Calculating Expansion Cracks Width for Concrete Highway -30°C to +32°C

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SUMMARY

The discussion focuses on calculating the required width of expansion cracks for a concrete highway slab measuring 12 meters in length, considering temperature variations from -30°C to +32°C. The coefficient of thermal expansion used is 1200e-6. The correct approach involves determining the maximum expansion due to temperature increase, which results in a calculated change in length of 0.1728 cm for the temperature rise from 20°C to 32°C. The consensus is to focus solely on expansion rather than contraction, emphasizing the importance of using Kelvin for thermal calculations.

PREREQUISITES
  • Understanding of thermal expansion concepts
  • Familiarity with the coefficient of thermal expansion (alpha)
  • Basic knowledge of temperature scales, particularly Celsius and Kelvin
  • Ability to perform calculations involving linear expansion formulas
NEXT STEPS
  • Study the principles of thermal expansion in materials
  • Learn about the implications of temperature variations on concrete structures
  • Research the use of Kelvin in thermodynamic calculations
  • Explore best practices for designing expansion joints in concrete construction
USEFUL FOR

Civil engineers, construction project managers, and anyone involved in the design and maintenance of concrete infrastructure will benefit from this discussion.

Trizz
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Homework Statement



A concrete highway is built of slabs 12 m long (20°C). How wide should the expansion cracks be (at 20°C) between the slabs to prevent buckling if the range of temperature is -30°C to +32°C?

________ cm


Homework Equations



change in L = L original * alpha * change in temp

sorry i don't have symbols and stuff

The Attempt at a Solution



I've tried it one way so far.

Basically, I just found the difference between the higher temp and the highway temp, and solved that as one equation. Then i found the difference between the lower temp and the highway temp, and solved that as one equation. Then i added the two answers.

change in L = 12m * 1200e-6(our teacher gave us this for alpha) * (32-20) = .1728

change in L = 12m * 1200e-6 * (20-(-30)) = .72

then i got .8928 cm


any help?
 
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I don't think you care about how much they shrink as compared to how much the slabs expand. The big ouch comes if the slabs expand and hit each other, while if they shrink then the gap just gets a little bigger. I don't see any constraint on the overall gap size itself. Also, you should be working in Kelvin!
 
It seems like you shouldn't add the two. Just find the largest value (as Mindscrape said we only care about expansion) and that should be how much space you should put between the slabs. So the real problem it seems is finding the maximum value of

|x - 20|

On the range 20 < x \le32
 
ok so something like...

change in L = 12m * 1200e-6 * 12 = .1728 cm

ohh and its 12 regardless if i work in celsius or kelvin

so does that seem better?
 
thanks guys i got it
 
Yeah, for changes in temperature the Kelvin scheme usually doesn't matter, though it did for the contraction equation you worked out because you should have got a negative when you had a positive. Get in the habit of working with Kelvin in thermo. You'll be sorry later if you don't.

Yep, that's good. Technically you should have in inequality in your answer.
 
Mindscrape said:
Yeah, for changes in temperature the Kelvin scheme usually doesn't matter, though it did for the contraction equation you worked out because you should have got a negative when you had a positive. Get in the habit of working with Kelvin in thermo. You'll be sorry later if you don't.

Yep, that's good. Technically you should have in inequality in your answer.

He actually would have gotten a negative delta T but he switched the final and initial temperatures.
 

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