Calculation of Stiffness in Structural elements
ANS.1.
Given ,
System has 3 DOF's
also,
L = 2h
ELb = ELc
Let us assume at each case, 1 Degree of freedom as unit value.
Case 1:-
Unit displacement u1 = 1
So,
The stiffness coefficient for column translation will be = `frac{12EI}{h^3}`
So for both the columns it will be multiplied by 2
Hence
Stiffness Coefficient for both column translation will be = 2 x `frac{12EI}{h^3}`
Stiffness Coefficient for beam rotation and both the ends will be `frac{EI}{h^2}`
Case 2:-
Unit rotation u2 = 1
So here,
Stiffness coefficient for column rotation at right end will be = `frac{6EI}{h^2}`
And
Stiffness Coefficient for column + Beam rotation at left end = `frac{4EI}{h}` + `frac{4EI}{L}`
As we have L = 2h
So equation will become
`frac{4EI}{h}` + `frac{4EI}{2h}` =`frac{6EI}{h}`
Case 3 :-
Unit rotation u3 = 1
Stiffness coefficient for column rotation at left end will be = `frac{6EI}{h^2}`
and
Stiffness coefficient of column and beam rotation at right End will be = `frac{4EI}{h}` + `frac{4EI}{L}`
As we have L = 2h
So equation will become
`frac{4EI}{h}` + `frac{4EI}{2h}` =`frac{6EI}{h}`
So the matrix equation will become like this
`[ [frac{24EIc}{h^3} , frac{6EIc}{h^2} , frac{6EIc}{h^2} ] , [frac{6EIc}{h^2} , frac{6EIc}{h} , frac{EIc}{h} ] , [ frac{6EIc}{h^2} , frac{EIc}{h} , frac{6EIc}{h} ] ]`
Hence we know
Fs = k.u ( eqn.1.)
so,
`( (fs) , (0) ,(0) )` = `frac{EIc}h^3` `[ [ 24 , 6h , 6h ] , [ 6h , 6h^2 , h^2 ] , [ 6h , h^2 , 6h^2 ] ]` `( (u1),(u2),(u3) )`
So now using the static condensation method to reduce the Shri unknown expression into two unknown expressions
Hence,
`( (u2) , ( u3) )` = - `[ [6h^2 , h^2] , [ h^2 , 6h^2 ] ]^-1` `[ [6h] ,[6h] ]` u1
= `- frac{6}{7h} [ [1] , [1] ]` u1
u2 = -6u1
u3 = - 7h x u1
Now placing the values in equation 1
we get ,
Fs = `(frac{24EIc}{h^3} - frac{EIc}{h^3} frac{6}{7h} [[6h , 6h ]][[1],[1] ]) u1`
Fs = `frac{96}{7} frac {EIc}{h^3} u1`
Hence we know that
for any unit displacement, stiffness will be equal to Fs
Or we can say that
Fs = k x u1
k = Fs/ u1
as u1 = 1 , so
k= Fs
Hence, Stiffness will be
k = `frac{96}{7} frac {EIc}{h^3} `
ANS.2.
2(a) :-
As given in figure,
The moment of inertia is infinite as the horizontal beam is having infinite moment of inertia and it is resting on a roller support therefore there will be no restrains in horizontal direction
the position of the beam can be described by using one coordinate only that is u(t)
As the given system has no restrains in horizontal direction so the degree of freedom will be of only single degree
We can see that the column has the fixed and at the bottom
So the stiffness coefficient will be equal to
`k = frac{3EI}{h^3}`
As there are no damping elements so the equation of motion will become
`M overset(°°)(u) + ku = 0`
Therefore the natural frequency will be
we know
Natural frequency = `omega = √k/m`
So placing the value of k, we get natural frequency as
`omega = √frac{3Ei}{Mh^3}`
2(b)
In this question also the horizontal beam has infinite moment of inertia and is resting on a support where two ends are fixed and one is simply supported, therefore there will be no restrains in horizontal direction.
The position of the beam can be described by using one coordinate only that is u(t) as beam having infinite stiffness will induce uniform translation in all directions.
Hence,it is a single degree of freedom structure.
