PRINT REVISIffffffffffffffffffffffffffffffffffffffffffffffff

A. Data Perencanaan

a. Mutu Baja (BJ) = 41 ; Fu = Mpa Fy = Mpa

b. Lebar bangunan (L) = m = cm

c. Tinggi bangunan (H) = + + = m

d. Panjang bangunan (B) = x m

e. Over stek = m

f. Kemiingan atap = 47^o ; cos α = sin α =

g. Bahan penutup atap = seng = kg/m² tan α =

h. Jarak antar kuda-kuda = m = cm h atap = m

i. Beban mati = Kg/m2

j. Beban hidup = Kg/m2

B. Menghitung jarak antar purlin

c a = mm

c = mm

1/2 c = mm

b 1/2 c +a = mm

a

47^o 6 = 6 purlin daerah a = 9

3.64865 = 3 purlin daerah c = 6

mm mm

# Menentukan panjang atap +

# Perencanaan purlin

Maka, jarak antar purlin : (sama dengan jarak maksimu sagrod) b

n

# Total purlin n = 2 (n+1)

= 2 ( 14 + 1 )

= 30 bh

# Cek jarak antar purlin

r = ( x 14 ) - 60

= cm r < b

< . . . Ok !!!

C. Menghitung beban yang bekerja pada purlin Desain Gording

Data perencanaan

a. Mutu Baja (BJ) = 41 ; Fu = Mpa Fy = Mpa

b. Lebar bangunan (L) = m = cm

c. Kemiringan atap = 47^o

d. Atap asbes = kg/m²

e. Jarak antar kuda-kuda = m = cm

f. Jarak antar purlin = m

g. Tekanan angin = kg/m²

0.731

1.2 120

25

250

12 1200

1.8 180

0.775 2.56

cm

1025.047 1085.047

410

=

= 0.775 m

=

b 1400 6000

cos α

410 250

12 1200

7.00

5 7

10850.466 mm =

2.56

10.85

14 =

1.4

0.682

10.850 m

1400 6000

=

77.503

77.503 1025.0466

300 175

1.072 6.4

4.75 2.25

2053 8798 4399 6452

Menggunakan profil Light Lip Chanel 75 20

A = cm² Zx = cm²

q = kg/m Zy = cm²

Ix = cm⁴ ix = cm ≤ 1 . . . Ok !!!

Iy = cm⁴ iy = cm . . . Ok !!!

# Menghitung beban yang bekerja pada gording a. beban mati

qbm sin α (y)

qbm cos α (x)

Gording = kg/m

Asbes = x 2.6 = kg/m

kg/m

Toeslagh = 10 % x = kg/m

= kg/m

= x cos α = x = kg/m

= x sin α = x = kg/m

= ¹/8 x x L² = ¹/8 x x 1.8² = kg/m

= ¹/8 x x ¹/2L² = ¹/8 x x 1² = kg/m

b. beban hidup

beban hidup akibat pekerja (P) = kg

Px

= x cos α = x = kg

= x sin α = x = kg

= ¹/4 x x L = ¹/4 x x 1.8 = kg.m

= ¹/4 x x ¹/2L = ¹/4 x x 0.9 = kg.m

c. beban angin

- Beban angin yang bekerja = kg/m²

- Kemiringan atap = 47^o ≤ 65^o

Koefisien angin tekan = ( 0,02 x α - 0,4 )

= harus positif

Koefisien angin hisap =

W Tekan (qwx) = x x 25 = kg/m

W Hisap (qwy) = x x 25 = kg/m

Angin Tekan ( Mx ) = ¹/8 x x 2² = kg.m

Angin Hisap ( My ) = ¹/8 x x 2³ = kg.m

73.135 16.455

1.684 0.452 Mbm y

qbm x qbm y

Px 68.200 30.690

0.775

4.159 4.460

100

qbm

Py

P 100 0.682

4.159 4.460 Mbm x

Px

M Px

M Py

68.200

P 100 0.731 73.135

125 4

0.775

3.560 1.984

0.391

5.544 0.554 5.544

Py

10.463 -7.750

10.463 4.237

25

0.54 -0.4

0.54 0.775

-7.750 -3.139

-0.40 Py

68.0

14.30 5.40 7.84 2.71 9.34

3.56 573

qbm

6.098

0.731 6.098

0.682 6.098 qbm

qbm x qbm y

qbm

qbm

d. Menghitung beban kombinasi

1. U = 1.4 D 2. U = 1.2 D + 0.5 L 3. U = 1.2 D + 1.6 L + 0.8 W 4. U = 1.2 D +1.3 W + 0.5 L

Jadi : Mux = = N.mm

Muy = = N.mm

Asumsi penampang kompak

Mnx → Zx x fy = x = N.mm

Mny → Zy x fy = x = N.mm

Untuk Mengatasi Masalah Puntir maka Mny dapat dibagi 2 sehingga

0.9 x 0.9 x

+ ≤ 1

≤ 1 . . . Ok !!!

e. menghitung lendutan (defleksi)

- beban yang menentukan adalah beban mati dan beban hidup qx = qbmx + qwx

= + = kg/m = kg/cm

qy = qbmy + qwy

= + = kg/m = kg/cm

Px = kg

Py = kg

x qx x l⁴ Px x l³

x E x Ix 48 x E x Ix

5 x x ⁴ x ³

x x 10⁶ x 48 x x 10⁶ x

= +

= cm

5 x x ⁴ x ³

x x 10⁶ x 48 x x 10⁶ x

= +

= cm δtotal = dx * cos α dy * sin α

= * + *

= cm

> = ^ + ^

cm > cm . . . Ok !!!

> . . . Ok !!!

Q Arah X Kombinasi Beban

2.358 ( kg/m )

5400 250

3575000 1350000

Mux

3575000 545152.306 +

1350000 268706.422

Q Arah Y

( kg/m ) 0.632 8.770 26.871 8.770 M Arah X M Arah y ( kg.m ) ( kg.m )

5.823 6.244

39.091 41.920

122.481 122.369

52.693 41.920

17.366 54.515 22.875

δx

384

5 +

4.159 10.463 14.622

384 2.1 573

0.1462 120

0.14622

4.460 7.750 12.210 0.1221

2.1 68.200

2.1 57758400000

117849316.6

≤ 1

0.169 0.2212

= 462067200000 151602058

+ +

68.200 73.135

=

573 120

0.391

0.731 68 126598348 + 126377919.6

54835200000 6854400000

0.00231 0.0184

=

0.02075

0.0024 2.000 0.02075 2.000

0.0209 240

0.00237

= 0.1221 120 + 73.135 120

384

0.017

0.017

0.00237 0.682 0.021 δy

2.1

L > δtotal

120 240 0.5

0.00033 0.002

=

68 Øb x.Mny

Øb x Mnx + Muy ≤ 1

54.515 26.871

545152.306 268706.422

14300 250

√(δ_𝑥 ^2+δ_𝑦^2)

√

= +

f. Design sagrod

beban yang bekerja = beban mati + beban hidup + beban angin/hujan pada sumbu X + beban purlin

a. beban mati

abm sin α (y)

abm cos α (x)

