Constructing a Deck with Beam supports

Hello all,

I am currently trying to figure out a cheap way to build a "bridge" that will go in my back yard. It will primarily be used to drive my 2000lb lawn mower over a small creek. I want to support it with 2 to beams at a 30 ft span as I will rest 2x8 or 2X10 pieces of wood on it. It will be flat and just have a small incline on each side to get on and off it. I need to figure out what beam size to use (preferably I-beam). Any formulas or any advice that applies to my situation helps a lot. Thank you in advance. I will also attach a sketch so you can visualize what I am discussing.

Here is attachment

Did you mean to have the I beams span the creek, instead of a 30' long 2x8?

I only plan to implement two I-beams one on each side of the creek, possibly even one in the center. The Wood will rest on top of these I beams

Ok, humor me for a minute, I just want to make sure I understand your intent.

You will have an I beam resting on the ground on each side of the creek.

The 2x8's (or possibly 2x10's) will be rigidly attached, at each end, to the top of the I beam, spanning the creek.

The I beam will not "span" anything, because it is sitting on the ground for its entire length.

As you drive across the bridge, the tires will likely ride on a single 2x8 on the left side,
and another 2x8 on the right side.
So, the entire 2000 pounds will be carried by two 30 long 2x8's, approximately 1000 pounds on each.

Right?

I am using sonotube and rebar to create concrete uprights and then I plan to rest the I beam on top of this. There will be one on each side of the creek . I will then place 15 or so 2X10's next to eachother resting upon the I beams.

Ok, one last question before we delve into the specifics of your design:
The span from I beam to I beam, over the creek, is approximately 30 feet.

Correct?

that is correct

Hi srw2104,
You have your materials bass-akwards. Beams need to run across the crick and wood side to side.

You say primarily for running your mower across the crick... What's the secondary purpose? Maybe running a pick-up load of firewood ovor it?

Your preliminary design sketch concerns me. Simple a thing as it might seem to be... there's a little more to it than finding a couple beams that will hold up. If you're set on building this check your local codes and make sure the DEP isn't going to have issue with it. There are beam bending calculators on this site and others. Most here will steer clear of offering you any specifics on size required etc. without knowing all the details and liability issues. I'll toss a suggestion for size if you decide to move forward with this but with all the other things to consider it won't mean that you couldn't end up in the crick with your mower on top of you.

Good Luck,
Bob

.........do you know of a more efficient way to accomplish this?

You are completely right. I wasn't thinking correctly. My design will have two beams that will span the 30 feet, one on each side....and then the wood will lay side to side on top of this. The I beams will be about 6 feet apart

Well, I guess the I-beam is intended to function as the bridge footing. I would think that four concrete pillars set into the ground might be more cost and functionally effective. I would turn the I-beams (W-flanges) perpendicular to lay accross the crick to function as the primary load bearing members. Then lay the wood from w-flange to w-flange. The math to verify the structure should then be reasonably easy to calculate.

The footing should be sunk into the ground, probably round and with enough surface area to distribute the loading on the soil or ground without over-loading the ground and moving around.

Just for fun - see [url]http://www.engineersedge.com/engineering/Engineering_Downloads/Engineering_Productivity_Misc/[/url] Bridge Design Manual

Maybe...
Depends on what you will actually run over this and how daring you are to take advice from some unknown on the internet... Also many other factors. Does crick flood much? What's your soil like at banks and yard? What sort of rock is there around this? Basically how stable are the creeksides? DEP and code issues need to be given consideration... or not... up to you it's your concern not mine.

We had a bridge in the township out for a few years. Wasn't that it couldn't have been finished quickly... it was an issue with some tree huggers about a small pool area in the crick just below the bridge where fish liked to hang out. The tree huggers were concerned that the rebuild would disturb the fishies hole they loved so much so all work was stopped. A couple years later they permitted the project to continue after digging out a new spot a little further down the crick and seeing that the fishies did indeed enjoy their new little pool.

If you have rocky sides to the crick you might get away with some big cut stones to sit your beams on. If it's soft dirt you'll need to watch that your support doesn't wash away if the crick floods. Might be as simple as laying a couple forty foot beams across the thing and slapping some boards on them. Can't know without seeing and being familiar with the turf wherever it is that you are.

Ah, you have refined your design... good.

