Calculation of Stiffness in Structural elements
1)Aim : To compute lateral stiffness of the one storey frame with an intermediate realistic stiffness of the beam.
Answer:
Case 1: . Unit displacement u1=1
Stiffness coefficient of Column translation (2 SIMILAR) 2 x `frac(12EI)(h^3)`
Stiffness coefficient of Beam rotation (2 ends) `frac(EI)(h^2)`
Case 2: . Unit rotation u2=1
. Stiffness coefficient of Column rotation at right end = `frac(6EI)(h^2)`
. Stiffness coefficient of Column + Beam rotation at left end = `frac(4EI)(h)+frac(4EI)(L)`
= `frac(4EI)(h)+frac(4EI)(2h)=frac(6EI)(h)`
Case 3: . Unit rotation u3=1
. Stiffness coefficient of Column rotation at left end = `frac(6EI)(h^2)`
. Stiffness coefficient of Column + Beam rotation at right end = `frac(4EI)(h)+frac(4EI)(L)`
= `frac(4EI)(h)+frac(4EI)(2h)=frac(6EI)(h)`
. fs = k x u
. fs = `frac(24EIc)(h^3)-frac(EIc)(h^3)frac(6)(7h)(6h,6h )[(1),(1)]u1=frac(96)(7)frac(EIc)(h^3)u1`
. 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 = `frac(96)(7)frac(EIc)(h^3)`
2)Aim: To determine the number of degrees of freedom for dynamic analysis and to establish the equation of motion and also calculate their natural frequencies for the given structures (a) and (b).
Answer: (a)
. Position of the element can be described by 1 coordinate only u(t).
. The system has single degree of freedom structure.
. The column has fixed end at bottom. So the stiffness coefficient will be as follow
k = `frac(3EI)(h^3)`
. The equation of motion shall be M`ddotu`+ku = 0.
. Natural frequency formula `omega= sqrtfrac(k)(m)`
`omega= sqrtfrac(3EI)(Mh^3)`
ANSWER: (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 = frac(12EI)(h^3)` `k2 = frac(12EI)(h^3)` `k3 = frac(3EI)(h^3)`
For parallel stiffness elements Ke = (k1+k2+k3)=`frac(27EI)(h^3)`
. The equation of motion shall be `Mddotu + ku = 0`
. Natural frequency formula `omega = sqrtfrac(k)(m)`
`omega = sqrtfrac(27EI)(Mh^3)`
3) Aim: To determine the natural period of system if a large mass 'M' is placed at the intersection of the beams at Point 'A' from the figure shown below.
Answer:
.Both elements have same material and section. It means they have same modulus of elasticity and second moment of inertia. But the lengths are different.
.When a mass 'M' is applied at the point, both elements will respond simultaneously. It means they are connected in parallel (as depicted in the sketch below)
Effective stiffness of the system Ke = Kb + Kc
`Kb = frac(48EI)(lb^3` `Kc = frac(3EI)(lc^3)`
`Ke = 3EI(frac(1)(lc^3)+frac(16)(lb^3))`
. Natural frequency formula `omega = sqrtfrac(k)(m)`
`omega = sqrtfrac(3EI)*frac(1)(lc^3)+frac(16)(lb^3)`
. Time period `T = frac(2pi)(omega)`
4) Aim: To determine an expression for the effective stiffness of the following systems:
Answer: (a)
. The stiffness coefficient of the beam element is `frac(3EI)(L^3)`
. The stiffness coefficient of the spring is K.
. The stiffness elements are in series. So,
`frac(1)(Ke)=frac(L^3)(3EI)+frac(1)(K)`
`frac(1)(Ke)=frac(KL^3+3EI)(3EIK`
`Ke=frac(3EIK)(KL^3+3EI`
Answer: (b)
. The stiffness coefficient of the both the column elements is `frac(3EI)(h^3)`
. 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'=frac(p)(costheta)`
Inclined component of deformation `'u'=u costheta`
Stiffness coefficient of spring,
`K=frac(p)(u cos^2theta`
`frac(p)(u)=Ks=K cos^2theta`
`costheta=frac(L)(sqrt(L^2+h^2`
`cos^2theta=frac(L^2)(L^2+h^2`
`Ks=Kfrac(L^2)(L^2+h^2`
. The effective stiffness of the system, `Ke=frac(6EI)(h^3)+frac(KL^2)(L^2+h^2)`
Answer:(c)
. The stiffness coefficient of the pinned-pinned beam is `K1=frac(48EI)(L^3`
. The effective stiffness coefficient for the system has to be formed by adding the 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 'm' is half of its right end nodal deformation.
`frac(mg)(2)=Kea,vert * 2 * Delta2`
`frac(mg)(Delta2)=4 * Kea.vert`
`K2=4 Kea,vert`
We assumed vertical axial stiffness in our calculation. But we have inclined axial stiffness in the given system. So we have to resolve the (Kea,vert) into (Kea).
`K2=frac(4 underset(ea)(K) h^2)(L^2+h^2`
. The effective stiffness shall be arrived as, `frac(1)(Kunderset(e)`=`frac(1)(Kunderset(1)`+`frac(1)(Kunderset2` `frac(1)(Kundersete`=`frac(1)((frac(48EI)(L^3))`+`frac(1)((frac(4Kea h^2)(L^2+h^2))`
`frac(1)(Kundersete`=`frac(1)((frac(48EI)(L^3))`+`frac(1)(frac(4EAh^2)((L^2+h^2)^1.5`
`Kunderset(e)`=`(frac(4AEh^2)((L^2+h^2)^1.5))frac(48EI)(L^3)`/`frac(4AEh^2)((L^2+h^2)^1.5`+`frac(48EI)(L^3)`