Pavement Analysis & Design 2nd Edition Solution Manual
April 2, 2018 | Author: Anonymous a19RxVjkk | Category: N/A
Short Description
Solutions Manual...
Description
SOLUTIONS MANUAL
PAVEMENT ANALYSIS AND DESIGN SECOND EDITION
YANG H. HUANG
SOLUTIONS MANUAL
PAVEMENT ANALYSIS AND DESIGN
YANG H. HUANG
Pearson Education, Inc. Upper Saddle River, New Jersey 07 458
Acquisitions Editor: Laura Fischer Supplement Editor: Andrea Messineo Executive Managing Editor: Vince O'Brien Managing Editor: David A. George Production Editor: Barbara A. Till Manufacturing Buyer: Ilene Kahn
•
© 2004 by Pearson Education, Inc .
.
Pearson Prentice Hall Pearson Education, Inc . Upper Saddle River, NJ 07458
All rights reserved. No part of this book may be reproduced in any form or by any means, without permission in writing from the publisher.
The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. Pearson Prentice Hall® is a trademark of Pearson Education, Inc. Printed in the United States of America
10 9 8 7 6 5 4 3 2 1
ISBN 0-13-184244-7 Pearson Education Ltd .. London Pearson Education Australia Pty. Ltd .. Sydney Pearson Education Singapore. Pte. Ltd. Pearson Education North Asia Ltd., Hong Kong Pearson Education Canada. Inc., Toronto Pearson Educaci6n de Mexico, S.A. de C.V. Pearson Education-Japan, Tokyo Pearson Education Malaysia. Pte. Ltd. Pearson Education, Inc., Upper Saddle Rivn; New Jersey
Contents 1
Introduction
1
2
Stresses and Strains in Flexible Pavements
2
3
KENLA YER Computer Program
11
4
Stresses and Deflections in Rigid Pavements
19
5
KENSLABS Computer Program
32
6
Traffic Loading and Volume
42
7
Material Characterization
49
8
Drainage Design
58
9
Pavement Performance
63
10
Reliability
70
11
Flexible Pavement Design
84
12
Rigid Pavement Design
91
13
Design of Overlays
102
lll
Chapter 1 Introduction 1-1.
The wheel follows
c o n f i g u r a t i o n , f tj r I
a
d u a 1 - t a n d e m a x 1e
1
i s
as
·1 ·~
I:. f
.l.~· t
-+
The 40 Kip load is applied over 8 tires. Thus, each tire bears 5000 Lbs (22.2 kN) load with 100 Psi (690 kPa) tire pressure. The contact area of each tire, Ac= Load of each 2 tire/tire pressure= 5000/100 = 50 in.2 (8.2 x 10 4 mm ). The dimension of the contact area is From Eq.1.7.
(Lecture Text) /so/0.5227 /Ac/0.5227 9.78 in. ( 24 8 mm).
L
width The most rectangle and figure : .... , -
=
0.6 L
=
5 .. 87
in (149 mm.)
./
I
realistic contact area consisting a two ~emi-circles as s h o wn .; 'i 11 f o 1 l o w i n g
~9.Jl1it.---+
10· ... ~ i
0
.
.~
......
.r:-f\l \J__U 1-2.
Freezing Index
Each tire imprint,if considered as a rectangular area,should have
:nle~g~~d~~ ~f 8~~~ ~'. ~; ~:~~ i~: ~ J/6 ,·,,,
= (32
- 24) x 30 + (32 + 3) x 31 + (32 - 14) x 31 + (32 - 16) x 28 + (32 - 22) x 31 + (32 -25) x. 30 = 1§2..L degree days.
Yes, this value is likely to be different because the last few da.ys in
October and tge first few days in May may have mean dl:J.ilY tempera.. tures lower than 32 F, so the degree days for these two months , may not be zero. ~
,._
Chapter 2 Stresses and Strains in Flexible Pavements C:Z-1. .fo l...·fion :
~ ::. I J Fie•... fi/. ( G; 4
%" ~
2.
2 · 2. ..
) I DO/.: ~
-=
~
I
s
o.1g· 1;
fiJ. x 1-l·
I o.
6
5". .J-4
6
~'
J.J~,,1.. ~ .. .1ir41'·
:
/'a .. ,(·,.,., •
S'O
0.61.r 1'2. 6 4f6
(¥).
6]J..-
12·
-
I
.2
2.
6
8 z.
2.
o.31
.I
. : . . - - - - ... / . .? 'i jO ooo
:
_,
2-b
,4c~
~
.--,--,---;-' y
~' ..,.4ooooo ps;
lb
IQt>.PS• '
{
l~,
:
~= ":'
1S· 7-s,- _,",, ·--/.-
- - - - - - - - · - - - - - _.Q ...
