Tigtening torque for M8X1.25 with 12.9 class 6

[elogesh]

Hai,

One of my colleague has asked the following doubt,

In one of the product assemblies, it was mentioned in the drawing to use bolt M8X1.25 bolt with class 8.8. The prescribed torque was 25 N-m.

But due to some reasons, they have used M8X1.25 bolt with class 12.9. The prescribed torque for this category bolt is 42 N-m. But they tighted the bolt with 25 N-m torque corresponding to class 8.8.

Colleague asked me what will be the effect.

I felt, the bolt will be subjected to full external load.

If it is tightened to full torque then most of the load is taken up by joint members. If it tightened to some percentage of full torque(say 50%), then bolt will experience most of the external load.

Whether am I right?

I am looking forward for your opinions.

Regards,
Logesh.E

 

[yates]

The clamping force produced by an installation torque of 25 Nm. is independent of the strength class of bolt used. If the designer felt that the clamping force produced by 25 Nm. of torque applied to a class 8.8 bolt, was sufficient - then this will also be the case for the 12.9 class bolt. The clamping force/torque relationship is only governed by thread geometry, bearing surface area and friction coefficient (type of lubrication)
'Full torque' does not have any significant meaning. The prescribed installation torque of 25 Nm. for 8.8 or 42 Nm. for class 12.9 only corresponds to a torque producing a certain stress in the bolt (80% of elastic limit is a typical criterion) The theory behind the prescibed installation torque is that it is desirable to produce the greatest clamping force possible without exceeding the elastic limit in the bolt (80% of this leaves a safety factor to take into account variations of friction coefficient or torque wrench accuracy)
To put it simply, the greater the clamping force the greater proportion of the load will be taken by the joint members rather than the bolt.

 

[Screwman]

In substituting a PC 12.9 for an 8.8 fastener you need to very aware of the possibility of a stress Corrosion failure. In this case, if the seating torque remains as designed for the PC 8.8 you should be safe, since sensitivity is directly proportional to the applied load (tension in the fastener combined with service load). SCC is one of the main causes of failures that I have seen in 12.9 fasteners.
Dick

 

[rmw]

Some bolt torques are determined to bring the bolt to within 80% of the elastic limit as stated by yates above, and that produces some "stretch" in the bolt, which is figured into the clamping forces on the joint. The same torque on a class 12.9 would not produce the same "stretch" on the higher strength bolt, and, hence might give a different characteristic with respect to the joint, and its clamping forces.

I am speaking in generalities regarding torque applied to bolts, and not specifically with respect to 25 Nm of torque on a M8X1.25 bolt (screw).

rmw

 

[israelkk]

rmw

When you say "stretch" I assume you mean deflection. If so, I believe your statement that " The same torque on a class 12.9 would not produce the same "stretch" on the higher strength bolt" is not accurate. If the tightning torque is the same the resultant force in the bolt is the same too where F=Torque/K/D (assuming K is the same for 8.8 and 12.9).

Therefore the stress and deflection is the same too because the modulus of elesticity is the same for both types of bolts.

elogesh

Tightning to 25Nm of both types of bolts will create the same clamping force on the joint. What really defines the rigidity of the assembly is the ratio between the stiffness of the clamped parts with respect to the stiffness of the bolt. The higher the ratio the better.

For the sake of the example lets assume that the stiffness of the clamped parts is 10 times higher of the bolt. Now let us assume that when we tightnem the bolt with 25Nm creates 1000 kgf force in the bolt and its length increase by 0.1 mm. Therefore theclamping force is 1000 kgf too on the clamped parts and the length/thickness of the clamped parts will decrease by 0.01 mm (10 timee stiffer).

Now here is the point! when one exerts a 1000 kgf force on the two clamped part that tried to separate them from each other the length/thickness of the clamped parts will increase by the 0.01 mm just before they are separated. therefore, the bolt length will increase just by extra 0.01 mm (not by extra 0.1 mm) resulting in 1100 kgf in the bolt. What we really achieved with the tightning of the bolt is a 10 times stiffer assembly compared to the same assembly with zero tighting torque.

If there was no tightning torque the two parts will move 0.1 mm apart from each other when a 1000 kgf is trying to separate them.

 

[CoryPad]

israelkk is correct and rmw is incorrect regarding "stretch" or elastic displacement.

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[rmw]

Cory,

What I said was not incorrect, but thank you for sending me back to the textbooks (and the internet) to research and review what I learned almost 40 years ago, and have taken for granted ever since.

