2-Lectures LEC 12

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    Chapter 5

    Deflection and StiffnessDeflection and Stiffness

    A. Aziz Bazoune

    03-Oct-07 Chapter 5 1

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    OutlineOutline

    1. Spring Rates2. Deflection in Tension, Compression & Torsion

    3. Deflection due to Bending

    4. Strain Energy5. Castiglianos Theorem

    6. Statically Indeterminate Problems

    7. Compression MembersLong Columns with Central Loading

    Intermediate Length Columns with Central Loading

    03-Oct-07 Chapter 5 2

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    55--10 Statically Indeterminate Problems10 Statically Indeterminate Problems

    A system in which the laws of statics are not sufficient to

    determine all the unknown forces or moments is said to bestatically indeterminate.

    03-Oct-07 Chapter 5 4

    The additional equations needed to solve for the unknownscome from equations pertaining to the deformation of the

    part.

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    Equilibrium:

    (a)

    The total force is resisted

    by a force F1 in spring 1plus the force F2 in spring2.

    1 2 0F F F F = =

    03-Oct-07 Chapter 5 5

    Since there are twounknowns and only oneequation , the system isstatically indeterminate.

    Figure 5-14

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    Write another equation.

    The two springs have the samedeformation.

    Thus we obtain the second equationas

    (b)or

    (c)

    1 2 = =

    1 2

    1 2

    F F

    k k=

    03-Oct-07 Chapter 5 6

    Solve Eq.(c) for F1 and substitute

    the result into (a) gives

    ( )1 1 2 2F k k F =

    2

    2

    1 2

    kF F

    k k=

    +

    1

    2

    1 2

    kF F

    k k=

    +and

    Notice that for othersituations relations betweendeformations may not be aseasy.

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    Procedure for general statically

    indeterminate problems

    ProcedureProcedure

    1. Choose redundant reaction (force or moment).2. Solve equation of equilibrium in terms of the redundants.3. Write equation for total energy U.

    03-Oct-07 Chapter 5 7

    4. Find expression for redundant reaction by taking .5. Solve resulting equation for reaction.6. Find rest of reactions using equilibrium equations.

    0i

    R =

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    ForFor thethe beambeam shown,shown, determinedetermine thethe supportsupport reactionsreactions usingusingsuperpositionsuperposition andand ProcedureProcedure 11 fromfrom SectionSection 55..1010..

    Example (Problem 5.57 Textbook)

    1. Choose RB as the

    redundant reaction.

    03-Oct-07 Chapter 5 8

    2. Force equilibriumgives

    0B C

    B C

    F R R wl

    R R wl

    = + =

    + =

    3. Moment equilibrium gives

    ( ) 02

    c B c

    lM R l a wl M

    = + =

    (b)

    (a)

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    Example (Problem 5.57 Textbook)

    4. By Superposition (You may use Castiglianos Theorem)

    03-Oct-07 Chapter 5 9

    4. By substituting

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    55--11 Compression Members11 Compression Members--GeneralGeneral

    03-Oct-07 Chapter 5 10

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    55--11 Compression Members11 Compression Members--GeneralGeneralColumnsColumns

    Straight, slender members loaded axially in compression.

    UsesUses

    03-Oct-07 Chapter 5 11

    .

    machine linkages, sign posts, support for highway overpasses.

    BucklingBucklingSudden large lateral deflection of a column due to a small increase in anexisting compressive load. This leads to instability and collapse of themember.

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    03-Oct-07 Chapter 5 12

    1. Long columns with central loading.2. Intermediate length columns with central loading3. Columns with eccentric loading.4. Struts or short columns with eccentric loading.

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    Critical LoadCritical Load

    The border between stabilityand instability occurs when anew equilibrium position is

    obtained:

    orP kl =

    ( ) 0P kl =

    03-Oct-07 Chapter 5 13

    for any displacement .

    This condition is referred toasneutral equilibrium.

    From the foregoingexpression we define thecritical loadas

    .

    crP k l= Figure 5-17

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    Physically, represents the load for which the system is on the vergeof buckling.

