Indeterminate Structures: Force Methods - Cheat Sheet
Your 30-minute revision companion for force (flexibility) method problems, three-moment equations, and redundant selection on the PRC Civil Engineer Licensure Exam. Covers propped cantilevers, continuous beams, and consistent deformation.
Degree of Indeterminacy & Redundant Selection
Formulas
DI = 3C + R – 3J (plane frame)
Meaning
C = closed loops, R = reaction components, J = pin/hinge points; counts how many forces are redundant
Watch out
Do NOT count hinges as separate supports — each hinge removes one constraint. Avoid confusing DI with number of supports.
When to use
First step: determine how many redundants must be released to get a determinate primary structure
DI = N – 2J + 3C (truss)
Meaning
N = number of members, J = joints, C = closed loops
Watch out
This is for trusses only; for frames use the 3-member equation.
When to use
For truss indeterminacy (less common in force methods but still tested)
Important facts
- A structure is statically determinate if DI = 0; indeterminate if DI > 0; unstable if DI < 0.
- Choose redundants wisely: preferably reactions at supports or moments at fixed ends (easier to compute deflections).
- Avoid choosing internal member forces as redundants unless necessary — they require cutting the member and is more tedious.
- The number of compatibility equations = DI; solve them simultaneously to find all redundants.
- A propped cantilever is DI = 1 (one redundant: the prop reaction).
Key definitions
Redundant
Example
In a propped cantilever, the prop is the redundant. In a two-span continuous beam, the interior support moment is the redundant.
A reaction or internal force whose removal does not cause instability; its removal gives a determinate primary structure.
Primary (Released) Structure
Example
A propped cantilever becomes a simple cantilever when the prop is removed.
The determinate structure obtained by removing all redundants; used as the base case for force method.
Flexibility Coefficient (δ_ij)
Example
δ_11 is the deflection at a released support caused by applying unit force at that same point.
Deflection at point i per unit redundant force at point j; inverse of stiffness.
Degree of Indeterminacy (DI)
Example
DI = 1 for a propped cantilever or two-span simple beam; DI = 3 for a fixed-fixed beam.
Number of unknown force(s) exceeding equilibrium equations; equals the number of redundants.
Diagrams to know
- Statically determinate beam (simply supported, cantilever) vs indeterminate (propped cantilever, fixed-fixed, continuous).
- Primary structure and redundants after release (sketch showing removed support or moment).
- Bending moment diagram for primary structure under applied load.
- Bending moment diagram for primary structure under unit redundant.
Method of Consistent Deformation (Flexibility Method)
Formulas
δ₀ + R₁δ₁₁ + R₂δ₁₂ + ... = Δ (compatibility for one released point)
Meaning
δ₀ = deflection at released point from applied loads; R = redundant force; δ_ij = flexibility; Δ = known settlement (0 for fixed support)
Watch out
Sign convention: deflections in same direction are positive; if δ₀ and R·δ_ii have opposite senses, one must be negative. Δ = 0 for a support with no settlement.
When to use
Always — this is the core compatibility equation; set up one equation per redundant.
δ = ∫(M·m)/(EI) dx (virtual work / unit-load method)
Meaning
M = bending moment from applied loads; m = bending moment from unit load at the point of interest; integrate over all segments
Watch out
Must break integral into segments where both M and m are continuous. Use conjugate beam or tables for standard cases (cantilever, simple span).
When to use
Computing deflections δ₀ and δ_ij; the most common approach for frames and beams.
δ = (1/EI)∫M·m dx ≈ (1/EI)∑(M_c·m_c·ΔL) (graphical / area method)
Meaning
Multiply bending moment area by ordinate of m diagram at the centroid of M; sum across segments
Watch out
Only works if EI is constant. Must align centroid of M with corresponding ordinate in m diagram carefully.
When to use
Quick hand calculation when M and m diagrams are composed of simple shapes (triangles, parabolas, rectangles).
Common values
Value
PL³/(3EI)
Symbol
δ
Quantity
Cantilever end deflection (point load P at free end, length L)
Value
wL⁴/(8EI)
Symbol
δ
Quantity
Cantilever end deflection (UDL w, length L)
Value
5wL⁴/(384EI)
Symbol
δ
Quantity
Simply supported beam center deflection (UDL w, span L)
Value
PL²/(2EI)
Symbol
θ
Quantity
Cantilever slope at free end (point load P)
Important facts
- Always compute δ₀ and δ_ii with consistent sign conventions: positive = downward (or direction of redundant force).
- For a cantilever of length L with free-end deflection: δ = wL⁴/(8EI) for UDL, δ = PL³/(3EI) for point load at free end.
- For a simple span with center load: δ_center = PL³/(48EI); with UDL: δ_center = 5wL⁴/(384EI).
- The flexibility coefficient δ_ii is always positive; it represents the stiffness inverse (smaller δ = stiffer structure).
- If DI > 1, set up a system of simultaneous compatibility equations; matrix form: {δ_ij}{R_j} = {−δ_0i}.
Key definitions
Compatibility Equation
Example
For a propped cantilever, the prop prevents downward deflection: δ_due_to_load + R_prop × (deflection_per_unit_prop) = 0.
Mathematical statement that the actual deflection at a released point equals the prescribed value (usually zero for a fixed support).
Virtual Work (Unit-Load Method)
Example
To find vertical deflection at a point, apply a unit vertical load there; to find slope, apply a unit moment.
Technique to find deflection by imagining a unit load at the desired point and integrating M×m/(EI) over the structure.
Δ (Support Settlement)
Example
If a support settles 10 mm downward, Δ = +0.01 m (or −0.01 m depending on sign convention; be consistent).
Known vertical displacement of a support (positive downward); default = 0 for idealized fixed/pinned supports.
Diagrams to know
- Bending moment diagrams: primary structure under applied load, and under unit redundant (separate sketches).
- Deflection curve qualitatively: concave down (sagging) vs concave up (hogging) based on moment sign.
- Virtual load diagram: unit load (or unit moment) applied at the point where deflection is desired.
Reactions or equations
For single redundant: R = (Δ − δ₀) / δ₁₁
Note
Rearrange from δ₀ + R·δ₁₁ = Δ; if Δ = 0, then R = −δ₀/δ₁₁ (negative sign often absorbed into sign convention).
Conditions
When δ₀ and δ₁₁ have opposite effects (load pulls down, redundant supports up), the equation gives the correct sign.
[δ_ij]{R_j} = −{δ_0i} (matrix form for multiple redundants)
Note
δ_ij = deflection at i due to unit force at j; δ_ii (diagonal) is always positive; off-diagonal terms δ_ij = δ_ji (reciprocal theorem).
Conditions
DI > 1; solve by Gaussian elimination or Cramer's rule.
Propped Cantilever (Common DI = 1 Case)
Formulas
R_prop = 3wL/8 (UDL, length L)
Meaning
Reaction at the prop (roller at free end); w = uniform load
Watch out
This is NOT wL/2 (which is the simple-beam reaction). The fixed end 'clamps' the beam, reducing the prop reaction.
When to use
Propped cantilever under uniformly distributed load — MEMORIZE this value for exams.
M_fixed = wL²/8 (hogging moment at fixed end for propped cantilever with UDL)
Meaning
Bending moment at the fixed support (fixed-end moment); negative by convention (sagging into the beam).
Watch out
The sign is hogging (negative), not sagging. Moment equilibrium: M_fixed = (1/2)wL² − (3wL/8)·L = wL²/8 (check by moments about the fixed end).
When to use
Always arises in propped cantilever under UDL; memorize alongside R_prop = 3wL/8.
R_prop = 5PL/(16L) = 5P/16 (point load P at midspan)
Meaning
Reaction when propped cantilever carries a point load at center.
Watch out
NOT 5P/8 or P/2. The factor 5/16 comes from the flexibility ratio: (5L³/48EI) / (L³/3EI) = 5/16.
When to use
Propped cantilever with central point load.
R_A = wL − R_prop (reaction at fixed end)
Meaning
Equilibrium of vertical forces: upward shear at fixed end + prop reaction = total load.
Watch out
Not the same as R_B in a simple beam; the fixed-end shear is LARGER because the fixed end also resists moment.
When to use
Always to check your answer and find the fixed-end reaction.
