2: Forces in Statically Determinate Beams and Plane Frames

2.2: Stability and Determinacy

2.2.1: General Concepts of Determinacy

A structure is statically determinate if the equations of static equilibrium are sufficient by themselves to determine all forces acting on and within a structure. Sometimes it is useful to distinguish between external determinacy and internal determinacy. A structure is externally statically determinate if its reactions can be completely determined using only the equations of equilibrium. A structure is internally statically determinate if all member forces (normal, shear and bending moments) can be completely determined using only the equations of statics.

The degree of indeterminacy of a structure is the excess number of unknown quantities (reactions and internal forces) over the number of independent equations of equilibrium available to solve for them. If the degree of indeterminacy is greater than 0, the structure is statically indeterminate. If it is equal to 0, the structure is statically determinate, but only if it is stable. If the degree is less than 0, then the structure is unstable (see below), and its statical determinacy is not very relevant.

To determine the degree of indeterminacy of a structure, we simply count the number of unknowns and compare to the number of equations available.

2.2.2: Determinacy of Beams and External Determinacy of 2D Frames

For beam structures that are single linear elements with one or more supports, we can take a somewhat simplified approach, one that deals with external determinacy only. Thats OK for beam structures as they will be internally determinate if they are externally determinate.

Number of Unknowns

The number of unknowns is simply the number of independent reaction components, $r$, in the entire structure:

If $r =$ the number of independent reaction components for the entire structure, where $r = \sum r_c$ over all of the supports in the structure, and $r_c$ for each support can be determined from the following table:

reaction components

Figure 2.1: Reaction Components

then the number of unknowns is:

\[r\]

Number of Equations

There are 3 equations available from the overall FBD. Each condition that specifies a known internal force value and that allows the beam to be cut into two separate parts at the condition adds 1 independent equation.

conditions

Figure 2.2: Internal Conditions

If $c$ is the total number of releases in the beam.

The number of equations is:

\[3 + c\]

Statical Determinacy

if $r < 3+c$, then the beam is unstable and determinacy does not matter very much.

if $r = 3+c$, then the beam is statically determinate, if it is stable.

if $r > 3+c$, then the beam is statically indeterminate. It may still be unstable.

Stability must be investigated separately from determinacy; one does not imply the other.

2.2.3: Determinacy of Beam and Frame Structures

The more general case works for beams and 2D Frames and also includes the evaluation of internal determinacy.

Number of Unknowns

Fig. 2.1 shows the most common support symbols and the independent reaction components for each; column $r_c$ shows the corresponding number of independent reaction components. As before, we can define:

$r$ The number of independent reaction components for the entire structure, $r = \sum r_c$ over all of the supports in the structure.

Members in a beam or frame structure are those portions that span between supports, or joints (connections to other members), or free ends. Each member has exactly two joints, one at either end. If the three internal force components – normal force (N), shear force (V) and bending moment (M) are known at a point in a member, then that member is completely determined. That is, we can use simple statics to determine the internal forces at any other point. It can thus be said that there are 3 unknowns associated with each member – 3 independent quantities that need be determined before everything is ‘known’. So, defining

$m$ The number of members in a structure.

Then we have, as the number of unknowns:

\[N_u = 3m + r\]

Number of Equations

To determine the number of independent equilibrium equations that are available to solve for the unknowns, we define:

$j$ The number of joints in the structure. Every member is connected to exactly two joints, one at each end.

$c$ The number of internal release conditions. A condition is a constraint that specifies that an internal force must be a particular numerical value (almost always zero) at a point. An example is an internal hinge or pin that constrains the bending moment to be 0.

It is possible to write 3 equations of equilibrium at each joint, and one more for each condition. We state without proof, then, that the number of independent equilibrium equations is:

\[N_e = 3j + c\]

Statical Determinacy

The degree of statical indeterminacy is the number of unknowns in excess of the number of equations:

\[N_u - N_e = (3m + r) - (3j + c)\]

To judge determinacy, we have the following three cases:

if $(3m+r) < (3j + c)$, then the system is unstable.

if $(3m+r) = (3j + c)$, then the system is statically determinate, if it is stable.

if $(3m+r) > (3j + c)$, then the system is statically indeterminate. It may also be unstable.

The latter two cases, $(3m+r)\ge(3j+c)$ are necessary, but not sufficient, for stability.

The degree of statical indeterminacy is given by $(3m+r)-(3j+c)$ where $m$ = number of members, $r$ = number of independent reaction components, $j$ = number of joints, and $c$ = number of conditions.
If the degree of statical determinacy is negative, the structure is unstable. If it is 0 or positive, the structure may be stable.
If a real structure is unstable, that is almost always a disaster.

