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Differential quadrature and transformation methods for vibration problems

Author :  S. Rajasekaran    Posted On :  27.08.2016 10:32 pm

Differential quadrature and transformation methods for vibration problems in relation to structural dynamics during earthquakes Abstract: In this page, natural frequencies are obtained for beam-like structures with different boundary conditions using the differential quadrature method and differential transformation methods. The results obtained are compared with those obtained from the finite element method and other numerical methods. Programs in MATLAB are given for solving beam problems by the differential quadrature method. The symbolic programming package MATHEMATICA is ideally suited to solve recursive equations of differential transformation methods.

Differential quadrature and transformation methods for vibration problems in relation to structural dynamics during earthquakes

Abstract: In this page, natural frequencies are obtained for beam-like structures with different boundary conditions using the differential quadrature method and differential transformation methods. The results obtained are compared with those obtained from the finite element method and other numerical methods. Programs in MATLAB are given for solving beam problems by the differential quadrature method. The symbolic programming package MATHEMATICA is ideally suited to solve recursive equations of differential transformation methods.

Key words: differential quadrature, harmonic quadrature, Lagrange interpolation, differential transformation, boundary conditions, natural frequency.

Introduction

Numerical solutions to free vibration analysis of beams and columns are obtained by the method of differential quadrature (DQ) and harmonic differential quadrature (HDQ) for various support conditions. The obtained results are compared with the existing solutions available from other numerical methods such as finite element method (FEM) and analytical results. In addition, this chapter also uses a recently developed technique, known as the differential transformation (DT) to determine the natural frequency of beams and columns. In solving the problem, governing differential equations are converted to algebraic equations using DT methods which must be solved together with applied boundary conditions. The symbolic programming package MATHEMATICA is ideally suited to solve such recursive equations by considering fairly large numbers of terms.

DQ method

These problems of free vibration of beams and columns either prismatic or non-prismatic, could easily be solved using the DQ method, which was introduced by Bellman and Casti (1971). With the application of boundary conditions as per Wilson’s method (Wilson 2002) the DQM method will also be straightforward and easy for engineers to use. Since the introduction of this method, applications of the DQ method to various engineering problems have been investigated and their success has shown the potential of the method as an attractive numerical analysis tool. The basic idea of the DQM method is to quickly compute the derivatives of a function at any grid point within its bounded domain by estimating the weighted sum of the values of the functions at a small set of points related to the domain. In the originally derived DQM, Lagrangian interpolation polynomial was used (Bert and Malik 1996, Bert et al. 1993, 1994). A recent approach of the original differential quadrature approximation, HDQ, was originally proposed by Striz et al. (1995). Unlike the DQ method, HDQM uses harmonic or trigonometric functions as the test functions. As the name of the test function suggested, this is called the HDQ method. All the problems in this chapter have demonstrated that the application of the DQ and HDQ methods will lead to accurate results with less computational effort and that there is a potential that the method may become alternative to conventional methods such as finite difference, finite element and boundary element.

Lagrangian interpolation

This interpolation technique is applied if the given points may or may not be equally spaced (see Fig. 15.1). The polynomial is an approximation to the function f (x), which coincides with the polynomial at (xi, yi).

The fourth order governing differential equation for free vibration of column with varying flexural rigidity ‘D’ (D = EI) and w (= the lateral deflection) may be written as

and the number of sampling points n > m.

A natural and often convenient choice for sampling point is that of equally spaced points or CGL mesh distribution as given by Eq. 15.8. For the sampling points, we adopt well accepted Chebyshev–Gauss–Lobatto mesh distribution given by Shu (2000) as

L’ is the length of the column, the column is divided into ‘ne’ (say 40) divisions or ne+1 (41) sampling points in the case of DQM and xi is the distance from the bottom end of the column.

HDQ method

The harmonic test function hi(ξ) used in the HDQ method is defined as

Higher order derivatives can be obtained using Eq. 15.9.

Transverse vibration of pre-tensioned cable

The transverse vibration of a taut cable or string has an equation of motion similar to the governing longitudinal vibration of a uniform rod. Consider a uniform elastic cable, having mass/unit length (ρ A), ρ being the mass density, to be stretched under tension T between two fixed points as shown in Fig. 15.2. Assuming small deflections and slopes, the equation of motion for transverse vibration is given by

where c =    T/ρA is the velocity of wave propagation along the cable and y is the transverse deflection of the string at any distance x. The above equation is also identical to the wave equation. The transverse deflection may be assumed as

A MATLAB program to find the natural frequency of transverse vibration of pretensioned string is shown below.

Program 15.1: MATLAB program for finding the natural frequency of lateral vibration of pre-tensioned string

STRINGVIB

% free vibration of a pretensioned cable by differential quadrature clc; close all;

ne=20;

n=ne+1;

nn=2*n;

no=4;

m=zeros(n,1);

x=zeros(n,1);

c=zeros(n,n,no);

d=zeros(n+2,n+2);

f=zeros(n+2,n+2);

%give length and mass density and area of the cable l=1;

dl=l/ne;

rho=7800;

ar=0.005;

ma=rho*ar;

%give tension in the cable T=4*ma;

format long; for i=1:n

x(i)=.5*(1-cos((i-1)*pi/ne)); end

d(1:n,1:n)=c(:,:,2)/l^2;

%application of boundary conditions d(n+1,1)=1.0;

d(n+2,n)=1.0;

d(1,n+1)=1.0;

d(n,n+2)=1.0;

din=inv(d); f(1:n,1:n)=-eye(n,n); ddf=din*f; [u,eu]=eig(ddf);

for i=1:n ww1(i)=u(i,3); ww2(i)=u(i,4); ww3(i)=u(i,5);

end

disp(‘ fundamental natural frequency\n’) wn1=sqrt(T/(eu(3,3)*ma)) disp(‘fundamental mode shape\n’)

ww1'

disp(‘ second natural frequency’) wn2=sqrt(T/(eu(4,4)*ma)) disp(‘second mode shape’)

ww2'

disp(‘ third natural frequency’) wn3=sqrt(T/(eu(5,5)*ma)) disp(‘ third mode shape’) ww3'

figure(1);

plot(x,ww1);

xlabel(‘x’);

ylabel(‘w’);

title(‘ fundamental mode shape’); figure(2);

plot(x,ww2);

xlabel(‘x’);

ylabel(‘w’);

title(‘ fundamental mode shape’); figure(3);

plot(x,ww3);

xlabel(‘x’);

ylabel(‘w’);

title(‘ fundamental mode shape’);

m=zeros(n,1);

c=zeros(n,n,4); for i=1:n

m(i,1)=1; for k=1:n

if ((k ==i )) jk=i;

else m(i,1)=m(i,1)*sin((x(i)-x(k))*pi/2);

format long end

end end m;

for i=1:n for j=1:n

if(j==i)

jk=i; else

c(i,j,1)=(pi*m(i,1))/(2.0*sin((x(i)-x(j))*pi/2)*m(j,1)); end

end end

for i=1:n c(i,i,1)=0.0;

for j=1:n if ((i==j)) jk=i;

else c(i,i,1)=c(i,i,1)-c(i,j,1);

end

end end o=2;

for i=1:n for j=1:n

if(j==i)

jk=i; else

c(i,j,o)=c(i,j,1)*(2.0*c(i,i,1)-pi/tan(((x(i)-x(j))*pi/2))); end

end end

for i=1:n c(i,i,2)=0.0;

for j=1:n if ((i==j)) jk=i;

else c(i,i,2)=c(i,i,2)-c(i,j,2);

end

end end

for o=3:no for i=1:n

for j=1:n c(i,j,o)=0.0;

for k=1:n c(i,j,o)=c(i,j,o)+c(i,k,1)*c(k,j,(o-1));

