Equations for Chapters 13 & 14 in Meyers & Chawla

A₀L₀ = A₁L₁

e_NH = A _NH [D_l G b / kT] (b/d)² (s/G)

e = C sⁿ

T(log t_r + C) = m e_C = A_C [D_gb G b / kT] (d/d)(b/d)³ (s/G)

ln t_r – (Q / kT ) = m e_HD = A _HD [D_l G b / kT] (s/G)

e_s t_r = k’

e = A [D_l G b / kT] (s/G)⁵

e_s = A’ exp (-Q_C / kT)

ln a_T = - C₁ (T –T_s) / [C₂ + T – T_s]

D = D₀ exp (-Q_D / kT)

if T_s = T_g, then C₁ = 17.5 and C₂ = 52K

s = s₀ exp (-t/t)

t = 3h / E [ t is relaxation time] J(t) = e(t) / s₀ = 1/E [ 1 – exp(-t/t)]

s = E e₀

de / dt = s₀ / 3h

Ds = s_max – s_min

s_a = (s_max + s_min) / 2

s_m = (s_max +s_min) / 2

stress ratio = R = s_max – s_min å n_i / N_i = 1

K_IC = Y s (p a)^1/2

da/dN = C (DK)^m

DK = Y Ds (pa)^1/2

Ds (N_f)^a = C C = A / (DK₀) ^m

s_a = s₀ [ 1 – s_m / s_uts]

De_e / 2 = s_a / E = (s_f’/E) (2N_f)^b De_p / 2 = s_a / E = e_f’ (2N_f)^c