Does anyone know if there is a place on the web that shows the load
carrying ability of different sizes of dimensional lumber over
different spans?

It sure be helpful if someone does.

Thanks

j walker

Try http://www.fpl.fs.fed.us/documnts/fp.../fplgtr113.htm

Everything you asked about - and more...
--
JeffB
remove no.spam. to email

j walker wrote:

Does anyone know if there is a place on the web that shows the load
carrying ability of different sizes of dimensional lumber over
different spans?

It sure be helpful if someone does.

Thanks

j walker

J,

Does anyone know if there is a place on the web that shows the load
carrying ability of different sizes of dimensional lumber over
different spans?

This is a good one. Be sure to select "US Species" if you are in the US.

http://www.cwc.ca/design/tools/calcs/SpanCalc_2002/

Anthony

Thanks for the links. Now I am off to find an engineer to interpret
the tables for me.

Thanks again.





On Sun, 10 Apr 2005 21:47:14 -0500, j walker wrote:


Does anyone know if there is a place on the web that shows the load
carrying ability of different sizes of dimensional lumber over
different spans?

It sure be helpful if someone does.

Thanks

j walker

Hi J Walker, --gotta love that...

Try these online calculators:

http://www.woodbin.com/calcs/index.htm

The one you may want is called the sagulator, about the 6th link down.

--
Mike Wenzloff
Wenzloff & Sons Cabinet Makers
(503) 359-4191
http://www.WenzloffandSons.com

"j walker" wrote in message
...
Thanks for the links. Now I am off to find an engineer to interpret
the tables for me.

Thanks again.





On Sun, 10 Apr 2005 21:47:14 -0500, j walker wrote:


Does anyone know if there is a place on the web that shows the load
carrying ability of different sizes of dimensional lumber over
different spans?

It sure be helpful if someone does.

Thanks

j walker

Thanks for this reply. This site is easy to use for shelves but I
found the terminology when applying the application to a beam to be
difficult.

Thanks again!



On Thu, 14 Apr 2005 00:58:45 GMT, "Mike Wenzloff" mwenz *@*
wenzloff.com wrote:

Hi J Walker, --gotta love that...

Try these online calculators:

http://www.woodbin.com/calcs/index.htm

The one you may want is called the sagulator, about the 6th link down.

j walker wrote:

Thanks for this reply. This site is easy to use for shelves but I
found the terminology when applying the application to a beam to be
difficult.

Just make the "depth" equal to the width of the beam and the "thickness"
equal to the height.

A shelf is just a "deep", "thin" beam. A beam is just a "thick" shelf that
is not very "deep".

Thanks again!



On Thu, 14 Apr 2005 00:58:45 GMT, "Mike Wenzloff" mwenz *@*
wenzloff.com wrote:

Hi J Walker, --gotta love that...

Try these online calculators:

http://www.woodbin.com/calcs/index.htm

The one you may want is called the sagulator, about the 6th link down.

--
--John
to email, dial "usenet" and validate
(was jclarke at eye bee em dot net)

I think that I use the term "beam" in a different manner.

I call a floor joist a beam and apparently that is incorrect.

This clarifies one thing; why the sagulator refers to a beam "standing
on end."

So, how do I get the sag in a floor joist?

What I am attempting to understand is how much sag should occur in a
2X8 floor joist if the span is 8 feet. The application is for a water
tank stand which will measure 8X8 and will hold an 1100 gallon water
tank. The rim joists are 2X12 and the floor joist ends rest on a
ledger nailed to the side on the 2X12. There are 9 joists in between
the 2X12.

On Sat, 16 Apr 2005 22:50:20 -0400, "J. Clarke"
wrote:

j walker wrote:

Thanks for this reply. This site is easy to use for shelves but I
found the terminology when applying the application to a beam to be
difficult.

Just make the "depth" equal to the width of the beam and the "thickness"
equal to the height.

A shelf is just a "deep", "thin" beam. A beam is just a "thick" shelf that
is not very "deep".

Thanks again!



On Thu, 14 Apr 2005 00:58:45 GMT, "Mike Wenzloff" mwenz *@*
wenzloff.com wrote:

Hi J Walker, --gotta love that...

Try these online calculators:

http://www.woodbin.com/calcs/index.htm

The one you may want is called the sagulator, about the 6th link down.

j walker wrote:

I think that I use the term "beam" in a different manner.

I call a floor joist a beam and apparently that is incorrect.

