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perancangan struktur beton
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1ENCV600101 - Perancangan Struktur Beton 1
Ch 3 Lentur: Balok dengan tulangan tekan
Sjahril A. RahimDepartemen Teknik Sipil FTUI
2014
Pengaruh tulangan tekan pada Kekuatan dan Perilaku
b
h
As
b
h d
As
As'd'
d
a1 c1 Cc1
T=Asfy
j1d=(d-a/2)
j2d
Cs
Cc2C
T=Asfy
a2 c1c2 s
cu
cu
a2 < a1Dgn tulangantekan
Tanpa tulangan tekan
s
s
Perbandingan
Tulangan tarik As Tulanga tekan As=0 Cc1=T T=Asfy Cc1 > Depth a1 > Lengan momen: j1d Mn=Asfy(j1d) j1 little >
Tulangan tarik As Tulangan tekan As C=Cc2+Cs=T T=Asfy Cc2 a2 Lengan momen: j2d Mn=Asfy (j2d) j2
Addition of comp. steel little effect on the usable ultimate moment
Peningkatan kapasitas momen akibat tulangan tekan
2Alasan untuk menyediakan tulangan tekan
Mengurangi defleksi beban yang berkelanjutan
Peningkatan daktilitas Perubahan kegagalan mode dari keruntukan
tekan menjadi keruntuhan tarik Memudahkan fabrikasi
Mengurangi defleksi beban yang berkelanjutan
Meningkatkan daktilitas Merubah moda keruntuhan dari tekan ke tarik
3Analysis of Beams with tension and Compression reinforcement
As
d=dt
d'
h
b
As'
As1
d=dt
d'As'
As2
d=dt(d-d') (d-a/2)
Cc
Cs
Cc
Cs
T1=As1fy T2=As2fy
0.003 0.85fc'
a/2fs'
fs=fyT
+a
0.85fc'
a/2
c a
sy
Analysis of Beams with tension and Compression reinforcement
The strain, stresses distribution and internal forces see figure b,c and d.The cross section was hypothetically divided in to two beams;Beam 1, consisting of the compression reinforcement at the top and sufficient at the bottom so that T1 = Cs, and beam 2, consisting of the concrete web and the remaining tensile reinforcement, as shown in Figure e and f.The strain and stress in the compression reinforcement:
003,0''
c
dcs sss Ef
If ys ' then ys ff '
s
yy E
fwhere
Case 1: Compression steel yields
The beam can be divided in to two imaginary beams, each with C=T:Beam 1: Consist of reinforcement in tension and compression and resists moment as steel force couple.
''1
'1
1'
1
ddfAMAA
fAfA
TC
yss
ss
ysys
s
As1
d=dt
d'As'
(d-d')
Cs=As'f's
T1=As1fy
Beam 1
Case 1: Compression steel yields
Beam 2: Consist of the concrete plus the remaining steel:
bafCAA
ffAAA
cc
ss
ys
sss
'
'1
12
85,0
Since C=T for beam 2, where T=(As-As)fy, the depth of the compression stress block, a, is
bfcfAA
a yss'85,0
'
As2
d=dt (d-a/2)
Cc
T2=As2fy
a
0.85fc'
a/2
4Case 1: Compression steel yields
The nominal moment capacity of beam 2 is:
2
)( '2adfAAM yssn
The total moment capacity of a beam with compression steel is
2
)()'( '' adfAAddfAM yssysn
Case 1: Compression steel yields
Correction due to concrete displaced by steel compression:
285,01'85,01 ''
'' adffAAddf
ffAM ycssy
y
csn
y
css f
fAA'
'1
85,01
and
bfc
ff
fAA
ay
y
css
'85,0
85,01'
'
where
Determination of Whether fs=fy in Compression Reinforcement
Tulangan tekan leleh, jika:
003,0''
c
dcs sss Ef
''
ys ff 'kemudian
MPa
MPafEf y
s
yys 000.200
'
Untuk mengkonfirmasi bahwa tulangan tekan leleh, tunjukkan
Determination of Whether fs=fy in Tension Reinforcement
Dengan kedalaman sumbu netral diketahui, asumsi kelelehan ketegangan baja tulangan dapat diperiksa. Dari segitiga serupa dalam distribusi regangan linear pada gambar, ekspresi berikut dapat diturunkan:
cuscus
ccdsehingga
ccd
Untuk mengkonfirmasi tulangan tarik leleh, tunjukkan
MPa
MPafEf y
s
yys 000.200
As
d=dt
d'
h
b
As'
0.003
c s
5Case 1: Compression steel does not yields
If the compression reinforcement does not yield, fs is not known, and different solution is required. Assuming that tensile steel yields, the internal forces in the beam are:
bafC
fAT
cc
ys
'85,0
Ignoring the correction to the compressive stress in the steel:'' )( ssss AEC 003,0'1 1'
ad
s
where
From the equilibrium:
ysssc
sc
fAadAEbaf
TCC
003,0'185,0 1''
Case 1: Compression steel does not yields
