Module 16 · Lesson 16.5
Kinematic indeterminacy and indeterminate trusses
Counting the other kind of unknown, and using the force method on a truss.
Why this matters
There are two ways to count what you do not know about a structure: how many forces statics cannot find, and how many displacements are free to happen. The first drives the force method, the second the stiffness method — and the choice between them is usually decided by whichever number is smaller. Then the force method, which solved the propped cantilever, transfers to a truss with almost no change.
By the end of this lesson you should be able to
- Count the kinematic indeterminacy of a frame
- Explain why it decides whether a stiffness or a flexibility approach is easier
- Apply the force method to a truss with one redundant member
- Recognise that member sizes now affect the forces, unlike in a determinate truss
What you should already know
- Static indeterminacy and the force method (this module)
- The unit-load method for trusses (Module 15)
- Method of joints (Module 4)
Static indeterminacy counts the forces that equilibrium cannot find. Kinematic indeterminacy counts the joint displacements that are free to occur — the unknowns the stiffness method solves for.
In a plane frame every joint has three degrees of freedom: two translations and a rotation. So:
kinematic indeterminacy = 3j − (restrained degrees of freedom)
A further reduction is common. If the members are taken as axially rigid — a standard assumption for frames, where axial deformations are tiny compared with bending ones — each such member removes one more degree of freedom, because the distance between its ends can no longer change.
The two counts are independent, and they point in opposite directions:
- A continuous beam on many supports is highly statically indeterminate but has few free rotations, so it is kinematically simple. Moment distribution — a stiffness method — handles it easily.
- A truss with one extra member is statically indeterminate to degree 1 but has two free translations at every joint, so it is kinematically complex. The force method is the easier route by hand.
Computers do not care — the stiffness method is systematic and needs no judgement, which is why it won. By hand, the count tells you which way to go.
Predict first
A portal frame has four joints and two fixed bases. What is its kinematic indeterminacy if axial deformations are ignored?
Now the truss. A truss with one member more than determinacy requires is statically indeterminate to the first degree, and the force method applies exactly as it did to the propped cantilever:
- 1.Release the redundant — cut one member, or remove one support.
- 2.Analyse the released truss under the real loads. Call these forces N. The cut member carries nothing.
- 3.Analyse it under a unit force in the redundant, applied as a self-equilibrating pair at the cut. Call these forces n, with n = 1 in the redundant itself.
- 4.Impose compatibility. In the real truss there is no gap at the cut, so the gap opened by the loads must be closed by the redundant:
R = −Σ(N n L/AE) ÷ Σ(n² L/AE)
- 5.The final force in every member is N + R·n.
The sums must include the redundant member itself. Its contribution to the numerator is zero (its N is zero) but to the denominator it is L/AE, which is exactly the flexibility of the member being restored.
Worked example
A three-bar indeterminate system
Given
- Three bars meet at a common joint and are anchored above it
- One bar is vertical, length 1000 mm; the other two are at 45° either side, so each is 1414 mm long
- All bars: A = 1000 mm², E = 200 000 N/mm²
- A vertical load of 100 kN hangs at the joint
Find
The force in each bar, by the force method.
Assumptions
- Pin joints throughout, so the bars carry axial force only
- Small deflections, so the geometry is unchanged
Practice
A plane frame has 4 joints and 6 restrained degrees of freedom at its supports. Ignoring axial deformation of its 3 members, what is its kinematic indeterminacy?
Practice
Three bars support a joint: one vertical of length 1000 mm and two at 45° of length 1414 mm, all with A = 1000 mm² and E = 200 000 N/mm². A load of 100 kN hangs at the joint. What force does the vertical bar carry, in kN?
Practice
For the same system, what force does each inclined bar carry, in kN?
Summary
- Kinematic indeterminacy = 3j − restrained dof, less one per axially rigid member
- It is the number of unknowns a stiffness method must solve for
- Continuous beams are statically complex but kinematically simple; trusses are the reverse
- Force method on a truss: R = −Σ(NnL/AE) ÷ Σ(n²L/AE), including the redundant
- Final force in each member = N + R·n
- In an indeterminate truss the member sizes change the forces, so design is iterative
This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint