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MSE Stresses

MSE Stresses. Steve Scott December 2002. M2: Shear Stresses due to in-plane Acceleration. Nominal stress = P / (T*(W-D)) Maximum Stress = K m * Nominal Stress K m = 3.

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MSE Stresses

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  1. MSE Stresses Steve Scott December 2002

  2. M2: Shear Stresses due to in-plane Acceleration • Nominal stress = P / (T*(W-D)) • Maximum Stress = Km * Nominal Stress • Km = 3 • M2 Mass: assume circle, diameter 15 cm, thickness 0.8 cm, quartz ( r = 2.65)  M =0.375 kg • Acceleration: assume 100g.  P = 100 * M * g = 367 Newtons • T = 8 mm, D = 3 mm, W = 15 cm. • Result: maximum stress = 0.92 MPa(http://www.stacieglass.com/cgi-bin/scf/sym_circ_hole_calc.cgi) • Quartz tensile strength: 48 MPA (http://www.goodfellow.com/csp/active/static/A/SI61.HTML) • Quartz Shear strength: 70 MPa • Quartz Compressive Strength: 1100 MPa • Conclusion: in-plane acceleration is not the problem. MSE stresses 9Dec2002.ppt

  3. M2: Compressive Stresses on Pin Sleeve due to In-Plane Acceleration Entire force of 367 Newtons is concentrated on 2 pins, each with cross-sectional area 3mm x 8mm. 367 N • Compressive stress = 367 / 2 / 0.003 / 0.008 = 7.6 Mpa MSE stresses 9Dec2002.ppt

  4. y x M2: PIN-reacting forces due to vertical applied load(e.g. due to edge contact with turret) s Vertical force: F + P1y + P2y = 0 Horizontal force: Fx = 0 = P1x + P2x Torque about C1:Fd + (R - s)P2x – (R - s)P1x = 0 P1 Pin-1 C1 R Solution: P1x = Fd / 2(R-s) P2x = -Fd / 2(R-s) What determines ratio of P1y to P2y ? Pin-2 P2 d F MSE stresses 9Dec2002.ppt

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