Question 1 of 5: Load Regimes, Time-Dependent Response, and Tension-Test Properties

Question 1: Load Regimes, Time-Dependent Response, and Tension-Test Properties (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a) — Loading regimes and time dependence (10 marks)

Static and dynamic load application (3 marks). A load is applied statically when it is brought on slowly enough that the acceleration of the mass it acts on is negligible, so no inertial force is generated and the material has time to reach equilibrium at every load increment. The stress state then depends only on the magnitude of the load, not on how quickly it arrived. Dead load on a bridge girder, the hydrostatic pressure inside a reservoir, stored grain in a silo, and a laboratory compression test run at a controlled 0.25 mm/min are all static applications. A load is applied dynamically when it changes fast enough that inertia, damping, and the strain-rate sensitivity of the material all matter. Truck axles crossing a pavement joint, wind gusts on a tall building, earthquake ground motion, pile-driving hammer blows, and blast or vehicle impact on a barrier are dynamic. The practical consequence is that most structural materials show a higher apparent strength and a lower apparent ductility as the strain rate rises, and that repeated dynamic application introduces fatigue, which no static analysis will reveal.

Time-dependent response (3 marks). A material has a time-dependent response when its deformation continues to change while the applied stress is held constant — equivalently, when its stress-strain curve is a function of the rate and the duration of loading rather than of the load alone. A purely elastic solid responds instantaneously and recovers instantaneously; a real civil-engineering material superimposes on that elastic response a delayed, partly recoverable (viscoelastic) component and a permanent, unrecoverable (viscous) component. The same physics appears in two dual forms: holding the stress constant and watching the strain grow is creep, while holding the strain constant and watching the stress decay is relaxation — prestress loss in a post-tensioned tendon is the second, long-term camber of the same girder is the first.

Creep (4 marks, first half). Creep is the time-dependent increase in strain under a sustained stress that is well below the short-term strength. A classical creep curve has three stages: a primary stage in which the strain rate decreases as the microstructure adjusts, a secondary or steady-state stage at an essentially constant minimum rate, and a tertiary stage in which damage accumulates, the rate accelerates, and rupture follows. In Portland-cement concrete the mechanism is migration of adsorbed water in the C-S-H gel, and the practical results are prestress loss, long-term deflection of slabs, and redistribution of moments in indeterminate frames; creep strain is commonly one to three times the elastic strain. Structural steel creeps negligibly at service temperature but creeps severely above roughly 400 °C, which is why fire protection governs steel-frame design. Timber floor joists carrying sustained load acquire a creep deflection that CSA O86 handles through a duration-of-load factor.

Viscous flow (4 marks, second half). Viscous flow is the part of the time-dependent deformation that is entirely unrecoverable: the material moves like a very stiff liquid, with a strain rate governed by the stress rather than a strain governed by the stress. For a Newtonian material the two are proportional through the viscosity, so that a constant deviator stress produces a strain that grows linearly and without limit in time. Asphalt binder is the standard civil example — at a summer pavement temperature it flows, which is precisely why an asphalt-concrete wearing course ruts under channelised truck traffic and why the same mix behaves almost elastically in winter. Fresh concrete flowing in a form, bitumen bleeding from an over-asphalted mix, and soft clay squeezing beneath an embankment are other examples. The distinction to keep straight is that elastic strain is instantaneous and fully recovered, viscoelastic (creep) strain is delayed and partly recovered, and viscous strain is delayed and never recovered.

Part (b) — Reading the tension-test curves (10 marks)

Given. The printed chart gives the engineering stress-strain traces of two metals tested in tension to fracture, with stress in ksi from 0 to 150 and strain in in/in from 0 to 0.14. Reading the chart:

Find. For each metal: the proportional limit, the 0.2 % offset yield stress, the ultimate strength, the modulus of resilience, the toughness, and a reasoned statement of which metal is the more ductile.

[Figure not reproduced: Figure 1.1 — the two traces redrawn from the examination chart. Green circles mark the proportional limits, the grey dashed lines are the 0.2 % offset construction, and the black circles are the offset yield points. Each curve stops at fracture. See the official exam paper.]

Approach. Take the slope of the initial straight line as the modulus of elasticity, take the proportional limit where the trace leaves that line, obtain the yield stress by intersecting the offset line with the curve, read the ultimate strength and the fracture strain directly, and get the two energy quantities as areas — the elastic triangle for resilience and the whole area under the curve for toughness.

