Structural Systems Application and Selection | B3.2

Students must be able toAnalyse and model the forces acting on and within the structure of existing products and be able to suggest how existing structures can be strengthened.

A structure is any system of interconnected parts designed to support loads and resist forces without unacceptable deformation or failure. Structures appear in every product: a chair's legs, a mobile phone's chassis, a bicycle frame, a bridge deck, and a skyscraper's core are all structures serving the same fundamental purpose.

Five types of stress act on structures:

1. Tension: A pulling force trying to elongate the material. Tensile stress is distributed uniformly across the cross-section. Examples: Suspension bridge cables; a rope in a tug-of-war; the bottom chord of a truss.
2. Compression: A pushing force trying to shorten the material. Like tension, compressive stress is distributed uniformly across the cross-section. Examples: Building columns; chair legs; the top chord of a truss; concrete in a dam.
3. Shear: A sliding force that causes one part to slide over an adjacent part. Unlike tension or compression, shear stress isn't spread evenly through the cross-section: it peaks right at the neutral axis and drops away to nothing at the outer surfaces. Shear takes three forms:
   • Single shear: One shear plane, e.g., a single bolt connecting two overlapping plates. τ = F / A
   • Double shear: Two shear planes, e.g., a pin in a clevis joint. τ = F / 2A
   • Punching shear: A punch forcing through a plate. τ = F / (π × d × t) where d is punch diameter and t is plate thickness.
4. Torsion: A twisting force (torque) that rotates one end relative to the other. The resulting stress is concentrated at the outer surface and falls away to almost nothing at the centre. Because that core material is doing so little work, a hollow shaft can match a solid one of the same outer diameter for torsional strength while cutting out a significant amount of weight. Examples: Car drive shafts; screwdriver blades; drill bits.
5. Bending (flexural stress): Produces tension on one face and compression on the opposite face of the beam, separated by the neutral axis (zero stress). Bending stress is maximum at the outermost fibres. This is why I-beams concentrate material at the flanges (top and bottom), which is where bending stress is highest, leaving the web thin to resist shear. Examples: Shelves sagging under books; a diving board; a car axle.

Load types:

• Dead loads: Permanent, constant weight: the structure itself plus fixed attachments. Does not change over time.
• Live loads: Variable, temporary loads: people, furniture, vehicles, stored goods. Must be estimated for the worst credible case.
• Environmental loads: Wind pressure, rain, snow accumulation, seismic ground motion, thermal expansion and contraction. Dynamic and unpredictable.
• Other loads: Foundation settlement, machinery vibration, impact (collision), blast.

Designers analyse structures by identifying all load types, determining how they combine in the worst-case scenario, and ensuring every structural member can carry its share of those combined loads with an adequate safety margin.

Students must be able toCalculate Young's Modulus using the formula E = σ / ε, and interpret stress-strain graphs identifying Young's Modulus, yield strength, ultimate strength and fracture.

Stress (σ) is the internal force per unit cross-sectional area that a material develops in response to an applied load:

σ = F / A    [Pa or MPa; note: 1 N/mm² = 1 MPa]

Where F = applied force (N) and A = cross-sectional area (m² for Pa; mm² for MPa). Use mm² consistently to work in MPa directly.

Strain (ε) is the fractional change in length caused by that stress:

ε = ΔL / L₀    [dimensionless: no units]

Where ΔL = change in length and L₀ = original length (use consistent units: both mm, or both m).

Young's Modulus (E), or stiffness, is the ratio of stress to strain in the elastic (linear) region of the stress-strain graph:

E = σ / ε    [GPa or MPa]

E is the slope of the straight-line portion of the stress-strain curve. A steeper slope = stiffer material. Steel: E ≈ 200 GPa. Aluminium: E ≈ 70 GPa. Rubber: E ≈ 0.01–0.1 GPa.

Reading the stress-strain graph (key points):

1. Elastic (linear) region: Stress and strain are proportional (Hooke's Law). The slope = E. Remove the load and the material returns to its original shape. The Young's Modulus is read from this region only.
2. Elastic limit / proportional limit: The stress at which the stress-strain relationship ceases to be linear. Below this point, deformation is fully reversible.
3. Yield point (yield strength, σ_y): The stress at which permanent (plastic) deformation begins. For many mild steels, there is a distinct upper yield point (sudden drop in load) followed by a lower yield point. Once passed, the material will not return to its original shape when unloaded.
4. Proof stress (for materials without a clear yield point): Aluminium alloys, titanium alloys, and copper do not show a well-defined yield point. The 0.2% proof stress (also called offset yield strength) is found by drawing a line parallel to the elastic region but offset 0.2% along the strain axis. Where this line intersects the stress-strain curve is the proof stress. This is the IB-standard method for these materials.
5. Ultimate Tensile Strength (UTS): The maximum stress on the graph: the peak of the curve. Beyond this point, the material begins to neck (thin locally) and load-carrying capacity decreases even as the material is still extending.
6. Necking: The localised reduction in cross-sectional area in ductile materials after UTS. The material thins at one point, and true stress actually increases there even though engineering stress (based on original area) decreases.
7. Fracture point: Where the material finally breaks. In ductile materials (mild steel, aluminium), there is significant plastic deformation between yield and fracture. In brittle materials (glass, cast iron, concrete in tension), fracture occurs with little or no plastic deformation: the curve drops sharply from near the elastic limit.

Effect of temperature: For plain carbon steels, Young's Modulus decreases as temperature increases. At elevated temperatures, steel softens and its stiffness falls: a critical consideration for fire-resistant structural design and high-temperature industrial equipment.

Worked example (calculating Young's Modulus):

A steel test piece, original length 50 mm and cross-sectional area 20 mm², is pulled with a force of 80 kN. It stretches 0.1 mm. Find E.

σ = F / A = 80,000 N / 20 mm² = 4,000 MPa

ε = ΔL / L₀ = 0.1 mm / 50 mm = 0.002

E = σ / ε = 4,000 MPa / 0.002 = 2,000,000 MPa = 200 GPa ✓ (consistent with steel)

Students must be able toIdentify why a structure has failed, including interpreting data from finite element analysis (FEA).

Structural failure occurs when stress in any part of a structure exceeds the material's capacity to resist it. Failure modes fall into four main categories:

1. Overloading: Applied loads exceed what the structure was designed for. Causes include unexpected live loads, dynamic impacts, or failure to account for cumulative loading.
2. Wrong material choice: A material with insufficient strength, toughness, or corrosion resistance for the operating environment. A mild steel fastener in a marine environment corrodes rapidly; a glass component in a high-impact application fractures brittlely.
3. Wrong size or shape: A section that is too thin for the load, or a geometry that concentrates stress at notches, holes, or sudden cross-section changes (stress concentrators). Fatigue cracks almost always initiate at stress concentrations.
4. Buckling: A mode of failure unique to slender members under compression. When a long, thin column or strut is loaded in compression, it can suddenly deflect laterally and collapse at a load far below its compressive strength. The critical buckling load depends on the member's slenderness (length-to-radius-of-gyration ratio), its cross-section shape, and its end conditions. Hollow tubes and I-sections resist buckling better than solid rods of the same area because they have a larger radius of gyration.

Case study (Quebec Bridge collapse, 1907):

During construction of what was to be the world's longest cantilever bridge (548 m main span), the south anchor arm collapsed, killing 75 of the 86 workers on site. The primary cause was the underestimation of the structure's own weight: the calculated weight was 30,857 tonnes but the actual weight was later found to be 36,408 tonnes. This led to compressive stresses in the lower chord members that exceeded the material's buckling resistance. Signs of buckling had been observed and reported for nearly a month before the collapse, but work continued. The bridge collapsed two hours after buckling was formally reported to the chief engineer.

Engineering lessons: Accurate weight calculations are non-negotiable. Warning signs of buckling must be acted on immediately. Clear lines of responsibility and authority to halt construction are essential. The disaster directly led to the reform of Canadian professional engineering standards.

Case study (Tacoma Narrows Bridge collapse, 1940):

The Tacoma Narrows Bridge was revolutionary in its slenderness: a depth-to-span ratio of 1:350 (compared to the typical 1:84 of contemporary suspension bridges) and a width-to-span ratio of 1:72, the narrowest of any comparable bridge. Its shallow plate girder sides acted as a solid wall to wind, rather than allowing air to pass through as an open truss would. During construction, the bridge already oscillated vertically in wind, earning the nickname "Galloping Gertie." Four months after opening, a 67 km/h gale induced a coupled bending-torsion oscillation (aeroelastic flutter): the roadway twisted at increasing amplitude until the deck tore apart.

