Chapter 3 Print Mander

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.

65Christopher 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


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



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.



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.



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)



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;



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



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



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)


Dead -986,7 0 0

Live -234,31 0 0



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)


1.2D+1.6L 0 86,2012 -68,7441.2D+1L+1E+0.2S 0 144,8652 -237,2220.9D+1E 0 110,7989 -209,764

Table 3. 7. Load combination of the column (2012 seismic load)


Table 3. 6. Load combination of the beam (2012 seismic load)


1.2D+1.6L 0 86,2012 -68,744

1.2D+1L+1E+0.2S 0 114,5452 -162,432

0.9D+1E 0 80,4789 -134,974


1.2D+1.6L -1558,936 0 0

1.2D+1L+1E+0.2S -1657,838 23,76 75,14

0.9D+1E -1127,518 23,76 75,14


1.2D+1.6L -1558,936 0 01.2D+1L+1E+0.2S -1877,85 213,11 747,290.9D+1E -1347,53 213,11 747,29


From the forces above, the beam-column joint can be designed. The design

forces of the beam column joint are shown as below;


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)


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



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.



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).



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.



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.



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.



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



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



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



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

column, the design is considered safe.

Figure 3. 13. Column reinforcement layout(Source: Personal)



1.3.1.3. Joint Strength Check and Panel Zone Detailing

Both the column and the joint will then connected in a beam-column joint.

This area is critical and needed to be detailed well, since poor detailing may result

in connection failure occurring before the formation of plastic hinge on the beam.

Several important notes which have to be done in this case; the joint

confinement and the beam rebar anchorage. The joint confinement has been

secured due to the close spacing of column stirrups, while longitudinal rebar

anchorage of the beam longitudinal section has to be considered more carefully

here.

To prevent connection failure, adequate bond force should be added into

the longitudinal bar. Since the bond force depends heavily on the development

and anchorage of the rebar, this length should be assessed. According to SNI

2847:2013 point 12.2.1, for D25 rebar, the minimum development length ld is 300

mm or stated as below;

ld=( f yΨ tΨ e

1.7 λ√ f c' )db (3.10)

where fy is the yielding strength of the longitudinal bars, Ψ t and Ψ e is factors

regarding the bar configuration, fc’ is the compressive strength of the concrete, λ

is the reduction factor for lightweight concrete, and db is the rebar diameter. By

taking fy as 390 MPa, Ψ t as 1.0, Ψ e as 1.0, fc’ as 30MPa, λ as 1.0 and db as

25mm, the minimum development length from this formula is 1047.12mm, which

means that the longitudinal tension bars should not be discontinued along this

length. Since this value greatly exceeds the width of the column, bend is required

to be formed.

To determine the horizontal length of beam rebar anchorage required in

the column, the formula ldh is used. This requirement is based on SNI 2847:2013

point 21.7.5.1, where the minimum value of ldh is the largest of 8db, 150mm, and;



ldh=f y db

5.4√ f c'(3.11)

from these three values, the minimum ldh are 200mm, 150mm, and 329.65mm.

Therefore, the minimum horizontal length of the anchorage is taken as 350mm.

The anchorage is taken as ldh = 700mm for the upper steel and 650 mm for the

lower steel because the rebar should be anchored near the end of the column side.

Since these numbers exceeds 350mm, the horizontal development length is fine.

Figure 3. 14. Minimum anchorage length of hooks and bends (Source: SNI 2847:2013)

from the figure above, it is stated that the minimum length of the vertical rebar

after the bend is 12 db. In this case, it corresponds with 300mm. the vertical

length of the anchorage is (300mm + 4db) =400mm.

To check the shear of the beam-column joint, SNI 2847:2013 section

21.7.4. The effective area of the joint A j is h = 800mm multiplied by b+2×200 =

800. Multiplying both will result in Aj = 640000mm2. This will give a value of

√ f c' A j as 3505.42 kN. Since this value is much larger than the shear acting on the

joint, the joint is considered safe.



1.3.2. SNI 2847:2002 Specimen Design

The second sample constructed in accordance to SNI 2847:2002 should

also be assessed as comparison to the newer SNI 2847:2002. This second

specimen will be designed based on the earthquake loading calculated by SNI

1726:2002.

1.3.2.1. Beam Design

If it is compared with the sample designed in accordance to SNI

2847:2013, smaller internal forces occur on the beam due to smaller seismic load.

In this case, the beam is only required to withstand 114.55 kNm of negative

moment. By using equation 3.2, with similar parameter with the beam designed

by using SNI 2847:2013, the required reinforcement is only As=690.97mm2.

However, similar with SNI 2847:2013, SNI 2847:2002 requires special moment

resisting frame to have a minimum reinforcement which follows equation 3.3.

According to this equation, the minimum reinforcement of the beam is 700mm2.

To fulfill this requirement, two D25 reinforcement (As=981.75mm2) is used for

both upper and lower part of the beam. This configuration will be able to carry

178.7 kNm of factored moment.

To design the shear reinforcement, the design shear force should be the

larger of the one caused by probable moment of the beam and the one gained from

the structural analysis. The structural analysis done earlier gives a design shear

value of 114.55 kN. By calculating the shear caused by the probable moment

according to equation 3.4. By calculating the over-strengthened moment capacity

of the beam using equation 3.6, a value of 221.13 kNm is gained.

Since the value of ½(qul) is the same with the previous sample (79.488

kN), the value of shear force due to the probable moment can be calculated. By

using equation 3.4, using MPrA and MPr

B as 221.13 kNm, l as 6 meters, and ½(qul)

as 79.488 kN, the shear caused by probable shear is Vc = 153.2 kN. Since this

value is larger than the one gained from structural analysis, this value is taken as

the design shear force.

To determine whether the concrete will contribute to the shear strength of

the beam, SNI 2847:2002 section 23.3.4.2 is checked. According to the section,



concrete contributes to the shear strength if the factored axial load acting on the

beam is less than Agf’c/20 and the shear contributed by the seismic load is less

than the half of the total shear force. The factored axial load acting on the beam

itself is 7.24 kN, which is less than Agf’c/20=330 kN. The shear contributed by

seismic force is 40.86 kN, which is less than a half of the shear calculated by

structural analysis (114.55 kN). Therefore, the concrete is able to contribute to the

shear strength. The concrete contribution to the shear strength itself is calculated

by using equation 2.48, which results in a value of 180.83 kN. This value is larger

than the design shear, so the stirrup spacing is determined by the limiting

regulation instead of strength consideration.

A pair of ∅12 bars is used as the stirrup of the beam. To determine the

spacing required. SNI 2847:2002 sections 23.3.3.2 requires that the maximum

spacing of stirrup is a quarter of the beam’s effective depth, eight times of the

longitudinal bar diameter, 24 times of the diameter of the stirrup bar, and 300mm.

These requirements will limit the beam spacing to121.875mm (a quarter of the

beam’s effective depth). Therefore, a spacing of 120mm is taken.

Figure 3. 15. Beam section designed in accordance to SNI 2847:2002 (Source: Personal)



1.3.2.2. Column Design

There are no major change between the SNI 2847:2002 and SNI

2847:2013 regarding the column regulation of special moment resisting frame.

This also apply for the confinement regulation, which requires the column to be

heavily confined. To accommodate transverse rebar, small longitudinal rebar

cannot be used. In this case, 16 D25 rebar is used as longitudinal reinforcement

and ∅12 rebar is used for confinement reinforcement, which is exactly similar

with the previous sample.

The column is expected to hold against an axial load of 1657.84 kN and a

moment of 75.14 kNm. According to the SNI 2847:2002 section 23.4.2.2, the

flexural strength of the column must be more than 1.2 the flexural strength of

the beams connected to it. This will result in a design moment of 3 times

178.7kNm times 1.2, which is equal to 643.32 kNm.

