Home | | **Structural Dynamics and Earthquake Engineering** | Modal response contribution using Chopraâ€™s method

The forced vibration of undamped system and the dynamic equation of motion may be written as
muË™Ë™ + ku = F (t )

**Modal response contribution
using Chopraâ€™s method (Chopra, 2002) **

The forced vibration of undamped
system and the dynamic equation of motion may be written as

*mu***Ë™Ë™*** *+* ku *=*
F *(*t *) --- ---
18.22

We now consider the common
loading case in which the force *F _{j}*(

where *Ï†** _{r}* is the
normalized eigenvector for the

Pre-multiplying both sides with *Ï†*_{n}^{T} and
utilizing the orthogonalization property of the modes we get

which is independent of how the modes are normalized.

Equation 18.26 may be viewed as
an expression of the distribution *F* of applied force in terms of force
distribution *F _{n}* associated with natural period. This
interpretation becomes apparently clear by considering the structure vibration
in the

Two
useful properties are to be noted:

1. The force
vector *F _{n}*

2. This
dynamic response in the *n*th mode is entirely due to the partial force
vector *F _{n}p*(

Example 18.4

Consider a five storey building
(rigid floor beams and slabs) with lumped mass *m* at each floor, and same
storeyed stiffness *k*. <*F*>= <0 0 â€“1 1 2). See Fig. 18.8.

Solution

Use the MATHEMATICA or MATLAB
package. Assume k = 1 *m* = 1.

Using the MATLAB program we get
five natural frequencies and five normalized mode shapes.

*Ï‰*_{1} = 0.2439; *Ï‰*_{2} =
0.6689; *Ï‰*_{3} = 1.0; *Ï‰*_{4} = 1.286;
*Ï‰*_{5} = 1.6850
rad/s

**Modal analysis for ****Î“**** f (t) **

The factor Î“* _{n}* which
multiplies the force

Let us consider the uncoupled
equation of motion

Hence *y _{n}* is
readily available once equation 18.30 has been solved for

which is the contribution of the *n*th
mode of modal displacement *u*(*t*). Substituting Eq. 18.30 to get
the equivalent static force as

The *n*th mode contribution
to any response quantity *R*(*t*) is determined by static analysis of
structures subjected to forces *f _{n}*(

The model analysis procedure just
presented is a special case of the one presented earlier. It has the advantage
of providing a basis for identifying and understanding the factors that
influence the relative modal contributions to the response. The above procedure
can be applied only if the *f* (*t*) for all the forces are the same.

**Interpretation of modal
analysis **

1. Determine
natural frequencies and modes.

2. Force
distribution is expanded into modal components {*F*}.

The rest of the procedure is explained in Table 18.1.

**Modal contribution factor **

The contribution *R _{n}*
of the

where *R ^{st}* is
the static value of

These modal contribution factors have three important
properties:

1. They are
dimensionless.

2. They are
independent of how modes are normalized.

3. The sum
of modal contribution factors over all modes is unity.

**Modal response and required number of modes**

Consider the displacement *D _{n}*(

*D _{n}*

The corresponding value *R _{n}*(

The algebraic system of *R _{no}*
is same as the modal static response

For a fixed *J, e _{j}*
depends on spatial distribution of

**Modal contributions**

For the five storey situation
three modes are required for base shear and two modes are required for roof
displacement.

Example 18.5

For the five storey building example

*F*(*t*)
=* Ff *(*t*); <* F *>=< 0 0 â€“ 1 1 2>

The system is undamped. Compute only the steady state
response.

Example 18.6

Figure 18.11 shows a shear frame
(i.e. rigid beams) and its floor masses and storey stiffness. This structure is
subjected to horizontal harmonic force at the top floor. *p*_{0} =
500 kN.

(a) Determine
the equation for steady state displacement of the structure.

(b) Determine
the direct solution of coupled equations.

(c) Determine
the modal analysis (neglect damping).

