Lecture eighteen:

Assistant professor Sahib M. Mahdi - 1 -

(( Manufacturing )) Example 21.1 Orthogonal Cutting

In a machining operation that approximates orthogonal cutting, the cutting tool has a

rake angle = 10o. The chip thickness before the cut to = 0.50 mm and the chip

thickness after the cut tc = 1.125 in. Calculate the shear plane angle and the shear

strain in the operation.

Solution: The chip thickness ratio can be determined from Eq. (21.2):

The shear plane angle is given by Eq. (21.3):

Finally, the shear strain is calculated from Eq. (21.4):

The main surfaces & planes in cutting process:

1 – The cutting surface in which the cutting force act along.

2 – The Rake face surface in which the friction force act along.

3 – The shear plane in which the shear force act along.

Velocity in Chip Formation:

The velocity of the chip is defined as the velocity with which the chip moves over the

rake face, the velocity with which the metal shears along the shear plane. The relationship

between these two velocities and the cutting velocity can be readily determined as following:

Again consider fig. ( 12 ) below:

Where: Vc = Vo ( The cutting velocity ).

Vs = The shear velocity.

Vf = The chip velocity.

= The Rake velocity.

= The shearing angle.

From trigonometric ( sine law ), it could be gotten that:

𝑽𝒔

𝒔𝒊𝒏 ( 𝟗𝟎 − 𝜶 ) =

𝑽𝒐

𝒔𝒊𝒏 [ 𝟗𝟎 − ( ∅ − 𝜶)] =

𝑽𝒇

𝒔𝒊𝒏 ∅

𝑽𝒔

𝒔𝒊𝒏( 𝟗𝟎 − 𝜶 ) =

𝑽𝒐

𝒄𝒐𝒔( ∅ − 𝜶 ) =

𝑽𝒇

𝒔𝒊𝒏 ∅

𝑽𝒔

𝐜𝐨𝐬 𝜶 =

𝑽𝒐

𝒄𝒐𝒔( ∅ − 𝜶 ) =

𝑽𝒇

𝒔𝒊𝒏 ∅

Lecture eighteen:

Assistant professor Sahib M. Mahdi - 2 -

Fig. ( 12 ): Schematic drawing showing the relation between cutting velocities.

Theories of Cutting process:

Merchant theory ( Shear angle Solution ):

Assumptions of this theory are:

1. Orthogonal cutting.

2. Minimum energy consumption is required.

3. Materials are formed by shearing.

4. The Tool used should be wider than work piece.

5. The stresses are uniformly distributed on the shear plane.

6. The total point is infinitely sharp.

7. The chips are being held in equilibrium condition.

Now consider fig. ( 13 ) below for the Forces diagram:

Fig. ( 13 ): Forces diagram - circle explanation.

90

90

Vo

Vf

Vs

90

90

Vo

Vf

Vs

90

90

Vo

Vf

Vs

Lecture eighteen:

Assistant professor Sahib M. Mahdi - 3 -

Fc , Ft are forces on the cutting surface.

Ff , Fn are forces on the rake plane.

Fs , FN are forces on the shear plane.

= The angle of friction.

From fig. ( 13 ), we will find that:

Ff = AE + ED = Fc . sin + Ft . cos

Fs = AQ HQ = Fc . cos Ft . sin

Fn = Fc . cos Ft . sin

FN = Ft . cos + Fc . sin

Fc = R . cos ( )

Fs = R . cos ( )

By dividing the cutting force ( Fc ) with the shear force ( Fs ), we obtain:

𝑭𝒄

𝑭𝒔 =

𝑹 . 𝒄𝒐𝒔 ( 𝜷 − 𝜶 )

𝑹 . 𝒄𝒐𝒔 ( ∅ + 𝜷 − 𝜶 ) ⟹ 𝑭𝒄 =

𝑭𝒔 . 𝒄𝒐𝒔 ( 𝜷 − 𝜶 )

𝒄𝒐𝒔 ( ∅ + 𝜷 − 𝜶 )

But the friction factor ( ) is equal to:

𝝁 = 𝒕𝒂𝒏 𝜷 = 𝑭𝒇

𝑭𝒏 =

𝑭𝒄 . 𝒔𝒊𝒏 𝜶 + 𝑭𝒕 . 𝒄𝒐𝒔 𝜶

𝑭𝒄 . 𝒄𝒐𝒔 𝜶 − 𝑭𝒕 . 𝒔𝒊𝒏 𝜶

The shearing stress ( s ) is equal to:

𝝉𝒔 ( 𝒔𝒉𝒆𝒂𝒓 𝒔𝒕𝒓𝒆𝒔𝒔 ) = 𝑭𝒔

𝑨

Where: A = Area of the shear plane

A = b . OA

But, from the beside schematic drawing:

𝑶𝑨 = 𝒕𝒐

𝒔𝒊𝒏 ∅ ⟹ 𝒕𝒐 = 𝑨𝑶 . 𝒔𝒊𝒏 ∅

𝑨𝒐 = 𝒕𝒉𝒆 𝒖𝒏𝒄𝒖𝒕 𝒄𝒉𝒊𝒑 𝒂𝒓𝒆𝒂 = 𝒃 . 𝒕𝒐

∴ 𝑨𝒐 = 𝒃 . 𝑨𝑶. 𝒔𝒊𝒏 ∅

Or, 𝑨𝒐 = 𝑨 . 𝒔𝒊𝒏 ∅

∴ 𝑨 = 𝑨𝒐

𝒔𝒊𝒏 ∅

Then the cutting force will become:

𝑭𝒄 = 𝑨𝒐 . 𝝉𝒔 . 𝒄𝒐𝒔 ( 𝜷 − 𝜶 )

𝒔𝒊𝒏 ∅ . 𝒄𝒐𝒔 ( ∅ + 𝜷 − 𝜶 )

Lecture seven:

4

But the Merchant theory assumes minimum energy required which means that the

differentiation of the cutting force ( Fc ) with respect to the shear angle ( ∅ ), we will obtain:

𝒅𝑭𝒄

𝒅∅ =

− 𝑨𝒐 . 𝝉𝒔 . 𝐜𝐨𝐬 ( 𝜷 − 𝒂 ). [ − 𝒔𝒊𝒏 ∅ . 𝐬𝐢𝐧 ( ∅ + 𝜷 − 𝜶 ) + 𝐜𝐨𝐬 ( ∅ + 𝜷 − 𝜶 ). 𝐜𝐨𝐬 ∅ ]

𝒔𝒊𝒏𝟐 ∅ . 𝒄𝒐𝒔𝟐 ( ∅ + 𝜷 − 𝜶 )

Equating the differentiation to zero:

𝒅𝑭𝒄

𝒅∅ = 𝟎

Then we get that:

cos ) = 0

cos ( ) = 0

∴ 𝟐 ∅ + 𝜷 − 𝜶 = 𝝅

𝟐

The effect of the Rake angle:

Rake angle is varies with the material being cut, since some materials slide more easily than

others, while some break up into small pieces. Brass, for instance, breaks up into small pieces,

and a rake angle of ( 0° ) is used. Aluminum, on the other hand, has a tendency to stick to the

face of the tool and requires a steep rake angle, usually in the region of ( 30° ).

Relative advantages of such rake angles are:

• Positive rake – helps reduce cutting force and thus cutting power requirement, larger

angle uses for soft material to avoid material stick on the tool.

• Negative rake – to increase edge-strength and life of the tool, and it is used for tough

materials using the cemented carbide cutting tools, it is necessary, due to the brittle

nature of the carbide, to give maximum support to the tip. To achieve this, a negative

rake is used.

• Zero rakes – to simplify design and manufacture of the form tools, used for

fragmented material such as brass.

The Uncut, Cut & width of Chip:

To understand well the terms mentioned above, we should look carefully to the figures below,

[fig. ( 14 )] shows the Straight turning cutting, where the width of chip represent ( b ) which

called also the depth of cut and it can be calculated from the below equation:

Fig. ( 14 ): The Straight turning.

Lecture seven:

5

𝒃 = 𝑫𝒐 − 𝑫𝟏

𝟐

The Uncut chip thickness ( to ) is called the feed per revolution ( Feed = mm / rev. ), so the Cut

chip thickness can be calculated from the formulas which are given in cutting lectures.

And [fig. ( 15 )] shows the Face or End face turning, where the width of chip represent ( b )

which called also the depth of cut and it can be calculated from the below equation:

𝒃 = 𝑳𝒐 − 𝑳𝟏

Fig. ( 15 ): The Face or End Face turning.

The Uncut chip thickness ( to ) is called the feed per revolution ( Feed = mm / rev. ), so the

Cut chip thickness can be calculated from the formulas which are given in cutting

lectures.

Steps to draw the Force Circle:

In order to draw the force circle diagram, we should follow the steps mentioned below

carefully. The force circle has its benefit to find the forces required or to find the related

angles and the most important to find the relationship between all the forces in terms of

other forces.

The steps are:

1 – For convenience forces R & R’ can be considered to be act at the tip tool point.

2 – Draw the force R to represent the diameter of a circle, this circle is the reference

circle and called the Force Circle. See fig. ( 16 – a ) below.

3 – The Force Circle is now can be used to draw any force acting along any surface

by decomposing the resultant force R into two perpendicular components ( such as ( Fc )

which is parallel to the cutting surface & ( Ft ) normal to it. See fig. ( 16 – b ) below.

4 – In the same way, R can be decomposed into ( Ff ) and ( Fn ) or into ( Fs ) & ( FN

), as shown in fig. ( 16 – c ).

5 – Then the relations in between the forces can be determined by using the

diagram shown in fig. ( 16– d ).

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Fig. ( 16 )

(d)(c)

(b)(a)

(d)(c) (d)(c)

(b)(a) (b)(a)

Lecture seven:

7

Example 21.2 Shear Stress in Machining

Suppose in Example 21.1 that cutting force and thrust force are measured during

an orthogonal cutting operation: Fc = 1559 N and Ft = 1271 N. The width of the

orthogonal cutting operation w = 3.0 mm. Based on these data, determine the

shear strength of the work material.

Solution: From Example 21.1, rake angle =10o, and shear plane angle =

25.4o. Shear force can be computed from Eq. (21.11):

The shear plane area is given by Eq. (21.8):

Thus the shear stress, which equals the shear strength of the work material, is

This example demonstrates that cutting force and thrust force are related to the

shear strength of the work material. The relationships can be established in a

more direct way Recalling from Eq. (21.7) that the shear force Fs = S As, the

force diagram of Figure 21.11 can be used to derive the following equations:

And

Lecture seven:

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These equations allow one to estimate cutting force and thrust force in an

orthogonal cutting operation if the shear strength of the work material is known.

Example 21.3 Estimating Friction Angle

Using the data and results from our previous examples, determine (a) the friction

angle and (b) the coefficient of friction.