UNDERSTAND IT. WORK IT OUT.

Learn: Concrete shear resistance — coefficient form

Concrete without calculated shear reinforcement still has a shear-resistance model based on concrete strength, longitudinal reinforcement and section size. This lesson evaluates the explicitly supplied main expression, including a compressive-stress contribution.

Beginner-friendlyFree · No accountTwo worked examples + separate practice
01

What the formula is saying

First take the cube root of 100ρfck, scale it by C and k, and add k1σcp. That produces a shear-stress expression; multiplying by bw d turns it into a force.

VRd,c = [C k (100ρl fck)^(1/3) + k1σcp] bw d

Read the symbols in plain language

C
Coefficient CRd,c

CRd,c for this empirical concrete shear branch; the material partial factor is normally included in the selected coefficient.

ratio / no unit

A dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.

k
Size-effect factor

Size-effect coefficient supplied after the code’s depth expression and upper cap have been applied.

ratio / no unit

A dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.

ρl
Longitudinal ratio

Longitudinal tensile reinforcement ratio appropriate to the shear section; supply it after the code’s ratio limit.

ratio / no unit

A dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.

fck
Concrete strength

Specified characteristic compressive cylinder strength, not cube strength; use the strength class and age required by the model.

MPa

One megapascal equals one N/mm² and 1000 kPa.

k1
Compression coefficient

Coefficient multiplying the mean axial compression contribution to this shear-stress branch.

ratio / no unit

A dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.

σcp
Mean compression

Mean compressive stress from the supplied axial force divided by concrete area, subject to the code limit.

N/mm²

One N/mm² equals one MPa.

bw
Web width

Web width. Multiplying stress by web width and effective depth converts this branch into a force.

mm

Millimetres measure length; 1000 mm = 1 m.

d
Effective depth

Distance from the extreme compression face to the centroid of tensile reinforcement; do not substitute the overall section depth.

mm

Millimetres measure length; 1000 mm = 1 m.

VRd,c
Result to find

Concrete shear resistance — coefficient form. This converted force still requires the separate minimum and applicability checks described above.

kN

Sort out the units first

Use fck in MPa, σcp in N/mm² and bw and d in mm. ρ is a decimal ratio: 1% means 0.01. The calibrated empirical constants require these units; divide the final newtons by 1000 for kN.

Assumptions before calculating

Use the first-generation EC2 teaching model and the supplied design coefficients. Material strengths, geometry, load situation and coefficients must be mutually compatible; selecting them from the adopted code and National Annex is outside this calculation.

This is a first-generation Eurocode teaching relationship or an explicitly simplified coefficient calculation. The numbers supplied here are exercise data, not a recommendation for any country. Check the adopted edition, relevant clause, National Annex, applicability conditions and all other limit states before any real design.

02

Let’s solve one together

Read the given values, follow each operation, then check what the result means.

Read the supplied values as one complete study case. Find VRd,c and explain the result in the stated output unit.

C · Coefficient CRd,c
0.12
k · Size-effect factor
1.8
ρl · Longitudinal ratio
0.01
fck · Concrete strength
30 MPa
k1 · Compression coefficient
0.15
σcp · Mean compression
0 N/mm²
bw · Web width
300 mm
d · Effective depth
550 mm

  1. Find the strength–reinforcement cube-root term

    Use the decimal reinforcement ratio and the calibrated MPa strength before taking the cube root.

    (100 × (0.01) × (30))^(1 ÷ 3) ≈ 3.107232506 empirical MPa basis

  2. Form the supplied shear-stress expression

    Add the concrete-size contribution and the supplied axial-compression contribution.

    (0.12) × (1.8) × (3.107232506) + (0.15) × (0) ≈ 0.6711622213 N/mm²

  3. Apply the effective web area

    Multiplying stress by web width and effective depth converts this branch into a force.

    (0.6711622213) × (300) × (550) ≈ 110741.7665 N

  4. Report the main-branch value

    This converted force still requires the separate minimum and applicability checks described above.

    (110741.7665) ÷ 1000 ≈ 110.7417665 kN
Answer110.7417665 kN

Does this worked answer make sense?

With all other values fixed, doubling section area bw d doubles the calculated force only if k and ρ are also held fixed. The concrete term grows with a cube root, not directly with fck.

A second worked example — different values

A second case uses different data. Predict which way the answer will change, then calculate it without reusing the first answer.

C · Coefficient CRd,c
0.12
k · Size-effect factor
1.7
ρl · Longitudinal ratio
0.012
fck · Concrete strength
35 MPa
k1 · Compression coefficient
0.15
σcp · Mean compression
1 N/mm²
bw · Web width
300 mm
d · Effective depth
600 mm

  1. Find the strength–reinforcement cube-root term

    Use the decimal reinforcement ratio and the calibrated MPa strength before taking the cube root.