Gording = kg/m

Zinkalium = x 3 = kg/m

Sagrod = kg/m

kg/m

Toeslagh = 10 % x = kg/m

= kg/m

= x cos α = x = kg/m

= x sin α = x = kg/m

= ¹/8 x x L² = ¹/8 x x 2² = kg m

= ¹/8 x x ¹/2 L² = ¹/8 x x 1² = kg m

b. beban hidup

beban hidup akibat pekerja (P) = kg

Px

= x cos α = x = kg

= x sin α = x = kg

= ¹/4 x x L = ¹/4 x x 2 = kg.m

= ¹/4 x x ¹/2 L = ¹/4 x x 1 = kg.m

c. Beban sendiri gurt/purlin

P = kg/m

Px = P . Cos α

= x

= kg/m Maka beban yang menentukan adalah Px = kg

# Kombinasi pembebanan

P = + Px

= +

= kg

# Dimensi sagrod

P π x D²

G

= cm² x D²

D² = Diameter yang digunakan min

D = cm 4 mm

182.838 5.532

0.110

4 3.14

= 4

= A 0.1098

A = = 175.658

1600

0.139855109 0.37397207 abm

3.560

Py P 250 0.731 182.838

170.500

qbm x

M Px Px 170.500 76.725

M Py Py 41.139

Mbm y qbm y 0.560

250

Py

qbm

Px P 250 0.682 170.500

3.56 3.56

qbm y qbm 7.564 0.731 5.532

Mbm x qbm x 5.158 2.089

qbm 7.564

qbm x qbm 7.564 0.682 5.158

0.775

6.876

6.876 0.688

1.332 1.984

5.158 170.500

175.658 0.682 2.42791416

Desain menurut AISC (America Institute Steel Construction)

# Lebar efektif (gelagar dalam) Jarak antar gelagar = m

B a1 b a2

tp

5 x m

Bo = 4 /2 + 4 /2 = 3.5 m = 350 cm

L 4

= cm

a1 = a2 = ¹/2 x

= cm

B2 = a1 + b + a2

= + + = cm Nilai B yang di pakai adalah

yang terkecil

a1 = a2 = 8 x tp B = cm

= 8 x 12

= 96 cm B3 = a1 + b + a2

= 96 + + 96 = cm

# WF yang digunakan a. beban mati

Slab beton bertulang t x = kg/m²

Plafon = kg/m²

Keramik = kg/m²

Spasi = kg/m²

= kg/m²

qDL = x

= kg/m

MDL = 1/24 x qDL x l²

x ²

= kg.m = kg.cm

344 3.5

350

1204

=

344

175.00

0.12 2400 288

11 24 175

7

700

350 35 350 735.0

1.8

35 227

162.54000 16254.000

1204 24

≤

B1 = 700

4

3.5

d1 d2

21 (tebal plat)

7 m

m 5

b. Beban yang bekerja

wt = kg/m2

qLa = wt x bo

= x

= kg/m

MLa = 1/8 x qDL x l² = kondisi saat pelaksanaan

x ²

= kg.m = kg.cm

c. Beban hidup

wt = kg/m2

qL = wt x l

= x

= kg/m

MLL = 1/24 x qDL x l²

x ²

= kg.m = kg.cm

Rumus taksiran AISC

x G

x

= cm3

Dicoba profil 19

A = cm² Ix = cm⁴

Wx = cm³ qbs = kg/m

mutu beton K-300

Ekomposit = + 20 % Ebeton *

= + √

= kg/m² 23500

# Angka ekivalen (n) Es

Ec

√ fc'

√

= diambil 9

Transformed Area Method

# Penentu garis netral

gn. Beton dc d = 12 cm

gn. Komposit

hs dt ds

gn. Baja Bs = 35 cm

2 B dc2 + 2 Fs dc - Fs hs = 0

2 dc2 + 2 dc - ( 12 + ) = 0

dc2 + dc - = 0

8.936174700210000 25 =

h 175.0 9 2300

40000 137

2100000

= w profil

24

= 350 7

8 2143.750

173.9

Ebaja

350

25 2106400

= h

= 4700210000

214375.000

250

250 3.5

100

100 3.5

350

0.2 6400

= 0.6616254.000001600 15.392

350

= 875

= 875 7

1786.458 178645.833

0.66 MDL

173.9 17.5

347.800 5130.050

350 12

173.9 38.889

-b + √ b² - 4ac 2a

= + √ ² - 4

2

= cm

ds = ( 12 + ) - ( 2 dc )

= -( 2 x )

= cm

Statis momen terhadap bagian atas slab beton ( Ac x Yc ) + ( As x Ys ) = ( At + Yt )

x x 6 ( 12 + ) = 12 x

9

+ = yt

yt = cm

GN Komposit didaerah Beton Ic Fc dc2

h

1 4 ^{2 2}

12 9

= + + +

= cm⁴

Analisa kekuatan berdasarkan metode pelaksanaan 1. Saat pelaksanaan

a. beban mati

Slab beton bertulang t x = kg/m²

bekisting = kg/m2 +

kg/m2

qDL= x

= +

= kg/m

MDL = 1/8 x qDL x l²

x ²

= kg.m = kg.cm

MDL = kg.m

Momen akibat beban kerja (beban pada saat pelaksanaan)

1 ² kondisi saat pelaksanaan

8

= kg.m

+

= kg/cm < kg/cm . . . Ok !!!

2. Saat Pemakaian a. beban mati

Slab beton bertulang t x = kg/m²

Plafon = kg/m²

Keramik = kg/m²

Spasi = kg/m²

= kg/m²

qDL = x + berat profil wf

= +

= kg/m

MDL = 1/24 x qDL x l²

x ²

= kg.m = kg.cm

MDL = kg.m

Momen akibat beban kerja (beban pada saat pelaksanaan)

1 ²

24

= kg.m

29.50 7.8535

x 100

412.106 1600

2737.87500

1786.458

= dc

100 3.5

7 2143.750

5130.050 38.889

7.854

17.50

7

6 7.8535 9

= 7334.68750 2143.750 2300

x 3.5 x

Wx

MDL + MLL x 100 Gs

82379.350

7334.69

MWL =

+ 175.00

h + Fs ds²

Is + +

=

-347.8 347.8 38.889

33083.580

x x x

Itotal

yt

173.9 5130.050

1400.000 407.233

13.793 175.0 12.00

9

x

2737.87500 273787.500

1204 137

303 3.5

1197.5

= 1197.5 7

8 7334.68750

= 40000 + 175.00 6

40000 2100 7195.770306

=

MWL = x 250

0.12 2400 288

11 24 21

16.035 13.793

+ 173.9 17.5 175.00 +

+ 173.9

288 15 303

733468.750

1060.5 137

+ berat profil wf

0.12 2400

344

344 3.5

1341

= 1341 7

24

+ x x ( 13.793 + )

= kg/cm < kg/cm . . . Ok !!!

+ x x ( 13.7929 + )

9

= kg/cm < <

σ

c

= σc/3.5

= 250/ 3.5

< kg/cm . . . Ok !!!

3. Perencanna shear connector

B C D

QE

QA

qt= qDL+ qLL qt x

= + 2 2

= kg/m = kg

qt QC = 0

4

= x

4

= kg

4. Gaya geser longitudinal St = Fs x ds

= x

= cm³

St x QA St x QA

x x

= kg/cm = kg/cm

5. Gunakan stud connector

Diameter = mm

Tinggi = cm

H 45

d 16

Q = 10 x H x d x √ fc'

= 10 x 45 x x √

= N

= kg

Q Q

qa qb

= =

= cm = cm

6. Cek Lendutan (saat konstruksi)

1 x x ⁴ 1 x x ⁴

x 10⁶ x x 10⁶ x

= cm = cm

+ <

cm < . . . Ok !!!

3670.978 19.096

17.5

2216 7

3878

Itotal qb = Itotal

= 2398.590 2398.590

71.4285714

7756 171.863

Gs = MDL Itotal+ MDL x ys

= 2737.87500 1786.458

GC = Mh x IDL + MtotalDL x ys

= 2737.87500 1786.458 100 17.5

82379.350 1600 82379.350 100

A E

QA

1341 875 =

2216

= 2216

7756

7

700

384 2.1

0.02605 S2 =

= qa

=

225.827 112.913

3878 82379.350 112.913

4.5 16

= = 2.8125

25.0 225.827

2398.590 82379.350

= 3670.978 = 3670.978

16

=

40000

16.256 32.511

36000

< L dTotal

0.875 0.875

0.08914 2.1 dBH = 3.500

32 16

dBM

384

11.975 700

40000

= QB =

173.900 13.793

S1

0.08914 0.0261 0.115

800