Lezione 4 di Costruzione di macchine

Carico su fatica con intaglio

Senza intaglio

La curva è quindi una interpolazione di punti, uniti ai carichi.

Ma come determino la curva σ/N?

Faccio ruotare il provino e lo carico ^{2 modi}.

• Da una sola estremità
• Da _{2 estremità}

Trave a sbalzo!

Da prove come questa trovo la tensione media!

σ_media = _{σmax + σmin} / 2

Contiamo su fatica

Sotto intaglio

Con intaglio la curva è quindi una interpolazione di punti, uniti al continuo.

Ma come determino la curva σ/N?

Faccio ruotare il provino e lo carico ^{2 modi}.

• Da una sola estremità
• Da 2 estremità

Trave a sbalzo.

Da prove come questa trovo la tensione media!

σ_media = (σmax + σmin) / 2

Altri test per la fatica

Posso fare anche altri test per la fatica:

Non qua consideriamo i dati provenienti da una prova di flessione rotante!

The main experimental standard conditions of Wöhler or S-N diagrams are:

• Rotating bending
• No average stress (σ_m = 0, corresponding to R = -1)
• Specimen with circular cross section
• Specimen diameter about 10 mm
• Specimen surface polished

Ma la prova di fatica è molto dispersiva, un po’ si usa questo grafico S-N-P ossia introduce la probabilità.

E per determinare il limite di fatica di un materiale e la % di successi o rotture si usa questo metodo:

M8 screws (Rm = 400 MPa)

Step, d = 10 MPa

x = failure o = run-out N = 5 x 10⁶

N = 7 A = 9 B = 15

σ_N(50%) = σ₀ + d NB - > 0.3 s = 1.62d Otherwise s = 0.53 d

σ_N(10%) = σ_N(50%) - 1.28·s σ_N(90%) = σ_N(50%) + 1.28·s

σ_D-1 Fatigue limit (of the material) in standard conditions with a 50% probability of failure (B50)

Characterization of the materials fatigue limit requires a huge experimental effort a relation between static strength and fatigue limit it is helpful in order to have an estimation of the fatigue strength.

Criteri per DR e limite fatica

Vediamo ora dei criteri per DR e limite fatica:

Bach criterion

σ_D-1 = 0.5R_m (R = -1 σ_min = -σ_max)

σ_D-1 = 0.3R_m (R = 0 σ_min = 0)

Fuchs criterion (1)

σ_D-1 = 0.5R_m (R_m < 1400 MPa)

σ_D-1 = 700 MPa (R_m ≥ 1400 MPa)

Ghisa

Cast iron σ_D-1 = 0.4 R_m → ottenuti da esperimenti.

Dopo schie vanno fatte delle approssimazioni, tipo:

Ginocchio → curva

_F: σ_F è lo sforzo di prova trattivo → σ_u = 0.9R_m

_G: σ_G → σ_D ⇒ limite fatica

N_G → N_D rottura

N_σ^b = B i.e. log(N) = log(B) - k log(σ_I)

k = frac{log(N_σ) - log(N_r)}{log(σ_f) - log(σ_D)} log(B) = log(N_σ) + frac{log(N_f) - log(N_r)}{log(σ_f) - log(σ_D)} log(σ_D)

σ_σ = AN^b i.e. log(σ_σ) = b log(A) + b log(N)

b = frac{log(σ_f) - log(σ_D)}{log(N_f) - log(N_r)} - frac{1}{k} log(A) = log(σ_σ) - frac{log(σ_D) - log(σ_τ)}{log(N_D) - log(N_r)} log(N_σ)

→ esprimere curva con “i”

→ esprimere curva con “b”

σ_a steel σ_m 0 _a = 60,000 psi

S-N diagram for two average stresses (σ_m=0, σ_m=400 MPa). The fatigue strength clearly decrease when the tension average stress is applied. An opposite behaviour, increasing of fatigue strength can be found if a compression stress is applied.

σ_a Experimental results

N constant

N = 2 × 10⁶ for steels

Goodman line

Goodman line, σ_D R_m σ_m

Strength verification:

Limit line equation:

σ_a σ_m

-------- + ------ ≤ 1

σ_D-1 R_m

σ_D σ_m

-------- + ------ = 1

σ_D-1 R_m

⇒ σ_D = σ_D-1 - σ_m ----- σ_D-1 ------------ R_m σ_a

S_s Yield (Langer) line

S_u Gerber line

Load line, slope r = S _u/S_s

Modified Goodman line

Soderberg line

ASME-elliptic line

0 S_st

Midrange stress σ_m

Haigh diagram drawing during the exam=> altri diagrammi

Fattori relativi al componente e alle condizioni di servizio

Factors related to the component:

• Size, C_S
• Notches, K_f
• Surface finish, C_F
• Surface treatments (mechanical, thermal or chemical, coatings...)

Factors related to service conditions:

• Loading mode, C_L
• Reliability, C_R
• Temperature
• Environment (moisture, corrosives...)

The fatigue limit of the component’s material σ^c_D-1 will be then:

σ^c_D-1 = σ_D-1(ΠC_i)

With σ_D-1 fatigue limit of the standard specimen

1) Effetto sui carichi -> “C_L”

The fatigue failure comes from the stress in a process zone instead of the maximum stress like in static. This process zone include the external skin and a fraction of the internal part under the skin.

On the same component, the effective stress in the process zone is higher if the load mode is tension/compression than in stresses distribution with gradient like in bending or torsion.

σ _M < σ _a,eff^f

In plane banding C_L ≍1;

In tension/compression C_L ≈0.6-0.85. It can be assumed a midrange C_L0.7. In some books (e.g. Shigley) is used C_L=0.85.

Bending

Fatigue failure comes if σ _a,eff= σ⁰D_-1

Gradient => σ _a,eff = α σ^f_d with α<1

Fatigue failure comes if α σ _a= σ⁰D_-1 (1)

Experimentally Fatigue failure comes if σ _a= σ^bD_-1 (3)

By comparison of (1), (3) => α σ^bD_-1 = σ⁰D_-1

Tension/compression

Fatigue failure comes if σ _a,eff= σ⁰D_-1

No gradient => σ _a,eff = σ _tc

Fatigue failure comes if α σ _a= σ⁰D_-1(2)

Experimentally Fatigue failure comes if σ _a= σ^tcD_-1(4)

By comparison of (2),(4) => σ ^tcD_-1= σ⁰D_-1

α = C_L and αK, by experiment α=0.6;0.85

σ ^b > σ ^tc

2) Effetto di scala -> “C_s”

Consider two components with circular cross section having different size but stressed with a gradient stress distribution (bending or torsion). Both components are stressed with the same maximum alternating stress. At this point, assuming that the process zone p is part of total cross section where the stress is higher than a fraction (e.g. 80%) of the maximum alternating stress. The process zone:

• Is lower in the smaller component;
• Is higher in the bigger component;
• Increases if the component size increase.

Assuming that the defect density as a function of the size is constant and taking in account that the majority of defect is usually localized in the neighbour of surface, there will be more defects in the component with larger process zone. It can be concluded that the solution in more probably in components higher stress. Consequently, if the size increase the fatigue strength decrease.