Calculation of Stiffness in Structural elements

1.Compute lateral stiffness of the one story frame with an intermediate realistic stiffnessof the beam.The system has e dof's as shown.Assume L=2h and Elb=Elc

• SOLUTION

• Stiffness matrix is determined by the method called stiffness matrix method of analysis

1. In this above image the degree of freedom (DOF) is 3 `Stiffness matrix is determined by the method called stiffness matrix method of analysis .
2. Now we have to determine the system coordinate and element coordinate

• Now we have to find stiffnes matrix by putting
• `u1= 1kn , u2 = u3 =0`

1. `[a1] = [ [-1/h ,-1/h],[0,0],[-1/h,-1/h]]`

• Now apply u2=1 and u1=u3=0

• `[a2] = [[0,-1],[-1,0],[0,0]]`

• Now apply u3=1 and u1=u2=0

• `[[0,0],[0,-1],[-1,0]]`

• 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]]`
• `[a^t]=[[-1/h,-1/h,0,0,-1/h,-1/h],[0,-1,-1,0,0,0],[0,0,0,-1,-1,0]]`

• Now we have to determine stiffness matrix

1. `[k 1]=EIdivh.[[4,2],[2,4]]`
2. `[k2]=[EI div h.[[2,1],[1,2]]]`
3. `[k3]=EIdivh.[[4,2],[2,4]]`

• Now stiffness matrix can be determine as

• `[k s]=EIdivh.[[4,2,0,0,0,0],[2,4,0,0,0,0],[0,0,2,1,0,0],[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][ks][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=EIdivh.[[24/h^2,6/h,6/h],[6/h,6,1],[6/h,1,6]]= EI divh^3.[[24,6h,6h],[6h,6h^2,h^2],[6h,h^2,6h^2]]`
• We know that,k.u=fs
• i.e. `EIdivh^3.[[24,6h,6h],[6h,6h^2,h^2],[6h,h^2,6^2]][[u1,u2,u3]] =[[fs,0,0]]`
• Now,by static condensation method,we get
• `[[u2,u3]]=[[6h^2,h^2],[h^2,6h^2]]^-1.[6h,6h].u1=-6/7h[[1,1]]u1` 2

Now,substitute eqn2 in first eqn of eqn1

i.e. , `fs=24(EI)divh^3.u1+ EIdivh^3(6h.u2+6h.u3)`

`fs=24EIdivh^3.u1+EIdivh^3.[6h,6h].[u1,u3]`

`fs=24EIdivh^3.u1-72EIdiv7h^3.u1`

`fs=96/7.EIdivh^3.u1`

• Hence ,lateral stiffness `k=96/7.EIdivh^3`

2. For the Following structure

• Determine the number of degrees of freedom for dynamic analysis
• Establish the equation of motion
• Calculate their natural frequencies

• Degree of freedom-1
• Equation of motion
• The standard equation of motion is given by : `mddotu + cddotu+ 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 k11=`3EIdivl^3`
• So the equation of motion can be written as
• `mddotu+(3EIdivl^3)u=P(t)`

Natural frequency,fn

fn=1/Tn

`w= 2pidivTn`

also, `w=sqrt(k/m)`

`w=sqrt(3EIdivml^3)`

`Tn=2pidivsqrt(3EIdivml^3`

`fn=1div2pisqrt(3EIdivml^3)`

Hence,fn=`infty`

b)

• Degree of freedom-4
• Equation of Motion

The standard equation of motion is given by `mddotu +cddotu+ku=p(t)`

But here,there is no damping and no external force hence.C=P(t)=O.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 and 3Ei/l^3 for the columnwith farend hinged k value for rotaion is

(6EI/l^2)

So the equation of motion can be written as

`m u + (24EIdivl^3 +3EIdivl^3)u4 + (6EI/l^3)(u1+u2+u3)=p(t)`

-Natural frequency,fn

fn=1/Tn---->Tn=`2pidivw`

`w = 2pidiv Tn`

Also, `w = sqrtk(k/m)`

But here,I value is `infty`so k value will be

3.

Natural period

• Tn=`2pidiv w`

`w=sqrt(k/m)`

`W = ku----> k= w/u`

`W = Mg ----> M=W/g`

`w=sqrt(w/udivw/g)`

`w = sqrt(g/u)`

Hence ,The natural time period ,Tn=(`2pi).sqrt(u/g)`

4. Determine an expression for the effective stiffness of the following systems

a) In this question one fixed beam and one spring is connected to the mass in series connection ,so the effective stiffness ca be determined as

Effective stiffness, `1divKeff=1divK1+1divK2`Where k1 is the stiffness of the beam and k2 is the stiffness of spring

`Keff= K1.K2div K1+K2`

`K1=3EIdivL^3`

K2=K

`Keff=(3EIdivL^3).Kdiv(3EIdivl^3)+K`

`Keff=3EI.Kdiv3EI+Kl^3`

b)In this equation3 components are conneected to the mass they are two column and one spring .These are connected in parallel Hence,the expression foe effective stiffnesss is

`Keff=K1+K2+K3`

k1=k2=`12EIdivh^3`

k3=k

keff=`12EIdivh^3+12EIdivh^3+K=@4EIdivh^3+K`

`Keff=(24EI)+Kh^3divh^3`

c)Here,Two componenets are connected to the mass which incli=udes one beam with flexural rigidity =(k1) and the other axial rigidity (k2)>These are connected in parallel

Keff=K1+k2

`K1=48EIdivl^3`

`K2=AEdiv sqrt(l^2+h^2)`

Keff=`48EIdivl^3+AEdiv sqrt(l^2+h^2`