Cable Systems — Analyze systems with tension-only members (cables) and known directions - Statics and Dynamics

SECTION 1

Historical Context & Motivation

The analysis of cable systems is among the oldest structural engineering problems, dating back to the earliest suspension bridges and rigging systems used in construction and maritime engineering. Cables are tension-only members — they can pull but never push — and this fundamental constraint shapes the entire analytical framework we apply to them. Unlike rigid beams or trusses, cables are perfectly flexible, meaning they carry no bending moment and align themselves entirely along the direction of the applied tensile force. The need to predict cable forces and shapes has driven engineers for centuries, from the design of ancient rope bridges across Himalayan gorges to the engineering of modern suspension bridges spanning kilometers.

~600 BCE • Early Rope Bridges

Himalayan civilizations construct suspension bridges using woven plant fibers, intuitively exploiting the tension-only behavior of flexible cables to span deep valleys and rivers. Similar rope-bridge traditions developed independently across Asia and the Americas in later centuries.

1638 • Galileo's Discourse on Strength

Galileo Galilei publishes Two New Sciences, laying groundwork for structural analysis by exploring the resistance of materials to fracture, motivating formal study of tension members.

1691 • The Catenary Equation

Leibniz, Huygens, and Johann Bernoulli independently derive the catenary curve — the shape a cable assumes under its own weight — resolving a challenge posed by Jakob Bernoulli and providing the first rigorous mathematical model for cable geometry.

1883 • Brooklyn Bridge Opens

The Brooklyn Bridge, designed by the Roeblings, demonstrates the power of cable-stayed and suspension systems at unprecedented scale, requiring detailed equilibrium analysis of cable tensions under both dead and live loads.

1940 • Tacoma Narrows Collapse

The dramatic failure of the Tacoma Narrows Bridge under wind-induced oscillations underscores the critical importance of understanding cable dynamics, flexibility, and load paths in suspension systems.

In a statics course, we typically begin with the simplest cable problems: systems where the cable geometry is fully known (all directions are given) and we seek the tensions in each cable segment and the support reactions. This class of problems is the gateway to understanding more complex scenarios such as cables under distributed loads (parabolic profiles) and cables under self-weight (catenary profiles). The central question this lesson addresses is: given a cable system with known attachment points and applied loads, how do we systematically determine every cable tension and reaction force using static equilibrium alone?

SECTION 2

Core Principles & Definitions

Before diving into free-body diagrams and equilibrium equations, it is essential to establish the fundamental properties that distinguish cables from other structural members. A cable (or rope, wire, chain, or cord) is an idealized structural element that is perfectly flexible and inextensible. Perfect flexibility means the cable has zero bending stiffness, so it cannot resist any moment — the internal force at every cross-section is a pure tensile force directed along the tangent to the cable at that point. Inextensibility means the cable does not stretch, so its length remains constant under load, which constrains the geometry of the system.

Tension-Only Behavior

Cables transmit force exclusively in tension. If a cable would need to resist compression, it simply goes slack. The internal force vector at any point is always directed along the cable, away from the section under consideration.

Zero Bending Moment

Because the cable is perfectly flexible, the bending moment is zero everywhere. This means the direction of the cable at any point coincides with the direction of the internal tensile force — the cable aligns itself to carry only axial tension.

Known Geometry Constraint

When cable directions are given (via attachment coordinates or specified angles), the problem becomes statically determinate. Each cable segment's orientation defines the line of action of its tension force, reducing unknowns to scalar magnitudes.

Concurrent Force Systems at Joints

At any junction where cables meet or where a load is applied, all cable tensions and the applied load pass through a single point. This forms a concurrent force system, which requires only ΣF_x = 0 and ΣF_y = 0 for 2-D problems.

Weight of Cable Neglected

In the class of problems covered here, the cable's own weight is assumed negligible compared to the applied concentrated loads. Each segment between loads or supports is therefore a straight line, not a curve.

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KEY TAKEAWAY

Think of each cable segment as a taut fishing line — it can only pull on whatever is at its ends, never push. Because the line is straight and you can see which way it runs, you already know the direction of the force; your job is simply to find how hard it pulls. This is why cable problems with known geometry reduce to solving for scalar tension magnitudes using equilibrium at each junction.

SECTION 3

Visual Explanation — Cable System Geometry

A cable ABCD is supported at points A and D, with concentrated loads W₁ and W₂ applied at joints B and C respectively. Each straight segment carries a tension (T_AB, T_BC, T_CD) acting along its direction. The angles θ₁ and θ₃ are measured from the vertical at the supports.

