Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Exam 1 Solutions, Rolling Resistance - Geometric Design of Highway Facilities | CME 402, Exams of Civil Engineering

Material Type: Exam; Class: Geometric Design of Highway Facilities; Subject: Civil & Materials Engineering; University: University of Illinois - Chicago; Term: Spring 2006;

Typology: Exams

2011/2012

Uploaded on 05/18/2012

koofers-user-3fl
koofers-user-3fl 🇺🇸

4.2

(5)

10 documents

1 / 6

Toggle sidebar

Often downloaded together


Related documents


Partial preview of the text

Download Exam 1 Solutions, Rolling Resistance - Geometric Design of Highway Facilities | CME 402 and more Exams Civil Engineering in PDF only on Docsity! University of Illinois at Chicago Department of Civil and Materials Engineering CME 402, Highway Geometric Design Mid-Term Test 1 March 2, 2006 SB + Time: 60 minutes Closed Book Exam. Total points: 100 Attempt all questions. Show all intermediate and final steps and results, The following equations may prove useful: Ve=att+ Vy d=Vot + 0.5at? (Vi? 2ad + (Vo)? Stopping Sight Distance: U oe —s SSD = Vt +[V?/2g@/gt+G Lg ty=2.5 sec a=3.4 m/s* g=9.81 m/s* Horizontal Curves: Ry=V"/[e*(e+f)] M, = R, [1-cos(28.65*SSD/R,) S = (Ry/28.65)* [cos (R, — Msy/R,)] L=RAga Vertical Curves: General equation: y = ax’ +bx +e A=(B2- gi) a=A/2L Offset: ¥ =| ax?|=|A[* x2/2L Page | of 6 1. Define each of the following terms as they have been used in the course lectures: \ S / [20 marks, recommended time budget 12 min] “4 Uo a. Factors influencing perception reaction time -DRweR AGE -URVER SUEEPINESS / ALERIVESS \ RR Accotge CoNSUMPTIO/ / — pisTehenonS b. Rolling Resistance Ps vehicle's Keo @& hes a Cortain resistance. te rolliry om He Po vemedt, Aue to Aefirwatiyn a6 He Kt as it rolls aut, Sct we between Hee and Pa Ueneh This is Ceprsrel os tle Rnomebic force Ro. c. Obstacle Envelope Ditewe betwee He edge of rhe Mod way Gud & toad -side abject, A mi mimun ghstacte \ Omvelope mor be needed st Onsure feguireol stopping sight Aistane @ aroun Curves, a d. Deflection Angle and Chord FoR Fiero Aft OF CURVES From oe kn pwr onk oow a curey ot Kugwi rod jus wl ofhie parts Con be Gun b? caloulatng He stevight lime dighee fo gmsther port and Crd Lie deflection. Oargh e- between Ho FEreb known iw we Va Steed pont, TLE curve Com Hen be stoked. ouch te Fle GI Nm gene TI® Ena rind ing “Survaring eqesorebet, stead of baving ta physely God iy ty, a Name: 4. Design a horizontal curve with a 40-degree central angle (A=40) for a two-lane road with 3-m lanes. The design speed is 110 km/h and superelevation limit is 6%. Give the radius and length of the curve that you would recommend (assume f, = 0.11). What distance must be cleared from the inside edge of the inside lane to provide adequate stopping sight , a distance? [25 marks, recommended time budget 13 min] AN “ Q=40 X aes @ Ba each. v Vi = 110 kon/p=30.56% tae & = AOE R? cP mM? Cran 5 vr7 g (e +f) _ 30,567 ~ 4s (O06+0u) ~ ou = Aesiqn CadieS Ch centerline) L= Ra= S65 (3 )I2920 ssD= tye + [07/24] (9 geede ) oe _ 205b™ = 3056(2%S) + 221) Ks) 7b4 £ 13724= QB7T we SSD M=RCI- cos (ARSC) “10,7. fru) ane DAUE ie le®, f Ap look at it backwe-ae Gre Bene lo 5: A vertical curve is designed for 90 km/h, and it has an initial grade of +2.5% and a final grade of -1.0%. The point of vertical tangent (end of vertical curve or EVC) is at station 34-480. It is known that a point on the curve at station 3+440 is at elevation 75 m. What is the stationing and elevation of the BVC (point of vertical curve or beginning of vertical = 7 curve)? What is the stationing and elevation of the high point on the curve? | t ¢ qa.” ( 9,7 -2.$ [25 marks, recommended time budget 15 min] — Ar ga f eg, ath c= BUC Azas-Cl) Lx ib ~ f = 2,5 = 0.0285 a Ae. LE -0.035 __ q arb, cake h™ ATR) =O, 00012 Y(% 2 z QV 34140 Bx y77° x. x ve 4 770, 001 Tx™ + Nl tl, wo 3444 c= 71.F06 ns) NS —0, 0001247 + 0.0 + 71. 806 7 \ —.* Page 6 of 6 a
Docsity logo



Copyright © 2024 Ladybird Srl - Via Leonardo da Vinci 16, 10126, Torino, Italy - VAT 10816460017 - All rights reserved