perhitungan gempa
TRANSCRIPT
Tahapan perhitungan
PERHITUNGAN BEBAN DAN GAYA GESER
Data : Kota Sukabumi (Wilayah Gempa 4)
Tanah= Keras
Daktail Penuh
L1= 425 cmH1= 500 cm
L2= 425 cmH2= 400cm
L3= 500 cmH3= 350 cm
K1= 45/45 cm
P.atap= 10 cm K2= 40/40 cm
P.lantai = 14 cmK3= 30/40cm
B1= 20/50 cm
B2= 25/50 cm
B3= 30/50 cm
2.1 Perhitungan Beban
Beton tanpa dinding geser :
T = 0,0731 x H (3/4)
= 0,0731 x 16,5 (3/4)
= 0,598Dari grafik dibawah ini didapat :
T = 0,598 dan C = 0,30 = 0,501
0,598
2.1.1 Beban Plat Atap / Lantai 4 (W4)
B1
100
500
400
250B2 100
500
400
200
B3 100
500
400
300
Berat atap (t =10 cm)
q1 = 17 x 23,5 x 0,1 x 2400 =95880Kg Balok melintang
q2 = 0,25 x 0,40 x 4,25 x 2400 x 12 = 12240 = 0,20 x 0,40 x 4,25 x 2400 x 12 = 9792 = 22032 kg Balok Memanjang
q3 = 0,30 x 0,40 x 23,5 x 5 x 2400 = 33840 kg
Berat langit-langit dan gantungannya
q4= 17 x 23,5 x 18 =7191 Kg Penutup aspal dan pasir
q5 = 17 x 23,5 x 0,01 x 1400 =5593 Kg Air Hujan ( t = 5 cm)
q6= 17 x 23,5 x 0,05 x 1000 =1997Kg
Beban Kolom (1/2 k3)
q7 = 0,30 x 0,40 x H x 2400 x 30 = 0,30 x 0,40 x 3,50 x 2400 x 30 = 15120Kg
Beban Hidup
q7 = 17 x 23,5 x 100 x 0,3 = 11985Kg +
Total W4 (q )= 211616 Kg2.1.2 Beban Lantai 3 (t = 14) B1
140
500
360
250
B2
140
500
360
200
B3
140
500
360
300 Berat plat lantai (t=14 cm)
q1 = 17 x 23,5 x 0,14 x 2400 =134232Kg Balok melintang
q2 = 0,25 x 0,36 x 4,25 x 2400 x 12= 11016 = 0,20 x 0,36 x 4,25 x 2400 x12= 8812,8
= 19828,8 kg Balok Memanjang q3 = 0,30 x 0,36 x 23,5 x 5 x 2400= 30456 Kg
Berat langit-langit dan gantungannya
q4= 17 x 23,5 x 18 = 7191Kg Beban Kolom (1/2 k4)
q7 = 0,40 x 0,40 x H4 x 2400 x 30 = 0,40 x 0,40 x 3,50 x 2400 x 30= 20160 Kg
Beban Kolom (1/2 k3)
q7 = 0,40 x 0,40 x H3 x 2400 x 30 = 0,40 x 0,40 x 4 x 2400 x 30= 23040 Kg
Keramik + adukan ( t = 5 cm)
q12= 17 x 23,5 x 5 x 24 kg/m2 = 47940 Kg
Beban Hidup
q14 = (17 x 23,5 x 250 x 0,3)= 29962,5Kg
Total W3 = 312810,3 kg2.1.3 Beban Lantai 2 (T=14 cm)
B1 = B4
140
500
360
250
B2
140
500
360
200
B3
140
500
360
300 Berat plat lantai (t=14 cm)
q1 = 17 x 23,5 x 0,14 x 2400 = 134232Kg
Balok melintang
q2 = 0,25 x 0,36 x 4,25 x 2400 x 12= 22032 = 0,20 x 0,36x 4,25 x 2400 x 12 = 8812,8
= 30844,8 kg Balok Memanjang
q3 = 0,30 x 0,36 x 23,5 x 5 x 2400=30456 Kg
Berat langit-langit dan gantungannya
q4= 17 x 23,5 x 18 =7191 Kg Beban Kolom (1/2 k3)
q7 = ,40 x 0,40 x H3 x 2400 x 30 = 0,40 x 0,40 x 4 x 2400 x 30 = 23040 Kg
Beban Kolom (1/2 k2)
q7 = 0,40 x 0,40 x H2 x 2400 x30 = 0,40 x 0,40 x 4 x 2400 x 30= 23040 Kg
Keramik + adukan ( t = 5 cm)
q12= 17 x 23,5 x 5 x 24 kg/m2 = 47940 Kg
Beban Hidup
q14 = (17 x 23,5 x 250 x 0,3)= 29962,5Kg
Total W2 = 326706,3kg
2.1.4 Beban Lantai 1 (T=14 cm)
B1 = B4
140
500
360
250
B2
140
500
360
200
B3
140
500
360
300 Berat plat lantai (t=14 cm)
q1 = 17 x 23,5 x 0,14 x 2400 = 134 232Kg
Balok melintang
q2 = 0,25 x 0,36 x 4,25 x 2400 x 12= 11016 = 0,20 x 0,36 x 4,25 x 2400 x 12 = 8812,8
= 198 288 kg Balok Memanjang
q3 = 0,30 x 0,36 x 23,5 x 5 x 2400= 30456 Kg
Berat langit-langit dan gantungannya
q4= 17 x 23,5 x 18 = 7191 Kg Beban Kolom (1/2 k2)q7 = 0,40 x 0,40 x H2x 2400 x30 = 0,40 x 0,40 x 4 x 2400 x 30= 23040 Kg
Beban Kolom (1/2 k1)
q7 = 0,45 x 0,45 x H1 x 2400 x 30 = 0,45 x 0,45 x 5 x 2400 x 30= 36450 Kg
Keramik + adukan ( t = 5 cm)
q12= 17 x 23,5 x 5 x 24 kg/m2 = 47940 Kg
Beban Hidup
q14 = (17 x 23,5 x 250 x 0,3 )= 29962,5 Kg
Total W1 = 507559,5kg
LantaiWi (kg)hi(m)Wi x hi
4211.616,016,53.491.664
3312.810,30134.066.534
2326.706,3092.940.357
1507.559,5052.537.798
wi = Wt1.358.692wi x hi13.036.352
Setiap Portal menerima gaya V sebanyak (Melintang) :
2.2 Perhitungan Beban Horizontal Terpusat Melintang
Total = F1 + F2 + F3 + F4 = 13354,94 KgKontrol :
F = V 13354,94 = 13354,94 .. OK
2.3 Perhitungan gaya geser tingkat (Qi)
Gbr 2-6 pembagian gaya geserQ4 = F4 = 3576,99 Kg
Q3 = Q4 + F3 = 3576,99 + 4165,91 = 7742,91KgQ2 = Q3 + F2 = 7742,91 + 3012,21 = 10755,12Kg Q1 = Q2 + F1 = 10755,12 + 2599,82 = 13354,94Kg2.4 Perhitungan Beban Horizontal Terpusat Memanjang
Setiap Portal menerima gaya V sebanyak (Memanjang) :
lantaiWi (kg)hi(m)Wi x hi
4211.616,016,53.491.664
3312.810,30134.066.534
2326.706,3092.940.357
1507.559,5052.537.798
wi = Wt1.358.692wi x hi13.036.352
Setiap Portal menerima gaya V sebanyak (Memanjang) :
Total = F1 + F2 + F3 + F4 = 16025,92 KgKontrol :
F = V 16025,92 = 16025,92 .. OK
2.3 Perhitungan gaya geser tingkat (Qi)
Q4 = F4 = 4292,39Kg
Q3 = Q4 + F3 = 4292,39 + 4999,09 = 9291,49KgQ2 = Q3 + F2 = 9291,49+ 3614,66 = 12906,14Kg Q1 = Q2 + F1 = 12906,14 + 3119,78 = 16025,92 KgPERHITUNGAN D -VALUE PORTAL MELINTANG3.1 Tingkat Satu
3.1.1 Kolom pinggir
= 2450,98 E
= 2733,75 E
D1.1 = D1.5= 0,482 . 2733,75 E = 1318,06 E
3.1.2 Kolom Tengah 1
= 2450,98 E
= 1960,78 E
= 2733,75 E
D1.2 = D1.4 = 0,58 . 2733,75 = 1.599,04E 3.1.2 Kolom Tengah 2
= 1960,78 E
= 1960,78 E
= 2733,75 E
D3 = 0,5633 . 2733,75 = 153980 E
Gaya geser setiap kolom pinggir :
Gaya geser setiap kolom tengah :
Gaya geser setiap kolom tengah 2 :
3.2 Tingkat Dua
3.2.1 Kolom pinggir
= 2450,98E
= 2450,98E
D.2.1= D.2.5= 0,53 . 2133,33 E = 1140,58E
3.2.2 Kolom tengah
k1 = k2 = 2450,98 E
= 1960,78 E
D2.2= D2.4 = 0,51. 2133,33 E = 1085 E
3.2.2 Kolom tengah 2
= 1960,78 E
= 1960,78 E
D3.2= 0,4789. 2133,33 E = 1021,71 E
Gaya geser setiap kolom pinggir :
Gaya geser setiap kolom tengah :
Gaya geser setiap kolom tengah :
3.3 Tingkat Tiga
3.3.1 Kolom pinggir
= 2450,98E
= 2450,98E
D3.1=D3.5 = 0,53 . 2133,33 E = 1140,58E
3.3.2 Kolom tengah
= 2450,98E
k3 = k4 = 1960,78 E
D3.2= D3.4= 0,51. 2133,33 E = 1085 E
3.3.3 Kolom tengah 2
= 1960,78 E
= 1960,78 E
D3.3= 0,4789. 2133,33 E = 1021,71 E
Gaya geser setiap kolom pinggir :
Gaya geser setiap kolom tengah :
Gaya geser setiap kolom tengah :
3.4 Tingkat Empat 3.4.1 Kolom pinggir
= 2450,98E
= 2450,98E
D4.1=D4.5 = 0,40 . 1828,57 E = 733,74E
3.4.2 Kolom tengah
= 2450,98E
=1960,78 E
D4.2 =D4.4 =0,51. 1828,57 E = 1084,50 E
3.4.3 Kolom tengah 2
=1960,78 E
=1960,78 E
D4.3= 0,4789. 1828,7 E = 875,75 E
Gaya geser setiap kolom pinggir :
Gaya geser setiap kolom tengah :
Gaya geser setiap kolom tengah :
PERHITUNGAN D-VALUE PORTAL MEMANJANG
5.1 Tingkat Satu
5.1.1.Kolom Pinggir
K1
KC
5.1.2 Kolom Tengah - 1
K1
K2
KC
5.1.3 Kolom Tengah - 2
K1
K2
KC
5.1.4Gaya Geser Pada Setiap Kolom Pinggir
5.1.5 Gaya Geser Pada Setiap Kolom Tengah
5.2 Tingkat Dua
5.2.1Kolom Pinggir
K1
Kc
K2
K2 = 1960,78 E
5.2.2 Kolom Tengah 1
K1 K2
Kc
K3 K4
5.2.3 Kolom Tengah 2
K1 K2
Kc
K3 K4
5.2.4 Gaya Geser Pada Setiap Kolom Pinggir
5.2.5 Gaya Geser Pada Setiap Kolom Tengah
5.3 Tingkat Tiga
5.3.1.Kolom Pinggir
K1
Kc
K2
K2 = 1960,78 E
5.3.2 Kolom Tengah 1
K1 K2
Kc
K3 K4
5.3.3 Kolom Tengah 2
K1 K2
Kc
K3 K4
5.2.4 Gaya Geser Pada Setiap Kolom Pinggir
5.2.5 Gaya Geser Pada Setiap Kolom Tengah
5.4 Tingkat Empat
5.4.1.Kolom Pinggir
K1
Kc
K2
5.4.2 Kolom Tengah -1
K1 K2
Kc
K3 K4
5.4.3 Kolom Tengah 2
K1 K2
Kc
K3 K4
5.3.3 Gaya Geser Pada Setiap Kolom Pinggir
5.3.4 Gaya Geser Pada Setiap Kolom Tengah
PERHITUNGAN MOMEN PORTAL MELINTANG
4.1 Perhitungan Momen Ujung Kolom
Rumus dari tabel Kyoshi Muto
GAB Bekerja Pada Titik Balok AB
Ma = GAB (1 y) h
Mb = GAB (yh)
Y = y0 + y1 + y2 + y3
1. Mencari Y0 Bangunan tingkat 4:
MENCARI Yo ( dari tabel Kyosi Muto )
TingkatKolom tepikolom tengahkolom tengah
K-YoK-YoK-Yo
10,8970,651,610,571,430,579
22,2980,502,070,501,840,492
32,2980,4652,070,4541,840,45
41,3400,3672,410,4211,840,392
2. Mencari y1 Bangunan Bertingkat 4Untuk mencari nilai y1, dapat ditentukan dengan cara menginterpolasikan data yang ada pada tabel 2. Nilai y1 tergantung harga k dan kekakuan balok atas + bawahnya.
Terbawah y1=0
Jika (k1+k2) > (k3+k4), maka dan harga y1 menjadi negatif.
Tabel y1 untuk kolom tepi dan kolom tengah
MENCARI Y1 ( dari tabel Kyosi Muto )
TingkatKolom tepikolom tengahkolom tengah
K1K21Y1K1+K2K3+K41Y1K1+K2K3+K41Y1
22450,982450,98104901,964901,96109803,929803,9210
32450,982450,98104901,964901,96109803,929803,9210
42450,982450,98104901,964901,96109803,929803,9210
3. Mencari y2 Bangunan Bertingkat 4
Untuk mencari nilai y2, dapat ditentukan dengan cara menginterpolasikan data yang ada pada tabel 3. Nilai y2 tergantung harga k, perbandingan kolom atas (ha) terhadap kolom lantai itu sendiri (h). Kolom teratas , yi = 0.
ha = tinggi kolom d atasnya
1 y2 = 0
Tabel y2 untuk kolom tepi dan kolom tengah
MENCARI Y2 ( dari tabel Kyosi Muto )
tingkatha (cm)h (cm)2kolom tepikolom tengahkolom tengah
K_y2K-YoK-Yo
14005000,80,8965601,613801,4345020
240040012,2977902,068001,8382350
33504000,8752,2977902,068001,8382350
4. Mencari y3 Bangunan Bertingkat 4
Untuk mencari nilai y3, ditentukan dengan cara menginterpolasikan data yang ada pada tabel 4 . Nilai y3 tergantung harga k, perbandingan kolom bawah (hb) terhadap kolom lantai itu sendiri (h).
Kolom terbawah , y3 = 0, 1 y3 = 0
Tabel y3 untuk kolom tepi dan kolom tengah
MENCARI Y3 ( dari tabel Kyosi Muto )
tingkathb (cm)h (cm)3kolom tepikolom tengahkolom tengah
K_y2K-YoK-Yo
25004001,252,2977902,0701,8382350
34003501,14292,29779402,0701,8382350
435035011,34002,4101,8382350
Resume y0, y1, y2 dan y3
RESUME
TingkatKolomY0Y1Y2Y3Y = Y0+Y1+Y2+Y3
1tepi0,650000,65
tengah10,570000,57
tengah20,579
2tepi0,500000,50
tengah10,500000,50
tengah20,50
3tepi0,4920000,492
tengah10,4540000,454
tengah20,450
4tepi0,3670000,367
tengah10,421
tengah20,3920000,392
4.1.1 Perhitungan Momen Ujung Kolom Lantai Satu
h = 500 cm = 5 m
Kolom Tepi
G11 = 2387,121y= 0,65Mbawah = G11 . y . h = (2387,121) . 0,65 . 5 = 7758,144 kg.m
Matas = G11 . (1 y) .h = (2387,121) . 0,35 . 5 = 4177,462 kg.m
Kolom Tengah 1G21 = 2895,995
y= 0,57
Mbawah = G21 . y . h = (2895,995). 0,57 . 5 = 8253,586kg.mMatas = G21.(1 y) . h = (2895,995) .0,43 . 5 = 6226,390kg.m
Kolom Tengah 2
G21 = 1889,102
y= 0,579
Mbawah = G21 . y . h = (1889,102). 0,579 . 5 = 8073,294kg.mMatas = G21.(1 y) . h = (1889,102) .0,421 . 5 = 5870,219kg.m4.1.2 Perhitungan Momen Ujung Kolom Lantai Dua
h = 400 cm = 4 m
Kolom Tepi
G12 = 1889,102y= 0,50Mbawah = G12 . y . h = (1889,102) . 0,50 . 4 = 3778,205kg.mMatas = G12 . (1 y) . h = (1889,102) . 0,50 . 4 = 3778,205kg.m Kolom Tengah
G22 = 1796,227y= 0,50Mbawah = G22 . y . h = (1796,227) . 0,50 . 4 = 2397,581kg.mMatas = G22 . (1 y) . h = (1796,227) . 0,50 . 4 = 2475,552kg.m
Kolom Tengah
G22 = 1218,283
y= 0,492
Mbawah = G22 . y . h = (1218,283) . 0,492 . 4 = 2397,581kg.mMatas = G22 . (1 y) . h = (1218,283) . 0,508 . 4 = 2475,552kg.m4.1.3 Perhitungan Momen Ujung Kolom Lantai Tiga
h = 400 cm = 4 m
Kolom Tepi
G13 = 1360,017y= 0,465
Mbawah = G12 . y . h = (1360,017) . 0,465 . 4 = 2529,631kg.mMatas = G12 . (1 y) . h = (1360,017) . 0,535 . 4 = 2910,436kg.m Kolom Tengah 1G23 = 1293,153y= 0,454Mbawah = G22 . y . h = (1293,153) . 0,454. 4 = 2348,366kg.mMatas = G22 . (1 y) . h = (1293,153). 0,546. 4 = 2824,246kg.m
Kolom Tengah 2
G33 = 1452,674
y= 0,45
Mbawah = G22 . y . h = (1452,674) . 0,45. 4 = 1452,674kg.mMatas = G22 . (1 y) . h = (1452,674). 0,55. 4 = 2614,814kg.m4.1.4 Perhitungan Momen Ujung Kolom Lantai Empat
h = 350 cm = 3,5 m
Kolom Tepi
G14 = 581,661 y= 0,367Mbawah = G12 . y . h = (581,661). 0,367 . 3,5 = 747,144kg.mMatas = G12 . (1 y) . h = (581,661). 0,633 . 3,5 = 1288,670 kg.m Kolom Tengah 1G24 = 859,7176y= 0,421Mbawah = G22 . y . h = (859,7176). 0,421 . 3,5 = 1266,794kg.mMatas = G22 . (1 y) . h = (859,7176). 0,579 . 3,5 = 1742,218kg.m Kolom Tengah 2
G24 = 694,2364
y= 0,392
Mbawah = G22 . y . h = (694,2364). 0,392 . 3,5 = 1266,794kg.mMatas = G22 . (1 y) . h = (694,2364). 0,608 . 3,5 = 952,492kg.m4.2 Perhitungan Momen Ujung Balok
4.2.1 Perhitungan Momen Ujung Balok Lantai Satu
a. Kolom Tepi
b. Kolom Tengah 1
= 4363,925 Kg.m
= 5454,918 Kg.m
c. Kolom Tengah 2
= 4133,900 Kg.m
= 4133,900 Kg.m
4.2.2 Perhitungan Momen Ujung Balok Lantai Dua
a. Kolom Tepi
b. Kolom Tengah 1
= 2640,361 Kg.m
= 3300,458Kg.m
c. Kolom Tengah 2
= 2545,183 Kg.m
= 2545,183Kg.m
4.2.3 Perhitungan Momen Ujung Balok Lantai Tiga
a. Kolom Tepi
b. Kolom Tengah 1
= 1818,238 Kg.m
= 2272,802 Kg.m
b. Kolom Tengah 2
= 2074,188 Kg.m
= 2074,188 Kg.m4.2.4 Perhitungan Momen Ujung Balok Lantai Empat
a. Kolom Tepi
b. Kolom Tengah 1
= 774,318 Kg.m
= 967,900 Kg.m
c. Kolom Tengah 2
= 738,668 Kg.m
= 738,668 Kg.m
BAB 6
PERHITUNGAN MOMEN PORTAL MEMANJANG
6.1 Perhitungan Momen Ujung Kolom
Tabel 1. Mencari Y0 Bangunan tingkat 5:
TingkatKolom tepiKolom tengah 1Kolom tengah 2
K-YoK-YoK-Yo
10.3750.80.750.650.750.65
20.5250.51.050.51.050.5
31.2820.4652.560.52.560.5
41.2820.452.560.4782.560.478
51.2820.3652.560.4282.560.428
Tabel 2
TingkatKolom tepikolom tengah 1kolom tengah 2
K1K21Y1K1+K2K3+K41Y1K1+K2K3+K41Y1
23125.003125.00106250.006250.00106250.006250.0010
33125.003125.00106250.006250.00106250.006250.0010
43125.003125.00106250.006250.00106250.006250.0010
53125.003125.00106250.006250.00106250.006250.0010
Tabel 3
tingkatha (cm)h (cm)2kolom tepikolom tengah 1kolom tengah 2
K_y2K-YoK-Yo
13506000.58330.37500.7500.750
235035010.52501.0501.050
335035011.28202.5602.560
435035011.28202.5602.560
Tabel 4
tingkathb (cm)h (cm)3kolom tepikolom tengah 1kolom tengah 2
K_y2K-YoK-Yo
25003501.42860.52501.0501.050
335035011.28202.5602.560
435035011.28202.5602.560
535035011.28202.5602.560
Kesimpulan :
tingkatkolomY0Y1Y2Y3Y = Y0+Y1+Y2+Y3
1tepi0.80000.8
tengah 10.650000.65
tengah 20.650000.65
2tepi0.50000.5
tengah 10.50000.5
tengah 20.50000.5
3tepi0.4650000.465
tengah 10.50000.5
tengah 20.50000.5
4tepi0.450000.45
tengah 10.4780000.478
tengah 20.4780000.478
5tepi0.3650000.365
tengah 10.4280000.428
tengah 20.4280000.428
6.1.1 Perhitungan Momen Ujung Kolom Lantai Satu
Tingkat 1
h = 500 cm = 5 m
Kolom tepi : G11 = 9302,41Kg ; y = 0,80
Mbawah = G11 x Yh = 9302,41x 0,8 x 5 = 37209,64 Kgm
Matas = G11 x (1-Y).h = 9302,41x 0,2 x 5 = 9302,41 Kgm
Kolom tengah 1 : G21 = 11740,17 Kg ; y = 0,65
Mbawah = G21 . y . h
= 38155,54 Kgm
Matas = G21 . (1 y) . h = 20545,29 Kgm
Kolom tengah 2 : G31 = 11993,46 Kg ; y = 0,65
Mbawah = G31 . y . h
= 38978,76 Kgm
Matas = G31 . (1 y) . h = 20988,56 Kgm
6.1.2 Perhitungan Momen Ujung Kolom Lantai Dua
Tingkat 2 h = 350 cm = 3,5 m
Kolom tepi : G12 = 7103,151 Kg ; y = 0,50
Mbawah = G12 . y . h = 12430,514 Kgm
Matas = G12 . (1 y) . h = 12430,514 Kgm
Kolom tengah 1: G22 = 12298,598 Kg ; y = 0,50
Mbawah = G22 . y . h
= 21522,547 Kgm
Matas = G22 . (1 y) . h = 21522,547 Kgm
Kolom tengah 2 : G32 = 12811,040 Kg ; y = 0,50
Mbawah = G32 . y . h
= 22419,320 Kgm
Matas = G32 . (1 y) . h = 22419,320 Kgm
6.1.3 Perhitungan Momen Ujung Kolom Lantai Tiga
Kolom tepi : G13 = 7358,913 Kg ; h = 350 ; y = 0,465
Mbawah = G13 . y . h
= 11976,630 Kgm
Matas = G13 . (1 y) . h = 13779,564 Kgm
Kolom tengah 1: G23 = 10902,593 Kg ; y = 0,50
Mbawah = G23 . y . h
= 19079,538 Kgm
Matas = G23 . (1 y) . h = 19079,538 Kgm
Kolom tengah 2 : G33 = 10902,593 Kg ; y = 0,50
Mbawah = G33 . y . h
= 19079,538 Kgm
Matas = G33 . (1 y) . h = 19079,538 Kgm
6.1.4 Perhitungan Momen Ujung Kolom Lantai Empat
Tingkat 4
h = 350 cm = 3,5 m
Kolom tepi : G14 = 6441,285 Kg ; y = 0,45
Mbawah = G14 . y . h
= 10145,024 Kgm
Matas = G14 . (1 y) . h = 12399,473 Kgm
Kolom tengah : G24 = 9543.083 Kg ; y = 0,478
Mbawah = G24 . y . h
= 15965,577 Kgm
Matas = G24 . (1 y) . h = 17435,212 Kgm
Kolom tengah 2 : G34 = 9543,083 Kg ; y = 0,478
Mbawah = G34 . y . h
= 15965,577 Kgm
Matas = G34 . (1 y) . h = 17435,212 Kgm
6.1.5 Perhitungan Momen Ujung Kolom Lantai Lima
Tingkat 5
h = 350 cm = 3,5 m
Kolom tepi : G15 = 5256,016 Kg ; y = 0,365
Mbawah = G15 . y . h
= 6714,560 Kgm
Matas = G15 . (1 y) . h = 11681,495 Kgm
Kolom tengah 1: G25 =7787,048 Kg ; y = 0,428
Mbawah = G25 . y . h
= 11664,998 Kgm
Matas = G25 . (1 y) . h = 15589,670 Kgm
Kolom tengah 2 : G35 = 7787,048 Kg ; y = 0,428
Mbawah = G35 . y . h
= 11664,998 Kgm
Matas = G35 . (1 y) . h = 15589,670 Kgm
6.2 Perhitungan Momen Ujung Balok
6.2 1 Perhitungan Momen Ujung Kolom Lantai Satu
a. Kolom Tepi
b. Kolom Tengah 1
=31307,115 Kgm
= 27393,714 Kgm
c. Kolom Tengah 2
=29983,660Kgm
=29983,660Kgm
6.2.2 Perhitungan Momen Ujung Kolom Lantai Dua
a. Kolom Tepi
b. Kolom Tengah 1
= 22957,39Kgm
= 20087,71 Kg.m
c. Kolom Tengah 2
= 22419,32 Kg.m
= 22419,32 Kg.m
6.2.3 Perhitungan Momen Ujung Kolom Lantai Tiga
a. Kolom Tepi
b. Kolom Tengah 1
=20351,511Kg.m
= 20351,511Kg.m
c. Kolom Tengah 2
= 17807,565Kg.m
= 20351,511Kg.m
6.2.4 Perhitungan Momen Ujung Kolom Lantai Empat
a. Kolom Tepi
b. Kolom Tengah 1
= 17813,757Kg.m
= 17813,757Kg.m
c. Kolom Tengah 2
=15587,032Kg.m
= 17813757Kg.m
6.2.5 Perhitungan Momen Ujung Kolom Lantai Lima
a. Kolom Tepi
b. Kolom Tengah 1
= 14535,83Kg.m
= 14535,83Kg.m
c. Kolom Tengah 2
=12718,84Kg.m
= 14535,83Kg.m
BAB 7
PERHITUNGAN MOMEN MAKSIMUM
7.1 Perhitungan Momen Maksimum Pada Portal Melintang
7.1.1Perhitungan Reaksi
a. Free Body EF = GH
Perhitungan reaksi perletakan
E F
MEF
MFE
q
VEF 500 VFE
VEF= 6.068,002 kgm
VFE= - VEF = -6.068,002 kN
Perhitungan Gaya Dalam
Mx =
EMBED Microsoft Equation 3.0
Mmax =
VGH q . x = 0
x =
Maka momen max =
EMBED Microsoft Equation 3.0
= 6.068,002 .2,528 - . 2400 .(5)2 1184,54
= -15.842,603 kgm
Dari perhitungan diatas dapat di tuliskan sebagai berikut :
Menghitung Momen Maximum dan Jarak
Free Body EF = Free Body GH
q=2.400,000Kg/m
L=5,000m
MEF=1.184,540Kgm
MFE=844,530Kgm
(1/2.ql2)=30.000,000Kg
VEF=6.068,002Kg
VFE=-6.068,002Kg
x=2,528m
VEF. X=15.341,937Kgm
(1/2.qx2)=30.000,000Kg
Mmax=-15.842,603Kgm
b. Free Body FG
Menghitung Momen Maximum dan Jarak
Free Body FG
q=2.400,000Kg/m
L=4,000m
MEF=844,530Kgm
MFE=1.184,540Kgm
(1/2.ql2)=19.200,000Kg
VEF=4.714,998Kg
VFE=-4.714,998Kg
x=1,965m
VEF. X=9.263,001Kgm
(1/2.qx2)=19.200,000Kg
Mmax=-10.781,529Kgm
c. Free Body IJ
Menghitung Momen Maximum dan Jarak
