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# Answer to Question #129714 in Mechanics | Relativity for Hamid

Question #129714
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.
1
2020-08-17T09:23:26-0400

As per the question,

for rectangular block,

length=125mm

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.

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