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AtholAthol J. CarrJ. CarrProfessorProfessor of Civil of Civil EngineeringEngineering

DepartmentDepartment of Civil of Civil andand Natural Natural ResourcesResources EngineeringEngineeringUniversity of Canterbury,University of Canterbury,

Christchurch, Christchurch, NewNew ZealandZealand..

Análisis Sísmico de Edificios de Hormigón Armado.Respuesta Dentro del Rango No Lineal

ACHISINAAsociación Chilena de Sismología e Ingeniería Antisísmica

Santiago de Chile, 2 al 6 de junio de 2008

Section 1 – Degrees of Freedom - Mass representation

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ReferencesClough, R.W. and Penzien, J., Dynamics of Structures, 2nd Edition, McGraw-Hill, 1993.Chopra, A.K., Dynamics of Structures, 2nd Edition, Prentice Hall, 2001.Carr, A.J., Ruaumoko User Manuals, Volume 1, Theory,Volume 2, 2D structuresVolume 3, 3D structuresVolume 4, Manuals for other programs in Ruaumoko suiteVolume 5, Appendices, Hysteresis loop dataDepartment of Civil Engineering, University of Canterbury, 2007Humar, J.L., Dynamics of Structures, Prentice Hall, 1990.Paz, Mario, Structural Dynamics . 3rd Edition, Van Nostrand Reinhold, 1991.Hurty, Walter C. and Rubinstein, Moshe F. Dynamics of Structures, Prentice Hall 1964. Craig, Roy R., Structural Dynamics, Wiley,1981

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Single Degree of Freedom Systems

Single Degree of freedom systems do not normally exist in real life.We live in a three-dimensional world and all mass is distributed resulting in systems that have an infinite number of degrees of freedom.There are, however, instances where a structure may be approximated as a single degree of freedom system.There are, in use in current engineering practice, many single degree of degree of freedom approximate solutions.The study of such systems is an integral step in understanding the responses of more complicated and realistic systems.

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For example:

This water towermay be consideredas a singledegree of freedomsystem when one considersvibration in one horizontaldirection only

3

5

Single Degree of Freedom SystemsThis structure may be simplified by assuming that the column has negligible mass along its length. This is reasonable assuming that the tube is hollow and that the mass of the tube is insignificant when compared with the mass of the water tank and water at the top.We can assume that the vertical displacements are small compared with the horizontal displacements of the tank and we ignore torsional rotations about the column axis.We can assume that the rotational inertia of the tank about the horizontal axes is insignificant when compared to the translational inertia of the tank. We will assume that the water mass in the tank does not rotate as the tank moves sideways. This means that we can consider that the tank is a point massWe will consider only the displacement in one horizontal direction at a time.

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Number of Degrees of Freedom

The number of coordinates required to uniquely define the inertia forcesThe number of coordinates required to uniquely define the accelerationsThe number of coordinates required to uniquely define the displacements of the massesEvery structure with distributed mass has an infinite number of degrees of freedom. The equation of dynamic equilibrium for distributed mass systems may be written as a Partial Differential Equation in space and time.Most engineers do not like solving Partial Differential EquationsFor most structural problems the solutions for Partial Differential Equations are not practical solutions.

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7

Number of Degrees of Freedom for a Beam with Distributed Mass

Given the beam with elastic properties E,A and I and mass ρ per unit length distributed along the x axis the vertical displacement u along the length is described by the partial differential equation

)t,x(P)x

)t,x(u(EI(x)

xt)t,x(u

)x(ρ 2

2

2

2

2

2

=∂

∂∂∂

+∂

∂

In this form the beam has an infinite number of dynamic degrees of freedom.

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Distributed Mass StructuresThe equation shown on the previous slide is approximate as the properties are assumed to be concentrated along the centreline whereas in real systems they are distributed across the cross section.

Connecting the PDE for one member to those of other members that meet at a common joint will be difficult. A transformation to a common coordinate system will be required, the displacements have to be continuous, the slopes (1st derivatives) will have to be the continuous, the moments (-EI times the 2nd derivatives) for all members will have to sum to the applied moment at the joint and the shears (EI times the 3rd derivatives) for all members will have to sum to the applied force.

