Universitas Indonesia Chapter 3 Research Methodology 3.1. General View In this research, the rotational stiffness of an exterior beam-column connection is assessed. The connection which is wanted to be assessed is an exterior joint in the second floor of a 6-storey rigid frame building, which is designed in accordance to the Indonesian SNI 2847:2013 and SNI 2847:2002 about structural reinforced concrete for building. The lateral force of the structure is then analyzed by referring to SNI 1726:2012 and SNI 1726:2002 about earthquake and seismic design of structures. The building which is used in this research is modeled by using ETABS 2013 software by CSi. ETABS is a finite element software which can be used to model reinforced concrete buildings. This software is capable in doing both structural analysis and structural design of the modeled building. From the building modeled by ETABS software, an exterior beam-column joint is taken into analysis. The detailing of the anchorage in this connection is then designed according to SNI 2847:2013 and SNI 2847:2002. The beam-column joint is then scaled into smaller specimen, so it can be tested in laboratory. 65 Christopher Kevinly
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Universitas Indonesia
Chapter 3
Research Methodology
1.1. General View
In this research, the rotational stiffness of an exterior beam-column
connection is assessed. The connection which is wanted to be assessed is an
exterior joint in the second floor of a 6-storey rigid frame building, which is
designed in accordance to the Indonesian SNI 2847:2013 and SNI 2847:2002
about structural reinforced concrete for building. The lateral force of the structure
is then analyzed by referring to SNI 1726:2012 and SNI 1726:2002 about
earthquake and seismic design of structures.
The building which is used in this research is modeled by using ETABS
2013 software by CSi. ETABS is a finite element software which can be used to
model reinforced concrete buildings. This software is capable in doing both
structural analysis and structural design of the modeled building.
From the building modeled by ETABS software, an exterior beam-column
joint is taken into analysis. The detailing of the anchorage in this connection is
then designed according to SNI 2847:2013 and SNI 2847:2002. The beam-column
joint is then scaled into smaller specimen, so it can be tested in laboratory.
There are two sets of activities which have to be done in this research; the
laboratory-scale experiment of beam-column joint sample and the numerical
modeling by using Fiber Model Analysis. Both experimental and numerical model
are tested in cyclic loading, which end product is the moment-rotation
relationship. The result which is gained from both procedures are then compared
and analyzed.
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66Christopher Kevinly
Experimental data processing
Laboratory testing on sample
Constructing Sample
Assign Sample Connection
Building Modelling
START
Numerical data processing
Running model
Sample modelling in Drain 2DX
FINISH
Conclusion
Comparing data from experiment and numerical calculation
Chart 3. 1. Research planning flow chart
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1.2. Building Modeling
As stated in the previous sub-chapter, the sample which is used in this
research is taken from a modeled structure instead of an existing one. The planned
following researches using different Indonesian standards are going to use the
internal forces of the same building model in order to create their samples, so
comparison between this research and the following researches in the future is
possible.
1.2.1. Preliminary Design
The building which is used will have 6 stories which have a story height of
3.5 meters. The building itself will has a similar beam span of 6 meters. The
building has 8 rows of column along the y axis and 6 rows of column along the x
axis. This building is assumed to be built on deep foundation system, so the
supports of the building are assumed to be fixed supports. The skeletal plan and
the 3D skeletal view of the building model can be seen on the figure below.
Figure 3. 1. Skeletal plan and 3D skeletal view of the building model (Source: Personal)
As a reinforced concrete structure, this building will contain both concrete
and steel reinforcement in it. The concrete which is used is fc’ 30MPa, which has
an unit weight of 24 kN/m3 and a modulus of elasticity of 25742.96 MPa. The
primary reinforcing steel is assigned as deformed rebar which is planned to be
BjTS40 rebar which has a yielding stress of 390MPa, while the stirrups
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reinforcing steel is assigned as plain steel bar BjTP24 which have 235MPa of
yielding stress. Both primary rebar and stirrups will have a modulus of elasticity
of 200000 MPa.
In this building, all beams are assumed to share the same stiffness. This
also applies on the columns. In this building, the column is designed to be a
800mm × 800mm rectangular column, while the beam is planned to be a 400mm
× 550mm beams.
1.2.2. Gravity Load Assignment
This building is planned to be an office building, which according to SNI
1727:2013 about minimum loading for structure has a distributed live load of 2.4
kN/m2. Office live load applies on all stories of the building except the roof,
which is planned to have a live load of 0.96 kN/m2. The slab in this structure will
be assumed as load, which thickness is assumed to be 15cm for all stories, which
is usually used in common office building except for roof, which uses 12cm slab
thickness. This slab’s weight, along with the self-weight of the skeletal elements
will be assumed as the dead load of the structure. Note that the slab along with its
live load will be modeled as shell, which will act as rigid diaphragm.
From the loading above, the forces acting on the beam of each floor can be
defined. The load itself is transferred to the beam through the slab, which
distribution will result in trapezoidal load (in this case, triangle load since width
and length of a slab section is similar).
Although the slab will be modeled as shell element, which will act as rigid
diaphragm, the gravity load transfer scheme to the beam should also be
determined, since it will be required in the probable moment calculation of the
beam.
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Figure 3. 2. Area load distribution to beam(Source: Personal)
The shaded area on the figure above is the area of slab which weight and
load acting on it contribute to the corresponding beam. The peak of the triangle
load will be;
q peak (kNm
)=6 (m)×w( kNm2
) (3.1)
where w is the area load due to live load or dead load. Referring to equation 3.1,
the qpeak for live load is 14.4 kN/m for every story except roof, where the roof will
have a qpeak of 5.76 kN/m.
In order to assign the dead load acting on the structure, the weight of the
slab is assumed to behave the same way with the live load. In order to determine
the area load, the thickness of the slab (15 cm and 12 cm) has to be multiplied
with the unit weight of the slab, which is taken as concrete unit weigt (24 kN/m).
This will yield to an uniform area load of 3.6 kN/m2 for intermediate stories and
2.88 kN/m2 for roof. By using equation (3.1), the qpeak for dead load can be
determined as 21.6 kN/m for intermediate stories and 17.28 kN/m for roof.
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1.2.3. Seismic Load Assignment
1.2.3.1. Seismic Load Assignment in Accordance to SNI 1726:2012
For taking seismic forces into account, the seismic loading assignment in
ETABS can be done. Since the preferred code which is used is the Indonesian SNI
1726:2012, which has similar calculation logarithm to ASCE 7-10, the ASCE
code is used to determine the seismic loading of the structure by static equivalent
approach. Although both SNI 1726 and ASCE 7-10 share the same calculation
method, the input of the spectra needs adjustment. It is assumed that this building
is to be built on Padang on class D soil, which has a value of S s of 1.345g and S1
of 0.599g.
The spectra which is resulted from ETABS by ASCE 7-10 is similar with
the spectra published by PUSKIM and ITB for Padang Region, so the seismic
calculation by ASCE 7-10 can be used. The calculation itself is done
automatically by ETABS software through auto lateral load feature. Note that the
dead load and a quarter of the live load is taken as the seismic weight.
Figure 3. 3. Response spectra of Padang City by ETABS for C class soil(By: Personal)
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Figure 3. 4. Response Spectra of Padang City published by PUSKIM and ITB for C class soil(Source: http://puskim.pu.go.id/)
1.2.3.2. Seismic Load Assignment in Accordance to SNI 1726:2002
To construct the second sample, the seismic loading of the second sample
should be assessed in accordance to SNI 1726:2002. Although the SNI 1726:2002
calculation mechanism is similar with UBC1997, auto lateral load cannot be used,
since the short period used by UBC is 0.1 second, while the short period used by
SNI is 0.2 second. Therefore, the equivalent lateral load should be calculated
manually.
Different with SNI 1726:2012, in the older SNI 1726:2002, the period of
the building have to be limited to ζn, where ζ is the ductility factor and n is the
story number of the building. In this case, ζ = 0.16 for seismic region 5 (Padang),
and n is taken as 6. This yields the limiting building period as 0.96 seconds. The
open-frame system itself has a period of 0.944 second, which does imply with the
requirement. The spectrum for seismic region 5 is shown as below;
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Figure 3. 5. Response spectrum for seismic area 5 (Source: SNI 1726:2002)
From the spectrum above, for T=0.944 second, can be known that C value of
0.529 can be used.
