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

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

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

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

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

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

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

Steve,

Thinking about your project a little more, I would choose to alter the
design rather than trying to add a stand-off. There are simpler and
stronger designs possible, such as this for example:

http://www.hi-techedu.com/products/hfc14.htm

And here are a couple of older cranes which might give you ideas:

http://www.tclayton.demon.co.uk/pics/canal/npb5.html
http://www.stroudwater.co.uk/cpsn/dudbridge%20crane.jpg

Note that both of these have a compression member which is at an angle
to the vertical. This is a nice, simple design. You just need to avoid
applying large moments to the beam as in your original proposal. In pure
compression 2" x 2" x 0.25" tubing should be fine.

The most critical part of the crane is likely to be the point at which
the top is attached to the wall. If you choose to use expanding bolts,
look up their load rating and use a good margin of safety (a factor of
10, say). Use the right size drill and make sure the wall isn't crumbly.
Should it fail, a joint like this will not fail in a nice way.

All the best,

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

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

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

"Christopher Tidy" wrote in message
...
Steve,

Thinking about your project a little more, I would choose to alter the
design rather than trying to add a stand-off. There are simpler and
stronger designs possible, such as this for example:

http://www.hi-techedu.com/products/hfc14.htm

I had thought of that, but the vertical beam joins to the sonotube at that
point, and just sits on top. The pushing sideways and the leverage would
cause the approximately 7' vertical corner support to push off the top of
the Sonotube, or come loose where it is only nailed together. This is going
to be the point of attatchment for the lower standoff, but I will make the
bracket 18" long, and at a right angle so I can wrap the whole corner, and
attatch to post, facia, etc, and increase the distribution of whatever side
load from the flex is. I don't think it will be that much. Certainly not
as much as mounting the bottom support of a stiff leg to it.

Seen lots of stifflegs loading boats. They are really strong, but all were
mounted into humongous concrete bases.

Steve


And here are a couple of older cranes which might give you ideas:

http://www.tclayton.demon.co.uk/pics/canal/npb5.html
http://www.stroudwater.co.uk/cpsn/dudbridge%20crane.jpg

Note that both of these have a compression member which is at an angle to
the vertical. This is a nice, simple design. You just need to avoid
applying large moments to the beam as in your original proposal. In pure
compression 2" x 2" x 0.25" tubing should be fine.

The most critical part of the crane is likely to be the point at which the
top is attached to the wall. If you choose to use expanding bolts, look up
their load rating and use a good margin of safety (a factor of 10, say).
Use the right size drill and make sure the wall isn't crumbly. Should it
fail, a joint like this will not fail in a nice way.

All the best,

Chris

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

Steve,

Thinking about your project a little more, I would choose to alter the
design rather than trying to add a stand-off. There are simpler and
stronger designs possible, such as this for example:

http://www.hi-techedu.com/products/hfc14.htm


I had thought of that, but the vertical beam joins to the sonotube at that
point, and just sits on top. The pushing sideways and the leverage would
cause the approximately 7' vertical corner support to push off the top of
the Sonotube, or come loose where it is only nailed together. This is going
to be the point of attatchment for the lower standoff, but I will make the
bracket 18" long, and at a right angle so I can wrap the whole corner, and
attatch to post, facia, etc, and increase the distribution of whatever side
load from the flex is. I don't think it will be that much. Certainly not
as much as mounting the bottom support of a stiff leg to it.

Yes, you would need to ensure that the bottom bearing can handle the
horizontal force if you go for the "stiff leg" type of crane. But if you
use your original design without a stand-off part way down, your bottom
bearing will carry a horizontal force. It would be a good idea for it to
be capable of carrying a horizontal force anyway. If you can resolve
these small problems, and you don't need to lift bulky loads to the full
height of the crane, I think the stiff leg is the way to go because it
will be strong and reliable. If you stick with your original design I
would put one stand-off at the very top and one a foot below it, as
opposed to the way they're shown in your diagram. This will mean that
the vertical column isn't being bent at all, which is a good thing. I
think this is more important than having a stand-off in the middle.
Also, make the stand-offs adjustable so you don't have to force the
thing together in the first place!

Best wishes,

Chris

In article ,
says...


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.

The original problem, which you've represented accurately
in your first sketch, is a simply supported beam with a
moment applied at one end. You've replaced the beam, which
was previously supported at both ends, with a cantilever.
The moment has become a concentrated force applied at the
opposite end. Extraordinary claims require extraordinary
proofs.

Or if anyone does the calculation by different means I'd like to
know your result, too.


Case 3e, Chapter 5, Roark's 6th edition - concentrated
moment on a simply supported beam. The angular deflection
at the end of the tube adjacent to the triangle is .0188
radian = 1.08 degrees; .68" deflection at the winch's
attachment point 36" from the vertical tube.

