Penulangan Tumpuan Arah X Tumpuan Tengah Penulangan Tumpuan Arah Y Tumpuan Tengah Penulangan Tumpuan Arah X

 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   perlu   min digunakan  perlu = 0,0065  As =  . b . d = 0,0065 . 1000 . 85 = 552,5 mm 2  Jumlah tulangan n n = 2 . . 4 1 d As  = 2 10 . 14 , 3 . 4 1 552,5 = 7,03  8 bh  Jarak tulangan s = n b = 8 1000 = 125 mm  Jarak maksimum s = 2 x h = 2 x 120 = 240 mm  As yang timbul = 8 x ¼ x 3,14 x 10 2 = 628 As ………ok  Jadi digunakan  10 – 160 mm

e. Penulangan Tumpuan Arah X

Tumpuan Tepi b = 1000 mm, d = 95 mm  Mu = Mtix = 328,896kg m = 328,896.10 4 N.mm  Mn = 8 . 10 896 , 328 4 x Mu   = 411,12.10 4  Rn = 2 .d b Mn = 2 4 95 1000 10 411,12. x = 0,455 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 . 455 , . 2 1 1 294 , 11 1 x = 0,0019   perlu  min digunakan  min = 0,0025  As =  . b . d = 0,0025 . 1000 . 95 = 237,5 mm 2  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

f. Penulangan Tumpuan Arah Y

Tumpuan Tepi b = 1000 mm, d = 68 mm  Mu = Mtiy = 279,561 kg m = 279,561.10 4 N.mm  Mn = 8 . 10 561 , 279 4 x Mu   = 349,451.10 4  Rn = 2 .d b Mn = 2 4 85 1000 10 349,451. x = 0,483 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 . 483 , . 2 1 1 294 , 11 1 x = 0,00203   perlu  min digunakan  min = 0,0025  As =  . b . d = 0,0025 . 1000 . 85 = 212,5 mm 2  Jumlah tulangan n n = 2 . . 4 1 d As  = 2 10 . 14 , 3 . 4 1 212,5 = 2,70  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 REKAPITULASI TULANGAN  Tulangan Lapangan Arah X =  10 – 240 mm  Tulangan Lapangan Arah Y =  10 – 240 mm  Tulangan Tumpuan Tengah arah X =  10 – 160 mm  Tulangan Tumpuan Tengah Arah Y =  10 – 160 mm  Tulangan Tumpuan Tepi Arah X =  10 – 240 mm  Tulangan Tumpuan Tepi Arah Y =  10 – 240 mm