Wood in Compression and Steel Test Methods: 16-Civ-B11 Structural Materials, Undated paper, Q5

Question 5: Wood in Compression and Steel Test Methods (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) — Stress–strain relationship, modulus of elasticity and failure stress (12 marks)

Given. A clear wood prism, 1 in × 1 in in cross-section and 4 in long, grain parallel to the length, loaded in compression parallel to the grain until it failed. Thirteen load–displacement readings were recorded:

Load (lb)	0	20	35	85	475	1650	2575	3825	4575	5325	5125	4575	4350
Displacement (in)	0	0.015	0.075	0.170	0.185	0.215	0.238	0.265	0.305	0.325	0.365	0.385	0.407

Find. The stress–strain curve, the modulus of elasticity from its straight portion, and the failure (maximum) stress.

Approach. Convert each load to a stress by dividing by the 1 in² cross-section and each displacement to a strain by dividing by the 4 in length, plot the result, fit a straight line to the linear run and take its slope as the modulus, extrapolating that line back to zero stress to correct for the seating toe, and read the failure stress as the peak of the curve.

Convert loads and displacements to stresses and strains. The cross-sectional area and the gauge length are $$A=1\ \text{in}\times 1\ \text{in}=1\ \text{in}^{2},\qquad L=4\ \text{in},$$ so that $$\sigma=\frac{P}{A}=P\ \ [\text{psi}],\qquad \varepsilon=\frac{\delta}{L}=\frac{\delta}{4}\ \ [\text{in/in}].$$ Because the area is exactly one square inch, the stress in psi is numerically equal to the load in pounds, and the thirteen strains are 0, 0.00375, 0.01875, 0.04250, 0.04625, 0.05375, 0.05950, 0.06625, 0.07625, 0.08125, 0.09125, 0.09625 and 0.10175 in/in.
Plot the curve and identify the straight portion. The plotted points fall into three clearly separated regions. The first four (0 to 85 lb over 0.170 in of travel) form an almost horizontal toe in which the load barely rises: this is the seating of the platens on the end grain and the take-up of the machine, not a property of the wood. The next four points, from 475 lb at ε = 0.04625 to 3825 lb at ε = 0.06625, form a straight line. Beyond 3825 lb the curve bends over towards the peak at 5325 lb and then falls away as the fibres buckle and crush.
Fit the straight run and take the modulus of elasticity as its slope. A least-squares line through those four points, $$E=\frac{\sum(\varepsilon_i-\bar{\varepsilon})(\sigma_i-\bar{\sigma})}{\sum(\varepsilon_i-\bar{\varepsilon})^{2}}=\frac{36.145}{2.1667\times 10^{-4}},$$ gives a slope of 166 800 psi and an intercept of −7 284 psi. As a two-point check, taking the first and last points of the run directly, $$E=\frac{3825-475}{0.06625-0.04625}=\frac{3350}{0.0200}=167\,500\ \text{psi},$$ which agrees with the fitted slope to better than half a per cent. $$\boxed{\,E\approx 1.67\times 10^{5}\ \text{psi}\ (1.15\ \text{GPa})\,}$$
Apply the toe correction. Extrapolating the fitted line back to zero stress locates the corrected origin of the strain axis: $$\varepsilon_0=-\frac{b}{E}=\frac{7\,284}{166\,821}=0.0437\ \text{in/in},$$ so the strain recorded before the specimen began to take load is 0.0437, and every strain should be reduced by that amount before it is quoted as a material strain. On that basis the corrected strain at failure is $$\varepsilon_f=0.08125-0.0437=0.0376\ \text{in/in}.$$ The toe correction does not change the modulus — a slope is unaffected by a shift of origin — but it is essential to any statement about strain, and the graph should be drawn with the corrected origin marked.
Read the failure stress. The maximum load carried was 5 325 lb, at a displacement of 0.325 in, after which the load fell steadily as the specimen crushed: $$\sigma_f=\frac{P_{\max}}{A}=\frac{5\,325}{1.0}$$ $$\boxed{\,\sigma_f=5\,325\ \text{psi}\ (36.7\ \text{MPa})\,}$$ This is squarely inside the 4 000 to 8 000 psi band expected for clear softwood loaded in compression parallel to the grain, so the strength result is sound.

