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    Mohr Circle & Buckling Task Example

    Uploaded by

    Muhammad Akbar Firmansyah
    The document is a homework assignment on mechanics of materials containing 4 problems. The summary provides the solutions to each problem in the following 3 sentences: Problem 1 calculates the buckling load of a column as 25680.6 kips about the x axis and 16785.4 kips about the y axis. Problem 2 finds the allowable load on a column is 55.1 kN. Problem 3 determines the maximum principal stress is 106.24 MPa at an angle of 68.52 degrees and the maximum in-plane shear stress is 60.373 MPa. Problem 4 finds the normal and shear stresses on a 30 degree element are -85.2 MPa and 43.5 MPa respectively.

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    © All Rights Reserved
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    Mohr Circle & Buckling Task Example

    Uploaded by

    Muhammad Akbar Firmansyah
    37 views7 pages
    The document is a homework assignment on mechanics of materials containing 4 problems. The summary provides the solutions to each problem in the following 3 sentences: Problem 1 calculates the buckling load of a column as 25680.6 kips about the x axis and 16785.4 kips about the y axis. Problem 2 finds the allowable load on a column is 55.1 kN. Problem 3 determines the maximum principal stress is 106.24 MPa at an angle of 68.52 degrees and the maximum in-plane shear stress is 60.373 MPa. Problem 4 finds the normal and shear stresses on a 30 degree element are -85.2 MPa and 43.5 MPa respectively.

    Original Description:

    contoh tugas materi Buckling Dan Lingkaran Mohr, Mekanika Bahan

    Original Title

    Mohr Circle & Buckling task example

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    The document is a homework assignment on mechanics of materials containing 4 problems. The summary provides the solutions to each problem in the following 3 sentences: Problem 1 calculates the buckling load of a column as 25680.6 kips about the x axis and 16785.4 kips about the y axis. Problem 2 finds the allowable load on a column is 55.1 kN. Problem 3 determines the maximum principal stress is 106.24 MPa at an angle of 68.52 degrees and the maximum in-plane shear stress is 60.373 MPa. Problem 4 finds the normal and shear stresses on a 30 degree element are -85.2 MPa and 43.5 MPa respectively.

    Copyright:

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

    Mohr Circle & Buckling Task Example

    Uploaded by

    Muhammad Akbar Firmansyah
    The document is a homework assignment on mechanics of materials containing 4 problems. The summary provides the solutions to each problem in the following 3 sentences: Problem 1 calculates the buckling load of a column as 25680.6 kips about the x axis and 16785.4 kips about the y axis. Problem 2 finds the allowable load on a column is 55.1 kN. Problem 3 determines the maximum principal stress is 106.24 MPa at an angle of 68.52 degrees and the maximum in-plane shear stress is 60.373 MPa. Problem 4 finds the normal and shear stresses on a 30 degree element are -85.2 MPa and 43.5 MPa respectively.

    Copyright:

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

    Nama : Muhammad Akbar Firmansyah

    NIM : 20/464364/SV/18683
    Teknologi Rekayasa Pelaksanaan Bangunan Sipil

    TUGAS MEKANIKA BAHAN

    1.

    Diket : Column : 3 ft
    Est : 20
    σY : 60 ksi
    Ditanya : the load it can support so that
    the column does not buckle
    Jawab :

     (𝐾𝐿)𝑥 = 0,5(3𝑓𝑡)
    (𝐾𝐿)𝑥 = 1,5 𝑓𝑡
    (𝐾𝐿)𝑥 = 18 𝑖𝑛𝑐

     (𝐾𝐿)𝑦 = 0,7(3𝑓𝑡/2)
    (𝐾𝐿)𝑦 = 1,05 𝑓𝑡
    (𝐾𝐿)𝑦 = 12,6 𝑖𝑛𝑐

     𝐼𝑥 = 29,1 𝑖𝑛^4
     𝐼𝑦 = 9,32 𝑖𝑛^4
    𝜋 2 𝐸𝐼𝑥
     (𝑃𝑐𝑟)𝑥 = (𝐾𝐿)𝑥 2
    𝜋 2 . 29𝑥103 𝑘𝑠𝑖. 29,1 𝑖𝑛4
    (𝑃𝑐𝑟)𝑥 =
    18 𝑖𝑛2
    (𝑃𝑐𝑟)𝑥 = 25680,6 𝑘𝑖𝑝
    𝜋 2 𝐸𝐼𝑥
     (𝑃𝑐𝑟)𝑦 = (𝐾𝐿)𝑥 2
    𝜋 2 . 29𝑥103 𝑘𝑠𝑖. 9,32 𝑖𝑛4
    (𝑃𝑐𝑟)𝑦 =
    12,6 𝑖𝑛2
    (𝑃𝑐𝑟)𝑦 = 16785,4 𝑘𝑖𝑝

