RC STAIR DESIGN
In accordance with EN1992-1-1:2004 incorporating Corrigenda January 2008 and the recommended
values
Tedds calculation version 1.0.03
Design summary
Description Unit Provided Required Utilisation Result
Bottom long. reinf. -Mid Span mm2 1847 1639 0.89 PASS
Bottom long. reinf. -Upper land. mm2 924 820 0.89 PASS
Bottom long. reinf. -Lower land. mm2 924 820 0.89 PASS
Allowable Actual Utilisation Result
Span-to-depth ratio 25.24 22.87 0.91 PASS
Shear capacity -Upper supp. kN 149.44 94.04 0.63 PASS
Shear capacity -Lower supp. kN 149.44 94.04 0.63 PASS
Stair geometry
Number of risers; Nsteps = 8
Going; G = 300 mm
Rise; R = 160 mm
Waist; hwaist = 300 mm
Breadth; b = 1400 mm
Length of the tread span; Lmid = (Nsteps – 1) G = 2100 mm
Overall height of stairs; hstairs = Nsteps R = 1280 mm
Angle of stairs; stairs = atan (R / G) = 28.07
Upper landing - Simple end support
Length of the upper landing; Lup = 2690 mm
Depth of the upper landing; hup = 300 mm
Width of the supporting element; wup = 250 mm
Distance to the centre of the support; ds,up = 125 mm
Lower landing - Simple end support
Length of the lower landing; Llow = 1590 mm
Depth of the lower landing; hlow = 300 mm
Width of the supporting element; wlow = 250 mm
Distance to the centre of the support; ds,low = 125 mm
Effective span – 5.3.2.2
Overall length; Ltotal = Llow + Lmid + Lup = 6380 mm
Clear distance between the supports faces; Ln = Ltotal - wup -wlow = 5880 mm
Effective span variable – cl.5.3.2.2(1); aup = min(0.5 hup, 0.5 wup) = 125 mm
Effective span variable – cl.5.3.2.2(1); alow = min(0.5 hlow, 0.5 wlow) = 125 mm
Length of the effective span – exp.5.8; Lspan = Ln + alow + aup = 6130 mm
Concrete details - Table 3.1.
Concrete strength class; C20/25
Characteristic compressive cylinder strength; fck = 20 N/mm2
Mean value of compressive cylinder strength fcm =fck + 8 N/mm2 = 28.0 N/mm2
Mean value of axial tensile strength fctm = 0.3 N/mm2 (fck/ 1 N/mm2)2/3 = 2.2 N/mm2
Secant modulus of elasticity of concrete Ecm = 22 kN/mm2[fcm/10 N/mm2]0.3 = 29962.0 N/mm2
Partial factor for concrete – Table 2.1N; c = 1.5
Density of concrete; conc = 25.00 kN/m3
Compression chord coefficient - cl.6.2.3(3); cw = 1
Compressive strength coefficient – cl.3.1.6(1); cc = 1.00
Design compr. concrete strength - exp.3.15 fcd = cc fck / c = 13.3 N/mm2
Effective strength factor – exp.3.21; =1
Effect. compr. zone height factor – exp.3.19; = 0.8
Ultimate strain - Table 3.1; cu2 = 0.0035
Shortening strain - Table 3.1; cu3 = 0.0035
Maximum aggregate size; hagg = 20 mm
Reinforcing steel details
Characteristic yield strength of reinforcement; fyk = 400 N/mm2
Partial factor for reinforcing steel - Table 2.1N; s = 1.15
Design yield strength of reinforcement; fyd = fyk / s = 348 N/mm2
Loading details
Self weight slab; gself,slab = hwaist / cos(stairs) (conc) b = 11.9 kN/m
Self weight steps; gself,steps = R / 2 (conc) b = 2.8 kN/m
Average self weight; gself,aver = gself,slab + gself,steps = 14.7 kN/m
Loading from finishes; gfin = 2.4 kN/m2
Imposed variable load; qk = 3.0 kN/m2
Permanent action factor; G = 1.35
Imposed action factor; Q = 1.50
Quasi-permanent value of variable action 2 = 0.30
Design load; FEd = G (gself,aver + gfin b) + Q qk b = 30.7 kN /m
Quasi-permanent load; FQP = 1.0 (gself,aver + gfin b) + 2 qk b = 19.3 kN /m
Midspan design
= 0.125
Design moment M = abs( FEd Lspan2) = 144.1 kNm
Nominal cover to reinforcement cnom = 25 mm
Diameter of bar for long. direction l =14 mm
Depth of reinforcement d = h - cnom - l / 2 = 268 mm
Redistribution ratio = 1.00
k1 = 0.44
k2 = 1.25 (0.6 + 0.0014 / cu2) = 1.25
K coefficient K = M / (b d2 fck) = 0.072
K' coefficient
K' = (2 cc / c) (1 - ( - k1) / (2 k2)) ( ( - k1) / (2 k2)) = 0.196
K < K' -Compression reinforcement is not required
Lever arm z = min(0.5 d [1 + (1 - 2 Kz / ( cc / c))0.5], 0.95 d)
z = 253 mm
Tension area required in longitudinal direction As,req = M / (fyd z) = 1639 mm2
Minimum area of reinforcement – exp.9.1.N As,min = max(0.26 fctm / fyk b d, 0.0013 b d)= 539
mm2
Maximum area of reinforcement - cl.9.2.1.1(3) As,max = 0.04 b h = 16800 mm2
Tension reinforcement check
Diameter of bars for longitudinal direction l = 14 mm
Number of bars in longitudinal direction N = 12
Diameter of bars for transverse direction t = 10 mm
Bar spacing in transverse direction s = 250 mm
Tension area provided in longitudinal direction As,prov = N l2 /4 = 1847 mm2
PASS - Tension reinforcement area is greater than area required
Tension secondary area required in transverse direction – cl.9.3.1.1(2)
As,req,t = 0.2 max(As,prov, As,req) = 369 mm2
Tension area provided in transverse direction As,prov,t = t2 /4 (b / s) = 440 mm2
PASS - Tension reinforcement in transverse direction is greater than area required
Minimum bar spacing - Section 8.2
Bar spacing (clear distance) in long. direction; sbar,l = (b - (2 cnom + N l)) / (N - 1) = 107 mm
kspc1 = 1
kspc2 = 5 mm
Minimum allowable bar spacing – cl.8.2(2); smin,l = max(kspc1 l, hagg + kspc2, 20 mm) = 25 mm
PASS - Actual bar spacing exceeds minimum allowable
Bar spacing (clear distance) in transv. direction; sbar,t = s - t = 240 mm
Minimum allowable bar spacing – cl.8.2(2); smin,t = max(kspc1 t, hagg + kspc2, 20 mm) = 25 mm
PASS - Actual bar spacing exceeds minimum allowable
Crack control - Section 7.3
Maximum crack width wk = 0.3 mm
Design value modulus of elasticity reinf – 3.2.7(4) Es = 200000 N/mm2
Mean value of concrete tensile strength fct,eff = fctm = 2.2 N/mm2
Stress distribution coefficient kc = 0.4
Non-uniform self-equilibrating stress coefficient k =min(max(1 + (300 mm - min(h, b)) 0.35 / 500 mm,
0.65), 1) = 1.00
Actual tension bar spacing sbar = (b -(2 (cnom + / 2))) / (N - 1) = 121 mm
Maximum stress permitted - Table 7.3N s = 303 N/mm2
Steel to concrete modulus of elast. ratio cr = Es / Ecm= 6.68
Distance of the Elastic NA from tension face y = (b h2 / 2 + As,prov (cr - 1) (h - d)) / (b h + As,prov
(cr - 1))
y = 147 mm
Area of concrete in the tensile zone Act = b y = 205977 mm2
Minimum area of reinforcement required - exp.7.1 Asc,min = kc k fct,eff Act / s = 601 mm2
PASS - Area of tension reinforcement provided exceeds minimum required for crack control
Quasi-permanent moment MQP = abs( FQP Lspan2) = 90.7kNm
Permanent load ratio RPL = MQP / M = 0.63
Service stress in reinforcement sr = fyd As,req / As,prov RPL = 194 N/mm2
Maximum bar spacing - Tables 7.3N sbar,max = 257 mm
PASS - Maximum bar spacing exceeds actual bar spacing for crack control
Deflection control - Section 7.4
Reference reinforcement ratio 0 = (fck / 1 N/mm2)0.5 / 1000 = 0.00447
Required tension reinforcement ratio = As,req / (b dmid) = 0.00437
Required compression reinforcement ratio ' = A's,req / (b dmid) = 0.00000
Structural system factor - Table 7.4N Kb = 1.00
Basic allow. span to depth ratio -exp.7.16(a)
stdbasic = Kb [11 + 1.5 (fck / 1 N/mm2)0.5 0 / +3.2 (fck / 1 N/mm2)0.5 (0 / - 1)3/2] = 17.917
Reinforcement factor - exp.7.17 Ks = min(As,prov / As,req 500 N/mm2 / fyk, 1.5) = 1.408
Long span supp. brittle partition factor-cl.7.4.2(2) F1 = 1.00
Allowable span to depth ratio stdallow = stdbasic Ks F1 = 25.236
Actual span to depth ratio stdactual = Lspan / dmid = 22.873
PASS-Span to effective depth is less than the maximum allowable
Upper landing support
Design moment at support M = 0.0 kNm
Nominal cover to reinforcement cnom = 25 mm
Diameter of bar for long. direction l =14 mm
Depth of reinforcement d = h - cnom - l / 2 = 268 mm
Calculated reinforcement at midspan cl. 9.3.1.2(1) As,span = Mmid / (fyd zmid) = 1639 mm2
Minimum area of reinforcement – exp.9.1.N As,min = max(0.26 fctm / fyk b d,0.0013 b d, 0.5
As,span)
As,min = 820 mm2
Maximum area of reinforcement - cl.9.2.1.1(3) As,max = 0.04 b h = 16800 mm2
Tension reinforcement check
Diameter of bars for longitudinal direction l = 14 mm
Number of bars in longitudinal direction N=6
Diameter of bars for transverse direction t = 10 mm
Bar spacing in transverse direction s = 300 mm
Tension area provided in longitudinal direction As,prov = N l2 /4 = 924 mm2
PASS - Tension reinforcement area is greater than area required
Tension secondary area required in transverse direction – cl.9.3.1.1(2)
As,req,t = 0.2 max(As,prov, As,req) = 185 mm2
Tension area provided in transverse direction As,prov,t = t2 /4 (b / s) = 367 mm2
PASS - Tension reinforcement in transverse direction is greater than area required
Minimum bar spacing - Section 8.2
Bar spacing (clear distance) in long. direction; sbar,l = (b - (2 cnom + N l)) / (N - 1) = 253 mm
kspc1 = 1
kspc2 = 5 mm
Minimum allowable bar spacing – cl.8.2(2); smin,l = max(kspc1 l, hagg + kspc2, 20 mm) = 25 mm
PASS - Actual bar spacing exceeds minimum allowable
Bar spacing (clear distance) in transv. direction; sbar,t = s - t = 290 mm
Minimum allowable bar spacing – cl.8.2(2); smin,t = max(kspc1 t, hagg + kspc2, 20 mm) = 25 mm
PASS - Actual bar spacing exceeds minimum allowable
Shear capacity check - Section 6.2
Shear coefficient; = 0.50
Design shear force; V = FEd Lspan = 94.0 kN
Shear stress; v = V / (b d) = 0.25 N/mm2
Design shear resist. coefficient. – cl.6.2.2(1) CRd,c = 0.18 / c = 0.120
