Frames, Cables, Arches and Structural Load Paths

Civil structural systems are not all trusses. Building frames develop shear and bending, flexible cables carry tension, arches can carry dominant compression, and pinned connections may turn a frame into a mechanism. A load path identifies how each applied action travels through slabs, beams, columns, braces and foundations. Drawing that path is as important as solving a small equation set. If a load path is interrupted—for example by an omitted connection, a discontinuous support or a temporary removal—the simplest equilibrium model may produce impossible reactions.

Engineering boundary: These exercises analyze idealized statics models for education. They do not certify structural strength, earthquake/wind resistance, geotechnical bearing, connection design, construction safety, or compliance with any jurisdiction’s building code. Real works require local standards, authoritative loading data and qualified professional checking.

Learning outcomes

• Formulate an explicit free-body model for frames, cables, arches and structural load paths.

• Use signed forces, consistent units and geometric measurements to construct equilibrium relations.

• Check results with an independently chosen equilibrium statement or limiting case.

• Distinguish what idealized statics establishes from what structural design and field verification must establish.

Lesson 4.1: Frames and multi-force rigid bodies

A frame member with three or more applied forces usually develops shear and bending, not merely axial force. Cutting at joints requires correct moments and interaction forces. Rigid frame joints transmit moments; pin idealizations do not.

Governing relationships

A fixed base in planar idealization can transmit Fx, Fy and a reaction moment. A member isolated between two pin joints without distributed action can be treated as a two-force member, but only under that strict condition.

Worked engineering example

A 3 m cantilever column subjected to 8 kN horizontal force at its top needs a base shear of 8 kN and a base moment of 24 kN·m, neglecting its self-weight.

Model audit and common error

Do not model a welded moment connection as frictionless pin without a justified reason.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving frames and multi-force rigid bodies. Mark coordinate axes, every restraint action, all known magnitudes and signed reference distances. Explain which connections or soil assumptions would invalidate the model. Report results in both a compact table and an independent equilibrium check.

Engineering discussion

This topic should be evaluated with the full load path rather than by its equation alone. For frames, cables, arches and structural load paths, the quantities that matter are the action itself, where it enters the idealized body, and which physical connection or foundation provides the balancing action. A model that omits one of these produces a formally neat answer without identifying the actual mechanism. Separating gross structural equilibrium from local member strength prevents that confusion. The analyst should first decide what the isolated body is, then measure or state geometry, then apply equilibrium. If an intermediate step changes the sign convention or datum, it should be flagged in the calculation record. Numerical solutions should preserve enough precision for checks but not imply more site accuracy than the input geometry supports.

Lesson 4.2: Flexible cables and sag sensitivity

An ideal cable carries axial tension only, and the cable shape depends on load distribution. For uniform load per horizontal projection, an ideal parabolic shape is appropriate; a cable under its own constant weight per actual cable length has a catenary shape.

Governing relationships

For horizontal span L, midspan sag f and uniform vertical load w per unit horizontal span, the parabolic model gives H=wL²/(8f). Approximate end tension is sqrt(H²+(wL/2)²).

Worked engineering example

For L=20 m, sag f=2 m, and w=1 kN/m horizontal, H=25 kN and support vertical reaction is 10 kN; end tension is sqrt(25²+10²)=26.93 kN.

Model audit and common error

Small sag increases horizontal force: a visually shallow cable is not necessarily a lightly loaded cable.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving flexible cables and sag sensitivity. Mark coordinate axes, every restraint action, all known magnitudes and signed reference distances. Explain which connections or soil assumptions would invalidate the model. Report results in both a compact table and an independent equilibrium check.

