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    Cek Kelangsingan Batang

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    regita tri cahyani
    1. The document provides output data from the SAP 2000 software for a steel girder profile including maximum and minimum values of internal forces. 2. It summarizes checks done on the girder for slenderness, compactness, bending strength, lateral buckling strength, shear strength, and tensile strength. 3. All checks show the girder's demand is less than its capacity and it is deemed adequate for the applied loads.

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    © All Rights Reserved
    Download as DOCX, PDF, TXT or read online from Scribd

    Cek Kelangsingan Batang

    Uploaded by

    regita tri cahyani
    48 views7 pages
    1. The document provides output data from the SAP 2000 software for a steel girder profile including maximum and minimum values of internal forces. 2. It summarizes checks done on the girder for slenderness, compactness, bending strength, lateral buckling strength, shear strength, and tensile strength. 3. All checks show the girder's demand is less than its capacity and it is deemed adequate for the applied loads.

    Original Description:

    perhitungan batang tarik - tekan

    Original Title

    973-1417-2-PB

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    © © All Rights Reserved

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    1. The document provides output data from the SAP 2000 software for a steel girder profile including maximum and minimum values of internal forces. 2. It summarizes checks done on the girder for slenderness, compactness, bending strength, lateral buckling strength, shear strength, and tensile strength. 3. All checks show the girder's demand is less than its capacity and it is deemed adequate for the applied loads.

    Copyright:

    © All Rights Reserved
    Download as DOCX, PDF, TXT or read online from Scribd
    Download as docx, pdf, or txt
    48 views7 pages

    Cek Kelangsingan Batang

    Uploaded by

    regita tri cahyani
    1. The document provides output data from the SAP 2000 software for a steel girder profile including maximum and minimum values of internal forces. 2. It summarizes checks done on the girder for slenderness, compactness, bending strength, lateral buckling strength, shear strength, and tensile strength. 3. All checks show the girder's demand is less than its capacity and it is deemed adequate for the applied loads.

    Copyright:

    © All Rights Reserved
    Download as DOCX, PDF, TXT or read online from Scribd
    Download as docx, pdf, or txt
    of 7

    Hasil Output SAP 2000

    Pu Vu Mu
    Profil
    Max Min Max Min Max Min
    Gelagar Melintang
    0 0 49.626 -49.626 94.276 -39.6179
    (GM)

    Data profil baja gelagar melintang (GM) 400.400.13.21


    Kuat Leleh Baja (fy) = 410 Mpa W = 171.68 kg/m
    Kuat Tarik Baja (Fu) = 550 Mpa lx = 66600 cm4
    Elastisitas Baja (€) = 200000 Mpa ly = 22400 cm4
    Angka Poisson (€) = 0.3 rx = 174.5 mm
    h = 400 mm ry = 101.2 mm
    bf = 400 mm L = 9000 mm
    tw = 13 mm Sx = 3330000 mm3
    tf = 21 mm Sy = 120000 mm3
    Luas Penampang (A) = 21870 mm2 L Jembatan = 31000 mm
    Modulus Geser (G) = 80000 Mpa S = mm

    1. Cek Kelangsingan Batang


    a. Sumbu X (λx)

    𝐾 ×𝐿 0.65 × 9000
    ≤ 200 = ≤ 200
    𝑟𝑥 174.5
    = 33.5243553 ≤ 200

    b. Sumbu Y (λy)
    𝐾 ×𝐿 0.65 × 9000
    ≤ 200 = ≤ 200
    𝑟𝑦 101.2
    = 57.80632411 ≤ 200
    Karena λy > λx, maka perhitungan dilakukan di sumbu Y
    2. Cek Kekompakan Batang
    a. Flens
    𝑏𝑓 250
    λf = < λr =
    2𝑡𝑓 √𝑓𝑦
    400 250
    = <
    2 × 21 √410
    = 9.523809524
    < 12.34661996 → 𝐾𝑜𝑚𝑝𝑎𝑘
    b. Web
    ℎ 665
    λw = < λr =
    𝑡𝑤 √𝑓𝑦
    400 665
    = <
    13 √410
    = 30.76923077
    < 32.84200909 → 𝐾𝑜𝑚𝑝𝑎𝑘

