Calculating bending stress on a pipe

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SUMMARY

This discussion focuses on calculating bending stress in 1" Schedule 40 PVC pipes, specifically comparing two designs for bending a 90-degree angle. Design A, which maintains a longer bending section of 7.75', exhibits less bending stress and is more durable for temporary shade structures, while Design B, with a shorter bending section of 5.75', is easier to transport but more prone to failure due to increased bending stress. Key calculations involve determining the radius of curvature and using the formula for bending strain, which is (half the pipe diameter) / (radius of curvature). Additionally, the discussion highlights the impact of creep on plastic materials, complicating long-term predictions of structural integrity.

PREREQUISITES
  • Understanding of bending stress and strain calculations
  • Familiarity with PVC material properties, specifically PVC 1120
  • Knowledge of creep behavior in plastics
  • Basic skills in using tools for bending and clamping PVC pipes
NEXT STEPS
  • Research "bending moment radius of curvature" for detailed calculations
  • Explore "plastics creep" to understand long-term material behavior
  • Learn about U-bolts and their applications in pipe construction
  • Investigate techniques for creating curved templates for bending PVC
USEFUL FOR

This discussion is beneficial for engineers, designers, and DIY enthusiasts involved in constructing temporary structures using PVC pipes, particularly those focused on optimizing design for durability and ease of transport.

dustball23
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TL;DR
I wish to compare the bending stress of 2 different designs.
  • In both designs, a PVC pipe is being bent 90 degrees, fixed at each end.
  • In both designs, the pipe is fixed at both ends.
  • In both designs, the pipe is 1" (inner diameter) Schedule 40 PVC pipe, 450 PSI. Material code PVC 1120. NSF pw-G ASTM D1785 PPFA 02199101H6BX 0934.
  • In design A, the length of the pipe that bends is 7.75'. (The actual pipe is 10' long but the ends will not bend much for two different reasons. The one end slides over a 1' pole to hold it in place. The other end slides 1.25' into a larger pipe. So the middle 7.75' must bend 90 degrees.
  • In design B, the 10' pipe is cut in half, and a 2' rigid outer sleeve joins them in the middle. So the part that bends (the 7.75' from the original design) is now only 5.75' - but really two separate sections of 2.875' each.
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Design A will last longer because there is less bending stress. Design A is a well known and well-tested design for temporary shade structures in the desert. Design B is easier to transport because the 10' pipes are cut in half, but will fail earlier because the pipes will get bent too much too early. How much additional bending stress is happening? (Typically, these structures are used 10 days a year, and will be reused every year until the pipes get too bent.)
 
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Since you know the shape, you can measure or calculate the radius of curvature. From that, and the diameter of the pipe, you can calculate the bending strain. The strain equals (half the pipe diameter) / (radius of curvature). The stress can be calculated from the strain. Good search term to learn more: bending moment radius of curvature.

HOWEVER, the above assumes that the material has a linear stress / strain curve. Plastics typically do not have linear stress / strain curves. The strain can be calculated exactly, while the stress is a function of the (nonlinear) stress / strain curve plus time due to creep.

dustball23 said:
Typically, these structures are used 10 days a year, and will be reused every year until the pipes get too bent.

Plastics are also subject to creep. If the pipes are taking a permanent bend, that is the result of creep. Creep gets worse with increasing stress. Good search terms to learn more: plastics creep. It is difficult to predict life when the failure mode is due to creep because you need specific creep curves for your specific material. Also, creep curves are typically plotted against stress, while you have a constant strain design. The stress decreases with time, while the strain stays the same.

Is it practical to change Design B such that the center sleeve is replaced with a center bracket curved to the same radius as Design A? You would need U-bolts or something to pull the ends of the pipe into place. The result would have the exact same pipe curvature as Design A.
 
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All very interesting points. That makes sense how the stress decreases with time (as the plastic creeps) and I was thinking the same thing. Maybe having bent pipes isn't so bad, since each year they will simply take more of a shape of the final structure anyway.

I'm not sure I have have the resources to create a precisely curved sleeve. Maybe I could take a heat gun, warm the sleeve, and bend it by hand to a best approximation of the curve. I'm also starting to doubt the curve need to be that long, I don't know why 2' was chosen by the designer of Design B. (My gut feeling is that by using too short of a piece - say a standard PVC coupler of a few inches - the stress would be too much and the design would buckle in that place.)
 
dustball23 said:
My gut feeling is that by using too short of a piece - say a standard PVC coupler of a few inches - the stress would be too much and the design would buckle in that place.
Your gut feeling is correct. If the fit to the sleeve is loose, then the pipe and the sleeve will be at angles instead of a smooth curve.

I think that the 2 foot length is about right. Your idea of a heat gun is good, and it is certainly easy enough to try. I suggest cutting a piece of wood to the desired curve to use as a bending template.

Here's another idea:
P8060022.JPG

Take a piece of wood about 1.5" thick and 2 feet long. Trace a piece of bent pipe to get the curve correct, cut the curve, then carve out a groove. Clamp the pipes in place with a pair of U-bolts. If you cannot find the right size U-bolts, just buy a piece of 1/4" threaded rod, cut to length, and bend into a U.
 

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