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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
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