So, at a very basic level, you have three problems to solve:
1. determine the size of planks required to carry a 2000# dynamic load across a 6' span
2. determine the size of a 30' I beam required to carry the weight of the planks plus a 2000# dynamic load mid-span
3. determine the surface area needed of a concrete footing to resist sinking or tilting in your soil

I do not believe this site has the necessary formulas for determining load ratings of dimensional lumber planks;
You may want to take a look at this site for engineering data on planks: [url]www.awc.org[/url]

To determine the predicted deflection of a 30' I beam mid-span under a 2000# static load,
you will want to become familiar with this formula on this site:
[url]http://www.engineersedge.com/beam_bending/beam_bending2.htm[/url]

Perhaps others can give advice on sizing the footings.

[QUOTE=dalecyr;662]I do not believe this site has the necessary formulas for determining load ratings of dimensional lumber planks;
You may want to take a look at this site for engineering data on planks: [url]www.awc.org[/url][/QUOTE]

[url]http://www.engineersedge.com/lumber.htm[/url]
[url]http://www.engineersedge.com/civil_engineering/lumber-structural-5x5.htm[/url]
[url]http://www.engineersedge.com/commercial_lumber_sizes.htm[/url]
[url]http://www.engineersedge.com/civil_engineering/concrete/concrete-column-volume-calculator.htm[/url]
[url]http://www.engineersedge.com/civil_engineering/concrete/concrete-footing-volume-calculator.htm[/url]

Dalecyr:

Do you know how I would approach solving for numbers 1 and 2 in your last post. Any formulas you can provide would be much appreciated.

"To determine the predicted deflection of a 30' I beam mid-span under a 2000# static load,
you will want to become familiar with this formula on this site: [URL]http://www.engineersedge.com/beam_bending/beam_bending2.htm[/URL]
"

I have a bit more time right now, let me enhance that last post.

It turns out that the question, how much will a beam carry,
is measured in degrees, rather than in absolutes (mostly).

A beam will (usually) start to deflect (bend) before it fails completely.
Some deflection is acceptable; even desirable.
But too much, and the beam collapses.

The equation noted above will help you predict the amount of deflection
of a particular beam, when under a specific stress (weight).

You will probably be safe with a deflection of less than [I]l[/I]/1200,
where [I]l[/I] is the length of the I beam in inches.

To use the equation, you will have to come up with the Moment of Inertia (I)
of the particular I beam under consideration. (look it up)
Your load (W) is known.
You do not care about Distance (x). (use half of the length)
Your length (l) is known.
For the Distance to the neutal axis just use half the height of the I beam under consideration.

I am always amazed that Kelly can pull up a table that I did not know exists on this site,
but if you can not find a table for MofI, you can solve for I,
by playing with the Moment of Inertia field in the equation.

Enter, oh, say... 150 for (I), enter your other parameters,
and see if you come up with an acceptable Maximum Deflection at Load (y).
You calculated the acceptable deflection from the equation above [i]l[/i]/1200.

It will take some playing around with the equation, but that is not a bad thing;
you will start to understand how each parameter affects the outcome.

You will still have to find an I beam with the correct (I),
but at least you will know what you are looking for.

Post your answer...

I did it a somewhat different way in that i plugged .3 for a deflection because I have a length of 30 feet which is 360in. L/1200 gives me 360/1200 = .3.....I then plugged in for all other variables. W = 3000lbs, L = 360, y = .3, and E = 30E6. I then solved for the moment inertia (Ixx) and got a final value of 324 in^4. So i can then use the I beam chart to find a beam with a Moment of inertia greater than or equal to 324. Is this correct? Becuase a moment of inertia of 324 seems a little high

[QUOTE=srw2104;669]I did it a somewhat different way in that i plugged .3 for a deflection because I have a length of 30 feet which is 360in. L/1200 gives me 360/1200 = .3.....I then plugged in for all other variables. W = 3000lbs, L = 360, y = .3, and E = 30E6. I then solved for the moment inertia (Ixx) and got a final value of 324 in^4. So i can then use the I beam chart to find a beam with a Moment of inertia greater than or equal to 324. Is this correct? Becuase a moment of inertia of 324 seems a little high[/QUOTE]



Yup, except I would go way bigger than 324 Moment of I. Also, .3 for deflection seems small for 30'..

what deflection seems more accurate?....and are you sure I should go bigger than a moment of inertia of 324?? I'm looking at a chart of W I beams and beams with a moment of inertia higher than 200 are beginning to get extremely pricey

I think 1/2 to 3/4 deflection should be OK if not a little more - others?

