# Bearing overload

## Main Question or Discussion Point

I am trying to determine rod bearing load when a 90 foot pound torque starter motor stalls. I have drawn the geometry in the picture below. I am figuring using a 10 tooth starter gear & a 153 tooth flywheel, so I calculate 15.3 x 90 ft lbs = 1377 ft lbs torque at the flywheel. So if I consider the distance traveled by the flywheel at 12 inches from it's center during a transition from 5 degrees BTC to TDC I get 1.04687 inches. During this same transition, the piston moves .00864. To calculate the effective leverage, I divide 1.04687 by .00864 to get 121.165, then I take 121.165 x 1377 lbs at 12 inches to get 166.844 lbs, and divide it by 2.704 square inches or rod bearing area to get 61,703 PSI on a rod bearing. Where did I go wrong? is the load really that high? A soft metal bearing could not endure that load could it?

[URL=http://s924.photobucket.com/user/tnoelle1/media/5c_zpsc411a418.jpg.html][PLAIN]http://i924.photobucket.com/albums/ad88/tnoelle1/5c_zpsc411a418.jpg[/URL][/PLAIN]

## Answers and Replies

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Baluncore
Science Advisor
2019 Award
I think you are wrongly assuming the starter motor is stalled by an infinite compression ratio.
The maximum cylinder pressure = (atmospheric pressure * compression ratio).
That limits the peak bearing pressure, = (peak pressure * piston area).
A “hydraulic lock” results when water gets into the cylinder, being incompressible, something will then fail.

I wasn't limiting the stall to just compression. I think the most common occurrence is due to ignition BTC causing the expanding burning charge to force the piston down while the starter motor is trying to force the piston up. The results of my calculations seem off somewhere however, because even the compressive strength of the rod would seem to be exceeded at 166,844 lbs.

SteamKing
Staff Emeritus
Science Advisor
Homework Helper
I think this is where teeth get stripped from the flywheel.

Ok, let's try this another way, a stock starter motor on a Chevy 350 has 90 ft lbs of torque. For modified high compression engines, and other hard starting engines, higher torque starters are available in 160,180,200,or even 250 ft lb levels. If an engine needs more than the stock 90 ft lb starter, how much load is being put on the bearings by the fully loaded starter?

Baluncore
Science Advisor
2019 Award
351tom said:
...how much load is being put on the bearings by the fully loaded starter?
The obvious theoretical answer is “infinite load”. That is because as the crank passes TDC, it reverses the direction of piston movement, at which point it has an infinite advantage.

There are two things here. The torque on the starter motor and the cylinder area * pressure at TDC. Those forces are opposed through the ring gear, crank and rod.
The load applied to the rod bearings will be the smaller of those two.

You must also consider the load characteristics of the battery, starter cables and switch. They may limit starter current and hence the available torque to below the starter rating.

A starter motor can destroy a motor when hydraulic lock prevents rotation. Repeated pre-ignition can also result in damage.

Disregarding the battery, cables , etc, and assuming the cylinder area * pressure at TDC is just enough to load the starter at 90 ft lbs, are my original calculations correct?

Baluncore
Science Advisor
2019 Award
No. Your calculations are wrong.
You fail to take the crank advantage at the instant before TDC. Instead you consider a broad average of the 5% before TDC.

Ok, would I be fairly close if I took the distance from 0.5 degrees BTC TO TDC? In this case, the flywheel moves 0.10472" and the piston moves 90 millionths of an inch, so the advantage is 1163.5 and multiplied by the 1377 lbs is 1,602,139 lbs on the rod and 592,507 psi on the rod bearings.

Baluncore
Science Advisor
2019 Award
351tom said:
Ok, would I be fairly close if I took the distance from 0.5 degrees BTC TO TDC?
No. In short, your simplistic model and analysis is inapplicable to the situation.
Any attempt to approach a limit without actually getting there involves a false assumption if the mechanism must then pass that limit. Ask yourself why you do not analyse it from 0.25% before to 0.25% after TDC.

You should also consider the expulsion of the bearing oil film, stretching of the main bearing bolts and stretching of the cylinder head bolts. Along with piston pressure, those will also limit the maximum bearing pressure.

The long connecting rod has a cross-section that is less than the crank bearing area. You should expect elastic compression of the connecting rod. If that compression exceeds the yield point, then the rod will bend and so begin the catastrophic failure of the engine. Again, elastic compression will reduce peak bearing pressure.

Your responses keep leading me to very high forces being developed. Would it be safe to say that if a stock 90 ft lb starter is not sufficient to crank the engine, then a higher torque starter is not a solution because at 90 ft lbs, the engine is already incurring damage?

