T6 T2 D2 D6 D1 D5 A8 A1 T1 T5 T7 D3 D4 A7 A B · PDF file4.1 Gaya Batang Akibat Beban Mati...

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

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

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

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°Pxq

Py

45°Wxq

Wy

45°W'x q

W'y

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

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

8

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.

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

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

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

A4A5

A6

A7

T1

T2

T3

T4

T5

T6

T7

B1A B

A8

45°

70

70

70

70

70

70

70

35 35

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

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

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

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

15

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

16

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 -

17

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


D1 = -916,121 Kg

D2 = -1448,514 Kg

D3 = -2048,663 Kg

D4 = +2048,663 Kg

D5 = +1448,514 Kg

D6 = +916,121 Kg

19

Tabel Gaya Batang Akibat Beban Berguna ( P = 70 Kg )



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


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



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


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



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


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

Ntp = d . s . σtp

= 2,7 . 1 . 2. 1600

= 8640 Kg


Ntp =

11746 ,486

6080

n = 1,93 ~ 2 buah


- 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


ρ2 = 30 mm

t = 60 mm

39

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

Pmax = 2404,865 Kg




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 Kg

Ntp = d . s . σtp

= 2,7 . 1 . 2. 1600

= 8640 Kg


Ntp =

11746 ,486

8640

n = 1,359 ~ 2 buah


- 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


ρ2 = 50 mm

t = 90 mm

6.4 Sambungan Paku untuk Batang Tegak ( T1 s/d T7 )

Pmax = 6133,652 Kg




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


Ntp =

6133,652

6720

n = 0,913 ~ 1 buah


- 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


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 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


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 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


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 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


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

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