Analisis Optimasi Profil Rangka Baja Dalam Perencanaan Bangunan Industri

(1)

TUGAS AKHIR BENTANG 20 m

= a = m H = m = m =

= L = m h = m = =

= =

z1 = m

20 m

DIMENSIONING GORDING

Mutu Profil 160 Mpa

Mpa

sin 20 0,342 20

cos 20 0,940 1,5

360 Mpa

7 Jarak Portal ( ) 6 Mutu Baut 360

5,250 Mutu Profil 240Mpa Mutu Las BAB III

Gaya angin 65kg/cm2

1,5 / cos = 1,596 10,0 / cos = 10,642

Page 1

z1 = m

z2 = m

m

+ ) / =

=

A. 1.

= × =

= = +

= ( =

= ( × × 2 / 8

= ( + q )

=

= ( × × 2 / 8

= ( + q )

8,7 9 buah

Asumsikan berat sendiri atap 4,00kg/m2

n = ( 1,596 10,642 1,400 Asumsikan jarak gording = 1,400

Maka banyak gording untuk bentang z1 + z2 ==> 1,5 / cos = 1,596

10,0 / cos = 10,642

5,600 kg/m

Berat sendiri gording q q kg/m

Dimensi Gording

Dimensi Gording Variasi Tanpa Ada Track Stang Muatan Tetap

Berat sendiri atap 1,4 4,00

MX1 q total × cos × L2 / 8 5,600 + q ) cos q total 5,600 + q ) kg/m

MY1 q total × sin × L2 / 8 5,600 + q ) sin

20 6,0

23,6803 4,2286 kg m

20 6,0

8,6189 1,5391 kg m

q c o s q s i n

q S b . Y

S b . X

(2)

TUGAS AKHIR BENTANG 20 m 2.

qa =

= × - =

= × × qa

= × ×

=

= × 2 / 8

= = 0

qa =

= × × qa

= × ×

=

= × 2 / 8

= = 0

3.

Muatan Angin 2a. Angin Datang

65 kg/cm2

0,4 0,000 qa datang Koefisien Jarak Gording

Koefisien = 0.02 × × 0.4 0,02 20

MX2a q_{a datang} × L2 / 8

0,000 6,0

0,000 1,4 65

0,000 kg/m

65 kg/cm2

Koefisien = -0,4

0,0000 kg m MY2a

2b. Angin Pergi

6,0 -163,8000 kg m -36,400 kg/m

MX2b q_{a datang} × L2 / 8 qa datang Koefisien Jarak Gording

-0,400 1,4 65

MY2b

Muatan Tak Terduga

-36,400

q d a t a n g

S b . Y S b . X

P = 1 0 0 k g

Sb.Y P q p e r g i

S b . Y S b . X

Page 2 =

= × × / 4

= =

= × × / 4

=

= ( + q ) + ( ) = ( + q )

= ( + q ) + ( 0 ) = ( + q )

= ( + q ) + ( ) = ( + q )

= ( + q ) + ( 0 ) = ( + q )

= ( + q ) + ( ) = ( + q )

= ( + q )

= = =

= ( + ) =

100 cos 20 6,0

MX3 P × cos × / 4

51,3030 kg m Kombinasi Momen

Akibat Muatan 1 + Muatan 2a (Angin Datang)

100 sin 20 6,0

140,954 kg m MY3 P × sin × / 4

kg m

MY Total 8,6189 1,5391 8,6189 1,5391 kg m

MX Total 23,6803

Akibat Muatan 1 + Muatan 2b (Angin Pergi)

MX Total 23,6803 4,2286 -163,800

23,6803

4,2286 0,0000 4,2286

Akibat Muatan 1 + Muatan 3 (Muatan Tak Terduga)

MX Total 23,6803 4,2286 140,954

-140,120

164,634

4,2286 kg m

MY Total 8,6189 1,5391 8,6189 1,5391 kg m

4,2286 kg m

MY Total 8,6189 1,5391 51,3030 59,9219 1,5391 kg m

Dicoba profil Channel Lips C 150×65×20×3.2

Data: q 7,51 kg/m Ix

kg m MY Total 59,9219 1,5391 kg m

Maka momen maksimum yang menentukan : MX Total 164,634 4,2286

cm

Wx 44,27 cm3 _Iy _{53,8 cm}332 cm4 _iy _{2,37139 cm}

4 _ix _5,89

kg/cm2

q total 5,600 7,510 13,110 kg/m 9,567 cm

2 _E _2100000

Wy 12,26 cm3 _F

Sb.Y

Sb.X

P cos P sin

= ( + ) =

q total 5,600 7,510 13,110 kg/m

(3)

( + × ) ( + × )

× × 4 × × 3

× =

× × 4 × × 3

× =

F = FX2 + FY2 = 2 + 2 =

F <

B.

Kontrol Tegangan

Terjadi = MX +

= 16463,4 422,86 7,5

MY

profil

WX WX

1600 kg/cm2

44,27 + 5992,19 12,26153,91 7,5

2100000

384 E × Ix 48 E × Ix

.. OK

Kontrol Lendutan

FX = 5 qTotal× cos × L 4

+ 1 P× cos × L3

1026,659 kg/cm2 _{1600 kg/cm}2

FY = 5 qTotal× cos × L 4

2100000 332

0,905 cm 48

= ₃₈₄5 0,131 cos _× 20₃₃₂ 600 + 1 100 cos 20 600

= ₃₈₄5 0,131 _2100000cos

+ 1 P× cos × L3

384 E × Iy 48 E × Iy

2,032 cm

100 cos 20 600

20 600 ₊ 1

0,905 2,032 2,224 cm

2100000 53,8

× 53,8 48

Direncanakan gading-gading kap memiliki 3 medan ekonomis

2,00

Fizin .. OK

Dimensi Gording Variasi Dengan Track Stang

600 ₌ _{3,333 cm}

180 Fizin = ₁₈₀L =

L y G a d in g K a p

Page 3 m

1.

= ( =

= ( × × 2 / 8

= ( + q )

=

= ( × × ( / 3 )2 _{/ 8}

= ( + q )

2. = = = = 3. =

= × × / 4

= 2,00

2,00 m

+ q ) kg/m MX1 q total × cos × L_X2 / 8 2,00 m

Muatan Tetap

q total 5,600

MY1 q total × sin × L_Y2 / 8

5,600 + q ) sin

6,0

23,6803 4,2286 kg m

5,600 + q ) cos 20

MX2a 0,0000

20 6,0

0,9577 0,1710 kg m

kg m

MY2b 0 kg m

2b. Angin Pergi

MX2b -163,8000

kg m

MY2a 0 kg m

6,0 140,954 kg m

Muatan Tak Terduga MX3 P × cos × _X / 4

100 cos 20

T ra c k S ta n g L y

L x

G a d in g K a p

S b .Y

S b .X

= =

= × × ( / 3

=

140,954 kg m MY3 P × sin × _Y / 4

) / 4 17,1010 kg m

100 sin 20 6,0

(4)

= ( + q ) + ( ) = ( + q )

= ( + q ) + ( 0 ) = ( + q )

= ( + q ) + ( ) = ( + q )

= ( + q ) + ( 0 ) = ( + q )

= ( + q ) + ( ) = ( + q )

= ( + q )

= = =

= ( + ) =

( + × ) ( + × )

