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

Tugas Akhir Perencanaan Struktur Gedung Sekolahan 2 Lantai BAB 5 Plat Lantai 99 n = s b = 100 1000 = 10 As yang timbul = 10. ¼ .  . 10 2 = 785 mm 2 As perlu …..…ok Dipakai tulangan  10 – 120 mm

5.6. Penulangan tumpuan arah y

Mu = 971,802 kgm = 971,802.10 6 Nmm Mn =  M u =  8 , 802 , 971 6 12,1475.10 6 Nmm Rn =  2 .dx b Mn    2 6 95 . 1000 10 . 1475 , 12 1,346 Nmm 2 m = 412 , 9 30 . 85 , 240 . 85 ,   c f fy  perlu =           fy Rn . m 2 1 1 . m 1 = . 412 , 9 1         240 346 , 1 . 412 , 9 . 2 1 1 = 0,00576   max   min , di pakai  perlu = 0,00576 As perlu =  perlu . b . dx = 0,00576 . 1000 . 95 = 547,58 mm 2 Tugas Akhir Perencanaan Struktur Gedung Sekolahan 2 Lantai BAB 5 Plat Lantai 100 Digunakan tulangan  10 As = ¼ .  . 10 2 = 78,5 mm 2 S = p erlu As b As . = 58 , 547 1000 . 5 , 78 = 143,358 ~ digunakan 120 mm n = s b = 100 1000 = 10 As yang timbul = 10. ¼ .  . 10 2 = 785 mm 2 As perlu …..…ok Dipakai tulangan  10 – 120 mm

5.7. Penulangan lapangan arah x

Mu = 400,154 kgm = 4,00154.10 6 Nmm Mn =  M u = 6 6 10 . 0019 , 5 8 , 10 . 00154 , 4  Nmm Rn =  2 .dx b Mn    2 6 95 . 1000 10 . 0019 , 5 0,554 Nmm 2 m = 412 , 9 30 . 85 , 240 . 85 ,   c f fy  perlu =           fy Rn . m 2 1 1 . m 1 =         240 554 , . 412 , 9 . 2 1 1 . 412 , 9 1 = 0,00233   max   min , di pakai  perlu = 0,0025 As =  perlu . b . dx Tugas Akhir Perencanaan Struktur Gedung Sekolahan 2 Lantai BAB 5 Plat Lantai 101 = 0,0025. 1000 . 95 = 237,5 mm 2 Digunakan tulangan  10 As = ¼ .  . 10 2 = 78,5 mm 2 S = p erlu As b As . = 5 , 237 1000 . 5 , 78 = 330,52 ~ 300 mm Jarak maksimum = 2 x h = 2 x 120 = 240 mm 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

5.8. Penulangan lapangan arah y

Mu = 400,154 kgm = 4,00154.10 6 Nmm Mn =  M u = 6 6 10 . 0019 , 5 8 , 10 . 00154 , 4  Nmm Rn =  2 .dx b Mn    2 6 95 . 1000 10 . 0019 , 5 0,554 Nmm 2 m = 412 , 9 30 . 85 , 240 . 85 ,   c f fy  perlu =           fy Rn . m 2 1 1 . m 1 Tugas Akhir Perencanaan Struktur Gedung Sekolahan 2 Lantai BAB 5 Plat Lantai 102 =         240 554 , . 412 , 9 . 2 1 1 . 412 , 9 1 = 0,00233   max   min , di pakai  perlu = 0,0025 As =  perlu . b . dx = 0,0025. 1000 . 95 = 237,5 mm 2 Digunakan tulangan  10 As = ¼ .  . 10 2 = 78,5 mm 2 S = p erlu As b As . = 5 , 237 1000 . 5 , 78 = 330,52 ~ 300 mm Jarak maksimum = 2 x h = 2 x 120 = 240 mm 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