    (100 × (0.012) × (35))^(1 ÷ 3) ≈ 3.476026645 empirical MPa basis

  2. Form the supplied shear-stress expression

    Add the concrete-size contribution and the supplied axial-compression contribution.

    (0.12) × (1.7) × (3.476026645) + (0.15) × (1) ≈ 0.8591094356 N/mm²

  3. Apply the effective web area

    Multiplying stress by web width and effective depth converts this branch into a force.

    (0.8591094356) × (300) × (600) ≈ 154639.6984 N

  4. Report the main-branch value

    This converted force still requires the separate minimum and applicability checks described above.

    (154639.6984) ÷ 1000 ≈ 154.6396984 kN
Answer154.6396984 kN
03

Now try your own values

Change a value or its unit. The same method will show your calculation, step by step.

CRd,c for this empirical concrete shear branch; the material partial factor is normally included in the selected coefficient.

Size-effect coefficient supplied after the code’s depth expression and upper cap have been applied.

Longitudinal tensile reinforcement ratio appropriate to the shear section; supply it after the code’s ratio limit.

Specified characteristic compressive cylinder strength, not cube strength; use the strength class and age required by the model.

Coefficient multiplying the mean axial compression contribution to this shear-stress branch.

Mean compressive stress from the supplied axial force divided by concrete area, subject to the code limit.

Web width. Multiplying stress by web width and effective depth converts this branch into a force.

Distance from the extreme compression face to the centroid of tensile reinforcement; do not substitute the overall section depth.

English, Arabic and Persian digits are supported. The steps convert inputs to the formula’s base units.

Results update only when you calculate. The lesson example above stays unchanged.

04

Your turn — check your understanding

Solve this separate case yourself. Use only the values below; the two worked examples use different data. Give the requested result in the selected unit.

C · Coefficient CRd,c
0.12
k · Size-effect factor
1.8
ρl · Longitudinal ratio
0.008
fck · Concrete strength
25 MPa
k1 · Compression coefficient
0.15
σcp · Mean compression
0.5 N/mm²
bw · Web width
250 mm
d · Effective depth
500 mm

Find: Learn: Concrete shear resistance — coefficient form

For repeating decimals, use at least four significant figures. Accepted rounding tolerance: 0.05% of the expected value; zero uses an absolute tolerance of 10⁻¹².

A hint, not the answer

First take the cube root of 100ρfck, scale it by C and k, and add k1σcp. That produces a shear-stress expression; multiplying by bw d turns it into a force.

Use fck in MPa, σcp in N/mm² and bw and d in mm. ρ is a decimal ratio: 1% means 0.01. The calibrated empirical constants require these units; divide the final newtons by 1000 for kN.

Show the full practice solution

Compare the steps with your work; revealing a solution does not mark the lesson complete.

  1. Find the strength–reinforcement cube-root term

    Use the decimal reinforcement ratio and the calibrated MPa strength before taking the cube root.

    (100 × (0.008) × (25))^(1 ÷ 3) ≈ 2.714417617 empirical MPa basis

  2. Form the supplied shear-stress expression

    Add the concrete-size contribution and the supplied axial-compression contribution.

    (0.12) × (1.8) × (2.714417617) + (0.15) × (0.5) ≈ 0.6613142052 N/mm²

  3. Apply the effective web area

    Multiplying stress by web width and effective depth converts this branch into a force.

    (0.6613142052) × (250) × (500) ≈ 82664.27565 N

  4. Report the main-branch value

    This converted force still requires the separate minimum and applicability checks described above.

    (82664.27565) ÷ 1000 ≈ 82.66427565 kN
Answer82.66427565 kN

Avoid the common trap

Do not omit the cube root or enter ρ as a whole percentage. Do not assume the returned main-branch force is the governing complete code resistance, and do not treat d in metres when selecting a millimetre-based size factor.

When this method applies — and when it does not

This is only the stated main branch, not the full VRd,c check: the minimum-resistance branch, limits on σcp, k selection, anchorage, axial stress and member-specific requirements remain outside it. The lesson restricts k to 1–2 and ρ to 0–0.02 for the stated first-generation teaching range.

For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.

One idea understood. Keep going.

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Sources & further reading

References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.

Reading focus: Concrete shear resistance — coefficient form. First-generation EN 1992 teaching: material properties and the relevant bending, shear, serviceability, detailing or prestress relationship. Read the applicability conditions as well as the expression.

Lesson updated: · Both examples and the separate practice case are checked against an independent high-precision numerical implementation. This verifies arithmetic for the stated model, not engineering certification.

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