Tulangan Baja TEORI DASAR

Tegangan setelah topping terpasang σ a2l σ a1l Mcorpek St + = Ft 2.881 Mpa ⋅ = ≤ σ a2l 0.994 − Mpa ⋅ = Fc 14.94 − Mpa ⋅ = σ b2l σ b1l Mcorpek Sb − = ≤ σ b2l 1.147 − Mpa ⋅ = Tegangan setelah prop dilepaskan Sebagai pelat komposit σ a3l σ a2l Mprop h Cbk − ⋅ I ck − = σ a3l 1.461 − Mpa ⋅ = Fc 14.94 − Mpa ⋅ = ≤ σ b3l σ b2l Mprop Sbk + = σ b3l 0.107 − Mpa ⋅ = Ft 2.881 Mpa ⋅ = ≤ c. Pada saat beban layan bekerja Sebagai pelat komposit σ a4l σ a3l Mll Msdl + h Cbk − I ck ⋅ − = σ a4l 2.054 − Mpa ⋅ = ≤ Fc 14.94 − Mpa ⋅ = σ b4l σ b3l Mll Sbk + Msdl Sbk + = σ b4l 1.213 Mpa ⋅ = Ft 2.881 Mpa ⋅ = ≤ Tegangan_beton_Saat_Layan MEMENUHI =

6. Check Kapasitas Momen

Dw 6 mm ⋅ = Diameter Tulangan Pratekan Dw Dia = Diameter tulangan Ds 6 mm ⋅ = As1 1 4 π ⋅ Ds 2 ⋅ = As1 28.274 mm 2 ⋅ = Luas per tulangan Banyaknya tul. tekan nc = Asc 0 mm 2 ⋅ = Luas Total tul. tekan Asc nc As1 ⋅ = Banyaknya tul. tarik np = Ast 0 mm 2 ⋅ = Luas Total tul. tarik Ast np As1 ⋅ = Selimut Beton dc 20 mm ⋅ = Lengan Tulangan Pratekan dp Ct e + = dp 180 mm ⋅ = d 221 mm ⋅ = Lengan Tulangan Non Pratekan d hsl dc − Dw − 0.5 Ds ⋅ − = fpe Peff Aps = Tegangan Tul. Pratekan Efektif ≥ fpe 780 Mpa ⋅ = 0.5 fpu ⋅ 812.5 Mpa ⋅ = maka : Nilai untuk p : 66 fpy fpu 0.9 = γ p 0.2 8 = β 1 0.85 fc 30 Mpa ⋅ ≤ if 0.65 fc 55 Mpa ⋅ ≥ if 0.85 0.008 fc Mpa 30 − ⎛⎜ ⎝ ⎞⎟ ⎠ ⋅ − 30 Mpa ⋅ fc 55 Mpa ⋅ ≤ if = β 1 0.824 = ρ p Aps bw dp ⋅ = Asc 0 cm 2 ⋅ = Ast 0 cm 2 ⋅ = ρ c Asc bw d ⋅ = ρ c = ρ t Ast bw d ⋅ = ρ t = ω c ρ c fy fc ⋅ = ω c = ω t ρ t fy fc ⋅ = ω t = fps fpu 1 γ p β 1 ρ p fpu fc ⋅ d dp ω t ω c − ⋅ + ⎡⎢ ⎣ ⎤⎥ ⎦ ⋅ − ⎡⎢ ⎣ ⎤⎥ ⎦ ⋅ = fps 1607.319 Mpa ⋅ = ρ p Aps bw dp ⋅ = ρ p 0.001 = ω p ρ p fps fc ⋅ = ω p 0.032 = Tps fps Aps ⋅ = Tps 227.229 kN ⋅ = a Tps 0.85 fctop ⋅ bw ⋅ = a 11.929 mm ⋅ = Mn Tps dp htop + a 2 − ⎡⎢ ⎣ ⎤⎥ ⎦ ⋅ Ast fy ⋅ d a 2 − ⎛⎜ ⎝ ⎞⎟ ⎠ ⋅ + Asc fy ⋅ dc a 2 − ⎛⎜ ⎝ ⎞⎟ ⎠ ⋅ + = φ Mn ⋅ 45.817 kN m ⋅ ⋅ = Mu 35.203 kN m ⋅ ⋅ = Momen_Kapasitas_Penampang MEMENUHI =