(d) If *k*
= 63 600 kN/m, *m* = 45 413 kg and *Ï‰* = 0.75 *Ï‰*_{1} split
the load into modal components and obtain the solution using Chopraâ€™s method.

Solution

Example 18.7

Figure 18.13 shows a mass-less
simply suported beam with three lumped masses and the following properties: *L*
= 3.81 m, *m* = 33 665 kg, *E* = 207.15 GPa.

We are interested in studying the
dynamic response of the beam to *F*(*t*) = *Ff*(*t*) where
<* F *>=<1 0 0>.

(a) Determine
the modal expansion of vector {*F*} that defines the spatial distribution
of force.

(b)For the bending moment M_{1}
at the location of U_{1} degrees of freedom determine the modal static
response

(c) Calculate
and tabulate modal contribution factors their cumulative values for various
numbers of modes included *J* = 1, 2, 3 and the error *e _{j}*
for static response. Comment on how the relative values of modal contribution
factors and the error

(d)Determine the peak value of (*M*_{1n})_{0}
modal response due to *F*(*t*)

Assume *t _{d}*
= 0.598 s which is the same as T

The distribution of the pulse *td* = *T*_{1}
the fundamental period of the system. For the shock spectrum of half cycle sine
wave *R _{d}* = 1.73, 1.14 and 1.06 for

(e) Comment
on how the peak modal response determined in part (d) depend on modal static
response, modal contributed factor *M*_{1n} and *R _{dn}*
and

(f) Is it
possible to determine the peak value of the total (considering all modes)
response from peak modal response? Justify your answer.

*E *= 207.15 GPa* m *= 33 665 kg

*L *= 3.81 m

*I *= 4.1623*
*Ã—* *10^{7}*
*mm^{4}

**Step 7 **Determine
modal static response as shown in Fig. 18.16. The value** **of moments due
to forces *F* is determined by the linear combination to the above three
load cases. The resultants are as given in Table 18.5.

Next we can determine

**Step 8 **Determine
modal contribution factors, their cumulative values and** **error.

The modal
contribution factors and error are given in Table 18.6

In the above the modal
contribution factor is largest for first mode and progressively decreases for
the second and third modes.

**Step 9 **Determine
response to the half cycle sine pulse. The peak modal** **response equation
is specialized for *R* = *M*_{1} to obtain (see Table 18.7)

**Step 10 **Comments

â€¢ For the
given force, the modal response decreases for higher modes. The decrease is
more rapid because of *R _{dn}*. It also decreases with mode

The peak value of the total
response cannot be determined from the peak modal response because the modal
peaks occur at different time instants. Square root of sum of squares (SRSS)
and complete quadratic combination (CQC) do not apply to pulse excitations.

**Program 18.1: MATLAB
program to find the ratio of dynamic shear to static shear in a multi-storey
building **

%program
to get modal components of the forces and calculate

%ratio of
dynamic shear to static shear

clc; close all;

% m=45413*[1 0;0 0.5];

m=[1 0 0 0 0;0 1 0 0 0;0 0 1 0
0;0 0 0 1 0;0 0 0 0 1]; disp(â€˜ mass matrixâ€™)

m

%you can give stiffness matrix disp(â€˜ stiffness
matrixâ€™)

% k=63600000*[2 -1 ;-1 1];

k=[2 -1 0 0 0;-1 2 -1 0 0;0 -1 2 -1 0;0 0 -1 2
-1;0 0 0 -1 1]; k

a=inv(k);

%or you
can given flexibility matrix directly

%a=[.75 .5
.25;.5 1 .5;.25 .5 .75];

disp(â€˜ flexibility matrixâ€™) a

c=a*m;

[ms,ns]=size(m);

par=zeros(ns,ns);

