Formula Sheet: Structural Analysis Methods

Determinacy and Stability

Determinacy of Structures

Degree of Indeterminacy (Beams and Trusses): \[i = r - e\]

• i = degree of static indeterminacy
• r = number of unknown reactions
• e = number of equilibrium equations available (typically 3 for 2D structures: ΣFx = 0, ΣFy = 0, ΣM = 0)
• If i = 0: statically determinate
• If i > 0: statically indeterminate to the i^th degree
• If i <>: unstable (insufficient reactions)

Truss Determinacy: \[i = m + r - 2j\]

• m = number of members
• r = number of reaction components
• j = number of joints
• If i = 0: statically determinate
• If i > 0: statically indeterminate
• If i <>: unstable

Frame Determinacy: \[i = 3m + r - 3j - c\]

• m = number of members
• r = number of reaction components
• j = number of joints
• c = number of condition equations (releases, hinges)
• Internal hinge releases 1 moment constraint
• Internal roller releases 1 force and 1 moment constraint

Stability Criteria

• Structure must have sufficient reactions to prevent rigid body motion
• Reactions must not be concurrent (pass through a single point)
• Reactions must not be parallel
• For trusses: members must be properly triangulated

Truss Analysis

Method of Joints

Equilibrium at Each Joint: \[\sum F_x = 0\] \[\sum F_y = 0\]

• Apply at each joint sequentially
• Begin at joints with ≤ 2 unknown member forces
• Tension forces pull away from joint (positive convention)
• Compression forces push into joint (negative convention)

Method of Sections

Equilibrium of Cut Section: \[\sum F_x = 0\] \[\sum F_y = 0\] \[\sum M = 0\]

• Cut through ≤ 3 members (for 2D truss)
• Apply equilibrium to either portion
• Take moments about strategic points to solve for individual member forces
• Useful for finding forces in specific members without analyzing entire truss

Zero-Force Members

• At a joint with 2 non-collinear members and no external load: both members are zero-force
• At a joint with 3 members (2 collinear, 1 at angle) and no external load: the non-collinear member is zero-force

Beam Analysis

Static Equilibrium Equations

\[\sum F_x = 0\] \[\sum F_y = 0\] \[\sum M = 0\]

• Used to determine support reactions for statically determinate beams
• Moment equilibrium can be taken about any point

Shear and Moment Relationships

Relationship between Load, Shear, and Moment: \[\frac{dV}{dx} = -w(x)\] \[\frac{dM}{dx} = V(x)\] \[\frac{d^2M}{dx^2} = -w(x)\]

• V(x) = shear force at position x
• M(x) = bending moment at position x
• w(x) = distributed load at position x (positive downward)
• x = distance along beam

Change in Shear: \[\Delta V = -\int_{x_1}^{x_2} w(x) \, dx\]

• Change in shear equals negative of area under load diagram
• Point load causes discontinuity (jump) in shear diagram

Change in Moment: \[\Delta M = \int_{x_1}^{x_2} V(x) \, dx\]

• Change in moment equals area under shear diagram
• Concentrated moment causes discontinuity (jump) in moment diagram

Shear and Moment Diagram Characteristics

• No load region (w = 0): V is constant, M is linear
• Uniform load region: V is linear, M is parabolic
• Linearly varying load: V is parabolic, M is cubic
• Maximum/minimum moment: occurs where V = 0
• Point of inflection: occurs where M = 0

Deflection of Beams

Double Integration Method

Governing Differential Equation: \[EI \frac{d^2y}{dx^2} = M(x)\]

• E = modulus of elasticity
• I = moment of inertia of cross-section
• y = deflection (positive upward)
• M(x) = bending moment at position x

First Integration (Slope): \[EI \frac{dy}{dx} = \int M(x) \, dx + C_1\]

• \(\frac{dy}{dx}\) = slope (θ)
• C₁ = constant determined from boundary conditions

Second Integration (Deflection): \[EI \cdot y = \iint M(x) \, dx \, dx + C_1 x + C_2\]

• C₁, C₂ = constants determined from boundary conditions
• Apply deflection and slope boundary conditions to solve for constants

Moment-Area Method

First Moment-Area Theorem: \[\theta_{B/A} = \frac{1}{EI} \int_A^B M(x) \, dx\]

• θ_B/A = change in slope from point A to point B (radians)
• Equals area under M/EI diagram between A and B
• Positive when M/EI diagram is positive (typically causing sagging)