5

9

Lumped Mass Approximation

The beam is modelled as a mass-less beam and in this example the mass is lumped at the quarter points along the beam.

The extra masses of ρL/8 located at the ends of the member do not move and hence have no effect on the response.

This beam now has only 3 degrees of freedom, we only have to write the equations of equilibrium of the three masses.

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Lumped Stiffness Approximation

Rigid

L/4L/4L/4L/4

ρ

Rotational Spring

The beam is modelled as being composed of 4 rigid links. Each link has a distributed mass of ρ per unit length but the displacements are defined as the vertical displacements of the 3 joints so the beam has only 3 degrees of freedom.

The appropriate rotational spring stiffnesses will be computed byenforcing the same vertical deflections at the joints due to, say,point loads applied at the joints, as would be obtained by applyingthe same point loads to the original flexible beam.

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Limited Displacement Shapes Approximation

Lxn

sina)x(uN

1nn

π= ∑

=

Use a Fourier Series to describe the variation of displacement along the beam

For 3 degrees of freedom this becomes

Lx3

sinaL

x2sina

Lx1

sina)x(u 321π

+π

+π

=

This approach has some advantages and disadvantages. The approachmay need a very large number of terms to model localized load effects.

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This is a Ritz type solution method where the solution is approximatedby a series of specified functions Φ which must satisfy the displacement boundary conditions. This was, in the past, very commonly used for problems in structural dynamics and structuralstability

)x(a)x(u n

N

1nnΦ= ∑

=

In this example the values of Φ would have to be zero at x=0 and x=L.

The success of this approach depends very much on the skill of theengineer in selecting functions that well represent the displacementsof the structure expected for the loading being applied.

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13

Finite Element Modelu3

u4

u1

u2

u1 =1

u2 =1

u3 =1

u4 =1

N1(x)

N2(x)

N3(x)

N4(x)

( )⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

=

4

3

2

1

4321

uuuu

)x(N)x(N)x(N)x(Nxu

The Ni(x) are cubic Hermitian interpolation functions derived from cubic polynomials alongthe length of the member.

The 4 displacements can uniquely define a cubic polynomial in x.

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To obtain the interpolation functions one starts with the cubic polynomial

( )⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

αααα

=

4

3

2

1

32 xxx1xu

The slope is

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

==

4

3

2

1

2

αααα

x3x210dxdu

)x(θ

The end displacements are

[ ]⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

αααα

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

αααα

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

4

3

2

1

4

3

2

1

2

32

4

3

2

1

A

L3L21.LLL1..1....1

uuuu

Inverting the matrix [A] gives {α} as a function of the enddisplacement degrees of freedom

[ ]⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

αααα

−

4

3

2

1

1

4

3

2

1

uuuu

A

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32

1 Lx

2Lx

31)x(N ⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛−=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=

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2 Lx

Lx

2Lx

L)x(N

32

3 Lx

2Lx

3)x(N ⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛−=

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4 Lx

Lx

L)x(N

where

( )⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

=

4

3

2

1

4321

uuuu

)x(N)x(N)x(N)x(Nxu

Substituting for {α} gives

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Differentiating the displacement twice with respect to x to get the curvature along beam

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

==χ

4

3

2

1

''4

''3

''2

''1 2

2

uuuu

)x(N)x(N)x(N)x(Ndx

ud)x(

where⎟⎠⎞

⎜⎝⎛ +=

Lx2

1L6

)x(N 2''1

⎟⎠⎞

⎜⎝⎛ +−=

Lx3

2L2

)x(N''2

⎟⎠⎞

⎜⎝⎛ −=

Lx2

1L6

)x(N 2''3

⎟⎠⎞

⎜⎝⎛ +−=

Lx3

1L2

)x(N''4

The curvature has a linear variation along the beam member.

