I'm afraid this might be a silly question but how can I calculate the maximum surface pressure on a (soft) aluminium plate, avoiding an imprint on the surface. In a construction I'm having a rod with in the center a threaded hole. In the Al plate there is a hole through which a bolt connects both parts. With the bolttorque I'm using now, the rod gives an imprint of approx. 1mm in the aluminium plate, which is not allowed. I would like to lower the torque to a value where no damage to the aluminium surface occurs. Of course I can do this by trail and error but for future cases I would like to be able to calculate the allowable surface pressure/torque.
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Maximum surface pressure on aluminium
- Thread starter ksnar
- Start date
Ksnar,
Here goes :-
Using Stress = Load /Area were Load = surface pressure
You need to limit the load to below that which the material yields (Aluminium approx = 172 MN/mm2)
To turn torque into load use (neglecting friction)
Torque = Load x pitch of screwthread/6.2832 (use SI units ie N & mm)
Simple approach but should give you a reasonable answer.
GB
Gb
Here goes :-
Using Stress = Load /Area were Load = surface pressure
You need to limit the load to below that which the material yields (Aluminium approx = 172 MN/mm2)
To turn torque into load use (neglecting friction)
Torque = Load x pitch of screwthread/6.2832 (use SI units ie N & mm)
Simple approach but should give you a reasonable answer.
GB
Gb
- Thread starter
- #3
You're looking at a thin sheet of metal. Is not simple like that.
If the loading is normal to the surface. The result of that loading will cause the plate to shear in the thin directions. Therefore what you have is the tension in the thin area directions. That is a simple thought.
You should check up elastic and plastic theary of thin plate on literature for the exact equation.
If the loading is normal to the surface. The result of that loading will cause the plate to shear in the thin directions. Therefore what you have is the tension in the thin area directions. That is a simple thought.
You should check up elastic and plastic theary of thin plate on literature for the exact equation.
- Thread starter
- #5
vtl: The aluminium sheet is approximately 15mm thick and the rod is made of plain carbon steel.
Diameter of the rod is 18mm with center hole M10.
Based on these dimensions I don't think that shear will be a problem, but thanks for the remark.
If you don't agree I would appreciate your reply.
ksnar
Diameter of the rod is 18mm with center hole M10.
Based on these dimensions I don't think that shear will be a problem, but thanks for the remark.
If you don't agree I would appreciate your reply.
ksnar
I thought yield strength of typical aluminum alloy plate, say Al 6061-T651, is about 260 N/mm^2 = 260 MPa (according to MIL-HDBK-5; slightly higher in matweb). So the allowable surface pressure will be S = (260 MPa)/FS, where FS = factor of safety against yield for your project.
Also, to compute preload P, see . Notice gjb1 is using only the first of the three terms in Eq. 3 therein, which is extremely conservative and might cause your preload to be too low relative to what is typically recommended for a bolted connection. However, in your case, for tapped holes, thread shear (stripping) strength of the internal (tapped) thread in weak material will often govern unless you have a long thread engagement length, which apparently you do.
Then, once P is obtained, your surface pressure sigma = P/A, where A = contact area. E.g., in your case, A = about 0.25(pi)(18^2 - 10^2) = 176 mm^2, or actually less considering chamfer on your threaded hole, which you should not neglect.
Also, to compute preload P, see . Notice gjb1 is using only the first of the three terms in Eq. 3 therein, which is extremely conservative and might cause your preload to be too low relative to what is typically recommended for a bolted connection. However, in your case, for tapped holes, thread shear (stripping) strength of the internal (tapped) thread in weak material will often govern unless you have a long thread engagement length, which apparently you do.
Then, once P is obtained, your surface pressure sigma = P/A, where A = contact area. E.g., in your case, A = about 0.25(pi)(18^2 - 10^2) = 176 mm^2, or actually less considering chamfer on your threaded hole, which you should not neglect.
trainguy,
You are absolutely correct, however there is scant data on allowable bearing stress for materials. MIL-HDBK-5 has some data, and ASTM has a test method, but they are for actual bearings elongating a circular hole, so it is not exactly equivalent. I have seen one technical journal article on the subject, but forgot to keep it (it was highly math-intensive, and not suitable for quick use). So, if anyone has good info sources, please share.
