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This calculator allows the user to assess the structural integrity of concrete beams to ensure compliance with the Australian Standard AS 3600. The calculation will identify the design capacities of concrete beams to meet flexural and shear design requirements to ultimate limit state design (ULS) methods.
❗ This calculation has been written in accordance with AS3600:2018.
Cross-section of a rectangular beam with symbols used in this calculator


Calculation

Technical notes and assumptions

  1. An axial capacity check of the beam is not included. "Axial force, N*" input value is used in the shear capacity calculation.
  1. "Bending moment, M*" input value assumes a sagging moment, i.e. where tension is at the bottom of the beam. M* shall be a positive value. A negative value for M* does not represent a hogging moment.


Inputs

Loads



N*
:0 kN​



V*
:300kN​



M*, kNm
:125​



Untitled
:0.00 kN​



Material Properties



f'c
:40MPa​



Ec
:32800MPa​



fsy
:500MPa​



Es
:200000MPa​



Section Geometry



B
:1200mm​



D
:1100mm​



c
:50mm​


Tensile (Bottom) Reinforcement:


db.st
:20mm​



n1
:9​


Shear Reinforcement:


db.sv
:20mm​



s
:125mm​



n
:4​



Shear Check - Method of Analysis



Method for calculating shear factor kv
:General​



dg (for general method)
:16mm​



Output

Material Properties



f'ct.f
:3.794733192202055MPa​



f'ct
:2.2768399153212333MPa​



Geometric Properties



Ast provided
:2827.4333882308138mm2​



Ag
:1.32e+6mm2​



Zx
:2.4e+8mm3​



Ixx
:1.3e+11mm4​



α2
:0.85​



γ
:0.77​



d
:1020.0mm​



dn
:17.324959486708416mm​



Shear Properties



kv
:0.25340793063291445​



bv
:1200.0mm​



dv
:918.0mm​



√f'c
:6.324555320336759​



εx
:3.8565504244731844E-4​


Design Checks

Flexural Checks

Flexural capacity:


ϕ (moment)
:0.8​



ϕMu
:1133.998754837186kN m​



Utilisation
:0.11022939793081774​

Minimum tension reinforcement check:


Ast min
:2,161mm2​



Ast provided
:2,827mm2​



Min tension reinf check
:OK​

Ductility check:


kuo
:0.044117543892814914​



ku <0.36
:PASS​





Shear Checks

Shear capacity:


ϕ (shear)
:0.7​



ϕVuc
:1235.868222015952kN​



ϕVus
:4445.796052135642kN​



ϕVu
:5681.664274151593kN​



Utilisation
:0.05280143027190692​

Minimum shear reinforcement check:


Shear reinf requirement
:REQUIRED​



Asv min
:151.78932768808224mm2​



Asv provided
:1256.6370614359173mm2​



Min tension reinf check
:OK​



Maximum shear strength check:


ϕVu.max
:8067.165709300708kN​



ϕVu < ϕVu.max
:PASS​


Explanation

Beams are an important structural element in reinforced concrete (RC) structures. They are designed to provide resistance to external loads that cause shear forces, bending moments and, in some cases, torsion across their length.
Concrete is strong in compression but weak in tension, so reinforcement is added to take the tensile stresses induced when beams are loaded. In a sagging beam the tensile stresses are along the bottom of the beam, so theoretically, no top reinforcement is required. However most beams will still have top reinforcement for constructability reasons, so that the ligs can hang from the top bars.

Longitudinal (flexural) reinforcement in a concrete beam

Shear reinforcement (also referred to as 'ligs', 'links' or 'fitments') in a concrete beam

The latest Australian Standard for concrete structures, AS3600-2018, has undergone significant updates and changes since the 2009 edition, mainly concerning:
  1. Stress-block configuration for the analysis and design of reinforced and prestressed members in bending
  2. Values of capacity reduction factor, Φ, for different member strengths
  3. Shear and torsional strength evaluation of concrete members
  4. Effective moment of inertia (I_ef) estimations for deflection calculations.
Despite the design equations for shear and torsion changing in AS3600-2018, the strength design procedure and limit state design philosophy remains unchanged:

Rd≥Ed ,whereRd=Design capcity (= ϕRu)Ed=Design action effect    R_d \ge E_d \space,\text{\\where} \\ R_d = Design \ capcity \ (=\ \phi R_u) \\ E_d = Design \ action \ effect\ \ \ \
For a given section of any structural member to be designed, 'Ed' is the ‘action effect’ or moment, shear, torsion or axial force due to the most critical combination of external service loads specified in AS1170.0, each multiplied by a corresponding load factor. 'Rd' is the computed ultimate resistance of a member at a particular section. And, 'ϕ' is the capacity reduction factor given in Table 2.2.2.

General Steps to RC Beam Design

Generally, the limit state checks for RC beam design include the following:
  1. Moment flexural capacity
  2. Shear capacity
  3. Deflection
  4. Stability
  5. Crack control
This calculator covers the first two checks - flexural and shear capacity. Deflection, stability checks and crack control require a more detailed examination by the engineer.

