What force is preventing car wheel bolts from being removed?

[lightarrow]

In my personal experience, for what it can count, the concept of "moment of a force" was taught to us at school, the concept of "couple of forces" only at university. The concept of "torque" its'not much used in Italy.

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[Merlin3189]

I started a response to the OP, but it went off in a math-linguistic direction while I was still drawing my diagrams of wheel bolts!

What bothered me about the original post, was the F arrow - not because I didn't think there should be a force there in that direction, nor because I thought there should be a torque or couple or moment. But friction always opposes movement between two objects in contact and I couldn't see any relative movement at that point.
I can see that the downward force on the wrench, applied to the bolt, is trying to move the bolt down. But I thought that force would be opposed, upto more than anything you are likely to apply, by the vertical reaction of the hub. Even if the hub did rotate, say because the wheel slipped, the bolt and its housing would move together. The force through the centre of the bolt perpendicular to the hub radius would still be reaction, not friction.

But, of course, it is friction that is stopping the bolt from turning in its housing. So we need to show where this friction occurs.

Despite previous discussion I've shown friction as a number of distinct arrows. It should be understood that friction occurs over the whole of the mating surfaces, so we could have an infinite number of little arrows, all pointing perpendicular to the radius at their point of action. Their sense will, as always with friction, be opposite to the direction of movement, or attempted movement until friction is overcome.

When we have linear (attempted) movement between two surfaces, all the little arrows point in the same direction - again opposite to the (attempted) movement. So the sum of them all is a single large arrow in that direction.
When the (attempted) movement is rotation, every little arrow has an opposite arrow on the other side of the disc, so the sum of all the little arrows is zero! There is no net force in any direction.

But all these little forces DO cause a net torque, moment, couple, or whatever you want to call it, about the centre of rotation.
Each little force multiplied by its radius give a turning force in the same direction, so these sum to a non-zero total.

As the previous discussion has said, we can't represent this torque with a single force, but we can represent it with a pair of equal and opposite forces acting at equal and opposite distances from the centre of rotation, ie a couple. (As shown on the left part of the diagram.)

Calculating (the limiting value of) this frictional torque is not quite so easy. I'll leave that* to the excellent mathematicians who've been arguing above.
But we shouldn't need to calculate: If we "torque" it up to 120 Nm when we fit the bolts, presumably that's about the torque we should need to reverse the process.

Here I've shown only the friction between the bolt head and the wheel/hub, but there will also be frictional torque on the threads. Again, since friction is always reactive, this torque will always add to the total frictional torque.

*(Literally on the back of an envelope, I got it to be $T = \frac {2} {3} \frac {(R^3 - R_0^3)} {(R^2 - R_0^2)} μ F_N$
where $F_N$ is the normal force between the mating surfaces - here the bolt head and the wheel/hub
$μ$ the coefficient of friction between these surfaces
$R$ is half the outside diameter of the bolt head and $R_0$ half the inside diameter of the bolt head
and T does not include friction elsewhere, such as between the bolt threads and the hub.. )

BTW

It's an active topic of discussion in the moderator's forum.

It was a bit rhetorical. I thought most of the participants would have read it. I intended to suggest that some should perhaps bear it in mind when thinking about their contributions here.

 

[A.T.]

When we have linear (attempted) movement between two surfaces, all the little arrows point in the same direction - again opposite to the (attempted) movement. So the sum of them all is a single large arrow in that direction.

If the arrows represent local friction forces, then they cannot point in the same direction, because friction is always parallel to the contact surface

When the (attempted) movement is rotation, every little arrow has an opposite arrow on the other side of the disc, so the sum of all the little arrows is zero! There is no net force in any direction.

Yes, but In the given case the friction on opposite sides don't have to be equal but opposite. They can produce a net torque and a net force.

 

[Merlin3189]

Disagree with first point - parallel vectors must point in the same direction, else they're not parallel.

Aggree with second point - I just tried to distinguish between the linear and rotational cases. Yes you could also have a mixture.

 

[A.T.]

Disagree with first point - parallel vectors must point in the same direction, else they're not parallel.

The point was that the friction forces are tangential to the curved contact surface, so they cannot all "point in the same direction".

 

[russ_watters]

The point was that the friction forces are tangential to the curved contact surface, so they cannot all "point in the same direction".

They don't have to be, and in this case they are not. @Merlin3189 correctly broke the problem apart into the linear force and the torque (a bunch of little tangential forces). Adding them together yields forces not tangential to the curves because the linear force is mostly not tangential. Or flipped over; the only places where the friction forces are tangential are along the horizontal line perpendicular to the vertical force...because that's the only place the vertical force is tangential to the curves.

Anyway, his explanation is why I'm opposed to showing the torque (moment) as anything but a torque. Showing it as a collection of linear forces adds assumptions and maybe even calculus for no reason...and opportunities for unnecessary confusion.

