Penulangan Lapangan Arah X Penulangan Lapangan Arah Y

 Jumlah tulangan n n = 2 . . 4 1 d As  = 2 10 . 14 , 3 . 4 1 5 , 237 = 3,025 = 4 bh  Jarak tulangan s = n b = 4 1000 = 250  240 mm  Jarak maksimum s = 2 x h = 2 x 120 = 240 mm  As yang timbul = 4 x ¼ x 3,14 x 10 2 = 314 As ………ok  Jadi digunakan  10 – 240 mm

b. Penulangan Lapangan Arah Y

b = 1000 mm, d = 85 mm  Mu = Mly = 409,292 kg m = 409,292.10 4 N.mm  Mn = 8 . 10 292 , 409 4 x Mu   = 511,615. 10 4  Rn = 2 .d b Mn = 2 4 85 1000 10 615 , 511 x x = 0,708 Nmm 2  m = 294 , 11 25 . 85 , 240 . 85 ,   c f fy   perlu =         fy m Rn x m . . 2 1 1 1 =         240 708 , . 294 , 11 . 2 1 1 294 , 11 1 x = 0,003   perlu   min digunakan  perlu = 0,003  As =  . b . d = 0,003 . 1000 . 85 = 255 mm 2  Jumlah tulangan n n = 2 . . 4 1 d As  = 2 10 . 14 , 3 . 4 1 255 = 3,24  4 bh  Jarak tulangan 1 m 2 s = n b = 4 1000 = 250  240 mm  Jarak maksimum s = 2 x h = 2 x 120 = 240 mm  As yang timbul = 4 x 14 x 3,14 x 10 2 = 314 As ………ok  Jadi digunakan  10 – 240 mm

c. Penulangan Tumpuan Arah X Tumpuan Tengah

b = 1000 mm, d = 95 mm  Mu = Mtx = 789,350 kg m = 789,350.10 4 N.mm  Mn = 8 . 10 350 , 789 4 x Mu   = 986,687.10 4  Rn = 2 .d b Mn = 2 4 95 1000 10 986,687. x = 1,093 Nmm 2  m = 294 , 11 25 . 85 , 240 . 85 ,   c f fy   perlu =         fy m Rn x m . . 2 1 1 1 =         240 294 , 11 . 093 , 1 . 2 1 1 294 , 11 1 x = 0,0046   perlu   min digunakan  perlu = 0,0046  As =  . b . d = 0,0046 . 1000 . 95 = 437 mm 2  Jumlah tulangan n n = 2 . . 4 1 d As  = 2 10 . 14 , 3 . 4 1 437 = 5,56  6 bh  Jarak tulangan s = n b = 6 1000 = 166  160 mm  Jarak maksimum s = 2 x h = 2 x 120 = 200 mm  As yang timbul = 6 x ¼ x 3,14 x 10 2 = 471 As ………ok  Jadi digunakan  10 – 160 mm

d. Penulangan Tumpuan Arah Y Tumpuan Tengah

b = 1000 mm, d = 85 mm  Mu = Mty = 877,056 kg m = 877,056.10 4 N.mm  Mn = 8 . 10 056 , 877 4 x Mu   = 1096,32.10 4  Rn = 2 .d b Mn = 2 4 85 1000 10 1096,32. x = 1,517 Nmm 2  m = 294 , 11 25 . 85 , 240 . 85 ,   c f fy   perlu =         fy m Rn x m . . 2 1 1 1 =         240 294 , 11 . 517 , 1 . 2 1 1 294 , 11 1 x = 0,0065