Deformations in timber structures according to Eurocode 5

[Ahmada]

Hello,

I have a question regarding a section in Eurocode 5 page 22, where it is stated that the psi_2 value should be ignored in load combination 6.16b when calculating the final deformations.

This means, if I interpret the section correctly, that the creep deformations are calculated with a characteristic value (6.16b without psi_2 becomes the same as 6.14b). This implies that the creep deformations are calculated to be as large as the initial deformations.

This is confusing to me, as I can't find any source where anyone does a calculation this way; everyone considers the psi_2 value when calculating creep deformations according to equation 6.16b.

Should one follow the Eurocodes and calculate creep deformations with characteristic values (6.16b without psi_2)? How do you calculate creep deformations?

 

[Ahmada]

Hi,

I have a question regarding a section in Eurocode 5 page 22, where it states that the psi_2 value should be ignored in load combination 6.16b. When the final deformations are calculated.

This means, if I interpret the section correctly, that creep deformations are calculated with a characteristic value (6.16b without psi_2 is the same as 6.14b). Which therefore means that the creep deformations are calculated to be as large as the initial deformations.

This confuses me, as I cannot find any source where someone makes a calculation in this way, everyone considers the psi_2 value when calculating creep deformations according to equation 6.16b.

Should one follow the Eurocodes and calculate creep deformations with characteristic values (6.16b without psi_2)? How do you calculate creep deformations?

Update: I have now received a response from another English forum that someone else reading the thread might find useful.

In large parts of Sweden (especially in Skåne where I operate), we do not need to follow the advice in Eurocode 5 to calculate with equation 6.16b without psi_2. This is because we have snow so few months of the year that the contribution to creep deformations due to snow loads becomes negligible.

So in this case, which is located in Skåne, we can consider psi_2 when calculating creep deformations.

Any structural engineer working in Sweden is welcome to give their perspective.

 

[witten]

I think you're interpreting the passage incorrectly. If you are going to calculate the deformation according to 2.2.3(3), i.e., load combination 6.16b, you should not omit psi 2.

However, if you calculate according to (5), you should omit psi 2, as the factor is already incorporated into expressions (2.4) and (2.5).

(In plain language, these equations state that you calculate the final deformation from the instantaneous deformation with characteristic values + a long-term deformation that is a product of the instantaneous deformation times psi 2 and the k-factor.)

I don't immediately see the point of using 2.2.3(3) instead of 6.16b. I don't think it's a simplification as it's portrayed, but I haven't thought deeply about this, so there might be good reasons for the option to exist. Feel free to share if you see the benefit of it.

 

[Ahmada]

I think you're interpreting the section incorrectly. If you are calculating the deformation according to 2.2.3(3), i.e., load combination 6.16b, you should not exclude psi 2.

But if you're calculating according to (5), you should exclude psi 2, as the factor is already embedded in expressions (2.4) and (2.5).

(In plain language, these equations are saying that you calculate the final deformation based on the instantaneous deformation with characteristic values + a long-term deformation that is a product of the instantaneous deformation times psi 2 and k-factor.)

I don't immediately see the point of calculating with 2.2.3(3) instead of 6.16b. I don't think it's a simplification as it's portrayed, but I haven't thought deeply about this, so there might be good reasons for the possibility. Feel free to share if you see the benefit of it.

Thank you very much for your response, it all "makes sense" now! Should one then calculate u_inst with eq. 6.16b (including psi_2) and then multiply with 1 + k_def to get the total deformation? I had the idea that you calculated "instant deformation" with equation 6.14b and creep deformation with 6.16b. But you can't do that if the (1+k_def) method is to be applied, instead, it must be the same load combination for "instant deformation" as for the creep.

Regarding why they use a more challenging "simplifying" method, I don’t know either. It might be easier if using software like MathCad perhaps?

 

[Ahmada]

Thank you very much for your reply, the whole "makes sense" now! Should one then calculate u_inst with eq 6.16b (including psi_2) and then multiply by 1 + k_def to get the total deformation? I thought you calculated "instant deformation" with equation 6.14b and creep deformation with 6.16b. But you can't do that if the (1+k_def) method is to be applied without using the same load combination for "instant deformation" as for creep.

Regarding why they use a more difficult "underlättrande" method, I don't know either, possibly it facilitates if you calculate with software like MathCad maybe?

Now I double-checked with Eurocode 5 and saw that it separates G and Q loads in the simplified expressions. Now it makes sense to me, I understand the connection. Thank you once again for the help.

 

[witten]

Glad you got it sorted Eurokod is unnecessarily complicated in many places.

 

[Ahmada]

It's great that you got it sorted Eurocode is unnecessarily complicated in many places.

I have to agree with that!