MX Total 23,6803 4,2286 0,0000

Kombinasi Momen

MX Total 23,6803 4,2286 -163,800

23,6803 4,2286 kg m

MY Total 0,9577 0,1710 0,9577 0,1710 kg m

MX Total 23,6803 4,2286 140,954

-140,120

164,634

4,2286 kg m

MY Total 0,9577 0,1710 0,9577 0,1710 kg m

4,2286 kg m

MY Total 0,9577 0,1710 17,1010 18,0587 0,1710 kg m

kg m MY Total 18,0587 0,1710 kg m

cm

Wx 21,92 cm3 _Iy _{20,6 cm}137 cm4 _iy _{1,89 cm}

4 _ix _4,88

Kontrol Tegangan

Terjadi = MX +

kg/cm2

q total 5,600 4,510 10,110 kg/m 5,747 cm

2 _E _2100000

Wy 6,22 cm3 _F

= 16463,4 422,86 4,5

MY

profil

WX WX

1600 kg/cm2

21,92 + 1805,87 6,2217,10 4,5

.. OK

Kontrol Lendutan

FX = 5 qTotal× cos × LX 4

+ 1 P× cos × LX3

1140,803 kg/cm2 _{1600 kg/cm}2

Page 4

× × 4 × × 3

× =

× × ( / 3 )4 _× _{× (} _{/ 3 )}3

× =

F = FX2 + FY2 = 2 + 2 =

F < = = = = = = = = = =

n = = + 1 =

2100000

384 E × Ix 48 E × Ix

Kontrol Lendutan

+ 1 P× cos × LX3

FY = 5 qTotal× cos × LY 4

2100000 137

2,027 cm 48

= ₃₈₄5 0,101 cos _× 20₁₃₇ 600 + 1 100 cos 20 600

= ₃₈₄5 0,101 _2100000cos

+ 1 P× cos × LY3

384 E × Iy 48 E × Iy

0,148 cm

100 cos 20 600

20 600 ₊ 1

2,027 0,148 2,032 cm

2100000 20,6

× 20,6 48

C 125×50×20×2.3

h 125 mm

Fizin .. OK

Kesimpulan : Profil gording yang di pakai adalah Channel Lips

600 ₌ _{3,333 cm}

180 Fizin = ₁₈₀L =

t 2,3 mm

F 5,747 cm2

b 50 mm

d 20 mm

Wy 6,22 cm3

Ix 137 cm4

q 4,51 kg/m

Wx 21,92 cm3

9 10 buah 8,74

Iy 20,6 cm4

Jumlah profil Channel Lips C yang diperlukan : ( Jumlah Lapangan Gording ) + 1

(5)

2 m

a = 1,400 m

b = 1,038 m

Berat sendiri atap = 4,00 kg/m2

Jarak batang tarik = 2 m

Berat sendiri gording (q) = 4,51 kg/m

Beban tak terduga = 100 kg

Beban yang bekerja

A. = × × × = kg

= × × = kg

= × = kg +

= kg

Tipe 1 = Atap 2,00 2,840

BAB II

DIMENSI BATANG TARIK (TRACK STANG)

Gording 2,00 4,511,04 4,00sin 20sin 20 3,085

Tipe 2 = Atap 2,00 1,4 4,00 sin 20

Beban Tak Terduga 100 sin 20

3,831 34,202

N1 40,127

a a

b

Page 5

A. = × × × = kg

= × × = kg

= × = kg +

= kg

= + = kg

:

× / 4

d

d = mm

Tipe 2 = Atap 2,00 1,4 4,00 sin 20 3,831

Gording 2,00 4,51 sin 20 3,085

Dibutuhkan Terjadi profil

NTotal N1 8 N2

34,202

N2 41,118

Beban yang dipikul oleh track stang :

369,069

0,542 cm

Ambil : 6

d ₁₆₀₀ 369,069

d NTotal

1600 /4 NTotal ₁₆₀₀ _kg/cm2

d2_/4

NTotal ₁₆₀₀ _kg/cm2

A

(6)

= a = m H = m = m =

= L = m h = m = =

= =

z1 = m

BAB III DIMENSIONING GORDING

160

Mpa

sin 20 0,342 25

cos 20 0,940 1,5

360 Mpa

7 Jarak Portal ( ) 6 Mutu Baut 360

5,250 Mutu Profil 240Mpa Mutu Las Dimensioning Gording

Gaya angin 65kg/cm2

1,5 / cos = 1,596 12,5 / cos = 13,302

Mutu Profil Mpa

Page 1

z1 = m

z2 = m

m

+ ) / =

=

A. 1.

= × =

= = +

= ( =

= ( × × 2 / 8

= ( + q )

=

= ( × × 2 / 8

= ( + q )

10,6 11 buah

Asumsikan berat sendiri atap 4,00kg/m2

n = ( 1,596 13,302 1,400 Asumsikan jarak gording = 1,400

Maka banyak gording untuk bentang z1 + z2 ==> 1,5 / cos = 1,596

12,5 / cos = 13,302

5,600 kg/m

Berat sendiri gording q q kg/m

Dimensi Gording

Dimensi Gording Variasi Tanpa Ada Track Stang Muatan Tetap

Berat sendiri atap 1,4 4,00

MX1 q total × cos × L2 / 8 5,600 + q ) cos q total 5,600 + q ) kg/m

MY1 q total × sin × L2 / 8 5,600 + q ) sin

20 6,0

23,6803 4,2286 kg m

20 6,0

8,6189 1,5391 kg m

q c o s q s i n

q S b . Y

S b . X

(7)

TUGAS AKHIR BENTANG 25 m 2.

qa =

= × - =

= × × qa

= × ×

=

= × 2 / 8

= = 0

qa =

= × × qa

= × ×

=

= × 2 / 8

= = 0

3.

65 kg/cm2

0,4 0,000 qa datang Koefisien Jarak Gording

Koefisien = 0.02 × × 0.4 0,02 20

MX2a q_{a datang} × L2 / 8

0,000 6,0

0,000 1,4 65

0,000 kg/m

65 kg/cm2

Koefisien = -0,4

0,0000 kg m MY2a

2b. Angin Pergi

6,0 -163,8000 kg m -36,400 kg/m

MX2b q_{a datang} × L2 / 8 qa datang Koefisien Jarak Gording

-0,400 1,4 65

MY2b

Muatan Tak Terduga

-36,400

q d a t a n g

S b . Y S b . X

P = 1 0 0 k g

Sb.Y P q p e r g i

S b . Y S b . X

Page 2 =

= × × / 4

= =

= × × / 4

=

= ( + q ) + ( ) = ( + q )

= ( + q ) + ( 0 ) = ( + q )

= ( + q ) + ( ) = ( + q )

= ( + q ) + ( 0 ) = ( + q )

= ( + q ) + ( ) = ( + q )

= ( + q )

= = =

= ( + ) =

100 cos 20 6,0

MX3 P × cos × / 4

51,3030 kg m Kombinasi Momen

100 sin 20 6,0

140,954 kg m MY3 P × sin × / 4

kg m

MY Total 8,6189 1,5391 8,6189 1,5391 kg m

MX Total 23,6803

MX Total 23,6803 4,2286 -163,800

23,6803

4,2286 0,0000 4,2286

MX Total 23,6803 4,2286 140,954

-140,120

164,634

4,2286 kg m

MY Total 8,6189 1,5391 8,6189 1,5391 kg m

4,2286 kg m

MY Total 8,6189 1,5391 51,3030 59,9219 1,5391 kg m

kg m MY Total 59,9219 1,5391 kg m

cm

Wx 44,27 cm3 _Iy _{53,8 cm}332 cm4 _iy _{2,37139 cm}

4 _ix _5,89

kg/cm2

q total 5,600 7,510 13,110 kg/m 9,567 cm

2 _E _2100000

Wy 12,26 cm3 _F

Sb.Y

Sb.X

P cos P sin

= ( + ) =

q total 5,600 7,510 13,110 kg/m

(8)

( + × ) ( + × )

× × 4 × × 3

× =

× × 4 × × 3

× =

F = FX2 + FY2 = 2 + 2 =

F <

B.

