# Failure in torsion

Why does a ductile rod in torsion fail at an angle perpendicular to its axis? I know the reason for brittle material failing in pure shear but I don't understand the reason for ductile material.

• materials are rarely 100% ductile Aug 16 '17 at 15:26
• ok but assuming it is.Can you give me the theoritical answer? Aug 16 '17 at 15:59
• Do a Mohr's circle and note the direction of the principal stresses. They are 45° from the axis of the rod (90° on Mohr's circle). Aug 16 '17 at 19:43

As a rule of thumb: When brittle materials are subjected to torsion they fail in the plane, where tension is at its highest, i.e. at a 45° angle. Ductile materials on the other hand fail in the plane of maximum shear stress.

Take a look at Mohr's Circle for pure shear. Maximum-Shear-Stress Theory states:

[...] yielding of the [ductile] material begins when the absolute maximum shear stress in the material reaches the shear stress that causes the same material to yield when it is subjected only to axial tension.
R.C. Hibbeler, Mechanics of Materials, p.525

for ductile materials: $$\tau_{max}=\frac{\sigma_y}{2}$$ As you can see in the graphic, $\tau_{max}$ is reached before $\sigma_y$, thus the plane of failure is perpendicular to the primary axis.

For brittle materials on the other hand, Maximum-Normal-Stress Theory says:

[...] a brittle material will fail when the maximum tensile stress, $\sigma_1$, in the material reaches a value that is equal to the ultimate normal stress the material can sustain […]
R.C. Hibbeler, Mechanics of Materials, p.528

for brittle materials: $$\tau_f=\sigma_f$$ So, under pure shear it fails in tension at a 45° angle.

The stresses of a rod in shear and in torsion are of the same kind. So they fail of the same reason.

There is a conversion factor for materials. It is sqrt(3). So with the same stress induced to a rod under torsion compared to one under tensile load is more likely to break by this factor.