Linearly Elastic System

Here the dynamics of structures problem (earth quake engineering) is studied for simple structures that can be considered as a system with lumped mass and a mass less, supporting structure. Structures which are linearly elastic as well as inelastic subjected to applied dynamic force or earthquake force is considered. The differential equation, for this can be solved in different ways.

We must need a knowledge about Newton 2nd law of motion, that is the definition of force, formula F = ma. It’s a function of time, acceleration is s/t2, s=ut+gt2/2. So a force is noted by P (t) distance by S (t), differential equations. We can start with a simple structure. A column fixed at base and loaded at the other end. It has been a single degrees of freedom system. When an earth quake force occurs, the structure may displace or vibrate, let displacement is S(t). By single degrees of freedom it meant this can move only in one direction.

The vibration must be damped for that a viscous damper is provided, usually by giving a shear wall. Thus, the forces acting are P(t)-resisting force =0. For a linear system the lateral force f and deformation S which is linear the formula is f =k S where k is lateral stiffness or force/ length. For the external force there is a resisting force according Newton third law, to every action there is an equal and opposite reaction. By linear we meant that when the force applied is removed the deformation also will disappear.

Md2S/dt2-ks = 0   k = stiffness= Force/ unit displacement, S = deformation

This is the differential equation governing the lateral displacement S (t).If some initial displacement is given S(0), then released and permitted to vibrate freely, the structure will oscillate about its initial equilibrium position.

The process by which vibration steadily diminishes in amplitude is called damping. Here the damping element is viscous damper. There are other methods also to damp the vibration.

Lateral Stiffness k

Stiffness k for the column is 12EI/L3, that is the force required for the unit displacement would be 12EI/L3, if the beam is rigid. For 2 columns is 24EI/L3, the first term of the stiffness matrix k11.

The n-by-n stiffness matrix of the structure shown in figure is found, from matrix stiffness method. Finally lateral stiffness of the frame is calculated. The structure has 3 degrees of freedom, one lateral movement and 2 angular rotations.

K = 96EIc/7L3

To get the first column of the 3 by 3 stiffness matrix, apply unit displacement S1=1, with S2=S3=0. The forces for this deflected shape to obtain shown in the figure.

The elements in the 2nd column of the stiffness matrix are calculated by applying S2 = 1 with S1 = S3 = 0 see fig.

For getting the 3rd column of stiffness matrix we apply S3=1 with S1=S2=0. For a frame with Ib = Ic

And for lateral force f the equilibrium equations are arranged in the matrix form

k11 = 2(12EI)/L3, k21 = 6EI/L2, k31 = 6EI/L2

k12 = 6EI/L2, k22 = 6EI/L, k32 = EI/L, see figure S1 is lateral displacement S2 and S3 are rotations

 


From the 2nd and 3rd equations, the joint rotations are expressed in terms of lateral deformation S1, see matrix equation below.


f = 96EIS1/7L3 by substituting equation 2nd into the first 3 equations of equation 1, dividing by S1 for k

Thus, the lateral stiffness k = 96EI/7L3, that is the force required to produce unit displacement to the frame

This is for elastic system. The frame can be analyzed by moment distribution method also.