Question

In a single pass drilling operation, a through hole of 15 mm diameter is to be drilled in a steel plate of 50 mm thickness. Drill spindle speed is 500 rpm, feed is 0.2 mm/rev and drill point angle is 118°. Assuming 2 mm clearance at approach and exit, the total drill time in seconds is
1. 35.1
2. 32.4 
3. 31.2
4. 30.1

Answer 1

Correct Answer - Option 1 : 35.1

Concept:

Machining time in drilling can be calculated by:

\({T_m} = \frac{{L + AP + OR + X}}{{f \times N}}\)

where L = thickness of the plate to be drilled

AP = approach

OR = over travel

X = necessary approach

The necessary approach can be calculated using:

\(X = \frac{D}{{2 \times \tan \beta }}\)

where D = diameter of the hole to be drilled, β = half drill point angle.

Calculation:

Given:

L = 50 mm, D = 15 mm N = 500 rpm, f = 0.2 mm/rev, AP = 2 mm, OR = 2 mm

2β = 118° ⇒ β = 59°

Necessary approach (X)

\(X = \frac{D}{{2 \times \tan \beta }} = \frac{{15}}{{2 \times \tan 59^\circ }} = 4.5\;mm\)

Machining time (T_m)

\({T_m} = \frac{{L + AP + OR + X}}{{f \times N}}\)

\({T_m} = \frac{{50 + 2 + 2 + 4.5}}{{0.2 \times 500}} = 0.585\;minute = 35.1\;seconds\)