structural dynamics free and force vibration undamped impulse load

structural dynamics free and force vibration undamped impulse load

1. Class #10

Structural Dynamics

Multi-Degree of Freedom Systems
Free & Forced Vibrations — Undamped
Impulse Load
Inter-story Shear Forces

Dr. Tesfaye Alemu

2. MDOF — Undamped Response to Forced Vibrations

— 2 Degrees of Freedom

Coupled equations of motion for the 2 DOF system are:

+= DY F,=0
mu ii,(t)— (k, +k Ju i(t)+ku,(t)+ p,()=0
0

rewriting :

mii, (1) (&X u(t)) ku, (t)+ p,(t)=

mi (0) + (ky + Ju (e) ku (1) - p,()=0
ml) kate (t)+ Ku (e)= pa(t)=0

Ir li, ( )

k, [u,(t)]

p(t) mlü,()] ¡920
u(t) volt)

k, [u,(t)—u, ()]

3. MDOF — Undamped Response to Forced Vibrations
— 2 Degrees of Freedom

Coupled equations of motion for the 2 DOF system are:

mit, (0), (k, tk Ju (e) ku, (1) = p(t)

m,ü i,(t)— ku (t)+k,u,(t)=+p,(t)
k=k+k

olla lol la) ve Rn

ky = k,

Pa(t)

[no {ii,(t) pa(t)
zus Me)
sli. O] rollo), O]

4. ml, ()+ (& + ky u,()—ku, ()= +P, (e)

m,ü,(t)- ku, (t)+k,w,(t)= +p, (0)
. ky =k tk,
7 „or cloro ve

ko =k,

Mi + Ku = P(t).....….............… Eqn()

[lee ol

5. Let us uncouple MDoF equations of motion

6. Modal Superposition

Letus assume u; = 6; For 2DoF, 4u=4,Z, +42,

u=0Z...... (2) o=[4, 9]
Multiply (2) by DM 4 (À |
T T D> On
us nl Initial conditions mapping
T from geometric to modal
A (3a) coordinate
"DM ,
Zoe RU S (3c)
EDDY DMD
= ; m _ D'Mü(0)
TAN N (3b) Z,(0) = oo (3d)

7. Mii + Ku = P(t)......... Egn(l)
MOZ + KOZ = Pi)... Eqn(5)
Multiply Eqn (5) by p”
MZ + KZ = Man. Egn(6)

Where: M=®'M® Modal Mass
K =®' K® Modal Stiffness
P(t) = 0" p(t) Modal Forces

8. MDOF — Undamped Response to Forced Vibrations
— 2 Degrees of Freedom

The two coupled equations of motion for the 2-DOF system:

mul, ()+ (& ” ky a ()-ku,() =+P (e)

m,ii,(t)— ku, ()+ ku,l) =+P, (e)

Become two uncoupled, independent equations of motion for
2 independent SDOF systems:

M,Z,(t)+ K,Z,(t)= Pe) M,Z,(t)+ K,Z,(t)=P,()

These are 2 decoupled SDOF equations of motion.
They can be solved independently for the motions of Z, and Z,

Z,(t) = response of a SDOF system containing modal mass M;
during the j" mode of vibration at frequency w;.

9. MDOF — Undamped Response to Forced Vibrations
— 2 Degrees of Freedom

The two uncoupled, independent equations of motion are:
MZ (c)+ K,Z,(1)=A() M,Z,(t)+ K,Z,(t)= Ple)
M,Z,(t)+a,, M,Z,(t)= P(t) M,Z,(t)+a,,M,Z,(t)= P, t

Divide by the Modal Masses, M, andM,

A B= 9 ()= Ae hae)

Z,(t)+@%, Z,(t)= Pole) _ P:()= Ap (0)+0,,p, (e)

2 2
2 mb, + MD

10. MDOF — Undamped Response to Forced Vibrations
— 2 Degrees of Freedom

The two uncoupled, independent equations of motion are:

Z,()+0,,Z,()= P'(t)= ÿ P()+6,:p,()
Z,()+o,Z,()= Bo (t)=4,p,(c)+0,,p,(0)

