Calculation of Stiffness in Structural elements

1.AIm:

Compute lateral stiffness of the one story frame with an intermediate realistic stiffness of the beam. The system has 3 DOFs as
shown. Assume L = 2h and El = El

Lateral stiffness of the one story frame:

• 
  This structure can be analyzed by any of the standard methods, including momentdistribution. Here we use the definition of stiffness influence coefficients to solve the
  problem.
• 
  Thus the 3 × 3 stiffness matrix of the structure is known and the equilibrium equations can bewritten. For a frame with Ib = Ic subjected to lateral force fS, they are,
  The degeer of freedom is 3.

Step-1:
U1=1,U2=0,U3=0
The stiffness matrix is calculated,
K11=24EIc/h^3
K12=6EIc/h^2
K13=6EIc/h^2

Step-2:
U1=0,U2=1,U3=0
The stiffness matrix is calculated,
K21=6EIc/h^2
K22=4EIc/h+4EIb/L
K23=2EIb/L

Step-3:
U1=0,U2=0,U3=1
The stiffness matrix is calculated,
K31=6EIc/h^2
K32=2EIb/L
K33=4EIc/h+4EIb/L
therefore,
K=matrix of[K11 K12 K13
K21 K22 K23
K31 K32 K33]
K=EIc/h^3matrix of[24 6h 6h {u1 ={fs
6h 6h^2 h^2 u2 0
6h h^2 6h^2] u3} 0}
The equation written as,
24u1+6hu2+6hu3=fs------1
6hu1+6h^2u2+h^2u3=0-----2
6hu1+h^2u2+6h^2u3=0-----3
The equation can be slove by two method are Elimination methid or static condansation method.
now we slove,
[ 6h^2 h^2 { 2 6h{U1
b c .

2a.Aim:
TO find the below solution
Determine the number of degrees-of-freedom for dynamic analysis
Establish the equation of motion
Calculate their natural frequencies
1. DOF =1
2. Equation of the motion
=mu+ku=0 (mu=mass&ku=stiffness of the frame)
stiffness of the frame=3EI/h^3
3. For natural frequency,wn= k/m
m=mass(M)
sub the value of m and k in wn formula
therefore,Natural of frequency of wn=root 3EI/h^3/m

2b.
1. DOF =1
2. Equation of the motion
=mu+ku=0 (mu=mass&ku=stiffness of the frame)
stiffness of the frame k=12EI/h^3+12EI/h^3+3EI/h^3
=27EI/h^3
3. For natural frequency,wn= k/m
m=mass(M)
sub the value of k in wn formula
therefore,Natural of frequency of wn=root (27EI/h^3/m)

3.AIM:
Consider the propped cantilever . The beams are made from the same steel section and have lengths as shown on the diagram.
Determine the natural period of this system if a large mass, M, is placed at the intersection of the beams at point A. The weight of the
beams in comparison with the mass M is very small.
so,
The lateral stiffness we need to add the both stiffness of beam
therefore we know,
For simple supported beam lateral stiffness is=48EI/L^3
For cantilever supported beam lateral stiffness is=3EI/L^3
Add both beam sstiffness,
K=(48EI/L^3+3EI/L^3)
K=51(EI/L^3)
The natural frequency,wn=root(k/m)
wn =root(51(EI/L^3)/m

4. Expression for the effective stiffness :
1. The effective stiffness of cantilever beam is =K=48EI/L^3
2. The effective stiffness of simply supported beam is =K=3EI/L^3
3. The effective stiffness of 2 simply supported beam is =K=3EI/L^3+3EI/L^3
=6EI/L^3