University of Florida/Eml4500/f08 Team Homework 3

Homework 3Edit

Derivation of Elemental Free-body Diagram with respect to the Global Coordinate SystemEdit




  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.




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 RelationshipEdit

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.

Closing the Loop between FEM and Statics: Virtual DisplacementEdit

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.

Infinitesimal displacement (related to virtual displacement)Edit

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 PointsEdit

PQ Line Segment Diagram






Equation for line perpendicular to PQ passing through P:


Determination of (XD,YD)Edit

Line Perpendicular to Point B:



Line Perpendicular to Point C:



Intersection at Point (XD,YD):












3-bar Truss SystemEdit

3-bar Truss System






Convenient Local Node Numbering:

3-bar Truss System Local Node Numbering

3-Bar Truss System

ΣFx = 0
ΣFy = 0
ΣMA = 0 (Trivial)

→ 2 equations, 3 unknowns → Statically Indeterminate

Question: How about MB? (3-D Explanation)


(for A' on line of action of \overline{F}


Back to 3-bar truss:

Node A is in equilibrium:



Ai' = any point on line of action of Fi



MatLab HomeworkEdit

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.

The blue dashed line represents the un-deformed truss, while the solid red line shows the truss after it has been deformed.


Eml4500.f08.wiki1.oatley 20:02, 8 October 2008 (UTC)

Eml4500.f08.wiki1.brannon 01:31, 8 October 2008 (UTC)

Eml4500.f08.wiki1.ambrosio 21:27, 7 October 2008 (UTC)

Eml4500.f08.wiki1.aguilar 20:45, 8 October 2008 (UTC)

Eml4500.f08.wiki1.schaet 19:42, 8 October 2008 (UTC)

Eml4500.f08.wiki1.handy 20:25, 8 October 2008 (UTC)