end end

end end y=c;

m=zeros(n,1);

c=zeros(n,n,4); for i=1:n

m(i,1)=1; for k=1:n

if ((k ==i )) jk=i;

else m(i,1)=m(i,1)*(x(i)-x(k));

format long end

end end

for i=1:n for j=1:n

if(j==i)

jk=i; else

c(i,j,1)=m(i,1)/((x(i)-x(j))*m(j,1)); end

end end

for i=1:n c(i,i,1)=0.0;

for j=1:n if ((i==j)) jk=i;

else c(i,i,1)=c(i,i,1)-c(i,j,1);

end

end end

for o=2:no for i=1:n

for j=1:n c(i,j,o)=0.0;

for k=1:n c(i,j,o)=c(i,j,o)+c(i,k,1)*c(k,j,(o-1));

end end end

end c(:,:,1); y=c;

OUTPUT

fundamental natural frequency

wn1 = 6.28318530717963

fundamental mode shape ans =

0

0.00733128915883

0.02911774918179

0.06459034487273

0.11203447410058

0.16833305876402

0.22868163719304

0.28673229986014

0.33532231670280

0.36772667715649

0.37911498526798

0.36772667715649

0.33532231670280

0.28673229986015

0.22868163719305

0.16833305876402

0.11203447410058

0.06459034487273

0.02911774918180

0.00733128915883

0

second natural frequency wn2 =

12.56637061435911

second mode shape

ans =

0

0.01384802941406

0.05484814152807

0.12024309852971

0.20220761642762

0.28495445387276

0.34458819094757

0.35438646074539

0.29557188373010

0.16900007820430

-0.00000000000001

-0.16900007820432

-0.29557188373010

-0.35438646074539

-0.34458819094755 -0.28495445387275 -0.20220761642761 -0.12024309852971 -0.05484814152806 -0.01384802941406 0

third natural frequency wn3 =

18.84955592153890 third mode shape

ans =

0

0.02026414001026

0.07989008567297

0.17170783853740

0.27374891998932

0.34314670851919

0.32560576874274

0.18817009067074 -0.03995789607568 -0.25873859745335 -0.34947339621276 -0.25873859745337 -0.03995789607570 0.18817009067073 0.32560576874275 0.34314670851920 0.27374891998931 0.17170783853740 0.07989008567298 0.02026414001026 0

Example 15.1

A string of length 1 m subjected to pre-tension of 156 N is under transverse vibration. Calculate the first three fundamental natural frequencies assuming mass density = 7800 kg/m3 and area of the cable = 0.005 m2.

Solution

The nth natural frequency is given by

ωn = 2 n π

The values obtained by DQ are

which agree with the true values. The mode shapes corresponding to these three frequencies are shown in Fig. 15.3 for first, second and third modes.

Lateral vibration of uniform Euler beams

The governing differential equation of lateral vibration of uniform beams is given as

Example 15.2

Find the fundamental three natural frequencies of a simply supported beam given that E = 200 GPa; I = 18.6 × 10–6; A = 2.42 × 10–4; ρ = 7800 kg/m3;

L = 20 m.

Solution

The closed form solutions for the problem are given by

EULERBEAMVIB

% free vibration of a beam by differential quadrature for different boundary conditions

clc; close all; ne=20; n=ne+1; nn=2*n; no=4; m=zeros(n,1); x=zeros(n,1);

c=zeros(n,n,no);

d=zeros(n+4,n+4);

f=zeros(n+4,n+4); iy=18.6e-6; ar=2.42e-4; ymod=200e9; l=20.0;

dl=l/ne;

mden=7800; format long; for i=1:n

x(i)=.5*(1-cos((i-1)*pi/ne)); end

d(1:n,1:n)=ymod*iy*c(:,:,4)/l^4; %application of boundary conditions %pinned - pinned

d(n+1,1)=1.0;

d(n+2:n+2,1:n)=c(1,1:n,2)/l^2;

d(n+3,n)=1.0;

d(n+4:n+4,1:n)=c(n,1:n,2)/l^2;

d(1,n+1)=1.0;

d(n,n+3)=1.0; for i=1:n

d(i,n+2)=d(n+2,i);

d(i,n+4)=d(n+4,i); end

%fixed – fixed

%d(n+1,1)=1.0;

