Lezione 6 di Costruzione di macchine

Safety factor

➢ The definition of Safety Factor is complicated by the fact that there could be two, or more, loading modes acting independently to promote failure.

➢ The definition of the safety factor depends on the way each loading mode increases the stresses as a consequence of:

• Higher request of performance
• Dimensional variations

➢ As a general rule, unless differently prescribed, minimum fatigue safety factor must be taken as

Haigh diagram and safety factor

Step 1: Haigh diagram drawing.

Step 2: Operating point P(σ_m^P, σ_a^P) is placed in the Haigh diagram. The Safety Factor SF is then calculated according to one of the following criteria according to the dependence of the stresses on the component performance.

The definition of Safety Factor is complicated by the fact that there could be two, or more, loading modes acting independently to promote failure. The definition of the safety factor depends on the way each loading mode increases the stresses as a consequence of:

• Higher request of performance
• Dimensional variations

As a general rule, unless differently prescribed, minimum fatigue safety factor must be taken as 3.

Step 1: Haigh diagram drawing.

Step 2: Operating point P(σ_m^P, σ_a^P) is placed in the Haigh diagram. The Safety Factor SF is then calculated according to one of the following criteria according to the dependence of the stresses on the component performance.

SF = ^σ_lim/_σ^Pa

Step 2: Operating point P(σ_a, σ_m) is placed in the Haigh diagram. The Safety Factor SF is then calculated according to one of the following criteria according to the dependence of the stresses on the component performance.

Se la retta passa per P ho retta di Goodman che mi dà il coefficiente di sicurezza.

Step 2: CASE A: average stress constant – alternate stress varying with performance.

Caso in cui sono a fatica media costante.

Step 2: CASE B: both average stress and alternate stress varying with performance (average stress proportional to alternate stress).

Caso in cui crescono entrambi uguali.

Dal punto in cui cade provino.

Altro modo per vedere Ja/Jm che cambiano o rimangono costanti.

Endurance Safety Factor: posso sapere questo.

Strength Safety Factor: o anche questo.

Poi faccio i conti con uno o con l'altro.

Carico multiassiale

Non complex multiaxial stress state: all principal stresses reach the maximum and minimum at the same time and principal directions do not change as a function of time.

Complex multiaxial stress state: the loads have not the same phase and consequently the principal directions are a function of time.

Intermediate case: the loads have the same phase but the principal directions are a function of time (not fixed) because midrange stresses have not the same principal directions of alternating stresses.

Multi-axial loading: combination of loading modes

Gough and Pollard Analysis

Material.

• Plan carbon and low alloyed steels¹
• Grey iron
• Ductile iron
• Aluminium alloy
• Copper
• Brass
• Bronze
• Titanium alloy TiAl6V4

¹ Steels for quench and tempering heat treating.

From Gough Pollard analysis, it has been found that the fatigue failure of ductile materials (e.g. steels) stressed by both alternating bending and alternating torsion comes if:

where σ_D-1 is the fatigue strength evaluated by a bending test. Moreover, the left side of this equation (1) can be seen as a Von Mises equivalent stress. If the further hypothesis that this result (eq. 1) is valid in general, even if σ_a=0, is added, it can be written:

⇒ Equivalent alternating stress: σ_aw = √(σ_a² + 3τ_a²) (2)

⇒ Equivalent midrange stress: σ_mw = √(σ_m² + 3τ_m²) (3)

Using the load coefficient (C_t) the relation between the fatigue limit measured with a tension/compression test (σ_D-1^⋅) and fatigue limit F measured with a bending (σ_D-1) is:

σ_D-1^⋅ = C_t • σ_D-1 = 0.6 • 0.85 • σ_D-1

where σ_D-1 is the fatigue strength evaluated with a bending test.

Assuming (C_t=0.85): σ_D-1^⋅ = 0.85 • σ_D-1

Consequently, the failure with only a tension/compression stress state comes:

Concluding, in order to compare the tension/compression stress state with the fatigue limit by bending test, the load coefficient (C_t=0.85) has to be applied as a correction factor.

On the basis of the last conclusion, the initial problem of the combination of bending, tension/compression, and torsion stress state can be analysed by adding a tension/compression stress state corrected by a load coefficient (C_t=0.85) to the bending stress in the equivalent alternating stress of Gough and Pollard:

Consequently the equivalent stresses are:

⇒ Equivalent alternating stress: σ_aw = √((σ_a^⋅/0.85)²+3(τ_a)²) (5)

⇒ Equivalent midrange stress: σ_mw = √((σ_m^⋅)²+3(τ_m)²) (6)

The safety factor in multiaxial fatigue loading is similar to the monoaxial case but it is based on the equivalent stresses. Taking into account a notched component:

⇒ Equivalent alternating stress: σ_aw = √((K_f^⋅σ_a^⋅/0.85)²+3(K_fs)² τ_a²) (7)

⇒ Equivalent midrange stress: σ_mw = √((K_f^⋅σ_m^⋅)²+3(K_fs)² τ_m²) (8)

⇒ Assuming a working point: P (σ_aw, σ_mw)

On the Haigh diagram, the fatigue stress of the component: σ_D-1^⋅ = C_s • C_f • σ_D-1^⋅

Where σ_D-1 is the fatigue strength evaluated by a bending test. The fatigue limit of the component has to be computed neglecting the load coefficient (C_t). The reason is that this effect has been already considered by the correction (C_t=0.85 in equation 6).

To the normal stress in tension/compression.