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Chapter 4 Problem Solutions

# Chapter 4 Problem Solutions - Determine which truck will...

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Unformatted text preview: Determine which truck will cause more Eavement damage. a1:=D.44 a2:= 0.2 a3z= [1.11 D1—3 D22=5 —S M: =10 M3.=1.CI SKI: + + SK= 3.4 Using linear interpolation for Truck A ungleleip := 0.2326 angle} Skip := 2.574 Totall SloFIE5:19.113,L := unglelEkip + ungleEE-kip Totall SlopE 8.59.1.le = 1.7966 Using linear interpolation for Truck B unglegktp := [1.01146 tandem=l3kip := 1.705 Totall ShpESALB := nugleShp + taudem-‘Bkip Tarall SlopESALB = 1.7196 Therefore. Truck A causes more damage. Problem 4.5 [Table 4.6] [given] (givem lE'II- 4.9] [Table 4.2, 4.3, and 4.4] Problem 4.11:1 Determine the reduction in pavement design life. SE:=3.E SD :=D.4U [given] PSI:=-—i..7 1&713:=1800363N T51 ;= :5 2R ;= _1_54j (for 95% reliability] [Table 4.5] C'BR:= 9 Current Design Life: 31351:: PSI — TSI JPSI = 2.2 [given] MR: 150001311 MR: 1.35): 10" Using Eq. 4}: 10110.1 ' leg: 7 _I 1031013}: zR-sa + 0.30:iog.:s:~:+ 1)) — 020+ i _ + 133109111111} — 00? [340+ ' _ 3.19 - 1:53 + 1} _ _ WIS = 313130.705): 1|C|I5 W13 N := — N = 12.333 before EDGE: before 18D013=134x 10* Tie" 13 Rafter := —2340 3'53 Maﬁa. = 9.452 iyears := Nbefcre — Maﬁa. Ajears = 1.336 Problem 4.13 Determine the prelmblility [reliability] that this pavement will last ED years belere reaching its terminal serviceability. 511:: 0.44 a} := [3.4-1 513:: 0.11 [Table 4.5] D1 :=i D3 := 6 D3:=1I:I [given] M]:=1.{> M3 := 1.0 [given] 315:: al-D1+ aE-DE-ME + aa-Di-Mi [Eli]. 4.9] 33': 5.94 N := II years Axle Leads [interpolating]: (Tables 4.2: 4.3. 4.4] nuglelﬂkip:=1.3332|30 tandem4'3'kjp := 1132-200 angle} Ekip := 2.164 SCI TeleleEqr := unglelﬂkip + tandemiﬂkip + ungleEEkip TeTAKleEqr = 913.13 W13 := ToTAxleEqr- 31-365 wig = 55.537765; 1.35 53 := 0.5 MR := 3000 2.1331 := 2.0 (given-J Using Eq. 4.? @1331 ‘ log: 1 _' tingle-'13]: zR-so + 9.3ejieg.:s:~:+ 1)) — n.2u+ ‘ _ + 2.32-lcg[MR] _ 3]}? 040+ - _ 5.19 - (53 + 1) _ _ ZR = 43.93 .773 R := 82.3 H. [interpolating from Table 4.5] Considering Ex. 4.5 determine which truck will cause more EﬂVEITIEllt damage? Axle load equivalency factors for truck A: 311.1g1elElripu:= Ill—Ii angle} Skip := 2.913 “J.”arﬁmlequrr__qL := unglelEkip + ungleEBkip TaTAxleEqrA = 3.09 13 kip ESAL Axle lead equivalency factors for truck B: ungleShp := DJIBE'. tandemﬂkip := “3.245 TaTAxleEqYB := ungleShp + tandemﬂkip TmAxleEqa-B = 5.217 18- kip ESAL Therefore. Tmck El causes more damage. Problem 4.21 Determine Estimated Daily Truck Trafﬁc. Problem 4.22 D := 11 N := :0 (given) ZR:=—1.282 s: := 000 5D :=0_35 Cd := 00 1:31:40 1:10 -= 1 EL "5 Ec:=4-1i]ﬁ 01331::1331— T31 k: 150 01:31:13 Using Eq. 4.13: ." . _" . , i '“E 30 _1 Sch.i_I}U"5—1.132,i 10501131: ZRsJ — "1.33110ch — 11:1 — 000— + {42"- 0.32-TSIJ-10‘ 1—1341 1.52410" - 215.631- D”"'5— —" 1+ — I, .025 {0+ 0"“; i E 'I 1: .' 1013:1102): 10' Axle lead equivalency factors from Tables 4.7", 4.8, and 4.9, for D = 11 313.1glell'JhLip:= 1.33 tends-11126111]: := 0.619 t1ip]e34l1:ip:= 0.593 TetAxleEq'.‘ := singlelﬂkip + taudemlﬁl-Cip + triple34lcip TetAxleEq'.‘ = 2.792 W 13 Daﬂ3rDeaignlaueT1'aﬁic := — 363 N-TetAxleEq‘r Daﬂ3.DegigﬂlaueTl-aﬁ3c = 31102 trucksiday {ier design lane] For 3 lanes and a conservative design, 1111;: 113a DailyDesignLaueTraﬂ'ic PDL TetalTlaﬂ'ic := TotalTlaﬂ'ic = r110 trucksidar {total for 3 lanes} ...
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Chapter 4 Problem Solutions - Determine which truck will...

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