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Unformatted text preview: Hw 2:2 2 SOLU’HONS‘ Problem # 1 Cam quozgz Aegixqﬁsz C11: 55%
3174;». = R7) + 37>
ELEVQSTA 232m; a «(ma WWW“;
ELgVQST/rzsmmcmg’m 2W
€162.90
b)
. ELEVPV, 2 (355A — dst‘Zis'casziisbsb)
ELCNW. 34359 —— Elma BVC = 34.55 +o;‘o_;s(LI_/7;) =' '36le 4,6131.
Ema/x: 49.481; X= % +20% 175.41” 4% Lam/(WC) + (31.x + £25: = 341.5% +,0{Z§L .ou(% +160) + .oe(ég+'w)2
2:. 3;; "4,31%: 34<§+.m€/L '54_'2.Qg(%2+ zmAwaw)
0:43.43 + mm + 31» .1352 0:19.0le ~/0.?3L, + Iago
L=. fiaiﬁh 0r L=1Hus P}. Problem # 1 cont.. (mm) 41939, 1499057 ‘ b ck
3E4 Nth/5‘0 is My; 13%, Sa H MW {.2 on +Le,(:cm30~+
43 ELEV(1‘N+$D>= ELEVW, éﬁiﬂzme(ZWKDQ := 002((é1w)
WW5“? Problem # 2 0) 'EPLVE ': 3553,
{5qu z 350 Md
Elia/cf : 350—401 I
t 561‘, = 358'3'—(350—4d) :35?3'—350++d.
l
l
i {’5 awarx by 9;: 4C3};
' 77
A : 612an —: 2—04) :67; Problem # 2 cont.. 5C 2 924%! 4m ﬁguﬂ) cum! 4 _ I: «(—4» W m Woww
. ._...————G ‘ ‘Hmb 0010th LWEI‘W ugfmﬂ OVYIOI SDI/vimj
WY cl: 9457—5 1/: 601 :— 2015xé: {245 WW5
1 “WW 1W3” 05/ m WW: 1‘3 aﬁcsgrwt. AUCWMMC mg a
i Em. M: 350+ 40/1) 1: 35mm A ' ﬁrm; oUé’tomw ’b/r‘m 6V0 £73 MW [WWW PDWWL 1‘5 . . ‘zw—m 1. «4w rfEP'rt’Ai/‘YICZC’ (9%, X 7; {:69 ,, 961, (so
I ,Problem # 2 cont.. Rom {up amoral, eal wwbbn Elm : Eta/5. + Cq7r+ AXZ
2b 35323:: 5 L“ ‘. ' 7’
. A (3m) Gwyn 551%) 501w?) for 1, ) £313.45 avgm; I. W It,wa Hm, We d5 ,‘24—5fcc6 v0 190.5155. :35: " 3mm?» (xii .ggvjc JﬁﬁrdsMDJH _ gm) 3x ._ 100' 1296"“ id) W'm'm whom 4MB)  This point E1: 15de Eva mmd EVL‘ mgu'w, +0“,
ngWCbL .chméut’fm am he/ Mal ‘ 3 9c: SEMIOMJIOvmg .sm(615+5o) :; [090.7%
Eavzx: iO‘lc '3 4" ("0‘045V19‘10)+ (6905)(;0¢r9)Z
r 52% 1900
Elie/5" M :3 " i 5121:1090 “" Problem # 3 ‘;A vertical curve with PVC at Station 43+OO and elevation of 426.00 ft is on a
:_'grade of 44.2% that intersects a +26% grade. The length of vertical curve iconnecttng the two‘grades is 850 ft. Find:
. . 3(a) The curve elevation at Station 48+60.00.
' ‘ (b) The station at which the slope of line tangentto the vertical curve is —2.7%. _ "(0) The station beyond the PVI atwhich the elevation of the curve is  “735'  L.= 550‘ = 8.5 si—a I
h 560' = SACps‘l’a " ‘ rave: 4am; TNT: 5i+5o /° 21.: 414.20 e’fz'é,
ﬁt. PVI= “+25
EL: 408MB 5 x2 +39: +Pvcdw = ( 2':(::SZ))(5.¢)Z+(—4.z )(6.o)
W +4ZéOO .1“; m» V. Zax Ho => ~27!" 2'<o.4»wo)(x) 42 = 2.'l +4.2.
X 2(o.4) ' 5f“: (45.1.00) + ( 3150) = = (0.4) X2 ’42): 4r 42$.OO = Lezso stat: 157.58 8.25
W . aAngarzxf {mace—47,15 __. 4.2: 2.lo‘zl3 g 1.8835(5lz
0'5 or 2.614,! sin Problem # 4 A vertical curve is to pass through a railroad at grade. The crossing must be at
station 37+40.00 and at an elevation of 574.00. The initial grade of the highway
gis_+1.8% and it meets a «3.4% grade at station 38+00 at an elevation of 577.73
ft. The rate of change along the vertical curve must not exceed 1.0% per station. l‘.(a) What length of curve (to the nearest foot) will meet these conditions?
 I (b) What is the station and elevation of the high point of the curve? K7— lbo LM’AE'. jV‘elev. F = 511.13  (cup) 0.8%) = 519.95 a = 517.13 + (0.6,) (3.4%) = '_ daE wleuﬁl = 61400513911 ._. ,5 ‘ clams deaf 514.00'510195 MG“) =" 2(Oab)(l.41SSQH) : 924$} s‘l‘a '   . (5'1) (1.4166‘l vi) l I ' _ _. (9241, = (926 >620 ‘ ‘ :11 __ ~3i —l8 ‘.Za>‘+5=0 X: a " 2(32’5i) =z 644.8 ‘' 24635 ,  _‘ f'ZL— _ 2(625) _OI‘HGO (9 «ex 035): s.(z§ij§fss) + 517.15— 1.3 ) = (Zlé.%§) = . Problem # 5 ':9A culvert is to be placed at the tow point of a vertical curve used in the design of
V Ta rural highway. The beginning of the vertical curve is at station 30+60.00 (A), ' _ 'ivvithgi = —2.5% and 92 = +1.6%. The length of vertical curve is 900 ft. 5: ’ f . i (a) What is the station of the culvert location? j '_ (b) ' If the elevation of station A is 438.00 and the culvert invert is to be 18.0
‘  a ft below the ﬁnished roadway, compute the elevation of the culvert
' invert. . .r_ , ._ . a 202.2315) a z LbrcZS) ‘
 2(a) = e .4878 5+4 Vt. em = (mm (s4ane>= 5X2  26x ‘i‘ 458. co Problem # 6 A new vertical curve of length 1200 ft is to pass through Point D, which is located
at the hi point of the curve. The percent grade for 92 that satisﬁes the design
constraints as shown is most nearly: ’PVI’ cz+~oo '
EH 4561.26 73:56.3 1.5 x 1— WC
. _ aw M‘a (E ) a ~ Iva(ta) 3,1.5 '1" 1.5 ‘3245 1... 4.50. 25 (gz—Is) + i5 (ia) ‘ scarier (arm r + y‘z‘ho — Z @345) Problem # 7 'Two tangents with 91 = +45% and 92 = +15% intersect at the PVl with a station
of,57+50 and elevation of 620.00. At station 55+70.00 it is desired to pass a
vertical curve through Point 8 at elevation 611.40. The length of vertical curve
(to. the nearest 5 ft) that will satisfy this constraint is: :‘?_'.I:,_“:Y5‘l__"3??(67+50)(66m): leo‘ = Leela fieW F = 97—000  ﬂamed.) = (,1 Leo‘ 51 2 620.00 (1.a)(i,57¢3 .5 @1501 z t l “33‘ _e9eu.6 25260} cqu'eu‘io . ’05 t 3.436” _ 2g s+0 ' ( 2(l.5)(3.43$l\+0 —_. Q5568 stat V (S'D' (3.455lll) . of Problem # 8 sta 82+45 9W PVC
sta 81 +60 PVI
elev 425.38 ft Find the elevation of the point of vertical curve (PVC). LVC
EPVC = EPVI + 01 = 425.38 ft + (3.6 (2.5 sta)
Sta.
2 434.38 ft Find the rate of change of grade per station.
R _ G2 — G1 H 4.4% « (—3.6%)
F‘ L _ 5 sta
= 1.60%/sta Problem # 9 sta 81 +50 Ev C
elev 638.42 ft BVC
sta 76+00
etev 646.12 ft PVI Marranging,
ﬂ; 2
(0.037 (550 ft)
638.42 f1: — 646.12 fl; [ 19.8 ft; = L
(0.037 (550 ft)?
L ~ 12.1 ft
= 1.68 sta (168 ft)
mtVL‘X ‘,
Z
ELE'VX = ELEVevJ G! ,1 + A’: x 2 2mm 444200 2. 95540
2 7 AZGL’&':3'g'—3r‘€f_7.+ egg/42 = 4464?, +3.gxg.g+ 7455
2L
\2I\ w, 7er><§~§2
2L
L1", 74XS5159‘Zf54'Q
2x :2 x 13’ Problem # 10 (a) L StaBVC = stav _ ,_2_ = (7420} — 14020 ft = 0+20 sta.
