A concrete bridge is built of 325-cm-long concrete slabs with an expansion joint between them. The slabs just touch on a ${115}^{0}F$day, the hottest day for which the bridge is designed. What is the gap between the slabs when the temperature is $0^{0}F$?

The process of thermal expansion occurs when objects are heated as well as expanded. In case the temperature of an object changes through $∆T$, as well as length changes through $∆L$and the length's fractional changes $∆L/L$are effective to the temperature's change $∆T$

The process of thermal expansion occurs when objects are heated as well as expanded. In case the temperature of an object changes through $∆T$ , as well as length changes through $∆L$and the length's fractional changes $∆L/L$are effective to the temperature's change $∆T$and the equation $(18.8)$is the form is:

$\frac{∆L}{L}=\alpha ∆T$ (1)

In this case, $\alpha$is the coefficient. Table 18.4 effectively gives the coefficient expression of linear for common materials, for instance $\alpha =1.2\times {10}^{-5\circ }{C}^{-1}$

The equation for $∆L$is

$∆L=\alpha L∆T$ (2)

In this case, L is the initial length. The temperature of the bridge effectively changes from $T_1=115°F$to $T_2=0°F$

Therefore, the change of temperature is:

$$\begin{gathered} ∆T=T_2-T_1 \\ =0°F-115°F \\ =-115°F \end{gathered}$$

Let's focus on the conversion from Fahrenheit to Celcius and the equation is:

$∆T_F=\frac{9}{5}∆Tc$ (3)

In terms of determining the temperature in Celcius and solving the equation (3) for Tc in the form of:

$∆T_c=\frac{5}{9}(∆T_F)$

Thus, plug the value for temperature into equation(4)

$$\begin{gathered} ∆T_c=\frac{5}{9}(∆T_F) \\ =\frac{5}{9}(-115°) \\ =-63.89°C \end{gathered}$$

Thus, plug the value for $L,∆Tc,\alpha$into Equation (2) in order to get $∆L$

$$\begin{gathered} ∆L=\alpha L∆T \\ =(325cm)(1.2\times {10}^{-5°}{C}^{-1})(-63.89°C) \\ =0.25cm \end{gathered}$$