Pier and Beam Garage Foundation???

[dablack]

Ok guys.  We will be moving forward with our plans in the next year or so.  I would really like to build a garage first.  Most likely something simple but with a small bathroom. 

SO!  Does anyone know of plans for a pier and beam foundation for a garage?  I know the loads are higher but I can handle that as long as it is in the plans. 

I much rather build a simple garage before building the house.  Maybe 20x20 or something like that.  We are planning on building the universal 20 by XX two story (most likely 20 x 40), and I think perfecting my skills on a garage would be a good idea. 

thanks
Austin

[cmsilvay]

Do you want to park a vechicle in there or are talking more of a workshop.

[jbos333]

I don't have any plans for you, but I do have a pier and beam garage that I have parked my truck in, approx. 12 x 24. So it is possible.

[dablack]

CMS,

It would start off as a workshop but later it would be an actual garage.  I'm thinking along the lines of an old looking carriage house with real doors (not fold up garage type doors).  So, eventually it would house a couple of cars.

jbos,

Tell me about your foundation and flooring. 

thanks
Austin

[jbos333]

Foundation consists of 12" sonotube piers on 6' centers length and widthwise (total of 15), 4x6 rough cut support beams spanning all piers length and widthwise and two additional rows 4x6 lengthwise between those on hangers. Double 2x6 yellow pine floor deck laid widthwise first layer, lengthwise top layer. This was built with salvaged materials so that's the reason for the 2x6 floor. But it seems plenty strong, I parked my truck in it for a long time, now my tractor/loader and snowmobile live there.

There is a "keywall" at the overhead door opening. This was built on a sloped site with the "rearmost" portion of the 24' length sitting on piers approx. 3 feet above grade.

Just keep in mind this was backyard engineered, so I can only say that it has worked for my use for about 20 yrs., YMMV.

[dablack]

jbos,

Thanks for the detailed info.  I'm surprised you remember after 20 years.  I have a hard time remembering what I did last year!

thanks again
Austin

[Squirl]

What kind of floor are you looking for?  Crushed gravel?

[jbos333]

Rest assured I did not remember! I went out and peeked underneath!

[dablack]

Squirl,

It would have to be wood.  If it didn't have a floor and I used crushed gravel then it I would just build a pole barn. 

thanks
Austin

[Squirl]

I was going to recommend a pole barn.  I guess the major issue is Load.  John sells a basic 20x30 plan that can be shortened to 20x20.  
On the homepage for country plans it has the link to the span calculator from AWC. I use this for my girder sizing: http://publicecodes.citation.com/st/ny/st/b400v07/st_ny_st_b400v07_5_sec002_par013.htm   Remembering that you are in TX, I don't think you have to worry about much of a snow load.  So you could probably easily build carriage doors on the sides not taking the roof load.
Using my jeep as an example, it weighs 4500 lbs. It is approximately 10' x 5' at the wheels. That is 90 lbs per square ft.  If you put 2 side by side in a 20 ft wide garage it would be 9000 lbs in a 20 x 12.5 ft area. It would be around 36 lbs per square foot.  I would build it in the 50-70lbs per square foot sizing.

[Don_P]

Dynamic non uniform vs. uniform static loads;
A vehicle bears on 4 or more point loads that can be anywhere. A point load has twice the bending moment of a uniform load. Then, double a rolling load, quadruple it if you plan on hitting the brakes hard. We've talked about wind and seismic being lateral loads, stopping a car is a big horizontal load.

[dablack]

Don,

I'm not a civil engineer (I'm a ME but haven't done any real engineering in SOME time), but knew it was a dynamic load in a very small area. 

Squirl,
You can't average the load over the size of the car.  My 71 F100 is about 4000 lbs (short bed with a mild 460) and that means about 1300 lbs on each front tire and about 700 lbs on each rear.  That 1300 lbs isn't even spread over a full square foot.  I'm running original 5.5" wide wheels so the contact patch is about 6" x 8" max.    1300 lbs spread over 48 sq inches is the same as 3900 lbs / sq ft.  Then I would most likely need to double it.  So we are talking about 8kpsf (8000 lbs / sq ft). 

I have ZERO idea how to take that number and determine what I need to do. 

[Don_P]

dablack, never tried this, I'm a carpenter so take it for what its worth.  

I believe you've taken the 1300 lb load and tried to derive a uniform load... I'm going to take the 1300 and double it and use that 2600 lbs as a point load on the decking in the center of a span between 2 joists.

I'm going to assume #2 southern yellow pine, base design values are Fb=1200 psi, forget deflection (1.6x10⁶), this isn't a house. Better pay attention to shear, Fv=175 psi, this is a short heavily loaded beam. So I'd check bending strength and horizontal shear.

There are some adjustments to Fb, bending strength, to think about. Wood can handle large overloads for short duration, C_D;
C_D- ten year = Fb X 1.0
C_D- 2 month (snow) = Fb X 1.15
C_D- 7 day = Fb X 1.25
C_D- 10 minutes = Fb X 1.6
C_D- impact = Fb X 2.0

You can multiply the result of that by 1.15 for repetitive member use.

