Design Of Cantilever Beam 1s2s5i
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DESIGN OF CANTILEVER BEAM (L.S.M)
1) Design a cantilever beam of span 3m subjected to u.d.l of 10KN/m. useM20 grade concrete and HYSD bars. Design as per L.S.M. Data: For M20 grade concrete, fck = 20N/mm². For HYSD bars, fy = 415N/mm². Super imposed load = 10KN/m. Span = 3m. xu max. = 0.48d. Breadth and depth of the beam: Assume depth, d = L/7 = 3000/7 = 428mm. let us adopt overall depth of the beam, D = 500mm. at the fixed end. The depth can be reduced to 200mm (min. is 150mm) at the free end. Where the B.M will be zero. Provide, b= D/2 = 500/2= 250mm. the effective span = clear span for cantilevers = 3m. eff. Cover = 40mm. d, eff. Depth = 500-40 =460mm.
Loads and B.M: Self weight of beam = (0.5+0.2)/2 *0.25*3*25 = 6.56KN/m.
This acts at = 3/3[(0.5+2*0.2)/(0.5+0.2)] = 1.29m from the fixed end. Taking a load factor of 1.5, Factored B.M at fixed end = (1.5*10*3²/2)+(1.5*6.56*1.29) = 80.15KN.m. Depth for B.M: M.R of a singly reinforced balanced section Mu = 0.36*fck*b*xu max.*(d-(0.42 xu max)) ---------------------- (1) 80.15*10^6= 0.36*20*250*0.48*d.*(d-(0.42 *0.48*d)) 80.15*10^6 = 689.8d² d=340.8mm <460mm (available depth) o.k. Depth provided is larger than the depth required for balanced section the section becomes U.R.section Steel for bending moment: To find steel area, Mu=0.87*fy*Ast(d-0.42xu) Where xu, depth of neutral axis for U.R.section This is given by Total compression = total tension 0.36fck*b*xu=0.87fy*Ast xu =[ (0.87fy*Ast)/(0.36fck*b)] =[ (0.87*415*Ast)/(0.36*20*250)] = 0.2006 Ast Substituting this in equation (1) Mu= 0.87*415*Ast (460-(0.42*0.2006Ast)) 80.15*10^6 =166083 Ast- 30.42Ast² Ast = 535mm² Min Ast= 0.85bd/fy =0.85*250*460/415 = 236mm² < 535 mm² o.k. Using 16mm-dia bars, no. of bars =535/201 =2.7 =3bars Provide 3 – 16 mm dia bars at the top
Check for deflection: Actual l/d = 3000/460 =6.52 Allowable l/d = (basic l/d)*β*γ*δ*λ β = 1, δ=1, λ=1 γ, modification factor for percentage of steel and stress in steel is given by, % steel= [(3*201)/ (250*460)] *100 =0.52% Stress in steel = 0.85*fy*(Ast required / Ast provided) =0.85*415*(2.67 / 3) = 214 N/mm² Ref. to fig 4 of IS-456,γ=1.4 Basic l/d= 7 for cantilevers (clause 23.2.1) Allowable l/d= 7*1.4 =9.8 > actual l/d
o.k.
Check for shear: Factored max. Shear force at the fixed end = 1.5(10*3) + 1.5(6.56) = 54.84 KN. τv , nominal shear = for beams of varying depth ,τv =[Vu-((Mu/d)*tanβ)]/bd τv =[54.84*10^3- ((80.15*10^6/460)*0.1)]/250*460 = 0.33 N/mm² For 0.52% steel and M20 grade concrete τc , permissible shear stress = 0.49 N/mm² > τv There fore provide minimum shear reinforcement. Using 8mm dia 2-legged vertical stirrups Spacing = Asv* (0.87fy)/(0.4b) = 2*50* (0.87*415)/(0.4*250) = 361mm Max. Spacing = 0.75 d = 0.75*460 =345 mm. There fore, provide 8mm dia 2-legged vertical stirrups @ 300mm c/c. through out . Check for development length: Inorder to develop full tensile strength at the face of , each of the 3-16dia bars must be embedded by a length equal to the development length. Ld= Φσs/4τbd
= Φ *(0.87fy)/(4*1.2*1.6) = 47.01Φ
500 mm
= 752mm. Check for cracking: Clear distance between bars = [(250-(2*25)-(3*16)]/2 = 76mm<180mm
250mm
(o.k)
Ast provided is 3*201 = 603 mm² Min. Ast = 236mm²
Ast provided (o.k.) Curtailment of longitudinal reinforcement: The reinforcement can be curtailed towards the free end of the cantilever as bending moment decreases very rapidly. Let us curtail 1-16Φ bar at x distance from free end where this is not required to resist bending moment. Effective depth of the beam at distance x =200-40 + [(460-160)/3000]*x = 160 +0.1 x .mm. Factored bending moment at x from free end = 80.15*10^6*x²/3000² = 8.91x² Area of the remaining 2-16Φ bars = 2*201 =402 mm² M.R. = 0.87fy*Ast(d-0.42xu) Where xu= (0.87*fy*Ast)/ (0.36*fck*b) = (0.87*415*2*201)/(0.36*20*250) = 80.63mm. There fore, 8.91x² = (0.87*415*402) [(160+0.1x)-80.63] = 11519928+14514.2x x = 2213 mm.& d = 381.3 mm Hence, 1-16Φ can be curtailed at (3000-2213) 787mm from the fixed end. However as per clause 26.2.3.1 , the reinforcement shall extend beyond the point at which it is no longer required. To resist flexure for a distance equal to the effective depth of the member or 12 Φ which ever is greater. Here d= 381.3mm 12Φ = 12*16 =192mm
Greater is 381.3 mm. Extend the for a length of 381.3mm from the theoretical cutoff point. Practical cutoff point = 787+381.3 = 1168.3
say 1170mm.
Min.distance at which any bar can be curtailed from the fixed end is Ld = 752mm< 1170mm(o.k.) Further ,the remaining 2-16Φ bars must be continued upto a length Ld beyond the Actual cutoff point in order to develop their full tensile strength at that section.
LINTEL CUM SUNSHADE (P) Design the lintel and sun shade over an opening 2m wide in a wall 300mm thickness. The length of the sun shade is 750mm.the live load on sunshade is 500N/m² and the density of masonry is 19KN/m³.The height of opening is 2m with the height of the floor being 3.2m.Use M20 concrete and Fe415 steel.
Design of sunshade: (1)Breadth and depth: The 750mm wide sunshade act as a cantilever with clear span = 0.75m It will be designed for a unit width and the same reinforcement will be provided in the entire width. Let the overall thickness of sunshade be 1/10 th of span = 750/10 =75mm (2) Loads and bending moment:
Dead load of sunshade = 0.075*25 = 1.875KN/m² Live load on sunshade = 0.5KN/m² Total load = 2.375 KN/m² For 1m wide strip, load per a meter = 2.375KN/m Wu,factored load = 1.5*2.375 =3.56KN/m Mu,factored bending moment = Wul²/2 = 3.56*0.75²/2 = 1.002KN-m. (3) Depth for bending moment M.R.of a singly reinforced balanced section = 0.36*fck*b*xumax(d-0.42xumax) 1.002*10^6 = 0.36*20*1000*0.48d(d-0.42*0.48d) d = 19.1mm
too small
adopt 75mm overall depth at the fixed end and 50mm overall depth at free end. Effective depth at fixed end = 75-15-6/2 =57mm > 19.1mm (O.K.) Depth provided is larger than the depth required for a balanced section. and here the section become under reinforced section. (4) Steel for B.M.: Mu= 0.87*fy*Ast*(d-0.42xu) Where xu = [(0.87*fy*Ast)/(0.36*fck*b)] = [(0.87*415*Ast)/(0.36*20*1000)] = 0.05Ast 1.002*10^6 =0.87*415*Ast*[57-0.42(0.05Ast)] Ast²-2707.5Ast+131842 =0 Ast = 50mm² However minimum Ast = 0.12%of b*D = 0.12/100*1000*(75+50/2)= 75mm² Therefore provide minimum Ast = 75mm² Using 6mmΦ, spacing = (28.5/75)*1000 = 380mm <3*d = 3*75 = 171mm <450mm Therefore provide 6mmΦ@ 170mmc/c at top of sunshade as shown in fig. also provide distribution steel in the transverse direction @0.12%of b*D = 75mm² Therefore provide 6mmΦ@170mm c/c.
(5) Check for deflection: Actual l/d = 750/57 = 13.16 Allowable l/d = (basic l/d)*β*γ*δ*λ Basic l/d = 7for cantilever slab γ, modification factor for %steel & stress in steel is given by, %steel = [[(1000/170)*28.3]/(1000*57)]*100 = 0.3% Stress in steel = 0.58fy*(Ast req)/(Ast provided) = 0.58*415*50/((1000/170)*28.3) = 72.3N/mm² γ, from fig. no 4 of IS 456= 2 Therefore allowable l/d = 7*2 = 14 > actual l/d
(safe.)
(6) check for shear: Factored S.F at the end of fixed end for 1m wide strip = [0.75*0.5+ ((0.075+0.05)/2)*0.75*25] Vu = 2.32KN τv, nominal shear stress = Vu/(b*d) = 2.32*10³/(1000*57) = 0.04n/mm² τc, shear strength of M20 grade concrete @0.3% steel Design shear strength of slab = k.τc = 1.3*0.38 >> τv (7) Check for development length:
(safe.)
In order to develop full tensile strength at the face of , extended the reinforcement for a length Ld in to the Ld = (Φ σs)/(4τbd) = 47Φ = 47*6 = 282mm Design of lintel: The clear span of lintel = 2m Assuming width of bearing on either side as 300mm Assume breadth of lintel = thickness of wall = 0.3m Assume depth of lintel, d = l/10 =200mm D, overall depth of lintel = 200+40 = 240mm Effective span is given by (1) c/c of bearings = 2+0.3/2+0.3/2 = 2.3m (2) clear span + d = 2+0.2 = 2.2m The least of the above is effective span, l =2.2m
Loads and B.M: The height of roof above the bottom of lintel =3.2-2 = 1.2m <1.25 (1.91) Therefore no arch action is possible. Sin 60 = h/l or h= l*sin 60 = 2.2*sin 60 = 1.91m Hence loading due to wall will be uniform.
Self weight of lintel = 0.3*0.24*1*25 = 1.8KN/m. Dead load of masonry above lintel = (1.2-0.24)*0.3*19 = 5.47KN/m. Super imposed load from floor slab = 15KN/m
(assumed)
Load from sunshade =n 0.75*1*0.5+ [(0.075+0.05)/2]*0.75*1*25 = 1.55KN/m Total load on lintel = 23.82KN/m Factored load = 1.5(23.82) = 35.73KN/m Max. B.M = wl²/8 = 35.73*2.2²/8 = 21.61KN-m. Depth for B.M: M.R of singly reinforced balanced section, Mu = 0.36*fck*xumax(d-0.42xumax) Where xu max = 0.48d B = 300mm Fck = 20N/mm² 21.61*10^6 = 0.36*20*300*0.48d*(d-0.42*0.48d) d =√(21.61*10^6/(0.138*20*300)) = 161.5mm Effective depth available = 200mm > d required
(o.k)
Depths provide is larger than depth required for a balanced section. And hence the section becomes under reinforced section. Steel for B.M: Mu = 0.87*fy*Ast(d- 0.42xu) Where xu = [(0.87*fy*Ast)/(0.36*fck*b)] = [(0.87*415*Ast)/(0.36*20*300)] = 0.167Ast. 21.61*10^6 = 0.87*415*Ast(200-0.42*0.167Ast) 21.61*10^6 = 72210Ast-25.27Ast² Ast = 340mm² Minimum Ast = 0.85bd/fy = 0.85*300*200/415 = 123mm²
γ for %steel = [(2*201)/(300*200)]*100 = 0.67% And stress in steel = 0.58*fy [Ast req/Ast prov] = 0.58*415*[340/102] = 203.6N/mm² γ from fig. 4 of IS456 = 1.3 Basic l/d = 20
for simply ed beams.