A floor joist is a beam.

This clarifies one thing; why the sagulator refers to a beam "standing
on end."

To the sagulator a beam is like a shelf stood on its edge in a manner of
speaking. Remember, the sagulator was intended for shelf deflection. The
math is the same for anything rectangular in cross section and of uniform
material properties, what is different is the words one might use to
describe different measurements. "On end" is I suspect careless
terminology--I suspect that "on edge" was meant.

To the sagulator, looking at your beam from the end, (i.e. so that you're
looking into the end grain in my pathetic attempt at ASCII art below)


(this is the top, the load you are supporting would be up here)
____________________ __
| | /|\
| | |
| | |
| | |
| | |
| | |
| | |
| | |
| | thickness
| | |
| | |
| | |
| | |
| | |
| | |
| | |
|____________________|_\|/
|----- depth ------|

And the "width" in the sagulator is the "length" of your beam (i.e. 8 feet,
only expressed in inches).

So, how do I get the sag in a floor joist?

What I am attempting to understand is how much sag should occur in a
2X8 floor joist if the span is 8 feet. The application is for a water
tank stand which will measure 8X8 and will hold an 1100 gallon water
tank. The rim joists are 2X12 and the floor joist ends rest on a
ledger nailed to the side on the 2X12. There are 9 joists in between
the 2X12.

If I read you correctly there are 9 2x8s supporting this tank, with the 2x8s
being supported on the ends by 2x12s.

To use the sagulator to get an approximation of the sag in the 2x8s, you
would plug 1100 pounds as the weight, uniform load (I'm assuming that the
tank covers the full span--if you want to be safe call it a "center load",
which will calculate a higher deflection), 96 inches as the width (8 ft x
12 inches), 18 as the depth (2 inches per beam x 9 beams), and 8 as the
thickness. You'd also of course have to pick the appropriate species of
wood. You'd then want to do the same calculation for the 2x12s to make
sure that they're up to the task.


On Sat, 16 Apr 2005 22:50:20 -0400, "J. Clarke"
wrote:

j walker wrote:

Thanks for this reply. This site is easy to use for shelves but I
found the terminology when applying the application to a beam to be
difficult.

Just make the "depth" equal to the width of the beam and the "thickness"
equal to the height.

A shelf is just a "deep", "thin" beam. A beam is just a "thick" shelf
that is not very "deep".

Thanks again!



On Thu, 14 Apr 2005 00:58:45 GMT, "Mike Wenzloff" mwenz *@*
wenzloff.com wrote:

Hi J Walker, --gotta love that...

Try these online calculators:

http://www.woodbin.com/calcs/index.htm

The one you may want is called the sagulator, about the 6th link down.

--
--John
to email, dial "usenet" and validate
(was jclarke at eye bee em dot net)

j walker wrote:

I think that I use the term "beam" in a different manner.

What you mean? Different from what?

I call a floor joist a beam and apparently that is incorrect.

Why you think that? It functions that way, a "joist" is a specific use.

This clarifies one thing; why the sagulator refers to a beam "standing
on end."

A beam "standing on end" is a column.

So, how do I get the sag in a floor joist?

I plugged in 1100 gal * 62.3 lb/ft^3 * 0.1337 lb/gal / 9 joists == 1020
lb load per joist. Putting in a span of 96", height of 7.5 and depth of
1.5 for an tuba8 and a long-leaf pine (SYP) for specie I got a
deflection of roughly 0.1". For a 2x10 it was closer to 0.05.

Depending on your actual orientation/layout, it might be better to
consider the load distributed on 8 instead of 9 intermediate joists.
That'd raise the load by 9/8 and increase the deflection to 0.15",
roughly for the 2x8.

HTH...

BTW, seems reasonable altho I didn't hand check the calcs returned by
the robot...

What you need to make sure of (at a bare minimum) is that you've got
adequate structural connection details throughout the entire
structure, including whatever the 2x12's of which you speak are resting
upon and what their span/spacing is to ensure they have enough
additional loading bearing capacity.

J. Clarke wrote:

j walker wrote:

I think that I use the term "beam" in a different manner.

I call a floor joist a beam and apparently that is incorrect.

A floor joist is a beam.

This clarifies one thing; why the sagulator refers to a beam "standing
on end."

To the sagulator a beam is like a shelf stood on its edge in a manner of
speaking. Remember, the sagulator was intended for shelf deflection. The
math is the same for anything rectangular in cross section and of uniform
material properties, what is different is the words one might use to
describe different measurements. "On end" is I suspect careless
terminology--I suspect that "on edge" was meant.

