# Undesired vibration of a beam under foot traffic

The other day I was in the waiting room of a doctor's office in a new building as part of UCLA medical complex. A new construction as part of multi million dollar extension of existing medical complex in west Los Angeles.

I felt a small tremor and because of California's proximity to San Andreas fault immediately suspected an earth-quake! When the tremor did not go on as I expected and died out I realized it must have been caused by the impact of steps of another patient. I paid more attention and noticed the vibrations are very noticeable specially when the stride of the person resonates with natural frequency of the floor.

Here is my question re. undesired vibration of beams and floors under foot traffic.

Consider we have a beam of steel WF 12x120 with 20 feet span with I(xx) = 1070 inch^4.

What is the deflection under the foot of a man weighing 180 lbs.
Lets assume he is wearing work-boots with negligible flexibility and when he lands on his foot he exert equivalent momentum as if he fell down 2 inches?
And what is the vibration of this beam if we assume 5% damping?

As a comparison static deflection of this beam under same load is. 0.001614 inch = $Wl^3/48EI$

• Just for comparison and to get a feeling of the magnitude of dynamic loading let's compare the static deflection to dynamic deflection calculated by @mg4w at first period or approximately t= 0.11s: $0.171/0.00.1612= 106.08$. or 100 times rounded down!! Aug 12 '16 at 0:41
• apologies there was a units error in the natural frequency calculation (now corrected). I don't work in US units very often ... the ratio you calculated (the Dynamic Amplification Factor) is approximately 50. Aug 12 '16 at 10:46

An exact theoretical analysis of this is rather involved so I will take an approximate approach.

We can idealise the beam as a single degree of freedom (SDOF) dynamic system. The theory behind this can be found in a variety of dynamics textbooks. For example: Biggs - Introduction to Structural Dynamics (1964).

Because the beam has uniform mass and stiffness we can obtain a Load-mass factor $K_{LM} = 0.49$ for a point load in the center of a simply supported beam. This converts the mass and load of a distributed beam into an equivalent SDOF spring-mass system by assuming a mode shape.

For the central point load it is straightforward to show that the stiffness $k = \frac{48EI}{L^3}$.

We can then calculate the circular natural frequency of the equivalent SDOF system: $$\omega_n = \sqrt{\frac{k}{K_{LM}M}}$$ where $M$ is the total mass of the beam.

And the damped natural frequency: $$\omega_d = \omega_n\sqrt{1-\zeta^2}$$ where, $\zeta$ is the damping ratio (in this case we take this as 0.05 since you specify 5% damping).

Taking the situation you describe, we can consider the system as having an initial velocity, which we will determine by conservation of momentum.

Velocity of the man when he impacts the beam: $$v_f = \sqrt{2 a d} = \sqrt{2(32.2 \frac{ft}{s^2})(\frac{2}{12}ft)}=3.3\frac{ft}{s}$$

If he 'sticks' to the beam the initial velocity of the beam-man system is found by conservation of momentum:

$$v_i (0.49 M+m)= v_f m \rightarrow v_i = v_f \frac{m}{m+0.49 M}$$

The displacement of a damped single degree of freedom system subject to an initial velocity is (see a dynamics text for derivation):

$$x(t)=\frac{v_i e^{-\zeta \omega_n t}}{\omega_d} \sin{\left(\omega_d t\right)}$$

so with $$v_i = 3.3 \frac{ft}{s}\frac{180lb}{180lb+0.49 (120\frac{lb}{ft})(20ft)}=1.59 \frac{in}{s}$$ $$k = 48 \frac{EI}{L^3} = \frac{48(29000 ksi)(1070in^4)}{\left[(20ft)(12\frac{in}{ft})\right]^3}=108 \frac{kips}{in}$$ $$\omega_n = \sqrt{\frac{k}{K_{LM}M+m}}=\sqrt{\frac{108\frac{kips}{in}(386\frac{in}{s^2})}{0.49 (2400lb)+180lb}} = 175 \frac{1}{s}$$ $$\omega_d = \omega_n\sqrt{1-\zeta^2} = 175\sqrt{1-0.05^2}=174.8\frac{1}{s}$$

and we can now calculate the beam mid-span displacement x(t): $$x(t)=\frac{v_i e^{-\zeta \omega_n t}}{\omega_d} \sin{\left(\omega_d t\right)} = \frac{1.59 \frac{in}{s} e^{-0.05 (175\frac{1}{s}) t}}{174.8\frac{1}{s}} \sin{\left(174.8\frac{1}{s} t\right)}$$

The response is: The frequency of the loading, for example due to walking, will have a large effect on the response. Especially if the frequency of loading is close to the natural frequency of the element (resonance).

For floor vibrations you may find the steelconstruction.info page on this topic interesting.

• Nice plot of graph. Can I ask you what you used? Aug 11 '16 at 17:55
• The graph was made using Mathematica Aug 12 '16 at 10:51

I am not an expert in these type calculations, but I do know it very much depends on the natural frequency of the floor as a system; not just assumptions on one of the supports. Typically knowing and avoiding the natural frequency at any energy input level is more important than knowing the energy inputs at various frequencies. It may be that the natural frequency of the floor system, at a particular location/area on the floor, coincided with the frequency of a person walking (~1 to 2Hz) in which case there would be very little dampening (as you may have witnessed).

This could be addressed by adding additional mass to the floor in the problem location in order to increase inertia and lower the natural frequency. To determine how much mass to achieve a natural frequency of 0.5Hz for example will require experiment and/or some serious calculations and/or FEA analysis. Maybe that is why there are so many of those oddly placed giant potted plants in modern architecture ;-)

• I used to practice in CA up to a few years ago and am familiar with the competing ideas in California's building code favoring less mass versus comfort! Mass is the most undesirable property of a building at the time of a quake, hence the bias for light weight wood frame buildings. The question is simple by design, assuming just a beam independent of the floor and the distributed mass of beam and slab. I want to focus on dynamic loading where a person walking can have an effective load several times his weight on a slab that is strong enough to support may be 100 people safely. Aug 11 '16 at 5:31