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. 3-30). 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.
(c) What is the maximum permissible torque T_{\max } if the allowable normal strain is \varepsilon_{a} = 0.9 × 10^{-3} ?
(d) If T = 4.0 kN.m and \varepsilon_{a} = 0.9 × 10^{-3}, what new outer diameter is required so that the tube can carry the required torque T (assume that the inner diameter of the tube remains at 60 mm)?
(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{T r}{l_{p}}=\frac{(4000 \mathrm{~N} \cdot \mathrm{m})(0.040 \mathrm{~m})}{\frac{\pi}{32}\left[(0.080 \mathrm{~m})^{4}-(0.060 \mathrm{~m})^{4}\right]}=58.2 \mathrm{~MPa}The maximum shear stresses act on cross-sectional and longitudinal planes, as shown by the stress element in Fig. 3-31a, where the x axis is parallel to the longitudinal axis of the tube.
The maximum tensile and compressive stresses are
These stresses act on planes at 45° to the axis (Fig. 3-31b).
(b) Maximum strains. The maximum shear strain in the tube is obtained from Eq. (3-30) \gamma=\frac{\tau}{G}. The shear modulus of elasticity is obtained from Table H-2, Appendix H, as G = 27 GPa. Therefore, the maximum shear strain is
Table H-2
Moduli of Elasticity and Poisson’s Ratios
| Material | Modulus of elasticity E | Shear modulus of elasticity G | Poisson’s ratio ν |
| Gpa | Gpa | ||
| Aluminum alloys 2014-T6 6061-T6 7075-T6 |
70-79 73 70 72 |
26-30 28 26 27 |
0.33 0.33 0.33 0.33 |
| Brass | 96–110 | 36–41 | 0.34 |
| Bronze | 96–120 | 36–44 | 0.34 |
| Cast iron | 83–170 | 32–69 | 0.2–0.3 |
| Concrete (compression) | 17–31 | 0.1–0.2 | |
| Copper and copper alloys | 110–120 | 40–47 | 0.33–0.36 |
| Glass | 48–83 | 19–35 | 0.17–0.27 |
| Magnesium alloys | 41–45 | 15–17 | 0.35 |
| Monel (67% Ni, 30% Cu) | 170 | 66 | 0.32 |
| Nickel | 210 | 80 | 0.31 |
| Plastics Nylon olyethylene |
2.1-3.4 0.7-1.4 |
0.4 0.4 |
|
| Rock (compression) Granite, marble, quartz Limestone, sandstone |
40-100 20-70 |
0.2-0.3 0.2-0.3 |
|
| Rubber | 0.0007–0.004 | 0.0002–0.001 | 0.45–0.50 |
| Steel | 190–210 | 75–80 | 0.27–0.30 |
| Titanium alloys | 100–120 | 39–44 | 0.33 |
| Tungsten | 340–380 | 140–160 | 0.2 |
| Wood (air dry) Douglas fir Oak Southern pine |
11-13 11-12 11-14 |
The deformed element is shown by the dashed lines in Fig. 3-28c.
The magnitude of the maximum normal strains [from Eq. (3-33) L_{b d}=\sqrt{2} h\left(1+\varepsilon_{\text {max }}\right) ] is
\varepsilon_{\max }=\frac{\gamma_{\max }}{2}=0.0011Thus, the maximum tensile and compressive strains are
\varepsilon_{t}=0.0011 \quad \varepsilon_{c}=-0.0011The deformed element is shown by the dashed lines in Fig. 3-31d for an element with sides of unit length.
(c) Maximum permissible torque. The tube is in pure shear, so the allowable shear strain is twice the allowable normal strain [see Eq. (3-32) \varepsilon_{\max }=\frac{\gamma}{2} ]:
\gamma_{a}=2 \varepsilon_{a}=2\left(0.9 \times 10^{-3}\right)=1.8 \times 10^{-3}From the shear formula [Eq. (3-13) \tau_{\max }=\frac{T r}{I_{P}}], we get
\tau_{\max }=\frac{T\left(\frac{d_{2}}{2}\right)}{I_{p}} \quad \text { so } \quad T_{\max }=\frac{\tau_{a} I_{p}}{\left(\frac{d_{2}}{2}\right)}=\frac{2\left(G \gamma_{a}\right) I_{p}}{d_{2}}where d_{2} is the outer diameter. Substituting numerical values gives
(d) New outer diameter of tube. We can use the previous equation but with T = 4.0 kN . m to find the required outer diameter d_{2}:
\frac{I_{p}}{d_{2}}=\frac{T}{2 G \gamma_{a}} \text { or } \frac{d_{2}^{4}-(0.06 \mathrm{~m})^{4}}{d_{2}}=\frac{\left(\frac{32}{\pi}\right) 4 \mathrm{~kN} \cdot \mathrm{m}}{2(27 \mathrm{~GPa})\left(1.8 \times 10^{-6}\right)}=0.41917 \mathrm{~m}^{3}Solving for the required outer diameter d_{2} numerically gives
d_{2} = 83.2 mm
