jerryv
Mechanical
- Nov 10, 2000
- 22
Were can I find information on Steel on Steel Friction Coefficients Values with different surface finishes?
Tek-Tips is the largest IT community on the Internet today!
Members share and learn making Tek-Tips Forums the best source of peer-reviewed technical information on the Internet!
-
Congratulations cowski on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!
Steel Friction Coefficients Values 5
- Thread starter jerryv
- Start date
- Status
-
3
- #2
Dear Jerryv
It depends on what you are going to use it for. In the simplest case you can find it in (for example) Machinery’s Handbook, Twentieth Ed, p.553, Table 1 “Coefficients of Static Friction for Steel on Various Materials.”
I should tell you, however, the friction coefficient is not a physical constant. It varies over a wide range depending on the contact pressure, shape of contact bodies, presents and type of lubricants and many many other parameters. In my personal opinion, it is useless parameter since it differs for each particular case in non-predictable manner. The friction force and the wear rate are much better characteristics of any contact.
Regards
Viktor
It depends on what you are going to use it for. In the simplest case you can find it in (for example) Machinery’s Handbook, Twentieth Ed, p.553, Table 1 “Coefficients of Static Friction for Steel on Various Materials.”
I should tell you, however, the friction coefficient is not a physical constant. It varies over a wide range depending on the contact pressure, shape of contact bodies, presents and type of lubricants and many many other parameters. In my personal opinion, it is useless parameter since it differs for each particular case in non-predictable manner. The friction force and the wear rate are much better characteristics of any contact.
Regards
Viktor
When I need to know friction coefficients for materials that I can't find or want a more accurate figure, I perform some simple experiments to determine the force required to initiate motion for the static coefficient and same for the dynamic coefficient where the specimen is moved at a constant velocity. Also if you are looking for "friction" of a rolling body the coefficients are not the same as for sliding.
-
1
- #5
Gianfranco
Mechanical
Hi Jerryv,
You have a very useful source of data on Marks' Standard Handbook, par. 3.2. and by the references cited on it.
In detail:
1. Coefficient of static friction fs for Steel on Steel are ranging from 0,013 to infinite (see pag. 3.25 and mainly on table 3.2.1 for different couples of steels.
2. Commonly fs=0,39 (grease-free in air). An excerpt of Marks' Handbook is availbale at
3. The effect of surface films strongly affects the fs. For instance for steel-steel:
fs=0,78 clean and dry without oxide film
fs=0,27 clean and dry with oxide film (by heating at 500°C)
4. The effect of surfaces finishings are shown on pag. 3.27. It is singular discovering the non linearity of Fs vs roughness; for instance:
Roughness fs
microinches) with mineral oil
2 0,128
7 0,189
20 0,360
50 0,372
65 0,378
Grit-Blasted 0,212
See for details: standards methods of measurements of fs and fd.
BTW, it is interesting to know that the Occupational Safety and Health Administration (OSHA) recommends a flooring surface with a coefficient of friction greater than 0.5. The Americans with Disabilities Act recommends higher coefficients of friction for some types of application. Accessible routes should have a coefficient of friction of 0.6 and ramps should be .80. Underwriters Laboratories (UL) standard also requires a 0.5 coefficient of friction (UL 410).
Regards to all.
Gianfranco.
You have a very useful source of data on Marks' Standard Handbook, par. 3.2. and by the references cited on it.
In detail:
1. Coefficient of static friction fs for Steel on Steel are ranging from 0,013 to infinite (see pag. 3.25 and mainly on table 3.2.1 for different couples of steels.
2. Commonly fs=0,39 (grease-free in air). An excerpt of Marks' Handbook is availbale at
3. The effect of surface films strongly affects the fs. For instance for steel-steel:
fs=0,78 clean and dry without oxide film
fs=0,27 clean and dry with oxide film (by heating at 500°C)
4. The effect of surfaces finishings are shown on pag. 3.27. It is singular discovering the non linearity of Fs vs roughness; for instance:
Roughness fs
microinches) with mineral oil
2 0,128
7 0,189
20 0,360
50 0,372
65 0,378
Grit-Blasted 0,212
See for details: standards methods of measurements of fs and fd.
BTW, it is interesting to know that the Occupational Safety and Health Administration (OSHA) recommends a flooring surface with a coefficient of friction greater than 0.5. The Americans with Disabilities Act recommends higher coefficients of friction for some types of application. Accessible routes should have a coefficient of friction of 0.6 and ramps should be .80. Underwriters Laboratories (UL) standard also requires a 0.5 coefficient of friction (UL 410).
