Answer to Question #129714 in Mechanics | Relativity for Hamid

Question

A rectangular block of metal of 50 mm x 25 mm cross-section and 125 mm length carries a tensile load of 100 kN along its length, a compressive load of 1.0 MN on its 50 x 125 mm faces and a tensile load of 400 kN on its 25 x 125 mm faces. If E = 208 GN/m² and Poisson’s ratio = 0.3, Find (a) change in volume of bar (b) increase required in 1.0 MN load to produce no change in volume.

Answer 1

As per the question,

for rectangular block,

length=125mm

breadth=50mm

height=25mm

Load along its length=100 KN

Load along its breadth= -1 MN=-1$\times10^3$ KN( since It is compressive load)

Load along its height=400 KN

Young's modulus=208GN/m$^2$ =208$\frac{\times 10^6}{10^6}$ KN/mm$^2$ =208KN/mm$^2$

poission's ratio=0.3

Stress in the x direction $\sigma_x$ =$\frac{load in x direction}{breadth\times height}$

=$\frac{100}{25\times 50}$

=80 N/mm$^2$

Stress in the y direction $\sigma_y$ =$\frac{load in y direction}{lenght\times breadth}$

= -$\frac{1000}{50\times125}$

=-100 N/mm$^2$

Stress in the z direction $\sigma_z$ =$\frac{load in z direction}{length\times height}$

=$\frac{100}{125\times50}$

=128 N/mm$^2$

Let the volume of block be V,

then V=$length\times breadth\times height$

=$125\times 50\times 25$

=156250 mm$^3$

Now the change in volume to original volume is given by,

$\frac{dV}{V}= \frac{(\sigma_x -\sigma_y+\sigma_z)(1-2u)}{E}$

=$\frac{(80-100+128)(1-2\times 0.3)}{208\times 10^3}$

=0.0002076

Now the change in volume,

dV =0.0002076$\times$ V

=0.0002076$\times$ 156250

=32.4375 mm$^3$

Hence the change in volume is 32.4375mm$^3$

(B) To keep the volume of the Block zero, we have to make the net load applied to zero.

This can be done by making the compressive 1 MN load to 0.5 MN.