 

[A.T.]

the only places where the friction forces are tangential are along the horizontal line perpendicular to the vertical force

Friction forces are always tangential to the contact surface, per definition.

 

[russ_watters]

Friction forces are always tangential to the contact surface, per definition.

What definition? Are you using "tangential" in place of "parallel"? Please explain, because what you are saying does not make sense.

 

[A.T.]

What definition?

Friction is the component of the contact force that is tangential to the contact surface.

 

[russ_watters]

Friction is the component of the contact force that is tangential to the contact surface.

Again: are you sure you don't mean "parallel"? Surfaces are, by definition, flat, so they don't have tangents.

...and they also aren't "a component of the contact force"; they are perpendicular to the contact force.

 

[jbriggs444]

Again: are you sure you don't mean "parallel"? Surfaces are, by definition, flat, so they don't have tangents.

...and they also aren't "a component of the contact force"; they are perpendicular to the contact force.

The "contact force" can include normal components. The pressure of a book on a table or the tension of a drop of glue hanging from the ceiling.

 

[russ_watters]

The "contact force" can include normal components. The pressure of a book on a table or the tension of a drop of glue hanging from the ceiling.

Ok, I was thinking from the usage that the "contact force" was only the normal component. And then friction is a separate parallel force. But if both are combined into a single resultant "contact force", ok...

...with that word usage issue aside, then...not sure where that leaves us. Still not sure about "parallel" vs "tangential"...

 

[PeterDonis]

Surfaces are, by definition, flat, so they don't have tangents.

First, surfaces don't have to be flat--what about the surface of a ball?

Second, flat surfaces do have tangents; they just happen to be parallel to the surface. So a simple solution to the conundrum you have set yourself here is to just interpret "tangential" to mean "parallel" for flat surfaces (which, as I understand it, is exactly what mathematicians in fact do).

 

[russ_watters]

First, surfaces don't have to be flat--what about the surface of a ball?

Second, flat surfaces do have tangents; they just happen to be parallel to the surface. So a simple solution to the conundrum you have set yourself here is to just interpret "tangential" to mean "parallel" for flat surfaces (which, as I understand it, is exactly what mathematicians in fact do).

Ok, so parallel = tangential in the case of a flat surface, which we have here.

So then a linear applied force is opposed by a linear friction force, parallel to the surface of the nut, right? And:

If the arrows represent local friction forces, then they cannot point in the same direction, because friction is always parallel to the contact surface.

In the case of a linear force applied to the nut, the elements of friction force are parallel to each other and pointed in the opposite direction (same direction as each other). Right?

 

[PeterDonis]

a linear applied force is opposed by a linear friction force, parallel to the surface of the nut, right?

It's a little complicated because there is not a single flat "surface of the nut"; this case is not as simple as the case of, say, pushing a heavy block along a rough floor.

First, the force exerted by the wrench socket on the hex head of the nut that causes the nut to turn is not parallel to the side of the hex head that a given point on the wrench is in contact with. If it were, the side of the hex head would just slide along itself, not rotate. The force exerted by the wrench at a given point on the hex head is at an angle to that side of the hex head.

Second, the friction force that opposes the motion of the nut is not exerted along the hex head; it's exerted along the threads. The thread surface is a helix, not a flat plane. The force exerted by the wrench on the hex head has to be transmitted through the nut to get the nut as a whole to move; and the friction along the threads opposes that motion. At a given point on the threads, the motion of the nut and the friction force are both collinear (and in opposite directions, yes), but the line along which they act is not in the same plane as the plane of rotation of the hex head; it is tilted at an angle that depends on the thread pitch. This line is tangent to the surface of the thread at the chosen point, but since the thread is not a flat plane, the word "parallel" is not really appropriate in this case.

 

[SWB123]

Let's say we want to change a wheel in a car. We want to remove bolts fastening the wheel using this tool:
View attachment 230237
I have also drawn a diagram of the forces in operation:
View attachment 230238

Now, from experience, I can say that the point of rotation of the wrench will be the blue point. Now, trying to determine the torque relative to that point leaves us with a net torque anticlockwise (friction gets canceled out).

Yet, we know that removing these bolts requires some effort. Therefore, there either are more forces in action or the axis of rotation I chose is incorrect (or both)

What am I missing?

If this is a 'real world' problem and not a sterile academic assignment then you are probably missing the fact that corrosion has actually fused the nut to the bolt; and, until you can apply enough force to actually shear the metallic bond then no other forces are going to come into play.

 

[Joseph M. Zias]

Just a somewhat humorous note. One time I had to remove the lug nuts to change a flat tire. I put the now standard right angled lug wrench on the lug and could not move the wrench. I ended up standing on it (my 170 lb) and with a flexing jump snapped the wrench into two pieces. Now those were tight lug nuts.

