B3.2 Structural Systems — Application and Selection

B3.2.1 Structures in Everyday Products

Structures are present in the design of everyday products.

In engineering and architecture, a structure is a system of interconnected parts designed to support weight and resist forces.

Structures can be found in cars, machinery, bridges, dams, buildings and everyday products such as chairs and tables. Each structure is designed to perform a specific function, remain stable, support loads and resist forces.

Structural Analysis

Structural analysis is used to understand how a structure responds to loads.

When analysing an existing product, designers and engineers consider:

• the loads acting on the structure,
• the stresses created within the structure,
• where the structure may deform or fail,
• how the structure could be strengthened.

Structural designs are assessed to ensure that they perform their intended function, withstand expected loads, include appropriate safety factors and comply with relevant standards and codes.

Modelling Structures

During the initial design stage, engineers often model structures before they are manufactured or constructed.

Computer-based methods such as Finite Element Analysis (FEA) can be used to predict how a structure responds to expected loads and to identify areas of high stress, possible deformation and potential failure.

Structural Assessment Throughout a Product's Life

During construction and throughout the life of a structure, engineers may continue to use:

• visual inspections,
• testing,
• structural analysis.

These assessments can identify potential issues or deficiencies and help maintain safety, structural integrity and regulatory compliance.

Forces and Stresses Within Structures

There are five basic types of stress that may occur within a structure:

1. tension,
2. compression,
3. shear,
4. torsion,
5. bending.

Structural analysis requires designers to understand where and how these stresses develop within a product.

1. Tension

Tension is produced by forces trying to pull apart or lengthen a material.

Tensile stress is distributed uniformly across the cross-section.

2. Compression

Compression is produced by forces trying to compress or shorten a material.

Compressive stress is distributed uniformly across the cross-section.

3. Shear

Shear stress is produced when forces try to slide one part of a material over another.

Shear may occur as:

• single shear,
• double shear,
• punching shear.

Unlike tensile and compressive stress, shear stress is not uniformly distributed across a cross-section. Its distribution depends on the shape of the cross-section.

Shear stress is typically:

• maximum at the neutral axis,
• zero at the outermost surfaces.

Example

A bolt connecting structural components may experience shear when the connected parts are forced in opposite directions.

4. Torsion

Torsion is stress induced by a twisting force, also known as torque.

Torsional stress:

• is highest at the outer surface,
• decreases towards the centre,
• tends towards zero at the centre.

Why Are Drive Shafts Often Hollow?

Because torsional stress is highest towards the outside of a shaft, material near the centre contributes less to torsional resistance.

A hollow drive shaft can therefore provide the same torsional strength as a solid shaft of the same outer diameter while using less material and therefore less weight.

5. Bending

Bending, also known as flexural stress, causes a structural member to curve.

Bending produces:

• tensile stress on one side,
• compressive stress on the opposite side.

Between these regions is the neutral axis, with maximum stresses at the outer surfaces.

Comparing the Five Types of Stress

Loads Acting on Structures

A load is an external force acting on a structure.

Loads create internal stresses such as tension, compression, shear, torsion and bending.

Load → external action on the structure

Stress → internal response produced within the material

1. Dead Loads

Dead loads are permanent, unchanging static loads.

They include the weight of the structure itself and permanent fixtures, such as walls and floors.

Dead loads are primarily vertical and may produce:

• compression,
• bending moments.

They are important when determining whether foundations and other structural components can support the total weight.

2. Live Loads

Live loads, also called imposed loads, are temporary or variable loads that are not permanently part of the structure.

Examples include:

• people,
• furniture,
• vehicles.

These loads are often primarily vertical and compressive. If unevenly distributed, they may also create bending stresses.

Unforeseen loads and misuse are typically considered through safety factors.

Dead Loads vs Live Loads

3. Environmental Loads

Environmental loads include:

• wind,
• rain,
• snow,
• seismic activity,
• flooding,
• thermal expansion and contraction.

They may create compression, tension, shear and bending depending on their magnitude, direction and distribution.

Example: Wind Load

Wind pressure against a building wall can create bending, inducing:

• compression on the windward side,
• tension on the leeward side.

Wind load → Bending → Compression + Tension

Example: Thermal Loads

Thermal expansion and contraction can create tensile and compressive stresses depending on how the structural components are constrained.

Example: Seismic Loads

Earthquakes can create a rapidly changing combination of:

• tension,
• compression,
• shear.

Their magnitude and direction can change throughout the seismic event.

4. Other Structural Loads

Settlement

Settlement is the downward vertical movement of soil or ground under an applied load as soil compresses or consolidates.

Hydrostatic Loads

Groundwater can create hydrostatic loads on subsurface structures such as foundations.

These loads act horizontally, normal to the structure, and increase in magnitude with depth.

Vibrational Loads

Vibration from machinery, wind or seismic activity can create combinations of:

• compression,
• tension,
• shear,
• torsion,
• bending.

Impact Loads

Impact can also create a complex combination of tensile, compressive and shear stresses that varies across the structure.

Analysing and Modelling Forces in Existing Products

When analysing an existing product, determine:

• what loads act on it,
• where those loads act,
• how loads are transferred through the structure,
• what stresses develop within its components,
• where structural weaknesses may exist.

A useful sequence is:

Product → Loads → Supports → Load Path → Internal Stresses → Potential Weaknesses

Step 1 — Identify the Structure

Ask:

• Which parts support the load?
• Which parts connect the structure?
• Where are the supports?
• Which components provide rigidity?

Step 2 — Identify the Loads

Consider:

• dead loads,
• live loads,
• environmental loads,
• impact,
• vibration,
• possible misuse.

Step 3 — Model Load Direction

Represent the loads using arrows showing their direction.

For example:

• downward weight,
• horizontal wind,
• rotational torque,
• side impact.

Step 4 — Identify Supports and Reactions

Identify where the product is supported and the reaction forces that oppose the applied loads.

Step 5 — Follow the Load Path

A load path describes how a load is transferred through the structural components.

For example:

Person → Chair seat → Frame → Legs → Floor

Step 6 — Identify Internal Stresses

Determine which components experience:

• tension,
• compression,
• shear,
• torsion,
• bending.

Step 7 — Identify Potential Weaknesses

Look for vulnerable areas such as:

• joints,
• connections,
• thin sections,
• unsupported spans,
• areas experiencing bending,
• areas subjected to impact or repeated loading.

Step 8 — Suggest Improvements

Propose a change that responds to the specific weakness identified.

Example Analysis: Chair

A person sitting on a chair creates a live load acting vertically downward. The chair itself creates a dead load.

