Equilibrium, Reactions, Supports and Static Determinacy

Global equilibrium is necessary for a structure to remain at rest, but its equations do not prove that a foundation has adequate capacity or a beam has sufficient strength. Pinned supports, rollers, fixed connections and contact surfaces restrain different degrees of freedom. The first audit is a boundary-condition audit: identify which movement is resisted and what reaction the idealized support can supply. A stable determinate beam usually has sufficient independent reactions to solve force and moment balance. Real construction tolerances, restraint stiffness and settlement complicate the idealized picture; they are introduced later in structural analysis.

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 equilibrium, reactions, supports and static determinacy.

• 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 2.1: Free-body diagrams and support reactions

Isolating the member eliminates hidden internal interactions but requires every actual external force and couple to remain. Support symbols are hypotheses about constraints, not decorative drawings.

Governing relationships

In two-dimensional equilibrium, ΣFx=0, ΣFy=0 and ΣMz=0. A simple pin transmits two force components; a frictionless roller normal to a horizontal surface transmits one vertical component.

Worked engineering example

A 6 m span with a 12 kN downward point load 2 m from A has RB=12×2/6=4 kN and RA=8 kN. Recompute moments about B to independently check.

Model audit and common error

Never assign a horizontal reaction to a frictionless horizontal roller.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving free-body diagrams and support reactions. 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 equilibrium, reactions, supports and static determinacy, 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 2.2: Uniform and variable loads on simply supported beams

A loading diagram may contain permanent dead weight, temporary construction loading and localized point forces. Reactions can be found by combining equivalents, then internal sections must use the original load distribution.

Governing relationships

For constant w on a simply supported beam of length L, RA=RB=wL/2. If a point force P is at a from A, RB=wL/2+Pa/L and RA=wL/2+P(L−a)/L.

Worked engineering example

For L=8 m, w=2 kN/m and P=10 kN at a=6 m, RB=8+7.5=15.5 kN and RA=8+2.5=10.5 kN. The reactions add to 26 kN.

Model audit and common error

Carry all loads at the correct geometric positions, not at the nearest support.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving uniform and variable loads on simply supported beams. 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 equilibrium, reactions, supports and static determinacy, 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 2.3: Determinacy versus stability

The count of unknown reaction components is a useful screening tool, but geometry determines whether their directions are independent. Three nominal support reactions may still form a mechanism when their lines of action are concurrent or parallel in an unsuitable manner.

Governing relationships

Compare independent equilibrium equations with independent reaction unknowns, then examine whether a rigid-body displacement is geometrically possible. The determinant of the reaction-equilibrium matrix offers an algebraic check.

Worked engineering example

A beam with a vertical roller at each end has only two vertical reactions and no horizontal restraint. It cannot resist arbitrary horizontal loading under the ideal support model, although its vertical reactions may be calculated.

Model audit and common error

Do not confuse the ability to solve one selected loading case with stable restraint under all allowed actions.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving determinacy versus 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 equilibrium, reactions, supports and static determinacy, 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 2.4: Internal cuts and action–reaction consistency

The free-body diagram of part of a beam makes internal axial force, shear and bending moment appear on the cut. Opposite cut faces carry equal and opposite internal actions. A globally balanced structure can still experience high internal demand.

Governing relationships

For the left segment x of a simply supported uniform beam, V(x)=RA−wx and M(x)=RA x−wx²/2 when 0≤x≤L, with the chosen convention.

Worked engineering example

For L=6 m and w=4 kN/m, RA=12 kN. At x=2 m, V=12−8=4 kN and M=24−8=16 kN·m. At midspan Mmax=wL²/8=18 kN·m.

Model audit and common error

Support reactions describe external equilibrium, while section forces describe how load is transferred internally.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving internal cuts and action–reaction consistency. 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 equilibrium, reactions, supports and static determinacy, 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

Bridge-deck support audit: derive reactions for a simplified span with two point actions and a continuous self-weight. Test whether each proposed support can deliver its modeled action and document missing settlement or uplift information.