    Clearly, the system is in stable equilibrium for and in unstableequilibrium for .

    crP P

    03-Oct-07 Chapter 5 14

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    55--11 Long Columns with Central Loading11 Long Columns with Central Loading

    Figure 5-18

    (a) Both ends rounded or pivoted

    03-Oct-07 Chapter 5 15

    (b) Both end fixe(c) One end free(d) One end rounded and pivoted

    and one end fixed

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    55--11 Long Columns with Central Loading11 Long Columns with Central Loading

    Referring to Figure 5-18 (a), assume a bar of length

    loaded by a force acting along the centroidal axis on

    rounded or pinned ends.

    The bar is bent in the positive direction. This

    P

    l

    y

    03-Oct-07 Chapter 5 16

    ,

    Remember Eq. (5-12)

    Equating between the two above equations gives

    2

    2

    d yM EI

    dx

    =

    2

    2 0

    d y Py

    dx EI

    + =

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    The previous equation represents a second order ordinary differentialequation generally known for simple harmonic motion . It can be writtenas

    where is known as the frequency of oscillations.

    The eneral solution of the revious differential e uation is

    2

    2

    2 0

    d yy

    dx

    + =

    2P EI =

    03-Oct-07 Chapter 5 17

    where A and B are constants of integration must determined from the

    B.Cs

    The first B.C. yields B=0, and the second leads to

    ( ) ( )sin cosy A x B x = +

    ( )( )

    0 0

    0

    y

    y l

    =

    =

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    The foregoing is satisfied if either or . The first of these corresponds to a condition of no buckling and yields a trivial

    solution. The second case is satisfied if

    or

    ( )sin 0A l =

    0A =

    ( ) ( )1,2,3,l n n = =

    ( )sin 0l =

    03-Oct-07 Chapter 5 18

    Solving for gives

    ( )1,2,3,l n nEI = =

    P

    ( )

    2 2

    2 1,2,3,n EI

    P nl

    = =

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    The value of has a physical significance, as it determines thesmallest value of for which a buckled shape or mode can occur understatic loading. Therefore, the critical load for a column with pinned/rolled end is

    P

    ( )2

    2 1,2,3,

    cr

    EIP n

    l

    = =

    Euler FormulaEuler Formula

    ( )1n=

    03-Oct-07 Chapter 5 19

    The preceding result, after L. Euler (1707-1783), is known as EulersFormula; the corresponding load is called the Euler Buckling Load.

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    Leonhard Euler

    Portrait by Johann Georg Brucker

    BornApril 15, 1707(1707-04-15)Basel, Switzerland

    DiedSeptember 18 [O.S. September 7] 1783St Petersburg, Russia

    Leonhard Paul Euler (pronounced Oiler; IPA [l]) (April

    15, 1707 September 18 [O.S. September 7] 1783) was apioneering Swiss mathematician and physicist, who spentmost of his life in Russia andGermany. He published more papers than any othermathematician of his time.[2]

    Euler made important discoveries in fields as diverse as

    calculus and graph theory. He also introduced much of themodern mathematical terminology and notation, particularlyfor mathematical analysis, such as the notion of a mathematical function.[3] He is alsorenowned for his work in mechanics, optics, and astronomy.

    http://en.wikipedia.org/wiki/Leonhard_Euler

    03-Oct-07 Chapter 5 20

    ResidencePrussiaRussiaSwitzerland

    Nationality Swiss

    Field Mathematics and physics

    InstitutionsImperial Russian Academy of Sciences

    Berlin Academy

    Alma mater University of Basel

    Religion Calvinist[1]

    u er s cons ere o e e preem nen ma ema c an o

    the 18th century and one of the greatest of all time. He isalsoone of the most prolific; his collected works fill 6080 quartovolumes.[4] A statement attributed to Pierre-Simon Laplaceexpresses Euler's influence on mathematics: "Read Euler,read Euler, he is the teacher (master) of us all".[5]

    Euler was featured on the sixth series of the Swiss 10-francbanknote[6] and on numerous Swiss, German, and Russianpostage stamps. The asteroid 2002 Euler was named in hishonor. He is also commemorated by the Lutheran Church ontheir Calendar of Saints on May 24.

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    QUESTIONS ?QUESTIONS ?

    03-Oct-07 Chapter 5 21