MATERIAL REDUCTION & SYMMETRY PLANES

Transcription

1 MATERIAL REDUCTION & SYMMETRY PLANES ISSUES IN SIMULATING A CONNECTING ROD

2 LIMITATIONS ON THE ANALYSES Recall from previous lectures that, in static stress analysis that we are subjecting the component to constraints and loads in such a way that the displacements, strains and stresses do not go beyond the linear scope. The engineer is interested in ensuring that the material provides the required service without failing. Keeping within the linear scope ensures that the assumptions of the governing equations remain valid. If these are exceeded, the results, no matter how beautiful the graphics you present, are invalid.

3 THE THREE QUANTITIES Recall that the Finite Element Analysis package already KNOWS the governing equations. It is however YOUR responsibility that the conditions stipulated by those equations remain valid to ensure reliability of your results. You specify constraints, loading and other conditions including contacts. You are looking for three things: Displacement vector, Strain tensor, Stress Tensor. The results provide you with these if you know where to look.

4 EQUIVALENT STRESSES & STRAINS Tensors are characterized by three scalars: the Principal invariants. Once these are known, any other information or components in any reference frame is a matter of easy computation; the heavy lifting is already done. For ductile materials such as steel, aluminium, etc, the second principal invariant of the deviatoric stress and strains, called Von-Mises equivalent stresses and strains are of pivotal importance.

5 COMPUTATION FORMULAE Stress as well as strains are second-order tensors; Here are the computation formulae for the principal Invariants of any second-order tensor T J 1 = tr T J 2 = tr T c = 1 2 tr2 T tr T 2 J 3 = det T Now the Von Mises stress, T VM = 3 J 2 is concerned about the second invariant of the deviatoric stress tensor. This tensor is traceless. Hence, tr 2 T = 0, so that only the second term matters. Next slide shows the contents of a Mathematica file that does this computation for Cartesian coordinates:

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7 VON MISES From here we can see that 3 times the second Invariant gives us, σ 2 x + σ 2 y + σ 2 z σ x σ y σ y σ z σ z σ x σ x σ y + 3τ 2 xy + 3τ 2 2 yz + 3τ zx The formula for the equivalent strain is similar.

8 We are going to examine the behavior of the shown connecting rod to the service condition of operation. Notice that the firing position of the piston sometimes puts transverse resultant forces on the connecting rod

9 PLANES OF SYMMETRY Our assumption here is that all forces acting on this component are restricted to the x-y plane. And, since this plane is also a plane of geometric symmetry, we can save computational cost by analyzing only half of the body since everything is repeated as a mirror image around that plane of symmetry. We therefore cut the body into two along that plane using the drop down Split bodies from the modify drop down, and we will proceed to hide half the body. Furthermore, we shall cut the big pin holder into two because we are not presently interested in that part of the connecting rod.

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11 SYMMETRY CONSTRAINTS The symmetry constraint here is that there will be no displacement in the z-direction. The holding constraint for this stydy is that the big pin holder is fixed in space. We will apply forces 200 N longitudinally downwards to simulate the firing push on the rod while we apply the same force laterally at the small pin.

12 STRAINS

13 STRESSES

14 DISPLACEMENTS

15 TWO MORE ISSUES From the results we have, it appears that the connecting rod as designed is OK. You can see that the large part of the connecting rod has a hole. How did the original designers of this part arrive at that decision? Suppose we had made the connecting rod solid from the start, would there be a clear indicator that a hole may be a good way to reduce material usage and cost? How can we know unless we subject a whole connecting rod without this hole to the same stresses and loading.

16 SAFETY FACTOR REPORT The minimum safety factor in the full connecting rod is 3.65 compared with 3.28 in the case where we had a material reduction How do we decide if this extra safety factor is worth the cast? We will have to compare it to the cost of manufacture and material input. There are instances in which the cost of reducing the material input can be completely lost in manufacturing costs. There is always a tradeoff and this aspect is not an exact science requiring feedbacks from the several sections of the industry and time.

17 THERMAL STRESSES Once temperature effects are large enough for us to consider the stresses induced by them, there are major changes in the governing equations. The simplest are the quasi static thermoelastic equations that make a number of assumptions including that of linearity and material isotropy. (See Boley & Wiener, Theory of Thermal Stresses) The number of unknowns increases to include the scalar temperature θ distribution in the material and on its boundary. The additional equation needed for this extra variable, temperature θ is the steady state equation of heat conduction: div grad θ + 1 k Q = 0 where Q is the rate of heat flux (assumed known).

18 THERMAL STRAINS The stress strain relations can remain unchanged provided we add the spherical thermal strain tensor to the strain computation: In this case, the total strain E t, the sum of strains caused by the applied structural load and that caused by thermal expansion now becomes, E t = E + αθi = 1 2 grad u + gradt u + αθi if we ignore higher order terms. For the case of infinitesimal strains α is the coefficient of linear expansion, and the temperature function is a defined as a field in the Euclidean point space. (This simply means that it has a value for each defined point in the space).

19 THERMAL STRESSES The stress strain relation, T = λi tre + 2μE, can now be written in terms of total strain, as T = λi tr E t αθi + 2μ E t αθi = [λ tre t 3λ + 2μ αθ]i + 2μE t This means heating, with the prevention strain, will cause stresses in the body: E t = 0 T = [ 3λ + 2μ αθ]i leading to a spherical stress tensor. A simple example of this is shown below:

20 THERMAL STRESSES It can happen that the stresses building up in a member that has been prevented from expanding causes serious damage. For example, consider a short bar that has been heated but not allowed to expand. The prevention of that expansion creates the stress we have calculated above. This build-up of stresses can have disastrous consequences The picture below shows rail tracks in the USA (Wikipedia) that became warped and caused many derailments and deaths. That is why rail tracks are in bits to make room for expansion.

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22 STRAIN COMPLETELY THERMAL What happens if in a thermoelastic situation, the total stresses equal to the thermal stresses? Will there still be any stresses in the system? In the equation, T = λi tr E t αθi + 2μ E t αθi E t = αθi This leads to T = 0. When the heated member is allowed to expand freely, there are no stresses as expected.

23 SOLVING THERMAL STRESS PROBLEMS Again, just like the elastostatics problems we have been looking at, the thermal stresses governing equations are already well known to your FEA solver. You duty as usual, is to ensure you can specify the constraints and boundary conditions correctly and can interpret the results appropriately. We will look at the same connecting rod problem again as an illustration of this problem and some of the peculiarities that arise:

24 THERMAL STRESSES IN THE CONNECTING ROD Furuhamma, S, Oya, Y and Sasaki, H, Temperature measurements of the connecting Rod, Piston pin and crank bearing pin of an automobile gasoline engine, JSME Vol 9, No 33, They report temperatures ranging from 115 to 140 at the small pin at rotational speeds of 1600 to 4500 rpm. 80 to 100 at the big pin in the crankshaft. Beginning with our now familiar sketch, we are in a position to simulate the simple connecting rod using the findings of Furuhamma et. al. as our guide:

25 THERMAL STRESSES IN THE CONNECTING ROD

26 SPLIT THE SINGLE BODY AT THE BIG PIN We extrude the three bodies and create a joining rather than a new body so that this figure gives us a single body. We take a z-x cutting plane and use the Modify menu to Split into two bodies after which we turn off the head of the connecting rod to obtain:

27 CONSTRAINTS & THERMAL LOADS We can now apply the constraints and thermal loads. In reality, we know that the thermal load here is sinusoidal as there is a firing once in 720 degrees of crank rotation. However, from the experiments quoted above, it is a good approximation to assume that the temperature at the small pin is about 140C and larger pin 90C. We assume an ambient of 60C in the engine. We fix the big pin end and simulate

28 THERMAL STRESS RESULTS

29 EQUIVALENT STRESS DISTRIBUTION

30 EQUIVALENT STRAINS

31 TOTAL DISPLACEMENT

32 BRAKE DISK ROTOR We will now work through the brake disk rotor simulation contained in the Autodesk example files. Samples section of your

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34 THERMAL STRESSES In your simulation, select Thermal stresses. Unlike the Autodesk instructions, when you edit the units in this study, select SI units rather than than the suggested US Units. Proceed to simply the model if you have not already done this in the model environment. Material for this simulation is Iron, Gray Cast ASTM A48 Grade 20

35 SIMPLIFYING THE MODEL In simplify, select Split Body and here use the XY plane from the Origin inside the browser to select the XY plane as shown. This gives you two bodies. Hide one of them and obtain the next body that you will split using the ZX plane as splitting tool:

36 SPLIT

37 SIMPLIFYING THE MODEL In simplify, select Split Body and here use the XY plane from the Origin inside the browser to select the XY plane as shown. This gives you two bodies. Hide one of them and obtain the next body that you will split using the ZX plane as splitting tool:

38 SIMPLIFIED BODY As a result of the bodies that have been turned off, the simplified body shown in the adjoining picture results. We are now able to apply constraints and loads on our way to simulating the thermal stresses

39 STRUCTURAL CONSTRAINTS Fixed all the structural constraints by preventing motion in the directions perpendicular to each cut surface. For the cut surface elements perpendicular to the x- direction, activate only x- movements as fixed. Do the same for the other surfaces. It is essential you get this right!

40 THERMAL LOADING Now we apply thermal loads just the same way as we have been applying structural loads: Set the type to applied temperature and specify 330K to the bearing end and 530K to the mating surface for the brake pad.

41 MESHING In Manage reduce the model based size from 10% to 2% for a finer mesh while allowing for mesh scaling per part. If the precheck is still not giving you the gree light to go on at this stage, try an automatic contact for the simulation and you can go ahead and simulate in the cloud.