Calculate shear strength A36 plate

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Discussion Overview

The discussion revolves around the design and calculations for a tank intended to hold diesel fuel, focusing on the structural integrity of the materials used, particularly A36 steel plate and 10 gauge steel. Participants explore the shear strength calculations, lifting mechanisms, and safety considerations related to the tank's construction.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant outlines the dimensions and weight of the tank, proposing calculations for shear strength based on the formula F = P x T x PSI, leading to a calculated force per hole.
  • Another participant corrects the initial shear strength calculations, noting that the allowable shear stress for A36 steel is lower than its tensile strength, suggesting a de-rating factor.
  • Concerns are raised about the design of lifting lugs, emphasizing the need for careful sizing to match shackles and the importance of using thicker materials for the lugs to withstand potential stresses.
  • A participant suggests that having multiple lifting lugs would stabilize the load and reduce shear stress on each lug.
  • There is a discussion about the potential risks of using thin gauge metal for the tank and the impact of corrosion over time.
  • One participant, identifying as a mechanical engineer, recommends lifting from the top to ensure that the lifting lugs experience axial stress.
  • A later reply expresses concern about the thread's continuation due to the inexperience of the original poster (OP) and the complexity of the topic.

Areas of Agreement / Disagreement

Participants express various viewpoints on the calculations and design considerations, with some corrections and suggestions made. However, there is no clear consensus on the best approach or final design, indicating ongoing debate and uncertainty.

Contextual Notes

Participants highlight the need for safety factors in design, the importance of material selection, and the potential for errors in initial calculations. There are also concerns about the long-term durability of the materials used.

Who May Find This Useful

Individuals interested in mechanical engineering, structural design, rigging, and safety considerations in construction may find this discussion relevant.

Hugh99
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I'm trying to build a tank to hold diesel fuel, it will be on skids to lift with a fork lift but I also want to extend the the sides to provide a way to raise the tank from two attachment points on each side. The sides will be 1/4 A36 steel plate and the rest of the tank will be 10 gauge (0.140625). The tank should weigh around 350 lbs and hold 119 gallons of diesel. Using maximums that's 119x7.5 lbs = 892.5 lbs, including the weight of the tank that is roughly 1250 lbs.

Assuming the holes at the tope of the side plate are 2 in diameter and the shear formula F = P x T X PSI
I was going to calculate P as 2 in x pi and then divide by the portion of the whole that would be in contact with the rigging, so assuming 1/4 of the perimeter is in contact with perimeter, = 1.57
F = 1.57 x .25 x 36,000
F = 14,130 lbs per hole, so I should be able to support over 28,000 pounds for the two holes.

Does my math look correct?
If I left the sides as 10 gauge, it looks like it could still support 7900 lbs per hole for total of 15,800 lbs so I may not need 1/4 plate on the sides.

Thanks
 

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Hugh99 said:
I'm trying to build a tank to hold diesel fuel, it will be on skids to lift with a fork lift but I also want to extend the the sides to provide a way to raise the tank from two attachment points on each side. The sides will be 1/4 A36 steel plate and the rest of the tank will be 10 gauge (0.140625). The tank should weigh around 350 lbs and hold 119 gallons of diesel. Using maximums that's 119x7.5 lbs = 892.5 lbs, including the weight of the tank that is roughly 1250 lbs.

Assuming the holes at the tope of the side plate are 2 in diameter and the shear formula F = P x T X PSI
I was going to calculate P as 2 in x pi and then divide by the portion of the whole that would be in contact with the rigging, so assuming 1/4 of the perimeter is in contact with perimeter, = 1.57
F = 1.57 x .25 x 36,000
F = 14,130 lbs per hole, so I should be able to support over 28,000 pounds for the two holes.

This is your first mistake. A36 steel has a minimum tensile strength of 36,000 psi. In shear, the allowable stress is much lower, typically 0.4 * 36,000 = 14,400 psi, which allows vagaries of shear stress distribution, provides for a minimal factor of safety on the load, etc. Even in bending, which provides partial tensile loads, the allowable stress should be kept below 0.6 * 36,000 = 21,600 psi. Many designers use higher safety factors depending on the service, sometimes 2:1 or 3:1 even.
Does my math look correct?
If I left the sides as 10 gauge, it looks like it could still support 7900 lbs per hole for total of 15,800 lbs so I may not need 1/4 plate on the sides.

Thanks
It's not a question of getting the arithmetic correct as knowing what you are doing with the numbers.

It would be better to have 2 lifting lugs per end to provide stability for the tank while it is being lifted. One lug per end could allow the tank to flip in the slings if not carefully handled.

You also need to check that plain plate can hold all that diesel fuel when it is suspended only at the ends. Such thin gage metal may appear attractive now, but what happens down the road once a little metal is lost due to corrosion?

It's better to have someone knowledgeable, like an engineer, take a look at what you want to build while it's still on paper.
 
Thanks for the information, I did not know you had to de-rate the PSI for shear stress, so that changes the calculations. There will be a steel tube substructure under the bottom plate for the forks to fit under and support the weight on the bottom, so I think that will be strong enough. Having multiple lifting lugs on each side make sense and as you said should stabilize the load, this should also decrease the shear stress of each lug.
 
Hugh99 said:
Thanks for the information, I did not know you had to de-rate the PSI for shear stress, so that changes the calculations. There will be a steel tube substructure under the bottom plate for the forks to fit under and support the weight on the bottom, so I think that will be strong enough. Having multiple lifting lugs on each side make sense and as you said should stabilize the load, this should also decrease the shear stress of each lug.
The lugs have to be carefully designed as well. If you make a lug with a 2" diameter hole and then use a 1" shackle to lift with, the stresses go up in the lug material.

Size the holes in the lifting lugs to fit one size shackle, which is rated to take the load of lifting a tank full of diesel. Each shackle size has a particular clearance between the bolt and the hoop. Don't ignore this clearance.

You must also check the pull out stress around the hole on the lug as well. Because of the potential beating the lugs are going to get, I would not recommend that they be constructed from thin gauge material. It's better, I think, to use a heavier plate for the lugs and weld them to the tank.
 
Hugh99 said:
I'm trying to build a tank to hold diesel fuel, it will be on skids to lift with a fork lift but I also want to extend the the sides to provide a way to raise the tank from two attachment points on each side. The sides will be 1/4 A36 steel plate and the rest of the tank will be 10 gauge (0.140625). The tank should weigh around 350 lbs and hold 119 gallons of diesel. Using maximums that's 119x7.5 lbs = 892.5 lbs, including the weight of the tank that is roughly 1250 lbs.

Assuming the holes at the tope of the side plate are 2 in diameter and the shear formula F = P x T X PSI
I was going to calculate P as 2 in x pi and then divide by the portion of the whole that would be in contact with the rigging, so assuming 1/4 of the perimeter is in contact with perimeter, = 1.57
F = 1.57 x .25 x 36,000
F = 14,130 lbs per hole, so I should be able to support over 28,000 pounds for the two holes.

Does my math look correct?
If I left the sides as 10 gauge, it looks like it could still support 7900 lbs per hole for total of 15,800 lbs so I may not need 1/4 plate on the sides.

Thanks
I am a mechanical engineer and have taught rigging to a fairly large group. I strongly suggest that you lift from the top so that your lifting lugs are primarily in axial stress as close to normal (90 degrees) as possible.
 
Because of the sizes and weights involved, and the inexperience of the OP, this thread should have been closed a while ago.
 

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