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
An axial capacity check of the beam is not included. "Axial force, N*" input value is used in the shear capacity calculation.
"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*
:500 kN
M*
:125 kN m
Material Properties
f'c
:40MPa
Ec
:32.8 GPa
fsy
:500 MPa
Es
:200 GPa
Section Geometry
B
:1.2 m
D
:1.1 m
c
:50 mm
Tensile (Bottom) Reinforcement:
db.st
:20 mm
n1
:9
Shear Reinforcement:
db.sv
:20 mm
s
:125 mm
n
:4
Shear Check - Method of Analysis
Method for calculating shear factor kv
:General
dg (for general method)
:16 mm
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
:17.3249594867083mm
Shear Properties
kv
:0.21695023483340845
bv
:1200mm
dv
:918.0mm
√f'c
:6.324555320336759
εx
:0.0005624938681049827
Design Checks
Flexural Checks
Flexural capacity:
ϕ (moment)
:0.8
ϕMu
:1133.998754837176kN m
Utilisation
:0.11022939793081832
Minimum tension reinforcement check:
Ast min
:2,161mm2
Ast provided
:2,827mm2
Min tension reinf check
:OK
Ductility check:
kuo
:0.0441175438928147
ku <0.36
:PASS
Shear Checks
Shear capacity:
ϕ (shear)
:0.7
ϕVuc
:1058.06436412564kN
ϕVus
:4445.796052135642kN
ϕVu
:5503.860416261314kN
Utilisation
:0.09084532713125064
Minimum shear reinforcement check:
Shear reinf requirement
:REQUIRED
Asv min
:151.7893276880822mm2
Asv provided
:1256.6370614359173mm2
Min tension reinf check
:OK
Maximum shear strength check:
ϕVu.max
:8067.16570930066kN
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
Stress-block configuration for the analysis and design of reinforced and prestressed members in bending
Values of capacity reduction factor, Φ, for different member strengths
Shear and torsional strength evaluation of concrete members
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:
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:
Moment flexural capacity
Shear capacity
Deflection
Stability
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.
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
Where there is no pre-stressing, (Muo)min is given by:
(Muo)min=1.2Zfct.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
For rectangular sections:αb=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≤0.36kuo=ku 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−2γkud)
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:
ϕVu=ϕ(Vuc+Vus+Pv)≤ϕVu.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 > 750mm
If fitments/ligatures are deemed required, the minimum shear cross-sectional area is calculated as per Cl. 8.2.1.7:
sAsv.min=fsy.f0.08fc′bvs=centre-to-centre spacing of shear reinforcement parallel to member longitudinal axisfsy.f=yield strength of shear reinforcement
Contribution to Shear Strength from Concrete (Vuc)
Vuc=kvbvdvfc′
The factor kv can be determined using the simplified method in accordance with Cl. 8.2.4.3 if the following is satisfied:
No prestress and no applied axial tension
Strength of concrete, f'c, is less than 65 MPa
Size of aggregates, kdg, is not less than 10mm
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.
Effective shear width (bv) and depth (dv) is determined as per Cl. 8.2.1.5 and 8.2.1.9:
bv=bw−kdΣddkdΣdd=factored diameters of prestressing ducts, if anydv=max(0.72D,0.9d)
Contribution to Shear Strength from Reinforcement (Vus)
For shear reinforcement placed at an inclination:
Vus=(sAsvfsy.fdv)(sin(αv)cot(θv)+cos(αv))αv=angle between the inclined shear reinforcement and the longitudinal tensile reinforcmentθv=angle between the axis of the concrete compression strut and the longitudinal axis of the member(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=(sAsvfsy.fdv)cot(θv)
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: