Learn: Maximum shear compression-strut resistance — study form
Adding more stirrups cannot increase shear resistance indefinitely: the inclined concrete compression strut can crush. This lesson evaluates the stated simplified strut-crushing ceiling for a supplied angle and concrete-strength reduction factor.
What the formula is saying
The numerator bw z ν fcd combines effective dimensions with reduced concrete strength. The angle denominator cot θ + tan θ reflects the geometry of the inclined strut; both terms must be retained.
Read the symbols in plain language
- bw
- Web width
Web width. Apply the specified strength reduction to the effective web-and-lever-arm area.
mmMillimetres measure length; 1000 mm = 1 m.
- z
- Lever arm
Perpendicular distance between the tensile and compressive resultants that form the resisting internal couple.
mmMillimetres measure length; 1000 mm = 1 m.
- ν1
- Strength reduction factor
Strength reduction factor. Apply the specified strength reduction to the effective web-and-lever-arm area.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- fcd
- Design concrete strength
Design concrete strength. Apply the specified strength reduction to the effective web-and-lever-arm area.
N/mm²One N/mm² equals one MPa.
- θ
- Strut angle
Angle between the concrete compression strut and the longitudinal member axis; the calculator converts degrees for trigonometry.
degAngles are entered in degrees; multiply by π/180 for trigonometric calculations in radians.
- VRd,max
- Result to find
Maximum shear compression-strut resistance — study form. Divide by the angular factor and convert newtons to kilonewtons.
kN
Sort out the units first
Use bw and z in mm, fcd in N/mm² and θ in degrees. ν is a dimensionless strength-reduction factor, not Poisson’s ratio here. The result before division by 1000 is N.
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.
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,max and explain the result in the stated output unit.
- bw · Web width
- 300 mm
- z · Lever arm
- 500 mm
- ν1 · Strength reduction factor
- 0.6
- fcd · Design concrete strength
- 20 N/mm²
- θ · Strut angle
- 45 deg
Compute the tangent of the strut angle
Convert the degree input to radians before evaluating the trigonometric function.
tan((45) × π ÷ 180) = 1Combine cotangent and tangent
The full strut geometry requires both reciprocal tangent and tangent contributions.
1 ÷ (1) + (1) = 2Form the reduced concrete force term
Apply the specified strength reduction to the effective web-and-lever-arm area.
(300) × (500) × (0.6) × (20) = 1800000 NCalculate the stated strut-crushing limit
Divide by the angular factor and convert newtons to kilonewtons.
(1800000) ÷ (2) ÷ 1000 = 900 kN
Does this worked answer make sense?
At 45°, tan θ and cot θ both equal 1, so the denominator is 2. For a positive acute angle their sum is at least 2, making 45° the maximum of this isolated angular expression.
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.
- bw · Web width
- 250 mm
- z · Lever arm
- 450 mm
- ν1 · Strength reduction factor
- 0.55
- fcd · Design concrete strength
- 25 N/mm²
- θ · Strut angle
- 30 deg
Compute the tangent of the strut angle
Convert the degree input to radians before evaluating the trigonometric function.
tan((30) × π ÷ 180) ≈ 0.5773502692Combine cotangent and tangent
The full strut geometry requires both reciprocal tangent and tangent contributions.
1 ÷ (0.5773502692) + (0.5773502692) ≈ 2.309401077Form the reduced concrete force term
Apply the specified strength reduction to the effective web-and-lever-arm area.
(250) × (450) × (0.55) × (25) = 1546875 NCalculate the stated strut-crushing limit
Divide by the angular factor and convert newtons to kilonewtons.
(1546875) ÷ (2.309401077) ÷ 1000 ≈ 669.8165232 kN
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
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.
- bw · Web width
- 350 mm
- z · Lever arm
- 550 mm
- ν1 · Strength reduction factor
- 0.6
- fcd · Design concrete strength
- 20 N/mm²
- θ · Strut angle
- 35 deg
Find: Learn: Maximum shear compression-strut resistance — study form
A hint, not the answer
The numerator bw z ν fcd combines effective dimensions with reduced concrete strength. The angle denominator cot θ + tan θ reflects the geometry of the inclined strut; both terms must be retained.
Use bw and z in mm, fcd in N/mm² and θ in degrees. ν is a dimensionless strength-reduction factor, not Poisson’s ratio here. The result before division by 1000 is N.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Compute the tangent of the strut angle
Convert the degree input to radians before evaluating the trigonometric function.
tan((35) × π ÷ 180) ≈ 0.7002075382Combine cotangent and tangent
The full strut geometry requires both reciprocal tangent and tangent contributions.
1 ÷ (0.7002075382) + (0.7002075382) ≈ 2.128355545Form the reduced concrete force term
Apply the specified strength reduction to the effective web-and-lever-arm area.
(350) × (550) × (0.6) × (20) = 2310000 NCalculate the stated strut-crushing limit
Divide by the angular factor and convert newtons to kilonewtons.
(2310000) ÷ (2.128355545) ÷ 1000 ≈ 1085.344977 kN
Avoid the common trap
Do not use only cot θ in the denominator. Do not confuse ν with steel slenderness or Poisson’s ratio, and do not assume satisfying this ceiling also verifies the actual shear reinforcement.
When this method applies — and when it does not
This exact expression omits a separate αcw multiplier, equivalent to taking it as 1. It does not choose code angle limits, adjust web width for ducts, derive ν or verify stirrup resistance. Mathematical angles require 0 < θ < 90°, but not every such angle is code-permitted.
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.
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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: Maximum shear compression-strut resistance — study 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.