Load Path

Person → Seat → Chair Frame → Legs → Floor

Internal Stresses

Possible stresses include:

• bending in the seat,
• compression in the legs,
• shear at joints and connections.

If the person moves or sits down suddenly, additional dynamic or impact loads may occur.

Potential Weaknesses

Weaknesses may occur at:

• seat-to-leg connections,
• joints in the frame,
• long unsupported members.

Possible Strengthening

The chair could potentially be strengthened by:

• adding bracing between the legs,
• reinforcing joints,
• using stronger connections,
• increasing the thickness of critical members.

Strengthening Existing Structures

There are four basic methods of strengthening a structure:

1. reinforcement with additional materials,
2. adding bracing,
3. increasing material thickness or density,
4. improving connections.

1. Reinforcement

Additional material can increase:

• strength,
• load-bearing capacity.

Reinforcement should be placed where structural analysis indicates additional strength is required.

2. Adding Bracing

Bracing can increase resistance to lateral forces such as:

• wind,
• seismic forces.

It can increase rigidity and reduce unwanted deformation.

3. Increasing Thickness or Density

Increasing the thickness or density of existing materials can improve their strength and ability to resist loads.

4. Improving Connections

Connections can be strengthened using:

• stronger bolts,
• welding.

Improved connections help transfer forces safely between structural components.

Comparing Strengthening Methods

From Analysis to Improvement

A strong structural analysis establishes the relationship:

Load → Stress → Weakness → Structural Improvement

Example

Observation: A horizontal load causes a rectangular frame to deform.

Analysis: The frame has insufficient resistance to lateral loading.

Structural effect: The members and joints are subjected to forces that distort the frame.

Improvement: Add bracing.

Explanation: The bracing increases rigidity and improves resistance to lateral forces.

Structural Analysis Checklist

When analysing an existing product, ask:

1. What is the structure designed to do?
2. What are its main structural components?
3. What loads act on the structure?
4. Where and in which direction do the loads act?
5. Where are the supports and reaction forces?
6. What is the load path through the product?
7. Where do tension, compression, shear, torsion or bending occur?
8. Where could structural failure occur?
9. How could the structure be strengthened?
10. Why would the proposed modification improve the structure?

B3.2.2 Young's Modulus

Young's Modulus is a measure of the stiffness of a material.

It describes how resistant a material is to elastic deformation when subjected to tensile or compressive stress.

Young's Modulus is represented by the symbol:

and is calculated using:

where:

• E = Young's Modulus,
• σ = tensile stress,
• ε = tensile strain.

Stress and Strain

Before calculating Young's Modulus, it is necessary to understand stress and strain.

Tensile Stress

Stress (σ) relates the applied load to the cross-sectional area of the material.

where:

• σ = stress,
• F = force,
• A = cross-sectional area.

Stress is measured in pascals (Pa).

In engineering calculations, dimensions are often measured in millimetres. In this case:

Therefore, stress is commonly expressed in MPa.

Tensile Strain

Strain (ε) measures the change in length relative to the material's original length.

where:

• ε = strain,
• ΔL = change in length,
• L₀ = original length.

Strain has no units because it is a ratio between two lengths measured using the same units.

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Change in length:

Therefore:

Calculating Young's Modulus

Within the elastic region:

Young's Modulus therefore compares:

how much stress is applied

with

how much elastic strain is produced

A material that requires a large stress to produce a small strain has a high Young's Modulus and is therefore stiff.

A material that produces greater strain for the same stress has a lower Young's Modulus and is more flexible.

Worked Example 1 --- Young's Modulus from Stress and Extension

An 18 mm square steel section is 3 m long. It is subjected to a stress of 21.6 MPa and extends by 0.2 mm.

Calculate Young's Modulus.

Step 1 --- Convert the original length

Step 2 --- Calculate strain

Step 3 --- Apply the Young's Modulus formula

Since:

then:

Worked Example 2 --- Calculate Stress First

A tensile force of 45 kN is applied to a circular specimen with a cross-sectional area of approximately 113.1 mm².

The resulting strain is 0.0015.

Calculate Young's Modulus.

Step 1 --- Convert force to newtons

Step 2 --- Calculate stress

Step 3 --- Calculate Young's Modulus

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The Stress-Strain Graph

A stress-strain graph shows the relationship between tensile stress applied to a material and the resulting strain.

The axes are:

• vertical axis (y) → Stress (σ),
• horizontal axis (x) → Strain (ε).

A typical stress-strain curve can provide information about many mechanical properties.

For B3.2.2, the most important are:

• Young's Modulus,
• yield strength,
• ultimate tensile strength,
• fracture.

1. Elastic Region

At the beginning of a typical stress-strain curve, stress and strain have a straight-line relationship.

In this region:

• deformation is elastic,
• stress is proportional to strain,
• the material returns to its original length when the load is removed.

This relationship is associated with Hooke's Law.

2. Young's Modulus on the Graph

Young's Modulus is represented by the slope of the straight-line elastic region.

or, when using two points from the linear region:

A graph with a steeper elastic slope represents a material with a higher Young's Modulus.

A graph with a shallower elastic slope represents a material with a lower Young's Modulus.

Steeper slope → Higher E → Stiffer material

Shallower slope → Lower E → More flexible material

Calculating E from a Stress-Strain Graph

Suppose a point in the linear elastic region has:

and:

Then:

3. Elastic Limit

The elastic limit, or proportional limit, marks the end of the linear elastic behaviour.

Up to this region, the material behaves elastically.

Beyond the elastic limit, the stress-strain relationship becomes non-linear and the material may begin to experience plastic deformation.

4. Yield Strength

The yield point occurs when the material begins to undergo significant plastic deformation.

Once the yield point has been passed:

• permanent deformation occurs,
• the specimen will not completely return to its original dimensions when the load is removed.

For structural applications, yield strength is particularly important because a structure may become unusable due to permanent deformation before it actually fractures.

Yield Point and Proof Stress

Not every material has a clearly defined yield point.

Low-carbon steels may display clearly identifiable yield behaviour, but materials such as:

• aluminium alloys,
• copper,
• titanium,
• magnesium,
• some polymers,
• medium and high-carbon steels,

may transition gradually from elastic to plastic deformation.

For these materials, engineers can use proof stress, also known as offset yield stress.

A common value is the 0.2% proof stress.

This provides a practical yield-like value for materials without an obvious yield point.

5. Plastic Region

After yielding, the material enters the plastic region.

In this region:

• deformation is permanent,
• removing the load will not return the material completely to its original dimensions,
• strain can continue to increase.

The curve eventually reaches its maximum stress.

