Consider the frame below where the column properties are E 3

Consider the frame below where the column properties are E= 30 x 106 psi, 1-150 in, and L = 12 ft. The rigid beam (El=o) has a mass 2.5m where m 1.0 lb-sec2/in. El=0 , Rigid 2.5m 1.51EI 1.5EI a) Determine the equation of motion for this system. b) Calculate the natural circular frequency. c) The frame is set into motion by initial conditions: uo-0 and ú6 ft/sec. Determine the expression describing the displacement time history and calculate displacement at t- 1 sec.

Solution

Solution:-

                =(K/M)

                K = K₁ + K₂

            K = 12 EI/(1.5 L)³ + 3 EI/L³

K =(12x 30x 10⁶ x 4.448 x 150)/(1.5 x 12 x 12)³ + (3 x 30 x 10⁶ x 4.448 x 150)/(12 x12)³    

K = 43943.9 N/inch

Where 1 inch = 0.0254 m

K =1.73 x 10⁶ N/m

Mass =2.5 x 175.13 = 437.82 kg

Where 1 lbf-s²/inch = 175.13 kg

(b) Natural circular frequency () = ((1.73 x10⁶ ) /( 437.82))

                                                      = 62.86 radian/second      Answer

(a) Equation of motion for system:-

u =u₀ cos t + u₀’/ x sin t

u = u₀ cos (62.86 t)   + 0.016 u₀’ sin (62.86 t)   Answer

Where u₀’ = velocity

(c) Displacement :-

When u₀ =0 ,   u₀’ = 6 feet/second ,   t = 1 second

u = u₀cos(62.86 t) + 0.016 x u₀’ sin(62.86 t)

u =0 x cos (62.86 x1) + 0.016 x 6 x cos (62.86 x 1)

u = 0.0854 feet     Answer