Calculate shear strength A36 plate

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SUMMARY

The forum discussion focuses on calculating the shear strength of an A36 steel plate for a diesel fuel tank design. The user initially calculated the shear force using the formula F = P x T x PSI, arriving at a potential load capacity of 28,000 lbs for two holes. However, experts clarified that the allowable shear stress for A36 steel is typically 14,400 psi, necessitating a reevaluation of the calculations. Recommendations included using multiple lifting lugs for stability and ensuring proper sizing of holes to accommodate shackles, emphasizing the importance of consulting a knowledgeable engineer for safety and design integrity.

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  • A36 steel properties and specifications
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This discussion is beneficial for mechanical engineers, structural designers, and anyone involved in the design and construction of load-bearing structures, particularly in the context of lifting and rigging applications.

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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