Kajian Perilaku Struktur Rangka Berpengaku Eksentrik Tipe-D Dengan Inovasi Pengaku Badan Pada Elemen Link

  LAMPIRAN A

• • Kurva Tegangan - Regangan Material

  ε (mm/mm) σ (MPa) 0,00 0,00 0,001774 373,00

  0,013938 373,00 0,028399 423,88 0,077690 486,34 0,105605 495,52 0,161324 498,28 0,225314 491,48

  Tabel A.1 Data Kurva Tegangan versus Regangan Gambar A.1 Kurva Tegangan versus Regangan

  100 200 300 400 500 600

  

0,05 0,1 0,15 0,2 0,25

teg an g an ( M P

  a) regangan (mm)

  LAMPIRAN B

• KONTUR TEGANGAN VON MISES PADA STRUKTUR AKIBATBEBAN MONOTONIK

Gambar B.1 Kontur Tegangan Von Mises Struktur Menggunakan Link Geser Standar AISC, pada Kondisi Beban Maksimum

  Gambar B.2 Kontur Tegangan Von Mises Struktur Menggunakan Link Geser dengan Pengaku Badan Diagonal, pada Kondisi Beban Maksimum

  Gambar B.3 Kontur Tegangan Von Mises Struktur Menggunakan Link Geser dengan Pengaku Badan Vertikal Diagonal, pada Kondisi Beban Maksimum

  LAMPIRAN C

• CONTOH PERHITUNGAN DISAIN LINK, BEAM OUTSIDE LINK, BRACING DAN COLUMN

I. DESAIN LINK GESER

  Gaya dalam maksimum yang bekerja: Mu = 52.63 kNm Vu = 138.96 kN Profil awal link geser : WF 200. 150. 6. 9

2 A = 39.01 cm

  rx = 8.30 cm ry = 3.61 cm

  4 Ix = 2690 cm

  4 Iy = 507 cm

  3 Sx = 277 cm

  3 Sy = 67.6 cm

  Parameter yang digunakan fy = 240 MPa fu = 370 MPa ʋ = 0.3

  5 Es = 2 x 10 MPa

  1. Cek kelangsingan penampang b 150

  Flenge : = = = 8.33

  2t f

  2.9 170 170

  = = = 10.97 λ

  � √240

  , penampang kompak λ f ≤ λ p

  200 ℎ −2 9

  Web: = = = 30.33

  6

  = = 0 0 ≤ 0.125

  ∅ ∅ 1365 1365

  Maka, = (1

  �1 − 1.54 � = − 0) = 88.11

  ∅ � √240

  , penampang kompak λ w ≤ λ p

2. Kapasitas penampang

  3

  2 Mp = Zx . f = 1.12 .277000 mm . 240 N/mm = 74457600 Nmm y

  Mp = 74.4576 kNm Mn = Mp Ø Mn = 0.9 . 74.4576 = 67.012 kNm Mu = 52.63 kNm Ø Mn > Mu , Memenuhi persyaratan Strength rasio = 52.63/67.012 = 0.785

   Analisa Geser 3.

  2 2 . 74.4576 = = = 372.288

  0.4 Vn = Vp = 0.6 . fy (h – 2tf).tw = 0.6 . 240. (200 – 2 . 9).6 = 157248 N = 157.25 kN

  ØVn = 0.9 . 157.25 = 141.523 kN Vu = 138.96 kN Strength rasio = 138.96/141.523 = 0.0.982 ok!!!

  Cek panjang link Mu = 52.63 kNm ; Mp = 74.4576 kNm Vu = 138.96 kN ; Vp = 157.25 kN Link geser

  1.6 1.6 . 74.4576 = = 0.76

  = = 760 157.25

  1.6 = 0.76

  = 0.4 ≤ Panjang link mencukupi

  ( ) (200

  ℎ − 2.9) Spasi stiffener = 30 . 6 = 30 . 6 = 143.6

  − −

  5

  5 II. DISAIN BALOK DI LUAR LINK

A. Desain Balok

  Gaya dalam maksimum yang bekerja: Vu = 76.74 kN; 1.1 . 1.5 Vu = 126.621 kN Mu = 50.27 kNm; 1.1 . 1.5 Mu = 82.95 kNm Balok didesain berdasarkan kekuatan link Vu = 1.1 Ry .Vn = 1.1 . 1.5 . 126.621 = 208.925 kN

  . 157.25 . 0.4 = 1.1 = 1.1 . 1.5 = 51.89

  ≤ 1.1 . 1.5

  2

  2 Maka diambil Mu = 82.95 kNm ; Vu = 208.925 kN Profil awal WF 200.150.6.9 1.

   Cek kelangsingan penampang

  2

  1365 √240

  (1 − 0) = 88.1

  λ w ≤ λ p

  penampang kompak!! Cek apakah sudah plastis

  ℎ ≤ 1.1� .

  = 5 +

  5 ( ℎ

  � )

  = 5 +

  

∅

  5 (6000 (200 − 2.9)

  � )

  2

  = 5.005 (200

  − 2.9)

  6 ≤ 1.1�

  5.005 . 200000 240

  = 71.04

  � =

  �1 − 1.54

  Flenge: =

  Web: =

  2

  =

  150 2 . 9

  = 8.33 =

  170 �

  =

  170 √240

  = 10.97 λf ≤ λp , penampang kompak

  ℎ

  1365 �

  =

  200 −2.9

  6

  = 30.33

  ∅

  =

  ∅ .

  = 00 ≤ 0.125 maka,

  =

2. Kapasitas Geser

  30.33 ≤ 71.04 kondisi plastis sempurna

  Vn = Vp = 0.6 . f (h – 2 t ) t = 0.6 . 240. (200 – 2.9).6 = 157.25 kN

  y f w

  ØVn = 0.9 . 157.25 = 141.53 kN Vu = 126.621 kN

  ≤ ØVn penampang mencukupi Strength rasio = 126.621/141.53 = 0.895 ok!!

3. Kapasitas lentur

  Analisa tahanan lentur nominal balok Lb = 5600 mm

  200000 = 1.76 . = 1.76 . 83 = 4217

  � �

  240 f = f - f = 240 – 0.3 . 240 = 168 MPa

  l y r

  200000 = = 76923.1

  = 2(1 + 2(1 + 0.3)

  ʋ)

  1

  1

  1

  1

  3

  3

  3

  3

• 2. . = + 2. . 150. 9

  (200 = �ℎ − 2. � . − 2.9)6

  3

  3

  3

  3 J = 86004 mm

  . 200000.76923.1.86004.3901 . . = =

  1 � � 2 277000

  2 = 18219.89

  2

  2

  ( ) (200

  ℎ − − 2.9)

  10

  = = 5070000 = 4.2

  10

  4

  4

  10

  277000

  4.2

  10 = 4 = 4 = 1.387

  2 � � �

  76923.1 . 86004 5070000 . �

  18219.89

  1

  2

  2

  = = 83 � � �1 + �1 +

  2 � � �1 + �1 + 1.387. 168

  168 = 126936 mm

  Lr

  L

  p ≤ L b ≤ L r

  − = (

  − � − � � � −

  Ma = momen di ¼ bentang = 6.17 kNm Mb = momen di ½ bentang = 65.32 kNm Mc = momen di ¾ bentang = 60.57 kNm

  12.5 = 2.5 + 3 + 4 + 3 12.5 . 96.34

  = 2.5 . 6.34 + 3 .6.17 + 4 .65.32 + 3 .60.57

  1.73 C b = M = S . f = 277000 . 168 = 46.536 kNm

  r x L

  M = 1.12 . S . f = 1.12 . 277000 . 168 = 52.12 kNm

  p x L 5600 −4217

  Mn = 1.73 �52.12 − (52.12 − 46.536) � �� = 95.278

  7022 −4217

  ØMn = 0.9 . 95.278 = 85.75 kNm Mu = 50.27 kNm ØMn

  ≥ Mu Strength rasio = 50.27/85.75 = 0.586 ok!!