Second Moment-Area Theorem: \[t_{B/A} = \frac{1}{EI} \int_A^B M(x) \cdot x \, dx\]

• t_B/A = tangential deviation of point B from tangent at point A
• Equals first moment of M/EI diagram area about point B
• Used to calculate deflections relative to tangent lines

Conjugate Beam Method

• Conjugate beam: fictitious beam with same length as real beam
• Load on conjugate beam: M/EI diagram from real beam
• Shear in conjugate beam: equals slope in real beam
• Moment in conjugate beam: equals deflection in real beam

Support Conversions (Real → Conjugate):

• Fixed support → Free end
• Free end → Fixed support
• Pin/roller support → Pin/roller support (same)
• Internal hinge → Internal hinge (same)

Superposition Method

• Total deflection = sum of deflections from individual loads
• Valid only for linear elastic behavior
• Use standard beam deflection formulas from tables

Virtual Work and Energy Methods

Principle of Virtual Work

Virtual Work Equation for Deflection: \[\Delta = \frac{1}{EI} \int_0^L m(x) \cdot M(x) \, dx\]

• Δ = deflection at point and in direction of virtual unit load
• m(x) = virtual moment due to unit load applied at point where deflection is sought
• M(x) = real moment due to actual loads
• L = length of member

Virtual Work for Rotation: \[\theta = \frac{1}{EI} \int_0^L m(x) \cdot M(x) \, dx\]

• θ = rotation at point where virtual unit moment is applied
• m(x) = virtual moment due to unit moment applied at point where rotation is sought

Virtual Work for Trusses

Truss Deflection: \[\Delta = \sum \frac{u \cdot N \cdot L}{AE}\]

• Δ = deflection at joint and in direction of virtual unit load
• u = virtual force in member due to unit load at point where deflection is sought
• N = real force in member due to actual loads
• L = length of member
• A = cross-sectional area of member
• E = modulus of elasticity
• Sum over all members

Virtual Work for Frames

Frame Deflection (Bending Only): \[\Delta = \sum \frac{1}{EI} \int_0^L m(x) \cdot M(x) \, dx\]

• Sum over all members in frame
• For frames with axial and shear deformations, add corresponding terms

Complete Frame Deflection (Including Axial and Shear): \[\Delta = \sum \left[ \frac{1}{EI} \int_0^L m \cdot M \, dx + \frac{1}{AE} \int_0^L n \cdot N \, dx + \frac{1}{GA} \int_0^L v \cdot V \, dx \right]\]

• n = virtual axial force
• N = real axial force
• v = virtual shear force
• V = real shear force
• G = shear modulus
• Shear and axial deformations often negligible for typical frames

Castigliano's Second Theorem

Deflection at Point of Applied Load: \[\Delta_i = \frac{\partial U}{\partial P_i}\]

• Δ_i = deflection at point where load P_i is applied, in direction of P_i
• U = total strain energy in structure
• P_i = applied load

Strain Energy in Beam (Bending): \[U = \int_0^L \frac{M^2}{2EI} \, dx\] Strain Energy in Truss: \[U = \sum \frac{N^2 L}{2AE}\]

• Sum over all members

For Deflection at Point with No Load:

• Apply dummy load Q at point where deflection is sought
• Calculate: \(\Delta = \frac{\partial U}{\partial Q}\)
• Set Q = 0 after differentiation

Influence Lines

Influence Line Definition

• Influence line: diagram showing variation of a response function (reaction, shear, moment) at a specific point as a unit load moves across the structure
• Ordinate at position x = value of response function when unit load is at position x
• Used to determine critical load positions for maximum response

Mueller-Breslau Principle

• The influence line for a response function is given by the deflected shape when the restraint corresponding to that function is removed and a unit displacement is applied
• For reaction: remove support, apply unit displacement in direction of reaction
• For shear: insert internal roller, apply unit relative translation
• For moment: insert internal hinge, apply unit relative rotation
• Deflected shape is proportional to influence line ordinates

Quantitative Influence Lines - Beams

Influence Line for Reaction at A:

• Simple span beam of length L
• At support A: ordinate = 1
• At support B: ordinate = 0
• Linear variation between supports
• For simply supported beam: \(IL_{R_A} = 1 - \frac{x}{L}\)

Influence Line for Shear at Section C (distance a from left support):

