structural element stiﬀness, mass, and damping...

Structural ElementStiffness, Mass, and Damping Matrices

CEE 541. Structural DynamicsDepartment of Civil and Environmental Engineering

Duke University

Henri P. GavinFall 2018

1 Preliminaries

This document describes the formulation of stiffness and mass matrices for structural elementssuch as truss bars, beams, plates, and cables(?). The formulation of each element involves thedetermination of gradients of potential and kinetic energy functions with respect to a set ofcoordinates defining the displacements at the ends, or nodes, of the elements. The potentialand kinetic energy of the functions are therefore written in terms of these nodal displacements(i.e., generalized coordinates). To do so, the distribution of strains and velocities within theelement must be written in terms of nodal coordinates as well. Both of these distributionsmay be derived from the distribution of internal displacements within the solid element.

1.1 Displacements

Figure 1. Displacements within a solid continuum.

2 CEE 541. Structural Dynamics – Duke University – Fall 2018 – H.P. Gavin

A component of a time-dependent displacement ui(x, t), (i = 1, · · · , 3) in a solid continuumcan be expressed in terms of the displacements of a set of nodal displacements, un(t) (n =1, · · · , N) and a corresponding set of “shape functions” ψin.

ui(x, t) =N∑n=1

ψin(x1, x2, x3) un(t) (1)

= Ψi(x) u(t) (2)u(x, t) = [Ψ(x)]3×N u(t) (3)

Engineering strain, axial strain εii, shear strain γij.

εii(x, t) = ∂ui(x, t)∂xi

≡ ui,i(x, t) (4)

γij(x, t) = ∂ui(x, t)∂xj

+ ∂uj(x, t)∂xi

≡ uj,i(x, t) + ui,j(x, t) (5)

(6)

Displacement gradient

∂ui(x)∂xj

=N∑n=1

∂

∂xjψin(x1, x2, x3) un(t) (7)

ui,j(x) =N∑n=1

ψin,j(x) un(t) (8)

Strain-displacement relations

εii(x, t) =N∑n=1

ψin,i(x) un(t) (9)

γij(x, t) =N∑n=1

(ψin,j(x) + ψjn,i(x)) un(t) (10)

Strain vectorεT(x, t) = ε11 ε22 ε33 γ12 γ23 γ13 (11)

ε(x, t) = [ B(x) ]6×N u(t) (12)

CC BY-NC-ND December 13, 2017, H.P. Gavin

http://creativecommons.org/licenses/by-nc-nd/3.0/

Structural Element Stiffness, Mass, and Damping Matrices 3

1.2 Geometric Deformation

In frame elements, nonuniform transverse displacements (u′y(x) 6= 0) induce longitudinaldeformation, u′x(x). These longitudinal deformation are called geometric deformations.

dx

dx yu (x) + u’(x) dxy

u (x)yu’(x)dxx

Figure 2. Transverse deformation u′y(x) and geometric deformation u′x(x).

Geometric deformation can be derived from the Pythagorean theorem and is quadratic inu′x(x) and u′y(x).

(dx− u′x(x)dx)2 + (u′y(x)dx)2 = (dx)2

(dx)2 − 2u′x(x)(dx)2 + (u′x(x))2(dx)2 = (dx)2 − (u′y(x)dx)2

2u′x(x)(dx)2 − (u′x(x))2(dx)2 = (u′y(x))2(dx)2

u′x(x) = 12(u′x(x))2 + 1

2(u′y(x))2

Since slender structural elements are much more stiff in axial deformation than in transversedeformation, and since elastic tensile strains are limited to within 0.002 for most structuralmaterials, (u′x)2 < (0.002)2 (u′y)2, the geometric portion of the deformation for structuralframe elements is approximated as

u′x(x) ≈ 12(u′y(x))2 . (13)

The accuracy of the approximation (13) can be quantified in terms of the strains in a displacedbar. Figure 3 shows a bar with four displacement coordinates and Figure 4 shows the errorsassociated with the finite strain approximation. In this figure, L1 is the true stretched lengthof the truss bar, L1 =

√(Lo + u3 − u1)2 + (u4 − u2)2, L2 uses the finite-strain approximation

shown above, L2 = Lo + (u3 − u1) + (1/(2Lo))(u4 − u2)2, and the logarithmic strain isln (L1/Lo). For a bar stretched to a level of linear strain (u3 − u1)/Lo that is significant formost metallic materials, the approximation (13) is accurate to within 0.001% for rotationsu′y < 0.1, at which point the total strain would be approximately 0.005, which is abouttwice the yield strain for most metals. For analyses of structures incorporating geometricnonlinearity but within the range of elastic behavior of metals, the approximation (13) forthe geometric portion of the deformation is accurate to within one part per million.




Lo

L

3u

u 4

u

u

1

2

Figure 3. A bar with four displacement coordinates in its original and displaced configurations.

10-3

10-2

10-1

100

10-4 10-3 10-2 10-1

stra

in

(u4-u2)/Lo = 0.050

L1 (Pythagorean Thm.)/Lo-1

L2 (approx)/Lo-1

log(L1/Lo)

10-7

10-6

10-5

10-4

10-3

10-4 10-3 10-2 10-1

rela

tive

erro

r, L

2/L 1

- 1

(u3-u1)/Lo

Figure 4. Finite-strain approximations and their associated errors.

1.3 Green-Lagrange Strain

Following the development from the previous section, we can consider the planar displace-ment of a short line segment of length dlo to a line segment of length dl. In the originalconfiguration, the line segment has a squared length of

dl2o = dx2 + dy2

and in the displaced configuration it has a squared length of

dl2 = (dx+ ux,xdx+ ux,ydy)2 + (dy + uy,xdx+ uy,ydy)2

= (dx)2 + (ux,x)2(dx)2 + (ux,y)2(dy)2 + 2(ux,x)(dx)2 + 2(ux,y)(dx)(dy) + 2(ux,xux,y)(dx)(dy) +(dy)2 + (uy,x)2(dx)2 + (uy,y)2(dy)2 + 2(uy,x)(dx)(dy) + 2(uy,y)(dy)2 + 2(uy,xuy,y)(dx)(dy)




dl o

+ u (x,y) dy

+ u (x,y) dx

u (x,y)

y,y

y,x

y

u (x,y)y

u (x,y)x

(x , y)

(x+dx , y+dy)

u (x,y)

+ u (x,y) dx

+ u (x,y) dyx,y

x,x

x

dl

Figure 5. Displacement of line segment dlo to line segment dl through planar displacementsux(x, y) and uy(x, y).

The difference between the squared lengths, dl2 − dl2o, is quadratic in dx and dy and can beexpressed in a “matrix form.”

l2 − l2o = 2[dx

dy

]T [ux,x + 1

2u2x,x + 1

2u2y,x

12ux,y + 1

2uy,x + 12ux,yuy,x

12ux,y + 1

2uy,x + 12ux,yuy,x uy,y + 1

2u2x,y + 1

2u2y,y

] [dx

dy

]

= 2[dx

dy

]T [εxx

12γxy

12γyx εyy

] [dx

dy

](14)

This “matrix” is called the Green-Lagrange strain tensor, or the Lagrangian strain tensor, orthe Green-Saint Venant strain tensor for plane strain. This relationship may be confirmedin two simple cases:

• dx = dlo, dy = 0, ux,x = ∆/lo, ux,y = 0, uy,y = 0, ⇒ dl2 − dl2o = 2lo∆ + ∆2.

• dx = dlo, dy = 0, uy,x = ∆/lo, ux,x = 0, uy,y = 0, ⇒ dl2 − dl2o = ∆2.

Generalizing this result to three dimensions, the strain-displacement equations for linearstrain and geometric deformation are:

εii = ui,i + 12u

2i,i + 1

2u2j,i + 1

2u2k,i = ∂ui

∂xi+ 1

2

(∂ui∂xi

)2

+ 12

(∂uj∂xi

)2

+ 12

(∂uk∂xi

)2

(15)

γij = ui,j + uj,i + ui,juj,j + uj,iui,i = ∂ui∂xj

+ ∂uj∂xi

+ ∂ui∂xj

∂uj∂xj

+ ∂uj∂xi

∂ui∂xi

(16)

In the following models, finite deformation effects are limited to εxx ≈ ux,x + (1/2)u2y,x for

the exclusive purpose of deriving geometric stiffness matrices. For metalic materials in whichstrains are typically around 0.001, this approximation does not compromise accuracy.




1.4 Stress-strain relationship (isotropic elastic solid)

σ11

σ22

σ33

τ12

τ23

τ13

= E

(1 + ν)(1− 2ν)

1− ν ν ν

ν 1− ν ν

ν ν 1− ν12 − ν

12 − ν

12 − ν

ε11

ε22

ε33

γ12

γ23

γ13

(17)

Stress vectorσT(x, t) = σ11 σ22 σ33 τ12 τ23 τ13 (18)

σ = [ Se(E, ν) ]6×6 ε (19)

1.5 Potential Energy and Stiffness

Consider a system comprising an assemblage of linear springs, with stiffness ki, each with anindividual stretch, di. The total potential energy in the assemblage is

U = 12∑i

kid2i

If displacements of the assemblage of springs is denoted by a vector u, not necessarily equalto the stretches in each spring, then the elastic potential energy may also be written

U(u) = 12uTKu

= 12

n∑i=1

uifi

= 12

n∑i=1

uin∑j=1

Kijuj

where K is the stiffness matrix with respect to the coordinates u. The stiffness matrix Krelates the elastic forces fi to the collocated displacements, ui.

f1 = K11u1 + · · ·+K1juj + · · ·+K1NuN

fi = Ki1u1 + · · ·+Kijuj + · · ·+KiNuN

fN = KN1u1 + · · ·+KNjuj + · · ·+KNNuN

A point force fi acting on an elastic body is the gradient of the elastic potential energy U

with respect to the collocated displacement ui

fi = ∂

∂uiU

The i, j term of the stiffness matrix may therefore be found from the potential energy functionU(u),

Kij = ∂

∂ui

∂

∂ujU(u) (20)




1.6 Strain Energy and Stiffness in Linear Elastic Continua

U(u) = 12

∫Ω

σ(x, t)Tε(x, t) dΩ (21)

= 12

∫Ω

ε(x, t)T Se(E, ν) ε(x, t) dΩ

= 12

∫Ω

u(t)TB(x)T Se(E, ν) B(x)u(t) dΩ

= 12 u(t)T

∫Ω

[B(x)T Se(E, ν) B(x)

]N×N

dΩ u(t) (22)

Elastic element stiffness matrix

fe = ∂U

∂u= Ke u

Ke =∫

Ω

[B(x)T Se(E, ν) B(x)

]N×N

dΩ (23)

1.7 Kinetic Energy and Mass

The impulse-momentum relationship states that∫f dt = δ(mu)

f = d

dt(mu)

f = d

dt

(∂

∂u

12mu

2)

f = d

dt

(∂

∂uT

),

where T is the kinetic energy, and is assumed to be independent of u(t).

Consider a system comprising an assemblage of point masses, mi, each with an individualvelocity, vi. The total kinetic energy in the assemblage is

T = 12∑i

miv2i

If displacements of the assemblage of masses are defined by a generalized coordinate vectoru, not necessarily equal to the velocity coordinates, above, then the kinetic energy may alsobe written

T (u) = 12 uTMu

= 12

n∑i=1

uin∑j=1

Mijuj




where M is the constant mass matrix with respect to the generalized coordinates u. Themass matrix M relates the inertial forces fi to the collocated accelerations, ui.

f1 = M11u1 + · · ·+M1juj + · · ·+M1N uN

fi = Mi1u1 + · · ·+Mijuj + · · ·+MiN uN

fN = MN1u1 + · · ·+MNjuj + · · ·+MNN uN

The i, j term of the constant mass matrix may therefore be found from the kinetic energyfunction T ,

Mij = ∂

∂ui

∂

∂t

∂

∂ujT (u) = ∂

∂ui

∂

∂ujT (u) (24)

1.8 Inertial Energy and Mass in Deforming Continua

T ( ˙u) = 12

∫Ωρ |u(x, t)|2 dΩ (25)

= 12

∫Ωρ u(x, t)Tu(x, t) dΩ

= 12

∫Ωρ ˙u(t)TΨ(x)T Ψ(x) ˙u(t) dΩ

= 12

˙u(t)T∫

Ωρ[Ψ(x)T Ψ(x)

]N×N

dΩ ˙u(t) (26)

Consistent mass matrix

∂T

∂ ˙u=

∫Ωρ[Ψ(x)T Ψ(x)

]N×N

dΩ ˙u(t) (27)

fi = d

dt

(∂T

∂ ˙u

)=

∫Ωρ[Ψ(x)T Ψ(x)

]N×N

dΩ ¨u(t) (28)

M =∫

Ωρ[Ψ(x)T Ψ(x)

]N×N

dΩ (29)




2 Bar Element Matrices

2D prismatic homogeneous isotropic truss bar.

Uniform uni-axial stress σT = σxx, 0, 0, 0, 0, 0T

Corresponding strain εT = (σxx/E) 1,−ν,−ν, 0, 0, 0T.Incremental strain energy dU = 1

2σTε dΩ = 12σxxεxx dΩ = 1

2Eε2xx dΩ

2.1 Bar Displacements

Figure 6. Truss bar element coordinates and displacements.

ux(x, t) =(

1− x

L

)u1(t) +

(x

L

)u3(t) (30)

= ψx1(x) u1(t) + ψx3(x) u3(t) (31)

uy(x, t) =(

1− x

L

)u2(t) +

(x

L

)u4(t) (32)

= ψy2(x) u2(t) + ψy4(x) u4(t) (33)

Ψ(x) =[

1− xL

0 xL

00 1− x

L0 x

L

](34)

[ux(x, t)uy(x, t)

]= Ψ(x) u(t) (35)




2.2 Bar Strain Energy and Elastic Stiffness Matrix

Strain-displacement relation

εxx = ∂ux∂x

+ 12

(∂uy∂x

)2

(36)

= ψx1,xu1 + ψx3,xu3 + 12 ψy2,xu2 + ψy4,xu42 (37)

=(− 1L

)u1 +

( 1L

)u3 + 1

2

(− 1L

)u2 +

( 1L

)u4

2(38)

=[− 1L

0 1L

0]u + 1

2

(− 1L

)u2 +

( 1L

)u4

2(39)

= B u + 12

(− 1L

)u2 +

( 1L

)u4

2(40)

B =[− 1L

0 1L

0]. (41)

Strain energy and elastic stiffness

U = 12

∫Ωεxx E εxx dΩ (42)

Ke =∫ L

x=0

[BT E B

]A dx (43)

= EA∫ L

x=0

1/L2 0 −1/L2 0

0 0 0 0−1/L2 0 1/L2 0

0 0 0 0

dx (44)

= EA

L

1 0 −1 00 0 0 0−1 0 1 0

0 0 0 0

(45)




2.3 Bar Kinetic Energy and Mass Matrix

T = 12

˙uT∫

Ωρ[Ψ(x)T Ψ(x)

]N×N

dΩ ˙u(t) (46)

M =∫ L

x=0ρ[Ψ(x)T Ψ(x)

]A dx (47)

= ρA∫ L

x=0

(1− x

L)2 0 (1− x

L)( xL

) 00 (1− x

L)2 0 (1− x

L)( xL

)( xL

)(1− xL

) 0 ( xL

)2 00 ( x

L)(1− x

L) 0 ( x

L)2

dx (48)

= 16ρAL

2 0 1 00 2 0 11 0 2 00 1 0 2

(49)

2.4 Bar Stiffness Matrix with Geometric Strain Effects

U = 12

∫ L

0εxx E εxx A dx (50)

= EA

2

∫ L

0

∂ux∂x

+ 12

(∂uy∂x

)22

dx (51)

= EA

2

∫ L

0

(∂ux∂x

)2

+ ∂ux∂x

(∂uy∂x

)2

+ 14

(∂uy∂x

)4 dx (52)

Substitute∂ux∂x

= − 1Lu1 + 1

Lu3 (53)

∂uy∂x

= − 1Lu2 + 1

Lu4 (54)

to obtainU = EA

2L

((u3 − u1)2 + 1

L(u3 − u1)(u4 − u2)2

)(55)

So,

∂U

∂u= EA

L

1 0 −1 00 0 0 0−1 0 1 0

0 0 0 0

u1

u2

u3

u4

+ EA(u3 − u1)L2

0 0 0 00 1 0 −10 0 0 00 −1 0 1

u1

u2

u3

u4

(56)

= Ke u + N

LKg u (57)




3 Bernoulli-Euler Beam Element Matrices

2D prismatic homogeneous isotropic beam element, neglecting shear deformation and rotatoryinertia.

3.1 Bernoulli-Euler Beam Coordinates and Internal Displacements

Consider the geometry of a deformed beam. The functions ux(x) and uy(x) describe thetranslation of points along the neutral axis of the beam as a function of the location alongthe un-stretched neutral axis.

Figure 7. Beam element coordinates and displacements.

We will describe the deformation of the beam as a function of the end displacements (u1, u2, u4, u5)and the end rotations (u3, u6). In a dynamic context, these end displacements will changewith time.

ux(x, t) =6∑

n=1ψxn(x) un(t)

uy(x, t) =6∑

n=1ψyn(x) un(t)

The functions ψxn(x) and ψyn(x) satisfy the boundary conditions at the end of the beam andthe differential equation describing bending of a Bernoulli-Euler beam loaded statically atthe nodal coordinates. In such beams the effects of shear deformation and rotatory inertiaare neglected. For extension of the neutral axis,

ψx1(x) = 1− x

L

ψx4(x) = x

L




and ψx2 = ψx3 = ψx5 = ψx6 = 0 along the neutral axis. For bending of the neutral axis,

ψy2(x) = 1− 3(x

L

)2+ 2

(x

L

)3

ψy3(x) =(x

L− 2

(x

L

)2+(x

L

)3)L

ψy5(x) = 3(x

L

)2− 2

(x

L

)3

ψy6(x) =(−(x

L

)2+(x

L

)3)L

and ψy1 = ψy4 = 0.

Ψ(x) =

[ 1 − xL

0 0 xL

0 0

0 1 − 3(xL

)2+ 2(xL

)3(xL

− 2(xL

)2+(xL

)3)L 0 3

(xL

)2− 2(xL

)3(

−(xL

)2+(xL

)3)L

](58)

[ux(x, t)uy(x, t)

]= Ψ(x) u(t) (59)

These expressions are analytical solutions for the displacements of Bernoulli-Euler beamsloaded only with concentrated point loads and concentrated point moments at their ends.Internal bending moments are linear within beams loaded only at their ends, and the beamdisplacements may be expressed with cubic polynomials.

3.2 Bernoulli-Euler Beam Strain Energy and Elastic Stiffness Matrix

In extension, the elastic potential energy in a beam is the strain energy related to the uni-form extensional strain, εxx. If the strain is small, then the extensional strain within thecross section is equal to an extension of the neutral axis, (∂ux/∂x), plus the bending strain,−(∂2uy/∂x

2)y.

εxx = ∂ux∂x− ∂2uy

∂x2 y

=6∑

n=1

∂

∂xψxn(x) un −

6∑n=1

∂2

∂x2ψyn(x) y un (60)

=6∑

n=1ψ′xn(x) un −

6∑n=1

ψ′′yn(x) y un

=6∑

n=1Bn(x, y) un

= B(x, y) u (61)

where

B(x, y) =[− 1L,

6yL2 −

12xyL3 ,

4yL− 6xy

L3 ,1L,−6yL2 + 12xy

L3 ,2yL− 6xy

L2

]. (62)




The elastic stiffness matrix can be found directly from the strain energy of axial strains εxx.

U = 12

∫Ωεxx E εxx dΩ (63)

Ke =∫ L

x=0

∫A

[B(x, y)T E B(x, y)

]dA dx. (64)

Note that this integral involves terms such as∫A y

2dA and∫A ydA in which the origin of the

coordinate axis is placed at the centroid of the section. The integral∫A y

2dA is the bendingmoment of inertia for the cross section, I, and the integral

∫A ydA is zero.

It is also important to recognize that the elastic strain energy may be evaluated separatelyfor extension effects and bending effects. For extension, the elastic strain energy is

U = 12

∫ L

x=0EA (εxx)2 dx

= 12

∫ L

x=0EA

( 6∑n=1

ψ′xn(x) un)2

dx

and the ij stiffness coefficient (for indices 1 and 4) is

Kij = ∂

∂ui

∂

∂uj

12

∫ L

x=0EA

( 6∑n=1

ψ′xn(x) un)2

dx

=∫ L

x=0EA ψ′xi(x) ψ′xj(x) dx. (65)

In bending, the elastic potential energy in a Bernoulli-Euler beam is the strain energy relatedto the curvature, κz.

κz = ∂2uy∂x2 =

6∑n=1

∂2

∂x2ψyn(x) un =6∑

n=1ψ′′yn(x) un

The elastic strain energy for pure bending is

U = 12

∫ L

x=0EI (κz)2 dx

= 12

∫ L

x=0EI

( 6∑n=1

ψ′′yn(x) un)2

dx

and the ij stiffness coefficient (for indices 2,3,5 and 6) is

Kij = ∂

∂ui

∂

∂uj

12

∫ L

x=0EI

( 6∑n=1

ψ′′yn(x) un)2

dx

=∫ L

x=0EI ψ′′yi(x) ψ′′yj(x) dx. (66)




3.3 Bernoulli-Euler Beam Kinetic Energy and Mass Matrix

The kinetic energy of a particle within a beam is half the mass of the particle, ρAdx, timesits velocity, u, squared. For velocities along the direction of the neutral axis,

ux(x) =6∑

n=1ψxn(x) ˙un ,

The kinetic energy function and the mass matrix may be by substituting equation (58) intoequations (26) and (29).

T = 12

˙uT∫

Ωρ[Ψ(x)T Ψ(x)

]N×N

dΩ ˙u(t) (67)

M =∫ L

x=0ρ[Ψ(x)T Ψ(x)

]A dx (68)

It is important to recognize that kinetic energy and mass associated with extensional velocitiesmay be determined separately from those associated with transverse velocities. The kineticenergy for extension of the neutral axis is

T = 12

∫ L

x=0ρA (ux)2 dx

= 12

∫ L

x=0ρA

( 6∑n=1

ψxn(x) ˙un)2

dx

and the ij mass coefficient (for indices 1 and 4) is

Mij = ∂

∂ ˙ui∂

∂ ˙uj12

∫ L

x=0ρA

( 6∑n=1

ψxn(x) ˙un)2

dx

=∫ L

x=0ρA ψxi(x) ψxj(x) dx. (69)

For velocities transverse to the neutral axis,

uy(x) =6∑

n=1ψyn(x) ˙un ,

the kinetic energy for velocity across the neutral axis is

T = 12

∫ L

x=0ρA (uy)2 dx

= 12

∫ L

x=0ρA

( 6∑n=1

ψyn(x) ˙un)2

dx

and the ij mass coefficient (for indices 2,3,5 and 6) is

Mij = ∂

∂ ˙ui∂

∂ ˙uj12

∫ L

x=0ρA

( 6∑n=1

ψyn(x) ˙un)2

dx

=∫ L

x=0ρA ψyi(x) ψyj(x) dx. (70)




3.4 Bernoulli-Euler Stiffness Matrix with Geometric Strain Effects

The axial strain in a Bernoulli-Euler beam including the geometric strain is

εxx = ∂ux∂x− ∂2uy

∂x2 y + 12

(∂uy∂x

)2

(71)

The potential energy with geometric strain effects is

U = 12

∫ L

x=0

∫Aεxx E εxx dA dx (72)

= 12

∫ L

0E

∫A

(∂ux∂x− ∂2uy

∂x2 y + 12

(∂uy∂x

)2)2

dx (73)

= 12

∫ L

0E

∫A

(u2x,x − 2ux,xuy,xxy + ux,xu

2y,x + u2

y,xxy2 − uy,xxu2

y,xy + 14u

4y,x

)dAdx (74)

Note that∫A ydA = 0 and

∫A y

2dA = I and neglect u4y,x so that

U = 12

∫ L

0EA

(u2x,x

)dx+ 1

2

∫ L

0EI

(u2y,xx

)dx+

∫ L

0EA

(ux,xu

2y,x

)dx . (75)

Substitute

uy,x =6∑

n=1ψ′yn(x) un (76)

uy,xx =6∑

n=1ψ′′yn(x) un (77)

ux,x =6∑

n=1ψ′xn un = N

EA(78)

and differentiate with respect to ui and uj to obtain,

Kij = EA∫ L

0ψ′xiψ

′xj dx+ EI

∫ L

0ψ′′yi(x)ψ′′yj(x) dx+N

∫ L

0ψ′yi(x)ψ′yj(x) dx (79)

so that,K = Ke + N

LKg (80)




3.5 Bernoulli-Euler Beam Element Stiffness and Mass Matrices

For prismatic homogeneous isotropic beams, substituting the expressions for the functionsψxn and ψyn into equations (65) - (70), or substituting equation (62) into equation (64) and(58) to equation (68) results in element stiffness matrices Ke, M, and Kg.

Ke =

EAL

0 0 −EAL

0 0

12EIL3

6EIL2 0 −12EI

L36EIL2

4EIL

0 −6EIL2

2EIL

EAL

0 0sym

12EIL3 −6EI

L2

4EIL

(81)

M = ρAL420

140 0 0 70 0 0

156 22L 0 54 −13L

4L2 0 13L −3L2

140 0 0sym

156 −22L

4L2

(82)

Kg = NL

0 0 0 0 0 0

65

L10 0 −6

5L10

2L2

15 0 − L10 −L2

30

0 0 0sym

65 − L

10

−2L2

15

(83)




4 Timoshenko Beam Element Matrices2D prismatic homogeneous isotropic beam element, including shear deformation and rotatoryinertia

Consider again the geometry of a deformed beam. When shear deformations are includedsections that are originally perpendicular to the neutral axis may not be perpendicular tothe neutral axis after deformation. The functions ux(x) and uy(x) describe the translation of

Figure 8. Deformation of beam element including shear deformation.

points along the neutral axis of the beam as a function of the location along the un-stretchedneutral axis. If the beam is not slender (length/depth < 5), then shear strains will contributesignificantly to the strain energy within the beam. The deformed shape of slender beams isdifferent from the deformed shape of stocky beams.

The beam carries a bending moment M(x) related to axial strain εxx and a shear force, Srelated to shear strain γxy. The potential energy has a bending strain component and a shearstrain component.

U = 12

∫Ω

σTε dΩ

= 12

∫Ωσxxεxx dΩ + 1

2

∫Ωτxyγxy dΩ

= 12

∫ L

0

∫A

M(x)yI

M(x)yEI

dA dx+ 12

∫ L

0

∫A

SQ(y)Ib(y)

SQ(y)GIb(y) dA dx

= 12

∫ L

0

M(x)2

EI2

∫Ay2 dA dx+ 1

2

∫ L

0

S2

GI2

∫A

Q(y)2

b(y)2 dA dx

= 12

∫ L

0

M(x)2

EIdx+ 1

2

∫ L

0

S2

G(A/α) dx (84)

where the shear area coefficient α reduces the cross section area to account for the non-uniformdistribution of shear stresses in the cross section,

α = A

I2

∫A

Q(y)2

b(y)2 dA .

For solid rectangular sections α = 6/5 and for solid circular sections α = 10/9 [2, 3, 4, 5, 8].




4.1 Timoshenko Beam Coordinates and Internal Displacements(including shear deformation effects)

The transverse deformation of a beam with shear and bending strains may be separated intoa portion related to shear deformation and a portion related to bending deformation,

uy(x, t) = u(b)y(x) + u(s)y(x) (85)

where

EIu′′(b)y(x) = M(x) = −M1

(1− x

L

)+M2

(x

L

)(86)

G(A/α)u′(s)y(x) = S(x) = 1L

(M1 +M2) (87)

It can be shown that the following shape functions satisfy the Timoshenko beam equations(equations (85), (86) and (87)) for transverse displacements.

ψy2(x) = 11 + Φ

[1− 3

(x

L

)2+ 2

(x

L

)3+(

1− x

L

)Φ]

ψy3(x) = L

1 + Φ

[x

L− 2

(x

L

)2+(x

L

)3+ 1

2

(x

L−(x

L

)2)

Φ]

ψy5(x) = 11 + Φ

[3(x

L

)2− 2

(x

L

)3+ x

LΦ]

ψy6(x) = L

1 + Φ

[−(x

L

)2+(x

L

)3− 1

2

(x

L−(x

L

)2)

Φ]

The term Φ gives the relative importance of the shear deformations to the bending deforma-tions,

Φ = 12EIG(A/α)L2 = 24α(1 + ν)

(r

L

)2, (88)

where r is the “radius of gyration” of the cross section, r =√I/A, ν is Poisson’s ratio. Shear

deformation effects are significant for beams which have a length-to-depth ratio less than 5.To neglect shear deformation, set Φ = 0. These displacement functions are exact for frameelements with constant shear forces S and linearly varying bending moment distributions,M(x), in which the strain energy has both a shear stress component and a normal stresscomponent,

U = 12

∫ L

0EI

( 6∑n=1

ψ′′(b)yn(x)un)2

dx+ 12

∫ L

0G(A/α)

( 6∑n=1

ψ′(s)yn(x)un)2

dx (89)

where the bending and shear components of the shape functions, ψ(b)yn(x) and ψ(s)yn(x) are:




ψ(b)y2(x) = 11 + Φ

[1− 3

(x

L

)2+ 2

(x

L

)3]

ψ(s)y2(x) = Φ1 + Φ

[1− x

L

]ψ(b)y3(x) = L

1 + Φ

[x

L− 2

(x

L

)2+(x

L

)3+ 1

2

(2xL−(x

L

)2)

Φ]

ψ(s)y3(x) = − LΦ1 + Φ

[12x

L

]ψ(b)y5(x) = 1

1 + Φ

[3(x

L

)2− 2

(x

L

)3]

ψ(s)y5(x) = Φ1 + Φ

[x

L

]ψ(b)y6(x) = L

1 + Φ

[−(x

L

)2+(x

L

)3+ 1

2

((x

L

)2)

Φ]

ψ(s)y6(x) = − L

1 + Φ

[12x

LΦ]

4.2 Timoshenko Beam Element Stiffness Matrices

The geometric stiffness matrix for a Timoshenko beam element may be derived as was donewith the Bernoulli-Euler beam element from the potential energy of linear and geometricstrain,

Kij = EA∫ L

0ψ′xi(x)ψ′xj(x) dx

+ EI∫ L

0ψ′′(b)yi(x)ψ′′(b)yj(x) dx

+ G(A/α)∫ L

0ψ′(s)yi(x)ψ′(s)yj(x) dx

+ N∫ L

0ψ′yi(x)ψ′yj(x) dx (90)

where the displacement shape functions ψ(x) are provided in section 4.1.




4.3 Timoshenko Beam Element Stiffness and Mass Matrices,(including shear deformation effects but not rotatory inertia)

For prismatic homogeneous isotropic beams, substituting the previous expressions for thefunctions ψxn(x) and ψ(b)yn(x), and ψ(s)yn(x) into equation (90) and (68), results in theTimoshenko element elastic stiffness matrices Ke, mass matrix M, and geometric stiffnessmatrix Kg

Ke =

EAL

0 0 −EAL

0 0

121+Φ

EIL3

61+Φ

EIL2 0 − 12

1+ΦEIL3

61+Φ

EIL2

4+Φ1+Φ

EIL

0 − 61+Φ

EIL2

2−Φ1+Φ

EIL

EAL

0 0sym

121+Φ

EIL3 − 6

1+ΦEIL2

4+Φ1+Φ

EIL

(91)

M =ρAL

840

280 0 0 140 0 0

312 + 588Φ + 280Φ2 (44 + 77Φ + 35Φ2)L 0 108 + 252Φ + 175Φ2 −(26 + 63Φ + 35Φ2)L

(8 + 14Φ + 7Φ2)L2 0 (26 + 63Φ + 35Φ2)L −(6 + 14Φ + 7Φ2)L2

280 0 0sym

312 + 588Φ + 280Φ2 −(44 + 77Φ + 35Φ2)L

(8 + 14Φ + 7Φ2)L2

(92)

Kg = N

L

0 0 0 0 0 0

6/5+2Φ+Φ2

(1+Φ)2L/10

(1+Φ)2 0 −6/5−2Φ−Φ2

(1+Φ)2L/10

(1+Φ)2

2L2/15+L2Φ/6+L2Φ2/12(1+Φ)2 0 −L/10

(1+Φ)2−L2/30−L2Φ/6−L2Φ2/12

(1+Φ)2

0 0 0sym

6/5+2Φ+Φ2

(1+Φ)2−L/10(1+Φ)2

2L2/15+L2Φ/6+L2Φ2/12(1+Φ)2

(93)




4.4 Timoshenko Beam Element Mass Matrix(including rotatory inertia but not shear deformation effects)

Consider again the geometry of a deformed beam with linearly-varying axial beam displace-ments outside of the neutral axis. The functions ux(x, y) and uy(x, y) now describe the

Figure 9. Deformation of beam element showing axial-direction displacements ux(x, y, t) outsidethe neutral axis.

translation of points anywhere within the beam, as a function of the location within thebeam. We will again describe these displacements in terms of a set of shape functions,ψxn(x, y) and ψyn(x), and the end displacements u1, · · · , u6.

ux(x, y, t) =6∑

n=1ψxn(x, y) un(t)

uy(x, t) =6∑

n=1ψyn(x) un(t)

The shape functions for transverse displacements ψyn(x) are the same as the shape functionsψyn(x) used previously. The shape functions for axial displacements along the neutral axis,ψx1(x, y) and ψx4(x, y) are also the same as the shape functions ψx1(x) and ψx4(x) usedpreviously. To account for axial displacements outside of the neutral axis, four new shapefunctions are derived from the assumption that plane sections remain plane, ux(x, y) =−u′(b)y(x)y.

ψx2(x, y) = −ψ′y2 y = 6(x

L−(x

L

)2)y

L

ψx3(x, y) = −ψ′y3 y =(−1 + 4x

L− 3

(x

L

)2)y

ψx5(x, y) = −ψ′y5 y = 6(−xL

+(x

L

)2)y

L

ψx6(x, y) = −ψ′y6 y =(

2xL− 3

(x

L

)2)y




Because ψyn, ψx1 and ψx4 are unchanged, the stiffness matrix is also unchanged. The kineticenergy of the beam, including axial and transverse effects is now,

T = 12

∫ L

x=0

∫ h/2

y=−h/2ρb(y)

( 6∑n=1

ψxn(x, y) ˙un)2

dy dx+ 12

∫ L

x=0ρA

( 6∑n=1

ψyn(x) ˙un)2

dx

and the mass matrix coefficients are found from

Mij = ∂

∂ ˙ui∂

∂ ˙ujT ( ˙u)

Evaluating equation (24) using the new shape functions ψx2, ψx3, ψx5, and ψx6, results in amass matrix incorporating rotatory inertia.

M = ρAL

13 0 0 1

6 0 0

1335 + 6

5r2

L211210L+ 1

10r2

L0 9

70 −65r2

L2 − 13420L+ 1

10r2

L

1105L

2 + 215r

2 0 13420L+ 1

10r2

L0

sym 13 0 0

1335 + 6

5r2

L2 − 11210L+ 1

10r2

L

1105L

2 + 215r

2

(94)

Beam element mass matrices including the effects of shear deformation on rotatory inertiaare more complicated. Refer to p 295 of Theory of Matrix Structural Analysis, by J.S.Przemieniecki (Dover Pub., 1985).




5 Coordinate Transformations for Bars and Beams

5.1 Beam Element Stiffness Matrix in Local Coordinates, K

3EIL3 ∆ 3EI

L2 θ6EIL2 θ

12EIL3 ∆

−→ −→ −→ −→EILθ 3EI

L2 ∆ 3EILθ 4EI

Lθ 6EI

L2 ∆

EILθ 2EI

Lθ 6EI

L2 ∆←− ←− ←− ←−3EIL3 ∆ 3EI

L2 θ6EIL2 θ

12EIL3 ∆

N1

V1

M1

N2

V2

M2

=

EAL 0 0 −EA

L 0 0

12EIL3

6EIL2 0 −12EI

L36EIL2

4EIL 0 −6EI

L22EIL

EAL 0 0

sym12EIL3 −6EI

L2

4EIL

u1

u2

u3

u4

u5

u6

f = K u




5.2 Beam Element Stiffness Matrix in Global Coordinates, K

Geometric relationship between u and u: u = T u

u1 = u1 cos θ + u2 sin θ u2 = −u1 sin θ + u2 cos θ u3 = u3

where

T =

c s 0−s c 0 0

0 0 1c s 0

0 −s c 00 0 1

c = cos θ = x2 − x1

L

s = sin θ = y2 − y1

L

The stiffness matrix in global coordinates is K = TT K T

K =

EAL c

2 EAL cs −EA

L c2 −EA

L cs

+12EIL3 s

2 −12EIL3 cs −6EI

L2 s −12EIL3 s

2 +12EIL3 cs −6EI

L2 s

EAL s

2 −EAL cs −EA

L s2

+12EIL3 c

2 6EIL2 c +12EI

L3 cs −12EIL3 c

2 6EIL2 c

4EIL

6EIL2 s −6EI

L2 c2EIL

EAL c

2 EAL cs

+12EIL3 s

2 −12EIL3 cs

6EIL2 s

symEAL s

2

+12EIL3 c

2 −6EIL2 c

4EIL

f = K u




5.3 Beam Element Consistent Mass Matrix in Local Coordinates, M

N1

V1

M1

N2

V2

M2

= ρAL

420

140 0 0 70 0 0

156 22L 0 54 −13L

4L2 0 13L −3L2

140 0 0sym

156 −22L

4L2

ü1

ü2

ü3

ü4

ü5

ü6

f = M ü




5.4 Beam Element Consistent Mass Matrix in Global Coordinates, M

Geometric relationship between u and u: u = T u

u1 = u1 cos θ + u2 sin θ u2 = −u1 sin θ + u2 cos θ u3 = u3

where

T =

c s 0−s c 0 0

0 0 1c s 0

0 −s c 00 0 1

c = cos θ = x2 − x1

L

s = sin θ = y2 − y1

L

The consistent mass matrix in global coordinates is M = TT M T

M = ρAL

420

140c2 −16cs −22sL 70c2 16cs 13sL+15s2 +54s2

140s2 22cL 16cs 70s2 −13cL+156c2 +54c2

4L2 −13sL 13cL −3L2

140c2 −16cs 22sLsym +156s2

140s2 −22cL+156c2

4L2

f = M u




6 2D Plane-Stress and Plane-Strain Rectangular Element Matrices

2D, isotropic, homogeneous element, with uniform thickness h.

Approximate element stiffness and mass matrices based on assumed distribution of internaldisplacements.

6.1 2D Rectangular Element Coordinates and Internal Displacements

Consider the geometry of a rectangle with edges aligned with a Cartesian coordinate system.(0 ≤ x ≤ a, 0 ≤ y ≤ b) The functions ux(x, y, t) and uy(x, y, t) describe the in-planedisplacements as a function of the location within the element.

Figure 10. 2D rectangular element coordinates and displacements.

Internal displacements are assumed to vary linearly within the element.

ux(x, y, t) = c1x

a+ c2

x

a

y

b+ c3

y

b+ c4

uy(x, y, t) = c5x

a+ c6

x

a

y

b+ c7

y

b+ c8




The eight coefficients c1, · · · , c8 may be found uniquely from matching the displacementcoordinates at the corners.

ux(a, b) = u1 , uy(a, b) = u2

ux(0, b) = u3 , uy(0, b) = u4

ux(0, 0) = u5 , uy(0, 0) = u6

ux(a, 0) = u7 , uy(a, 0) = u8

resulting in internal displacements

ux(x, y, t) = xy u1(t) + (1− x)y u3(t) + (1− x)(1− y) u5(t) + x(1− y) u7(t) (95)uy(x, y, t) = xy u2(t) + (1− x)y u4(t) + (1− x)(1− y) u6(t) + x(1− y) u8(t) (96)

where x = x/a (0 ≤ x ≤ 1) and y = y/b (0 ≤ y ≤ 1) so that

Ψ(x, y) =[xy 0 (1− x)y 0 (1− x)(1− y) 0 x(1− y) 00 xy 0 (1− x)y 0 (1− x)(1− y) 0 x(1− y)

](97)

and [ux(x, y, t)uy(x, y, t)

]= Ψ(x, y) u(t) (98)

Strain-displacement relations

εxx = ∂ux∂x

= 1a

∂ux∂x

εyy = ∂uy∂y

= 1b

∂uy∂y

γxy = ∂ux∂y

+ ∂uy∂x

= 1b

∂ux∂y

+ 1a

∂uy∂x

so that

εxxεyyγxy

=

y/a 0 −y/a 0 −(1− y)/a 0 (1− y)/a 00 x/b 0 (1− x)/b 0 −(1− x)/b 0 −x/bx/b y/a (1− x)/b −y/a −(1− x)/b −(1− y)/a −x/b (1− y)/a

u1

u2

u3

u4

u5

u6

u7

u8

or

ε(x, y, t) = B(x, y) u(t)




6.2 Stress-Strain relationships

6.2.1 Plane-Stress

In-plane behavior of thin plates, σzz = τxz = τyz = 0For plane-stress elasticity, the stress-strain relationship simplifies to

σxxσyyτxy

= E

1− ν2

1 ν 0ν 1 00 0 1

2(1− ν)

εxxεyyγxy

(99)

orσ = Spσ ε (100)

6.2.2 Plane-Strain

In-plane behavior of continua, εzz = γxz = γyz = 0For plane-strain elasticity, the stress-strain relationship simplifies to

σxxσyyτxy

= E

(1 + ν)(1− 2ν)

1− ν ν 0ν 1− ν 00 0 1

2 − ν

εxxεyyγxy

(101)

orσ = Spε ε (102)

6.3 2D Rectangular Element Strain Energy and Elastic Stiffness Matrix

V = 12

∫A

σ(x, y, t)Tε(x, y, t) h dx dy (103)

= 12 u(t)T

∫A

[B(x, y)T Se(E, ν) B(x, y)

]8×8

h dx dy u(t) (104)

Elastic element stiffness matrix

Ke =∫A

[B(x, y)T Se(E, ν) B(x, y)

]8×8

h dx dy (105)

6.4 2D Rectangular Element Kinetic Energy and Mass Matrix

T ( ˙u) = 12

∫Aρ |u(x, y, t)|2 h dx dy (106)

= 12

˙u(t)T∫Aρ[Ψ(x, y)T Ψ(x, y)

]8×8

h dx dy ˙u(t) (107)

Consistent mass matrix

M =∫Aρ[Ψ(x, y)T Ψ(x, y)

]8×8

h dx dy (108)




6.5 2D Rectangular Plane-Stress and Plane-Strain Element Stiffness and Mass Matrices

6.5.1 Plane-Stress stiffness matrix

Ke = Eh12(1−ν2) ·

4c+ kA kB −4c+ kA/2 −kC −2c− kA/2 −kB 2c− kA kC

kB 4/c+ kD kC 2/c− kD −kB −2/c− kD/2 −kC −4/c+ kD/2

−4c+ kA/2 kC 4c+ kA −kB 2c− kA −kC −2c− kA/2 kB

−kC 2/c− kD −kB 4/c+ kD kC −4/c+ kD/2 kB −2/c− kD/2

−2c− kA/2 −kB 2c− kA kC 4c+ kA kB −4c+ kA/2 −kC

−kB −2/c− kD/2 −kC −4/c+ kD/2 kB 4/c+ kD kC 2/c− kD

2c− kA −kC −2c− kA/2 kB −4c+ kA/2 kC 4c+ kA −kB

kC −4/c+ kD/2 kB −2/c− kD/2 −kC 2/c− kD −kB 4/c+ kD

where c = b/a and

kA = (2/c)(1− ν)kB = (3/2)(1 + ν)kC = (3/2)(1− 3ν)kD = (2c)(1− ν)

6.5.2 Plane-Strain stiffness matrix

Ke = Eh12(1+ν)(1−2ν) ·

kA + kB 3/2 −kA + kB/2 6ν − 3/2 −kA/2 − kB/2 −3/2 kA/2 − kB 3/2 − 6ν

3/2 kC + kD 3/2 − 6ν kC/2 − kD −3/2 −kC/2 − kD/2 6ν − 3/2 −kC + kD/2

−kA + kB/2 3/2 − 6ν kA + kB −3/2 kA/2 − kB 6ν − 3/2 −kA/2 − kB/2 3/2

6ν − 3/2 kC/2 − kD −3/2 kC + kD 3/2 − 6ν −kC + kD/2 3/2 −kC/2 − kD/2

−kA/2 − kB/2 −3/2 kA/2 − kB 3/2 − 6ν kA + kB 3/2 −kA + kB/2 6ν − 3/2

−3/2 −kC/2 − kD/2 6ν − 3/2 −kC + kD/2 3/2 kC + kD 3/2 − 6ν kC/2 − kD

kA/2 − kB 6ν − 3/2 −kA/2 − kB/2 3/2 −kA + kB/2 3/2 − 6ν kA + kB −3/2

3/2 − 6ν −kC + kD/2 3/2 −kC/2 − kD/2 6ν − 3/2 kC/2 − kD −3/2 kC + kD

where c = b/a and

kA = (4c)(1− ν)kB = (2/c)(1− 2ν)kC = (4/c)(1− ν)kD = (2c)(1− 2ν)




6.5.3 Mass matrix

The element mass matrix for the plane-stress and plane-strain elements is the same.

M = ρabh

36

4 0 2 0 1 0 2 00 4 0 2 0 1 0 22 0 4 0 2 0 1 00 2 0 4 0 2 0 11 0 2 0 4 0 2 00 1 0 2 0 4 0 22 0 1 0 2 0 4 00 2 0 1 0 2 0 4

(109)

Note, again, that these element stiffness matrices are approximations based on an assumeddistribution of internal displacements.

7 Element damping matrices

Damping in vibrating structures can arise from diverse linear and nonlinear phenomena.

If the structure is in a fluid (liquid or gas), the motion of the structure is resisted by thefluid viscosity. At low speeds (low Reynolds numbers), this damping effect can be takento be linear in the velocity, and the damping forces are proportional to the total rate ofdisplacement (not the rate of deformation). If the fluid is flowing past the structure at highflow rates (high Reynolds numbers), the motion of the structure can interact with the flowingmedium. This interaction affects the dynamics (natural frequencies and damping ratios) ofthe coupled structure-fluid system. Potentially, at certain flow speeds, the motion of thestructure can increase the transfer of energy from the flow into the structure, giving rise toan aero-elastic instabililty.

Damping can also arise within structural systems from friction forces internal to the structure(the micro-slip within joints and connections) inherent material viscoelasticity, and inelas-tic material behavior. In many structural systems, a type of damping in which dampingstresses are proportional to strain and in-phase with strain-rate are assumed. Such so-called“complex-stiffness damping” or “structural damping” is commonly used to model the damp-ing in soils. Fundamentally, this kind of damping is neither elastic nor viscous. The force-displacement behavior does not follow the same path in loading and unloading, behaviorbut instead follows a “butterfly” shaped path. Nevertheless, this type of damping is com-monly linearized as linear viscous damping, in which forces are proportional to the rate ofdeformation.

