Ok..  I am building an arbor over my patio.  I have checked my local
building codes no permit or inspection is required.  I still want to
ensure I have a safe and structurally sound design.  My problem lies
in the wood I have chosen, Ipe. I cannot find a span table for this
wood, it is very heavy (69lb/ft^3) and very strong (modulus of
elasticity 3,140,000).  The design is straightforward: two main beams
spanning 14ft.  Between the two beams will be 11 joists 10 ft long. No
roof, no additional loads, but the structure itself will weigh nearly
3000 lbs.  So the question: what size do I need for the main beams?
Most importantly: What is the calc for this (be specific and
detailed).

Jeff

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Request for Question Clarification by redhoss-ga on 09 Feb 2006 06:35 PST

I will need to know the height/width of the beam and how the beam is
oriented (hopefully longest dimension vertical). Also, do you know the
yield strength of the wood.

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Clarification of Question by jscott82-ga on 09 Feb 2006 07:06 PST

The height/width of the beam is open.  The only stipulation is it
would need to be composed of standard dimentional lumber.  I was
thinking of a pair of 2x12.  But again this is open.  The best
technical data I can find for Ipe is http://ipe-wood.com/tech.html. 
Let me know if this is not sufficient.

Wow Jeff, that is some seriously bad boy lumber. Here are the calcs:

First we need the section modulus and moment of inertia for a 2x12.
(finished dimensions 1.5 x 11.5)

S (section modulus) = bd^2 / 6 = 1.5 x 132.25 / 6 = 33.06 in^3
I (moment of inertia) = bd^3 / 12 = 1.5 x 1520.875 / 12 = 190.1 in^4

Next we need a reasonable weight to use. We will assume 2x12 beams and 2x6 joists:

2x12x14' weight = 2(1.5 x 11.5 x 14 x 69 / 144) = 232#
2x6x10' weight = 11(1.5 x 5.5 x 10 x 69 /144) =   435#
TOTAL                                             667#

PSF = 667 / 140 sq ft = 4.76   say 5 PSF

The maximum bending moment is found by:

M = (wl^2 / 8) x 144 sq in per sq ft
Where w is applied load in # per ft = 5 x 5 = 25

M = 25 x 196 x 144 / 8 = 88,200 in#

The bending stress is:

s = M / S = 88,200 / 33.06 = 2,668 psi

Compare this to the bending strength of 22,560 psi (from website) and
you are very safe. Safety factor = 25,360 / 2,668 = 9.5

Now we will look at deflection:

D = (5wl^4 / 384 E I) x 1728 cubic inches per cubic ft
Where E is the modulus of elasticity = 3,140,000 psi

D = (5 x 25 x 38,416 / 384 x 3,140,000 x 190.1) x 1728 = .036 inch
Again this number is very safe.

Just out of curiosity I ran the numbers for a 2x6 beam. To keep from
boring you further (and me also) here are the results:

S = 7.56 in^3
I = 20.8 in^4

s = 11,666 psi 
Now we have reduced the safety factor to 25,360 / 11,666 = 2.17
I would be a little nervous with this number.

D = 0.33 inch

Obviously your first inclination to use 2x12 beams is a very sound
choice. You could certainly use 2x8 beams, but the question then
becomes more about aesthetics. The 2x12 beams appear more massive and
that may be the look you want.

Hopefully you have been able to follow all of this. However, please
ask for a clarification if there is something you don't understand.

Good luck with your project, Redhoss

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Clarification of Answer by redhoss-ga on 09 Feb 2006 20:24 PST

No problem. The equations would be:

Maximum bending moment:
M = (P x l) / 4
Where P is the point load in lbs.

Deflection:
D = (P x l^3) / 48EI

These formulas can all be found in the AISC Manual of Steel Construction.

redhoss,
Thanks that is exacty what I am looking for.  I have run into this
same problem with other materials and will keep the math for future
projects.  This realy helps alot.  Not part of my initial question but
if you would be so kind.  How would the bending stress equation change
if it were a point load?  Thanks Again.

Redhoss,

>M = (wl^2 / 8) x 144 sq in per sq ft
>Where w is applied load in # per ft = 5 x 5 = 25

>M = 25 x 196 x 144 / 8 = 88,200 in#

Is there a unit error here?  You've multiplied 25 pounds/ft by 196
ft^2, which gives you an answer in pounds-ft.  To convert to
pounds-inches, multiply by 12, not 144.

The resulting bending stress is quite a bit smaller; the "safety
factor" would be positively enormous.  Note that the usual bending
stresses allowed (Fb) under code for joists made of stuff like Douglas
Fir are on the order of 1000 psi -- I'm not sure how to compare the
Ipe-wood 25,400 psi number to this, but in any case the arbor will
stay up just fine.

The deflection calculation looks right (I checked against some beam tables).

I get 972 psi bending stress for 2x6 main beams, which is within the
code value for Douglas Fir, so I assume it would be for Ipe.

And this makes intuitive sense as well.  If I dropped a big load on a
14' 2x6, and the thing sagged 1/3 of an inch, I'd trust it to hold.  I
sure wouldn't let anyone walk on it, though.

And so my reservation with all these calculations is that they're for
a dead load only.  Usually one has to design for a live load as well
(folks standing on the thing).  What does code say about arbors?

Finally, and I'm much less sure of this, is modulus = bd^2 / 6
correct?  If I integrate:

integral(-d/2, d/2) of b*x*dx

I get

b*d^2/4

If I try the integral(-d/2, d/2) of b*x^2*dx, I get bd^3/12, just as
you wrote, for moment of inertia.  But perhaps there is a convention
of which I am unaware.

Very good iainmcclatchie-ga. I did use the wrong factor to convert to
in-lbs. I was lucky in this case that it made no difference in the
outcome. In my defense, I used "w" (lbs/ft) rather than "W" (lbs) and
didn't quite compensate. Thanks for keeping me honest. Yes, I would
normally apply a live load. However, Jeff stated in his question, "no
additional loads". I have never seen any code requirements for arbors.
The value I used for section modulus is correct. What you have derived
is the "polar section modulus" usually denoted as "J".