Regards to all.
Gianfranco.
Dear gainfrancoIn most engineering and physical situations, friction effects are described by a constant coefficient of Coulomb friction
f=F/P (1)
where P is the normal force and F is the frictional force.
Although it is well-established that contact between two bodies is limited to only a few microscopic high points (asperities), it is customary to calculate stresses by assuming that the forces are distributed over the total (apparent) area A. Such an approximation, however, is not far from reality in metalworking where the actual and apparent contact areas are almost the same due to high contact pressures. Thus the stress normal to interface P/A, the shearing (frictional) stress at the interface F/A so that Eq. (1) becomes
Friction coefficient=(shear stress)/(normal stress) (2)
Equation (2) reveals that if the friction coefficient is constant, the ratio of the shear and normal stresses should be the same along the entire contact length.
The above analysis was for sliding friction at the interface, as in our first encounter with friction in elementary physics. At the other extreme, we can envision a situation where the interface has a constant shear strength tau. The most usual case is sticking friction, where there is no relative motion at the interface. For sticking friction tau = the flow stress in shear. With von Mises’ yield criterion, the coefficient of friction under sticking conditions is
Friction coefficient=(tau/normal stress)=(the flow shear strength/the ultimate strength stress)=((the ultimate strength/SQART(3))/(the ultimate strength) = 0.577 (3)
Therefore, the value of the friction coefficient defined by Eq. (3) should be considered as the limiting value so that if friction coefficient ³ 0.577 no relative motion can occur at the interface.
Therefore, the value in your references are doubtfull.
Regards
Viktor
f=F/P (1)
where P is the normal force and F is the frictional force.
Although it is well-established that contact between two bodies is limited to only a few microscopic high points (asperities), it is customary to calculate stresses by assuming that the forces are distributed over the total (apparent) area A. Such an approximation, however, is not far from reality in metalworking where the actual and apparent contact areas are almost the same due to high contact pressures. Thus the stress normal to interface P/A, the shearing (frictional) stress at the interface F/A so that Eq. (1) becomes
Friction coefficient=(shear stress)/(normal stress) (2)
Equation (2) reveals that if the friction coefficient is constant, the ratio of the shear and normal stresses should be the same along the entire contact length.
The above analysis was for sliding friction at the interface, as in our first encounter with friction in elementary physics. At the other extreme, we can envision a situation where the interface has a constant shear strength tau. The most usual case is sticking friction, where there is no relative motion at the interface. For sticking friction tau = the flow stress in shear. With von Mises’ yield criterion, the coefficient of friction under sticking conditions is
Friction coefficient=(tau/normal stress)=(the flow shear strength/the ultimate strength stress)=((the ultimate strength/SQART(3))/(the ultimate strength) = 0.577 (3)
Therefore, the value of the friction coefficient defined by Eq. (3) should be considered as the limiting value so that if friction coefficient ³ 0.577 no relative motion can occur at the interface.
Therefore, the value in your references are doubtfull.
Regards
Viktor
Gianfranco
Mechanical
Hi all,
I personally have appreciated very much the point of view of vicktor an expert on metal-cutting Let me say shortly:
1. We are talking about of friction under static (dry or not) or dinamic/sliding (dry or not) conditions.
2. All data shown in my previous communication are excerpt from the Marks' Handbook par 3.2. one of the most important engineering reference book.
3. Vickor says that the values of the friction coefficients reported on Marks' Handbook (greater than 0,577, I suppose), are doubtfull. Vicktor, is your thought summarized correctly?
4. As a consequence of (3), we have the hazard of consulting and using one of the most known reference book affected of mistakes, or of unreliable data, or that the sources, used by the Prof. Emeritus D. Fuller (author of sec. 3.2 of the book)expert on "Friction and Fluid Film Bearings", author of Theory and Practice of Lubrication for Engineers, are not sure. But, up to now, I have not any errata corrige for this section or data. Please, Vickor could you check the content of sec. 3.2 of the Marks' Handbook.
4. Anyway, I'll appreciate further pointings out about others references or obs from the experts in friction attending at the forums.
Thanks to all.
Gianfranco
I personally have appreciated very much the point of view of vicktor an expert on metal-cutting Let me say shortly:
1. We are talking about of friction under static (dry or not) or dinamic/sliding (dry or not) conditions.
2. All data shown in my previous communication are excerpt from the Marks' Handbook par 3.2. one of the most important engineering reference book.
