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![BOUNDARY CONDITIONS
Referring to Fig., we find that there are displacement boundary
conditions and surface-loading conditions. If u is specified on part of
the boundary denoted by Su, we have
U= 0 OR Sx
We can also consider boundary conditions such as u = a, where a
is a given displacement.
We now consider the equilibrium of an elemental tetrahedron ABCD,
shown in Fig., where DA, DB, and DC are parallel to the X-, y-, and
z-axes, respectively, and area ABC, denoted by dA, lies on the
surface. If n = [nx, ny, nz]T is the unit normal to dA, Consideration of
equilibrium along the three axes directions gives
These conditions must be satisfied on the boundary, St, where the
tractions are applied. In this description, the point loads must be
treated as loads distributed over small, but finite areas.](https://image.slidesharecdn.com/finiteelementmethod-170409063521/75/Finite-element-method-7-2048.jpg)
![ The first step is to model the bar as a stepped shaft, consisting of a
discrete number of elements, each having a uniform cross section.
Specifically, let us model the bar using four finite elements. A simple
scheme for doing this is to divide the bar into four regions, as shown in
Fig.
In the finite element model, every element connects to two nodes.
However, cross-sectional area, traction, and body forces can differ in
magnitude from element to element.
Better approximations are obtained by increasing the number of
elements.
It is convenient to define a node at each location where a point load is
applied.
In a one-dimensional problem, every node is permitted to displace only
in the ±x direction. This, each node has only one degree of freedom
(dof).
The five-node finite element model in Fig. has five dofs.
The displacements along each dof are denoted by Q1, Q2, Q3... Q5. In
fact, the column vector Q = [Q1. Q2 .. 'Q5]T is called the global](https://image.slidesharecdn.com/finiteelementmethod-170409063521/75/Finite-element-method-22-2048.jpg)
![ The shape functions N1 and N2 are shown in Figs. Respectively.
The graph of the shape function N1 in Fig.(a) is obtained from equation
N1(ξ) by noting that N1 = 1 at ξ = -1, N1 = 0 at ξ = 1.
and N1 is a straight line between the two points.
Similarly, the graph of N2 in Fig. 3.7(b) is obtained from Eq. N2(ξ).
Once the shape functions are defined, the linear displacement field within
the element can be written in terms of the nodal displacements q1 and q2
as
u= N1q1 + N2q2
or, in matrix notation, as u=Nq
Where N = [N1 N2] and q=[q1 q2]T
In these equations, q is referred to as the element displacement
vector.
It is readily verified from equation of u that u=q1 at node 1 and u=q2 at
node 2, and that u varies linearly.
It may be noted that the transformation from x to ξ in eq of u can be
written in terms of N1 and N2 as
x= N1x1 + N2x2](https://image.slidesharecdn.com/finiteelementmethod-170409063521/75/Finite-element-method-27-2048.jpg)
More Related Content
Finite element method
- 1. Finite Element Method ComputerAided Design
- 2. Finite Element Method FEM: Method for numerical solution of field problems. Description - FEM cuts a structure into several elements (pieces of the structure). - Then reconnects elements at “nodes” as if nodes were pins or drops of glue that hold elements together. - This process results in a set of simultaneous algebraic equations. Number of degrees-of-freedom (DOF) Continuum: Infinite FEM: Finite (This is the origin of the name, Finite Element Method)
- 4. Approximate Numerical Method Inmechanics of solids, our problem is to determine the displacement u of the body shown. It satisfying the equilibrium equations. stresses are related to strains, which, in turn, are related to displacements. Solution of this set of equations is generally referred to as an exact solution which are available for simple geometries and loading conditions. For problems of complex geometries and general boundary and loading conditions, obtaining such solutions is an almost impossible task. Approximate solution methods usually employ potential energy or variational methods, which place less stringent conditions on the functions.
- 5.
- 6. The stressesacting on the elemental volume dV are shown in Fig. When the volume dV shrinks to a point, the stress tensor is represented by placing its components in a (3 x 3) symmetric matrix. However, we represent stress by the six independent components as in where σx, σy, σz are normal stresses and τyz, τxz, τzy are shear stresses. Let us consider equilibrium of the elemental volume shown in Fig. First we get forces on faces by multiplying the stresses by the corresponding areas. Writing ΣFx=0, ΣFy = 0, ΣFz = 0 and recognizing dV = dx dy dz, we get the equilibrium equations
- 7. BOUNDARY CONDITIONS Referringto Fig., we find that there are displacement boundary conditions and surface-loading conditions. If u is specified on part of the boundary denoted by Su, we have U= 0 OR Sx We can also consider boundary conditions such as u = a, where a is a given displacement. We now consider the equilibrium of an elemental tetrahedron ABCD, shown in Fig., where DA, DB, and DC are parallel to the X-, y-, and z-axes, respectively, and area ABC, denoted by dA, lies on the surface. If n = [nx, ny, nz]T is the unit normal to dA, Consideration of equilibrium along the three axes directions gives These conditions must be satisfied on the boundary, St, where the tractions are applied. In this description, the point loads must be treated as loads distributed over small, but finite areas.
- 8. STRESS-STRAIN RELATIONS Forlinear elastic materials, the stress-strain relations come from the generalized Hooke's law. For isotropic materials, the two material properties are Young's modulus (or modulus of elasticity) E and Poisson's ratio v. Considering an elemental cube inside the body, Hooke's law gives The shear modulus (or modulus of rigidity), G, is given by From Hooke's law relationships note that Substituting for and so on into we get inverse relations.
- 9. D isthe symmetric (6 x 6) material matrix given by Special Cases:- One dimension:- In one dimension, we have normal stress u along x and the corresponding normal strain ϵ, Stress-strain relations are simply Two dimensions:- In two dimensions, the problems are modeled as plane stress and plane strain. Plane Stress:- A thin planar body subjected to in-plane loading on its edge surface is said to be in plane stress. A ring press fitted on a shaft, Here stresses in z direction are set as zero. The Hooke's law relations then give us which is used as σ= Dϵ.
- 10. Plane Strain:If a long body of uniform cross section is subjected to transverse loading along its length, a small thickness in the loaded area, as shown in Fig, can be treated as subjected to plane strain, Here ϵz , ϒzx, ϒ yz are taken as zero. Stress σ z may not be zero in this case. The stress-strain relations can be obtained directly from D here is a (3 X 3) matrix, which relates three stresses and three strains.
- 12. The totalpotential energy Π (Capital Pi) of an elastic body, is defined as the sum of total strain energy (U) and the work potential: Π = Strain energy + Work potential (U) (WP) For linear elastic materials., the strain energy per unit volume in the body is ½*σTϵ For the elastic body shown in Fig. 1.1, the total strain energy U is given by The work potential WP is given by The total potential for the general elastic body shown in Fig.
- 13. Rayleigh-Ritz Method(Potential Energy Approach) Forcontinua, the total potential energy II in previous Eq. can be used for finding an approximate solution. The Rayleigh-Ritz method involves the construction of an assumed displacement field, say, The functions Øi are usually taken as polynomials. Displacements u, v, w must be kinematically admissible. That is, u, v, w must satisfy specified boundary conditions. Introducing stress-strain and strain-displacement relations, and substituting above equations into potential energy equations gives
- 14. where r= number of independent unknowns. Now, the extremum with respect to ai (i = 1 to r) yields the set of r equations- Principle of Minimum Potential Energy:- For conservative systems, of all the kinematically admissible displacement fields, those corresponding to equilibrium extremize the total potential energy. If the extremum condition is a minimum. the equilibrium state is stable.
- 15. GALERKIN'S METHOD Galerkin'smethod uses the set of governing equations in the development of an integral form. it is usually presented as one of the weighted residual methods. For our discussion, let us consider a general representation of a governing equation on a region V: Lu = P For the one-dimensional rod considered, the governing equation is the differential equation is We may consider L as the operator operating on u. The exact solution needs to satisfy governing eq. at every point x. If we seek an approximate solution ũ it introduces an error ϵ(x), called the residual: The approximate methods revolve around setting the residual relative to a weighting function wi to zero:
- 16. GALERKIN'S METHOD in elasticity Let us turn our attention to the equilibrium equations In elasticity. Galerkin’s method requires is an arbitrary displacement consistent with the boundary conditions of u. if is a unit normal at a point x on the surface.
