CELE Structural Theory & Analysis — Indeterminate Structures: Displacement MethodsCheat Sheet

Formula

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

Formula

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)

Term

Redundant

Example

In a propped cantilever, the prop is the redundant. In a two-span continuous beam, the interior support moment is the redundant.

Definition

A reaction or internal force whose removal does not cause instability; its removal gives a determinate primary structure.

Term

Primary (Released) Structure

Example

A propped cantilever becomes a simple cantilever when the prop is removed.

Definition

The determinate structure obtained by removing all redundants; used as the base case for force method.

Term

Flexibility Coefficient (δ_ij)

Example

δ_11 is the deflection at a released support caused by applying unit force at that same point.

Definition

Deflection at point i per unit redundant force at point j; inverse of stiffness.

Term

Degree of Indeterminacy (DI)

Example

DI = 1 for a propped cantilever or two-span simple beam; DI = 3 for a fixed-fixed beam.

Definition

Number of unknown force(s) exceeding equilibrium equations; equals the number of redundants.

Formula

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.

Formula

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.

Formula

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).

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)

Term

Compatibility Equation

Example

For a propped cantilever, the prop prevents downward deflection: δ_due_to_load + R_prop × (deflection_per_unit_prop) = 0.

Definition

Mathematical statement that the actual deflection at a released point equals the prescribed value (usually zero for a fixed support).

Term

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.

Definition

Technique to find deflection by imagining a unit load at the desired point and integrating M×m/(EI) over the structure.

Term

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).

Definition

Known vertical displacement of a support (positive downward); default = 0 for idealized fixed/pinned supports.

Note

Rearrange from δ₀ + R·δ₁₁ = Δ; if Δ = 0, then R = −δ₀/δ₁₁ (negative sign often absorbed into sign convention).

Equation

For single redundant: R = (Δ − δ₀) / δ₁₁

Conditions

When δ₀ and δ₁₁ have opposite effects (load pulls down, redundant supports up), the equation gives the correct sign.

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).

Equation

[δ_ij]{R_j} = −{δ_0i} (matrix form for multiple redundants)

Conditions

DI > 1; solve by Gaussian elimination or Cramer's rule.

Formula

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.

Formula

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.

Formula

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.

Formula

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.

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

Term

Propped Cantilever

Example

A building facade beam fixed to a column at one end and resting on a prop (support) at the other end.

Definition

A beam fixed at one end and supported (pinned or roller) at the other free end; DI = 1, statically indeterminate.

Term

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.

Definition

Negative bending moment causing concave-up curvature (sagging region above the neutral axis tends to go in compression).

Note

Shear goes from +R_A to −R_prop; sign change indicates where max sagging moment occurs.

Equation

Shear at any section: V(x) = R_A − wx (where x = 0 at fixed end, x = L at prop)

Conditions

UDL w over the whole span; V = 0 at x = R_A/w = (5wL/8)/w = 5L/8.

Note

At x = L (prop), M = 0 (simple support). At x = 5L/8 (zero shear), M = max sagging moment.

Equation

Moment at any section: M(x) = M_fixed + R_A·x − (1/2)wx²

Conditions

UDL w; M_fixed = −wL²/8 (negative = hogging); integrate shear to get moment.

Formula

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).

Formula

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.

Formula

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.

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

Term

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).

Definition

Bending moment diagram of a single span if it were simply supported (ignoring continuity); used to compute the load terms in three-moment equation.

Term

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.

Definition

A support within the beam (not at the ends); moment at an interior support is typically unknown in a continuous beam.

Term

Load Term (RHS of Three-Moment)

Example

For two equal UDL spans: RHS = −6(wL³/4 + wL³/4) = −3wL³/2.

Definition

The quantity −6(A₁x̄₁/L₁ + A₂x̄₂/L₂), which encodes the effect of applied loads on the moments at three consecutive supports.

Note

The equation automatically enforces moment continuity (same moment on left and right).

Equation

At an interior support: sum moments about that support from left and right spans must balance the unknown moment M there.

Conditions

Three-moment equation is derived from moment equilibrium at the support.

Note

Always compute reactions from beam equilibrium (vertical and moment) AFTER solving for moments.

Equation

Reaction at an interior support: R_B = (sum of end moments)/L + midspan load term

Conditions

For symmetrical loading, reactions simplify.

Formula

[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.

Formula

{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}.

Formula

[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.

Term

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.

Definition

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.

Term

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).

Definition

Deflection at point i due to unit load at j equals deflection at j due to unit load at i: δ_ij = δ_ji.

Note

For 3 or more unknowns, Cramer's rule becomes tedious; use Gaussian elimination or software.

Equation

Cramer's rule for 2 redundants: R₁ = −det([δ₀, δ₁₂; δ₀₂, δ₂₂]) / det([δ_ij])

Conditions

2×2 system; numerator uses the load vector in place of the first column.

Formula

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.

Formula

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.

Formula

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.

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

Term

Fixed End

Example

A beam welded to a column (moment-resisting joint) at both ends; both ends are fixed.

Definition

A support that prevents both vertical displacement and rotation; develops both a reaction and a bending moment.

Term

Fixed-End Moment

Example

In a fixed-fixed UDL beam, the fixed-end moment = −wL²/12 at each end.

