BAB 1 PERHITUNGAN PANJANG BATANG
Transcript of BAB 1 PERHITUNGAN PANJANG BATANG
1
BAB 1
PERHITUNGAN PANJANG BATANG
1.1 Perhitungan Secara Matematis
Panjang Batang Bawah
B1 = B2 = B3 = B4 = B5 = B6 = B7 = B8
B = 30
8= 3,75 m
Panjang Batang Tegak
T1 = T7
T1 = tan α . B1
= tan 45° . 3,75
= 3,75 m
T2 = T6
T2 = tan α .( B1+B2 )
= tan 45° .( 3,75+3,75 )
= 7,5 m
B2 B3 B4 B5 B6 B7 B8
D1
D2
D3 D4
D5
D6
30
A1
A2
A3
A4A5
A6
A7
T1
T2
T3
T4
T5
T6
T7
B1A B
A8
45°
A1 T1
B1
A1
A2
T1
T2
B1 B2
2
T3 = T5
T3 = tan α . ( B1+B2+B3 )
= tan 45° . ( 3,75+3,75+3,75 )
= 11,25 m
T4
T4 = tan α .( B1+B2+B3+B4 )
= tan 45° .( 3,75+3,75+3,75+3,75 )
= 15 m
Panjang Batang Atas
A1 = A2 = A3 = A4 = A5 = A6 = A7 = A8
A = 3,752 + 3,752
= 5,303
Panjang Batang Diagonal
D1 = D6
D1 = 3,752 + 3,752
= 5,303
D2 = D5
D2 = 7,52 + 3,752
= 8,385
A1
A2
A3
T1
T2
T3
B1 B2 B3
A1
A2
A3
A4
T1
T2
T3
T4
B1 B2 B3 B4
A1 T1
B1
T1
B2
D1
T2
B3
D2
3
D3 = D4
D3 = 11,252 + 3,752
= 11,859
Tabel Panjang Batang
No A ( m ) B ( m ) T ( m ) D ( m )
1 5,303 3,75 3,75 5,303
2 5,303 3,75 7,5 8,385
3 5,303 3,75 11,25 11,859
4 5,303 3,75 15 11,859
5 5,303 3,75 11,25 8,385
6 5,303 3,75 7,5 5,303
7 5,303 3,75 3,75 -
8 5,303 3,75 - -
T3
B4
D3
4
BAB 2
PERHITUNGAN GORDING
2.1 Gording
Data – data :
o Bentang rangka atap = 30 m
o Jarak kuda – kuda ( λ ) = 3 m
o Berat atap genteng biasa = ±24 Kg/m
o Jarak gording = 5,303 m
o Beban angin ( W ) = 70 Kg/m2
o Beban Berguna ( P ) = 70 Kg
2.2 Perencanaan Dimensi Gording
Dicoba gording INP.30
Data Profil F = 69,1 Cm2
G = 54,2 Kg/m
Ix = 9800 Cm4
Iy = 451 Cm4
Wx = 653 Cm3
Wy = 72,2 Cm3
2.3 Pembebanan Gording
a. Beban Mati
- Berat sendiri gording = 1 × 54,2 = 54,2 Kg/m
- Berat penutup atap = ( a × berat sendiri atap × 1 )
= 5,303 × 24 × 1
= 127,272 Kg/m
o q1 = 54,2 + 127,272
= 181,472 Kg/m
o Brancing 10 % . q1 = 10 % . 181,472
q2 = 18,147 Kg/m
o q total = q1 + q2
= 181,472 + 18,147
= 199,619 Kg/m
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b. Beban Berguna ( P ) = 70 Kg
c. Beban Angin
o Angin tekan
c = 0,02 α – 0,4
= 0,02. 45 – 0,4
= 0,5
o Angin Isap = c’ = - 0,4
o Beban angin tekan
W = c × w × a × 1
= 0,5 × 70 × 5,303 × 1
= 185,605 Kg/m
o Beban angin isap
W’ = c ‘ × w × a × 1
= -0,4 × 70 × 5,303 × 1
= -148,484 Kg/m
2.4 Momen Pada Gording
a. Akibat beban Mati
qy = q cos α
= 199,619 cos 45°
= 141,152 Kg/m
qx = q sin α
= 199,619 sin 45°
= 141,152 Kg/m
Mqy = 1
8 . qy .λ
2 Mqx =
1
8 . qx . λ
2
= 1
8 . 141,152 . 3
2 =
1
8 . 141,152 . 3
2
= 158,796 Kg/m = 158,796 Kg/m
45°qxq
qy
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b. Akibat beban berguna
Py = P cos α
= 70 cos 45°
= 49,497 Kg
Px = P sin α
= 70 sin 45°
= 49,497 Kg
MPy = 1
4 . Py .λ MPx =
1
4 . Px .λ
= 1
4 . 49,497 . 3 =
1
4 . 49,497 . 3
= 37,123 Kg/m = 37,123 Kg/m
c. Akibat beban angin
o Angin tekan
Wy = W = 185,605 Kg/m
Wx = 0
MWy = 1
8 . wy . λ
2
= 1
8 . 185,605 . 3
2
= 208,806 Kg/m
MWx = 0
o Angin isap
Wy’ = W’ = -148,484 Kg/m
Wx’ = 0
MWy’ = 1
8 . wy’ . λ
2
= 1
8 .-148,484. 3
2
= - 167,045 Kg/m
MWx’ = 0
45°Pxq
Py
45°Wxq
Wy
45°W'x q
W'y
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2.5 Kombinasi Momen
Arah Beban Mati Beban Hidup Beban ngin Kombinasi
( 1 ) ( 2 ) Tekan ( 3 ) Isap ( 4 ) ( 1+2 ) (1+2+3) (1+2+4)
X 158,796 37,123 0 0 195,919 195,919 195,919
Y 158,796 37,123 208,806 - 167,045 195,919 404,725 28,874
Catatan :ambil nilai yang terbesar
2.6 Kontrol Terhadap Tegangan
Data :
