boussias 1.docx

7/27/2019 Boussias 1.docx

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Experimental Methods in Seismic Engineering

Prof. S. Bousias

Department of Civil Engineering University of Patras

MEEES - Academic Year 2012/2013

Astrid Aubry

Homework # 1

Exercise #2

Question a)

The size effects are related with the effect that length scale produces: generally the reduction

in the specimen size leads to an increase in strength properties. For example, the paper of

Wallace and Krawinkler, Small Scale Model Test of Structural Steel Assemblies, shows a

comparison between a prototype and replica model (at scale 1:12.5) of 3 beam-column

assemblies in term of beam-load deflection. This one of the results

Left: Model Speciment; Right Prototype

Question b)

The main conclusion of the tests by Abrams is that at the small scale the bond strength is

difficult to reproduce (simulation of bond). Since, this parameter mainly affects the response ofthe concrete-beam RC assemblies, if it is not improved in the future tests, Abrams

recommended that the minimum usable scale factor for testing isolated RC components in

flexure should be one-quarter.


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Question c)

By comparing some results shown in the 2 papers, it may appear that strength and stiffness

degradation is more significant for scaled model in RC beam-column joints than for scaled steel

assemblies. In the former this behavior may be inferred to the bond deterioration, that it is not

well reproduced in the small scale model.

Left: Small Scale 1:12 RC Beam Column Joint; Right Large Scale 3:4 RC Beam Column Joint;

for Steel Assemblies

In fact, one of the conclusion of Wallace and Krawinkler is that in their study the reduced-scale

model tests led to the same major conclusions on

structural behavior as prototype tests (by considering that

some localize failure modes could not be reproduced, and

those should be studied using full scale component tests) .

Strain rate effects:


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In general an increase in strain rate leads to an increase of the yield strength.

In the case of Structural Steel Assemblies tests, the authors say that strain rate effect are of

relatively small and predictable importance in dynamic model tests, except for some localized

failure modes. While in the Concrete Beam Column joints, the author says that in the tests they

apply different loading histories (see the left picture) to the each scaled models, leading me to

understand that for concrete the strain rate is of more significant importance than for

structural steel assemblies. It is also worth to note that it is function of the similitude that one

wants to do (dynamic with pseudostatic) and of the kind of test that one is doing (comparing

dynamic with pseudo-static tests).

References:

Small-Scale Model Tests of Structural Steel Assemblies by B.J. Wallace and H.Krawinkler;

Scale Relation for Reinforced Concrete Beam-Column Joints by Abrams.


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Exercise #3

The relevant geometric quantities are reported in the following figure:

The following notation will be used: the subscript 1 and 2 refers to the large and small model,

respectively.

The geometric scale is:Ge0m.Model1 : Geom.Model2 = 2 : 1

In the table on the left Hint is an average value:1 Scale 2 Scale G. Scale

D est [mm ] 16.00 8.00 D est,1 /Dest,2 2

D int [mm ] 13.50 6.75 D int,1 /Dint,2 2

H washer [mm] 3.00 1.50 H wash,1 /Hwash,2 2

H Nut [mm ] 12.60 6.30 H Nut,1 /HNut,2 2H Int [mm ] 450.00 225.00 H Int,1 /HInt,2 2

H [mm ] 25.00 12.50 H 1 /H2 2

B [mm ] 750.00 375.00 B 1 /B2 2

L [mm ] 500.00 250.00 L1 /L2 2

1 Scale 2 Scale G. Scale

H 1story[mm ] 450.00 225.00 H 1story,1 /H1story,2 2

H 2story[mm ] 450.00 220.00 H 2story,1 /H2story,2 2.045455

H 3story[mm ] 450.00 230.00 H 3story,1 /H3story,2 1.956522

H Top [mm ] 108.000 54.000 H Top,1 /HTop,2 2

H tot [mm] 1458.00 729.00 H ToT,1 /HToT,2 2


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Therefore, these are the scale factors for all other physical quantities which affect dynamic

response:

the left table was constructed under

the condition that the two models

were built by using the same material

= 1.

Even though, by measurement of the

slab and rod masses and dimensions

in the lab, I got:

Quantity G. Case Using the same Material

Length L 2.00

Area A = 2

L 4.00

Volume V = 3

L 8.00

Second Moment of Area I = 4

L 16.00

Young's Modulus E 1.00

Strain e = 1.00 1.00

Stress s = E 1.00

Mass Density =E / (L a) 1.00

Mass m = 3

L 8.00

Acceleration a= 1/LE/ 0.50

Velocity v= (La)0.5

1.00

G. Acceleration g = 1.00 1.00

Time, Period t = (L / a)0.5

2.00

Frequency w = 1/L (E / )0.5

0.50

Impuse i = 3

L (E )0.5

8.00

Energy e = E 3

L 8.00

Critical Damping x

= 1.00 1.00

Gravitational Forces fg = 3

L 8.00

Deflection d = L 2.00

1 Scale 2 Scale

H [mm ] 25 12.5

B [mm ] 750 375

L [mm ] 500 250

mslab [Kg ] 6.500 0.850

slab [Kg / m3] 693.33 725.33

=slab,1 / Slab,2 0.955882353

1 Scale 2 Scale

L [mm] 502.000 582.000

D int [mm ] 13.5 6.75

mrod [Kg ] 0.610 0.180

rod [Kg / m

3

] 8489.24 8642.75=rod,1 / rod,2 0.982237716


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Eigenvectors and Eigenfrequencies:Under the hypothesis of:

1. Mass concentrated in the center of mass of each story;2. Center of mass = center of stiffness;3. Bending stiffness of the slab >> Bending Stiffness of the rod

We have that the problem is simplified to the evaluation of eigenvalues and eigenvectors of this

model:

For both systems, we get:

( ) or ( ) Where are the eigenvectors (or modal shapes) associated to eigenvalues (square of angular frequency) of the matrix (thesubscript , ).In order to define , I consider the mass of the rods and slabs forboth model (neglecting the washer and nut masses):

*note that m 8.

long direction short direction

Lslab [mm] 434 284

Eslab [MPa]

Islab [mm4] 976563 651042

(EI/L)Slab 22501440 22924002

Hrod

[mm] 394 394

Erod[MPa]

Irod[mm4] 1630 1630

2*(EI/L)rod 1738916 1738916

(EI/L)Slab/2*(EI/L)rod 13 13

10000

210000

1 Scale 2 Scale m

m1 [Kg ] mslab+4*mrod 8.687 1.128 7.70

m2 [Kg ] mslab+4*mrod 8.687 1.128 7.70

m3 [Kg ] mslab+4*mrod/2 7.594 0.989 7.68


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Where the values of k1, k2, k3 are

expressed in the previous sketchand their values are:

*note that k = L = 2.

The important things to point out are:

H is the free length of the rod:

The moment of inertia Iof the cross section of the rod is evaluated considering the innerdiameter measured in the lab.

Therefore, eigenvalues and the eigenvectors of are:Modes 1 Scale 2 Scale f

f1 [Hz = 1/T] 12.9 25.3 0.510

f2 [Hz = 1/T] 35.8 70.2 0.510

f3 [Hz = 1/T] 50.9 98.8 0.515

1 Scale 2 Scale 1 Scale 2 Scale 1 Scale 2 Scale

u1 0.3331 0.333 -0.7288 -0.729 -0.5621 -0.5628

u2 0.5954 0.5953 -0.2685 -0.2699 0.7306 0.7308

u3 0.7312 0.7313 0.6299 0.6291 -0.3877 -0.3862

1st Mode 2nd Mode 2nd Mode

1 Scale 2 Scale

HInterStory [mm] 450 225

2 HWasher [mm ] 6 3

2 HNut [mm ] 25.2 12.6

HSlab [mm ] 25 12.5

H [mm] 394 197

1 Scale 2 Scale k = L

k1 [N/m] 269115 134558 2

k2 [N/m] 269115 134558 2

k3 [N/m] 269115 134558 2


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If we compare the obtained fwith the one evaluated in the scale factortable, we see that we

are very closed to the required 0.5. The not exact mach is due to the fact that the m 8 (it is

almost equal).

I dont compare the modal shapes because these are not absolute quantities but they depends

on a coefficient a.It is worth to note that the acceleration that we will measure with the test will be characterized

by a = 0.5. It means that in the small model the acceleration will be the double of the one

measured on the large scale.

If our requirement is that a=1, the scale factortable changes:

In order to satisfy similitude, since we

cannot increase the mass density in the

small scale model, we add a mass so that weincrease its effective mass density.

By assuming that the m = 8 , we have to

double the mass on small scale model. While

if we consider the real ratio over each floor,

we have:

Therefore, now the natural frequencies of

the two models are:

By adding a mass on the small scale model we satisfy all the requirement ofscale factortable.

Quanti ty G. Case Re qui re d Provi de

Length L 2.00 2.00

Area A = 2

L 4.00 4.00

Volume V = 3

L 8.00 8.00

Second Moment of Area I = 4

L 16.00 16.00

Young's Modulus E 1.00 1.00

Strain e = 1.00 1.00 1.00

Stress s = E 1.00 1.00

Mass Density =E / (L a) 0.50 1.00

Mass m = 3

L 4.00 8.00

Acceleration a= 1/

LE/ 1.00 1.00

Velocity v= (La)0.5

1.41 1.41

G. Acceleration g = 1.00 1.00 1.00

Time, Period t = (L / a)0.5

1.41 1.41

Frequency w = 1/L (E / )0.5

0.71 0.50

Impuse i = 3

L (E )0.5

5.66 8.00

Energy e = E 3

L 8.00 8.00

Critical Damping x = 1.00 1.00 1.00

Gravitational Forces fg =

3

L 4.00 8.00Deflection d = L 2.00 2.00

1 Scale 2 Scale m Dm2

m1 [Kg ] mslab+4*mrod 8.687 1.128 7.70 1.043

m2 [Kg ] mslab+4*mrod 8.687 1.128 7.70 1.043

m3 [Kg ] mslab+4*mrod/2 7.594 0.989 7.68 0.909

Modes 1 Scale 2 Scale f

f1 [Hz = 1/T] 12.9 18.3 0.707

f2 [Hz = 1/T] 35.8 50.6 0.707

f3 [Hz = 1/T] 50.9 72.0 0.707

boussias 1.docx

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