A cylindrical metal specimen 12.7 mm (0.5 in.) in diameter and 250 mm (10 in.) long is to be
subjected to a tensile stress of 28 MPa (4000
psi); at this stress level the resulting deformation
will be totally elastic.

If, in addition, the maximum permissible diameter
decrease is 1.2 103 mm (4.7 105 in.)
when the tensile stress of 28 MPa is applied,
which of the metals that satisfy the criterion in
part (a) are suitable candidates? Why?

*( part a was : If the elongation must be less than 0.080
mm (3.2 103 in.), which of the metals in
Table 6.1 are suitable candidates? Why? )

Steel, Titanium and Tungsten

Explanation:

Given:-

- The diameter of the cylindrical specimen, d = 12.7 mm

- The length of the cylindrical specimen, L = 250 mm

- The tensile stress in the specimen, σ = 28 MPa

- Assume elastic deformation of the material up-to the specified tensile stress

Solution:-

- To determine or select suitable materials that are subjected to the given stress ( σ ) and an elongation constraint ( Δl ) or contraction of diameter ( Δd ). We need material intrinsic properties and relate them to our analysis.

- We studied two material intrinsic properties that is the Elastic Modulus ( E ) and Poisson ratio ( v ).

- Since, the question poses a constraint on the diameter. Note that the axis of stress applied ( axial ) and the contraction dimension are orthogonal to each other. In such case, we will determine the poisson ratio ( v ) of the suitable material as follows:

- We will first determine the permissible strain in the diametrical direction ( εx ). It is the ratio of allowable contraction of diameter ( Δd ) to the initial diameter ( d ):

                              εx = Δd / d

                              εx = ( -1.2*10^-3 )  / ( 12.7 )

                              εx = -0.00009

- The poisson ratio ( v ) of a material is defined as the negative ratio of the transversal strain ( εx ) to the axial strain ( εz ). The axial strain ( εz ) is determined as the ratio of elongation in length ( Δl ) to the initial length ( L ) of the specimen as follows:

                              εz = Δl / L

                              εz = ( 0.08 )  / ( 250 )

                              εz = 0.00032

- The poisson ratio ( v ) is expressed as:

                             v = - [ εx / εz ]

                              v = [ 0.00009 / 0.00032]

                              v = 0.2812

- Hooke's law gives us a linear relation between the applied stress ( σ ) and the engineering strain in the direction of applied stress. Since, the specimen is subjected to tensile stress and abides by the Hooke's law ( Elastic deformation ). The modulus of elasticity ( E ) is given as :

                              E = σ / εz

                              E = (28*10^6 ) / 0.00032

                              E = 87.5 GPa

- As per maximum elongation allowed the modulus of elasticity calculated above is the minimum value that must be satisfied by the selected material. We will use a list of materials ( ASTM ) standard and list a few that meet the elongation criteria as follows:

                  Copper, Nickel , Steel, Titanium , Tungsten

- From the above selected metals the most closely associated poisson ratio ( v ) calculated above would be:

                   Metals                         poisson ratio ( v ) range

                   Copper                                     0.33

                   Nickel                                       0.31

                   Steel                                    0.27–0.30

                   Titanium                              0.265–0.34

                   Tungsten                                  0.27

- The suitable materials would be:

                         Steel, Titanium and Tungsten

answer:

rawr

explanation: