Question

On a day that the temperature is 21.0°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).) (a) What is the stress in the cement on a hot day of 33.0°C? N/m2

Answer 1

1*10^6 N/m^2

Step-by-step explanation:

Coefficient of Linear Expansion for Concrete = α = 1.2 x 10^-5 (°C)^-1

Change in temperature = ΔT

ΔT= T2 - T1

ΔT = 33 - 21

ΔT = 12°C

ΔL = α * L(i) * ΔT

ΔL = (1.2 x 10^-5 (°C)^-1) * L(i) * (12°C)

ΔL = 1.44 x 10^-4

Stress = F / A

Strain = ΔL / L

Strain = (1.44*10^-4) * (L) / L

Strain = 1.44*10^-4

Y = Stress / Strain

Stress = Y * Strain

Stress = (7.00*10^9 N/m^2) * (1.44*10^-4)

Stress = 1*10^6 N/m^2

Thus, the stress in the cement on a hot day of 33° is 1*10^6 N/m^2