0
q'J,5 wer
NBoND
400000 8000 10000 (11) E 1 (13) LOAD 4.5 75 (14) CR CP 1 (19) NPT 0 13.5 0 6.75 (20) XW YW XPT 1 15 (25) NOLAY ITENOL 2 0 (26) LAYNO NCLAY 6 (27) ZCNOL 0 0 0 0 0.01 (28) RCNOL XCNOL YCNOL SLD DELNOL 0. 5 (29) RELAX 145 135 0 (30) GAM 0.5 0.6 (31) K2 KO 8000 8000 (33) PHI Kl
1 (1) NPROB PROBLEM 3.9 3 0 1 1 (3) MATL NDAMA NPY NLG 0.001 (4) DEL 2 1 80 1 0 (5) NL NZ ICL NSTD NUNIT 8 (6) TH (7) PR 0.5 0.5 0 1
I 1-.
3 /. I J 3 ' -3.a83 p.>
;,,, -'
O. o 2a~6 ,':.,, J
o. 0 208 7 ;11, _,
~. o Za "JZ /"1,
CJ> J '3.51
0
.J
0
(39) TYME 4.57E-06
4.381:-06
0.0000046
(42)
~
1 (1) NPROB PROBLEM 3.10 3 1 1 1 (3) MATL NDAMA NPY NLG 0.001 (4) DEL 2 0 80 9 0 (5) NL NZ ICL NSTD NUNIT 8 (6) TH 0.5 0.5 (7) PR 1
(9) NBOND
10000 (11) E (13) LOAD 6 75 (14) CR CP 1 (16) NR 0 (17) RC 0.1 (36) DUR 1 1 (37) NVL LNV 11 (38) NTYME 0.001 0.003 0.01 0
qr'JJ
0
0.03
0.1
2 I. o '7 :Je'1r->
wer:
0.3
1
3
10
30
100
(39) TYME
0.113 70 (41) BETA TEMPREF 2.04E-06 2.12E-06 2.37E-06 2.92E-06 3.81E-06 4.19E-06 4.49E-06 4.57E-06 0.0000046 0.0000046 (42) CREEP 70 (44) TEMP 1 1 (46) NLBT NLTC 1 (47) LNBT 2 (48) LNTC 100000 (49) TNLR 0.0796 3.291 0.854 (50) FTl FT2 FT3 1.365E-09 4.477 (51) FT4 FT5
be 2
(1) NPROB N?J,tjHA ..ShD'-J}.::l 1 PRO!!IJiM 3. l~ 1 ~ (3) MATL NDAMA NPY NLG 0.001 (4) DEL 3 0 80 9 0 (5) NL NZ ICL NSTD NUNIT 6 8 (6) TH 0.4 0.35 0.45 (7) PR 1
.fo
9-e-I:..
I'~
4.38E-06
-fa J)ow1~
4w-c 7° ~x.. --:4-'7.s,-/-e S-l->"qi.., e:>, o ~3..>J ~?4 ~Jrti qtJff: ro r:/~:ftr4/!r.>~4) $"t->q;,, = q e;-08801D4HJ4je '74-1-/-o
=
(9) NBOND
(11) E 740000 23000 11000 2 (13) LOAD 4.52 70 (14) CR CP 3 (19) NPT 48 13.5 0 0 0 3.375 0 6.75 (20) XW YW XPT 1 1 ( 46) NLBT NLTC 1 (47) LNBT 3 (48) LNTC 200000 (49) TNLR 0.0796 3.291 0.854 (50) FTl FT2 FT3 1.365E-09 4.477 (51) FT4 FT5
(1) NPROB 3 Problem 3-12 (a) 2 1 1 1 ( 3) MATL ND.AMA NPY NLG 0.001 (4) DEL 7 2 80 9 0 (5) NL NZ ICL NSTD NUNIT (6) TH 2 2 2 2 2 2 0.35 0.35 0.35 0.35 0.35 0.35 1
~
0.45
(7) PR
(9) NBOND
120000 100000 80000 68000 40000 20000 5000 0 (13) LOAD 6 100 (14) CR CP 1 (16) NR 0 (17) RC 7 15 (25) NOLAY ITENOL 1 0 2 0 3 0 4 0 5 0 6 0 7 1 1 3 5 7 9 11 13 (27) ZCNOL 0 0 0 0 0.01 (28) RCNOL XCNOL YCNOL SLD DELNOL 0.5 (29) RELAX 135 135 135 135 135 135 115 (30) GAM 0.5 0.6 (31) K2 KO 0.5 0.6 (31) K2 KO 0.5 0.6 (31) K2 KO 0.5 0.6 (31) K2 KO 0.5 0.6 (31) K2 KO 0.5 0.6 (31) K2 KO 6.2 1110 178 0.8 (31) K2 K3 K4 KO 0 10000 (33) PHI Kl 0 10000 (33) PHI Kl 0 10000 (33) PHI Kl 0 10000 (33) PHI Kl 0 10000 (33) PHI Kl
(11) E
(26) LAYNO NCLAY
0 10000 (33) PHI K1 1827 7682 3020 (33) EMIN EMAX K1 0 1 ( 46) NLBT NLTC 7 (48) LNTC 1000 (49) TNLR 1.365E-09 4.477 (51) FT4 FT5 Problem 3-12 (b) 2