• Simple span beam of length L
• Left of C: \(IL_{V_C} = \frac{b}{L}\) where b = L - a
• Right of C: \(IL_{V_C} = -\frac{a}{L}\)
• Discontinuity (jump) of magnitude 1.0 at section C

Influence Line for Moment at Section C (distance a from left support):

• Simple span beam of length L
• At section C: ordinate = \(\frac{ab}{L}\) where b = L - a
• Linear from zero at left support to maximum at C
• Linear from maximum at C to zero at right support

Using Influence Lines

Response Due to Point Load P: \[R = P \cdot y\]

• R = response (reaction, shear, or moment)
• P = magnitude of point load
• y = ordinate of influence line at position of load P

Response Due to Distributed Load w: \[R = \int w(x) \cdot y(x) \, dx\]

• For uniform load w: R = w × (area under influence line)
• y(x) = ordinate of influence line at position x

Maximum Response:

• Place loads where influence line has same sign (all positive or all negative)
• For concentrated loads: place at maximum ordinate(s)
• For uniform loads: cover entire positive or negative region

Influence Lines for Trusses

• Construct influence line for member force by analyzing truss with unit load at various panel points
• Between panel points, influence line ordinate varies linearly
• Maximum force occurs when loads positioned to maximize \(\sum P_i y_i\)

Approximate Analysis Methods

Portal Method (Building Frames - Lateral Loads)

Assumptions:

• Interior columns carry twice the shear of exterior columns at each level
• Points of inflection occur at mid-height of columns
• Points of inflection occur at mid-span of beams

Distribution of Lateral Load: \[V_{\text{exterior}} = \frac{H}{2n}\] \[V_{\text{interior}} = \frac{H}{n}\]

• H = total lateral load at level
• n = number of bays
• V_exterior = shear in exterior column
• V_interior = shear in each interior column

Cantilever Method (Building Frames - Lateral Loads)

Assumptions:

• Structure behaves as vertical cantilever
• Axial stress in columns varies linearly with distance from centroid of all columns
• Points of inflection at mid-height of columns and mid-span of beams

Axial Force Distribution: \[P_i = \frac{M \cdot d_i}{\sum d_i^2}\]

• P_i = axial force in column i
• M = overturning moment at level
• d_i = distance of column i from centroid of columns
• Tension on windward side, compression on leeward side

Approximate Analysis for Vertical Loads on Frames

Assumptions for Gravity Load Analysis:

• Points of inflection at 0.1L from each end of beams (L = span)
• Points of inflection at mid-height of columns
• Insert hinges at inflection points to create determinate structure
• Analyze resulting statically determinate structure

Slope-Deflection Method

Slope-Deflection Equations

General Slope-Deflection Equation (No Settlement): \[M_{AB} = \frac{2EI}{L} \left( 2\theta_A + \theta_B - \frac{3\Delta}{L} \right) + M_{AB}^{FEM}\] \[M_{BA} = \frac{2EI}{L} \left( 2\theta_B + \theta_A - \frac{3\Delta}{L} \right) + M_{BA}^{FEM}\]

• M_AB = moment at end A of member AB (acting on member)
• M_BA = moment at end B of member AB (acting on member)
• θ_A = rotation at end A (positive clockwise)
• θ_B = rotation at end B (positive clockwise)
• Δ = relative transverse displacement of B with respect to A (positive when B moves in positive y direction relative to A)
• L = length of member AB
• M_AB^FEM = fixed-end moment at A due to loads on span
• M_BA^FEM = fixed-end moment at B due to loads on span
• EI = flexural rigidity (constant along member)

Simplified Form (No Support Settlement, Δ = 0): \[M_{AB} = \frac{2EI}{L} \left( 2\theta_A + \theta_B \right) + M_{AB}^{FEM}\] \[M_{BA} = \frac{2EI}{L} \left( 2\theta_B + \theta_A \right) + M_{BA}^{FEM}\] For Far End Fixed (θ_B = 0): \[M_{AB} = \frac{4EI}{L} \theta_A + M_{AB}^{FEM}\] For Far End Hinged (M_BA = 0): \[M_{AB} = \frac{3EI}{L} \theta_A + M_{AB}^{FEM} - \frac{M_{BA}^{FEM}}{2}\]

Fixed-End Moments (Common Load Cases)

