How to stop a weight from falling

1. Aug 3, 2016

ikeatwork

Hello, new member here so first post, I apologize for any unintentional faux pas.

Anyway...

I have a problem which I have been asked to sort out in my work place. We have a rig which is used to move a pneumatic ram up and down. Currently it is on an electrical winch system and I have been tasked in finding a way to make this safe for people working under it by having a secure safety system in place just in case the winch fails. The easiest way that I can think of doing this is by having a pin pass through the two outer edges of the RSJ it is attached to and go under a protruding piece of steel work of the movable section, as shows in the image below:

What I need to do is stop it dead in its tracks, if the pin bends and jams under the movable part of the rig that is fine, it can be sorted later. What it cannot do is break the pin upon impact. The numbers I have gathered from varying sources are below, all i want to know is what is the thickness of the pin that i need?

O.k. I am assuming a weight of 500kg, this is a lot and much more than what it actually weighs but if i overestimate then I have a little wiggle room.

The pin will have the following characteristics:
• A density of 7.85g/cm3
• A length of 250mm
• The impact point on the pin will be 20mm in from the end which is unsupported by the RSJ
• The bend point will be where the pin goes into the first upright of the RSJ at 50mm from the end.
• The steel has a yield of 250MPa
• An ultimate strength of 400MPa
• I think the modulus of elasticity is 30.3Mpsi

It has to withstand a weight of 500kg falling 110mm. The fall will have the following characteristics:

• Time to fall: 0.149779026187 seconds
• Velocity: 1.46883048716m/s
• kinetic energy: 539 joules.. or 539kN worked out using the following formula:
• mass x gravity x height = 500 x 9.8 x 0.11 = 539
If anyone could help I would appreciate it. If any of the numbers do not seem right in your eyes please tell me and I shall investigate further. I think I have everything i need to do the maths but I can't figure out how to do the maths. Every example I have seen has the impact in between the supports and not outside them, so just to clarify I have attached another diagram below which should make it clear what I am after:

On a separate note, would the shape of the pin in question make a difference? I.e. a hex shaped pin, or a more square one rather than a cylinder? Also I read somewhere that boring a hole through the center of the pin could strengthen it for this type of impact due to the larger surface area, is this true?

2. Aug 3, 2016

jack action

It depends on the amount of deformation desired (reference). Normally, you would want the pin deformation to stay in the elastic zone of the material. So, the transverse displacement $\Delta x$ can be found with the definition of the Young Modulus:
$$\Delta x = \frac{Fl}{AG}$$
Where $F$ is the shear force, $l$ is the initial length, $A$ is the shear area and $G$ is the Young Modulus. Also:

$$F = \frac{\tau_y}{S_f}A$$
Where $\tau_y$ is the shear yield stress (that we do not wish to exceed to stay in the elastic zone) and $S_f$ is the safety factor.

The amount of work done ($F \Delta x$) is equal to the energy input:
$$mgh = F \Delta x$$
$$mgh = \frac{F^2 l}{AG}$$
$$mgh = \left(\frac{\tau_y A}{S_f}\right)^2\frac{l}{AG}$$
Or:
$$A = \frac{S_f^2 Gmgh}{\tau_y^2 l}$$
That would be the pin shear area needed.

To be noted:

• The force is actually an average force during the displacement $\Delta x$, so it may have higher peaks. A higher safety factor may be required without testing.
• This assumes that only the pin deforms. If there is more deformation involved i.e. the part that hits the pin (and it will deform as well), this adds up as a safety margin ($F$ will actually be smaller since both $\Delta x$ combined will be larger).

The shape has an effect on the deformation pattern. If the pin is a triangle where the object hits a corner first, the «initial» area will be small, thus it will deform a lot, creating a flat spot more easily. Deformation is good (increased $\Delta x$), but the area of the deformed part must be great enough to withstand the shear stress. This is why a safety factor could be needed. If your pin is square but crooked, the part could land on the corner first and result may be different than if it lands perfectly on an edge.
I cannot stipulate if this is true or not, but if there is a hole in the center, the part will more easily deform, thus increasing $\Delta x$. This requires a deeper analysis and testing before attempting such design.