staEvc = (0+20) + L = (0+20) + 1400 ft = 14+20 sta. . . L
elevationgvc = elevationv + 91 1400 ft
= 226.88 ft + (0.02) ( 2 ) = 240.88 ft y = 1) x2 + 9193 + elevationBVC 4 ? (—2)
= $2 * 256 + 240.88 ft = 0.2142st2 — 23: + 240.88 ft
At y = 240 ft, calculate 3:. 0.214293:2 — 22: + 240.88 = 240
2:1 = 8.871 sta. (887.1 ft)
:52 = 0.462 sta, (46.2 ft)
stal = (0+20) + 46.2 ft = 04662 sta stag = (0+20) + 887.1 ft = 9+07.1 sta Between stations U+66.2 and 94—0721, the curve drops"
below the flood plain. The stations dropping below this
50year ﬂood plain are 1+00, 2+00, 3+00, 4+00, 5+00, 6+00, 7+00, 8+00
and 9+00
(b) 907.113; —— 662 ft = 840.9 ft (C) increasing the length of curve will decrease the
length of the portion that will be submerged because
elevations on the curve will generally increase. (d) Decreasing the length of curve will increase the
length of the portion that will submerge. (e) When the road is ﬂooded, the embankment and
the subgrade may deteriorate, which in turn degrades
the pavement, regardless of pavement material. De—
pending on its importance, the highway may have to be
rerouted. In addition, providing an adequate drainage
system, paving the shoulders, and using highquality,
plantmixed hot bituminous pavement material can help. 15 Problem # 11 59mm:
1. Determine the elevation at station 38+00.00.
Note that station 384410.00 is on Curve 1.
Compute the P.V.C. station on Curve 1.
P.\’.C..~sta s F’A'LI stn  '/2 L, = 42+00  mumo; = 35+00
Compute the tangent elevation at station 38400.
Tan. elev. = P.V.l.I elev.  g, at distance
= 436.56 a — 4.5% x I 42 st:  3s stal = 418.56 {x
W om statiOn on Curve 1. >
e = (1.15; (g:  g.) a (14 Sta/6) [3.0'}  H.523 ll
=  13.13 ft (neglect. the sign)
Compute the tangent offset at sfhtion 38+00.
y = (4e I L“) X"
g l4 2 13.13 t't/(l.400 ﬁt"! [(38+0m  (35+0011" = 2.41 ft Compute the elevation at station 33100.
(a) Y,,.....  415.5611  2.41 It  $15.15“, THE CORRECT ANSWER IS (A) 2. Determine the station at the high point on Curve 1.
Compute the rate of change of grade. r. for Curve 1.
r = (g;  g.)/ L = [3.0%  «4.51:» I 14 stn a o.5351;: Ism
Compute the distance from the P.V.C.I to the high point on Curve 1.
X" = g‘/r =  (+4.5ﬁ)/(0.5357ﬁ Ista l = 8.4002 sta
Compute the P.V.C. station on Curve 1.
P.V.C., = P.V.I.,  '1‘: L = (42+00)  'A (14400) = 35+00
Compute the station of the high point on Curve 1.
High point station = P.V.C.. + X.“
= (354410.00) 1 (8440.02)
L E) = 43+4D.D2 which is most nearly. Am 3. Determine the elevation at the high point of Curve 1.
Compute the tangent elevation at station 43440.02 on the g. tangent extended.
Tan. elet‘. = 436.56 it. + (4.51%) (43.4002 Sta  42.0000 sta) I $42.86 ﬂ:
Compute the tangent offset. y“, at the high point on Curve 1.
y...” e H e I unc‘ = (4 x 13.13 a “1.400)”; a: [(43+4o.021  (35+oo.00) I“ = 18.91 1179‘
Compute the vertical curve elevation at the high point on Curve 1. ‘\ ‘ Curve etan"...~ a: tangent elevation  tangent offset
LL.) Curve elem"...  442,86  18.9l II Problem # 12 “ 1.5% 5°]:
BVC *2 evo
elev 562 ft sta (68+00) sta (sown) PV!
sla (64+00) (a) L0 = (68+00) ~ (60+00) = 800 ft = 92—91 2
y ( 21,0)? +91$+ynvc _ 2.5 — (~15)
“ (2)(8 3123.)
