refresh my memory:
i'm looking for the section modulus for a boom section to understand
allowable bending stress.
sx= bd(squared) ?
but this is for rectangular sections right?
how do you calculate this for an oval section?
gf.

"DSK" wrote in message
.. .
Nav wrote:
Ad_hominem?

Yep. Couldn't happen to a nicer guy, too.

After the way you have acted, to get all nose-in-the-air about my poking
fun at your obsessive behavior is rather funny.


Are you sure? The geometry _is_ defined.

You're right. Lesson 1- always look the problem over thoroughly.


For the record, the compression on the boom would be the weight
multiplied by the cosine of the angle.


Are you saying it is not 118 lbs in the topping lift case?

Are you saying it's not the cosine of the angle formed by the topping
lift? Tell you what, go down the hall and ask one of the engineering
profs... if any will speak to you...

I gather that you have never heard of a "Free Body Diagram"? Freshman
engineering stuff. That is the way to solve such problems. If you
don't believe me, ask Scout.


But I'm not asking Scout. I'm asking _you_ to solve this freshman
problem -if you can.

Well, it is only a few minutes to draw up a free body diagram. I did a
rough one earlier, but it will take longer to do it on the comuter and
post it. Another job for tomorrow....

I wonder why the boom vang situation bothers you so much, or why you are
*so* sure that a solid vang cannot lift a heavy weight. I've seen it
done several times on several different boats, so obviously it can.

Hey, I got an idea... let's have a little wager on it... no wait, you
don't pay up when you lose...

Fresh Breezes- Doug King

gonefishiing wrote:
refresh my memory:
i'm looking for the section modulus for a boom section to understand
allowable bending stress.
sx= bd(squared) ?

No, IIRC it's the integral of the solid cross section area distance from
the axis. That's why triangular sections have the greatest rigidty for
their cross section area, and square sections are more rigid than oval
or round.

I can refer you to a pretty good text book.

but this is for rectangular sections right?
how do you calculate this for an oval section?

Personally, I don't. I look it up! But this is also not foolproof, you'd
be amazed how many mfg'rs fudge their specs (or maybe they can't do math).

Hope this helps.

Fresh Breezes- Doug King

"DSK" wrote in message
. ..
gonefishiing wrote:
refresh my memory:
i'm looking for the section modulus for a boom section to understand
allowable bending stress.
sx= bd(squared) ?

No, IIRC it's the integral of the solid cross section area distance from
the axis. That's why triangular sections have the greatest rigidty for
their cross section area, and square sections are more rigid than oval
or round.

MEANING TO THE EXTREME OUTER FIBERS IN BOTH AXIS?

I can refer you to a pretty good text book.

but this is for rectangular sections right?
how do you calculate this for an oval section?

Personally, I don't. I look it up! But this is also not foolproof, you'd
be amazed how many mfg'rs fudge their specs (or maybe they can't do math).

YEAH BEEN THERE RECENTLY: IT IS ALSO CALLED EXPENSIVE WHEN YOU DISCOVERED
THEIR PUBLISHED ERRORS.

Hope this helps.

Fresh Breezes- Doug King

No, IIRC it's the integral of the solid cross section area distance from
the axis. That's why triangular sections have the greatest rigidty for
their cross section area, and square sections are more rigid than oval
or round.


gonefishiing wrote:
MEANING TO THE EXTREME OUTER FIBERS IN BOTH AXIS?

No, all fibers along the tension/compression axis. That's why it's an
integral.


but this is for rectangular sections right?
how do you calculate this for an oval section?

Personally, I don't. I look it up! But this is also not foolproof, you'd
be amazed how many mfg'rs fudge their specs (or maybe they can't do math).


YEAH BEEN THERE RECENTLY: IT IS ALSO CALLED EXPENSIVE WHEN YOU DISCOVERED
THEIR PUBLISHED ERRORS.

I hope nobody got hurt. This is called "letting your customers do your
failure mode testing." It's very popular with software firms, too

Regards
Doug King

DSK wrote:

gonefishiing wrote:

refresh my memory:
i'm looking for the section modulus for a boom section to understand
allowable bending stress.
sx= bd(squared) ?


No, IIRC it's the integral of the solid cross section area distance from
the axis.

Perhaps term you are groping for is "section moment"? You can look it up
on the web.

That's why triangular sections have the greatest rigidty for
their cross section area, and square sections are more rigid than oval
or round.

