Merencana Baja Final

MERENCANA

KONSTRUKSI

BAJA

DIMAS FATCHUR R.

NIM 095534057

DOSEN PEMBIMBING:

DWIARIANTO SYAHIRUL A.,ST

S1 PEND. TEKNIK BANGUNAN

JURUSAN TEKNIK SIPIL

FAKULTAS TEKNIK

UNIVERSITAS NEGERI SURABAYA

2013

Direncanakan menggunakan gording profil baja Light Lip Chanel

, dan memiliki data - data profil sebagai berikut :

Profil CNP 150 50 20 4.5

A = 11.72 cm2 lx = 368 cm4

W = 9.2 kg/m ly = 35.7 cm4

A = 150 mm ix = 5.6 cm

B = 50 mm iy = 1.75 cm

C = 20 mm Zx = 49 cm3

t = 4.5 mm Zy = 10.5 cm3

1. Pembebanan Pada Gording

A. Beban Mati

- Beban Gording = 9.2 kg/m

- Beban Atap = 2.32 kg/m (Brosur)

Berat Total = 11.52 kg/m

- Berat Pengikat 10% = 1.152 kg/m +

qd = 12.67 kg/m

Gording ditempatkan tegak lurus bidang penutup atap dan beban mati Px

bekerja Vertikal, P diuraikan pada sumbu X dan Sumbu Y Sehingga diperoleh :

Gambar gaya yang bekerja

qdx = qd x cos α = 12.67 = 11.91 kg/m

qdy = qd x sin α = 12.67 = 4.33 kg/m

B. Beban Hidup Terpusat

Beban berguna atau beban hidup adalah beban terpusat yang bekerja ditengah

tengah bentang gording. Beban ini diperhiutngkan kalau ada orang yang bekerja

diatas gordng. Besarnya beban hidup dambil dari PPIUG 1987,

P = 100 kg

GORDING

0.940

0.342

x

x

x x x

Gambar gaya yang bekerja

Px = P cos α = 100 0.9397 = 93.97 Kg

Py = P Sin α = 100 0.342 = 34.20 Kg

C. Beban Angin

Beban angin diperhitungkan dengan menganggap adanya tekanan positif (Tiup

dan tekanan negatif (Hisap), yang bekerja tegak lurus pada bidang atap. Menurut

PPIUG 1987, tekanan tiup harus diambil minimal 25 kg/m2. Dalam perencanaan

ini, besarnya tekanan angin (w) diambil sebesar 25 kg/m2

Jenis Perencanaan Gedung = - Gedung Tertutup

- Gedung terbuka

Koefisien angin

- Kondisi 1

Koefisien angin tekan = -0.02α - 0.4 = 0.02. 20 - 0.4 =

Koefisien angin hisap = 0.4

- Kondisi 2

Koefisien angin tekan = 0.8

0

x

x

Koefisien angin hisap = 0

- Kondisi 3



Tekanan angin

- Kondisi 1

>Tekanan angin tekan = Koef. x tekanan angin x jrk gording

= 0 25 1.6

= 0 Kg/m

>Tekanan angin hisap =Koef. x tekanan angin x jrk gording

= 0.4 25 1.6

= 16 Kg/m

- Kondisi 2


= 0.8 25 1.6

= 32 Kg/m


= 0 25 1.6

= 0 Kg/m

- Kondisi 3


= 1.2 25 1.6

= 48 Kg/m


= 0.4 25 1.6

= 16 Kg/m

Dari ketiga kondisi angin diatas, maka diambil tekanan angin yang terbesar

untuk mewakili = 48 Kg/m. karena tekanan angin tegak lurus bidang

atap maka :

qwx = 48 Kg/m

qwy = 0 ( karena tegak lurus sumbu gording)

2. Perhitungan beban kombinasi

1. Kombinasi M + H ( 1.2D+1.6L )

- akibat beban merata atap ( Beban mati)

1.2 qx = 1.2 11.91 = 14.29 kg/m = 0.14 kg/cm

1.2 qy = 1.2 4.33 = 5.20 kg/m = 0.05 kg/cm

- akibat beban terpusat orang (beban hidup)

1.6 Px = 1.6 93.97 = 150.35 kg

1.6 Py = 1.6 34.20 = 54.72 kg

2. Kombinasi M + A ( 0.9D+1.3W )


x x

x x

x x

x x

x x

x x

x

x

x

x

x

0.9 qx = 0.9 11.91 = 10.72 kg/m = 0.11 kg/cm

0.9 qy = 0.9 4.33 = 3.90 kg/m = 0.04 kg/cm

- akibat beban merata angin

1.3 qx = 1.3 48.00 = 62.40 kg/m = 0.62 kg/cm

1.3 qy = 0 kg/m =

3. Kontrol Lendutan

1. Kombinasi M + H

fx = 5 (qux atap) x L + 1 Px . L

384 E . Ix 48 E . Ix

5 0.14 x 600 + 1 150.35

384 368 48

0.31 + 0.875

1.19 cm

fy = 5 (quy atap) x (L/3) + 1 Py . (L/3)

384 E . Iy 48 E . Iy

5 0.05 x 200 + 1 54.72

384 35.7 48 2100000 36

0.01 + 0.122

0.14 cm

f ' = fx + fy f ijin = L/250

= 1.19 0.14 = 600 250

= 1.20 cm < = 2.4 cm

Karena f ' yang terjadi lebih kecil dari f ijin Maka dalam kombinasi 1

dinyatakan KUAT

2. Kombinasi M + A

fx = 5 (qux atap+ quxangin) x Lx

384 E . Ix

5 0.107 0.624 600

384 368

1.60 cm

fy = 5 (quy atap+ quy angin) x (L/3)

384 E . Iy

5 0.039 0 200

384 35.7

0.01 cm

f ' = fx + fy f ijin = L/250

= 1.60 0.01 = 600 250

= 1.60 cm < = 2.4 cm

2100000

600

2100000 2100000 368

200

2100000

2100000

x

x

x

= x

= x

=

=

= x

= x

=

=

x

x

x

x

+ /

= x

= x

=

( ) +

x

= x

= x

=

( ) +

x

+ /

4 3

4 3

4

4

4

4


dinyatakan KUAT

4. Perhitungan Momen Kombinasi

1 . Kombinasi M + H

Mx = Atap= 1 qux atap L

8

= 0.125 14.29 6 = 64.3 kg.m

Orang= 1 Pux Orang L

4

= 0.25 150.4 6 = 225.5 kg.m +

Mx1 = 289.8 kgm

My = Atap= 1 quy atap (L/3)

8

= 0.125 5.20 6 /3) = 2.6 kg.m

Orang= 1 Puy Orang (L/3)

4

= 0.25 54.72 6 /3) = 27.36 kg.m +

My1 = 29.96 kgm

Kontrol tegangan

Mx My < σ

Wx Wy

< 1600

kg/cm2 < 1600 kg/cm2

Karena f ' yang terjadi lebih kecil dari f ijin Maka dalam kombinasi

1 dinyatakan KUAT

2 . Kombinasi M + A


8

= 0.125 10.72 6 = 48.23 kg.m

angin = 1 qux angin L

8

= 0.125 62.4 6 = 280.8 kg.m +

Mx1 = 329.0 kgm

28982.83 2996.21

49 10.5

656.72

x

2

2

2

2 (

(

+

+

2

2

2

2

x x

x x

x x

x x

x x

x x

x x

x x

x x

x x

x x

x x

My = Atap 1 quy atap L/3

8

= 0.125 3.901 2 = 1.95 kg.m +

My1 = 1.950 kgm

Kontrol tegangan

Mx My < σ

Wx Wy

< 1600



2 dinyatakan KUAT

5. KONTROL KEKUATAN PROFIL

A. Penampang Profil

Sayap

bw < 170 bw = lebar gording

2tf fy tf = tebal gording

50 < 170 h = tinggi gording

9 15.49 tw = tebal gording

5.56 < 10.97

OK

Badan

h < 1680

tw fy

150 < 1680

4.5 15.49

33.33 < 108.4

OK

B. Lateral Buckling

Jarak pengaku lateral ( Sagrod )

