379 
1907-8.] On the Cohesion of Steel, and Yield Points. 
they represent the actual behaviour in its complete aspect. They give, 
however, a fair working hypothesis. 
Let K be the cohesion of the metal, expressed in the same units as the 
stresses. 
Then /xK is the additional frictional resistance due to cohesion. If c is 
the axial stress corresponding with the yield point of the material in com- 
pression, then at this load we have 
c/2 . cos <p = /x . c/2 . (1 - sin <p) + /x . K. 
Or 
W 
II 
| COS <p — fX . (1 
- sin p) j> 
Putting: sin </> = 
IX 
and cos <p — 
1 
O r 
n/1 + ’ 
Jl+f* 2 ’ 
the expression becomes 
FI 
II 
■Th 
)• 
In the case of a tensile load which gives rise to an axial stress, t , — 
The normal stress on a surface of sliding is t .sin 2 (45° + ^/2) 
— t/ 2 . (1 + sin (p ). 
The decrease in frictional resistance along this surface due to the 
normal stress is fi.t/ 2.(1 + sin <p). 
The tangential stress along the surface is t . sin (45° + (p/2 ) . cos (45° + (p/2) 
= t/2 . cos (p . 
But in this case the normal stress acts in the direction opposite to the 
cohesion ; the latter force must be the greater if the metal is unbroken. 
Hence, if the yield point in tension corresponds with the axial stress t, we 
have 
t/2 . cos <p = [x . K - fx . t/2 . (1 + sin <p). 
Or 
For steel, taking as before 0 = 10°, yU = 0T76, we obtain 
K = 2-384 c = 3*384 t. 
The ratio t/c is obviously the same as that found in paragraph 2. 
If it be the case that fracture of a metal bar cannot occur until the 
normal tensile stress on a plane cross section becomes equal to the cohesion, 
we can see, either by finding the value of a which makes K = £ in the 
tension equation, or by noting that the surface of least total cohesion is 
the surface of greatest normal stress, that the bar must begin to break 
