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CIVL 7/8111 Time-Dependent Problems - 2-D Equations 1/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations Swiss Mathematician and physicist discovered the wave equation in three space dimensions.

Leonhard Euler (1707 – 1783) was a pioneering Swiss mathematician and physicist. He made important discoveries in fields as diverse as infinitesimal calculus and graph theory. He is also renowned for his work in mechanics, , optics, astronomy, and music theory

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations The wave equation is an important second-order linear partial for the description of – as they occur in physics – such as waves, waves and water waves.

It arises in fields like , electromagnetics, and fluid dynamics.

A solution of the wave equation in two dimensions with a zero- boundary condition along the entire outer edge. CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 2/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations Consider the motion of a rectangular membrane (in the absence of gravity) using the two-dimensional wave equation. Plots of the spatial part for modes are illustrated below.

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations Consider the motion of a rectangular membrane (in the absence of gravity) using the two-dimensional wave equation. CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 3/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations The governing that describe the pro- pagation of waves in situations involving two independent variables appear typically as 2u k  u xyt,,   f xyt,,  0 in  t 2

ugst,on1 u kstuqst ,,on   n 2 uxy,,0 cxy , in uxy,,0 dxy,in t

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations

Where  is the interior domain, and 1 and 2 form the boundary of the domain.

s

1 u kstuqst ,,   n

2u kuxyt ,,   fxyt,,  0 t 2 ugst ,

2 CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 4/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations The physical constants are k and , and s is a linear mea- sure of the position on the boundary.

s

1 u kstuqst ,,   n

2u kuxyt ,,   fxyt,,  0 t 2 ugst ,

2

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations

The type of boundary condition specified on 2 results from a local balance between internal and external forces.

s

1 u kstuqst ,,   n

2u kuxyt ,,   fxyt,,  0 t 2 ugst ,

2 CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 5/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations This equation is similar in form to the parabolic initial- presented in the previous section with the very important change in the time derivative term from a first to a , and with the addition of a second initial condition on the velocity.

With these changes, the problem changes from a parabolic initial-boundary value problem to a hyperbolic initial boundary value problem.

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations The basic steps of discretization, interpolation, elemental formulation, assembly, constraints, solution, and computation of derived variables are presented in this section as they relate to the two-dimensional hyperbolic initial-boundary value problem.

The Galerkin method, in connection with the corresponding weak formulation to be developed, will be used to generate the finite element model.

Discretization - Referred to the material in Chapter 3. CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 6/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations The Galerkin Finite Element Method Interpolation - The solution is assumed to be expressible in

terms of the nodally based interpolation functions ni(x, y) introduced and discussed in Section 3.2. In the present setting, these interpolation functions are used with the semidiscretization.

N1 uxyt(,,)  uii () tn (, xy ) i 1

The ni(x, y) are nodally based interpolation functions and can be linear, quadratic, or as otherwise desired.

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations The Galerkin Finite Element Method Elemental Formulation - The starting point for the elemental formulation is the weak formulation of the initial- boundary value problem.

The first step in developing the weak formulation is to multiply the differential equation by an arbitrary test

function v(x, y) vanishing on 1.

The result is then integrated over the domain  to obtain

2u vku    fd 0   2  t CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 7/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations The Galerkin Finite Element Method Elemental Formulation - Using the two-dimensional form of the divergence theorem to integrate the first term by parts, the results after rearranging are: 2u vkud  v d   2 t vkudn  vfd 

where  = 1 + 2

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations The Galerkin Finite Element Method

Elemental Formulation - Recalling that v vanishes on 1 and that ku/x = kn u = q - hu on 2, it follows that 2u vkud  v d   2 t vq hud  vfd 

