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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.

MSE Stresses

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MSE Stresses

Steve Scott

December 2002

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

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

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