Question 2.3: Three metal balls are suspended by three wires of equal leng...
Three metal balls are suspended by three wires of equal length arranged in sequence as shown in Figure 2.32. The masses of the balls, starting at the top, are 2 kg, 4 kg, and 3 kg, respectively. In the same order, beginning at the top, the wires have diameters 2 mm, 1.5 mm, and 1 mm, respectively. (a) Determine the highest stressed wire, and (b) by changing the location of the balls, optimize the mass locations to achieve a system with minimum stresses.
Given: Dimensions and arrangement of steel balls.
Find: Stresses in each wire; lowest-stress configuration.
Assume: Neglect weights of wires.
We must find the internal force within each wire and then divide by the wire’s cross-sectional area to find the normal stress in each wire. For each wire, the internal force will equal the mass this wire must support times the acceleration of gravity. For example, the top wire, a, must support 2 + 4 + 3 kg, so its internal axial force is 88.3 N. We tabulate these calculations:
| P_i(N) | A_i(m^2) | \sigma i (MPa) | |
| Wire a | 88.3 | 3.14\times 10^{-6} | 28.1 |
| Wire b | 68.7 | 1.77 \times 10^{-6} | 38.8 |
| Wire c | 29.4 | 0.79\times 10^{-6} | 37.2 |
The wire subjected to the highest stress is wire b.
To achieve a minimum stress system, we recognize that stress is inversely proportional to cross-sectional area. Hence, since A_a > A_b > A_c , wire a should carry the largest load (which it must), and wire b and wire c should support as little load as possible. This leads us to the following configuration (Figure 2.33)
| \sigma _i (MPa) | |
| Wire a | 28.1 |
| Wire b | 27.7 |
| Wire c | 24.8 |
In the configuration of part (a), the total stress in the three-wire system is 104 MPa; in part (b), the total stress is 80.6 MPa.