Free Body IJ = Free Body KL
q=2.400,000Kg/m
L=5,000m
MEF=849,020Kgm
MFE=739,790Kgm
(1/2.ql2)=30.000,000Kg
VEF=6.021,846Kg
VFE=-6.021,846Kg
x=2,509m
VEF. X=15.109,429Kgm
(1/2.qx2)=30.000,000Kg
Mmax=-15.739,591Kgm
d. Free Body JK
Menghitung Momen Maksimum dan Jarak
Free Body JK
q=2.400,000Kg/m
L=4,000m
MEF=739,790Kgm
MFE=849,020Kgm
(1/2.ql2)=19.200,000Kg
VEF=4.772,693Kg
VFE=-4.772,693Kg
x=1,989m
VEF. X=9.491,081Kgm
(1/2.qx2)=19.200,000Kg
Mmax=-10.448,709Kgm
e. Free Body MN
Menghitung Momen Maximum dan Jarak
Free Body MN = Free Body OP
q=2.400,000Kg/m
L=5,000m
MEF=684,340Kgm
MFE=591,620Kgm
(1/2.ql2)=30.000,000Kg
VEF=6.018,544Kg
VFE=-6.018,544Kg
x=2,508m
VEF. X=15.092,863Kgm
(1/2.qx2)=30.000,000Kg
Mmax=-15.591,477Kgm
f. Free Body NO
Menghitung Momen Maximum dan Jarak
Free Body NO
q=2.400,000Kg/m
L=4,000m
MEF=591,620Kgm
MFE=684,340Kgm
(1/2.ql2)=19.200,000Kg
VEF=4.776,820Kg
VFE=-4.776,820Kg
x=1,990m
VEF. X=9.507,504Kgm
(1/2.qx2)=19.200,000Kg
Mmax=-10.284,116Kgm
g. Free Body QR
Menghitung Momen Maximum dan Jarak
Free Body QR = Free Body ST
q=2.400,000Kg/m
L=5,000m
MEF=437,880Kgm
MFE=410,720Kgm
(1/2.ql2)=30.000,000Kg
VEF=6.005,432Kg
VFE=-6.005,432Kg
x=2,502m
VEF. X=15.027,172Kgm
(1/2.qx2)=30.000,000Kg
Mmax=-15.410,708Kgm
h. Free Body RS
Menghitung Momen Maximum dan Jarak
Free Body RS
q=2.400,000Kg/m
L=4,000m
MEF=410,720Kgm
MFE=437,880Kgm
(1/2.ql2)=19.200,000Kg
VEF=4.793,210Kg
VFE=-4.793,210Kg
x=1,997m
VEF. X=9.572,859Kgm
(1/2.qx2)=19.200,000Kg
Mmax=-10.037,861Kgm
i. Free Body UV
Menghitung Momen Maximum dan Jarak
Free Body UV = Free Body WX
q=1.400,000Kg/m
L=5,000m
MEF=270,690Kgm
MFE=196,980Kgm
(1/2.ql2)=17.500,000Kg
VEF=3.514,742Kg
VFE=-3.514,742Kg
x=2,511m
VEF. X=8.823,865Kgm
(1/2.qx2)=17.500,000Kg
Mmax=-8.946,825Kgm
j. Free Body VW
Menghitung Momen Maximum dan Jarak
Free Body VW
q=1.400,000Kg/m
L=4,000m
MEF=196,980Kgm
MFE=270,690Kgm
(1/2.ql2)=11.200,000Kg
VEF=2.781,573Kg
VFE=-2.781,573Kg
x=1,987m
VEF. X=5.526,533Kgm
(1/2.qx2)=11.200,000Kg
Mmax=-5.870,477Kgm
7.2 Perhitungan Momen Maksimum Pada Portal Memanjang
a. Free Body FG = IJ
Perhitungan reaksi perletakan
F G
MFG
MGF
q
VFG 400 VGF
VFG= 4.822,775
VGF= - VFG = -4.822,775 kN
Perhitungan Gaya Dalam
Mx =
EMBED Microsoft Equation 3.0
Mmax =
VFG q . x = 0
x =
Maka momen max =
EMBED Microsoft Equation 3.0
= 4.822,775 .2,009 - . 2400 .(2,009)2 1011,64
= -10.520,324 kgm
Dari perhitungan diatas dapat di tuliskan sebagai berikut :
Menghitung Momen Maximum dan Jarak
Free Body FG = Free Body IJ
q=2.400,000Kg/m
L=4,000m
MAB=1.011,640Kgm
MBA=920,540Kgm
(1/2.ql2)=19.200,000Kg
VAB=4.822,775Kg
VBA=-4.822,775Kg
x=2,009m
VAB. X=9.691,316Kgm
(1/2.qx2)=19.200,000Kg
Mmax=-10.520,324Kgm
b. Free Body BC = CD
Menghitung Momen Maximum dan Jarak
Free Body GH = Free Body HI
q=2.400,000Kg/m
L=5,000m
MEF=920,540Kgm
MFE=1.011,640Kgm
(1/2.ql2)=30.000,000Kg
VEF=5.981,780Kg
VFE=-5.981,780Kg
x=2,492m
VEF. X=14.909,038Kgm
(1/2.qx2)=30.000,000Kg
Mmax=-16.011,502Kgm
