Kantilever

FAKULTAS TEKNIK SIPIL DAN TUGAS STRUKTUR BAJA HALAMAN: 49 PERENCANAAN BAGIAN STRUKTUR: NO. GAMBAR: NRP: 21409159

UNIVERSITAS KRISTEN PETRA KANTILEVER 21409169

PERHITUNGAN

BAB VII

KANTILEVER

VII.1. Perhitungan Kantilever

Digunakan profil WF 100×50×5×7

Data – data :

h = 100 mm tf = 7 mm Iy = 14.8 cm4 Sx = 37.5 cm3 Zy = 8750

mm3

b = 50 mm A = 1185 mm2 ix = 39.8 mm Sy = 5.91 cm3 q = 9.3

kg/m

tw = 5 mm Ix = 187 cm4 iy = 11.2 mm Zx = 44157.3 mm3

Pembebanan :

a. Beban mati :

reaksi gording (Pd) = 13.695 × 6 = 82.17 kg (lihat Bab II -

Gording)

berat sendiri kantilever (qd) = 9.3 kg/m

b. Beban hidup :

beban hidup gording = 100 kg (lihat Bab II -

Gording)

berat talang = 0.2 × 0.2 × 6 × 1000 = 240 kg +

beban hidup total (Pl) = 340 kg

Kombinasi pembebanan :

FAKULTAS TEKNIK SIPIL DAN TUGAS STRUKTUR BAJA HALAMAN: 49 PERENCANAAN BAGIAN STRUKTUR: NO. GAMBAR: NRP: 21409159

UNIVERSITAS KRISTEN PETRA KANTILEVER 21409169

PERHITUNGAN

Pu = 1.2D+1.6L = 1.2 × 82.17 + 1.6 × 340 = 642.604 kg

Pu = 1.2D+0.5L+1.3W = 1.2 × 82.17 + 0.5 × 340 = 268.604 kg

Pu = 1.4D = 1.4 × 82.17 = 115.038 kg

Pu = 642.604 kg

Perhitungan momen lentur :

Mux = 642 .604 ×1+1. 2×0. 5×9. 3×( 1

cos 150 )2

= 694.72 kgm = 6.9472 kNm

Perhitungan gaya normal tarik :

Nu = Pu sin α +1.2 × q × L × cos α × sin α

= 642 .604 ×sin 150+1. 2×9 . 3×1×cos150×sin 150 = 169.12 kg

Perhitungan gaya geser :

Vu = 642.604 + 1.2 x 9.3 x 1 / cos 15º

= 654.16 kg

Cek kapasitas penampang profil WF 100×50×5×7

a. Terhadap momen lentur :

Sayap Badan

= b/t =50/7 = 7.143 = h/tw =86/5 = 17.2

p =170/√ fy =170/√240=10.97 p =1185/√ fy =1185/√240 = 76.49

< p (penampang kompak) < p (penampang kompak)

Mnx = Mpx = Zx×Fy

= 44 .1573×103 mm3×240

Nmm2

= 10597 . 792×103 Nmm

= 10597 .752 Nm

ø Mnx = 9537.977 > Mux …(OK).

b. Terhadap lateral torsional buckling :

Lp = 1.76 iy √ Efy = 1.76 x 11.2 x √ 2 .1×105

240 = 583.089 mm

fL = fy – fr = 240 – 70 = 170 MPa

E = 2.1 x 105 Mpa

PERHITUNGAN

NRP : 2140916421409167

BAGIAN STRUKTUR:

KANTILEVER

Halaman : 50TUGAS STRUKTUR BAJAFAKULTAS TEKNIK SIPIL DAN

PERENCANAANUNIVERSITAS KRISTEN PETRA

G =

E2 (1+μ) =

2. 1×105

2 (1+0 .3 ) = 80769.231 MPa

J = Σ

bt3

3 =

(2×100×73+86×53 )3 = 29058.667 mm4

A = 1185 mm2

X1 =

πSx √ EGJA

2 =

π37500 √ 2 .1×105×80769 .23×29057 .667×1185

2

= 45271.593 Mpa

Iw = 0.25 Iy h2 = 0.25 x 14.8 x 104 x 862 = 0.0274 x 1010 mm6

X2 = 4

IwIy (

SxG J )2 = 4 x

0 .0274×1010

148000 (

3750080769 . 23×29057 .667 )2 = 0.189x10-5 mm4

Lr = iy

X1

fL √1+√1+ X2×fL2

= 11.2 (

45271 . 593170 )√1+√1+0.189×10−5×1702

= 4246.353 mm

L = 1000 mm

Lp < L < Lr maka termasuk bentang menengah.

Mr = Sx (fy – fr) = 37500 x (240 – 70) = 6.375 kNm

Cb = 2.3

Mn = Cb ( Mr + (Mp – Mr)

Lr−LLr−Lp )

= 2.3 ( 6.375 + (10.967 – 6.375)

4246 .353−10004246 .353−583. 089 )

= 24.022 kNm

Mnx (21.62 kNm) > Mux (6.9472 kNm) …(OK).

c. Terhadap gaya geser :

kn = 5 + 5/(a/h)2

= 5 + 5/(1000/86)2

= 5.03698

PERHITUNGAN

NRP : 2140916421409167

BAGIAN STRUKTUR:

KANTILEVER



Cek : h/tw ≤1 .1√ kn . E

fy

17.2 ≤ 73.0267 termasuk plastic buckling in shear.

ø Vn = 0.9 x 0.6 x fy x Aw

= 0.9 x 0.6 x 240 x 100 x 6

= 77.76 kN > Vu (6.513 kN) …(OK).

d. Terhadap gaya normal tarik:

Kondisi leleh: Kondisi fraktur:

øNn=ø×A×fy øNn=ø×Ag× fu

= 0 .85×1185×240 = 0 .75×1185×370

= 241740 N = 328837.5 N

ø Nn = 241740 N ≥ Nu = 1684.3 N ...(OK).

e. Rumus interaksi:

Nu2 φ Nn

+( Muxφ bMnx

+ Muyφ bMny )≤1

..........(SNI 2002 pasal 11.3 hal 75)

1684 .32×241740

+(64590 . 9×24022

+0)≤1

0.3 < 1…(OK).

Hasil perhitungan:

Untuk kantilever digunakan profil WF 100×50×5×7, karena telah memenuhi syarat

rumus interaksi dari SNI 2002 pasal 11.3 hal 75.

PERHITUNGAN

NRP : 2140916421409167

BAGIAN STRUKTUR:

KANTILEVER



Kantilever

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