Text

C-CSS-099.20-063, Rev 1, Shield Building Design Calculation. Part 5 of 7
	ML15280A304
Person / Time
Site:	Davis Besse
Issue date:	09/03/2014
From:	; FirstEnergy Nuclear Operating Co
To:	; Advisory Committee on Reactor Safeguards
Shared Package
ML15280A293	List: L-15-310, C-CSS-099.20-069, Rev 0, Shield Building Laminar Cracking Limits; ML15280A287; ML15280A303; ML15280A304; ML15280A305; ML15280A306; ML15280A308; ML15280A309; ML15280A310; ML15280A311;
References
	L-15-310, TAC ME4640 C-CSS-099.20-063, Rev 1
	Download: ML15280A304 (107)
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{{#Wiki_filter:Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-056 Rev. 001 105 of 212 000 CALCULATION OF CIRCUMFERENTIAL STRESSES AND CRACK WIDTH: C5 Region Moment curvature relations presented in this section are derived using strain compatibility (no slip among concrete and rebar) and equilibrium concepts. The figure shown below and the following assumptions are used for the derivations: Plane sections before bending remain plane after bending The stress strain behavior of concrete and steel is known Neutral axis + c <r-* Section strain stress Internal Forces External demand Concrete (sc) and steel (es) strains are calculated using strain compatibility as follows: 'cm -y c-dt -s. The concrete and steel stresses are calculated using appropriate constitutive laws, as those given below. In general any type of constitutive behavior can be used. Conservatively, the concrete tensile capacity is ignored. iooofcec(i - 250- ec) if ee < 0.002 unconfined concrete model per Park and Paulay 1975 fc\\l.2-100-£r\\ if <<*> 0.002 4000 fc [psi] 3000 2000 woo 0.002 0.003 Stress strain curve used for concrete

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-056 Rev. 001 106 of 212 000 Es£si fy <f Es£si>fy Elastoplastic reinforcing steel model 60" 0.002 0.003 Stress strain curve used for reinforcing steel The concrete resultant (C) and its moment arm (yh) are calculated by integrating the concrete stress /c(sc) that acts on each infinitesimal concrete area b(d)dy. Note that b(d) is the width of the concrete section a location d. yQ = Max(0,c - h) c= [ fc(ec}b(y)dy 1 Distance from the neutral axis to the first concrete fiber under compression. Internal resultant of concrete compression Location of concrete compression measured from the top concrete fiber f.=fsj.Asi Internal steel forces Equilibrium of the reinforced concrete section is verified as follows i = 0 Pure compression capacity Pure tension capacity j =0

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-056 Rev. 001 107 of 212 000 MATERIAL PROPERTIES fc a 4000psi fy - 60ksi 7C - 150pcf a =0.0000055 Ec m 57000 Ifc psi = 3605ksi l/= £cu =- 29000ksi 0.25 Jy

=

0.003 £t = 7.5-Ifc psi = 0.00013 Rebar Configuration h m 2.5ft Shell thickness dRl-= 15 0.33-1.56 1.33-1.56 1.33-1.56 2 in Concrete compressive strength Rebar yielding stress (note that this values is only to define stress-strain curve) Concrete unit weight Coefficient of thermal expansion (1/F) Concrete modulus of elasticity Rebar modulus of elasticity Concrete Poisson's ratio Yielding strain Concrete crushing strain Concrete tensile cracking strain Modular ratio

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-056 Rev. 001 108 of 212 000 Stresses in Reqion C5 for maximum DCR Demand of moments and membrane forces f 2.14 -72.90 "] 2.87 -100.82 R1M:= P:= RIM -kip = <-72.9^ -100.8 -3.8 -94.6 -41.4 -41.7 -74.7 -74.7) kip M:= RIM kip-ft = '2.1 N 2.9 -7.6 3.4 1.2 1.9 2.1 K -' J -7.58 -3.80 3.40 -94.58 1.19 -41.35 1.91 -41.73 2.12 -74.71 {2.12 -74.71 j Parameters for crack width calculation (ACI SP20) tb - 4in dw := ~- d'w := dw kipft

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-056 Rev. 001 109 of 212 000 LC101 k = 0 Pc := Pk = -72.9kip M£ d:=dRJ = 15 \\26) 'm As:=AsRl = guess g^= h = -5ft Given PNiAc, e^, d,A$,fc \\ = -6/ c <0 PNy{c,£ci,d,As,fc} =PC MN^c,£cj,d,As,fc)=Mc Converged results c = -17.8ft £ci = -0.00038 PN<^c,£ci,d,As,fc'j = -72.9 kip MN(f{c, £ci, d,As,fc) = 2.7-kip-ft JMcm, '= Jc[for*c) fMsOF k := £s{c' dmax

eci)' Es fhdsIF k:= £s{c'dmm'Sci)Es Maximum crack width wMmaxk = "NA"

= M, = 2.1-kip-ft Demands acting on the section 2.1 in dmax := max(d^ = 26-in dm-m := min(d) = 4-in M £ci:= -0.0003 7 kip MNi^c, £ci, d, As,fc) = 6. Ikipft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand fh/lcm = °PS' maximum concrete compressive stress fMsOF " ~J2-3ksi O.F rebar stress (tension is negative) fMsIF =-Ul-ksi L.F rebar stress (tension is negative) 1

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev.: C-CSS-099.20-056 Rev. 001 110 of 212 000 LC201 - LC206 = M = 2.9-kip-ft Demands acting on the section &= dRl" guess 4^ 15 in ,26, -0.2-h = -0.5/? (2.1\\ 2.1 \\2.1j 2 in Given PN(f{c, eci, d,As,fc} =-189.5 kip c<0 PNirfc,£ci,d,As,fc)=Pc MNcf{c,£ci,d,As,fc)=Mc )=26-in ^= -0.0003 MNi^c,£ci,d,As,fc} = 60.7-kip-ft =4in Converged results c =-18.4ft £ci =-0.00052 ,£ci,d,As,fc)=-100.8kip = 2.9-kipft 'Mem JMcmk = °PSI

= es{c'dmin<eci)Es fMsOFk = ~17-M fMs!Fk = 4-ksi Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative)

I.F rebar stress (tension is negative) Maximum crack width 'Mmaxk-="NA" wMmaxk="NA"in

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-056 Rev. 001 111 of 212 000 LC301 - LC302 £=2 ^=Pk = -3.8kip j^ Negtive moment switch OF and IF M d := dn r = (4) 15 &*<= AsRl = f2.t 2.1 \\26J \\2.1j guess ^.= 20-h = 50ft Given P<Ac,eci,d,As,fc\\ = 443.2kip c> 0 MV{c,eci,d,As,fc)=Mc Converged results c = 0.5ft £ci= 0.00002 Py{c,eci,d,As,fc)=-3.8kip jk/^m C,£"_-, u,^4,y ^, i = /.O' Kip'jt J Mein j J c I ciyJ c ) JKdctfi fMsOF. := £s\\c' dmin < £ci)' Es ^MsOF fMsIF k'= £s\\c>dmax'£ci)Es fhisIFk Maximum crack width ve "pjA"

= M

= -7.6-kip ft Demands acting on the section ^.= (-M) =7.6-kip-ft 2 &o*ii>>/= max(") = 26in ^gnmi<= min(d) = 4in M*j= 0-0003 M<f{c,eci,d,As,fc)=4.4-kipft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand = 82.5psi maximum concrete compressive stress = 0.2-ksi O.F rebar stress (tension is negative) = -2.3-ksi I.F rebar stress (tension is negative) wMmaxk= "NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-056 Rev. 001 112 of 212 000 LC401-404 JL=3 Lv=P,, = 6kip &= dRl = f 4^ 15 in /&k= AsRl " [26) guess c := -0.2-h = -0.5ft AAA J 'zi> 2.1 Given PN(Ac,£ci,d,As,fc\\=-189.5kip c <0 PN^c,eci,d,As,fc)=Pc Converged results c =-14.3ft eci =-0.00048 PN^c,eci,d,As,fc) = -94.6kip MNiAc, £ci,d, As,fc \\ = 3.4-kip-ft JMcm k -= Ac {£ci 'Jc) JMcm, fMsOF k := £s{c' dmax > £ci)' Es f] fMsIF k:= £s{C'dmin'£cijEs fl Maximum crack width k HsOF idsIF t = m = 3.4-kipft Demands acting on the section 2 aW= -°0003 MNif(c, ecj, d,As,fc) = 60.7-kip-ft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand = Opsi maximum concrete compressive stress = -16.1ksi O.F rebar stress (tension is negative) = -14.3-ksi I.F rebar stress (tension is negative) wMmaxk="NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev.: C-CSS-099.20-056 Rev. 001 113 of 212 000 LC501-502 = dRl = M = 1.2-kip-ft Demands acting on the section 75 in

= A sRl 2.1 2

in guess ^.= -0.2h = -0.5ft Given PNi^c, eci, d,As,fc j = -189.5 kip c<0 MN^c,eci,d,As,fc)=Mc Converged results c = -18.2ft eci =-0.00021 = -4l.4 kip MN^c,sci,d,As,fc) = 1.2-kip-ft JMem '~ Jc \\ ci'J c } fMsOFk:= £s{c>dmax<eci)Es fMsIFk-= £s(c>dmin'eci)Es Maximum crack width

=

"NA" fhMFk = = max(d) = 26-in

=

min(d) = 4-in MNy(c, ecj, d,A$,fc^ = 60.7-kip-ft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) 'Mmaxk="m"'m

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-056 Rev. 001 114 of 212 000 LC601

= 5

£&<= Pk = ~4L 7kip Mes-^ Mk = L9'kiP'fi Demands acting on the section &=dRl = 15 \\26j = AsRl 2.1 \\Z.l guess g^ h = -25ft Given PNi^c, £ci, d,As,fc }=-37.9 kip c<0 ^=-0.0002 MNi^c,eci,d,As,fc) = 0.8-kip-ft Converged results c = -lift eci = -0.00021 PN^c, eci, d,As,fc ) = -41.7 kip MN<f(c,£ci,d,As,fc) = 19-kip-ft JMcm 1=

c \\£ci'J c )

fjvisOF k := £s\\c' dmax £ci)' Es fMsIFk= £s{c'dmin>£ci)-Es Maximum crack width NA := c = ~6-2ksi Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) h-NA h-NA-ti. R=l

wMmax,

=

0.000091-R ksi -5 in,0in

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-056 Rev. 001 115 of 212 000 LC701-LC702 /wga.:= P = -74.7kip J^-= M, = 2.1-kip-ft Demands acting on the section = dRl 15 in i= AsR1 2.1 2 in = max(d) = 26 in &,*<<*= min(d) = 4-in guess: c\\=-10-h = -25ft Given PN<4lc,£ci,d,AsJl^ = -15l.6kip c,eci,d,As,fc) = 3.2kiP-ft c <0 MN¥{c,£ci>d,As,fc)=Mc Converged results c = -18.5ft £ci = -0.00039 PNv{c,eci,d,As,fc)= -74.7 kip fMcm f\\lsOF. := £s{c' dmax * £ci)Es fMslFk:= £s{c'dmin'£ci)Es Maximum crack width Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) h-NA h-NA-t R= 1 b wMmaxk := max 1-R in 2 m -5 in,0in

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev.: C-CSS-099.20-056 Rev. 001 116 of 212 000 LC801 LC803 k=7 I £= dRl guess Given Converged c = -75.5ft ^=Pk = -74.7 kip M^ 15 K26) p.7^ 2.1 £f= -0.2h = -0.5ft PNy(c,£ci,d,As,fc) = -189.5 kip c<0 PNy(c,£ci,d,As,fc)=Pc

esults sci =-0.00039 PN(f{c,£ci, MNi^c,£ci fMcmk=h fMsOFk =

fMslFk=t Maximum wMmaxk d,As,fc) =-74.7 kip d,As,fc) = 2.1-kip-ft

(£crfc)

fMcm,

£s\\c > dmax' £cij' Es fhlsOF crack width "SIM"

= M

= 2.1kipft Demands acting on the section 2 in 8mG>>s= max(") = 26-in $Mtm<= m>n(d) = 4-in ^=-0.0003 MN<fi(c,£ci,d,AsJ'c) = 60.7-kip-ft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand = Opsi maximum concrete compressive stress = -12.6ksi O.F rebar stress (tension is negative) k = -11.4ksi I.F rebar stress (tension is negative) wMmaxk="NA"-i"

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-056 Rev. 001 117 of212 000 Summarv of Circumferential Stresses and Crack width in Region C5 Maximum Concrete stress

o N 0 82.5 0 0 0 0 0 MAXIMUM DCRM Psi fMsOF = Maximum OF rebar stress f-12.3^ -17 0.2 -16.1 -7 -7.2 -12.6 -126)

to-Maximum IF rebar stress f-tu) fhisIF =

-15.4 -2.3 -14.3 -6.3 -6.2 -11.4 y-11.4;

to-Maximum crac wMmax =

( <<NA" "NA" "NA" "NA" "NA" 0.001496 0.005081 i "NA" m

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 118 Of 212 000 CALCULATION OF MERIDIONAL STRESSES AND CRACK WIDTH: C5 Region Moment curvature relations presented in this section are derived using strain compatibility (no slip among concrete and rebar) and equilibrium concepts. The figure shown below and the following assumptions are used for the derivations: Plane sections before bending remain plane after bending The stress strain behavior of concrete and steel is known Neutral "axis ~ + c Section strain stress Internal Forces External demand Concrete (ec) and steel (es) strains are calculated using strain compatibility as follows: £cm y e-di 'cm ip= c The concrete and steel stresses are calculated using appropriate constitutive laws, as those given below. In general any type of constitutive behavior can be used. Conservatively, the concrete tensile capacity is ignored. woo-fcec\\i - 250-ec) if ec < 0.002 Unconfinecl concrete model per Park and Paulay 1975 fc-(l.2-100-£^) if sr> 0.002 4000 /f[psi] 3000 200C WOO 0.002 0.003 Stress strain curve used for concrete

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 119 of 212 000 Es£si fy Elastoplastic reinforcing steel model 60-fy 0.002 0.003 Stress strain curve used for reinforcing steel The concrete resultant (C) and its moment arm (yb) are calculated by integrating the concrete stress /c(ec) that acts on each infinitesimal concrete area b(d)dy. Note that b{6) is the width of the concrete section a location d. y =Max(0,c-h) C= fc{sc\\b(y)dy yb=c~~c' Distance from the neutral axis to the first concrete fiber under compression. Internal resultant of concrete compression Location of concrete compression measured from the top concrete fiber Fi =fsi'Asi Internal steel forces Equilibrium of the reinforced concrete section is verified as follows i = o Pure compression capacity i = 0 Mc = Pure tension capacity Mt = j=0

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 120 of 212 000 MATERIAL PROPERTIES fc a 4000psi fy=60ksi yc = 150pcf a =0.0000055 E a 57000 If -psi = 3605-ksi Es a 29000ksi v=0.25 ecu:= 0.003 £t = 7.5lfc-psi = 0.00013 Ec n = =8 Ec Rebar Configuration h 2.5ft Shell thickness 1.56\\2 Concrete compressive strength Rebar yielding stress (note that this values is only to define stress-strain curve) Concrete unit weight Coefficient of thermal expansion (1/F) Concrete modulus of elasticity Rebar modulus of elasticity Concrete Poisson's ratio Yielding strain Concrete crushing strain Concrete tensile cracking strain Modular ratio

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 121 of 212 000 Stresses in Reaion C5 for maximum DCR Demand of moments and membrane forces r 8.49 2.44^ 10.48 2.55 -14.61 1.40 10.41 2.22 4.76 1.63 5.89 1.65 7.66 2.04 7.66 2.04 RIM := P:=R1M -kip = 2.6 1.4 2.2 1.6 1.7 2 (o) kip M := RIMV-kip-ft = 8.5 10.5 -14.6 10.4 4.8 5.9 7.7 7.7 kip-ft Parameters for crack width calculation (ACI SP20) w =

4. in h

2 =dW

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 122 of 212 000 LC101 k-=0 Pc := P = 2.4kip Mc := M = 8.5-kip-ft Demands acting on the section fa ^

dRl = \\26\\in As'

sRl n guess ^= o.2-h = 0.5ft Given Pi^c, £ci,d,As,fc) = 0.6kip c>0 P<fi(c,£ci,d,As,fc)=Pc My{c,eci,d,As,fc)=Mc 2 i 6\\in dmax := max(d) = 26in dmin '= min £ci:= 0.0001 = 4-in Mcf(c,eci,d,As,fc)=30.5-kip-ft Converged results c = 0.6 ft eci = 0.00003 PN?(c,eci,d,As,fc)=-2.5kip MN<f{c,£ci,d,As,fc) = 3.4-kip-ft fMcmk:=fc{£ci'fc) fMsOFk := £s{c'dmax' £cijEs fMsIFk := £s{c< dmin * £cij'Es Maximum crack width wMmaxk '= "NA" fMcmk = ">9.4psi = ~2ksi Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) wMmaxk =

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev.: C-CSS-099.20-063 Rev. 001 123 of 212 000 LC201 - LC206 &T dRl guess Given Converged c = 0.6ft / A \\ / I I in A

A ( \\26)m s'" sR1~{l.6 ^:= 0.2-h = 0.5ft P<f(c,eci,d,As,fc) = 0.3kiP c>0 P^c,eci,d,As,fc)=Pc M^c,ed,d,As,fc)=Mc results sci = 0.00003 P(f{c,Eci,d,AsJc) = 2.6kip M(fi{c,eci,

JMcm,

~

fM,OFk-w4> Maximum wMmaxk := d,As,fc) = W.5-kip-ft fc(£crfc) fMa = £s\\c'dmax'£cij'Es flMst

£s\\c

dmin' £ci)' Es J*Msl crack width "NA"
= m

= 10.5-kip-ft Demands acting on the section Yin ^memj.= max{d) = 26-in ^m>>K= m'n(d) = 4-in &oi}= 0.0003 Mp(c,eci,d,As,fc)=90-kipft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand n = 135.3psi maximum concrete compressive stress -)F = -2.6-ksi O.F rebar stress (tension is negative) p = 0.4-ksi I.F rebar stress (tension is negative) wMmaxk = "NA"-'n

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 124 of 212 000 LC301 - LC302 /C

2 P
= P 14 kit?

A' AAA AM0A ^ A' Negtive moment switch OF and IF Mm £= dRi 126 h" i^:= A5<<y = L 6 guess: ^= 0.2-h = 0.5ft Given P(f(c,£ci,d,As,/c) = O.J^p c> 0 My{c,£ci,d,As,fc)=Mc Converged results c = 0.5ft eci = 0.00005 P<(c,eci,d,As,fc) = 1.4kip My{c,£ci,d,AsJc)= 14.6-kip-ft fMcmk-fc{£c,fc)

fMcm, fMsOF k:= £s{c'dmax'£ci)'Es flMsOF fMsIF k := £s{c-dmin * £a)' Es 4fj/F Maximum crack width wMmaxk:= "NA" r^:= m

= -14.6-kip-ft Demands acting on the section

.= {-M)k = 14.6-kip-ft

\\ 2 yin $An<Mto'= max(d) = 26-in 4<<tmK= mln(d) =4-in &$ir om03 My{c,£ci,d,As,fc)=90-kip-ft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand = 188.8psi maximum concrete compressive stress

= -4.3-ksi O.F rebar stress (tension is negative) = 0.5-ksi I.F rebar stress (tension is negative) c wMmaxk="NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 125 of 212 000 LC401-404 fc=3 &K=pk = Z2kiP M* Ak~ dRl ~ \\26)m &"' AsR1 " [l.6j guess ^= o.Zh = 0.5ft Given Pi^c, £ci,d,As,fc) = 0.3 kip c>0 Py(c,eci,d,As,fc)=Pc Mv{c,eci,d,As,fc)=Mc \\£&4 Converged results c = 0.6ft sci = 0.00003 P^C£ci,d,As,fc)=2.2kip My(c.eci.d,As,fc) = 10.4kip-ft JMcm k *~ Jc \\fci'J c) JMcm ^ fMsOF k := £s{c' dmax > £ci)' Es ffAsOF fMsIFk:= £s\\c'dmin'£cij'Es fhtsIF Maximum crack width wMmaxk := "NA"

= Mk =

10.4-

'"2 Am

= 134.6psi = -2.7-ksi = 0.4-ksi kip-ft m>>J= ma /£a*/~ M<p(c Demands acting on the section x(d) = 26-in AmitiK~ min(d) = 4-in 0.0003 ,ed,d,As,fc)=90-kip-ft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) wMmaxk="NA"-in

Attachment E Reinforcement LC501-502 J&;= dRl" guess Given Converged C = 0.7/7 / Concrete Stresses and Crack Width &= 0.2-h = 0.5ft P<^c,sci,d,As,fc) = 0.3kip c>0 P^c,eci,d,As,fc)=Pc M<f(c,ecl,d,As,fc)=Mc esults eci = 0.00002 P^c,eci,d M(f{c,£cix JMcm

" -A fMsOFt-'

wt-Maximum ,As,fc)=1.6kip >.As,fc)=4.8-kip-fi

{£ci'fc) fMcmk = 61JP

£s[c'dmax>£ci)-Es fMsOFk = -hk rs[c>dmin'£ci)Es fMsIFk = a2^ crack width "NA" Calc. No: C-CSS-099.20-063 Rev. 001 Sheet No: 126 of 212 Sheet Rev.: 000 4.8-kip-ft Demands acting on the section /§[owwv:= max(d) = 26-in j^jti^-= min(d) = 4-in &>>}= °-0003 Mif[c,eci,dAs,fc)=90-kipft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand 5; maximum concrete compressive stress si O.F rebar stress (tension is negative) sb I.F rebar stress (tension is negative) wMmaxk = "NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 127 of 212 000 LC601 o) 1.6 1.6 2 in guess Given £j= 0.2-h = 0.5ft P<p(c,£ci,d,As,fc) = 0.6kip c>0 Pv{c.eci,d,As,fc)=Pc Mcp(c,£ci,d,As,fc)=Mc Converged results c = 0.6ft £ci = 0.00002 PN(f(c,£ci,d,As,fc)=-L8kip MN<f(c, £ci,d,As,/c) = 2.3-kip-ft fMcmk:=fc{£ci>fc) fMsOF k'= £s[c'dmax'£ci)'Es fMsIFk~£s{c'dmin'eci)Es Maximum crack width NA:= c Demands acting on the section

=

max(d) = 26-in

= min(d) = 4-in if= °-000]

M<p(c,£ci,d,As,fc)=30.5-kip-fi Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) ksi h-NA h-NA-t, in.Oin wMmaxk = Oin

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 128 of 212 000 LC701-LC702 ^=Mk= 7.7-kip-ft '-=dRl = \\26[in V= 1.6 sR, n guess ^:= 0.2-h = 0.5ft Given Pip[c.eci,d.As,fc) = 0.3 kip c>0 P<f(c,eci,d,As,fc)=Pc Mp(c,eci,d,As,fc)=Mc Converged results c = 0.6ft eci = 0.00002 Pif{c,eci,d,As,fc)=2kip M<p[c,eci,d,As,fc) = 7.7-kip-ft -fc{£cvfc)

= £s{c'dmax'eci)-Es JMcm, '= *c [ £ci'J c Maximum crack width

= c

=

max lb A 0.000091-l R in 2 in 2 in Demands acting on the section ,:= max(d) = 26-in J^mmA'= m"2(a0 = ^'JW g= 0.0003 M<f(c,eci,d,As,fc) = 9 Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) h-NA h-NA-tt. ksi -5 in,0in wMmaxk = °-i

Attachment E Reinforcement / Concrete Stresses and Crack Width LC801 - LC803 guess G/ven P^c,£iCi-,ii,A5,/(:j = 0.3kip c>0 P<f{c,eci,d,As,fc)=Pc M¥{c,eci,d,As,fc)=Mc \\'):=Find{c,ecl) Converged results c = 0.6ft £ci = 0.00002 P<f{c,£ci,d,As,fc)=2kip M(fiic,eci,d,As,fc) = 7.7-kip-ft fMcm, := fc {£ci'fc) fMcm , = 98-9Psi K A. fMsOF

= £s[c'dmax' £ci)'Es fhtsOF

. = ~^-8lcsi fMsIF, := £s[c'dmin-£a)-Es fhfsIF. = 03ksi Maximum crack width wMmaxk := "NA" Calc. No: C-CSS-099.20-063 Rev. 001 Sheet No: 129 of 212 Sheet Rev.: 000 dp-ft Demands acting on the section 9lm^= max(d) = 26-in Mntki^-~ min(d) = 4-in My(c,eci.d,As,fc)=90-kip-ft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) wMmaxk="NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 130 of 212 000 Summarv of Meridional Stresses and Crack width in Region C5 Maximum Concrete stress 'lO9.4\\ 135.3 Maximum OF rebar stress MAXIMUM DCRM JMcm 188.8 134.6 61.1 76 98.9 98.9 Psi fMsOF = -2 -2.6 -4.3 -2.7 -1 -1.4 -1.8 \\ ksi Maximum IF rebar stress '0.4 0.4 0.5 0.4 0.2 0.3 0.3 V wMmax = Maximum crack width "NA" "NA" "NA" "NA" 0 0 in

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 131 of 212 000 CALCULATION OF CIRCUMFERENTIAL STRESSES AND CRACK WIDTH: D1 Region Moment curvature relations presented in this section are derived using strain compatibility (no slip among concrete and rebar) and equilibrium concepts. The figure shown below and the following assumptions are used for the derivations: Plane sections before bending remain plane after bending The stress strain behavior of concrete and steel is known Neutral Section strain stress Internal Forces External demand Concrete (ec) and steel (es) strains are calculated using strain compatibility as follows: 'cm c-di -y c <p = c The concrete and steel stresses are calculated using appropriate constitutive laws, as those given below. In general any type of constitutive behavior can be used. Conservatively, the concrete tensile capacity is ignored. WOOfc ec-(l - 250-ec) if ec < 0.002 ijnconfined concrete model per Park and Paulay 1975 fc-[l.2-100-er] if s,> 0.002 fc 4000-3000-0.003 Stress strain curve used for concrete

Attachment E Reinforcement / Concrete Stresses fsi{£si) = Esesi fy $ Es£si>fy fy [ksi] Calc. No: and Crack Width Sheet No: Sheet Rev.: Elastoplastic reinforcing steel model 60 41 20 0 / / / ' / i i 0 0.001 0.002 0.003 Stress strain curve used for reinforcing steel C-CSS-099.20-063 Rev. 001 132 of 212 000 The concrete resultant (C) and its moment arm (yb) are calculated by integrating the concrete stress /c(e0) that acts on each infinitesimal concrete area b(6)dy. Note that b(6) is the width of the concrete section a location d. y0 =Max(0,c-h) C= fc(ec)-b{y)dy

7b y

-C-L[C f le).b(y).y yo Fi=fsi-Asi Distance from the neutral axis to the first concrete Internal resultant of concrete compression iy Location of concrete compression measured from Internal steel forces Equilibrium of the reinforced concrete section is verified as follows n i = 0 Pure compression capacity Pure tension capacity i=0 j=0 j=o fiber under compression. the top concrete fiber I

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 133 of 212 000 MATERIAL PROPERTIES fc 4000psi f =60ksi 7c = 150pcf a =0.0000055 Ec = 57000 Ifc -psi = 3605-ksi V e 29000ksi 0.25 Jy Es

=

0.003 £, = 7.5 \\fr -psi = 0.00013 1 E V c Rebar Configuration h m 2.0ft Shell thickness dR]:= 20) A sRJ 0.79),, \\in 1.00) 2 Concrete compressive strength Rebar yielding stress (note that this values is only to define stress-strain curve) Concrete unit weight Coefficient of thermal expansion (1/F) Concrete modulus of elasticity Rebar modulus of elasticity Concrete Poisson's ratio Yielding strain Concrete crushing strain Concrete tensile cracking strain Modular ratio

Attachment E Reinforcement / Concrete Stresses and Crack Width Caic. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 134 of 212 000 Stresses in Region D1 for maximum DCR Demand of moments and membrane forces -47.12 100.11^ 39.34 38.51 -18.26 -31.08 58.17 42.37 34.78 29.40 -0.90 17.14 -1.37 29.94 41.64 39.77 R1M:= P:=R1M -kip = \\ 100.1^ 38.5 -31.1 42.4 29.4 17.1 29.9 39.8 kip M := RIMf-kip-ft = J r-47.l\\ 39.3 -18.3 58.2 34.8 -0.9 -1.4 . 41.6 j kip-ft Parameters for crack width calculation (ACI SP20) tb := 4m A

= 2-tKb dw = - - lb

Attachment E Reinforcement / Concrete Stresses and Crack Width LC101 k:=0 P

= P

= 100.1 kip M

= M

-47.1-kipft For negtive moment, switch OF and IF rebar d:= dnj

Yin A, := guess ^:= o.2-/i = 0.4ft Given P<f(c,£ci.d,As,fc) = 12.5kip c>0 P<p(c,£ci,d,As,fc)=Pc M?(c,£ci,d,As,fc)=Mc /Wl I / \\

=

Fjnrfl cfi^j-l Converged results c = 7.6/i £ci = 0.00021 Pp(c,£ci,d,As,fc) = W0.1 kip M?(c,£ci,d,As,fc)= 47.1-kipft fMcmk-=fc{eCvfc) fMcmk = 807.8psi

JMsOF,

= £s\\c' min'£ci) s

MsOF k~

sl fMsIF k := £s{c'dmax* sci)'Es ffrfsIFk=-°-2-ksi Maximum crack width w "A/4 " MtnoXt Calc. Sheet Sheet

=

me £c0= No: C-CSS-099.20-063 Rev. 001 No: 135 of 212 Rev.: 000 Demands acting on the section x{d) = 20-in dmin := mjw(^) = 4-in 0.0003 ,£ci.d,As,fc)=43.9-kipft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) wMmaxk="NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width LC201 - LC206 Jp 1 P^:= Pk = 38.5 kip jjfa= k A A (4\\- A (°A.2 a:=flpi= yin A

= AK1

=\\ ym K1 \\2o) s sR1 y i) guess ^= 0.2-h = 0.4ft Given P(f{c, £ci,d,A$,fc) = 6.4 kip c>0 Mc'£ci'd'As'fc)=Pc M^c,eci,d,As,fc)=Mc Converged results c = 0.7ft eci = 0.00021 P<f(c,Eci,d,AsJc)=38.5kip M<f(c,eci,d,As,fc)=39.3.kiPft JMcm, '= *c [ £ci'J c j JMcm , =

MsOF k'~

£s\\c' max' £ci)' s JMsOF k fMsIFk := £s{c'dmin* £cij'Es ?MsIF f = Maximum crack width wMmaxk := "NA" Calc. No: C-CSS-099.20-063 Rev. 001 Sheet No: 136 of 212 Sheet Rev.: 000 = 39.3-kip-ft Demands acting on the section /dMUUl^= max(d) = 20-in j(himiA= """W) = ^"'n /&V- °-0003 M<f{c.eci,d,As,fc)= 47.5-kip.fi Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand 795.4psi maximum concrete compressive stress = -7.5-ksi O.F rebar stress (tension is negative) 3.4-ksi I.F rebar stress (tension is negative) wMmaxk="NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack LC301 - LC302 k=2 iP^=pk=-3Llkip m For negtive moment, switch OF and IF rebar faj= (-M)k = 18.3-kip-ft d-d -(4\\in (A sRl i AsRln ( guess gg= o.2-h = 0.4ft Given Pt/^c, £ci, d,As,fc)= 8.7 kip c>0 P*(c,eci,d,As,fc)=Pc M<f{c,£ci,d,As,fc)=Mc lsJ:=Find(C'e^ Converged results c = 0.1ft £ci = 0.00009 PNy(c,£ci,d,As,fc)=-33.8kip MN^c.eci,d,As,fc) = 15.6-kip-ft fMcmk=fc{£cvfc) fM fMsOF k := ss[c

dmin' £ci)' Es

^M. fMsIF k:= £s[c'dmax'£cij'Es f\\f. Maximum crack width wMmaxk~ "NA" Width 'wT Mk - 'O.2 [in 0.8) <OFk = ~- Calc. No: Sheet No: Sheet Rev.: C-CSS-099.20-063 Rev. 001 137 of 212 000 = -18.3-kip-ft Demands acting on the section /^g)iBf:= max(d) = 20-in /$uw*ii/\\:= min(d) = 4-in £*,*= 0.0002 My{c,£ci, .7psi \\2-ksi 5.2-ksi d,As,fc)=29.6-kipft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) IJF rebar stress (tension is negative)

Attachment E Reinforcement / Concrete Stresses and Crack Width LC401-404 A:= dRl: guess Given P^=Pk = 42.4 kip M*-=Mk = 58- -{4\\in A -A -(°'8]in2 g£= 0.2-h = 0.4ft P<f(c,£ci,d,As,fc)=6.4kip c>0 Py{c,eci,d,As,fc)=Pc Mp(c,eci,d,As,fc)=Mc Converged results c = 0.6ft £ci = 0.00034 PtAc'£ci Mcf(c,£cl 'Mem,

~

fM,OFt-wt> Maximum wMmaxk := d.As.fc)= 42.4 kip ,d,As,fc)=58.2-kip-ft fc(£c,fc) fMcmk^239.6pi -4M-~4to fMsOFk = -^-k £s{c-dmin-£ci)-Es fMsIFk = 4A'ksi crack width "MA" Calc. Sheet Sheet 2-kip-ft Mtf{c i si No: C-CSS-099.20-063 Rev. 001 No: 138 of 212 Rev.: 000 Demands acting on the section uvc(d) = 20-in 4smmf:= TOn(^) = ^'" 0.0005 Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) wMmaxk="NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width LC501-502 k=4 L^=pk = 29-4kiP M*--=Mk = 3-(4 \\ (°^^\\ 2 guess jg* 0.2-h = 0.4ft Given Pi^c,£ci,d,As,fc) = 6.4kip c> 0 Py{c,eci,d,As,fc)=Pc M<f(c,£ci,d,As,fc)=Mc Converged results c = 0.7ft eci = 0.00019 P?(c,£ci,d,As,fc)=29.4kip M<p(c,£ci,d,As,fc)=34.8-kip-ft fMcmk--fc{£cvfc) fMcmk = ?33^ fftisOF,:= £s[c'dmax'£ci)'Es $MsOF =S-4-k fMsIFk := es{c'dmin'£ciJEs $MsIFk = 2-8'ksi Maximum crack width wMmaxk:= "NA" Calc. Hoc C-CSS-099.20-063 Rev. 001 Sheet No: 139 of 212 Sheet Rev.: 000 t.8-kip-ft Demands acting on the section $4X0^-= max(d) = 20-in ^mi^= min(d) = 4-in /£^= 0.0003 M<p[c,£d,d,As,fc)=47.5-kipft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand i maximum concrete compressive stress si O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) wMmaxk = "NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 140 of 212 000 LC601 &= 5 &^= Pk = 17J kiP M*:= Mk 9-kipft For negtive moment, switch OF and IF rebar = 0.9-kip-ft 2 dn1 = \\ \\-in \\20 guess £5= o.2-h = 0.4ft Given P<p[c, Sd, d,As,fc) = 12.5 kip c>0 P<f(c,eci.d,As,fc)=Pc Converged results c = 7.9ft £ci = 0.00002 Pi^{c,£ci,d,AsJc) = 17.1 kip M<f{c,£ci,d,As,fc)= 0.9-kip-ft fMcm k:= fc{£ci>fc) fMsOF k:= £s{c'dmin-sci)'Es fMsIF k:= £s\\c'dmax'£ci}'Es Maximum crack width NA:= c wMmaxk := rmx 0.000091-R- in .2 in Demands acting on the section ,:= max(d) = 20-in = 4-in ^= 0.0003 Mip[c,eci.d.As.fc)=43.9-kip-ft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) i.F rebar stress (tension is negative) R;= h-NA h-NA-tt. ksi -5 in,0in R-0.9 wMmaxk °-i

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 141 of 212 000 LC701-LC702 - Mk = -1.4-kip-ft For negtive moment, switch OF and IF rebar d := 20 A

=

5 fAsRl AsRl \\ I 0) n.2 [in 0.8 guess g^= 0.2-h = 0.4ft Given Pifi{c, eci,d,As,fc) = 12.5 kip c>0 Mc'eci'd'As'fc)=Pc Mv{c,sci,d,As,fc)=Mc Converged results c = 8.9ft eci = 0.00003 P<f{c,sci,d,As,fc)= 29.9 kip M<f{c,eci,d,As,fc) = 1.4-kip-ft fMcmk:=fc{sci'fc) fMsOF k := £s[c' dmin * £cijEs fMsIF k := £s{c' dmax < sci)' Es Maximum crack width wMmaxk ~ tU A 0.000091-l R in 2 in Demands acting on the section -= max(d) = 20-in T °-0003 Mp(c.eci,d,As,fc)=43.9-kip-ft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) h-NA h-NA-tb R = ksi -5 in,0in

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev.: C-CSS-099.20-063 Rev. 001 142 of 212 000 LC801 - LC803 k'.= 7 P

= P

= 39.8 kip M

= Af

= 41.6-1 <Sk d/?7~l 2Q \\'m s-~ AsRl ~\\ j \\'ln fa** 9UeS$: ^=0.2-h = 0.4ft Given P<fl(c,£ci,d,As,fc) = 6.4kip c>0 Mc'£ci'd'As'fc)=Pc M(fi[c,£ci,dAs,fc)=Mc

= Find(c,£cj\\

Converged results c = 0.7ft £ci = 0.00022 P^£,vd,As,fcy-39Xkip M<p(c,£ci,d,As,fc)=41.6-kipft fMcmk := fc{£cffc) fMcmft = 846.6psi fMsOFk = 4c-dmax-£c^s fMsOFk=^ksi JMslF -= £s\\c' min'£ci) s JMsIF , ->-ksi Maximum crack width wMmaxk:= "NA" dp-ft Demands acting on the section Mp(c,£ci,d,As,fc)=47.5-kip-fi Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative)

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 143 of 212 000 Summarv of Circumferential Stresses and Crack width in Region D1 MAXIMUM DCRM Maximum Concrete stress 807.8 795.4 344.7 1239.6 733.4 65.2 111.9 846.6 ) JMcm Maximum OF rebar stress 4.9 ^ -7.5 Ps< fMsOF = \\ -5.2 -17.1 -8.4 0.5 0.8 -8.4 ksi Maximum IF rebar stress -0.2 3.4 -36.2 4.4 2.8 0.4 0.7 3.5 ksi Maximum crack width "NA"^ "NA" "NA" "NA" "NA" 0 0 "NA"

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 144 of 212 000 CALCULATION OF MERIDIONAL STRESSES AND CRACK WIDTH: D1 Region Moment curvature relations presented in this section are derived using strain compatibility (no slip among concrete and rebar) and equilibrium concepts. The figure shown below and the following assumptions are used for the derivations: Plane sections before bending remain plane after bending The stress strain behavior of concrete and steel is known 4 i r Neutral axis 1 r j& I f>2

fs: \\ rM l

Section strain stress Internal Forces External demand Concrete (ec) and steel (e ) strains are calculated using strain compatibility as follows: 'cm .y c-^. -cm c <p = c The concrete and steel stresses are calculated using appropriate constitutive laws, as those given below. In general any type of constitutive behavior can be used. Conservatively, the concrete tensile capacity is ignored. fc(ec) = 11000-fc-ec\\l - 250-ec) if ec < 0.002 \\fc \\1.2 - 100-£,} if £r > 0.002 Unconfined concrete model per Park and Paulay 1975 4000 3000 2000 1000 0 - / 1 1 0 0.001 0.002 0.003 Stress strain curve used for concrete

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 145 of 212 000 Esesi fy if' Elastoplastic reinforcing steel model 60" 40-20-0 0.001 0.002 0.003 s Stress strain curve used for reinforcing steel The concrete resultant (C) and its moment arm fj^j are calculated by integrating the concrete stress /c(ec) that acts on each infinitesimal concrete area b(d)dy. Note that b{d) is the width of the concrete section a location d. y0 =Max(0,c-h) C = fc(ec)-b(y)dy yt>=c~~c' Fi=fsiAsi Distance from the neutral axis to the first concrete fiber under compression. Internal resultant of concrete compression Location of concrete compression measured from the top concrete fiber Internal steel forces Equilibrium of the reinforced concrete section is verified as follows n i = 0 Pure compression capacity i=0 / =

Pure tension capacity Mt =

j=0 5-

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 146 of 212 000 MATERIAL PROPERTIES fc = 4000psi fy = 60ksi 7C

ISOpcf a =0.0000055 E

= 57000 If -psi = 3605-ksi Es = 29000ksi i/= 0.25 ecu:= 0.003 £t = 7.5-Ifc psi = 0.00013 Ec Q = Ec~ Rebar Configuration h m 2.0ft Shell thickness dR1 := l \\ 20 j i.oo Concrete compressive strength Rebar yielding stress (note that this values is only to define stress-strain curve) Concrete unit weight Coefficient of thermal expansion (1/F) Concrete modulus of elasticity Rebar modulus of elasticity Concrete Poisson's ratio Yielding strain Concrete crushing strain Concrete tensile cracking strain Modular ratio

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 147 of 212 000 Stresses in Reaion D1 for maximum DCR Demand of moments and membrane forces f-14.94 44.20 47.28 44.97 43.07 39.27 57.72 39.27 27.19 18.33 -2.20 20.02 -3.38 24.23 R1M:= {31.96 26.07) / P:= RIM -kip = 44.2s* 45 39.3 393 18.3 20 24.2 kip M := RIM'kip-ft = \\ -14.9s" 47.3 43.1 57.7 27.2 -2.2 -3.4 32 kipft Parameters for crack width calculation (ACI SP20) tb := 4in ._ A w 2 d'w := dw

Attachment E Reinforcement / Concrete Stresses LC101 k:=0 Pc:= Pk = 44.2 kip and Crack Width Mc := M = -14.9-kip-ft For negtive moment, switch OF and IF rebar )fc= (-M)k = 14.9-kipft d:=dRi={2or A*:= A A \\ guess c^= o.2-h = 0.4ft Given Pi^c,eci,d,As,fc) = c>0 P<f(c,eci,d.As,fc) = Converged results c = 2ft eci = 0.00007 Pip(c, eci, d,As.fc)= 44.2 kip M<f(c,eci.dAs.fc)= 14.9-kip-ft

>Mcm

'~ Jc\\ cN c) fMsOFk-=<c<*minMEs fMsIF^^-^M^ Maximum crack width wMmaxk- "NA" Calc. Sheet Sheet No: No: Rev C-CSS-099.20-063 Rev. 001 148 of 212 000 Demands acting on the section

[oJ*a dmax:=max{d) 12.5 kip Pc fMcmk = 290.5psi fMs0F k

1.8-ksi fMsIF = 0-4'ksi eci '- M(f{c = 20-in ^rnin := min(d) = 4-in 0.0003

£ci-d,AsJc)=43.9-kip-ft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative)

I.F rebar stress (tension is negative) wMmaxk="NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 149 of 212 000 LC201 - LC206 fa=l ^p^ 45 kip J*\\. /&.' Rl \\ ->r\\\\ln s sRl \\20J guess ^= 0.2-h = 0.4ft Given Pi^c,eci,d,As,fc) = 6.4k c>0 Pcf{c,eci,d,As,fc)=Pc Converged results c = 0.7ft £ci = 0.00026 Pp(c,eci,d,As,fc)=45kip M^{c,eci,d,As,fc)=47.3-kipft fMcmk-=fc{£cvfc) fMsOF j := £s\\c' dmax * £ci) 'Es ftdsIFk := £s{c'dmin'£ci)'Es Maximum crack width wMmaxk-= "NA" MaS= Mk = 47.3-kip-ft - (°'8\\ in2

- =

ip M(Ac fMcmk = 958JPsi fMsOF = ~9.6-&.H fMsIF =4^si Demands acting on the section iax(d)=20-in jtj^^^ min(d) = 4-in 0.0003 ,£ci,d,As,fc)=47.5-kip-ft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) wMmaxk="NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width LC301

LC302 guess

^= 0.2-h = 0.4ft Given Pif{c, £ci,d,As,fc) = 6.4kip c>0 Pt{c,eci,d,As,fc)=Pc MV{c,eci,d,As,fc)=Mc

= Findl c,£c\\

Converged results c = 0.7ft sci = 0.00024 Pp(c,eci,d,As,fc)= 39.3 kip M<f(c,eci,d,As,fc)=43.1-kip-fi fMcmk--fc{£c,fc) fMcmk = ^ fMsOFk := £s{c-dmax- £ci}-£s ffAsOFk = ~9A'k fMsIFk ~ £s{c'dmin'£ci)-Es ftfsIFk = 3-&ksi Maximum crack width w = "NA" Calc. No: C-CSS-099.20-063 Rev. 001 Sheet No: 150 of 212 Sheet Rev.: 000 3.1-kipft Demands acting on the section /$kWMl/= max(d) = 20-in $j&imr}= min(d) =4-in &>>}= °-0003 Mif{c,eci,d,AsSc) =47.5-kipft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand i maximum concrete compressive stress si O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) wMmaxk = "NA"-'m

Attachment E Reinforcement / Concrete Stresses and Crack Width LC401-404 k=3 &r dRiz guess Given Convergec c = 0.6ft Pje,= Pr 39.3 kip MoJ.= Mk - y20) At-sRl ~ y j J lr c~ 0.2-h = 0.4ft P<fi(c,eci,d,As,fc)=6.4kip c>0 P^c,eci,d,As,fc)=Pc M<p(c,eci,d,As,fc)=Mc I results sci = 0.00034 Mf{c,eci

JMcm,

"~ fMsOFk-Maximum maxk d,As,fc)= 39.3 kip ,d,As,fc)=57.7-kip-ft fc{£ci'fc) fMcmk = I2A = £s{c

dmax * £cij' Es

$MsOF ~ ~>> crack width "NA" Caic. Sheet Sheet = 57.7-kip-ft 3.7psi No: C-CSS-099.20-063 Rev. 001 No: 151 of 212 Rev.: 000 Demands acting on the section uuc(d) = 20-in $mm<~ mi/2(l2O = 4'^n 0.0003 ,eci,d,As,fc)=47.5-kipft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) wMmaxk="NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev.: C-CSS-099.20-063 Rev. 001 152 of 212 000 LC501-502 L=4 guess Given Converged c = 0.6ft Pe/:=Pk = 18.3 kip ^:=Mk = 27.2-i {4\\' A^-A -{°'%n <L ££= 0.2-h = 0.4ft P^c,sci.d.As,fc) = 6.4kip c>0 P4ceci,d,As,fc)=Pc Mp(c,eci,d,As,fc)=Mc results eci = 0.00076 pAc'£cv M<f(c,eci, fMcmk-fMsOFk-- fMsIF,= d,As,fc) = 18.3 kip d,AsJc)=27.2-kipft fc{£cvfc) fMcmk = 602.3psi

£s[c-dmax' £ci)'Es

^MsOFk = ~S-6'ksi £s{c'dmin'£cifEs fMsIF k = Lg-ksi Maximum crack width wMmaxk := "NA" dpft Demands acting on the section ^1= max(d) = 20-in &mm<~ m(W(^) = ^'in gtf* 0.0003 M(f(c,eci.d.As,fc)=47.5-kip-fi Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) wMmaxk "NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 153 of 212 000 LC601 For negtive moment, switch OF and IF rebar d:= dR7 = '**N

hRl, guess

^= o.2h = 0.4ft Given P<f{c.eci,d,As,fc) = 12.5kip c>0 P?(c,£ci.d,As,fc)=Pc Converged results c = 4.2ft £ci = 0.00002 P<f(c,£ci,d,As,fc)=20kip M<fi(c,£ci,d,As,fc)=2.2-kip-ft ffAsOFk '*' £s\\c'dmin* £ci)'Es fMsIF k := £s[c

dmax * £cij' Es Maximum crack width NA:=

c wMmaxk := h=-M in 2 in 0.8 Demands acting on the section

= max{d) = 20-in Mif[c,£ci,d,AsJc)=43.9-kipft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative)

I.F rebar stress (tension is negative) h-NA h-NA-tb R = 0.9 ksi -5 in.Oin

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 154 of 212 000 LC701-LC702 = Pk = 24- -= Mk = -3.4-kip-fi For negtive moment, switch OF and IF rebar d\\ dni = guess Given [in P<f{c, c>0

Pdc, 2-h

£ci £ci (A A \\ = 0.4ft .d,As,fc) = ,d,As,fc) = \\ sRl j 4.6kip ? Mtf(c,£ci,d,As,fc)=Mc Converged results c = 3.5ft £ci = 0.00003 MN<f(c,£ci,d,As,fc) = 0.3-kip-ft

= £s{c>dmin£cij-Es fMsIFk := £s[c'dmax> £ci)Es Maximum crack width Demands acting on the section h-NA h-NA-tb wMmaxk := "^

tb A 0.000091-l R in 2 ksi -5 in,0in r a0001 Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) R = 0.8

Attachment E Reinforcement / Concrete LC801 - LC803 7r "7 O D ^>> 9UeSS: ^0.2-* = I ^ I CB

=

Fj Converged results c = 0.6ft eci = 0.00018 Stresses and Crack Ukip As'-- 0.4ft Tf\\ nd{^£ci Pif{c,sci,d,As.fc) =26.1 kip Mc'£cvd'As'fc) = fMcmk--=fc(£ci-fc) fMsOFk:~ £s\\c'dmax' fMslF k= £s(^min^ Maximum crack width w "f>JA " AfftUXXt 32-kip-ft £ci}-Es (o. " AsRl = ^ j = 6.4 kip = Pc fMa ?MsC ffAsl Width 8\\in2 nk = 68i W =~ Fk = 2" Calc. Sheet Sheet = 32-kip-ft 8.1-ksi j-ksi No: C-CSS-099.20-063 Rev. No: 155 of Rev.: Demands acting on the section 0.0003 ,eci,d,As,fc)=47.5-kip-fi Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) wMmaxk="NA"-in 001 212 000

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 156 of 212 000 Summarv of Meridional Stresses and Crack width in Reqion D1 MAXIMUM DCRM

Mcm n Concrete stress

' 290.5 " 958.7 885.9 1243.7 602.3 87.4 112.7 Maximum OF Maximum IF -ebar stress rebar stress Maximum cr ( 1.8 -9.6 -9.4 -18.2 -8.6 0.6 0.7 fMsIF = <0.4\\ 4 3.6 4.2 1.9 0.4 0.4 ,2.5, ksi wMmax = ("NA"\\ "NA" "NA" "NA" "NA" 0 0 K"NA",

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 157 of 212 000 CALCULATION OF CIRCUMFERENTIAL STRESSES AND CRACK WIDTH: D2 Region Moment curvature relations presented in this section are derived using strain compatibility (no slip among concrete and rebar) and equilibrium concepts. The figure shown below and the following assumptions are used for the derivations: Plane sections before bending remain plane after bending The stress strain behavior of concrete and steel is known Neutral axis c Section strain stress Internal Forces External demand Concrete (ec) and steel (es) strains are calculated using strain compatibility as follows: -cm -y -cm c

c The concrete and steel stresses are calculated using appropriate constitutive laws, as those given below. In general any type of constitutive behavior can be used. Conservatively, the concrete tensile capacity is ignored. 1000fc ed(l - 250-£c) if £c < 0.002 UncOnfined concrete model per Park and Paulay 1975 fc\\l.2-100-£r) if £r> 0.002 4000- /c[psi] 3000-2000-1000-0.003 Stress strain curve used for concrete

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 158 of 212 000 Es£si Elastoplastic reinforcing steel model fy ifEs£si 60-40-fy 20-0.002 0.003 Stress strain curve used for reinforcing steel The concrete resultant (C) and its moment arm <yb) are calculated by integrating the concrete stress /c(ec) that acts on each infinitesimal concrete area b{6)dy. Note that b(6) is the width of the concrete section a location d. y =Max(0,c - h) C= fc{ec)-b(y)dy Distance from the neutral axis to the first concrete fiber under compression. Internal resultant of concrete compression 1 f yb=c fc(ec\\-b(y)-y dy Location of concrete compression measured from the top concrete fiber Fi -hi Asi Internal steel forces Equilibrium of the reinforced concrete section is verified as follows n i = 0 Pure compression capacity j=o Pure tension capacity Mf = j =0 i = 0 - -d 2 J 'A--**

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 159 of 212 000 MATERIAL PROPERTIES fc = 4000psi ~fc = ISOpcf a =0.0000055 E

57000 [f~pTi = 3605-ksi L>

V £CU = 29000ksi 0.25 Es

= 0.003

£t = 7.5-Ifc -psi = 0.00013 Ec n = =8 Ec Rebar Configuration h = 2.0ft Shell thickness dRl:= 0.79 1-1.56 2 in Concrete compressive strength Rebar yielding stress (note that this values is only to define stress-strain curve) Concrete unit weight Coefficient of thermal expansion (1/F) Concrete modulus of elasticity Rebar modulus of elasticity Concrete Poisson's ratio Yielding strain Concrete crushing strain Concrete tensile cracking strain Modular ratio

Attachment E Reinforcement / Concrete Stresses and Crack Width Stresses in Reqion D2 for maximum OCR Demand of moments and membrane forces R1M:= (-4.68 8.53 42.89 36.00 43.36 36.32 61.85 36.19 32.81 17.00 -2.77 6.20 -3.51 -1.49 y40.38 30.80) P:= RIM -kip = 36 36.3 36.2 17 6.2 -1.5 K3O.8J Parameters for crack width calculation (ACI SP20) tb = 4in A := 2-iwb h dw := ~ " lb d'w := dw Calc. No: Sheet No: Sheet Rev.: (n) kip M := RIM' kip-ft = C-CSS-099.20-063 Rev. 001 160 of 212 000 (-4.7^ 42.9 43.4 61.8 32.8 -2.8 -3.5 \\40Aj kip-ft

