Learn: Masonry compression resistance
A masonry wall’s design compressive strength must be applied to an effective area and reduced for the specified wall behaviour. This equation evaluates the supplied compression-resistance model using a reduction factor Φ.
What the formula is saying
Area times fd gives the reference compressive force. Multiplying by Φ reduces that force for the effects represented by the supplied factor, such as the chosen slenderness and eccentricity model.
Read the symbols in plain language
- Φ
- Reduction factor
Reduction factor. The supplied factor accounts for the wall behaviour represented in the selected model.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- A
- Loaded area
Loaded area. Multiply the appropriate wall area by its already established design compressive strength.
mm²Square millimetres measure area; 1 mm² = 10⁻⁶ m².
- fd
- Design masonry strength
Design masonry strength. Multiply the appropriate wall area by its already established design compressive strength.
N/mm²One N/mm² equals one MPa.
- NRd
- Result to find
Masonry compression resistance. Report this compression component in kilonewtons without treating it as a full wall approval.
kN
Sort out the units first
A is in mm² and fd in N/mm², so the force is N before division by 1000 for kN. Φ is a decimal reduction factor from above zero to 1; it is not an angle.
Assumptions before calculating
The supplied effective area and reduction factor must belong to the wall section and adopted first-generation EC6 method being checked. fd is already a design strength.
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 NRd and explain the result in the stated output unit.
- Φ · Reduction factor
- 0.8
- A · Loaded area
- 200000 mm²
- fd · Design masonry strength
- 2.5 N/mm²
Find the reference compression force
Multiply the appropriate wall area by its already established design compressive strength.
(200000) × (2.5) = 500000 NApply the wall reduction factor
The supplied factor accounts for the wall behaviour represented in the selected model.
(0.8) × (500000) = 400000 NConvert the wall resistance component
Report this compression component in kilonewtons without treating it as a full wall approval.
(400000) ÷ 1000 = 400 kN
Does this worked answer make sense?
The reduced resistance cannot exceed A fd when 0 < Φ ≤ 1. Halving Φ halves the force without changing the masonry material strength itself.
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.
- Φ · Reduction factor
- 0.65
- A · Loaded area
- 180000 mm²
- fd · Design masonry strength
- 2 N/mm²
Find the reference compression force
Multiply the appropriate wall area by its already established design compressive strength.
(180000) × (2) = 360000 NApply the wall reduction factor
The supplied factor accounts for the wall behaviour represented in the selected model.
(0.65) × (360000) = 234000 NConvert the wall resistance component
Report this compression component in kilonewtons without treating it as a full wall approval.
(234000) ÷ 1000 = 234 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.
- Φ · Reduction factor
- 0.75
- A · Loaded area
- 240000 mm²
- fd · Design masonry strength
- 2.5 N/mm²
Find: Learn: Masonry compression resistance
A hint, not the answer
Area times fd gives the reference compressive force. Multiplying by Φ reduces that force for the effects represented by the supplied factor, such as the chosen slenderness and eccentricity model.
A is in mm² and fd in N/mm², so the force is N before division by 1000 for kN. Φ is a decimal reduction factor from above zero to 1; it is not an angle.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Find the reference compression force
Multiply the appropriate wall area by its already established design compressive strength.
(240000) × (2.5) = 600000 NApply the wall reduction factor
The supplied factor accounts for the wall behaviour represented in the selected model.
(0.75) × (600000) = 450000 NConvert the wall resistance component
Report this compression component in kilonewtons without treating it as a full wall approval.
(450000) ÷ 1000 = 450 kN
Avoid the common trap
Do not apply the material partial factor again to fd. Do not use gross area when the method requires an effective or net area, or assume Φ = 1 without checking the wall model.
When this method applies — and when it does not
The calculator does not derive Φ, establish effective height/thickness, verify eccentricity, openings, concentrated loads, lateral stability or execution quality. These can govern the wall separately.
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: Masonry compression resistance. Unreinforced masonry material and resistance relationships. Unit groups, mortar type, permitted coefficients and reduction factors must correspond to the selected model.
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.