c1=ea/l; c2=12e*i/(l^3); c3=6ei/(l^2); c4=2ei/l; c=cos(theta); s=sin(theta); K=[]; Verification

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Rose Welch
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1 UNIVERSITY OF ILLINOIS AT URBANA-CHAMPAIGN Civil and Environmental Engineering CEE Methods of Structural Analysis Fall Semester, 2 Problem 1: Writing the functions Laboratory #6 Solutions There were 3 functions that had to be completed for this laboratory. They were: 1) feeldof.m 2) feasmbl.m 3) cong2.m Also, the script file lab6.m had to be completed. The file feaplyc.m was given in working order and no modifications were necessary. The code for this laboratory will be presented in the following order: 1) cong2.m 2) feaplyc.m 3) feeldof.m 4) feasmbl.m Following the presentation of a code, verification for that function will be provided. Program 1: cong2.m >> type cong2.m function K=cong2(E,I,L,A,theta) ================================================================= this function determines the global stiffness matrix contribution of a frame member input variables E = Young's modulus A = area L = length I = moment of inertia theta = angle between global and local coordinate frames internal variables c1 = temp variable for generating the terms in K c2 = temp variable for generating the terms in K c3 = temp variable for generating the terms in K c4 = temp variable for generating the terms in K c = cosine of theta s = sine of theta k1 = temp variable component of K k2 = temp variable component of K k3 = temp variable component of K k4 = temp variable component of K k5 = temp variable component of K k6 = temp variable component of K k7 = temp variable component of K output variables K = global stiffness matrix contribution ================================================================= define some common constants c1=e*a/l; c2=12*e*i/(l^3); c3=6*e*i/(l^2); c4=2*e*i/l; c=cos(theta); s=sin(theta); k1=(c1*c^2)+(c2*s^2); k2=(c1-c2)*c*s; k3=c3*s; k4=(c1*s^2)+(c2*c^2); k5=c3*c; k6=2*c4; k7=c4; define the elemental stiffness matrix in global coordinates K=[]; K=[ k1 k2 -k3 -k1 -k2 -k3;... k2 k4 k5 -k2 -k4 k5;... -k3 k5 k6 k3 -k5 k7;... -k1 -k2 k3 k1 k2 k3;... -k2 -k4 -k5 k2 k4 -k5;... -k3 k5 k7 k3 -k5 k6]; Verification The verification for this function was completed as part of Laboratory #3. The reason why we require verifications is so that we can use a function from a previous laboratory without having to check that function s output. Please view the solutions to Laboratory #3 if you wish to see this verification.

2 Program 2: feasmbl.m >> type feasmbl.m function [kk]=feasmbl(kk,k,index) MATLAB M-file: feasmbl.m Purpose: Assembly of element matrices into the system matrix Synopsis: [kk]=feasmbl(kk,k,index) Variable Description: Input Variables kk - system matrix before assembly k - element matrix index - d.o.f. vector associated with an element Internal Variables edof = number of dof's in index i = counter for which dof in index being added in j = counter for which dof in index being added in ii = temp variable for dof matching counter i jj = temp variable for dof matching counter j Output Variables kk - system matrix after assembly edof=length(index); for i=1:edof ii=index(i); for j=1:edof jj=index(j); kk(ii,jj)=kk(ii,jj)+k(i,j); Verification We provided you with this function and therefore you can assume that it was verified by us. Though this is a dangerous assumption, it is valid for this case. Program 3: feeldof.m >> type feeldof.m MATLAB M-file: feeldof.m Purpose: Compute system dofs associated with each element Synopsis: [index]=feeldof(nd,nnel,ndof) Variable Description: Input Variables nd - nodes connected to the element nnel - number of nodes per element ndof - number of dofs per node Internal Variables start = first dof corresponding to the given node i = counter for number of nodes per element j = counter for number of dof per node k = indicates position in index being filled Output Variables index - system dof vector associated with element "iel" k=; for i=1:nnel start=(nd(i)-1)*ndof; for j=1:ndof k=k+1; index(k)=start+j; Verfication >> nd=[5 2];nnel=2,ndof=6; nnel = 2 >> nd=[5 2];nnel=2;ndof=6; >> index=feeldof(nd,nnel,ndof) index = function [index]=feeldof(nd,nnel,ndof)

