Contoh Soal:kapal konstruksi memanjang mpembujur adalah searah sumbu Xpelintang adalah searah sumbu Y

Y 5.6 m

X b

a

nama bagian simbol formula nilai satuan keteranganpanjang bagian - - 8400 mm

lebar bagian - - 5600 mmjarak pembujur * b - 700 mm * tergantung perencanaanjarak pelintang * a - 2100 mm

plate thickness as built ta - 12 mm

material factor k - 1 mm

corrosion allowance acc. to K 1.70 mm for ta > 10 mm,

nominal plate thickness t 10.30 mmlateral load p - 100 Pa

0.10beban tekan - 60 MPa

60- 30 MPa

30t - 30 MPa

30aspect ratio a a/b 3 -

number of single plate… n 8 -correction factor F1 - 1 - for stiffeners sniped at both ends

modulus young E - 2.06E+05 for steel

nominal yield point - 235 for hull structural steels according to Section 2, B.2.

safety factor S - 1.1 - in general

tK 0.1 ta / (k)0.5 + 0.5

for ta 10 mm tK = 1.5 mm

ta - tK

KN/mm2

sx

N/mm2 sy

N/mm2

N/mm2

S long. stiffener + 1

N/mm2

ReH N/mm2

£

load case nama bagian simbol formula nilai satuan keteranganedge stress ratio y - 1 - 1buckling factor K 4 -

b - c 1.13 - 1.25

reference stress 40.14099

reference degree of slenderness 1.20979 - 0.673 to be usedweighting factor F2 - 0 -

reductions factor 0.76419 - 1

F2 0 : 0.76419 -

F 1.40346 -

0.47284 -

load caseedge stress ratio y - 1 - 1

buckling factor K 1.23457 -- c 1.13 - 1.25

reference stress 40.14099

b reference degree of slenderness 2.17762 - 0.673 to be used

reductions factor 0.46649 - 1- 1.87 - 1.87

weighting factor F2 0.88099 -

F 3.20721 -

0.17980 -

F2 > 0 0.243976 -

F3 2.25E-04 -

load case

- 4.89556 - 1

t buckling factor K 8.4794 -

reference stress 40.14099

b reference degree of slenderness 0.83092 -

reduction factor 1.01093 - 1weighting factor F2 - 0

a > sx sx 8.4/(y + 1.1) 1 y 0

1.25 - 0.12 yse 0.9 . E (t/b)2 kN/m2

a b y .sx l (ReH/K .se)^0.5

kx c(1/l - 0.22/l2)

sx > 0 (compressive stress)

kpx kpx = kx

0.5(1+ 0.34(l - 0.2)+l2)

kk 1/(F+(F2-l2)0.5) for l > 0.2

a

F1(1+1/a2)2*2.1/(y+1.1) 1 y 01.25 - 0.12 y

sy y sy se 0.9 . E (t/b)2 kN/m2

l (ReH/K .se)^0.5

ky c(1/l - 0.22/l2) sy y sy lp lp = l 1.225 lp

(lp2+0.5 - K/0.91)/(lp2 - 0.5)a b

sy > 0 (compressive stress)

0.5(1+ 0.34(l - 0.2)+l2)

kk 1/(F+(F2-l2)0.5) for l > 0.2

kpy (1 - F22)ky + F22kk

sx and sy > 0 (compressive stress) (kpx kpy)5

Kt (5.34 - 4/a2) a

Kt . 30.5

se 0.9 . E (t/b)2 kN/m2

l (ReH/K .se)^0.5

kt 0.84/l

a b

t

£

£

³

£

t £

£

££

³

t £

³ ³

³³ ³

³

exponents e1 1.34104 -

e2 1.00354 -

e3 1.19054 -

proof of single plate fields

S 1.1 66

60 179.585

0.76419

235 0.26123 must be < 1 1).e1 1.34104

S 1.1 33

30 57.334

0.24398

235 0.57444 must be < 1 2).e2 1.00354

F3 0.00023 2178

60 0.039438660027161630

S 1.1 8.88525362210658E-06 must be < 1 3).

