3 Determine the natural period of the system shown below usi

3. Determine the natural period of the system shown below using L-100 in, E 108 (lb), W= 3000 lb, and k= 2000 lb/in. If the weight W has an initial displacement of uo = 1.0 in and initial velocity uo-20 in/sec, determine the displacement and velocity 1.0 second later. EI, L V-y

Solution

Solution:-

K₁ = 12 EI/ L³

    = 12 x 10⁸/(100)³ = 1200 lb/inch

Total K = k₁ + k + k

              = 1200 +2000 + 2000 = 5200 lb/inch

K = 5200 x 175.12 = 910.624 x 10³ N/m

M = 3000 lb

M = 3000 x 0.4535

      = 1360.5 kg

= (K/M)

    = (910.624 x 10³/(1360.5)) = 25.87 radian/second

Natural period of system(T) =2/

= 2/(25.87)

= 0.243 second   Answer

Given u₀ = 1 , u₀’(velocity) = 20 inch/second , t =1 second

u = u₀cos (t) + u₀’/ sin(t)

u = 1 x cos (25.87 x1) + 20/25.87 xsin(25.87 x1)

(Displacement) u = 1.23 inch Answer

Veocity after 1 second

Du/dt = - u₀ sin (t)x + u₀’cos (t)

= -1 x sin (25.87) x 25.87 + 20 cos(25.87)

  Velocity = 6.7 inch/second   Answer