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 2.1

Given: A simply supported 4.0 m beam carries 1.50 kN/m across the span and a 5.0 kN point load at x=0.80 m. Determine end reactions.

Solution: Uniform action = wL=6.000 kN at L/2. Taking moments about A yields RB=wL/2+Pa/L=4.000 kN. Vertical balance gives RA=wL/2+P(L−a)/L=7.000 kN. Check RA+RB=11.000 kN matches wL+P.

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 2.2

Given: A simply supported 4.5 m beam carries 1.75 kN/m across the span and a 6.3 kN point load at x=1.12 m. Determine end reactions.

Solution: Uniform action = wL=7.875 kN at L/2. Taking moments about A yields RB=wL/2+Pa/L=5.513 kN. Vertical balance gives RA=wL/2+P(L−a)/L=8.662 kN. Check RA+RB=14.175 kN matches wL+P.

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 2.3

Given: A simply supported 5.0 m beam carries 2.00 kN/m across the span and a 7.6 kN point load at x=1.50 m. Determine end reactions.

Solution: Uniform action = wL=10.000 kN at L/2. Taking moments about A yields RB=wL/2+Pa/L=7.280 kN. Vertical balance gives RA=wL/2+P(L−a)/L=10.320 kN. Check RA+RB=17.600 kN matches wL+P.

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 2.4

Given: A simply supported 5.5 m beam carries 2.25 kN/m across the span and a 8.9 kN point load at x=1.93 m. Determine end reactions.

Solution: Uniform action = wL=12.375 kN at L/2. Taking moments about A yields RB=wL/2+Pa/L=9.303 kN. Vertical balance gives RA=wL/2+P(L−a)/L=11.973 kN. Check RA+RB=21.275 kN matches wL+P.

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 2.5

Given: A simply supported 6.0 m beam carries 2.50 kN/m across the span and a 10.2 kN point load at x=2.40 m. Determine end reactions.

Solution: Uniform action = wL=15.000 kN at L/2. Taking moments about A yields RB=wL/2+Pa/L=11.580 kN. Vertical balance gives RA=wL/2+P(L−a)/L=13.620 kN. Check RA+RB=25.200 kN matches wL+P.

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 2.6

Given: A simply supported 6.5 m beam carries 2.75 kN/m across the span and a 11.5 kN point load at x=2.93 m. Determine end reactions.

Solution: Uniform action = wL=17.875 kN at L/2. Taking moments about A yields RB=wL/2+Pa/L=14.113 kN. Vertical balance gives RA=wL/2+P(L−a)/L=15.262 kN. Check RA+RB=29.375 kN matches wL+P.

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 2.7

Given: A simply supported 7.0 m beam carries 3.00 kN/m across the span and a 12.8 kN point load at x=3.50 m. Determine end reactions.

Solution: Uniform action = wL=21.000 kN at L/2. Taking moments about A yields RB=wL/2+Pa/L=16.900 kN. Vertical balance gives RA=wL/2+P(L−a)/L=16.900 kN. Check RA+RB=33.800 kN matches wL+P.

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 2.8

Given: A simply supported 7.5 m beam carries 3.25 kN/m across the span and a 14.1 kN point load at x=4.12 m. Determine end reactions.

Solution: Uniform action = wL=24.375 kN at L/2. Taking moments about A yields RB=wL/2+Pa/L=19.942 kN. Vertical balance gives RA=wL/2+P(L−a)/L=18.532 kN. Check RA+RB=38.475 kN matches wL+P.

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 2.9

Given: A simply supported 8.0 m beam carries 3.50 kN/m across the span and a 15.4 kN point load at x=4.80 m. Determine end reactions.

Solution: Uniform action = wL=28.000 kN at L/2. Taking moments about A yields RB=wL/2+Pa/L=23.240 kN. Vertical balance gives RA=wL/2+P(L−a)/L=20.160 kN. Check RA+RB=43.400 kN matches wL+P.

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 2.10

Given: A simply supported 8.5 m beam carries 3.75 kN/m across the span and a 16.7 kN point load at x=1.70 m. Determine end reactions.

Solution: Uniform action = wL=31.875 kN at L/2. Taking moments about A yields RB=wL/2+Pa/L=19.277 kN. Vertical balance gives RA=wL/2+P(L−a)/L=29.297 kN. Check RA+RB=48.575 kN matches wL+P.

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 2.11

Given: A simply supported 9.0 m beam carries 4.00 kN/m across the span and a 18.0 kN point load at x=2.25 m. Determine end reactions.

Solution: Uniform action = wL=36.000 kN at L/2. Taking moments about A yields RB=wL/2+Pa/L=22.500 kN. Vertical balance gives RA=wL/2+P(L−a)/L=31.500 kN. Check RA+RB=54.000 kN matches wL+P.

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 2.12

Given: A simply supported 9.5 m beam carries 4.25 kN/m across the span and a 19.3 kN point load at x=2.85 m. Determine end reactions.

Solution: Uniform action = wL=40.375 kN at L/2. Taking moments about A yields RB=wL/2+Pa/L=25.977 kN. Vertical balance gives RA=wL/2+P(L−a)/L=33.697 kN. Check RA+RB=59.675 kN matches wL+P.

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. Sketch and label the unknown reactions for a pin–roller supported bridge span. Explain why one reaction component is absent.

Guidance: Use the governing equation and a labeled diagram from Unit 2; 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 6 m beam carries 12 kN at x=4 m. Find both end reactions and check moments about each end.

Guidance: Use the governing equation and a labeled diagram from Unit 2; 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 nominal three-reaction planar structure can still be geometrically unstable.

Guidance: Use the governing equation and a labeled diagram from Unit 2; report direction, magnitude, units and model assumptions. Numerical assessment answers should be obtained from the worked methods, not from inspection of figures alone.

Q4. Describe why a simply supported beam can meet global equilibrium while its bending demand exceeds material capacity.

Guidance: Use the governing equation and a labeled diagram from Unit 2; 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 2](../simulations/simulation-02.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 2: Equilibrium, reactions, supports and static determinacy

Global equilibrium is necessary for a structure to remain at rest, but its equations do not prove that a foundation has adequate capacity or a beam has sufficient strength. Pinned supports, rollers, fixed connections and contact surfaces restrain different degrees of freedom. The first audit is a boundary-condition audit: identify which movement is resisted and what reaction the idealized support can supply. A stable determinate beam usually has sufficient independent reactions to solve force and moment balance. Real construction tolerances, restraint stiffness and settlement complicate the idealized picture; they are introduced later in structural analysis.

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 equilibrium, reactions, supports and static determinacy.

• 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 2.1: Free-body diagrams and support reactions

Isolating the member eliminates hidden internal interactions but requires every actual external force and couple to remain. Support symbols are hypotheses about constraints, not decorative drawings.

Governing relationships

In two-dimensional equilibrium, ΣFx=0, ΣFy=0 and ΣMz=0. A simple pin transmits two force components; a frictionless roller normal to a horizontal surface transmits one vertical component.

Worked engineering example

A 6 m span with a 12 kN downward point load 2 m from A has RB=12×2/6=4 kN and RA=8 kN. Recompute moments about B to independently check.

Model audit and common error

Never assign a horizontal reaction to a frictionless horizontal roller.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving free-body diagrams and support reactions. 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 equilibrium, reactions, supports and static determinacy, 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 2.2: Uniform and variable loads on simply supported beams

A loading diagram may contain permanent dead weight, temporary construction loading and localized point forces. Reactions can be found by combining equivalents, then internal sections must use the original load distribution.

Governing relationships

For constant w on a simply supported beam of length L, RA=RB=wL/2. If a point force P is at a from A, RB=wL/2+Pa/L and RA=wL/2+P(L−a)/L.

Worked engineering example

For L=8 m, w=2 kN/m and P=10 kN at a=6 m, RB=8+7.5=15.5 kN and RA=8+2.5=10.5 kN. The reactions add to 26 kN.

Model audit and common error

Carry all loads at the correct geometric positions, not at the nearest support.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving uniform and variable loads on simply supported beams. 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 equilibrium, reactions, supports and static determinacy, 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 2.3: Determinacy versus stability

The count of unknown reaction components is a useful screening tool, but geometry determines whether their directions are independent. Three nominal support reactions may still form a mechanism when their lines of action are concurrent or parallel in an unsuitable manner.

Governing relationships

Compare independent equilibrium equations with independent reaction unknowns, then examine whether a rigid-body displacement is geometrically possible. The determinant of the reaction-equilibrium matrix offers an algebraic check.

Worked engineering example

A beam with a vertical roller at each end has only two vertical reactions and no horizontal restraint. It cannot resist arbitrary horizontal loading under the ideal support model, although its vertical reactions may be calculated.

Model audit and common error

Do not confuse the ability to solve one selected loading case with stable restraint under all allowed actions.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving determinacy versus 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 equilibrium, reactions, supports and static determinacy, 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 2.4: Internal cuts and action–reaction consistency

The free-body diagram of part of a beam makes internal axial force, shear and bending moment appear on the cut. Opposite cut faces carry equal and opposite internal actions. A globally balanced structure can still experience high internal demand.

Governing relationships

For the left segment x of a simply supported uniform beam, V(x)=RA−wx and M(x)=RA x−wx²/2 when 0≤x≤L, with the chosen convention.

Worked engineering example

For L=6 m and w=4 kN/m, RA=12 kN. At x=2 m, V=12−8=4 kN and M=24−8=16 kN·m. At midspan Mmax=wL²/8=18 kN·m.

Model audit and common error

Support reactions describe external equilibrium, while section forces describe how load is transferred internally.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving internal cuts and action–reaction consistency. 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 equilibrium, reactions, supports and static determinacy, 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

Bridge-deck support audit: derive reactions for a simplified span with two point actions and a continuous self-weight. Test whether each proposed support can deliver its modeled action and document missing settlement or uplift information.

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 2.1

Given: A simply supported 4.0 m beam carries 1.50 kN/m across the span and a 5.0 kN point load at x=0.80 m. Determine end reactions.

Solution: Uniform action = wL=6.000 kN at L/2. Taking moments about A yields RB=wL/2+Pa/L=4.000 kN. Vertical balance gives RA=wL/2+P(L−a)/L=7.000 kN. Check RA+RB=11.000 kN matches wL+P.

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 2.2

Given: A simply supported 4.5 m beam carries 1.75 kN/m across the span and a 6.3 kN point load at x=1.12 m. Determine end reactions.

Solution: Uniform action = wL=7.875 kN at L/2. Taking moments about A yields RB=wL/2+Pa/L=5.513 kN. Vertical balance gives RA=wL/2+P(L−a)/L=8.662 kN. Check RA+RB=14.175 kN matches wL+P.

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 2.3

Given: A simply supported 5.0 m beam carries 2.00 kN/m across the span and a 7.6 kN point load at x=1.50 m. Determine end reactions.

Solution: Uniform action = wL=10.000 kN at L/2. Taking moments about A yields RB=wL/2+Pa/L=7.280 kN. Vertical balance gives RA=wL/2+P(L−a)/L=10.320 kN. Check RA+RB=17.600 kN matches wL+P.

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 2.4

Given: A simply supported 5.5 m beam carries 2.25 kN/m across the span and a 8.9 kN point load at x=1.93 m. Determine end reactions.

Solution: Uniform action = wL=12.375 kN at L/2. Taking moments about A yields RB=wL/2+Pa/L=9.303 kN. Vertical balance gives RA=wL/2+P(L−a)/L=11.973 kN. Check RA+RB=21.275 kN matches wL+P.

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 2.5

Given: A simply supported 6.0 m beam carries 2.50 kN/m across the span and a 10.2 kN point load at x=2.40 m. Determine end reactions.

Solution: Uniform action = wL=15.000 kN at L/2. Taking moments about A yields RB=wL/2+Pa/L=11.580 kN. Vertical balance gives RA=wL/2+P(L−a)/L=13.620 kN. Check RA+RB=25.200 kN matches wL+P.

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 2.6

Given: A simply supported 6.5 m beam carries 2.75 kN/m across the span and a 11.5 kN point load at x=2.93 m. Determine end reactions.

Solution: Uniform action = wL=17.875 kN at L/2. Taking moments about A yields RB=wL/2+Pa/L=14.113 kN. Vertical balance gives RA=wL/2+P(L−a)/L=15.262 kN. Check RA+RB=29.375 kN matches wL+P.

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 2.7

Given: A simply supported 7.0 m beam carries 3.00 kN/m across the span and a 12.8 kN point load at x=3.50 m. Determine end reactions.

Solution: Uniform action = wL=21.000 kN at L/2. Taking moments about A yields RB=wL/2+Pa/L=16.900 kN. Vertical balance gives RA=wL/2+P(L−a)/L=16.900 kN. Check RA+RB=33.800 kN matches wL+P.

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 2.8

Given: A simply supported 7.5 m beam carries 3.25 kN/m across the span and a 14.1 kN point load at x=4.12 m. Determine end reactions.

Solution: Uniform action = wL=24.375 kN at L/2. Taking moments about A yields RB=wL/2+Pa/L=19.942 kN. Vertical balance gives RA=wL/2+P(L−a)/L=18.532 kN. Check RA+RB=38.475 kN matches wL+P.

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 2.9

Given: A simply supported 8.0 m beam carries 3.50 kN/m across the span and a 15.4 kN point load at x=4.80 m. Determine end reactions.

Solution: Uniform action = wL=28.000 kN at L/2. Taking moments about A yields RB=wL/2+Pa/L=23.240 kN. Vertical balance gives RA=wL/2+P(L−a)/L=20.160 kN. Check RA+RB=43.400 kN matches wL+P.

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 2.10

Given: A simply supported 8.5 m beam carries 3.75 kN/m across the span and a 16.7 kN point load at x=1.70 m. Determine end reactions.

Solution: Uniform action = wL=31.875 kN at L/2. Taking moments about A yields RB=wL/2+Pa/L=19.277 kN. Vertical balance gives RA=wL/2+P(L−a)/L=29.297 kN. Check RA+RB=48.575 kN matches wL+P.

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 2.11

Given: A simply supported 9.0 m beam carries 4.00 kN/m across the span and a 18.0 kN point load at x=2.25 m. Determine end reactions.

Solution: Uniform action = wL=36.000 kN at L/2. Taking moments about A yields RB=wL/2+Pa/L=22.500 kN. Vertical balance gives RA=wL/2+P(L−a)/L=31.500 kN. Check RA+RB=54.000 kN matches wL+P.

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 2.12

Given: A simply supported 9.5 m beam carries 4.25 kN/m across the span and a 19.3 kN point load at x=2.85 m. Determine end reactions.

Solution: Uniform action = wL=40.375 kN at L/2. Taking moments about A yields RB=wL/2+Pa/L=25.977 kN. Vertical balance gives RA=wL/2+P(L−a)/L=33.697 kN. Check RA+RB=59.675 kN matches wL+P.

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. Sketch and label the unknown reactions for a pin–roller supported bridge span. Explain why one reaction component is absent.

Guidance: Use the governing equation and a labeled diagram from Unit 2; 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 6 m beam carries 12 kN at x=4 m. Find both end reactions and check moments about each end.

Guidance: Use the governing equation and a labeled diagram from Unit 2; 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 nominal three-reaction planar structure can still be geometrically unstable.

Guidance: Use the governing equation and a labeled diagram from Unit 2; report direction, magnitude, units and model assumptions. Numerical assessment answers should be obtained from the worked methods, not from inspection of figures alone.

Q4. Describe why a simply supported beam can meet global equilibrium while its bending demand exceeds material capacity.

Guidance: Use the governing equation and a labeled diagram from Unit 2; 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 2](../simulations/simulation-02.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. [Trusses, joints, sections and idealization](units/unit-03.md)