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Home My WebLink AboutRetaining wall analysisProject Job Ref. ° Carmel Office Building Wall - 1701 E. 116th Street, Carmel IN 23-026 Section Sheet nolrev. South Retaining Wall 1 Badger Engineering Cale. by Date Chk'd by Date ApWd by Dace 9830 Michigan Rd., Suite D ccb 4/21/2023 ccb 4/22/2023 RETAINING WALL ANALYSIS In accordance with International Building Code 2018 Tedds pk:ulation version 2.9.71 Analysis summary Design summary Overall design utilisation 1,325 Overall design status Fail Description Unit Cana A lied F o S Result Slidina stability If 1231196 1 10957 2.117 PASS Overturning stability Ib ft/ft 1135731 156538 12.401 PASS Bearing pressure Ipsf 12500 12211 11.131 PASS Design summary Descri lion Unit Provided Re wired Utilisation Result Stem pO rear face - Flexural reinforcement inz/ft 0,663 0.625 0,943 PASS Stem pO - Shear resistance lb/ft 17788 8830 0,496 PASS Base top face - Flexural reinforcement inz/ft 0,295 0,618 1,760 FAIL Base bottom face - Flexural reinforcement inz/ft 0,295 0.518 1,760 FAIL Base - Shear resistance lb/ft 23622 8650 0.366 PASS Key - Flexural reinforcement inz/ft 0,295 0,518 1.760 FAIL Key - Shear resistance Wit 23622 8199 0.347 PASS Transverse stem reinforcement inz/ft 0.589 0.432 0.733 PASS Transverse base reinforcement inz/ft 0.589 0.518 0.880 PASS Retaining wall details Stem type Cantilever Stem height h:tem =10 ft Stem thickness =18 g a = Angle to rear face of stem = 90 deg Stem density ystem = 150 pcf ,unnno, Toe length 6 = 3.5 ft •• 5(�p N C 9q %,, Heel length Ira = 8 It �Q��GtSTF4 OOCa,�� Base thickness time = 24 in No. Key position pkey = 3.5 It _ PE60910326 - Key depth dkey = 3 ft ST E _ Key thickness tkay = 24 in Base density yeaae = 150 pcf V• Height of retained soil h,at = 8 ft , hill Angle of soil surface p = 0 deg Depth of cover dwvu =1.75 It Depth of excavation dam = O.667 ft Height of water hwatu = 0 ft Water density yw = 62 pcf Retained soil properties Soil type Medium dense well graded sand Moist density ym, = 135 pcf Saturated density yu = 145 pcf Project Job Ref. = a Carmel Office Building Wall - 1701 E. 116th Street, Carmel IN 23-026 Section Sheet no./rev. South Retaining Wall 2 Badger Engineering Cala by Date Chk'd by Date App' Date 9830 Michigan Rd., Suite D ccb 4/21/2023 Gcb 4/22/2023 Z� Effective angle of internal resistance ¢r = 30 deg Effective wall friction angle Sr = 0 deg Base soil properties Soil type Dense gravel Soil density yb = 111 pcf Cohesion Cb = 0 psf Effective angle of internal resistance ¢b = 36 deg Effective wall friction angle Sb =18 deg Effective base friction angle Sbb = 24 deg Gross allowable bearing pressure ciaan-gress = 2500 psf Loading details Dead surcharge load SurchargeD = 25 psf Live surcharge load Surcharges = 1000 psf Calculate retaining wall geometry Base length Base height Saturated soil height Moist soil height Length of surcharge load - Distance to vertical component