Question 8.4: The rotor shaft of a helicopter drives the rotor blades that...
The rotor shaft of a helicopter drives the rotor blades that provide the lifting force to support the helicopter in the air (Fig. 8-24a). As a consequence, the shaft is subjected to a combination of torsion and axial loading (Fig. 8-24b).
For a 50-mm diameter shaft transmitting a torque T = 2.4 kN·m and a tensile force P = 125 kN, determine the maximum tensile stress, maximum compressive stress, and maximum shear stress in the shaft.
The stresses in the rotor shaft are produced by the combined action of the axial force P and the torque T (Fig. 8-24b). Therefore, the stresses at any point on the surface of the shaft consist of a tensile stress \sigma_{0} and shear stresses τ_{0}, as shown on the stress element of Fig. 8-24c. Note that the y axis is parallel to the longitudinal axis of the shaft.
The tensile stress \sigma_{0} equals the axial force divided by the cross-sectional area:
The shear stress \tau_{0} is obtained from the torsion formula (see Eqs. 3-11 and 3-12 of Section 3.3):
\tau_{\max}=\frac{Tr}{I_{P}} (3-11)
\tau_{\max}=\frac{16T}{\pi d^{3}} (3-12)
\tau_{0}=\frac{Tr}{I_{P}}=\frac{16T}{\pi d^{3}}=\frac{16(2.4_\ kN\cdot m)}{\pi(50_\ mm)^{3}}=97.78_\ MPaThe stresses \sigma_{0} and \tau_{0} act directly on cross sections of the shaft.
Knowing the stresses \sigma_{0} and \tau_{0}, we can now obtain the principal stresses and maximum shear stresses by the methods described in Section 7.3. The principal stresses are obtained from Eq. (7-17):
\sigma_{1,2}=\frac{\sigma_{x}+\sigma_{y}}{2}\pm \sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right)^{2}+\tau^{2}_{xy}} (d)
Substituting \sigma_{x}=0, \sigma_{y}=\sigma_{0} = 63.66 MPa, and \tau_{xy}=-\tau_{0} = -97.78 MPa, we get
\sigma_{1,2} = 32 MPa ± 103 MPa
or
\sigma_{1} = 135 MPa \sigma_{2} = -71 MPa
These are the maximum tensile and compressive stresses in the rotor shaft.
The maximum in-plane shear stresses (Eq. 7-25) are
\tau_{\max}=\sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right)^{2}+\tau^{2}_{xy}} (e)
This term was evaluated previously, so we see immediately that
\tau_{\max} = 103 MPa
Because the principal stresses \sigma_{1} and \sigma_{2} have opposite signs, the maximum in-plane shear stresses are larger than the maximum out-of-plane shear stresses (see Eqs. 7-28a, b, and c and the accompanying discussion). Therefore, the maximum shear stress in the shaft is 103 MPa.
(\tau_{\max})_{x_{1}}=\pm \frac{\sigma_{2}}{2} (\tau_{\max})_{y_{1}}=\pm \frac{\sigma_{1}}{2} (\tau_{\max})_{z_{1}}=\pm \frac{\sigma_{1}-\sigma_{2}}{2} (7-28a,b,c)