Allowable l/d = 20*1.3 = 26 rel="nofollow"> actual l/d
(o.k)
Check for shear: Factored S.F = Vu = wu.l/2 = 35.73*2/2 = 35.73KN. τv = Vu/bd = [(35.73*10³)/(300*200)] = 0.6N/mm². τc, permissible shear stress for M20 concrete with 0.67% steel = 0.53N/mm². τc <τv < τcmax=2.8N/mm² Provide shear reinforcement Shear to be resisted by steel = Vu-(τc*b*d) = 35.73*10³-(0.53*300*200) Vus = 3.93*10³N. = 3.93KN. Using 6mmΦ 2-legged vertical stirrups, spacing = (Asv*0.87*fyd)/Vus (2*28.3*0.87*415*200)/3.93*10³ = 1040mm. Spacing of minimum shear reinforcement = (Asv*0.87*fy)/(0.4*b) = (2*28.3*0.87*415)/(0.4*300) = 170mm Max. Spacing = 0.75*d = 0.75*200 = 150mm Provide 6mmΦ 2-legged vtl. stirrups@150mm c/c Check for development length: 1.3*(M1/V) +Lo>= Ld Ld = (Φ*σs)/(4*τbd) = (Φ*0.87*fy)/(4*τbd) = (16*0.87*415)/(4*1.2*1.6) = 47 Φ = 752mm M1 = M.R = 0.87*fy*Ast(d-0.42xu) = 0.87*415*2*201*(200-(0.42*0.167*402)) = 24.94KN-m Vu = 35.73KN. Lo = bs/2-25-3Φ+8Φ+100
= 150-25-3*16+8*16+100 = 305mm 1.3(M1/V)+Lo = 1.3*[(24.94*10^6)/(35.73*10³)]+305 = 1212mm > Ld
(safe)
T- Beam floor A hall of internal dimensions of 6m*15m has beans spaced at 3m c/c. the beams are ed by wall around 300mm thick design the T-beam roof for a live load intensity of 1.5KN/m² use M20mix and Fe415 steel.
ly/lx = 6/3 = 2 Therefore the slab is designed as one way slab Design of intermediate slab Assume width of beam = 300mm. Thickness of slab from deflection criteria: lx/d = 26γ Assume γ = 1.2 d = lx/ (26*1.2) = 3000/(26*1.2) = 96mm Assuming a clear cover of 15mm for 10mmΦ bars. Overall depth, D = 96+15+10/2 = 120mm. Provide D = 120mm. Effective depth provide = 120-20 = 100mm.
Loads and B.M: Dead load of slab = 0.12*25 = 3KN/m² Live load on slab = 1.5KN/m² Assume floor finish = 1KN/m² Total load = 5.5KN/m² Factored load = 1.5*5.5 = 8.25KN/m² Consider 1m wide strip: Effective span for interior span is ( as per cl.22.2.b) As the width of (here beam) = 300mm > 1/12 of clear span = (1/12)*2700 = 225mm The eff. Span for intermediate spans is clear span i.e. l =2700mm. for intermediate span, -ve B.M at = (wd*l²/12)+(wl*l²/9) = (4*2.7²/12) + (1.5*2.7²/9) = 3.645KN-m +ve B.M at mid span = (wd*l²/16) + (wl*l²/12) = (4*2.7²/16) + (1.5*2.7²/12) = 2.73KN-m. Factored B.Ms: At = 1.5*3.465 = 5.47KN-m At mid span = 1.5*2.73 = 4.1KN-m Check the depth for B.M: Mu = 0.36*fck*b*xumax(d-0.42xumax) Where xumax = 0.48d 5.46*10^6 = 0.36*20*1000*0.48d*(d-0.42*0.48d) d = 44.52mm < d provide
(o.k)
the depth provide is larger than the depth required for a balanced section, the section becomes U.R section Steel for B.M: For U.R sections, Ast for –ve B.M = (0.5*fck*b*d/fy)*[1-√(1-((4.6*Mu)/(fck*b*d²)))] = (0.5*20*1000*100/415)*[1-√(1-((4.6*5.47*10^6)/(20*1000*100²)))]
= 157mm². Ast for +ve B.M = (0.5*fck*b*d/fy)*[1-√(1-((4.6*Mu)/(fck*b*d²)))] = 0.5*20*1000*100/415)*[1-√(1-((4.6*4.1*10^6)/(20*1000*100²)))] = 116mm²
(50/144)*1000 = 347mm
But the maximum spacing < 450mm < 5*d = 500mm Provide 8mmΦ 340mm c/c both at top and bottom as shown in fig…. Check for deflection: Actual l/d ratio = 2700/100 = 27 Allowable l/d = (basic l/d)* β*γ*δ*λ Basic l/d =26 fot continuous slabs % of steel = [((1000/300)*50)/(1000*100)]*100 = 0.17% Stress in steel = 0.58*(Ast req/Ast prov) = 0.58*415*(116/167) = 167N/mm² γ from fig. 4 of IS 456 = 2 allowable l/d = 26*2 = 52 rel="nofollow"> allowable l/d
(o.k)
Check for shear: S.F coefficients: Shear force at interior s Due to dead load = 0.50wd*l = 0.5*4*2.7 = 5.4KN. Due to live load = 0.60wl*l = 0.6*1.5*2.7 = 2.43KN. Total S.F = 7.83KN. Factored S.F = Vu = 1.5*7.83 = 11.75KN. τv, nominal shear stress = Vu/(bd)= 11.75*10³/(1000*100) = 0.12N/mm² τc, permissible shear stress for 0.17% steel&M20 concrete = 0.296N/mm² τc, for slabs = ks τc = 1.3*0.296 = 0.38N/mm² (where ks = 1.3 for D< 150mm) > τv
(safe)
Check for development length: At points of inflection Ld <= (M1/V) + Lo Ld = (Φ*σs)/(4*τbd) = (8*0.87*415)/(4*1.6*0.8) = 564mm M1 = M.R = 0.87*fy*Ast(d-0.42*xu) xu = (0.87*fy*Ast)/(0.36*fck*b) M1 = 0.87*fy*Ast(d-0.42*((0.87*fy*Ast)/(0.36*fck*b)))\ = 0.87*fy*Ast(d-((fy*Ast)/(fck*b))) Ast = (1000/300)*50 = 166mm² M1 = 0.87*415*166(100-((415*166)/(20*1000))) = 5.78KN-m Shear force at point of inflection assume point of inflection at (0.15l) from the face of the . S.F = shear force at the face of the ------------ (0.15l)*wu = 11.75-(0.15*2.7)*8.25 = 8.4KN. Lo = 12Φ = 12*8 = 96mm d = 100mm Grater of the above is 100mm Lo = 100mm
(M1/V)+Lo = (5.78*10^6/8.4*10³)+100 = 787mm.
> Ld
(safe)
Design of beam: (1) breadth and depth of the beam: Clear span of the beam = 6m Assume depth as 1/15 of L = (1/15)*6000 = 400mm Overall depth = 400+50eff cover = 450mm. Let breadth of the beam = 300mm Effective span: (1) clear span +c/c of bearings = 6000+300/2+300/2 = 6300mm (2) clear span +d = 6000+400 = 6400mm The least of the above is l = 6300mm Effective flange width = (lo/6)+bw+6Df = (6300/6)+300+(6*120) = 2070mm <3000mm Therefore take bf = 2070mm (3) loads and B.M:
Load from slab = (1.5+1+3)*3 = 16.5KN/m Self weight of rib = 0.3*0.33*1*25 = 2.5KN/m Total load per meter run of the beam = 19KN/m Max. B.M = wu*l²/8 = 28.5*6.3²/8 = 141.4KN-m Depth of N.A : Assume N.A lies in flange b = bf Total comp. = total ten. 0.36*fck*b*xu = 0.87*fy* Ast xu = (0.87*fy*Ast)/(0.36*fck*bf)
Steel: Equating M.R to design B.M, Mu = 0.87*fy*Ast(d- 0.42xu) Ast = (0.5*fck*b*d/fy)*[1-√(1-((4.6*Mu)/(fck*b*d²)))] = (0.5*20*2070*400/415)*[1-√(1-((4.6*141.6*10^6)/(20*2070*400²)))] = 1005mm². xu = (0.87*415*1005)/(0.36*20*2070) = 24.34mm< Df Therefore our assumption is correct. Minimum Ast = (0.85*bw*d)/fy = (0.85*300*400)/415 = 246mm². Provide 1005mm². Using 16mmΦ, no of bars = 1005/201 = 5 bars. Provide 5-16mmΦ bars at bottom. Check for deflection: Actual l/d = 6300/400 = 15.75 Allowable l/d = (basic l/d)*β*γ*δ*λ % of steel = [(5*201)/(2070*400)] = 0.12% Stress in steel = 0.58*fy*(Ast req/Ast prov) = 240N/mm² Γ from fig. 4 of IS 456 = 2 λ modification factor for flanged section- related to ratio of (web width/ flange width) = 300/2070 = 0.145 λ = 0.8
(from fig. 6 of IS 456)
Basic l/d = 20 for simply ed beams
Allowable l/d = 20*2*0.8 = 32
> actual l/d (safe)
Check for shear: Cross section for shear lies at d from the face of the Vu , shear at the critical section = wu*l/2 – (wu*d) = 28.5*6/2-(28.5*0.4) = 74.1KN. τv, nominal shear stress = Vu/bw*d = 74.1*10³/(300*400) = 0.62 N/mm² τc, permissible shear stress for % of steel = [(5*201)/(300*400)]*100 = 0.84% and M20 grade concrete is = 0.58N/mm² < τv τc<τv <τcmax =2.8N/mm² there fore provide shear reinforcement . using 8mm Φ 2-legged vertical stirrups spacing is (1) [Asv*0.87*fy*d]/Vus =[ 2*50*0.87*415*400]/(4.5*1000) = 3209 mm (2) Asv*0.87*fy/0.4b
= 2*50*0.87*415/(0.4*300) = 301 mm
(3) 0.75d
= 0.75*400 =300 mm.
Vus, shear to be resisted by steel = Vu-Vuc =74.1-(0.58*300*400/1000) = 4.5 KN. The least of the above is 300 mm Provide 8mm Φ 2-legged vertical stirrups @ 300mm c/c through out.
Check for development length: Ld <= 1.3(M1/V) + Lo Ld = (Φ*0.87*fy)/(4*τbd) = (16*0.87*415)/(4*1.6*0.8) = 1128 mm. M1= 0.87*fy*Ast(d-(fy*Ast/fck*b)) = 0.87*415*5*201(400-(415*5*201/20*300)) = 119.9 KN-m. V= 28.5*6/2 = 85.5 KN. Lo = (bs/2)-25-3Φ+8Φ+100 = 150-25-3*16+8*16+100 = 305 mm. (1.3M1/V)+Lo = 1.3(119.9*10^6/85.5*10³)+305 = 2128 mm
>Ld
(O.K)
Design of a section subjected to B.M.,S.F. and Torsion Moment. 2) Design a rectangular beam 300 mm wide and 600 mm deep subjected to a BM of 100 kN-m, shear force of 25 kN and twisting moment of 40 kN-m. Use M20 grade
concrete
and
HYSD
bars.Design 300 mm
as
per
LSD
method.
600 mm
A). Mu = 80 kN-m
Vu = 70 kN
Tu = 40 kN-m
Assume effective cover = 50mm effective depth = d = 600 – 50 = 550m Equivalent shear Ve = Vu + 1.6Tu/b = 70 + 1.6 * 40/0.3 = 283.3 kN. ve equivalent nominal shear stress =
Ve 283.3 * 10 3 1.72 N / mm 2 bd 300 * 450
Assuming % tension steel as 0.25% c for M20 concrete = 0.36 N/mm2 (table 19) cmax for M20 concrete = 2.8 N/mm2 c<ve<cmax Torsion reinforcement is required in the form of longitudinal steel and transverse reinforcement.
Longitudinal steel Equivalent B.M = Me1 = Mu + Mt. l D /b 1 600 / 300 40 70.6kN .m. 1. 7 1. 7
Where, M t Tu Me1 = 80 +70.6
= 150.6 kN –m.
Mt < Mu No steel is required on compression face. (cl.: 41.4.2.1) Check depth Me1 = 0.36 fck b xumax (d-0.42 xumax) Xumax = 0.48d 150.6*106 = 0.36 *20*300*0.48d( d- 0.42 *0.48d) d = 438 mm
< d available (o.k)
Since d available is greater than drequired for a balanced section,the section will be designed as U.R.Beam. Ast
0.5 fckbd fy
1 1
4.6 M e1 fckbd 2
0.5 * 20 * 300 * 550 4.6 * 150.6 * 10 6 1 1 415 20 * 300 * 550 2
=849.5 mm2 min Ast
0.85bd 0.85 * 300 * 550 338 Ast (O.K ). fy 415
Using16 bars no. of bars = 849.5/201 = 4.22 5 Provide 5 – 16 at the bottom. Transverse Reinforcement: Assuming 10 - 2 legged vertical stirrups Spacing is given by 1). From strength requirement (as per clause 41.4.3) Asv
Tu Sv VuSv b1d 1(0.87 fy ) 2.5d 1(0.87 fy )
b1 d1 157.08
b – [2 (25) + 2 (16/2)]
=
300 – 66 = 234
=
600 – (50) * 2 = 500
40 * 10 6 * Sv 70 * 10 3 * Sv 234 * 500 * 0.87 * 415 2.5 * 500 * 0.87 * 415
=
0.947 Sv + 0.155 Sv.
=
1.10Sv
Sv
=
142 mm
Asv
(ve - c) b Sv 0.87 fy
157.08
2).
=
157.08
(1.72 0.52) * 300 * Sv 0.87 * 415
c based on 5- 16 mm dia. Bars % steel = 5*201*100/300*550 = 0.61 For 0.61% steel and M 20 grade concrete, c = 0.52 N/mm2 Sv
=
282.15 mm.
