5.Truss Deflection
2. Statical Determinacy and Kinematic Stability
Instability due to improper support. A beam or frame is kinematically unstable if the support conditions are such that the whole structure is allowed to move as a mechanism.
Examples of improper support and insufficient support are shown below.
Improper or insufficient support conditions.
Instability due to improper connection. A beam or frame is kinematically unstable if the internal connection conditions are such that part of or the whole structure is allowed to move as a mechanism. Examples of improper connections are shown below.
M
V
T
T
V M
Improper internal connections.
Statical determinacy. A stable beam or frame is statically indeterminate if the number of force unknowns is greater than the number of equilibrium equations. The difference between the two numbers is the degree of indeterminacy. The number of force unknowns is the sum of the number of reaction forces and the number of internal member force unknowns. For reaction forces, a roller has one reaction, a hinge has two reactions and a clamp has three reactions as shown below.
Reaction forces for different supports.
To count internal member force unknowns, first we need to count how many members are in a frame. A frame member is defined by two end nodes. At any section of a member there are three internal unknown forces, T , V, and M. The state of force in the member is completely defined by the six nodal forces, three at each end node, because the three internal forces at any section can be determined from the three equilibrium equations taken from a FBD cutting through the section as shown below, if the nodal forces are known.
Internal section forces are functions of the nodal forces of a member.
Thus, each member has six nodal forces as unknowns. Denoting the number of members by M and the number of reaction forces at each support as R, the total number of force unknowns in a frame is then 6M+ΣR.
x
T
V M
On the other hand, each member generates three equilibrium equations and each node also generates three equilibrium equations. Denoting the number o f nodes by N, The total number of equilibrium equations is 3M+3N.
FBDs of a node and two members.
Because the number of members, M, appears both in the count for unknowns and the count for equations, we can simplify the expression for counting unknowns as shown below.
Counting unknowns against available equations.
The above is equivalent to considering each member having only three force unknowns.
The other three nodal forces can be computed using these three nodal forces and the three member equilibrium equations. Thus, a frame is statically determinate if 3M+ΣR= 3N.
If one or more hinges are present in a frame, we need to consider the conditions generated by the hinge presence. As shown in the following figure, the presence of a hinge within a member introduces one more equation, which can be called the condition of construction. A hinge at the junction of three members introduces two conditions of construction. The other moment at a hinge is automatically zero because the sum of all moments at the hinge (or any other point) must be zero. We generalize to state that the conditions of construction, C, is equal to the number of joining members at a hinge, m, minus one, C=m-1. The conditions of construction at more than one hinges is ΣC.
Nodal Equilibrium
Member Equilibrium
Member Equilibrium
Number of unknowns=6M+ΣR Number of equations= 3M+3N
Number of unknowns=3M+ΣR Number of equations= 3N
Since the conditions of construction provide additional equations, the available equation becomes 3N+C. Thus, in the presence of one or more internal hinges, a frame is statically determinate if 3M+ΣR= 3N+ΣC.
Presence of hinge introduces additional equations.
Example 1. Discuss the determinacy of the beams and frames shown.
Solution. The computation is shown with the figures.
M T V
M=0 T V
M T V
M T V
M T V
M T V
M=0 T V
M=0 T V M
T V
M T V
=0
=0
Counting internal force unknowns, reactions, and available equations.
For frames with many stories and bays, a simpler way of counting of unknowns and equations can be developed by cutting through members to produce separate “trees” of frames, each is stable and determinate. The number of unknowns at the cuts is the number of degree of indeterminacy as shown in the example that follows.
Example 2. Discuss the determinacy of the frame shown.
R=3
R=2
Number of unknowns = 3M+ΣR =7, Number of equations = 3N+ΣC = 7 Statically determinate.
R=1 R=3
Number of unknowns 3M+ΣR =15, Number of equations = 3N+C = 13 Indeterminate to the 2nd degree.
Number of unknowns = 3M+ΣR =21, Number of equations = 3N+C = 20 Indeterminate to the 1st degree.
R=3 R=3 R=2 R=1
R=3
R=1 Number of unknowns = 3M+ΣR =10, Number of equations = 3N+C= 10 Statically determinate.
Number of unknowns = 3M+ΣR =12, Number of equations = 3N+ΣC= 11 Indeterminate to the 1st degree.
R=1 R=3
R=2
M=1, N=2, C=0 M=1, N=2, C=1
M=3, N=4, C=1 M=5, N=6, C=2
M=2, N=3, C=1
M=2, N=3, C=2
R=3 Number of unknowns = 3M+ΣR =8,
Number of equations = 3N+ΣC = 6 Indeterminate to the 2nd degree.
Multi-story multi-bay indeterminate frame.
We make nine cuts that separate the original frame into four “trees” of frames as shown.
Nine cuts pointing to 27 degrees of indeterminacy.
We can verify easily that each of the stand-alone trees is stable and statically determinate, i.e. the number of unknowns is equal to the number of equations in each of the tree problems. At each of the nine cuts, three internal forces are present before the cut. All together we have removed 27 internal forces in order to have equal numbers of unknowns and equations. If we put back the cuts, we introduce 27 more unknowns, which is the degrees of indeterminacy of the original uncut frame.
This simple way of counting can be extended to multi-story multi-bay frames with hinges: simply treat the conditions of construction of each hinge as “releases” and subtract the ∑C number from the degrees of indeterminacy of the frame with the hinges removed. For supports other than fixed, we can replace them with fixed supports and counting the “releases” for subtracting from the degrees of indeterminacy.
Example 3. Discuss the determinacy of the frame shown.
Indeterminate frame example.
1
2 3
4
5 6
7 8 9
Solution. Two cuts and five releases amounts to 2x3-5=1. The frame is indeterminate to the first degree.
Short cut to count degrees of indeterminacy.
1 2
C=1 C=2
C=2