Overview of statically indeterminate structures
If unknowns are more than the equations ΣFx = 0, ΣFy = 0, ΣM = 0, equilibrium alone is not enough. Degree of indeterminacy = (unknown reactions + internal unknowns) − (equations). For a plane beam: r − 3. For a truss: m + r − 2j.
- Simple beam: r = 3, degree 0.
- Propped cantilever or two-span beam: r = 4, degree 1.
- Beam fixed at both ends: r = 6, degree 3.
To solve we add compatibility conditions: for example, the deflection at an extra support must be zero. They use E, I and the length, so the answer depends on stiffness EI, unlike a determinate structure.
Advantages: smaller moments and sag, a second path for load if one part fails (safer), saving material. Disadvantages: settlement, temperature changes and fabrication errors create stress, and the analysis is harder.
Methods of solving
Force (flexibility) method: remove the redundant, solve the remaining determinate structure, find the gap it leaves, then apply the redundant force that closes the gap (compatibility). Displacement (stiffness) methods: slope-deflection and moment distribution take joint rotations as unknowns; computers use the matrix stiffness method.
Steps of the force method: (1) choose a redundant, (2) remove it, (3) find the deflection from the real loads, (4) find the deflection from a unit redundant, (5) write compatibility, (6) solve, (7) add the effects.
Statically indeterminate beams and frames
Standard results for an even load w:
- Propped cantilever (fixed one end, roller at other), length L: prop reaction 3wL/8, fixed-end moment wL²/8.
- Fixed-fixed beam: end moments wL²/12 (top fibres stretched), mid-span moment wL²/24, mid sag wL⁴/(384EI), which is 5 times less than a simple beam.
- Two equal spans l (continuous): end reactions 3wl/8, middle reaction 5wl/4, moment at the middle support wl²/8.
Frames: a portal frame with both feet fixed has 6 reactions, degree 3. The stiff corners share the moment between beam and columns, so the beam moment is smaller than in a pin-ended frame.
Support settlement Δ in a fixed beam adds end moments of 6EIΔ/L²; a determinate beam is not affected at all.
Key formulas and definitions
- Degree of indeterminacy (plane beam) = r − 3
- Truss: degree = m + r − 2j
- Propped cantilever (even load w): R_prop = 3wL/8, M_fixed = wL²/8
- Fixed-fixed: M_end = wL²/12, M_mid = wL²/24, δ = wL⁴/(384EI)
- Two equal spans l: R_mid = 5wl/4, R_end = 3wl/8, M_mid support = wl²/8
- Simple beam: M_max = wL²/8, δ = 5wL⁴/(384EI)
- Settlement in a fixed beam: ΔM = 6EIΔ/L²
Worked examples
1. Find the degree of indeterminacy of a beam with one fixed end and two rollers.
Reactions r = 3 + 1 + 1 = 5. Degree = 5 − 3 = 2.
2. A propped cantilever 4 m long carries 6 kN/m. Find the prop reaction and the fixed-end moment.
R_prop = 3wL/8 = 3 × 6 × 4 / 8 = 9 kN. M_fixed = wL²/8 = 6 × 16 / 8 = 12 kN·m.
3. A beam 5 m long is fixed at both ends and carries 12 kN/m. Find the end and mid-span moments.
M_end = wL²/12 = 12 × 25 / 12 = 25 kN·m. M_mid = wL²/24 = 12.5 kN·m.
4. A continuous beam has two equal spans of 3 m under 8 kN/m. Find the middle reaction.
R_mid = 5wl/4 = 5 × 8 × 3 / 4 = 30 kN. Each end reaction = 3wl/8 = 9 kN. Check: 9 + 30 + 9 = 48 = 8 × 6.
5. Compare the largest moment in a simple beam and a fixed-fixed beam of the same span L = 6 m under 10 kN/m.
Simple: wL²/8 = 10 × 36 / 8 = 45 kN·m. Fixed-fixed: wL²/12 = 30 kN·m. The fixed beam has 1.5 times smaller maximum moment.
6. A truss has 10 bars, 6 joints and 3 reactions. Find the degree of indeterminacy.
m + r − 2j = 10 + 3 − 12 = 1. It is indeterminate to the first degree.
Common mistakes
- Thinking equilibrium alone solves an indeterminate beam. Compatibility with EI is needed.
- Forgetting that fixed ends give 3 reactions each, so r = 6 in a plane.
- Assuming extra supports are always better. A settling support can overstress the structure.
- Mixing up the end moment (wL²/12) and the mid-span moment (wL²/24) of a fixed beam; end moments are bigger.