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# Re: Ridge beam/joist analysis

• To: <seaint(--nospam--at)seaint.org>
• Subject: Re: Ridge beam/joist analysis
• From: "Jordan Truesdell, PE" <seaint(--nospam--at)truesdellengineering.com>
• Date: Tue, 29 Jun 2004 16:05:39 -0400

Good call, Nels. Serves me right for not looking at both sides of my connection. Also your pencil analogy is great - I'll break that one out next time I have to explain it.
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Craig - one way to consider the incongruity is to note that in the local axis of the board, you will have both shear and axial stresses. If you resolve only the axial stress into vertical and horizontal components, you will end up with a net horizontal thrust at either end. By resolving both shear and axial forces into components and then summing them, you should find the horizontal reactions to be zero.
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I know that I'd solved this before, but I let myself type before I finished re-solving it in my head. The roller analogy fails spectacularly because the vertical reaction force is applied at a distance r (roller radius) from the actual support point, and depending on the angle of the board there is a moment generated about that support by the vertical reaction.
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If you've been imagining a board sliding off a 24 in 12 pitch roof, that's a bit of a different scenario: Using a point (okay, line) contact at the roof and wall, the resisting can be looked at as global (stable) or rafter-local (unstable). Why the difference? In the rafter-local frame of reference, there is a normal (shear) and parallel (axial) force at the point of contact (the vector sum of which is the global vertical force). Now the normal force is stable, but the parallel force is resisted by friction. At a u_s of 0.5, the system is stable at any applied vertical load up to about 7 in 12 (cos^-1(0.5) = (90-pitch angle) but go higher, and the parallel force overcomes friction and the whole thing starts to slide. This is the case I was thinking about.
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Nels example avoids this stability problem by providing a parallel-axis constraint, preventing the pencil from sliding across the string. Make that string frictionless, and you'll be picking up the pencil off the floor. That doesn't invalidate his example - it still provides the demonstration that there is no external net force in the system, and that - as long as you know that instability exists - you can design the structure below the bearing points for vertical reactions only.
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Jordan

At 02:18 PM 6/28/2004 -0700, you wrote:
```
```Jordan,
|
| Cut a horizontal seat for the rafter at each end, and the rafter won't
| roll -- there will be no force due to gravity that will displace the
rafter
| laterally -- even if each end of the rafter bears on a roller on the
| horizontal seat.  There are other reasons, of course, for making a secure
| connection each end of the rafter, but a lateral force due to gravity is
not
| one of the reasons.
|
| Try this: hang a pencil from two threads, one each end.  Regardless of
which
| end of the pencil is higher, or if it is horizontal, there is no lateral
| restraint needed to keep both of the threads vertical.
|
| Nels
|
| Nels Roselund
| Structural Engineer
| South San Gabriel, CA
| njineer(--nospam--at)att.net
```
```

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