Question 3.6: A circular tube with an outside diameter of 80 mm and an ins...
A circular tube with an outside diameter of 80 mm and an inside diameter of 60 mm is subjected to a torque T = 4.0 kN⋅m (Fig. 3-26). The tube is made of aluminum alloy 7075-T6.
(a) Determine the maximum shear, tensile, and compressive stresses in the tube and show these stresses on sketches of properly oriented stress elements.
(b) Determine the corresponding maximum strains in the tube and show these strains on sketches of the deformed elements.
(a) Maximum stresses. The maximum values of all three stresses (shear, tensile, and compressive) are equal numerically, although they act on different planes. Their magnitudes are found from the torsion formula:
\tau _{max} =\frac{Tr}{I_{P} }=\frac{\left(4000 N⋅m\right)\left(0.040 m\right) }{\frac{\pi }{32}\left[\left(0.080 m\right)^ {4} -\left(0.060 m \right) ^{4} \right] }=58.2 MPaThe maximum shear stresses act on cross-sectional and longitudinal planes, as shown by the stress element in Fig. 3-27a, where the x axis is parallel to the longitudinal axis of the tube.
The maximum tensile and compressive stresses are
\sigma _{t}=58.2 MPa \sigma _{c}= – 58.2 MPa
These stresses act on planes at 45° to the axis (Fig. 3-27b).
(b) Maximum strains. The maximum shear strain in the tube is obtained from Eq. (3-31). The shear modulus of elasticity is obtained from Table I-2, Appendix I (available online), as G=27 GPa. Therefore, the maximum shear strain is
The deformed element is shown by the dashed lines in Fig. 3-27c.
The magnitude of the maximum normal strains (from Eq. 3-33) is
\epsilon _{max} =\frac{\gamma }{2} (3-33)
\epsilon _{max} =\frac{\gamma _{max} }{2} = 0.0011Thus, the maximum tensile and compressive strains are
\epsilon _{t} = 0.0011 \epsilon _{c} = -0.0011
The deformed element is shown by the dashed lines in Fig. 3-27d for an element with sides of unit length.
