Theory of linear elasticity. I assume you have learned the elements of linear

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1 Physial Metallurgy Frature mehanis leture 1 In the next two letures (Ot.16, Ot.18), we will disuss some basis of frature mehanis using ontinuum theories. The method of ontinuum mehanis is to view a solid as a ontinuous distribution of material partiles. ah material partile onsists of many atoms. The material partile represents the olletive behavior of many atoms. Following topis will be disussed. Trouble with linear elasti theory of strength. The Griffith approah. Frature energy. nergy release rate. Appliations of frature mehanis. The problem to be solved in frature mehanis. A body is subjet to a load. What is the magnitude of the load that will ause the body to frature? Let us begin with a body of a silia glass, whih deforms elastially by small strains. A proedure has been taught in strength of materials probably goes as follows. We first determine the imum stress in the body. Then we determine the strength of the material. The body is supposed to frature when the imum stress in the body reahes the strength of the material. I will first review this proedure, so that we agree exatly what this proedure is. I will then explain why this proedure is diffiult to y in pratie. Theory of linear elastiity. I assume you have learned the elements of linear elastiity. Imagine a body subjet to an ied stress,. At eah material partile, the state of stress is a tensor, with 6 omponents. In the body, the state of stress varies from one material partile to another. Thus, the state of stress in a body is desribed by a field. The field is determined by solving the boundary-value problem. A list of notes of solving boundary-value problems in elastiity are following: Governing equations. Geometri relations. Fore balane equations. Constitutive law. Boundary onditions. Tration or displaement boundary onditions. Approximate solutions. Beams and plates. xat analytial solutions. Few problems an be solved exatly. Theory of lastiity, S. P. Timoshenko and J. N. Goodier, MGraw-Hill, New York. Numerial solutions. Finite element methods. Commerial software suh as ABAQUS. Maximum stress in a body. Solving boundary-value problem is a big task by itself, but is not the subjet of this lass. Let s say we already have the solution. That is, we know all six omponents of stress at every material partile in the body, ( x, x, x ), ( x, x, x ) What do we do with this massive amount of data? We are interested in 1

2 prediting the onditions of frature of the body. For example, if the ied stress is small enough, the body will not frature. How small is small enough? Here is the proedure taught in strength of materials. From the field of stress we determine the imum stress in the body,. The imum omponent of stress at eah material partile is determined by priniple stresses, solved by an eigenvalue problem. Then we look for the largest value of the priniple stress by omparing all material partiles in the body. Today, all this proedure is embodied in ommerial finite element software. So you know the imum omponent of stress of the body,. Stress onentration fator. The equations in elastiity are linear, so that the imum stress in the body is proportional to the ied stress. = C, where C is a dimensionless number. The basi phenomenon that the stress is higher at some material partile in a body than others is known as stress onentration. The number C is known as the stress onentration fator. When a irular hole is embedded in a muh larger plate, this boundary-value problem is solved analytially by Timoshenko and Goodier. The imum stress ours at the surfae of the hole. The stress onentration fator is = 3. Consider an ellipti hole in an infinite plate subjet to a remote stress. This boundary-value problem an still be solved analytially (Stresses in a plate due to the presene of raks and sharp orners, C.. Inglis, 1913). The imum stress given by is a = 1+, b where a and b are semi-axes of the ellipse. The stress onentration fator depends on the shape of the hole, haraterized by the ratio a/b. When a=b, the hole is irular, and the stress onentration fator is 3. When the ellipse is very elongated, a>>b, the stress onentration fator is very large. Denote the radius of urvature at the tip of the ellipse as ρ, after some math you an find ρ = b / a. You an express the above formula in terms of a and ρ, namely, a. ρ This formula may be used to estimate the stress onentration fator for a flaw of

3 some other shapes, where a is interpreted as the overall size of the flaw, and ρ is the radius at the root of the flaw. Strength of a material. Now we have alulated the imum stress the body. Will the body sustain this stress? The theory of elastiity will not answer this question. You need to find the strength of the material somewhere else. For example, you may estimate the theoretial strength of a material by S th =. 10 For a glass, Young s modulus is 70 GPa, so that the estimated theoretial strength is 7 GPa. Prof. Ju Li has told you this value is about two orders of magnitude higher than the strength measured in a bulk sample. You an determine the strength of the material experimentally. You pull a sample until it breaks, then reord the stress that breaks the sample. We all this stress as the experimental strength of the material. The experimental values of vlass are on the order of S ~100 MPa. Design for strength based on linear elastiity. What is the imum load that an be sustained by a body? We now summarize the proedure as follows. Calulate the stress field by solving boundary-value problems. Loate the imum stress in the body. Assume the material has a definite strength. That is, the same material has the same strength, independent of the shape of the body. Measure the strength S using a simple sample of the material, suh as a tensile bar. Make sure the imum stress in the body is below the strength of the material. in Why is this proedure hard to use in pratie? The imum stress in a body is sensitive to the shape of the flaw. The shape of the flaw is a body is seldom known in pratie. The proedure assumes that the body is linearly elasti everywhere, whih is never true. The proedure assumes that the strength of a material is independent of the sample used in experiments. In reality, strengths measured from different samples are different, beause eah sample has different flaws. That is, the proedure is hit in both ways: the imum stress is impossible to alulate, and the strength is impossible to measure. 3

5 The elasti energy redues. To determine the amount of the redution, one has to solve the boundary-value problem. This tough elastiity problem is for professional elastitians. Look how ompliated the stress field must be near the rak. Griffith used the elastiity solution of Inglis, beause a rak is just a speial ase of an ellipse when b/ a 0. This part of the Griffith paper is diffiult to read, and is not very interesting. In the end he made small errors. An alternative approah is to invoke linearity and dimensional onsiderations. For a linearly elasti problem, the stress field is linearly proportional to the ied stress. The elasti energy per unit volume is proportional to /. The elasti energy in an infinite sheet is infinite. However, we are interested in the differene in elasti energy between the raked sheet and the unraked sheet. Note that the rak length a is the only length sale in the boundaryvalue problem. Consequently, the differene in elasti energy between the two sheets takes the form β a, where β is a numerial value. Thus, from very basi onsiderations, we get nearly everything exept for a pure number. This number must be determined by solving the elastiity boundary value problem. The solution turns out to be β = π. You an find the solution of the full problem in Timoshenko and Goodier. Relative to the unraked sheet, in the raked sheet the ombined surfae energy and elasti energy is U( a) =+ 4aγ π a. The rak length, a, is the thermodynami variable. The surfae energy density γ and the ied stress are taken to be onstant for the time being. As expeted, when the rak length inreases, the surfae energy inreases, but the elasti energy dereases. Critial rak size. Plot the free energy as a funtion of the rak length. The free energy first goes up, reahes a peak, and then goes down. Beause there is no minimum free energy, the rak annot reah equilibrium. The free energy reahes the peak at the rak length γ a* =. π Let us examine the physial signifiane of this partiular rak length. Let a be the length of the rak pre-existing in the sheet. Distinguish two situations. 5

6 Crak healing. If a< a*, the surfae energy prevails over the elasti energy. To redue the free energy, the rak length must derease. The rak does so by healing, i.e., forming atomi bonds one by one, like losing a zipper. In reality rak healing is not often observed. This is not beause the thermodynamis is wrong, but beause surfaes are not flat to the atomi dimension, so that atoms annot meet aross the gap and form bonds. Several examples show that rak healing happens. Adhesives. Soft material an heal readily by flows. Wafer bonding. If the surfaes are indeed made flat, they will join. Sintering. At elevated temperatures, atoms an diffuse, so that the two surfaes hange shape and an join. Crak growth. If a> a*, the elasti energy prevails over the surfae energy. To redue the free energy, the rak must grow. The rak does so by breaking atomi bonds one by one, like opening a zipper. This is the situation studied in this ourse. The Griffith experiments. The main predition of the Griffith theory an be written as γ a =. π He performed several experiments to asertain various parts of the equation. xperiment 1. Confirm that a = onstant, independent of the size of the rak. Start with several glass sheets (large spherial bulbs atually). Introdue a rak in eah sheet. Measure the strength of eah sheet. Two important points: (1) The rak introdued is in the mm to m range, muh longer than any natural flaws in the sheets, so that the natural flaws are negligible. In this way Griffith irumvented the unertainties assoiated with the natural flaws. () The introdued raks in different sheets have different lengths, and the measured strengths are also different. His data onfirmed that a = onstant. xperiment. Confirm that the onstant is indeed γ / π. Young s modulus for the glass used by Griffith was = 6 GPa. The surfae energy inferred from the measured breaking strength is γ = 1.75 J/m. Griffith needed an independent measurement of the surfae energy. He did the reeping fiber experiment. The value he obtained was γ = 0.54 J/m. The agreement was fair. xperiment 3. Measure strengths of glass fibers. For a fixed pre-existing rak size a, there is a ritial stress: γ =. π a 6

7 This is the stress needed to frature the sample. This relation shows that the frature strength depends on the rak size. Beause different samples have different rak sizes, the frature strength is not a material property. The measured strength has large satter. Take representative values γ = 1 J/m, 11 = 10 N/m, strength is 50 MPa. This orresponds to the experimental strength. a 6 = 10 m, the Sine the Griffith paper, the siene of frature was born. 7

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