Engineering lessons: Wind is not just a static pressure load; it can induce dynamic resonance. Torsional stiffness and aerodynamic stability are as important as vertical strength. Wind tunnel testing of scale bridge models became mandatory following this failure. Modern long-span bridges use open truss girders, aerodynamic deck profiles, and tuned mass dampers to prevent flutter.

Case study (Genoa Morandi Bridge collapse, 2018):

A 210 m section of the cable-stayed Morandi Bridge collapsed, killing 43 people. The failure was caused by decades of corrosion to steel cables inside the bridge's distinctive reinforced concrete pylons, combined with deterioration of the prestressed concrete. The original design embedded the stay cables inside concrete shrouds, a design that prevented inspection or replacement of the steel. Corrosion progressed invisibly until the cable system no longer had sufficient capacity to carry the bridge's dead load.

Engineering lessons: Safety factors degrade over time as materials corrode and fatigue. Infrastructure must be designed for inspectability and maintainability: hidden structural elements are a design failure, not just an operational problem. The replacement bridge (Genoa San Giorgio, 2020), designed by Renzo Piano, incorporates continuously operating monitoring robots that inspect every structural element for corrosion and cracking.

Finite Element Analysis (FEA):

FEA is a computational method that divides a complex structure into thousands of tiny elements (triangles or tetrahedra in 2D/3D). The software applies loads and constraints, then calculates stress, strain, and displacement at each element. The results are displayed as colour maps (stress contour plots): regions in red/orange indicate the highest stress, green and blue indicate lower stress. Designers use FEA to:

• Identify stress concentrations before any physical prototype is built.
• Optimise cross-section shapes and thicknesses to eliminate over-stressed regions without adding unnecessary material.
• Predict where fatigue cracks will initiate under cyclic loading.
• Verify that a proposed design meets safety factor requirements.

Interpreting FEA output: a region shown in red that coincides with a geometric feature (hole, fillet, notch) is a stress concentration: the designer should increase the radius of the fillet, add a gusset plate, or choose a stronger material for that region. A large region of uniform low stress (blue) indicates over-engineered material that could be removed to reduce weight.

The Hyatt Regency Walkway Collapse

The deadliest structural failure in US history.

Read case study →

Students must be able toInterpret simple force diagrams for a given structure.

Engineers represent forces and their effects using standardised diagrams. Three types are essential for structural analysis:

1. Free Body Diagrams (FBDs)

A free body diagram isolates a single object (or a section of a structure) and shows all external forces acting on it as vectors: arrows indicating direction and magnitude. FBDs are the starting point for every structural calculation.

Rules for drawing FBDs:

• Draw the object in isolation: remove all surrounding context.
• Show every external force: applied loads (arrows at the point of application), reactions at supports (upward at supports, horizontal at pinned supports if required).
• Label each force with its magnitude and direction.
• Show the coordinate system (x–y axes).

2. Force polygons (tip-to-tail vector addition)

When multiple forces act on a point, their resultant (combined effect) can be found graphically. Draw each force vector to scale, placing the tail of each arrow at the tip of the previous one. The resultant is the vector from the starting point to the final tip. If the forces are in equilibrium, the polygon closes (the last tip meets the first tail): this is the graphical equivalent of ΣF = 0.

3. Support reactions

Before drawing shear force and bending moment diagrams, the reactions at all supports must be found using the two equilibrium equations:

ΣF = 0    (sum of all vertical forces = 0)

ΣM = 0    (sum of moments about any point = 0)

Two support types:

• Roller support: Provides a reaction force perpendicular to the surface only (vertical for a horizontal beam). It cannot resist horizontal forces. One unknown: R_vertical.
• Pinned support: Provides reactions in any direction: both vertical and horizontal components. Two unknowns: R_vertical and R_horizontal.

Worked example (finding support reactions): A 6 m simply-supported beam carries a 10 kN point load at 2 m from the left (pinned) support A. Find reactions R_A and R_B (roller at B).

Take moments about A: ΣM_A = 0 → R_B × 6 = 10 × 2 → R_B = 20/6 = 3.33 kN

ΣF_vertical = 0 → R_A + R_B = 10 → R_A = 10 − 3.33 = 6.67 kN

4. Shear force diagrams (SFD)

A shear force diagram plots the internal shear force at every cross-section along the beam. Sign convention: shear that tends to cause clockwise rotation of the left segment is positive (+). The diagram is built by working from left to right, adding each load or reaction encountered. Point loads cause vertical jumps in the diagram; uniformly distributed loads (UDL) cause linear slopes.

5. Bending moment diagrams (BMD)

A bending moment diagram plots the internal bending moment at every cross-section. The critical rule: maximum bending moment occurs where shear force equals zero. The BMD is the area under the SFD. Point loads produce triangular shapes; UDLs produce parabolic curves. The BMD is essential for sizing the beam: the peak moment location is where the beam is most likely to fail in bending.

Uniformly distributed loads (UDL): Expressed in kN/m. A UDL of 5 kN/m over 4 m is equivalent to a point load of 20 kN at the midpoint of the span for the purpose of calculating support reactions. The total UDL force = w × L where w is load intensity (kN/m) and L is span length (m).

Cantilever beams: Fixed at one end, free at the other. The fixed support must resist both vertical force and bending moment. The maximum bending moment in a cantilever occurs at the fixed end, not in the middle. Examples: Stadium roof canopies, balconies, aircraft wings. A UDL on a cantilever of length L and intensity w gives: maximum bending moment M_max = w × L² / 2 at the fixed end.

Students must be able toCalculate SFs using the formula SF = Ultimate Load (Stress) / Allowable Load (Stress); calculate maximum intended loads for given structures; and design structures with an SF.

The safety factor (also called factor of safety, FOS) is the ratio of a structure's ultimate strength to the maximum stress it is designed to carry in service:

SF = Ultimate Load (or Stress) / Allowable Load (or Stress)

Rearranged to find allowable working stress:

σ_working = UTS / SF

And maximum working load:

F_working = σ_working × A = (UTS / SF) × A

Why use a safety factor? Real structures operate in conditions that are imperfect, unpredictable, and changing. The SF absorbs uncertainty across nine categories:

1. Certainty of loads: Actual loads may exceed estimates due to accidents, misuse, or unforeseen conditions.
2. Design life: A structure designed for 50 years must remain safe as materials fatigue and degrade.
3. Manufacturing quality: Real materials have flaws, inconsistencies in composition, and surface defects that reduce strength below laboratory values.
4. Consequences of failure: Catastrophic failure (loss of life, uncontained release of hazardous materials) warrants a higher SF than failure of a non-critical component.
5. Environmental influences: Corrosion, UV degradation, temperature cycling, and chemical attack reduce material strength over time.
6. Criticality: Whether the component is part of a redundant system (failure of one does not cause total collapse) or a single point of failure.
7. Repairability: A buried pipeline cannot easily be inspected or repaired: a higher SF is warranted compared to a visible, accessible structural member.
8. Certainty of material properties: Properties from material databases are average values; actual strength may be ±15% of the published figure.
9. Statutory and code requirements: Industry standards (ISO, ASTM, EN, AS) mandate minimum SF values for specific applications. These are legal requirements, not suggestions.

Typical safety factors by application (from A3.2):

Worked example 1 (calculating SF):

A steel rod (diameter 12 mm) fails at a load of 90 kN. It is designed to carry a working load of 30 kN. What is the SF?

A = π × d² / 4 = π × 144 / 4 = 113.1 mm²

UTS = 90,000 / 113.1 = 795.8 MPa

σ_working = 30,000 / 113.1 = 265.3 MPa

SF = 795.8 / 265.3 = 3.0

Worked example 2 (calculating maximum working load, from the MD):

A 16 mm diameter steel rod has UTS = 590 MPa and SF = 4. Find the maximum working load.

A = π × 16² / 4 = 201.1 mm²

σ_working = 590 / 4 = 147.5 MPa

F_working = 147.5 × 201.1 = 29,662 N ≈ 29.7 kN

Safety factors degrade over time (the Genoa Morandi lesson):

The Morandi Bridge was designed with an adequate SF at opening in 1967. Over 50 years, corrosion of the embedded steel cables progressively reduced their cross-sectional area and tensile strength, effectively lowering the actual SF year by year. By 2018, the SF for the corroded cables had fallen below 1, and the bridge collapsed. The lesson: the SF at the time of construction is not the SF in service. Infrastructure monitoring, regular inspection, and maintenance are required to keep the actual SF above the design SF throughout the structure's intended life.