If I changed the last word in my first paragraph above to "characteristics" instead of "forces" it might make my point more clear.

What israelkk said in response to my post is partially correct, with respect to the forces, but not with respect to the modulus of elasticity being equal for the two bolts.

I do not dispute that the force on the joint would be the same, given the same torque and "k" factors. The "stretch," a term commonly used in industry for elongation of bolts, however is another matter.

First lets deal with the modulus of elasticity (E), or Young's Modulus, which is the ratio between an applied stress, and the strain (elongation) that results. As a ratio, it is the function of the slope of the straight line portion of the stess strain curve, before plastic deformation occurs. (Everything herein pertains only to elastic deformation, or "stretch")

There are a plethera of sources for E for a variety of materials, steel in the case of the bolt material of this thread, and they are generally denoted as "average," "typical," or "representative", depending on the source. They vary from reference to reference, and sometimes even within the same reference.

Shigley, in the 1963 edition of "Mechanical engineering Design" states the average E for carbon steel as 28.5 X 10*6, and 27.6 X 10*6 for 18-8 SS in Appendix A, table A-1. This same table states the "average" E for gray cast iron (CI) as 14.5, while table A-4 gives a range of E's for gray cast iron, ranging from 11-18 X 10*6 psi.

Van Vlack, in "Elements of Material Science" second edition, states the value of the E for a variety of carbon and stainless steels, as well as gray (and white) CI in appendix C as 30 X 10*6 (across the board). But in an example problem in the text, chapter 1, page 3, he states the average E for steel as 29.5 X 10*6. In another place on page 142, he states the average E for steel as 30 X 10*6. Which is it?

He goes further (in explaining this average value) in the same paragraph to state that the E for iron crystals range from 18-41 X 10*6 psi, depending upon orientation with respect to a uniaxial force.

In a psu.edu reference listed at the end, E ranges from 28.6 to 30 X 10*6.

So, we can see from the literature that even the "average" E for steel is not consistant. E is generally given as an average in order to compare it with the average E of other materials, such as Al, CU, Ti, etc.

E is a function of the modulus of rigidity, (G) or shear modulus, and poissons ratio, (v). So, as these vary within a given metallurgy, E varies. The modulus of elasticity, E, is defined as E=2G(1+v).

G and v are affected by tensile strength which is generally associaated with hardness in steels. Strength is a measure of the ability of a material to resist stress, so it follows that as a material is rated higher in strength, the resistance to stress is going to increase, and the corresponding elongation due to a specific stress is going to be less, resulting in a different stress/strain ratio, and hence a different curve, and a different E.

Table A-4 in Shigley, which is for CI, rather than steel, shows how E increases incrementally from 11-18 X 10*6 psi through a range of 6 types of CI, (ASTM 20, 25, 30, 35, 40, & 50) as hardness ranges from 110 - 250 BHN, and tensile strength ranges from 20 - 50 K psi.

Frankly, I never found such a stand alone table for carbon steel, but it is obvious from many listings of various carbon steel alloys that as tensile strength (and hardness) increase, elongation (in 2 in), and reduction in area of the fracture point decrease, indicating that the corresponging E, G, and v values are not equal.

Now, back to my comments in my post above. A good reference I found that gives a comparison of the relative strengths (as well as hardness) of Metric 8.8 and 12.9 bolts is:

http://euler9.tripod.com/bolt-database/22.html

A good place for me to make my point with visual aids is on the fifth page down, figure 2.2.2 of:

http://www.utm.edu/departments/engin/lemaster/Machine Design/Lecture 01.pdf

Two observations about the values presented on this page.

First, it is clear that the slopes of the stress strain relationship of a variety of steels presented (again, pertaining to the elastic region only) is not identical, indicating that the E for these steels is, in fact, different.

Second, in order to re-state my case, lets look at the two different curves for the same metal, specifically in this case, SAE 3140, one in a 190K psi strength, and the other in a 240K psi strength.

Note first that the slope of the elastic portion of the curve is different.

But for the point that I made, imagine, if you will, that the 8.8 and 12.9 bolts in question in the original post of the thread are of this SAE 3140 material. Let the 190K value represent the E curve for the 8.8 and the 240K value represent the E curve for the 12.9.

