hitungan baja.xls

Data - data Konstruksi :Kuda - kuda type = D

= 30 ˚Bentang Kuda - kuda ( B ) = 20 mPanjang Gudang ( L ) = 44 mJarak antar kolom = 4.5 mPenutup atap = Seng gelombangSifat dinding = TertutupTinggi kolom ( h ) = 4 m

DETAIL KUDA-KUDA

RENCANA ATAP

Kemiringan atap ( α )

P1 X 12

A

P2

D

P3

E

P4

G

P5

I

P6

K

P7

M

P8

O

P9 X 12

B

A1

V1 D1

B1 B8 C F H J L N P B2 B3 B4 B5 B6 B7

V2

V3

V4

V5

V6

V7

D2

D3 D4

D5

D6

A2

A3

A4 A5

A6

A7

A8

2000

G G

G

G G

G

G G

G

G G

G

G G

G

G G

G

G G

G

G G

G

G G

G

G G

G

KK

KK

KK

KK

KK

KK

KK

KK

KK

N

4000

2000

200

200

200 200

PERHITUNGAN PANJANG BATANG :

Batang bawah ( B ) =B : L : X → X = 8 bh

L = 20 m

B =20

= 2.5 m8

Batang atas ( A ) =A : B : Cos α → Cos 30 ˚ = 0.866

B = 2.5 m

A =2.5

= 2.887 m0.866

Batang vertikal ( V ) =V1 = V7 = 2.5 . tg 30 ˚

= 2.5 . 0.577= 1.443 m

V2 = V6 = 2.5 . tg 30 ˚= 2.5 . 0.577 . 2= 2.885 m

V3 = V5 = 2.5 . tg 30 ˚= 2.5 . 0.577 . 3= 4.328 m

V4 = 2.5 . tg 30 ˚= 2.5 . 0.577 . 4= 5.770 m

Batang diagonal ( D ) =D1 = D6 = ( 2.5 ² + 1.443 ²

= ( 6.250 + 2.081= ( 8.331= 2.886 m

D2 = D6 = ( 2.5 ² + 2.886 ²= ( 6.250 + 8.331= ( 14.581= 3.818 m

D3 = D4 = ( 2.5 ² + 4.328 ²= ( 6.250 + 18.727= ( 24.977= 4.998 m

) ½) ½

) ½

) ½) ½

) ½

) ½) ½

) ½

PERHITUNGAN DIMENSI GORDING

Dicoba memakai gording C 12H = 120 mmB = 55 mmD = 7 mmWx = 60.7 mmWy = 11.1 mmBerat profil = 13.4

# Berat gording dan atap …………………. ( K1 )gording = 13.4seng gelombang = 10.0 +

Q = 23.4Baut 10 % = 2.34 +

25.7 → Q total = 26

Qy = Q total . Cos 30 ˚= 26 . 0.866= 22.516

= ⅛ . Qy . L ²= 0.125 . 22.516 . 4.5 ²= 56.9936

= ⅛ . Q total . Sin α . L ²= 0.125 . 26 . . 4.5 ²= 0.125 . 26 . 0.5 . 20.25= 32.9063

# Beban angin pada atap …………………. ( K2 )Beban angin = 40kemiringan atap = 30 ˚

dipihak angin C1 = 0.02 . α - 0.4= 0.02 . 30 - 0.4= 0.6 - 0.4 = 0.2

dibelakang angin C2 = -0.4

Dari tabel pembebanan pengaruh angin ditentukan :beban dipihak angin W1 = C1 . A . 40

= 0.2 . 2.887 . 40= 23.09469

beban dibelakang angin W2 = C2 . A . 40

Kg.m¹

Kg.m¹Kg.m¹Kg.m²Kg.m²Kg.m² Kg.m²

Kg.m¹

Mx₁

Kg.m¹

My₁Sin 30˚

Kg.m¹

Kg.m²

Kg.m¹

= -0.4 . 2.887 . 40= -46.1894

= ⅛ . W1 . L ²= 0.125 . 23.0947 . 4.5 ²= 58.4584

= ⅛ . W2 . L ²= 0.125 . -46.189 . 4.5 ²= -116.917

= 0

# Beban hidup …………………. ( K3 )untuk beban ini ditentukan P = 100 Kg terpusat di tengah batang

= ¼ . 100 . . 4.5= 0.25 . 100 . 0.866 . 4.5= 97.425

= ¼ . 100 . . ( 4.5 / 2 )= 0.25 . 100 . 0.5 . 2.25= 28.125

Kombinasi pembebanan ( P ditiadakan saat angin bekerja )

Kombinasi K1 + K2Mx₁ + Mx₂ = 56.9936 + 58.4584

= 115.452Mx₁ + Mx₂ = 56.9936 + -116.917

= -59.923My₁ + My₂ = 32.9063 + 0

= 32.906My₁ + My₂ = 32.9063 + 0

= 32.906Total Mx = 115.452 Kg.m¹

-59.923 Kg.m¹ +55.529 Kg.m¹

Total My = 32.906 Kg.m¹32.906 Kg.m¹ +65.813 Kg.m¹

Kombinasi K1 + K3Mx₁ + Mx₃ = 56.9936 + 97.425

= 154.419My₁ + My₂ = 32.9063 + 28.125

= 61.031

Kg.m¹

Mx₂

Kg.m¹

Mx₂

Kg.m¹

My₂

Mx₃ Cos 30˚

Kg.m¹

My₂ Sin 30˚

Kg.m¹

Kg.m¹

Kg.m¹

Kg.m¹

Kg.m¹

Total Mx = 154.419 Kg.m¹

Total My = 61.031 Kg.m¹

TEGANGAN YANG TERJADI

σ =Mx

+My

Wx Wy

=154.419 . 10 ²

+61.031 . 10 ²

60.7 11.1= 254.396 + 549.831= 804.227 Kg.cm²

σ < σ Baja BJ 37804.227 Kg.cm² < 1600

OOOOKKKKKEEEEYYY

KONTROL LENDUTANe = 2.1 . 10 ⁶

Fy =4

xQ . .

+1

xP . Cos α .

384 E . I x 48 E . I x

=4

x26 . 0.866 . 450 ⁴

+1

x100 . 0.866 . 450 ³

384 2.1 . 10 ⁶ . 364 48 2.1 . 10 ⁶ . 364

=3693

+7891

293529.6 36691.2= 0.013 + 0.215 = 0.228

Fx =4

xQ . .

+1

xP . Sin α .

384 E . I x 48 E . I x

=4

x26 . 0.500 . 450 ⁴

+1

x100 . 0.500 . 450 ³

384 2.1 . 10 ⁶ . 43.21 48 2.1 . 10 ⁶ . 43.21

=2132

+4556

34844.544 4355.568= 0.061 + 1.046 = 1.107

F = ( Fx ² + Fy ²= ( 1.107 ² + 0.228 ²

JADI, kombinasi terbesar yang menentukan adalah kombinasi K1 + K3

Kg.cm²

JADI, Memakai Baja BJ 37

Cos α L⁴ L³

Sin α L⁴ L³

) ½) ½

= ( 1.226 + 0.052= ( 1.278= 1.130

F izin = 1x L =

1x 450 = 1.8

250 250

JADI, F < F izin1.130 < 1.8

OOOOKKKKKEEEEYYYPERHITUNGAN PEMBEBANAN

> Data - data Konstruksi :Berat gording = 13.2 Kg.m¹

= 30 ˚Bentang Kuda - kuda ( B ) = 25 mJarak antar kolom = 4.5 mPenutup atap = Seng gelombang : 10 Kg.cm²Penggantung Plafond = 18 Kg.cm²Tinggi kolom ( h ) = 5 mJarak gording = A

