Constructing a Deck with Beam supports

[srw2104]

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

[Kelly_Bramble]

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.

[dalecyr]

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.)

[Kelly_Bramble]

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.

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.)

[srw2104]

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?

[dalecyr]

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...

[srw2104]

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

[RWOLFEJR]

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.

[dalecyr]

+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.

[CCR5600Design]

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

[PinkertonD]

It seems it would have been easier to just put larger diameter tires on the mower and lawn cart

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.

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

[srw2104]

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