%force
vector

%s=[0;500000] s=[0;0;0;-1;2];
su=0;

for i=1:ns su=su+s(i);

end

%imposed
frequency

%omimp=21.376;

disp(â€˜ imposed frequencyâ€™)
omimp=0.15

%eigen values and eigen vectors
[V,D]=eig(c);

for i=1:ms e(i)=1/D(i,i);

end Qh=max(e)+0.001; Ql=0;

for i=1:ms for j=1:ms

if e(j) > Ql & e(j) < Qh kk=j;

Qh=e(j); else

end end Ql=Qh;

Qh=max(e)+0.001;

om1(i)=e(kk);

omega(i)=sqrt(e(kk)); for l=1:ms

p1(l,i)=V(l,kk); end

end

%Normalizing the mode shape
L=p1'*m*p1;

%develop modal matrix for i=1:ms

for j=1:ms p(i,j)=p1(i,j)/sqrt(L(j,j));

end end

disp(â€˜ Natural frequencies in
rad/secâ€™) disp(omega)

disp(â€˜ normalized modal vectorâ€™)
disp(p)

disp(â€˜ check pT m p=Iâ€™) pâ€™*m*p

%for earthquake analysis
%s=[m(1,1);m(2,2);m(3,3);m(4,4);m(5,5)] gamma=pâ€™*s;

for i=1:ns par(i,i)=gamma(i);

end

%modal
contribution of forces

disp(â€˜ modal contribution of
forcesâ€™) ee=m*p*par

disp(â€˜ dynamic magnification
factorsâ€™) for i=1:ns

rdn(i,i)=1/(1-(omimp/omega(i))^2);
end

rdn ust=a*ee; u=ust*rdn; for
i=1:ns

dis(i)=0; for j=1:ns

dis(i)=dis(i)+u(i,j); end

end

%disp(â€˜ amplitude of
displacementsâ€™); dis;

fo=k*disâ€™;

sum=0; for i=1:ns

sum=sum+fo(i); end

ratio=sum/su;

disp(â€˜ ratio of dynamic base
shear to static shearâ€™) ratio

OUTPUT mass matrix

a =

1.0000 1.0000 1.0000 1.0000 1.0000

1.0000 2.0000 2.0000 2.0000 2.0000

1.0000 2.0000 3.0000 3.0000 3.0000

1.0000 2.0000 3.0000 4.0000 4.0000

1.0000 2.0000 3.0000 4.0000 5.0000

imposed frequency

omimp =

0.1500

Natural frequencies in rad/sec

0.2846 0.8308 1.3097 1.6825 1.9190

normalized modal vector

â€“0.1699 â€“0.4557 â€“0.5969 â€“0.5485 0.3260

â€“0.3260 â€“0.5969 â€“0.1699 0.4557 â€“0.5485

â€“0.4557 â€“0.3260 0.5485 0.1699 0.5969

â€“0.5485 0.1699 0.3260 â€“0.5969 â€“0.4557

â€“0.5969 0.5485 â€“0.4557 0.3260 0.1699

check pT m p=I

ans =

1.0000 â€“0.0000 â€“0.0000 â€“0.0000 0.0000

â€“0.0000 1.0000 0.0000 0.0000 0.0000

â€“0.0000 0.0000 1.0000 0.0000 â€“0.0000

â€“0.0000 0.0000 0.0000 1.0000 â€“0.0000

0.0000 0.0000 â€“0.0000 â€“0.0000 1.0000

modal contribution of forces

ee =

0.1096 â€“0.4225 0.7386 â€“0.6851 0.2594

0.2104 â€“0.5534 0.2102 0.5692 â€“0.4364

0.2941 â€“0.3023 â€“0.6788 0.2122 0.4748

0.3539 0.1575 â€“0.4034 â€“0.7455 â€“0.3625

0.3851 0.5086 0.5640 0.4072 0.1352

dynamic magnification factors

rdn =

1.3845 0 0 0 0

0 1.0337 0 0 0

0 0 1.0133 0 0

0 0 0 1.0080 0

0 0 0 0 1.0061

ratio of dynamic base shear to
static shear

ratio =

1.5039

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