Indeterminacy & Stability of a Structure

 

Indeterminacy & Stability of a Structure

Statically Determinate Structures

Conditions of equilibrium are sufficient to analyse the structure. Bending moment and shear force is independent of the cross-sectional area of the components and flexural rigidity of the members. No stresses are caused due to temperature change. No stresses are caused due to lack of fit or differential settlement.

Statically Indeterminate Structures

Additional compatibility conditions are required. Bending moment and shear force depends upon the cross-sectional area and flexural rigidity of the members. Stresses are caused due to temperature variation. Stresses are caused due to lack of fit or differential settlement.

Important Terms

1. Stable/Unstable: A stable structure is one that will not collapse when disturbed. Stability may also be defined as "The power to recover equilibrium." In general, there are may ways that a structure may become unstable, including the buckling of compression members, yielding/rupture of members, however, for linear structural analysis, the main concern is instability caused by insufficient reaction points or poor layout of structural members.
2. Internally Stable: In internally stable structure is one that would maintain its shape if all the reactions supports were removed. A structure that is internally unstable may still be stable if it has sufficient external support reactions. An example is shown below in Figure.3. External Determinacy: The ability to calculate all of the external reaction component forces using only static equilibrium. A structure that satisfies this requirement is externally statically determinate. A structure for which the external reactions component forces cannot be calculated using only equilibrium is externally statically indeterminate.

   4. Internal Determinacy: The ability to calculate all of the external reaction component forces and internal forces using only static equilibrium. A structure that satisfies this requirement is internally statically determinate. A structure for which the internal forces cannot be calculated using only equilibrium is internally statically indeterminate. Typically if one talks about 'determinacy', it is an internal determinacy that is meant.

   5. Redundant: Indeterminate structures effectively have more unknowns than can be solved using the three equilibrium equations (or six equilibrium equations in 3D). The extra unknowns are called redundant.

   6. Degree of Indeterminacy: The degree of indeterminacy is equal to the number of redundant. An indeterminate structure with 2 redundant may be said to be statically indeterminate to the second degree or "2º S.I."

   Static Indeterminacy

   If a structure cannot be analyzed for external and internal reactions using static equilibrium conditions alone then such a structure is called indeterminate structure.

   (i) D_S = D_Se + D_Si

   Where,

   D_S = Degree of static-indeterminacy

   D_Se = External static-indeterminacy

   D_Si = Internal static-indeterminacy

   External static indeterminacy:

   It is related with the support system of the structure and it is equal to number of external reaction components in addition to number of static equilibrium equations.

   (ii) D_Se = r_e - 3 For 2D

   D_Se = r_e – 6 For 3D

   Where, r_e = total external reactions

   Internal static indeterminacy:

   It refers to the geometric stability of the structure. If after knowing the external reactions it is not possible to determine all internal forces/internal reactions using static equilibrium equations alone then the structure is said to be internally indeterminate.

   For geometric stability sufficient number of members are required to preserve the shape of rigid body without excessive deformation.

   (iii) D_Si = 3C - r_r …… For 2D

   D_Si = 6C - r_r …… For 3D

   where, C = number of closed loops.

   and

   r_r = released reaction

   (iv) r_r = ∑(m_j - 1) …… For 2D

   r_r = 3∑(m_j - 1) ……. For 3D

   where m_j = number of member connecting with J number of joints.

   and J = number of hybrid joint.

   (v) D_s = m + r­_e – 2j ….. For 2D truss

   D_Se = r_e - 3 & D_Si = m – (2j – 3)

   (vi) D_S = m + r_e – 3j ….. For 3D truss

   D_Se = r_e – 6 & D_Si = m – (3j - 6)

   (vii) D_S = 3m + r_e – 3j - r_r ….. 2D Rigid frame

   (viii) D_s = 6m + r­_e – 6j - r_r ….. 3D rigid frame

   (ix) D_S = (r_e – 6) + (6C – r_r) ….. 3D rigid frame

   Kinematic Indeterminacy

   It the number of unknown displacement components are greater than the number of compatibility equations, for these structures additional equations based on equilibrium must be written in order to obtain sufficient number of equations for the determination of all the unknown displacement components. The number of these additional equations necessary is known as degree of kinematic indeterminacy or degree of freedom of the structure.

   A fixed beam is kinematically determinate and a simply supported beam is kinematically indeterminate.

   (i) Each joint of plane pin jointed frame has 2 degree of freedom.