Common values
Value
3wL/8
Symbol
R_prop
Quantity
Propped cantilever UDL: prop reaction
Value
wL²/8
Symbol
M_fixed
Quantity
Propped cantilever UDL: fixed-end moment
Value
5wL/8
Symbol
R_A
Quantity
Propped cantilever UDL: fixed-end shear
Value
5L/8 from fixed end
Symbol
x₀
Quantity
Zero-shear location in propped cantilever UDL
Important facts
- Propped cantilever is the canonical DI = 1 problem; appears in nearly every structural theory exam.
- The prop reaction is always LESS than the simple-beam reaction (wL/2) because the fixed end carries negative moment.
- The shear diagram changes sign (zero crossing) between the prop and the free end; find it from V(x) = wL − 3wL/8 − wx.
- The maximum positive moment (sagging) occurs where shear = 0; max negative moment is at the fixed end.
- For point loads, use the flexibility method: δ₀ and δ₁₁ depend on load position; tables or hand integration required.
Key definitions
Propped Cantilever
Example
A building facade beam fixed to a column at one end and resting on a prop (support) at the other end.
A beam fixed at one end and supported (pinned or roller) at the other free end; DI = 1, statically indeterminate.
Hogging Moment
Example
In a propped cantilever, the fixed end always has a hogging moment; the interior moments may sag or hog depending on load.
Negative bending moment causing concave-up curvature (sagging region above the neutral axis tends to go in compression).
Diagrams to know
- Bending moment diagram: negative (hogging) at fixed end, reaches maximum positive (sagging) at zero-shear location, returns to zero at prop.
- Shear diagram: positive at fixed end (R_A), decreases linearly with slope −w, crosses zero, reaches −R_prop at prop.
- Deflection curve: concave down (hogging) near fixed end, concave up (sagging) near prop; zero slope at fixed end (fixed support), zero deflection at prop.
Reactions or equations
Shear at any section: V(x) = R_A − wx (where x = 0 at fixed end, x = L at prop)
Note
Shear goes from +R_A to −R_prop; sign change indicates where max sagging moment occurs.
Conditions
UDL w over the whole span; V = 0 at x = R_A/w = (5wL/8)/w = 5L/8.
Moment at any section: M(x) = M_fixed + R_A·x − (1/2)wx²
Note
At x = L (prop), M = 0 (simple support). At x = 5L/8 (zero shear), M = max sagging moment.
Conditions
UDL w; M_fixed = −wL²/8 (negative = hogging); integrate shear to get moment.
Three-Moment Equation for Continuous Beams
Formulas
M_A·L₁ + 2M_B(L₁ + L₂) + M_C·L₂ = −6(A₁·x̄₁/L₁ + A₂·x̄₂/L₂)
Meaning
M = bending moment at supports A, B, C (three consecutive); L = span lengths; A = area of simple-beam M diagram; x̄ = distance from left end to centroid of A; negative RHS = sagging area convention
Watch out
The RHS term is −6(A₁x̄₁/L₁ + A₂x̄₂/L₂), NOT +6. The negative sign accounts for sagging moment areas creating hogging at interior supports. Get the load term RIGHT or the entire answer is wrong.
When to use
Continuous beam; write one equation per interior support. Simply supported ends: M = 0. Fixed end: unknown (write a second three-moment equation or use fixed-end conditions).
For UDL w on both spans: 6(A·x̄/L) = 6·(wL³/12)·(L/2)/L = wL³/4 (per span)
Meaning
Simplified form for uniform load: area of M diagram = wL³/12, centroid at L/2 from either end, so term = wL³/4.
Watch out
The value is wL³/4 (NOT wL²/4 or other variants). Memorize: UDL span term = wL³/4.
When to use
Quick calculation when all spans carry the same UDL.
For central point load P on span: 6(A·x̄/L) = 6·(PL²/8)·(L/2)/L = 3PL²/8 (term for that span)
Meaning
For a point load at midspan: triangular M diagram of area PL²/8, centroid at 2L/3 from one end, so 6·(PL²/8)·(2L/3) / L = PL²/4. (Note: standard is 3PL²/8 for general loading tables.)
Watch out
The factor depends on load position. For central load: 3PL²/8 (if at midspan). Off-center loads require integration or tables.
When to use
Continuous beam with concentrated loads.
