modal analysis - solved example

8/18/2019 Modal Analysis - Solved example

1/11

Widener University

Earthquake Response of Structures

Time History Response

Three Story Shear Frame

SYSTEM INFORMATION:

Moment of Inertia: ≔ I ⋅T

42875 42875 42875 in4

Young's Modulus of Concrete (fc=7000psi): ≔ E ⋅5072.2 ksi

Story Heights: ≔L ⋅T

12 10.67 10.67 ft

GLOBAL STIFFNESS MATRIX:

≔k1 2621.89 kip

in

≔k2 3733.13 kip

in

≔k3 3733.13 kip

in

≔ K +k1 k2 −k2 0

−k2 +k2 k3 −k30 −k3 k3

⎡⎢

⎣

⎤⎥

⎦

= K

6355.02 −3733.13 0

−3733.13 7466.26 −3733.130 −3733.13 3733.13

⎡

⎢⎣

⎤

⎥⎦

kip

in

MASS MATRIX:

≔m1 ―――2165.74 kip

g

=m1 67313.257 slug Mass of First Floor

≔m2 ―――2165.74 kip

g

=m2 67313.257 slug Mass of Second Floor

≔m32165.74 kip

g

=m3 67313.257 slug Mass of Third Floor

≔ M m1 0 00 m2 00 0 m3

⎡⎢

⎣

⎤⎥

⎦

= M 67313.257 0 0

0 67313.257 00 0 67313.257

⎡⎢⎣

⎤⎥⎦slug

Mansoor IllahiPage 1 of 11


2/11

NATURAL FREQUENCIES:

≔ D ⋅ M −1

K = D1132.916 −665.509 0−665.509 1331.017 −665.509

0 −665.509 665.509

⎡⎢⎣

⎤⎥⎦

1

s

2

Eigenvalues: ≔λ ⋅eigenvals D1

s

2

≔λ sort λ

=λ106.752924.024

2098.666

⎡⎢⎣

⎤⎥⎦

1

s

2

Number of Modes: ≔n 3 ≔i ‥1 n

Natural Circular Frequency: ≔ωi

‾λi

=ω10.33230.39845.811

⎡⎢⎣

⎤⎥⎦

― d

s

Natural Period: ≔T i

―⋅2

ωi

=T 0.6080.2070.137

⎡⎢⎣

⎤⎥⎦s

Mode Shapes: ≔ϕi

eigenvec⎛ , D λi⎞

=ϕ1

0.3850.5930.707

⎡⎢⎣

⎤⎥⎦

=ϕ2

−0.756−0.237

0.611

⎡⎢⎣

⎤⎥⎦

=ϕ3

0.53−0.769

0.357

⎡⎢⎣

⎤⎥⎦

Eigenvector Matrix:

≔Φ augment⎛ ,,ϕ1

ϕ2

ϕ3⎞ =Φ

0.385 −0.756 0.530.593 −0.237 −0.7690.707 0.611 0.357

⎡

⎢⎣

⎤

⎥⎦



3/11

:

Time Variables:

Time Increment: ≔∆t ⋅0.01 s =∆t 0.01 s

Number of data Points: ≔ N 8900 = N 8900

Range of Data Points: ≔ j ‥1 − N 4

Starting Time: ≔tstart ⋅0 s =tstart 0 s

Time Step Array: ≔t j

+tstart ⋅ j ∆t

Seismic Data:www.strongmotioncenter.org

“Record of Fri Apr 18, 2008 04:37:37.0 CDT (Avol1 v7.5 7/07 CSMIP) ”“CSMIP Preliminary Processing (Origin: To be determined) ”

“02407-K4180-08109.01 Start time: 4/18/08, 09:37:37.0 UTC (GPS) ”“Station No. 2407 38.494N, 90.281W K2 s/n 4180 (3 Channels) ”

“St. Louis - Jefferson Barracks VAMC USGS ”“Chan 1: 90 Deg ”

“Record of Fri Apr 18, 2008 04:37:37.0 CDT Fri Apr 18, 2008 04:37:37.0 CDT ”“Hypocenter: To be determined. ML: To be determined. ”

“Instr Period = .0049 sec, Damping = .700, Sensitivity = 1.25 v/g (Nominal)”

“ Record length = 89.000 sec. ”

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎦

8900 points of accel data equally spaced at .010 sec, in cm/sec2.

