PERENCANAAN BALOK 1. Balok 30 x 60
90
Gambar 4.14. Distribusi beban nominal statik ekuivalen arah-x
Gambar 4.15. Distribusi beban nominal statik ekuivalen arah-y
4.4. PERENCANAAN BALOK 4.4.1. Balok 30 x 60
Data – data balok : Tinggi balok h
: 600 mm Lebar balok b
: 300 mm Selimut Beton p
: 20 mm Diameter tul. Utama
: 20 mm Diameter tul. Sengkang
: 10 mm Mutu tulangan fy
: 240 Mpa
91
Mutu Beton fc : 24 Mpa
Gaya rencana yang dipakai adalah gaya maksimum pada batang : P
= 89150 N Vu
= 149820 N Tu
= 25733283.1 Nmm Mu
= 184604805 Nmm Penulangan
Longitudinal d = h – selimut beton – sengkang – ½ Ø tulangan
= 600 – 40 – 10 – 222 = 539 mm
Penulangan Pada Momen Mu
= 184604805 Nmm Mu Mr = Ø . Mn
Mu = Ø . Mn Mn
= φ
Mu
0,85 . fc’ . a . b . d – ½ a = φ
Mu
0,85 . 24 . 300 . 539 – ½ a = 8
, 184604805
6120 . a . 539 – ½ a = 230756006.3
539 a – ½ a
2
=
6120 3
. 230756006
539 a – ½ a
2
= 37705.22979
x 2
92
1078a – a
2
= 75410.45958
a
2
– 1078a + 75410.45958 = 0
a
12
= 2
. .
4
2
c a
b b
− ±
−
a
12
= 2
45958 ,
75410 .
1 .
4 1078
1078
2
− −
± −
−
= 2
8383 ,
301641 1162084
1078 −
±
= 2
1617 ,
860442 1078
±
=
2 6
, 927
1078 ±
= 75,2
mm
C = T 0,85 . fc’ . a . b
= As . fy 0,85 . 24 . 75,2 . 300 = As . 240
460224 = As
. 240
As =
240 460224
As =
1917.6 mm
2
As tulangan D22 = ¼ . 3,14 . D
2
= ¼ . 3,14 . 22
2
= 379,94 mm
2
Dipakai 6 Ø 22 As = 2279,64 mm
2
93
ρ =
d b
As .
= 539
. 300
6 ,
1917
= 0.012 ρ
max
= 0,75
. ρ
b
= 0,75
.
⎟⎟ ⎠
⎞ ⎜⎜
⎝ ⎛
+ fy fc
600 600
. 240
. .
85 ,
1
β
= 0,75
. ⎟
⎠ ⎞
⎜ ⎝
⎛ + 240
600 600
. 240
85 ,
. 24
. 85
,
= 0,0387
ρ
min
= fy
4 ,
1 =
240 4
, 1
= 0.0058 ρ
min
ρ
aktual
ρ
maks
0.0058 0.012 0.0387 Dipakai 6 Ø 22
As = 2279,64 mm
2
Kontrol spasi
=
2 22
. 6
20 300
− −
= 74 mm Penulangan
geser Vu
= 149820 N d = h – selimut beton – sengkang – ½ Ø tulangan
= 600 – 40 – 10 – 222 = 539 mm
Vc = 16 .
fc
. b
w
. d
94
= 16
. 24 . 300 . 539
= 132027,4971
N Vs
= 6
, .vc
Vu φ
−
= 6
, 4971
, 132027
. 6
, 149820
−
= 6
, 49826
, 79216
149820 −
= 117672,5029
N • Vu ½ . Ø . Vc
149820 ½ . 0,6 . 132027,24913 149820 39608,24913
Tidak memenuhi Syarat • ½ . Ø . Vc
≤ Vu ≤ Ø . Vc ½ . 0,6 . 132027,4971
≤ 149820 ≤ 0,6 . 132027,4971 39608,24913
≤ 149820 ≥ 79216,49826 Tidak Memenuhi Syarat
• Ø . Vc ≤ Vu
0,6 . 132027,4971 ≤ 149820
79216,49826 ≤ 149820 memenuhi syarat
• Vs 13 .
fc
. b
w
. d 117672,5029 13 . 24 . 300 . 539
117672,5029 264054,9943 memenuhi syarat Dimensi sudah memenuhi syarat
S maks = d2
95
= 539 2 =
269,5 mm
→ dipakai 150 mm Penulangan
geser minimum
Av =
3 1
. fy
s b
w
.
=
3 1
.
240 150
. 300
= 62,5
mm
2
Jadi dipakai
Ø 10 – 150
4.5. PERENCANAAN KOLOM 4.5.1. Kolom K4 60x60