As per given figure the stiffness will be ,
For `k1` ,
Stiffness coefficient will be `k1 = frac(12EI)(h^3)`
For `k2` ,
Stiffness coefficient will be `k2 = frac(12EI)(h^3)`
For `k3` ,
Stiffness coefficient will be `k3 = frac(3EI)(h^3)` as it is pinned
Now,
For parallel stiffness elements
`Ke = k1 + k2 + k3`
So,
`Ke = frac(27EI)(h^3)`
As there are no damping elements so the equation of motion will become
`M overset(°°)(u) + ku = 0`
Therefore the natural frequency will be
we know
Natural frequency = `omega = √k/m`
So placing the value of k, we get natural frequency as
`omega = √frac{27EI}{Mh^3}`
ANS.3.
Given,
Length of fixed component acting as column = Lc
Length of component acting as beam = Lb
hence, the effective stiffness for the system will be
Ke = kb + kc
For kb ,
`kb = frac(48EI)((lb)^3)`
For kc,
`kc = frac(3EI)((lc)^3)`
Now,
Ke = `frac(48EI)((lb)^3)` +
`frac(3EI)((lc)^3)`
Ke = `3EI(frac(1)((lc)^3) + frac(16)((lb)^3 ) )`
Hence natural frequency will be
Natural frequency =`omega = frac(3EI(frac(1)((lc)^3) + frac(16)((lb)^3 )))(M)`
Also, time period will be `T = frac(2π)(omega)`
Hence by putting the value of Omega we can get the time period also.
ANS.4.
4(a).
As given figure has one end fixed and the load will be transfered from other end.
so ,
Stiffness coefficient for beam element will be = `frac(3EI)(L^3)`
Now, stiffness Coefficient for spring given is K.
The given figure is in a series,so the stiffness Coefficient will be in a series.
Therefore,
`frac(1)(Ke) = frac(L^3)(3EI) + frac(1)(K)`
`frac(1)(Ke) = frac(Kl^3 + 3EI)(3EIK)`
`Ke = frac(3EIK)(Kl^3 + 3EI)` = ans.
4(b).
From the given figure,
Stiffness Coefficient for column will be `frac(3EI)(h^3)`
Stiffness Coefficient for spring
As we can see that the spring is attached to inclined position and for that we have to resolve the components for the spring to find out the stiffness of the spring.
So,
let the load be P acting there and deformation be u
for inclined component of load P = `frac(P)(costheta)`
for inclined component of deformation u = `ucostheta`
So,
Stiffness Coefficient for spring will be `K = frac(P)(ucos^2theta)`
Hence,
`K = frac(P)(ucos^2theta)`
`frac(P)(u) = Ks`
where,
`Ks = Kcos^2theta`
also we know,
`costheta= frac(L)(√L^2+ h^2)`
So `cos^2theta = frac(L^2)(L^2+h^2)`
putting the value of `cos^2theta` in value of Ks
therefore,
`Ks = Kfrac(L^2)(L^2+ h^2)`
Hence the effective stiffness for the system will be
Ke = `frac(3EI)(h^3)` + `frac(3EI)(h^3)` + Kfrac(L^2)(L^2+ h^2)`
Ke = `frac(6EI)(h^3) + frac(KL^2)(L^2 +H^2 )` = ans.
4(c).
As per given figure,
Stiffness coefficient for pinned beam will be `k1 = frac(48EI)(L^3)`
Also force acting at right end is `frac(mg)(2)`
Deformation caused by strut at point m will be half of it's right end nodal deformation.
So, `frac(mg)(2) = K(ea,vert) × 2 × Delta2`
`frac(mg)(Delta2) = 4K(ea,vert)`
`K2 = 4 K(ea,vert)`
Now,resolving K(ea,vert) into K(ea)
`K2 = frac(4k(ea)h^2)(L^2+h^2)`
So, effective stiffness will be
`frac(1)(Ke) = frac(1)(K1) + frac(1)(K2)`
`frac(1)(Ke) = frac(1)(frac(48EI)(L^3)) + frac(1)(frac(4k(ea)h^2)(L^2+h^2))`
`frac(1)(Ke) = frac(1)(frac(48EI)(L^3)) + frac(1)(frac(4EAh^2)((L^2+h^2)^1.5))``Ke = frac((frac(4EAh^2)((L^2+h^2)^1.5)) × frac(48EI)(L^3))((frac(4EAh^2)((L^2+h^2)^1.5))+ frac(48EI)(L^3)` = ans.