Ned Simmons

"Christopher Tidy" wrote


Yes, you would need to ensure that the bottom bearing can handle the
horizontal force if you go for the "stiff leg" type of crane. But if you
use your original design without a stand-off part way down, your bottom
bearing will carry a horizontal force. It would be a good idea for it to
be capable of carrying a horizontal force anyway. If you can resolve these
small problems, and you don't need to lift bulky loads to the full height
of the crane, I think the stiff leg is the way to go because it will be
strong and reliable. If you stick with your original design I would put
one stand-off at the very top and one a foot below it, as opposed to the
way they're shown in your diagram. This will mean that the vertical column
isn't being bent at all, which is a good thing. I think this is more
important than having a stand-off in the middle. Also, make the stand-offs
adjustable so you don't have to force the thing together in the first
place!

Best wishes,

Chris


One of the design criteria is that when the load is at the top, right before
it hits the winch kickout switch, the load needs to be swung into the porch
over the 36" high handrail. A stiffleg wouldn't make it in to where the
load could be landed on the deck before it hit the side of the deck. This
needs to be lifted to its highest possible point, and then swung about 150
degrees over the rail and inside the railing, where the load would be hand
unloaded, or lowered to a waiting dolly.

Your suggestion would work perfectly IF I didn't have to have that extra 60
degrees of swing.

Steve

"Ned Simmons" wrote

Case 3e, Chapter 5, Roark's 6th edition - concentrated
moment on a simply supported beam. The angular deflection
at the end of the tube adjacent to the triangle is .0188
radian = 1.08 degrees; .68" deflection at the winch's
attachment point 36" from the vertical tube.

Ned Simmons


This is a HF winch with two attachment points. The one closest to the end
would be about 2'6" from the vertical, and the other, about 8" closer to the
vertical than that. I really don't want to put the winch all the way out,
but only as far as needed to clear the rail when I swing the load into the
deck.

I am going to have to make two new hangers, as HF gives you two for 2" round
stock. I CAN slide the winch out so that the drum actually goes PAST the
tip of the horizontal, but don't want to do that until after testing. At
the point of sliding it out, the hanger would hit the diagonal coming down.

Just gonna have to build this puppy, hook it up, get to a safe place, and
watch what happens.. Just like in real life. Can't really do a test until
failure, as some of the components are holding up my roof and
deck..............

Steve

Ned Simmons wrote:

snip

Case 3e, Chapter 5, Roark's 6th edition - concentrated
moment on a simply supported beam. The angular deflection
at the end of the tube adjacent to the triangle is .0188
radian = 1.08 degrees; .68" deflection at the winch's
attachment point 36" from the vertical tube.

Hi Ned,

When I read your post, at first I disagreed with you, then I agreed, and
now I partly agree. After some thought I am of the opinion that neither
of us have got it quite right.

My cantilever would be fine if the horizontal member was attached half
way up the vertical member, but the model becomes less accurate the
further it is moved from this position. Your model is fine in the
absence of the diagonal tension member (Steve's flat bar). With the
tension member present it isn't a moment which is applied to the
vertical member, but a point load instead.

I did some sketches and free body diagrams:

http://www.mythic-beasts.com/~cdt22/davit_calc2.jpg

I now think that the correct model is a simply supported beam under the
action of a point load F, where F is applied distance A from one end and
B from the other. In this case F = 3W, where W is the load on the
suspension point, A = 14' and B = 1'. What do you think?

I don't have a formula for this case. Perhaps if your book gives a
formula for this case, giving the deflection of the point at which the
load is applied, you could run a quick calculation and get an answer for
Steve? My guess is that it will come out with an answer a good bit less
than my original suggestion of 8", but I'd be interested to know.

I suspect that whatever model we use, we aren't going to get much closer
than an order of magnitude figure unless we know exactly how Steve will
construct it. Nevertheless I believe it would be wise to avoid putting
the long vertical member under a bending load.

Best wishes,

Chris

Chris,

My cantilever would be fine if the horizontal member was attached half
way up the vertical member,

Based on what?


but the model becomes less accurate the
further it is moved from this position.

It is not so much a matter of accurate and inaccurate, it is about right
and wrong. There are approximations and judgement calls all through
this kind of work, but one must be clear on the details before trying to
be creative.


Your model is fine in the
absence of the diagonal tension member (Steve's flat bar). With the
tension member present it isn't a moment which is applied to the
vertical member, but a point load instead.

Actually, the model Ned cited is not "fine" in that case. It is however
reasonable, and much more so than a non-related cantelever beam.


I did some sketches and free body diagrams:

http://www.mythic-beasts.com/~cdt22/davit_calc2.jpg

What about the vertical component of the tension in the beam FBD on the
left?



I don't have a formula for this case.

Sorry, somebody has to say it: if you need a "formula" for something
like that, then you should not be giving advice on structural mechanics.

Bill

What about the vertical component of the tension in the beam FBD on the
left?

Make that horizontal.

Bill Schwab wrote:
Chris,

My cantilever would be fine if the horizontal member was attached half
way up the vertical member,


Based on what?

Based on the fact that the deflection would then be symmetrical about
the horizontal member, so the root of the cantilever would not be rotated.

It is not so much a matter of accurate and inaccurate, it is about right
and wrong.

So what's right then? In my opinion engineering is all about accuracy.
My first model may have been fairly inaccurate, but that's the way it
goes: you build one model, think about it and discuss it, then build a
better one.

