Answer to Question #170093 in Mechanical Engineering for ahmed

Question

Draw a sketch for a hand lever of effective length= 600 mm , hand force = 100 N

tensile stress = 7000 N/m^2 , shear sress = 6000 N/m^2

Determine ;

shaft diameter in the boss and in the bearing ,

boss diameter and length ,

width and thickness of arm .

Answer 1

P x L = $\pi$ /16 x q x d³

P - hand force; L - effective length; q - shear sress; d - shaft diameter in the boss

100 N x 0.6 m = $\pi$ /16 x 6000 N/m² x d³

d = 0.370 m = 370 mm 

The diameter of boss d₂ = 1.6 x d = 1.6 x 370 mm = 592 mm

Lenght of boss l₁ = (2 x P x L) / {t₂ x $\sigma$ x (d + t₂)}

$\sigma$ - tensile stress; t₂ = thickness of boss = 0.3d = 0.3 x 0.37 m = 0.111 m

l₁ = (2 x 100 N x 0.6 m) / {0.111 m x 7000 N/m² x (0.370 m + 0.111 m)} = 0.321 m = 321 mm

P x (L² + l²)^1/2 = $\pi$ /16 x q x d₁³

l = 2 x l₁ = 2 x 0.321 m = 0.642 m; d₁ - shaft diameter in the bearing

100 N x ((0.6 m)² +(0.642 m)²)^1/2 = $\pi$ /16 x 6000 N/m² x d₁³

d₁ = 0.422 m = 422 mm