Engineering discussion

This topic should be evaluated with the full load path rather than by its equation alone. For frames, cables, arches and structural load paths, the quantities that matter are the action itself, where it enters the idealized body, and which physical connection or foundation provides the balancing action. A model that omits one of these produces a formally neat answer without identifying the actual mechanism. Separating gross structural equilibrium from local member strength prevents that confusion. The analyst should first decide what the isolated body is, then measure or state geometry, then apply equilibrium. If an intermediate step changes the sign convention or datum, it should be flagged in the calculation record. Numerical solutions should preserve enough precision for checks but not imply more site accuracy than the input geometry supports.

Lesson 4.3: Arches, horizontal thrust and load line

Arches redirect loading partly through compression and may generate significant horizontal support thrust. A funicular arch under its design loading may carry little bending, but other loading patterns and support movements introduce bending.

Governing relationships

For an ideal three-hinged parabolic arch with span L, central rise f and uniform vertical load w, horizontal thrust H=wL²/(8f), obtained from zero bending moment at the crown hinge.

Worked engineering example

A symmetric 30 m arch with rise 5 m and w=3 kN/m has H=3×900/(40)=67.5 kN. Global vertical support reaction is wL/2=45 kN at each end.

Model audit and common error

An arch solution is valid only with its specified hinge, geometry and loading assumptions.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving arches, horizontal thrust and load line. Mark coordinate axes, every restraint action, all known magnitudes and signed reference distances. Explain which connections or soil assumptions would invalidate the model. Report results in both a compact table and an independent equilibrium check.

Engineering discussion

This topic should be evaluated with the full load path rather than by its equation alone. For frames, cables, arches and structural load paths, the quantities that matter are the action itself, where it enters the idealized body, and which physical connection or foundation provides the balancing action. A model that omits one of these produces a formally neat answer without identifying the actual mechanism. Separating gross structural equilibrium from local member strength prevents that confusion. The analyst should first decide what the isolated body is, then measure or state geometry, then apply equilibrium. If an intermediate step changes the sign convention or datum, it should be flagged in the calculation record. Numerical solutions should preserve enough precision for checks but not imply more site accuracy than the input geometry supports.

Lesson 4.4: Load-path continuity and construction detailing

A system can satisfy an attractive free-body calculation and still be unbuildable because forces cannot cross an absent or weak connection. A useful load-path diagram traces each force to ground and identifies the member or interface responsible for every transfer.

Governing relationships

At a floor-to-beam-to-column interface, the tributary floor load enters the beam, beam end reactions load columns and columns send resultant forces into the foundation. Compatibility and deformation redistribute load when the system is indeterminate.

Worked engineering example

A precast plank bearing on a girder transfers vertical reaction through contact. If the bearing seat is removed during construction, the intended gravity load path does not exist even though a drawing still shows nominal support reactions.

Model audit and common error

Equilibrium of individual elements must agree with the complete structural assembly.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving load-path continuity and construction detailing. Mark coordinate axes, every restraint action, all known magnitudes and signed reference distances. Explain which connections or soil assumptions would invalidate the model. Report results in both a compact table and an independent equilibrium check.

Engineering discussion

This topic should be evaluated with the full load path rather than by its equation alone. For frames, cables, arches and structural load paths, the quantities that matter are the action itself, where it enters the idealized body, and which physical connection or foundation provides the balancing action. A model that omits one of these produces a formally neat answer without identifying the actual mechanism. Separating gross structural equilibrium from local member strength prevents that confusion. The analyst should first decide what the isolated body is, then measure or state geometry, then apply equilibrium. If an intermediate step changes the sign convention or datum, it should be flagged in the calculation record. Numerical solutions should preserve enough precision for checks but not imply more site accuracy than the input geometry supports.

Integrated civil case study

Footbridge cable-and-arch concept: draw two plausible load paths and compare resulting support horizontal thrust, internal moments and sensitivity to changing sag or rise. This is a mechanics study, not a buildable design.

Worked problem studio

Each problem below is a separate numerical case. Work it first without looking at the answer, then compare the force and moment checks. Changing dimensions can reveal near-singular geometry or support reactions that switch direction.

