Calculation of Stiffness in Structural elements

1)

calculation of stiffness coefficients:

step 1:

• taking unit displacement as unit at U1, i.e U1=1
• stiffness coefficient of columns translation, k=2*{(12EI/h^3)}=(24EI)/(h^3)
• stiffness coefficients of beam rotation, k=(EI)/(h^2)

step 2:

• unit rotation U2=1
• stiffness coefficient of column rotation at right end, k=6EI/(h^2)
• stiffness coefiicient of column+beam rotation @ left end={4EI/h}+{4EI/h}=6EI/h

step 3:

• unit rotation @ U3=1
• stiffness coefficient of column @ left end=6EI/(h^2)
• stiffness coefficient of column+beam rotation @ right end={4EI/h}+{4EI/h}=6EI/h
• stiffness (K)

=

=

• F_s=K.U

by static condensation method,we are reducing the 3 unknown expressions into 2 unknown equation

=

• F_s=K.U==

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 follows

k=(96/7)(EI/h^3)

2)

a)

• position of element can be described by only one coordinate u(t)
• it has single degree of freedom
• the column is fixed at right end. so, the stiffness is as follows

k=(3EI)/(h^3)

• equation of motion shall be
• natural frequency

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
• it has single degree of freedom
• the two columns are fixed. so, the stiffness coefficient is as follows

K1=(12EI)/(h^3)

K2=(12EI)/(h^3)

• one column is pinned fixed end.so, the stiffness coefficient is as follows

K3=(3EI)/(h^3)

• columns looks like parallel.so, for parallel stiffness elements Ke=(K1=+K2+K3)=(27EI)/(h^3)
• equation of motion will be

• natural frequency

3)

• both elements are same in material, section, modulus of elasticity and second moment of inertia and different in lengths
• when load applies at the point, both will react. so they are in parallel as represented in below figure

• stiffness along length Lc

Kc=(3EI)/(Lc^3)

• stiffness along length Lb

Kb=(48EI)/(Lb^3)

• effective stiffness

Ke=Kb+Kc

Ke=3EI((1/Lc^3)+(16/Lb^3))

• natural frequency formula ω=`squaroot`of k/m

ω='squaroot' of [{3EI((1/Lc^3)+(16/Lb^3)}/{m}]

• time period t=[2`pi`]/[`omega`]

4)

a)

• stiffness coefficient of beam element is 3EI/L^3
• stiffness coefficient of the spring is K
• stiffness elements are in series. so,

b)

• stiffness coefficient of the both the column elements is 2*[3EI/h^3]
• stiffnes coefficient of the spring is K, but it is inclined. so, the horizontal force(P) and deformation (u) has to be resolved into inclined components to arrive coefficient of the spring

• inclined component of load 'p'=p/cosθ

inclined component of deformation'u'=u cosθ

stiffness coefficient of spring,

k=p/u

k=p/u cos^2 θ

[p/u]=Ks=K cos^2 θ

cosθ=L/('squaroot' of L^2 +h^2)

cos^2 θ=L^2/(L^2+h^2)

Ks=K*[L^2/(L^2+h^2)]

• effective stiffness of the system

Ke={6EI/h^3}+{KL^2/(L^2+h^2)}

c)

• stiffness coefficient of pinned-pinned beam is K1=48EI/L^3
• effective stiffness=relative axial stiffness provided by the inclined strut connected at 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*2*△2

K2=4*Kea,vert

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

K2={(4Kea*h^2)/(L^2+h^2)}

• the effective stiffness shall be arrived as, (1/Ke)=(1/K1)+(1/K2)