Your model is fine in the absence of the diagonal tension member
(Steve's flat bar). With the tension member present it isn't a moment
which is applied to the vertical member, but a point load instead.


Actually, the model Ned cited is not "fine" in that case. It is however
reasonable, and much more so than a non-related cantelever beam.


I did some sketches and free body diagrams:

http://www.mythic-beasts.com/~cdt22/davit_calc2.jpg


What about the vertical component of the tension in the beam FBD on the
left?

We can't include it as we know nothing about the kinds of joints Steve
intends to employ in the structure. I assumed that there are pin joints
at each end of the beam because this makes the structure statically
determinate. We know so little about the structure that this is only
assumption one can make. Granted, I did not draw a pin joint. That was
an honest mistake.

I don't have a formula for this case.


Sorry, somebody has to say it: if you need a "formula" for something
like that, then you should not be giving advice on structural mechanics.

In which case, why are there big books of formulae published for
structural engineers to use? Like Roark's formula book which Ned
mentioned? Probably I could have figured out the formula from first
principles if I'd chosen to spend an hour or two on it, but I didn't
want to. Structural engineers don't begin all their calculations from
scratch.

So okay, I am not a structural engineer. I'm someone who took a few
structural engineering courses a while back. But I'm the only one who
actually tried to help Steve in the first place. Others then joined the
discussion, which I feel is a good thing if everyone remains reasonable,
but now it seems some traditional Usenet insults are being thrown in.

And no one here has yet claimed to be a professional structural engineer...

Chris

Chris,

My cantilever would be fine if the horizontal member was attached
half way up the vertical member,



Based on what?


Based on the fact that the deflection would then be symmetrical about
the horizontal member, so the root of the cantilever would not be rotated.

Not good enough.


It is not so much a matter of accurate and inaccurate, it is about
right and wrong.


So what's right then? In my opinion engineering is all about accuracy.

No - engineering is more about managing inaccuracy. Before you can
learn to do that, you must become proficient with the type of problem
solving that appears in undergraduate texts.


My first model may have been fairly inaccurate,

tactOff
Chris, it was nonsense.
/tactOff


but that's the way it
goes: you build one model, think about it and discuss it, then build a
better one.

Not in this case. You clearly lack understanding of the basics that go
into such a model.



What about the vertical component of the tension in the beam FBD on
the left?

We can't include it as we know nothing about the kinds of joints Steve
intends to employ in the structure. I assumed that there are pin joints
at each end of the beam because this makes the structure statically
determinate. We know so little about the structure that this is only
assumption one can make. Granted, I did not draw a pin joint. That was
an honest mistake.

Again, nonsense. You called it a tension member, hence you were assuming
frictionless pins and negligable weight. Otherwise known as a two force
member, and it's item one in any engineering statics course.


I don't have a formula for this case.



Sorry, somebody has to say it: if you need a "formula" for something
like that, then you should not be giving advice on structural mechanics.


In which case, why are there big books of formulae published for
structural engineers to use? Like Roark's formula book which Ned
mentioned? Probably I could have figured out the formula from first
principles if I'd chosen to spend an hour or two on it,

Time wasted if the FBD is incorrect.


but I didn't
want to. Structural engineers don't begin all their calculations from
scratch.

Clearly the tables exist (I consult them at times myself), and they
exist to save time, not to escape learning. My statement stands.



So okay, I am not a structural engineer. I'm someone who took a few
structural engineering courses a while back.

If you still have interest, get a Schaum's outline on statics and start
solving problems. It will eventually come back to you.


But I'm the only one who
actually tried to help Steve in the first place. Others then joined the
discussion, which I feel is a good thing if everyone remains reasonable,
but now it seems some traditional Usenet insults are being thrown in.

And no one here has yet claimed to be a professional structural engineer...

Wisely so. A pro won't touch something like this - responsibility w/o
compensation. Some time ago, a poster suggested that the OP needs to
find a structural engineer, which was very good advice.

Bill

Bill Schwab wrote:

snip

My first model may have been fairly inaccurate,


tactOff
Chris, it was nonsense.
/tactOff

Bill, you have yet to propose a model of your own. You could have
proposed a model in the time you spent criticising mine.

but that's the way it goes: you build one model, think about it and
discuss it, then build a better one.


Not in this case. You clearly lack understanding of the basics that
go into such a model.

An easy insult to offer, but as I asked, where is your model?

What about the vertical component of the tension in the beam FBD on
the left?


We can't include it as we know nothing about the kinds of joints Steve
intends to employ in the structure. I assumed that there are pin
joints at each end of the beam because this makes the structure
statically determinate. We know so little about the structure that
this is only assumption one can make. Granted, I did not draw a pin
joint. That was an honest mistake.


Again, nonsense. You called it a tension member, hence you were assuming
frictionless pins and negligable weight. Otherwise known as a two force
member, and it's item one in any engineering statics course.

Consider my above answer retracted. Your question does not make it clear
which member you're talking about, and now you've changed the force
you're talking about from vertical to horizontal.