235

30 57.1576766497729

S 1.1 237.568344967712

1.01093

235 0.183398140127156 must be < 1 4).e3 1.19054

cek buckling :

1). + 2). - 3). + 4). 1.0

= 1.01905916894972 buckling

1 + kpx4

1 + kpy4

1 + kpx . kpy . kt2

|sx| . S

|sx| kpx . ReH

kpx

ReH ( |sx| . S / kpx . ReH )e1

|sy| . S

|sy| kpy . ReH

kpy

ReH ( |sy| . S / kpy . ReH )e2

sx . sy . S2

sx sx . sy . S2 / ReH2

sy

F3( sx . sy . S2 / ReH2 )

ReH

| t | | t | . S . 30.5

kt . ReH

kt

ReH (| t | . S . 30.5 / kt . ReH)e3

£

Merancang ukuran pembujur

tebal pelat awal ta 12 mmtebal pelat t 10.30 mm

jarak pembujur * b 700 mmjarak pelintang * a 2100 mm

lateral load p 100

0.1

lebar efektif bm 534.933 mm

pembujur direncanakan menggunakan profil L dengan ukuran sbb:

ukuran pembujur * bm 534.933 ta 12hw 150 tw 9bf 75 tf 9

* direncanakan

angle-, tee and bulb setions:

web: hw/tw 16.666667 ukuran terpenuhi

60

flange: bi = bf - tw 66bi/tf 7.3333333 ukuran terpenuhi

19.5

gambar pembujur: Y

bm

129

150lebar efektif boleh diambil sejarak 6 kali tebal pelat web (pak eko budi jatmiko)

X75

9

Pa

KN/m2

kpx . b

hw/tw 60.0 (k)0.5

60.0 (k)0.5

bi/tf 19.5 (k)0.5

19.5 (k)0.5

£

£

no. nama bagian hor ver A d A d

mm mm mm1 pelat pengikut 534.933 12 6419.200 165 1059168 174762713.9 77030.3972 pelat bilah 9 150 1350.000 84 113400 9525600 25312503 pelat hadap 75 9 675.000 4.5 3037.5 13668.75 4556.25

8444.200 1175605.5 184301982.65 2612836.6

z1 = 139.22047017181 mmz2 = h - z1 31.779529828192 mm

Ix = 186914819.29827

23246473.997837

Wst = 731492.06811784

lateral bucklingnama bagian simbol formula nilai satuan keterangan

1

uniformly distributed… 60 for long. Stiffener

moment of inertia… Ix - 1.87E+04

modulus young E - 2.06E+05 for steel

ideal buckling force 8.62E+07 N for long. Stiffener

bending moment M1 12862.5 Nmm for long. Stiffener- a/b - 3 -- m1 - 1.47 - a/b >= 2.0- m2 - 0.49 - for long. Stiffener

(m1/a2 + m2/b2) 1.333E-06 -

8.034 -

- -52.751 - 00 -

nominal yield point - 235 for hull structural steels

- 60

- 30

t - 30

1 1.0285714285714

2 2.2618447379837 for long. Stiffener

3 0

A d2 h*v3/12

mm2 mm3 mm4 mm4

S Ax =

S2 / S1

S3 + S4 mm4

INA = Ix - z12S1 mm4

INA / z2 mm3

(sa + sb) S/ReH

sa sa = sx N/mm2

cm4

N/mm2

FKix p2/a2 .E.Ix.104

p . b . a2/24 . 103

(ReH E (m1/a2 + m2/b2))0.5

t1 t - t(ReH E (m1/a2 + m2/b2))0.5

ReH N/mm2

sx N/mm2

sy N/mm2

N/mm2

sy 2 ta/b

sx p2/a2 (Ax + bta)

