Structural Engineering Tools

Reinforced concrete shear wall to SIA 262

Checks a reinforced concrete wall loaded in its plane, to SIA 262:2025: bending with axial force and shear, detailing, buckling of a strip out of the plane and, for earthquake, capacity design and the detailing of the plastic zone. Every step shows its formula and the clause it comes from.

Basic SIA 262:2025

Part of the Basic plan

This module is included in the Basic plan and above, with its calculation note.

About this check

clauses, worked example and questions

Checks a reinforced concrete wall loaded in its plane, to SIA 262:2025: bending with axial force and shear, detailing, buckling of a strip out of the plane and, for earthquake, capacity design and the detailing of the plastic zone. Every step shows its formula and the clause it comes from.

What it checks

  • Design strengths: fcd = ηfc · ηt · fck/γc, clause 2.4.2.3, equation (2), with ηfc from 4.2.1.2, equation (26), and fyd = fyk/γs, equation (4).
  • Axial force and bending in the plane, checks 1 and 2: Nd ≤ NRd,c, the whole section at εc2d, and Md ≤ [M_Rd], with the compression zone x found from N(x) = Nd (4.2.1.5).
  • Shear of the web, checks 3 and 4: Vd ≤ VRd,s = Asw/s · z · fyd · cot α, 4.3.3.4.3, equation (45), and Vd ≤ VRd,c = bw0 · z · kc · fcd · sin α · cos α, 4.3.3.4.6, equation (47), with z = 0.9 · d (4.3.3.4.2) and α between αmin and 45° (4.3.3.3.2, equation (42)).
  • Tension chord, check 5: Md/z − Nd/2 + FtVd/2 ≤ As,E · fyd (4.3.3.4.10), with FtVd = Vd · cot α, 4.3.3.4.11, equation (51).
  • Detailing: bw0 ≥ 150 mm cast in place (5.5.4.1, Table 22); 0.6 % ≤ ρv,tot ≤ 8 % (5.5.4.3, 5.5.4.5); sv ≤ min(2 · bw0 ; 300 mm); ρh ≥ 25 % of ρv (5.5.4.11); cross-ties with st ≤ min(15 · Ømin ; bw0 ; 300 mm), 5.5.4.7, equation (117), and Øt ≥ Ømax/3 (5.5.4.9).
  • Out of the plane, on a strip of 1 m: nd ≤ nRd,c, then e0d from αi (4.1.3.2.3, equation (17)) and d/30 (4.3.7.5, equation (75)), e1d = m1d/nd, equation (76), e2d = χd · lcr²/c, equation (77), with χd from equation (78), and md = nd · ed ≤ [m_Rd] (4.3.7.3).
  • Earthquake, ductile behaviour (4.3.9.3): q from Table 15, the concrete between C20/25 and C50/60 (4.3.9.3.2), the checks in the plane under the seismic actions, and the shear by capacity design, V+d = ε · κ · Vd, 4.3.9.4.4, equations (95) to (97), with M+Rd from fy,m (4.3.9.3.3), against a web with αmin = 40° and kc = 0.4.
  • Plastic zone (5.7.1): hpl, νd ≤ 0.4, the ratios of the boundary elements and the web between 0.3 % and 3 %, the bar spacings, the web and boundary element dimensions (5.7.1.2, 5.7.1.3), and the confinement ωwd, 5.7.1.4, equations (118) and (119), for μφ = 2 · q − 1.

What you enter and what you get

For the wall, you enter lw, bw0, constant thickness or thickened ends, and cast in place or precast. Then the cover cv, the nE end bars ØE in pairs at sE, the web bars Øv at most sv apart, the horizontal bars Øh at sh, the materials, the resistance factors and ηt. In the plane: Nd, Md and Vd, the condition of the web, and α chosen or entered. Then the cross-ties Øt at st. Out of the plane: nd, m1d, lcr and l. For earthquake: the behaviour, the ductility class of the steel, the seismic actions, hw, hs, the number of storeys n, the ends of the wall, lc, the hoops Øw at scx with s0 and nt, and T1 and TC.

You get the bar layout used, every resistance and check with its utilisation and clause, and the warnings on detailing. With the Basic plan, it prints as an A4 calculation note.

Assumptions and limits

  • The wall is checked at the section you enter the actions for, usually the base. A wall in tension is outside this module.
  • Wall and bars are symmetric: both ends alike, and the sign of Md does not matter.
  • The web bars are spread evenly between the end bars, at sv or a little less; the spacing used is reported.
  • Out of the plane, the most loaded metre is checked with the estimated curvature, creep and shrinkage included.
  • In capacity design the concrete is kept at fcd: 4.3.9.3.3 names the steel only. The hoops are taken outside the vertical bars, and ωwd leaves out the anchorage lengths.
  • Not checked: the crack requirements of 4.4.2 for the horizontal bars, and the elastic zones above the plastic zone (4.3.9.4.3).

Worked example

A wall 5,000 mm long and 250 mm thick in fck = 30 N/mm², with eight Ø20 end bars at each end, Ø10 web bars at 200 mm, Ø12 horizontal bars at 150 mm and Ø8 cross-ties at 125 mm, under Nd = 2,500 kN, Md = 6,000 kNm and Vd = 900 kN, with nd = 500 kN/m and m1d = 5 kNm/m over lcr = 2.8 m, and designed for ductile behaviour as a six-storey wall 18 m high under Nd = 2,000 kN, Md = 5,000 kNm and Vd = 600 kN, with 700 mm boundary elements. These are the module's own results for its worked example, computed when this page was built.

