Theory of Structures – Complete Study Notes

Comprehensive chapter-wise notes covering every aspect of Theory of Structures — from static determinacy and indeterminacy through influence line diagrams, arches, trusses, matrix methods of analysis, and structural dynamics. All formulae, diagrams, solved examples, IS code references, and exam-focused tables included.

GATE ESE / IES SSC JE State PSC RRB JE

Ch 1 · Determinacy & Indeterminacy Ch 2 · Influence Line Diagrams Ch 3 · Rolling Loads Ch 4 · Arches Ch 5 · Suspension Bridges Ch 6 · Methods of Structural Analysis Ch 7 · Trusses Ch 8 · Matrix Methods Ch 9 · Structural Dynamics Ch 10 · Suspended Cables Quick Revision

1Determinacy and Indeterminacy►

1.1 Introduction

A structure is statically determinate if all reaction forces and internal forces can be found using the three static equilibrium equations alone (ΣFx = 0, ΣFy = 0, ΣM = 0). A statically indeterminate (or hyperstatic) structure has more unknowns than equilibrium equations — the extra unknowns are the degree of static indeterminacy (DSI), also called redundants.

⭐ Key formula: DSI = Total unknowns − Equations of equilibrium available. Redundants must be found using compatibility (force method) or stiffness (displacement method).

1.2 Degrees of Static Indeterminacy (DSI) — Beams & Frames

For beams and frames:
DSI = (m + r) − 3j [pin-jointed frames; m = members, r = reactions, j = joints]

For rigid frames:
DSI = 3m + r − 3j − releases

Where: m = no. of members, r = no. of external reactions, j = no. of joints
Internal hinge introduces 1 condition equation (reduces DSI by 1)

External Indeterminacy

Number of reactions minus the number of equilibrium equations:

External DSI = r − 3 (for 2D plane structures)
External DSI = r − 6 (for 3D space structures)

Internal Indeterminacy

Arises from closed rings, redundant members, or moment-resistant joints:

Internal DSI (rigid frames) = 3 × number of closed loops
Internal DSI (trusses) = m − (2j − 3) for plane truss
Internal DSI (space truss) = m − (3j − 6)

1.3 Quick Classification Table

Structure Type	Reactions (r)	DSI Formula	Example DSI
Simply supported beam	3	r − 3 = 0	0 (Determinate)
Propped cantilever	4	r − 3 = 1	1
Fixed-fixed beam	6	r − 3 = 3	3
Fixed-pinned beam	5	r − 3 = 2	2
Two-hinged arch	4	r − 3 = 1	1
Fixed arch	6	r − 3 = 3	3
Portal frame (fixed base)	6	3m+r−3j = high	6

1.4 Degree of Kinematic Indeterminacy (DKI)

Also called degree of freedom (DOF) — the number of independent displacements/rotations that define the deformed shape of the structure.

DKI = 3j − r (2D frames; each joint has 3 DOF: u, v, θ)
DKI = 6j − r (3D space frames)
Axial deformation neglected: DKI = 2j − r (for beams; only rotations count if inextensible)

Structure	DOF per Joint	DKI (approx)	Notes
Simply supported beam	Rotation at each end	2	2 rotations (slopes)
Fixed-fixed beam	—	0	Both ends fully fixed; no unknown displacements
Propped cantilever	Rotation at prop	1	1 slope at roller end
Continuous beam (n spans)	Rotation at interior joints	n − 1	Interior slopes only

1.5 Stability of Structures

A structure must be stable (will not collapse or move as a rigid body) before it can be classified as determinate or indeterminate.

Unstable (mechanism): DSI < 0, or geometry places all reactions concurrent / parallel
Determinate: DSI = 0 (all unknowns solvable by equilibrium alone)
Indeterminate: DSI > 0 (compatibility equations needed)

⚠️ DSI = 0 does NOT guarantee stability — geometry matters! Example: Three parallel reaction forces (all vertical) cannot resist a horizontal load → unstable even if DSI = 0.

1.6 Solved Examples

Example 1 — Fixed-End Beam

Fig. 1.1 — Fixed-Fixed Beam under UDL (DSI = 3)

r = 6 (RA, HA, MA at A; RB, HB, MB at B)
DSI = r − 3 = 6 − 3 = 3
∴ Three redundants (e.g., MA, MB, and HA taken as redundants)

Fixed End Moments (standard result for UDL on fixed-fixed beam):
MAB = +wL²/12 (hogging at A)
MBA = −wL²/12 (hogging at B)
RA = RB = wL/2

Example 2 — Pin-Jointed Plane Truss

Given: 7 members (m), 5 joints (j)
r = 3 (pin at one end + roller at other)
DSI = m + r − 2j = 7 + 3 − 2×5 = 10 − 10 = 0
∴ Statically Determinate truss

📝 GATE Tip: For rigid frames with internal hinges, each hinge releases 1 condition equation. Subtract 1 from DSI for every internal hinge: DSI_net = DSI_gross − n_hinges.

2Influence Line Diagrams (ILD)►

2.1 Concept of Influence Lines

An Influence Line is a graph showing the variation of a structural response function (reaction, shear force, bending moment) at a fixed point as a unit concentrated load moves across the structure. It is a powerful tool for determining the most critical position of moving loads.

⭐ Key distinction: A bending moment diagram (BMD) shows forces everywhere for a fixed load position. An influence line shows one response at a fixed point as the load moves.

2.2 Müller-Breslau Principle

The Müller-Breslau principle states: "The influence line for any response function is the same as the deflected shape of the structure when the restraint corresponding to that function is removed and a unit displacement/rotation is applied in its direction."

For ILD of a reaction: remove the corresponding support, apply unit displacement at that point
For ILD of shear at section C: cut the beam at C, apply unit relative shear displacement
For ILD of moment at section C: insert a hinge at C, apply unit relative rotation

ℹ️ Müller-Breslau is especially useful for indeterminate structures — the deflected shape IS the ILD (qualitative shape). For determinate structures, both the qualitative and quantitative ILD are linear (straight lines).

2.3 ILD for Simply Supported Beam

Fig. 2.1 — ILD for Reaction R_A (triangle, peak = 1 at A)

Fig. 2.2 — ILD for BM at Midspan C (peak = L/4)

2.4 Key ILD Ordinates — Simply Supported Beam (Span L, section at distance 'a' from A)

Response	ILD Shape	Max Ordinate	Location of Peak
Reaction R_A	Triangle	1.0	At A
Reaction R_B	Triangle	1.0	At B
SF at C (left)	Two triangles (positive and negative)	b/L (pos), a/L (neg)	Between A&C / C&B
BM at C	Triangle (peak at C)	ab/L	At C; a+b=L
BM at midspan	Triangle	L/4	At midspan

2.5 Maximum Effect Using ILD

For a single concentrated load P:
Max response = P × (max ILD ordinate)

For a UDL of intensity w (length ℓ):
Max response = w × (area of ILD under load)

For UDL over full span: Max BM at centre = w × (L/4) × (L/2) = wL²/8 ✓

2.6 ILD for Indeterminate Structures

For indeterminate beams, ILD ordinates are curved (not straight lines). Müller-Breslau principle gives the qualitative shape; ordinates must be computed from compatibility equations or standard tables.

📝 GATE/ESE Tip: For a propped cantilever, the ILD for the prop reaction has a cubic parabola shape. The peak ordinate at mid-span is 5/16 × L (for reaction at prop end). Always sketch qualitatively using Müller-Breslau — it earns partial marks.

3Rolling Loads (Moving Loads)►

3.1 Classification of Rolling Loads

Single concentrated load — locomotive, crane wheel
Series of concentrated loads — train bogies, truck axles
Uniformly Distributed Load (UDL) — longer than, equal to, or shorter than the span
Equivalent UDL — used for design; replaces the actual series of loads for convenience

3.2 Absolute Maximum Bending Moment (AMBM)

The Absolute Maximum BM is the largest bending moment anywhere in the structure due to any possible position of the moving load system.

⭐ Key rule for a single concentrated load P on span L: AMBM = PL/4, occurring at midspan when load is at midspan.

For a Series of Concentrated Loads

The AMBM occurs under one of the heavy loads, when the load and the resultant of all loads on the span are equidistant from midspan (the midspan bisects the distance between the load and the resultant).

Criterion: place load such that midspan = midpoint of (load position, resultant position)

Fig. 3.1 — Series of rolling loads; R = resultant, AMBM criterion: midspan bisects load–resultant distance

3.3 Maximum Shear Force Due to Rolling Loads

Absolute Maximum SF = Maximum reaction = sum of all loads × (span − distance of resultant from nearer end) / span
Occurs when the leading (heaviest) load reaches the support.

3.4 UDL Shorter than Span

Condition	For Maximum BM at C	For Maximum SF at C
UDL length d < span L	Load must straddle C symmetrically; centroid at C	Load placed from C to the nearer support (A or B)
UDL length d = span L	UDL covers full span; BM at mid = wL²/8	SF at end = wL/2
UDL length d > span L	Cover full span; surplus load beyond span not considered	Same as full span

3.5 Equivalent Uniformly Distributed Load (EUDL)

The EUDL is a hypothetical UDL that produces the same maximum bending moment or shear as the actual series of concentrated loads.

EUDL for BM: w_e = 8 × M_max / L²
EUDL for SF: w_e = 2 × F_max / L

📝 EUDL values for standard railway loading (BG, MG) are tabulated in Railway Bridge Rules and are directly used in design of railway bridges. ESE often asks about this in bridge engineering context.

4Arches►

4.1 Introduction

An arch is a curved structural member that carries loads primarily through compression. The horizontal thrust generated at the supports reduces bending moments dramatically compared to a beam of the same span and loading — this is the fundamental advantage of arches.

⭐ Key benefit: For a parabolic arch under UDL, bending moment everywhere = 0 (pure compression). This is called a funicular arch.