Transverse reinforcement as per Cl. 26.5.1.7 of IS 456 3).
Spacing , Sv = x1 x1
= 234 + 16 + 10 = 260 mm
y1
= 600 – 2 (50) + 2(16/2) + 2(10/2) = 526 mm.
4)
x1 y1 260 526 196.5 4 4
5)
300 mm. Provide stirrups at 140 c/c As the depth is greater than 450mm provide side face reinforcement @ 0.1% of
C.S area (Cl.25.5.1.7(b)) =
0.1 * 300 * 600 180mm 2 100
Using 12 , no of bars
=
180 = 1.59 2. 113
Provide 2-12 on each side, and also provide 2-12 at top as anchor bars
Reinforcement details
1) Design a cantilever beam of span 3m subjected to u.d.l of 10KN/m. useM20 grade concrete and HYSD bars. Design as per L.S.M. Data: For M20 grade concrete, fck = 20N/mm². For HYSD bars, fy = 415N/mm². Super imposed load = 10KN/m. Span = 3m. xu max. = 0.48d. Breadth and depth of the beam: Assume depth, d = L/7 = 3000/7 = 428mm. let us adopt overall depth of the beam, D = 500mm. at the fixed end. The depth can be reduced to 200mm (min. is 150mm) at the free end. Where the B.M will be zero. Provide, b= D/2 = 500/2= 250mm. the effective span = clear span for cantilevers = 3m. eff. Cover = 40mm. d, eff. Depth = 500-40 =460mm.
Loads and B.M: Self weight of beam = (0.5+0.2)/2 *0.25*3*25 = 6.56KN/m.
This acts at = 3/3[(0.5+2*0.2)/(0.5+0.2)] = 1.29m from the fixed end. Taking a load factor of 1.5, Factored B.M at fixed end = (1.5*10*3²/2)+(1.5*6.56*1.29) = 80.15KN.m. Depth for B.M: M.R of a singly reinforced balanced section Mu = 0.36*fck*b*xu max.*(d-(0.42 xu max)) ---------------------- (1) 80.15*10^6= 0.36*20*250*0.48*d.*(d-(0.42 *0.48*d)) 80.15*10^6 = 689.8d² d=340.8mm <460mm (available depth) o.k. Depth provided is larger than the depth required for balanced section the section becomes U.R.section Steel for bending moment: To find steel area, Mu=0.87*fy*Ast(d-0.42xu) Where xu, depth of neutral axis for U.R.section This is given by Total compression = total tension 0.36fck*b*xu=0.87fy*Ast xu =[ (0.87fy*Ast)/(0.36fck*b)] =[ (0.87*415*Ast)/(0.36*20*250)] = 0.2006 Ast Substituting this in equation (1) Mu= 0.87*415*Ast (460-(0.42*0.2006Ast)) 80.15*10^6 =166083 Ast- 30.42Ast² Ast = 535mm² Min Ast= 0.85bd/fy =0.85*250*460/415 = 236mm² < 535 mm² o.k. Using 16mm-dia bars, no. of bars =535/201 =2.7 =3bars Provide 3 – 16 mm dia bars at the top
Check for deflection: Actual l/d = 3000/460 =6.52 Allowable l/d = (basic l/d)*β*γ*δ*λ β = 1, δ=1, λ=1 γ, modification factor for percentage of steel and stress in steel is given by, % steel= [(3*201)/ (250*460)] *100 =0.52% Stress in steel = 0.85*fy*(Ast required / Ast provided) =0.85*415*(2.67 / 3) = 214 N/mm² Ref. to fig 4 of IS-456,γ=1.4 Basic l/d= 7 for cantilevers (clause 23.2.1) Allowable l/d= 7*1.4 =9.8 > actual l/d
o.k.
Check for shear: Factored max. Shear force at the fixed end = 1.5(10*3) + 1.5(6.56) = 54.84 KN. τv , nominal shear = for beams of varying depth ,τv =[Vu-((Mu/d)*tanβ)]/bd τv =[54.84*10^3- ((80.15*10^6/460)*0.1)]/250*460 = 0.33 N/mm² For 0.52% steel and M20 grade concrete τc , permissible shear stress = 0.49 N/mm² > τv There fore provide minimum shear reinforcement. Using 8mm dia 2-legged vertical stirrups Spacing = Asv* (0.87fy)/(0.4b) = 2*50* (0.87*415)/(0.4*250) = 361mm Max. Spacing = 0.75 d = 0.75*460 =345 mm. There fore, provide 8mm dia 2-legged vertical stirrups @ 300mm c/c. through out . Check for development length: Inorder to develop full tensile strength at the face of , each of the 3-16dia bars must be embedded by a length equal to the development length. Ld= Φσs/4τbd
= Φ *(0.87fy)/(4*1.2*1.6) = 47.01Φ
500 mm
= 752mm. Check for cracking: Clear distance between bars = [(250-(2*25)-(3*16)]/2 = 76mm<180mm
250mm
(o.k)
Ast provided is 3*201 = 603 mm² Min. Ast = 236mm²
Greater is 381.3 mm. Extend the for a length of 381.3mm from the theoretical cutoff point. Practical cutoff point = 787+381.3 = 1168.3
say 1170mm.
Min.distance at which any bar can be curtailed from the fixed end is Ld = 752mm< 1170mm(o.k.) Further ,the remaining 2-16Φ bars must be continued upto a length Ld beyond the Actual cutoff point in order to develop their full tensile strength at that section.
LINTEL CUM SUNSHADE (P) Design the lintel and sun shade over an opening 2m wide in a wall 300mm thickness. The length of the sun shade is 750mm.the live load on sunshade is 500N/m² and the density of masonry is 19KN/m³.The height of opening is 2m with the height of the floor being 3.2m.Use M20 concrete and Fe415 steel.
Design of sunshade: (1)Breadth and depth: The 750mm wide sunshade act as a cantilever with clear span = 0.75m It will be designed for a unit width and the same reinforcement will be provided in the entire width. Let the overall thickness of sunshade be 1/10 th of span = 750/10 =75mm (2) Loads and bending moment:
Dead load of sunshade = 0.075*25 = 1.875KN/m² Live load on sunshade = 0.5KN/m² Total load = 2.375 KN/m² For 1m wide strip, load per a meter = 2.375KN/m Wu,factored load = 1.5*2.375 =3.56KN/m Mu,factored bending moment = Wul²/2 = 3.56*0.75²/2 = 1.002KN-m. (3) Depth for bending moment M.R.of a singly reinforced balanced section = 0.36*fck*b*xumax(d-0.42xumax) 1.002*10^6 = 0.36*20*1000*0.48d(d-0.42*0.48d) d = 19.1mm
too small
adopt 75mm overall depth at the fixed end and 50mm overall depth at free end. Effective depth at fixed end = 75-15-6/2 =57mm > 19.1mm (O.K.) Depth provided is larger than the depth required for a balanced section. and here the section become under reinforced section. (4) Steel for B.M.: Mu= 0.87*fy*Ast*(d-0.42xu) Where xu = [(0.87*fy*Ast)/(0.36*fck*b)] = [(0.87*415*Ast)/(0.36*20*1000)] = 0.05Ast 1.002*10^6 =0.87*415*Ast*[57-0.42(0.05Ast)] Ast²-2707.5Ast+131842 =0 Ast = 50mm² However minimum Ast = 0.12%of b*D = 0.12/100*1000*(75+50/2)= 75mm² Therefore provide minimum Ast = 75mm² Using 6mmΦ, spacing = (28.5/75)*1000 = 380mm <3*d = 3*75 = 171mm <450mm Therefore provide 6mmΦ@ 170mmc/c at top of sunshade as shown in fig. also provide distribution steel in the transverse direction @0.12%of b*D = 75mm² Therefore provide 6mmΦ@170mm c/c.
(5) Check for deflection: Actual l/d = 750/57 = 13.16 Allowable l/d = (basic l/d)*β*γ*δ*λ Basic l/d = 7for cantilever slab γ, modification factor for %steel & stress in steel is given by, %steel = [[(1000/170)*28.3]/(1000*57)]*100 = 0.3% Stress in steel = 0.58fy*(Ast req)/(Ast provided) = 0.58*415*50/((1000/170)*28.3) = 72.3N/mm² γ, from fig. no 4 of IS 456= 2 Therefore allowable l/d = 7*2 = 14 > actual l/d
(safe.)
(6) check for shear: Factored S.F at the end of fixed end for 1m wide strip = [0.75*0.5+ ((0.075+0.05)/2)*0.75*25] Vu = 2.32KN τv, nominal shear stress = Vu/(b*d) = 2.32*10³/(1000*57) = 0.04n/mm² τc, shear strength of M20 grade concrete @0.3% steel Design shear strength of slab = k.τc = 1.3*0.38 >> τv (7) Check for development length:
(safe.)
In order to develop full tensile strength at the face of , extended the reinforcement for a length Ld in to the Ld = (Φ σs)/(4τbd) = 47Φ = 47*6 = 282mm Design of lintel: The clear span of lintel = 2m Assuming width of bearing on either side as 300mm Assume breadth of lintel = thickness of wall = 0.3m Assume depth of lintel, d = l/10 =200mm D, overall depth of lintel = 200+40 = 240mm Effective span is given by (1) c/c of bearings = 2+0.3/2+0.3/2 = 2.3m (2) clear span + d = 2+0.2 = 2.2m The least of the above is effective span, l =2.2m
Loads and B.M: The height of roof above the bottom of lintel =3.2-2 = 1.2m <1.25 (1.91) Therefore no arch action is possible. Sin 60 = h/l or h= l*sin 60 = 2.2*sin 60 = 1.91m Hence loading due to wall will be uniform.
Self weight of lintel = 0.3*0.24*1*25 = 1.8KN/m. Dead load of masonry above lintel = (1.2-0.24)*0.3*19 = 5.47KN/m. Super imposed load from floor slab = 15KN/m
(assumed)
Load from sunshade =n 0.75*1*0.5+ [(0.075+0.05)/2]*0.75*1*25 = 1.55KN/m Total load on lintel = 23.82KN/m Factored load = 1.5(23.82) = 35.73KN/m Max. B.M = wl²/8 = 35.73*2.2²/8 = 21.61KN-m. Depth for B.M: M.R of singly reinforced balanced section, Mu = 0.36*fck*xumax(d-0.42xumax) Where xu max = 0.48d B = 300mm Fck = 20N/mm² 21.61*10^6 = 0.36*20*300*0.48d*(d-0.42*0.48d) d =√(21.61*10^6/(0.138*20*300)) = 161.5mm Effective depth available = 200mm > d required
(o.k)
Depths provide is larger than depth required for a balanced section. And hence the section becomes under reinforced section. Steel for B.M: Mu = 0.87*fy*Ast(d- 0.42xu) Where xu = [(0.87*fy*Ast)/(0.36*fck*b)] = [(0.87*415*Ast)/(0.36*20*300)] = 0.167Ast. 21.61*10^6 = 0.87*415*Ast(200-0.42*0.167Ast) 21.61*10^6 = 72210Ast-25.27Ast² Ast = 340mm² Minimum Ast = 0.85bd/fy = 0.85*300*200/415 = 123mm²
γ for %steel = [(2*201)/(300*200)]*100 = 0.67% And stress in steel = 0.58*fy [Ast req/Ast prov] = 0.58*415*[340/102] = 203.6N/mm² γ from fig. 4 of IS456 = 1.3 Basic l/d = 20
for simply ed beams.
Allowable l/d = 20*1.3 = 26 rel="nofollow"> actual l/d
(o.k)
Check for shear: Factored S.F = Vu = wu.l/2 = 35.73*2/2 = 35.73KN. τv = Vu/bd = [(35.73*10³)/(300*200)] = 0.6N/mm². τc, permissible shear stress for M20 concrete with 0.67% steel = 0.53N/mm². τc <τv < τcmax=2.8N/mm² Provide shear reinforcement Shear to be resisted by steel = Vu-(τc*b*d) = 35.73*10³-(0.53*300*200) Vus = 3.93*10³N. = 3.93KN. Using 6mmΦ 2-legged vertical stirrups, spacing = (Asv*0.87*fyd)/Vus (2*28.3*0.87*415*200)/3.93*10³ = 1040mm. Spacing of minimum shear reinforcement = (Asv*0.87*fy)/(0.4*b) = (2*28.3*0.87*415)/(0.4*300) = 170mm Max. Spacing = 0.75*d = 0.75*200 = 150mm Provide 6mmΦ 2-legged vtl. stirrups@150mm c/c Check for development length: 1.3*(M1/V) +Lo>= Ld Ld = (Φ*σs)/(4*τbd) = (Φ*0.87*fy)/(4*τbd) = (16*0.87*415)/(4*1.2*1.6) = 47 Φ = 752mm M1 = M.R = 0.87*fy*Ast(d-0.42xu) = 0.87*415*2*201*(200-(0.42*0.167*402)) = 24.94KN-m Vu = 35.73KN. Lo = bs/2-25-3Φ+8Φ+100
= 150-25-3*16+8*16+100 = 305mm 1.3(M1/V)+Lo = 1.3*[(24.94*10^6)/(35.73*10³)]+305 = 1212mm > Ld
(safe)
T- Beam floor A hall of internal dimensions of 6m*15m has beans spaced at 3m c/c. the beams are ed by wall around 300mm thick design the T-beam roof for a live load intensity of 1.5KN/m² use M20mix and Fe415 steel.
ly/lx = 6/3 = 2 Therefore the slab is designed as one way slab Design of intermediate slab Assume width of beam = 300mm. Thickness of slab from deflection criteria: lx/d = 26γ Assume γ = 1.2 d = lx/ (26*1.2) = 3000/(26*1.2) = 96mm Assuming a clear cover of 15mm for 10mmΦ bars. Overall depth, D = 96+15+10/2 = 120mm. Provide D = 120mm. Effective depth provide = 120-20 = 100mm.
