I am making a simple spreadsheet that will calculate the capacity of a double angle bolted connection, similar to what is done in AISC SCM 14th ed., Table 10-1. In the description of that table, it says that these tables check bolt shear, bolt bearing on the angles, shear yielding of the angles, shear rupture of the angles, and block shear rupture of the angles. So far I believe that my spreadsheet correctly calculates the bolt shear and shear rupture of the angles because it agrees with the tables when these modes govern. For bolt bearing/block shear rupture I am using equation J3-6a. For shear yielding of the angles I am using the information in Section G4 of the Specification. For this post let's say (4) rows 3/4" A325 bolts, threads included, standard sized holes, with 1/4" thick angles. The allowable load capacity according to the SCM Table 10-1 is 67.1 kips. Can someone explain to me how to calculate this capacity? I would really appreciate any help. I have also attached the spreadsheet that I am working on. I am not expecting anyone to pour through it but have attached it if it helps to see what I am doing.
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AISC SCM 14th ed., Table 10-1, All Bolted Double Angle Connections
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Note that this table changed in the 15th edition and will almost certainly change again in the 16th. This is due to the subject of the user note in Section 3.10 of the 14th. I haven't run numbers, but I'm pretty sure the change will result in strengths that are nontrivially lower than in the 14th Table 10-1.
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Thank you for the response 271. There is a copy of the 15th edition in my office, maybe I will take a look at that. I am curious what the change is. Reading in the Commentary of the 14th edition, Section J3.10 it says "While the effective strength of some bolts in the connection may be less than others, the connection has enough ductility to allow all of the bolts to reach their individual effective strengths." I wonder if that is what has changed. Thanks.
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- #6
Nutte - Thanks for the response, but I am still not sure how this is. Looking at section J4.3 of the Spec (2010) Block Shear Strength is defined as when there are failure surfaces in tension and failure surfaces in shear. I don't think that applies in this situation where the bolts are in a single row. The Block Shear equation would essentially just degenerate to the shear yielding and shear rupture equations (except the shear yielding would have a factor of safety of 2). You are probably right, I just am not getting it through my thick skull yet.
JAE - Thanks for the tip. I will add that check. I think that is sort of what the bottom part of Table 10-1 does so I can check it from there.
JAE - Thanks for the tip. I will add that check. I think that is sort of what the bottom part of Table 10-1 does so I can check it from there.
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- #7
I reached out to AISC solutions and they sent me this email (SPOILER ALERT: I was wrong in my previous post).
Me - I am trying to make designing with steel easier for my firm and am making a spreadsheet that does basically what Table 10-1 in the SCM 14th edition does, get the capacity of an all bolted double angle connection (at least for the angles and bolts). I have a bolt shear check and a shear rupture check for the angles that I believe are working correctly because when they govern my capacity agrees with that of the Table 10-1. I also have an excessive bearing check and a shear yielding of the angles check but I think one of these aren't working properly because many times I do not get the correct capacity out of table 10-1. Lets just take an example of (4) 3/4" A325 bolts in standard holes, threads included, in 1/4" thick angles. On page 10-21 of the SCM 14th edition it says the ASD capacity is 67.1 kips. Can you tell me where that number is coming from? What is the governing limit state? What equation in the Spec I should be looking at and maybe any other pertinent information about it. I greatly appreciate it.
AISC Steel Solutions - This seems to be a popular activity lately. We have received several questions from people trying to recreate the tables. I am sort of surprised that engineers do not do this more often. We get questions with some regularity involving misinterpretations of the tables. Often engineers do not read the language that we provide to describe what is or is not considered in the tables. Sometimes even when they do, they still are left with misconceptions about what the tables can and cannot be relied upon to do. Occasionally trying to derive table values at least on a somewhat random basis would go a long way towards avoiding such potential problems. I know of instances where engineers have incorrectly assumed for decades that Manual tables checked things that they have never checked. Obviously this could have real world consequences. Trying to derive table values also could provide a quality check for the engineers to ensure that they are applying provisions of the Specification properly.
The discussion that introduces the Table 10-1 states, “Tabulated bolt and angle available strengths consider the limit states of bolt shear, bolt bearing on the angles, shear yielding of the angles, shear rupture of the angles, and block shear rupture of the angles.” These are the limit states that are considered.
67.1 kips is the block shear strength of the angles. It is calculated as:
Agv= 2(0.25)(3(4-1)+1.25) = 5.13
Anv= 2(0.25)(3(4-1)+1.25 – (4-0.5)(0.875)) = 3.59
Ant= 2(0.25)( 1.25 – 0.5(0.875)) = 0.406
MIN(0.6(58)(3.59)+58(0.406), 0.6(36)(5.13)+58(0.406)) = MIN(148, 134.4) = 134.4
ASD = 134.4/2 = 67.18 kips about 67.1 kips. The program carries all figures until the end and then rounds. I have not.
Please let me know if you have any further questions.
Me - I am trying to make designing with steel easier for my firm and am making a spreadsheet that does basically what Table 10-1 in the SCM 14th edition does, get the capacity of an all bolted double angle connection (at least for the angles and bolts). I have a bolt shear check and a shear rupture check for the angles that I believe are working correctly because when they govern my capacity agrees with that of the Table 10-1. I also have an excessive bearing check and a shear yielding of the angles check but I think one of these aren't working properly because many times I do not get the correct capacity out of table 10-1. Lets just take an example of (4) 3/4" A325 bolts in standard holes, threads included, in 1/4" thick angles. On page 10-21 of the SCM 14th edition it says the ASD capacity is 67.1 kips. Can you tell me where that number is coming from? What is the governing limit state? What equation in the Spec I should be looking at and maybe any other pertinent information about it. I greatly appreciate it.
AISC Steel Solutions - This seems to be a popular activity lately. We have received several questions from people trying to recreate the tables. I am sort of surprised that engineers do not do this more often. We get questions with some regularity involving misinterpretations of the tables. Often engineers do not read the language that we provide to describe what is or is not considered in the tables. Sometimes even when they do, they still are left with misconceptions about what the tables can and cannot be relied upon to do. Occasionally trying to derive table values at least on a somewhat random basis would go a long way towards avoiding such potential problems. I know of instances where engineers have incorrectly assumed for decades that Manual tables checked things that they have never checked. Obviously this could have real world consequences. Trying to derive table values also could provide a quality check for the engineers to ensure that they are applying provisions of the Specification properly.
The discussion that introduces the Table 10-1 states, “Tabulated bolt and angle available strengths consider the limit states of bolt shear, bolt bearing on the angles, shear yielding of the angles, shear rupture of the angles, and block shear rupture of the angles.” These are the limit states that are considered.
67.1 kips is the block shear strength of the angles. It is calculated as:
Agv= 2(0.25)(3(4-1)+1.25) = 5.13
Anv= 2(0.25)(3(4-1)+1.25 – (4-0.5)(0.875)) = 3.59
Ant= 2(0.25)( 1.25 – 0.5(0.875)) = 0.406
MIN(0.6(58)(3.59)+58(0.406), 0.6(36)(5.13)+58(0.406)) = MIN(148, 134.4) = 134.4
ASD = 134.4/2 = 67.18 kips about 67.1 kips. The program carries all figures until the end and then rounds. I have not.
Please let me know if you have any further questions.
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