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Shear force, bending moment, determinate/indeterminate structures and core analysis methods.
| Concept | Description | Key Point |
|---|---|---|
| Structural Analysis | Determination of internal forces, stresses, and deformations | Ensures safety & serviceability under loads |
| Statically Determinate | Reactions & internal forces found from equilibrium alone | Equations of statics are sufficient |
| Statically Indeterminate | Additional compatibility equations needed | Degree of indeterminacy = unknowns − equations |
| Stability | Structure maintains equilibrium under small disturbances | Unstable structures collapse under any load |
| Ductility | Ability to undergo large plastic deformation before failure | Gives warning before collapse |
| Redundancy | Multiple load paths available | Improves robustness & safety |
| Load Type | Symbol | Description |
|---|---|---|
| Dead Load (DL) | G | Self-weight of structure + permanent fixtures |
| Live Load (LL) | Q | Movable/variable loads (people, furniture, vehicles) |
| Wind Load | W | Dynamic pressure from wind; depends on terrain & height |
| Seismic Load | E | Inertial forces due to earthquake ground motion |
| Snow Load | S | Weight of accumulated snow on roof surfaces |
| Impact Load | Sudden application of load (vehicles, dropping loads) | |
| Accidental Load | A | Explosion, fire, vehicle collision (ULS check) |
| Structure | Load Transfer | Primary Forces |
|---|---|---|
| Truss | Axial members | Tension / Compression only |
| Beam | Flexural bending | Shear + Bending Moment |
| Frame | Beam-column combo | Axial + Shear + Moment |
| Arch | Curved compression | Primarily compression |
| Cable | Tension-only | Catenary / parabolic profile |
| Plate/Shell | 2D surface loading | Bending + Membrane forces |
┌─────────────────────────────────────────────────────────────────┐
│ DEGREE OF STATIC INDETERMINACY (DSI) │
├─────────────────────────────────────────────────────────────────┤
│ │
│ GENERAL FORMULA: │
│ DSI = (m + r) − 2j (for 2D pin-jointed trusses) │
│ │
│ Where: │
│ m = number of members │
│ r = number of reaction components │
│ j = number of joints │
│ │
│ RESULT INTERPRETATION: │
│ ┌────────────────┬──────────────────────────────────────────┐ │
│ │ DSI = 0 │ Statically determinate & stable │ │
│ │ DSI > 0 │ Statically indeterminate (redundant) │ │
│ │ DSI < 0 │ Unstable (mechanism) │ │
│ └────────────────┴──────────────────────────────────────────┘ │
│ │
│ EXAMPLE: Simple triangular truss (3 members, 3 joints, 3 rxns) │
│ DSI = (3 + 3) − 2(3) = 0 → Determinate ✓ │
│ │
│ EXAMPLE: Warren truss (11 members, 7 joints, 3 rxns) │
│ DSI = (11 + 3) − 2(7) = 0 → Determinate ✓ │
│ │
│ FOR BEAMS / FRAMES: │
│ DSI = (3m + r) − 3j (rigid joints, 2D) │
│ DSI = (6m + r) − 6j (rigid joints, 3D) │
│ │
│ KINEMATIC INDETERMINACY (DKI): │
│ DKI = Total DOF − (reaction components) │
│ For each rigid joint in 2D: 3 DOF (Δx, Δy, θ) │
│ For each rigid joint in 3D: 6 DOF │
└─────────────────────────────────────────────────────────────────┘m + r ≥ 2j is necessary but not sufficient for stability.┌─────────────────────────────────────────────────────────────────┐
│ EQUATIONS OF STATICAL EQUILIBRIUM │
├─────────────────────────────────────────────────────────────────┤
│ │
│ 2D STRUCTURES: │
│ ────────────── │
│ ΣFx = 0 (Sum of horizontal forces = 0) │
│ ΣFy = 0 (Sum of vertical forces = 0) │
│ ΣMz = 0 (Sum of moments about any point = 0) │
│ │
│ 3D STRUCTURES: │
