4 ms·

I'm not a structural engineer, but would there be any advantage to the larger bolt spreading the shear load over a larger area of the wood? Imagine a needle-thi

by jleader 12y ago

I'm not a structural engineer, but would there be any advantage to the larger bolt spreading the shear load over a larger area of the wood? Imagine a needle-thin (but incredibly strong) bolt; it might just tear through the wood fibers under load. I've seen rings (like http://www.portlandbolt.com/products/others/splitrings.html http://www.portlandbolt.com/products/others/splitrings.html) used to spread the load over a larger area of the wood, with a bolt keeping the wood members pressed against the ring.

msandford 12y ago
In reality what you want to do is get a lot of clamp force which will then allow you to use the friction between the two members to carry a lot of the load as well. Handily in that case the load is transferred over a large area meaning that the joint could be as strong as the wood on either side of it. Here's a nice reference for torque vs clamp force for a range of fasteners: http://www.spaenaur.com/pdf/sectionD/D48.pdf http://www.spaenaur.com/pdf/sectionD/D48.pdf If you use a 1/2-13 fastener and torque to 75 ft-lbs it'll provide nearly 5 tons of clamp force (provided that said clamp force doesn't cause the wood to get squished). Now standard friction equation applies: F = u * n And wood-on-wood is between 0.25 and 0.5 so let's be conservative and use 0.25. 9000 lbs * 0.25 = 2250 lbs of frictional force keeping the one wooden member from sliding down the other one. http://www.physlink.com/Reference/FrictionCoefficients.cfm http://www.physlink.com/Reference/FrictionCoefficients.cfm Here you can see the shear strength of a variety of bolts. You'll want to look at the second page under "single shear" as that's the application we're talking about. https://nucor-fastener.com/Files/PDFs/TechDataSheets/TDS_013_Shear_Strength.pdf https://nucor-fastener.com/Files/PDFs/TechDataSheets/TDS_013... If you look here a 2x6 as a vertical column is only good to support about 900lbs so basically all of these equations and speculations are a joke. Grade 2, 1/4" bolts would have 2000lbs of shear strength each, so the only reason to go bigger is for a better interface to the lumber. But that can be largely negated so long as you have a decent amount of clamp force, for 900lbs you would need 3600lbs of clamp force at most. Which you can get with a single 5/16" fastener, and probably a pair of large washers to spread the force out to prevent crushing the wood. https://courses.cit.cornell.edu/arch264/calculators/example7.1/index.html https://courses.cit.cornell.edu/arch264/calculators/example7...
- tomsaffell 12y ago
  The 900 lb limit you mention is for an 8.5' long 2x6 loaded in compression, where buckling is the likely failure mode, hence the 900lb rating. Looking at the photos, theirs are only a few feet long. A 4' long unbraced 2x6 is rated at 4247 lbs in the page you linked. Additional to that, the design may also rely on the tensional strength of the member, which is greater still.

by jleader 12y ago

msandford 12y ago

In reality what you want to do is get a lot of clamp force which will then allow you to use the friction between the two members to carry a lot of the load as well. Handily in that case the load is transferred over a large area meaning that the joint could be as strong as the wood on either side of it. Here's a nice reference for torque vs clamp force for a range of fasteners: http://www.spaenaur.com/pdf/sectionD/D48.pdf http://www.spaenaur.com/pdf/sectionD/D48.pdf If you use a 1/2-13 fastener and torque to 75 ft-lbs it'll provide nearly 5 tons of clamp force (provided that said clamp force doesn't cause the wood to get squished). Now standard friction equation applies: F = u * n And wood-on-wood is between 0.25 and 0.5 so let's be conservative and use 0.25. 9000 lbs * 0.25 = 2250 lbs of frictional force keeping the one wooden member from sliding down the other one. http://www.physlink.com/Reference/FrictionCoefficients.cfm http://www.physlink.com/Reference/FrictionCoefficients.cfm Here you can see the shear strength of a variety of bolts. You'll want to look at the second page under "single shear" as that's the application we're talking about. https://nucor-fastener.com/Files/PDFs/TechDataSheets/TDS_013_Shear_Strength.pdf https://nucor-fastener.com/Files/PDFs/TechDataSheets/TDS_013... If you look here a 2x6 as a vertical column is only good to support about 900lbs so basically all of these equations and speculations are a joke. Grade 2, 1/4" bolts would have 2000lbs of shear strength each, so the only reason to go bigger is for a better interface to the lumber. But that can be largely negated so long as you have a decent amount of clamp force, for 900lbs you would need 3600lbs of clamp force at most. Which you can get with a single 5/16" fastener, and probably a pair of large washers to spread the force out to prevent crushing the wood. https://courses.cit.cornell.edu/arch264/calculators/example7.1/index.html https://courses.cit.cornell.edu/arch264/calculators/example7...

tomsaffell 12y ago
The 900 lb limit you mention is for an 8.5' long 2x6 loaded in compression, where buckling is the likely failure mode, hence the 900lb rating. Looking at the photos, theirs are only a few feet long. A 4' long unbraced 2x6 is rated at 4247 lbs in the page you linked. Additional to that, the design may also rely on the tensional strength of the member, which is greater still.