FEA as an aid to Design Andrei Lozzi

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FEA as an aid to Design Andrei Lozzi 2017 1. Too Much Information The

FEA as an aid to Design Andrei Lozzi 2017 1. Too Much Information The FEA analysis of a typical part under load can initially overburden us with information. We therefore need to analyse and separate the various effects that loads have on components. 2. Multiple Load Paths FEA can help us separate the different paths that a load can take from where it is applied to where it is resisted, and how the load may be divided between paths. Furthermore which path can be more effectively reinforced to reduce stress and or deflection. 3. Select cross sections. For loads that are predominantly tension, compression, bending or torsion there are well known cross sections that are well suited for each, possibly reducing sizes and mass. 3. Stress Concentration. Wherever there is a change of shape there will be some sort of stress redistribution. There will be increase in stress at concave corners and a reduction at convex corners. We can limit the concentration using established design techniques, but never reduce it to zero. 1

A single load path from the point of application of a load, to the

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 effective load path is a straight line from the load to the reaction. Point or points where the reaction to the load takes place Point where the load is applied to the component. The ‘load’ may be some combination of forces and moments. The component is presumed to be contained within this envelope 2

A sequence of load paths from the point of application of a load, to

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.

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,

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 The applied external force is divided between the two load paths in proportion to the ratio of their stiffness to the total stiffness. FT 5

A force F is shown applied to an edge of the ‘angle’ L section

A force F is shown applied to an edge of the ‘angle’ L section at right, and reproduced in plan view below l 1 F l 2 l 1 l 2 The applied force is somehow divided between the two legs of the L section - l 1 & l 2. What resistance does that force meet ? The aligned leg - l 1 has to bend about the neutral axis shown, it has a relatively large second moment of area I 1, hence it will be stiff. Leg l 2 has a small I 2 hence will be relatively flexible. 6

l 1 An Example of reinforcing the less rigid load path on the stiffness

l 1 An Example of reinforcing the less rigid load path on the stiffness and stress level of a part. F l 2 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 as shown on slide 6. Von Mises 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 There are two loads paths for F to reach the base plate, through l 1, or through l 1, l 2 and the web, to the plate. Obviously the second path is the less stiff. 7

An Example of reinforcing the more rigid load path on the stiffness and stress

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 a web as shown on slide 6. Von Mises stress is reduced from 87 to 70 N/mm^2 at the outer corner. Stress at the free edge has decreased 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: Placing the webs in just about the least advantageous

We make the following observations: Placing the webs in just about the least advantageous position reduce 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%. Your challenge may be to add webs and holes to a point when there will be diminishing return for the amount of work required. To decide what is a better modification and what is not and when to stop, prepare a table of design changes and outcomes. 9

Selecting shapes for a component This is the applied force This is the load

Selecting shapes for a component This is the applied force This is the load path Assuming that we have a single load path through your part (slide 2). you have to invent a shape for the part, with a cross section that generates the least stresses, for the least amount of material. This is the invention process that will test your understanding and creativity, as a designer. If we have two load paths as represented on slide 3 & 4, the vertical path would be subjected to bending and compression while the horizontal one predominantly to bending and torsion. There is a perpendicular component of the applied force that will cause a bending moment, which will reach a maximum at the reaction end. There is a larger component that will cause a compressive - buckling load, with the greatest effect near the centre of the part. 10

Axial torque axial tension bending moment axial compression These components are shown as elements

Axial torque axial tension bending moment axial compression These components are shown as elements of approximately equal crosssectional area. Sections suitable for different loads Tube of max practical diameter bar or any section I beam tube or barrel of practical diameters

Relative merits of common sections, when used to carry bending moments, torsion and other

Relative merits of common sections, when used to carry bending moments, torsion and other loads. I beam U channel Box section An I beam and a box section, made to the same cross-sectional area, can be equally effective in bending. The box will also be better in torsion and in compression than the I beam, but will close off a volume for most other purposes. The channel section will be about 7% less effective than the I beam, in bending about the axis shown, but may be easier to manufacture than the box section, and can provide an open volume for other purposes. But it is relatively ineffective in compression and torsion. 12

A real example – This ‘wheel upright’ is connected to the chassis by the

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, about 1 -2 mm thick. 13

Stress and deflection is shown here. What would you do to principally reduce deflection

Stress and deflection is shown here. What would you do to principally reduce deflection while actually trying to reduce mass ? 14

SOME HINTS - when working on the FEA design problems Search for a reasonable

SOME HINTS - when working on the FEA design problems Search for a reasonable solution: Possibly less than 6 webs, bolt holes closest to most heavily loaded weld , or built into the web, minimize or place no part into bending. Some impractical solutions: An infinity of webs will make a stiff joint but a very impractical one, likewise an infinity of hold down bolt holes. Put your best solution up front, otherwise we will not pick out what we think to be the best and reward you for it. We have to presume that you could not tell the difference. Make table of attributes as you try different features or shapes, the differences between good and bad may not be very large. After a 15% improvement may be very significant 15

In general, to reduce stress concentration, add material where the stress is high, and

In general, to reduce stress concentration, add material where the stress is high, and remove material where the stress is low – very simple, sometimes hard to see how that can be done. 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. 16