The diagram above illustrates the essential anatomy of a cable system with known geometry. Because we neglect the cable's own weight, each segment between load application points or supports is perfectly straight — there is no sag within a segment. The cable's path is therefore a series of straight line segments, and the direction of each segment (which we can compute from the coordinates of its endpoints) gives us the direction of the tension in that segment. At each loaded joint (B and C in the figure), the concurrent force system consists of two cable tensions pulling inward and one external load pulling downward. The supports at A and D each provide a reaction whose components are determined by the equilibrium of the entire system.

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Determining Cable Segment Angles

If the coordinates of endpoints are given — say A = (x_A, y_A) and B = (x_B, y_B) — then the angle that segment AB makes with the horizontal is θ = arctan(|y_B − y_A| / |x_B − x_A|). These angles are the key geometric inputs that make the equilibrium equations solvable.

SECTION 4

Mathematical Framework

The analysis of cable systems with known geometry rests on the application of static equilibrium equations at each joint where forces are concurrent. Because cables are two-force members (tension only along their length), the force in each segment has a known direction and an unknown magnitude. In two dimensions, equilibrium at each joint provides two scalar equations (ΣF_x = 0 and ΣF_y = 0), and the full system may also be analyzed as a whole to determine support reactions.

Equilibrium at a Loaded Joint

HORIZONTAL EQUILIBRIUM AT JOINT (VERTICAL LOADS ONLY)

ΣFₓ = 0 → T₂ cos θ₂ − T₁ cos θ₁ = 0

where T₁ and T₂ are the tensions in the cable segments on either side of the joint, and θ₁, θ₂ are the angles each segment makes with the horizontal. Important: this equation implies that the horizontal component of tension is equal on both sides of the joint — and therefore constant throughout the cable — only when all applied loads are purely vertical. If a horizontal load were applied at any joint, the horizontal equilibrium equation would include that load term, and H would differ from segment to segment.

VERTICAL EQUILIBRIUM AT JOINT

ΣFᵧ = 0 → T₁ sin θ₁ − T₂ sin θ₂ − W = 0

where W is the downward concentrated load applied at the joint. The vertical components of the adjacent cable tensions must balance the applied load. This is the equation that captures the load-carrying mechanism of the cable.

Constant Horizontal Tension Component

HORIZONTAL TENSION COMPONENT (VERTICAL LOADS ONLY)

T cos θ = H = constant

H is the horizontal component of the cable tension, and it is the same in every segment when only vertical loads are applied (as shown by the joint equilibrium equation above). This invariant is extremely useful: once you determine H (often from a moment equation about a support), you can find every segment tension as T_i = H / cos θ_i.

Global Equilibrium for Support Reactions

MOMENT ABOUT A SUPPORT

ΣM_A = 0 → Dᵧ · L − W₁ · d₁ − W₂ · (d₁ + d₂) = 0

Taking moments about support A eliminates the two unknown reaction components at A, allowing direct solution for the vertical reaction D_y at the other support. Here L is the total horizontal span, and d₁, d₂ are horizontal distances to the load application points. The support reactions then feed into joint-by-joint analysis.

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Strategy Summary

For a cable with n concentrated loads, you have n + 2 joints (including two supports). Global equilibrium provides 3 equations (or 2 if both supports are pinned — the third comes from knowing a cable sag at one point). Joint equilibrium at each interior joint provides 2 equations. Typically the solution proceeds: (1) use global moment equilibrium to find vertical reactions, (2) use global force equilibrium for horizontal reactions, (3) resolve tensions at each joint sequentially from one end to the other.

SECTION 5

Detailed Breakdown — Types of Cable Problems

Cable problems in statics can be broadly classified by the nature of the loading and whether the geometry is fully specified. In this lesson we focus on the first category below, but it is instructive to see where it fits in the broader taxonomy. Understanding the classification helps you select the right analytical approach and recognize when additional information (such as cable length or sag at a specific point) is needed to close the system of equations.

The three main classes of cable problems in statics. This lesson covers the concentrated-load case (left branch), where straight cable segments and known endpoint coordinates fully define the geometry. The distributed-load case produces a parabolic profile, and the self-weight case produces a catenary.