3. Aug 3, 2016

Baluncore

Welcome to PF.
An alternative solution to a movable pin would be a spring loaded cam attached to the travelling load. The cam would pinch the rail when the load was being lowered. A chord would need to be pulled to release the cam when the load was to be lowered.
Some lifting / lowering equipment fitted with locking cams automatically apply the cam when tension in the lifting wire falls below the value set by a spring.

4. Aug 3, 2016

Ketch22

I am a little confused by your description. Are you using the winch to move the ram and the ram provides a separate function or is the ram somehow supplemented by the winch and the ram provides the primary function? Also Are you talking about a regular work process or is this a safety procedure for maintenance persons?
As to the calculations notice that the kinetic energy developed increases rapidly with fall height. If you can reduce that the gains are significant. Another idea if in fact you are talking about the ram and not the winch. A standard maintenance item for workers with Hydraulic and pneumatic rams is to use a steel channel that fits around the pushrod. The piston is extend to slightly past the desired position, the channel is then placed around the ram. This is locked in place with a non load bearing pin or keeper. The ram is then retracted to alleviate the load and also reduce potential fall height to 0. This style preventer can often be placed by a single person and support high loads for maintenance.

5. Aug 3, 2016

rootone

Instead of a pin you could use a flat surface made of iron or similar.
No way any small falling object will break through that.

6. Aug 4, 2016

ikeatwork

The ram is not part of the lifting mechanism at all, that is all done by electric pulley. It's a safety procedure for regular use of the device the ram is used to apply loads on a vertical surface and quite regularly needs to be raised above head height and it is not uncommon for people to have to work under it. It's the type of thing that if the pulley failed due to a bolt sheering or cable fraying then a hardhat would do little to protect the person it fell upon.

Sorry about this, but you lost me there. I will be honest, to me to work out a part of one of the equations I have to go back to an earlier equation but then I go back and it is asking me for a number I need to work out using the equations that are towards the end. I.e in the first equation it asks to use A and in the last equation it tells you how to work out A but you need to have some of the numbers worked out previously to have this equation make sense. Am i doing this utterly wrong? Am i utterly over thinking this? All i need to know is if i need a pin of 8mm, 10mm, 12mm or 15mm diamter steel rod to do the job. I only picked 500kg as the falling weight to make sure it is definitely safe for what is falling as i am unable to physically weigh it and I know it is a lot less than 500kg, which would mean that if it did fall on a pin designed to withstand 500kg then the pin would not deform. I just want it stopped dead in its tracks. Again sorry for this.

7. Aug 5, 2016

Terrysv

Hi
I suggest a wedge as shown below. The unit could be moved to a desired height and protect the product when in use.
Black = RSJ
Red = wraparound collar.
Pink = set screw
Blue = wedge
Green = permanent magnet

8. Aug 5, 2016

CWatters

Ideally there should be some sort of interlock so that people cannot work under the weight unless the safety device is in place. Perhaps a gate lock that is only energised when the safety bolt is fully engaged. Should also fail safe in the event of a power cut etc

9. Aug 6, 2016

jack action

Only the last equation is needed: $A = \frac{S_f^2 Gmgh}{\tau_y^2 l}$. You should know all of these variables.

