Question

1. A traffic signal on the four-leg signalized intersection is designed based on the information below. Please answer the following questions. The startup lost time per phase is 2.0 seconds, and the clearance lost time per phase is 1.5 seconds. The deceleration rate of the vehicles is 3.4 m/s², and the average length of the vehicles is 6.0 m. The lane width of normal lane is 3.3 m, and the turn bay width is 2.8 m. Please ignore the median width and shoulders. Assume there is no sidewalk.

Approach	Grade	Number of Approaching Lane	# of Left- Turn Bay	LT (veh/h)	TH (veh/h)	RT (veh/h)	Speed Limit (km/h)	Displayed Green Time	Effective Red Time
EB	+ 2.0 %	3	1	300	500	200	80	30	35
WB	- 2.0 %	2	0	220	300	250	80	30	35
NB	- 3.5 %	2	1	200	320	200	70	25	30
SB	+ 3.5 %	3	1	250	300	150	70	25	30

1.1 Compute a displayed yellow signal of all approaches for safe operation of the intersection.

1.2 Compute an all-red signal length of all approaches for safe operation of the intersection. Assume that the number of lanes on leaving direction from the intersection is three for all. Consider the number of lanes on right side of driving direction for this problem.  

1.3 Compute the capacity of through movement of all approaches. A saturation flow rate for through movement is 1800 veh/h. If the multiple cycle lengths are available from calculation, please use the longest cycle length for all approaches to determine the capacity.

2. A flexible pavement is designed with an initial PSI of 4.5 and terminal PSI of 2.0. The subgrade soil was tested and provided CBR value of 6. The relationship between CBR and resilience modulus is given as Mr=1500*CBR. The structural number of this pavement is 4.5 and the overall standard deviation is 0.35, and the reliability is 99%. The original pavement was designed for the following daily traffic:
   • 1400 single axles with 8kips
   • 30 single axles with 40 kips
   • 200 tandem axles with 40 kips

When the axle passes of a single axle with 40 kips and a tandem axle with 40 kips increased by 40% into this highway traffic during the entire service period, how many years will be reduced in the pavement service life?

Answer 1

1.1 Yellow change interval is calculated by the eauation

where Y = yellow time (seconds)

t = perception time = driver reactiontime for starting and stopping = assume = 1 .5sec

V = 85th percentile approach speed in mph

a = deceleration rate in ft/sec2 = 3.4 m/sec2 =

g = approach grade expressed as a decimal, value is negative if downgrade

So we an calculate yellow time for each

Approach	Grade	Number of	# of Left- Turn Bay	LT	TH	RT	Speed	Speed	Displayed	Effective	deceleration (ft/sec2)	Yellow time
Approaching	(veh/h)	(veh/h)	(veh/h)	Limit	Limit	Green	Red
Lane				(km/h)	mph	Time	Time
EB	2.00%	3	1	300	500	200	80	49.71	30	35	11.154856	4.596623503
WB	-2.00%	2	0	220	300	250	80	49.71	30	35	11.154856	4.976083661
NB	-3.50%	2	1	200	320	200	70	43.50	25	30	11.154856	4.688073099
SB	3.50%	3	1	250	300	150	70	43.50	25	30	11.154856	4.102989153

so displayed yellow time = 5 seonds

Red interval is calculated as follows

where R = all red nterval = intersection width (feet), L = length of vehicle ft, V = 85th percentile speed in mph.

= (6 *3.3) * 3.2804 = 64.96, becaue 1 meter = 2.38084

L = length of vehicle ft = 6m = 6*3.28084 = 19.685

W + L = 19.685 + 64.96 = 84.465

R = 84465/ 1.47 * 49.71 = 84.465/73.0737 = 7.198 -1

Red clearance time is computed for all movements below

Approach	Grade	Number of	# of Left- Turn Bay	Speed	Speed	Displayed	Effective	deceleration (ft/sec2)	Yellow time	Red clearance interval (ft)
Approaching	Limit	Limit	Green	Red
Lane	(km/h)	mph	Time	Time
EB	2.00%	3	1	80	49.71	30	35	11.154856	4.596623503	0.158426971
WB	-2.00%	2	0	80	49.71	30	35	11.154856	4.976083661	0.158426971
NB	-3.50%	2	1	70	43.50	25	30	11.154856	4.688073099	0.323916538
SB	3.50%	3	1	70	43.50	25	30	11.154856	4.102989153	0.323916538

Since computed red time is less than 1, assume Red clearnace time is assumed to be equal to 1 second