Structural Engineering Tools

Reinforced concrete tie in two-way bending to SIA 262

Checks a reinforced concrete tie under an axial tension and moments about both axes, on a section of any shape with its bars at their own coordinates, to SIA 262:2025: tension resistance, resistance in the direction of the moment, cracking and bar stresses. 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 tie under an axial tension and moments about both axes, on a section of any shape with its bars at their own coordinates, to SIA 262:2025: tension resistance, resistance in the direction of the moment, cracking and bar stresses. Every step shows its formula and the clause it comes from.

What it checks

  • Section: Ac, Ix and Iy about the axes through the centroid G, the bars As, the ratio ρ = As/Ac and the smallest clear cover c.
  • Design strengths: ηfc = (40/fck)^(1/3) ≤ 1, clause 4.2.1.2, equation (26); fcd = ηfc · ηt · fck/γc, equation (2); fyd = fyk/γs, equation (4).
  • Check 1, axial force: |Nd| ≤ NRd. In tension NRd,t = As · fyd: all the bars yield, the concrete takes nothing. In compression NRd,c follows clause 4.2.1.5.
  • Check 2, oblique bending with axial force (clause 4.2.1.5): Md = √(Mdx² + Mdy²) ≤ MRd(Nd, β). The module finds the inclination θn of the neutral axis and the depth x for which N(x) = Nd, and gives MRd,x and MRd,y parallel to the design moment, with the strain εs of the most stretched bar.
  • Cracking: the largest tension σct,max of the uncracked section, a plane strain on Ac + (αs − 1) · As (clause 4.1.4.2.1), against fct,eff = kt · fctm, clause 4.4.1.3. The upper value fctk0,95 = 1.3 · fctm (equation (8), clause 4.4.1.4) can be chosen where late cracking is unfavourable.
  • Bar stresses in service, Table 17: σs ≤ fyd − 80 N/mm², and for the quasi-permanent load case σs ≤ σs,adm = √(9 · Es · fctm · wnom/Øs) ≤ fyk, clause 4.4.2.3.4, equation (102).

What you enter and what you get

You choose the shape of the tie section (rectangle, circle, or a polygon drawn by clicking its corners) and click each bar into the drawing, or place a row of bars at once; bars can be dragged, removed or edited in a table. You enter the concrete and steel properties, the resistance factors and ηt, then the sense of the axial force (tension for a tie), Nd, Mdx and Mdy. For serviceability: the tensile strength for cracking, the requirement level, wnom from Table 17, N, Mx, My and the load case.

You get both checks with their utilisation, the state of the tie in service, cracked or not, with the stresses of the most stretched and the most compressed bar, and the interaction contour of the two moments. With the Basic plan, it prints as an A4 calculation note.

Assumptions and limits

  • N acts at the centroid G of the gross concrete section; its magnitude is entered with the sense chosen.
  • Mdx is positive when it compresses the top of the drawn section (+y), Mdy when it compresses the right side (+x).
  • The check is radial: Nd and the direction of the moment are kept.
  • In the cracked section the concrete in tension is ignored. Tension stresses are positive, compression negative.
  • Crack widths are not computed.

Worked example

The module's worked example is a compressed section: 400 × 600 mm in fck = 30 N/mm² with eight Ø20 bars (three at the top, three at the bottom, one on each side at mid-depth), under Nd = 1,500 kN in compression, Mdx = 200 kNm and Mdy = 80 kNm, and a quasi-permanent service state of N = 1,000 kN, Mx = 120 kNm and My = 50 kNm at the increased requirement level with wnom = 0.5 mm. These are the module's own results for its worked example, computed when this page was built.

Inputs
InputValueRef.
Shape of the sectionRectangle
Width b400 mm
Depth h600 mm
Axial force NCompression
Bars in the section{"b":[[50,50,20],[200,50,20],[350,50,20],[50,550,20],[200,550,20],[350,550,20],[50,300,20],[350,300,20]],"p":[],"v":[]}
Characteristic compressive strength fck30 N/mm²§ 3.1.2.2.2
Ultimate compressive strain εc2d3.0 ‰§ 4.2.1.5
Mean tensile strength fctm2.9 N/mm²§ 3.1.2.2.4
Modulus of elasticity Ecm33600 N/mm²§ 3.1.2.3.3
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 force Nd1500 kN
Design moment about the x axis Mdx200 kNm
Design moment about the y axis Mdy80 kNm
Tensile strength for crackingMean tensile strength (4.4.1.3)§ 4.4.1.3
Requirement levelIncreased§ 4.4.2.2.3
Nominal crack width wnom0.5 mmTable 17
Axial force in the load case N1000 kN
Moment about the x axis Mx120 kNm
Moment about the y axis My50 kNm
Load caseQuasi-permanentTable 17
Calculation: Concrete section
StepValueRef.
Gross concrete areaAc = 240,000 mm²
Second moment of area about xabout the x axis through GIx = 7,200,000,000 mm⁴
Second moment of area about yabout the y axis through GIy = 3,200,000,000 mm⁴
Calculation: Reinforcement
StepValueRef.
Barsns = 8Ø20
Total area of the barsΣ π·Ø²/4As = 2,513 mm²
Reinforcement ratioAs/Acρ = 1.05 %
Smallest clear cover of the barsc = 40 mm
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) + min(fyd ; Es·εc2d)·Asthe whole section at εc2dNRd,c = 5,842 kN§ 4.2.1.5
Resistance in centred tensionAs·fydall the bars yield, no concreteNRd,t = 1,093 kN
Calculation: Check 1 — axial force: |N_d| ≤ N_Rd
StepValueRef.
Axial compression against the resistance|Nd|/NRdη = 0.257 · Pass26 % of the capacity used§ 4.2.1.5
Calculation: Section at the ultimate limit state
StepValueRef.
Resultant design moment√(Mdx² + Mdy²)Md = 215.4 kNm
Direction of the design momentatan2(Mdx, Mdy)the side the moment compresses, measured from +x towards +yβ = 68.2 °
Inclination of the neutral axis to the x axisgenerally not perpendicular to β: the section and its bars are not alike in all directionsθn = 133.7 °
Depth of the compression zonefrom the most compressed corner, across the neutral axis; N(x) = Ndx = 332 mm§ 4.2.1.5
Force in the compressed concretefcd·A(0.85·x)Fc = 1,590 kN
Resistance about xΣ Fi·yiMRd,x = 395.2 kNm
Resistance about yΣ Fi·xiMRd,y = 158.1 kNm
Bending resistance in the direction of Md√(MRd,x² + MRd,y²), (MRd,x ; MRd,y) parallel to (Mdx ; Mdy)MRd = 425.6 kNm§ 4.2.1.5
Calculation: Check 2 — oblique bending with axial force: M_d ≤ M_Rd
StepValueRef.
Design moment against the resistance in its directionMd/MRdη = 0.506 · Pass51 % of the capacity used§ 4.2.1.5
Strain of the most stretched baryieldsεs = 2.73 ‰
Calculation: Uncracked section
StepValueRef.
Modular ratio of the barsEs/Ecmαs = 6.10
Largest concrete tension, uncrackedplane strain ε0 + κy·x + κx·y on Ac + (αs − 1)·Astension positiveσct,max = 3.59 N/mm²§ 4.1.4.2.1
Size factor1/(1 + 0.5·t), t = hn/3 in mhn = 710 mm, the depth of the section across its neutral axiskt = 0.894§ 4.4.1.3
Effective tensile strengthkt·fctmfct,eff = 2.59 N/mm²§ 4.4.1.3
Calculation: State under N = 1,000 kN, M_x = 120.0 kNm, M_y = 50.0 kNm
StepValueRef.
State of the sectioncracked (σct,max > fct,eff)
Compressed part of the concretethe concrete in tension is ignoredAcc/Ac = 0.830
Inclination of the neutral axis to the x axisθn = 135.3 °
Largest concrete compressioncompression is negativeσc = -12.31 N/mm²
Stress of the most stretched barbar 1σs = 19.5 N/mm²
Stress of the most compressed barbar 6σs′ = -64.6 N/mm²
Calculation: Check 1 — no yielding of the bars
StepValueRef.
Bar stress against fyd − 80 N/mm²σs/(fyd − 80)η = 0.055 · Pass5 % of the capacity usedTable 17
Calculation: Check 2 — bar stress under the quasi-permanent load
StepValueRef.
Admissible bar stress√(9·Es·fctm·wnom/Øs) ≤ fykØs = largest bar, 20 mmσs,adm = 366 N/mm²§ 4.4.2.3.4 (102)
Bar stress against the crack limitσs/σs,admη = 0.053 · Pass5 % of the capacity usedTable 17

How this module is verified

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

Questions

How do I check a tie?

Choose Tension for the axial force and enter its magnitude for Nd and N. The resistance in centred tension is then NRd,t = As · fyd.

Where do I get the crack width of the tie?

Not here. In the governing direction, the module Concrete tie in bending and axial force gives it.

Which plan includes this module?

The Basic plan, with its calculation note.

Related

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