This equation is the required weak formulation for the two- dimensional diffusion problem. CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 8/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations The Galerkin Finite Element Method Elemental Formulation - Substituting the approximation of

u(x, y, t) into the weak formulation and taking v = nk, k = 1, 2, ... yields: nnnn kikiku ku d  ii   xxyy

nundnhund   kiikii  

nfd nqd kk  k  1, 2,   2

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations The Galerkin Finite Element Method

- - Elemental Formulation – Where  and 2 represent the elemental areas approximating , and the collection of the

elemental edges approximating 2,respectively. nnnn kikiku ku d  ii   xxyy

nundnhund   kiikii  

nfd nqd kk  k  1, 2,   2 CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 9/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations The Galerkin Finite Element Method Elemental Formulation – This N x N set of linear algebraic equations can be written as N  AkiuBuFt i ki i k ( ) k  1,2,..., N i 1

nnnn A kikkdnhnd ki ki  k i xxyy

Bnnd ki k i 

Fnfdnqd kk  k  2

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations The Galerkin Finite Element Method Elemental Formulation – Note that assembly is contained implicitly within the formulation. N  AkiuBuFt i ki i k ( ) k  1,2,..., N i 1

nnnn A kikkdnhnd ki ki   k i xxyy

Bnnd ki k i 

Fnfdnqd kk  k  2 CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 10/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations The Galerkin Finite Element Method Elemental Formulation – In terms of the corresponding elementally based interpolations

TT uxyeee(, )Nu u N The finite element model can be expressed as Au Bu F

' B= r AkGG a  G ee e

' F= fGG q ee

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations The Galerkin Finite Element Method Elemental Formulation – In terms of the corresponding elementally based interpolations

TT uxyeee(, )Nu u N The finite element model can be expressed as

NNTT  NN ke kkdA  xxyy Ae  rNN  T dA aNN  T ds e  e  Ae  2e

fN fdA qN qds e  e  Ae  2e CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 11/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations The Galerkin Finite Element Method Elemental Formulation – The initial conditions for the system of first-order differential equations are obtained from the initial conditions prescribed for the original initial- boundary value problem.

Generally u(0) is determined by evaluating the function c(x, y) at the nodes to obtain:

T uu(0) 0123ccc... cNN 1 c

th where ci = c(xi, yi) with (xi, yi) the coordinates of the i node.

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations The Galerkin Finite Element Method Elemental Formulation – The initial conditions for the system of first-order differential equations are obtained from the initial conditions prescribed for the original initial- boundary value problem.

Generally ů(0) is determined by evaluating the function d(x, y) at the nodes to obtain

 T uu(0) 0123ddd... dNN 1 d

th where di = d(xi, yi) with (xi, yi) the coordinates of the i node. CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 12/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations The Galerkin Finite Element Method Constraints - The constraints arise from the boundary

conditions specified on 1.

Generally the values of the constraints are determined from the q function with the constrained value of u at a node on

1 being taken as the value of q at that point.

These constraints are then enforced on the assembled equations, resulting in the final global constrained set of linear first-order differential equations.

 Mu  Ku f uu(0)  0 uu(0)  0

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations The Galerkin Finite Element Method Solution - The system of equations is precisely the same in form and character as the corresponding equations developed in Section 4.2.2 for the one-dimensional wave problem.

The analytical method as well as the numerical methods using the central difference and Newmark algorithms can be used for integrating the above set of equations. CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 13/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations The Galerkin Finite Element Method Solution - The central difference algorithm will be conditionally stable with the critical time step depending on the maximum eigenvalue of the associated problem (K - 2M)v = 0.

The Newmark algorithm will be unconditionally stable for  = 0.5 and  = 0.25( + 0.5)2.

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations The Galerkin Finite Element Method Derived variables - In a physical situation governed by a wave equation, the derived variables are usually the

internal forces computed according to Fe = keue per element for each time step. CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 14/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1 Consider the two-dimensional wave problem shown below.

2 2  x y Tp 2 0 cos cos  t 22  y

(1,1) (-1,1)  xt,1,  0

 xy,,0  0  1,yt , 0  1,yt ,  0

x  xy,,0  0 t

(-1,-1) (1,-1)  xt,1, 0

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1 Discretization. Due to symmetry, we will model the top right-most quadrant of the membrane using four equally- sized 4-noded quadrilaterals.

y 2 2  p0 x y  2  cos cos  t T 22 

  1,yt , 0   x T  xt,1,0

 xy,,0  xy,,0  0  0 t CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 15/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1

Elemental Formulation – This N x N set of linear algebraic equations can be written as:

N  Aki iBFtk ki i k ()  1,2,..., N i 1

nn  nn Adki ki ki    xx  yy

Bnnd ki k i 

Fnfd kk 

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1

Elemental Formulation – In terms of the corresponding elementally based interpolations

TT eee(,xy )NN The finite element model can be expressed as

NNTT  NN ke dA  xx  yy Ae 

rNN T dA e  Ae

fN fdA e  Ae CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 16/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1 Interpolation - In matrix notation, the distribution of the function over the element in local (s, t) coordinates is:

TT e xy, eeNN 11st  Nst1 ,  t 4 (1,1) (1, 1) 11st  4 3 Nst,  2 4 s (0,0) 11st  Y Nst,  3 4 1 2 (1,1) (1, 1) 11st  Nst4 ,  X 4

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1 Rectangular Elements - The elemental interpolations for a rectangular element are:

4 3 4 3 N 1 N2

12 12

4 3 N4 4 3 N3

12 12 CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 17/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1 Rectangular Elements - The shape functions are visually deceiving. There is no in directions parallel to any side; however, there is a twist due to the xy term in the element representation.

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1 Interpolation - This integrals may be transformed into the

local coordinate space using as = x - x0 and bt = y - y0, for a rectangular element 2a x 2b in size: 11  , Fxydxdy, Fx00 asy, bt dsdt    st, Ae 11

where the Jacobian (,)/(s, t) has a value of Ae/4.

The resulting global system matrix ke is:

2211 2112   1 ba2211 1221  k  e 6 ab1122 1221  1122  2112  CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 18/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1

Rectangular Elements - Evaluation of ke

If the element is square, then a = b, then ke becomes: 2211 2112   1 ba2211 1221  k  e 6 ab1122 1221  1122  2112 

4121   1 14 12 k   e 6 2141   12  14

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1

The elemental me matrices are: 11  , mNN T dA Fx asy, bt dsdt e   00 st, Ae 11

where the Jacobian (,)/(s, t) has a value of Ae/4.

The resulting global system stiffness matrix me is:

4212  A 2421 m  e  e 36 1242  2124 CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 19/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1

Transforming the integral into the non-dimensional coordinates (s, t) yields: 11 A fNNf T e ds dt e  11 4

42ffff1234 2  Ae 242ffff1234 fe   36 ff12242 ff 3 4  224ff12 f 3  f 4 The resulting 4 x 1 elemental load vector contributes to the global system equations at those locations corresponding to the four nodes defining the element.

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1 (0, 1) (½, 1) (1, 1) For element 1: 9 8 7 (0, ½) (½, ½) (1, ½) 1 4121  4 3 6  1 14 122 1 1  (0, 0) (1, 0) k1  A1  6 2141 3 4 1 (½, 0) 2 5  12  14 4

2 0.05089 1 42121 f1      0.02659 2 1 24212 f2 1 2   m  f    f1   1 f 2 1 0.02659 144 12423 3   3    0.01389 2124 f4 2  4 1  4 CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 20/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1 (0, 1) (½, 1) (1, 1) For element 2: 9 8 7 (0, ½) (½, ½) (1, ½) 2 4121  4 3 6  1 14 125 1 2  (0, 0) (1, 0) k2  A2  6 2141 6 4 1 (½, 0) 2 5  12  14 3

0.04539 2 42122 f2 2      0.01329 5 1 24215 f5 1 0   m  f    f2   2 f 2 0.02371 144 12426 6 0  6    0.00694 2124 f3 1  3 2  3

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1 (0, 1) (½, 1) (1, 1)

9 8 7 For element 3: 3 (0, ½) (½, ½) (1, ½) 4 4121  4 3 6  1 14 123 1  (0, 0) (1, 0) k3  A3  6 2141 8 4 1 (½, 0) 2 5  12  14 9

0.04048 3 42124 f4 2      0.01185 6 1 24213 f3 1 1   m  f    f3   3 f 2 0.01185 144 12428 8 0  7    0.00347 2124 f9 0  9 3  8 CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 21/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1 (0, 1) (½, 1) (1, 1) For element 4: 9 8 7 4 (0, ½) (½, ½) (1, ½) 3 4121  4 3 6  1 14 126 1  (0, 0) (1, 0) k4  A4  6 2141 7 4 1 (½, 0) 2 5  12  14 8

0.04539 3 42123 f3 1     f 0.02371 8 1 24216 6 1 0  m  f   f4   4 f 02 0.01329 144 12427 7  9 f  0.00694 2124 8 4 0  8 4

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1

The assembled KG, MG, and F matrices are:

4-1-2-1000001 -18-2-2-1-2000 2

-2 -2 16 -2 -2 -2 -2 -2 -2 3  -1 -2 -2 8 0 0 0 -2 -1 4 1 K  0-1-204-1000 G 6 5 0 -2 -2 0 -1 8 -1 -2 0 6 00-200-14-10 7 0 0 -2 -2 0 -2 -1 8 -1 8  00-2-1000-149 CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 22/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1

The assembled KG, MG, and F matrices are:

42 1 200000 1 0.05089 28 4 121000  2 0.07197 14164141413 0.10178  21 4 800012  4 0.07197 1  MG  02 1 042000 F  0.01329 5  144  01 4 0282106 0.01880  00 1 002420 7 0.00347  00 4 1012828 0.01880  00 1 2000249 0.01329

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1 Applying nonhomogeneous boundary conditions that

{5, … , 9} = {0, 0, 0, 0, 0}; the constrained equations are:  KMFGG   4-1-2-1000001 42 1 2000001 0.05089   -1 8 -2 -2 -1 -2 0 0 0  28 4 121000  0.07197 2 2   -2 -2 16 -2 -2 -2 -2 -2 -2 3 1416414141 01. 0178  3    -1 -2 -2 8 0 0 0 -2 -1 4 21 4 8000124 0.07197 1     0-1-204-10005 02 1 042000 0.01329 6 0 144 5 0-2-20-18-1-20 01 4 028210 0.01880 06 6   00-200-14-1007 00 1 0024207 0.00347   0 0 -2 -2 0 -2 -1 8 -1 0 00 4 101282  0.01880 8 8  0   0 0 -2 -1 0 0 0 -1 4 9 00 1 2000 2 4 9 0.01329

  59,,   0 CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 23/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1 Assembly. With both the boundary conditions essential, BT=0 and bt=0. It follows that the assembled equations are:  KMFGG 

 4-1-2-11 42121 0.05089   1 -1 8 -2 -22  2 8 4 1  0.07197   2    6-2 -2 16 -23 144 1 4 16 4 3 0.10178     -1 -2 -2 84 2 1 4 8 4 0.07197

Let assume the  = 1 for this example.

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1 For the Newmark algorithm, the critical time step is associated with the largest eigenvalue of the (K – 2M) = 0 system.

2 hcr  max

The eigenvalue problem is:

2 KMGG0

4-1-2-1 4212  1 -1 8 -2 -22 2 8 4 1  0 6-2 -2 16 -2 144 1 4 16 4  -1 -2 -2 8 2 1 4 8 CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 24/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1

The eigenvalues can be found using a variety of available solution techniques. In Matlab, use [V,D] = eig(A).

