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The Global Forces are determined by the following matrix
6X1 6X6 6X1
Global Force Global Stiffness Global Displacement
Column Matrix Matrix Column Matrix
|How to solve in alternative ways:|
This is the global diagram. The forces are chosen in the direction parallel to each of the members. Both of the two elements that make up the global picture are purely two force members.
This is the global diagram broken down into its element components. The forces can be chosen along the axis of to make finding the solution easier to solve. By breaking it into its elements one can solve using the Finite Element Analysis method.
The above figure can also be solved using statics. The figure below breaks down the problem for solving it using statics.
Because all forces must equal zero the equations can be formed below. There are two unknown forces so two equations must be formed.
Solve the above equations for F1 and F2 then solve the system of equations
An example of a statically indeterminate problem is shown below.
Before the global matrix can be constructed, you first need to find the displacements in the elements themselves. To do this for each element, the following equation may be used:
4X4 4X1 4X1
where: =Element stiffness matrix for element e
=Element displacement matrix of element e
=Element force matrix
Element Stiffness Matrix:
The element stiffness matrix for the equation above is calculated with the matrix shown (below).
where: Axial stiffness of bar element "e"
and =Director cosines of the axis with respect to the (x,y) coordinates for element e.
Director cosines relate the angle at which the element is on to the traditional x,y axis. These values are needed to calculate the stiffness matrix k.
Model of the bar turss system (continued)
θ(1) = 30°
l(1) = cos(1)(θ) = cos(1)(30°) =
m(1) = sin(1)(θ) = sin(1)(30°) =
k(1)= = [k4 × 4, where and = number of element
Note: this matrix is symetircal with respect to the diagnol
= [k4 × 4
Observe: Absolute values for all the coeficients are the same [k4 × 4, where . All that differs are the positive/negative signs in front of the magnitudes of the coeficients.
Element F-D relation:
Global F-D Relation("free-free" structure)
there are degrees of freedom (6 forces are acting upon the sturcture)
Global Force Displacement Relationship
- Note: The following matrix formulas show that there is a difference between the notation for the stiffness and force matrices , , and , . For example, is the global stiffness matrix, and is the element stiffness matrix.
The matrix form for the example problem in progress is as follows:
|The Shorthand Notation for|
For general applications, the shorthand can be written as follows:
The and subscripts are not arbitrary, however, so written in scalar form;
|The Global Matrix Formula and the Shorthand Notation for|
The matrix formula for the element matrices is written as follows:
|The Element Matrix Formula and the Shorthand Notation for|
is the matrix formula for the following variables in local coordinates.
is the element stiffness matrix.
is the element displacement matrix.
is the element force matrix
In order to express the element matrices in global matrices to solve for the displacements, there is an assembly process.
You must identify the correspondence between the element displacement degrees of freedom and the global displacement degrees of freedom.
For the current example, the global displacement degrees of freedom are:
At the element level:
Then it is necessary to identify the global - element degrees of freedom. In other words, determine the element displacement degrees of freedom that are actually equivalent to one another if the separate elements were to be put back together, as well as which are in turn actually describing a global displacement degree of freedom:
at Global Node 1
at Global Node 2
at Global Node 3
Conceptual Step of Assembly: Topology of K
After calculation of the local stiffness matrices K1 and K2, they can be combined following the conceptual step of assembly method called: The Topology of K as follows:
Within the intersection of the two matrices, the terms that would overlap there are simply added together. (See 9/15/08 for the matrix completed for the current example.)
The resulting global stiffness matrix is:
A sample calculation of the values of K:
|Numerical Values for the global stiffness matrix|
Using the data below, numerical values for the global stiffness matrix was calculated:
First in fractional format:
Then in decimal format:
Step 4: Introduction of Boundary Equations
In this step eliminating known degrees of freedom will reduce the global Force Displacement Relationship. By knowing the displacements for global nodes 1 and 3 are zero because of the boundary conditions expressed in the problem statement. The boundary conditions have been applied to the global stiffness matrix below.
The global stiffness matrix reduces to a 6X2 matrix when applying the boundary equations. The top and bottom two rows are also eliminated due to the principle of virtual work. The Force Displacement relation then becomes the equation shown below.
To solve for the displacement of global node 2. The equation above needs to be multiplied on each side by K-1, and the values for the stiffness matrix and the force matrix inserted. This solves d3 and d4 to be 4.352 and 6.1271 respectively.
|How to find the inverse of the global stiffness matrix|
In order to solve for the displacements of the example, each element is considered:
The displacements are found by solving the 4x4 matrix, after the eliminations discussed previously. This method provides the solutions:
The internal forces for nodes 1 and 3 will be zero because they are fixed points. (See "Boundary Conditions") Therefore, the matrix will reduce to:
Where the known external forces will be plugged in.
To continue, for example, using specifically element 1, the now known displacements from the previous step, can be plugged back in for the variable labels, and the reactions at the supports can be found. This is the last step.
