Struktur Beton i

STRUKTUR BETON I(SNI 2847-2002)

OLEH :

MUDJI IRMAWANBAMBANG PISCESA

TABLE OF CONTENTS (1)

Design Method And Strength Requirement General Principles Of Strength Design Design For Flexure Design For Flexure And Axial Load Design For Slenderness Effect Shear Torsion Shear Friction


Bracked, Corbel And Beam Ledges Deep Flexural Members One Way Slab System Two Way Slab System Footings


Additional : Design For Biaxial Loading Shear In Slabs Walls Structural Plain Concrete Unified Design Provision

REFERENCES

PCA NOTES-ON ACI 318-99 SNI 2847-2002 NAWY MCGREGOR JACK C. MCCORMAC CHUKIAWANG-CHARLE S G.SALMON

DESIGN METHOD AND STRENGTH REQUIREMENTS (1)

Design Strength >= Required Strength (U)Design Strength = Strength Reduction Factor (f)

x Nominal StrengthStrength Reduction Factor :1. Faktor yang memperhitungkan kemungkinan

terjadinya kekuatan elemen struktur dibawah kekuatan yang diinginkan akibat variasi dari material penyusunnya dan juga akibat variasi dimensinya.


Strength Reduction Factor :2.Faktor yang memperhitungkan ketidak

akuratan persamaan desain (Design Equation).3.Faktor yang memperhitungkan Derajad

Daktilitas dan Keandalan yang diperlukan pada elemen yang dibebani.

4.Faktor yang memperhitungkan kepentingan dari elemen pada struktur.


Nominal Strength : Kekuatan elemen strutkur atau potongan-penampang yang dihitung dengan menggunakan asumsi dan persamaan kekuatan dari metode desain kekuatan (Strength Design method) sebelum diaplikasikan Strength Reduction Factor.


Required Strength (U) : Load Factor x Service Load Effect.

Load Factor : Faktor overload yang terjadi akibat kemungkinan terjadinya variasi pembebanan yang berlebih pada beban kerja (Service Load).

Service Load : Beban spesifik yang diatur oleh peraturan pembebanan (Unfactored).


Notation : Required Strength (U):Mu = Factored Flexural MomentPu = Factored Axial LoadVu = Factored Shear ForceTu = Factored Torsional Moment


Notation : Nominal Strength :Mn = Nominal Flexural Moment StrengthPn = Nominal Axial Load Strength At Given

Eccentricity.Vn = Nominal Shear StrengthTn = Nominal Torsional Moment Strength


Notation : Nominal Strength :fMn = Design Flexural Moment StrengthfPn = Design Axial Load Strength At Given

Eccentricity.fVn = Design Shear StrengthfTn = Design Torsional Moment Strength


Strength Requirement :Design Strength > Required Strength (U)Strength Reduction Factor (f) x Nominal

Strength > Load Factor x Service Load EffectExample :

f Mn > Mu f Pn > Pu f Vn > Vu


Nominal Strength:Mn=As.fy.(d-a/2)f.Mn=f[As.fy.(d-a/2)

Required Strength (U):Mu=1.2Md+1.6Ml

Strength Design Requirement :f.Mn>Muf[As.fy.(d-a/2)>1.2Md+1.6Ml


Required Strength :U=1.4DU=1.2D+1.6LU=1.2D+1.6L+0.5(A or R)U=1.2D+1.0L+1.6W+0.5(A or R)U=0.9D+1.6WU=1.2D+1.0L+1.0EU=0.9D+1.0EU=1.4(D+F)U=1.2(D+T)+1.6L+0.5(A or R)


Keterangan :D=Beban MatiL=Beban HidupA=Beban AtapR=Beban HujanW=Beban AnginE=Beban GempaT=Beban Kombinasi Rangkak,Susut Dan Perbedaan Penurunan


Strength Reduction Factor (f):Flexure Without Axial Load : 0.8Axial Compression And Axial Compression W/ Flexure :Members With Spiral Reinforcement : 0.7Others Member : 0.65Shear And Torsion : 0.75

GENERAL PRINCIPLES OF STRENGTH DESIGN (1)

Stress-Strain Curves Of Concrete


Stress-Strain Reinforcement


Design Assumption #1Strain in Reinforcement And Concrete Shall Be

Assumed Directly Proportional From The Neutral Axis


Design Assumption #2Maximum usable strain at extreme concrete

compression fiber shall be assumed equal to eu =0.003


Design Assumption #3Stress in reinforcement fs below the yield strength fy shall be taken as Es times the steel strain es . For strains greater than fy/Es, stress in reinforcement shall be considered independent of strain and equal to fy.


