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 :
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(9) NBOND
400000 10000 10000 (11) E 1 (13) LOAD 4.5 75 (14) CR CP 3 (19) NPT 0 13.5 0 0 0 4.5 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 K1 Cjl],.$ 1 1 (46) NLBT NLTC 1 (47) LNBT 3 (48) LNTC 36500 (49) TNLR 0.0796 3.291 0.854 (50) FT1 FT2 FT3 1.365E-09 4.477 (51) FT4 FT5
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2
1
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(11) E 400000 50000 40000 30000 20000 10000 1 (13) LOAD 4.5 75 (14) CR CP 3 (19) NPT (20) XW YW XPT 0 13,5 0 0 0 4.5 0 6.75 4 15 (25) NOLAY ITENOL 2 0 3 0 4 0 5 0 (26) LAYNO NCLAY 5 7 9 11 (27) ZCNOL 0 0 0 0 0.01 (28) RCNOL XCNOL YCNOL SLD DELNOL 0.5 (29) RELAX (30) GAM 145 135 135 135 135 130 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 8000 (33) PHI K1 Ci?.$KJeY: 0 8000 (33) PHI Kl 0 8000 (33) PHI Kl 0 8000 (33) PHI Kl 1 1 (46) NLBT NLTC 1 (47) LNBT 6 (48) LNTC 36500 (49) TNLR .0796 3.291 .854 (50) FTl FT2 FT3 1.365E-09 4.477 (51) FT4 FT5
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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
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(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
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(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
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0.45
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(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
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2
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(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
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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
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1
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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
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Chapter 5 KENSLABS Computer Program 5-1. (a) Given
cr = 400 psi and h = 7 in, find bending moment per in. width of slab
cr = 6M/h2 , so M = crh 2/6 =(400 x 49)/6 = 3267 lb-in.fin. of slab. (b) Given:
E1 =7 x 105 psi, E2 = 4 x10 6 psi and two bonded layer with h1
=3 in. and
h2 = 5 in. Find bending stress at top and bottom. With E1/E 2 = 0.175, from Eq. 5.8, 2
. = 0.175x3(0.5x3 + 7) + 0.5(7) = 3 .849 m. d 0.175x3 + 7
F
E
5 9, rom q..
le= 0.175[_!_(3 ~ + 3(0.5x3 + 7 -3.849) 2 ] + _!_(7) + 7(3.849 - 0.5x7")2 12 12 3
= 11.750 + 28.583 + 0.853 = 41.186 in
3
Bending stress at bottom of PCC
f = 3267x3 .849 = 305 _3 si 2 41.186 p Bending stress at top of HMA fi = 3267x(l0 - 3.849) xO.l ? 5 = SS.4 psi 41.186 5-2.
Same as 5-1, but the two layers are unbonded. From Eqs. 5.12, the modulus of rigidity of each layer
R = 7x105x(3)3 =l.875x106 lb-in.2 1 2 12(1 - (0 .4 ) ] 6 R 2 = 4 xl0 x(7 =1.1696x10 8 lb-in. 2 2 12(1 - (0 .15 ) ]
Y
R1 + R2 = 1.1884x108
The moment carried by each layer is proportional to R,
M 1 =(1.875x106/1.1884x10 8 )x3267
=51.545 in.-lb/in.
M2 = 3267 - 51.545 = 3215.455 in.-lb/in. f1 = (51.545 x 6)/(3) 2 = 34.4 psi 393.7 psi f 2 = (3215.455 x 6)/(7) 2 = ____...,.
1
(1) NPROB
Problem 5-3 0
0
1 9 1
0
1
1
( 3) NFOUND NDAMA NPY NLG
( 4) NSLAB NJOINT 6 0 0 0 0 1 0 0 0 0 0 1
(5) NX NY JONOl JON02 JON03 JON04 0 1 0 0 9 6 0 0 2 1 2 (6) NI.AYER NNCK NOTCON NGAP NPRINT INPUT NBOND NTEMP NWT NCYCLE NATl NAT2 NSX NSY MDPO NUNIT UL TC CL -24 150 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) (9) X's and then Y's 0 12 28 44 60 76 92 108 120 0 12 28 44 60 72 • 8 0 .15 4000000 (10) T PR YM /thdt:::. ..S-
An..s w-e.,.. : a.t.
-.>J86s- ,PS~
1
(1) NPROB
Problem 5.4 0
0
1 7 1
0
1
( 4) NSLAB NJOINT 6 0 0 0 0 1 0 0 0 0 0 0
1
( 3) NFOUND NDAMA NPY NLG
1
(1) NPROB
(5) NX NY JONOl JON02 JON03 JON04 0 1 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 O. 000005 0. 001 1 (7) TEMP GAMA(1) GAMA(2) PMR(l) PMR.(2) CT DEL FMAX 17. 61 0 17. 61 0 (8) Fl(l) F1(2) F2 (1) F2 (2) (9) X's and then Y's 0 12 24 36 54 78 120 0 17.6 36 54 78 120 10 0.15 4000000 (10) T PR YM 2 (12) NUDL 0 (13) NCNF 0 0 (14) NNMX NNMY 1 0 10.44 0 7.19 79.932 (15) LS XLl XL2 YLl YL2 QQ 1 0 10.44 14.005 21.195 79.932 (15) LS XLl XL2 YL1 YL2 QQ 1 (22) FSAF 0 200 (24) HAS SUBMOD
Problem 5.5 0
0
1 7 1
0
1
1
(3) NFOUND NDAMA NPY NLG
( 4) NSLAB NJOINT 7 0 0 0 0 (5) NX NY JONOl JON02 JON03 JON04 1 0 0 0 0 0 0 0 1 0 0 7 7 0 0 1 2 2 (6) NLAY.ER 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 (1) Fl (2) F2 (1) F2 (2) (9) X's and then Y's 0 7 14 26 48 83 118 0 7 14 26 48 83 118 10 0.15 4000000 (10) T PR YM 1 (12) NUDL 0 (13) NCNF 0 0 (14) NNMX NNMY (15) LS XLl XL2 YL1 YL2 QQ 1 0 5.22 3.405 10.595 79. 932 43 (19) NODSX 1 8 15 22 29 36 (20) NODSY 6 7 2 3 4 5 1 (22) FSAF 1 200 (24) NAS SUBMOD 0
-IZ..8.
9-o7
1
(1) NPROB
Problem 5 . 6 0
0
1
1
(3) NFOUND NDAMA NPY NLG
1 0 (4) NSLAB NJOINT 7 7 0 0 0 0 (5) NX NY JONOl JON02 JON03 JON04 1 1 0 0 0 0 0 0 0 1 0 0 0 7 0 0 1 2 2 (6) NJ.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) 0 8 18 30 54 84 120 0 8 17.6 30 54 84 120 (9) X's and then Y's 10 0.15 4000000 (10) T PR YM 2 (12) NUDL a'/..S weY: 0 (13) NCNF 0 0 (14) NNMx: NNMY 1 0 5.22 0 7.19 79.932 (15) LS XLl XL2 YLl YL2 QQ 79.932 (15) LS XL1 XL2 YLl YL2 QQ 1 0 5.22 14.005 21.195 1 2 3 4 5 6 7 (20) NODSY 1 (22) FSAF 200 (24) NAS SUBMOD 0
1 (1) NPROB Problem 5. 7 0
0
1
1
(3) NFOUND NDAMA NPY NLG
1 0 ( 4) NSLAB NJOINT 7 7 0 0 0 0 (5) NX NY JONOl JON02 JON03 JON04 1 1 0 0 0 0 0 0 0 1 0 0 7 7 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(1) 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 10 20 40 60 90 120 0 10 20 40 60 90 120 (9) X's and then Y's 8 0.15 4000000 (10) T PR YM 1 (12) NUDL - 299. 3Zo p.s_;,, 0 (13) NCNF 0 0 (14) NNMx: NNMY 14.15 99.985 (15) LS XLl :µ.2 YLl YL2 QQ 1 13.975 26.025 5.85 1 8 15 22 29 36 43 (19) NODSX 1 2 3 4 5 6 7 (20) NODSY 1 (22) FSAF 0 100 (24) NAS SUBMOD
4,,.swer =
1
NPROB
(1)
Problem 5.8 0
0
1
1.
( 3) NFOUND NDAMA NPY NLG
1 0 ( 4) NSLAB NJOINT 0 0 0 0 (5) NX NY JONOl JON02 JON03 JON04 7 7 1 1 0 0 0 0 0 0 0 1 0 0 0 7 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) 0 10 20 40 60 90 120 0 12 24.15 36 60 90 120 (9) X's and then Y's 8 0.15 4000000 (10) T PR YM 2 (12) NUDL 0 (13) NCNF 0 0 (14) NNMX NNMY 1 13.975 26.025 0 8.3 99.985 (15) LS XLl XL2 YLl YL2 QQ 1 13.975 26.025 20 28.3 99.985 ( 15) LS XLl XL2 YLl YL2 QQ 1 2 3 4 5 6 7 (20) NODSY (22) FSAF 1 0 100 (24) NAS SUBMOD
1 (1) NPROB Problem 5.9 0
0
1 8 1
0
1
1
( 3) NFOUND
NDAMA NPY NLG
( 4) NSLAB NJOINT 6 0 0 0 0 1 0 0 0 0 0 1
(5) NX NY JONOl JON02 JON03 JON04 0 1 0 0 0 6 0 0 1 1 2 ( 6) NLAYER NNCK NOTCON NGAP NPRINT INPUT NBOND NTEMP NWT NCYCLE NATl NAT2 NSX NSY MDPO NUNIT UL TC CL 13.5 150 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) (9) x· s and then y• s 0 12 30 54 90 138 186 240 0 12 24 48 84 132 9 0.15 4000000 (10) T PR YM 1
(12) NUDL
(13) NCNF 0 (14) NNMX NNMY 0 4.744 0 9.487 99.997 2 3 4 5 6 (20) NODSY (22) FSAF 200 (24) NAS SUBMOD
0
0 1 1 1 0
-1.>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<7. J l::J.5--<-> 0 (13) NCNF 0 0 (14) NNMX NNMY (15) LS XLl XL2 YLl YL2 QQ 1 108 120 0 12 83.333 83.333 1 108 l.20 72 84 (15) LS XLl XL2 YLl YL2 QQ (22) FSAF 1 0 300 (24) NAS SUBMOD 29000000 0.3 (34) YMSB PRSB "u . ,,. (35j SPCONl. SPC'ON2 SCKV .BD as WJ GDC NNAJ 0 0 1cnnnnn 1 1.2 0.01 0
-38
.
