Wedge Anchor Torque Spec vs Ultimate Tension Spec

[Factotum]

TL;DR Summary

I have been reviewing the specifications for Red Head Trubolt wedge anchors. I am confused by what appears to be a contradiction between the specified installation torque, the resulting installation axial tensile force, and the anchors ultimate tension specification. I want to understand what I am misinterpreting.

[Mentor Note: See post #16 for a corrected/updated version of this post]
https://www.physicsforums.com/threa...vs-ultimate-tension-spec.1056964/post-6965453

I have been reviewing the specifications for Red Head Trubolt wedge anchors. I am confused by what appears to be a contradiction between the specified installation torque, the resulting installation axial tensile force, and the anchors ultimate tension specification. Per my calculations, the installation torque would result in a tensile force greater than the ultimate tension specification for the anchor. Since the allowable tension specification is 25% of the ultimate tension specification, the installation tensile force would far exceed the allowable tension specification. If my understanding and calculations are correct, there is a good chance that applying the installation torque to the anchor will cause the concrete to fracture.

I figure one of two things is happening. Either, I am incorrectly calculating the installation axial tension force, or I am misinterpreting the ultimate tension specification.

Per my understanding, the ultimate tension specification conveys the force needed to pull the anchor from the concrete. An event that usually results in fracturing and dislodging a cone of concrete along with the anchor.

I used the following equation to calculate the tensile force resulting from the applied installation torque.

F = T/(K*D)​

Where

• F is the tensile force.
• T is the torque applied to the anchor nut.
• K is the friction of the bolt contact. (0.2 for steel)
• D is the diameter of the bolt.

In the Red Head Trubolt specification document, I am looking specifically at the specifications for a 3/8” anchor embedded 3 inches into a 2000 PSI concrete slab. Below are the specification values I obtained.

• Installation Torque 25 ft. lbs.
• Ultimate Tension 3,480 lbs.

I calculated the installation tensile force as follows.

F = 25 ft. lbs. / (0.2 * (0.375 in./12 in. per ft.)) = 4000 lbs.

Again, if my understanding and calculations are correct, the installation tensile force of 4,000 lbs. exceeds the ultimate tension specification of 3,480 lbs.

Please let me know where I am going wrong.

Thank you.

 

[jack action]

Reverse engineering the product, the needed friction coefficient to reach the given ultimate tension force with the given torque is 0.23.

0.2 is usually for steel on steel.

Hot-dip galvanized bolts can use 0.25 (source). According to your linked source, it seems to be a possible finish for your bolt.

 

[Dullard]

I may be confused, but:

Any equation which relates torque to tension (in a bolt) and doesn't (somehow) include thread pitch may be missing something.

 

[Lnewqban]

Welcome, @Factotum !

Could you do a simple wedge calculation disregarding friction, which is the maximum clamping force possible?
Your mechanical advantage should be π x diameter / pitch.

Please, see:
http://emweb.unl.edu/negahban/em223/note16/note16.htm

 

[jrmichler]

Per my understanding, the ultimate tension specification conveys the force needed to pull the anchor from the concrete. An event that usually results in fracturing and dislodging a cone of concrete along with the anchor.

I read it the same way.

Again, if my understanding and calculations are correct, the installation tensile force of 4,000 lbs. exceeds the ultimate tension specification of 3,480 lbs.

True, and different cases. The tension specification is the maximum external force on the fastener. If that force is too large, the fastener comes flying out along with a cone of concrete.

The installation tensile force is the force of the nut clamping the object being clamped to the concrete. Since the clamping force is only pulling the object against the concrete, it is not trying to pull out a concrete cone. The installation tensile force needs to be greater than the maximum external force applied to that object. If the installation tensile force was less than the external force, the object would pull free due to elastic strain in the bolt.

A quick look at a bolt torque tension table verifies your tension calculation.

Any equation which relates torque to tension (in a bolt) and doesn't (somehow) include thread pitch may be missing something.

In typical threaded joint calculations, the effect of thread pitch is so small compared to the effect of friction that it can be ignored. This is not the case with ball screws, which have much lower friction and larger thread pitch.

 

[Factotum]

Hot-dip galvanized bolts can use 0.25 (source). According to your linked source, it seems to be a possible finish for your bolt.

Hi Jack action, thank you for your reply.

You are correct. Carbon steel with hot-dipped galvanizing is one type of Trubolt wedge anchor. Other types are carbon steel with zinc plating, 304 stainless steel, and 316 stainless steel. In the ITW Red Head specification, the table that indicates the 25 ft. lb. installation torque value for a 3/8 in. anchor also specifies all the aforementioned anchor types. I interpret this as meaning that the 25 ft. lb. installation torque specification applies to all the anchor types.

From what I could find, the appropriate nut factors are:

• Zinc plated carbon steel 0.22 (source)
• Hot dipped galvanized steel 0.25 (source)
• Stainless Steel 0.35 (source)

My use of a 0.2 nut factor may be a little low, but the ultimate tension specification seems to apply to all anchor types. I'm hoping the manufacturer will shed some light on the issue.

Any equation which relates torque to tension (in a bolt) and doesn't (somehow) include thread pitch may be missing something.

Hi Dullard, thank you for your reply.

I agree with jrmichler's assessment. From what I’ve researched, the friction component dominates the influence of thread pitch.

Could you do a simple wedge calculation disregarding friction, which is the maximum clamping force possible?
Your mechanical advantage should be π x diameter / pitch.

Hi Lnewqban, thank you for your reply.

The force applied at the edge of a 3/8-16 bolt due to an applied torque of 25 ft. lb. is:

​

Ft = T / r

where,

• Ft = The force tangential to the bolt's threads in pounds resulting from the applied torque.
• T = The applied torque in ft. lb.
• r = radius of bolt in feet.

Calculation:
Ft = 25 ft lb / (0.375 in./2 * 1 ft./12 in.) = 1600 lb

The mechanical advantage is calculated as follows:

MA = π * D / P

where,

• MA = The mechanical advantage
• D = The thread diameter of the bolt in inches.
• P = The thread pitch of the bolt in inches.

Calculation:
MA = π * 0.375 in / (1/16 in) = 18.85

The axial force on the bolt (ignoring friction) is calculated as follows:

Fa = MA * Ft

where,
Fa = The axial force on the bolt in pounds.
MA = The mechanical advantage.
Ft = The force tangential to the bolt threads in pounds.

Calculation:
Fa = 18.85 * 1600 lb = 30,160 lb

I found this article that indicates that only ~10% of the applied torque results in bolt tension. The rest is consumed by friction. Using this fact, I can attempt to take friction into account.

Recalculating the axial tension force.
Ft = (0.1 * 25 ft lb) / (0.375 in. / 2)(1 ft / 12 in) = 160 lb
Fa = 18.85 * 160 lb = 3,016 lb

The result is significantly lower than the 4000 lbs. I calculated previously, but it is still close to the ultimate tension spec. value of 3480.

The installation tensile force is the force of the nut clamping the object being clamped to the concrete. Since the clamping force is only pulling the object against the concrete, it is not trying to pull out a concrete cone. The installation tensile force needs to be greater than the maximum external force applied to that object. If the installation tensile force was less than the external force, the object would pull free due to elastic strain in the bolt.

Hi jrmichler, thank you for your reply.

I am trying to understand what you are trying to convey here. Per my understanding, the installation tensile force is required to wedge (i.e., compress) the anchors retention clip against the wall of the concrete hole. The resulting compressive force increases the friction between the anchors retention clip and the wall of the concrete hole. It is this frictional force that will oppose the installation tensile force, as well as any applied external force pulling axially on the anchor. Ultimately, the concrete must bear the entire axial load of the anchor, i.e., the installation tensile force plus any external axial load placed on the anchor. See Figure 1 below.