For example, [V,D] = eig(MG\KG) gives eigenvalues D:

22 2 2 125.19332 34.28571  3  34.28571  4  63.37811

The eigenvectors V are: 0.66667 0.89443 0.09022 0.66667  0.47140 0.00000 0.70350 0.47140 VV12 V 3  V 4  0.33333 0.44721 0.04511 0.33333 0.47140 0.00000 0.70350 0.47140 

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1

For the example the largest eigenvalue is 63.37811, the critical time step is: 2 2 hcr    0.2512sec. max 63.37811 CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 25/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1 Newmark's Method– The equations to be solved at the first step can be written as:

222 MKMhhc100101 hh  F  10 h  0  ch  10 h 20 1 c   hc  1 2  10 0 10 1 2  h  c 1 2 2

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1 Newmark's Method– Evaluating the differential equation at t = 0 yields

0 00    1 MFK00 00MFK 

47.02041 11.75510 2.93878 11.75510 0.05089   11.75510 23.51020 5.87755 2.93878 0.07197 0   2.93878 5.87755 11.75510 5.87755 0.10178  11.75510 2.93878 5.87755 23.51020 0.07197  1.00000  0.70711   0.50000 0.70711 CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 26/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1 Newmark's Method– Taking  = 0.25,  = 0.5, and h = 0.01 seconds yields:

0.02779 0.01388 0.00694 0.01388 0.12723  2 0.01388 0.05559 0.02777 0.00694 250.17993 MKh h F  10 0.00694 0.02777 0.11118 0.02777 0.25446  0.01388 0.00694 0.02777 0.05559 0.1799 3

 00  00 At step 1: 0   0   222 MKMhhc100101 hh  F

0.02779 0.01388 0.00694 0.01388 0.12723 0.01388 0.05559 0.02777 0.00694 0.17993 25  10 M0.25h 10 0.00694 0.02777 0.11118 0.02777 0.25446  0.01388 0.00694 0.02777 0.05559 0.1799 3

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1

Newmark's Method– The solution for 1 is

5 T 1  10 4.9994 3.5351 2.4997 3.5351

Now solve for the velocity at t = h   0 00   00   0 10 h  0  ch  10h 20  00  0 4.99935 1.00000  3.53507 0.70711 1  5  120ch 10 ch2  h h 2.49968 0.50000 3.53507 0.70711 T  0.01000 0.00707 0.005 0.00707 CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 27/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1 Newmark's Method– The equation for acceleration at

t = 0 is 0 00   hc   c   10 0 101 10 1 h2  h2   0 00  Now solve for the acceleration at t = h 4.99935  1.00000 3.53507  0.70711   5  1210 ch  h 2.49968  0.50000 3.53507  0.70711

 T 1  0.99974 0.70692 0.49987 0.70692

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1 Newmark's Method – For n = 2:

4 T 2  10 1.99948 1.41385 0.99974 1.41385  T 2  0.01999 0.01414 0.01000 0.01414  T 2  0.99896 0.70637 0.49948 0.70637

For n = 3: 4 T 3  10 4.49786 3.18047 2.24893 3.18047

 T 3  0.02998 0.02120 0.01499 0.02120  T 3  0.99766 0.70546 0.49883 0.70546 CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 28/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1

The values of 1, 2, and 3 for 0 < t < 10 sec. are show below:

1

2

3

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1

The velocity values of 1, 2, and 3 for 0 < t < 10 sec. are:

 1

 2  3 CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 29/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1

The acceleration values of 1, 2, and 3 for 0 < t < 10 sec. are:

 1  2  3

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1

 x,,yt (,,)x yt   t CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 30/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1

2 x,,yt (,,)x yt   t 2

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1 Consider the two-dimensional wave problem shown below.

y

 xt,1, 0 (1,1) ft(0,0, ) (-1,1)  

 4 2 ft() uyt1, , 0 t uyt1, ,  0 0.1 t x  xy,,0  0

(-1,-1)  xt,1, 0 (1,-1) CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 31/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 2 Discretization. Due to symmetry, we will model the top right-most quadrant of the membrane using four equally- sized 4-noded quadrilaterals.

y 2 2 ft() t 2

 1,yt , 0  xt,1,0 x

 xy,,0  xy,,0  0  0 t

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 2

Rectangular Elements - Evaluation of ke

As in Example 1, all the ke are:

4121 1 4121 2  1 14 122 1 14 125 k1  k2  6 2141 3 6 2141 6  12 14 12 14   4   3

4121 3 4121 3  1 14 126 1 14 128 k3  k4  6 2141 7 6 2141 9  12 14 12 14   8   4 CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 32/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 2

The elemental me matrices are: 11  , mNN T dA Fx asy, bt dsdt e   00 st, Ae 11

where the Jacobian (,)/(s, t) has a value of Ae/4.

The resulting global system stiffness matrix me is:

4212  A 2421 m  e  e 36 1242  2124

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 2

Rectangular Elements - Evaluation of me

As in Example 1, all the me are:

42121 42122   1 24212 1 24215 m1  m2  144 12423 144 12426   2124 2124 4 3

42123 42123   1 24216 1 24218 m3  m4  144 12427 144 12429   2124 2124 8 4 CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 33/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 2

Transforming the integral into the non-dimensional coordinates (s, t) yields: 11 A fNNf T e ds dt e  11 4

42ffff1234 2  Ae 242ffff1234 fe   36 ff12242 ff 3 4  224ff12 f 3  f 4 The resulting 4 x 1 elemental load vector contributes to the global system equations at those locations corresponding to the four nodes defining the element.

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 2 (0, 1) (½, 1) (1, 1) For element 1: 9 8 7 (0, ½) (½, ½) (1, ½)

f1 1 0 4 3 6    1 f2 0 0 (0, 0) (1, 0) f      1 (½, 0) 2 5 f3 0 0 f   4 1 0 t 0.1 0 t 0.1

0.02778 1 0 1   0.01389 2 0 2 f1     0.00694 3 0 3 0.01389 0 t 0.1 4 t 0.1 4 CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 34/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 2 (0, 1) (½, 1) (1, 1) For elements 2 to 4: 9 8 7 4 3 (0, ½) (½, ½) (1, ½) 2 f2 0 0 4 3 6    f 0 0 5 2 5   (0, 0) (1, 0) f   f2   f6 0 0 6 1 (½, 0) 2 5 f 0 0 3 2  3

3 3 f3 0 0 f3 0 0       f6 0 0 6 f8 0 0 8 f   f3   f   f4   f7 0 0 7 f9 0 0 9 f 0 0 f 0 0 8 3  8 4 4  4

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 2

The assembled KG, MG, and F matrices are:

4-1-2-1000001 -18-2-2-1-2000 2

-2 -2 16 -2 -2 -2 -2 -2 -2 3  -1 -2 -2 8 0 0 0 -2 -1 4 1 K  0-1-204-1000 G 6 5 0 -2 -2 0 -1 8 -1 -2 0 6 00-200-14-10 7 0 0 -2 -2 0 -2 -1 8 -1 8  00-2-1000-149 CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 35/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 2

The assembled KG, MG, and F matrices are:

42 1 200000 1 28 4 121000 2

14164141413  21 4 8000124 1 M  02 1 042000 G 144 5 01 4 0282106 00 1 002420 7 00 4 1012828  00 1 2000249

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 2

The assembled KG, MG, and F matrices are:

0.02778 1 0 1 0.01389  0  2  2

0.00694 3 0 3   0.01389 4 0 4   F  0.00000 5 F  0 5 0.00000  0  6  6 0.00000 7 0 7     0.00000 8 0 8   0.00000 t0.1 9 0 t 0.1 9 CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 36/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 2 Applying nonhomogeneous boundary conditions that

{5, … , 9} = {0, 0, 0, 0, 0}; the constrained equations are:  KMFGG  4-1-2-1000001 42 1 2000001 0.02778   -1 8 -2 -2 -1 -2 0 0 0  284121000 0.01389 2 2   -2 -2 16 -2 -2 -2 -2 -2 -2 3 1416414141 0.00694  3    -1 -2 -2 8 0 0 0 -2 -1 4 21 4 8000124 0.01389 1  1   0-1-204-100005 02 1 0420005 0.00000 6 144  0-2-20-18-1-20 0 01 4 028210 0.00000 6 6   00-200-14-1007 00 1 0024207 0.00000   0 0 -2 -2 0 -2 -1 8 -1 0 004101282 0.00000 8 8   0   0 0 -2 -1 0 0 0 -1 4 9 00 1 2000 2 4 9 0.00000 t 0.1

  59,,   0

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 2 Assembly. With both the boundary conditions essential, BT=0 and bt=0. It follows that the assembled equations are:  KMFGG

 4-1-2-11 42121 0.02 7778   11-18-2-22 2841  0.013889   2    6144-2 -2 16 -23 1 4164 3 0.006944      -1 -2 -2 84 2 1 498 4 0.01388 t 0.1  4-1-2-11 42121 0   11-18-2-22 2841  0  2   6144-2 -2 16 -23 1 4 16 4 3 0    -1 -2 -2 84 2 1484 0 t 0.1 CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 37/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 2 For the Newmark algorithm, the critical time step is associated with the largest eigenvalue of the (K – 2M) = 0 system.

2 hcr  max

The eigenvalue problem is the same as Example 1:

2 KMGG0

4-1-2-1 4212  11-1 8 -2 -2 2 8 4 1  0 6-2 -2 16 -2 144 1 4 16 4  -1 -2 -2 8 2 1 4 8

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 2

The eigenvalues can be found using a variety of available solution techniques. In Matlab, use [V,D] = eig(A).

For example, [V,D] = eig(MG\KG) gives eigenvalues D:

22 2 2 125.19332 34.28571  3  34.28571  4  63.37811

The eigenvectors V are: 0.66667 0.89443 0.09022 0.66667  0.47140 0.00000 0.70350 0.47140 VV12 V 3  V 4  0.33333 0.44721 0.04511 0.33333 0.47140 0.00000 0.70350 0.47140  CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 38/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 2

For the example the largest eigenvalue is 63.37811, the critical time step is: 2 2 hcr    0.2512sec. max 63.37811

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 2 Newmark's Method – The equations to be solved at the first step can be written as:

222 MKMhhc100101 hh  F  10 h  0  ch  10h 20 1 c   hc  1 2  10 0 10 1 2  h  c 1 2 2 CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 39/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 2 Newmark's Method – Evaluating the differential equation at t = 0 yields

0 00    1 MFK00 00MFK 

47.02041 11.75510 2.93878 11.75510 0.027778   11.75510 23.51020 5.87755 2.93878 0.013889 0   2.93878 5.87755 11.75510 5.87755 0.006944  11.75510 2.93878 5.87755 23.51020 0.013889 1.00000  0.00000   0.00000 0.00000

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 2 Newmark's Method – Taking  = 0.25,  = 0.5, and h = 0.01 seconds yields:

0.02779 0.01388 0.00694 0.01388 6.94444  2 0.01388 0.05559 0.02777 0.00694 273.47222 MKh h F  10 0.00694 0.02777 0.11118 0.02777 1.73611  0.01388 0.00694 0.02777 0.05559 3.4722 2

 00  00 At step 1: 0   0   222 MKMhhc100101 hh  F

0.02779 0.01388 0.00694 0.01388 6.94444 0.01388 0.05559 0.02777 0.00694 3.47222 27  10 M0.25h 10 0.00694 0.02777 0.11118 0.02777 1.73611  0.01388 0.00694 0.02777 0.05559 3.4722 2 CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 40/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 2

Newmark's Method – The solution for 1 is

T 1  4.996E-05 1.284E-08 6.595E-12 1.284E-08

Now solve for the velocity at t = h   0 00   00   0 10 h  0  ch  10h 20

0 00 

T  9.9914E-03 2.5670E-06 1.3191E-09 2.5670E-06

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 2 Newmark's Method – The equation for acceleration at

t = 0 is 0 00   hc   c   10 0 101 10 1 h2  h2   0 00  Now solve for the acceleration at t = h

 T 1  0.998295.1341E-04 2.6381E-07 5.1341E -04 CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 41/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 2 Newmark's Method – For n = 2:

T 2  1.9966E-04 1.0259E-07 7.9076E-11 1.0259E-07  T 2  0.01995 1.5385E-05 1.3177E-08 1.5385E-05  T 2  0.99316 2.0501E-03 2.1078E-06 2.0501E-03

For n = 3: T 3  4.4859E-04 4.2268E-07 4.8051E-10 4.2268E-07  T 3  0.02984 4.8633E-05 6.7109E-08 4.8633E-05  T 3  0.98464 4.5996E-03 8.6787E-06 4.5996E-03

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 2 Newmark's Method – For n = 11, t > 0.1 sec.

0.027778 1 0 1   0.013889 2 0 2 F   F   0.006944 3 0 3   0.013889 t 0.1 4 0 t 0.1 4

3 T 11  10 5.81661 0.06193 0.00055 0.06193

 3 T 11  10 97.58691 2.19325 0.02862 2.19325  T 11  0.19688 0.05772 0.00126 0.05772 CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 42/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1

The values of 1, 2, and 3 for 0 < t < 10 sec. are show below:

1

2

3

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1

The velocity values of 1, 2, and 3 for 0 < t < 10 sec. are:

 1

 2  3 CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 43/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1

The acceleration values of 1, 2, and 3 for 0 < t < 10 sec. are:

 1

 2

 3

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1

 x,,yt (,,)x yt   t CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 44/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations - Example 1

2 x,,yt (,,)x yt   t 2

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations

Closure - Time-dependent problems are inherently more difficult and expensive to solve than their corresponding steady-state counterparts.

The expense of generating the global matrices is higher for the time-dependent problems because of the necessity of computing the mass matrices.

The main extra expense, however, is in solving the resulting time-dependent global equations. CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 45/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations

Closure - For an analytical approach to the solution, additional expense is incurred in terms of having to determine eigenvalues and eigenvectors.

The actual amount of expense depends on the specific form of the stiffness and mass matrices and the algorithm used, but in any case it is significantly in excess of the expense of solving the single set of linear algebraic equations associated with the steady-state problem.

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations

Closure - For a time domain integration technique, the additional expense is clearly related to the number of time steps necessary to trace out the desired time history.

In addition to several matrix multiplications and additions, each step can involve the solution of a set of linear algebraic equations.

In some instances this expense can be minimized by using a decomposition that can be reused for the computation of the solution at each new time. CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 46/47

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations

Closure - In this regard recall that the Euler and central difference algorithms require that the size of the time step not exceed a value proportional to the inverse of the largest eigenvalue.

For large systems this critical step size can be very small resulting in many applications of the algorithm to trace out the time history.

TIME-DEPENDENT PROBLEMS Two-Dimensional Wave Equations

Closure - The unconditionally stable Crank-Nicolson and Newmark algorithms, on the other hand, can be used with arbitrary step size that has been chosen so as to accurately integrate the lower modes, with significant improvement in the expense relative to the conditionally stable Euler and central difference algorithms.

There are of course other algorithms available that are specifically tailored to address other numerical issues. CIVL 7/8111 Time-Dependent Problems - 2-D Wave Equations 47/47

End of Chapter 4d