│ ────────────── │
│ ΣFx = 0, ΣFy = 0, ΣFz = 0 │
│ ΣMx = 0, ΣMy = 0, ΣMz = 0 │
│ │
│ CONCURRENT FORCE SYSTEM (2D): ΣFx = 0, ΣFy = 0 │
│ PARALLEL FORCE SYSTEM (2D): ΣFy = 0, ΣM = 0 │
│ GENERAL COPLANAR (2D): ΣFx = 0, ΣFy = 0, ΣM = 0 │
│ │
│ FREE BODY DIAGRAM (FBD) STEPS: │
│ ┌──────────────────────────────────────────────────────────┐ │
│ │ 1. Isolate the body (cut free from all supports) │ │
│ │ 2. Show ALL external forces (applied loads + reactions) │ │
│ │ 3. Show internal forces at cut sections │ │
│ │ 4. Choose a convenient coordinate system │ │
│ │ 5. Apply equilibrium equations to solve unknowns │ │
│ └──────────────────────────────────────────────────────────┘ │
└─────────────────────────────────────────────────────────────────┘| Support Type | Symbol | Reactions | Restrained DOF |
|---|---|---|---|
| Roller | △ or ○ | 1 (normal to surface) | 1 translation |
| Pin / Hinge | △ | 2 (Fx, Fy) | 2 translations |
| Fixed / Built-in | ▬▬▬ | 3 (Fx, Fy, Mz) | All (2 trans + 1 rot) |
| Hinged roller | △ with rotation | 2 (Fx, Fy) | 2 translations; free rotation |
| Link / Cable | ──○── | 1 (along link) | 1 (along link direction) |
| Rocker | curved surface | 1 (normal) | 1 translation |
| Quantity | Positive (+) | Negative (−) |
|---|---|---|
| Axial Force | Tension (pulling away) | Compression (pushing in) |
| Shear Force | Clockwise on left face | Counter-clockwise on left face |
| Bending Moment | Sagging / hogging (varies) | Hogging / sagging (convention) |
| Deflection | Upward | Downward |
| Slope / Rotation | Counter-clockwise | Clockwise |
┌─────────────────────────────────────────────────────────────────┐
│ SUPPORT REACTION CALCULATION — EXAMPLE │
├─────────────────────────────────────────────────────────────────┤
│ │
│ Simply supported beam, span L = 6 m: │
│ Point load P = 20 kN at 2 m from A (left support) │
│ UDL w = 5 kN/m over full span │
│ │
│ A (Pin) [UDL = 5 kN/m] B (Roller) │
│ △──────────▲──────────────────────────────────△ │
│ 2m P=20kN │
│ │
│ ΣMB = 0 (take moment about B): │
│ RA × 6 − 20 × 4 − (5 × 6) × 3 = 0 │
│ RA × 6 − 80 − 90 = 0 │
│ RA = 170 / 6 = 28.33 kN │
│ │
│ ΣFy = 0: │
│ RA + RB − 20 − (5 × 6) = 0 │
│ 28.33 + RB − 20 − 30 = 0 │
│ RB = 50 − 28.33 = 21.67 kN │
│ │
│ Check: ΣMA = 0 │
│ −21.67 × 6 + 20 × 2 + 30 × 3 = −130 + 40 + 90 = 0 ✓ │
└─────────────────────────────────────────────────────────────────┘| # | Assumption | Implication |
|---|---|---|
| 1 | Members are straight | No bending, only axial forces |
| 2 | Joints are frictionless pins | No moment transfer at joints |
| 3 | Loads applied at joints only | Members carry axial force (T/C) only |
| 4 | Self-weight neglected (or split to joints) | Simplifies analysis |
| 5 | Material is linear elastic | Hooke's law applies |
| Truss Type | Configuration | Use Case |
|---|---|---|
| Pratt | Diagonals slope toward center | Bridges (verticals in compression) |
| Warren | Equilateral triangles | Bridges, roof trusses |
| Howe | Diagonals slope away from center | Roof trusses (verticals in tension) |
| K-Truss | K-shaped vertical subdivisions | Long-span railway bridges |
| Fink | Sub-divided Warren | Roof trusses with variable pitch |
| Baltimore | Sub-divided Pratt | Heavy-duty railway bridges |
┌─────────────────────────────────────────────────────────────────┐
│ METHOD OF JOINTS │
├─────────────────────────────────────────────────────────────────┤
│ │
│ PROCEDURE: │
│ 1. Find support reactions using equilibrium │
│ 2. Select a joint with ≤ 2 unknowns │
