Modeling of Mechanical (Lumped Parameter) Elements

Equivalent Spring Coefficients:Select an equivalent displacement, find the total strain energyand equate to that of the equivalent element.(a) springs connected by leversV s = ½ {K 1 X 2 1 + K 2 X 2 22} = ½ K eq X eq[N.m=J]Let X eq = X 1 , then X 2 = θL 2 , X 1 = θL 1so then X 2 = X 1 (L 2 / L 1 ) andK eq = K 1 + K 2 (L 2 / L 1 ) 2 [N/m](b) torsional springs on geared shaftsV s =½ {K 1 θ 1 2 + K 2 θ 2 2 } = ½ K eq θ eq 2[N.m=J]since θ1N1 = θ2 N2, where N is the number of teethand select θ eq = θ1 , thenK eq = K 1 + K 2 (N 1 / N 2 ) 2[N.m/rad](c) coupled torsional and translationalspringsV s = ½ {K x 2 + K t θ 2 } = ½ K eq X eqLet Xeq = X , and since X = R θ , then2K eq = K + K t / R 2[N/m], orLet X eq = θ , then the equivalent torsional spring isKteq = K R 2 + K t [N.m/rad]Note: in (a) thru (c) one of the spring ends is connected to ground.MEEN 617 Notes: Handout 1 / © Luis San Andrés 2008 1.4