Tugas besar albayyinah putri

TUGASBESAR KONSTRUKSI KAYU 2013
Politeknik Negeri Jakarta – AlbayyinahPutri
3112120002 – 2 SIPIL1 SORE
PERMASALAHAN
Di suatutaman kota akandibuatjembatanyangdigunakanuntukmenyeberangisungai dengan
lebarsungai 7.5 meterlebarjembatan1.5meter. PembuatanjembatanakanmenggunakanKayukelas
II,mutu A.Kontruksi tersebut merupakankonstruksitidakterlindungsertamenerimabebantetapdan
tidaktetap.
α = 44o
cos α = 0, 719
sin α = 0, 694
A
H
B
C D E F G
I J K L M N
1250 1250 1250 1250 1250 1250
1200
A1
A2
H1
C1
I1
D1
J1
E1
K1
F1
L1
G1
M1
I2 D2 K2 F2 B2C2 J2 E2 L2 G2 B1A3

3112120002 – 2 SIPIL1 SORE
PERHITUNGAN REAKSI PERLETAKAN
*Asumsi bebansudahtermasuk bebanmanusiadanbebankonstruksi sendiri
Rangka jembatandigunakanuntukpejalankaki disebuahtamankota(pedestrian)
Alasjembatanmenggunakanpapandanbalok
Rc (AkibatP) =
𝛴𝑃
2
=
300
2
= 150 kg
Rc (Akibatq) =
𝑞𝐿
2
=
60 .1,5
2
= 45 kg +
P = 195 kg
ANALISIS GAYA BATANG DENGAN METODE
KESEIMBANGAN TITIK BUHUL

VA =
𝑃1.7,5+P2.6,25+ P2.6,25 + P3.5 + P4.3,75 + P4.3,75 + P5.2,5 + P6.1,25+ P7.0
7,5
=
97,5.7,5 + 195.6,25 + 195.5 + 195.3,75 + 195.2,5 + 195.1,25 + 0
7,5
=
731,25 + 1218,75 + 975 + 731,25 + 487,5 + 243,75
7,5
VA =
4387,5
7,5
= 585 KG

3112120002 – 2 SIPIL1 SORE
 Titik Buhul H
ΣH = 0
H1 = 0
ΣV = 0
A3 = 0
 Titik Buhul A
ΣV = 0
A3 + AV2 + P1 + VA = 0
0 + 0,694A2 + 97,5 + 585 = 0
0,694A2 = -487,5
A2 = - 702,449 kg
ΣH = 0
A1 + AH2 = 0
A1 + 0,719A2 = 0

3112120002 – 2 SIPIL1 SORE
A1 + 0,719(- 702,449) = 0
A1 = 505,061 kg
 Titik Buhul C
ΣV = 0
C2 – P2 = 0
C2 = 195 kg
ΣH = 0
A1 – C1 = 0
C1 = 505,061 kg
 Titik Buhul I
ΣV = 0
C2 + I2V + A2V – H1 = 0
195 + 0,694 I2 + (-702,449)(0,694) = 0
195 + 0,694 I2 – 487,499 = 0
0,694 I2 = 292,499
I2 = 421,468 kg
ΣH = 0
I1 + I2H – A2H – H1 = 0
I1 + 0,719(421,468) – (-702, 449)(0,719) – 0 = 0

3112120002 – 2 SIPIL1 SORE
I1 + 421,468 + 505,060 = 0
I1 = -926,528 kg
 Titik Buhul J
ΣH = 0
J1 – I1 = 0
J1 = -926,528 kg
ΣV = 0
J2 = 0
 Titik Buhul D
ΣV = 0
I2V + J2 – P3 + D2V = 0
421,468(0,694) + 0 – 195 + 0,694D2 = 0
292,498 – 195 + 0,694D2 = 0
0,694D2 = -97,498
D2 = -140,487 kg
ΣH = 0
D2H + D1 – I2H – C1 = 0
(-140,487)(0,719) + D1 – (421,468)(0,719) – 505,061 = 0
-101,010 + D1 – 303,035 – 505,061 = 0
D1 = 909,106 kg

3112120002 – 2 SIPIL1 SORE
 Titik Buhul E
ΣV = 0
E2 – P4 = 0
E2 = 195 kg
ΣH = 0
E1 – D1 =
E1 = 909,106 kg
 Titik Buhul K
ΣV = 0
E2 + K2V + D2V = 0
195 + 0,694K2 + (0,694)(-140,487) = 0
195 + 0,694K2 – 97,497 = 0
K2 = -140,494 kg
ΣH = 0
K1 + K2H – J1 – D2H = 0
K1 + (-140,494)(0,719) – (-926,528) – (-140,487)(0,719) = 0
K1 – 101,015 + 926,528 + 101,010 = 0
K1 = -926,528 kg

3112120002 – 2 SIPIL1 SORE
TABEL GAYA-GAYA BATANG
NO. BATANG PANJANG (m) TARIK + (kg) TEKAN – (kg) DIMENSI
1 A1 1,25 505,061 -
2 A2 1,73 - 702,449
3 A3 1,20 0 -
4 H1 1,25 0 -
5 C1 1,25 505,061 -
6 C2 1,20 195 -
7 I1 1,25 - 926,528
8 I2 1,73 421,468 -
9 D1 1,25 909,106 -
10 D2 1,73 - 140,487
11 J1 1,25 - 926,528
12 J2 1,20 0 -
13 E1 1,25 909,106 -
14 E2 1,20 195 -
15 K1 1,25 - 926,528
16 K2 1,73 - 140,494
17 F1 1,25 505,061 -
18 F2 1,73 421,468 -
19 L1 1,25 - 926,528
20 L2 1,20 0 -
21 G1 1,25 505,061 -
22 G2 1,20 195 -
23 B1 1,20 0 -
24 B2 1,73 - 702,449
25 M1 1,25 0 -

3112120002 – 2 SIPIL1 SORE
ANALISA PENDIMENSIAN BATANG TARIK DAN
BATANG TEKAN
 BATANG TEKAN
σtk =
𝑃.𝜔
𝐴.𝑏𝑟
P = -926,528 kg = -0,927 Ton
I min = 50Ptk.lk2
= 50 . 0,927 . 1,252
= 72,421
I min =
1
12
𝑏ℎ3
72,421 =
1
12
𝑏ℎ3
72,421 =
1
12
𝑏4
 asusmsi b= h
b4
= 72,421 x 12
b4
= 869,052
b = √869,0524
= 5.43 ≈ 6
I min = 72,421
1
12
bh3
= 72,421
1
12
6 . h3
= 72,421
h3
= 72,421 x 2
h3
= 144, 842
h = √144,823
= 5,25 ≈ 12
*menggunakan kayu dimensi 6/12
σtk =
𝑃.𝜔
𝐴.𝑏𝑟
ω  λ
λ 
𝑙𝑘
𝑖 𝑚𝑖𝑛
 √
𝐼 𝑚𝑖𝑛
𝐴 𝑏𝑟
50 Ptklk2
(Dari PKKI)

3112120002 – 2 SIPIL1 SORE
I min= √
𝐼 𝑚𝑖𝑛
𝐴𝑏𝑟
= √
1
12
𝑏ℎ
bh3
= √
1
12
6(12)3
6𝑥12
= √
864
72
= 3,464 cm
λ =
𝑙𝑘
𝑖 𝑚𝑖𝑛
=
125𝑐𝑚
3,464𝑐𝑚
= 36,085 ≈ 36
ω = 1,32
σtk =
𝑃.𝜔
𝐴𝑏𝑟
=
𝑃.𝜔
𝑏.ℎ
=
926,528𝑥1,32
6𝑥12
= 16,986 𝒌𝒈 𝒄𝒎⁄
∴ 𝜎𝑡𝑘 𝑖𝑧𝑖𝑛 = 150. 𝑔. 𝑓𝑎𝑘𝑡𝑜𝑟
= 150.(0,8).
5
4
.
2
3
= 100  KuatTekan izin

3112120002 – 2 SIPIL1 SORE
#Pengecekan Kekuatan Batang
A). UntukBatang I1, J1, K1, L1
P = 926,528
I min =
1
12
𝑏ℎ3
=
1
12
6(12)3
= 864
i min = √
864
6𝑥12
= √2√3 = 𝟑, 𝟒𝟔𝟒𝒄𝒎
λ =
𝑙𝑘
𝑖 𝑚𝑖𝑛
=
125
3,464
= 𝟑𝟔, 𝟎𝟖𝟓 ≈ 𝟑𝟔
ω = 1,32
σtk // =
926,528 𝑥 1,32
6 𝑥 12
= 𝟏𝟔, 𝟗𝟖𝟔
𝒌𝒈
𝒄𝒎 𝟐⁄
𝟏𝟔, 𝟗𝟖𝟔
𝒌𝒈
𝒄𝒎 𝟐⁄ < 𝟏𝟎𝟎
𝒌𝒈
𝒄𝒎 𝟐⁄ (𝑶𝑲!)
B). Untuk Batang A2 dan B2
P = 702,449
i min = 𝟑, 𝟒𝟔𝟒𝒄𝒎
λ =
𝑙𝑘
𝑖 𝑚𝑖𝑛
=
173
3,464
= 𝟒𝟗, 𝟗𝟒𝟐 ≈ 𝟓𝟎
ω = 1,50
σtk // =
702,449 𝑥 1,50
6 𝑥 12
= 𝟏𝟒, 𝟔𝟑𝟒
𝒌𝒈
𝒄𝒎 𝟐⁄
𝟏𝟒, 𝟔𝟑𝟒
𝒌𝒈
𝒄𝒎 𝟐⁄ < 𝟏𝟎𝟎
𝒌𝒈
𝒄𝒎 𝟐⁄ (𝑶𝑲!)
C). UntukBatang D2 dan K2

3112120002 – 2 SIPIL1 SORE
P = 140,487
i min = 𝟑, 𝟒𝟔𝟒𝒄𝒎
λ =
𝑙𝑘
𝑖 𝑚𝑖𝑛
=
173
3,464
= 𝟒𝟗, 𝟗𝟒𝟐 ≈ 𝟓𝟎
ω = 1,50
σtk // =
140,487 𝑥 1,50
6 𝑥 12
= 𝟐, 𝟗𝟐𝟕
𝒌𝒈
𝒄𝒎 𝟐⁄
𝟐, 𝟗𝟐𝟕
𝒌𝒈
𝒄𝒎 𝟐⁄ < 𝟏𝟎𝟎
𝒌𝒈
𝒄𝒎 𝟐⁄ (𝑶𝑲!)
 BATANG TARIK
A). UntukBatang D1 dan E1
P = 909,106 kg
σtr =
𝑃
𝐴𝑛𝑒𝑡𝑡𝑜
Anetto= Abr x c
σtr //izin= 150(
2
3
.
5
4
)(0,8)
= 100
𝑘𝑔
𝑐𝑚2⁄
σtr //
𝑃
≤ σtr // izin
Anetto = Abrx C
= (6x12) x (1-0,2)
= 72 x 0,8 = 57,6 cm2

3112120002 – 2 SIPIL1 SORE
σtr =
𝑃
=
909,106
57,6
= 𝟏𝟓, 𝟕𝟖𝟑
𝒌𝒈
𝒄𝒎 𝟐⁄
15,783
𝑘𝑔
𝑐𝑚2⁄ < 100
𝑘𝑔
𝑐𝑚2
B). Untuk C1, F1, G1
P = 505,061
σtr =
𝑃
=
505,061
57,6
= 𝟖, 𝟕𝟔𝟖
𝒌𝒈
𝒄𝒎 𝟐⁄
8,768
𝑘𝑔
𝑐𝑚2⁄ < 100
𝑘𝑔
𝑐𝑚2
C). Untuk I2 dan F2
P = 421,468
σtr =
𝑃
=
421,468
57,6
= 𝟕, 𝟑𝟏𝟕
𝒌𝒈
𝒄𝒎 𝟐⁄
7,317
𝑘𝑔
𝑐𝑚2⁄ < 100
𝑘𝑔
𝑐𝑚2
D). Untuk C2, E2, G2
P = 195
σtr =
𝑃
=
195
57,6
= 𝟑, 𝟑𝟖𝟓
𝒌𝒈
𝒄𝒎 𝟐⁄
3,385
𝑘𝑔
𝑐𝑚2⁄ < 100
𝑘𝑔
𝑐𝑚2
SAMBUNGAN BAUT
 Kelas kuat 2, Gol. 2 sambungan baut tampang 2  λ = 4,3
 α =
5
4
; β =
5
6

3112120002 – 2 SIPIL1 SORE
 b3 = 6cm  dimensi kayu 6
8⁄
 b1 = plat = 0,3d = 0,45cm
 𝑑 =
𝑏
𝜆
=
6
4,3
= 1,395 ≈ 1,4 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑏𝑎𝑢𝑡
 Kekuatan sambungan dinaikkan 25%, karena dalam kondisi terlindung & dapat
mengering dengan cepat.
𝑆̅1 = 100. 𝑑. 𝑏3 (1 - 0,6 sinα)
= 100.1,4.6 (1 - 0,6 sin0)
= 840
𝑆̅2 = 200. 𝑑. 𝑏1 (1 - 0,6 sinα)
= 200.1,4.0,45 (1 - 0,6 sin0)
= 126
𝑆̅3 = 430. 𝑑2
(1 - 0,35 sinα)
= 430.1,42 (1 - 0,35 sin0)
= 842,8
JUMLAH BAUT
𝑁 =
𝑃
𝑠
 𝑆̅ = (840 + 840 x 25%)
5
4
.
5
6
= 1093,75

3112120002 – 2 SIPIL1 SORE
1. Batang A1 =
505,061
1093,75
= 0,462 ≈ 2 𝑏𝑢𝑎ℎ 𝑏𝑎𝑢𝑡
2. Batang A2 =
702,449
1093,75
= 0,642 ≈ 2 𝑏𝑢𝑎ℎ 𝑏𝑎𝑢𝑡
3. Batang A3 =
0
1093,75
= 0 ≈ 2 𝑏𝑢𝑎ℎ 𝑏𝑎𝑢𝑡
4. Batang H1 =
0
1093,75
5. Batang C1 =
505,061
1093,75
= 0,462 ≈ 2 𝑏𝑢𝑎ℎ 𝑏𝑎𝑢𝑡
6. Batang C2 =
195
1093,75
= 0,178 ≈ 2 𝑏𝑢𝑎ℎ 𝑏𝑎𝑢𝑡
7. Batang I1 =
926,528
1093,75
= 0,847 ≈ 2 𝑏𝑢𝑎ℎ 𝑏𝑎𝑢𝑡
8. Batang I2 =
421,468
1093,75
= 0,385 ≈ 2 𝑏𝑢𝑎ℎ 𝑏𝑎𝑢𝑡
9. Batang D1 =
909,106
1093,75
= 0,831 ≈ 2 𝑏𝑢𝑎ℎ 𝑏𝑎𝑢𝑡
10.Batang D2 =
140,487
1093,75
= 0,128 ≈ 2 𝑏𝑢𝑎ℎ 𝑏𝑎𝑢𝑡
11.Batang J1 =
926,528
1093,75
= 0,847 ≈ 2 𝑏𝑢𝑎ℎ 𝑏𝑎𝑢𝑡
12.Batang J2 =
0
1093,75
13.Batang E1 =
909,106
1093,75
= 0,831 ≈ 2 𝑏𝑢𝑎ℎ 𝑏𝑎𝑢𝑡
14.Batang E2 =
195
1093,75
= 0,178 ≈ 2 𝑏𝑢𝑎ℎ 𝑏𝑎𝑢𝑡
PERHITUNGAN KEKUATAN KOMPONEN JEMBATAN
1. PerhitunganLantai Jembatan

3112120002 – 2 SIPIL1 SORE
𝑃 =
1
2
𝑞𝑙
195 =
1
2
. 𝑞. 1,5
390 = 0,15𝑞
𝑞 =
390
1,5
= 260𝑘𝑔𝑚
𝑞 = 𝑊𝑡𝑜𝑡𝑎𝑙. 𝑏
260 = 𝑊𝑡𝑜𝑡𝑎𝑙.1,25
𝑊𝑡𝑜𝑡𝑎𝑙 = 208
𝑘𝑔
𝑚⁄ = 2,08
𝑘𝑔
𝑐𝑚⁄
= 208.1,25
= 260
𝑘𝑔
𝑚⁄ = 2,6
𝑘𝑔
𝑐𝑚⁄
𝑀 𝑚𝑎𝑘𝑠 =
1
8
𝑞𝑙2
=
1
8
.260. 1,52
= 73,125
𝑘𝑔
𝑚⁄ = 7312,5
𝑘𝑔
𝑐𝑚⁄
Menghitungkekuatanmultipleks
 Trial 1
Misalkanmenggunakanmultipleks9mm
𝑡 = 9𝑚𝑚 = 0,9𝑐𝑚
I =
1
12
𝑏ℎ3 =
1
12
.125.0,93 = 7,594𝑐𝑚4
𝜔 =
1
6
𝑏ℎ2 =
1
6
.125.0,92 = 16,875𝑐𝑚3
 Cekterhadapkekuatan
𝜎//𝑖𝑧𝑖𝑛 = 150
𝑘𝑔
𝑐𝑚2⁄ (𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒𝑘𝑠 𝑘𝑎𝑦𝑢 𝑘𝑒𝑙𝑎𝑠 1)
𝜎// =
𝑀
𝑊
=
7312,5
16,875
= 433.333
𝑘𝑔
𝑐𝑚2⁄

3112120002 – 2 SIPIL1 SORE
𝜎// > 𝜎//𝑖𝑧𝑖𝑛(𝑡𝑖𝑑𝑎𝑘 𝑑𝑖𝑖𝑧𝑖𝑛𝑘𝑎𝑛)
Tindakan yang harus dilakukan untuk mengatasinya:
1. Nilai ℓ diperkecil
2. Tebal mutipleksditambah
 Trial 2
ℓ = 0,75𝑚
Misalkanmenggunakanmultipleks12mm
𝑡 = 1,2 𝑐𝑚
1
8
𝑞𝑙2
=
1
8
.260. 0,752
= 18,281
𝑘𝑔
𝑚⁄ = 1828,125
𝑘𝑔
𝑐𝑚⁄
𝐼 =
1
12
𝑏ℎ3 =
1
12
.125.1,23 = 18𝑐𝑚4
𝜔 =
1
6
𝑏ℎ2 =
1
6
.125.1,22 = 30𝑐𝑚3
𝜎//𝑖𝑧𝑖𝑛 = 150 𝑘𝑔
𝑐𝑚2⁄
𝜎// =
𝑀
𝑊
=
1828,125
30
= 60,937
𝑘𝑔
𝑐𝑚2⁄
𝜎// < 𝜎//𝑖𝑧𝑖𝑛(𝑑𝑖𝑖𝑧𝑖𝑛𝑘𝑎𝑛)
 Cekterhadaplendutan
𝛿𝑖𝑧𝑖𝑛 =
1
400
. 𝑙 =
1
400
. 75 = 0,1875𝑐𝑚
𝛿 =
2,5 . 𝑞 .ℓ4
384 EI

3112120002 – 2 SIPIL1 SORE
=
2,5 . 2,6 .754
384 .125000 .18
= 0,238𝑐𝑚
= 0,339 cm
𝛿 > 𝛿𝑖𝑧𝑖𝑛(𝑡𝑖𝑑𝑎𝑘 𝑑𝑖𝑖𝑧𝑖𝑛𝑘𝑎𝑛)
 Trial 3
ℓ = 0,75𝑚
Misalkanmenggunakanmultipleks15 mm
𝑡 = 1,5 𝑐𝑚
1
8
𝑞𝑙2
=
1
8
.260. 0,752
= 18,281
𝑘𝑔
𝑚⁄ = 1828,1
𝑘𝑔
𝑐𝑚⁄
𝐼 =
1
12
𝑏ℎ3 =
1
12
.125.1,53 = 35,156𝑐𝑚4
𝜔 =
1
6
𝑏ℎ2 =
1
6
.125.1,52 = 46,875𝑐𝑚2
𝜎//𝑖𝑧𝑖𝑛 = 150
𝑘𝑔
𝑐𝑚2⁄ (𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒𝑘𝑠 𝑘𝑎𝑦𝑢 𝑘𝑒𝑙𝑎𝑠 1)
𝜎// =
𝑀
𝑊
=
1828,1
46,875
= 38,999
𝑘𝑔
𝑐𝑚2⁄

3112120002 – 2 SIPIL1 SORE
1
400
. 𝑙 =
1
400
. 75 = 0,1875𝑐𝑚
𝛿 =
2,5 . 𝑞 .ℓ4
384 EI
=
2,5 . 2,6 .754
384 .125000 .35,156
= 0,122𝑐𝑚
𝛿 < 𝛿𝑖𝑧𝑖𝑛(𝑑𝑖𝑖𝑧𝑖𝑛𝑘𝑎𝑛)
Jadi, tebal multipleks yang digunakan ukuran 15 mm dengan ℓ = 0,75 m.
2. PerhitunganGelagarAnak
𝑀𝑖𝑠𝑎𝑙𝑛𝑦𝑎 𝑘𝑎𝑦𝑢 𝑦𝑎𝑛𝑔 𝑑𝑖𝑔𝑢𝑛𝑎𝑘𝑎𝑛 5
10⁄ 𝑘𝑎𝑟𝑒𝑛𝑎 𝑠𝑎𝑙𝑎ℎ 𝑠𝑎𝑡𝑢 𝑠𝑖𝑠𝑖𝑛𝑦𝑎 𝑑𝑖𝑘𝑒𝑡𝑎𝑚 𝑚𝑎𝑘𝑎 𝑢𝑘𝑢𝑟𝑎𝑛𝑛𝑦𝑎 𝑚𝑒𝑛𝑗𝑎𝑑𝑖 5
9⁄
𝐼 =
1
12
𝑏ℎ3 =
1
12
.5.93 = 303,75𝑐𝑚4
𝜔 =
1
6
𝑏ℎ2 =
1
6
.5. 92 = 67,5𝑐𝑚3
= 208.0,75
= 156
𝑘𝑔
𝑚⁄ = 1,56
𝑘𝑔
𝑐𝑚⁄
1
8
𝑞𝑙2
=
1
8
.156. 1,562
= 30,469
𝑘𝑔
𝑚⁄ = 3046,9
𝑘𝑔
𝑐𝑚⁄
𝜎//𝑖𝑧𝑖𝑛 = 100
𝑘𝑔
𝑐𝑚2⁄

3112120002 – 2 SIPIL1 SORE
𝜎// =
𝑀
𝑊
=
3046,9
67,5
= 45,139
𝑘𝑔
𝑐𝑚2⁄
1
400
. 𝑙 =
1
400
. 125 = 0,3125𝑐𝑚
𝛿 =
5𝑞𝑙4
384𝐸𝐼
=
5 . 1,56 .1254
384 .100000 .303,75
= 0,163𝑐𝑚
𝛿 < 𝛿𝑖𝑧𝑖𝑛(𝑑𝑖𝑖𝑧𝑖𝑛𝑘𝑎𝑛)
Jadi, untuk gelagar anak dapat menggunakan kaso ukuran 5/10.
3. PerhitunganRangka Batang
TEGANGANYANG DIPERKENANKANUNTUK KAYUMUTU A KELAS KUAT II :
𝝈̅ 𝒕𝒌 𝑰𝑰
KELAS KUAT
KI II
𝝈̅ 𝒍𝒕(
𝒌𝒈
𝒄𝒎 𝟐⁄ ) 100
𝝈̅ 𝒕𝒓//
(
𝒌𝒈
𝒄𝒎 𝟐⁄ ) 85
𝝈̅ 𝒕𝒌⊥
(
𝒌𝒈
𝒄𝒎 𝟐⁄ ) 25
𝝉̅//(
𝒌𝒈
𝒄𝒎 𝟐⁄ ) 12
𝑓̅ =
1
700
. 1,25 = 0,179
𝜔 =
1
6
𝑏ℎ2
=
1
6
6. 122
= 144𝑐𝑚2
 𝑃 =
1
2
𝑞𝑙
195 =
1
2
. 𝑞. 1,5
390 = 1,5𝑞
𝑞 = 260
 𝑞 = 𝑊𝑡𝑜𝑡𝑎𝑙 . 𝑏
260 = 𝑊𝑡𝑜𝑡𝑎𝑙 . 1,5