4. Kombinasi Lentur dan geser

• 0.625

  ≤ 1.35 ∅ ∅ 0.895+ 0.625 . 0.586

  ≤ 1.35 1.26 < 1.35 …….ok!!

B. Desain Bresing

  Gaya-gaya dalam maksimum yang bekerja Nu = 146.78 kN ; 1.25 . 1.5 Nu = 275.213 kN Mu = 30.567 kNm ; 1.25 . 1.5 Mu = 57.30 kNm Bresing didesain berdasarkan kekuatan link: Nu = 1.25 . 1.5 Vn link = 1.25 . 1.5 . 157.25 = 294.84 kN Mu = 1.25 Ry.0.5Vn.e = 58.97 kNm < 1.25 . 1.5 Mu Maka diambil Nu = 294.84 kN Mu = 58.97 kNm Coba penampang WF 200.150.6.9

1. Cek Kelangsingan Penampang

  150

  Flange: = = = 8.33

  2

  2.9 170 170

  = = = 10.97

  � √240

  penampang kompak λ f ≤ λ p

  (200 ℎ −2.9)

  Web: = = = 30.33

  6

  294.84 = = = 0.3705

  ≥ 0.125 . 0.85 . 3901 . 240

  ∅ ∅

  

500 665

  Maka, = . � �2.33 − � , � = 70.23; 42.92 = 70.23

  ∅

� �

  penampang kompak λ w ≤ λ p

2. Analisa komponen tekan

  kcx = kcy = 1 Analisa tekuk bresing L = 5.97 m = 5970 mm fy = 240 MPa E = 200000 MPa Arah x: Lkx = 5970 mm

  5970 = = = 71.93

  ≤ 200 ‼

  83

  1 1 240 = = . 71.93. = 0.793

  � � 200000

  0.25

  cx

  ≤ λ ≤ 1.25

  1.43

  1.43 = = 1.34

  = 1.6 )

  1.6 − (0.67 − (0.67 . 0.793)

  240 = . = 3901. = 698686.6

  = 698.687

  1.34 Arah y: Lky = 5970 mm

  5970

  = = = 88.31 < 200 ok!!

  67 .6

  1 1 240 = . . = . 88.31. = 0.974

  � � 200000

  0.25

  cy

  ≤ λ ≤ 1.25

  1.43

  1.43 = = 1.51

  = 1.6 )

  1.6 − (0.67 − (0.67.0.974)

  240 = . = 3901. = 620026.5

  = 620.03

  1.51 Nn = min (Nnx; Nny) = Nny = 620.03 kN ØNn = 0.85 . 620.03 = 527.03 kN ØNn

  ≥ Nu Nu = 294.84 Strength rasio = 294.84/527.03 = 0.56 ok!!

3. Kapasitas Lentur

  Analisa tahanan lentur nominal bresing Lb = 5970 mm

  200000 = 1.76 . . = 1.76 . 36.1 . = 1834

  � �

  240 = = 240

  − − 0.3 . 240 = 168

  200000 = = 76923.1

  = 2(1 2 (1 + 0.3)

  − ʋ)

  1

  1

  1

  1

  3

  3

  3

  3

  . + 2. . = . (200 + 2. . 150. 9 = �ℎ − 2. � . − 2.9). 6

  3

  3

  3

  3

  4

  = 86004 200000 . 76923.1 . 86004 . 3901 = . = .

  � �

  1

  2 277000

  2 = 18219.89

  2

  2

  ( ) (200 ℎ − 2 − 2 . 9)

  10

  = . = 5070000. = 4.2

  10

  4

  4

  10

  277000

  4.2

  10

  −4

  = 4 . = 4 = 1.387 2 � � � .

  10 76923.1 . 86004 5070000

  � _{1 2}

18219.89

₂

  −4 = . = 36.1 . 168 ₂ � ��1 + �1 + � � �1 + �1 + 1.387 10

  

168

  = 7022 mm

  Lr Lp

  ≤ Lb<Lr −

  = � − � − � � ��

  − Ma = momen di ¼ bentang = 10.52 kNm Mb = momen di ½ bentang = 52.76 kNm Mc = momen di ¾ bentang = 50.48 kNm

  12.5 . = = 1.39

  2.5 + 3 + 4 + 3 = . = 277000 . 168 = 46536000

  = 46.536

  = 1.12 . = 1.12 . 277000 . 168 = 51120320 = 51.12

  5970 − 1834

  = 1.39 �51.12 − (51.12 − 46.536) � � = 65.98

  7022 − 1834�

  ØMn = 0.9 x 65.98 = 59.38 kNm Mu = 58.97 kNm ØMn > Mu ok!! Strength rasio = 58.97/59.38 = 0.99 ok!!

4. Persamaan interaksi aksial momen (Mn hanya arah x)

  8

• 9

  ∅ ∅ 157.25

  8

• 0.85 . 620.03

  9

  56.36 M . M

  ux = Rangka tak bergoyang = δ b nt

  Mnt = 56.36 kNm =

  ≥ 1 �1 − �

  Cm = 1 . 3901 . 240

  = = = 2600.67

  2

  2

  0.6

  1 = = 1.064

  � � �1 − 157.25 2600.67

  Mux = 1.064 x 56.36 = 59.97 kNm

  157 .25

  8

  51.12

• Maka,

  ≤ 1

  0.85 . 698.687

  9

  56.36

   Desain Kolom C.

  Gaya dalam kolom yang bekerja Mu = 60.71 kNm; 1.1 . 1.5 Mu = 100.172 kNm Nu = 350.82 kN; 1.1 . 1.5 Nu = 578.85 kN Vu = 78.13 kN; 1.1 . 1.5 Vu = 128.92 kN Pembesaran = 1.1 . Ry Vn link : Nu = Vu = 1.1 . 1.5 . 157.25 = 259.46 kN

  . 157.25 . 0.4 = 1.1 = 1.1 . 1.5 = 51.893

  2

  2 Untuk menghindari strongth beam weak column, maka diambil Mu kolom = 6/5 Mn balok = 1.2 . 85.75 = 102.9 kNm Maka diambil: Mu max. = 102.9 kNm Vu max. = 157.25 kN Nu max. = 578.85 kN Sx min = Mu/(1.12 . fy) = 382812.5 mm Profil awal: WF 300.200.8.12

1. Cek kelangsingan penampang

  200

  Flange: = = = 8.33

  2 2 .12 170 170

  = = = 10.97

  √240 � penampang kompak ≤

  ,

  (300 − 2 . 12)

  Web: � = 34.5

  � =

  8 = ℎ

  3

  259.46 . 10 = = = 0.176 > 0.125

  . 0.85 . 7238 . 240 ∅ ∅

  500 665

  Maka, = . � �2.33 − � , � = (59.31, 42.93) = 59.31

  � ∅ �

  penampang kompak ≤

  ,

2. Kapasitas Geser

  Cek apakah sudah plastis .

  ℎ ≤ 1.1 . �

  5

  5 = 5 + = 5 + = 5.024

  2

  2

  �4000 � � � � 300

  − 2.12 � ℎ 300 5.024 . 200000

  < 1.1 .

  � 8 240 37.5 < 71.175 ok!! Plastis sempurna Vn = Vp = 0.6 . fy (h – 2tf) tw Vp = 0.6 . 240 .(300 – 2 . 12) = 317.952 kN ØVn = 0.9 . 317.952 = 286.157 kN Vu = 157.25 kN ØVn > Vu Strength rasio = 157.25/286.157 = 0.55

  3. Kapasitas Lentur

  Analisa tahanan lentur nominal kolom

  L = 3800 mm b

  200000 = 1.76 . . = 1.76 . 47.1 . = 2393

  � �

  240 <L

  L b p

  Mn = Mp = Zx . fy = 771.000 . 240 = 185.04 kNm ØMn = 0.9 . 185.04 = 166.536 kNm Mu = 102.9 kNm ØMn > Mu Strength rasio = 102.9/166.536 = 0.62 ok!!

  4. Kombinasi Lentur dan Geser + 0.625 .

  ≤ 1.35 ∅ ∅ 0.62 + 0.625 . 0.555 < 1.35 1.12 < 1.35

  5.Analisis komponen tekan

  Struktur Rangka Bergoyang Arah x: Mencari nilai k

  ∑ � � =

  ∑ � � Kolom L = 3800 mm

  c1

  L = 3800 mm

  c2

  4

  4 I = 11300 x 10 mm c1

  4

  4 I = 11300 x 10 mm c2

  Balok L = 4600 mm

  b1

  L = 4600 mm

  b2

  4

  4 I = 2690 x 10 mm b1

  4

  4 I = 2690 x 10 mm b1 ₄ 11300

  10

  2

  3800 ₄ = 5.085

  =

  2690

  10

  2

  4600

  GB = 0 k = 1.46 (dari tabel)

  cx

  Arah y: Kolom L = 3800 mm

  c1

  L = 3800 mm

  c2

  4

  4 I = 1600 x 10 mm c1

  4

  4 I = 1600 x 10 mm c2

  Balok L = 4000 mm

  b1

  L = 4000 mm

  b2

  4

  4 I = 507 x 10 mm b1

  4

  4 I = 507 x 10 mm b1 ₄ 1600

  10

  2

  3800 ₄ = 3.322

  =

  507

  10

  2

  4000

  = 0

  GB

  k = 1.38 (dari tabel)

  cy

  Arah x: L = 1.46 x 3800 = 5548 mm

  kx

  5548 = = = 44.38 < 200

  ‼ 125

  1 1 240 = . . = . 44.38 . = 0.489

  � � 200000

  0.25 ≤ λcx < 1.25

  1.43

  1.43 = = 1.124

  =

  1.6

  1.6 − (0.67 . − (0.67 . 0.489)

  240 = . = 7238 . = 1545.48

  1.124 Arah y: L = 1.38 . 3800 = 5244 mm

  ky

  5244 = = = 111.34 < 200, ok

  ‼

  47.1

  1 1 240 = . . � = . 111.34 . = 1.23

  � 200000

  0.25 < < 1.25

  1.43

  1.43 = = 1.84

  = 1.6 )

  1.6 − (0.67 − (0.67 . 1.23)

  240 = . = 7238 . = 944.086

  1.84 = min ; = 944.086

  � � = = 0.85 . 944.086 = 802.473

  ∅ N = 578.85 kN

  u

  ØN > N

  n u

  Strength rasio = 578.85/802.473 = 0.72

6. Persamaan Interaksi aksial momen

  8

• 9

  ∅ ∅ 578.85

  8 + . 9 0.9 . 185.04

  ∅ . 802.473 = .

  = = 185.04 kNm

  M nt

  1 =

  ≥ 1

  ∑

  �1 − �

  ∑

  = 1736.55 �

  . 7238 . 240 = = = 7264.607

  �

  2

  2

  0.489

  1 = = 1.03

  1736 .55

  �1 − �

  7264 .607

  = . = 1.03 . 102.9 = 105.99 Maka,

  578.85 8 105.99 .

• 0.85 . 1545.48

  ≤ 1

  9 0.9 . 185.04

  1.0 ≤ 1.0 ok!!