General arrangement -sketch pressures relate to bearing check Ibasa = hba + fstern + I� =13 ft hbsse = ibasa + dker = 5 ft h at = hwater + d� er =1,75 ft hme i = hret - hmw = 8 it Isu = Ihew = 8 it xwr V = Ibaaa - Iheei / 2 = 9 ft i Project Job Ref. WINE Carmel Office Building Wall - 1701 E. 1161h Street. Carmel IN 23-026 A Section Sheet no./rev. r� a� t(Mv � "°" South Retaining Wall 3 Badger Engineering caic. by Date ChWd by Date App'd by oat 9830 Michigan Rd., Suite D cob 4/21/2023 cob 4/22/2023 Z 2 Effective height of wall - Distance to horizontal component - Distance to horizontal component above key Area of wall stem Distance to vertical component Area of wall base - Distance to vertical component Area of saturated soil hea = heaw + deow + brat =14,75 ft xscr_n = heff / 2 - dkey = 4.375 ft xsvr_h_a = (heff - dkey) / 2 = 5.875 It Astern = hstem x tstem = 15 112 xstam = hoe + tstem / 2 = 4,25 ft Abase = lease x lease + dkey x tkey = 32 ft2 xbasa = (lbase2 x tease / 2 + dkey x tkey x (pkey + tkay / 2)) / Abase = 6.125 ft Asat = heat x Ineei =14 ft2 - Distance to vertical component xsat_ff = Ibasa - (heat x I� / 2) / Asat = 9 ft - Distance to horizontal component xsat h = (hsat + hbase) / 3 - dkey = -0.75 It - Distance to horizontal component above key xsai_b_a = (heat + tease) / 3 = 1,25 It Area of water Awater = hsat x Inem =14 ft2 - Distance to vertical component xwater a = Ibase - (hsat x Inee? / 2) / Asat = 9 ft - Distance to uplift component xwatar_a = 2 x Ibase / 3 = 8.667 ft - Distance to horizontal component xwata _h = (heat + hose) / 3 - dkey = -0.75 It - Distance to horizontal component above key xwaiar_h_a = (hsat + tease) / 3 =1.25 It Area of moist soil Amast = hmeut x Ineeq = 64 ft' - Distance to vertical component - Distance to horizontal component above key Area of base soil - Distance to vertical component Distance to horizontal component Area of excavated base soil - Distance to vertical component - Distance to horizontal component Using Coulomb theo Passive pressure coefficient From IBC 2018 c1.1807.2.3 Safety factor xmow v = lease - (hmo;st x Inee? / 2) / Amdst = 9 R xpass_v = lease - (da„er x hoax (lease -hoe ! 2)) / Apass =1.75 ft xoass_n = isst aer + hose) / 3- dkey = -0.75 it Aexc = hpass x hoe = 3.792 ft2 xex� v = (base - (hpaw x Itoex (Ibse - lum / 2)) / Aexc = 1.75 ft xexc_n = (hpass + hease) / 3- dkey = -0.972 ft 0.333 Kr = sin(90 - qb)2 / (sin(90 + Sb) x [1 - J[sin(Qb + Sb) x sin(Qb) / (sin(90 + sb))]]2) = 8.022 Load combination 1 1.0 x Dead + 1.0 x Live + 1.0 x Lateral earth Sliding check Vertical forces on wall Wall stem Fstem = Astem x ystem = 2250 pit Wall base Fosse = Abase ybase = 4800 plf Surcharge load Fsu<_k = (SurchargeD + 0 x Surchargeb) x Ihea = 200 pit Saturated retained soil Feet_v = Aeat x (yet - yw) =1158 pit Water Fwater_v = A, ester x yw = 872 pit Project Job Ref. MW Carmel Office Building Wall - 1701 E. 116th Street. Carmel IN 23-026 section Sheet no./rev. South Retaining Wall 4 Badger Engineering caw. by Date Chkd by Date np Date 9830 Michigan Rd., Suite D ccb 4/21/2023 ccb 4/22/2023 1 do Water uplift F„acare = (heat + tbase) x Ibass / 2 x yw =1519 plf Moist retained soil Fu st „= Aaoist x ym = 8640 plf Base soil Foxe „= Aexc x yb = 422 pif Total Ftwa v = Fstem + Fbase + Fsw v + Fsat v + Fwatw v - Fwater e + Fm st_v + Fa.o_v = 16823 pit Horizontal forces on wall Surcharge load Saturated retained soil Water Moist retained soil Total Check stability against sliding Base soil resistance Base friction Resistance to sliding Factor of safety Overturning check Vertical forces on wall Wall stem Wall base Surcharge load Saturated retained soil Water Water uplift Moist retained soil Fsu,_n = KA x (Surchargeo + Surcharges) x hen = 5040 plf Fsat_h = KA x (yar - yw) x (heat + hbase)2 / 2 = 628 pif Fwater h = Yw x (hwatw + drover + hbase)Z / 2 =1419 pit Fmmst_n = KA x yrw x ((hen - hsat - hbase)2 / 2 + (hen - hsat - hbase) x (hsai + hbaw)) = 3870 pit Ftd.�_n = Fsarn + Fsat_h + Fwater_n + F„,Rst_h =10957 pit m Fex� n = Ka x cos(Sb) x Yb x (hpass + hbase)2 / 2 =15705 plf F Aon = FtDw v x tan(Sbb) = 7490 pit Frest = Fexc h + Fmwon = 23196 Of FoSss = Fmi / Fwai_n = 2.117 > 1.5 PASS - Factor of safety against sliding is adequate Fstaa, = Astem x ystem = 225D plf Fbase = Abase x ybew = 4800 pit F ery = (Surchargeo + 0 x Surcharges.) x Ineei = 200 pit Peaty = Asat x (ysr - yw) =1158 pit Fwater_v = Awatw x yw = 872 pif Fvaeer_a = (heat + tbaw) x Imse / 2 x yw =1519 pif Fo %st_v = Aa,wst x yrw = 8640 pif Base soil Fexc_v = Aexc x yb = 422 plf Total Ftaasy = Fstem + Fbase + Fear v + Fsat_v + F„ater_v - Fwater_e + F,,,dst_v + Faxo_v = 16823 pit Horizontal forces on wall Surcharge load Saturated retained soil Water Moist retained soil Base soil Total Overturning moments on wall Surcharge load Saturated retained soil Fsw_n = KA x (Surchargeo + Surcharges) x (hee - dxey) = 4015 plf Fsat_n = W. x (Yu - yw) x (heat + tbase)2 / 2 = 194 plf Fwater h = Yw x (h„ater + drover + tbaee)2 / 2 = 438 pit Fatasi h = KA x y,nr x ((hex - hsat - hbase)2 / 2 + (herc - hsat - hoses) x (hsat + tbase)) _ 2790 pit Fexc_h = max(-Kp x cos(Sb) x yb x (hpass + hbase)2 / 2,-(Fsat_h + Fmdst_n + Fwat. n + Fsw_n)) _ -7436 pif Ftaai_n = Few h + Fsat h + Fw ern + Fmoist_h + Fexc_h = 0 plf Msw or = Fsw n x xsw n a = 23586 Ib_fUft Meat or = Fsat h x xsat n a = 242 Ib fl/fl Project Job Ref. Carmel Office Building Wall - 1701 E. 116th Street. Carmel IN 23-026 Section Sheet wJrev. South Retaining Wall 5 Badger Engineering calc. by Dale Chk'd by Date App' Date 9830 Michigan Rd., Suite D ccb 4/21 /2023 ccb 4/22/2023 ZZA Z Z Water Mwatw oT x Xmtr h_a + FH,rer " x xwaw" =13708 Ib_ff/ft Moist retained soil Mmmst_OT = Fmouth x xmmst_h_a =11771 Ib_ft/ft Base soil Men oT = Fexe h x xexo h = 7230 Ib_ft/ft Total Mierai_oT = Ms- oT + Meal oT + MwaterpT + Mnxat_OT + Mexe_oT = 56538 Ib_tUft Restoring moments on wall Wall stem Mua,n_R = Fstem x xstem = 9562 Ib_ft/ft Wall base Mmsa_R = Feria x xwa = 29400 Ib_ft/ft Saturated retained soil Msat R = Fsat v x xsat v = 10420 Ib_ft/ft Water Mwate,_R = Fwatar_v x xwater_v = 7850 Ib_ft/ft Moist retained soil Mn st_R = Fa st_v x xm ist_v = 77760 lb_ft/ft Base soil Mexe_R = Fex_v x xexe_v = 738 Ib_tf/ft Total Mimai_R = Mste,n_R + Moase_R + Msat_R + Mwata< R + Mn-ost_R + Mexe R=135731 Ib fgft Check stability agains4 overturning Factor of safely FoSm = Mmtai R / Mtowi_oT = 2.401 > 1.5 PASS - Factor of safety against overturning is adequate Bearing pressure check Vertical forces on wall Wall stem Fstem = Astm x ym m = 2250 plf Wall base Fbaea = Aoase x yoaw = 4800 plf Surcharge load Fsw_v = (Surchargeo + Surchargel-) x Inem = 8200 pit Saturated retained soil Fsat_v = Asat x (yw - yw) =1158 pit Water Fwat" v = Awat" x yw = 872 pit Moist retained soil Finast_w = Ammst x yav = 8640 pit Base soil Fpass_w = Apass x yb = 681 pit Total Fto m_v = Fstem + Fbasa + Fsw_v + Fsat_v + Fwate v + Fmotst_v + Fpass_v = 26601 pit Horizontal forces on wall Surcharge load Fsw_h = KA x (Surchargeo + Surcharges) x (hen - dkey) = 4015 pif Saturated retained soil Fsat_h = KA x (ysr - yw) x (hset + loasep / 2 =194 plf Water Fwatet_h = yw x (hwat" + dwvu + tbase)2 / 2 = 438 pit Moist retained soil Fnast_h = KA x ymr x ((hen - hwt - tt a )z / 2 + (hen - hw - hbaw) x (hsm + toase)) _ 2790 pit Base soil Fpass n = max(-Kp x cos(Sb) x yb x (doover + hbasa)z / 2,-(Fwt_h + Fnaat_h + Fwater h + Fsa _h)) _ -7436 pit Total Ft«m_h = max(Fsw_h + Fsat_h + Fwater_h + Fmotst_h + Fpasa_h - Fwat_v x tan(Sbb), 0 plf) = 0 pit Moments on wall Wall stem Mstem = Fstem x xstea, = 9562 lb Will Wall base Moasa = Fbasa x xoase = 29400 Ib_Wft Surcharge load Msw = F= v x xsw_v - Fsu,_h x xwh_a = 50214 Ib_ft/ft Saturated retained soil Msat = Fsat_v x xsat_v - Fsat_h x xsat_h_a =10178 Ib_ft/ft Project Job Ref. Carmel Office Building Wall - 1701 E. 116th Street. Carmel IN 23-026 Section Sheet no./rev. South Retaining Wall 6 Badger Engineering Cale. by Date Chk'd by Date APO by Dat 9E3o Michigan Rd., Suite D ccb 4/21/2023 Ccb 4/22/2023 2 Water IVI"tw = Fv atai_v x xv wt v - Fwaze<_h x xwata- h_a = 7302 Ib_it/ft Moist retained soil Mmbsi = Finast_v x xm«sty - Faoath x Xmoist_h_a = 65989 Ib_ft/ft Base soil Mori= = Fpass_v x xpass_v - Fpass_h x xpass_h = -4385 Ib_ft/ft Total Motor = Mstem + Mbasa + MW + Msw + M„ w + Maoist + Mpass=168261 Ib_f /ft Check bearing pressure Distance to reaction Eccentricity of reaction Loaded length of base Bearing pressure at toe Bearing pressure at heel Allowable bearing capacity Factor of safety RETAINING WALL DESIGN In accordance with ACI 318-14 x=Mt«w/Fmtaiv=6.325ft e= x-Ibasa/2=-0.175ft Iwaa = Ibasa =13 ft qbe = Fwtw_v / loose x (1 - 6 x e / Ibase) = 2211 psf ghaei = Ftow_v / Ibasa x (1 + 6 x e / Ibasa) =1881 psf garmv = gaim sms = 2500 psf FoSbp = gaaon / max(gw, ghew) =1,131 PASS - Allowable bearing pressure exceeds maximum applied