Question 4 of 7: Sag curve under a bridge, and a 1200 m curve profile

Question 4: Sag curve under a bridge, and a 1200 m curve profile (20 marks — (a) 10, (b) 10)

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) a sag curve joining $-6\ \%$ to a level grade, whose PVI lies 1 m below the curve, passing under a structure with only 3.5 m of clearance; (b) a 1200 m sag curve from $-2.4\ \%$ to $+3.2\ \%$ with its PVI at 10+300.000, elevation 153.20 m.

Find. (a) the speed limit the underpass sight distance permits; (b) the curve elevation and slope at every 100 m, plus the low point.

Part (a) — sight distance under the structure

Approach. Convert the 1 m offset into a curve length, apply the AASHTO undercrossing sight-distance pair with the truck-eye and object heights, and read the resulting $S$ back to a tabulated speed on the governing downgrade.

1. Recover the curve length from the offset. The algebraic difference is $A=|G_2-G_1|=6.0\ \%$, and the PVI-to-curve offset of a symmetrical parabola is $e_m=AL/800$, so $$ L=\frac{800\,e_m}{A}=\frac{800(1.00)}{6.0}=133.33\ \text{m}. $$
2. Choose the correct sight-distance model. A sag curve under a structure is not limited by headlights but by the soffit cutting the sight line. AASHTO measures from a truck driver’s eye at $h_1=2.40$ m to an object at $h_2=0.60$ m, so the effective clearance is $$ C-\frac{h_1+h_2}{2}=3.50-\frac{2.40+0.60}{2}=2.00\ \text{m}. $$
3. Test the $S\lt L$ branch, then reject it. With $L=A S^{2}/\!\left[800\left(C-\frac{h_1+h_2}{2}\right)\right]$, $$ S=\sqrt{\frac{800(2.00)(133.33)}{6.0}}=188.56\ \text{m}\;\gt\;L, $$ so that branch contradicts itself.
4. Solve the consistent branch. For $S\ge L$, $L=2S-\dfrac{800\left(C-\frac{h_1+h_2}{2}\right)}{A}$, hence $$ 133.33=2S-\frac{800(2.00)}{6.0}=2S-266.67 \quad\Longrightarrow\quad\boxed{S=200.0\ \text{m}} $$ and $200\ \text{m}\ge133.33\ \text{m}$ checks out.
5. Read 200 m back to a posted speed. Descending the $-6\ \%$ approach with $f=0.30$ at 90 km/h, $$ \text{SSD}=\frac{25.000^{2}}{2(9.81)(0.30-0.06)}+2.5(25.000) =132.73+62.50=195.23\ \text{m}\;\lt\;200\ \text{m}, $$ while 100 km/h needs 240.43 m and fails. In the opposite direction the vehicle is climbing, so taking $G=0$ conservatively, 100 km/h needs 205.06 m — still just over the 200 m available — and 90 km/h needs 168.68 m. Both directions therefore land on $$ \boxed{\text{posted speed}=90\ \text{km/h}} $$

Part (b) — elevations and slopes at 100 m intervals

Approach. Station the PVC, write the parabola and its derivative in one place, then tabulate.

1. Locate the PVC and PVT. The PVI is at mid-length, so the PVC is 600 m back at station $10+300.000-600=9+700.000$, and the PVT 600 m ahead at 10+900.000. Their tangent elevations are $$ \text{el}_{PVC}=153.20+0.024(600)=167.60\ \text{m},\qquad \text{el}_{PVT}=153.20+0.032(600)=172.40\ \text{m}. $$
2. Write the curve equation once and use it everywhere. With $x$ measured from the PVC in metres and grades as decimals, $$ \text{el}(x)=167.60-0.024x+\frac{0.032-(-0.024)}{2(1200)}x^{2} =167.60-0.024x+2.3333\times10^{-5}x^{2}, $$ $$ \frac{d(\text{el})}{dx}=-0.024+\frac{0.056}{1200}x =-0.024+4.6667\times10^{-5}x. $$
3. Evaluate at every 100 m. Substituting $x=0,100,\dots,1200$ gives the table below; as a spot check, at $x=300$ m $\text{el}=167.60-7.200+2.100=162.500$ m and the slope is $-0.024+0.014=-0.010$, i.e. $-1.000\ \%$.
4. Locate the low point, because that is where the drainage inlet goes. The slope vanishes at $$ x_{low}=\frac{-G_1L}{G_2-G_1}=\frac{0.024(1200)}{0.056}=514.286\ \text{m} \;\Rightarrow\;\text{station }10+214.286, $$ $$ \text{el}_{low}=\boxed{161.4286\ \text{m}} $$ Note that this is not at the 500 m computation point: the tabulated minimum of 161.433 m at $x=500$ m is 4 mm high, which matters for a gutter grade even though it is invisible on the profile.