Uniformly Distributed Load w over entire span L: \[M_{AB}^{FEM} = -\frac{wL^2}{12}\] \[M_{BA}^{FEM} = +\frac{wL^2}{12}\] Concentrated Load P at midspan: \[M_{AB}^{FEM} = -\frac{PL}{8}\] \[M_{BA}^{FEM} = +\frac{PL}{8}\] Concentrated Load P at distance a from A (b from B, a + b = L): \[M_{AB}^{FEM} = -\frac{Pab^2}{L^2}\] \[M_{BA}^{FEM} = +\frac{Pa^2b}{L^2}\] Triangular Load (zero at A, w at B): \[M_{AB}^{FEM} = -\frac{wL^2}{30}\] \[M_{BA}^{FEM} = +\frac{wL^2}{20}\] Sign Convention:

• Negative FEM causes compression on top fiber (sagging)
• Positive FEM causes tension on top fiber (hogging)

Solution Procedure

1. Calculate fixed-end moments for all loaded members
2. Write slope-deflection equations for each member end
3. Apply equilibrium equations at each joint: \(\sum M = 0\)
4. Solve system of equations for unknown rotations (and displacements if applicable)
5. Substitute rotations back into slope-deflection equations to find member end moments
6. Calculate shears and reactions using statics

Moment Distribution Method

Distribution Factor

Distribution Factor at Joint: \[DF_{AB} = \frac{K_{AB}}{\sum K}\]

• DF_AB = distribution factor for member AB at joint A
• K_AB = stiffness factor for member AB
• ΣK = sum of stiffness factors of all members meeting at joint A
• Sum of all DFs at a joint = 1.0 (for non-fixed joints)
• DF = 0 for fixed support

Stiffness Factor

Member with Far End Fixed: \[K = \frac{4EI}{L}\] Member with Far End Hinged (Pinned): \[K = \frac{3EI}{L}\] Relative Stiffness:

• If E is constant throughout structure, can use \(K = \frac{I}{L}\) (far end fixed) or \(K = \frac{3I}{4L}\) (far end pinned)

Carry-Over Factor

Far End Fixed: \[COF = +0.5\]

• Moment applied at one end induces moment of +0.5 times magnitude at fixed far end

Far End Hinged: \[COF = 0\]

• No moment carried over to hinged end

Moment Distribution Procedure

1. Calculate fixed-end moments for all loaded members
2. Calculate distribution factors at all joints
3. Start with all joints locked; determine unbalanced moments at each joint
4. Release joints one at a time (or simultaneously):
   • Distribute unbalanced moment to connecting members using DFs
   • Carry over half the distributed moment to far ends (if fixed)
5. Repeat process until moments converge (unbalanced moments negligible)
6. Sum all moments at each member end to obtain final moments

Unbalanced Moment at Joint: \[M_{\text{unbal}} = -\sum M_{\text{FEM}}\]

• Negative sign because distribution moments oppose unbalanced moment

Distributed Moment to Member AB: \[M_{\text{dist}} = DF_{AB} \times M_{\text{unbal}}\] Carried-Over Moment: \[M_{\text{CO}} = COF \times M_{\text{dist}}\]

Modified Stiffness for Symmetric Structures

Member at Line of Symmetry (Symmetric Loading):

• Rotation at symmetry line = 0
• Treat as fixed end: \(K = \frac{4EI}{L}\)

Member at Line of Symmetry (Antisymmetric Loading):

• Moment at symmetry line = 0
• Treat as hinged end: \(K = \frac{3EI}{L}\)

Matrix Analysis - Stiffness Method

Member Stiffness Matrix

Local Member Stiffness Matrix for 2D Beam Element: \[ [k] = \begin{bmatrix} \frac{AE}{L} & 0 & 0 & -\frac{AE}{L} & 0 & 0 \\ 0 & \frac{12EI}{L^3} & \frac{6EI}{L^2} & 0 & -\frac{12EI}{L^3} & \frac{6EI}{L^2} \\ 0 & \frac{6EI}{L^2} & \frac{4EI}{L} & 0 & -\frac{6EI}{L^2} & \frac{2EI}{L} \\ -\frac{AE}{L} & 0 & 0 & \frac{AE}{L} & 0 & 0 \\ 0 & -\frac{12EI}{L^3} & -\frac{6EI}{L^2} & 0 & \frac{12EI}{L^3} & -\frac{6EI}{L^2} \\ 0 & \frac{6EI}{L^2} & \frac{2EI}{L} & 0 & -\frac{6EI}{L^2} & \frac{4EI}{L} \end{bmatrix} \]

