Question 3.6: A large pendulum is composed of a 10.0 kg ball suspended by ...
A large pendulum is composed of a 10.0 kg ball suspended by an aluminum wire having a diameter of 1.00 mm and a length of 6.30 m. The aluminum is the alloy 7075-T6. Compute the elongation of the wire due to the weight of the 10 kg ball.
Objective Compute the elongation of the wire.
Given Wire is aluminum alloy 7075-T6; diameter = D = 1.00 mm.
Length = L = 6.30 m; ball has a mass of 10.0 kg.
Analysis The force on the wire is equal to the weight of the ball, which must be computed from w = m · g. Then, the stress in the wire must be checked to ensure that it is below the proportional limit. Finally, because the stress will then be known, Equation (3–8) will be used to compute the elongation of the wire.
\delta = \frac{\sigma L}{E} (3-8)
Results Force on the wire: F = w = m · g = (10.0 kg)(9.81 m/s2) = 98.1 N.
Axial tensile stress: σ = F/A
A = \frac{ \pi D^{2}}{4} = \frac{ \pi(1.00 mm)^{2}}{4} = 0.785 mm²
\sigma = \frac{F}{A} = \frac{98.1 N}{0.785 mm^{2}} = 125 MPa
Appendix A–14 lists the yield strength of 7075-T6 aluminum alloy to be 503 MPa. The stress is well below the proportional limit.
The stress can be considered to be steady for a slow-moving pendulum, and the aluminum is ductile, having 11% elongation. From Table 3–2, the design stress can be computed from
| A–14 Typical properties of aluminum alloys .^{a} | |||||||
| Ultimate strength, s_{u} | Yield strength, s_{y} | Shear strength, s_{us} | |||||
| Alloy and temper | ksi | MPa | ksi | MPa | percent elongation | ksi | MPa |
| Alloys in wrought form | |||||||
| 1100-H12 | 16 | 110 | 15 | 103 | 25 | 10 | 69 |
| 1100-H18 | 24 | 165 | 22 | 152 | 15 | 13 | 90 |
| 2014-0 | 27 | 186 | 14 | 97 | 18 | 18 | 124 |
| 2014-T4 | 62 | 427 | 42 | 290 | 20 | 38 | 262 |
| 2014-T6 | 70 | 483 | 60 | 414 | 13 | 42 | 290 |
| 3003-0 | 16 | 110 | 6 | 41 | 40 | 11 | 76 |
| 3003-H12 | 19 | 131 | 18 | 124 | 20 | 12 | 83 |
| 3003-H18 | 29 | 200 | 27 | 186 | 10 | 16 | 110 |
| 5154-0 | 35 | 241 | 17 | 117 | 27 | 22 | 152 |
| 5154-H32 | 39 | 269 | 30 | 207 | 15 | 22 | 152 |
| 5154-H38 | 48 | 331 | 39 | 269 | 10 | 28 | 193 |
| 6061-0 | 18 | 124 | 8 | 55 | 30 | 12 | 83 |
| 6061-T4 | 35 | 241 | 21 | 145 | 25 | 24 | 165 |
| 6061-T6 | 45 | 310 | 40 | 276 | 17 | 30 | 207 |
| 7075-0 | 33 | 228 | 15 | 103 | 16 | 22 | 152 |
| 7075-T6 | 83 | 572 | 73 | 503 | 11 | 48 | 331 |
| Casting alloys—permanent mold castings | |||||||
| 204.0-T4 | 48 | 331 | 29 | 200 | 8 | – | – |
| 206.0-T6 | 65 | 445 | 59 | 405 | 6 | – | – |
| 356.0-T6 | 41 | 283 | 30 | 207 | 10 | – | – |
| TABLE 3–2 Design stress guidelines: Direct normal stresses. | ||
| Manner of loading | Ductile material | Brittle material |
| Static | \sigma_{d} = s_{y} /2 | \sigma_{d} = s_{u} /6 |
| Repeated | \sigma_{d} = s_{y} /8 | \sigma_{d} = s_{u} /10 |
| Impact or shock | \sigma_{d} = s_{y} /12 | \sigma_{d} = s_{u} /15 |
\sigma_{d} = s_{y} /2 = 503 MPa/2 = 251 MPa
Therefore, the wire is safe.
Elongation: For use in Equation (3–8), all data are known except the modulus of elasticity, E. The footnote of Appendix A–14 lists the value of E = 72 GPa = 72 × 10^{9} Pa. Then,
\delta = \frac{\sigma L}{E} = \frac {(125 MPa)(6.30 m)}{72 GPa} = \frac{(125 \times 10^{6} Pa)(6.30 m)}{72 \times 10^{9} Pa }
\delta = 10.9 \times 10^{-3} m =10.9 mm
Comment What do you see around you right now that has a dimension similar to 10.9 mm (0.429 in.)? Measure the thickness of one of your fingers. Certainly, the design of the system containing the pendulum in this example problem would have to take this deflection into account.
Added Note This problem is modeled after the “Foucault pendulum,” found in many science museums and universities. The oscillation of the pendulum allows the observation of the rotation of
the earth on its access.