Common values
Value
wL³/4
Symbol
Load term per span
Quantity
Three-moment UDL term (one span)
Value
3PL²/8
Symbol
Load term (midspan P)
Quantity
Three-moment central point load term
Value
−wL²/8
Symbol
M_interior
Quantity
Two-span uniform beam interior moment
Important facts
- Three-moment equation relates moments at three consecutive supports; write one equation per interior support (DI equations).
- Simply supported ends: M = 0; fixed ends: use a second three-moment equation or boundary conditions (slope = 0).
- The load term RHS has a negative sign by convention (sagging areas → hogging interior moments). Do NOT drop this sign.
- For a uniform two-span beam (equal L, equal w): M_B = −wL²/8 (at the interior support).
- Support reactions follow from moment diagrams: R = (M_left − M_right)/L + distributed load contribution.
- Always check equilibrium: sum of all reactions = sum of applied loads; sum of moments about any point = 0.
Key definitions
Simple-Beam Moment Diagram
Example
For a UDL on a 6 m span, the simple-beam M diagram is a parabola with peak = wL²/8 = 3w (where w in kN/m).
Bending moment diagram of a single span if it were simply supported (ignoring continuity); used to compute the load terms in three-moment equation.
Interior Support
Example
In a three-span continuous beam A–B–C–D, B and C are interior supports; A and D are end supports.
A support within the beam (not at the ends); moment at an interior support is typically unknown in a continuous beam.
Load Term (RHS of Three-Moment)
Example
For two equal UDL spans: RHS = −6(wL³/4 + wL³/4) = −3wL³/2.
The quantity −6(A₁x̄₁/L₁ + A₂x̄₂/L₂), which encodes the effect of applied loads on the moments at three consecutive supports.
Diagrams to know
- Bending moment diagram of the continuous beam (sketch showing hogging at interior supports, sagging in spans).
- Simple-beam moment diagrams for each span (dashed lines, for reference only — not the final answer).
- Shear force diagram (derived from slopes of M diagram; discontinuities at supports and under concentrated loads).
- Deflection curve (qualitative; concave up at interior hogging moments, concave down in sagging regions).
Reactions or equations
At an interior support: sum moments about that support from left and right spans must balance the unknown moment M there.
Note
The equation automatically enforces moment continuity (same moment on left and right).
Conditions
Three-moment equation is derived from moment equilibrium at the support.
Reaction at an interior support: R_B = (sum of end moments)/L + midspan load term
Note
Always compute reactions from beam equilibrium (vertical and moment) AFTER solving for moments.
Conditions
For symmetrical loading, reactions simplify.
Multiple Redundants & Simultaneous Equations
Formulas
[δ_ij] = [[δ₁₁, δ₁₂, δ₁₃, ...], [δ₂₁, δ₂₂, δ₂₃, ...], ...] (flexibility matrix)
Meaning
δ_ij = deflection at point i due to unit force at redundant j; symmetrical matrix (δ_ij = δ_ji by reciprocal theorem).
Watch out
Ensure consistent sign convention: all deflections and redundants in the same direction (e.g., downward positive). The diagonal terms δ_ii are always positive; off-diagonal can be positive or negative.
When to use
DI > 1; set up a system of compatibility equations in matrix form.
{δ₀} = {deflection at each released point from applied loads} (load vector)
Meaning
δ₀ᵢ = deflection at point i caused by the actual applied loads on the primary structure.
Watch out
Sign: if δ₀ is downward and the redundant force opposes downward motion, include the negative sign in the equation setup.
When to use
Right-hand side of the matrix equation: [δ_ij]{R_j} = −{δ₀}.
[δ_ij]{R_j} = −{δ₀} (matrix form: deflection matrix × redundants = negative load deflections)
Meaning
Simultaneous compatibility equations for DI > 1; solve by Gaussian elimination, Cramer's rule, or matrix inversion {R} = −[δ_ij]⁻¹{δ₀}.
Watch out
Invert the flexibility matrix (or solve the system); the negative sign is critical. Common error: forgetting the inversion or getting the sign of {R} wrong.
When to use
Fixed-fixed beam (DI = 3), two-span beam with interior support moment as redundant, frames with multiple redundants.
Important facts
- Number of compatibility equations = DI; always solvable if the structure is stable.
- The flexibility matrix is symmetric and positive-definite (all eigenvalues > 0).