2
1
1
(3) MATL NDAMA NPY NLG
0. 001 (4) DEL 2 2 80 9 0 (5) NL NZ ICL NSTD NUNIT 12 (6) TH (7) PR 0.35 0. 45 1
(9) NBOND
10000 3020 (11) E 0 (13) LOAD 6 100 (14) CR CP 1 (16) NR 0 (17) RC 2 15 (25) NOLAY ITENOL 1 0 2 1 (26) LAYNO NCLAY 4 13 (27) ZCNOL 0 0 0 0 0.01 (28) RCNOL XCNOL YCNOL SLD DELNOL 0.5 (29) RELAX 135 115 (30) GAM 0.5 0.6 (31) K2 KO 6.2 1110 178 0.8 (31) K2 K3 K4 KO 10000 10000 (33) PHI K1 1827 7682 3020 (33) EMIN EMAX K1 0 1 (46) NI.BT NLTC 2 (48) LHTC 1000 (49) TNLR 1.365E-09 4.477 (51) FT4 FTS Prob1em 3-12 (c) 2
2
1
0. 001 2
2
12 0.35 1
1
(3) MATL NDAMA NPY NLG
(4) DEL 80 9 0 (6)
(5) NL NZ ICL NSTD NUNIT
TH
0.45
(7)
PR
(9) NBOND
10000 3020 (11) E 0 (13) LOAD 6 100 (14) CR CP 1 (16) NR 0
(17) RC
2
15
1
0
(25) NOLAY ITENOL 2 1 (26) LAYNO NCLAY 6 13 (27) ZCNOL 0 0 0 0 0.01 (28) RCNOL XCNOL YCNOL SLD DELNOL 0.5 (29) RELAX 135 115 (30) GAM 0.5 0.6 (31) K2 KO 6.2 1110 178 0.8 (31) K2 K3 K4 KO 40 10000 (33) PHI K1 1827 7682 3020 (33) EMIN EMAX K1 0 1 (46) NLBT NLTC 2 (48) LNTC 1000 (49) TNLR 1.365E-09 4.477 (51) FT4 FTS
2
(1) NPROB 3-13 (a) 3 1 1 1 (3) MATL ND.AMA NPY NLG 0.001 (4) DEL 2 2 80 9 0 (5) NL NZ ICL NSTD NUNIT 10 (6) TH 0.5 0.5 (7) PR PROBLEM
1
(9)
0
0
NBOND
(11) E (13) LOAD 100 (14) CR CP
0
10 1 0
(16) NR (17) RC (36) DUR
0.1
1 2 (37) NVL LNV 7 (38) NTYME 0.01 0.03 0.1 0.3 1 0 0 ( 41) BETA TEMPREF 1.02E-06 1.06E-06 1.21E-06 2
0
0
( 41)
3
10
1.59E-06
(39) TYME 2.68E-06
0.000105 0.000289 0.000732 0.00125 0 0 (44) TEMP 1 1 (46) NI.BT NLTC 1 (47) LNBT 2 (48) LNTC 60000 (49) TNLR 0.0796 3.291 0.854 (50) FTl FT2 FT3 1.365E-09 4.477 (51) FT4 FT5 PROBLEM 3-13 (b) 3 1 1 1 (3) MATL ND.AMA NPY NLG 0.001 (4) DEL 2 2 80 9 0 (5) NL NZ ICL NSTD NUNIT 10 (6) TH 0.5 0.5 (7) PR 1 0 0
4.84E-06
9.27E-06
(42) CREEP
BETA TEMPREF
0.00195
0.00359
0.00732
(42) CREEP
//. 6 z
(9) 0
NBOND (11) E (13) LOAD
10 1
100 (14) CR CP (16) NR 0 (17) RC 0.2 (36) DUR 2 1 2 (37) NVL LNV 7 (38) NTYME 0.01 0.03 0.1 0.3 0
0
1.02E-06 0
0
1
3
10
(39) TYME
( 41) BETA TEMPREF
1.06E-06
1.21E-06
1.59E-06
2.68E-06
4.84E-06
9.27E-06
(42) CREEP
( 41) BETA TEMPREF
0.000105 0.000289 0.000732 0.00125 0 0 (44) TEMP 1 0 ( 46) NI.BT NLTC 2 (47) LNBT 60000 (49) TNLR 0.0796 3.291 0.854 (50) FTl FT2 FT3
0.00195
0.00359
0.00732
(42) CREEP
1
(1) NPROB
PROBLEM 3-14 4 1 2 2 (3) MATL ND.AMA NPY NLG 0. 001 (4) DEL 6 0 80 9 0 (5) NL NZ ICL NSTD NUNIT 6 2 2 2 2 (6) TH 0.4 0.35 0.35 0.35 0.35 0.45 (7) PR 1
(9) NBOND
0 0