To the sagulator, looking at your beam from the end, (i.e. so that you're
looking into the end grain in my pathetic attempt at ASCII art below)


(this is the top, the load you are supporting would be up here)
____________________ __
| | /|\
| | |
| | |
| | |
| | |
| | |
| | |
| | |
| | thickness
| | |
| | |
| | |
| | |
| | |
| | |
| | |
|____________________|_\|/
|----- depth ------|

And the "width" in the sagulator is the "length" of your beam (i.e. 8
feet, only expressed in inches).

So, how do I get the sag in a floor joist?

What I am attempting to understand is how much sag should occur in a
2X8 floor joist if the span is 8 feet. The application is for a water
tank stand which will measure 8X8 and will hold an 1100 gallon water
tank. The rim joists are 2X12 and the floor joist ends rest on a
ledger nailed to the side on the 2X12. There are 9 joists in between
the 2X12.

If I read you correctly there are 9 2x8s supporting this tank, with the
2x8s being supported on the ends by 2x12s.

To use the sagulator to get an approximation of the sag in the 2x8s, you
would plug 1100 pounds as the weight,


My apologies I misread "gallons" as "pounds"--that should be 8800 pounds (1
gallon weighs 8 pounds to a good approximation--8.33 to be exact).

uniform load (I'm assuming that the
tank covers the full span--if you want to be safe call it a "center load",
which will calculate a higher deflection), 96 inches as the width (8 ft x
12 inches), 18 as the depth (2 inches per beam x 9 beams), and 8 as the
thickness. You'd also of course have to pick the appropriate species of
wood. You'd then want to do the same calculation for the 2x12s to make
sure that they're up to the task.


On Sat, 16 Apr 2005 22:50:20 -0400, "J. Clarke"
wrote:

j walker wrote:

Thanks for this reply. This site is easy to use for shelves but I
found the terminology when applying the application to a beam to be
difficult.

Just make the "depth" equal to the width of the beam and the "thickness"
equal to the height.

A shelf is just a "deep", "thin" beam. A beam is just a "thick" shelf
that is not very "deep".

Thanks again!



On Thu, 14 Apr 2005 00:58:45 GMT, "Mike Wenzloff" mwenz *@*
wenzloff.com wrote:

Hi J Walker, --gotta love that...

Try these online calculators:

http://www.woodbin.com/calcs/index.htm

The one you may want is called the sagulator, about the 6th link down.


--
--John
to email, dial "usenet" and validate
(was jclarke at eye bee em dot net)

Duane Bozarth wrote:
....
I plugged in 1100 gal * 62.3 lb/ft^3 * 0.1337 lb/gal / 9 joists == 1020
lb load per joist. Putting in a span of 96", height of 7.5 and depth of
1.5 for an tuba8 and a long-leaf pine (SYP) for specie I got a
deflection of roughly 0.1". For a 2x10 it was closer to 0.05.
....

That of course, assumes uniformly distributed load...I just noticed
after posting the "stand" part...how the stand is configured determines
how the load is transferred to the floor and whether the floor can
support the concentrated load if it has small contact points is yet
another question...

"Duane Bozarth" wrote in message

What you need to make sure of (at a bare minimum) is that you've got
adequate structural connection details throughout the entire
structure, including whatever the 2x12's of which you speak are resting
upon and what their span/spacing is to ensure they have enough
additional loading bearing capacity.

snip of much good advice

.... and the 'crown' always goes up.

--
www.e-woodshop.net
Last update: 4/17/05

"J. Clarke" wrote:
....

If I read you correctly there are 9 2x8s supporting this tank, with the
2x8s being supported on the ends by 2x12s.

To use the sagulator to get an approximation of the sag in the 2x8s, you
would plug 1100 pounds as the weight,

My apologies I misread "gallons" as "pounds"--that should be 8800 pounds (1
gallon weighs 8 pounds to a good approximation--8.33 to be exact).

But the calculation is on each joist so the load/joist is only 1/9 (or
1/8 if one wants to discount an end and be a little more conservative)
so the weight to plug into the robot is roughly 1000 lb (or 1150 lb if
use 8 instead of 9).

Duane Bozarth wrote:

....
What you need to make sure of (at a bare minimum) is that you've got
adequate structural connection details throughout the entire
structure, including whatever the 2x12's of which you speak are resting
upon and what their span/spacing is to ensure they have enough
additional loading bearing capacity.

