Question 2 of 5: Horizontal curve on a four-lane divided urban expressway

Question 2: Horizontal curve on a four-lane divided urban expressway (20 marks: a 5, b 10, c 5)

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. A 4 km circular curve of 750 m centreline radius on a four-lane divided urban expressway, posted at 100 km/h, carrying 52,200 vehicles per day, in a cut with 3:1 back slopes.

Find. (a) whether 750 m is an adequate radius; (b) the clear-zone width on the inside and the outside of the curve; (c) a recommended spiral length.

Approach. Test the radius against the point-mass equation at the posted speed, read the clear zone from the straight-section table and adjust the outside for curvature, then size the spiral by the three standard criteria and take the largest.

1. (a) Compare the radius against the minimum for the posted speed. The point-mass relation printed on the appendix formula sheet is $$R_{\min} = \frac{V^{2}}{127\,(e + f_s)}$$ An urban facility takes $e_{\max} = 0.06$ (snow and ice, frequent stopped traffic on the curve), and the appendix side-friction table gives $f_s = 0.11$ at 100 km/h. Hence $$R_{\min} = \frac{100^{2}}{127\,(0.06 + 0.11)} = \frac{10\,000}{21.59} = \boxed{463\ \text{m}}$$ The built radius of 750 m is 1.6 times this, so the curve is safe at the posted speed.
2. Confirm the margin the other way round, through the friction actually demanded. Rearranging the same relation at the built radius with the design superelevation still at its maximum, $$f_{\text{demand}} = \frac{V^{2}}{127R} - e = \frac{10\,000}{127 \times 750} - 0.06 = 0.105 - 0.060 = \boxed{0.045}$$ Only 0.045 of the 0.11 available side friction is called on — a 41 % utilisation, so there is a comfortable reserve against a wet surface or an over-speeding driver.
3. Establish how much over-speed the curve tolerates, since this is an expressway. Repeating the first step at higher speeds with their own tabulated side-friction values, $$R_{\min}(110) = \frac{12\,100}{127(0.06+0.09)} = 635\ \text{m}, \qquad R_{\min}(120) = \frac{14\,400}{127(0.06+0.07)} = 872\ \text{m}$$ The 750 m radius therefore satisfies a 110 km/h design speed but not 120 km/h. This is the reason the question can quote its relative slope “at 120 km/h” and still be answered at 100: the geometry of the curve itself will not support a 120 km/h design speed, so the posted 100 km/h remains the design speed for the spiral calculation.
4. (b) Read the straight-section clear zone. The appendix clear-zone table is entered with a design speed of 100 km/h, a design ADT above 6,000 vehicles per day, and a cut section with a 3:1 back slope. That cell reads $$\text{clear zone (straight)} = 6.0\ \text{to}\ 6.5\ \text{m}$$ measured from the edge of the travelled way. A 3:1 back slope in cut is a traversable, recoverable slope, which is why the cut columns carry real numbers where the corresponding 3:1 fill column carries only a footnote.
5. Apply the curve-adjustment factor to the outside of the curve only. The second appendix table gives a horizontal-curve adjustment factor for a 750 m radius at a 100 km/h design speed of $$K_{cz} = 1.2$$ (the tabulated values at 700 m and 900 m are both 1.2, so no interpolation is needed). The note beneath that table is explicit that the factor applies to the outside of curves only, and that curves flatter than 900 m need no adjustment at all. Hence $$\boxed{\text{outside: } 6.0 \times 1.2 \ \text{to}\ 6.5 \times 1.2 = 7.2\ \text{to}\ 7.8\ \text{m}} \qquad \boxed{\text{inside: } 6.0\ \text{to}\ 6.5\ \text{m}}$$ A vehicle that leaves the road on the outside of a curve departs along the tangent and therefore travels further from the pavement edge before it stops, which is the whole physical content of the factor. On the inside, the departure path curves back toward the road, so no widening is warranted. For design, take 7.8 m outside and 6.5 m inside — the upper end of each band, since this is a high-volume expressway.
6. (c) Establish the design superelevation before the spiral can be sized. The runoff criterion needs a value of e, and at $R = 750\ \text{m}$ the point-mass equation demands none at all: $V^{2}/127R = 0.105$ is less than the 0.11 of side friction available, so a flat curve would technically stand up. Superelevation is nevertheless provided on any curve sharper than the normal-crown threshold, and it is distributed between the maximum and zero in proportion to the sharpness. Using the linear (AASHTO Method 1) distribution, $$e = e_{\max}\frac{R_{\min}}{R} = 0.06 \times \frac{463}{750} = 0.037 \rightarrow \text{adopt } e = 0.040$$ which is a practical 0.5 % increment and errs on the safe side.
7. Size the spiral by the superelevation-runoff criterion. The relative-slope criterion sets the length over which the outer edge can be raised without an objectionable edge profile. On a divided highway each carriageway rotates about its median edge, so the rotated width is the two lanes of one direction: $$L_s = \frac{w\,e}{s} = \frac{(2 \times 3.75)(0.040)}{0.004} = \frac{7.50 \times 0.040}{0.004} = \boxed{75.0\ \text{m}}$$
8. Size the spiral by the two remaining criteria and take the largest. The travel-time criterion requires at least two seconds on the spiral, $$L_s = \frac{2V \times 1000}{3600} = \frac{2 \times 100 \times 1000}{3600} = 55.6\ \text{m}$$ and the comfort (rate of change of lateral acceleration) criterion uses the spiral parameter from the formula sheet, $$A = \sqrt{0.03577\,V^{3}} = \sqrt{0.03577 \times 10^{6}} = 189.1, \qquad L_s = \frac{A^{2}}{R} = \frac{35\,770}{750} = 47.7\ \text{m}$$ The superelevation runoff governs, so $$\boxed{L_s = 75\ \text{m}}$$ is recommended at each end of the circular arc, leaving 4000 − 2(75) = 3850 m of circular arc. Runoff governing is the expected outcome on a flat curve: the comfort and travel-time criteria both shrink as the radius grows, while the runoff length depends only on the width being rotated and the superelevation reached.