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

Lateral stiffness:

The ability of a body to resist lateral deflection when a lateral force is applied. This is also called as Storey Stiffness, wherein the lateral deflection is storey drift and lateral force is storey shear.

Solution:

Given deg of freedom is 3

Now,

Unit displacement U1=1

Stiffness coefficient of column translation (2 similar)

2 x12EI/h^3

Stiffness coefficient of Beam rotation (2 ends)

EI/h^2

Now ,

Unit Rotation U2=1

Stiffness coefficient of column rotation at right end

6EI/h^2

Stiffness coefficient of Column+Beam rotation at left end

4EI/h+4EI/2h=6EI/h

Now ,

Unit Rotation U3=1

Stiffness coefficient of column rotation at left end

6EI/h^2

Stiffness coefficient of Column+Beam rotation at right end

4EIh+4EIL">4EI/h+4EI/L

4EIh+4EI2h=6EIh"> 4EI/h+4EI/2h=6EI/h

Therefore,

24EIc/h^3 6EIc/h^2 6EIc?h^2

6EIc/h^2 6EIc/h EIc/h

6EIc/h^2 EIc/h 6EIc/h

fs=k⋅u">fs=k⋅u

fs=k⋅u" 6ELc/h^3* 24 6h 6h u1 fs

fs=k⋅u"> 6h 6h^2 h^2 u2 = 0

fs=k⋅u" 6h h^2 6h^2 u3 0

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

u2

= - * 6h^2 h^2 ^-1 6h -6/7h 1 u1

u3 h^2 6h^2 6h u1 = 1

fs=24EIc/h^3 -EIc/h^3 6/7h<6h.6h> 1 u1 =96EIc/7h^3 u1

1

Stiffness is the amount of force required for unit displacement. So interchanging the translation u1 to the left side, we arrive the stiffness coefficient as follow.

k=96EIc/7h^3

RESULT:

Stiffness coefficient k=96EIc/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

AIM:

Determine the number of degrees-of-freedom for dynamic analysis

Establish the equation of motion

Calculate their natural frequencies

SOLUTION:

a ) Position of the element can be described by 1 coordinate only u(t)

The system has single degree of freedom structure.

..

The Equation of motion is, Mu..+ku=0">Mu+ku=0

The column has fixed end at bottom. So the stiffness coefficient will be as follow

k=3EI/h^3

Applying the value of K in above equation

Mu+3EI/h^3(1)=0

The Natural Frequency, ω=√3EI/Mh^3

b) Position of the element can be described by 1 coordinate only u(t) Because Beam with infinite stiffness will induce uniform translation in all columns

The system has single degree of freedom structure

Two columns with both end fixed & One column has pinned-fixed ends. So the stiffness coefficient will be as follow

k1=12EI/h^3 K2=12EI/h^3 K3=3EI/h^3

For parallel stiffness calulating the sum of K1,K2,K3

Ke=(K1+K2+K3)= (12EI/h^3+12EI/h^3+3EI/h^3)

Ke=27EI/h^3

Natural frequency formula, ω=√k/m

Where, k = stiffness

m = mass

substitue the value of k in ω

ω=√27EI/Mh3

RESULT:

ω=√27EI/Mh3

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.

AIM:

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:

we need to add the lateral stiffness of both cantilever beam and simply supported beam, k=kb+kck=kb+kc

therefore, k=48EI/lb^3+3EI/lc^3=3EI{16/lb3+1/lc3}

Natural frequency formula, ω=√k/m

Where, k = stifness

m = mass

Now, subtitute k value in ω =sq of 3EI(16/lb^3+1/lc^3)/M

RESULT:

ω =sq of 3EI(16/lb^3+1/lc^3)/M

Question 4

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

AIM:

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

Solution:

a) The stiffness coefficient of the beam element is 3EI/L^3

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

1/Ke=KL^3+3EI/3EIK

Ke=3EIK/KL^3+3El

b)

Inclined component of load, P=p/cosθ

Inclined component of deformation, u=u⋅cosθ

Stiffness coefficient of spring, K=p/u.cos^2θ

ks=K⋅L^2+h^2

• The effective stiffness of the system, Ks=6EI/h^3+KL^2/L^2+h^2

c) K2=4Keah^2/L^2+h^2