Lb = L/3 = 150 cm= 1500 mm

Lp = 1.76 x iy x E

fy

= 1.76 1.75

= 91.11 cm= 911 mm

671.74

32902.65 195.03

49 10.5

2100000

2400

2

2

+

+

x x

Lb > Lp Bukan Bentang Pendek

fL = fy - fr

240 70 = 170 Mpa

= 3038 + 4283 + 941.6

= 8262 mm4

3.14 8262 1172

49000

= Mpa

Iw = Iy .h2 = 150 4.5

4 4

= mm

= 4 2E+09 = 1.163504 mm2/N2

80000 8262 35.7

= 10.5 1 + 1+ 1.164 170

= 10577 mm= 1058 cm > Lb= 150

Bentang Menengah

Mpx = fy * Zx = 240 49000 = Nmm

Mpy = fy * Zy = 240 10500 = Nmm

Mrx = Sx * ( fy-fr )= 170 = Nmm

Mry = Sy * ( fy-fr )= 170 = Nmm

Mux = kgm

Ma= kgm

Mb= kgm

72.46

289.83

49000

10500

11760000

2520000

8330000

1785000

289.83

12611.59

170

200000 80000

2

12611.59

49000

1889444813

357000

𝐿𝑟 = 𝑍𝑦𝑋1

𝑓𝐿1 + 1 + 𝑋2𝑓𝐿2

𝑋1 =𝜋

𝑆

𝐸𝐺𝐽𝐴

2

𝑋2 = 4𝑆

𝐺𝐽

2𝐼𝑤

𝐼𝑦

𝐽 = 2.1

3. 𝑏. 𝑡3 +

1

3. ℎ𝑡 − 2𝑡 . 𝑡3 +

2

3. 𝑎 − 𝑡 . 𝑡3

2

- =

= x x x

2 x

𝑀𝑛 = 𝐶𝑏 𝑀𝑟 + (𝑀𝑝 − 𝑀𝑟)(𝐿𝑟 − 𝐿)

(𝐿𝑟 − 𝐿𝑝)≤ 𝑀𝑝

2 x - ) (

Mc = kgm

= 1.316

= Nmm

Mpx = Zx . fy = 49 2400

= kg.cm= Nmm

= Nmm

Mny = Zy . fy = 1/4 tf. bf 2 . fy

= 0.25 0.45 5 2400

= kg.cm= Nmm

6. Kontrol Interaksi momen

Mux Muy

φ Mnx φ Mny

0.9 0.9

+ = 0.31 < 1

217.37

15198706.23

3256865.62

11760000

0.21 0.10

OK

117600

6750

2898282.67 299620.59

15198706.23 3256865.62

675000

+

+

2

𝐶𝑏 =12.5 𝑥 𝑀𝑢𝑥

(2.5 𝑥 𝑀𝑢𝑥 + 3 𝑥 𝑀𝑎 + 4 𝑥 𝑀𝑏 + 3 𝑥 𝑀𝑐)

𝑀𝑛𝑥 = 𝐶𝑏 𝑀𝑟𝑥 + (𝑀𝑝𝑥 − 𝑀𝑟𝑥)(𝐿𝑟 − 𝐿)


𝑀𝑛𝑦 = 𝐶𝑏 𝑀𝑟𝑦 + (𝑀𝑝𝑦 − 𝑀𝑟𝑦)(𝐿𝑟 − 𝐿)


Pembebanan

a. Beban Mati:

Berat Gording = 9.2 Kg/m

Berat Atap = 2.32 Kg/m

= 11.52 Kg/m

Pengikat 10% = 1.152 Kg/m +

qD = 12.67 Kg/m

b. Beban Hidup

Py = P Sin α

= 100 Sin 20 = 34.2 Kg

Pmax = + Py

= 13 Sin 20 6

= Kg

Dimensi Sagrod

Jumlah Lapangan, n= 3

tan α= x = 1.6 = 0.8

y 2

α =

R = n. Pmax = 3 42.87

Sin 38.66

= Kg

Luas Perlu = 205.9 = 0.129 cm2

1600

d perlu = 4

3.14

= cm= 4.049 mm

Jadi digunakan diameter Sagrod ɸ 8 (As = 0.502 mm2)

0.40

SAGROD

3

qy . Ly

3

34.20

42.87

38.66

Sin α

205.877

0.129

x

x x +

α

h1

h2

a = 20

6.5 6.5

26

tekanan angin = 1.2 x 25 30 kg/m2

tg a = h1 / 13 tg a = h2/6.5

h1 = tg 20 13 h2 = tg 20 6.5

h1 = 0.364 13 h2 = 0.36 6.5

4.732 m 2.37 m

Perhitungan gaya yang bekerja

F1 = 4.732 2.37 6.5 30

= kg

F2 = 6.5 2.37 30

= 230.7 kg

Ftot = F1 + F2

= 692 231 923 kg

Luas Batang = 922.66 1600 = 0.5767 cm2

Jadi digunakan diameter ɸ 10 dengan ( As = 0.7854 cm2)

Kontrol tegangan = 922.7

0.7854

= 1175 kg/cm2 < 1600 kg/cm2

OKE

2

692.00

IKATAN ANGIN

1/2 x

/

= +

x

=

x + ( )

x x

7

13 m 13 m

Horizontal Bracing - Kolom (Ikatan Gigi Anjing)

F = 13 7 25 0.9

= 2048 Kg

Luas Perlu = F

1600

= 2047.5 = 1.2797 cm2

1600

d perlu = 1.280 4

= cm= 12.77 mm

Digunakan diameter ɸ 13 dengan ( As = 1.327 mm2)

Kontrol tegangan = 2048

1.33

= 1543 kg/cm2 < 1600 Kg/cm2

OKE

Cek Profil Regel

Digunakan Profil CNP 125 50 20 4.5

A = 7.01 cm2 lx = 238 cm4

W = 5.5 kg/m ly = 28.3 cm4

A = 125 mm ix = 3.9 cm

B = 50 mm iy = 1.87 cm

C = 20 cm Zx = 38 cm3

t = 4.5 mm Zy = 8.19 cm3

q

F

REGEL

1.28

3.14

x x x

x x x

m

Pembebanan

Beban q = Beban Profil CNP + Berat Diagonal

= 5.5 + 1.04 2.236

= 5.5 + 0.4651

= 5.97 Kg/m

Beban horizontal (F) = 13 7 25 0.9

= 2047.5 Kg

Dari Perhitungan SAP diperoleh M = Kgm

Kontrol tegangan

M < σ

W

< 1600


OKE

Kontrol Lendutan

f = 5 M . L

48 E . I

5 39106 x 100 f ijin = L/250

48 28.3 = 600 250

cm < = 2.4 cm

OKE

38

1029.11

391.06

2100000

0.69

39106

= x

= x

=

2

/

x x x

/ ) (

/6 ) ( 2

Beban Mati

- Beban atap Berat atap x jarak kolom x panjang lereng atap total

= 5.71 6 = Kg

- Beban Gording Berat Profil gording x jarak kolom x jumlah gording

= 9.2 6 11 = Kg

- Berat Gabungan Berat atap + Berat Gording

= 521 + 607.2 = Kg

- Berat Perangkai 10% dari berat gabungan

= Kg

- Berat Mati Total Berat gabungan + Berat Perangkai

= = Kg

Beban Terpusat (Pd) = 1241 Kg = kg/m

11

Beban Hidup

Akibat orang bekerja ( P ) = 100 kg

Beban Angin

- Kondisi 1 ( Tertutup )

q1 = 0.02 20 0.4 x 25 1.6

= 0 25 1.6 = 0 kg/m ( Tekan )

q2 = 0.4 25 1.6 = 16 kg/m ( Hisap )

q3 = 0.9 25 1.6 = 36 kg/m ( Tekan )

q4 = 0.4 25 1.6 = 16 kg/m ( Hisap )

- Kondisi 2 ( Terbuka )

q1 = 0.8 25 1.6 = 32 kg/m ( Tekan )

q2 = 0


q1 = 0.4 25 1.6 = 16 kg/m ( Hisap )

q2 = 1.2 25 1.6 = 48 kg/m ( Hisap )

