Paul,
The principle you illustrate is true for thin (or relatively thin) walls around a hole, but for large solid blocks and small holes there are sometimes other considerations. Here is a snip out of a conversation thread I found on the web that attempts to expound on that concept (it is important to note that their conversation is about whether a hole will shrink or expand in a block that is frozen, whereas our question is whether the hole will shrink or expand when the block is heated):
In general, with ordinary materials, the hole will contract. There even was someone giving specific number of how much they will contract which apply to only a linear (straight line, rail road tie). Although many people have everyday observations about getting one pipe into another, or bearings into holes which they are bigger, the lessons learned from those are not to be applied to situations where thick walls are greater than the small hole. To further confuse, most are talking about outer diameter and not inner diameter. For reference, lets bring up Mechanics of Materials chapter of Thick Cylinder Walls, specifically applied to Cylinders with holes in the center.
Look at figure 10.1 and figure 10.2 and observe that under thin cylinders (small thickness, large hole) we only consider the linear contractions along the circumference, leading to a contraction coefficient of sigma_H = rho*D/2/t, we call this hoop stress. This is the formula which is very intuitive.
But our question is for a thick cylinder wall. Lets look at Figure 10.2, and understand the sigma_H = A + B/r2 (hoop stress), and the radial stress is sigma_r = A-B/r2. Where r is the center line of the material. From the area of between the hole and the center line, the metal is under extreme compression, especially where the hole is. The metal is under tensile stress on the outside. This tensile stress makes the object want to contract its outer diameter. This is the reason the outside of the metal can fit inside another hole. The large compression forces near the hole, counteract the effect of the entire object wanting to contract. Lets again focus on, the hole in the middle, not the entire object shrinking. For most metals (all?), I would agree that the hole would shrink, but this occurs under the condition that tensile strength being stronger and a non symmetric (not equally opposite) of compression strength. So while this will apply very accurately for most metals, doing the same for concrete (which has almost no tensile strength) the hole will not contract. In fact, if the thickness of the metal is extremely larger than the hole (approaches infinity), than the accuracy of the thin wall cylinder formula is off by an infinate amount; refer to Figure 10.7. For instance your drill a 1 in hole inside a 2 inch (diameter) block, the approximation of the others' formulas is off by 12%.
Lets View Figure 10.8, and reiterate that if the Hoop stress coefficient is a Positive B, and the Radial stress coefficient is Negative B. These are usually equal and opposite, but B_hoop is not guaranteed to be the same magnitude as B_radial.
Edit- Reference for figures:
http://www.ewp.rpi.edu/hartford/users/papers/engr/ernesto/poworp/Project/4. Supporting_Material/Books/32658_10.pdf
This reference
http://courses.washington.edu/me354a/Thick Walled Cylinders.pdf has actual data of normal metals, showing that what other people said to be correct (but not for the reasons they stated). In fact, metals have tensile strength much greater than compression strength.
TLDR; The hole will contract due to mechanics of material in steel having about 1.5x higher tensile strength over compression strength. If we have a material which has higher compression strength, the hole will not contract.
[and the response]
Unless I'm missing something, your sources say nothing about thermal contraction, they're about deformation under mechanical stress.
Look at figure 10.1 and figure 10.2 and observe that under thin cylinders (small thickness, large hole) we only consider the linear contractions along the circumference, leading to a contraction coefficient of sigma_H = rho*D/2/t, we call this hoop stress. This is the formula which is very intuitive, and every one in this article is citing.
You're confusing stress and strain. And that's not a rho, that's a p for the internal pressure, as you can see from where the formula is given the first time, in the previous chapter. Not sure why they (apparently) used a lower case p when the diagram labels it as P.
Where r is the center line of the material. From the area of between the hole and the center line, the metal is under extreme compression, especially where the hole is.
...The metal where the hole is is under radial stress equal to the internal pressure, as is clearly marked, because otherwise it wouldn't be in equilibrium. And r doesn't represent the "center line", or any particular line. It's the independent variable in the equations given for the radial and circumferential stress. The stresses each vary continuously over the radius.
[and the reply to the response]
My wordy interpretation was not fully accurate, but my take home points still stand. From my sources, there were many Figures, and the formulas didn't use nomenclature the same as the formulas. Thank you for pointing that out.
I was the only person in this discussion to point out that linear contraction does not apply for thick wall cylinders. Perhaps I got too wordy, and lacked emphasis to my important points; distracting our readers by formulas.
You point out that I'm concerned with stresses. this is because DUE to contraction of different circumferences. The compression force on the inner radius counteracts the contraction. Say we have a thin piece of metal, were know it to have a linear contraction of x per meter. if the inner radius is very different than the outer radius (a thick cylinder), then we must apply hooks law and consider that the metal will act like a spring, with out outer wall being stretch, and the inner wall being compressed. That is why I mentioned all these formulas, some are additive to the contraction, some are subtractive. I needed to explain these pressures so you know which situation to ignore them, and when to just consider the linear thermal expansion/contraction. For example a hole in a concrete slab, for which compression forces dominate the thermal contraction forces. I'm not deleting these posts, and I appreciate you being observant and looking into my sources in a true AskScience manner, but I implore you to accept the points of which were valid while you point out that a rho is a 'p'.
The top rated comment mentioned how much metal will contract while refering to a pipe around another pipe, but yet quoted linear contraction coefficient. Perhaps we should consider coefficient of the arclength of the circumference of the thin walled cylinder and its ratio to diameter. Because any contraction will be along the circumference, (which corresponds to a diameter contraction ratio of 1/pi), ultimately affecting the radius, but not as much as the linear contraction itself.
NOTE: The above clip was very slightly edited for clarity here.
And with all that said, Paul, you're certainly correct anyway, because we're not actually talking about a small hole in a very large block (which is what the above referenced thread was specifically talking about). So your argument does apply much more strongly since we're discussing a very large hole (actually 4 very large holes) in a relatively small block. And in that case the holes will certainly grow as the block is heated.
I should have probably just said that the clearance between the piston and cylinder will shrink as temperatures rise (and which according to the above tech talk is arguably true to some degree), and also because the cylinder diameter would not grow as rapidly as the piston diameter, if only because the cylinder has full-flood cooling jackets surrounding it and the piston is cooled mostly by conduction transfer to the cylinder wall through the oil film (and also by a small amount of oil splash from underneath), so the piston is bound by the law of thermal conduction to run hotter than the cylinder wall whenever the engine is under load.
But the short answer is, you're right of course...the cylinder diameter will grow with temperature. I do stand corrected, and thank you for the lump on the head!