Physics/engineering question
Steve B
sheet steel decking at bargain basement prices, but want to make some A
frame trusses both to hold up the sheets, and to form a hanger for a hoist
that will lift no more than 500#. Mostly for things too big for me to load
and unload from my pickup. For bigger things, I am building a dock.
I am thinking of using 3 x 3 x .120" square tubing for the main two trusses,
then make some smaller ones out of 2" square, as they will only be holding
up sheeting. The center of the peak will only be about two feet vertical.
Between the two 3 x 3 trusses, I want to make a 8' I beam dolly so I can
lift something up, and load or unload it without having to back the truck
up.
I would weld it directly to the top of the containers, either at some strong
points there, or lay a long beam or I beam to get a strong base.
Do you think this would be safe for center point lifting of no more than
500#? Or should I beef up the center load section? I also have a little
height, so could vee it down additionally to peaking it, and add that
horizontal compression stiffener. I can also add several diagonal
stiffeners in there just for GP.
I tend to overengineer things, but I like to make it about 3x swl.
Thanks
Steve
visit my blog at http://cabgbypasssurgery.com watch for the book
A fool shows his annoyance at once, but a prudent man overlooks an insult.
Jim Wilkins
> I want to start covering the 16' between my containers. I found a bunch of
> sheet steel decking at bargain basement prices, but want to make some A
> frame trusses both to hold up the sheets, and to form a hanger for a hoist
> that will lift no more than 500#. Mostly for things too big for me to load
> and unload from my pickup. For bigger things, I am building a dock.
> Steve
I use pressure treated wood for load-bearing beams, sized from this
table:
http://www.awc.org/pdf/wsdd/C2B.pdf
http://www.awc.org/technical/spantables/tutorial.htm
http://picasaweb.google.com/KB1DAL/Firewood#5287788504883032706
Some day the wood will be reused in a new tool shed. I buy PT because
that area floods, and to get southern yellow pine (SYP) which is a
very strong wood.
I do have textbooks on steel and wood structural design, which explain
shear, concentrated loads, end fastening, etc. There's more to
designing a beam than the load rating, they can fail in many non-
obvious ways, like the Minnesota highway bridge.
http://en.wikipedia.org/wiki/I-35W_Mississippi_River_bridge
Since I'm an amateur at this I test them for deflection and proof load
with a load cell.
jsw
Existential Angst
news:ee2c6b50-de9e-4c67...@k39g2000yqd.googlegroups.com...
http://picasaweb.google.com/KB1DAL/Firewood#5287788504883032706
===========================================
Is a load cell where you put the ends of yer beam on two 2x4s laid on the
floor, jump up and down on the middle of the beam, and see if the middle
touches the floor?
If so, then I have a load cell!! :)
--
EA
jsw
Jim Wilkins
wrote:> ...
>
> Is a load cell where you put the ends of yer beam on two 2x4s laid on the
> floor, jump up and down on the middle of the beam, and see if the middle
> touches the floor?
>
> If so, then I have a load cell!! :)
> --
> EA
Yeah, if the 2x4s let the beam deflect more than it should in service.
The chart I referenced uses 1/180 of the length as the worst-case
deflection limit.
I test 2x4s in the store when no one is looking by laying one end on
the lowest shelf, the weaker way, and stepping on the middle. The ones
that make noise go back on the rack.
For wider planks that might be able to support 1500 Lbs that isn't
really a good enough test but it may find hidden cracks.
The orange channel iron in my photo was bent from previous pallet rack
service. I was able to straighten it with a jack in the middle of two
pieces which were chained together at the ends. They also have to be
restrained from twisting by crosspieces. That was an easy way to load
a beam to over a ton. Be careful, it's also a crossbow.
If you know the unloaded axle weights of your vehicles you can roughly
calibrate jack force to the reading of a spring scale on the handle.
jsw
Bob Engelhardt
will be in the geometry of them. It would be silly to ignore that
aspect & choose beams based on their "naked" strength.
My $.02,
Bob
Jim Wilkins
Do you sense a disconnect between designing a truss and the questions
asked?
jsw
Existential Angst
=============================================
Be nice, now.
Speaking of design, tho, mebbe just buy bitty I beams from a steel
warehouse.
They come as small as 2x3" overall cross-section, and are stronger'n'hell,
and not super-expensive -- proly less than "engineered" wood trusses, etc.
Actually, I don't even see "I" beams listed... a " 3 x 5.7 " is an "S"
beam, 3" tall, 5.7 lbs/ft, and 2.33" wide.... and did I say
stronger'n'hell?
A 20 ft length would be about 120 lbs, and proly about $60. Most warehouses
will cut.
Also, structural channel is a possibility, for more compactness.
--
EA
jsw
Jim Wilkins
wrote:> ....