Inputs
InputValueRef.
Length of the wall lw5000 mm
Thickness of the wall (web) bw0250 mm
Ends of the wallConstant thickness
ConstructionCast in place§ 5.5.4.1
Clear cover to the vertical bars cv30 mm
End bars at each end nE8
Diameter of the end bars ØE20 mm
Spacing of the pairs of end bars sE100 mm
Diameter of the vertical web bars Øv10 mm
Largest spacing of the vertical web bars sv200 mm
Diameter of the horizontal bars Øh12 mm
Spacing of the horizontal bars sh150 mm
Characteristic compressive strength fck30 N/mm²§ 3.1.2.2.2
Ultimate compressive strain εc2d3.0 ‰§ 4.2.1.5
Characteristic yield strength fyk500 N/mm²
Modulus of elasticity Es205000 N/mm²
Resistance factor for concrete γc1.5§ 2.4.2.6
Resistance factor for steel γs1.15§ 2.4.2.6
Duration factor ηt1.0 — both conditions of 4.2.1.3 met§ 4.2.1.3
Design axial compression Nd2500 kN
Design bending moment in the plane Md6000 kNm
Design shear force in the plane Vd900 kN
Condition of the web αminNormal case — αmin = 30°, kc = 0.55§ 4.3.3.3.2
Inclination of the compression field αChosen for the largest resistance§ 4.3.3.3.2
Ties holding the vertical barsCross-ties through the wall§ 5.5.4.10
Diameter of the ties Øt8 mm
Vertical spacing of the ties st125 mm
Design axial compression per metre nd500 kN/m
First-order moment per metre, out of the plane m1d5 kNm/m
Buckling length lcr2.8 m
Height for the inclination l2.8 m§ 4.1.3.2.3
Design for earthquakeDuctile behaviour — capacity design (4.3.9.3)§ 4.3.9.1.2
Ductility class of the reinforcing steelB§ 4.3.9.3.5
Axial compression Nd2000 kN
Bending moment in the plane Md5000 kNm
Shear force in the plane Vd600 kN
Height of the wall hw18 m
Height of the lowest storey hs3 m
Number of storeys n6§ 4.3.9.4.4
Ends of the wallFree ends — boundary elements§ 5.7.1.2
Length of each boundary element lc700 mm
Diameter of the hoops Øw8 mm§ 5.7.1.4
Vertical spacing of the hoops scx100 mm
Distance from the base to the first hoop s050 mm
Legs across the thickness in one boundary element nt5
Fundamental period of the building T10.8 s
Corner period of the design spectrum TC0.5 s
Calculation: Concrete section
StepValueRef.
Area of the wall sectionlw·bw0Ac = 1,250,000 mm²
Slenderness of the sectiona wall is much longer than it is thick: it bends easily out of its planelw/bw0 = 20.0
Calculation: Vertical reinforcement
StepValueRef.
End bars at each end4 pairs, area 2,513 mm²nE = 8Ø20
Vertical web bars21 on each facenv = 42Ø10
Spacing of the web bars usedweb length shared evenly, at most svsv = 196 mm
Total area of the vertical barsΣ π·Ø²/4As = 8,325 mm²
Ratio of vertical reinforcementAs/Acρv,tot = 0.67 %
Ratio of the web, both faces2·(π·Øv²/4)/(sv·bw0)ρv = 0.32 %
Calculation: Horizontal reinforcement
StepValueRef.
Area per unit height, both faces2·π·Øh²/4/shash = 1.508 mm²/mm
Ratio of horizontal reinforcementash/bw0ρh = 0.60 %
Calculation: Design strengths
StepValueRef.
Factor for the brittleness of strong concrete(40/fck)^(1/3) ≤ 1ηfc = 1.000§ 4.2.1.2 (26)
Design compressive strength of the concreteηfc·ηt·fck/γcfcd = 20.0 N/mm²§ 2.4.2.3 (2)
Design yield strength of the barsfyk/γsfyd = 434.8 N/mm²§ 2.4.2.5 (4)
Calculation: Axial resistance of the section
StepValueRef.
Resistance in centred compressionfcd·(Ac − As) + σs·As, σs = min(fyd ; Es·εc2d)the whole section at εc2dNRd,c = 28,453 kN§ 4.2.1.5
Calculation: Check 1 — axial force: N_d ≤ N_Rd,c
StepValueRef.
Axial compression against the resistanceNd/NRd,cη = 0.088 · Pass9 % of the capacity used§ 4.2.1.5
Calculation: Section at the ultimate limit state under N_d
StepValueRef.
Depth of the compression zonefrom N(x) = Nd, εc2d at the compressed endx = 874 mm§ 4.2.1.5
Relative depthx/lw = 0.175
Force in the compressed concretefcd·A(0.85·x)Fc = 3,716 kN
