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Solution: Suppose there are two spherical bodies of radius R1 and R2 which are in contact under the action of an axial force F. As we know if there is a force applied over a surface, stress is generated and that stress deforms the object if it exceeds the yield stress for a particular material. We calculate the stress as the ratio of force applied over the area of application. Here, we have two spheres. Suppose they are idea spheres with no deformation. When they come into contact with each other, the nature of contact established is limited to a certain point. We can say, that the area of a point is zero. So, if a force is applied over a point contact, there will be infinite stress generated at the contact zone, due to which there will be a certain deformation of one or both bodies until the two objects can resist the deformation caused by reduced stress which is the consequence of increased contact area. Suppose the force F acts in the z-direction and the plane perpendicular to this axis, called x-y plane, is located at the contact surface of both spheres as depicted in the figure. If you look at the deformed zone from the z-axis, it will look like a circle and if you look at the same zone in the y-z plane, it looks like an ellipse. If the radius of the circle of deformation is r then it will be equal to the semi-major axis of the ellipse and the height of penetration will be equal to the semi-minor axis. That means, r=a. In this situation, (1) the radius of the contact area, r = a is given by, where, F = force applied R1 = radius of sphere 1 R2 = radius of sphere 2 = poison's ratio for sphere 1 = poison's ratio for sphere 2 E1 = modulus of elasticity of sphere 1 E2 = modulus of elasticity of sphere 2 If and, then (2) maximum contact pressure, Pmax: The maximum contact pressure, Pmax occurs at the center of the circular contact area. When we look at the deformation zone, in the y-z plane, we can see that a semi-elliptic pressure distribution is developed. In this distribution, the maximum pressure is given by, (3) Principal Stresses Three principal stresses along the axes x,y and z are generated such that the stresses along x and y-axes are the same due to symmetry. So, equation (1) and z which can be given by, equation (2) (4) Principal Shear Stress The principal shear stress is given by, equation (3) and Using the equation, we can calculate the principal stresses and the maximum shear and plot the graph on MATLAB. Since the question does not provide the data about the force applied and other related situations, I can only show you its nature.
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