• Relates member end forces to end displacements: {F} = [k]{d}
• DOFs: axial displacement, transverse displacement, rotation at each end (6 total)
• Local coordinate system aligned with member axis

Local Member Stiffness Matrix for 2D Truss Element: \[ [k] = \frac{AE}{L} \begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \]

• Only axial deformations considered (2 DOFs per node)
• Simplified from beam element with I = 0

Coordinate Transformation

Transformation from Local to Global Coordinates: \[[K] = [T]^T [k] [T]\]

• [K] = member stiffness matrix in global coordinates
• [k] = member stiffness matrix in local coordinates
• [T] = transformation matrix

Transformation Matrix for 2D Beam Element: \[ [T] = \begin{bmatrix} \cos\theta & \sin\theta & 0 & 0 & 0 & 0 \\ -\sin\theta & \cos\theta & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & \cos\theta & \sin\theta & 0 \\ 0 & 0 & 0 & -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{bmatrix} \]

• θ = angle of member axis measured counterclockwise from global x-axis

Global Stiffness Matrix Assembly

Structure Stiffness Equation: \[\{F\} = [K_s] \{D\}\]

• {F} = global force vector (applied loads and reactions)
• [K_s] = global structure stiffness matrix
• {D} = global displacement vector
• Assemble [K_s] by adding member stiffness contributions at appropriate DOF locations

Solution Procedure

1. Number all nodes and define global coordinate system
2. Determine member properties (E, A, I, L) and orientation (θ)
3. Formulate member stiffness matrices in local coordinates
4. Transform member stiffness matrices to global coordinates
5. Assemble global structure stiffness matrix [K_s]
6. Apply boundary conditions (eliminate rows/columns for restrained DOFs or use partitioning)
7. Solve for unknown displacements: {D} = [K_s]^-1{F}
8. Calculate member end forces using member stiffness relations
9. Calculate reactions by substituting displacements into eliminated equations

Partitioned System

\[ \begin{Bmatrix} F_k \\ F_r \end{Bmatrix} = \begin{bmatrix} K_{kk} & K_{kr} \\ K_{rk} & K_{rr} \end{bmatrix} \begin{Bmatrix} D_k \\ D_r \end{Bmatrix} \]

• Subscript k: known (free) DOFs
• Subscript r: restrained DOFs
• F_k = known applied forces
• F_r = unknown reactions
• D_k = unknown displacements
• D_r = known displacements (typically zero)

Solve for Unknown Displacements: \[\{D_k\} = [K_{kk}]^{-1} \left( \{F_k\} - [K_{kr}]\{D_r\} \right)\] Calculate Reactions: \[\{F_r\} = [K_{rk}]\{D_k\} + [K_{rr}]\{D_r\}\]

Plastic Analysis

Plastic Hinge Concept

• Plastic hinge: location where cross-section reaches full plastic moment capacity M_p
• At plastic hinge, section can rotate freely while maintaining moment = M_p
• Mechanism: forms when sufficient plastic hinges develop to transform structure into kinematic chain

Plastic Moment Capacity: \[M_p = F_y \cdot Z\]

• M_p = plastic moment capacity
• F_y = yield stress
• Z = plastic section modulus

Shape Factor: \[f = \frac{Z}{S} = \frac{M_p}{M_y}\]

• f = shape factor
• S = elastic section modulus
• M_y = yield moment = F_yS
• For rectangular sections: f ≈ 1.5
• For I-shaped sections: f ≈ 1.10 to 1.15

Number of Plastic Hinges Required

Degree of Indeterminacy Plus One: \[n = i + 1\]

• n = number of plastic hinges needed to form mechanism
• i = degree of static indeterminacy
• For determinate structures (i = 0): one plastic hinge creates mechanism

Virtual Work Method for Collapse Load

Principle of Virtual Work: \[W_{\text{external}} = W_{\text{internal}}\] \[\sum P_i \Delta_i = \sum M_p \theta_i\]

• P_i = applied load (factored by collapse load multiplier λ)
• Δ_i = virtual displacement at load application point
• M_p = plastic moment at hinge location
• θ_i = virtual rotation at plastic hinge

Collapse Load Multiplier: \[\lambda = \frac{\sum M_p \theta_i}{\sum P_i \Delta_i}\]

• Express Δ and θ in terms of assumed mechanism geometry
• Assume unit virtual displacement at convenient location