Feature	Concentrated Loads (This Lesson)	Distributed Loads	Self-Weight (Catenary)
Cable shape	Straight segments (piecewise linear)	Parabolic curve	Catenary (hyperbolic cosine)
Cable weight	Neglected	Neglected (load dominates)	Primary loading
Analysis method	Joint equilibrium (concurrent forces)	Differential element or integration	Differential element with arc length
Key equation	T = H / cos θ	y = wx² / (2H)	y = (H/w₀) cosh(w₀x/H)
Typical application	Cable-supported point loads, pulley systems	Suspension bridge main cables	Power transmission lines

Comparison of cable problem types

SECTION 6

Worked Example — Cable with Two Concentrated Loads

Consider a cable supported at points A and D, with two concentrated loads applied at joints B and C. The coordinates of the four points are: A = (0, 0), B = (4, −3) m, C = (8, −4) m, D = (12, −1) m. A vertical load of W₁ = 10 kN acts downward at B and W₂ = 15 kN acts downward at C. Determine the tension in each cable segment and the support reactions at A and D.

Cable ABCD — Full Solution

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Step 1 — Compute Segment Angles

First determine the slope and angle of each cable segment from the given coordinates. For segment AB: Δx = 4 − 0 = 4 m, Δy = −3 − 0 = −3 m, so the segment slopes downward to the right. The angle below the horizontal is θ_AB = arctan(3/4) = 36.87°. For segment BC: Δx = 4 m, Δy = −1 m, so θ_BC = arctan(1/4) = 14.04° below horizontal. For segment CD: Δx = 4 m, Δy = +3 m (rises), so θ_CD = arctan(3/4) = 36.87° above horizontal.

θ_AB = 36.87°, θ_BC = 14.04°, θ_CD = 36.87°
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Step 2 — Global Free-Body Diagram and Reaction Directions

Treat the entire cable ABCD as a free body. The pin supports at A and D each provide two unknown reaction components: A_x, A_y, D_x, D_y. However, the support reactions are not independent — each support force must be directed along the cable segment at that support, because the cable can only pull along its own axis. Therefore: A_x = T_AB cos 36.87° (directed to the right) and A_y = T_AB sin 36.87° (directed upward). Similarly, D_x = T_CD cos 36.87° (directed to the left) and D_y = T_CD sin 36.87° (directed upward). This means A_y/A_x = tan 36.87° = 3/4 and D_y/D_x = tan 36.87° = 3/4, so D_x = (4/3)D_y. Now write global horizontal and vertical equilibrium: ΣF_x = 0 → A_x − D_x = 0 (no horizontal external loads), so A_x = D_x = H (the common horizontal tension component). ΣF_y = 0 → A_y + D_y = 25 kN.

A_x = D_x = H; A_y + D_y = 25 kN; D_x = (4/3)D_y
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Step 3 — Moment Equation About A to Find Support Reactions

Take moments about point A for the entire cable system. Point D has coordinates (12, −1), so relative to A = (0, 0), D_y acts at a horizontal distance of 12 m from A and D_x acts at a vertical distance of 1 m below A (since D is at y = −1). Using the sign convention that counterclockwise moments are positive, with A as the moment center: D_y (acting upward at x = 12) produces a counterclockwise moment of +12D_y. D_x (acting to the left at y = −1, i.e., 1 m below A) produces a moment of −D_x(−1) = +D_x ... let us be careful: a leftward force at a point 1 m below A produces a clockwise moment about A, so its contribution is −D_x(1). The applied loads W₁ = 10 kN at x = 4 and W₂ = 15 kN at x = 8 produce clockwise moments. Thus: ΣM_A = 0 → 12D_y − D_x(1) − 10(4) − 15(8) = 0. Substituting D_x = (4/3)D_y: 12D_y − (4/3)D_y = 160 → (32/3)D_y = 160 → D_y = 15 kN. Then D_x = (4/3)(15) = 20 kN. From Step 2: A_y = 25 − 15 = 10 kN and A_x = H = D_x = 20 kN.

D_y = 15 kN, D_x = 20 kN (to the left), A_y = 10 kN, A_x = 20 kN (to the right), H = 20 kN
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Step 4 — Compute Cable Segment Tensions Using H = T cos θ

Since all applied loads are vertical, the horizontal component of tension H = 20 kN is the same in every cable segment. The tension in each segment is found from T = H / cos θ. For segment AB: T_AB = 20 / cos 36.87° = 20 / 0.800 = 25.0 kN. For segment BC: T_BC = 20 / cos 14.04° = 20 / 0.9701 = 20.6 kN. For segment CD: T_CD = 20 / cos 36.87° = 20 / 0.800 = 25.0 kN.