=2.887

= 0.64 mL 4.5

> Pembebanan

beban gording= 4 . 13.2 . 4.5 = 237.600 Kg.cm²

beban atap dan plafond= 28 . 2.887 . 4.5 = 363.741 Kg.cm²

beban kuda - kuda= 27 . 2.887 . 4.5 = 350.751 Kg.cm²

sambungan dan perlengkapan= 30 % 350.751 = 105.225 Kg.cm²

beban hidup = 100.000 Kg.cm² +P = 1157.32 Kg.cm²

muatan tak terduga 10 % = P x 10 %= 1157.32 x 10 %= 115.732

P total = 115.732 + 1157.317= 1273.049

RA = RB = P1 + P2 + P3 + P4 + P5 + P6 + P7 + P8 + P9

) ½) ½


= 636.524 + 1273.05 + 1273.05 + 1273.05 + 1273.05 + 1273.05 + 1273.05 +2

1273.05 + 636.524

=10184.390

2

= 5092.195 Kg

> Perhitungan pembebanan anginBeban angin = 40kemiringan atap = 30 ˚

dipihak angin C1 = 0.02 . α - 0.4= 0.02 . 30 - 0.4= 0.6 - 0.4 = 0.2


Dari tabel pembebanan pengaruh angin ditentukan :beban tekan angin W1 = C1 . 40

= 0.2 . 40= 8

beban hisap angin W2 = C2 . 40= -0.4 40= -16

Pengaruh angin tekan ( W1 ) = 8 (titik buhul)

A =2.887

x 4.5 x 8 = 51.96305 Kg2

D = 2.887 x 4.5 x 8 = 103.9261 Kg

E = 2.887 x 4.5 x 8 = 103.9261 Kg

G = 2.887 x 4.5 x 8 = 103.9261 Kg

I =2.887

x 4.5 x 8 = 51.96305 Kg2

Pengaruh angin hisap ( W2 ) = -16 (titik buhul)

I =2.887

x 4.5 x -16 = -103.926 Kg2

K = 2.887 x 4.5 x -16 = -207.852 Kg

Kg.m²

Kg.m¹

Kg.m¹

M = 2.887 x 4.5 x -16 = -207.852 Kg

O = 2.887 x 4.5 x -16 = -207.852 Kg

B =2.887

x 4.5 x -16 = -103.926 Kg2

MUATAN TETAP PER BATANG

BATANG SKALA BESAR GAYA BATANG ( Kg )

NAMA PANJANG ( cm ) GAYA ( Kg) TARIK ( + ) TEKAN ( - )

A1 21.77 1000 0 21770A2 18.66 1000 0 18660A3 15.55 1000 0 15550A4 12.44 1000 0 12440A5 12.44 1000 0 12440A6 15.55 1000 0 15550A7 18.66 1000 0 18660A8 21.77 1000 0 21770B1 18.85 1000 18850 0B2 18.85 1000 18850 0B3 16.16 1000 16160 0B4 13.46 1000 13460 0B5 13.46 1000 13460 0B6 16.16 1000 16160 0B7 18.85 1000 18850 0B8 18.85 1000 18850 0V1 0 1000 0 0V2 1.55 1000 0 1550V3 3.11 1000 0 3110V4 9.33 1000 0 9330V5 3.11 1000 0 3110V6 1.55 1000 0 1550V7 0 1000 0 0D1 3.11 1000 0 3110D2 4.11 1000 0 4110D3 5.38 1000 0 5380D4 5.38 1000 0 5380D5 4.11 1000 0 4110D6 3.11 1000 0 3110

REAKSI GAYA SIMPUL A & B

REAKSI

PANJANG GAYA ANGIN GAYA ANGIN KOMBINASI

TARIK TEKAN TARIK TEKAN TARIK TEKAN

( + ) ( - ) 1 : 1000 1 : 1000 ( + ) ( - )

RA 10.62 7.029 10620 7029 3591 0

RB 10.62 2.650 10620 2650 7970 0

RAV 5.31 5.310 5310 5310 0 0

RAH 9.2 4.6 9200 4600 4600 0

ANGIN KIRI TEKAN



A1 8.614 1000 0 8614A2 7.295 1000 0 7295A3 5.967 1000 0 5967A4 4.639 1000 0 4639A5 5.303 1000 0 5303A6 5.303 1000 0 5303A7 5.303 1000 0 5303A8 5.303 1000 0 5303B1 11.47 1000 11470 0B2 11.47 1000 11470 0B3 9.192 1000 9192 0B4 6.89 1000 6890 0B5 4.59 1000 4590 0B6 4.59 1000 4590 0B7 4.59 1000 4590 0B8 4.59 1000 4590 0V1 0 1000 0 0V2 1.131 1000 0 1131V3 2.640 1000 0 2640V4 3.970 1000 0 3970V5 0 1000 0 0V6 0 1000 0 0V7 0 1000 0 0D1 2.638 1000 0 2638D2 3.506 1000 0 3506D3 4.594 1000 0 4594D4 0 1000 0 0D5 0 1000 0 0

D6 0 1000 0 0

ANGIN KANAN HISAP



A1 10.62 1000 10620 0A2 10.62 1000 10620 0A3 10.62 1000 10620 0A4 10.62 1000 10620 0A5 9.29 1000 9290 0A6 11.95 1000 11950 0A7 14.590 1000 14590 0A8 17.27 1000 17270 0B1 0 1000 0 0B2 0 1000 0 0B3 0 1000 0 0B4 0 1000 0 0B5 4.6 1000 0 4600B6 9.19 1000 0 9190B7 13.8 1000 0 13800B8 13.8 1000 0 13800V1 0 1000 0 0V2 0 1000 0 0V3 0 1000 0 0V4 7.96 1000 7960 0V5 5.311 1000 5311 0V6 2.64 1000 2640 0V7 0 1000 0 0D1 0 1000 0 0D2 0 1000 0 0D3 0 1000 0 0D4 9.2 1000 9200 0

D5 7.01 1000 7010 0D6 5.32 1000 5320 0

GAYA RANGKA BATANG AKIBAT ANGIN ( KOMBINASI )

BATANG

ANGIN TEKAN ( Kg ) ANGIN HISAP ( Kg ) KOMBINASI ( Kg )


( + ) ( - ) ( + ) ( - ) ( + ) ( - )

A1 0 8614 10620 0 2006 0A2 0 7295 10620 0 3325 0A3 0 5967 10620 0 4653 0A4 0 4639 10620 0 5981 0A5 0 5303 9290 0 3987 0A6 0 5303 11950 0 6647 0A7 0 5303 14590 0 9287 0A8 0 5303 17270 0 11967 0B1 11470 0 0 0 11470 0B2 11470 0 0 0 11470 0B3 9192 0 0 0 9192 0B4 6890 0 0 0 6890 0B5 4590 0 0 4600 -10 0B6 4590 0 0 9190 0 4600B7 4590 0 0 13800 0 9210B8 4590 0 0 13800 0 9210V1 0 0 0 0 0 0V2 0 1131 0 0 0 1131V3 0 2640 0 0 0 2640V4 0 3970 7960 0 3990 0V5 0 0 5311 0 5311 0V6 0 0 2640 0 2640 0V7 0 0 0 0 0 0D1 0 2638 0 0 0 2638

D2 0 3506 0 0 0 3506D3 0 4594 0 0 4594D4 0 0 9200 0 9200 0D5 0 0 7010 0 7010 0D6 0 0 5320 0 5320 0

GAYA RANGKA BATANG ( KOMBINASI )AKIBAT BEBAN TETAP DAN BEBAN ANGIN

BATANG

BEBAN TETAP ( Kg ) BEBAN ANGIN ( Kg ) KOMBINASI ( Kg )


( + ) ( - ) ( + ) ( - ) ( + ) ( - )

A1 0 21770 2006 0 0 19764A2 0 18660 3325 0 0 15335A3 0 15550 4653 0 0 10897A4 0 12440 5981 0 0 6459A5 0 12440 3987 0 0 8453A6 0 15550 6647 0 0 8903A7 0 18660 9287 0 0 9373A8 0 21770 11967 0 0 9803B1 18850 0 11470 0 30320 0B2 18850 0 11470 0 30320 0B3 16160 0 9192 0 25352 0B4 13460 0 6890 0 20350 0B5 13460 0 -10 0 13450 0B6 16160 0 0 4600 11560 0B7 18850 0 0 9210 9640 0B8 18850 0 0 9210 9640 0V1 0 0 0 0 0 0V2 0 1550 0 1131 0 2681V3 0 3110 0 2640 0 5750V4 0 9330 3990 0 0 5340V5 0 3110 5311 0 2201 0

V6 0 1550 2640 0 1090 0V7 0 0 0 0 0 0D1 0 3110 0 2638 0 5748D2 0 4110 0 3506 0 7616D3 0 5380 0 4594 0 9974D4 0 5380 9200 0 3820 0D5 0 4110 7010 0 2900 0D6 0 3110 5320 0 2210 0

