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Perhitungan Manual Balok Anak
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Perhitungan Manual Balok Anak
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ANALISA BALOK ANAK
Bab Balok Anak
perencanaan balok
Lx
Ly
A.
P1
× =½×
Mmax
= P1
P
b/2
=½
P1
P1
qa
×
P=¼× P
( b/2 – 1/3
×
=¼(½
P1
×
b/2
×
=(¼ P1
b
P
×
× b
×
Mmax
qeq
b ) ( b/3 )
×
qa )
×
= Meq-max
1/24 × qa × b2 = ⅛ × qeq × b2 qeq
Lx=b
= ⅓ × qa × b
= ⅓ × qa × L x
qeq
Ly-b
b/2
x
P1+P2
P1
= (P1 + P2)
×
Dima Dimana na : P1
P3
P2
Ly/2 – P1 (Ly/2 Ly - ⅔ =½
⋅
b/2
P
⋅
b/2
P4
⋅
P1+P2
b/2) – P2 (Ly –
b
b/3 )
Meq max = ⅛ × qeq × b Lx=b
×
b
×
b/3
P2
P ⋅ ( Ly 2
=
Meq-max = ⅛ Mmax
⋅
− b)
qeq
⋅
Ly2
⋅
Lx2/Ly2)
= Meq-max
qeq = ½ × qa × Lx (1 - ⅓
4.2. 35/55
Lx = 350 – (35/2 + 25/2) = 320 cm
Ly = 500 – (35/2 + 35/2) = 465 cm 0 4 / 5 2
5 5 / 5 3
5 5 / 5 3
3.50
0 0 . 5
3.50
Dari data perhitungan pembebanan plat didapat : qd = 402 kg/m2 ql = 100 kg/m2
Beban mati (qd)
Berat sendiri balok= 0.25 x 0.28 x 2400 Beban mati plat = 2
×
½
×
402
×
= 168
3.20 × (1 - ⅓
⋅
(3.20/4.65)2) = 1083.33 kg/m2 qd
Beban
×
100
×
3.20
×
(1 - ⅓
⋅
(3.20/4.65) 2)
= 269.48 kg/m2 Beban
= 1251.33 kg/m2
hidup (ql)
Beban hidup plat = 2 × ½
kg/m2
berfaktor (qu)
qu = 1.2 x qd + 1.6 x ql = 1.2 x 1251.33 + 1.6 x 269.48 = 1520.81 kg/m2
Koefisien momen dan gaya lintang (sesuai SNI 03-2847-2002 Pasal
10.3.3.5) -1 / 16
A
- 1 / -1 11 / 1 1
- 1 / -1 10 / 1 1 +1 / 14
+1 / 14
B
C
- 1 / 1- 11 / 1 1 +1 / 14
- 1 / 1- 11 / 1 0 +1 / 14
D
E
-1 / 11 +1 / 14
Momen o
Bentang Ujung
M A
qu ⋅ l 2
=
M A-B
=
16 qu ⋅ l 2
=
=
14 qu ⋅ l 2
=
1520 .81 x 4.65 2 16 1520 .81 x 4.65 2 14
1520 .81 x 4.65 2
MB
=
•
Bentang Dalam
MB
10
qu ⋅ l 2
=
MB-C
MC
=
Contoh
=
11 qu ⋅ l 2
=
=
16 qu ⋅ l 2
=
11
10
11 1520 .81 x 4.65 2 16
1520 .81 x 4.65 2 11
2055.232
kgm
=
2348.837
kgm
=
3288.371
kgm
=
2989.429
kgm
=
2055.232
kgm
=
2989.429 kgm
Perhitungan Balok Anak Atap :
Data-data : - b
ρ b =
1520 .81 x 4.65 2
=
= 250 mm
d = 400 – (40 + 8 + 1/2 * 12) = 346 mm
-
h
= 400 mm
Tulangan Utama = D 12 mm
-
fc’
= 35 MPa
Tulangan Sengkang = D 8 mm
-
fy
= 400 MPa
0.85β 1 fc ' 600
= + 600 fy
fy
0.85 x 0.81 x 35 400
ρ max = 0.75 ρ b = 0.75 x 0.036 = 0.027 ρ min
=
1.4 fy
1.4 =
400
= 0.0035
600 = 0.036 600 + 400
F
m
=
fy 0.85 fc '
=
400 0.85 x 35
= 13.45
• Tumpuan
Mn
=
Rn
=
ρperlu
=
A
2055.232 × 9,81 × 1000 Mu = = 25202282.4 Nmm φ 0.8 Mn bd
2
=
25202282.4 250 × 346
= 0.842 N/mm2
2
1 1 − 1 − 2m × Rn = 1 1 − 1 − 2 × 13.45 × 0.842 13.45 m fy 400
= 0.0021
ρpakai Aspakai
× = ρ⋅ b ⋅ = 1.33
0.0021
= 0.0028 <
d = 0.0035 × 250
×
ρmin
346
= 302.75 mm2 Maka dipasang tulangan 4-D12
0 6 0 4 4 3 2-D12
250
Lapangan A – B
Mn =
Rn
ρperlu
2348.837 × 9.81 × 1000 Mu = = 28802613.71 N-mm φ 0 .8
=
Mn bd
2
=
28802613.7 1 250 × 346
2
= 0.0.962 N/mm2
1 2 × 13.45 × 0.962 1 2m × Rn 1 1 − − 1 1 − − = = 13.45 m fy 400
= 0.0024
ρpakai Aspakai
= 1.33
×
0.0024
ρ⋅
⋅
d = 0.0035× 250
=
b
= 0.0032 <
×
ρmin
346
= 302.75 mm2 Maka dipasang tulangan
2-D12
0 6 0 4 4 3
4-D12
250
Tumpuan B
Mn =
3288.371 × 9.81 × 1000 Mu = = 40323649.39 N-mm φ 0.8
Rn
ρperlu
=
=
Mn
=
bd2
40323649.3 9 250 × 346
2
= 1.347 N/mm2
1 1 − 1 − 2m × Rn = 1 1 − 1 − 2 × 13.45 × 1.347 13.45 m fy 400 = 0.0034
ρpakai Aspakai
= 1.33
×
0.0034
ρ⋅
⋅
d = 0.0045 × 250
=
b
= 0.0045 >
×
ρmin
346
= 389.25 mm2 Maka dipasang tulangan 4-D12
0 6 0 4 4 3 2-D12
250
o
Tumpuan
Mn
=
Rn
=
ρperlu
=
B
2989.429 × 9,81 × 1000 Mu = = 36657873.11 Nmm φ 0.8 Mn bd
=
2
36657873.11 250 × 346
2
= 1.225 N/mm2
1 1 − 1 − 2m × Rn = 1 1 − 1 − 2 × 13.45 × 1.225 13.45 m fy 400
= 0.0031
ρpakai Aspakai
= 1.33
×
0.0031
ρ⋅
⋅
d = 0.004 × 250
=
b
= 0.004 >
×
ρmin
346
= 346 mm2 Maka dipasang tulangan 4-D12
0 6 0 4 4 3 2-D12
250
Lapangan B - C
Mn
=
Rn
=
ρperlu
=
2055.232 × 9.81 × 1000 Mu = = 25202282.4 N-mm φ 0.8 Mn bd
=
2
1 − m 1
25202282.4 250 × 346
1−
2
2m × Rn fy
= 0.842 N/mm2
=
1 − 13.45 1
= 0.0021
ρpakai Aspakai
= 1.33
×
0.0021
ρ⋅
⋅
d = 0.0035 × 250
=
b
= 302.75 mm2
= 0.0028 <
×
ρmin
346
1−
2 × 13.45 × 0.842 400
Maka dipasang tulangan
2-D12
0 6 0 4 4 3
4-D12
250
Tumpuan
Mn
=
Rn
=
ρperlu
=
C
2989.429 × 9,81 × 1000 Mu = = 36657873.11 Nmm φ 0.8 Mn bd
=
2
36657873.1 1 250 × 346
2
= 1.225 N/mm2
1 1 − 1 − 2m × Rn = 1 1 − 1 − 2 × 13.45 × 1.225 13.45 m fy 400
= 0.0031
ρpakai Aspakai
= 1.33
×
0.0031
ρ⋅
⋅
d = 0.004 × 250
=
b
= 0.004 >
×
ρmin
346
= 346 mm2 Maka dipasang tulangan 4-D12
0 6 0 4 4 3 2-D12
250
×
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