    3. Kontrol Lentur
    a. Menentukan batasan momen plastic (Mp)
    1
    𝐾𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑎 𝑃𝑢𝑛𝑡𝑖𝑟 (𝐽) = × 𝑏 × 𝑡3
    3
    1
    = × 400 × 213
    3
    = 1234800 𝑚𝑚4

    1
    𝐼𝑤 = × 𝑓𝑦 × ℎ2
    4
    1
    = × 410 × 4002
    4
    = 16400000 𝑚𝑚6
    𝜋 𝐸×𝐺×𝐽×𝐴
    𝑋1 = ×√
    𝑆𝑥 2

    𝜋 200000 × 80000 × 16400000 × 31870


    = ×√
    3330000 2

    = 13872.2882 𝑀𝑃𝑎
    𝑆𝑥 2 𝐼𝑤
    𝑋2 = 4 × ( ) ×
    𝐺×𝐽 𝐼𝑦
    2
    3330000 16400000
    =4×( ) ×
    (80000⁄1000) × 1234800 224000000
    2
    = 0.000332791 𝑚𝑚 ⁄𝑁

    𝑍𝑥 = 𝑏 × 𝑡𝑓(𝑑 − 𝑡𝑓) + 1⁄4 𝑡𝑤 × (𝑑 − 2𝑡𝑓)2

    = 400 × 21(400 − 21) + 1⁄4 13 × (400 − 2 × 21)2


    = 3600133 𝑚𝑚3
    𝑀𝑛 = 𝑀𝑝 = 𝑓𝑦 × 𝑍𝑥
    = 410 × 4600133
    = 1476054530 𝑁𝑚𝑚

    b. Menentukan kuat nominal penampang dengan pengaruh tekuk lateral.


    Kontrol penampang termasuk bentang pendek, menengah, atau panjang.
    𝐿 = 9000 𝑚𝑚

    𝐸
    𝐿𝑝 = 1.76𝑟𝑦 × √
    𝑓𝑦

    200000
    = 1.76(101.2) × √
    410

    = 3933.835994 𝑚𝑚
    𝑓𝑙 = 𝑓𝑦 − 𝑓𝑟
    = 410 − 70
    = 340 𝑀𝑃𝑎
    𝑋1
    𝐿𝑟 = 𝑟𝑦 ( ) × √1 + √1 + 𝑋2 × 𝑓𝑙 2
    𝑓𝑙
    13872.2882
    = 101.2 ( ) × √1 + √1 + 0.000332791 × 3402
    340
    = 11142.73539 𝑚𝑚
    Jadi, Karena Lp < L < Lr, maka penampang termasuk batang menengah.

    c. Menentukan kuat lentur plastis Mp


    12.5𝑀𝑚𝑎𝑥
    𝐶𝑏 = < 2.3
    2.5𝑀𝑚𝑎𝑥 + 3𝑀𝑎 + 4𝑀𝑏 + 3𝑀𝑐
    12.5(94.276)
    = < 2.3
    2.5(94.276) + 3(94.276) + 4(39.6179) + 3(94.276)
    = 1.227785363 < 2.3
    Nilai Cb yang digunakan adalah nilai yang terkecil, yaitu 1.227785363.
    𝑀𝑟 = 𝑆𝑥 × (𝑓𝑦 − 𝑓𝑟)
    = 3330000 − (410 − 70)
    = 1132200000 𝑁𝑚𝑚
    𝐿 − 𝐿𝑝
    𝑀𝑛 = 𝐶𝑏 (𝑀𝑝 − (𝑀𝑝 − 𝑀𝑟) × ) ≤ 𝑀𝑝
    𝐿𝑟 − 𝐿𝑝

    = 1.227785363 (1476054530 − (1476054530 − 1132200000)