PERHITUNGAN PROFIL KUDA - KUDA

Batang atas A

Nama batang = A1, A2, A3, A4, A5, A6, A7, A8Gaya batang Max ( P ) = 19764 KgPanjang batang = 3.608 m → 360.8 cm

Dicoba menggunakan profil 100 . 100 . 10 Baut = 14 mmIx = 177.0F = 19.2ix = 3.04 cms = 28.2 2.82 cm

e total = 2.82 + 0.5 = 3.32 cm

iy = 2 ( Ix + F )= 2 ( 177.0 + 19.2 ( 3.32 )= 777.26 cm²

iy = ( 777.26 / 2 . F ) ½= ( 777.26 / 2 . 19.2 ) ½= ( 20.2412 ) ½= 4.499 cm

maka ix = 3.04 cm adalah i minimum ( i min )e = 2.1 . 10 ⁶

cm⁴cm²

mm →

(e)²)²

λ =LK

=360.8

= 118.684i min 3.04

= 3.14 ² . 2.1 . 10 ⁶= 20705160 = 1469.917

118.684 ² 14085.942

v kt=

1469.917= 521.247 Kg.cm²

s 2.82

P=

19764= 514.688 Kg.cm²

2 F 2 . 19.2

σ < σ kt514.688 Kg.cm² < 521.247 Kg.cm²

OOOOKKKKKEEEEYYY

Profil 100 . 100 . 10 bisa dipakai.

Batang bawah B

Nama batang = B1, B2, B3, B4, B5, B6, B7, B8Gaya batang Max ( P ) = 30320 KgPanjang batang = 3.13 m → 313 cmTegangan izin = 1600 Kg.cm²

Dicoba menggunakan profil 90 . 90 . 9 Baut = 14 mmIx = 116.0F = 15.5 → 2 F = 31ix = 2.74 cms = 25.4 2.54 cm

F efk = 31 - 2 ( 1.5 . 0.6 )= 29.2

λ =LK

=312.5

= 114.051i min 2.74

P=

30320= 1038.356 Kg.cm²

F efk 29.2

σ < Ŧ1038.356 Kg.cm² < 1600 Kg.cm²

OOOOKKKKKEEEEYYY

v kt = π² . e Kg.cm²

x²

σ kt =

Tegangan yang timbul ( σ )

σ =

cm⁴cm²

mm →

cm²

σ =


Batang diagonal D

BATANG TARIKNama batang = D4, D5, D6 Gaya batang Max ( P ) = 3820 KgPanjang batang Max = 5.409 m → 540.9 cmTegangan izin = 1600 Kg.cm²

Dicoba menggunakan profil 80 . 80 . 10 Baut = `14 mmIx = 87.5F = 15.1 → 2 F = 30.2ix = 2.41 cms = 23.4 2.34 cm

λ =LK

=540.9

= 224.440i min 2.41

λ g = π ( e / 0.7 Te ) ½= 3.14 ( 2.1 . 10 ⁶ / 0.7 . 2400 )= 111.016

λ = 224.440 = 2.021693 ≥ 1λ g 111.016

untuk λ ≥ 1 maka w = 2.381 . λ s= 2.381 . 2.021693 = 4.81365

F efk = 30.2 - 2 ( 1.5 . 1.5 )= 25.7

F efk . б =

25.7 . 1600= 8542.371 Kg

w 4.814

σ > P 8542.371 Kg > 3820 Kg

OOOOKKKKKEEEEYYY


BATANG TEKANNama batang = D1, D2, D3 Gaya batang Max ( P ) = 9974 KgPanjang batang Max = 5.409 m → 540.9 cmTegangan izin = 1600 Kg.cm²

cm⁴cm² cm²

mm →

λ s =

cm²

σ =


λ =LK

=540.9

= 197.409i min 2.74

F efk = 31 - 2 ( 1.5 . 1.5 )= 26.5

P=

9974= 321.742

F efk 31

σ < Ŧ321.742 Kg.cm² < 1600 Kg.cm²

OOOOKKKKKEEEEYYY


Batang vertikal V

BATANG TARIKNama batang = V5, V6 Gaya batang Max ( P ) = 2201 KgPanjang batang Max = 6.247 m → 624.7 cmTegangan izin = 1600 Kg.cm²

Dicoba menggunakan profil 80 . 80 . 12 Baut = 20 mmIx = 102.0F = 17.9 → 2 F = 35.8ix = 2.39 cms = 24.1 2.41 cm

λ =LK

=624.7

= 261.381i min 2.39

F efk = 35.8 - 2 ( 1.5 . 1.5 )= 31.3

P=

2201= 70.319

cm⁴cm² cm²

mm →

cm²

σ =

cm⁴cm² cm²

mm →

cm²

σ =

F efk=

31.3= 70.319

σ < Ŧ70.319 Kg.cm² < 1600 Kg.cm²

OOOOKKKKKEEEEYYY


BATANG TEKANNama batang = V2, V3, V4 Gaya batang Max ( P ) = 5340 KgPanjang batang Max = 6.247 m → 624.7 cmTegangan izin = 1600 Kg.cm²


λ =LK

=624.7

= 261.381i min 2.39

λ g = π ( e / 0.7 Te ) ½= 3.14 ( 2.1 . 10 ⁶ / 0.7 . 2400 )= 111.016

λ = 261.381 = 2.354447 ≥ 1λ g 111.016

untuk λ ≥ 1 maka w = 2.381 . λ s= 2.381 . 2.354447 = 5.60594

F efk = 35.8 - 2 ( 1.5 . 1.5 )= 31.3

F efk . б =

31.3 . 1600= 8933.383 Kg

w 5.606

σ > P8933.383 Kg > 5340 Kg

OOOOKKKKKEEEEYYY


σ =

cm⁴cm² cm²

mm →

λ s =

cm²

σ =

PERHITUNGAN BAUT PADA TIAP TITIK BUHUL

Profil seluruh kuda-kuda 100 . 100 . 10tebal plat = 10 mmdiameter baut = 14 mmdiameter lubang = 14 + 1 = 15 mmtegangan geser izin б = 1600 Kg.cm²

Ng = 0.25 . π . d ² . б= 0.25 . 3.14 . 1.5 ² . 1600= 2826 Kg

Ntp = d . t . бt . б Ng > Ntp= 1.5 . 1.0 . 0.9 . 1600 2826 > 2160= 2160 OOOOKKKKKEEEEYYY

JUMLAH BAUT TITIK BUHUL A = B

Batang A1

n =gaya A1

=19764

= 6.99363 → 5 bhNg 2826

Batang B1

n =gaya B1

=30320

= 10.729 → 7 bhNg 2826

JUMLAH BAUT TITIK BUHUL D = O

Batang A1 dan A2

n =gaya A1 - gaya A2

=19764 - 15335

= 1.567 → 3 bhNg 2826

Batang D1

n =gaya D1

=5748

= 2.034 → 3 bhNg 2826

Batang V1

n =gaya V1

=0

= 0 → 3 bhNg 2826

JUMLAH BAUT TITIK BUHUL E = M

Batang A2 dan A3


=15335 - 10897

= 1.570 → 3 bhNg 2826

Batang D2

n =gaya D2

=7616

= 2.695 → 3 bhNg 2826

Batang V2

n =gaya V2

=2681

= 0.949 → 3 bhNg 2826

JUMLAH BAUT TITIK BUHUL E = K

Batang A3 dan A4


=10897 - 6459

= 1.570 → 3 bhNg 2826

Batang D3

n =gaya D3

=9974

= 3.529 → 3 bhNg 2826

Batang V3

n =gaya V3

=5750

= 2.035 → 3 bhNg 2826

JUMLAH BAUT TITIK BUHUL I

Batang A4

n =gaya A4

=6459

= 2.286 → 3 bhNg 2826

Batang A5

n =gaya A5

=8453

= 2.991 → 3 bhNg 2826

Batang V4

n =gaya V4

=5340

= 1.890 → 3 bhNg 2826

JUMLAH BAUT TITIK BUHUL C = P

Batang B1 dan B2

n =gaya B1 - gaya B2

=30320 - 30320

= 0.000 → 3 bhNg 2826

Batang V1

n =gaya V1

=0

= 0.000 → 3 bhNg 2826

JUMLAH BAUT TITIK BUHUL F = N

Batang B2 dan B3


=30320 - 25352

= 1.758 → 3 bhNg 2826

Batang D1

n =gaya D1

=5748

= 2.034 → 3 bhNg 2826

Batang V2

n =gaya V2

=2681

= 0.949 → 3 bhNg 2826

JUMLAH BAUT TITIK BUHUL H = L

Batang B3 dan B4


=25352 - 20350

= 1.770 → 3 bh0 2826

Batang D2

n =gaya D2

=7616

= 2.695 → 3 bhNg 2826

Batang V3

n =gaya V3

=5750

= 2.035 → 3 bhNg 2826

JUMLAH BAUT TITIK BUHUL J

Batang B4

n =gaya B4

=20350

= 7.201 → 5 bhNg 2826

Batang B5

n =gaya B5

=13450

= 4.759 → 4 bhNg 2826

Batang V4

n =gaya V4

=5340

= 1.890 → 3 bhNg 2826

Batang D3

n =gaya D3

=9974

= 3.529 → 3 bhNg 2826

Batang D4

n =gaya D4

=3820

= 1.352 → 3 bhNg 2826

MENENTUKAN KOPELjumlah pelat koppel = 8 bhtebal pelat ( + ) = 8 mmgaya max = 9803 Kglebar pelat = 18 cm