Bending resistance under NdΣ Fi·(lw/2 − zi)MRd = 13,282 kNm§ 4.2.1.5
Calculation: Check 2 — bending with axial force: M_d ≤ M_Rd
StepValueRef.
Design moment against the bending resistance under NdMd/MRdη = 0.452 · Pass45 % of the capacity used§ 4.2.1.5
Calculation: Shear resistance of the web
StepValueRef.
Effective depthcompressed end to the centroid of the bars in the tension halfd = 4,337 mm§ 4.3.3.4.2
Lever arm of the internal forces0.9·dz = 3,903 mm§ 4.3.3.4.2
Width of the webbw0 = 250 mm
Shear reinforcement (horizontal bars)2·π·Øh²/4/shboth facesAsw/s = 1.508 mm²/mm
Smallest inclination of the compression fieldαmin = 30 °§ 4.3.3.3.2
Reduction factor for the concrete strengthkc = 0.55§ 4.2.1.7
Inclination of the compression field usedαmin ≤ α ≤ 45°, for the largest min(VRd,s ; VRd,c)α = 30.0 °§ 4.3.3.3.2 (42)
Calculation: Check 3 — horizontal bars: V_d ≤ V_Rd,s
StepValueRef.
Resistance of the horizontal barsAsw/s·z·fyd·cotαVRd,s = 4,432 kN§ 4.3.3.4.3 (45)
Design shear against the horizontal barsVd/VRd,sη = 0.203 · Pass20 % of the capacity used§ 4.3.3.4.3 (45)
Calculation: Check 4 — compression field: V_d ≤ V_Rd,c
StepValueRef.
Resistance of the compression fieldbw0·z·kc·fcd·sinα·cosαVRd,c = 4,648 kN§ 4.3.3.4.6 (47)
Design shear against the compression fieldVd/VRd,cη = 0.194 · Pass19 % of the capacity used§ 4.3.3.4.6 (47)
Calculation: Check 5 — tension chord with the shear
StepValueRef.
Chord force from the shearVd·cotαhalf to each chordFtVd = 1,559 kN§ 4.3.3.4.11 (51)
Tension chord: end bars of one endFt,d/Ft,Rdη = 0.976 · Pass98 % of the capacity used§ 4.3.3.4.10
Calculation: Check 1 — thickness of the wall
StepValueRef.
Thickness of a reinforced walltmin/bw0η = 0.600 · Pass60 % of the capacity used§ 5.5.4.1 (Table 22)
Calculation: Check 2 — vertical reinforcement
StepValueRef.
Minimum ratio of vertical reinforcement0.6 %/ρv,totη = 0.901 · Pass90 % of the capacity used§ 5.5.4.3
Maximum ratio of vertical reinforcementρv,tot/8 %η = 0.083 · Pass8 % of the capacity used§ 5.5.4.5
Spacing of the vertical web barssv/smaxη = 0.655 · Pass65 % of the capacity used§ 5.5.4.3
Calculation: Check 3 — horizontal reinforcement
StepValueRef.
Horizontal against vertical reinforcement0.25·ρv/ρhη = 0.133 · Pass13 % of the capacity used§ 5.5.4.11
Calculation: Check 4 — cross-ties
StepValueRef.
Smallest vertical barØmin = 10 mm
Vertical spacing of the tiesst/smaxη = 0.833 · Pass83 % of the capacity used§ 5.5.4.7 (117), 5.5.4.10
Diameter of the ties(Ømax/3)/Øtη = 0.833 · Pass83 % of the capacity used§ 5.5.4.9, 5.5.4.10
Calculation: Strip of 1 m of the web
StepValueRef.
Thicknessbw0 = 250 mm
Vertical bars per metre, each faceπ·Øv²/4·1000/svas = 400 mm²/m
Effective depthbw0 − cv − Øv/2d = 215 mm
Depth of the compressed barscv + Øv/2d′ = 35 mm
Resistance in centred compressionnRd,c = 5,332 kN/m§ 4.2.1.5
Calculation: Check 1 — axial force: n_d ≤ n_Rd,c
StepValueRef.
Axial compression per metre against the resistancend/nRd,cη = 0.094 · Pass9 % of the capacity used§ 4.2.1.5
Calculation: Slenderness and eccentricities
StepValueRef.
Slendernesslcr/i, i = bw0/√12for information: the larger λ, the more the deflection mattersλ = 38.8
Inclination of the wall0.01/√l, between 1/300 and 1/200 (l in m)limited to 1/200αi = 0.00500 rad§ 4.1.3.2.3 (17)
Eccentricity from the inclinationαi·lcr/2 = 7.0 mm§ 4.3.7.5 (75)
Minimum eccentricityd/30 = 7.2 mm§ 4.3.7.5 (75)
Eccentricity due to imperfectionsthe larger of the twoe0d = 7.2 mm§ 4.3.7.5 (75)
First-order eccentricitym1d/nde1d = 10.0 mm§ 4.3.7.6 (76)
Curvature of the wall2·fyd/(Es·(d − d′))creep and shrinkage includedχd = 0.02357 1/m§ 4.3.7.8 (78)
Integration constantπ²c = 9.87§ 4.3.7.12 (81)
Eccentricity from the deflectionχd·lcr²/ce2d = 18.7 mm§ 4.3.7.7 (77)