BTW, don't neglect the weight of the tank and other structural material
as well as whatever other dead load there is plus there should be a
safety margin as well.

And, for the supporting 2x12 and remaining portion of the structure,
don't fail to include the other loads they're required to support as
well--which may not be simply determined w/o someone much more expert
looking at the design...

Remember the weight of this tank/contents is additional load beyond
other structural loads.

Duane Bozarth wrote:

"J. Clarke" wrote:
...

If I read you correctly there are 9 2x8s supporting this tank, with the
2x8s being supported on the ends by 2x12s.

To use the sagulator to get an approximation of the sag in the 2x8s,
you would plug 1100 pounds as the weight,

My apologies I misread "gallons" as "pounds"--that should be 8800 pounds
(1 gallon weighs 8 pounds to a good approximation--8.33 to be exact).

But the calculation is on each joist so the load/joist is only 1/9 (or
1/8 if one wants to discount an end and be a little more conservative)
so the weight to plug into the robot is roughly 1000 lb (or 1150 lb if
use 8 instead of 9).

Note that rather than dividing the load I multiplied the width to encompass
all members, so for the method I decribed 8800 would be the load to apply.

--
--John
to email, dial "usenet" and validate
(was jclarke at eye bee em dot net)

On Sun, 17 Apr 2005 09:58:42 -0500, j walker wrote:


What I am attempting to understand is how much sag should occur in a
2X8 floor joist if the span is 8 feet. The application is for a water
tank stand which will measure 8X8 and will hold an 1100 gallon water
tank. The rim joists are 2X12 and the floor joist ends rest on a
ledger nailed to the side on the 2X12. There are 9 joists in between
the 2X12.


I would recommend that you have an engineer do your load calculation
and framing plan for this.

Keep in mind a couple of things:

The liquid volume alone will weigh about 9160 Lbs. before you add
anything in for what will hold the water, temporary point loading
during servicing, base level allowance for live load, etc.

Let's say that your tank and water come in at around 10,000 Lbs.
(we'll leave out the other stuff for now).

You are proposing to carry this on eight joists which will rest on a
ledger. Assuming the ledger to be 2X stock (1-1/2" net) you will have
36" to rest your 10,000 Lb. load on (1-1/2 x 1-1/2 x 16 = 36).

That is about 278 Lbs. per Sq. In.

Let's think about the fact that you are using a 2 x 12 rim joist and
butting 2 x8 joists to it on a ledger.

The 2 x 12 is 11-1/4" and the 2 x 8 is 7-1/4", leaving 4" for the
ledger. Unless you specify that the ledger must be ripped from a 2 x
6 to the full 4" height, you may wind up with a 2 x 4 ledger sitting
1/2" above the top plate. I wouldn't want to see the fasteners handle
that load all by their lonesome. Neither would I want to see the
space shimmed with typical shim stock.

I would want to make sure that the plate was doubled and that the
joists rested directly over the vertical framing members below the
plate, and that the load was transmitted all the way to the foundation
in the same fashion.

It seems to me that the amount of sag is not the real problem. That
is not to say that it is not a problem at all. I would glue and screw
1/2" min. ply (not CDX) to the underside of the joists and glue and
screw the subfloor on the top, creating a sort of torsion box on the
cheap.

The other thing that I would consider is that you are attempting to
hold a lot of water on a framing system that will decline in its
ability to hold the weight if it is subjected to water damage.

Anything that holds water will eventually leak.


These sorts of considerations are what make people like me pick up the
phone and transfer the liability to my friendly local engineering
firm.



Tom Watson - WoodDorker
tjwatson1ATcomcastDOTnet (email)
http://home.comcast.net/~tjwatson1/ (website)

On Sun, 17 Apr 2005 15:41:33 -0400, Tom Watson
wrote:


You are proposing to carry this on eight joists which will rest on a
ledger. Assuming the ledger to be 2X stock (1-1/2" net) you will have
36" to rest your 10,000 Lb. load on (1-1/2 x 1-1/2 x 16 = 36).


Tom,

Why isn't the calculation above (1-1/2 + 1-1/2) x 8 = 24





Tom Watson - WoodDorker
tjwatson1ATcomcastDOTnet (email)
http://home.comcast.net/~tjwatson1/ (website)

On Sun, 17 Apr 2005 18:31:52 -0500, j walker wrote:


Why isn't the calculation above (1-1/2 + 1-1/2) x 8 = 24

Because you have eight joists with two ends apiece.