Worked problem 4.1

Given: A parabolic cable spans 12 m with sag 1.20 m and vertical load 1.00 kN per horizontal metre. Compute horizontal thrust and end tension.

Solution: H=wL²/(8f)=15.000 kN; vertical end reaction V=wL/2=6.000 kN; end tension sqrt(H²+V²)=16.155 kN. Changing the distributed-load basis to weight per cable length requires a different model.

Review check: State the units and verify whether the result reflects the applied load path. Identify one assumption whose failure would require a new analysis.

Worked problem 4.2

Given: A parabolic cable spans 14 m with sag 1.35 m and vertical load 1.15 kN per horizontal metre. Compute horizontal thrust and end tension.

Solution: H=wL²/(8f)=20.870 kN; vertical end reaction V=wL/2=8.050 kN; end tension sqrt(H²+V²)=22.369 kN. Changing the distributed-load basis to weight per cable length requires a different model.

Review check: State the units and verify whether the result reflects the applied load path. Identify one assumption whose failure would require a new analysis.

Worked problem 4.3

Given: A parabolic cable spans 16 m with sag 1.50 m and vertical load 1.30 kN per horizontal metre. Compute horizontal thrust and end tension.

Solution: H=wL²/(8f)=27.733 kN; vertical end reaction V=wL/2=10.400 kN; end tension sqrt(H²+V²)=29.619 kN. Changing the distributed-load basis to weight per cable length requires a different model.

Review check: State the units and verify whether the result reflects the applied load path. Identify one assumption whose failure would require a new analysis.

Worked problem 4.4

Given: A parabolic cable spans 18 m with sag 1.65 m and vertical load 1.45 kN per horizontal metre. Compute horizontal thrust and end tension.

Solution: H=wL²/(8f)=35.591 kN; vertical end reaction V=wL/2=13.050 kN; end tension sqrt(H²+V²)=37.908 kN. Changing the distributed-load basis to weight per cable length requires a different model.

Review check: State the units and verify whether the result reflects the applied load path. Identify one assumption whose failure would require a new analysis.

Worked problem 4.5

Given: A parabolic cable spans 20 m with sag 1.80 m and vertical load 1.60 kN per horizontal metre. Compute horizontal thrust and end tension.

Solution: H=wL²/(8f)=44.444 kN; vertical end reaction V=wL/2=16.000 kN; end tension sqrt(H²+V²)=47.237 kN. Changing the distributed-load basis to weight per cable length requires a different model.

Review check: State the units and verify whether the result reflects the applied load path. Identify one assumption whose failure would require a new analysis.

Worked problem 4.6

Given: A parabolic cable spans 22 m with sag 1.95 m and vertical load 1.75 kN per horizontal metre. Compute horizontal thrust and end tension.

Solution: H=wL²/(8f)=54.295 kN; vertical end reaction V=wL/2=19.250 kN; end tension sqrt(H²+V²)=57.606 kN. Changing the distributed-load basis to weight per cable length requires a different model.

Review check: State the units and verify whether the result reflects the applied load path. Identify one assumption whose failure would require a new analysis.

Worked problem 4.7

Given: A parabolic cable spans 24 m with sag 2.10 m and vertical load 1.90 kN per horizontal metre. Compute horizontal thrust and end tension.

Solution: H=wL²/(8f)=65.143 kN; vertical end reaction V=wL/2=22.800 kN; end tension sqrt(H²+V²)=69.018 kN. Changing the distributed-load basis to weight per cable length requires a different model.

Review check: State the units and verify whether the result reflects the applied load path. Identify one assumption whose failure would require a new analysis.

Worked problem 4.8

Given: A parabolic cable spans 26 m with sag 2.25 m and vertical load 2.05 kN per horizontal metre. Compute horizontal thrust and end tension.

Solution: H=wL²/(8f)=76.989 kN; vertical end reaction V=wL/2=26.650 kN; end tension sqrt(H²+V²)=81.471 kN. Changing the distributed-load basis to weight per cable length requires a different model.

Review check: State the units and verify whether the result reflects the applied load path. Identify one assumption whose failure would require a new analysis.

Worked problem 4.9

Given: A parabolic cable spans 28 m with sag 2.40 m and vertical load 2.20 kN per horizontal metre. Compute horizontal thrust and end tension.

Solution: H=wL²/(8f)=89.833 kN; vertical end reaction V=wL/2=30.800 kN; end tension sqrt(H²+V²)=94.967 kN. Changing the distributed-load basis to weight per cable length requires a different model.

Review check: State the units and verify whether the result reflects the applied load path. Identify one assumption whose failure would require a new analysis.

Worked problem 4.10

Given: A parabolic cable spans 30 m with sag 2.55 m and vertical load 2.35 kN per horizontal metre. Compute horizontal thrust and end tension.

Solution: H=wL²/(8f)=103.676 kN; vertical end reaction V=wL/2=35.250 kN; end tension sqrt(H²+V²)=109.505 kN. Changing the distributed-load basis to weight per cable length requires a different model.

Review check: State the units and verify whether the result reflects the applied load path. Identify one assumption whose failure would require a new analysis.

Worked problem 4.11

Given: A parabolic cable spans 32 m with sag 2.70 m and vertical load 2.50 kN per horizontal metre. Compute horizontal thrust and end tension.

Solution: H=wL²/(8f)=118.519 kN; vertical end reaction V=wL/2=40.000 kN; end tension sqrt(H²+V²)=125.087 kN. Changing the distributed-load basis to weight per cable length requires a different model.

Review check: State the units and verify whether the result reflects the applied load path. Identify one assumption whose failure would require a new analysis.

Worked problem 4.12

Given: A parabolic cable spans 34 m with sag 2.85 m and vertical load 2.65 kN per horizontal metre. Compute horizontal thrust and end tension.

Solution: H=wL²/(8f)=134.360 kN; vertical end reaction V=wL/2=45.050 kN; end tension sqrt(H²+V²)=141.711 kN. Changing the distributed-load basis to weight per cable length requires a different model.

Review check: State the units and verify whether the result reflects the applied load path. Identify one assumption whose failure would require a new analysis.

Unit assessment (with guidance)

Q1. Distinguish a bending frame from an ideal pin-jointed truss using a real walkway example.

Guidance: Use the governing equation and a labeled diagram from Unit 4; report direction, magnitude, units and model assumptions. Numerical assessment answers should be obtained from the worked methods, not from inspection of figures alone.

Q2. For w=1.5 kN/m, L=16 m and sag 2 m compute the parabolic cable horizontal force.

Guidance: Use the governing equation and a labeled diagram from Unit 4; report direction, magnitude, units and model assumptions. Numerical assessment answers should be obtained from the worked methods, not from inspection of figures alone.

Q3. Explain why a shallow arch or cable can create unexpectedly large horizontal support thrust.

Guidance: Use the governing equation and a labeled diagram from Unit 4; report direction, magnitude, units and model assumptions. Numerical assessment answers should be obtained from the worked methods, not from inspection of figures alone.

Q4. Trace the load path from a precast slab panel through girder, column, footing and supporting ground.

Guidance: Use the governing equation and a labeled diagram from Unit 4; report direction, magnitude, units and model assumptions. Numerical assessment answers should be obtained from the worked methods, not from inspection of figures alone.

Interactive experiment

Open [simulation 4](../simulations/simulation-04.html) and vary at least three parameters. Create an observation table from the CSV. Describe one change for which the response is nonlinear, ill-conditioned or physically infeasible.

End-of-unit checklist

• External forces and reactions identified on isolated body.

• Dimensions, datums, load units and sign convention documented.

• Vertical/horizontal resultant and moment balance checked separately.

• Real-world limits and uncertainty recorded before any design interpretation.

Unit 4: Frames, cables, arches and structural load paths

Civil structural systems are not all trusses. Building frames develop shear and bending, flexible cables carry tension, arches can carry dominant compression, and pinned connections may turn a frame into a mechanism. A load path identifies how each applied action travels through slabs, beams, columns, braces and foundations. Drawing that path is as important as solving a small equation set. If a load path is interrupted—for example by an omitted connection, a discontinuous support or a temporary removal—the simplest equilibrium model may produce impossible reactions.

Engineering boundary: These exercises analyze idealized statics models for education. They do not certify structural strength, earthquake/wind resistance, geotechnical bearing, connection design, construction safety, or compliance with any jurisdiction’s building code. Real works require local standards, authoritative loading data and qualified professional checking.

Learning outcomes

• Formulate an explicit free-body model for frames, cables, arches and structural load paths.

• Use signed forces, consistent units and geometric measurements to construct equilibrium relations.

• Check results with an independently chosen equilibrium statement or limiting case.

• Distinguish what idealized statics establishes from what structural design and field verification must establish.

Lesson 4.1: Frames and multi-force rigid bodies

A frame member with three or more applied forces usually develops shear and bending, not merely axial force. Cutting at joints requires correct moments and interaction forces. Rigid frame joints transmit moments; pin idealizations do not.

Governing relationships

A fixed base in planar idealization can transmit Fx, Fy and a reaction moment. A member isolated between two pin joints without distributed action can be treated as a two-force member, but only under that strict condition.

Worked engineering example

A 3 m cantilever column subjected to 8 kN horizontal force at its top needs a base shear of 8 kN and a base moment of 24 kN·m, neglecting its self-weight.

Model audit and common error

Do not model a welded moment connection as frictionless pin without a justified reason.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving frames and multi-force rigid bodies. Mark coordinate axes, every restraint action, all known magnitudes and signed reference distances. Explain which connections or soil assumptions would invalidate the model. Report results in both a compact table and an independent equilibrium check.

Engineering discussion

This topic should be evaluated with the full load path rather than by its equation alone. For frames, cables, arches and structural load paths, the quantities that matter are the action itself, where it enters the idealized body, and which physical connection or foundation provides the balancing action. A model that omits one of these produces a formally neat answer without identifying the actual mechanism. Separating gross structural equilibrium from local member strength prevents that confusion. The analyst should first decide what the isolated body is, then measure or state geometry, then apply equilibrium. If an intermediate step changes the sign convention or datum, it should be flagged in the calculation record. Numerical solutions should preserve enough precision for checks but not imply more site accuracy than the input geometry supports.

Lesson 4.2: Flexible cables and sag sensitivity

An ideal cable carries axial tension only, and the cable shape depends on load distribution. For uniform load per horizontal projection, an ideal parabolic shape is appropriate; a cable under its own constant weight per actual cable length has a catenary shape.

Governing relationships

For horizontal span L, midspan sag f and uniform vertical load w per unit horizontal span, the parabolic model gives H=wL²/(8f). Approximate end tension is sqrt(H²+(wL/2)²).

Worked engineering example

For L=20 m, sag f=2 m, and w=1 kN/m horizontal, H=25 kN and support vertical reaction is 10 kN; end tension is sqrt(25²+10²)=26.93 kN.

Model audit and common error

Small sag increases horizontal force: a visually shallow cable is not necessarily a lightly loaded cable.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving flexible cables and sag sensitivity. Mark coordinate axes, every restraint action, all known magnitudes and signed reference distances. Explain which connections or soil assumptions would invalidate the model. Report results in both a compact table and an independent equilibrium check.

Engineering discussion

This topic should be evaluated with the full load path rather than by its equation alone. For frames, cables, arches and structural load paths, the quantities that matter are the action itself, where it enters the idealized body, and which physical connection or foundation provides the balancing action. A model that omits one of these produces a formally neat answer without identifying the actual mechanism. Separating gross structural equilibrium from local member strength prevents that confusion. The analyst should first decide what the isolated body is, then measure or state geometry, then apply equilibrium. If an intermediate step changes the sign convention or datum, it should be flagged in the calculation record. Numerical solutions should preserve enough precision for checks but not imply more site accuracy than the input geometry supports.

Lesson 4.3: Arches, horizontal thrust and load line

Arches redirect loading partly through compression and may generate significant horizontal support thrust. A funicular arch under its design loading may carry little bending, but other loading patterns and support movements introduce bending.

Governing relationships

For an ideal three-hinged parabolic arch with span L, central rise f and uniform vertical load w, horizontal thrust H=wL²/(8f), obtained from zero bending moment at the crown hinge.

Worked engineering example

A symmetric 30 m arch with rise 5 m and w=3 kN/m has H=3×900/(40)=67.5 kN. Global vertical support reaction is wL/2=45 kN at each end.

Model audit and common error

An arch solution is valid only with its specified hinge, geometry and loading assumptions.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving arches, horizontal thrust and load line. Mark coordinate axes, every restraint action, all known magnitudes and signed reference distances. Explain which connections or soil assumptions would invalidate the model. Report results in both a compact table and an independent equilibrium check.

Engineering discussion

This topic should be evaluated with the full load path rather than by its equation alone. For frames, cables, arches and structural load paths, the quantities that matter are the action itself, where it enters the idealized body, and which physical connection or foundation provides the balancing action. A model that omits one of these produces a formally neat answer without identifying the actual mechanism. Separating gross structural equilibrium from local member strength prevents that confusion. The analyst should first decide what the isolated body is, then measure or state geometry, then apply equilibrium. If an intermediate step changes the sign convention or datum, it should be flagged in the calculation record. Numerical solutions should preserve enough precision for checks but not imply more site accuracy than the input geometry supports.

Lesson 4.4: Load-path continuity and construction detailing

A system can satisfy an attractive free-body calculation and still be unbuildable because forces cannot cross an absent or weak connection. A useful load-path diagram traces each force to ground and identifies the member or interface responsible for every transfer.

Governing relationships

At a floor-to-beam-to-column interface, the tributary floor load enters the beam, beam end reactions load columns and columns send resultant forces into the foundation. Compatibility and deformation redistribute load when the system is indeterminate.

Worked engineering example

A precast plank bearing on a girder transfers vertical reaction through contact. If the bearing seat is removed during construction, the intended gravity load path does not exist even though a drawing still shows nominal support reactions.

Model audit and common error

Equilibrium of individual elements must agree with the complete structural assembly.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving load-path continuity and construction detailing. Mark coordinate axes, every restraint action, all known magnitudes and signed reference distances. Explain which connections or soil assumptions would invalidate the model. Report results in both a compact table and an independent equilibrium check.

Engineering discussion

This topic should be evaluated with the full load path rather than by its equation alone. For frames, cables, arches and structural load paths, the quantities that matter are the action itself, where it enters the idealized body, and which physical connection or foundation provides the balancing action. A model that omits one of these produces a formally neat answer without identifying the actual mechanism. Separating gross structural equilibrium from local member strength prevents that confusion. The analyst should first decide what the isolated body is, then measure or state geometry, then apply equilibrium. If an intermediate step changes the sign convention or datum, it should be flagged in the calculation record. Numerical solutions should preserve enough precision for checks but not imply more site accuracy than the input geometry supports.

Integrated civil case study

Footbridge cable-and-arch concept: draw two plausible load paths and compare resulting support horizontal thrust, internal moments and sensitivity to changing sag or rise. This is a mechanics study, not a buildable design.

Worked problem studio

Each problem below is a separate numerical case. Work it first without looking at the answer, then compare the force and moment checks. Changing dimensions can reveal near-singular geometry or support reactions that switch direction.

Worked problem 4.1

Given: A parabolic cable spans 12 m with sag 1.20 m and vertical load 1.00 kN per horizontal metre. Compute horizontal thrust and end tension.

Solution: H=wL²/(8f)=15.000 kN; vertical end reaction V=wL/2=6.000 kN; end tension sqrt(H²+V²)=16.155 kN. Changing the distributed-load basis to weight per cable length requires a different model.

Review check: State the units and verify whether the result reflects the applied load path. Identify one assumption whose failure would require a new analysis.

Worked problem 4.2

Given: A parabolic cable spans 14 m with sag 1.35 m and vertical load 1.15 kN per horizontal metre. Compute horizontal thrust and end tension.

Solution: H=wL²/(8f)=20.870 kN; vertical end reaction V=wL/2=8.050 kN; end tension sqrt(H²+V²)=22.369 kN. Changing the distributed-load basis to weight per cable length requires a different model.

Review check: State the units and verify whether the result reflects the applied load path. Identify one assumption whose failure would require a new analysis.

Worked problem 4.3

Given: A parabolic cable spans 16 m with sag 1.50 m and vertical load 1.30 kN per horizontal metre. Compute horizontal thrust and end tension.

Solution: H=wL²/(8f)=27.733 kN; vertical end reaction V=wL/2=10.400 kN; end tension sqrt(H²+V²)=29.619 kN. Changing the distributed-load basis to weight per cable length requires a different model.

Review check: State the units and verify whether the result reflects the applied load path. Identify one assumption whose failure would require a new analysis.

Worked problem 4.4

Given: A parabolic cable spans 18 m with sag 1.65 m and vertical load 1.45 kN per horizontal metre. Compute horizontal thrust and end tension.

Solution: H=wL²/(8f)=35.591 kN; vertical end reaction V=wL/2=13.050 kN; end tension sqrt(H²+V²)=37.908 kN. Changing the distributed-load basis to weight per cable length requires a different model.

Review check: State the units and verify whether the result reflects the applied load path. Identify one assumption whose failure would require a new analysis.

Worked problem 4.5

Given: A parabolic cable spans 20 m with sag 1.80 m and vertical load 1.60 kN per horizontal metre. Compute horizontal thrust and end tension.

Solution: H=wL²/(8f)=44.444 kN; vertical end reaction V=wL/2=16.000 kN; end tension sqrt(H²+V²)=47.237 kN. Changing the distributed-load basis to weight per cable length requires a different model.

Review check: State the units and verify whether the result reflects the applied load path. Identify one assumption whose failure would require a new analysis.

Worked problem 4.6

Given: A parabolic cable spans 22 m with sag 1.95 m and vertical load 1.75 kN per horizontal metre. Compute horizontal thrust and end tension.

Solution: H=wL²/(8f)=54.295 kN; vertical end reaction V=wL/2=19.250 kN; end tension sqrt(H²+V²)=57.606 kN. Changing the distributed-load basis to weight per cable length requires a different model.

Review check: State the units and verify whether the result reflects the applied load path. Identify one assumption whose failure would require a new analysis.

Worked problem 4.7

Given: A parabolic cable spans 24 m with sag 2.10 m and vertical load 1.90 kN per horizontal metre. Compute horizontal thrust and end tension.

Solution: H=wL²/(8f)=65.143 kN; vertical end reaction V=wL/2=22.800 kN; end tension sqrt(H²+V²)=69.018 kN. Changing the distributed-load basis to weight per cable length requires a different model.

Review check: State the units and verify whether the result reflects the applied load path. Identify one assumption whose failure would require a new analysis.

Worked problem 4.8

Given: A parabolic cable spans 26 m with sag 2.25 m and vertical load 2.05 kN per horizontal metre. Compute horizontal thrust and end tension.

Solution: H=wL²/(8f)=76.989 kN; vertical end reaction V=wL/2=26.650 kN; end tension sqrt(H²+V²)=81.471 kN. Changing the distributed-load basis to weight per cable length requires a different model.

Review check: State the units and verify whether the result reflects the applied load path. Identify one assumption whose failure would require a new analysis.

Worked problem 4.9

Given: A parabolic cable spans 28 m with sag 2.40 m and vertical load 2.20 kN per horizontal metre. Compute horizontal thrust and end tension.

Solution: H=wL²/(8f)=89.833 kN; vertical end reaction V=wL/2=30.800 kN; end tension sqrt(H²+V²)=94.967 kN. Changing the distributed-load basis to weight per cable length requires a different model.

Review check: State the units and verify whether the result reflects the applied load path. Identify one assumption whose failure would require a new analysis.

Worked problem 4.10

Given: A parabolic cable spans 30 m with sag 2.55 m and vertical load 2.35 kN per horizontal metre. Compute horizontal thrust and end tension.

Solution: H=wL²/(8f)=103.676 kN; vertical end reaction V=wL/2=35.250 kN; end tension sqrt(H²+V²)=109.505 kN. Changing the distributed-load basis to weight per cable length requires a different model.

Review check: State the units and verify whether the result reflects the applied load path. Identify one assumption whose failure would require a new analysis.

Worked problem 4.11

Given: A parabolic cable spans 32 m with sag 2.70 m and vertical load 2.50 kN per horizontal metre. Compute horizontal thrust and end tension.

Solution: H=wL²/(8f)=118.519 kN; vertical end reaction V=wL/2=40.000 kN; end tension sqrt(H²+V²)=125.087 kN. Changing the distributed-load basis to weight per cable length requires a different model.

Review check: State the units and verify whether the result reflects the applied load path. Identify one assumption whose failure would require a new analysis.

Worked problem 4.12

Given: A parabolic cable spans 34 m with sag 2.85 m and vertical load 2.65 kN per horizontal metre. Compute horizontal thrust and end tension.

Solution: H=wL²/(8f)=134.360 kN; vertical end reaction V=wL/2=45.050 kN; end tension sqrt(H²+V²)=141.711 kN. Changing the distributed-load basis to weight per cable length requires a different model.

Review check: State the units and verify whether the result reflects the applied load path. Identify one assumption whose failure would require a new analysis.

Unit assessment (with guidance)

Q1. Distinguish a bending frame from an ideal pin-jointed truss using a real walkway example.

Guidance: Use the governing equation and a labeled diagram from Unit 4; report direction, magnitude, units and model assumptions. Numerical assessment answers should be obtained from the worked methods, not from inspection of figures alone.

Q2. For w=1.5 kN/m, L=16 m and sag 2 m compute the parabolic cable horizontal force.

Guidance: Use the governing equation and a labeled diagram from Unit 4; report direction, magnitude, units and model assumptions. Numerical assessment answers should be obtained from the worked methods, not from inspection of figures alone.

Q3. Explain why a shallow arch or cable can create unexpectedly large horizontal support thrust.

Guidance: Use the governing equation and a labeled diagram from Unit 4; report direction, magnitude, units and model assumptions. Numerical assessment answers should be obtained from the worked methods, not from inspection of figures alone.

Q4. Trace the load path from a precast slab panel through girder, column, footing and supporting ground.

Guidance: Use the governing equation and a labeled diagram from Unit 4; report direction, magnitude, units and model assumptions. Numerical assessment answers should be obtained from the worked methods, not from inspection of figures alone.

Interactive experiment

Open [simulation 4](../simulations/simulation-04.html) and vary at least three parameters. Create an observation table from the CSV. Describe one change for which the response is nonlinear, ill-conditioned or physically infeasible.

End-of-unit checklist

• External forces and reactions identified on isolated body.

• Dimensions, datums, load units and sign convention documented.

• Vertical/horizontal resultant and moment balance checked separately.

• Real-world limits and uncertainty recorded before any design interpretation.

1. [Centroids, second moments and composite cross sections](units/unit-05.md)