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 El_b = El_c

STEP 1:

• unit displacement u1 = 1
• stiffnesss coefiicient of column translation (2 similar) = 2 * 12EI / h^3
• stifness coefficient of beam rotation (2ends) = EI / h^2

STEP 2:

• unit displacement u2 = 1
• stifness coefficient of column rotation @ right end = 6EI / h^2
• stiffness coefficient of column + beam rotation @ left end = 4EI/h + 4EI/L
• 4EI/h + 4EI/2h = 6EI/h

STEP 3:

• unit displacement u2 = 1
• stifness coefficient of column rotation @ right end = 6EI / h^2
• stiffness coefficient of column + beam rotation @ right end = 4EI/h + 4EI/L
• 4EI/h + 4EI/2h = 6EI/h

by applying the static condensation method, we are deducing the 3 unknoen expression into 2 unknown equation,

stifness amount of force required for unit displacement so inter changing the translation u1 to the left side, we arrive the stiffness coefficient as follows,

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

STEPS :

• position of the elemnt can be described by 1 coordinate omly u(t)
• the system has single degree of freedom structure
• the column has fixed end at bottom, so the stiffness coefficient will be as follows,
• K = 3EI/h^2
• The equation of motion shall be Mu + Ku = 0
• Natural frequency formula

STEPS :

• Position of the element can be described by 1 coordinate only u(t) because beam with infinite stiffness will induce uniform trnaslation columns
• the system has single degree of freedom structure.
• two columns with both end fixed & one column has pinned fixed end the stifness coefficient will be as follows ,
• 

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.

STEPS :

• Both elements have same material and section , it means they have same modolus of elasticity and second moment of inertia . but the lengths are different
• when a mass M is applied at the point both elemnts will respond simultaneously
• It means they are connected in parallel
• 

Question 4

Determine an expression for the effective stiffness of the following systems:

STEP 1:

• An expression for the effective stiffness of fig (a) is
• K = 3EI / L^3

STEP 2:

• An expression for the effective stiffness of fig (b) is
• K = 3EI/L^3 + 3EI/L^3 = 6EI/L^3

STEP 3 :

• An expression for the effective stiffness of fig (c) is
• K = 12EI/L^3 + 12EI/L^3 = 24EI/L^3