University of Florida/Eml4500/f08 Team Homework 3

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Homework 3[edit]

Derivation of Elemental Free-body Diagram with respect to the Global Coordinate System[edit]

HW-3Picture1.jpg


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 .

HW-3Picture2.jpg

= 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]

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 Displacement[edit]

Two-bar truss system:

HW-3Picture3.jpg

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?

HW-3Picture4.jpg
Note: The displacement of node 2 is in the direction of vector D.


Infinitesimal displacement (related to virtual displacement)[edit]

Infinitesimaldisp1.JPG

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]

Pqpicture.JPG

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):

Simplification:

3-bar Truss System[edit]

3bartruss.JPG

3-bar Truss System

Convenient Local Node Numbering:

3bartrusslocalnodes.JPG

3-bar Truss System Local Node Numbering


3-Bar Truss System



Hw3-1.JPG



ΣFx = 0
ΣFy = 0
ΣMA = 0 (Trivial)

→ 2 equations, 3 unknowns → Statically Indeterminate


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

Hw3-2.JPG

(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 Homework[edit]

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.



Plot for truss.GIF

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



Authors[edit]

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)