- For hand calculation of 2×2 or 3×3 systems, Cramer's rule is faster than Gaussian elimination.
- Always check: number of equations = number of unknowns (redundants).
- After solving for redundants, use equilibrium to find all reactions, shears, and moments.
Key definitions
Flexibility Matrix
Example
For a fixed-fixed beam with three redundants (two fixed-end moments + one internal release), the 3×3 flexibility matrix encodes all pairwise deflections.
A square matrix [δ_ij] where each entry is the deflection (at point i) caused by a unit force (at point j); inverse of stiffness matrix.
Reciprocal Theorem (Maxwell's Law)
Example
The deflection at point A caused by a unit load at point B equals the deflection at B caused by a unit load at A (both on the primary structure).
Deflection at point i due to unit load at j equals deflection at j due to unit load at i: δ_ij = δ_ji.
Diagrams to know
- Primary structure with all redundants released (e.g., fixed-fixed beam treated as simple span with end moments as redundants).
- Moment diagrams: (1) primary structure under applied loads, (2) unit moment at first redundant, (3) unit moment at second redundant, etc.
- Final moment diagram combining all contributions: M_final = M₀ + R₁·m₁ + R₂·m₂ + ... (superposition).
Reactions or equations
Cramer's rule for 2 redundants: R₁ = −det([δ₀, δ₁₂; δ₀₂, δ₂₂]) / det([δ_ij])
Note
For 3 or more unknowns, Cramer's rule becomes tedious; use Gaussian elimination or software.
Conditions
2×2 system; numerator uses the load vector in place of the first column.
Fixed-Fixed Beam (DI = 3 Canonical Case)
Formulas
M_A = −wL²/12, M_B = wL²/12 (fixed-end moments, UDL, symmetrical)
Meaning
Both ends fixed and symmetric; negative at left (hogging), positive is wrong — use hogging sign convention.
Watch out
Memorize as ±wL²/12 (NOT wL²/8 or wL²/6). The negative sign at the left end indicates hogging moment (concave up, tension on top).
When to use
Fixed-fixed beam with UDL; memorize for quick answer checks.
R_A = R_B = wL/2 (reactions at fixed ends, UDL, symmetrical)
Meaning
By symmetry, each end takes half the total load.
Watch out
Reactions are unchanged; moments are different. Fixed ends support large moments but don't change vertical reactions.
When to use
Check your answer for fixed-fixed UDL beam; reactions are the same as simple beam.
For central point load P on fixed-fixed: M_A = −PL/8, M_B = −PL/8 (both hogging)
Meaning
Central load on fixed-fixed; both fixed ends develop hogging moments of equal magnitude.
Watch out
Both moments are negative (hogging), not one positive. The interior maximum moment is different from the end moments.
When to use
Symmetric concentrated load on fixed-fixed beam.
Common values
Value
wL²/12
Symbol
M_end
Quantity
Fixed-fixed beam UDL: end moments
Value
wL²/24
Symbol
M_center
Quantity
Fixed-fixed beam UDL: center moment
Value
wL/2
Symbol
R
Quantity
Fixed-fixed beam UDL: reactions
Important facts
- Fixed-fixed beam is DI = 3 (three redundants: two end moments + one constraint); typically solved via three-moment or consistent deformation.
- The internal shear and moment diagrams differ from the simple-beam case due to the fixed-end moments.
- Maximum internal moment in a fixed-fixed UDL beam is often at midspan (positive, sagging), not at the ends.
- For asymmetrical loading or spans, use the three-moment equation or consistent deformation to find end moments.
- Always compute the maximum moment and the location of zero shear; they may not coincide in a fixed-fixed beam.
Key definitions
Fixed End
Example
A beam welded to a column (moment-resisting joint) at both ends; both ends are fixed.
A support that prevents both vertical displacement and rotation; develops both a reaction and a bending moment.
Fixed-End Moment
Example
In a fixed-fixed UDL beam, the fixed-end moment = −wL²/12 at each end.
The bending moment at a fixed support, arising from applied loads and continuity constraints.
Diagrams to know
- Bending moment diagram: negative (hogging) at both ends, positive (sagging) in the interior; symmetric for symmetric loading.
- Shear force diagram: symmetric, zero at center, linear variation.
