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445.204 Introduction to Mechanics of Materials (재료역학개론) Chapter 10: Deflection of beam (Ch. 12 in Shames) Myoung-Gyu Lee, 이명규 Tel. 880-1711; Email: [email protected] TA: Chanmi Moon, 문찬미 Lab: Materials Mechanics lab.(Office: 30-521) Email: [email protected]

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  • 445.204

    Introduction to Mechanics of Materials

    (재료역학개론)

    Chapter 10: Deflection of beam

    (Ch. 12 in Shames)

    Myoung-Gyu Lee, 이명규

    Tel. 880-1711; Email: [email protected]

    TA: Chanmi Moon, 문찬미

    Lab: Materials Mechanics lab.(Office: 30-521)

    Email: [email protected]

  • Contents

    - Differential equations for deflection of symmetric beams- Statically indeterminate beams- Superposition methods- Energy method

    2

  • Differential equations

    • Recall pure bending

    • For general loading, the above equation gives local radius of curvature of the neutral surface

    • Normal stress (or normal strain) results in deflection of the beam

    • Radius of curvature as a function of deflection, from analytic geometry

    zz

    z

    EIRM

    =

    ( )

    22

    3/22

    1

    1

    d vdx

    R dvdx

    κ= = +

    2

    2

    1 d vR dx=

    Simplification/small deformation

  • Differential equations

    • Basic differential equations

    • From

    • Therefore,

    zz

    z

    EIRM

    =2

    2

    1 d vR dx=&

    2

    2z

    zz

    Md vdx EI

    =2

    2zz zd vEI Mdx

    =or

    zy

    dM Vdx

    = and y ydV

    wdx

    = −

    2

    2z

    yd M wdx

    = −

    22 2

    2 2 2z

    zz yd Md d vEI w

    dx dx dx

    = = −

    4

    4zz yd vEI wdx

    = −or

  • Differential equations: example 12.1

    2

    2z

    zz

    Md vdx EI

    =

    • Boundary conditions

    0dvvdx

    = = at x L=

  • Differential equations: example 12.2

    ( ) ( )1

    1

    nn

    n

    P x aP x a dx C

    n

    +−− = +∫

    +

  • Differential equations: example 12.2

  • Differential equations: example 12.2

    Boundary conditions

  • Differential equations: example 12.2

  • Differential equations: example 12.4

    Pin connected beam

    Q: Find max. deflection

    Solution step

    1. Apply moment equilibrium at the position of pinAnd determine the reaction force at A2. Obtain bending moment along the beam axisNote: divide the region at the pin position3. Apply boundary conditionsBCs. 0 deflection at x=0 & x=9, zero slope at x=9

    Apply patch condition.

  • Differential equations: example 12.4

    Pin connected beam

    Q: Find max. deflection

    Solution step

    1. Apply moment equilibrium at the position of pinAnd determine the reaction force at A2. Obtain bending moment along the beam axisNote: divide the region at the pin position3. Apply boundary conditionsBCs. 0 deflection at x=0 & x=9, zero slope at x=9

    Apply patch condition.

  • Differential equations: example 12.4

    Pin connected beam

    Q: Find max. deflection

  • Statically indeterminate problem: example 12.5

    Solution step

    1. Calculate bending moment as a function of x with unknown R1 (considered as redundant constant)

    2. Obtain deflection and slope for each region between 0~5, and 5~10

    3. Apply proper boundary condition at x=0, x=10, and patch conditions at x=5

    4. Obtain additional unknowns for R2 and M2 at the wall (or x=10)

  • Statically indeterminate problem: example 12.5

  • Superposition method

    Homework!!!!Derive the left equations!

  • Superposition method

  • Energy method

    • Strain energy related 1) normal stress, 2) transverse shear stress• For normal stress

    • For shear stress resulting from Vy

    zxx

    zz

    M yI

    σ = −

    2 2 2 2* 2

    0 02 22 2 2L Lxx z z

    V A Azz zz

    M y MU dv dA dx y dA dxE EI EI

    σ = = =∫ ∫ ∫ ∫ ∫

    2*

    0 2L z

    zz

    MU dxEI

    = ∫

    2*

    2xy

    VU dvGτ

    = ∫

  • Energy method

    • Shear stress may vary across the cross section as a function of both y and z

    • Therefore, instead of determining the exact distribution of shear stress, use “tabulated shape factors”

    • The shape factors correlate the energy computed using actual shear stress distribution with that of a cross section with equal area but with a constant distribution of shear stress

    2 2*

    2 2xy actual xy simplified

    xy xyyV VU dv dvG G

    τ τ

    τ τα

    −> −>

    = =∫ ∫

    2 2*

    0 202 2

    LL y y

    y yA

    V VU dA dx dx

    GA GAα α

    = =∫ ∫ ∫

    xy simplified

    yVAτ −>

    =

  • Energy method – Example 12.11

  • Energy method

  • Energy method- Example 12.12

  • Homework by yourself

    Solve examples 12.3, 12.6, 12.8, 12.13, 12.14

    22

    슬라이드 번호 1ContentsDifferential equationsDifferential equationsDifferential equations: example 12.1Differential equations: example 12.2Differential equations: example 12.2Differential equations: example 12.2Differential equations: example 12.2Differential equations: example 12.4Differential equations: example 12.4Differential equations: example 12.4Statically indeterminate problem: example 12.5Statically indeterminate problem: example 12.5Superposition methodSuperposition methodEnergy methodEnergy methodEnergy method – Example 12.11Energy methodEnergy method- Example 12.12Homework by yourself