CE 382 L10 - Approximate Analysis

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    Approximate Analysis ofStatically Indeterminate

    Structures

    Every successful structure mustbe capable of reaching stableequilibrium under its appliedloads, regardless of structuralbehavior. Exact analysis of indeterminate structures involvescomputation of deflections andsolution of simultaneousequations . Thus, computer programs are typically used.

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    To eliminate the difficultiesassociated with exact analysis,preliminary designs of indeter-minate structures are often basedon the results of approximateanalysis.

    Approximate analysis is basedon introducing deformationand/or force distribution

    assumptions into a staticallyindeterminate structure, equalin number to degree of indeter-minacy, which maintains stable

    equilibrium of the structure .

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    No assumptions inconsistentwith stable equilibrium areadmissible in any approximateanalysis .

    Uses of approximate analysisinclude:

    (1) planning phase of projects,when several alternative designs

    of the structure are usuallyevaluated for relative economy;

    (2) estimating the various

    member sizes needed to initiatean exact analysis;

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    (3) check on exact analysisresults;

    (4) upgrades for older structuredesigns initially based onapproximate analysis; and

    (5) provide the engineer with a

    sense of how the forcesdistribute through the structure.

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    In order to determine the reac-tions and internal forces for indeterminate structures usingapproximate equilibrium me-

    thods, the equilibrium equationsmust be supplemented byenough equations of conditionsor assumptions such that the

    resulting structure is stable andstatically determinate .

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    The required number of suchadditional equations equalsthe degree of static indeter-minacy for the structure , witheach assumption providing anindependent relationshipbetween the unknownreactions and/or internalforces .

    In approximate analysis, theseadditional equations are basedon engineering judgment of appropriate simplifying assump-tions on the response of thestructure.

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    Approximate Analysis of

    a Continuous Beam forGravity Loads

    Continuous beams and girdersoccur commonly in building floor systems and bridges. In theapproximate analysis of con-tinuous beams, points ofinflection or inflection point(IP) positions are assumedequal in number to the degreeof static indeterminacy .

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    For continuous beam struc-tures, the degree of staticindeterminacy in bending ( Ib)equals

    number of bendingreactions (vertical andmoment support reactions)

    C = number of equationsof condition in bending

    b bRI N C 2=

    bRN =

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    Each inflection point positionintroduces one equation of

    condition to the static equilibri-um equations . Three strategiesare used to approximate thelocation of the inflection points:

    1. qualitative displacementdiagrams of the beamstructure,

    2. qualitative bending momentdiagrams ( preferred methodfor students ), and

    3. location of exact inflectionpoints for some simplestatically indeterminatestructures.

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    Approximate analysis of con-tinuous beams using the quali-tative deflection diagram isbased on the fact that the elasticcurve ( deflected shape ) of acontinuous beam can generallybe sketched with a fair degree of accuracy without performing anexact analysis . When the elasticcurve is sketched in this manner,the actual magnitudes of deflec-tion (displacements and rota-tions) are not accurately por-trayed, but the inflection pointlocations are easily estimatedeven on a fairly rough sketch.

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    Qualitative bending momentdiagrams can also be used tolocate inflection points .Bending moments in spans withno loading are linear or constant;with point loading the spanbending moment equations arepiecewise linear; and withuniform loading the moments arequadratic. Remember, internalbending moments at interior support locations adjacent to oneor two loaded spans is negative .

    Recall that zero momentlocations correspond to theinflection point locations .

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    From the total set of inflectionpoints, select the needednumber to achieve a solutionby statics .

    In the case of beams , there willnormally be enough inflectionpoints to reduce the structureto a statically determinatestructure and typically thereare more inflection points thanthe degree of indeterminacy .

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    With the inflection points located

    (equal in number to the degreeof static indeterminacy), theanalysis can proceed on thebasis of statics alone. Since an

    inflection point is a zero momentlocation , it may be thought ofas an internal hinge forpurposes of analysis .

    Some examples to guide thelearning and practice aregiven on the following pages.

    Both the elastic curve andbending moment diagrams aregiven.

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    Fixed-Fixed Beam Subjectedto a Uniform Load

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    Fixed-Fixed Beam Subjectedto a Central Point Force

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    Propped Cantilever BeamSubjected to a Uniform Load

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    Propped Cantilever BeamSubjected to aCentral Point Force

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    When considering problems thatdo not match the exact valuesgiven, some useful guides are:

    Inflection points move

    towards positions of reducedstiffness,

    No more than one inflectionpoint can occur in an un-loaded span, and

    No more than two inflectionpoints will occur in a loaded

    span .

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    Example Approximate Analysisof a Continuous Beam

    L L

    qEI = constant

    IP1

    IP2

    L/3L

    0.1 < 0.25

    QualitativeDeflection Diagram

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    Qualitative BendingMoment Diagram

    -M

    0.5 M

    L/3

    L

    L/3 2L/3 L (1- )LV1 V2R1

    M1

    R 2 R3

    FBD through IPs

    q q

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    3

    2 3

    q(1 )LR2

    q(1 )LV R

    2

    =

    = =

    From the last FBD:

    From the middle FBD:

    2

    1 2

    1

    2L q( L)V LV

    3 2

    3q LV4

    = + +

    =

    2M 0=

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    From the first FBD:

    y 2 1 2F 0 R V q L V= = +

    2qL

    R (2 5 )4

    = +

    2

    1 1 1L q L

    M 0 M V3 4

    = = =

    y 1 1 3q LF 0 R V 4= = =

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    Shear Force Diagram

    R1

    R2

    -R 3

    R2 + R 1

    M1

    -2M1

    0.125q[(1- )L]2

    (1- )L/2

    Bending Moment Diagram

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    Frame Example

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    Trusses with Double

    DiagonalsTruss systems for roofs, bridgesand building walls often containdouble diagonals in each panel ,which makes each panelstatically indeterminate .

    Approximate analysis requiresthat the number of assump-tions introduced must equalthe degree of indeterminacy sothat only the equations of equi-librium are required to performthe approximate analysis.

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    Since one extra diagonal exists

    in each double diagonal panel,one assumption regarding theforce distribution between thetwo diagonals must be made in

    each panel . If the diagonals areslender , it may be assumed thatthe diagonal members are onlycapable of resisting tensile

    forces and that diagonalssubjected to compression canbe ignored since they aresusceptible to buckling , i.e.,

    assume very small bucklingload and ignore post-bucklingstrength .

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    Such an assumption is illustratedin Fig. DD.1(a). In Fig. DD.1(a),the total panel shear is assumedto be resisted by the tensiondiagonal as shown. Compres-sion diagonals are assumed toresist no loading . With thisassumption, the truss of Fig.DD.1 is statically determinate.

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    28Fig. DD.1 Truss with Double

    Diagonal Panels

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    The assumption discussed in the

    previous paragraph is generallytoo stringent , i.e., the compres-sion diagonals can resist aportion of the panel shear .

    Figures DD.1(b) and (c) showtwo different assumptionsregarding the ability of the

    compression diagonals to resistforce.

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    Figure DD.1(b) shows the shear (vertical) components of thediagonal members assuming thatthe compression and tensiondiagonals equally resist thepanel shear .

    Figure DD.1(c) shows thevertical force distribution amongthe compression and tensiondiagonals based on the tensiondiagonal resisting twice the forceof the compression diagonal or two-thirds of the panel shear .

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    Any reasonable assumptioncan be made .

    The compression diagonalassumptions for double diagonaltrusses can be mathematicallysummarized as:

    C = T; 0 1

    If P S = Panel Shear , then

    T(1+ ) = P S

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    Once the diagonal memberforces are determined, theremaining member forces inthe truss can be calculatedusing simple statics , i.e., themethod of sections and/or themethod of joints.