.L~
~JVVVUV
1 (1) NPROB Problem 5.13 0
0
1
1
( 3) NFOUND ND.AMA NPY NLG
2 1 (4) NSLAB NJOINT 6 . 6 0 1 0 0
(5) NX NY JON01 JON02 JON03 JON04
6
6
(5j l'l"'X 1'.-X JONOl. JON02 JON03 JO:N04
1
34
1
0
0
0
0
0
0
0
0
0
0
1
0
0
12
0
0
0
2
2
1 ( 6) NLAn:R NNCK NOTCON NGAP
NPRINT INPUT NBOND NTEMP NWT NCYCLE NAT1 NAT2 NSX NSY MDPO NUNIT UL TC CL
0 150 0 500 0 0.000005 0.001 10 (7) TEMP GAMA(l) GAMA(2) PMR.(1) PMR(2) CT D~T. FMA..~ 17.61 0 17.61 0 (8) F1(1) 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 15 30 42 54 72 (9) X's and then Y's 10 0.15 4000000 (10) T PR YM 0 (12) NUDL 1 (13) NCNF Q.IJ.jW~Y: 0 0 (14) NNMX NNMY 34 12000 (16) NN FF 1 7 13 19 25 31 37 43 49 55 61 67 (19) NODSX 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
1
(1) NPROB
Problem 5.14 0
0
2 6 6 1
1
1
1
(3) NFOUND ND.AMA NPY NLG
( 4) NSLAB NJOINT 8 0 1 0 0 (5) NX NY JON01 JON02 JON03 JON04 8 1 0 0 0 (5) NX NY JON01 JON02 JON03 JON04 45 0 0 0 0 0 0 0 1 0 0 12 0 0 0 2 2 1 (6) NI.AYER 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 0.00110 (7) TEMP GAMA(l) GAMA(2) PMR.(1) PMR(2) CT DEL FMAX 17.61 0 17.61 0 (8) Fl(1) F1(2) F2(1) F2(2) O 30 60 90 108 120 O 6 18 30 42 54 66 72 (9) X's and then Y's o 12 30 60 90 120 0 6 18 30 42 54 66 72 (9) X's and then Y's 10 0.15 4000000 (10) T PR YM 0 (12) NUDL 1 (13> NCNF Q,l'J.jy..Jer: at:, N~cle -ZZ 2 PS-<.J 0 0 (14) NNMX NNMY 45 12000 (16) NN FF 1 9 17 25 33 41 49 57 65 73 81 89 (19) NODSX 1 (22) FSAF 300 (24) NAS SUBMOD 0 29000000 0.3 (34) YMSB PRSB 0 0 1500000 1 l.2 0.25 0 8 (35) SPCON1 SPCON2 SCKV BD BS WJ GDC NNAJ 0 1 1 1 1 1 1 0 (36) BARNO
9-.>
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 ~<..
ac
0
(13)
7S
NCNF
7. '7- ?>i._
7 z:.
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7/ 0 0 (14) NNMX NNMY 1 108 120 0 12 83.333 (15) LS XLl XL2 YLl YL2 QQ (15) LS XLl XL2 YLl YL2 QQ 1 108 120 72 84 83.333 57 43 44 45 46 47 48 49 50 51 52 53 54 55 56 58 59 67 68 69 70 71 72 73 74 75 60 61 62 63 64 65 66 76 77 78 79 80 81 82 83 84 (18) NP 1 (22) FSAF 16000 0.4 (28) YMS PRS 29000000 0.3 (34) YMSB PRSB 0 0 1500000 1 12 0.25 0 14 (35) SPCONl SPCON2 SCKV BD BS WJ GDC NNAJ 0 1 1 1 1 1 1 1 1 1 1 1 1 0 (36) BARNO
2
NPROB
(1)
PROBLEM 5 .18 (1st step) 1 0 1 1 ( 3) NFOUND ND.AMA NPY NLG
2 1 ( 4) NSLAB NJOINT 7 8 0 1 0 0 (5) NX NY JONOl JON02 JON03 JON04 6 8 1 0 0 0 (5) NX NY JONOl JON02 JON03 JON04 1 49 0 0 11 0 0 0 1 10 0 0 0 0 0 0 2 1 2 (6) NI.AYER NNCK NOTCON NGAP NPRINT INPUT NBOND NTEMP NWl' NCYCLE NAT! NAT2 NSX NSY MDPO NUNIT UL TC CL 0 150 0 0 0 0.000005 0.001 1 (7) TEMP GAMA(l) GAMA(2) PMR(l) PMR.(2) CT DEL FMAX 0 0 0 0 (8) Fl(l) F1(2) F2(1) F2(2) 0 60 110 128 146 164 180 O 16 41 66 82 98 121 144 (9) X's and then Y's 0 16 40 80 120 180 0 16 41 66 82 98 121 144 (9) X's and then Y's 7 0.15 4000000 (10) T PR YM 0 (12) NUDL 0
(13) NCNF
(14) NNMX NNMY 17 25 29 33 41 49 50 1 (22) FSAF 5000 0.45 (28) YMS PRS 29000000 0.3 (34) YMSB PRSB 0 0 1500000 1 12 0. 25 PROBLEM 5.18 (2nd step) 0
1
2 7
6 1
0
0 1
1
1
51
0
52
0
53
54
(18) NP
(35) SPCONl SPCON2 SCKV BD BS WJ GDC NNAJ
( 3) NFOUND ND.AMA NPY NLG
(4j NSLAB NJOnn·
8 8 49
0 1 0
0 0
1 0 0
11
1
0 0 0
(5) NX NY JONOl JON02 JON03 JON04 (5) NX NY JONOl JON02 JON03 JON04 0 0 10 0 0 0 0 0 0 1 1 2 (6) NI.AYER NNCK NOTCON
NGA.P NPRINT INPUT NBOND NTEMP NWT NCYCLE NAT1 N..:&..T2 NSX N'SY MDPO l-tl"l.r.IT t."L TC CL
0 150 0 0 0 0.000005 0.001 1 (7) TEMP GAMA(l) GAMA(2) PMR(l) PMR(2) CT DEL FMAX 0 0 0 0 (8) Fl (1) Fl (2) F2 (1) F2 (2) O 60 110 128 146 164 180 O 16 41 66 82 98 121 144 (9) X's and then Y's 0 16 40 80 120 180 0 16 41 66 82 98 121 144 (9) X's and then Y's 7 0.15 4000000 (10) T PR YM 8 (12) NUDL 0
(13) NCNF
P?
0 (14) NNMX NNMY 90.007 ( 15) LS XLl XL2 YLl YL2 QQ 123.983 132.017 5.532 0 90. 007 (15) LS XLl XL2 YLl YL2 QQ 123.983 132.017 13.234 18.766 90. 007 ( 15) LS XLl XL2 YLl YL2 QQ 123.983 132.017 72.484 78.016 90. 007 (15) LS XLl XL2 YL1 YL2 QQ 1 123.983 132.017 85.984 91.516 (15) LS XL1 XL2 YL1 YL2 QQ 1 171.967 180 0 5.532 90.007 1 171. 967 90.007 (15) LS XL1 XL2 YL1 YL2 QQ 180 13.234 18.766 90.007 {15) LS XL1 XL2 YL1 YL2 QQ 1 171.967 180 72.484 78.016 1 171. 967 180 85.984 91.516 90. 007 (15) LS XL1 XL2 YL1 YL2 QQ 17 25 29 33 41 49 50 51 52 53 54 (18) NP 1 (22) FSAF l)vwe"7' -: 4.t. rJo:Je 5000 0.45 (28) '!MS PRS 4-':J 29000000 0.3 (34) YMSB PRSB 0 0 1500000 1 12 0.25 0 (35) SPCONl SPCON2 SCKV BD BS WJ GDC NNAJ 0 0
1 1 1
zs
1
(1) NPROB
PROBLEM 5 • 19 1 2 2 3 ( 3) NFOUND NDAMA NPY NLG
1 9 1
0 9
(4) NSLAB NJOINT 0 0 0 0 0 0 3 0 0 0
(5) NX NY JON01 JON02 JON03 JON04 0 1 0 0 0 9 0 0 1 2 2 (6) NI.AYER NNCK NOTCON NGAP NPRINT INPUT NBOND NTEMP NWT NCYCLE NAT1 NAT2 NSX NSY MDPO NUNIT UL TC CL 0 0 0 600 0 0.000005 0.001 1 (7) TEMP GAMA(1) GAMA(2) PMR(1) PMR(2) CT DEL FMAX 14.75 0 17.61 0 (8) F1(1) F1(2) F2(1) F2(2) 0 12 24 36 48 72 96 132 168 0 10 20 39 58 77 · 97 120 144 (9) X's and then Y's 8 0.15 4000000 (10) T PR YM 4 4 8 (12) NUDL 0 (13) NCNF 0 0 (14) NNMX NNMY 1 0 4.042 0 5.567 100 (15) LS XL1 XL2 YL1 YL2 QQ 1 0 4.042 14.433 20 100 (15) LS XLl XL2 YL1 YL2 QQ 1 0 4.042 77 82.567 100 (15) LS XLl XL2 YL1 YL2 QQ 1 0 4.042 91.433 97 100 (15) LS XL1 XL2 YL1 YL2 QQ 1 19.958 28.042 0 5.567 100 (15) LS XL1 XL2 YLl YL2 QQ 1 19.958 28.042 14.433 20 100 (15) LS XL1 XL2 YLl YL2 QQ 1 19.958 28.042 77 82.567 100 (15) LS XLl XL2 YL1 YL2 QQ 1 19.958 28.042 91.433 97 100 (15) LS XL1 XL2 YLl YL2 QQ 1 0 4.042 0 5.567 100 (15) LS XL1 XL2 YL1 YL2 QQ 1 0 4.042 14.433 20 100 (15) LS XL1 XL2 YL1 YL2 QQ 1 0 4.042 77 82.567 100 (15) LS XLl XL2 YL1 YL2 QQ 1 0 4.042 91.433 97 100 (15) LS XLl XL2 YL1 YL2 QQ 1 43.958 52.042 0 5.567 100 (15) LS XL1 XL2 YL1 YL2 QQ 1 43.958 52.042 14.433 20 100 (15) LS XL1 XL2 YLl YL2 QQ 1 43.958 52.042 77 82.567 100 (15) LS XLl XL2 YL1 YL2 QQ 1 43.958 52.042 91.433 97 100 (15) LS XLl XL2 YLl YL2 QQ 1 19 37 (18) NP 1 2 3 4 5 6 7 8 9 (20) NODSY 1 0.8 (22) FSAF 5000 0.4 (28) YMS PRS 3650 3650 3650 (37) TNLR 3650 3650 3650 (37) TNLR 1
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(1) NPROB
PROBLEM 5. 20
(first step) 2 0 1 1 ( 3) NFOUND NDAMA NPY NLG 2 1 ( 4) NSLAB NJOINT 8 9 0 1 0 0 (5) NX NY JON01 JON02 JON03 JON04 8 9 1 0 0 0 (5) NX NY JON01 JON02 JON03 JON04 1 68 0 32 0 0 0 1 1 10 0 0 1 0 1 0 2 1 2 ( 6) NI.AYER NNCK NOTCON NGAP NPRINT INPUT NBOND NTEMP NWT NCYCLE NAT1 NAT2 NSX NSY MDPO NUNIT UL TC CL -20 150 135 0 0 0.000005 0.001 1 (7) TEMP GAMA(1) GAMA.(2) PMR(1) PMR(2) CT DEL FMAX 0 0 0 0 (8) F1 (1) F1 (2) F2 (1) F2 (2) 0 40 80 120 140 160 170 180 0 26 52 62 72 82 92 118 144 (9) X's and then Y's 0 10 20 40 60 100 140 180 O 26 52 62 72 82 92 118 144 (9) X's and then Y's 10 0.15 4000000 (10) T PR YM 0 (12) NUDL 0 (13) NCNF 0 0 (14) NNMX NNMY 1 (19) NODSX 27 0.05 36 0.05 45 0.05 54 0.05 63 0.05 72 9 0.05 18 0.05 0.05 81 0.05 0.05 99 0.05 108 0.05 117 0.05 126 0.05 135 90 0.05 144 0.05 0.05 65 66 0.05 67 68 0.05 69 0.05 0.05 64 75 76 0.05 77 0.05 0.05 70 0.05 0.05 73 74 0.05 0.05 71 0.05 0.05 78 0.05 0.05 80 79 (21) NG CURL 1 (22) FSAF 2 30 (30) NL MAXIC 8 (31) TH 20000 5000 (32) E 0.3 0.4 (33) PRBF 29000000 0.3 (34) YMSB PRSB 0 0 1000000 1 12 0.125 0 0 (35) SPCON1 SPCON2 SCKV BD BS WJ GDC NNAJ PROBLEM 5.20 (second step) 2 0 1 1 ( 3) NFOUND NDAMA NPY NLG 2 1 ( 4) NSLAB NJOINT 8 9 0 1 0 0 (5) NX NY JON01 JON02 JON03 JON04 8 9 1 0 0 0 (5) NX NY JON01 JON02 JON03 JON04 1 68 0 0 0 1 0 0 0 10 0 0 1 0 0 0 1 1 2 ( 6) NLAYER NNCK NOTCON NGAP NPRINT INPUT NBOND NTEMP NWT NCYCLE NAT1 NAT2 NSX NSY MDPO NUNIT UL TC CL 0 150 135 0 0 0.000005 0.001 1 (7) TEMP GAMA(1) GAMA(2) PMR(1) PMR(2) CT DEL FMAX 0 0 0 0 (8) F1(1) F1(2) F2(1) F2(2) 0 40 80 120 140 160 170 180 0 26 52 62 72 82 92 118 144 (9) X's and then Y's 0 10 20 40 60 100 140 180 0 26 52 62 72 82 92 118 144 (9) X's and then Y's 10 0.15 4000000 (10) T PR YM 1 (12) NUDL 0 (13) NCNF o o (14; h""NMX h""NM"x 1 170 180 67 77 200 (15) LS XL1 XL2 YL1 YL2 QQ 1 (19) NODSX 1 (22) FSAF 2 30 (30) NL MAXIC 8 (31) TH 20000 5000 (32) E 0.3 0.4 (33) PRBF 29000000 0.3 (34) YMSB PRSB 0 0 1000000 1 12 0.125 0 0 (35) SPCONl SPCON2 SCKV BD BS WJ GDC NNAJ
