Finite Element Truss Problem 6.0 Trusses Using FEA We started this series of lectures looking at truss problems. We limited the discussion to statically determinate structures and solved for the forces in elements and reactions at supports using basic concepts from statics. Next, we developed some basic one dimensional finite elements concepts by looking at springs. We developed a basic system of node numbering that allows us to solve problems involving several springs by simply adding the stiffness matrices for each spring. In the next two lectures, we will extend the basic one dimensional finite element development to allow us to solve generalized truss problems. The technique we will develop is a little more complex than that originally used to solve truss problems, but it allows us to solve problems involving statically indeterminate structures. 6.1 Local and Global Coordinates We can extend the one dimensional finite element analysis by looking at a one dimensional problem in a two dimensional space. Below we have a finite element (could be a spring) that attaches to nodes 1 and 2. The 〉′′〈yx,coordinates are the local coordinates for the element and 〉〈yx,are the global coordinates. The local coordinate system looks much like the one dimensional coordinate system we developed in the last lecture. We can convert the displacements shown in the local coordinate system by looking at the following diagram. We will let 1q′ and 2q′ represent displacements in the local coordinate system and q1, q2, q3, and q4 represent displacements in the x-y (global) 1 2 x’ y’ Local coordinate system x y Global coordinate System Figure 1 - Local and global coordinate systemscoordinate system. Note that the odd subscripted displacements are in the x direction and the even ones are in the y direction as shown in the following diagram. In a previous lecture we looked at the deformation of springs by looking at the displacements at the ends of the springs. Here we are going a step farther into finite element development by looking at the strain energy of the element. The element could be a spring but in this case we will generalize and look at it as any solid material element. The only restriction we will place on the element is that the deformation is small compared to its total length. We know from Hook’s law that the force is directly proportional to the deformation. xkF∆= (6.1) We can compute the energy by integrating over the deformation 2021kQxdxkuQ==∫ (6.2) where LAEk = the element stiffness, A = the cross sectional area of the element, E = Young’s modulus for the material, and L = the length of the element. Q is the total change in length of the element. Note that we are assuming the deformation is linear over the element. All equal length segments of the element will deform the same amount. We call this a constant strain deformation of the element. We can rewrite this change in length as q1’ q2’ ? Deformed element θsin2qq1 q2 q4 q3 θcos1qUn-deformed element Figure 2 - The deformation of an element in both local and global coordinate systems.)('1'2qqQ −= (6.3) Substituting this into equation (6.2) gives us 212)(21qqku′−′= (6.4) or )2(21211222qqqqku′+′′−′= (6.5) Rewriting this in vector form we let ′′=′21qqq (6.6) and −−=′1111LAEk (6.7) With this we can rewrite equation (6.5) as: qkquT′′′=21 (6.8) We can do the indicated operations in (6.8) to see how the vector notation works. We do this by first expanding the terms then doing the multiplication. { }′′−−′′=212111112 qqqqLAEu (6.9) { }′′′+′−′−′=2121212 qqqqqqLAEu (6.10) ( ))()(2122211qqqqqqLAEu′−′′+′−′′= (6.11) ( )212221212qqqqqqLAEu′′−′+′′−′= (6.12))2(2222121qqqqLAEu′+′′−′= (6.13) Which is the same as equation (6.5). Equation (6.7) is the stiffness matrix for a one dimensional problem. It bears very close resemblance to equation (5.7) used in our one dimensional spring development. [ ]−−=kkkkK (5.7) −−=′1111LAEk (6.7) 6.2 Two Dimensional Stiffness Matrix We know for local coordinates that ′′=′21qqq (6.6) and for global coordinates (See Figure 2) =4321qqqqq (6.14) We can transform the global coordinates to local coordinates with the equations θθ sincos211qqq +=′ (6.15) and θθ sincos432qqq +=′ (6.16) This can be rewritten in vector notatio n as: qMq=′ (6.17) where=scscM0000, (6.18) θcos=c , and θsin=s . Using qkquT′′′=21 (6.8) we can substitute in equation (6.17) []qMkMquTT′=21 (6.19) Now we will let MkMkT′= (6.20) and doing the multiplication, k our stiffness matrix for global two dimensional coordinates becomes −−−−−−−−=22222222scsscscsccscscsscscsccscLAEk (6.21) where: E = Young’s modulus for the element material A = the cross sectional area of the element L = the length of the element θcos=c θsin=s 6.3 Stress Computations The stress can be written as εσE= (6.22) where εis the strain, the change in length per unit of length. We can rewrite this as:LqqE12′−′=σ (6.23) In vector form we can write the equation as { }′′−=2111qqLEσ (6.24) From our previous discussion, we know that in local coordinates ′′=′21qqq (6.25) and in global coordinates =4321qqqqq (6.26) From equation (6.17) we know that qMq=′ (6.17) where =scscM0000 (6.18) Substituting this in to the equation (6.24) yields { }qMLE11−=σ (6.27) Now we multiply M by the vector { }qscscLE−−=σ (6.28) total deformation length of