%d(n+2:n+2,1:n)=c(1,1:n,1)/l;

%d(n+3,n)=1.0

%d(n+4:n+4,1:n)=c(n,1:n,1)/l;

%d(1,n+1)=1.0;

%d(n,n+3)=1.0;

%for i=1:n

%d(i,n+2)=d(n+2,i);

%d(i,n+4)=d(n+4,i);

%end

%fixed - free

%d(n+1,1)=1;

%d(n+2:n+2,1:n)=c(1,1:n,1)/l;

%d(n+3:n+3,1:n)=c(n,1:n,2)/l^2;

%d(n+4:n+4,1:n)=c(n,1:n,3)/l^3;

%for i=1:n

%d(i,n+1)=d(n+1,i);

%d(i,n+2)=d(n+2,i);

%d(i,n+3)=d(n+3,i);

%d(i,n+4)=d(n+4,i);

%end

din=inv(d);

f(1:n,1:n)=eye(n,n);

ddf=din*f;

[u,eu]=eig(ddf); for i=1:n

ww1(i)=u(i,5);

ww2(i)=u(i,6);

ww3(i)=u(i,7); end

sprintf(‘ fundamental natural frequencies\n’) wn1=sqrt(1/(mden*ar*eu(5,5))) wn2=sqrt(1/(mden*ar*eu(6,6))) wn3=sqrt(1/(mden*ar*eu(7,7)))

sprintf(‘ fundamental mode shape\n’) ww1'

figure(1);

plot(x,ww1);

xlabel(‘x’);

ylabel(‘w’);

title(‘ fundamental mode shape’); figure(2);

plot(x,ww2);

xlabel(‘x’);

ylabel(‘w’);

title(‘ second mode shape’); figure(3);

plot(x,ww3);

xlabel(‘x’);

ylabel(‘w’);

title(‘ third mode shape’);

OUTPUT

ans =

fundamental natural frequencies

wn1 =   34.63827370993050

wn2 =  1.385530948382473e+002

wn3 =  3.117444633857984e+002

To find natural frequency and mode shape given variation of D = EI for Euler beam with axial load

In this problem, (D = EI) and P are known and w and ω are unknown values that can be found by solving as an eigenvalue problem as explained below. Assuming c(:,:,m) is the mth derivative , i.e.      C m , the governing equation may be written as

Boundary conditions

Since it is a fourth order differential equation, four boundary conditions should be given. The boundary conditions will be applied as follows.

Clamped–pinned

w = 0 at x = 0; G[n + 1, 1] = 1.0

w = 0 at x = 0; G[n + 2, 1: n] = c(1, 1: n, 1)/L

w = 0 at x = L; G[n + 3, n] = 1

w = 0 at x = L; G[n + 4, 1: n] = c(n, 1: n, 2)/L              ----15.35

Clamped–clamped

w = 0 at x = 0; G[n + 1, 1] = 1.0

w = 0 at x = 0; G[n + 2, 1: n] = c(1, 1: n, 1)/L w = 0 at x = L; G[n + 3, n] = 1

w = 0 at x = L; G[n + 4, 1: n] = c(n, 1: n, 2)/L          --- 15.36

Pinned–pinned

w = 0 at x = 0; G[n + 1, 1] = 1.0

w = 0 at x = 0; G[n + 2, 1: n] = c(1, 1: n, 2/L2) w = 0 at x = L; G[n + 3, n] = 1

