Question 1 of 8: Suspension Cable — Wire Count and Materials Testing

Question 1: Suspension Cable — Wire Count and Materials Testing (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.

Given. Cable length $L=20$ ft $=240$ in; total tensile load $P=20{,}000$ lb; wire diameter $d=3/16$ in (1080 steel); allowable stress $=70\%$ of yield; maximum allowable cable elongation $\Delta L_{max}=\tfrac12$ in $=0.50$ in; $E=30\times10^6$ psi; $\sigma_y=100{,}000$ psi. Wires equally loaded (parallel, isostrain).

Find. (a) Number of wires $n$ required. (b) Experimental method for yield strength and Poisson's ratio.

Approach

Two independent limits govern the wire count — an allowable-stress limit (70% of $\sigma_y$) and an allowable-elongation limit (Hooke's law, $\Delta L = \sigma L/E$); each gives an allowable per-wire stress, and the lower of the two (the more restrictive) sets the required wire count.

1. Wire cross-sectional area. $$A_{wire}=\frac{\pi}{4}d^2=\frac{\pi}{4}(0.1875)^2=\boxed{0.02761\ \text{in}^2}.$$
2. Stress-limit allowable stress. $$\sigma_{allow,stress}=0.70\,\sigma_y=0.70\times100{,}000=70{,}000\ \text{psi}.$$
3. Elongation-limit allowable stress. From $\Delta L=\sigma L/E$, $$\sigma_{allow,elong}=\frac{E\,\Delta L_{max}}{L}=\frac{30\times10^6\times0.50}{240}=\boxed{62{,}500\ \text{psi}}.$$ Since $62{,}500 < 70{,}000$ psi, the elongation limit governs.
4. Required number of wires. With all $n$ wires equally loaded, $\sigma=P/(nA_{wire})\le\sigma_{allow,elong}$, so $$n\ge\frac{P}{\sigma_{allow,elong}\,A_{wire}}=\frac{20{,}000}{62{,}500\times0.02761}=11.59\ \Rightarrow\ \boxed{n=12\ \text{wires}}.$$

(b) Experimental measurement of yield strength and Poisson's ratio. Machine a standard round tensile specimen from the same 1080-steel stock and load it in a calibrated universal testing machine at a slow, controlled strain rate. Mount an axial extensometer (or strain gauge) along the gauge length to record load vs. axial strain; from the resulting stress–strain curve, locate the yield strength by the 0.2% offset method (draw a line parallel to the elastic slope, offset by $\varepsilon=0.002$, and read the stress where it intersects the curve) since 1080 steel does not show a sharp yield point. To obtain Poisson's ratio, bond a second strain gauge transverse to the loading axis (or use a biaxial rosette / diametral extensometer) at the same location; within the elastic region, record the transverse strain $\varepsilon_{lat}$ simultaneously with the axial strain $\varepsilon_{ax}$, and take $\nu=-\varepsilon_{lat}/\varepsilon_{ax}$ from the slope of a plot of $\varepsilon_{lat}$ vs. $\varepsilon_{ax}$ over several load increments below the proportional limit.