This gives you your theoretical point load.

The most important thing that I said in the post, and why I repeated
it, is that you should get an engineer in on this.

Clarke and Bozarth have responded with math that does not take into
consideration the stresses and loads that are involved in this.

What you are describing is an unusual load situation that is beyond
the capabilities of most onsite guys. I would want to be very
particular in my inquiry that the stresses could be absorbed all the
way down to the foundation level.

You are contemplating sitting a huge mass of potential energy in a
position that can do great damage if things go wrong.

Good luck with it.


Tom Watson - WoodDorker
tjwatson1ATcomcastDOTnet (email)
http://home.comcast.net/~tjwatson1/ (website)

On Sun, 17 Apr 2005 18:31:52 -0500, j walker wrote:


Why isn't the calculation above (1-1/2 + 1-1/2) x 8 = 24

I didn't want to address this in the previous post because I was
trying to be polite.

Upon further reflection and given the potential ramifications of
misunderstanding, I thought it best to inform you that 1.5 x 1.5 x 8 =
18.

If you are making these kind of mistakes, for gawd's sake don't do
your own engineering.



Tom Watson - WoodDorker
tjwatson1ATcomcastDOTnet (email)
http://home.comcast.net/~tjwatson1/ (website)

"J. Clarke" wrote:
....
Note that rather than dividing the load I multiplied the width to encompass
all members, so for the method I decribed 8800 would be the load to apply.

Sorry, I missed that...

Tom Watson wrote:

....
Clarke and Bozarth have responded with math that does not take into
consideration the stresses and loads that are involved in this.
....

I certainly added beyond the simple deflection calculation that he
wasn't looking at anywhere near the whole picture...

On Sun, 17 Apr 2005 20:00:52 -0500, Duane Bozarth
wrote:

Tom Watson wrote:

...
Clarke and Bozarth have responded with math that does not take into
consideration the stresses and loads that are involved in this.
...

I certainly added beyond the simple deflection calculation that he
wasn't looking at anywhere near the whole picture...


You and Clarke were jerking each other off about deflection when the
primary problem was related to the perimeter loading.

This is the same kind of crap that goes on in the endless 'lecktricity
threads.

The only reason that the real engineers, who lurk on this newsgroup,
don't respond and smack your head, is that they would be giving
professional advice in a situation where they might be liable but
could never profit.

The OP has a situation that could result in damage to the structure
and injury to the occupants if things go wrong.

You and Clarke should sit down and have a nice cup of "shut the ****
up" when the situation is such that it could result in harm to another
person.

I don't have a problem with you generally, Bozarth, but in this
situation you could get someone hurt.


Tom Watson - WoodDorker
tjwatson1ATcomcastDOTnet (email)
http://home.comcast.net/~tjwatson1/ (website)

In article , Tom Watson wrote:
On Sun, 17 Apr 2005 18:31:52 -0500, j walker wrote:


Why isn't the calculation above (1-1/2 + 1-1/2) x 8 = 24

I didn't want to address this in the previous post because I was
trying to be polite.

Upon further reflection and given the potential ramifications of
misunderstanding, I thought it best to inform you that 1.5 x 1.5 x 8 =
18.

True, but irrelevant: he said (1.5 + 1.5) x 8 = 24, which it does.

If you are making these kind of mistakes, for gawd's sake don't do
your own engineering.

Ahem....

--
Regards,
Doug Miller (alphageek at milmac dot com)

Nobody ever left footprints in the sands of time by sitting on his butt.
And who wants to leave buttprints in the sands of time?

On Mon, 18 Apr 2005 01:34:07 GMT, (Doug Miller)
wrote:

In article , Tom Watson wrote:
On Sun, 17 Apr 2005 18:31:52 -0500, j walker wrote:


Why isn't the calculation above (1-1/2 + 1-1/2) x 8 = 24

I didn't want to address this in the previous post because I was
trying to be polite.

Upon further reflection and given the potential ramifications of
misunderstanding, I thought it best to inform you that 1.5 x 1.5 x 8 =
18.

True, but irrelevant: he said (1.5 + 1.5) x 8 = 24, which it does.

If you are making these kind of mistakes, for gawd's sake don't do
your own engineering.

Ahem....


Douglas:

Did I have to make the obvious point that the calculation was without
merit?