= 0.25:1:2 — 1.59: + 562 ft m2 — 1.53: + 562a; Elevations of the road at ESQft intervals can be cal
culated by inserting values of a; = {3.5, 1.0, 1.5, . .., 8.0
stations in the preceding equation. ye4+so = (0.25)(4.5 8133.)? — (1.5)(45 sta) + 562 a». = 560.31 ft y65+go = (0.25) (5 sta)2 — (1.5) (5.0mm) + 562 ft 31654.00 = (0.25) (6 sta)2 — (1.5)(6 sta) + 562 ft = 562.0 I I7 Problem # 12 cont. 0?) sta {64+50) sta (66+00) The elevation of the centerline pile (pile C) is found
ﬁgom Sol. 5.1, less 3 ft. There is a. 12% superelevation,
so for each 105; run, the rise is 1.2 ft. For the ﬁrst bent at 64+50, pile C top = 560.31 ff: — 3.0 ft = 557.31 ft
pile B top = 557.31 ft — 1.2 ft = 556.11 ﬁ:
pile A top = 557.31 ft — (2)(1.2 ft) = 554.91ft
pile D top : 557.31 ft + 1.2 ft = 558.51 ft
pile E top = 557.31 ft + (2X12 ft) = 559.71 ft bent at A B C D E
station 64+50 554.91 556.11 557.31 558.51 559.71
65+00 555.35 556.55 557.75 558.95 560.15
65+50 555.91 557.11 558.31 559.51 560.71
66+00 556.6 557.8 559.0 560.2 561.4 Problem # 13 (a) The curve passes through point E at Sta 75+2£
and elevation 314 ft. sta (75+20) EVC sta (74+oo) G
elev 310 ft not to scale EG = elevationg — elevationv + Idgli
= 314ft w 310 ft; + K7520 ft — 7400 ft)(~—0.013)[
= 5.56 ft EF 2 eleVatidnE — elevationv  dgz
= 314 ft — 310 ft — [(7520 ~— 7400) (01318)] = 1.84 ft
The iength of curve is F9. +1 L0 = 2d HEP ‘EE _ 1 EF
{5.56 ft + 1
= (2)(120ft) ———————1'34 ft
5.56 ft _ 1
V 1.84 ft
= 890.13 ft 1 q Problem # 13 cont. (b) L0 = 1300 ft. LC
Stagvo : stay + 2— = (74+00) — LOO ft 2
= 67 +50 sta. LC elevationBvc = stay + .91 = 310 ft + (13020 ft) (0.013) = 318.45 ft __ 92 — .91 2 ,
y ._ ( 2L0 ) x + 91:1: + elevatlonng _ 1.8 — (—1.3)
_ (2)(13 sta)
= 0.119%2  1.3:: + 318.45 ft {£2 — 1.33: + 318.45 ft At point E, (up; = (75+20) — (67+50) = 7.70% w; = (o.1192)(7.7 sta)2 — (1.3)(7.7 sta) + 318.45 ft
= 315.51 ft The difference in elevation is 315.51 ft — 314fb = 1.51 ft Problem # 13 cont. (C) The grate inlet should be pierced at the low point
on the curve. y : 0.119232 — 1.311: + 318.45 ft; At the 10w point, dy/da: = 0. ﬂ = 0.2384x — 1.3 = 0
dz: x = 5.453 sta (543.3 ft)
Stagrate inlet 2 StaBVG + 3 = (67+50) “}‘ (5+45.3) = 72+95.3 sta elevationgrate inlet = (0.1192) (5.453 stay2
— (1.3)(5.453 $123) + 318.45 ft = 314.91 ft 7.1 Problem # 14 A —2.9% street grade and +21% street grade intersect at station 18+40 at
elevation 1140.40. it is necessary to pass an overhead structure at station 19+.10 with a 14.5 ft clearance requirement. The iowest steel girder support is at .‘an elevation of 1160.00. Compute the maximum length of parabotic curve  .jreqqired to meet the given conditions. (HMO) —(\e+4o)' = 70 0.“! 5+6! afﬁaw, E c “(90.00 445' = Meso' E73150. F = “40.40 + (o.1)(z.t%) = H4t.6’l
_‘j._,.: aorta = ii4o.4o +(o.1)(2‘I7é)= 1138.31 : dw.e aw . q =
.; .: I: Ia'w""cTEZJ_—’ . IF 2.4(90 ._.. 2(o.1)(I.4oI+‘i+I) ___ 3,5140 5h I. 0.40144"! ) or. 1:45.50—03831 __ (L 4
it4€.So—u4l.91 ' " 0‘ a 71% Problem # 15 L= 950' = 8.55% 3‘ (z>= 840.00 «aw—:3 —~ W? W]: +3‘ 840.00 + (4.2)( = (“4269 >63er 5o(l.6)+58.1s = 826, .oo‘ zCas)
W “m5+u8 0°54”) (6 0532 +5 46.05) + 818.15 = w Aaagtbriow " 45““ 1 . 850. 50 23 apt—r £953
26 Maw/m. 4% «LW cwv ~C4‘L .»_,, 2,, .
C (WC PW I: , 4ﬂu ‘ 10¢, ,450%' _—o
,u «19.,
k X. 45750 = 43% $§+ G, 7’4503g ’
4r2«5'o = 45053; ,_
4Qa§0 :— 43’23. __ AQJO  45—025 ; "4‘0" + A :ELL
7 2L1 A
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A 25/ AZ
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91/ + 91:9
A 2A' r
"Sam —\— \rYLg'YZ—
k . ’27
5:93: «\ + L1 3
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 Spring '11
 Moses

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