Static Equilibrium - · PDF fileStatic equilibrium: Sum of vertical and horizontal forces must...

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Transcript of Static Equilibrium - · PDF fileStatic equilibrium: Sum of vertical and horizontal forces must...

  • Static EquilibriumStatic equilibrium requires a balance of forces and a balance of moments.

    0 0 0

    0 0 0

    ===

    ===

    zyx

    zyx

    MMM

    FFF

    Example 1:

    A painter stands on a ladder that leans against the wall of a house at an angle of 20o. Assume that the painter is a midheight of the ladder. Calculate the minimum coefficient of friction for static equilibrium.

  • 0)( 0

    21

    1221

    =++====

    PmmPFPPPPF

    personladdery

    x

    ++

    =+

    +=

    1)(

    1

    )(21

    gmmP

    gmmP plpl

    0)20(2

    )()20()20( 21 =+ SinLgmmLCosPLSinP pl

    1763.0=

    Static equilibrium: Sum of vertical and horizontal forces must be zero.

    Taking the moment about point O and setting it equals to zero:

  • Load Analysis 3-D

    Newton First Law: A body at rest tend to remain at rest and a body in motion at constant velocity will tend to maintain that velocity unless acted upon by an external force.

    Newton Second Law: The time rate of change of momentum of a body is equal to the magnitude of applied force and acts in the direction of the force.

    Newton Third Law: when two particles interact, a pair of equal and opposite reaction forces will exist at their contact point. This force pair will have the same magnitude and act along the same direction line but have opposite sense.

  • Newtons second law can be written for a rigid body in two forms, one for linear forces and one for moments or torques

    GG HMmaF&=

    = F=Force; m=mass, a= accelerationMG=moment about the center of gravity

    GH& =time rate of change of the moment or the angular momentum about the CG

    zzyyxx maFmaFmaF ===

    kIjIiIH zzyyxxG ++=

    yxyxzzz

    xzxzyyy

    zyzyxxx

    IIIM

    IIIM

    IIIM

    )(

    )(

    )(

    =

    =

    =

    If the x, y and z axes are chosen to coincide with the principal axes of inertia of the body. Where Ix, Iy and Iz are the principal centroidal mass moments of inertia (second moments of mass). In 3-D

    Eulers Equations

  • For 2-D:

    yyxx maFmaF == zzz IM =Where z is the angular acceleration

    Moment of a Force about an axis

  • 3-D Equilibrium Example:Two transmission belts pass over sheaves welded to an axle supported by bearings at B and D. The radius at A = 2.5 and at C = 2. The axle rotates at constant speed. Find T and the reaction forces at B, D. Assume that bearing at D exerts no axial thrust and neglect weights of sheaves and axle.

  • 8in

    6in6in

  • x

    y

    A B C D33.75lb 33.75lb67.5lb

    33.75lb

    33.75lb

    202.5lb-in

  • x

    z

    A B C D70lb 28lb42lb

    28lb

    42lb

    336lb-in

  • DefinitionsCentroid of Area (Center of Gravity of an area): Point that defines the geometric center of the area

    ( )yx,

    A

    Axx

    A

    Ayy AA

    ==

    Q axis-y therespect towith

    Areaan ofMoment First

    Q axis- x therespect towith

    Areaan ofMoment First

    y

    x

    AxAx

    AyAy

    A

    A

    ===

    ===

    IbVQ

    =

  • Example 2:

    Calculate the center of gravity of the rectangle:

    (a)Without a hole

    (b)With a hole of dimensions cand d

    a) Without a hole

    22

    22

    1

    1

    aba

    baa

    xA

    Axx

    A

    Axx

    bba

    bab

    yA

    Ayy

    A

    Ayy

    n

    iiA

    n

    iiA

    =

    ===

    =

    ===

  • b) With a hole

    dcba

    dccebaa

    xA

    Axx

    A

    Axx

    dcba

    dcdfbab

    yA

    Ayy

    A

    Ayy

    n

    iiA

    n

    iiA

    ===

    +

    ===

    )2

    (2

    )2

    (2

    1

    1

    Second Moment or Moment of Inertia of an Area

    ==A

    yA

    x AxIAyI 22

    +==A

    yxO IIAJ 2

    Rectangular moments of inertia

    Polar moments of inertia

  • Radius of Gyration

    2222

    2

    2

    yxOOO

    yy

    xx

    rrrArJ

    ArI

    ArI

    +==

    =

    =

    Example 3:Find the moment of inertia of the circular area about the x and y axes, the polar moment of inertia and the radius of gyration about the x and y axes. CosrxSinry == yCosrArea = 2

    = Cosry

    2

    4

    42

    44

    2

    2

    4224

    rIIJrI

    rCosSinrI

    yxzy

    x

    =+=

    =

    ==

  • Parallel-Axis Theorem

    2' yxx dAII +=

    Example 4:

    ( )4

    '

    224

    2'

    4

    2516

    r4 4

    4

    r.I

    r rdAII

    rI

    x

    yxx

    x

    =

    +

    =+=

    =

  • IMy

    =

  • Mass Moment of Inertia

    It is the product of the elements mass and the square of the elements distance from the axis.

    ( )( )( )

    +=

    +=

    +=

    amz

    amy

    amx

    myxI

    mzxI

    mzyI

    22

    22

    22

  • Classification with respect to the method of application:

    (a)Normal tensile

    (b)Normal compressive

    (c)Shear

    (d)Bending

    (e)Torsion

    (f) Combined

    Load Classification and Sign Convention

  • The sign convention that will be used here is as follows:

    deflectionx

    => 02

    2

  • Distributed loads, Shear Force and

    Bending Moment in Beams

  • 4

    4

    3

    3

    2

    2

    2

    2

    xEI-qq

    xV

    xEIVV

    xM

    xEIM

    EIM

    x

    ==

    ==

    ==

  • Successive Integration Method

    1

    2

    1

    2600)(

    600)(

    )()(

    CxxV

    CxxxV

    xxqxV

    qxV

    +=

    +=

    =

    =

    2/3m

    =650N1050N=

    10

  • NVthenmxFor

    xxV

    CVxFor

    CxxV

    xFor

    1050)2(_____2__

    1502

    600)(

    150__150500350)1(__1__2

    600)(

    21__

    22

    2

    2

    ==

    +=

    ====

    +=

  • 0)2(____2__500__550)1(____1__

    150100)(

    1502

    600)()(

    1502

    600)(

    21__

    4

    43

    2

    2

    =====

    ++=

    +==

    +=

  • Normal Stress and Strain:Where F: force, normal to the cross-sectional area,

    A0: original cross-sectional area 0AF

    =

    1Pa = 1 N.m-2; 1MPa = 106Pa; 1GPa=109Pa

    Normal strain or tensile strain or Axial strain

    00

    0

    ll

    lll =

    =

  • Hookes LawWhen strains are small, most of materials are linear elastic.

    E

    Normal: =

    Youngs modulus

    EAlFl

    o

    o

    =

    Springs: the spring rate

    o

    o

    lEA

    lFk =

    =

  • Torsion Loading resulting from the twist of a shaft.

    lr

    zrZ

    =,

    lrG

    zrGG zz

    == ,,

    G = Shear Modulus of Elasticity

    Shear strain

    Twist Moment or Torque

    ==A zA z

    Arl

    GArT 2,,

    = A ArJ 2Area Polar Moment

    of InertiaJGlT

    lJGT

    =

    = or

    JrT

    z

    ,Thus: JrT o

    Max

    Angular spring rate:

    lGJTka

    ==

  • We can obtain the normal and shear stresses from flexure and shear formulas

    Where: is the normal stress acting on the cross section, M is the bending moment, y is the distance from the neutral axis and I is the moment of inertia of the cross sectional area with respect to

    the neutral axis. is the shear stress at any point in the cross section, V is the shear force,Q is the first moment of the cross sectional area outside of the point in the cross section where the stress is being found, and b is the width of the cross section.

    IMy

    =

    Maximum Stresses in Beams

    IbVQ

    =

  • The normal stresses obtained from the flexure formula have their maximum values at the farthest distance from the neutral axis. The normal stresses are calculated at the cross section of maximum bending moment.

    The shear stress obtained from the shear formula usually have their highest value at the neutral axis. The shear stresses are calculated at the cross section of maximum shear force. In most circumstances, these are the only stresses that are needed for design purposes. However to obtain a more complete picture of the stresses, we will need to determine the principal stresses and maximum shear stresses at various points in the beam.

  • Consider a simple rectangular beam below, and a cross section to the left of the load. Points A and E are at the top and bottom of the beam. Point C is in the midheight of the beam and points B and Dare in between.If Hookes law applies (linear elasticity), the normal and shear stresses at each of these five points can be readily calculated from the flexure and shear formulas.All the elements of vertical and horizontal faces, are in plane stress, because there is no stresses acting perpendicular to the plane of the figure.

    Beams of Rectangular Cross Section

  • Stresses in a beam of rectangular cross section:

    (a) simple beam with points A, B, C, D, and E on the side of the beam;

    (b) normal and shear stresses acting on stress elements at points A, B, C, D, and E;

    (c) principal stress; and (d) maximum shear stresses.

    Points A and E elements are in uniaxialcompressive and tensile stresses respectively. Point C (neutral axis) element is in pure shear. Points B and Delements have both normal and shear stresses.

  • We may use either the