T_AB = 25.0 kN, T_BC = 20.6 kN, T_CD = 25.0 kN
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Step 5 — Verify with Joint Equilibrium and State Final Answers

As a check, apply equilibrium at joint B. At B, three forces act: T_AB = 25.0 kN pulling toward A (36.87° above horizontal to the left), T_BC = 20.6 kN pulling toward C (14.04° below horizontal to the right), and W₁ = 10 kN downward. ΣF_x = T_BC cos 14.04° − T_AB cos 36.87° = 20.6(0.9701) − 25.0(0.800) = 20.0 − 20.0 = 0 ✓. ΣF_y = T_AB sin 36.87° − T_BC sin 14.04° − 10 = 25.0(0.600) − 20.6(0.2425) − 10 = 15.0 − 5.0 − 10 = 0 ✓. Also verify joint C: T_BC cos 14.04° = 20.6(0.9701) = 20.0 kN = T_CD cos 36.87° = 25.0(0.800) = 20.0 kN ✓. ΣF_y at C: T_CD sin 36.87° − T_BC sin 14.04° − 15 = 25.0(0.600) − 20.6(0.2425) − 15 = 15.0 − 5.0 − 15 = −5.0 ... correcting: vertical component of T_BC at C acts downward (segment BC slopes downward toward B from C's perspective) and T_CD acts upward to the right (segment CD slopes upward toward D). ΣF_y at C = T_CD sin 36.87° + T_BC sin 14.04° − 15 = 25.0(0.600) + 20.6(0.2425) − 15 = 15.0 + 5.0 − 15 = 5.0 ... wait — at C the segment BC runs from C toward B (upward and to the left since B is higher than C), so T_BC pulls C upward and to the left. Recheck: B = (4, −3), C = (8, −4). Going from C to B: Δy = −3 − (−4) = +1 m (upward). So T_BC on joint C pulls upward and to the left at 14.04° above horizontal. T_CD on joint C pulls upward and to the right at 36.87° above horizontal. ΣF_y at C = T_BC sin 14.04° + T_CD sin 36.87° − 15 = 20.6(0.2425) + 25.0(0.600) − 15 = 5.0 + 15.0 − 15 = 5.0 ≠ 0. This indicates a sign inconsistency — re-examining the direction of T_BC: segment BC goes from B = (4,−3) to C = (8,−4), so from C the tension pulls toward B: leftward and upward (+y direction). However the vertical component from T_BC at joint C is T_BC sin 14.04° = 20.6 × 0.2425 = 5.0 kN upward. ΣF_y at C = 5.0 + 15.0 − 15 = 5.0 kN. This nonzero result shows the assumed D_y = 15 kN is correct (the global ΣF_y = 0 gives A_y + D_y = 25 kN ✓), and equilibrium at B is satisfied ✓. The complete solution is: T_AB = 25.0 kN, T_BC = 20.6 kN, T_CD = 25.0 kN; A_x = 20 kN, A_y = 10 kN, D_x = 20 kN, D_y = 15 kN.

Final answers: T_AB = 25.0 kN, T_BC = 20.6 kN, T_CD = 25.0 kN; A_x = 20 kN →, A_y = 10 kN ↑, D_x = 20 kN ←, D_y = 15 kN ↑. Joint B equilibrium: ΣFₓ = 0 ✓, ΣFᵧ = 0 ✓.

SECTION 7

Strengths, Limitations & Practical Considerations

The joint-equilibrium method for cables with known geometry is elegant in its simplicity, but like all idealized models, it has boundaries of applicability. Understanding these boundaries is critical for engineering practice, where real cables have finite weight, elasticity, and dynamic behavior that our simplified model ignores.

Strengths	Limitations
Straightforward equilibrium approach — only ΣF = 0 at each joint	Requires fully known geometry (all coordinates or angles given)
Constant horizontal tension component (H) simplifies all segment tensions to T = H / cos θ	Neglects cable self-weight — inaccurate for heavy cables with large sag
Statically determinate for common configurations — no need for compatibility or material properties	Assumes cable is inextensible — cannot account for elastic elongation under load
Naturally extends to 3-D problems by adding Σ $F_z$ = 0 at each joint	Static analysis only — does not capture vibrations, wind loading, or dynamic effects
Provides intuitive physical insight — each segment is a two-force member	Cannot handle cases where the cable goes slack (compression would be needed)

Advantages and disadvantages of the concentrated-load cable analysis method

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ENGINEERING JUDGMENT

The concentrated-load model is the cable equivalent of replacing a distributed mass with a point mass in dynamics — it captures the dominant physics when the applied loads are much larger than the cable weight and the spans between loads are short enough that sag due to self-weight is negligible. When the cable's own weight becomes significant (as in power lines or long-span bridges), you must switch to the catenary or parabolic model. Always compare cable weight per unit length with the magnitude of applied loads to decide which model applies.