7500 7500 7500 7500 3020 (11) E 10000 10000 10000 10000 12340 (11) E 0 (13) LOAD 5.35 100 (14) CR CP 1 (16) NR 0 (17) RC 1 (13) LOAD 4.52 70 (14) CR CP 3 (19) NPT 0 13.5 0 0 0 3.375 0 6.75 (20) XW YW XPT 5 15 (25) NOLAY ITENOL 0 3 0 4 0 5 0 6 1 2 (26) LAYNO NCLAY 7 9 11 13 15 (27) ZCNOL 0 0 6.75 0 0.01 (28) RCNOL XCNOL YCNOL SLD DELNOL 0.5 0.5 (29) RELAX 145 135 135 135 135 125 (30) GAM 0.5 0.6 (31) K2 KO 0.5 0.6 (31) K2 KO 0.5 0.6 (31) K2 KO 0.5 0.6 (31) K2 KO 6.2 1110 178 0.8 (31) K2 K3 K4 KO 0 7500 (33) PHI K1 0 7500 (33) PHI K1 0 7500 (33) PHI K1 0 7500 (33) PHI K1 1827 7682 3020 (33) EMIN EMAX K1 0 10000 (33) PHI K1 0 10000 (33) PHI K1 0 10000 (33) PHI K1 0 10000 (33) PHI K1 7605 17002 12340 (33) EMIN EMAX K1 0.1 (36) DUR 1 1 (37) NVL LNV 11 (38) NT!ME 0.001 0.003 0.01 0.03 0.1 0.3 1 3 10 30 100 (39) TYME 0 .113 70 (41) BETA TEMPREF 3.7E-07 5.2E-07 8.6E-07 1.45E-06 0.0000025 0.000004 0.0000062 0.0000086 0.000012 0.000016 0.000019 (42) CREEP 60 (44) TEMP 80 (44) TEMP 1 1 ( 46) NLBT NLTC 1 (47) LNBT Qn.jwer: 3 (48) LNTC 25. 15 9125 18250 (49) TNLR 9125 18250 (49) TNLR 0.0796 3.291 0.854 (50) FT1 FT2 FT3 1.365E-09 4.477 (51) FT4 FT5
jeqrJ
Chapter 4 Stresses and Deflections in Rigid Pavements
+-1 ca.)
..,.,
~:.
6'! I I
I
tI
C'
t::k,: - ~ 0 f
- --------~A
G' I
+ I
8''
,-k_,,S)pci
6:: (
,!\)':: 0 . '4r
10
A'
\.() I
~~
4XUIG
I
_/
>f
• 6 ·; I of
0.- \, -: 15 X LO
II
?f·µ,·r
6 o't .At
N. E, . O ~b l z. 00 0 l
V'Vt.1tt.i.7
2. o,S'"
/~ IJt11tJi~r ~
IV= cSo
zs : :
t::S.f
~ ..J.1..Ye ~u;re_
.>, 27 q»:t.s.
2. OS°
/ Iz ~ 'O == .ss.
b)e;t:.K.J
3 S8
2. c;5 t.in~· t~
-fo,PJ ~c;
s, "l.7 : ~ 2
I
a..t
:t?
M::: /oox(~.1)z.>1Z9o::. ~9-.>-Z- /'v;-1o/f";.,_ /~_...t!:)OO
0 --,,.-17
0
/o ckkr"'~ ?'-he
1;, ~
l'l?t>Mtte'7/._
)d"'f,·Jve;J;">".J q17~~.f10':) I ~ ~;J~,,-lq.J.~"J
0
~:f -?-~ -/J~f
~ ?'.,&.:,~
IHllft
7'o:
q_s
.S/;c.v.J"> /";;; +-~ k-Fi ~"1uY-c_ Nok -1-~ 7-J.;e rr9">t -1wo ..;,~~$ .S-1,,-~+t:/k.J 6e-lw~'? µ5,·7,-~ ~ ~..J,~ 6/of!.ks arA.. ~Y-e v-e~y hH)-e -eff~-l °"> ~6e &lock' ~,,t, Al= 2So /1-:::. loax (.S°8.1fJt. ZS"o = 8/f',?~ 1~.-1~;,,
=-
~~l
4.15. 5.53
8.~o
i/ 1ocro/ 13.~
o.3L,o.4L
---r-
II'\
L
;iJ
12. ( I -
ce "c....rc tr
t'l"f
l_·-_E_h_~__
4
o.t, L
t"4~10'
V~) K
r0 :
psi
0,15
) 4_,,10". ~~
8.ZS? (15) LS XLl XL2 YLl YL2 QQ
1 (1) NPROB Problem 5 .10 1 (3) NFOUND NDAMA NPY NLG (4) NSLAB NJOINT 9 0 1 0 0 (5) NX NY JONOl JON02 JON03 JON04 9 1 0 0 0 (5) NX NY JONOl JON02 JON03 JON04 46 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 2 2 (6) NIAYER NNCK NOTCON NGAP NPRINT INPUT NBOND NTEMP NWT NCYCLE NATl NAT2 NSX NSY MDPO NUNIT UL TC CL 0 150 0 500 0 0.000005 0.001 1 (7) TEMP GAMA(l) GAMA(2) PMR(l) PMR.(2) CT DEL FMAX 17.61 0 17.61 0 (8) F1(1) F1(2) F2(1) F2(2) 0 30 60 90 108 120 0 12 24 42 66 78 90 117 144 (9) x• s and then Y's 0 12 30 60 90 120 0 12 24 42 66 78 90 117 144 (9) x• s and then Y's 10 0.15 4000000 (10) T PR YM 2 (12) NUDL 0 (13) NCNF 0 0 (14) NNMX NNMY 1 108 120 0 12 83.333 ( 15) LS XLl XL2 YLl YL2 QQ 1 108 120 72 84 83.333 (15) LS XL1 XL2 YL1 YL2 QQ 1 (22) FSAF 0 300 (24) NAS SUBMOD 29000000 0.3 (34) YMSB PRSB 0 0 1500000 1 12 0.25 0 0 ( 35) SPCONl SPCON2 SCKV BD BS WJ GDC NNAJ 0
