Statically Inde. Structures

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    ANALYSIS OF INDETERMINATESTRUCTURES

    1.IntroductionA structure of any type is classified as statically

    indeterminate when the number of unknown reaction or

    internal forces exceeds the number of equilibrium

    equations available for its analysis.

    What is statically DETERMINATE structure? (Fig.1)

    No. of unknown = 3

    No. of equilibrium equations = 3

    3 = 3 thus statically determinate

    Fig.1

    No. of unknown = 4

    No. of equilibrium equations = 3

    43 thus statically Indeterminate

    Fig.2

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    Fig.3

    No. of unknown = 6

    No. of equilibrium equations = 5

    6 5 thus statically Indeterminate

    2.Why we study indeterminate structure ? Most of the structures designed today are statically indeterminate

    Reinforced concrete buildings are considered in most cases as

    statically indeterminate structures since the columns & beams are

    poured as continuous member through the joints & over the supports

    More stable compare to determinate structure or in another wordsafer.

    In many cases more economical than determinate.

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    Contrast

    Determinate Structure Indeterminate Structure

    PL3 /48EIP

    PL3 /192EIP

    Considerable compared to

    indeterminate

    structure

    DEFLECTION

    PL/4

    P PL/8 PSTRESS

    High moment caused thicker

    member & more material needed

    Less moment, smaller cross

    section& less material needed

    TEMPERATUREP

    No effect & no stress would be

    Developed in the beam

    Serious effect and stress would

    be developed in the beam

    P

    3

    Generally smaller than

    determinate structure

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    EQUILIBRIUM

    Equilibrium issatisfied when the

    reactive forces

    hold the structure

    at rest

    The force -

    displacement

    requirement depend

    upon the way the

    material responds,

    which assumedlinear-elastic

    response

    FORCE-DISPLACEMENT

    COMPATIBILITY

    Compatibility is

    satisfied when thevarious segments

    of the structure fit

    together without

    intentional breaks

    or overlaps

    Methods of Analyses

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    Methods of Analysis

    Two different Methods are available

    Force method

    known as consistent deformation, unit load method,

    flexibility method, or the superposition equations method.

    The primary unknowns in this way of analysis are forces

    Displacement method Known as stiffness matrix method

    The primary unknowns are displacements

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    Force method of analysisThe deflection or slope at any point on a structure as a result of a

    number of forces, including the reactions, is equal to the algebraic

    sum of the deflections or slopes at this particular point as a result

    of these loads acting individually

    General Procedure Indeterminate to thefirst degree

    1 Compatibility equation

    is needed

    Choosing one of the

    support reaction as a redundant

    The structure become statically

    determinate & stable

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    Downward displacementB at B calculated (load action)

    BB upward deflection per unit force at

    B

    Compatibility equation0 =B + ByBB

    Reaction By known

    Now the structure is statically

    determinate

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    General Procedure

    Indeterminate to the first

    degree

    1 Compatibility equation isneeded

    Choosing MA at A as a

    redundant

    The structure become statically

    determinate & stable

    Rotation A at A caused by load P is

    determined

    AA rotation caused by unit couple

    moment applied at A

    Compatibility equation

    0 = A + MA AA

    Moment MA known

    Now the structure is statically determinate 8

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    General Procedure Indeterminate to the 2nd degree

    2 Compatibility equations needed

    Redundant reaction B & C

    Displacement B &C caused by load P1 & P2 are determined

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    BB & BC Deflection per unit

    force at B are determined

    CC &CB Deflection per unit

    force at C are determined

    1

    Compatibility equations

    0 =B + ByBB + CyBC

    0 =C + ByCB + CyCC

    Reactions at B & C are known

    Statically determinate structure

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    Maxwells Theorem

    The displacement of a point B on a structure due to a unit load

    acting at a point A is equal to the displacement of

    point A when the unit load is acting at point B is:

    fBC=fCB

    The rotation of a point B on a structure due to a unit moment

    acting at a point A is equal to the rotation of point A when the unitmoment is acting at point B the is

    BC=CB

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    Procedure for Analysis

    Determine the degree of statically indeterminacy

    Identify the redundants, whether its a force or a moment, that

    would be treated as unknown in order to form the structurestatically determinate & stable

    Calculate the displacements of the determinate structure at the

    points where the redundants have been removed

    Calculate the displacements at these same points in thedeterminate structure due to the unit force or moment of each

    redundants individually

    Workout the compatibility equation at each point where there is

    a redundant & solve for the unknown redundants

    Knowing the value of the redundants, use equilibrium to

    determine the remaining reactions

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    Determine the reaction at B

    Indeterminate to the 1st degree

    thus one additional equationneeded

    Lets take B as a redundant

    Determine the deflection at point B

    in the absence of support B. Using

    the momentarea method

    B

    = 1/EI[300x6/2{(x6)+6}]

    B = 9000 kNm3/EI

    Determine the deflection caused by

    the unit load at point B

    Example

    CA

    B

    50kN300kNm

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    BB =1/EI[12x12/2xx12]

    BB = 576m3/EI Compatibility equation

    0=B

    +BY

    BB

    0= 9000 /EI + BY

    576 /EI

    BY

    =15.6kN

    The reaction at B is known now so

    the structure is staticallydeterminate & equilibrium

    equations can be applied to get

    the rest of the unknowns

    AB

    1

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    90kN

    C

    ED

    BA

    60kN

    C

    BA

    CC

    C ED

    A B

    1.671.34

    0.440.56

    1.0

    2.67

    60kN90kN

    68.33 81.67

    C ED BA

    245230205

    C

    60kN 90kN

    EDAB

    C

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    A

    2.5

    20kN8kN / m

    B C

    3 2.5

    B

    8kN / m20kN

    A CB

    B1 =RB BBA C

    B

    RB

    Compatibility Equation

    B =B1 =RBBB

    RB = B / BB