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Fatigue Strength (6.4, 6.7-6.8, 6.11). MAE 316 – Strength of Mechanical Components NC State University Department of Mechanical and Aerospace Engineering. Fatigue Strength (6.8). Up to now, we have designed structures for static loads. t. w. P. P. P. ( σ max is also constant). d. t.
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Fatigue Strength(6.4, 6.7-6.8, 6.11) MAE 316 – Strength of Mechanical Components NC State University Department of Mechanical and Aerospace Engineering Fatigue Strength
Fatigue Strength (6.8) • Up to now, we have designed structures for static loads. t w P P P (σmaxis also constant) d t Fatigue Strength
Fatigue Strength (6.8) • What if loading is not constant? • Even if σmax≤ Sy, failure could occur if enough cycles are applied. P t Fatigue Strength
Fluctuating Stresses (6.11) • If σmin = - σmax, this is known as “fully-reversed” loading. σmax σmin σ t Fatigue Strength
S-N Diagram (6.4) Sf (fatigue strength) - stress level at which a corresponding number of cycles (N) will lead to failure (crack initiation) Se (endurance limit) - stress level below which failure will never occur Fatigue Strength
Endurance Limit (6.7) • The simplest design rule to prevent fatigue failure is • This is a valid concept, but not quite so simple in reality. • Se is determined experimentally. • Simple approximate Se formulas exist for steel, but must be used carefully – better to have actual data. • where Sut = ultimate strength and Se’ = unmodified, laboratory determined value Fatigue Strength
Endurance Limit (6.7) • For real design we will modify Se’ to account for the surface finish, stress concentration, temperature, etc. • These effects decrease the effective endurance limit. Fatigue Strength
Predicting Fatigue Life (6.8) • High-cycle fatigue life (N > 1000 cycles) • Typical S-N diagram for steel Sl’ Se’ (log Sf) (log N) Fatigue Strength
Fatigue strength fraction Fatigue strength
Example • Find Sf of 1020 hot-rolled steel if the required life is 250,000 cycles, bending loads. • Given: Sut = 57 ksi for 1020 steel • Note: For steel, Sl’ = 0.9Su (bending), 0.75Su (axial), and 0.72Su (torsion). • What is the life if Sf = 40 ksi? Fatigue Strength
High Cycle Fatigue(6.9-6.10) MAE 316 – Strength of Mechanical Components NC State University Department of Mechanical and Aerospace Engineering High Cycle Fatigue
Modified Endurance Limit (6.9) • Modified endurance limit is defined as • ka = surface finish factor = aSutb Table 6-2 Surface finish factors ka High Cycle Fatigue
Modified Endurance Limit (6.9) • kb= size factor • Axial loading • kb = 1 • Bending and torsion • kb = 0.879d-.107 (0.11 in ≤ d ≤ 2 in) • kb = 0.91d-.157 (2 < d < 10 in) • kb = 1.241d-.107 (2.79 ≤ d ≤ 51 mm) • kb = 1.51d-.157 (51 < d < 254 mm) • d is the diameter of the round bar or the equivalent diameter (de) of a non-rotating or non-circular bar (Table 6-3). High Cycle Fatigue
Modified Endurance Limit (6.9) • kc= loading factor • 1 (bending) • 0.85 (axial) • 0.59 (torsion) • kd = temperature factor • If endurance limit (Se’) is known, oruse equation • If Se’ is not known, use kd = 1 and temperature-corrected tensile strength (Sut) (see Example 6-5 in textbook) High Cycle Fatigue
Modified Endurance Limit (6.9) • ke = reliability factor Table 6-5 Reliability factors ke High Cycle Fatigue
Modified Endurance Limit (6.9) • kf = miscellaneous-effects factor • Corrosion • Electrolytic plating • Metal Spraying • Cyclic frequency • Frettage corrosion • If none of the above conditions apply, kf = 1 High Cycle Fatigue
Fatigue Stress Concentration Factor (6.10) • Kf= fatigue stress concentration factor • Kf = 1 + q(Kt – 1) • q = notch sensitivity • Kt= stress concentration factor • Kfcan be used to reduce Se (multiply Se by 1/Kf) or to modify the nominal stress (σmax = Kfσnom). High Cycle Fatigue
Fatigue Stress Concentration Factor (6.10) Figure 6-20 Notch sensitivity for bending and axial Figure 6-21 Notch sensitivity for torsion High Cycle Fatigue
Example • For the plate shown below, find the maximum allowable load F for the plate to have infinite life. • Given: 1018 cold-drawn steel, Sy = 373 MPa, Sut = 442 MPa t = 10 mm w =60 mm F F F d = 12 mm t High Cycle Fatigue