Calculation of Stiffness in Structural elements
Ques:1
`EI_c/h^3``[[24,6h,6h],[6h,6h^2,h^2],[6h,h^2,6h^2]]{[u_1],[u_2],[u_3]}={[f_s],[0],[0]}`
`{[u_2],[u_3]}=-[[6h^2,h^2],[h^2,6h^2]]^-1 [[6h],[6h]]u_1=-6/7h[[1],[1]]u_1`
`f_s=[24EI_c/h^3-EI_c/h^3 6/7h([6h,6h])u_1=-6/7h[[1],[1]]u_1`
`k=96/7EI_c/h^3`
Ques:2
a.
No of Degree of Freedom
In this structure the column portion is pinned and the beam portion is roller supported and also the force is acted laterally to the column that is the force is acted horizontally u(t).
So one force reaction is obtained at the pinned support therefore this structure has Single Degree Of Freedom
Equation of Motion
Since there is no dampers
`mddotu+kdotu=u(t)`
Natural Frequencies
`omega=sqrt(k/m)`
Stiffness k=`sqrt(3EI/mh^3)` since pinned
therefore
`omega=sqrt((3EI)/(mh^3))`
b.
No of Degree of Freedom(DOF)
In this structure two columns were fixed and one column is pinned and force at single coordinates which act as Single DOF
Equation of Motion
Since there is no dampers
`mddotu+kdotu=u(t)`
Natural Frequencies
`omega=sqrt(k/m)`
Stiffness at the two fixed Column is considered as `k_1, k_2`
`k_1, k_2 = 12EI/h^3`
Stfiffness at the pinned column `k_3=3EI/h^3`
K=`k_1+k_2+k_3`
k=`27EI/h^3`
therefore
`omega=sqrt(27EI/(mh^3))`
Ques:3
Natural Frequecy
T=`2pi/omega`
Propped Cantilever Beam is the Beam fixed at one end and a simple supoort is given at the other end
Stiffness need to get find out since natural frequency `omega` is needed
let `k_b` be the stiffnes of the simply supported beam and `k_c` be the stiffness of the propped cantilever beam
`k_b=48EI/l_b^3`
`k_c=3EI/l_c^3`
K=`k_b+k_c`
K= `48EI/l_b^3+3EI/l_c^3`
`omega=sqrt((48EI/l_b^3+3EI/l_c^3)/m`
T`=(2pi/omega)/(sqrt((48EI/l_b^3+3EI/l_c^3)/m)`
Ques:4
a.
Effective Stiffness=K
In this structure one end is fixed and the other end is free with mass and spring acting at the free end
So Stiffness is `K_1=3EI/h^3`
Since the spring is added to the structure and also in Series
Stiffness `1/K=1/k_1+1/k`
`1/K=l^3/(3EI)+1/k`
`1/K=(kl^3+3EI)/(3EIk)`
`K=(3EIk)/(kl^3+3EI)`
b.
Effective Stiffness=K
In this structure two columns are pinned
So Stiffness is
`k_1,k_2=3EI/h^3`
spring is acted inclined so angle `theta` is considered
spring stiffness k=k`cos^2theta`
`costheta= L/(sqrt(L^2+h^2)`
`cos^2 theta= L^2/(L^2+h^2)`
`k=kL^2/(L^2+h^2`
Stiffness K= `3EI/h^3+3EI/h^3+kL^2/(L^2+h^2)`
c.
Effective Stiffness=K
Since the members are in series
`1/K=1/k_1+1/k_2`
The straight beam is considered as `k_1`
The inclined one is considered as `k_2`
`k_1=48EI/L^3`
Since the member for `k_2` is inclined
Force at end = mass/2
`k_2/2=2kcos^2 theta`
`k_2=4kh^2/(L^2+h^2)`
`1/K=L^3/(48EI)+(L^2+h^2)/(4kh^2)`