What the hell does 1.5 + 1.5 have to do with anything?

The thrust of the inquiry was geared towards determining the load on a
specific area, which would be calculated by L X W.

Have you become so reflexive in your posting that you ignore the
fundamentals of the argument?

BTW - that was a rhetorical question.

Ahem...





Tom Watson - WoodDorker
tjwatson1ATcomcastDOTnet (email)
http://home.comcast.net/~tjwatson1/ (website)

In article , Tom Watson wrote:
On Mon, 18 Apr 2005 01:34:07 GMT, (Doug Miller)
wrote:

In article , Tom Watson
wrote:
On Sun, 17 Apr 2005 18:31:52 -0500, j walker wrote:


Why isn't the calculation above (1-1/2 + 1-1/2) x 8 = 24

I didn't want to address this in the previous post because I was
trying to be polite.

Upon further reflection and given the potential ramifications of
misunderstanding, I thought it best to inform you that 1.5 x 1.5 x 8 =
18.

True, but irrelevant: he said (1.5 + 1.5) x 8 = 24, which it does.

If you are making these kind of mistakes, for gawd's sake don't do
your own engineering.

Ahem....


Douglas:

Did I have to make the obvious point that the calculation was without
merit?

I don't know whether it had merit or not; didn't see what led up to it. But it
was calculated correctly.

--
Regards,
Doug Miller (alphageek at milmac dot com)

Nobody ever left footprints in the sands of time by sitting on his butt.
And who wants to leave buttprints in the sands of time?

Duane Bozarth wrote:

Tom Watson wrote:

...
Clarke and Bozarth have responded with math that does not take into
consideration the stresses and loads that are involved in this.
...

I certainly added beyond the simple deflection calculation that he
wasn't looking at anywhere near the whole picture...

I didn't respond with any "math". I was simply explaining how to use the
sagulator.

--
--John
to email, dial "usenet" and validate
(was jclarke at eye bee em dot net)

Tom Watson wrote:
You are contemplating sitting a huge mass of potential energy in a
position that can do great damage if things go wrong.

Good luck with it.


Now THAT's funny!

-Phil Crow

Tom Watson wrote:

....a diatribe...

I also told him connection and support was an issue...if you didn't
think it was strong enough, sorry...

j walker wrote:

On Sun, 10 Apr 2005 21:47:14 -0500, j walker wrote:

I am beginning to wish I hadn't asked the original question. Somebody
out there needs to quit reading between the lines and assuming what
they don't know.

I gave a brief view of the project simply because I thought perhaps I
had confused someone attempting to help.

Looking back I should have merely ignored that and left well enough
alone

OR

given a detailed word picture of the platform as it has existed for
several years, complete with the 3 X 4 angle braces on all sides but
one and that one being reinforced with a 4X4X3/8 el piece of steel. I
probably should have mentioned that the 1100 gallon capacity will only
rarely be reached and only by manual override and that normally the
limit is 800 gallons at which time the water flows out the overflow
pipe. And finally it stands alone on the farm, filled by a windmill
and should a catastrophe occur it isn't likely to hurt anyone.

Famous last words. Remember Murphy's Law. If it _can_ fail, the day it
_does_ the family of the meanest trial lawyer in the country will be
standing under it.

And
just in case someone worries about the tank sinking in the ground
please be advised that it sits proudly on 8 inches of 4000 psi
concrete and the soil under the concrete was properly prepared.

I am a farmer not an engineer but I am not an idiot about how to build
something and have it stand up for a while.

Thanks to those who were helpful.



Does anyone know if there is a place on the web that shows the load
carrying ability of different sizes of dimensional lumber over
different spans?

It sure be helpful if someone does.

Thanks

j walker

--
--John
to email, dial "usenet" and validate
(was jclarke at eye bee em dot net)

IFIRC: The US Department of Agriculture has all sorts of preengineered
plans for things like the tank and stand/tower you are trying to build
online, hosted by the Extension Services, at various Universities. Try
a google search, or just call your extension agent and see if they can
help you.

Note 2, I haven't run the calcs, (and won't, I am a licensed engineer
and don't want to risk the liability), but the 2X12 perimeter is
probably undersized for the loads.

You might want to discuss your project with a Professional Engineer, the
cost shouldn't be too high (it is a pretty simple problem and shouldn't
involve too much of their time). Depending on where you are building
the tank, the Building Department will require sealed drawings before
they will give you a permit.

j walker wrote:
Does anyone know if there is a place on the web that shows the load
carrying ability of different sizes of dimensional lumber over
different spans?