A, 1 9 2

MES D ba

| are Orthonormalized mode shapes

11. Multi-Degree of Freedom Systems
Free Vibrations — Undamped

Set > P(t)=0

12. MDOF — Undamped Response to Forced Vibrations
— 2 Degrees of Freedom

The two coupled equations of motion for the 2-DOF system:

mii, (t) + (k, +k, Ju, (t)—k,w,(t)=0

m,ii,(t)— kyu, ()+ ku,l) =0

Become two uncoupled, independent equations of motion for
2 independent SDOF systems:

M,Z\(t)+K,Z,(t)=0 M,Z,(t)+K,Z,(t)=0

These are 2 decoupled SDOF equations of motion.
They can be solved independently for the motions of Z, and Z,

Z,(t) = response of a SDOF system containing modal mass M;
during the j" mode of vibration at frequency w,. ”

13. MDOF — Undamped Response to Forced Vibrations
— 2 Degrees of Freedom

The two uncoupled, independent equations of motion are:

MZ (t)+K,Z,(t)=0 M,Z,(t)+K,Z,(t)=0
MZ, (c)+ Dr, MZ, (t)= 0 M,Z,(t)+ oy, M,Z,(t)= 0

Divide by the Modal Masses, M, andM,

20+0iz(0=0

Z,()}+03,Z,())=0

14. MDOF — Undamped Response to Forced Vibrations
— 2 Degrees of Freedom

SAV: Sum of Absolute Values :

Ya = Dz itor

jl

Ure =|, Zany |+

n=number of modes

B.2...| and u, =

B.2,,.|*

PoZ,

2uax

SRSS : Square Root Sum of the Squares: (2 dim ensional systems)

n 5
Uno = D n=number of modes
j=l

Wig, = iZ F haz. +

2

EPA UA

and u, =
wax

mx

COC: Complete Quadratic Combination: (3 dimensional systems)

Ying = y bz... lez.) n=number of modes
jel k=l

822(1+ 7%" a poe

= =damping ratio de
ik 2P 2 2
(1-12 f +47, (tr) o, Cer

15. Example 2: MDOF Free Vibration Example

Calculate by modal superposition, the free vibration
response of Example 1. The initial conditions are as follows:
u, = Up = 0.02m. Assume the system is undamped

Solution:
n n n u m = mis
itial co itio i u

m 0 1 0
Mass Matrix M= = 20000 kg
0 m o 1

Stiffness Matrix X = 2k —k =18x10° 2 ll Re
k ok -1 1

16. @, =18.54rad/sec

Two natural frequency AS e
, = 48.

0.61803 — ee)
1 1

Mode Shapes ¢=|¢, #1= |

ModalMass M = p M

— |27639.2 0.2
M = kg
0.2 72360.4

17. Initial condition in modal coordinates

D'Mu(0)

Z,(0) = —
©) D'MD

0 20000 | 0.02
27639.3

[0.618 i

20000 0 Le

z,(0) | 0.023416m

20000 0 70.02
[-1.618 1
> 0 20000 | 0.02
x 72360.7

| 0.003416m

18. The free vibration response for each modal coordinate for
undamped structure is

2.6)= z,(0\Cos(w2)]+ 0 sito, 0]

Thus

z(t) 23.416cos ot
= mm
Z,(t) —3.416cos 0,t

19. The displacement in geometric coordinates system are
obtained by superposition of modal coordinate displacements

u = PZ

Thus

0 0.618 -1.618 | 23.416cos at
17) =
1 1 —3.416cos @,t

14.471 cos 0,1 +5.527 cos @,t
u(t) = mm
23.416cos @,t —3.416cos @,t

20. Example 3: MDOF Forced Vibration Example

The two story structure given in Example 1 is acted up on at
floor level by horizontal triangular impulsive forces as
shown below. For this structure, determine the maximum
floor displacement and maximum shear forces in the
columns.

75KN

Pa(t)

25KN |
ji Time (sec)

ye Od

Impulsive Load Functions

21. MDOF — Undamped Response to Forced Vibrations
— 2 Degrees of Freedom - Impulse Load Example:

75KN

25KN

Impulsive Load Functions

P(t)
m, li, ()] u,(t)

Pa(t)

= 0.1

mi, (c)]

k, [u,(t)] k, [u,(t)—u, ()]

Time (sec)

Can use a
Response Spectra
To obtain the
MDOF response
to any
forcing function:
- Harmonic
- Non-harmonic
- Impulse
- Blast
- Aperiodic

a(t) Z

u;(t)

Zo =R, (Zu )

