Impact force on shear pin
Impact force on shear pin
(OP)
I am working on a shear pin design in a sawmill application. A sorter mechanism that holds lumber travels vertically up and down an HSS column. The sorter cradle and full load of lumber could fall a distance of approximately 12" if the hydraulics fail. A steel pin through the HSS column stops the fall of the cradle (see attached sketch). Using Blodgett's formula (Section 2.8-3, paragraph 5), I get a huge spring constant (K) as the shear deformation/deflection of the pin + column shortening are infintesimal (0.0128"). This produces an astronomical force of 1065 kips based on 12" fall & 24 kip body weight. Half of that goes to each column (532.5 k) and is way in excess of the column capacity. What am I missing? Is there a better way to determine the impact force for this application? Thanks.






RE: Impact force on shear pin
Obviously - F=ma and a = delta distance/ delta time
Trying to find the acceleration (or de-acceleration) is the key. To determine this you need to know how fast/slow and how far it takes to stop the load. It is not instantaneous - as we all know. Because then acceleration would be infinite. That is why it is easier to fall on soft soil versus concrete. That is why people often use springs in these apps. It allows the energy (force) to dissipate over a longer distance.
Determining "a" or "-a" is almost impossible without a bit of testing. If it was easy, Ford, Chevy, BMW, etc wouldn't spend so much money wrecking cars every year....
RE: Impact force on shear pin
An accurate yet relatively easy experiment like something on mythbusters may solve the problem, huh? See if you can talk the mill owner into that! :)
RE: Impact force on shear pin
You're confusing shear pins with solid stops, and you have essentially a solid stop. The solid stop leads to small time durations to stop and small deflections or deformations which lead to very high impact loads or stresses. Shear pins intentionally shear, absorbing some of the energy in their shearing failure, over some distance and time in the process, and thus the forces are much smaller to stop the fall.
I would have to know much more about how this system is structured and operates, but a few thoughts...
1. Put pressure actuated check valves right at the lifting cylinders, so that if you break a hydraulic hose, the cylinders are held right there.
2. Lower the 3" solid stop pin so you can put in a couple smaller shear pins above the solid stop, and let them act as the operating stops for the cradle guides, or as pins that really do fail in shear if the whole sorter falls. They are replaceable fail-safe parts.
3. I assume the inner plates on the lifting guide are wear plates against the HSS column? Under the outer plates (castings?) with the lifting and load carrying bosses, just outside of your downward force arrows, install vert. plates about the same thickness as the plates with the bosses. Cut an inverted 'Vee' notch in these plates a couple inches high, which will engage a shear pin (maybe your 3" stop pin) shortly above the bottom edge of this new plate. Then if a failure occurs your system will rip these new plates at the tip of the 'Vee,' thus absorbing energy in the process, over some time and over some travel, and decreasing the magnitude of the forces involved. Again, these new plates are fail-safe replaceable parts.
4. Does your 3" pin weaken the HSS column? It will be driven down in the column wall in a failure and cause buckling and bearing failure of the column walls. You said... 533k/{(2x3")(.25" wall thick.)} = 355ksi bearing stress. That's a pretty substantial bearing stress. Whatever you do, you don't want to cause a column failure or you may have an even bigger failure on your hands. Some system which absorbs energy and lessens the magnitudes of the forces is called for to prevent this.
RE: Impact force on shear pin
RE: Impact force on shear pin
RE: Impact force on shear pin
In fact there are some helically wound springs made out of a tapered piece of spring steel bar stock of some length. The bar is wound up/twisted, in height, helically and around itself to make four or five wound layers in the diameter direction. Then the t&b bearing surfaces are ground flat. When the spring is compressed in height the wedge shaped (tapered) bars work against each other frictionally, and this frictional action increases with reduced height because the diametrical layers must expand in dia. w.r.t. each other. These are very effective shock absorbers w.r.t. heavy loads and fairly short travel. There are also hydraulic snubbers which I believe would take these kinds of loads. These convert hydraulic fluid flow to heat to absorb energy, and the fluid flow orifice decreases in size with piston travel to require more force per inch of travel, prior to hitting an internal solid stop. I would have to do some digging to find literature and cap'ys. on these.
You could also make a heavy compression spring, maybe with a 6.5" I.D. to fit over/around the HSS column, sit on a solid stop; and gag it down so that it would take some predetermined load (half the column cap'y.?) without deflecting downward; but then travel and pick up more load to dampen the impact, but not overload the column. Therein lies the spring design problem for this spring. Maybe two off-the -shelf heavy compression springs, one under each boss could be made to work in a similar fashion. But, I gotta know more about his entire system to comment further.
RE: Impact force on shear pin