A metal cube of initial dimension L_1 = 25.4mm, L_2 = 25.4mm, L_3 = 25.4mm has n

Question

A metal cube of initial dimension L_1 = 25.4mm, L_2 = 25.4mm, L_3 = 25.4mm has normal stresses applied to its 3 sides: sigma_1 = 344.7 MPa sigma_2 = -206.8 MPa sigma_3 unknown The steel material has sigma_Y = 413.7MPa and sigma = 827 epsilon^0.2 (MPa) The new specimen dimensions are: L_f3 = 21 mm (in direction of sigma_3) L_f2 = 23mm (in direction of sigma_2) Define the new dimension L_f1 (in direction of sigma_1) assuming plastic deformation Define the normal stress sigma_3 assuming plastic deformation

Explanation / Answer

Here the steps to be followed as -

1) We know stress = Force / Area - we can find the force on 1.

2) We know stress = Force / Area - we can find the force on 2.

Now assuming Poisson ratio = -(change in length / original length) / young's modulus - change in length for 1 can be found out .

Also , Youmg's modulus = stress / strain - Stress in 3 is obtained .

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