Figure 3. 16. Factored interaction diagram of the column designed based on 2002 standards(Source: Personal)

By using the interaction diagram shown on the figure above, it is

known that for an axial force of 1657.84 kN, the moment of 643.32 kNm is still

within the envelope. The current longitudinal reinforcement of 16 D25 rebar

also covers just 1.23% of the total cross section, which is more than 1% and

less than 6%. Therefore, this longitudinal reinforcement configuration will be

used.



Besides of bending moment, the beam is also expected to bear shear.

According to SNI 2847:2002 point 23.4.5, design shear is calculated by

calculating the shear caused by probable moment of both the beam and the

column. This is similar with the one done in the previous sample.

Figure 3. 17. Interaction diagram of column designed on 2002 standards (overstrengthened)(Source: Personal)

From the over-strengthened interaction diagram above, it is known with a

compressive strength of 1657.84 kN, the flexural capacity of the column is 1549

kNm. This value is then taken as the probable moment of the column. By

multiplying it by 2 and deviding it by its clear span (2.95m), shear force of

1050.17 kN is gained. Since this value is much larger from the shear calculated

through structural analysis, this value is taken as the design shear.

The configuration of the stirrups is regulated by the SNI 2847:2002 section

23.4.4. This regulation requires the stirrup spacing to be less than the least of a

quarter of the smallest member width, six times the diameter of longitudinal bars,

or the value gained from equation 3.7. These require the maximum stirrup spacing

to be smaller than 200mm, 150mm, and 161.333mm. However, to provide proper

confinement, the spacing of stirrups is taken as 110mm.

Similar with the newer SNI, transverse reinforcement configuration should

also follow SNI 2847:2002 section 23.4.4.1, where the minimum area of

reinforcement should be more than the value gained from equation 3.8 and 3.9. By



using s = 110mm, fc’ = 30 MPa, fyt = 235 MPa, bc = 720mm, 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.

This configuration is similar with the one designed by using SNI 2847:2013. If

this configuration is used to accommodate the shear force, the spacing required is

182.43mm, which is more than 110mm. Therefore, 110mm spacing is used.

Figure 3. 18. Column reinforcement layout designed by 2002 standards(Source: Personal)

1.3.2.3. Joint Strength Check and Panel Zone Detailing

After designing the beam and column, the connection between both are

checked. Note that the older and newer standards have the same requirement for

beam anchorage. But since the beam uses smaller rebar, the development length

should be recalculated by using equation 3.10. By taking fy as 390 MPa, Ψ t as

1.0, Ψ e as 1.0, fc’ as 30MPa, λ as 1.0 and db as 22mm, the minimum development



length from this formula is 921.47mm. This value exceeds the width of the

column, so bend should be made to provide adequate anchorage.

The bend requirement itself is similar with the one from SNI 2847:2013.

Smaller rebar means that smaller anchorage length is required. However, the

requirement for special moment resisting frame is that the bend should be made so

the vertical part of the bar is close to the column end. Therefore, the same

anchorage is used, as ldh for the upper steel is taken as 700mm and the lower steel

as 650mm. For the vertical part of the rebar, 12 db of length is required, which is

equal to 300mm for D25 bars. The bend radius itself is equal to 4db, which is

equal to 100mm. adding both will result in vertical development of 400mm.

taking the similar length with the previous model; the vertical development of the

rebar is taken as 400mm.

To check the shear of the beam-column joint, SNI 2847:2002 section has

similar requirement with SNI 2847:2013. The effective area of the joint A j is h =

800mm multiplied by b+2×200 = 800. Multiplying both will result in Aj =

640000mm2. This will give a value of √ f c' A j as 3505.42 kN. Since this value is

much larger than the shear acting on the joint, the joint is considered safe.

1.3.3. Full-Scale Sample Configuration

After doing the design of the joint elements, the samples which are used to

do this experiment can be designed. For doing the test, half of the beam span will

be taken. If the span measured from the column centre is 6000mm, half of it will

be taken. The beam length is now 3000mm measured from the middle of the

column. Since 400mm of the beam is in the column, the clear span of the beam

sample is 2600mm. Both samples have the same dimension.

