How do I determine the yield and tensile strength of a specific diameter of bolt?

In most cases, the strength of a given material used to make a fastener has strength requirements or parameters described as pounds per square inch (psi) or thousands of pounds per square inch (ksi). This is helpful when analyzing what grade of material should be used for a given application, but this doesn’t tell us the actual strength of that diameter of material. In order to calculate the actual strength values of a given diameter, you would use the following formulas:

Note: the formulas below do not depend on the finish of the fastener.

Ultimate Yield Strength

y * a = s

Take the minimum yield in psi of the ASTM grade (see our Strength Requirements by Grade Chart for this value), multiplied by the stress area of the specific diameter (see our Thread Pitch Chart). This formula will give you the ultimate yield strength of that size and grade of bolt.

Example: What is the ultimate yield strength of a 3/4″ diameter F1554 Grade 36 rod?

36,000 psi * 0.334 in^2 = 12,024 lbs
This is the minimum requirement for F1554 grade 36. In other words, a 3/4″ diameter F1554 grade 36 anchor rod will be able to withstand 12,024 pounds force (lbf) without yielding.

Ultimate Tensile Strength

y * a = s

Take the minimum tensile strength in psi of the ASTM grade, multiplied by the stress area of the diameter. This formula will give you the ultimate tensile strength of that size and grade of bolt.

Example: What is the ultimate tensile strength of a 3/4″ diameter F1554 Grade 36 rod?

58,000 psi * 0.334 in^2 = 19,372 lbs
This is the minimum requirement for F1554 grade 36. In other words, a 3/4″ diameter F1554 grade 36 anchor rod will be able to withstand 19,372 pounds force (lbf) without breaking.

Shear Strength

First, find the ultimate tensile strength using the formula above. Take that value and multiply it by 60% (0.60). It is important to understand that this value is only an estimate. Unlike tensile and yield strengths, there are no published shear strength values or requirements for ASTM specifications. The Industrial Fastener Institute (Inch Fastener Standards, 7th ed. 2003. B-8) states that shear strength is approximately 60% of the minimum tensile strength. For more information, please see our FAQ on bolt shear strength considerations.

Written October 10, 2011 by
Anthony PorrecoAnthony Porreco

Phone: 800.599.6926


  1. i was curious if you could tell me how much weight a 5/8 inch by 12 inch lag bolt can hold up i have a tree stand 21 feet in the air so i would like to tell people this so they feel more comfortable thank you for your time

    1. @Ryan Mozingo – Unfortunately, the value that you are looking for is impossible to calculate with any accuracy. We have an FAQ that addresses the difficulty of calculating strength values for lag bolts.

      If you are looking for a generic calculation, I can offer a strength calculation based on a mild steel rod. A 5/8″ lag bolt has a minor diameter of 0.471″, which calculates to a stress area of 0.17497 sq. in. Applying the formula for calculating tensile strength for a mild steel bolt, we get a value of 10,500 lbs. for the ultimate tensile strength of a 5/8″ mild steel bolt. The shear strength, which I think you are concerned about would be about 60% of that value. Unfortunately, there is no way of verifying what grade your bolt is, or what steel your bolt is made out of, so there is absolutely no way of verifying the validity of the above information. I would consult an engineer and have your bolts tested if you are concerned about safety.

    1. @Ranjeet – there is not a one-size-fits-all formula for proof stress. Many ASTM specifications have published tables for proof load values, but not all do. Normally, the proof load values are about 90-92% of yield,but that can vary.

  2. What is the shear strength required for an expansion bolt into concrete with a load of 3,500 # axial load. In what book is it shown in. can I download it on any website. What would be that website or manufacturer be?

  3. Example 1: Area of rectangular cross section
    width (w) = 0.505 in.
    thickness (t) = 0.5 in.
    Area (a) = W x T
    AREA (A) = 1.5 X 0.5
    AREA (A) = 0.75

    LOAD =12500 LBS
    AREA = 0.2 IN. 2
    Tensile = 12500 x 5
    tensile = 62500
    Must be an easier way of computing this Tensile Stength of Steel in question.
    most of the time I deal with ASTM OR API CODES


    1. @Ravish – Yes it is possible, but not common, that a bolt can pass the tensile test but fail the proof load test. Typically speaking, the 0.2% offset is only used when testing for yield strength, not for testing proof stress. Proof stress is a simple pass or fail test where the full size bolt is tensioned to a predetermined proofload value, held for 10 seconds, then measured to see if it elongated. Yield strength differs in that the bolt or test coupon is pulled to failure, and the yield is calculated (using the 0.2% offset method) along with the tensile, elongation and reduction of area. In order to calculate using the 0.2% offset method, you must first secure data from which a stress-strain diagram may be drawn. More detailed information can be found by looking at ASTM F606 Section or by contacting an accredited test lab.

  4. Frequently people are concerned about threads pulling through the nut (shearing the threads). If nut and bolt are made from the same material,and thread engagement is at least 55% the bolt will pull into (tensil) before the thread shears.(General Electric standard 12A1200),
    Tensil strength of any material is available in many metalurgical sites on the internet. You only need to calculate the cross section area of the bolt X tensil strength of material bolt is made of. That will tell you how many bolts are needed to support the load + safety factor

  5. I am trying to calculate the yield strength of that 5/8″ bolt and I’m not seeing how you are doing your math. If I take the area (pie x r sqd) I get 3.14 x .312 x.312 =1.962 x 36,000psi tensile = 70,650 yield. what am I doing wrong?
    Thanks Ted

    1. @Ted – The error you are making is a common one. When calculating the strength of a fastener, the tensile stress area you use is the area of the minor(root) diameter of the threads, in this case 5/8-11 is 0.226 sq in. So 36,000 x 0.226= 8,136lbs. If instead, you are trying to calculate the strength of a full sized bar with no threads, then you did the math correctly.

  6. I have a welded test coupon that is made of x70 pipe. the cross section of the coupon is .250x.250. I am using a tensile tester to measure tensile strength. what is the formula to see what tensile the material actually is?

    1. @Erinosho – There are conversion charts available to convert hardness to tensile for steel and steel alloy products, but I do not know if they are accurate for Titanium alloys. I can happily send you a copy if you are interested.

  7. how to calculate the strength of the steel for tension member without using tensile load method?
    any other method to calculate strength of steel with or without using IS(indian standard) code book?

    1. @Adhi – I am sorry, I do not completely understand your question. We can help in calculating tensile strength values for fasteners using the above methodology. If you have some specifics about what you are trying to calculate, please email me and perhaps I can assist.

  8. Sy = 58 Ksi ER316L rode yield strength
    Pr=2.94Ks Hydrostatic test pressure
    0 = 1.13in Bolt diameter
    l=1.375 in Bolt’s head size
    T (in) Weld fillet throat dimension
    P (lbs) Load applied on the bolt by the internal pressure
    SVP, comment calculer la gorge du boulons apres soudage.

  9. Im looking for a simple steal tensile strength calculator for metal tubing/axles. Specifically 35mm diameter 3mm thick 1000 mm long. Looking for the minimum tensile strength that would be needed to be applied in the center in order to distort/bend the tube/axle.

  10. I’m hanging a porch swing into a 7foot long 2×12 that is supporting my balcony. The total weight it needs to support is 700lbs and my mounting hardware has two bolt holes for each chain. Spread over four galvanized lag bolts, what’s the minimum diameter and length of a lag bolt I need?

    1. @Matt – from a bolt strength standpoint, anything 1/4″ diameter or larger would support the weight. However, the weakest point will be where the bolt screws into the wood and the strength will be limited by the pull out value of the wood. That is information we do not have. Apologies.

  11. I’m standing a 5000lb tare on top of another 5000 lb tare. Each has a 115 ton capacity. They are supporting a 100 ton rod shaped stator field. The field can move in the axial direction some which is why I’d like to bolt the two tare’s together. My question is what size bolts would I need to be able to support the stress of the field moving in the axial direction?

  12. I have a question.
    What would be the tensile strength of a fastener Shoulder bolt: .500 dia shaft 3/8-16 thread?



    1. @Al – I think shoulder bolts are made per the same standard as regular socket head cap screws (ASTM A574). If that is true, your 3/8-16 shoulder bolt would have an ultimate minimum tensile strength of 13,900lbs. You may want to confirm that your bolts were made to the A574 standard, we don’t usually handle shoulder bolts, so I am not certain.

  13. If i have a Cabinet that weights 200 to 2000 pounds and we are using 3/8″ grade 2 bolts, will we have any problem?

    thank you

    1. @Shaira – You should be OK, but it depends a bit on exactly how the bolts are being used and how the cabinets are attached to the framing. Your best bet would be to consult an engineer to confirm.

  14. Hello, I am building a 125” long by 4’ dining room table out of laminated slabs 3” thick (5 slabs side by side to make 4’) and I am going to drill 5, 1/2” holes divided equally all the way through them I am planing on using steel threaded rods to secure the table top together and I was wandering what would be the best steel to support this kind of load to prevent sagging in the middle. Thank you.

    1. @Dave – I am sorry, we we don’t have the expertise to make any recommendations like that. We do have strength and grade summaries on our website, and are happy to give to any technical information we have, but we do not have any engineers on staff.

  15. Thank you for a very concise explanation on calculating yield and tensile strength.

    Now, onto my question:

    I want to hang an elephant over my bed with a single lag bolt screwed into my ceiling. I don’t know how much the elephant weighs nor do I know if it will just be into drywall or a rafter. What lag bolt do I need (don’t use a safety factor – I like to live dangerously)?

    Yes, feel free to moderate my comment and delete my sarcasm (or whole post for that matter), but I do thank you for the information.

    1. @Pat – The biggest question would be Asian or African elephant? Let’s go with African, as they are the larger of the two. Males can weigh up 14,000lbs, so we’ll use that as a worst case scenario. A 3/4″ lag bolt, made to ASTM A307A would have an ultimate tensile strength of 14,874lbf, so it would support the weight by itself. However, the limiting factor will be the material it is screwed into, and the wood will typically yield before the lag bolt will, so you may have a problem using only one lag bolt. Not to mention that I doubt your roof/ceiling was designed to support that kind of weight. My suggestion would be to rig up a sling like they did in Smokey and the Bandit II, that supported Charlotte the elephant, and she was pregnant.

  16. 6″ dia pipe. 1/2″ wall 15″ tall capped on both ends. in Static state, approx. wt. to withstand in compression. stated in tons

  17. I am trying to determine the sheer strength (which I know is an approximation) of three different diameter Grade 5 hitch pins: 3/4 inch, 1 inch, and 1 1/4 inch. Thank you for your assistance.

    1. @Don – Using the area (3.14 x r^2) and multiplying that by the approximate shear strength (60% of tensile), I get the following values: 3/4″=31,824lbs. 1″=56,520lbs. 1-1/4″=77,301lbs.

  18. how can I calculate the strength of rear cover of dc motor whose material is aluminium…give mi hint about the formula’s required for calculation…if thickness of rear cover reduces will it affect the strength? plz guide me….

    1. @Vani- Apologies, but we do not have any information regarding this kind of material. Our expertise is limited to steel and stainless steel fasteners.

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