Figure 5.1 — the stress–strain curve from the recorded data, with the seating toe, the straight run used for the modulus, the toe-corrected origin and the peak marked. Stress in psi is numerically equal to load in pounds because the specimen area is exactly 1 in².

Check: the modulus computed above, 1.67 × 10⁵ psi, is about six times lower than the 1.0 to 1.9 × 10⁶ psi that clear softwood shows in compression parallel to the grain, and the corrected strain at failure, 3.8 per cent, is several times the 0.5 to 1 per cent normally measured. Both discrepancies have the same cause: the recorded displacement is crosshead travel over a 4 in specimen, not the extension of a gauge length measured by an extensometer, so it contains the elastic deformation of the loading train and the local crushing of the end grain as well as the strain in the specimen. The 0.170 in of travel absorbed before the load reaches 85 lb is direct evidence of it. The number obtained is therefore an apparent modulus of the specimen-plus-machine system and should be reported as such; it is the quantity the question's data permit, and the data have not been adjusted to force agreement with the book value. The failure stress, which depends only on the load cell and the measured area, is unaffected and is reliable.

Quantity	Value
Cross-sectional area / gauge length	1.00 in² / 4.00 in
Straight run used for the modulus	475 to 3 825 lb (ε = 0.04625 to 0.06625)
Modulus of elasticity (apparent)	1.67 × 10⁵ psi = 1.15 GPa
Toe correction (corrected strain origin)	ε₀ = 0.0437 in/in
Maximum (failure) load	5 325 lb at 0.325 in
Failure stress	5 325 psi = 36.7 MPa
Corrected strain at failure	0.0376 in/in

Part (b) — Significance and use of the three steel tests (8 marks)

i. Tension test (ASTM A370 / ASTM E8; CSA G40.20 for structural steel). A machined or full-section coupon is pulled to fracture in a testing machine while load and extension are recorded, giving the stress–strain diagram and from it the yield strength (by the 0.2 per cent offset or by the upper/lower yield point), the tensile strength, the percentage elongation over a 50 mm or 8 in gauge length, and the reduction of area. Its significance is that it is the primary test from which the design strength of the steel comes: every resistance equation in CSA S16 is written in terms of $F_y$ and $F_u$, and the mill certificate that accompanies a shipment of steel is essentially a record of tension tests. Its uses go beyond strength. The elongation and reduction of area are the acceptance measures of ductility, which is what allows moment redistribution, plastic design and the energy dissipation that seismic design depends on; the ratio $F_u/F_y$ is specified as a minimum in seismic applications to guarantee strain hardening; and the shape of the curve distinguishes a hot-rolled structural steel, with its sharp yield plateau, from a cold-worked or heat-treated product that has none.

ii. Charpy V-notch impact test (ASTM E23; CSA G40.21 category T steels). A 10 mm square bar carrying a standard 2 mm deep V-notch is broken by a swinging pendulum at a specified temperature, and the energy absorbed in fracture is reported in joules, along with the lateral expansion and the percentage of shear (fibrous) fracture on the broken surface. Its significance is that it measures toughness in the presence of a notch and at a controlled temperature, which is exactly the combination the tension test cannot address. A structural steel that is thoroughly ductile in a slow tension test at room temperature can fail in a brittle, near-zero-energy manner when it contains a notch, is loaded rapidly, and is cold — the mechanism behind the Liberty ship and bridge fractures that led to the test being adopted. Its use is to establish the ductile-to-brittle transition temperature and to qualify steel for service: the specification calls for a minimum absorbed energy (commonly 27 J) at the lowest anticipated service temperature, which in Canada may be −45 °C for a northern bridge. It is mandatory for fracture-critical members, for thick plate, and for members that are welded, since welding introduces both notches and residual tension.

