PROBLEM 1.2

Find the minimum diameter of a steel wire, which is used to raise a load of 6000 N, if the stress in the rod is not to exceed 105 MN/m² .

Sol:

As per given data:

            Load           P = 6000 N

            Stress        σ = 105 MN/m²    = 105 x 10 ⁶ N/m²            (1 MN = 10⁶ N)

                                  = 105 N/mm²                                        ( 10⁶ N/m² = 1 N/mm²)

 Let       D = The diameter of the rod in mm

Area      A = (π/4) D²

We know that, Stress  σ = Load/Area = P/A

                                    105 = 6000/(π/4) D²

                                     D² = 6000 x 4 / π x 105

                                     D   = 72.75