3112120002 – 2 SIPIL1 SORE
𝑊𝑡𝑜𝑡𝑎𝑙 = 173,333
 𝑞 = 𝑊𝑡𝑜𝑡𝑎𝑙 . 𝑏
= 173,333.1,5
= 260 𝑘𝑔𝑚
= 2, 6 kgcm
 𝐼 =
1
12
. 6. 123
= 864𝑐𝑚4
 𝑀 𝑚𝑎𝑥 =
1
8
𝑞𝑙2
=
1
8
. 260 .1,252
50, 781 ≈ 5078,125
 𝜎̅𝑙𝑡 =
𝑀
𝑊
=
5078,125
173,333
= 29, 297 < 100
 𝑄 =
1
2
𝑞𝑙 =
1
2
. 260.1,25 = 162, 5𝑘𝑔
 𝒯// =
3
2
.
𝑄
𝑏.ℎ
=
3
2
.
162,5
6.12
= 2,257 < 12
Untuk konstruksi rangka batang yang tidak terlindung, 𝑓𝑚𝑎𝑥 <
1
700
𝑙
f = lendutan
l = jarak bentang
 𝑓𝑚𝑎𝑥 =
5
384
.
𝑞.𝑙4
𝐸𝐼
=
5
384
.
2,6 .1254
100000.864
= 𝟎, 𝟎𝟗𝟔 < 𝟎, 𝟏𝟕𝟗 (𝑨𝑴𝑨𝑵!)
LAMPIRAN GAMBAR
 Rangka Batang Sambungan Baut
A
H
I J K L NM
BC D E F G
1250 1250 1250 1250 1250 1250
1200
TAMPAK SAMPING
Skala 1:50

3112120002 – 2 SIPIL1 SORE
Sambungan Baut
42
84
42
20
42
84
42
60
80
60
80
268
208
Kayu 80x60mm
Baut Ø 14mm
Plat
20
42
84
42
268
220
80
221
60
80
60
80
60
80
Baut Ø 14mm
Plat
Kayu 80x60mm

3112120002 – 2 SIPIL1 SORE
42
84
42
20
42
84
42
356
359
210210
60
80
60
80
60
80
60
80
60
80
Kayu 80x60mm
Baut Ø 14mm
Plat
DETAIL I,K,M
Skala 1:10
42
84
42
20
42
84
42
60
80
356
268
80
138 80 138
60
80
60
Kayu 80x60mm
Plat
Baut Ø 14mm
Multiplek 12mm
12
DETAIL C,E,G
Skala 1:10
42
84
42
20
42
84
42
60
80
356
268
13880138
60
80
60
Kayu 80x60mm
Plat
Baut Ø 14mm
DETAIL J,L
Skala 1:10
42
84
42
20
42
84
42
60
80
60
80
268
208
Kayu 80x60mm
Baut Ø 14mm
Plat
DETAIL N
Skala 1:10

3112120002 – 2 SIPIL1 SORE
42
84
42
20
42
84
42
356
359
210 210
60
80
60
80
60
80
60
80
60
80
Kayu 80x60mm
Baut Ø 14mm
Plat
DETAIL D,F
Skala 1:10
8442
84
42
208
268
220
80
221
60
80
60
80
60
80
Baut Ø 14mm
Plat
Kayu 80x60mm 42
DETAIL B
Skala 1:10

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