Arches, Cables and Matrix Methods
Three-hinged and two-hinged arches, horizontal thrust and Eddy's theorem, cables under point loads and UDL, cable tension and sag, and stiffness and flexibility matrix methods.
Arches and cables carry load mainly by axial force: arches in compression, cables in tension. Matrix methods are the general way computers analyse any structure.
Three-hinged arch
Hinges at both supports and at the crown make it statically determinate (4 reactions, 3 equilibrium equations + 1 hinge condition: M = 0 at the crown).
Parabolic arch with UDL w over the full span L and rise h:
- Horizontal thrust H = wL²/(8h).
- Bending moment is zero everywhere: the parabola is the funicular shape for a UDL.
Eddy's theorem: the bending moment at any section = H × (vertical distance between the arch axis and the linear arch, the line of thrust).
Moment at a section x: M = M_beam(x) − H y(x), where M_beam is the simply supported beam moment and y the arch height.
Two-hinged arch
Indeterminate to degree 1 (H is the redundant), found by energy methods: H = ∫ M_beam y ds/(EI) ÷ ∫ y² ds/(EI). For a parabolic two-hinged arch with UDL over the full span, also H = wL²/(8h). A rise in temperature increases H.
Cables
A cable carries only tension and takes the funicular shape of its loads.
- At any point: H × (sag at that point) = M_beam at that point (same equation as the arch).
- UDL over the span (parabolic cable, supports level, central sag d): H = wL²/(8d); vertical reaction V = wL/2; maximum tension T_max = √(H² + V²) at the supports.
- Length of a parabolic cable ≈ L + 8d²/(3L).
Matrix methods
- Flexibility (force) method: {Δ} = [F]{P}; unknowns are redundant forces. Matrix size = Ds.
- Stiffness (displacement) method: {P} = [K]{Δ}; unknowns are joint displacements. Matrix size = Dk. Used in most software.
- Properties: [K] and [F] are symmetric (Maxwell), with positive diagonal terms; [K] = [F]⁻¹ for the same coordinates.
- Stiffness matrix of a beam element (rotations only, both ends): (EI/L)[[4, 2], [2, 4]].
Worked example
A cable spans 100 m with a central sag of 10 m and carries 2 kN/m.
- H = 2 × 100²/(8 × 10) = 250 kN; V = 100 kN.
- T_max = √(250² + 100²) = 269.3 kN.
Practice
A three-hinged parabolic arch of span 20 m and rise 4 m carries a UDL of 16 kN/m over the whole span. What is the horizontal thrust in kN?
Show answer and explanation
Answer: 200H = wL²/(8h) = 16 × 400/32 = 200 kN.
For the arch above, what is the bending moment at the quarter-span section (kN·m)?
Show answer and explanation
Answer: 0A parabolic arch under a full-span UDL is funicular: the moment is zero everywhere.
A cable of span 60 m with central sag 6 m carries 3 kN/m. What is the maximum tension in kN? (Give 1 decimal place.)
Show answer and explanation
Answer: 242.3H = 3 × 3600/48 = 225 kN and V = 90 kN, so T_max = √(225² + 90²) = √58,725 = 242.3 kN.
The size of the stiffness matrix of a structure equals its:
Show answer and explanation
Answer: BThe stiffness method's unknowns are joint displacements, so the matrix is Dk × Dk.