PERMFILE117687

• Permit No: C1981022
• IBM Index Class Name: Permit File
• Doc Date: 10/10/2003
• Doc Name: Sedimentation Pond E Embankment Wall Design ‎(Sept 2003)‎
• Section_Exhibit Name: Exhibit 2.05-E4 Part 11
• Media Type: D
• Archive: No

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CLIENT: Oxbow Mining JOB NO: DATE: 6/3/2003
.TASK: Wall Reinforcing BY: Christian Questad
DESCRIPTION: LowerBaz/Dowel CHKD:
Service Moment (k-Kips) = 30 81
Bending Moment, Mu (k-kips) = 5238
Sheaz (w/o ACI 350 L.F.), V u (k) = 0.00
Axial Force(Tension), Pu (k) _ -000
Load Factor (For bending) _ '170
Load Factor for Shear Steel (ACI 350) = 13
Compression flange width, be (in) = 12 0
Thickness of slab (in) = 24
Cover (in) = 2
Is [he bar on the outside or inside?.outstde --
Depth to reinforcing, d (in) = 2150
f c (psi) = 4000 ` dto Max=. 0.0214.
fy (psi) = 60000" Therefore As max = 5.52 sq in
Es (Psi) = 29000000
Ec (psi) = 3604997
n = Es/Ec = 8
Allowable Stress for Direct Tension (ksi) _ . 18.0
Beta1 = 0.85 - ~.
phi, for Bending = 0.9
phi, for Shear and Torsion = 0.85 ,
=Thickness -Cover - 1/2 Baz Diameter
Shear Ca aci y and Reinforcement (Rasesi nn AC131R.113.2) Am~mec Nn Torsinn
Note: For slabs & walls, shear reinforcement is not required as long as Vu < phi*Vc
V u *d/Mu (max of 1.0) = 0.00
1.9*sart(fc) +2500*As/(bd) * Vu d/Mu = 120.17
phi*Vd(phi*sgrt(fc)*bd) = 1.900 Wall nr Slab thirknecc is adegi ate
No Shear Reinforcement is required for Beam
a Stimtp Spacing = " `. 5 ' in ~
~ Max spacing of shear reinforcement (in) = 10.75 Non-Seismic loads -ACI 318-11.5.4 ~
rn Max spacing of shear reinforcement (in) = 10.75 Seismic requirements -See ACI 318-21.3.3 ~
~ Min. area of one leg of shear reinforcement = 0.025 (50 Bw s / fyy2 -See ACI 3I8-1 I.5.5.3 y
m Req'd Area of One Leg of Shear Stirrups, Av = noon ~
Remiired Stimm Rar Size fnr Reams o # 3
Rending Ru = Mu/(phi*bc*d^2) _
Rho = (.85*f c/fy)* { 1-[ 1-(2*Ru/(.85 *f c))]^.5 ] _
As Required for beri in =Rho*bc*d =
Min (fir bendine) _
Baz
As Total
y
(Z
ok r
Jl M'ICC
125.9 Use Ru = 1259
0.0021
OS52 in^2
0.860 Per ACI 318-99 Sect 10.5.1
` As Required for Direct Tension (in^2) = 0.000
As supplied for bending (in^2) = 0.942
\. Number of bars = 1.2000
M
~~
~cktng Location of C.G. of cra 'or
(from compression fiber) (in) _
fs (ksi)= Ms/(As*moment arm) _
Ae = 2*(dc-cover (2 max)tll2 Baz Dia)*bc =
A = Ae/number of bars =
Z = fs*((d~cover (2 max)+ 1/2 Bar dia)*A)^(1!3) _
File:lower.xJs
4.62
19.65 Z< 95 for extreme exposures
60 Z< 115 for hydraulic structtves
50.00 Z< 145 for exterior exposure
483 Z< 175 interior ex sure
Rev 101001 Print Date: 6/3/2003