Question

A solid circular shaft is subjected to bending moment of 3000 N-m and a torque of 10000 N-m the shaft is made of 45 C 8 steel having ultimate tensile stress of 700 MPa and an ultimate shear stress of 500 MPa. Assuming a factor of safety as 6. determine the diameter of the shaft.

Answer 1

made of 45 C 8 steel having ultimate tensile stress of 700 MPa and an ultimate shear stress of 500 MPa. Assuming a factor of safety as 6. determine the diameter of the shaft.

Answer 2

Given :

M = 3000 N-m = 3 × 10⁶N-mm ;

T = 10 000 N-m = 10 × 10⁶ N-mm ;

σtu = 700 MPa = 700 N/mm² ;

τu = 500 MPa = 500 N/mm²

Explanation:

Now we know that  allowable tensile stress,

                     σt or σb  = σtu

OS = 700/6

= 116.7 N/mm2

shear stress,

                                     τ = τ u

FOS

                                         500/6

                                   83.3 N/mm2

Let d be the diameter of the shaft in mm.

Now , according to maximum shear stress theory,

equivalent twisting moment =

                               Te = √M² + T²

= √(3 × 106 )2 + (10 × 106 )2

= 10.44 × 106 N-mm

Hence (Te),

                               =10.44 × 106

                               = π * τ*d3

                    d3 = 10.44 × 106 / 16.36

                         = 0.636 × 106 or

                         = 86 mm

Answer = 86mm