Real World Beam Failure Analysis

On a recent theatrical show, during the "load-out",
it was noted that a "beam" failed sometime during the show.

Later that week I was shown the "beam", which is actually a section of schedule 40 pipe, 4 feet long.
The pipe was spanning a grid "well" (see drawing), a distance of 10 inches.

In attempting to determine the nominal stress to show the strain as indicated,
I'm thinking that this formula applies:

[url]http://www.engineersedge.com/beam_bending/beam_bending2.htm[/url]

even tho the "beam" is pipe.

Am I thinking correctly?


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Hi dalecyr,
Very interesting question!

I think you are right. I would have used the same formulas but this doesn't mean that it is right :grin:

I'm curious to know what the other members think.

What is different between a pipe and a "classic beam" is the internal stress.
As you see, the pipe is subject at the following stress:
- bending: the moment of inertia of a pipe is different from a beam
- transverse shear: the shear factor for a pipe is different from a beam.

You are right to use that calculator. Being in the temporary works game, specifically scaffold design, I work with "tube" quite a bit. PierArg is correct in noting that the section properties between "beam" and "pipe" are quite different. In bending, tube is rubbish. Once the FOS has been applied, we can only put a maximum of 1.12 kN.m on a piece of 48.3 mm (outside dia.) tube (UK). Even without the FOS, it's not much better. In addition, there will be different results between a suspended load as opposed to a directly applied load. It may not have been a bending failure but a buckling failure.

Do you have a link to the show?

Hi Cake,
what do you mean for FOS? Maybe "force"?
(i'm sorry but i don't understand the abbreviation)

I'd like to point out that the "pipe" is not recommended for that application (IMHO).
You need a beam shape with a high moment of inertia when bended...such as a rectangular shape with the long side oriented in the vertical direction.
Or just a beam with the "I" section.
In this way the bending is limited.

Factor of safety. Sorry. I'm so used to using this abbreviations, I sometimes forget that not everybody knows what they mean. Damn you, international internet!

Thanks Cake!G:J

What do you think about my opinion?
Should have you used a pipe for that application?

Over a shorter span to decrease bending and increase shear (pipe can do quite well in shear), it would have been fine. A beam or just a larger diameter pipe would have been better.

"On a recent theatrical show, during the "load-out",
it was noted that a "beam" failed sometime during the show."

Please describe the "failure"

In my understanding, the term "beam" has to do with the loading, not the shape. If the loading is along the axis, its a column. If the loading is perpendicular to the axis, its a beam. Think of the columns and beams in a building. Material of any cross-sectional shape can be subjected to those loads. The moment of inertia factor in the formulas accounts for the effects of the different shapes and how they respond to the applied loading.

And yes, you are using the right formula.

[QUOTE=zeke;5843]"On a recent theatrical show, during the "load-out",
it was noted that a "beam" failed sometime during the show."

Please describe the "failure"[/QUOTE]

Hi Zeke.

By "failure" I specifically mean that the pipe was permanently deformed.
The 4' long pipe was "bent" about 3/4" at the center point,
resulting in the ends lifting off the grid, as shown in the right hand half of the drawing.

Doom:
This method of rigging is standard practice in the U.S. in this industry.
However, a 4" schedule 40 pipe is normally used in that location in this house.
I don't know why the 1.5" pipe was used this time.
(I was not in the building at the time,
but I have been talking to the other riggers who installed / deinstalled that point.)

The load is a self-climbing chain motor, with a load attached under that.
The chain motor capacity is nominally 1 ton (2000 pounds),
with a brake capacity of (typically) 2-3 times that.

The show maintained that the static load was less than 1200 pounds.
Motor startup / stopping typically increases the static load by 2-3 times.

Industry standards for FOS for this configuration is 5 times the dynamic load minimum.


So, I have a question and two concerns.
The question is 'how much force does it take to permanently deform that pipe by that much'.
One concern is that we are much closer to the failure point than we realized.
The other concern is that the show may not know the load amount accurately.

Joe:
That is my understanding also, so I'm off to find (actually, re-check) the E and I for 1.5" schd 40 pipe.
The last answer I got didn't make sense...

[QUOTE=zeke;5843]"On a recent theatrical show, during the "load-out",
it was noted that a "beam" failed sometime during the show."

Please describe the "failure"[/QUOTE]

Hi Zeke.

By "failure" I specifically mean that the pipe was permanently deformed.
The 4' long pipe was "bent" about 3/4" at the center point,
resulting in the ends lifting off the grid, as shown in the right hand half of the drawing.

Doom:
This method of rigging is standard practice in the U.S. in this industry.
However, a 4" schedule 40 pipe is normally used in that location in this house.
I don't know why the 1.5" pipe was used this time.
(I was not in the building at the time,
but I have been talking to the other riggers who installed / deinstalled that point.)

The load is a self-climbing chain motor, with a load attached under that.
The chain motor capacity is nominally 1 ton (2000 pounds),
with a brake capacity of (typically) 2-3 times that.

The show maintained that the static load was less than 1200 pounds.
Motor startup / stopping typically increases the static load by 2-3 times.

Industry standards for FOS for this configuration is 5 times the dynamic load minimum.


So, I have a question and two concerns.
The question is 'how much force does it take to permanently deform that pipe by that much'.
One concern is that we are much closer to the failure point than we realized.
The other concern is that the show may not know the load amount accurately.

Joe:
That is my understanding also, so I'm off to find (actually, re-check) the E and I for 1.5" schd 40 pipe.
The last answer I got didn't make sense...

"The load is a self-climbing chain motor, with a load attached under that.
The chain motor capacity is nominally 1 ton (2000 pounds),
with a brake capacity of (typically) 2-3 times that."

The calculation is not for a static load but the fact that it is braking makes it dynamic and energy methods have to be employed to get the correct pipe in addition to considering the startup load which is derived from 5x nominal torque. It is NOT 5x the weight.

[QUOTE=zeke;5859]

The calculation is not for a static load but the fact that it is braking makes it dynamic and energy methods have to be employed to get the correct pipe in addition to considering the startup load which is derived from 5x nominal torque. It is NOT 5x the weight.[/QUOTE]

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?

Just did a rough sums and this what I came up with:-

I=5.98cm^4
I/D = 3.07cm
O/D = 3.81cm
F = 90kN (9 metric ton based on a rough conversion)
Max BM = 5.63kN.m
Max Shear = 45kN
Def = ~2.37mm

Look about right? Our nearest equivalent section can only take 1.3kN.m so that would be a big no no.

Please forgive me Cake!
And the torsional stress? You said there's a motor...the part of the beam between the motor and the rope is subject to torque when the motor runs.

Or not?

I'll have to look in to that a bit later. This is a sneaky post from work and I've got a job on at the minute that is being a bit of a female dog.

[QUOTE=Cake of Doom;5874]This is a sneaky post from work and I've got a job on at the minute that is being a bit of a female dog.[/QUOTE]
what does it mean?

As I told you I speak just an elementary english ;?

Mea culpa. Posted from my work computer whilst nobody else is looking.

As far as I can tell the beam isn't fixed at any end so I don't think torsional stress should be a factor.

I agree, there's no torsional stress when the ends are not fixed...but how does it turn?
Where is applied the motor?I suppose there is a motor joined at the end (right or left side)

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

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"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.

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

What I have drawn is just a simple scheme.

[QUOTE=PierArg;5884]I thought the motor was on the top, as I drew.

What I have drawn is just a simple scheme.[/QUOTE]

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.

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

[ATTACH=CONFIG]602[/ATTACH]

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).

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 "Dynamic Load = Load x 1.5"

(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?

In thinking about this some more,
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[I] is [/I]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...

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

I'd sure like to see the math / table / whatever on that.

See image below

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Thank you.

Kelly:
Any chance of adding this equation to the site?

[QUOTE=dalecyr;5929]Kelly:
Any chance of adding this equation to the site?[/QUOTE]

I think we do...

[URL]http://www.engineersedge.com/beam_bending/beam_bending15.htm[/URL]

Needs an associated calculator...

"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.