 

[AZFIREBALL]

Point of clarification:
As shown in the above image, when the lug nut is tightened there is a frictional interface between the lug nut and the lug bolt threads. The tighter the nut the greater the friction between the two.

The lug nut should be tightened just enough to take full advantage of this frictional force without over-stressing the lug bolt’s tensile properties. This is achieved by applying the proper tightening torque using the proper wrench for the job as per the car’s specifications.Also note the bevel on the end of the lug nut. It is designed to fit into a corresponding taper in the wheel rim. This causes the wheel to be centered on the lug bolts and removes any sloppiness (play) between the wheel rim and the hub. There must be some sloppiness to allow the wheel to be fitted over the multiple lug bolts. However, this sloppiness can not be allowed in the working interface between the wheel rim and hub because, under the right conditions, any slight movement between the two will ,over time, loosen the lug nut. Note: Most cars have right hand threaded lug bolts fitted to one side of the vehicle and left hand threaded lugs fitted to the other side to help prevent loosening of the lug nuts. As long as the lug nut is clamping the rim to the hub (it is tight), there is also friction at the rim-lug nut-hub interface (On the tapered portion and at the very end of the lug nut).

So, to remove the lug nut, one must therefore overcome the frictional forces on the threads and between the lug nut-rim tapered interface using a moment or torque. (Foot-pounds).
With corrosion these forces can be immense.

 

[berkeman]

Most cars have right hand threaded lug bolts fitted to one side of the vehicle and left hand threaded lugs fitted to the other side to help prevent loosening of the lug nuts.

Is that true? I've never had a car like that.

 

[AZFIREBALL]

I may have been stretching it a little when I said “most”. I was thinking US cars in the 50’s and 60’s.

 

[berkeman]

I was thinking US cars in the 50’s and 60’s.

Ah, interesting. I wonder how many lug bolts had to get stripped before they standardized on the same threads on both sides of the car.

I know the countershaft sprocket bolt on most motorcycles has left-hand threads (for the reason you mention). But usually the folks who work on motorcycles know to expect that.

 

[rude man]

I've always wondered how those little cotter pins can guarantee a wheel in place. Seems to me a loose wheel, or a misaligned hub grinding a pin, could easily pop it off. I keep getting reassured that such friction torque cannot exist.

 

[AZFIREBALL]

Normally a cotter-pin is not used on wheel lug bolts. They ARE used however to insure the large nut holding the front wheel bearings onto the spindle do not come off.

The cotter-pin is a safety device that prevents the bearing ‘jam’/preload nut from coming completely off the spindle should wear or disintegration of the wheel bearings take place. Once the nut becomes loose there is little or no torque force available to unscrew the nut farther or cause shearing of the cotter-pin by the nut. Plus, there is a large/thick washer, keyed to the spindle with a tab, to prevent rotation, that isolates the nut from the outer bearing race rotation, should it occur.

 

[Tom.G]

See pg 5 of:
https://www.ellsworth.com/globalass...g-brochure-molykote-lubrication-solutions.pdf

ppg 10-12 of:
https://www.fastenal.com/content/feds/pdf/Article - Bolted Joint Design.pdf

 

[AZFIREBALL]

Tom.G:

The links you supplied provides all the info. one would need to understand the relationship between torque and bolt tightening.

One key point to consider in applying torque to a fastener using a given specification
is weather the torque is to be applied to a DRY or WET connection.

It is plain to see that applying the torque specified for a DRY (none lubricated bolt and nut) connection to a WET (lubricated bolt and nut) connection will result in too much torque being applied, thus possibly elongating (or breaking) the bolt or distorting the threads.

I think the torque spec. for tightening the lug bolts on a automobile are generally specified for a DRY connection.

To clarify, applying 50 foot-pounds of torque to a WET connection does not stress the bolt the same as applying 50 foot-pounds to a DRY connection because of the difference in the coefficient of frictional between a WET and DRY connection. In a DRY connection, a great deal of the torque applied is used in overcoming the high frictional component.

 

[Tom.G]

To clarify, applying 50 foot-pounds of torque to a WET connection does not stress the bolt the same as applying 50 foot-pounds to a DRY connection because of the difference in the coefficient of frictional between a WET and DRY connection.

If you are talking about twisting the bolt to failure, I agree. If you are talking about elongating the bolt to failure by applying too much tension, then the opposite is true; i.e. for the same torque, the lower friction of a wet thread will allow more nut travel along the threads than a dry thread will.

Two different things to worry about!

 

[AZFIREBALL]

for the same torque, the lower friction of a wet thread will allow more nut travel along the threads than a dry thread will.

...thereby tending to elongate the bolt.

 

[Dadface]

I think there must be a certain amount of bolt elongation (and compression of the system being bolted) in order to get a adequate clamping. The elastic /mechanical properties of the system parts should be such that they can withstand the optimum forces required.

 

[AZFIREBALL]

Dadface:

You are correct. Generally speaking, to get maximum benefit from a bolted joint the tightening torque should be that which places the tensional stress in the bolt just below the Yield Point of the bolt’s material. Any tighter than that causes the bolt’s thread root diameter tensional stresses to enter, what is known as the ‘plastic’ region, where the bolt becomes permanently elongated (distorted). Another consideration is the clamping system (all the bolts together) must be robust enough that no one element should never experience loads that causes it to exceed its Yield Point during the service life of the joint. Additionally, to reduce what is called ‘Fatigue Failures”, the stresses induced by the assembly process (preload on the bolt) should always be greater than the stresses it will see during normal operation.

 

[Mister T]

I'm not sure where this thread is going. Is there something that's not as obvious as it seems to me? An additional tangential force on the nut could have been added to the original diagram and that would have 'explained' everything, I think.

The original diagram is a free-body diagram of the wrench. But the thread title is a question about the bolt. Wondering about forces on one object while referring to a free-body diagram of a different object can be a source a confusion.

 

[lightarrow]

Just a somewhat humorous note. One time I had to remove the lug nuts to change a flat tire. I put the now standard right angled lug wrench on the lug and could not move the wrench. I ended up standing on it (my 170 lb) and with a flexing jump snapped the wrench into two pieces. Now those were tight lug nuts.

I would rather say that was a very economic wrench
Last time I had to use that ("cross like") wrench was 3 months ago on highway. To remove a bolt I had to jump on it and it suddenly rotated > 90º kicking painfully on my leg. .. Next time I'll use motocross shin guards

 

[sophiecentaur]

I would rather say that was a very economic wrench
Last time I had to use that ("cross like") wrench was 3 months ago on highway. To remove a bolt I had to jump on it and it suddenly rotated > 90º kicking painfully on my leg. .. Next time I'll use motocross shin guards

I think that's the problem when we aren't constantly playing with our vehicles - as when we were 'boy racers'. Joints dry out and a tiny amount of corrosion can change cause a lot of stiction.
I must say, tyre shops these days are well enough behaved when it comes to torquing up the nuts correctly. Way back they just used the pneumatic wrench and ZZZZZZZped up the nuts as hard as they'd go.
I was impressed to find that the L wrench, supplied with my Landrover actually has a telescopic extending section to increase the torque when needed. There's posh.

 

[Tom Rauji]

I'm a "gear head" and have been for over 50 years. Let me make a few points:

Wheels are either hub centric where the hub locates the wheel, or lug centric where the lug locates the wheel.

While the lug nut or bolt is held in place by friction, most of the locking is caused by distortion of the fastener. I actually lubricate fasteners on my wheels, like I do any fastener with a torque specification. My present drag race car puts at least 16,000 ft-lb total on the rear axles. None of my vehicles ever loosen a lug. They do not loosen connecting rod bolts either, even though the rod bolts are pre-lubed in assembly and bathed in oil. As long as the fastener is always under tension or "stretch", it won't back out.

So while friction holds it, the primary concern in a non-locked fastener is not allowing the fastener to reach zero tension.

By the way, I'm very selective with rear wheels in my drag car. I won't use a wheel with a standard tapered lug centric nut (with are all standard thread direction and always have been on all of my vehicles since a 1950's Plymouth I owned). I use shank type nuts with thick wheels that close fit the nut shanks. This keeps most of the force as shear right at the hub surface. A regular taper nut (lug centric) spaced out away from the hub would greatly increase bending on the stud and I'd likely have failures.

I disagree with the notion friction is the critical parameter for locking the fastener. The critical parameter is the fastener stretch and elasticity, which can never be allowed to go to zero. While friction holds it, it is the stretch that prevents loosening. I'll continue to lube and properly torque my fasteners, because I want proper stretch.

 

[sophiecentaur]

I'll continue to lube and properly torque my fasteners

I guess that the instructions are not aimed at someone doing the job 'properly'. If there is a chance of a nut being under-tightened then a bit of sticktion under a dry nut may be 'statistically' safer than having a slippy nut. Your comment about con rod bolts applies all over a car but, in general, the more critical a nut is, the less likely that an amateur will be touching it. Almost anyone vaguely strong enough may be up for changing a wheel, even if they have no mechanical knowledge.

 

[Javier Lopez]

< I actually lubricate fasteners on my wheels>
The attaching force = area*coefficient of friction (steel to steel) /sin(angle) * force used to attach the bolt
The angle is the arctangent of screw thread mm/turn/2 divided by the diameter
The area =area of bolt inside the wheel holder

regards