Analisis Balok Tampang T
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Transcript of Analisis Balok Tampang T
CONTOH HITUNGAN ANALISIS
BALOK BERTULANG PENAMPANG T
b = dicari/dihitung Titik berat tulangan tarik 5D19:
h = 500 mm ds =(3*50+2*100)/5 = 70 mm
hf = 100 mm
bw = 200 mm fc' = 25 MPa 1 = 0.85
ds' = 50 mm fy = 400 MPa
ds = 70 mm Es = 200000 MPa y = 0.002
d = 430 mm DL = 15 kN/m'
L = 7200 mm LL = 4 kN/m2
L1= 1800 mm
L2 = 1800 mm
b
3,6 m 3,6 m 2,4 m
7,2 m
50 cm
= lebar efektif balok tampang T
lebar lajur beban utk satu balok
b
100 50
50 50
200
500
2D19
5D19
POT. I ‐ I
b
100 50
50 50
200
500
5D19
2D19
POT. II ‐ II
50
ds
Lebar efektif pelat untuk balok T:bw = 200 mm
b1 = 8*hf = 800 mm
b1 = 1/2*L1 = 900 mm digunakan nilai b1 terkecil, b1 = 800 mm
b2 = 8*hf = 800 mm
b2 = 1/2*L2 = 900 mm digunakan nilai b2 terkecil, b2 = 800 mm
jadi: b = bw+b1+b2= 1800 mm
syarat lain: b < 1/4*L= 1800 mm digunakan nilai b = 1800 mm
Jadi untuk analisis penampang digunakan ukuran tampang sbb:
a. Di daerah momen positif (sisi atas atau pelat lantai tertekan), digunakan tampang T dg ukuran:
b = 1800 mm hf = 100 mm
bw = 200 mm ds = 70 mm
h = 500 mm ds' = 50 mm
d = 430 mm
b. Di daerah momen negatif (sisi atas atau pelat lantai tertarik), digunakan tampang persegi dg ukuran:
b = 1800 mm ds = 70 mm
bw = 200 mm ds' = 50 mm
h = 500 mm d = 430 mm
Analisis struktur: Lebar jalur beban = 2 m
DL = 15 kN/m'
LL = 2*4 = 8 kN/m'
qu = 1,2*DL + 1,6*LL= 30.8 kN/m'
SFD (kN) 73.92
98.56 RAV = 98.56 kN
RBV = 197.12 kN
123.2 XSF=0 = 3.2 m, dari A.
BMD (kNm) 88.704
M(+)max = 157.696 kNm
M(‐)max = 88.704 kNm
157.696
Analisis kuat lentur nominal penampang T (daerah momen positif):
Kondisi balance:
cb = 600/(600+fy)*d = 258 mm > hf = 100 mm
berarti g.n. berada di badan.
ab = b1*cb = 219.30 mm
Asf = 0.85*fc'*hf*(b‐bw)/fy = 8500 mm2
Asb = Asf +(0.85*fc'*bw*ab)/fy = 10830.063 mm2
As max = 0.75*Asb = 8122.5469 mm2
DLLL
7,2 m 2,4 m
A B C
Cek tulangan terpasang:
As' = 2D19 = 566.77 mm2
As = 5D19 = 1416.925 mm2
As1 = As ‐ As' = 850.155 mm2 < As max= 8122.547 mm2 OK
> As min1= 268.750 mm2
> As min2= 301.000 mm2 menentukan
Ok
Asumsi: blok tekan beton berada di sayap di atas tulangan atas (desak) atau a < ds',
dan tulangan atas mengalami tegangan tarik dan mencapai tegangan leleh, maka
Cc = Ts' + Ts
a = (As+As')*fy/(0.85*fc'*b) = 20.745 mm < ds'= 50 mm
c = a/0.85 = 24.405 mm
regangan tulg atas = 0.00315 > reg.leleh = 0,002. OK, asumsi benar.
Mn
regangan tarik baja tulg bawah =
0.050 >> reg.leleh =0.002
Cc = 0.85*fc'*b*a = 793478 N
Ts' = As'*fy = 226708 N
Ts = As * fy = 566770 N Cek: Cc = Ts' + Ts = 793478 N, OK
Mn = Cc * (d‐a/2) ‐ Ts'*(d‐ds')
= 246816339 Nmm
= 246.816 kNm
Mr = .Mn = 0.8*Mn = 197.4531 kNm > Mu = 157.696 kNm, OK.
Analisis kuat lentur nominal penampang persegi (daerah momen negatif):
Kondisi balance:
cb = =600/(600+fy)*d = 258 mm
ab = =1*cb = 219.30 mm
Asb = =(0.85*fc'*bw*ab)/fy = 2330.0625 mm2
As max = 0.75*Asb = 1747.5469 mm2
Cek tulangan terpasang:
As' = 2D19 = 566.77 mm2
As = 5D19 = 1416.925 mm2
As1 = As ‐ As' = 850.155 mm2 < As max= 1747.547 mm2 Ok
> As min1= 537.500 mm2 Ok
> As min2= 2418.750 mm2
(utk yg ini mestinya tulg tarik pd plat ikut diperhitungkan)
posisi tulg atas
Cc
Ts'
Ts
Asumsi I: tulangan tekan 2D19 dan tarik 5D19 sudah leleh
a = (AS1*fy) / (0.85*fc'*bw) = 80.015 mm
c = a/0.85 = 94.135 mm
cek regangan tulg tekan = 0.0014065 < y = 0.002, belum leleh
asumsi salah.
Asumsi II: tulangan tekan 2D19 belum leleh, tulangan tarik 5D19 sudah leleh
disusun persamaan kuadrat dalam c yang didasarkan pada persyaratan Fh = 0:Cc + Cs = Ts
Cc = 0.85*fc*b*(1*c)Cs = As'*fs = As' * (s*Es) = As'*Es*(c‐ds')/c*0.003Ts = As * fy
didapat nilai c = 106.86 mm regangan baja tulg tekan:
a = 90.831 mm s' = 0.001596
fs' = 319.259 MPa
Momen nominal (Mn):
Mn1 = 0.85*fc'*a*bw*(d‐a/2) = 148461828 Nmm
Mn2 = As'*fs'*(d‐ds') = 68759607 Nmm
Mn = Mn1 + Mn2 = 217221435 Nmm
217.221 kNm
Mr = f.Mn = 0.8*Mn = 173.777 kNm > Mu= 88.704 kNm
OK
KESIMPULAN:
Balok dengan dimensi dan tulangan penampang tsb
mampu menahan beban tsb dg aman !