# Force to compress flexible elastic tubing

How would one work out the force required to compress flat, using a flat bar on top pressing down, a thick walled (15mm OD 10mm ID) piece of flexible plastic (PVC) tubing i.e. so reducing the x-sectional area inside the tube to 0?

Just looking for any way to approximate the real situation without having to use an FEA simulation.

At the crushed configuration we can assume we are dealing with a flat beam. Width= L of pipe, span = 15pi.

Height = 5mm.

for simplicity we assume three plastic hinges ( we will refine this by assuming more hinges, such as one at either edge of pressure block if need be), one at either crushed edge of the pipe, one at the center of horizontal flatly deflected top of the pipe right under the middle of back block.

Why do we assume this ultimate failure configuration? because we think this is the lowest energy needed, of course we later can try other collapse models.

The two plastic hinges on the side need to be enough to counter the resistance of these hinges M_p.

$$M_p = L5^2/4 \times \sigma_{plastic yield}$$

And the force needed is $$M_p/7.5\times \pi$$

We need to add to above the force needed to crush the pipe at the center which is equal to previous force/2: effective length is double.

This is the first collapse mechanism. now we can assume other modes and compare the results, in a few iterations we get close to required precision.

• Hi Kamran, thanks for this good idea. Where did you get the formula for Mp from? May 19 '18 at 9:35
• M_p is when the section is stressed past yield point and you have two equal rectangular stress blocks for compression and tension. so you get a moment arm of H/2 multiplied by BH/2 which is stress block area, so you get M_p = BH^2/4. May 19 '18 at 15:38