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From the Principle of Virtual Work it can be shown that the forces acting on the ends of the member can be related to the displacements at the member as

[ ] [ ]⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

=

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

∫4

3

2

1

4

3

2

1

''4

''3

''2

''1

L

0

T''4

''3

''2

''1

4

3

2

1

uuuu

k

uuuu

dxNNNNEINNNN

ffff

where

[ ]

⎥⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−−−

−

−

=

LEI4

LEI6

LEI2

LEI6

LEI6

LEI12

LEI6

LEI12

LEI2

LEI6

LEI4

LEI6

LEI6

LEI12

LEI6

LEI12

k

22

2323

22

2323

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This method also falls within the Ritz family of solution methods butdiffers in that here the functions are specified only over each elementrather than over the whole structure as was required in the classicalRitz method.

This means that the solution functions are easier to specify and mesh refinement is used to increase the number of degrees of freedom along with the number of elements.

The displacement functions selected for the elements must be continuous up to one derivative below the highest derivative that occurs in the strain energy expression. In this case energy is curvature*moment. The curvature is related to the second derivative of u with respect to x so the functions chosen must be continuous up to the first derivative, the slope, from element to element.

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Example using Limited Displacement FunctionsColumn with E,I,ρ constant and p(x,t)=p(t)Let u(x,t) = Φ(x)Y(t)Assume

L2x

cos1)x(π

−=φ

Seems reasonable, no displacement or slope at x=0and no curvature at x=L. Actually a good approximation to the 1st mode

L 2x

sinL 2

)x(x

' ππ=φ=

∂φ∂

L 2x

cosL4

)x(x 2

2''

2

2 ππ=φ=

∂φ∂

Inertia force at a point along the column = mass * acceleration

)t(Y)x(tY

)x(tu

)t,x(f 2

2

2

2

i ρφ−=∂∂

ρφ−=∂∂

ρ−=

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The bending moment at a point along the column =EI*curvature

)t(Y)x(EI)t(Yx

EIxu

EI)t,x(m ''2

2

2

2

φ=∂φ∂

=∂∂

=

Virtual Work = real force*virtual displacement= real moment*virtual curvature= real stress*virtual strain

Total Virtual Work = integral of elements of work throughout system

Principle of Virtual WorkInternal Virtual Work = External Virtual Work

Let the virtual displacement be δu, therefore

Y)x(u Y)x(u Y)x(u '''''' δφ=δδφ=δδφ=δ

Internal virtual work = elastic moments*virtual curvatureExternal virtual work = applied forces*virtual displacement

+inertia forces*virtual displacement

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Therefore

∫ ∫∫ δφρφ−+δφ=δφφL

0

L

0

L

0

'''' dx Y)x()).t(Y)x(( dx Y)x().t,x(pdx Y)x().t(Y)x(EI

Assume δY is not equal to zero then

∫∫∫ φ=φ+φρL

0

L

0

2''L

0

2 dx )x().t,x(p)t(Y dx ))x(EI( )t(Y dx ))x((

Let∫ φρ=L

0

2* dx ))x((M

∫ φ=L

0

2''* dx ))x(EI(K

∫ φ=L

0

* dx )x().t(p)t(p

This is an equivalent single degree of freedom system

)t(p)t(YK)t(YM *** =+

Then

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L227.0)4

23

(Ldx)L2x

cos1(dx)x(M 2L

0

L

0

2* ρ=π

−ρ=π

−ρ=ρφ= ∫∫

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4L

0

222

22

L

0

''*

LEI

04.3LEI

32dx

L2x

cos)4L

EI(dx))x(EI(K =π

=ππ

=φ= ∫∫

)t(p L364.0)2

1(L)t(pdx)L2x

cos1()t(pdx)x()t(p)t(pL

0

L

0

* =π

−=π

−=φ= ∫∫

This result shows that the equivalent one degree of freedom structure could be considered as a cantilever with a stiffness of approximately 3EI/L3

having a mass of 0.227 of the total mass located at the top and a dynamic load of 0.364 of the total load acting as a point load at the top of the cantilever.

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Summary of Generalized Single Degree of Freedom SystemsLet )t(Y)x()t,x(u φ=

2i

N

1ii

L

0

2* mdx ))x()(x(M φ+φρ= ∑∫=

2i

N

1ii

L

0

2* cdx ))x()(x(cC φ+φ= ∑∫=

2i

N

1ii

L

0

2'L

0

2''L

0

2* kdx ))x(GA(x)(dx ))x(EI(x)(dx ))x(k(x)(K φ+φ+φ+φ= ∑∫∫∫=

∫ ∑ φ+φ==

L

0i

N

1ii

* pdx )x().t(p)t(p

Where the mi ci ki pi represent point masses, dampers, springs and forcesThe Φi are the values of Φ at the point where these point masses etc. act.The GA term represents shear stiffness associated with shear deformationand k(x) are distributed Winkler-like springs

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Generalized Single Degree of Freedom Systems – Matrix Systems

[M]{x} [C]{x} [K]{x} {p(t)}+ + =

{x} { }Y(t)= φ

T T T T{ } [M]{ }Y(t) { } [C]{ }Y(t) { } [K]{ }Y(t) { } {p(t)}φ φ + φ φ + φ φ = φ

* * * *M Y(t) C Y(t) K Y(t) P (t)+ + =

Given the equation of dynamic equilibrium for the multi-degree of freedom system

Assume that the displacement varies in the structure as the assumed shape {φ} . This again a Ritz solution, the shape function is known!

Then substituting for {x} and its derivatives and pre-multiplying by {φ}T gives

Which is equivalent to the generalized single degree of freedom system as

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Mass Representation for a Complete Structure

Lumped Mass Models. The mass is lumped at points in the structure and is usually only associated with the displacement degrees of freedom. These are usually in the form of point masses. Note: a point mass has no rotational inertia.

Consistent Mass Models. The mass is associated with the structural members and the inertia forces are consistent with the displacement variations throughout the members. All degrees of freedom in the structure have mass and the form of the mass matrix is similar to that of the stiffness matrix for the structure.

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Lumped Mass Models

In the lumped mass model the mass is lumped at points in the structure.

Not all joints (nodes) in the structure need have mass and in some cases if a rigid floor diaphragm model is assumed then there may be as few as 3 inertia terms associated with each floor, two horizontal translations at the centre of mass and a rotation about the vertical axis. For some modeling arrangements not all translational degrees of freedom at a joint need have the same mass.

The structure mass matrix [M] is a diagonal matrix where the terms associated with the rotational degrees of freedom are usually zero

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Consistent Mass Models

The variation of displacements throughout a member as a function of the joint displacements are specified by the displacement patterns assumed for that element.

The variation of accelerations in the member follow the same patterns and from these local accelerations the local inertia forces are computed.

The equivalent joint forces may be then obtained from Virtual Work relationships in a similar way that the joint forces associated with member applied forces are found.

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Consistent Mass Models

The structure’s mass matrix [M] is obtained by summing the member mass matrices [m] by using the Direct Stiffness Method in the same way that the structure’s stiffness matrix [K] is obtained form the member stiffness matrices [k].

The structure’s mass matrix is usually a sparse symmetric matrix with the same skyline form as the stiffness matrix.

As the inertial and elastic forces follow the same displacement functions, the natural frequencies computed from the consistent mass model are bounded with the true values always being a lower bound. This may lead to a bias in computed time-history responses.

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Lumped Mass Matrix for 2D Beam Member

L um ped

1 . . . . .. 1 . . . .. . 0 . . .LM

2 . . . 1 . .. . . . 1 .. . . . . 0

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥ρ⎡ ⎤ = ⎢ ⎥⎣ ⎦⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

Given a beam with mass ρ per unit length where the mass is assumed to lumped equally as point masses at each end of the beam.

The mass matrix is

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Consistent Mass Matrices

•If a lumped mass model is used, the bounds on the frequencies disappear, some frequencies may be too high, some frequencies may be too low.

•This often gives a more realistic set of time-history results when the structure is subjected to an earthquake excitation.

•If one is carrying out a frequency analysis of the structure then the bounded results associated with the consistent mass model may bemore useful.

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Consistent Mass Matrix for 2D Beam Member (i)u5

u6

u2

u3

u2 =1

u3 =1

u5 =1

u6 =1

N2(x)

N3(x)

N5(x)

N6(x)

u1u4 x

y

u1 =1

u4 =1N4(x)

N1(x)The Ni(x) are linear and cubic Hermitianinterpolation functions derived from linear and cubic polynomials.

