Question

1). Two spherical balls are contacting each other due to a pair of compressive forces F = 50 N. The bigger ball with a radius of R2 = 20 mm is made of a steel with elastic modulus of 210 GPa and Poisson's ratio of 0.3. The smaller ball with a radius of R1 = 5 mm is treated as a rigid body. According to Hertz contact theory, what is the pressure distribution along a radial path in the contact area? 2). An elastic-plastic material has the Young's modulus of 190 GPa and Poisson's ratio of 0.3. Its plastic work-hardening behaviour can be described by a power law: 0 0 = Ce" for € 2 E with n = 0.1 and Oy = 300 MPa. Provide 5 sets of data in correct order as the data lines for the key word *Plastic in Abaqus. The last entry of the yield stress should be 450 MPa.

Answer 1

1 given that, F = 50N R₂ = 20mm D2 = 40mm E = 210Gipa = 2108 103 N/mm² u 0.3 R, - mm , D = 10mm L (Diameter) Contact area - ana² a = 3 3F (1-82/E 8 Ya, You 2. 3x50 8 (€ 1-0.373)/210x10 o + Yo 40 3 150 0.91 X DO XIto 5X210X103 Beogn 0.65*10-3 0.086 mm 3F 3 x 50 Pmax an az 2X3.14 x bigo (0.086)2 0 3 229.49 N/mm² pressure distribution Brosch 3229.49 N/mm²

= 1906 Pa = 190x10% e giren. Yong's modulus = 0.3 Mpa n - o.1 300 MPa we know, E = 3k(1-24) 190 = 3k (1-2 *03) k 190 = 158.33 Gpa. 3 X 0.4 Here, o 1 Bulk modulus is represented by 'C' c = k = 158.33 Gpa Gpa = 158. 33x103 MP a Nowe, given, E ay A E - = C En when ay = 300MPa E » 1.57X10² 190x 103 CE2 158.33x10°(1.57X103)001 83.0 X 10 MPa Νσο, ay -3 = 350 mpa E > 350 1.84 x 10 190x103 ascen153.33 X (1.84X10") x103 = = 84.358103 0.1 3. МРО

when 3 4- oo MP4 £ $ чоо 19ojo - это хто = 1 33 х х (эох 163) 8.46 x to3 4 2o when ay $ > 420 - 9.21х16 3 1qox to2 a = can 15.33 x 102 85.9ox to Mea х (а. 1x163) 1 when 3 = 450 $ 7 4SO 2.36 x13 190х 103 = cen 58.33 xo'x (2-36 x 3) 86.49 х 10 lo 2 MPa о E Зоом Ра 30 м Р 1.гэх 103 3 1. 34 x 10 -3 9. 10 х10 2.21 1 3 Мра 83. от хомро 84. 2x x 103 MPA 25.46 Kip Mpa 85,40 x 102 MP4 26, 42 x 102 м а 420 MPa D -2 MPa 3. 36 х 10