- 17. The integrationby parts formula is where a and (J are functions of (x, y, z). For multidimensional problems, above Eq is usually referred to as the Green-Gauss theorem or the divergence theorem. Using this formula, integrating Eq. by parts, and rearranging terms, we get
- 18. One-Dimensional Problems Thetotal potential energy and the stress-strain and strain-displacement relationships are now used in developing the finite element method for a one- dimensional problem. For the one-dimensional problem, the stress, strain, displacement, and loading depend only on the variable x. so variables are as Furthermore, the stress-strain and strain-displacement relations are
- 19. For one-dimensionalproblems, the differential volume dV can be written as dV = Adx The loading consists of three types: the body force F, the traction forte T, and the point load Pi. These forces are shown acting on a body in Fig. Body Force:- A body force is a distributed force acting on every elemental volume of the body and has the units of force per unit volume. traction Force:- A traction force is a distributed load acting on the surface of the body. For the one-dimensional problem considered here, however, the traction force is defined as force per unit length. Point Load:- Finally, Pi is a force acting at a point i and ui is the x displacement at that point.
- 21. 1-D FINITE ELEMENT MODELING The finite element modeling of a one-dimensional body is considered in Section. The basic idea is to discretize the region and express the displacement field in terms of values at discrete points. Linear elements are introduced first. Stiffness and load concepts are developed using potential energy and Galerkin approaches. Boundary conditions are then considered.
- 22. The firststep is to model the bar as a stepped shaft, consisting of a discrete number of elements, each having a uniform cross section. Specifically, let us model the bar using four finite elements. A simple scheme for doing this is to divide the bar into four regions, as shown in Fig. In the finite element model, every element connects to two nodes. However, cross-sectional area, traction, and body forces can differ in magnitude from element to element. Better approximations are obtained by increasing the number of elements. It is convenient to define a node at each location where a point load is applied. In a one-dimensional problem, every node is permitted to displace only in the ±x direction. This, each node has only one degree of freedom (dof). The five-node finite element model in Fig. has five dofs. The displacements along each dof are denoted by Q1, Q2, Q3... Q5. In fact, the column vector Q = [Q1. Q2 .. 'Q5]T is called the global
- 24. SHAPE FUNCTIONS Considera typical finite element e in Fig. In the local number scheme, the first node will be numbered 1 and the second node 2. The notation X1 = x-coordinate of node 1 X2 =x-coordinate of node 2 is used. We define a natural or intrinsic coordinate system, denoted by ξ as
- 25. From Fig.we see that ξ = -1 at node 1 and ξ = 1 at node 2. The length of an element is covered when ξ changes from -1 to 1. We use this system of coordinates in defining shape functions, which are used in interpolating the displacement field. Now the unknown displacement field within an element will be interpolated by a linear distribution. Fig.3.6 To implement this linear interpolation, linear shape functions will be introduced as
- 27. The shapefunctions N1 and N2 are shown in Figs. Respectively. The graph of the shape function N1 in Fig.(a) is obtained from equation N1(ξ) by noting that N1 = 1 at ξ = -1, N1 = 0 at ξ = 1. and N1 is a straight line between the two points. Similarly, the graph of N2 in Fig. 3.7(b) is obtained from Eq. N2(ξ). Once the shape functions are defined, the linear displacement field within the element can be written in terms of the nodal displacements q1 and q2 as u= N1q1 + N2q2 or, in matrix notation, as u=Nq Where N = [N1 N2] and q=[q1 q2]T In these equations, q is referred to as the element displacement vector. It is readily verified from equation of u that u=q1 at node 1 and u=q2 at node 2, and that u varies linearly. It may be noted that the transformation from x to ξ in eq of u can be written in terms of N1 and N2 as x= N1x1 + N2x2
- 28. Strain - displacementmatrix The strain--displacement relation as Upon using the chain rule of differentiation, we obtain From the relation between x and ξ, we have Also, since So, can be written as
- 29. where the(1 x 2) matrix B, called the element strain-displacement matrix, is given by Note: Use of linear shape functions results in a constant B matrix and, hence, in a constant strain within the element. The stress, from Hooke's law, is
- 30. Element stiffness matrix Consider the strain energy term Substituting for and in above eq. In the finite element model, the cross- sectional area of element e, denoted by Ae, is constant.
- 31. Also, Bis a constant matrix. Further, the transformation from x to ξ in shape function equation. where -1 <ξ<1, and le = IX2 – X1l is the length of the element. The element strain energy Ue is now written as where Ee is Young's modulus of element e. Noting that and substituting for B from Eq. we get
- 32. This equationis of the form where the element stiffness matrix ke is given by observe that Ke is linearly proportional to the product Ae Ee and inversely proportional to the length le.
- 33. PROPERTIES OF K(Stiffness Matrix) 1. The dimension of the global stiffness K is (N x N), where, N is the number of nodes. This follows from the fact that each node has only one degree of freedom. 1. K is symmetric. 2. K is a banded matrix. That is. all elements outside of the band are zero.