Total eccentricitye0d + e1d + e2ded = 35.9 mm§ 4.3.7.4 (74)
Design moment per metrend·edmd = 17.9 kNm/m§ 4.3.7.3 (73)
Calculation: Check 2 — bending out of the plane with second order: m_d ≤ m_Rd
StepValueRef.
Depth of the compression zonex = 38 mm§ 4.2.1.5
Bending resistance per metre under ndmRd = 88.5 kNm/m§ 4.2.1.5
Design moment against the resistance under ndmd/mRdη = 0.203 · Pass20 % of the capacity used§ 4.3.7.3
Calculation: Behaviour factor
StepValueRef.
Behaviour factor for the seismic analysissteel of ductility class B: the analysis to SIA 261 must use at most this qq = 3.0§ 4.3.9.3.5 (Table 15)
Calculation: Check 1 — concrete strength for ductile behaviour
StepValueRef.
Characteristic compressive strengthfckη = 0.667 · Pass67 % of the capacity used§ 4.3.9.3.2
Calculation: Wall to which the rules apply
StepValueRef.
Slenderness of the wallmore than 2: the rules of 4.3.9.4 and 5.7.1 applyhw/lw = 3.60§ 4.3.9.4.1
Calculation: Axial resistance of the section
StepValueRef.
Resistance in centred compressionfcd·(Ac − As) + σs·As, σs = min(fyd ; Es·εc2d)the whole section at εc2dNRd,c = 28,453 kN§ 4.2.1.5
Calculation: Check 2 — axial force: N_d ≤ N_Rd,c
StepValueRef.
Axial compression against the resistanceNd/NRd,cη = 0.070 · Pass7 % of the capacity used§ 4.2.1.5
Calculation: Section at the ultimate limit state under N_d
StepValueRef.
Depth of the compression zonefrom N(x) = Nd, εc2d at the compressed endx = 777 mm§ 4.2.1.5
Relative depthx/lw = 0.155
Force in the compressed concretefcd·A(0.85·x)Fc = 3,303 kN
Bending resistance under NdΣ Fi·(lw/2 − zi)MRd = 12,393 kNm§ 4.2.1.5
Calculation: Check 3 — bending with axial force: M_d ≤ M_Rd
StepValueRef.
Seismic moment against the bending resistance under NdMd/MRdη = 0.403 · Pass40 % of the capacity used§ 4.2.1.5
Calculation: Capacity design of the shear
StepValueRef.
Mean yield strength with overstrength1.1·fyk (class B)fy,m = 550 N/mm²§ 4.3.9.3.3
Bending resistance with overstrengthsection under Nd with fy,m on the bars, fcd on the concretethe concrete is kept at fcd: 4.3.9.3.3 names the steel onlyM+Rd = 14,307 kNm§ 4.3.9.3.3
Factor for the overstrengthM+Rd/Mdε = 2.86§ 4.3.9.4.4 (96)
Factor for the higher modes1.5 (n > 5)κ = 1.50§ 4.3.9.4.4 (97)
Magnified design shearε·κ·VdV+d = 2,575 kN§ 4.3.9.4.4 (95)
Calculation: Shear resistance of the plastic zone
StepValueRef.
Effective depthcompressed end to the centroid of the bars in the tension halfd = 4,337 mm§ 4.3.3.4.2
Lever arm of the internal forces0.9·dz = 3,903 mm§ 4.3.3.4.2
Width of the webbw0 = 250 mm
Shear reinforcement (horizontal bars)2·π·Øh²/4/shboth facesAsw/s = 1.508 mm²/mm
Smallest inclination of the compression fieldlevel 1, plastic deformations of the chordsαmin = 40 °§ 4.3.9.3.3, 4.3.3.3.2
Reduction factor for the concrete strengthkc = 0.40§ 4.2.1.7
Inclination of the compression field usedαmin ≤ α ≤ 45°, for the largest min(VRd,s ; VRd,c)α = 40.0 °§ 4.3.3.3.2 (42)
Calculation: Check 4 — horizontal bars: V+_d ≤ V_Rd,s
StepValueRef.
Resistance of the horizontal barsAsw/s·z·fyd·cotαVRd,s = 3,050 kN§ 4.3.3.4.3 (45)
Design shear against the horizontal barsV+d/VRd,sη = 0.844 · Pass84 % of the capacity used§ 4.3.3.4.3 (45)
Calculation: Check 5 — compression field: V+_d ≤ V_Rd,c
StepValueRef.
Resistance of the compression fieldbw0·z·kc·fcd·sinα·cosαVRd,c = 3,844 kN§ 4.3.3.4.6 (47)
Design shear against the compression fieldV+d/VRd,cη = 0.670 · Pass67 % of the capacity used§ 4.3.3.4.6 (47)
Calculation: Plastic zone at the base
StepValueRef.
Height of the plastic zonehpl ≥ lw and ≥ hw/6the detailing below applies over this heighthpl = 5,000 mm§ 5.7.1.1
Calculation: Check 6 — axial force in the plastic zone
StepValueRef.
Normalised axial forceνd/0.4η = 0.200 · Pass20 % of the capacity used§ 5.7.1.1