4.2 Types of Arches

Type	Hinges	DSI	Key Feature
Three-Hinged Arch	2 ends + 1 crown	0 (Determinate)	Temperature change, settlement → no secondary stresses
Two-Hinged Arch	2 ends only	1	Most common; H found from compatibility
Hingeless (Fixed) Arch	0	3	Highest efficiency; sensitive to temp & settlement

4.3 Analysis of Three-Hinged Arch

Fig. 4.1 — Three-Hinged Arch: Horizontal thrust H acts inward at both supports

Step 1 — Find RA, RB using ΣM = 0 (same as a simply supported beam):
RA = P·b/L RB = P·a/L (where a, b are distances of load from A and B)

Step 2 — Use hinge condition at crown (ΣM_crown = 0 for either half):
ΣM about crown for left half: RA × (L/2) − H × h = 0
∴ H = RA × L / (2h)

Step 3 — Normal thrust and radial shear at any section:
N = H·cos θ + V·sin θ (compression, +ve)
Q = V·cos θ − H·sin θ (radial shear)

4.4 Parabolic Arch under UDL (Pure Compression)

For a parabolic arch with equation y = 4h·x(L−x)/L², subjected to a uniform load w per unit horizontal length, the horizontal thrust H = wL²/(8h) and bending moment at every section = 0.

Arch rib equation (parabola, origin at A):
y = (4h/L²) · x · (L − x)

Horizontal thrust H = wL² / (8h)

BM at any section x: M = M_beam − H·y = 0 (for parabolic arch under UDL) ✓

4.5 Two-Hinged Arch — Finding Horizontal Thrust

For a two-hinged arch, the horizontal thrust H is the redundant. Using the compatibility condition (horizontal displacement at B = 0):

H = ∫(M₀·y·ds/EI) / ∫(y²·ds/EI)

For a parabolic two-hinged arch under UDL w:
H = 5wL² / (8h) × (1/(1 + 16h²/5L²)) ≈ 5wL²/(8h) for flat arches (h/L < 1/5)

Rise-to-span ratio: h/L typically 1/4 to 1/6 for economy

4.6 Temperature Effects in Arches

Two-hinged arch — increase in H due to temperature rise ΔT:
H_T = (α·ΔT·L) / (2 × ∫y²·ds/EI) [approximate for flat parabolic arch]

Three-hinged arch: temperature rise does NOT induce additional stresses (determinate)

⚠️ Critical: In indeterminate arches (2-hinged, fixed), temperature change and support settlement cause secondary stresses. This is a major disadvantage and a frequent GATE/ESE question.

4.7 Eddy's Theorem (Bending Moment in Arches)

BM at any section of arch = BM in the corresponding simply supported beam (M₀) − H × y
where y = rise of arch rib above the springing line at that section

If arch rib follows the funicular polygon for the loading: M = 0 everywhere (pure compression)

📝 Exam Angle: The linear arch (or pressure line) is the locus of the point of application of the resultant force at every section. If the pressure line coincides with the arch rib → no bending. If pressure line lies outside the middle third → tensile stress in arch.

5Suspension Bridges►

5.1 Introduction

A suspension bridge consists of a flexible cable draped between towers and anchored at both ends. The deck (stiffening girder or truss) is hung from the cable by vertical hangers/suspenders. The cable carries loads entirely in tension — unlike an arch (compression), making it ideal for long spans where steel is excellent in tension.

⭐ World's longest suspension spans: Akashi-Kaikyō Bridge (Japan) ~1991 m main span. In India: Bandra–Worli Sea Link is a cable-stayed (not pure suspension) bridge.

5.2 Components of a Suspension Bridge

Main cable — high-tensile steel wire bundles; primary load-carrying element (tension only)
Towers (pylons) — carry compressive force and transfer cable tension to foundations
Anchorages — massive concrete blocks or rock anchors; resist horizontal cable pull
Hangers/Suspenders — vertical cables connecting main cable to stiffening girder
Stiffening girder or truss — distributes concentrated loads; provides aerodynamic stability
Deck (roadway) — supported by stiffening structure

Fig. 5.1 — Suspension Bridge: Towers, main cable, hangers, stiffening girder, and anchorages

5.3 Cable Shape Under UDL (Suspended Cable)

Under uniformly distributed load per unit horizontal length, the cable takes a parabolic shape. Under self-weight (per unit arc length), the true shape is a catenary — but for most bridge cables, the parabola approximation is acceptable.

Cable equation (parabola, origin at lowest point):
y = wx² / (2H) where H = horizontal tension (constant throughout cable)

Sag d (at midspan): d = wL² / (8H)
∴ H = wL² / (8d)

Cable tension at any point:
T = √(H² + V²) where V = w·x (vertical component at distance x from midspan)

Maximum tension T_max (at supports) = H × √(1 + 16d²/L²)

5.4 Stiffening Girder

The stiffening girder prevents the cable from changing shape under unsymmetrical concentrated loads. It distributes concentrated loads over a larger portion of the cable. Without a stiffening girder, the bridge would oscillate (as in the famous Tacoma Narrows Bridge collapse, 1940 — aerodynamic flutter).