Design Assumption #4Tensile strength of concrete shall be neglected in

flexural calculation of reinforced concrete.

Design Assumption #5Relationship between concrete compressive stress

distribution and concrete strain shall be assumed to be rectangular, trapezoidal, parabolic, or any other

shape that results in prediction of strength in substantial agreement with results of

comprehensive tests.


Design Assumption #5


Design Assumption #6Parabolic Stress-

Strain distribution in concrete may be considered satisfied by an

equivalent rectangular

concrete stress distribution.

30MPa 58MPa

65.005.07

30'85.01

cf


Dalam menganalisa struktur beton bertulang ada tiga kondisi kegagalan yang harus diketahui :•Balanced Strain ConditionKondisi dimana kegagalan elemen struktur terjadi pada saat beton serat terluar tertekan mengalami kehancuran bersamaan dengan terjadinya leleh pada tulangan baja tarik terluar.•Over-Reinforced ConditionKondisi dimana kegagalan elemen struktur terjadi pada saat beton serat terluar tertekan mengalami kehancuran tetapi leleh belum terjadi pada tulangan tarik terluar. (Getas)


Dalam menganalisa struktur beton bertulang ada tiga kondisi kegagalan yang harus diketahui :•Under-Reinforced ConditionKondisi dimana kegagalan elemen struktur terjadi pada saat tulangan tarik terluar mengalami pelelehan tetapi beton pada serat tertekan paling luar belum mengalami kegagalan. (Daktail)


Balanced Strain Condition



fyfy

600

600

200000/003.0

003.0



fyfy

f c

600

60085.0 '1

Untuk menjamin sebuah struktur berada dalam kondisi under reinforced maka :

fyfy

f cbal 600

60085.075.075.0 '1

max


Minimum Reinforcement Of Flexural Member

dbf

dbf

fA w

yw

y

cs

4.13 'min


Stress Strain Diagram In 3 Condition :

DESIGN FOR FLEXURE (1)

Single Reinforced Concrete Beam


Force Equilibrium :


Moment Equilibrium :


A nominal strength coefficient of resistance Rn is obtained when both sides of Eq. (2) are divided by bd2:

When b and d are preset, r is obtained by solving the quadratic equation for Rn:


Equation (3) can be used to determine the steel ratio r given Mu or vice-versa if the section properties b and dare known. Substituting Mn = Mu/f into Eq. (3) and dividing each side by fc :


Graphics Rn Vs r


Example 1 (Analyzing) :Calculate the Moment Nominal of The Concrete Section :

d=350mm

h=400mm

b=250mm

cover=50mm

3D19

Mutu Beton (f’c) : 35 MPaMutu Baja (f’c) : 400 MPa


Example 1 (Analyzing) :22 8501925.03 mmAs

mmbf

fAa

c

ys 71.452503585.0

400850

85.0 '

kNmM

M

adfAM

n

n

ysn

229.111

2

71.45350400850

2

T

C

d

a

d-a/2


Example 2 (Design) :Design the concrete section of the beam, which is simple supported and were loaded as below :

L=6m

Ql=1.5t/m;Qd=1t/m

kNmMMM

kNmtmlqM

kNmtmlqM

ldu

ll

dd

76.15815.666.11.442.16.12.1

15.6675.665.18

1

8

1

1.445.4618

1

8

1

22

22

Spesification :Mutu Beton (f’c)=35 MPaMutu Baja (fy)=400 MPa


Example 2 (Design) :#1 Determine maximum reinforcement ratio (rmax) for material strength f’c=35MPa And fy=400MPa :

65.0814.01

65.005.07

303585.01

65.005.07

30'85.01

cf

0272.0

400600

600

400

35814.085.075.0

600

60085.075.075.0

max

max

'1max

fyfy

f cbal

Calculating b1

Calculating rmax :


Example 2 (Design) :#2 Compute bd2 required :

32

'22

23008696625.88.0

100000076.158

625.83585.0

4000272.05.014000272.0

85.0

5.01

mmR

Mbd

MPaR

f

ff

bd

M

bd

MR

n

u

n

c

yy

unn


Example 2 (Design) :#3 Size member so that bd2 > bd2 required :

mmd

mmb

37.303250

23008696

250

Minimum Beam Depth (h) = d + cover Minimum Beam Depth (h) = 303.37 + 50 = 353.37 mmUse minimum Beam Depth (h) = 400 mmTherefore :d = h – cover = 400 – 50 = 350 mm


Example 2 (Design) :#4 Using the 400 mm. beam depth, compute a revised value of r :

MPabd

MR

bd

MR

un

un

48.63502508.0

100000076.15822

2

0035.00185.00272.0

400

4.14.1

3585.0

48.6211

400

3585.00272.0

4.1

85.0

211

85.0

minmax

minmax

min'

'max

fy

fyf

R

f

f

c

n

y

c


Example 2 (Design) :#5 Compute As Required :As = r x b x dAs = 0.0185 x 250 x 350As = 1618.75 mm2Pakai 6 D19 (As=1701 mm2)

#6 CrossCheck The Moment Nominal with Moment Ultimate :

kNmMkNmM

kNma

dfAM

nu

ysn

071.17876.158

071.1782

71.4535040017018.0

2


Example 2 (Design) :#7 Ilustrated The Section With Reinforcement :

d=350mm

b=250mm

h=400mm

cover=50mm


Concrete Section With Compression Reinforcement



2'

'85.0

1

'

1

adfAAM

fAAabf

TC

yssn

yssc

c



''

''

'

'

2

2

2

ddfAM

or

ddfAM

fAfA

TC

ysn

ssn

yssss

s



''2

'

21

ddfAa

dfAAM

MMM

ssyssn

nnn

21

'

TTT

CCC

TC

sc

Force Equilibrium :

Solving The Force Equilibrium We Achieved That :

bf

fAAa

bf

fAfAa

c

yss

c

ssys

'' 85.0

';

85.0

'


Design Procedure :#1 Calculate c < 0.75 cb :

dfy

cc b 600

60075.075.0

#2 Calculate As1 with c determined from above :

y

cs f

cbfA '185.0

#3 Calculate Mn1 :

21

adfAM ysn


Design Procedure :#4 Calculate Mn-Mn1 :

#5 If Compression Reinforcement is Required Then Calculate T2 :

#6 Control The Yield Of Compression Reinforcement :

0

0

1

1

nn

nn

MM

MM (Compression Reinforcement Is Required)

(Compression Reinforcement Is Not Required)

'' 12 dd

MMTC nn

s

fyc

dfs

fyc

dfs

600'

1

600'

1 (Compression Reinforcement Yield)

(Compression Reinforcement Not Yield)


Design Procedure :#7 Calculate The Required Compression Reinforcement :

#8 Calculate The Additional Tension Reinforcement :

#9 Calculate Required Reinforcement :As = As1 + AssAs’ = As’

#10 Check Required Strength :fMn > Mu

cs

ss ff

CA

'85.0'

''

yss f

TA 2


Example 3 (Design) :Design the concrete section below if the factored moment applied to the concrete section is 300 kNm (Mu) :

d=550mm

h=600mm

b=300mm

cover=50mm


cover=50mm


Example 3 (Design) :#1 Calculate c < 0.75 cb :

mmdfy

cc b 5.247550400600

60075.0

600

60075.075.0

#2 Calculate As1 with c determined from above c=85mm:

2' 1543400

3008535814.085.085.0mm

f

cbfA

y

cs

#3 Calculate Mn1 :

kNmM n 134.3182

85813.055040015431


Example 3 (Design) :#4 Calculate Mn-Mn1 :

#5 If Compression Reinforcement is Required Then Calculate T2 :

#6 Control The Yield Of Compression Reinforcement :

0686.56314.3188.0

3001 kNmM

Mn

u

Ndd

MMTC nn

s 113372500

1000000686.56

'' 1

2

MPafyMPafs 40024760085

501

(Compression Reinforcement Not Yield)

(Compression Reinforcement Is Required)


Example 3 (Design) :#7 Calculate The Required Compression Reinforcement :

#8 Calculate The Additional Tension Reinforcement :

#9 Calculate Required Reinforcement :As = 1543 + 283 = 1826 mm2 ( 7 D 19 = 1988 mm2)As’ = 521 mm2 (2 D 19 = 568 mm2)

2

'

5213585.0247

113372

85.0'

'' mm

ff

CA

cs

ss

22 283400

113372mm

f

TA

yss


Example 3 (Design) :#10 Check The Moment Required :

kNmMkNmM

M

ddfAa

dfAfAM

un

n

ssssysn

300325148.70190.3378.0

505502475682

26.7055024756840019888.0

''2

'

mmbf

fAfAa

c

ssys 378.733003585.0

2475684001988

85.0

'

'


Example 3 (Design) :Design the concrete section below if the factored moment applied to the concrete section is 300 kNm (Mu) :

d=550mm

h=600mm

b=300mm

cover=50mm


cover=50mm


T-Beam Concrete Section

bw

hd

beff

hf



bw 2bo bw bw2bo

beff beff beff

hf



Lo



No Effective Width

Spandrel Center T-Beam

1 beff < bw + 1/12 Lo beff < 1/4 Lo

2 beff < bw + 6 hf beff < bw + 16 hf

3 beff < bw + bo beff < bw + 2bo

Single T-Beam

1 beff < 4 bw

2 Flange Thickness (hf) > 1/2 bw


T-Beam Concrete Section Neutral Axis x < hf


T-Beam Concrete Section Neutral Axis x < hf

abfC effc '85.0

ys fAT

Force Equilibrium :

effc

ys

bf

fAa

'85.0

2

adfAM ysn


T-Beam Concrete Section Neutral Axis x > hf


T-Beam Concrete Section Neutral Axis x > hf

fweffc hbbfC '2 85.0

ys fAT

Force Equilibrium :

feff

e

wc

ys hb

b

bf

fAa

1'85.0

22 21f

n

hdC

adCM

abfC wc '1 85.0


T-Beam Concrete Section (Example 1) Analyze Mn :

bw=300mm

d=510mm

beff

hf=120mm

h=600mm

Specification :f’c = 40 MPafy = 400 MpaLo = 6000 mmBo = 2000 mm

10 D19


T-Beam Concrete Section (Example 1) :#1 Calculate Beff Of T-Beam :

mmbbb

mmhbb

mmLb

oweff

fweff

oeff

4300200023002

22201201630016

1500600025.04

1

Diambil beff = 1500 mm


T-Beam Concrete Section (Example 1) :#2 Calculate a Of T-Beam :

mmhfmmbf

fyAsa

effc

120235.2215004085.0

4002835

'85.0

kNmM

adfAM

n

ysn

565

2

235.225104002835

2

#3 Calculate Mn Of T-Beam :



bw=300mm

d=510mm

beff

hf=120mm

h=600mm


10 D25


T-Beam Concrete Section (Example 2) :#1 Calculate beff Of T-Beam :

mmbbb

mmhbb

mmLb

oweff

fweff

oeff

4300200023002

22201201630016

1000400025.04

1



T-Beam Concrete Section (Example 2) :#2 Calculate a Of T-Beam :

mmhfmmbf

fyAsa

effc

120375.14410002085.0

5004908

'85.0

mmb

b

bf

fyAsa

eff

e

wc

76.2011300

1000

3002085.0

50049081

'85.0


T-Beam Concrete Section (Example 2) :#3 Calculate Mn Of T-Beam :

kNmM

M

adC

hdCM

kNabfC

kNhbbfC

hdC

adCM

n

n

fn

wc

fweffc

fn

57.106397.4206.642

2

76.2015101028

2

1205101428

22

102876.2013002085.085.02

142812030010002085.085.01

22

21

'

'

21



bw=300mm

d=510mm

beff

hf=120mm

h=600mm


10 D19

d’=20mm

5 D19


T-Beam Concrete Section (Example 3) :#1 Calculate Beff Of T-Beam :

mmbbb

mmhbb

mmLb

oweff

fweff

oeff

4300200023002

22201201630016

1500600025.04

1



T-Beam Concrete Section (Example 3) :#2 Calculate a Of T-Beam with assumption that compression and tension steel yield :

'12.11

15004085.0

40014172835

'85.0

'dhmm

bf

fAAa f

effc

yss

Therefore the first assumption that the compression steel is yield is not true because a < d’, this means that the compression steel is tension steel. Analyzing the beam using single reinforced concrete :

