6 thoughts on “3.4”

  1. Just a addition point worth mentioning to possibly simplify the method of adding the spring coefficients.

    For springs in series, use the same method of adding capacitors in series and you will arrive at the same equivalent stiffness coefficient. Example: for part a) k1+(k2^-1+k3^-1)^-1 = k1+ k2*k3/(k2+k3). Intuitively, this can be understood by comparing a springs behavior under a load as storing mechanical energy in the same respect as a capacitor under an electrical load stores electrical energy. Thus the same series addition can be applied.

    For springs in parallel, you simply add their values together... like adding capacitors in parallel in an electric circuit. Hints why in the above example you add k1 to the inverse sum of the inverses.

    1. You are absolutely correct. In fact each of the mechanical elements has an electrical (and fluidic) counterpart.

      Spring -> Capacitor
      Damper -> Resistor
      Mass -> Inductor

      The analysis of mechanical systems and electric circuits can both be reduced to the same fundamental idea. In fact there are methods of analysis which are completely agnostic of the type of system you are investigating, they focus entirely on how each element stores, transfers, or dissipates energy.

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