Plastic Analysis Procedures

Mechanism Method:

1. Identify possible collapse mechanisms (beam, panel, combined)
2. For each mechanism, apply virtual work to find collapse load
3. Critical mechanism gives lowest (governing) collapse load
4. Verify equilibrium and yield conditions satisfied

Equilibrium Method:

1. Assume locations of plastic hinges
2. Write equilibrium equations using M_p at hinge locations
3. Solve for collapse load
4. Check that moment nowhere exceeds M_p
5. Check that assumed hinges actually form (positive rotations)

Theorems of Plastic Analysis

Lower Bound Theorem (Static/Safe Theorem):

• If any load factor λ can be found for which equilibrium is satisfied without M exceeding M_p anywhere, then λ ≤ λ_collapse
• Provides safe (conservative) estimate

Upper Bound Theorem (Kinematic/Unsafe Theorem):

• If any mechanism is assumed and λ calculated from virtual work, then λ ≥ λ_collapse
• Provides unsafe (unconservative) estimate
• True collapse load is lowest of all kinematically possible mechanisms

Uniqueness Theorem:

• If both equilibrium and mechanism conditions are satisfied, the solution is unique and gives exact collapse load

Cable Analysis

Cable Under Concentrated Loads

Equilibrium at Support: \[\sum F_x = 0, \quad \sum F_y = 0, \quad \sum M = 0\]

• Solve for horizontal and vertical reactions
• Horizontal component of tension H is constant along cable

Tension in Cable Segment: \[T = \sqrt{H^2 + V^2}\]

• T = tension in cable
• H = horizontal component of tension (constant)
• V = vertical component of tension (varies)

Cable Geometry: \[\tan\theta = \frac{V}{H}\]

• θ = angle of cable from horizontal

Cable Under Uniformly Distributed Load

Cable Equation (Parabolic Shape): \[y = \frac{wx^2}{2H} + C_1 x + C_2\]

• y = vertical position of cable
• x = horizontal position
• w = uniform load per unit horizontal length
• H = horizontal component of tension (constant)
• C₁, C₂ = constants from boundary conditions

Maximum Sag (Symmetric Cable, Span L, Supports at Same Level): \[h = \frac{wL^2}{8H}\]

• h = sag at midspan
• L = horizontal span

Horizontal Tension Component: \[H = \frac{wL^2}{8h}\] Maximum Tension (at Support): \[T_{\max} = \sqrt{H^2 + \left(\frac{wL}{2}\right)^2}\] Approximate Cable Length (Small Sag): \[s \approx L \left(1 + \frac{8h^2}{3L^2}\right)\]

• s = cable length
• Valid for h/L <>

Catenary (Cable Under Own Weight)

Catenary Equation: \[y = \frac{H}{w_0} \cosh\left(\frac{w_0 x}{H}\right)\]

• w₀ = weight per unit length of cable
• Measured from lowest point of cable

Parameter c: \[c = \frac{H}{w_0}\]

• Characteristic length parameter

Cable Length: \[s = c \sinh\left(\frac{x}{c}\right)\]

• For small sags, catenary approximates parabola

Three-Hinged Arches

Analysis Procedure

• Three-hinged arch is statically determinate (i = 0)
• Hinges typically at supports and crown (or other interior point)
• Use 3 equilibrium equations plus moment = 0 at internal hinge

Horizontal Thrust at Supports:

• Take moment about internal hinge from one side
• Solve for horizontal reaction H
• H is same at both supports (for symmetric arch)

General Procedure:

1. Apply \(\sum M = 0\) at internal hinge to find horizontal thrust H
2. Use \(\sum F_x = 0\) and \(\sum F_y = 0\) to find vertical reactions
3. Cut arch at any section; apply equilibrium to find internal forces (N, V, M)

Moment at Any Section: \[M = M_{\text{beam}} - H \cdot y\]

• M_beam = moment at section if arch were simply supported beam
• y = vertical height of arch centerline at section
• For properly shaped arch under uniform load, M can be zero everywhere if arch follows funicular shape

Funicular Arch

• Arch shape that carries load purely in axial compression (M = 0 everywhere)
• For uniform load w: funicular shape is parabolic
• For point loads: funicular shape is linear segments between loads

Parabolic Arch Under Uniform Load: \[y = \frac{4h}{L^2}x(L-x)\]

• h = rise at crown
• L = span
• x = horizontal distance from left support
• For this shape, M = 0 throughout arch