- Deflection curve: symmetric, zero at both ends (fixed), maximum deflection at center (downward).
Reactions or equations
Shear at any section x (UDL): V(x) = R_A − wx = (wL/2) − wx; zero at x = L/2 (center).
Note
Maximum moment occurs at zero shear; use M(x) = M_A + R_A·x − (1/2)wx² to find it.
Conditions
Symmetric UDL on fixed-fixed; shear is the same as simple beam.
Moment at center (UDL fixed-fixed): M_center = M_A + R_A(L/2) − (1/2)w(L/2)²
Note
The center moment is positive (sagging), much smaller than the simple-beam value (wL²/8), because the fixed ends absorb load via negative moments.
Conditions
Substituting M_A = −wL²/12, R_A = wL/2, x = L/2: M_center = −wL²/12 + wL²/4 − wL²/8 = wL²/24.
Settlement, Sinking Support & Temperature Effects
Formulas
δ₀ + R·δ₁₁ = Δ (compatibility with support settlement)
Meaning
Δ = known downward displacement (positive) at the released support; set equal to the actual displacement.
Watch out
Δ ≠ 0; common error is setting Δ = 0 when a support has moved. Always read the problem: 'support settles 10 mm' means Δ = −0.01 m (negative if downward and we take up as positive) or +0.01 m (if downward is positive).
When to use
When a support settles, sinks, or is intentionally displaced by a known amount (e.g., from construction error, soil subsidence).
ΔL_thermal = α·ΔT·L (axial thermal strain in a member, free expansion)
Meaning
α = coefficient of thermal expansion, ΔT = temperature change, L = member length; unrestricted expansion/contraction.
Watch out
If the structure is determinate, thermal effects cause no additional stresses (free expansion). Only indeterminate structures develop thermal stresses.
When to use
When a structure experiences temperature change and one or more directions are restrained (indeterminate problem).
Thermal force = −(α·ΔT·L) / δ₁₁ (redundant force due to temperature change)
Meaning
Treat thermal expansion like a 'load': δ_thermal = α·ΔT·L acts like a displacement, and the redundant resists it.
Watch out
Set up: (−α·ΔT·L) + R·δ₁₁ = 0 (if no settlement); R = (α·ΔT·L) / δ₁₁. Sign: positive ΔT (rise) → expansion; if restrained, R opposes the expansion (compression).
When to use
Indeterminate structure with fixed/restrained ends, subject to temperature rise or drop.
Common values
Value
12 × 10⁻⁶ /°C
Symbol
α
Quantity
Steel thermal expansion coefficient
Value
10 × 10⁻⁶ /°C
Symbol
α
Quantity
Concrete thermal expansion coefficient
Important facts
- Support settlement is a boundary condition; set Δ = known value in the compatibility equation.
- If multiple supports settle (unequal), use relative settlement: Δ_relative = Δ_A − Δ_B for the compatibility equation between A and B.
- Thermal effects only create stresses in indeterminate structures; in determinate structures, members expand/contract freely with no additional stress.
- The axial thermal strain (α·ΔT) is dimensionless; multiply by E to get thermal stress in a fully restrained member: σ_thermal = E·α·ΔT.
- Temperature changes in indeterminate trusses require the same force method: release one member, compute thermal expansion δ_thermal, then solve for the member force.
Key definitions
Support Settlement (Differential Settlement)
Example
Foundation sinking 5 cm due to soil consolidation; this is Δ = −0.05 m (taking upward as positive) or +0.05 m (downward positive).
A known downward (or upward) displacement of a support; if all supports settle equally, no extra stresses (relative movement matters).
Thermal Strain
Example
A steel beam at 20°C heated to 60°C experiences ε_thermal = 12×10⁻⁶ (1/°C) × 40 (°C) = 480×10⁻⁶ (0.048%), so a 10 m beam expands 4.8 mm.
Strain due to temperature change: ε_thermal = α·ΔT; free expansion ΔL = α·ΔT·L.
Thermal Stress
Example
A fixed-fixed beam heated: the ends cannot expand freely, so the beam develops compressive stress σ = E·α·ΔT.
Stress developed in a restrained structure due to thermal strain; occurs only if the structure is indeterminate.
Diagrams to know
- Moment diagram for the redundant force due to settlement.
- Final moment diagram combining primary structure loads + settlement effect.
- Qualitative deformed shape before and after settlement (exaggerated).
Reactions or equations
For settlement: δ₀ + R·δ₁₁ = Δ → R = (Δ − δ₀) / δ₁₁
Note
If Δ > 0 (downward) and δ₀ + R·δ₁₁ represents upward, the equation balances; solve for R (may be positive or negative).
Conditions
Single redundant; Δ is the prescribed settlement (known value).
For thermal: (−α·ΔT·L) + R·δ₁₁ = 0 → R = (α·ΔT·L) / δ₁₁
Note
Thermal stress: σ = R / A = E·α·ΔT (if fully restrained, no deflection).
Conditions
Restrained member; positive ΔT → expansion → compressive force (negative R if compression).
Quick Formulas & Standard Cases
Formulas
Cantilever deflection (free end, point load P at tip): δ = PL³/(3EI)
Meaning
Standard formula; used to compute flexibility coefficients.
Watch out
If load is not at the free end (offset by a), use: δ = Pa²(3L − a)/(6EI).
When to use
Propped cantilever, consistent deformation setup.
Cantilever slope (free end, point load P at tip): θ = PL²/(2EI)
Meaning
Slope (rotation) at the free end.
Watch out
Slope is PL²/(2EI), NOT PL²/(3EI) (which is deflection coefficient).
When to use
When the released redundant is a slope (moment release) instead of deflection.
Simple span deflection (center, UDL w): δ = 5wL⁴/(384EI)
Meaning
Maximum downward deflection at midspan of a simply supported beam under UDL.
Watch out
For point load at center: δ = PL³/(48EI) (different coefficient).
When to use
Computing δ₀ for a continuous beam when the primary structure is a simple span.
Simple span deflection (center, point load P at center): δ = PL³/(48EI)
Meaning
For concentrated load at midspan.
Watch out
Coefficient is 1/48 (UDL is 5/384 ≈ 0.013; point load 1/48 ≈ 0.021).
When to use
Continuous beam with concentrated interior loads.
Important facts
- Always use EI = E × I in SI units; E in Pa (or MPa if I in mm⁴), I in m⁴ (or mm⁴ if E in MPa).
- For standard deflections, consult a table or use integration: δ = ∫M·m/(EI) dx.
- The method of virtual work (unit-load method) is the most flexible approach for non-standard cases.
Must remember
- 1. DI = number of redundants to release; set up one compatibility equation per redundant.
- 2. Propped cantilever under UDL: R_prop = 3wL/8, M_fixed = −wL²/8 (memorize exactly).
- 3. Three-moment equation RHS = −6(A₁x̄₁/L₁ + A₂x̄₂/L₂); DO NOT drop the negative sign or get the load terms wrong.
- 4. UDL load term in three-moment = wL³/4 per span; central point load term = 3PL²/8.
- 5. Consistency equation: δ₀ + R·δ_ij = Δ; if Δ = 0 (fixed support), then R = −δ₀/δ_ij (note the negative sign).
- 6. Flexibility coefficient δ_ii is always positive; it represents how much the point deflects per unit redundant force.
- 7. Simply supported ends: M = 0 in three-moment; fixed ends: unknown (requires second equation).
- 8. For multiple redundants, use matrix form [δ_ij]{R} = −{δ₀}; solve by Cramer's rule (2×2) or Gaussian elimination (3×3+).
- 9. Support settlement: set Δ = known value in compatibility; thermal effects: use δ_thermal = α·ΔT·L as an imposed 'deflection'.
- 10. Always check equilibrium (vertical and moment) after solving for redundants; this catches sign errors immediately.
Last minute tips
- PROPPED CANTILEVER: If you see 'cantilever with a prop at the free end,' immediately think DI = 1, R_prop = 3wL/8 (UDL), and use consistency: δ_cantilever_UDL = wL⁴/(8EI) on the RHS. Most common exam question.
- THREE-MOMENT EQUATION: Write it DOWN in full every time: M_A·L₁ + 2M_B(L₁+L₂) + M_C·L₂ = −6(...load terms...). The negative sign kills half the exam answers; memorize it as part of the formula.
- SIGN CONVENTION: Choose one (e.g., downward deflection + compression = positive) and STICK to it throughout. Mixing signs is the #1 cause of wrong answers. Draw a convention box on your scratch paper.
