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• Miriam Ontiveros

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Mechanical Design of Machine Elements

Quiz 3: Fully Reversed Loading in Fatigue Suppose you have a 4340 steel plate in fully reversed axial loading . The plate is 1⁄4” thick, 2” high, and 5” long. The bar is machined, with factor a = 2.70 ksi and exponent b = -0.265. The bar is operating at 600 degrees F. The program required reliability is 99.99.

What is the fully adjusted endurance limit of this bar?

1.

Determine fully adjusted endurance limit Se’ = 100 ksi as Sut > 200 ksi ka = 2.70(260)-0.265 = 0.62 kb = 1 (axial load) kc = 0.85 (axial load) kd = 0.90 (Reference Sut on ‘Effect of temperature on strength’ chart) ke = 0.70 (reference chart for reliability of 99.99)

Se = (0.62)(1)(0.85)(0.9)(0.70)(100) = 33.22 ksi

Mechanical Design of Machine Elements

Quiz 3: Fully Reversed Loading in Fatigue Suppose you have a 4340 steel plate in fully reversed axial loading, with an axial load varying from 15000 lbf to -15000 lbf. The plate is 1⁄4” thick, 2” high, and 5” long. A hole is in the center of the plate, and is 1/4“ in diameter. The stress concentration factor chart is in the Quiz 3 References. The fully adjusted endurance limit was found to be Se = 25 ksi.​ The fatigue strength fraction f is 0.76. The notch sensitivity is 0.95. 2) What is the life in cycles of this bar? 5.00“

+

2.00“

1.00“ 0.25“

2.50“

2) Step 1: Determine stress at critical location. σnom = F/A = (15000lbf)/[(2 in - 0.25 in) * 0.25 in] = 34,286.71 psi Step 2: add in stress concentration factor Kt = 2.70 (from graph, d/H = (0.25/2) = 0.125) Kf = 1+q(Kt-1) = 1+0.95(2.70-1) = 2.62 σmax = Kf*σnom = 89.66 ksi

Step 3: Compare to endurance limit 89.66 ksi > 25 ksi Step 4: Calculate life a = (f*Sut)2/Se = [(0.76*260)2]/25 = 1561.8 ksi b = -(1/3)log[(f*Sut)2/Se ] = -0.30 N = (σnom/a)1/b =(89.66/1561.8)(1/-0.30) = 14002.0 cycles

F

Ductile to Brittle Transition Temperature 4340 Steel

Mil-HDBK-5J, Jan 2003

Theoretical Stress-Concentration Factor Kt

By Cdang [CC0], via Wikimedia Commons https://commons.wikimedia.org/wiki/File%3AKt_plaque_percee_symetrique_traction.svg

MIl-HDBK-5J, Jan 2003 Size Factor kb Torsion and bending Range kb = 0.11 < d < 2.01 in 0.879d^(-0.107) 2 < d < 10 in 0.91d^(-0.157) Axial kb = 1

Loading Factor kc Loading Type kc Bending 1 Axial 0.85 Torsion 0.59