Note also that this assumes that the distance $l$ is a for rigid bar. Looking at your drawing, if we take the pin as a beam, the displacement $\Delta x$ can be greater and it would be represented by $y_{x_1 = a}$ (see reference) or:
$$\Delta x = \frac{Fa}{6EI}(2aL+3a^2 - a^2) = \frac{Fa^2 (L+a)}{3EI}$$
In this case:
$$mgh= \frac{F^2 a^2 (L+a)}{3EI}$$
Or:
$$F^2 = \frac{3EImgh}{a^2(L+a)}$$
In your particular case where the pin is round, $I = \frac{\pi r^4}{4} = \frac{A^2}{4\pi}$. Thus:
$$F^2 = \frac{3EA^2mgh}{4\pi a^2(L+a)}$$
The shear stress is $\tau = \frac{F}{A}$, thus we can determine that:
$$\tau = \sqrt{\frac{3Emgh}{4\pi a^2(L+a)}}$$
Or:
$$\tau_y \gt S_f \sqrt{\frac{3Emgh}{4\pi a^2(L+a)}}$$
Furthermore, in this case (beam), we have to consider maximum stress due to bending moment (see reference) where $n = r$ and $r = \sqrt{\frac{A}{\pi}}$ for a round beam, thus:
$$\sigma_{max} = \frac{n}{I}M_{max} = \frac{\sqrt{\frac{A}{\pi}}}{\frac{A^2}{4\pi}}Fa = \sqrt{\frac{16\pi a^2}{A^3}F^2}$$
$$\sigma_{max} = \sqrt{\frac{16\pi a^2}{A^3}\frac{3EA^2mgh}{4\pi a^2(L+a)}} = \sqrt{\frac{12Emgh}{A(L+a)}}$$
Or, if we set the maximum stress to the tensile yielding stress + the safety factor ($\sigma_{max} = \frac{\sigma_y}{S_f}$):
$$A = \left(\frac{S_f}{\sigma_y}\right)^2\frac{12Emgh}{L+a}$$
I did not check, but this should give you a smaller and more realistic value.

With the previous, you get the cross-sectional area of the pin. You can find the pin diameter from there.

10. Aug 6, 2016

Nidum

(1) It would be better to arrange the design so that the load can't actually fall at all or at least not very far .

(2) Common arrangement is to have a selection of holes so that pin can be placed in best location .

(3) Ideal arrangement is to have just one or a small number of pin locations and to lower the load under control until it seats on the pin .

(4) Putting a pin through plain holes in an RSJ is not good practice . Ideally have some reinforcement to maintain the strength of the beam and to provide some seating area for the pin .

Last edited: Aug 6, 2016
11. Aug 9, 2016

ikeatwork

The rig can be raised over 3m, there is a hole every 100mm.

See above.

Not possible due to the nature of the work.

What would you suggest? I am looking into the wedge system described in Terrysv's post. I am attempting to work out the formulae in jack actions posts. I have thought of having a u shaped steel pin going through two of the holes in the RSJ to the stress of anything falling is spread over the two points and there is less chance of the steel bending, I think. Another option is to have an adjustable chain looping over the top of the rig and down under the movable part to catch it but then you are replying on a chain and don't know if there are any weak links. Any other options tend to get more complicated. I am after a simple solution to a problem so that if it breaks then it is 1. easy and 2. cheap to fix.

The current safety device is not adequate at all really. I want myself and others in my work place to feel safe when working around it.

12. Aug 9, 2016

Nidum

(1) The simplest solution of all is just a system of props . Very commonly used is a selection of lengths of railway sleeper . May need a pair of opposing wedges to get small height adjustments and lock the prop in place .

(2) Anecdote - I once designed a high speed high loading hydraulic arrester device . A prototype had to be impact tested to validate the design . Not many suitable test facilities in UK but something suitable was found in the Railway Research Centre at Derby . They had a very big drop hammer , instrumentation and experience of conducting this sort of test . I was intrigued to find that the drop hammer probably dated from the 1920's at the latest and would originally have been steam driven . It had now been adapted for modern work with electric hoist and greatly extended slides to give longer drops . They just used pieces of sleeper and bits of RSJ as safety props .

(3) If you decide to stay with pins then ideally they need to run in tubes welded between the RSJ flanges . Probably easier in practice to weld drilled square blocks rather than round tubes .

(4) I would be tempted to at least look at a device which bolted through any set of adjacent holes . Work of a couple of minutes to bolt in place and unbolt afterwards . Doing it this way you could have a catcher system of immense strength and functionally better shaped than a round pin .

Actual catcher piece could then be a solid rectangular block faced with a thick buffer layer of industrial rubber and optionally you could have have some metal slip pieces that could be put in to reduce the gap between load and catcher to zero .

A difficulty is that you would need to work out a configuration that allows you to bolt the parts in place without needing to work under the temporarily unsupported load .

