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| Input | Value | Ref. |
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| Length of the wall lw | 5000 mm | |
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| Thickness of the wall (web) bw0 | 250 mm | |
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| Ends of the wall | Constant thickness | |
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| Construction | Cast in place | § 5.5.4.1 |
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| Clear cover to the vertical bars cv | 30 mm | |
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| End bars at each end nE | 8 | |
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| Diameter of the end bars ØE | 20 mm | |
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| Spacing of the pairs of end bars sE | 100 mm | |
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| Diameter of the vertical web bars Øv | 10 mm | |
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| Largest spacing of the vertical web bars sv | 200 mm | |
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| Diameter of the horizontal bars Øh | 12 mm | |
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| Spacing of the horizontal bars sh | 150 mm | |
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| Characteristic compressive strength fck | 30 N/mm² | § 3.1.2.2.2 |
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| Ultimate compressive strain εc2d | 3.0 ‰ | § 4.2.1.5 |
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| Characteristic yield strength fyk | 500 N/mm² | |
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| Modulus of elasticity Es | 205000 N/mm² | |
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| Resistance factor for concrete γc | 1.5 | § 2.4.2.6 |
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| Resistance factor for steel γs | 1.15 | § 2.4.2.6 |
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| Duration factor ηt | 1.0 — both conditions of 4.2.1.3 met | § 4.2.1.3 |
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| Design axial compression Nd | 2500 kN | |
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| Design bending moment in the plane Md | 6000 kNm | |
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| Design shear force in the plane Vd | 900 kN | |
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| Condition of the web αmin | Normal case — αmin = 30°, kc = 0.55 | § 4.3.3.3.2 |
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| Inclination of the compression field α | Chosen for the largest resistance | § 4.3.3.3.2 |
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| Ties holding the vertical bars | Cross-ties through the wall | § 5.5.4.10 |
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| Diameter of the ties Øt | 8 mm | |
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| Vertical spacing of the ties st | 125 mm | |
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| Design axial compression per metre nd | 500 kN/m | |
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| First-order moment per metre, out of the plane m1d | 5 kNm/m | |
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| Buckling length lcr | 2.8 m | |
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| Height for the inclination l | 2.8 m | § 4.1.3.2.3 |
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| Design for earthquake | Ductile behaviour — capacity design (4.3.9.3) | § 4.3.9.1.2 |
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| Ductility class of the reinforcing steel | B | § 4.3.9.3.5 |
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| Axial compression Nd | 2000 kN | |
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| Bending moment in the plane Md | 5000 kNm | |
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| Shear force in the plane Vd | 600 kN | |
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| Height of the wall hw | 18 m | |
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| Height of the lowest storey hs | 3 m | |
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| Number of storeys n | 6 | § 4.3.9.4.4 |
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| Ends of the wall | Free ends — boundary elements | § 5.7.1.2 |
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| Length of each boundary element lc | 700 mm | |
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| Diameter of the hoops Øw | 8 mm | § 5.7.1.4 |
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| Vertical spacing of the hoops scx | 100 mm | |
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| Distance from the base to the first hoop s0 | 50 mm | |
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| Legs across the thickness in one boundary element nt | 5 | |
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| Fundamental period of the building T1 | 0.8 s | |
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| Corner period of the design spectrum TC | 0.5 s | |
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Calculation: Concrete section| Step | Value | Ref. |
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| Area of the wall sectionlw·bw0 | Ac = 1,250,000 mm² | |
