I got some 2" x 2" x .250" steel square receiver stock today. Sheesh. $4
per foot.

Anyway, I am making a figure 4 davit. The long leg will be 15 long. The
short horizontal will be 3 foot.

It will mount at the top and the bottom. I am considering putting in a
standoff about half way up to work against the flexing.

The winch weighs about 25#, and I intend to lift no more than 200# with this
800# capacity winch.

The horizontal will be one foot down from the top with a diagonal flatbar
going from the top out to the end of the horizontal to help transfer the
load to the top of the upright.

How much flex can I expect? Do I need the standoff in the middle? I will
probably put one on anyhow just to be sure.

Thanks in advance.

Steve

You don't mention what the loading will be. 2" square heavy tubing is
pretty light weight for any kind of beam.
Better to pay a structural engineer for your design than to depend on
this E-world for advice, which may range from qualified professional
engineer to the Village Idiot.
Bugs

SteveB wrote:
I got some 2" x 2" x .250" steel square receiver stock today. Sheesh. $4
per foot.

Anyway, I am making a figure 4 davit.

snip details

How much flex can I expect? Do I need the standoff in the middle? I will
probably put one on anyhow just to be sure.

Hi Steve,

I couldn't find any pictures of a "figure 4" davit online, but I think I
understand what you're talking about.

I had a bit of time this morning so I did a quick calculation. It is
about the most rudimentary calculation possible, but if I've understood
your structure correctly I believe it gives a useful answer. It ignores
all but one mode of deflection, which I believe will be the most
significant, and makes many assumptions. Naturally these assumptions are
a matter of discretion and people are welcome to discuss them. But there
wouldn't be any point in doing complex calculations without a lot more
knowledge of the project.

Here's what I did. If the diagrams don't match what you intended, let me
know. I assumed that you're going to have a pivot at the top and bottom
to allow the davit to rotate. You may not, but I doubt it will make much
difference because, either way, the anchor points are unlikely to be
highly rigid in a torsional sense.

http://www.mythic-beasts.com/~cdt22/davit_calc.jpg

The calculation suggests that the winch suspension point will be
deflected downwards about 8 inches by a 200 lb load. This is only an
order of magnitude figure, but it is way too much. It means that your
davit would be very bouncy, your top and bottom anchor points could be
damaged, and it might cause failure in an unexpected way.

You might want to rethink the need for the long, vertical column.
Presumably the davit will be supported by a wall or gantry of some kind?
Can you place the bearing carrying the vertical load just below the
horizontal member? If not, a stand-off in the middle would reduce the
deflection to about one-eighth the value I calculated (i.e., 1 inch),
but I still wouldn't be entirely happy with this. A stand-off at the
level of the horizontal member would be better.

Comments welcome!

Best wishes,

Chris

"Christopher Tidy" wrote in message
...
SteveB wrote:
I got some 2" x 2" x .250" steel square receiver stock today. Sheesh.
$4 per foot.

Anyway, I am making a figure 4 davit.

snip details

How much flex can I expect? Do I need the standoff in the middle? I
will probably put one on anyhow just to be sure.

Hi Steve,

I couldn't find any pictures of a "figure 4" davit online, but I think I
understand what you're talking about.

I had a bit of time this morning so I did a quick calculation. It is about
the most rudimentary calculation possible, but if I've understood your
structure correctly I believe it gives a useful answer. It ignores all but
one mode of deflection, which I believe will be the most significant, and
makes many assumptions. Naturally these assumptions are a matter of
discretion and people are welcome to discuss them. But there wouldn't be
any point in doing complex calculations without a lot more knowledge of
the project.

Here's what I did. If the diagrams don't match what you intended, let me
know. I assumed that you're going to have a pivot at the top and bottom to
allow the davit to rotate. You may not, but I doubt it will make much
difference because, either way, the anchor points are unlikely to be
highly rigid in a torsional sense.

http://www.mythic-beasts.com/~cdt22/davit_calc.jpg

The calculation suggests that the winch suspension point will be deflected
downwards about 8 inches by a 200 lb load. This is only an order of
magnitude figure, but it is way too much. It means that your davit would
be very bouncy, your top and bottom anchor points could be damaged, and it
might cause failure in an unexpected way.

