cee 254-2 design for shear

CEE254 Spring2015

NagiAboShadi,PhD,SE,PEng 1

Design for ShearDirect Shear (One-way Shear)

Vn Vu

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Example of One-way Shear Cracks

CEE254 Spring2015


Shear Strength

Vn = (Vc +Vs)

Where:=0.75Vc =ShearstrengthbyconcreteVs =Shearstrengthbyreinforcement

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Thevaluesof(fc)shallnotexceed100psi3

Concrete Strength for Shear

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Thevaluesof(fc)shallnotexceed100psi

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CEE254 Spring2015


Steel Reinforcing Strength for Shear

Where:Av = Total reinforcement cross sectional area of ties resisting shearfyt = Yield strength of tie reinforcements = Tie spacing through the longitudinal axis of the beam

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Shear Ties

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CEE254 Spring2015


Example 1Determine the shear strength adequacy for the beamsection of example 2.1. The beam clear span is 20 ft.Dead load is1.5 klf and the live load is 0.7 klf. The concretestrength, fc = 4,000 psi and steel yield strength, fy = 60,000psi.

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Solution:DL=1.5klfLL=0.7klfWu =1.4D=1.4x1.5=2.1Wu =1.2D+1.6L=1.2x1.5+1.6x0.7=2.92k/ftRu =20x2.92/2=29.2kipsVu =29.2 (16"/12x2.92)=25.3kipsSheartobetakenatthecriticalcolumnsection.Atddistancefromthefaceofthesupport. Vc =0.75x(2x10"x13.54,000)/1,000=12.8kipsVs =0.75(2x0.2x60x13.5/14)=17.36kipsVn =fVc +fVs =12.8+17.36=30.16kipsVn >Vu OK

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CEE254 Spring2015


Punching ShearTwo-way ShearPunchingshearcausessuddenfailureinconcretememberssuchasreinforcedconcretefootingsandreinforcedconcreteslabs

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Examples of Punching Shear Cracks

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bx = Cx +dby= Cy+dbo = 2(bx +by)Ac= bo xd

vu=

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vc = ShearFactor

= Strengthreductionfactor =0.75ShearFactor =minimumofthefollowingshearfactors:SFactoro = 2 SFactorc = 2 c = o = SFactormax = 4.0

s =40 forinteriorcolumns=30 foredgecolumns=20 forcornercolumns

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CEE254 Spring2015


Example2A gravity reinforced concrete footing of (86x86x26 deep)supports a 24x24 reinforced concrete column. The concretespecified strength at 28 days is equal to 3,000 psi. The columnultimate axial load is equal to equal to 680 kips. Which of thefollowing statements is correct?

A)ThefootingissafeagainstpunchingshearB)Tomakethepunchingshearsafe,thespecifiedconcretestrength

needstobeaminof3,500psiC)Tomakethepunchingshearsafe,thefootingtotalthickness

needstobeaminof30D)Tomakethepunchingshearsafe,thefootingtotalthickness

needstobeaminof40

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Solution:SlabColumnanalysis:Slabeffectivedepth=30 3.5=26.5inbx =Cx +d=24+26.5=50.5inby =Cy +d=24+26.5=50.5inbo =2(bx +by)=202inAc =bo xd=5,353inUltimateshearstress:vu =Vu /Ac =680/5,353=0.127ksi

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CEE254 Spring2015


Shearcapacity:s =40c =Long/ShortColDim=1.00o =bo /d=10.92Factorc =2+4/c =6.00Factoro =2+s/o =5.66Factormax =4.00Factor(Controls)=4.00

4.0 3,000 219 0.75219 164 > vu Safe

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Example3A twoway 10thick slab is supported by columns at 30 feeton both directions while the column size is 16x16. Thespecified concrete strength for the slab and columns is 4,000psi. The occupancy of the floor supported by the concrete slab isoffice use. The weight of the additional miscellaneousmechanical and ceiling is 5 psf. The ultimate shear stress andshear strength are equal to:

A) 226psiand189psiB) 189psiand226psiC) 340psiand226psiD) 226psiand340psi

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CEE254 Spring2015


Solution:Tributaryarea=30x30=900ft2ReductionFactor,R=(900 150)0.08=60Rmax =40%LLReduced =(1.0 0.4)50psf=30psfPartitionLoad=15psfLL=30psf+15psf=45psfDL=10.5/12x150pcf+5psf=136.25psfWu=1.4D=1.4x0.136=0.191ksiWu =1.2D+1.6L=(1.2x0.136ksf)+(1.6x0.045ksf)=0.235ksfPu =30x30x0.235ksf =211.50kips

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Slab-Column analysisSlabeffectivedepth=10.5 1.25=9.25inbx =Cx +d=25.25inby =Cy +d=25.25inbo =2(bx +by)=101inAc =bo xd=934.25inUltimateshearstress:vu =Pu /Ac =0.226ksi

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CEE254 Spring2015


Shearcapacity:s =40c =Long/ShortColDim=1.00o =bo /d=10.92Factorc =2+4/c =6.00Factoro =2+s /o =5.66Factormax =4.00Factor(Controls)=4.00

4.0 4,000 253 0.75253 189.75

CEE254 Spring2015


Testing for Torsion

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Stress due to Shear and Torsion

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Aoh =Beamcrosssectionalareameasuredtothecenterlineofthetie(in2)Ph =Beamperimeteratthecenterlineofthetie(in)

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1)Selecttheconcretesectiondimensionsothatforsolidsections:

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ForHollowSections:

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Where: =Strengthreductionfactor=0.75Vu =UltimateshearstrengthTu =UltimatetorsionAoh =BeamcrosssectionalareameasuredtothecenterlineofthetiePh =Beamperimeteratthecenterlineofthetie(in)

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CEE254 Spring2015


2)Strengthdesign Tn Tu

Cot (ACI19.11.6.3.6)

Where: = 45o fornonprestressed concreteAo = 0.85Aohs = Tiespacing

2 At = CrosssectionalareaofonelegfortorsionAv = Crosssectionalareaofalllegsforshear

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Example 4The concrete beam shown has a concrete strength of 5,000 psi andis used as a pedestrian bridge. The weight of the additional deadload is 20 psf. An unbalanced uniform load may be calculated byapplying step loading. Which of the following statements iscorrect?

A) The beam dimensions are adequate for the applied torsion.B) The beam dimensions are inadequate for the applied torsion.C) The ACI318 does not cover case for beams exposed to torsion

and shear.D) None of the above.

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CEE254 Spring2015


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Solution:SelfWeight=(12'x1'+4'x4)0.15kcf =4.2k/ftDL=4.2+0.02ksf x12'=4.44k/ftLLStep loading =0.1ksf x12'/2=0.6k/ftwu =1.4D=1.4x4.44=6.22k/ftwu =1.2D+1.6L=1.2x4.44+1.6x0.6=6.29k/ftVu =Shearatcriticalsection(dfromfaceofsupport)Vu =6.29(15' 5')=62.9kipsTorsionAnalysiswuTorsion =1.6x0.6k/ft =0.96k/ftMT =0.96x3'=2.88kft/ftTu =2.88(15 5)=28.8kft

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CEE254 Spring2015


Forsolidsectionbw =48'', d=57''Ph =(48 4 0.5)x2+(60 4 0.5)x2=198inAoh =(48 4 0.5)(60 4 0.5)=2,414.25in2

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24psi

CEE254 Spring2015


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Example5ThemosteconomicaltiestobeusedforExample4are:A)#4@14''(2Legs)B)#4@8''(4Legs)C)#5@14''(4Legs)D)#5@14''(2Legs)Solution:Ao =0.85Aoh =2,052in2

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Tu = Tn =184,680(At /S)=28.8x12At /S=0.0019

At /S=0.02 ControlsTakeS=8inches, At =8x0.02=0.16in2 #4has0.2in2TakeS=14inches, At =14x0.02=0.28in2 #5has0.31in2

, 0.02

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Shear-friction Strength

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Example of Corbel Shear Cracks

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11.6.4 Shear-friction design methodWhereshearfrictionreinforcementisperpendiculartotheshearplane,Vn shallbecomputedby

Vn =Avf fy Vn =Avf fy (sin+cos )

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Corbel Example

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Beam Ledge Example

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cee 254-2 design for shear

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