Four displacements can uniquely define a cubic polynomial in x and two displacements uniquelydefine a linear polynomial.

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Consistent Mass Matrix for 2D Beam Member (ii)Beam member displacement interpolation

( )( ) [ ]{ }

1

2

x 1 4 3

y 2 3 5 6 4

5

6

uu

u x N (x) . . N (x) . . uN u

u x . N (x) N (x) . N (x) N (x) uuu

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎧ ⎫ ⎡ ⎤⎪ ⎪ ⎪ ⎪= =⎨ ⎬ ⎨ ⎬⎢ ⎥

⎪ ⎪ ⎣ ⎦⎪ ⎪⎩ ⎭⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

where

2 3

2x xN (x) 1 3 2L L

⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2 3

3x x xN (x) L 2L L L

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − +⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠2 3

5x xN (x) 3 2L L

⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2 3

6x xN (x) LL L

⎛ ⎞⎛ ⎞ ⎛ ⎞= − +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

1xN (x) 1L

= −

4xN (x)L

=

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Consistent Mass Matrix for 2D Beam Member (iii)Acceleration along beam member

Inertia force along member with mass per unit length along beam of ρ

[ ]{ }ix x

iy y

f (x) u (x).u

f (x) u (x).⎧ ⎫ ⎧ ⎫ρ⎡ ⎤⎪ ⎪ ⎪ ⎪= − = − ρ⎨ ⎬ ⎨ ⎬⎢ ⎥ρ⎪ ⎪ ⎪ ⎪⎣ ⎦⎩ ⎭ ⎩ ⎭

Let the Virtual Displacement be δu(x) i.e.

( )( ) [ ]{ }

1

2

x 1 4 3

y 2 3 5 6 4

5

6

uu

u x N (x) . . N (x) . . uN u

u x . N (x) N (x) . N (x) N (x) uuu

⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎧ ⎫ ⎡ ⎤⎪ ⎪ ⎪ ⎪= =⎨ ⎬ ⎨ ⎬⎢ ⎥

⎪ ⎪ ⎣ ⎦ ⎪ ⎪⎩ ⎭⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭

( )( ) [ ]{ }

1

2

x 1 4 3

y 2 3 5 6 4

5

6

uu

u x N (x) . . N (x) . . uN u

u x . N (x) N (x) . N (x) N (x) uuu

δ⎧ ⎫⎪ ⎪δ⎪ ⎪⎪ ⎪⎧ ⎫δ δ⎡ ⎤⎪ ⎪ ⎪ ⎪= = δ⎨ ⎬ ⎨ ⎬⎢ ⎥δ δ⎪ ⎪ ⎣ ⎦⎪ ⎪⎩ ⎭⎪ ⎪δ⎪ ⎪δ⎪ ⎪⎩ ⎭

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Consistent Mass Matrix for 2D Beam Member (iv)Internal Virtual Work = Real Force multiplied by Virtual Displacement

TLx ix

Internaly iyx 0

u (x) f (x)W dx

u (x) f (x)=

δ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪= ⎨ ⎬ ⎨ ⎬δ⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭∫

SubstitutingTL

x ixInternal

y iyx 0

u (x) f (x)W dx

u (x) f (x)=

δ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪= ⎨ ⎬ ⎨ ⎬δ⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭∫

{ } [ ] [ ] [ ]{ }L

T TInternal

x 0

W u N(x) N(x) u dx=

= δ ρ∫

{ } [ ] [ ] [ ] { }L

T TInternal

x 0

W u N(x) N(x) dx u=

= δ ρ∫

As the nodal displacements are not functions of position along the beam these can be taken outside the integral

18

35

Consistent Mass Matrix for 2D Beam Member (v)Let {f} be the nodal forces at the member ends due to the accelerations{ü} at the member ends.

