technical ceramics and glasses only, using a ‘Tree’ stage (materials such as foam have artificially low values of Gc because they are mostly air).
1James Joule (1818–1889), English physicist, did not work on fracture or on surfaces, but his demonstration of the equivalence of heat and mechanical work linked his name to the unit of energy.
2Waloddi Weibull (1887–1979) was a Swedish engineer and mathematician whose technical interests were stimulated by his early naval career with the coast guard, where he developed the use of sound waves from explosive charges to investigate the nature of the seabed, something the oil industry can thank him for.
The probability distribution that bears his name had previously been associated with describing the statistics of particle-size distributions.