Knuckle Joint

KNUCKLE JOINT

• PROBLEM: Design a knuckle joint to transmit 150KN. The design stresses may be taken as 80 Mpa in tension, 60 Mpa in shear and 150 Mpa in compression.

MAJOR COMPONENTS OF KNUCKLE JOINT (1) (2) (3) (4) (5)

Single eye end Double eye end Knuckle pin Collar Taper pin / Split pin

Knuckle Joint Assembly Knuckle Joint Assembly

Single Eye End Single Eye End

Double Eye End Double Eye End

Knuckle Pin Knuckle Pin

Collar

Collar

TaperTaper Pin / Split Pin Pin / Split Pin

Knuckle Joint Assembly Knuckle Joint Assembly

(1) Failure of the Solid rod in tension:

(1) Failure of the of Solid rod in tension: Let d = Diameter the rod Let d =transmitted Diameter of(P), the rod The load 3 2 load transmitted 150The x 10 = (Π/4) x d2 x σ(P), = (Π/4) x d x 80 t 3 2 2 150 x 10 = (Π/4) x d x σ = (Π/4) x d x 80 d = 15.45 say 16 mm. t

= 15.45 say 16 =mm. So,dRod Diameter 16mm So, Rod Diameter = 16mm

(2) Failure of the Knuckle Pin in shear: Because of tensile load P, knuckle pin may fail in double shear. dp = Knuckle pin diameter τ = Allowable shear stress P = 2 x (Π/4) x dp2 x τ 15000 = 2 x (Π/4) x dp2 x 60 So, dp = 12.61mm

Generally, pin diameter is taken equal to rod diameter. So, dp = d = 16 mm

(3) Failure of the single eye or rod in tension: Generally, outer diameter of eye D1 and thickness t of single eye end is taken from the dimension of pin diameter. So, D1 = 2 dp = 2 x 16 = 32mm t = 1.2d = 1.2 x 16 = 19.2mm say 20mm

Now, P = (D1-dp) x t x σ t 15000 = (32-16) x 20 x σ t σ t = 46.88 N/mm2 ≤ [σ t ]

(4) Failure of the single eye hole and knuckle pin crushing: P = dp x t x σ c 15000 = 16 x 20 x σ c σc = 46.88 n/mm2 So, σ c ≤ [σ c] So, the eye end and pin is safe in crushing failure.

(5) Failure of the fork end in tensile: P = 2 x C x t1 x σ t t1= t/2 = 20 / 2 = 10mm And C = 2 t1 = t =20 taken, 15000 = 2 x 20 x 10 x σ t σ t = 37.5 N/mm2 So, σ t ≤ [σ t] So, the fork is safe in tensile failure.

(6) Failure of the fork eye end at pin hole section in tensile: P = 2 x (D1 – dp ) x t1 x σ t 15000 = 2 x ( 32-16) x 10 x σ t σ t = 46.88 N/mm2 So, σ t ≤ [σ t] So, the fork eye end at pin hole section is safe in tensile failure.

(6) Failure of the single eye end in double shear: P = 2 x { (D1 – dp )/2 } x t x τ 15000 = 2 x { ( 32-16) / 2 } x 20 x τ τ = 46.88 N/mm2 So, τ≤ [τ] So, the single eye end is safe in double shear.