Mx = 195,919 Kg.m = 19591,9 Kg.cm
My = 404,725 Kg.m = 40472,5 Kg.cm
Wx = 653 cm3
Wy = 72,2 cm3
σ = 1600 Kg/cm2
𝑀𝑥
𝑊𝑥+𝑀𝑦
𝑊𝑦 ≤ σ
19591,9
653+
40472 ,5
72,2
≤ σ
590,564 Kg/cm2 ≤ 1600 Kg/cm
2 Aman !
2.7 Kontrol Terhadap Lendutan
Data :
qx = 141,152 Kg/m = 1,41152 Kg/cm2
qy = 141,152 Kg/m = 1,41152 Kg/cm2
Px = 49,497 Kg
Py = 49,497 Kg
Ix = 9800 cm4
Iy = 451 cm4
λ = 3 m = 300 cm
E = 2,1 x 106 Kg/cm2
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o Lendutan Arah Sumbu x
δx = 5
384.𝑞𝑥 .𝜆⁴
𝐸 .𝐼𝑥+
1
48.𝑃𝑥 .𝜆³
𝐸 .𝐼𝑥
= 5
384.
1,41152 .300⁴
2,1𝑥106 .9800+
1
48.
49,497.300³
2,1𝑥106 .9800
= 0,008605 cm
o Lendutan Arah Sumbu y
δy = 5
384.𝑞𝑦 .𝜆⁴
𝐸 .𝐼𝑦+
1
48.𝑃𝑦 .𝜆³
𝐸 .𝐼𝑦
= 5
384.
1,41152 .300⁴
2,1𝑥106 .451+
1
48.
49,497.300³
2,1𝑥106 .451
= 0,186583 cm
δ = 𝛿𝑥2 + 𝛿𝑦2
= 0,0086052 + 0,1865832
= 0,18678 cm
δ ≤ 1
250 . λ
≤ 1
250 . 300
0,18678 cm ≤ 1,2 cm Aman !
Maka INP.30 aman terhadap tegangan dan lendutan yang akan terjadi.
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BAB 3
PERHITUNGAN RANGKA KUDA-KUDA
3.1 Perhitungan Kuda-Kuda
Berat sendiri kuda-kuda = 2 + 0,66 L
= 2 + 0,66 . 30
= 21,8 Kg/m2
Berat total = 𝐿 .𝑇
2 × 21,8
= 30 . 15
2 × 21,8
= 4905 Kg
Berat sendiri gording INP.30 = 54,2 Kg/m2
Jumlah gording = 10 buah
Berat atap = ±24 Kg/m2
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3.2 Menentukan Beban Mati Vertikal
Berat gording = jumlah gording b.s gording jarak kuda-kuda
= 10 54,2 3
= 1626 Kg
Berat sendiri atap = berat atap batang atas kuda-kuda jarak kuda-kuda
= 24 2 (21,212) 3
= 3054,528 Kg
Berat rangka keseluruhan + Berat sendiri gording + Berat sendiri atap
G = 4905 + 1626+ 3054,528
= 9585,528 Kg
Brancing 10 % . G = 10 % . 9585,528
= 958,5528 Kg
G total = G + Brancing 10 %
= 9585,528 + 958,5528
= 10544,081 Kg
Beban mati per titik tumpu = Gtotal
8
= 10544 ,081
8
= 1318,010 Kg
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3.3 Menentukan beban berguna( P )
o Beban berguna ( P ) = 70 Kg
3.4 Menentukan beban angin
w = 70 Kg
o Koefisien angin tekan ( c ) = 0,5
o Koefisien angin isap ( c’ ) = -0,4
Tiap titik simpul tengah menerima beban, yaitu :
Angin tekan
W = λ jarak gording c w
= 3 5,303 0,5 70
= 556,815 Kg
Angin isap
W’ = λ jarak gording c’ w
= 3 5,303 -0,4 70
= -445,452 Kg
B2 B3 B4 B5 B6 B7 B8
D1
D2
D3 D4
D5
D6
30
A1
A2
A3
A4A5
A6
A7
T1
T2
T3
T4
T5
T6
T7
B1A B
A8
45°
70
70
70
70
70
70
70
35 35
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Tiap titik simpul menerima beban yaitu :
Tepi bawah ( di titik A ) = W
2=
556,815
2 = 278,408 Kg
Tepi bawah ( di titik B ) = W′
2=
−445,452
2 = - 222,726 Kg
Angin Kiri
Angin Kanan
B2 B3 B4 B5 B6 B7 B8
D1
D2
D3 D4
D5
D6
30
A1
A2
A3
A4A5
A6
A7
T1
T2
T3
T4
T5
T6
T7
B1A B
A8
45°
556,815
556,815
556,815
-445,452
-445,452
-445,452
278,408 -222,726
278,408 -222,726
B2 B3 B4 B5 B6 B7 B8
D1
D2
D3 D4
D5
D6
30
A1
A2
A3
A4A5
A6
A7
T1
T2
T3
T4
T5
T6
T7
B1A B
A8
45°
-445,452
-445,452
-445,452
-222,726
-222,726
556,815
556,815
556,815
278,408
278,408
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BAB 4
PERHITUNGAN GAYA BATANG
4.1 Gaya Batang Akibat Beban Mati dengan Cara Cremona
RAH = 0
∑ H = 0
Beban simetris karena beban kiri dan beban kanan sama
∑MA=∑MB = 0
= P1+P2+P3+P4+P5+P6+P7+P8+P9
2
= 659,005+1318 ,010+ 1318,010+ 1318,010+ 1318 ,010+ 1318,010+ 1318,010+1318 ,010+659,005
2
RAV = RBV = 5272,04 Kg ( ↑)
Cek : ∑V = 0
RAV + RBV – ( P1+P2+P3+P4+P5+P6+P7+P8 ) = 0
10544,08 – 10544,08 = 0 OK
A1
A2
A3
A4A5
A6
A7
T1
T2
T3
T4
T5
T6
T7
B1 B2 B3 B4 B5 B6 B7 B8
D1
D2
D3 D4
D5
D6
A B
30
A8
45°
1318,010
1318,010
1318,010
1318,010
1318,010
1318,010
1318,010
659,005 659,005
30.0000
RA RB
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Gaya batang yang diperoleh akibat beban mati dengan cara cremona :
Batang atas ( A )
A1 = - 6523,817 Kg
A2 = - 5591,843 Kg
A3 = - 4659,869 Kg
A4 = - 3727,895 Kg
A5 = - 3727,895 Kg
A6 = - 4659,869 Kg
A7 = - 5591,843 Kg
A8 = - 6523,817 Kg
Batang bawah ( B )
B1 = + 4613,035 Kg
B2 = + 4613,035 Kg
B3 = + 3954,030 Kg
B4 = + 3295,025 Kg
B5 = + 3295,025 Kg
B6 = + 3954,030 Kg
B7 = + 4613,035 Kg
B8 = + 4613,035 Kg
Batang tegak ( T )
T1 = + 0
T2 = - 659,005 Kg
T3 = + 1318,010 Kg
T4 = + 3954,030 Kg
T5 = + 1318,010 Kg
T6 = + 659,005 Kg
T7 = + 0
Batang diagonal ( D )
D1 = - 931,929 Kg
D2 = - 1473,580 Kg
D3 = - 2083,957 Kg
D4 = + 2083,957 Kg
D5 = + 1473,580 Kg
D6 = + 931,929 Kg
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Tabel Gaya Batang Akibat Beban Mati
No. Batang Gaya Batang ( Kg )
Tarik ( + ) Tekan ( - )
A1 - 6523,817
A2 - 5591,843
A3 - 4659,869
A4 - 3727,895
A5 - 3727,895
A6 - 4659,869
A7 - 5591,843
A8 - 6523,817
B1 4613,035 -
B2 4613,035 -
B3 3954,030 -
B4 3295,025 -
B5 3295,025 -
B6 3954,030 -
B7 4613,035 -
B8 4613,035 -
T1 0 -
T2 - 659,005
T3 1318,010 -
T4 3954,030 -
T5 1318,010 -
T6 659,005 -
T7 0 -
D1 - 931,929
D2 - 1473,580
D3 - 2083,957
D4 2083,957 -
D5 1473,580 -
D6 931,929 -
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4.2 Gaya Batang Akibat Beban Berguna ( P = 70 Kg )
Reaksi Perletakan
Beban yang digunakan adalah beban kiri dan beban kanan sama.
RAH = 0
∑ H = 0
∑MA=∑MB = 0
= P1+P2+P3+P4+P5+P6+P7+P8+P9
2
= 35+70+ 70+ 70+ 70+ 70+ 70+70+35
2
RAV = RBV = 280 Kg ( ↑)
Cek : ∑V = 0
RAV + RBV – ( P1+P2+P3+P4+P5+P6+P7+P8 ) = 0
560 – 560 = 0 OK !
A1
A2
A3
A4A5
A6
A7
T1
T2
T3
T4
T5
T6
T7
B1 B2 B3 B4 B5 B6 B7 B8
D1
D2
D3 D4
D5
D6
A B
A8
45°
30
RA RB
35 kg
70 kg
70 kg
70 kg
70 kg
70 kg
70 kg
70 kg
35 kg
18
Gaya batang yang diperoleh akibat beban berguna dengan cara cremona :
Batang atas ( A )
A1 = -346,482 Kg
A2 = -296,985 Kg
A3 = -247,487 Kg
A4 = -197,9899 Kg
A5 = -197,9899 Kg
A6 = -247,487 Kg
A7 = -296,985 Kg
A8 = -346,482 Kg
Batang bawah ( B )
B1 = +4534,848 Kg
B2 = +4534,848 Kg
B3 = +3887,052 Kg
B4 = +3239,038 Kg
B5 = +3239,038 Kg
B6 = +3887,052 Kg
B7 = +4534,848 Kg
B8 = +4534,848 Kg
Batang tegak ( T )
T1 = 0
T2 = - 647,795 Kg
T3 = +1295,674 Kg
T4 = +1943,386 Kg
T5 = +295,674 Kg
T6 = +647,795 Kg
T7 = 0
Batang diagonal ( D )
D1 = -916,121 Kg
D2 = -1448,514 Kg
D3 = -2048,663 Kg
D4 = +2048,663 Kg
D5 = +1448,514 Kg
D6 = +916,121 Kg
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Tabel Gaya Batang Akibat Beban Berguna ( P = 70 Kg )
No. Batang Gaya Batang ( Kg )
Tarik ( + ) Tekan ( - )
A1 - 6413,199
A2 - 5496,948
A3 - 4580,736
A4 - 3664,571
A5 - 3664,571
A6 - 4580,736
A7 - 5496,948
A8 - 6413,199
B1 4534,848 -
B2 4534,848 -
B3 3887,052 -
B4 3239,038 -
B5 3239,038 -
B6 3887,052 -
B7 4534,848 -
B8 4534,848 -
T1 0 -
T2 - 647,795
T3 1295,674 -
T4 1943,386 -
T5 1295,674 -
T6 647,795 -
T7 0 -
D1 - 916,121
D2 - 1448,514
D3 - 2048,663
D4 2048,663 -
D5 1448,514 -
D6 916,121 -
20
4.3 Gaya Batang Akibat Beban Angin Kiri dengan Cara Cremona
Reaksi Perletakan
∑ H = 0
= RAH + 278,408sin45° + 556,815sin45° + 556,815sin45° + 556,815sin45° +
278,408sin45° - 222,726sin45° - 445,452sin45° - 445,452sin45° -
445,452sin45° - 222,726sin45° = 0
RAH = - 315,62 Kg ( ← )
∑MB = 0
= RAV×30 + 278,408sin45°×0 - 278,408cos45°×30 + 556,815sin45°×3,75 –
556,815cos45°×26,25 + 556,815sin45°×7,5 – 556,815cos45°×22,5 +
556,815sin45°×11,25 – 556,815cos45°×18,75 + 278,408sin45°×15 –
278,408cos45°×15 + (-222,726sin45°×15) + (-222,726cos45°×15) + (-
445,452sin45°×11,25) + (-445,452cos45°×11,25) + (-445,452sin45°×7,5) +
(-445,452cos45°×7,5) + (-445,452sin45°×3,75) + (-445,452cos45°×3,75) = 0
RAV = 1417,4195 Kg ( ↑ )
∑MA = 0
= -RBV×30 + (-222,726sin45°×0) - (-222,726cos45°×30) + (-
445,452sin45°×3,75) - (-445,452cos45°×26,25) + (-445,452sin45°×7,5) - (-
445,452cos45°×22,5) + (-445,452sin45°×11,25) - (-445,452cos45°×18,75) +
(-222,726sin45°×15) – (-222,726cos45°×15) + 278,408sin45°×15 +
278,408cos45°×15 + 556,815sin45°×11,25 + 556,815cos45°×11,25 +
556,815sin45°×7,5 + 556,815cos45°×7,5 + 556,815sin45°×3,75 +
556,815cos45°×3,75 + 278,408sin45°×0 = 0
RBV = 1417,4195 Kg ( ↑ )
30
A1
A2
A3
A4A5
A6
A7
T1
T2
T3
T4
T5
T6
T7
B1 B2 B3 B4 B5 B6 B7 B8
D1
D2
D3 D4
D5
D6 A8
RA RB
RAH
12 W
12 W
W
W
12 W'
12 W'
W'
W'
W'
W
21
∑V = 0
= RAV + RBV - 278,408cos45° - 556,815cos45° - 556,815cos45° -
556,815cos45° - 278,408cos45° - 222,726cos45° - 445,452cos45° -
445,452cos45° - 445,452cos45° - 222,726cos45° = 0
= 2834,839 – 2834,839 = 0 →OK !!!
Gaya batang yang diperoleh akibat beban angin kiri dengan cara cremona :
Batang atas ( A )
A1 = + 55,682 Kg
A2 = + 55,682 Kg
A3 = + 55,682 Kg
A4 = + 55,682 Kg
A5 = - 445,452 Kg
A6 = - 445,452 Kg
A7 = - 445,452 Kg
A8 = - 445,452 Kg
Batang bawah ( B )
B1 = + 2598,603 Kg
B2 = + 2598,603 Kg
B3 = + 2204,375 Kg
B4 = + 1811,147 Kg
B5 = + 1102,437 Kg
B6 = + 787,455 Kg
B7 = + 472,473 Kg
B8 = + 472,473 Kg
Batang tegak ( T )
T1 = 0
T2 = + 393,728 Kg
T3 = + 787,455 Kg
T4 = + 236,237 Kg
T5 = + 629,964 Kg
T6 = + 314,982 Kg
T7 = 0
22
Batang diagonal ( D )
D1 = - 556,815 Kg
D2 = - 880,402 Kg
D3 = -1245,076 Kg
D4 = + 996,061 Kg
D5 = + 704,321 Kg
D6 = + 445,452 Kg
Tabel Gaya Batang Akibat Beban Angin Kiri
No. Batang Gaya Batang ( Kg )
Tarik ( + ) Tekan ( - )
A1 55,682 -
A2 55,682 -
A3 55,682 -
A4 55,682 -
A5 - 445,452
A6 - 445,452
A7 - 445,452
A8 - 445,452
B1 2598,603 -
B2 2598,603 -
B3 2204,375 -
B4 1811,147 -
B5 1102,437 -
B6 787,455 -
B7 472,473 -
B8 472,473 -
T1 - -
T2 393,728
T3 787,455
T4 236,237
T5 629,964
T6 314,982
T7 - -
D1 - 556,815
D2 - 880,402
D3 - 1245,076
D4 996,061 -
D5 704,321 -
D6 445,452 -
23
4.4 Gaya Batang Akibat Beban Angin Kanan dengan Cara Cremona
Reaksi Perletakan
∑ H = 0
= RAH + 278,408sin45° + 556,815sin45° + 556,815sin45° + 556,815sin45° +
278,408sin45° - 222,726sin45° - 445,452sin45° - 445,452sin45° -
445,452sin45° - 222,726sin45° = 0
RAH = - 314,983 Kg ( ← )
∑MA = 0
= -RBV×30 + 278,408sin45°×0 + 278,408cos45°×30 – 556,815sin45°×3,75 +
556,815cos45°×26,25 – 556,815sin45°×7,5 + 556,815cos45°×22,5 –
556,815sin45°×11,25 + 556,815cos45°×18,75 – 278,408sin45°×15 +
278,408cos45°×15 – (-222,726sin45°×15) – (-222,726cos45°×15) – (-
445,452sin45°×11,25) – (-445,452cos45°×11,25) – (-445,452sin45°×7,5) – (-
445,452cos45°×7,5) – (-445,452sin45°×3,75) – (-445,452cos45°×3,75) = 0
RBV = 1417,4195 Kg ( ↑ )
∑MB = 0
= RAV×30 – (-222,726cos45°×30) – (-445,452sin45°×3,75) – (-
445,452cos45°×26,25) – (-445,452sin45°×7,5) – (-445,452cos45°×22,5) – (-
445,452sin45°×11,25) – (-445,452cos45°×18,75) – (-222,726sin45°×15) – (-
222,276cos45°×15) – 278,408sin45°×15 – 278,408cos45°×15 –
556,815sin45°×11,25 – 556,815cos45°×11,25 – 556,815sin45°×7,5 –
556,815cos45°×7,5 – 556,815sin45°×3,75 – 556,815cos45°×3,75 –
278,408sin45°×0 = 0
RAV = 1417,4195 Kg ( ↑ )
A1
A2
A3
A4A5
A6
A7
T1
T2
T3
T4
T5
T6
T7
B1 B2 B3 B4 B5 B6 B7 B8
D1
D2
D3 D4
D5
D6 A8
RA RB
RAH
30
12 W
12 W
W
12 W'
12 W'
W'
W'
W' W
W
24
∑V = 0
= RAV + RBV + (-222,726cos45°) + (-445,452cos45°) + (-445,452cos45°) + (-
445,452cos45° + (-222,726cos45°) – 278,408cos45° - 556,815cos45° -
556,815cos45° - 556,815cos45° - 278,408cos45° = 0
= 2834,839 – 2834,839 = 0 →OK !!!
Gaya batang yang diperoleh akibat beban angin kanan dengan cara 24remona :
Batang atas ( A )
A1 = - 445,452 Kg
A2 = - 445,452 Kg
A3 = - 445,452 Kg
A4 = - 445,452 Kg
A5 = + 55,683 Kg
A6 = + 55,683 Kg
A7 = + 55,683 Kg
A8 = + 55,683 Kg
Batang bawah ( B )
B1 = - 2362,368 Kg
B2 = - 2362,368 Kg
B3 = - 2047,385 Kg
B4 = - 1732,403 Kg
B5 = - 1023,693 Kg
B6 = - 629,965 Kg
B7 = - 236,237 Kg
B8 = - 236,237 Kg
Batang tegak ( T )
T1 = 0
T2 = - 314,982 Kg
T3 = - 629,965 Kg
T4 = + 236,236 Kg
T5 = + 787,455 Kg
T6 = + 393,728 Kg
T7 = 0
25
Batang diagonal ( D )
D1 = + 445,453 Kg
D2 = + 704,322 Kg
D3 = + 996,062 Kg
D4 = - 1245,076 Kg
D5 = - 880,402 Kg
D6 = - 556,815 Kg
Tabel Gaya Batang Akibat Beban Angin Kanan
No. Batang Gaya Batang ( Kg )
Tarik ( + ) Tekan ( - )
A1 - 445,452
A2 - 445,452
A3 - 445,452
A4 - 445,452
A5 55,683 -
A6 55,683 -
A7 55,683 -
A8 55,683 -
B1 - 2362,368
B2 - 2362,368
B3 - 2047,385
B4 - 1732,403
B5 - 1023,693
B6 - 629,965
B7 - 236,237
B8 - 236,237
T1 - -
T2 - 314,982
T3 - 629,965
T4 236,236 -
T5 787,455 -
T6 393,728 -
T7 - -
D1 445,453 -
D2 704,322 -
D3 996,062 -
D4 - 1245,076
D5 - 880,402
D6 - 556,815
29
BAB 5
PERHITUNGAN PROFIL KUDA-KUDA
5.1 Profil Batang Atas ( A )
Profil siku dobel ( ┘└ )
- Beban maksimum ( P max ) = 13382,468 Kg ( Batang Tekan )
- Panjang batang ( Lk ) = 5,303 m = 530,3 cm
- σ = 1600 Kg/cm2
- Tebal plat penyambung ( S ) = 10 mm = 1 cm
Fperlu = Pmax
σ + 2,5 Lk
2
= 13382 ,468
1600 + 2,5 . 5,303
2
= 78,669 cm2
Dicoba dimensi profil ┘└ 140.140.15
- F = 40 × 2 = 80 cm2 Fperlu = 78,669 cm
2
- Ix = Iy = 723
- ix = iy = 4,25
- e = 4,00
Y
Y
e
30
a. Kontrol as bahan ( sb x – x )
λx = 𝐿𝑘𝑥
𝑖𝑥 =
530,3
4,25 = 124,776 125
x = 0,271
σdx = x . σ
= 0,271 × 1600
= 433,6 Kg/cm2
P = σdx . F
= 433,6 × 80
= 34688 Kg Pmax = 13382,468 Kg → Ok !
b. Kontrol as bebas bahan ( sb y – y )
1. Seluruh profil
Iy fiktif = { Iy + F ( e + ½ .s )2 }× 2
= { 723 + 40 ( 4 + ½ . 1 )2
× 2
= 3066 cm4
iy fiktif = 0,9 .3066
80
= 5,873
λy fiktif = Lky
iy fiktif
= 530,3
5,873
= 90,29 ~ 90,3
y fiktif = 0,481
P = y fiktif . F . σ
= 0,481 . 80 . 1600
= 61568 Kg Pmax = 13382,468 Kg → Ok!
31
2. Satu profil / Profil tunggal
x = y’ . y1
y1 = x
y’
= 0,271
0,481
= 0,563
λy1 = 83 ( Tabel Tekuk )
Lky1 = λy1 . iy
= 83 . 4,25
= 352,75
n = L
Lky 1
= 530,3
352,75
= 1,5 ~ 5 medan
Jumlah Pelat Kopel = n + 1 = 5 + 1 = 6 buah
5.2 Profil Batang Bawah ( B )
- Beban maksimum ( Pmax ) = 11746,486 Kg ( tarik )
- Panjang Batang ( Lkx ) = 3,750 m = 375 cm
- σ = 1600 Kg/cm2
Fnetto = Pmax
σ
= 11746 ,486
1600
= 7,342 cm2
Fbruto = Fnetto
0,85
= 7,342
0,85
= 8,638
Dicoba Profil └ 150.150.16 → F = 45,7 cm2 F brutto
32
5.3 Profil Batang Diagonal ( D )
Profil siku dobel ( ┘└ )
- Beban maksimum ( P max ) = 2404,865 Kg ( Batang Tekan )
- Panjang batang ( Lk ) = 5,303 m = 530,3 cm
- σ = 1600 Kg/cm2
- Tebal plat penyambung ( S ) = 10 mm = 1 cm
Fperlu = Pmax
σ + 2,5 Lk
2
= 2404 ,865
1600 + 2,5 . 5,303
2
= 71,807 cm2
Dicoba dimensi profil ┘└ 130.130.16
- F = 39,3 × 2 = 78,6 cm2 Fperlu = 71,807 cm
2
- Ix = Iy = 605
- ix = iy = 3,92
- e = 5,37
Y
Y
e
33
a. Kontrol as bahan ( sb x – x )
λx = 𝐿𝑘𝑥
𝑖𝑥 =
530,3
3,92 = 135,281 136
x = 0,229
σdx = x . σ
= 0,229 × 1600
= 366,4 Kg/cm2
P = σdx . F
= 366,4 × 78,6
= 28799,04 Kg Pmax = 2404,865 Kg → Ok !
b. Kontrol as bebas bahan ( sb y – y )
1. Seluruh profil
Iy fiktif = { Iy + F ( e + ½ .s )2 }× 2
= { 605 + 39,3 ( 5,37 + ½ . 1 )2 × 2
= 3918 cm4
iy fiktif = 0,9 .3918
78,6
= 6,698
λy fiktif = Lky
iy fiktif
= 530,3
6,698
= 79,17 ~ 80
y fiktif = 0,588
P = y fiktif . F . σ
= 0,588 . 78,6 . 1600
= 73946,88 Kg Pmax = 2404,865 Kg → Ok!
34
2. Satu profil / profil tunggal
x = y’ . y1
y1 = x
y’
= 0,229
0,588
= 0,389
λy1 = 104 ( Tabel Tekuk )
Lky1 = λy1 . iy
= 104 . 3,92
= 407,68
n = L
Lky 1
= 530,3
407,68
= 1,3 ~ 3 medan
Jumlah Pelat Kopel = n + 1 = 3 + 1 = 4
5.4 Profil Batang Tegak ( T )
- Beban maksimum ( Pmax ) = 6133,652 Kg ( tarik )
- Panjang Batang ( Lkx ) = 7,5 m = 750 cm
- σ = 1600 Kg/cm2
Fnetto = Pmax
σ
= 6133 ,652
1600
= 3,834 cm2
Fbruto = Fnetto
0,85
= 3,834
0,85
= 4,511
Dicoba Profil └ 110.110.10 → F = 21,2 cm2 F brutto
35
DAFTAR REKAPITULASI DIMENSI PROFIL
Nama Batang Nomor Batang Dimensi
Batang Tepi Atas A1 s/d A8 ┘└ 140.140.15
Batang Tepi Bawah B1 s/d B8 └ 150.150.16
Batang Diagonal D1 s/d D6 ┘└ 130.130.16
Batang Tegak T1 s/d T7 └ 110.110.10
36
BAB 6
PERHITUNGAN SAMBUNGAN PAKU KELING
6.1 Sambungan Paku untuk Batang Atas ( A1 s/d A8 )
Pmax = 13382,468 Kg
Sambungan profil ┘└ 140.140.15
a. Menentukan ∅ . P.K
∅ = 2 × Tebal Rata −rata plat yang disambung
2
= 2 × 10+15
2
= 25 mm
d = 25 + 1 = 26 mm
b. Jumlah Paku Keling
Ngs = 2 π d²
4 × τ
= 2 π 2,6²
4 × 0,8 × 1600
= 13591,786
Ntp = d . s . σtp
= 2,6 . 1 . 1,6. 1600
= 6656 Kg
Ambil yang paling kecil n = Pmax
Ntp =
13591,786
6656
n = 2,042 ~ 3 buah
c. Penempatan Paku Keling
- Cukup tempat
t ≥ 2 d
52 mm
- Cukup kuat
t ≥ 3 d
78 mm
37
ρ1 ≥ 2 d
52 mm
ρ2 ≥ 1 ½ d
39 mm
- Cukup Rapat
t ≤ 4,5 d
≤ 117 mm
ρ1 ≤ 3 d
≤ 78 mm
ρ2 ≤ 3 d
≤ 78 mm
Ambil → ρ1 = 60 mm
ρ2 = 40 mm
t = 80 mm
6.2 Sambungan Paku untuk Batang Bawah ( B1 s/d B8 )
Pmax = 11746,486 Kg
Sambungan profil ┘└ 150.150.16
a. Menentukan ∅. P.K
∅ = 2 × Tebal Rata −rata plat yang disambung
2
= 2 × 10+16
2
= 26 mm
d = 26 + 1 = 27 mm
b. Jumlah Paku Keling
Ngs = 2 π d²
4 × τ
= 2 π 2,7²
4 × 0,8 × 1600
= 14657,415
38
Ntp = d . s . σtp
= 2,7 . 1 . 2. 1600
= 8640 Kg
Ambil yang paling kecil n = Pmax
Ntp =
11746 ,486
6080
n = 1,93 ~ 2 buah
c. Penempatan Paku Keling
- Cukup tempat
t ≥ 2 d
38 mm
- Cukup kuat
t ≥ 3 d
57 mm
ρ1 ≥ 2 d
38 mm
ρ2 ≥ 1 ½ d
28,5 mm
- Cukup Rapat
t ≤ 7 d
≤ 189 mm
ρ1 ≤ 3 d
≤ 57 mm
ρ2 ≤ 3 d
≤ 57 mm
Ambil → ρ1 = 40 mm
ρ2 = 30 mm
t = 60 mm
39
6.3 Sambungan Paku untuk Batang Diagonal ( D1 s/d D6 )
Pmax = 2404,865 Kg
Sambungan profil ┘└ 130.130.16
a. Menentukan ∅. P.K
∅ = 2 × Tebal Rata −rata plat yang disambung
2
= 2 × 10+16
2
= 26 mm
d = 26 + 1 = 27 mm
b. Jumlah Paku Keling
Ngs = 2 π d²
4 × τ
= 2 π 2,7²
4 × 0,8 × 1600
= 14657,415 Kg
Ntp = d . s . σtp
= 2,7 . 1 . 2. 1600
= 8640 Kg
Ambil yang paling kecil n = Pmax
Ntp =
11746 ,486
8640
n = 1,359 ~ 2 buah
c. Penempatan Paku Keling
- Cukup tempat
t ≥ 2 d
54 mm
- Cukup kuat
t ≥ 3 d
81 mm
ρ1 ≥ 2 d
54 mm
ρ2 ≥ 1 ½ d
40,5 mm
40
- Cukup Rapat
t ≤ 4,5 d
≤ 121,5 mm
ρ1 ≤ 3 d
≤ 81 mm
ρ2 ≤ 3 d
≤ 81 mm
Ambil → ρ1 = 60 mm
ρ2 = 50 mm
t = 90 mm
6.4 Sambungan Paku untuk Batang Tegak ( T1 s/d T7 )
Pmax = 6133,652 Kg
Sambungan profil ┘└ 110.110.10
a. Menentukan ∅. P.K
∅ = 2 × Tebal Rata −rata plat yang disambung
2
= 2 × 10+10
2
= 20 mm
d = 20 + 1 = 21 mm
b. Jumlah Paku Keling
Ngs = 2 π d²
4 × τ
= 2 π 2,1²
4 × 0,8 × 1600
= 8866,831 Kg
Ntp = d . s . σtp
= 2,1 . 1 . 2. 1600
= 6720 Kg
41
Ambil yang paling kecil n = Pmax
Ntp =
6133,652
6720
n = 0,913 ~ 1 buah
c. Penempatan Paku Keling
- Cukup tempat
t ≥ 2 d
42 mm
- Cukup kuat
t ≥ 3 d
63 mm
ρ1 ≥ 2 d
42 mm
ρ2 ≥ 1 ½ d
31,5 mm
- Cukup Rapat
t ≤ 7 d
≤ 147 mm
ρ1 ≤ 3 d
≤ 63 mm
ρ2 ≤ 3 d
≤ 63 mm
Ambil → ρ1 = 50 mm
ρ2 = 40 mm
t = 70 mm
42
BAB 7
PERHITUNGAN PELAT KOPEL
7.1 Pelat Kopel untuk Batang Atas ( A1 s/d A8 )
o Dengan profil ┘└ 140.140.15
o P max = 13382,468 Kg
o Lk = 5,303 m = 530,3 m
o d = 26 mm
o ex = ey = 4,00
t = 80 mm = 9 cm
ρ1 = 60 mm = 6 cm
ρ2 = 40 mm = 4 cm
o b = 280 mm = 28 cm
hn = 2 . ex + δ
= 2. 4,00 + 0,8
= 8,8 cm
F = 2 × 40 = 80 cm2
Ix = Iy = 723 cm4
Z1 = ex + ½ s
= 4 + ½ . 1
= 4,5 cm
Diambil 5 Medan
L1 = Lk
5 =
530,3
5 = 106,06 cm
D = 1,5 % × Pmax
= 1,5 % × 13382,468
= 200,737 Kg
δ = F . Z1
= 80 . 4,5
= 360 cm2
43
L = 𝐷 .𝛿 .𝐿1
𝐼𝑥 =
200,737 .360 .106,06
723 = 10600,913 Kg
M = 𝐿 .ℎ𝑛
𝑒 =
10600 ,913 .8,8
6 = 15548,006
Tegangan Pelat
τ = D.δ
b .Ix =
200,737 . 360
28 .723 = 3,569 Kg/cm
2
Tegangan Potongan Tunggal
R = ¼ . 𝐿² +𝑚²
= ¼ . 10600,913² + 15548,006²
= 4704,518 Kg
Kontrol Tegangan
τ = R
¼𝜋 .𝑑² ≤ τ
= 4704,518
¼𝜋 .2,6² ≤ 0,8 . 1600
= 886,091 ≤ 1280 Kg/cm2 ( Aman )
7.2 Pelat Kopel untuk Batang Bawah ( B1 s/d B8 )
o Dengan profil ┘└ 150.150.16
o P max = 11746,486 Kg
o Lk = 3,75 m = 375 m
o d = 27 mm
o ex = ey = 4,29
t = 60 mm = 6 cm
ρ1 = 40 mm = 4 cm
ρ2 = 30 mm = 3 cm
o b = 150 mm = 15 cm
hn = 2 . ex + δ
= 2. 4,29 + 0,8
= 9,38 cm
44
F = 45,7 cm2
Ix = Iy = 949 cm4
Z1 = ex + ½ s
= 4,29 + ½ . 1
= 4,79 cm
Diambil 3 Medan
L1 = Lk
3 =
375
3 = 125 cm
D = 1,5 % × Pmax
= 1,5 % × 11746,486
= 176,197 Kg
δ = F . Z1
= 45,7 . 4,79
= 218,903 cm2
L = 𝐷 .𝛿 .𝐿1
𝐼𝑥 =
176,197 .218,903 .125
949 = 5080,355 Kg
M = 𝐿 .ℎ𝑛
𝑒 =
5080,355 .9,38
6 = 7942,288 Kg
Tegangan Pelat
τ = D.δ
b .Ix =
176,197 . 218,903
15 .949 = 2,709 Kg/cm
2
Tegangan Potongan Tunggal
R = ¼ . 𝐿² +𝑚²
= ¼ . 5080,355² + 7942,288²
= 2357,037 Kg
Kontrol Tegangan
τ = R
¼𝜋 .𝑑² ≤ τ
= 2357,037
¼𝜋 .2,7² ≤ 0,8 . 1600
= 411,669 ≤ 1280 Kg/cm2 ( Aman )
45
7.3 Pelat Kopel untuk Batang Diagonal ( D1 s/d D6 )
o Dengan profil ┘└ 130.130.16
o P max = 2404,865 Kg
o Lk = 5,303 m = 530,3 m
o d = 19 mm
o ex = ey = 3,80
t = 60 mm = 6 cm
ρ1 = 40 mm = 4 cm
ρ2 = 30 mm = 3 cm
o b = 130 mm = 13 cm
hn = 2 . ex + δ
= 2. 3,80 + 0,8
= 8,4 cm
F = 2 × 39,3 = 78,6 cm2
Ix = Iy = 605 cm4
Z1 = ex + ½ s
= 3,80 + ½ . 1
= 4,3 cm
Diambil 3 Medan
L1 = Lk
3 =
530,3
3 = 176,767 cm
D = 1,5 % × Pmax
= 1,5 % × 2404,865
