Calculation of Stiffness in Structural elements

Q1

Aim:- To calculate the lateral stiffness of the frame shown

step 1:- consider unit deformation at u1

• stiffness coefficient of the column is 12EI/h^3
• stiffness coefficient of the beam rotation is EI/h^3

step 2:- consider unit deformation at u2

• stiffness coefficient of the column at u2 is 6EI/h^2
• stiffness coefficient at u2 because of the beam rotation and column rotation is 4EI/h + 4EI/ L

step 3:- consider unit deformation at u3

• stiffness coefficient of the column at u3 is 6EI/h^2
• stiffness coefficient at u2 because of the beam rotation and column rotation is 4EI/h + 4EI/ L

• write the stiffness matrix

• then we have fs=k.u

after solving the matrix form and equations stiffness can be determined

Q2

Aim:- To determine the degree of freedom, establish the equation of motion and calculate natural frequencies

(A)

• from the frame given it can be inferred by the observation that the frame has a single degree of freedom i.e., horizontal deformation
• the given column is hinged at the bottom
• so the stiffness of the column

K= 3EI/ `h^3`

• EQUATION OF MOTION

mu+fd+fs=p(t) neglect fd since there is no dynamic element

mu+fs=p(t)

mu+(3EI/ `h^3)u =p(t)

• NATURAL FREQUENCY

=`√ K/M

=``√3EI/ `h^3M

(B)

• from the frame given it can be inferred by the observation that the frame has a single degree of freedom i.e., horizontal deformation
• given 2 columns are fixed at the end and 1 column is hinged
• since the translation in the beam is uniform, the translation in the column will occur simultaneously
• so for parallel columns, equivalent stiffness is Ke= K1+ K2+K3
• so the stiffness of the column

K1= 12EI/ `h^3` K2= 12EI/ `h^3` K= 3EI/ `h^3`

• equivalent stiffness Ke= 27EI/ `h^3`

• EQUATION OF MOTION

mu+fd+fs=p(t) neglect fd since there is no dynamic element

mu+fs=p(t)

mu+(27EI/ `h^3)u =p(t)

• NATURAL FREQUENCY

=`√ K/M

=``√27EI/ `h^3M

Q3

Aim:- to determine the natural period of the system given below

• stiffness of the horizontal member with length Lb is Kb=48EI/ `Lb^3
• stiffness of the lateral member with length Lc is Kc= 3EI/ `Lc^3

Equivalent stiffness Ke= Kb+ Kc

Ke= 48EI/ `Lb^3 +3EI/ `Lc^3

• Natural frequency od the system = √ K/M
• time period of the system = 2`π`/ `ω`

Q4

Aim:- to write the effective stiffness of the frames shown

(A)

• stiffness of the beam is Kb=3EI/ `L^3
• stiffness of the spring is given as k

since the members are in series equivalent stiffness id is written as

1/Ke = 1/Kb +1/K

1/Ke=1/3EI/ `L^3 +1/K

(B)

• the beam is rigid because the flexural rigidity is given as infinity
• the stiffness of the column elements is 3EI/ `L^3
• the stiffness of the inclined spring is k/cos^2 θ

equivalent or effective stiffness = Kc+ K

Ke= 3EI/ `L^3 + 3EI/ `L^3 + k/cos^2 θ

Ke= 6EI/ `L^3 + k/cos^2 θ

(C)

• stiffness of the beam which is pinned at both ends is given as Kb=48EI/ `L^3
• stiffness of the inclined strut is derived from the equation

stiffness = load/deformation

since the elements are connected in series

equivalent stiffness= 1/Ke = 1/Kb +1/K