To be able to calculate the lateral load, the seismic weight of the building
itself should be assessed. The seismic load is divided into lumped mass per story,
which includes the self-weight of the structure and its live load. The base shear
should be known first before. By using the ETABS software, the story weight
along with the corresponding lateral force can be seen as below;
Story Story Weight (DL+LL) (kN)
Wz (kNM) Story Shear (kN)
Story 6 12458,835 261635,535
1395,442188
Story 5 19619,55 343342,125
1831,227116
Story 4 19619,55 274673,7 1464,981693
Story 3 19619,55 206005,275
1098,73627
Story 2 19619,55 137336,85 732,4908465
Story 1 19619,55 68668,425 366,2454232
Sigma 110556,585 1291661,9 6889,12353
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1 7Table 3. 1. Story Weight and Story Shear
From the story shear value above, the lateral load can be inputted into
ETABS software. 100% of the force goes along the y direction and 30% to the x
direction.
Figure 3. 6. Lateral load input on ETABS
1.2.4. Load Combination
In this model, there will be three load combinations which will be
assessed; 1.2D+1.6L; 1.2D+L+E+0.2S; and 0.9D+E. D is the dead load acting on
the structure, L is the live load acting on the structure, and E is the earthquake
force acting on the structure, which is gained from equivalent lateral load
approach. These three combinations are taken because these three combinations
have the most possibility of giving the maximum internal forces of the structure.
Note that these combinations are extracted from SNI 1727:2013.
1.2.5. Analysis Result
By running the analysis on ETABS 2013 software, the internal forces of
the frame can be determined. In this case, only the internal forces acting on joint
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F4 at story 1 will be assessed, since this joint is planned to be the sample of this
experiment.
Figure 3. 7. Location of the joint assessed
The internal force of each load types and cases are shown on the table
following this paragraph. All the three load cases are evaluated, and the design
will based on the largest forces acting on the member.
Loading Axial Load (kN) Shear (kN) Moment (kNm)
Dead 0 44,021 -34,46
Live 0 20,86 -17,12
Seismic (2002) 0 40,86 -103,96
Seismic (2012) 0 71,18 -178,75
Table 3. 2. Internal forces calculated on beam (Max)
Loading Axial Load (kN) Shear (kN) Moment (kNm)
Dead -986,7 0 0
Live -234,31 0 0
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Seismic (2002) -239,488 23,76 75,14
Seismic (2012) -459,5 213,11 747,29
Table 3. 3. Internal forces calculated on column (Max)
Table 3. 4,Load combination of the beam (2002 seismic load)
Table 3. 5. Load combination of the column (2002 seismic load)
From the forces above, the beam-column joint can be designed. The design
forces of the beam column joint are shown as below;
Loading Axial Load (kN) Shear (kN) Moment (kNm)
Beam 0 114.55 -162.43
Column -1657,84 23,76 75,14
Table 3. 8. Ultimate internal forces of the members corresponding to the joint (2002 Seismic load)
Loading Axial Load (kN) Shear (kN) Moment (kNm)
Beam 0 144,87 -237,22
Column -1877,85 213,11 747,29
Table 3. 9. Ultimate internal forces of the members corresponding to the joint (2012 Seismic load)
1.3. Specimen Design
1.3.1. SNI 2847:2013 Specimen Design
As soon as the connection sample has been assigned, the corresponding
members of it should be designed in the design; layout of the beam-column joint
to be designed and checked will be in two dimensional plane.
As stated before, the connection which is assessed is the exterior
connection of the second story of the building. This joint and its corresponding
members is designed by using SNI 2847:2013 as a beam-column joint of a special
moment resisting frame.
1.3.1.1. Beam Design
The first member to be designed is the beam, along with its reinforcing
bars. In this beam design, the beam is expected to undergo plastic hinge at 2 times
of the beam depth from the column. In the preliminary design, the dimension of
the beam is taken as 550mm tall and 400 mm wide. The concrete cover which is
planed is 40mm, the stirrups is planned to be ∅12 bars, and the longitudinal bars
is planned to be D25 bars. This will give an effective depth value of 485.5mm.
This beam is expected to bear 237.22kNm of unfactored negative moment
on the column face. In order to simplify the design, the reinforcing bars
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configuration will be designed according to the largest moment (negative moment
on column face) for both positive and negative moment design. This will result in
the same reinforcing bar for both upper and lower part of the beam. In bending
capacity calculation, ϕ=0.9 is used.
