# Heating a steel plate with a hole in it

#### Nikitin

1. Homework Statement
If a steel plate with a hole of 10 cm in diameter is heated 35 degrees kelvin, what will the new diameter be? $\alpha_L = 13 \cdot 10^{−6} K^{−1}$

2. Homework Equations
$\Delta L = \alpha_L \cdot \Delta T \cdot L$
3. The Attempt at a Solution
If I understand the length-expansion formula correctly, all the dimensions of the substance should change by a factor $\Delta L /L$ (why?).

Then the diameter will then increase in size to $L+\Delta L$? Uhm, how so? Considering there is a void inside the plate, and the rest of the plate is expanding, shouldn't the plate also expand into the hole? But according to the formula, the plate should expand in the same way as if the hole did not exist!

Can somebody explain how this works? I'm new to thermodynamics.

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#### BvU

Homework Helper
Consider the difference between a) making a ring from a steel rod, then heating, and
b) heating a rod of the exact same length, then making a ring of it.

#### Nikitin

There is no difference.

But my problem here is, that according to the formula the plate from the OP will NOT expand inwards, despite there being a void there. It will only expand outwards. This makes no sense to me.

#### BvU

Homework Helper
As you say, there is no difference.
Like the rod, the inner ring around the hole wants to get longer. Going inward would make it shorter.

#### Chestermiller

Mentor
To expand on what BvU is saying, the circumference of the hole is one of the "lines" in the steel plate. Since all lines in the steel plate must increase in length by the same percentage, so must the circumference of the hole. If the circumference gets larger, so must the diameter of the hole.

Chet

#### BvU

Homework Helper
There is no difference.

But my problem here is, that according to the formula the plate from the OP will NOT expand inwards, despite there being a void there. It will only expand outwards. This makes no sense to me.
Sense, intuition, gut feeling... Very important in science. A matter of having the right mental model in your head. So let me try another tack:

Draw (or imagine) a 2D grid of 1 cm x 1 cm squares. Our simplest possible model for a solid. The grid points are the Fe atoms, the 1 cm connecting lines represent the distances between neighboring atoms. They are like springs at equilibrium: compress and the atoms repel each other more, expand and they attract more.

Cut out a circle around the center. Or color the grid points and strings within such a circle.

Now for some reason the equilibrium state of the springs shifts to a little bigger distance between atoms when the temperature goes up. The expansion coëfficiënt tells us how much.

In physics (and some other areas as well) it's nice to exaggerate 'a bit' and wonder what if...
So let's exaggerate the expansion coëfficiënt and the temperature rise and extend all connecting lines to 11 mm. Draw or imagine again. What happened to the circle ? Without any tension, shear, warping, whatever, everything one-dimensional got 10% bigger: diameter of the circle, circumference, you name it.

2D is too simple for reality, I hear someone mutter. Well, now it gets interesting. Use our imagination or do a lot of ("3D") drawing, doodling with steel balls and springs, or whatever.

Do you agree that a 10% increase in all connecting springs lets the diameter of a cut-out cylinder go up by 10% as well?

So why did your senses tempt you to think of a kind of swelling inwards? (And you are definitely not the only one. If physics were a democratic process, you'd probably win the elections).

Suppose, just suppose, there is a fixed layer above and a fixed layer below our 3D grid. (Fixed z)With the same springs vertically welded to 1 cm grid points between fixed layers and our springy 'solid'. Suppse we heat up this contraption - without the fixed layers giving in even the slightest bit - so much that all springs are 11 mm again. This time we get quite a bit of swelling inwards, just as you intuitively expected!

Only, that wasn't the situation described in the exercise....

#### kuruman

Homework Helper
Gold Member
And here is the math behind what @Chestermiller posted. The change in size $S$ (1, 2, or 3 D) is given by
$$\Delta S=\alpha S_0 \Delta T$$. The new size is $$S=S_0 +\Delta S=(1+\alpha \Delta T)S_0.$$So if you have two objects of different size $S$ and you heat them, they will both expand by the same percentage, while their ratio will remain the same. It's like taking a photograph of the plate and then enlarging it by a factor of $(1+\alpha \Delta T)$. If the diameter of the hole in one picture is one-half the side of the plate, it will be one-half the side of the plate in the enlarged picture.

#### DEvens

Gold Member
If I understand the length-expansion formula correctly, all the dimensions of the substance should change by a factor $\Delta L /L$ (why?).
The level of answer you need to this "why" is dependent on your understanding of solid state physics. But try this.

Think of a solid as a bunch of atoms held together by springs. When the material gets hotter, the atoms vibrate faster. Depending on a *lot* of details (a LOT of details) this often makes the material get larger. Maybe because as the atoms vibrate the springs are stretched on average.

As a mental image that's probably OK for this level of question. Just be aware of the huge pile of details I am omitting. It was a year of university, it was in 1980, and I only got a B in the class. And it was mostly about x-ray techniques anyway.

Then the diameter will then increase in size to $L+\Delta L$? Uhm, how so? Considering there is a void inside the plate, and the rest of the plate is expanding, shouldn't the plate also expand into the hole? But according to the formula, the plate should expand in the same way as if the hole did not exist!

Can somebody explain how this works? I'm new to thermodynamics.
It's because the steel expands uniformly. Go back to the picture of atoms on springs. All the springs get longer by the same fraction. It's just a scale factor. (It would be very different if the heating were not uniform or the material of the plate was not uniform.)

So this

becomes this.

Note that, in terms of "number of intersections" the geometry is exactly the same after the stretch. 9 wide by 14 high. Each rectangle the same ratio. And the round hole is exactly the same number of intersections wide and tall as before. Only now each intersection is farther apart. So the hole gets bigger in exactly the same ratio.

Note that it means that any shape you draw on, or cut out of, the plate will also grow by the same scale.

"Heating a steel plate with a hole in it"

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