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