And no one here has yet claimed to be a professional structural
engineer...


Wisely so. A pro won't touch something like this - responsibility w/o
compensation. Some time ago, a poster suggested that the OP needs to
find a structural engineer, which was very good advice.

So here we have three guys who know a little about structural
engineering having an argument. If we'd worked together we could
possibly have been helpful, but I do not believe we are helping Steve
now, so I will not continue this discussion.

I do, however, stand by my recommendation that Steve should try to avoid
subjecting his 2" x 2" x 0.25" vertical column to a bending moment. I
think he should either place one stand-off at the very top (15' level)
and one at the level of the horizontal member (14' level), or change the
design to something stronger like the "stiff leg" style of crane. People
are welcome to agree or disagree with this opinion.

Chris

For the engineers in the group, and those who seem to know of these things
..........

If you would like, e mail me at , and I will fax or
send you a jpeg of my drawing, and you can get it just right.

I DO believe this thing will work, and I am not going to use it to lift
things much heavier than a couple of suitcases or a few bags of groceries.

So far, I'm into this about $200, so it is not just a frivolous widget.
But, failure can cause heavy damage to my structure. I think if I stay less
than 200#, probably more like 100 most of the time, I shouldn't have a
problem. I would just like to be sure, and hear from someone who can
calculate such things.

Steve

SteveB wrote:
For the engineers in the group, and those who seem to know of these things
.........

If you would like, e mail me at , and I will fax or
send you a jpeg of my drawing, and you can get it just right.

I DO believe this thing will work, and I am not going to use it to lift
things much heavier than a couple of suitcases or a few bags of groceries.

So far, I'm into this about $200, so it is not just a frivolous widget.
But, failure can cause heavy damage to my structure. I think if I stay less
than 200#, probably more like 100 most of the time, I shouldn't have a
problem. I would just like to be sure, and hear from someone who can
calculate such things.

Hi Steve,

As Bugs suggested, I suspect this is the best you'll get from this
group. There probably aren't many true structural engineers who hang out
here, and those who do may have reasons for not wanting to get involved
(ranging from boredom with structures through to liability). There may
be other Usenet groups where you'll find better help, but I can't
suggest any. There is also the regrettable tendency for people to be
non-constructive in their discussions on Usenet.

As has been demonstrated, one of the problems with your proposed
structure is that it isn't as simple as it looks. Its behaviour is more
complicated and less easy to predict than I thought, and probably than
Ned and Bill thought, too. One of my main reasons for suggesting the
"stiff leg" type of crane is that the analysis of the structure will be
much more straightforward and less open to dispute. I would attack the
problem by making some changes to the design (you should still be able
to use the materials you've bought) rather than trying to find someone
who's a professional structural engineer.

Best wishes,

Chris

Christopher Tidy wrote:
Bill Schwab wrote:

snip

My first model may have been fairly inaccurate,



tactOff
Chris, it was nonsense.
/tactOff


Bill, you have yet to propose a model of your own. You could have
proposed a model in the time you spent criticising mine.

Responsibility w/o compensation. Nice try, but I'm not going on the
record with a solution.


but that's the way it goes: you build one model, think about it and
discuss it, then build a better one.


Not in this case. You clearly lack understanding of the basics that
go into such a model.

An easy insult to offer, but as I asked, where is your model?

It is not an insult; it is a professional opinion.


What about the vertical component of the tension in the beam FBD on
the left?



We can't include it as we know nothing about the kinds of joints
Steve intends to employ in the structure. I assumed that there are
pin joints at each end of the beam because this makes the structure
statically determinate. We know so little about the structure that
this is only assumption one can make. Granted, I did not draw a pin
joint. That was an honest mistake.



Again, nonsense. You called it a tension member, hence you were
assuming frictionless pins and negligable weight. Otherwise known as
a two force member, and it's item one in any engineering statics course.


Consider my above answer retracted. Your question does not make it clear
which member you're talking about, and now you've changed the force
you're talking about from vertical to horizontal.

The identity of the member would be clear to an engineer; it would be
clear to just about anyone since I used the unique name _you_ gave it.
The h/v thing was a typo, no doubt the result of my distraction with
trying to find a nice way to put this.



So here we have three guys who know a little about structural
engineering having an argument. If we'd worked together we could
possibly have been helpful, but I do not believe we are helping Steve
now, so I will not continue this discussion.

I think you will find the level of knowledge of the "three guys" varies
just a bit. Hopefully the OP has figured out that your advice is not to
be taken at face value.


Toward taking my leave of this thread, I will address some of your other
comments here.

As has been demonstrated, one of the problems with your proposed
structure is that it isn't as simple as it looks. Its behaviour
is more
complicated and less easy to predict than I thought, and probably than
Ned and Bill thought, too.

Speak for yourself. You haven't come close to the complex part
(searching for the weak link in the chain so to speak). The deformation
analysis is at the level of an undergrad homework problem.




I would attack the
problem by making some changes to the design (you should still be able
to use the materials you've bought) rather than trying to find someone
who's a professional structural engineer.

More bad advice.


To the group at large, I offer an apology for the nature of this
discussion, but I have learned to respect you. To not point out the
flaws would be unfair to you, and potentially dangerous to the OP. I
trust you will do the same should anyone give me bad advice about
clamping or feed rates.

Bill

"Christopher Tidy" wrote

I would attack the
problem by making some changes to the design (you should still be able to
use the materials you've bought) rather than trying to find someone who's
a professional structural engineer.

Best wishes,

Chris


Thanks for your input, Chris. This is going to be mounted at a corner, and
needs to swing in through the handrail, so a stiffleg is out because it
would hit the building.

I have built a lot of "stuff", and most of it has worked. Anything that
really needed to be engineered, I built from plans okayed by an engineer and
provided by the builder.

However, as we all know, we do make Rube Goldberg things for ourselves with
varying results. One of the major things to have is a positive solution
finding attitude. It may take a prototype or six, and some modifications,
but I have found that a workable solution to most problems is always found.

That said, there are all different types. I am sure you saw the back yard
swing ..... as it was intended ....... as an engineer would look at it
........ etc. Sometimes, it's just like the big bucks fighting to the death
while the little forkyhorn runs in and mates with the doe. A lot gets done
by us little forkyhorns.

I like what Henry Ford said, "Whether you think you can or can't, you're
right."

Intellectualizing gives me a headache and takes too much time. I like
getting it done, and then ruminating afterward over some cold beer.

Steve

Bill,

I feel I must enter this thread one more time.

What about the vertical component of the tension in the beam FBD on
the left?

We can't include it as we know nothing about the kinds of joints
Steve intends to employ in the structure. I assumed that there are
pin joints at each end of the beam because this makes the structure
statically determinate. We know so little about the structure that
this is only assumption one can make. Granted, I did not draw a pin
joint. That was an honest mistake.

Amid your stream of criticism and confusion between horizontal and
vertical, I gave the answer to a different question: the question "Why
is there no vertical force shown on the left-hand FBD acting at the
junction of the horizontal and vertical members?" I stand by the answer
I gave to this question. The answer to your question is very simple. I
believe the free body diagram on the left to be complete and correct. No
horizontal force is shown acting on the top end of the beam because the
upper support provides a reaction against this (see the roller support
in my diagram). If you disagree with this, please elaborate.

You called it a tension member, hence you were
assuming frictionless pins and negligable weight. Otherwise known as
a two force member, and it's item one in any engineering statics course.

I agree with your description of the tension member. I don't see what
you're complaining about here.

snip

As has been demonstrated, one of the problems with your proposed
structure is that it isn't as simple as it looks. Its behaviour
is more
complicated and less easy to predict than I thought, and probably than
Ned and Bill thought, too.

Speak for yourself. You haven't come close to the complex part
(searching for the weak link in the chain so to speak). The deformation
analysis is at the level of an undergrad homework problem.

If it's so simple, I still don't understand why you don't go ahead and
solve it. From my recollection of undergraduate structures questions,
this might have been a difficult second year "bonus" question. The kind
which looks simple but isn't. We are also presented with the problem of
having to model a vague structure from scratch; we haven't been given an
idealised structure on which to perform the mathematics.

I would attack the
problem by making some changes to the design (you should still be able
to use the materials you've bought) rather than trying to find someone
who's a professional structural engineer.

More bad advice.

Why? Steve is unlikely to hire a structural engineer. Some structures
are easier to analyse than others. Some structures are designed so that
their behaviour is easier to predict. I've seen older bridges which are
built with a roller support at one end. Why? So that the structure is
statically determinate and easier to model. Changing a design so that
its behaviour is more predictable is a sensible thing to do. If you
disagree with this, please explain why.

To the group at large, I offer an apology for the nature of this
discussion, but I have learned to respect you. To not point out the
flaws would be unfair to you, and potentially dangerous to the OP. I
trust you will do the same should anyone give me bad advice about
clamping or feed rates.

You seem devoted to accusing me of offering bad advice without
explaining why my advice is bad, or attempting to offer better advice of
your own. Your refusal to solve what you claim is an easy problem is, to
me, suspicious. It's all too convenient. The risk involved in offering
Steve a solution is not great. You're not a practising structural
engineer, so you needn't worry about your reputation, and if you say
that your advice is offered without warranty I don't believe you have
any legal worries. The risk doesn't concern me, and it doesn't appear to
concern Ned either.

I tried to shed some light on Steve's problem in good faith. Perhaps the
model I proposed was somewhat flawed, but you just comdemned it as
"nonsense". At no point did you attempt to explain why it was nonsense,
or how I might refine the model to give a better answer. You never
explain why my advice is "bad". Furthermore, some of your questions and
comments barely make sense.

I'm of the opinion that you're attempting to make me look a fool, and I
don't appreciate it. An engineer contacted me by e-mail this afternoon
and offered some suggestions about how to refine my model. Why didn't
you take this approach?

I am also intrigued to learn that you work in an anaesthesiology lab:
http://needle.anest.ufl.edu/anest4/bills/

Chris

Construction and installation of this major project will take place in
Spring of 2006. I will keep all appraised. Photos will be available as
fabrication progresses, at installation, testing, and finish.

