- #1
Turbine
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Hi guys,
Working on a 50-ton hydraulic press design using a standard H-frame design, like this:
I'm working on the design of the vertical members. Similar to the photo above, the verticals have 1.25" diameter holes drilled for pins that support the horizontal bed. The problem: while determining the size of steel to use, I'm finding that the hole stresses require a much thicker steel than what I see being used in commercial machines.
Specifically, in order to support the maximum 50-ton load, that equates to 25,000 lbf per vertical support. If I use two 1.25" pins per side (4 total) to support the horizontal bed, that works out to 12,500 lbf of force per hole.
I've modeled a 1"x4" A36 flat bar at some length with 1.25" dia. holes space 5" apart. The base is fixed, there is a bearing force of 12,500 lbf per hole (so 25,000 lbf total), and a reaction force of 25,000 lbf along the top of the bar. Running SolidWorks Simulation, I get the following:
The max stress is in the top hole, 27,600 psi. While this is OK, if I use, say, 1/2" plate, it yields. Commercial units I've seen use 1/2" or less. Many only use a single pin per side. I just don't get it. Am I doing something wrong here? Are the manufactures not taking into account the hole stresses? Are they not as important as I'm suspecting?
Any feedback would be greatly appreciated. Thank you!
Working on a 50-ton hydraulic press design using a standard H-frame design, like this:
I'm working on the design of the vertical members. Similar to the photo above, the verticals have 1.25" diameter holes drilled for pins that support the horizontal bed. The problem: while determining the size of steel to use, I'm finding that the hole stresses require a much thicker steel than what I see being used in commercial machines.
Specifically, in order to support the maximum 50-ton load, that equates to 25,000 lbf per vertical support. If I use two 1.25" pins per side (4 total) to support the horizontal bed, that works out to 12,500 lbf of force per hole.
I've modeled a 1"x4" A36 flat bar at some length with 1.25" dia. holes space 5" apart. The base is fixed, there is a bearing force of 12,500 lbf per hole (so 25,000 lbf total), and a reaction force of 25,000 lbf along the top of the bar. Running SolidWorks Simulation, I get the following:
The max stress is in the top hole, 27,600 psi. While this is OK, if I use, say, 1/2" plate, it yields. Commercial units I've seen use 1/2" or less. Many only use a single pin per side. I just don't get it. Am I doing something wrong here? Are the manufactures not taking into account the hole stresses? Are they not as important as I'm suspecting?
Any feedback would be greatly appreciated. Thank you!