BAB 1 PERHITUNGAN PANJANG BATANG
A4
A5
A6
A3
T4 A2 T3
D4
D3
T5
D2
A7 D5 T6
T2 A1
T1
A8 T7
45°
A
D6
D1
B1
B2
B4
B3
B5
B6
30
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
A1
T1
T1
= tan α . B1 = tan 45° . 3,75 = 3,75 m
T2
= tan α .( B1+B2 ) = tan 45° .( 3,75+3,75 ) = 7,5 m
B1
T2 = T6 A2
T2 A1
T1 B1
B2
1
B7
B8
B
T3 = T5 A3
T3
= tan α . ( B1+B2+B3 ) = tan 45° . ( 3,75+3,75+3,75 ) = 11,25 m
T4
= tan α .( B1+B2+B3+B4 ) = tan 45° .( 3,75+3,75+3,75+3,75 ) = 15 m
A2 T3
T2 A1
T1 B1
B2
B3
T4
A4
A3
T4 A2 T3
T2 A1
T1 B1
B2
B4
B3
Panjang Batang Atas A1 = A2 = A3 = A4 = A5 = A6 = A7 = A8 A A1
= 3,752 + 3,752
T1
= 5,303 B1
Panjang Batang Diagonal D1 = D6
D1
T1
D1
= 3,752 + 3,752 = 5,303
B2
D2 = D5 D2
= 7,52 + 3,752
D2 T2
= 8,385
B3
2
D3 = D4 D3
T3
= 11,252 + 3,752 = 11,859
D3
B4
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
-
-
3
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 q2
o
q total
= 10 % . 181,472 = 18,147 Kg/m = q1 + q2 = 181,472 + 18,147 = 199,619 Kg/m 4
b. Beban Berguna ( P )
=
70
Kg
qy
= q cos α
c. Beban Angin o
Angin tekan c
=
0,02 α – 0,4
=
0,02. 45 – 0,4
=
0,5
= c’ = - 0,4
o
Angin Isap
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
= 199,619 cos 45° = 141,152 Kg/m
qy qx
45°
qx q
= q sin α = 199,619 sin 45° = 141,152 Kg/m
Mq y
1
= 8 . qy .λ2
Mq x
1
1
= 8 . qx . λ2 1
= 8 . 141,152 . 32
= 8 . 141,152 . 32
= 158,796 Kg/m
= 158,796 Kg/m
5
b. Akibat beban berguna Py
= P cos α = 70 cos 45° = 49,497 Kg
Py 45°
Px
Px
q
= P sin α = 70 sin 45° = 49,497 Kg
MPy
1
= 4 . Py .λ
MPx
1
1
= 4 . Px .λ 1
= 4 . 49,497 . 3
= 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 1
MWy = 8 . wy . λ2 1
= 8 . 185,605 . 32
Wy 45°
Wx
o
= 208,806 Kg/m MWx = 0
q
Angin isap Wy’
= W’ = -148,484 Kg/m
Wx’
=0 1
MWy’ = 8 . wy’ . λ2 1
= 8 .-148,484. 32
W'y W'x
45°
= - 167,045 Kg/m
q
MWx’ = 0
6
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
3
Wx
= 653 cm
Wy
= 72,2 cm3
σ
= 1600 Kg/cm2
𝑀𝑥 𝑊𝑥
+
𝑀𝑦 𝑊𝑦
19591,9 653
+
40472 ,5 72,2
590,564 Kg/cm2
≤
σ
≤
σ
≤
1600 Kg/cm2
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
=3m
E
= 2,1 x 106 Kg/cm2
= 300 cm
7
Aman !
o Lendutan Arah Sumbu x
δx
= =
5 384 5 384
. .
𝑞𝑥 .𝜆⁴ 𝐸.𝐼𝑥
+
1 48
.
𝑃𝑥 .𝜆³ 𝐸.𝐼𝑥
1,41152 .300⁴ 2,1𝑥10 6 .9800
+
1
.
49,497.300³
48 2,1𝑥10 6 .9800
= 0,008605 cm o Lendutan Arah Sumbu y
δy
= =
5 384 5 384
. .
𝑞𝑦 .𝜆⁴ 𝐸.𝐼𝑦
+
1 48
.
1,41152 .300⁴ 2,1𝑥10 6 .451
𝑃𝑦 .𝜆³ 𝐸.𝐼𝑦
+
1
.
49,497.300³
48 2,1𝑥10 6 .451
= 0,186583 cm
δ
= 𝛿𝑥 2 + 𝛿𝑦 2 = 0,0086052 + 0,1865832 = 0,18678 cm
δ
≤ ≤
1 250 1 250
.λ . 300
0,18678 cm ≤ 1,2 cm
Aman !
Maka INP.30 aman terhadap tegangan dan lendutan yang akan terjadi.
8
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
9
10
3.2 Menentukan Beban Mati Vertikal Berat gording
= jumlah gording b.s gording = 10
jarak kuda-kuda
54,2 3
= 1626 Kg Berat sendiri atap
= berat atap batang atas kuda-kuda = 24
2 (21,212)
jarak kuda-kuda
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
11
3.3 Menentukan beban berguna( P ) o Beban berguna ( P )
= 70 Kg 70
70
A4
A5
70
70 A6
A3
T4
70 A2 T3
70 D4
D3
A7
T5
D5
D2
35 T1
35
D6
D1
A8 T7
45°
A
T6
T2 A1
70
B1
B2
B3
B4
B5
B6
30
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
=3
5,303
c
w
0,5 70
= 556,815 Kg Angin isap W’
=λ
jarak gording
=3
5,303
c’
-0,4 70
= -445,452 Kg
12
w
B7
B8
B
Tiap titik simpul menerima beban yaitu : Tepi bawah ( di titik A ) = Tepi bawah ( di titik B ) =
W
=
2 W′
=
2
556,815 2
= 278,408 Kg
−445,452 2
= - 222,726 Kg
Angin Kiri 278,408
A4
556,815
556,815
556,815
-222,726
-445,452
A5
T4 A2
D4
D3
T3
T5
D2
A7
278,408 T1
-445,452
D5 T6
T2 A1
-445,452
A6
A3
D6
D1
A8 T7
45°
A
B1
B2
B4
B3
B5
B7
B6
B8
-222,726 B
30
Angin Kanan -222,726
-445,452
-445,452
278,408
A4
556,815
A5
556,815
A6
A3
T4
-445,452
A2 T3
D4
D3
D2
-222,726
D5 T6
T2 A1
T1
556,815
A7
D6
D1
B1
B2
278,408
A8 T7
45°
A
T5
B3
B4
B5 30
13
B6
B7
B8
B
BAB 4 PERHITUNGAN GAYA BATANG
4.1 Gaya Batang Akibat Beban Mati dengan Cara Cremona
1318,010
1318,010
1318,010
A4
A5
1318,010
1318,010 A6
A3
1318,010
1318,010 T4
A2 T3
659,005
D4
D3
D2
T5
A7
T6
T2 A1
T1
659,005
D5
D6
D1
A8 T7
45°
A
B1
B2
B3
B4
B5
B6
B7
B8
B
30
30.0000 RA
RB
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
14
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 15
Tabel Gaya Batang Akibat Beban Mati No. Batang A1 A2 A3 A4 A5 A6 A7 A8 B1 B2 B3 B4 B5 B6 B7 B8 T1 T2 T3 T4 T5 T6 T7 D1 D2 D3 D4 D5 D6
Gaya Batang ( Kg ) Tarik ( + ) 4613,035 4613,035 3954,030 3295,025 3295,025 3954,030 4613,035 4613,035 0 1318,010 3954,030 1318,010 659,005 0 2083,957 1473,580 931,929
16
Tekan ( - ) 6523,817 5591,843 4659,869 3727,895 3727,895 4659,869 5591,843 6523,817 659,005 931,929 1473,580 2083,957 -
4.2 Gaya Batang Akibat Beban Berguna ( P = 70 Kg ) 70 kg
70 kg
70 kg
A4
A5
70 kg
70 kg A6
A3
70 kg
70 kg T4
A2
T3
35 kg
D4
D3
D2
T5
A7
T6
T2 A1
D6
D1
A8 T7
45°
A
T1
35 kg
D5
B1
B2
B4
B3
B5
B6
B7
B8
B
30 RA
RB
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 ) 560
–
560
=0 = 0 OK !
17
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 18
Tabel Gaya Batang Akibat Beban Berguna ( P = 70 Kg ) No. Batang A1 A2 A3 A4 A5 A6 A7 A8 B1 B2 B3 B4 B5 B6 B7 B8 T1 T2 T3 T4 T5 T6 T7 D1 D2 D3 D4 D5 D6
Gaya Batang ( Kg ) Tarik ( + ) 4534,848 4534,848 3887,052 3239,038 3239,038 3887,052 4534,848 4534,848 0 1295,674 1943,386 1295,674 647,795 0 2048,663 1448,514 916,121
19
Tekan ( - ) 6413,199 5496,948 4580,736 3664,571 3664,571 4580,736 5496,948 6413,199 647,795 916,121 1448,514 2048,663 -
4.3 Gaya Batang Akibat Beban Angin Kiri dengan Cara Cremona 1 2
1 2
W
W'
W
W' A4
A5 W'
W A6
A3 W A2 1 2
D2
W A1
RAH
T1 B1
D1 B2
D3
W' A7
T5 D5
T2
1 2
T6 D6 T7
B3
RA
T3
T4 D4
B4
B5 30
B6
B7
W'
A8 B8 RB
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 ( ↑ ) 20
∑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
21
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 A1 A2 A3 A4 A5 A6 A7 A8 B1 B2 B3 B4 B5 B6 B7 B8 T1 T2 T3 T4 T5 T6 T7 D1 D2 D3 D4 D5 D6
Gaya Batang ( Kg ) Tarik ( + ) 55,682 55,682 55,682 55,682 2598,603 2598,603 2204,375 1811,147 1102,437 787,455 472,473 472,473 393,728 787,455 236,237 629,964 314,982 996,061 704,321 445,452 22
Tekan ( - ) 445,452 445,452 445,452 445,452 -
556,815 880,402 1245,076 -
4.4 Gaya Batang Akibat Beban Angin Kanan dengan Cara Cremona 1 2
1 2
W'
W
W
W' A4
A5 W
W' A6
A3 W' A2 1 2
D2
W' A1 RAH
T1 B1
D1
D3
W A7
T5 D5
T2
B2
1 2
T6 D6 T7
B3
RA
T3
T4 D4
B4
B5 30
B6
B7
W
A8 B8 RB
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 ( ↑ ) 23
∑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
24
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 A1 A2 A3 A4 A5 A6 A7 A8 B1 B2 B3 B4 B5 B6 B7 B8 T1 T2 T3 T4 T5 T6 T7 D1 D2 D3 D4 D5 D6
Gaya Batang ( Kg ) Tarik ( + ) 55,683 55,683 55,683 55,683 236,236 787,455 393,728 445,453 704,322 996,062 25
Tekan ( - ) 445,452 445,452 445,452 445,452 2362,368 2362,368 2047,385 1732,403 1023,693 629,965 236,237 236,237 314,982 629,965 1245,076 880,402 556,815
26
27
28
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
-
σ
= 1600 Kg/cm2
-
Tebal plat penyambung ( S )
= 10 mm
Fperlu
= =
Pmax σ
= 530,3 cm
= 1 cm
+ 2,5 Lk2
13382 ,468 1600
+ 2,5 . 5,3032
= 78,669 cm2 Dicoba dimensi profil ┘└ 140.140.15 -
F
= 40 × 2 = 80 cm2 Fperlu = 78,669 cm2
-
Ix = Iy
= 723
-
ix = i y
= 4,25
-
e
= 4,00
Y
e
Y 29
a. Kontrol as bahan ( sb x – x ) λx x
σdx
=
𝐿𝑘𝑥
=
𝑖𝑥
530,3 4,25
= 124,776 125
= 0,271 =
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
P
= 0,481 =
y fiktif . F . σ
= 0,481 . 80 . 1600 = 61568 Kg Pmax = 13382,468 Kg → Ok!
30
2. Satu profil / Profil tunggal x
=
y1
=
y’ .
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
31
5.3 Profil Batang Diagonal ( D ) Profil siku dobel ( ┘└ ) -
Beban maksimum ( P max )
= 2404,865 Kg ( Batang Tekan )
-
Panjang batang ( Lk )
= 5,303 m
-
σ
= 1600 Kg/cm2
-
Tebal plat penyambung ( S )
= 10 mm
Fperlu
= =
Pmax σ
= 530,3 cm
= 1 cm
+ 2,5 Lk2
2404 ,865 1600
+ 2,5 . 5,3032
= 71,807 cm2 Dicoba dimensi profil ┘└ 130.130.16 -
F
= 39,3 × 2 = 78,6 cm2 Fperlu = 71,807 cm2
-
Ix = Iy
= 605
-
ix = i y
= 3,92
-
e
= 5,37 Y
e
Y
32
a. Kontrol as bahan ( sb x – x ) λx x
σdx
=
𝐿𝑘𝑥
=
𝑖𝑥
530,3 3,92
= 135,281 136
= 0,229 =
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
P
= 0,588 =
y fiktif . F . σ
= 0,588 . 78,6 . 1600 = 73946,88 Kg Pmax = 2404,865 Kg → Ok!
33
2. Satu profil / profil tunggal x
=
y1
=
y’ .
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 34
DAFTAR REKAPITULASI DIMENSI PROFIL
Nama Batang
Nomor Batang
Batang Tepi Atas
A1 s/d A8
Batang Tepi Bawah
B1 s/d B8
Batang Diagonal
D1 s/d D6
Batang Tegak
T1 s/d T7
35
Dimensi
┘└ 140.140.15 └ 150.150.16
┘└ 130.130.16 └ 110.110.10
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×
=2×
Tebal Rata −rata plat yang disambung 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 = d . s . σtp
Ntp
= 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 ≥2d 52 mm - Cukup kuat t ≥3d 78 mm
36
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×
=2×
Tebal Rata −rata plat yang disambung 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
37
= d . s . σtp
Ntp
= 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 ≥2d 38 mm - Cukup kuat t ≥3d 57 mm
ρ1 ≥ 2 d 38 mm
ρ2 ≥ 1 ½ d 28,5 mm -
Cukup Rapat t
≤7d ≤ 189 mm
ρ1 ≤ 3 d ≤ 57 mm
ρ2 ≤ 3 d ≤ 57 mm Ambil → ρ1 = 40 mm
ρ2 = 30 mm t
= 60 mm
38
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×
=2×
Tebal Rata −rata plat yang disambung 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 = d . s . σtp
Ntp
= 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 ≥2d 54 mm - Cukup kuat t ≥3d 81 mm
ρ1 ≥ 2 d 54 mm
ρ2 ≥ 1 ½ d 40,5 mm
39
-
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×
=2×
Tebal Rata −rata plat yang disambung 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
40
Ambil yang paling kecil n =
Pmax Ntp
=
6133 ,652 6720
n = 0,913 ~ 1 buah c. Penempatan Paku Keling - Cukup tempat t ≥2d 42 mm - Cukup kuat t ≥3d 63 mm
ρ1 ≥ 2 d 42 mm
ρ2 ≥ 1 ½ d 31,5 mm -
Cukup Rapat t
≤7d ≤ 147 mm
ρ1 ≤ 3 d ≤ 63 mm
ρ2 ≤ 3 d ≤ 63 mm Ambil → ρ1 = 50 mm
ρ2 = 40 mm t
= 70 mm
41
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