c. Free Body FG = IJ
Menghitung Momen Maximum dan Jarak
Free Body KL = Free Body NO
q=2.400,000Kg/m
L=4,000m
MEF=1.012,240Kgm
MFE=987,190Kgm
(1/2.ql2)=19.200,000Kg
VEF=4.806,263Kg
VFE=-4.806,263Kg
x=2,003m
VEF. X=9.625,066Kgm
(1/2.qx2)=19.200,000Kg
Mmax=-10.587,174Kgm
d. Free Body GH = HI
Menghitung Momen Maximum dan Jarak
Free Body LM = Free Body MN
q=2.400,000Kg/m
L=5,000m
MEF=987,190Kgm
MFE=1.012,240Kgm
(1/2.ql2)=30.000,000Kg
VEF=5.994,990Kg
VFE=-5.994,990Kg
x=2,498m
VEF. X=14.974,960Kgm
(1/2.qx2)=30.000,000Kg
Mmax=-16.012,230Kgm
e. Free Body KL = NO
Menghitung Momen Maximum dan Jarak
Free Body PQ = Free Body ST
q=2.400,000Kg/m
L=4,000m
MEF=816,720Kgm
MFE=373,520Kgm
(1/2.ql2)=19.200,000Kg
VEF=4.910,800Kg
VFE=-4.910,800Kg
x=2,046m
VEF. X=10.048,315Kgm
(1/2.qx2)=19.200,000Kg
Mmax=-9.968,405Kgm
f. Free Body LM = MN
Menghitung Momen Maximum dan Jarak
Free Body QR = Free Body RS
q=2.400,000Kg/m
L=5,000m
MEF=373,520Kgm
MFE=816,720Kgm
(1/2.ql2)=30.000,000Kg
VEF=5.911,360Kg
VFE=-5.911,360Kg
x=2,463m
VEF. X=14.560,074Kgm
(1/2.qx2)=30.000,000Kg
Mmax=-15.813,446Kgm
g. Free Body PQ
Menghitung Momen Maximum dan Jarak
Free Body UV = Free Body XY
q=2.400,000Kg/m
L=4,000m
MEF=531,830Kgm
MFE=498,810Kgm
(1/2.ql2)=19.200,000Kg
VEF=4.808,255Kg
VFE=-4.808,255Kg
x=2,003m
VEF. X=9.633,048Kgm
(1/2.qx2)=19.200,000Kg
Mmax=-10.098,782Kgm
h. Free Body QR
Menghitung Momen Maximum dan Jarak
Free Body VW = Free Body WX
q=2.400,000Kg/m
L=5,000m
MEF=498,810Kgm
MFE=531,830Kgm
(1/2.ql2)=30.000,000Kg
VEF=5.993,396Kg
VFE=-5.993,396Kg
x=2,497m
VEF. X=14.966,998Kgm
(1/2.qx2)=30.000,000Kg
Mmax=-15.531,812Kgm
i. Free Body UV
Menghitung Momen Maximum dan Jarak
Free Body Z-aa = Free Body ac-ad
q=1.400,000Kg/m
L=4,000m
MEF=257,270Kgm
MFE=197,130Kgm
(1/2.ql2)=11.200,000Kg
VEF=2.815,035Kg
VFE=-2.815,035Kg
x=2,011m
VEF. X=5.660,301Kgm
(1/2.qx2)=11.200,000Kg
Mmax=-5.796,969Kgm
j. Free Body VW
Menghitung Momen Maximum dan Jarak
Free Body ab-ac = Free Body ac-ad
q=1.400,000Kg/m
L=5,000m
MEF=197,130Kgm
MFE=257,270Kgm
(1/2.ql2)=17.500,000Kg
VEF=3.487,972Kg
VFE=-3.487,972Kg
x=2,491m
VEF. X=8.689,963Kgm
(1/2.qx2)=17.500,000Kg
Mmax=-9.007,167Kgm
EMBED AutoCAD.Drawing.17
EMBED AutoCAD.Drawing.17
EMBED AutoCAD.Drawing.17
EMBED AutoCAD.Drawing.17
kc
k1
kc
k2
k1
K3
K5
K3
K3
K5
K4
K6
kc
K2
kc
k1
kc
k4
K2
K3
k1
K5
K4
K6
kc
k1
K3
K2
k4
kc
k1
kc
K2
K4
K6
kc
k1
K3
K2
k4
kc
k1
kc
K2
dari nilai momen atas (Ma) dari kolom bawah
dari nilai momen bawah (Mb) dari kolom atas
Mki
Mka
kki
kki
A
B
yh
( 1-y ) h
EMBED AutoCAD.Drawing.17
EMBED Microsoft Equation 3.0
EMBED Microsoft Equation 3.0
EMBED Microsoft Equation 3.0
EMBED Microsoft Equation 3.0
EMBED Microsoft Equation 3.0
EMBED Microsoft Equation 3.0
EMBED Microsoft Equation 3.0
EMBED Microsoft Equation 3.0
EMBED Equation.3
EMBED AutoCAD.Drawing.17
EMBED AutoCAD.Drawing.17
K2
K3
kc
60Analisis Gempa Statik Ekivalen / UPI / 0900759
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