Analysis of Statically Indeterminate ?· – Indeterminate to the first degree ... redundant & solve…

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  • Analysis of Statically Indeterminate Structures

    Force Method of Analysis

    Chapter 10

  • Methods of AnalysisTwo 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 method

    The primary unknowns are displacements

    2

  • 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

    3

  • Force Method of Analysis General Procedure

    Indeterminate to the first degree

    1 Compatibility equation is needed

    Choosing one of the support reaction as a redundant

    The structure become statically determinate and stable

    Downward displacement B at B calculated (load action)

    BB upward deflection per unit force at B

    Compatibility equation

    0 = B + ByBB Reaction By known

    Now the structure is statically determinate

    4

  • Force Method of Analysis General Procedure

    Indeterminate to the first degree

    1 Compatibility equation is needed

    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

    5

  • Force Method of Analysis 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

    6

  • Force Method of Analysis

    General Procedure BB & BC Deflection per unit force

    at B are determined

    CC & CB Deflection per unit force at C are determined

    Compatibility equations

    0 = B + ByBB + CyBC0 = C + ByCB + CyCC

    Reactions at B & C are known

    Statically determinate structure

    7

  • Maxwells Theorem The displacement of a point B on a

    structure due to a unit load acting at a point C is equal to the displacement of point C when the unit load is acting at point B the is

    fBC=fCB

    The rotation of a point B on a structure due to a unit moment acting at a point C is equal to the rotation of point C when the unit moment is acting at point B the is

    BC=CB

  • Force Method of Analysis 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 structure statically 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 the determinate 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

    9

  • Beam Deflections and Slopes

    3 2

    3 2

    max max

    36

    3 2

    Px Lx

    EI

    PL PLatx L

    EI EI

    20

    2

    0 0max max

    2

    2

    Mx

    EI

    M L M Latx L

    EI EI

    Positive (+)

  • Beam Deflections and Slopes

    4 3 2 2

    4 3

    max max

    4 624

    8 6

    wx Lx L x

    EI

    wL wLatx L

    EI EI

    3 2

    3

    max

    2

    max

    4 3 , 048 2

    2 48

    0 or 2 16

    P Lv x L x x

    EI

    L PLatx

    EI

    L PLatx

    EI

    Positive (+)

  • Beam Deflections and Slopes

    3 2 3

    4

    max

    3

    max

    224

    5

    2 384

    24

    wxv x Lx L

    EI

    L wLatx

    EI

    wL

    EI

    2 2 2 , 06

    6 6L R

    Pbxv L b x x a

    LEI

    Pab L b Pab L a

    LEI LEI

    Positive (+)

  • Beam Deflections and Slopes

    3 2 3

    3 2 2 3

    3 3

    9 24 16 02384

    8 24 17384

    2

    3 7

    128 384L R

    wx LL Lx x xEI

    wLx Lx L x L

    EI

    L x L

    wL wL

    EI EI

    2 20

    2

    0max

    0 0

    3 26

    9 2

    6 3L R

    M xv x Lx L

    EIL

    M L

    EI

    M L M L

    EI EI

    Positive (+)

  • Example 1 Determine the reaction at B

    Indeterminate to the 1st degree thus one additional equation needed

    Lets take B as a redundant

    Determine the deflection at point B in the absence of support B. Using the moment-area method

    Determine the deflection caused by the unit load at point B

    39000 .B

    kN m

    EI

    14

    506=

    300kN.m

    B.M.D

    BA

    50kN

    1 6 2300 . 6 6

    2 3B

    EI

  • Example 1

    Compatibility equation

    0 = B + ByBB

    By = 15.6kN

    The reaction at B is known now so the structure is statically determinate & equilibrium equations can be applied to get the rest of the unknowns

    3576BB

    mf

    EI

    15

    1 12 212 12

    2 3BBf

    EI

    112=

    12m

    B.M.D

    BA

    19000 576

    0 yBEI EI

    B.M.D

  • Example 2 Determine the moment at A

    Indeterminate to the 1st degree thus one additional equation needed

    Lets take MA as a redundant

    Determine the slope A at point A ignoring the fixation at A. Using the moment-area method

    1

    1 10 10 333.3320

    2 3d

    EI EI

    A

    20k.ft

    B.M.D

    B1 1 333.33.

