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Rectangular RC Section Design Theory - Eurocode 2

The theory behind this rectangular reinforced-concrete section calculator: how the ULS bending capacity is found from the rectangular stress block and strain compatibility, how shear is checked with the variable-strut-inclination method, and how serviceability stresses and crack width are controlled to Eurocode 2 (EN 1992-1-1).

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A rectangular reinforced-concrete section - a beam or a one-way slab strip - is checked to Eurocode 2 (EN 1992-1-1) at the ultimate limit state for bending and shear, and at the serviceability limit state for stress and crack width. This page sets out the method behind the calculator.

Materials and design strengths

Concrete and steel are reduced from characteristic to design strengths by partial factors. For concrete the design compressive strength carries the long-term factor αcc\alpha_{cc} (often 0.85):

fcd=αccfckγC,fyd=fykγSf_{cd} = \alpha_{cc}\,\frac{f_{ck}}{\gamma_C}, \qquad f_{yd} = \frac{f_{yk}}{\gamma_S}

with γC=1.5\gamma_C = 1.5 and γS=1.15\gamma_S = 1.15. The concrete strain at the extreme compression fibre is limited to εcu2=3.5\varepsilon_{cu2} = 3.5‰ (for grades up to C50/60).

ULS bending - the rectangular stress block

At bending failure the concrete compression is idealised as a uniform (rectangular) stress block of intensity ηfcd\eta f_{cd} over a depth λx\lambda x, where xx is the neutral-axis depth (λ=0.8\lambda = 0.8, η=1.0\eta = 1.0 up to C50/60). For a singly-reinforced section with tension steel AsA_s at effective depth dd, force equilibrium gives the lever arm and the moment of resistance:

ηfcdbλx=Asfyd    z=dλx2,MRd=Asfydz\eta f_{cd}\,b\,\lambda x = A_s f_{yd} \;\Rightarrow\; z = d - \tfrac{\lambda x}{2}, \qquad M_{Rd} = A_s f_{yd}\,z

It is convenient to work with the dimensionless moment K=MEd/(bd2fck)K = M_{Ed}/(b d^{2} f_{ck}). If KKbalK \le K_{bal} (about 0.167 for the recommended 15% redistribution limit) the section is singly reinforced and the lever arm follows in closed form:

z=d[0.5+0.25K/1.134]0.95dz = d\left[\,0.5 + \sqrt{0.25 - K/1.134}\,\right] \le 0.95\,d

If K>KbalK > K_{bal} the concrete alone cannot carry the moment, so compression reinforcement is added; the calculator sizes both layers automatically and reports whether the section is singly or doubly reinforced.

ULS shear - variable strut inclination

Shear is checked by the truss analogy of §6.2. A member without shear reinforcement has resistance VRd,cV_{Rd,c}:

VRd,c=[CRd,ck(100ρlfck)1/3+0.15σcp]bwd  vminbwdV_{Rd,c} = \Big[C_{Rd,c}\,k\,(100\,\rho_l f_{ck})^{1/3} + 0.15\,\sigma_{cp}\Big] b_w d \ \ge\ v_{\min}\,b_w d

where k=1+200/d2.0k = 1 + \sqrt{200/d} \le 2.0 and ρl=Asl/(bwd)0.02\rho_l = A_{sl}/(b_w d) \le 0.02. If VEd>VRd,cV_{Ed} > V_{Rd,c}, designed links are required and the resistance is the lesser of the link capacity and the concrete-strut limit, with the strut angle θ\theta chosen in the range 21.8θ4521.8^\circ \le \theta \le 45^\circ:

VRd,s=Aswszfywdcotθ,VRd,max=αcwbwzν1fcdcotθ+tanθV_{Rd,s} = \frac{A_{sw}}{s}\,z\,f_{ywd}\cot\theta, \qquad V_{Rd,\max} = \frac{\alpha_{cw}\,b_w z\,\nu_1 f_{cd}}{\cot\theta + \tan\theta}

A flatter strut (larger cotθ\cot\theta) needs fewer links but raises the compression in the concrete strut, so VRd,maxV_{Rd,\max}sets the lower bound on θ\theta.

