Learn: Punching shear design stress
A concentrated column load can punch through a slab around a control perimeter. This calculation spreads the design shear, adjusted by a supplied eccentricity factor, over the idealized vertical area u d.
What the formula is saying
The denominator is perimeter times effective slab depth, not the horizontal area enclosed by the perimeter. The factor β accounts for the specified nonuniformity model before the average stress is calculated.
Read the symbols in plain language
- β
- Load eccentricity factor
Load eccentricity factor. The supplied eccentricity factor scales the shear demand before it is spread over the perimeter.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- VEd
- Design shear
Design shear. The supplied eccentricity factor scales the shear demand before it is spread over the perimeter.
NNewtons measure force; 1000 N = 1 kN.
- ui
- Control perimeter
Chosen code control perimeter around the loaded region; this is a length, not the loaded area.
mmMillimetres 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.
mmMillimetres measure length; 1000 mm = 1 m.
- vEd
- Result to find
Punching shear design stress. Divide adjusted demand by the vertical area; comparison with resistance is a separate step.
N/mm²
Sort out the units first
Use V in N and both u and d in mm. The result is N/mm², equal to MPa. A column load given in kN must be converted or entered with the correct selected unit.
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 vEd and explain the result in the stated output unit.
- β · Load eccentricity factor
- 1.15
- VEd · Design shear
- 800000 N
- ui · Control perimeter
- 6000 mm
- d · Effective depth
- 250 mm
Adjust the design shear
The supplied eccentricity factor scales the shear demand before it is spread over the perimeter.
(1.15) × (800000) = 920000 NFind the vertical control area
Multiply perimeter length by effective depth to obtain the idealized shear-transfer area.
(6000) × (250) = 1500000 mm²Calculate punching shear demand stress
Divide adjusted demand by the vertical area; comparison with resistance is a separate step.
(920000) ÷ (1500000) ≈ 0.6133333333 N/mm²
Does this worked answer make sense?
For the same adjusted force, doubling u or d halves the stress. Increasing β increases demand; it does not improve the slab’s resistance.
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.
- β · Load eccentricity factor
- 1.4
- VEd · Design shear
- 600000 N
- ui · Control perimeter
- 5000 mm
- d · Effective depth
- 220 mm
Adjust the design shear
The supplied eccentricity factor scales the shear demand before it is spread over the perimeter.
(1.4) × (600000) = 840000 NFind the vertical control area
Multiply perimeter length by effective depth to obtain the idealized shear-transfer area.
(5000) × (220) = 1100000 mm²Calculate punching shear demand stress
Divide adjusted demand by the vertical area; comparison with resistance is a separate step.
(840000) ÷ (1100000) ≈ 0.7636363636 N/mm²
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.
- β · Load eccentricity factor
- 1.15
- VEd · Design shear
- 900000 N
- ui · Control perimeter
- 6500 mm
- d · Effective depth
- 260 mm
Find: Learn: Punching shear design stress
A hint, not the answer
The denominator is perimeter times effective slab depth, not the horizontal area enclosed by the perimeter. The factor β accounts for the specified nonuniformity model before the average stress is calculated.
Use V in N and both u and d in mm. The result is N/mm², equal to MPa. A column load given in kN must be converted or entered with the correct selected unit.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Adjust the design shear
The supplied eccentricity factor scales the shear demand before it is spread over the perimeter.
(1.15) × (900000) = 1035000 NFind the vertical control area
Multiply perimeter length by effective depth to obtain the idealized shear-transfer area.
(6500) × (260) = 1690000 mm²Calculate punching shear demand stress
Divide adjusted demand by the vertical area; comparison with resistance is a separate step.
(1035000) ÷ (1690000) ≈ 0.6124260355 N/mm²
Avoid the common trap
Do not use the column perimeter automatically if the check requires another control perimeter. Do not use overall thickness instead of effective depth, or treat the calculated demand stress as a resistance.
When this method applies — and when it does not
The control perimeter, effective depth, net shear and β must already be established for the actual geometry. Openings, edge and corner columns, soil reaction deductions, maximum punching stress and reinforcement resistance are not determined here.
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: Punching shear design stress. 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.