6. Ultimate Tensile Strength

The Ultimate Tensile Strength (UTS) is the maximum engineering stress reached during the tensile test.

On the stress-strain graph:

UTS = highest point on the engineering stress-strain curve

After this point, a ductile specimen may begin to neck.

7. Necking

Necking is a localised reduction in cross-sectional area.

It occurs in ductile materials as the specimen continues to elongate after reaching UTS.

The reduction becomes concentrated in one area of the specimen.

Eventually, fracture occurs in this region.

Both:

• elongation,
• reduction in area,

can be used as measures of ductility.

8. Fracture

Fracture is the point where the tensile specimen breaks.

It represents the final failure of the specimen during the tensile test.

On the stress-strain graph, fracture occurs at the end of the curve.

Reading a Typical Stress-Strain Graph

The sequence for a typical ductile material can be interpreted as:

Linear elastic region → Elastic limit → Yield → Plastic deformation → UTS → Necking → Fracture

Feature What It Represents What the Designer Learns

Elastic region Reversible deformation Material returns to original dimensions

Young's Modulus Slope of linear elastic region Stiffness

Elastic limit End of linear elastic behaviour Limit of proportional behaviour

Yield strength Beginning of significant permanent deformation Practical structural limit

Plastic region Permanent deformation Material will not fully recover

UTS Maximum engineering stress Maximum stress reached

Necking Local reduction in cross-sectional area Specimen approaching fracture

Fracture Specimen breaks Final failure

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Interpreting Stress-Strain Graphs

When given a stress-strain graph, use a systematic approach.

Step 1 --- Check the Axes

Identify:

• stress and its units,
• strain and how it is expressed.

Strain may be shown as:

• a decimal,
• a percentage.

Step 2 --- Find the Linear Elastic Region

Locate the initial straight section.

Its slope represents Young's Modulus.

Step 3 --- Identify Yield Strength

Find the point where significant plastic deformation begins.

If there is no obvious yield point, the graph may use proof stress.

Step 4 --- Identify UTS

Find the maximum stress on the engineering stress-strain curve.

This is the Ultimate Tensile Strength.

Step 5 --- Identify Fracture

Follow the curve to its endpoint.

This represents fracture or breaking of the specimen.

Interpreting Two Materials

Stress-strain graphs can also be used to compare materials.

If Material A has a steeper elastic slope than Material B:

Therefore Material A is stiffer.

However, the graph must be analysed further before making conclusions about:

• yield strength,
• UTS,
• ductility,
• fracture behaviour.

Engineering Stress and True Stress

Most conventional stress-strain graphs use engineering stress.

Engineering stress uses the specimen's original cross-sectional area:

During a tensile test, however, the cross-sectional area changes, particularly during necking.

True stress uses the actual changing cross-sectional area:

True stress-strain curves provide a more accurate representation during significant plastic deformation and may be useful in applications such as:

• metal forming,
• crash simulations.

Temperature and Young's Modulus

Young's Modulus is affected by temperature.

As temperature rises, Young's Modulus will generally decrease.

The material therefore tends to become:

• less stiff,
• more flexible,
• less resistant to elastic deformation.

This occurs because increased thermal energy causes greater atomic vibration and affects the resistance of interatomic bonds to deformation.

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Worked Exam-Style Questions

Question 1

A material experiences a tensile stress of 160 MPa and a tensile strain of 0.0008 while remaining within its elastic region.

Calculate Young's Modulus.

Answer

Question 2

A 500 mm long specimen extends by 0.25 mm while subjected to a tensile stress of 100 MPa.

Calculate Young's Modulus.

Step 1 --- Calculate strain

Step 2 --- Calculate Young's Modulus

Question 3

A stress-strain graph shows that a material reaches:

• 150 MPa at a strain of 0.001 in its linear region,
• yield at 260 MPa,
• maximum stress at 410 MPa,
• fracture after the maximum point.

Identify:

a) Young's Modulus\ b) Yield strength\ c) Ultimate tensile strength

Answer

a) Young's Modulus

b) Yield strength

c) Ultimate tensile strength

The fracture point would be identified at the endpoint of the curve.

B3.2.2 Analysis Checklist

When solving a Young's Modulus or stress-strain graph question:

1. Identify the stress.
2. Identify or calculate the strain.
3. Ensure strain is expressed as a decimal, not a percentage.
4. Confirm the data point is within the linear elastic region when calculating E.
5. Use:

1. State the appropriate units for E, usually MPa or GPa.
2. Identify the yield strength.
3. Identify the UTS as the maximum engineering stress.
4. Identify fracture at the end of the curve.
5. Interpret what each point means for the material's behaviour.

B3.2.3 Structural Failure

Structures can fail due to:

• overloading,
• inappropriate material choice,
• inadequate size,
• inappropriate shape.

Structural failure occurs when a structure or one of its components can no longer perform its intended function safely.

All structures must therefore be designed with careful consideration of the forces and loads that will act on them.

Failure to correctly assess these forces can result in:

• excessive deformation,
• instability,
• buckling,
• collapse,
• increased costs,
• injury.

Causes of Structural Failure

The four key factors for B3.2.3 are:

1. Overloading
2. Material choice
3. Size
4. Shape

These factors may act individually or in combination.

1. Overloading

Overloading occurs when the forces acting on a structure exceed the load that its components can safely withstand.

Loads may be underestimated during design or may become greater than originally expected.

This can cause:

• excessive stress,
• excessive deformation,
• buckling,
• fracture,
• collapse.

Buckling

Buckling is particularly associated with long, thin structural members subjected to compression.

If a critical compressive load is exceeded, the member becomes unstable and deforms laterally.

This can cause the member to lose its ability to support the load and may lead to collapse.

Compression + slender member + critical load exceeded → Buckling

2. Material Choice

A structure can fail if the selected material does not have suitable properties for the conditions it experiences.

When investigating material choice, consider whether the material can adequately resist the:

• stresses,
• deformation,
• environmental conditions,
• repeated loading,

experienced by the structure.

FEA data can help determine whether the stress or deformation predicted in a component is acceptable for the selected material.

3. Size

The dimensions of a structural component affect its ability to withstand loads.

Structural problems may occur when components are:

• too thin,
• too slender,
• inadequately sized for the applied load.

Reducing the dimensions of a structural component may reduce its structural capacity.

For example, a long, thin member under compression may be particularly vulnerable to buckling.

4. Shape

The geometry of a structure affects:

• how loads are carried,
• how forces are distributed,
• structural stiffness,
• stability.

A structure can therefore fail even if it contains strong materials if its geometry does not adequately resist the loads acting on it.