= 36,073 Kg
δ = F . Z1
= 78,6 . 4,3
= 337,98 cm2
L = 𝐷 .𝛿 .𝐿1
𝐼𝑥 =
36,073 .337,98 .176,767
605 = 3562,206 Kg
46
M = 𝐿 .ℎ𝑛
𝑒 =
3562,206 .8,4
6 = 4987,088
Tegangan Plat
τ = D.δ
b .Ix =
36,073 . 337,98
13 .605 = 1,550 Kg/cm
2
Tegangan Potongan Tunggal
R = ¼ . 𝐿² +𝑚²
= ¼ . 3562,206 ² + 4987,088²
= 1532,163 Kg
Kontrol Tegangan
τ = R
¼𝜋 .𝑑² ≤ τ
= 1532,163
¼𝜋 .1,9² ≤ 0,8 . 1600
= 540,390 ≤ 1280 Kg/cm2 ( Aman )
7.4 Pelat Kopel untuk Batang Tegak ( T1 s/d T7 )
o Dengan profil ┘└ 110.110.10
o P max = 6133,652 Kg
o Lk = 7,5 m = 750 m
o d = 21 mm
o ex = ey = 3,07
t = 70 mm = 7 cm
ρ1 = 50 mm = 5 cm
ρ2 = 40 mm = 4 cm
o b = 110 mm = 11 cm
hn = 2 . ex + δ
= 2. 3,07 + 0,8
= 6,94 cm
F = 21,2 cm2
Ix = Iy = 239 cm4
47
Z1 = ex + ½ s
= 3,07 + ½ . 1
= 3,57 cm
Diambil 5 Medan
L1 = Lk
5 =
750
5 = 150 cm
D = 1,5 % × Pmax
= 1,5 % × 6133,652
= 92,005 Kg
δ = F . Z1
= 21,2 . 3,57
= 75,684 cm2
L = 𝐷 .𝛿 .𝐿1
𝐼𝑥 =
92,005 .75,684 .150
239 = 4370,276 Kg
M = 𝐿 .ℎ𝑛
𝑒 =
4370,276 .6,94
6 = 5054,953 Kg
Tegangan Pelat
τ = D.δ
b .Ix =
92,005 . 75,684
11 .239 = 2,649 Kg/cm
2
Tegangan Potongan Tunggal
R = ¼ . 𝐿² +𝑚²
= ¼ . 4370,276² + 5054,953²
= 1670,551 Kg
Kontrol Tegangan
τ = R
¼𝜋 .𝑑² ≤ τ
= 1670,551
¼𝜋 .2,1² ≤ 0,8 . 1600
= 482,316 ≤ 1280 Kg/cm2 ( Aman )
48
BAB 8
LENDUTAN KONSTRUKSI
Batang F1
( cm2 )
F = 2.F1
( cm2 )
L/F Gaya Batang
(Kg) Sin α
L/F.B.Sinα
(Kg/cm2)
No
Batang L (m)
A1 5,303 40 80 13,258 13382,468 0,707 125439,306
A2 5,303 40 80 13,258 11534,243 0,707 108115,143
A3 5,303 40 80 13,258 9686,057 0,707 90791,345
A4 5,303 40 80 13,258 7837,918 0,707 73467,988
A5 5,303 40 80 13,258 7392,466 0,707 69292,585
A6 5,303 40 80 13,258 9546,605 0,707 89484,205
A7 5,303 40 80 13,258 11088,791 0,707 103939,740
A8 5,303 40 80 13,258 12937,016 0,707 121263,903
B1 3,75 45,7 91,4 8,206 11746,486 0 0
B2 3,75 45,7 91,4 8,206 11746,486 0 0
B3 3,75 45,7 91,4 8,206 10045,457 0 0
B4 3,75 45,7 91,4 8,206 8345,21 0 0
B5 3,75 45,7 91,4 8,206 7636,5 0 0
B6 3,75 45,7 91,4 8,206 8628,537 0 0
B7 3,75 45,7 91,4 8,206 9620,356 0 0
B8 3,75 45,7 91,4 8,206 9620,356 0 0
D1 5,303 39,3 78,6 13,494 2404,865 0,707 22943,033
D2 8,385 39,3 78,6 21,336 3802,496 0,707 57358,949
D3 11,859 39,3 78,6 30,176 5377,696 0,707 114730,089
D4 11,859 39,3 78,6 30,176 5128,861 0,707 109421,336
D5 8,385 39,3 78,6 21,336 3626,415 0,707 54702,846
D6 5,303 39,3 78,6 13,494 2293,502 0,707 21880,601
T1 3,75 21,2 42,4 17,689 0 0,707 0
T2 7,5 21,2 42,4 35,377 393,728 0,707 9847,743
T3 11,25 21,2 42,4 53,066 3401,139 0,707 127602,783
T4 15 21,2 42,4 70,755 6133,653 0,707 306828,539
T5 11,25 21,2 42,4 53,066 3243,648 0,707 127602,783
49
Lendutan yang terjadi :
∑ × L/F × B × Sin α = 1744560,66 Kg/cm2
E = 2,1 × 106
L = 30 m = 3000 cm
δ = ∑ × L/F × B × Sin α
𝐸 ≤
1
500 . L
= 1744560 ,66
2,1 × 106 ≤
1
500 . 3000
= 0,831 ≤ 6,0 → Aman !
Jadi lendutan yang ditinjau aman
T6 7,5 21,2 42,4 35,377 1621,782 0,707 9847,743
T7 3,75 21,2 42,4 17,689 0 0,707 0
∑ 1744560,66