Ebook A first course in the finite element method (4th edition) Part 2

... Da, daaaa! There they are The blocks you wanted Again, our diagonal block was a scalar, so this was easy What would have happened if it was a matrix? Do you have to compute that inverse and apply ... of the values on its vertices Otherwise, we can try a quadrature rule to approximate the integrals It is important at this stage to note that the choice of an adequate quadrature rule has to take ... are as many unknowns as there are equations here Note that in the general case, the values uj are not nodal values, since an arbitrary basis of a linear space has nothing to with nodes or evaluations...

... solutions 7 12 24 28 32 35 44 48 50 71 71 72 73 75 80 91 102 10 7 11 4 14 1 14 1 14 4 14 5 14 7 15 1 15 3 16 0 viii The Finite Element Method Further topics in the finite element method 5 .1 The variational ... Consider the equation (2 .13 ) −u = f, < x < 1, 21 Weighted residual and variational methods Table 2.3 The integrals for the Galerkin method i, j vi Lvj dx and 10 15 21 15 10 10 5 84 10 10 35 14 84 15 10 5 ... 24 c0 c1 = , giving c0 = 12 , c1 = 16 The approximate solution is then u 1 (x) = x (1 − x) (1 + 2x) 12 Example 2.6 The final method to be considered is the Galerkin method In this method the integral...

... + 32 (1 + 2U2 + U3 )U2 − 32 (U2 + U3 )U3 = − , − 32 (U2 + U3 ) + 32 (U2 + 2U3 )U3 = − Solving these yields U 22 = 49 , U 32 = 19 21 2 The Finite Element Method Just as in Example 5.5, the positive ... after the conditions U11 = 0, U21 = 29 , U31 = 89 , U41 = 2, U 12 = 1, U 42 = are enforced, becomes ⎤ ⎥ ⎢ ⎢ U 22 ⎥ + ⎣ U 32 ⎦ ⎡ which yields U 22 = 11 , U 32 = 17 29 46 29 29 46 29 ⎡ ⎤ ⎡ ⎤ 0 2 ... Performing the integrations, ⎤ ⎡ 156 22 h 54 −13h kh ⎢ 4h2 13h −3h2 ⎥ ⎥ + EI ⎢ ke = ⎣ 156 22 h ⎦ h3 420 sym 4h2 fe = fh [−6 12 −h h dξ, 4EI N N h2 i j ⎡ ⎢ ⎢ ⎣ 12 − 12 −6h 12 6h 4h2 sym ⎤ 6h 2h2 ⎥ ⎥,...