Question

A CATER TOWER OF MASS 1500kg IS SUSPENDED BY FOUR PIPES OF A COLLECTIVE STIFFNESS OF 3000N/m• IF THE DAMPING OF THE STRUCTURE IS 360 DE TERMINE THE DISPLACEMENT AND VELOCID OF THE MASS AT te a sess. FIGURE GNEN: Xo = 200 mm to = - 15 mm/s 7771)

Answer 1

Given, ma 1500 kg. K z 3000 N/m c - 3000 s/m. V=n. 2 200 mm 2 0.2 o 2 – 15 mmls : - m, 0.015 m/s. t 22s. Natural frequency cwn) - v tom 3ooo 1500 & Won - 1.4142 ratss.

Damping ratio (3) am w 3000 2x1500 x 1.4142 20.7071 Damped frequency (p) & od 1752 Wo 2 1.4142 V1 + 0.707,2 > Wp = 1.73203 radys. Now, displacement at any time it is geven by +&+?wy v (t) =ė swt fu coswort ab ta as, -6.707 x 1.4142x2) Vitze 0.2 xcos (1-7322) + 60015)+(0.7071x1.414240 •2) rein (1732x2) 1.732 – 0.03 0254 m. vlt) u(t) D - 30.254 mm Ans.

To find the relocity at t=2s, Differentiating the displacement equation with respect to t ů (t) a esot Ev.wp. sin wat + ( + 3 wro) cos(pt)] & è sot (arw). . [lows upt + ut 3 who sro (wors] i at taas, Últ ale 0704* *4142 »2 (-0.2 31732 x sin (1-732*2) + (-0.015+0.7071*1-414220.2) * cos (1.732*2)744 0.7074x1.4142+2 (-0.7071x1.41.42) × [0.2 x Cos (1.732x2) -0.015) +0.70 4x 1.4142× 0.2 1.732 sin (1.732x2)

.24 241 0 ، 0S = (1) ق ( m / s 1، 52 = (1) ډ