This can be reduced to the quadratic equation in a, given by:
0)'003,0()003,0()85,0( 1''2' dAEafAAEabf ssysssc
Once the depth of the stress block, a, is known, the nominalmoment capacity of the section is
)'(2
ddCadCM scn
Determination of Whether fs=fy in Tension Reinforcement
Dengan kedalaman sumbu netral diketahui, asumsi kelelehan ketegangan baja tulangan dapat diperiksa. Dari segitiga serupa dalam distribusi regangan linear pada gambar, ekspresi berikut dapat diturunkan:
cuscus
ccdsehingga
ccd
Untuk mengkonfirmasi tulangan tarik leleh, tunjukkan
MPa
MPafEf y
s
yys 000.200
As
d=dt
d'
h
b
As'
0.003
c s
Pengecekan regangan tarik net, tb
h ddt
d'
As
As' c
Dengan kedalaman sumbu netral diketahui, strain regangan tarik net pada tulangan tarik terjauh dapat di ceck,. Dari segitiga serupa dalam distribusi regangan linear pada gambar, ekspresi berikut dapat diturunkan:
cut
tcu
t
t
ccdsehingga
ccd
t
Untuk menjamin potongan balok daktail, pastikan bahwa regangan tarik net:t 0.004
untuk memenuhi SNI 2847:2013, Pasal 10.3.5
6Upper Limit on Reinforcement in Beams
Under-reinforced beams fails in a ductile manner Over-reinforced beams in a brittle manner SNI section 10.3.5 attempts to prevent non-ductile failures by limiting the strain t at nominal flexural strength condition shall be greater than or equal to 0,004:
004,0t
new
As
d=dt
d'
h
b
As'Cs
Cc
0.003 0.85fc'
a/2fs'
fs=fyT
c a
s=0.004
Upper Limit on Double Reinforcement in Beams
Strain compatibility: tscu
cu dc
tcu
cu dc004.0
Equilibrium: C=T
ysy
s
ys
cu
cu
y
c
yycsc
ffifff
ffif
ff
bdbdfbdfbcfCabf
''
max
'max
'
1max
max'
1'
'
'
004.085.0
''85.085.0
Tulangan minimum pada komponen struktur lentur
ACI 318/SNI 2847:2013 Pasal 10.5.1 menentukan luas minimum tulangan longitudinal berikut untuk potongan balok yang di lentur positif,
y
ww
y
cs f
dbdbf
fA 4.1
25.0 'min,
fc dan fy dalam Mpa.bw = lebar badan balok, d = tinggi efektif balok
(10.3)
Ties for Compression Reinforcement
As the ultimate load is approach the compression steel in a beam may buckle, causing the surface layer of concrete to spall off, possibly leading to failure. For this reason it is necessary to enclose compression reinforcement with closed stirrup or ties. Please refer to SNI 9.11 for details.
7Analysis procedure of a Beam with Compression Reinforcement: Compression
reinforcement yields Assuming that fs=fy and fs=fy and divide the beam
into two components Compute a for beam 2 Check if compression reinforcement steel yields Check if fs=fy and whether the section is tension-
controlled Check if As As,min Compute Mn
Analysis procedure of a Beam with Compression Reinforcement: Compression reinforcement yields
Assume fs=fyfs=fy
'1 ss AA 12 sss AAA
bffAA
ac
yss'
'
85,0)(
1
Beam 1 Beam 2Y
cus cdc
''
1ac
Analysis procedure of a Beam with Compression Reinforcement: Compression reinforcement yields
1
ys ff '
Y2
Check yielding ofcompression reinforcement ys
'
s
yy E
f
cus ccd
ys
Analysis procedure of a Beam with Compression Reinforcement: Compression reinforcement yields
2
yy
cs f
bdbdff
A 4,14
'
min,
min,ss AA Increase AsN
YBeam 1 Beam 2
Check for minimum reinforcement
''1 ddfAM ysn
2
)( '2adfAAM yssn
21 nnn MMM 3
8Analysis procedure of a Beam with Compression Reinforcement: Compression reinforcement yields
cut
t ccd
3
t0.005
0.005
9Analysis procedure of a Beam with Compression Reinforcement: Compression reinforcement Does
Not Yields3
''
'
'85,0
ssss
cus
cc
AECc
cdbafC
)'(2
ddCadCM scn
4
Analysis procedure of a Beam with Compression Reinforcement: Compression reinforcement Does
Not Yields
cut
t ccd
4
t0.005
0.005
10
Latihan 2 Analisis
d'
b
dh
2D28 mm
6D28 mm
Figure 1
Case 2: d=80 mm
Rectangular beam with compression reinforcementCompute Mn for the beam shown in Figure 1, where fc=20 MPa, fy=400 MPa, b=300 mm, d=810 mm, dt= 935 mm, d=80 mm and h=900 mm.
