Question 1 of 7: Directional transition around a protected area

Question 1: Directional transition around a protected area (20 marks — (a) 8, (b) 8, (c) 4)

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 four-lane freeway with 4.0 m lanes and 3.0 m shoulders is to be curved through $\Delta=60^\circ$ about a PI at 2+120.000, keeping a 10 m statutory buffer from a protected area whose nearest boundary is 80 m from the PI, with $e=0.06$ and no grade.

Find. (a) the largest radius that fits; (b) whether 120 km/h can be retained and, if not, the recommended speed; (c) the stations of the PC and PT.

Approach. Convert the environmental buffer into an upper bound on the external distance $E$, invert $E=R[\sec(\Delta/2)-1]$ for the radius, then test that radius against both the superelevation–side-friction limit and the sight distance the trees leave, and station the curve from the PI.

1. Fix the half-width of the graded roadway. The statute measures to the edge of the highway including the shoulder, so the controlling offset from the alignment centreline is $$ h=\tfrac{1}{2}nw+s=\tfrac{1}{2}(4)(4.0)+3.0=11.0\ \text{m}. $$
2. Turn the environmental clearance into a limit on the external distance. Along the bisector the mid-point of the centreline sits $E$ from the PI, and the inside shoulder edge a further $h$ beyond it. That edge must stop $b$ short of the area, so $$ E+h\le d-b \quad\Longrightarrow\quad E_{max}=80-10-11.0=59.00\ \text{m}. $$
3. Invert the external-distance formula. With $E=R\left[\sec(\Delta/2)-1\right]$ and $\sec 30^\circ-1=0.154701$, $$ R_{max}=\frac{E_{max}}{\sec(\Delta/2)-1}=\frac{59.00}{0.154701} =\boxed{381.38\ \text{m}} $$ Any flatter (larger-radius) curve would swing the pavement into the 10 m buffer, so this is the largest radius the site allows — and, because capacity for speed grows with radius, it is also the radius to adopt.
4. Test the adopted radius against the 120 km/h design speed. Using $f_s=0.09$ at 120 km/h from the friction table and $V=33.333\ \text{m/s}$, $$ R_{req}=\frac{V^{2}}{g\,(f_s+e)}=\frac{33.333^{2}}{9.81(0.09+0.06)} =755.09\ \text{m}. $$ The site allows only 381.38 m, so a reduced design speed is unavoidable.
5. March down the tabulated speeds until the radius suffices. At 100 km/h, $f_s=0.12$ and $R_{req}=27.778^{2}/[9.81(0.18)]=436.97\ \text{m}$, still more than 381.38 m. At 90 km/h, $f_s=0.13$ and $R_{req}=25.000^{2}/[9.81(0.19)]=335.32\ \text{m}\lt 381.38\ \text{m}$, so the curvature criterion permits 90 km/h.
6. Now check the trees, because the question warns that they obstruct sight. A driver in the innermost lane runs on a path of radius $R_v=R-1.5w=381.38-6.0=375.38\ \text{m}$, i.e. 6.0 m from the PI side of the centreline, so the clear middle ordinate to the edge of the protected area is $$ M_s=d-(E_{max}+1.5w)=80-(59.00+6.00)=15.00\ \text{m}. $$
7. Convert that middle ordinate into an available sight distance. Inverting the exam-sheet relation $M_s=R_v-R_v\cos\!\left[\dfrac{90\,S}{\pi R_v}\right]$ gives $$ S=\frac{\pi R_v}{90}\,\cos^{-1}\!\left(1-\frac{M_s}{R_v}\right) =\frac{\pi(375.38)}{90}\cos^{-1}(0.960041)=212.95\ \text{m}. $$
8. Read that back to a speed. On the level curve ($G=0$, $t_{pr}=2.5$ s) the demand is $\text{SSD}=V_0^{2}/(2gf)+V_0t_{pr}$: at 100 km/h, $27.778^{2}/[2(9.81)(0.29)]+2.5(27.778)=205.06\ \text{m}$, which fits inside 212.95 m; at 110 km/h it grows to 246.34 m, which does not. Sight distance therefore permits 100 km/h.
9. Take the governing value. Curvature caps the section at 90 km/h and sight distance at 100 km/h, so $$ \boxed{V_{design}=90\ \text{km/h}} $$ is the recommended reduced design speed and posted limit. At 90 km/h the sight demand is only 168.68 m against 212.95 m available, so the trees are not critical once the curvature limit is respected.
10. Station the curve. With $R=381.38$ m and $\Delta=60^\circ$, $$ T=R\tan(\Delta/2)=381.38\tan 30^\circ=220.19\ \text{m},\qquad L=\frac{\pi\Delta R}{180}=\frac{\pi(60)(381.38)}{180}=399.38\ \text{m}. $$ Working from the PI, $$ \text{PC}=(2+120.000)-220.19=\boxed{1+899.81},\qquad \text{PT}=\text{PC}+L=\boxed{2+299.19}. $$