CELE Structural Theory & Analysis — Indeterminate Structures: Force MethodsStudy Notes

A structure is statically indeterminate when the number of unknown reactions and internal forces exceeds the number of independent equilibrium equations available (three in 2D: ΣFx = 0, ΣFy = 0, ΣM = 0). The degree of indeterminacy (DI) quantifies how many redundants must be removed to make a structure determinate.

For a structure with r reactions and m internal moment equations from releases:
DI = r + m - 3 (for 2D structures)
DI = r + m - 6 (for 3D structures)

Alternatively, for beams and frames:
DI = (number of unknown reactions + number of internal cuts) - (number of equilibrium equations)

A redundant is any reaction force or internal force that you can remove without causing the structure to become unstable. Removing all redundants produces the primary (or released) structure, which is always statically determinate. For example:

• In a propped cantilever (fixed at one end, supported on a roller at the free end), the prop reaction is the single redundant. Remove it → you have a simple cantilever, which is determinate.
• In a two-span continuous beam (simply supported at ends A and C, continuous over interior support B), the moment at B is one redundant. In fact, the reaction at B is also a redundant; typically we choose the moment MB as the primary unknown.
• In a fixed-fixed beam, there are three redundants: the two end moments and one of the end reactions (once you have those, the other reaction follows from equilibrium).

The key strategy: (1) identify the degree of indeterminacy, (2) select that many redundants, (3) remove them to obtain the determinate primary structure, (4) solve using compatibility of deformations.

The method of consistent deformation solves for redundants by enforcing that the actual displacements in the original (indeterminate) structure match the geometry and boundary conditions. The procedure has three core steps:

Step 1: Release the redundants and find the primary-structure response to applied loads.

Remove all redundants from the structure to obtain the primary (determinate) structure. Apply all the actual loads (not the redundants) to this primary structure and calculate the displacement at each point where a redundant was removed. Denote the displacement caused by the applied loads at the location of redundant i as δ₀ᵢ (the subscript 0 means 'due to applied loads').

For example, if the redundant is a prop reaction RB at the free end of a cantilever, then δ₀ is the downward deflection of that end under the applied load. If the redundant is a support moment, δ₀ is the rotation at that support due to the applied loads.

Step 2: Find the flexibility coefficients.

Apply a unit load (or unit moment) at each location where a redundant was removed, and calculate the resulting displacement at that same location. The flexibility coefficient δᵢⱼ is the displacement at location i due to a unit force (or moment) at location j.

For a single redundant, we denote this as δ₁₁, which is the deflection at the redundant's location caused by a unit value of that redundant. For the propped cantilever example, δ₁₁ is the upward deflection (per unit prop force) at the free end.

For multiple redundants, δᵢⱼ represents the displacement at location i due to a unit redundant at location j. The set of all δᵢⱼ forms the flexibility matrix.

Step 3: Enforce compatibility.

The original structure has fixed or known boundary conditions. At each location where a redundant was removed, the actual displacement must equal the physical constraint. If the location is a fixed support, the displacement is zero. If there is a known settlement Δ, the actual displacement equals Δ.

The compatibility equation is:

δ₀ + R₁·δ₁₁ + R₂·δ₁₂ + ... = Δ

For one redundant:
δ₀ + R₁·δ₁₁ = Δ
R₁ = (Δ - δ₀) / δ₁₁

For multiple redundants, you obtain a system of linear equations:

δ₀₁ + R₁·δ₁₁ + R₂·δ₁₂ + ... + Rₙ·δ₁ₙ = Δ₁
δ₀₂ + R₁·δ₂₁ + R₂·δ₂₂ + ... + Rₙ·δ₂ₙ = Δ₂
...
δ₀ₙ + R₁·δₙ₁ + R₂·δₙ₂ + ... + Rₙ·δₙₙ = Δₙ

Once the redundants are known, all other internal forces and reactions follow from equilibrium applied to the determinate primary structure with the redundants included as external loads.

Key principle: The flexibility method treats redundants as unknowns and uses the condition that actual displacements (deformations) must be compatible with the structural geometry and supports. This is why it is also called the method of compatible deformations.

The three-moment equation (also called Clapeyron's equation) is a powerful relationship that directly connects the bending moments at three consecutive supports of a continuous beam. It is derived from the force method but is expressed in a form that requires no explicit calculation of flexibility coefficients for each individual problem.