≔ Edata ⋅READEXCEL ,“.\CE 634 Project Data (Raw).xlsx” “Data2!A1:H1113” ―m

s

2

= Edata

… 0 0 0 0 0 00 0.001 0.002 0.002 0.002 0.003

−0.002 −0.004 −0.004 −0.002 −0.002 −0.003−0.005 −0.002 0.006 0.01 0.002 −0.011

0.019 0.023 0.016 −0.001 −0.012 0−0.013 −0.04 −0.051 −0.038 −0.016 0.016

⋮

⎡⎢⎢

⎢⎢⎢

⎣

⎤⎥⎥

⎥⎥⎥

⎦

―ms

2



4/11

Table data

≔n per_row 8 8 data points per row

≔nrow N

n per_row=nrow 1112.5 ≔nrow 1112

≔TableT

Edata Transpose the data

≔ag‖‖‖‖‖

‖

←ag Table1

for ∊ j ‥1 −nrow 1‖‖ ←ag stack ,ag Table

j

ag

Stack the columns from Table to form avector containing the ground acceleration

Create the "unity" vector that applies the ground accelerationto the horizontal translational degree of freedom. All DOF arehorizontal translation DOF, therefore:

≔oneT

1 1 1

GROUND ACCELERATION VS. TIME PLOT:

-0.015

-0.01

-0.005

0

0.005

0.01

0.015

0.02

-0.025

-0.02

0.025

1.99 2.98 3.97 4.96 5.95 6.94 7.93 8.92 9.91 10.9 11.8912.8813.8714.8615.8516.8417.8318.8219.810.01 1 20.8

ag g

t s



5/11

Specify Initial Conditions

Initial Displacement: ≔x0 ⋅T

0 0 0 in

Initial Velocity: ≔v0 ⋅T

0 0 0 ―nsec

Transform initial conditions to modal coordinates

≔q0i

―――

⋅⋅T⎛ϕ

i⎞ M x0

⋅⋅T⎛ϕ

i⎞ M ϕ

i

=q0

000

⎡⎢⎣

⎤⎥⎦

in

≔q' 0i ―――

⋅⋅T

ϕi

M v0

⋅⋅T⎛ϕi⎞ M ϕ

i

=q' 0

000

⎡⎢⎣

⎤⎥⎦

in

s

Calculate modal stiffnesses

≔ K mi

⋅⋅T

Φ K Φ,i i

= K m

598.8195183.255

11772.336

⎡⎢⎣

⎤⎥⎦

kip

in

Calculate modal mass terms

≔ M mi ⋅⋅T

Φ M Φ ,i i = M m

67313.257

67313.25767313.257

⎡

⎢⎣

⎤

⎥⎦

slug

Define modal damping ratios

Modal Damping Ratio: ≔ζmT

0.03 0.04 0.05

≔Cmi

⋅⋅⋅2 ωi

M mi

ζmi

=Cm

3477.413641.125697.5

⎡⎢⎣

⎤⎥⎦

⋅lbf s

in

Transform modal damping matrix to natural coordinates for use in Part 2:

≔ξξi

⋅⋅2 ζmi

―

ωi

M mi

≔ξ

ξξ1

0

⋅lug s

0

⋅lug s

――0

⋅lug sξξ

2――

0

⋅lug s

――0

⋅lug s――

0

⋅lug sξξ

3

⎡⎢⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎥⎦



6/11

≔Cnat ⋅⋅⋅⋅ M Φ ξT

Φ M =Cnat

15522.18 −7236.219 −483.009−7236.219 17193.175 −7575.451

−483.009 −7575.451 10100.734

⎡⎢⎣

⎤⎥⎦

⋅bfs

in

Inertial forces in modal coordinates

≔ Γ i

⋅⋅T

ϕi

M one = Γ 9453.152

−2144.033662.142

⎡⎢⎣

⎤⎥⎦

⋅bf s2

in

≔ Γ normi

Γ i

M mi

= Γ normi

1.685−0.382

0.118

⎡⎢⎣

⎤⎥⎦

≔ P ,i j

⋅− Γ i

ag j

= P ,3 2472

19139.083 ⋅b ft

s

2

Calculate modal displacements using central difference method.Calculate the starting acceleration

≔a0i

⋅―1

M mi

⎛⎝

−− P ,i 1

⋅Cmi

q' 0i

⋅ K mi

q0i⎞⎠

=a0

000

⎡⎢⎣

⎤⎥⎦

in

s

2

≔qmi

−−q0i

⋅∆t q' 0i

――

⋅a0i

∆t2

2fictitious displacement q-1.

Calculate the central difference coefficients for each mode.

