Friction, Sliding, Overturning and Stability

Foundation contact and retaining systems can fail by sliding, overturning, loss of bearing contact or internal structural distress. These mechanisms are different; passing one check does not prove the system safe. A rigid-body sliding calculation is governed by the maximum friction or other available resistance, whereas overturning depends on the relative location of stabilizing and destabilizing resultants. Soil–structure interaction, groundwater, passive pressure, anchorage and applicable factors require specialist treatment beyond introductory statics.

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 friction, sliding, overturning and stability.

• 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 6.1: Coulomb friction with unilateral contact

A dry contact model bounds tangential friction by μN. In static equilibrium the friction magnitude adjusts to demand up to its upper limit. A compressive normal reaction N cannot become negative when contact cannot carry tension.

Governing relationships

For a simple horizontal base without other vertical loads, N=W and static friction may satisfy |F|≤μsN. At impending slip, |F|=μsN under the idealized model.

Worked engineering example

A 60 kN gravity load on a level base with coefficient μ=0.4 offers at most 24 kN nominal friction resistance. A 10 kN horizontal demand requires 10 kN static friction, not automatically 24 kN.

Model audit and common error

Friction coefficient values depend on interface condition, moisture, contamination and testing; do not treat classroom numbers as design guarantees.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving coulomb friction with unilateral contact. 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 friction, sliding, overturning and stability, 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 6.2: Sliding checks and load combinations

Sliding equilibrium compares driving and resisting horizontal actions. If external anchorage, passive pressure or geometry contributes resistance, distinguish which mechanisms are reliably mobilized and compatible.

Governing relationships

For an isolated rigid block with H acting horizontally and μW available, the nominal ratio against sliding is μW/H for H>0. A code-based check may use different resistance and load factors.

Worked engineering example

With W=80 kN, μ=0.35 and H=20 kN, nominal available friction is 28 kN and ratio is 1.40. That simple ratio does not certify geotechnical safety.

Model audit and common error

Do not add incompatible peak resistances from different movement states without engineering justification.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving sliding checks and load combinations. 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 friction, sliding, overturning and stability, 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 6.3: Overturning moments and contact resultant

Overturning resistance is a moment balance around a specified toe. A nominal positive stabilizing moment does not alone ensure compression across the entire base; resultant eccentricity can create partial contact.

Governing relationships

For base width B, downward load W centered at B/2 and lateral force H at height h, stabilizing moment about the toe is WB/2, overturning moment is Hh. Resultant eccentricity e=M/W in a simple vertical-resultant model.

Worked engineering example

For B=2 m, W=100 kN, H=20 kN at h=3 m, resisting moment WB/2=100 kN·m and overturning moment Hh=60 kN·m. Nominal toe ratio is 1.67.

Model audit and common error

Local bearing pressures, soil capacity and load factors are beyond the friction-and-moment calculation.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving overturning moments and contact resultant. 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 friction, sliding, overturning and stability, 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 6.4: Retaining-wall pressures and conceptual stability

A retained-soil wall experiences lateral stress that often increases with depth, producing a triangular resultant for simple idealizations. Real earth-pressure distribution depends on soil strength, groundwater, wall movement, surcharge and drainage.

Governing relationships

Under a uniform effective earth-pressure gradient increasing from 0 at ground to pmax at depth H, total lateral resultant per metre of wall is P=pmax H/2, acting H/3 above the base.

Worked engineering example

With H=4 m and base lateral pressure pmax=24 kPa, the ideal triangular resultant is 48 kN per metre wall, at 1.333 m above the base. Units: kPa×m=kN/m.

Model audit and common error

This pressure shape is a teaching assumption; real wall design requires geotechnical investigation and appropriate earth-pressure theory.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving retaining-wall pressures and conceptual 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 friction, sliding, overturning and stability, 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

Conceptual retaining wall: compare horizontal sliding tendency, overturning about the toe, and effect of a higher groundwater level. Identify missing geotechnical inputs before any actual wall 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 6.1

Given: A simple rigid block has W=80.0 kN, base width B=2.00 m, horizontal push H=14.0 kN at height h=1.50 m and assumed friction coefficient μ=0.25. Compare nominal sliding and toe-overturning actions.

Solution: Nominal available friction μW=20.000 kN versus demand H=14.000 kN. Toe resisting moment WB/2=80.000 kN·m; overturning moment Hh=21.000 kN·m. Comparisons 1.429 and 3.810 are unfactored classroom ratios, not design factors of safety.