'235.2215004085.0

4002835

'85.0dhfmm

bf

fyAsa

effc


T-Beam Concrete Section (Example 3) :From the calculation before a < hf and a > d’ this means that actualy the top reinforcement is in compression but not yielding. #3 Calculate the stress in compression reinforcement :

65.0778.0

65.005.07

304085.0

65.005.07

3085.0

1

1

'1

cf

mma

x 57.28778.0

235.22

1


T-Beam Concrete Section (Example 3) :#3 Calculate the stress in compression reinforcement :

MPax

dfs 978.179600

57.28

201600

'1'

mm

bf

fAfAa

effc

ssys 23.1715004085.0

99.17914174002835

'85.0

'

#4 Calculate the new value of a :

kNmM

M

ddfAa

dfAfAM

n

n

ssssysn

5678215.12639.441

2

205101791417

2

23.1751017914174002835

''2

'



bw=300mm

d=510mm

beff

hf=100mm

h=600mm


10 D25

5 D19


T-Beam Concrete Section (Example 4) :#1 Calculate beff Of T-Beam :

mmbbb

mmhbb

mmLb

oweff

fweff

oeff

4300200023002

22201201630016

1000400025.04

1



T-Beam Concrete Section (Example 4) :#2 Calculate a Of T-Beam with assumption that compression and tension steel yield :

'103

10002085.0

50014174908

'85.0

'dhmm

bf

fAAa f

effc

yss

Therefore the first assumption that the compression steel is yield is might be true because a > d’, #3 Calculate the stress in compression reinforcement :85.01

mma

x 176.12185.0

103

1

MPafyMPax

dfs 50097.500600

176.121

201600

'1'


T-Beam Concrete Section (Example 4) :#4 Since the compression steel is yielding therefore the first assumption is right, therefore the a value before is used :

kNmM

M

ddfAa

dfAfAM

n

n

ssssysn

69,113825.35444.784

2

205105001417

2

176.12151050014175004908

''2

'

DESIGN FOR SHEAR (1)

Shear Strength Of Concrete Section



Perlawanan geser yang terjadi setelah retak miring :1. Perlawanan geser beton yang belum retak, Vcz.2. Gaya ikat (interlock) antara aggregat atau

transfer geser antar permukaan.3. Aksi pasak (dowel action), Vd.4. Aksi pelengkung (arch action), Deep Beam.5. Perlawanan tulangan geser bila ada, Vs.


Shear Strength Of Concrete SectionTypes of Reinforcement



Fungsi tulangan geser (Sengkang Begel):1. Memikul sebagian gaya geser, Vs.2. Melawan pertumbuhan geser miring dan ikut

menjaga terpeliharanya lekatan/geseran antar aggregat.

3. Mengikat batang tulangan memanjang untuk tetap diposisinya.

4. Aksi pasak pada beton dan aksi ikatan (confinement) sengkang meningkatkan kekuatan.



Shear nominal strength (Vn) of concrete section is a combination of shear strength from concrete (Vc) and sehar strength from shear reinforcement (Vs) :

Vn=Vc+VsWhere Vc :

If accurate calculation were used then :

dbfVc wc'6

1

dbfdbbM

dVfV wcw

u

uwcc '' 3.0120

7

1



Where :

bd

Asw 1

u

u

M

dV

If there is an axial compression load acting in the concrete section therefore Vn can be calculated as follows:

dbfA

NV wc

g

uc '6

1

141

Nu/Ag is in MPa.



If accurate calculation were used (Axial Compression Load) :

Where :

g

uwcw

m

uwcc A

Ndbfdb

M

dVfV

3.013.0120

7

1''

8

4 dhNMM uum



If there is an axial tension load acting in the concrete section therefore Vn can be calculated as follows:

Nu/Ag is in MPa.

For circular concrete section area for Vc can be calculated from diameter multiplied by effective width (d=0.8h).

06

13.01 '

dbf

A

NV wc

g

uc



Minimum shear reinforcement area Avmin :

Minimum shear force acquired from shear reinforcement :

y

wv f

sbA

3min

dbV ws 3

1min


Design Of Shear Reinforcement

Design of shear reinforcement is divided into several categories, which each category is corresponds with the shear forces acting on the concrete section. This category is divided as below :Condition 1 :

cu VV 5.0Shear reinforcement is not required.Condition 2 :

cc VVuV 5.0Smax < d/2 or Smax < 600 mm, minimum shear reinforcement need to be chekced.