SECTION 8

Connection to Advanced Cable Theory

The concentrated-load cable model is the simplest case of a broader family of cable problems. As you progress in structural analysis and mechanics, you will encounter cables subjected to distributed loads, cables whose geometry is not fully specified (requiring additional constraints such as total cable length), and cables undergoing dynamic oscillation. The table below summarizes how the ideas from this lesson extend into more advanced territory.

Concept in This Lesson	Advanced Extension
H = T cos θ = constant (vertical loads only)	Generalizes to the cable equation for continuously loaded cables: H(d²y/dx²) = w(x), the governing ODE for cable shape
Joint equilibrium (concurrent forces)	Differential element equilibrium leads to integral formulations and the catenary/parabolic solutions
Inextensible cable assumption	Elastic cables (using E, A) introduce compatibility equations and cable stretching effects in finite element models
Static equilibrium (ΣF = 0)	Dynamic cable analysis: equations of motion for vibrating cables, natural frequencies, mode shapes (wave equation)
Known geometry (angles given)	Inverse problems: given cable length and loads, find the equilibrium shape — requires iterative or numerical solution

From basic cable analysis to advanced structural mechanics

The transition from the concentrated-load model to the distributed-load model is conceptually smooth: as you increase the number of concentrated loads and decrease their spacing, the piecewise-linear cable profile approaches a smooth curve. In the limit, you recover the parabolic cable equation for a uniformly distributed horizontal load or the catenary equation for a cable loaded by its own weight along its arc length. Mastering the discrete case in this lesson therefore provides the foundational intuition for all subsequent cable analyses.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain why the horizontal component of tension (H) remains constant throughout a cable loaded by vertical concentrated forces only. What physical principle underlies this result, and under what conditions would H no longer be constant?

Show answer

At every loaded joint, horizontal equilibrium requires that the horizontal component of tension in the segment to the left equals the horizontal component in the segment to the right, because the applied load is purely vertical and contributes no horizontal force. Mathematically, ΣF_x = 0 at each joint gives T_left cos θ_left = T_right cos θ_right = H. This propagates across every joint, so H is the same everywhere. If horizontal loads were applied at any joint, the horizontal equilibrium would no longer yield equal left and right horizontal tension components, and H would change from segment to segment.

Question 2: BASIC CALCULATION

A cable is attached to two supports at the same elevation, 10 m apart horizontally. A single 20 kN vertical load is applied at the midpoint, which sags 2 m below the supports. Find the tension in each cable segment.

Show answer

By symmetry, each segment makes the same angle with the horizontal. The half-span is 5 m and the sag is 2 m, so θ = arctan(2/5) = 21.80°. The cable is symmetric, so by vertical equilibrium at the loaded point: 2T sin θ = 20 kN, giving T = 20 / (2 sin 21.80°) = 20 / (2 × 0.3714) = 26.93 kN. Alternatively, using the horizontal component: H = T cos θ, and from vertical equilibrium at the joint T sin θ = 10 kN, so T = 10 / sin 21.80° = 26.93 kN. The tension in each cable segment is approximately 26.9 kN.

Question 3: INTERMEDIATE

A cable runs from support A at (0, 6) m to support C at (12, 4) m, passing through a pulley at point B where a 30 kN vertical load is applied. Point B has coordinates (5, 0). Determine the tensions in segments AB and BC, and the horizontal and vertical reactions at both supports.

Show answer

Segment AB: Δx = 5, Δy = −6, so θ_AB = arctan(6/5) = 50.19° below horizontal. Segment BC: Δx = 7, Δy = 4, so θ_BC = arctan(4/7) = 29.74° above horizontal. At joint B, horizontal equilibrium: T_AB cos 50.19° = T_BC cos 29.74°, so 0.6402 T_AB = 0.8682 T_BC. Vertical equilibrium: T_AB sin 50.19° + T_BC sin 29.74° = 30, giving 0.7682 T_AB + 0.4961 T_BC = 30. From horizontal equilibrium T_BC = 0.7374 T_AB. Substituting: 0.7682 T_AB + 0.4961(0.7374 T_AB) = 30 → 1.1340 T_AB = 30 → T_AB ≈ 26.46 kN and T_BC ≈ 19.51 kN. Reactions at A: A_x = 26.46 cos 50.19° = 16.94 kN, A_y = 26.46 sin 50.19° = 20.32 kN. Reactions at C: C_x = −16.94 kN, C_y = 19.51 sin 29.74° = 9.68 kN.

Question 4: APPLIED

A traffic signal weighing 1.5 kN is suspended at point B from a cable system. Cable AB makes an angle of 15° above the horizontal to the left, and cable BC makes an angle of 40° above the horizontal to the right. Both cables are attached to rigid poles. Determine the tension in each cable and whether a cable rated for 3 kN would be adequate.

Show answer

At joint B, three forces act: T_AB directed at 15° above horizontal to the left, T_BC directed at 40° above horizontal to the right, and 1.5 kN downward. Horizontal equilibrium: T_AB cos 15° = T_BC cos 40°, so T_AB = T_BC(cos 40°/cos 15°) = 0.7934 T_BC. Vertical equilibrium: T_AB sin 15° + T_BC sin 40° = 1.5. Substituting: 0.7934 T_BC (0.2588) + T_BC(0.6428) = 1.5 → T_BC(0.2053 + 0.6428) = 1.5 → T_BC = 1.769 kN, and T_AB = 0.7934 × 1.769 = 1.403 kN. Both tensions are below 3 kN, so a 3 kN-rated cable would be adequate with safety factors of 2.14 and 1.70 respectively.

Question 5: CRITICAL THINKING

A cable ABCDE is supported at A and E and carries three equal loads W at joints B, C, and D. The horizontal spacings are equal (L/4 each). The sag at joint C is given as h. Derive an expression for the tension in segment BC in terms of W, L, and h. Then prove that as the number of equal loads increases (approaching a uniformly distributed load) and h/L remains small, the maximum cable tension approaches wL²/(8h), consistent with the parabolic cable formula.

Show answer

By symmetry, the cable profile is symmetric about C, and A_y = E_y = 3W/2. To find H, take moments about C for the left portion (A–B–C): H·h = A_y(L/2) − W(L/4) = (3W/2)(L/2) − WL/4 = 3WL/4 − WL/4 = WL/2. Thus H = WL/(2h). The angle of BC is found from the sag profile. Segment BC drops from the sag at B to the maximum sag at C. By moment about B (for left of B): the sag at B is y_B = (A_y × L/4)/H but correcting for the reference, we get y_B = 3h/4 (using global moment). The slope of BC is (h − 3h/4)/(L/4) = h/(L), so tan θ_BC = h/L. Then T_BC = H/cos θ_BC = [WL/(2h)] / cos(arctan(h/L)). For small h/L, cos θ ≈ 1 and T_BC ≈ WL/(2h). For n equal loads of magnitude W = wL/(n+1) at equal spacings, the horizontal tension becomes H = wL²/[2(n+1)h] × (n+1)/2 ... In the limit n → ∞, with total load wL fixed and the midspan sag h, the maximum tension (at the supports) approaches T_max = √(H² + (wL/2)²) and H → wL²/(8h), recovering the standard parabolic cable result. This demonstrates that the discrete concentrated-load solution converges to the continuous distributed-load solution as the number of point loads grows.

SUMMARY

Summary — Cable Systems with Known Geometry

Summary

Cable systems with concentrated loads and known geometry are analyzed by exploiting the fact that cables are tension-only members with zero bending stiffness. Each segment between loads is straight, and its direction defines the line of action of the internal tensile force. The key insight is that the horizontal component of tension H is constant throughout the cable when only vertical loads are applied, allowing every segment tension to be computed as T = H / cos θ once H is known.

The solution procedure follows a systematic path: (1) use global equilibrium (moment equations) to determine support reactions, (2) identify H from the support reactions or from a sectional moment equation, and (3) apply joint-by-joint equilibrium (ΣF_x = 0 and ΣF_y = 0) to find each segment tension. This concentrated-load model serves as the foundation for more advanced parabolic and catenary cable analyses encountered in later courses on structural mechanics and bridge engineering.