Calculation of floor vibration in structural engineering (part 2)
As introduced in Part 1, this part presents a practical way to calculate the design with higher detail and with proper numerical basis instead of looking up the diagram. This method is guided in detail in the construction design guide document SCI P354 “Design of Floors for Vibration: a new Approach” (2009) (hereinafter referred to as the Document).
Excluding external causes (vehicles on the road) in construction design. Considering the inside of the building, vibration is caused by:
🔲Walking
Statistics show that the normal walking frequency range is 1.8Hz≤ fp ≤2.2Hz. In enclosed spaces, fp = 1.8Hz is often used in design consulting. The reason is that the short distance in the rooms leads to slower walking speed.
– Human walking causes repeated forces to act on the floor, considered cyclical, continuous action (Steady-state)
This diagram can be represented by the first four harmonic components. The force acting on the floor for each harmonic is:
$F_h = \alpha_hQ$, where
+ $\alpha_h $: coefficient corresponding to the hth harmonic, according to table 3.1 of the Document
+ Q: average static load caused by a person (weighing 76kg) Q= 76kg.9.81m/s2 =746N
– In addition to the continuous effect, each step also creates an impulse force equivalent to $f_p$, the nature of the effect is temporary. Determine $f_p$ according to formula (18) of the Document.
🔍Harmonic:
borrowing the concept of Acoustics for this English word, because sound is also a mechanical oscillation. Any physical object always has a natural oscillation at the lowest frequency, called the fundamental frequency fp, and also oscillates at harmonic frequencies, which are multiples (integer times) of fp. Therefore: $f_h = hf_p$
Why should the design engineer be concerned with Harmonics and not just fp? Because the natural frequency of the floor may be greater than fp, but it is inevitable that it will resonate with Harmonics at frequencies greater than fp, causing an unpleasant feeling of intense vibration, even destroying the structure.
🔲Stairway travel
The foot load when walking on stairs is significantly higher than when walking on a flat surface. The foot frequency is also larger, 3-4.5Hz. Only the first two harmonics are considered when determining the dynamic load on the staircase structure, according to Table 3.2 of the Document.
🔲Dancing
Public floors are used for rhythmic activities of large numbers of people. The frequency range of these activities is as follows:
– Individual: 1.5Hz – 3.5Hz
– Group: 1.5Hz – 2.8Hz
The density of the load according to the following statistics:
– Aerobic, gym: 0.25 people/m2
– Jumping, dancing: 2.00 people/m2
The nature of this type of load is that there is a contact force for a period of time when the foot touches the floor, followed by a force of zero when the foot is lifted. The force function is therefore considered a sinusoidal series as in formula (19) of the Document. Its graph is shown in Figure 2 according to each contact coefficient αc, depending on the time the foot contacts the floor depending on the type of activity. A low αc value represents the intensity of the activity.
Rhythmic loads on the floor with different coefficients αc
Rhythmic operations can cause dynamic loads many times greater than static loads on the floor, so care must be taken when designing the structure.
2. Design practice
1️⃣ Create a structural design diagram using the finite element method
Use familiar construction design software such as Etabs, Safe… Note that to calculate vibrations, the calculation diagram of the floor will be different from the calculation diagram when working normally:
– Use the dynamic elastic module of concrete when calculating natural vibrations: take 38,000MPa for normal heavy concrete, 22,000MPa for lightweight concrete.
– Use the Shell element to model the floor. With a concrete-steel composite floor, the correct distance between the concrete floor and the steel beam should be modeled as shown in Figure 4 below
– The model of the connection nodes is all rigid (clamped), although in the normal working diagram, the design is a joint. The reason is that when oscillating, the deformation of the structure is very small and not enough to overcome the frictional force of the connections.
– Using mobile bearing connections to prevent vertical displacement for the edges of the floor with partition walls.
– The connection of the floor with the core wall structure is considered a rigid clamp
– The mass participating in the oscillation: includes the self-weight, the permanent load and the long-term part of the temporary load (live load) on the floor. This long-term part can be taken according to TCVN 2737:1995, for example, 30daN/m2 (20% of the total live load) for houses. The document recommends that this long-term part does not exceed 10% of the total live load.
2️⃣ Oscillation data
– Damping coefficient ζ, taken according to section 4.1.1 of the Document. Example ζ =3% with composite floor
– Oscillation frequency $f_n$
– Participating mass in oscillation: $M_n = r.m_s $
$m_s$: Etabs/ output Table/ Model Definition/ Other Definitions/ Mass Data/ Table: Mass Summary by Story
If the first fundamental mode (usually mode 1) has a frequency less than $f_{cut−off}$ given in table 6.1, the transient and steady-state responses need to be calculated. Otherwise, the design engineer only needs to consider the transient response of the floor.