(5) PS: Energy methods are useful when sizing pins to withstand impact loading .

Last edited: Aug 9, 2016
13. Aug 9, 2016

Terrysv

Hi Ike 2 ideas
1. Another option is to use seat belt material. 2” nylon material has a breaking strength is approximately 2722kg. Secure the material from above in place of a chain.
Long ago I saw where the material was folded in half and had many rows of cross stitching. The idea is that one or several rows of cross stitching burst, absorbing energy as the weight is falling.

2. Make a wedge with a pin to be placed in a hole below the rig. Instead of bending the pin the wedge will be jammed into the 20mm space.

14. Aug 10, 2016

Tom.G

Along with some means of keeping it in place under vibration (and gravity), it seems one of the better ideas. Still have to figure out how to position it safely.

15. Aug 10, 2016

Terrysv

The larger mass is below the pin to let gravity hold the wedge in the proper rotation.
A cotter pin could also be used to secure the unit.
If the rig fell a secondary problem may see the rig itself be torn apart and having a second fall. The drawing below shows a series of cuts in the wedge. The idea is each tab is to bend down on top of each other as they absorb energy from the rig falling.

16. Aug 11, 2016

tkyoung75

These days it is quite a feat to allow any of your workers to work under a suspended load.
We would not be designing a pin (or a wedge like those which used to be used as brakes on conveyor take-up towers) to withstand a dynamic "impact" load. It is just too dangerous, and can fail.
Any interlock would be used to ensure that people cannot work under it when it is suspended. That would be a feedback saying there is load on the cylinder, don't open gate.
Safe solution is to not work under the suspended load. This is achieved by building a structure to support the load, engineered, designed 'rated' certified (cock on the block). I would consider going down to the local mechanic and see what they do when they work under cars, maybe ask them how they would feel about standing underneath the car while you drop it 110mm onto a pin. Sorry!?!

17. Aug 11, 2016

ikeatwork

As i have said before, not working under the load is not an option. In the job that it is used for the rising part of the rig is used to raise an lower items used to perform the tests which need to be done. I just need to make it as safe as possible for the person or persons in the position of having to adjust the test apparatus during the test. I looked at the rig and i just thought of the worst case scenario and a way to minimize damage to personnel. The load is already supported by the electric winch and pulley system, what i am after is something to make it safe if that system fails for some reason.

18. Aug 11, 2016

Tom.G

A different approach.

Here is a concept sketch for automatic braking. The spring strength and brake pads must be sized such that each one alone can stop the load. For redundancy, use two concentric springs on each side (wound in opposite directions), each strong enough to stop the load. If the implementation is easier, use leaf springs instead of coil springs. Of course the braking end of the leaf spring will have to be held down but allowed to slide.

Last edited: Aug 11, 2016
19. Aug 11, 2016

tkyoung75

1 J = 1 Nm, Not 1kN. Energy and Force are not the same. To convert 539 Joules (kinetic / Potential), you need to calculate over a distance to allow conversion to Force (I.e. pin displacement / elastic deformation). We would need to see that calculation to check your impact load.

Moving the mass of the pin away from its centre of mass increases the Section Modulus. This helps with bending, not shear (thats why structural engineers use I-beams). As mentioned by jack, It will also affect your deformation, and therefore the load as it moves through the pin. I think jack gave you your formulas.

Be specific in your design. Quote sources of 'weights' etc. Design to data, then add safety factors in accordance with "standard safety factor". This will avoid over specifying your design, and provide clarity regarding how safe it is.

This is a good insight into what you are working on. Don't forget (below) you still need to look at bushing your holes in the RSJ..

The image below might help.

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Last edited: Aug 11, 2016
20. Aug 25, 2016

ikeatwork

Thank you for all your help on this matter. After weeks of going through the different options I have narrowed down the final design something akin to Terrysv's two ideas with the wedges, the one with the bracket surrounding it and the one with the slots cut out. I have found the process of asking you all quite rewarding in a number of ways and has sent me down some paths of learning that I wasn't entirely prepared for. Thank you all, I think I shall stick around and see what else I can learn.