As both members are attached to each other at the ends they will both
elongate the same amount. One member cannot stretch further than the other.

Now, taking into account e=PL/EA (explained earlier by Tom Pettit). We know
that e is the same for both, and L is the same for both, and assuming E is
the same for both, we get:

P1/A1 = P2/A2 and therefore P1/P2 = A1/A2

Whe P1 = load on member 1
P2 = load on member 2
A1 = cross-sectional area of member 1
A2 = cross-sectional area of member 2

If we have have the cross-sectional areas of both members we can work out
how the load is distributed between them. As we also have the strengths of
the two members we can calculate the total load at failure. You don't want
to go anywhere near that loading.

Ed

PS I've just realized we've been given the areas for both earlier. So here
goes:

P1/P2 = 1.00/0.667 = 1.5

If the total load is P,

P = P1 + P2 = 1.5xP2 + P2 = 2.5P2
or
P = P1 + P2 = P1 + 0.667xP1 = 1.667P1

Now we substitute the strength of each member for the load (remember the
strength is the load at failure).

For P1

S1 = A1 x 80,000 = 80,000 lbs (I hope that's right, I'm used to SI units)

Therefo P = 133,400 lbs

For P2

S2 = A2 x 120,000 = 80,000 (Obviously designed to take the same load)

Therefo P = 200,000 lbs

From this the failure load would be 133,400 lbs. When the Weaker member
fails, the stronger one suddenly receives the whole load (and it can only
take 80,000 lbs before failing). By juggling the figures you should be able
to reduce the area of the stronger member without chnging the strength of
the whole thing.

From a safety point of view, I wouldn't go within half of that load.

I'd be really grateful if somebody could check this. And I'm sure there will
be plenty of questions.

On 13/1/06 9:21 am, in article
, "Alan Baker"
wrote:

In article , "Highflyer"
wrote:

"Alan Baker" wrote in message
...
In article ,
"Dick" wrote:

okay, got pounds per Square Inch and elongation differences and the
failure
sequence in my noodle.

For simplicity, let's use 120,000 and 80,000 psi numbers.

So if the 120,000 # piece is .667 square inches
and the 80,000# piece is 1.0 square inch,,,are the elongations the
same???

Thanks, Dick

Ummm...

Think about it for a moment. To be different strengths, the two steels
must differ in chemical composition, right? Do you think that that
wouldn't also lead to different elongation under the same load?

--
Alan Baker

No it wouldn't lead to a different elongation under the same load. The
alloy materials are such a small percentage of the metal that generally the
bulk metal overrides and different alloys have, for all practical purposes,
the same youngs modulus. In this case, both alloys were 4130 so there isn't
even a difference in the alloy. If they have the same cross section area
and the same load they will have the same elongation. The heat treatment
doesn't change the slope of the stress strain curve. It merely moves the
yield point closer to the ultimate rupture point.

But then the piece without the heat treatment is going to yield at an
earlier point and thus elongate differently, isn't it?


Highflyer
Highflight Aviation Services
Pinckneyville Airport ( PJY )

Vancouver, British Columbia
"If you raise the ceiling 4 feet, move the fireplace from that wall
to that wall, you'll still only get the full stereophonic effect
if you sit in the bottom of that cupboard."