Design procedure of a Beam with Compression Reinforcement:
1. Calculate the maximum moment that can be resisted by the undereinforced section with =max, or 0.005 to ensure =0.90. The corresponding tensile area is As=bd, and Mn=Asfy(d-a/2), with a=(Asfy)/(0.85fcb)
2. Find the excess moment, if any, that must be resisted, and set M2=Mn, as calculated in step 1, M1=Mu/-M2. Now As from step 1 is defined as As2, ie., that part of the tension steel area in the double reinforced beam that works with the compression force in the concrete, As-As=As2.
3. Tentatively assume that fs=fy. Then As=M1/(fy(d-d). Alternatively, if the compression reinforcement is known not to yield, go to step 6.
4. Add an additional amount of tensile steel As1=As. Thus, the total tensile steel reinforcement area As is As2 from step 2 plus As1.
5. Analyze the doubley reinforced beam to see if fs=fy; that is, check the tensile reinforcement ratio against cy.
Design procedure of a Beam with Compression Reinforcement:
6. If
11
M2=Mn
21 MMM u
ss AA 2
sy
6
1
c=a/1
cus cdc
''
f s=fy
)'(1'
ddfMA
ys
As1=As
As=As1+As2
2
Design procedure of a Beam with Compression Reinforcement:
yn
Analysis procedure of a Beam with Compression Reinforcement: Compression reinforcement yields
Assume fs=fyfs=fy
'1 ss AA 12 sss AAA
bffAA
ac
yss'
'
85,0)(
3
Beam 1 Beam 2Y
cus cdc
''
1ac
2
Analysis procedure of a Beam with Compression Reinforcement: Compression reinforcement yields
3
ys ff '
Y4
Check yielding ofcompression reinforcement ys
'
s
yy E
f
cus ccd
ys
Analysis procedure of a Beam with Compression Reinforcement: Compression reinforcement yields
4
yy
cs f
bdbdff
A 4,14
'
min,
min,ss AA Increase AsN
YBeam 1 Beam 2
Check for minimum reinforcement
''1 ddfAM ysn
2
)( '2adfAAM yssn
21 nnn MMM 5
12
Analysis procedure of a Beam with Compression Reinforcement: Compression reinforcement yields
cut
t ccd
5
t0.005
0.005
13
Analysis procedure of a Beam with Compression Reinforcement: Compression reinforcement Does
Not Yields
9
Y
10
y
ww
y
cs f
dbdbff
A 4,14
'
min,
min,ss AA Increase AsN
Y
Check for minimum reinforcement
Check for tension control
cus ccd
ys
1ac
Analysis procedure of a Beam with Compression Reinforcement: Compression reinforcement Does
Not Yields10
''
'
'85,0
ssss
cus
cc
AECc
cdbafC
)'(2
ddCadCM scn
11
Analysis procedure of a Beam with Compression Reinforcement: Compression reinforcement Does
Not Yields
cut
t ccd
11
t0.005
0.005
14
Daftar referensi
1. James K Wight, James McGregor : Reinforced Concrete, Mechanics and Design, Sixth Edition, Pearson, 2012
2. ________________, Persyaratan beton struktural untuk bangunan gedung, SNI 2847:2013, Badan Standarisai nasional
3. ________________, Building Code requirements for Structural Concrete, ACI 318-2011, American Concrete Institute
Alternative determination of Whether fs=fy in Tension Reinforcement
The tension steel will yield if a tension or balanced failure occurs. If fs=fy, the balance conditions:
yc
yysss
fbdffffAA
600600
'85,0/
1
''
Substituting =As/bd and =As/bd gives
yy
c
by
s
fff
ff
60060085,0' '1
'
ys
bysys
ffffff
)/'()/'( '' If then
Old
Upper Limit on Tension Reinforcement in Beams with Compression Steel
b)'(75,0)'(
where
yy
cb ff
f600
60085,0''
1
yy
c
by
s
fff
ff
60060085,0' '1
'
or
Old
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