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If we simplify the whole bridge into 2D thin beam with a constant section size, no internal damping and subject only to small vertical deflections, then the natural frequency is determined by simple harmonic motion: $$n_0 = \frac{1}{2 \pi} \sqrt{ \frac{ k } { m } }$$ Where $n_0$ is the natural frequency, $k$ is the ratio between restorative force and ...

6

In case Eurocodes do not provide enough information, some sources exist. In the case of elastic critical moment for lateral-torsional buckling, an NCCI (Non-contradictory, complementary information) document exists. The document code is SN 003, and one version (maybe not the latest) can be accessed here. Hopefully, this will cover your current needs. In ...

3

As @Wasabi and @AndyT mentioned in the comments, this means the maximum value $k$ can have is $2$, so if $d<200$ then $k=2$. This can be confirmed in the section on shear design for slabs from the document 'How to Design Concrete Structures to Eurocode 2'. Table 7 from this document shows: which has $V_{Rd,c}$ unchanged for effective depths $d<200$ ...

3

Here's a possible answer. I found this document (not sure of the exact source), which contains a related derivation: In a simple harmonic motion problem, $$n_0=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}$$ where $k$ is the elastic stiffness and $m$ is the mass undergoing vibration. $$k=\frac{\text{load}}{\text{deflection}}=\frac{F}{\delta}$$ where $F$ is force and ...

3

I know it's tagged Eurocodes, but another answer for the Americans: specification section F of the AISC manual (pg. 16.1-44 of the 13th edition) delineates the equations applicable to lateral-torsional buckling. For the usual case of doubly symmetric I shapes bent about the major axis (and normal, "compact" flanges), it's section F2 (pg. 16.1-47). However, ...

2

There are three options: 1) Using paragraph 6.3.2.4 "Simplified assessment methods form beams with restraints in buildings" Which gives a conservative way to design a beam resistant to lateral torsion buckling. 2) Looking in the Apendices and Supplementary informations of the National Annex for the country in question. In the Belgian annex, for example, ...

2

For a flat bar - there is no local buckling, only global buckling. For a plate you follow the rules for plated sections. In Europe that would be EC3-1-5. The local design code should define the limits at which a flat bar becomes a plate. (I can't remember off the top of my head where in the eurocodes this would be found.)

2

It isn't clear from your sketch whether the end supports are fixed or pinned, so I'm going to assume they were pinned. However, this is going to be a qualitative answer, so that doesn't actually matter too much. Assuming pinned supports, the original slab was therefore basically a simply-supported beam, with no moment at the ends and positive longitudinal ...

2

One important thing to remember about the strut-and-tie method is that as a lower-bound method it is based on the plasticity theory. That is, the principle is to find a safe and statically admissible stress distribution, and if any such distribution is possible the structure will not collapse at this load. This principle only makes sense for the ultimate ...

1

I am not sure which version of Eurocode you have this (I have access to an older version circa 2002), and 6.10a and b have a slightly different form. In any case I will answer and I will later update my answer. Equation A: Dominant secondary actions and combination of transients loads First of all: equation a is not only one equation. It represents all ...

1

In your sketch, there seems to be no off-center loading so no torque. But if we had a slab sitting on say the lip of the lower flange then we had torque, $$t=w*x (\text{w is slab weight}/ ft \ , \ x= offset),$$ This torque would rotate the fix- fix (fix for rotation )beam by this equation: $$\theta_{max-at- center}= \frac{t*L^2}{8*G*j}$$ G = Shear ...

1

You've calculated the Euler buckling force, not the compressive strength. The part of the code you want is EN 1993-1-1 section 6.3.1, which specifies how to account for imperfections and safety factor in order to calculate the allowable load. The size of the imperfections depend on whether the profile is cold formed or hot finished and the relevant safety ...

1

The root problem of your question is that you're asking us to make some engineering judgement calls and calculations for a building where we don't know all the relevant details. You could however calculate the slenderness of the existing column and then propose a section that has this slenderness or less. The slenderness parameter that is commonly used here ...

1

There are two major effects that make the elasticity modulus of concrete vary with time: hardening and creep. In a Eurocode context $E_{cm}(t)$ refers to the hardening of concrete until it reaches its 28-day strength. This is section 3.1.3(3). If you need to calculate a crack width at an early stage, you should probably use $E_{cm}(t)$ instead of $E_{cm}$, ...

1

Looking at section "7.3.3 - Control of cracking without direct calculation", we have equations (7.6) and (7.7) which define the maximum bar diameter for crack control: \begin{align} \phi_s &= \phi_s^*\cdot\dfrac{f_{ct,eff}}{2.9}\cdot\dfrac{k_c h_{cr}}{2(h-d)} &&\text{Bending (part of section in compression)}\tag{7.6} \\ \phi_s &= \phi_s^*\...

1

They refer to the factors defined in section 7.3.4 on calculation of crack widths (and whatever corresponding national annex that applies to your case). Section 7.3.3 is intended to be an easy to use summary of section 7.3.4 so that is where you should look for any additional information. Personally I never use section 7.3.3 because the situations where all ...

1

I haven't seen any method of calculation to determine the cross section classification of a plate but SCI advisory note AD 391 & also design manual for steel frame for SAP 2000 all indicate/assume the cross section classification of a plate as class 3. See below link for AD 391: http //www.newsteelconstruction.com/wp/wp-content/uploads/TechPaper/AD391....

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