BAB 1 PERHITUNGAN PANJANG BATANG

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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 A4 A5 A6 A7 T1 T2 T3 T4 T5 T6 T7 B1 A B A8 45 ° A1 _T1 B1 A1 A2 T1 T2 B1 B2

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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 _{+ 3,75}2 = 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

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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

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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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5 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

q

y = q cos α = 199,619 cos 45° = 141,152 Kg/m

q

x = q sin α = 199,619 sin 45° = 141,152 Kg/m M

q

y = 1 8 . qy .λ 2 M

q

x = 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° qx q qy

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6

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°

Px

q

Py

45°

Wx

q

Wy

45°

W'x

_q

W'y

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7 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

1600 Kg/cm2 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

49,497.300³ 2,1𝑥106_.451 = 0,186583 cm

δ

=

𝛿𝑥2 _{+ 𝛿𝑦}2 = 0,0086052 _{+ 0,186583}2 = 0,18678 cm

δ

≤

1 250

. λ

≤

1 250

. 300

0,18678 cm ≤ 1,2 cm Aman !

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9 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

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11 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

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12 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 A4 A5 A6 A7 T1 T2 T3 T4 T5 T6 T7 B1 A B A8 45 ° 70 70 70 70 70 70 70 35 35

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13

 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 A4 A5 A6 A7 T1 T2 T3 T4 T5 T6 T7 B1 A 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 A4 A5 A6 A7 T1 T2 T3 T4 T5 T6 T7 B1 A 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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14 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 A4 A5 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 A4 A5 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

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 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 )

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 -

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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 A4 _A5 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 1 2 W 1 2 W W W 1 2 W' 1 2 W' W' W' W' W

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∑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

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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

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 -

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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 A4 _A5 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 1 2 W 1 2 W W 1 2 W' 1 2 W' W' W' W' W W

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∑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

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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

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

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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 cm2 - Ix = Iy = 723 - ix = iy = 4,25 - e = 4,00 Y Y e

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30 a. Kontrol as bahan ( sb x – x ) λx = 𝐿𝑘𝑥 𝑖𝑥 = 530,3 4,25 = 124,776  125 x = 0,271

σ

dx = x .

σ

= 0,271 × 1600 = 433,6 Kg/cm2 P =

σ

dx . 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 yfiktif = 0,481 P = _yfiktif . F . σ = 0,481 . 80 . 1600 = 61568 Kg  Pmax = 13382,468 Kg → Ok!

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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

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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 cm2 - Ix = Iy = 605 - ix = iy = 3,92 - e = 5,37 Y Y e

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33 a. Kontrol as bahan ( sb x – x ) λx = 𝐿𝑘𝑥 𝑖𝑥 = 530,3 3,92 = 135,281  136 x = 0,229

σ

dx = x .

σ

= 0,229 × 1600 = 366,4 Kg/cm2 P =

σ

dx . 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 yfiktif = 0,588 P = _yfiktif . F . σ = 0,588 . 78,6 . 1600 = 73946,88 Kg  Pmax = 2404,865 Kg → Ok!

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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 ) = 6133,652Kg ( 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

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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

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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)

2 = 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 × 10+16

2

=

26 mm

d = 26 + 1 = 27 mm

Ngs = 2 π d² 4 × τ = 2 π 2,7²

4 × 0,8 × 1600 = 14657,415

(38)

38

₂ = 30 mm t = 60 mm

(39)

39

6.3 Sambungan Paku untuk Batang Diagonal ( D1 s/d D6 )

Pmax = 2404,865 Kg

Sambungan profil

┘└

2

=

2 × 10+16

2

= 26 mm

d = 26 + 1 = 27 mm

Pmax = 6133,652 Kg

Sambungan profil

┘└

2

=

2 × 10+10

2

=

20 mm

d = 20 + 1 = 21 mm

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)

₂ = 40 mm t = 70 mm

(42)

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)

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

= _{¼𝜋 .𝑑²}R

≤

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

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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

τ

= _{¼𝜋 .𝑑²}R

≤

τ

= 2357,037 ¼𝜋 .2,7²

≤

0,8 . 1600 = 411,669 ≤ 1280 Kg/cm2

( Aman )

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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)

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

R = ¼ . 𝐿² + 𝑚² = ¼ . 3562,206 ² + 4987,088² = 1532,163 Kg Kontrol Tegangan

τ

= _{¼𝜋 .𝑑²}R

≤

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 = 6133,652Kg 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

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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

τ

= R ¼𝜋 .𝑑²

≤

τ

= 1670,551 ¼𝜋 .2,1²

≤

0,8 . 1600 = 482,316 ≤ 1280 Kg/cm2

( Aman )

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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

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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

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