CEE 30 nd Midterm, Fall 009 CEE 30 nd Midterm Examination (50 minute) Pleae write your name on thi cover. Pleae write you lat name on all other exam page You are allowed to ue one 8.5 by heet of note.. There are 6 quetion worth a total of 00 point. Each quetion lit the point value for that quetion. Pleae work quietly and repect other people pace. Carefully read each quetion and enure that you anwer what i aked. If you need to untaple page, I have a tapler to retaple them at the end of the examination. Name (firt, lat):
CEE 30 nd Midterm, Fall 009 True or Fale ( point) A traight ection of road ha a ag vertical curve of length 500. Another flat ection of road ha a horizontal curve of length 500. If one were to meaure each curve along the centerline of the road ON THE ACTUA OAD SUFACE (rather than ue tationing), which meaurement would be longer? vertical In ituation where the topping ight ditance (SSD) i greater than the vertical curve length, i the lited curve length in the deign table (Table 3. and 3.3 in your text) more than or le than the actual required vertical curve length baed on actual geometry? Circle one: More than actual curve length e than actual curve length Are the following tatement true or fale? Sag vertical curve deign differ from cret vertical curve deign in the ene that ight ditance on a ag vertical curve i governed by nighttime condition. True In horizontal curve deign, v i meaured to the center of the vehicle travelled path. True The critical lane group ha the highet traffic intenity. True
CEE 30 nd Midterm, Fall 009 Short Anwer (5 point each) Calculate the number of equivalent ingle axle load for an empty double axle delivery truck with a front axle weight of 0,000 pound and a rear axle weight of 5,000 pound. 0,000 8,000 4 5,000 8,000 4 0.0ESA per bu At leat how many feet away from an obtacle mut a driver tart to brake (on a level grade) in order to top before hitting the object if the vehicle i travelling 6 mph? How many additional feet away from the obtacle mut the driver tart to brake if the driver i travelling along a 6% downgrade? Aume the driver i initially travelling at 6 mile per hour. Given: = 6 mph, = 0 mph, and a =. / (6 580/ 3600) d 369.5 a. The driver mut tart to brake at leat 369.5 ahead of the top line. (6 580/ 3600) d a gg. 3. 0.06 446. 0 Given G = 6% = 0.06 and g = 3. / d d d 446.0 369.5 76.95 If a 6% downgrade i conidered, the ditance i 76.95 longer than the ditance calculated for (). What i the lane group capacity if the aturation flow rate i 800 vehicle per hour, the lane group effective green time i 5 econd, and the cycle length i 75 econd? =800*5/75=600 veh/hour
CEE 30 nd Midterm, Fall 009 Quetion (5 point) The Pocono aceway in Pennylvania conit of three turn a diagramed below. Turn data are given in the table below. Uing tandard deign aumption, what i the deign peed (to the nearet mph) for the Pocono aceway baed on horizontal curve geometry only (you mut perform calculation for all 3 curve to get full credit). Note that thi i the deign peed for a typical automobile and not for a race car. Aume v = and the coefficient of ide friction = 0.55 in all cae and for all peed. A Turn North A Turn Turn 3 A 3 Turn Number Curve ength Superelevation Angle 800 0.5% A = 60 degree 750 4. % A = 90 degree 3 675 4.9 % A 3 = 30 degree Note : Angle meaured are a indicated in the picture and are NOT Δ (delta). You need to firt calculate the radiu for each turn and then ue the baic uperelevation equation to determine the velocity. Do thi for all three turn. The lowet of the three calculated velocitie would be the controlling deign peed and would, therefore, be the deign peed of the entire race track. ecognize that the central angle of the curve (Δ) i equal to 80 minu the angle given in the table. Alo, you need to convert the velocity you obtain from /econd to mph, which involved dividing by.47. Turn Find curve radiu 80 80 800 38. 97 80 60 80 Find deign peed v
CEE 30 nd Midterm, Fall 009 v g f e 38.97 3. 0.55 0.05 38. g f e.47.47 5 Turn Find curve radiu 80 80 750 477. 46 80 90 80 Find deign peed v g f e 477.46 3. 0.55 0.4 45. g f e.47.47 9 Turn 3 Find curve radiu 80 80 675 57. 83 80 30 80 Find deign peed v g f e 57.83 3. 0.55 0.49 39. g f e.47.47 4 v v mph mph mph Deign Speed = the lowet of the three rounded down to the nearet mph = 38 mph
CEE 30 nd Midterm, Fall 009 Quetion (0 point) You are deigning the vertical alignment of an eat-wet portion of S 58 through Maryville. An equal tangent cret vertical curve mut go over an exiting north-outh Olympic oil pipeline. According to afety regulation, the top of the pipeline mut be at leat 6 below the centerline roadway urface. Known grade, tationing and elevation are given in the drawing below. Deign the curve for the highet poible deign peed without violating the pipeline 6 cover requirement. eport the longet poible curve length, and the aociated deign peed rounded down to the nearet 5 mph (be careful with unit in your calculation!). Profile iew G =.5% Y =? PI Station = 0+00 Elevation = 335 G = -4.5% PC 6 minimum Oil Pipeline Diameter = Station = 9+00 Elevation of center = 34 PT There are two principal way you can olve thi problem. Either one i fine, although the firt way i horter and perhap le prone to math error. Method : Determine uing vertical curve offet A G G.5 4.5 7 At tation 9+00, the elevation of the roadway mut be at leat: Elevation of pipeline (34 ) + half the diameter of the pipe (0.5 ) + 6 = 330.5 Alo realize that the PI i at the half-way point on the vertical curve, or /. Thi make tation 9+00 = PI tation 00. O / 00. Ue the offet equation: Y A 00 x Note that the offet i the elevation of G at tation 9+00 minu the roadway elevation and that the elevation of G at tation 9+00 i the PI elevation 00(G ): Y 335 00 0.05 330.5
CEE 30 nd Midterm, Fall 009 Y 0 A 7 x 00 00 00 0.035 0.5 00 0,000 0.5 57.4 _0,000 57.4 0.5 00 0,000 olve quadratic and get = 556.7 or 7.9. Since 7.9 i too hort (the curve would not even extend to tation 9+00 and it would alo not be the ONGEST curve one could deign), chooe = 556.7. Method : Determine uing the equation for a vertical curve (working in tation and percent grade) At PC, y = c. Therefore, c = 335 (elevation of PI) /(G ) = 335.5 At PC, b = G =.5 Anywhere, a G G 4.5.5 7 The point on the curve you know i right above the pipline: tation 9+00, elevation 330.5 (ee method for a determination of the elevation). The tation (9+00) i actually /. 3.5 Ue the point and the equation for the curve to olve for : y ax 3.5 bx c 330.5.5 335.5 330.5 0 0 0 3.5 4 3.5 0.5 3.5 0.5 0.875 5.5 3.5.5.5 335.5 olve quadratic and get = 5.567 or 0.79 tation Since 0.79 tation i too hort (the curve would not even extend to tation 9+00 and it would alo not be the ONGEST curve one could deign), chooe = 5.567 tation or 556.7. Determine the deign peed (extra credit), full point were given if you got thi far K=/A o K=79.5 Adequate SSD for 45 mph require K of 6, and for 50 mph require K of 84, o maximum deign peed i 45 mph.
tunnel wall tunnel wall tunnel wall CEE 30 nd Midterm, Fall 009 Quetion 3: (0 point) At milepot 3.44, Aurora Avenue (S 99) enter the Battery Street Tunnel in the outhbound direction. There i a horizontal curve entering thi tunnel with the dimenion hown below. Determine the deign peed for thi tunnel to the nearet 5 mph increment baed on thi horizontal curve (remember to round down). The outhbound direction ha two lane of travel. Cro Section iew Plan iew Northbound Southbound 4.. wide PT = 409. =.69º No uperelevation 4. 0. 0. = 409. PC Northbound Southbound You need to check both the horizontal curve SSD and the uperelevation. I realize you can get the right anwer even if you ignore uperelevation but doing o doe not repreent good engineering practice. You really have to look at it. Firt, check SSD M = ½ inide lane + ditance to wall = 5 + = 7 v = 409 4 0 5 = 390 SSD v 90 co v M v 390 390 7 = co 48 90 390 For full credit, you need to check uperelevation for 0 mph: 0.47 99. v g f e 3. 0.7 0 4 If f= 0.7, you only need 99.4 but your actual v i 390 o the curve i okay for 0 mph in term of uperelevation (or lack of uperelevation). The limiting factor i SSD and the deign peed i 0 mph. 48 (.) 33 (.5), Other value of f were accepted if aumption tated.
CEE 30 nd Midterm, Fall 009 Quetion 4 (0 point) A ignal i being retimed o that it aociated interection will operate at 80% capacity. Aume a lot time of 4 econd per phae. The interection operate in phae: Phae : eatbound and wetbound traffic go at the ame time. Phae : northbound and outhbound traffic go at the ame time. The calculated aturation flow and meaure flow are hown below. Phae ane Group Saturation Flow Meaured Flow Eatbound 000 veh/hr 00 veh/hr Wetbound 000 veh/hr 00 veh/hr Northbound 800 veh/hr 600 veh/hr Southbound 600 veh/hr 600 veh/hr eport the following:. The critical lane group for phae and phae.. The minimum neceary cycle length (rounded up to the nearet 5 econd). 3. The optimal cycle length uing Webter practical C opt equation (rounded up to the nearet 5 econd). Critical ane Group Thee are imply the lane group in each phae with the highet v/ ratio Critical lane group are wetbound and outhbound C min 8 0.80 Cmin 8 00 600 0.80 000 600 Or C min = 30 econd (rounded up to the nearet 5 econd) C opt.5 8 5 C opt 40 00 600 000 600 Or C opt = 40 econd (rounded up to the nearet 5 econd)