1241.391128.53 112.85

112.85

Pembebanan Rafter

15.217 521.33

607.20

112.85

1128.53

x x

x x

x x

x -

x x

x x

x x

) ( x

x x

+

x x

x x

=

=

=

=

=

+0.9 -0.4

TABLE: Element Forces - Frames

Frame StationOutputCaseCaseType P V2 M3

Text m Text Text Kgf Kgf Kgf-m

1 0 MH Combination-1856.37 -1975.56 -7713.54

1 0.934 MH Combination-1842.05 -1936.21 -5886.16

1 0.934 MH Combination-1787.33 -1785.86 -5886.16

1 1.267 MH Combination-1782.23 -1771.84 -5294.06

1 1.267 MH Combination -1735.9 -1644.56 -5294.06

1 1.729 MH Combination-1728.82 -1625.1 -4538.55

1 2.534 MH Combination-1716.48 -1591.19 -3243.96

1 2.534 MH Combination-1670.15 -1463.92 -3243.96

1 3.459 MH Combination-1655.98 -1424.99 -1908.9

2 0 MH Combination-1655.98 -1424.99 -1908.9

2 0.676 MH Combination-1645.62 -1396.53 -955.61

2 0.676 MH Combination-1544.58 -1118.9 -955.61

2 1.729 MH Combination-1528.43 -1074.53 199.84

2 2.276 MH Combination-1520.05 -1051.51 780.73

2 2.276 MH Combination-1473.73 -924.24 780.73

2 3.459 MH Combination-1455.59 -874.42 1844.49

3 0 MH Combination 50.53 138.82 7.276E-12

3 0.692 MH Combination 61.13 167.95 -106.1

3 1.383 MH Combination 71.73 197.09 -232.35

4 0 MH Combination-1455.59 -874.42 1844.49

4 0.417 MH Combination -1449.2 -856.85 2205.6

4 0.417 MH Combination-1348.15 -579.22 2205.6

4 1.729 MH Combination-1328.04 -523.96 2929.36

4 2.017 MH Combination-1323.62 -511.84 3078.44

4 2.017 MH Combination -1277.3 -384.56 3078.44

4 3.459 MH Combination -1255.2 -323.85 3589

5 0 MH Combination -1255.2 -323.85 3589

5 0.159 MH Combination-1252.77 -317.17 3639.83

5 0.159 MH Combination-1151.72 -39.55 3639.83

5 1.729 MH Combination-1127.65 26.61 3649.99

5 1.809 MH Combination-1126.43 29.95 3647.75

5 1.809 MH Combination-1080.11 157.22 3647.75

5 3.459 MH Combination-1054.81 226.72 3330.99

6 0 MH Combination-1054.81 -226.72 3330.99

6 1.65 MH Combination-1080.11 -157.22 3647.75

6 1.65 MH Combination-1126.43 -29.95 3647.75

6 1.729 MH Combination-1127.65 -26.61 3649.99

6 3.3 MH Combination-1151.72 39.55 3639.83

6 3.3 MH Combination-1252.77 317.17 3639.83

6 3.459 MH Combination -1255.2 323.85 3589

7 0 MH Combination -1255.2 323.85 3589

7 1.441 MH Combination -1277.3 384.56 3078.44

7 1.441 MH Combination-1323.62 511.84 3078.44

7 1.729 MH Combination-1328.04 523.96 2929.36

7 3.041 MH Combination-1348.15 579.22 2205.6

7 3.041 MH Combination -1449.2 856.85 2205.6

7 3.459 MH Combination-1455.59 874.42 1844.49

8 0 MH Combination-1455.59 874.42 1844.49

8 1.183 MH Combination-1473.73 924.24 780.73

8 1.183 MH Combination-1520.05 1051.51 780.73

8 1.729 MH Combination-1528.43 1074.53 199.85

8 2.783 MH Combination-1544.58 1118.9 -955.6

8 2.783 MH Combination-1645.62 1396.53 -955.6

8 3.459 MH Combination-1655.98 1424.99 -1908.9

9 0 MH Combination-1655.98 1424.99 -1908.9

9 0.924 MH Combination-1670.15 1463.92 -3243.96

9 0.924 MH Combination-1716.48 1591.19 -3243.96

9 1.729 MH Combination-1728.82 1625.1 -4538.55

9 2.191 MH Combination -1735.9 1644.56 -5294.06

9 2.191 MH Combination-1782.23 1771.84 -5294.06

9 2.524 MH Combination-1787.33 1785.85 -5886.14

9 2.524 MH Combination-1842.05 1936.21 -5886.14

9 3.459 MH Combination-1856.37 1975.56 -7713.54

10 0 MH Combination-3150.25 -1068.74 -4.366E-12

10 1.167 MH Combination-3097.96 -1068.74 1246.86

10 2.333 MH Combination-3045.67 -1068.74 2493.73

11 0 MH Combination-3045.67 -1068.74 2493.73

11 1.167 MH Combination-2993.38 -1068.74 3740.59

11 2.333 MH Combination-2941.09 -1068.74 4987.46

12 0 MH Combination-2941.09 -1068.74 4987.46

12 1.167 MH Combination -2888.8 -1068.74 6234.32

12 2.333 MH Combination-2836.51 -1068.74 7481.19

13 0 MH Combination-3150.25 1068.74 0

13 1.167 MH Combination-3097.96 1068.74 -1246.86

13 2.333 MH Combination-3045.67 1068.74 -2493.73

14 0 MH Combination 71.73 -197.09 -232.35

14 0.692 MH Combination 61.13 -167.95 -106.1

14 1.383 MH Combination 50.53 -138.82 -2.642E-12

15 0 MH Combination-3045.67 1068.74 -2493.73

15 1.167 MH Combination-2993.38 1068.74 -3740.59

15 2.333 MH Combination-2941.09 1068.74 -4987.46

16 0 MH Combination-2941.09 1068.74 -4987.46

16 1.167 MH Combination -2888.8 1068.74 -6234.32

16 2.333 MH Combination-2836.51 1068.74 -7481.19

Mu = Kgm (Momen)

Nu = Kg (Geser)

Vu = Kg (Aksial)

Ma = kgm= kgcm

Mb = kgm= kgcm

Ms = kgm= kgcm

Dari hasil perhitungan SAP 2000, diperoleh Profil WF dengan data Sbb :

WF 300 x 150 x 6.5 x 9

A = 46.78 cm2 tf = 9 mm Zx = 481 cm3

W = 36.7 kg/m Ix = 7210 cm4 Zy = 67.7 cm3

a = 300 mm Iy = 508 cm4

bf = 150 mm tw = 6.5 mm

iy = 3.29 cm ix = 12.4 cm

Kontrol Lendutan

f ijin = L = 1383 = 3.84 cm

360 360

f = 5 1383 0.1

48 2.1 E6 7210

f = 1.892 cm < 3.84 cm OK

Kontrol Tekuk

untuk arah x :

kcx = 1.2 (jepit-rol tanpa putaran sudut)

L = 1383 cm

Lkx = 150 cm

= 150 = 12.1

12.4

= 3.14 46.78

12.1

= kg

Cek Profil Rafter

1844.49

2.1 E6

6619110.95

7713.54

1856.38

1975.56

7713.54

3649.99

771354

364999

184449

184449 364999771354

𝑓 =5

48.

