Tensile Strength Question Continued

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

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
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."

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???

It depends on their length.

The elongation is: e=PL/EA where e=elongation, P=load, L=length of
piece, A=cross sectional area of piece, and E=elastic constant.

All steels have approximately (withing a few percent) the same elastic
constant, so the tensil strength of a steel plays no part in it's
stiffness, only it's strength. In your case, if the lengths of both
pieces are the same, then the smaller cross section will deflect the
most. All this is dependent on the geometry of loading being able to
accomodate different deflections.

tom pettit

"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.

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."

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."

--
Alan Baker
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."

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."

Alan Baker wrote:
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?

Surprisingly enough, that is not true. A treatable steel can be
hardened and annealed over a broad range of tensile strengths.
The bulk composition of the alloy does not change, but the
crystaline structure and the compositions of those crystals
changes.

Crudely stated, metals consist of microscopic crystals that butt
up against each other. The simplest heat treatable alloy,
carbon steel, is just iron and carbon. When fully annealed
(softest) all of the carbon is collected into tiny crstals
of graphite and the bulk of the material is pure iron crystals.
In that form, it really isn't steel at all, it is just wrought iron.
Heat treating will cause some of the carbon to combine chemically
with the iron and form carbides. Then you have steel and there
will be some pure iron crystals, some pure graphite crystals,
and some carbide crystals but the bulk proportion of
carbon atoms to iron atoms will remain the same.

The different hardnesses depend partly on how much carbide is
formed and partly on the geometry of the atoms in those crystals.
The geometry within the crystals can be manipulated by heating
to a temperature at which the geometry (packing) changes and then
cooling it so fast that they 'freeze' in that state.

All this happens without any change in the bulk composition of the
alloy. Which, I grant you is quite counter-intuitive.

Do you think that that
wouldn't also lead to different elongation under the same load?


The relationship between elongation and applied load depends on
a property called Young's modulus. It depends almost entirely on
the bulk composition of the material and is independent of the
strength of the material. Heat treating an alloy to a different
strength
does not affect its modulus.

Iron has the highest modulus of any of the natural elements which is
why really big stars become supernovae. More relevent to the
present discussion, small differences in the compositon of a
steel have almost no effect on their moduli. Thus all steel alloys
that are 99% iron have esentially the same modulus as pure
iron.

Even the 18-8 stainless steels that are less than 65% iron have
moduli that about 90% that of pure iron.

Neat, eh?

--

FF

Several posters have given good info. The elastic (Young's) modulus
does not change with heat treatment, so the 80 ksi steel stretches the
same as the 120 ksi for the same dimensions. The easiest way to solve
the problem is to calculate how much load is needed to stretch each
individual piece a set amount. Then for each increment of displacement,
the total load needed will be the sum of each, since they are attached
and must displace the same amount as load is applied. This is true up
to the yield point of the first piece to reach yield. After that things
get more involved, with plastic deformation, etc.
With different cross sectional areas, they will not stretch the same,
and the yield stress may be less than the sum of the two pieces. Each
pice must be analyzed separately, just as all parts of a structure must
be. Often times I see peolple adding steel straps to beef up aluminum
structure. This is not the best way as the higher modulus of the steel
means that it usually carries most of the load. In order to make a
situation like this work, you need to match the load versus elongation
of each piece to get the most out of it. You just have to do the
calculations!
By the way, since you didn't give the yield strengths, I can't
calculate which one will hit yield first. But to answer your question,
assuming that the two straps are attached by the
same bolts, they must stretch the same under load. This means that
since the 80 ksi steel has a larger area, it takes more load to stretch
it a set amount. But since it has a larger area, its psi is lower so it
may hit yield at the same time as the 120 ksi steel. It does carry 50%
more load than the 120 ksi material. For every 100 lbs load, 60 lbs
goes thru the 80 ksi, and 40 lbs through the 120ksi.


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