Calculation of Stiffness in Structural elements
Question 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 :
System has degree of freedom 3 that is one lateral displacement and 2 joint rotations
Elements Ki1 of the stiffness matrix are determined imposing displacements U1=1, U2=0, U3=0
Elements Ki2 of the stiffness matrix are determined imposing displacements U1=1, U2=1, U3=0
Elements Ki3 of the stiffness matrix are determined imposing displacements U1=1, U2=0, U3=1
The equilibrium equations are written in 3 X 3 stiffness matrix of the structure with Ib = Ic which is subjected to lateral force Fs is written as
EIc/`h^3``[[24, 6h, 6h] , [6h, 6h^2, h^2] , [6h, h^2, 6h^2]]` `[[U1] , [U2] , [U3]]` = `[[Fs] , [0] , [0]]`
Joint rotations are expressed in terms of lateral displacement :
`[[U2] , [U3]]` = -`[[ 6h^2, h^2] , [h^2, 6h^2]]^-1` `[[6h] , [6h]] *U1` = -6/7h `[[1] , [1]] *U1`
Substituting above equation in the 1st row of the matrix
Fs = ((24EIc/`h^3`) - (EIc/`h^3`) * (6/7h) * `[[6h, 6h]]`*`[[1],[1]]`) U1 = 96/7 (EIc / `h^3`) U1
Lateral stiffness of the frame k = 96/7 (EIc/`h^3`)
Question 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)
i) Degree of freedom
DOF = 4 ( 3 rotations and 1 displacement)
DOF = 3J - (R+M)
= 3(3) -(3+2) = 9 - 5 = 4
ii) Equation of motion
`m(ddot u)` + k(u) = u(t)
Where k = `(48EI)/h^3`
iii) Natural frequencies
Frequency = 1/Time period
Time period = 2`pi`/`omega`
`omega` = `sqrt (K/m)`
Where k = `(48EI)/h^3`
b)
i) Degree of freedom
DOF = 6 ( 3 rotations and 3 displacements)
DOF = 3J - (R+M)
= 3(6) - (8 + 4) = 18 - 12 = 6
ii) Equation of motion
`m(ddot u)` + k( u ) = u ( t )`
where k = `(12E(I+I))/h^3` + `(48 EI)/h^3`
iii) Natural frequencies
Frequency = 1/Time period
Time period = 2`pi`/`omega`
`omega` = `sqrt (K/m)`
Frequency = `sqrt (K /m)` x `1/ (2pi)`
K `= (12E(I+I))/h^3` + `(48 EI)/h^3`
Question 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 :
Stiffness of propped cantilever beam is 4EI/L
Time period = `(2pi) / omega`
`omega` = `sqrt (k/m)`
Time period = `(2pi) / sqrt(k/m)`
= `(2pi) / (sqrt(((4EI)/L)/(m)))`
Question 4:
Determine an expression for the effective stiffness of the following systems:
Solution :
a)
`m(ddot u)` + `K ( bar u)` = mg + u(t)
`bar u` = `delta` spring + `delta` beam
F = K (`bar u`)
Ks `delta` spring = Kb `delta` beam
Ks = K spring and Kb = K beam
`(F) / K = ((Fs)/(Kb)) + (( Fs)/(Ks))`
`K = (KsKb ) / (Ks + Kb)`
b)
`m(ddot u)` + `K ( bar u)` = u(t)
`bar u` = `delta` spring + `delta` beam
F = K (`bar u`)
`(F) / (bar U) = ( Fs) / (deltas) + (Fb) / (deltab)`
K = Ks + Kb
c)
`m (ddot u) + K (baru) = mg + u(t)`
F = K (`bar u`)
F1 = `(AE) / L`U1
F2 = `((48EI) / (L/2)^3) U2`
`F/K = ((F1)/((AE)/L ) + (F2)/((48EI)/(0.5L))^3)`