《塑性加工模拟及自动控制》课件：SimulationsMaterials2

1、Simulations of plastic processing: elasto-plastic deformationPart 2: elastic deformationTeacher: 法法 QQ号：2306847727， office 523 A区Content Origin of Elastic deformation Hooke law 胡克定律 Applications 应用 Material are made of atoms Atoms are bounded by electrical forces Atoms cannot be seen as rigid balls,

2、 but can be modeled like points linked by springs: When external forces are applied to the material, the inter-atomic space changes There is an equilibrium between the electric forces and the external forces.Origin of Elastic deformationOrigin of Elastic deformation Electrical force between atoms: t

3、he interatomic potentialInteratomic space (when no external forces are applied)Origin of Elastic deformation For small variations around equilibrium, the interatomic potential is modeled as a linear relationship, so the force is linear and proportional to the displacement, as the force of a spring:

4、F = k*lOrigin of Elastic deformation This first linear simplification makes force proportional to the displacement of the interatomic space. So there is a linear relation between applied force (pressure in fact) and strain: =E E is call the Youngs modulus. This simple relation is right in uniaxial d

5、eformation, along the direction of the force.Origin of Elastic deformation In case of elastic deformation, the volume changes. When a force (or pressure) is applied in one direction, there is a deformation in the lateral direction:beforeWith applied forceOrigin of Elastic deformation The lateral str

6、ain depend on the material. Its measure give the Poisson ratio noted (Greek letter nu) yy=-/ExxbeforeWith applied forceXYHooke law For elastic deformation, we have to use the Hooke law ij = Cijkl : kl , C is the stiffness tensor (刚度张量) C is a tensor of rank 4, but we write it as a matrix 6*6, which

7、is for isotropic materials C = and = 2+ 0 0 0 2+ 0 0 0 2+ 0 0 0 0 0 0 2 0 0 0 0 0 0 2 0 0 0 0 0 0 2xx yy zz xy xz yzGangduHooke law ij = ijkk+2ij ij = Kijkk + 2(ij 1/3*ijkk) with K: Bulk modulus (or incompressibility), =K-2/3* Lam parameter (lamda)=G: shear modulus (or rigidity, mu), ij=1 if i=j, el

8、se 0 the Kronecker delta normal stress: xx = (xx+yy+zz)+2xx = K*V+2(xx-V/3) Shear stress: xy = 2xyHooke law ij = Sijkl : kl , S is the compliance tensor (柔度张量) ij=1/E*(1+)ij-ijkk E: Young modulus, = Poissons ratio (nu) S = 1/E*1 - - 0 0 0 - 1 - 0 0 0 - - 1 0 0 0 0 0 0 (1+) 0 0 0 0 0 0 (1+) 0 0 0 0 0

9、 0 (1+) roudu zhangliangHooke law Relationships between K, , , E, : = E/2(1+) K = E/3(1-2) = +2/3* = E/(1-2)(1+) =K-2/3* E = 9K/(3K+) = (9+6)/(3-) = (3K-2)/2(3K+) = /2(-/3)Hooke law Uniaxial traction along X: yy = zz = - xx V= xx + yy + zz = xx*(1-2v)/E xx = (V)+2xx=EV/(1-2 )(1+) + 2E/2(1+)*xx = xx

10、/(1+) + E/(1+)*xx xx(1+-)/(1+) = E/(1+)xx xx = E xxApplications Compute the elastic deformation during a compression test of Mg alloy Compute the change of volume for Mg during a tensile test e = 0.2/E = 200/45000 = 0.0044 xx +yy +zz = -*zz/E -*zz/E + 1/E*zz = (1-2*0.35)*200/45000 = 0.00133Applications in uniaxial deformation:V= xx +yy +zz = xx*(1-2)/E Metal