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Case Study --- Quebec Bridge Collapse, 1907

The Quebec Bridge collapse demonstrates the consequences of underestimating structural loads and inadequately designing structural members.

During construction, the central span collapsed.

The investigation found that:

• the weight of the bridge had been significantly underestimated,
• the required strength of structural members had therefore been assessed incorrectly,
• the lower compression chords near the main pier were inadequately designed,
• buckling was observed before collapse.

The bridge eventually collapsed after the lower compression chords became unstable.

Failure Analysis

Underestimated weight → Inadequate member design → Excessive compression → Buckling → Collapse

This failure also demonstrated the importance of:

• rigorous design checks,
• adequate supervision,
• responding to evidence of structural distress,
• clear engineering responsibility.

Case Study --- Tacoma Narrows Bridge, 1940

The Tacoma Narrows Bridge was exceptionally slender for a suspension bridge of its time.

Its main span had:

• a very shallow stiffening girder,
• a narrow roadway,
• unusually slender proportions.

These characteristics made the bridge susceptible to wind-induced oscillation.

Wind produced aeroelastic flutter and increasingly severe twisting of the bridge deck.

The bridge eventually collapsed after torsional oscillation developed during strong winds.

Failure Analysis

Slender geometry → Susceptibility to wind → Aeroelastic flutter → Torsional oscillation → Structural failure

At the time, engineers had limited understanding of these aerodynamic effects and limited access to wind-tunnel testing.

The failure subsequently influenced:

• suspension bridge design,
• wind-load analysis,
• aerodynamic stability assessment,
• development of more sophisticated analytical methods.

Case Study --- Sampoong Department Store, 1995

The Sampoong Department Store collapse resulted from a combination of:

• design changes,
• construction errors,
• increased loading,
• reduced structural capacity,
• inadequate engineering evaluation,
• negligence.

Changes included:

• adding an additional above-ground floor,
• removing some columns to accommodate escalators,
• using thinner columns,
• increasing column spacing,
• installing a heavy rooftop air-conditioning unit.

These modifications simultaneously:

• increased the dead load,
• reduced structural capacity.

Cracking was observed before the collapse but was ignored or dismissed.

Failure Analysis

Increased load + reduced column capacity + design/construction problems → Structural distress → Collapse

Comparing the Failure Examples

Case Important Cause Structural Effect

Quebec Bridge Underestimated weight and inadequate compression members Buckling and collapse

Tacoma Narrows Bridge Extremely slender geometry and wind interaction Torsional oscillation and collapse

Sampoong Department Store Increased loads and reduced structural capacity Progressive structural distress and collapse

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Finite Element Analysis --- FEA

Finite Element Analysis (FEA) is a computerised method used to test virtual models under different loading conditions.

FEA can be used to:

• test new product designs,
• analyse existing products,
• simulate service conditions,
• predict structural behaviour,
• identify critical regions before physical manufacture.

FEA is particularly useful for stress calculations.

How FEA Works

FEA divides a complex virtual model into many smaller, simpler regions called finite elements.

These elements are connected through nodes.

Instead of analysing the entire complex structure as one object, the software analyses the behaviour of these interconnected elements.

The results are then combined to predict the behaviour of the complete structure.

Complex model → Finite elements → Nodes → Numerical analysis → Structural results

What Can Be Simulated?

The structural component of FEA can simulate how an object responds to service loads and conditions.

Loads may result from:

• external forces,
• pressure,
• temperature changes,
• vibration.

Static and dynamic loads can be simulated to investigate behaviour such as:

• extension,
• bending,
• twisting.

FEA can provide information about:

• stress,
• strain,
• displacement,
• stiffness,
• strength.

Interpreting FEA Data

The syllabus requires you to be able to use FEA information to help identify why a structure may fail.

FEA results are often presented using:

• contour plots,
• colour maps,
• numerical scales,
• displacement plots,
• Safety Factor simulations.

The important skill is not simply recognising an FEA image.

You must be able to interpret the data shown.

1. Read the Result Type

First identify what the FEA plot is showing.

It may represent:

• stress,
• strain,
• displacement,
• Safety Factor,
• another structural quantity.

2. Read the Legend and Units

Contour plots use a scale to show the magnitude of the result across the model.

In the examples described in the source material:

• red identifies high-stress or danger zones,
• blue identifies low-stress or safer zones,
• intermediate colours show the transition between them.

However, the numerical legend or scale is what gives the colours their meaning.

3. Locate Critical Regions

Look for areas where the FEA result indicates particularly high values.

These may identify:

• stress concentrations,
• highly loaded members,
• vulnerable joints,
• critical welds,
• areas of excessive displacement.

Ask:

Where is the highest value located?

and then:

Why might that location be critical?

4. Interpret Stress Contour Plots

Stress contour plots show how stress is distributed through the structure.

Look for:

• areas of high stress,
• areas of low stress,
• sudden changes in stress,
• localised stress concentrations.

A local region of high stress may indicate a possible weakness requiring further investigation.

FEA can, for example, be used to determine peak stresses around welds and help assess fatigue resistance.

5. Interpret Displacement Plots

Displacement plots show how much parts of the structure move when loads are applied.

High displacement may indicate:

• insufficient stiffness,
• excessive deformation,
• a possible structural problem.

The important question is whether the predicted movement is acceptable for the intended design.

6. Interpret Safety Factor Results

Safety Factor simulations compare structural capacity with the stress or load experienced by the structure.

A Safety Factor result can help identify regions with a lower margin against failure.

When analysing the result, identify where the lowest Safety Factor occurs.

That location may be more critical than regions with a larger Safety Factor.

From FEA Data to a Failure Diagnosis

A useful sequence is:

Load condition → FEA result → Critical region → Structural behaviour → Possible cause of failure

For example:

Applied load → High stress around connection → Local stress concentration → Connection vulnerable → Possible failure associated with connection design

The FEA result provides evidence for the diagnosis.

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FEA and Structural Failure

FEA can help investigate whether failure is associated with the four syllabus factors.

Overloading

Evidence may include:

• very high predicted stress,
• excessive displacement,
• low Safety Factor,
• highly loaded structural members.

Material Choice

FEA may show that the predicted stresses or deformation are inappropriate for the material selected.

Size

FEA may identify excessive stress or displacement in structural members whose dimensions are inadequate for the simulated loading.

Shape

FEA can show how geometry affects the distribution and concentration of stresses.

Particular shapes or transitions may produce localised regions requiring further investigation.

Example FEA Interpretation

Imagine an FEA stress plot of a bracket under load.

The legend shows:

• maximum stress = 280 MPa,
• minimum stress = 5 MPa.