    10A

    d

    L EI

    233.33 .A

    k ft

    EI

    B

    A Ad1

    Elastic Curve

  • Example 2 Determine rotation caused by the unit

    moment applied at A

    Compatibility equation

    0 = A +MAAA

    MA = -10k.ft

    The moment at A is known now so the structure is statically determinate

    2

    1 10 2 10 33.331

    2 3d

    EI EI

    17

    33.33 3.330 AM

    EI EI

    B

    A AAd2

    Elastic Curve

    1 B.M.D

    BA

    12 1 33.33 3.33.10

    AA

    d ft

    L EI EI

  • Example 3

  • Example 3

  • Example 3

  • Example 3

  • Neglect the axial load

    The end moments at A&B will be considered as redundants

    From Table inside the front cover

    1(375)A

    EI

    1(291.7)B

    EI

    Example 4

  • -

    3.33BA

    EI

    13.33

    20(2/3)10

    6.67

    6.67AA

    EI

    6.67BB

    EI

    3.33AB

    EI

    375 6.67 3.330 A BM M

    EI EI EI

    291.7 3.33 6.670 A BM M

    EI EI EI

    45.8 .AM k ft 20.8 .BM k ft

    Example 4

  • 24Example 4

  • Example 5Determine the reaction at the support for the beam shown, EI is Constant.Choose the internal moment at internal support as the redundant

    Compatibility Equations

  • Example 5

  • Example 5

  • Example 6Determine the support reactions on the frame shown EI is Constant.

  • Example 6

  • Example 6

  • Example (Additional)The frame, shown in the photo is used to support the bridge deck.

    Assuming EI is constant, a drawing of it along with the dimensions and

    loading is shown. Determine the support reactions.

  • Example 7Determine the moment at the fixed support A for the frame shown EI is

    Constant.

  • Example 7

  • Example 7

  • Example 7

  • Problem 1

  • Problem 2

  • Problem 3

  • Force Method of Analysis: TrussExample8

    Determine the force in member AC.

    Assume EA is the same for all the members

  • 44

    Example 8

  • Force Method of Analysis: Truss

  • 46

    Example 9

    Determine the force in each member if the turnbuckle on member AC is used to shorten the member by 0.5in. Each member has a cross-section area of 0.2 in2 & E=29(106)psi

  • 47

    Example 9

  • Example 9

  • Example 10The beam shown is supported by a pin at A and two pin-connected bar at B.

    Determine the force in member BD. Take E=29(103), I=800 in4 for the beam

    and A=3 in2 for each bar.

  • Example 10

  • Example 10

  • Example 11The simply supported beam shown in the photo is to be designed to support a

    uniform load of 2 kN/m. Determine the force developed in member CE. Neglect

    the thickness of the beam and assume the truss members are pin connected to the

    beam. Also, neglect the effect of axial compression and shear in the beam. The

    cross-sectional area of each strut is 400 mm2, and for the beam I=20(106)mm4 .

    Take E=200 GPa

  • Example 11

  • Example 11

  • Example 11

  • Example 11

  • Example 11

  • Symmetric Structures

  • Anti-symmetric Structures

  • Transformation of Loading

  • Influence Lines for Statically Indeterminate Beams

    Reaction at A

  • 1

    Scale Factor 1

    y DA

    AA

    AA

    A ff

    f

    Reaction at A

    Influence Lines for Statically Indeterminate Beams

  • 1

    Scale Factor 1

    E DE

    EE

    EE

    V ff

    f

    Shear at E

    Influence Lines for Statically Indeterminate Beams

  • Moment at E

    1

    Scale Factor 1

    E DE

    EE

    EE

    M f

    Influence Lines for Statically Indeterminate Beams

  • Example 11

  • Example 11

  • Example 11

  • Example 11

  • Influence Line of a continuous Beam

    4m 8m

    A B C

    G D E F

    Draw the Influence Line of1. Reaction at A, B and C2. Shear at G and E3. Moment at G and E

