Determinacy, Indeterminacy and Stability
Contents
Introduction
Before beginning to analyse a structure, it is important to know what kind of structure it is. Different types of structures may need to be analysed using different methods. For example, structures that are determinate may be completely analysed using only static equilibrium, whereas indeterminate structures require the use of both static equilibrium and compatibility relationships to find the internal forces. In addition, real structures must be stable. This means that the structure can recover static equilibrium after a disturbance. There is no point analyzing a structure that is not stable.
This section will explain the concepts of determinacy, indeterminacy and stability and show how to identify determinate, indeterminate and stable structures.
Important Terms
- Stable/Unstable
- A stable structure is one that will not collapse when disturbed. Stability may also be defined as "The power to recover equilibrium."^{[1]} In general, there are may ways that a structure may become unstable, including buckling of compression members, yielding/rupture of members, or nonlinear geometric effects like P-Delta; however, for linear structural analysis, the main concern is instability caused by insufficient reaction points or poor layout of structural members.
- Internally Stable^{[2]}
- 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 in Figure 1.
- External Determinacy
- The ability to calculate all of the external reaction component forces using only static equilibrium.^{[2]} 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.
- Internal Determinacy
- The ability to calculate all of the external reaction component forces and internal forces using only static equilibrium.^{[2]} 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 internal determinacy that is meant.
- 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 redundants.^{[2]}
- Degree of Indeterminacy
- The degree of indeterminacy is equal to the number of redundants.^{[2]} An indeterminate structure with 2 redundants may be said to be statically indeterminate to the second degree or "2º S.I."
External Determinacy
If a structure is externally determinate, then all of the reactions may be calculated using equilibrium alone. To calculate external determinacy, the following equations are used:^{[2]}
$ r < 3 + e_c_c $ Statically unstable externally ~Eq. 1 $ r = 3 + e_c $ Statically determinate externally ~Eq. 2 $ r > 3 + e_c $ Statically indeterminate externally ~Eq. 3
- where $ r $ is the number of reaction components, and $ e_c $ is the number of equations of condition. Both of these are described in detail below.
The degree of indeterminacy is given by the following equation:^{[2]}
- $ i_e = r - (3 + e_c) $ ~Eq. 4
Reaction Components
In the equations above, $ r $ is equal to the total number of reaction components as follows:
For multiple reaction points, $ r $ is the sum of all the components for all the reaction points in the structure.
Equations of Condition
Additionally, $ e_c $ is the "number of equations of condition." These are release conditions within the structure that provide extra equilibrium equations beyond the three for global equilibrium.^{[3]}
For example, if an internal hinge is added to the structure, as shown in Figure 2, then there is one equation of condition. If there was no internal hinge in this example, then the structure would be indeterminate and it would not be possible to find the reaction forces or the internal forces (since is has four reaction components). The addition of the hinge provides an additional equilibrium condition which forces the internal moment to be equal to 0 at point B ( $ \sum {M_B} = 0 $ ). This may be seen if the structure is split into two free body diagrams as shown in the lower part of Figure 2. At point B, there are three internal force components that exist in equal and opposite action/reaction pairs on either side of point B:
- Axial force - $ B_x^{AB} $ and $ B_x^{BC} $
- Shear force - $ B_y^{AB} $ and $ B_y^{BC} $
- Moment - $ M_B^{AB} $ and $ M_B^{BC} $
- So, $ M_B^{AB} = M_B^{BC} = 0 $, because they are action reaction pairs.
Therefore, only one extra equilibrium equation is possible due to the introduction of the hinge: either $ \sum {M_B^{AB}} = 0 $ or $ \sum {M_B^{BC}} = 0 $ but not both because the two equations are not independent. So, for each internal hinge in a structure, there is a single equation of condition: $ e_c = 1 $.
For a structure with an internal roller, such as that shown in Figure 3, both the force transfer in the direction of the roller and the moment are equal to zero at the location of the roller. This provides two extra equilibrium equations, and therefore two equations of condition. For the structure shown in Figure 3, the extra equations are:
- $ \sum {M_B^{AB}} = 0 $ ( or $ \sum {M_B^{BC}} = 0 $ )
- and
- $ \sum {B_x^{AB}} = 0 $ ( or $ \sum {B_x^{BC}} = 0 $ )
So, for each internal roller, there are two equations of condition: $ e_c = 2 $.
If there are additional members that frame into a single internal hinge, then there is an additional equation of condition for each additional member.^{[2]} For example, for three members connected at a hinge, then there are two extra independent equilibrium equations that are added to the system (because in an equal and opposite action/reaction pair, there can only be two sides). So, for a hinge connection with multiple elements, $ e_c = n - 1 $ where $ n $ is equal to the number of members connected to the hinge.^{[2]}
Similarly, for a roller connection with multiple members, each additional member adds two equations of condition, so $ e_c = 2 * (n - 1) $.^{[2]}
In summary:
Internal
Support TypeEquations
of ConditionHinge $ e_c = n - 1 $ ~Eq. 5 Roller $ e_c = 2 * (n - 1) $ ~Eq. 6
- where $ n $ is the number of members connected to the hinge or roller.
WARNING: This method of determining external determinacy is not valid for indeterminate structures which contain closed loops.
Internal Determinacy
If a structure is internally determinate, then all of the reactions and internal forces may be calculated using equilibrium alone. Internal determinacy is generally much more important than external determinacy in structural analysis. To calculate internal determinacy, the following equations are used:^{[2]}
$ 3m + r < 3j + e_c $ Statically unstable ~Eq. 7 $ 3m + r = 3j + e_c $ Statically determinate ~Eq. 8 $ 3m + r > 3j + e_c $ Statically indeterminate ~Eq. 9
- where $ m $ is the total number of members in the structure, $ r $ is the number of reaction components, $ j $ is the total number of joints in the structure, and $ e_c $ is the number of equations of condition. The meaning of $ r $ and $ e_c $ are the same as for #External Determinacy above. The definition of members and joints will be discussed below.
The degree of indeterminacy is given by the following equation:^{[2]}
- $ i_e = 3m + r - (3j + e_c) $ ~Eq. 10
Members and Joints
There is no specific way that a structure must be split into members and joints for the purposes of the determinacy analysis. Any division of the structure is okay as long as the members and joints are consistent with each other; however, joints should be placed at least at the following locations:
- Free ends
- Reactions
- Intersections of three or more elements
For an example of how to calculate the numbers of members and joints, see Figure 4.
Stability
An unstable structure generally cannot be analysed. Therefore, it is useful to know if a structure is stable or unstable before a structural analysis is conducted. There are four main ways that a structure may be geometrically unstable. These apply only to linear geometric stability and not to instability caused by buckling, member yielding or nonlinear geometry.
- There are not enough reactions^{[3]}: This will generally be clear from an application of the determinacy equations (Eq. 7-9).
- The reactions are parallel^{[4]}: All of the reaction components point in the same direction. An example of such a situation is shown in Figure 5. In this example, the horizontal equilibrium $ \sum {F_x} = 0 $ cannot be solved and there will be a net horizontal force on the system with no resistance.
- The reactions are concurrent^{[4]}: All of the reaction components meet at a point. An example of such a situation is shown in Figure 6. Effectively, the system is free to rotate as a rigid body around the point that the reaction components meet at.
- There is an internal collapse mechanism^{[3]}: This is any situation in which there is an internal mechanism in the system that will cause it to deform between the supports. In some such situations, this will be clear from the use of the determinacy equations, but in others, it may not. In all such cases, though, the instability will become clear during the structural analysis because it will be impossible to solve for all of the internal forces. An example internal collapse mechanism is shown in Figure 7.
Example Problem
Determine whether the following structures shown in Figure 8 are externally determinate, internally determinate, externally indeterminate, internally indeterminate or unstable. If a structure is indeterminacy, determine how many degrees of indeterminacy it has.
a) External Determinacy:
- $ i_e = r - (3+ e_c) $
- $ r = 4, \; e_c = 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 $ i_e = 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 a fixed-end support this case would be considered a stable, statically determinate structure.
References
- ↑ http://www.colorado.edu/engineering/CAS/courses.d/Structures.d/IAST.Lect23.d/IAST.Lect23.pdf
- ↑ ^{2.00} ^{2.01} ^{2.02} ^{2.03} ^{2.04} ^{2.05} ^{2.06} ^{2.07} ^{2.08} ^{2.09} ^{2.10} ^{2.11} Kassimali, A. (2011). Structural Analysis: SI Edition (4th ed.). Stamford, CT: Cengage Learning.
- ↑ ^{3.0} ^{3.1} ^{3.2} Leet, K.M., Uang, C-M., Gilbert, A.M. (2011). Fundamentals of Structural Analysis (4th ed.). New York, NY: McGraw Hill.
- ↑ ^{4.0} ^{4.1} Hibbeler, R.C. (2012). Structural Analysis (8th ed.). Upper Saddle River, NJ: Pearson Prentice Hall.