CHAPTER (6)

Drilling

Dr. Ahmed Abou El-Wafa

6.1 Introduction

Drilling is a process used extensively by which

through or blind holes are originated or

enlarged in a workpiece. Drilling is considered

a roughing operation and, therefore, the

accuracy and surface finish in drilling are

generally not of much concern. If high

accuracy and good finish are required,

drilling must be followed by some other

operation such as reaming, boring, or

grinding. The most commonly employed

drilling tool is the twist drill, which is available

in diameters ranging from 0.15 to 80 mm.

Drill

f

v

Workpiece

Drilling operation

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Referring to the terminology of twist drill, the helix angle of the twist

drill is the equivalent of the rake angle of other cutting tools. The

standard helix is 30°, which, together with a point angle of 118°, is

suitable for drilling steel and CI.

-Nomenclature of twist drill

6.2 Drilling with Twist Drills

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Friction makes matters worse in two ways. In addition to the usual friction

between the chip and the drill, friction also results from rubbing between

the outside of the drill and the hole wall. This generates heat, which

causes the temperature of the drill and work to rise. Delivery of cutting

fluid to the drill point is difficult because the chips are flowing in the

opposite direction. Because of chip removal and heat, a twist drill is

normally limited to a hole depth no greater than four times its

diameter.

Some twist drills are made with internal holes running through their

length which cutting fluid can be pumped into the hole near the drill point.

An alternative approach with twist drills that do not have fluid holes is to

periodically withdraw the drill from the hole to clear the chips before

proceeding deeper.

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N = 1000 v / π.D

Where:

N : Spindle speed (r.p.m)

v : Cutting speed (m/min.)

D : Drill diameter (mm)

and: fr = N . f

Where:

f : Feed (mm/rev.)

fr : Feed rate (mm/min.)

6.3 Cutting Conditions in Drilling

6.3.1 Machining time

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- Machining (drilling) time, Tm (min.), can be calculated

based on whether the hole is through or blind type:

A

tw.p

drill

Through hole Blind hole

r

m f

AtT

Where: t : Work thickness (mm) A: Approach allowance accounts for

drill point angle, and it can be given by: A = 0.5 D . tan (90 - /2)

: Drill point angle.

r

m f

dT

Where: d : Hole depth (mm)

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d

w.p

drill

6.3.2 Metal (Material) removal rate (MRR)

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fDMRR r

2 )mm3/min(.

This equation is valid only after the drill reaches full diameter

and excludes the initial approach of the drill into the work.

SURFACESPEED

FEED, mm/rev (in./rev)DRILL DIAMETER

RPMWORKPIECEMATERIAL

m/min ft/min 1.5 mm(0.060 in.)

12.5 mm(0.5 in.)

1.5 mm 12.5 mm

Aluminum alloysMagnesium alloysCopper alloysSteelsStainless steelsTitanium alloysCast ironsThermoplasticsThermosets

30-12045-12015-6020-3010-206-20

20-6030-6020-60

100-400150-40050-20060-10040-6020-6060-200100-20060-200

0.025 (0.001)0.025 (0.001)0.025 (0.001)0.025 (0.001)0.025 (0.001)0.010 (0.0004)0.025 (0.001)0.025 (0.001)0.025 (0.001)

0.30 (0.012)0.30 (0.012)0.25 (0.010)0.30 (0.012)0.18 (0.007)0.15 (0.006)0.30 (0.012)0.13 (0.005)0.10 (0.004)

6400-25,0009600-25,0003200-12,0004300-64002100-43001300-4300

4300-12,0006400-12,0004300-12,000

800-30001100-3000400-1500500-800250-500150-500500-1500800-1500500-1500

Note: As hole depth increases, speeds and feeds should be reduced. Selection of speeds andfeeds also depends on the specific surface finish required.

General recommendations for speeds and feeds in drilling

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6.3.3 Cutting power

Power=

60000

TN2kW

Where:T : The torque required to operate a drill (N.m), and has been found

experimentally to be:

T = C. f 0.75 D1.8 N.m

C : A constant depending on the w.p. material, found from the following Table:

Carbon tool steel

Mild Steel

Cast iron

Soft brass

AluminumMaterial being

drilled

0.4 0.36 0.07 0.084 0.11 C

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6.4 Operations Related to Drilling

- Reaming: Reaming is used to slightly enlarge a hole, to provide a better tolerance, and to improve its surface finish. The tool is called a reamer.

- Tapping: This operation is performed by a tap and is used to provide internal screw threads on an existing hole.

- Counterboring ( الأسطواني :(التخويشis the operation of enlarging the end of a hole cylindrically, as for the recess for a counter-sunk rivet. The tool used is known as counter-bore.

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6.4 Operations Related to Drilling (cont.)

- Countersinking ( المخروطي :(التخويشis the operation of making a coneshaped enlargement of the end of a hole, as for the recess for a flat head screw. This is done for providing a seat for counter sunk heads of the screws so that the latter may flush with the main surface of the work.

- Centering: Also called centerdrilling, this operation drills a starting hole to accurately establish its location for subsequent drilling. The tool is called centerdrill.

- Spotfacing: This is the operation of removing enough material to provide a flat surface around a hole to accommodate the head of a bolt or a nut. A spot-facing tool is very nearly similar to the counter-bore.

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6.5 Drill Presses

(a) Upright drill press

These machines are used for

machining holes up to 50 mm in

diameter in relatively small-size

work. It has a wide range of spindle

speeds and feeds. Therefore, they

are employed not only for drilling

from solid material, but also for core

drilling, reaming, and tapping

operations.

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(b) Radial drill press These machines are especially designed for drilling, counterboring, countersinking, reaming, and tapping holes in heavy and bulky WPs that are inconvenient or impossible to machine on the upright drilling machines. They differ from upright drill presses in that the spindle axis is made to coincide with the axis of the hole being machined by moving the spindle in a system of polar coordinate to the hole, while the work is stationary.

6.5 Drill Presses (Cont.)

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(c) Gang drill pressThe spindles are arranged in a row, and each spindle is driven by its own motor. The gang machine is in fact several upright drilling machines having a common base and single worktable. They are used for consecutive machining of different holes in one workpiece, or for the machining of a single hole with different cutting tools.- A related machine is the multiple-spindle drilling machine which is used to drill a number of holes in a job simultaneously and to reproduce the same pattern of holes in a number of identical pieces in a mass production work. This machine has several spindles and all the spindles holding drills are fed into the work simultaneously. Feeding motion is usually obtained by raising the worktable.

6.5 Drill Presses (Cont.)

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6.6 Work Holding Devices

Examples of the drilling jigs

1- Vise (for small workpieces) 2- Worktable surface, using standard tee slots (for large workpieces) 3- Drilling Jigs (for mass production).

Drilling jigs are special devices designed to hold a particular workpiece and guide the cutting tool. Jigs enable work to be done without previously laying out the workpiece. Drilling using jigs is, therefore, accurate and quicker than standard methods. However, larger quantities of workpieces must be required to justify the additional cost of the equipment.

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Example 6.1

A drilling operation is to be used to drill 10 mm diameter hole to a certain depth. It takes 4.5 min. to perform the drilling operation. The cutting conditions are N = 3000 rev/min at a feed = 0.2 mm/rev. To improve the surface finish in the hole, it has been decided to increase the speed by 20% and decrease the feed by 25%. How long will it take to perform the operation at the new cutting conditions.

Given: D = 10 mm (blind hole), Tm1 = 4.5 min., N1 = 3000 rpm, f1 = 0.2 mm/rev.N2 = N1 + 0.2 N1 , f2 = f1 – 0.25 f1

Required: Tm2

Solution

N2 = 3000 + 0.2 * 3000 = 3000 * 1.2N2 = 3600 rpm

f2 = 0.2 - 0.25 * 0.2 = 0.2 * 0.75

f2 = 0.15 mm/rev. 15

fr2 = N2 f2

fr2 = 540 mm/min.

From 1st step:

Tm1 = d / fr1

4.5 = d / (3000 * 0.2)

d = 2700 mm.

From 2nd step:

Tm2 = d / fr2

= 2700 / 540

Tm2 = 5 min.

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Example 6.2

A 15 mm drill is drilling mild steel (C = 0.36) at 80 m/min. If the feed

is 0.2 mm/rev., calculate the torque, and the input power if frictional

losses are equivalent to 25% of the cutting power. Find also the

volume of metal removed.

Solution

Given: D = 15 mm, C = 0.36, v = 80 m/min., f = 0.2 mm/rev., and frictional losses = 25% of the cutting power.

Required: (a) Torque (T) (b) Input power (Pinp) (c) Metal Removal Rate (MRR)

(a) 8.175.0 DfCT

= 0.36 * (0.2)0.75 * (15)1.8

T = 14.1 N.m

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(b) Cutting power Pcut = 60000

TN2

N = 1000 v / (π D)

N = 1000 * 80 / (π * 15) = 1697.65

r.p.m

Pcut = 60000

1.14*65.1697*2= 2.5 kW

Input power (Pinp) = Pcut + 0.25 Pcut = 1.25 * Pcut

Pinp = 1.25 * 2.5

Pinp = 3.13 kW

(c) Metal Removal Rate (MRR) = 4

fD r

2

fr = N . f = 1697.65 * 0.2 = 339.53 mm/min.

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53.339*15*MRR

2

MRR = 59999.9 ≈ 60000 mm3/min. 18

Web site:

http://www.staff.zu.edu.eg/awafa/

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