- LOAD TERMS: For three-moment, UDL term = wL³/4 (NOT wL²/4), point load = 3PL²/8 (NOT PL²/4). Write them on a flashcard; they appear in almost every continuous-beam problem.
- MATRIX SETUP: For DI > 1, lay out [δ_ij] and {δ₀} clearly before solving. Invert (or row-reduce) slowly; one arithmetic error cascades. Use Cramer's rule for 2×2; it's faster and less error-prone than full Gaussian elimination by hand.
Comparison tables
Statically Determinate vs Indeterminate Structures
| Property | Determinate | Indeterminate |
|---|---|---|
| Equilibrium alone | Sufficient to find all reactions/moments | Insufficient; compatibility required |
| Number of unknowns | = number of equilibrium equations (3 for 2D) | > number of equilibrium equations; DI > 0 |
| Support reactions | Unique, do not depend on geometry or material | Depend on EI, span lengths, load distribution |
| Deflections | Computed after finding reactions (not needed for statics) | Part of the solution (compatibility equations) |
| Redundants | None; structure is fully constrained | DI redundants; remove to get determinate primary structure |
| Support settlement effect | No change in reactions (if all settle equally) | Creates additional stresses and moments |
Propped Cantilever vs Fixed-Fixed Beam vs Continuous Beam
| Aspect | Propped Cantilever | Fixed-Fixed Beam | Continuous Beam (2-span) |
|---|---|---|---|
| DI | 1 (one redundant: prop reaction) | 3 (two end moments + one slope) | 1 (one redundant: interior moment) |
| Typical redundant | R_prop (vertical reaction) | M_A, M_B (fixed-end moments) | M_B (interior support moment) |
| UDL prop/end reaction | R_prop = 3wL/8, R_fixed = 5wL/8 | R = wL/2 (symmetrical) | R_A = wL/2 − |M_B|/L, etc. |
| UDL end moment | M_fixed = −wL²/8 (hogging) | M_end = −wL²/12 (hogging) | M_A = 0 (simple end), M_B = −wL²/8 (interior) |
| Primary structure | Cantilever (remove prop) | Simple span (release end moments) | Two simple spans (release interior support) |
| Typical exam question | Find prop reaction and fixed-end moment | Find end moments and maximum moment | Find interior support moment and reactions |
Three-Moment Load Terms for Common Cases
| Loading | Simple-Beam M Diagram | Area A | Centroid x̄ | 6Ax̄/L Term |
|---|---|---|---|---|
| UDL w over span L | Parabola, peak = wL²/8 | wL³/12 | L/2 from either end | wL³/4 |
| Point load P at center | Two triangles, peak = PL/4 | PL²/8 | 2L/3 from left end | 3PL²/8 (standard) |
| Point load P at distance a from left | Two triangles (asymmetric) | Pab/2L (where b = L−a) | Varies | Requires integration or tables |
| Triangular load (0 at left, w at right) | Cubic curve | wL³/16 | 3L/4 from left | 9wL³/32 (approx.) |
Method Comparison: Force (Flexibility) vs Displacement (Stiffness)
| Aspect | Force Method (Flexibility) | Displacement Method (Stiffness) |
|---|---|---|
| Primary unknown | Redundant forces/moments | Nodal displacements/rotations |
| Number of unknowns | = DI (typically small for hand calc) | = DOF (often large for frames) |
| Key equation | Compatibility: δ₀ + R·δ_ij = Δ | Equilibrium: [k]{Δ} = {F} |
| Matrix to invert | Flexibility [δ_ij] (small, hand-friendly) | Stiffness [k] (large, typically software) |
| Hand calculation suitability | Best for DI ≤ 3 | Tedious for DI > 2; use software |
| Primary structure required? | Yes (must be determinate) | No (assemble global stiffness) |
Diagrams
Force Method Decision Tree
Step-by-step flowchart guiding the force method from identifying indeterminacy through final answer. Shows decision points for choosing redundants, solving single vs multiple redundant systems, and verification checks.
Indeterminate Structures: Force Methods Overview
Hierarchical mind map showing the main topics and subtopics within force methods. Central node branches into fundamentals, single/multiple redundant cases, the three-moment equation, and special considerations.