perhitungan tugas baja2 revie fix

8/12/2019 Perhitungan Tugas Baja2 REVIE Fix

1/49

STRUKTUR BAJA II

Perencanaan Gedung Bertingkat Metode LRFD

Ir. R. Pandaleke, M

>> PRADESAIN PENAMPANG

1. BALOK

L = 8 m

Lb = 5.6 m

a. Mutu baja BJ-37, fy = 240 MPa

b. Beban terfaktor,

Beban Mati :Berat plat beton bertulang, t plat = 12 cm

W plat = 1612.8 kg/m

Berat finishing, = 100 kg/m

W finishing = 560 kg/m

Berat sendiri profil, W profil = 94.3 kg/m

qDL = 2267.1 kg/m

Beban Hidup :

Fungsi bangunan kantor, qL = 250 kg/m

qLL = 1400 kg/m

Kombinasi Beban :

qU = 4960.52 kg/m

c. Momen nominal,

Mu = 39684.16 kgm

= 0.9 Mn = 44093.5111 kgm

d. Persyaratan Tekuk Lokal

1. Flens p = 11.0

2. Web p = 108.4

e. Perencanaan DimensiMp = Mn = 44093.5 kgm

Zx = 1837.2 cm

Sx = 1670.2 cm

Lb

L

= 1,2 +1,6

=18 .qU.

=

= 170

=1680

=

=1,1

qU

L

REVIE. O. F. WANTALANG

09 02 11


2/49

STRUKTUR BAJA II


Ir. R. Pandaleke, M

f. Desain Penampang Direncanakan

Maka dicoba memakai profil WF, h = 386 mm

b = 299 mm

tw = 9 mm

tf = 14 mm

r = 22 mm

W = 94.3 kg/m

g. Pemeriksaan Kelayakan Dimensi

Tekuk lokal,

1. Flens, = 10.7

< p OK

2. Web, = 42.9

< p OK

Maka profil yang direncanakan berpenampang = KOMPAK

h. Cek Kuat Lentur Terfaktor,

hw = 35.8 cm

hf = 37.2 cm

Zx = 1845.6 cm

Mn = 44293.5 kgm

Mn > Mn OK

Maka dicoba menggunakan penampang profil WF mm

400 x 300

400 x 300

2.

= 2. =

=.

4 +.. = = .


09 02 11


3/49

STRUKTUR BAJA II


Ir. R. Pandaleke, M

2. KOLOM Lt.1

L = 8 m

Lb = 5.6 m

h = 4 m


E = 200000 MPa

b. Beban terfaktor,


P plat = 12902.4 kg

Berat profil balok, = 1282.48 kg


W finishing = 4480 kg

PDL = 18664.88 kg

Beban Hidup :


PLL = 11200 kg

Beban Aksial Ultimate :

PU = 29864.9 kg

c. Desain Penampang Direncanakan


b = 200 mm

tw = 8 mm

tf = 12 mm

r = 13 mm

rx = 86.2 mm

ry = 50.2 mm

Ag = 6350 mm

W = 49.9 kg/m

d. Periksa Kelangsingan Penampang

1. Flens, = 8.3

r = 16.1

2. Web, = 25.0

r = 42.9

200 x 200

OK

OK

Lb

L

= 1,2 +1,6

Pu

h

2.

=250 2.<

=665 <


09 02 11


4/49

STRUKTUR BAJA II


Ir. R. Pandaleke, M

Kondisi tumpuan sendi-sendi, k = 1

1. Arah sumbu kuat x

x = 46.404

cx = 0.512

x = 1.137

Pn = 133981.8 kg

= 0.85

Pn = 113884.5 kg

OK

2. Arah sumbu lemah y

y = 79.681

cy= 0.879

y= 1.414

Pn = 107780.9 kg

= 0.85

Pn = 91613.8 kg

OK

Maka dicoba menggunakan penampang profil WF mm

0.326=

< 10.262=

< 1

200 x 200

=.

=

.

0,25 < < 1,2 = 1,431,60,67.

= .

=.

=

.

0,25 < < 1,2 = 1,431,60,67.

= .


09 02 11


5/49

STRUKTUR BAJA II


Ir. R. Pandaleke, M

3. KOLOM Lt.2

L = 8 m

Lb = 5.6 m

h = 3.5 m


E = 200000 MPa

b. Beban terfaktor,


P plat = 12902.4 kg

Berat profil balok, = 11.56 kg


W finishing = 4480 kg

PDL = 17393.96 kg

Beban Hidup :


PLL = 11200 kg

Beban Aksial Ultimate :

PU = 28594.0 kg

c. Desain Penampang Direncanakan


b = 150 mm

tw = 7 mm

tf = 10 mm

r = 11 mm

rx = 64.9 mm

ry = 37.5 mm

Ag = 4014 mm

W = 31.5 kg/m

d. Periksa Kelangsingan Penampang

1. Flens, = 7.5

r = 16.1

2. Web, = 21.4

r = 42.9

150 x 150

OK

OK

Lb

L

= 1,2 +1,6

Pu

h

2.

=250 2.<

=665 <


09 02 11


6/49


7/49

STRUKTUR BAJA II


Ir. R. Pandaleke, M

>> PRADESAIN PEMBEBANAN PORTAL

1. Akibat Beban Hidup

Besar beban hidup untuk fungsi bangunan kantor, qL = 250 kg/m

Besar daerah tinjauan, Lb = 5.6 m

qLL = 1400 kg/m

2. Akibat Beban Mati

Berat sendiri profil baja yang digunakan, W profil = 94.3 kg/m

Berat sendiri plat beton bertulang, t plat = 12 cm

W plat = 1612.8 kg/m


W finishing = 560 kg/m

qDL = 2267.1 kg/m

Berat Atap, P = 1481.98 kg

3. Akibat Beban Horizontal (Quake)

Beban horizontal pada titik, D = 2500 kg

G = 3000 kg


09 02 11


8/49

STRUKTUR BAJA II


Ir. R. Pandaleke, M

Lt. hx (m) xe (mm)

71.5

13 71.5

147.4

1 4

>> PERIKSA DEFLEKSI / SIMPANGAN ANTAR TINGKAT (SPRMK)

75.9

2 7.5 26.8

87.5

100

OK

OK

= 5,5 = 0,025 = 1 > GAMBAR DIAGRAM FREE BODY (dari hasil perhitungan pada SAP)

A B

27390.61 kg 78031.41 kg

97.671653.965 kgm

2056.174185.56

1602.57 kg 1602.57

qL qD

qL qDqL qD

qL qD

H1

H2

V1


10/49

STRUKTUR BAJA II


Ir. R. Pandaleke, M

>> MENCARI NILAI k

G WF 400 x 300 mm H WF 400 x 300 mm I

WF 250 x 250 mm WF 250 x 250 mm WF 250 x 250 mm h2

D WF 400 x 300 mm E WF 400 x 300 mm F

WF 250 x 250 mm WF 250 x 250 mm WF 250 x 250 mm h1

L L

WF 250 x 250 mm Ix = 4720 cm L = 800 cmIy = 1600 cm h1 = 400 cm

WF 250 x 250 mm Ix = 20400 cm h2 = 350 cm

Iy = 6750 cm

WF 400 x 300 mm Ix = 33700 cm

Iy = 6240 cm

GA = 1 GD = 1.5 GG = 0.3

GB = 1 GE = 0.8 GH = 0.2

GC = 1 GF = 1.5 GI = 0.3

A B C

G =


09 02 11


11/49

STRUKTUR BAJA II


Ir. R. Pandaleke, M

Hasil perhitungan kemudian di plot kedalam nomogram,

Komponen Kolom

k AD = 0.8 k DG = 0.72

k BE = 0.76 k EH = 0.67

k CF = 0.8 k FI = 0.72


09 02 11


12/49

STRUKTUR BAJA II


Ir. R. Pandaleke, M

Hasil perhitungan kemudian di plot kedalam nomogram,

Komponen Balok

k DE = 0.78 k GH = 0.61

k EF = 0.78 k HI = 0.61


09 02 11


13/49

STRUKTUR BAJA II


Ir. R. Pandaleke, M

1. Kolom Lt.1

78031.41 kg fy = 240 MPa

E = 200000 MPa

kg WF 250 x 250

h = 250 mm

b = 250 mm

tw = 9 mm

tf = 14 mm

kg r = 16 mm

rx = 108.2 mm

ry = 62.9 mm

78031.41 kg Ag = 9218 mm

W = 72.36 kg/m

L = 4000 mm

Nu = 78031.41 kg

Vu = 3346.15 kg

Mu = 7530.154 kgm

a. Aksi Kolom

Menghitung rasio kelangsingan maksimum

k = 0.76

= 48.3

c = 0.53

= 1.15

Nn = 192292.724 kg

= 0.85

Nn = 163448.815 kg

Maka digunakan persamaan,

b. Aksi Balok

Periksa apakah profil baja yang digunakan kompak atau tidak,

p = 10.973

OK

Ny = 221232 kg

b= 0.9

b.Ny = 199108.8 kg

0.477 > 0,2

8.929

0.392 > 0,125

kgm

kgm

>> DESAIN BALOK KOLOM METODE LRFD

3346.15

3346.15

7530.154

5854.456

=.

=1 . .

= 1,431,60,67.

= .

0,25 < < 1,2

=

+

89 .

.+

.

2.=

= 170

= .

.=

h

b

tw

tf


09 02 11


14/49

STRUKTUR BAJA II


Ir. R. Pandaleke, M

sehinggap ,

OK

maka dimensi profil yang digunakan,

fr = 115 MPa

G = 80000 MPa

J = 511279.3333 mm

Sx = 864 cm

X1 = 22326.90861 N/mm

Iy = 3650 cm

Iw = 5.70313E+11 mm

X2 = 2.789E-05 N/mm

Lr = 16657.327 mm

Sehingga,

Maka, Zx = 936889 mm

Mp = 22485.336 kgm

Mr = 10800 kgm

Cb = 2.23

Mn = 48606.948 kgm

Mn dipakai = 22485.336 kgm

b = 0.9

b.Mnx = 20236.802 kgm

c. Perbesaran Momen

Rasio kelangsingan yang diperhitungkan dalam perbesaran momen hanya sumbu lenturnya,

dalam hal ini sumbu lentur adalah sumbu x, sehingga :

kx = 0.76

Cm = 0.289

Nel = 2305015.229 kg

b = 0.299

b = 1

Mux = 7530.154 kgm

28.096

Lp = 3207.540 mm

p = 62.552

27.778

KOMPAK

42.926

= 500 2,33 .

665

=

=790.

= . 1 . 1 + 1 + 2 .

= 13 . .

1 = . .. .

2

2 = 4. .

.

= .

4

< <

= . = .()

= . + .

. =

= 0,60,4. 121 = =...

= 1

1

= .


09 02 11


15/49

STRUKTUR BAJA II


Ir. R. Pandaleke, M

d. Periksa terhadap persamaan :

karena momen lentur terjadi pada arahxmaka persamaan di atas menjadi,

e. Kontrol terhadap geser :

Aw = 1998 mm

Vm = 28771.2 kg

= 0.9

Vm = 25894.08 kg

mampu memikul beban-beban yang diberikan.

0.808

Jadi profil WF, 250 x 250

0.129

+

89 .

.+

.

+

89 .

. 1 1

= 0,6.. =

= 1


09 02 11


16/49

STRUKTUR BAJA II


Ir. R. Pandaleke, M

2. Kolom Lt.2

39737.48 kg fy = 240 MPa

E = 200000 MPa

kg WF 250 x 250

h = 250 mm

b = 250 mm

tw = 9 mm

tf = 14 mm

kg r = 16 mm

rx = 108.2 mm

ry = 62.9 mm

39737.48 kg Ag = 9218 mm

W = 72.36 kg/m

L = 3500 mm

Nu = 39737.48 kg

Vu = 2653.9 kg

Mu = 4859.801 kgm

a. Aksi Kolom


k = 0.67

= 37.3

c = 0.41

= 1.08

Nn = 204921.510 kg

= 0.85

Nn = 174183.284 kg


b. Aksi Balok


p = 10.973

OKNy = 221232 kg

b= 0.9

b.Ny = 199108.8 kg

2653.94859.801 kgm

2653.94428.832 kgm

0.228 > 0,2

8.929

0.200 > 0,125

=.

=1 . .

= 1,431,60,67.

= .

0,25 < < 1,2

=

+

89 .

.+

.

2.=

=170

= .

.=

h

b

tw

tf


09 02 11


17/49

STRUKTUR BAJA II


Ir. R. Pandaleke, M

sehinggap ,

OK


fr = 115 MPa

G = 80000 MPa

J = 511279.3333 mm

Sx = 472 cm

X1 = 40869.59542 N/mm

Iy = 1600 cm

Iw = 2.5E+11 mm

X2 = 8.323E-06 N/mm

Lr = 29538.896 mm

Sehingga,

Maka, Zx = 936889 mm

Mp = 22485.336 kgm

Mr = 5900 kgm

Cb = 2.25

Mn = 50177.530 kgm


b = 0.9

b.Mnx = 20236.802 kgm

c. Perbesaran Momen

Rasio kelangsingan yang diperhitungkan dalamperbesaran momen hanya sumbu lenturnya,

dalam hal ini sumbu lentur adalah sumbu x, sehingga :

kx = 0.67

Cm = 0.235

Nel = 3873782.851 kg

b = 0.238

b = 1

Mux = 4859.801 kgm

27.778

KOMPAK

Lp = 3207.540 mm

21.673

p = 68.759 42.926

= 500 2,33 .

665

=

=790.

= . 1 . 1 + 1 + 2 .

= 13 . .

1 = . .. .

2

2 = 4. .

.

= .

4

< <

= . = .()

= . + .

. =

= 0,60,4. 121 = =...

= 1

1

= .


09 02 11


18/49

STRUKTUR BAJA II


Ir. R. Pandaleke, M


karena momen lentur terjadi pada arahxmaka persamaan di atas menjadi,


Aw = 1998 mm

Vm = 28771.2 kg

= 0.9

Vm = 25894.08 kg


0.442

0.102


+

89 .

.+

.

+

89 .

. 1 1

= 0,6.. =

= 1


09 02 11


19/49

STRUKTUR BAJA II


Ir. R. Pandaleke, M

3. Balok

12855.82 kg 20108.34 kg fy = 240 MPa

E = 200000 MPa

3799.76 3799.76 kg WF 400 x 300

h = 386 mm

b = 299 mm

2600.335 kg 31610.41 kg tw = 9 mm

tf = 14 mm

r = 22 mm

rx = 167 mm

ry = 72.1 mm

Ag = 12010 mm

W = 94.3 kg/m

L = 8000 mm

Nu = 3799.76 kg

Vu = 20108.34 kg

Mu = 31610.409 kgm

a. Aksi Kolom


k = 0.61

= 67.7

c = 0.75

= 0.70

Nn = 413993.594 kg

= 0.85

Nn = 351894.555 kg


b. Aksi Balok


p = 10.973

OK

Ny = 288240 kg

b= 0.9

b.Ny = 259416 kg

10.679

0.015 < 0,125

0.011 < 0,2

=.

=1 . .

= 1,25.

= .

> 1,2

=

2.=

=170

= .

.=

qL qD

2.+

.+

.

h

b

tw

tf


09 02 11


20/49

STRUKTUR BAJA II


Ir. R. Pandaleke, M

sehinggap ,

OK


fr = 115 MPa

G = 80000 MPa

J = 633964.6667 mm

Sx = 1740 cm

X1 = 14091.25417 N/mm

Iy = 6240 cm

Iw = 2.32434E+12 mm

X2 = 1.754E-04 N/mm

Lr = 13922.014 mm

Sehingga,

Maka, Zx = 1845561 mm

Mp = 44293.464 kgm

Mr = 21750 kgm

Cb = 2.1

Mn = 73039.252 kgm


b = 0.9

b.Mny = 39864.118 kgm

c. Perbesaran Momen

Rasio kelangsingan yang diperhitungkan dalamperbesaran momen hanya sumbu lenturnya,

dalam hal ini sumbu lentur adalah sumbu y, sehingga :

ky = 0.61

Cm = 1.000

Nel = 517491.992 kg

b = 1.007

b yang dipakai = 1.007

Muy = 31844.230 kgm

67.684

Lp = 3676.688 mm

p = 104.075

42.889

KOMPAK

=1680 1 2,75..

=

=790.

= . 1 . 1 + 1 + 2 .

= 13 . .

1 = . .. .

2

2 = 4. .

.

= .

4

< <

= . = .()

= . + .

. =

= 0,60,4. 121 = =...

= 1

1

= .


09 02 11


21/49

STRUKTUR BAJA II


Ir. R. Pandaleke, M


karena momen lentur terjadi pada arah y maka persamaan di atas menjadi,


Aw = 3222 mm

Vm = 46396.8 kg

= 0.9

Vm = 41757.12 kg


0.804


0.482

2.+

.+

.

< 12.+ . < 1

= 0,6.. =

= 1


09 02 11


22/49

STRUKTUR BAJA II


1. Sambungan pada titik D

a. Baut pada sisi kolom

DATA :

WF 250 x 250 Pu = 1Mu = 2

fu =

WF 400 x 300 fu =

h =

b =

b siku =

digunakan profil h siku =

t siku =

d baut =

d lubang =

> 38.1 s =

> 19.05 s1 =

> 38.1 s' => 19.05 s1' =

Kekuatan tarik nominal terfaktor baut,

Rn = 10

=

.Rn = 7

Luas pengganti, lebar bagian tertarik,

a =

Letak garis netral,

Momen inersia luas pengganti,

Gaya tarik maksimum pada baut,

Gaya tarik maksimum,

R maks =

R maks < .Rn

Ab =

c =

I = 2523

ft =

>> SAMBUNGAN BAUT BALOK KOLOM (diperkaku)

=14 . .

