A knuckle joint is to be design to connect two mild steel bars under a tensile load of 150 KN. The allowable stresses are 75Mpa in tension, 50Mpa in shear and 150 Mpa in crushing. (Assume empirical relations as Diameter of knuckle pin$ d_1 = d$ , Outer diameter of eye $d_2 = 2d$, diameter of knuckle pin head and collar $d_3 = 1.5d$, thickness of single eye t = 1.25d, thickness of fork $t_1 = 0.75d$, thickness of pin head $t_2 = 0.5d$)

1. Draw neat sketch of knuckle joint.
2. Find the diameter of the rod (d).
3. Using empirical find all dimensions.
4. With neat sketches for failure cross section areas check all components under different failures.

Subject: Machine Design -I

Topic: Design against static Loads, Bolted and welded joints under eccentric loading. Power Screw

Difficulty: Medium

1. Draw a neat sketch of knuckle joint:

1. FInd the diameter of the rod(d):

$d=\sqrt \frac{4P}{\sigma_t \pi}=\sqrt \frac{4(150*10^3)}{\pi *75}=50.462 or 52mm$

1. Using empirical find all dimensions:

Diameter of knuckle pin:

$d_1=d=52mm$

Outer diameter of eye:

$d_2=2d=104mm$

Diameter of knuckle pin head and collar:

$d_3=1.5d=78mm$

Thickness of single eye:

$t=1.25d=656mm$

Thickness of fork:

$t_1=0.75d=39mm$

Thickness of pin head:

$t_2=0.5d=26mm$

1. Check for components under different failures:

Stresses in single eye/eye:

Tensile stress:

$\sigma_t=\frac{p}{t(d_2-d)}=44.37MPa$

Crushing stress:

$\sigma_c=\frac{p}{td}=44.37MPa$

Shear stress:

$\tau=\frac{p}{t(d_2-d)}=44.37MPa$

Stresses in double eye/fork:

Tensile stress:

$\tau_t=\frac{p}{2t_1(d_2-d)}=36.98MPa$

Crushing stress:

$\sigma_c=\frac{p}{2t_1d}=36.98MPa$

Shear stress:

$\tau=\frac{p}{2t_1(d_2-d)}=36.98MPa$