- 9-S
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1 (1) NPROB Problem 5 . 21 1 0 1 1 (3) NFOUND ND.AMA NPY NLG 2 1 (4) NSLAB NJOINT 6 6 0 1 0 0 (5) NX NY JONOl JON02 JON03 JON04 6 6 1 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) 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) Fl(l) F1(2) F2(1) F2(2) 0 30 60 90 108 120 0 15 30 42 54 (9) X's and then Y's 72 0 12 30 60 90 120 0 15 30 42 54 (9) X's and then Y's 72 10 0.15 4000000 (10) T PR YM 0 (12) NUDL 1 (13) NCNF 0 0 (14) NNMX NNMY 34 12000 (16) NN FF 34 (18) NP 1 7 13 19 25 31 37 43 49 61 67 (19) NODSX 55 1 (22) FSAF 15000 0.4 (28) YMS PRS 29000000 0.3 (34) YMSB PRSB 0 0 1500000 1 12 0.25 0 0 (35) SPCONl SPCON2 SCKV BD BS WJ GDC NNAJ
1 (1) NPROB Problem 5. 22 1 0 1 1 ( 3) NFOUND ND.AMA NPY NLG 2 1 ( 4) NSLAB NJOINT 0 1 0 0 (5) NX NY JONOl JON02 JON03 JON04 6 8 6 8 1 0 0 0 (5) NX NY JONOl JON02 JON03 JON04 1 45 0 0 1 0 0 0 0 1 0 0 12 0 0 0 2 2 1 (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 10 (7) TEMP GAMA(l) GAMA(2) PMR(l) llMR.(2) CT DEL liMAX 17.61 0 17.61 0 (8) Fl(l) Fl(2) F2(1) F2(2) 0 30 60 90 108 120 0 6 18 30 42 54 (9) X's and then Y's 66 72 (9) X's and then Y's 72 66 • 0 12 30 60 90 120 0 6 18 30 42 54 10 0.15 4000000 (10) T PR YM 0 (12) NUDL 1 (13) NCNF 0 0 (14) NNMX NNMY 45 12000 (16) NN FF 45 (18) NP 1 9 17 25 33 49 41 57 65 73 81 (19) NODSX 89 1 (22) FSAF 15000 0.4 (28) YMS PRS 29000000 0.3 (34) YMSB PR.SB 0 0 1500000 1 12 0.25 0 8 ( 35) SPCONl SPCON2 SCKV BD BS WJ GDC NRAJ 0 1 1 1 1 1 1 0 (36) BARNO
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SINGLE AXLE LOAD
EALF (TABLE 6.4) 1.5 5 7.5 10 14 17 18.5 19.5 21 23 25 28
NUMBER
.0001 .005 . 02695' .0877 . 36 . .796 1.12 1. 375 1.83 2.58 3.53 5.39
ESAL
0 3188 2843 9942 3111 1899 1078 423 598 144 6 6
0
15.94 76.61885 871.9134 1119.96 1511.604 1207.36 581.625 1094.34 371.52 21.18 32.34
TANDEM AXLE LOAD
EALF (TABLE 6.4)
3 9 15 .., 1 ........ Z7 30.75 31. 15 33 35'
-:,1 39 41 43 4-5 ~8 w;-...,
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NUMBER
ESAL
0
7
0
2631 2541 "'""lL""' ".JO"' 3103 703 141 503 388 280 247 183 160 51 64
.036 1 A 0 • ..L,U
.426 .72925 .831 .971 1. 23 1. 53 1. 89 2.29 2.76 3.27 4.17 5.63
COMBINED ESAL FOR 24 DAY PERIOD ESAL DURING FIRST YEAR (365.25 DAYS)
7
0 0
91.476 349.576 1321.878 512.66275 117.171 488.413 477.24 428.4 466.83 419.07 441.6 166.77 266.88 39.41 12491. 778 190109
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EALF Table 6.7
Number Table P6.7
ESAL col C x col D
Single Axle 5 7.5 10 14 17 18.5 19.5 21 23 25 28
3188 2843 9942 3111 1899 1078 423 598 144 6 6
0.006 0.0265 0.082 0.341 0.802 1.1425 1.4275 1.955 2.85 4.015 6.29
19.128 75.3395 815.244 1060.851 1522.998 1231.615 603.8325 1169.09 410.4 24.09 37.74 6970.328
Tandem Axle 3 9 15 21 27 30.75 31.75 33 35 37 39 41 43 45 48 52
Es/:}L
( /::.r
7 2631 2541 2362 3103 703 141 503 388 280 247 183 160 51 64 7
0.0003 0.009 0.065 0.257 0.736 1.271 1.446 1.705 2.175 2.73 3.385 4.145 5.015 6.005 7.73 10.6
0.0021 23.679 165.165 607.034 2283.808 893.513 203.886 857.615 843.9 764.4 836.095 758.535 802.4 306.255 494.72 74.2
Sum
9915.2071
Total Sum
16885.5351
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p7.3 x
5
254 228 197 185 165 137 115 91
5.056905 5.060698 5. 127105 5.204120 5 .457882 5.346353 5.444045 5.498311
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p7.9 log x
J.aJ,eD
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....
Constant 5.m954 Std Err of Y Est .0740420 R Squared(Adj,Raw) .8558775 .8738928 No. of Observations 9 7 Degrees of Freedom
L1 ~::: ~ c E0
« {'£
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f z,. ~:i,
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.4417041/
Std Err of Coef_
-~
s
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1+.6
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= S":J9; ZZ92J~ ,,_~/
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Regression Output:
+ ~ Po;.11,
4
JI. (
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Coefficient(s) Std Err of Coef.
o-71/
= J},J-o e
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"
Regression Output:
Coefficient(s) Std Err of Coef.
J'1~
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logy 278
L~ H,t1. =12~ k, -r- I~ f~e
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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
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50.11872
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39.81071
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Chapter 8 Drainage Design
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8-4. (a) Can geotextiles be used as a filter for soil C in Figure P8.3? Why? Soil C is a clayed soil with 100% passing #200 sieve. Geotextiles cannot be used as a filter because the soil is too fine and will cause clogging. (b) If geotextiles are used for soil B, what AOS do you recommend? Soil B has 65% passing #200 sieve with Das = 0.25 mm or #60. If woven fabric is used, AOS
~
Pas or AOS
~
#60
(c) If geotextiles are used for soil A, what AOS do you recommend? Soil C has less than 50% passing #200, so AOS
~
B x Das
Cu =0 60 /0 10 = 0.9/0.3 = 3, which is between 2 and 4, so B = 0.5Cu = 1.5 Das= 2 mm, AOS
~
1.5 x 2 =3 mm or AOS
~
#8
8-5. (a) Estimate the permeability of soil B in Figure P8.3 by Eq. 8.3 From Figure P8.3, 010 = 0.0052 mm
From Table 8.4, Ck= 5 to 8
From Eq. 8.3, k = Ck 010 so k= 5 (0.0052)2 = 1.352 x 104 mm/s
lft )(60x60x24secJ ( 304.8mm lday
= 0.0383 fUday (lower limit) k= 8 (0.0052) 2 = 2.163 x 104 mmls = 0.0613 tuday (upper limit) (b) Given Ydrr 110 lb/ft3 , estimate the permeability of soil B by Eq. 8.5a Assume Gs= 2.7, from Eq. 8.5b, n = 1-
110 = 0.3471 62.4x2.7
From Figure P8.3, percent passing #200 (0.075 mm) is P200 = 65% From Eq. 8.5a, k
= 6.214xl 0 5 (0.0052)1. 478 (0.3471) 6 ·654 (65)0 .597
= = 0.0189 fUday
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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
qr(!
t:> lr/.q· /'> -c
(14) CR CP 24
~
96
(17) RC
1 (1) NPROB PROBLEM 9.6 2 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
~ E3
9000 (11) E 500000 ~35000 0 (13) LOAD 6 79.5775 (14) CR CP 7 (16) NR 0 8 12 24 48 72 96 (17) RC 1 15 (25) NOLAY ITENOL 0 (26) LAYNO NCLAY 2 7 (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
z::,
'\ I< I
~
ooo f'><..
Baoo
1
(1) NPROB
PROBLEM 9. 7 2 0 1 1 (3) MATL ND.AMA NPY NLG 0.001 (4) DEL 8 1 80 1 0 (5) NL NZ ICL NSTD NUNIT 4 2 2 2 2 2 2 ( 6) TH 0.35 0.35 0.35 0.35 0.35 0.35 0 1
0.35
0.45
(7) PR
(8) zc (9) NBOND
500000 35000 35000 35000 35000 35000 35000 9000 (11) E 0 (13) LOAD 6 79.5775 (14) CR CP 7 · (16) NR 0 8 12 24 48 72 96 (17) RC 6 15 (25) NOLAY ITENOL 2 0 3 0 4 0 5 0 6 0 7 0 (2 6) LAYNO NCLAY 5 7 9 11 13 15 (27) ZCNOL 3 0 0 0.25 0.01 (28) RCNOL XCNOL YCNOL SLD DELNOL 0.5 (29) RELAX 145 135 135 135 135 0 135 135 (30) GAM 0.5 0.6 (31) K2 KO 0.5 0.6 (31) K2 KO S-oo., ooa 0.5 0.6 (31) K2 KO 0. 5 0. 6 ( 31) K2 KO 0.5 0.6 (31) K2 KO j'>-<... zooo 0.5 0. 6 (31) K2 KO 1:::3 0 8000 (33) PHI Kl 0 8000 (33) PHI Kl ~ ()Ot7 0 8000 (33) PHI Kl 0 8000 (33) PHI Kl 0 8000 (33) PHI Kl 0 8000 (33) PHI Kl
.
F.>,I..
t:.,
k,
1
.
