Any mechanical or structural engineers here?
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Cargo Master
Joined: Jul 2008
Posts: 2,586
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Any mechanical or structural engineers here?
I am trying to find the right mix of stress equations for loading a bolted or pinned assembly in-plane (shear force on the bolts/pins).
I have the plate stress area down perpendicular to the load (end surface area minus the diameter of the hole). However, I'm stuck on the stress of the parallel force centered on the hole....
My question is I am trying to find the load capacity of the metal around a hole. I don't care about the bolts or pins (way more shear capacity than the application).
If I have my variables right - Proof Load is the same as Yield Force (maximum force before deformation), and Ultimate Stress and Tensile Stress are the same (point at which the metal fails).
That having been said, the variables I have are:
Steel Yield Force/Proof Load = 22,000PSI
Steel Ultimate Stress/Tensile Strength = 40,000PSI
Plate thickness is 3/8"
Hole diameter is 7/16"
Plate Width is 4" - hole is centered in the plate (2-7/32" either side of the hole)
What I want to do is get the thickness of metal between the hole and edge of the plate (in-line with the force) to withstand 7000lbs+ force at breaking, or around 4000lbs+ force yield. How far in do I drill the hole?
Any help on calculating this??? Thanks!
I have the plate stress area down perpendicular to the load (end surface area minus the diameter of the hole). However, I'm stuck on the stress of the parallel force centered on the hole....
My question is I am trying to find the load capacity of the metal around a hole. I don't care about the bolts or pins (way more shear capacity than the application).
If I have my variables right - Proof Load is the same as Yield Force (maximum force before deformation), and Ultimate Stress and Tensile Stress are the same (point at which the metal fails).
That having been said, the variables I have are:
Steel Yield Force/Proof Load = 22,000PSI
Steel Ultimate Stress/Tensile Strength = 40,000PSI
Plate thickness is 3/8"
Hole diameter is 7/16"
Plate Width is 4" - hole is centered in the plate (2-7/32" either side of the hole)
What I want to do is get the thickness of metal between the hole and edge of the plate (in-line with the force) to withstand 7000lbs+ force at breaking, or around 4000lbs+ force yield. How far in do I drill the hole?
Any help on calculating this??? Thanks!
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Cargo Master
Joined: Jul 2008
Posts: 2,586
Likes: 62
More thoughts....
The area of the sides of the holes, and stress they can tolerate, is:
4" (plate width) -7/16" (hole diameter) = 4-.4375 = 3.5625"
(3.5625)(.375 plate thickness) = 1.3359375in.sq = tensile stress area
22,000lbs x 1.3359375in.sq = 29,390.625lbs = yield strength
in.sq
40,000lbs x 1.3359375in.sq = 53,437.5lbs = tensile strength
in.sq
According to those numbers, the only limiting factor is how much force it would take to blow out the back side of the hole (inline with the force). The sides are more than adequate.
I can't calculate the shear area parallel to the hole at the width of the hole as a valid figure because that force is spread out radially, not a straight line, from the pin in the hole. The edge of the hole (holding back the load) would mostly be in compression, also, not shear or tension (though the area near the full diameter on the bottom through the top/opposite end as the force would be in tension, if any).
The area of the sides of the holes, and stress they can tolerate, is:
4" (plate width) -7/16" (hole diameter) = 4-.4375 = 3.5625"
(3.5625)(.375 plate thickness) = 1.3359375in.sq = tensile stress area
22,000lbs x 1.3359375in.sq = 29,390.625lbs = yield strength
in.sq
40,000lbs x 1.3359375in.sq = 53,437.5lbs = tensile strength
in.sq
According to those numbers, the only limiting factor is how much force it would take to blow out the back side of the hole (inline with the force). The sides are more than adequate.
I can't calculate the shear area parallel to the hole at the width of the hole as a valid figure because that force is spread out radially, not a straight line, from the pin in the hole. The edge of the hole (holding back the load) would mostly be in compression, also, not shear or tension (though the area near the full diameter on the bottom through the top/opposite end as the force would be in tension, if any).
Thread Starter
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Cargo Master
Joined: Jul 2008
Posts: 2,586
Likes: 62
Thats a valid thought - make the metal as thick as I can.
I test-fit the clevis I am thinking of using ( 3/8" grade 70) and a 5/16" safety hook. The clearance I have left is 1/2" between the hook and pin in the clevis. I don't think 1/2" of metal there is going to do the trick. I may have to get some other kinds of attaching devices - a clevis that is longer, for example. Or I can stack them so there is more room for the hook.
