Reinforcement of a Concrete Beam Calculator

About Concrete Beam Reinforcement Calculation

The online reinforced concrete beam calculator performs reinforcement and strength calculations for a rectangular concrete beam with either simple supports or cantilever support. For a specified slab load, span, section dimensions, concrete class, and reinforcement parameters, it determines the bending moment, shear force, required reinforcement cross-section, suitable bar diameter, and strength utilization. These values are the main parameters used when assessing the strength and deflection behaviour of a reinforced concrete beam.

The calculation is carried out step by step. First, the applied load is combined with the self-weight of the beam, then the internal forces, effective depth, and required reinforcement area are determined. After a standard bar diameter has been selected, the calculator checks the section geometry again, verifies whether the bars fit within the beam, and evaluates bending and shear resistance.

Guidelines and recommendations

Calculation sequence

Loads and units. The cross-section dimensions and beam length are entered in millimetres. Before the structural calculation, the length is converted to metres. A distributed load is used in kN/m and a point load in kN. If the load is entered in kg/m or kg, the calculator converts it to force using a gravitational acceleration of 9.81 m/s2.

q = qkg · 9.81 / 1000

P = Pkg · 9.81 / 1000

Self-weight. The self-weight is calculated automatically from the actual section dimensions. The density of reinforced concrete is taken as 2500 kg/m3.

g = b · h · 2500 · 9.81 / 109

In this formula, b and h are in mm, and the resulting g is obtained in kN/m. For a distributed load, the self-weight is added to the external load.

Bending moment and shear force

Simply supported beam. For a uniformly distributed load acting over the full span, the standard expressions for the maximum bending moment at mid-span and the support reaction are used.

MEd = (q + g) · L2 / 8

VEd = (q + g) · L / 2

Cantilever beam. For a uniformly distributed load, the maximum internal forces occur at the fixed end.

MEd = (q + g) · L2 / 2

VEd = (q + g) · L

Point load. For a simply supported beam, the point load is assumed to act at mid-span. For a cantilever, it is assumed to act at the free end. The beam self-weight remains a uniformly distributed load.

MEd = g · L2 / 8 + P · L / 4

MEd,c = g · L2 / 2 + P · L

Load factors. The calculator does not additionally apply the partial factors γG and γQ to the entered external load. If a design load combination is required, it should be formed before the load is entered into the calculator.

Concrete and reinforcing steel

Design basis. The calculation relationships follow the approaches of EN 1992-1-1, Eurocode 2: Design of concrete structures. Part 1-1: General rules and rules for buildings. Related documents for actions are EN 1990, Eurocode: Basis of structural design, and EN 1991-1-1, Eurocode 1: Actions on structures. Part 1-1: General actions, densities, self-weight and imposed loads for buildings. Concrete exposure classes are related to EN 206, Concrete: Specification, performance, production and conformity.

Design strength of concrete. A material factor of γc=1.5 is used for all concrete classes. The basic design compressive strength is obtained by dividing the characteristic strength fck by 1.5. For classes up to C50/60, an additional coefficient of αc=1.00 is used; for classes C55/67 and above, αc=0.95 is used.

fcd = fck / 1.5

Concrete stress block parameters. For classes C8/10-C50/60, the calculation uses εcu=3.5‰, ω=0.810, and k2=0.416. For higher-strength concrete, these parameters are reduced:

• C55/67: εcu=3.2‰, ω=0.754, k2=0.403.
• C60/75: εcu=3.0‰, ω=0.694, k2=0.380.
• C70/85: εcu=2.8‰, ω=0.642, k2=0.371.
• C80/95: εcu=2.8‰, ω=0.605, k2=0.366.
• C90/105: εcu=2.8‰, ω=0.590, k2=0.366.

Tensile strength of concrete. The minimum reinforcement calculation uses fctm values that depend on the concrete class. For C8/10-C25/30, the values are respectively 1.2, 1.6, 1.9, 2.2, and 2.6 MPa; for C30/37-C50/60 they are 2.9, 3.2, 3.5, 3.8, and 4.1 MPa; and for C55/67-C90/105 they are 4.2, 4.4, 4.6, 4.8, and 5.0 MPa.

Reinforcing steel. For B500A, B500B, and B500C, the characteristic yield strength is taken as fyk=500 MPa, the material factor as γs=1.15, and the modulus of elasticity as Es=200000 MPa. The resulting design yield strength is therefore fyd=434.783 MPa.

Concrete cover and effective depth

Concrete cover. In the simplified selection by exposure condition, the calculator uses 20 mm for an indoor environment with normal humidity, 25 mm for an indoor environment with increased humidity, 30 mm for outdoor exposure, and 40 mm for contact with soil.

Exposure classes. The calculator uses the following fixed values: X0 = 20 mm; XC1 = 30 mm; XC2, XC3, XC4 = 35 mm; XD1, XD2, XD3 = 50 mm; XA1 = 25 mm; XA2 = 30 mm; XA3 = 40 mm; XF = 40 mm; XS = 50 mm.

Effective depth. The calculation is made to the centre of the longitudinal reinforcing bar. Therefore, after the bar diameter has been determined, the effective depth is reduced by the concrete cover and half the bar diameter.

d = h - z1 - φ1/2

For the second reinforcement layer, the bar centre position is determined in the same way:

d2 = z2 + φ2/2

Iterative recalculation. The bar diameter affects the effective depth, while the effective depth affects the required reinforcement area. The calculation therefore starts with an assumed diameter of 12 mm, then selects a diameter, recalculates the effective depth, and repeats the calculation. Up to 12 iterations are performed.

Required area of the main reinforcement

Normalized bending moment. After the design moment has been determined, a dimensionless parameter describing the utilization of the concrete section is calculated:

αm = MEd / (αc · fcd · b · d2)

The internal calculation continues only while the value remains within the valid range of the adopted model. The limiting condition is αm / (ω/k2) ≤ 0.25.

Internal lever arm. For a valid section, the coefficient ζ is determined:

ζ = 0.5 + √(0.25 - αm / (ω/k2))

The calculated area of tensile reinforcement is then determined from the bending moment equilibrium:

As,calc = MEd / (fyd · ζ · d)

Minimum reinforcement. At the same time, the minimum longitudinal reinforcement area is calculated. The larger of the two code-based expressions is used:

As,min = max(0.26 · fctm/fyk; 0.0013) · b · d

Final requirement. The main reinforcement must satisfy both the bending calculation and the minimum reinforcement requirement.

As,req = max(As,calc; As,min)

When the second reinforcement layer is calculated

Limit of singly reinforced behaviour. If the additional reinforcement layer is enabled, the calculator determines the limit beyond which part of the action is assigned to the second group of bars. For B500 steel, the design yield strain is approximately 2.174‰.

ξlim = εcu / (εcu + 2.174)

αm,lim = ω · ξlim · (1 - k2 · ξlim)

If αm ≤ αm,lim, no additional calculated reinforcement is required. If the limit is exceeded and the second layer is enabled, its required area is calculated.

Second-layer coefficient. For this calculation branch, a tabulated relationship between the coefficient ξ and αm is used. The value is interpolated within the range from 0.04 to 0.40. The ratio d2/d is limited to the range 0.04-0.16.

An additional coefficient depends on the steel class: 3 is used for B500A, 5 for B500B, and 10 for B500C. The resulting value is limited to a maximum of 1.

ks = min((ξ - d2/d) · kclass; 1)

Additional reinforcement area. The second reinforcement layer carries the part of the moment above the singly reinforced limit. Its area is also not allowed to be less than the calculated As,min.

As2 = max(As,min; (MEd - Mlim) / (ks · fyd · (d-d2)))

For a simply supported beam under the loads considered here, the main tension reinforcement is located at the bottom of the section. For a cantilever, the main working reinforcement is located at the top.

Selecting the reinforcement bar diameter

Standard diameter series. After the required reinforcement area has been determined, the calculator does not return an arbitrary bar diameter. It checks the following diameters in sequence: 6, 8, 10, 12, 14, 16, 18, 20, 22, 25, 28, 32, 36, and 40 mm.

For each diameter, the total cross-sectional area of the specified number of bars is calculated:

As,prov = n · π · φ2 / 4

Selection principle. The first diameter in the series for which the provided reinforcement area is not less than the required area is selected. The result is therefore always rounded up to the next available diameter. If even 40 mm bars do not provide the required area with the selected number of bars, the calculator indicates that the number of bars must be increased.

Arrangement of reinforcement within the section

Minimum clear spacing. After selecting the diameter, the calculator checks whether all bars fit within the beam width. The clear spacing between adjacent bars is taken as not less than 20 mm and at the same time not less than the bar diameter.

aclear = max(20 mm; φ)

Required width. The check includes concrete cover on both sides, the diameters of all bars, and the clear spaces between them.

breq = 2z + nφ + (n-1) · aclear

The reinforcement is considered to fit only when breq ≤ b. The main and additional reinforcement layers are checked separately.

Maximum reinforcement. For each selected group of bars, an upper limit equal to 4% of the full concrete cross-sectional area is used.

As,max = 0.04 · b · h

Bending resistance

Provided reinforcement. After the bar diameter has been selected, the calculation is repeated using the actual provided reinforcement area. The calculator determines the largest bending moment MRd that this reinforcement can resist within the adopted calculation model.

Numerical search. The value of MRd is found by repeatedly bisecting the search interval. Up to 80 iterations are performed, so the result is effectively independent of any manual search increment.

Strength utilization. The final percentage is the ratio of the design bending moment to the calculated bending resistance.

ηM = MEd / MRd · 100%

A value up to 100% means that the selected longitudinal reinforcement passes the bending check. A value above 100% means that the calculated bending resistance has been exceeded.

Shear check

Longitudinal reinforcement ratio. The shear check uses the actual provided area of the main longitudinal reinforcement. The reinforcement ratio is limited to a maximum value of 0.02.

ρl = min(0.02; Asl / (b · d))

Size factor. The effective depth d is used in millimetres:

k = min(2; 1 + √(200/d))

Concrete shear resistance. A coefficient of CRd,c=0.12 is used. The calculator determines the main resistance value and the minimum limit, then uses the larger result.

vRd,c,1 = 0.12 · k · (100 · ρl · fck)1/3

vmin = 0.035 · k3/2 · √fck

VRd,c = max(vRd,c,1; vmin) · b · d

The contribution from longitudinal compressive stress is taken as zero in this check. Shear utilization is evaluated separately from bending strength.

ηV = VEd / VRd,c · 100%

FAQs

Why does the beam self-weight affect the result even when the external load is unchanged?

The beam self-weight is calculated from the section width and height and is added automatically to the applied load. Increasing the beam dimensions can therefore increase its resistance while also increasing the distributed load acting on it.

Why does the calculator select a different bar diameter when the number of bars changes?

The required reinforcement area is distributed between the specified number of bars. A larger number of bars may provide the required area with a smaller diameter, but the calculator also checks whether the complete group fits within the beam width with the required clear spacing.

Why does enabling additional reinforcement not always result in a second reinforcement layer?

The second reinforcement layer is used only when the calculated limit for singly reinforced behaviour is exceeded. If the main reinforcement can provide the required resistance within the adopted model, the required additional reinforcement area remains zero.

What does a strength utilization of 70% or 110% mean?

The percentage is the ratio of the design action to the calculated resistance. For example, 70% means that approximately seven tenths of the available resistance is being used, while 110% means that the calculated resistance is exceeded by approximately one tenth.

Why can the bending check pass while the shear check does not?

These are independent checks. Bending resistance is evaluated from the bending moment and longitudinal reinforcement, while shear resistance depends on the concrete, effective depth, and longitudinal reinforcement ratio, so the utilization percentages can differ substantially.