Learn: Timber design strength
Timber strength depends on load duration and moisture conditions as well as the characteristic grade. This teaching equation adjusts the relevant characteristic strength by kmod and divides by the timber material partial factor.
What the formula is saying
The modification factor belongs to the chosen service class and load-duration class. Apply it to the specific characteristic property, such as bending or compression parallel to grain, before applying γM.
Read the symbols in plain language
- kmod
- Modification factor
kmod modifies strength for load duration and service class; do not use kdef, which concerns deformation.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- fk
- Characteristic strength
Statistically defined characteristic property for the selected material/model. Do not replace it with a mean or design value.
MPaOne megapascal equals one N/mm² and 1000 kPa.
- γM
- Material factor
Supplied material/resistance partial factor in the divisor. Select its code clause and National Annex before real design.
ratio / no unitA dimensionless ratio has no physical unit; 0.01 as a ratio is 1% when the percent option is selected.
- fd
- Result to find
Timber design strength. Division gives the design strength for this property and this stated combination of conditions.
MPa
Sort out the units first
fk and fd are MPa. kmod and γm are dimensionless. A strength perpendicular to grain is a different property from strength parallel to grain, even though their units match.
Assumptions before calculating
Use compatible timber product, strength class, grain direction, service class and duration class. The supplied factors follow the selected first-generation EC5 teaching convention, not an automatic product approval.
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 fd and explain the result in the stated output unit.
- kmod · Modification factor
- 0.8
- fk · Characteristic strength
- 24 MPa
- γM · Material factor
- 1.3
Apply the duration-and-service modifier
Modify the characteristic strength for the specified load duration and service conditions.
(0.8) × (24) = 19.2 MPaApply the timber partial factor
Division gives the design strength for this property and this stated combination of conditions.
(19.2) ÷ (1.3) ≈ 14.76923077 MPa
Does this worked answer make sense?
At fixed fk and γm, a larger kmod increases fd. A lower design strength does not necessarily mean a different timber grade; it can reflect a different duration or moisture condition.
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.
- kmod · Modification factor
- 0.6
- fk · Characteristic strength
- 24 MPa
- γM · Material factor
- 1.3
Apply the duration-and-service modifier
Modify the characteristic strength for the specified load duration and service conditions.
(0.6) × (24) = 14.4 MPaApply the timber partial factor
Division gives the design strength for this property and this stated combination of conditions.
(14.4) ÷ (1.3) ≈ 11.07692308 MPa
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.
- kmod · Modification factor
- 0.9
- fk · Characteristic strength
- 28 MPa
- γM · Material factor
- 1.3
Find: Learn: Timber design strength
A hint, not the answer
The modification factor belongs to the chosen service class and load-duration class. Apply it to the specific characteristic property, such as bending or compression parallel to grain, before applying γM.
fk and fd are MPa. kmod and γm are dimensionless. A strength perpendicular to grain is a different property from strength parallel to grain, even though their units match.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Apply the duration-and-service modifier
Modify the characteristic strength for the specified load duration and service conditions.
(0.9) × (28) = 25.2 MPaApply the timber partial factor
Division gives the design strength for this property and this stated combination of conditions.
(25.2) ÷ (1.3) ≈ 19.38461538 MPa
Avoid the common trap
Do not use concrete or steel partial factors for timber. Do not confuse kmod, a strength modifier, with kdef, a deformation factor, or choose the wrong grain-direction strength.
When this method applies — and when it does not
Size effects, system factors, instability, notches, connections, fire, creep and other strength modifications are not included unless already part of the supplied model. Do not transfer the example factors to an unrelated timber product.
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: Timber design strength. Timber design values and bending relationships: distinguish the strength modifier kmod from the deformation factor kdef, and apply the correct service class and load duration.
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.