Steel plate below glass rail shoe

[crss]

I have to design the connection for a glass rail shoe to a wood beam. I used a steel plate on top of the wood beam and the glass rail shoe has screws to this plate. The shoe connection to the steel plate comes from the glass rail manufacturer specs. I basically designed the connection of the steel plate to the wood beam to resist 50plf along the top of the rail. Shoe and steel plate are continuous. My steel plate is 5/16" thick x 3.5" wide. The lags are at 12"o.c. 3/4" from each edge. I am stuck on how to check the adequacy of the plate and the bearing perpendicular to the wood.

 

[Ron]

I'm confused by your description of what you need to check further. The steel plate is only serving a shear function. Its bearing function is only the weight of the glass and rail...insignificant. Am I missing something?

 

[racookpe1978]

What are the wood beam dimensions?

3/4 inch in from the edge of a wood beam is very small for a lag bolt, which works by trying to split the beam down the grain. A through bolt at least tries to compress (squeeze together) the wood fibers rather than spread them further apart.

 

[crss]

Ron, a plan checker threw these calcs for checking the bearing perpendicular to the wood and bending of the steel plate. he uses formulas that look like masonry calcs with j and k factors. It didn't make any sense to me at all. I just found his contact info this morning and will be calling him to clarify his calculations.

Racookpe, the lags are 1/4" dia and working in tension.

 

[crss]

Spoke to plan checker. He says that my calculation for T and C is not correct if I just use the moment at the base divided by the moment arm (??). I must use the formula 2M/jkd2b. And on top of that he says that the wood beam below the glass rail needs to be designed using a factor of 4x the load because of section 2407.1.1. I interpret this section as the components that hold the glass rail assembly together but not carried through the design of the beam at the edge of the deck. Section 1607.7 already confuses all by saying that a 200# load anywhere should be used for residential work. This makes sense if it is a guardrail connected with posts at 4'o.c.. If it is a continuous connection along the bottom rail, 200# anywhere does not make sense at all. Nothing will work. I see glass rails by CR Laurence installed everywhere and never saw any outrageous connection device along the bottom of those. I used 50plf for my force. My T=C=1050lbs per foot with a 2" moment arm. What am I missing??

 

[dhengr]

Crss:
We can’t see it from here, so we have no idea what you are talking about. So.... the first thing you are missing is a good sketch showing us what you are dealing with. That should include loads, sizes, dimensions, reasonable proportions, materials, etc., etc.

 

[crss]

here is the detail I have on my plans. next post will have the page where I noted my calcs and the plan checker added his comments.

 

[crss]

page with calcs and plan checker's comments

 

[JLNJ]

The tension is taken by one of the fasteners, but the other fastener is not really in compression and you can't simply use the distance between them. The compression is likely some sort of quasi-triangular distribution on the wood under the steel plate. To get the depth of the compression block, you need to use the modular ratio, similar to the design of reinforced masonry. This actually may help your fastener force, but the compression in the wood may control.

 

[crss]

I realized that after thinking over. I was only thinking of the tension. And I also realized that the reason he finds the wood failing is because he is using 200lbs concentrated load instead of the 50plf I used. When the load is distributed uniformly, the wood is ok below the plate.
Thanks!