You are absolutely correct, however there is scant data on allowable bearing stress for materials. MIL-HDBK-5 has some data, and ASTM has a test method, but they are for actual bearings elongating a circular hole, so it is not exactly equivalent. I have seen one technical journal article on the subject, but forgot to keep it (it was highly math-intensive, and not suitable for quick use). So, if anyone has good info sources, please share.
gjb1,
I am surprised to see you suggest that friction may be ignored when trying to estimate the bolt tensions resulting from an applied torque. In fact, once there is any reasonable load in the bolt, more than 80% of the torque can be taken up in friction.
When I look at my fastener handbook, I see that for a 6mm dia 'commercial low tensile' bolt with 1mm pitch, the manufacturers suggest that a preload of 2940 N requires a torque of 3500 Nmm, where the non-friction calculation would say 468 Nmm. ie for that particular bolt at that load, 87% of the torque is taken up by friction.
For a 20mm dia and 2.5mm pitch the figures are 35800 N, 143000 Nmm with friction, 14244 without friction, 90% frictional torque.
For a very rough estimate of bolt load, I would multiply the torque from your formula by 8 at least.
I am surprised to see you suggest that friction may be ignored when trying to estimate the bolt tensions resulting from an applied torque. In fact, once there is any reasonable load in the bolt, more than 80% of the torque can be taken up in friction.
When I look at my fastener handbook, I see that for a 6mm dia 'commercial low tensile' bolt with 1mm pitch, the manufacturers suggest that a preload of 2940 N requires a torque of 3500 Nmm, where the non-friction calculation would say 468 Nmm. ie for that particular bolt at that load, 87% of the torque is taken up by friction.
For a 20mm dia and 2.5mm pitch the figures are 35800 N, 143000 Nmm with friction, 14244 without friction, 90% frictional torque.
For a very rough estimate of bolt load, I would multiply the torque from your formula by 8 at least.
Vonlueke/Austim,,
The aluminium was described as soft hance my conservatism using a value of 172MPa (25KSI).
Point regarding friction is valid . However, I was pointing out that the equation given takes no account of friction (ie simple approach).
If you need to take friction into account the equation becomes
For motion in direction of load
Torque (assists) = Load x (6.3ur-p/6.3r+up)x r
For motion opposite load
Torque (resists) = Load x (p +6.3ur/6.3r-up)x r
Where r = bolt rad, u=friction co-efficient p=bolt pitch.
Thanks for the feedback.
gjb
The aluminium was described as soft hance my conservatism using a value of 172MPa (25KSI).
Point regarding friction is valid . However, I was pointing out that the equation given takes no account of friction (ie simple approach).
If you need to take friction into account the equation becomes
For motion in direction of load
Torque (assists) = Load x (6.3ur-p/6.3r+up)x r
For motion opposite load
Torque (resists) = Load x (p +6.3ur/6.3r-up)x r
Where r = bolt rad, u=friction co-efficient p=bolt pitch.
Thanks for the feedback.
gjb
- Thread starter
- #11
Meanwhile I made some tests (calculations I made were including friction) and found different results. Reading vonlueke's remarks I realised that the rod indeed has a chamfer both on the treaded hole as the OD, which could be the cause of the different results.
The aluminium we use is AlSi7Mg (LM25) in the as cast condition, so with minimum yield even as low as 80 N/mm2. Material certificates show however values varying between 85 and 140 N/mm2. For my case I can allow some imprintment of the steel rod in the aluminium (up to 0.3 mm) so I took yield a little higher than minimum and friction factor 0.2 for the tread.
For a rod OD20 and bolthole 13 in the aluminium (for the M12 bolt) i found with torque 75 Nm an imprintment of 0.15 to 0.3 mm on several samples. This is acceptable for the construction.
The aluminium we use is AlSi7Mg (LM25) in the as cast condition, so with minimum yield even as low as 80 N/mm2. Material certificates show however values varying between 85 and 140 N/mm2. For my case I can allow some imprintment of the steel rod in the aluminium (up to 0.3 mm) so I took yield a little higher than minimum and friction factor 0.2 for the tread.