In materials in which stress depends on strain and strain rate, a Voigt viscoelasticity modelmay be assumed, in which stress is proportional to both strain ε and strain-rate ε,

σ = [ Se(E, ν) ] ε + [ Sv(η) ] ε




The internal virtual work of real viscous stresses Svε moving through virtual strains δε is

δW ( ˙u) =∫

Ωσ(x, t)T δε(x, t) dΩ (110)

=∫

Ωε(x, t)T Sv(η) δε(x, t) dΩ

=∫

Ω˙u(t)TB(x)T Sv(η) B(x)δu(t) dΩ

= ˙u(t)T∫

Ω

[B(x)T Sv(η) B(x)

]N×N

dΩ δu(t) (111)

Given a material viscous damping matrix, Sv, a structural element damping matrix can bedetermined for any type of structural element, through the integral in equation (111), as hasbeen done for stiffness and mass element matrices earlier in this document. In doing so, itmay be assumed that the internal element displacements ui(x, t) (and the matrices [Ψ] and[B]) are unaffected by the presence of damping, though this is not strictly true. Further,the parameters in Sv(η) are often dependent of the frequency of the strain and the strainamplitude. Damping behavior that is amplitude-dependent is outside the domain of linearanalysis.

7.1 Rayleigh damping matrices for structural systems

In an assembled model for a structural system, a damping matrix that is proportional tosystem’s mass and stiffness matrices is called a Rayleigh damping matrix.

Cs = αMs + βKs

RTCsR =

2ζ1ωn1

. . .2ζNωnN

= α

1

. . .1

+ β

ωn

21

. . .ωn

2N

(112)

where ωn2j is an eigen-value (squared natural frequency) and the columns of R are mass-

normalized eigen-vectors (modal vectors) of the generalized eigen-problem

[Ks − ωn2jMs ]rj = 0 . (113)

From equations (112) it can be seen that the damping ratios satisfy

ζj = α

21ωnj

+ β

2ωnj

and the Rayleigh damping coefficients (α and β) can be determined so that the dampingratios ζj have desired values at two frequencies. The damping ratios modeled by Rayleighdamping can get very large for low and high frequencies. Rayleigh damping grows to ∞ asω → 0 and increases linearly with ω for large values of ω. Note that the Rayleigh dampingmatrix has the same banded form as the mass and stiffness matrices. In other words, withRayleigh damping, internal damping forces are applied only between coordinates that areconnected by structural elements.




7.2 Caughey damping matrices for structural systems

The Caughey damping matrix is a generalization of the Rayleigh damping matrix. Caugheydamping matrices can involve more than two parameters and can therefore be used to providea desired amount of damping over a range of frequencies. The Caughey damping matrix foran assembled model for a structural system is

Cs = Ms

j=n2∑j=n1

αj(M−1s Ks)j

where the index range limits n1 and n2 can be positive or negative, as long as n1 < n2. Aswith the Rayleigh damping matrix, the Caughey damping matrix may also be diagonalizedby the real eigen-vector matrix R. The coefficients αj are related to the damping ratios, ζk,by

ζk = 12

1ωk

j=n2∑j=n1

αjω2jk

The coefficients αj may be selected so that a set of specified damping ratios ζk are obtainedat a corresponding set of frequencies ωk. If n1 = 0 and n2 = 1, then the Caughey damp-ing matrix is the same as the Rayleigh damping matrix. For other values of n1 and n2 theCaughey damping matrix loses the banded structure of the Rayleigh damping matrix, imply-ing the presence of damping forces between coordinates that are not connected by structuralelements.

Structural systems with classical damping have real-valued modes rj that depend only onthe system’s mass and stiffness matrices (equation (113)), and can be analyzed as a sys-tem of uncoupled second-order ordinary differential equations. The responses of the systemcoordinates can be approximated via a modal expansion of a select subset of modes. Theconvenience of the application of modal-superpostion to the transient response analysis ofstructures is the primary motivation

7.3 Rayleigh damping matrices for structural elements

An element Rayleigh damping matrix may be easily computed from the element’s mass andstiffness matrix C = αM+βK and assembled into a damping matrix for the structural systemCs. The element damping is presumed to increases linearly with the mass and the stiffness ofthe element; larger elements will have greater mass, stiffness, and damping. System dampingmatrices assembled from such element damping matrices will have the same banding as themass and stiffness matrices; internal damping forces will occur only between coordinatesconnected by a structural element. However, such an assembled damping matrix will not bediagonizeable by the real eigenvectors of the structural system mass matrix Ms and stiffnessmatrix Ks.




7.4 Linear viscous Damping elements

Some structures incorporate components designed to provide supplemental damping. Thesesupplemental damping components can dissipate energy through viscosity, friction, or inelas-tic deformation. In a linear viscous damping element (a dash-pot), damping forces are linearin the velocity across the nodes of the element and the forces act along a line between thetwo nodes of the element. The element node damping forces fd are related to the elementnode velocities vd through the damping coefficient cd[

fd1

fd2

]=[

cd −cd

−cd cd

] [vd1

vd2

]

The damping matrix for a linear viscous damper connecting a node at (x1, y1) to a node at(x2, y2) is found from the element coordinate transformation,

C6×6 =[c s 0 0 0 00 0 0 c s 0

]T [cd −cd

−cd cd

] [c s 0 0 0 00 0 0 c s 0

]

where c = (x2 − x1)/L and s = (y2 − y1)/L. Structural systems with supplemental dampingcomponents generally have non-classical system damping matrices.

References[1] Clough, Ray W., and Penzien, Joseph, Dynamics of Structures, 2nd ed. (revised), Com-

puters and Structures, 2003.

[2] Cowper, G.R., “Shear Coefficient in Timoshenko Beam Theory,” J. Appl. Mech.,33(2)(1966):335-346

[3] Dong, S.B, Alpdogan, C., and Taciroglu, E. “Much ado about shear correction factors inTimoshenko beam theory,” Int. J. Solids & Structures, 47(2010):1651-1655.

[4] Gruttman, F., and Wagner, W., “Shear correction factors in Timoshenko’s beam theoryfor arbitrary shaped cross-sections,” Comp. Mech. 27(2001):199-207.

[5] Kaneko, T., “An experimental study of the Timoshenko’s shear coefficient for flexurallyvibrating beams,” J. Phys. D: Appl. Phys. 11 (1978): 1979-1988;

[6] Paz, Mario, Structural Dynamics Theory and Computation, Chapman & Hall, 2000.

[7] Przemieniecki, J.S., Theory of Matrix Structural Analysis, Dover, 1985. ?

[8] Rosinger, H.E., and Ritchie, I.G., “On Timoshenko’s correction for shear in vibratingisotropic beams,” J. Phys. D: Appl. Phys., 10 (1977): 1461–1466.



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