Floor load formula 1

[BobNeff]

I'm having some trouble, does anyone have a formula for figuring out floor load on standard construction. I have 2x12 joists 14" on center and the longest span being 19', with 3/4" pine floor boards. If this information is not sufficent please inform.

Thanks Bob

 

[boo1]

Missing information:
Joist wood type and grade
Loading (residential second floor? ??PSF)

14" OC joists is not standard construction
3/4" pine floor is not standard construction

Formulas:
Bending Fb=Mc/I
Shear Fv=3V/2db
Deflection D=

 

[BobNeff]

Joist wood type: fir, grade unknown assume lowest
Loading: this requires aliitle explaining.
The floor is the second floor of an old lumber yard I bought. And to help with the bills I wanted to utilize the second floor as warehouse space. I was wondering how to figure out what the floor is rated at so I may deside what will and what won't go up there. The building was built sometime in the 1930's, which I think is the reason for the funny joist spacing. The room is about 48'x110'with a 19'aisle in the middle. I have a drawing if would like to see that to get a better idea of the colum spacing. Its in .dwg format that I drew up quickly in case someone asked for it.

 

[RockEngineer]

Work the equations backwords to find an allowable load. Most likely deflection will govern since your span is 19 ft. For most floors the live load deflection is limited to L/360 which would be .6 in for your joists.
E for all grades of Hem-Fir can be taken at 1,300,000 psi

defl = (5*w*L^4)/(364*E*I) or

w= (364*E*I)*defl/(5*L^4)

w would be in lb/ft. Account for 14" spacing and reslove units and you have an allowable uniform load based on live load deflection.

Then check to make sure your bending stress is not too high for the live load plus dead load. A conservative for an old growth 2x12 is 1100 psi.

 

[BobNeff]

Thats exactly what I need. What value should I use for "I"?

 

[TFL]

Are you sure your joists are not at 16" o.c.? 14 1/2" would be the clear distance between joists at 16" o.c. which is standard.

It really sounds like you are out of your element. you might want to consider finding someone to do this for you.

from statics i of a rectangle I= (1/12)(b)(h^3)

 

[BobNeff]

Thanks for the advice TFL, worth about as much as I paid, I'll be fine.

 

[TFL]

that is the danger of trying to get random people on the internet to do your work for you .

By the way i was just trying to be helpful. so the sarcasm is not appreciated. There is much more that goes into your simple prolem then you think. if you search previous post you will find many with a similiar questions. all of them usually have someone who suggests they should hire someone to do it for them. usually alot ruder then i was

 

[UcfSE]

That should probably be 384, not 364 in the above post by RockEngineer, unless I am mistaken.

Bob,

You should hire someone who is knowledgeable. Honestly, we cannot design things for you and you should know that. We are here to help engineers with problems, not help someone save a buck over hiring a professional when needed. There is much more to figuring your floor load than you seem to think. Have you evaluated your walls yet? How about foundations? Connections? In short, there is no simple floor load formula. In the future don't respond so rudely. It isn't our fault you don't know what to do.

 

[JAE]

I was out of town and just read this post - I'd have to second what UcfSE says - there is a lot more to this than using some applicable equations and assuming some wood properties of old lumber.

In fact, Bob, I would suggest that you are putting yourself in serious difficulty should the floor get loaded beyond its capacity and fail. Here's some thoughts:

1. Your purchase of the existing building and the use of it is most likely putting you under the local building codes and your use of the floor for storage, etc. would be dependent upon a proper evaluation of the floor loading capacity.

2. If you aren't licensed in structural engineering in your locale, then you are adding to your liability by performing engineering calculations to determine if the floor is safe for a particular load. This could be construed as practicing engineering without a license (assuming you don't have a license - forgive me if you do).

3. Even if you have a license, it sounds like you aren't well versed in the structural field, so this again may be violating the engineering laws in your area. (your handle says "mechanical").

4. Should something fall down and injure someone, you would be held liable for the damages.

5. Many insurance carriers require mezzanine and other floors to be load rated to qualify for insurance. I've checked a number of mezzanines just for this purpose.

So all of my wordiness above - to simply suggest that you go out and hire a structural engineeer to evaluate the whole structural system, not just the floor.

 

[RockEngineer]

Thanks UcfSE. I knew I made a mistake in the equation as soon as I hit send. It is 384 not 364.

Even though I provided the basic equations, I agree with the others that you are taking a liability on yourself. I know that many rural areas of the country do not enforce building codes but you do need to be careful and conservative. It is your liability in the end and your decision what you do.

A dollar spent early on in planning to get knowledgeable help can save thousands of dollars and lives later on. My two cents worth.

 

[TFL]

thanks to everyone who said it much more eloquently then I

 

[OkieDokie]

TFL, I get a bigger laugh out of this post the way you said it.