A concrete cylinder of diameter 150 mm and length 300 mm when subjected to an axial compressive load of 248…

Question

A concrete cylinder of diameter 150 mm and length 300 mm when subjected to an axial compressive load of 248 kN if result in an increase of diameter by 0.127 mm and an decrease of 0.3 mm compute the value of Poisson ratio and modulus of elasticity.

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let's compute the Poisson's ratio and modulus of elasticity.

Convert all given values to SI units (meters and Newtons).
Original diameter, $D = 150 \text{ mm} = 0.150 \text{ m}$
Original length, $L = 300 \text{ mm} = 0.300 \text{ m}$
Compressive load, $F = 248 \text{ kN} = 248 \times 10^3 \text{ N}$
Increase in diameter, $\Delta D = 0.127 \text{ mm} = 0.127 \times 10^{-3} \text{ m}$
Decrease in length, $\Delta L = 0.3 \text{ mm} = 0.3 \times 10^{-3} \text{ m}$
Calculate the longitudinal strain ( $\epsilon_L$ ) and lateral strain ( $\epsilon_D$ ).
Longitudinal strain is the change in length divided by the original length. Since it's a decrease, we use a negative sign for the change in length.
$\epsilon_L = \frac{\Delta L}{L} = \frac{-0.3 \times 10^{-3} \text{ m}}{0.300 \text{ m}} = -0.001$
Lateral strain is the change in diameter divided by the original diameter. Since it's an increase, it's positive.
$\epsilon_D = \frac{\Delta D}{D} = \frac{0.127 \times 10^{-3} \text{ m}}{0.150 \text{ m}} \approx 0.00084667$
Compute Poisson's ratio ( $\nu$ ).
Poisson's ratio is the negative ratio of lateral strain to longitudinal strain.
$\nu = -\frac{\epsilon_D}{\epsilon_L} = -\frac{0.00084667}{-0.001} = 0.84667$
Rounding to three significant figures:
$\boxed{\text{Poisson's ratio} \approx 0.847}$
Calculate the cross-sectional area ( $A$ ) of the cylinder.
$A = \frac{\pi D^2}{4} = \frac{\pi (0.150 \text{ m})^2}{4} = \frac{\pi \times 0.0225 \text{ m}^2}{4} \approx 0.017671 \text{ m}^2$
Calculate the compressive stress ( $\sigma$ ).
Stress is the force divided by the cross-sectional area.
$\sigma = \frac{F}{A} = \frac{248 \times 10^3 \text{ N}}{0.017671 \text{ m}^2} \approx 14033900 \text{ Pa}$
Compute the modulus of elasticity ( $E$ ).
The modulus of elasticity (Young's modulus) is the ratio of stress to the magnitude of longitudinal strain.
$E = \frac{\sigma}{|\epsilon_L|} = \frac{14033900 \text{ Pa}}{0.001} = 14033900000 \text{ Pa}$
Converting to GigaPascals (GPa):
$E = 14.0339 \text{ GPa}$
Rounding to three significant figures:
$\boxed{\text{Modulus of elasticity} \approx 14.0 \text{ GPa}}$

A concrete cylinder of diameter 150 mm and length 300 mm when subjected to an axial compressive load of 248 kN if result in an increase of diameter by 0.127 mm and an decrease of 0.3 mm compute the value of Poisson ratio and modulus of elasticity.

Answer

Step-by-step solution

Convert all given values to SI units (meters and Newtons).

Calculate the longitudinal strain ( $\epsilon_L$ ) and lateral strain ( $\epsilon_D$ ).

Compute Poisson's ratio ( $\nu$ ).

Calculate the cross-sectional area ( $A$ ) of the cylinder.

Calculate the compressive stress ( $\sigma$ ).

Compute the modulus of elasticity ( $E$ ).

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More Physics Questions

A concrete cylinder of diameter 150 mm and length 300 mm when subjected to an axial compressive load of 248 kN if result in an increase of diameter by 0.127 mm and an decrease of 0.3 mm compute the value of Poisson ratio and modulus of elasticity.

Answer

Step-by-step solution

Convert all given values to SI units (meters and Newtons).

Calculate the longitudinal strain (ϵL\epsilon_LϵL​) and lateral strain (ϵD\epsilon_DϵD​).

Compute Poisson's ratio (ν\nuν).

Calculate the cross-sectional area (AAA) of the cylinder.

Calculate the compressive stress (σ\sigmaσ).

Compute the modulus of elasticity (EEE).

Try one yourself

Still stuck on this one?

More Physics Questions

Calculate the longitudinal strain ( $\epsilon_L$ ) and lateral strain ( $\epsilon_D$ ).

Compute Poisson's ratio ( $\nu$ ).

Calculate the cross-sectional area ( $A$ ) of the cylinder.

Calculate the compressive stress ( $\sigma$ ).

Compute the modulus of elasticity ( $E$ ).