Similar with the beam sample assignment, the column sample height is

taken as the half of the story height of the column. Since there will be two

columns connected in the joint, the overall height of the sample can be taken as

the story height of the building, which is 3500mm.



Figure 3. 19. Full-Scale sample with reinforcement(Source: Personal)

1.4. Laboratory Scale Experiment

As one of the way of measuring the semirigid behavior of the connection,

laboratory scale experiment is done. In general, there will be two main parts of the

experiment, sample construction and preparation, and the sample testing itself.

1.4.1. Preparation and Sample Construction

In this research, a sample is made in the laboratory. The sample itself will

has a scale factor of 0.4. In this scaled sample, the entire dimension is reduced by

half of the initial dimension, without compromising the quality of the material.

For example, D13 BjTS40 rebar is used instead of D32 BjTS40 rebar, and 320mm

× 320mm column is built using fc’ 30MPa concrete instead of 800mm × 800mm



column using fc’ 30MPa concrete. For the longitudinal rebar of the column made

by SNI 2847:2002, D25 BjTS40 is replaced with D10 BjTS40.

The rebar being used in the full-scale sample is then reduced to the half of

the initial size. The beam, which has been reduced into a 220mm x 160mm, will

has D10 and ∅5 rebar to substitute the D25 and ∅12 for the sample designed by

SNI 2847:2013 and D10 and ∅5 rebar to substitute D25 and ∅12 rebar initially

installed. Note that ∅5 steel is not available in SNI standard, so the steel found in

the market should be tested in the laboratory.

Figure 3. 20. Scaled specimen (Source: Personal)



Figure 3. 21. Scaled beam specimen (2013 standards design) (Source: Personal)

Figure 3. 22. Scaled Column Specimen (2013 standards design)(Source: Personal)



Figure 3. 23. Scaled beam specimen (2002 standards design) (Source: Personal)

Figure 3. 24. Scaled Column Specimen (2002 standards design)(Source: Personal)

Dry-mix concrete is used in the making of the sample. This is used due to

its simplicity and accuracy, since making a mix design personally is rather

inaccurate and risky. The dry-mixed concrete which is used will first be tested in



the laboratory first before being used in the sample. Fc’30 MPa dry-mixed

concrete will be used in this experiment.

In order to detect the presence of plastic hinge, strain gauges should be

installed on the reinforcing bars. These strain gauges should be installed in the

areas where rebar yielding is expected to occur. In this case, strain gauges will be

placed on the flexural bar, not more than 50mm from the column face. This gauge

placement is done because this reinforcing bar is expected to experience the

highest stress, since that area will has the largest bending moment during the test.

The other strain gauge will be located on the development of the beam’s

rebar inside the column. Only the top will be installed with strain gauge, since that

part will undergo major tension, which may cause bond slippage. The lower rebar

development of the beam is not equipped with strain gauge because the

compressive stress is not as high as the top rebar due to the existence of concrete

compressive strength.

As shear force may become dominant as soon as the occurrence of plastic

hinge on the beam, strain gauges should also be installed on the beam stirrups.

The stirrup which is equipped with strain gauge is the first beam stirrup from the

column face. The strain gauge installation plan is shown on the figure below;

Figure 3. 25. Proposed strain gauge installation locations(Source: Personal)



Curing process is done by using fiber sags which has been saturated with

water. The curing process itself is impossible to be done in the curing tub due to

the large dimension of the sample. The sag’s humidity should always be

controlled, since dry sag may cause the heat caused from the hydration process is

not absorbed and causes cracking on the sample. Therefore, the sag should be kept

wet by showering it twice a day; at 11AM in the morning and at 6PM in the

evening and stored indoor.

The sample which has been cured for 28 days is then fitted into the loading

frame to be loaded. The sample itself will be loaded in semi-cyclic sequence,

which load is increased for each loading. The loading itself is done on the end of

the beam by hydraulic jack. The deformation of the beam and column is assessed

by using displacement dials. The experiment layout is shown as below;

Figure 3. 26. Experimental setup (Source: Personal)



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;



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



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. 32. Accelerometer Placement (Source: Personal)

The mode shape from the first three modes indicates great deflection on

the half-end of the beam. Therefore, three accelerometers will be installed along

the half-end of the beam. Another accelerometer will also be installed in the joint

of the column and two more accelerometers are installed on the mid-span of each

upper and lower part of the column.

In order to trigger the excitation, two methods are proposed. The first

method is by hitting the end of the beam with a rubber hammer. The second

method uses a vibration generating device to pull the end of the beam and release

it to create free vibration. This device consists of a cable hoop which is joined to a

single steel bar (rod). This rod which has a loop / hole is attached to a bearing bar.

To initiate vibration, the cable hoop is placed around the tip of the sample. The

hydraulic jack deflects the beam tip so the rod can be attached to the bearing bar.

The hydraulic jack is then lifted, and the bearing bar is pulled fast, releasing the

rod and causing the beam to vibrate freely.



Figure 3. 33. Schematic view of the vibration generator(Source: Personal)

1.5. Numerical Analysis

The numerical analysis in this experiment is done by using fiber model

analysis through Drain-2DX software. This is done by splitting the model into

‘layers’ of fiber to be analyzed numerically.

The Structure will be simplified as a beam and columns with two pinned

support. The beam-column joint panel is taken as the rigid zone. The beam which

is analyzed is the scaled beam, so the span of the beam is 840mm. This beam is

going to be discretized into 2 elements. The two columns which have a single

height of 690mm. The column itself is modeled as one element for each column

parts. In total, there are 4 elements and 5 nodes, where the load is placed on Node

4.



Figure 3. 34. Structure modeling for numerical analysis (Source: Personal)

The beam will consist of two elements; Element 3 which has a length of

760mm and Element 4 which has a length of 80mm. Element 3 is discretized into

7 sections which has a length of 108.57mm for each section, while element 4 will

not be discretized. Element 3 will start at Node 2, Element 3 and Element 4 will

be connected with Node 4, while Element 4 will end at Node 5

Figure 3. 35. Beam modeling for numerical analysis (Source: Personal)

Each column part will be discretized into 5 parts; each part will have a

thickness of 138mm. Element 1 (lower column) will start at Node 1 and end at

Node 2, while Element 2 (upper column) will start at Node 3 and end at Node 2.

Note that both Node 1 and Node 3 will be restrained for translations.



Figure 3. 36. Column modeling for numerical analysis (Source: Personal)

In this analysis, both the beam and the column are analyzed through fiber

model analysis. The beam cross section of the sample is split into 10 parts of

concrete fibers; 2 parts will consist of fully unconfined concrete and 8 parts will

consist of combination of confined and unconfined concrete. 2 layers of steel will

also be added into the analysis. The two unconfined layers will have a thickness

of 20.8mm for each of the layers, and the rest of the concrete fibers will have a

thickness of 22.25mm.

The column will be divided into layers. 8 layers of concrete and 5 layers of

steel are assigned. The thickness of intermediate layer is 46.4mm and the

thickness of unconfined outer layer is 20.8mm.

Fiber Center of Gravity (mm) Area Fiber Type

1 99.5 3360 Unconfined

2 85.6 314.16 Steel

3 77.875 934.5 Unconfined

4 77.875 2625.5 Confined


6 55.625 2625.5 Confined




8 33.375 2625.5 Confined


10 11.125 2625.5 Confined

11 -11.125 2625.5 Confined

12 -11.125 934.5 Unconfined

13 -33.375 2625.5 Confined

14 -33.375 934.5 Unconfined

15 -55.625 2625.5 Confined

16 -55.625 934.5 Unconfined

17 -77.875 2625.5 Confined

18 -77.875 934.5 Unconfined

19 -85.6 314.16 Steel

20 -99.5 3360 Unconfined

Table 3. 10. Fiber determination of beam cross section (2013 Sample)(Source: Personal)