Baluncore
Science Advisor
2019 Award
Not necessarily.

There are two things here. Is the starter motor capable of turning the motor, and how fast can it turn it. Power = torque * RPM. The starter must turn the engine fast enough to start after some oil has drained from worn piston rings. (This is especially true of diesel engines).

During starting, the ignition timing should prevent an advanced spark occurring, so we can ignore pre-ignition.
The greatest starter motor torque will be required somewhere in the last half of the compression stroke.

I define BDC as -90° and TDC as +90°.
The cylinder pressure rises as the reciprocal of the remaining cylinder volume. P1*V1 = P2*V2.
Advantage follows a 1/Cos function. The crank arm lever advantage has a minimum of unity near 0° and increases as it approaches TDC. Starter torque will be a maximum where the combination of those two effects peaks. That peak pressure will occurs close to +60° and will be about 87 psi.

If you work out the pressure * area and then force to the crank, you can apply the crank throw and pinion to ring gear tooth count ratio to compute maximum starter motor torque required to turn the motor.

Baluncore
Science Advisor
2019 Award
Here is a first try at computing torque needed to turn the engine against compression.
I don't trust it yet. It needs to be checked.
Code:
         deg   x       y      h      vol     psi    force  advantage  ft.lb
T.D.C  90.0  0.000   1.750  7.500   4.398  161.7  2032.0  1.633E+16  0.000
80.0  0.304   1.723  7.465   4.833  147.1  1849.1  5.759E+00  3.060
70.0  0.599   1.644  7.363   6.117  116.3  1461.0  2.924E+00  4.763
60.0  0.875   1.516  7.199   8.186   86.9  1091.8  2.000E+00  5.203
50.0  1.125   1.341  6.979  10.939   65.0   817.0  1.556E+00  5.005
40.0  1.341   1.125  6.716  14.245   49.9   627.4  1.305E+00  4.581
30.0  1.516   0.875  6.422  17.949   39.6   497.9  1.155E+00  4.110
20.0  1.644   0.599  6.108  21.886   32.5   408.3  1.064E+00  3.657
10.0  1.723   0.304  5.790  25.893   27.5   345.2  1.015E+00  3.240
0.0  1.750   0.000  5.477  29.817   23.9   299.7  1.000E+00  2.857
-10.0  1.723  -0.304  5.182  33.530   21.2   266.5  1.015E+00  2.502
-20.0  1.644  -0.599  4.911  36.929   19.3   242.0  1.064E+00  2.168
-30.0  1.516  -0.875  4.672  39.940   17.8   223.8  1.155E+00  1.847
-40.0  1.341  -1.125  4.467  42.516   16.7   210.2  1.305E+00  1.535
-50.0  1.125  -1.341  4.298  44.632   15.9   200.2  1.556E+00  1.227
-60.0  0.875  -1.516  4.167  46.276   15.4   193.1  2.000E+00  0.920
-70.0  0.599  -1.644  4.074  47.447   15.0   188.4  2.924E+00  0.614
-80.0  0.304  -1.723  4.019  48.147   14.8   185.6  5.759E+00  0.307
B.D.C -90.0  0.000  -1.750  4.000  48.381   14.7   184.7  1.633E+16  0.000

After seeing those small torque requirement numbers, it's hard to imagine how compression could cause a starter motor to stall. Even using an extreme 14:1 compression ratio, a stock starter could easily handle it;
14:1
deg____x______y______h_____vol___psi_____force___ft. lbs
70 __ 0.599__1.644__ 7.363 __ 5.177__120.0 __1508.0__ 4.92
65__ 0.756 __1.578__ 7.278 __ 6.246__ 111.6__1402.9__ 5.77
60__ 0.875___1.516__7.199 __7.238___81.6 ___1025.4__4.89

11:1
65__ 0.756___1.578__7.278___7.188__98.9___1243.4 __ 5.12

Baluncore
Science Advisor
2019 Award
I still don't believe it. I must have made a mistake somewhere. Your numbers come out close to mine, and I know you did not copy my equations, that's why I did not post them. The conversion from inch_pound to foot_pound appears not to be the problem.

It seems very logical that the peak starter torque should be somewhere in the 60° to 70° range.

Because the starter is a DC motor, the maximum torque is when stalled. Without load it will have a maximum RPM determined by supply voltage. In between those extremes, torque will fall linearly as RPM rises.

Direct injection diesels without turbo chargers can have a compression ratio up to 22:1. The starters used on diesel engines often have an additional gear reduction stage within the starter. Maybe that compensates for their higher compression ratios and permits standard starter components to be used.

256bits
Gold Member
Hey guys. Just a few points to add.

