Theory of Structures builds directly on Solid Mechanics to analyse complete structural systems — from checking whether a structure is statically determinate, through influence lines for moving loads, arches and suspension bridges, the classical displacement/force methods of indeterminate analysis, truss analysis, matrix methods used by modern structural software, and finally structural dynamics and suspended cables. Every formula, diagram, solved example, IS code reference and exam-pattern table is included.

Learning Objectives

After studying this chapter you will be able to:

• Compute the degree of static indeterminacy (DSI) and kinematic indeterminacy (DKI) for beams, frames and trusses, and assess structural stability
• Construct influence line diagrams using the Müller-Breslau principle and use them to find the maximum response due to moving loads
• Determine the absolute maximum bending moment (AMBM) and equivalent UDL (EUDL) for rolling-load systems such as trains and truck axles
• Analyse three-hinged, two-hinged and fixed arches — find horizontal thrust, apply Eddy's theorem, and evaluate temperature and settlement effects
• Analyse suspension bridges and suspended cables under point loads, UDL (parabolic cable) and self-weight (catenary)
• Apply slope-deflection, moment distribution (Hardy Cross), Kani's method and the three-moment equation to indeterminate beams and frames
• Analyse trusses by the method of joints and method of sections, identify zero-force members, and compute deflections by the virtual work (unit-load) method
• Assemble stiffness/flexibility matrices for beam and bar elements and formulate SDOF/MDOF equations of motion for structural dynamics problems

Exam Relevance

Concept Map

Where This Fits

Prerequisite: Solid Mechanics (SFD/BMD and beam-deflection formulae are used throughout this module). Leads to: Steel Structures and RCC/PCC Design, which use these analysis results to size real structural members.

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 (\(\Sigma F_x = 0, \Sigma F_y = 0, \Sigma M = 0\)). 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.

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: \(DSI = 3m + r - 3j - \text{releases}\), where m = no. of members, r = no. of external reactions, j = no. of joints. An 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

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: \(u, v, \theta\)); \(DKI = 6j - r\) (3D space frames). If axial deformation is neglected: \(DKI = 2j - r\) (for beams; only rotations count if inextensible).

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)

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.

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

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

2.4 Maximum Effect Using ILD

For a single concentrated load P: Max response \(= P \times\) (max ILD ordinate). For a UDL of intensity w (length ℓ): Max response \(= w \times\) (area of ILD under load). For a UDL over the full span: Max BM at centre \(= w \times (L/4) \times (L/2) = wL^2/8\) — matching the standard SS-beam result ✓

2.5 ILD for Indeterminate Structures

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

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.

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 the load such that midspan = midpoint of (load position, resultant position).

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. It occurs when the leading (heaviest) load reaches the support.

3.4 UDL Shorter than 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.

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.

4.2 Types of Arches

4.3 Analysis of Three-Hinged Arch

Step 1 — Find \(R_A, R_B\) using \(\Sigma M = 0\) (same as a simply supported beam): \(R_A = P\cdot b/L,\ R_B = P\cdot a/L\) (where \(a, b\) are distances of the load from A and B). Step 2 — Use the hinge condition at the crown (\(\Sigma M_{crown} = 0\) for either half): taking moments about the crown for the left half, \(R_A \times (L/2) - H \times h = 0\), so \(H = R_A \times L/(2h)\). Step 3 — Normal thrust and radial shear at any section: \(N = H\cos\theta + V\sin\theta\) (compression, +ve), \(Q = V\cos\theta - H\sin\theta\) (radial shear).

4.4 Parabolic Arch under UDL (Pure Compression)

For a parabolic arch with equation \(y = 4h\cdot x(L-x)/L^2\), subjected to a uniform load \(w\) per unit horizontal length, the horizontal thrust \(H = wL^2/(8h)\) and bending moment at every section = 0. BM at any section \(x\): \(M = M_{beam} - H\cdot y = 0\) (for a 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): \(H = \int(M_0 \cdot y\cdot ds/EI) / \int(y^2\cdot ds/EI)\). For a parabolic two-hinged arch under UDL \(w\): \(H = (5wL^2/(8h))\times 1/(1 + 16h^2/(5L^2)) \approx 5wL^2/(8h)\) for flat arches (\(h/L < 1/5\)). Rise-to-span ratio \(h/L\) is typically 1/4 to 1/6 for economy.

4.6 Temperature Effects in Arches

For a two-hinged arch, the increase in H due to temperature rise \(\Delta T\): \(H_T = (\alpha\cdot\Delta T\cdot L)/(2\times \int y^2\cdot ds/EI)\) [approximate for a flat parabolic arch]. A three-hinged arch is determinate, so temperature rise does NOT induce additional stresses.

4.7 Eddy's Theorem (Bending Moment in Arches)

BM at any section of an arch = BM in the corresponding simply supported beam (\(M_0\)) \(- H \times y\), where \(y\) = rise of the arch rib above the springing line at that section. If the arch rib follows the funicular polygon for the loading: \(M = 0\) everywhere (pure compression).

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.

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

5.3 Cable Shape Under UDL (Suspended Cable)

Under a 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.

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 a two-hinged stiffening girder: the cable takes up UDL corresponding to dead load → treated like a simple cable. Live load induces bending in the stiffening girder and changes cable tension: BM in stiffening girder \(=\) BM in equivalent beam \(- H_{live} \times y\) (similar to arch analysis).

5.5 Comparison: Arch vs Suspension Bridge

6.1 Classification of Methods

Principle of superposition: for a linearly elastic structure (small deflections, Hooke's law obeyed), the response — reaction, internal force, or displacement — to a combination of loads equals the algebraic sum of the responses to each load acting separately. It is the foundation of the force/flexibility method (combining the primary structure with each redundant's effect), of influence-line construction, and of the method of consistent deformation. It does not hold when the response is non-linear (large deflections, \(P\)-\(\Delta\) effects, or inelastic material).

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:

Where: \(M_{FAB}, M_{FBA}\) = Fixed End Moments (FEM) due to applied loads; \(\theta_A, \theta_B\) = slopes at A and B (positive = clockwise); \(\psi\) = chord rotation \(= \Delta/L\) (sway); \(EI\) = flexural rigidity; \(L\) = member length.

Standard Fixed End Moments

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 a joint for member \(ij\): \(DF_{ij} = K_{ij}/\Sigma K\) (sum over all members meeting at the 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).

6.4 Kani's Method

An extension of moment distribution that handles sway frames more systematically. It introduces rotation contributions (u) and shear contributions (v) that are iterated until convergence. Rotation contribution: \(u_{ij} = -\tfrac{1}{2} \times DF_{ij} \times (M'_{ij} + \Sigma u_{ki})\), 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. It relates moments at three successive supports:

\(A_1\) = area of the free BMD in span 1; \(\bar a_1, \bar b_2\) = distance of the centroid of the free BMD from the 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 the column area group)
• Inflection points at mid-height of columns and mid-span of beams
• More accurate for slender tall frames

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.

7.2 Classification of Trusses

7.3 Methods of Analysis

Method of Joints

Apply equilibrium (\(\Sigma F_x = 0, \Sigma F_y = 0\)) 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 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, and apply \(\Sigma F_x = 0, \Sigma F_y = 0, \Sigma M = 0\). Best for finding force in specific members without solving the whole truss. For maximum efficiency: choose the moment centre at the intersection of the other two unknown forces → one equation in one unknown.

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.

7.5 Deflection of Trusses — Virtual Work Method

Deflection at any joint in the direction of a virtual unit load: \(\delta = \Sigma(F\cdot u\cdot L/(AE))\), where \(F\) = actual force in the member (due to real loads), \(u\) = force in the member due to a unit virtual load at the point and direction of the desired deflection, \(L\) = member length, \(A\) = cross-section area, \(E\) = Young's modulus.

8.1 Overview

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

8.2 Stiffness Method — Procedure

1. Number all degrees of freedom (free DOF first, then restrained DOF)
2. Develop the local stiffness matrix \([k]\) for each element
3. Transform to global coordinates: \([K_{global}] = [T]^T[k][T]\)
4. Assemble the global stiffness matrix \([K]\) by adding contributions at each DOF
5. Apply boundary conditions (zero displacement at supports)
6. Solve: \(\{d\} = [K_{ff}]^{-1}\{F_f\}\) for the free DOF
7. Compute member forces: \(\{f\} = [k][T]\{d\} - \{f_0\}\)

8.3 Local Stiffness Matrix — Beam Element

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

DOF order: \([v_1, \theta_1, v_2, \theta_2]\) (Rotations \(\theta\) 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 \([u_i, u_j]\):

For an inclined member at angle \(\theta\) to the global X-axis:

where \(c = \cos\theta,\ s = \sin\theta\).

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: \([f]\{F\} = \{\delta_0\} - \{\delta_{compat}\}\), where \(\{F\}\) = redundant forces, \(\{\delta_0\}\) = displacements in the released (determinate) structure under applied loads, and \(\{\delta_{compat}\}\) = prescribed displacements at redundant positions (usually zero for rigid supports).

8.6 Transformation Matrix [T]

For a member inclined at angle \(\theta\): \(\{f_{local}\} = [T]\{F_{global}\}\), where

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)
• The 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

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: \(m\ddot x + c\dot x + kx = F(t)\), where \(m\) = mass (kg), \(c\) = damping coefficient (N·s/m), \(k\) = stiffness (N/m), \(x, \dot x, \ddot x\) = displacement, velocity, acceleration, and \(F(t)\) = applied force (function of time).

9.3 Free Vibration — Undamped

Natural circular frequency: \(\omega_n = \sqrt{k/m}\) rad/s. Natural frequency: \(f_n = \omega_n/(2\pi)\) Hz (cycles/sec). Natural time period: \(T_n = 1/f_n = 2\pi\sqrt{m/k} = 2\pi\sqrt{\delta_{st}/g}\) seconds, where \(\delta_{st}\) = static deflection under weight \(mg\). Response: \(x(t) = A\cos(\omega_n t) + B\sin(\omega_n t)\), with \(A = x_0\) (initial displacement) and \(B = v_0/\omega_n\) (initial velocity / \(\omega_n\)).

9.4 Free Vibration — Damped

Critical damping coefficient: \(c_c = 2m\omega_n = 2\sqrt{km}\). Damping ratio: \(\zeta = c/c_c\).

Cases: \(\zeta < 1\) → Underdamped: oscillatory decay, \(x(t) = e^{-\zeta\omega_n t}[A\cos(\omega_d t) + B\sin(\omega_d t)]\). \(\zeta = 1\) → Critically damped: fastest return to equilibrium without oscillation. \(\zeta > 1\) → Overdamped: slow non-oscillatory return.

Damped natural frequency: \(\omega_d = \omega_n\sqrt{1 - \zeta^2}\). For structural engineering: \(\zeta\) typically 2–5% (steel: 1–2%; RC: 5%; soil: 10–30%).

9.5 Forced Vibration — Harmonic Excitation

Applied force: \(F(t) = F_0\cdot\sin(\Omega t)\) (\(\Omega\) = forcing frequency). Steady-state amplitude: \(X = (F_0/k) \times D\), where \(D\) = Dynamic Magnification Factor (DMF) \(= 1/\sqrt{(1-r^2)^2 + (2\zeta r)^2}\) and \(r\) = frequency ratio \(= \Omega/\omega_n\).

At resonance (\(r = 1\), undamped): \(D \to \infty\) (theoretically); limited by damping in practice. With damping \(\zeta\): \(D_{max} = 1/(2\zeta)\) at \(r \approx \sqrt{1 - 2\zeta^2} \approx 1\) for small \(\zeta\). Phase angle: \(\phi = \tan^{-1}[2\zeta r/(1 - r^2)]\).

9.6 Multi-Degree of Freedom (MDOF) Systems

Equation of motion: \([M]\{\ddot x\} + [C]\{\dot x\} + [K]\{x\} = \{F(t)\}\). Free undamped vibration → Eigenvalue problem: \(([K] - \omega^2[M])\{\phi\} = \{0\}\); characteristic equation \(\det([K] - \omega^2[M]) = 0\). Roots \(\omega_1^2, \omega_2^2, \ldots, \omega_n^2\) → natural frequencies; corresponding eigenvectors \(\{\phi_1\}, \{\phi_2\}, \ldots\) → 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 \(\zeta\).

Design base shear (IS 1893): \(V_B = A_h \times W\), where \(A_h = \dfrac{Z}{2} \times \dfrac{I}{R} \times \dfrac{S_a}{g}\); \(Z\) = 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), \(S_a/g\) = spectral acceleration coefficient (from the IS 1893 spectrum, depends on \(T_n\) and soil type).

9.8 Natural Frequencies — Standard Structural Cases

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 \(P_1, P_2, \ldots\) at known horizontal positions: (1) Assume the sag (\(d\)) or the position of the lowest point; (2) Take moments about one support to find vertical reactions; (3) Use the 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) \(\tan\theta(x) = V(x)/H\) → slope of the cable at \(x\); (6) Cable tension \(T = \sqrt{H^2 + V^2}\).

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 the origin at the lowest point (both supports at the same level): \(y = wx^2/(2H) \to\) parabola. Sag at centre: \(d = wL^2/(8H)\), so \(H = wL^2/(8d)\).

Tension at any point: \(T(x) = \sqrt{H^2 + (wx)^2}\). Maximum tension (at supports): \(T_{max} = \sqrt{H^2 + (wL/2)^2} = H \times \sqrt{1 + 16d^2/L^2} \approx H \times (1 + 8d^2/L^2)\) for small sag. Length of cable: \(\ell \approx L[1 + (8d^2)/(3L^2) - (32d^4)/(5L^4) + \ldots]\) (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 \(w_0\)), it takes the shape of a catenary. Catenary equation (origin at the lowest point): \(y = (H/w_0)[\cosh(w_0 x/H) - 1]\), where \(H\) = horizontal tension (constant). Arc length \(s = (H/w_0)\cdot\sinh(w_0 x/H)\).

Tension at any point: \(T = H\cdot\cosh(w_0 x/H) = H + w_0 y\). Maximum tension \(T_{max}\) (at supports) \(= H + w_0\,y_{support}\).

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 the position of the 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: \(y_A = w\cdot x_A^2/(2H)\), \(y_B = w\cdot x_B^2/(2H)\), with \(x_A + x_B = L\) (span) and \(y_B - y_A = h\). Solving: \(x_B = L/2 + hH/(wL)\), \(x_A = L/2 - hH/(wL)\). \(H\) is found from additional information (e.g., a specified sag at mid-span).

10.6 Anchor Cable / Guy Wire

For a straight inclined cable (taut — no sag) at angle \(\alpha\) to horizontal: Tension \(T = \text{load}/\sin\alpha\) (if the load is vertical). For an anchor cable supporting a mast against lateral force \(F\): \(T = F/\cos\alpha\) (where \(\alpha\) is the angle between the 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: \(E_{eq} = E/(1 + w^2 L^2 E/(12 T^3))\), where \(E\) = actual Young's modulus of the cable material, \(w\) = cable self-weight per unit length, \(L\) = horizontal projected length, \(T\) = average cable tension. As \(T \to \infty\) (highly tensioned): \(E_{eq} \to E\) (sag effect negligible, cable acts like a rod).

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