tam 251 equation sheetmechref.engr.illinois.edu/sol/formulasheet.pdftam 251 equation sheet page 2...

Click here to load reader

Post on 22-Jan-2021

4 views

Category:

Documents

0 download

Embed Size (px)

TRANSCRIPT

  • TAM 251 EQUATION SHEET

    Main Equations

    Stress σavg =F

    Aτavg =

    V

    A

    Strain �eng =δ

    L0�true = ln

    LfL0

    γ =δxLy

    +δyLx

    Constitutive Relations σ = E� τ = Gγ

    Material Properties (Isotropic) ν = − �lat�long

    G =E

    2(1 + ν)

    Thermal Expansion �th = α∆T δth = αL0∆T

    Axial Loading δ =F L0EA

    σ =F

    A

    Torsion φ =T L0GJ

    τ =T ρ

    Jγ =

    φρ

    L0

    Stiffness and Flexibility kaxial =EA

    L0=

    1

    faxialktorsion =

    GJ

    L0=

    1

    ftorsion

    Bending σ =Mc

    I

    Thin-Walled Pressure Vessels σh =pr

    tσa =

    pr

    2t

    Transverse Shear τ =V Q

    I tq =

    V Q

    I

    Miscellaneous

    Distributed Loads, Shear,& Bending Moments

    dV

    dx= −w dM

    dx= V

    Inclined Plane: Normal Stress σn = σx cos2 θ + 2τxy sin θ cos θ + σy sin2 θ

    Inclined Plane: Shear Stress τn,s = (σy − σx) sin θ cos θ + τxy(cos2 θ − sin2 θ

    )

    Tresca Criterion

    |σ1| = σyield, |σ2| = σyieldwhen σ1, σ2 have thesame sign

    |σ1 − σ2| = σyieldwhen σ1, σ2 have theopposite sign

    Von-Mises Criterion σ21 − σ1σ2 + σ22 = σ2yield

    TAM 251 Equation Sheet Page 1 Apr. 2019TAM 251 Equation Sheet Page 1 Apr. 2019TAM 251 Equation Sheet Page 1 Apr. 2019

  • Stress Transformations

    Plane Stress

    σx′ =σx + σy

    2+σx − σy

    2cos(2θ) + τxy sin(2θ)

    σy′ =σx + σy

    2− σx − σy

    2cos(2θ) − τxy sin(2θ)

    τx′y′ = −σx − σy

    2sin(2θ) + τxy cos(2θ)

    Mohr’s Circle σavg =σx + σy

    2R =

    √(σx − σy

    2

    )2+ τ 2xy

    Principal Stresses σ1 = σavg + R σ2 = σavg − R τmax = R

    Plane Orientations tan 2θp =2τxy

    σx − σytan 2θs = −

    σx − σy2τxy

    Moments and Geometric Centroids

    Q = yA Ix =

    ∫A

    y2 dA Jo =

    ∫A

    ρ2dA y =1

    A

    ∫A

    ydA

    Rectangle x

    y

    b

    h Ix =1

    12bh3

    Circle

    ry

    x Ix =π

    4r4 Jz =

    π

    2r4

    Semicircle x

    y

    x`

    ryIx =

    π

    8r4 Ix′ =

    8− 8

    )r4 y =

    4r

    Parallel Axis Theorem Ic = Ic′ + Ad2cc′

    Buckling

    Critical Load Pcr =π2EI

    (Le)2

    pinned-pinned Le = L

    pinned-fixed Le = 0.7L

    fixed-fixed Le = 0.5L

    fixed-free Le = 2L

    TAM 251 Equation Sheet Page 2 Apr. 2019TAM 251 Equation Sheet Page 2 Apr. 2019TAM 251 Equation Sheet Page 2 Apr. 2019

  • Beam Deflection

    DiagramMax. Deflection Slope at End Elastic Curve

    ymax θ y(x)

    y P

    xymax

    L

    −P L3

    3EI−P L

    2

    2EIy(x) =

    P

    6EI

    (x3 − 3Lx2

    )

    y

    x

    w

    ymaxL

    −wL4

    8EI−wL

    3

    6EIy(x) = − w

    24EI

    (x4 − 4Lx3 + 6L2x2

    )

    y

    ymaxx

    L

    M

    −ML2

    2EI−MLEI

    y(x) = − M2EI

    x2

    yw

    ymaxx

    L

    − 5wL4

    384EI± wL

    3

    24EIy(x) = − w

    24EI

    (x4 − 2Lx3 + L3x

    )

    yL2

    PL2

    ymaxx −

    P L3

    48EI± P L

    2

    16EI

    For 0 ≤ x ≤ L2

    y(x) =P

    48EI

    (4x3 − 3L2x

    )

    by

    x

    P

    ymax

    a

    A B

    Lxm

    For a > b

    −P b(L2 − b2

    )3/29√

    3EIL

    at xm =

    √L2 − b2

    3

    θA = −P b(L2 − b2

    )6EIL

    θB = +P a

    (L2 − a2

    )6EIL

    For x < a :

    y(x) =P b

    6EIL

    [x3 − x

    (L2 − b2

    )]For x = a :

    y = −P a2 b2

    3EIL

    yM

    ymax

    A B

    Lxm

    x

    − ML2

    9√

    3EI

    at xm =L√3

    θA = −ML

    6EI

    θB = +ML

    3EI

    y(x) =M

    6EIL

    (x3 − L2x

    )

    TAM 251 Equation Sheet Page 3 Apr. 2019TAM 251 Equation Sheet Page 3 Apr. 2019TAM 251 Equation Sheet Page 3 Apr. 2019