Post on 28-Nov-2015
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
TABLE OF CONTENTS (2)
Bracked, Corbel And Beam Ledges Deep Flexural Members One Way Slab System Two Way Slab System Footings
TABLE OF CONTENTS (3)
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.
DESIGN METHOD AND STRENGTH REQUIREMENTS (2)
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.
DESIGN METHOD AND STRENGTH REQUIREMENTS (3)
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.
DESIGN METHOD AND STRENGTH REQUIREMENTS (4)
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).
DESIGN METHOD AND STRENGTH REQUIREMENTS (5)
Notation : Required Strength (U):Mu = Factored Flexural MomentPu = Factored Axial LoadVu = Factored Shear ForceTu = Factored Torsional Moment
DESIGN METHOD AND STRENGTH REQUIREMENTS (6)
Notation : Nominal Strength :Mn = Nominal Flexural Moment StrengthPn = Nominal Axial Load Strength At Given
Eccentricity.Vn = Nominal Shear StrengthTn = Nominal Torsional Moment Strength
DESIGN METHOD AND STRENGTH REQUIREMENTS (7)
Notation : Nominal Strength :fMn = Design Flexural Moment StrengthfPn = Design Axial Load Strength At Given
Eccentricity.fVn = Design Shear StrengthfTn = Design Torsional Moment Strength
DESIGN METHOD AND STRENGTH REQUIREMENTS (8)
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
DESIGN METHOD AND STRENGTH REQUIREMENTS (9)
DESIGN METHOD AND STRENGTH REQUIREMENTS (10)
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
DESIGN METHOD AND STRENGTH REQUIREMENTS (11)
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)
DESIGN METHOD AND STRENGTH REQUIREMENTS (12)
Keterangan :D=Beban MatiL=Beban HidupA=Beban AtapR=Beban HujanW=Beban AnginE=Beban GempaT=Beban Kombinasi Rangkak,Susut Dan Perbedaan Penurunan
DESIGN METHOD AND STRENGTH REQUIREMENTS (13)
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
GENERAL PRINCIPLES OF STRENGTH DESIGN (2)
Stress-Strain Reinforcement
GENERAL PRINCIPLES OF STRENGTH DESIGN (3)
Design Assumption #1Strain in Reinforcement And Concrete Shall Be
Assumed Directly Proportional From The Neutral Axis
GENERAL PRINCIPLES OF STRENGTH DESIGN (4)
Design Assumption #1Strain in Reinforcement And Concrete Shall Be
Assumed Directly Proportional From The Neutral Axis
GENERAL PRINCIPLES OF STRENGTH DESIGN (5)
Design Assumption #2Maximum usable strain at extreme concrete
compression fiber shall be assumed equal to eu =0.003
GENERAL PRINCIPLES OF STRENGTH DESIGN (6)
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.
GENERAL PRINCIPLES OF STRENGTH DESIGN (7)
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.
GENERAL PRINCIPLES OF STRENGTH DESIGN (8)
Design Assumption #5
GENERAL PRINCIPLES OF STRENGTH DESIGN (9)
Design Assumption #5
GENERAL PRINCIPLES OF STRENGTH DESIGN (10)
Design Assumption #5
GENERAL PRINCIPLES OF STRENGTH DESIGN (11)
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
GENERAL PRINCIPLES OF STRENGTH DESIGN (12)
Design Assumption #6
GENERAL PRINCIPLES OF STRENGTH DESIGN (13)
Design Assumption #6
GENERAL PRINCIPLES OF STRENGTH DESIGN (14)
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)
GENERAL PRINCIPLES OF STRENGTH DESIGN (15)
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)
GENERAL PRINCIPLES OF STRENGTH DESIGN (16)
Balanced Strain Condition
GENERAL PRINCIPLES OF STRENGTH DESIGN (17)
Balanced Strain Condition
fyfy
600
600
200000/003.0
003.0
GENERAL PRINCIPLES OF STRENGTH DESIGN (18)
Balanced Strain Condition
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
GENERAL PRINCIPLES OF STRENGTH DESIGN (19)
Minimum Reinforcement Of Flexural Member
dbf
dbf
fA w
yw
y
cs
4.13 'min
GENERAL PRINCIPLES OF STRENGTH DESIGN (20)
Stress Strain Diagram In 3 Condition :
DESIGN FOR FLEXURE (1)
Single Reinforced Concrete Beam
DESIGN FOR FLEXURE (2)
Force Equilibrium :
DESIGN FOR FLEXURE (3)
Moment Equilibrium :
DESIGN FOR FLEXURE (4)
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:
DESIGN FOR FLEXURE (5)
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 :
DESIGN FOR FLEXURE (6)
DESIGN FOR FLEXURE (7)
Graphics Rn Vs r
DESIGN FOR FLEXURE (8)
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
DESIGN FOR FLEXURE (9)
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
DESIGN FOR FLEXURE (10)
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
DESIGN FOR FLEXURE (11)
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 :
DESIGN FOR FLEXURE (12)
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
DESIGN FOR FLEXURE (13)
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
DESIGN FOR FLEXURE (14)
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
DESIGN FOR FLEXURE (15)
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
DESIGN FOR FLEXURE (16)
Example 2 (Design) :#7 Ilustrated The Section With Reinforcement :
d=350mm
b=250mm
h=400mm
cover=50mm
DESIGN FOR FLEXURE (17)
Concrete Section With Compression Reinforcement
DESIGN FOR FLEXURE (18)
Concrete Section With Compression Reinforcement
DESIGN FOR FLEXURE (19)
Concrete Section With Compression Reinforcement
2'
'85.0
1
'
1
adfAAM
fAAabf
TC
yssn
yssc
c
DESIGN FOR FLEXURE (20)
Concrete Section With Compression Reinforcement
''
''
'
'
2
2
2
ddfAM
or
ddfAM
fAfA
TC
ysn
ssn
yssss
s
DESIGN FOR FLEXURE (20)
Concrete Section With Compression Reinforcement
''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 FOR FLEXURE (21)
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 FOR FLEXURE (22)
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 FOR FLEXURE (23)
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
DESIGN FOR FLEXURE (24)
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
Mutu Beton (f’c) : 35 MPaMutu Baja (f’c) : 400 MPa
cover=50mm
DESIGN FOR FLEXURE (25)
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
DESIGN FOR FLEXURE (26)
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)
DESIGN FOR FLEXURE (27)
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
DESIGN FOR FLEXURE (28)
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
'
'
DESIGN FOR FLEXURE (29)
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
Mutu Beton (f’c) : 35 MPaMutu Baja (f’c) : 400 MPa
cover=50mm
DESIGN FOR FLEXURE (30)
T-Beam Concrete Section
bw
hd
beff
hf
DESIGN FOR FLEXURE (31)
T-Beam Concrete Section
bw 2bo bw bw2bo
beff beff beff
hf
DESIGN FOR FLEXURE (32)
T-Beam Concrete Section
Lo
DESIGN FOR FLEXURE (33)
T-Beam Concrete Section
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
DESIGN FOR FLEXURE (34)
T-Beam Concrete Section Neutral Axis x < hf
DESIGN FOR FLEXURE (35)
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
DESIGN FOR FLEXURE (36)
T-Beam Concrete Section Neutral Axis x > hf
DESIGN FOR FLEXURE (37)
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
DESIGN FOR FLEXURE (38)
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
DESIGN FOR FLEXURE (39)
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
DESIGN FOR FLEXURE (40)
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 :
DESIGN FOR FLEXURE (41)
T-Beam Concrete Section (Example 2) Analyze Mn :
bw=300mm
d=510mm
beff
hf=120mm
h=600mm
Specification :f’c = 20 MPafy = 500 MpaLo = 4000 mmBo = 2000 mm
10 D25
DESIGN FOR FLEXURE (42)
T-Beam Concrete Section (Example 2) :#1 Calculate beff Of T-Beam :
mmbbb
mmhbb
mmLb
oweff
fweff
oeff
4300200023002
22201201630016
1000400025.04
1
Diambil beff = 1000 mm
DESIGN FOR FLEXURE (43)
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
DESIGN FOR FLEXURE (44)
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
DESIGN FOR FLEXURE (45)
T-Beam Concrete Section (Example 3) Analyze Mn :
bw=300mm
d=510mm
beff
hf=120mm
h=600mm
Specification :f’c = 40 MPafy = 400 MpaLo = 6000 mmBo = 2000 mm
10 D19
d’=20mm
5 D19
DESIGN FOR FLEXURE (46)