gaya batang yang dipikul pelat koppel ( D ) :D = 0.015 . P

= 0.015 . 9803 = 147.045 Kg

Ng = 0.25 . π . d ² . б= 0.25 . 3.14 . 2 ² . 1600= 5024 Kg

Ntp = d . t . бt . б= 1.8 . 0.8 . 2 . 1600= 4608

Ng > Ntp maka, Ntp lah yang menentukan

n =D

=147.045

= 0.032 → 2 bh

n =Ntp

=4608

= 0.032 → 2 bh

PERHITUNGAN KOLOM

> Data - data Konstruksi :Berat gording = 13.2

= 30 ˚Bentang Kuda - kuda ( B ) = 25 mJarak bagin Kuda - kuda = 3.1 mJarak antar kolom = 4.5 mPenutup atap = Seng gelombang : 10 Kg.cm²Penggantung Plafond = 18 Kg.cm²Tinggi kolom ( h ) = 5 mJarak gording = A

=3.608

= 0.80 mL 4.5

> Pembebanan






Perhitungan reaksi kolom : PP P P

P PP P

RA RA20

∑MA = 0


.

- RB . 25 + P . 21.9 + P . 19 + P . 15.6 + P . 13 +

P . 9.4 + P . 6 + P . 3.1 + = 0

- RB . 25 + P ( 25 + 21.9 + 19 + 15.6 + 13 + 9.4 + 6 + 3.1 )=0

- RB . 25 + P ( 113 )

RB =P . 113

=113

P = 4.5 P25 25

RA = RB = 4.5 . P= 4.5 . 1362.09= 6129.412

Pendimensian kolomtinggi kolom ( h ) = 5 m → 500 mmб kolom Bj 37 = 1600 Kg.cm²N = 6129.412

Dicoba menggunakan profil WF 350 . 250t 1 = 9 mmt 2 = 14 mmA = 101.5i x = 14.6 cm²i y = 6 cm²Wx = 1280Wy = 292

arah tegak lurus sumbu xk = 1

λ x =L ky

=k L

=1 . 500

= 34.2466i x i x 14.6

λ g = π ( e / 0.7 Te ) ½= 3.14 ( 2.1 . 10 ⁶ / 0.7 . 2400 )= 111.016

λ x=

34.247= 0.308

λ g 111.016

untuk 0.183 < λ s < 1 maka

Wx =1.41

=1.41

= 1.0981.593 - λ s 1.593 - 0.308

`

nx =π² E A

=3.14 2.1 . 10 ⁶ 101.5

=2E+09

= 292.342N . 6129.412 . 34.247 7E+06

cm²

cm³cm³

λ s =

² .λ x² ²

nx=

292.342= 1.003 → 1

nx - 1 292.342 -1

Bx = 0.6 + 0.4 . Mx1/Mx2= 0.6 + 0.4 . 0= 0.6

Mx = 0.125 . N . L= 0.125 . 6129.412 . 500= 383088.263


λ y =L ky

=k L

=1 . 500

= 83.3333i y i y 6

λ g = π ( e / 0.7 Te ) ½= 3.14 ( 2.1 . 10 ⁶ / 0.7 . 2400 )= 111.016

λ y=

83.333= 0.751

λ g 111.016


Wy =1.41

=1.41

= 1.6741.593 - λ s 1.593 - 0.751

`

ny =π² E A

=3.14 2.1 . 10 ⁶ 101.5

=2E+09

= 49.373N . 6129.412 . 83.333 4E+07

ny=

49.373= 1.021 → 1

ny - 1 49.373 -1

Cek persamaan interaksia )

WxN

+ Bxny

xMx

< бA ny - 1 Wx

1.0986129.412

+ 0.6 x 1383088.263

< 1600 Kg.cm²101.5 1280

245.860 Kg.cm² < 1600 Kg.cm²

b )Wx

N< б

A

1.6746129.412

< 1600 Kg.cm²101.5

λ s =

² .λ x² ²

101.083 Kg.cm² < 1600 Kg.cm²

c ) N+

Mx< б

A Wx6129.412

+383088.263

< 1600 Kg.cm²101.5 1280

359.676 Kg.cm² < 1600 Kg.cm²

Profil WF 350 . 250 bisa dipakai.

Pelat dasar kolom pada pondasi1 kip = 4.536 . 10 ² Kg

= 6.452

Beban aksial ( P ) = 6129.412 = 1351.281P ' C = 3 ksi

Fp = 0.35 . P ' C= 0.35 . 3= 1.05 kgi

A perlu = P / FP= 1351.281 / 1.05= 1286.935

Profil yang digunakan adalah WF 350 . 250A = 101.5d = 34 cmb = 25 cm

menentukan lebar ( B ) dan panjang ( N ) pelat dimana harus memenuhi syarat :

B x N ≥ 5590.486 cm²

untuk pemilihan B dan N sedemikian hingga perbedaan ( selisih ) diantara keduanya mendekati :

0.92 d - 0.80 b0.92 . 34 - 0.8 . 25 = 11.28 cm

dicoba berbagai gaya B dan N sebagai berikut :

B x N ≥ 5590.486 cm²

B N Was selisih75 76 5700 175 78 5694 3

1 m² cm²

cm²

71 80 5680 970 83 5740 13

gunakan B = 70 dan N = 83tekanan aktual adalah

Fp =P

=6129.412

= 1.055 Kg.cm²B . N 70 . 83

hitung M dan N

M =N - 0.95 d

=83 - 0.95 . 34

= 25.35 cm2 2

N =N - 0.80 d

=70 - 0.80 . 25

= 25 cm2 2

hitung tebal pelat ( tp )tp = N ( FP / 0.25 Fy ) ½

= 25 ( 1.055 / 0.25 . 2400 )½= 1.048 → 1 cm

70 . 1 . 83

Menentukan jumlah baut

pada kolom digunakan profil WF . 350 . 250 . 9 . 14diameter baut = 2 ( 14 + 9 / 2 ) = 23 mm

diameter lubang = 23 + 1 = 24 mme1 = 2 d

= 2 . 24 = 48s = 3 d

= 3 . 24 = 72

Ng = 0.25 . π . d ² . б= 0.25 . 3.14 . 2.4 ² . 1600= 7234.56 Kg

Ntp = d . t . б= 2.4 . 0.9 . 1600= 3456


n =P

=6129.412

= 1.774 → 2 bhNtp 3456

Cek kekuatan baut :

jadi, pelat landasan yang dipakai adalah

untuk mengetahui kuat atau tidaknya baut menahan suatu beban harus memenuhi syarat berikut :б ≤ 0.8 x 1600 Kg.cm²

P

6129.4122 . 0.25 . 3.14 2.4 ² . 3

225.931 Kg.cm² ≤ 0.8 x 1600 Kg.cm²

225.931 Kg.cm² ≤ 1280 Kg.cm²OOOOKKKKKEEEEYYY

2 . 1/4 . π . d² . 3

G G

G

G G

G

G G

G

G G

G

G G

G

G G

G

G G

G

G G

G

G G

G

G G

G

KK

KK

KK

KK

KK

KK

KK

KK

KK

N

4000

2000

200

200

200 200

1706.8

30.24

0.00342064884091075

66.2876075362953

179.572623046875245.86023058317

5074.831014.966

1118.8143477589932.6887786732796

1065.53747405618

34.5= 250 ( 250 + 14 / 2 ) > 1.5 diameter lubang= 33000

l 2

Data - data Konstruksi :Kuda - kuda type = D

= 30 ˚Bentang Kuda - kuda ( B ) = 25 mPanjang Gudang ( L ) = 40 mJarak antar kolom = 4.5 mPenutup atap = Asbes GelombangSifat dinding = TertutupTinggi kolom ( h ) = 5 m

DETAIL KUDA-KUDA

RENCANA ATAP

PERHITUNGAN PANJANG BATANG :

Batang bawah ( B ) =B : L : X → X = 8 bh

L = 25 m

B =25

= 3.125 m8

Batang atas ( A ) =A : B : Cos α → Cos 30 ˚ = 0.866

B = 3.125 m

A =3.125

= 3.609 m0.866


P1 X 12

A

P2

D

P3

E

P4

G

P5

I

P6

K

P7

M

P8

O

P9 X 12

B

A1

V1 D1

B1 B8 C F H J L N P B2 B3 B4 B5 B6 B7

V2

V3

V4

V5

V6

V7

D2

D3 D4

D5

D6

A2

A3

A4 A5

A6

A7

A8

2000

G G

G

G G

G

G G

G

G G

G

G G

G

G G

G

G G

G

G G

G

G G

G

G G

G

KK

KK

KK

KK

KK

KK

KK

KK

KK

N

4000

2000

200

200

200 200

Batang vertikal ( V ) =V1 = V7 = 3.125 . tg 30 ˚

= 3.125 . 0.577= 1.803 m

V2 = V6 = 3.125 . tg 30 ˚= 3.125 . 0.577 . 2= 3.606 m

V3 = V5 = 3.125 . tg 30 ˚= 3.125 . 0.577 . 3= 5.409 m

V4 = 3.125 . tg 30 ˚= 3.125 . 0.577 . 4= 7.212 m

Batang diagonal ( D ) =D1 = D6 = ( 3.125 ² + 1.803 ²

= ( 9.766 + 3.251= ( 13.017= 3.608 m

D2 = D5 = ( 3.125 ² + 3.608 ²= ( 9.766 + 13.017= ( 22.783= 4.773 m

D3 = D4 = ( 3.125 ² + 5.409 ²= ( 9.766 + 29.261= ( 39.027= 6.247 m

PERHITUNGAN DIMENSI GORDING

Dicoba memakai gording C 12H = 120 mmB = 55 mmD = 7 mmWx = 60.7 mmWy = 11.1 mmBerat profil = 13.4

# Berat gording dan atap …………………. ( K1 )gording = 13.4asbes gelombang = 10.3 +

Q = 23.7Baut 10 % = 2.37 +

26.1 → Q total = 26

Qy = Q total . Cos 30 ˚= 26 . 0.866= 22.516

= ⅛ . Qy . L ²= 0.125 . 22.516 . 4.5 ²= 56.9936

= ⅛ . Q total . Sin α . L ²= 0.125 . 26 . . 4.5 ²= 0.125 . 26 . 0.5 . 20.25= 32.9063

# Beban angin pada atap …………………. ( K2 )Beban angin = 40kemiringan atap = 30 ˚

dipihak angin C1 = 0.02 . α - 0.4= 0.02 . 30 - 0.4= 0.6 - 0.4 = 0.2

) ½) ½

) ½

) ½) ½

) ½

) ½) ½

) ½

Kg.m¹

Kg.m¹Kg.m¹Kg.m²Kg.m²Kg.m² Kg.m²

Kg.m¹

Mx₁

Kg.m¹

My₁Sin 30˚

Kg.m¹

Kg.m²


Dari tabel pembebanan pengaruh angin ditentukan :beban dipihak angin W1 = C1 . A . 40

= 0.2 . 3.609 . 40= 28.86836

beban dibelakang angin W2 = C2 . A . 40= -0.4 . 3.609 . 40= -57.73672

= ⅛ . W1 . L ²= 0.125 . 28.868 . 4.5 ²= 73.073

= ⅛ . W2 . L ²= 0.125 . -57.73672055427 . 4.5 ²= -146.146

= 0

# Beban hidup …………………. ( K3 )untuk beban ini ditentukan P = 100 Kg terpusat di tengah batang

= ¼ . 100 . . 4.5= 0.25 . 100 . 0.866 . 4.5= 97.425

= ¼ . 100 . . ( 4.5 / 2 )= 0.25 . 100 . 0.5 . 2.25= 28.125

Kombinasi pembebanan ( P ditiadakan saat angin bekerja )

Kombinasi K1 + K2Mx₁ + Mx₂ = 56.993625 + 73.073037

= 130.067Mx₁ + Mx₂ = 56.993625 + -146.146

= -89.152My₁ + My₂ = 32.90625 + 0

= 32.906My₁ + My₂ = 32.90625 + 0

= 32.906Total Mx = 130.067 Kg.m¹

-89.152 Kg.m¹ +40.914 Kg.m¹

Total My = 32.906 Kg.m¹32.906 Kg.m¹ +65.813 Kg.m¹

Kombinasi K1 + K3Mx₁ + Mx₃ = 56.993625 + 97.425

= 154.419My₁ + My₂ = 32.90625 + 28.125

= 61.031

Total Mx = 154.419 Kg.m¹

Total My = 61.031 Kg.m¹

Kg.m¹

Kg.m¹

Mx₂

Kg.m¹

Mx₂

Kg.m¹

My₂

Mx₃ Cos 30˚

Kg.m¹

My₂ Sin 30˚

Kg.m¹

Kg.m¹

Kg.m¹

Kg.m¹

Kg.m¹

JADI, kombinasi terbesar yang menentukan adalah kombinasi K1 + K3

TEGANGAN YANG TERJADI

σ =Mx

+My

Wx Wy

=154.419 . 10 ²

+61.031 . 10 ²

60.7 11.1= 254.396 + 549.831= 804.227 Kg.cm²

σ < σ Baja BJ 37804.227 Kg.cm² < 1600

AMAN

KONTROL LENDUTANe = 2.1 . 10 ⁶

Fy =4

xQ . .

+1

xP . Cos α .

384 E . I x 48 E . I x

=4

x26 . 0.866 . 450 ⁴

+1

x100 . 0.866 . 450 ³

384 2.1 . 10 ⁶ . 364 48 2.1 . 10 ⁶ . 364

=3693

+7891

293529.6 36691.2= 0.013 + 0.215 = 0.228

Fx =4

xQ . .

+1

xP . Sin α .

384 E . I x 48 E . I x

=4

x26 . 0.500 . 450 ⁴

+1

x100 . 0.500 . 450 ³

384 2.1 . 10 ⁶ . 43.21 48 2.1 . 10 ⁶ . 43.21

=2132

+4556

34844.544 4355.568= 0.061 + 1.046 = 1.107

F = ( Fx ² + Fy ²= ( 1.107 ² + 0.228 ²= ( 1.226 + 0.052= ( 1.278= 1.130

F izin = 1x L =

1x 450 = 1.8

250 250

JADI, F < F izin1.130 < 1.8

OKEY

Kg.cm²

JADI, Memakai Baja BJ 37

Cos α L⁴ L³

Sin α L⁴ L³

) ½) ½

) ½) ½

PERHITUNGAN PEMBEBANAN

> Data - data Konstruksi :Berat gording = 13.4 Kg.m¹

= 30 ˚Bentang Kuda - kuda ( B ) = 25 mJarak antar kolom = 4.5 mPenutup atap = asbes gelombang : 10.3 Kg.cm²Penggantung Plafond = 18 Kg.cm²Tinggi kolom ( h ) = 5 mJarak gording = A

=3.609

= 0.80 mL 4.5

> Pembebanan






muatan tak terduga 10 % = P x 10 %= 1370.7179 x 10 %= 137.072

P total = 137.072 + 1370.718= 1507.790

RA = RB = P1 + P2 + P3 + P4 + P5 + P6 + P7 + P8 + P9= 753.8948441 + 1507.79 + 1507.79 + 1507.79 + 1507.79 + 1507.79 + 1507.79 +

