`9EXAMPLE 6-1 Design of Vertical Stirrups WSD v.s. SDMReinforced Concrete DesignHome work: Select the stirrup spacing for the beam shown below.Shear &amp; Diagonal Tension #2Shear Design Summary WSD v.s. SDM Design Examples Location of Max. Shear for the Design of Beams Shear Span Deep BeamMongkol JIRAVACHARADETf c' = 280 ksc, and fy = 4,000 ksc Use DB10 stirrups.Show your results on a scaled sketch. PL = 5 tons PD = 2 tons A A 2.5 m 4.0 m 2.5 m Section A-A PL = 5 tons PD = 2 tons wL = 3 t/m wD = 2 t/m 40 cmd = 53 cmSURANAREE UNIVERSITY OF TECHNOLOGYINSTITUTE OF ENGINEERING SCHOOL OF CIVIL ENGINEERINGShear2_03Shear Design SummaryWSDShear: V = VDL + VLLShear Diagram : (ton) 29.5 w = 5 t/m 17 10 4m 2.5 mWSDSDMShear: Vu = 1.4 VDL + 1.7 VLL Vn = Vu /  Concrete: Vc = 0.29 fc b d Steel: Vs = V - Vc Spacing: s = Av fs d / Vs Min. Stirrup: smax = Av fy / 3.5 b Chk. light shear: V  0.795 fc b d smax  d/2  60 cm Chk. heavy shear: V  1.32 fc b d smax  d/4  30 cmConcrete: Vc = 0.53 fc b d Steel: Vs = Vn - Vc Spacing: s = Av fy d / Vs Min. Stirrup: smax = Av fy / 3.5 b Chk. light shear: Vs  1.1 fc b d smax  d/2  60 cm  Chk. heavy shear: Vs  2.1 fc b d smax  d/4  30 cmShear2_022.5 m-10 -17Assume column width = 0.3 m @ critical section V = 29.5 ­ 5(0.15+0.53) Shear strength of concrete Vc = 0.29(280)1/2(40)(53)/1,000 Required shear strength of steel Vs = 26.10 ­ 10.29 Check lightly shear rein. 0.795(280)1/2(40)(53)/1,000-29.5 = 26.10 ton = 10.29 ton = 15.81 ton = 28.20 ton &gt; V OKShear2_04WSD Select RB9 : Av = 2(0.636) = 1.27 cm2, fs = 1,200 ksc Stirrup spacing : s = 1.27(1.2)(53)/15.81 = 5.11 cm &lt; [53/2=26.5 cm] &lt; 60 cm USE Stirrup RB9 @ 0.05 m Shear @ x = 2.5 m, V = 10 ton Use min. stirrup : s = 1.27(2,400)/3.5(40) = 21.8 cm &lt; [53/2=26.5 cm] &lt; 60 cm USE Stirrup RB9 @ 0.20 m Select RB9 : Av = 2(0.636) = 1.27 cm2, fy = 2,400 ksc Stirrup spacing : s = 1.27(2.4)(53)/30.00 = 5.38 cm &lt; [53/2=26.5 cm] &lt; 60 cm USE Stirrup RB9 @ 0.05 m Shear @ x = 2.5 m, Vu/ = 15.8/0.85 = 18.6 ton  Use min. stirrup : s = 1.27(2,400)/3.5(40) = 21.8 cm &lt; [53/2=26.5 cm] &lt; 60 cm USE Stirrup RB9 @ 0.20 mSDM[email protected][email protected][email protected][email protected][email protected][email protected]2.5 m3.7 m2.5 mShear2_052.5 m3.7 m2.5 mShear2_07Shear Diagram : (ton) 46.85wu = 1.4(2)+1.7(3) = 7.9 t/m Pu = 1.4(2)+1.7(5) = 11.3 ton wu = 7.9 t/m 27.1 15.8 4m 2.5 mSDMLocation of Maximum Shear for the Design of BeamsACI 11.1.3.1 ­ For nonprestressed members, sections located less than a distance d from face of support shall be permitted to be designed for Vu computed at a distance d.2.5 m-15.8 -27.1 d -46.85 = 41.48 ton = 18.80 ton = 30.00 ton = 39.02 ton &gt; Vs OKShear2_06d d dAssume column width = 0.3 m @ critical section Vu = 46.85 ­ 7.9(0.15+0.53) Shear strength of concrete Vc = 0.53(280)1/2(40)(53)/1,000 Required shear strength of steel Vs = 41.48/0.85 ­ 18.80 Check lightly shear rein. 1.1(280)1/2(40)(53)/1,000Shear force diagram for designShear2_08Typical support conditions for locating Vud Critical section Vu Vu Vu dEXAMPLE 6-2 More Detailed Design of Vertical Stirrups SDM The simple beam supports a uniformly distributed service dead load of 2 t/m, including its own weight, and a uniformly distributed service live load of 2.5 t/m. Design vertical stirrups for this beam. The concrete strength is 250 ksc, the