Where the subscripts 1 and 2 denote the reactions at the supports , and the subscripts 3 and 4 denote the internal forces. (As labeled on the free body diagrams for element 1 above.)
The solutions of the reactions and internal forces for element 1 are as follows:
Take note of the fact that each of the internal forces and reactions for the x and y directions will sum to zero, providing an algebraic proof that the element is in equilibrium.
|Proof that the|
For the single truss element 1, there is no moment along the truss (the truss is assumed to be massless with no external moment applied). To prove this we create a coordinate system such that the x-axis runs along the length of the truss and the y-axis is perpendicular to the truss. For the truss to be in equilibrium, the force components from P and from element 2 acting perpendicular to the truss must be equal and opposite otherwise a rotation about point 1 would occur. Because these forces are equal and opposite while acting at the exact same point on element 1, there can therefore be no moment created by either force. Also because these perpendicular forces counteract one another, there is no perpendicular force component at point 1. Next the force component from P, element 2, and element 1's ground support that act along the truss are also in equilibrium. These forces cannot however create a moment along the truss as they are acting along the length of the truss.
|Proof that node 2 is in equilibrium|
Solving a Two-Bar Truss System using MatLab
The same system can be solved using MatLab.
The following code solves for the parameters: "Stiffness matrices, displacements, and reactions."
% Two bar truss example clear all; e = [3 5]; A = [1 2]; P = 7; L=[4 2]; alpha = pi/3; beta = pi/4; nodes = [0, 0; L(1)*cos(pi/2-alpha), L(1)*sin(pi/2-alpha); L(1)*cos(pi/2-alpha)+L(2)*sin(beta),L(1)*sin(pi/2-alpha)-L(2)*cos(beta)]; dof=2*length(nodes); conn=[1,2; 2,3]; lmm = [1, 2, 3, 4; 3, 4, 5, 6]; elems=size(lmm,1); K=zeros(dof); R = zeros(dof,1); debc = [1, 2, 5, 6]; ebcVals = zeros(length(debc),1); %load vector R = zeros(dof,1); R(4) = P; % Assemble global stiffness matrix K=zeros(dof); for i=1:elems lm=lmm(i,:); con=conn(i,:); k_local=e(i)*A(i)/L(i)*[1 -1; -1 1] k=PlaneTrussElement(e(i), A(i), nodes(con,:)) K(lm, lm) = K(lm, lm) + k; end K R % Nodal solution and reactions [d, reactions] = NodalSoln(K, R, debc, ebcVals) results=; for i=1:elems results = [results; PlaneTrussResults(e, A, ... nodes(conn(i,:),:), d(lmm(i,:)))]; end format short g results
Here are the calculaled results from MatLab:
k_local = 0.75 -0.75 -0.75 0.75 k = 0.5625 0.32476 -0.5625 -0.32476 0.32476 0.1875 -0.32476 -0.1875 -0.5625 -0.32476 0.5625 0.32476 -0.32476 -0.1875 0.32476 0.1875
k_local = 5 -5 -5 5 k = 2.5 -2.5 -2.5 2.5 -2.5 2.5 2.5 -2.5 -2.5 2.5 2.5 -2.5 2.5 -2.5 -2.5 2.5 K = 0.5625 0.32476 -0.5325 -0.32476 0 0 0.32476 0.1875 -0.32476 -0.1875 0 0 -0.5625 -0.32476 3.0625 -2.1752 -2.5 2.5 -0.32476 -0.1875 -2.1752 2.6875 2.5 -2.5 0 0 -2.5 2.5 2.5 -2.5 0 0 2.5 -2.5 -2.5 2.5 R = 0 0 0 7 0 0 d = 0 0 4.352 6.1271 0 0 reactions = -4.4378 -2.5622 4.4378 -4.4378 results = 1.7081 5.1244 8.5406 5.1244 17.081 0.6276 1.8828 3.138 1.8828 6.276
It can be observed that the values that were calculated previously are similar to the ones calculated using Matlab.
The two "k_local" that MatLab calculated are our k(1) and k(2) respectively.
The two lowercase k's (k) from MatLab are our local stiffness matrices k(1) and k(2).
The Uppercase k (K) found in the Matlab Results is equivalent to our global stiffness matrix K.
The "R" matrix is the known forces matrix.
The "d" matrix is the displacement matrix.
The "reaction" matrix is for the reaction forces.
Contributing Team Members
--Eml4500.f08.jamama.justin 18:26, 26 September 2008 (UTC)
Cedric Adam--Eml4500.f08.jamama.adam 18:29, 26 September 2008 (UTC)
William Mueller Eml4500.f08.jamama.mueller 19:07, 26 September 2008 (UTC)
Chris Alford--Eml4500.f08.jamama.chris 20:28, 26 September 2008 (UTC)
Megan Alvarez - --EML4500.f08.jamama.megan 21:00, 26 September 2008 (UTC)
Arunas Janulevicius --Eml4500.f08.jamama.jan 22:42, 26 September 2008 (UTC)