0
2 6 6 1
1
1
1 (1) NPROB Problem 5.11
0 0 1 1 (3) NFOUND ND.AMA NPY NLG 2 1 ( 4) NSLAB NJOINT 6 14 0 1 0 0 (5) NX NY JONOl JON02 JON03 JON04 6 14 1 0 0 0 (5) NX NY JONOl JON02 JON03 JON04 1 46 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 2 2 (6) NIAYER NNCK NOTCON NGAP NPRINT INPUT NBOND NTEMP NWT NCYCLE NATl NAT2 NSX NSY MDPO NUNIT UL TC CL 0 150 0 500 0 0.000005 0.001 1 (7) TEMP GAMA(l) GAMA(2) PMR.(1) PMR(2) CT DEL FMAX 17.61 0 17.61 0 (8) Fl(l) F1(2) F2(1) F2(2) 0 30 60 90 108 120 0 6 18 30 42 54 66 78 90 102 114 126 138 144 (9) X's and then Y's 66 78 90 0 12 30 60 90 120 0 6 18 30 42 54 102 114 \ 126 138 144 (9) x• s and then Y's QIJ.)WeY: 4):. N~ 72- tfo~2. '7 f'.P-' 10 0.15 4000000 (10) T PR YM - !Jo. 8 z.b P~ 2 (12) NUDL 0 (13) NCNF ?I e>. 0 2 7 ;>]. 0 0 (14) NNMX NNMY 1 108 120 0 12 83.333 (15) LS XLl XL2 YLl YL2 QQ
78
z
J
(15) LS XLl XL2 YL1 YL2 QO 1 108 120 72 84 83.333 1 (22) FSAF 0 300 (24) HAS SUBMOD 29000000 0.3 (34) YMSB PRSB 0 14 (35) SPCONl SPCON2 SCKV BD BS WJ GDC NNAJ 0 0 1500000 1 12 0.25 1 1 1 1 0 (36) BARNO 0 1 1 1 1 1 1 1 1
1
(1) NPROB
Problem 5.12 0
0
2 6 6 1
1
1
1
(3) NFOUND NDAMA NPY NLG
( 4) NSLAB NJOINT 9 0 1 0 0 (5) NX NY JONOl JON02 JON03 JON04 9 1 0 0 0 ( 5) NX NY JONOl JON02 JON03 JON04 46 0 0 0 0 0 0 0 l 0 0 0 0 0 0 1 2 2 (6) NLAYER NNCK NOTCON NGAP NPRINT INPUT NBOND NTEMP NWT NCYCLE NATl NAT2 NSX NSY MDPO NUNIT UL TC CL 0 150 0 500 0 0.000005 0.001 1 (7) TEMP GAMA(l) GAMA(2) PMR(1) PMR(2) CT DEL FMAX 17.61 0 17.61 0 (8) F1(1) F1(2) F2(1) F2(2) 117 144 (9) X's and then Y's 0 30 60 90 108 120 0 12 24 42 66 78 90 144 (9) X's and then Y's 0 12 30 60 90 120 0 12 24 42 66 78 90 117 10 0.15 4000000 (10) T PR YM \ 2 (12) NUDL Z NCNF Q,l'J.jy..Jer: at:, N~cle -ZZ 2 PS-
1
Bf;
(1) NPROB
Problem 5.15 0
0
1
1
( 3) NFOUND NDAMA NPY NLG
2 1 (4) NSLAB NJOINT 6 6 0 1 0 0 (5) NX NY JONOl JON02 JON03 JON04 6 6 l 0 0 0 (5) NX NY JONOl JON02 JON03 JON04 1 34 0 0 1 0 0 0 0 1 0 0 12 0 0 0 2 2 1 ( 6) NLAYER NNCK NOTCON NGAP NPRINT IN.PUT NBOND NTEMP NWT NCYCLE NATl NAT2 NSX NSY MDPO NUNIT UL TC CL 0 150 0 500 0 0.000005 0.001 l. (7) TEMP GAMA(1) GAMA(2) PMR(1) PMR(2) CT DEL FMAX 17.61 0 17.61 0 (8) Fl(l) F1(2) F2(1) F2(2) 0 30 60 90 108 120 0 15 30 42 54 72 (9) X's and then Y's 0 12 30 60 90 120 0 l.5 30 42 54 72 (9) X's and then Y's 10 O.l.5 4000000 (l.0) T PR YM 0 (12) NUDL l. (l.3) NCNF ps~ 0 0 (l.4) NNMX NNMY 34 12000 (16) NN FF 34 (18) NP 1 7 13 l.9 25 31 37 43 49 55 61 (19) NODSX 67 1 (22) FSAF 300 (24) NAS SUBMOD 0 29000000 0.3 (34) YMSB PRSB 0 0 1500000 l. 12 0.25 0.01 0 (35) SPCON1 SPCON2 SCKV BD BS WJ GDC NNAJ