w = 0 at x = L; G[n + 4, 1: n] = c(n, 1: n, 2)/L2     --- 15.37

Clamped–free

w = 0 at x = 0; G[n + 1, 1] = 1.0

w = 0 at x = 0; G[n + 2, 1: n] = c(1, 1: n, 1)/L w = 0 at x = L; G[n + 3, n] = c(n, 1: n, 2)/L2

dD w ′′ + Dw ′′′ = Pw at x = L ; G [n + 4,1: n] dx

= αnnc(n, 1: n, 2)/L2 + Dnnc(n, 1: n, 3)/L3

Pc (n, 1: n, 1)/L                --- 15.38

1Wilson’s method (Wilson, 2002) of applying boundary conditions

In general, the boundary conditions are given by

[G]1{w} = [E]1{w}                         --- 15.39

Combining governing equations and boundary conditions, we get

The above equation has both equilibrium and equation of geometry. Solving Eq. 15.41 is an eigenvalue problem; one will be able to obtain the natural frequency.

Program 15.3: MATLAB program for solving free vibration problem of non-prismatic beam with or without axial load

EULERVIB

% free vibration of non-prismatic Euler beams with or without axial load %using differential quadrature method

clc;

ne=50;

n=ne+1;

nn=2*n;

no=4;

m=zeros(n,1);

x=zeros(n,1);

xi=zeros(n,1);

c=zeros(n,n,no);

d=zeros(n+4,n+4);

e=zeros(n+4,n+4);

z=zeros(n+4,1);

f=zeros(n+4,1);

alp=zeros(n,n);

bet=zeros(n,n);

zz=zeros(n,1);

ki=zeros(n,n);

eta=zeros(n,n);

const=1.0;

l=12;

ymod=200e09;

rho=7800; format long; for i=1:n

xi(i)=.5*(1-cos((i-1)*pi/ne));

%mi(i)=0.000038*(1-xi(i)^2/2);

ar(i)=1/rho;

mi(i)=0.000038;

ki(i,i)=ymod*mi(i);

end

for i=1:n

alp(i,i)=0;

bet(i,i)=0;

for j=1:n

alp(i,i)=alp(i,i)+c(i,j,1)*ki(j,j)/l;

bet(i,i)=bet(i,i)+c(i,j,2)*ki(j,j)/l^2;

end

end

d=zeros(n+4,n+4);

% free vibration of the beam

% axial load on the beam t=+ if it is compressive t=- if it is tensile

% weight of the beam / unit length

t=520895.0;

d(1:n,1:n)=2.0*alp*c(:,:,3)/l^3+bet*c(:,:,2)/l^2+ki*c(:,:,4)/l^4+eta+t*c(:,:,2)/

l^2;

% boundary conditions

% clamped - free

% d(n+1,1)=1.0;

% d(n+2:n+2,1:n)=alp(n,n)*c(n,1:n,2)/l^2+ki(n,n)*c(n,1:n,3)/l^3+t*c(n,1:n,1)/

l;

% d(n+3:n+3,1:n)=c(1,1:n,1)/l;

% d(n+4:n+4,1:n)=ki(n,n)*c(n,1:n,2)/l^2;

% d(1,n+1)=1.0;

% for i=1:n

% d(i,n+2)=d(n+2,i);

% d(i,n+3)=d(n+3,i);

% d(i,n+4)=d(n+4,i); % end

% pinned - pinned d(n+1,1)=1.0;

d(n+2:n+2,1:n)=ki(n,n)*c(n,1:n,2)/l^2;

d(n+3:n+3,1:n)=ki(1,1)*c(1,1:n,2)/l^2;

d(n+4,n)=1.0;

d(n,n+4)=1.0;

d(1,n+1)=1.0;

d(n+4,n)=1.0;

for i=1:n d(i,n+2)=d(n+2,i);

d(i,n+3)=d(n+3,i); end

e=zeros(n+4,n+4); for i=1:n

e(i,i)=rho*ar(i); end

din=inv(d);

z=din*e;

[ev,euv]=eig(z); for i=1:n

zz(i)=ev(i,5); end

omega=sqrt(1/euv(5,5)); sprintf(‘ natural frequency\n’) omega

figure(1);

plot(xi,zz) xlabel(‘ x/L ’) ylabel(‘ z’)

title (‘ fundamental mode shape ’)

Example 15.3

Find the buckling load of a pinned–pinned column given E = 200 GPa; I = 0.000 038 m4; mass density = 7800 kg/m3, A = 1/7800; span = 12 m. Find the natural frequency if the axial load is tension of magnitude 300 000 N.