-
(1) NPROB
PROBLEM 9.8 1 0 1 1 (3) MATL ND.AMA NPY NLG 0.001 (4) DEL 3 1 80 1 0 (5) NL NZ ICL NSTD NUNIT 8 12 (6) TH (7) PR 0.35 0.35 0.45 0
1
(8)
zc
C9>' NBOND
240000 0 (13) 6 80 5 (16) 0 12
19200
4800
E
(14)
CR CP
NR
24
& = /ao
36
48
,Ps(.; .J
o. aZ.<77/ .q
(11)
LOAD
~. o J
(17) RC
cz.
=~
Zif. z 7
s ec7"k4l PJ(..
.J
1YJ
J
c
/~o7 J
J
tt::J,o/S-)oJ
4~ <7.o 112_$-
' l".l, _,
AR~~ = '
f I t- .:<. (
o.
~
d,4cif.7
:
c8.~~s
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..
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w. ,}'
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q,,.:t..
w, >
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cl z.
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=/&»7 (
)<
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k= /e;.7(~) .::),iOO~
\1.12 .I
.
w, "
K
z So, o
pc:.<-
=z'"· <7
pt:-t,...
-
='"I 7 ( <>,o•).J" 0., I ) =ZJ8, Z t:J,
10.7 (tl.4~8) ~
K .=
-::. ZJ'j.8
6,01>
~y~~1~ I<.= Z.$3,7 ,Pc:.~.; S4.f Z85'p<.it.. <:riv~?
I,
c, = 9: 7 x
.3
p 5 ;_,,
/o"'
£;-.
pt:)#J
'7. ~8
= I~ [ 1-(o.ISJ"J x 223, 7 x ( z.<:;)-t- =._)co, 8 49-, 7 ~
h - 7.
-? 4-' 1~
/'0
s~ y
;,,,3
6
,~~
8
,
J
"I-
/0.
/:'f
0 ,,,
'F"!-:
,.,,,",,, hJ-. 9.4!J-J/ r. r>"~ Ef. '7-~7,,
40 :::
1
~
Ee ::: 2[/-(o.'f-.2._ ] x T
"
'f' ;". ~.J8c _, 4., =~.1.>c.,
?.3'9 ./ _;_ =
l/.j-
c/z:::
~ J68" c/J = a.07$ cl~
~c~ cl~ = j('Jf)o( ~)
W~
.
r~-G
PC<.
<;~ ~
'7-11
cl~
I, :.
4- '~, ) ~ = 6 ~Y.).-7 .
3 ( /
I=
Fyr;w,
~
)0 ~ -Je~~>'e CyqcJ<,~
s- ~ b
.f"o/..>oo ;Ft
6
~~ -Ff://~eJo 7-t
r
'f;4 ~ ~ 9.S+c:t.. o.f21J
F/iTULT
= (Zo)
- (). / 'f'? 2 (
<}-(!;,
~ 9.
.;;-r
I
3~J'97~ -r ( 6. '''6-o.87~/ >< 1- +- /,JJ89~h)~(a./xzoi) ~ 2./,!>-16~3 - /, 21.f
~ / 9. S' 'f-1Z.
~-I~
It?= ~o 0c
R..e7ui'7e.~ : C1'4
Sa J1..1.J.1d")
S6,.t = 2~
I)
,~60
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0,0.J"~
/J
~.,~ o~
t- o. o
J -
1o ft8 { 1)
0. t:J 8 43
, 1
~
'FroM ~ ~·~' ~
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=
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¢,()ZZ} - 0.
o,~t:Jo362
o/ 3)X
x /,IJ<>o]
( b, o ;'cinf.s pe">"' l Jb. j JCl,",,,.f.; p-e~
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) 1l l - ei,o't-19 l )
t I)
o
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,;:o)-
)/-jfl1
.Se
Chapter 10 Reliability 10-1. Given:
f(x) = [
~: 1/6
x=O Find: expectation and variance of x
x =1 x =2
1 1 I E[x] = Ox - +Ix-+ 2x -
From Eq. 10.2a,
3
2
6
5
=- 6
<:>,
B33
t2
10-2. Given:
f(t)= te
2
=0
when t 2 0 when t
Find: Mean and variance of x t2
E[t]=
J t f(tX!t =rX)Jo t e -2 dt 00
2
-oo
From mathematical handbook, we can find the following definite integral
i
oo
0
2
x 2 e-x dx
J;C =4
so E[t] must be transformed to the same form, or
E(t)=2v'iJ:(}iJe-(~f{k)=2F2( =~ =1.253
:J
From Eq. 10.8
v[t]= E[t 2 ]-(E[tD2 12
E[t2
t2
]= fooo t3e -2 dt =~ fooo t2e -2 d(t2)
From mathematical handbook, we can find the following integral
I
eax
xeaxdx = - 2 a
(ax -1)
With x = t 2 and a= _ _!.
2
00
When t = oo, the value is O because the exponential term predominates. When t = 0, the value is -2. so E[t2] = 0 - (-2) = 2. V[x] = 2 - (1.253)2 = 0.429
10-3. Given: f(x, y)
x= 1 x= 2 x= 3
[ % %
=
112
y= 2 y= 3 y =4
Find: Cov(x, y), p(x, y), V[x + y]
From Eq. 10.2a
I 1 1 E[x] =Ix-+ 2x - + 3x - = 2.25 4 4 2
I 1 1 E[Y ] =2x-+3x-+4x-=3.25
4
4
2
From Eq. 10.15b
Cov[x,y ]= (1- 2.25)(2 - 3.25)x.!. + (2 - 2.25)(3- 3.25)x.!. +
4
4
(3- 2.25)( 42 - 3.25)x.!. = 0.6875 2 From Eq. 10.7b V[x] = (1 - 2.25)2 x 0.25 + (2 - 2.25)2 x 0.25 + (3 - 2.25) 2 x 0.5 = 0.6875 V[y]
=(2 -
F
rom
E
q.
3.25) 2 x 0.25 + (3 - 3.25)2 x 0.25 + (4 - 3.25) 2 x 0.5 10 8 ·1
10-4. Given:
f(x, y)
(x )0.6875 -1 ,y - .Jo.6875x0.6875 - -
P
V[x, y] = 0.6875 + 0.6875 + 2 x 0.6875
From Eq. 10.21
=0.6875
=6xy(2-x-y)
x:S:x:S: 1,
=O
otherwise
=2.75
O:S: y:S:1
Find:
Cov[x, y] and p[x,y]
Note Cov[x, y] = E[x, y] -E[x] E[y], so it is necessary to compute E[x, y], E[x], and E[y].
E[x,y] = =
J~J~xy(12xy- 6x 2y-6xy 2)dxdy
J~J~ (12x 2y 2 -
3 2 2 3 6x y - 6x y )dxdy
2 2 E[x] = J~J~ x(12xy- 6x y- 6xy )dxdy
= s~s; (12x 2y =
2 2 3 6x y- 6x y )dxdy
6 y- 2x y 2]l dy il[4x y- -x 4 4
3
3
0
=[2y2
0
-~y2 -~y3]1 =2-~-~=2_ 4
3
0
Because of symmetry of x and y, E[y]
4
3
12
=E[x] =7/12
1 7 7 1 Cov[x y]=---x-=-, 3 12 12 144
-
.
[
]
Cov[x, y]
Since p x,y = ~V[x]V[y], it is necessary to determine V[x] and V[y]. Because V[x] = V[y]
= E[x2] -(E[x])2 , E[x2] must be determined first.
E[x ]= J~J~ x (12xy - 6x y- 6xy )dxdy 2
2
=
2
2
f:J: (12x y-6x y- 6x y )dxdy 4
3
3 2
1 fl[ 4 6 5 6 4 2] =Ji 3x y--x y--x y dy 0
5
4
= [ ~ y2 _ ~ y2 _ ~ y3
I
0
=~ _ ~ _ ~ =~
Because of symmetry of x and y, E[y] = E[x] = 215
1 Cov[x,y] - -144 5 [ ] p x,y = "'V[x]V[y] =--=4=3===43==--43
-x720 720
2
10-5. Given t = (0.25 + 0.125logn)
7x5EP) [ (
_ a2
]~V3 {I EEP EP
t = 6.5 in., n = 100,000, P = 9000 lb, E = 10,000 psi, a= 6 in., and Ep =400,000 psi. The coefficients of variation of all the above variables are 1O.~ Find: Reliability oft based on Taylorseries expansion. (a) Variance oft due ton After substituting the values of all variables, except n, the following equation is obtained.
t = 6.03266(0.25 + 0.125logn)
~ = 6.03266x0.125x loge 8n
n =
v[t].
6
7.5408xl0- x0.4343 = 3.275xl0-6
=(:J
v[n]
J
= (3.275xl0- 6 (100,000x0.1) 2 = l.0726xl0- 3 (b) Variance oft due to P
t = 0.25585J5.6993xl0- 6 P
2
-
~=
0 25585 · (5.6993xl0- 6 P 2 8P 2 = 6.36x10- 4
v[t1 = (6.36xlo- 4 )
36 36 }-
-
05 · xl
l.3986P
(9000x0.1) 2 = 0.328
(c) Variance oft due to E
t = 0.0119 VE J4.616xl0 10 E- 2
-
36 2
~ = 0.0119(4.616xl0 10 E- 2 00
36
3 r.s x.!_(E)3
+ 0.0119 VEG)4.616xl0 10 E- 2 -36f
05
(-9.232x10 10 E- 3 )
=l.76314xl0- 4 _ 6.21xlo- 3 (-9.232x10- 2 )= -3.9699x10- 4 v[tk = (-3.9699xlo- 4 )
(10,000x0.1) 2 = 0.1576
(d) Variance oft due to a
t = 0.2559 J461.64 - a
2
~= 0.2559x0.5(461.64- a 2 }-
aa
0
.5(-2a)= -7.442x10- 2
V[tl = (-7.442x10- 2 J(6x0.1) 2 = 0.1994xlo- 3
(e) Variance oft due to Ep I
t = 388.917(EP
f3
~=- 388.917 (E f~ =-4.399xl0-6
aa
3
p
V[tkP =(-4.399xl0- 6 J(400000x0.1) 2 =0.031 Variance oft
V[t]= v[t]n + v[t]p + V[t]E + V[t]a + v[t]E p = 0.0011+0.328 + 0.1576 + 0.002 + 0.031=0.5197 Standard deviation of t
st=
.JVftl = .Jo.5297 = 0.121
E[1J~co25 + 01251ogn{ (~;J2 -a ]~ 2
~ [ro.25 + 0.12510g(lOOOOOil
( 75x9000 1tX10,000
= 0.875x20.63 lx0.2924 = 5 .28 in.
z = 6.5 - 5.28=1.692 0.721 From Tabbie 10.1 Area under normal curve= 0.5 + 0.455 = 0.955 Reliability= 95.5%
J2 - 36]3
10,000 400000
/o-~
i=
&1v-e1?
= Io~ ooo
J?
= ~ooo
P
(o.z.>+-a.12S~n)[Jc?SP)2-_c:l)~ 1TE V~
± / c; co a = / Jc; coo
:t
9do
c = I qooo r /., «:>o q_
6 ±
==
r=.~ :;
Cr
By
= J7c.
ee:
Gr1ile11
q,,4,_
~... /oo
) I,
q~
~ ~ ,P~~
..>.