t1 20.5ta/b

£

³

nama bagian simbol formula nilai satuan keterangan

nominal lateral load 1+2+3 3.2904161665552 for long. Stiffener

numerical imperfection 2.8 10 mma/250 8.4b/250 2.8

deformation of stiffener w1 9.21E-05 mm for long. Stiffenerw wo + w1 2.80 mm

elastic support ….. 192.8560 for long. Stiffener

bending moment… Mo 4.19E+06 Nmmcf - pz 189.5656 0

section modulus… Wst - 731.49206811784

5.7432501210582safety factor S - 1.1 - in general

cek lateral buckling 0.3077343622688 1 tidak buckling

torsional bucklingnama bagian simbol formula nilai satuan keterangan

1- ef hw + tf/2 154.5 mm

web area Aw hw tw 1350

flange area Af bf tf 675

sectoral moment of … 15608.90566

3.507219

1.684719

St. Vernant's moment.. 5.191938

polar moment of…. Ip 2623.741875

degree of fixation e 1.0092442705091 -

- 433.74730346209

reference degree of… 0.7360639318142

- f 0.8620259242673

- - 0.7430886941089

- - 0.5417901117178

- - 0.4486631056719

- 0.76

cek torsional buckling 0.3681084084084 1 tidak buckling

pzx N/mm2

wo a/250 wox b/250

p b a4/(384 107 E Ix)

cfx FKix p2/a2 N/mm2

Fki pz w/(cf - pz)

cm3

sb (Mo + M1)/(Wst 103) N/mm2

(sa + sb) S/ReH

sx S / (kT ReH)

mm2

mm2

Iw Af bf2 ef2 / 12 106 {(Af+2.6Aw)/(Af+Aw)} cm4

IT 1 (hw tw3 / 3 104) (1 - 0.63 tw/hw) cm4

IT 2 (bf tf3 / 3 104) (1 - 0.63tf/bf) cm4

IT IT 1 + IT 2 cm4

((Aw hw2)/3 + Af ef2) 10-4 cm4

1+10-4(a4/(Iw(b/t3 + 4hw/3tw3))0.5

sKiT E/Ip (p2 Iw 102/a2 e + 0.385 IT) N/mm2

lT (ReH/sKiT)0.5

0.5 (1 + 0.34(lT - 0.2) + lT2)

f2

lT2

(f2 - lT2)0.5

kT 1 / (f + (f2 - lT2)0.5) for lT > 0.2

sx S / (kT ReH)

£³ £

³

£

£

£

Merancang ukuran pelintangtebal pelat awal ta 12 mm

tebal pelat t 10.30 mmjarak pembujur * b 700 mmjarak pelintang * a 2100 mm

lateral load p 100

0.1

lebar efektif am 512.351 mm

pelintang direncanakan menggunakan profil T dengan ukuran sbb:

ukuran pelintang * bm 512.351 ta 12hw 700 tw 14bf 200 tf 14

* direncanakan

angle-, tee and bulb setions:

web: hw/tw 50 ukuran terpenuhi

60

flange: bi = (bf - tw)/2 93bi/tf 6.6428571 ukuran terpenuhi

19.5Y

gambar pelintang: bm

1214

700

X200

14

Pa

KN/m2

kpy a

hw/tw 60.0 (k)0.5

60.0 (k)0.5

bi/tf 19.5 (k)0.5

19.5 (k)0.5

£

£

no. nama bagian hor ver A d A d

mm mm mm1 pelat pengikut 512.351 12 6148.207 720 4426708.93 3187230432.07 73778.4822242 pelat bilah 14 700 9800.000 364 3567200 1298460800 400166666.673 pelat hadap 200 14 2800.000 7 19600 137200 45733.333333

18748.207 8013508.93 4485828432.07 400286178.48

z1 = 427.42802000722 mmz2 = h - z1 298.57197999278 mm

Iy = 4886114610.5522

1460916353.8258

Wst = 4893012.2440195

lateral bucklingnama bagian simbol formula nilai satuan keterangan

1

uniformly distributed… 30 for trans. Stiffener

moment of inertia… Iy - 4.89E+05

modulus young E - 2.06E+05 for steel

ideal buckling force 3.17E+08 N for trans. Stiffener

bending moment M1 8.23E+05 Nmm for trans. Stiffener- a/(n b) - 0.375 -- m1 - 0.49 - a/(n b) < 0.5- m2 - 0.023 - for trans. Stiffener

(m1/a2 + m2/b2) 1.580E-07 -

2.766 -

- 1.515 - 02 -

nominal yield point - 235 for hull structural steels

- 60

- 30

t - 30

120.315712685551 1.3750367164062

2 0.4149407676336 for trans. Stiffener

3 0.0367323320304

A d2 h*v3/12

mm2 mm3 mm4 mm4

S Ax =

S2 / S1

S3 + S4 mm4

INA = Iy - z12S1 mm4

INA / z2 mm3

(sa + sb) S/ReH

sa sa = sy N/mm2

cm4

N/mm2

FKiy p2/(n b)2 .E.Iy.104

p . a . (n b)2/cs. 8 . 103

(ReH E (m1/a2 + m2/b2))0.5

t1 t - t(ReH E (m1/a2 + m2/b2))0.5

ReH N/mm2

sx N/mm2

sy N/mm2

N/mm2

sx1 sx(1 + Ax/(b ta))sx1 2 ta/a

sy p2/(n b)2 (Ay + ata)

t1 20.5ta/a

£

³

nama bagian simbol formula nilai satuan keterangan

nominal lateral load 1+2+3 1.8267098160703 for trans. Stiffener

numerical imperfection 8.4 10 mmn b/250 22.4a/250 8.4

factor accounting …. - 1 - for simply supported stiff.

deformation of stiffener w1 2.67E-03 mm for trans. Stiffenerw wo + w1 8.403 mm

elastic support ….. 99.6961 for trans. Stiffener

bending moment… Mo 4.97E+07 Nmmcf - pz 97.8694 0

section modulus… Wst - 4893.0122440195

10.321797765368safety factor S - 1.1 - in general

cek lateral buckling 0.1887403299656 1 tidak buckling

torsional bucklingnama bagian simbol formula nilai satuan keterangan

1- ef hw + tf/2 707 mm

sectoral moment of … 4665257.33333

63.219930666667

17.486597333333

St. Vernant's moment.. 80.706528

web area Aw hw tw 9800

flange area Af bf tf 2800

polar moment of…. Ip 300024.38666667

degree of fixation e 1.0000141758031 -

- 6473.360098655

reference degree of… 0.191

- f 0.5165418389224

- - 0.2668154713573

- - 0.0363026305379

- - 0.4801175281319

- 1.0034

cek torsional buckling 0.1399564217566 1 tidak buckling

pzy N/mm2

wo n b/250 woy a/250

cs

5 a p (n b)4/(384 107 E Iy cs2)

cfy cs FKiy p2/(n b)2 N/mm2

Fki pz w/(cf - pz)

cm3

sb (Mo + M1)/(Wst 103) N/mm2

(sa + sb) S/ReH

sy S / (kT ReH)

Iw bf3 tf ef2 / 12 106 cm4

IT 1 (hw tw3 / 3 104) (1 - 0.63 tw/hw) cm4

IT 2 (bf tf3 / 3 104) (1 - 0.63tf/bf) cm4

IT IT 1 + IT 2 cm4

mm2

mm2

((Aw hw2)/3 + Af ef2) 10-4 cm4

1+10-4(b4/(Iw(a/t3 + 4hw/3tw3))0.5

sKiT E/Ip (p2 Iw 102/b2 e + 0.385 IT) N/mm2

lT (ReH/sKiT)0.5

0.5 (1 + 0.34(lT - 0.2) + lT2)

f2

lT2

(f2 - lT2)0.5

kT 1 / (f + (f2 - lT2)0.5) for lT > 0.2

sy S / (kT ReH)

£³ £

³

£

£

£

Pengecekan kekakuan Penegar

Momen inersia ( I ) dari penegar tidak boleh kurang dari:

I =

nama bagian simbol nilai satuan keterangan

momen inersia I 18691.48 for long. Stiffener

- 488611.5 for trans. Stiffenerc 4 - if both ends are simply supported

section modulus W 731.4921 for long. Stiffener

- 4893.012 for trans. Stiffenerunsupported span l 2.1 m for long. Stiffener

- 0.7 m for trans. Stiffener

Kekakuan Pembujur

I = 6144.533

18691.48 > 6144.533

kekakuan pembujur: terpenuhi

Kekakuan Pelintang

I = 13700.43

488611.5 > 13700.43

kekakuan pelintang: terpenuhi

c W l cm4

cm4

cm4

cm3

cm3

cm4

cm4

abg