Condition 3 :

Smax < d/2 or Smax < 600 mm, minimum shear reinforcement need to be chekced.Condition 4 :

minscc VVVuV

dbfVVuVV wccsc 'min 3

1

Smax < d/2 or Smax < 600 mm, shear reinforcement need to be calculated as follows :

s

dfAVVVV yv

scusperlu ;



Condition 5 :

Smax < d/4 or Smax < 600 mm, shear reinforcement need to be calculated as follows :

s

dfAVVVV yv

scusperlu ;

dbfVVudbfV wccwcc '' 3

2

3

1



Condition 6 :

Enlarge the concrete cross section.

dbfVVu wcc '3

2


Design Of Shear Reinforcement (Example 1)

L=6m

Ql=6t/m;Qd=4t/m Spesification :Mutu Beton (f’c)=35 MPaMutu Baja (fy)=400 MpaLebar Balok = 300 mmTinggi Balok = 500 mm

Desain tulangan geser yang diperlukan untuk memikul beban geser yang terjadi.

Ra=Rb=43.2 ton = 423.36 kN

423.36 kN

423.36 kN

Bidang Gaya Lintang



#1 Check All Shear Boundary Condition before calculating the shear reinforcement :Condition 1 :

Condition 2 :

kNdbfVc wc 133450300356

1

6

1'

kNVV cu 9.4913375.05.05.0

kNVukN

VVuV cc

8.999.49

5.0




Condition 4 :

kNVs 454503003

1min

kNVukN

VVVuV scc

5.1334513375.08.99min

kNVukN

dbfVVuVV wccsc

5.29926613375.05.133

3

1'min




Condition 6 (Concrete Section Is Satisfied) :

kNVukN

dbfVVudbfV wccwcc

75.49853213375.05.299

3

2

3

1''

kNkN

dbfVVu wcc

75.4986.423

3

2'



#2 Drawing The Shear Boundary Condition :

Vu=423.36 kN

V=498.75kN

IV=299.5kN

III=133kN

II=99.8kN

X1=0.878m

X2=2.057m

X3=2.3m

X4=3m



#3 Calculate each boundary distance :

mX

mQlQd

VuX

mQlQd

VuX

mQlQd

VuX

3

300.28.586.12.392.1

8.9936.423

6.12.1

133

057.28.586.12.392.1

13336.423

6.12.1

133

878.08.586.12.392.1

5.29936.423

6.12.1

5.299

4

3

2

1



#4 Calculate required shear reinforcement in each condition :Condition II :

Use 2 Leg of f10, As = 157 mm2,. Use s = 200 mm.

Since smax < d/2 = 450/2 =225 mm or smax <600mm.Therefore s = 200mm is adequate.

kNdbV ws 453

1min

22min 15750

4003

200300

3mmAvmm

f

sbA

y

wv

kNVkNs

dfAV s

yvs 453.141

200

450400157min



#4 Calculate required shear reinforcement in each condition :Condition III :

Use 2 Leg of f10, As = 157 mm2,. Use s = 200 mm.

Since smax < d/2 = 450/2 =225 mm or smax <600mm.Therefore s = 200mm is adequate.

kNdbV ws 453

1min

22min 15750

4003

200300

3mmAvmm

f

sbA

y

wv

kNVkNs

dfAV s

yvs 453.141

200

450400157min



#4 Calculate required shear reinforcement in each condition :Condition IV :

Therefore in condition IV use s = 100 mm, since smax < d/2 = 450/2 = 225 mm and smax < 600 mm, s =100 mm is adequate.

kNVV

V cusperlu 26.266

75.0

8.995.299

mmV

dfAs

s

yv 10626.266

450400157



#4 Calculate required shear reinforcement in each condition :Condition V :

Therefore in condition IV use s = 50 mm, since smax < d/3 = 450/4 = 112.5 mm and smax < 300 mm, s =50 mm is adequate.

kNVV

V cusperlu 73.431

75.0

8.996.423

mmV

dfAs

s

yv 6573.431

450400157



#5 Drawing the shear reinforcement with the spacing shown below :

3m

0.878m 1.179m

0.943m

S=50mm

S=100mm

S=200mm



#4 Drawing the cross section with longitudinal and shear reinforcement :Condition V

f10-50mm

b=300mm

h=500mm

4D 19

10D 19

Struktur Beton i

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