3. Vickor says that the values of the friction coefficients reported on Marks' Handbook (greater than 0,577, I suppose), are doubtfull. Vicktor, is your thought summarized correctly?
4. As a consequence of (3), we have the hazard of consulting and using one of the most known reference book affected of mistakes, or of unreliable data, or that the sources, used by the Prof. Emeritus D. Fuller (author of sec. 3.2 of the book)expert on "Friction and Fluid Film Bearings", author of Theory and Practice of Lubrication for Engineers, are not sure. But, up to now, I have not any errata corrige for this section or data. Please, Vickor could you check the content of sec. 3.2 of the Marks' Handbook.
4. Anyway, I'll appreciate further pointings out about others references or obs from the experts in friction attending at the forums.
Thanks to all.
Gianfranco
Dear Gainfranko
Thank you for understanding what I was pointing out. Unfortunately I cannot type formulas on this screen so I tried my best to explain the point qualitatively. The problem with friction coefficients more than 0,577 is that they violate the von Mises criterion for stresses. In other works – the metal rather deforms that slides because the force holding the contact is greater than that needed for deformation. SQUART(3) is valid for most of metals. I think that von Mises criterion holds for other substances also.
Now we should consider why the friction coefficient obtained experimentally exceeds sometimes 0.577. The reason for that is adhesion which creates the force which adds to the friction force so the overall force is so great that the friction coefficient exceeds the critical value. In ‘pure’ sense of friction (I would say, in Coulomb friction) there is no adhesion, diffusion chemical reactions, etc. In reality, we have all this stuff so the wear due to friction is not pure abrasion wear but rather a combination of abrasion, adhesion, diffusion, and chemical wear. It all depends on the relative velocity, contact stresses, lubricant and many other factors.
In conclusion I would like to point out that Column friction model can be used only to the first, thus very rough approximation at earlier stages of design. To understand the nature of contact processes at the interface, the energy balance of this interface should be considered. I’ve done a lot of work on the Tribology of Metal Cutting. This year my chapter “Tribology of Metal Cutting” is going to be published book in “Surface Modification and Processing: Physical and Chemical Tribological Methodologies.” by Marcel Dekker. Then a book “Physics of Strength and Fracture Control” is scheduled for publishing by CRC Press.
Regards
Viktor
Thank you for understanding what I was pointing out. Unfortunately I cannot type formulas on this screen so I tried my best to explain the point qualitatively. The problem with friction coefficients more than 0,577 is that they violate the von Mises criterion for stresses. In other works – the metal rather deforms that slides because the force holding the contact is greater than that needed for deformation. SQUART(3) is valid for most of metals. I think that von Mises criterion holds for other substances also.
Now we should consider why the friction coefficient obtained experimentally exceeds sometimes 0.577. The reason for that is adhesion which creates the force which adds to the friction force so the overall force is so great that the friction coefficient exceeds the critical value. In ‘pure’ sense of friction (I would say, in Coulomb friction) there is no adhesion, diffusion chemical reactions, etc. In reality, we have all this stuff so the wear due to friction is not pure abrasion wear but rather a combination of abrasion, adhesion, diffusion, and chemical wear. It all depends on the relative velocity, contact stresses, lubricant and many other factors.
In conclusion I would like to point out that Column friction model can be used only to the first, thus very rough approximation at earlier stages of design. To understand the nature of contact processes at the interface, the energy balance of this interface should be considered. I’ve done a lot of work on the Tribology of Metal Cutting. This year my chapter “Tribology of Metal Cutting” is going to be published book in “Surface Modification and Processing: Physical and Chemical Tribological Methodologies.” by Marcel Dekker. Then a book “Physics of Strength and Fracture Control” is scheduled for publishing by CRC Press.
Regards
Viktor
-
1
- #9
Very interesting points.
Viktor, I think I appreciate what you are getting at, but I am not sure that I agree completely with your statement.
I wonder whether your statement is really more relevant for cutting simulations, rather than for connections. For the situation in which the normal forces and shear forces come from the same source, then I think your logic may be correct.
However, take the following example (a fairly common situation):
Bodies A and B are held together by bolts, with significant normal forces, but not within the range of material yield. Body B is fixed to ground, and body A is then loaded with a shearing force. For this situation, the shearing force is independent of the normal force.
Let us presume that the normal forces from the bolts do not result in stresses greater than 50% of yield. Let us also presume that the surfaces are very rough, resulting in an approximate coefficient of friction of 1.0. For this situation, the shear stresses could be as high as 50% of yield (resulting in a shear force of the same value as the normal forces). Von Mises is not violated in this situation, as the connection will slip due to shear stress before the underlying material yields.
