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

Question 1: Load Application, 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) — Static and dynamic loading, and the time-dependent response of materials (10 marks)

Static load application (3 marks). A load is applied statically when it is brought onto the structure slowly enough that the rate of loading has no measurable effect on the response, and when it is then held essentially constant. Two conditions are implied. First, the acceleration of the mass of the structure is negligible, so no inertial force appears in the equation of equilibrium and the internal forces follow directly from statics. Second, the strain rate is low enough that the measured stiffness and strength are the quasi-static values quoted in the code. The self-weight of a bridge girder, the weight of the fill on a buried culvert, the permanent equipment load on a plant floor and the water pressure on the upstream face of a dam are all static. A laboratory compression test on a concrete cylinder is deliberately run as a static test: CSA A23.2-9C requires the load to be applied at 0.15 to 0.35 MPa per second precisely so that rate effects do not contaminate the reported strength.

Dynamic load application (3 marks). A load is dynamic when it varies rapidly enough with time that the inertia and the damping of the structure participate in carrying it, so that the response must be found from the equation of motion rather than from statics alone, and when the strain rate is high enough to change the material's own behaviour. Vehicle impact on a bridge deck, the wheel loads of a truck at highway speed on a pavement, wind gusts and vortex shedding on a slender stack, earthquake ground motion, blast and pile-driving hammer blows are the standard civil examples. Two consequences matter to the materials engineer. The first is amplification: a suddenly applied load can produce roughly twice the deflection of the same load applied gradually, which is why the bridge codes add a dynamic load allowance to the truck load. The second is a change in the material response itself — concrete and steel both show a higher apparent strength and a lower ductility at high strain rate, and a steel that is perfectly ductile in a slow tension test can fracture in a brittle manner under impact at low temperature, which is exactly what the Charpy test of Q.5(b) is designed to detect.

Time-dependent response (4 marks). A material has a time-dependent response when its strain is not a unique function of the current stress: the strain continues to change while the stress is held constant, and the stress relaxes while the strain is held constant. An ideal elastic solid has no time dependence, because Hooke's law fixes a one-to-one relation between stress and strain and the deformation is recovered instantly on unloading. Real construction materials are viscoelastic to some degree — concrete, asphalt, wood and polymers markedly so, steel appreciably so only at elevated temperature. The engineering significance is that the deformation the designer must check is not the one measured at first loading. The two mechanisms named in the question sit at the two ends of this behaviour.

Creep is the gradual increase of strain under a sustained stress, at a decreasing rate, with a part of the strain recovered slowly on unloading and a part left permanently. In concrete it arises from the migration of adsorbed water in the calcium–silicate–hydrate gel and from micro-cracking; the creep strain of a normal structural concrete under service stress typically reaches two to three times the initial elastic strain after several years. It is the reason a prestressed girder loses prestress and continues to camber for years after transfer, the reason a long-span flat slab must be checked for long-term deflection using a multiplier on the immediate deflection, and the reason CSA A23.3 requires sustained-load deflections to be computed separately. Creep in a timber roof member under permanent snow load, and creep in the steel of a fire-exposed member above about 400 °C, are the other classic civil examples.

Viscous flow is the continued deformation of a material at a rate proportional to the applied shear stress, with no recovery whatsoever on unloading: the material behaves as a liquid, so all of the deformation is permanent. Asphalt binder is the textbook example. At high pavement temperature the binder is essentially a Newtonian liquid, the shear strain rate is proportional to the shear stress, and repeated wheel loads accumulate unrecovered strain in the wheelpath — that accumulation is rutting. It is precisely because binders flow viscously that they are graded by viscosity or by the Superpave high-temperature performance grade, and why a stiffer binder is specified for slow-moving or standing traffic. Fresh concrete before setting and a soft clay under a sustained embankment load are the other examples usually offered. The practical distinction to state is that creep is delayed but partly recoverable deformation in a solid, whereas viscous flow is unrecoverable flow of a material that has no equilibrium shape at all.

Part (b) — Tension-test properties of metals A and B (10 marks)

Given. The printed stress–strain chart for two metals tested in tension to fracture, with stress in ksi on the vertical axis (gridlines at 0, 50, 100 and 150) and strain in in/in on the horizontal axis (0.00 to 0.14). Metal A is the solid curve and metal B the dashed curve. Reading the chart:

Find. For each metal: the proportional limit, the 0.002 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. See the official exam paper.]

Approach. Take the proportional limit and the elastic modulus from the slope and the end of the initial straight portion of each curve, draw a line of that slope from ε = 0.002 to find the offset yield, read the ultimate strength and fracture strain directly, then obtain the modulus of resilience as the triangular area under the elastic line and the toughness as the whole area under the curve to fracture.

1. Read the proportional limit and the modulus of elasticity for each metal. The proportional limit is the stress at which the initial straight portion of the curve ends. Metal A leaves its straight line at about 50 ksi, reached at a strain of about 0.0025; metal B leaves its straight line at about 45 ksi, reached at a strain of about 0.0045. The modulus of elasticity is the slope of that straight portion, $$E=\frac{\sigma_{PL}}{\varepsilon_{PL}}$$ so that $$E_A=\frac{50}{0.0025}=20\,000\ \text{ksi},\qquad E_B=\frac{45}{0.0045}=10\,000\ \text{ksi}.$$ $$\boxed{\ \sigma_{PL,A}\approx 50\ \text{ksi},\quad E_A\approx 20\times 10^{3}\ \text{ksi};\qquad \sigma_{PL,B}\approx 45\ \text{ksi},\quad E_B\approx 10\times 10^{3}\ \text{ksi}\ }$$ Metal A is twice as stiff as metal B, and the two moduli are the orders of magnitude expected of a steel and of an aluminium alloy respectively.
2. Construct the 0.002 offset line and read the yield stress. The offset yield stress is defined as the ordinate of the point where a line of slope $E$, drawn from a strain of 0.002 at zero stress, cuts the stress–strain curve: $$\sigma=E\,(\varepsilon-0.002).$$ For metal A the construction line $\sigma=20\,000(\varepsilon-0.002)$ meets the curve where the curve has already turned over, at $$\varepsilon_y\approx 0.0052\ \text{in/in},\qquad \sigma_{y,A}\approx 63\ \text{ksi}.$$ For metal B the flatter line $\sigma=10\,000(\varepsilon-0.002)$ meets its curve at $$\varepsilon_y\approx 0.0071\ \text{in/in},\qquad \sigma_{y,B}\approx 51\ \text{ksi}.$$ $$\boxed{\ \sigma_{y,A}\approx 63\ \text{ksi},\qquad \sigma_{y,B}\approx 51\ \text{ksi}\ }$$ Both offset yields lie above the corresponding proportional limits, as they must: the offset construction is only meaningful once the curve has begun to depart from the elastic line.
3. Read the ultimate strength. The ultimate strength is the highest ordinate reached before fracture. Neither curve shows a falling branch — each simply stops — so on this chart the ultimate strength coincides with the fracture stress: $$\boxed{\ \sigma_{u,A}\approx 132\ \text{ksi}\ \ \text{at}\ \varepsilon_f=0.079;\qquad \sigma_{u,B}\approx 73\ \text{ksi}\ \ \text{at}\ \varepsilon_f=0.117\ }$$ Metal A is therefore the stronger metal by a factor of about 1.8, both at yield and at ultimate.
4. Compute the modulus of resilience. Resilience is the elastic strain energy stored per unit volume up to the proportional limit — the area of the triangle under the straight portion: $$U_r=\tfrac12\,\sigma_{PL}\,\varepsilon_{PL}=\frac{\sigma_{PL}^{2}}{2E}.$$ Substituting for each metal, $$U_{r,A}=\frac{(50)^2}{2(20\,000)}=0.0625\ \text{ksi},\qquad U_{r,B}=\frac{(45)^2}{2(10\,000)}=0.10125\ \text{ksi}.$$ Expressing these in the more familiar units of energy per unit volume, $$\boxed{\ U_{r,A}\approx 62.5\ \text{in}\cdot\text{lb/in}^{3},\qquad U_{r,B}\approx 101\ \text{in}\cdot\text{lb/in}^{3}\ }$$ The weaker metal has the greater resilience, which is not a contradiction: resilience rewards a high elastic limit combined with a low stiffness, and metal B trades only 5 ksi of proportional limit for half the modulus.
5. Compute the toughness. Toughness is the total energy absorbed per unit volume up to fracture, that is the whole area under the stress–strain curve: $$U_t=\int_{0}^{\varepsilon_f}\sigma\,d\varepsilon .$$ Integrating the two digitised curves numerically by the trapezoidal rule gives $$U_{t,A}\approx 8.2\ \text{ksi},\qquad U_{t,B}\approx 7.5\ \text{ksi},$$ that is $$\boxed{\ U_{t,A}\approx 8\,200\ \text{in}\cdot\text{lb/in}^{3},\qquad U_{t,B}\approx 7\,500\ \text{in}\cdot\text{lb/in}^{3}\ }$$ A quick check by the Ludwik approximation $U_t\approx \tfrac23\sigma_u\varepsilon_f$ gives 7.0 and 5.7 ksi respectively, the right order and the right ranking, which confirms the numerical areas. Metal A is the tougher material: it gives up some fracture strain but more than makes it up in stress.
6. Decide which metal is the more ductile, and why. Ductility is the ability to sustain permanent deformation before fracture, and it is measured by the strain at fracture (or by the percentage elongation and reduction of area computed from it), not by the area under the curve and not by the strength. Comparing the two fracture strains, $$\frac{\varepsilon_{f,B}}{\varepsilon_{f,A}}=\frac{0.117}{0.079}=1.48,$$ so metal B stretches about 48 per cent further before it breaks, and the corresponding elongations over the gauge length are about 11.7 per cent for B against 7.9 per cent for A. $$\boxed{\ \text{Metal B is the more ductile: }\varepsilon_{f,B}=0.117>\varepsilon_{f,A}=0.079\ }$$ The point worth making in the answer is that B is the more ductile metal while A is simultaneously the stronger and the tougher one — ductility and toughness are different properties, and a material can have more of one and less of the other.