Solution

By trial and error giving various axial loads the natural frequency is calculated for each axial load and the load at which natural frequency becomes imaginary is the buckling load. For the problem, buckling load is 520 895 N which agrees with (π 2 E I/L2) = 520 895 N.

When the axial load = 0 the natural frequency corresponds to the Euler beam simply supported conditions without axial load and is given by ωn = 188.9 rad/s. When the axial load is negative, i.e. tension say P = –300 000 N, the natural frequency is 237.19 which corresponds to the true value

Example 15.4

A cantilever beam shown in Fig. 15.4 is analysed for free vibration. The data E = 200 GPa; mass density = 7800 kg/m3 L = 4.572 m. The width of the flange = 0.203 m and thickness of flange and web are 0.0178 and 0.0114 m respectively.

Solution

The natural frequency is obtained as 195.628 rad/s which agrees with 191 rad/s obtained by Wekezer (1987).

Vibration of Timoshenko beam by DQ method

The governing differential equations for free vibration of Timoshenko beam (including shear deformation and rotary inertia) have been given in Eq. 13.112 as

where [I] is the unit matrix and I is the moment of inertia. Equation 15.45 is written as

λ[ B ]{q} = [ D ]{q}                        --- 15.46

where [B] is of size nn = 2 × n where n is the number of discrete points. Applying boundary conditions (clamped clamped conditions)

B[nn + 1, 1] = 1.0

B[nn + 2, n] = 0

B[nn + 3, n + 1] = 1.0

B[nn + 4, 2n] = 1.0

Using Wilson’s method

B[1, nn + 1] = 1.0

B[n,nn + 2] = 1.0

B[n + 1, nn + 3] = 1.0

B[2n, nn + 4] = 1.0

Now [B] and [D] matrices of size nt + 4 because of the application of boundary conditions. Solving as an eigenvalue problem λ and hence ω can be determined.

Example 15.5

Find the natural frequency of a clamped clamped beam for the following conditions: L = 10 m, E = 200 GPa, ρ = 7800 kg/m3, assume unit width (k = 0.83).

TIMOSHENKOVIB

% free vibration analysis of timoshenko beam clc; close all;

ne=20;

no=2;

n=ne+1;

nn=2*n;

nt=nn+4;

m=zeros(n,1);

x=zeros(n,1);

c=zeros(n,n,no);

d=zeros(nt,nt);

e=zeros(nt,nt);

l=10;

e=2e11;

g=.8e11;

ar=2;

ir=0.667;

ak=0.83;

rho=7800;

15.5 Mode shape for (a) lateral deflection; and (b) φ.

for i=1:n x(i)=0.5*(1-cos((i-1)*pi/ne));

d(1:n,1:n)=ak*g*ar*c(:,:,2)/l^2; d(1:n,n+1:nn)=-ak*g*ar*c(:,:,1)/l; d(n+1:nn,1:n)=-ak*g*ar*c(:,:,1)/l; d(n+1:nn,n+1:nn)=ak*g*ar*eye(n,n)-e*ir*c(:,:,2)/l^2; % %boundary conditions for fixed end d(nn+1,1)=1.0;

d(nn+2,n)=1.0;

d(nn+3,n+1)=1.0;

d(nn+4,nn)=1.0;

d(1,nn+1)=1.0;

d(n,nn+2)=1.0;

d(n+1,nn+3)=1.0;

d(nn,nn+4)=1.0;