Statement of the Three-Moment Equation:

For three consecutive supports A, B, and C of a continuous beam, with spans L₁ (from A to B) and L₂ (from B to C), the equation is:

MA·L₁ + 2MB·(L₁ + L₂) + MC·L₂ = -6·(A₁·x̄₁/L₁ + A₂·x̄₂/L₂)

Where:

• MA, MB, MC = bending moments at supports A, B, C (positive = sagging/tension on bottom)
• L₁, L₂ = span lengths
• A₁, A₂ = areas of the simple-span bending moment diagrams for each span (due to applied loads only)
• x̄₁, x̄₂ = distances from the left end to the centroids of A₁ and A₂, respectively
• The term 6·(A₁·x̄₁/L₁ + A₂·x̄₂/L₂) is the "load term" for the three consecutive spans

Key observations:

1. If either end (A or C) is a simple support or free (no moment is transmitted), set that moment to zero: MA = 0 or MC = 0.
2. If either end is fixed, the support moment is unknown and must be included in the equation.
3. The load term on the right side depends on the type of load (UDL, point load, triangular, etc.) and its position.
4. Apply the equation at each interior support to generate as many equations as there are unknown support moments.

Load Terms for Common Loadings:

For a span of length L with load, the simple-span moment diagram has area A and centroid at distance x̄ from the left. The load term is 6Ax̄/L.

• Uniformly distributed load w across the full span: A = wL³/12, x̄ = L/2, so 6Ax̄/L = 6·(wL³/12)·(L/2)/L = wL³/4
• Concentrated point load P at distance a from the left: A = Pa(L-a)/2, x̄ = (L-a)/2 (from left), so 6Ax̄/L = 3Pa(L-a)·(L-a)/(2L) = 3Pa(L-a)²/(2L) (if a = L/2, this becomes 3PL²/8)
• Concentrated moment M applied at any point in the span: The moment does not produce a simple-span moment diagram (it creates a discontinuity), so this term appears only in the 'load term' of the equation.

Application Procedure:

1. Identify all interior supports where the moment is unknown.
2. Apply the three-moment equation at each interior support, treating the two adjacent spans.
3. For each unknown interior support moment, write one equation.
4. Solve the system of linear equations simultaneously.
5. Once all support moments are known, use equilibrium to find all reactions and internal forces.

Advantages:

• Direct approach; no need to compute flexibility coefficients manually for each problem.
• Applicable to variable-moment regions or non-uniform loads within the force-method framework.
• Systematic for multi-span beams.

Limitations:

• Applies only to continuous beams (not to other indeterminate structures like frames).
• Requires careful computation of the load terms A and x̄.
• For structures with supports that settle, the equation is modified (see Example 3 below).

The force method (and the three-moment equation as its special case for continuous beams) appears throughout professional engineering practice and the PRC licensure examination. Understanding how to classify problems and choose the right approach is essential.

A. Propped Cantilevers:

A propped cantilever is a cantilever with an additional support (roller or pin) at or near the free end. It is statically indeterminate to the first degree (DI = 1). The prop reaction is the single redundant.

Standard results (memorize these):

• Under UDL w over length L:

  • Prop reaction: Rprop = 3wL/8
  • Fixed-end reaction: Rfixed = 5wL/8
  • Fixed-end moment: Mfixed = wL²/8 (hogging)

• Under point load P at distance a from fixed end:

  • Prop reaction: Rprop = P·a²(3L - 2a) / (2L³)
  • Fixed-end moment: Mfixed = P·a(L - a) / L

For an exam problem on a propped cantilever:

1. Identify the load type and position.
2. Choose the prop reaction as the redundant.
3. Use the method of consistent deformation or the standard formulas.
4. Calculate the fixed-end moment and reaction from equilibrium.

B. Continuous Beams (Two or More Spans):

Continuous beams are among the most common structures in building and bridge design. They can have:

• Simple supports at the ends (pinned or roller).
• One or more interior supports (pinned or roller, typically).
• Fixed supports at one or both ends (less common in practice, but possible).

Degree of indeterminacy: DI = number of interior supports (for a beam with simply supported ends).

For example:

• Two-span beam (one interior support): DI = 1, one unknown (the interior moment)
• Three-span beam (two interior supports): DI = 2, two unknowns (moments at both interior supports)

Using the three-moment equation:

1. Write one equation at each interior support.
2. Substitute known boundary moments (zero for simple ends, unknown for fixed ends).
3. Substitute load terms for each span.
4. Solve the system of linear equations.
5. Use equilibrium to find all reactions and internal forces.

C. Fixed-Fixed Beams:

A fixed-fixed beam is supported by fixed connections at both ends. It has DI = 3 (the two end moments and one of the end reactions are redundant).

For a fixed-fixed beam under a central point load P:

• The two end moments are equal in magnitude: MA = MC = PL/8 (hogging at both ends)
• The two end reactions are equal: RA = RC = P/2
• The maximum positive moment at mid-span is PL/8 (sagging)
• The shear is constant in each half: V = ±P/2

For a fixed-fixed beam under UDL w:

• End moments: MA = MC = wL²/12 (hogging)
• End reactions: RA = RC = wL/2
• Maximum positive moment at mid-span: wL²/24 (sagging)

Deriving these requires the method of consistent deformation. The results are standard and should be memorized.

D. Frames and Other Indeterminate Structures:

For frames (rigid jointed structures), the degree of indeterminacy is higher. The force method still applies but requires more computational effort. Choose redundants strategically:

• For a simple rectangular portal frame: DI = 3 (typically, choose the three moment components or one moment and two reactions).
• For a three-hinge frame: remove one hinge to create a statically determinate arch-like structure.

