# Required force to compress flexible elastic tubing with fluid inside at pressure P

How can one calculate the force needed to completely flatten a flexible plastic (PVC) tubing with an outer diameter of 4.1mm and an inner diameter of 3mm using a flat surface pressing down from above? This process would reduce the internal cross-sectional area of the tube to zero, and the tube contains a fluid at a given pressure P.

• Is there a pressure regulator? Or just a fixed volume? Apr 18, 2023 at 13:43
• Is it supposed to be done just locally? If yes, how long is the section to be compressed and how long is the tube in total. Also is the fluid gas or liquid? Apr 18, 2023 at 16:06
• It entirely depends on where the fluid is going to go when being squeezed out. If it's in a sealed tube with nowhere to go, the applied force will build until it exceeds the stress required to fail the pvc tube May 18, 2023 at 18:29
• This is exactly what happens in peristalic pumps, where the rollers squeeze the tube (sometimes the term "crush" is used) to make a series of "pillows". I'd use that as a possible line of search in the literature Jan 14 at 21:27

A quick rough estimate is to consider the geometry of the deformation and calculate the ultimate force that pushes the pipe, which is working like an arched beam, beyond its plastic section modulus to create plastic hinges and flatten the pipe.

lets call the length and points on the flattenned figure starting from the left to right as the following.

• L is the length of pipe
• left edge A
• the black block left side B
• the black block right side C
• the right edge of flattened pipe D
• thickness of the pipe 4.1-3=1.1 mm
• span of flat pipe 4.1*3.14= 12.88 mm
• $$I_{plastic}=BH^2/4 =L*1.1^2/4$$

The Force F is the sum of the pressure of the liquid and the plastic moment at the 4 points ,A,B,C,D to crete 4 hinges. Let's fist calculate just the bending.

$$F* AB= I_{plastic} *\sigma_{y_{plastic}}$$ $$F=\frac{I_{plastic} *\sigma_{y_{plastic}}}{AB}$$ Now we add the component of the liquid pressure $$F= \frac{I_{plastic} *\sigma_{y_{plastic}}}{AB} +(L*3\pi*P)$$ This force will bend the pipe at A and D and bring the to a flat position the points B and C (a bit of distortion is left)

@kamran F=Iplastic∗σyplasticAB+(L∗3π∗P)

For unit conversion LP, the result is end up with N/mm. It should be N only please help to solve this issue. Thanks in advance.