For a stiffened suspension bridge with two-hinged stiffening girder:
The cable takes up UDL corresponding to dead load → treated like a simple cable.
Live load induces bending in stiffening girder + changes cable tension.
BM in stiffening girder = BM in equivalent beam − H_live × y (similar to arch analysis)

5.5 Comparison: Arch vs Suspension Bridge

Feature	Arch	Suspension Bridge
Primary stress	Compression	Tension
Horizontal force	Outward thrust (pushes abutments)	Inward pull (anchorages resist)
Material advantage	Concrete, masonry (compression)	High-tensile steel wire (tension)
Span range	Up to ~500 m (steel arch)	Up to 2000+ m
Foundation requirement	Must resist large horizontal thrust	Need heavy anchorage blocks

6Methods of Structural Analysis►

6.1 Classification of Methods

Category	Methods	Unknowns	Best for
Force (Flexibility) Method	Method of consistent deformation, Three-moment equation, Column analogy	Forces (redundants)	Low DSI structures
Displacement (Stiffness) Method	Slope deflection, Moment distribution, Kani's method	Displacements (rotations/deflections)	Low DKI structures
Matrix Methods	Stiffness matrix, Flexibility matrix	Matrices of forces or displacements	Computer analysis (all structures)
Approximate Methods	Portal method, Cantilever method	Assumed inflection points	Preliminary design of tall frames

6.2 Slope Deflection Method

A displacement method that expresses end moments in terms of slopes (rotations) and deflections at the joints.

Slope-Deflection Equations:
M_AB = M_FAB + (2EI/L)(2θ_A + θ_B − 3ψ)
M_BA = M_FBA + (2EI/L)(2θ_B + θ_A − 3ψ)

Where:
M_FAB, M_FBA = Fixed End Moments (FEM) due to applied loads
θ_A, θ_B = slopes at A and B (positive = clockwise)
ψ = chord rotation = Δ/L (sway)
EI = flexural rigidity
L = member length

Standard Fixed End Moments

Loading	M_FAB	M_FBA
UDL w over full span L	+wL²/12	−wL²/12
Central point load P at L/2	+PL/8	−PL/8
Point load P at distance 'a' from A (b=L−a)	+Pab²/L²	−Pa²b/L²
Triangular load (w at B, 0 at A)	+wL²/20	−wL²/30
Support settlement Δ (no load)	+6EIΔ/L²	+6EIΔ/L²

6.3 Moment Distribution Method (Hardy Cross Method)

An iterative displacement method that distributes unbalanced moments at joints, cycling until convergence. Does not require solving simultaneous equations — ideal for manual computation.

Stiffness factor (K):
K = 4EI/L (far end fixed)
K = 3EI/L (far end pinned or free)

Distribution Factor (DF) at joint for member ij:
DF_ij = K_ij / ΣK (sum over all members meeting at joint)

Carry Over Factor (COF):
COF = 0.5 (far end fixed)
COF = 0 (far end pinned/free)

Procedure:
1. Lock all joints → compute FEMs
2. Unlock joint → distribute unbalanced moment using DF
3. Carry over half to far ends
4. Repeat until convergence (typically 3–4 cycles suffices for GATE)

Solved Example — Propped Cantilever (Moment Distribution)

Propped cantilever AB (fixed at A, roller at B), span L, UDL w:
FEM_AB = +wL²/12, FEM_BA = −wL²/12
DF at B: K_BA = 3EI/L (pinned end) → DF = 1.0
Unbalanced moment at B = −wL²/12
Distribution at B = +wL²/12 × 1.0 = +wL²/12
Carry over to A = 0 (pinned end) → none back to B
Carry over to A = +wL²/24 (from B)... No, COF=0 for pinned B.

Final moment: M_AB = wL²/12 + wL²/24 = wL²/8 (hogging at A) ✓
M_BA = 0 ✓ (roller end)

6.4 Kani's Method

An extension of moment distribution that handles sway frames more systematically. Introduces rotation contributions (u) and shear contributions (v) that are iterated until convergence.

Rotation contribution: u_ij = −(1/2) × DF_ij × (M'_ij + Σu_ki)
where M'_ij = fixed-end moment contribution
Final moment = 2u_ij + u_ji + FEM_ij

6.5 Three-Moment Equation (Clapeyron's Theorem)

Used for analysis of continuous beams. Relates moments at three successive supports:

M_A·L₁/I₁ + 2M_B(L₁/I₁ + L₂/I₂) + M_C·L₂/I₂
= −6[A₁·ā₁/(I₁·L₁) + A₂·b̄₂/(I₂·L₂)]

A₁ = area of free BMD in span 1
ā₁, b̄₂ = distance of centroid of free BMD from left and right supports

6.6 Approximate Methods for Lateral Load Analysis

Portal Method (low-rise frames)

Assumes inflection points at midpoints of all beams and columns
Interior columns carry double the shear of exterior columns
Quick but less accurate for tall buildings (10+ storeys)

Cantilever Method (tall frames)

Assumes columns behave like flanges of a cantilever (axial stress proportional to distance from centroid of column area group)
Inflection points at mid-height of columns and mid-span of beams
More accurate for slender tall frames

ℹ️ For GATE: Slope deflection and moment distribution are the most frequently tested analysis methods. Know FEM tables thoroughly — they appear in almost every question involving indeterminate beams.

7Trusses►

7.1 Introduction and Assumptions

A truss is an assembly of straight members connected at their ends through frictionless pin joints to form a stable triangulated structure. Members carry only axial forces (tension or compression) — no bending or shear.

ℹ️ Assumptions: (1) Members are straight (2) Loads apply only at joints (3) All joints are frictionless pins (4) Members are weightless (self-weight neglected or distributed to joints). In reality joints are rigid — secondary moments exist but are usually small.

7.2 Classification of Trusses

Type	Description	Application
Simple truss	Built by adding 2 members and 1 joint at a time from a basic triangle	Most common; determinate
Compound truss	Two or more simple trusses connected	Large spans; may be determinate
Complex truss	Does not follow simple/compound rules	Special structures; analysis by matrix
Pratt truss	Vertical members + diagonal members in tension under downward load	Railway/road bridges
Howe truss	Diagonals in compression; verticals in tension	Timber roofs
Warren truss	Diagonals only (no verticals); alternating tension-compression	Light bridges
K-truss	K-shaped diagonals; reduced unsupported length	Long-span bridges

7.3 Methods of Analysis

Method of Joints

Apply equilibrium (ΣFx = 0, ΣFy = 0) at each joint in sequence. Start from a joint with no more than 2 unknowns.

Convention: Assume all unknown member forces are tensile (+ve).
If the result is positive → member is in Tension (T)
If the result is negative → member is in Compression (C)

Method of Sections

Pass an imaginary plane through no more than 3 members, isolate one portion of the truss, apply ΣFx = 0, ΣFy = 0, ΣM = 0. Best for finding force in specific members without solving the whole truss.

For maximum efficiency: choose moment centre at the intersection of the other two unknown forces → one equation in one unknown.

Fig. 7.1 — Pratt Truss: verticals in compression, inclined diagonals in tension under downward loads

7.4 Zero-Force Members

Zero-force members carry no load under a given loading condition. Identifying them simplifies analysis significantly.

Rule 1 (2-member joint, no external load):
If only 2 members meet at an unloaded joint AND they are not collinear → both are zero-force members.

Rule 2 (3-member joint, no external load):
If 2 of the 3 members are collinear AND the third is at an angle → the non-collinear member is zero-force.

📝 Zero-force members are not useless — they provide stability against buckling, resist loads under different loading combinations, and maintain geometric form. Never remove them from the actual structure!

7.5 Deflection of Trusses — Virtual Work Method

Deflection at any joint in the direction of a virtual unit load:
δ = Σ(F·u·L / AE)

Where:
F = actual force in member (due to real loads)
u = force in member due to unit virtual load at the point and direction of desired deflection
L = member length, A = cross-section area, E = Young's modulus

8Matrix Method of Structural Analysis►

8.1 Overview

Matrix methods provide a systematic, computer-oriented approach to structural analysis. There are two formulations:

Method	Matrix	Unknowns	Based on
Stiffness Method	Stiffness matrix [K]	Joint displacements {d}	Equilibrium: [K]{d} = {F}
Flexibility Method	Flexibility matrix [f]	Redundant forces {F}	Compatibility: [f]{F} = {δ}

⭐ The stiffness method is almost universally used in modern software (STAAD.Pro, ETABS, SAP2000) because it is easier to automate — the global stiffness matrix is assembled from individual member stiffness matrices.

8.2 Stiffness Method — Procedure

Number all degrees of freedom (free DOF first, then restrained DOF)
Develop local stiffness matrix [k] for each element
Transform to global coordinates: [K_global] = [T]^T[k][T]
Assemble global stiffness matrix [K] by adding contributions at each DOF
Apply boundary conditions (zero displacement at supports)
Solve: {d} = [K_ff]⁻¹{F_f} for free DOF
Compute member forces: {f} = [k][T]{d} − {f₀}

8.3 Local Stiffness Matrix — Beam Element

For a 2D beam element with 4 DOF (v_i, θ_i, v_j, θ_j) and flexural rigidity EI, length L:

[k] = (EI/L³) ×
┌ 12 6L −12 6L ┐
│ 6L 4L² −6L 2L² │
│ −12 −6L 12 −6L │
└ 6L 2L² −6L 4L² ┘

DOF order: [v₁, θ₁, v₂, θ₂]
(Rotations θ positive clockwise; transverse displacement v positive upward)

8.4 Local Stiffness Matrix — Bar/Truss Element

For a bar element (axial only), 2 DOF per node: [u_i, u_j]
[k] = (AE/L) × [ 1 −1 ]
[−1 1 ]

For inclined member at angle θ to global X-axis:
[k_global] = (AE/L) ×
┌ c² cs −c² −cs ┐
│ cs s² −cs −s² │
│ −c² −cs c² cs │
└ −cs −s² cs s² ┘

c = cos θ, s = sin θ

8.5 Flexibility Matrix and Method

The flexibility matrix [f] has elements f_ij = deflection at DOF i due to unit force at DOF j (with all other loads zero). It is the inverse of the stiffness matrix: [f] = [K]⁻¹.

Compatibility equations: [f]{F} = {δ₀} − {δ_compat}
{F} = redundant forces
{δ₀} = displacements in the released (determinate) structure under applied loads
{δ_compat} = prescribed displacements at redundant positions (usually zero for rigid supports)

8.6 Transformation Matrix [T]

For a member inclined at angle θ:
{f_local} = [T] × {F_global}
[T] = [ cos θ sin θ 0 0 ]
[−sin θ cos θ 0 0 ]
[ 0 0 cos θ sin θ ]
[ 0 0 −sin θ cos θ ]

Global stiffness: [K_g] = [T]^T[k_L][T]

8.7 Properties of Stiffness and Flexibility Matrices

Both matrices are symmetric (Maxwell's reciprocal theorem)
Stiffness matrix is positive definite (for a stable structure)
Stiffness matrix diagonal elements are always positive
Before applying boundary conditions, the global stiffness matrix is singular (rigid body motion possible)
After applying boundary conditions (eliminating restrained DOF), it becomes non-singular and invertible

📝 GATE Tip: The stiffness matrix for a simply supported beam (2 DOF: rotations θ_A, θ_B) is:
[k] = (2EI/L) × [2 1; 1 2] — know this by heart. It appears in slope deflection and matrix methods alike.

9Structural Dynamics►

9.1 Introduction

Structural dynamics deals with the response of structures to time-varying loads — earthquakes, winds, blasts, machinery vibrations, and moving vehicles. The distinguishing feature is the role of inertia (mass × acceleration) and damping forces in addition to static stiffness forces.

9.2 Single Degree of Freedom (SDOF) System

Equation of motion for SDOF:
mẍ + cẋ + kx = F(t)

Where:
m = mass (kg)
c = damping coefficient (N·s/m)
k = stiffness (N/m)
x, ẋ, ẍ = displacement, velocity, acceleration
F(t) = applied force (function of time)

Fig. 9.1 — SDOF spring-mass-damper system; mẍ + cẋ + kx = F(t)

9.3 Free Vibration — Undamped

Natural circular frequency: ω_n = √(k/m) rad/s
Natural frequency: f_n = ω_n / (2π) Hz (cycles/sec)
Natural time period: T_n = 1/f_n = 2π√(m/k) = 2π√(δ_st/g) seconds

where δ_st = static deflection under weight mg

Response: x(t) = A·cos(ω_nt) + B·sin(ω_nt)
A = x₀ (initial displacement), B = v₀/ω_n (initial velocity / ω_n)

9.4 Free Vibration — Damped

Critical damping coefficient: c_c = 2mω_n = 2√(km)
Damping ratio: ζ = c / c_c

Cases:
ζ < 1 → Underdamped: oscillatory decay → x(t) = e^(−ζω_nt)[A·cos(ω_dt) + B·sin(ω_dt)]
ζ = 1 → Critically damped: fastest return to equilibrium without oscillation
ζ > 1 → Overdamped: slow non-oscillatory return

Damped natural frequency: ω_d = ω_n√(1 − ζ²)

For structural engineering: ζ typically 2–5% (steel: 1–2%; RC: 5%; soil: 10–30%)

9.5 Forced Vibration — Harmonic Excitation

Applied force: F(t) = F₀·sin(Ωt) (Ω = forcing frequency)

Steady-state amplitude:
X = (F₀/k) × D
D = Dynamic Magnification Factor (DMF) = 1 / √[(1−r²)² + (2ζr)²]
r = frequency ratio = Ω/ω_n

At resonance (r = 1, undamped):
D → ∞ (theoretically); limited by damping in practice.
With damping ζ: D_max = 1/(2ζ) at r ≈ √(1 − 2ζ²) ≈ 1 for small ζ

Phase angle: φ = tan⁻¹[2ζr / (1 − r²)]

9.6 Multi-Degree of Freedom (MDOF) Systems

Equation of motion: [M]{ẍ} + [C]{ẋ} + [K]{x} = {F(t)}

Free undamped vibration → Eigenvalue problem:
([K] − ω²[M]){φ} = {0}
Characteristic equation: det([K] − ω²[M]) = 0
Roots ω₁², ω₂², ... ω_n² → natural frequencies
Corresponding eigenvectors {φ₁}, {φ₂}... → mode shapes

9.7 Response Spectrum Method (Seismic)

Used in IS 1893:2016 for earthquake design. The response spectrum gives the maximum response (displacement, velocity, acceleration) of SDOF systems with various time periods T_n for a given earthquake record and damping ratio ζ.

Design base shear (IS 1893):
V_B = Ah × W
Ah = (Z/2) × (I/R) × (Sa/g)

Z = seismic zone factor (II: 0.10, III: 0.16, IV: 0.24, V: 0.36)
I = importance factor (1.0–1.5)
R = response reduction factor (3–5 for RC frames)
Sa/g = spectral acceleration coefficient (from IS 1893 spectrum, depends on T_n and soil type)

📝 GATE/ESE Tip: Know the DMF formula and its behaviour at r = 0 (D=1), r = 1 (resonance), r → ∞ (D→0). The DMF graph is a standard question. For seismic, memorise the zone factor values for all four zones.

9.8 Natural Frequencies — Standard Structural Cases

Structure	ω_n	T_n
Mass m on spring k	√(k/m)	2π√(m/k)
Simple pendulum length L	√(g/L)	2π√(L/g)
Cantilever, mass at tip (EI, L)	√(3EI/mL³)	2π√(mL³/3EI)
Simply supported beam, central mass (EI, L)	√(48EI/mL³)	2π√(mL³/48EI)
Fixed-fixed beam, central mass	√(192EI/mL³)	2π√(mL³/192EI)

10Suspended Cables►

10.1 Introduction

A suspended (flexible) cable is a structural element that can only carry tension — it has no flexural or shear stiffness. Its shape adjusts to the applied loading so that it is always in pure tension (funicular shape). Cables are fundamental to suspension bridges, cable-stayed bridges, aerial ropeways, power lines, and guy wires.

10.2 Cable Under Concentrated Loads

For a cable with two supports at the same level, with concentrated loads P₁, P₂, ... at known horizontal positions:
1. Assume the sag (d) or the position of lowest point
2. Take moments about one support to find vertical reactions
3. Use condition at the point of known sag (usually one known point) to find H
4. At any point x: V(x) = vertical shear = sum of vertical forces to the left
5. Tan θ(x) = V(x)/H → slope of cable at x
6. Cable tension T = √(H² + V²)

Fig. 10.1 — Cable carrying two concentrated loads; H is constant throughout; T = √(H² + V²)

10.3 Cable Under UDL — Parabolic Cable

When a cable carries a uniformly distributed load per unit horizontal length (e.g., from a bridge deck hung from it), it takes a parabolic shape.

Taking origin at lowest point (both supports at same level):
y = wx² / (2H) → parabola

Sag at centre: d = wL² / (8H)
∴ H = wL² / (8d) (horizontal tension — constant along cable)

Tension at any point: T(x) = √(H² + (wx)²)

Maximum tension (at supports): T_max = √[H² + (wL/2)²]
T_max = H × √(1 + 16d²/L²) ≈ H × (1 + 8d²/L²) for small sag

Length of cable: ℓ ≈ L[1 + (8d²)/(3L²) − (32d⁴)/(5L⁴) + ...] (series expansion for small sag)

10.4 Cable Under Self-Weight — Catenary

When a cable carries only its own weight (uniform per unit arc length w₀), it takes the shape of a catenary.

Catenary equation (origin at lowest point):
y = (H/w₀)[cosh(w₀x/H) − 1]

where H = horizontal tension (constant)

Arc length s = (H/w₀)·sinh(w₀x/H)

Tension at any point: T = H·cosh(w₀x/H) = H + w₀y
Maximum tension T_max (at supports) = H + w₀y_support

Note: For large sag-to-span ratios, catenary must be used. For d/L < 1/8, parabola is an adequate approximation.

10.5 Supports at Different Levels

When the two supports of a cable are at different elevations (height difference h), the lowest point of the cable may lie outside the span. The analysis uses two unknowns (H and position of lowest point) and two equations (moments about each support).

Let support A be at height y_A, support B at height y_B above the lowest point.
y_A = w·x_A² / (2H), y_B = w·x_B² / (2H)
x_A + x_B = L (span), y_B − y_A = h
Solving: x_B = L/2 + h·H/(wL), x_A = L/2 − h·H/(wL)
H is found from additional information (e.g., specified sag at mid-span).

10.6 Anchor Cable / Guy Wire

For a straight inclined cable (taut — no sag) at angle α to horizontal:
Tension T = load / sin α (if the load is vertical)

For an anchor cable supporting a mast against lateral force F:
T = F / cos α (where α is the angle between cable and horizontal)

10.7 Cable Stiffness and Equivalent Modulus (Ernst Formula)

A sagging cable is less stiff than a straight cable because part of the axial deformation goes into changing the sag rather than elastic strain. This is captured by the equivalent modulus:

E_eq = E / (1 + w²L²E / 12T³)

where:
E = actual Young's modulus of cable material
w = cable self-weight per unit length
L = horizontal projected length
T = average cable tension

As T → ∞ (highly tensioned): E_eq → E (sag effect negligible, cable acts like a rod)

📝 Ernst's equivalent modulus is critical for cable-stayed bridges where cable sag is significant for longer cables. This appears in ESE optional and advanced structural analysis papers.

★Quick Revision – Formulae & Mnemonics►

Determinacy at a Glance

Plane Truss: DSI = m + r − 2j
Rigid Frame: DSI = 3m + r − 3j (count conditions for internal hinges)
Space Truss: DSI = m + r − 3j
Ext. DSI (2D): r − 3 | Ext. DSI (3D): r − 6
DKI (2D frame): 3j − r | DKI (continuous beam): = interior joints (if axial neglected)

ILD — Key Ordinates (Simply Supported Beam, Span L, Section at distance a from A)

R_A : peak = 1 at A, zero at B (linear)
SF at C (left side): +b/L (between A and C), −a/L (between C and B)
BM at C: peak = ab/L at C; a + b = L
BM at midspan: peak = L/4 (unit load at midspan)

Arch Formulae

3-hinged arch: H = RA × (L/2) / h (from crown moment condition)
2-hinged arch UDL: H = 5wL²/(8h) [approx for flat arch]
Eddy's Theorem: M_arch = M_beam − H·y
Funicular arch (parabola + UDL): M = 0 everywhere ✓
Linear arch (pressure line) outside middle third → tension → avoid in masonry arches

Slope Deflection & Moment Distribution

SDE: M_AB = M_FAB + (2EI/L)(2θ_A + θ_B − 3ψ)
DF = K_ij / ΣK | COF = 0.5 (fixed far end), 0 (pinned far end)
K = 4EI/L (far end fixed) | K = 3EI/L (far end pinned)
FEM: UDL → ±wL²/12 | Central P → ±PL/8 | P at 'a' → +Pab²/L², −Pa²b/L²

Matrix Method

Beam element [k] size: 4×4 (2 DOF/node: v, θ)
Bar element [k] = (AE/L)[1,−1;−1,1]
Inclined bar: c=cosθ, s=sinθ; [k_global] = AE/L × [c² cs −c² −cs; ... ]
Global K assembled by direct stiffness method; singular before BCs applied
[K]·{d} = {F} → solve for unknown displacements after applying BCs

Structural Dynamics

ω_n = √(k/m); T_n = 2π/ω_n = 2π√(m/k)
ζ = c/(2√km); ω_d = ω_n√(1−ζ²)
DMF D = 1/√[(1−r²)²+(2ζr)²]; r = Ω/ω_n
Resonance: r=1 → D=1/(2ζ)
Cantilever ω_n = √(3EI/mL³); SSB central mass ω_n = √(48EI/mL³)

Suspended Cables

Parabolic (UDL per horizontal length): y = wx²/(2H); d = wL²/(8H)
H = wL²/(8d); T_max = H√(1+16d²/L²)
Catenary (UDL per arc length): y = (H/w₀)[cosh(w₀x/H)−1]
T = H+w₀y; Use catenary when d/L > 1/8
Ernst equivalent modulus: E_eq = E/(1 + w²L²E/12T³)

Mnemonics

STIFF vs FLEX:
Stiff people Stand (Stiffness → displacements as unknowns)
Flex people Force (Flexibility → forces as unknowns)

FEM sign convention:
"First letter clockwise = positive" → M_AB is positive if it makes a clockwise moment on the near end A

Truss zero-force rule:
"Two members, no load, not in line → both are ZERO"
"Three members, two in line, third is alone → third is ZERO"

Arch types memory: 3-2-0 hinges → 0-1-3 DSI
(3 hinged → DSI 0; 2 hinged → DSI 1; 0 hinged/fixed → DSI 3)

Cable shape memory:
"Horizontal load → catenary; Horizontal UDL (from deck) → parabola"

Damping ratio typical values:
Steel 1–2% | RCC 5% | Soil 10–30% → "Steel Stays, RCC Rocks, Soil Soaks"

Exam-Angle Comparison Table

Topic	GATE Focus	ESE Focus	SSC JE Focus
Determinacy	Numerical DSI/DKI computation	Concept + numerical	Identify type only
ILD	Ordinates for SS beams; propped cantilever	ILD + rolling load combination	Shape and peak values
Arches	3-hinged arch: H, BM, normal thrust	2-hinged arch; temperature effects	Types; H formula
Structural Analysis	SDE, MDM, Kani's method	All methods + frames with sway	FEM values only
Trusses	Method of joints & sections; zero-force	Virtual work deflection; redundant	Basic force calculation
Matrix Methods	Stiffness matrix assembly; 2-member	Full stiffness/flexibility method	Not usually tested
Dynamics	ω_n, T_n, DMF, resonance; SDOF	MDOF eigenvalue, response spectrum	Basic T_n formula
Cables	Sag, H, T_max; catenary vs parabola	Suspension bridge analysis, Ernst	H formula; sag-span relation