Penulangan tumpuan arah y Penulangan lapangan arah x Penulangan lapangan arah y

Tugas Akhir Perencanaan Struktur Bank Pasar 2 Lantai 101 BAB 5 Plat Lantai m = 2942 , 11 25 . 85 , 240 . 85 ,   c f fy  perlu =       fy Rn . m 2 1 1 . m 1 = . 2942 , 11 1       240 1862 , 1 . 2942 , 11 . 2 1 1 = 0,005089   max   min , di pakai  perlu = 0,006024 As perlu =  perlu . b . dx = 0,005089 . 1000 . 95 = 483,46 mm 2 Digunakan tulangan  10 As = ¼ .  . 10 2 = 78,5 mm 2 S = per lu As b As . = 46 , 483 1000 . 5 , 78 = 162,37 ~ 160 mm n = s b = 160 1000 = 6,25 As yang timbul = 6,25. ¼ .  . 10 2 = 490,63 mm 2 As perlu …..…ok Dipakai tulangan  10 – 160 mm

5.6. Penulangan tumpuan arah y

Mu = 1089,99 kgm = 10,8999.10 6 Nmm Tugas Akhir Perencanaan Struktur Bank Pasar 2 Lantai 102 BAB 5 Plat Lantai Mn =  M u = 6 6 10 . 625 , 13 8 , 10 . 8999 , 10  Nmm Rn =  2 .dy b Mn   886 , 1 85 . 1000 10 . 625 , 13 2 6  Nmm 2 m = 294 , 11 25 . 85 , 240 . 85 ,   c f fy  perlu =       fy Rn . m 2 1 1 . m 1 =       240 886 , 1 . 294 , 11 . 2 1 1 . 294 , 11 1 = 0,008242   max   min , di pakai  perlu = 0,008242 As perlu =  perlu . b . d = 0,008242 . 1000 . 85 = 700,57 mm 2 Digunakan tulangan  10 As = ¼ .  . 10 2 = 78,5 mm 2 S = per lu As b As . = 57 , 700 1000 . 5 , 78 = 112,05 ~ 110 mm n = s b = 110 1000 = 9 As yang timbul = 9. ¼ .  . 10 2 = 706,5 mm 2 As perlu …ok Dipakai tulangan  10 – 110 mm Tugas Akhir Perencanaan Struktur Bank Pasar 2 Lantai 103 BAB 5 Plat Lantai

5.7. Penulangan lapangan arah x

Mu = 342,57 kgm = 3,4257.10 6 Nmm Mn =  M u = 6 6 10 . 282 , 4 8 , 10 . 4257 , 3  Nmm Rn =  2 .dx b Mn    2 6 95 . 1000 10 . 282 , 4 0,4745 Nmm 2 m = 294 , 11 25 . 85 , 240 . 85 ,   c f fy  perlu =       fy Rn . m 2 1 1 . m 1 =       240 4745 , . 294 , 11 . 2 1 1 . 294 , 11 1 = 0,001999   max   min , di pakai  min = 0,0025 As =  min . b . dx = 0,0025. 1000 . 95 = 237,5 mm 2 Digunakan tulangan  10 As = ¼ .  . 10 2 = 78,5 mm 2 S = per lu As b As . = 5 , 237 1000 . 5 , 78 = 330,52 ~ 330 mm Jarak maksimum = 2 x h = 2 x 120 = 240 mm n = s b = 240 1000 Tugas Akhir Perencanaan Struktur Bank Pasar 2 Lantai 104 BAB 5 Plat Lantai = 4,2  5 As yang timbul = 5. ¼ .  . 10 2 = 392,5 mm 2 As …ok Dipakai tulangan  10 – 240 mm

5.8. Penulangan lapangan arah y

Mu = 498,28 kgm = 4,9828.10 6 Nmm Mn =  M u = 6 6 10 . 23 , 6 8 , 10 . 9828 , 4  Nmm Rn =  2 .dy b Mn    2 6 85 . 1000 10 . 23 , 6 0,8623 Nmm 2 m = 294 , 11 25 . 85 , 240 . 85 ,   c f fy i  perlu =       fy Rn m m . . 2 1 1 . 1 = . 294 , 11 1       240 8623 , . 294 , 11 . 2 1 1 = 0,00367   max   min , di pakai  perlu = 0,00367 As =  perlu b . d = 0,00367 . 1000 . 85 = 311,95 mm 2 Digunakan tulangan  10 As = ¼ .  . 10 2 = 78,5 mm 2 S = p er lu As b As . = 95 , 311 1000 . 5 , 78 = 251,643 ~ 250 mm Jarak maksimum = 2 x h = 2 x 120 = 240 mm Tugas Akhir Perencanaan Struktur Bank Pasar 2 Lantai 105 BAB 5 Plat Lantai n = s b = 240 1000 = 4,2  5 As yang timbul = 5. ¼ .  . 10 2 = 392,5 mm 2 As …ok Dipakai tulangan  10 – 240 mm

5.9. Rekapitulasi Tulangan