You might want to rethink the need for the long, vertical column.
Presumably the davit will be supported by a wall or gantry of some kind?
Can you place the bearing carrying the vertical load just below the
horizontal member? If not, a stand-off in the middle would reduce the
deflection to about one-eighth the value I calculated (i.e., 1 inch), but
I still wouldn't be entirely happy with this. A stand-off at the level of
the horizontal member would be better.

Comments welcome!

Best wishes,

Chris


You are correct about the top and bottom mounting. I believe I will add a
middle standoff to keep flex down. I already have the materials for this,
and will build and test it this spring. It will either work or fail on the
first test load of 200#.

We'll see, and I'll keep everyone posted. It is merely a device to haul
groceries and supplies up without having to traverse thin steep steps. It
doesn't have to hold a lot, and I am thinking it will suffice.


Steve

SteveB wrote:
"Christopher Tidy" wrote in message
...

SteveB wrote:

I got some 2" x 2" x .250" steel square receiver stock today. Sheesh.
$4 per foot.

Anyway, I am making a figure 4 davit.

snip details

How much flex can I expect? Do I need the standoff in the middle? I
will probably put one on anyhow just to be sure.

Hi Steve,

I couldn't find any pictures of a "figure 4" davit online, but I think I
understand what you're talking about.

I had a bit of time this morning so I did a quick calculation. It is about
the most rudimentary calculation possible, but if I've understood your
structure correctly I believe it gives a useful answer. It ignores all but
one mode of deflection, which I believe will be the most significant, and
makes many assumptions. Naturally these assumptions are a matter of
discretion and people are welcome to discuss them. But there wouldn't be
any point in doing complex calculations without a lot more knowledge of
the project.

Here's what I did. If the diagrams don't match what you intended, let me
know. I assumed that you're going to have a pivot at the top and bottom to
allow the davit to rotate. You may not, but I doubt it will make much
difference because, either way, the anchor points are unlikely to be
highly rigid in a torsional sense.

http://www.mythic-beasts.com/~cdt22/davit_calc.jpg

The calculation suggests that the winch suspension point will be deflected
downwards about 8 inches by a 200 lb load. This is only an order of
magnitude figure, but it is way too much. It means that your davit would
be very bouncy, your top and bottom anchor points could be damaged, and it
might cause failure in an unexpected way.

You might want to rethink the need for the long, vertical column.
Presumably the davit will be supported by a wall or gantry of some kind?
Can you place the bearing carrying the vertical load just below the
horizontal member? If not, a stand-off in the middle would reduce the
deflection to about one-eighth the value I calculated (i.e., 1 inch), but
I still wouldn't be entirely happy with this. A stand-off at the level of
the horizontal member would be better.

Comments welcome!

Best wishes,

Chris



You are correct about the top and bottom mounting. I believe I will add a
middle standoff to keep flex down. I already have the materials for this,
and will build and test it this spring. It will either work or fail on the
first test load of 200#.

You'd be better off putting the stand-off just 1 foot down from the top
(i.e., at the same level as the horizontal member). This will mean that
the vertical column only has to support a compressive load. The
compressive load will be well within the capabilities of the column. It
is the bending of the long vertical column that is the major weakness in
your design. Fix this problem and the rest of the design will most
likely be fine.

Let us know how it goes.

Best wishes,

Chris

"Christopher Tidy" wrote in message
news:43A1978C.5030906@


You'd be better off putting the stand-off just 1 foot down from the top
(i.e., at the same level as the horizontal member). This will mean that
the vertical column only has to support a compressive load. The
compressive load will be well within the capabilities of the column. It is
the bending of the long vertical column that is the major weakness in your
design. Fix this problem and the rest of the design will most likely be
fine.

Let us know how it goes.

Best wishes,

Chris


Yes, the horizontal leg will be mounted to a beam, and be about a foot long.
Therefore, it will have a top and bottom vertical attachment pin/rod, and
two standoffs, one a foot from the top, and one about half way down. The
top foot from the horizontal to the piece is there to attach the diagonal 2"
x 1/4" flat bar leg.

I think it's going to be strong enough, but the addition of the middle
standoff leg will insure flex will be directed to a solid standoff.

We'll see.

Now I have to wait for the snow to melt and for it to warm up to pour the
SonoTube base.

