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.

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Netta00 Netta00 · Answer 1 · 2019-08-02T09:28:09+02:00

Answer:

[tex]\Delta a=-0.166 cm [/tex]

[tex]\Delta b=-0.249 cm [/tex]

[tex]\Delta c=-0.332 cm [/tex]

Explanation:

Given that E=600 KPa

Poisson ratio=0.45

We know that for hydroststic stress ,strain given as

[tex]\varepsilon =\dfrac{\sigma}{E}(2\mu -1)[/tex]

Here given that [tex]\sigma =-50 KPa[/tex]

Now by putting the values

[tex]\varepsilon =\dfrac{50}{600}(2\times 0.45 -1)[/tex]

[tex]\varepsilon =-0.00833[/tex]

Negative sign indicates that dimensions will reduces due to compressive stress

We know that strain given as

[tex]\varepsilon =\dfrac{\Delta L}{L}[/tex]

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

So [tex]\Delta a=-0.00833\times 20 cm [/tex]

[tex]\Delta a=-0.166 cm [/tex]

[tex]\Delta b=-0.00833\times 30 cm [/tex]

[tex]\Delta b=-0.249 cm [/tex]

[tex]\Delta c=-0.00833\times 40 cm [/tex]

[tex]\Delta c=-0.332 cm [/tex]