In order to design the reinforcement of the beam, the equation 2.44 is used
to calculate the minimum reinforcing bar needed on the section. The equation 2.44
itself can be rearranged into;
( f y2
1.7 f c' b )A s
2−( f y d) A s+M n /ϕ=0 (3.2)
by arranging the equation into this form, the value of minimum reinforcement can
be calculated by solving the polynomial equation 3.2. After doing the calculation,
there are two values of As; 1963.495mm2 and 23,917.29mm2. From these two
values, the smaller value is taken, since it can be rationally accepted.
To fulfill the minimum reinforcement requirement, 4 D25 (1963.495mm2)
bars are needed. This configuration should also be checked due to SNI 2847:2013
point 21.5.2.1, which states that the maximum reinforcement area is 2.5% and the
minimum area of reinforcement is stated as below;
A smin=
1.4 bw d
f y
(3.3)
where bw is the width of the beam web, d is the effective depth of the beam, and f y
is the yielding strength of the rebar. The planned As (two times 1963.495mm2)
covers 1.785% of the gross area of the section, and the minimum longitudinal
reinforcement required by SNI is 700 mm2. Note that the rebar area used complies
with both requirements, so it is fine to use this configuration.
Since the installed rebar area is more than the minimum rebar area, the
flexural strength of the beam should be calculated. By using equation 2.40, the
nominal moment capacity of the beam will be 343.03 kNm, while reduced
moment capacity is 308.73 kNm.
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Other than flexural capacity, the shear capacity of the beam should also be
assessed based on the ultimate shear occurred on the beam. Note that the ultimate
shear force occurs on the beam is 144.86kN. In design of shear reinforcement in
joint of special moment resisting frame, according to SNI 2847:2013 point
21.5.4.2., the shear capacity contributed by concrete is neglected. This is done to
accommodate plastic hinge condition where the concrete will have negligible
shear capacity due to heavy cracking.
Besides of the ultimate shear gained from structural analysis, the
controlling shear force should also be determined by calculating the shear force
resulting from probable moment of the beam. This moment is gained from the
flexural capacity of the beam, calculated by using 1.25fy and no reduction factor
(ϕ=1.0). The value of shear force resulting in the beam due to the probable
moment is stated as;
V c=MPr
A MPrB
l+qu l2
(3.4)
Where Vc is the maximum shear force may act on the column face, MApr is
the probable moment of beam at point A, MBpr is the probable moment of beam at
point B, l is the beam span, and qu is the ultimate factored distributed load
(1.2qDL+1.0qLL) (SNI 2847:2013).
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Figure 3. 8. Loading cases for shear design of beams based on probable moment of the beam. (Source: SNI 2847:2013)
The probable moment of the beam being designed is taken as its flexural
capacity by using 1.25 fy as its yielding strength and not inducing any reduction
factor. In this case, its compression block may expand due to increased tensile
yielding strength of the rebar. Substituting fy with 1.25fy on equation 2.37 will
yield;
0.85 f c' ba=A s 1.25 f y
a=1.25 A s f y
0.85 f c' b
a=1.47A s f y
f c' b
(3.5)
From this increased compressive block, the moment capacity of the beam
can be determined from equation 2.32. Note that T of the column is now equals to
1.25fyAs, which alters the result of the initial equation.
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M n=1.25 A s f y (d−a2 )
M n=1.25 A s f y (d−
1.47A s f yf c' b
2)
M n=1.25 A s f y (d−A s f y
1.36 f c' b ) (3.6)
Inserting As as 1963.495mm2, fy as 390 MPa, d as 485.5mm, fc’ as 30 MPa, and b
as 400mm, will result in Mn of 419,808,767.7 Nmm or 419.81 KNm.
The value of ½(qul) is basically is the value of the shear force due to the
factored gravity load in the end of the beams. To determine the value of qu, the qLL
is taken as 14.4/2 kN/m or 7.2 kN/m and the qDL as 21.6/2 kN/m + (0.4m × 0.55m
× 24 kN/m3) or 16.08 kN. Note that the triangle loads on qDL and qLL have to be
multiplied by ½ to make it equivalent with distributed load. The value of qu
according to [18] is then 1.0×7.2 + 16.08×1.2 kN/m or 26.496 kN/m. The value of
½(qul) is then 79.488 kN.
By using the equation 3.4 and taking MApr and MB
pr as 419.81 KNm, l as 6
meters, and ½(qul) as 79.488 kN, the Vc is 219.425 kN. Therefore, the design of
stirrups of the beam should be done based on this shear force value.
In order to calculate the shear capacity of the beam, equation 2.52 is used.
