Slide 1

Free Vibration Without Damping

System with Translation DOFSystem with Rotation DOF

SDOF System Spring Having Translation Stiffness (in x-x, y-y, z-z )

Is already studied but just for the reference

Free vibration without damping

MATHEMATICAL MODEL Where, m = mass in kg k = stiffness in N/m x = displacement in m By considering vertical system

k

W

k = W

(When the mass moves downward)(When the mass moves upward)

6

Calculation of angular natural frequency from equation of motion So our basic equation of motion is

We assume the solution as

Putting these values in equation of motion

The eqn. must hold true for all values of t, therefore =natural frequency

8The complete solution of Equation of motion is obtained by method of operator. The solution to the above equation can be obtained Using method of operator.

Let the operator be D , and

Above equation can be written as, (mD2+k)=0

9 x(t) = A(cos + i sin) + B(cos - i sin) x(t) = (A +B) cos + (A - B) i sin The above Equation is the complete solution of free vibration in single degree of freedom system without damping.

10 Since x=Acos(t) is a solution of eqn of motionWe can also assume that,

is also a solution.

Now differentiating above eqn. twice we get

11

Therefore the equation of motion becomes

Displacement of undamped system is as shown:

Example:-x(t) = 5 sin(1t) Compare it with x(t) = Asin ( t) Here, A = amplitude= 5

= Natural frequency= 1 rad/sec

So one cycle will be completed in 2 time

13

T(sec)5-5SDOF System Spring Having Rotational Stiffness (Rocking Stiffness)

Connected at CG of Base Area of the Block

Consider the block having mass m and

Mass Moment of Inertia Mmz about centroid Z-axis passing through centroid C .A rotational spring having rotational stiffness k is attached to the support point O of the block (centre of the base point)

The block is constrained such that it cannot move either in X or Y direction.

But it can only rotate in XY plane about Z axis

(perpendicular to the plane of the paper) passing through O. The centroid C is at a distance h from the centre of the base of the block O. The DOF of the system is rotation at point O.The centroid C is at a distance h from the centre of the base of the block O

The DOF of the system is rotation at point O

Fig. Undamped SDOF System with Rotational Spring attached at Centre of Base of Block

Static Equilibrium

Consider the position of the mass at rest i.e.

mean position of the mass.

The gravity force mg is taken care of by reaction R at point O.

Forces acting on the system

(a) System at rest (b) Rotation about O (c) FBDFig. Undamped SDOF System with Rotational Spring attached at centre of Base of Block Free Vibration

System under Motion:Let us disturb the block so as to cause it to rotate slightly about point O and then release the block. The block shall start oscillating about the mean position (i.e. position at rest) at point O in X-Y plane.

Consider that any instant of time t, the block position is rotated anti-clockwise by an angle

Due to rotation, centroid point C'. Rotation at O induces rotation and translation h at centroid C'

Forces acting on the systemRotational Inertia = Mmz acting clockwise (opposite to direction of motion) at centroid C'

Translational Inertia force (along direction perpendicular to line OC) = mh (opposite to direction of motion).

Resisting moment offered by spring (clockwise) at O = ( k) Moment due to self-weight (anticlock wise) at O = mg . h sin = mgh (For to be small sin = )

Considering equilibrium of forces at DOF location i.e. at point O, we get

Y = 0 mg R = 0 or mg = RMz = 0 .. .. (Mmz + mh . h ) + k mgh = 0 ..(Mmz + mh2) + (k mgh) = 0 ..Mmoz + (k mgh ) = 0Here Mmoz = Mmz + mh2 is the mass moment of inertia of the block about Zaxis passing through support point OSolution of this equation gives: Motion of the block

= Acospt + Bsinpt

Here p represents rotational natural frequency of the system and

constants A & B are evaluated using initial conditions

For all practical real life cases, it is seen that term mgh is negligible compared to k and Can be conveniently be ignored without any loss of accuracy in the frequency value.Hence the natural frequency becomes:

..

Second solution consider the same eq. let be represented asSubstituting equation (b)in equation (a), equation becomes

Since exponent is not zero, therefore for solution to exist

This gives two values of s Denotingthe solution taken the formUsing De Moivres theorem, the equation is rewritten as

The values of constants A & B are obtained using initial conditions.

This is the equation of motion for Rocking about Z-axis i.e. DOF

For system with rocking DOF

, replacing all parameters with parameters SDOF System Spring Having Rotational Stiffness (Torsional Stiffness)

Connected At CG of Base Area of the BlockConsider a block having mass m and mass moment of inertia Mmy about Y-axis passing through CG of the Base AreaA Torsional spring having rotational stiffness k is attached to the block at the CG of the base area point OThe block is constrained such that it can neither move in X nor in Z direction But

It can only rotate in XZ plane about Y-axis passing through O

we get equation of motion as

Mmoy = Mmy is the mass moment of inertia of the block about Y-axis passing through support point OThis is the equation of motion of an undamped SDOF system with Torsional DOF at point OSolution of this equation gives: Motion of the block p is represents Torsional natural frequency of the system

Constants A & B are evaluated using initial conditions

UnitsForce and Weight kNMass kgLength in mTime in secondG gravity m/sq secE N/sq MeterPressure N/sq MeterDensity kg/cum

Problem 1

A RCC Block having length (along Z-axis) L=4000

mm , Width B=2000 mm and Height H=500 mm is

supported by a rotational spring (Rocking about Z-

axis) having stiffness of k = 3*106 Nm/rad attached

at centre point of base of block, point O as shown in

Fig. Consider the system as undamped. Density of

concrete is 25 kn/m3 (2500 kg/m3 .)

Find the natural frequency of the system

a) considering that the system performs only rocking motion about Z-axis passing through CG of the base area point O

b) If the applied spring at point O is in Rocking direction about X-axis and the system performs only rocking motion about X-axis passing through centre point O

c) If the applied spring at point O is in Torsional direction about Y-axis and the system performs only torsional motion about Y-axis passing through centre point O

Rocking Natural Frequency (about Z axis )Rocking Natural Frequency (about X axis ) c) Torsional motion about Y - axis Mass of the block m=2500* 4*2*0.5 = 10000 kg Rotational stiffness of spring (Rocking about X-axis) k = 3*10^6 Nm/rad Height of centroid C above center of base O , h= 0.5*H = 0.5*0.5 = 0.25 m Since point C and point O lie on the same vertical line , Mass moment of inertia of the block about Y - axis passing point O = Mmoy

Rocking Natural Frequency (about X axis )