Figure 3. 37. Beam section modeling by layer approach (2013 Sample)(Source: Personal)



2 85.6 157.08 Steel


4 77.875 2625.5 Confined




6 55.625 2625.5 Confined


8 33.375 2625.5 Confined


10 11.125 2625.5 Confined

11 -11.125 2625.5 Confined

12 -11.125 934.5 Unconfined

13 -33.375 2625.5 Confined

14 -33.375 934.5 Unconfined

15 -55.625 2625.5 Confined

16 -55.625 934.5 Unconfined

17 -77.875 2625.5 Confined

18 -77.875 934.5 Unconfined

19 -85.6 157.08 Steel


Table 3. 11. Fiber determination of beam cross section (2002 Sample)(Source: Personal)

Figure 3. 38. Beam section modeling by layer approach (2002 Sample)(Source: Personal)



2 134.2 392.7 Steel



3 116 1948.8 Unconfined

4 116 12899.2 Confined


6 69.6 12899.2 Confined

7 69.6 157.08 Steel


9 23.2 12899.2 Confined

10 0 157.08 Steel

11 -23.2 12899.2 Confined

12 -23.2 1948.8 Unconfined

13 -67.1 157.08 Steel

14 -69.6 12899.2 Confined

15 -69.6 1948.8 Unconfined

16 -116 12899.2 Confined

17 -116 1948.8 Unconfined

18 -134.2 392.7 Steel


Table 3. 12. Fiber determination of column cross section (2013 Sample)(Source: Personal)

Figure 3. 39. Column section modeling by layer approach (2013 Sample)(Source: Personal)





2 134.2 392.7 Steel

3 116 1948.8 Unconfined

4 116 12899.2 Confined


6 69.6 12899.2 Confined

7 69.6 157.08 Steel


9 23.2 12899.2 Confined

10 0 157.08 Steel

11 -23.2 12899.2 Confined

12 -23.2 1948.8 Unconfined

13 -67.1 157.08 Steel

14 -69.6 12899.2 Confined

15 -69.6 1948.8 Unconfined

16 -116 12899.2 Confined

17 -116 1948.8 Unconfined

18 -134.2 392.7 Steel


Table 3. 13. Fiber determination of column cross section (2002 Sample)(Source: Personal)

Figure 3. 40. Column section modeling by layer approach (2002 Sample)(Source: Personal)



After defining the layers of the beam and column, the material properties

and stress-strain relationship should be stated. Four properties should be stated;

confined concrete, unconfined concrete, longitudinal reinforcement steel and

transverse reinforcement stirrups steel.

The concrete property which is inputted into Drain-2DX program includes

the maximum compressive strength of the concrete, the modulus of elasticity of

the concrete, and ultimate strain of the concrete for both confined and unconfined

concrete. The stress-strain relationship itself will be expressed by using Kent-

Park’s curve for unconfined concrete and Mander’s stress-strain curve for

confined concrete. By using these curves, there are five curves which are required

to be calculated; unconfined concrete, confined concrete for the beam and column

of the 2002 and 2013 samples respectively.

For the unconfined concrete, the relationship for the stress and strain can

be found by using

f c=30,000 ε0+7,500,000 ε 02 (3.12)

while the declining part of the stress-strain relationship for the unconfined

concrete is stated as

f c=50.1−10,050 ε 0 (3.13)

In order to input the stress-strain relationship above to Drain-2DX, the

properties of the unconfined concrete itself is inputted as on the tables below;

Property Value

Compressive Strength 30 MPa

Strain at Ultimate Stress 0.002

Spalling Strain 0.0044

Table 3. 14. Material properties of concrete being used(Source: Personal)

Stress Strain



12,158 0,0005

27,590 0,0015

29,53125 0,00175

30 0,002

0 0,0044Table 3. 15. Stress and strain point to be inputted for unconfined concrete property

(Source: Personal)

After determining the property of the unconfined concrete, the stress-strain

of confined concrete should be determined. Since reinforcement configuration

will result in different confinement behavior, each factor which influences the

stress-strain relationship is shown for each sample.

