Question

A surface grinding operation has been performed on a Cast Iron plate having dimensions 300 mm (length) \( \times \) 10 mm (width) \( \times \) 50 mm (height). The grinding was performed using an alumina wheel having a wheel diameter of 150 mm and wheel width of 12 mm. The grinding velocity used is 40 m/s, table speed is 5 m/min, depth of cut per pass is 50 μm and the number of grinding passes is 20. The average tangential and average normal force for each pass is found to be 40 N and 60 N respectively. The value of the specific grinding energy under the aforesaid grinding conditions is \(\underline{\hspace{2cm}}\) J/mm³ (round off to one decimal place).

Hint:

The specific grinding energy is influenced by the tangential force, grinding velocity, wheel dimensions, and the number of grinding passes.

Answer 1

Correct Answer: 38 - 39

The specific grinding energy \( E_g \) is calculated using the formula: \[ E_g = \frac{F_{\text{tangential}} \cdot v_{\text{grinding}}}{b \cdot d \cdot N} \] Where:
- \( F_{\text{tangential}} = 40 \ \text{N} \),
- \( v_{\text{grinding}} = 40 \ \text{m/s} \),
- \( b = 12 \ \text{mm} \) (width of the grinding wheel),
- \( d = 50 \ \text{μm} = 0.05 \ \text{mm} \) (depth of cut per pass),
- \( N = 20 \) (number of passes).
Substituting the values: \[ E_g = \frac{40 \times 40}{12 \times 0.05 \times 20} = \frac{1600}{12 \times 1} = 38.0 \ \text{J/mm}^3 \] Thus, the specific grinding energy is: \[ \boxed{38.0 \ \text{J/mm}^3} \]