FE Civil Transportation Eng. practice problems

Uphill is positive grade, so $G = +0.05$ and the denominator grows, shortening the braking distance. Reaction distance $= 1.47 \times 55 \times 2.5 = 202.1$ ft. The denominator factor is $a/32.2 + G = 0.3478 + 0.05 = 0.3978$, so braking $= 55^2/(30 \times 0.3978) = 3{,}025/11.93 = 253.5$ ft. Total SSD $= 202.1 + 253.5 = 455.6 \approx 456$ ft. The 202 ft choice is the reaction distance only. The 492 ft choice used a level grade ($G = 0$): $202.1 + 290 = 492$. The 541 ft choice used a downgrade ($G = -0.05$), wrongly making stopping harder.

1. Reaction distance:
   $d_r = 1.47 \times 55 \times 2.5 = 202.1 \text{ ft}$
2. Braking distance with $G = +0.05$:
   $d_b = \frac{55^2}{30\left(\dfrac{11.2}{32.2} + 0.05\right)} = \frac{3{,}025}{30 \times 0.3978} = \frac{3{,}025}{11.93} = 253.5 \text{ ft}$
3. Total SSD:
   $SSD = 202.1 + 253.5 = 455.6 \approx 456 \text{ ft}$

FE Reference Handbook 10.6, p. 306

$PHF = V/(4 \times V_{15}) = 2{,}400/(4 \times 700) = 2{,}400/2{,}800 = 0.857$. The 0.292 option divided $V_{15}$ by $V$ and forgot the factor of 4 in the wrong place: $700/2{,}400 = 0.292$. The 1.167 option inverted the ratio: $2{,}800/2{,}400$. The 0.750 option used $V_{15} = 800$ by mistake. PHF is always between 0.25 and 1.00.

1. Peak hour factor:
   $PHF = \frac{V}{4 \times V_{15}} = \frac{2{,}400}{4 \times 700} = \frac{2{,}400}{2{,}800} = 0.857$

FE Reference Handbook 10.6, p. 306

Test the $S \leq L$ case first: $L = AS^2/2{,}158 = 4 \times 700^2/2{,}158 = 1{,}960{,}000/2{,}158 = 908 \approx 909$ ft. Check the assumption: $S = 700 \leq L = 908$ is TRUE, so this branch is valid and governs. Now check the $S > L$ branch: $L = 2S - 2{,}158/A = 1{,}400 - 539.5 = 860.5$ ft, which would require $S > L$ (i.e., $700 > 860.5$) -- that is FALSE, so the $S > L$ branch is invalid here. The minimum length is therefore about 909 ft. The 863 ft choice is approximately the invalid $S > L$ branch value ($2S - 2{,}158/A = 860.5$, rounded). The 1,400 ft choice is just 2S, dropping the second term. The 540 ft choice is the term $2{,}158/A$ alone.

1. Apply the $S \leq L$ crest formula:
   $L = \frac{AS^2}{2{,}158} = \frac{4 \times 700^2}{2{,}158} = \frac{1{,}960{,}000}{2{,}158}$
2. Evaluate and check the assumption ($S = 700 \leq L$):
   $L = 908.2 \approx 909 \text{ ft} \quad (700 \leq 908 \checkmark)$

FE Reference Handbook 10.6, p. 308

$A = |g_2 - g_1| \times 100 = |0.03 - (-0.02)| \times 100 = 0.05 \times 100 = 5$. Then $K = L/A$ rearranges to $L = K \times A = 160 \times 5 = 800$ ft. The 160 ft choice reported $K$ as if it were $L$. The 32 ft choice computed $K/A = 160/5$ instead of $K \times A$. The 320 ft choice used $A = 2$ (only one grade magnitude): $160 \times 2$.

1. Algebraic difference:
   $A = |g_2 - g_1| \times 100 = |0.03 - (-0.02)| \times 100 = 5$
2. Curve length:
   $L = K \times A = 160 \times 5 = 800 \text{ ft}$

FE Reference Handbook 10.6, p. 307

$M = R[1 - \cos(I/2)] = 1{,}000[1 - \cos(20\degree)] = 1{,}000(1 - 0.9397) = 1{,}000 \times 0.0603 = 60.3$ ft. The 234 ft choice used $\cos(I) = \cos(40\degree) = 0.766$ instead of $\cos(I/2)$: $1{,}000(1 - 0.766) = 234$. The 64.2 ft choice is the external distance $E = R[1/\cos(I/2) - 1] = 1{,}000(1/0.9397 - 1) = 64.2$, not the middle ordinate. The 30.2 ft choice halved the correct answer.