Property Value

Unconfined Compressive Strength 30 MPa

Ineffective Confinement Area 9335.335 mm2

Center-center Stirrup Spacing 50 mm

Clear Stirrup Spacing 40 mm

Effective Confinement Area 8602.058 mm2

ke 0.4115ρx 0.0044ρy 0.0133

f’lx 0.429 MPa

f’ly 1.268 MPa

Confined Compressive Strength 33 MPa

Strain at Confined Compressive Strength 0.003

Maximum Strain (First Hoop Fracture) 0.0393

Table 3. 16. 2013 beam properties which influences stress-strain relationship(Source: Personal)



The Mander’s stress-strain curve will result in an equation of

f c=19207.32 ε0

0.74612+ε 0

1.74612

3.944×10−5

(3.14)

which will result in a graph below;

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.0450.0005.000

10.00015.00020.00025.00030.00035.000

Stress-Strain relationship of Concrete (2013 Beam)

ConfinedUnconfined

Strain

Stre

ss (M

Pa)

Figure 3. 41. Stress and strain curve of the concrete used for 2013 beam(Source: Personal)

From the figures above, the input values for confined concrete can be

inputted.

Stress Strain

12,158 0,0005

27,590 0,0015

31,010 0,002

33 0,003

8,382 0,0393Table 3. 17. Stress and strain point to be inputted for confined concrete property of 2013 beam

(Source: Personal)



With similar manner, stress-strain relationship of the confined concrete of

the column sample designed by using 2013 standard can also be determined.

Property Value






ke 0.866ρx 0.0141ρy 0.0141

f’lx 2.869 MPa

f’ly 2.869 MPa

Confined Compressive Strength 45.9 MPa



Table 3. 18. 2013 column properties which influences stress-strain relationship(Source: Personal)

From the figures above, the stress-strain graph, equation, and input values will be

stated respectively as;

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.050.0005.000

10.00015.00020.00025.00030.00035.00040.00045.00050.000

Stress-Strain relationship of Concrete (2013 Column)

ConfinedUnconfined

Figure 3. 42. Stress and strain curve of the concrete used for 2013 column(Source: Personal)



f c=8319.847 ε0

0.3232+ε 0

1.3232

0.0014885

(3.15)

Stress Strain

12,158 0,0005

27,590 0,0015

33,049 0,002

45,9 0,0073

32,911 0,0444Table 3. 19. Stress and strain point to be inputted for confined concrete property of 2013 column

(Source: Personal)

The confined concrete stress-strain relationship of the beam sample

designed in accordance to the 2002 standard can be calculated by using similar

manner;

Property Value






ke 0.323ρx 0.0044ρy 0.00666

f’lx 0.334 MPa

f’ly 0.5055 MPa




Table 3. 20. 2002 beam properties which influences stress-strain relationship(Source: Personal)



The stress-strain equation, stress-strain curve, and the input value for

Drain-2DX are shown below;

f c=25779.7 ε0

1.00143+ε0

2.00143

0.000058544

(3.16)

0 0.005 0.01 0.015 0.02 0.025 0.030.0005.000

10.00015.00020.00025.00030.00035.000

Stress-Strain relationship of Concrete (2002 Beam)

ConfinedUnconfined

Strain

Stre

ss (M

Pa)

Figure 3. 43. Stress and strain curve of the concrete used for 2002 beam(Source: Personal)

Stress Strain

12,158 0,0005

27,590 0,0015

30,715 0,002

31,300 0,00243

5,523 0,02725

Table 3. 21. Stress and strain point to be inputted for confined concrete property of 2002 beam(Source: Personal)

With similar manner, stress-strain relationship of the confined concrete of

the column sample designed by using 2002 standard can also be determined.



Property Value






ke 0.858ρx 0.0141ρy 0.0141

f’lx 2.8425 MPa

f’ly 2.8425 MPa




Table 3. 22. 2002 column properties which influences stress-strain relationship(Source: Personal)

The stress-strain equation, stress-strain curve, and the input value for

Drain-2DX are shown below;

f c=8399.891 ε0

0.3263+ε0

1.3263

0.0014394

(3.17)

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.050.000

10.000

20.000

30.000

40.000

50.000

Stress-Strain relationship of Concrete (2002 Column)

ConfinedUnconfined

Strain

Stre

ss (M

Pa)

Figure 3. 44. Stress and strain curve of the concrete used for 2002 column(Source: Personal)



cc Strain

12,158 0,0005

27,590 0,0015

32,994 0,002

45,600 0,0072

32,394 0,0447

Table 3. 23. Stress and strain point to be inputted for confined concrete property of 2002 column(Source: Personal)

Beside of the compressive stress-strain relationship is assessed, the tensile

stress-strain relationship should also be checked. By using equation 2.6 and taking

fc’ as 30 MPa, the modulus of rupture of the concrete being used is 2.74 MPa.

From zero stress to the ultimate stress, the modulus of elasticity of 25,742.96 MPa

is taken, while after the ultimate stress has been reached, the crushing strain shall

be 20 times the strain at the ultimate stress. The crushing stress of the concrete is

usually taken as 0.001 MPa.

0 0.0005 0.001 0.0015 0.002 0.00250

0.5

1

1.5

2

2.5

3

Tensile Stress-Strain relationship of Concrete

Strain

Stre

ss (M

Pa)

Figure 3. 45. Concrete tensile stress-strain relationship (Source: Personal)

Stress Strain



2.74 0,00010645

0.001 0,00213Table 3. 24. Stress and strain point to be inputted for concrete in tension

(Source: Personal)

For the steel property, the yield stress and ultimate stress of the steel

should be inputted, along with their corresponding strain. The stress-strain

diagram of the longitudinal rebar, along with its properties to be inputted into

Drain-2DX is shown as below;

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.160

100

200

300

400

500

600

Stress-Strain of Longitudinal Rebar

Strain

Stre

ss (M

Pa)

Figure 3. 46. Stress-strain relationship of steel used for longitudinal rebar(Source: Personal)

Property Value

Yielding Stress 390 MPa

Ultimate Stress 560 MPa

Pre-Yield Modulus of Elasticity 200,000 MPa

Yielding Strain 0.00195

Ultimate Strain 0.16

Table 3. 25. Material properties of steel being used as longitudinal rebar(Source: Personal)

Stress Strain



390 MPa 0.00195

560 MPa 0.16Table 3. 26. Stress and strain point to be inputted for longitudinal rebar property

(Source: Personal)

The material property of the stirrups along with its stress-strain diagram

should also be shown in the same manner.

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.20

50100150200250300350400

Stress-Strain of Transverse Rebar

Strain

Stre

ss (M

Pa)

Figure 3. 47. Stress-strain relationship of steel used for transverse rebar(Source: Personal)

Property Value

Yielding Stress 235 MPa

Ultimate Stress 380 MPa

Pre-Yield Modulus of Elasticity 200,000 MPa

Yielding Strain 0.001175

Ultimate Strain 0.2

Table 3. 27. Material properties of steel being used as transverse rebar(Source: Personal)

Stress Strain

235 MPa 0.001175

380 MPa 0.2Table 3. 28. Stress and strain point to be inputted for transverse rebar property



(Source: Personal)

1.6. Result Processing and Comparison

After both moment-rotation graphs are gained, the function for both

graphs should be determined. These functions are then compared to each other, so

that the difference between rotational stiffness gained from experimental and

numerical approach can be gained.



REFERENCES

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[2] ASTM C 39. 2015. Standard Test Method for Compressive Strength of

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[3] Hibbler, R.C. 2011. Structural Analysis 8th Edition. Singapore: Pearson

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http://www.scielo.br/

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