SLS - stresses and crack width

At service load the section is analysed cracked-elastic, using the modular ratio αe=Es/Ecm\alpha_e = E_s/E_{cm} to transform the steel into equivalent concrete. The neutral-axis depth follows from first moments of the transformed area, and the stresses are checked against the §7.2 limits - the concrete compression against 0.6fck0.6 f_{ck} (characteristic combination) and the steel tension against 0.8fyk0.8 f_{yk}.

Crack width is controlled either by limiting bar spacing/diameter (the deemed-to-satisfy tables) or by direct calculation (§7.3.4):

wk=sr,max(εsmεcm)w_k = s_{r,\max}\,(\varepsilon_{sm} - \varepsilon_{cm})

where sr,maxs_{r,\max} is the maximum crack spacing and (εsmεcm)(\varepsilon_{sm} - \varepsilon_{cm}) the mean strain difference between steel and concrete, including tension stiffening. The result is compared with the limit wmaxw_{\max} (commonly 0.3 mm) for the exposure class.

Minimum and maximum reinforcement (§9.2), detailing and the National Annex values for αcc\alpha_{cc}, CRd,cC_{Rd,c} and the crack limits always govern the final design.

Frequently asked questions

At the ultimate limit state the concrete compression is represented by a rectangular stress block of intensity η·f_cd (η = 1.0 up to C50/60) over a depth λx (λ = 0.8). Equating the concrete force to the tension-steel force A_s·f_yd gives the neutral-axis depth x and the lever arm z = d − λx/2, so the moment of resistance is M_Rd = A_s·f_yd·z. Using the dimensionless moment K = M_Ed/(b·d²·f_ck), if K ≤ K_bal (≈0.167) the section is singly reinforced; if K is larger, compression steel is added.

The real parabola-rectangle concrete stress distribution at failure is replaced, for design, by an equivalent rectangular block of uniform stress η·f_cd acting over a depth λx measured from the most-compressed fibre, where x is the neutral-axis depth. For concrete up to C50/60, λ = 0.8 and η = 1.0; both reduce for higher grades. This simplification gives the same compression force and almost the same lever arm as the true distribution, so the bending capacity comes out in a simple closed form.

A singly-reinforced section relies on the concrete alone to balance the tension steel in compression. When the applied moment exceeds the balanced capacity - the dimensionless moment K = M_Ed/(b·d²·f_ck) is greater than K_bal (about 0.167 for the usual ≤15% redistribution) - the concrete compression zone would have to be too deep, over-straining the section. Compression reinforcement is then added in the top of the section to carry the excess, making it doubly reinforced.

Eurocode 2 models a beam with shear links as a truss: the links are vertical ties and the concrete between diagonal cracks forms compression struts. The designer may choose the strut angle θ between 21.8° and 45° (cot θ from 2.5 to 1.0). A flatter strut (cot θ = 2.5) needs fewer links because each crossed link is more effective, but it raises the compression in the strut, so V_Rd,max (the strut-crushing limit) caps how flat θ can be. The shear resistance is the link capacity V_Rd,s, checked against V_Rd,max.

Crack width can be controlled in two ways. The simplest is to satisfy deemed-to-satisfy rules - limiting bar diameter and spacing from tables for the relevant steel stress. Alternatively the crack width is calculated directly as w_k = s_r,max·(ε_sm − ε_cm), where s_r,max is the maximum crack spacing and (ε_sm − ε_cm) is the difference between the mean steel and concrete strains allowing for tension stiffening. The result must not exceed the limit w_max (typically 0.3 mm) for the exposure class.

The design strengths are the characteristic strengths divided by partial material factors. For concrete f_cd = α_cc·f_ck/γ_C with γ_C = 1.5 (and α_cc commonly 0.85 for long-term effects). For reinforcement f_yd = f_yk/γ_S with γ_S = 1.15. These are the recommended values for persistent and transient design situations; accidental situations use lower factors, and the National Annex may adjust them.

Ready to check your own section? Get the moment capacity, shear resistance, SLS stresses and crack width for any rectangular RC beam or slab.

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