Question 11.1: An aluminium shaft (σyp = 330 MPa) of circular cross-section...
An aluminium shaft \left(\sigma_{y p}=330 MPa \right) of circular cross-section with 20 mm diameter is subjected to an axial load, P = 50.0 kN. Using the octahedral shear stress criterion of failure by yielding, determine
(a) the torque T that can be applied to initiate yielding;
(b) the torque T that can be applied to the shaft, if it is designed with a factor of safety of 1.75 for both P and T against the initiation of yielding.
We first calculate the stresses produced due to P and T as follows:
\sigma_{x x}=\frac{P}{A}=\frac{50\left(10^3\right)}{\pi(20)^2 / 4} N / mm ^2=159.15 MPa
and \tau_{x y}=\frac{16 T}{\pi d^3}=\frac{16 T}{\pi(20)^3}=6.366\left(10^{-4}\right) T MPa
where T is in Nmm.
(a) Failure theory to be followed is octahedral shear stress, using which we can write by considering \sigma_{y y}=\sigma_{z z}=\tau_{y z}=\tau_{z x}=0 from Eq. (11.44) and using the values of I_1 \text { and } I_2 :
\sqrt{I_1^2-3 I_2}=\sigma_{ yp } (11.44)
\sigma_{x x}^2+3 \tau_{x y}^2=\sigma_{ yp }^2 (1)
Therefore,
159.15^2+3 \tau_{x y}^2=330^2
or \tau_{x y}=166.9 MPa =6.366\left(10^{-4}\right) T
Thus,
T = 262.17 Nm
The torque that can be applied to the shaft without causing yielding is: 262.17 Nm
(b) If same factor of safety is used, then Eq. (1) becomes
\sigma_{x x}^2+3 \tau_{x y}^2=\left\lgroup \frac{\sigma_{ yp }}{\text { factor of safety }} \right\rgroup^2
\Rightarrow 159.15^2+3 \tau_{x y}^2=\left\lgroup \frac{330}{1.75} \right\rgroup^2
\Rightarrow \tau_{x y}=58.4 MPa =6.366\left(10^{-4}\right) T
⇒ T = 91737.4 Nmm = 91.74 Nm
Hence, the applied torque is 91.74 Nm