4 Finite Element Method for Trusses

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Moris Fletcher
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1 4 Finite Element Method for Trusses To solve the system of linear equations that arises in IPM, it is necessary to assemble the geometric matrix B a. For the sake of simplicity, the applied force vector f is constant and independent of the design. A bar, r, connects the nodes i and j. It has length l r, area a r, and an unit vector t r pointing from node i to node j. The components of t r in 2D and 3D are shown in Figure 4.1, while the bar magnitudes are illustrated in Figure 4.2. DIM denotes the dimension of the problem, i.e., 2 or 3 for two or three dimensions, respectively, and is denoted with the superscript 2D or 3D. Figure 4.1: Axis definition and direction cosines for 2D and 3D. Figure 4.2: Bar magnitudes for a bar, r. The unit vector for r for 2D and 3D are: [ ] cos α r t 2D cos α r r =, t 3D r = cos β r. (4-1) cos β r cos γ r a Although the stiffness matrix K is not explicitly used in this work, its assembly steps will also be shown.

2 Chapter 4. Finite Element Method for Trusses 37 The displacement vector for r is: [ u r = u r,i u r,j ], (4-2) where the components of the nodal displacements are: [ ] [ ] and u 2D r,i = u 3D r,i = u r,ix u r,iy u r,ix u r,iy u 2D r,j = u 3D r,j = u r,jx u r,jy u r,jx u r,jy, (4-3). (4-4) The elongation of bar r is: u r,iz u r,jz where B 2D r = δ r = (u r,j u r,i ) t r t r,x t r,y t r,x t r,y = B T r u r, ; B3D r = t r,x t r,y t r,z t r,x t r,y (4-5). (4-6) The nodal force vector in r, can be given by: t r,z f r = B r n r, (4-7) where n r is the axial force in the bar. The relationship between axial force and elongation is obtained from Hooke s law: n r = σ r a r = Eϵ r a r = Eδ ra r l r = D a,r δ r, (4-8) where σ is the axial stress, ϵ r is the corresponding strain, E is the Young s modulus, and: D a,r = Ea r. (4-9) l r Using (4-5) and (4-8) in (4-7), we find that: f r = k r u r, (4-10) where k r is known as the element stiffness matrix and can be defined as: k r = B r D a,r B T r. (4-11) Matrix k r can also be expressed, in 2D and 3D, in terms of its components:

3 Chapter 4. Finite Element Method for Trusses 38 k 2D r = D a,r t 2 r,x t r,x t r,y t 2 r,x t r,x t r,y t r,x t r,y t 2 r,y t r,x t r,y t 2 r,y t 2 r,x t r,x t r,y t 2 r,x t r,x t r,y, k 3D r t r,x t r,y t 2 r,y t r,x t r,y t 2 r,y t 2 r,x t r,x t r,y t r,x t r,z t 2 r,x t r,x t r,y t r,x t r,z t r,x t r,y t 2 r,y t r,y t r,z t r,x t r,y t 2 r,y t r,y t r,z t r,x t r,z t r,y t r,z t 2 r,z t r,x t r,z t r,y t r,z t 2 r,z = D a,r t 2 r,x t r,x t r,y t r,x t r,z t 2 r,x t r,x t r,y t r,x t r,z t r,x t r,y t 2 r,y t r,y t r,z t r,x t r,y t 2 r,y t r,y t r,z. t r,x t r,z t r,y t r,z t 2 r,z t r,x t r,z t r,y t r,z t 2 r,z (4-12) The element displacement vector of each bar is obtained from the global displacement vector, u: u r = C r u, (4-13) where C r is a 2 DIM N n matrix of zeros and ones that selects the components of u that correspond to u r. The global force vector is: f = C T r f r, (4-14) where f r is local force vector of each bar. 4.1 Geometric Matrix Assembly To calculate B, we combine equations (4-7) and (4-14): f = B n, (4-15) ] B = [C T1 B 1 C TNb B Nb. (4-16) Each product C T i B i corresponds to a column of B and contains at most 2DIM non zeros, related to the direction cosines. 4.2 Stiffness Matrix Assembly The stiffness matrix is obtained by multiplying Equation (4-10) by C T i and adding over all bars: f = Ku, (4-17) K = K r ; K r = C T r k r C r ; (4-18)

4 Chapter 4. Finite Element Method for Trusses 39 K is the global stiffness matrix; K r and k r are the element stiffness matrices expressed in global and local coordinate systems, respectively. This element by element assembling process is not the only way to obtain the stiffness matrix. Equation (4-8) can be rewritten as: n = D a δ, (4-19) where D a is a diagonal matrix that contains the elements D a,r, and the elements of δ can be obtained from the global displacement vector: B T 1 C 1 δ =. u B T N b C Nb (4-20) = B T u. Combining Equations (4-20), (4-21), and (4-15), one obtains: n = D a B T u, (4-21) and: f = BD a B T u. (4-22) Therefore: K = BD a B T. (4-23) It is important to compare the stiffness matrix (Equation(4-23)) with the first of the normal equations presented in IPM. This Equation (3-22) is repeated below for the sake of clarity: M = BD ipm B T. (4-24) They only differ in the middle diagonal matrix. Therefore, the same solvers used in a finite element analysis can be applied to solve the first of the normal equations. Also M can be assembled as a sum of element matrices in similar way as in the case of the stiffness matrix: M = M r ; M r = C T r m r C r, (4-25) m r = B r D ipm B T r. (4-26) Therefore, an element by element iterative solver can be used to solve the normal equations. This solver is very suitable to be used in cases where the truss presents a large number of bars, since the global stiffness matrix does not need to be assembled. Another interesting aspect regarding the finite element analysis is the

5 Chapter 4. Finite Element Method for Trusses 40 application of boundary conditions. For trusses with a low number of bars, an intuitive approach to apply boundary conditions to K is to eliminate the columns and rows that correspond to the restrained degrees of freedom. However, when the size of the problem is large this technique is not appropriate because it involves restructuring the matrix and moving large sections of data. A more computationally efficient method is shown in the next chapter.

Stress analysis of a stepped bar

Stress analysis of a stepped bar Problem Find the stresses induced in the axially loaded stepped bar shown in Figure. The bar has cross-sectional areas of A ) and A ) over the lengths l ) and l ), respectively.