What I intend to do is fully mount everything, and leave the middle standoff
out. I will then make a 200# load and measure deflection. I will then
install the standoff as a CYA feature. I am going to make a public guess on
record that the deflection in the middle of 15' and the lifting point 2'6"
off the vertical will be less than one inch. We'll see if I am even close.

Thanks for all the help, input, and insights into areas I had never thought
of.

Steve

SteveB wrote:
Thanks for your input, Chris. This is going to be mounted at a corner, and
needs to swing in through the handrail, so a stiffleg is out because it
would hit the building.

I get a better idea what you need now. I'm sorry this thread has turned
into a flame war about mathematics with which none of use are entirely
familiar. It does muddy the waters.

I can see that a stiff leg is out of the question. Is your building
wooden, brick, stone or what? 200 lb is not an enormous load and I think
it might be possible to attach the davit to the building without the
long vertical column. If you do a Google image search for "wall crane"
and "wall hoist" you'll see pictures of cranes which are attached
directly to the structure of a building. Some of these are quite hefty.

As I suggested (but it got lost in the argument somewhere) I think the
weakest feature of your design is that the load will be attempting to
bend the vertical column. I'm not a structural engineer, but my
intuition is generally good and this is fairly clear I think. If you
took a long length of box section, clamped one end to the bed of a
pick-up and leant on the other end, it would deflect quite a bit. But if
you push on the end of the section, as if you were trying to push the
pick-up along, it would be much harder to buckle it. So basically you
could build something like one of those wall cranes, but have a column
below which carries compression only. I did a little ASCII sketch (turn
the fixed font on now):

-|\ -|\
| \ | \
| \ | \
-|---\ |---\
| |
| |
| -|
| |
| |
| |
| |
_|__________________________|_______
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

My preference is for the design on the left, as it avoids bending the
column. But I do think my original suggestion of an 8" deflection
without stand-offs is inaccurate. Given Ned's estimate, plus perhaps a
little extra flexibility in the mountings, 1" to 2" seems more
reasonable. But that's still a bit bouncy for my liking as flexing can
cause problems such as pivots jamming, etc.

I'd be interested to see pictures and drawings. Feel free to post them
online or e-mail them to the above address (remove NOSPAM). Good luck
and let us know how you get on.

Best wishes,

Chris

Chris,

Bill,

I feel I must enter this thread one more time.

What about the vertical component of the tension in the beam FBD
on the left?


We can't include it as we know nothing about the kinds of joints
Steve intends to employ in the structure. I assumed that there are
pin joints at each end of the beam because this makes the structure
statically determinate. We know so little about the structure that
this is only assumption one can make. Granted, I did not draw a pin
joint. That was an honest mistake.


Amid your stream of criticism and confusion between horizontal and
vertical, I gave the answer to a different question: the question "Why
is there no vertical force shown on the left-hand FBD acting at the
junction of the horizontal and vertical members?" I stand by the answer
I gave to this question. The answer to your question is very simple. I
believe the free body diagram on the left to be complete and correct. No
horizontal force is shown acting on the top end of the beam because the
upper support provides a reaction against this (see the roller support
in my diagram). If you disagree with this, please elaborate.

I don't know how to be any more plain about it: you are wrong. The
horizontal reaction is one applicable force, but that does not change
the behavior of two-force members. Your FBD is incorrect. It's not
"less accurate" - it's wrong.

Bill

"Christopher Tidy" wrote in message
...
SteveB wrote:
Thanks for your input, Chris. This is going to be mounted at a corner,
and needs to swing in through the handrail, so a stiffleg is out because
it would hit the building.

I get a better idea what you need now. I'm sorry this thread has turned
into a flame war about mathematics with which none of use are entirely
familiar. It does muddy the waters.

I can see that a stiff leg is out of the question. Is your building
wooden, brick, stone or what? 200 lb is not an enormous load and I think
it might be possible to attach the davit to the building without the long
vertical column. If you do a Google image search for "wall crane" and
"wall hoist" you'll see pictures of cranes which are attached directly to
the structure of a building. Some of these are quite hefty.

As I suggested (but it got lost in the argument somewhere) I think the
weakest feature of your design is that the load will be attempting to bend
the vertical column. I'm not a structural engineer, but my intuition is
generally good and this is fairly clear I think. If you took a long length
of box section, clamped one end to the bed of a pick-up and leant on the
other end, it would deflect quite a bit. But if you push on the end of the
section, as if you were trying to push the pick-up along, it would be much
harder to buckle it. So basically you could build something like one of
those wall cranes, but have a column below which carries compression only.
I did a little ASCII sketch (turn the fixed font on now):

-|\ -|\
| \ | \
| \ | \
-|---\ |---\
| |
| |
| -|
| |
| |
| |
| |
_|__________________________|_______
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

My preference is for the design on the left, as it avoids bending the
column. But I do think my original suggestion of an 8" deflection without
stand-offs is inaccurate. Given Ned's estimate, plus perhaps a little
extra flexibility in the mountings, 1" to 2" seems more reasonable. But
that's still a bit bouncy for my liking as flexing can cause problems such
as pivots jamming, etc.