yield strength of the flexural reinforcement is 4,000 ksc.DL = 2 t/m LL = 2.5 t/mwu = 1.4(2) + 1.7(2.5) = 7.05 t/md = 64 cmBeam loaded near bottom Critical section Vu Vu Critical section Beam supported by tension force dBeam column jointL = 10 mwLu = 1.7(2.5) = 4.25 t/m wuL/2 = 7.05(10)/2 = 32.25 t/m30 cmwLuL/8 = 4.25(10)/8 = 5.31 t/m32.25/0.85 = 37.94 ton Vu/ Diagram : 5.31/0.85 = 6.25 tonBeam with concentrated load close to supportShear2_09 Shear2_11Shear at Midspan of Uniformly Loaded BeamsIn normal building, the dead load is always present over the full span, the live load may act over the full span, or over part of the span. LL full span DL full span w L Vu = u 2assume column width = 0.40 cm Vu /  at d = 37.94 ­ (0.84/5)(37.94 ­ 6.25) = 32.62 ton  Shear strength of concrete Vc = 0.53 fc b d = 0.53 250 (30)(64) /1,000 = 16.09 ton 37.94 t84 cmCritical section 32.62 t Required Vs 16.09 t Vc 8.05 t 6.25 t MidspanMax. shear @ ends LL half span DL full span Vu =Vu/w Lu L 8 Support0.5VcMax. shear @ midspan w L Vu = u 2 w L Vu = Lu 8Is the cross section large enough?  Vn,max = Vc + 2.1 fc b d = 16.09 + 2.1 250 (30)(64) /1,000 = 79.84 &gt; 32.62 ton  Vc + 1.1 fc b d = 16.09 + 1.1 250 (30)(64) /1,000 = 55.6 &gt; 32.62 tonShear2_10Shear force envelop :OK smax  d / 2  60 cmShear2_12Minimum stirrup : (ACI 11.5.6.3) USE RB9 : Av = 2(0.636) = 1.27 cm2, fy = 2400 ksc  A v,min = 0.2 fc Rearranging gives smax = smax = A v fy  0.2 fc b A v fy 3.5b = bs fy = 1.27(2,400) = 32 cm 0.2 250 (30) (ACI Eq. 11-13) s=15 cm @ x = 140 cm [email protected] [email protected] s=29 cm @ x = 239 cm [email protected]but not less than1.27(2,400) = 29 cm 3.5(30)20 cm[email protected] cm 1 cm[email protected] cm[email protected] cmUse smax = 29 cm &lt; [d/2 = 64/2 = 32 cm] &lt; 60 cm Compute stirrup resuired at d from support s= A v fy d Vu /  - VcSupport500 cm OK OKMidspanRB9 @ 0.11 m : [email protected] = 142 cm &gt; 140 cm = 1.27(2.4)(64) = 11.8 cm 32.62 - 16.09 RB9 @ 0.15 m : [email protected] = 247 cm &gt; 239 cm RB9 @ 0.29 m : 247 + [email protected] = 479 cmUse [email protected] Change spacing to s = 15 cm where this is acceptable, and then to the maximum spacing of 29 cm. Compute Vu/ where s can be increased to 15 cm.  Vu A v fyd 1.27(2.4)(64) = + Vc = + 16.09 = 29.1 ton  s 15Shear2_13Shear2_1537.94 t84 cmCritical section 32.62 t 29.1 t Vc = 16.09 t 8.05 t 6.25 tShear Span (a = M /V )Distance a over which the shear is constant P PVu/aa0.5Vc 500 cm x 37.94 - 29.1 × 500 = 140 cm from support x= 37.94 - 6.25 Change s to 29 cm, compute Vu/ Vu =  1.27(2.4)(64) + Vc = + 16.09 = 22.82 ton s 29 37.94 - 22.82 x= × 500 = 239 cm from support 37.94 - 6.25 A v fy dSupportMidspanShear DiagramV = +P + V = -P M = VaMoment Diagram+Shear2_14Shear2_16Crack Pattern in Several Lengths of Beam Mark 1 2 3 4 5 6 7/1 8/1 10/1 9/1Span (m) 0.90 1.15 1.45 1.70 1.95 2.35 3.10 3.60 4.70 5.80a/d 1.0 1.5 2.0 2.5 3.0 4.0 5.0 6.0 8.0 7.0DEEP BEAMShear2_17Brunswick Building. Note the deep concrete beams at the top of the ground columns. These 168-ft beams, supported on four columns and loaded by closely spaced fascia columns above, are 2 floors deep. Shear stresses and failure mechanisms were studied on a small concrete model. (Chicago, Illinois)Shear2_19Variation in Shear Strength with a/d for rectangular beamsFlexural moment strengthInclined cracking strength, VcShear-compression strength Failure moment = VaDeep beamsShear-tension and shear-compression failuresFlexural failuresDiagonal tension failures0123 a/d4567Shear2_18Shear2_20Deep BeamsDeep beams are structural elements loaded as beams in which a significant amount of the load is transferred to the supports by a compression thrust joining the load and the reaction.Design Criteria for Shear in Deep BeamsBasic Shear Strength: whereVn  Vu Vn = Vc + VsWhen shear span a = M /V to depth ratio &lt; 2 Mechanism:Location for Computing Factored Shear:(a) Simply Supported BeamsUse both horizontal and vertical may prevent cracks(Critical section located at distance z from face of support) Compressive struts - z = 0.15Ln  d for uniform loading - z = 0.50a  d for concentrated loading (b) Continuous Beams Critical section located at face of supportIf unreinforced, large cracks may open at lower midspan.Shear2_21Limitation on Nominal Shear StrengthVn,max = 2.7 fc b dShear2_23Definition of Deep BeamACI 10.7.1 ­ Deep beams are members loaded on one face and supported on the opposite face so that compression struts can develop between the loads and the supports, and have either: (a) clear spans, Ln, equal to or less than four times the overall member depth; orShear Strength of Concrete, Vc M  V d   Vc =  3.5 - 2.5 u   0.50 fc + 176  u  b d  1.6 fc b d Vu d   Mu   where 1.0  3.5 - 2.5 Mu  2.5 VudIf some minor unsightly cracking is not tolerated, the designer can use h Ln / h  4 Ln (b) regions with concentrated loads within twice the member depth from the face of the support. P x h x &lt; 2h Simplified method:  Vc = 0.53 fc b dShear Reinforcement, Vs A  1 + Ln / d  Vs =  v   +  sv  12  Av = sv =Shear2_22A vh  11 - Ln / d     fy d sh  12   Avh = sh =( .2), F ( .),( .2) F ( .)Shear2_24Minimum Shear Reinforcementmaximum sv  maximum sh  and minimum A vh = 0.0015 b sh d  30 cm 5 d  30 cm 5(b)F0.50a = 0.5(1.20) = 0.60FF&lt; [d = 0.90a = 1.20 ].F (c)F 0.60 . F FMu 102(60) = = 0.67 Vu d 102(90) 3.5 - 2.5 Mu = 3.5 - 2.5(0.67) = 1.83 &lt; 2.5 Vu d OKF1.7 LL = 1.7(60) = 102FFminimum A v = 0.0025 b sv V d v c = 1.83 0.50 fc + 176  u  Mu  w =Shear2_254(10.18) = 0.0129 35(90)Shear2_275.6 60 F d = 90 . F f c= 280 F1.20 mFF F ./ .260 tF3.6 fy = 4,00060 tF 35 ./ .2. 176 ( 0.0129 )  v c = 1.83 0.50 280 +  0.67   = 1.83[8.37 + 3.39] = 21.5./ .21.20 mUpper limit: v c = 1.6 fc = 1.6 240 = 24.8 kg/cm290 cmVc = vc bw d = 21.5(35)(90)/1,000 = 67.84DB36(d)FRequired Vn = Vu5 cm 40 cm[email protected] = 3.5 m 3.6 m5 cm 40 cm35 cm=102 = 120 ton 0.85(a)FFLn/d = 360/90 = 4 VnShear2_26 Vn,max = 2.7 fc b d = 2.7 280(35)(90) /1,000 = 142&gt; 120 FOKF&gt; Vc (120 &gt; 67.8)Shear2_28(e)Av  1 + Ln / d  Avh  11 - Ln / d  Vs +  = fd s  12  s2  12     y Ln/d = 4 : Vs = 120 ­ 67.8 = 52.2 b = 35.fy = 4,000./ .2A v  5  A vh  7  52.2 + = = 0.145 sv  12  sh  12  4.0(90)     min Av = 0.0015 b sv min Avh = 0.0025 b sh max sv = d/5 =18 max sh = d/5 = 18. .sh = 182 2FDB12FFF.min Avh = 0.0025(35)(18) = 1.58FFAvh = 2(1.13) = 2.262&gt; 2.2OKShear2_29FAvhA v  5  2.26  7   +   = 0.145 s  12  18  12  Av 12 = [0.145 - 0.0732] = 0.172 s 5 DB12: Av = 2(1.13) = 2.262F FDB12s = 2.26/0.172 = 13.1 12. &lt; [d/5 = 18 F.]OK.[email protected] 90 cm[email protected] 4DB36[email protected] = 3.6 m 40 cm 3.6 m 40 cm35 cm Shear2_30`

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