I test-fit the clevis I am thinking of using ( 3/8" grade 70) and a 5/16" safety hook. The clearance I have left is 1/2" between the hook and pin in the clevis. I don't think 1/2" of metal there is going to do the trick. I may have to get some other kinds of attaching devices - a clevis that is longer, for example. Or I can stack them so there is more room for the hook.
Thread Starter
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Cargo Master
Joined: Jul 2008
Posts: 2,586
Likes: 62
I did a few more calculations on the bearing capacity of the steel in the hole, based on the 22000lb proof load/yield strength (force before metal deformation).
The hole surface area for the bearing load is the diameter times the thickness = .375 x .4375 = .164063. Then you take that number and multiply it by the proof load/yield strength and get 3609.375lbs where deformation starts. That is a bit short of my 4000lb number, but then again upping the pressure capacity of the bearing member to the tensile strength/ultimate load of the steel at 40,000PSI the load goes to a little over 6500lbs before failure. That still gives me some margin.
What those numbers still don't answer is how much force it would take to blow out the thin metal between the hole and the edge of the plate. Those numbers just say that a block of steel with the given surface area has roughly those numbers associated with the load and failure points.
Let me see if this shows up OK....
The hole surface area for the bearing load is the diameter times the thickness = .375 x .4375 = .164063. Then you take that number and multiply it by the proof load/yield strength and get 3609.375lbs where deformation starts. That is a bit short of my 4000lb number, but then again upping the pressure capacity of the bearing member to the tensile strength/ultimate load of the steel at 40,000PSI the load goes to a little over 6500lbs before failure. That still gives me some margin.
What those numbers still don't answer is how much force it would take to blow out the thin metal between the hole and the edge of the plate. Those numbers just say that a block of steel with the given surface area has roughly those numbers associated with the load and failure points.
Let me see if this shows up OK....
Fleet Owner
Joined: Aug 2010
Posts: 21,255
Likes: 1,657
From: Reynoldsburg, Ohio
I was being facetious so anything I posted that's useful was pure accident! 
What are you building?
What are you building?
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Cargo Master
Joined: Apr 2003
Posts: 3,476
Likes: 224
From: Frequently frozen MN
You might get some help here. I don't know that, but they seem to be discussing a lot of complex issues.
Mechanical Engineering Forum
Also wonder if some of the places that make things like step bumpers don't have some standard load rules. Maybe a smaller place would even tell you what they are
Good Luck,
hj
Mechanical Engineering Forum
Also wonder if some of the places that make things like step bumpers don't have some standard load rules. Maybe a smaller place would even tell you what they are
Good Luck,
hj
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Thread Starter
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Cargo Master
Joined: Jul 2008
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Well I ended up changing things up a bit.
I found a 5/8" shackle and measured that. I wasn't getting a whole lot more of a gap between the pin and anything attached to it for the size increase, the only real benefit would be a larger diameter = higher bearing area = the hole supporting the pin/shackle/load could have a larger "footprint".
So to boost the amount of metal between the hole and the edge of the plate I swapped the shackle idea for a loop of cable and a bolt. I put 1-1/2" of steel between the hole and edge of the plate with a 5/8" grade 8 bolt. I figured up that a 1/2" hole and bolt would give me enough on the bearing pressure on the hole, so I bumped it up to the 5/8" for some head room.
I still don't know what the load capacity is of the metal between the hole and the bolt/pin, but I am a full inch thicker now than I would have been with the 3/8" shackle from before. That should be enough.
I found a 5/8" shackle and measured that. I wasn't getting a whole lot more of a gap between the pin and anything attached to it for the size increase, the only real benefit would be a larger diameter = higher bearing area = the hole supporting the pin/shackle/load could have a larger "footprint".
So to boost the amount of metal between the hole and the edge of the plate I swapped the shackle idea for a loop of cable and a bolt. I put 1-1/2" of steel between the hole and edge of the plate with a 5/8" grade 8 bolt. I figured up that a 1/2" hole and bolt would give me enough on the bearing pressure on the hole, so I bumped it up to the 5/8" for some head room.
I still don't know what the load capacity is of the metal between the hole and the bolt/pin, but I am a full inch thicker now than I would have been with the 3/8" shackle from before. That should be enough.
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