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 6.2

Given: A simple rigid block has W=86.0 kN, base width B=2.15 m, horizontal push H=16.0 kN at height h=1.65 m and assumed friction coefficient μ=0.27. Compare nominal sliding and toe-overturning actions.

Solution: Nominal available friction μW=23.220 kN versus demand H=16.000 kN. Toe resisting moment WB/2=92.450 kN·m; overturning moment Hh=26.400 kN·m. Comparisons 1.451 and 3.502 are unfactored classroom ratios, not design factors of safety.

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 6.3

Given: A simple rigid block has W=92.0 kN, base width B=2.30 m, horizontal push H=18.0 kN at height h=1.80 m and assumed friction coefficient μ=0.29. Compare nominal sliding and toe-overturning actions.

Solution: Nominal available friction μW=26.680 kN versus demand H=18.000 kN. Toe resisting moment WB/2=105.800 kN·m; overturning moment Hh=32.400 kN·m. Comparisons 1.482 and 3.265 are unfactored classroom ratios, not design factors of safety.

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 6.4

Given: A simple rigid block has W=98.0 kN, base width B=2.45 m, horizontal push H=20.0 kN at height h=1.95 m and assumed friction coefficient μ=0.31. Compare nominal sliding and toe-overturning actions.

Solution: Nominal available friction μW=30.380 kN versus demand H=20.000 kN. Toe resisting moment WB/2=120.050 kN·m; overturning moment Hh=39.000 kN·m. Comparisons 1.519 and 3.078 are unfactored classroom ratios, not design factors of safety.

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 6.5

Given: A simple rigid block has W=104.0 kN, base width B=2.60 m, horizontal push H=22.0 kN at height h=2.10 m and assumed friction coefficient μ=0.33. Compare nominal sliding and toe-overturning actions.

Solution: Nominal available friction μW=34.320 kN versus demand H=22.000 kN. Toe resisting moment WB/2=135.200 kN·m; overturning moment Hh=46.200 kN·m. Comparisons 1.560 and 2.926 are unfactored classroom ratios, not design factors of safety.

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 6.6

Given: A simple rigid block has W=110.0 kN, base width B=2.75 m, horizontal push H=24.0 kN at height h=2.25 m and assumed friction coefficient μ=0.25. Compare nominal sliding and toe-overturning actions.

Solution: Nominal available friction μW=27.500 kN versus demand H=24.000 kN. Toe resisting moment WB/2=151.250 kN·m; overturning moment Hh=54.000 kN·m. Comparisons 1.146 and 2.801 are unfactored classroom ratios, not design factors of safety.

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 6.7

Given: A simple rigid block has W=116.0 kN, base width B=2.90 m, horizontal push H=26.0 kN at height h=2.40 m and assumed friction coefficient μ=0.27. Compare nominal sliding and toe-overturning actions.

Solution: Nominal available friction μW=31.320 kN versus demand H=26.000 kN. Toe resisting moment WB/2=168.200 kN·m; overturning moment Hh=62.400 kN·m. Comparisons 1.205 and 2.696 are unfactored classroom ratios, not design factors of safety.

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 6.8

Given: A simple rigid block has W=122.0 kN, base width B=3.05 m, horizontal push H=28.0 kN at height h=2.55 m and assumed friction coefficient μ=0.29. Compare nominal sliding and toe-overturning actions.

Solution: Nominal available friction μW=35.380 kN versus demand H=28.000 kN. Toe resisting moment WB/2=186.050 kN·m; overturning moment Hh=71.400 kN·m. Comparisons 1.264 and 2.606 are unfactored classroom ratios, not design factors of safety.

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 6.9

Given: A simple rigid block has W=128.0 kN, base width B=3.20 m, horizontal push H=30.0 kN at height h=2.70 m and assumed friction coefficient μ=0.31. Compare nominal sliding and toe-overturning actions.

Solution: Nominal available friction μW=39.680 kN versus demand H=30.000 kN. Toe resisting moment WB/2=204.800 kN·m; overturning moment Hh=81.000 kN·m. Comparisons 1.323 and 2.528 are unfactored classroom ratios, not design factors of safety.

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 6.10

Given: A simple rigid block has W=134.0 kN, base width B=3.35 m, horizontal push H=32.0 kN at height h=2.85 m and assumed friction coefficient μ=0.33. Compare nominal sliding and toe-overturning actions.