I think you may be confusing a truss with a section. For bending in
plane, the most rigid section per unit weight is close to a T
section with the top under compression and the bulb (or smaller bottom
plate) at the bottom in tension. If the direction can be up and down
this becomes the familiar I beam. For compression, a circle (tube) is
generally best as again, it places most material away from the axis for
all planes containing the axis. This important as the failure mode is
almost always buckling and not material compression. For a combination
of compression and bending the best solution lies between these cases so
we see egg shapes and ovals but the object is always to place as much
material from the axis as needed to contain the stress well below a
yield stress.

Hope this helps.

Cheers

That's why triangular sections have the greatest rigidty for
their cross section area, and square sections are more rigid than oval
or round.


Nav wrote:
I think you may be confusing a truss with a section.

Nope.

... For bending in
plane

In what plane?

... the most rigid section per unit weight

Did I say "per unit weight"? No I said cross section area.

... is close to a T
section with the top under compression and the bulb (or smaller bottom
plate) at the bottom in tension.

You're describing an asymmetric I-beam.

Now, think for a moment... have you seen any sailboat booms shaped like
I-beams?

No, you haven't. But if you've looked at pictures of fancy hi-dollar
racing yachts, you have seen triangular section booms.

Why is that, Navvie?


Maybe you better go call up Bruce Farr and tell him he's been getting it
wrong. He'll probably offer you a job.

DSK

You may be confused. Perhaps you would like to re-read your earlier
post? I'll highlight it for you:


DSK wrote:
That's why triangular sections have the greatest rigidty for
their cross section area, and square sections are more rigid than
oval or round.

You did say most rigid didn't you? I'd say that rigid means best able to
resist bending -but perhaps you have some other use of the term? It's
strange but engineering texts say that the section that best resists
bending is an I or T section. But Doug is always right so we must all be
wrong!

Now, since you've neatly drawn and posted a diagram why haven't you
actually shown us where your cosine is or what the forces on the boom
are. I gave you a solution but you've not explained why it's wrong.
C'mon share your engineering expertise!

I'd say this is a smokscreen and you are trying to wriggle off the hook.

Cheers

Nav wrote:
You may be confused. Perhaps you would like to re-read your earlier
post? I'll highlight it for you:


DSK wrote:
That's why triangular sections have the greatest rigidty for
their cross section area, and square sections are more rigid than
oval or round.


You did say most rigid didn't you? I'd say that rigid means best able to
resist bending -but perhaps you have some other use of the term? It's
strange but engineering texts say that the section that best resists
bending is an I or T section.


No, engineering texts say that "I" or "T" sections have the greatest
rigidity *in a defined plane*.

... But Doug is always right

Nope, I've been wrong several times. But this isn't one of them.

... so we must all be
wrong!

Who's "we" kimosabe?

BTW do you know what a cosine is? If you have the slightest clue about
any sort of engineering ... or even high school geometry... then my
diagram should be relatively easy to understand for you...

The compression force on the boom is the cosine of what angle?
What is that strange symbol on the mast represent? I believe earlier you
claimed the force on the mast had to be the exact same as the weight
suspended from the boom? Care to backpedal on that one, or don't you
understand what you yourself said?

DSK

DSK wrote:



No, engineering texts say that "I" or "T" sections have the greatest
rigidity *in a defined plane*.

Yes that's what I said ealier! What is it with your inability to read?
Anyway, now you know a triangle section isn't _always_ the strongest for
a given amount of material. You must be so happy to have learnt
something new! Hooray!


... But Doug is always right


Nope, I've been wrong several times. But this isn't one of them.

... so we must all be wrong!


Who's "we" kimosabe?

BTW do you know what a cosine is? If you have the slightest clue about
any sort of engineering ... or even high school geometry... then my
diagram should be relatively easy to understand for you...

The compression force on the boom is the cosine of what angle?

No, no, you silly man you need to re-read your posts. It was _you_ who
keeps saying it's a cosine -not me! Are you now realizing that you were
wrong and it's not a cosine after all? You do know what a cosine is
don't you?

Cheers

gonefishiing wrote:

refresh my memory:
i'm looking for the section modulus for a boom section to understand
allowable bending stress.
sx= bd(squared) ?
but this is for rectangular sections right?
how do you calculate this for an oval section?
gf.


The shape of the section is taken care of by it's moment of inertia.
The bending stress is the bending moment times the distance from the
neutral axis divided by the moment of inertia of the section.

Stress = M Y /I

This is known as the flexure formula.

Cheers