= .

=

=

7

=.

3 +.

3

= 2 .

= .

=

3 > > 151,5 > 1 > 4 + 100

s

s1

b

h

s' s1'

s

s1

h


23/49

STRUKTUR BAJA II


b. Baut pada sisi balok

WF 250 x 250 DATA :

WF 400 x 300 d baut =

d lubang =

> 38.1 s => 19.05 s1 =

> 38.1 s' =

> 19.05 s1' =

WF 400 x 300

Sambungan tipe tumpu, tinjau terhadap kekuatan geser,

m =

r1 =

Rn =

=

.Rn =

Tinjau terhadap kekuatan tumpu,

tp =

Rn =

.Rn =

Perhitungan resultan gaya pada baut,

n =

M = 2

x y x y M Mx My Rx Ry P = Rv(mm) (mm) (mm ) (mm ) (kg mm) (kg mm ) (kg mm ) (kg) (kg) (kg)

1 0 283 0 80089 2600335.000 0 735894805 2203.331 0 1606.978

2 0 223 0 49729 2600335.000 0 579874705 1736.193 0 1606.978

3 0 163 0 26569 2600335.000 0 423854605 1269.056 0 1606.978

4 0 103 0 10609 2600335.000 0 267834505 801.919 0 1606.978

5 0 103 0 10609 2600335.000 0 267834505 801.919 0 1606.978

6 0 163 0 26569 2600335.000 0 423854605 1269.056 0 1606.978

7 0 223 0 49729 2600335.000 0 579874705 1736.193 0 1606.978

8 0 283 0 80089 2600335.000 0 735894805 2203.331 0 1606.978

0 333992

Jadi digunakan baut dengan,

pada sisi kolom 8 bh 12.7 60 mm, jarak tepi baut

pada sisi balok 8 bh 12.7 60 mm, jarak tepi baut

No. Baut

333992

P =

mm, jarak antar baut


= .1. .

= 2,4. ..

= 1 =

= = =

s

s1

P

M

s

s1

s' s1'

3 > > 151,5 > 1 > 4 + 100


24/49

STRUKTUR BAJA II


2. Sambungan pada titik E kiri


DATA :

WF 250 x 250 Pu =

Mu =

fu =WF 400 x 300 fu =

h =

b =

b siku =


t siku =

d baut =

d lubang =

> 38.1 s =

> 19.05 s1 =

> 38.1 s' =

> 12.7 s1' =


Rn =

=

.Rn =


a =

Letak garis netral,




R maks =

R maks < .Rn

c =

I = 20

ft =

Ab = =14 . .

=

= 7

=.

3 + .

3

= 2 .

= .

=

3 > > 151,5 > 1 > 4 + 100

s

s1

b

h

s' s1'

s

s1

h

= .


25/49

STRUKTUR BAJA II



WF 250 x 250 DATA :

WF 400 x 300 d baut =

d lubang =

> 57.15 s => 28.58 s1 =

> 57.15 s' =

> 28.58 s1' =

WF 400 x 300


m =

r1 =

Rn =

=

.Rn =


tp =

Rn =

.Rn =


n =

M = 31

x y x y M Mx My Rx Ry P = Rv

(mm) (mm) (mm ) (mm ) (kg mm) (kg mm ) (kg mm ) (kg) (kg) (kg)

1 30 313 900 97969 31610409.000 948312270 9894058017 13399.611 1284 1256.7712 30 233 900 54289 31610409.000 948312270 7365225297 9974.790 1284 1256.771

3 30 153 900 23409 31610409.000 948312270 4836392577 6549.969 1284 1256.771

4 30 73 900 5329 31610409.000 948312270 2307559857 3125.149 1284 1256.771

5 30 73 900 5329 31610409.000 948312270 2307559857 3125.149 1284 1256.771

6 30 153 900 23409 31610409.000 948312270 4836392577 6549.969 1284 1256.771

7 30 233 900 54289 31610409.000 948312270 7365225297 9974.790 1284 1256.771

8 30 313 900 97969 31610409.000 948312270 9894058017 13399.611 1284 1256.771

9 30 313 900 97969 31610409.000 948312270 9894058017 13399.611 1284 1256.771

10 30 233 900 54289 31610409.000 948312270 7365225297 9974.790 1284 1256.771

11 30 153 900 23409 31610409.000 948312270 4836392577 6549.969 1284 1256.771

12 30 73 900 5329 31610409.000 948312270 2307559857 3125.149 1284 1256.771

13 30 73 900 5329 31610409.000 948312270 2307559857 3125.149 1284 1256.771

14 30 153 900 23409 31610409.000 948312270 4836392577 6549.969 1284 1256.771

15 30 233 900 54289 31610409.000 948312270 7365225297 9974.790 1284 1256.77116 30 313 900 97969 31610409.000 948312270 9894058017 13399.611 1284 1256.771

14400 723984




P =

No. Baut

738384



= .1.

.

= 2,4. ..

= 1 =

= = =

s

s1

P

M

s

s1

s' s1'

3 > > 151,5 > 1 > 4 + 100


26/49

STRUKTUR BAJA II


3. Sambungan pada titik E kanan


DATA :

WF 250 x 250 Pu =

Mu =

fu =WF 400 x 300 fu =

h =

b =

b siku =


t siku =

d baut =

d lubang =

> 38.1 s =

> 19.05 s1 =

> 38.1 s' =

> 19.05 s1' =


Rn =

=

.Rn =


a =

Letak garis netral,




R maks =

R maks < .Rn

Ab =

c =

I = 20

ft =

=14 . .

= .

=

= 7

= .

3 + .

3

= 2 .

= .

