Question

On a day that the temperature is 10.0°C, a concrete walk is poured in such a way that the ends of the walk are unable to move. Take Young's modulus for concrete to be 7.00 109 N/m2 and the compressive strength to be 2.00 109 N/m2. (The coefficient of linear expansion of concrete is 1.2 10-5(°C−1).)
What is the stress in the cement on a hot day of 42.0°C? N/m2

Answer 1

Answer:

The stress is  $stress = 2688000 \ N$

Explanation:

From the question we are told that

    The first temperature is  $T_1 = 10 ^o \ C$

    The  young modulus is  $Y = 7.00 *10^9\ N/m^2$

    The compressive strength is  $\sigma = 2.00 *10^{9} \ N/m^2$

     The coefficient of  linear expansion is  $\alpha = 1.2 *10^{-5} \ ^o C ^{-1}$

     The  second temperature is  $T_2 = 42.0^o \ C$

Generally the change in length of the concrete is mathematically represented as

      $\Delta L = \alpha * L * [T_2 - T_1 ]$

=>  $\frac{\Delta L}{L} = \alpha * [T_2 - T_1 ]$

=> $strain = \alpha * [T_2 - T_1 ]$

Now  the young modulus is  mathematically represented as

        $Y = \frac{stress}{strain}$

=>     $7.00 *10^9 = \frac{stress}{\alpha(T_2 - T_1 ) }$

=>   $stress = \alpha (T_2 - T_1 ) * 7.00 *10^9$

=>   $stress = 1.2* 10^{-5} (42 - 10 ) * 7.00 *10^9$

=>   $stress = 2688000 \ N$