Calculation of Stiffness in Structural elements

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  

 

Answer: 

assuming unit value as a 1

so, in 1st case,

unit diaplacement u1=1

 

stiffness coefficient of column translation 

        2X12EI/h3.....(2 similar) 

 

stiffness coefficient of beam rotation

   EI/h2.......(two ends)

 

2nd case

unit rotation u2 =1

 

stiffness coefficient of column rotation at right end 

6EI/h2

 

stiffness coefficient of column + beam rotation at left end

4EI/h + 4EI/L

 

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

 

3rd case 

unit rotation u3=3 

 

stiffness coefficient of column rotation at left end 

6EI/h2

 

stiffness coefficient of column + beam rotation at right end 

4EI/h + 4EI/L

4EI/h + 4EI/2h  = 6EI/h

 

[24EIC/h3   6EIc/h2   6EIc/h2

 6EIC/h2     6EIC/h    EIc/h

 6EIc/h2     EIc/h       6Eic/h ]

 

Fs = k.u

 

EIc/h3  [24     6h        6h     { u1}          { fs }

             6h     6h2      h2     { u2 }  =     { 0 }

             6h     h2       6h2]   { u3 }         { 0 }

 

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

{ u2 }        [ 6h2    h2   ] -1  [6h]                    [1]

            =                                    u1 = - 6/7h          u1

{ u3 }        [  h2     6h2 ]      [6h]                    [1]

 

 

 

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

 

Answer: 

position of the element can be described by one coordinate only 

the system has single degree of freedom structure

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

 K= 3EI/h3

 

the equation of motion shall be Mu + ku = 0

 

natural frequency formula i =  `under root k/m`  under root 3EI/mh3

 

 

 

 

Answer: 

position of the element can be described by one coordinate only

because beam with infinite stiffness will induced uniform translation in all columns 

the system has single degree of freedom structure 

two columns with both end fixed and one column has pinned fixed ends, so the stiffness coefficient will be as follow 

 

k1 = 12EI/h3       k2= 12Ei/h3      k3= 3EI/h3

 

for parallel stiffness elements Ke = ( K1 + K2 + K3 ) = 27EI/h3

 

the equation of motion shall be Mu + Ku = 0

 

natural frequency formula (i) = under root of k/m 

    i = under root of 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.

 

Answer : 

Both elements have same materials ans section it means they have same modulus of elasticity and second moment of inertia. But the length are diffrent 

When mass M is applied at the point both elements will respond simultaneopusly it means they are connected in parallel ( as depiicted in the sketch below ) 

 

Effective stiffness of the system Ke= Kb + Kc 

     Kb = 48EI/tb3             

     Kc = 3EI/ Ic3

    Ke = 3EI ( 1/Ic3 + 16/Ib3 ) 

 

natural frequency formula (i) = under root of k/m 

     i = under root of [ 3EI(1/Ic3 +16/Ib3)/ M ]

 

Time period T = 2`pi`/(i) 

 

 

 

Question 4

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

Answer 

A)

the stiffness coefficient of the beam elements is 3EI/L3

the stiffness coefficient of the spring is K 

the stiffness elements are in series so, 

 

1/Ke = L3/3EI + 1/K

1/Ke = KL3+3EI/3EIK

Ke = 3EIK/KL3+3EI

 

B) 

the stiffness of coefficient of the both the column elements is 3EI/h3

 

The stiffness coefficient of the spring is K, but is inclined. so the horizontal force (P) and deformation (u) has to be resolved into inclined components to arrive actual coefficient of the spring 

 

Inclined component of load P = p/cos0

 

Inclined component of deformation u = u cos0

 

stiffness coefficient of spring 

K= p/ucos20

 

p/u = Ks = Kcos20

 

cos0 = L/ under root of L2 + h2

 

cos20 = L2/L2 + h2 

 

Ks = K L2/L2 + h

 

the effective stiffness of the system, Ke = 6EI/h3 + KL2/L2+h2

 

 

 

C)

The stiffness coefficient of the pinned-pinned beam is K1= 48EI/L3

 

the effective stiffness coefficient for the system has to be formed by adding the relative axial stiffness provided by the inclined strut connected to the right end 

 

force acting at the right end is mg/2 and deformation caused by the strut at the point of m is half of its right end nodal deformation 

 

mg/2 = Kea.vert X 2 X ^2

 

mg/^2 = 4 Kea.vert 

 

K2 = 4 Kea.vert 

 

we assumed vertical axial stiffness in our calculation but we have an inclined axial stiffness in the given system. so we have to resolve the (Kea.vert) into (Kea)

 

K2 = 4Keah2/L2+h2

 

the effective stiffness shall be arrived as 1/Ke = 1/k1 + 1/k2

 

1/Ke = 1/(48Ei/L3)  + 1/(4Keah2/L2+h2)

 

1/Ke = 1/(48EI/L3) + 1/(4EAh2/(L2+h2)1.5)

 

Ke = {4 AEH2/(L2+h2)3/2) 48EI/L3} / {4EAH2/(L2+h2)3/2 + 48EI/L3