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 161 of 212 000 LC101 k;=0 Pc:= P =8.5 tip Mc:= M

-4.7-kip-ft For negtive moment, switch OF and IF rebar d := dgj

Yin As := (L6\\ 2 guess ^= 0.2-h = 0.4ft Given P<fi(c,£d.d,As,fc) = 13.3 kip c>0 Pp(c,£ci,d,As,fc)=Pc Mp(c,£ci,d,AsJc)=Mc

= Findie, £r\\

Converged results c = 1.4ft £ci = 0.00002 Pcp[c,£ci,d,As,fc)=8.5kip M?{c,£ci,d,As,fc)=4.7-kip-fi fMcm k:=fc{ £ci'fc) fMcm k = 77-5Psi fMsOFk := £s\\c'dmin'£ci)'Es fMsOFk = °-4ksl fMsIF , := £s[c'dmax' £ci)'Es fMsIF, = ~0--*5' Maximum crack width wMmaxk:= "NA" Demands acting on the section x := max(d) = 20-in ^min := m'n^ ~ ^'ln £ci:= 0.0003 M<p(c,£ci,d,As,fc)=44.4-kip-ft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) wMmaxk = "NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width LC201 - LC206 fc= l pe:= Pk = 36 kip ^:= jl/ft = 42.9- ^=dR1%0Jm As--=AsRl={]6y* Mm guess: ^= 0.2-h = 0.4ft Given Pif^c, eci, d,As,fc) = -9 kip c>0 Pv(c,eci,d,As,fc)=Pc

=: Pindices

] Converged results c = 0.8ft £ci = 0.00021 Pp(c,£ci,d,As,fc)=36kip M<f{c,eci,d,As,fc)=42.9-kipft fMcm . := ft {£ci'fc) fMcm , = 805-8Psi k k fMsOFk := £s{c'dmax-£ci)-Es fMsOF, = ~7A-ksl fMsIFk := £s{c-dmin<£ci)-Es fhfsIFk = 3-5'ksi Maximum crack width wMmaxk-= "NA" Calc. No: C-CSS-099.20-063 Rev. 001 Sheet No: 162 of 212 Sheet Rev.: 000 kip-ft Demands acting on the section mt/:= max(d) = 20-in Mmmii= min{d) = 4-in &&= 0-0003 M<dc,eci,d,As,fc)=57.8-kip-ft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) wMmaxk = "NA"-in

Attachment E Calc. 1 Reinforcement / Concrete Stresses and Crack Width Sheet Sheet LC301 - LC302 guess £i= Given Piplc = 36.3kip M>>/:= Mk ~ 43.4-kip-ft (0-8\\ 2 in fiw;= ^-sRl = 0.2h = 0.4ft

,£ci,d,As,fc)=-5.6kip c>0 Mif{

(&* -eci'd'As<fc)=Pc C'tci'd-As-fc)=Mc j:= Find(c,£cl) Converged results c = 0.8ft £ci = 0.00022 PNi^c.e^d.A, MNt^c,£ci,d,A fMcmk-=fc{eC fMsOFk:= £s{c f\\tsIFk:= es{c- ,fc)=-8.9kip s,fc)=9.6-kipft -fc) fMcmk = max'sci) s JMsOFk dmin<£ci)-Es fMsIFkz Maximum crack width wMmaxk~ "NA" 814.6psi = -7.5-ksi = 3.5-ksi <<Jo: C-CSS-099.20-063 Rev. 001 No: 163 of 212 Rev.: 000 Demands acting on the section ax(d) = 20-in ]bnm^= min(d) = 4-in -- 0.0002 c,£ci,d,As,fc)=38.9-kip-ft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) wMmaxk = "NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width LC401-404 tor* ^pk=362kip M*:=Mk = R1 ~ {20) m ^~ sR1 ~ {1.6) m guess c\\= o.2-h = 0.4ft Given Pirfc, £ci,d,As,fc) = -9 kip c>0 P<f(c,eci,d,As,fc)=Pc M<f(c,eci,d,As,fc)=Mc Converged results c = 0.6ft £ci = 0.00032 P^{c,eci,d,As,fc)=36.2kiP Mif{c,£ci,d,As,fc^ = 61.8-kip-ft fMcmk ~ fc{£ci'fc) fMcmk = ]182/ fMsOF^ ^.^W£i fMsOFk = fMsIFk ~ £s{c>dmin>£ci)Es ffAslF k = 4Ski Maximum crack width wMmaxk:= "NA" Calc. No: C-CSS-099.20-063 Rev. 001 Sheet No: 164 of 212 Sheet Rev.: 000 61.8-kip-ft Demands acting on the section ^MUt^= max{d) = 20-in ^atm^= min^) = 4^n Mp(c,£ci,d,AsJc)=57.8-kipft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand 7pSi maximum concrete compressive stress ksi O.F rebar stress (tension is negative) i I.F rebar stress (tension is negative) wMmaxk="NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width LC501-502 k=4 i guess Given Converged c = 0.6ft f*!* ~A -{°'%n \\20) sR1 {l.6J &=0.2-h = 0.4ft PV{c,eci,d,As,fc)=-9kip c>0 Pv(c,£ci,d,As,fc)=Pc results eci = 0.00017 P*(c.eci, M<f(c,£ci,

JMcm,

" J Maximum

^M^ := UAs,fc) = 17kip d,As,fc)=32.8-kipft Cc{£ci'fc) fMcmk = 645APi £s{c-dmax- £cij-Es flAsOF k = ~8Sk £s{c-dmin-£ci}-Es fMslFk =1Zksi crack width "NA" Calc. No: C-CSS-099.20-063 Rev. 001 Sheet No: 165 of 212 Sheet Rev.: 000 l.8-kip-ft Demands acting on the section ^^^1= max(d) = 20-in 4sm>>>c= minW = 4-in &>>}= °-0003 M<p(c,£ci,d,As,fc)=57.8-kip-fi Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress a O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) wMmaxk="NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width LC601 L=5 ^=pk = 6-2kiP M>>--=Mk = -Z8kl For negtive moment, switch OF and IF rebar ^:= {~M)k = 2.8-kipft f 4\\ 'AsRl^ yAsR1oj -{L6\\in guess ^= o.2-h = 0.4ft Given Pt^c£ci,d,As.fc) = 13.3 kip c>0 P<f{c,£ci,d,As,fc)=Pc M(p(c,£ci,d,As,fc)=Mc Converged results c = 1.7ft sci = 0.00001 P?(c,£ci,d,As,fc)=6.2kip Mif(c,£ci,d,As,fc)= 2.8-kip-ft fMcm k'-=fc{ £ci'fc ) fMcm k = 46 fltfsOF k := £s\\c' dmin * £ci)' Es ^MsOF k ~ C fhfsIF k := es{c' dmax * £a)' Es flMsIF k = 0 Maximum crack width NA:=c wMmaxk

r'* 0.000091-R- I in .2 ij in ( fMsIF)k ksi Calc. Sheet Sheet No: C-CSS-099.20-063 Rev. 001 No: 166 of 212 Rev.: 000 p-ft Demands &maMS~ ma .9psi ksi h-NA -NA-tb in,0in acting on the section 0.0003 '£ci d,As,fc)=44.4-kipft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) R ?'8 Mr]

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev.: C-CSS-099.20-063 Rev. 001 167 of 212 000 LC701-LC702 For negtive moment, switch OF and IF rebar d:= dK1 = yin Kl \\20 A.n; V "J 1.6) 2 08i guess: ^= 0.2-h = 0.4ft Given P<p(c, £ci,d,As,fc) = 13.3 kip c>0 Pp(c,eci,d,As,fc)=Pc My{c,eci,d,As,fc)=Mc Converged results c = 0.3ft £ci = 0.00002 P^c,eci,d,As,fc)=-1.5kip M<p(c,£ci,d,As,fc)=3.5-kipft

'Mem

'~ Jc \\ ci'J c) fMsOFk:= £s{c-dmin'£ci)-Es fMsIFk'-= £s[c'dmax'£ci)-Es Maximum crack width wMmaxk := max tu A 0.000091-l R in 2 in Demands acting on the section = 00003 c) =44.4-kip-ft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) h-NA h-NA-tb ksi -5 in.Oin

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 168 of 212 000 LC801 - LC803 k=7 S^=pk = 30-8kiP M* (4\\ (0.8^ &i=dRl = [20\\ln As:=AsR]=[16 9UeSS: ^=0.2-k = 0.4ft Given P(f{c,eci,d,As,fc)=-9kip c>0 P<f{c,eci,d,As,fc)=Pc M<f(c,eci,d,As,fc)=Mc Converged results c = 0.7ft eci = 0.0002 P(f(c,£ci,d,As,fc)= 30.8 kip M?{c,eci,d,As,fc)=40.4-kipft 'Mem i,'~ c{ ci'J c)

Mcm
MsOFk'~ £s\\c' max'£ci) s
MsOF fMsIF,

= £s[c-dmin'£ci)'Es

^MsIF Maximum crack width w = "NA "

= m

= 40.4-kipft Demands acting on the section 'in dmw:= rnaxid) = 2Oin /dMmi^= min(d) = 4'in &>>}= °-0003 Mv{c,eci,d,AsJc)=57.8-kiPft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand = 768.3psi maximum concrete compressive stress = -7.8-ksi O.F rebar stress (tension is negative) = 3.1ksi I.F rebar stress (tension is negative) l wMmaxk = "NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 169 of 212 000 Summarv of Circumferential Stresses and Crack width in Reqion D2 Maximum Concrete stress 77.5

805.8 MAXIMUM DCRM 814.6 1182.7 645.4 46.9 98.1 768.3 ) Psi fMsOF = Maximum OF rebar stress 0.4 -7.4 -7.5 -15 -8.5 0.3 -0.2 -7.8 Maximum IF rebar stress ksi V 3.5 3.5 4.5 2.2 0 -3.8 3.1 ksi wMrru2x = Maximum crack width "NA" "NA" "NA" "NA" 0 0 "NA" in

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 170 of 212 000 CALCULATION OF MERIDIONAL STRESSES AND CRACK WIDTH: D2 Region Moment curvature relations presented in this section are derived using strain compatibility (no slip among concrete and rebar) and equilibrium concepts. The figure shown below and the following assumptions are used for the derivations: Plane sections before bending remain plane after bending The stress strain behavior of concrete and steel is known Neutral "axis ~ Section strain stress Internal Forces External demand Concrete (ec) and steel (es) strains are calculated using strain compatibility as follows: 'cm -y c-dt cm ec

c The concrete and steel stresses are calculated using appropriate constitutive laws, as those given below. In general any type of constitutive behavior can be used. Conservatively, the concrete tensile capacity is ignored. 1000fc ec-[l - 250-ec) if £c < 0.002 unconfined concrete model per Park and Paulay 1975 fc-(l.2-100-£r) if £r> 0.002 fe 4000-3000-2000-iooc-0.003 Stress strain curve used for concrete

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 171 of 212 000 \\Es£si \\fy r Elastoplastic reinforcing steel model y f**-*a>fy 60-40-L [ksi] 20" 0.002 0.003 Stress strain curve used for reinforcing steel The concrete resultant (C) and its moment arm (yb) are calculated by integrating the concrete stress/c(ec) that acts on each infinitesimal concrete area b{6)dy. Note that b{6) is the width of the concrete section a location d. yo =Max(0,c-h) C= fc(ec)-b(y)dy Distance from the neutral axis to the first concrete fiber under compression. Internal resultant of concrete compression i f y -c-.\\ f le \\-b(y)-y dy Location of concrete compression measured from the top concrete fiber F- =f -A i Internal steel forces Equilibrium of the reinforced concrete section is verified as follows n M =C-\\- i=0 F.-\\--d; Pure compression capacity Mc = j=0 Pure tension capacity j=0

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 172 of 212 000 MATERIAL PROPERTIES fc = 4000psi fy=60ksi 7c = 150pcf a =0.0000055 Ec = 57000 Ifc -psi = 3605-ksi Es = 29000ksi u= 0.25 /v ecu:= 0.003 St = 7.5-Ifc psi = 0.00013 Ec Rebar Configuration h m 2.0ft Shell thickness dR]:= l.56 Concrete compressive strength Rebar yielding stress (note that this values is only to define stress-strain curve) Concrete unit weight Coefficient of thermal expansion (1/F) Concrete modulus of elasticity Rebar modulus of elasticity Concrete Poisson's ratio Yielding strain Concrete crushing strain Concrete tensile cracking strain Modular ratio

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 173 of 212 000 Stresses in Region D2 for maximum DCR Demand of moments and R1M:= (-10.09 43.20 41.31 59.15 31.57 -6.18 -8.00 K 36.06 36.04^ 41.45 37.41 37.37 19.94 20.36 28.80 28.71, membrane forces P:=R1M{1) kip = (36^ 41.5 37.4 37.4 19.9 20.4 28.8 ,28.7, kip M := RIM kip-ft = 43.2 41.3 59.1 31.6 -6.2 -8 36.1 j kip-ft Parameters for crack width calculation (ACI SP20) tb := 4in

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 174 of 212 000 LC101 k:=0 Pc := Pk = 36kip M£ := A*fc = -10.1-kip-ft For negtive moment, switch OF and IF rebar d := dR] = guess Given Convergec c = 2.2ft )k = lO.lkip-ft \\ri^ AsR10 ^:= 0.2-/i = 0.4/r c>0 I results £ci = 0.00005 Pf{c,eci, MAc'£a k fMsOFk-Maximum wMmaxk := d,As,fc)=36kip ,d,As,fc)= 10.1-kip-ft fc(£crfc) fMcmk = 209.6Psi = £s{c'dmin.'£cij-Es fMsOFk = L3ksi £s{c'dmax' £cij-Es fttsIF k = 9'44si crack width "NA" Demands acting on the section

