Question

Determine handle diameter , lever arm cross-secion , and journal diameter if :

effort = 400 N

bending sterss = 50 N/mm^2

bear stress = 40 N/mm^2

Answer 1

Let D = Diameter of the handle

Bending moment, M = P × L = 400 × 450 = 180 × 10³ N-mm

The twisting moment depends upon the point of application of the effort.

T = 400 × 100 = 40 × 10³ N-mm

We know that equivalent twisting moment,

$T_e = \sqrt{(M²+T²)} = 184.4 × 10³ N-mm$

We also know that equivalent twisting moment (T_e

),

$184.4 × 10³ = \dfrac{π}{16} × Torque × D³$

$D³ = \dfrac{184.4× 10³}{10.8} = 17.1 × 10³$

D = 25.7 mm