iii. Bend test (ASTM E290; CSA G30.18 for reinforcing bars, CSA W59 for welds). A specimen is bent through a specified angle around a mandrel of specified diameter, usually 180 degrees, and the outside of the bend is examined for cracking. Its significance is that it is a simple, qualitative but very searching test of ductility and soundness in a severely strained surface: the outer fibre is taken to strains of the order of 10 to 20 per cent in one operation, so it exposes surface defects, laminations, inclusions, embrittlement from over-heating and poor weld fusion that a tension test on a machined coupon can miss entirely. Its uses are threefold. It is the standard qualification test for welders and welding procedures, where face, root and side bends are required to remain crack-free. It is a mandatory mill test for reinforcing steel, since a bar must be capable of being bent cold around a standard pin to form stirrups and hooks without cracking. And it is used as a fabrication-suitability check on plate that must be cold-formed into bends, where the acceptable mandrel diameter is itself the specification.

Topic. Compression parallel to the grain in wood, machine-compliance artefacts in a load–deformation record, and the three standard mechanical acceptance tests for structural steel.

Key relations. $\sigma=P/A$; $\varepsilon=\delta/L$; $E=\Delta\sigma/\Delta\varepsilon$ over the linear run; toe correction $\varepsilon_0=-b/E$ from the least-squares intercept $b$; series compliance $1/k_{\text{measured}}=1/k_{\text{specimen}}+1/k_{\text{machine}}$, which is why crosshead travel always overstates strain; typical clear softwood compression parallel to grain: $F_c\approx 4\,000$ to $8\,000$ psi, $E\approx 1.0$ to $1.9\times 10^{6}$ psi.

Why this works. Wood is a bundle of hollow cellulose tubes aligned with the grain. Loaded along that axis it is being loaded along the axis of the tube walls, so the stiffness and the strength are high and are governed by the cell wall material itself; loaded across the grain it merely crushes the tubes flat, and both properties fall by an order of magnitude. That anisotropy is why the question specifies the grain direction, and why CSA O86 tabulates entirely different design values for $F_c$ parallel and perpendicular to the grain. The failure mode in the parallel direction is buckling of the individual fibres, which is a progressive rather than a sudden mechanism — hence the rounded peak and the gentle falling branch in the plotted curve, quite unlike the sharp fracture of a metal. The toe correction works because the machine and the crushed end grain behave as a spring in series with the specimen: once the seating is complete, that spring contributes a constant offset, so the straight portion is displaced but not tilted, and extrapolating it back to zero stress recovers the true origin.

Common pitfall. The main trap is to compute the modulus from the origin to the first substantial load point, which includes the whole seating toe and yields a modulus several times too low; the slope must be taken from the straight run only. The second is to quote the resulting modulus as the modulus of elasticity of wood without noticing that it is an order of magnitude below every published value — a sanity check against the reference range should be automatic. On the steel tests, the recurring error is to describe the Charpy test as measuring toughness in general: it measures notch toughness at a stated temperature, and a result is meaningless unless the test temperature is quoted with it.

Source. Mamlouk & Zaniewski, Materials for Civil and Construction Engineers, 4th ed., Ch. 10 (Wood — anisotropy, compression parallel and perpendicular to grain, test methods) and Ch. 3 (Steel — tension, impact and bend testing); ASTM D143 (wood test methods), ASTM A370/E8 (tension), ASTM E23 (Charpy V-notch), ASTM E290 (bend); CSA O86 Engineering Design in Wood and CSA G40.21 for the Canadian toughness categories.

Question 5 of 5: Wood in Compression and Steel Test Methods

Question 5: Wood in Compression and Steel Test Methods (20 marks)

Part (a) — Stress–strain relationship, modulus of elasticity and failure stress (12 marks)

Part (b) — Significance and use of the three steel tests (8 marks)