3 This is exactly what you would expect. Since there are 6 dof s per node, the dof s corresponding to node 5 would be 25 to 3. We know this because node 1 would be 1-6, node 2 would be 7-12, node 3 would be 13-18, and node 4 would be which makes node 5 dof s As mentioned, node 2 would have dof s 7-12 which is exactly what was derived in this case. Additionally, the dof s linearly increase by 1 as you go from dof to dof at a node. This behavior is captured here. Finally, the first six positions in index should correspond to node nd(1)=5 because ndof equals 6. Likewise, the next six positions in index should correspond to nd(2)=2 for the same reason. If there was a third node per element (nnel=3), the 13 th to the 18 th position of index would correspond to nd(3). To demonstrate the generality of the program, let s change nnel to 3. >> nd=[2 1 3];nnel=3;ndof=2; >> index=feeldof(nd,nnel,ndof) index = For this case, we should have got index with six terms where the first two terms are the 2 dof s for node 2 in a 2 dof per node system. These values would be 2*2-1=3 and 2*2-=4. Likewise, for the 3 rd and 4 th components of index would correspond to the 2 dofs at node 1. These values are 1 and 2. Finally, the 3 rd node s contribution would be dof s 5 and 6 and would appear in the 5 th and 6 th position of index. This is exactly what happened therefore we will assume that the function works perfectly. Program 4: feasmbl.m >> type feasmbl.m function [kk]=feasmbl(kk,k,index) MATLAB M-file: feasmbl.m Purpose: Assembly of element matrices into the system matrix Synopsis: [kk]=feasmbl(kk,k,index) Variable Description: Input Variables kk - system matrix before assembly k - element matrix index - d.o.f. vector associated with an element Internal Variables edof = number of dof's in index i = counter for which dof in index being added in j = counter for which dof in index being added in ii = temp variable for dof matching counter i jj = temp variable for dof matching counter j Output Variables kk - system matrix after assembly edof=length(index); for i=1:edof ii=index(i); for j=1:edof jj=index(j); kk(ii,jj)=kk(ii,jj)+k(i,j); Verification >> nd=[2 1];nnel=2;ndof=3; >> index=feeldof(nd,nnel,ndof) index = >> temp=rand(6,6); >> k=temp+temp' gives k as a symmetric matrix k = >> kk=zeros(8,8); >> kk=feasmbl(kk,k,index) kk = Columns 1 through

4 Column 8 >> kk=zeros(8,8); >> kk(1,1)=1; >> kk=feasmbl(kk,k,index) kk = Columns 1 through Column 8 >> index=[ ]; >> kk=zeros(7,7); >> temp=rand(5,5); >> k=temp+temp' k = >> kk=feasmbl(kk,k,index) kk = Several different tests were completed here to verify that the function worked and that the function was general. For all these examples, adding a random matrix to the transpose of that matrix generated a random but symmetric matrix. In the first example, a six by six matrix k was generated. The matrix kk was created as an zeros matrix of size 8,8. According to index, the 2 nd quadrant of k was to be placed in kk(4:6,4:6). By following the drawn arrow, you can see that is exactly what happened. Likewise, the 4 th quadrant of k was to be placed in kk(1:3,1:3). Again, following the arrow you can see this is what exactly happened. We knew where each quadrant should be placed in kk by knowing index. Looking at index, we can see that the first three components of index equal 4, 5 and 6. Thus, k(1:3,1:3) should be placed in kk(index(1:3),index(1:3)) or kk(4:6,4:6). To test that each element s stiffness contribution would be added into kk, the above example was repeated but with kk not initially being a matrix of all zeros. Instead, kk(1,1) was specified to be 1.. Therefore, after the function feasmbl.m was run, the kk(1,1) term should be 1. greater this time than the previous time. This is exactly what happens. Instead of.3983, kk(1,1) equaled As a test of the generality of the function, a 5x5 k matrix was specified. Though this case is unlikely based on the elements used in this class, it will show that the function can handle any situation meeting the input requirements. The index specified in this case was [ ]. Thus, after feasmbl.m was run, k(2:5,2:5) should be kk(1:4,1:4) assuming kk was initially a zeros matrix. Based on each one of these tests being passed by the function, we can safely assume that this function works as desired and is therefore verified. Problem 2: lab6.m After the 4 functions for this laboratory were completed, the lab6 script file also needed to be finished. The given structure had to be defined and then the script file could be run to calculate the displacements in the structure. >> type lab Laboratory #6 MATLAB M-file: lab6.m Problem description Solution of static 2D frame structure Find deflections and stresses of the frame shown in the figure

5 NOTE: This program is a script file, i.e. to run this program, one must type the following at the MATLAB prompt; >> lab6 This program is also problem specific, i.e. this program will only be applicable for lab6 and no other assignment. Before this program will run though, the user must make the correct changes to the program so to model the assigned problem. Internal Variable descriptions k = element stiffness matrix kk = system stiffness matrix ff = system force vector index = a vector containing system dofs associated with each element x = vector of the x coordinate values y = vector of the y coordinate values el = elastic modulus area = area xi = moment of inertia of cross-section nel = number of elements nnel = number of nodes per element ndof = number of dofs per node nnode = total number of nodes sdof = total number of dofs nodes = nodal connectivity matrix for each element bcdof = a vector containing dofs associated with boundary conditions bcval = a vector containing boundary condition values associated with the dofs in 'bcdof' leng = element length beta = angle between the local and global axis for an element nd = vector of the nodes of a specific element fsol = displacements vector control input data nel=6; nnel=2; ndof=3; nnode=(nnel-1)*nel+1; sdof=nnode*ndof; total system dofs nodal coordinates numer of elements number of nodes per element number of dofs per node total number of nodes in system x(1)=; y(1)=; x, y coord. values of node 1 in terms of the global axis x(2)=; y(2)=2; x, y coord. values of node 2 in terms of the global axis x(3)=; y(3)=4; x, y coord. values of node 3 in terms of the global axis x(4)=; y(4)=6; x, y coord. values of node 4 in terms of the global axis x(5)=; y(5)=8; x, y coord. values of node 5 in terms of the global axis x(6)=12; y(6)=8; x, y coord. values of node 6 in terms of the global axis x(7)=24; y(7)=8; x, y coord. values of node 7 in terms of the global axis material and geometric properties el=3; area=2; xi=8/12; nodal connectivity nodes(1,1)=1; nodes(1,2)=2; nodes(2,1)=2; nodes(2,2)=3; nodes(3,1)=3; nodes(3,2)=4; nodes(4,1)=4; nodes(4,2)=5; nodes(5,1)=5; nodes(5,2)=6; nodes(6,1)=6; nodes(6,2)=7; applied constaints elastic modulus cross-sectional area moment of inertia of cross-section bcdof(1)=1; transverse deflection at node 1 is constrained bcval(1)=; whose described value is bcdof(2)=2; vertical deflection at node 1 is constrained bcval(2)=; whose described value is bcdof(3)=3; rotation deflection at node 1 is constrained bcval(3)=; whose described value is bcdof(4)=19; transverse deflection at node 7 is constrained bcval(4)=; whose described value is bcdof(5)=21; rotation deflection at node 7 is constrained bcval(5)=; whose described value is initialization to zero ff=zeros(sdof,1); initialization of system force vector kk=zeros(sdof,sdof); initialization of global stiffness matrix index=zeros(nel*ndof,1); initialization of index vector applied nodal force

6 ff(2)=-75; load applied at node 7 in the negative y direction loop for elements (DO NOT MODIFY THIS PROGRAM BEYOND THIS POINT) for iel=1:nel nd(1)=nodes(iel,1); nd(2)=nodes(iel,2); x1=x(nd(1)); y1=y(nd(1)); x2=x(nd(2)); y2=y(nd(2)); leng=sqrt((x2-x1)^2+(y2-y1)^2); loop for the total number of elements 1st connected node for the (iel)-th element 2nd connected node for the (iel)-th element x and y coordinates of 1st node x and y coordinates of 2st node element length dx=x2-x1; dy=y2-y1; if(dx==) compute the angle between the local and global axes if (dy>) beta=pi/2; else beta=-pi/2; else beta=atan2(dy,dx); index=feeldof(nd,nnel,ndof); k=cong2(el,xi,leng,area,beta); kk=feasmbl(kk,k,index); [kk,ff]=feaplyc(kk,ff,bcdof,bcval); fsol=kk\ff; print fem solutons extract system dofs for the element compute element matrix assemble into system matrix apply the boundary conditions solve the matrix equation to find nodal displacements Problem 3: MaStAn2 and Comments The two structures given with the assignment were inputted into MaStAn2 and a first order elastic analysis completed. We have included the displacement results below as well as the deformed shape, shear Y, axial and moment Z plots for each case. Figure 1 Displacement Report ************ MASTAN2 v1. ************ Time: 13:8:2 Date: 1/15/2 Problem Title: Laboratory #6, Figure 1 Solution ************** ############################## Results of Structural Analysis ############################## num=1:1:sdof; store=[num' fsol] Results >> lab6 store = print displacements General Information: Structure Analyzed as: Planar Frame Analysis Type: First-Order Elastic Analytical Results: (i) Displacements at Step # 1, Applied Load Ratio = 1. Deflections Node X-disp Y-disp Z-disp

7 1.e+.e+.e e-2-2.5e-4.e e-2-5.e-4.e e-2-7.5e-4.e e-5-1.e-3.e e e-2.e+ 7.e e-1.e+ Rotations (radians) Node X-rot Y-rot Z-rot 1.e+.e+.e+ 2.e+.e e-3 3.e+.e e-3 4.e+.e e-4 5.e+.e e-3 6.e+.e e-3 7.e+.e+.e+ ##################################### End of Results of Structural Analysis #####################################

8 12.e+.e e-3 13.e+.e+.e+ ************ MASTAN2 v1. ************ Time: 15:38:3 Date: 1/15/2 ##################################### End of Results of Structural Analysis ##################################### Problem Title: Laboratory #6, Figure 2 Solution ************** ############################## Results of Structural Analysis ############################## General Information: Structure Analyzed as: Planar Frame Analysis Type: First-Order Elastic Analytical Results: (i) Displacements at Step # 1, Applied Load Ratio = 1. Deflections Node X-disp Y-disp Z-disp 1.e+.e+.e e-2-2.5e-4.e e-2-5.e-4.e e-2-7.5e-4.e e-5-1.e-3.e e e-2.e e e-1.e e e-2.e e-5-1.e-3.e e-2-7.5e-4.e e-2-5.e-4.e e-2-2.5e-4.e+ 13.e+.e+.e+ Rotations (radians) Node X-rot Y-rot Z-rot 1.e+.e+.e+ 2.e+.e e-3 3.e+.e e-3 4.e+.e e-4 5.e+.e e-3 6.e+.e e-3 7.e+.e e-18 8.e+.e e-3 9.e+.e e-3 1.e+.e e-4 11.e+.e e-3