│ 3. Draw FBD of the joint │
│ 4. Apply ΣFx = 0, ΣFy = 0 │
│ 5. Assume members in tension (+ve); if result is −ve → compress│
│ 6. Move to the next joint with ≤ 2 unknowns │
│ 7. Repeat until all members are solved │
│ │
│ IDENTIFYING ZERO-FORCE MEMBERS: │
│ ┌──────────────────────────────────────────────────────────┐ │
│ │ Rule 1: If a joint has only 2 non-collinear members │ │
│ │ and NO external load → both members are zero. │ │
│ │ │ │
│ │ Rule 2: If a joint has 3 members where 2 are collinear │ │
│ │ and NO external load → the 3rd member is zero. │ │
│ │ │ │
│ │ Rule 3: If a joint has 2 members and external load acts │ │
│ │ along one member → the other member is zero. │ │
│ └──────────────────────────────────────────────────────────┘ │
│ │
│ C (no external load) │
│ / \ │
│ / \ ← Both are ZERO FORCE members │
│ / \ │
│ A───────B │
└─────────────────────────────────────────────────────────────────┘┌─────────────────────────────────────────────────────────────────┐
│ METHOD OF SECTIONS │
├─────────────────────────────────────────────────────────────────┤
│ │
│ BEST FOR: Finding force in a SPECIFIC member without solving │
│ the entire truss. │
│ │
│ PROCEDURE: │
│ 1. Find support reactions │
│ 2. Cut the truss with a section passing through the target │
│ member (max 3 members cut for 2D) │
│ 3. Draw FBD of either portion │
│ 4. Apply equilibrium equations (ΣFx, ΣFy, ΣM) │
│ 5. Choose equation that eliminates other unknowns │
│ │
│ STRATEGY FOR CHOOSING MOMENT CENTER: │
│ ┌──────────────────────────────────────────────────────────┐ │
│ │ • Take moment at intersection of 2 unknown forces │ │
│ │ → solves the 3rd force directly │ │
│ │ • Use ΣFy = 0 when 2 unknowns are horizontal │ │
│ │ • Use ΣFx = 0 when 2 unknowns are vertical │ │
│ └──────────────────────────────────────────────────────────┘ │
│ │
│ EXAMPLE: Pratt truss, find force in member CD │
│ ───────────────────────────────────── │
│ A ─── B ─── C ─── D ─── E │
│ | / | / | / | / | │
│ | / | / | / | / | │
│ F ─── G ─── H ─── I ─── J │
│ │
│ Cut section through BD, CD, CI (3 members) │
│ Take left portion: Apply ΣMH = 0 to find force in BD │
│ Apply ΣFy = 0 to find force in CD │
│ Apply ΣMG = 0 to find force in CI │
└─────────────────────────────────────────────────────────────────┘| Feature | Method of Joints | Method of Sections |
|---|---|---|
| Solves | All member forces | Specific member(s) |
| Approach | Joint by joint (2 eqns each) | Cut section (3 eqns total) |
| Best For | Small trusses, all forces needed | Large trusses, specific members |
| Max Unknowns | 2 per joint (2D) | 3 per cut (2D) |
| Speed | Slower for large trusses | Fast for targeted members |
| Zero-Force ID | Easy to identify at joints | Cannot directly identify |
┌─────────────────────────────────────────────────────────────────┐
│ BEAM TYPES & LOADING NOTATION │
├─────────────────────────────────────────────────────────────────┤
│ │
│ SIMPLY SUPPORTED BEAM: Pin at one end, Roller at other │
│ CANTILEVER BEAM: Fixed at one end, Free at other │
│ OVERHANGING BEAM: Supports inside, extends beyond │
│ CONTINUOUS BEAM: More than 2 supports (indeterminate)│
│ FIXED BEAM: Fixed at both ends (indeterminate) │
│ PROPPED CANTILEVER: Fixed + Roller (indeterminate) │
│ │
│ SIGN CONVENTION (Left Segment Rule): │
│ ────────────────────────────── │
│ │
│ SHEAR FORCE (V): │
│ Upward force on LEFT face → POSITIVE (+) │
│ │
│ ↑ +V │
│ ─────┤ │
│ │
│ BENDING MOMENT (M): │
│ Sagging (concave up) → POSITIVE (+) │
│ Hogging (concave down) → NEGATIVE (−) │
│ │
│ Sagging (+M): Hogging (−M): │
│ ╲ ╱ ╱ ╲ │
│ ╲──╱ ╱──╲ │
│ │
│ RELATIONSHIPS BETWEEN LOAD, SHEAR, MOMENT: │
│ w(x) = dV/dx (load = rate of change of shear) │
│ V(x) = dM/dx (shear = rate of change of moment) │
│ d²M/dx² = w(x) (moment-curvature relationship) │
└─────────────────────────────────────────────────────────────────┘| Loading | SFD Shape | BMD Shape |
|---|---|---|
| No load (0) | Constant / horizontal | Linear (sloped) |
| Point load | Vertical jump by P | Slope change (kink) |
| UDL (w) | Linear slope = −w | Parabolic (degree 2) |
| UVL (triangular) | Parabolic | Cubic (degree 3) |
| Point moment | No change | Vertical jump by M |
| Concentrated couple | No change in V | Discontinuity = M₀ |
| Beam | Loading | Max Moment |
|---|---|---|
| Simply Supported | Central point load P | PL/4 (at center) |
| Simply Supported | UDL w over full span | wL²/8 (at center) |
| Cantilever | Point load P at free end | PL (at fixed end) |
| Cantilever | UDL w over full span | wL²/2 (at fixed end) |
| Fixed-Fixed | Central point load P | PL/8 (at ends & center) |
| Fixed-Fixed | UDL w over full span | wL²/12 (at supports) |
| Simply Supported | Two equal loads P at L/3 | PL/3 (at midspan) |
┌─────────────────────────────────────────────────────────────────┐
│ SFD & BMD — CANTILEVER WITH UDL + POINT LOAD │
├─────────────────────────────────────────────────────────────────┤
│ │
│ Cantilever: L = 4 m, UDL w = 10 kN/m, Point load P = 20 kN │
│ at free end (B). Fixed at A. │
│ │
│ B (free end) A (fixed end) │
│ ○──────────────────────────┬┬ │
│ │ P=20kN w=10 kN/m ││ │
│ ↓ ▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼▼││ │
│ │
│ FIXED END REACTIONS: │
│ VA = 20 + 10×4 = 60 kN (upward) │
│ MA = 20×4 + 10×4×2 = 160 kN·m (CCW, hogging) │
│ │
│ SHEAR FORCE (from free end B, x from B): │
│ ┌────────────────────────────────────────────────────────┐ │
│ │ At B (x=0): V = −20 kN (downward point load) │ │
│ │ At x: V = −20 − 10x │ │
│ │ At A (x=4): V = −20 − 40 = −60 kN │ │
│ │ SFD: Linear from −20 to −60 kN │ │
│ └────────────────────────────────────────────────────────┘ │
│ │
│ BENDING MOMENT (from free end B, x from B): │
│ ┌────────────────────────────────────────────────────────┐ │
│ │ At B (x=0): M = 0 │ │
│ │ At x: M = −20x − 10x²/2 = −20x − 5x² │ │
│ │ At A (x=4): M = −80 − 80 = −160 kN·m │ │
│ │ BMD: Cubic (parabolic + linear) from 0 to −160 kN·m │ │
│ └────────────────────────────────────────────────────────┘ │
│ │
│ POINT OF MAX SHEAR: At A → |Vmax| = 60 kN │
│ POINT OF MAX MOMENT: At A → |Mmax| = 160 kN·m │
└─────────────────────────────────────────────────────────────────┘┌─────────────────────────────────────────────────────────────────┐
│ FRAME CLASSIFICATION │
├─────────────────────────────────────────────────────────────────┤
│ │
│ 1. BRACED FRAME (Non-Sway / Braced): │
│ • Lateral stability provided by bracing or shear walls │
│ • Approximate analysis: assume pins at beam-column joints │
│ • Axial forces in beams ≈ zero │
│ • Columns carry axial load + end moments from beams │
│ │
│ 2. UNBRACED FRAME (Sway / Rigid): │
│ • Lateral stability through rigidity of connections │
│ • Joints are rigid — transfer moment │
│ • More complex: must consider sway effects │
│ │
│ PORTAL METHOD (Approximate — for lateral loads): │
│ ───────────────────────────────────────── │
│ Assumptions: │
│ • Inflection points at mid-height of columns │
│ • Inflection points at mid-span of beams │
│ • Interior columns carry twice the shear of exterior columns │
│ • Horizontal shear is distributed equally to exterior cols │
│ and shared by interior columns │
│ │
│ CANTILEVER METHOD (Approximate — for tall frames): │
│ ───────────────────────────────────────────── │
│ Assumptions: │
│ • Inflection points at mid-height of columns │
│ • Inflection points at mid-span of beams │
│ • Axial stress in columns is proportional to distance from │
│ the centroid of the column group │
└─────────────────────────────────────────────────────────────────┘| Quantity | Formula |
|---|---|
| General Equation | M_AB = (2EI/L)[2θA + θB − 3δ/L] + MF_AB |
| Fixed End Moment (UDL) | MF = wL²/12 (at both ends) |
| Fixed End Moment (Point Load) | MF = Pab²/L² at A; Pa²b/L² at B |
| Sway Modification | Add −6EIδ/L² to each end moment |
| Equilibrium | ΣM at each joint = 0 (solve for θ, δ) |
| Step | Operation | Description |
|---|---|---|
| 1 | Calculate Stiffness (k) | k = 4EI/L (far end fixed), 3EI/L (far end pinned) |
| 2 | Distribution Factor (DF) | DF = k_member / Σk_at_joint |
| 3 | Fixed End Moments (FEM) | Calculate FEMs for each span |
| 4 | Release & Distribute | Apply opposite of unbalanced moment × DF |
| 5 | Carry Over (CO) | CO = ½ of distributed moment to far end |
| 6 | Repeat | Cycle until moments converge (≤ 1-2% change) |
| 7 | Sum | Final moment = FEM + Σ distributed + Σ carry-over |
┌─────────────────────────────────────────────────────────────────┐
│ MOMENT DISTRIBUTION — CONTINUOUS BEAM EXAMPLE │
├─────────────────────────────────────────────────────────────────┤
│ │
│ Two-span continuous beam: AB = 4 m, BC = 6 m │
│ UDL on AB: w = 20 kN/m, Point load on BC: P = 40 kN at center │
│ All supports are simple pins (except B is continuous). │
│ │
│ STIFFNESS: │
│ k_AB = 4EI/4 = EI (B is continuous, treat as fixed) │
│ k_BC = 4EI/6 = 2EI/3 (B is continuous) │
│ Σk at B = EI + 2EI/3 = 5EI/3 │
│ │
│ DISTRIBUTION FACTORS AT B: │
│ DF_AB = (EI)/(5EI/3) = 3/5 = 0.6 │
│ DF_BC = (2EI/3)/(5EI/3) = 2/5 = 0.4 │
│ │
│ FIXED END MOMENTS: │
│ FEM_AB = −wL²/12 = −20(16)/12 = −26.67 kN·m │
│ FEM_BA = +26.67 kN·m │
│ FEM_BC = −PL/8 = −40(6)/8 = −30.00 kN·m │
│ FEM_CB = +30.00 kN·m │
│ │
│ UNBALANCED MOMENT AT B = 26.67 − 30.00 = −3.33 kN·m │
│ │
│ DISTRIBUTION: │
│ M_AB_distributed = +3.33 × 0.6 = +2.00 kN·m │
│ M_BC_distributed = +3.33 × 0.4 = +1.33 kN·m │
│ │
│ CARRY OVER (½ of distributed to far end): │
│ CO to A = +2.00/2 = +1.00 kN·m │
│ CO to C = +1.33/2 = +0.67 kN·m │
│ │
│ (Repeat cycles until convergence...) │
└─────────────────────────────────────────────────────────────────┘3EI/L instead of 4EI/L and no carry-over occurs to the pinned end. The Slope-Deflection method gives the same results but requires solving simultaneous equations.| Feature | Influence Line | Shear/Moment Diagram |
|---|---|---|
| Represents | Variation of ONE quantity due to unit load moving across span | Effects of FIXED loading at all positions |
| X-axis | Position of moving unit load | Position along the beam |
| Y-axis | Value of reaction/shear/moment at a fixed point | Value of S/M at that section |
| Loading | Unit load (1 kN) at varying positions | Actual loads at fixed positions |
| Use | Finding max response under moving loads | Design for fixed loads |
| Function | Remove Restraint | Apply |
|---|---|---|
| Vertical Reaction | Remove vertical support | Unit vertical displacement |
| Shear at Section | Cut + allow vertical slip (no rotation) | Unit relative vertical displacement |
| Moment at Section | Insert hinge | Unit relative rotation |
| Axial Force | Cut + allow axial slip | Unit relative axial displacement |
┌─────────────────────────────────────────────────────────────────┐
│ INFLUENCE LINES — SIMPLY SUPPORTED BEAM (L = 10 m) │
├─────────────────────────────────────────────────────────────────┤
│ │
│ 1. INFLUENCE LINE FOR REACTION RA: │
│ ──────────────────────────── │
│ IL(RA): Straight line from 1.0 at A to 0.0 at B │
│ Equation: RA = (L − x)/L = 1 − x/L │
│ At x = 3m: RA = 0.7 │
│ At x = 7m: RA = 0.3 │
│ │
│ 2. INFLUENCE LINE FOR SHEAR AT SECTION C (4 m from A): │
│ ──────────────────────────────────────────── │
│ Left of C: V_C = RA − 1 = (L−x)/L − 1 = −x/L │
│ Right of C: V_C = RA = (L−x)/L │
│ │
│ Shape: Two straight lines │
│ At A (x=0): V_C = 0 │
│ Just left of C (x=4): V_C = −4/10 = −0.4 │
│ Just right of C (x=4): V_C = 6/10 = +0.6 │
│ At B (x=10): V_C = 0 │
│ Max positive ordinate = 0.6 (just right of C) │
│ Max negative ordinate = −0.4 (just left of C) │
│ │
│ 3. INFLUENCE LINE FOR MOMENT AT SECTION C (4 m from A): │
│ ──────────────────────────────────────────────── │
│ Left of C: M_C = RA × 4 = 4(L−x)/L │
│ Right of C: M_C = RB × 6 = 6x/L │
│ │
│ Shape: Two straight lines meeting at C │
│ At A: M_C = 0 │
│ At C: M_C = 4 × 6 / 10 = 2.4 (maximum ordinate) │
│ At B: M_C = 0 │
│ │
│ USING INFLUENCE LINES WITH MULTIPLE LOADS: │
│ ──────────────────────────────────────── │
│ For a system of loads P1, P2, ..., Pn at positions x1..xn: │
│ Response = Σ (Pi × yi) where yi = IL ordinate at xi │
│ │
│ For UDL w over a region [a, b]: │
│ Response = w × (Area under IL between a and b) │
└─────────────────────────────────────────────────────────────────┘┌─────────────────────────────────────────────────────────────────┐
│ VIRTUAL WORK PRINCIPLES │
├─────────────────────────────────────────────────────────────────┤
│ │
│ PRINCIPLE OF VIRTUAL WORK (PVW): │
│ External Virtual Work = Internal Virtual Work │
│ Σ(P × δ) = Σ(F × δL) + Σ(T × θ) + Σ(M × φ) │
│ │
│ ──────────────────────────────────────────────────────────── │
│ UNIT VIRTUAL LOAD METHOD (for deflections): │
│ ──────────────────────────────────────────────────────────── │
│ │
│ For TRUSSES: │
│ δ = Σ (n × N × L) / (A × E) │
│ │
│ Where: │
│ δ = deflection at desired point/direction │
│ n = force in member due to UNIT virtual load at desired point │
│ N = force in member due to REAL loads │
│ L = length of member │
│ A = cross-sectional area of member │
│ E = Young's modulus of elasticity │
│ │
│ For BEAMS (bending only): │
│ δ = ∫ (m × M) / (E × I) dx │
│ │
│ Where: │
│ m = bending moment due to UNIT virtual load │
│ M = bending moment due to REAL loads │
│ E = Young's modulus │
│ I = moment of inertia of cross-section │
│ │
│ STEPS: │
│ 1. Calculate real forces/moments (N or M) due to actual loads │
│ 2. Apply unit virtual load at the point where deflection is │
│ needed (in the direction of desired deflection) │
│ 3. Calculate virtual forces/moments (n or m) │
│ 4. Apply the formula and sum/integrate │
│ 5. If result is +ve → deflection in direction of unit load │
│ If result is −ve → deflection opposite to unit load │
└─────────────────────────────────────────────────────────────────┘| Theorem | Statement |
|---|---|
| Maxwell-Betti | Work done by load P1 acting through displacement δ12 due to P2 = Work done by P2 acting through δ21 due to P1 |
| Reciprocal Deflections | δ_ij = δ_ji (deflection at i due to unit load at j = deflection at j due to unit load at i) |
| Application | Only need to compute deflection in one direction — the reciprocal is automatically known |
| Theorem | Formula | Use For |
|---|---|---|
| First | δ_i = ∂U/∂P_i | Deflection under load P_i |
| Second | P_i = ∂U/∂δ_i | Load for given displacement |
(n, N, L, A) in columns for systematic computation.┌─────────────────────────────────────────────────────────────────┐
│ MAXIMUM DEFLECTION FORMULAS │
├─────────────────────────────────────────────────────────────────┤
│ │
│ SIMPLY SUPPORTED BEAM: │
│ ────────────────────── │
│ Central point load P: │
│ δ_max = PL³ / (48EI) at center │
│ │
│ UDL w over full span: │
│ δ_max = 5wL⁴ / (384EI) at center │
│ │
│ Moment at one end M: │
│ δ_max = ML² / (9√3 EI) at x = L/√3 │
│ │
│ Two equal loads P at L/3: │
│ δ_max = 23PL³ / (648EI) at center │
│ │
│ CANTILEVER BEAM: │
│ ──────────────── │
│ Point load P at free end: │
│ δ_max = PL³ / (3EI) at free end │
│ │
│ UDL w over full span: │
│ δ_max = wL⁴ / (8EI) at free end │
│ │
│ Moment M at free end: │
│ δ_max = ML² / (2EI) at free end │
│ │
│ FIXED-FIXED BEAM: │
│ ──────────────── │
│ Central point load P: │
│ δ_max = PL³ / (192EI) at center │
│ │
│ UDL w over full span: │
│ δ_max = wL⁴ / (384EI) at center │
└─────────────────────────────────────────────────────────────────┘| Method | Principle | Best For |
|---|---|---|
| Double Integration | EI(d²y/dx²) = M(x) | Simple loading, standard cases |
| Macaulay's Method | Singular functions for discontinuous loads | Point loads at any position |
| Moment-Area (Mohr's) | Area of M/EI diagram = slope change | Sketching deflected shape, angles |
| Conjugate Beam | Elastic load = M/EI diagram | Finding slopes & deflections by beam analogy |
| Virtual Work (Unit Load) | δ = ∫mM/EI dx | Complex loading, trusses, frames |
| Castigliano's | δ = ∂U/∂P | Energy methods, varying loads |
| Superposition | Sum of standard cases | Combination of simple loadings |
| Theorem | Statement |
|---|---|
| First | Change in slope between A and B = Area of M/EI diagram between A and B |
| Second | Deflection at B relative to tangent at A = Moment of M/EI diagram between A and B, taken about B |
┌─────────────────────────────────────────────────────────────────┐
│ MACAULAY'S METHOD (STEP FUNCTION METHOD) │
├─────────────────────────────────────────────────────────────────┤
│ │
│ Used for discontinuous loading on beams. │
│ │
│ STEP FUNCTION NOTATION: │
│ ⟨x − a⟩ⁿ = 0 for x < a │
│ ⟨x − a⟩ⁿ = (x − a)ⁿ for x ≥ a │
│ │
│ INTEGRATION RULES: │
│ ∫⟨x − a⟩ⁿ dx = ⟨x − a⟩ⁿ⁺¹ / (n + 1) [n ≥ 0] │
│ ∫⟨x − a⟩⁻¹ dx = ⟨x − a⟩⁰ [n = −1] │
│ │
│ EXAMPLE: Simply supported beam, L = 6 m │
│ UDL w = 10 kN/m over full span │
│ Point load P = 20 kN at x = 2 m from A │
│ │
│ Reaction RA = (wL/2 + Pb/L) [from equilibrium] │
│ = (10×6/2 + 20×4/6) = 30 + 13.33 = 43.33 kN │
│ │
│ BENDING MOMENT EXPRESSION: │
│ M(x) = RA·x − w·x²/2 − P·⟨x−2⟩¹ │
│ M(x) = 43.33x − 5x² − 20⟨x−2⟩ │
│ │
│ EI·d²y/dx² = M(x) = 43.33x − 5x² − 20⟨x−2⟩ │
│ │
│ Integrate once (slope): │
│ EI·dy/dx = 43.33x²/2 − 5x³/3 − 20⟨x−2⟩²/2 + C₁ │
│ │
│ Integrate again (deflection): │
│ EI·y = 43.33x³/6 − 5x⁴/12 − 20⟨x−2⟩³/6 + C₁x + C₂ │
│ │
│ Apply BCs: y = 0 at x = 0 → C₂ = 0 │
│ y = 0 at x = 6 → solve for C₁ │
│ │
│ Then find maximum deflection by setting dy/dx = 0 │
└─────────────────────────────────────────────────────────────────┘span/250 for general serviceability or span/360 for sensitive finishes. For cantilevers, the limit is often span/125. Always check deflection under service loads (unfactored), not ultimate loads.