# Background

Trying to determine how much weight a post can support without necking when a monitor is attached to an articulated arm: a cantilever problem.

# Problem

There are three objects involved in this problem:

• a support post that has two sections bolted together;
• a monitor; and
• an articulated arm that extends the weight away from the post.

## Monitor

The monitor, which is mounted to a post:

• weighs 12kg; and
• has dimensions of 449.6mm x 690.2mm x 83mm

## Post

The post is:

Where the poles attach together a black plastic ring holds a mounting bracket. The mounting bracket wraps around the top post and hides the seam.

## Arm

The articulated arm has:

• a weight of 3.6kg;
• a maximum extent of 50cm; and
• an appearance as follows:

.

# Problem

Stainless steel has a typical yield strength of 520MPa and ultimate strength of 860MPa.

An engineer from the pole's manufacturer said that the pole will not support the weight without necking. This conclusion was reached before the engineer knew any of the weights or measurements involved.

# Attachment

The objective is to attach the articulated arm to the bracket attached to the pole. This will either be accomplished using four bolts on the plate, or having the two plates welded together.

# Questions

1. Is the post strong enough to support the monitor at the maximum extent of the articulated arm without buckling, necking, breaking, or otherwise becoming structurally unsound?
2. How would you perform the calculation?

Thank you!

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The crucial detail here, that's missing from your description, is how the arm is going to be mounted onto the pole. The manual for the arm only gives two options for mounting it: a wood stud or a flat concrete surface. Obviously this would be something different from those two options, so it needs to be specified in detail. –  Keenan Pepper Mar 8 '11 at 0:54
@Keenan: I was presuming the plates would be secured sufficiently to hold the weight. That was a rather large and erroneous assumption. Thank you. –  Dave Jarvis Mar 8 '11 at 1:12
the plate at the right end of the beam has a lot of holes, similar to the plate at the pole. This two might fit together in soem holes, if not, a drill will help. –  Georg Mar 8 '11 at 13:10

You have 3 obvious weak points in the system.

1. Where the (longer) arm is attached to the existing bracket
2. Where the bracket attaches to the pole
3. Where the pole is attached to the ground / base.

Notation I'll use:

• $W$ = weight of monitor
• $L$ = maximum arm length
• $=> M = WL$ = moment at point of pole attachement
• $x$ = distance between attachment points. E.g. width of bracket, width of base etc.
• $\sigma$ = yield stress of material
• $A$ = area

1 Arm - bracket I'm going to assume you'll be able to make this bracket strong enough. If you need to check, you just need to balance moments: $M = x * \sigma * A_{bolts}$

2 Bracket attachment to pole This is the most difficult calculation and I'm not sure if it's possible to solve analytically. It also depends critically on the details of the attachment bracket.

The moment in the pole is $M$. This is constant through the length of the pole (assuming pole is perpendicular to the arm) from the attachment point to the base. I'm assuming the top of the pole is unattached to anything.

This moment cause a compressive force along the front of the pole. With a wide, thin-walled pole the most likely failure will be by buckling. The calculations for buckling are difficult and often not possible to solve analytically - normally FEA would be used instead.

This is compounded by the second load on the pole. The moment is conveyed to the pole via the attachment bracket, pulling on the top of the bracket and pushing at the bottom. This means we have a (large) compressive force applied to the front of the pole at the bottom of the bracket - at the point that is already under compression due to the moment. This compressive force would eventually cause failure by buckling the pole at this point.

So the attachment to the pole will fail by causing the pole to buckle under the bottom of the bracket. Calculating the force to do this is very difficult (probably requires FEA) and depends on the precise details of the pole and the bracket itself.

3 Pole attachment to base The calculation is exactly as for 1 - $M$ must be counteracted by forces in the attachment spread over the base width $x$ of the attachment.

Conclusion It's going to be difficult (but certainly not impossible) to make the attachments strong enough, due to $L$ being much bigger than $x$, and $W$ being large. Having said that, decent bolts should be plenty strong enough.

The thing that'll be difficult to calculate is the weight required to caused buckling of the pole. Since buckling is a sudden failure (switching from one stability mode to another), you won't get much warning of this before failure, so it's difficult to predict just from suck-it-and-see (i.e. it'll be fine, right up to the point where it smashes through your desk).

If I were going to try this I'd probably try to get an identical pole & bracket and test to destruction. Much more fun, too! Failing that, you might get somewhere by seeing how steady it looks with the monitor not extended too far... (though watch out for sudden failure as above!).

You can mitigate the buckling effect by:

• Making the bracket wider, to spread the load on the pole
• Adding a sleeve to the pole under the bracket (i.e. make the pole stringer and stiffer)
• Fill the pole with something strong under compression (eg concrete) that will help to prevent the pole buckling inwards

UPDATE I've just looked in a bit more detail at the picture of the bracket itself, and I've realised that's another weak point ...! The bracket appears to be made of a few fairly thin sheets of metal, joined by small bolts (in sliding joints). This isn't designed to carry a large moment (monitor on long arm). I'd expect the sliding joints to slide, and wouldn't be surprised if the bracket itself were to buckle and collapse. Also the height of the bracket is quite small, meaning the forces applied to the pole (to transmit the moment) are large.

Given that, my next recommendation would be to build a new bracket! You've talked about welding some pieces together already, so I'm guessing this should be possible for you. Design a new bracket that is explicitly designed to carry a large moment, and to attach to the specific arm that you have.

Maybe something like:

• Foot-long tube that fits snugly around the pole
• Foot-long plate, with mounting holes for the arm, welded the full length to the tube (I guess this would be slightly easier to use a C-section to fit more easily to the tube?)
• If you want to be able to change the height, drill a couple of holes in the tube and screw bolts through to press against the pole. Put these at 90$^o$ to the arm mounting.
• This should work better, but if the pole itself is too weak it may still fail.

Hope this helps. Key thing with all this waffle from me is it's just some general ideas without knowing the full details of the system. I may be being pessimistic about the properties of the system, but without seeing the thing for real it's impossible to say. I just hope I've given you some appropriate things to consider.

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Thank you for a thorough analysis. The bracket fits the pole quite snugly, so adding a sleeve under the bracket would be possible if the sleeve is thinner than paper. Filling the pole is not an option (hidden wires inside the pole; needs to disassemble for moving). Making the bracket taller to spread the load would be difficult and expensive. Thanks for the ideas, though, I appreciate it. –  Dave Jarvis Mar 9 '11 at 18:16
Lastly, for what it is worth, I will likely move the bracket down so that it covers the seam between the top and bottom pole, reducing the stress at the connection point and thereby distributing some of the weight to the bottom pole. –  Dave Jarvis Mar 9 '11 at 18:21
@ Dave: (without knowing the full details of the joint) - I think the weakest point in the system won't be the joint between the arm and bracket, but between the bracket and pole. Clearly the arm is designed to be bolted to the wall, so whatever bolts would be used for that must be strong enough to make the equivalent join to the bracket. Assuming the holes in the bracket are same / similar distance apart as those in the bracket - if they're significantly closer together the forces will be larger. –  Sam Davies Mar 10 '11 at 9:38
Thank you, Sam. I opted to simplify the problem and saved up to purchase a new, thicker post. imgur.com/zK8gJP0,dTYrsn1,s1aH3z4,zHQgZ4w,CzwBYL1#0 –  Dave Jarvis Aug 28 '13 at 0:30

I used to make robotic hands and the main problem always was the torque generated by the end load.

At the end of this arm you have an extremely heavy TV, which increases the force on the poor little column the farther away it is. Since, torque is proportional to the radius. For a small radius the bolts will tolerate it quite well, but there will be a point where the torque exceeds the manufacturers limits of the force per unit cross sectional area (i.e. pressure) of the pole that will lead to the necking...

You can actually find this out by using the shear modulus and calculating the amount of weight $m_{\text{TV}}g$ times the perpendicular distance $r$ of the pole to the pole. Using the radius of the bolts you will get something like

$\text{Strain} = \frac{\text{stress}}{\text{Shear Modulus for the material}}$

$\frac{\delta l}{L} = \frac{\frac{F}{A}}{\text{Shear Modulus}}$

Since, you already know the max pressure you don't have to worry about anything. Just use,

$r = \frac{\text{The manufacturer's pressure rating}}{m_{\text{TV}} g}$

The result would be the maximum radius possible for your TV without buckling.

That aside, what I would do to prevent necking is either get a thicker pole (too expensive) or get another pole and bind it with the first one (just make sure that the restraints for the arm reach around them both. This is crucial) and then ground it firmly with the first one on a broad base (making sure the entire thing doesn't tip over either).

This will work because you've essentially increased the crossectional area of the poll. Kinda like the old stories about how sticks that stick together are harder to break. If this still doesn't work then add a 2 more poles and interlock them with restraints on a broad base. Do note that if I were you I would give em a nice weld and turn this into an awesome home DIY project. So, take everything with a grain of salt.

I think that should solve your problem.

Take care,

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@Anna: Thank you. A thicker pole is not feasible (notice that the pole goes through the top of the desk---nearly to the floor). The pole is locked in place by two long, thick, pieces of wood secured to the back of the desk. A second (or third) pole is also not feasible for the same reason as a bigger pole: the desk is finished. –  Dave Jarvis Mar 8 '11 at 18:01
@Anna: I used the following calculation: bit.ly/gm3eNp at WolframAlpha, but the answer was meaningless to me. Any idea what I've done wrong? –  Dave Jarvis Mar 8 '11 at 18:19
@Dave: You did $g^2$ instead of just $g$. Wait do you have any more photos? Maybe we can figure something out without ruining your expensive desk. The best thing in my mind would be to perhaps stuff this poll with a smaller poll that just fits in there? Or maybe something like prestressed concrete (you know just pour concrete into the thing with someone's help and stand back, wait for it to cool)? So, that it's essentially a compressive force that you've got to worry about. –  Anna Mar 9 '11 at 7:10
@Anna: I did not do g^2. I used r = 520 MPa / (14 kg * (9.8 m/s/s)). Unless you mean a different value for g than 9.8m/s^2. A smaller post won't work. Take a look at how the posts are assembled in the PDF: omnimount.box.net/shared/static/v4hvzsg7r1.pdf ... Wires go up the inside of the post (as you can see in the picture), so pouring concrete would be at odds to one of the pole's purposes (hide wires). I can bolt supporting rods to the outside of the pole, though, with some care and difficulty. –  Dave Jarvis Mar 9 '11 at 9:28
Down-voting for the concerns Georg raises in comments to his answer. There's no explanation (or reason as far as I can see) to go from $strain = stress/Young modulus$ to $r = pressure rating / m.g$. The answer makes no mention of buckling and doesn't look at how the arm is attached to the pole. –  Sam Davies Mar 9 '11 at 17:41