1. Middle ordinate:
   $M = R\left[1 - \cos\!\left(\frac{I}{2}\right)\right] = 1{,}000\left[1 - \cos(20\degree)\right]$
2. Evaluate:
   $M = 1{,}000(1 - 0.9397) = 60.3 \text{ ft}$

FE Reference Handbook 10.6, p. 309

Rearrange $0.01e + f = V^2/(15R)$ for the minimum radius: $R = V^2/[15(0.01e + f)] = 60^2/[15(0.08 + 0.12)] = 3{,}600/[15 \times 0.20] = 3{,}600/3.0 = 1{,}200$ ft. The 2,000 ft option used only friction ($0.01e + f = 0.12$): $3{,}600/(15 \times 0.12) = 2{,}000$. The 240 ft option forgot to convert $e$: used $0.01e = 0.08$ but then added $e = 8$ giving an inflated denominator. The 600 ft option doubled the denominator factor by using $0.01e + f = 0.40$.

1. Rearrange for radius:
   $R = \frac{V^2}{15(0.01e + f)}$
2. Substitute:
   $R = \frac{60^2}{15(0.08 + 0.12)} = \frac{3{,}600}{15 \times 0.20}$
3. Evaluate:
   $R = \frac{3{,}600}{3.0} = 1{,}200 \text{ ft}$

FE Reference Handbook 10.6, p. 309

Convert: $v = 35 \times 1.467 = 51.3$ ft/sec. On an upgrade, gravity aids stopping, so the denominator grows: $2a + 64.4G = 2(10) + 64.4(0.04) = 20 + 2.576 = 22.58$. $y = 1.0 + 51.3/22.58 = 1.0 + 2.27 = 3.27 \approx 3.3$ sec. The 3.6 sec option used a level grade ($G = 0$): $1.0 + 51.3/20 = 3.57$. The 3.9 sec option reversed the sign (treated the upgrade as a downgrade): $1.0 + 51.3/17.42 = 3.95$. The 2.6 sec option used mph in the numerator: $1.0 + 35/22.58 = 2.55$.

1. Convert speed:
   $v = 35 \times 1.467 = 51.3 \text{ ft/sec}$
2. Effective deceleration term (upgrade, +G):
   $2a + 64.4G = 2(10) + 64.4(0.04) = 20 + 2.58 = 22.58$
3. Yellow interval:
   $y = 1.0 + \frac{51.3}{22.58} = 1.0 + 2.27 = 3.27 \approx 3.3 \text{ sec}$

FE Reference Handbook 10.6, p. 306

The only term that differs between the two crosswalks is $0.27 N_{\text{ped}}$. The difference in that term is $0.27(25 - 5) = 0.27 \times 20 = 5.4$ sec, so B needs about 5.4 sec more. The start-up constant (3.2) and the walking time ($L/S_p$) are identical for both. The option that they are equal because the walking time term $L/S_p$ dominates the calculation is wrong because the volume term, while smaller, is not zero. The option that Crosswalk B requires 5 times the green time of Crosswalk A wrongly assumes green time scales linearly with pedestrian count -- only one of three terms depends on $N_{\text{ped}}$. The option that Crosswalk A requires more green time because fewer pedestrians walk more cautiously invents a behavior not in the formula.

1. Identical terms: $3.2 + L/S_p$ are the same for both crosswalks.
2. Difference comes only from the volume term:
   $\Delta G_p = 0.27(N_B - N_A) = 0.27(25 - 5) = 5.4 \text{ sec}$
3. Crosswalk B requires about 5.4 sec more green time.

FE Reference Handbook 10.6, p. 306

First find speed from Greenshields: $S = S_f - (S_f/D_j)D = 50 - (50/160)(40) = 50 - 0.3125 \times 40 = 50 - 12.5 = 37.5$ mph. Then flow $V = S \times D = 37.5 \times 40 = 1{,}500$ veh/hr/ln. The 2,000 veh/hr/ln choice used the free-flow speed without the density reduction: $50 \times 40 = 2{,}000$. The 1,000 veh/hr/ln choice used the optimum speed $S_f/2 = 25$: $25 \times 40$. The 500 veh/hr/ln choice used only the speed-reduction term ($12.5 \times 40$) as the speed.