E. Support Settlements and Temperature Changes:

When supports settle or the structure experiences differential temperature effects, the compatibility equation is modified:
δ₀ + R·δ₁₁ = Δ (where Δ is the known settlement or displacement, not zero)

For the three-moment equation with settlement:
MA·L₁ + 2MB·(L₁ + L₂) + MC·L₂ = -6·(load terms) - 6EI·(1/L₁ + 1/L₂)·Δ

Temperature changes create curvature changes (related to the coefficient of thermal expansion and the temperature difference across the depth). These are handled similarly, with the "load term" replaced by the equivalent curvature term.

Success in solving indeterminate structure problems requires both conceptual understanding and systematic problem-solving skills. Here is a structured approach for the PRC examination:

Step 1: Classify the Problem

• Identify the structure type: propped cantilever, continuous beam, fixed-fixed beam, frame, etc.
• Count the degree of indeterminacy. For a beam: DI = (number of unknown reactions) - (number of equilibrium equations) = r - 3.
• Determine the number of redundants = DI.

Step 2: Choose a Solution Method

• For continuous beams with simply supported or fixed ends: use the three-moment equation (fastest).
• For propped cantilevers or other single-redundant structures: memorize or use the standard formulas, or apply consistent deformation if needed.
• For frames or more complex structures: use the method of consistent deformation.

Step 3: Apply the Method

• Force method (consistent deformation):

  1. Release each redundant and calculate δ₀ (displacement due to applied loads).
  2. Apply unit redundant and calculate δᵢⱼ (flexibility coefficients).
  3. Write compatibility equations: δ₀ + Σ(Rᵢ·δᵢⱼ) = Δ.
  4. Solve for redundants.

• Three-moment equation:

  1. Write the equation at each interior support.
  2. Substitute boundary conditions (zero moments at simple supports).
  3. Calculate load terms from the simple-span diagrams.
  4. Solve the system of equations.

Step 4: Verify and Complete

• Use equilibrium (ΣF = 0, ΣM = 0) to find remaining unknowns.
• Sketch the bending moment and shear diagrams to check for physical reasonableness.
• Verify total loads and reactions add up correctly.

Common Examination Pitfalls to Avoid:

1. Sign errors in flexibility coefficients:

   • Always be consistent with your sign convention. If the redundant points upward and the load deflection is downward, they oppose each other.
   • δ₀ and δ₁₁ should have opposite signs in the compatibility equation δ₀ + R·δ₁₁ = 0.

2. Incorrect load terms in the three-moment equation:

   • Memorize: UDL over full span → wL³/4; central point load → 3PL²/8.
   • If the load doesn't cover the full span or is not at mid-span, recalculate A and x̄ carefully.
   • Don't confuse the moment area (A) with the bending moment value.

3. Forgetting boundary conditions:

   • Simple supports: moment = 0.
   • Fixed supports: moment is unknown (include it in the equations).
   • Overhangs: if the overhang is free, the end moment is zero (unless a concentrated moment is applied).

4. Arithmetic errors in solving systems of equations:

   • For a large system, use substitution or matrix methods carefully.
   • Check your answer by substituting back into the original equations.

5. Failing to complete the solution:

   • If asked for reactions, moments, and diagrams, calculate all of them.
   • Always sketch the bending moment diagram; it provides a sanity check.

Examination Techniques:

1. Time management: The three-moment equation is faster than the general force method for continuous beams. Recognize when each applies.
2. Diagram sketches: Quickly sketch the primary structure and the M-diagram to visualize the problem. A picture prevents many errors.
3. Dimensional analysis: Check that your final units are correct (kN for force, kN·m for moment).
4. Symmetry: Exploit symmetry to reduce the number of unknowns (e.g., MB = MC for a three-span beam with equal spans and identical loading).
5. Known results: Memorize the standard formulas for propped cantilevers and fixed-fixed beams under common loads. These save time in the exam.

Key Formulas to Memorize:

Propped cantilever (UDL): Rprop = 3wL/8, Mfixed = wL²/8
Fixed-fixed beam (UDL): Mend = wL²/12, Mmax(interior) = wL²/24
Fixed-fixed beam (central point load P): Mend = PL/8
Three-moment equation: MA·L₁ + 2MB·(L₁ + L₂) + MC·L₂ = -6·(A₁x̄₁/L₁ + A₂x̄₂/L₂)
Compatibility (1 redundant): δ₀ + R·δ₁₁ = Δ

Practice Strategy:

1. Solve problems of increasing complexity: start with two-span beams, then three-span, then frames.
2. Solve each problem two ways if possible (e.g., by the three-moment equation and then by consistent deformation) to verify.
3. Keep a list of the load terms (A, x̄) for various loading patterns; refer to it until memorized.
4. Sketch the final M-diagram for every problem to visualize the structural response.