If there are four shear reinforcement bars are planned, where each bar is a ∅12 bar
which has a yielding strength of 235 MPa, depth of the beam is 485.5mm, and ϕ
for shear is taken as 0.75, equation 2.52 will give a maximum spacing of
176.42mm. According to SNI 2847:2013 point 21.5.3.2., the minimum spacing of
stirrups must not exceed one-fourth of the effective depth of the beam, six times
the diameter of longitudinal reinforcement, or 150mm. In this case, it will be the
smallest of 121.375mm and 150mm. therefore, the maximum stirrups spacing
allowed is 121.375mm. 120mm of stirrups spacing is taken to simplify
implementation.
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Figure 3. 9. Beam reinforcement configuration
(Source: Personal)
1.3.1.2. Column Design
In the column design, the design axial load is 1877.85kN. According to
SNI 2847:2013 point 21.6.1, the factored axial load should be less than 0.1AgFc’.
Therefore, an 800 x 800mm (Ag = 640,000mm2) column with 16 D25 rebar is
taken. The design moment according to SNI 2847:2013 point 21.6.22 is taken as
1.2 of the unfactored nominal moment capacity of the beams connected to the
column, which is 1.2 times 3 beams times 343.03 kNm or 1224 kNm. The design
shear of the column itself is based on the maximum force may occur on the
column. This design shear force can be designed based on the shear caused by the
probable moment as much as the moment capacity of the column with the
corresponding axial force, with no reduction factor and fy = 1.25 fys. The
maximum shear also comes from dividing the over-strengthened moment capacity
of the beam by the half-span of the beam. The shear calculated by this method is
then compared to the shear acquired from the structural analysis.
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Figure 3. 10. Options of calculating column shear (Source: Seismic Design of Reinforced Concrete Special Moment Frames: A Guide for Practicing Engineers,
Jack P. Moehle et al., 2008)
Figure 3. 11. Interaction diagram of column (excluding reduction factor and increased fy)(Source: Personal)
From the interaction diagram above, it is known that moment capacity of
the column with no reduction factor and increased fy for 1877.85kN of
compressive force is 1504.5 kNm. By multiplying it by two and dividing it by its
clear length (2.95m), the maximum shear force caused by these moments is
1020kN. If the probable moment from the beam is assessed, by using Mpr =
419.81 KNm (gained from subchapter 3.3.1.1), the shear force resulted from this
moment can be determined by subtracting the longitudinal bar force with the
upper column shear (refer section 2.4.2). The beam probable moment will cause
rebar force to be 864.7 kN. The column’s shear gained from the structural analysis
is 213.11 kN. Since the value generated from the column’s probable moment is
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larger than the value generated from the structural analysis and the beam’s
probable moment, 1020kN is taken as the design shear.
To design the longitudinal reinforcing bar of the column, the rebar
configuration has to be stated in the first place. A moment-axial load interaction
diagram is then made by using ETABS 2013 software.
According to SNI 2847:2013, the area of longitudinal reinforcement must
not be less of 1% Ag and more than 6% Ag and the minimal number of reinforcing
bar is 6 bars. In that case, 16 D25 bars is taken as the longitudinal reinforcement.
The reinforcing bars make up 1.23.% of the cross-sectional area of the column.
The confinement rebar (stirrups) is taken as ∅12 bars, and the concrete cover
taken is 40mm.
From the configuration above, an interaction diagram of axial force and
bending moment of the column can be made. By using ETABS 2013, the
interaction diagram (factored) of the column configuration above is shown as
below.
Figure 3. 12. Interaction diagram of column (factored)(Source: Personal)
From the interaction diagram shown, it is known that the axial load
(1877.85 kN) and the bending moment (1224 kNm) are inside the interaction
envelope. Therefore, the proposed longitudinal bar configuration can be used.
To design the transverse reinforcement (confinement stirrups), several
perimeters should be assessed. To find the minimum transverse reinforcement
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needed, the spacing of it should be assessed before determining the number of the
stirrups needed.
The spacing should also be checked in accordance to SNI 2847:2013. Two
spacing should be considered, the spacing nearby the connection with the length
of lo, and the spacing outside the lo length. In this case, both spacing are taken as
the same. The regulation requires the maximum stirrups spacing to be the least of
one fourth of minimum structural component dimension, six times of the
longitudinal rebar diameter, or less than the formula stated below;
so=100+( 350−hx
3 ) (3.7)
where hx is the distance between stirrups in millimeters, where So may not exceed
150mm and it is not necessary to take So less than 100mm. From these three
criteria, the maximum stirrups spacing due to dimension, rebar diameter, and
equation 3.7 respectively are 200mm, 192mm, and 161.333mm. However, to
provide proper confinement, the spacing of stirrups is taken as 110mm.