> Also, structural channel is a possibility, for more compactness.
The steel in my photo is 3" channel, two pieces bolted together back
to back. A single one twists when loaded. That Harbor Freight 1 ton
trolley didn't fit until I turned the wheels down. It may not be good
for a ton any more, but neither are the channels if I apply a
reasonable safety factor to their (measured) yield strength.
jsw
ATP*
"Steve B" <pittma...@hotmail.com> wrote in message
news:temlf7-...@news.infowest.com...
with it. 16 feet is a long unsupported span. The more overlap you can get on
the containers the better.
cavelamb
?
Overlap onto the relatively flexible top skin?
Why?
The load stresses would be taken by the vertical sides - not the top.
My shop roof was built using 10" purlin to free span 30 feet.
A bit more expensive in terms of metal, but more than offset
by lower labor cost.
--
Richard Lamb
Steve B
"cavelamb" <cave...@earthlink.net> wrote in message
news:uaydnYgeB9dYzrTR...@earthlink.com...
You do have a definite point about the top skin. Plus, it is slightly domed
shaped to make the water run off. I think I shall probably be better served
by running four poles to ground, and having the majority of the weight
supported by them. The top is not very strong, and the sides at that point
don't look that hefty, either.
I knew when it came time to cover that 16' x 40' span, it was going to be
fun.
Jim Wilkins
> ...
>
> You do have a definite point about the top skin. Plus, it is slightly domed
> shaped to make the water run off. I think I shall probably be better served
> by running four poles to ground, and having the majority of the weight
> supported by them. The top is not very strong, and the sides at that point
> Steve
If you can reduce the span with posts you are within the AWC table I
mentioned and can support around 3/4 ton with fairly inexpensive wood
beams.
jsw
ATP
container and get secured at the ends. That would reduce deflection as
compared to a smaller overlap.
cavelamb
That's nonsense.
BIG waste of material.
--
Richard Lamb
ATP
Steve B
"ATP" <walter...@unforgiven.com> wrote in message
news:4c2d31e2$0$21703$607e...@cv.net...
When I try to comprehend engineering questions, I like to exaggerate them
greatly.
If I heard you correctly, I should make the beam 32' long, that is 8' per
container, and 16' inbetween, and anchor at the end points only. I'm no
engineer, but common sense tells me that a 16' length of anything will lift
more than a 32' length of the same anything. And bend less.
You did say secure it on the ends. If I actually were to do this, wouldn't
securing it at the 16' point kill the elastic modulus that lets the beam sag
at those points? And the ends welded like that to the outsides would
actually hold down the ends that would tend to spring up when lifting causes
the beam to go into a U formation if it was supported at the inner container
wall points? I mean at 8' in from each end, and at each end.
And wouldn't you end up with a slightly stronger assembly if you actually
DID use the 32' beam welded in four places instead of a 16' welded in two?
And definitely a 32' beam welded on only the ends.
Or, I could just exaggerate the problem, and put poles way out, and use a
100' beam supported on the ends ................. <g>
James Waldby
> "ATP" <walter...@unforgiven.com> wrote :
>> "cavelamb" <cave...@earthlink.net> wrote :
>>> ATP* wrote:
>>>> "Steve B" <pittma...@hotmail.com> wrote:
>>>>> I want to start covering the 16' between my containers. I found a
>>>>> bunch of sheet steel decking at bargain basement prices, but want to
>>>>> make some A frame trusses both to hold up the sheets, and to form a
>>>>> hanger for a hoist that will lift no more than 500#. Mostly for
>>>>> things too big for me to load and unload from my pickup. For bigger
>>>>> things, I am building a dock.
>>>>>
>>>>> I am thinking of using 3 x 3 x .120" square tubing for the main two
>>>>> trusses, then make some smaller ones out of 2" square, as they will
>>>>> only be holding up sheeting. The center of the peak will only be
>>>>> about two feet vertical. Between the two 3 x 3 trusses, I want to
>>>>> make a 8' I beam dolly so I can lift something up, and load or
>>>>> unload it without having to back the truck up.
>>>> I would look around for a heavy, deep I beam cheap and bridge the
>>>> containers with it. 16 feet is a long unsupported span. The more
>>>> overlap you can get on the containers the better.
>>> Overlap onto the relatively flexible top skin? Why?
>>> The load stresses would be taken by the vertical sides - not the top.
>> Ideally the beam would overlap all the way to the far sides of each
>> container and get secured at the ends. That would reduce deflection as
>> compared to a smaller overlap.
>
> When I try to comprehend engineering questions, I like to exaggerate
> them greatly.
>
> If I heard you correctly, I should make the beam 32' long, that is 8'
> per container, and 16' inbetween, and anchor at the end points only.
> I'm no engineer, but common sense tells me that a 16' length of anything
> will lift more than a 32' length of the same anything. And bend less.
>
> You did say secure it on the ends. If I actually were to do this,
> wouldn't securing it at the 16' point kill the elastic modulus that lets
> the beam sag at those points? And the ends welded like that to the
> outsides would actually hold down the ends that would tend to spring up
> when lifting causes the beam to go into a U formation if it was
> supported at the inner container wall points? I mean at 8' in from
> each end, and at each end.
Some of what you write in above paragraph is unclear to me (such as
what "the 16' point" is). For clarity, let case A = 32' beam laid
across without fastenings; case B = 32' beam laid across, fastened
at ends (ie at outer walls of each container); case C = 32' beam
laid across, fastened at 8' intervals (ie, at each wall of each
container). Suspend a 10 ton weight from the center of each beam.
Suppose that in case A the center drops two inches. Then the ends
each rise about four inches. The lower edge of the beam between the
containers is in tension and the upper edge is in compression.
In case B, the center drops an inch and the 4'-from-end points each
rise an inch. Over the walls (ie, at 8' intervals) the beam height
remains fixed. Over the roofs, the lower edge of the beam is in
compression and the upper edge is in tension.
In case C, about the same thing happens as in case B. Ie, welding
at the inner walls has little effect (assuming flex at such welds)
> And wouldn't you end up with a slightly stronger assembly if you
> actually DID use the 32' beam welded in four places instead of a 16'
> welded in two? And definitely a 32' beam welded on only the ends.
>
> Or, I could just exaggerate the problem, and put poles way out, and use
> a 100' beam supported on the ends ................. <g>
What ATP wrote precludes having the beam ends out beyond the far sides
of each container, but if they were, I think the effect is complicated
and depends on beam cross-section; whether the ends are supported;
whether there are welds, and at which walls; length beyond walls;
and of course on the crush strength of the container walls.
--
jiw
Jim Wilkins
> "ATP" <walter_mun...@unforgiven.com> wrote in message
> > "cavelamb" <cavel...@earthlink.net> wrote in message
> >> ATP* wrote:
> >>> "Steve B" <pittmanpir...@hotmail.com> wrote in message
> >>>...
> > Ideally the beam would overlap all the way to the far sides of each
> > container and get secured at the ends. That would reduce deflection as
> > compared to a smaller overlap.
Think of the forces if you simplify the problem by cutting the 32'
beam in half. Now each half is like a diving board.
Hang 1000 Lbs on one end in the center. The far end pulls up on the
outer wall with 1000 Lbs. The leverage ratio at the inner wall is 2:1
and it has to support 2000 Lbs, twice the applied load. The total
added load under the container is still 2000-1000=1000.
If you don't cut the center the added inner wall load is less, but
still more than with a 16' beam. You can't really calculate the 32'
beam very well without knowing how much the container walls yield or
the whole box tips.
So the 32' cantilevered beam applies a greater downward force on the
inner walls and costs "more", how much depends on what you can find
surplus. If you look up beam load formulas you'll see that a beam
"fixed" at the columns is stiffer than one that's only supported and
free to pivot, but you would have to figure it out both ways and
compare costs to see which is better for you.
jsw
James Waldby
...
>For clarity, let case A = 32' beam laid across
> without fastenings; case B = 32' beam laid across, fastened at ends (ie
> at outer walls of each container); case C = 32' beam laid across,
> fastened at 8' intervals (ie, at each wall of each container). Suspend
> a 10 ton weight from the center of each beam.
That should have read, "fastened at ends and 8' from ends" - see below.
> Suppose that in case A the center drops two inches. Then the ends each
> rise about four inches. The lower edge of the beam between the
> containers is in tension and the upper edge is in compression.
Also was wrong, see below.
> In case B, the center drops an inch and the 4'-from-end points each rise
> an inch. Over the walls (ie, at 8' intervals) the beam height remains
> fixed. Over the roofs, the lower edge of the beam is in compression and
> the upper edge is in tension.
...
I said 32' beam when defining the cases but was thinking 24', ie,
with a wrong span of 8' rather than 16' between the containers,
so numbers were all wrong.
There are a couple of beam calculation programs you can download
and try different beam sizes. Eg, atlas and beamboy. With the
latter, you can calculate the effect of 1 ton at the center of a
32', 3"x3"x.12" tube supported at 8' from the ends [center sags
5.1", ends rise 7.7"] or supported at the ends and 8' from the
ends [center sags 2.2", 4'-from-end points rise about 0.3"].
This is with I=1.91 in^4 moment of inertia, 30*10^6 psi modulus
of elasticity, and c=1.5"
Of course, you said you'd use an I beam dolly between
trusses, so the above isn't like your plan, it's just an
example of the sort of thing you can compute with beamboy.
--
jiw