Kontrol Tegangan

Terjadi = MX +

= 16463,4 422,86 7,5

MY

profil

WX WX

1600 kg/cm2

44,27 + 5992,19 12,26153,91 7,5

2100000

384 E × Ix 48 E × Ix

.. OK

Kontrol Lendutan

FX = 5 qTotal× cos × L 4

+ 1 P× cos × L3

1026,659 kg/cm2 _{1600 kg/cm}2

FY = 5 qTotal× cos × L 4

2100000 332

0,905 cm 48

= ₃₈₄5 0,131 cos _× 20₃₃₂ 600 + 1 100 cos 20 600

= ₃₈₄5 0,131 _2100000cos

+ 1 P× cos × L3

384 E × Iy 48 E × Iy

2,032 cm

100 cos 20 600

20 600 ₊ 1

0,905 2,032 2,224 cm

2100000 53,8

× 53,8 48

Direncanakan gading-gading kap memiliki 3 medan ekonomis

2,00

Fizin .. OK

Dimensi Gording Variasi Dengan Track Stang

600 ₌ _{3,333 cm}

180 Fizin = ₁₈₀L =

Page 3 m

1.

= ( =

= ( × × 2 / 8

= ( + q )

=

= ( × × ( / 3 )2 _{/ 8}

= ( + q )

2. = = = = 3. =

= × × / 4

= 2,00

2,00 m

+ q ) kg/m MX1 q total × cos × L_X2 / 8 2,00 m

Muatan Tetap

q total 5,600

MY1 q total × sin × L_Y2 / 8

5,600 + q ) sin

6,0

23,6803 4,2286 kg m

5,600 + q ) cos 20

MX2a 0,0000

20 6,0

0,9577 0,1710 kg m

kg m

MY2b 0 kg m

2b. Angin Pergi

MX2b -163,8000

kg m

MY2a 0 kg m

6,0 140,954 kg m

Muatan Tak Terduga MX3 P × cos × _X / 4

100 cos 20

T ra c k S ta n g L y

L x

G a d in g K a p

S b .Y

S b .X

= =

= × × ( / 3

=

140,954 kg m MY3 P × sin × _Y / 4

) / 4 17,1010 kg m

100 sin 20 6,0

(9)

= ( + q ) + ( ) = ( + q )

= ( + q ) + ( 0 ) = ( + q )

= ( + q ) + ( ) = ( + q )

= ( + q ) + ( 0 ) = ( + q )

= ( + q ) + ( ) = ( + q )

= ( + q )

= = =

= ( + ) =

( + × ) ( + × )

MX Total 23,6803 4,2286 0,0000

Kombinasi Momen

MX Total 23,6803 4,2286 -163,800

23,6803 4,2286 kg m

MY Total 0,9577 0,1710 0,9577 0,1710 kg m

MX Total 23,6803 4,2286 140,954

-140,120

164,634

4,2286 kg m

MY Total 0,9577 0,1710 0,9577 0,1710 kg m

4,2286 kg m

MY Total 0,9577 0,1710 17,1010 18,0587 0,1710 kg m

kg m MY Total 18,0587 0,1710 kg m

cm

Wx 21,92 cm3 _Iy _{20,6 cm}137 cm4 _iy _{1,89 cm}

4 _ix _4,88

Kontrol Tegangan

Terjadi = MX +

kg/cm2

q total 5,600 4,510 10,110 kg/m 5,747 cm

2 _E _2100000

Wy 6,22 cm3 _F

= 16463,4 422,86 4,5

MY

profil

WX WX

1600 kg/cm2

21,92 + 1805,87 6,2217,10 4,5

.. OK

Kontrol Lendutan

+ 1 P× cos × LX3

1140,803 kg/cm2 _{1600 kg/cm}2

Page 4

× × 4 × × 3

× =

× × ( / 3 )4 _× _{× (} _{/ 3 )}3

× =

F = FX2 + FY2 = 2 + 2 =

F < = = = = = = = = = =

n = = + 1 =

2100000

384 E × Ix 48 E × Ix

Kontrol Lendutan

+ 1 P× cos × LX3

FY = 5 qTotal× cos × LY 4

2100000 137

2,027 cm 48

= ₃₈₄5 0,101 cos _× 20₁₃₇ 600 + 1 100 cos 20 600

= ₃₈₄5 0,101 _2100000cos

+ 1 P× cos × LY3

384 E × Iy 48 E × Iy

0,148 cm

100 cos 20 600

20 600 ₊ 1

2,027 0,148 2,032 cm

2100000 20,6

× 20,6 48

C 125×50×20×2.3

h 125 mm

Fizin .. OK

Kesimpulan : Profil gording yang di pakai adalah Channel Lips

600 ₌ _{3,333 cm}

180 Fizin = ₁₈₀L =

t 2,3 mm

F 5,747 cm2

b 50 mm

d 20 mm

Wy 6,22 cm3

Ix 137 cm4

q 4,51 kg/m

Wx 21,92 cm3

11 12 buah 10,6

Iy 20,6 cm4

Jumlah profil Channel Lips C yang diperlukan : ( Jumlah Lapangan Gording ) + 1

(10)

2 m

a = 1,400 m

b = 0,898 m

Berat sendiri atap = 4,00 kg/m2

Jarak batang tarik = 2 m

Berat sendiri gording (q) = 4,51 kg/m

Beban tak terduga = 100 kg

Beban yang bekerja

A. = × × × = kg

= × × = kg

= × = kg +

= kg

Tipe 1 = Atap 2,00 2,458

BAB II

Gording 2,00 4,510,9 4,00sin 20sin 20 3,085

34,202

N1 39,745

Tipe 2 = Atap 2,00 1,4 4,00 sin 20

3,831

a a

b

Page 5

A. = × × × = kg

= × × = kg

= × = kg +

= kg

= + = kg

:

× / 4

d

d = mm

Tipe 2 = Atap 2,00 1,4 4,00 sin 20

0,599 cm

Ambil : 6

d ₁₆₀₀ 450,922

d NTotal

1600 /4 NTotal ₁₆₀₀ _kg/cm2

d2_/4

NTotal ₁₆₀₀ _kg/cm2

A

450,922

3,831

Gording 2,00 4,51 sin 20 3,085

Dibutuhkan Terjadi profil

NTotal N1 10 N2

34,202

N2 41,118

Beban yang dipikul oleh track stang :

(11)

TUGAS AKHIR BENTANG 30 m

=

= = H = = =

= L = = = =

=

1 =

=

20 0,364

5,459554

G 65

2

1,5 / = 1,596

15,0 / = 15,963

BAB III DIMENSIONING GORDING30

160

M

20 0,342 30

20 0,940 1,5

360 M

7 J P () 6 M B 360

5,250 M P 240M M L

P 1

1 =

=

2 =

+ ) / = "

=

A. 1.

= =

= = +

= ( =

= ( 2 / 8

= ( + )

=

= ( 2 / 8

= ( + )

20 6,0

8,6189 1,5391

M

2_{/ 8}

5,600 + )

20 6,0

23,6803 4,2286

M

2_{/ 8}

5,600 + )

5,459554

5,600

B

D

D ! #

M

B 1,4 4,00

12,5 13 $

A $ 4,00

2

%( 1,596 15,963 1,400

A & % 1,400

M $ $ 1 + 2 ==>

1,5 / % 1,596

15,0 / % 15,963

q c o s q s i n

q S b . Y

S b . X

(12)

TUGAS AKHIR BENTANG 30 m

2. + , = = - - = = - - + , = - -= =

= - 2 / 8

= = 0 + , = = - - + , = - -= =

= - 2 / 8

=

= 0

3.

