Memory has been initialized. Room for 2425 values. FEMOD1 A program for solving one dimensional boundary value problems using the finite element method. Program maxima: Elements = 100 Nodes = 501 Nodes per element = 6 Materials = 4 Quadrature points = 10 Point loads = 5 Last modified on 17 May 1999 Enter command (HELP for menu) FEMOD1 commands: Help Print list of commands. Quit Stop the program. Set BC, ELEM, EXACT, GAUSS, MAT, NODE, POINT, TITLE. Solve Solve the system. Type BC, ELEM, ERROR, EXACT, GAUSS, MAT, NODE, POINT, SOL, SYS, TITLE. # To make a comment. PREP Input the data in an orderly fashion. PROC Solve the system. POST Print out the solution data. Enter command (HELP for menu) Set up the exact solution U and derivative. Enter a formula for the exact solution U. Enter a formula for dUdX or RETURN to leave it blank. Enter command (HELP for menu) Enter a title for the problem Set up the material functions. Set the functions K(x), C(x), B(x) and F(x) in -d/dx K dU/dx + C dU/dx + B U = F + P Enter the number of materials, up to 4 (Simple problems use 1 material.) For material 1 enter formulas for K(x), C(x), B(x) and F(x): Enter formula for K function. Enter formula for C function. Enter formula for B function. Enter formula for F function. Set up the elements. Set up element 1 Enter polynomial degree (between 1 and 5 Enter Y if all the elements have been defined. If there are more elements in a row, like this one, type the number of the LAST such element. between 2 and 100 If there are no more elements, type 1 Enter element number. Enter Y if all the elements have been defined. Choose a quadrature rule order between 1 and 10 Set up the nodes. Enter the location of the LEFT boundary node. Define equally spaced subintervals by giving an element, and its rightmost X coordinate. The left element is 1 The left X value is 0.0000000E+00 Enter right element between 1 and 5 Enter X coordinate of rightmost node in element 5 Set up the boundary conditions. LEFT boundary condition. BC type 1 has the form "U=V1", 2 has the form "K dU/dx = V1 U + V2". Enter 1 or 2, to choose boundary condition. Enter V1 in condition "U = V1" RIGHT boundary condition. BC type 1 has the form "U=V1", 2 has the form "K dU/dx = V1 U + V2". Enter 1 or 2, to choose boundary condition. Enter V1 in condition "U = V1" Set up the point loads. (Simple problems have 0 point loads). Enter a node number, between 1 and 6 where a point load applies, and the value of the point load there. (0, 0.0 to quit) Linear Elements Boundary conditions: At the left boundary point, the boundary condition is of type 1 U = 0.0000000E+00 At the right boundary point, the boundary condition is of type 1 U = 1.000000 Element Degree Material Node1 Node2 Node3 Node4 Node5 Node6 1 1 1 1 2 2 1 1 2 3 3 1 1 3 4 4 1 1 4 5 5 1 1 5 6 The quadrature rule is of of order 2 I Weight(I), X(I) 1 0.0000000E+00 0.0000000E+00 2 0.0000000E+00 0.0000000E+00 Node X coordinate 1 0.000000E+00 2 0.200000 3 0.400000 4 0.600000 5 0.800000 6 1.00000 There are no point loads! Material functions. There is 1 material. Each material is defined by functions K(x), C(x), B(x) and F(x), which appear in - d/dx K(x) dU/dx + C(x) dU/dx + B(x) U = F(x) + P Functions for material 1 K(x) = 1 C(x) = 0 B(x) = 0 F(x) = -2 Enter command (HELP for menu) Set up and solve the linear system: 1: Compute the bandwidth of the stiffness matrix GK. Lower bandwidth = 1 Upper bandwidth = 1 2: Assemble the stiffness matrix GK and right hand side GF. Enter 0 to display no more element matrices or number between 0 and 5 or enter 6 for matrices 0 to 5 EKF_WRITE: Element matrix and force vector for element 1 1 2 1 5.000 -5.000 -0.2000 2 -5.000 5.000 -0.2000 EKF_WRITE: Element matrix and force vector for element 2 2 3 2 5.000 -5.000 -0.2000 3 -5.000 5.000 -0.2000 EKF_WRITE: Element matrix and force vector for element 3 3 4 3 5.000 -5.000 -0.2000 4 -5.000 5.000 -0.2000 EKF_WRITE: Element matrix and force vector for element 4 4 5 4 5.000 -5.000 -0.2000 5 -5.000 5.000 -0.2000 EKF_WRITE: Element matrix and force vector for element 5 5 6 5 5.000 -5.000 -0.2000 6 -5.000 5.000 -0.2000 3: Modify the system for boundary conditions. 4: Modify the right hand side GF for point loads. 5: Factor the stiffness matrix. 6: Solve the system GK * U = GF. Enter command (HELP for menu) Linear Elements The solution has been computed. Linear Elements X U computed U exact dUdx computed dUdx exact 0.000000E+00 0.000000E+00 0.000000E+00 0.200000 0.000000E+00 0.200000 0.400000E-01 0.400000E-01 0.600000 0.400000 0.400000 0.160000 0.160000 1.00000 0.800000 0.600000 0.360000 0.360000 1.40000 1.20000 0.800000 0.640000 0.640000 1.80000 1.60000 1.00000 1.00000 1.00000 1.80000 2.00000 Exact solution information. U(X): x*x U'(X): 2*x Error computations Element L2-norm Energy-norm 1 0.298143E-02 0.516398E-01 2 0.298143E-02 0.516398E-01 3 0.298145E-02 0.516398E-01 4 0.298146E-02 0.516398E-01 5 0.298141E-02 0.516398E-01 Total 6.6666943E-03 0.1154700 Enter command (HELP for menu) System matrix and right hand side Column 1 2 3 4 RHS Row 1 1.00000 0.000000E+00 0.000000E+00 2 0.000000E+00 10.0000 -5.00000 -0.400000 3 -5.00000 10.0000 -5.00000 -0.400000 4 -5.00000 10.0000 -0.400000 5 -5.00000 4.60000 Column 5 6 RHS Row 4 -5.00000 -0.400000 5 10.0000 0.000000E+00 4.60000 6 0.000000E+00 1.00000 1.00000 Enter command (HELP for menu) Enter "YES" to confirm you want to quit. FEMOD1 is stopping now.