1908-9.] 
End Friction with Hard Plates. 
435 
of friction at the extremities may be considered as equivalent to a shearing 
stress having the value /xV sin 2 /3', together with a normal stress equal to 
jul'c' sin /3' cos /3'. The new tangential stress acts in the direction opposite 
to that due to the crushing load, and tends to strengthen the material. The 
new normal stress increases the pressure between the sliding surfaces, and 
therefore increases the internal frictional resistance to sliding by the 
amount jul/jl'c' sin /3' cos /S', again tending to strengthen the bar. 
The total resistance to sliding is 
fxc sin 2 (3' + /xK + /x/xV sin S' cos (3' + /xV sin 2 f3'. 
And the inclination of the surface of sliding is such that 
c sin /3' cos [3' - (/xc' sin 2 (3' 4- /x/xV sin (3' cos / 3 ' + /xV sin 2 (3') 
is a maximum. This occurs when 
tan 2/3' = = cot (</> + <£') ; 
that is, when, 
I3' = (i5° - 4,' 12. 
Since /a and K are, by assumption, constants for any particular material, 
the ratio of c to c is given by 
c sin (3 cos (3 — /x sin 2 (3 
c sin f3' cos (3' - /x sin 2 (3' - /x^' sin (3 ' cos (3' - fj! sin 2 f3’ 
|{cos (/> - /x( 1 - sin </>)} 
J[(l - /x/x r ) cos (</> + <J>') ~(fjL + /i'){l - sin (cf) + </>')}] 
cos (f>'( 1 — sin (/>) 
1 — sin (cf) + (f>') 
Or, more conveniently for purposes of calculation, 
c V f + /x 2 — /x 
c v /(l + y 2 )( 1 + /A 2 ) - (^ + /x') 
If the length of the bar is relatively less than in fig. 2, sliding takes 
place as in fig. 3 or fig. 5. The conditions for each portion are the same as 
in fig. 2, so that the inclination of the sliding surfaces should remain 
constant. If the relative length of the bar is much greater than in fig. 2, 
bending occurs, and the above results no longer hold good. 
4. Cast Iron. 
For cast iron the value of as determined from Morin’s experiments on 
sliding friction, varies from 0T5 to 0T6. The corresponding values of (j> are 
8° 32' and 9° 5'. Hence (3 varies from 40° 44' to 40° 28'. 
