Design of a Knuckle Joint for an Axial Tensile Load of 18 kN

A knuckle joint is used to connect two rods subjected to tensile loads, allowing for slight flexibility or angular movement. It consists of three main parts: the single eye end, double eye end (fork end), and knuckle pin. Below is the step-by-step design procedure for a knuckle joint to carry an axial tensile load of 18 kN.

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Step 1: Selection of Material

• Material: Mild steel

• Properties:

  • Ultimate tensile strength ($S_{ut}$) = 435 MPa

  • Yield strength ($S_{yt}$) = 246 MPa

  • Yield shear strength ($S_{ys}$) = 154 MPa

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Step 2: Selection of Factor of Safety

• Factor of Safety (n): 2

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Step 3: Determination of Allowable Stresses

1. Design Tensile Strength ($S_{dt}$):

   $$S_{dt}=\frac{S_{yt}}{n}=\frac{246}{2}=123\text{MPa}$$

2. Design Shear Strength ($S_{ds}$):

   $$S_{ds}=\frac{S_{ys}}{n}=\frac{154}{2}=77\text{MPa}$$

3. Design Bearing Strength ($S_{dbr}$):

   $$S_{dbr}=S_{dt}=123\text{MPa}$$

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Step 4: Empirical Relations for Dimensions

The dimensions of the knuckle joint are initially estimated using empirical relations:

1. Diameter of rod ($d$): 15 mm

2. Diameter of knuckle pin ($d_p$): $d_p=d=15\text{mm}$

3. Diameter of eye or fork ($d_1$): $d_1=2d=30\text{mm}$

4. Thickness of each eye of fork ($t_1$): $t_1=1.2d=18\text{mm}$

5. Thickness of eye end ($t_2$): $t_2=0.75d=11.25\text{mm}\approx 12\text{mm}$

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Step 5: Design of Knuckle Pin

5a: Check for Shear Stress

The knuckle pin is subjected to double shear. The shear stress ($\tau$) is calculated as:

$$\tau =\frac{W}{2\cdot \frac{\pi }{4}{d}_p^2}$$$$\tau =\frac{18,000}{2\cdot \frac{\pi }{4}\cdot 15^2}=50.92\text{MPa}$$

Since $50.92\text{MPa}<77\text{MPa}$, the pin is safe in shear.

5b: Check for Bending Stress

The knuckle pin is subjected to bending stress. The bending stress (${\sigma }_b$) is calculated as:

$${\sigma }_b=\frac{32\cdot W\cdot l}{\pi \cdot d_p^3}$$

Assume $l=20\text{mm}$ (distance between the centers of the eyes):

$${\sigma }_b=\frac{32\cdot 18,000\cdot 20}{\pi \cdot 15^3}=231.35\text{MPa}$$

Since $231.35\text{MPa}>123\text{MPa}$, the pin is unsafe in bending. Increase the diameter of the knuckle pin to $d_p=20\text{mm}$.

Recalculate bending stress:

$${\sigma }_b=\frac{32\cdot 18,000\cdot 20}{\pi \cdot 20^3}=97.4\text{MPa}$$

Since $97.4\text{MPa}<123\text{MPa}$, the pin is now safe in bending.

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Step 6: Design of Eye End

6a: Check for Tensile Stress

The eye end is subjected to tensile stress across the slot. The tensile stress (${\sigma }_t$) is calculated as:

$${\sigma }_t=\frac{W}{(d_1-d_p)\cdot t_1}$$$${\sigma }_t=\frac{18,000}{(30-20)\cdot 18}=100\text{MPa}$$

Since $100\text{MPa}<123\text{MPa}$, the eye end is safe in tension.

6b: Check for Shear Stress

The eye end is subjected to shear stress. The shear stress ($\tau$) is calculated as:

$$\tau =\frac{W}{(d_1-d_p)\cdot t_1}$$$$\tau =\frac{18,000}{(30-20)\cdot 18}=100\text{MPa}$$

Since $100\text{MPa}>77\text{MPa}$, the eye end is unsafe in shear. Increase $d_1$ to 34 mm.

Recalculate shear stress:

$$\tau =\frac{18,000}{(34-20)\cdot 18}=71.42\text{MPa}$$

Since $71.42\text{MPa}<77\text{MPa}$, the eye end is now safe in shear.

6c: Check for Crushing Stress

The eye end is subjected to crushing stress. The crushing stress (${\sigma }_{cr}$) is calculated as:

$${\sigma }_{cr}=\frac{W}{d_p\cdot t_1}$$$${\sigma }_{cr}=\frac{18,000}{20\cdot 18}=50\text{MPa}$$

Since $50\text{MPa}<123\text{MPa}$, the eye end is safe in crushing.

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Step 7: Design of Fork End

7a: Check for Tensile Stress

The fork end is subjected to tensile stress across the slot. The tensile stress (${\sigma }_t$) is calculated as:

$${\sigma }_t=\frac{W}{(d_1-d_p)\cdot 2t_2}$$$${\sigma }_t=\frac{18,000}{(34-20)\cdot 2\cdot 12}=53.57\text{MPa}$$

Since $53.57\text{MPa}<123\text{MPa}$, the fork end is safe in tension.

7b: Check for Shear Stress

The fork end is subjected to shear stress. The shear stress ($\tau$) is calculated as:

$$\tau =\frac{W}{(d_1-d_p)\cdot 2t_2}$$$$\tau =\frac{18,000}{(34-20)\cdot 2\cdot 12}=53.57\text{MPa}$$

Since $53.57\text{MPa}<77\text{MPa}$, the fork end is safe in shear.

7c: Check for Crushing Stress

The fork end is subjected to crushing stress. The crushing stress (${\sigma }_{cr}$) is calculated as:

$${\sigma }_{cr}=\frac{W}{d_p\cdot 2t_2}$$$${\sigma }_{cr}=\frac{18,000}{20\cdot 2\cdot 12}=37.5\text{MPa}$$

Since $37.5\text{MPa}<123\text{MPa}$, the fork end is safe in crushing.

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Final Design Specifications:

1. Rod:

   • Diameter ($d$) = 15 mm

2. Knuckle Pin:

   • Diameter ($d_p$) = 20 mm

3. Eye End:

   • Diameter ($d_1$) = 34 mm

   • Thickness ($t_1$) = 18 mm

4. Fork End:

   • Thickness ($t_2$) = 12 mm

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This design ensures the knuckle joint can safely carry an axial tensile load of 18 kN.