Derivation of Elemental Free-body Diagram with respect to the Global Coordinate System[edit | edit source]
axial displacement of element 'e' at local node 'i' axial force of element 'e' at local node 'i'
We want to find the relationship between and , and and .
These relationships can be expressed in the form:
Consider the displacement vector of the local node 1 denoted by .
= axial displacement of node 1 is the prthagonal projection of the displacement vector of node 1 on the axis of element 'e'.
Here we can see that is a 1x1 scalar.
We do this for node 2 as well.
which leads us to:
where this is the equation we set out to derive,
Similarly, (same argument):
Where P is a 2x1 matrix, T is a 2x4 matrix, and f is a 4x1 matrix.
This relationship is the same as saying
Recall the axial Fd relationship:
where k is a 2x2 matrix, q is a 2x1 matrix, and P is a 2x1 matrix.
Goal: We want to have k(e)d(e)=f(e) so "move" Te from right side to the left side by pre multiplying equation by T(e)-1, the inverse of T(e). Unfortunately, T(e) is a rectangular matrix of the size 2x4 and cannot be inverted. Only square matricies can be inverted. To solve this issue, transpose T.
Ans:
where k is on p6.1, k hat is on 12-2, and T is on 12-5.
Justification for (1): PVW see 10-1 for 1st applications of PVW, reduction of global FD relationship.
Remember: Why not solve as follows?
Used mathcad to try to solve K-1 and could not, due to singularity of K., i.e. the determinant of K=0 and thus K is not invertible. Recall that you need to computer to find K-1.
Why? For an unconstrained structure system, there are 3 possible rigid body mostion in 2-D (2 translational and 1 rotational).
HW: Find the eigenvalues of K and make observations about the number of eigenvalues.
Dyanmic eigenvalue problem Kv = λMv
Where K is the stiffness matrix, M is the mass.
Zero evaluation corresponds to zero stored elastic energy which means rigid body motion.
Using the Global Free-body Diagram Relationship[edit | edit source]
Using d3, and d4 in and using boundary conditions to eliminate rows in the global displacement matrix we get:
Note: We really only need to do the computations for rows 1,2,5, and 6 to get F1,F2,F5,and F6 because computation from rows 3 and 4 gives the applied load which is already known.
MATLAB Coding Showing Computations
K=
d=
MatLab Code:
>> K = [-.5625,(-3*sqrt(3))/16;(-3*sqrt(3))/16,-.1875;3.0625,-2.1752;-2.1752,2.6875;-2.5,2.5;2.5,-2.5];
>> d = [4.352;6.1271];
>> f = K*d
f =
-4.4378
-2.5622
0.0003
7.0001
4.4377
-4.4377
This gives us: F=
Closing the Loop between FEM and Statics: Virtual Displacement[edit | edit source]
Two-bar truss system:
Since our two-bar truss system is statically determinant because we can view element 1 and 2 as two force bodies. By statics, the reactions are known and therefore, so are the member forces , and .
We then compute axial displacement degrees of freedom. This is the amount of extension within the bars.
(Node 1 is fixed)
(Node 2 is fixed)
How do we back out the displacement DOFs of node 2 from above results?
Note: The displacement of node 2 is in the direction of vector D.
Statics reason that previous method cannot be done with arcs of a compass
(if alpha is small) (first order)
Infinitesimal displacement (related to virtual displacement)[edit | edit source]
Global Free-Body Diagram showing Infinitesimal Displacements
Find (x,y) coordinates of B and C:
Point B:
Point C:
We now have two unknowns, (XD,YD)
We need the equations for line and
Method for the Determination of the Slope between Two Arbitrary Points[edit | edit source]
PQ Line Segment Diagram
Equation for line perpendicular to PQ passing through P:
Part of this assignment required the team to develop a MATLAB code that plots the deformed and un-deformed configurations of the two-bar truss system solved in class. The code is based on two models that were created by Dr. Vu-Quoc and X.G. Tan.
MATLAB Code
%HW 3%Assignment:%Write a matlab code to plot the initial underformed %configuration of the two-bar truss system solved in class %and then superpose the deformed configuration on the same figure. functiontruss_plotdof=2;%number of dofs per noden_node=3;%number of nodesn_elem=2;%number of elementstotal_dof=2*n_node;%total number of dofs%*******************************************************************% UNDEFORMED BEAM%*******************************************************************%coordinates of undeformed positionposition(:,1)=[0;0];position(:,2)=[2*sqrt(3);2];position(:,3)=[2*sqrt(3)+sqrt(2);2-sqrt(2)];%set up the nodal coordinate arrays for undeformed beamsfori=1:n_nodex(i)=position(1,i);y(i)=position(2,i);end%define elements%element 1node_connect(1,1)=1;node_connect(2,1)=2;%element 2node_connect(1,2)=2;node_connect(2,2)=3;%*******************************************************************% DEFORMED BEAM %*******************************************************************%the deformed beam variables are denoted with a '2'%coordinates of deformed position%(NOTE: node 1 and node 2 remain undeformed)position2(:,1)=[0;0];position2(:,2)=[2*sqrt(3)+4.352;2+6.1271];position2(:,3)=[2*sqrt(3)+sqrt(2);2-sqrt(2)];%set up the nodal coordinate arrays for deformed beamsfori=1:n_nodex2(i)=position2(1,i);y2(i)=position2(2,i);end%define elements%element 1node_connect2(1,1)=1;node_connect2(2,1)=2;%element 2node_connect2(1,2)=2;node_connect2(2,2)=3;%*******************************************************************% CONNECT BEAMS AND PRINT %*******************************************************************fori=1:n_elem;%undeformed beamnode_1=node_connect(1,i);node_2=node_connect(2,i);xx=[x(node_1),x(node_2)];yy=[y(node_1),y(node_2)];%deformed beamnode_1_deformed=node_connect2(1,i);node_2_deformed=node_connect2(2,i);xx_deformed=[x2(node_1_deformed),x2(node_2_deformed)];yy_deformed=[y2(node_1_deformed),y2(node_2_deformed)];%plotholdonplot(xx,yy,'--b')plot(xx_deformed,yy_deformed,'-r')endend
The blue dashed line represents the un-deformed truss, while the solid red line shows the truss after it has been deformed.