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    Fe Strength of Materials 0930

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    fem mece3381
    This document provides an overview of mechanics of materials and fundamentals of engineering concepts. It covers topics such as: 1) The relationship between shear and moment diagrams, with the slope of the moment diagram equaling the ordinate of the shear diagram. 2) Analyzing composite cross sections using a transformed section approach. 3) Calculating stresses and transforming stresses to new coordinate systems, including determining principal stresses and shear stresses. 4) Finding principal planes and shear planes using equations relating normal and shear stresses.

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    Fe Strength of Materials 0930

    Uploaded by

    fem mece3381
    16 views31 pages
    This document provides an overview of mechanics of materials and fundamentals of engineering concepts. It covers topics such as: 1) The relationship between shear and moment diagrams, with the slope of the moment diagram equaling the ordinate of the shear diagram. 2) Analyzing composite cross sections using a transformed section approach. 3) Calculating stresses and transforming stresses to new coordinate systems, including determining principal stresses and shear stresses. 4) Finding principal planes and shear planes using equations relating normal and shear stresses.

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    This document provides an overview of mechanics of materials and fundamentals of engineering concepts. It covers topics such as: 1) The relationship between shear and moment diagrams, with the slope of the moment diagram equaling the ordinate of the shear diagram. 2) Analyzing composite cross sections using a transformed section approach. 3) Calculating stresses and transforming stresses to new coordinate systems, including determining principal stresses and shear stresses. 4) Finding principal planes and shear planes using equations relating normal and shear stresses.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    16 views31 pages

    Fe Strength of Materials 0930

    Uploaded by

    fem mece3381
    This document provides an overview of mechanics of materials and fundamentals of engineering concepts. It covers topics such as: 1) The relationship between shear and moment diagrams, with the slope of the moment diagram equaling the ordinate of the shear diagram. 2) Analyzing composite cross sections using a transformed section approach. 3) Calculating stresses and transforming stresses to new coordinate systems, including determining principal stresses and shear stresses. 4) Finding principal planes and shear planes using equations relating normal and shear stresses.

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    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 31

    Mechanics of Materials

    Fundamentals of Engineering
    Review Course

    Fall, 2020

    Dr. S. Moorthy (moorthy@lsu.edu)


    -w

    -w dx
    V SLOPE OF M DIAGRAM EQUALS THE
    POSITIVE OF THE ORDINATE OF THE
    SHEAR DIAGRAM

    Vdx CHANGE IN M EQUALS THE POSITIVE OF THE


    AREA UNDER THE SHEAR DIAGRAM
    MC
    MC
    FC

    FC

    IF THERE ARE CONCENTERATED FORCES/MOMENTS ACTING AT A POINT,

    𝑉 + ∆𝑉 = 𝑉 + 𝐹𝐶 ↑
    𝑀 + ∆𝑀 = 𝑀 + 𝑀𝐶 ( )
    1350

    8 12 16
    V
    6.75’ -250

    -850
    For a composite X-section
    30
    Steel, Est=30x103 ksi, n2= 3 =10,sall=25 ksi
    6n2=60”
    6”

    1” 0.818” 1”
    60P
    60P
    6” 6”

    3
    1” Timber, Esim=3x103 ksi, n1=3=1,sall=4 ksi 1”

    Original X-section Transformed X-section

    1
    60 × 2 + 6 × 4 60 × 13 1 × 63
    2
    𝑦= = 0.818, 𝐼 = + 60 × (0.818 − 0.5) + + 6 × (4 − 0.818)2 = 89.82 𝑖𝑛4
    60 + 6 12 12

    60𝑃 × (6 − 0.818) 60𝑃 × (0.818)


    𝑛1 =4 𝑃 = 1.15 𝑘𝑖𝑝, 𝑛2 = 25 𝑃 = 4.57 𝑘𝑖𝑝
    89.82 89.82

    𝑃 = 1.15 𝑘𝑖𝑝 (𝑡𝑖𝑚𝑏𝑒𝑟 𝑔𝑜𝑣𝑒𝑟𝑛𝑠)


    34

    -9
    -22
    STRESS TRANSFORMATION

    𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦 𝜎𝑥 − 𝜎𝑦
    𝜎= + cos 2𝜃 + 𝜏𝑥𝑦 sin 2𝜃 , 𝜏 = − sin 2𝜃 +𝜏𝑥𝑦 cos(2𝜃)
    2 2 2
    y
    56.2 MPa y’

    -38.2 MPa
    =
    x
    x’
    25o

    86.2 MPa

    𝜎𝑥 = −60 (𝑀𝑃𝑎), 𝜎𝑦 = 90 𝑀𝑃𝑎 , 𝜏𝑥𝑦 = 30 (𝑀𝑃𝑎)

    −60 + 90 −60 − 90
    𝜎𝑥 ′ = + cos 2 × −25 + 30 sin −50 = −56.2 (𝑀𝑃𝑎)
    2 2

    −60 + 90 −60 − 90
    𝜎𝑦 ′ = + cos 2 × 65 + 30 sin 130 = 86.2 (𝑀𝑃𝑎)
    2 2

    −60 − 90
    𝜏𝑥′𝑦′ = − sin −50 +𝜏𝑥𝑦 cos −50 = −38.2 (𝑀𝑃𝑎)
    2
    PRINCIPAL (NORMAL) VALUES/PLANES

    𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦 𝜎𝑥 − 𝜎𝑦
    𝜎= + cos 2𝜃 + 𝜏𝑥𝑦 sin 2𝜃 , 𝜏 = − sin 2𝜃 +𝜏𝑥𝑦 cos(2𝜃)
    2 2 2
    y
    95.2 MPa

    =
    x
    10.8

    125.2 MPa

    𝜎𝑥 = −60 (𝑀𝑃𝑎), 𝜎𝑦 = 90 𝑀𝑃𝑎 , 𝜏𝑥𝑦 = 30 (𝑀𝑃𝑎)

    30
    tan(2𝜃𝑃 ) = 2𝜃𝑃 = −21.8 𝑜𝑟 180 + −21.8 𝜃𝑝 = −10.9 𝑜𝑟 79.1
    −60 − 90
    2
    −60 + 90 −60 − 90
    𝜎1 = 𝜎𝜃=−10.9 = + cos 2 × −10.9 + 30 sin −21.8 = −95.2(𝑀𝑃𝑎)
    2 2

    −60 + 90 −60 − 90
    𝜎2 = 𝜎𝜃=79.1 = + cos 2 × 79.1 + 30 sin 158.2 = 125.2(𝑀𝑃𝑎)
    2 2

    −60 − 90
    𝜏𝜃=−10.9,79.1 = − sin 2𝜃 +𝜏𝑥𝑦 cos 2𝜃 = 0
    2
    PRINCIPAL SHEAR VALUES/PLANES
    𝜎𝑥 + 𝜎𝑦 𝜎𝑥 − 𝜎𝑦 𝜎𝑥 − 𝜎𝑦
    𝜎= + cos 2𝜃 + 𝜏𝑥𝑦 sin 2𝜃 , 𝜏 = − sin 2𝜃 +𝜏𝑥𝑦 cos(2𝜃)
    2 2 2
    y

    34.1o
    =
    x 110.2 MPa

    15 MPa
    15 MPa

    𝜎𝑥 = −60 (𝑀𝑃𝑎), 𝜎𝑦 = 90 𝑀𝑃𝑎 , 𝜏𝑥𝑦 = 30 (𝑀𝑃𝑎)

    −60 − 90
    −
    tan(2𝜃𝑠 ) = 2 2𝜃𝑠 = 68.2 𝑜𝑟 180 + 68.2 𝜃𝑝 = 34.1 𝑜𝑟 124.1
    30

    −60 + 90 −60 − 90
    𝜎𝜃=34.1 = + cos 2 × 34.1 + 30 sin 68.2 = 15 (𝑀𝑃𝑎)
    2 2

    −60 + 90 −60 − 90
    𝜎𝜃=124.1 = + cos 2 × 124.1 + 30 sin 248.2 = 15 (𝑀𝑃𝑎)
    2 2

    −60 − 90
    𝜏1/2 = 𝜏𝜃=34.1/124.1 = − sin 2𝜃 +𝜏𝑥𝑦 cos 2𝜃 = ±110.2 (𝑀𝑃𝑎)
    2
    The End

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