I'd be interested to see pictures and drawings. Feel free to post them
online or e-mail them to the above address (remove NOSPAM). Good luck and
let us know how you get on.

Best wishes,

Chris


The problem with the drawing I provided is that it is waaaaaaaaaaaaaay
out of scale. The top triangle is 1 foot x 3 foot x the hypotenuse.

Your idea of putting the second standoff at the base of the triangle won't
work, because of the following configuration:

Starting from the base it goes something like this -from ground up, the
first six feet is sonotube, next foot is joisting, next eight foot is column
hooked directly to top of deck joisting, top foot is beam sitting on top of
eight foot column. There's no straight continuous piece in the whole deal.

That is why I wanted the majority of the weight to be transferred vertically
to the base of concrete in Sonotube. The standoffs are just to keep the
load from swinging as I swing it in over the handrail.

My main concern was how much the 14' vertical would flex under a 200# load.

Steve

In article ,
says...
Ned Simmons wrote:

snip

Case 3e, Chapter 5, Roark's 6th edition - concentrated
moment on a simply supported beam. The angular deflection
at the end of the tube adjacent to the triangle is .0188
radian = 1.08 degrees; .68" deflection at the winch's
attachment point 36" from the vertical tube.

Hi Ned,

When I read your post, at first I disagreed with you, then I agreed, and
now I partly agree. After some thought I am of the opinion that neither
of us have got it quite right.

My cantilever would be fine if the horizontal member was attached half
way up the vertical member, but the model becomes less accurate the
further it is moved from this position.

I don't doubt you can find *some* case where your model is predictive,
but a model that only works in one specific instance isn't very useful.

Your model is fine in the
absence of the diagonal tension member (Steve's flat bar). With the
tension member present it isn't a moment which is applied to the
vertical member, but a point load instead.

In your original sketch you made the assumption that the triangle was a
rigid body. Considering that the triangle is only 1 foot tall, and the
rest of the tube is 14 feet, that seems like a reasonable
simplification. In other words, I'm confident that the deflection I
calculated is quite accurate, erring on the conservative side, as a
result of my using the full 15 feet as the beam's length. In hindsight,
it may have been more accurate to treat the beam as 15' long, but figure
the slope at 14'.

Of course there is also a vertical compression load in the vertical tube
equal in magnitude to the suspended load, but its contribution to
deflection will be negligible, so I ignored it.


I did some sketches and free body diagrams:

http://www.mythic-beasts.com/~cdt22/davit_calc2.jpg

I now think that the correct model is a simply supported beam under the
action of a point load F, where F is applied distance A from one end and
B from the other. In this case F = 3W, where W is the load on the
suspension point, A = 14' and B = 1'. What do you think?

I think you could solve the problem this way as well, and the result
would be within a few percent of the way I figured it. You would still
have to calculate the slope of the vertical tube at the attachment point
of the horizontal.


I don't have a formula for this case. Perhaps if your book gives a
formula for this case, giving the deflection of the point at which the
load is applied, you could run a quick calculation and get an answer for
Steve? My guess is that it will come out with an answer a good bit less
than my original suggestion of 8", but I'd be interested to know.

I suspect that whatever model we use, we aren't going to get much closer
than an order of magnitude figure unless we know exactly how Steve will
construct it.

Unless we've completely misinterpreted Steve's description, the only
thing I can think of that would have a signficant effect on the results
of the deflection calculations would be a change in the way the
structure is supported, e.g., adding another support near the horizontal
member, as you've suggested. In that case the structure would be
statically indeterminate, and the problem would become one appropriate
for the end of a first sememster structures course instead of one
introduced near the beginning.

Ned Simmons

In article ,
says...


Hi Steve,

As Bugs suggested, I suspect this is the best you'll get from this
group. There probably aren't many true structural engineers who hang out
here, and those who do may have reasons for not wanting to get involved
(ranging from boredom with structures through to liability).

For the record, since I'm an ME, but not a registered structural
engineer, and am not privy to all the construction details of this
device, I've deliberately avoided making a judgement on its safety.


As has been demonstrated, one of the problems with your proposed
structure is that it isn't as simple as it looks. Its behaviour is more
complicated and less easy to predict than I thought, and probably than
Ned and Bill thought, too.

Why would you assume that? It's a very simple problem - the only thing
that makes it appear out of the ordinary is that it's more easily worked
by looking at the vertical member's slope rather than the more commonly
solved for deflection.


Ned Simmons

Ned: I could post further responses to what you've said, but apart from
a few minor points I agree with you. Were we face-to-face we could
discuss the merits of different models and likely reach an agreement on
a good one fairly quickly, but it's rather hard via Usenet. Usenet does
tend to be a little combative.

Bill: I am none the wiser. If you believe my FBD to be incorrect, please
explain clearly where you believe I should add or remove forces and why.
Better still, draw your own FBD and post it. Then I'll be glad to
discuss it with you.