External Virtual Work

{ } { }

T1 1

2 2

T3 3E x t e r n a l

4 4

5 5

6 6

u fu fu f

W u fu fu fu f

δ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪δ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪δ⎪ ⎪ ⎪ ⎪= = δ⎨ ⎬ ⎨ ⎬δ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪δ⎪ ⎪ ⎪ ⎪δ⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭

Equating External Virtual Work and Internal Virtual Work and assuming that the Virtual Displacement {δu} is not null

36

Consistent Mass Matrix for 2D Beam Member (vi)

2 2

Consistent

2 2

140 . . 70 . .. 156 22 L . 54 13 L. 22 L 4 L . 13 L 3 L LM

420 70 . . 140 . .. 54 13 L . 156 22 L. 13 L 3 L . 22 L 4 L

⎡ ⎤⎢ ⎥−⎢ ⎥⎢ ⎥−ρ

=⎡ ⎤ ⎢ ⎥⎣ ⎦⎢ ⎥⎢ ⎥−⎢ ⎥

− − −⎣ ⎦

{ } [ ] [ ] [ ] { } { }L

TConsistent

x 0

f N(x) N(x) dx u M u=

= ρ = ⎡ ⎤⎣ ⎦∫Then

where

If the mass per unit length ρ is not constant but varies along themember as ρ(x) then this is easily accommodated in the integral alongthe member length.

19

37

Consistent Mass Matrix for 2D Beam Member (vii)

[ ]

3 2 3 2

2 2

3 2 3 2

2 2

AE AE. . . .L L

12EI 6EI 12EI 6EI. .L L L L

6EI 4EI 6EI 2EI. .L L L Lk

AE AE. . . .L L

12EI 6EI 12EI 6EI. .L L L L

6EI 2EI 6EI 4EI. .L L L L

−⎡ ⎤⎢ ⎥⎢ ⎥

−⎢ ⎥⎢ ⎥⎢ ⎥−⎢ ⎥⎢ ⎥= ⎢ ⎥−⎢ ⎥⎢ ⎥⎢ ⎥− − −⎢ ⎥⎢ ⎥

−⎢ ⎥⎢ ⎥⎣ ⎦

This mass matrix is consistent, in an energy sense, with the stiffness matrix for the same beam member in that they are both based on the same displacement functions

38

Consistent Mass Matrices - NotesThe Consistent Mass matrix can be further modified to account for the fact that the mass of the beam is not actually concentrated along the centreline of the beam element but is actually distributed across the beam cross-section. This means that as the beam cross-section rotates there is a moment generated at that cross-section. This rotatory inertia term can be incorporated into the formulation at no extra cost. This is only likely to be of significance for sections with a wide cross section or with heavy flanges. Most computer programs ignore this effect. The Consistent Mass matrix can also incorporate shear deformations as well as the flexural deformations covered in thepreceding pages. Again, usually ignored, but can be of significance in structural wall models where a large part of themember deformation is a shear deformation.Most current structural dynamics programs use only a Lumped Mass model.

20

39

Consistent Mass MatricesThe same approach is used to form the consistent mass matrix forany finite element in 1, 2 or 3 dimensional space using the displacement functions that are used for forming the stiffness matrix of the element.The mass matrix will have the same degrees of freedom as does the associated stiffness matrix for that element.For finite elements that are not based on displacement functions, i.e. Equilibrium, Hybrid Stress and Mixed model finite elements then either an appropriate formulation must be used or the mass matrix might be approximated by that for a similar Displacementfinite element with the same external degrees of freedom.Special members, such as contact elements, springs etc may be represented with the consistent mass of a truss element or, moreusually, by using a local lumped mass approximation.

40

Structural Mass matrix

To form the total mass matrix for the structure then the member mass matrices, either lumped member mass matrices or consistent member mass matrices are combined using the Direct Stiffness Method in exactly the same way that the member stiffness matrices are combined to form the structure’s stiffness matrix.If only Lumped mass models are used then the structure’s mass matrix will be a diagonal matrix.If Consistent mass models are used then the structure’s mass matrix will have the same skyline form as that of the structure’s stiffness matrix.If all degrees of freedom have some mass then the structure’s mass matrix is symmetric and positive definite.If some degrees of freedom do not have mass then the structure’s mass matrix is symmetric and non-negative definite.

21

41

Multi-storey FramesFor a static analysis this frame has almost 1000 degrees of freedom, assuming 6 kinematic degrees of freedom at each joint.The stiffness matrix is very sparse, assuming only frame members (beams and columns), there are a maximum of 42 non-zero coefficients in any row or column of the matrix. Using a finite element model for the floor slabs will increase the connectivity but will still only have a small effect on the density of non-zero coefficients in the stiffness matrix.

42

Multi-storey Frames•If a consistent mass model is used then there are almost a 1000 inertial degrees of freedom with the mass matrix having the sameform as the stiffness matrix.•If a lumped mass matrix is assumed then the number of degrees of freedom is halved, there being only 3 acceleration degrees offreedom per node and the matrix is a diagonal matrix.•If the floors are modelled as rigid diaphragms in their own planes then there are 3 degrees of freedom for the in-plane motions plus 1 degree of freedom for each node that has mass in the vertical direction. i.e 7*(3+24)=189 degrees of freedom•If the vertical inertia at each node is ignored (ETABS model) then there are only the 3 degrees of freedom at each floor (two horizontal accelerations and a rotational acceleration about thevertical axis) giving a total of 21 degrees of freedom for the whole structure.

22

43

Floor DiaphragmsLet dA = dx.dyThe total mass M

∫ ρ=A

dA)y,x(M

Centre of mass

∫

∫ρ=

ρ=

A0

A0

/M )dA.y).y,x((y

/M )dA.x).y,x((x

Moment of Inertia about origin

∫

∫=

=

A

2yy

A

2xx

dAy)y,x(ρI

dAx)y,x(ρI

Note:

yyxxzz

22

a

2

azz

222

III

dA)yx)(y,x(ρdAr)y,x(ρI

yxr

+=

+==

+=

∫∫

44

Floor DiaphragmsOf more interest are the moments of inertia about the Centre of Gravity

∫ ∫

∫ ∫=+−ρ=−ρ=

=+−ρ=−ρ=

A A

20yy

200

220yy0

A A

20xx

200

220xx0

yM-I dA).yyy2y).(y,x( dA.)yy).(y,x(I

xM-I dA).xxx2x).(y,x( dA.)xx).(y,x(I

Similarly

2zzyy0xx0zz0

20

20

azz0

MrIIII

dA))yy()xx)((y,x(I

−=+=

−+−ρ= ∫

These rotational moments of inertia are required for many analyses in programs such as ETABS which use rigid floor diaphragm models.

23

45

Floor Diaphragmsu3

u1

u2

Centre of Mass

a

bMass of floor be M and the Polar Moment of inertia be I with uniform mass ρ per unit area.

⎟⎟⎠

⎞⎜⎜⎝

⎛ +=ρ+ρ=ρ=

12ba

M12ba

12ab

I abM2233

Inertia forces at the Centre of Mass due to accelerations at theCentre of Mass

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

3

2

1

3

2

1

uuu

I...M...M

FFF

The mass matrix of the floor is a diagonal matrix.

46

Floor Diaphragms

u3

x

yCentre of Mass

a

b

u1

u2

Reference origin

Suppose that the reference origin does not coincide with the Centre of Mass.

There are now moments generated about the reference origin due to translational accelerations and the rotations generate translations at the Centre of Mass

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++−

−−=

⎪⎭

⎪⎬

⎫

⎪⎩

⎪⎨

⎧

3

2

1

223

2

1

uuu

))yx(MI(MxMyMxM.My.M

FFF

The mass matrix of the floor is no longer a diagonal matrix.

24

47

Floor DiaphragmsPrograms such as ETABS use the centre of mass as the reference point for the computation of responses to earthquake loadings asthe use of a diagonal mass matrix simplifies the computation.Most design codes of practice require that analyses be carried out for different eccentricities of mass, to allow for uncertainty of mass distribution and for difficult in assessing torsional effects in the building.If the output results are those of the centre of mass of the floor, as in early versions of ETABs, the program user finds it very difficult to compare the results of the different analyses because a different reference point is used for each analysis.Today, most computer programs will use a user defined reference centre with all output referred to this point. However, the program will still use the centre of mass as its internal reference point for the computation and then transform the solution to that at the users’ reference point, simplifying the computation yet making the results easier for the user to understand.