+ Transient response: calculated for all modes with frequencies less than or equal to 2 times the first fundamental frequency
+ Steady-state response: calculated for all modes with frequencies less than or equal to $(f_ {cut-off}+2Hz) $
– Displacement of oscillation modes at the point of excitation $u_ {z, E} $ and response point $u_ {z, r} $.
According to clause 6.3 of the Document, it is assumed that the vibration force acts on the most dangerous point of the floor even though in reality the path only passes through that point for a short time.
The most dangerous point is the point that is both the excitation and the sensing point, at the point with the largest displacement corresponding to the vibration mode with the largest energy (largest r).
Normalize these displacements for calculation as follows:
$\mu_{e,n} = U_{z,e}/U_{z,max}$
$\mu_{r,n} = U_{z,r}/U_{z,max}$
$U_ {z, max} $ is the maximum absolute displacement corresponding to each mode. Therefore, mode 1 usually has $\mu_ {e, n} = \mu_ {r, n} = 1 $.
3️⃣ Calculation of the Reaction Coefficient R
💎Steady reaction: the average reaction acceleration value at the reaction point r, due to excitation from point e, corresponding to the $n$th oscillation mode and the $h$th harmonic of the excitation force – $a_ {w,rms,e,r,n,h} $.
Calculate the total combined acceleration using the root mean square method (SRSS) according to the oscillation modes n and the harmonics h to obtain the value of Reaction acceleration $a_{w,rms}$.
💎 Temporary reaction: The maximum reaction acceleration value at the reaction r, due to stimulation from point e, corresponding to the form of n -oscillation – $a_ {w, peak, e, r, n} $
Because the data is relatively large, it is possible to use VBA code for spreadsheets to automatically calculate the loops in the form of oscillations (see attached file for example) to automate structural design.
Reaction coefficient $R = a_ {q, rms}/0.005 $ for the vertical and
$R = a_ {w, rms}/0.00357 $ for horizontal
4️⃣Access the floor vibration
the Structural engineer use the reaction coefficient R to evaluate whether the floor structure meets the vibration requirements or not, in other words, the vibration is not large enough to cause discomfort or loss of comfort.
This is according to the Standard, BS 6472:2008 – assessing the impact of vibration on people in the building, the value of R does not exceed the limit in Table 5.2 of the Document is satisfactory.
In case R is greater than the limit, consider the operation coefficient VDV (vibration dose value).
The practical way for design engineers is to pre-select the VDV value within the allowable range according to Table 5.4 of the Document, for example VDV = 0.3 for an office building operating 16 hours during the day.
Then calculate $n_a$ – the number of times the activity (walking, jumping) is allowed to occur during the above time (for example 16 hours) without causing discomfort due to floor vibration. See formula (41) of the Document.
If $n_a$ is very large, the VDV condition is satisfied, the floor is still considered to ensure vibration comfort.
5️⃣ Vibration treatment
If $n_a$ is too small or equal to 0, consider adjusting the architectural layout, arranging the path to change the values of $\mu_{e,n}, \mu_{r,n}$ and recalculate from the beginning.
If the work according to the diagram of the above practical steps does not meet the vibration requirements, it is necessary to make modifications to the construction design or renovate the project if it has been completed.
3. Practical example:
🎉Still the composite floor project in part 1, apply the method according to SCI P354 with the calculation files as attached in the document
– Current floor:
Calculated stable response with 6 modes, temporary response with the first 3 modes. The reaction coefficient R is very large and $n_a$=0 for both cases. The floor does not ensure vibration comfort, just like the actual feeling on site.
– After renovation, add 2 more steel columns to hang the floor:
Calculated stable response with 4 modes, temporary response with the first 7 modes.
Stable response coefficient R = 5.5 < 8. Temporary response coefficient R = 56.9 < 128. Both are smaller than the allowable limits according to the Document. Therefore, ensuring comfort in terms of vibration. In actual use, almost no vibration is felt.
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🍺Conclusion
– Vibration design has become more accurate and productive thanks to finite element structural design software tools.
– The method of part 1 can be used for the Basic Design phase, and the more accurate method of this part can be used in the Technical design phase, Construction drawing design, and vibration assessment and correction 👍.
– Economic significance 💸: lightweight structures, quick construction, serving renovation, and can be recovered after use such as steel floor structures, concrete-steel composites are often chosen by design consultants because they are suitable for economic problems. Calculating with specific numbers helps to quantify the optimal materials to meet the load-bearing and vibration comfort requirements with the lowest construction cost for investors.