    9000 − 3933.835994
    × ) ≤ 1476054530
    11142.73539 − 3933.835994
    = 1515585022 ≤ 1476054530
    ∅𝑀𝑛 = 0.9 × 1515585022
    = 1364026520
    Kontrol kekuatan penampang berdasarkan Mn
    𝑀𝑢 ≤ ∅𝑀𝑛
    1364026520
    94.276 ≤
    1000000
    94.276 ≤ 1364.02652 → 𝐴𝑀𝐴𝑁‼!
    4. Kontrol Kuat Geser

    ℎ 𝐸
    𝐾𝑒𝑡𝑒𝑏𝑎𝑙𝑎𝑛 min 𝑤𝑒𝑏 = ≤ 2.24√
    𝑡𝑤 𝑓𝑦

    400 200000
    = ≤ 2.24√
    13 410

    = 30.76923077 ≤ 49.47332368 → 𝐴𝑀𝐴𝑁‼!


    Maka, kuat geser nominal = 0.6 × 𝑓𝑦 × 𝐴𝑤
    𝐴𝑤 = 𝑡𝑤 × ℎ
    = 13 × 400
    = 5200 𝑚𝑚2
    𝑉𝑛 = 0.6 × 𝑓𝑦 × 𝐴𝑤
    = 0.6 × 410 × 5200
    = 1279200 𝑁
    ∅𝑉𝑛 = 0.9 × 𝑉𝑛
    = 0.9 × 1279200
    = 1151280 𝑁

    𝑉𝑢 ≤ ∅𝑉𝑛
    1151280
    49.626 ≤
    1000
    49.626 ≤ 1151.28 → 𝐴𝑀𝐴𝑁‼!

    5. Parameter Kelangsingan
    Ditinjau berdasarkan sumbu terlemah (Y)

    𝐿×𝐾 𝑓𝑦
    λcy = ×√
    𝜋𝑟𝑦 𝐸

    9000 × 0.65 410


    = ×√
    𝜋(101.2) 200000

    = 0.832774931
    Nilai koefisien tekuk (ω) diambil berdasarkan:
    Untuk λcy ≤ 0.25 =1
    1.43
    Untuk 0.25 ≤ λcy ≤ 1.2 = 1.6−0.67λcy

    Untuk λcy ≥ 1.2 = 1.25 λcy2

    1.43
    Karena Untuk 0.25 ≤ λcy ≤ 1.2, maka nilai ω = 1.6−0.67λcy = 1.372307116

    6. Tahanan Nominal Tekan


    𝑁𝑛 = 𝐴𝑔 × 𝑓𝑐𝑟
    𝑓𝑦
    = 𝐴𝑔 × ( )
    ω
    410
    = 21870 × ( )
    1.372307116
    = 6534033.011 𝑁
    ∅𝑁𝑛 = 0.85 × 𝑁𝑛
    = 0.85 × 6534033.011
    = 5553928.059 𝑁

    𝑁𝑢 = 𝑃𝑢 ≤ ∅𝑁𝑛
    5553928.059
    0 ≤
    1000
    0 ≤ 5553.928059 → 𝐴𝑀𝐴𝑁‼!

    7. Tahanan Nominal Tarik


    a. Batas leleh (Tn)
    𝑇𝑛 = ∅𝐴𝑔 × 𝑓𝑦
    = 0.85 × 21870 × 410
    = 7621695 𝑁
    𝑃𝑢 ≤ 𝑇𝑛
    7621695
    0 ≤
    1000

    0 ≤ 7621.695 𝐾𝑁 → 𝐴𝑀𝐴𝑁‼!
    b. Batas putus
    Nilai b/h = 400/400 >2/3
    = 1 > 2/3
    Makan nilai reduksi, U = 0.9

    Batas putus, 𝑇𝑛 = ∅𝐴𝑛 × 𝑈 × 𝑓𝑢


    = (0.9 × 21870) × 0.9 × 550
    = 9743085 𝑁
    𝑃𝑢 ≤ 𝑇𝑛
    9743085
    0 ≤
    1000

    0 ≤ 9743.085 𝐾𝑁 → 𝐴𝑀𝐴𝑁‼!

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