talonturbo said:
I'm a final year Mechanical Engineering student.
And I earned my mechanical engineering degree twenty-two years ago. Trust me, your education is just beginning.
When you get out in the real world, you'll discover that that things are always more complicated than they look now: tolerances stack up in ways you didn't anticipate; parts deform and wear, changing the geometry you though you analyzed; and of course, users will do things you didn't plan for. On the other hand, you'll also learn what to safely ignore or approximate, so you can put your work into what's important.
Anyhow, enough of this "When I was a lad..." drivel.
It's not at all difficult to put a couple hundred inch-pounds of torque on the pivot of a hard-use knife like a Strider. On the Strider STR shows, the point where the tang and lock meet is about a half inch from the pivot axis; that 200 in-lb torque translates to 400 pounds at the point. Now, what's the area that force is applied to? Let's take a guess, based on STR's observations: about 3mm = 0.12" high, by perhaps 1mm = 0.04" wide, for a total area of 0.005 sq in, for an overall pressure of 80 kpsi.
Now, here's the fun part: since the interface is angled, both surfaces will deform, but there will be a peak load near one of the outside corners; my semi-professional guess would be about half the load over probably a quarter of the area: 200 lb over .00125 sq in, or 160 kpsi. On brand-new sharp-edged hardened parts, it could be even higher.
If memory serves, 6Al4V titanium has a yield strength of 130kpsi, and a tensile strength of 145kpsi or so. Still think that titanium lock can't shear?
So, for the record, I'm four-square behind STR here.
When a properly-fitted lock engages, it'll "wipe" any crud from the locking surfaces, so Strider's explanation sounds to me to be a cop-out. All I can think of is, he wants to maximize the pressure at the interface so that titanium's galling properties will make the lock less likely to slip. But that's just a guess...
As for the question of why you want a wide lock bar (other than appearances), note that the contact point is about a half inch below the pivot,
and about a quarter inch
behind it, so there's a significant upward component to the load. A thin lockbar like STR drew would be at risk to fail by bending upward!
Having spent a lot of time on a factory floor, I'm inclined to believe STR's theory of fitted stop pins; shop tolerances being what they are, there has to be
some way to adjust the fit, and the stop pin is the most obvious way to do so.