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.
Expert's answer

As per the question,

for rectangular block,




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}"


=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}"


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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