,..,,
-
O<=>D
~
le>,
t::
l.b
.4~ ~OD -+- 4d.; ~o ::: 44-o; ..aoo q~ 3 ~., DOD p..$~ 7 ~ '-:I ~J /6, ~ ~ Z,, I> ?'~ ~lkw~,,-- /efulf,1
p (lb) 9900 9900 9900 9900 9900 9900 9900 9900 8100 8100 8100 8100 8100 8100 8100 8100 9900 9900 9900 9900 9900 9900 9900 9900 8100 8100 8100 8100 8100 8100 8100 8100
n
110000 110000 110000 110000 110000 110000 110000 110000 110000 110000 110000 110000 110000 110000 110000 110000 90000 90000 90000 90000 90000 90000 90000 90000 90000 90000 90000 90000 90000 90000 90000 90000
Fr6NI
~ '9ao
= ~. " <=111-/..
b
:?
i.Jf~ Excel .5,P~~s~el:.. ~">t.. 4>".t::1"''7'1J-...
tj>6C1f'.5
E [t.]
tt)_
_
e:n
0
E (psi) 11000 11000 11000 11000 9000 9000 9000 9000 11000 11000 11000 11000 9000 9000 9000 9000 11000 11000 11000 11000 9000 9000 9000 9000 11000 11000 11000 11000 9000 9000 9000 9000
Y.'f9 / 3Z -
~I, v £ t ]
Ep (PSI)
6.6 6.6 5.4 5.4 6.6 6.6 5.4 5.4 6.6 6.6 5.4 5.4 6.6 6.6 5.4 5.4 6.6 6.6 5.4 5.4 6.6 6.6 5.4 5.4 6.6 6.6 5.4 5.4 6.6 6.6 5.4 5.4
440000 360000 440000 360000 440000 360000 440000 360000 440000 360000 440000 360000 440000 360000 440000 360000 440000 360000 440000 360000 440000 360000 440000 360000 440000 360000 440000 360000 440000 360000 440000 360000 Sum
= Z8.
9
9-: 8 .%
~;,.:J,:,.;m:cl;
(t)"2 (sq. in.)
27.69251 31.65647 28.64632 32. 7468 37.43425 42.79265 38.26863 43.74646 17.58416 20.10118 18.53796 21 .19152 24.22492 27.69251 25.05929 28.64632 27.01126 30.8777 27.94161 31.94122 36.51335 41 .73993 37.3272 42.67028 17.15158 19.60668 18.08192 20.6702 23.62897 27.01126 24.44282 27.94161 920.5796
.>: 3 Jl.f
t
E[ i"'J = '7:z.o,r7~6/)2 =. Z8. 768. 768 - ( S.J/¥..)z =IC), sz ~ st = ~~J z7
o. 7 2
A~-e-t b~
t (in.)
5.262368 5.626408 5.352226 5. 722482 6.118354 6.541609 6.186164 6.614111 4.193347 4.483434 4.305574 4 .603425 4.921882 5.262368 5.005926 5.352226 5.197236 5.556771 5.285982 5.651656 6.042628 6.460645 6.109599 6.532249 4.141446 4.427943 4.252284 4.546449 4.860964 5.197236 4.943968 5.285982 170.0449
)'Jere.
'""
s. J1<;-~
~ == 6 ·->-f. ~' 5'l-
'·$ ,
R-e./,'.c/ 6 .-1#·ly -
a (in.)
Vt&7>iq6/e.r
=
e::>,
= /. ~ 31. fro'? /.S = c:J. 9 ~ 0
7
/a.
J
10-7. Find reliability based on n by Taylor's series expansion 2
t
= (0.25+0.125logn)[
8logn _
at
8
r
IEEP ( ~:) -a2J 3)VEP IL.G
_ _ _ _ _ _ _8_ _ _ _ _ _ =1.
r (~:) 2 2J V{rEEP r( Ep _a
11..G
2
75x9000 ) nxl0,000
3
2
75 ) (2p) 8t ( ;£
8x6.5(
nxl0,000
6
21
2
1t
3
400,000 2
) (9000)
2_ 21 6
= -1.0388 xl0-3 3 3 10,000
400,000
400,000 10 000 10
[ ( 75x9000 )' -6:
]3
nxl0,000
=-2.873xl0- 4 + 9.349x10- 4 = 6.476x 10-4 _a_Io_g_n = _ _ _ _S_ta____
aa
[
2 (7n5EP) _a21
3
JIEEP VEP
8x6.5x6
-r . . .--(
2 75x9-000 )_6 2] nxl0,000
= 3 3
326
3 10,000
1tXlO,OOO
Sx _ (75x9000) 65
= ---;=(=75=x9=000=)2=_=62 +
-
75
75x9000 ) ( nxl 0,000
__ 8x_6_.5 3 400,000 3 10,000 4
2
10,000 400,000
0 1215 .
8
-t
ologn =
3
[ (:)'-a' J(EE,' V[logn1
~( o~~n
8 -x6.5 =--;:======3==-------=7.183x 10-6 2 75 9000 2 2 · -6 Jv10 ooox4oo 000 x [( 7r:Xl0,000 ' '
J
J' {tC[tD'
2 2 V[lognl = (1.326) + (6.5x0.1) = 0. 7429 V[logn]P = (-1.0388x10-3
J + (9000x0.1)2 = 0.8741
J
2 V[logn]E = {6.476x10- 4 + (10,000x0.1) = 0.4194 2 2 V[lognl = (0.1215) +(6x0 1) = 0.005314 V[log n ]Ep = {7.183x10-6
J+ (400,000x0.1)2 = 0..08255
V[logn] = 0.7429 + 0.8741+0.4194 + 0.005314 + 0.08255 = 2.1242 S[logn]= .Jv[Iogn] = 1.4574 E[logn]=
8 65 x · 2 75x9000 _ 62 xxl0,000
( l
J J
-2=6.62 3
10,000 400,000
log nT = log I 00,000 = 5 · lognT -E[logn] 5-6.62 z= = =-1.1116 S[log n] 1.4572 Reliability =86. 75%
t
p (lb)
9900 9900 9900 9900 9900 9900 9900 9900 8100 8100 8100 8100 8100 8100 8100 8100 9900 9900 9900 9900 9900 9900 9900 9900 8100 8100 8100 8100 8100 8100 8100 8100
7.15 7.15 7.15 7.15 7.15 7.15 7.15 7.15 7.15 7.15 7.15 7.15 7.15 7.15 7.15 7.15 5.85 5.85 5.85 5.85 5.85 5.85 5.85 5.85 5.85 5.85 5.85 5.85 5.85 5.85 5.85 5.85
E (psi) 11000 11000 11000 11000 9000 9000 9000 9000 11000 11000 11000 11000 9000 9000 9000 9000 11000 11000 11000 11000 9000 9000 9000 9000 11000 11000 11000 11000 9000 9000 9000 9000
a (in.)
6.6 6.6 5.4 5.4 6.6 6.6 5.4 5.4 6.6 6.6 5.4 5.4 6.6 6.6 5.4 5.4 6.6 6.6 5.4 5.4 6.6 6.6 5.4 5.4 6.6 6.6 5.4 5.4 6.6 6.6 5.4 5.4
E U:;t"] = 21~ ~ZS: 6. 7z3 -r. {" ,,~ IY)
c:c. / r
// .:>. ~ /
V
r .f& n] J.. (!
S--2:1" - /2. zib z-::::. 5- b.723 = l.~8~
Ep (PSI)
440000 360000 440000 360000 440000 360000 440000 360000 440000 360000 440000 360000 440000 360000 440000 360000 440000 360000 440000 360000 440000 360000 440000 360000 440000 360000 440000 360000 440000 360000 440000 360000 Sum
log n
(log n)"2
7.567168 6.948153 7 .406546 6. 797923 6.228677 5.696265 6.138478 5.611901 10.00615 9.229329 9.693206 8.936631 8.229006 7.567168 8.057272 7.406546 5.827683 5.321216 5.696265 5.1983 4.732554 4.296944 4.658754 4.227919 7 .823215 7 .187633 7.567168 6.948153 6.369187 5.827683 6.228677 5.696265 215.128
57.26204 48.27683 54.85692 46.21175 38.79642 32.44743 37 .68091 31.49344 100.1231 85.18051 93.95824 79.86337 67.71654 57.26204 64.91964 54.85692 33.96189 28.31534 32.44743 27 .02233 22.39707 18.46373 21.70399 17.8753 61 .2027 51.66207 57.26204 48.27683 40.56654 33.96189 38.79642 32.44743 1517.269
E[ (1.") ,,/): ir'f~b2 =- o/-7. 'f!S'
= ~7-1.f-JS -
z ( 6.Ji.3') = Z ~ Z
/
h
10-9. Given
Nr = f1 (ct tf2 f1
=0.00462,
f2
=2.69,
C[f1]
C[f2]
et =0.0021,
=30%
=5%
C[Et]
=10%
Find Expectation and variance of Nt based on the first order Taylor series expansion. First order approximation of expectation or mean is E[g] = E[µ] E[Nt]
= 0.00462 (0.0021r2·59 = 73,769
(a)
V(Nr l t
aNf
= f1 (- f2 Xct tfz-l = -0.00462x2.69x(0.002lt 3 ·69 = -94,494,180
OE1 2
V(Nr
2 l =(oNr J V(E 1 ]= (- 94,494,180 )2(0.002lx0.1) = 3.94xl0 8 OE t
t
(b) v[Nr Jr1 ( ' : :r 1
J;
V(N r le,
(e, tr, ; (o.0021t 2·69
; ( 8;;r 1
=
r
;
0.1597x!0 8 2
2
V(f1 ] ; (0. l 597xl0 8 ) ( 0 .00462x0 .3 )
4.899xl0 8
= 4.55xl0 5
V[Nr
12 =( ':::
r ]= V[f2
= 37.45xl0
5
(4.55xl0 )2(2.69x0.05)2
8
(d) Cross - Product
fi{fJ = 0.00462x0.3 =0.001386 .Jv[f2 ] = 2.69x0.05 = 0.1345 Cov[f1,f2 ]= -0.867x0.001386x0.1345 = -1.616x10- 4 BNr ( BNr1
J(
BNr 8Nr2
Jcov[f1,f2]= 0.1597x10 8 x4.55xl0 5 x(-1.616xl0- 4 )
= -11.75x10 8 V[Nr ]= 3.94xl0 8 + 4.899x10 8 + 37.45xl0 8
= 2.28xl0 9
-
2x(l 1.75xl0 8 )
10-10. Serviceability index PSI= P1 - 1.91 log(1 + SV)- 1.38 RD 2 - 0.01
JC
(a) The expected value of PSI using second order Taylor series 1
Efg] = g(µ)+2
n
a2
i=l
Oxi
L -{ v[xi]
oPSI = _ 1.91 (-1-) = _ 1.91 (-1-) = _ 0 323 oSV l+SV lnlO 1+1.572 In 10 . 2
8 PSI=
asv 2
1.91 (-1-) = 1.91 (-1-) = O 125 (1+sv) 2 InIO (1+1.572)2 1n10 ·
oPSl
02
oRD
oRD 2
- - = -1.38x2xRD = 1.38x2x0.284 = -0.784
PS I = -1.38x2 = -2. 76
1 oPSI = -0.01 (!) _l_ = -0.01(!)--=0.000456
ac
02
JC,
2
2
.fl20
1 PSI =-0.01(_!_)(-_!_)--=0.01(_!_) 2
ac
2
2
.J2i
4
l
J12o
3
=0.0000019
E[PSI] = 4.2 - 1.91 log(1 +1.572) - 1.38 x (0.284) 2 - o.01 J120 + 0.5(0 x 0.06 + 0.125 x 0.32 -2.76 x 0.003 + 0.0000019 x 0) =3.211 (b) Variance
V(gl =
~(!, )'vrx.J
V[PSI] = 1 x 0.06 + (-0.323) 2 x 0.32 + (-0.784) 2 x 0.003 + (0.000456) 2 x O= 0.09522 (c) Terminal serviceability index for 95 % reliability From Table 8.1, u
=1.645 for 95% reliability
(PSI)T -E[PSI]
~V[PSI]
= -1.
