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Homework #6 solution

# Homework #6 solution - Problem 3.2 Determine the...

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Unformatted text preview: Problem 3.2? Determine the sugereletratien reguired at the design sgeed. r'JrlseI cemgute the degree at currreI length at currreI and statiening at the PC annd PT. ‘u’ := 100 R := 1000 J. := 33 [given] stap|:=11251|3 r5 ;= 0.20 g:=32.2 Since the racetrack is single-lane. := R = 1000 Salve fcr required superelet‘aticn . 2 tut-1.46? . e+f3=;-}1—fS-e: rqu.3.34] e = [1.413 salve fer degree at cur-re 13mm D := —R D = 5.?3 degrees Ing. 3.35] use this and Equaticn 3.33 tc salve fer length cf cur-re rauuu : a: — L= _-e-_~. [Eq 3.39] r-D ran L ;= 103'; L: 523.5 ft calculate tangent length It's—1". tilt.“ I T := R-tan: 2 Edeg :: T = 251943 IIEI:|. 3.3I3] Shape I= StEPI — T Shape = StEPC = staFr-I- := stapc + L staFr-I- = 113755.55 staF.T = 112346365 Pmbmm331 Determine the design sgeed used. Since the ramp is a single 12-feet lane. center efreadwa‘y is center eftraveled path 1.:= 31] L := 528 11:13:: 13.4 [given] using L and ﬂ. sewe fer R : L-1E3EI L: —-R-_". R := R: 333.3 [EIZ|. 3.33] 131] :3. R; := R = 333.3 since Me and RV are knewn. we can use Equatien 3.43 te ﬁnd SSE] IR,- :'i I: Fit- - W133": SSE] := SID-deg laces H SSE] = 250.1 11 [El 343] frern Table 3.1. SSE] fer 35 mifh is 25th - cuwe is designed fer 35 mifh Alternative Selutien since Ms and Re are knewn. we can sewe Equatien 3.42 te ﬁnd SSD f fassse ‘9 MS: 1 — ces: F. -deg :: [EC]. 342] SSE] = 250.1 ft frem Table 3.1. SSE] fer 35 mifh is 251] ft - curve is designed fer 35 mifh Cernering Check 1'.i":=35-1.21|3|3? 1'.i"=51.3 fer 35 nw'h. f3 := [1.155 [Table 3.5] g := 32.2 2 ‘u’ e:=——fS e=E|.E|5 93s Se this cemhinatien efspeed. radius. and superelevatien is OK Problem 3.34 Determine the design sgeed used te design the curve. e := [1.116 {given} . . . 12 since the read Is feur-lane Wlil'l 12-11 lanes. 11.13:: 52 — 12 — E ME = 34 11 tr'_.,.r EU mi-“h Ru := 1341] [Table 3.5} see ;= are [Table 3-1} r” Eggssg TH 11:13:: 1 — ees: -deg :: ME = 30.134 11 (Ea. 3.42} i *1 Jr} this is less than the required distance. try again tr'_.,.r TE] mi-“h Ru := 2051] [Table 3.5} see ;= ran [Table 3-1} r” Eggssg TH 11:13:: 1 — ees: -deg :: ME = 32.4113 11 (Ea. 3.42} i *1 Jr} this is less than the required distance. try again tr'_.,.r EU mi-“h R133: 305” [Table 3.5} fan-see ‘1‘} 11:13:: 1 — ees_ _ -deg __ ME = 33.?65 11 (Ea. 3.42} i *1 Jr} this rounds in 34 11. therefore the design speed is 31] mifh Determine the statiening and elevatien ef the P‘U’Cs and P‘U'Ts. Problem 3'9 EU mifh design speed Kc : 151 [Table 3.2] Ks 3: 135 [Table 3.3] A ;= In — 2| A = 2 calculate lengths cf crest and sag cuwes L3 ;= KS-A L3 = 212 Calculate staticn and elevaticn cf PVT fcr crest cur-5e. staFm: = 1+51 A-LE 20E] elevpyﬂ-c := 1m] — eleupV-I-E = 95.98 ﬁ Parr: ;= n + 302 Parr: = 302 staWTc = s + [:2 Calculate staticn ancl elevaticn cf PVT and PVC fcr sag cuwe. staFmS = 41+E‘l elevFllul-I-S := 1m: — [DUE-4000] eleupV-I-E = 20 ft L L PUTS ;= 3': + 4mm] + 33 PUTS = 42s? staWTs = 42 + s? A-LS elevFllugcS := 20 + 2m] eleanulCS = 22.?2 ft F'ch ;= PWS — LS F'ch = 4015 mama = 4n + 15 ...
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