I think you'll move the truck to another board at least once every 2 months so 1.15
1200 x 1.15 C_D=1380... 1380 x 1.15 C_r= 1587 psi

You can multiply Fv x C_D
175 psi x 1.15 = 201.25 psi

http://www.windyhilllogworks.com/Calcs/beamclc_ctrptld.htm
I plugged in;
2600 lb load
16" span
5.5" width
3.5" depth
Fb 1587 psi
E 1.6
Fv 201.25 psi

With a 4x6 deck it passed up to 4400 lbs, couldn't get a 2x deck to pass. Shear transfer between 2 layers would take enough nails between layers to resist the max shear output if both layers are across the joists.

anyway, there's my take on the deck  

And yup, I've driven across a 2x12 spanning several feet  

edit; playing a bit more
I tightened the joist spacing to 8" o.c. and passed with 2x10 decking.

[dablack]

Don,

Thanks so much.  I knew there was info out there on how to calculate this but I had no idea where.  I was about to start looking up the civil engineers that I knew in college.

I'm really surprised that a 2x10 decking would make it even with a 8" spaced joist. 

I guess the next question is, what size do the joists need to be?  4x6 like jbos?

thanks again,
Austin

[Don_P]

Quite welcome Austin,
I'm not saying looking up your buddies would be a bad idea  
The NDS is the reference for wood design, awc.org, their free download library has a good bit of info. Most of my calcs are using the formulas that can be found in their Beam Design Formulas free pub. The design values are in the NDS Supplement, also a free download.

I dug a little more last night and found rolling wheel load formulas in the AISC manual, I'd be happy to scan and email that page. Long story short, if both wheels are in a span we need to find the cg and work the moment by trial. If we can keep a single wheel on a span then it is using the same formula as in the calc above, Pl/4. For those following the discussion, the moment formula for a uniformly loaded beam is wl/2... the bending force in a beam is twice as great for a concentrated load at midspan compared to a uniformly distributed load.

Think about moment as if you were pulling on a torque wrench, if the handle were 1' long and you hung a 100 lb weight on the end it would produce 100 ft lbs of torque on the bolt, if you hung 10 lb weights every tenth of a foot from the handle end back to the bolt it would only produce 50 ft/lbs of torque. If the lever were 10' long and the 100 lb weight was on the end, moment=1000 ft-lbs. All the rest of this is to find the amount of wood and shape of beam to safely resist that bending moment. In wood were running about a 2 to 1 safety margin on the weakest stick likely to be in the grade, ultimately the best sticks won't fail fail till about 5 times the safe load. In the lab I've seen sticks break at the design number... grading oopses happen, don't get froggy   Using "common sense" I slid a steel beam off a delivery stack across some double stacked studs last week. It hit the ground surrounded by splinters like it didn't even know they were there.

On to joists;
If there are 4 girder lines the joist spans are roughly 80" and we'll never have more than 1 wheel at midspan so I'll try that first. I'm also going to count on half the load being distributed to adjoining joists so 1300 lbs at midspan on a 80" joist. Plugging those into the same calc with the same design values and a 4x6 works, as does a 2x10. Notice bending controls, max moment is 2166 ft-lbs.

Lets put both wheels symmetrically on the joist, I'm going to call the stance 5' wide. Mmax= Pa, a is the distance from the support to the wheel, ~1'... 1300x1= 1300 ft-lbs, the case above controls, we're good.

Girder;
Try an 80" span. Better put the axle on it and double it, 5200 lbs... double 2x12 just fails in bending as does a triple 2x10. If you calc down to exact bearing lengths I think these are the minimums. 16 piers.


Now I need help, if we're rolling 4000 lbs into the garage at 15 feet per second and stop the car right now. That horizontal force is delivered to the floor and the piers. force= mass x acceleration in seconds squared. Horizontal force = 4000 lbs x 15 x (1x1)? This part is why I think perimeter piers are nuts. If that force is delivered along the length of a wall the overturning stress is very low.

[dablack]

Ok Don,

It has been a while but I will do my best.  I will explain it first and calculate it later.

We are entering the garage with a certain amount of energy.  Since we are moving we have Kinetic Energy (KE).  KE=1/2 M V V (one half * mass * velocity * velocity)

Now to stop the truck we need to scrub off this energy.  So we put on the brakes and use friction between the tires and floor to burn off the energy.  How much energy we put into the floor depends on how fast we were going when we started stopping (V) and how far it took us to stop.  Think about it this way, if there was ice on the floor and I just locked up the brakes and drove through the other side of the garage, I didn't really put much energy into the floor. 

So, when we are talking energy and friction the formula is uMGd.  u is the coefficient due to friction.  With good rubber on a good road it is close to 1.  I have no idea what it would be between stained wood and rubber.  M is mass, G is gravity, and d is distance. 

