FEA aid to design

```FEA as an aid in Design
Andrei lozzi 2014
1. Applying FEA to a fairly complex design can initially overburden us with
information. We therefore need a method of analysing and separating the
various effects that a components or assembly is subjected to, when loaded
by some combination of forces and moments.
2. FEA can help us determine the different paths the load takes from its
application to its reaction, and how the load is divided between the paths.
Subsequently if we need to strengthen or stiffen the part, we can choose
the more effective path to reinforce.
3. Depending on the pattern of the stresses created, indicative of the
component being subjected to tension, compression, bending or torsion, or
a combination, we can change the cross sections to better suit the pattern.
4. Thirdly we can see the effect of changes in shape on the stress distribution.
That is, we can quantitatively assess stress concentration, that principally
occur at concave corners. We can control stress concentration using well
established design techniques.
1
A single load path from the point of application of a load,
to the point where reaction to that load takes place.
If at all possible, the most
to the reaction.
applied to the component.
combination of forces and
moments.
The component is
presumed to be contained
within this envelope
Point or points where the
reaction to the load takes place
2
A sequence of load paths from the point of application of a load,
to the point where reaction to that load takes place.
Here the path is not the shortest, but
is made up of two indirect paths.
The stresses created within those
paths will be some complex
combination due to bending, shear,
torsion and other.
This may make arriving at an
effective overall shape of the
component somewhat difficult to
conceive.
3
Multiple load paths from the point of application of a load, to the
reaction.
In general, the load will be shared
between the paths according to their
relative stiffness.
Reinforcing the less stiff path will have
a lesser outcome than augmenting
the more rigid one.
4
FT
A simple example of parallel load paths.
Here for the spring at left, we can say because it
has fewer coils, of thicker wire and of smaller coil
diameter, will be stiffer than the one at right. Let
the relative stiffness be:
KL = 10 KR
If the applied force F causes an extension of Δ,
the force carried by the left and right springs are:
FR = Δ KR
FL = Δ KL = Δ 10 KR
Where
FT = FL + FR = Δ 11 KR
The fraction of the force carried by either spring is
proportional to the relative stiffness of that
spring.
FL / FT = 10 / 11
FR / FT = 1/ 11
FT
The applied external force is divided between the
two load paths in proportion to the ratio of their
stiffness to the total stiffness.
5
A force is shown applied to an edge of
the ‘angle’ L section at right, and
reproduced in plan view below
l1
l2
l1
l2
The applied force is somehow divided between the
two legs of the L section - l1 & l2 .
What resistance does that force meet ? The aligned leg
- l1 has to bend about the neutral axis shown, it has a
relatively large second moment of area I1, hence it will
be stiff. Leg l2 has a small I2 hence will be relatively
flexible.
6
An Example of reinforcing the more flexible load path
on the stiffness and stress level of a part.
Here a web has been added to the more
flexible of the two legs of this ‘angle’
section. We compare here the stress and
deflection with a similar connection
without the web shown here
vMises stress at the free edge has
increased from 65 to 70 N/mm^2.
Stress have been reduced from 87
to 73 N/mm^2 at the outer corner.
Max deflection has been reduced
from 0.494 to 0.447 mm
7
An Example of reinforcing the more rigid load path on
the stiffness and stress level of a part.
Here a web has been added to the more
rigid of the two legs of this ‘angle’ section.
As before we compare here the stress and
deflection with a similar connection
without this web.
vMises stress is reduced from 87 to
70 N/mm^2 at the outer corner.
Stress at the free edge has
increased from 65 to 28 N/mm^2.
Max deflection has been reduced
from 0.494 to 0.307 mm.
8
We make the following observations:
deflection but by only 10%.
Placing similar sizes webs in a relatively advantageous position reduced
deflection by 40%.
Placing the hold-down bolt holes, close to the most heavily loaded welds,
reduces deflection by a further 10%.
9
In general, to reduce stress concentration,
add material where the stress is high, and
remove material where the stress is low
– very simple.
Examples of the means that may be used to
control stress concentration.
In (a) and (b) above the large fillet radii and
notches in the low stress areas, help to smooth
the force field (stress levels). Adding notched in
the high stress areas has the opposite effect.
At left the additional small holes before and after
the large one, helps to redistribute stress as if a
smoother elliptical hole were present.
10
A real example – This ‘wheel upright’ is connected to the chassis by the 4 bolt holes at
top and the 2 at the bottom. The wheel is supported by 2 bearings in the centre. The
wheel transmits forces and moments through those bearings and a forces come from
the brake calliper. All these loads are then passed on to the chassis. The upright shown
here, is made in two halves in Al alloy, welded together at the mid-plane and is hollow,