Torsion

132 CHAPTER 5. TORSION
Next consider the case of a material under pure plane shear, i.e. all stress components vanish except for
τ12. The equivalent stress become σeq = √ 3 τ12, and the strength criterion τ12 ≤ σallow/ √ 3. This important
result implies that the allowable shear stress is related to the allowable axial stress as
τallow = σallow
√ . (5.42)
3
This result is found to be in excellent agreement with experimental measurements, providing an experimental
verification of Von Mises strength criterion. The following examples describe various applications of this
criterion.
Pressure vessel
In section 2.6, a pressure vessel subjected to internal pressure was shown develop both hoop stresses σh =
pR/e and axial stresses σa = σh/2. Since all other stress components vanish the equivalent stress become
σeq = [σ 2 h + σ2 a − σhσa] 1/2 . Von Mises criterion now implies
√ 3
2
pR
e ≤ σallow, or p ≤ 2
√ 3
eσallow
R
eσallow
≈ 1.15 . (5.43)
R
It is interesting to note that if the strength criterion was erroneously applied by taking into account the sole
hoop stress (the maximum stress component) the safe service internal pressure would be p ≤ eσallow/R, a
more stringent condition.
Propeller shaft
Consider an aircraft propeller connected to a homogeneous, circular shaft of radius R. The engine applies a
torque M1 to the shaft and the propeller exerts a thrust N1. The corresponding stresses are
respectively. Von Mises criterion then requires
τ = 2M1
N1
, and σ = , (5.44)
πR3 πR2
( N1
πR2 )2 + 3( 2M1
1/2 )2 ≤ σallow
πR3 (5.45)
for safe service load conditions. It is convenient to rewrite the criterion in a non dimensional form as
N1
2
+ 12
M1
2 ≤ 1. (5.46)
πR 2 σallow
πR 3 σallow
Fig. 5.8 shows the geometric interpretation of the criterion: safe loads correspond to combinations inside an
ellipse in the non-dimensional load space, non-dimensional axial force N1/(πR 2 σallow) versus non-dimensional
torque M1/(πR 3 σallow).
Shaft under torsion and bending
Consider a circular shaft subjected to both bending and torsion, as would occur, for instance, in a cantilever
shaft with a tip pulley. Let M3 and M1 be the applied bending moment and torque, respectively. The
corresponding stresses are
σ = 2M3
, and
πR3 respectively. Von Mises criterion then requires
2M1
τ = ,
πR3 (5.47)
( 2M3
πR2 )2 + 3( 2M1
1/2 )2 ≤ σallow
πR3 (5.48)