Frame with 6 DOFs. v m. determining stiffness, k k = F / water tower deflected water tower dynamic response model

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1 CE 533, Fall 2014 Undaped SDOF Oscillator 1 / 6 What is a Single Degree of Freedo Oscillator? The siplest representation of the dynaic response of a civil engineering structure is the single degree of freedo (SDOF) oscillator. A degree of freedo is a displaceent coponent. For exaple, the 2D frae shown below has six degrees of freedo since each of the top joints can translate horizontally, translate vertically and rotate Frae with 6 DOFs A structure whose dynaic response (due to ground shaking) lends itself well to a single degree of freedo response odel (aka a SDOF oscillator) is a water tower, depicted below. Although the tank of the water tower rotates and displaces vertically (slightly), these degrees of freedo can be ignored for a siple analysis. v tank support water tower deflected water tower dynaic response odel The dynaic response odel of the tower considers the following (see sketch below): the horizontal displaceent of the center of ass of the tank, as a function of tie = v(t) the ass of the tank including water (and neglecting the ass of the support) luped at a point = the force per inch or elastic stiffness required to deflect the center of ass = k v F deterining stiffness, k k = F /

2 CE 533, Fall 2014 Undaped SDOF Oscillator 2 / 6 The equation of otion for a SDOF Oscillator The dynaic response of a SDOF oscillator is usually derived using a block on rollers with luped ass attached to a support with a spring of stiffness k. The displaceent fro the at rest position (no tension or copression in the spring) is denoted v(t). Note that when the block is displaced in the positive direction the spring exerts a force on the ass = kv in the negative direction. +v k at rest: v with +'ve v: kv The velocity and acceleration of the ass are expressed as follows: v dv velocity, in / s dt v d 2 v acceleration, in / dt 2 s2 Newton's 2 nd Law says that the su of the forces on an object is equal to the ass of the object ties its acceleration: F = a. Applying Newton s 2 nd law to the block: F a kv v v kv 0 The equation above is called the equation of otion for the SDOF oscillator. The solution of the equation of otion has the for: v(t) = G e st, where G and s are coplex nubers.

3 CE 533, Fall 2014 Undaped SDOF Oscillator 3 / 6 Coplex Nubers Coplex nubers are a tool that siplifies certain types of calculations, for exaple calculating the roots of a linear differential equation with constant coefficients (our case). A coplex nuber is a cobination of a real nuber and an iaginary nuber. Iaginary nubers, when squared, give a negative result. The unit iaginary nuber is i", where i 1. Iaginary nubers were developed in the 1500 s and detractors called the new nubers fictitious or iaginary. The ter has stuck. (European atheaticians also resisted the concept of negative nubers until the 1700s). The coplex nuber G can be represented two ways: in Cartesian coordinates: G G R ig I and in polar coordinates: G G cos ig sin Iaginary G I G G R Real Euler s identity relates the trigonoetric functions and the coplex exponential function: cos i sin e i Therefore, the coplex nuber G can be expressed a third way: G Ge i

4 CE 533, Fall 2014 Undaped SDOF Oscillator 4 / 6 Solution of the equation of otion To solve the equation of otion, we will substitute v Ge st into the equation of otion. The velocity and acceleration ters are calculated by differentiating v: v sge st v s 2 Ge st The equation of otion (fro Page 2): v kv 0 becoes: (s 2 Ge st ) k(ge st ) 0 collecting ters: (s 2 Ge st ) k(ge st ) 0 (s 2 k)ge st 0 Since Ge st 0 is a trivial solution, s 2 k 0 Dividing by : s 2 k 0, We set: 2 k,where undaped natural circular frequency in rad/sec The resulting equation has two roots, s 1 and s 2 : s s 1 2, s 2 2 s 1 i, s 2 i

5 CE 533, Fall 2014 Undaped SDOF Oscillator 5 / 6 The solution to the equation of otion now becoes: v(t) G 1 e s1t G 2 e s 2t v(t) G 1 e it G 2 e it Expressing the coplex constants G1 and G2 in Cartesian coordinates: v(t) (G 1R ig 1I )e it (G 2 R ig 2 I )e it Expressing the coplex exponential in trigonoetric for (using Euler s identity): v(t) (G 1R ig 1I )(cost isint) (G 2 R ig 2I )(cost isint) Expanding: v(t) G 1R cost ig 1R sint ig 1I cost G 1I sint G 2 R cost ig 2 R sint ig 2 I cost G 2I sint Collecting like ters: v(t) (G 1R G 2 R )cost (G 1I G 2 I )sint i[(g 1I G 2I )cost (G 1R G 2 R )sint] Since the response v(t) is real, the iaginary coponents ust = 0. Therefore: G 1I G 2I 0, and G 1R G 2R 0 The equations above iply that: the iaginary parts of G 1 and G 2 are equal and opposite, so G 1I G I and G 2I G I the real parts of G 1 and G 2 are equal, so G 1R G 2 R G R Therefore: G 1 G R ig I G 2 G R ig I So: now (G 1R G 2 R ) G R G R 2G R and (G 1I G 2 I ) G I (G I ) 2G I v(t) 2G R cost 2G I sint,

6 CE 533, Fall 2014 Undaped SDOF Oscillator 6 / 6 2G R and 2G I are two real nubers which we ll renae A and B, respectively. Therefore: v(t) Acost Bsint The initial conditions lead to: (1) : v(t 0) v(0) Acos(0) Bsin(0) A A v(0) (2) : v(0) A sin(0) B cos(0) B B v(0) An alternative way to express v(t) can be derived by letting : A cos and B sin v( t) cos cost sin sint cos( t ) (trigonoetric identity) v( t) cos( t ) where : A B ( cos ) ( sin ) (cos sin ), or A B sin B B 1 B and :, tan, or tan cos A A A