Hex Standoff inserted into baseplate thread failure
Hex Standoff inserted into baseplate thread failure
(OP)
looking for help on how to conduct an analysis with the following scenario:
A 0.25in thick vertical aluminum baseplate (plate a) has 4x tapped thru all #6 holes of spaced out in a rectangle pattern with dimensions of 4.5in width 3.5in height.
Each one of these tapped holes has a hex standoff inserted into the #6 tapped holes at a depth of about .2in.
The standoffs are 6-32 thread 1in long 0.25in hex width stainless steel male / female standoffs from McMaster Carr.
At the female end of these standoffs, there is a second plate (plate b) which is bolted into all four standoffs.
This plate weighs 5lb.
Plate b is secured to the standoffs using 6-32 pan head screws which insert about 0.25 into the standoff threads.
I am trying to figure out how to calculate the max weight plate b can be before the standoffs rip out of the tapped holes.
To me, it seems to be a two part analysis of the bending stress a point load at distance 1in causes and then the thread failure due to this point load.
I tried looking up beams and bending stress but most examples consider beams as a fixed end to the wall.
This lead me to trying to use max bending stress of M*c/I but it did not account for thread rip out.
I also tried looking into tensile strength and shear strength but all these consider the shear force acting on a bolt with a nut on the other end.
Tensile was only in the axial direction. Shear was only shown when two plates were clamped together by the bolt.
Thanks in advance
A 0.25in thick vertical aluminum baseplate (plate a) has 4x tapped thru all #6 holes of spaced out in a rectangle pattern with dimensions of 4.5in width 3.5in height.
Each one of these tapped holes has a hex standoff inserted into the #6 tapped holes at a depth of about .2in.
The standoffs are 6-32 thread 1in long 0.25in hex width stainless steel male / female standoffs from McMaster Carr.
At the female end of these standoffs, there is a second plate (plate b) which is bolted into all four standoffs.
This plate weighs 5lb.
Plate b is secured to the standoffs using 6-32 pan head screws which insert about 0.25 into the standoff threads.
I am trying to figure out how to calculate the max weight plate b can be before the standoffs rip out of the tapped holes.
To me, it seems to be a two part analysis of the bending stress a point load at distance 1in causes and then the thread failure due to this point load.
I tried looking up beams and bending stress but most examples consider beams as a fixed end to the wall.
This lead me to trying to use max bending stress of M*c/I but it did not account for thread rip out.
I also tried looking into tensile strength and shear strength but all these consider the shear force acting on a bolt with a nut on the other end.
Tensile was only in the axial direction. Shear was only shown when two plates were clamped together by the bolt.
Thanks in advance
Talk To Other Members
RE: Hex Standoff inserted into baseplate thread failure
The minor diameter is about 0.11 inches and the hex has a, roughly, 0.125 inch distance from center to prying edge; then you have an 8:1 mechanical advantage from the length of the standoff. The flats are 0.25 apart, but there is usually a chamfer that reduces the distance to the pivot even more so I expect an even higher ratio, 10:1 is easy to get to.
The split between resistance by pure bending on the 0.11 diameter neck and the tension load from the 8:1 prying depends on how embedment takes place at the pivot. This will change dramatically with relative hardness.
Because this interaction is difficult to determine, it's faster just to get a selection of standoffs and drill and tap the same alloy and temper of aluminum plate. I expect you will find out more about the situation than you expect.
Note that the same issue occurs at the female end of the standoffs except that, because there isn't so much continuity through the screw, it will mainly be prying over bending at the start eventually transitioning to bending. To determine the total effect, duplicate the intended use and test it.
Just an estimate, but unless this item is going to be attached to a concrete building that is not in an earthquake zone, those are very tiny screws for this amount of weight.