Question 1.6: An antenna and receiver are suspended on a steel wire from a...
An antenna and receiver are suspended on a steel wire from a helicopter to measure the effects of wind turbines on a local radar installation (see Fig.1-25). Obtain a formula for the maximum stress in the wire, taking into account the weight of the wire itself. Calculate the maximum stress in the wire in MPa using the following numerical properties: L_{1} = 6 m , L_{2} = 5 m, d = 9.5 mm; antenna weight is W_{1} = 380 N; receiver weight is W_{2} = 700 N. Note that the weight density \gamma of steel is 77.0 kN/m³ (from TableI-1 in Appendix I).
| Table I-1 | ||||
| Weights and Mass Densities | ||||
| Material | Weight density γ | Mass density p | ||
| lb/ft³ | kN/m³ | slugs/ft³ | kg/m³ | |
| Aluminum alloys 2014-T6, 7075-T6 6061-T6 |
160–180 175 170 |
26–28 28 26 |
5.2–5.4 5.4 5.2 |
2600–2800 2800 2700 |
| Brass | 520–540 | 82–85 | 16–17 | 8400–8600 |
| Bronze | 510–550 | 80-86 | 16-17 | 8200–8800 |
| Cast iron | 435–460 | 68–72 | 13–14 | 7000–7400 |
| Concrete
Plain |
145 |
23 |
4.5 |
2300 |
| Copper | 556 | 87 | 17 | 8900 |
| Glass | 150-180 | 24-28 | 4.7–5.4 | 2400–2800 |
| Magnesium alloys | 110-114 | 17-18 | 3.4-3.5 | 1760–1830 |
| Monel (67% Ni, 30% Cu) | 550 | 87 | 17 | 8800 |
| Nickel | 550 | 87 | 17 | 8800 |
| Plastics
Nylon |
55–70 |
8.6–11 |
1.7–2.2 |
880–1100 |
| Rock
Granite, marble, quartz |
165–180 |
26–28 |
5.1–5.6 |
2600–2900 |
| Rubber | 60–80 | 9-13 | 1.9-2.5 | 960–1300 |
| Sand, soil, gravel | 75–135 | 12-21 | 2.3-4.2 | 1200–2200 |
| Steel | 490 | 77.0 | 15.2 | 7850 |
| Titanium | 280 | 44 | 8.7 | 4500 |
| Tungsten | 1200 | 190 | 37 | 1900 |
| Water, fresh sea |
62.4 63.8 |
9.81 10.0 |
1.94 1.98 |
1000 1020 |
| Wood (air dry)
Douglas fir |
30–35 |
4.7–5.5 |
0.9–1.1 |
480–560 |
Use the following four-step problem-solving approach.
1. Conceptualize : A free-body diagram of the suspended instrument package is shown in Fig.1-26. The antenna (W_{1} ) and receiver (W_{2} ) weights are concentrated forces at specified locations along the wire; the weight of the wire is a uniformly distributed axial force expressed as w(x)=\gamma A, where A is the cross-sectional area of the wire. Cutting the wire at some point x leads to upper and lower free-body diagrams (Fig.1-27); either can be used to find the internal axial force N(x) at the location of the cut section. The internal axial force in the wire is a maximum at the point at which it is attached to the helicopter (x = 0).
2. Categorize : Start by solving for the reaction force R at the top of the wire and then cut the wire a short distance below the support to find N_{max}. The wire is prismatic, so simply divide N_{max} by cross-sectional area A to find the maximum axial normal stress \sigma_{\max }.
3. Analyze :
Reaction force R: Use the free-body diagram in Fig.1-26 to obtain
The minus sign indicates that reaction force R acts in the (-x) direction, or upward in Figs.1-26 and 1-27.
Internal axial forces N(x) in hanging wire: The internal axial force in the wire varies over the length of the wire. Cutting through the wire in upper and lower segments (the lower segment is cut in Fig.1-27) gives
N(x)=W_{1}+W_{2}+w\left(L_{1}+L_{2}-x\right) \quad 0 \leq x \leq L_{1}N(x)=W_{2}+w\left(L_{1}+L_{2}-x\right) \quad L_{1} \leq x \leq L_{1}+L_{2}
Internal force N(x) is shown as a pair of forces acting away from the cut section in accordance with a deformation sign convention in which the wire is initially assumed to be in tension and that tension is positive. The maximum force in the wire is at x = 0: N_{\max }=N(0)=W_{1}+W_{2}+w\left(L_{1}+L_{2}\right).
Formula for maximum stress in the wire: The cross-sectional area A of the wire is constant, so dividing N_{max} by A leads to a formula for maximum stress in the wire:
\sigma_{\max }=\frac{N_{\max }}{A}=\frac{W_{1}+W_{2}+w\left(L_{1}+L_{2}\right)}{A}=\frac{W_{1}+W_{2}}{A}+\gamma\left(L_{1}+L_{2}\right)Numerical calculations: The cross-sectional area of the wire is
A=\frac{\pi}{4} d^{2}=\frac{\pi}{4}(9.5 mm )^{2}=70.88 mm ^{2}Solving for maximum normal stress gives
\sigma_{\max }=\frac{W_{1}+W_{2}}{A}+\gamma\left(L_{1}+L_{2}\right)=\frac{380 N +700 N }{70.88 mm ^{2}}+77.0 \frac{ kN }{ m ^{3}}(6 m +5 m )=16.08 MPa4. Finalize: If the weight of the wire is ignored, the maximum normal stress is reduced to 15.24 MPa, which is a decrease of more than 5%. Although the stresses are low here, eliminating the self-weight of the wire from the stress calculation is not recommended.