Ultimate Flexural Capacity

The flexural (also referred to as 'bending' or 'moment') capacity of a beam cross-section is determined using the Rectangular Stress Block method (Cl. 8.1.2). The stress distribution in concrete under bending is curved in reality, however it can be converted to an equivalent rectangular stress block by the use of reduction factors α2 and γ.
The strain and equivalent rectangular stress block of a typical beam section


Factor ku is not defined explicitly in the standards and must be computed via mechanics. It is the ratio of the depth of the neutral axis and the centroid of tensile reinforcement, taken from the extreme compressive fibre.

α2=0.85−0.0015fc′≥0.67γ=0.97−0.0025fc′≥0.67ku=Astfsyα2fc′γbd+Ascfsy\alpha_2=0.85-0.0015f'_c \ge 0.67 \\ \gamma = 0.97-0.0025 f'_c \ge 0.67\\k_u = \frac{A_{st}f_{sy}}{\alpha_{2}f'_c\gamma bd+A_{sc}f_{sy}}
The ultimate flexural strength at critical sections should not be less than the minimum required strength in bending (Muo)min, given by Eq. 8.1.6.1 (1):

(Muo)min=1.2[Z(fct.f′+Pe/Ag)+Pee]Z=section modulus of the uncracked cross-sectionfct.f′=characteristic flexural tensile strength of the concretePe=total effective prestress force allowing for all losses of prestresse=eccentricity of the prestressing force\small{(M_{uo})_{min}=1.2[Z(f'_{ct.f}+P_e/A_g)+P_ee]} \\ \small{Z= \text{section modulus of the uncracked cross-section}} \\ \small{f'_{ct.f} = \text{characteristic flexural tensile strength of the concrete}} \\ \small{P_e = \text{total effective prestress force allowing for all losses of prestress}} \\ \small{e = \text{eccentricity of the prestressing force}}
Where there is no pre-stressing, (Muo)min is given by:

(Muo)min=1.2Zfct.f′(M_{uo})_{min} = 1.2Zf'_{ct.f}
For reinforced concrete sections, the above requirement is deemed to be satisfied if the total area of the provided tensile reinforcement satisfies the following:

Ast≥[αb(D/d)2fct.f′/fsy]bwd\small{A_{st}\geq [\alpha_b(D/d)^2f'_{ct.f}/f_{sy}]b_wd}

For rectangular sections:αb=0.2\small{\text{For rectangular sections:}} \\\small{\alpha_b=0.2}
To finalise your ultimate bending capacity limit state, you must check your section is ductile. A ductile section ensures the reinforcement will yield before the concrete crushes, which is deemed a less dangerous failure mechanism since the failure occurs relatively slow compared to a sudden brittle failure. As per Cl 8.1.5, a section is ductile if:

kuo≤0.36  kuo = ku factor for a section in pure bendingk_{uo} \le 0.36\ \\\normalsize\small\ k_{uo}\ =\ k_u\text{ factor\ for\ a\ section\ in\ pure\ bending}
Once the reduction factor ϕ is determined from Table 2.2.2, the reduced moment flexural capacity is thus given as:

ϕMu=ϕAstfsy×(d−γkud2)\phi M_u = \phi A_{st}f_{sy} \times (d - \frac{\gamma k_ud}{2})

Ultimate Shear Capacity

Section 8.2 outlines methods for calculating strength of beams in shear. Torsion is omitted from explanations, but it must be checked if the concrete member sees in-plane rotation.
The total shear capacity of a section (Vu) is the combination of the shear strength contributed by concrete (Vuc), shear reinforcement (Vus) and vertical component of prestressing (Pv), limited by Vu.max:

ϕ Vu = ϕ(Vuc+Vus+Pv) ≤ϕ Vu.max\phi\ V_u\ =\ \phi(V_{uc}+V_{us}+P_v)\ \le\phi\ V_{u.max}
Truss analogy of shear resistance: concrete strut in compression and shear reinforcement (ligs) in tension

Shear fitments/ligatures are required if any of the following is true:

V∗ >ϕ(Vuc+Pv)or;Overall beam depth, D > 750mmV^*\ >\phi(V_{uc}+P_v)\hspace{0.5cm}\text{or;}\\\text{Overall\ beam\ depth,\ D\ > 750mm}
If fitments/ligatures are deemed required, the minimum shear cross-sectional area is calculated as per Cl. 8.2.1.7:

Asv.mins=0.08fc′ bvfsy.fs=centre-to-centre spacing of shear reinforcement parallel to member longitudinal axisfsy.f=yield strength of shear reinforcement\large\frac{A_{sv.min}}{s}=\frac{0.08\sqrt{f'_c}\ b_v}{f_{sy.f}}\\\normalsize\small \\ s=\text{centre-to-centre\ spacing\ of\ shear\ reinforcement parallel\ to\ member\ longitudinal\ axis}\\\normalsize\small \\ f_{sy.f}=\text{yield\ strength\ of\ shear\ reinforcement}

Contribution to Shear Strength from Concrete (Vuc)

Vuc=kvbvdvfc′V_{uc}=k_vb_vd_v\sqrt{f'_c}
The factor kv can be determined using the simplified method in accordance with Cl. 8.2.4.3 if the following is satisfied:
  1. No prestress and no applied axial tension
  2. Strength of concrete, f'c, is less than 65 MPa
  3. Size of aggregates, kdg, is not less than 10mm
  4. Yield strength of longitudinal reinforcement is less than 500 MPa.
For most design scenarios, the above conditions will be satisfied. If not, general method must be used as per Cl. 8.2.4.2.