Loads and B.M: Dead load of slab = 0.12*25 = 3KN/m² Live load on slab = 1.5KN/m² Assume floor finish = 1KN/m² Total load = 5.5KN/m² Factored load = 1.5*5.5 = 8.25KN/m² Consider 1m wide strip: Effective span for interior span is ( as per cl.22.2.b) As the width of (here beam) = 300mm > 1/12 of clear span = (1/12)*2700 = 225mm The eff. Span for intermediate spans is clear span i.e. l =2700mm. for intermediate span, -ve B.M at = (wd*l²/12)+(wl*l²/9) = (4*2.7²/12) + (1.5*2.7²/9) = 3.645KN-m +ve B.M at mid span = (wd*l²/16) + (wl*l²/12) = (4*2.7²/16) + (1.5*2.7²/12) = 2.73KN-m. Factored B.Ms: At = 1.5*3.465 = 5.47KN-m At mid span = 1.5*2.73 = 4.1KN-m Check the depth for B.M: Mu = 0.36*fck*b*xumax(d-0.42xumax) Where xumax = 0.48d 5.46*10^6 = 0.36*20*1000*0.48d*(d-0.42*0.48d) d = 44.52mm < d provide
(o.k)
the depth provide is larger than the depth required for a balanced section, the section becomes U.R section Steel for B.M: For U.R sections, Ast for –ve B.M = (0.5*fck*b*d/fy)*[1-√(1-((4.6*Mu)/(fck*b*d²)))] = (0.5*20*1000*100/415)*[1-√(1-((4.6*5.47*10^6)/(20*1000*100²)))]
= 157mm². Ast for +ve B.M = (0.5*fck*b*d/fy)*[1-√(1-((4.6*Mu)/(fck*b*d²)))] = 0.5*20*1000*100/415)*[1-√(1-((4.6*4.1*10^6)/(20*1000*100²)))] = 116mm²
(50/144)*1000 = 347mm
But the maximum spacing < 450mm < 5*d = 500mm Provide 8mmΦ 340mm c/c both at top and bottom as shown in fig…. Check for deflection: Actual l/d ratio = 2700/100 = 27 Allowable l/d = (basic l/d)* β*γ*δ*λ Basic l/d =26 fot continuous slabs % of steel = [((1000/300)*50)/(1000*100)]*100 = 0.17% Stress in steel = 0.58*(Ast req/Ast prov) = 0.58*415*(116/167) = 167N/mm² γ from fig. 4 of IS 456 = 2 allowable l/d = 26*2 = 52 rel="nofollow"> allowable l/d
(o.k)
Check for shear: S.F coefficients: Shear force at interior s Due to dead load = 0.50wd*l = 0.5*4*2.7 = 5.4KN. Due to live load = 0.60wl*l = 0.6*1.5*2.7 = 2.43KN. Total S.F = 7.83KN. Factored S.F = Vu = 1.5*7.83 = 11.75KN. τv, nominal shear stress = Vu/(bd)= 11.75*10³/(1000*100) = 0.12N/mm² τc, permissible shear stress for 0.17% steel&M20 concrete = 0.296N/mm² τc, for slabs = ks τc = 1.3*0.296 = 0.38N/mm² (where ks = 1.3 for D< 150mm) > τv
(safe)
Check for development length: At points of inflection Ld <= (M1/V) + Lo Ld = (Φ*σs)/(4*τbd) = (8*0.87*415)/(4*1.6*0.8) = 564mm M1 = M.R = 0.87*fy*Ast(d-0.42*xu) xu = (0.87*fy*Ast)/(0.36*fck*b) M1 = 0.87*fy*Ast(d-0.42*((0.87*fy*Ast)/(0.36*fck*b)))\ = 0.87*fy*Ast(d-((fy*Ast)/(fck*b))) Ast = (1000/300)*50 = 166mm² M1 = 0.87*415*166(100-((415*166)/(20*1000))) = 5.78KN-m Shear force at point of inflection assume point of inflection at (0.15l) from the face of the . S.F = shear force at the face of the ------------ (0.15l)*wu = 11.75-(0.15*2.7)*8.25 = 8.4KN. Lo = 12Φ = 12*8 = 96mm d = 100mm Grater of the above is 100mm Lo = 100mm
(M1/V)+Lo = (5.78*10^6/8.4*10³)+100 = 787mm.
> Ld
(safe)
Design of beam: (1) breadth and depth of the beam: Clear span of the beam = 6m Assume depth as 1/15 of L = (1/15)*6000 = 400mm Overall depth = 400+50eff cover = 450mm. Let breadth of the beam = 300mm Effective span: (1) clear span +c/c of bearings = 6000+300/2+300/2 = 6300mm (2) clear span +d = 6000+400 = 6400mm The least of the above is l = 6300mm Effective flange width = (lo/6)+bw+6Df = (6300/6)+300+(6*120) = 2070mm <3000mm Therefore take bf = 2070mm (3) loads and B.M:
Load from slab = (1.5+1+3)*3 = 16.5KN/m Self weight of rib = 0.3*0.33*1*25 = 2.5KN/m Total load per meter run of the beam = 19KN/m Max. B.M = wu*l²/8 = 28.5*6.3²/8 = 141.4KN-m Depth of N.A : Assume N.A lies in flange b = bf Total comp. = total ten. 0.36*fck*b*xu = 0.87*fy* Ast xu = (0.87*fy*Ast)/(0.36*fck*bf)
Steel: Equating M.R to design B.M, Mu = 0.87*fy*Ast(d- 0.42xu) Ast = (0.5*fck*b*d/fy)*[1-√(1-((4.6*Mu)/(fck*b*d²)))] = (0.5*20*2070*400/415)*[1-√(1-((4.6*141.6*10^6)/(20*2070*400²)))] = 1005mm². xu = (0.87*415*1005)/(0.36*20*2070) = 24.34mm< Df Therefore our assumption is correct. Minimum Ast = (0.85*bw*d)/fy = (0.85*300*400)/415 = 246mm². Provide 1005mm². Using 16mmΦ, no of bars = 1005/201 = 5 bars. Provide 5-16mmΦ bars at bottom. Check for deflection: Actual l/d = 6300/400 = 15.75 Allowable l/d = (basic l/d)*β*γ*δ*λ % of steel = [(5*201)/(2070*400)] = 0.12% Stress in steel = 0.58*fy*(Ast req/Ast prov) = 240N/mm² Γ from fig. 4 of IS 456 = 2 λ modification factor for flanged section- related to ratio of (web width/ flange width) = 300/2070 = 0.145 λ = 0.8
(from fig. 6 of IS 456)
Basic l/d = 20 for simply ed beams
Allowable l/d = 20*2*0.8 = 32
> actual l/d (safe)
Check for shear: Cross section for shear lies at d from the face of the Vu , shear at the critical section = wu*l/2 – (wu*d) = 28.5*6/2-(28.5*0.4) = 74.1KN. τv, nominal shear stress = Vu/bw*d = 74.1*10³/(300*400) = 0.62 N/mm² τc, permissible shear stress for % of steel = [(5*201)/(300*400)]*100 = 0.84% and M20 grade concrete is = 0.58N/mm² < τv τc<τv <τcmax =2.8N/mm² there fore provide shear reinforcement . using 8mm Φ 2-legged vertical stirrups spacing is (1) [Asv*0.87*fy*d]/Vus =[ 2*50*0.87*415*400]/(4.5*1000) = 3209 mm (2) Asv*0.87*fy/0.4b
= 2*50*0.87*415/(0.4*300) = 301 mm
(3) 0.75d
= 0.75*400 =300 mm.
Vus, shear to be resisted by steel = Vu-Vuc =74.1-(0.58*300*400/1000) = 4.5 KN. The least of the above is 300 mm Provide 8mm Φ 2-legged vertical stirrups @ 300mm c/c through out.
Check for development length: Ld <= 1.3(M1/V) + Lo Ld = (Φ*0.87*fy)/(4*τbd) = (16*0.87*415)/(4*1.6*0.8) = 1128 mm. M1= 0.87*fy*Ast(d-(fy*Ast/fck*b)) = 0.87*415*5*201(400-(415*5*201/20*300)) = 119.9 KN-m. V= 28.5*6/2 = 85.5 KN. Lo = (bs/2)-25-3Φ+8Φ+100 = 150-25-3*16+8*16+100 = 305 mm. (1.3M1/V)+Lo = 1.3(119.9*10^6/85.5*10³)+305 = 2128 mm
>Ld
(O.K)
Design of a section subjected to B.M.,S.F. and Torsion Moment. 2) Design a rectangular beam 300 mm wide and 600 mm deep subjected to a BM of 100 kN-m, shear force of 25 kN and twisting moment of 40 kN-m. Use M20 grade
concrete
and
HYSD
bars.Design 300 mm
as
per
LSD
method.
600 mm
A). Mu = 80 kN-m
Vu = 70 kN
Tu = 40 kN-m
Assume effective cover = 50mm effective depth = d = 600 – 50 = 550m Equivalent shear Ve = Vu + 1.6Tu/b = 70 + 1.6 * 40/0.3 = 283.3 kN. ve equivalent nominal shear stress =
Ve 283.3 * 10 3 1.72 N / mm 2 bd 300 * 450
Assuming % tension steel as 0.25% c for M20 concrete = 0.36 N/mm2 (table 19) cmax for M20 concrete = 2.8 N/mm2 c<ve<cmax Torsion reinforcement is required in the form of longitudinal steel and transverse reinforcement.
Longitudinal steel Equivalent B.M = Me1 = Mu + Mt. l D /b 1 600 / 300 40 70.6kN .m. 1. 7 1. 7
Where, M t Tu Me1 = 80 +70.6
= 150.6 kN –m.
Mt < Mu No steel is required on compression face. (cl.: 41.4.2.1) Check depth Me1 = 0.36 fck b xumax (d-0.42 xumax) Xumax = 0.48d 150.6*106 = 0.36 *20*300*0.48d( d- 0.42 *0.48d) d = 438 mm
< d available (o.k)
Since d available is greater than drequired for a balanced section,the section will be designed as U.R.Beam. Ast
0.5 fckbd fy
1 1
4.6 M e1 fckbd 2
0.5 * 20 * 300 * 550 4.6 * 150.6 * 10 6 1 1 415 20 * 300 * 550 2
=849.5 mm2 min Ast
0.85bd 0.85 * 300 * 550 338 Ast (O.K ). fy 415
Using16 bars no. of bars = 849.5/201 = 4.22 5 Provide 5 – 16 at the bottom. Transverse Reinforcement: Assuming 10 - 2 legged vertical stirrups Spacing is given by 1). From strength requirement (as per clause 41.4.3) Asv
Tu Sv VuSv b1d 1(0.87 fy ) 2.5d 1(0.87 fy )
b1 d1 157.08
b – [2 (25) + 2 (16/2)]
=
300 – 66 = 234
=
600 – (50) * 2 = 500
40 * 10 6 * Sv 70 * 10 3 * Sv 234 * 500 * 0.87 * 415 2.5 * 500 * 0.87 * 415
=
0.947 Sv + 0.155 Sv.
=
1.10Sv
Sv
=
142 mm
Asv
(ve - c) b Sv 0.87 fy
157.08
2).
=
157.08
(1.72 0.52) * 300 * Sv 0.87 * 415
c based on 5- 16 mm dia. Bars % steel = 5*201*100/300*550 = 0.61 For 0.61% steel and M 20 grade concrete, c = 0.52 N/mm2 Sv
=
282.15 mm.
Transverse reinforcement as per Cl. 26.5.1.7 of IS 456 3).
Spacing , Sv = x1 x1
= 234 + 16 + 10 = 260 mm
y1
= 600 – 2 (50) + 2(16/2) + 2(10/2) = 526 mm.
4)
x1 y1 260 526 196.5 4 4
5)
300 mm. Provide stirrups at 140 c/c As the depth is greater than 450mm provide side face reinforcement @ 0.1% of
C.S area (Cl.25.5.1.7(b)) =
0.1 * 300 * 600 180mm 2 100
Using 12 , no of bars
=
180 = 1.59 2. 113
Provide 2-12 on each side, and also provide 2-12 at top as anchor bars
Reinforcement details
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