To determine the required area of the transverse reinforcement of the
column, SNI 2847:2013 point 21.6.4.4 requires the confinement reinforcement
area to be more than the larger of
A sh=0.3sbc f c
'
f yt (( Ag
Ach)−1) (3.8)
or
A sh=0.09sbc f c
'
f yt
(3.9)
where Ash is the minimum confinement reinforcement area, s is the spacing
between stirrups, bc is the width of the column area which is bounded with the
stirrups, fc’ is the concrete compressive strength, fyt is the stirrups steel yielding
strength, Ag is the gross area of the column, and Ach is the area bounded by the
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outer part of the confinement stirrups. By using s = 100mm, fc’ = 30 MPa, fyt =
235 MPa, bc = 708mm, Ag = 640,000mm2 and Ach = 518,400mm2, the minimum
reinforcement gained from equation 3.8 and 3.9 respectively are 699.63mm2 and
894.79mm2. To fulfill the transverse reinforcement requirement, eight ∅12 bars
(Ash = 904.78mm2) should be used. Note that confinement bars are applied on
both sides.
According to the reinforcement configuration, the transverse
reinforcement will consist of eight ∅12 bars a segment. Therefore, Av = Ash will be
904.78mm2. By using equation 2.52 and assuming that no shear strength
contributed by concrete, the maximum spacing gained for 1020 kN of shear force
is 153.32 mm. Since the maximum spacing is more than the spacing used in the
To ensure sample quality before testing, UPV (Ultrasonic Pulse Test) is
done. This is done to make sure that the sample is perfectily fit to be tested,
without any major cracking or flaw which may alter the experiment result. The
sample is considered good if the pulse velocity of the UPV test exceeds 3.5 km/s.
[19]
The sample lifting should be carefully done, since improper handling of
the sample may cause cracking and alter the experiment result. Therefore, the
lifting point of the sample will be properly arranged. Two lifting points are
assigned to lift the sample and fit it on the testing frame. The first point is right
below the column, while the second point is on the midspan of the beam. No
major collision is allowed to occur during the lifting and fitting process to prevent
damage on the sample.
1.4.2. Sample Testing
Loading is done on the end of the beam element, which is done by using
hydraulic jack. The sample is planned to be loaded by semicyclic loading schame,
which scheme is similar with the loading pattern being used in [10]. The loading
scheme itself is a deflection based scheme. In the first 3 cycles, loading is given as
0.25 Δy. For the 4th to the 6th cycle, the load is raised to 0.5 Δy. For the 7th to 9th
cycle, the load is raised to 0.75 Δy. For the 10 th to 12th cycle, the load is raised to
Δy. For the 13th to the 17th cycle, the load is taken as (n-11) Δy, where n is the
number of cycle and Δy is the yielding displacement of the beam section. The
yielding displacement itself should be determined by referring the load-
displacement relationship of the beam, park [20] proposes several definitions of
yield displacement as below;
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Figure 3. 27. Several approaches to determine yield displacement (Source: Evaluation of Ductility of Structures and Structural Assemblages from Laboratory Testing, R. Park,
1988)
0 2 4 6 8 10 12 14 160
1
2
34
5
6
7
Cycle
Disp
lace
men
t (∆y
)
Figure 3. 28. Proposed Loading Scheme
For each cycle, the cracking occurred on the sample is assessed and
marked, so the cracking development of the sample can be known better. After
each loading sequence, vibration test is done in order to assess the stiffness of the
overall structure. According to a research conducted by Q.-B. Bui [21], the
stiffness degradation will cause a decrease of natural frequency of the beam-
column sample. The vibration test itself is done by excite the sample with a free
vibration from the tip of the beam. To know the placement of accelerometer, the
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mode shape of the sample should be assessed first. By using SAP2000 software,
the first three modes shape of the sample is shown as below;
Figure 3. 29. Mode 1 of the sample (Source: Personal)
Figure 3. 30. Mode 2 of the sample (Source: Personal)
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Figure 3. 31. Mode 3 of the sample (Source: Personal)