1.3 Thermal expansion  (Page 4/10)

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Check Your Understanding Does a given reading on a gasoline gauge indicate more gasoline in cold weather or in hot weather, or does the temperature not matter?

The actual amount (mass) of gasoline left in the tank when the gauge hits “empty” is less in the summer than in the winter. The gasoline has the same volume as it does in the winter when the “add fuel” light goes on, but because the gasoline has expanded, there is less mass.

Thermal stress

If you change the temperature of an object while preventing it from expanding or contracting, the object is subjected to stress that is compressive if the object would expand in the absence of constraint and tensile if it would contract. This stress resulting from temperature changes is known as thermal stress    . It can be quite large and can cause damage.

To avoid this stress, engineers may design components so they can expand and contract freely. For instance, in highways, gaps are deliberately left between blocks to prevent thermal stress from developing. When no gaps can be left, engineers must consider thermal stress in their designs. Thus, the reinforcing rods in concrete are made of steel because steel’s coefficient of linear expansion is nearly equal to that of concrete.

To calculate the thermal stress in a rod whose ends are both fixed rigidly, we can think of the stress as developing in two steps. First, let the ends be free to expand (or contract) and find the expansion (or contraction). Second, find the stress necessary to compress (or extend) the rod to its original length by the methods you studied in Static Equilibrium and Elasticity on static equilibrium and elasticity. In other words, the $\text{Δ}L$ of the thermal expansion equals the $\text{Δ}L$ of the elastic distortion (except that the signs are opposite).

Calculating thermal stress

Concrete blocks are laid out next to each other on a highway without any space between them, so they cannot expand. The construction crew did the work on a winter day when the temperature was $5\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ . Find the stress in the blocks on a hot summer day when the temperature is $38\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$ . The compressive Young’s modulus of concrete is $Y=20\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{9}\phantom{\rule{0.2em}{0ex}}{\text{N/m}}^{2}$ .

Strategy

According to the chapter on static equilibrium and elasticity, the stress F / A is given by

$\frac{F}{A}=Y\frac{\text{Δ}L}{{L}_{0}},$

where Y is the Young’s modulus of the material—concrete, in this case. In thermal expansion, $\text{Δ}L=\alpha {L}_{0}\text{Δ}T.$ We combine these two equations by noting that the two $\text{Δ}L’\text{s}$ are equal, as stated above. Because we are not given ${L}_{0}$ or A , we can obtain a numerical answer only if they both cancel out.

Solution

We substitute the thermal-expansion equation into the elasticity equation to get

$\frac{F}{A}=Y\frac{\alpha {L}_{0}\text{Δ}T}{{L}_{0}}=Y\alpha \text{Δ}T,$

and as we hoped, ${L}_{0}$ has canceled and A appears only in F / A , the notation for the quantity we are calculating.

Now we need only insert the numbers:

$\frac{F}{A}=\left(20\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{9}\phantom{\rule{0.2em}{0ex}}{\text{N/m}}^{2}\right)\left(12\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\text{/}\text{°}\text{C}\right)\left(38\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}-5\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}\right)=7.9\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}{\text{N/m}}^{2}.$

Significance

The ultimate compressive strength of concrete is $20\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}{\text{N/m}}^{2},$ so the blocks are unlikely to break. However, the ultimate shear strength of concrete is only $2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}{\text{N/m}}^{2},$ so some might chip off.

Check Your Understanding Two objects A and B have the same dimensions and are constrained identically. A is made of a material with a higher thermal expansion coefficient than B . If the objects are heated identically, will A feel a greater stress than B ?

Not necessarily, as the thermal stress is also proportional to Young’s modulus.

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process whereby the degree of hotness of a body (or medium) changes
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Q=mcΔT
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where The letter "Q" is the heat transferred in an exchange in calories, "m" is the mass of the substance being heated in grams, "c" is its specific heat capacity and the static value, and "ΔT" is its change in temperature in degrees Celsius to reflect the change in temperature.
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what was the temperature of the soft drink when it was removed ?
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15 degree Celsius
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15 degree
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ok I think is just conversion
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15 degree Celsius to Fahrenheit
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0 degree Celsius = 32 Fahrenheit
Salim
15 degree Celsius = (15×1.8)+32 =59 Fahrenheit
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I dont understand
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the question said you should convert 15 degree Celsius to Fahrenheit
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To convert temperatures in degrees Celsius to Fahrenheit, multiply by 1.8 (or 9/5) and add 32.
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what is d final ans for Fahrenheit and Celsius
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it said what is temperature change in Fahrenheit and Celsius
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the 15 is already in Celsius
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So the final answer for Fahrenheit is 59
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what is d final ans for Fahrenheit and Celsius
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mechanical stiffness and small size
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d=dQ+w