Choosing Steel Thickness for a bracket

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Discussion Overview

The discussion revolves around determining the appropriate thickness of steel for a bracket that will be part of a robotic arm designed to lift and move heavy loads. Participants explore calculations related to tensile stress, yield strength, and the mechanical requirements of the arm's design, including the forces exerted by hydraulic cylinders.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant outlines the calculations for tensile stress and the required thickness of the steel plate based on a worst-case load of 2 metric tons, concluding that a thickness of 0.5" would be sufficient.
  • Another participant questions the clarity of the bracket's intended support role and requests additional details about the application.
  • A later reply provides context about the robotic arm's design, noting that it is controlled by hydraulic cylinders and is intended to regularly hold and move 2 metric tons.
  • Further clarification is provided regarding the hydraulic cylinder's specifications, indicating it can exert a push/pull force of approximately 2.4/1.8 metric tons, although the exact lifting capacity of the arm remains uncertain.

Areas of Agreement / Disagreement

Participants express uncertainty about the exact forces acting on the brackets and the implications for their design. There is no consensus on the adequacy of the proposed bracket thickness or the overall design considerations for the robotic arm.

Contextual Notes

Participants highlight the need to consider the forces at the attachment points of the hydraulic cylinder, which may influence the bracket's design and required thickness. The discussion reflects varying assumptions about the mechanical loads and stresses involved.

roboemperor
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Just need someone to double check my work.

So the bracket is going to be cut out from a steel plate in a weird, rounded out triangular shape, and then a hole will be drilled through it. For simplicity though, let's just say the bracket is a ring with one half of it welded onto a wall. OD of the ring is 1.25", ID is 0.765", which results in a thickness of 0.2425".

The equation for tensile stress is σ = Fn / A.

Since the bracket is going to be pushed and pulled, the cross sectional area is:
(0.2425" * H) * 2, where H is the height of the ring when it is placed on the table, which is the thickness of the steel plate I need to cut the brackets out from. Equation simplifies to 0.485H.

The worst case scenario this bracket has to endure is 2 metric tons, which translates to 4410lbs

The yield strength of the steel plate I intend to use is 36,000psi. As I understand, 0.6 represents the ratio of maximum bending stress to yield stress for the material, so I'm guessing it is the same for maximum tensile stress? Which makes the maximum allowable stress 21600psi.

So if we solve for H,
H = 4410 / (0.485 * 21600) = 0.421", so if I get a steel plate that is 0.5" thick to cut my brackets out from I should be fine.
 
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roboemperor said:
Just need someone to double check my work.

So the bracket is going to be cut out from a steel plate in a weird, rounded out triangular shape, and then a hole will be drilled through it. For simplicity though, let's just say the bracket is a ring with one half of it welded onto a wall. OD of the ring is 1.25", ID is 0.765", which results in a thickness of 0.2425".

The equation for tensile stress is σ = Fn / A.

Since the bracket is going to be pushed and pulled, the cross sectional area is:
(0.2425" * H) * 2, where H is the height of the ring when it is placed on the table, which is the thickness of the steel plate I need to cut the brackets out from. Equation simplifies to 0.485H.

The worst case scenario this bracket has to endure is 2 metric tons, which translates to 4410lbs

The yield strength of the steel plate I intend to use is 36,000psi. As I understand, 0.6 represents the ratio of maximum bending stress to yield stress for the material, so I'm guessing it is the same for maximum tensile stress? Which makes the maximum allowable stress 21600psi.

So if we solve for H,
H = 4410 / (0.485 * 21600) = 0.421", so if I get a steel plate that is 0.5" thick to cut my brackets out from I should be fine.
A 1/2" bracket is a pretty hefty member.

It's not clear what this bracket is supposed to support. Why don't you provide some extra details?
 
The device is a large complex robotic arm controlled by multiple hydraulic cylinders, which in turn is controlled by a FPGA board. The arm is designed to regularly hold and move 2 metric tons of weight.

These brackets will be welded directly onto the arm and the hydraulic cylinders will regularly push/pull the brackets to change the arm's orientation.

I chose size 4" schedule 80 pipes for the arm's skeleton.
 
roboemperor said:
The device is a large complex robotic arm controlled by multiple hydraulic cylinders, which in turn is controlled by a FPGA board. The arm is designed to regularly hold and move 2 metric tons of weight.

These brackets will be welded directly onto the arm and the hydraulic cylinders will regularly push/pull the brackets to change the arm's orientation.

I chose size 4" schedule 80 pipes for the arm's skeleton.
The 2 tonnes is at the arm end I presume.
The hydraulic cylinder is attached to the arm somewhere between both ends of the arm.
You need to find out what force the hydraulic cylinder will be acting with at its attachment points.
 
I phrased my comment wrong. My bad.

The bottleneck here is the cylinder. I got the strongest one available (3000 operating psi) for its size which equates to about 2.4/1.8 metric tons of push/pull. I don't know how much weight the arm can lift, but it's not too important right now. I'm trying to make the arm utilize the full power of what these cylinders can offer.
 

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