Thermodynamics
PV = C would be for an isothermal process, where the temperature of the gas as it is compressed remains constant.

PV^gamma = C is for an adiabatic process where the compression of the gas is quick enough so that the heat produced does not have time to transfer from gas to the cylinder walls.
For air gamma = 1.4

If you use model the compression as adaibatic rather than isothemal, the maximum cylinder pressure at TDC would be about 25 times that at BDC for a 10:1 compression ratio.

Starter Motor torque.
I do not know how that starter motors are rated but let us assume that maximum torque is at stall minimum velocity=0, and minimum torque=0 at free running maximum velocity.
One can plot these two points on a graph of torque vs velocity( angular ) and join them with a straight line to obtain a simple motor curve that we innocently will assume the motor to follow.

Assuming the starter motor maximum velocity is 1000 rpm as an example and that 90 ft-lbs is required to overcome stiction and all that to initiate turning the petrol engine over, and that the turning friction requires 45 ft-lbs. Matching 45 ft-lbs on the curve and we see that this starter will rotate at 500 rpm while cranking the engine.

If we now replace the starter with a beefier model capabable of twice the starting torque of 180 ft-lbs at stall but still having a free running velocity of 1000rpm, graph that and match the turning torque of 45 ft-lbs, we find a different rotational speed while cranking greater than 500 rpm. We conclude that larger starters are not necessary chosen to overcome the torque of the petrol engine but could be selected to turn it over at a faster rate while cranking.

Multiple cylinder engines
Calculating the torque necessary to overcome one cylinder does not translate into a multiple of the cylinders. It could, but friring and timing issues reduce the multplication. If you have a 4 cylinder engine and all cylinders are arranged so that all compress the air and fire at the same time then the torque to turn that engine over would be the multiple. Most engine are not built that way which adds another complication to multiple engines.

I hope that adds yo the analysis.
So far it has been interesting.
thanks

Chronos
Science Advisor
Gold Member
Consider shock loading, a common cause of bearing failure. Loads are usually averaged in engineering calculations which does not always reflect actual conditions.

Baluncore
Science Advisor
2019 Award
256bits. I agree with all your points.
While starting a multi-cylinder engine, the expanding air after TDC might be expected to cancel the compression on another cylinder at that time. This would be true if not for gas leakage past the rings, or with beta = 1.4 for air that would be a thermal loss during both the compression and expansion strokes.
A compression testing gauge on a good engine gives a pressure that is very close to the compression ratio * atmospheric pressure. That suggests to me that beta is closer to 1.0 during starting, probably because the compressed air is losing heat to the cold cylinder walls faster than the engine is being turned.

Chronos. Yes, shock loading can do funny things. With the exception of failure such as hydraulic lock, I see no likely cause for shock to the connecting rod bearings. Lack of lubrication might be a problem initially. The original fear suggested in this thread was that the infinite advantage at TDC could generate an infinite load on the rod bearings. I think it is now clear that the gas pressure limits that bearing load.

A lot of good points made. I found some real world data on a similar size Ford engine I was troubleshooting a while back. Along with the torque characteristics assumptions of 256bits & the observed pressures by Baluncore, I put together the following plot;
the cranking speed was 185 rpm so if I use the 15.3:1 starter/flywheel ratio, & 90 ft lb starter, that puts the starter rpm at 2831. The engine had a 10.4 compression ratio, so, at 65 degrees it works out to 5.226 ft lbs load at the starter vs. 6.044 ft lbs for a 14:1 compression. The starter rpm at 14:1 compression works out to 2804. That works out to a engine rpm difference of only 2 rpm. If I look at using a 200 ft lb starter, the engine rpm difference is only 6 or 7 rpm. We have not been considering other resistance to turning the engine like drag of the rings or compressing valve springs, but regardless, this shows the difference in starter load due to compression between a 10.4cr engine & a 14cr engine is minimal.

Chronos
Science Advisor
Gold Member
Shock is a concern with every load cycle. A typical bearing endures an enormous number of shock loads over its life cycle. The number of shock loads can be enormous [as in many millions] in a typical reciprocating application. Lubrication is the usual suspect. You need an exotic lubricant to reliably provide efficient lubrication under such conditions. Even the best lubricants can only perform reliably for a few months under heavy loads.

