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design/layout for weight bearing mortise

Started by bigmish, August 05, 2006, 10:57:30 PM

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bigmish

I have a some questions about mortise design/layout. I understand that a regular mortise looks like A in the picture below. However, I am designing a shed where the beams will be carrying significant weight. As I understand it there are basically two options for this joint: B) this joint seems unnecessarily complex to layout and cut, and C) this seems like the easiest way to lay this joint out.

Is there any disadvantage to using the joint in C? Why do people use the joint in b, is it just aesthetic?



Further, I'm intending to vertically align the beams. The following image seems to be the way to do this if I used joint C, is this correct? A few notes on how it seems as though this would be done:
- one tenon seems like it would have to be short to allow the other tenon to pass by it
- one tenon would have to have rabbit to allow the other tenon face to get flush to the post




Any answers to these questions or related comments would be helpful.
Thanks, Mischa

Raphael

I have you joint "C" all over my house as it is square ruled.
  Joint "C" is used when the beam is housed ~½" into the beam, joint "B" would be used where you need more bearing for the beam usually a full inch, possibly more.  The sloped shape allows maximum conservation of fibers in the post and channels load from above more effectively, a 90° reduction of an inch or more at the top of the joint would create a stress point (line).
  In the post diagram, the 'beam' with the shorter tennon is usually a girt (a beam parallel to the joists and most often the same size as the joists) and the one with the longer tennon is a true beam which carries the joists and therefore the majority of the floor load.  Since the girt is narrower it's housing doesn't extend all the way across the post and there isn't need of a rabbet on the beam.

  In my house frame the 8x10 Beams are housed to 7.5" measured from the far side of the 8" posts, their tennons are 7-3/8" long and 9.5" deep, measured from the top of beam.  So the housing into my post is as close to ½" as my post is to 8" and any variation from the sawmill doesn't affect my meaurements.  The mortice for the girt extends all the way to the mortice for the beam allowing a 4-3/8" tennon on the girt (enough to hold a 3/4" or 13/16" peg).  My girts are 4.5x7" and my floor joists are 5x7"; the girts are smaller to minimize the housing in the post (more conservation of fibers).  This works because the girts support half the floor load of the joists.
  I'd say the two full housings in your example make that post look really undersized.  You say the shed needs to carry significant weight on the beams, how much weight are we talking about?

What did you use to produce nice 3D drawings?
... he was middle aged,
and the truth hit him like a man with no parachute.
--Godley & Creme

Stihl 066, MS 362 C-M & 24+ feet of Logosol M7 mill

bigmish

Thanks for the response Raphael. I modeled up what I understand you to be saying but it looks a bit off to me, can you look a the following two pictures and tell me if I understood you right?

Note: the second image is too small to read due to the Forum's size restrictions, please see (Links to off site pictures are not allowed) for a larger readable version.






bigmish

Also, in answer to your question, I use sketchup (http://www.sketchup.com/).

It's a really great FREE program that it really well designed and very useful for anyone working in dimensions. It's also very learnable; I was able to understand the program well enough after watching a few video tutorials (also free online) and playing with it for a couple spare hours. I highly recommend it.

.M

Raphael

After looking at you latest picture I realise I got one detail wrong.  The housing for my main beam does extend all the way across the face of the post so there is no reduction at the end of the beam...  Much easier to cut and that much more bearing for the beam.

Also your girt and joist should be rotated 90°, you want them deeper than they are wide as they are stronger that way, and the reductions at their ends to fit into the housings usually come back ~1" and the angle out at 45°.   Here's a shot of basically the same area in my frame, you can see the reduction fairly well at the end of the joist and girt.



... he was middle aged,
and the truth hit him like a man with no parachute.
--Godley & Creme

Stihl 066, MS 362 C-M & 24+ feet of Logosol M7 mill

Tom

Is this what you were trying to do?  It is still within Forum restrictions.  :)




bigmish

Tom, I'd like to post an image that large but when I tried to upload it into my gallery it kept saying it was too large untill I shrunk it down to 300 pixel width. How did you manage?

Thanks, M

Furby

I'm guessing you didn't compress the image, that's why you had to shrink the pixels down.
It was probly the file size that was stopping you in this case.
Did you read the photo posting thread? Look under the "help" button at the top of the page.

Tom

Yes, Furby is right.  :)

There are two things that you manipulate to fix a picture for the WEB and the Forum.

One is picture size and that is controlled by Pixels.  There are so many wide and so many tall that make the picture be the size square you want. (think of pixels as inches in a square foot)  The reason for the limitation is that some people have small screens and big pictures  out-grow their screens and destroy the integrity of the size of the forum page.


The other is Kilobytes.  That is the data processing term for the number of individual building blocks that it takes to make a letter, or a word or a picture, etc.  You think of it as density.    The density can be "thinned" until the number of building blocks becomes so few that the picture is no longer discernible. You don't want to go that far, but, you need to get it thinned of useless building blocks.

By controlling the combination of the two, you can get a real good large picture with little density.

XAT, the program that I use, even lets you "optimize" specific areas of a picture.  That comes in real handy. :D

mark davidson

just wanted to mention that the sloped shoulder goes faster with the circular saw than the square shoulder.
and really not that much different to lay out.... the bottom point of the housing is the critical point in both situations...
you should watch your post when bringing beams in at the same elevation, do you have enough material left in the post to support what is going on above?

Jim_Rogers

Quote from: Raphael on August 07, 2006, 04:35:05 AM
Also your girt and joist should be rotated 90°, you want them deeper than they are wide as they are stronger that way, and the reductions at their ends to fit into the housings usually come back ~1" and the angle out at 45°.   

Raphael is completely correct in stating that your joists want to be deeper than they are wide. If you are intending on storing heavy loads then this design will need careful review of bearing surfaces, joinery as well as sizes of timbers.

Reductions on the end of timbers are used when you are creating a frame that is made out of rough sawn timbers. If the frame has been planed, and the timbers are somewhat true, then reductions are not needed.
The reduction is always on the opposite side of the timber from the reference face, measured away from the reference edge. On horizontal timbers the reference face is the top so that all flooring will lay flat. And timbers on the perimeter of the building have the adjacent face to the outside. Inner floor joists should have the reference face up and the adjacent face towards one side or end of the building so that standard distances can be measured and laid out correctly so that all joists will lay flat, level, and parallel to each other at the necessary spacing to support the load.

Also one thing to consider when placing two mortises in one post at the same elevation is that there maybe not enough post left to support the roof load coming down it.

Here is a cross section drawing based on your dimensions of what is left of the post:



If you can't read the print it says that what's left of the post is around 52 % of the total cross section.
So what you'll have to look at is the total load coming down the post from the roof at this corner post, and see if it exceeds the amount of area that can support that load.
You'll need to know the type of wood you're using and the pounds per sq inch it can support parallel with the grain from a NDS chart for that type.

For example the compression allowed in the NDS for EWP (eastern white pine) parallel to the grain Fc for grade #2 Posts and Timbers is 500 psi. So 500 x 64 = 32,000 lbs. But you have now reduced that by 48 % to 15,360 lbs.

You'll have to look at the roof area and all other loads from above that will be supported by this post at that cross section point to make sure you're not exceeding 15,360 lbs.

That may seem like a large number but when you start adding up all the loads this post will carry it may not be all that much......

Also, you're going to have to look at the loads being transmitted to the post through the bearing surfaces of the joists and tie beam tenons at the post.

Just some things to thing about.

Jim Rogers
Whatever you do, have fun doing it!
Woodmizer 1994 LT30HDG24 with 6' Bed Extension

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