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  • Statically Indeterminate Structures

    [email protected]

  • • Statically Determinate Problems(静定问题)

    • Statically Indeterminate Problems(超静定问题)

    • Degrees of Indeterminacy(超静定次数)

    • Advantages & Disadvantages of Indeterminate Structures

    (超静定结构的优缺点)

    • Redundancy & Basic Determinate System(冗余约束与 基本静定系)

    • General Ideas for Analyzing Indeterminate Structures

    (超静定问题一般解法)

    Contents

    2

  • • Force Method Solution Procedure(力法)

    • Statically Indeterminate Bars(超静定拉压杆)

    • Assembly Stress(装配应力)

    • Thermo Stress(热应力)

    • Statically Indeterminate Shafts(超静定扭转轴)

    • Statically Indeterminate Beams(超静定弯曲梁)

    • Moment-Area Theorems with Indeterminate Beams(图乘 法求解超静定梁)

    • Combined Indeterminate Structures (组合超静定结构)

    Contents

    3

  • • Problems can be completely solved via static equilibrium

    • Number of unknowns (forces/moments) = number of

    independent equations from static equilibration

    0 ? 0 ?

    ?0

    x Ax

    y Ay

    ByA

    F F F F

    FM

         

           

      

    Statically Determinate Problems

    4

    A B

  • A B C

    • Problems cannot be solved from static equilibrium alone

    • Number of unknowns (forces/moments) > number of

    independent equations from static equilibration

    ? 0 ?

    0 ?

    ?0 ?

    Ax

    x Ay

    y By

    CxA

    Cy

    F F F

    F F

    FM F

        

         

        

      

    Statically Indeterminate Problems

    5

  • • For a co-planer structure, there are at most three

    equilibrium equations for each portion of the structure. If

    there is a total of n portions and m unknown reaction

    forces from supports:

    3 Statically determinate

    3 Statically

    - 3 Degrees of indetermi

    indet

    n

    erminate

    acy

    m n

    m n

    m n 

     

     

    Statically determinate Degrees of Indeterminacy = 2

    Degrees of Indeterminacy of Structures

    6

  • • Advantages:

    Advantages & Disadvantages of Indeterminate Structures

    - Redistribution of reaction forces / internal forces

    - Smaller deformation

    - Greater stiffness as a whole structure

    • Disadvantages:

    - Thermal and residual stresses due to temperature change

    and fabrication errors

    7

  • • Redundancy: unnecessary restraints without which the

    static equilibrium of a structure still holds.

    • Basic determinate system: the same structure as of a

    statically indeterminate system after replacing redundant

    restraints with extra constraining loads

    Redundancy & Basic Determinate System

    8

  • • Basic determinate structure: obtained via replacing

    redundant restraints with extra constraining loads.

    How to Analyze Statically Indeterminate Structures?

    • Equilibrium: is satisfied when the reaction forces at

    supports hold the structure at rest, as the structure is

    subjected to external loads

    • Deformation compatibility: satisfied when the various

    segments of the structure fit together without intentional

    breaks or overlaps

    • Deformation-load relationship: depends on the manner

    the material of the structure responds to the applied loads,

    which can be linear/nonlinear/viscous and elastic/inelastic;

    for our study the behavior is assumed to be linearly elastic

    9

  • • The method of consistent deformations or force method

    was originally developed by James Clerk Maxwell in 1874

    and later refined by Otto Mohr and Heinrich Műller-

    Breslau.

    Christian Otto Mohr

    (1835 – 1918)

    James Clerk Maxwell

    (1831 – 1879)

    Heinrich Műller-Breslau

    (1851 - 1925)

    Force Method (Method of Consistent Deformations)

    10

  • • Making the indeterminate structure determinate

    Force Method Solution Procedure

    • Obtaining its equations of equilibrium

    • Analyzing the deformation compatibility that must be

    satisfied at selected redundant restraints

    • Substituting the deformation-load relationship into

    deformation compatibility, resulting in complementary

    equations

    • Joining of equilibrium equations and complementary

    equations

    • Types of Problems can be addressed include (a) bar

    tension/compression; (b) shaft torsion; and (c) beam

    bending 11

  • Statically Indeterminate Bars

    • For an axially loaded bar with both ends fixed, determine

    the reaction forces.

      12

    1

    2

    0 BB

    B

    A

    P R LR L L

    EA EA

    L R P

    L

    L R P

    L

         

     

     

    • Solution:

    12

  • • Determine the reaction forces at the bar ends.

     

     

     

    6

    3

    3 64

    3 1

    6

    3

    6

    400 10

    600 10

    150 10 400 10 0

    600 10

    250 10

    900 10

    250 10

    577kN

    900 577 323kN

    B

    B

    i i

    i i B

    B

    B

    A

    R

    R

    F L L

    EA E R

    R

    R

    R

     

         

              

        

      

        

     

     

       

    • Solution:

    Sample Problem

  • • Solution:

    Sample Problem

    • Determine the reaction forces, if any, at the bar ends. E = 200 Gpa.

     

     

     

    4 3

    1

    6

    3

    3 6

    9 3

    6

    3

    6

    4.5 10

    400 10

    600 10

    150 10 400 10

    200 10 600 10

    250 10

    900 10

    250 10

    115.4kN

    900 115.4 784.6kN

    i i

    i i

    B

    B

    B

    B

    B

    A

    F L L

    EA

    R

    R

    R

    R

    R

    R

     

       

         

           

         

      

        

     

     

       

    14

  • Sample Problem

    • What is the deformation of the

    rod and tube when a force P is

    exerted on a rigid end plate as

    shown?

    • Solution:

    1 2

    1 2 1 2

    1 1 2 2

    1 1 2 2 1 2

    1 1 2 2 1 1 2 2

    1 1 2 2

    ,

    P P P

    PL P L L L

    E A E A

    E A P E A P P P

    E A E A E A E A

    PL L

    E A E A

             

        

       

    15

  • • Given: tension/compression rigidities E1A1 = E2A2 = EA,

    E3A3, external load F. Find: internal forces in bars.

    1. Make the truss determinate by

    replacing extension/contraction

    restraint along Bar 3 with FN3 F

    

    A

    1 23

    l

    1 2

    1 2 3

    0

    0 ( )cos

    x N N

    y N N N

    F F F

    F F F F F

      

        

     A F

    FN3FN1 FN2

    Sample Problem

    • Solution:

    2. Static equilibrium at joint A

    16

  • 4. Displacement-force relationship

    3. Displacement compatibility at joint A

    Due to symmetry

    21 ll  

     cos31 ll 

    1 1

    cos

    NF ll EA 

      33

    3 3

    AE

    lF l N

    

    A

    1 23

    l

    A*

    17

  • 5. Complementary equation

    21 3 1 3 1 3

    1 1 3 3 3 3

    cos cos cos cos

    N N N N

    F l F l EA l l F F

    E A E A E A   

           

    6. Joint of force equilibrium and complementary equations

    1 2 3 3

    1 2 2

    1 2 3

    32 3

    1 3

    3 3 3 3

    2cos cos

    ( )cos

    cos 1 2 cos

    N N

    N N

    N N N

    N

    N N

    F F F

    E A F F

    EA F F F F

    F FEA EAF F

    E A E A

     

     

              

     