Solution: Nominal available friction μW=44.220 kN versus demand H=32.000 kN. Toe resisting moment WB/2=224.450 kN·m; overturning moment Hh=91.200 kN·m. Comparisons 1.382 and 2.461 are unfactored classroom ratios, not design factors of safety.

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 6.11

Given: A simple rigid block has W=140.0 kN, base width B=3.50 m, horizontal push H=34.0 kN at height h=3.00 m and assumed friction coefficient μ=0.25. Compare nominal sliding and toe-overturning actions.

Solution: Nominal available friction μW=35.000 kN versus demand H=34.000 kN. Toe resisting moment WB/2=245.000 kN·m; overturning moment Hh=102.000 kN·m. Comparisons 1.029 and 2.402 are unfactored classroom ratios, not design factors of safety.

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 6.12

Given: A simple rigid block has W=146.0 kN, base width B=3.65 m, horizontal push H=36.0 kN at height h=3.15 m and assumed friction coefficient μ=0.27. Compare nominal sliding and toe-overturning actions.

Solution: Nominal available friction μW=39.420 kN versus demand H=36.000 kN. Toe resisting moment WB/2=266.450 kN·m; overturning moment Hh=113.400 kN·m. Comparisons 1.095 and 2.350 are unfactored classroom ratios, not design factors of safety.

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. Explain why static friction does not always equal μN.

Guidance: Use the governing equation and a labeled diagram from Unit 6; report direction, magnitude, units and model assumptions. Numerical assessment answers should be obtained from the worked methods, not from inspection of figures alone.

Q2. With W=75 kN and μ=0.4, calculate ideal maximum base friction. What information is still missing for design?

Guidance: Use the governing equation and a labeled diagram from Unit 6; report direction, magnitude, units and model assumptions. Numerical assessment answers should be obtained from the worked methods, not from inspection of figures alone.

Q3. A wall has 100 kN weight over a 2 m base and 25 kN lateral action at 3 m. Compare opposing toe moments.

Guidance: Use the governing equation and a labeled diagram from Unit 6; 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 a positive nominal overturning ratio not prove satisfactory soil bearing pressure?

Guidance: Use the governing equation and a labeled diagram from Unit 6; 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 6](../simulations/simulation-06.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 6: Friction, sliding, overturning and stability

Foundation contact and retaining systems can fail by sliding, overturning, loss of bearing contact or internal structural distress. These mechanisms are different; passing one check does not prove the system safe. A rigid-body sliding calculation is governed by the maximum friction or other available resistance, whereas overturning depends on the relative location of stabilizing and destabilizing resultants. Soil–structure interaction, groundwater, passive pressure, anchorage and applicable factors require specialist treatment beyond introductory statics.

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 friction, sliding, overturning and stability.

• 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 6.1: Coulomb friction with unilateral contact

A dry contact model bounds tangential friction by μN. In static equilibrium the friction magnitude adjusts to demand up to its upper limit. A compressive normal reaction N cannot become negative when contact cannot carry tension.

Governing relationships

For a simple horizontal base without other vertical loads, N=W and static friction may satisfy |F|≤μsN. At impending slip, |F|=μsN under the idealized model.

Worked engineering example

A 60 kN gravity load on a level base with coefficient μ=0.4 offers at most 24 kN nominal friction resistance. A 10 kN horizontal demand requires 10 kN static friction, not automatically 24 kN.

Model audit and common error

Friction coefficient values depend on interface condition, moisture, contamination and testing; do not treat classroom numbers as design guarantees.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving coulomb friction with unilateral contact. 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 friction, sliding, overturning and stability, 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 6.2: Sliding checks and load combinations

Sliding equilibrium compares driving and resisting horizontal actions. If external anchorage, passive pressure or geometry contributes resistance, distinguish which mechanisms are reliably mobilized and compatible.

Governing relationships

For an isolated rigid block with H acting horizontally and μW available, the nominal ratio against sliding is μW/H for H>0. A code-based check may use different resistance and load factors.

Worked engineering example

With W=80 kN, μ=0.35 and H=20 kN, nominal available friction is 28 kN and ratio is 1.40. That simple ratio does not certify geotechnical safety.

Model audit and common error

Do not add incompatible peak resistances from different movement states without engineering justification.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving sliding checks and load combinations. 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 friction, sliding, overturning and stability, 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 6.3: Overturning moments and contact resultant

Overturning resistance is a moment balance around a specified toe. A nominal positive stabilizing moment does not alone ensure compression across the entire base; resultant eccentricity can create partial contact.

