But he is a chipper fellow. It doesn't offset his tenuous grasp of
logic and engineering - still, it must count for something.
Do you know that TV show NUMB3RS? Nick would be a great "guess" star.
While he was running off half-cocked and making stupefying assumptions,
the stars of the show could proceed to resolve things in a rational
manner. Kind of a Point/Counterpoint thing.
Nick, this is not a mental exercise. We're not playing with numbers
because we're bored. It's a real guy's house, with a real question
about the structure. With out real information about snow loads,
earthquake, wind loading, all of that fun stuff you ignore, and which
can _easily_ supercede any occupancy or dead load, the numbers are
imaginary. They have no meaning. So what's the point in _guessing_ at
The fact is, you really don't care whether it's a real situation of not
- a real house or not. You see no difference because your first love
is juggling numbers. To each, his own. But please don't go selling
your "solution" as anything other than a mental exercise.
On 21-Mar-2006, email@example.com wrote:
Estimating its strength doesn't answer his question. The bafflegab
you post might mislead the naive into believing it is an answer.
You're being irrisponsible and can't bring yourself to admit it.
Most of the other responses have been of the "you haven't provided
enough information" sort, because - he hasn't provided enough information.
Like the response below? :-)
Article: 797996 of alt.home.repair
Subject: Re: Beam advice
Date: 19 Mar 2006 07:55:38 -0500
Organization: Villanova University
Depends on the kind of wood and the load and its distribution.
Like, what's the load on the beam? With bending moment M = WL/8 in-lb and
a total uniform load W in pounds and L = 14x12" and S = M/f = bd^2/6 in^3
and f = 1000 psi and d = 11.25" and b = 5" (3x1.5+0.5, if half the plywood
grain runs lengthwise), W = 8bd^2f/(6L) = 8x5x11.25^2x1000/(6x14x12) = 5022
pounds. You might make it stronger by substituting some metal for plywood.
On 20-Mar-2006, firstname.lastname@example.org wrote:
Your numbers are useless in the absence of load information and
details on the beam support. There are two sides to the general
problem of building a structure, P and R - loads and resistance.
You are focussing on only one.
I had a very similar situation where I removed a chimney. I kick and
kick and kick myself for not hiring a structural engineer before and
DURING the work. Hired one after the fact, engineer thinks the work
may be ok but I do not. We have had many rain leaks (even after
replacing the roof twice) and now I may have to rebuild a portion of an
exterior wall because I think there has been movement. My theory is
that the chimney was providing stiffness to the structure, like a
column. In our case, the chimney was also supporting wood beams which
were difficult to remedy. I think the structure is moving differently
and somehow letting water in. Don't be a big dumby like I was. I
have spent 5x the cost of the engineer and still have water dripping
into my house because of movement. Also, if you hire a structural
engineer make sure you have them come to the site and approve the work
before it gets covered up!
Structural engineering, which he also teaches and writes books about.
He's the lead engineer on a $20 million building now. He uses lots of
Unistrut in his work.
We talked about that ("page 61") and I asked where the curve came from.
He said Unistrut has several failure modes. I guessed the curve came from
the main twisting one with the open side up, vs the local buckling one for
short lengths, with a safety factor. Where do you think it came from?
Perhaps you can answer this question: "Exactly how much perturbation is
needed to cause torsional longitudinal failure as a function of load?"
You might say "Exactly zero, at full load," or "Exactly zero, at the
derated load on page 61," but neither of us would believe that :-)
Then again, from a quantum-mechanical point of view, there is an
infinitestimal but finite probability that every molecule in a 10'
strut will suddenly decide to be somewhere else for an instant...
On 21-Mar-2006, email@example.com wrote:
Writes books? - give us his name.
He'd better brush up on his engineering skills...
Guesses are not what you need - you need knowledge.
Tell your idiot friend to look up "lateral-torsional buckling" in a real
engineering book - e.g. page 203 of "Structural Steel Design". Lambert Tall
ed. 2nd edition Ronald Press Company, NY.
Lateral-torsional buckling is critical in slender beams - and a ten foot long
unistrut supported only at the ends qualifies. Open channel sections
are more sensitive to it than closed sections (as I have already pointed out).
To prevent this, you have to either brace the beam along the length or limit
the load and that's what the load reduction factors are for. The OP was
asking for a closet rod unsupported along its length, hence only the latter
approach is appropriate.
You're an blowhard that thinks he know more than he does.
Yes, esp after the FIT exam, with questions about dam design, crystallography,
electronic controls, and so on. In PA, an electrical PE can legally design
bridges and skyscrapers. And pediatricians can do brain surgery...
What's your point? That there are stupid regulations with lots of
loopholes in existence?
Or are you suggesting that since you don't know anything about the
specific design requirements of the original question, you can
concentrate on Beam Theory 101 to "offset" the lack of information?
Gee, my car's not starting, but I don't know anything about engines.
Let me make sure the tires are at the right pressure...
Just give up on Nick. He's always long on equations but completely
lacking in any common sense or practical real world experience. This
whole thing is laughable. He tells the guy how much weight a bunch of
2X12's can hold, as if that is the solution to his problem. Like the
guy already knows the load that's going to be on the beam, how it's
distributed, what structural support there is at the endpoints, etc.
Laughable, yes, but I do get concerned that someone who doesn't know
any better will believe his spiel.
He reminds me of a kid that is _so_ proud of their newfound knowledge,
forgetting that with knowledge comes responsibility. He's
irresponsible with his knowledge, and careless with his assumptions.
Many people on usenet are. Maybe I just expect more from him, I don't
After reading your additional information and viewing the photo, I would
have to say that I feel all the stronger than an engineer is not optional.
I can't express this in any other way, but to say:
It would be extremely foolish not to pay the cost of an engineer. Keep
in mind that if it fails and you did not use an engineer, your home
insurance will likely be voided.
First off, I am a structural engineer and second, forget about the
insurance costs or voided policy. If the building fails and someone
is inside, you may lose lives.
To the OP: Would the cost of an engineering opinion be worth the cost
of losing a family member?
I followed most of this thread and while I understand the concepts of
engineering design and code, most of the thread is valid. I will say,
rather than get stuck on "too" much theory, I'd follow the local
building codes for this situation (I didn't look at any pictures tho).
And besides the sizing of the beam, the number of fasteners or
location in the flitch beam and the ends will also matter
(inotherwords, what good is a correct beam size if the connections
fail). And I don't think live load reductions will come into play
here tho I'm not up on all local codes but I've used the UBC codes
some time ago.
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