Hertz Theory of Elastic Contact Conforming Contact Surfaces of the two bodies ‘fit’ exactly or even closely together without deformation Eg: Flat Slider bearings, journal bearings etc. Non-Conforming Contact Bodies which have dissimilar profiles When brought into contact without deformation, “point contact” or “line contact” occurs Eg: Contact between races and balls in a ball or roller bearing, Contact between a cam and follower, etc. On application of normal load, contact area is small Stresses are highly concentrated in the vicinity of contact Effect is unaffected by the shape of the bodies away from contact area Elastic Half Space When the dimensions of the body are large compared to the contact area, stresses in this region are unaffected by the shape of the contacting bodies far away from the contacting region The stresses may be calculated to a good approximation by considering each body as a semi-infinite elastic solid bounded by a plane surface: i.e. an elastic half space Bodies of any profile may be regarded as semi-infinite in extent and having a plane surface If the loading is along a narrow strip, a state of plane strain is assumed Justified when the thickness of the solid is large compared with the width of the loaded region Classical theory of elasticity can now be used to determine the contact pressure and subsurface stresses Hertz Theory (1882) When two spheres are in contact with a normal force P, Hertz made the hypothesis that the contact area is, in general elliptical For calculating local deformations, each contacting body can be regarded as an elastic half space, loaded over a small elliptical region of its plane surface The significant dimensions of the contact area must be small compared to the dimensions of each body and with the relative radii of curvature of each surface Strains in the contact region are considered to be sufficiently small so as to lie within the scope of the linear theory of elasticity The surfaces are assumed to be frictionless so that only a normal pressure is transmitted between them Two solids of general Shape, after deformation Separation between the two surfaces is given by, h = Ax 2 + By 2 Within the contact region, u z1 + u z 2 + h = δ 1 + δ 2 u z1 + u z 2 = δ − Ax 2 − By 2 (δ = δ 1 + δ 2 ; h = S1 + S 2 ) Outside the contact region, u z1 + u z 2 < δ − Ax 2 − By 2 Hertz Theory (Contd…) Assumptions: The surfaces are continuous and non-conforming: a<<R Strains are small: a<<R Each solid can be considered as an elastic half space: a<<R1,2 , a<<l The surfaces are frictionless: qx = qy = 0 The two surfaces do not touch or interfere outside the loaded region P(x,y) is required to be found. Two Solids of revolution Contact area is circular with radius, a Boundary conditions for displacements within the contact, uz1 + uz 2 = δ − (1 / 2 R ) r 2 (1 / R ) = (1 / R1 ) + (1 / R2 ) Pressure distribution proposed by Hertz, 2 2 1 1− ν 1 1− ν 2 = + Writing , * E E1 E2 p = p0 {1 − ( r / a ) } 2 1/ 2 a = π p0 R /( 2 E * ) Radius of the contact circle is, And the mutual approach of distant points in the two solids is given by, * δ = π ap0 /( 2 E ) Total load compressing the solids is given by, a P= ∫ 0 In terms of total load, 3PR a= * 4E 2 p ( r ) 2π rdr = p0π a 2 3 1/ 3 9P a δ = = *2 R 16 RE 2 ; 2 1/ 3 6 PE 3P p0 = = 3 2 2 2π a π R *2 ; Before applying this equation, Uniqueness of the boundary conditions are to be checked Ensure that the surfaces do not touch or interfere outside the contact region 1/ 3 Stress and pressure distribution Two-dimensional contact of Cylindrical Bodies Separation between corresponding points on the unloaded surface is given by h = z1 + z2 = Ax 2 = 1 2 (1 R1 + 1 R2 ) x 2 = (1 R ) x 2 Within the contact area u z1 + u z 2 = δ − Ax 2 = δ − 1 2 1 2 (1 / R) x 2 Outside the contact region, u z1 + u z 2 < δ − 12 (1 / R ) x 2 Due to the inherent drawback of 2D problems, relationship with reference to surface gradients are generally obtained ∂ u z1 ∂ u z 2 + = − (1 / R ) x ∂x ∂x Two-dimensional contact of Cylindrical Bodies Solving the equations, { p ( x) = p0 1 − ( x / a ) 4 PR a= * πE } 2 1/ 2 1/ 2 1 1 − ν 12 1 − ν 22 = + ∗ E E1 E2 2 P PE p0 = = π a πR * 1/ 2 π a2 E* P= 4R 1 1 1 = + R R1 R2 P is the load (per unit length (thickness)) on the cylinders = 4 π pm Stresses in 2D contact σ x p0 = − a { (a 2 + 2z ( p0 2 σz= − a + z2 a { ) 2 )( a 2 + z − 2z − 1/ 2 ( p0 τ1= − z − z 2 a2 − z 2 a (τ 1 ) max = 0.30 p0 ) 2 − 1/ 2 at ) − 1/ 2 } z = 0.78a } Example Two mating steel spur gears are 20 mm wide, and the tooth profiles have radii of curvature at the line of contact of 10 mm and 15 mm for the first and second gear respectively. A force of 250 N is transmitted between them. (a) Compute the maximum contact pressure and the width of contact. (b) How deep below the surface does the maximum shear stress occur, and what is its value? For steel, modulus of elasticity and Poisson’s ratio are 200 GPa and 0.3 respectively.

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