Cutting Forces in manufacturing technology

cutting force

1. CUTTING FORCES and POWER Fig: (a) Forces acting on a cutting tool during 2-dimensional cutting. Note that the resultant force, R must be collinear to balance the forces. (b) Force circle to determine various forces acting in the cutting zone.

2. Knowledge of the cutting forces and power involved in machining operations is important for the following reasons: Machine tools can be properly designed to minimize distortion of the machine components, maintain the desired dimensional accuracy of the machined part, and help select appropriate tool holders and work-holding devices. The work piece is capable of withstanding these forces without excessive distortion. Power requirements must be known in order to enable the selection of a machine tool with adequate electric power. CUTTING FORCES and POWER 2

3. CUTTING FORCES and POWER Cutting force, Fc, acts in the direction of cutting speed, V, and supplies energy required for cutting. Thrust force, Ft , acts in a direction normal to cutting velocity, perpendicular to WP. The resultant force, R can be resolved into two components : Friction force: F, along the tool-chip interface N ormal force: N, perpendicular to it. F = R sin β N = R cos β R is balanced by an equal and opposite force along the shear plane and is resolved into a shear force, Fs , and a normal force, Fn Fs = Fc cos Ø – Ft sin Ø Fn = Fc sin Ø + Ft cos Ø 3

4. Coefficient of Friction Coefficient of friction between tool and chip: Friction angle related to coefficient of friction as follows: 4 The ratio of F to N is the coefficient of friction, μ, at the tool-chip interface, and the angle β is the friction angle. The coefficient of friction in metal cutting generally ranges from about 0.5 to 2.

5. Shear Stress Shear stress acting along the shear plane: where A s = area of the shear plane, Shear stress = shear strength of work material during cutting 5

6. CUTTING FORCES & POWER Thrust Force If the thrust force is too high or if the machine tool is not sufficiently stiff, the tool will be pushed away from the surface being machined. This movement will, in turn, reduce the depth of cut, resulting in lack of dimensional accuracy in the machined part, As the rake angle increases and/or friction at the rake face decreases, this force can act upward. This situation can be visualized by noting that when μ = 0 (that is, β = 0), the resultant force, R, coincides with the normal force, N. In this case, R will have a thrust-force component that is upward. 6

7. The Power consumed/ work done per sec in cutting : The Power consumed/ work done per sec in shear : The Power consumed/ work done per sec in friction : The total Power required : Power required in Metal cutting 90

8. Specific Energy Specific Energy, u t ,is defined as the total energy per unit volume of material removed . Where wt v c is the MRR Units for specific energy are typically N‑m/mm 3 Therefore is simply the cutting force to the projected area of cut. If u f and u s be specific energy for friction and specific energy for shearing, then As the rake angle increases, the frictional specific energy remains more or less constant, where as the shear specific energy rapidly reduced . 91

9. Approximate Specific-Energy Requirements in Cutting Operations 92

10. Cutting Forces Fig: (a) Forces acting on a cutting tool during 2-dimensional cutting. Note that the resultant force, R must be collinear to balance the forces. (b) Force circle to determine various forces acting in the cutting zone. 10

11. The following is a circle diagram. Known as Merchant’s circle diagram, which is convenient to determine the relation between the various forces and angles. In the diagram two force triangles have been combined and R and R’ together have been replaced by R. the force R can be resolved into two components F c and F t . Fc and Ft can be determined by force dynamometers. The rake angle (α) can be measured from the tool, and forces F and N can then be determined. The shear angle (  ) can be obtained from it’s relation with chip reduction coefficient. Now F s & Fn can also be determined. Merchant’s Circle Diagram ∅ Work Tool Chip Clearance Angle F t F c F N F n F s α α β ( β - α ) R 77

12. Merchant’s Circle Diagram Work Tool Chip Clearance Angle F t F c F N F n F s α α β ∅ ( β - α ) R 78

13. Clearance Angle The procedure to construct a Merchant’s circle diagram Work Tool Chip F t F c F N F n F s α α β ∅ R 79

