Question

Design the rectangular key for a shaft of 50 mm diameter. The shearing and crushing stresses for the key

material are 42 MPa and 70 MPa.​

Answer 1

Answer:

material are 42 MPa and 70 MPA

Answer 2

Given:

d= 50 mm

τ = 42 Mpa = 42 N/mm²:

σ= 70 Mpa = 70/mm²

Explanation:

width of key, w= d/4                                 w  = 12.5 mm

and thickness of key as d/6                     t  = 8.3 mm

The length pf key is obtained by considering the key in shearing and crushing.

Let                      l = Length of key.

Considering shearing of the key. We know that shearing strength(or torque transmitted) of the key,

           T= l × w ×τ× (d/2)

             = l×17×42×(50/2)     =  13125 l N-mm

and torsional shearing strength of the shaft,

T= (π/16)×τ×d³   = (π/16)×42(50)³

                           =1.03×10³×² N-mm

 From Eqs. we have

         l = 1.03×(10^6/13125)

           = 79.25 mm

N ow considering crushing the key. We know that shearing strength of the key,

 T = l×(t/2)×σ×(d/2)

     = l×(8.3/2)×70×(50/2)×

     = 7262.5 l N-mm

From Eqs.  we get

         l =1.03×(10^6/8750)

           = 141.8 mm

Taking larger of the two values, we have length of key,

          l= 141.8 = 142

Answer = 142