It sure be helpful if someone does.

Thanks

j walker

On Sun, 10 Apr 2005 21:47:14 -0500, j walker wrote:

I am beginning to wish I hadn't asked the original question. Somebody
out there needs to quit reading between the lines and assuming what
they don't know.

I gave a brief view of the project simply because I thought perhaps I
had confused someone attempting to help.

Looking back I should have merely ignored that and left well enough
alone

OR

given a detailed word picture of the platform as it has existed for
several years, complete with the 3 X 4 angle braces on all sides but
one and that one being reinforced with a 4X4X3/8 el piece of steel. I
probably should have mentioned that the 1100 gallon capacity will only
rarely be reached and only by manual override and that normally the
limit is 800 gallons at which time the water flows out the overflow
pipe. And finally it stands alone on the farm, filled by a windmill
and should a catastrophe occur it isn't likely to hurt anyone. And
just in case someone worries about the tank sinking in the ground
please be advised that it sits proudly on 8 inches of 4000 psi
concrete and the soil under the concrete was properly prepared.

I am a farmer not an engineer but I am not an idiot about how to build
something and have it stand up for a while.

Thanks to those who were helpful.



Does anyone know if there is a place on the web that shows the load
carrying ability of different sizes of dimensional lumber over
different spans?

It sure be helpful if someone does.

Thanks

j walker

j walker wrote in news:l2r8615of6gcsqpv0i7adhvq1r8nchua8d@
4ax.com:

snip
I am a farmer not an engineer but I am not an idiot about how to build
something and have it stand up for a while.


Farmers who _are_ idiots tend not to survive all that long. Tough life,
even when you know what you are doing.

Thank you. Success in your work!

Patriarch

"j walker" wrote in message
...
On Sun, 10 Apr 2005 21:47:14 -0500, j walker wrote:

I am beginning to wish I hadn't asked the original question. Somebody
out there needs to quit reading between the lines and assuming what
they don't know.

Well, when the details are missing, some assumptions are likely going to
have to be made to develop a reasonably-complete answer.

I gave a brief view of the project simply because I thought perhaps I
had confused someone attempting to help.

Looking back I should have merely ignored that and left well enough
alone

OR

given a detailed word picture of the platform as it has existed for
several years,

That probably would have been helpful

complete with the 3 X 4 angle braces on all sides but
one and that one being reinforced with a 4X4X3/8 el piece of steel. I
probably should have mentioned that the 1100 gallon capacity will only
rarely be reached and only by manual override and that normally the
limit is 800 gallons at which time the water flows out the overflow
pipe. And finally it stands alone on the farm, filled by a windmill
and should a catastrophe occur it isn't likely to hurt anyone. And
just in case someone worries about the tank sinking in the ground
please be advised that it sits proudly on 8 inches of 4000 psi
concrete and the soil under the concrete was properly prepared.

I am a farmer not an engineer but I am not an idiot about how to build
something and have it stand up for a while.

Well, you're talking about a non-trivial amount of load and not everybody
(even farmers) is prepared to take the amount of static loads, dynamic
loads, shear stresses, beam deflections, etc into account when designing a
structure. Since your original question, which dealt with deflection of a
beam, involves (relatively) elementary structural analysis, someone in the
know about engineering of a structure is likely to assume the worst which is
that you have little to no knowledge in the area.

Good luck with whatever you end up with.

todd

j walker wrote:

On Sun, 10 Apr 2005 21:47:14 -0500, j walker wrote:

I am beginning to wish I hadn't asked the original question. Somebody
out there needs to quit reading between the lines and assuming what
they don't know.
....

Well, you see what frequently (usually?) happens on usenet...

What fun would it be if there weren't anybody around guessing? But, it
is a two-way street--the info one gets usually can only be as good as
the info regarding the question supplied. W/ a general description, the
imagination of respondents is unfettered, and as you note, there are
some who will attempt to make the most dire prediction they can imagine.

If you happen back by one more time, where are you farming and what
type?

I'm in SW KS on dryland wheat/milo raising feeder calves over the winter
for the feedlots...and wishing we'd get some of the rain that's been
scattered around but missed us (but don't need the 2" diam hail that
covered the ground in N central part of the state last night)

On Tue, 19 Apr 2005 11:12:13 -0500, Duane Bozarth
wrote:

j walker wrote:

On Sun, 10 Apr 2005 21:47:14 -0500, j walker wrote:

I am beginning to wish I hadn't asked the original question. Somebody
out there needs to quit reading between the lines and assuming what
they don't know.
...