Best wishes,

Chris

SteveB wrote:

The problem with the drawing I provided is that it is waaaaaaaaaaaaaay
out of scale. The top triangle is 1 foot x 3 foot x the hypotenuse.

Your idea of putting the second standoff at the base of the triangle won't
work, because of the following configuration:

Starting from the base it goes something like this -from ground up, the
first six feet is sonotube, next foot is joisting, next eight foot is column
hooked directly to top of deck joisting, top foot is beam sitting on top of
eight foot column. There's no straight continuous piece in the whole deal.

That is why I wanted the majority of the weight to be transferred vertically
to the base of concrete in Sonotube. The standoffs are just to keep the
load from swinging as I swing it in over the handrail.

My main concern was how much the 14' vertical would flex under a 200# load.

Okay, got it. I thought the Sonotube was something you were going to
install, rather than an existing part of your house. So you have what we
in Britain would call a "balcony" (an outdoor platform on the first
floor) and you want to lift stuff onto it? From your description I can
see the problem of having a stand-off in the position I suggested, and I
can see why you don't want to change the design.

Ned's suggestion of a deflection of an inch or so at the winch
attachment point probably isn't far off. But I would be more concerned
about the deflection in the middle of the vertical column: if it's close
to the wall, it might jam. Similarly the bending of the column might
cause the top and bottom bearings to jam. And of course you might just
want to lift more than 200 lbs sometimes!

If it were me, my gut reaction would be to use something stouter than 2"
x 2" x 0.25" box section in order to resist the bending. I would likely
use something like 4" x 0.25" round tube. Some might say this is
overkill, but it gives a good margin of safety without doing about
calculations, allows you to lift heavy stuff without worrying about it,
and I think round would look neat. I know a place where I could get some
4" x 0.25" round pretty cheap, but it all depends on what you can get
locally.

Best wishes,

Chris

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

Chris,

Ned: I could post further responses to what you've said, but apart from
a few minor points I agree with you. Were we face-to-face we could
discuss the merits of different models and likely reach an agreement on
a good one fairly quickly, but it's rather hard via Usenet. Usenet does
tend to be a little combative.

Bill: I am none the wiser. If you believe my FBD to be incorrect, please
explain clearly where you believe I should add or remove forces and why.
Better still, draw your own FBD and post it. Then I'll be glad to
discuss it with you.

Please note that I did describe the problems in each case. However, I
will try this from another angle.

Going all the way back to Newton's Laws and how to apply them, a free
body diagram is a diagram of a free body. A free body is one that has
been isolated from its surroundings, with all interacting bodies
represented by the forces and moments they exert _on the free body_.
Some people treat force fields (magnetic, electric, gravitational)
separately, but it's the same deal: didn't draw the earth? Put in the
weight of the object facing toward the center of the earth (down for
this type of work), and add support reactions.

Never show internal forces on an FBD.

One of the most insightful things I have ever heard on this front came
the late John Wesley Hoover: "take it to pieces". Sadly you can't hear
him bellow that, but do your best to imagine it (add a slight southern
US accent for character). Pick any one or more bodies of a system, draw
them and represent each missing item by the forces/moments it exerts on
the system. Again, never show forces or moments internal to the system.
Feel free to cut any member; when you do, add the forces/moments the
part you removed exerts on the part you kept. I've repeated myself a
bit, but it deserves repeating.

You cannot correctly reason about mechanics w/o a full understanding of
this stuff. Sometimes you will get lucky and miss something that does
not completely change the behavior; there is also risk of missing
something that leads to dangerously incorrect results.

For drill, grab any statics book or relevant Schaum's outline and tackle
the example problems; follow along on one or two, then compare your
independent solution with theirs. In particular, look for problems
suitable to the method of sections; that tends to pound on the rules,
and is very useful for getting at specific information and/or reducing
the coupling of equations one would suffer with a naive decomposition
into the simplest possible FBDs.

From there, review technical beam theory. The FBD of a section of bent
beam and the cut used to get at shear stresses across a vertical section
through a beam will further reinforce how this works. Likewise,
computer shear and bending moment by cutting the beam is good
reinforcement. On the latter, pay particular attention to distributed
loads: they can be reduced to resultants, but only _after_ cutting the
beam. Doing otherwise leaves the support reactions unchanged but alters
the internal shear and bending moment.

Re the top/left diagram we discussed, the "tension member" must be
represented by its tension in the direction of the member. You can
treat that as horizontal and vertical compontents, or as magnitude and
direction. I prefer the latter because it helps in scanning for errors.

I suspect you drew an FBD of the pin supporting the load. Either way,
start there, summing forces to zero vertically and then horizontally.
Out of that you will get the tension in the "tension member" and the
horizontally directed compressive force in the horizontal member.

Then draw an FBD of the beam itself, representing the members above by
the forces they exert on the beam. It is the diagonal tension member
where you have problems. Remember, for a weightless member with smooth
pins, the tension/compression acts along the line between the pins. If
its weight is significant, then draw a free body, including its weight
and unknown horizontal and vertical components of reactions the pins
apply to the member. To start, ignore the weight of the tension member
and treat it as a two-force member.

Bill