It seems to me that any axial tension force, as long as it's not too sudden, will act to wedge the anchor's retention clip against the wall of the concrete hole. Consequently, it shouldn't matter whether the axial tension force is created by the external load force or by torquing on the anchor nut.

I can understand the need to initially apply the installation tensile force by torquing the anchor nut to the specified installation torque, but after doing so, it seems that the anchor nut torque should be adjusted so that it only applies enough clamping force to securely attached the fixture to the concrete. A reduction in the axial tension force introducted by torquing the anchor nut, will allow for a larger external axial load force to be applied.

Figure 1: Force on the concrete

All that being said, I still don't understand why my calculated installation tensile force is greater than the ultimate tensile force of the anchor. I emailed the manufacture and their initial response was very vague. I responded asking if they would provide me the range of possible installation tensile force values that could occur when the installation torque value of 25 ft. lbs. is applied.

 

[Tom.G]

It seems that the flange (bottom plate) of the fixture is distributing the force of bolt tension at the concrete surface; thus the installation torque has no effect on the extraction (concrete failure) from an externally applied load.

Cheers,
Tom

(A case of a picture is worth a thousand words?)

 

[Factotum]

Hi Tom.G, thank you for your reply.

Here is my hypothesis.

First, assume there is no external load force. Tightening the anchor nut will force the anchor clip against the concrete. The resulting friction prevents the anchor from sliding out of the anchor hole. As the anchor nut is tightened, the stress in the concrete and the anchor increase. The concrete experiences stress at two places: where the anchor clip engages the concrete and where the fixture engages the concrete. See Figure 1.

Figure 1: Forces due to tightening anchor nut

If more and more torque is applied to the anchor nut, either the anchor or concrete will fail. One possibility is that the anchor begins to deform plastically and finally snaps. Another possibility is that the concrete cracks allowing the anchor to slide within the anchor hole. See Figure 2. These are only two of many failure modes.

Figure 2: Anchor clip slides in hole due to crack

Second, let’s say the anchor nut is not tightened to the point of failure, but close. If an external tensile force is added, the concrete and anchor will experience more stress. The concrete will experience more stress at the anchor clip boundary. The anchor will experience more tensile stress. See Figure 3.

Figure 3: Forces due to anchor nut tightening plus external force on fixture

In this case, the lateral forces in the concrete at the anchor clip will be larger due to the additional tension force on the anchor. I'm thinking the larger lateral forces may play a role in the concrete cracking and the anchor being pulled out.

 

[Tom.G]

Could be. My comments were aimed more at the conical failure pull-out discussion. That part of the conversation was considering that the bolt tightening tension was directly added to any vertical extraction loading.

 

[Factotum]

What prompted me to examine the Red Head Trubolt anchor specification was my desire to know how much clamping force would be applied to a fixture when the anchor installation torque was applied. I need to design a wedge shim to allow a railing post to be installed on a 11° ramp. One idea is to cast the wedges from polyurethane. I wanted to know the clamping force of the anchors so I could calculate how much compression force the polyurethane wedges would be subjected to.

I did some online research to find the clamping force equation, F = T/(K*D). I decided to do a sanity check on the equation by comparing its results with the Red Head Trubolt anchor specification. Unfortunately, the clamping force I calculated using the Red Head Trubolt anchor installation torque seemed quite high and not consistent with the ultimate tensile force specification. My first thought is that I was not doing something right.

Rather than go directly to the manufacturer to rectify the inconsistency, I chose to post to this forum as it could be that I was missing something obvious. When no glaring issues with my calculation was identified in this post, I brought the inconsistency to the attention of Red Head. Perhaps they could shed some light on my problem. Unfortunately, Red Head would not directly address my question. Instead, I was essentially told to follow the specification and everything will be fine.

Unfortunately, the specification does not give me any idea of what clamping force I can expect. At least, I don’t know how to extract it from the information provided.

So, I am still at square one. How do I gain confidence that the F = T/(K*D) equation is accurate? Well, at least somewhat accurate. All my research indicates that calculating clamping force from torque has an uncertainty of ± 25%.

Can you provide me with your assessment of the F = T/(K*D) equation and the fact that I calculated a clamping force of 4000 lbs. for an applied torque of 25 ft. lbs.? Do you believe that a 4000 lb. clamping force is in the ballpark?

Thank you for your help.

 

[Factotum]

Also, is there any way I can edit my original post? I've determined that the K in the F = T/(K*D) equation is actually referred to as the torque coefficient.

 

[berkeman]

Also, is there any way I can edit my original post? I've determined that the K in the F = T/(K*D) equation is actually referred to as the torque coefficient.

It's better to post the correction as a new reply, to keep from confusing folks who read the thread going forward. Go ahead and post a reply with the correction, and use the Report link in your post to ask the Mentors to update the initial post with a note pointing to the correction. Thanks.

 

[Dullard]

If I understand your concern (from post #10), there may be a "Gordian" solution:

Perform the "Anchor Setting" and "Final Assembly" as 2 independent operations. The wedge need not be present when installation torque is applied. I understand that this may be more/less feasible depending on the circumstances.

 

[jrmichler]

Can you provide me with your assessment of the F = T/(K*D) equation and the fact that I calculated a clamping force of 4000 lbs. for an applied torque of 25 ft. lbs.?

Search torque tension chart. That will bring up charts of torque vs tension for different types and sizes of bolts, and also different assumptions on thread friction. Look for a bolt of similar type and size. You might have to extrapolate to find the tension for your specific torque. Any discrepancies between different charts are normally due to different assumptions about thread friction.

 

[Dullard]

Where it really, really, really matters (like mounting turbopumps on Space Shuttle Main Engines) it is common to measure (ultrasound) the length of the "stretched" bolt.

 

[Factotum]

Correction to original post!

In my original post, I referred to K in the equation F = T/(K*D) as “the friction of the bolt contact. (0.2 for steel)”. This is incorrect. Although friction is a dominant factor in the determination of K, characteristics of the bolt thread also play a role. Hence, it is incorrect to refer to K as having to do with friction alone. See this resource for a more detailed explanation of K.

In the aforementioned resource, K is referred to as the “torque coefficient”. I have also seen it referred to as the “nut factor” in other publications. I am going to use “torque coefficient”. As such, the equation should be as follows.

F = T/(K*D)​

Where​

• F is the tensile force.
• T is the torque applied to the anchor nut.
• K is the torque coefficient.
• D is the diameter of the bolt.

​

 

[Factotum]

It's better to post the correction as a new reply, to keep from confusing folks who read the thread going forward. Go ahead and post a reply with the correction, and use the Report link in your post to ask the Mentors to update the initial post with a note pointing to the correction. Thanks.

Thank you berkeman. I have done as you recommended.

 

[Factotum]

If I understand your concern (from post #10), there may be a "Gordian" solution:

Perform the "Anchor Setting" and "Final Assembly" as 2 independent operations. The wedge need not be present when installation torque is applied. I understand that this may be more/less feasible depending on the circumstances.

Thanks for the suggestion. My goal is to know how much force is being applied to the wedge so I can use it to select a suitable wedge material. To do so, I need to determine the amount of clamping force exerted by whatever torque is applied; even if that torque is less than the installation torque.

Search torque tension chart. That will bring up charts of torque vs tension for different types and sizes of bolts, and also different assumptions on thread friction. Look for a bolt of similar type and size. You might have to extrapolate to find the tension for your specific torque. Any discrepancies between different charts are normally due to different assumptions about thread friction.

Thanks for the input. I have looked at several bolt torque charts. Based on the provided clamping loads and recommended tightening torque, I am confident that I am correctly calculating the clamping force for a given torque.

 

[Factotum]

Thank you all for your help. I now feel confident I can perform the calculations I require. There is a lot of uncertainty, but that's engineering .