Compare FEA&Formulas (Simple)
The whole point of using
formulas is to be able to predict the behaviour of an object. We could
use simple formulas for simple shapes or we can turn to Finite Element
Analysis (FEA) for complex shapes. Here, we applyboth methods to a
simple cantilever to see how they compare to the real thing.
Lecture Notes: empty.pdf empty.one
Lecture Video: Empty
Beam BendingStandard formulas have been derived for common arrangments of loaded beams. See Beam Bending tables here. An example is shown at right. For a simple cantilever beam, where a weight W is applied at the end of the cantilever of length L; Maximum Bending Moment = WL Therefore, from the bending stress theory; where; σ is the bending stress
(Ivonaff uses fb)
Maximum Deflection = WL3/(3EI) which occurs at the end of the beam. |
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Predicting Stress and Deflection
In this exercise we will use two methods to predict the behaviour of a cantilever beam made of aluminium.
The two methods are:
1. Empirical formulas. (As listed above). These can be reviewed in MEM30006A Stresses.
2. Finite Element Analysis. A computer calculated approximation.
Both the maximum stress and the maximum deflection will be compared.
In class..
Calculations:
How to calculate I
- by formula
- by AutoCad
- by Inventor
Setting up formulas in Excel.
Adding
Forces in Excel (Mathematical addition of
forces)
SIMPLY SUPPORTED BEAM
This steel beam is loaded with weights and the dimensions measured.
Note that this is a truss beam turned sideways - in the weak direction.
The material is mild steel (hot dip galvanised).
You will need to model this in INVENTOR, so take all the necessary dimensions to build the frame. Take care when measuring the flat bar on each side since these are doing all the work (use a micrometer to get accurate measurement).
Calculating a maximum safe load
Flat depth: h = 19.3, 19.35, 19.67, 19.25. 19.25, 19.25 Average = 19.345
Flat breadth: b = 6.78, 6.875, 6.5, 6.875, 6.83 Average = 6.772
Beam length: L = 2167
Beam width: BW = 140
Second moment of Area: Ixx = bh3/12 = 6.772*19.345^3/12 = 4085.5 mm4
Maximum allowable stress: 100 MPa
Maximum allowable bending moment from bending stress equation;
where;
σ is the bending stress = 100
MPa
M - the moment about
the neutral axis = ?
y -
the distance to the neutral axis = 19.345/2
= 9.6725 mm
Ix -
the second
moment of area about the neutral axis x = 4085.5
mm4
This is for one flat, but we have two flats, so total = 4085.5 * 2 = 8171 mm4
Re-arrange to find max allowable bending moment M
M = σ Ix / y = 100*8171/9.6725 = 84476 Nmm
Max Bending Moment: M = WL/4
so W = 4M/L = 4*84476/2167 = 156 N (about 16 kg)
Maximum load applied at the end of the beam is 16 kg.
Weight of beam is about 7kg, so we can add about 10kg.
You must use the actual dimensions, not these!