645
(PSI)T -3.211 .Jo.04522
=-1. 645
or
(PSl)T
=2. 703
2 2 10-11. Given g = l - r + +r + ( rX ) Find V[g] when r=1, C[r]=0.2 1+ r 3 + 6r + r 2 4 + r 2 6 + r
V[r] = (0.2 x 1)2 = 0.04
og
-=
Or
(1+r)(-2)-(1-2r)(l) (3 + 6r + r 2)(1)-(2 + r)(6 + 2r) (4 + r 2)(6 + r)(l)- r(4+12r + 3r 2 ) + +---------(l+r) 2 (3+6r+r 2) 2 [(4+r 2)(6+r)] 2
For r = 1,
ag = -~ + 0.14 + 0.0131 =-0.5969 Or
zr 4
V[g] = (
2
V[ r l = (--0.5969) (0.04) = 0.01425
10-12. Given g =
1 - 2r - s 1+ rs
+
r+s 3 + 6s + r
2
+
rs ( r + s) 2
,
Find V[g] when r = v[r] =10, s = V[s] =20
0g (l+rs)(-2)-(1-2r-s)(s) (3+6s+r 2 )(1)-(r+s)(2r) (r+s) 2 (s)-rs(2r+2s) _ -= + + 0r (l+rs) 2 (3+6s+r 2) 2 (r+s) 2 (r+s) 2 (1+200)(-2)-(1-20-20)(20)
-'------'--'"-'---'-----''--'----~+
(1+20xl0)
2 (3+6x20+10 2 )-(30)x2xl0 +...;__ (10_+ 20) x20-20xl0(2xl0 + 2x20) ___;__ _ _ _.;___ _ _
~
2
3+120+100)
2
2
(30) (30)
= 0.00936- 0.00758 + 0.00741 = 0.00919
0g (1+rs)(-1)-(1-2r- s)(r) (3 + 6s + r -= + 2 Os
= -201-(-39)(10) + 223-180 2 (223)
900xl0- 200(40 + 20) 900x900
= 0.001839
=(Zr v[rJ+(:)' v[s]
V[g] = (0.00919) 2x10 + (0.001839) 2 x20 = 0.00139
(r + s) 2 r - rs(2s + 2r)
+...;_--'-~---'-~~
+-~~~~~-
= 0.004678 + 0.000865-0.003704
V(g)
)(1)- (r + s)x6
(3+6s+r 2 ) 2
(l+rs)
(201)2
2
(r+s) 2 (r+s) 2
2
Chapter 11 Flexible Pavement Design 11-1 Find moisture content at Points A and B. From Eq. 11.2
~
ft
3
A=
2ft 2
fa-
ft.
3fe
~
u
sa,,4_ /'=
u = S + ap
=-z Yw
S = -z Yw - ap
IZo~cf
C/'t'j
tr= /oo r~f'
'""wT
At Point A with sand, a= 0, S = -z Yw = -7 x 62.4 = -463.8 psf = - 3.033 psi From Fig. P11.1
w= 10 %
At Point B with clay, a= 0.5, P = 5 x 120 + 2 x 100 = 800 psf S = -3 x 62.4 --0.5 x 800 = -587.2 psf = -4.08 psi From Fig. 11.1
w=21 %
11-2. Estimate the coefficients of permanent deformation (a, b, and m) in the Ohio State Model in Eq. 11.16 based on the test results in Figs. 11.4 and 11.5.
Let A= {
~:
r
or log A =log a -
blog(~: J
From Fig. 11.5, when A= 0.0002, MR/crd
=104 , when A = 0.0006, MR/ad = 3000
Log(0.0002) =log a - b log(104 ) or
-3.69897 =log a - 4 b
Log(0.0006) =log a - b 109(3000) or
-3.22185 =log a - 3.47712 b
Solving a and b
a= 0.89324, b = 0.9125
Fig. 11.4 shows a straight line relationship between log(cp/N), thus m is a constant, which is equal to the slope of the straight line From Fig. 11.4 5
m = log(2xl0- )- log(2xl0logl0,000-log70
11-3. Given:
Nt
7
)
= _ 0 79 -
=5 x 1o-e (ctf3, ct= 0.0005, and C[Nt] =0.8.
Find: Percent area cracked if the pavement is subjected to 5000 thermal cycles.
Nt= 5 x 10-e(0.0005f3
=40,000
V[Nt] = (0.8 x 40,000) 2 = 1.024 x 109 Cracking Index
c = n/ Nt = 5000140,000 = 0.125 2
V(c]=[- ( n Nr
V[logc]=
S1ogc
f]
2
V[Nr]=[-
0.1~ 86 V[c ]= c
t
SOOO (40,000
188
0.125
f]
9
xl.024xl0 =0.01
~2 x0.01=0.1207
= .J0.1207 = 0.3474
loge= log(0.125)= -0.903 Z= 0+0.903 0.3474 From Table 10.1, Area under normal curve = 0.5 + 0.495 or Reliability= 99.5 % Or Area cracked= 100 - 99.5 = 0.5 %
(b}
B.J =- /. >-t o. f1 /if ;;5 11' ~
53 :::
J
J: 31 > A
(3 t :; (') .tfk3 J(. I/ =0
:. I- 7'1-'l7b /
J.f}¥ ;3 +- "· oZ.$-¥z.
J )( .f
-o. 170,;. )C.
-o oz.7?J-
>/ ( f)
-t- o. '1JI J
.
+ ~"'7°}7JX z._)
(&j j,
-
t;, o }f?
.,j,,
't
' V~
= I, 71 of;/:,-- Q.: I. .Jll'f'l-Z.J' j .
~I. 71
1""
br ...,.je:r-
fL ..)v~ce.. p.>~
9--. 2- )4. Io
~. ~ ""10 J'
,
I/,;;___
[,,,,,c:/~~-
11-S.
(t:t)
1=S~L =('4z:>r~ (T)( ~ )((7-)(Z>)(L)CJ(,~)(''()
'F>-a,.,, '=1: t.Jo
C,..111~,., : (A2'7)a C 7) = 18 8S~ (D) ( L)
= o.y..
a. o ~) .::., 0
o.S-2.
X
o
t-z.. ( .f"~e ~6/-e?
( c5-ee Jqb/~ 6. IS'_),
( <.;-) C'() .::
f; : :. I
20
Jt-
~
.fr°'? EJ-: 6.33
= 2 9 ~ 7~
-
2 9, 78 >< o,
4,.,~
X
365°
=
9"-z6
6. 10)
X /() &
Cb)
JI.I,,~ ?} C5 4SL
••
u,,/-r-e.,.,4-1_ Cfj'JT"-e"j~ bq.>e
cf
= 9-: z' )(
/o:
J/. JI
-J--f;,e s~6.s:A-1uf1~"'J 7tr l;..q.
e"Jq/s/fi~
;:;r
~re~
i.5
erspfuJt JI.
o,,
€>F
?:
11.1z
tf'7..../.,
/C)
~
I •' H-1-1.e; +o I
I~ I
<::je./erPJ1',-;e. 11
°..f
/cy~ ..[
,w;,'X
'-¥ f'q1)-q~
-7iJ< #i4
HM/;
T-i} I/. I z. > ~ co,.,61;,~ -Tl>.,-~K~J..5 d:_F 1-iNP.J
<1% ~h'JGlh-;p·~.:1... Cf->?41Jl
/f?1';-c,.
t's
~$.5~.H'J'J~ "- ~l~IWJ'I.-, H H~
4.F'
$u/,sl,·"1.,,.f,~"J ;fq~~
::
I 2 -
11.s--
I 2
=(
10-
Z
2
t:f
* 8,s -
z)x I t:JS-3
/'J.
2 ,,., '
u.se Z '~ a:f' H HA 1 -161~1<~
kt~
I?
11.
~.1,:f'~ .:j~tl
'~-
' :I~ vx .J,5 ;~ ~F >IH/-1/ -:JiJ,c_k,,~$S .::>:f" ~u/5'.:f'i~~
= (Io- 3,5)
4.Jj>4:fIt &4J~ 11-6.
v-lz.tt.f
&,.A/Jevc .J,.,,_
d~J-' 3 r4 .tfo--J ,,
cK, ""
= I IJ :=
/,().f.] :: 7 ;'1 .
:
s-cf1<.1'
I ~t vi-fo-;1. ::
X,
J~u
rlrwlf:
~hrr
I:::
3~dJZ.l'
I ~ (/ fi!../
~:::::; CH
/l'---' u. ;;> J
>
. ·_
/ J't lf'-f~~ (
o-'Z A4
.J' r4
~t.
{ I f-0.0 SJ) - I
1FJ/I)
( /./:1~1)
cffiJ( /.?
lJ
J';eJ')
_
o.osr (J.+ /). ()JJ°') .::
I.I'
I
-,.
_o.O~
::
-
=..?tJ d10 /:J'/ll
.lt /{ ArS'•/ )
U.
'
?s~,,
11.s- /"'.
~~W'J
'I:F
l'11e= lo
,,
f'~I'>') )-~
hY~l .. u..>~ F1JiJY--e.S
CC.)
JtJ-
)t .
-rA.1
f 1'21'1 cl'~.< : 3C, :; ..?tJoco
JJ-. ..
S:36
~ I~ ~OD /&o
ft? =
//.tr :
/;
l(
-
;ft
/ / . 10
0
~OD
-_'5"-
~
/
=~,.<1 lroo :=
a s. . --.z . .!( /() 0
I
~24 /;,,
11-7
·C(.Z..:.
SN= ;/.
CJ.LlfJ
Ap 11,.>Lo- ~
o,
o<:Jlb
8' x 4.'f8 + 2 XO. c.j/l == 1-:6
/(A/Jr :7~"T p4k,..,,,~
~.k~
Cr?!
.:/A~
¢~L
J.
/l)c~,,,.->f
#cAh
-
//.,-"4'~>/
~~Jul .fa f
' /YJ_ :=
~,,('~
rt'.A· 4~"' /
/:unlAp .PT&"'dvb ( 7~4. //.
//J
J.
.Ph·. ,,.,,cdv~~
Mwilf.. NR {Pfi)
JllN
.2'}
yoo
p/J
,J/:,Jt)IJ
/iM.
4.liCO
Ut IJ. ~/J
.ro f
tJ.()D]I/
() .P/)
.z..c-q
11.f/?.
SODPD
lJ.b () I rf'1
f'/111
J .rt]oo
/).().,(,lr./J
JVM
1.f100
~.tJ1:;-8<1
j,,iL/
JfiS°OD ·
o.o 1V1'16
1/116.
/9KOo
o.o 1z.fy
/1-f;e_ =
/1
~ol. I o · ~
.1lfrr. l1t.f.
.21.l..OO
0.0108>
.'l
()./)/) yf/y
AfCV.
&},Z./'"IJO
O.OD94'/
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..2u01>
(). () I)
z
tJ. il)J.t.
.,-)'()"1 /'•
°fat6/e I~. /
I
·- /.~ 104'/
kJ;.;-~ Z
v
;J
= I./ 6
J
~ qy~
C~.,, Yf:. > q .5 s ~ ~w >7 b:J- -J-?-e. C71J.$5 ~-f.ch'!r o.Jr7::. a,B7/ &J'Y 88 ~o
t.; r;cJer
~-e
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'7 c>-l'Y)"fl
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J]ir-.; ,._"'(/' :
_6v_,,fJ'
,,,,.,,/,-e~rA.·y
\4' -k.A?....:vic">• .
JM4 ·· !'I~ : -'f".r'j( I 0 v~/',:/>" G:'78.. . .
-
6'~ "" I evt d'i-1 .i. .J &1-h
A
4.
~.J.
z-fl.