    (𝑃𝑐𝑟)𝑦
     𝜎𝑐𝑟 = 𝐴
    16785,4
    𝜎𝑐𝑟 =
    4,43 𝑖𝑛2
    𝜎𝑐𝑟 = 3789,0 𝑘𝑠𝑖

     𝑃𝑐𝑟 = 16785,4 𝑘𝑖𝑝

    2.

    Diket : Eal : 70 Gpa


    σY : 215 Mpa
    A : 7,5(10-3) m2
    Ix : 61,3(10-6) m4
    Iy : 23,2(10-6) m4
    F.S. : 3.0.
    Column : 8 m
    (W6x15)
    Ditanya : The largest allowable load P that can be applied.
    Jawab :
     (𝐾𝐿)𝑥 = 2. (8 𝑚)
    (𝐾𝐿)𝑥 = 16 𝑚

     (𝐾𝐿)𝑦 = 0,7. (8 𝑚)
    (𝐾𝐿)𝑦 = 5,6 𝑚
    𝜋 2 𝐸𝐼𝑥
     (𝑃𝑐𝑟)𝑥 = (𝐾𝐿)𝑥 2
    𝜋 2 . 70𝑥109 𝑁/𝑚2 . 61,3𝑥10−6 𝑚4
    (𝑃𝑐𝑟)𝑥 =
    16 𝑚2
    (𝑃𝑐𝑟)𝑥 = 165263,8 𝑁
    (𝑃𝑐𝑟)𝑥 = 165 𝑘𝑁
    𝜋 2 𝐸𝐼𝑥
     (𝑃𝑐𝑟)𝑦 = (𝐾𝐿)𝑥 2
    𝜋 2 . 70𝑥109 𝑁/𝑚2 . 23,2𝑥10−6 𝑚4
    (𝑃𝑐𝑟)𝑦 =
    5,6 𝑚2
    (𝑃𝑐𝑟)𝑦 = 510586,4 𝑁
    (𝑃𝑐𝑟)𝑦 = 0,5 𝑀𝑁

    By comparison, as P is increased the column will buckle about the x-x axis. The
    allowable load is therefore.

    𝑃𝑐𝑟
     𝑃𝑎𝑙𝑙𝑜𝑤 = 𝐹.𝑆.
    165
    𝑃𝑎𝑙𝑙𝑜𝑤 =
    3,0
    𝑃𝑎𝑙𝑙𝑜𝑤 = 55,1 𝑘𝑁

    𝑃𝑐𝑟
     𝜎𝑐𝑟 = 𝐴
    165 𝑘𝑁
    𝜎𝑐𝑟 =
    7,5(10−3 )𝑚2
    𝜎𝑐𝑟 = 22 𝑀𝑝𝑎 < 215 Mpa
    𝐸𝑢𝑙𝑒𝑟 ′ 𝑠𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑐𝑎𝑛 𝑏𝑒 𝑎𝑝𝑙𝑙𝑖𝑒𝑑
    3.

    Diket : σx : -31 MPa


    σY : 85 MPa
    τxy : 54 MPa
    Ditanya : Principal Stress, dan sudut ?
    In Plane Maximum shear stress, dan tegangan beserta sudutnya ?
    Jawab :
    Orientation of Elements
    54
     tan 2𝜃𝑝 = (−31−85)/2
    54
    tan 2𝜃𝑝 =
    −68
    tan 2𝜃𝑝 = −0,931034483