T-Beam Concrete Section (Example 3) :#1 Calculate Beff Of T-Beam :
mmbbb
mmhbb
mmLb
oweff
fweff
oeff
4300200023002
22201201630016
1500600025.04
1
Diambil beff = 1500 mm
DESIGN FOR FLEXURE (47)
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
DESIGN FOR FLEXURE (48)
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
DESIGN FOR FLEXURE (49)
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
'
DESIGN FOR FLEXURE (50)
T-Beam Concrete Section (Example 4) Analyze Mn :
bw=300mm
d=510mm
beff
hf=100mm
h=600mm
Specification :f’c = 20 MPafy = 500 MpaLo = 4000 mmBo = 2000 mm
10 D25
5 D19
DESIGN FOR FLEXURE (51)
T-Beam Concrete Section (Example 4) :#1 Calculate beff Of T-Beam :
mmbbb
mmhbb
mmLb
oweff
fweff
oeff
4300200023002
22201201630016
1000400025.04
1
Diambil beff = 1000 mm
DESIGN FOR FLEXURE (52)
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'
DESIGN FOR FLEXURE (53)
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
DESIGN FOR SHEAR (2)
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.
DESIGN FOR SHEAR (3)
Shear Strength Of Concrete SectionTypes of Reinforcement
DESIGN FOR SHEAR (4)
Shear Strength Of Concrete Section
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.
DESIGN FOR SHEAR (4)
Shear Strength Of Concrete Section
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
DESIGN FOR SHEAR (5)
Shear Strength Of Concrete Section
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.
DESIGN FOR SHEAR (6)
Shear Strength Of Concrete Section
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
DESIGN FOR SHEAR (7)
Shear Strength Of Concrete Section
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
DESIGN FOR SHEAR (8)
Shear Strength Of Concrete Section
Minimum shear reinforcement area Avmin :
Minimum shear force acquired from shear reinforcement :
y
wv f
sbA
3min
dbV ws 3
1min
DESIGN FOR SHEAR (9)
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.
DESIGN FOR SHEAR (10)
Design Of Shear Reinforcement
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 ;
DESIGN FOR SHEAR (11)
Design Of Shear Reinforcement
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
DESIGN FOR SHEAR (12)
Design Of Shear Reinforcement
Condition 6 :
Enlarge the concrete cross section.
dbfVVu wcc '3
2
DESIGN FOR SHEAR (13)
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
DESIGN FOR SHEAR (14)
Design Of Shear Reinforcement
#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
DESIGN FOR SHEAR (15)
Design Of Shear Reinforcement
#1 Check All Shear Boundary Condition before calculating the shear reinforcement :Condition 3 :
Condition 4 :
kNVs 454503003
1min
kNVukN
VVVuV scc
5.1334513375.08.99min
kNVukN
dbfVVuVV wccsc
5.29926613375.05.133
3
1'min
DESIGN FOR SHEAR (16)
Design Of Shear Reinforcement
#1 Check All Shear Boundary Condition before calculating the shear reinforcement :Condition 5 :
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'
DESIGN FOR SHEAR (17)
Design Of Shear Reinforcement
#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
DESIGN FOR SHEAR (18)
Design Of Shear Reinforcement
#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
DESIGN FOR SHEAR (19)
Design Of Shear Reinforcement
#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
DESIGN FOR SHEAR (20)
Design Of Shear Reinforcement
#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
DESIGN FOR SHEAR (21)
Design Of Shear Reinforcement
#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
DESIGN FOR SHEAR (22)
Design Of Shear Reinforcement
#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
DESIGN FOR SHEAR (23)
Design Of Shear Reinforcement
#5 Drawing the shear reinforcement with the spacing shown below :
3m
0.878m 1.179m
0.943m
S=50mm
S=100mm
S=200mm
DESIGN FOR SHEAR (24)
Design Of Shear Reinforcement
#4 Drawing the cross section with longitudinal and shear reinforcement :Condition V
f10-50mm
b=300mm
h=500mm
4D 19
10D 19