I think I appreciate your rationale from the metal-cutting standpoint, but for many connection-interface issues, I don't think it applies.
Please correct me if my logic is in error.
Brad
Viktor, I think I appreciate what you are getting at, but I am not sure that I agree completely with your statement.
I wonder whether your statement is really more relevant for cutting simulations, rather than for connections. For the situation in which the normal forces and shear forces come from the same source, then I think your logic may be correct.
However, take the following example (a fairly common situation):
Bodies A and B are held together by bolts, with significant normal forces, but not within the range of material yield. Body B is fixed to ground, and body A is then loaded with a shearing force. For this situation, the shearing force is independent of the normal force.
Let us presume that the normal forces from the bolts do not result in stresses greater than 50% of yield. Let us also presume that the surfaces are very rough, resulting in an approximate coefficient of friction of 1.0. For this situation, the shear stresses could be as high as 50% of yield (resulting in a shear force of the same value as the normal forces). Von Mises is not violated in this situation, as the connection will slip due to shear stress before the underlying material yields.
I think I appreciate your rationale from the metal-cutting standpoint, but for many connection-interface issues, I don't think it applies.
Please correct me if my logic is in error.
Brad
Dear Bradh
I would not agree with you point that my considerations are valid only for metal cutting. I guess they are valid universally. OK, lets consider your example:
“Let us presume that the normal forces from the bolts do not result in stresses greater than 50% of yield. Let us also presume that the surfaces are very rough, resulting in an approximate coefficient of friction of 1.0. For this situation, the shear stresses could be as high as 50% of yield (resulting in a shear force of the same value as the normal forces). Von Mises is not violated in this situation, as the connection will slip due to shear stress before the underlying material yields.”
We cannot assume that because: friction coefficient=(friction force)/normal force. If we multiply the nominator and denominator by the apparent contact area we have the ratio of tangential and normal stresses regardless of their levels (50% of yield or greater). The maximum of this ratio is at the yield point and equal to 0.577. When the normal stress is lower that according to the Mohr’s circle of stress, this ratio is lower. Therefore, it is not possible have friction coefficient 1.0 for the indicated conditions.
Now the ball is in your court.
Regards
Viktor
I would not agree with you point that my considerations are valid only for metal cutting. I guess they are valid universally. OK, lets consider your example:
“Let us presume that the normal forces from the bolts do not result in stresses greater than 50% of yield. Let us also presume that the surfaces are very rough, resulting in an approximate coefficient of friction of 1.0. For this situation, the shear stresses could be as high as 50% of yield (resulting in a shear force of the same value as the normal forces). Von Mises is not violated in this situation, as the connection will slip due to shear stress before the underlying material yields.”
We cannot assume that because: friction coefficient=(friction force)/normal force. If we multiply the nominator and denominator by the apparent contact area we have the ratio of tangential and normal stresses regardless of their levels (50% of yield or greater). The maximum of this ratio is at the yield point and equal to 0.577. When the normal stress is lower that according to the Mohr’s circle of stress, this ratio is lower. Therefore, it is not possible have friction coefficient 1.0 for the indicated conditions.
Now the ball is in your court.
Regards
Viktor
Viktor,
My volley . . .
Let me first state that, given your background, I expected this to ultimately end with me hitting myself on the head and realizing where my logic is wrong.
But first, let me reclarify an earlier statement. I didn't state that your statements were ONLY valid for metal cutting. I will restate my contention more specifically--I thought your contention holds for the situation in which material yield is exceeded (obviously metal-cutting is an example of this). If there is no yielding of the materials at the interface, then I contend that the statement holds.
I went about calculating the math behind my example, and quickly realized that my given example is garbage. I had been doing my tensor calculations in my head, and screwed something up.
So I redefined my problem . . .
Same situation as above, friction coefficient=1.0, but with the resulting normal stress=20% of yield. When I do this, my resulting peak normal/shear combination no longer exceeds yield.
My calculations are based on: neglecting poisson effects, resulting in a stress state of:
Sxx= -0.2, tauyz= 0.2, all others zero. (.2 meaning normalized to yield--20% of Sy)
This situation then (if I did my math right) results in a Von Mises stress of 0.6 of yield, so for the situation in which the normal stresses are about or below 20% of yield, the coefficient of friction can indeed be 1.0, and not violate Von Mises.
However, before I get too smug about my statements (even if my math is correct), I think this may have been the very point that you were making originally, that being that this approximation breaks down for many problems.
