Ground vibrations can come from a large number of different sources, such as rail and vehicular traffic, jackhammering, bulldozing, pavement milling, vibratory compaction, pile driving, and blasting.
The magnitude of these vibrations can vary from being imperceptible, to causing a minor nuisance, to possibly even causing extreme damage. Generally speaking, the main factors to consider are the strength of the source vibrations, duration of exposure, distance from the source, type and condition of the affected building, and the local soil type. Our building stock and infrastructure are aging. More and more there will be a need to carry out construction activities near these aging buildings. Consequently, there will likely be a rise in the number of complaints, and insurance claims, due to construction-induced ground vibrations in the future.
Ground vibrations are typically quantified in terms of Peak Particle Velocity (PPV), usually expressed in millimetres per second (mm/s) or inches per second (in/sec). Stronger vibrations have a higher PPV. Vibrations can damage buildings directly as a result of the movement in the walls or floors, or the damage can be indirect as a result of soil movements that were caused by the vibrations. In Canada, there is no national standard for ground vibrations, but standards from other countries generally agree that ground vibrations should be limited to 5 mm/s to prevent direct damage to small residential buildings. This 5 mm/s threshold is based on empirical data and probabilities. The threshold is intended to ensure there is a very low risk of cosmetic cracking in drywall, plaster, and old unreinforced masonry. There is no guarantee that damage cannot occur at lower levels, but that would be very rare. There's also no guarantee that exceeding the 5 mm/s threshold will cause damage to a building. In fact, most buildings can tolerate stronger vibrations, especially buildings that are well-constructed, in good condition, and/or are only briefly exposed to the vibration.
Whether or not an activity will exceed a vibration threshold depends on the distance from the source, local soils, and the strength of the source vibrations. Weak and soft soils are good at dissipating vibrations, while hard soils or rock can transmit vibrations a long way. Just think of how you would feel the vibrations in your hands if you swung a steel pipe at concrete wall as hard as you could, and how different your hands would feel if you swung a pillow instead. But remember, the 5 mm/s threshold is for direct damage to the building, so it might not be applicable if the building sits on a soil that is susceptible to vibrations, making the building vulnerable to indirect damage.
Susceptibility to indirect damage depends on the local soil and the duration of exposure to vibrations. Susceptible soils include sand, gravel, and some fine-grained soils known as loess. These soils can be consolidated by vibrations, which can in turn result in foundation movement and cracking. Imagine the particles in a loosely compacted sand are like a house of cards. The vibration makes the particles move around and fall into a more compact configuration, like when a house of cards collapses. The amount of consolidation that can take place depends on how well these soils were compacted beforehand and how long they're exposed to the vibrations. There are no standards limiting vibrations to prevent soil consolidation because more research into the problem is needed. However, most vibration standards at least recognize that problems can occur with loose granular soils and recommend that a geotechnical engineer be consulted. Limited available empirical data on damage caused by vibration-induced soil consolidation shows that some buildings in loess soils have been damaged at vibration levels below 1 mm/s. Therefore, both the soil characteristics and the ground vibrations need to be considered together when investigating vibration damage.
Building occupants are typically much more sensitive to vibrations than the structures themselves. Vibrations of just 0.3 mm/s could be sufficient to disturb some homeowners and vibrations of about 1 mm/s are likely to cause complaints. A vibratory roller compactor working in an area with hard soils might cause complaints more than 400 feet away! Therefore, the sensitivity of the occupants is frequently an issue with vibration damage claims. Often, when building owners perceive nuisance vibrations they go looking for damage afterwards. They then find a number of cracks that they hadn't ever paid any attention to before. In many cases, most or all of the damage reported was actually pre-existing.
Cracking caused by ground vibrations may be indistinguishable from other common types of damage, like cracking from thermal and moisture movements or small foundation settlements. Therefore, a vibration damage investigation involves a lot more than a visual examination. Additional information needs to be considered, including the local soil conditions, type of building involved, general condition of the building, and the duration, frequency, and source location of the vibrations. Documentation of pre-existing conditions can be the most valuable information for an investigation.
If you look hard enough, you'll find cracks in any house. If you're a homeowner, it's a good idea to looking for cracks once in a while and document them with photos. You should also do this as soon as you see any major construction work going on nearby. That way, if damage does occur, you will have some documentation of existing conditions from when the construction work started. That record of pre-existing conditions will help in determining the extent of the damages.
If you're a contractor, the best way to prevent complaints is to inform nearby property owners about the ground vibrations they can expect and to conduct pre-construction surveys to document any existing cracks. Nuisance vibrations are more tolerable when they are expected, and if you do the pre-construction survey then owners won't be discovering pre-existing cracks while the construction work is going on. And if damage does occur, that record of pre-existing conditions will help limit the costs to just the vibration damage.
Dowding, C.H. (2000). Construction Vibrations. Second edition. Prentice Hall, Upper Saddle River, NJ.
Svinkin, M.R. (2012). "The necessity of condition surveys for structural protection against pile driving effects," Proceedings of the 9th International Conference on Testing and Design Methods for Deep Foundations, T. Matsumoto, ed., Kanazawa e-Publishing, Japan, 429-439.
Svinkin, M.R. (2015). "Tolerable Limits of Construction Vibrations," Practice Periodical on Structural Design and Construction. 10.1061/(ASCE)SC.1943-5576.0000223.
Wiss, J.F. and Parmalee, R.A. (1974). "Human Perception of Transient Vibrations," Journal of the Structural Division, ASCE, Vol. 100, No. S74, pp. 773-787.
Monday, 28 November 2016
Thursday, 29 September 2016
|Okay, maybe it CAN be worse than Paint Your Wagon.|
It's that time again. Time to solve some math problem involving birthdays. I've previously solved the Classic Birthday Problem, as well as the Near-Match, First Match (and First Near-Match), Same Birthday as You, and Matched Couples (and larger groupings) variations.
I'm going to call this the Cover All Birthdays variation. We're looking for the probability that a group of people has every day of the year covered for birthdays.
Let's start by looking at a simpler problem. Suppose that we're tossing rings at some pegs. In this scenario we're assuming that every peg has an equal probability of being hit and that every ring must land on a peg. For any given numbers of n rings and k pegs, what's the probability that every peg is hit at least once? It is impossible to hit k pegs with fewer than k rings, so there is an implicit assumption in this analysis that n ≥ k.
With only one peg, the problem is trivial because all rings necessarily land on that one peg. There is only one possible outcome irrespective of n: the probability of hitting all pegs is 100%.
Since rings can land on any peg, including on the same peg multiple times, the total number of possible outcomes in this ring toss scenario is equal to k to the n-th power. To calculate the probability of hitting all pegs we need to count all of the ways that we could have at least one lone peg. So let's look at a two-peg situation now. There are 2 to the n-th power possible outcomes with the ring tosses. If we have one lone peg, that means there's just one peg left to hit. We already know that there's only one way to hit just one peg. But now, either peg could be the one that gets all the rings, so with 2 pegs there are 2*1 = 2 possible ways to have a lone peg. That means there are 2^n - 2 ways of hitting each peg at least once. The probability of covering all the pegs is therefore [2^n - 2] / [2^n] = 1 - 2^(1-n).
|Number of ways to throw n rings at 2 pegs, hitting all pegs at least once.|
|Probability of hitting all pegs at least once when throwing n rings at 2 pegs.|
What happens when we add in a third peg? Again, we could have all the rings land on one peg. This time there are 3 different pegs that could get all the rings, so there are 3 * 1 = 3 ways to hit only one peg. We already figured out that there were 2^n - 2 ways of covering 2 pegs, but now there are three pegs we could choose the two from, or C(3,2) = 3 ways to pick two out of three pegs. That gives us a total of 3 * (2^n - 2) ways of covering exactly two out of three pegs. We're left with 3^n - 3*(2^n-2) - 3 ways of covering all the pegs, out of a total 3^n permutations. Therefore, the probability of covering all three pegs is:
|Number of ways to throw n rings at 3 pegs, hitting all pegs at least once.|
|Probability of hitting all pegs at least once when throwing n rings at 3 pegs.|
Let's now define a function F[n,k] as the number of ways of hitting each of the k pegs at least once from n tosses. That recursive relationship looks like this:
|Recursive relationship defining the number of ways you can throw n rings hitting all k pegs at least once.|
|Probability of throwing n rings and hitting all k pegs at least once.|
As you keep adding pegs you'll see that the equations quickly start to get lengthy. Using the recursive relationship we could take a brute force calculation approach to evaluate F[n,k], but that quickly gets to be very inefficient as k increases. We'd like a different form of the equation. If we expand and simplify for k = 4 and k = 5, we get the following:
|Expanded and simplified expressions for F[n,k] and P[n,k] for k = 1 to 5.|
Just for fun, let's plot those out as well so you can see how many tosses you need to have a reasonable chance of hitting all of the pegs.
|Tabulated values of the probability of throwing n rings and hitting all k pegs at least once|
|Graph showing the probability of hitting all pegs at least once with n tosses.|
In order to have at least 50% chance of hitting all of the pegs, you need only 2 tosses when there are 2 pegs, 5 tosses for 3 pegs, 7 tosses for 4 pegs, and 10 tosses for 5 pegs.
Now, going back to the expanded equations we wrote out, you might've spotted a pattern. We can define F[n,k] in closed-form using binomial coefficients and summation notation as follows:
|Solution to the recursive relation for F[n,k].|
Okay, so how does this relate to the Cover All Birthdays problem? Well if instead of n rings we have n randomly selected people and instead of k pegs we have 365 possible days a person could be born on, the problem is modeled exactly the same way. There are 365^n permutations of birthdays in a group of n people and there are F[n,365] ways of covering each possibility at least once:
|Number of birthday combinations in a group of n people where every day of the year is covered at least once.|
Dividing by 365^n to get the probability of covering all birthdays, we can make the following graph:
|Probability of covering all 365 birthdays in a group of randomly selected people.|
|Size of group required to achieve the given threshold probabilities of covering all 365 birthdays.|
For n < 365 we know the probability of covering all the birthdays is exactly zero. For n = 365, there's less than 1 in 6.873×10¹⁵⁶ chance of covering every day of the year. If we double that to n = 730, we increase the probability by a factor of nearly 3.45×10¹²⁸, but that's still a negligibly small chance of about 1 in 1.993×10²⁸. Chances of covering all birthdays are vanishingly small until approximately n = 1400. We need at least n = 2287 to exceed a 50% chance of covering all birthdays and at least n = 3828 to give us better than a 99% chance.