Calculation of Stiffness in Structural elements
1
Unit Displacement u1= 1
Stiffness coefficient of coloumn translation (2 similar) 2* 12EI/h^3
Stiffness coefficient of beam rotation (2 ends ) EI/h^2
Unit rotation u2= 1
Stiffness coefficient of coloumn rotation @rigth end 6EI/h^2
Stiffness coefficient of coloumn + beam rotation @ left end
4EI/h+4EI/L
4EI/h + 4EI/2h = 6EI/h
Unit rotation u2= 1
Stiffness coefficient of coloumn rotation @left end 6EI/h^2
Stiffness coefficient of coloumn + beam rotation @ right end
4EI/h+4EI/L
4EI/h + 4EI/2h = 6EI/h
By applying the condition we can get the equation is
fs = k.u
By applying the static condensation method,we are deducting the 3 unknown expression into 2 unknown equation
Stiffness is the amount of force required for thr unit displacement so interchanging the translation u1 to the left side,we arrive the stiffness coefficient s follows
k= 96EIc/7h^3
2
Position of the element can be described by 1 coordinate only u(t) because the beam with infinite stiffness will induce uniform translation in all the coloumns
The system has single degree of freedom structure
Two coloumns with both the ends fixed and one coloumn has pinned-fixed ends so the stifness coefficient will be as follows
k1 =12EI/h^3
k2 =12EI/h^3
k3 =3EI/h^3
For parallel stiffness elements ke = k1+k2+k3
=27EI/h^3
The equation of motion shall be Mu+Ku =0
Natural frequency formula = √k/m = √27EI/Mh^3
3
Both elements have same materials and section.It means they have same modulus of elasticity and second moment of inertia.But the lengths are different
When a mass M is applied at the point, both the elements will respond simultaneously.It means they are connected in parallel (as depicted in the sketch below )
Effective stiffness of the system ke= kb+kc
kb=48 EI/lb^3 kc=3EI/lc^3
ke=3EI(1/lc^3 +16/lb^3)
Natural frequency formula = √k/m = √3EI(1/lc^3 +16/lb^3)/M
Time period T=2π/Natural frequency
4
The stiffness coefficient of the beam elements is 3EI/L^3
The stiffness coefficient of the spring is k
The stiffness elements are in the series,so
1/ke =L^3/3EI +1/k
1/ke=KL^3+3EI/3EIK
ke=3EIK/kL^3+3EI
The stiffness coefficient of the both the coloumn elements is 3EI/h^3
The stiffness coefficient of the spring is k, but is inclined,so the horizontal force (p) and the deformation (U) has to be resolved into the inclined components to arrive actual coefficient of the spring
Inclined component of load P =p/cosθ
Inclined component of deformation u =ucosθ
Stiffness coefficient of spring
K= p/ucos^2θ
p/u=ks=Kcos^2θ
cosθ=L/√L^2+h^2
ks=K L^2/L^2+h^2
The effective stiffness of the system ke= 6EI/h^3+KL^2/L^2+h^2
The stiffness coefficient of the pinned-pinned beam is k1=48EI/L^2
The effective stiffness coefficient for the system ha to be formed by adding the relative axial stiffness provided by the inclined strut connected at the rigth end
Force acting at the rigth end is mg/2 and deformation caused by the strut at the point of "m" is half of its nodal deformation
mg/2=kea.vert*2*`delta`δ2
mg/δ2=4 kea.vert
k2=4 kea.vert
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=4kea h^2/L^2+h^2
The effective stiffness shall be arrived as 1/ke=1/k1+1/k2