The maximum stress is concentrated around a connection.

A correct interpretation would be:

Observation:\ The highest stress occurs around the connection.

Evidence:\ The FEA predicts a maximum stress of 280 MPa in this region.

Interpretation:\ The connection is the most highly stressed region in the simulated load case.

Diagnosis:\ This region should be investigated as a possible structural weakness.

To determine whether failure is actually predicted, the designer would need to compare this result with the relevant allowable limit or material/design criterion.

FEA in Product Design

FEA data can assist designers with decisions involving:

• weight minimisation,
• material selection,
• manufacturing processes,
• cost,
• structural strengthening.

Complex connections can be analysed to determine:

• force flow,
• peak stresses around welds,
• fatigue resistance.

FEA may also be used to verify simplified design methods.

FEA in the Automotive Industry

FEA can be used to simulate vehicle collisions.

A simulation may show:

• distributed forces,
• twisting,
• buckling,
• movement,
• failure of structural elements.

The results can inform decisions about where strengthening may be required.

This allows designers to investigate crash scenarios before manufacturing and physically testing every design iteration.

Modelling Defects

FEA can also investigate defects in components.

For example, an internal flaw in a casting may be modelled as:

• a crack,
• a void.

A sharp crack representation may be appropriate when investigating:

• fracture mechanics,
• crack propagation.

A void may be more appropriate when investigating the overall behaviour of a component containing a larger manufacturing defect.

Advantages and Limitations of FEA

Advantages

FEA can:

• analyse complex structures,
• simulate multiple loading conditions,
• reduce development time,
• reduce the need for some physical prototypes,
• identify critical regions before manufacture,
• assist design optimisation.

It can also be used to assess existing structures and investigate whether continued service or proposed repairs may be acceptable.

Limitations

FEA does not automatically guarantee a correct result.

Results may be inaccurate if the analysis uses:

• incorrect assumptions,
• inappropriate modelling techniques,
• flawed input data.

These problems can lead to incorrect interpretations of:

• stress,
• strain,
• displacement,
• other critical parameters.

FEA and Physical Testing

FEA can reduce the cost and time associated with manufacturing and testing physical prototypes.

However, computer simulation may only inform a design to a certain point.

Physical testing may still be required to confirm the performance and specifications of the final product.

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How to Identify Why a Structure Failed

When given a failed structure, photograph, case study, test result or FEA output, use the following process.

Step 1 --- Identify the Load

Ask:

• What forces were acting?
• Was the load static or dynamic?
• Was it expected or unexpected?
• Could the structure have been overloaded?

Step 2 --- Identify the Failure Location

Look for:

• buckling,
• cracking,
• excessive deformation,
• fractured members,
• damaged joints,
• highly stressed regions in FEA.

Step 3 --- Use the Evidence

Use available:

• dimensions,
• loads,
• material data,
• FEA values,
• displacement data,
• Safety Factor data,
• observations.

Do not diagnose failure only from appearance.

Step 4 --- Connect the Evidence to a Cause

Consider:

• Overloading --- Was the load too high?
• Material choice --- Was the material appropriate?
• Size --- Were the structural members adequately sized?
• Shape --- Did the geometry respond adequately to the load?

Step 5 --- Explain the Failure Mechanism

State how the identified cause produced structural failure.

For example:

Excessive compressive load → slender member becomes unstable → buckling → loss of load-bearing capacity

Exam-Style FEA Question

An FEA simulation of a structural component shows a highly stressed region around a welded connection and much lower stress throughout the rest of the component.

Question

What can be concluded from the FEA result?

Answer

The FEA indicates that the welded connection is a stress concentration and is the most highly stressed part of the simulated structure.

This makes the connection an important area to investigate when assessing possible structural failure.

The numerical stress value should be compared with an appropriate allowable material or design limit before concluding that failure will occur.

Exam-Style Failure Question

A long, thin structural member bends sideways while carrying a large compressive load.

Question

Identify the likely failure mechanism.

Answer

The likely mechanism is buckling.

The member is slender and subjected to compression. If the critical compressive load is exceeded, it can become unstable, deform laterally and lose its ability to support the load.

B3.2.3 Analysis Checklist

When identifying why a structure failed:

1. Identify the applied loads.
2. Identify where failure occurred.
3. Look for evidence of overloading.
4. Consider whether the material choice was appropriate.
5. Consider whether structural members were adequately sized.
6. Consider whether the shape or geometry contributed to the problem.
7. If FEA is provided, identify the result type.
8. Read the legend, numerical values and units.
9. Locate areas of high stress, displacement or low Safety Factor.
10. Relate the FEA evidence to the observed or predicted failure.
11. Distinguish between simulation evidence and your failure conclusion.
12. Remember that FEA results depend on the quality of the model and input data.

B3.2.4 Force Diagrams

Forces acting on a structure, or within structural components such as beams and trusses, can be represented diagrammatically.

Force diagrams simplify a real structure so that the forces acting on it can be identified and interpreted.

The most common type is a Free Body Diagram (FBD).

Free Body Diagrams

A Free Body Diagram (FBD) isolates an object from its surroundings and represents the external forces acting on it.

The real object is simplified so that the forces can be analysed clearly.

Forces are represented using arrows.

An arrow communicates:

• the direction of the force,
• the line of action of the force,
• and, when drawn to scale or labelled, its magnitude.

For example:

        ↑ 50 N
        │
        ●
        │
        ↓ 50 N

The two forces are equal and opposite.

Therefore:

and the object is in translational equilibrium.

Resultant Force

The resultant force is the single force that represents the combined effect of all the forces acting on an object.

It may be written as:

If two forces act in opposite directions, their magnitudes can be subtracted.

For example:

50 N ←  ●  → 80 N

Therefore:

acting to the right.

Unbalanced forces → Resultant force ≠ 0

If all forces balance:

the object has no resultant translational force.

Equilibrium

For a structure to remain stable, it must be in equilibrium.

In simple terms:

The forces and turning effects acting on the structure must balance.

For translational equilibrium:

and:

For rotational equilibrium:

where M represents moment.

Forces Acting at Angles

Forces do not always act horizontally or vertically.

A force acting at an angle has:

• a horizontal component,
• a vertical component.

For example:

           ↗ F
          /
         /
        ●

The force can be resolved into:

          ↑ Fy
          │
          ●────→ Fx

where:

and:

depending on how the angle is defined.

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Loads on Beams

Two common ways of representing loads on beams are:

• concentrated or point loads,
• uniformly distributed loads (UDL).

Point Load

A concentrated load or point load acts over such a small length that it is treated as acting at a single point.