.i-.....4 · /'11t. ~ .JiFo?J
' -rA1c.l.-1.I'n
=~.107
£
,HI' ffi.,.,-d
U-'I
II· y'6./_
~.6(2,~4
/JI\/
C!..7(3
~j
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~-'I
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~.6 .b.ei.;-e :
;;;.v11-. =- /. :i )( 10 t ..r.;-Cb 2
l _
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t/s
ro /Jiltn~.rr
d?.foCfl/'h·.6k
&>t..3 :: 3"' ~- .;.; -+
I I -11.
o/"·
l
;-1-H A :
tJ/i/'v'1~
=
{
I
r.n ·
s-11/"'- = a~ ~, + di'{ .lJo{
J.f'.11
~
lN"Aj"''.:f" tl.rr/j -= CJ. J'o ( 7 ~ (, t/. ~)
AAJ'llTt.J
/11'{ ::
:. CJ.
'
d3 = tl.Uj(/f/ Jl-'l)-IJ.if .l:J 42
}) . / ;.
.a., = o. ~v. /I~ =d.,r)(/o.r /rr ·
. .
JV. 1 /Z: c
-
=
C',,(4(/;- _.lfll,=t
))..:;_ x r CA
6
I/}
/
s. I
ti N1 = /I, '[)-,
t- .Ii'- 'PL.. /71, -t 4.7 'i>; />1 J J'./ ; . ~- r(x 31- 6 )( CJ. 11 4- 0. icr.J )/,. lJs .( o. 8.
I
7 ""
. R:5~)!,. 6 '])3 :: g. 9 ;/,, ~~J.'-e,,"" ~hi -= ..> x 10 .. H~ =..s: ~~ r.s '° 4;.::L ~f>s.z ~ i 7 PDl'>J ;i;_y;_ l),.Z.J' SN fe>- 1>?<:1;,, /,,;"ff: j:qve"'e" t :: 1- 2 . . ; "tr. G-1v~'J Pe =2.J"fa 4: ~: ~ aZ. .% , ~~= Sx10" (~oz.>+- ~o0o<;J = O, 12 6 >< /O ,, -f"ro"? F1?. 11. ZS'
-
...5N -fo;Y .s~t./-ey P17"C,. :::: 2' St-
Chapter 12 Rigid Pavement Design 12-1. Given: a concrete pavement on untreated subbase without concrete shoulders. h 10 in., k 150 pci, and n 2 x 106 repetitions.
=
=
=
Estimate: Allowable comer deflection. From Eq. 12.9
~C 2 n·1 % erosion damage = 100 L..J--
N-1
When damage is 100 %, allowable number of repetitions N = C2n C2= 0.06 for pavements with concrete shoulders, so
N = 0. 06 x 2 x 106 = 0.12 x 106 From Eq. 12.7
log N
= 14.524 - 6.777(C 1P - 9.0)0·103
C1 = 1 for untreated subbase log (0.12x106 )= 14.524 - 6.777(P - 9.0)0·103 P
=34.0946
From Eq. 12.8
p2 p = 268.7 hk:0.73
p=
Phk
0 73 · =
268.7
34.0946xl0x(l50)
0 73 · =
268.7
si _ 7 014 p
7 014 w =p= · = 0.04676 in. k 150 12-2. Same as Problem 12-1 but with concrete shoulders. C2
=0. 94, so N = 0. 94 x 2 x 106 = 1. 88 x 106
Log(1.88 x 106 ) = 14.524 - 6. 777(P - 9.0) 0·103 P = 15.7485 From Eq. 12.8 p = 4.767 f'..S".A-
or
w:::: ~767/1So
12-3. Total number of trucks during design period on design lane
=2500 x 365 x 20 x 0.35 x 0.5=3,193,750
=o.~318 ;~.
Calculation of Pavement Thickness Project_......P_R>_J, ___ }e_,.,, _____!Z._-_3 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __
B ___
Trial thickness_ _ _ Subbase-subgrade k
~
Modulus of rupture, MR
~
Load safety factor, LSF
I·/
in
Doweled joints:
y e s / no _ _
pci
Concrete shoulder:
Yes _ _ no.../
psi
Design period
Fatigue analysis Axle load kips 1
Multiplied by LSF 2
Expected repetitions
Allowable repetitions
~
years
Erosion analysis
Fatigue damage
Allowable repetitions
Y.
5
4
3
Y. 7
6
8. Equivalent stress :Z 'rZ... 10. Erosion 9. Stress ratio factor
Single Axles
factor
:Z~$--o
~09000
0.3
~, o'° a
CJ~4-
o.JlZ-
2B
30.8
·z&7lf'
I~ c6CJ
1.5"-I
7J..
zx.6
bOb~
-4-S:ae:sD
I
Z-if'
Zb·f-
/~a,oO'D
7.Z...
z.ir. -z.,
7"!4-b tt;j,8zJ
13..> 1-. ft::,
X'Ot>~o
6,"Z,
3,
zo
i2.o
-zo'J,
I~
r1 ..8
Jb
t 7.lo
15J~x'a 4--J ~. t>£ z....
Tandem Axles
o/8
S-z.~
Erosion damage
I
-
J 6oa~c8()
-
4C>OOOt>oo ,
z.o$ 13. factor ~. '3Z..O
11. Equivalent stress
Erosion
12. Stress ratio
factor
/ t0,, 600001. /
~.3
i.rg. 'f'
53,S-39-
-
l.Lociaot>
4-o
4-
-
-Z~4'°
']b ]Z
31.b
.>J. t.>.> .Sl.f. '733
-
,<-;/co aoo
J,>:Z,
76,bSO
7J8
3o.B
;019,~br-
1-Cf
Z.b.4--
zr61J9
-
-
-
-z.. ~ 3 ~?/
7~~0
~
Total
1-4Z.9 z' z_,,,, ;.a
s--~ooo
7, .ztj~()CJ<::J
-
3o. 6,bo
6,4-
Z,/Ot:)e>~O
2.1
z ./
I 2," "~ t)O 0
/,0 .1>,k,
~~z:,oC
o.~
-
39,7
Total
I
7· 9
12-4. Determine the thickness of a concrete pavement using the PCA procedure and check the result by the simplified procedure. Trial thickness = 8.5 in. K = 150 pci MR= 650 psi LSF 1.2
Doweled joints: yes Shoulders: yes Design period = 20 years
=
Total no. of trucks on the design lane during the design period= 3460 x 365 x 20 =25,258,000
Single Axle
-
30 28 26 24 22 20 18 16 14 12
36 33.6 31.2 28.8 26.4 24 21.6 19.2 16.8 14.4
11,383 21,500 45,024 131,784 198,562 413,059 636,157 804,871 1,207,306 4,604,105
110,00C 300,00( 1,750,00C
11.4 7.17 2.57
680,00C 1,200,00C 2,000,00C 5,000,00C 15,000,00C
1.67 1.79 2.25 2.63 1.32
Eq. Stress= 191 Stress Ratio F = O.29 Erosion Factor= 2.32
680,00C 1,300,00C 3,000,00Cl 8,000,00Cl
4.43 5.66 6.75 6.74
Eq. Stress 166 Stress Ratio F = 0.26 Erosion Factor= 2.45
Tandem Axles ..
52 48 44 40 36 32 28 24 20 16
62.4 57.6" 52.8 48 43.2 38.4 33.6 28.8 24 19.2
30,100 73,607 202,609 539,026 1,422,816 2,621,269 3,066,199 1,834,863 2,173,809 ? 51? 71:.A
Total
21.1
=
33.2 -~
For h = 8.0 in., the fatigue damage is greater than 100 %, so h = 8.5 in. For the simplified procedure, from Table 12.15, the allowable ADTT is 5, 700 for an 8-in. slab with medium support, which is greater than the actual ADTT of 3460, so the thickness required is 8 in.
-
12-5. Same as Problem 12-4 but without concrete shoulder Trial thickness = 9.5 in. Doweled joints: yes K =150 pci Concrete shoulders: no MR= 650 psi Design period = 20 years LSF = 1.2 Total no. of trucks on the design lane during the design period= 25,258,000 From Table 12.15, a thickness of 9.5 in. is required. A 9-in. slab is not adequate because an allowable ADTT of 2700 is smaller than the actual ADTT of 3460. Trial thickness = 9.5 in. Load Multiplie< Expected Fatiaue Analvsis kips by LSF Repetition Allowable RI % Damage
Erosion Analvsis Allowable RJ% Damage
Single Axle 30 28 26 24
22 20 18 16 14 12
36 33.6 31.2 28.8 26.4 24 21.6 19.2 16.8 14.4
11,383 40,000 21,500 120,000 45,024 400,000 131,784 3,000,000 198,562 413,059 636,157 804,871 1,207,306 4,604,105
28.5 17.92 11.26 4.4
1,400,000 2,000,000 3,400,000 5,800,000 10,000,000 21,000,000 60,000,000
0.81 1.08 1.32 2.27 1.99 1.97 1.06
0.75
900,000 1,400,000 2,100,000 4,000,000 8,000,000 20,000,000 80,000,000
3.34 5.26 9.65 13.48 17.79 13.11 3.83
Eq. Stress= 200 Stress Ratio F =0.308 Erosion Factor =2.59
Tandem Axles '
52 48 44 40 36 32 28 24
20 16
62.4 57.6 52.8 48 43.2 38.4 33.6 28.8 24 19.2
30,100 73,607 202,609 539,026 1,422,816 2,621,269 3,066,199 1,834,863 2, 173,809 2 512 756
4,000,000
Total
62.83 ,.---!!'~
....
76.96 -----~
Eq. Stress =183 Stress Ratio F=0.282 Erosion Factor=2. 775
.A
rJ.f.5U"'J-e
V~· 5"
1 =
j1e41r_J
~
'~-S {<1. 4 ?-J
-iJ =/. "98
~?::: t-~/o
1
· o,6 A.PSI= o,c8 (~. .>) .: o. 289
~p.5I
6
X/O
=Z-o,Z89= I· 7 II
6 J--t1. W : 6.11f7t- rJ 87~ l-s:; .:.2!.l : 6. /JZ.~ ~>' W1.<,::.. /. 3/;,X/O ~r 18 ' ~~ 3 I
f:IJ
. /. 36 ~ 10' = S x /o"
f
=
~ /.l..72
..!"} /, D 'f
1#.5"~~
y = I
[ ( /, ()
= 6, ~.
~
z...
'I
'I-') - I
z.
J
J
(!, o
o "r
I . z.J Z-
je-9>'..!
y-eqJ'.S
~ b =f"X/o 6 [ ( /,dlf),,,Z..- I = /,38>1ts =: /. 3 ~,.. ,0" J.
w,~ .A
7k
,
.
~
t:1..G./tJ4)
'E5AL. ~f / ..JSx/a~
is
.s~,~f~ ~b?-fJ)~>
-J-~~"' +Ix ~J)oW46J~ l:56l of / . .3., X /t:J' So ,P~;;·6r::l ~.;
'' Z
fa"r>".S JS ~sfado?'~
q
~'fiYW/4•1~~
12-7. Given a concrete pavement with Sc= 650 psi, J = 3.3, ~PSI = 4.5 -2.0 = 2.5, and poor drainage with 5% of time near saturation. Determine W 18 by Eq. 12.21 without the reliability term and compare the result with Figure 11.25. From Table 12.20, C = 0.9 for poor drainage with 5% of time near saturation. With Dse= 8 in., MR= 5000 psi, Ese = 30,000 psi, from Figure 12.18, the composite modulus of subgrade reaction k = 350 pci. From Eq. 12.21
log W18 =7.35log(8+1)- 0.06 +
1og(-2.5J 3
7+
l + l.624xl0 (8+1)8.46
1
650x0.9 s0·75 -1.132) (4.22 - 0.32x2.0 )log -------'~.____ ___,____
215.63x3.2 g0.75 -(
w18 = 7.01368- 0.06 W1a =8,597A13 log
18.:2J0.25 4xl0 350
0.069621 + 0.050309 = 6.9343678
By assuming a reliability of 50% and drawing a line along the scale of overall standard deviation, from the AASHTO design chart, W 1a = 8.5 x 106 . Both solutions check very closely.
12-8. Given D = 8.5 in. and the subgrade modulus, MR, for each month Determine the effective modulus, kerr, by assuming k = MR /19.4 Ur= (D0.75 _ 0 _39 ko.25)3.24 The computed k and Ur are listed below:
MR
Month JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC
(psi)
k (pci)
15,900 27,300 38,700 50,000 900 1,620 2,340 3,060 3,780 4,500 4,500 4,500
819.6 1407.2 1994.8 2577.3 46.4 83.5 120.6 157.7 194.8 232.0 232.0 232.0
Sum
Ur (%) 31.2 21.8 16.4 12.9 86.4 75.5 68.5 63.2 59.0 55.6 55.6 55.6 601.8
50.15 = [8.5°· 75 - 0.39 (kett) 0·25] 3·24 or kett = 305 pci
Ur= 60.18/12 = 50.15
12-9. Same as Problem 12-8 except that the relationship between MR and k is
with E = Ee = 4 x106 ,
Vt=
0.45, and h = D = 8.5 in.
The computed k and Ur are listed below:
Month JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC
Ur= 853.9/12 = 71.16
MR (psi)
k (pci)
15,900 27,300 38,700 50,000 900 1,620 2,340 3,060 3,780 4,500 4,500 4,500
353.0 725.7 1155.6 1626.2 7.7 16.8 27.4 39.2 52.0 65.6 65.6 65.6
Ur (%)
47.3 33.4 25.1 19.5 115.3 103.6 95.6 89.4 84.4 80.1 80.1 80.1 853.9
Sum 71.16 = [8.5°· 75 - 0.39 (kett) 025] 3 ·24 or kett = 105 pci
I Z-10. {_cz,)
ue.fe,-,.,,,,,",;-e -!-);/<-/
= 7,,,.. r.O<:::~.,
nv
11 '-1
-r:: = 4-x/a c<:
..>
6
c
;:>5~
=-
0c ;
R~ 1Sfa
J. c;f,
c4.,:.
.q,,..i .a.ps~:..
.,-1;.. ~k~..>.$
v ::
l/j.F Skb
')(./0
6 ,
.
9.
s
I~.
/.x:rY per-
...)
2-~r /4r;-e
I
Wl,e~/ ~c:L. 4<.J-e- -lo Ci:?'>.5..f;'/c .f,,~
lb,
k: 75;
41?-l
~~ ;:;~
12.,
z. z.
2JS ?5~
=
"'/~'-
;'?~
= J ~ .O~Q
-f>'"fff,<.. c5~
= 9. 5
D
"'E5AL .::: /o
z. '?
f>-o,-,,, ;=,/'-'re / z. 1?
Z,,
(b} v~krm;,,~ l'J<.N'/?6€-~ of ,<.)~• (i:p..J.e,.,
~ = t'), 3_,
I
6,s-o ;~>~ _, J::.
M117;,.,VW')
a<"
QMIJ(..t'1i
7
...Ske/ )':),,,,-:J •
x
:foJlo"";"'r ..;h-re-e C¥ik~,~ :
_saJ,·sfy
#'JU$l
-i-J,e
< 8 .ft , cw<~ o Y.._ ;,., ~ 4 ~
<5s < 6z~~a P..SG
&111--e,,., ~
o<.c:::
~$
:ic
=-
~.at, ~ _.. .fsr P.f~ ~-== s;< /;6/
o...z::
J
/a-'-/ 0~
(
L,~..>kl'J'e Ca::,r~ ~-t~f'<~f~~ J;i,k
IZ.
ep= f ,:, _
=- ~-•z.>1~. z:~~~J~ k1z,.z..,J FrOIV) t:~ 1Z. J3 ( i>ef5~ ~ X = & {t.) c ~.S ~·~j~' r;.
~,'r]
-.:::
/. o&, Z t:J
( 9)
=
z
1
1i,
--
I
( I + µ;;;; ) -I
-
'2'f6S"
o.~.>ff (
.=.
J, '+.$7(
z.3s
i>,u7
~. 9..,.,7
GJ-:
fro,.,, Pm11-J
..>5'9
(lr ~)
( 6::,.;~-1..
J t" {;e>
(~.~¥-)'
..,.. f"'IO -
1.1.3
~.
o,,
2
~
(I+ e>,~zs-./ I
/o
CW=
~~o (l+~)/,07~
85'4-
•
'
:'\
e>J~ (.. l +- I ooo.,,. ~ 04o;JI:,)
'°·
0
}'-
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I~ o.~186 - I ::: ~. 38
Q.
)
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z.-3
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m4..X/ 1J7'1n:J
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Gt.
e.;..
-;:YlJM
=
f:..
Cl'1~'-'? t ~.f' Ste e I ~..x
12 I
( &.>~.::/_, e;~
"J 3
x :-
-Ft'
3 ..>ft
/, ,.3 {" (). "'l..5 . t> S'-76 1.~4,z( J·S-5'7) ( /.-1,58) ( /. &z5) ·
lf)~X
CJ. S )~' 'Z/7 2. 8 71 b
-
-
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"""'' WJ"HJ
( /, 3') a.?$-9
-
I
0/ /0
.
~ 1z7 3 x ()• I/i•
t:?
~
l/Jc
( /, Z
k
Z-
>< 12)(1
o:f Jg
z >< 9• .,;;),../ro.tz.s)
h4rs
Pe'!"
= I}. 8,
12-ft )<1'1-C ~
N,_,,,z
= ~."I z73 x o.S-3 x IZ.,, 12.,, '1· >,/('6,6z..f) =2 3, 6
u~
.q,.
~~~"'"'? ~ 2}'- b;y$ ~, /2-f't
/'91")-'e ·
Calculation of Pavement Thickness
~-----------~
Project _ _ ,/)._'ro_b_}2_n?__1_z_-_l_I_______
a·. s-
Trial thickness _ _ _ _ __
Doweled joints:
yes
Subbase-subgrade k___ ; J"l_c:::>_ _ pci
Concrete shoulder:
Yes- - n o 7
Modulus of rupture, MR_6_J-o __ psi
Design period
in
z.o
/
no
years
Load safety factor, LSF__ J._Z....._ Fatigue analysis Axle load kips
Multiplied by LSF
Expected repetitions
Allowable repetitions
3
4
2
1
Fatigue damage
r.
Erosion analysis Allowable repetitions
s
10. Erosion
9. Stress ratio factor o.Jb
Single Axles
7
3o 28
3b 'JJ..b
z_b
'JI. -z_,
z'f'
2-~·8
uJz,,
z_Z-
z.b. 4-
3%6
_360,.o"D
~o
2-~0
$z.'1-'7
4o•o.oo 0
(~ }/:,
z../.b
) ~ 70.t 1~079-
t \f-
1.i;. "/_,, 1b.~
IV
I
'f. '/"
"71194-;J
r.
7
6
8. Equivalent stress 2} f
Erosion damage
factor z,7_.3
Z,fdD
':),5
b/O,, ooo
9-Z'f
~OtJO
t:-+-
£-.J_-~oo't)
o.o O•/
9o0
z'Jo•O
3.:7
jI
q.o 0, ()0 f.)
o.J
5"·1
O./
1~3
z , /000&0 , z, 'f co,, t:Jl!JO
o, -z.,...
/.&i~DOO
,i..Z
2 'f_ /I)
f-2
-
(14'0
-
1 'r ooo. ()trt:
-
"3 '9 0 00, ootl>
-
-
0 ,Z..,
If!)./ o-1 d,O
-
-
11. Equivalent stres~ 12. Stress ratio
Tandem Axles
.>z.
h'J.:4-
t:r17
4-'r 4-0
l"7~,,e1~0
J-7.6
60/ J.t; 70
fZ.
4o(fh
J~,~o.ooo
4g-.o
/C>, 765'
31:>
4-J. z,.,
z8, '1''5
3z.. -i-5
?l?·'t-
j-i,,3.)0
~J.h
1,1,z.35'
VJ-' -z,,.o
zff.~
If'
hifl.f' 1-~. 'f-13
lb
/tf.Z/
..ta,183
dl,
j /,
x~~oo
-
zo g 13. Erosion factor o.J-zfactor Z • 90 e>J O· I 4-30000 CJ, 7 /
0
/,~oOO
/, 'joo,coo
-
J~~~
-
f?.
11) 00.
t>CI'{)
Zt>, tt>OO, 00 0
-
/~~~c.oO
-
',2,;
0
c.4-
I!)."
a.7 0,7
0.-;, 0
-
Total
oJ~ooo
-
ZS::J
Total
3.J
Calculation of Pavement Thickness Project
_p_"4_~_6_k_"? __l_;(_-_I_~_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ 7_____ k ---<-J-'J-c'---<=> __
Trial thickness _ _ _
in.
Subbase-subgrade
pci
Modulus of rupture. MR Load safety factor, LSF
6
$"b
Doweled joints: Concrete shoulder:
Design period _2.0 _ _ years
psi
I· Z..
Fatigue analysis Axle load, kips
Multiplied by LSF
1
2
Expected repetitions
3
Fatigue, percent
Allowable repetitions
Damage, percent
4
5
6
7
9. Stress ratio factor
z. z.,~
Erosion analysis
Allowable repetitions
8. Equivalent stress
Single Axles
4
yes no - .yes _ _ no _ _
z.lf- 'i? L>.) f>Z/
7Jo :z.. kcri>
10. Erosion factor
/ ..f?;!.
6 <7-e:J
z.rfa-
CJ. I
o. z_O 6. z."if 0, I
~
o~
4.o
/e:>O~
to~/~
/.
/.. C> d&J! t:J·..:rO
:z,s-tr0~
c;_ tJ.
2,bZ-
11. Equivalent stress
Tandem Axles
13. Erosion factor - - - - -
12. Stress ratio factor
Total
7
.0
Total
Chapter 13 Design of Overlays
13-1. (a.) YJ;fh 1>1eq'J ;CXf>i~rq.J">-e of ~ --';:: q,1!.
fr:,_s~ 45 &-1~1? ~l"Y4...
I;;;._
~ =
;::
/~, 0 .
tt:>.J9- ,,;,,
H/'1.11 OV<*rky
2;
1J,
'°°°' ~-Q ,
/.$ )1l. 7c X6.4-
=16 Bc;.::i
( [
J- [
j
Pf~
Er, 1J.10
4
°. ~ z
01
~.
Ta.~ (
-
~. ~ ~ J'i
P s /.7-L = /~ .f~ '9. ?a?; w6~ /s ~.,
..f-/,;>e
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c-Q
I a ~o oao #
;
e>
5;v e~
So
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z
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J
q. 15,'9 J6
8-:to
} )( $'~ ~~
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sl.9J;:f}_y
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;:
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:: 1.5 x /ox &. 4Z.
s.:
=
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Gr.s
cf;e~x. by
4r?
+ z_x
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(
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ct-
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1
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he
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==
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fe
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= S-, 6 9-
<:1~4..- ""'-' -/-e,,,,peJ'lt:f..J.~;.-<:
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;: "·coo..)(,
:::
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oyterl~y :H;.-~/(~$
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Upper Saddle River, NJ 07458 www.prenhall.com
11
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