1. MEGN 424: Computer Aided Engineering Truss Structural Stability & Rigid Body Motion

2. Truss Structures
Two-force members connected by a ball and socket joint (i.e., elements do not transmit bending moments)
Each member of a truss structure is only subject to axial tensile or compressive forces (i.e., a two-force member).
Each truss member has a specific cross sectional area and material properties.
Each truss member can only stretch/contract along the axis of the truss.
When viewing the mesh and results in SWS, the elements are represented by cylinders regardless of the actual cross section.  
A curved structural member is modeled with a number of straight elements.

3. Stability of Truss Structures
Unstable
– rotates freely without resistance (and collapses)
Stable
– deforms with resistance

4. Stability Equations Stability equation for a 2D Truss Structure :
where
e – number of elements
n – number of joints (nodes)

5. 1D Truss Element
Element e 1 2 u1 u2 F1 F2 ke
For a 1D truss element, the element stiffness matrix is:
A node only has 1 degree of freedom (1 DOF).
In a 1D truss analysis two conditions may be applied to a node:
The axial translation may be fixed (specified displacement – may be zero or non-zero)
Apply a load in the axial direction (+ or -)

6. 2D Truss Element
Nodes 2
x y x y
u1y
u2x
u2y
F1x
F1y
F2x
F2y q
u1x ke
node 1
node 2
Node 1
Node 2
4x x x1
The element stiffness matrix is now a 4x4 instead of a 2x2, because each node can now move in the x- or y- direction. In this case each node has 2 DOFs.
In a 2D truss analysis, two conditions may be applied to a node:
The x-translation or y-translation may be fixed (specified displacement – may be zero or non-zero)
Apply a load in the x-direction or y-direction (+ or -)

7. Standard Boundary Condition Symbols
Fixed pin/node constraints:
1) Pin: All Translations fixed
2) Roller: one Translation fixed y
Examples
Tx,Ty Ty x z
Translation fixed in x & y
Translation fixed in y,

8. Standard Boundary Condition Symbols
Fixed boundary constraints:
1)Fully fixed: all boundary
translations
2)Roller: boundary rolls
without friction y
1) Plate bottom
edge fixed x z
Tx,Ty
2) Right side of
sheet on rollers Tx
Tx,Ty,Rz
Note, a fully fixed condition (including Rotation about z-axis) can be applied to a pin/node on a Beam not a Truss

9. The Golden Rules of Finite Elements
Elements must be connected at nodes.
Forces and boundary conditions (fixed translations) can only be applied at nodes.
The application of a nodal force in a DOF that is fixed at the same node has no effect on the displacements and stresses in the system.
EXAMPLE: If the x-translation DOF is fixed and a force is applied in the x-direction at a node, the overriding nodal constraint is the fixed translation, and the applied force will have no effect on displacements or stresses anywhere in the system. However the reaction force at the fixed node will be changed.

10. FE Truss Model – Modeling Errors1 2 3 4 5 6 a) F
Model a) – ERRORS
Force F, cannot be applied
at the mid-side of element 2
2. Element 6 is only connected
to elements 1 and 5. 1 2 3 4 5 6 7 b)
F/2
Model b) - CORRECTED
Force F/2 applied at
nodes 4 and 5.
2. Element 6 is divided into
elements 6 and 7, that are
joined to elements 3 and 4
through the common
node 2.
Notation
Black dots indicate nodes
Small numbers are node IDs
Larger numbers are elements IDs

11. Boundary Condition Sets (BCSs) or Restraints
For the FE equations to have a solution (you must be able to compute [K]-1), the BCS must be valid!
A valid or allowable BCS must prevent all types of rigid body motion.
In 2D problems there are a total of 3 types of rigid body motion which must prevented:
Two in-plane translations (usually in the x-y plane), and
Rotation perpendicular to the plane (usually rotation about the z-direction)
In 3D problems there is a total of 6 types of rigid body motion which must be prevented:
Three translations (along the x, y and z directions), and
three rotations (around the x, y and z directions).

12. Rigid Body Motion
Computers approximate forces as a very small number instead of exactly 0. This will cause rigid body motion even though we don’t expect a force in that direction.
From the 2D Truss Worksheet
Fx applied (approx. 0)
Fy applied

13. Rigid Body Motion ExampleF
R1=F
R2=0
R3=Fy
Fixtures and applied force F are in the x-direction, so we don’t expect displacement in the y- or z-directions.
But computers will approximate the y- and z-forces as a very small number, instead of exactly 0, which will cause rigid body (sometimes called out of plane) motion in y- and z-directions.
Therefore, we must apply fixtures in all three directions to prevent rigid body motion.

14. Example of 2D BCS’s Motions ? NO NO NO => VALID Motions ? NO NO YES
=> NOT VALID!
Force

15. Condition Number Check
When a BCS is valid, the FE linear equations:
[K]{u} = {F} can be solved.
Previously we have said that these equations cannot be solved when det[K] = 0; however, in practice this condition needs to be defined in a more precise normalized manner.
This is because the calculations are usually being performed approximately (by computer) on very large matrices.
We therefore test a normalized quantity called the Condition Number (CN) of [K] which is defined by:
If CN indicated determinant is close to 0 then we conclude that [K]–1 either cannot be computed or the computation is NOT ACCURATE!

16. Accurate & Valid Boundary Condition Sets
Example – Truss bridge with supports at its two ends (at pts B and D) C H D B x y z

17. Alternative BCS BCS 1): x- and y-translation fixed at B and D
Not Accurate / VALID C H D B
Alternative BCS C H D B
BCS 2): y-translation fixed at B and D
Accurate / Not VALID C H D B
BCS 3): y-translation fixed at B and x- and y- translation at D
Accurate / VALID
In reality, the actual Boundary Conditions may be somewhere between BCS 1) and 3), because the rollers may not be completely frictionless (effect of dirt, rust etc.)

18. Distributed Loading Given a load/unit length e.g., 40 lbf/foot
And span length e.g., 15 feet
 total load = 600 lbf
Given a number of elements
e.g., 3 elements
 200 lbf per element!
15 ft

19. Distributed Loading (cont.)
First, consider each element separately…you have 200 lbf per element, so split the force evenly and apply 100 lbf at each node
Now put the elements back together…
100
100
200
100 
100
200
100
100
100
100