Question

i. A solid circular shaft is subjected to a bending moment of 3000 N-m and torque of 1000 N-m. The shafts are made of 45 C8 steel having ultimate tensile stress of 700 MPa and ultimate shear stress of 500 MP: Assuming | factor of safely as 6, determine the diameter of shafts.

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Answer 1

Answer:

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Answer 2

Answer:

The answer is 86 mm.

Explanation:

Given: M= 3000-m = 3×106 N-mm ;

          T= 10000 N-m = 10×10⁶ N-mm ;

          σtu = 700 MPa = 700 N/mm² ;

          ζυ = 500 MPa = 500 N/mm²

We know that the allowable tensile stress,

σt or σb = σtu

FOS = 700/6

       = 116.7 N/mm²

and allowable shear stress,

ζ = ζu

FOS = 500/6

       = 83.3 N/mm²

Let d= diameter of the shaft in mm.

According to the maximum shear stree thjeory, equivalent twisting moment,

Te = √M²+T²

    = √(3×106)²+(10×106)²

   = 10.44 × 106 N-mm

We also know that equivalent twisting moment (Te),

 10.44×106 = π*ζ*d³

16

d³ = 10.44×106/16.36

   =0.636×106

  = 86mm

Hence the answer is 86mm.