2.2.4: Determinacy Examples

Fig. 2-3, below, shows a number of examples of determinacy calculations.

examples

Fig. 2-3: Examples of Determinacy Calculation

Observations

  1. Internal hinges (conditions) such as those shown in examples, iv, v, vii, and viii are not usually counted as joints, and thus we usually count just one member continuously through the hinge. However, they can be counted as joints, in which case the one member becomes two, for consistency (every member has exactly two joints). This adds 1 to m and 1 to j and thus does not change the important result. This is illustrated in examples iv and v, which are otherwise identical, and in example vii.

  2. The determinacies given are overall – they include both member forces and reaction determination. It is possible for a structure to be indeterminate, overall, and yet still be statically determinate with respect to the reactions (i.e., externally statically determinate). This is illustrated in example viii.

  3. If we define $c’$ as the number of conditions at which it is possible to separate the structure into two complete halves, then $c’=1$ in example viii. The minimum number of reaction components necessary for stability is $r_{min}=3+c’$ and thus $r_{min}=4$ for this example. If $r=r_{min}$, as it is in this case, the structure is statically determinate externally.

The minimum number of independent external reaction components require for stability is $r_{min} = 3 + c'$, where $c'$ = the number of conditions at which the structure may be separated into two complete parts.
If $r = r_{min}$, the structure is statically determinate externally, if it is stable.

2.2.5: Stability of Beam and Frame Structures

A structure is stable if it cannot undergo any small rigid-body geometry changes. It is important to recognize instability as it is vitally important that no civil engineering structures are unstable. An unstable structure will lead to either collapse, large deformations, or very high member forces.

It is always possible to precisely determine the stability of a structure by setting up all the possible equilibrium equations in matrix form. If the rank of the resulting matrix is less than the number of rows, then the set of equations has no unique solution and the structure is unstable.

Someday there will be an example of the matrix method, but not today.

This method is laborious and impractical for hand calculation, so we look for something easier to apply. The method we will choose involves looking for mechanisms - i.e. small geometry changes that are otherwise consistent with all the displacement constraints in the structure. Only when we can’t find a mechanism can we have some confidence that the structure is stable.

You can prove a structure is unstable by finding a mechanism. You can't use this method to prove a structure is stable, however -- you can only fail to find a mechanism. You have to try all possible mechanisms, and that number may be very large for larger structures.

Stability Example 1

Example 1

Fig. 2-4: Stability Example 1.

Consider the first example of Fig. 2-4. Counting the members, joints, reactions and conditions tells us that the structure is statically determinate, if it is stable. But is it stable? We must look for mechanisms.

Start at one edge of the structure and proceed segment by segment, looking for valid displacements. Start with member a-b; it is constrained against horizontal and vertical displacement at a, but is free to rotate about a, and must stay straight from a to b. Point b can move anywhere vertically on the line through b (remember, small displacements); let it move downwards.

Now point b on b-c-d must also move downwards, as there is a constraint that both sides of b must displace the same amount. However, there is no constraint against relative rotation at b, so we can let b-c-d rotate relative to a-b. b-c-d is also constrained against vertical displacement at c and must remain straight. This can be accomplished by allowing point d to displace upwards, along the vertical line through its original location.

Point d on d-e, now, must also move upward the same amount, and it can be letting d-e rotate clockwise about e.

So, we have found a mechanism - a set of small, consistent, geometry changes. Therefore, that beam is unstable.

Stability Example 2

Example 2

Fig. 2-5: Stability Example 2

Now consider the beam structure in Fig. 2-5. All of the supports and counts are the same as the previous example, but this time the supports at c and e are displaced vertically. Again, try to find a mechanism.

Start with a-b and imagine a rotation about a. Point b must remain on the vertical line through b – that line is perpendicular to the radius from a to b.

Now consider a rotation of b-c-d about c. For this to happen, point b must remain along the inclined line through b, as that line is perpendicular to the radius from c to b. Point b, then has to lie on both lines, and the only point of intersection of those two lines is at the original location of b. Therefore, point b cannot move,so that is not a possible mechanism.

This does not prove that the structure is stable, you must look for other possible mechanism. For this structure, you will not find a mechanism, and you may eventually conclude that the structure is stable.

Stability Example 3

Example 3

Fig. 2-6: Stability Example 3

Now see what happens when we change the pin at c to a roller, as shown in Fig. 2-6. Starting with a-b, we allow b to move along the vertical line through b. b on b-c-d can also move the same amount along the same line by allowing point c to move to the right as b-c-d rotates counter clockwise. It can do that, as c is a roller and allows the horizontal displacement. Point d moves upwards and to the right as a result of the rotation of b-c-d. d-e can rotate about e while e moves rightward (remember that the lengths of the members must not change).

So, we have found a mechanism, so this structure is unstable. :-( But you could have also told that by counting.