𝐿2

𝐸 . 𝐼𝑥. 𝑀𝑠 − 0.1 𝑀𝑎 − 𝑀𝑏

2

2

x

EANcrbx

ix

Lkxx

x x 2

2

( )) ( - - x 2

untuk arah y :

kcy = 0.8 (jepit-Sendi)

L = 300 9 = 159 cm

2

Lky = 300 159 = 141 mm = 14.1 cm

= 14.1 = 4.286

3.29

= 3.14 46.78

4.286

=

Maka dipakai lx karena lx > ly

= 12.1 2400

3.142 2.1 E6

= 0.13

λc < 0.25 ,maka ω = 1

Pn = Ag fy = 46.78 2400

w

= kg

0.85

= 0.021 < 0.2

Batang Dianggap Tidak Bergoyang Maka : (sumbu X )

Cmx = 1

112272

112272

1975.56

1

2.1 E6

52734202.06

E

fyxc

22yEANcrby x x

iy

Lkyy

2

2

y

EANcrby

x x 2

2 2

cPn

Pu

x

x

1

)(1

Ncrbx

Vu

CmxSbx

+

-

Sbx =

1 -

Sbx = 1.00 > 1.00

Sbx = 1

Mux = Mutx . Sbx

Mux = 1 = 7716 kgm

Batang Dianggap Tidak Bergoyang Maka : (sumbu Y )

Cmy = 1

Sby =

1 -

Sby = 1 > 1

Sby = 1

Muy = Muty . Sby

Muy = 1 = 0 kgm

Kontrol kekuatan Profil

Penampang Profil

Sayap





8.33 < 10.97

OK

Badan

h < 1680

tw fy

300 < 1680

6.5 15.49

46.15 < 108.4

OK

1975.56

52734202.06

0

1

6619110.95

1975.56

7713.54

1

1

)(1

Ncrby

Vu

CmySby

Lateral Buckling

Jarak pengaku lateral ( Rib )

Lb = 160 cm

Lp = 1.76 x iy x E

fy

= 1.76 3.29

= 171.3 cm

Bila Lb < Lp ,maka Mnx = Mpx

Mnx = Mpx = Zx . fy = 481 2400

= kg.cm

Mny = Zy . fy = 1/4 tf. bf 2 . fy

= 0.25 0.9 15 2400

= kg.cm

Interaksi Momen

1.7 0.9 0.9

+ +

0.013 < 1

Kontrol Kuat Rencana Geser

300 < 1100

6.5 15.49

46.15 < 71

Plastis

Vn = 0.6 fy Aw

= 0.6 2400 30 0.9 0.65

= 0.6 2400 18.33

= Kg

Vu < ФVn

1976 < 0.9

1976 < 23756

OK

0.0070.005

26395.2

26395.2

0

1154400

121500

112272 1154400 121500

7715.8428941975.56 0

2.1 E6

2400

x x

x

x x x 2

12

bMny

Muy

bMnx

Mux

cPn

Pu

x x x 2. + +

fytw

h 1100

x x -(2. )) (

x x

Mu = Kgm= Kgcm

Du = Kg

Asumsi Diameter Baut ɸ 0.5 inch = 12.7 mm

Tebal Pelat penyambung 10 mm

fu = 8070

fy = 6210

Kontrol Kekuatan Baut

Perhitungan Ruv yang Diterima Setiap Baut

Ruv = Du = 1856 = 309.4 kg

n 6

Perhitungan Kuat Geser Baut

f Rnv = 0.75 0.5 fu Ab n

= 0.75 0.5 8070 1.27 1

= kg

Perhitungan Kuat Tumpu Baut

f Rn = 2.4 d tp fu 0.75

= 2.4 1.27 1 8070 0.75

= kg

Perhitungan Kuat tarik Baut

f Rnt = 0.75 0.75 fu Ab

= 0.75 0.75 8070 1.266

= 5747 kg

Rumus Interaksi Geser dan Kuat Tarik Baut

309.4 + Rut < 1

3832 5747

Rut = T =5283 kg

SAMBUNGAN RAFTER 1

7713.54

1856.38

3831.62

771354

18448.02

1)()( 22 nt

ut

nv

uv

R

R

R

R

. . . .

. . . .

. . . .

. . . .

. . .

. . .

Kontrol Momen Sambungan

Letak Garis Netral a:

100

100

250

150

150

a = Σ T = 5283 6

fy B 6210 15

= 0.34 cm

d1 = 15 0.34 = 14.66 cm

d2 = 14.66 25 = 39.66 cm

d3 = 39.66 10 = 49.66 cm

Sdi = 104 cm

f Mn = 0.9 fy a2

B + S T di

= 0.9 6210 0.116 15 + 104 5283 2

= kgcm > kgcm

OK

7713541103566

2

2

d1

d2

d3

a

2T

2T

2T

-

+

+

. . .

. . . . .

Mu = Kgm= Kgcm

Du = Kg

Asumsi Diameter Baut ɸ 0.5 inch = 12.7 mm

Tebal Pelat penyambung 10 mm

fu = 8070

fy = 6210

Kontrol Kekuatan Baut

Perhitungan Ruv yang Diterima Setiap Baut

Ruv = Du = 1055 = 175.8 kg

n 6


f Rnv = 0.75 0.5 fu Ab n

= 0.75 0.5 8070 1.27 1

= kg

Perhitungan Kuat Tumpu Baut

f Rn = 2.4 d tp fu 0.75

= 2.4 1.27 1 8070 0.75

= kg


f Rnt = 0.75 0.75 fu Ab

= 0.75 0.75 8070 1.266

= 5747 kg

Rumus Interaksi Geser dan Kuat Tarik Baut

175.8 + Rut < 1

3832 5747

Rut = T =5484 kg

SAMBUNGAN RAFTER 2

3330.99 333099

1054.81

3831.62

18448.02

1)()( 22 nt

ut

nv

uv

R

R

R

R

. . . .

. . . .

. . . .

. . . .

. . .

. . .

Kontrol Momen Sambungan

Letak Garis Netral a:

100

100

250

150

150

a = Σ T = 5484 6

fy B 6210 15

= 0.353 cm

d1 = 15 0.353 = 14.65 cm

d2 = 14.65 25 = 39.65 cm

d3 = 39.65 10 = 49.65 cm

Sdi = 103.9 cm

f Mn = 0.9 fy a2

B + S T di

= 0.9 6210 0.125 15 + 104 5484 2

= kgcm > kgcm

OK

2

2

1145189 333099

d1

d2

d3

a

2T

2T

2T

-

+

+

. . .

. . . . .

Beban Mati

- Beban atap Berat atap x jarak kolom x panjang lereng atap total

= 5.71 6 = Kg

- Beban Gording Berat Profil gording x jarak kolom x jumlah gording

= 9.2 6 11 = Kg

- Berat Gabungan Berat atap + Berat Gording

= 521 + 607.2 = Kg

- Berat Perangkai 10% dari berat gabungan

= Kg

- Berat Mati Total Berat gabungan + Berat Perangkai

= = Kg

Beban Terpusat (Pd) = 1241 Kg = kg/m

11

Beban Hidup

Akibat orang bekerja ( P ) = 100 kg

Beban Angin

- Kondisi 1 ( Tertutup )

q1 = 0.02 20 0.4 x 25 1.6

= 0 25 1.6 = 0 kg/m ( Tekan )

q2 = 0.4 25 1.6 = 16 kg/m ( Hisap )

q3 = 0.9 25 1.6 = 36 kg/m ( Tekan )

q4 = 0.4 25 1.6 = 16 kg/m ( Hisap )


q1 = 0.8 25 1.6 = 32 kg/m ( Tekan )

q2 = 0


q1 = 0.4 25 1.6 = 16 kg/m ( Hisap )

q2 = 1.2 25 1.6 = 48 kg/m ( Hisap )

1241.391128.53 112.85

112.85

Pembebanan Rafter Rangka

15.217 521.33

607.20

112.85

1128.53

x x

x x

x x

x -

x x

x x

x x

) ( x

x x

+

x x

x x

=

=

=

=

=

+0.9 -0.4

Perhitungan batang tekan

Gaya max pada batang A1: Pmax = Kg

Lalu Dicoba Profl Double Siku 70 70 7 dengan data data

Sebagai Berikut

Ix =Iy = 42.4 cm4 w = 7.38 kg/m

Imin = 1.37 cm ix = iy= 2.12 cm

A = 9.4 cm2 wx = wy = 8.43 cm3

e = 1.97 cm t plat= 6 mm

Syarat I : pemeriksaan terhadap sumbu bahan (x-x)

λx = lkx 320 150.9 ( PPBBI, hal 12 )

ix 2.12

ωx = 3.236

Tegangan yang terjadi :

