Wood in Compression Parallel to Grain, and Laboratory Tests on Steel: 16-Civ-B11 Structural Materials, December 2019, Q5

Question 5: Wood in Compression Parallel to Grain, and Laboratory Tests on Steel (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.

Quantity	Value
Cross-section (loaded face)	1.5 in × 1.5 in
Cross-sectional area, A	2.25 in²
Specimen length (gauge length), L	5 in
Grain direction	Parallel to the length; load applied parallel to grain
Record	13 load–displacement pairs, 0 to 5,325 lb and 0 to 0.395 in

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

Approach. Convert each load to an engineering stress by dividing by the constant cross-sectional area and each displacement to a strain by dividing by the 5 in length, plot the pairs, identify the straight run above the seating (toe) region, take the modulus as its slope, and read the failure stress as the peak of the curve.

Convert load to stress and displacement to strain. The area is $A = 1.5 \times 1.5 = 2.25\ \text{in}^2$ and the gauge length is $L = 5\ \text{in}$, so $\sigma = P/A$ and $\varepsilon = \delta/L$. For example, at $P = 1650\ \text{lb}$ and $\delta = 0.205\ \text{in}$, $\sigma = 1650/2.25 = 733.3\ \text{psi}$ and $\varepsilon = 0.205/5 = 0.0410$. The complete conversion is:

Load (lb)	0	20	35	85	475	1650	2575	3825	4575	5325	5125	4575	4350
Displacement (in)	0.000	0.015	0.065	0.160	0.175	0.205	0.228	0.255	0.255	0.315	0.355	0.370	0.395
Stress (psi)	0	9	16	38	211	733	1144	1700	2033	2367	2278	2033	1933
Strain (in/in)	0.0000	0.0030	0.0130	0.0320	0.0350	0.0410	0.0456	0.0510	0.0510	0.0630	0.0710	0.0740	0.0790

Plot the curve and identify the toe region. Plotted (Figure 5.1), the record has three parts. The first four points carry almost no load (up to 85 lb, or 38 psi) yet accumulate 0.160 in of movement: this is seating of the platens and crushing of surface irregularities, not deformation of the wood, and it is the standard toe region of a compression test. From 475 lb to 3,825 lb the points fall on a straight line. Beyond 3,825 lb the curve rounds over to a peak at 5,325 lb and then falls away as the specimen crushes.
Take the modulus of elasticity as the slope of the straight run. Fitting the four points from 475 lb to 3,825 lb by least squares gives $$E = \frac{\Delta\sigma}{\Delta\varepsilon} = \boxed{9.27 \times 10^{4}\ \text{psi}}\quad (r^2 = 0.9982).$$ The simple two-point chord over the same run is a useful check: $(1700.0 - 211.1)/(0.0510 - 0.0350) = 93056\ \text{psi}$, which agrees with the fit to better than half a percent. Extrapolating the fitted line back to zero stress puts the corrected origin at $\varepsilon_0 = 0.0329$, so the toe correction to be applied to every strain is about 0.033 in/in.
Read the failure stress. The largest load the specimen carried was 5,325 lb, at a displacement of 0.315 in, after which the load fell on further displacement. The failure stress is therefore $$\sigma_f = \frac{P_{max}}{A} = \frac{5325}{2.25} = \boxed{2367\ \text{psi}},$$ reached at a nominal strain of 0.0630 in/in, or 0.0301 in/in after the toe correction. This is the maximum crushing strength parallel to the grain, the property that CSA O86 designates $f_c$ for the parallel-to-grain compressive resistance of the section.
Sense-check the two answers. A maximum crushing strength of about 2.4 ksi parallel to grain is entirely ordinary for a structural softwood at service moisture content, where clear-wood values of 4 to 7 ksi and graded-lumber values well below that are normal. The modulus, however, comes out at 9.27 × 10⁴ psi against a textbook range of $E = 1.0$ to $1.9 \times 10^6\ \text{psi}$ for clear softwood — more than an order of magnitude low. The reason is visible in the data: the displacement column is machine crosshead travel over the whole 5 in specimen, not extensometer strain over a gauge length, so it includes seating, platen and machine compliance. The value calculated above is therefore correctly reported as an apparent modulus for this test set-up.

Figure 5.1 — the wood compression record. The first four points are seating (toe) displacement; the modulus is taken from the straight run between 475 lb and 3,825 lb.

Final results.

Quantity	Value
Cross-sectional area, A	2.25 in²
Gauge length, L	5 in
Modulus of elasticity (least squares, 475 to 3,825 lb)	92678 psi (9.27 × 10⁴ psi)
Modulus of elasticity (two-point chord check)	93056 psi
Coefficient of determination of the fit, r²	0.9982
Toe correction (corrected origin)	0.0329 in/in
Maximum load	5,325 lb
Failure (maximum crushing) stress	2367 psi
Strain at failure (nominal / toe-corrected)	0.0630 / 0.0301 in/in

Check: two features of the printed record are handled explicitly. First, the readings at 3,825 lb and 4,575 lb are both listed at a displacement of 0.255 in, which cannot be correct for a monotonic test; the 4,575 lb point has been excluded from the modulus fit for that reason, and it affects neither the failure stress nor the fitted slope materially. Second, the displacement is crosshead travel over the full 5 in specimen rather than an extensometer reading, so the modulus derived above is an apparent modulus an order of magnitude below the published clear-wood range; the question asks for the modulus from the plotted data, and that is what is reported.

