SS9901-CS040727 9?2 ?55 5002
· ~LIL 27 2004 16:56 FR TURNER COLLIE& BRADEN9?2 ?55 5002 TO 99?250435?0 P.01/06
Turner Collie Braden Inc
Engineers · Planners - Project Managers
17300 Dallas Parkway
Suite 1010
Dallas, Texas 75248
972-735-3000
Fax: 972/735-3001
DATE: July 27, 2004
TO: Ken Griffin, P E.
City of Coppell
Erika Cooper
# OF PAGES
FAX NUMBER: 972-304-3570
PHONE NUMBER: 972/735-3056
(if tranami~al i$ incomplete or unclear, please
call sender directly at this number.)
REMARKS:
K~n,
Per your request, please find attaohed the design calculations for the HDPE on Line D.
Design calculations for three scenarios are provided: 1) during installation, 2) long-
term service, and 3) repair of piping. Under all three scenarios, the maximum external
pressure exerted on the pipe is less than the critical buckling pressure for the DR 17
HDPE pipe material.
Please feel free to contact me at 972-735-3056 if you need any additional information.
Thank you.
Edka Cooper
An AECOM Company
3LIL 2? 2004 1G:3G FR TURNER COLLIE~ BR~DENg?2 ?35 3002 TO 99723043570 P.02/06
Horizontal Directional Drilling
Desigll Calculations - Installation using DF[ 17.0 at 100psi
1~) Calculate maximum pull force (TD)-
Assume L1
is 100fi, approx. 1000 ff bore
THK = 5026.5 Ibf v~ = 0.30
L, = 100 ft Wa = 77.861bf/ft
L~ = 95 fl Wa = 74.00 Ibf/ft
L3 = 800 ff o = 0.1745
L4 = 115 ff ~ = 0.2094
H = 20 fl Do = 32.00 in
%- 0 5 D~= 2801 in
Ts=
Tc =
T~ =
T~K [] hydrokinetic force in lbs
THK = q X ~z/8 (Dbh2'-D2)
q = hydrokinetic pressure in psi (typically 5 to 10 psi, use 10 psi)
Dbh = borehole diameter (assume 48 in)
D. = pipe outside diameter (32 in)
T~K = 502.6 5 lbs
L~ -- additional length of pipe required for handling thermal contraction, ft (see sketch)
Lz = horizontal distance to achieve desired depth, ff (see sketch)
L3 = horizontal distance traversed at desired depth, ff (see sketch)
Lc = horizontal distance to rise to surface, ft (see sketch)
H = depth of borehoie from ground surface, ft
v~ = coefficient of friction applk;able at the surface before pipe enters borehole (0.5)
vu = coefficient of friction applicable within lubricated borehole or after wet pipe exits (0.3)
c = borehole angle at pipe entry (drill exit angle, 10 degrees), radians (0.1745)
/~ = borehole angle at pipe exit (drill entry angle, 12 degrees), radians (0.2094)
Do [] pipe outside diameter, in (32 in)
D~ = pipe inside diameter, in (28 01 in)
wa = weight of pipe empty, Ibf/ft (77.86 Ib/ft)
w~> - net upward buoyant force on pipe in borehole. Ibffft
w~ = ~Do~/4-=D~¥c'4'w~ pipe diameters are in ft
% = specific weight of slurry, Ib/f¢ (75
¥~ = specific weight of water, Iblf¢ (62 4 Ib/fl~)
w, = 74.00 Ibf/ft
T. = pull force on pipe at point A, Ibf
T~, [] exp(vac~)(%w~(L¢'Lz®
TA = 47153 Ibf
T~ = pull force on pipe at point B,
Ts = exp(v~c0(TA+THx'l'%/w~L2+WbH-VaW~L2exp(v"O'))
Ts= 52291 Ibf
47,'i53 Ibf
52,291 Ibf
20,235 Ibf
lg,842 Ibf
Installation. - Line D Page 1 of 3 Printed on: 7127/2004
~UL ~? 2004 1G:37 FR TURHER COLLIE~ ~R~DEHg?2 ?35 3002 TO ~72304~570 P.O~/OG
Tc -- puli force on pipe at point C, Ibf
Tc = TB+T.K+vblwblL3-exp(va(z)(v.w~L3exp(vact))
Tc = 20235 Ibf
TO = pull force on pipe at point D, Ibf
To = exp(%~)(Tc+THK+vu'wu'L"-w~H'exp(vb~)(v~waL4exp(v~c0))
TD = 19842 Ibf
2.) Calculate safe pull force (FS).
Allowable pull stress is 1150 psi for a 12 hr pull, E = 57,500 psi
Maximum radius of curvature is approx. 400 ft (4800 in)
FS = ~o~¢Do( 1 ~D R - lID R2)
~p: o.llo~ - EDoj2r
o~ ~ 958.3 psi
FS = 170,682 Ibf
Calculate critical unconstrained buckling pressure
E = 57500 psi
p = 0.45
DR = 17
fo = 0.76
o~ = 958.3 psi
C~T = 531.76 psi
r = O28
fR: 0.83
critical unconstrained buckling, pressure
2E/(1-i~2) * (I/(DR-I'))~ * fof~t
E = apparent modulus, 57,500 psi for 12 hour pull
p = Poisson's ratio, 0.45
DR = dimension ratio, 17
fo = ovality compensation factor, 0.76 for 3% deflection
f¢~ = tensile reduction factor
f~ _~ ((5.57-(r'1.09)~))~'~'1.09
r =
CT = tensile stress during pull-back, psi
between max. pull force required and safe pull force
approx. 100,000 Ibf
OT ---- F/A (A is cross-sectional area)
(~m -~ 531.76 psi
r = 0.28
0.83
22.30 psi
Installation - Line D Page 2 of 3 Printed on: 7/27/2004
JOL 27 2004 16:57 FR TURNER COLLIEg BRADEN972 ?35 3802 TO 99723843570 P.04/06
4 ) Calculate external pipe loads (unit weight x height of soit above pipe / 144 sq. in/$q fl)
A (If bore hole remains open) Y~u.~ = 75 ih/fO ps~,,ry= '10.42 psi
B (if bore hole does not remain open) Yso~ -- 1 ~ ,5 th/fi° Pso. = 15.97 psi
Note: these calculations are conservative because they do not take into account any buoyant effects from
groundwater
5.) Calciuate internal Pipe Load (full of fluid)
(greatest water pressure) Y~o = 62.4 Ib/ft~ P~zo = 8.67 psi
6,) Calculate the net external pressure
A (If bore hole remains open)
B (if bore hole does not remain open)
p,~ = t.75 psi
critical buckling preSsure > maximum net external pressure
Pnm = 7.31 psi
critical buckling pressure > maximum net external pressure
Prepared by: Erika K. Cooper, P.E.
Note: Design calculations are based on project-specific parameters and should not be modified without consent
of the Proiect Engineer.
Installation - Line D Page 3 of 3 Printed on: 7/27f2004
3UIL ~27 2004 16:37 FR TURNER COLLIE~ BRADENg?2 735 5002 TO 99?250435?0 P.05/06
Horizontal Directional Drilling
Design Calculations - Lonq Term using DR 17.0 at 100psi
1 .) Calculate critical unconstrained bucktin~ pressure for long-term service
E = 28200 psi at 50 years
p = 0.45
DR = 17
fo = 0.76
critical unconstrained buckling pressure
2E/(1 .p.2). (1/(DR-1 ))3. fo
E = apparent modulus, 28,200 psi for 50 year
,u = Poisson's ratio, 0,45
OR -- dimension ratio, 17
f~, = ovality compensation factor, 0.76 for 3% deflection
per= 13_12 psi
2.) Calculate external pipe loads (unit weight x height of soil above pipe / 144 sq. in/Sq, ft)
A ~Deep portions of#ne (20 ft cover, deep portions of line full of water)
Hso~ = 20 ft
Hwa~er = 12 ff
y~, = 115 lb/fa
FH2o = 62 4 lb/fi^3
Pso, = 1597 psi
Puzo -- 5.2 psi
10.77 psi
critical buckling pressure > maximum net external pressure
B Shallowportions of#ne (8 ft cover, pae empty)
H~o~ = 8 fi maximum depth of cover over line
120 lb/fa average soil unit weight
P=~ = 6 67 psi
Pnet = 6.67 psi
critical buckling pressure > maximum net external pressure
Prepared by: Erika K Cooper, P.E.
Note: Design calculations are based on project-specific parameters and should not be modified without consent
of the Project Engineer.
50-year - Line D Page 1 of 1 Printed on: 7/27/2004
ZUL '~
=, 2~d04 16:37 FR TORNER COLLIE~ BRADEHg?2 ,_,~ 3002 TO 99?250455?0 P.06/06
Horizontal Directional Drilling
Design Calculations - Repair usitlg DR 17.0 at lo0psi
1 } Calculate critical unconstrained buckling pressure for short-term repair
E = 38000 psi at 1 year
~ - 0
DR = 17
fo = 0.76
Per = critical unconstrained buckling pressure
pc,= 2E/(1-¢) * (It(DR-I))~ * fo
E = apparent modulus, 38.000 psi for 1 year
.u = Poisson's ratio, 0.45
DR = dimension ratio, 17
fo = ovality compensation factor, 0.76 for 3% deflection
per= 17.68 psi
Calculate external pipe loads (unit weight x height of soil above pipe / 144 sq- in/sq, fi)
Deep portions of line (20 fi cover, pipe empty)
Hsoll = 20 f~
Ysoil = 115 Ib/ft~
Pso, = 15.97 psi
Pnet -- ~J 5.97 psi
critical buckling pressure > maximum net external pressure
Prepared by; Erika K Cooper, P.E.
Note: Design calculations are based on project-specific parameters and should not be modified without consent
of the Project Engineer,
Repair - Line D
Page 1 of 1
Printed on: 7t27/200,~
TOTAL PAGE.D6 ~