To summarize my points, then:
1) the coefficient of friction can exceed .577 without violating Von Mises criterion.
2) this can only happen for stresses significantly below yield of the contacting materials (<10-25% material yield).
3) Most joining operations (bolts, etc.) exist in a state of stress near yield, so that my problem described above is not relevant to numerics of most real-world problems.
Essentially, at this point I think I am in agreement with what you said. However, if I am still wrong, or I have missed something, please let me know.
Oh, and the ball just hit in my court and went of bounds.
Point, Viktor.
Best regards,
Brad
My volley . . .
Let me first state that, given your background, I expected this to ultimately end with me hitting myself on the head and realizing where my logic is wrong.
But first, let me reclarify an earlier statement. I didn't state that your statements were ONLY valid for metal cutting. I will restate my contention more specifically--I thought your contention holds for the situation in which material yield is exceeded (obviously metal-cutting is an example of this). If there is no yielding of the materials at the interface, then I contend that the statement holds.
I went about calculating the math behind my example, and quickly realized that my given example is garbage. I had been doing my tensor calculations in my head, and screwed something up.
So I redefined my problem . . .
Same situation as above, friction coefficient=1.0, but with the resulting normal stress=20% of yield. When I do this, my resulting peak normal/shear combination no longer exceeds yield.
My calculations are based on: neglecting poisson effects, resulting in a stress state of:
Sxx= -0.2, tauyz= 0.2, all others zero. (.2 meaning normalized to yield--20% of Sy)
This situation then (if I did my math right) results in a Von Mises stress of 0.6 of yield, so for the situation in which the normal stresses are about or below 20% of yield, the coefficient of friction can indeed be 1.0, and not violate Von Mises.
However, before I get too smug about my statements (even if my math is correct), I think this may have been the very point that you were making originally, that being that this approximation breaks down for many problems.
To summarize my points, then:
1) the coefficient of friction can exceed .577 without violating Von Mises criterion.
2) this can only happen for stresses significantly below yield of the contacting materials (<10-25% material yield).
3) Most joining operations (bolts, etc.) exist in a state of stress near yield, so that my problem described above is not relevant to numerics of most real-world problems.
Essentially, at this point I think I am in agreement with what you said. However, if I am still wrong, or I have missed something, please let me know.
Oh, and the ball just hit in my court and went of bounds.
Point, Viktor.
Best regards,
Brad
Dear Brand
Firs of all, thank you for deep thoughts. I think you’ve made your point and I should agree with it. Actually, it gave me a lot of to think on the essence of friction coefficient. On one had, it should be independent on the normal load thus its graph in sigma-tau coordinate system should be represented by a straight line. On the other hand (or whatever part), this line should pass through point 0.577. If we connect the coordinate origin with this point then…..all the most interesting begins. If you have a chance to read
Dieter G. Mechanical Metallurgy, Third Edition, McGraw-Hill, p. 539, Section Friction and Lubrication, you might find a lot of to think about the friction coefficient and it real meaning. Reading this Section, I am more and more convinced that it is not properly understood and thus applied in everyday practice.
Regards
Viktor
Firs of all, thank you for deep thoughts. I think you’ve made your point and I should agree with it. Actually, it gave me a lot of to think on the essence of friction coefficient. On one had, it should be independent on the normal load thus its graph in sigma-tau coordinate system should be represented by a straight line. On the other hand (or whatever part), this line should pass through point 0.577. If we connect the coordinate origin with this point then…..all the most interesting begins. If you have a chance to read
Dieter G. Mechanical Metallurgy, Third Edition, McGraw-Hill, p. 539, Section Friction and Lubrication, you might find a lot of to think about the friction coefficient and it real meaning. Reading this Section, I am more and more convinced that it is not properly understood and thus applied in everyday practice.
Regards
Viktor
Viktor,
I will try to find the book. I think rather than making my points, I managed to convince myself (and hopefully many others) of the general validity of your statements.
This was a very enjoyable interchange (for me anyway). Thank you for making me think, and rethink, some aspects of this which I have never really given a large amount of knowledge to. It was fun to walk through the implications of this.
Best regards,
Brad
I will try to find the book. I think rather than making my points, I managed to convince myself (and hopefully many others) of the general validity of your statements.
This was a very enjoyable interchange (for me anyway). Thank you for making me think, and rethink, some aspects of this which I have never really given a large amount of knowledge to. It was fun to walk through the implications of this.
Best regards,
Brad
- Status
Similar threads
- Locked
- Question