Tensile Strength Question

Help an old mind come to the correct conclusion G.

The situation is (2) steel straps side by side and fully in contact with a
bolt hole at each end . One is 4130 with about 80,000 psi tensile and the
other a heat treated 4130 to a 124,000 psi tensile strength.

Is the combination strength 80,000 or 124,000 or
204,000 psi tensile??

Thanks, Dick

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"Dick" wrote in message
...
Help an old mind come to the correct conclusion G.

The situation is (2) steel straps side by side and fully in contact with a
bolt hole at each end . One is 4130 with about 80,000 psi tensile and the
other a heat treated 4130 to a 124,000 psi tensile strength.

Is the combination strength 80,000 or 124,000 or
204,000 psi tensile??

Thanks, Dick



GOD NO

Tensile strength is stated in PSI per unit area, usually a square inch.
If you think about this for a minute, ask the question.
How can 2 pieces of metal, one that will carry at most
124KSI, carry 204 KSI if each of them will have failed
way before reaching that load??
You can't add them up. You need to look at the applied
load, figure the differences in elongation and what it will
do to the joint. This isn't particularly hard, you just need to
look at the charts and do the math. There are many good
books out there......

pbc76049 wrote:

FASTEST RACE GLASS ON THE PLANET
"Dick" wrote in message
...
Help an old mind come to the correct conclusion G.

The situation is (2) steel straps side by side and fully in contact with a
bolt hole at each end . One is 4130 with about 80,000 psi tensile and the
other a heat treated 4130 to a 124,000 psi tensile strength.

Is the combination strength 80,000 or 124,000 or
204,000 psi tensile??

Thanks, Dick

GOD NO

Tensile strength is stated in PSI per unit area, usually a square inch.
If you think about this for a minute, ask the question.
How can 2 pieces of metal, one that will carry at most
124KSI, carry 204 KSI if each of them will have failed
way before reaching that load??
You can't add them up. You need to look at the applied
load, figure the differences in elongation and what it will
do to the joint. This isn't particularly hard, you just need to
look at the charts and do the math. There are many good
books out there......

i.e.: pick one material or the other and size it to carry the desired load.

Richard

The 124 KSI strap will have less elongation than the 80 KSI
strap and will therefore take more of the applied load, with the
combination failing much closer to the 124 KSI figure than the
collective 204 KSI limit. When the 124 strap fails, the 80 strap will
stretch and fail almost instantly.
That's how I'd see it, anyway.

Dan

Dick wrote:
Help an old mind come to the correct conclusion G.

The situation is (2) steel straps side by side and fully in contact with a
bolt hole at each end . One is 4130 with about 80,000 psi tensile and the
other a heat treated 4130 to a 124,000 psi tensile strength.

Is the combination strength 80,000 or 124,000 or
204,000 psi tensile??

Thanks, Dick



Probably none of the above. I don't know of any way to simply "combine"
the yield strengths of two different materials. Typically, you would
look at the Emod of both materials, compute n, use that to convert the
area of the weakest material to be equivalent to that of the strongest
material, determine epsilon under the maximum load you expect to see,
and then convert that back to sigmas for both materials and then
determine if you have exceeded the yield strength of the weaker material.


Matt

Matt Whiting wrote:
Dick wrote:

Help an old mind come to the correct conclusion G.

The situation is (2) steel straps side by side and fully in contact
with a bolt hole at each end . One is 4130 with about 80,000 psi
tensile and the other a heat treated 4130 to a 124,000 psi tensile
strength.

Is the combination strength 80,000 or 124,000 or
204,000 psi tensile??

Thanks, Dick


Probably none of the above. I don't know of any way to simply "combine"
the yield strengths of two different materials. Typically, you would
look at the Emod of both materials, compute n, use that to convert the
area of the weakest material to be equivalent to that of the strongest
material, determine epsilon under the maximum load you expect to see,
and then convert that back to sigmas for both materials and then
determine if you have exceeded the yield strength of the weaker material.


Matt

Dick,

Problems like this need to be solved by deflection methods, as long as
the loads are less than ~70% of the yield strength (in the elastic
region )and the strap areas are equal, then the load divided by the
cross sectional area of the two straps will be the total stress. This
works because as long as the load on each strap is in the elastic region
the Modulus of Elasticity is very close for all steels, 29 x 10**6psi
if memory serves. The ultimate tensile strength of material only comes
into play after the "elastic limit" has been reached. At which point
your part is on its way to breaking. It then has started to work
harden, though it may not break, it is not the same anymore. If the
areas are not equal, the elastic stretch of each strap will be different
and a more complicated "free body diagram" would be needed to even get
close.
Ask any 2nd or 3rd year Mechanical Engineering student to borrow their
"Shigley" stress textbook. It will have examples of the more complicated
situations and problems as well. There may be other issues such as
stress risers that may cause other problems. Heck you could even have a
student turn the whole thing into a pretty cool Senior Project.

Good Luck,

Mike Butler

On 10 Jan 2006 11:55:37 -0800, wrote:

The 124 KSI strap will have less elongation than the 80 KSI
strap and will therefore take more of the applied load, with the
combination failing much closer to the 124 KSI figure than the
collective 204 KSI limit. When the 124 strap fails, the 80 strap will
stretch and fail almost instantly.
That's how I'd see it, anyway.

Dan

Youngs modulus is virtually independent of alloy and heat treat
condition. The load vs elongation will be "the same" if they are the
same size. ASSUMING the straps are of the same size, if the load
remains below the yield of the weaker member, you doubled your load
capacity based on that one. The higher one does virtually no good.

Youngs modulus is virtually independent of alloy and heat treat
condition. The load vs elongation will be "the same" if they are the
same size.

I was wondering when someone was going to point this out.

The slope of the stress /strain curve is independent of heat treatment. Both
stretch the same amount with the same force applied. The stronger stuff just
stretches further before it breaks.

Thanks ,

That makes much more sense!

Cam.


Youngs modulus is virtually independent of alloy and heat treat
condition. The load vs elongation will be "the same" if they are the
same size.

I was wondering when someone was going to point this out.

The slope of the stress /strain curve is independent of heat treatment. Both stretch the same
amount with the same force applied. The stronger stuff just stretches further before it breaks.

pbc76049 (removethis) wrote:
FASTEST RACE GLASS ON THE PLANET
"Dick" wrote in message
...
Help an old mind come to the correct conclusion G.

The situation is (2) steel straps side by side and fully in contact with a
bolt hole at each end . One is 4130 with about 80,000 psi tensile and the
other a heat treated 4130 to a 124,000 psi tensile strength.

Is the combination strength 80,000 or 124,000 or
204,000 psi tensile??

Thanks, Dick



GOD NO

Tensile strength is stated in PSI per unit area, usually a square inch.

Huh?

Tensile strength, like pressure, is stated in units of force per
unit area, for example pounds per square inch (PSI). In this
context tensile 'strength' is a misnomer, it should be called
tensile stress, or more correctly, yield stress in tension or
rupture stress in tension.

If you want to know how much force it take to to yield (or rupture)
a part in pure tension, neglecting the effects of fatigue, stree risers
etc, you multiply the tensile strength by the smallest cross-sectional
of the piece that is bearing the full load. For a bolt that will be be
area at the thread root, but keep in mind that threads are a
significant
stress riser.

Thus if you have a rod with a cross sectional area of one square inch
and a yield strength of 100,000 psi then hanging a weight in excess
of 50 tons on that rod will yield it.

Maybe that is what you meant, but it was not particularly clear.

--

FF