Yes, shock loads-this seems to bring us back to the original premise; that it's not high compression that requires high torque starters, but rather ignition occurring well befofe top dead center.
Back in the 60's & early 70's before electronic ignitions were available, there was no provision to retard the spark at start-up, but the stock engines didn't require all that much advance to begin with, so they were able to get by. Today, however, circle track racers for example, often run 42 degree locked out distributors, but they usually push start the cars & don't flip the ignition on until the engine rpm is well above typical starting speeds. Others, however, who don't have the option of push starting use starters to start their engines. While aftermarket ignition systems often have a high degree of starting retard circuitry built in them, many do not, & it seems there are an ample number of high torque starter motor manufacturers ready to sell them high torque starters to get the job done.
So in the case of ignition well before top dead center, the original calculation of nearly 62,000 PSI (which is probably about 10 times the load normally seen at full power) on the rod bearings could (depending on the exact time the ignition fires) be true!

Baluncore
Science Advisor
2019 Award
Shock is not really a problem with big end bearings. The bearing has a huge contact area with a pressurised oil supply. It is much more likely that a shock will bend the connecting rod than the bearing will be damaged or it's oil film lost.

Gone are the days when cars had a spark advance adjustment on the steering column. If you had to turn the motor with a crank handle you would definitely want it to fire closer to 2° past TDC.

Spraying ether into the air intake of an old diesel makes for an easier cold start. But in modern diesels, ether spray only incites them to run backwards.

These new fangled self starters aren't all they are cranked up to be.

Shock is not really a problem with big end bearings. The bearing has a huge contact area with a pressurised oil supply. It is much more likely that a shock will bend the connecting rod than the bearing will be damaged or it's oil film lost.

Not during cranking, because no oil pressure is yet developed, and the soft babbit layer gets squeezed like a mashed potato.

Baluncore
Science Advisor
2019 Award
“squeezed like a mashed potato” is an emotional exaggeration. “the soft babbit layer”, as you refer to it, is so thin that it can not move sideways. It is only a surface coating on the steel shell. Pressures in the system all come down to bearing area or member cross section. The cross section of the rod is the bottle neck for plastic deformation at low RPM.

The big end bearing has a greater area than the section of the rod. That is so because when running, the high surface velocity of the big end bearing requires a large area oil film. When starting the surface velocity is very low so lubrication is non-critical. The oil pump does not have to provide bearing pressure because the film that supports the load will build up during cranking or immediately the engine starts. The film is the tail of an oil wedge. Oil is deliberately circulated to keep the oil and bearing cool.

Babbitt bearings were replaced with thin shell bearings over half a century ago. That was done to reduce weight and to facilitate maintenance. It also meant that only the inner surface needed to be plated with the much more expensive corrosion resistant low friction bearing material. Extrusion of thin shell bearing material is not an issue.

“squeezed like a mashed potato” is an emotional exaggeration. “the soft babbit layer”, as you refer to it, is so thin that it can not move sideways........................................Extrusion of thin shell bearing material is not an issue.
From the Clevite Engine Bearing failure analysis guide;
For many years, nearly all camshaft bearings were manufactured with a lining of babbitt. Babbitt is a soft
slippery material made up primarily of lead and tin and is quite similar to solder. As a bearing surface layer,
babbitt possesses the desirable properties necessary to survive under adverse conditions such as foreign
particle contamination, misalignment and marginal lubrication on start up.
The trend in modern engines has been toward higher operating temperatures and higher valvetrain loads. Babbitt
is limited in its ability to survive under these conditions due to its relatively low strength. When babbitt cam bearings
are installed under these demanding conditions, the lining may extrude or fatigue. Fatigue can be identified by
craters in the bearing surface where sections of lining material have flaked out.

While this is speaking specifically of cam bearings, another section about rod & main bearings titled "Foreign Particles in Lining" describes the same effects;

DAMAGING ACTION
Dust, dirt, abrasives and/or metallic particles present in the
oil supply embed in the soft babbitt bearing lining, displacing
metal and creating a high-spot.

Pressures in the system all come down to bearing area or member cross section. The cross section of the rod is the bottle neck for plastic deformation at low RPM.
ok, so the rod has a smaller cross section, but the bearing lining material has much less compressive strength
When starting the surface velocity is very low so lubrication is non-critical.
This would only be true in an extremely lightly loaded situation. Now add the overload condition as In the case of using the starter to force the piston up against the expanding combustion gases (as when a high amount of static timing is used).
From;
http://machinedesign.com/bearings/detect-oil-film-bearing-failure
Generally, boundary lubrication occurs when the shaft is running at speeds less than the minimum speed required for a full oil-film development. This can happen, for example, during starts and stops. Boundary lubrication permits metal-to-metal contact and this is when smaller particulates will embed into the babbitt or abrade the babbitt material.

The oil pump does not have to provide bearing pressure because the film that supports the load will build up during cranking or immediately the engine starts.
Again, this would only be true if the ignition were retarded enough at start up to prevent the types of loads that tax the stock starter motors, otherwise, the load has already occurred, then later during cranking, or ever later after start up the film will eventually build up.