# Thread: Real World Beam Failure Analysis

1. The motor is suspended from the beam using a cable. Think of how a "normal" hoist works, then turn it upside down.

2b0396a3-888f-4c04-b1d3-faf14dd78ed1.jpg  Reply With Quote

2. "Ok, fair enough.
But I'm showing that it would take well over 200,000 pounds of force to bend a schedule 40 pipe 3/4 of an inch over a span of 10 inches."

You were assuming elastic deformation which it was NOT. It failed at a much smaller load and subsequently deformed 3/4 inch

To do this problem you need an accurate assessment of the dynamic load ( dragging the load up a chain under the worst conditions, namely startup when the motor accelerates the load.

You need the maximum motor startup torque which may be as much as 3 to 5X the nominal motor design torque.
Next you write the equation
T*2*Pi=F*x
which is the energy equation
x= the distance the load climbs for one revolution of the motor shaft
T= maximum torque delivered by motor
F= the maximum force applied to the chain that lifts the load and is the pipe design force.
On top of this you need a factor of safety of at least 3.
Not so simple. DO NOT use the weight of the load for the design calculation.  Reply With Quote

3. I thought the motor was on the top, as I drew.

What I have drawn is just a simple scheme.  Reply With Quote

4. Originally Posted by PierArg I thought the motor was on the top, as I drew.

What I have drawn is just a simple scheme.
Now I see where you were getting the Torque from! Put the motor over the load and you got it. The motor then winds the cable in to it's self (like a man climbing a rope) to lift both the load and the motor.  Reply With Quote

5. Thanks! Now i've understood.
I supposed the motor was on the top because there is only one cable.  Reply With Quote

6. pipe failure .jpg

Point of clarification:

There is no torsion load on the pipe.
As pictured above, the motor hangs from a wire rope.
The wire rope is pulled up (by a rigger) between two C channel beams,
and the pipe is inserted into a loop of the wire rope, as shown.

The motor then climbs its own chain, lifting the load attached underneath (not shown).  Reply With Quote

7. Zeke:

Thanks for your guidance on this problem.

And thanks for that formula, I haven't used that one before.
We rarely have the information available to use it though.

However, what we do have is a industry standard rule of thumb
to calculate the dynamic load when using a specific brand / model hoist.
(The standard hoist in this industry)

That rule of thumb has been verified in a lab and in the field (many times)
by direct measurements using a load cell.

(it is actually a range of 1.25 to 1.5,
but we always use the higher value of 1.5)

So the dynamic load is a relatively known quantity.

Using a maximum weight of 1200 pounds as claimed by the show,
the potential dynamic load is 1800 pounds.

Using a FOS of 5 (industry standard for the pipe as used in this application),
the pipe design force is 9000 pounds.

Since I am (now) not assuming elastic deformation,
where do I go from here?  Reply With Quote

I don't see how the motor start/stop dynamic load can possibly deform that pipe in that configuration.

That type of deformation would have to come from a shock load.
For example, while the point was being let down to the deck,
maybe it caught on something, the chain went slack,
then the load slipped and fell, creating a shock load on the pipe / rigging.

A shock load can get pretty high very quickly.
And shock loads are not common, but they are not rare in this industry either.

A shock load is easy to calculate;
the equation is well known and relatively simple,
and the parameters are usually readily available.

But what I am more interested in
is how to calculate the stress needed to deform that pipe in this configuration.
The "beam" is so short that I'm not sure the 'beam bending' equation is the correct one to use.
The specific situation is almost more of a 'crush resistance' calculation.

If the beam bending equation is the correct one to use,
I'm getting a number so high that other elements would fail long before the pipe would.

And Zeke hinted that the beam bending equation is not the right way to go about it anyway.
I'll do the work, make the calculations, look up the numbers.
I don't want anybody to solve the math or just provide the solution.
But I sure could use a hand being pointed in the right direction...  Reply With Quote

9. The load required to reach yield (assuming 30 ksi), is about 4000 lbs. That is when your pipe will permanently deform.  Reply With Quote

10. I'd sure like to see the math / table / whatever on that.  Reply With Quote

11. See image below

8hjchY3.jpg  Reply With Quote

12. Thank you.  Reply With Quote

13. Kelly:
Any chance of adding this equation to the site?  Reply With Quote

14. Originally Posted by dalecyr Kelly:
Any chance of adding this equation to the site?
I think we do...

http://www.engineersedge.com/beam_be..._bending15.htm

Needs an associated calculator...  Reply With Quote

15. "Ok, fair enough.
But I'm showing that it would take well over 200,000 pounds of force to bend a schedule 40 pipe 3/4 of an inch over a span of 10 inches.

(I=.3)

Does that sound right?"

No, what you did was use elastic theory to get the deflection.
That deflecrttion was the resuly of a yield load and most of it was plastic deformation after yield.

Your problem is simple, it was simply underdeigned

To do it right,

1 find the maximum starting torque the motor can put from which you get the corresponding tension in the wire rope.

Apply a factor of safety of at least 2.5

Use beam theory to keep teh flexural stress less than the yield stress using the design force.  Reply With Quote

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