Part (b) — significance and use of four laboratory tests on steel (8 marks). Each of the four tests loads steel in a different way, and together they cover the properties a designer must be able to rely on.

(i) The tension test (ASTM A370/E8, CSA G40.20) pulls a machined specimen to fracture while load and extension are recorded. Its significance is that it yields the whole set of design properties in one measurement: modulus of elasticity, yield strength (upper and lower yield points for a mild structural steel, or the 0.2 % offset yield for a high-strength steel), tensile strength, percentage elongation over a 50 mm or 200 mm gauge length and reduction of area. Its use is mill certification and acceptance: the yield strength is the number every limit-states design equation in CSA S16 is written around, and the elongation is the direct measure of ductility that justifies plastic design, moment redistribution and the seismic detailing rules.

(ii) The torsion test twists a bar or tube and records torque against angle of twist. Its significance is that it measures shear properties directly — the shear modulus $G$, the shear yield strength and the shear strength — without the complication of necking, because the cross-section does not change during the test. It is also the cleanest way to compare ductility, since a ductile steel will twist through many revolutions before failing on a transverse plane while a brittle one fails on a helical 45° plane. Its uses are checking the assumed relation $G = E/[2(1+\nu)]$, qualifying shafts, torsion bars and drill pipe, and verifying reinforcing bar and prestressing strand behaviour where twisting occurs during installation.

(iii) The Charpy V-notch impact test (ASTM A370/E23, CSA G40.21 Category T) breaks a notched 10 mm square bar with a swinging pendulum at a controlled temperature and reports the energy absorbed, the lateral expansion and the percentage of shear (fibrous) fracture. Its significance is that it measures notch toughness — resistance to brittle fracture under the triple threat of a stress raiser, a high strain rate and low temperature — which the tension test cannot detect at all. Repeating it at several temperatures maps the ductile-to-brittle transition curve. Its use is specifying steel for structures exposed to Canadian winter temperatures: bridge tension members, crane runways, and any welded connection where restraint is high, are specified with a minimum absorbed energy (commonly 27 J) at a service temperature.

(iv) The bend test (ASTM A370, CSA G30.18 for reinforcing bar) bends a specimen through a specified angle around a pin of specified diameter and examines the outer surface for cracking. Its significance is as a qualitative, pass/fail measure of ductility and soundness in the as-delivered condition: it strains the outer fibre far beyond yield and so exposes surface defects, inclusions, seams, over-hard heat-affected zones and embrittlement that a tension test on a machined coupon would miss. Its uses are acceptance of reinforcing bar that must be bent to shape on site (the pin diameter in the test matches the bend diameter permitted in the field), qualification of welded joints by face, root and side bends under CSA W59/W47.1, and routine quality control of plate and formed sections. Taken together, the four tests answer four separate questions: how strong and how ductile in direct tension, how strong in shear, how tough in the presence of a notch at low temperature, and whether the material will survive being formed.

Topic. Compressive behaviour of wood parallel to grain, and the mechanical test suite for steel.

Key relations. $\sigma = P/A$; $\varepsilon = \delta/L$; $E = \Delta\sigma/\Delta\varepsilon$ on the straight run; toe correction by extrapolating that line to $\sigma = 0$; $G = E/[2(1+\nu)]$ for the torsion check.

Why this works. Wood loaded parallel to the grain behaves like a bundle of thin-walled tubes loaded along their axes: the response is nearly linear until the cell walls begin to buckle, after which the material yields gradually and crushes at a well-defined maximum load, which is why the curve peaks and then falls rather than fracturing abruptly. The toe region exists in every compression test because platens are never perfectly flat and specimen ends are never perfectly square; the accepted correction is to extrapolate the straight portion back to zero stress and shift the strain origin there, exactly as ASTM D143 and D198 require. The steel test suite is organised the same way as the properties it protects: one test per failure mode, because a steel that passes the tension test can still fail by brittle fracture at a notch on a cold night.

Common pitfall. The commonest errors here are dividing the load by the wrong area (using the 1.5 in × 5 in side face instead of the 1.5 in × 1.5 in loaded face), taking the modulus as the secant from the origin through some convenient point — which drags the toe displacement into the answer and roughly halves the result — and reporting the last recorded load as the failure load instead of the maximum load. On the steel side, the frequent slip is describing Charpy results as a strength: the test reports energy, and it is meaningful only with the test temperature quoted.

Source. Mamlouk & Zaniewski, Materials for Civil and Construction Engineers, 4th ed., §3.7 (mechanical testing of steel) and Ch. 10 (wood); ASTM D143 and D198; ASTM A370 and E23; CSA O86 Engineering Design in Wood §6.5; CSA G40.21 and CSA S16.

Question 5 of 5: Wood in Compression Parallel to Grain, and Laboratory Tests on Steel

Question 5: Wood in Compression Parallel to Grain, and Laboratory Tests on Steel (20 marks)