# Linear Expansion Problem: Finding Optimal Crack Width for Highway Slabs

• yossup
In summary, if the temperature range is -30C to 43C, the cracks between the slabs should be 12m wide to prevent buckling.
yossup
linear expansion problem...urgent!

## Homework Statement

A concrete highway is built of slabs 12m long (20 degrees C). How wide should the expansion cracks between the slabs be (at 15 C) to prevent buckling if the range of temperature is - 30 C to 43 C?

## Homework Equations

change in length = (coefficient) (initial length) (change in temperature)

coefficient = 12 * 10^(-6)

## The Attempt at a Solution

So first I solved the length of the slabs at 15 C =

coefficient (12m) (-5 degrees C) = -7.2 * 10^(-4)

so (12m - 7.2 * 10^(-4)) = 11.99m

so at 15 degrees C, the slab is 11.99m

Then I solved how much it would expand/contract at 43C/-30C

basically

change of length from 15C to 43C = .00403m

change of length from 15c to -30c = -.00648

since in an expansion crack, there are slabs on both sides which would BOTH expand, I thought the answer would be .00403m + .00403m but it's wrong.

I've tried

.00403m + .00648m

.00403*2 + .00648*2

but these are all wrong.

What am I missing? No matter how much I think about it...I don't get what about my calculations are wrong. Thanks!

Last edited:
yossup said:
A concrete highway is built of slabs 12m long (20 degrees C). How wide should the expansion cracks between the slabs be (at 15 C) to prevent buckling if the range of temperature is - 30 C to 43 C?

change in length = (coefficient) (initial length) (change in temperature)

coefficient = 12 * 10^(-6)since in an expansion crack, there are slabs on both sides which would BOTH expand, I thought the answer would be .00403m + .00403m but it's wrong.

I've tried

.00403m + .00648m

.00403*2 + .00648*2

but these are all wrong.

Hi yossup!

Why are you going down to -30C?

The question only involves expansion from 15C to 43C.

(and you only need to consider one slab, not two)

so the answer is simply =>

(coefficient_concrete) (11.99m) (28C) ?

why only one slab? don't you need to consider two since if two slabs next to each other both expand for example, .1m. then if they are expanding towards each other...in order for them not to buckle, there would have to be .2m of space between them.

yossup said:
why only one slab? don't you need to consider two since if two slabs next to each other both expand for example, .1m. then if they are expanding towards each other...in order for them not to buckle, there would have to be .2m of space between them.

ah … but imagine ten slabs with 9 gaps of .01m = 120.09m total.

If they each expand by .01m, then they take up 120.1m …

so it's only one slab that has to be accounted for at both ends …

and if the highway is infinitely long, you don't even have to bother about that!

## What is linear expansion problem?

Linear expansion problem refers to the phenomenon where an object expands or contracts in length when it is exposed to changes in temperature. This change in length is directly proportional to the original length of the object and the change in temperature.

## What causes linear expansion?

The main cause of linear expansion is the increase in kinetic energy of the particles in an object when it is heated. This increase in energy causes the particles to vibrate more and take up more space, resulting in the expansion of the object.

## How is linear expansion measured?

Linear expansion is measured using the coefficient of linear expansion, which is a constant value that represents the change in length of an object per unit change in temperature. This value varies for different materials and is usually measured in units of 1/°C or 1/°F.

## What are some real-life applications of linear expansion?

Linear expansion is used in various fields such as construction, engineering, and manufacturing. It is used to design and construct structures that can withstand changes in temperature without causing damage. It is also used in the calibration of thermometers and other temperature-measuring devices.

## How can linear expansion be controlled or compensated for?

One way to control or compensate for linear expansion is by using materials that have a low coefficient of linear expansion, such as steel or concrete. Another method is to use expansion joints or gaps in structures to allow for expansion and contraction. Additionally, in precision instruments, the coefficient of linear expansion can be taken into account during the design process to minimize its effects.

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