Your design is really simple in that you’re trying to support the loading with two w-flanges. Ultimately strength characteristics, costing of materials and weight are what make truss designs attractive. The difference being that there will be increased manufacturing/fabrication costs associate with a truss type structural bridge design.

A truss design will distribute the loading such that smaller individual structural components (smaller moments of I) can be used.

Ok Kelly, now you have me confused.

Increasing M of I represents an increase in stiffness, resulting in less deflection.
If .3" is small, increasing M of I more than 324 will result in even smaller deflection.
Right?

If we use M of I of 90, on a 30' span, with a 2,000# load, the deflection is .72".

srw2104: I am sure you realize we are talking about the capacity of a single beam.
You will have two beams, effectively doubling the load carrying capacity of the bridge
over the capacity of a single beam.
(There are other considerations, but we will get to those later.)

Yes,

What I meant and failed to articulate by saying "increasing M of I" is that for FOS the structrual w-flange M of I should be bigger. I should have said you need a FOS (Factor of Safety) of some number (1.5 maybe) and increase the M of I to accomplish this.

[QUOTE=dalecyr;673]Ok Kelly, now you have me confused.

Increasing M of I represents an increase in stiffness, resulting in less deflection.
If .3" is small, increasing M of I more than 324 will result in even smaller deflection.
Right?

If we use M of I of 90, on a 30' span, with a 2,000# load, the deflection is .72".

srw2104: I am sure you realize we are talking about the capacity of a single beam.
You will have two beams, effectively doubling the load carrying capacity of the bridge
over the capacity of a single beam.
(There are other considerations, but we will get to those later.)[/QUOTE]

yes, i do realize this is a calculation for a single beam. Do you think that since the load is shared between the two supports, I can divide it in half and use a value of W to be equal to that of 1500 instead of 3000?

I am probably more conservative than others;
I would use a load (W) of 2000#, which you state is the expected load.

This will give a FOS of (approx. (not quite)) 2:1.

But I would like to see the opinions of others on the forum...

okay...than you for the help dalecyr and kelly. It is much appreciated

I'd go with 40 foot W14 x 22 twice (I of 199 each) and some rough cut 2" hardwood planks. If the load will indeed be limited to a mower and/or 2,000 lbs. for your own private residence. The 14" beam is lightest you can get with a high enough moment and light = less steel = less dollars. Or as Kelly mentioned you could look into trusses. Beams with all the useless parts cut out. Space the beams maybe a foot wider than your mowers width. and overhang the boards by a foot each side. Sink some big cut stones in the ground a few feet off the bank each side. Drop the beams on the rocks and poke a bunch of holes for stainless carriage bolts. Further you sink the stones the less need for a ramp. Thrwo some rip rap along the edges of the crick and call it a day.

+1,
and Bob has addressed a few more issues that we had not gotten to yet.
Like making the beam long enough to provide a point of attachment to the footing;
reducing the need for a ramp by sinking footings into the ground;
treating the side of the creek to prevent erosion in that area;
using rough cut lumber to increase M of I on the planks;
placement of the beams relative to the load distribution and beam supports.

Do remember tho, this is a generic answer to a somewhat theorectical problem;
if you change the "problem", the answer no longer applies.

For example, you have not stated that there could be any lateral load at all.
Although the I of a W14x22 is 199 vertically, it is only 7 horizontally.
Run an I of 7 through the equation, and think about what happens when
some debris gets caught under the bridge and starts to make a dam.

There are ways to mitigate those forces.
But then, you did not ask that question. :)

It seems it would have been easier to just put larger diameter tires on the mower and lawn cart to increase ground clearance and just drive through the crick...

:)

Ron

[QUOTE=CCR5600Design;690]It seems it would have been easier to just put larger diameter tires on the mower and lawn cart[/QUOTE]

Or, my personal favorite, save a bunch of money with a couple of yards of concrete, some rebar and corrugated steel drain pipe (if needed) and make a concrete crossing. Have it flow through the pipe for normal flow and over the ford for high flow.

Much less cost, no cranes, unless of course two men can carry a 40 foot W14 x 22 on their shoulders. :D

Dave
Generally, I will not give you the answer to your question, but I *will* guide you into discovering how to solve this yourself.

I will absolutely consider that as one of my options...thank you!