Calculation of Stiffness in Structural elements

1.AIM

To Compute lateral stiffness of the one story frame with an intermediate realistic stiffness of the beam. The system has 3 Degree of freedom as shown. Assume L = 2h and Elb = Elc

SOLUTION

• By using Moment distribution method
• Given the Degree of freedom is 3

Now,

• The Unit Displacement U1=1, Where U2=U3
• Stiffness coefficient of Column translation (2 similar)

2 x `frac{12EI}{h^3}`

• Stiffness coefficient of Beam rotation (2 ends)

`frac{EI}{h^2}`

Now,

• Unit rotation U2 = 1

• Stiffness coefficient of Column rotation at right end is

`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}`

Now,

• 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}`

Therefore,

`[[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}]]`

`fs=k*u`

`frac{ELc}{h^3}[[24,6h,6h],[6h,6h^2,h^2],[6h,h^2,6h^2]]{[u1],[u2],[u3]}={[fs],[0],[0]}`

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

`{[u2],[u3]}=-[[6h^2,h^2],[h^2,6h^2]]^-1 [[6h],[6h]]u1=-frac{6}{7h}[[1],[1]]u1`

`fs=(frac{24EIc}{h^3} -frac{EIc}{h^3}frac{6}{7h}(:6h *6h:)[[1],[1]]u1)=frac{96EIc}{7h^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{96EIc}{7h^3}`

2.AIM

• Determine the number of degrees-of-freedom for dynamic analysis
• Establish the equation of motion
• Calculate their natural frequencies

SOLUTION

a]Ans

• 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, `Mddotu+ku=0` ......(eq-1)`

• The column has fixed end at bottom. So the stiffness coefficient will be as follow `k=frac{3EI}{h^3}`

Now, we have to substitute the K in the (eq-1)

• Equation of Motion, `Mddotu+ku=0`

`Mddotu+frac{3EI}{h^3} u=0`

`Mddotu+frac{3EI}{h^3} (1)=0`

• Natural frequency formula, `omega=sqrtfrac{k}{m}` .....(eq-2)

Where, k = stifness

m = mass

• by substituting the value of k in (eq-2), we get

Thus, the Natural Frequency, `omega=sqrtfrac{3EI}{Mh^3}`

SOLUTION

b]Ans

• 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
• Stiffness of the 1st column, `k1=frac{12EI}{h^3}`
• Stiffness of the 1st column, `k1=frac{12EI}{h^3}`
• Stiffness of the 1st column, `k1=frac{3EI}{h^3}`
• For parallel stiffness element, `ke=(k1+k2+k3)=(frac{12EI}{h^3}+frac{12EI}{h^3}+frac{3EI}{h^3})=frac{27EI}{h^3}`
• the Equation of motion is, `Mddotu+ku=0`

Now, we have to substitute the 'K' in above equation

• Equation of Motion, `Mddotu+ku=0`
• `Mddotu+frac{27EI}{h^3} u=0`
• Natural frequency formula, `omega=sqrtfrac{k}{m}`

Where, k = stifness

m = mass

• by substituting the value of k in `omega`, we get

Thus, the Natural Frequency, `omega=sqrtfrac{27EI}{Mh^3}`

3.AIM

To Determine the natural period of this system if a large mass, M, is placed at the intersection of the beams at point A

SOLUTION

• 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
• The two beams contribute to the overall stiffness of the system in a parallel manner.
• This implies that the systems stiffness can be found from the direct addition of the stiffnesses of the individual beams.

Lateral stiffness of simply supported beam, `kb=frac{48EI}{lb^3}`

Lateral stiffness of cantilever beam, `kc=frac{3EI}{lc^3}`

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

therefore, `k=frac{48EI}{lb^3}+frac{3EI}{lc^3}=3EI{frac{16}{lb^3}+frac{1}{lc^3}}`

• Natural frequency formula, `omega=sqrtfrac{k}{m}`

Where, k = stifness

m = mass

Now, subtitute k value in `omega`

`omega=sqrtfrac{3EI(frac{16}{lb^3}+frac{1}{lc^3})}{M}`

• The natural period of the system can be found from this natural frequency using the expression, `Tn=frac{2pi}{omega}`
• `Tn=frac{2pi}{sqrtfrac{3EI(frac{16}{lb^3}+frac{1}{lc^3})}{M}}`
• `Tn=frac{2pi*sqrtM}{sqrt{3EI(frac{16}{lb^3}+frac{1}{lc^3})`

4.AIM

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

SOLUTION

a.Ans

• 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}`

b.Ans

• 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}=K*cos^2theta`

where, `frac{p}{u}=ks`

therefor, above equation became as, `ks=K*cos^2theta`

now, find the value of `cos^2theta`

`costheta=frac{L}{sqrt(L^2+h^2)}`

`Cos^2theta=frac{L^2}{L^2+h^2}`

substitute value of `cos^2theta` in `ks`, the `ks` equation becomes

`ks=K*frac{L^2}{L^2+h^2}`

• The effective stiffness of the system, `Ks=frac{6EI}{h^3}+frac{KL^2}{L^2+h^2}`

c.Ans

• 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 `frac{mg}{2}` and deformation caused by the strut at the point of '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 an inclined axial stiffness in the given system. So we have to resolve the (kea,vert) into (Kea).

`k2=frac{4*kea*h^2}{L^2+h^2}`

• The effective stiffness shall be arrived as,

`frac{1}{ke}=frac{1}{k1}+frac{1}{k2}`

`frac{1}{ke}=frac{1}{frac{48EI}{L^3}}+frac{1}{frac{4*kea*h^2}{L^2+h^2}}`

`frac{1}{ke}=frac{1}{frac{48EI}{L^3}}+frac{1}{frac{4*EA*h^2}({l^2+h^2}^frac{3}{2})}`

`ke=frac{(frac(4*AE*h^2)((L^2+h^2)^frac{3}{2}})*frac{48EI}{L^3}){frac{4*AE*h^2}({L^2+h^2}^frac{3}{2})+frac{48EI}{L^3}`