Question

Figure Q1 illustrates a simple pressure relief valve system, which consists of a rigid L-shaped beam, hinged at a point where the horizontal part of the beam has length 2L and the vertical part has length L. A spring of stiffness k is attached midway along the horizontal part of the beam, and a damper with damping coefficient c is attached to the vertical part of the beam, at a distance 0.75L from the hinge, O. The pressure relief valve is attached at the opposite end of the horizontal part of the beam from the hinge, as illustrated. This relief valve is subject to a pressure, PO), over a cross-sectional area, A. The beam has total mass 3m, and the moment of inertia of the beam about the hinge has been found to be 16 = 3ml?. For this system, show that the equation of motion is given by the following equation [3 marks] (a) - + (b) 30 k 2 ë+ @= APO) 16m 3m 3ml The system has the properties m= 1.5 kg, L = 1.2 m, c = 4 Ns/m, k = 810 N/m and A = 0.1 m², The pressure applied is zero until a sudden increase to a pressure of 800 Pa. Considering this as a step input to the system, using a Laplace Transform approach, derive an expression for the response (t) of the system to this input. [12 marks) Calculate the damping ratio, &, that is required to ensure that the maximum overshoot of the system is no greater than 10% of the steady-state value. [5 marks) (c) 2L and Plt) Area, A Figure Q1

Answer 1

Given: Io = 3m 2² 0.25L A² 0.75 L.O DA5L A By tasque Method PLU) A.2L REOT 2 Too 3m 3ML 2. c. 10.75L) Iö + (0.75 L O + KLO PLE) A.2L 3mit öt cx a tot Kto = P(E) A. 2X 16 ot 3c é tko 2 A.PH) 16m b> mzl.5kg t2 1.2 m Pan C²4 N/m 800 Pa K2810 N/m A201² P ₂ 800 Pa Step Suput time

for second order system Gric) ROS Сck) an K(X+ 2gon) -Unit feedback C) G(K) Wn R(K) I+G(K) K²+2zWnkton Chadeerste Since Input is unit step. R(K) = P * for the Given fr 27 Wn K 3m 23 n 2 3.c. 16m We 2 81082 2X3x 1180 - 3x4 X 183 16x1.5 wn 180 Wnz 8180 sad/sec 41780 [320.0186 n2 13.416 rad/sec Q331 an R(K) (+6, 3)²+ on 1 (1-3%) CCK) P. on А Bstc kccktwon) + coñ (1-3995 (5+3n) & On (137) After simplifying Аз Р. B2-P daw (1-2) 2 t we get, Ce-230mP

Cod CCB :P (K+3wn) 3 (k+3wnt wat j-25 16+ Applying j Gnieode laplace an both side ķ we get, -zont e e(t)= (1 (UF-3) Colkat) +3 Sto (241)P Substituting Values e(t) = (CS1-0.0186)C63 byt + 356efcb)2 we get 0.0186--1180-Xt VI -0.01867 bd (1-3 wna 129.97 gad/sec 2 cat B2002 oldali ē 0.9998 -0.2495t 60.9998) Cos (29.974)+0.0186 SMC179974

c) -RATI 7.O.S. = e X100 for y.O.S. 20 to e X100 -3 र 2 2.30 25 3 20.733 -- 3-2 0.537 - 03377 इ20.5912

To ensure that maximum overshoot is not greater than 10% the damping ratio must be 0.5912.