σx = ω . P 3.236 7458

2 F 2 9.4

= 1283.79 kg/cm2

< 1600 kg/cm2

OK

σ = P

2 A 2 9.4

= 396.721 kg/cm2 < 1284 kg/cm2 OK

Syarat II : Pemeriksaan terhadap sumbu bebas bahan ( y - y )

Kelangsingan batang dicari setelah dibagi dengan medan ganjil

Iy = Lk 320 106.7

3 3

𝛿

Jarak antar 2 batang karena pelat simpul

a = e + 1/2𝛿 h = 2e +𝛿

= 1.97 + 1/2 x 0.6 = 2 x 1.97 + 0.6

= 2.27 cm = 4.54 cm

Momen inersia dari susunan profil ganda :

Iy-y = 2 ( Iy + F .a2 )

Rangka Batang

7458.36

7458.36

x

x =

= x

= =

2 ( )

= =

x x

= 106.67 9.4 2.27

= 310.208 cm4

Jari jari minimum (iy )

= 310.2 = 4.06 cm

9.4

Menentukan kelangsingan sebelum plat kopel

λy = 320 78.78 = 79

4.06

Setelah dipasang plat kopel

ly = 320 78.78

4.06

λl = ly 78.78 57.5

imin 1.37

Sehingga angka kealngsingan ideal didapat sbb:

= 79 2 57.5

2

= 97.7111

Ternyata 1.2 λl = 117.3 > 1.2 57.5

117.3 > 69

vy = 1.934 (PPBBI hal 25)

σy = Wy. P 1.934 7458

2 F 2 9.4

= 767.26 kg/cm2

< 1600 kg/cm2

OK

σkerja= P

2 .F 2 9.4

= 396.721 kg/cm2

< kg/cm2

OK

Jadi, Profil yang digunakan masih dalam kategori KUAT

7458.36

767.3

2 ( x ) +

𝑖𝑦 = 𝐼𝑦−𝑦

2 𝐹

=

=

= =

𝜆𝑖𝑦 = 𝜆𝑦2 +𝑚

2. 𝜆𝑙2 1/2

+ 2 2 1/2

( ) .

x

x =

= x

Kontrol Pemakaian Pelat Kopel

Dmax= 0.02 N

= 0.02 7458 = 149

Lmax = Dmax x li 149 = kg

h 4.54

(N.C) - (L.b) = 0 maka ; b= h.n

= 4.54 1 4.54

= 2588.33 4.54 = 839 kg

14

Kontrol pelat kopel

M = 1/2 L .b = 1/2 2588 4.54 = kgm

W = 1/6 s c2

= 1/6 0.6 14 = cm3

σmax = M 5876 = 300 kg/cm2

W 19.6

τ max = 3 L

2 S c

= 3 = 462 kg/cm2

2 0.6 14

= 300 + 3 x 462

= 854.842 kg/cm2

< 1600 kg/cm2

OK

Jadi besi pelat dengan tebal 6 mm memenuhi syarat untuk dijadikan

pelat kopel

Perhitungan Batang Tarik

Gaya max pada batang yang terjadi = kg

Lk = 310 cm

σ = 0.75 x σijin

0.75 x 1600 = 1200 kg/cm2

F perlu = P = = cm2

σ

Tegangan yang bekerja :

σ = P = 6897 = 367 kg/cm2

< kg/cm2

F netto 18.8 OK

Kontrol kelangsingan

λ = Lk = 310 = 226 < 240

i min 1.37

Jadi Profil diatas memenuhi syarat

2588.33

1200

5.75

1600

19.6

2588.33

6896.65

6896.65

78.78

5875.5138

x =

x

= x

𝑁 = 𝐿.𝑏

𝐶

x

x x

x x

=

x

x x

𝜎𝑖 = 𝜎𝑚𝑎𝑥2 + 3. 𝑇𝑚𝑎𝑥2

2 2

Perhitungan kuat baut

fu = 4100

diameter baut = 1/2 inch = 1.27 mm


f Rnv = 0.8 0.5 fu Ab

= 0.8 0.5 4100 1.27

= kg


f Rnt = 0.8 0.75 fu Ab

= 0.8 0.75 4100 1.27

= kg

Dalam Perhitungan Jumlah baut digunakan kuat baut yang terkecil, antara

kuat geser dan kuat tarik

Simpul A

Perhitungan Jumlah Baut:

S1 = P = = 2.15 = 3 baut

f

S2 = P = = 1.21 = 2 baut

f

S3 = P = = 2.15 = 3 baut

f

Simpul B

S1 = P = = 3.74 = 4 baut

f

S2 = P = = 3.46 = 4 baut

f

1946.7

2920.0

1946.7

4178.66

2348.45

1946.7

4178.66

1946.7

7274.5

1946.7

6741.48

1946.7

Perhitungan Sambungan Rangka

S1

S2

S3

S1

S2

x x x

x x x

x x x

x x x

Simpul C

S1 = P = = 2.64 = 3 baut

f

S2 = P = = 0.44 = 2 baut

f

S3 = P = = 0.73 = 2 baut

f

S4 = P = = 2.15 = 3 baut

f

Simpul D

S1 = P = = 0.59 = 2 baut

f

S2 = P = = 2.88 = 3 baut

f

S3 = P = = 2.43 = 3 baut

f

S4 = P = = 0.44 = 2 baut

f

Simpul D

S1 = P = = 0.73 = 2 baut

f

S2 = P = = 2.43 = 3 baut

f

S3 = P = = 2.43 = 3 baut

f

S4 = P = = 0.73 = 2 baut

5143.97

1946.7

1143.68

1946.7

5597.21

1946.7

862.7

1946.7

1423.64

1946.7

4178.66

1423.64

1946.7

4733.9

1946.7

4733.9

1946.7

4733.9

1946.7

862.7

1946.7

1423.64

1946.7

S1

S3 S2

S4

S1

S2 S3

S4

S1

S2 S3

S4

S5

f

S5 = P = = 1.21 = 2 baut

f

1946.7

2348.45

1946.7

Direncanakan menggunakan profil baja Light Lip Chanel , dan memiliki

data - data profil sebagai berikut :

Profil CNP 125 50 20 4.5

A = 10.59 cm2 lx = 238 cm4

W = 8.32 kg/m ly = 33.5 cm4

A = 125 mm ix = 4.74 cm

B = 50 mm iy = 1.78 cm

C = 20 mm Zx = 38 cm3

t = 4.5 mm Zy = 10.1 cm3

1. Pembebanan

A. Beban Mati

- Beban Profil = 8.32 kg/m

- Beban Atap = 15.5 kg/m (Brosur)

Berat Total = 23.82 kg/m

- Berat Pengikat 10% = 2.382 kg/m +

qd = 26.2 kg/m

B. Beban Angin

Tekanan Angin 25 kg/m2

- Kondisi



- Kondisi 3

>Tekanan angin tekan = Koef. x tekanan angin x jrk Profil

= 0.9 25 1.3

= 29.25 Kg/m

>Tekanan angin hisap =Koef. x tekanan angin x jrk Profil

= 0.4 25 1.3

= 13 Kg/m

Dari ketiga kondisi angin diatas, maka diambil tekanan angin yang terbesar

untuk mewakili = 29.25 Kg/m.

qwx = 29.25 Kg/m

qwy = 0 Kg/m

2. Perhitungan beban kombinasi

Kombinasi M + A ( 0.9D+1.3W )


Cladding wall

x x

x x

x

x x x

0.9 qx = 0.9 26.20 = 23.58 kg/m = 0.24 kg/cm

0.9 qy = 0.9 0.00 = 0.00 kg/m = 0.00 kg/cm

- akibat beban merata angin

1.3 qx = 1.3 29.25 = 38.03 kg/m = 0.38 kg/cm

1.3 qy = 0 kg/m =

3. Kontrol Lendutan

2. Kombinasi M + A

fx = 5 (qux atap+ quxangin) x Lx

384 E . Ix

5 0.236 0.38 600

384 238

2.08 cm

fy = 5 (quy atap+ quy angin) x (L/3)