%boundary conditions for ssd end

%d(nn+1,1)=1.0;

%d(nn+2,n)=1.0;

%d(nn+3:nn+3,n+1:nn)=c(1:1,1:n,1)/l;

%d(nn+4:nn+4,n+1:nn)=c(n:n,1:n,1)/l;

%d(1,nn+1)=1.0;

%d(n,nn+2)=1.0;

%for i=1:n

%d(n+i,nn+3)=d(nn+3,n+i);

%d(n+i,nn+4)=d(nn+4,n+i);

%end

e(1:n,1:n)=-rho*ar*eye(n,n); e(n+1:nn,n+1:nn)=rho*ir*eye(n,n); e(nn+1:nt,1:nt)=0;

ddi=inv(d);

f=ddi*e;

[evv,ev]=eig(f);

disp(‘ natural frequency’); wn1=sqrt(1/ev(5,5)) figure(1);

for i=1:n z(i)=evv(i,5);

end plot(x,z) xlabel(‘ x ‘) ylabel(‘ z’)

title (‘ mode shape of deflection’) figure(2);

for i=1:n z(i)=evv(n+i,5);

end plot(x,z) xlabel(‘x’)

ylabel(‘ phi’)

title(‘ mode shape of phi’)

OUTPUT

natural frequency wn1 =

5.292070310268033e+002

DT method

The concept of the DT method was first introduced some 30 years ago by Pukhov (Chai and Wang 2006) Since then, DT has been used with success in structural mechanics. The concept of the DT method is readily available in Russian literature. For a function w(x), DT exists as

Equation 15.49 is obviously a Taylor series expansion of the function w(x) about x = 0. The differential technique essentially converts a differential equation into an algebraic equation, similar to integral transform methods such as Laplace and Fourier transform. The final resulting algebraic equations are solved together with boundary conditions. The transformation can be given by a simple formula as

Transverse vibration of pre-tensioned cable

The governing equation for the transverse vibration of a pre-tensioned cable

nonlinear equation in terms of ‘a’ and linear in terms of ‘c’. Equation 15.59 may be written as A * c = 0, and since c 0, A must be zero where A is the coefficient of c in f (c, a). It should be noted that a fairly a large number of terms are needed for convergence of the natural frequency coefficient. Figure 15.6 shows the convergence of a for different numbers of terms in the summation of Eq. 15.59 and a is obtained as 9.8696. Hence natural frequency is obtained as

Program 15.5: MATHEMATICA program for finding the natural frequency of vibration of a pre-tensioned cable

Free vibration analysis of Euler beam

The governing equation is given by Eq. 15.24 as

Boundary conditions

Case 1 Simply supported at both ends

y(0) = 0; y(0) = 0; y(1) = 0; y (1) = 0 The DT equivalents are

Y[0] = 0; Y[1] = c; Y[2] = 0; Y[3] = d

And

since c and d are not zero, for a non-trivial solution to exist the determinant of the matrix must be zero, i.e.

aa × dd cc × bb = 0            --- 15.68

where aa, bb, are the coefficients of c and d in the equation S = 0 and cc, dd are the coefficients of c and d in the equation T = 0. The root of Eq. 15.68 is the solution for the problem.

For a beam with simply supported ends a = 97.4091 leads to natural frequency asy(1) = 0.

Program 15.6: MATHEMATICA program for finding the natural frequency of vibration of an Euler beam

Natural frequency of Euler beam subjected to axial load

The governing differential equation for an Euler beam subjected to axial load is given by

where EI is flexural rigidity, P is axial compressive load, ρ is mass density/ unit volume, L is span of the beam and y = lateral deflection.

