Question

A rectangular block having dimensions 20 cm X 30 cm X 40 cm is subjected to a hydrostatic stress of -50 kPa (i.e. under compression). Calculate the change in length of each side. Data: Young's modulus of the block E = 600 kPa, Poisson ratio v=0.45.

Answer 1

Answer:

$\Delta a=-0.166 cm$

$\Delta b=-0.249 cm$

$\Delta c=-0.332 cm$

Explanation:

Given that E=600 KPa

Poisson ratio=0.45

We know that for hydroststic stress ,strain given as

$\varepsilon =\dfrac{\sigma}{E}(2\mu -1)$

Here given that $\sigma =-50 KPa$

Now by putting the values

$\varepsilon =\dfrac{50}{600}(2\times 0.45 -1)$

$\varepsilon =-0.00833$

Negative sign indicates that dimensions will reduces due to compressive stress

We know that strain given as

$\varepsilon =\dfrac{\Delta L}{L}$

Lets take a=20 cm,b=30 cm,c=40 cm.

So $\Delta a=-0.00833\times 20 cm$

$\Delta a=-0.166 cm$

$\Delta b=-0.00833\times 30 cm$

$\Delta b=-0.249 cm$

$\Delta c=-0.00833\times 40 cm$

$\Delta c=-0.332 cm$