So, these two things have to be equal in order for the friction to stop the truck.  1/2mvv = uMGd

So at this point I'm stuck.  I have no idea what u is and I have no idea what the distance would be so.....we are going to have to estimate what u is so we can calculate distance.  I'm going to try and find what u is between wood and a tire. 

[dablack]

Ok, so the best I can find is u = 1 for rubber on dry concrete and = .30 for wet concrete.  We will just assume for now that rubber on dry wood is the same as wet concrete and use u = .30. 

So now we can calculate if I'm able to stop in time (before going through the wall).  Sorry but metric is easier in physics so that is what I'm going to use.  Your 15ft/sec now turns into 5 meters/ sec.  So now we have 1/2mvv = uMGd and we can plug in some numbers.  NOTE:  we have M on both sides of the equation so they cancel out.  Thats right boys and girls.  Mass doesn't matter when calculating stopping distance.  So:  1/2 vv = uGd

Lets plug in:  1/2*5*5 = .30 * 9.8 * d      Now we solve for D (hope you remember your algebra)  D= (1/2*5*5)/ (.30 * 9.  D = 4.25 meters or about 12 feet.  So if I enter my garage at 5 meters per second and hit the brakes, it will take me about 12 feet to stop (assuming we have our u correct).

Now after all that I can tell you what the force due to the fricton is.  Ff (force due to friction) is just Force due to gravity times u.  Force due to gravity is also known as weight.  So Ff=Weight*u.  We will assume our above u is correct and say that Ff=4000lbs*.30     Ff=1200lbs.  This is a max and when I have the brakes locked up. 

Does that answer your question?  Yeah, we are finally talking about stuff I know.....

Austin

[Don_P]

 
So, did the stopping distance formula do anything we needed? Can I strip it down to just Ff=Weight*u?

What I was trying to figure out was the bracing requirement. After writing my rant on piers it occured to me that there is some form of ramp tieing this thing to the ground... as long as it is well connected to the ground and structure it takes care of that concern. 

but since I have your ear let's check my thinking on braces. If the horizontal force I'm trying to resist is 1200 lbs and I want to use a 45 degree angled brace between girder and pier I'm assuming the way to calculate that force is to divide 1200 lbs by the sine of 45*, or .7071 So, 1200/.7071=~1700 lbs. We have 16 piers, if each is braced they need to be attached with connections good for a bit over 100 lbs to resist the vehicle stopping force?

[dablack]

Ok, I want to make sure I understand what you are saying.  First we have a horizontal girder and a vertical pier.  Then you place a brace (support) between the girder and pier at 45 degrees.  If that is correct then I'm going to have to look into this a little.  I know how to resolve an angled force into its verticle and horizontal portions, but we are take a force (along the girder) and splitting it.  The force then hits the pier via the girder and via the brace.  I can't remember how to split that force between the two members.  Obviously the horizontal force on the pier can't be greater than what was in the girder to begin with (1200 lbs).  So I'm going to have to look into it.  I'm thinking some of the force will stay in the girder and some will go down the support.  I just don't know how much goes where. hmmmmm

[Don_P]

Yes, you're understanding the scenario I was exploring. The horizontal force on the pier is the same 1200 lbs. below as above. I was looking into the force running diagonally down the axis of the brace to understand the connection requirement between pier and brace.

This is a graphic from a simple online truss design program. Pier, girder and brace form a trussed assembly. Multiply the forces X10 for our example. I've set it up with the yellow point fixed in location but free to rotate in place, the red pin is allowed to slide horizontally as needed, a flexing pier top. I have the load moving to the right at 1200 lbs, the braking vehicle.


This is the solved graphic showing the axial load in the brace, and the reactions, there is your 1200 lbs horizontal on the base of the pier.


What I was trying to get at is that the connection requirement between brace and girder or pier is a good bit higher than the load. If the brace is shallower than a 45 it gets worse.

[dablack]

I see what you are doing now.  Again, I need to look into it more, but I was assuming that the 1200 lb force would be to the left of the girder and brace attachment point (the red dot) but still pointing to the right (like you have it).  I think the horizontal force will be shared between the girder and brace.  So there would be less than 1200 in the girder after the attachment point.  The problem I'm having is figuring out how to calculate the % that goes down into the brace. 

We just adopted two children (we have two of our own as well) and I'm finishing up putting a 3rd row seat in my wife's old explorer.  I'm also finishing up remolding the kid's bathroom.  As much as I love physics problem (I really do), it might take me a day or so to crack my physics or statics book to find the answer.  I might be over thinking it.  Oh, yeah.  WE also have to go get a tree today!  It is 5:30am and I better get to work on this stuff!

[Don_P]

Hey, take your time. Good luck on the tree hunt, the tractor trailers full have quit rolling out of here this week. I've got some more design resources to reread as well, we'll get back round tuit when your ready .