1. Speed at $D = 40$:
   $S = 50 - \frac{50}{160}(40) = 50 - 12.5 = 37.5 \text{ mph}$
2. Flow:
   $V = S \times D = 37.5 \times 40 = 1{,}500 \text{ veh/hr/ln}$

FE Reference Handbook 10.6, p. 312

$VMT = ADT \times \text{days} \times \text{length} = 5{,}000 \times (365 \times 5) \times 3 = 5{,}000 \times 1{,}825 \times 3 = 27{,}375{,}000$ vehicle-miles. Rate per 100 million VM $= A \times 100{,}000{,}000/VMT = 27 \times 100{,}000{,}000/27{,}375{,}000 = 98.6$. The 0.99 choice used the per-one-million factor instead of per hundred million: $27 \times 1{,}000{,}000/27{,}375{,}000$. The 32.9 choice is the correct rate divided by 3 (e.g., counting the 3-mile length twice in the VMT). The 493 choice used only 1 year instead of the 5-year study period, dividing the VMT by 5 and raising the rate fivefold ($98.6 \times 5 \approx 493$).

1. Vehicle-miles traveled over 5 years:
   $VMT = 5{,}000 \times (365 \times 5) \times 3 = 27{,}375{,}000$
2. Crash rate per 100 million VM:
   $R = \frac{27 \times 100{,}000{,}000}{27{,}375{,}000} = 98.6$

FE Reference Handbook 10.6, p. 313

$f_{HV} = 1/[1 + P_T(E_T - 1)] = 1/[1 + 0.20(3.0 - 1)] = 1/[1 + 0.20 \times 2] = 1/1.40 = 0.714$. The 0.625 distractor used $E_T = 3.0$ without subtracting 1: $1/[1 + 0.20 \times 3] = 1/1.60$. The 1.400 distractor reported the denominator itself instead of its reciprocal. The 0.600 distractor computed $1 - P_T \times E_T = 1 - 0.60$.

1. Heavy-vehicle factor:
   $f_{HV} = \frac{1}{1 + P_T(E_T - 1)} = \frac{1}{1 + 0.20(3.0 - 1)}$
2. Evaluate:
   $f_{HV} = \frac{1}{1 + 0.40} = \frac{1}{1.40} = 0.714$

FE Reference Handbook 10.6, p. 311

LOS is read directly from density. Segment 1 at 24 pc/mi/ln falls in the $> 18$ to 26 band, which is LOS C. Segment 2 at 40 pc/mi/ln falls in the $> 35$ to 45 band, which is LOS E. The 'LOS B / LOS D' choice shifts both grades one letter too good. The 'both at LOS D' choice wrongly lumps both into D. The 'LOS C / LOS F' choice pushes Segment 2 to F, but F only begins above 45 pc/mi/ln.

1. Segment 1: 24 pc/mi/ln is in the $> 18$ to 26 band.
2. That band is LOS C.
3. Segment 2: 40 pc/mi/ln is in the $> 35$ to 45 band, which is LOS E.

FE Reference Handbook 10.6, p. 311

Base contribution: $0.14 \times 10 \times 1.0 = 1.40$. Subbase: $0.12 \times 8 \times 1.0 = 0.96$. Remaining SN for the surface: $5.0 - 1.40 - 0.96 = 2.64$. Then $D_1 = 2.64/a_1 = 2.64/0.44 = 6.0$ in. The 11.4-in choice solved $D_1 = 5.0/0.44$ without subtracting the lower layers. The 2.6-in choice left the remaining SN of 2.64 as a thickness, forgetting to divide by $a_1$. The 7.7-in choice forgot the subbase, dividing $5.0 - 1.40 = 3.60$ by $0.44 \times 1.06$.

1. Base contribution:
   $a_2 D_2 m_2 = 0.14 \times 10 \times 1.0 = 1.40$
2. Subbase contribution:
   $a_3 D_3 m_3 = 0.12 \times 8 \times 1.0 = 0.96$
3. Surface thickness:
   $D_1 = \frac{5.0 - 1.40 - 0.96}{0.44} = \frac{2.64}{0.44} = 6.0 \text{ in}$

FE Reference Handbook 10.6, p. 314

Daily ESALs $= 300(1.00) + 150(7.9) = 300 + 1{,}185 = 1{,}485$ ESALs/day. Over the design life: $1{,}485 \times 365 \times 20 = 1{,}485 \times 7{,}300 = 10{,}840{,}500 \approx 10.8 \times 10^6$ ESALs. The $2.2 \times 10^6$ choice counted only the 18-kip axles over the design life: $300 \times 7{,}300 = 2.19 \times 10^6$. The $8.7 \times 10^6$ choice used the heavy axles only: $1{,}185 \times 7{,}300 = 8.65 \times 10^6$. The $0.54 \times 10^6$ choice forgot to multiply by the 20-year life and used only 365 days: $1{,}485 \times 365 = 542{,}025$.