1. Part (b) I — modulus of elasticity and proportional limit. The initial portion of each trace is straight through the origin, so $$E=\frac{\sigma}{\varepsilon}\Big|_{\text{initial straight line}}$$ Metal A stays straight to $\sigma=52\ \text{ksi}$ at $\varepsilon=0.0026$ and Metal B to $\sigma=45\ \text{ksi}$ at $\varepsilon=0.0041$, which gives $E_{A}=52/0.0026=20\,000\ \text{ksi}$ and $E_{B}=45/0.0041\approx11\,000\ \text{ksi}$. The proportional limit is the stress at which the trace first departs from that line: $$\boxed{\sigma_{PL,A}\approx52\ \text{ksi},\qquad \sigma_{PL,B}\approx45\ \text{ksi}}$$ Metal A is roughly twice as stiff as Metal B, which is already the signature of a steel tested alongside a light alloy.
2. Part (b) II — yield stress at the 0.002 offset. Draw a line of slope $E$ starting at $\varepsilon=0.002$ on the strain axis, $$\sigma_{\text{offset}}(\varepsilon)=E\,(\varepsilon-0.002)$$ and take the yield stress where it cuts the measured curve. For Metal A the intersection falls at $\varepsilon\approx0.0051$, and for Metal B at $\varepsilon\approx0.0068$, so $$\boxed{\sigma_{y,A}\approx62\ \text{ksi},\qquad \sigma_{y,B}\approx53\ \text{ksi}}$$ Neither metal shows a sharp yield plateau, which is exactly the situation the offset definition exists to handle.
3. Part (b) III — ultimate strength. The ultimate (tensile) strength is the highest engineering stress the specimen carries. Both traces rise monotonically to fracture, so the ultimate strength is the stress at the end of each curve: $$\boxed{\sigma_{u,A}\approx131\ \text{ksi},\qquad \sigma_{u,B}\approx74\ \text{ksi}}$$ Metal A is about 1.8 times as strong as Metal B.
4. Part (b) IV — modulus of resilience. Resilience is the elastic strain energy stored per unit volume up to the proportional limit, i.e. the area of the triangle under the straight portion: $$U_{r}=\tfrac{1}{2}\,\sigma_{PL}\,\varepsilon_{PL}=\frac{\sigma_{PL}^{2}}{2E}$$ Substituting, $U_{r,A}=52^{2}/(2\times20\,000)=0.0676\ \text{ksi}$ and $U_{r,B}=45^{2}/(2\times10\,976)=0.0922\ \text{ksi}$. Since $1\ \text{ksi}\times1\ \text{in/in}=1000\ \text{in}\cdot\text{lb/in}^{3}$, $$\boxed{U_{r,A}\approx68\ \frac{\text{in}\cdot\text{lb}}{\text{in}^{3}},\qquad U_{r,B}\approx92\ \frac{\text{in}\cdot\text{lb}}{\text{in}^{3}}}$$ Note the reversal: the weaker metal stores more elastic energy per unit volume, because resilience rewards a low modulus as much as a high elastic limit.
5. Part (b) V — toughness. Toughness is the total energy absorbed per unit volume to fracture, the whole area under the engineering stress-strain curve: $$U_{T}=\int_{0}^{\varepsilon_{f}}\sigma\,d\varepsilon$$ Evaluating the area by the trapezoidal rule over the plotted traces gives $U_{T,A}\approx8.08\ \text{ksi}$ and $U_{T,B}\approx7.60\ \text{ksi}$, that is $$\boxed{U_{T,A}\approx8{,}100\ \frac{\text{in}\cdot\text{lb}}{\text{in}^{3}},\qquad U_{T,B}\approx7{,}600\ \frac{\text{in}\cdot\text{lb}}{\text{in}^{3}}}$$ A quick sanity check is the rectangle approximation $U_{T}\approx\tfrac{1}{2}(\sigma_{y}+\sigma_{u})\varepsilon_{f}$, which returns $7.6$ and $7.4\ \text{ksi}$ respectively — the right size, and it shows why the two toughnesses are close despite the large difference in strength.
6. Part (b) VI — which metal is more ductile. Ductility is the amount of permanent strain the material accepts before it breaks, so it is measured by the strain at fracture (or, on a broken specimen, by percent elongation and reduction of area). Metal A fractures at $\varepsilon_{f}=0.079$ and Metal B at $\varepsilon_{f}=0.117$, so $$\boxed{\text{Metal B is the more ductile: }11.7\ \%\text{ elongation against } 7.9\ \%\text{, a ratio of about }1.5}$$ Metal B also spends far more of its curve on the flat post-yield plateau, which is the practical meaning of ductility in a structure: it gives visible warning, allows moment redistribution, and tolerates fabrication strain.

Collecting the six required quantities for both metals: