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

Tugas Akhir Perencanaan Struktur Gedung Perpustakaan 2 Lantai

5.5. Penulangan tumpuan arah x

Mu = 1867,514 kgm = 1,8675.10 7 Nmm Mn =  M u =  8 , 10 . 8675 , 1 7 2,3344.10 7 Nmm Rn =  2 .d b Mn    2 7 94 . 1000 10 . 3344 , 2 2,6419 Nmm 2 m = 294 , 11 25 . 85 , 240 . 85 ,   c f fy  perlu =       fy Rn . m 2 1 1 . m 1 = . 294 , 11 1       240 6419 , 2 . 294 , 11 . 2 1 1 = 0,012   max   min , di pakai  = 0,012 As =  . b . d = 0,012 . 1000 . 94 = 1128 mm 2 Digunakan tulangan  12 = ¼ .  . 12 2 = 113,04 mm 2 S = per lu As A 1000 . = 1128 1000 . 04 , 113 = 100,21 ~ 110 mm Jarak maksimum = 2 x h = 2 x 120 = 240 mm As yang timbul = 10. ¼ .  .12 2 = 1130,4 mm 2 As…..…ok Dipakai tulangan  12 – 110 mm Tugas Akhir Perencanaan Struktur Gedung Perpustakaan 2 Lantai

5.6. Penulangan tumpuan arah y

Mu = 1396,102 m = 1,3961.10 7 Nmm Mn =  M u =  8 , 10 . 3961 , 1 6 1,7451.10 7 Nmm Rn =  2 .d b Mn    2 7 82 . 1000 10 . 7451 , 1 2,5953Nmm 2 m = 294 , 11 25 . 85 , 240 . 85 ,   c f fy  perlu =       fy Rn . m 2 1 1 . m 1 = . 294 , 11 1       240 5953 , 2 . 294 , 11 . 2 1 1  = 0,012   max   min , di pakai  = 0,012 As =  . b . d = 0,012 . 1000 . 82 = 1128 mm 2 Digunakan tulangan  12 = ¼ .  . 12 2 = 113,04 mm 2 S = per lu As A 1000 . = 1128 1000 . 04 , 113 = 100.21 ~ 110 mm Jarak maksimum = 2 x h = 2 x 120 = 240 mm As yang timbul = 10. ¼ .  . 12 2 = 11304 mm 2 As…..…ok Dipakai tulangan  12 – 110 mm Tugas Akhir Perencanaan Struktur Gedung Perpustakaan 2 Lantai

5.7. Penulangan lapangan arah x

Mu = 870,248 kgm = 8,7025.10 6 Nmm Mn =  M u = 6 6 10 . 878 , 10 8 , 10 . 7025 , 8  Nmm Rn =  2 .d b Mn    2 6 94 . 1000 10 . 878 , 10 1,2312 Nmm 2 m = 294 , 11 25 . 85 , 240 . 85 ,   c f fy  perlu =       fy Rn . m 2 1 1 . m 1 =       240 2312 , 1 . 294 , 11 . 2 1 1 . 294 , 11 1 = 0,0053   max   min , di pakai  perlu = 0,0053 As =  perlu . b . d = 0,0053. 1000 . 94 = 498,2 mm 2 Digunakan tulangan  12 = ¼ .  . 12 2 = 113,04 mm 2 S = per lu As A 1000 . = 2 , 498 1000 . 04 , 113 = 226,89 ~ 200 mm Jarak maksimum = 2 x h = 2 x 120 = 240 mm As yang timbul =5. ¼ .  . 12 2 = 565,2 As ….…ok Dipakai tulangan  12 – 200 mm Tugas Akhir Perencanaan Struktur Gedung Perpustakaan 2 Lantai

5.8. Penulangan lapangan arah y