=

3 > > 151,5 > 1 > 4 + 100

s

s1

b

h

s' s1'

s

s1

h


27/49

STRUKTUR BAJA II



WF 250 x 250 DATA :

WF 400 x 300 d baut =

d lubang =

> 47.64 s => 23.82 s1 =

> 47.64 s' =

> 23.82 s1' =

WF 400 x 300


m =

r1 =

Rn =

=

.Rn =


tp =

Rn =

.Rn =


n =

M = 25



1 30 313 900 97969 25587096.000 767612880 8008761048 10846.336 1040 1136.5992 30 233 900 54289 25587096.000 767612880 5961793368 8074.110 1040 1136.599

3 30 153 900 23409 25587096.000 767612880 3914825688 5301.883 1040 1136.599

4 30 73 900 5329 25587096.000 767612880 1867858008 2529.657 1040 1136.599

5 30 73 900 5329 25587096.000 767612880 1867858008 2529.657 1040 1136.599

6 30 153 900 23409 25587096.000 767612880 3914825688 5301.883 1040 1136.599

7 30 233 900 54289 25587096.000 767612880 5961793368 8074.110 1040 1136.599

8 30 313 900 97969 25587096.000 767612880 8008761048 10846.336 1040 1136.599

9 30 313 900 97969 25587096.000 767612880 8008761048 10846.336 1040 1136.599

10 30 233 900 54289 25587096.000 767612880 5961793368 8074.110 1040 1136.599

11 30 153 900 23409 25587096.000 767612880 3914825688 5301.883 1040 1136.599

12 30 73 900 5329 25587096.000 767612880 1867858008 2529.657 1040 1136.599

13 30 73 900 5329 25587096.000 767612880 1867858008 2529.657 1040 1136.599

14 30 153 900 23409 25587096.000 767612880 3914825688 5301.883 1040 1136.599

15 30 233 900 54289 25587096.000 767612880 5961793368 8074.110 1040 1136.59916 30 313 900 97969 25587096.000 767612880 8008761048 10846.336 1040 1136.599

14400 723984




P =


No. Baut

738384


= .1.

.

= 2,4. ..

= 1 =

= = =

s

s1

P

M

s

s1

s' s1'

3 > > 151,5 > 1 > 4 + 100


28/49

STRUKTUR BAJA II


4. Sambungan pada titik F


DATA :

WF 250 x 250 Pu =

Mu =

fu =WF 400 x 300 fu =

h =

b =

b siku =


t siku =

d baut =

d lubang =

> 38.1 s =

> 19.05 s1 =

> 38.1 s' =

> 19.05 s1' =


Rn =

=

.Rn =


a =

Letak garis netral,




R maks =

R maks < .Rn

Ab =

c =

I = 20

ft =

=14 . .

= .

=

=7

= .

3 + .

3

= 2 .

= .

=

3 > > 151,5 > 1 > 4 + 100

s

s1

b

h

s' s1'

s

s1

h


29/49

STRUKTUR BAJA II



WF 250 x 250 DATA :

WF 400 x 300 d baut =

d lubang =

> 47.64 s => 23.82 s1 =

> 47.64 s' =

> 23.82 s1' =

WF 400 x 300


m =

r1 =

Rn =

=

.Rn =


tp =

Rn =

.Rn =


n =

M = 11



1 0 313 0 97969 11958988.000 0 3743163244 10340.458 0 1847.3212 0 233 0 54289 11958988.000 0 2786444204 7697.530 0 1847.321

3 0 153 0 23409 11958988.000 0 1829725164 5054.601 0 1847.321

4 0 73 0 5329 11958988.000 0 873006124 2411.672 0 1847.321

5 0 73 0 5329 11958988.000 0 873006124 2411.672 0 1847.321

6 0 153 0 23409 11958988.000 0 1829725164 5054.601 0 1847.321

7 0 233 0 54289 11958988.000 0 2786444204 7697.530 0 1847.321

8 0 313 0 97969 11958988.000 0 3743163244 10340.458 0 1847.321

0 361992




No. Baut

361992



P =

= .1.

.

= 2,4. ..

= 1 =

= = =

s

s1

P

M

s

s1

s' s1'

3 > > 151,5 > 1 > 4 + 100


30/49

STRUKTUR BAJA II


5. Sambungan pada titik G


DATA :

WF 250 x 250 Pu =

Mu =

fu =WF 400 x 300 fu =

h =

b =

b siku =


t siku =

d baut =

d lubang =

> 38.1 s =

> 19.05 s1 =

> 38.1 s' =

> 19.05 s1' =


Rn =

=

.Rn =


a =

Letak garis netral,




R maks =

R maks < .Rn

Ab =

c =

I = 25

ft =

=14 . .

= .

=

=7

=.

3 + .

3

= 2 .

= .

=

3 > > 151,5 > 1 > 4 + 100

s

s1

b

h

s' s1'

s

s1

h


31/49

STRUKTUR BAJA II



WF 250 x 250 DATA :

WF 400 x 300 d baut =

d lubang =

> 38.1 s => 19.05 s1 =

> 38.1 s' =

> 19.05 s1' =

WF 400 x 300


m =

r1 =

Rn =

=

.Rn =


tp =

Rn =

.Rn =


n =

M = 1



1 0 283 0 80089 1462081.000 0 413768923 1238.859 0 1594.5532 0 223 0 49729 1462081.000 0 326044063 976.203 0 1594.553

3 0 163 0 26569 1462081.000 0 238319203 713.548 0 1594.553

4 0 103 0 10609 1462081.000 0 150594343 450.892 0 1594.553

5 0 103 0 10609 1462081.000 0 150594343 450.892 0 1594.553

6 0 163 0 26569 1462081.000 0 238319203 713.548 0 1594.553

7 0 223 0 49729 1462081.000 0 326044063 976.203 0 1594.553

8 0 283 0 80089 1462081.000 0 413768923 1238.859 0 1594.553

0 333992




No. Baut

333992


P =


= .1.

.

= 2,4. ..

= 1 =

= = =

s

s1

P

M

s

s1

s' s1'

3 > > 151,5 > 1 > 4 + 100


32/49

STRUKTUR BAJA II


6. Sambungan pada titik H kiri


DATA :

WF 250 x 250 Pu =

Mu =

fu =WF 400 x 300 fu =

h =

b =

b siku =


t siku =

d baut =

d lubang =

> 38.1 s =

> 19.05 s1 =

> 38.1 s' =

> 19.05 s1' =


Rn =

=

.Rn =


a =

Letak garis netral,




R maks =

R maks < .Rn

I = 20

ft =

Ab =

c =

=14 . .