1. Daily ESALs:
   $300(1.00) + 150(7.9) = 300 + 1{,}185 = 1{,}485 \text{ ESALs/day}$
2. Total over the design life:
   $1{,}485 \times 365 \times 20 = 1{,}485 \times 7{,}300 = 10{,}840{,}500$
3. Rounded:
   $\approx 10.8 \times 10^6 \text{ ESALs}$

FE Reference Handbook 10.6, p. 314

Compacted volume $= 0.90 \times$ bank volume, so bank volume $=$ compacted volume $/ 0.90 = 9{,}000/0.90 = 10{,}000$ yd^3. Soil shrinks when compacted, so you must dig MORE bank material than the finished fill volume. The 8,100 yd^3 option multiplied by 0.90 instead of dividing: $9{,}000 \times 0.90$. The 9,000 yd^3 option ignored shrinkage entirely. The 9,900 yd^3 option added 10% instead of dividing by 0.90.

1. Relationship:
   $V_{\text{compacted}} = 0.90 \times V_{\text{bank}}$
2. Solve for bank volume:
   $V_{\text{bank}} = \frac{9{,}000}{0.90} = 10{,}000 \text{ yd}^3$

FE Reference Handbook 10.6, p. 315

Apply the average end area method to each 100-ft span and add. Span 1 ($A_1$ to $A_2$): $V = 100(120 + 200)/2 = 16{,}000$ ft^3. Span 2 ($A_2$ to $A_3$): $V = 100(200 + 160)/2 = 18{,}000$ ft^3. Total $= 34{,}000$ ft^3, then $34{,}000/27 = 1{,}259$ yd^3. The 1,037 yd^3 distractor collapsed the two spans into a single 200-ft span using only the first and last areas: $200(120 + 160)/2 = 28{,}000$ ft^3 $\to 1{,}037$ yd^3. The 1,333 yd^3 distractor wrongly applied the prismoidal formula treating $A_2$ as a mid-section: $200(120 + 4 \times 200 + 160)/6 = 36{,}000$ ft^3 $\to 1{,}333$ yd^3. The 593 yd^3 distractor used only the first span.

1. Span 1 to 2:
   $V_{12} = \frac{100(120 + 200)}{2} = 16{,}000 \text{ ft}^3$
2. Span 2 to 3:
   $V_{23} = \frac{100(200 + 160)}{2} = 18{,}000 \text{ ft}^3$
3. Total and convert:
   $V = \frac{16{,}000 + 18{,}000}{27} = \frac{34{,}000}{27} = 1{,}259 \text{ yd}^3$

FE Reference Handbook 10.6, p. 315

Fraction to zone 1 = 120/(120 + 80) = 120/200 = 0.60, so T = 800 × 0.60 = 480 trips. The 320 trips choice is the trips to zone 2 (80/200 × 800). The 400 trips choice splits evenly. The 120 trips choice reports the raw weight, not the trips.

1. Fraction to zone 1:
   $\frac{120}{120 + 80} = 0.60$
2. Trips:
   $T = 800 \times 0.60 = 480\text{ trips}$

FE Reference Handbook 10.6, p. 312

Rigid pavement design uses the modulus of subgrade reaction k (pressure per unit deflection) to describe how the subgrade supports the slab. The structural number (SN) is for flexible pavement; Marshall stability and penetration grade are asphalt-mix and binder properties.

1. Rigid pavement rests on the subgrade as a slab.
2. Subgrade support is measured by the modulus of subgrade reaction k.
3. The structural number applies to flexible pavement instead.

If no warrant is met, a signal is not justified, because installing one can increase rear-end crashes and cause unnecessary delay. The "install anyway" option wrongly assumes signals always help. The claim that warrants do not apply to existing intersections is false (warrants apply broadly). The "a signal must be installed" option inverts the purpose of a warrant analysis.

1. No warrant met means conditions do not justify a signal.
2. Unwarranted signals can increase crashes and delay.
3. So a signal should not be installed.

FE Reference Handbook 10.6, MUTCD, Traffic Signal Warrants