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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 torque coefficient. (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.

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  • $\begingroup$ K is not the coefficient of friction of the threaded bolt-nut inteface. K is the torque coefficient that is tabulated, or can be computed from a heuristic. See Alternative Torque Formula in engineeringlibrary.org/reference/… $\endgroup$
    – Phil Sweet
    Commented Nov 2, 2023 at 0:07
  • $\begingroup$ @ Phil Sweet. Thank you for your input and the excellent resource. I have corrected my original question to indicate that K is the torque coefficient. Despite incorrectly referring to the torque coefficient, do you believe that my calculated tensile force of 4000 lbs. is reasonable? Thanks. $\endgroup$
    – SLax
    Commented Nov 7, 2023 at 5:42
  • $\begingroup$ K isn't 0.2. Use the formula in the reference to compute K $\endgroup$
    – Phil Sweet
    Commented Nov 7, 2023 at 11:04
  • $\begingroup$ The Red Head Trubolt anchor is available: zinc plated steel, hot galvanized steel, 304 SS, and 316 SS. I assumed the friction coeff between threads and bearing surfaces to be the same. I estimated the friction coeffs as follows: zinc plated steel 0.17, hot galvanized steel 0.19, 304 SS 0.14, and 316 SS 0.14. Using these values, I calculated the torque coefficients as 0.22, 0.25, 0.19, and 0.19, respectively. Using the calculated torque coefficients, I calculate the installation tensile force as 3595 lbs, 3255 lbs, 4260 lbs, and 4260 lbs, respectively. $\endgroup$
    – SLax
    Commented Nov 10, 2023 at 23:45

1 Answer 1

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Your force equation is wrong.

F=T/(K∗D) - this equation doesn't turn torque into axial stress, which needs to convert the torque into axial force via the wedge force of a simple machine using the pitch angle of the threads. It might be the force at the surface of the bolt tangent to the radius.

As a practical matter, 25 ft pounds might break a tiny bolt, but certainly isn't much for a normal sized bolt. Car lug nuts are tightened to 148 ft-pounds. 25 foot pounds is a 1/4 inch ratchet type of torque, and no way are you lifting a car with a quarter inch ratchet wrench. Sometimes we have to take off our engineer hats and put on mechanic hats to see how things make sense.

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  • $\begingroup$ Thank you for your response. You say the force equation is wrong. Here is a link to a calculator on the Engineers Edge website that gives the same axial tension force that I calculated. See engineersedge.com/calculators/torque_calc.htm. I realize this is an estimate, but is it in the ballpark? Also, I don't believe the force constraint is about breaking the bolt. It is about overcoming the tensile strength of the concrete and causing the concrete to fracture. $\endgroup$
    – SLax
    Commented Nov 1, 2023 at 17:38
  • $\begingroup$ it is not in the ballpark, 25 ft pounds of torque doesn't make enough force to life a car. hat's a 160x advantage on a 12 inch wrench, $\endgroup$
    – Tiger Guy
    Commented Nov 1, 2023 at 18:21
  • $\begingroup$ Secondly, I have never heard of 2000 psi concrete. The junkiest stuff I have ever heard of is 3500 psi, which is used for bulk pours and sometimes delivered for foundations until I send the truck back a few times. If you are installing a two post lift or industrial equipment of any kind using wedge bolts, consider 4000 psi the absolute minimum - this is if you can verify the strength with test cylinders, and you measure the actual pour thickness in the installation areas. $\endgroup$
    – Phil Sweet
    Commented Nov 1, 2023 at 18:22
  • $\begingroup$ This is why you want the flippin' bolts to hold - google.com/… $\endgroup$
    – Phil Sweet
    Commented Nov 1, 2023 at 18:23
  • $\begingroup$ What does lifting a car have to do with my question? Also, I am talking about what appears to be an issue with the wedge anchor specification. It is the wedge anchor specification that refers to 2000 psi concrete. $\endgroup$
    – SLax
    Commented Nov 1, 2023 at 18:42

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