o b
t
= 80 mm
= 9 cm
ρ1
= 60 mm
= 6 cm
ρ2
= 40 mm
= 4 cm
= 280 mm = 28 cm = 2 . ex + δ
hn
= 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 Lk
=
D
= 1,5 % × Pmax
5
=
530,3
L1
5
= 106,06 cm
= 1,5 % × 13382,468 = 200,737 Kg δ
= F . Z1 = 80 . 4,5 = 360 cm2
42
L
=
M
=
𝐷 .𝛿 .𝐿1 𝐼𝑥
𝐿 .ℎ𝑛 𝑒
=
200,737 .360 .106,06 723
10600 ,913 .8,8
=
6
= 10600,913 Kg
= 15548,006
Tegangan Pelat
τ
D.δ
= b .Ix =
200,737 . 360 28 .723
= 3,569 Kg/cm2
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
o b
hn
t
= 60 mm
= 6 cm
ρ1
= 40 mm
= 4 cm
ρ2
= 30 mm
= 3 cm
= 150 mm = 15 cm = 2 . ex + δ = 2. 4,29 + 0,8 = 9,38 cm
43
= 45,7 cm2
F
Ix = Iy = 949 cm4
Z1
= ex + ½ s = 4,29 + ½ . 1 = 4,79 cm
Diambil 3 Medan Lk
=
D
= 1,5 % × Pmax
3
=
375
L1
3
= 125 cm
= 1,5 % × 11746,486 = 176,197 Kg δ
= F . Z1 = 45,7 . 4,79 = 218,903 cm2
L
=
𝐷 .𝛿 .𝐿1 𝐼𝑥
M
=
=
𝐿 .ℎ𝑛 𝑒
176,197 .218,903 .125 949
=
5080 ,355 .9,38 6
= 5080,355 Kg
= 7942,288 Kg
Tegangan Pelat
τ
D.δ
= b .Ix =
176,197 . 218,903 15 .949
= 2,709 Kg/cm2
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 )
44
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
o b
t
= 60 mm
= 6 cm
ρ1
= 40 mm
= 4 cm
ρ2
= 30 mm
= 3 cm
= 130 mm = 13 cm = 2 . ex + δ
hn
= 2. 3,80 + 0,8 = 8,4 cm = 2 × 39,3 = 78,6 cm2
F
Ix = Iy = 605 cm4
Z1
= ex + ½ s = 3,80 + ½ . 1 = 4,3 cm
Diambil 3 Medan Lk
=
D
= 1,5 % × Pmax
3
=
530,3
L1
3
= 176,767 cm
= 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
45
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/cm2
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
o b
hn
t
= 70 mm
= 7 cm
ρ1
= 50 mm
= 5 cm
ρ2
= 40 mm
= 4 cm
= 110 mm = 11 cm = 2 . ex + δ = 2. 3,07 + 0,8 = 6,94 cm
F
= 21,2 cm2
Ix = Iy = 239 cm4
46
Z1
= ex + ½ s = 3,07 + ½ . 1 = 3,57 cm
Diambil 5 Medan Lk
=
D
= 1,5 % × Pmax
5
=
750
L1
5
= 150 cm
= 1,5 % × 6133,652 = 92,005 Kg δ
= F . Z1 = 21,2 . 3,57 = 75,684 cm2
L
=
𝐷 .𝛿 .𝐿1 𝐼𝑥
M
=
=
𝐿 .ℎ𝑛 𝑒
92,005 .75,684 .150 239
=
4370 ,276 .6,94 6
= 4370,276 Kg
= 5054,953 Kg
Tegangan Pelat
τ
D.δ
= b .Ix =
92,005 . 75,684 11 .239
= 2,649 Kg/cm2
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 )
47
BAB 8 LENDUTAN KONSTRUKSI Batang
F = 2.F1
L (m)
( cm2 )
( cm2 )
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
Batang
L/F
48
Gaya Batang
L/F.B.Sinα
F1
No
(Kg)
Sin α
(Kg/cm2)
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
Lendutan yang terjadi : ∑ × L/F × B × Sin α = 1744560,66 Kg/cm2
δ
E
= 2,1 × 106
L
= 30 m = 3000 cm
= =
∑ × L/F × B × Sin α 𝐸 1744560 ,66 2,1 × 106
= 0,831
≤
≤
1 500 1
500
.L
. 3000
≤ 6,0 → Aman !
Jadi lendutan yang ditinjau aman
49
50
