能量释放率理论

1、Fracture and Damage Mechanics Chapter Three Energy release rate theoryChapter Three Energy release rate theory Energy release rate G is another important concept in fracture mechanics. There is a relationship between G and K. One is known. Another can also be known. Sometime, it is difficult to obta

2、in K factor. However, it is perhaps easy to calculate G.§3-1 Concept of energy release rate1. Basic concepts(1) Strain energy Work done by the external force is changed to the strain energy stored in the elastic deformation. The strain energy can be released and the elastic deformation then dis

3、appears. Strain energy density w: strain energy per unit volume., (for linear elastic body)Total strain energy U (Internal force potential): total energy stored in the volume V(2) External force potential UP: the negative value of virtual work done by the external force. Assume that the body force i

4、s and the surface force is on the stress boundary. The external force potential is (3) Total potential P:2. Energy release rate G The body force , the surface force on and the displacement on are given. Assume that the crack size is changed from a to a+Da. Accordingly, the displacement, strain, stre

5、ss, stain energy density, internal force potential, external force potential and total potential are also changed. The total potential is changed to . is the increment caused by the crack growth . Assume that the plate thickness is t. denotes the single surface area increment. The energy release rat

6、e G is defined asIf the plate thickness t is a constant, the energy release rate G is3. Constant force and constant displacement conditions(1) Constant force conditionA plate with a crack is applied by a constant force F as shown in Fig. 3.1(b). The external force virtual work W and external force p

7、otential UP, respectively, are , The total strain energy U (internal force potential) is Then, the total potential P isNow, ，. The energy release rate G under the constant force condition can be written as ，for constant F.In this case, , the strain energy in the body in fact increases rather than re

8、leases with the crack growth. G can not be called as the strain energy release rate. (2) Constant displacement conditionAfter a displacement d occurs, the plate is clamped. This is the constant displacement condition as shown in Fig. 3.1(c). In this case, there isW0，, for constant displacement d.It

9、is seen that only for the constant displacement condition, G can be called as the strain energy release rate. Since W=0, the energy needed by the crack growth comes from the release of strain energy stored in the body. That is the strain energy stored in the body decreases with the crack growth. Foe

10、 the constant force condition, the increment of external work is in which a part is used to increase strain energy dU while the other part is used for crack growth. However, the values of G for two cases are equal. Constant force case: ，. C is the compliance of the plate. Constant displacement case:

11、，It is seen that the values of G for two cases are equal.§3-2 Relation between G and K The relationship between G and K is one of the most important relations in fracture mechanics.1. Mode I crackBy consideration of stress and energy fields ahead of the crack tip, the relation between K and G c

12、an be established. Assume that a segment of crack length is closure. To this end, a distributed stress syy is applied. This distributed stress syy is the stress field in the vicinity in the crack tip for q=0, r=x, i.e.In addition, for the plane strain case, the displacement produced in the crack clo

13、sure process can be known from the asymptotic solution of displacement field in the vicinity of the crack tip. The point o is taken as the original point. For , , the displacement along y-axis is Assume that the plate thickness is 1. The applied force F in the area isAs shown in Fig. 3.2, the total

14、crack closure displacement is . Since the actual work done by the external force sy in the length dx is equal to the strain energy, there isThe crack closure length is a so that the change of system strain energy isThe change of system strain energy by closing the unit crack area isOn the other hand

15、, the crack growth case is similar to the crack closure case. The energy for crack growth is equal to the energy for crack closure. Therefore, in the case of constant displacement, the dissipative energy for the unit area increment of crack area, i.e. GI, can also be written asThe expression of and

16、can be inserted into the above eq. to givei.e., for plane strain caseFor the plane stress case, there is , for plane stress case.2. Mode II crack For the mode II crack, the relative slide displacement of crack surfaces along the x-axis by the shear stress is 2u. The energy release rate GII is The sh

17、ear stress txy and displacement u are respectively known as, , , , We have, for plane strain case, for plane stress case.3. Mode III crack For the mode III crack, the relative displacement by the anti-plane shear stress along the z-axis is 2w. The energy release rate GIII can also be written as The

18、anti-plane shear stress and anti-plane displacement w are known as, , , , We have For the complex crack, KI¹0, KII¹0, KIII¹0, the total energy release rate is§3-3 Bi-cantilever beam problem It is a mode I crack problem.1. Long crack case: When , the crack segment is equivalent to

19、 a cantilever beam. , (rectangular section), , For the plane stress problem, there isThe stress intensity factor can be known as2. Short crack case: In this case, the crack segment is equivalent to a dumpy beam (短粗梁). The shear deformation of the crack segment and the deformation in the uncracked se

20、gment must be considered. The displacement d can be divided into three parts. d1 is produced by bend and d2 by shear. It is known that, d3 is the body displacement induced by the deformation of the uncracked segment. A normalized compliance coefficient is introduced. The coefficients , and can be de

21、termined by theory or experiment. If the normalized compliance is known, the displacement d can be determined.Total potential of the system isThe energy release rate isFor the plane stress problem, the stress intensity factor can be determined fromthat3. Determination of a1, a2 and a3 by experiment

22、There are three specimens. The crack lengths are , and respectively. Three F versus d curves can be obtained by test. F and d are the generalized force and displacement. In light ofwe can have three equationsThe coefficients , and now can be known. Then, we calculate the energy release rate G and th

23、e stress intensity factor K. §3-4 Determination of SIF by experiment The method in the last section can be applied to the general case.Consider a Mode I crack problem. The crack can be a central or edge crack. F and d denote the generalized force and displacement respectively. For the different

24、 crack sizes , , , , we can obtain a group of F versus d straight lines by experiment. For a constant applied force F, we can have the displacement values , , , . The total potential can be calculated for different crack sizes., for , Finally, a group of data (Pi, ai) is obtained. The Pi versus ai c

25、an be plotted. For the Mode I and plane stress problem, in view ofthe stress intensity factor iswhere is the slope at the point a. Another treatment: Assume that the compliance l(a) of the system depends on the crack size a. The generalized displacement q and force F are related by l as When F=const

26、ant, the derivative is For a constant applied force F, we can have the displacement values , , , from Fig. 3-8. By using , a group of data can be obtained. The versus curve can be depicted. By using the slope at the point a, we can calculate the stress intensity factor for the crack size a and the a

27、pplied load F. §3-5 An infinite plate with a central crack under uniaxial tension This problem has been solved in chapter two. Now we resolve it by the energy approach. Case (a) can be divided into case (b) plus case (c).Case (b) is the same as no crack case, , such thatCase (a) is equal to cas

28、e (c).1. Solution for case (c) For the case (c), assume that the displacement distribution on the upper crack surface is an ellipse. The relation is substituted to giveWhen the distance r is very small, r2 can be ignored. On the crack surface near the crack tip, the displacement distribution is a pa

29、rabola. It has been known in chapter two that the displacement asymptotic solution isOn the upper crack surface, , we have For plane stress, , , the stress intensity factor can be obtained thatandThe displacement remains determined.2. The expression of energy release rate G In what follows, the expr

30、ession of energy release rate G is derived from its definition. Then, the displacement can be determined and the solution for the problem can be known. The system total potential is Note that is related to .The early result isA differential equation can be obtained that, Bernoulli equationThe general form of the Bernoulli equation isThe solution can be found in the mathematic handbook. The result isC is an integral constant. When , the displacement . This requires thatC=0. Finally, the solution is, , It is identical to the early result in chapter two. §3-6 M