Calculation of Stiffness in Structural elements
Ans1. In frist case
.Unit displacement u1=1
.Stiffness coefficient of column translation (2 similar)
`2*(12EI)/h^3`
.stiffnes coefficient of Beam rotation (2 ends)
`(EI)/h^2`
At second case.
.Unit rotation u2=1
.Stiffness coefficient of column rotation at right end
`(6EI)/h^2`
.Stiffness cofficient of column + beam rotation at left end
`(4EI)/h+(4EI)/L`
L=2h
then ,
`(4EI)/h+(2EI)/h`
`=(6EI)/h`
Third case:
.Unit rotation u3=1
.Stiffness coefficient of column rotation at left end
`(6EI)/h^2`
.Stiffness cofficiebt of column + Beam rotation at right end.
`(4EI)/h+(4EI)/L`
L=2h
`(4EI)/h+(2EI)/h`
`=(6EI)/h`
`[[[(24EIc)/h^3,(6EIc)/h^2,(6EIc)/h^2]],[[(6EIc)/h^2,(6EIc)/h,(EIc)/h]],[[(6EIc)/h^2,(EIc)/h,(6EIc)/h]]]`
`fs=k*u`
now,
`(EIc)/h^3*[[[24,6h,6h,]],[[6h,6h^2,6h^2]],[[6h,h^2,6h^2]]]``((u1),(u2),(u3))=((fs),(0),(0))`
By applying static conensation method , we are deducing the 3 unknown expression into 2 unknown equation.
`((u2),(u3))=-[[6h^2,h^2],[h^2,6h^2]]*[[6h],[6h]]*u1=-6/(7h)*[[1],[1]]*u1`
`fs=((24EIc)/h^3=(EIc)/h^3*6/(7h)*(6h,6h)*[[1],[1]])u1`
`=96/7*(EIc/h^3)*u1`
Stiffness is the amount of force required for unit displacement, So intechanging the translation u1 to the left side, we arrive the stiffness coefficient as follow.
`k=96/7*(EIc)/h^3`
Ans2.
a). Figure 1:
. Position of the element can be described by 1 coordinate only u(t)
. The system has single degree of freedom structure.
. The column has fixed end at bottom. so the stiffness cofficient will be as follow
`k=(3EI)/h^3`
. The equation of motion shall be `Mu' + ku=0`
.. Natural frequency formula `omega`=`sqrt(k/m)`
`omega=sqrt((3EI)/(Mh^3))`
b). Figure 2:
. Position of the element can be described by 1 coordinate only u(t) Because Beam with infinite stiffness will induce uniform translation in all columns
. The system has single degree of freedom structure.
. Two columns with both end fixed & One column has pinned-fixed ends. So the stiffness coefficient will be as follow
`k1=(12EI)/h^3` `k2=(12EI)/h^3` `k3=(3EI)/h^3`
For parallel stiffness elements `ke=(k1+k2+k3)=(27EI)/h^3`
Ans3. both element have same material and section. It means they have same modulus of elasticity and second moment of inertia. But the length are defferent.
. When a mass 'M' is applied at the point, both elements will respond simultaneously. it means they are connected in parallel ( as depictedin the sketch)
So,
.Effective stiffness of the system `ke=kb+kc`
`kb=(48EI)/(lb)^3` `kb=(3EI)/(lc)^3`
`ke=3EI(1/(lc)^3+16/(lb)^3)`
.Natural frequency formula `omega``=sqrt(k/m)` `omega=sqrt[[3EI(1/(lc^3)+16/(lb^3))]/M]`
. time period `T=(2pi)/omega`
Ans4. The stiffness coefficient of the both the column element is `(3EI)/h^3`
. The stiffness coefficient of the spring is K, but is inclined. So the horizontal force (P) and deformation (u) has to be resolved into inclined components to arrive actual coefficient of the spring.
Inclined component of load 'P' = `P/CosO/`
Inclined component of deformation 'u' = u `cosO/`
stiffness coefficient of spring.
`K=P/(ucos^2O/)`
`p/u=ks=Kcos^2O/`
b).
`(mg)/2=kea,vert X 2 X Delta2`
`(mg)/(Delta2)=2kea,vert`
`k2=4kea,veart`
We assumed vertical axial stiffness in our calculation. But we have an inclined axial stiffness in the given system. So we have to resolve the (kea,vert) into (kea).
`k2=(4 kea h^2)/(L^2+h^2)`
.The effective stiffness shall be arrived as, `1/(ke)=1/(k1)+1/(k2)`
c).
`1/(ke)=1/((48EI)/L^3)+1/((4kea h^2)/(L^2+h^2)`
`1/(ke)=1/(((48EI)/(L^3)))+1/(((4EAh^3)/(L^2+h^2)^15)`
`ke=(((4AEh^2)/((L^2+h^2)^(3/2)))*(48EI)/L^3)/(((4AEh^2)/((L^2+h^2)^(3/2)))+(48EI)/L^3`