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| Slenderness of the sectiona wall is much longer than it is thick: it bends easily out of its plane | lw/bw0 = 20.0 | |
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Calculation: Vertical reinforcement| Step | Value | Ref. |
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| End bars at each end4 pairs, area 2,513 mm² | nE = 8Ø20 | |
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| Vertical web bars21 on each face | nv = 42Ø10 | |
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| Spacing of the web bars usedweb length shared evenly, at most sv | sv = 196 mm | |
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| Total area of the vertical barsΣ π·Ø²/4 | As = 8,325 mm² | |
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| Ratio of vertical reinforcementAs/Ac | ρv,tot = 0.67 % | |
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| Ratio of the web, both faces2·(π·Øv²/4)/(sv·bw0) | ρv = 0.32 % | |
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Calculation: Horizontal reinforcement| Step | Value | Ref. |
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| Area per unit height, both faces2·π·Øh²/4/sh | ash = 1.508 mm²/mm | |
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| Ratio of horizontal reinforcementash/bw0 | ρh = 0.60 % | |
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Calculation: Design strengths| Step | Value | Ref. |
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| Factor for the brittleness of strong concrete(40/fck)^(1/3) ≤ 1 | ηfc = 1.000 | § 4.2.1.2 (26) |
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| Design compressive strength of the concreteηfc·ηt·fck/γc | fcd = 20.0 N/mm² | § 2.4.2.3 (2) |
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| Design yield strength of the barsfyk/γs | fyd = 434.8 N/mm² | § 2.4.2.5 (4) |
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Calculation: Axial resistance of the section| Step | Value | Ref. |
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| Resistance in centred compressionfcd·(Ac − As) + σs·As, σs = min(fyd ; Es·εc2d)the whole section at εc2d | NRd,c = 28,453 kN | § 4.2.1.5 |
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Calculation: Check 1 — axial force: N_d ≤ N_Rd,c| Step | Value | Ref. |
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| Axial compression against the resistanceNd/NRd,c | η = 0.088 · Pass9 % of the capacity used | § 4.2.1.5 |
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Calculation: Section at the ultimate limit state under N_d| Step | Value | Ref. |
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| Depth of the compression zonefrom N(x) = Nd, εc2d at the compressed end | x = 874 mm | § 4.2.1.5 |
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| Relative depth | x/lw = 0.175 | |
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| Force in the compressed concretefcd·A(0.85·x) | Fc = 3,716 kN | |
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| Bending resistance under NdΣ Fi·(lw/2 − zi) | MRd = 13,282 kNm | § 4.2.1.5 |
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Calculation: Check 2 — bending with axial force: M_d ≤ M_Rd| Step | Value | Ref. |
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| Design moment against the bending resistance under NdMd/MRd | η = 0.452 · Pass45 % of the capacity used | § 4.2.1.5 |
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Calculation: Shear resistance of the web| Step | Value | Ref. |
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| Effective depthcompressed end to the centroid of the bars in the tension half | d = 4,337 mm | § 4.3.3.4.2 |
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| Lever arm of the internal forces0.9·d | z = 3,903 mm | § 4.3.3.4.2 |
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| Width of the web | bw0 = 250 mm | |
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| Shear reinforcement (horizontal bars)2·π·Øh²/4/shboth faces | Asw/s = 1.508 mm²/mm | |
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| Smallest inclination of the compression field | αmin = 30 ° | § 4.3.3.3.2 |
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| Reduction factor for the concrete strength | kc = 0.55 | § 4.2.1.7 |
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| 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) |
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Calculation: Check 3 — horizontal bars: V_d ≤ V_Rd,s| Step | Value | Ref. |
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| Resistance of the horizontal barsAsw/s·z·fyd·cotα | VRd,s = 4,432 kN | § 4.3.3.4.3 (45) |
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| Design shear against the horizontal barsVd/VRd,s | η = 0.203 · Pass20 % of the capacity used | § 4.3.3.4.3 (45) |
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Calculation: Check 4 — compression field: V_d ≤ V_Rd,c| Step | Value | Ref. |
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| Resistance of the compression fieldbw0·z·kc·fcd·sinα·cosα | VRd,c = 4,648 kN | § 4.3.3.4.6 (47) |
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| Design shear against the compression fieldVd/VRd,c | η = 0.194 · Pass19 % of the capacity used | § 4.3.3.4.6 (47) |
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Calculation: Check 5 — tension chord with the shear| Step | Value | Ref. |