384 E . Iy

5 0 0 200

384 33.5

0.00 cm

f ' = fx + fy f ijin = L/250

= 2.08 0.00 = 600 250

= 2.08 cm < = 2.4 cm


dinyatakan KUAT

4. Perhitungan Momen Kombinasi

2 . Kombinasi M + A


8

= 0.125 23.58 6 = 106.1 kg.m

angin = 1 qux angin L

8

= 0.125 38.03 6 = 171.1 kg.m +

Mx1 = 277.2 kgm

My = Atap 1 quy atap L/3

8

= 0.125 0 2 = 0 kg.m +

My1 = 0.000 kgm

2100000

2100000

x

x

x

= x

= x

=

( ) +

x

= x

= x

=

( ) +

x

+ /

4

4

4

4

2

2

2

2

2

2

x x

x x

x x

x x

x x

x x

Kontrol tegangan

Mx My < σ

Wx Wy

< 1600



2 dinyatakan KUAT

5. KONTROL KEKUATAN PROFIL

A. Penampang Profil

Sayap

bw < 170 bw = lebar Profil

2tf fy tf = tebal Profil

50 < 170 h = tinggi Profil

9 15.49 tw = tebal Profil

5.56 < 10.97

OK

Badan

h < 1680

tw fy

125 < 1680

4.5 15.49

27.78 < 108.4

OK

B. Lateral Buckling

Jarak pengaku lateral ( Sagrod )

Lb = L/3 = 150 cm= 1500 mm

Lp = 1.76 x iy x E

fy

= 1.76 1.78

= 92.67 cm= 926.7 mm

Lb > Lp Bukan Bentang Pendek

fL = fy - fr

240 70 = 170 Mpa

27723.06 0.00

38 10.1

729.55

2100000

2400

+

+

- =

= 3038 + 3524 + 941.6

= 7503 mm4

3.14 7503 1059

38000

= Mpa

Iw = Iy .h2 = 125 4.5

4 4

= mm

= 4 1E+09 = 0.582016 mm2/N2

80000 7503 33.5

= 10.1 1 + 1+ 0.582 170

= 10005 mm= 1001 cm > Lb= 150

Bentang Menengah

Mpx = fy * Zx = 240 38000 = Nmm

Mpy = fy * Zy = 240 10100 = Nmm

Mrx = Sx * ( fy-fr )= 170 = Nmm

Mry = Sy * ( fy-fr )= 170 = Nmm

Mux = kgm

Ma= kgm

Mb= kgm

Mc = kgm

= 1.316

170

200000 80000

2

14730.95

335000

1216070938

38000

14730.95

9120000

2424000

38000 6460000

10100 1717000

277.23

69.31

277.23

207.92

𝐿𝑟 = 𝑍𝑦𝑋1

𝑓𝐿1 + 1 + 𝑋2𝑓𝐿2

𝑋1 =𝜋

𝑆

𝐸𝐺𝐽𝐴

2

𝑋2 = 4𝑆

𝐺𝐽

2𝐼𝑤

𝐼𝑦

𝐽 = 2.1

3. 𝑏. 𝑡3 +

1

3. ℎ𝑡 − 2𝑡 . 𝑡3 +

2

3. 𝑎 − 𝑡 . 𝑡3

2

= x x x

2 x

𝑀𝑛 = 𝐶𝑏 𝑀𝑟 + (𝑀𝑝 − 𝑀𝑟)(𝐿𝑟 − 𝐿)


𝐶𝑏 =12.5 𝑥 𝑀𝑢𝑥

(2.5 𝑥 𝑀𝑢𝑥 + 3 𝑥 𝑀𝑎 + 4 𝑥 𝑀𝑏 + 3 𝑥 𝑀𝑐)

(𝐿𝑟 𝐿)

2 x - ) (

= Nmm

Mpx = Zx . fy = 38 2400

= kg.cm= Nmm

= Nmm

Mny = Zy . fy = 1/4 tf. bf 2 . fy

= 0.25 0.45 5 2400

= kg.cm= Nmm

6. Kontrol Interaksi momen

Mux Muy

φ Mnx φ Mny

0.9 0.9

+ = 0.26 < 1

11778981.4

9120000

3130729.266

6750 675000

2772306.00 0.00

91200

11778981.4 3130729.266

0.26 0.00

OK

+

+

2

𝑀𝑛𝑥 = 𝐶𝑏 𝑀𝑟𝑥 + (𝑀𝑝𝑥 − 𝑀𝑟𝑥)(𝐿𝑟 − 𝐿)


𝑀𝑛𝑦 = 𝐶𝑏 𝑀𝑟𝑦 + (𝑀𝑝𝑦 − 𝑀𝑟𝑦)(𝐿𝑟 − 𝐿)


Mu = Kgm

Nu = Kg

Vu = Kg

Ma = kgm= kgcm

Mb = kgm= kgcm

Ms = kgm= kgcm

Dari hasil perhitungan SAP 2000, diperoleh Profil WF dengan data Sbb :

WF 300 x 150 x 6.5 x 9

A = 46.78 cm2 tf = 9 mm Zx = 481 cm3

W = 36.7 kg/m Ix = 7210 cm4 Zy = 67.7 cm3

a = 300 mm Iy = 508 cm4

bf = 150 mm tw = 6.5 mm

iy = 3.29 cm ix = 12.4 cm

Kontrol Lendutan

f ijin = L = 700 = 1.94 cm

360 360

f = 5 700 0.1

48 2.1 E6 7210

f = 0.238 cm < 1.94 cm OK

Kontrol Tekuk

untuk arah x :

kcx = 1.2 (jepit-rol tanpa putaran sudut)

L = 700 cm

Lkx = 233 cm

= 233 = 18.79

12.4

= 3.14 46.78

18.79

= kg

Cek Profil Kolom

759.77

1094.33

3394.05

759.77 75977

638.37 63837

716.89 71689

71689 63837

2.1 E6

2743281.26

75977

𝑓 =5

48.

𝐿2

𝐸 . 𝐼𝑥. 𝑀𝑠 − 0.1 𝑀𝑎 − 𝑀𝑏

2

2

x

EANcrbx

ix

Lkxx

x x 2

2

( )) ( - - x 2

untuk arah y :

kcy = 0.8 (jepit-Sendi)

L = 300 9 = 159 cm

2

Lky = 300 159 = 141 mm = 14.1 cm

= 14.1 = 4.286

3.29

= 3.14 46.78

4.286

= kg

Maka dipakai lx karena lx > ly

= 18.79 2400

3.142 2.1 E6

= 0.202

λc < 0.25 ,maka ω = 1

Pn = Ag fy = 46.78 2400

w

= kg

0.85

= 0.036 < 0.2

Batang Dianggap Tidak Bergoyang Maka : (sumbu X )

Cmx = 1

2.1 E6

52734202.06

1

112272

3394.05

112272

E

fyxc

22yEANcrby x x

iy

Lkyy

2

2

y

EANcrby

x x 2

2 2

cPn

Pu

x

x

1

)(1

Ncrbx

Vu

CmxSbx

Sbx =

1 -

Sbx = 1.001 > 1

Sbx = 1.001

Mux = Mutx . Sbx

Mux = 1.001 = 760.7 kgm

Batang Dianggap Tidak Bergoyang Maka : (sumbu Y )

Cmy = 1

Sby =

1 -

Sby = > 1

Sby = 1

Muy = Muty . Sby

Muy = 0 = 0 kgm

Kontrol kekuatan Profil

Penampang Profil

Sayap





8.33 < 10.97

OK

Badan

h < 1680

tw fy

300 < 1680

6.5 15.49

46.15 < 108.4

OK

2743281.26

1

3394.05

759.77

1

3394.05

52734202.06

1.0001

1.0001

1

)(1

Ncrby

Vu

CmySby

0.5

0.5

x

x

Lateral Buckling

Jarak pengaku lateral ( Rib )

Lb = 700 233.3 cm

3

Lp = 1.76 x iy x E

fy

= 1.76 3.29

= 171.3 cm

Bila Lb < Lp ,maka Mnx = Mpx

Mnx = Mpx = Zx . fy = 481 2400

= kg.cm

Mny = Zy . fy = 1/4 tf. bf 2 . fy

= 0.25 0.9 15 2400

= kg.cm

Interaksi Momen

1.7 0.9 0.9

+ +

0.01 < 1

Kontrol Kuat Rencana Geser

300 < 1100

6.5 15.49

46.15 < 71

Plastis

Vn = 0.6 fy Aw

= 0.6 2400 30 0.9 0.65

= 0.6 2400 18.33

= Kg

Vu < ФVn

3394 < 0.9

3394 < 23756

OK

0

2.1 E6

2400

1154400

121500

3394.05 760.7111692

26395.2

26395.2

112272 1154400 121500

0.009 0.001 0

x x

x

x x x 2

12

bMny

Muy

bMnx

Mux

cPn

Pu

x x x 2. + +

fytw

h 1100

x x -(2. )) (

x x

=

Panjang Plat Dasar L 40 cm

Lebar Plat Dasar B 40 cm

fc' beton 30 Mpa

Mu ( Momen ) 77215 kgcm

Du ( Geser ) 162.66 kg

Pu (Aksial) 3464.84 kg

Perhitungan Tegangan Yang Bekerja Akibat Adanya Eksentrisitas

e = M = = 22.29 cm

P

A = 40 40 = 1600 cm2

W = 1/6 B L2

1/6 40 402 = cm3

σ = P + M

A W

= +

σ makσ = 9.404 kg/cm2

σ min = 5.073 kg/cm2

Jadi, q = 9.404 kg/cm2

Perhitungan Momen Yang Bekerja

M = 1/2 x q x L2

= 1/2 9.404 10 470.22 kgcm

Perhitungan Tebal Pelat Baja

σ = 4 M t = 4 M

t2

σ plat

= 4 x 470.2

= 1.08423 ~ 1.2 cm

1600

Base Plate

77215

3464.84

3464.84 77215

1600 10666.67

10666.67

x

= x x

x x 2

=

Perhitungan Tegangan Yang Bekerja Pada Angker

10

σ min = σ max

x B - x

x

5.07 20

5.07 9.404

7.008 cm

y = B - x

= 20 - 7.008 12.99 cm

S min = 1.5 d = 1.5 ( 2 x tf )

= 1.5 2 0.9

= 2.7 cm

1/3 x = 2.34 cm < S min

1/3 y = 4.33 cm > S min

r = 20 - 1/3 y 20 4.33 = 15.67 cm

C = 40 4.33 2.336 = 33.33 cm

Angker

σ min + σ max

σmin B

20

P

M

C 1/3x

1/3

s max

s

min

x y

+

x =

=

x

=

- -

-

-

=

=

T = M - P r 15.7

C

= kg = Pu

Perencanaan Diameter Angker

fy = 4100

fu = 6210

- Leleh: Pu =

Ag perlu = = 0.19 cm2

0.9 4100

- Putus: Pu =

Ag perlu = = 0.196 cm2

0.75 0.75 6210

- A baut perlu = = 0.239 cm2

0.7 4100

Direncanakan untuk tiap sisi A baut perlu = 0.239 2 sisi 0.12 cm2

diameter angker baut 0.5 inch = 1.27 cm

dengan luasan baut, ( A = 1.26613 cm2 )

Jumlah baut = 0.120 / 1.266

= 0.094 ~ 2 buah

Jadi, Dipasang 2 1.27 ; (A= 2.532 cm2) untuk tiap sisi

Abaut = 2.532 > Aperlu = 0.12 OK

Perencanaan Panjang Angker

Diameter = 12.7 mm, dengan luasan = 253 mm2

Mutu Baja (fy) = 4100 kg/cm2 410 Mpa

fc' = 30 Mpa

I = 0.02 . A . Fy

= 0.02 253.2 410 = 379.1 mm

30

dan tidak kurang dari = 0.06x db x fy

= 0.06 12.7 410 = 312.4 mm

Jadi, Dipasang angkur dengan panjang 400 mm

fc'

686.3083684

77215 3464.84

33.4

686.3084

f fy Ag

686.3084

f 0.75 fu Ag

686.30837

= x -

/ =

x

x x

x

D

x x

x x

Data Teknis :

- Dimensi :: 400 400 Pu : kg

- fc' : 30 Mpa : N

- fy : 360 Mpa Mu : kgm

- lo : 75 cm Du : Kg

Diasumsikan :

- Diameter Tul. Utama = 16 mm

- Diameter Tul. Sengkang 10 mm

Perhitungan Tul. Longitudinal Kolom

Menentukan tinggi efektif

d = h - s -ds - du/2

= 400 40 10 16 mm

d' = s + Ds + du/2

= 40 10 16 = mm

Eksentrisitas

e = Mu 0.223 m = 222.9 mm

P

Perhitungan Pn perlu

0.1 Ag . fc = 0.1 400 400 30

= N = 480 kN

Digunakan faktor reduksi ɸ = 0.65

a = =

0.85 f'c b 0.85 30 400

= mm

223 400 47.06

= 2 2 = 224.1 mm2

360 342 66

As = As'

Kontrol luas tulangan

Ast = As + As' = 224.1 224.1 = 448.1 mm2

Ast min = 1% Ag = 0.01 400 400 = 1600 mm2

maka, digunakan tulangan:

Kolom Pedestal

3464.84

772.15

47.06

342

66

772.15

3464.84

480000

Pn perlu 480000

480000

34648.4

162.66

As =Pn perlu. e −

h2 + a

2

fy. d − d′

= =

x x x

x x x x

- +

( ) - x

+

( ) x x

- - - /2

+ + /2

=

x

Ast = 10 16 mm2

dengan Rasio = 1.256 % . Ag OK

( rasio tul. Longitudinal harus lebih dari 1% sampai 8%. Pasal 12.9.1)

Kontrol tulangan Leleh atau tidak

x 41.7 mm

0.85. 0.85 30

∈w= 0.003

= 342 41.7 =

= 0.022 =

Karena nilai ∈s > ∈y , maka

Tulangan termasuk kedalam kondisi leleh dan, fs = fy

Perhitungan Gaya dalam

T = As .fy = 360

Cc= 0.85 fc a b

0.85 30 47.06 400 = N

Cs = AS .fy

= 2010 360 = N

Pnb

7E+05 N

KN

= 7E+05 342 400 5E+05 400 47.06 400 342

2 2 2 2

= Nmm = KNmm

eb= Pnb = 224 mm

Mnb

e < eb

222.9 < 224 Kolom Runtuh Tekan

Pola Keturunan Tekan

An = Ag - Ast =

Pn0 = An . 0.85 . Fc' + Ast. Fy

0.85 30 360

N = 4752 Kn

Pn

480000

723456

480000 723456 480000

Cs +Cc - T

723456

160000 2009.6 157990.4

107294.12

480

107294117.6 107294.12

723456

xAs fy 1004.8 360

0.85.β1 fc' b 400

2009.6

0.003

41.7 200000

0.0018

360

2009.6

480

157990.4 2009.6

4752211.2

Pn0

D =

∈𝑠=𝑑 − 𝑥

𝑥. ∈𝑤 ∈𝑦=

𝑓𝑦

𝐸𝑠

= x x

= =

- x

x

x x x =

x

=

=

= + - =

=

+ + - - - .

=

=

1+ Pn0 - 1 e

Pnb eb

=

1 + - 1 223 N = 481.3 Kn

224

Kontrol Keamanan

ɸ. Pn = 0.65

ɸ. Pn = N > Pu = N

Kolom Aman

Cek Lebar Perlu

b = 400 mm ɸTu = 16 mm

Sb = 40 mm n = 3

ɸS = 10 mm

(b - (2. Sb) - (2. ɸS) - (n.ɸTu)) / (n - 1) > 40

126 > 40 OK

Tulangan Geser

Kontrol dimensi kolom

Vu max <

Vu max < 0.75 . 5/6 30 400

< N OK

Menentukan Vc (SNI pasal 13.3.1.2)

= 1 + 1/6 . 30 400 342

14 40 40

= N

Ø.Vc= 0.75 . = N

1/2 .Ø.Vc= = N

Cek Vc

Vc tidak perlu diambil lebih dari

= 0.3 30 400 342

= N > N OK

Penulangan Geser Sejauh lo

Vumax = N < N

(Maka, Digunakan tulangan Geser minimum)

- Smax = 342 171 mm

2 2

Ø.Vc= 100461.82

224785.3376 133949.0921

1626.60

133949.0921 100461.8191

0.5. 0.75 . 133949.09 50230.90953

342

1626.6 468302.7867

1626.6

133949.0921

481309.3496

312851.0773 34648.4

4752211.2

480000

481309.3496

4752211.2

=

=

∅ .5

6. 𝑓𝑐′. 𝑏. 𝑑

𝑉𝑐 = 1 +𝑁𝑢

14𝐴𝑔.1

6. 𝑓𝑐′. 𝑏. 𝑑

x x

x x x

d = =

x

0.3 . 𝑓𝑐′. 𝑏. 𝑑 x x x

- Smax = 600 mm

Dari perhitungan jarak tulangan, dipilih jarak yang terkecil =

Smax = 171 mm dan dambil S = 150 mm

75 30 400 150

1200 360

57.1 mm2

dipakai tulangan diameter 10

Av = 2 . As= 2 x 78.5 = 157 mm2 > 57 mm2

Jadi untuk Sengkang digunakan tulangan 10 - 150

mm

=

𝐴𝑣 =75. 𝑓𝑐′𝑏. 𝑠

1200. 𝑓𝑦 =

x x x

x

Tipe Pondasi : Tiang Pancang

Kedalaman Pond.: 1400 cm

Dimensi Pile : 25 cm 25 cm

Konfigurasi Pile

- Jarak antar tiang

2.5D < S < 3D

2.5 x 25 < S < 3D x 25

62.5 cm < S < 75 cm

diambil S = 62.5 cm

- Jarak Pancang ke Tepi

1.5D < S' < 2 D

1.5 25 < S' < 2 25

37.5 < S' < 50

diambil S' = 37.5 cm

Perhitungan P ijin Tiang

- Menurut Bahan

P ijin = 50 Ton = Kg ( Dari Brosur )

- Menurut Tanah

Data Tanah

Kedalam Tiang : 1400 cm = 14 m

Rata konus (Cn) 37

JHP 1406

= 25 25 37 4 25 1406

= 35828 kg = 35.83 Ton

Ket :

Cn = Tekanan konus rata2 sepanjang 8D dan 4D ( kg/cm2 )

JHP = JHP pada kedalam tanah tiang ( kg/cm )

A = Luas (cm2)

K = Keliling ( cm2 )

Dari dua Metode diatas Maka diambil P ijin tiang yang terkecil 35.83 Ton

Pondasi

50000

3 5

𝑃 𝑡𝑖𝑎𝑛𝑔 =𝐴 . 𝐶𝑛

3+

𝐾. 𝐽𝐻𝑃

5

x x

x x x x ) ( ) ( +

Pembebanan Pondasi

1.05

1.05

0.4

0.75

0.5

1.05

- Berat tanah :

1.05 1.05 0.75 0.75 0.4 0.4 2100 = 1484.4 Kg

- Berat Pilecap :

1.05 1.05 0.5 2400 = 1323 Kg

- Berat Sloof :

0.2 0.3 6 2400 = 864 Kg

- Beban Aksial Kolom

Aksial Dari SAP 2000 Kg +

Berat Total : Kg

Jumlah Pile yang dibutuhkan :

n = Berat Total = 0.199 buah = 1 buah

P ijin Pile

Efisiensi tiang

= 1 - arc tg 25 0 1 0 1

62.5 90 1 1

= 1 - (arc tg( 0 )

= 1 - 0 = 1

Sehingga Pijin (satu tiang)

- efisiensi x P tiang = 1 35828 = 35828 kg

7136

7136

35828.33333

3464.84

0.4

′𝜂 = 1 − 𝑎𝑟𝑐𝑇𝑔𝐷

𝑆

𝑚 − 1 . 𝑛 + 𝑛 − 1 . 𝑚

90. 𝑚. 𝑛

) ( ) ( x x x x x -

x x

x x

x

x

) ( ) ( x + x

x x

x

=

P maks = kg

Mx = 0.13 = kgm

= 3465 + 0.125

1 1 0.02

= 6930 kg

Cek

P ijin tiang (stlh efisiensi) > Pmaks

> OK

433.11

35828.33 6929.68

3464.84

3464.84 433.11

400

400

400

1050

Data Perencanaan :

𝜏 ijin = 2.2 kg/cm2

𝛾 tanah = 2.1 t/m3

M = kgm= kgcm

Berat vol. beton = 2400 kg/m3 = 2.4 t/m3

f'c = 30 Mpa

fy = 360 Mpa

Pembebanan Pilecap :

Berat tanah diatas Pilecap (A3)

- Volume tanah x 𝛾 tanah =

0.4 0.65 1.05 2.1 = 0.57 ton

- Berat Pilecap (A2)

Volume x Berat volume btn =

1.05 0.4 1.05 2.4 = 1.06 ton

- Berat Sloof (A1)

Volume x Berat volume btn =

0.2 0.3 6 2.4 = 0.86 ton +

P tot = 2.5 ton= Kg

Cek Tegangan Ijin Tanah

105 105 1/6 105 105

= 0.226 + 0.4

= 0.627 kg/cm2

< 2.200 kg/cm2

OK

105 105 1/6 105 105

= 0.226 - 0.4

= -0.174 kg/cm < 2.2 kg/cm2

OK

Pilecap

1050

772.15 77215

2495.7

2495.7 77215

2495.7 77215

𝜏𝑚𝑎𝑥 =𝑃

𝐴+

𝑀

𝑊 x x x 2 + =

𝜏𝑚𝑖𝑛 =𝑃

𝐴−

𝑀

𝑊 x x x 2 - =

x x x

x x x

x x x

Perhitungan Momen yang bekerja:

𝜏 = qu = 0.627 kg/cm2

Mu = 1/2 qu .t

= 1/2 0.627 32.5 = 330.9 kgcm Nmm

Perhitungan tinggi efektif (d)

asumsi tulangan = 19

dx = 400 40 1/2 19 = 351 mm

dy = 400 40 19 1/2 19 = 332 mm

Perhitungan batas kebutuhan tulangan

ρb = 0.85 . f'c . Β1

fy

= 0.85 30 0.85

360 600 360

ρmax = 0.75 . ρb

0.75 . 0.038

ρmin = 1.4 / fy

1.4 / 360 =

m = =

0.85 . 30

=

Perhitungan Arah X

Rn =

=

0.8 . 1000 350.5

=

ρperlu =

= 1 1 - 1 - 2 14.12

14.12

=

360

0.000001

0.0003

0.038

0.0282

0.00389

fy 360

0.85 . f'c

14.12

Mu

φ . b .dx2

33091.0

0.0003

600

33091

600

600 + fy

2

fy

Rnm

m

211

1

2

x x

+

x

x . .

=

. .

- - x

x - - -

t

Karena ρperlu < ρmin maka yang akan dipakai adalah

As = ρ . b . dx

= 1000 350.5

= 1363 mm2

digunakan tulangan diameter 19 - 200

As pasang = mm2

> As perlu = mm2

Perhitungan Arah Y

Rn =

=

0.8 . 1000 331.5

=

ρperlu =

= 1 1 - 1 - 2 14.12

14.12

=

Karena ρperlu < ρmin maka yang akan dipakai adalah

As = ρ . b . dy

= 1000 331.5

= 1289 mm2

digunakan tulangan diameter 19 - 200

As pasang = mm2

> As perlu = mm2

1289.2

360

0.000001

ρmin= 0.00389

0.00389

1416.9

1363.1

Mu

φ . b .dy2

33091.0

0.0004

0.0004

ρmin= 0.00389

0.00389

1416.9

x x

fy

Rnm

m

211

1

2

x x

x x

Merencana Baja Final

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