A circular rod of length 25 cm and diameter 8mm is made of g

A circular rod of length 25 cm and diameter 8mm is made of grade 1045 steel (E_steel=207 GPa, S_y= 414 Mpa). Calculate the stress and strain in the rod, and its extension, when it is subjected to 5kN of tension; At what force would the rod begin to yield? By what amount would the rod have to be stretched beyond its original length in order to yield?

Solution

Given data

   Circular rod length =25 cm =0.25m

   Circular rod diameter d = 8mm =0.008m

   E steel= 207 GPa, Sy=414 Mpa

To find

1)Stress,strain,extension of rod when P= 5KN

2)At what force would the rod begin to yield?

3)By what amount would the rod have to be stretched beyond its original length in order to yield?

Solution 1

                            Cross sectional area A= d^2/4

                                                                    = (*(0.008)^2)/4 =5.027 x 10^-5 m^2

                               Tensile stress = F/A

                                                             = 5000/5.027x10^-5

                                                              = 9.95 x10^7 Pa

                                                             =99.5 Mpa

                                    Strain = /E = (99.5 x 10^6)/( 207 x10^9) = 4.81 x10^-4

                                                                Strain = 4.81 x10^-4

                                    Elongation L = L = (4.81 × 10–4)(0.25) =120 × 10^-6 m

                                                        Elongation L = 120 × 10–6 m =120 m

Solution 2   The rod begins to yield when the force has magnitude

                                         F = A = SyA = 414 × 10^6 *5.027 × 10–5

                                                                F = 20.8kN

Solution 3 The rod begins to yield when it has been stretched by amount

                                                     = /E=Sy/E= 414 x 10^6/ 207 x 10^9 =2 x 10^-3

                                                    L = L= 2 x10^-3 *0.25 =5 x 10^-4 m =500 x 10^-6

                                                   L = 500 m