Governing relationships

For base width B, downward load W centered at B/2 and lateral force H at height h, stabilizing moment about the toe is WB/2, overturning moment is Hh. Resultant eccentricity e=M/W in a simple vertical-resultant model.

Worked engineering example

For B=2 m, W=100 kN, H=20 kN at h=3 m, resisting moment WB/2=100 kN·m and overturning moment Hh=60 kN·m. Nominal toe ratio is 1.67.

Model audit and common error

Local bearing pressures, soil capacity and load factors are beyond the friction-and-moment calculation.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving overturning moments and contact resultant. 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 friction, sliding, overturning and stability, 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 6.4: Retaining-wall pressures and conceptual stability

A retained-soil wall experiences lateral stress that often increases with depth, producing a triangular resultant for simple idealizations. Real earth-pressure distribution depends on soil strength, groundwater, wall movement, surcharge and drainage.

Governing relationships

Under a uniform effective earth-pressure gradient increasing from 0 at ground to pmax at depth H, total lateral resultant per metre of wall is P=pmax H/2, acting H/3 above the base.

Worked engineering example

With H=4 m and base lateral pressure pmax=24 kPa, the ideal triangular resultant is 48 kN per metre wall, at 1.333 m above the base. Units: kPa×m=kN/m.

Model audit and common error

This pressure shape is a teaching assumption; real wall design requires geotechnical investigation and appropriate earth-pressure theory.

Interpretation task

Prepare a free-body sketch of a civil infrastructure element involving retaining-wall pressures and conceptual 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 friction, sliding, overturning and stability, 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

Conceptual retaining wall: compare horizontal sliding tendency, overturning about the toe, and effect of a higher groundwater level. Identify missing geotechnical inputs before any actual wall 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 6.1

Given: A simple rigid block has W=80.0 kN, base width B=2.00 m, horizontal push H=14.0 kN at height h=1.50 m and assumed friction coefficient μ=0.25. Compare nominal sliding and toe-overturning actions.

Solution: Nominal available friction μW=20.000 kN versus demand H=14.000 kN. Toe resisting moment WB/2=80.000 kN·m; overturning moment Hh=21.000 kN·m. Comparisons 1.429 and 3.810 are unfactored classroom ratios, not design factors of safety.

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 6.2

Given: A simple rigid block has W=86.0 kN, base width B=2.15 m, horizontal push H=16.0 kN at height h=1.65 m and assumed friction coefficient μ=0.27. Compare nominal sliding and toe-overturning actions.

Solution: Nominal available friction μW=23.220 kN versus demand H=16.000 kN. Toe resisting moment WB/2=92.450 kN·m; overturning moment Hh=26.400 kN·m. Comparisons 1.451 and 3.502 are unfactored classroom ratios, not design factors of safety.

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 6.3

Given: A simple rigid block has W=92.0 kN, base width B=2.30 m, horizontal push H=18.0 kN at height h=1.80 m and assumed friction coefficient μ=0.29. Compare nominal sliding and toe-overturning actions.

Solution: Nominal available friction μW=26.680 kN versus demand H=18.000 kN. Toe resisting moment WB/2=105.800 kN·m; overturning moment Hh=32.400 kN·m. Comparisons 1.482 and 3.265 are unfactored classroom ratios, not design factors of safety.

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 6.4

Given: A simple rigid block has W=98.0 kN, base width B=2.45 m, horizontal push H=20.0 kN at height h=1.95 m and assumed friction coefficient μ=0.31. Compare nominal sliding and toe-overturning actions.

Solution: Nominal available friction μW=30.380 kN versus demand H=20.000 kN. Toe resisting moment WB/2=120.050 kN·m; overturning moment Hh=39.000 kN·m. Comparisons 1.519 and 3.078 are unfactored classroom ratios, not design factors of safety.

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 6.5

Given: A simple rigid block has W=104.0 kN, base width B=2.60 m, horizontal push H=22.0 kN at height h=2.10 m and assumed friction coefficient μ=0.33. Compare nominal sliding and toe-overturning actions.

Solution: Nominal available friction μW=34.320 kN versus demand H=22.000 kN. Toe resisting moment WB/2=135.200 kN·m; overturning moment Hh=46.200 kN·m. Comparisons 1.560 and 2.926 are unfactored classroom ratios, not design factors of safety.

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 6.6

Given: A simple rigid block has W=110.0 kN, base width B=2.75 m, horizontal push H=24.0 kN at height h=2.25 m and assumed friction coefficient μ=0.25. Compare nominal sliding and toe-overturning actions.

Solution: Nominal available friction μW=27.500 kN versus demand H=24.000 kN. Toe resisting moment WB/2=151.250 kN·m; overturning moment Hh=54.000 kN·m. Comparisons 1.146 and 2.801 are unfactored classroom ratios, not design factors of safety.

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 6.7

Given: A simple rigid block has W=116.0 kN, base width B=2.90 m, horizontal push H=26.0 kN at height h=2.40 m and assumed friction coefficient μ=0.27. Compare nominal sliding and toe-overturning actions.

Solution: Nominal available friction μW=31.320 kN versus demand H=26.000 kN. Toe resisting moment WB/2=168.200 kN·m; overturning moment Hh=62.400 kN·m. Comparisons 1.205 and 2.696 are unfactored classroom ratios, not design factors of safety.

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 6.8

Given: A simple rigid block has W=122.0 kN, base width B=3.05 m, horizontal push H=28.0 kN at height h=2.55 m and assumed friction coefficient μ=0.29. Compare nominal sliding and toe-overturning actions.

Solution: Nominal available friction μW=35.380 kN versus demand H=28.000 kN. Toe resisting moment WB/2=186.050 kN·m; overturning moment Hh=71.400 kN·m. Comparisons 1.264 and 2.606 are unfactored classroom ratios, not design factors of safety.

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 6.9

Given: A simple rigid block has W=128.0 kN, base width B=3.20 m, horizontal push H=30.0 kN at height h=2.70 m and assumed friction coefficient μ=0.31. Compare nominal sliding and toe-overturning actions.

Solution: Nominal available friction μW=39.680 kN versus demand H=30.000 kN. Toe resisting moment WB/2=204.800 kN·m; overturning moment Hh=81.000 kN·m. Comparisons 1.323 and 2.528 are unfactored classroom ratios, not design factors of safety.

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 6.10

Given: A simple rigid block has W=134.0 kN, base width B=3.35 m, horizontal push H=32.0 kN at height h=2.85 m and assumed friction coefficient μ=0.33. Compare nominal sliding and toe-overturning actions.

Solution: Nominal available friction μW=44.220 kN versus demand H=32.000 kN. Toe resisting moment WB/2=224.450 kN·m; overturning moment Hh=91.200 kN·m. Comparisons 1.382 and 2.461 are unfactored classroom ratios, not design factors of safety.

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 6.11

Given: A simple rigid block has W=140.0 kN, base width B=3.50 m, horizontal push H=34.0 kN at height h=3.00 m and assumed friction coefficient μ=0.25. Compare nominal sliding and toe-overturning actions.

Solution: Nominal available friction μW=35.000 kN versus demand H=34.000 kN. Toe resisting moment WB/2=245.000 kN·m; overturning moment Hh=102.000 kN·m. Comparisons 1.029 and 2.402 are unfactored classroom ratios, not design factors of safety.

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 6.12

Given: A simple rigid block has W=146.0 kN, base width B=3.65 m, horizontal push H=36.0 kN at height h=3.15 m and assumed friction coefficient μ=0.27. Compare nominal sliding and toe-overturning actions.

Solution: Nominal available friction μW=39.420 kN versus demand H=36.000 kN. Toe resisting moment WB/2=266.450 kN·m; overturning moment Hh=113.400 kN·m. Comparisons 1.095 and 2.350 are unfactored classroom ratios, not design factors of safety.

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. Explain why static friction does not always equal μN.

Guidance: Use the governing equation and a labeled diagram from Unit 6; report direction, magnitude, units and model assumptions. Numerical assessment answers should be obtained from the worked methods, not from inspection of figures alone.

Q2. With W=75 kN and μ=0.4, calculate ideal maximum base friction. What information is still missing for design?

Guidance: Use the governing equation and a labeled diagram from Unit 6; report direction, magnitude, units and model assumptions. Numerical assessment answers should be obtained from the worked methods, not from inspection of figures alone.

Q3. A wall has 100 kN weight over a 2 m base and 25 kN lateral action at 3 m. Compare opposing toe moments.

Guidance: Use the governing equation and a labeled diagram from Unit 6; 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 a positive nominal overturning ratio not prove satisfactory soil bearing pressure?

Guidance: Use the governing equation and a labeled diagram from Unit 6; 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 6](../simulations/simulation-06.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. [Virtual work, equilibrium checks and energy ideas](units/unit-07.md)