21507.79 + 753.89484411

=12062.318

2

= 6031.159 Kg


> Perhitungan pembebanan anginBeban angin = 40kemiringan atap = 30 ˚

dipihak angin C1 = 0.02 . α - 0.4= 0.02 . 30 - 0.4= 0.6 - 0.4 = 0.2


Dari tabel pembebanan pengaruh angin ditentukan :beban tekan angin W1 = C1 . 40

= 0.2 . 40= 8

beban hisap angin W2 = C2 . 40= -0.4 40= -16

Pengaruh angin tekan ( W1 ) = 8 (titik buhul)

A =3.609

x 4.5 x 8 = 64.953811 Kg2

D = 3.609 x 4.5 x 8 = 129.90762 Kg

E = 3.609 x 4.5 x 8 = 129.90762 Kg

G = 3.609 x 4.5 x 8 = 129.90762 Kg

I =3.609

x 4.5 x 8 = 64.953811 Kg2

Pengaruh angin hisap ( W2 ) = -16 (titik buhul)

I =3.609

x 4.5 x -16 = -129.9076 Kg2

K = 3.609 x 4.5 x -16 = -259.8152 Kg

M = 3.609 x 4.5 x -16 = -259.8152 Kg

O = 3.609 x 4.5 x -16 = -259.8152 Kg

B =3.609

x 4.5 x -16 = -129.9076 Kg2

Kg.m²

Kg.m¹

Kg.m¹

MUATAN TETAP PER BATANGBATANG SKALA BESAR GAYA BATANG ( Kg )

NAMA PANJANG ( cm ) GAYA ( Kg) TARIK ( + ) TEKAN ( - )A1 13.815 1000 0 13815A2 9.330 1000 0 9330A3 6.220 1000 0 6220A4 3.110 1000 0 3110A5 3.110 1000 0 3110A6 6.220 1000 0 6220A7 9.330 1000 0 9330A8 13.815 1000 0 13815B1 18.8534 1000 18853.4 0B2 18.8534 1000 18853.4 0B3 16.160 1000 16160 0B4 13.467 1000 13466.7 0B5 13.467 1000 13466.7 0B6 16.160 1000 16160 0B7 18.8534 1000 18853.4 0B8 18.8534 1000 18853.4 0V1 0 1000 0 0V2 1.555 1000 0 1555V3 3.110 1000 0 3110V4 9.330 1000 0 9330V5 3.110 1000 0 3110V6 1.555 1000 0 1555V7 0 1000 0 0D1 3.110 1000 0 3110D2 4.114 1000 0 4114.1D3 5.387 1000 0 5386.7D4 5.387 1000 0 5386.7D5 4.114 1000 0 4114.1D6 3.110 1000 0 3110

REAKSI GAYA SIMPUL A & B

REAKSI PANJANG GAYA ANGIN GAYA ANGIN KOMBINASI

TARIK TEKAN TARIK TEKAN TARIK TEKAN( + ) ( - ) 1 : 1000 1 : 1000 ( + ) ( - )

RA 7.082 4.687 7082.2 4686.6 2395.6 0

RB 7.084 1.768 7083.6 1767.7 5315.9 0

RAV 3.5411 3.585 3541.1 3584.8 0 43.7

RAH 6.196 3.067 6195.8 3066.7 3129.1 0

ANGIN KIRI TEKANBATANG SKALA BESAR GAYA BATANG ( Kg )

NAMA PANJANG ( cm ) GAYA ( Kg) TARIK ( + ) TEKAN ( - )A1 5.743 1000 0 5742.9A2 3.096 1000 0 3095.6A3 3.094 1000 0 3094.2A4 3.093 1000 0 3092.8A5 3.535 1000 0 3535.4A6 3.535 1000 0 3535.4A7 3.535 1000 0 3535.4A8 3.535 1000 0 3535.4B1 7.652 1000 7651.9 0B2 7.652 1000 7651.9 0B3 6.128 1000 6128.4 0B4 4.595 1000 4595.1 0B5 3.062 1000 3061.8 0B6 3.062 1000 3061.8 0B7 3.062 1000 3061.8 0B8 3.062 1000 3061.8 0V1 0 1000 0 0V2 0.880 1000 0 879.6V3 1.765 1000 0 1764.9V4 2.650 1000 0 2650.1V5 0 1000 0 0V6 0 1000 0 0V7 0 1000 0 0D1 1.759 1000 0 1759.2D2 3.057 1000 0 3057D3 3.062 1000 0 3061.8D4 0 1000 0 0D5 0 1000 0 0D6 0 1000 0 0

ANGIN KANAN HISAPBATANG SKALA BESAR GAYA BATANG ( Kg )

NAMA PANJANG ( cm ) GAYA ( Kg) TARIK ( + ) TEKAN ( - )A1 7.082 1000 7082.2 0A2 7.082 1000 7082.2 0A3 7.082 1000 7082.2 0A4 7.082 1000 7082.2 0A5 6.1969 1000 6196.9 0A6 11.503 1000 11502.8 0A7 9.732 1000 9732.3 0A8 11.511 1000 11511.3 0B1 0 1000 0 0B2 0 1000 0 0B3 0 1000 0 0B4 0 1000 0 0B5 3.069 1000 0 3069.1B6 6.128 1000 0 6128.4B7 9.205 1000 0 9204.9B8 9.205 1000 0 9204.9V1 0 1000 0 0V2 0 1000 0 0V3 0 1000 0 0V4 5.364 1000 5364.3 0V5 3.541 1000 3541.1 0V6 1.766 1000 1766.3 0V7 0 1000 0 0D1 0 1000 0 0D2 0 1000 0 0D3 0 1000 0 0D4 6.133 1000 6133.3 0D5 6.124 1000 6123.5 0D6 3.552 1000 3552.4 0

GAYA RANGKA BATANG AKIBAT ANGIN ( KOMBINASI )

BATANGANGIN TEKAN ( Kg ) ANGIN HISAP ( Kg ) KOMBINASI ( Kg )

TARIK TEKAN TARIK TEKAN TARIK TEKAN( + ) ( - ) ( + ) ( - ) ( + ) ( - )

A1 0 5742.9 7082.2 0 1339.3 0A2 0 3095.6 7082.2 0 3986.6 0A3 0 3094.2 7082.2 0 3988 0A4 0 3092.8 7082.2 0 3989.4 0A5 0 3535.4 6196.9 0 2661.5 0A6 0 3535.4 11502.8 0 7967.4 0A7 0 3535.4 9732.3 0 6196.9 0A8 0 3535.4 11511.3 0 7975.9 0B1 7651.9 0 0 0 7651.9 0B2 7651.9 0 0 0 7651.9 0B3 6128.4 0 0 0 6128.4 0B4 4595.1 0 0 0 4595.1 0B5 3061.8 0 0 3069.1 -7.3 0B6 3061.8 0 0 6128.4 0 3066.6B7 3061.8 0 0 9204.9 0 6143.1B8 3061.8 0 0 9204.9 0 6143.1V1 0 0 0 0 0 0V2 0 879.6 0 0 0 879.6V3 0 1764.9 0 0 0 1764.9V4 0 2650.1 5364.3 0 2714.2 0V5 0 0 3541.1 0 3541.1 0V6 0 0 1766.3 0 1766.3 0V7 0 0 0 0 0 0D1 0 1759.2 0 0 0 1759.2D2 0 3057 0 0 0 3057D3 0 3061.8 0 0 3061.8D4 0 0 6133.3 0 6133.3 0D5 0 0 6123.5 0 6123.5 0D6 0 0 3552.4 0 3552.4 0

GAYA RANGKA BATANG ( KOMBINASI )AKIBAT BEBAN TETAP DAN BEBAN ANGIN

BATANG

BEBAN TETAP ( Kg ) BEBAN ANGIN ( Kg ) KOMBINASI ( Kg )


( + ) ( - ) ( + ) ( - ) ( + ) ( - )

