Trusses, Joints, Sections and Idealization

An ideal truss uses straight two-force members joined at frictionless pins with loads applied only at joints. Under those restrictions members carry axial tension or compression, allowing efficient load transfer with less material than a deep solid member. The restrictions are strong: roofs often carry purlin loads between joints, connections have eccentricity, and out-of-plane instability may govern. Statics can determine nominal member forces, but not member capacity, buckling, corrosion or fabrication safety. A civil engineer must distinguish an educational truss calculation from construction design.

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 trusses, joints, sections and idealization.

• 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 3.1: Two-force members and truss geometry

A force at each end of an isolated pin-connected member must be collinear with the member when no other load or couple acts on the member. This result justifies labeling each ideal truss member with one unknown axial magnitude.

Governing relationships

Planar joint equilibrium gives ΣFx=0 and ΣFy=0 at each joint. Member directions follow measured joint coordinates. Tension is conventionally positive away from the joint.

Worked engineering example

At a symmetric roof-truss apex with vertical load 30 kN and two equal diagonals at 45° above horizontal, each diagonal must deliver 30/(2sin45°)=21.21 kN vertical-supporting compression magnitude.

Model audit and common error

If a panel is loaded between joints, the ideal two-force model may be invalid for that member.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving two-force members and truss geometry. 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 trusses, joints, sections and idealization, 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 3.2: Method of joints and zero-force members

The method of joints proceeds through joints with at most two unresolved member forces. Zero-force deductions can simplify a truss, but they apply only under their stated collinearity and load assumptions.

Governing relationships

For a joint with exactly two non-collinear members and no external force or support reaction, both member forces are zero. With three members, two collinear and no external force, the non-collinear member is zero.

Worked engineering example

A truss panel contains two diagonal members meeting at an unloaded lower joint and no vertical support there. If they are non-collinear and there are only two members, equilibrium forces both to zero in the ideal truss configuration.

Model audit and common error

Do not declare a member redundant without inspecting load applications, connection moments and geometric assumptions.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving method of joints and zero-force members. 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 trusses, joints, sections and idealization, 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 3.3: Method of sections and cut-force verification

To find selected forces deep inside a truss, cut through no more than three unknown members in a planar determinate model and equilibrate one side. Taking moments about an intersection can eliminate two unknown forces.

Governing relationships

Use ΣFx=ΣFy=ΣM=0 for one side of a cut, assigning trial directions to cut forces. A negative solved tension indicates compression under the chosen convention.

Worked engineering example

On a 4 m span with a simply supported 12 kN midspan roof load, reactions are 6 kN each. A section isolating an outer triangle must reproduce the 6 kN external vertical support through its cut-member force components.

Model audit and common error

Draw cut forces on one side only; otherwise internal forces are counted twice.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving method of sections and cut-force verification. 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 trusses, joints, sections and idealization, 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 3.4: Planar determinacy and member stability

For a simple planar pin-jointed truss, the count m+r=2j is a necessary but not sufficient condition for static determinacy; geometry can still be unstable. Compression-member buckling must be evaluated separately using member stiffness and end restraint.

Governing relationships

Here m is number of truss members, r support-reaction components and j joints. A simple determinate truss typically has m=2j−3 with three independent external reaction components.

Worked engineering example

A four-joint rectangle without a diagonal has four perimeter members, so m=4 while 2j−3=5. With ordinary pin joints the assembly is a mechanism until stabilized by a suitable diagonal or other restraint.

Model audit and common error

A high member force is not evidence of safe capacity; this chapter solves demand only.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving planar determinacy and member stability. 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 trusses, joints, sections and idealization, 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

Lightweight roof truss: document node locations, member labels, support types and roof-load transfer to joints. Compare two truss heights and explain how geometry changes axial force even under identical total gravity load.

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 3.1

Given: A symmetric two-member pin-jointed roof apex supports 12 kN downward. Each diagonal makes 25° with horizontal. Find diagonal axial-force magnitude.

Solution: At the apex each diagonal contributes an upward vertical component |N|sinθ. Balance 2|N|sinθ=P gives |N|=14.197 kN, compression in the usual roof arrangement. The ideal calculation assumes load applied at the joint, no joint couple and identical member directions.

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 3.2

Given: A symmetric two-member pin-jointed roof apex supports 15 kN downward. Each diagonal makes 28° with horizontal. Find diagonal axial-force magnitude.

Solution: At the apex each diagonal contributes an upward vertical component |N|sinθ. Balance 2|N|sinθ=P gives |N|=15.975 kN, compression in the usual roof arrangement. The ideal calculation assumes load applied at the joint, no joint couple and identical member directions.

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 3.3

Given: A symmetric two-member pin-jointed roof apex supports 18 kN downward. Each diagonal makes 31° with horizontal. Find diagonal axial-force magnitude.

Solution: At the apex each diagonal contributes an upward vertical component |N|sinθ. Balance 2|N|sinθ=P gives |N|=17.474 kN, compression in the usual roof arrangement. The ideal calculation assumes load applied at the joint, no joint couple and identical member directions.

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 3.4

Given: A symmetric two-member pin-jointed roof apex supports 21 kN downward. Each diagonal makes 34° with horizontal. Find diagonal axial-force magnitude.

Solution: At the apex each diagonal contributes an upward vertical component |N|sinθ. Balance 2|N|sinθ=P gives |N|=18.777 kN, compression in the usual roof arrangement. The ideal calculation assumes load applied at the joint, no joint couple and identical member directions.

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 3.5

Given: A symmetric two-member pin-jointed roof apex supports 24 kN downward. Each diagonal makes 37° with horizontal. Find diagonal axial-force magnitude.

Solution: At the apex each diagonal contributes an upward vertical component |N|sinθ. Balance 2|N|sinθ=P gives |N|=19.940 kN, compression in the usual roof arrangement. The ideal calculation assumes load applied at the joint, no joint couple and identical member directions.

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 3.6

Given: A symmetric two-member pin-jointed roof apex supports 27 kN downward. Each diagonal makes 40° with horizontal. Find diagonal axial-force magnitude.

Solution: At the apex each diagonal contributes an upward vertical component |N|sinθ. Balance 2|N|sinθ=P gives |N|=21.002 kN, compression in the usual roof arrangement. The ideal calculation assumes load applied at the joint, no joint couple and identical member directions.

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 3.7

Given: A symmetric two-member pin-jointed roof apex supports 30 kN downward. Each diagonal makes 43° with horizontal. Find diagonal axial-force magnitude.

Solution: At the apex each diagonal contributes an upward vertical component |N|sinθ. Balance 2|N|sinθ=P gives |N|=21.994 kN, compression in the usual roof arrangement. The ideal calculation assumes load applied at the joint, no joint couple and identical member directions.

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 3.8

Given: A symmetric two-member pin-jointed roof apex supports 33 kN downward. Each diagonal makes 46° with horizontal. Find diagonal axial-force magnitude.

Solution: At the apex each diagonal contributes an upward vertical component |N|sinθ. Balance 2|N|sinθ=P gives |N|=22.938 kN, compression in the usual roof arrangement. The ideal calculation assumes load applied at the joint, no joint couple and identical member directions.

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 3.9

Given: A symmetric two-member pin-jointed roof apex supports 36 kN downward. Each diagonal makes 49° with horizontal. Find diagonal axial-force magnitude.

Solution: At the apex each diagonal contributes an upward vertical component |N|sinθ. Balance 2|N|sinθ=P gives |N|=23.850 kN, compression in the usual roof arrangement. The ideal calculation assumes load applied at the joint, no joint couple and identical member directions.

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 3.10

Given: A symmetric two-member pin-jointed roof apex supports 39 kN downward. Each diagonal makes 52° with horizontal. Find diagonal axial-force magnitude.

Solution: At the apex each diagonal contributes an upward vertical component |N|sinθ. Balance 2|N|sinθ=P gives |N|=24.746 kN, compression in the usual roof arrangement. The ideal calculation assumes load applied at the joint, no joint couple and identical member directions.

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 3.11

Given: A symmetric two-member pin-jointed roof apex supports 42 kN downward. Each diagonal makes 55° with horizontal. Find diagonal axial-force magnitude.

Solution: At the apex each diagonal contributes an upward vertical component |N|sinθ. Balance 2|N|sinθ=P gives |N|=25.636 kN, compression in the usual roof arrangement. The ideal calculation assumes load applied at the joint, no joint couple and identical member directions.

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 3.12

Given: A symmetric two-member pin-jointed roof apex supports 45 kN downward. Each diagonal makes 58° with horizontal. Find diagonal axial-force magnitude.

Solution: At the apex each diagonal contributes an upward vertical component |N|sinθ. Balance 2|N|sinθ=P gives |N|=26.532 kN, compression in the usual roof arrangement. The ideal calculation assumes load applied at the joint, no joint couple and identical member directions.

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. Under which assumptions does a truss member act as a two-force element?

Guidance: Use the governing equation and a labeled diagram from Unit 3; report direction, magnitude, units and model assumptions. Numerical assessment answers should be obtained from the worked methods, not from inspection of figures alone.

Q2. A roof apex is loaded 20 kN with diagonals at 30°. Find each axial magnitude; classify axial action for a typical symmetric roof arrangement.

Guidance: Use the governing equation and a labeled diagram from Unit 3; report direction, magnitude, units and model assumptions. Numerical assessment answers should be obtained from the worked methods, not from inspection of figures alone.

Q3. Use an unloaded truss joint to state one valid zero-force member rule and its exception.

Guidance: Use the governing equation and a labeled diagram from Unit 3; report direction, magnitude, units and model assumptions. Numerical assessment answers should be obtained from the worked methods, not from inspection of figures alone.

Q4. Why does m+r=2j fail to guarantee stability for every planar truss geometry?

Guidance: Use the governing equation and a labeled diagram from Unit 3; 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 3](../simulations/simulation-03.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 3: Trusses, joints, sections and idealization

An ideal truss uses straight two-force members joined at frictionless pins with loads applied only at joints. Under those restrictions members carry axial tension or compression, allowing efficient load transfer with less material than a deep solid member. The restrictions are strong: roofs often carry purlin loads between joints, connections have eccentricity, and out-of-plane instability may govern. Statics can determine nominal member forces, but not member capacity, buckling, corrosion or fabrication safety. A civil engineer must distinguish an educational truss calculation from construction design.

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 trusses, joints, sections and idealization.

• 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 3.1: Two-force members and truss geometry

A force at each end of an isolated pin-connected member must be collinear with the member when no other load or couple acts on the member. This result justifies labeling each ideal truss member with one unknown axial magnitude.

Governing relationships

Planar joint equilibrium gives ΣFx=0 and ΣFy=0 at each joint. Member directions follow measured joint coordinates. Tension is conventionally positive away from the joint.

Worked engineering example

At a symmetric roof-truss apex with vertical load 30 kN and two equal diagonals at 45° above horizontal, each diagonal must deliver 30/(2sin45°)=21.21 kN vertical-supporting compression magnitude.

Model audit and common error

If a panel is loaded between joints, the ideal two-force model may be invalid for that member.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving two-force members and truss geometry. 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 trusses, joints, sections and idealization, 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 3.2: Method of joints and zero-force members

The method of joints proceeds through joints with at most two unresolved member forces. Zero-force deductions can simplify a truss, but they apply only under their stated collinearity and load assumptions.

Governing relationships

For a joint with exactly two non-collinear members and no external force or support reaction, both member forces are zero. With three members, two collinear and no external force, the non-collinear member is zero.

Worked engineering example

A truss panel contains two diagonal members meeting at an unloaded lower joint and no vertical support there. If they are non-collinear and there are only two members, equilibrium forces both to zero in the ideal truss configuration.

Model audit and common error

Do not declare a member redundant without inspecting load applications, connection moments and geometric assumptions.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving method of joints and zero-force members. 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 trusses, joints, sections and idealization, 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 3.3: Method of sections and cut-force verification

To find selected forces deep inside a truss, cut through no more than three unknown members in a planar determinate model and equilibrate one side. Taking moments about an intersection can eliminate two unknown forces.

Governing relationships

Use ΣFx=ΣFy=ΣM=0 for one side of a cut, assigning trial directions to cut forces. A negative solved tension indicates compression under the chosen convention.

Worked engineering example

On a 4 m span with a simply supported 12 kN midspan roof load, reactions are 6 kN each. A section isolating an outer triangle must reproduce the 6 kN external vertical support through its cut-member force components.

Model audit and common error

Draw cut forces on one side only; otherwise internal forces are counted twice.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving method of sections and cut-force verification. 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 trusses, joints, sections and idealization, 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 3.4: Planar determinacy and member stability

For a simple planar pin-jointed truss, the count m+r=2j is a necessary but not sufficient condition for static determinacy; geometry can still be unstable. Compression-member buckling must be evaluated separately using member stiffness and end restraint.

Governing relationships

Here m is number of truss members, r support-reaction components and j joints. A simple determinate truss typically has m=2j−3 with three independent external reaction components.

Worked engineering example

A four-joint rectangle without a diagonal has four perimeter members, so m=4 while 2j−3=5. With ordinary pin joints the assembly is a mechanism until stabilized by a suitable diagonal or other restraint.

Model audit and common error

A high member force is not evidence of safe capacity; this chapter solves demand only.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving planar determinacy and member stability. 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 trusses, joints, sections and idealization, 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

Lightweight roof truss: document node locations, member labels, support types and roof-load transfer to joints. Compare two truss heights and explain how geometry changes axial force even under identical total gravity load.

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 3.1

Given: A symmetric two-member pin-jointed roof apex supports 12 kN downward. Each diagonal makes 25° with horizontal. Find diagonal axial-force magnitude.

Solution: At the apex each diagonal contributes an upward vertical component |N|sinθ. Balance 2|N|sinθ=P gives |N|=14.197 kN, compression in the usual roof arrangement. The ideal calculation assumes load applied at the joint, no joint couple and identical member directions.

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 3.2

Given: A symmetric two-member pin-jointed roof apex supports 15 kN downward. Each diagonal makes 28° with horizontal. Find diagonal axial-force magnitude.

Solution: At the apex each diagonal contributes an upward vertical component |N|sinθ. Balance 2|N|sinθ=P gives |N|=15.975 kN, compression in the usual roof arrangement. The ideal calculation assumes load applied at the joint, no joint couple and identical member directions.

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 3.3

Given: A symmetric two-member pin-jointed roof apex supports 18 kN downward. Each diagonal makes 31° with horizontal. Find diagonal axial-force magnitude.

Solution: At the apex each diagonal contributes an upward vertical component |N|sinθ. Balance 2|N|sinθ=P gives |N|=17.474 kN, compression in the usual roof arrangement. The ideal calculation assumes load applied at the joint, no joint couple and identical member directions.

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 3.4

Given: A symmetric two-member pin-jointed roof apex supports 21 kN downward. Each diagonal makes 34° with horizontal. Find diagonal axial-force magnitude.

Solution: At the apex each diagonal contributes an upward vertical component |N|sinθ. Balance 2|N|sinθ=P gives |N|=18.777 kN, compression in the usual roof arrangement. The ideal calculation assumes load applied at the joint, no joint couple and identical member directions.

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 3.5

Given: A symmetric two-member pin-jointed roof apex supports 24 kN downward. Each diagonal makes 37° with horizontal. Find diagonal axial-force magnitude.

Solution: At the apex each diagonal contributes an upward vertical component |N|sinθ. Balance 2|N|sinθ=P gives |N|=19.940 kN, compression in the usual roof arrangement. The ideal calculation assumes load applied at the joint, no joint couple and identical member directions.

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 3.6

Given: A symmetric two-member pin-jointed roof apex supports 27 kN downward. Each diagonal makes 40° with horizontal. Find diagonal axial-force magnitude.

Solution: At the apex each diagonal contributes an upward vertical component |N|sinθ. Balance 2|N|sinθ=P gives |N|=21.002 kN, compression in the usual roof arrangement. The ideal calculation assumes load applied at the joint, no joint couple and identical member directions.

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 3.7

Given: A symmetric two-member pin-jointed roof apex supports 30 kN downward. Each diagonal makes 43° with horizontal. Find diagonal axial-force magnitude.

Solution: At the apex each diagonal contributes an upward vertical component |N|sinθ. Balance 2|N|sinθ=P gives |N|=21.994 kN, compression in the usual roof arrangement. The ideal calculation assumes load applied at the joint, no joint couple and identical member directions.

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 3.8

Given: A symmetric two-member pin-jointed roof apex supports 33 kN downward. Each diagonal makes 46° with horizontal. Find diagonal axial-force magnitude.

Solution: At the apex each diagonal contributes an upward vertical component |N|sinθ. Balance 2|N|sinθ=P gives |N|=22.938 kN, compression in the usual roof arrangement. The ideal calculation assumes load applied at the joint, no joint couple and identical member directions.

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 3.9

Given: A symmetric two-member pin-jointed roof apex supports 36 kN downward. Each diagonal makes 49° with horizontal. Find diagonal axial-force magnitude.

Solution: At the apex each diagonal contributes an upward vertical component |N|sinθ. Balance 2|N|sinθ=P gives |N|=23.850 kN, compression in the usual roof arrangement. The ideal calculation assumes load applied at the joint, no joint couple and identical member directions.

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 3.10

Given: A symmetric two-member pin-jointed roof apex supports 39 kN downward. Each diagonal makes 52° with horizontal. Find diagonal axial-force magnitude.

Solution: At the apex each diagonal contributes an upward vertical component |N|sinθ. Balance 2|N|sinθ=P gives |N|=24.746 kN, compression in the usual roof arrangement. The ideal calculation assumes load applied at the joint, no joint couple and identical member directions.

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 3.11

Given: A symmetric two-member pin-jointed roof apex supports 42 kN downward. Each diagonal makes 55° with horizontal. Find diagonal axial-force magnitude.

Solution: At the apex each diagonal contributes an upward vertical component |N|sinθ. Balance 2|N|sinθ=P gives |N|=25.636 kN, compression in the usual roof arrangement. The ideal calculation assumes load applied at the joint, no joint couple and identical member directions.

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 3.12

Given: A symmetric two-member pin-jointed roof apex supports 45 kN downward. Each diagonal makes 58° with horizontal. Find diagonal axial-force magnitude.

Solution: At the apex each diagonal contributes an upward vertical component |N|sinθ. Balance 2|N|sinθ=P gives |N|=26.532 kN, compression in the usual roof arrangement. The ideal calculation assumes load applied at the joint, no joint couple and identical member directions.

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. Under which assumptions does a truss member act as a two-force element?

Guidance: Use the governing equation and a labeled diagram from Unit 3; report direction, magnitude, units and model assumptions. Numerical assessment answers should be obtained from the worked methods, not from inspection of figures alone.

Q2. A roof apex is loaded 20 kN with diagonals at 30°. Find each axial magnitude; classify axial action for a typical symmetric roof arrangement.

Guidance: Use the governing equation and a labeled diagram from Unit 3; report direction, magnitude, units and model assumptions. Numerical assessment answers should be obtained from the worked methods, not from inspection of figures alone.

Q3. Use an unloaded truss joint to state one valid zero-force member rule and its exception.

Guidance: Use the governing equation and a labeled diagram from Unit 3; report direction, magnitude, units and model assumptions. Numerical assessment answers should be obtained from the worked methods, not from inspection of figures alone.

Q4. Why does m+r=2j fail to guarantee stability for every planar truss geometry?

Guidance: Use the governing equation and a labeled diagram from Unit 3; 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 3](../simulations/simulation-03.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. [Frames, cables, arches and structural load paths](units/unit-04.md)