-/655. 9
l. (1) NPROB Problem 5.16 0 0 1 l. ( 3) NFOUND NDAMA NPY NLG l. 0 (4) NSLAB NJOINT 9 6 0 0 0 0 (5) NX NY JONOl. JON02 JON03 JON04 l. 22 0 0 0 0 0 0 0 l. 0 0 9 6 0 0 l. 2 2 (6) NLAn:R NNCK NOTCON NGAP NPRINT INPUT NBOND NTEMP NWT NCYCLE NATl. NAT2 NSX NSY MDPO NUNIT UL TC CL 0 l.50 0 500 0 0.000005 0.001 l. (7) TEMP GAMA(1) GAMA(2) PMR(1) PMR(2) CT DEL FMAX 17.61 0 l.7.61 0 (8) Fl(l) F1(2) F2(l.) F2(2) 0 14 28 41 54 72 96 l.32 168 0 10 20 29.5 38.75 48 (9) X's and then Y's 8 0.15 4000000 (l.0) T PR YM 1 (l.2) NUDL tJ~e 0 (l.3) NCNF 0 0 (14) NNMX NNMY 1 37.5 44.5 23.5 35.5 150 (15) LS XLl XL2 YLl YL2 QQ l. 7 l.3 19 25 31 37 43 49 (19) NODSX l. 2 3 4 5 6 (20) NODSY l. (22) FSAF 0 150 (24) NAS SUBMOD
cit
22
1 (1) NPROB Probl.em 5.17 1 0 1 1 ( 3) NFOUND NDAMA NPY NLG 2
l
(4j NSLAB NJOINT
6 14 0 1 0 0 (5) NX NY JONOl JON02 JON03 JON04 6 14 1 0 0 0 (5) NX NY JONOl JON02 JON03 JON04 1 46 0 0 42 0 0 0 0 1 0 0 0 0 0 0 1 2 2 (6) NI.AYER NNCK NOTCON NGAP NPRINT INPUT NBOND NTEMP NWT NCYCLE NATl NAT2 NSX NSY MDPO NUNIT UL TC CL 0 150 0 500 0 0.000005 0.001 1 (7) TEMP GAMA(l) GAMA(2) PMR(l) PMR.(2) CT DEL FMAX 17.61 0 17.61 0 (8) Fl(l) F1(2) F2(1) F2(2) 54 66 114 0 30 60 90 108 120 0 6 18 30 42 78 90 102 126 138 144 (9) X's and then Y's 0 12 30 60 90 120 0 6 18 30 42 54 66 78 90 102 114 126 138 144 (9) X's and then Y's A,,,? weY Ala:/e -308 10 0.15 4000000 (10) T PR YM 2 (12) NUDL -13S:z.69 ~
:
=
7 ,t
;,,
1J
c!1 =
/.< /,,
...z.q
E
,' .,..
0·-360
/I I· ...-
o. 366
Ib
o.h3
84.5 6/4
p.6;24
.. -
(. (717~
18
/. S-1
;(,,()
A
1.~
47. 2
I· 52a
..2,/.4
,z.23 2
9
-3.l{,O
r;.
I
4. ~s
cl ..;,
...
.J4
? .03 4.07
12.
.( 'g
S.:3~
~.q
30
6'.97
/1 2
7, 7
//. !J
7 (} ·3
p.121
~
~b
18
o,eg
j.;( ?
N/11_ .R6
E:AlF'
q()
0.18'7
12
A
-=-t-t
.3
>"· 8S5 /?
q,q8 3
t).
IZ
1)
EAL
~
L
t' ~I
t=i n· I
~
-a6 s.,. 20 "' .z,q9
.,_ ID
6
/
I>
- (Ca. /. .J6.2-
CJ']<
~)" ~ :=.
I
2; I')
7)( 36S;i(j-
I g .> ~o-6 /
Chapter 7 Material Characterization
c.,ip.. ,
i)Z11,,1roJ11 /1''11.11'
7
/#A•./l.O.t•J
J;t.· ( 1""1)
1 /10 )
(IO "'1)
8
{ /'.II)
S". 8
10• .Jy>1-
/. ./
.-UJ.
Jo 60
'J.r
~
II·
z.
6
6. {)
JO
'l rl...e
)',..j,.n/ a"J.J '71
,6'.t,f·u{
y
;,A
2. ' (
A..t JI ',F =.>o \ _ /A.,a ~ 1 :...,"T .
ju-w .:
z.t- - .YJ I -;-
=
--·------
,LL
2-o - rur,
13..J. =
l-
-..)A(;-
:
"·'!!.)
~. l y
.
x / ('#.11})
:I- IA
/ . (J L
% t1a..J,.,-1 t
~
:
/t'P-
';-JI..
= . lvv - JY, J- 1-17 / : ;Ir X
/.Zo .
~,/-
p .-
_!1#7 .
~u1'/ l~•,q_
t
./~//. /lod.