9 Questions 1) Program Comparison The displacements calculated by both programs are equivalent. Any small differences that exist can be attributed to round-off error. Also, note that displacements are presented in different formats. MASTAN2 presents the displacements with respect to the node and direction where as the lab6 program just gives the displacements with respect to the global degrees of freedom. 2) Discussion of Diagrams The deformed shape of the frame in figure 2 are symmetric about the centerline of the frame. The deformed shape of figure 1 match the deformed shape for the left-hand side of the frame in figure 2. For the frame in figure 2, the displacements at the centerline (location of the symmetric plane) match the boundary conditions applied to that point on the symmetric frame of figure 1. This indicates that the boundary conditions applied to the frame of figure 1 were sufficient and necessary for the behavior of the symmetric structure to match the behavior of the full structure. From the results of the frame in figure 2, it should be noted that the displacements in the y-direction are of the same magnitude and direction on opposite sides of the centerline (symmetric plane). However, the displacements in the x-direction are of the same magnitude but are opposite in direction (i.e. change in sign). This behavior is expected and necessary for the deformed shape of the full structure to form "mirror" images on opposite sides of the centerline. In regards to the axial force diagram, it is evident that the axial loads are equal on opposite sides of the centerline. However, this is not always the case for the shear Y diagram. The member crossing the line of symmetry has shears that have equal magnitude but opposite signs across the line of symmetry. The columns have equal magnitudes and signs. This differs from the moment diagram. Here, the columns have equal magnitude but opposite signs across the line of symmetry. The member crossing the line of symmetry has moments that are mirrored across the line. This is expected. In regards to the magnitude of values, if the structure is geometrically and load symmetric, then the values must be equal across the line of symmetry. The sign differences are based on the properties of moment and shear. When the load is passed on the line of symmetry, shear must change signs. The discontinuity is possible. However, when crossing the line of symmetry, the moment must remain continuous. By the slope of the moment diagram equaling the magnitude of the shear diagram at a given point, the moment diagram must simply be mirrored across the line giving the same signs and magnitudes. 3) Changing the load If the load applied to the structure is cut in half, then the resulting displacements will also be halved. This occurs because we are assuming linear elastic behavior and therefore

10 superposition is possible. Superposition is defined as, "Under the principle (of superposition), the response of a structure to the application of a system of forces is identical to the summation of the responses of the same structure to the separate application of every force of the system" (p.11, class text). Problem 4: Meaning of the Stiffness Matrix The componenet K(i,j) is the force in DOF direction i due to a unit displacement in DOF direction j, all other displacements equal to zero. A column of the stiffness matrix [K] represents the forces induced in each DOF of the element due to a unit displacement in the direction of one DOF, all other displacements equal to zero. These induced forces are the set of equilibrium forces resulting from the prescribed unit displacement. Bonus Problem Below is the lab6.m script file and the resulting output corresponding to the bonus analysis of figure 2 using the script file. type lab Laboratory #6 MATLAB M-file: lab6.m Problem description Solution of static 2D frame structure Find deflections and stresses of the frame shown in the figure NOTE: This program is a script file, i.e. to run this program, one must type the following at the MATLAB prompt; >> lab6 This program is also problem specific, i.e. this program will only be applicable for lab6 and no other assignment. Before this program will run though, the user must make the correct changes to the program so to model the assigned problem. Internal Variable descriptions k = element stiffness matrix kk = system stiffness matrix ff = system force vector index = a vector containing system dofs associated with each element x = vector of the x coordinate values y = vector of the y coordinate values el = elastic modulus area = area xi = moment of inertia of cross-section nel = number of elements nnel = number of nodes per element ndof = number of dofs per node nnode = total number of nodes sdof = total number of dofs nodes = nodal connectivity matrix for each element bcdof = a vector containing dofs associated with boundary conditions bcval = a vector containing boundary condition values associated with the dofs in 'bcdof' leng = element length beta = angle between the local and global axis for an element nd = vector of the nodes of a specific element fsol = displacements vector control input data nel=12; nnel=2; ndof=3; nnode=(nnel-1)*nel+1; sdof=nnode*ndof; total system dofs nodal coordinates numer of elements number of nodes per element number of dofs per node total number of nodes in system x(1)=; y(1)=; x, y coord. values of node 1 in terms of the global axis x(2)=; y(2)=2; x, y coord. values of node 2 in terms of the global axis x(3)=; y(3)=4; x, y coord. values of node 3 in terms of the global axis x(4)=; y(4)=6; x, y coord. values of node 4 in terms of the global axis x(5)=; y(5)=8; x, y coord. values of node 5 in terms of the global axis x(6)=12; y(6)=8; x, y coord. values of node 6 in terms of the global axis x(7)=24; y(7)=8; x, y coord. values of node 7 in terms of the global axis x(8)=36; y(8)=8; x, y coord. values of node 8 in terms of the global axis x(9)=48; y(9)=8; x, y coord. values of node 9 in terms of the global axis x(1)=48; y(1)=6; x, y coord. values of node 1 in terms of the global axis x(11)=48; y(11)=4; x, y coord. values of node 11 in terms of the global axis x(12)=48; y(12)=2; x, y coord. values of node 12 in terms of the global axis x(13)=48; y(13)=; x, y coord. values of node 13 in terms of the global axis material and geometric properties el=3; area=2; xi=8/12; elastic modulus cross-sectional area moment of inertia of cross-section

11 nodal connectivity nodes(1,1)=1; nodes(1,2)=2; nodes(2,1)=2; nodes(2,2)=3; nodes(3,1)=3; nodes(3,2)=4; nodes(4,1)=4; nodes(4,2)=5; nodes(5,1)=5; nodes(5,2)=6; nodes(6,1)=6; nodes(6,2)=7; nodes(7,1)=7; nodes(7,2)=8; nodes(8,1)=8; nodes(8,2)=9; nodes(9,1)=9; nodes(9,2)=1; nodes(1,1)=1; nodes(1,2)=11; nodes(11,1)=11; nodes(11,2)=12; nodes(12,1)=12; nodes(12,2)=13; applied constaints bcdof(1)=1; transverse deflection at node 1 is constrained bcval(1)=; whose described value is bcdof(2)=2; vertical deflection at node 1 is constrained bcval(2)=; whose described value is bcdof(3)=3; rotation deflection at node 1 is constrained bcval(3)=; whose described value is bcdof(4)=37; transverse deflection at node 13 is constrained bcval(4)=; whose described value is bcdof(5)=38; vertical deflection at node 13 is constrained bcval(5)=; whose described value is bcdof(6)=39; rotation deflection at node 13 is constrained bcval(6)=; whose described value is initialization to zero ff=zeros(sdof,1); initialization of system force vector kk=zeros(sdof,sdof); initialization of global stiffness matrix index=zeros(nel*ndof,1); initialization of index vector applied nodal force ff(2)=-15; load applied at node 7 in the negative y direction loop for elements (DO NOT MODIFY THIS PROGRAM BEYOND THIS POINT) for iel=1:nel loop for the total number of elements nd(1)=nodes(iel,1); nd(2)=nodes(iel,2); x1=x(nd(1)); y1=y(nd(1)); x2=x(nd(2)); y2=y(nd(2)); leng=sqrt((x2-x1)^2+(y2-y1)^2); 1st connected node for the (iel)-th element 2nd connected node for the (iel)-th element x and y coordinates of 1st node x and y coordinates of 2st node element length dx=x2-x1; dy=y2-y1; if(dx==) compute the angle between the local and global axes if (dy>) beta=pi/2; else beta=-pi/2; else beta=atan2(dy,dx); index=feeldof(nd,nnel,ndof); k=cong2(el,xi,leng,area,beta); kk=feasmbl(kk,k,index); [kk,ff]=feaplyc(kk,ff,bcdof,bcval); fsol=kk\ff; print fem solutons num=1:1:sdof; store=[num' fsol] lab6 store = extract system dofs for the element compute element matrix assemble into system matrix apply the boundary conditions solve the matrix equation to find nodal displacements print displacements

CONTENT. A. Plane truss (2D) B. Space truss (3D) C. Observed steps of running programming... 02

CONTENT. A. Plane truss (2D) B. Space truss (3D) C. Observed steps of running programming... 02 CONTENT A. Plane truss (2D)... 01 B. Space truss (3D)... 01 C. Observed steps of running programming... 02 D. Details... 06 1. Part 1: Algorithm and program for truss structures by fem... 06 1.1 : Algorithm...