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| Chord force from the shearVd·cotαhalf to each chord | FtVd = 1,559 kN | § 4.3.3.4.11 (51) |
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| Tension chord: end bars of one endFt,d/Ft,Rd | η = 0.976 · Pass98 % of the capacity used | § 4.3.3.4.10 |
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Calculation: Check 1 — thickness of the wall| Step | Value | Ref. |
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| Thickness of a reinforced walltmin/bw0 | η = 0.600 · Pass60 % of the capacity used | § 5.5.4.1 (Table 22) |
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Calculation: Check 2 — vertical reinforcement| Step | Value | Ref. |
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| Minimum ratio of vertical reinforcement0.6 %/ρv,tot | η = 0.901 · Pass90 % of the capacity used | § 5.5.4.3 |
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| Maximum ratio of vertical reinforcementρv,tot/8 % | η = 0.083 · Pass8 % of the capacity used | § 5.5.4.5 |
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| Spacing of the vertical web barssv/smax | η = 0.655 · Pass65 % of the capacity used | § 5.5.4.3 |
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Calculation: Check 3 — horizontal reinforcement| Step | Value | Ref. |
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| Horizontal against vertical reinforcement0.25·ρv/ρh | η = 0.133 · Pass13 % of the capacity used | § 5.5.4.11 |
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Calculation: Check 4 — cross-ties| Step | Value | Ref. |
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| Smallest vertical bar | Ømin = 10 mm | |
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| Vertical spacing of the tiesst/smax | η = 0.833 · Pass83 % of the capacity used | § 5.5.4.7 (117), 5.5.4.10 |
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| Diameter of the ties(Ømax/3)/Øt | η = 0.833 · Pass83 % of the capacity used | § 5.5.4.9, 5.5.4.10 |
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Calculation: Strip of 1 m of the web| Step | Value | Ref. |
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| Thickness | bw0 = 250 mm | |
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| Vertical bars per metre, each faceπ·Øv²/4·1000/sv | as = 400 mm²/m | |
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| Effective depthbw0 − cv − Øv/2 | d = 215 mm | |
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| Depth of the compressed barscv + Øv/2 | d′ = 35 mm | |
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| Resistance in centred compression | nRd,c = 5,332 kN/m | § 4.2.1.5 |
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Calculation: Check 1 — axial force: n_d ≤ n_Rd,c| Step | Value | Ref. |
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| Axial compression per metre against the resistancend/nRd,c | η = 0.094 · Pass9 % of the capacity used | § 4.2.1.5 |
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Calculation: Slenderness and eccentricities| Step | Value | Ref. |
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| Slendernesslcr/i, i = bw0/√12for information: the larger λ, the more the deflection matters | λ = 38.8 | |
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| Inclination of the wall0.01/√l, between 1/300 and 1/200 (l in m)limited to 1/200 | αi = 0.00500 rad | § 4.1.3.2.3 (17) |
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| Eccentricity from the inclination | αi·lcr/2 = 7.0 mm | § 4.3.7.5 (75) |
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| Minimum eccentricity | d/30 = 7.2 mm | § 4.3.7.5 (75) |
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| Eccentricity due to imperfectionsthe larger of the two | e0d = 7.2 mm | § 4.3.7.5 (75) |
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| First-order eccentricitym1d/nd | e1d = 10.0 mm | § 4.3.7.6 (76) |
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| Curvature of the wall2·fyd/(Es·(d − d′))creep and shrinkage included | χd = 0.02357 1/m | § 4.3.7.8 (78) |
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| Integration constantπ² | c = 9.87 | § 4.3.7.12 (81) |
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| Eccentricity from the deflectionχd·lcr²/c | e2d = 18.7 mm | § 4.3.7.7 (77) |
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| Total eccentricitye0d + e1d + e2d | ed = 35.9 mm | § 4.3.7.4 (74) |
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| Design moment per metrend·ed | md = 17.9 kNm/m | § 4.3.7.3 (73) |
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Calculation: Check 2 — bending out of the plane with second order: m_d ≤ m_Rd| Step | Value | Ref. |
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| Depth of the compression zone | x = 38 mm | § 4.2.1.5 |
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| Bending resistance per metre under nd | mRd = 88.5 kNm/m | § 4.2.1.5 |
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| Design moment against the resistance under ndmd/mRd | η = 0.203 · Pass20 % of the capacity used | § 4.3.7.3 |
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Calculation: Behaviour factor| Step | Value | Ref. |