bearing pressure Tedtls cakuladon version 2.9.17 Concrete details Compressive strength of concrete Concrete type Reinforcement details Yield strength of reinforcement Modulus of elasticity or reinforcement Compression -controlled strain limit f a = 4000 psi Normal weight fy = 60000 psi Es = 29000000 psi e y = 0.002 Cover to reinforcement Front face of stem cst = 1.5 in Rear face of stem ev = 2 in Top face of base cbt = 2 in Bottom face of base cbb = 3 in From IBC 2018 c1.1605.2 Basic load combinations Load combination no.1 1.4 x Dead Load combination no.2 1.2 x Dead + 1.6 x Live + 1.6 z Lateral earth Load combination no.3 1.2 x Dead + 1.0 x Earthquake + 1.0 x Live + 1.6 x Lateral earth Load combination no.4 0.9 x Dead + 1.0 x Earthquake + 1.6 x Lateral earth 0 Project Carmel Office Building Wall - 1701 E. 116th Street. Carmel IN Job Ref. 23-026 Section Sheet no.lrev. South Retaining Wall 7 Badger Engineering Cal by Date Chkd by Date App'd Detail 9830 Michigan Rd., Suite D ccb 4/21/2023 Ccb 4/22/2023 ? 60�r brce-Car81w0'm No.l-dryri 9eNng rtOnixi-CgmH Jm lb.1 -fi�Ml Lmey deals - CanlnKvn No. i - IFW Ii a 0.01 �e I 25 — 0ag0sg6 Tr - 'm Hal 7fT�" 1 3f " SJ �x 0 0 6lmrfam-Co cm NO2-kp% oKdN mevN-CorrSrYbn Nat -ems Ml L•aerg eMah-Cw+G' .00 tbt-IiWO' Ilow I I 0.58 8J 378 -n E2 A.E — 13.5 ]) - Y...i 156 �I % 351 7 Project Job Ref. MINE Carmel Office Building Wall -1701 E. 116th Street. Carmel IN 23-026 Section Sheet no/rev. South Retaining Wall 8 Badger Engineering Cale. by Date ChWdby Dale AN" Date 9830 Michigan Rd., Suite D ccb 4/21/2023 ccb 4/2212023 ((�/JAB/��d%y- Lwd nJ htih - CartirSm l:e.] - LpLB � 0 58 0 s9 e Loaana 4ek - CmiN,c'm Ne 1- f ipNR � O.SB oo ow Check stem design at base of stem Depth of section SFax taro-Cwbma Noa -Npa SFea bro-Garb' San No.4-fipaM Rectangular section in Flexure -Section 22.3 Design bending moment combination 2 Depth of tension reinforcement Compression reinforcement provided Area of compression reinforcement provided Tension reinforcement provided Area of tension reinforcement provided Maximum reinforcement spacing - CIA 1.7.2 Depth of compression block h=18in eartre �a-c«mrocm w.a-� nm eeNrg rto�t-CarSnaOon NOA-per RM M = 37012 Ib_ff/ft d=h-cB,-�s</2=15.625in No.4 bars @ 8" do A pro = n x os? / (4 x ssf) = 0.295 in2/ft No.6 bars @ 8" C/c As .p o = n x Q�2 / (4 x su) = 0.663 in2/ft min(18 in, 3 x h) = 18 in PASS - Reinforcement is adequately spaced a = Aspaw x fy / (0.85 x fc) = 0.975 in e Check stem design at base of stem Depth of section SFax taro-Cwbma Noa -Npa SFea bro-Garb' San No.4-fipaM Rectangular section in Flexure -Section 22.3 Design bending moment combination 2 Depth of tension reinforcement Compression reinforcement provided Area of compression reinforcement provided Tension reinforcement provided Area of tension reinforcement provided Maximum reinforcement spacing - CIA 1.7.2 Depth of compression block h=18in eartre �a-c«mrocm w.a-� nm eeNrg rto�t-CarSnaOon NOA-per RM M = 37012 Ib_ff/ft d=h-cB,-�s</2=15.625in No.4 bars @ 8" do A pro = n x os? / (4 x ssf) = 0.295 in2/ft No.6 bars @ 8" C/c As .p o = n x Q�2 / (4 x su) = 0.663 in2/ft min(18 in, 3 x h) = 18 in PASS - Reinforcement is adequately spaced a = Aspaw x fy / (0.85 x fc) = 0.975 in e Project Job Ref. EMNOINIM Carmel Office Building Wall - 1701 E. 116th Street. Carmel IN 23-026 low Section Sheet no.lrey. South Retaining Wall 9 Badger Engineering caic. by Date Chk'd by Date App'd by Da 9830 Michigan Rd., Suite D ccb 4/21/2023 ccb 4/22/2023 Neutral axis factor - cl.22.2.2.4.3 pi = min(max(0.85 - 0.05 x (fc - 4 ksi) Depth to neutral axis c = a / pt = 1.147 in Strain in reinforcement et = 0.003 x (d - c) / c = 0.037885 Section is in the tension controlled zone Strength reduction factor Of = min(max(0.65 + 0.25 x(et - cry) / 0.003, 0.65). 0.9) = 0.9 Nominal flexural strength Mn = Atpm x fr x (d - a / 2) = 50157 Ib_ft/ft Design flexural strength Wn = 0 x Mn = 45142 Ib_ft/ft M/pMn=0.820 PASS - Design flexural strength exceeds factored bending moment By iteration, reinforcement required by analysis Asr.des = 0.54 Witt Minimum area of reinforcement - cl.9.6.1.2 A .r n = max(3 x J(fe x 1 psi), 200 psi) x d / fy = 0.625 in2/ft PASS - Area of reinforcement provided is greater than minimum area of reinforcement required Rectangular section in shear -Section 22.5 Design shear force V = 8830 Ib/fl Concrete modification factor - cl.19.2A R =1 Nominal concrete shear strength - egn.22.5.5.1 Ve = 2 x X x VC x 1 psi) x d = 23717 Ib/ft Strength reduction factor @s = 0.75 Design concrete shear strength - cl.11.6.1.1 OVe = Qs x Vc =17788 Ib/ft V / �Vc = 0.496 PASS - No shear reinforcement is required Horizontal reinforcement parallel to face of stem Minimum area of reinforcement - c1.11.6.1 Asxre� = 0.002 x tsts,n = 0.432 in2/fl Transverse reinforcement provided 19o.4 bars @ 8" do each face Area of transverse reinforcement provided Asxpw = 2 x n x Qsx2 / (4 x ssx) = 0.589 in2/ft PASS - Area of reinforcement provided is greater than area of reinforcement required Check base design at toe Depth of section Rectangular section in flexure -Section 22.3 Design bending moment combination 2 Depth of tension reinforcement Compression reinforcement provided Area of compression reinforcement provided Tension reinforcement provided Area of tension reinforcement provided Maximum reinforcement spacing - el.7.7.2.3 h=24in M = 15411 Ib_ft/ft d=h-rkb-Qbb/2=20.75in No.4 bars @ 8" do Ab POW= n x 4bt2 / (4 x sw) = 0.295 in2/ft NoA bars @ 8" c/c Abb.PM = n x Obb2 / (4 x sbb) = 0.295 in2/ft smsx = min(18 in, 3 x h) =18 in PASS - Reinforcement is adequately spaced Depth of compression block a = Abbprw x fY / (0.85 x fc) = 0.433 in Neutral axis factor - cl.22.2.2.4.3 pi = min(max(0.85 - 0.05 x (fc - 4 ksi) / 1 ksi, 0.65), 0.85) = 0.85 Depth to neutral axis c = a / pi; = 0,51 in Strain in reinforcement et = 0.003 x (d - c) / c = 0.119165 Section is in the tension