Assessment 2_updated

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CVEN3401 ± Term 2 2023 ± Highway Geometric Design Page 1 14/07/2023 Highway Geometric Design Assignment Question 1 (3 marks) A driver is travelling at 95 km/h on a road. An object is spotted on the road 146 m ahead and the driver is able to come to a stop just before hitting the object. Assuming 2.5 seconds for reaction time and 0.36 for longitudinal deceleration, determine the grade of the road. (3 decimal places) Question 2 (3 marks) A car has a drag coefficient 0.34, a frontal area of 2.20 m 2 ͕ Ă ǁĞŝŐŚƚ ŽĨ ϭϭϱϬϬ E ĂŶĚ ʌ с ϭ͘ϮϮϱ ŬŐͬŵ 3 . It is travelling at a speed 156 km/h. The engine develops maximum power of 135 hp. What is the maximum grade this vehicle could ascend on a paved surface while its engine is developing maximum power? (3 decimal places) Question 3 (3 marks) At a specific instance, a car is travelling on a paved surface at 145 ݇݉ / ݄ with ܥ ܦ = 0.35, ܣ ݂ = 1.90 ݉ 2 , ܹ = 6000 ܰ and ߩ = 1.225 ݇݃ / ݉ 3 . Its engine is producing 120 ݄݌ of power and the coefficient of losses between the motor and the wheels is 90%. What will the ĐĂƌ͛Ɛ ŵĂdžŝŵƵŵ ĂĐĐĞůĞƌĂƚŝŽŶ ƌĂƚĞ ďĞ ƵŶĚĞƌ ƚŚĞƐĞ conditions on a level road? (Use the relationship ( ܨ ௉ൈఌ ) where ܨ ݁ is the force generated by the engine in N, P is the power in watts, V is the vehicle speed in ݉ /s and ߳ is the coefficient of losses between motor and wheels). (3 decimal places) Question 4 (3 marks) An engineering student is driving on a level roadway and sees a construction sign 180 ݉ ahead in the middle of the roadway. The student strikes the sign at a speed of 60 ݇݉ / ݄ . If the student was travelling at 90 ݇݉ / ݄ when the sign was first ƐƉŽƚƚĞĚ͕ ǁŚĂƚ ǁĂƐ ƚŚĞ ƐƚƵĚĞŶƚ͛Ɛ associated perception/reaction time? How far back should the student have first observed the sign to be able to stop safely at a comfortable deceleration rate before hitting the sign? (3 decimal places) Question 5 (12 marks) A tunnel at level grade has a design speed of 120 km/h and curves of 1000 ݉ radius. The tunnel has one lane in each direction. Each lane is 4 ݉ wide and the sidewalk is 2 ݉ wide. (3 decimal places) a. Determine an appropriate superelevation rate for the circular curve. b. Check if the available sight distance exceeds the SSD. c. If the answer is no in part b, determine what the posted speed limit should be to ensure safe stopping. CVEN3401: Sustainable Transport and Highway Engineering School of Civil and Environmental Engineering Term 2, 2023
RT 2 5s d 0.36 V 95km h 95to 26.39 mls Stopping sight distance 254 2752 95 952 ssn 25427521 36 o_o icy object spotted on the road 146m ahead 146 25 25427599 G 8.351 p 1.225kglm3 fair density1 G 0.34 Af 22om2 Engine horsepower 135hp lhp 746w 1351 pt oo7iowlw Nms 1ikinamttic of vehicles Eractwe 11 RatRr ma Max grade to ascend on road given engineoperate at max.power 1 Power to over ome RA RA pen Af v2 RA ixnzsxosaxz zxl.mn 19853NBA Rav Rn 19853x1156kmlhx 860 2967W 2Power to oven come Rv Rr hw
fn o_o111 an fri o 0111 15 o0197 Rr o_o197xnsoo 226.409N 1 Rrxv Br 226.409xnix 9811.0567w 3 Power to over one Ra Ra WG Running horsepower overcome Ra 10 860 98110567 38.647 90038.647 11500G G 7.829 i The maximum gradenisveni.ie can ascend on pavedsurface is 7829 V 145km h 40.278 mls 0.35 Af 1.9ni W 6000N E p wikglm Engine 120hp 746 89520w 1Nmlsl coefficient of losses 90 On a level road 2 out of 3 resistance i Air resistance 2 rolling resistance 1 Air resistant e RA RA IPGA V2 RA ixi 2250.35 1 9x402782 660.7 N
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2 Rolling resistance I Rrl Rr few fi o 0111til f o 01 11 4 o 019 Rr o 019xbooo 114.028N Total resistance 660790 114028 774.818N Reina 774818x 40.278 31208.12 w Erie generated by engine Fe 8 0 2ooo2979N Extra tone provided for an elevation 2ooo2979 774818 1225 480N F ma 1225480 ta a 3.270ms 2 i The car's maximum acceleration rate is 3 27ms 2 V 60km h 16.667mls U 90 kmlh 25mls 1 Reaction time distance ISR SR R x V 3.6 2 Breaking distance SB SB U u2 zgldto.ci G
lal Perception1 reaction time Distance tuned 180m Assume d 0 36 a wrong to Table5 3 of Aumn Guide to road design Part 3 By SSD YY V2_u2 zgldto.nu 180m Tfmlh lot qx 2 9811036tooixoi RT 5234s Distance t stop safely SSD 523 90km h Got 2 2 9811036 01 219327 m a Determine appropriate super elevation rate for cinuiar curve By linear distribution method Vemax G 127Rlemaxtfmax Assume the roadis a rural highway f Ǔ Ì Grounded For design speed of Rokmlh emi 6 fmax one 1120kmlhit6 6 o 04 Grounded O 045 4.5 f o045 o_o68 lfe.se max use emax to calculate 7 Hf also sfmax reduce designspeed Determine super elevation development length Le To Sro Le Le in
Le maxllrr Lrgl R rate of rotation 5 Assume e 3 ez 4.5 as calculated Ri when speed Ookmlh R 3.5 35 When speed 80km h R 25 Lrr 027810.03 1 o 045 GR relative grade hnie.ge assume as 3 ez 45 WR 4m When speed C80km h Gr i When speed s 80km h Gr 9 0.3 Lrg 413 1 45 11 loon 03 Le loom 0Lmax for 2lanes 107 Lmax Lrg Lrg loom
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cven 3401 Q5 b Is amiable sight distant exceeds SSD Avalíable sight distance I Ru radius of centre t inner lane Rv R lane width Rv 1000 998m 2 m middle ordinate m inner lane road shoulder m 2 4m 3 SSD Arc length of inner lane Rv 4 2cosy R 2cos 1191 10.263 SSD arc length of inner lane 998 10 263 178.766m Required Going distance SSD Rjf Euz gldto Assume d 0.36 R 25s SSD 25 20 120 12 29811036 0 240.643 Therefore he auakablesightdstne does not exceeds SSD i Determine what t postedspeed limit should be Since the required stopping distance 240 643m round up 245m 245 2 0 lot 12 2981xlo 36 01 V 6 274km h The posted speed limit should be less tin design speed 126274km h We can set it as 120 kmlhr
Question 6 (15 marks) A two-lane highway (one 3.6 ݉ lane in each direction) with a design speed of 100 ݇݉ / ݄ goes from normal crown with 2% cross-slope to 6% superelevation by means of a spiral transition curve. The spiral curve is 50 ݉ long. If the superelevation is attained by rotating the road section around its centreline, draw a cross section of the road at 10, 20 and 40 ݉ from TS. (horizontal distances, and elevated heights must all be labelled in the drawing) Question 7 (15 marks) A roadway goes from tangent alignment to a 245 ݉ curve (spiral-circular-spiral) by means of a 85 ݉ long circular and two similar 80 ݉ long spiral transition curves. The deflection angle between the tangents is 45 ° . Use formulas to compute ܺ ǡ ܻ ǡ ݌ǡ ܭǤ Assume that the station of the PI measured along the back tangent is 150+000, and compute the stations of the TS, SC, CS and ST. (3 decimal places) Prepare a table for the first spiral curve giving coordinates, spiral angles, deflection angles and chords (from TS) for full stations and +20 points. Question 8 (8 points) You are asked to design a horizontal curve (with the smallest possible radius) for a two-lane road. The road has 3.6 ݉ lanes. Due to expensive excavation, it is determined that a maximum of 12 ݉ can be cleared ĨƌŽŵ ƚŚĞ ƌŽĂĚ͛Ɛ ĐĞŶƚƌĞůŝŶĞ ƚŽǁĂƌĚ ƚŚĞ ŝŶƐŝĚĞ lane to provide for stopping sight distance. Also, local guidelines dictate a maximum superelevation of 8%. What is the highest possible design speed for this curve? Question 9 (15 points) A vertical curve passes under a new 20 m wide railway bridge which is to be constructed as shown below. The design speed of the vertical curve is ͺͲ݇݉Ȁ݄ݎ . The bridge centerline is to be located 60 m to the right of the PVI which has an elevation of 100.0 m above sea level. a. Draw a sketch of the tangents of the vertical alignment and design curve to show your understanding of the problem. Document the critical station points of the curve and the respective elevations and please indicate stations and elevations in your drawing. b. Determine the bridge clearance from the paved surface if the elevation of the bottom of the bridge is 105.5 m. You may neglect the thickness of the pavement. Vertical Alignment ʹ Longitudinal profile Pkehnon 100 35G Mekuatm PVI elevation t Gz ii elevation Not35Gz Gz G Kinda 10.1 I A
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c. Comment on the adequacy of length of vertical curve on both the directions based on the design criteria: (a) stopping sight distance and (b) comfort and aesthetics. If the length of the curve is not sufficient suggest an improvement strategy. Question 10 (15 points) Two level sections of an east-west highway (G=0) are to be connected. Suppose it is necessary to keep the entire alignment within the 1300 ݉ that separate the two-level sections. It is determined that the crest and sag curves should be connected (i.e., the EVC of the crest and the BVC of the sag) with a constant- grade section that has the lowest grade possible. In solving this problem, use the design speed of 110 ݇݉ / ݄ , determine for the crest and sag vertical curves, the stationing and elevation of the BVCs and EVCs given that the westernmost level section ends at station 0+000.00 and elevation 44 and the elevation of the easternmost level section is 18. (3 decimal places) Question 11 (8 marks) Calculate the net earthwork required for a 400m long rural road section. The cross-sectional measured end areas at every 100m chainage are presented in the figure below. Assume that the material present on site is common earth with 10% reduction in volume. Cross-Section at every 100m chainage Fill = 5m 2 Cut = 18m 2 C.S 0+400 1.3m Fill = 35m 2 C.S 0+600 2.3m 2.2m Fill = 20m 2 C.S 0+500 0.8m Cut = 40m 2 C.S 0+700 2.4m Cut = 40m 2 C.S 0+800 2.5m XL 10 XL
Qn 5 18 ig too 5 20 25251 5 g o o 2750 00 0 0 35 ioiiiii 4894 5106 35 35 35 489 to8 o o o o_o 0 40 4014080801 ⼼⼼ 4000 cs cs to 23 Totals 457 97 5052.2 48.94 it 2.3a24 Net Earthworks 元去 1 Hs yield Goo_yoo ⼆⼀ v47 4570 97 5052211 10 24m3 2 3 0 600 0 047 Need 24m more soil for the construction t 648.94 108 8 1 61 9 it3 1.3 08 400500 O 021 O 021 461.9
V look mlh c 2 e 6 Spiralcurve Sro 50m By x so 5 x nnm By y c_ y 2 1 o.in At iifkreit.ci By 只靠 s not By a inner lane wont stouter lane on elevation 2 1.2 x3.6 m Atkzom s o 024 24 aeui cxmnerlanew.intsxontriane 2 xn 2 4 xi 6 o.mn At 40m Y 4 S 4.8 ekwnon Ctinnerianetsx outer lane 2 3.6 4 8 x3.6 ozaom 2448 1584 i 52 ǐ i is no 20 40 1.2 24 46 6 oEIEoifii.EE a i OE I Ě on ǚ
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Lc 85m Lsp 80m 450 i i Tteoi 85m.is 450 1 CurveRadius 2 Central Angle By a and L R 信⼼ 0 0 9 0.404 rad 23.170 0 450 0 45x o 0.785rad Let o o205 3 Spiral Angle 0 0 785 20 0 0s 05 019rad 10.90 0s os 20 Sub 20into 0 When L 80m 0 0.185 2 l fi.io 0785 k 80 4 Einaz 7971om le Ro YS 6R Ǜ 336 ǜ spp 85 R0 6xzniiim.fi qxap 5.064m 0 40 p 0.785 P ziiq 1 269 IE ࡣ? iqz 39 952m
T Rtpltn tk 1210.19 1 269 㕧些 1 39.952 127.476m Station value TS PI ITS 1150 01 10 127 4761 149 872 524 sc TS th SI 1149872524 10 080 149 952.524 CS Sctlc Cs 149 952 524 10 0851 150 37.52 51 Cstls ST 1150 37.524 10to801 150 112524 l X y os dlcrtci chord o o o o o 20 R Ǒ o 08 o_o nrad 0.23 20 40 39.99 0.63 o 048 0.90 3999 60 59 93 214 0.107 205 5997 80 7971 5.06 0190 363 7987 iii ǜ m LE Rea oc Os iq c.ru
i emax 8 如何 在 桃 is 3on i n_n T.it Stopping sight distance I Required Stopping dstaniel SSD 2 Assume G 0 d 0.36 m 25 SSD 40 im 45.3m 2 Avahabustppiy distance m R__z m 24 102 SSD Rv 01 o zcu I 28 106 SSD 49.5me 40218 1273010 to 8.84 around 9 emax 8
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Pkehnon 100 35G Mekuatm PVI elevation t Gz ianiiiii elevation Not35 G G G G PVI elevation 100m given yiitbxtc station in bin Pvci a ⾃若 elevation 100 35G C elevation at Pvc station 70 c 100 35G PUT elevation 100 35Gz station 70
I 10 1 10 1 Ǜ iii Elevation 00 Elevation to 1 g stay 70 1 fiiiiustniotooo ss tzso.cn G 0 25 25 125.5m s a L K zoolhi K 125 52 20010 65 125.5 114.4 A 1.223 A 1 Gz Gill.in 1Gz 01 Gz l.tn
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I 10 1 10 1 Ǜ ī iiiii x iy in PVI Ekunnniloo 5 Elevation to 1 g station 0 060 1 li ǜ kustmiotooo look Gi 0 a 4.36Zxió5 100 Y 4318xiit 100 Atrzom y 1o_o v09 Bridge clearance 5.4m
d other direction SSD 2 80 802 25410 36 1.223 128m ss D a L i ok K 802 129610.49.5 10 06 Ckll44 i ok If the length of curve is not sufficient one of the strategy is reduu the design speed so tat driver could have a longer avalíable stopping distance
go 0 G Ekfm Lc 1 inut Lt 1 Ekpm Let u tls Boom yt tys 4418 Yc ys yt GA 0 26m A c 1G 01 G As 10 Gl G G 26 o G Lc tl.tt s 1300 20 Ls.li Lc CS Ls cs lootoi tlt 0 his t 1300 2 K 2004 2 A2 A l G z G l IG0 1 G s s s D 12 25 0 to.mn 2087m G 3 16 Lc Kfi_ 2001 A Lc 275.7 417.4 44 6 u is 417.4 44_ Assumption wrong
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Ls is k 2001 A 1 0 G1 G ssn 0 2541 SSD 221.45 76.39 9 442_ KA 176 39t 9.44iih_ 2001065 439 91 44 A Á 1152 78 qn ǜ 54G 130 t 200 0 02327 tzooxq 4G G 152 78 quiiy 134654 t 焱品 G G Ahmfn u u 7241286G 377711170.98Gt 1394ooboo 72 Nooo G l 254Gt a 44 ⼆⽇ 2 7241286Gt 377711670.98Gt139400600.72 100001 33025 6604G t 8872G 2377447 7241206Git 377711670.98Gt 1394ooboo 72 33v2 a4125476oooo G 2377440000 25777714G 877048329Gt 2516840601 0 G 30 9 or GE 3.16
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Less Ls s less k zoojiik S 2081 Lc KA zo ā p2 4 2001 yi A Lc 97 306G Lsss kizoolhti Ii 217.757 K zoo 1 6 Ǘ stn1 330.98m Lai KA 176 39 ta.in I 2001065ttai ࡃ? Aiiiiiii ǜ ii 18823936.87G 500X7406187 429Gt Jòox 364241103.1 15637207.44Gt 434366 8732G2 596643062 9 16573418.42G G 18823936.875 3703093715 Gt 1.82 10 4343668732G2 32210625.86Gt 596643062.9 G 1Lctlsl 1300 2 gqyf8823936875 3703093715 Gt 1.82 10 f Boo 4343668732G2 32210625.86Gt 596643062.9 1.8 t 61090439 8345 6237 874.933Gt 58056949878 55
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564676935.2GF 112935387oG 4 MX 0G 0 8.37xi 7 76t1o 1.55i_ G 61090439 8345 6837380874933Gt 24 10 1129353870GE2258767740G 838xih 1.674 1.552 G 2 30 U 2 30x97 306 224 m 4 s s L s 122.6 L s s s U Boo_ 224 1226 953m Bin En crust Otooo otzz4 Bug Eksg 011177 0 1300 44 41.422 19 484 18
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