M./0

M123 245 2657 891: 2

-36,400 6,0

-163,8000 6:;

-36,400 6: < ;

M=/0 +

,>,?, @A -B

2_{/ 8}

+

,>, ?, @A K

C7DEF E74 J282 6G C89E4:

-0,400 1,4 65

65 6: < H ;

2

KC7DEF E74I -0,4

0,0000 6:;

M./J

2KL A4:E4M78:E

M=/J +

,>,?, @A -B

2_{/ 8}

0,000 6,0

0,000 1,4 65

0,000 6: < ;

0,4 0,000

+

,>, ?, @A K

C7DEF E74 J282 6G C89E4:

KC7DEF E74I 0.02 -N-O LP 0,02 20

M123 24Q4:E4

22L A4:E4R23 24:

65 6: < H ;

2

q d a t a n g

S b . Y S b . X

P = 1 0 0 k g

Sb.Y P q p e r g i

S b . Y S b . X

PS TUV

=

= - - / 4

= =

= - - / 4

=

W

= ( + +) + ( ) = ( + +)

= ( + +) + ( 0 ) = ( + +)

W

= ( + +) + ( ) = ( + +)

= ( + +) + ( 0 ) = ( + +)

W = ( + +) + ( ) = ( + +) = ( + +) + ( ) = ( + +) = ( + +) = ( + +) = = = = = = = = = = ( + ) =

6: < H ;

2

+3C3 2X 5,600 7,510 13,1106:<;

9,567H;

2 _E _2100000

Y Z 12,26

H;

3 _F

H ;

Y[ 44,27

H;

3 _IZ 53,8

H;

4 EZ 2,37139H ;

332H;

4

E [ 5,89

D23 2\ + 7,516: < ; I[

6:;

M]^ _ ?,` 59,9219 1,5391

6:;

M2 62

;C;7 4;26F E;1 ;

Z24:;7 47431 624\ M

a^_?,` 164,634 4,2286

M]^ _ ?,` 8,6189 1,5391 51,3030 59,9219 1,5391

6:;

DEHCK2b 8CDEXc d2447XBEbF C 150-ef -gO- hLg

A6E K23

i1 2324jki 1 23 2 4h(M1 23245 2657891:2)

Ma

^ _ ?,` 23,6803 4,2286 140,954

-140,120

164,634

4,2286 6:;

M]^ _ ?,` 8,6189 1,5391 8,6189 1,5391

6:;

4,2286 6:;

A6E K23i1 2324jki 1 23 2 4gK(A4:E4M78:E)

Ma^ _ ?,` 23,6803 4,2286 -163,800

23,6803

4,2286 0,0000 4,2286 6:;

M]^ _ ?,` 8,6189 1,5391 8,6189 1,5391

6:;

Ma^ _ ?,` 23,6803

51,3030 6:;

KC;KE42F EiC;74

A6E K23i1 2324jki 1 23 2 4g 2(A4:E4R23 24:)

100 F E4 20 6,0

140,954 6:;

M.l P

-F E4N-m<P

100 HCF 20 6,0

M=l P

-HCFN-m<P

Sb.Y

Sb.X

P cos P sin

= ( + ) =

+3C3 2X 5,600 7,510 13,1106:<;

(13)

TUGAS AKHIR BENTANG 30 m

( + n ) ( + n )

o

n n 4 n n 3

n

=

n n 4 n n 3

n

=

F = Fp

2 _{+ F}

q

2 ₌ ₂ ₊ _{2 =}

F <

B.

Dr s tuv wu wxwuy wzr uy {ywzruyxw|} t}r ~rxr} t zwut xu }r

2,00

F

OK

Dr } turs zr uyws r wrtuywuswv x wuy

600 ₌ _3,333

v }

180 F =

L ₌

180

0,905 2,032 2,224 v }

2100000 53,8

n 53,8 48

2,032 v}

100 v 20 600

20 600 ₊ 1

= 5 0,131 v

384 2100000

+ 1 Pnvn

3

384 E n

48 E n

Fq =

5 T

nv n

4

2100000 332 0,905 v}

48

= 20 600 100 v 20 600

+ 1

n 332

5 0,131 v

384 2100000

384 E n 48 E n

OK

Kus~tu z wu

Fp =

5 T

nv n

4

+ 1 Pnvn

3

1026,659xyv }

2 ₁₆₀₀ xyv } 2 o 1600 xyv } 2

44,27 + 5992,19 12,26153,91 7,5

= 16463,4 422,86 7,5

Mq o p p

Kus~tywuy wu

T

= M p

+

P¡ ¢£¤ }

1.

= ( =

= ( n n 2 / 8

= ( + )

=

= ( n n( / 3 )

2 _{/ 8}

= ( + ) 2. = = = = 3. =

= n n / 4

= 140,954 xy} 6,0

Mwwuwxtszy w

M¥ ¦ P

nv n§ p / 4

100 v 20

xy}

M¨©ª 0

xy}

2« Auyr u¬tsy r

M¥ ©ª -163,8000

xy}

M¨© 0

xy}

Mwwu®uyr u

2w Auyr uwwuy

M¥ © 0,0000

20 6,0

0,9577 0,1710 xy}

M¨¯

w~nr un q

2_{/ 8}

5,600 + ) ru

6,0

23,6803 4,2286 xy}

5,600 + ) v 20

+ ) xy}

M¥¯

w~nvn p

2_{/ 8}

2,00}

Mwwutw|

w~ 5,600

2,00

2,00} T ra c k S ta n g

L y

L x

G a d in g K a p

S b .Y

S b .X

= =

= n n( / 3

= 17,1010 xy} ) / 4

100 r u 20 6,0

140,954 xy}

M¨ ¦ P nrun§ q / 4

(14)

°

= ( + ±) + ( ) = ( + ±)

= ( + ±) + ( 0 ) = ( + ±)

°

= ( + ±) + ( ) = ( + ±)

= ( + ±) + ( 0 ) = ( + ±)

° = ( + ±) + ( ) = ( + ±) = ( + ±) + ( ) = ( + ±) = ( + ±) = ( + ±) = = = = = = = = = = ( + ) = ( + ² ) ( + ² ) ³ ´µµ OK

K ¶ ·¸ ¹ ¶

F = 5 T 4 ₊ 1 P 3

1140,803º»¼ ½¾

2 ₁₆₀₀ º»¼ ½¾ 2 ³ 1600 º»¼ ½¾ 2

21,92 + 1805,87 6,2217,10 4,5

= 16463,4 422,86 4,5

M¿ ³ À ÁÂ ÃÄ ÅÆ Ç È Ç È

KÉÊ¶Ë É·Ì Í»Î Ê»Î Ê

ÀT

ÏÂÐÑÒ Å = M È

+

º»¼ ½¾

2

±¶É¶Î· 5,600 4,510 10,110º»¼ ¾

5,747½¾

2 _E _2100000

ÇÓ 6,22 ½¾

3 _F

½¾

Ç Ô 21,92

½¾

3 _IÓ 20,6

½¾

4 ÕÓ 1,89½¾

137½¾

4

Õ Ô 4,88

DÎ¶Î Ö ± 4,51º»¼¾ IÔ

º»¾

M¿

×ÃØÑÆ 18,0587 0,1710

º»¾

MÎºÎ¾É¾Í Ê¾ÎºÙÕ¾ Ú¾ÓÎ Ê»¾Í ÊÍÊ¶ ÚºÎ Ê

Ö M

È×ÃØÑÆ 164,634 4,2286

M¿×ÃØÑÆ 0,9577 0,1710 17,1010 18,0587 0,1710

º»¾

DÕ ½ÉÛÎÜ ËÉÝÕ·Þ ßÎ ÊÊÍ·¸ ÕÜÙ C 125²àá²âá²âµã

AºÕÛÎ¶äÚÎ¶Î Ê åæ

äÚÎ¶ÎÊ ã(M

ÚÎ¶Î ÊÌ ÎºÌÍË¹ Ú»Î)

MÈ×ÃØÑÆ 23,6803 4,2286 140,954

-140,120

164,634

4,2286 º»¾

M¿×ÃØÑÆ 0,9577 0,1710 0,9577 0,1710

º»¾

4,2286 º»¾

AºÕÛÎ¶äÚÎ¶Î ÊåæäÚÎ¶ÎÊâ Û(AÊ» ÕÊçÍË» Õ)

MÈ×ÃØÑÆ 23,6803 4,2286 -163,800

23,6803 4,2286 º»¾

M¿×ÃØÑÆ 0,9577 0,1710 0,9577 0,1710

º»¾

MÈ×ÃØÑÆ 23,6803 4,2286 0,0000

KÉ¾ÛÕÊÎÙÕä É¾ÍÊ

Pè éêë

² ² 4 ² ² 3

²

=

² ²( / 3 )

4

² ²( / 3 )

3

²

=

F = FÈ

2 _{+ F}

¿

2 ₌ ₂ ₊ _{2 =}

F < = = = = = = = = = =

Ê = = 13 + 1 = 14

ÛÚÎ ß 12,5

IÓ 20,6

½¾

4

JÚ¾ ·Î ßÜ ËÉÝÕ·Þ ßÎÊÊÍ·¸ ÕÜÙÞÓÎÊ»¹ ÕÜÍ Ë·ÚºÎ ÊÖ ( JÚ¾·Îß¸ÎÜÎ Ê»Î Êì ÉË¹ ÕÊ») + 1 ÇÓ 6,22

½¾

3

IÔ 137

½¾

4

± 4,51º»¼¾

Ç Ô 21,92

½¾

3

¶ 2,3¾¾

F 5,747½¾

2

Û 50¾¾

¹ 20¾¾

C 125²àá²âá²âµã

ß 125

¾¾

FÅí Åî

´µ µ OK

KÍÙÕ¾ÜÚ·Î ÊÖç ËÉÝÕ·»É Ë¹ ÕÊ»ÓÎ Ê»¹ ÕÜÎºÎÕÎ¹Î·Î ßÞßÎ ÊÊÍ·¸ÕÜÙ

600 ₌ _3,333

½¾

180 FÅíÅî =

L ₌

180

2,027 0,148 2,032 ½¾

2100000 20,6

² 20,6 48

0,148 ½¾

100 ½ÉÙ 20 600

20 600 ₊ 1

= 5 0,101 ½ÉÙ

384 2100000

+ 1 P²½ÉÙï²¸ ¿

3

384 E ²ðÓ 48 E ²ðÓ

F¿ =

5 ±T

ÃØÑÆ ²½ÉÙï²¸ ¿ 4 2100000 137 2,027 ½¾ 48

= 20 600 100 ½ÉÙ 20 600

+ 1

² 137

5 0,101 ½ÉÙ

384 2100000

384 E ²ð Ô 48 E ²ð Ô

KÉÊ¶Ë É·¸ÍÊ¹ Ú¶Î Ê

FÈ =

5 ±T

ÃØÑÆ

²½ÉÙï²¸ È

4

+ 1 P²½ÉÙï²¸ È 3 h C x X Y b C y d t

(15)

TUGAS AKHIR BENTANG 30 m

2 ñ

ò = 1,400 ñ

ó = 0,759 ñ

Bô õ òö÷ô øùú õúòö òû = 4,00 üýþñ

2

Jò õòüóòö òøýö òõ ú ü = 2 ñ

Bô õ òö÷ô øùú õúýÿ õùú øý( ) = 4,51 üýþñ

Bô óòøö òüöôõùýò = 100 üý

Bô óòøòøýóô üôõò

A. = = üý

= = üý

= = üý +

= üý

T

Gÿ õ ùú øý 2,00

BAB II

3,831 34,202

N1 39,364

A 2,00 1,4 4,00 20

Bôóò ø òüôõùý ò 100 ÷ úø 20

4,51 ÷ úø 20 3,085

0,76 4,00 ÷ úø 20

Túûô Aö òû 2,00 2,077

a a

b

P

A. = = üý

= = üý

= = üý +

= üý

= + = üý

13 : "

573,9

/ 4

ù 0,676

ù = ññ

573,8932

NT

" 1600 üýþñ

2

ù

2_/4

NT

" 1600 üýþñ

2

A

Dúóö üò ø T

NT N1 12 N2

Túûô

0,651 ñ

Añ óú 8

ù

532,776 1600

ù

NT

1600þ

34,202

N2 41,118

Bô óòøòøýùúû ú üÿ ô ö õ òü÷ö òøý

Bôóò ø òüôõùý ò 100 ÷ úø 20

532,776

3,831 Gÿ õ ùú øý 2,00 4,51 ÷ úø 20 3,085

Aö òû 2,00 1,4 4,00 ÷ úø 20

(16)

Tugas Akhir Bentang 20 m

A.

=

= kg

Lk = 0,7 L = 0,7 × =

=

: q = =

h = =

199 mm iX 16,7 cm

396 mm IX 20000

Data 57 kg/m WX 1010 cm3

2,45 cm4 Lk 200 = Lk cm

Coba profil WF 400×200×7×11

Kolom

Ukuran kolom AC dan DF disamakan (Frame 1, 2, 3, 4)

Mmax 1290146 kg cm

Untuk komponen struktur tertekan (kolom) yang lebih menentukan adalah faktor tekuk

Menurut peraturan PPBBI 1983 untuk struktur seperti tampang pada gambar di samping :

700 490 cm

Dmax 2347,07

Nmax 2863,04

= 490₂₀₀ 200 imin

imin perlu =

L

h tw tf

Page 1

h = =

b = =

tw = =

tf = =

×

Y g

= 1 s

= s 1

Y =

×

X g

= 1 s

= s 1

X =

Maka : 1,0622.381 s

2 _untuk

Faktor tekuk untuk 0,18

= 1,41 untuk 0,183 < s

s = = 29,42943 =_111,0721 0,265

< 1 1.593 - s

111,072

OK

iX 16,7

g = E =

29,4294 200 ..

0.7 fy 0,721000002400 =

Maka : 2,318

Arah x

X = Lk = 490 =

< 1 1.593 - s

2.381 s2 untuk

0,18

= 1,41 untuk 0,183 < s

= 0,985 111,0721

Faktor tekuk untuk

s = = 109,375

g =

..

OK

iY 4,48 = 109,375 200

2100000 _{= 111,072}

0.7 fyE = 0,7 2400

Kontrol Kelangsingan Arah y

Y = Lk = 490

cm

11 mm A 72,16 cm2

7 mm iY 4,48

199 mm iX 16,7 cm

396 mm IX 20000 cm4

b

h tw

(17)

Tugas Akhir Bentang 20 m

( × - ) × ( / 2 - ) 2

=

× ×

0

X = X =

2_× _×

× 2

N nX

A nX - 1

- 1

= < =

N

kg/cm2 _.. _OK

= Y + M_WX

X

809,821 kg/cm2 _profil ₁₆₀₀

1319,497

OK

× 1290146

72,16 + 0,6 × 603,149603,149 1010

= 1,062 × 2863,04

+ X × MX

WX

= X

603,149 N X2 = 2863,04 29,4294294

2100000 72,16 ₌

nX = 2_{E A}

0,6 MX2

29,429429 --> 1,062

1290146 kg cm

X = 0,6 + 0,4 MX1 =

Kontrol Tegangan Desak

Momen pada ujung : MX1 =

MX2 =

OK

tw IX 0,7 20000 = 928 kg/cm

2 _..