Well, you see what frequently (usually?) happens on usenet...

What fun would it be if there weren't anybody around guessing? But, it
is a two-way street--the info one gets usually can only be as good as
the info regarding the question supplied. W/ a general description, the
imagination of respondents is unfettered, and as you note, there are
some who will attempt to make the most dire prediction they can imagine.

If you happen back by one more time, where are you farming and what
type?

I'm in SW KS on dryland wheat/milo raising feeder calves over the winter
for the feedlots...and wishing we'd get some of the rain that's been
scattered around but missed us (but don't need the 2" diam hail that
covered the ground in N central part of the state last night)


We have a small cow-calf operation down in AL and we could probably
spare some rain for you. We are approaching 6 days without rain. It
was in January the last time that happened. The pasture is beginning
to dry out and the grass toughen up some which the cows will like
better. I am still feeding some hay and usually we are finished by
March 10.

Sold some calves last week and prices are still good so I certainly
can't complain. Slaughter cows are higher than we have ever seen.

On Mon, 18 Apr 2005 21:53:46 -0500, j walker wrote:

On Sun, 10 Apr 2005 21:47:14 -0500, j walker wrote:

I am beginning to wish I hadn't asked the original question. Somebody
out there needs to quit reading between the lines and assuming what
they don't know.

I gave a brief view of the project simply because I thought perhaps I
had confused someone attempting to help.

Looking back I should have merely ignored that and left well enough
alone

OR

given a detailed word picture of the platform as it has existed for
several years, complete with the 3 X 4 angle braces on all sides but
one and that one being reinforced with a 4X4X3/8 el piece of steel. I
probably should have mentioned that the 1100 gallon capacity will only
rarely be reached and only by manual override and that normally the
limit is 800 gallons at which time the water flows out the overflow
pipe. And finally it stands alone on the farm, filled by a windmill
and should a catastrophe occur it isn't likely to hurt anyone. And
just in case someone worries about the tank sinking in the ground
please be advised that it sits proudly on 8 inches of 4000 psi
concrete and the soil under the concrete was properly prepared.

I am a farmer not an engineer but I am not an idiot about how to build
something and have it stand up for a while.

Thanks to those who were helpful.



Does anyone know if there is a place on the web that shows the load
carrying ability of different sizes of dimensional lumber over
different spans?

It sure be helpful if someone does.

Thanks

j walker


I've got a water tower on my property that stands off by itself and
I'm thinking of putting an 1100 gallon tank on it.

It has a foundation rated at 4000 psi (which is stronger than the mix
for any road in my state) and all loads bear directly. I haven't the
slightest idea of the framing sizes or how they align with the load.

I'm wondering if 2 x 8 joists will hold the load without there being a
problem with deflection.

The 2 x 8 's will bear on steel angle braces that are set who knows
how many inches apart, by god knows what kind of fasteners, but these
are fastened to 2 x 12 rim joists.

BTW - I have no idea how the holding tank engages the framing at the
footprint. It might be perimeter loaded or it might hit at eight
points distributed in some way that is not privy to me.

Thank you so much for helping me out with this.

Here's a span calculator for joists and rafters that ought to do what
you need: http://www.awc.org/calculators/span/calc/timbercalcstyle.asp




I've got a water tower on my property that stands off by itself and
I'm thinking of putting an 1100 gallon tank on it.

It has a foundation rated at 4000 psi (which is stronger than the mix
for any road in my state) and all loads bear directly. I haven't the
slightest idea of the framing sizes or how they align with the load.

I'm wondering if 2 x 8 joists will hold the load without there being a
problem with deflection.

The 2 x 8 's will bear on steel angle braces that are set who knows
how many inches apart, by god knows what kind of fasteners, but these
are fastened to 2 x 12 rim joists.

BTW - I have no idea how the holding tank engages the framing at the
footprint. It might be perimeter loaded or it might hit at eight
points distributed in some way that is not privy to me.

Thank you so much for helping me out with this.

On Wed, 20 Apr 2005 18:46:01 -0400, Roger
wrote:

Here's a span calculator for joists and rafters that ought to do what
you need: http://www.awc.org/calculators/span/calc/timbercalcstyle.asp

Now, by Gosh, that' funny.