For a rod OD20 and bolthole 13 in the aluminium (for the M12 bolt) i found with torque 75 Nm an imprintment of 0.15 to 0.3 mm on several samples. This is acceptable for the construction.
ksnar,
I have used all the information given, applied it to the formulas given and it works out you are that you require approx 275Mpa of resistant force to hold the 75Nm torque hence the deformation in the Al plate. However the 0.2 friction co-efficient appears optimistic as it's likely to be at least double that.
A frictiuon co-eff of 0.4 would bring the resistant force to 137.5MPa which would agree with your material certs as you would be around the point of yield for the material (hence the deformation).
gjb
I have used all the information given, applied it to the formulas given and it works out you are that you require approx 275Mpa of resistant force to hold the 75Nm torque hence the deformation in the Al plate. However the 0.2 friction co-efficient appears optimistic as it's likely to be at least double that.
A frictiuon co-eff of 0.4 would bring the resistant force to 137.5MPa which would agree with your material certs as you would be around the point of yield for the material (hence the deformation).
gjb
- Thread starter
- #13
gjb1,
This discussion has taken me deeper into the matter as I anticipated, I appreciate all support.
To understand you correct: do you mean that with friction co-efficient 0.4 you calculate that the pressure of the rod on the Al-plate is 137.5 MPa?
For calculation of the boltforce, torque and resulting pressure I used the "simplified" equation: M=F*mu*D/1000, with F=boltforce [N], mu=total friction co-eff (0.2), D=nominal bolt diameter (12mm), M=torque(75Nm) and calculated F=31250N and with area 181mm²(OD20-OD13) found pressure 172 N/mm². I concluded since this lies above the 0.2% yield of the certificates it explains the deformation.
Are you of the opinion that my approach is too simplified, especially the bolt torque equation or is it similair to what you did.
By the way I checked the referenced document of GD Euler (Eq.3) but was not sure what Ut and Uc co-efficients to take, therefore my simplified method.
ksnar
This discussion has taken me deeper into the matter as I anticipated, I appreciate all support.
To understand you correct: do you mean that with friction co-efficient 0.4 you calculate that the pressure of the rod on the Al-plate is 137.5 MPa?
For calculation of the boltforce, torque and resulting pressure I used the "simplified" equation: M=F*mu*D/1000, with F=boltforce [N], mu=total friction co-eff (0.2), D=nominal bolt diameter (12mm), M=torque(75Nm) and calculated F=31250N and with area 181mm²(OD20-OD13) found pressure 172 N/mm². I concluded since this lies above the 0.2% yield of the certificates it explains the deformation.
Are you of the opinion that my approach is too simplified, especially the bolt torque equation or is it similair to what you did.
By the way I checked the referenced document of GD Euler (Eq.3) but was not sure what Ut and Uc co-efficients to take, therefore my simplified method.
ksnar
Knsar,
ksnar,
My apologies I made a mistake in my earlier calculations and therefore the numbers are incorrect. I calculated an incorrect slip factor
Using
Torque = F x Slip Factor (usually 0.25 for s.s on s.s)x D
Therefore F = 25KN and Pressure = 138MPa.
Hope this closes this out.
gjb
ksnar,
My apologies I made a mistake in my earlier calculations and therefore the numbers are incorrect. I calculated an incorrect slip factor
Using
Torque = F x Slip Factor (usually 0.25 for s.s on s.s)x D
Therefore F = 25KN and Pressure = 138MPa.
Hope this closes this out.
gjb
- Thread starter
- #16
Guest
ksnar, please excuse me. My previous post is misleading. Let me explain why. By the way, what you call "mu", above, is usually named K. Your "simplified" formula, above, is Eq. 1 in euler9. Since you already are assuming, e.g., K = 0.20, you therefore don't need Eq. 3. Eq. 3 is if you don't already have K. So, as you did, you could use K = 0.20 (or maybe K = 0.25) in Eq. 1, i.e., P = T/(K*D), which you called M = F*K*D, above. I advise you don't use the name mu for K, because mu is not equal to K. Also, K is usually not called "slip factor"; it's called torque coefficient (or occasionally "nut factor"
. Regards.
. Regards.- Thread starter
- #18
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