Steve

Chris,

Here's what I did. If the diagrams don't match what you intended, let me
know. I assumed that you're going to have a pivot at the top and bottom
to allow the davit to rotate. You may not, but I doubt it will make much
difference because, either way, the anchor points are unlikely to be
highly rigid in a torsional sense.

http://www.mythic-beasts.com/~cdt22/davit_calc.jpg

The calculation suggests that the winch suspension point will be
deflected downwards about 8 inches by a 200 lb load. This is only an
order of magnitude figure, but it is way too much. It means that your
davit would be very bouncy, your top and bottom anchor points could be
damaged, and it might cause failure in an unexpected way.

I fear you are mixing models in a way that does not work. You compute
support reactions based on the ends being fixed horizontally, and then
model the deflection based on a cantelever beam. An FBD of the beam
itself will have (using your top right setup) loads from the bar and
cable (which you might want to reduce to an applied moment).

The compression of the column will move its neutral axis, but I imagine
that the most useful starting point will be to get an angular deflection
at the end of the beam and multiply that by 3 ft to get an estimate of
the load deflection.

Bill

Bill Schwab wrote:

I fear you are mixing models in a way that does not work. You compute
support reactions based on the ends being fixed horizontally, and then
model the deflection based on a cantelever beam.

The cantilever beam only represents part of the vertical member: the 14'
from the horizontal to the ground. I have made an imaginary cut 14' from
the ground. This point can move horizontally, but has a moment and shear
force applied to it. With a bit of spatial twisting and turning in my
mind, I believe this 14' section can be modelled as a cantilever. The
fact that the root moves horizontally while the tip is fixed doesn't matter.

An FBD of the beam
itself will have (using your top right setup) loads from the bar and
cable (which you might want to reduce to an applied moment).

I agree. But the cantilever model takes account of the moment and shear
force, does it not? I used the cantilever model so that I didn't have to
calculate the value of the moment and shear force.

The compression of the column will move its neutral axis

I chose to ignore the compressive load as I think it will produce a
small deflection compared to that resulting from the mode I've modelled.
This is only intended to be an order of magnitude model and I think it
serves this purpose, showing that the structure isn't stiff enough.

Best wishes,

Chris

In article ,
says...
Bill Schwab wrote:

I fear you are mixing models in a way that does not work. You compute
support reactions based on the ends being fixed horizontally, and then
model the deflection based on a cantelever beam.

The cantilever beam only represents part of the vertical member: the 14'
from the horizontal to the ground. I have made an imaginary cut 14' from
the ground. This point can move horizontally, but has a moment and shear
force applied to it. With a bit of spatial twisting and turning in my
mind, I believe this 14' section can be modelled as a cantilever. The
fact that the root moves horizontally while the tip is fixed doesn't matter.


But you've got the cantilever backwards. Your calculations
would predict the deflection at the base of the column as a
result of F3 if point Z were fixed (moment connection) and
if the base were free, which is not the case at all.

Bill is right about solving this by calculating the angular
deflection at Z.

Ned Simmons

Ned Simmons wrote:
In article ,
says...

Bill Schwab wrote:


I fear you are mixing models in a way that does not work. You compute
support reactions based on the ends being fixed horizontally, and then
model the deflection based on a cantelever beam.

The cantilever beam only represents part of the vertical member: the 14'
from the horizontal to the ground. I have made an imaginary cut 14' from
the ground. This point can move horizontally, but has a moment and shear
force applied to it. With a bit of spatial twisting and turning in my
mind, I believe this 14' section can be modelled as a cantilever. The
fact that the root moves horizontally while the tip is fixed doesn't matter.



But you've got the cantilever backwards. Your calculations
would predict the deflection at the base of the column as a
result of F3 if point Z were fixed (moment connection) and
if the base were free, which is not the case at all.

I don't believe this matters. The formula gives the deflection of the
cantilever tip relative to the root. It doesn't matter which moves. You
can build a system of guide rails such that a pin joint is at the fixed
end of the cantilever, and the root is allowed to move transversely, but
not rotate. In this case the formula is still valid, and this is the way
in which I'm using it here.

If anyone has a counter argument or proof I'd be very interested to hear
it. Or if anyone does the calculation by different means I'd like to
know your result, too.

Chris

Chris,

I fear you are mixing models in a way that does not work. You compute
support reactions based on the ends being fixed horizontally, and then
model the deflection based on a cantelever beam.


The cantilever beam only represents part of the vertical member: the 14'
from the horizontal to the ground. I have made an imaginary cut 14' from
the ground. This point can move horizontally, but has a moment and shear
force applied to it. With a bit of spatial twisting and turning in my
mind, I believe this 14' section can be modelled as a cantilever. The
fact that the root moves horizontally while the tip is fixed doesn't
matter.

That might end up being true, but I do not think you can hand-wave it
that way. With its max deflection somewhere in the middle of the beam,
it is more likely that it will function as two half(more or less)-length
cantelevers, which puts quite a hurt on the cubic term in the deflection.



An FBD of the beam itself will have (using your top right setup) loads
from the bar and cable (which you might want to reduce to an applied
moment).


I agree. But the cantilever model takes account of the moment and shear
force, does it not? I used the cantilever model so that I didn't have to
calculate the value of the moment and shear force.

My gut tells me that the deflection will be quadratic in length, vs.
cubic as for the cantelever. My gut has been known to be wrong.



The compression of the column will move its neutral axis


I chose to ignore the compressive load as I think it will produce a
small deflection compared to that resulting from the mode I've modelled.
This is only intended to be an order of magnitude model and I think it
serves this purpose, showing that the structure isn't stiff enough.

I think it will be a factor only in the stress calculations anyway, but
thought I'd mention it. If your deflection is correct, the thing will
tear itself apart anyway =:0


Regards,

Bill

Bill Schwab wrote:
Chris,

I fear you are mixing models in a way that does not work. You
compute support reactions based on the ends being fixed horizontally,
and then model the deflection based on a cantelever beam.



The cantilever beam only represents part of the vertical member: the
14' from the horizontal to the ground. I have made an imaginary cut
14' from the ground. This point can move horizontally, but has a
moment and shear force applied to it. With a bit of spatial twisting
and turning in my mind, I believe this 14' section can be modelled as
a cantilever. The fact that the root moves horizontally while the tip
is fixed doesn't matter.


That might end up being true, but I do not think you can hand-wave it
that way. With its max deflection somewhere in the middle of the beam,
it is more likely that it will function as two half(more or less)-length
cantelevers, which puts quite a hurt on the cubic term in the deflection.

I take your point here. The root of the cantilever may rotate a bit so
that the maximum deflection is somewhere in the middle of the beam. But
I believe that the cantilever model will give a useful result for a
10-minute, one-side-of-paper calculation. Whether it is more like a
single cantilever or two half-length cantilevers will depend on Steve's
joints and anchor points. When we know so little, there will inevitably
be some hand waving. I still think it's a fair model based on the
information we have. It is only intended to give an order of magnitude
result.

Chris

!\
! \
! \
-!----
!
!
!
!
!
-!
!
!
!
!
!

Here is an attempt to draw this thing. At top, it will pin into a pad eye
at the end of a receiver plate/saddle coming horizontal from the beam. At
bottom, it will rest entirely on washers on a plate mounted to concrete.

From the top, at one foot down, the horizontal goes out. The diagonal in
this drawing is not to scale, as the top angle would be much greater. The
hoist's outer attatchment will go about 2/6" from the vertical.

The diagonal would be the square root of 10 in length.

Each short horizontal standoff would be about a foot long with a pad eye.

I anticipate that stability will be from a solid mounting at the bottom, and
a beam saddle with extended baseplate to make a pad eye for the top
attatchment. The middle standoff will just stiffen it up a bit, and help
keep it straight when I swing it in towards the porch.

HTH. These things are hard to describe in words so as to communicate
exactly.

Steve

After all this hullabaloo, I will mention that I am a retired
registered professional engineer. I try to make constructive comments,
but will not attempt to suggest any design when all the facts are not
given.
If the loading is 200#, there are many models of pickup cranes
available that are guaranteed for 1/2 ton or more. Buy one of them.
If you want to use the material you bought, I would suggest an open web
steel joist construction to fit the dimensions of your ?boat? plain
tubing will deflect just as discussed.

____________________________________
!____________________________________!
\ \ ! \ \ / /! !
\ \ ! \ \ / / ! !
\ \ ! !\ \ / / ! !
\ \ ! ! \ \ / / ! !
\ \ ! ! \ \ / / ! !

Well, you get the general idea. Properly designed & executed, the joist
structure would pick up a Volkswagen. Look at a commercial construction
crane for inspiration.
Bugs