ForAsvs<Asv.mins: kv=2001000+1.3dv≤0.1ForAsvs≥Asv.mins:kv=0.15\text{For}\hspace{0.2cm}\large\frac{A_{sv}}{s}<\frac{A_{sv.min}}{s}:\ k_v=\frac{200}{1000+1.3d_v}\normalsize\leq0.1\\\text{For}\hspace{0.2cm}\large\frac{A_{sv}}{s}\geq\frac{A_{sv.min}}{s}:{k_v\normalsize=0.15}
Effective shear width (bv) and depth (dv) is determined as per Cl. 8.2.1.5 and 8.2.1.9:

bv=bw−kdΣdd kdΣdd=factored diameters of prestressing ducts, if anydv=max(0.72D,0.9d) b_v=b_w-k_d\Sigma{d_d}\ \normalsize\small\\\hspace{0.1cm}k_d\Sigma{d_d}=\text{factored diameters of prestressing ducts, if any}\\d_v = max(0.72D, 0.9d)\

Contribution to Shear Strength from Reinforcement (Vus)
For shear reinforcement placed at an inclination:

Vus=(Asvfsy.fdvs)(sin(αv)cot(θv)+cos(αv))αv=angle between the inclined shear reinforcement and the longitudinal tensile reinforcment θv=angle between the axis of the concrete compression strut and the longitudinal axis of the member(Cl. 8.2.4)\large{V_{us}= \left( \frac{A_{sv}f_{sy.f}d_v}{s} \right)(sin(\alpha_v)cot(\theta_v)+cos(\alpha_v))}\\\normalsize\small{\alpha_v = \text{angle between the inclined shear reinforcement and the longitudinal tensile reinforcment}}\\\ \small{\theta_v = \text{angle between the axis of the concrete compression strut and the longitudinal axis of the member}\hspace{0.1cm}\text{{(Cl.\ 8.2.4)}}}
The strut angle, θv can generally be taken as 36° as per Cl 8.2.4.3 (simplified method) or using a more rigorous analysis as per Cl. 8.2.4.2.2 or 8.2.4.2.3 (general method). The strut angle, θv is a variable parameter that varies with the magnitude of applied load. A smaller θv means the strut crosses more shear ligs and therefore a smaller area of shear ligs is needed to get the same shear capacity as a greater θv. Though, looking at the truss analogy figure above, a larger θv means more of the applied shear force is taken by the vertical tie (shear ligs) then the horizontal tie (longitudinal reinforcement). This calculator determines θv directly, with the general method, though limits θv ≥ 36° as per the simplified method to ensure the required shear reinforcement is not under-conservative.
For shear reinforcement placed perpendicular to the longitudinal axis of the member (𝛼v = 90°), as per Cl. 8.2.5.2:

Vus=(Asvfsy.fdvs)cot(θv)\large{V_{us}= \left( \frac{A_{sv}f_{sy.f}d_v}{s} \right)cot(\theta_v) }

Comparison against Shear Strength Limited by Web Crushing (Vu.max)
The sum of contribution to shear strength by concrete and steel (Vuc + Vus) is limited by web crushing as per Cl. 8.2.3.3, which defines Vu.max:

Vu.max=0.55[fc′bvdv(cot(θv)+cot(αv)1+cot2(θv))]+Pv...8.2.3.3(1)Vu.max=0.55[fcp′bvdv(cot(θv)+cot(αv)1+cot2(θv))]+Pv...8.2.3.3(2)V_{\text{u.max}}=0.55\left[ f'_cb_vd_v \left( \frac{cot(\theta_v)+cot(\alpha_v)}{1+cot^2(\theta_v)} \right) \right] + P_v \hspace{2cm}...8.2.3.3(1)\\V_{\text{u.max}}=0.55\left[ f'_{cp}b_vd_v \left( \frac{cot(\theta_v)+cot(\alpha_v)}{1+cot^2(\theta_v)} \right) \right] + P_v \hspace{1.7cm} ...8.2.3.3(2)
The total shear capacity (Vu) is calculated as below, with reduction factor again taken from Table 2.2.2:

ϕVu=ϕ min(Vuc+Vus,Vu.max)\phi V_u = \phi \space\text{min}(V_{uc} + V_{us}, V_{u.max})


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