     2𝜃𝑝2 = −42,95˚
     𝜃𝑝2 = −21,47˚

     2𝜃𝑝 = 180˚ + 2𝜃𝑝2


    2𝜃𝑝 = 137,045˚
     𝜃𝑝 = 90˚ + 𝜃𝑝2
    𝜃𝑝 = 68,52˚

    Principal Stress
    𝜎𝑥+𝜎𝑦 𝜎𝑥−𝜎𝑦 2
     𝜎1,2 = ± √( ) + 𝜏𝑥𝑦 2
    2 2

    −31+85 −31−85 2
     𝜎1 = + √( ) + 542
    2 2
    𝜎1 = 106,24 𝑀𝑃𝑎
    −31+85 −31−85 2
     𝜎2 = − √( ) + 542
    2 2
    𝜎2 = −52,24 𝑀𝑃𝑎
    𝜎𝑥+𝜎𝑦 𝜎𝑥+𝜎𝑦
     𝜎𝑥 ′ = 2 + 2 𝑐𝑜𝑠2𝜃 + 𝜏𝑥𝑦20
    𝜎𝑥 ′ = −52,246 𝑀𝑃𝑎

    Orientation of Elements
    −(𝜎𝑥−𝜎𝑦)/2
     tan 2𝜃𝑝𝑠 = 𝜏𝑥𝑦
    −31 − 85
    −
    tan 2𝜃𝑝𝑠 = 2
    54
    tan 2𝜃𝑝𝑠 = 1,0740

     2𝜃𝑠2 = 47,04˚
     𝜃𝑠2 = 23,522˚

     2𝜃𝑠1 = 180˚ + 2𝜃𝑠2


    2𝜃𝑠1 = 227,045˚
     𝜃𝑠1 = 90˚ + 𝜃𝑠2
    𝜃𝑠1 = 113,522˚

    Max in-Plane Shear Stress

    𝜎𝑥+𝜎𝑦 2
     𝜏𝑚𝑎𝑥 = √( 2 ) + 𝜏𝑥𝑦 2

    −31 + 85 2
    𝜏𝑚𝑎𝑥 = √( ) + 542
    2
    𝜏𝑚𝑎𝑥 = 60,373 𝑀𝑃𝑎

    Average of Normal Stress


    𝜎𝑥+𝜎𝑦
     𝜎𝑎𝑣𝑔 = 2
    −31 + 85
    𝜎𝑎𝑣𝑔 =
    2
    𝜎𝑎𝑣𝑔 = 27 𝑀𝑝𝑎
    4.

    Diket : σx : -55 MPa


    σY : 120 MPa
    τxy : 43 MPa
    Ditanya : Principal Stress, dan sudut ?
    In Plane Maximum shear stress dan tegangan dengan sudut ?
    Tegangan normal dan geser pada sudut 35,6 ?
    Jawab :
    Construction of the Circle
    𝜎𝑥+𝜎𝑦
     σavg = 2
    −55 + 120
    σavg =
    2
    σavg = 32,5 𝑀𝑃𝑎
    55+32,5 2
     R = √( 2 ) + 432
    R = 97,5

    Lingkaran Mohr
    Stresses on 30˚ Element (check)
    43
     ϕ = 𝑡𝑎𝑛−1 87,5 = 33,5˚
     ψ = 60˚ − 33,5˚ = 26,5˚

    counterclockwise 30˚
     σx ′ = 2 − 97,5. cos (26,5˚) = −85,2 𝑀𝑃𝑎
     τx ′ y ′ = 97,5. sin (26,5˚) = 43,5 𝑀𝑃𝑎

    clockwise 30˚
     σx ′ = 2 − 97,5. cos (26,5˚) = 89,2 𝑀𝑃𝑎
     τx ′ y ′ = 97,5. sin (26,5˚) = −43,5 𝑀𝑃𝑎 (check)

    Tegangan normal dan geser pada sudut 36,6˚


    43
     ϕ = 𝑡𝑎𝑛−1 87,5 = 33,5˚
     ψ = 73,2˚ − 33,5˚ = 39,7˚

    counterclockwise 30˚
     σx ′ = 2 − 97,5. cos (39,7˚) = −73 𝑀𝑃𝑎
     τx ′ y ′ = 97,5. sin (39,7˚) = 62,27 𝑀𝑃𝑎

    clockwise 30˚
     σx ′ = 2 − 97,5. cos (39,7˚) = 77 𝑀𝑃𝑎
     τx ′ y ′ = 97,5. sin (39,7˚) = −62,27 𝑀𝑃𝑎 (check)

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