Question

A steel cubic block of side 200 mm is subjected to hydrostatic pressure of 250 N/mm². The elastic modulus is \( 2 \times 10^5 \) N/mm² and Poisson ratio is 0.3 for steel. The side of the block is reduced by \(\underline{\hspace{2cm}}\) \(\text{mm (round off to two decimal places).}\)

Hint:

The reduction in dimensions due to pressure is related to the material's elastic properties and the applied pressure. Poisson's ratio plays a key role in this calculation.

Answer 1

Correct Answer: 0.08 - 0.12

To solve the problem of determining the reduction in the side of a steel cubic block due to hydrostatic pressure, we first need to understand the relationship between stress, strain, and changes in dimensions.

The hydrostatic pressure \( P \) applied on the block is 250 N/mm².

The Elastic Modulus \( E \) is \( 2 \times 10^5 \) N/mm² and the Poisson ratio \( \nu \) is 0.3 for the steel.

The volumetric strain \( \epsilon_v \) under hydrostatic pressure is given by:

\( \epsilon_v = \frac{3(1-2\nu)P}{E} \)

Substitute the known values:

\( \epsilon_v = \frac{3(1-2 \times 0.3) \times 250}{2 \times 10^5} \)

\( \epsilon_v = \frac{3 \times 0.4 \times 250}{2 \times 10^5} \)

\( \epsilon_v = \frac{300}{2 \times 10^5} \)

\( \epsilon_v = 0.0015 \)

Volumetric strain \( \epsilon_v \) also relates to change in volume \(\Delta V\) per unit volume: \( \epsilon_v = \frac{\Delta V}{V} \), where \( V \) is the original volume of the cube.

For a cubic shape, if the original side length is \( a = 200 \) mm, the change in side length can be related to volumetric strain by:

\( \Delta a = a \times \frac{\Delta V}{3V} = a \times \frac{\epsilon_v}{3} \)

Computing \( \Delta a \):

\( \Delta a = 200 \times \frac{0.0015}{3} \)

\( \Delta a = 200 \times 0.0005 \)

\( \Delta a = 0.10 \) mm

Thus, the side of the block is reduced by 0.10 mm.