~ /...!}, = ~ ~l" I~'~--~~;:,~-~- . -.~--·. .. ~- ~ -~ .- -/J, :
/IJ. dr--~7/~~~-t--~~~l~~_J
/
1000
10
2?u.gr,041
J.
~ = 1- .r ~ I- _tJ. 4/Y// :: Cl: .s-..rl'.3_ /t ~ - f..S • . /;Liz. )( ~. 'fl/1 ,.,.,,0-.;-.: . o. J (' 6 L .J. 4"f>l. ll 10- ..r · .
~r------r---+V·~--.---~~~_.:___,..,,
2..o
,f
/.k~
'" 0
7-10.
No. OF
PERMANEXT STRAIN
REPETITION
(10~-5)
1 10
2.:1n
5.813 12.773 13.93 19.715 36.231 73.506
100 200
1000 10000 100000
PROBLEM 7-10 .79.4J282
I
50.11872
_J
39.81071
I
....,_I
.......
JL622.77
·r
I 1
25.11886
....
1
_, •
19.95262
.J
z
15.8489J
-1
!()
(
0
< ~
f-
12.58925
zrz
10
(/)
I
I
63.:95 73
I
\
'
! \
!
1
~
I
w
-l I ...., 6.309573 I 5.011672 i 7.943282
-1:
~
er
w
a.
i
3.9Ei J7i
1
3.162277
j
2.5111'385
/
-4
1.995262 I •P
C
-
• 28, fO..,
S =
! I
\
r - -1 e.6.e
1~0-..i---,-:-----.-----r------,,-----_J ,. '1
r
1
, ODOO n
100.0
0. 2 ~ ")
NIJMl3ER OF LOAD k£PDI \10h
1 ... e.bz cl.-= 1-S
1 E
s
=
I - o. 2c:l s 2 ·b '2 ~
---4. /'2
o. -z95
~
--
/
o. 7 o 5
=
i=:'
I -
o. 187
-
Chapter 8 Drainage Design
--
Q:: \J ~ v::.~~
,i;;.J
8. 1..
.:x. ~ J.S:_19._,~
lo ...
~.
.3
8
~
D~-:..
I/ 'l 0....,..
"J.,{'-lc.-... - /OOm..,.__
/ Io. 001 :. 5-4 ~~ O.i
Io. 0?1~
: 10
D"'° ~I D10 ~ ~ 'Z.~ D.C)/O.~
~
'!:>
D~ ~ ?=. o . o7A ./l'IJW'\ D.
z:z.'6 c o. 07.A
~B (~) ~~ (~)
~~ o. co1 I o.01tc:, =
o.~"'
~-·
{-)~~ 0.001
,
~
fJIJI
J
Io. ooozz.d:>
,,,,~.;c ~ ~ -~ . o. 0~7~ I 0. OOlic?
Nit:~ J 4..>.S''?J-
7/~Ye /q~ 4-
3 ~ .f"c/;,,,;fc
a..
Fo.,- eirw- w}-~
::.
:ft~
I< ( H-H•)
t?..5 )< (zj-15") - - - - -- - ::.
'},.,
=Z 3
0 ,"S-~
L.,i.. =- 3. $' ( zj-IS)::::. 3o. 4' f-c
i. zo,,,
':. z. ~'-
8..s
=
/~IJY
o,6
5
frl"'?
o,
o 2.. Inn?.> _.J;.H-J
/c:f:ly.
-r 120x10)/1z :::::- I Lf-8 ?>f F-ro1t1 ~~ Y·t.5, q__ =
{),.,
e.; Jr;;; > ::. ~. o ~7 :f't- ~cfqyI ;-c- ~ ~ /'J .
;
,
c4
Z. 9
F~ E'~ g. .?~
z.
4:t.y ==
Sf. = 18 7;: .:
~h>""I h ~ ?J', I 8>
t1.,.." = -
-
/, .
'£
f'J~
L
Kl'i
o. o/8Z
447
a. a 7 -4r ><
c::r.o~
S/1 z.
I, ..S-
o '>"
/,.> ")/.
-
-
= /.of;
, ..
"
'-
0.25 ')(..
~(
o. ~~ ~>".