It is represented by a single force arrow.

             ↓ 10 kN
             │
     ────────┼────────

Examples could include a load transferred to a beam through a particular connection or support point.

Uniformly Distributed Load --- UDL

A Uniformly Distributed Load (UDL) acts evenly over a specified length.

It is normally expressed as force per unit length, for example:

A UDL may be represented as:

       ↓   ↓   ↓   ↓   ↓
       ↓   ↓   ↓   ↓   ↓
     ─────────────────────
          3 kN/m

The arrows indicate that the load is distributed across the beam rather than concentrated at one point.

Support Reactions

When a structure applies a force to its supports, the supports provide reaction forces.

These reactions help keep the structure in equilibrium.

A simply supported beam may be represented as:

              ↓ Load
              │
        ──────┼──────
       ▲             ○
       ↑             ↑
      RA             RB

The downward load is balanced by upward reactions at the supports.

Pinned Support

A pinned (hinged) support can provide a reaction whose components may act horizontally and vertically as required to maintain equilibrium.

It prevents translation at the support but allows rotation in the idealised model.

A reaction may therefore be represented using components such as:

and:

Roller or Rocker Support

A roller or rocker support allows movement along the supporting surface.

Its reaction acts perpendicular to the supporting surface.

For a horizontal surface, the reaction is therefore vertical.

        Beam
    ────────────
          ○
          ↑ R

Roller or rocker supports are useful in structures such as bridges because they can allow movement caused by temperature variations.

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Forces Within a Beam

A beam subjected to transverse loading tends to bend.

When a simple beam bends:

• one side is placed in compression,
• the opposite side is placed in tension,
• between them is the neutral axis.

For a typical simply supported beam bending downward:

               ↓ LOAD
        ─────────────────
          COMPRESSION
        -----------------
          NEUTRAL AXIS
        -----------------
            TENSION

The upper layers shorten and are therefore in compression.

The lower layers extend and are therefore in tension.

At the neutral axis, longitudinal bending stress is zero.

Stress Distribution Through a Beam

Bending stress varies through the depth of the beam.

The greatest tensile and compressive bending stresses occur at the outer surfaces, furthest from the neutral axis.

The stress reduces towards the neutral axis.

At the neutral axis:

for longitudinal bending stress.

This is important when interpreting why beam cross-sectional shape affects structural performance.

Beam Cross-Section and Force Resistance

The location of material relative to the neutral axis affects resistance to bending.

An I-beam places a large amount of material away from the neutral axis.

It consists of:

• two horizontal flanges,
• a vertical web.

     ━━━━━━━━━━━  ← flange
          │
          │        ← web
          │
     ━━━━━━━━━━━  ← flange

This arrangement gives an efficient resistance to bending for its cross-sectional area.

A box beam has two webs and two flanges and provides greater resistance to twisting or torsion, making it useful where torsional forces are important.

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Trusses

A truss is a framework consisting of straight structural members connected at joints.

Trusses are commonly used in:

• bridges,
• roofs,
• buildings,
• other lightweight structural frameworks.

Idealised trusses are often based on triangles.

Triangles are structurally useful because their geometry allows loads to be transferred through members while maintaining a stable shape.

Forces in Truss Members

For the idealised pin-jointed trusses considered here:

• loads act at the joints,
• members carry axial forces,
• members are therefore primarily in either tension or compression.

This makes force diagrams particularly useful for interpreting trusses.

Identifying Tension

If the force in a member acts away from a joint, the member is being treated as being in tension.

←────────●────────→
       JOINT

The member is resisting a stretching action.

Forces pulling away from the joint → Tension

Identifying Compression

If the force in a member acts towards a joint, the member is being treated as being in compression.

────────→●←────────
       JOINT

The member is resisting a compressive action.

Forces pushing towards the joint → Compression

Interpreting a Simple Truss

When examining a truss force diagram:

1. Identify the external loads.
2. Identify the support reactions.
3. Identify the direction of the forces at the joints.
4. Determine which members are in tension.
5. Determine which members are in compression.
6. Check whether the forces are consistent with equilibrium.

A truss distributes loads through multiple members and transfers them towards its supports.

Force Polygons

When several forces act at different angles, their vectors can be arranged tip-to-tail.

If the final vector returns to the starting point, the force polygon closes.

A closed force polygon indicates:

and therefore translational equilibrium.

If the polygon does not close, the gap represents the resultant force.

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Shear Force Diagrams

Internal forces within a beam can vary along its length.

A Shear Force Diagram (SFD) represents the magnitude of the internal shear force at different positions along a beam.

A simple example may look like:

Shear Force

 +3 kN ──────────┐
                 │
  0 ─────────────┼──────────────
                 │
 -3 kN           └──────────────

Changes in the shear force diagram correspond to loads and reactions acting on the beam.

For a beam in equilibrium, the shear force diagram returns to zero after all loads and reactions have been accounted for.

Bending Moment Diagrams

A Bending Moment Diagram (BMD) represents how bending moment changes along a beam.

Moment is the turning effect of a force about a point:

where:

• M = moment,
• F = force,
• d = perpendicular distance from the force to the point.

Moment is commonly measured in:

or:

Interpreting the Bending Moment Diagram

For a typical simply supported beam, the bending moment may:

• start at zero at one support,
• increase towards the loaded region,
• reach a maximum,
• return to zero at the other support.

The source material highlights an important relationship:

Maximum bending moment occurs where the shear force diagram equals zero.

These diagrams represent different but related information about the same loaded structure.

Cantilever Force Diagrams

A cantilever beam is fixed at one end and unsupported at the other.

FIXED                     FREE
████│──────────────────────
    │                 ↓ Load

Examples include:

• balconies,
• some bridge structures,
• stadium roofs,
• aircraft wings.

Because the cantilever is supported at only one end, the fixed support must resist both forces and the turning effect created by the load.

For a point load (F) acting at a distance (L) from the fixed end:

The largest bending moment occurs at the fixed end.

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Interpreting Force Diagrams --- Step by Step

When given a force diagram in an IB question, use the following process.

Step 1 --- Identify the Structure

Determine what is being represented:

• object,
• beam,
• cantilever,
• truss,
• joint.

Step 2 --- Identify External Loads

Look for arrows representing applied forces.

Record:

• magnitude,
• direction,
• position.

Step 3 --- Identify the Supports

Determine whether the diagram contains:

• pinned/hinged supports,
• roller/rocker supports,
• fixed supports.

Then identify the possible reaction directions.

Step 4 --- Identify Reactions

Look for forces generated by the supports.

Ask:

Which applied forces are these reactions balancing?