Static

R,=

= obtain from
Response Spectra

Dynamic Displacement

amplification factor
21

22. MDOF — Undamped Response to Forced Vibrations
— 2 Degrees of Freedom - Impulse Load Example:

A 20
i A Response Spectra |
al 19 |
H 2 i L
h 2 75KN 1.2 | |
: 2 R | |
+
ha E d l cl Fie)
j E à
| S Po(t) = :
2 | 25KN i %
4 3 [ I
h, E 2 0.05 0.1 0.2 0.5 1.0 2.0 5 10
y = IT
Time (sec Dynamic load factor for an undamped
system acted upon by a triangular force.
À =
(t) alt) — obtain from
7 a 2 static = Response Spectra
m, lii,(¢ )l u,(t) = u,(t) p p
\ m, lii, (o) R Dynamic Displacement

d ~~ amplification facto!

sli. (o)] k Le, 0-0, O] Arr

23. 75KN

Impulsive Load
Functions

From previous
25KN

2-DOF example 1:

©, =343.77(rad/s)

Oy, = 18.54(rad/s)

i. = 2 0.339(sec/cycle)
o,

4 _ ASA

Ty, 0.339(sec)

R, =0.9

Time (sec

MDOF — Undamped Response to Forced Vibrations
— 2 Degrees of Freedom - Impulse Load Example:

2.0
A Response Spectra

T
T
i
|
y
I
t
1
T
|

005 01 02 05 1.0
tglT

Dynamic load factor for an undamped
system acted upon by a triangular

orce.

obtain from
Response Spectra

— Dynamic Displacement
R= Brent
amplification factor

Zn =R, aa )

7 =

Static —

23

24. MDOF — Undamped Response to Forced Vibrations
— 2 Degrees of Freedom - Impulse Load Example:

75KN

Impulsive Load
Functions

From previous
2-DOF example:1

25KN

Ox, = 2356.23(rad/s)
Oy, = 48.54(rad/s) oz
Ty, = SE eee er)
Ny
ta __Olsee _ 0.773
Ty, 0.124(sec)
R,, =1.45

207 A Response Spectra
T T
1.6 | T
| | I
i |
1.2 + +
| |
R t
ds l (| F(t)
N
0.4 i 5 t
i T
o | i
0.05 0.1 0.2 0.5 1.0 2.0 5 10
talT
ime (sec) Dynamic load factor for an undamped
system acted upon by a triangular force.
Z.. . = obtain from
Static Response Spectra
R = Dynamic Displacement
=

amplification factor

Z a= R, en)

24

25. MDOF — Undamped Response to Forced Vibrations
— 2 Degrees of Freedom

The two uncoupled, independent equations of motion are:

Z,()+0,,Z,()= P'(t)= ÿ P()+6,:p,()
Z,()+o,Z,()= Bo (t)=4,p,(c)+0,,p,(0)

A, 1 9 2
O, 1 b 2

d=10? 0.37175 -0.60150 | | From previous
à 0.60150 0.37175 | | 2-DOF example:

Where ¢= |

| are Orthonormalized mode shapes

26. MDOF — Undamped Response to Forced Vibrations
— 2 DOF - Impulse Load Example

Obtain Zmax Dynamic from the two uncoupled equations of motion:

Z,(t)+ 0%, Z i(t)= P'(t)=4.r,0)+ P20)

For static condition, acceleration = 0 :

0+343.77Z, (e) = (0.37175 x 50000+ 0.60150*75000) x10~

zZ. =1.853m
Zu. Dnamic = Ra, XZ), = 0-9X1.853m
Z = 1.668m

1: Dynamic

27. MDOF — Undamped Response to Forced Vibrations
— 2 DOF - Impulse Load Example

Obtain Zmax Dynamic from the two uncoupled equations of motion:

Z,(t) | Oy, Z,(t)= AUET #20 | P(t)

For static condition, acceleration = 0 :

0+ 2356.23Z, (e) = (-0.60150x 50000 + 0.37175 * 75000) x 10”
Z;. =—0.00931m

Static

Zn. = Ry, x Zu, = 1.45x(-0.00931)
Z =-0.0135m

2Max_Dynamic

28. MDOF — Undamped Response to Forced Vibrations
— 2 DOF - Impulse Load Example

Obtain Umax Dynamic TOM Zmax Dynamic: (Sum of Absolute Values)

US es m 7 I Zu en | ar 0 222 ee |
= |(0.0037175)(1.668m) +|(- 0.006015)- 0.0135m)

= 0.0.0062m +|0.00008 1m] = 0.00628m = 6.28mm
First Second

Mode Mode
UN A ls Zier smu a VAR |
Woy. rage = |(0.006015)1.668m) + |(0.0037175/- 0.0135)
Woy. ps =0.01003m + 0.00005: = 0.010087 =10.08mm

IMax_ Dynami

IMax_Dynamic

28

29. MDOF — Undamped Response to Forced Vibrations
— 2 DOF - Impulse Load Example

Obtain Uyax Dynamic OM Zmax Dynamic: (Square Root Sum of Squares)

(SRSS, ASCE 7-05,
Sect. 12.9.3)

= (7 cn | + 2...

- = 0. QUES 668m) +[(—0.006015)(—0.0135m)P
21 Mode

Imax_Dyna;

U prune 20.0060 mm

EEE lo, ar A E

= y[(0. 15). P +[(0.003715)(- 0.0135)
[0.0060 5)1,668m)P +[(0.003715-0.0135)]
_ = 0.010033m = 10.033mm

30. MDOF — Undamped Response to Forced Vibrations
— 2 Degrees of Freedom - Inter-story Shear Forces

Vg ie lo, Au

(i= Story j= Mode )

Vi total = 1 y (21

Fi
(ASCE 7-05, Sect 12.9.3)

Vi =, lá, dr
V,= Zu (6, =d0
Vi _totat = VV, i ar Vv, BF

Vy = Zi (, -4,)

À puto _
m, lit, ()] u,(t) m, lii, ()] lt) Dn. ( Gee k,

k, [u,(t)] k,[u,(t)—u,(1)] Va _roiat = (V,,) + vy

31. MDOF — Undamped Response to Forced Vibrations
— 2 Degrees of Freedom - Inter-story Shear Forces

Va Tota à
Define: (ASCE 7-05, Sect 12.9.3)

t
E 18 Story Shear Force
SASS Vile ie lo, dk;

Vi =z, | 11 0

V,, =(1.668m)(0.0037175-0)2x18x10°)
V,, =223.23kN

V,= Zu (2 z %o2 Jo

a — Vi, +(—0.0135\—0.006015—0)(2x18 10°
Vi, +2.9233kN

(223.23kN Y +(2.9233kN)
31

" Total
V, rorar = 223.25KN

32. MDOF — Undamped Response to Forced Vibrations
— 2 Degrees of Freedom - Inter-story Shear Forces

V,

2_Total

Total
SRSS

Vy

=

Define: (ASCE 7-05, Sect 12.9.3)
2nd Story Shear Force
7 Zn (o, pk

Story Mode )

Va =2 lé, en,

= (1.668)0.006015 - 0.0037175/18x10°)
= 68.98KN

v,

7

1

0.0135)0.0037175- (-0.006015) 18x 10°)
2.365KN

Tor = N (68.98KN) +(=
V, =69.02kN

2_Total

33. V, = 68.98KN

Y,» =2.365kN ;

MDOF — Undamped Response to Forced Vibrations

— 2 Dearees of Freedom - Inter-story Shear Forces
Va_rout = 69.02KN

P,(t)

Hi li, 9) ul)
k, lu, (a)

Vi =2.92kN

m,[ii,(t)]
k, [u,(t)—u, ()]

Vi_roit = 223.25KN

a(t)

Define: (asce 7-05, sect 12.3)
Story Shear Force

Vie (6. br k,

i = story number

j = mode number

qe ‘ond
Mode Mode

V tomar = V (18.825k)° + (—2.32k)}
Vi ro =1 8.968(kip)

4st ‘ond
Mode Mode

Vy rorar = (7.0617k) ar (9.8 12k)
V rows = 12-089(kip) 33

34. MDOF — Undamped Response to Forced Vibrations

— 2 Dearees of Freedom - Inter-story Shear Forces
Vo à = 68.98KN Vy py = 69.02KN

V,:=2.365kN : 5
| | Define: (asce 7-05, sect 12.3)

Story Shear Force
VA = Zoe. (6. “bi k,

i = story number

Va =2.92kN Y tot = 223-25KN j = mode number

For the 1°! Story:
1st mode shear force
AU NDS N dominates
la pr(t)
m, li, ( )] u,(t) m, li, ()] a(t)

k, [u,(t)] k, [u,(t)—u, ()]