c IB)
.Ji:.) /2-
Chapter 9 Pavement Performance 9-1. Given PSI= Ao+ A1 log(RI) and Table P9.1 Derive equations and find Ao and A1 E =:L(PSR -PSl) 2 = :L[PSR -Ao-A1 log(Rl)] 2
-c3E =-2 :L[PSR -Ao- A1 log(RI)] = 0 8A 0
PSR = A 0 + A 1log(RI) .................................................... (1)
-BE =-2 :L{[PSR -Ao-A1 log(RI)] log(RI)} = 0 8A 1
I(PSR) log( RI) = Ao I log( RI) + A1 :L[log(Rl)]2 .... :. ............... (2)
Eqs. 1 and 2 can be used to solve Ao and A1
PSR =(1 + 2 + 3 + 4+ 5)/5 =2. 5 log(RI) = (log 800 + log 300 + log 200 + log 150 + log 80)/5 = (2.903 + 2.477 + 2.301 + 2.176 + 1.903)/5 = 11. 76/5 = 2.352 From Eq. 1 2.5 =Ao+ 2.352 A1 .................................. (3) From Eq. 2 1 x 2.903 + 2 x 2.477 + 2.5 x 2.301 + 3 x 2.176 + 4 x1.903 =11.76 Ao+ [(2.903) 2 + (2.477) 2 + (2.301 )2 + (2.176) 2 + (1.903) 2 ] A1 27,75 = 11.76 Ao+ 28.214 A1 ..................................... (4) Solve Eqs. 3 and 4 27.75 - 2.5 x 11.76 = (28.214 - 2.352 x 11.76) A 1 -1.65 = 0.554 A1
or A1 = -2.98
Ao= 2.5 - 2.352 x (-2.98) = 9.5 PSI= 9.5 - 2.98 log(RI)
9-2. Given Table P9.2 Develop equation for PSI Section log (1+ SV) No. Ri
-2
RD R2
.Jc+P
PSR
R1·R1
R2- R1
D1-D1
PSR- PSR
D1
1 2 3 4 5
0.580 0.833 1.076 1.250 1. 756
0.0036 0.0100 0.0121 0.0256 0 .0361
0 1 3.606 4.796 5.568
4.3 3.8 3.2 2.4 1.1
Average
1.099
0.0175
2.994
2.96
-0.519 -0.0139 -2.994 -0.266 -0.0075 -1.994 -0.023 -0.0054 0.612 0.151 0.0081 1.802 0.657 0.0186 2.574
1.34 0.84 0.24 -0.56 -1.86
From Eq. 9.9 2.96 = AQ + 1.099 A1 + 0.0175 A2+ 2.994 81 ................ (1) From Eq. 9.11 2 2 2 [(-0.519) + (-0.266) + (-0.023) + (0.151) 2 + (0.657) 2 ] A 1 + [(-0.519) (-0.0139) +(-0.266)(-0.0075) + (-0.023)(-0.0054) + (0.151 )(0.0081) + (0.657)(0.0186)] A2 + [(-0.519)(-2.994) +(-0.266)(-1.994) + (-0.023)(0.612) + (0.151)(1.802) + (0.657)(2.574)]81
=(-0.519)(1.34) +(-0.266)(0.84) +
(-0.023)(0.24) + (0.151)(-0.56) + (0.657)(-1.86) or 0. 7951 A1 + 0.0228 A2+ 4.033 8 1
=-2.231
.................. (2)
From Eq. 9.12 2 0.0228 A1 + [(-0.0139) + (-0.0075) 2 + (-0.0054)2 + (0.0081 )2 + 2 (0.0186) ] A2 + [(-0.0139)(-2.994) + (-0.0075)(-1.994) + (-0.0054)(0.612) + (0.0081)(1.802) + (0.0186)(2.547)] 81 = (-0.0139)(1.34) + (-.0075)(0.84) + (-0.0054)(0.24) + (0.0081)(-0.56) + (0.0186)(-1.86) or 0.0228 A1 + 0.00069 A2 + 0.1152 8 1= -0.0654 ............... (3) From Eq. 9.13
4.033 A1 + 0.1152 A2 + [(-2.994) 2 + (-1.994) 2 + (0.612)2 + (1.802) 2 + (2.574) 2 ] 81 = [(-2.994)(1.34) + (-1.994)(0.84) + (0.612)(0.24) + (1.802) (-0.56) + (2.574)(-1.86)]
or
4.033A1+0.1152A2+23.18781=-11.337 ............... (3) From Eqs. 2, 3, and 4
From Eq. 1 Ao= 2.96-1.099 (-1.7) + 0.0175 (-38.09)-2.994 (-0.004)
PSI
Ao= 5.51
=$'.SJ - ).70
}~(/r5v.)-38.09 RD~ ~col;- jC-:P
9-3. Derive Eq. 9.29
v
Average speed= -
F =ma
v
2
v
y2
y2
µ = - ............ (9.29)
a=-=-
2S
2S
Time to stop = -
2S
2gS
v 9-4. Given volume of beads= 2 in. 3 , diameter of patch= 10 in., and SN 40 = 40 Determine SN20 and SNso
2 MTD = I -1t(l0 )2 4
= 0.0255 in.
From Eq. 9.33 PNG = 0.157 (0.0255r0.47 = 0.881
From Eq. 9.31
0 881 40 = SN 0 exp(- · x4o) 100
SNo = 56.9
When V = 20 mph SN20 = 56.9 x exp (-0.881 x 0.2) = 47.7 When V = 60 mph SNso = 56.9 x exp (-0.881 x 0.6) = 33.5
1 (1) NPROB PROBLEM 9.5 1 0 1 1 (3) MA.TL NDAMA NPY NLG 0.001 (4) DEL 3 1 80 1 0 (5) NL NZ !CL NSTD NUNIT 4 12 (6) TH (7) PR 0.35 0.35 0.45 0 1
(8) zc (9) NBOND
500000
35000
0
(13) LOAD
6
79.5775 (16) NR 8 12
7
0
9000
(11) E
48
72
/~f lr/.q· /'> -c
View more...
Comments