A1 0 13815 1339.3 0 0 12475.7A2 0 9330 3986.6 0 0 5343.4A3 0 6220 3988 0 0 2232A4 0 3110 3989.4 0 0 -879.4A5 0 3110 2661.5 0 0 448.5A6 0 6220 7967.4 0 0 -1747.4A7 0 9330 6196.9 0 0 3133.1A8 0 13815 7975.9 0 0 5839.1B1 18853.4 0 7651.9 0 26505.3 0B2 18853.4 0 7651.9 0 26505.3 0B3 16160 0 6128.4 0 22288.4 0B4 13466.7 0 4595.1 0 18061.8 0B5 13466.7 0 -7.3 0 13459.4 0B6 16160 0 0 3066.6 13093.4 0B7 18853.4 0 0 6143.1 12710.3 0B8 18853.4 0 0 6143.1 12710.3 0V1 0 0 0 0 0 0V2 0 1555 0 879.6 0 2434.6V3 0 3110 0 1764.9 0 4874.9V4 0 9330 2714.2 0 0 6615.8V5 0 3110 3541.1 0 431.1 0V6 0 1555 1766.3 0 211.3 0V7 0 0 0 0 0 0D1 0 3110 0 1759.2 0 4869.2D2 0 4114.1 0 3057 0 7171.1D3 0 5386.7 0 3061.8 0 8448.5D4 0 5386.7 6133.3 0 746.6 0D5 0 4114.1 6123.5 0 2009.4 0D6 0 3110 3552.4 0 442.4 0

PERHITUNGAN PROFIL KUDA - KUDA

Batang atas A

Nama batang = A1, A2, A3, A4, A5, A6, A7, A8Gaya batang Max ( P ) = 12475.7 KgPanjang batang = 3.609 m → 360.9 cm

Dicoba menggunakan profil 100 . 100 . 10 Baut = 14 mmIx = 177.0F = 19.2ix = 3.04 cms = 28.2 2.82 cm

e total = 2.82 + 0.5 = 3.32 cm

iy = 2 ( Ix + F )= 2 ( 177.0 + 19.2 ( 3.32 )= 777.26 cm²

iy = ( 777.26 / 2 . F ) ½= ( 777.26 / 2 . 19.2 ) ½= ( 20.2412 ) ½= 4.499 cm

maka ix = 3.04 cm adalah i minimum ( i min )e = 2.1 . 10 ⁶

λ =LK

=360.9

= 118.702i min 3.04

= 3.14 ² . 2.1 . 10 ⁶= 20705160 = 1469.473

118.702 ² 14090.198

v kt=

1469.473= 521.090 Kg.cm²

s 2.82

P=

12475.7= 324.888 Kg.cm²

2 F 2 . 19.2

σ < σ kt324.888 Kg.cm² < 521.090 Kg.cm²

OOOOKKKKKEEEEYYY


cm⁴cm²

mm →

(e)²)²

v kt = π² . e Kg.cm²

x²

σ kt =

Tegangan yang timbul ( σ )

σ =

Batang bawah B

Nama batang = B1, B2, B3, B4, B5, B6, B7, B8Gaya batang Max ( P ) = 26505.3 KgPanjang batang = 3.13 m → 313 cmTegangan izin = 1600 Kg.cm²


F efk = 31 - 2 ( 1.5 . 0.6 )= 29.2

λ =LK

=312.5

= 114.051i min 2.74

P=

26505.3= 907.716 Kg.cm²

F efk 29.2

σ < Ŧ907.716 Kg.cm² < 1600 Kg.cm²

OOOOKKKKKEEEEYYY


Batang diagonal D

BATANG TARIKNama batang = D4, D5, D6 Gaya batang Max ( P ) = 746.6 KgPanjang batang Max = 6.247 m → 624.7 cmTegangan izin = 1600 Kg.cm²

Dicoba menggunakan profil 80 . 80 . 10 Baut = `14 mmIx = 87.5F = 15.1 → 2 F = 30.2ix = 2.41 cms = 23.4 2.34 cm

λ =LK

=624.7

= 259.218i min 2.41

λ g = π ( e / 0.7 Te ) ½= 3.14 ( 2.1 . 10 ⁶ / 0.7 . 2400 )= 111.016

λ = 259.218 = 2.3349666 ≥ 1λ g 111.016

cm⁴cm²

mm →

cm²

σ =

cm⁴cm² cm²

mm →

λ s =

untuk λ ≥ 1 maka w = 2.381 . λ s= 2.381 . 2.3349666 = 5.559556

F efk = 30.2 - 2 ( 1.5 . 1.5 )= 25.7

F efk . б =

25.7 . 1600= 7396.275 Kg

w 5.560

σ > P 7396.275 Kg > 747 Kg

OOOOKKKKKEEEEYYY


BATANG TEKANNama batang = D1, D2, D3 Gaya batang Max ( P ) = 8448.5 KgPanjang batang Max = 6.247 m → 624.7 cmTegangan izin = 1600 Kg.cm²


λ =LK

=624.7

= 227.998i min 2.74

F efk = 31 - 2 ( 1.5 . 1.5 )= 26.5

P=

8448.5= 272.532

F efk 31

σ < Ŧ272.532 Kg.cm² < 1600 Kg.cm²

OOOOKKKKKEEEEYYY


cm²

σ =

cm⁴cm² cm²

mm →

cm²

σ =

Batang vertikal V

BATANG TARIKNama batang = V5, V6 Gaya batang Max ( P ) = 431.1 KgPanjang batang Max = 5.409 m → 540.9 cmTegangan izin = 1600 Kg.cm²


λ =LK

=540.9

= 226.334i min 2.39

F efk = 35.8 - 2 ( 1.5 . 1.5 )= 31.3

P=

431.1= 13.773

F efk 31.3

σ < Ŧ13.773 Kg.cm² < 1600 Kg.cm²

OOOOKKKKKEEEEYYY


BATANG TEKANNama batang = V2, V3, V4 Gaya batang Max ( P ) = 6615.8 KgPanjang batang Max = 7.212 m → 721.3 cmTegangan izin = 1600 Kg.cm²


λ =LK

=721.3

= 301.778i min 2.39

λ g = π ( e / 0.7 Te ) ½= 3.14 ( 2.1 . 10 ⁶ / 0.7 . 2400 )= 111.016

λ = 301.778 = 2.7183368 ≥ 1λ g 111.016

untuk λ ≥ 1 maka w = 2.381 . λ s= 2.381 . 2.7183368 = 6.47236

F efk = 35.8 - 2 ( 1.5 . 1.5 )= 31.3

F efk . б =

31.3 . 1600= 7737.518 Kg

w 6.472

σ > P7737.518 Kg > 6616 Kg

OOOOKKKKKEEEEYYY


cm⁴cm² cm²

mm →

cm²

σ =

cm⁴cm² cm²

mm →

λ s =

cm²

σ =

PERHITUNGAN BAUT PADA TIAP TITIK BUHUL

Profil seluruh kuda-kuda 100 . 100 . 10tebal plat = 10 mmdiameter baut = 14 mmdiameter lubang = 14 + 1 = 15 mmtegangan geser izin б = 1600 Kg.cm²

Ng = 0.25 . π . d ² . б= 0.25 . 3.14 . 1.5 ² . 1600= 2826 Kg

Ntp = d . t . бt . б Ng > Ntp= 1.5 . 1.0 . 0.9 . 1600 2826 > 2160= 2160 OOOOKKKKKEEEEYYY

JUMLAH BAUT TITIK BUHUL A = B

Batang A1

n =gaya A1

=12475.7

= 4.414614 → 5 bhNg 2826

Batang B1

n =gaya B1

=26505.3

= 9.379 → 7 bhNg 2826

JUMLAH BAUT TITIK BUHUL D = O

Batang A1 dan A2


=12475.7 - 5343.4

= 2.524 → 3 bhNg 2826

Batang D1

n =gaya D1

=4869.2

= 1.723 → 2 bhNg 2826

Batang V1

n =gaya V1

=0

= 0 → 2 bhNg 2826

JUMLAH BAUT TITIK BUHUL E = M

Batang A2 dan A3


=5343.4 - 2232

= 1.101 → 2 bhNg 2826

Batang V2

n =gaya V2

=2434.6

= 0.862 → 2 bhNg 2826

JUMLAH BAUT TITIK BUHUL E = K

Batang A3 dan A4


=2232 - -879.4

= 1.101 → 3 bhNg 2826

Batang D3

n =gaya D3

=8448.5

= 2.990 → 3 bhNg 2826

Batang V3

n =gaya V3

=4874.9

= 1.725 → 2 bhNg 2826

Batang D2

n =gaya D2

=7171.1

= 2.538 → 3 bhNg 2826

JUMLAH BAUT TITIK BUHUL I

Batang A4

n =gaya A4

=-879.4

= -0.311 → 2 bhNg 2826

Batang A5

n =gaya A5

=448.5

= 0.159 → 2 bhNg 2826

Batang V4

n =gaya V4

=6615.8

= 2.341 → 3 bhNg 2826

JUMLAH BAUT TITIK BUHUL C = P

Batang B1 dan B2


=26505.3 - 26505.3

= 0.000 → 2 bhNg 2826

Batang V1

n =gaya V1

=0

= 0.000 → 2 bhNg 2826

JUMLAH BAUT TITIK BUHUL F = N

Batang B2 dan B3


=26505.3 - 22288.4

= 1.492 → 2 bhNg 2826

Batang D1

n =gaya D1

=4869.2

= 1.723 → 2 bhNg 2826

Batang V2

n =gaya V2

=2434.6

= 0.862 → 2 bhNg 2826

Batang D2

n =gaya D2

=7171.1

= 2.538 → 3 bhNg 2826

JUMLAH BAUT TITIK BUHUL H = L

Batang B3 dan B4


=22288.4 - 18061.8

= 1.496 → 2 bh0 2826

Batang V3

n =gaya V3

=4874.9

= 1.725 → 2 bhNg 2826

JUMLAH BAUT TITIK BUHUL J

Batang B4

n =gaya B4

=18061.8

= 6.391 → 7 bhNg 2826

Batang B5

n =gaya B5

=13459.4

= 4.763 → 5 bhNg 2826

Batang V4

n =gaya V4

=6615.8

= 2.341 → 3 bhNg 2826

Batang D3

n =gaya D3

=8448.5

= 2.990 → 3 bhNg 2826

Batang D4

n =gaya D4

=746.6

= 0.264 → 3 bhNg 2826

MENENTUKAN KOPELjumlah pelat koppel = 8 bhtebal pelat ( + ) = 8 mmgaya max = 5839.1 Kglebar pelat = 18 cm

gaya batang yang dipikul pelat koppel ( D ) :D = 0.015 . P

= 0.015 . 5839.1 = 87.586 Kg

Ng = 0.25 . π . d ² . б= 0.25 . 3.14 . 2 ² . 1600= 5024 Kg

Ntp = d . t . бt . б= 1.8 . 0.8 . 2 . 1600= 4608


n =D

=87.586

= 0.019 → 2 bhNtp 4608

PERHITUNGAN KOLOM

> Data - data Konstruksi :Berat gording = 13.4

= 30 ˚Bentang Kuda - kuda ( B ) = 25 mJarak bagin Kuda - kuda = 3.125 mJarak antar kolom = 4.5 mPenutup atap = Asbes Gelombang 10.3 Kg.cm²Penggantung Plafond = 18 Kg.cm²Tinggi kolom ( h ) = 5 mJarak gording = A

=3.609

= 0.802 mL 4.5

> Pembebanan







Perhitungan reaksi kolom : PP P P

P PP P

RA RA25

∑MA = 0- RB . 25 + P . 21.9 + P . 19 + P . 15.6 + P . 13 +

P . 9.4 + P . 6 + P . 3.1 + = 0

- RB . 25 + P ( 25 + 21.9 + 19 + 15.6 + 13 + 9.4 + 6 + 3.1 )=0

- RB . 25 + P ( 112.50 )

RB =P . 112.50

=112.50

P = 4.5 P25 25

RA = RB = 4.5 . P= 4.5 . 1370.72= 6168.231

Pendimensian kolomtinggi kolom ( h ) = 5 m → 500 mmб kolom Bj 37 = 1600 Kg.cm²N = 6168.231

Dicoba menggunakan profil WF 350 . 250t 1 = 9 mmt 2 = 14 mmA = 101.5i x = 14.6 cm²i y = 6 cm²Wx = 1280Wy = 292


λ x =L ky

=k L

=1 . 500

= 34.2466i x i x 14.6

λ g = π ( e / 0.7 Te ) ½= 3.14 ( 2.1 . 10 ⁶ / 0.7 . 2400 )= 111.016

λ x=

34.247= 0.308

λ g 111.016

cm²

cm³cm³

λ s =

.


Wx =1.41

=1.41

= 1.0981.593 - λ s 1.593 - 0.308

`

nx =π² E A

=3.14 2.1 . 10 ⁶ 101.5

=2E+09

= 290.502N . 6168.231 . 34.247 7E+06

nx=

290.502= 1.003 → 1

nx - 1 290.502 -1

Bx = 0.6 + 0.4 . Mx1/Mx2= 0.6 + 0.4 . 0= 0.6

Mx = 0.125 . N . L= 0.125 . 6168.231 . 500= 385514.408920323


λ y =L ky

=k L

=1 . 500

= 83.3333i y i y 6

λ g = π ( e / 0.7 Te ) ½= 3.14 ( 2.1 . 10 ⁶ / 0.7 . 2400 )= 111.016

λ y=

83.333= 0.751

λ g 111.016


Wy =1.41

=1.41

= 1.6741.593 - λ s 1.593 - 0.751

`

ny =π² E A

=3.14 2.1 . 10 ⁶ 101.5

=2E+09

= 49.062N . 6168.231 . 83.333 4E+07

ny=

49.062= 1.021 → 1

ny - 1 49.062 -1

Cek persamaan interaksia )

WxN

+ Bxny

xMx

< бA ny - 1 Wx

1.0986168.231

+ 0.6 x 1385514.4089

< 1600 Kg.cm²101.5 1280

247.417 Kg.cm² < 1600 Kg.cm²

² .λ x² ²

λ s =

² .λ x² ²

b )Wx

N< б

A

1.6746168.231

< 1600 Kg.cm²101.5

101.723 Kg.cm² < 1600 Kg.cm²

c ) N+

Mx< б

A Wx6168.231

+385514.40892032

< 1600 Kg.cm²101.5 1280

361.954 Kg.cm² < 1600 Kg.cm²

Profil WF 350 . 250 bisa dipakai.

Pelat dasar kolom pada pondasi1 kip = 4.536 . 10 ² Kg

= 6.452

Beban aksial ( P ) = 6168.231 = 1359.839P ' C = 3 ksi

Fp = 0.35 . P ' C= 0.35 . 3= 1.05 kgi

A perlu = P / FP= 1359.839 / 1.05= 1295.085

Profil yang digunakan adalah WF 350 . 250A = 101.5d = 34 cmb = 25 cm

menentukan lebar ( B ) dan panjang ( N ) pelat dimana harus memenuhi syarat :

B x N ≥ 5590.486 cm²

1 m² cm²

cm²

untuk pemilihan B dan N sedemikian hingga perbedaan ( selisih ) diantara keduanya mendekati :

0.92 d - 0.80 b0.92 . 34 - 0.8 . 25 = 11.28 cm

dicoba berbagai gaya B dan N sebagai berikut :

B x N ≥ 5590.486 cm²

B N Was selisih75 76 5700 175 78 5694 371 80 5680 970 83 5740 13

gunakan B = 70 dan N = 83tekanan aktual adalah

Fp =P

=6168.231

= 1.062 Kg.cm²B . N 70 . 83

hitung M dan N

M =N - 0.95 d

=83 - 0.95 . 34

= 25.35 cm2 2

N =N - 0.80 d

=70 - 0.80 . 25

= 25 cm2 2

hitung tebal pelat ( tp )tp = N ( FP / 0.25 Fy ) ½

= 25 ( 1.062 / 0.25 . 2400 )½= 1.052 → 1 cm

70 . 1 . 83

Menentukan jumlah baut

pada kolom digunakan profil WF . 350 . 250 . 9 . 14diameter baut = 2 ( 14 + 9 / 2 ) = 23 mm

diameter lubang = 23 + 1 = 24 mme1 = 2 d

= 2 . 24 = 48s = 3 d

= 3 . 24 = 72

Ng = 0.25 . π . d ² . б= 0.25 . 3.14 . 2.4 ² . 1600= 7234.56 Kg

jadi, pelat landasan yang dipakai adalah

Ntp = d . t . б= 2.4 . 0.9 . 1600= 3456


n =P

=6168.231

= 1.785 → 2 bhNtp 3456

Cek kekuatan baut :untuk mengetahui kuat atau tidaknya baut menahan suatu beban harus memenuhi syarat berikut :б ≤ 0.8 x 1600 Kg.cm²

P

6168.2312 . 0.25 . 3.14 2.4 ² . 3

227.362 Kg.cm² ≤ 0.8 x 1600 Kg.cm²

227.362 Kg.cm² ≤ 1280 Kg.cm²OOOOKKKKKEEEEYYY

2 . 1/4 . π . d² . 3

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