91,163 kg/cm2 _0.58 _profil

= 2347,0700 543,774 =

= Dmax SX

1,1

2 2

543,774 cm3

1,1 _{+ 0,7} 39,6

tw ( h/2 - tf)2

2 2

= 19,9 1,1 ) × ( 39,6

Kontrol Tegangan Geser SX = ( b × tf ) ( h - tf) +

Page 2

N A

= < =

N A

= < =

i = 2 + 3 2

= 2_{+ 3 ×} ₃

= =

B.

=

= kg

= 816,8296 cm3

profil 1600

WX Perlu = M = 1306927

Dmax 2030,21

Nmax 2080,28

Ukuran kolom GC, CH, HF dan FI disamakan (Frame 5, 6, 7, 8)

Balok GC dan FI digunakan profil WF yang dipotong seperti terlihat pada gambar berikut :

Mmax 1306927 kg cm

kg/cm2 _..

OK

Balok

1328,911 kg/cm2 _profil ₁₆₀₀

kg/cm2 _.. _OK

1319,50 91,163

91,970 kg/cm2 _profil ₁₆₀₀

= 2,318 × 2863,04_72,16

kg/cm2 _.. _OK

= Y

1369,342 kg/cm2 _profil ₁₆₀₀

+ 1290146

72,16 1010

= 2,318 × 2863,04

= Y + M_WX

X

G

C

K olom P rofil W F

(18)

Tugas Akhir Bentang 20 m

Lk = 0,5 L = =

=

: q = =

h = =

b = =

tw = =

tf = =

× 111,072

cm2

cm Untuk keadaan tertekan maka :

Menurut peraturan PPBBI 1983 untuk struktur seperti tampang pada gambar di atas :

1223,8 611,9 cm

= 1,230

s = = 136,5848

0.7 fy 0,721000002400 =

g = E =

= 136,585

11 mm A 72,16

200 ..

OK

Y = Lk_i

Y = 4,48

611,9

7 mm iY 4,48 cm

199 mm iX 16,7

396 mm IX 20000 cm4

Data 57 kg/m

612 3,0595 WX cm 200 200 200 imin

imin perlu = Lk =

= Lk

1010 cm3

b h t w t f P P L

Page 3

× Y g

= 1 s

= s 1

Y =

×

X g

= 1 s

= s 1

X =

( × - ) × ( / 2 - ) 2

=

< 1 < 1 111,072

1.593 - s s = = 36,75075 =_111,0721 0,331

1,1

2 2

543,774 cm3

1,1 _{+ 0,7} 39,6

tw ( h/2 - tf)2

2 2

= 19,9 1,1 ) × ( 39,6

Maka : 1,117

= 111,072

OK

iX 16,7 36,7508

2.381 s2 untuk

= 1,41 untuk 0,183 < s

g = _{0.7 fy}E = _0,72100000₂₄₀₀

Maka : 3,600

Arah x

X = Lk = 611,9 =

1.593 - s

2.381 s2 untuk

200 ..

0,18

= 1,41 untuk 0,183 < s

= 1,230 111,0721

Faktor tekuk untuk

s = = 136,5848

0.7 fy 0,721000002400 =

g = E =

(19)

Tugas Akhir Bentang 20 m

× ×

N A

= < =

N A

= < =

N A

= < =

i = 2 + 3 2

= 2_{+ 3 ×} ₃

= 1333,208 kg/cm2 =

profil 1600

kg/cm2 _.. _OK

1326,19 78,855

103,796 kg/cm2 _profil ₁₆₀₀

kg/cm2 _..

OK

= 3,600 × 2080,28_72,16

kg/cm2 _..

OK

= Y

1322,816 kg/cm2 _profil ₁₆₀₀

= 2080,28_72,16 + 1306927₁₀₁₀

kg/cm2 _..

OK

1600

= + M_WX

X

1326,194 kg/cm2

profil

+ 1306927

72,16 1010

= 1,117 × 2080,28

= X + M_WX

X

OK

tw IX 0,7 20000 = 928 kg/cm

2 _..

78,855 kg/cm2 _0.58 _profil

= 2030,2100 543,774 =

= Dmax SX

(20)

Tugas Akhir Bentang 20 m

(21)

(22)

(23)

(24)

Tugas Akhir Bentang

! #

A.

=

= kg

Lk = 0,7 L = 0,7 × =

=

: q = =

h = =

Kolom

Mmax 1610486 kg cm

700 490 cm

Dmax 2804,69

Nmax 3191,00

= 490₂₀₀ 200 imin

imin perlu = ₂₀₀Lk

= Lk

cm

2,45

cm4

200 mm iX 16,8 cm

400 mm IX 23700

L

h tw tf

Page 1

h = =

b = =

tw = =

tf = =

×

Y g

= 1 s

= s 1

Y =

×

X g

= 1 s

= s 1

X =

cm4

200 mm iX 16,8 cm

400 mm IX 23700

Y = Lk = 490

cm

13 mm A 84,12 cm2

8 mm iY 4,55

g =

.. OK

iY 4,55 = 107,692 200

2100000 _{= 111,072}

0.7 fyE = 0,7 2400

= 0,970 111,0721

Faktor tekuk untuk

s = = 107,6923

< 1 1.593 - s

2.381 s2 untuk

0,18

= 1,41 untuk 0,183 < s

Maka : 2,262

Arah x

X = Lk = 490 =

111,072

OK

iX 16,8

g = E =

29,184 200 ..

0.7 fy 0,721000002400 =

s = = 29,18404 =_111,0721 0,263

< 1 1.593 - s

2.381 s2 untuk

= 1,41 untuk 0,183 < s

Maka : 1,060

b

h tw

(25)

Tugas Akhir Bentang

$ % &

( × - ) × ( / 2 - ) 2

=

× ×

0

X = X =

2_× _×

× 2

N nX

A nX - 1

- 1

= < =

N

tw ( h/2 - tf)2

2 2

= 20 1,3 ) × ( 40

1,3

2 2

642,976 cm3

1,3 _{+ 0,8} 40

MX2 =

OK

tw IX 0,8 23700 = 928 kg/cm

2 _..

95,113 kg/cm2 _0.58 _profil

= 2804,6900 642,976 =

= Dmax SX

0,6 MX2

29,184038 --> 1,060

1610486 kg cm

X = 0,6 + 0,4 MX1 =

= X

641,506 N X2 = 3191,00 29,184038

2100000 84,12 ₌

nX = 2_{E A}

= 1,060 × 3191,00

+ X × MX

WX

× 1610486

84,12 + 0,6 × 641,506641,506 1185

.. OK

= Y + M_WX

X

856,917 kg/cm2 _profil ₁₆₀₀

1399,268 kg/cm

OK

2

Page 2

N A

= < =

N A

= < =

i = 2 + 3 2

= 2_{+ 3 ×} ₃

= =

B.

=

= kg

= Y + M_WX

X

+ 1610486

84,12 1185

= 2,262 × 3191,00

= 2,262 × 3191,00_84,12

kg/cm2 _.. _OK

= Y

1444,855 kg/cm2 _profil ₁₆₀₀

kg/cm2 _.. _OK

1399,27 95,113

85,795 kg/cm2 _profil ₁₆₀₀

Mmax 1627199 kg cm

kg/cm2 _..

