Structural Analysis — Complete Formula Sheet

Three Conditions of Equilibrium (2D) ⭐ GATE

ΣFx = 0  |  ΣFy = 0  |  ΣM = 0

For any 2D free body — applies to the whole structure and to any cut portion.

Support Reactions — Number of Unknowns

Simply Supported Beam — Reactions

Pin at A, roller at B, span L. Point load W at distance a from A (b = L − a):

R_A = Wb/L  |  R_B = Wa/L

Full-span UDL w: R_A = R_B = wL/2

Differential Relationships

dV/dx = −w(x) — slope of SFD = negative load intensity

dM/dx = V — slope of BMD = shear force

d²M/dx² = −w(x)

Maximum BM occurs where V = 0 (or changes sign).

Integral Relationships

V₂ − V₁ = −∫w dx — change in SF = area under load diagram

M₂ − M₁ = ∫V dx — change in BM = area under SFD

Load Type → Diagram Shape

Key BM Values ⭐ GATE

SS beam, central point load W: M_max = WL/4 (at midspan)

SS beam, full UDL w: M_max = wL²/8 (at midspan)

SS beam, eccentric load W at a from A: M at load = Wab/L

Cantilever, point load W at free end: M_max = WL (at fixed end, hogging)

Cantilever, full UDL w: M_max = wL²/2 (at fixed end, hogging)

Fixed beam, central point load W: M at supports = WL/8 | M at midspan = WL/8

Fixed beam, full UDL w: M at supports = wL²/12 | M at midspan = wL²/24

Propped cantilever, full UDL w: M at fixed end = wL²/8 | M_{max sag} = 9wL²/128 at 5L/8 from fixed end

Euler-Bernoulli Beam Equation

EI · d²y/dx² = M(x)

E = Young’s modulus [N/mm² or GPa] | I = second moment of area [mm⁴ or m⁴] | EI = flexural rigidity

Standard Maximum Deflection Formulae

Deflection ratio (same w, L, EI): Cantilever : SS : Fixed = 48 : 5 : 1 (for UDL)

Macaulay’s Method

EI·d²y/dx² = M(x) written with Macaulay brackets ⟨x − a⟩

⟨x − a⟩ = 0 when x < a | ⟨x − a⟩ = (x − a) when x ≥ a

Integrate keeping brackets intact. Apply BCs to find C₁ and C₂.

Partial UDL (a to b): Add compensation term: −w⟨x−a⟩²/2 + w⟨x−b⟩²/2

Mohr’s Theorems (Moment-Area Method)

Theorem 1: Change in slope = Area of M/EI diagram between two points

θ_AB = (1/EI) × Area of BMD from A to B

Theorem 2: Tangential deviation = First moment of M/EI diagram about the point

t_A/B = (1/EI) × Area of BMD × distance of centroid from A

DSI — Rigid Frames and Beams

DSI = (3m + r) − (3j + c)

m = members | r = reactions | j = joints | c = condition equations (internal hinges)

DSI = 0 → Determinate | DSI > 0 → Indeterminate | DSI < 0 → Mechanism

DSI — Pin-Jointed Trusses

DSI = m + r − 2j

m = members | r = reactions | j = joints

Internal Hinge Condition Equations

Single internal hinge: c = 1

n members meeting at a common hinge: c = n − 1

Each internal hinge reduces DSI by 1.

DSI — Quick Reference ⭐ GATE

Fixed End Moments — All Cases

Sign convention: FEM_AB positive (anticlockwise on near end), FEM_BA negative (clockwise on far end) for downward loading.

Slope-Deflection Equations

M_AB = (2EI/L)(2θ_A + θ_B − 3ψ) + FEM_AB

M_BA = (2EI/L)(2θ_B + θ_A − 3ψ) + FEM_BA

θ_A, θ_B = joint rotations (positive = clockwise)

ψ = chord rotation = δ/L (sway displacement / member length)

Modified SDE — Far End Pinned

When far end B is a pin/roller (M_BA = 0):

M_AB = (3EI/L)(θ_A − ψ) + FEM_AB − FEM_BA/2

Effective stiffness = 3EI/L | Carry-over to far end = 0

Far End Stiffness and Carry-Over ⭐ GATE

Distribution Factor

DF_member = K_member / ΣK_{all members at joint}

ΣDF at any joint = 1.0 (always)

DF at fixed end = 0 | DF at pin/roller end = 1.0

MDM Process Summary

1. Lock all joints → apply FEMs

2. Release one joint → distribute out-of-balance moment by DFs → carry over (COF × distributed moment) to far ends

3. Repeat until convergence

4. Sum all rows per column → final end moments

5. Check: ΣM = 0 at every internal joint

Sway Frame — Two-Cycle Method

Actual moments = (No-sway moments) + k × (Sway moments)

k = −P / P’

P = prop force in no-sway case | P’ = prop force in sway-only case

Sway FEM (both ends fixed): −6EIΔ/L² = −6EIψ/L

Sway FEM (pin at base, fixed at top): −3EIΔ/L²

Strain Energy

Bending: U = ∫M²dx / 2EI

Axial (truss member): U = F²L / 2AE

Shear: U = ∫V²dx / 2GA (usually neglected)

Castigliano’s Second Theorem

δ_i = ∂U / ∂P_i

Deflection at any point = partial derivative of strain energy w.r.t. load at that point.

For no load at point of interest: apply dummy load P = 0, differentiate, then set P = 0.

Unit Load Method (Virtual Work)

Beam deflection: δ = ∫(mM/EI) dx

Truss deflection: δ = Σ(f·F·L / AE)

M, F = real force effects | m, f = virtual effects from unit load at point of interest

Horizontal Thrust

H = M₀_C / h

M₀_C = free BM at crown (as for equivalent SS beam) | h = rise of arch at crown

Full UDL w: H = wL²/8h

Central point load W: H = WL/4h

Bending Moment at Any Section

M = M₀ − Hy

M₀ = free BM at that section | y = arch height above springing line at that section

M_C = 0 always (crown hinge condition) ✓

Parabolic Arch Geometry

y = 4hx(L−x)/L²

Slope: tan φ = dy/dx = 4h(L−2x)/L²

At crown (x = L/2): φ = 0 | At springing (x = 0): tan φ = 4h/L

Normal Thrust and Radial Shear ⭐ GATE

N = V sinφ + H cosφ (axial compression)

Q = V cosφ − H sinφ (radial shear)

V = net vertical force on left of section | H = horizontal thrust | φ = arch slope at section

Under full UDL on parabolic arch: Q = 0 everywhere, N = H/cosφ

Key Results — Parabolic Arch Under Full UDL

BM = 0 everywhere | Q = 0 everywhere | Pure axial compression

N at crown = H = wL²/8h

N at springing = √(V_A² + H²) = (wL/2)√(1 + L²/16h²)

ILD Ordinates — Simply Supported Beam (span L, section C at distance a from A, b = L−a)

Response Under Loads Using ILD

Point loads: F = Σ(P_i × η_i)

UDL of intensity w: F = w × (area of ILD over loaded length)

Maximum positive response → load where ILD is positive

Maximum negative response → load where ILD is negative

Müller-Breslau Principle ⭐ GATE

The ILD for any response = deflected shape of the released structure under a unit deformation corresponding to that response.

ILD for M_C → insert hinge at C, apply unit rotation → triangular shape

ILD for V_C → insert shear release at C, apply unit relative displacement → two straight lines

ILD for R_A → remove support at A, apply unit upward displacement → straight line

Absolute Maximum BM Under Moving Load Train ⭐ GATE

Occurs under one of the concentrated loads when:

Midspan bisects the distance between that load and the resultant of all loads

Place (load + resultant) symmetrically about midspan → each at d/2 from midspan

where d = distance between the load and the resultant

Governing Equation

[K]{D} = {F}

[K] = global stiffness matrix | {D} = nodal displacements | {F} = nodal forces

Solution: {D} = [K]⁻¹{F} (after applying boundary conditions)

Beam Element Stiffness Matrix (4×4) ⭐ GATE

DOF order: [v_i, θ_i, v_j, θ_j]

(EI/L³) ×

    ┌  12    6L   -12    6L  ┐
    │  6L    4L²  -6L    2L² │
    │ -12   -6L    12   -6L  │
    └  6L    2L²  -6L    4L² ┘

Key entries: k₂₂ = 4EI/L (near end stiffness) | k₂₄ = 2EI/L (carry-over stiffness)

Stiffness Coefficients — Definition

K_ij = force at DOF i when unit displacement is applied at DOF j (all other DOFs = 0)

f_ij = displacement at DOF i due to unit force at DOF j

[K] = [f]⁻¹

Properties of Stiffness Matrix ⭐ GATE

Kinematic Indeterminacy (DKI)

DKI = Total DOF − Restrained DOF

DKI = size of reduced stiffness matrix [K_ff]

Beam joint: 2 DOF (v and θ) | Frame joint: 3 DOF (u, v, θ) | Truss joint: 2 DOF (u and v)

Determinacy ⭐ GATE

DSI = m + r − 2j

= 0: Determinate | > 0: Indeterminate | < 0: Mechanism

Zero-Force Member Rules ⭐ GATE

Rule 1: Two members meet at a joint with no external load → both are zero-force members.

Rule 2: Three members at a joint, two collinear, no load → the third (non-collinear) member is zero-force.

Method of Sections — Moment Shortcut

To find force in one specific member: cut through that member + 2 others.

Take moments about the intersection of the other two cut members → direct solution for the required force.

Maximum 3 cut members per section (3 equilibrium equations available).

Truss Element Stiffness (Global, 4×4)

(AE/L) × (c = cosθ, s = sinθ):

    ┌  c²    cs   -c²   -cs ┐
    │  cs    s²   -cs   -s² │
    │ -c²   -cs    c²    cs │
    └ -cs   -s²    cs    s² ┘

Deflection Ratios (Same w, L, EI — UDL)

Max BM Ratios (Same w, L — UDL)

Propped Cantilever Under UDL — Key Values

R_B (prop) = 3wL/8 | R_A = 5wL/8 | M_A = wL²/8 (hogging)

Point of contraflexure at x = 3L/4 from fixed end

M_{max sag} = 9wL²/128 at x = 5L/8 from fixed end

ILD Peak Ordinates — Simply Supported Beam

Arch Thrust — Summary

MDM Stiffness and DF Summary