Calculation of Stiffness in Structural elements
AIM: Calculation of Stiffness in Structural elements
TASK 1: 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 Elb = Elc
SOLUTION:
Here, the degree of freedom is 3 hence the size of the stiffness matrix will be 3X3.
Let Assume, `EI_b= EI_c=EI`
- The first step is to determine the system coordinates and element coordinates.
- Next, we have to find out the Displacementtransformation matrix [a].
- For this apply unit force corresponding to each system coordinates and then find the displacement transformation matrix for each element coordinates.
1. Apply`u_1= 1kN and u_2=u_3=0`
The first column of transformation matrix:
`[a_1]=((-1/h),(-1/h),(0),(0),(-1/h),(-1/h))`
2. Apply `u_2= 1kN and u_1=u_3=0`
The second column of transformation matrix:
`[a_2]=((0),(-1),(-1),(0),(0),(0))`
3. Apply `u_3= 1kN and u_1=u_2=0`
The Third column of transformation matrix:
`[a_3]=((0),(0),(0),(-1),(-1),(0))`
Hence, the displacement transformation matrix can be written as:
`[a]=[[-1/h,0,0],[-1/h,-1,0],[0,-1,0],[0,0,-1],[-1/h,0,-1],[-1/h,0,0]]`
Now, `[a^T]=[[-1/h,-1/h,0,0,-1/h,-1/h],[0,-1,-1,0,0,0],[0,0,0,-1,-1,0]]`
Next we have to find out the element stiffness matrix:
`[k_(e1)]= (EI)/h*[[4,2],[2,4]]`
`[k_(e2)]= (EI)/(2h)*[[4,2],[2,4]]=(EI)/h*[[2,1],[1,2]]`
`[k_(e3)]= (EI)/h*[[4,2],[2,4]]`
Now Element Matrix for the structure can be determined as:
`[k_s]= (EI)/h*[[4,2,0,0,0,0],[2,4,0,0,0,0],[0,0,2,1,0,0],[0,0,1,2,0,0],[0,0,0,0,4,2],[0,0,0,0,2,4]]`
The stiffness matrix, `k=[a^T][k_s][a]`
`k= [[-1/h,-1/h,0,0,-1/h,-1/h],[0,-1,-1,0,0,0],[0,0,0,-1,-1,0]](EI)/h*[[4,2,0,0,0,0],[2,4,0,0,0,0],[0,0,2,1,0,0],[0,0,1,2,0,0],[0,0,0,0,4,2],[0,0,0,0,2,4]][[-1/h,0,0],[-1/h,-1,0],[0,-1,0],[0,0,-1],[-1/h,0,-1],[-1/h,0,0]]`
`k=(EI)/h*[[-6/h,-6/h,0,0,-6/h,-6/h],[-2,-4,-2,-1,0,0],[0,0,-1,-2,-4,-2]][[-1/h,0,0],[-1/h,-1,0],[0,-1,0],[0,0,-1],[-1/h,0,-1],[-1/h,0,0]]`
`k=(EI)/h*[[24/h^2,6/h,6/h],[6/h,6,1],[6/h,1,6]]=(EI)/h^3*[[24,6h,6h],[6h,6h^2,h^2],[6h,h^2,h^2]]`
we know that, `k*u=f_s`
i.e., `(EI)/h^3*[[24,6h,6h],[6h,6h^2,h^2],[6h,h^2,h^2]]* [[u_1],[u_2],[u_3]]=[[f_s],[0],[0]]` --------- `eq^n 1`
Now, by Static condensation method, we get
`[[u_2],[u_3]]=[[6h^2,h^2],[h^2,h^2]]^-1* [[6h],[6h]]*u_1=6/(7h)[[1],[1]]u_1` ----------`eq^n 2`
Now, substitute `eq^n 2` in `eq^n 1`
i.e., `f_s=(24(EI))/h^3*u_1+(EI)/h^3(6h*u_2+6h*u_3)`
`f_s=(24(EI))/h^3*u_1+(EI)/h^3*[[6h,6h]]*[[u_2],[u_3]]`
`f_s=(24(EI))/h^3*u_1+(EI)/h^3*[[6h,6h]]*(-6/(7h)*[[1],[1]]*u_1)`
`f_s=(24(EI))/h^3*u_1-(72EI)/(7h^3)*u_1`
`f_s=96/7*(EI)/h^3*u_1`
We know that, Stiffness is defined as the force required per unit deformation.
