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_b = El_{c .}

_{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 { u2 = 6h{U1

h^2 6h^2] u3} U1}

[u2=-[[ 6h^2 h^2 6h{U1 =6/7h[1 u1

u3] h^2 6h^2]-1 (inverse) 6h 1]

fs=24EIc/h^3-EIc/h^3*6/7h(6h 6h)1by1)u1

=96/7EIc/h^3u1

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=`root`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=`root`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