Design shear resist. size factor. – cl.6.2.2(1) Ksh = min(1 + (200 mm / d)1/2, 2) = 1.86
Design shear resist. factor. – cl.6.2.2(1) 1 = As,prov / (b d) = 0.00246
Design shear resist. factor. – exp.6.3.N vmin = (0.035 (N)1/2 / mm) Ksh 3 / 2 fck1/2= 0.40 N/mm2
Minimum design shear resistance – exp.6.2(b) VRd,cmin = (vmin) b d = 149.4 kN
Design shear resist. without reinf. – exp.6.2(a) VRd,c1 = [CRd,c Ksh (100 N2 / mm4 1 fck)1/3] b d
VRd,c1 = 142.8 kN
VRd,c = max(VRd,c1, VRd,cmin) = 149.4 kN
PASS - Design force is less than maximum allowable- No shear reinforcement is required;
Lower landing support
Design moment at support M = 0.0 kNm
Nominal cover to reinforcement cnom = 25 mm
Diameter of bar for long. direction l =14 mm
Depth of reinforcement d = h - cnom - l / 2 = 268 mm
Calculated reinforcement at midspan cl. 9.3.1.2(1) As,span = Mmid / (fyd zmid) = 1639 mm2
Minimum area of reinforcement – exp.9.1.N As,min = max(0.26 fctm / fyk b d,0.0013 b d, 0.5
As,span)
As,min = 820 mm2
Maximum area of reinforcement - cl.9.2.1.1(3) As,max = 0.04 b h = 16800 mm2
Tension reinforcement check
Diameter of bars for longitudinal direction l = 14 mm
Number of bars in longitudinal direction N=6
Diameter of bars for transverse direction t = 8 mm
Bar spacing in transverse direction s = 300 mm
Tension area provided in longitudinal direction As,prov = N l2 /4 = 924 mm2
PASS - Tension reinforcement area is greater than area required
Tension secondary area required in transverse direction – cl.9.3.1.1(2)
As,req,t = 0.2 max(As,prov, As,req) = 185 mm2
Tension area provided in transverse direction As,prov,t = t2 /4 (b / s) = 235 mm2
PASS - Tension reinforcement in transverse direction is greater than area required
Minimum bar spacing - Section 8.2
Bar spacing (clear distance) in long. direction; sbar,l = (b - (2 cnom + N l)) / (N - 1) = 253 mm
kspc1 = 1
kspc2 = 5 mm
Minimum allowable bar spacing – cl.8.2(2); smin,l = max(kspc1 l, hagg + kspc2, 20 mm) = 25 mm
PASS - Actual bar spacing exceeds minimum allowable
Bar spacing (clear distance) in transv. direction; sbar,t = s - t = 292 mm
Minimum allowable bar spacing – cl.8.2(2); smin,t = max(kspc1 t, hagg + kspc2, 20 mm) = 25 mm
PASS - Actual bar spacing exceeds minimum allowable
Shear capacity check - Section 6.2
Shear coefficient; = 0.50
Design shear force; V = FEd Lspan = 94.0 kN
Shear stress; v = V / (b d) = 0.25 N/mm2
Design shear resist. coefficient. – cl.6.2.2(1) CRd,c = 0.18 / c = 0.120
Design shear resist. size factor. – cl.6.2.2(1) Ksh = min(1 + (200 mm / d)1/2, 2) = 1.86
Design shear resist. factor. – cl.6.2.2(1) 1 = As,prov / (b d) = 0.00246
Design shear resist. factor. – exp.6.3.N vmin = (0.035 (N)1/2 / mm) Ksh 3 / 2 fck1/2= 0.40 N/mm2
Minimum design shear resistance – exp.6.2(b) VRd,cmin = (vmin) b d = 149.4 kN
Design shear resist. without reinf. – exp.6.2(a) VRd,c1 = [CRd,c Ksh (100 N2 / mm4 1 fck)1/3] b d
VRd,c1 = 142.8 kN
VRd,c = max(VRd,c1, VRd,cmin) = 149.4 kN
PASS - Design force is less than maximum allowable- No shear reinforcement is required;
