Answer to Question #169792 in Mechanical Engineering for Milan

Question #169792

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.



1
Expert's answer
2021-03-08T03:09:44-0500


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

1/2keqy2 = 1/2k1x12 + 1/2k2x22 = 1/2k1y2cos230 + 1/2k2y2sin230

Therefore, keq = 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

keq = 3k1/4 + k2/4 = 617 338 kN/m

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


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