 I am Mr. P. P. Mitragootri, Associate Professor in Mechanical Engineering Department at Valchand Institute of Technology, Sholapur. In the earlier session, we have studied. Hello friends, I am Mr. P. P. Mitragootri, Associate Professor in Department of Mechanical Engineering at Valchand Institute of Technology, Sholapur. In the earlier session, we have discussed merchant circle diagram and how to graphically represent forces acting along shear plane and forces acting along tool chip interface and how to graphically calculate values of these forces. But graphical representation sometimes may be values may vary in graphical representation depending upon the scale. So analytical method of calculation of forces will be better to calculate the forces more accurately. So in this particular session, we desire to study analytical method of calculation of forces outcome of this learning will be calculation of forces acting along the shear plane and calculation of forces acting along tool chip interface during orthogonal cutting by analytical method for this what we desire to know is the required data before that assumption is important. Again assumption is same that is orthogonal cutting which is obviously two dimensional cutting only two forces acting will be considered, two dimensional cutting will be considered and two dimensional forces will be considered for analysis. In this case again required data will be cutting force denoted by letter fc, thrust force denoted by letter ft and cutting force and thrust force are measured by tool dynamometer cutting force acts in the direction of tool travel and thrust force acts in a direction perpendicular to cutting force and both are measured by tool dynamometer and measured by measured in Newton. Again during metal cutting and rake angle of the cutting tool which is denoted by letter alpha is recorded from the tool that we are employing for cutting because we are knowing the specifications of the tool and shear angle which is denoted by letter phi shall be known to us. But many a times shear angle is not given instead of that chip thickness before cut and chip thickness after cut is given, chip thickness before cut is denoted by letter t and chip thickness after cut is denoted by letter tc. So, from that chip thickness before cut and chip thickness after cut we can calculate value of chip thickness ratio r which is equal to t upon tc and once we know r we can calculate shear angle phi by the formula shear angle phi is equal to tan inverse of r cos alpha divided by 1 minus r sin alpha. And for calculation of forces we have to draw triangle of forces cutting force thrust force and resultant and now in this particular diagram we have shown ab as a cutting force acting along the direction of tool travel, bc as a thrust force acting perpendicular to the cutting force ac represents resultant of these forces ac as already said is a resultant. Now in this with this particular diagram analytically we are going to derive the formulae for calculation of tool force f and normal to the tool force an tool force f act along the tool phase and normal to the tool force acts perpendicular to that. So, we have represented these two forces in the diagram and tool force f obviously make an angle alpha with a vertical tool force f is represented by line ad and normal to the tool force that is n is represented by line cd. Now further construction is to draw the line from point a making an angle alpha that is rake angle with ab that is cutting force that line is extended. Now after that we have to draw the line from point c parallel to ad cutting the line which has been drawn earlier that line will cut the earlier line at point l. So, c l will be parallel to ad and c l will be equal to ad means c l will also represent tool force. Now after that from point b we have to draw a line parallel to c l cutting the extended line drawn from point a making an alpha with cutting force and line drawn from b will cut that line at point j. So, b j will be parallel to c l. Now next one more construction we have to carry out that is from point b we have to draw a perpendicular bk to the line c l from this we can very easily understand that angle kbc can be none other than 90 minus alpha and angle kcb will be alpha. Now from this diagram tool force f is equal to ad equal to c l c l is equal to lk plus kc which can be represented as jb plus kc. Now from the triangle abj jb is equal to fc sin alpha which you can very easily understand and kc will be bc sin 90 minus alpha, but bc is nothing but fc sin 90. So, fc sin 90 minus alpha is equal to ft cos alpha. Therefore, tool force f can be represented in terms of cutting force thrust force and angle alpha as f is equal to fc sin alpha plus ft cos alpha. Then similarly normal to the tool force can be represented from the diagram as aj minus jl, but from the diagram we can understand that aj is fc cos alpha and jl is equal to kb and kb is ft cos of 90 minus alpha. So, ft is sin alpha. So, normal to the tool force n also is represented in terms of cutting force thrust force and rake angle. As fc cos alpha minus ft sin alpha after analytically calculating the formulas for tool force and normal to the tool force we have to calculate formulae for shear force and normal to the shear force which are acting along shear plane. For that we again have to draw a triangle of forces cutting force thrust force and resultant force this triangle abc as shown in the diagram. Now after that we have to draw a line from point A making an angle phi extended line making an angle phi with cutting force or line ab. Then we have to draw a normal to that line from point C cutting that line at point D. So, cd will represent normal to the shear force and ad will be magnitude of shear force fs. Then we have to draw a line from point B parallel to line cd cutting the line at point G. So, bg will be parallel to cd then from point B we have to draw a perpendicular to line cd which will cut the line cd at point H. So angle HBC in this case will be 90 minus phi and angle SCB will be phi. Now from the diagram shear force can be represented as fs which is ag minus gd but ag is equal to fs fc cos phi and gd is equal to bh ft sin phi. So, shear force fs is represented in terms of cutting force and thrust force and shear angle as fs is equal to fc cos phi minus ft sin phi. Similarly normal to the shear force fn can be represented as dh plus sc that is cd cd is equal to dh plus sc but dh is equal to bg bg is equal to fc sin phi and hc is equal to ft cos phi. So, fn that is normal to the shear force can be represented in terms of cutting force thrust force and shear angle as fn is equal to fc sin phi plus ft cos phi. So, all the analytical formulas calculated in this session have been represented in the next slide. You can understand that that is tool forces are represented in terms of cutting force thrust force and rake angle and forces acting along shear plane are represented in terms of cutting force thrust force and shear angle. Hope you have understood the analytical method of calculating the forces for further reading I will recommend machine tool engineering by J. R. Nakpal and textbook of production engineering by P. C. Sharma. Thank you.