controlled zone Strength reduction factor of = min(max(0.65 + 0.25 x(et - ey) / 0.003, 0.65), 0.9) = 0.9 Project Job Ref. ' Carmel Office Building Wall - 1701 E. 116th Street. Carmel IN 23-026 Section SheelnoJrev. " N, lv I <" "" South Retaining Wall 10 Badger Engineering catc.by Date Chk'dby Date APP'dby Date 9830 Michigan Rd., Suite D ccb 4/21 /2023 ccb 4/2212023 of 2 Nominal Oexural strength Mn = AbbPmv x fy x Design flexural strength OMn = Of x Mn = 27214 Ib_fUft M / QMn = 0.566 PASS - Design flexural strength exceeds factored bending moment By iteration, reinforcement required by analysis Abb.des = 0.166 in2/ft Minimum area of reinforcement - cL7.6.1.1 Abeam = 0.0018 x In = 0.518 inz/ft FAIL - Area of reinforcement provided is less than minimum area of reinforcement required Rectangular section in shear -Section 22.5 Design shear force V = 8650 Ib/ft Concrete modification factor - cl.19.2.4 X =1 Nominal concrete shear strength - egn.22.5.5.1 V, = 2 x X x J(fe x 1 psi) x d = 31496 Ib/ft Strength reduction factor 03 = 0.75 Design concrete shear strength - cl.7.6.3.1 OVA = Os x Vc = 23622 Ib/ft V / ¢Vc = 0.366 Check base design at heel Depth of section Rectangular section in flexure -Section 22.3 Design bending moment combination 2 Depth of tension reinforcement Compression reinforcement provided Area of compression reinforcement provided Tension reinforcement provided Area of tension reinforcement provided Maximum reinforcement spacing - cl,7.7.2.3 Depth of compression block Neutral axis factor - c1.22.2.2.4.3 Depth to neutral axis PASS - No shear reinforcemenP is required h=24in M = 37828 Ib_ft/ft d=h-rat-Qbt/2=21,75in NoA bars @ 8" do Abb.PM = it x Qbb2 / (4 x sbb) = 0.295 Will:NoA bars @ 8" do AbrPrw = n x Obt2 / (4 x sbf) = 0.295 Will: s = min(18 in, 3 x h) =18 in PASS - Reinforcement is adequately spaced ,w a = AbrPx fy / (0.85 x fe) = 0.433 in c=a/pt=0.51 in Strain in reinforcement et = 0.003 x (d - c) / c = 0.125052 Section is in the tension controlled zone Strength reduction factor Of = min(max(0.65 + 0.25 x(et - ey) / 0.003, 0.65), 0.9) = 0.9 Nominal flexural strength Mn = Auprw x fy x (d - a / 2) = 31711 Ib_f /ft Design flexural strength ¢Mn = Of x Mn = 28540 Ib_ft/ft M/¢Mn=1.325 FAIL - Design flexural strength is less than factored bending moment By iteration, reinforcement required by analysis Abrdes = 0.392 in2/ft Minimum area of reinforcement - cl.7.6.1.1 Abrrdn = 0.0018 x h = 0.518 in2/fl FAIL - Area of reinforcement provided is less than minimum area of reinforcement required Rectangular section in shear -Section 22.5 Design shear force V = 8641 Ib/ft Concrete modification factor - clA 924 7 =1 Project Job Ref. WINE Carmel Office Building Wall -1701 E. 116th Street, Carmel IN 23-026 ow Section Sheet no./rev. South Retaining Wall 11 Badger Engineering Cato. by Date Chk'd by Date App' Date 9a3e Michigan Rd., Suite D ceb 4/21/2023 ccb 4/22/2023 25az Nominal concrete shear strength - egn.22.5.5.1 Vc = 2 x 7 x J(1 c x 1 psi) x d = 33014 Iblft Strength reduction factor 05 = 0.75 Design concrete shear strength - cl.7.6.3.1 ¢Vc _ ¢s x Vc = 24761 Ib/ft V / OVA = 0,349 PASS • No shear reinforcement is required Check key design Depth of section h = 24 in Rectangular section in flexure -Section 22.3 Design bending moment combination 2 M = 13470 Ib_f ift Depth of tension reinforcement d = h- con - Qk / 2 = 20.75 in Compression reinforcement provided No.4 bars @ 8" Gc Area of compression reinforcement provided Akprw = a x W / (4 x sk) = 0.295 in2/ft Tension reinforcement provided No.4 bars @ 8" Gc Area of tension reinforcement provided Akprov = a x W / (4 x sk) = 0.295 in2/ft Maximum reinforcement spacing - cl.7.7.2.3 s� = min(18 in, 3 x h) =18 in PASS - Reinforcement is adequately spaced Depth of compression block a = Aepror x fy / (0.85 x fc) = 0.433 in Neutral axis factor - cl.22.2.2.4.3 (if = min(max(0.85 - 0.05 x (fc - 4 ksi) / 1 ksi, 0.65), 0.85) = 0.85 Depth to neutral axis c = a / or = 0.51 in Strain in reinforcement cr = 0.003 x (d - c) / c = 0.119165 Section is in the tension controlled zone Strength reduction factor of = min(max(0.65 + 0.25 x(et - ea) / 0.003, 0.65), 0.9) = 0.9 Nominal flexural strength Mn= Akprov x fr x (d - a / 2) = 30238 Ib_f /ft Design flexural strength Wn = of x Mn = 27214 Ib_tUft M / OMn = 0,495 PASS - Design flexural strength exceeds factored bending moment By iteration, reinforcement required by analysis Ardas = 0.145 in2/ft Minimum area of reinforcement - cl.7.6.1.1 Ake = 0.0018 x In = 0.518 in2/ft FAIL - Area of reinforcement provided is less than minimum area of reinforcement required Rectangular section In shear -Section 22.5 Design shear force V = 8199 Ib/ft Concrete modification factor - c1.192.4 l =1 Nominal concrete shear strength - egn.22.5.5.1 Vc = 2 x R x d(fc x 1 psi) x d = 31496 Ib/ft Strength reduction factor os = 0.75 Design concrete shear strength - cl.7.6.3.1 ¢Ve _ ¢s x Vc = 23622 Ib/ft V/QVc=0.347 PASS - No she reinforcement is required Transverse reinforcement parallel to base Minimum area of reinforcement - c1.7.6.1.1 Anxreq = 0,0018 x tb� = 0.518 in2lft Transverse reinforcement provided No.4 bars @ 8" Gc each face Area of transverse reinforcement provided Anxpmv = 2 x it x �b2 / (4 x sbx) = 0.589 in2/ft PASS - Area of reinforcement provided Is greater than area of reinforcement required "Dal Project Job Ref. A000A�MV="w ANOW Carmel Office Building Wall - 1701 E. 116th Street, Carmel IN 23-026 Section Sheet m./fev. South Retaining Wall 12 Badger Engineering catc.by Date chk!dby Date aped Da 9830 Michigan Rd., Suite D ccb 4/21/2023 ccb 4/22/2023 C/S/A= Z '. •.•No 4m.aer. xe-sa�ae�e� a..Ae�s xo wnar.,E F Reinforcement detalls No.4 bars � 8" cic No.4 bars ca 8" cic horizontal reinforcement parallel to face of stem No.6 bars (� 8" cIc 0.4 bars Q 8" cic 4 bars � 8" cic No.4 bars (cD 8" crc Reinforcement details No.4 bars � 8" do transverse reinforcement In base