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| Behaviour factor for the seismic analysissteel of ductility class B: the analysis to SIA 261 must use at most this q | q = 3.0 | § 4.3.9.3.5 (Table 15) |
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Calculation: Check 1 — concrete strength for ductile behaviour| Step | Value | Ref. |
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| Characteristic compressive strengthfck | η = 0.667 · Pass67 % of the capacity used | § 4.3.9.3.2 |
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Calculation: Wall to which the rules apply| Step | Value | Ref. |
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| Slenderness of the wallmore than 2: the rules of 4.3.9.4 and 5.7.1 apply | hw/lw = 3.60 | § 4.3.9.4.1 |
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Calculation: Axial resistance of the section| Step | Value | Ref. |
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| Resistance in centred compressionfcd·(Ac − As) + σs·As, σs = min(fyd ; Es·εc2d)the whole section at εc2d | NRd,c = 28,453 kN | § 4.2.1.5 |
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Calculation: Check 2 — axial force: N_d ≤ N_Rd,c| Step | Value | Ref. |
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| Axial compression against the resistanceNd/NRd,c | η = 0.070 · Pass7 % of the capacity used | § 4.2.1.5 |
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Calculation: Section at the ultimate limit state under N_d| Step | Value | Ref. |
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| Depth of the compression zonefrom N(x) = Nd, εc2d at the compressed end | x = 777 mm | § 4.2.1.5 |
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| Relative depth | x/lw = 0.155 | |
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| Force in the compressed concretefcd·A(0.85·x) | Fc = 3,303 kN | |
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| Bending resistance under NdΣ Fi·(lw/2 − zi) | MRd = 12,393 kNm | § 4.2.1.5 |
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Calculation: Check 3 — bending with axial force: M_d ≤ M_Rd| Step | Value | Ref. |
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| Seismic moment against the bending resistance under NdMd/MRd | η = 0.403 · Pass40 % of the capacity used | § 4.2.1.5 |
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Calculation: Capacity design of the shear| Step | Value | Ref. |
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| Mean yield strength with overstrength1.1·fyk (class B) | fy,m = 550 N/mm² | § 4.3.9.3.3 |
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| 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 only | M+Rd = 14,307 kNm | § 4.3.9.3.3 |
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| Factor for the overstrengthM+Rd/Md | ε = 2.86 | § 4.3.9.4.4 (96) |
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| Factor for the higher modes1.5 (n > 5) | κ = 1.50 | § 4.3.9.4.4 (97) |
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| Magnified design shearε·κ·Vd | V+d = 2,575 kN | § 4.3.9.4.4 (95) |
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Calculation: Shear resistance of the plastic zone| Step | Value | Ref. |
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| Effective depthcompressed end to the centroid of the bars in the tension half | d = 4,337 mm | § 4.3.3.4.2 |
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| Lever arm of the internal forces0.9·d | z = 3,903 mm | § 4.3.3.4.2 |
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| Width of the web | bw0 = 250 mm | |
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| Shear reinforcement (horizontal bars)2·π·Øh²/4/shboth faces | Asw/s = 1.508 mm²/mm | |
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| Smallest inclination of the compression fieldlevel 1, plastic deformations of the chords | αmin = 40 ° | § 4.3.9.3.3, 4.3.3.3.2 |
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| Reduction factor for the concrete strength | kc = 0.40 | § 4.2.1.7 |
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| 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) |
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Calculation: Check 4 — horizontal bars: V+_d ≤ V_Rd,s| Step | Value | Ref. |
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| Resistance of the horizontal barsAsw/s·z·fyd·cotα | VRd,s = 3,050 kN | § 4.3.3.4.3 (45) |
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| Design shear against the horizontal barsV+d/VRd,s | η = 0.844 · Pass84 % of the capacity used | § 4.3.3.4.3 (45) |
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Calculation: Check 5 — compression field: V+_d ≤ V_Rd,c| Step | Value | Ref. |
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| Resistance of the compression fieldbw0·z·kc·fcd·sinα·cosα | VRd,c = 3,844 kN | § 4.3.3.4.6 (47) |
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| Design shear against the compression fieldV+d/VRd,c | η = 0.670 · Pass67 % of the capacity used | § 4.3.3.4.6 (47) |
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Calculation: Plastic zone at the base| Step | Value | Ref. |
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| Height of the plastic zonehpl ≥ lw and ≥ hw/6the detailing below applies over this height | hpl = 5,000 mm | § 5.7.1.1 |
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Calculation: Check 6 — axial force in the plastic zone| Step | Value | Ref. |
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| Normalised axial forceνd/0.4 | η = 0.200 · Pass20 % of the capacity used | § 5.7.1.1 |
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Calculation: Check 7 — longitudinal reinforcement| Step | Value | Ref. |
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| Ratio in one boundary elementAs,c/(lc·bw) | ρc = 1.53 % | |
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| Boundary element: at least 0.3 %0.3 %/ρc | η = 0.197 · Pass20 % of the capacity used | § 5.7.1.1 |
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| Boundary element: at most 3 %ρc/3 % | η = 0.509 · Pass51 % of the capacity used | § 5.7.1.1 |
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| Ratio of the webAs,w/((lw − 2·lc)·bw0) | ρw = 0.33 % | |
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| Web: at least 0.3 %0.3 %/ρw | η = 0.905 · Pass90 % of the capacity used | § 5.7.1.1 |
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| Web: at most 3 %ρw/3 % | η = 0.111 · Pass11 % of the capacity used | § 5.7.1.1 |
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Calculation: Check 8 — horizontal reinforcement| Step | Value | Ref. |
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| Horizontal ratio: at least 0.3 %0.3 %/ρh | η = 0.497 · Pass50 % of the capacity used | § 5.7.1.1 |
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| Horizontal against vertical ratio0.25·ρv/ρh | η = 0.133 · Pass13 % of the capacity used | § 5.7.1.1 |
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Calculation: Check 9 — spacing of the bars| Step | Value | Ref. |
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| Vertical web barssv/smax | η = 0.785 · Pass79 % of the capacity used | § 5.7.1.1 |
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| Horizontal barssh/smax | η = 0.600 · Pass60 % of the capacity used | § 5.7.1.1 |
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| End barssE/smax | η = 0.400 · Pass40 % of the capacity used | § 5.7.1.1 |
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Calculation: Check 10 — thickness of the web| Step | Value | Ref. |
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| Web between the boundary elementsbw0,min/bw0 | η = 0.600 · Pass60 % of the capacity used | § 5.7.1.3 |
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Calculation: Check 11 — boundary elements| Step | Value | Ref. |
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| Thickness of the boundary elementsbw,min/bw | η = 0.800 · Pass80 % of the capacity used | § 5.7.1.2 |
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| Length of the boundary elementslc,min/lc | η = 0.777 · Pass78 % of the capacity used | § 5.7.1.2 |
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Calculation: Check 12 — confinement of the boundary elements| Step | Value | Ref. |
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| Confined core across the wallbw − 2·cv + Øwbetween hoop centrelines, hoops outside the vertical bars; cover to the hoops cv − Øw = 22 mm | bcz = 198 mm | |
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| Confined core along the walllc − cv + Øw/2 | bcy = 674 mm | |
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| Bars in one boundary element | nc = 10 (5 rows of up to 2) | |
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| Vertical spacing of the hoopsscx/smax | η = 0.833 · Pass83 % of the capacity used | § 5.7.1.4 |
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| First hoop above the bases0/50 | η = 1.000 · Pass100 % of the capacity used | § 5.7.1.4 |
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| Diameter of the hoops0.35·Øsl,max/Øw | η = 0.875 · Pass88 % of the capacity used | § 5.7.1.4 |
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| Every other row held, along the wallnt,req/nt | η = 0.600 · Pass60 % of the capacity used | § 5.7.1.4 |
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| Distance between held bars, along the wallscy/200 | η = 0.843 · Pass84 % of the capacity used | § 5.7.1.4 |
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| Distance between held bars, across the wallscz/200 | η = 0.990 · Pass99 % of the capacity used | § 5.7.1.4 |
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| Curvature ductility needed2·q − 1 (T1 ≥ TC)with q = 3.0 of Table 15 | μφ = 5.00 | § 5.7.1.4 |
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| Normalised axial force−Nd/(Ac·fcd), compression positive | νd = 0.080 | |
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| Mechanical ratio of the vertical web barsρv·fyd/fcd | ωv = 0.070 | |
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| Efficiency of the confinement(1 − √(scy² + scx²)/(2·bcz))·(1 − √(scz² + scx²)/(2·bcy)) | αs = 0.422 | § 5.7.1.4 (119) |
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| Mechanical volumetric ratio of the hoopsAw·(2·bcy + nt·bcz)/(bcy·bcz·scx)·fyd/fcdwithout the anchorage lengths | ωwd = 0.191 | § 5.7.1.4 (118) |
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| Confinement of the boundary elementωwd,req/ωwd | η = 0.310 · Pass31 % of the capacity used | § 5.7.1.4 (118) |
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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