Block shear with odd bolt patterns

[jwkilgore]

We analyze a lot of existing lattice steel structures (electrical transmission industry) with odd bolt patterns in angle connections.

Assuming minimum end, edge, and bolt-bolt distances are met, the basic block strength is A[sub]t[/sub]*F[sub]y[/sub] + 0.6*A[sub]v[/sub]*F[sub]u[/sub].

See the attached drawing. In the top and middle connections, the obvious failure paths would be the dashed lines.

Top connection: A quick analysis would just use the "s" lengths in the shear path length calculation, a more precise calculation would use the slightly longer "ss" lengths.
Mid connection: Shear path is straight, ignoring the first hole in the second gage line. Tension path length would be f + g[sub]2[/sub] + s[sup]2[/sup]/4g[sub]2[/sub] - 1.5 holes, similar to a basic net-section calculation.

The bottom connection is the question. At what point does the failure path transition from a straight path (1) to one that passes through the second hole (2)? The s[sup]2[/sup]/4g solution for the tension path has been around for over a hundred years, but I can't find anything on a "reverse slope" scenario, where the slope is in shear+compression. The conservative answer is to just assume the strength along both lines is the same (0.6*F[sub]u[/sub]) and use the shortest path. The detail was drawn specifically such that Path 2 is slightly shorter (accounting for the extra bolt hole) than path 1. But are we being too conservative? Does it have a higher strength due to the "reverse slope" section?

I'm working on a spreadsheet where we can rapidly input the bolt pattern information and it'll check all the possible failure paths to get the controlling failure block. As of now we're using the basic shear strength value for the sloped line in compression, (0.6*F[sub]u[/sub]),and not taking any credit for the compression component.

 

[Eng16080]

the basic block strength is At*Fy + 0.6*Av*Fu

This is mostly correct, although you should include shear yielding and the Ubs factor in your expression (per Section J4.3).
I would use: Rn = 0.60*min(Fu*Anv, Fy*Agv) + Ubs*Fu*Ant
According to the code, "Where the tension stress is uniform, Ubs=1; where the tension stress is nonuniform, Ubs=0.5." Unless the bolt locations (or resultant) happen to align with the centroid of the angle, I would probably use Ubs=0.5.

See the attached drawing. In the top and middle connections, the obvious failure paths would be the dashed lines.

I mostly agree, although depending on the length of the angle leg going into the page, it's possible that the failure path could continue straight up and go through the other leg. I suppose you can probably rule this out, but based on the drawing alone, it's unclear.

The bottom connection is the question. At what point does the failure path transition from a straight path (1) to one that passes through the second hole (2)? The s2/4g solution for the tension path has been around for over a hundred years, but I can't find anything on a "reverse slope" scenario, where the slope is in shear+compression.

While I don't know for sure, I would assume that path (2) cannot occur. I assume that a block shear failure must result in a clean break of the material in the direction of the load. Whereas path (2) would leave material to resist the load in compression, I don't think this would occur. At that point, wouldn't the remaining chunk that's in compression just fail along path (1)? Per some rather extensive research that I've done on block shear failures, I never came across this topic.

I'm working on a spreadsheet where we can rapidly input the bolt pattern information and it'll check all the possible failure paths to get the controlling failure block.

I recently created something similar, with the key feature being bolts in any arbitrary location. There isn't much guidance on what to do for sloped failure surfaces. One analysis option in the tool I created is to base the strength of a sloped segment on the tension rupture strength for slope to gage ratios up to 1. Beyond that ratio, the strength is based on shear yielding/rupture using the area from hole to hole. Another option is to use the lesser of tension or shear, although this is probably overly conservative.

While the spacing/gage formula s2/4g can be used for a sloped failure surface, I found there were some cases where the equation could possibly give non-conservative results. I forget the specifics now but I think it was mostly an issue with slope to gage ratios exceeding 1. Essentially, the equation could imply a greater tension failure area than what is actually there.

There is also a recent paper titled "New Design Approach to High-Strength Steel Staggered-Hole Bolted Connection Failing in Block Shear" by Viet Binh Pham which gets into this in greater depth. I only skimmed the paper, but it seems that they propose using a combination of the tension and shear components along a sloped path to estimate the block shear strength. Their approach is probably more accurate in terms of actual behavior of the material (and less conservative than the approaches I mentioned above).

Anyway, good luck! Hopefully you have an easier time than I did creating this.

 

[jwkilgore]

This is mostly correct, although you should include shear yielding and the Ubs factor in your expression (per Section J4.3)...

The structures I'm working on aren't under AISC 360, they're under ASCE 10. It uses slightly different equations aimed purely at angle members with either one or two legs connected. The equations account for bending due to eccentric connections, etc. The shear-lag function is applied elsewhere. Also, these are all old/historic structures designed using completely different codes and typically full-scale tested to failure.

... it's possible that the failure path could continue straight up and go through the other leg...

That possibility is calculated separately as a net-section capacity. Each capacity is calculated independently (bolt shear, bolt bearing, edge/end tearout, block shear, net section, etc.) then the lowest is used. The calculation I'm discussing is purely for block shear.

While I don't know for sure, I would assume that path (2) cannot occur. I assume that a block shear failure must result in a clean break of the material in the direction of the load. Whereas path (2) would leave material to resist the load in compression, I don't think this would occur. At that point, wouldn't the remaining chunk that's in compression just fail along path (1)? Per some rather extensive research that I've done on block shear failures, I never came across this topic.

The key is that nothing is restraining the angle vertically... if it failed along that sloped path the angle would just kick upwards as it moves to the right. But yes, it would be combined shear/compression along the shear plane until it tears.

I had never considered this scenario either, but look at the top detail again. What sent me down this rabbit hole was trying to code in a 2-bolt connection similar to the left two bolts on that detail. A straight horizontal shear failure path from the center hole would pass through the edge of the left hole, so the conservative method is to subtract the full bolt diameter. Since we're subtracting the full diameter, we can also use the full diagonal distance for the shear path, but that puts the shear failure on a "reverse slope" like we're discussing. It will have to kick upwards as the angle gets pulled to the right.

So that failure path can definitely occur in the top connection. If g[sub]2[/sub] is slightly more, it can still occur. If g[sub]2[/sub] is more than s (1:1 slope), then it probably cannot occur. Back to my original question, at what point does it change to "cannot occur"? What strength can I assign to the sloped path (assuming varies and increasing as slope increases) that can be directly compared to the strength of the straight path to pick the lower value?

While the spacing/gage formula s2/4g can be used for a sloped failure surface, I found there were some cases where the equation could possibly give non-conservative results. I forget the specifics now but I think it was mostly an issue with slope to gage ratios exceeding 1. Essentially, the equation could imply a greater tension failure area than what is actually there.

You are right, it's around a ratio of 1. For a regular sloped path along the tension plane I check both shear and tension then use the lower value for that segment. I go all the way to calculating strength per unit thickness because F[sub]u[/sub]/F[sub]y[/sub] ratios vary by material. Shear is 0.6*F[sub]u[/sub]*(sqrt(s[sup]2[/sup]+g[sub]2[/sub][sup]2[/sup])-hole). Tension is F[sub]y[/sub]*(g[sub]2[/sub] + s[sup]2[/sup]/4g[sub]2[/sub] - hole).

There is also a recent paper titled "New Design Approach to High-Strength Steel Staggered-Hole Bolted Connection Failing in Block Shear" by Viet Binh Pham which gets into this in greater depth. I only skimmed the paper, but it seems that they propose using a combination of the tension and shear components along a sloped path to estimate the block shear strength. Their approach is probably more accurate in terms of actual behavior of the material (and less conservative than the approaches I mentioned above).

My first ASCE download for the year. I also skimmed it and will dig in later. Mainly it's a modification of the positive sloped section I was discussing in the last paragraph, and appears to be more precise and less conservative as you observed. I may use it to refine my calculations along the sloped path, but probably won't go quite to his level of precision.

 

[Eng16080]

The structures I'm working on aren't under AISC 360, they're under ASCE 10.

Fair enough. I'm not familiar with ASCE 10, although it seems similar to AISC 360 in checking block shear.

That possibility is calculated separately as a net-section capacity.

Makes sense. The program I created checks a net-section failure as part of the block shear failure check. I tend to forget that it's a separate calculation (although fundamentally the same thing, at least per AISC 360).

The key is that nothing is restraining the angle vertically... if it failed along that sloped path the angle would just kick upwards as it moves to the right.

Yeah, I see what you're saying. It seems obvious this would occur per the top detail. To account for this, I can see the calculation potentially becoming a lot more complicated. A few general thoughts (for which I don't know the answers):
[ol 1]
[li]Does it make sense to use the component of the tension force acting in the direction of the reverse slope portion when looking at that failure plane? Seems like it might.[/li]
[li]I would guess that the capacity of a combined shear/compression failure might be greater than a combined shear/tension failure even if there's no resistance to "kick upwards".[/li]
[/ol]

My program can (somewhat) account for the situation you're describing. The algorithm sorts the fastener locations into rows, with rows being defined in the direction of the load. Fasteners that are perfectly aligned in the direction of the load end up in the same row, but also, if they're only offset a small distance apart perp. to the direction of load, they go in the same row as well. Another (more complicated) option allows fasteners to be placed in the same row if the slope (gage/spacing) between adjacent fasteners is less than some predetermined value. This got rather complicated to implement, but anyway, in your upper example, my program would (depending on the options used) probably place the three fasteners in the same row, and later on, it would arrive at a failure path going through all three holes of that row.

So that failure path can definitely occur in the top connection. If g2 is slightly more, it can still occur. If g2 is more than s (1:1 slope), then it probably cannot occur. Back to my original question, at what point does it change to "cannot occur"? What strength can I assign to the sloped path (assuming varies and increasing as slope increases) that can be directly compared to the strength of the straight path to pick the lower value?

I think this will end up being something similar to that spacing/gage formula. If the strength is simply a function of the path length, then the point where the failure path (per your last example) changes from being a straight line to going through the last bolt will be the point where the dashed line segments of option 2 equal the length of the dashed line of option 1 (minus the holes).

Within your spreadsheet/program, from the 2nd hole, you would essentially check every viable combination of paths and then use the one with the least capacity. So, from that hole, you would check a path straight up (net tension through the other angle leg), a path to the left (option 1), a path to the final bolt, and then from there, to the left (option 2). Essentially, you do this at every bolt hole so that you create a failure path or paths (not to go down the rabbit hole further) of least resistance which contain all bolt holes (not necessarily going through them but containing them). That is at least how my program works with the exception that it doesn't backtrack to revisit previous rows, meaning that it wouldn't check option 2 in this case. It would only check that scenario if somehow those three holes were in the same row (which is unlikely based on the geometry depicted.)

This stuff is somewhat complicated to explain, but hopefully this is some help to you!

 

[jwkilgore]

I think this will end up being something similar to that spacing/gage formula. If the strength is simply a function of the path length, then the point where the failure path (per your last example) changes from being a straight line to going through the last bolt will be the point where the dashed line segments of option 2 equal the length of the dashed line of option 1 (minus the holes).

This is my current plan; assume the strength along the reverse slope is the basic shear strength and pick the shortest path. I know this is quite conservative, and I'll investigate how conservative later.

Within your spreadsheet/program, from the 2nd hole, you would essentially check every viable combination of paths and then use the one with the least capacity. So, from that hole, you would check a path straight up (net tension through the other angle leg), a path to the left (option 1), a path to the final bolt, and then from there, to the left (option 2). Essentially, you do this at every bolt hole so that you create a failure path or paths (not to go down the rabbit hole further) of least resistance which contain all bolt holes (not necessarily going through them but containing them). That is at least how my program works with the exception that it doesn't backtrack to revisit previous rows, meaning that it wouldn't check option 2 in this case. It would only check that scenario if somehow those three holes were in the same row (which is unlikely based on the geometry depicted.)

It'll be something similar. Currently the spreadsheet requires the engineer to figure out the path and input the values accordingly, but staggered block shear is a complicated subject and hard for some people to understand. I want to make it more automated for typical connections. I'm still working through data entry, data storage for calc documentation, and path-choosing algorithms. The path will always incorporate the bolts in the first gage line closest to the heel. So it'll look at all possible shear paths from the end of the member to the first bolt on the first gage, pick weakest. Then look at all paths from first bolt to second bolt, pick weakest. All the way to the last bolt in that gage line, then transition to looking at all possible tension paths to the edge.

I truly do appreciate the feedback and discussion.

 

[ProgrammingPE]

While the spacing/gage formula s2/4g can be used for a sloped failure surface, I found there were some cases where the equation could possibly give non-conservative results. I forget the specifics now but I think it was mostly an issue with slope to gage ratios exceeding 1. Essentially, the equation could imply a greater tension failure area than what is actually there.

For a section in tension with staggered bolts, once s²/(4g) is greater than the hole width, skipping the staggered bolt has a lower capacity than taking the diagonal path. This is because the decreased area that you get for passing through an additional bolt hole does not make up for the increased diagonal area.

For 5/8" diameter bolts (3/4" wide holes after accounting for add'l 1/16" for damage), spaced 3" o.c., when the stagger dimension exceeds 3", the staggered bolts should be ignored.

s²/(4g) = (3 in)²/(4*(3 in)) = 3/4"​

If there was a diagonal block shear piece in the corner, I would likely treat it as part of the tension area using s²/(4g) up until it reaches the point discussed above where skipping a bolt would control the tension rupture case. For larger s values, I would switch to treating it as part of the shear area using the full diagonal length. This will be conservative since it will have a combination of shear and tension, and the tension capacity is higher. This should keep you within accepted methods even though there isn't specific guidance for how to handle diagonal block shear.

I would also ignore any benefit of the compression component for your "reverse slope" condition and use shear capacity along the diagonal length, although I agree that this is conservative. Same thing with the combined shear and tension component in your first example where the shear length is offset a small amount. Using von Mises criterion is probably the accurate way to capture the combined shear and tension/compression.

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[Eng16080]

If there was a diagonal block shear piece in the corner, I would likely treat it as part of the tension area using s²/(4g) up until it reaches the point discussed above where skipping a bolt would control the tension rupture case. For larger s values, I would switch to treating it as part of the shear area using the full diagonal length.

This is roughly the approach I took.

This will be conservative since it will have a combination of shear and tension, and the tension capacity is higher.

If using the AISC 360 block shear equation, if the connection is eccentric and a Ubs factor of 0.5 is being used, it's possible this won't be the case. Otherwise, I think you're right.

 

[ProgrammingPE]

If using the AISC 360 block shear equation, if the connection is eccentric and a Ubs factor of 0.5 is being used, it's possible this won't be the case.

That's a good point. If Ubs is less than 1, then you need to use the minimum of the shear and tension values since either could still control.

(It's probably also worth mentioning then that the AISC commentary gives you the option of using Ubs = (1 - e/l), which may help.)