176 APPLIED MECHANICS 
split into two or more pieces. The ball therefore gives way finally by 
tearing. Fig. 250 shows a ball which has split into two pieces by a 
erushing — load re: . 
acting in the 
direction of the 
arrows. The two 
cones referred to 
above are seen 
adhering, one to 
one half of the 
ball and theother ‘Fria. 260. Fig. 251. 
to the other. 
Fig. 251 shows a typical fracture of a cast-iron roller tested in com- 
pression between two hard flat plates. Here wedges form by shearing, 
and these finally split the roller. In the case illustrated, the splitting | 
was probably caused by the bottom wedge. 
In actual fractures by shearing in tension and compression tests the 
inclination of the plane of fracture to the axis is never 45°, although at 
that angle the shear stress produced by the external load is greatest. 
The reason why the fracture does not take place at 45° is that the resistance 
to sliding at an oblique section is affected by the normal stress on that sec- 
tion. The theory of the effect of the normal stress on an oblique section 
in altering the inclination of the shear fracture is as follows. Re- 
ferring to Fig. 182, p. 138, and using the notation of Art. 140, the 
external load P causes a shear force p sin 6 cos @ along CD per unit area 
of CD, and also a force p cos? 6 normal to CD per unit area of CD. Ifs 
is the normal cohesive force per unit area holding together the parts on 
opposite sides of CD, then the resultant normal force on CD per unit 
area is s +p cos® 6, where the upper sign applies to a compression test, and 
the lower sign to a tension test. Assuming that the resistance to sliding 
along CD per unit area is proportional to s +p cos? 0, and that it is equal 
to (s +p cos? 6), where p= tan > is a coefficient of resistance to sliding, © 
then p sin 6 cos 0 = u(s +p cos? 4), and this reduces to 
Bi 2s cos 
P~ sin (2044) Fain 
Hence p will be a minimum when sin(20F ¢) is a maximum, that is, when 
204% p=90°, or when cot 20= 4p, where the upper sign applies to a 
compression test, and the lower sign to a tension test. It follows from 
the above that the value of # in a compression test is the complement of — 
its value in a tension test of the same material. 
167. Fracture by Tension in Torsion Tests.—When a cylindrical 
specimen is subjected to simple torsion there is a pure shear stress in the ~ 
material in planes perpendicular to the axis, and also in planes containing — 
the axis, and in Art. 142, p. 140, it was shown that at any point in the 
material there is also a pure tensile stress equal to the shear stress at 
that point, the direction of this tensile stress being at 45° to the shear 
stresses. It may therefore be expected that when a specimen of a 
material whose resistance to direct tension is less than its resistance to — 
pure shear is subjected to torsion it will give way in tension. Cast-iron 
is such a material, and the form of the fracture of a hollow cylindrical 
he oe 