So, When `u_1=1, f_s=k`
Hence, Lateral Stiffness `k=96/7*(EI)/h^3`
TASK 2: For the following structures:
- Determine the number of degrees-of-freedom for dynamic analysis
- Establish the equation of motion
- Calculate their natural frequencies
SOLUTION:
a) 
- Degree of freedom: 1 (one Translation)
- Equation of Motion:
The standard equation of motion is given by: `mddotu+Cdotu+ku=P(t)`
But here, there is no damping and no external force hence, C=0.
So the equation of motion becomes: `mddotu+ku=P(t)`
Now, we have to find the stiffness value i.e. k
Here, the far end is hinged, so the stiffness value `k=(3EI)/l^3`
So the equation of motion can be written as `mddotu+((3EI)/l^3)u=P(t)`
- Natural Frequency:`(f_n)`
`f_n=1/T_n` (where `T_n=(2pi)/omega`)
Then, `omega=(2pi)/T_n`
Also, `omega=sqrt(k/m)`
i.e. `omega=sqrt((3EI)/l^3)/m`
`omega=sqrt((3EI)/(ml^3)`
`T_n=(2pi)/sqrt((3EI)/(ml^3))`
`T_n=2pisqrt((ml^3)/(3EI))`
Hence, `f_n=1/(2pi)*sqrt((3EI)/(ml^3))`
But here I value is `oo`
Hence, `f_n=oo`
b) 
- Degree of freedom: 4 (Three rotations at beam joints and One Translation)
- Equation of Motion:
The standard equation of motion is given by: `mddotu+Cdotu+ku=P(t)`
But here, there is no damping and no external force hence, C=P(t)=0.
So the equation of motion becomes: `mddotu+ku=0`
Now, we have to find the stiffness value i.e. k
k value is `(12EI)/l^3` for the columns with fixed ends & `(3EI)/(l^3)` forcolumn with far end hinged.
k value for rotation is `(6EI)/l^2`
So the equation of motion can be written as `mddotu+((24EI)/(l^3)+(3EI)/(l^3))u_4+((6EI)/(l^2))(u_1+u_2+u_3)u=P(t)`
- Natural Frequency:`(f_n)`
`f_n=1/T_n` (where `T_n=(2pi)/omega`)
Then, `omega=(2pi)/T_n`
Also, `omega=sqrt(k/m)`
But here I value is `oo`.
So k value will be `oo`.
Hence, `f_n=oo`
TASK 3: Consider the propped cantilever shown in the figure below. 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.
SOLUTION:
Natural period of system `T_n=(2pi)/omega`
where `omega=sqrt (k/m)`
`k=W/u` and mass(M) = `W/g`
Substituting the above relation in the circular frequency of vibration `(omega)`
`omega= sqrt((W/u)/(W/g)`
`omega=sqrt(g/u)`
Finally, `T_n=sqrt(u/g)*2pi`
`T_n=2sqrt u`
Where u = Displacement at joint
TASK 4: Determine an expression for the effective stiffness of the following systems:
a) 
SOLUTION:
There are 2 stiffness components connected to the mass. One is fixed cantilever beam. Another spring, both are connected in series with the mass.
So the effective stiffness of the system with components in series.
`1/(k_e)=1/(k_1)+1/(k_2)`
k1= Stiffness of beam
k2= Stiffness of spring = K
`k_1=(3EI)/L^3`
`1/(k_e)=L^3/(3EI)+1/(K)`
`k_e=(3EIK)/(3EI+L^3K)`
b) 
SOLUTION:
There are 3 stiffness components connected to the mass. Two of them are fixed end columns. Another spring. All are connected in parallelly to the mass.
So the effective stiffness of the system with components in parallel.
`k_e=k_1+k_2+k_3`
k1= Stiffness of Beam-1
k2= Stiffness of Column-2
k3= Stiffness of Spring = K
`k_e=(24EI)/h^3+(24EI)/h^3+K`
`k_e=(48EI+Kh^3)/h^3`
c) 
SOLUTION:
There are 2 stiffness components connected to the mass. One is fixed beam with bending stiffness. Another Inclined fixed beam with axial stiffness. Both are connected in parallel to the mass.
So the effective stiffness of the system with components in parallel.
`k_e=k_1+k_2`
k1= Bending Stiffness of Beam
k2= Axial Stiffness of inclined Beam
`k_e=(48EI)/L^3+(EA)/sqrt (h^2+L^2)`