Calculating bending stress for pipes

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Discussion Overview

The discussion revolves around calculating bending stress for pipes used in a robotic structure that relies on hydraulics. Participants explore the necessary parameters for determining the appropriate pipe size and thickness, considering factors such as load-bearing capacity and fatigue under cyclic loading. The conversation includes theoretical calculations and practical applications related to pipe mechanics.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant seeks guidance on calculating the maximum load a pipe can withstand under bending stress, providing specific dimensions and material properties.
  • Another participant explains the bending stress equation and emphasizes the importance of calculating the moment of inertia (I) and the distance to the neutral axis (y) for the pipe.
  • There is a discussion about the significance of the factor of safety in bending stress calculations, with a ratio of 0.6 to yield strength being mentioned as a guideline for design.
  • Some participants express confusion about the rationale behind using the 0.6 factor, leading to clarifications regarding its origin from the AISC Steel Construction Manual.
  • Concerns are raised about the implications of using ultimate tensile strength versus yield strength in design considerations, with a focus on ensuring safety in structural applications.
  • One participant suggests considering different pipe schedules for pressurized applications, indicating a need for further exploration of material specifications.

Areas of Agreement / Disagreement

Participants generally agree on the need to calculate bending stress and consider factors such as yield strength and safety ratios. However, there is some disagreement and confusion regarding the application of the 0.6 factor and the distinction between yield strength and ultimate tensile strength, indicating that the discussion remains unresolved on these points.

Contextual Notes

Some assumptions about material properties and loading conditions are not fully explored, and there are references to specific equations and factors that may depend on additional context or definitions not provided in the discussion.

roboemperor
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I'm making a robotic structure using hydraulics and pipes, and I need to choose what size pipe and thickness I will be using.

I've searched online to the best of my ability but all I got were equations for I-beams.

I've taken statics in college so I know how to calculate all the loads in a multi-jointed strucuture. I only need to figure out how to calculate whether or not the structure will hold in the worst-case-scenario and if it will build up fatigue.

To clarify, I will be choosing the pipe's diameter, length, thickness, material type (mild steel), and the load, and I need to calculate whether the resulting structure will break or develop fatigue. For now assume the structure is a simple cantilever until I can get a better grasp on the concept.

Any help is appreciated.
 
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It's not entirely clear what you are planning to do here. Are the pipes intended to be load-bearing members, in addition to supplying hydraulic fluid to the robot mechanism?

The bending stress equation, σ = My / I, works for pipes like I-beams. You need to know how to calculate I and y for the pipe.

All structures which undergo cyclic loading are subject to developing fatigue. The trick is designing the mechanism to keep the max. cyclic stresses low enough that a decent fatigue life can be expected.
 
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The pipes are the main load bearing members. The pipes make up the skeleton of the structure, and I will be welding various things onto the pipe. The reason why I am using pipes instead of round bars is to make the structure lighter. No fluid will be flowing through the pipes. The hydraulic cylinders will be making the pipes move, like muscles on a bone, and their fluid intake/outtake is handled by flexible hoses.

Do you mind giving me a full solution of a simple example so I can dissect it in an attempt to fully understand it? It would make my life a whole lot easier.

Lets consider a pipe that is 300mm long, outer diameter is 40mm, inner diameter is 30mm. One end of the pipe is firmly attached to a wall, and the other end has a load. If the pipe is made out of a material with a yield strength of 36,000 psi (250 MPa), what is the maximum load allowed? Feel free to change some values if it would make things easier.

Thank you for helping me out on this. Its been a while since I've done anything related to moments and shear, so I am attempting to relearn it right now.
 
Last edited:
roboemperor said:
The pipes are the main load bearing members. The pipes make up the skeleton of the structure, and I will be welding various things onto the pipe. The reason why I am using pipes instead of round bars is to make the structure lighter. No fluid will be flowing through the pipes. The hydraulic cylinders will be making the pipes move, like muscles on a bone, and their fluid intake/outtake is handled by flexible hoses.

Do you mind giving me a full solution of a simple example so I can dissect it in an attempt to fully understand it? It would make my life a whole lot easier.

Lets consider a pipe that is 300mm long, outer diameter is 40mm, inner diameter is 30mm. One end of the pipe is firmly attached to a wall, and the other end has a load. If the pipe is made out of a material with a yield strength of 36,000 psi (250 MPa), what is the maximum load allowed? Feel free to change some values if it would make things easier.

Thank you for helping me out on this. Its been a while since I've done anything related to moments and shear, so I am attempting to relearn it right now.
Like I said before, bending stress is bending stress; only the cross section properties change.

For a pipe L = 300 mm cantilevered with a concentrated load P at the free end,
max. Bending Moment at the fixed end is M = PL = 0.3 P N-m

For a pipe with OD = 40 mm; ID = 30 mm,
I = (π / 64) ⋅ (OD4 - ID4) = (π / 64) ⋅ (0.044 - 0.034) = 8.59×10-8 m4

y = 40 mm / 2 = 20 mm = 0.02 m

σ = M y / I = 0.3 P ⋅ 0.02 / (8.59 × 10-8) = 69,850 ⋅ P Pascals

For pure bending, σ < 0.6 ⋅ σy , or σ < 0.6 ⋅ 250 MPa; σ < 150 MPa max.

This figure for max. bending stress will of course be reduced depending on the amount of fatigue life you wish for the piece to have.

σ < 150 MPa

69850 P < 150 × 106 Pa

P < 2150 N (≈ 483 lbs)

Pipe usually doesn't have the same yield stress as things like bars, I-beams etc. Ordinary Grade A pipe has a min. yield stress of 30,000 psi, while Grade B pipe goes to 35,000 psi. Check the grade of pipe and its material when designing.

Also, bending is not the only consideration when designing. If you have compressive loads applied to the pipe, buckling must also be considered when evaluating the max. load.
 
SteamKing said:
For pure bending, σ < 0.6 ⋅ σy , or σ < 0.6 ⋅ 250 MPa; σ < 150 MPa max.

Sorry for the 2 month delay. This is the only part I don't understand. Why is there a 0.6?
 
roboemperor said:
Sorry for the 2 month delay. This is the only part I don't understand. Why is there a 0.6?
0.6 represents the ratio of maximum bending stress to yield stress for the material. It comes from the AISC Steel Construction Manual, and it represents the lowest factor of safety permitted for bending stress in steel construction. You never want to make the maximum bending stress = yield stress for the material.
 
SteamKing said:
0.6 represents the ratio of maximum bending stress to yield stress for the material. It comes from the AISC Steel Construction Manual, and it represents the lowest factor of safety permitted for bending stress in steel construction. You never want to make the maximum bending stress = yield stress for the material.

Thank you. My understanding was the ultimate tensile strength was the dangerous one and the yield stress was the optimal one. Thank you for clarifying before I made a huge mistake.
 
roboemperor said:
Thank you. My understanding was the ultimate tensile strength was the dangerous one and the yield stress was the optimal one. Thank you for clarifying before I made a huge mistake.
Ultimate strength is reached just before failure. You never want to design to ultimate strength, unless you want a piece to fail, such as a safety link.
 
roboemperor said:
Lets consider a pipe that is 300mm long, outer diameter is 40mm, inner diameter is 30mm. One end of the pipe is firmly attached to a wall, and the other end has a load. If the pipe is made out of a material with a yield strength of 36,000 psi (250 MPa), what is the maximum load allowed? Feel free to change some values if it would make things easier.

You may want to include the pipe (schedule) 20, "40, standard" ,"80 standard for pressurized pipe"

http://www.engineersedge.com/pipe_schedules.htm
 

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