15.19.1Pin roller support

The boundary conditions are

y(0) = y(0) = y(1) = y(1) = 0         --- - - 15.80

This can be interpreted in terms of DT as

Y[0] = 0; Y[2] = 0; Y[1] = c; Y[3] = d  --- ---- 15.81a

where aa, bb are the coefficients of c and d in Eq. 15.81b and cc and dd are the coefficients of c and d in Eq. 15.81c. For a numerical solution to exist, the determinant || A || has to be zero and hence a can be found once β is given. One has to include more terms, say, than 35 in Eq. 15.79 for accuracy.

Example 15.6

Find the buckling load of a pinned–pinned column given E = 200 GPa; I = 0.000 038 m4; mass density = 7800 kg/m3, A = 1/7800; span = 12 m. Find the natural frequency if the axial load is tension of magnitude 300 000 N.

Solution

By trial and error giving various axial loads the natural frequency is calculated for each axial load and the load at which natural frequency becomes imaginary is the buckling load. For the problem, the buckling load is (β = 9.862) P = 520 895 N, which agrees with (π2 EI/L2) = 520 895 N.

When the axial load = 0 the natural frequency corresponds to the Euler beam simply supported conditions without axial load and a is given by a =

97.4091 for which ωn = 188.9 rad/s.

When the axial load is negative, i.e. tension, say P = – 300 000 N (β = – 5.864), the value of a = 153.5 obtained from DT from which natural frequency is 237.19 which corresponds to the true value

Similarly, other boundary conditions could be tackled.

Program 15.7: MATHEMATICA program for finding the natural frequency of an Euler beam subjected to axial load

Natural frequency of a Timoshenko beam

The governing differential equation for free vibration of Timoshenko beams (including shear deformation and rotary inertia into account) is given in Eq. 15.42a and 15.42b. Combining these two equations and identifying y = w = lateral deflection of the beam, the governing differential equation is given by

For clamped beam the following boundary conditions are to be applied: y = 0 and φ = 0 both at x = 0 and x = 1. In the absence of incorporating the correct boundary conditions, assuming the shear deformation at the ends of the beam are negligible, we can apply the boundary conditions as

y = 0 and y = 0, both at x = 0 and x = 1.

Example 15.7

Find the natural frequency of a clamped-clamped beam for the following conditions. L = 10 m, E = 200 GPa, ρ = 7800 kg/m3. Assume unit width (k = 0.83).

Program 15.8: MATHEMATICA program for finding the natural frequency of a Timoshenko beam

Summary

The DQ and HDQ methods are applied to solve for natural frequency of strings and, beams with or without axial load. For finding the natural frequencies, unlike Rayleigh–Ritz methods, DQ and HDQ methods do not need the construction of an admissible function that satisfies the boundary conditions a priori. Accurate results are obtained for the problems even with a small number of discrete points used to discretize the domain. This approach is convenient for solving problems governed by higher order differential equations, and matrix operations could be performed using MATLAB with ease. It is also easy to write algebraic equations in the place of differential equations and to apply boundary conditions. It is also explained in this chapter how the Lagrange multiplier method is used to convert rectangular matrix to square matrix by incorporating boundary conditions using Wilson’s method. Results with high accuracy are obtained in all study cases and DQM and HDQ methods are computationally efficient. DQM and HDQ methods are straightforward so the same procedures can be easily employed for handling problems with the other boundary conditions.

In this chapter, the DT method is also highlighted and its usefulness demonstrated by solving stability analysis of fully supported prismatic and non-prismatic piles. It is also shown in this chapter how DT can be used to convert differential equation to a set of algebraic equations of recursive nature. It is also shown that, together with boundary conditions, these equations are solved for natural frequency of various types of problems. A fairly large number of terms, say 35 to 40, are required for convergence. DT is efficient and easy to implement, particularly in symbolic program packages such as MATHEMATICA. Mode shape also could be obtained using Eq. 15.48. It is expected that DQ, HDQ and DT will be more promising for further development into efficient and flexible numerical techniques for solving practical engineering problems in future.

Tags : Civil - Structural dynamics of earthquake engineering
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