Definition

The bending moment at a fixed support, arising from applied loads and continuity constraints.

Note

Maximum moment occurs at zero shear; use M(x) = M_A + R_A·x − (1/2)wx² to find it.

Equation

Shear at any section x (UDL): V(x) = R_A − wx = (wL/2) − wx; zero at x = L/2 (center).

Conditions

Symmetric UDL on fixed-fixed; shear is the same as simple beam.

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.

Equation

Moment at center (UDL fixed-fixed): M_center = M_A + R_A(L/2) − (1/2)w(L/2)²

Conditions

Substituting M_A = −wL²/12, R_A = wL/2, x = L/2: M_center = −wL²/12 + wL²/4 − wL²/8 = wL²/24.

Formula

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).

Formula

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).

Formula

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.

Value

12 × 10⁻⁶ /°C

Symbol

α

Quantity

Steel thermal expansion coefficient

Value

10 × 10⁻⁶ /°C

Symbol

α

Quantity

Concrete thermal expansion coefficient

Term

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).

Definition

A known downward (or upward) displacement of a support; if all supports settle equally, no extra stresses (relative movement matters).

Term

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.

Definition

Strain due to temperature change: ε_thermal = α·ΔT; free expansion ΔL = α·ΔT·L.

Term

Thermal Stress

Example

A fixed-fixed beam heated: the ends cannot expand freely, so the beam develops compressive stress σ = E·α·ΔT.

Definition

Stress developed in a restrained structure due to thermal strain; occurs only if the structure is indeterminate.

Note

If Δ > 0 (downward) and δ₀ + R·δ₁₁ represents upward, the equation balances; solve for R (may be positive or negative).

Equation

For settlement: δ₀ + R·δ₁₁ = Δ → R = (Δ − δ₀) / δ₁₁

Conditions

Single redundant; Δ is the prescribed settlement (known value).

Note

Thermal stress: σ = R / A = E·α·ΔT (if fully restrained, no deflection).

Equation

For thermal: (−α·ΔT·L) + R·δ₁₁ = 0 → R = (α·ΔT·L) / δ₁₁

Conditions

Restrained member; positive ΔT → expansion → compressive force (negative R if compression).

Formula

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.

Formula

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.

Formula

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.

Formula

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.

Values

• Sufficient to find all reactions/moments
• Insufficient; compatibility required

Property

Equilibrium alone

Values

• = number of equilibrium equations (3 for 2D)
• > number of equilibrium equations; DI > 0

Property

Number of unknowns

Values

• Unique, do not depend on geometry or material
• Depend on EI, span lengths, load distribution

Property

Support reactions

Values

• Computed after finding reactions (not needed for statics)
• Part of the solution (compatibility equations)

Property

Deflections

Values

• None; structure is fully constrained
• DI redundants; remove to get determinate primary structure

Property

Redundants

Values

• No change in reactions (if all settle equally)
• Creates additional stresses and moments

Property

Support settlement effect

Values

• 1 (one redundant: prop reaction)
• 3 (two end moments + one slope)
• 1 (one redundant: interior moment)

Property

DI

Values

• R_prop (vertical reaction)
• M_A, M_B (fixed-end moments)
• M_B (interior support moment)

Property

Typical redundant

Values

• R_prop = 3wL/8, R_fixed = 5wL/8
• R = wL/2 (symmetrical)
• R_A = wL/2 − |M_B|/L, etc.

Property

UDL prop/end reaction

Values

• M_fixed = −wL²/8 (hogging)
• M_end = −wL²/12 (hogging)
• M_A = 0 (simple end), M_B = −wL²/8 (interior)

Property

UDL end moment

Values

• Cantilever (remove prop)
• Simple span (release end moments)
• Two simple spans (release interior support)

Property

Primary structure

Values

• Find prop reaction and fixed-end moment
• Find end moments and maximum moment
• Find interior support moment and reactions

Property

Typical exam question

Values

• Parabola, peak = wL²/8
• wL³/12
• L/2 from either end
• wL³/4

Property

UDL w over span L

Values

• Two triangles, peak = PL/4
• PL²/8
• 2L/3 from left end
• 3PL²/8 (standard)

Property

Point load P at center

Values

• Two triangles (asymmetric)
• Pab/2L (where b = L−a)
• Varies
• Requires integration or tables

Property

Point load P at distance a from left

Values

• Cubic curve
• wL³/16
• 3L/4 from left
• 9wL³/32 (approx.)

Property

Triangular load (0 at left, w at right)

Values

• Redundant forces/moments
• Nodal displacements/rotations

Property

Primary unknown

Values

• = DI (typically small for hand calc)
• = DOF (often large for frames)

Property

Number of unknowns

Values

• Compatibility: δ₀ + R·δ_ij = Δ
• Equilibrium: [k]{Δ} = {F}

Property

Key equation

Values

• Flexibility [δ_ij] (small, hand-friendly)
• Stiffness [k] (large, typically software)

Property

Matrix to invert

Values

• Best for DI ≤ 3
• Tedious for DI > 2; use software

Property

Hand calculation suitability

Values

• Yes (must be determinate)
• No (assemble global stiffness)

Property

Primary structure required?