= .

=

= 7

= .

3 + .

3

= 2 .

= .

=

3 > > 151,5 > 1 > 4 + 100

s

s1

b

h

s' s1'

s

s1

h


33/49

STRUKTUR BAJA II



WF 250 x 250 DATA :

WF 400 x 300 d baut =

d lubang =

> 57.15 s => 28.58 s1 =

> 57.15 s' =

> 28.58 s1' =

WF 400 x 300


m =

r1 =

Rn =

=

.Rn =


tp =

Rn =

.Rn =


n =

M = 31



1 30 313 900 97969 31267393.000 938021790 9786694009 13254.206 1270 1262.9842 30 233 900 54289 31267393.000 938021790 7285302569 9866.550 1270 1262.984

3 30 153 900 23409 31267393.000 938021790 4783911129 6478.893 1270 1262.984

4 30 73 900 5329 31267393.000 938021790 2282519689 3091.237 1270 1262.984

5 30 73 900 5329 31267393.000 938021790 2282519689 3091.237 1270 1262.984

6 30 153 900 23409 31267393.000 938021790 4783911129 6478.893 1270 1262.984

7 30 233 900 54289 31267393.000 938021790 7285302569 9866.550 1270 1262.984

8 30 313 900 97969 31267393.000 938021790 9786694009 13254.206 1270 1262.984

9 30 313 900 97969 31267393.000 938021790 9786694009 13254.206 1270 1262.984

10 30 233 900 54289 31267393.000 938021790 7285302569 9866.550 1270 1262.984

11 30 153 900 23409 31267393.000 938021790 4783911129 6478.893 1270 1262.984

12 30 73 900 5329 31267393.000 938021790 2282519689 3091.237 1270 1262.984

13 30 73 900 5329 31267393.000 938021790 2282519689 3091.237 1270 1262.984

14 30 153 900 23409 31267393.000 938021790 4783911129 6478.893 1270 1262.984

15 30 233 900 54289 31267393.000 938021790 7285302569 9866.550 1270 1262.98416 30 313 900 97969 31267393.000 938021790 9786694009 13254.206 1270 1262.984

14400 723984




No. Baut

738384



P =

= .1.

.

= 2,4. ..

= 1 =

= = =

s

s1

P

M

s

s1

s' s1'

3 > > 151,5 > 1 > 4 + 100


34/49

STRUKTUR BAJA II


7. Sambungan pada titik H kanan


DATA :

WF 250 x 250 Pu =

Mu =

fu =WF 400 x 300 fu =

h =

b =

b siku =


t siku =

d baut =

d lubang =

> 38.1 s =

> 19.05 s1 =

> 38.1 s' =

> 19.05 s1' =


Rn =

=

.Rn =


a =

Letak garis netral,




R maks =

R maks < .Rn

I = 20

ft =

Ab =

c =

=14 . .

= .

=

= 7

= .

3 + .

3

= 2 .

= .

=

3 > > 151,5 > 1 > 4 + 100

s

s1

b

h

s' s1'

s

s1

h


35/49

STRUKTUR BAJA II



WF 250 x 250 DATA :

WF 400 x 300 d baut =

d lubang =

> 57.15 s => 28.58 s1 =

> 57.15 s' =

> 28.58 s1' =

WF 400 x 300


m =

r1 =

Rn =

=

.Rn =


tp =

Rn =

.Rn =


n =

M = 29



1 30 313 900 97969 29241060.000 877231800 9152451780 12395.247 1188 1220.6092 30 233 900 54289 29241060.000 877231800 6813166980 9227.132 1188 1220.609

3 30 153 900 23409 29241060.000 877231800 4473882180 6059.018 1188 1220.609

4 30 73 900 5329 29241060.000 877231800 2134597380 2890.904 1188 1220.609

5 30 73 900 5329 29241060.000 877231800 2134597380 2890.904 1188 1220.609

6 30 153 900 23409 29241060.000 877231800 4473882180 6059.018 1188 1220.609

7 30 233 900 54289 29241060.000 877231800 6813166980 9227.132 1188 1220.609

8 30 313 900 97969 29241060.000 877231800 9152451780 12395.247 1188 1220.609

9 30 313 900 97969 29241060.000 877231800 9152451780 12395.247 1188 1220.609

10 30 233 900 54289 29241060.000 877231800 6813166980 9227.132 1188 1220.609

11 30 153 900 23409 29241060.000 877231800 4473882180 6059.018 1188 1220.609

12 30 73 900 5329 29241060.000 877231800 2134597380 2890.904 1188 1220.609

13 30 73 900 5329 29241060.000 877231800 2134597380 2890.904 1188 1220.609

14 30 153 900 23409 29241060.000 877231800 4473882180 6059.018 1188 1220.609

15 30 233 900 54289 29241060.000 877231800 6813166980 9227.132 1188 1220.60916 30 313 900 97969 29241060.000 877231800 9152451780 12395.247 1188 1220.609

14400 723984




No. Baut

738384


P =


= .1.

.

= 2,4. ..

= 1 =

= = =

s

s1

P

M

s

s1

s' s1'

3 > > 151,5 > 1 > 4 + 100


36/49

STRUKTUR BAJA II


8. Sambungan pada titik I


DATA :

WF 250 x 250 Pu =

Mu =

fu =WF 400 x 300 fu =

h =

b =

b siku =


t siku =

d baut =

d lubang =

> 38.1 s =

> 19.05 s1 =

> 38.1 s' =

> 19.05 s1' =


Rn =

=

.Rn =


a =

Letak garis netral,




R maks =

R maks < .Rn

I = 20

ft =

Ab =

c =

=14 . .

= .

=

=7

= .

3 + .

3

= 2 .

= .

=

3 > > 151,5 > 1 > 4 + 100

s

s1

b

h

s' s1'

s

s1

h


37/49

STRUKTUR BAJA II



WF 250 x 250 DATA :

WF 400 x 300 d baut =

d lubang =

> 38.1 s => 19.05 s1 =

> 38.1 s' =

> 19.05 s1' =

WF 400 x 300


m =

r1 =

Rn =

=

.Rn =


tp =

Rn =

.Rn =


n =

M = 4



1 0 313 0 97969 4859801.000 0 1521117713 4202.075 0 1679.3032 0 233 0 54289 4859801.000 0 1132333633 3128.063 0 1679.303

3 0 153 0 23409 4859801.000 0 743549553 2054.050 0 1679.303

4 0 73 0 5329 4859801.000 0 354765473 980.037 0 1679.303

5 0 73 0 5329 4859801.000 0 354765473 980.037 0 1679.303

6 0 153 0 23409 4859801.000 0 743549553 2054.050 0 1679.303

7 0 233 0 54289 4859801.000 0 1132333633 3128.063 0 1679.303

8 0 313 0 97969 4859801.000 0 1521117713 4202.075 0 1679.303

0 361992




No. Baut

361992



P =

= .1.

.

= 2,4. ..

= 1 =

= = =

s

s1

P

M

s

s1

s' s1'

3 > > 151,5 > 1 > 4 + 100


38/49

STRUKTUR BAJA II


Ir. R. Pandaleke, MT

1. Tumpuan pada titik A

Beban Kolom

Pu = 27390.61 kg

Mu = 1653.965 kgm

Vu = 97.67 kgPlat Tumpuan

fy = 240 MPa

fu = 370 MPa

B = 450 mm

L = 450 mm

t = 25 mm

Kolom Pedestial

fc' = 25 MPa

I = 500 mm

J = 500 mm

Kolom Baja WF

ht = 250 mm

bf = 250 mm

tw = 9 mm

tf = 14 mm

Angkur Baut

fu = 825 MPa

fy = 400 MPa

d = 19.05 mm

nt = 4 bh

nc = 4 bh

f = 175 mm

La = 500 mm

Eksentrisitas Beban

e = 60.384 mm

L/8= 56.250 mm

e > L/8 OK

h = 236 mm

n = nt + nc = 8 bh

>> PERHITUNGAN TUMPUAN KOLOM

293et =

ec = 57 mm

mm

ht

h

P

Mu

Vu

La

L

J

B I

a a

f

=

= = + 2

=

2

REVIE. O. F. WANTALANGIE

09 02 11 047


39/49

STRUKTUR BAJA II



Tahanan Tumpu Beton

Gaya tarik pada angkur baut,

Pt = 5328.549 kg

Gay tekan total pada plat tumpuan,

Puc = 32719.159 kg

Panjang bidang tegangan tekan beton,Y = 321 mm

Luas plat tumpuan baja, A1 = B.L = 202500 mm

Luas penampang kolom pedestial, A2 = I.J = 250000 mm

Tegangan tumpu nominal,

fcn = 23.611 MPa

fcn = 1,70.fc' = 42.5 MPa

Tegangan tumpu nominal beton yang digunakan, fcn = 23.611 MPa

Faktor reduksi kekuatan tekan beton, = 0.65

Tegangan tumpu beton yang diijinkan, .fcn = 15.347 MPa

Tegangan tumpu maksimum yang terjadi pada beton,fcu = 4.530 MPa

Kontrol tegangan, OK

Kontrol Dimensi Plat Tumpuan

Lebar minimum plat tumpuan yang diperlukan,

Bp min = 132.830 mm

Lebar plat yang digunakan, B = 450 mm

Kontrol, OK

Panjang bagian plat tumpuan jepit bebas,

Modulus penampang plastis plat,

Z = 70312.5 mm

Momen yang terjadi pada plat akibat beban terfaktor,

Mup = 10237237.98 Nmm

Faktor reduksi kekuatan lentur, = 0.9Tahanan momen nominal plat, Mn = fy.Z = 16875000 Nmm

Tahanan momen plat, .Mn = 15187500 Nmm

Kontrol, OK

a =

fcu1 =

106.250

3.031

mm

MPa

= .

= +

= 3. 2

= 0,85. . 21

= 2. . .

min = 0,5...

= 0,95. 2

1 = 1 .

=14 . .

=12 ..1.+ 13 .. 1 .

.


09 02 11 047


40/49

STRUKTUR BAJA II



Gaya Tarik Pada Angkur Baut


Tu1 = 13321.372 N

Luas penampang angkur baut,

Ab = 285.022957 mm

Faktor reduksi kekuatan tarik, = 0.9

Tahanan tarik nominal angkur baut baut, Tn = 0,75.Ab.fu = 176357.9546 N

Tahanan tarik angkur baut, .Tn = 158722.1592 N

Kontrol, OK

Gaya Geser Pada Angkur Baut

Gaya geser pada angkur baut,

Vu1 = 122.088 N

Jumlah penampang geser, m = 1

Faktor pengaruh ulir pada bidang geser, r1 = 0.4

Luas penampang baut,Ab = 285.022957 mm

Faktor reduksi kekuatan geser, = 0.75

Tahanan geser nominal, Vn = r1.m.Ab.fu = 94057.576 N

Tahanan geser angkut baut, .Vn = 70543.18186 N

Kontrol, OK

Gaya Tumpu Pada Angkur Baut

Gaya tumpu pada angkur baut, Ru1 = Vu1 = 122.088 N

Tahanan tumpu nominal, Rn = 2,4.d.t.fu = 422910 N

Tahanan tumpu, .Rn = 317182.5 N

Kontrol, OK

Kombinasi Geser dan Tarik

Konstanta tegangan untuk baut mutu tinggi, f1 = 807 MPa

f2 = 621 MPa


Tegangan geser akibat beban terfaktor,

fuv = 0.043 MPa

Kuat geser angkur baut, .r1.m.fu = 247.5 MPa

Kontrol, OK

Gaya tarik akibat beban terfaktor, Tu1 = 13321.372 NTahanan tarik angkur baut, .f1.Ab = 172510.145 N

Kontrol, OK

1 =

=

1

4 . .

