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

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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

0 100 200 300 400 500 600

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

teg

an

g

an

(

M

P

a)

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LAMPIRAN B

•

KONTUR TEGANGAN VON MISES PADA STRUKTUR

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

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LAMPIRAN C

•

CONTOH PERHITUNGAN DISAIN LINK, BEAM OUTSIDE LINK,

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I. DESAIN LINK GESER

Gaya dalam maksimum yang bekerja:

Mu = 52.63 kNm

Parameter yang digunakan 3

fy = 240 MPa

fu = 370 MPa

ʋ = 0.3

Es = 2 x 105

1. Cek kelangsingan penampang

(9)

Web: 𝜆𝜆_𝑤𝑤 = ℎ

2. Kapasitas penampang

, penampang kompak

Strength rasio = 52.63/67.012 = 0.785

3. Analisa Geser

𝑉𝑉𝑛𝑛 =

(10)

Cek panjang link

Mu = 52.63 kNm ; Mp = 74.4576 kNm

Vu = 138.96 kN ; Vp = 157.25 kN

Link geser

𝑒𝑒 = 1.6 𝑀𝑀𝑝𝑝

𝑉𝑉𝑝𝑝 =

1.6 . 74.4576

157.25 = 0.76 𝑚𝑚 = 760 𝑚𝑚𝑚𝑚

𝑒𝑒 = 0.4≤1.6𝑀𝑀𝑝𝑝

𝑉𝑉𝑝𝑝 = 0.76

Panjang link mencukupi

Spasi stiffener = 30 . 6− (ℎ𝑤𝑤)

5 = 30 . 6−

(200−2.9)

5 = 143.6 𝑚𝑚𝑚𝑚

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

𝑀𝑀𝑢𝑢 = 1.1 𝑅𝑅𝑦𝑦𝑉𝑉𝑛𝑛

.𝑒𝑒

2 = 1.1 . 1.5

157.25 . 0.4

2 = 51.89 𝑚𝑚𝑁𝑁𝑚𝑚 ≤1.1 . 1.5 𝑀𝑀𝑢𝑢

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Profil awal WF 200.150.6.9

Flenge:𝜆𝜆_𝑓𝑓 = 𝑏𝑏

2. Kapasitas Geser

penampang kompak!!

Cek apakah sudah plastis

ℎ

5.005 . 200000

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30.33 ≤ 71.04 kondisi plastis sempurna

Vn = Vp = 0.6 . fy (h – 2 tf) tw

ØVn = 0.9 . 157.25 = 141.53 kN

= 0.6 . 240. (200 – 2.9).6 = 157.25 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

5070000 = 1.387

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4. Kombinasi Lentur dan geser 𝑀𝑀𝑢𝑢

∅𝑀𝑀𝑛𝑛 + 0.625

𝑉𝑉𝑢𝑢

∅𝑉𝑉𝑛𝑛 ≤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

Flange:𝜆𝜆_𝑓𝑓 = 𝑏𝑏

2𝑜𝑜_𝑓𝑓 = 150

2.9 = 8.33

𝜆𝜆𝑝𝑝 =170_�𝑓𝑓_𝑦𝑦 =_√170₂₄₀ = 10.97

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Web:𝜆𝜆_𝑤𝑤 = ℎ

2. Analisa komponen tekan

penampang kompak

kcx = kcy = 1

Analisa tekuk bresing

L = 5.97 m = 5970 mm

200000= 0.793

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Arah y:

200000= 0.974

0.25 ≤ λcy

3. Kapasitas Lentur

Analisa tahanan lentur nominal bresing

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𝑀𝑀𝑝𝑝 = 1.12 𝑆𝑆𝑥𝑥.𝑓𝑓𝐿𝐿 = 1.12 . 277000 . 168 = 51120320 𝑁𝑁𝑚𝑚𝑚𝑚= 51.12 𝑚𝑚𝑁𝑁𝑚𝑚

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C. Desain Kolom

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

𝑀𝑀𝑢𝑢 = 1.1 𝑅𝑅𝑦𝑦𝑉𝑉𝑛𝑛

.𝑒𝑒

2 = 1.1 . 1.5

157.25 . 0.4

2 = 51.893 𝑚𝑚𝑁𝑁𝑚𝑚

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

Flange:𝜆𝜆_𝑓𝑓 = 𝑏𝑏

2𝑜𝑜_𝑓𝑓 = 200

2 .12 = 8.33

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𝜆𝜆𝑓𝑓 ≤ 𝜆𝜆𝑝𝑝 , penampang kompak

2. Kapasitas Geser

Cek apakah sudah plastis

ℎ

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

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Strength rasio = 157.25/286.157 = 0.55

3. Kapasitas Lentur

Analisa tahanan lentur nominal kolom

Lb

L

= 3800 mm

𝐿𝐿𝑝𝑝 = 1.76 .𝑟𝑟𝑦𝑦.�_𝑓𝑓𝐸𝐸

𝑦𝑦 = 1.76 . 47.1 .�

200000

240 = 2393 𝑚𝑚𝑚𝑚

b<L

Mn = Mp = Zx . fy = 771.000 . 240 = 185.04 kNm

p

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 𝑁𝑁𝑢𝑢

∅𝑁𝑁𝑛𝑛 + 0.625 .

𝑉𝑉𝑢𝑢

∅𝑁𝑁𝑛𝑛 ≤1.35

0.62 + 0.625 . 0.555 < 1.35

1.12 < 1.35

5.Analisis komponen tekan

Struktur Rangka Bergoyang

Arah x:

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Ic2 = 1600 x 104 mm4

200000= 0.489

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Lky

200000= 1.23

0.25 <𝜆𝜆_{𝑜𝑜𝑦𝑦} < 1.25

Strength rasio = 578.85/802.473 = 0.72 u

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𝛿𝛿𝑚𝑚 =

1

�1− _{∑ 𝑁𝑁}∑ 𝑁𝑁𝑢𝑢

𝑜𝑜𝑟𝑟𝑚𝑚�

≥ 1

� 𝑁𝑁𝑢𝑢 = 1736.55 𝑚𝑚𝑁𝑁

� 𝑁𝑁𝑜𝑜𝑟𝑟𝑚𝑚 = 𝐴𝐴𝑔𝑔

.𝑓𝑓_𝑦𝑦

𝜆𝜆𝑜𝑜𝑥𝑥2

= 7238 . 240

0.4892 = 7264.607 𝑚𝑚𝑁𝑁

𝛿𝛿𝑚𝑚 =

1

�1− 1736 .55

7264 .607�

= 1.03

𝑀𝑀𝑢𝑢𝑥𝑥 = 𝛿𝛿𝑚𝑚 .𝑀𝑀𝑛𝑛𝑜𝑜 = 1.03 . 102.9 = 105.99 𝑚𝑚𝑁𝑁𝑚𝑚

Maka,

578.85

0.85 . 1545.48+

8 9 .

105.99

0.9 . 185.04≤ 1