≔ Ai

−⋅4 M mi

⋅⋅2 K mi

∆t2

+⋅Cmi

∆t ⋅2 M mi

≔ Bi

−⋅Cmi

∆t ⋅2 M mi

+⋅Cmi

∆t ⋅2 M mi

≔Ci

⋅2 ∆t2

+⋅Cmi

∆t ⋅2 M mi



7/11

Calculate the modal response

Revised Number of Data Points: ≔ N 3246 problem is the datais cut off

≔q ‖‖‖‖‖‖‖‖‖‖‖‖

‖

←n N for ∊i ‥1 3

‖‖‖

‖

←q,i 1

q0i

←q,i 2

++⋅ Ai

q0i

⋅ Bi

qmi

⋅Ci

P ,i 1

for ∊i ‥1 3‖‖

‖

for ∊ j , ‥3 4 −n 1‖‖

←q,i j

++⋅ Ai

q,i − j 1

⋅ Bi

q,i − j 2

⋅Ci

P ,i − j 1

q

-0.016

-0.012

-0.008

-0.004

0

0.004

0.008

0.012

0.016

0.02

-0.024

-0.02

0.024

10 15 20 25 30 35 40 450 5 50

q,1 j

in

q,2 j

in

q,3 j

in

t j

s

Plot Time-Histories for the Modal Displacements



8/11

Calculate response in natural (displacement) coordinates

＝xk

⋅Φ qk

≔k ‥1 n =n 3 index for number of nodes

For time series calculations =i

1

23

⎡

⎢⎣

⎤

⎥⎦

Displacement: ≔x ∑i

⋅Φ q

Natural Displacement Response

-0.026

-0.017

-0.009

0

0.0090.017

0.026

0.034

-0.043

-0.034

0.043

10 15 20 25 30 35 40 450 5 50

x ,1 j in

x,2 j

in

x,3 j

in

t j

s

Plot Mode Shapes:

In order to plot the mode shapes, we add a "node" at the base of the frame.

≔ψi

stack ⎛ ,0 ϕi⎞ =ψ

1

00.3850.5930.707

⎡⎢⎢

⎣

⎤⎥⎥

⎦

and generate story heights Zero line for plot

≔h

⋅0 ftL

1

+L1

L2

++L1

L2

L3

⎡⎢

⎢⎢⎢

⎣

⎤⎥

⎥⎥⎥

⎦

≔ z

0

000

⎡

⎢⎢

⎣

⎤

⎥⎥

⎦



9/11

Mode Shapes

7

10.5

14

17.5

21

24.5

28

31.5

0

3.5

35

-0.6 -0.45 -0.3 -0.15 0 0.15 0.3 0.45 0.6-0.9 -0.75 0.75

h ft

h ft

h ft

h ft

ψ1

ψ2

ψ3

z



10/11

Determine Interstory Drift:First Story Drift: ≔x1 =max x

10.026 in

Second Story Drift: ≔x2 =−max x2

x1 0.012 in

Third Story Drift: ≔x3 =−max x3

x2 0.03 in

Maximum Story Drift: ≔∆X max max ,,x1 x2 x3 =∆X max 0.03 in

≔x =x1x2x3

⎡⎢

⎣

⎤⎥

⎦

0.0260.0120.03

⎡⎢⎣

⎤⎥⎦in

Determine Interstory Force:

≔ F ⋅diag K ∆X max = F 192.014225.589112.795

⎡⎢⎣

⎤⎥⎦kip

Calculate the Relative Acceleration Response:(Using Central Difference Method)

≔q'' ‖‖‖‖‖‖‖‖‖‖‖‖‖‖‖‖‖‖‖‖‖

‖‖‖‖‖

‖

←n N for ∊i ‥1 3

‖‖‖

‖

←q,i 1

q0i

←q,i 2

++⋅ Ai

q0i

⋅ Bi

qmi

⋅Ci

P ,i 1

for ∊i ‥1 3‖‖‖‖

←q'' ,i 1

―――――⎛ +−qmi ⋅2 q0i q ,i 2⎞

∆t2

for ∊i ‥1 3‖‖

‖

for ∊ j , ‥3 4 −n 1‖‖

←q,i j

++⋅ Ai

q,i − j 1

Bi

q,i − j 2

⋅Ci

P ,i − j 1

for ∊i ‥1 3‖‖‖

‖‖

for ∊ j ‥2 −n 2‖‖

‖←q'' ,i j ―――――――

⎛ +−q,i − j 1

⋅2 q,i j

q,i + j 1

⎞

∆t2

for ∊i ‥1 3‖‖

←q'' ,i −n 1

q'' ,i −n 2

q''



11/11

Convert Modal Acceleration to Natural Coordinates:

≔x'' r ∑k

⋅Φ q''

For time series calculations:

Absolute Acceleration:

≔x'' r1 =max⎛

x'' r1 ⎞

37.898 in

s

2

≔x'' r2 =−max⎛

x'' r2 ⎞

x'' r1 −9.213 in

s

2

≔x'' r3

=−max⎛

x'' r

3 ⎞x''

r232.091

in

s

2

Maxima:

≔∆Amax max ,,x'' r1 x'' r2 x'' r3 =∆Amax 0.098 g

Maximum Absolute Acceleration in Terms of Gravity at EachLevel for each instant of time

-0.08

-0.06

-0.04

-0.02

0

0.02

0.04

0.06

0.08

-0.12

-0.1

0.1

18 27 36 45 54 63 72 810 9 90

x'' r ,1 jg

x'' r ,2 jg

x'' r ,3 jg

t j

s


modal analysis - solved example

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