Answer to Question #169792 in Mechanical Engineering for Milan

Question

The arm AD of the excavator shown in Fig. 1.93 can be approximated as a steel tube of outer diameter 10 in., inner diameter 9.5 in., and length 100 in. with a viscous damping coefficient of 0.4. The arm DE can be approximated as a steel tube of outer diameter 7 in., inner diameter .5 in., and length 75 in. with a viscous damping coefficient of 0.3. Estimate the equivalent spring constant and equivalent damping coefficient of the excavator, assuming that the base AC is fixed.

Answer 1

Consider the equivalent strain energy which is the sum of the strain energies from these two axially loaded springs:

1/2k_eqy² = 1/2k₁x₁² + 1/2k₂x₂² = 1/2k₁y²cos²30 + 1/2k₂y²sin²30

Therefore, k_eq = 3k₁/4 + k₂/4

For the axially loaded spring:

1 in = 0.0254 m

k₁ = A₁E₁/L₁ = ($\pi$ /4((0.254 m)² - (0.241 m)²) x 200 x 10⁹ Pa) / 2.54 m = 397 753 kN/m

k₁ = A₂E₂/L₂ = ($\pi$ /4((0.1778 m)² - (0.127 m)²) x 200 x 10⁹ Pa) / 1.905 m = 1 276 096 kN/m

k_eq = 3k₁/4 + k₂/4 = 617 338 kN/m

Similarly, c_eq = 3c₁/4 + c₂/4 = 3(0.4)/4 + 0.3/4 = 0.375