7. Kapasitas Retak

fr 0.7 fc Mpa ⋅ ⋅ = fr 4.033 Mpa ⋅ = Tegangan tarik retak Kondisi tegangan pada tepi bawah HCS akibat beban layan total : fakt Peff − Ac Peff e ⋅ Sb − Mslb Sb + Mcorpek Sb − Mprop Sbk + Msdl Sbk + Mll Sbk + = fakt 1.213 Mpa ⋅ = Mcr fr fakt − Sbk ⋅ Mts + = Mcr 48.727 kN m ⋅ ⋅ = φ Mn ⋅ Mu 1.301 = Rasio_Penampang_Retak MEMENUHI = 67

8. Pemeriksaan Geser

Lebar badan bw 1200 mm ⋅ = φ 0.85 = Faktor reduksi Kuat tekan beton fc 33.2 Mpa ⋅ = Tegangan leleh tul. fy 390 Mpa ⋅ = Gaya-gaya geser : Saat beban layan belum bekerja geser hanya ditahan oleh HCS saja Vu dl 1.2 1 2 Qslb Qtop + ⋅ Lsl ⋅ ⎡⎢ ⎣ ⎤⎥ ⎦ ⋅ = Vu dl 14.083 kN ⋅ = Saat beban layan bekerja geser ditahan oleh pelat komposit Qd Qslb Qtop + Qsdl + = Vll 1 2 Qll ⋅ Lsl ⋅ = Vll 9 kN ⋅ = Vd 14.736 kN ⋅ = Vd 0.5 Qd ⋅ Lsl ⋅ = Vu x 1.2 Vd Qd x ⋅ − 1.6 Vll Qll x ⋅ − ⋅ + = Mu x 1.2 Vd x ⋅ 0.5 Qd ⋅ x 2 ⋅ − ⋅ 1.6 Vll x ⋅ 0.5Qll x 2 ⋅ − ⋅ + = x1 0.1m = x3 50 Dw ⋅ = x2 0.5hsl = x2 0.125 m = x3 0.3 m = Vu1 Vu x1 = Vu1 31.014 kN ⋅ = Mu1 Mu x1 = Mu1 3.155 kN m ⋅ ⋅ = Vu2 Vu x2 = Vu2 30.746 kN ⋅ = Mu2 Mu x2 = Mu2 3.927 kN m ⋅ ⋅ = Vu3 Vu x3 = Vu3 28.875 kN ⋅ = Mu3 Mu x3 = Mu3 9.144 kN m ⋅ ⋅ = Persyaratan Geser menurut ACI : 0.4 fpu ⋅ 650 Mpa ⋅ = dapat menggunakan metoda sederhana sebagai berikut : fpe 780 Mpa ⋅ = Vu dp ⋅ Mu 1 ≤ vc1 1 20 fc Mpa ⋅ 4.8 Vu1 dp ⋅ Mu1 ⋅ + ⎛⎜ ⎝ ⎞⎟ ⎠ Mpa ⋅ = Vu1 dp ⋅ Mu1 1.769 = vc1 8.782 Mpa ⋅ = vc2 1 20 fc Mpa ⋅ 4.8 Vu2 dp ⋅ Mu2 ⋅ + ⎛⎜ ⎝ ⎞⎟ ⎠ Mpa ⋅ = Vu2 dp ⋅ Mu2 1.409 = vc2 7.053 Mpa ⋅ = vc3 1 20 fc Mpa ⋅ 4.8 Vu3 dp ⋅ Mu3 ⋅ + ⎛⎜ ⎝ ⎞⎟ ⎠ Mpa ⋅ = Vu3 dp ⋅ Mu3 0.568 = vc3 3.017 Mpa ⋅ = λ 1 = untuk beton normal λ 6 fc Mpa ⋅ ⋅ 0.96 Mpa ⋅ = vc1 8.782 Mpa ⋅ = ≥ 0.4 λ ⋅ fc Mpa ⋅ ⋅ 2.305 Mpa ⋅ = ≤ vc 0.4 λ ⋅ fc Mpa ⋅ ⋅ vc1 0.4 λ ⋅ fc Mpa ⋅ ⋅ if vc1 otherwise = vc 2.305 Mpa ⋅ = 68 Saat beban layan belum bekerja Vc vc bw dp ⋅ ⋅ = dp 180 mm ⋅ = Vc 497.832 kN ⋅ = Vu dl 14.083 kN ⋅ = φ Vc ⋅ 423.157 kN ⋅ = Saat beban layan telah bekerja Vc vc bw htop dp + ⋅ htop be ⋅ + [ ] ⋅ = Vc 739.834 kN ⋅ = Vu1 31.014 kN ⋅ = φ Vc ⋅ 628.859 kN ⋅ = Tulangan_Geser_Vertikal TI DAK PERLU DI PASANG =

9. Check Terhadap Defleksi a. Pada kondisi awal pelat prategang saja

Δ pi Pi − e ⋅ Lsl 2 ⋅ 8 Eci ⋅ I c ⋅ = Δ pi 3.945 − mm ⋅ = ke atas Defleksi akibat berat sendiri Δ bs 5 384 Qslb Lsl 4 ⋅ Eci I c ⋅ ⋅ = Δ bs 3.316 mm ⋅ = Dengan menerapkan faktor jangka panjang untuk defleksi bersih pada waktu ereksi, diperoleh : Δ 1 0.966 − mm ⋅ = Δ 1 1.85 Δ bs ⋅ 1.8 Δ pi ⋅ + =

b. Pada kondisi akhir pelat komposit

Defleksi akibat topping Δ top 1 48 P Lsl 3 ⋅ Ec I ck ⋅ ⋅ = Δ top 0.835 mm ⋅ = Defleksi akibat SDL Δ sdl 5 384 Qsdl Lsl 4 ⋅ Ec I ck ⋅ ⋅ = Δ sdl 0.58 mm ⋅ = Defleksi akibat Beban Hidup LL Δ L 5 384 Qll Lsl 4 ⋅ Ec I ck ⋅ ⋅ = Δ L 1.739 mm ⋅ = Lsl 360 16.667 mm ⋅ = syarat defleksi maksimum akibat beban hidup : Jadi total defleksi jangka panjang yang terjadi adalah : Δ 2 2.2 Δ pi ⋅ 2.4 Δ bs ⋅ + 2.3 Δ top ⋅ + 3 Δ sdl ⋅ + 3 30 Δ L ⋅ + = ke bawah Δ 2 4.505 mm ⋅ = Jadi defleksi total Δ tot Δ 2 Δ 1 − 70 Δ L + = Δ tot 6.688 mm ⋅ = Lsl 240 25 mm ⋅ = syarat defleksi maksimum Δ tot 6.688 mm ⋅ = 69 Defleksi_yang_terjadi MASI H MEMENUHI PERSYARATAN = Catatan : - Nilai negatif menunjukkan defleksi ke atas dan, - Nilai positif menunjukkan defleksi ke bawah - Defleksi memenuhi persyaratan bila defleksi max. lebih besar dari defleksi total yg terjadi

10. Pengecekan Geser Horizontal

Av1 0 mm 2 ⋅ = Lebar bid kontak bv bw = s 0mm = Tinggi efektif d dp htop + = ρ v Av1 bv s ⋅ = ρ v ⋅ = Vu ≤ Vnh dari data ETABS Akibat gempa Vg 10.54 kN ⋅ = Vu1 Vu1 Vg + = Vu1 41.554 kN ⋅ = φ 0.6 ⋅ bv ⋅ d ⋅ Mpa ⋅ 140.76 kN ⋅ = Vnh 3.5 bv ⋅ dp ⋅ Mpa ⋅ 756 kN ⋅ = Untuk_menahan_Geser_Horizontal TI DAK PERLU DI PASANG TULANGAN GESER =

11. Tulangan Lentur di Daerah Tumpuan

Mtop 1 11 Qtop ⋅ Lsl 2 ⋅ = Mtop 4.713 kN m ⋅ ⋅ = Mlle 1 11 Qll ⋅ Lsl 2 ⋅ = Mlle 9.818 kN m ⋅ ⋅ = Msdle 1 11 Qsdl ⋅ Lsl 2 ⋅ = Msdle 3.273 kN m ⋅ ⋅ = fctop 18.675 Mpa ⋅ = fc 33.2 Mpa ⋅ = Mue 1.2 Msdle Mtop + ⋅ 1.6 Mlle ⋅ + = Mue 25.292 kN m ⋅ ⋅ = Akibat gempa dari data ETABS Mg 17.716 kN ⋅ m = Mue 37.117 kN m ⋅ ⋅ = Mue 1.2 Msdle Mtop + ⋅ 1.0 Mlle ⋅ + 1.0 Mg ⋅ + = Mu Mue = Mu 37.117 kN m ⋅ ⋅ = Dtump 9 mm ⋅ = fy 400 Mpa ⋅ = dtop hsl dc − 0.5 Dtump ⋅ − = dtop 225.5 mm ⋅ = As1 1 4 π ⋅ Dtump 2 ⋅ = As1 0.636 cm 2 ⋅ = jarak tulangan lentur : Sptump 150 mm ⋅ = As bw Sptump As1 ⋅ = As 508.938 mm 2 ⋅ = faktor reduksi lentur : φ 0.9 = β 1 0.85 = ρ max 0.75 β 1 0.85 fc ⋅ fy ⋅ 87000 87000 fy psi + ⋅ ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ⋅ ρ max 0.027 = = ρ min 0.18 = ρ As bw dtop ⋅ = ρ 0.188 ⋅ = 70 a fy As ⋅ 0.85 fctop ⋅ bw ⋅ = a 10.687 m m ⋅ = Mn As fy ⋅ dtop a 2 − ⎛⎜ ⎝ ⎞⎟ ⎠ ⋅ = Mu φ 41.241 kN m ⋅ ⋅ = Mn 44.818 kN m ⋅ ⋅ = Tumpuan TULANGAN TERPASANG MEMENUHI =

12. Perhitungan Tulangan TransferLateral

Mtop 1 14 Qtop ⋅ Lsl 2 ⋅ = Msdl 1 14 Qsdl ⋅ Lsl 2 ⋅ = Mtop 3.703 kN m ⋅ ⋅ = Msdl 2.571 kN m ⋅ ⋅ = Mll 1 14 Qll ⋅ Lsl 2 ⋅ = Mll 7.714 kN m ⋅ ⋅ = Mu 15.429 kN m ⋅ ⋅ = Mu 1.2 Msdl ⋅ 1.6 Mll ⋅ + = dari data ETABS Mg 17.52 kN ⋅ m = Akibat gempa Mu 28.32 kN m ⋅ ⋅ = Mu 1.2 Msdl ⋅ 1.0 Mll ⋅ + 1.0Mg + = Faktor Reduksi Lentur φ 0.9 = ρ min 0.2 ⋅ = Rasio Tulangan Minimum Asmin 600 mm 2 ⋅ = Luas Tulangan Minimum Terpasang Asmin ρ min bw ⋅ hsl ⋅ = Diameter Tulangan Terpasang Dstr 8 mm ⋅ = Kuat Leleh Tulangan fy 400 Mpa ⋅ = Spasi Tulangan Sptr 150 mm ⋅ = As 0.25 π ⋅ Dstr 2 ⋅ bw Sptr ⋅ = Luas Tulangan Terpasang As 402.124 mm 2 ⋅ = a As fy ⋅ 0.85 fctop ⋅ bw ⋅ = Lebar Stress Beton a 8.444 mm ⋅ = jd hsl 18.5 mm ⋅ − 0.5 Ds ⋅ − 0.5 a ⋅ − = Lengan Momen jd 224.278 mm ⋅ = Momen Nominal Terpasang Mn φ As ⋅ fy ⋅ jd ⋅ = Mn 32.467 kN m ⋅ ⋅ = Mu 28.32 kN m ⋅ ⋅ = Tulangan MAMPU MENAHAN BEBAN LAYAN =

13. Tulangan Stek Sambungan Antar HCS

Mu 28.32 kN m ⋅ ⋅ = Momen Ultimit Mu φ 0.9 = Faktor Reduksi Lentur Dstek 12 mm ⋅ = Diameter Tulangan Terpasang fystek 400 Mpa ⋅ = Kuat Leleh Tulangan Spstek 120 mm ⋅ = Spasi Tulangan As 0.25 π ⋅ Dstek 2 ⋅ bw Spstek ⋅ 1.6 ⋅ = a As fystek ⋅ 0.85 fctop ⋅ bw ⋅ = As 1.81 10 3 × mm 2 ⋅ = Luas Tulangan Total Lebar Stress Beton a 37.999 mm ⋅ = jd 47 mm ⋅ = Lengan Momen Tulangan Lentur jd htop 0.5 Ds ⋅ − = 71