14. Set up x-y axis labeled with forces, and the origin in the centre of the page. The cutting force (Fc) is drawn horizontally, and the tangential force (Ft) is drawn vertically. (Draw in the resultant (R) of Fc and Ft. Locate the centre of R, and draw a circle that encloses vector R. If done correctly, the heads and tails of all 3 vectors will lie on this circle. Draw in the cutting tool in the upper right hand quadrant, taking care to draw the correct rake angle (α) from the vertical axis. Extend the line that is the cutting face of the tool (at the same rake angle) through the circle. This now gives the friction vector (F). 14 The procedure to construct a Merchant’s circle diagram

15. The procedure to construct a Merchant’s circle diagram A line can now be drawn from the head of the friction vector, to the head of the resultant vector (R). This gives the normal vector (N). Also add a friction angle ( β ) between vectors R and N. Therefore, mathematically, R = F c +F t = F + N. Draw a feed thickness line parallel to the horizontal axis. Next draw a chip thickness line parallel to the tool cutting face. Draw a vector from the origin (tool point) towards the intersection of the two chip lines, stopping at the circle. The result will be a shear force vector (Fs). Also measure the shear force angle between Fs and Fc . Finally add the shear force normal (Fn) from the head of Fs to the head of R. Use a scale and protractor to measure off all distances (forces) and angles. 15

16. Frictional Force System Relationship of various forces acting on the chip with the horizontal and vertical cutting force from Merchant circle diagram Ft Fc F N α α β ( β - α ) R α α α (90- α ) (90- α ) O A C B G E D ∅ Work Tool Chip Clearance Angle F t F c F N F n F s α α β ( β - α ) R 82

17. Shear Force System Also: Relationship of various forces acting on the chip with the horizontal and vertical cutting force from Merchant circle diagram ∅ Work Tool Chip Clearance Angle F t F c F N F n F s α α β ( β - α ) R ∅ Ft Fc A O Fn Fs α α ( β - α ) R B C D E ∅ ∅ (90- ∅ ) (90- ∅ ) 83

18. Relationship of various forces acting on the chip with the horizontal and vertical cutting force from Merchant circle diagram ∅ Work Tool Chip Clearance Angle F t F c F N F n F s α α β ( β - α ) R 84 F t = R Sin ( β - α ) F c = R Cos ( β – α )

19. The Merchant Equation Of all the possible angles at which shear deformation can occur, the work material will select a shear plane angle  that minimizes energy, given by Derived by Eugene Merchant. Based on orthogonal cutting, but validity extends to 3-D machining. 19

20. Assuming that the shear angle adjusts itself to minimize the cutting force, or that the shear plane is a plane of maximum shear stress. (21.3) β is the friction angle and is related to the coefficient of friction, μ, at the tool – chip interface (rake face): From Eq (21.3), as the rake angle decreases and/or the friction at the tool–chip interface increases, the shear angle decreases and the chip becomes thicker, Thicker chips mean more energy dissipation because the shear strain is higher. As work done during cutting is converted into heat, temperature rise is also higher. 20 The Merchant Equation

21. Ernest and Merchant gave the relation Theory of Ernst and Merchant (1944) Assumptions of the theory: Tool edge is sharp. The work material undergoes deformation across a thin shear plane. There is uniform distribution of normal and shear stress on the shear plane. The work material is rigid and perfectly plastic. The shear angle ∅ adjusts itself to give minimum work. The friction angle β remains constant and is independent of ∅. The chip width remains constant . M. Eugene Merchant 95

22. What the Merchant Equation Tells Us To increase shear plane angle Increase the rake angle Reduce the friction angle (or coefficient of friction) 22

23. <--------TUGAS------> NIM Ganjil NIM Genap Cutting Force (Fc) = 115 N Tangential Force (Ft) = 75 N Rake angle ( α ) = + 30° Tebal chip = 1 cm Cutting Force (Fc) = 110 N Tangential Force (Ft) = 80 N Rake angle ( α ) = + 35° Tebal chip = 1 cm Diketahui data-data pemotongan sebagai berikut: 1 . Dengan menggunakan Merchant Circle , carilah friction vector (F), Normal vector (N ), shear angle (Φ ), Shear Force (Fs), Normal Force (Fn), Friction angle ( β ), koefisien gesek ( μ )!Bandingkan antara metode grafis dengan metode analitis. 2. Apabila v=25mm/s, vs=15mm/s, dan vc=12mm/s, Berapa Daya Pemotongan? 3. Apabila MRR = 150 mm/s, Berapa spesific energy -nya?