OK

Balok

1408,933 kg/cm2 _profil ₁₆₀₀

profil 1600

WX Perlu = M = 1627199

Dmax 2182,74

Nmax 2622,79

= 1016,9992 cm3

G

C

Kolom Profil W F

(26)

Tugas Akhir Bentang

' ( )

Lk = 0,5 L = =

=

: q = =

h = =

b = =

tw = =

tf = =

745 3,7245 WX cm 200 200 200 imin

imin perlu = Lk =

= Lk

1185 cm3

400 mm IX 23700 cm3

Data 66 kg/m

8 mm iY 4,55 cm4

200 mm iX 16,8

= 163,714

13 mm A 84,12

200 .. OK

Y = Lk_i

Y = 4,55

744,9

0.7 fy 0,721000002400 =

g = E = 111,072

cm2

cm4

Untuk keadaan tertekan maka :

1489,8 744,9 cm b h t w t f P P L

Page 3

× Y g

= 1 s

= s 1

Y =

×

X g

= 1 s

= s 1

X =

( × - ) × ( / 2 - ) 2

=

0.7 fy 0,721000002400 =

g = E =

= 1,474 111,0721

Faktor tekuk untuk

s = = 163,7143

1.593 - s

2.381 s2 untuk

200 ..

0,18

= 1,41 untuk 0,183 < s

=

0.7 fy 0,721000002400

Maka : 5,173

Arah x

X = Lk = 744,9 =

= 111,072

OK

iX 16,8 44,3657

2.381 s2 untuk

= 1,41 untuk 0,183 < s

g = E

tw ( h/2 - tf)2

2 2

= 20 1,3 ) × ( 40

Maka : 1,181

1,3

2 2

642,976 cm3

1,3 _{+ 0,8} 40

< 1 < 1 111,072

1.593 - s s = = 44,36569 =_111,0721 0,399

(27)

Tugas Akhir Bentang

* + ,

× ×

N A

= < =

N A

= < =

N A

= < =

i = 2 + 3 2

= 2_{+ 3 ×} ₃

= =

= Dmax SX

OK

tw IX 0,8 23700 = 928 kg/cm

2 _..

74,022 kg/cm2 _0.58 _profil

= 2182,7400 642,976 =

+ 1627199

84,12 1185

= 1,181 × 2622,79

= X + M_WX

X

= + M_WX

X

1409,996 kg/cm2

profil

= 2622,79_84,12 + 1627199₁₁₈₅

kg/cm2 _..

OK

1600

= 5,173 × 2622,79_84,12

kg/cm2 _..

OK

= Y

1404,343 kg/cm2 _profil ₁₆₀₀

1415,813 kg/cm2

profil 1600

kg/cm2 _.. _OK

1410,00 74,022

161,282 kg/cm2 _profil ₁₆₀₀

kg/cm2 _..

OK

(28)

Tugas Akhir Bentang

- . /

(29)

Tugas Akhir Bentang

0 1 2

(30)

Tugas Akhir Bentang

3 4 5

(31)

Tugas Akhir Bentang

6 7 8

(32)

Tugas Akhir Bentang

30 m

A.

=

= kg

Lk = 0,7 L = 0,7 × =

=

: q = =

h = =

200 mm iX 18,6 cm

450 mm IX 33500

2,45

cm4

imin perlu = ₂₀₀Lk

= Lk

Kolom

Mmax 1975429 kg cm

700 490 cm

Dmax 3326,04

Nmax 3554,82

= 490₂₀₀ 200 imin

cm

L

h tw tf

Page 1

h = =

b = =

tw = =

tf = =

×

Y g

= 1 s

= s 1

Y =

×

X g

= 1 s

= s 1

X =

Maka : 1,0402.381 s

2 _untuk

= 1,41 untuk 0,183 < s

s = = 26,32993 =_111,0721 0,237

< 1 1.593 - s

111,072

OK

iX 18,6

g = E =

26,3299 200 ..

0.7 fy 0,721000002400 =

Maka : 2,394

Arah x

X = Lk = 490 =

< 1 1.593 - s

2.381 s2 untuk

0,18

= 1,41 untuk 0,183 < s

= 1,003 111,0721

Faktor tekuk untuk

s = = 111,3636

g =

..

OK

iY 4,4 = 111,364 200

2100000 _{= 111,072}

0.7 fyE = 0,7 2400

Y = Lk = 490

cm

14 mm A 96,76 cm2

9 mm iY 4,40

200 mm iX 18,6 cm

450 mm IX 33500 cm4

b

h tw

(33)

Tugas Akhir Bentang 30 m

( × - ) × ( / 2 - ) 2

=

× ×

0

X = X =

2_× _×

× 2

N nX

A nX - 1

- 1

= < =

N

kg/cm2 _.. _OK

= Y + _WMX

X

835,192 kg/cm2 _profil ₁₆₀₀

1364,885

OK

× 1975429

96,76 + 0,6 × 813,763813,763 1489

= 1,040 × 3554,82

+ X × MX

WX

= X

813,763 N X2 = 3554,82 26,32993

2100000 96,76 ₌

nX = 2_{E A}

0,6 MX2

26,32993 --> 1,040

1975429 kg cm

X = 0,6 + 0,4 MX1 =

MX2 =

OK

tw IX 0,9 33500 = 928 kg/cm

2 _..

89,438 kg/cm2 _0.58 _profil

= 3326,0400 810,7445 =

= Dmax SX

1,4

2 2

810,7445 cm3

1,4 _{+ 0,9} 45

tw ( h/2 - tf)2

2 2

= 20 1,4 ) × ( 45

Page 2

N A

= < =

N A

= < =

i = 2 + 3 2

= 2_{+ 3 ×} ₃

= =

B.

=

= kg

= 1245,0336 cm3

profil 1600

WX Perlu = M = 1992054

Dmax 2347,41

Nmax 3237,53

Mmax 1992054 kg cm

kg/cm2 _..

OK

Balok

1373,648 kg/cm2 _profil ₁₆₀₀

kg/cm2 _.. _OK

1364,88 89,438

87,934 kg/cm2 _profil ₁₆₀₀

= 2,394 × 3554,82_96,76

kg/cm2 _.. _OK

= Y

1414,616 kg/cm2 _profil ₁₆₀₀

+ 1975429

96,76 1489

= 2,394 × 3554,82

= Y + _WMX

X

G

C

Kolom Profil W F

(34)

Tugas Akhir Bentang 30 m

Lk = 0,5 L = =

=

: q = =

h = =

b = =

tw = =

tf = =

× 111,072

cm2

cm Untuk keadaan tertekan maka :

1755,9 877,95 cm

= 1,796

s = = 199,5341

0.7 fy 0,721000002400 =

g = E =

= 199,534

14 mm A 96,76

200 .. OK

Y = Lk_i

Y = 4,4

878

9 mm iY 4,40 cm

200 mm iX 18,6

450 mm IX 33500 cm4

Data 76 kg/m

878 4,38975 WX cm 200 200 200 imin

imin perlu = Lk =

= Lk

1489 cm3

b h t w t f P P L

Page 3

× Y g

= 1 s

= s 1

Y =

×

X g

= 1 s

= s 1

X =

( × - ) × ( / 2 - ) 2

=

< 1 < 1 111,072

1.593 - s s = = 47,17625 =_111,0721 0,425

1,4

2 2

810,7445 cm3

1,4 _{+ 0,9} 45

tw ( h/2 - tf)2

2 2

= 20 1,4 ) × ( 45

Maka : 1,207

= 111,072

OK

iX 18,6 47,1762

2.381 s2 untuk

= 1,41 untuk 0,183 < s

g = _{0.7 fy}E = _0,72100000₂₄₀₀

Maka : 7,684

Arah x

X = Lk = 878 =

1.593 - s

2.381 s2 untuk

200 ..

0,18

= 1,41 untuk 0,183 < s

= 1,796 111,0721

Faktor tekuk untuk

s = = 199,5341

0.7 fy 0,721000002400 =

g = E =

(35)

Tugas Akhir Bentang 30 m

×

N A

= < =

N A

= < =

N A

= < =

i = 2 + 3 2

= 2_{+ 3 ×} ₃

= 1382,559 kg/cm2 =

profil 1600

kg/cm2 _.. _OK

1378,23 63,123

257,100 kg/cm2 _profil ₁₆₀₀

kg/cm2 _..

OK

= 7,684 × 3237,53_96,76

kg/cm2 _..

OK

= Y

1371,306 kg/cm2 _profil ₁₆₀₀

= 3237,53_96,76 + 1992054₁₄₈₉

kg/cm2 _..

OK

1600

= + M_WX

X

1378,229 kg/cm2

profil

+ 1992054

96,76 1489

= 1,207 × 3237,53

= X + _WMX

X

OK

tw IX 0,9 33500 = 928 kg/cm

2 _..

63,123 kg/cm2 _0.58 _profil

= 2347,4100 810,7445 =

= Dmax SX

(36)

(37)

(38)

(39)

(40)

OUTPUT SAP2000 BENTANG 20 M

TABLE: Element Forces - Frames

Frame Station OutputCase CaseType P V2 M3 FrameElem ElemStation

9: ;< = 9: ; < 9:;< > ?@ >? @ > ? @A= 9:;< = B CDEFG B

DH=IJ KL<J HK AMC NO P QR

BOMC P SO

ASP Q SSTABM BAB

C B Q PUDEFG B DH=IJ KL<J HK AMC NO P QR RVMP BO AQ NOMPRO BAB QPU B ODEFG B DH=IJ KL<J HK AMC NO P QR AO QS P QQ AQRRUP BVBAB O B CDEFG M DH=IJ KL<J HK AM NS QPC R AM QR O PCO ASP Q SSTABM BAB C B Q PUDEFG M DH=IJ KL<J HK AM NS QPC R AB NRQPCO O Q QMPOQBAB QPU B ODEFG M DH=IJ KL<J HK AM NS QPC R AB QQVPCO B MVC B PRSBAB O B B CDEFG B DH=IJ KL<J HK BB PVO Q MP NV AQP B NQTABM BBAB C B B B PRDEFG B

DH=IJ KL<J HK B MN PNV QUR P B B AMO C PVBBAB

BPR B B B PRDEFG B DH=IJ KL<J HK BRCPN S QNO AMO C PVBBAB BPR B B

BPU VSMO

DEFG B

DH=IJ KL<J HK BUOPMU

R Q MPC R

AQUB P MN BBAB B PU VSMO B B BPU VSMODEFG B DH=IJ KL<J HK AB MRNPUV AB MO O PUN QCVQPVBBBAM C B B M P NDEFG B DH=IJ KL<J HK AB BRNPCS ABCCBP QV RRSUPURBBAM B P MCQOQ B B M P NDEFG B DH=IJ KL<J HK AB BQ SPCV A VS NPU RRSUPURBBAM B P MCQOQ B B RP MDEFG B DH=IJ KL<J HK ABC BVP BO ASROP MO UUVSPUNBBAM M P SCQOQ B B RP MDEFG B DH=IJ KL<J HK ABC COPM ASBRP QN UUVSPUNBBAM M P SCQOQ B B UP SDEFG B DH=IJ KL<J HK ANVC P MV AMVQ P BS 6231,86 11-2 4,00373

11 5,6 COMB1 Combination -878,32 -260,27 6231,86 11-2 4,00373

11 7 COMB1 Combination -761,4 60,96 6371,38 11-2 5,40373

11 8,4 COMB1 Combination -632,51 415,07 6015,14 11-2 6,80373

11 0 COMB2 Combination 11,97 32,89 2,592E-11 11-1 0

11 1,4 COMB2 Combination 128,89 152,51 -129,78 11-1 1,4

11 1,59627 COMB2 Combination -2080,28 -2030,21 -13069,27 11-2 0

11 2,8 COMB2 Combination -1979,75 -1927,36 -10687,34 11-2 1,20373

11 7 COMB2 Combination -1593,09 -1469,82 -3622,34 11-2 5,40373

15 0 COMB1 Combination -2863,04 2347,07 -9,095E-13 15-1 0

15 3,5 COMB1 Combination -2863,04 1843,07 -7332,73 15-1 3,5

15 7 COMB1 Combination -2863,04 1339,07 -12901,46 15-1 7

15 0 COMB2 Combination -2087,34 -1720,67 6,366E-12 15-1 0

15 7 COMB2 Combination -2087,34 736,33 3445,19 15-1 7

(41)

OUTPUT SAP2000 BENTANG 20 M

16 0 COMB2 Combination 11,97 32,89 0 16-1 0

16 1,4 COMB2 Combination 140,86 387 -270,9 16-1 1,4

16 1,59627 COMB2 Combination -1248,59 -1277,58 3093,91 16-2 0

(42)

OUTPUT SAP2000 BENTANG 25 M

TABLE: Element Forces - Frames

Text m Text Text Kgf Kgf Kgf-m Text m

1 0 COMB1 Combination -2821,34 1394,07 -5,23E-12 1-1 0

1 7 COMB1 Combination -2821,34 -1062,93 -1158,99 1-1 7

1 0 COMB2 Combination -3191 -2804,69 9,322E-12 1-1 0

1 3,5 COMB2 Combination -3191 -2300,69 8934,43 1-1 3,5

1 7 COMB2 Combination -3191 -1796,69 16104,86 1-1 7

11 0 COMB1 Combination 11,97 32,89 -1,01E-10 11-1 0

11 7 COMB1 Combination -1320,73 -520,88 7222,04 11-2 5,40373

11 12,6 COMB2 Combination -1622,24 -1018,27 1317 11-2 11,00373

(43)

OUTPUT SAP2000 BENTANG 25 M

15 0 COMB1 Combination -3191 2804,69 7,049E-12 15-1 0

15 3,5 COMB1 Combination -3191 2300,69 -8934,43 15-1 3,5

15 7 COMB1 Combination -3191 1796,69 -16104,86 15-1 7

15 7 COMB2 Combination -2821,34 1062,93 1158,99 15-1 7

(44)

OUTPUT SAP2000 BENTANG 30 M

TABLE: Element Forces - Frames

Text m Text Text Kgf Kgf Kgf-m Text m

1 0 COMB1 Combination -3507,91 1003,75 4,093E-12 1-1 0

1 7 COMB1 Combination -3507,91 -1453,25 1573,24 1-1 7

1 7 COMB2 Combination -3554,82 -2318,04 19754,29 1-1 7

17 0 COMB1 Combination -3554,82 3326,04 0 17-1 0

17 0 COMB2 Combination -3507,91 -1003,75 0 17-1 0

19 4,2 COMB2 Combination -3009,29 -2095,25 -14140 19-2 2,60373

19 4,2 COMB2 Combination -2997,32 -2062,36 -14140 19-2 2,60373

(45)

OUTPUT SAP2000 BENTANG 30 M

21 2,8 COMB1 Combination -3137,55 -2246,04 -17155,89