What I said was that if the joint was designed to require the fastener to be elongated (stretched) to within a certain percentage (we will use 90%, proof strength here) of the elastic limit, then the approximate force on this joint would be 190 x 0.9 = 171K psi with the 8.8 bolt. The elongation of the bolt can be read directly from the graph. (I used the proof strength of 90% rather than 80% of elastic limit so as to be in the elastic region in order to be able to make my point.)


Now, if a stronger 12.9 bolt of this same material is substituted, and torqued to the same value so as to produce the same force on the joint, no elongation occurs at all, because elongation in this material does not begin until approx 220K psi of stress.

So, to restate my case, if a joint requires that the bolt be "stretched" for joint integrity reasons, substituting a higher strength fastener, and torqueing it to the same torque as the lower strength bolt will not produce the same aamount of stretch, (if any at all in this case.)

Two types of joints that require "stretching" of the fasteners that occur to me right away are (1) a joint where the fastener and the joint material have a different coefficient of thermal expansion over a given temperature range, and the "stretching" of the bolt insures that tension is always maintained on the joint throughout the temperature range, and (2) a joint that might have a gasket material that was expected to further compress (with process induced force or temparature) after being placed in service, and in order to maintain tension on the joint upon returning to the original temperature, so the bolts are "stretched" to a predetermined value.

It is clear from my exercise, that while the value of E given in a lot of references hovers around 30 X 10*6, that this value of E can not be taken carte blanche for any material for the purposes of a statement such as the one made by israelKK above.

Other interesting references I found on this topic include:

http://www.stelfast.com/pages/techdata.htm

http://www.stbf.com/studchart.htm

http://www.personal.psu.edu/users/l/d/ldl130/experiment2.htm

http://www.ubka.uni-karlsruhe.de/cg...geo/4&search=/2003/bau-geo/4&format=1&page=40

http://www.ubka.uni-karlsruhe.de/cg...geo/4&search=/2003/bau-geo/4&format=1&page=41

I thouroughly enjoyed the mental exercise. Thanks!!

rmw

 

[CoryPad]

rmw,

You are correct that elastic constants vary along crystallographic directions. Nearly all materials used by society are polycrystals, so the elastic response is governed by the aggregate of all crystals, which leads to an "average" value.

E is a function of the modulus of rigidity, (G) or shear modulus, and poissons ratio, (v). So, as these vary within a given metallurgy, E varies. The modulus of elasticity, E, is defined as E=2G(1+v).

Young's modulus is not a function of the other elastic constants. You have shown the relationship amongst the three elastic constants for a cubic material. In actuality, for isotropic materials (like metallic polycrystals), there are only two independent elastic constants (the Lam[é] constants): [μ] and [λ], which can be used to derive all others (Young's modulus, shear modulus, Bulk modulus, and Poisson's ratio). This information may not be in introductory books like Van Vlack, so try Deformation and Fracture Mechanics of Engineering Materials by Hertzberg or Mechanical Behavior of Materials by Bowman.

G and v are affected by tensile strength which is generally associaated with hardness in steels. Strength is a measure of the ability of a material to resist stress, so it follows that as a material is rated higher in strength, the resistance to stress is going to increase, and the corresponding elongation due to a specific stress is going to be less, resulting in a different stress/strain ratio, and hence a different curve, and a different E.

This paragraph is rife with errors. A material's elastic constants are NOT affected by tensile strength. Elastic constants can change based on chemical composition, number of phases present, and crystallographic texture. Tensile strength can change because of these factors too. However, elastic and plastic responses are not functions of each other.

A good place for me to make my point with visual aids is on the fifth page down, figure 2.2.2 of:

http://www.utm.edu/departments/engin/lemaster/Mach...

Two observations about the values presented on this page.

First, it is clear that the slopes of the stress strain relationship of a variety of steels presented (again, pertaining to the elastic region only) is not identical, indicating that the E for these steels is, in fact, different.

This figure is a true stress-true strain graph that doesn't even show elastic response so it cannot be used to verify elastic property variation. The abscissa would need to be enlarged between zero and 0.2 (by about 10 times) to determine the elastic properties.

To go back to the original question, screws of property classes 8.8 and 12.9 will have essentially the same elastic constants because they have essentially the same chemical composition (> 99% Fe according to ISO 898-1), phases present (tempered martensite according to ISO 898-1), and crystallographic texture (essentially random due to austenitizing, quenching and tempering of a polycrystal).

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[Screwman]

Cory,
Great post.
Thanks for taking the time to clear up what was turning into an incorrect situation.

Dick