  • Influence Line of RA

  • Influence Line of RA

  • 1

    3

    1 1

    2

    3

    2 2

    0 8

    5.33

    12

    0 4

    18.67 64

    6

    xA

    xA

    x

    x xf

    EI EI

    x

    x xf

    EI EI EI

    4m 8mA

    B C

    D

    16/EI8/EI 4/EI

    5.33/EI

    x1y=x1/2x2

    y=x2

    18.67/EI

    64/EI

    Influence Line of RA

  • 1

    3

    1 1

    2

    3

    2 2

    0 8

    5.33

    12

    0 4

    18.67 64

    6

    xA

    xA

    x

    x xf

    EI EI

    x

    x xf

    EI EI EI

    Point fxA fxA/fAA

    A 64/EI 1.0

    G 28/EI 0.4375

    B 0 0

    D -14/EI -0.2188

    E -16/EI -0.25

    F -10/EI -0.1562

    C 0 0

    Influence Line of RA

  • 0

    8 12 1 0

    3

    2 8

    c

    B A

    AB

    M

    R R x

    R xR

    Point x RA RB

    A 12 1.0 0

    G 10 0.4375 0.5939

    B 8 0 1

    D 6 -0.2188 1.078

    E 4 -0.25 0.875

    F 2 -0.1562 0.485

    C 0 0 0

    Influence Line of RBUsing equilibrium conditions for the influence line of RB

  • 0

    4 8 1 8 0

    18 2

    B

    A C

    AC

    M

    R R x

    x RR

    Point x RA RC

    A 12 1.0 0

    G 10 0.4375 -0.0312

    B 8 0 0

    D 6 -0.2188 0.1406

    E 4 -0.25 0.375

    F 2 -0.1562 0.6719

    C 0 0 1

    Influence Line of RCUsing equilibrium conditions for the influence line of RC

  • 1

    Check

    0y

    A B C

    F

    R R R

  • Influence Line of VG

  • Influence Line of VE

  • Influence Line of MG

  • Influence Line of ME

  • Influence Line of ME

  • Live Load Pattern in Continuous Beams

    LL

    DL

  • Support Reactions

    Influence Line for positive reaction at support 1

    Load pattern for maximum positive reaction at support 1

  • Span Positive Moment

    Influence line for positive moment at 7

    Load pattern for maximum positive moment at 7

  • Support Negative Moment

    Influence line for negative moment at support 2

    Load pattern for maximum negative moment at support 2

  • Internal Shear

    Influence line for positive shear at 7

    Load pattern for maximum positive shear at 7

  • Moment EnvelopesThe moment envelope curve defines the extreme boundary values of

    bending moment along the beam due to critical placements of design

    live loading.

    Example of Three Span Continuous Beam

  • Moment Envelopes

    Dead Load Only

    BMD due to DL

  • Moment Envelopes

    Live Load Arrangement

    Max. Positive Moment at span 1&3(Min. Moment at span 2)

    Max. Positive Moment at span 2

    (Min. Moment at span 1 & 3)

  • Moment Envelopes

    Live Load Arrangement

    Max. negative Moment at support 2

    Max. negative Moment at support 3

  • Moment Envelopes

    Live Load Arrangement

    Max. positive Moment at support 2

    Max. positive Moment at support 3

  • Moment Envelopes

    Max. +M at span 2Max. +M at spans 1&3

    DL only

    Max. -M at support 2 Max. -M at support 3

    Max. +M at support 2 Max. +M at support 3

    Summarize all critical cases

  • Moment EnvelopesMoment Diagram for All Cases

    Solving each of these cases and combining the results with the dead

    only case, results in six different moment diagrams.

    The dead load only case has been left as a heavier line.

  • Moment EnvelopesMoment Diagram for All Cases

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