Calculation: Check 7 — longitudinal reinforcement
StepValueRef.
Ratio in one boundary elementAs,c/(lc·bw)ρc = 1.53 %
Boundary element: at least 0.3 %0.3 %/ρcη = 0.197 · Pass20 % of the capacity used§ 5.7.1.1
Boundary element: at most 3 %ρc/3 %η = 0.509 · Pass51 % of the capacity used§ 5.7.1.1
Ratio of the webAs,w/((lw − 2·lc)·bw0)ρw = 0.33 %
Web: at least 0.3 %0.3 %/ρwη = 0.905 · Pass90 % of the capacity used§ 5.7.1.1
Web: at most 3 %ρw/3 %η = 0.111 · Pass11 % of the capacity used§ 5.7.1.1
Calculation: Check 8 — horizontal reinforcement
StepValueRef.
Horizontal ratio: at least 0.3 %0.3 %/ρhη = 0.497 · Pass50 % of the capacity used§ 5.7.1.1
Horizontal against vertical ratio0.25·ρv/ρhη = 0.133 · Pass13 % of the capacity used§ 5.7.1.1
Calculation: Check 9 — spacing of the bars
StepValueRef.
Vertical web barssv/smaxη = 0.785 · Pass79 % of the capacity used§ 5.7.1.1
Horizontal barssh/smaxη = 0.600 · Pass60 % of the capacity used§ 5.7.1.1
End barssE/smaxη = 0.400 · Pass40 % of the capacity used§ 5.7.1.1
Calculation: Check 10 — thickness of the web
StepValueRef.
Web between the boundary elementsbw0,min/bw0η = 0.600 · Pass60 % of the capacity used§ 5.7.1.3
Calculation: Check 11 — boundary elements
StepValueRef.
Thickness of the boundary elementsbw,min/bwη = 0.800 · Pass80 % of the capacity used§ 5.7.1.2
Length of the boundary elementslc,min/lcη = 0.777 · Pass78 % of the capacity used§ 5.7.1.2
Calculation: Check 12 — confinement of the boundary elements
StepValueRef.
Confined core across the wallbw − 2·cv + Øwbetween hoop centrelines, hoops outside the vertical bars; cover to the hoops cv − Øw = 22 mmbcz = 198 mm
Confined core along the walllc − cv + Øw/2bcy = 674 mm
Bars in one boundary elementnc = 10 (5 rows of up to 2)
Vertical spacing of the hoopsscx/smaxη = 0.833 · Pass83 % of the capacity used§ 5.7.1.4
First hoop above the bases0/50η = 1.000 · Pass100 % of the capacity used§ 5.7.1.4
Diameter of the hoops0.35·Øsl,max/Øwη = 0.875 · Pass88 % of the capacity used§ 5.7.1.4
Every other row held, along the wallnt,req/ntη = 0.600 · Pass60 % of the capacity used§ 5.7.1.4
Distance between held bars, along the wallscy/200η = 0.843 · Pass84 % of the capacity used§ 5.7.1.4
Distance between held bars, across the wallscz/200η = 0.990 · Pass99 % of the capacity used§ 5.7.1.4
Curvature ductility needed2·q − 1 (T1 ≥ TC)with q = 3.0 of Table 15μφ = 5.00§ 5.7.1.4
Normalised axial force−Nd/(Ac·fcd), compression positiveνd = 0.080
Mechanical ratio of the vertical web barsρv·fyd/fcdωv = 0.070
Efficiency of the confinement(1 − √(scy² + scx²)/(2·bcz))·(1 − √(scz² + scx²)/(2·bcy))αs = 0.422§ 5.7.1.4 (119)
Mechanical volumetric ratio of the hoopsAw·(2·bcy + nt·bcz)/(bcy·bcz·scx)·fyd/fcdwithout the anchorage lengthsωwd = 0.191§ 5.7.1.4 (118)
Confinement of the boundary elementωwd,req/ωwdη = 0.310 · Pass31 % of the capacity used§ 5.7.1.4 (118)

How this module is verified

Verified clause by clause against SIA 262:2025 and against independent calculations used on real projects, before release.

Questions

Why is the seismic shear of the wall raised so much?

Capacity design protects the wall against a shear failure once the plastic zone forms. In the worked example ε = 2.86 and κ = 1.5 raise Vd = 600 kN to V+d = 2,575 kN.

When does the wall need no plastic zone detailing?

With non-ductile behaviour (4.3.9.2), the wall is designed for the seismic actions with a small q (Table 14) and needs nothing more. The rules of 4.3.9.4 and 5.7.1 apply where hw/lw exceeds 2. Boundary elements are not needed where both ends are tied into sufficiently reinforced transverse walls (5.7.1.2).

Which plan includes this module?

The Basic plan, with its calculation note.

Related

Norm: SIA 262:2025 · Page updated: 5 November 2026