concrete blocks are laid out next to each other on a highway without any space between them, so they cannot…

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 $t_i = 5.0^{circ}c$. find the thermal stress in the blocks on a hot summer day when the temperature is $t_f = 38^{circ}c$? the compressive youngs modulus of concrete is $y = 20\times10^{9} n/m^{2}$ and thermal expansion coefficient is $alpha_{concrete}=10\times10^{-6} ^{circ}c^{-1}$. $t_i = 5.0^{circ}c$ $t_f = 38.0^{circ}c$ $(\frac{f}{a}) = y(\frac{delta l}{l_0})$ $= yalphadelta t$ $= 20\times10^{9}\frac{n}{m^{2}}\times10\times10^{-6}\times(38 - 5)$ $\frac{n}{m^{2}}$ $\frac{n}{m^{2}}rightarrow(\frac{f}{a}) = stress$ (prop) strain $\frac{f}{a}=y(\frac{delta l}{l_0})$ $(\frac{delta l}{l_0})=alphadelta t$ $\frac{n}{m^{2}} = pa$

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 $t_i = 5.0^{circ}c$. find the thermal stress in the blocks on a hot summer day when the temperature is $t_f = 38^{circ}c$? the compressive youngs modulus of concrete is $y = 20\times10^{9} n/m^{2}$ and thermal expansion coefficient is $alpha_{concrete}=10\times10^{-6} ^{circ}c^{-1}$. $t_i = 5.0^{circ}c$ $t_f = 38.0^{circ}c$ $(\frac{f}{a}) = y(\frac{delta l}{l_0})$ $= yalphadelta t$ $= 20\times10^{9}\frac{n}{m^{2}}\times10\times10^{-6}\times(38 - 5)$ $\frac{n}{m^{2}}$ $\frac{n}{m^{2}}rightarrow(\frac{f}{a}) = stress$ (prop) strain $\frac{f}{a}=y(\frac{delta l}{l_0})$ $(\frac{delta l}{l_0})=alphadelta t$ $\frac{n}{m^{2}} = pa$

Answer

Explanation:

Step1: Identify the formula for thermal - stress

The formula for thermal stress $\sigma$ is $\sigma = Y\alpha\Delta T$, where $Y$ is Young's modulus, $\alpha$ is the coefficient of thermal expansion, and $\Delta T$ is the change in temperature.

Step2: Calculate the change in temperature

$\Delta T=T_f - T_i$. Given $T_i = 5.0^{\circ}C$ and $T_f=38^{\circ}C$, then $\Delta T=38 - 5=33^{\circ}C$.

Step3: Substitute the values into the formula

We know that $Y = 20\times10^{9}\ N/m^{2}$ and $\alpha=10\times10^{- 6}\ ^{\circ}C^{-1}$. Substituting these values and $\Delta T = 33^{\circ}C$ into $\sigma = Y\alpha\Delta T$, we get $\sigma=(20\times10^{9})\times(10\times10^{-6})\times33$. First, calculate $(20\times10^{9})\times(10\times10^{-6})=20\times10^{3}=2\times10^{4}$. Then, $\sigma=(2\times10^{4})\times33 = 6.6\times10^{5}\ Pa$.

Answer:

$6.6\times10^{5}\ Pa$