1 .

1 =

=14 . .

1 .

1 .

= .

. 1. .

1 . 1.


09 02 11 047


41/49

STRUKTUR BAJA II



Kuat tarik angkur baut, ft = 0,75.fu = 618.750 MPa

Batas tegangan kombinasi, f1-r2.fuv = 806.919 MPa

Kontrol, OK

OK

Kontrol Panjang Angkur Baut

Panjang angkur yang ditanam, La = 500 mmKuat tekan beton, fc' = 25 MPa

Tegangan leleh angkur baut, fy = 400 MPa

Panjang minimum angkur yang ditanam,

L min = 381 mm

Kontrol, OK

1 2. 2

= 4. .


09 02 11 047


42/49

STRUKTUR BAJA II



2. Tumpuan pada titik B

Beban Kolom

Pu = 78031.41 kg

Mu = 4185.56 kgm

Vu = 2056.17 kg

Plat Tumpuan

fy = 240 MPafu = 370 MPa

B = 450 mm

L = 450 mm

t = 35 mm

Kolom Pedestial

fc' = 25 MPa

I = 500 mm

J = 500 mm

Kolom Baja WF

ht = 250 mm

bf = 250 mm

tw = 9 mm

tf = 14 mm

Angkur Baut

fu = 825 MPa

fy = 400 MPa

d = 19.05 mm

nt = 5 bh

nc = 5 bh

f = 175 mm

La = 500 mm

Eksentrisitas Beban

e = 53.639 mm

L/10 = 45.000 mm

e > L/10 OK

h = 236 mm

n = nt + nc = 10 bh

et = 293 mm

ec = 57 mm

ht

h

P

Mu

Vu

La

L

J

B I

a a

f

=

= = + 2

= 2


09 02 11 047


43/49


44/49

STRUKTUR BAJA II





Tu1 = 30360.344 N


Ab = 285.022957 mm




Kontrol, OK



Vu1 = 2056.170 N







Kontrol, OK





Kontrol, OK



f2 = 621 MPa



fuv = 0.721 MPa


Kontrol, OK


Kontrol, OK

1 =

=

1

4 . .

1 .

1 =

=14 . .

1 .

1 .

= .

. 1. .

1 . 1.


09 02 11 047


45/49

STRUKTUR BAJA II





Kontrol, OK

OK





L min = 381 mm

Kontrol, OK

1 2. 2

= 4. .


09 02 11 047


46/49

STRUKTUR BAJA II



3. Tumpuan pada titik C

Beban Kolom

Pu = 29991.36 kg

Mu = 5854.456 kgm

Vu = 3346.15 kg

Plat Tumpuan

fy = 240 MPafu = 370 MPa

B = 450 mm

L = 450 mm

t = 25 mm

Kolom Pedestial

fc' = 25 MPa

I = 500 mm

J = 500 mm

Kolom Baja WF

ht = 250 mm

bf = 250 mm

tw = 9 mm

tf = 14 mm

Angkur Baut

fu = 825 MPa

fy = 400 MPa

d = 19.05 mm

nt = 3 bh

nc = 3 bh

f = 175 mm

La = 500 mm

Eksentrisitas Beban

e = 195.205 mm

L/6 = 75.000 mm

e > L/6 OK

h = 236 mm

n = nt + nc = 6 bh

et = 293 mm

ec = 57 mm

ht

h

P

Mu

Vu

La

L

J

B I

a a

f

=

= = + 2

= 2


09 02 11 047


47/49

STRUKTUR BAJA II



Tahanan Tumpu Beton


Pt = 5834.497 kg

Gay tekan total pada plat tumpuan,

Puc = 35825.857 kg

Panjang bidang tegangan tekan beton,Y = 321 mm

Luas plat tumpuan baja, A1 = B.L = 202500 mm

Luas penampang kolom pedestial, A2 = I.J = 250000 mm

Tegangan tumpu nominal,

fcn = 23.611 MPa

fcn = 1,70.fc' = 42.5 MPa

Tegangan tumpu nominal beton yang digunakan, fcn = 23.611 MPa

Faktor reduksi kekuatan tekan beton, = 0.65

Tegangan tumpu beton yang diijinkan, .fcn = 15.347 MPa

Tegangan tumpu maksimum yang terjadi pada beton,fcu = 4.960 MPa

Kontrol tegangan, OK

Kontrol Dimensi Plat Tumpuan

Lebar minimum plat tumpuan yang diperlukan,

Bp min = 145.443 mm

Lebar plat yang digunakan, B = 450 mm

Kontrol, OK

Panjang bagian plat tumpuan jepit bebas,

Modulus penampang plastis plat,

Z = 70312.5 mm

Momen yang terjadi pada plat akibat beban terfaktor,

Mup = 11209268.06 Nmm

Faktor reduksi kekuatan lentur, = 0.9Tahanan momen nominal plat, Mn = fy.Z = 16875000 Nmm

Tahanan momen plat, .Mn = 15187500 Nmm

Kontrol, OK

a = 106.250 mm

fcu1 = 3.318 MPa

= .

= +

= 3. 2

= 0,85. . 21

= 2. . .

min = 0,5...

= 0,95. 2

1 = 1 .

=14 . .

=12 ..1.+ 13 .. 1 .

.


09 02 11 047


48/49

STRUKTUR BAJA II





Tu1 = 19448.322 N


Ab = 285.022957 mm




Kontrol, OK



Vu1 = 5576.917 N







Kontrol, OK





Kontrol, OK



f2 = 621 MPa



fuv = 1.957 MPa


Kontrol, OK


Kontrol, OK

1 =

=

1

4 . .

1 .

1 =

=14 . .

1 .

1 .

= .

. 1. .

1 . 1.


09 02 11 047


49/49

STRUKTUR BAJA II





Kontrol, OK

OK





L min = 381 mm

Kontrol, OK

1 2. 2

= 4. .

perhitungan tugas baja2 revie fix

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