Step 5 --- Check Equilibrium

Consider:

and, where relevant:

You do not always need to calculate the values to interpret whether the diagram represents balanced or unbalanced forces.

Step 6 --- Identify Internal Structural Effects

Depending on the structure, look for:

• tension,
• compression,
• shear,
• bending.

For a truss, identify whether individual members are in tension or compression.

For a beam, identify the compression side, tension side and neutral axis.

Step 7 --- Interpret What the Diagram Means

Do not simply list the arrows.

Explain the structural behaviour.

For example:

The downward point load causes the simply supported beam to bend. The supports provide upward reaction forces. Under this bending condition, the upper region of the beam is in compression and the lower region is in tension.

This is interpretation, rather than simple identification.

Worked Interpretation 1 --- Simply Supported Beam

Consider:

                 ↓ 10 kN
                 │
          ───────┼───────
         ▲               ○
         ↑               ↑
        5 kN            5 kN

Interpretation

A downward 10 kN point load acts on the beam.

The two supports each provide an upward 5 kN reaction.

Therefore:

The vertical forces are balanced.

The beam is in vertical equilibrium.

The downward loading also produces bending within the beam.

Worked Interpretation 2 --- Unbalanced Forces

40 N ←  ●  → 65 N

The horizontal resultant is:

to the right.

Therefore the forces are not balanced.

The object has a resultant force of:

acting to the right.

Worked Interpretation 3 --- Truss Joint

Consider an isolated joint:

        ↖
         \
          ●────→
         /
        ↙

If the arrows representing member forces point away from the joint, those members are interpreted as being in tension.

If the arrows point towards the joint, they are interpreted as being in compression.

The combined forces at the joint must balance for the joint to remain in equilibrium.

::: page-break :::

Additional Context --- Bending Stress

When a beam bends, the bending stress at a particular position can be related to:

• the bending moment,
• the distance from the neutral axis,
• the geometry of the beam cross-section.

The relationship is:

where:

• σ = bending stress,
• M = bending moment,
• y = distance from the neutral axis,
• I = second moment of area of the beam section.

The equation helps explain why the maximum bending stress occurs at the outer surfaces:

As y increases, bending stress increases.

At the neutral axis:

therefore:

Second Moment of Area

The second moment of area (I) describes how the cross-sectional area is distributed relative to the neutral axis.

A larger second moment of area generally reduces bending stress for the same bending moment and distance from the neutral axis:

This explains why beam geometry can significantly affect bending performance.

The source material notes that, in current IB examinations, the second moment of area will be provided when required.

Common Interpretation Errors

They are related but do not show the same quantity.

B3.2.4 Force Diagram Checklist

When interpreting a simple force diagram, ask:

1. What structure or component is represented?
2. What loads are acting?
3. What is the magnitude of each force?
4. In which direction does each force act?
5. Where does each force act?
6. What supports are present?
7. What reaction forces are shown or implied?
8. Are the horizontal forces balanced?
9. Are the vertical forces balanced?
10. Are the turning effects balanced?
11. Is the structure in equilibrium?
12. Are structural members in tension or compression?
13. Is the beam experiencing bending or shear?
14. What does the diagram tell you about the behaviour of the structure?

B3.2.5 Safety Factors

A Safety Factor (SF), also called a Factor of Safety (FOS), provides a margin between the load or stress a structure is expected to experience in service and the load or stress that would cause unacceptable structural failure.

In its simplest form:

Safety Factor describes the extent to which the design strength exceeds the expected service requirement.

For B3.2.5, the required relationship is:

Safety Factor is a ratio, so it has no units.

Ultimate, Allowable and Intended Loads

It is important to distinguish between three related ideas.

Ultimate Load or Stress

The ultimate load is the load associated with the limiting strength used in the safety-factor calculation.

Similarly, an ultimate stress represents the corresponding limiting material stress.

The source material commonly relates this to Ultimate Tensile Strength (UTS).

Allowable Load or Stress

The allowable load is the maximum load permitted in normal design use after the Safety Factor has been applied.

The corresponding allowable stress, also called working stress, can be calculated using:

or:

Intended or Service Load

The intended load is the load the structure is expected to carry in service.

The design must provide sufficient capacity so that the intended load remains within the allowable design limit.

A useful design relationship is:

Calculating Safety Factor

The syllabus formula is:

This formula can be used with either loads or stresses, but the numerator and denominator must represent the same type of quantity.

For loads:

For stresses:

Worked Example 1 --- Calculate SF

A structural component has an ultimate load of 90 kN and an allowable load of 30 kN.

Calculate the Safety Factor.

Interpretation

The ultimate capacity used in the calculation is three times the allowable load.

Rearranging the Safety Factor Formula

The same relationship can be rearranged depending on the unknown quantity.

To Find Safety Factor

To Find Allowable Load or Stress

To Find the Required Ultimate Capacity

         ULTIMATE
        ──────────
        SF × ALLOWABLE

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Calculating Maximum Intended Loads

A common design problem is to determine the maximum load that may safely be applied to a component.

If the ultimate load is known:

Worked Example 2 --- Maximum Allowable Load

A component has an ultimate load capacity of 48 kN and is designed with:

Calculate the maximum allowable load.

Therefore, the design load should not exceed 12 kN under the assumptions of this calculation.

Designing a Structure with a Safety Factor

Sometimes the expected load and required Safety Factor are known, and the designer must determine the minimum structural capacity required.

Use:

Worked Example 3 --- Required Design Capacity

A structural support is expected to carry a maximum intended load of:

The required Safety Factor is:

The required ultimate capacity is:

Therefore, the structure must be designed around an ultimate capacity of at least 24 kN under the stated design assumptions.

Safety Factor Using Stress

Safety Factor can also be applied to stress.

The source material gives the relationship:

Therefore:

where:

• UTS = Ultimate Tensile Strength,
• σworking = allowable or working stress,
• FOS = Factor of Safety.

Worked Example 4 --- Allowable Stress

A steel has:

and:

Calculate the allowable working stress.

Since:

this can also be written as:

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Designing a Tensile Rod with a Safety Factor

Safety Factor calculations can be combined with the stress equation:

This allows a designer to determine the required cross-sectional area of a structural component.

For a circular rod:

where:

• A = cross-sectional area,
• d = diameter.

The calculation sequence is:

Material strength → Apply SF → Allowable stress → Required area → Required dimensions

Worked Example 5 --- Maximum Working Load of a Steel Rod

A steel rod has:

• diameter = 16 mm,
• UTS = 590 MPa,
• Safety Factor = 4.

Calculate the maximum working load.

Step 1 --- Calculate Cross-Sectional Area

Step 2 --- Calculate Allowable Stress

Since:

then:

Step 3 --- Calculate Maximum Working Load

From:

rearrange:

Therefore:

Rounded appropriately:

Worked Example 6 --- Minimum Diameter of a Steel Rod

A steel rod must carry a tensile load of:

The steel has:

and the required Safety Factor is:

Calculate the minimum rod diameter.

Step 1 --- Calculate Allowable Stress

Step 2 --- Calculate Required Area

From:

rearrange:

Convert the load:

Therefore:

Step 3 --- Calculate Diameter

For a circular section:

Therefore:

Substituting:

So the theoretical minimum is approximately:

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Choosing an Appropriate Safety Factor

The size of the Safety Factor is not arbitrary.

The source material identifies several considerations.

Certainty of Loads

If the loads acting on a structure are difficult to predict, greater contingency may be required.

Design Life

Structures expected to remain in service for long periods may experience:

• ageing,
• wear,
• deterioration,
• changing conditions.

Quality of Manufacture

Manufacturing variation and defects can affect structural performance.

Higher confidence in manufacturing quality reduces uncertainty.

Consequences of Failure

The consequences of structural failure may range from inconvenience to catastrophic loss.

Where failure could cause severe injury or major damage, the design margin becomes particularly important.

Environmental Influences

Structures may be affected by:

• corrosion,
• weather,
• temperature,
• moisture,
• other environmental exposure.

These factors can reduce structural capacity over time.

Criticality of the Component

Some components are more important to the overall safety of a structure than others.

Failure of a critical component may lead to much greater consequences.

Ease of Repair or Maintenance

Components that are difficult to:

• inspect,
• maintain,
• repair,
• replace,

may require additional consideration during design.

Certainty of Material Properties

Material properties are not perfectly identical in every manufactured component.

Variation and uncertainty must therefore be considered.

Statutory and Code Requirements

Engineering structures may be required to comply with:

• regulations,
• standards,
• engineering codes.

These may specify required design approaches and safety margins.

Safety Factor Selection Summary

Consideration Question for the Designer

Certainty of loads How predictable are the loads?

Design life How long must the structure remain safe?

Manufacturing quality How consistent is production?

Consequences of failure What happens if the component fails?

Environment Will corrosion, temperature or weather affect it?

Component criticality How important is this component to overall safety?

Maintenance Can it be inspected and repaired easily?

Material certainty How reliable are the material properties?

Codes What do regulations and standards require?

UTS, Yield Strength and Safety Factors

The source material makes an important distinction between brittle and ductile materials.

Brittle Materials

For brittle materials, UTS may be used as the limiting strength because yielding may not provide a useful warning before failure.

A simplified relationship is:

Ductile Materials

For ductile materials, yield strength is generally more relevant when permanent deformation is considered unacceptable.

A component may still be physically intact after yielding but may no longer satisfy its design requirements.

Therefore:

may be the more appropriate design relationship.

Safety Factor Throughout a Structure's Life

A Safety Factor established during initial design does not mean that a structure will retain the same safety margin forever.

Structural capacity can change over time.

Material Degradation

Materials may deteriorate because of:

• corrosion,
• environmental exposure,
• wear,
• ageing,
• other degradation mechanisms.

If structural strength decreases while service loads remain unchanged:

Changes in Use

The actual loading of a structure may also change.

For example, a bridge may experience:

• heavier vehicles,
• more vehicles,
• different traffic patterns,

than were originally assumed.

Therefore:

Original design assumptions may no longer represent current service conditions.

Monitoring and Reassessment

Structural safety therefore requires more than an initial calculation.

During the life of a structure, engineers may need to:

• inspect,
• monitor,
• reassess,
• maintain,
• repair,
• strengthen,

the structure.

::: page-break :::

Case Study --- Genoa Morandi Bridge

The Morandi Bridge collapse illustrates how structural capacity can deteriorate during the life of a structure.

The source material identifies important issues including:

• corrosion of steel cables,
• deterioration of concrete,
• reduction in structural strength,
• concerns about maintenance.

The bridge eventually suffered catastrophic structural failure.

Safety Factor Interpretation

The important principle is:

If structural capacity decreases while the loads remain similar or increase, the effective margin against failure becomes smaller.

A simplified representation is:

If the numerator decreases because of deterioration:

This demonstrates why inspection and maintenance are essential for ageing infrastructure.

Designing with Safety Factor --- A General Method

When asked to design a structural component with a specified Safety Factor, use the following sequence.

Step 1 --- Identify the Intended Load

Determine the maximum expected service load.

Step 2 --- Identify the Required Safety Factor

For example:

Step 3 --- Determine the Required Capacity

For a load-based problem:

or, for a material-stress problem:

Step 4 --- Determine the Required Size

Use the appropriate structural relationship.

For axial stress:

therefore:

Step 5 --- Convert Area into Component Dimensions

For a circular rod:

Therefore:

Step 6 --- Check the Design

Confirm that:

and that the intended load does not exceed the allowable design load.

Exam-Style Questions

Question 1 --- Calculate SF

A structural member has an ultimate capacity of 120 kN and an allowable load of 40 kN.

Calculate its Safety Factor.

Answer

Question 2 --- Calculate Maximum Intended Load

A component can withstand an ultimate load of 75 kN and must have:

Calculate the maximum allowable load.

Answer

Question 3 --- Design for an SF

A support is intended to carry:

It must have:

Determine the required ultimate load capacity.

Answer

Question 4 --- Calculate Allowable Stress

A material has an ultimate stress of:

The required Safety Factor is:

Calculate the allowable stress.

Answer

Question 5 --- Select Between Two Designs

A product must carry a maximum intended load of 20 kN with:

Two designs are available:

• Design A: ultimate load = 50 kN
• Design B: ultimate load = 65 kN

Determine which design satisfies the required Safety Factor.

Required Capacity

Design A

Does not meet the required SF of 3.

Design B

Meets the requirement.

Therefore:

is acceptable according to the stated Safety Factor criterion.

::: page-break :::

Common Calculation Errors

B3.2.5 Calculation Checklist

Before finishing a Safety Factor problem, check:

1. What is the intended or allowable load?
2. What is the ultimate load or limiting strength?
3. What Safety Factor is required?
4. Are you working with loads or stresses?
5. Are the units compatible?
6. Do you need:

1. Or:

1. Or:

1. Do you need to calculate stress using:

1. Do you need to calculate the dimensions of the component?
2. Is UTS or yield strength the appropriate limiting property in the stated problem?
3. Does the final design actually achieve the required Safety Factor?