   (ii) Each joint of space pin jointed frame has 3 degree of freedom.

   (iii) Each joint of plane rigid jointed frame has 3 degree of freedom.

   (iv) Each joint of space rigid jointed frame has 6 degree of freedom.

   Degree of kinematic indeterminacy is given by:

   1. D_k = 3j - r_e ………. For 2D Rigid frame when all members are axially extensible.
   2. D_k = 3j - r_e - m ………. For 2D Rigid frame if 'm' members are axially rigid / inextensible.
   3. D_k = 3(j + j’) - r_e – m + r_r …… For 2D Rigid frame when J' = Number of Hybrid joints is available.
   4. D_k = 6(j + j’) - r_e – m + r_r ….. For 3D Rigid frame
   5. D_k = 2(j + j’) - r_e – m + r_r ….. For 2D Pin jointed truss.
   6. D_k = 3(j + j’) - r_e – m + r_r …… For 3D Pin jointed truss.

   Examples

   Notations used in examples

   i_e is degree of Indeterminacy

   e_c is the number of equations of condition,

   Internal support type	Equations of conditions
   Hinge	ec=n-1
   Roller	ec=2*(n-1)

   where n is the number of members connected to the hinge or roller.

   1. Determination of the Number of Members and Joints

   
   2.  Instability due to Parallel Reactions
   
   3. Instability due to Concurrent Reactions
   
   4. Instability due to an Internal Collapse Mechanism
   
   5. Mixed up
   

   a) External Determinacy:

   i_e=r−(3+e_c)  

   r=4,ec=1 (The hinge on the left at the pin does not provide any additional equations of condition).

   Therefore,

   i_e=0.

   Then, is this structure statically determinate? No, it is unstable because if we take a free-body diagram of the left side of the beam, and take a sum of moments about the center hinge, the sum of moments will be non-zero due to the vertical reaction at the left pin (but we know that it has to be zero due to the existence of the pin).

   Internal Determinacy:

   i_e=(3m+r)−(3j+e_c)

   m=2,r=4,j=3,e_c=1 (Again, the hinge on the left at the pin does not provide any additional equations of condition).

   Therefore,

   3m+r=10, 3j+e_c=10, and ie=0.

   Then, is this structure statically determinate? No, it is unstable due to the same reason above.

   b) External Determinacy:

   r=3,e_c=0.

   Therefore,

   i_e=0.

   Then is this structure statically determinate? No, because the reactions are concurrent through the pin on the right.

   Internal Determinacy:

   m=2,r=3,j=3,e_c=0.

   Therefore,

   3m+r=9 and 3j+e_c=9,

   so the structure appears internally determinate, but it is still unstable due to the concurrent reactions.

   c) External Determinacy:

   r=3,e_c=0.

   Therefore,

   i_e=0.

   Since there are no sources of instability, this structure is externally statically determinate.

   Internal Determinacy:

   m=6,r=3,j=6,e_c=0.

   Therefore,

   3m+r=21 and 3j+e_c=18,

   so this structure is internally statically indeterminate to three degrees (or "3º S.I.").

   d) External Determinacy:

   r=5,e_c=2.

   Therefore,

   i_e=0.

   Since there are no sources of instability, this structure is externally statically determinate.

   Internal Determinacy:

   m=5,r=5,j=6,e_c=2.

   Therefore,

   3m+r=20 and 3j+e_c=20,

   so this structure is internally statically determinate (or "S.D.").

   e) External Determinacy:

   r=7,e_c=2. (Due to the three members connected to the internal hinge)

   Therefore,

   i_e=2.

   This structure can be described as 2 degrees externally statically indeterminate.

   Internal Determinacy:

   m=3,r=7,j=4,e_c=2.

   Solving,

   3m+r=16 and 3j+e_c=14,

   Again, this structure is found to be 2 degrees internally statically indeterminate.

   f) External Determinacy:

   r=4,e_c=2.

   Therefore,

   i_e=−1.

   Due to the design of the structure, the internal roller cannot be supported and the structure is classified as unstable.

   Internal Determinacy:

   m=2,r=4,j=3,e_c=2.

   Solving,

   3m+r=10 and 3j+e_c=−11,

   We can safely say that this structure is unstable, both by the equations of determinacy and by understanding how the structure will bend under loading.

   However, if the right-hand pin were fixed-end support this case would be considered a stable, statically determinate structure.