= max(d) = 20-in

^m[n '- min(d) = 4-in ed := 0.0003 Mif(c,eci,d,As,fc)=44.4-kip-ft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) wMmcak="NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width LC201 - LC206 L-=1 L*=pk = 4L5kiP M*s=b &= dRl = [20)in As := AsR1 = [l 6)in guess j-= o.Zh = 0.4ft Given P<{(c,eci,d,As,fc) = -9kip c>0 M<f{c,eci,d,As,fc)=Mc Converged results c = 0.8ft £ci = 0.00021 Pif{c,eci,d,As,fc)=41.5kip M<f(c,£ci,d,As,fc)=43.2-kipft JMcm

~ Jc[£ci'J c)

JMcm , ~ k k JMsOFk -= £s[c' max-£ci) s JMsOFk ?MsIFk := £s{c'dmin<£a)'Es ^MslFk = Maximum crack width wMmaxk:= "NA" <k=43.2-796.3psi = -6.2-ksi 3.6-ksi Calc. Sheet Sheet kipft No: C-CSS-099.20-063 Rev. 001 No: 175 of 212 Rev.: 000 Demands acting on the section 1^:= max(d) = 20-in {bati^- min(d) = 4-in M(Ac 0.0003 .£ci.d.As.fc)=57.8-kip-ft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) l.F rebar stress (tension is negative)

Attachment E Reinforcement / Concrete Stresses and Crack Width LC301 - LC302 &2 Pe,= Pk = 37.4kiP jfa -(4)in -A (°-8) \\ / V s guess c\\= 0.2-h = 0.4ft Given P(f{c,£ci,d,As,fc)=-9kip c>0 P?(c,eci,d,As,fc)=Pc Converged results c = 0.8ft eci = 0.0002 P<f(c,£ci,d,As,fc)=37.4kip M<p(c,£ci,d,As,fc)=41.3-kipft

Mcm,
" *c \\ £ci'J c)

JMcm JMs0F k -~ £s\\c-amax * £ci)' fis JMsOF

MsIF. -= £s[c' min * £ci)'

s JMslF - Maximum crack width -Mk=41.3 in d^ = 769 psi = -6.4-ksi = 3.4-ksi Calc. Sheet Sheet kip-ft /£*>>j= Mp(c No: C-CSS-099.20-063 Rev. 001 No: 176 of 212 Rev.: 000 Demands acting on the section 0.0003 ,£-d,rf,A5>/c)=J7.5-^>-/r Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) wMmaxk="NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width LC401-404 L=3 &T dRl guess Given Pe.= Pk = 37.4kip M*-=Mk = (4] (0.8\\ 2 \\20j A*t=AsRl=[16Jin £f= 0.2-h = 0.4ft Pp(c,£ci,d,AsJc)=-9kip c>0 M?{c,eci,d,As,fc)=Mc Converged results c = 0.7ft sci = 0.00031 Mc'eci M<f{c,£cl JMcm. Maximum wMmaxk := d,As,fc)= 37.4 kip ,d,As,fc)=59.1-kipft fc{£c-rfc) Jfcfa^-M* = £s\\c' dmax * £ci) Es fltfsOF k=~]3'

es{c-dmin>sci)-Es fMsIFk = 4-4'ki crack width "NA" Calc. No:

C-CSS-099.20-063 Rev. 001 Sheet No: 177 of 212 Sheet Rev.: 000 59.1-kipft Demands acting on the section 4^^= max(d) = 20-in 3*0^= min(d) = 4-in gtf 0.0003 M<p(c,eci,d,As.fc)=57.8-kip-ft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand >psi maximum concrete compressive stress 5-ksi O.F rebar stress (tension is negative) i II.F rebar stress (tension is negative) wMmaxk="NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 178 of 212 000 LC501-502 k=4 guess Given P^=Pk = 19.9kiP M*=Mk = ~ yiOJ '"~ sR1 ~ {].6J m £g= 0.2-h = 0.4ft P<f(c,£ci,d,As,fc)=-9kip c>0 Converged results c = 0.7ft £ci = 0.00016 Mc'£ci Mif{c,ecl fMcmk-= fMsOFk-fMstFk-- Maximum wMmaxk '~ d,As,fc)=19.9kip ,d,As,fc)=31.6-kipft fc{£cffc) fMcmk = 614-5 = £s[c, dmax, ecij Es fjrfsOF k = ~^'A 1 £s\\c' dmin > £ci)' Es ffdsIF k = 2'3'k crack width TOT 31.6-kip-ft psi ksi K Demands acting on the section max(d) =20-in 4shhik= m^n(^) =4-in c,eci,d,As,fc)=57.8-kip-ft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) wMmaxk="NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 179 of 212 000 LC601 A:= 5 £e^= pk = 20A kiP M^= Mk = "*2## For negtive moment, switch OF and IF rebar v:= (-M)k = 6.2-kip-ft '.:= dR1 =

sRl 1.6 0.8 2

in guess ^:= 0.2-h = 0.4ft Given P<f(c,£ci,d,As,fc) = 13.3 kip c>0 Pv(c,£ci,d,As,fc)=Pc M?(c,eci,d,As,fc)=Mc Converged results c = 2.1 ft sri = 0.00003 P?(c,eci,d,As,fc)=20.4kip Mif{c.eci.d.AsJc)= 6.2kip-ft fMcm k:= fc{£ci>fc) fMsOF k := £s{c-dmin * £ci)' Es fMsIFk-= £s{c'dmax'£cij-Es Maximum crack width NA~c Demands acting on the section

= max(d) = 20- in M<p(c,£ci,d,As,fc)=44.4-kip'ft fMsIFk = °-2-ksi h-NA h-NA-tL wMmaxk :=

\\ 0.000091-l R-in 2 in -5 in.Oin Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) R = 0.3

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 180 of 212 000 LC701-LC702 i^-= Mk = kip-ft For negtive moment, switch OF and IF rebar = Pk = 28.8 kip

= (~M)k = 8-kip-ft u

guess Given [ir K20j &:= 0.2-h Pv{c, c>0 Pf{c MiAc £cv £cv

£ci

= = 0.4ft d,A^ d,A^ .d.As-fc sRlj A (1-6) . 2 [in K0-8J = 4.8 kip = Pc )=Mc Converged results c = 2.3ft eci = 0.00004 MN^c,eci,d,As,fc)=0.9-kip-ft fMcmk:=fc{£ci'fc) fMsOFk:=es{c-dmin'eci)-Es fMsIF k:= es{c'dmax'£cij'Es Maximum crack width Demands acting on the section max(d) = 20-in r °-0001

= min(d) = 4-in Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative)

I.F rebar stress (tension is negative) h-NA h-NA-tb R = 0.4 wMmaxk := ksi -5 in,0in

Attachment E Reinforcement / Concrete Stresses and Crack Width LC801 - LC803 &=7 S^=pk = 28-^p M^=**k (4\\ (0.8~\\ 2 R1 \\20) s sR1 {I.6J 9U6SS: ^0.2-h = 0.4ft Given Pif{c, sci, d,As,fc) = -9 kip c>0 P<<{c,eci,d.As,fc)=Pc M<f(c,eci,d,AsJc)=Mc Converged results c = 0.7ft eci = 0.00018 P¥{c,eci,d,As,fc)=28.7kip Mcf(c,eci,d,As,fc)=36.1-kipft fMcmk-=fc{£c-rfc) fMcm^' JMsOF -= £s\\c

max' £ci)'

s JMsOF - fMsIF i := £s{c

dmin * £ci)' Es

^MsIF k~2'8 Maximum crack width w

=

"NA" Calc. Sheet Sheet = 36.1-kip-ft M<p(c .8PSI -,6-ksi ksi No: C-CSS-099.20-063 Rev. 001 No: 181 of 212 Rev.: 000 Demands acting on the section 0.0003 ,eci,d,As,fc)=57.8-kipft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) wMmaxk="NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 182 of 212 000 Summarv of Meridional Stresses and Crack width in Region D1 Maximum Concrete stress

209.6 796.3 MAXIMUM DCRM 769 1128.2 614.5 123.1 167 v 684.8 j Psi Maximum OF rebar stress 1.3

-6.2 -6.4 -13.5 -7.2 0.8 1 -6.6 \\ ksi ftislF = Maximum IF rebar stress 3.6 3.4 4.4 2.3 0.2 0.3 wMmax = Maximum crack width "NA" "NA" "NA" "NA" 0 0 "NA")

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 183 of 212 000 CALCULATION OF CIRCUMFERENTIAL STRESSES AND CRACK WIDTH: D3 Region Moment curvature relations presented in this section are derived using strain compatibility (no slip among concrete and rebar) and equilibrium concepts. The figure shown below and the following assumptions are used for the derivations: Plane sections before bending remain plane after bending The stress strain behavior of concrete and steel is known Neutral axis A Section strain stress Internal Forces External demand Concrete (ec) and steel (e ) strains are calculated using strain compatibility as follows: -cm -y The concrete and steel stresses are calculated using appropriate constitutive laws, as those given below. In general any type of constitutive behavior can be used. Conservatively, the concrete tensile capacity is ignored. iooo-fc ec-(; - 250- sc) if ec < O.OO2 Uncorrfjned concrete model per Park and Paulay 1975 fc -{1.2 - 100-sr) if er > 0.002 /c[psi] 40O0 3000 2000 1000 0 / 1 1 1 0.001 0.002 0.003 Stress strain curve used for concrete

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 184 of 212 000 Es£si /> $ Es-esi>fy Elastoplastic reinforcing steel model [ksi] 0.002 0.003 Stress strain curve used for reinforcing steel The concrete resultant (C) and its moment arm (yb) are calculated by integrating the concrete stress /c(ec) that acts on each infinitesimal concrete area b{d)dy. Note that b(6) is the width of the concrete section a location d. y =Mox(0,c-h) C = fc(ec)-b(y)dy Distance from the neutral axis to the first concrete fiber under compression. Internal resultant of concrete compression 1 f yb = c fc(ec)-b(y)-y dy Location of concrete compression measured from the top concrete fiber F; = LrAt.; Internal steel forces I it SI Equilibrium of the reinforced concrete section is verified as follows M=C-l M7-4 Pure compression capacity j=o Pure tension capacity j=o

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 185 of 212 000 MATERIAL PROPERTIES fc = 4000psi fy=60ksi 7C

150pcf a= 0.0000055 Ec = 57000 If c -psi = 3605-ksi Es a 29000ksi u= 0.25 fv ecu:= 0.003 et = 7.5-lfc-psi = 0.00013 Ec Rebar Configuration h = 2.0ft Shell thickness do! :=

I \\in \\20) 1.56 \\j \\in 2-1.56) Concrete compressive strength Rebar yielding stress (note that this values is only to define stress-strain curve) Concrete unit weight Coefficient of thermal expansion (1/F) Concrete modulus of elasticity Rebar modulus of elasticity Concrete Poisson's ratio Yielding strain Concrete crushing strain Concrete tensile cracking strain Modular ratio

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 186 of 212 000 Stresses in Reaion 03 for maximum DCR Demand of moments and membrane forces / R1M:= -5.08 -4.63 42.33 -72.32 41.79 -71.20 74.77 18.86 41.02 0.00 -3.00 -1.14 -3.83 -13.40 37.52 -53.58 P:= RIM' -kip = -72.3 -71.2 18.9 0 -1.1 -13.4 V kip M := RIM -kip-ft = 42.3 41.8 74.8 41 -3 -3.8 kip-ft Parameters for crack width calculation (ACI SP20) tb := 4in d'w := dw

Attachment E Reinforcement / Concrete Stresses and Crack Width LC101 k:=0 Pc := Pk = -4.6kip Mc := Mk = -5.1-kip-ft For negtive moment, switch OF and IF rebar J^joJ^ (~M), = 5.1-kip-ft d:=dRl = \\20\\-in V= AsRlQ) {3A\\ 2 guess ^:= o.2h = 0.4ft Given P(f{c,eci,d,As,fc) = -5.6kip c>0 Py{c,eci,d,As,fc)=Pc M?(c,eci,d,As,fc)=Mc Converged results c = 0.3ft £-i = 0.00002 P(f{c,eci,d,As,fc) =-4.6kip M<p[c,sci,d,As,fc)= 5.1-kip-ft fMcm,'-=fc{£ci'fc) fMcm, = 94-3Psi k k JMsOF

= £s\\c'dmin * £ci)'Es

^MsOF , = ~0-2ksi JMslFk := £s[c'dmax' £ci)'Es JMsIFk ~ ~3-7'ksl Maximum crack width Calc. Sheet Sheet vc:= ma £ci~ Mipic No: C-CSS-099.20-063 Rev. 001 No: 187 of 212 Rev.: 000 Demands acting on the section x(d)=20-in dmin:=min(d)=4-in ,eci,d,As,fc)=60.1-kip-fi Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) wMmaxk = "NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack LC201 - LC206 L=1 S*rpk = 3kiP f 4 V fl-guess gj= 0.2-h = 0.4ft Given P^c,£d,d.As,fc) = -50.9kip c>0 P?(c,eci,d,As,fc)=Pc My{c,eci,d,As,fc)=Mc Converged results c = O.lft eci = 0.00004 P<f(c,Sci,d,As,fc)=-72.3 kip M<p(c,£ci,d,As,fc)=42.3-kip-ft JMcm

~
c\\£c'vJ c)
M fMsOF k:= £s[c' dmax * £ci)' Es

$M fMsIFk:= £s{c'dmin'£ci)'Es ?M Maximum crack width w = "NA" Calc. No: C-CSS-099.20-063 Rev. 001 Width Sheet No: 188 of 212 Sheet Rev.: 000 ^:= m = 42.3-kipft Demands acting on the section Vin jh>>a>>,:= max(d) = 2O'in ^mm^-= min(d) = 4-in &ei}= 0.0003 M(f{c,£ci,d,As,fc)=87.2-kipfr Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand = I46.5psi maximum concrete compressive stress 0F =-2l.7-ksi O.F rebar stress (tension is negative) rp =-3.5-ksi I-F rebar stress (tension is negative) wMmaxk = "NA"-'ln

Attachment E Reinforcement / Concrete Stresses and Crack LC301 - LC302 L,2 P^Pk = -7Uk, k /&:= dRl = [20Jm ^'= AsR1 = {3 1 guess: c^ 0.2-h = 0.4ft Given Pifi{c,£ci,d,As,fc) =-33.5 kip c>0 P<f{c,eci,d,As,fc)=Pc M<p(c,eci,d,As,fc)=Mc Converged results c = O.lft £ci = 0.00004 PNy{c,£ci,d,As,fc) =-72.1 kip MNV{c,£ci,d,As,fc)=4hkip-ft fMcmk--=fc{£ci'fc) fli fMsOF k £s[c' dmax * £ci)' Es f?>4 fltfsIF k := £s[c

dmin * £ci)' Es

^M Maximum crack width wMmaxk := "NA" Calc. No: Width Sheet No: Sheet Rev.: C-CSS-099.20-063 Rev. 001 189 of 212 000 ^:= m = 41.8-kip-ft Demands acting on the section \\'in &auMJ= max(d) = 20- in jhttiut}= min(d) = 4-in /£t^= 0.0002 Mifi{c,£ci,

mk = I48.4Psi

,OFk = -2L4ksi slFk=-3A-ksi d,As.fc)=58.5-kipft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) wMmaxk="NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack LC401-404 k=3 &T dRl = guess Given Converged c = 0.7ft P = P = {>>> Mc'£ci c> 0 PtAc'£ci My{c,Ecl results £ci = 0.0003 k Maximum wMmaxk ~

.-V/c) =

>-*s-fc) fc[£ci'fc) 18.9kiP J a sR1 U = 0.4ft d,As,fc)=-50.9 kip d,As,fc)=Pc -d'As'fc)=Mc ~-18.9 kip = 74.8-kip-ft z £s[c'dmax-£ci)'Es ^Msi £s{c-dmin *£ci) s JMsi crack width "NA" Width lv=Mk = 74.8- %2 Mm n= 1117.5 psi k 2F = -13.5-ksi ,F = 4.3-ksi Calc. Sheet Sheet kipft M<fi[c No: C-CSS-099.20-063 Rev. 001 No: 190 of 212 Rev.: 000 Demands acting on the section iwc(d) = 20-in ^MtHu<= """(^O = 4-in 0.0003 ,eci,d,As,fc)=87.2-kip-ft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) wMmaxk="NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width LC501-502 fc=4 i &= dRl = guess Given Converged c = 0.6ft ^=Pk = OkiP ^Mk = (4\\in A.-A -(L6\\in2 \\20) m sR] {3.1) " &:= 0.2-h = 0.4ft P^c,eci,d,As,fc)=-50.9kip c>0 Pv(c,eci,d,As,fc)=Pc M<f(c,£ci,d,As,fc)=Mc results sci = 0.00016 PAc'£cv Mip[c,£ci, fMcmk=. fMsOFk--= fltfsIF.

=

i,As,fc)=0kip d,As.fc)=41-kip-ft fc{£ci'fc) fMcmk = 609-7 £s{c

dmax' £ci) Es

^MsOF fc = ~9 £s{c'dmin'£a)-Es ff4slF k = L9'k Maximum crack width wMmaxk

=

"NA" Calc. No: C-CSS-099.20-063 Rev. 001 Sheet No: 191 of 212 Sheet Rev.: 000 41-kipft Demands acting on the section /£kMB^:= max(d) = 20-in j^mkiK= min^) = 4^n JSgfP' 0.0003 M^c,ectd,AsJc)=87.2kip-ft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand psi maximum concrete compressive stress isi O.F rebar stress (tension is negative) si I.F rebar stress (tension is negative) w

"N£L".in

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 192 of 212 000 LC601 Demands acting on the section 20 1.6 3.1 2 in ^ max(d) = 20-in = min{d) = 4-in Mathcad has a difficulty to converge, solve this case by hand as: fMcm k := °Psi fMcm k = ° Psi

= -5.9ksi J

t

= -7.4ksi 1

maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) Check: AsRl kAsRl Q = ~30 kiP fMsOFkAsRl /(j - dmax ^ ~ d Maximum crack width NA:= c h-NA h-NA-t R = 1.3 b wMmaxk tb A 0.000091-l R in 2 in ksi -5 in,0in

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 193 of 212 000 LC701-LC702

= P

= -13.4 kip ]$#}= M = -3.8-kip-ft Demands acting on the section Mathcad has a difficulty to converge, solve this case by hand as: fMcmk~°Psi k:=-]Oksi

= ~15ksi maximum concrete compressive stress O.F rebar stress (tension is negative)

I.F rebar stress (tension is negative) Check: AsRl kAsRl Q = ~54-6 kiP ~ ~ dmin Maximum crack width = c h-NA h-NA-t, R = A 0.000091-l R-in 2 in ksi -5 in, Oin

Attachment E Reinforcement / Concrete Stresses and Crack Width LC801

LC803 k=7 s^^-szm jt^*B-ju

&;--dRi-\\ 20yn A^-AsRi-\\3]vin 3^ss: ^02.h = 04fi Given P<fi{c,£ci,d,As,fc) = -50.9kip c>0 Mcp(c,£ci,d,As,fc)=Mc ."= Findi c, £ l Converged results c = 0.2ft £ci = 0.00007 P¥{c,£ci,d,As,fc)=-53.6 kip Mp(c,£ci,d,As,fc)=37.5-kip-ft fMcmk := fc{£ci'fc) fMcm = 2?<<^?8B fMsOF - := £5(c * ^max' ^50' £5 fMsOF k=--l7-4-ks ffrlsIFk := £s\\c'dmin'£ci)'Es fhtsIFk ~ ~]-8'kst Maximum crack width w

"NA" Calc. No: C-CSS-099.20-063 Rev. 001 Sheet No: 194 of 212 Sheet Rev.: 000 'kip-ft Demands acting on the section 3^= max(d) = 20-in 4mi>>i<= min(^> =4-in Mp(c,£ci,d,As,fc)=87.2-kip-ft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress i O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) wMmaxk="NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 195 of 212 000 Summarv of Circumferential Stresses and Crack width in Reqion D3 MAXIMUM DCRM JMcm n Concrete stress 94.3 1 146.5 148.4 1117.5 609.7 0 0 i 276.3 Psi fMsOF = Maximum OF Maximum IF Maximum crack ebar stress rebar stress ' -0.2 ^ -21.7 -21.4 -13.5 -9 -5.9 -10 K-17.4y flUsIF = f-3.7\\ -3.5 -3.4 4.3 1.9 -7.4 -15 K-1.8J ksi wMmax = ( "NA" N ISt" "NA" "NA" "NA" 0.002067 0.008614 L W

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 196 of 212 000 CALCULATION OF MERIDIONAL STRESSES AND CRACK WIDTH: D3 Region Moment curvature relations presented in this section are derived using strain compatibility (no slip among concrete and rebar) and equilibrium concepts. The figure shown below and the following assumptions are used for the derivations: Plane sections before bending remain plane after bending The stress strain behavior of concrete and steel is known j i Neutral 1 f, JL Section strain stress Internal Forces External demand Concrete (ec) and steel (es) strains are calculated using strain compatibility as follows: £cm

e. =

y c-di The concrete and steel stresses are calculated using appropriate constitutive laws, as those given below. In general any type of constitutive behavior can be used. Conservatively, the concrete tensile capacity is ignored. 1000-fc ec-[1 - 250- £c) if £c < 0.002 fc-[l.2-100-£r) if £r> 0.002 Unconfined concrete model per Park and Paulay 1975 fc 4000-3000-2000-1000-0.003 Stress strain curve used for concrete

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 197 of 212 000 Esesi fy if* Elastoplastic reinforcing steel model fy 60~ 40-20-0.001 £ 0.002 0.003 Stress strain curve used for reinforcing steel The concrete resultant (C) and its moment arm (yb) are calculated by integrating the concrete stress/c(ec) that acts on each infinitesimal concrete area b{6)dy. Note that b(6) is the width of the concrete section a location d. yo =Max(O,c - h) C= fc(ec)-b(y)dy yb=c~~c' Distance from the neutral axis to the first concrete fiber under compression. Internal resultant of concrete compression Location of concrete compression measured from the top concrete fiber Internal steel forces Equilibrium of the reinforced concrete section is verified as follows M =C-\\- i = 0 i=0 Pure compression capacity M_ = j=0 Pure tension capacity M,= j =0

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 198 of 212 000 MATERIAL PROPERTIES fc m 4000psi fy=60ksi 7c = 150pcf a =0.0000055 Ec = 57000 Ifcpsi = 3605-ksi Es m 29000ksi = 0.25 ecu:= 0.003 £t 9 7.5-\\fc-psi = 0.00075 Ec n = =8 Ec Rebar Configuration h = 2.0ft Shell thickness dR1 := i

K20, 7.56V 2

\\in 1.56 Concrete compressive strength Rebar yielding stress (note that this values is only to define stress-strain curve) Concrete unit weight Coefficient of thermal expansion (1/F) Concrete modulus of elasticity Rebar modulus of elasticity Concrete Poisson's ratio Yielding strain Concrete crushing strain Concrete tensile cracking strain Modular ratio

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 199 of 212 000 Stresses in Reqion D3 for maximum DCR Demand of moments and membrane forces R1M:= (29.73 25.90^ 69.94 29.99 66.81 27.80 89.69 27.80 47.57 14.95 19.02 16.27 24.91 22.10 J8.04 22.10j P:= RIM -kip = (25.9^ 30 27.8 27.8 15 16.3 22.1 K22.1) (o) kip M:= RIM(kip-ft = (29.7^ 69.9 66.8 89.7 47.6 19 24.9 ,58 ) kip-ft Parameters for crack width calculation (ACI SP20) tb := 4in h 2 b dw := dw

Attachment E Reinforcement / Concrete Stresses and Crack Width LC101 k:=0 d:=dR1 = guess Given Pc := Pk = 25.9kip M£ := A^ = 29.7-kipft {20jln s'~ sR1~{l.6jm m ^:= 0.2-h = 0.4ft P<fi(c,£ci,d,As,fc)=-7.9kip c>0 P<f(c,£ci,d,As,fc)=Pc M?(c,£ci,d,As,fc)=Mc Converged results c = 0.8ft £ci = 0.00014 Mc'£ci fMsOFk: Maximum wMmaj:,

=

d,As,fc)= 25.9 kip ,d,As,fc)=29.7-kipft = £s[c, dmax, £c-j Es ffdsOF k=-4 z Ss{c>dmin>£ci)-Es fMsIFk = 2-3'ksi crack width "NA" Calc. Sheet Sheet No: No: Rev [ u := max(d) eci:= Mirfc C-CSS-099.20-063 Rev. 001 200 of 212 000 Demands acting on the section = 20-in dmin := m^n^) =4-in 0.0003

£ci-d,As,fc)=58.6-kip-ft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative)

I.F rebar stress (tension is negative) wMmaxk="NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width LC201 - LC206 fc=l Pg/:=Pk = 30kip ^:=M J J (4\\ (}A 2 &rdRl = \\?n\\m A^= AsRl = L, \\ln \\20) {1.6J guess ^= o.2-h = O.4ft Given P<f{c,£ci,d,As,f'c) = -7.9kip c>0 P?{c,eci,d,As,fc)=Pc Converged results c = 0.6ft £ci = 0.00036 P<fi(c,eci,d,As,fc)=30kip My(c,eci,d,As,fc)=69.9-kip-fi

Mcm k'~ Jc\\ecvJ c)
Mcm

. ~

MsOF k'=

£s[c' max' £ci) s ->MsOF k~ fMsIFk := £s{c'dmin<sci)'Es flUsIFk = Maximum crack width w = "NA" Calc. Sheel Sheet <k = 69.9-kiP-ft No: C-CSS-099.20-063 Rev. 001 No: 201 of 212 Rev.: 000 Demands acting on the section jjmut;-= max(d) = 20-in ^m^= min{d) = 4-in Mif{c 1306.6psi = -20.3-ksi 4.3-ksi 0.0003 ,eci,d,As,fc)=58.6-kiPft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) wMmaxk "NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width LC301 - LC302 &=2 ^P^ 27.8 kip ^=Mk = (4 V (L6~) 2 guess £j= o.2-h = 0.4ft Given P<^c,£ci,d,As,fc) = -7.9kip c>0 P¥{c,sci,d,As,fc)=Pc M<p(c,eci,d,As,fc)=Mc Converged results c = 0.6ft £ci = 0.00034 P<f{c,eci,d,As,fc)=27.8kip M<p(c,£ci,d,As,fc) = 66.8kipft fMcmk-=fc{£ci<fc) frb^-JMW JMsOF

= £s\\c'dmax'£ci)'E's JMsOF i,~~

JMsIF

= £!rlc>°mjn'5cij JMsIF k=

sl Maximum crack width wMmaxk := "NA" Calc.

Sheet Sheet 66.8-kip-ft No: C-CSS-099.20-063 Rev. 001 No: 202 of 212 Rev.: 000 Demands acting on the section O.OOOi Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) IF rebar stress (tension is negative) wMmaxk="NA"-in

Attachment E Reinforcement / Concrete Stresses and LC401-404 (4\\ /$U:= dRl = I 2Q I *m Ak= AsR guess c^= o.2-h = 0.4ft Given P(fi^c,£c^,d,As,fc ) = -7. i c>0 M<f(c,eci,d,As,fc)=Mc Converged results c = 0.5ft £ci = 0.00047 P(f(c,eci,d,As,fc) = 27.8 kip M<f{c, £ci,d, As,fc) = 89.7-kip-ft JMcm k'~ Jc\\£ci'* c) fMsOF k-= <^dmax'£chEs fMsIFk '= £s{c<dmin>£ci)-Es Maximum crack width w '= "NA " Calc. No: C-CSS-099.20-063 Rev. 001 Crack Width Sheet No: 203 of 212 Sheet Rev.: 000 M^:= M = 89.7-kip-ft Demands acting on the section (1.6\\ 2 j=\\ \\-in tijMa#/-= maxW = 20-in g^^.^ min(d) = 4-in \\].6J &>>!= °-0003 >kip M<p(c,£ci,d,As.fc)=58.6-kip-ft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand fMcm = 1656.8psi maximum concrete compressive stress fin OF =-29.4-^5! O.F rebar stress (tension is negative) fM IF = 5-ksi UF rebar stress (tension is negative) wMmaxk="NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 204 of 212 000 LC501-502 f 4 ] f 1.6 Jtfc~ R] l \\'ln >w*a'= sRl = guess c := 0.2-h = 0.4ft Given P!^c,eci,d,As,fc) = -7.9kip c>0

= Find(c,£c^

Converged results c = 0.5ft eci = 0.00024 P<p(c,£ci,d,As,fc) = 15kip Mip(c,£ci,d,As,fc} =47.6-kipft fMcmk := fc{£ci'fc) fUcm fMsOF k := £s\\c' dmax * £ci)' ^5 ^MsOl ?MsIF k:= £s{c' dmin' £ci)' Es ?MsIF Maximum crack width wMmaxk:= "NA" = M = 47.6-kip-ft \\ 2 ^= Mif{c = 912.2psi y =-15.4-ksi = 2.5-ksi Demands acting on the section 0.0003 ,£ctd,As,fc)=58.6-kipft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) wMmaxk="NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width LC601 Jti= 5 £a*:= Pk = 16J kiP M*~ Mk = 19-kiP-fi (4 s] (1.6) 2 J&= dRl = [2ojln ^= AsR1 = [l 6jm d' guess c^= 0.2-h = 0.4ft Given Pif(c.£ci,d,As,f'c) = -7.9kip c>0 P<f{c,£ci,d,As,fc)=Pc M<fi(c,£ci,d,As,fc)=Mc Converged results c = 0.7ft £ci = 0.00009 P<f{c,eci,d,As.fc) = 16.3 kip M^£ci,d,As,fc)-l,kipft fMcm := fc {£ci-fc) fUcm, = 349JPsi fMsOF

= £s\\Csdmax'£ci)'Es

^MsOFk = "*** fMsIFk := £s\\c'dmin'£cV'Es fltfsIFk = l-4'ksi Maximum crack width NA:= c h-NA h-NA-tb Mmaxk 3i 0.000091-

/?

in .2 \\ in '(-fMsOF) ] 5 ksi in,0in Calc. No: C-CSS-099.20-063 Rev. 001 Sheet No: 205 of 212 Sheet Rev.: 000 Demands acting on the section ^:= max(d) = 20-in ^jmm<= min(d) = 4'ln §^ 0.0003 M<p(c,£ci,d.As,fc)=58.6-kipft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative)

Attachment E Reinforcement / Concrete Stresses and Crack Width LC701-LC702 ^~ &a*:= Pk = 22AkiP MeS= Mk = 24-9-kiP-fi -(4)in -A -(L6]in2 R1 {20) sR1 {l.6j guess ^= 0.2-h = 0.4ft Given P<^c,eci,d,AsJc) = -7.9 kip c>0 P^c,eci,d,As,fc)=Pc Mp(c.eci,d,As,fc)=Mc Converged results c = 0.8ft £ci = 0.00012 P^c,sci,d>As,fc)= 22.1 kip M?{c,£ci,d,As,fc)= 24.9-kip-ft \\ fMcmk-=fc{£cvfc) fMcmk=4^P> fMsOFk :~ £s{c'dmax'£ci)'Es fMsOFk = ~*-ksi fMslF, := £s\\c'dmin'£ci)'Es JMsIFk = '-9'ksi Maximum crack width Mmaxk 3\\ 0.000091-R- \\in .2 y m Calc. No: Sheet No: Sheet Rev. Demands C-CSS-099.20-063 Rev. 001 206 of 212 000 acting on the section &HWH- "^^ M 2°-in J^m^-= mil%W " 44n &^= o-0003 M^c,£a i h-NA h-NA '(-fMsOF) 1 k 5 ksi in. -H Oin ,d,As,fc)=58.6-kip-ft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) I.F rebar stress (tension is negative) R-1.4 wMmaxk

Attachment E Reinforcement / Concrete LC801 - LC803 /6£\\:= dRl ~\\ Pn 9U6SS: a-<<M-Given P(Ac,£ci,d c>0 ?Ac-£cvd Converged results c = 0.5ft £ci = 0.0003 Stresses and Crack >.lkip A fl. 0.4ft ,As,fc)=-7.9kip >As'fc)=Pc P^c,eci,d,As,fc) = 22.1 kip Mv(c,eci,d,As,fc) = fMcmk-=fc{£ci>fc)

JMsOF,

= £s\\c'dmax' fMsIF k:= £s{c'dmin-£ 58-kip-ft fMc

£a)'Es fMs ci)'Es ^Ms Maximum crack width wMmaxk:= "NA" Width Calc. Sheet Sheet l^:= Mk = 58-kipft 6). 2 mk = 1099psi £&> Qp =-17.6-ksi IF k = 3.4-ksi No: C-CSS-099.20-063 Rev. 001 No: 207 of 212 Rev.: 000 Demands acting on the section iax(d) = 20-in /bating min(d) = 4-in 0.0003

Sci,d.As.fc)=58.6-kip-ft Neutral axis location Maximum compression strain Verification that P converges to demand Verification that M converges to demand maximum concrete compressive stress O.F rebar stress (tension is negative) il.F rebar stress (tension is negative) wMmaxk="NA"-in

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 208 of 212 000 Summarv of Meridional Stresses and Crack width in Region D1 MAXIMUM DCRM Maximum Concrete stress 540.9 1306.6 1252.6 1656.8 912.2 349.3 453.3 1099 fMsOF = Maximum OF rebar stress -20.3 -19.6 -29.4 -15.4 -3.2 -4 Maximum IF rebar stress v ksi fMsIF '2.3 4.3 4 5 2.5 1.4 1.9 3.4 \\ K ksi Maximum crack width "NA" "NA" "NA" "NA" 0 0 "NA" in

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 209 of 212 000 Algorisim and functions used in stress eviuation MAIN EQUATIONS USED IN THIS ATTACHMENT Concrete width as a function of d, (see figure below) b := lft Rebar strain £S{c'di'£cm):= c-d{ 'cm Rebar stress fysign(esi) if \\Es-esi\\ >f Rebar force at location d, Fs{Asi'fsi) ~fsi'Asi Concrete stress-strain relationship 1000-fc ec- (1 - 250- ec) if 0<ec<0.002 fc\\l.2-100-sc) if (ec> 0.002) 0 if £c<0 Concrete compression (*max(0,c) fc max(0,ch) Location of concrete compression C rmax(0,c) max(0,c-h) 'cm rmax(0,c) l "f >'./, max(0,c-h)

Attachment E Reinforcement / Concrete Stresses and Crack Width Calc. No: Sheet No: Sheet Rev. C-CSS-099.20-063 Rev. 001 210 of 212 000 Force equilibrium n <r-last{d\\ for is O..n fsi j=0 Moment equilibrium for BE 0..n

Attachment E Reinforcement / Concrete Stresses and Crack Width Location of neutral axis kd guess c := 0.1-h Given Range Variables ^:= 0.00007,0.00005.. 0.003 d-t := 0,1ft.. h Compressive Strains Used in Calculations Ae - 0.00005 i := 1.. 15 ecm

=

Solution for c < 0 Force equilibrium PN^ecm,drAb,fc):= n 4-last(ds)

ie 0..n fsi <<-/,M Calc. No: Sheet No: Sheet Rev.: (i-])-Ae+ 0.000001 C-CSS-099.20-063 Rev. 001 211 of 212 000}}

v t e Letter Sequence Other
TAC:ME4640, Steam Generator Tube Integrity (Approved, Closed)
Initiation Request, Request, Request, Request, Request, Request, Request, Request, Request, Request, Request Acceptance, Acceptance Supplement, Supplement, Supplement, Supplement, Supplement, Supplement, Supplement, Supplement, Supplement, Supplement, Supplement, Supplement, Supplement, Supplement Administration Withholding Request Acceptance Meeting Questions 28 February 2011 17 February 2011 17 March 2011 21 March 2011 18 March 2011 5 April 2011 30 March 2011 20 April 2011 2 May 2011 19 May 2011 20 June 2011 11 July 2011 12 July 2011 21 July 2011 21 July 2011 27 July 2011 11 August 2011 12 September 2011 9 September 2011 22 September 2011 11 October 2011 8 November 2011 22 November 2011 2 December 2011 2 December 2011 2 December 2011 27 December 2011 13 December 2011 21 December 2011 29 December 2011 12 July 2011 12 July 2011 12 July 2011 23 January 2012 30 January 2012 19 January 2012 1 February 2012 9 February 2012 27 January 2012 21 February 2012 22 February 2012 2 March 2012 27 February 2012 22 February 2012 24 February 2012 16 February 2012 27 February 2012 15 May 2012 28 February 2012 2 April 2012 ... further results Answers 18 March 2011 23 March 2011 23 March 2011 15 April 2011 20 April 2011 5 May 2011 24 May 2011 3 June 2011 24 June 2011 17 June 2011 22 July 2011 26 August 2011 17 August 2011 16 September 2011 7 October 2011 21 October 2011 9 November 2011 23 November 2011 13 January 2012 5 April 2012 14 June 2012 11 July 2012 16 August 2012 20 November 2012 21 January 2013 12 February 2013 28 February 2013 14 March 2013 19 April 2013 31 January 2014 11 March 2014 3 July 2014 29 July 2014 16 September 2014 28 October 2014 28 January 2015 15 April 2011 29 April 2011 9 March 2012 25 May 2012 Results Approval, Approval, Approval, Approval, Approval, Approval, Approval, Approval, Approval, Approval, Approval, Approval, Approval, Approval Other: L-11-114, Drawing No. M-042B, Rev. 35, Sampling System Sh. 2 Local Grab Samples, L-11-131, Drawing No. LR-OS041A2, Revision 1, Operational Schematic Emergency Diesel Generator Systems, L-11-334, Reply to Request Additional Information for the Review of the License Renewal Application Amendment No. 21, L-12-337, Review of the Safety Evaluation Report with Open Items Related to the License Renewal, L-12-341, Firstenergy Nuclear Operating Cos Initial Status Report in Response to March 12, 2012 Commission Order Modifying Licenses with Regard to Requirements for Reliable Spent Fuel Pool Instrumentation (Order Number EA-12-051), L-12-444, Submittal of Contractor Equivalent Margins Assessments for Reactor Vessel Welds (Nonproprietary Versions), L-12-456, Notification of Closure of Commitments Related to the Review of the License Renewal Application, L-13-257, Notification of Completion of a License Renewal Commitment Related to the Review of License Renewal Application TAC No. ME4640) and License Renewal Application Amendment No. 45), L-13-330, License Renewal Application Amendment No. 46 - Annual Update, L-13-341, Review of the Safety Evaluation Report Related to the License Renewal of Davis-Besse Nuclear Power Station, L-14-085, License Renewal Application (TAC No. ME4640) Amendment No. 48, L-14-206, License Renewal Application Amendment No. 50 - Annual Update, L-15-120, Notification of Completion of License Renewal Commitments Related to the Review of the Davis-Besse Nuclear Power Station, Unit No. 1, License Renewal Application and License Renewal Application Amendment No. 55, L-15-139, License Renewal Reactor Vessel Internals Inspection Plan, L-15-214, License Renewal Application Amendment No. 59 - Annual Update, L-15-309, License Renewal Application Amendment No. 60, L-15-310, C-CSS-099.20-069, Rev 0, Shield Building Laminar Cracking Limits, ML111050091, ML11110A089, ML11110A091, ML11110A092, ML11110A093, ML11110A094, ML11110A095, ML11110A105, ML11110A106, ML11110A107, ML11122A014, ML11126A017, ML11126A018, ML11126A019, ML11126A020, ML11126A021, ML11126A022, ML11126A023, ML11126A024, ML11126A025, ML11126A026, ML11126A032, ML11126A033, ML11126A034, ML11126A035, ML11126A036, ML11126A037, ML11126A038, ML11126A039, ML11126A040, ML11126A041, ML11126A042, ML11126A043... further results
MONTHYEAR2011-02-17 [Table View]

ML15280A304

Text

dRl = \\26\\in As'

-47.1-kipft For negtive moment, switch OF and IF rebar d:= dnj

[oJ*a dmax:=max{d) 12.5 kip Pc fMcmk = 290.5psi fMs0F k

57000 [f~pTi = 3605-ksi L>

-4.7-kip-ft For negtive moment, switch OF and IF rebar d := dgj

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