Created by GARY NUGROHO
Perhitungan Balok Dua Tumpuan dengan Metode SRPMK (Sistem Rangka Pemikul Momen Khusus) SNI 2847 - 2013
h
PROPERTIES :
b
L Momen Tumpuan KIRI
Momen Tumpuan KANAN
Momen LAPANGAN
Dimensi Balok : b= h= L= Ln =
300 600 7500 7200
mm mm mm mm
b= h= tp =
300 mm 300 mm 120 mm
(Kolom) (Kolom)
Hasil dari Analisa Struktur :
fy
=
fys
=
172,09 137,63 171,41 11,75 158712,90 5,08 18,64 40 19 10 12 540,5 59,5 25
kNm = kNm = kNm = kNm = N kN/m kN mm mm mm mm mm mm N/mm2
2 300 N/mm 2 240 N/mm
172090000 137630000 171410000 11751200
Nmm Nmm Nmm Nmm
D Tul. Lentur Ø Tul. Geser
d'
= = = = = = = = = = = = = =
d
Momen tumpuan kiri Momen lapangan Momen tumpuan kanan Tu Vu Q (1,2D + 1,6L) P (1,2D + 1,6L) Tebal selimut beton (cover) Diameter tulangan lentur (D) Diameter tulangan geser (Ф) Diameter tulangan puntir d = (h-cover-Ф-(0,5.D)) d' = h - d f c'
b cover
β1 = 0,85
untuk fc' 17 s.d. 28 N/mm2
β1 = 0,85 - 0,05*((fc' - 28)/7) untuk fc' > 28 N/mm
Pasal 10.2.7.3 2
β1 = 0,87 Asmin = (1,4*bw*d)/fy ρmax =
Pasal 10.5.1
=
2 756,70 mm
Pasal 21.5.2.1
=
0,0250
=
0,0047
=
14,12
ρmin = 1,4/fy m = fy/(0,85*fc')
SYARAT STRUKTUR LENTUR SRPMK
Created by GARY NUGROHO
Pu ≤ (Ag*fc')/10
=
450 KN OK
Ln ≥ 4*d
=
2162 mm OK
bw > a). 0,3*h b). 250 mm
= =
180 mm 240 mm OK
bw ≤ a). Bkolom b). 0,75*(b*h)kolom
=
300 mm
=
450 mm OK
PENULANGAN PUNTIR BALOK Ph
Luasan Acp dan keliling Pcp
Ph
Luasan Aoh dan Keliling Ph
Acp = b*h
=
Pcp = 2*(b+h)
=
Aoh = [b-2*cover-0,5*Ф]*[h-2*cover-0,5*Ф]
=
Ph = 2*[(b-2*cover-0,5*Ф)+(h-2*cover-0,5*Ф)] Momen Puntir Nominal Tn = Tu/φ ((φ*√(fc'))/3)*(Acp2/Pcp)
Tu-max = Cek Pengaruh Momen Puntir Tu-min = ((φ*√(fc'))/12)*(Acp2/Pcp) = 6E+06 Nmm
=
2 180000 mm
1800 mm 2 110725 mm 1460 mm
=
15668266,7 Nmm
=
22500000,0 Nmm
< <
Tu 11751200 OK
Nmm
Ruas Kiri ≤ Ruas Kanan Ruas Kiri = √((Vu/(b*d))2+((Tu*Ph)/(1,7*Aoh2))2) Vc = (1/6)*√(fc')*b*d Ruas kanan = φ*[(Vc/(b*d))+((2*√(fc'))/3)]
=
0,979
=
135125
=
3,125 OK
Tulangan Puntir untuk Geser Tn = ((2*A0*At*fyv)/s)*cot Ɵ A0 = 0,85*Aoh At/s = Tn/(2*Ao*fyv*cot Ɵ) Tulangan Puntir untuk Lentur
= =
2 94116,25 mm 2 0,347 mm
Created by GARY NUGROHO
A1-min = ((5*√(fc')*Acp)/12*fy)-(At/s)*Ph*(fyv/fy)
=
2 405,10 mm 2 844,90 mm
Apakai =
=
2 422,45 mm
=
2 452,39 mm
A1 = (At/s)*Ph*(fyv/fy)*cot2 Ɵ
As pasang =
n (buah) =
=
4
D12
PENULANGAN LENTUR TUMPUAN Tumpuan Kiri = Tumpuan Kanan Dipakai δ 2 Rn = ((1-δ)*Mu)/(φ*b*d )
= =
0,5 1,09
ρδ = (1/m)*(1-√(1-((2*m*Rn)/fy)))
=
0,0037
ρ' = (δ*Mu)/(φ*fy*(d-d')*b*d)
=
0,0041
ρ = ρδ + ρ'
=
0,0078
As pelu = As perlu + 0,25*As puntir As pasang = n (buah) = As' perlu = As' perlu + 0,25*As puntir As' pasang = n (buah) =
ρ*b*d 6
D19 ρ'*b*d
4
D19
= = =
1268,12 mm2 1479,35 mm2 2 1701,17 mm
= = =
662,55 mm2 873,77 mm2 2 1134,11 mm
Kontrol spasi tulangan Syarat = S ≥ 25 mm Pasal 7.6.1 S = (b-(2*cover)-(2*Ф)-(n*D))/(n-1) = 17,20 mm Tidak OKSyarat Tidak Terpenuhi Pasang satu lapis Tidak OK(Tidak Bisa, harus pasang dua (2) lapis) Pasang dua lapis (Perlu dipasang dua (2) lapis) nTotal = = 6 buah nDasar =
=
5 buah
nSisa =
=
1 buah
y-1 = cover+Ф+(0,5*D)
=
59,5 mm
y-2 = cover+Ф+D+25+(0,5*D)
=
103,5 mm
y-0 = ((nDasar*y-1)+(nSisa*y-2))/nTotal
=
66,8 mm
d = h - y-0
=
533,2 mm
a = (As*fy)/(0,85*fc'*b)
=
80,1 mm
φMn = 0,8*As*fy*(d-(a/2)) ( Syarat φMn > Mu )
=
Kontrol spasi tulangan Syarat = S ≥ 25 mm S = (b-(2*cover)-(2*Ф)-(n*D))/(n-1) OK Syarat Terpenuhi
=
201339505 Nmm (OK)
26,25 mm
Created by GARY NUGROHO
PENULANGAN LENTUR LAPANGAN Dipakai δ Rn = ((1-δ)*Mu)/(φ*b*d2) ρδ = (1/m)*(1-√(1-((2*m*Rn)/fy)))
= = =
0,5 0,87 0,0030
ρ' = (δ*Mu)/(φ*fy*(d-d')*b*d)
=
0,0033
ρ = ρδ + ρ'
=
0,0062
As pelu = As perlu + 0,25*As puntir As pasang = n (buah) =
ρ*b*d 5
D19
= = =
As' perlu = ρ'*b*d = As perlu + 0,25*As puntir = As' pasang = n (buah) = 3 D19 = Kontrol spasi tulangan Syarat = S ≥ 25 mm S = (b-(2*cover)-(2*Ф)-(n*D))/(n-1) = OK Syarat Terpenuhi Pasang satu lapis OK (Bisa dipasang satu (1) lapis) Pasang dua lapis (Tidak Perlu Pasang dua (2) lapis) nTotal = =
1011,52 mm2 1222,75 mm2 2 1417,64 mm 529,88 mm2 741,10 mm2 2 850,59 mm
26,25 mm
5 buah
nDasar =
=
5 buah
nSisa =
=
0 buah
y-1 = cover+Ф+(0,5*D)
=
59,5 mm
y-2 = cover+Ф+D+25+(0,5*D)
=
103,5 mm
y-0 = ((nDasar*y-1)+(nSisa*y-2))/nTotal
=
59,5 mm
d = h - y-0
=
540,5 mm
a = (As*fy)/(0,85*fc'*b)
=
66,7 mm
φMn = 0,8*As*fy*(d-(a/2)) ( Syarat φMn > Mu )
=
Kontrol spasi tulangan Syarat = S ≥ 25 mm S = (b-(2*cover)-(2*Ф)-(n*D))/(n-1) OK Syarat Terpenuhi Kontrol Balok T Lebar efektif 1). be = L/4 2). be = b + 8*tp Dipakai yang terkecil As (Lapangan) n (buah) = a = (As*fy)/(0,85*fc'*b) x = a/0,85 Syarat x < tp
5
D19
172547768 Nmm (OK)
=
26,25 mm
=
1875 mm
= = = = =
1260 mm 1260 mm 1417,64 mm2 66,71 mm 78,49 mm (OK) (dipakai balok T palsu)
Created by GARY NUGROHO
PENULANGAN GESER TUMPUAN 1). Momen Tumpuan Negatif As pasang = n (buah) = 6 a = (As*(1,25*fy))/(0,85*fc'*b)
D19
