 Hello friends, in earlier two sessions we have studied Merchant Circle Diagram graphical method of calculating the forces acting along tool chip interface and along shear plane. Now in this session we are going to solve numerical problem on this merchant circle diagram by analytical method. I am Mr. P. P. Mitrabotri, Associate Professor in Department of Mechanical Engineering at Vulturen Stop Technology. Outcome of this session will be, learners will be able to solve problem for calculation of shear force and tool force during orthogonal cutting using various formulae derived in earlier session. So I will read the problem which is available on the screen during orthogonal turning with single point carbide tool having rake angle of 10 degrees at a feed rate of 0.2 mm per revolution. Here I will take a pause and I will focus your attention towards the word orthogonal turning in the problem statement. Orthogonal we are carrying out orthogonal cutting and in that orthogonal turning and when we are carrying out orthogonal turning feed rate is uncut chip thickness. So here we need not require uncut chip thickness measurement and I will start reading the problem further following observations were required. Cheap thickness after cut 0.44 mm cutting force Fc and thrust force Ft are measured by plate tool diameter as 400 Newton and 240 Newton respectively. And we are required to calculate chip thickness ratio or shear angle phi tool force and normal to the tool force Fn and respectively and shear force and normal to the shear force Fs and Fn. For that given data is uncut chip thickness is feed rate that is given. Cheap thickness after cut Tc is given cutting force Fc is given thrust force Ft is given rake angle alpha that is 10 degree is given. For calculation of shear force and normal to the shear force we require shear angle phi but shear angle phi is not given. But uncut chip thickness and chip thickness after cut is given from that we can calculate shear angle. Step 1 will be to calculate chip thickness ratio chip thickness ratio is represented by later r. r is nothing but uncut chip thickness divided by chip thickness after cut chip thickness after cut. As it is given 0.2 divided by 0.44 which comes out to be 0.45 it does not have any unit as it is a ratio. Now once we know r we can calculate shear angle phi by the formula phi is equal to 10 inverse of r cos alpha where alpha is rake angle divided by 1 minus r sin alpha. We know value of r we know value of alpha as 10. So by substituting these values equation comes out to be tan inverse of 0.45 into cos of 10 degrees divided by 1 minus 40.45 into sin of 10 degrees. By calculating it with the help of calculator value of phi comes out to be 25.89 degrees. Now we know all the entities so we will start calculation of forces acting along the tool face. The forces acting along the tool face are tool force f and normal to the tool force n. And as derived earlier value of f is represented in terms of cutting force thrust force and rake angle. And formula for calculation is fc sin alpha plus ft cos alpha and n is calculated as fc cos alpha minus ft sin alpha. So we will first of all calculate f we know fc as 400 newton alpha is 10 degrees. We know ft as 240 newton and alpha is 10 degrees. So we have substituted these values in this by calculating it with the help of calculator tool force f comes out to be 305 newton. So first value we have calculated now we have to calculate value of normal to the tool force which is denoted by formula fc cos alpha minus ft sin alpha. So fc is 400 newton cos of 10 degrees ft that is thrust force is 240 newton into sin of 10 degrees. By calculating it value comes out to be 252 newton you can confirm it. So we have calculated tool force f and normal to the tool force n by the formulas we have derived in earlier session. Now next step will be calculation of shear force fs and normal to the shear force fn. And for that angle phi is known which is to be which is 25.89 fc is known as 400 newton ft is known which is 240 newton. Now formula for calculation of fs is fc cos phi minus ft sin phi. So in this we can put value of fc as 400 newton cos of 25.89 minus ft 240 newton into sin of 25.89 degrees. By calculating this it comes out to be 352 newton so shear force acting along the shear plane is 352 newton. Then we have to calculate normal to the shear force fn which is given by the formula fc sin phi plus ft cos phi. Now in this value of fc is 400 newton value of phi is 25.89 degrees plus value of ft is 240 newton and value of phi as already stated is 25.89. By calculating this it comes out to be 505 newton. So in all our results can be represented as tool forces f305 newton normal to the tool force n252 newton then shear force fs352 newton and normal to the shear force 505 newton. And chip thickness ratio 0.45 shear angle phi 25.89 degrees. So this is total solution of our problem in what steps, what steps we have followed to solve this problem all of you might have understood well. Machine tool engineering by G. R. Nakpal and textbook of production engineering by P. C. Sharma. Thank you all.