What is the necessary gap width for thermal expansion of concrete sidewalks?

So, the minimum gap width necessary to account for an increase in temperature of 27 C° would be 0.00078 m, as calculated earlier.
  • #1
Crimsonangel
5
0
Concrete sidewalks are always laid in sections, with gaps between each section. For example, the drawing shows three identical 2.4 m sections, the outer two of which are against immovable walls. The two identical gaps between the sections are provided so that thermal expansion will not create the thermal stress that could lead to cracks. What is the minimum gap width necessary to account for an increase in temperature of 27 C°?

http://www.webassign.net/CJ/p12-20.gif (here is a picture)

I used the equation (deltaL=alpha*Lo*deltaT). So I knew alpha was 12*10^-6 and Lo was 2.4m and deltaT was 27 C°. I got deltaL to be 7.776*10^-4 m. So the concrete expanded and how has a length of 2.4007776 m. After that I wasn't sure what to do. I thought it would be 3 times 7.776*10^-4 but it wasn't. Could you please help? Thanks so much!
 
Physics news on Phys.org
  • #2
umm... maybe I'm being dumb but it seems like since it expands both ways with a total of .0007776 m that means that each way it goes half of that, and since there are two around each gap, I think the gap is just .00078m.
 
  • #3
Crimsonangel said:
Concrete sidewalks are always laid in sections, with gaps between each section. For example, the drawing shows three identical 2.4 m sections, the outer two of which are against immovable walls. The two identical gaps between the sections are provided so that thermal expansion will not create the thermal stress that could lead to cracks. What is the minimum gap width necessary to account for an increase in temperature of 27 C°?

http://www.webassign.net/CJ/p12-20.gif (here is a picture)

I used the equation (deltaL=alpha*Lo*deltaT). So I knew alpha was 12*10^-6 and Lo was 2.4m and deltaT was 27 C°. I got deltaL to be 7.776*10^-4 m. So the concrete expanded and how has a length of 2.4007776 m. After that I wasn't sure what to do. I thought it would be 3 times 7.776*10^-4 but it wasn't. Could you please help? Thanks so much!

The 3 times 7.776*10^-4m gives you the total expansion of the three sections. You have two gaps that can be filled in by that increased length.
 
  • #4
my temperature for this question is 37 degrees Celsius...and i got deltaL to be 0.001066...i'm still not quite sure what to do with that
 
  • #5
jalpabhav said:
my temperature for this question is 37 degrees Celsius...and i got deltaL to be 0.001066...i'm still not quite sure what to do with that

I assume that is for one section. You have three sections that expand, and two gaps for them to expand into. The total expansion cannot be more than the total gap space.
 

Related to What is the necessary gap width for thermal expansion of concrete sidewalks?

1. What is linear thermal expansion?

Linear thermal expansion is the tendency of a material to expand or contract in one dimension (length) when heated or cooled.

2. What causes linear thermal expansion?

Linear thermal expansion is caused by the increase in molecular vibrations within a material when it is heated. This causes the molecules to take up more space and thus the material expands.

3. How is linear thermal expansion measured?

Linear thermal expansion is typically measured using the coefficient of linear thermal expansion (CLTE), which is the change in length per unit length per degree temperature change. It is expressed in units of 1/°C or 1/°F.

4. What factors can affect linear thermal expansion?

The main factors that can affect linear thermal expansion are the material's composition, temperature change, and length/dimension of the material. Other factors such as pressure and impurities can also have an impact.

5. Why is linear thermal expansion important?

Linear thermal expansion is important to consider in many industries, such as construction, engineering, and manufacturing. It allows for the design and construction of structures and machines that can withstand changes in temperature without causing damage or malfunction.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
7K
  • Introductory Physics Homework Help
Replies
6
Views
21K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
10K
  • Introductory Physics Homework Help
Replies
2
Views
7K
  • Introductory Physics Homework Help
Replies
2
Views
6K
Back
Top