Mpr- = As*(1,25*fy)*(d-(a/2))
= =
1701,17 mm2 100,07 mm
=
308,21 kNm
= =
1134,11 mm2 66,71 mm
=
215,68 kNm
2). Momen Tumpuan Positif As' pasang = n (buah) = 4 a = (As*(1,25*fy))/(0,85*fc'*b)
D19
Mpr+ = As*(1,25*fy)*(d-(a/2))
Gaya geser total pada muka tumpuan (pada muka kolom s.d. 2h) V total = (Qu*Ln)/2 = Ve, A = Ve, B =
((Mpr++Mpr-)/Ln)+((Qu*Ln)/2) (-(Mpr++Mpr-)/Ln)+((Qu*Ln)/2)
Ve, maks =
18,288 kN
=
91,05 kN
=
-54,48 kN
=
91,05 kN
Syarat nilai Vc = 0, apabila a). Ve, maks > 0,5 V total
(OK)
b). Pu < 0,1(Ag*fc') , …. Dimana Ag = b*h (mm2)
(OK)
Nilai Vc =
0
Vu tumpuan = Vn Vs = Vn/φ Vs max = (2/3)*b*d*√(fc')
=
91051,0403 N
=
121401,387 N
=
533166,667 N
Syarat Vs > Vs max Direncanakan As Geser (Av) untuk Ф10 n= 2 Av = 0,25*π*Ф s = (Av*fy*d)/Vs
(OK) = =
2 buah 2 157,08 mm
=
165,57 mm
Kontrol smax, tidak boleh lebih besar dari yang terkecil : smax = = = = dipakai s =
2*h d/4 6*D 150
Pasal 21.5.3.2 = = = = =
1200 133 114 150 110
mm mm mm mm mm
sepanjang 2h Dipasang 2Ф10 - 110 = 1200 mm ,dari muka kolom dimana tulangan geser (Ф) pertama dipasang sejarak 50 mm dari muka KOLOM.
Created by GARY NUGROHO
PENULANGAN GESER LAPANGAN Tulangan geser (Ф) di luar sendi plastis (di luar 2h) Vu lapangan Vn -=(2*h*Qu)
=
84955,0403 N
=
40537,50 N
pers.1
=
50671,88 N
pers.2
=
101343,75 N
Stop
[φVc+φVsmin] =
=
141881,25 N
Stop
[φVc+φ(1/3).√fc'.b.d] =
=
304031,25 N
Stop
Tulangan geser minimum smax = d/2
=
φVsmin = 0,75*(1/3)*b*d 0,5*φVc = 0,5*(0,75*(1/6)*√(fc')*b*d) φVc = 0,75*(1/6)*√fc'*b*d
Av = (b*s)/(2*fy) Direncanakan As Geser (Av) untuk Ф10 n= 2 Av = n*(0,25*π*d ) s= KONTROL LENDUTAN hmin balok = L/21
=
270,25 mm 2 168,91 mm
= = =
2 buah 2 157,08 mm 200,00 mm
=
357 mm OK, h desain terpenuhi
KONTROL RETAK z = fx*3√(dc*A) fx = 0,6*fy
=
dc = (cover+Ф+(0,5*D))
=
A = (2*dc*b)/n dasar As tumpuan z= Syarat z < 30 MN/m
= =
2 180,0 N/mm
59,5 mm 2 7140,0 mm 13,5 MN/m (OK)
Created by GARY NUGROHO
HASIL PERHITUNGAN TUMPUAN TUMPUAN TUMPUAN
LAPANGAN LAPANG ANAN LAPANG
As
As'
As'
As'
As'
As
As
h h
As
tp tp
POSISI
POTONGAN
bb Dimensi Tul. Atas Tul. Puntir Tul. Bawah Sengkang
bb
300 x 600 6D19 (5 + 1) 4Ф12 4D19 2Ф10 - 110
300 x 600 3D19 4Ф12 5D19 (5 + 0) 2Ф10 - 200
As Tumpuan Ln/4
As' Lapangan Ln/2
As Tumpuan Ln/4
Ln/4 As' Tumpuan
Ln/2 As Lapangan
Ln/4 As' Tumpuan
Sengkang di luar sendi plastis sejarak Ln-(2* 2h) Ln
Sengkang di sendi plastis sejarak 2h
50 mm Sengkang di sendi plastis sejarak 2h
50 mm
L
