Ramp-to-Failure Testing

Abstract

Engineers are constantly confronted with time issues. Applying a constant stress and waiting for failure can be very time-consuming. Thus, it is only natural to ask the question—does a rapid time-zero test exist that can be used on a routine sampling basis to monitor the reliability of the materials/devices? The answer to this question is often yes and it is called the ramp-to-failure test. While the test is destructive in nature (one has to sacrifice materials/devices), it is generally much more rapid than conventional constant-stress time-to-failure tests. The relative quickness of the test also enables the gathering of more data and thus the gathering of better statistics.

Problems

• 1. Capacitor dielectrics were randomly selected and ramp-to-breakdown tested at 105 °C, using a linear ramp rate of R = dE/dt = 0.5 MV/cm/s. During ramp-to-breakdown testing, it was determined that the breakdown distribution could be approximated by a normal distribution with: (E_bd)₅₀ = 12 MV/cm and σ = 1.0 MV/cm. Determine the expected time-to-failure for 1 % of the capacitors at an operational field of 5 MV/cm at 105 °C. Assume an exponential acceleration factor with γ = 4.0 cm/MV.

  Answer: TF(1 %) = 2.1 years

• 2. Turbine blades were randomly selected and ramped-to-rupture at 700 °C using a tensile stress with a linear ramp rate of R = dσ_stress/dt = 5 MPa/min. During ramp-to-rupture testing, it was determined that the rupture distribution could be approximated by a normal distribution with: (σ_rupture)₅₀ = 250 MPa with a standard deviation of σ = 25MPa. Determine the expected time-to-failure for 1 % of the turbine blades at an operational tensile stress of 15 MPa at 700 °C. Assume a power-law acceleration factor with n = 4.0 and (because of small cracks) no yield point.

  Answer: TF(1 %) = 0.4 years

• 3. Analyze Problem 1 except this time, assume a power-law acceleration with n = 40.

Answer: TF(1 %) = 4,297 years

• 4. Analyze Problem 2, except this time assume an exponential acceleration with γ = 0.06/MPa.

Answer: TF(1 %) = 0.3 years

• 5. Steel pipes were randomly selected for pressurizing-to-rupture testing. Using a linear ramp rate of R = dP/dt = 5 kpsi/min, the rupture data tended to obey a Weibull distribution with (P_rupture)₆₃ = 200 kpsi and a Weibull slope of β = 10. Determine the expected time-to-failure for 1 % of the pipes at an operational pressure of 5 kpsi. Assume that the stress in the cylindrical pipes is directly proportional to the pressure and assume a power-law acceleration factor with an exponent of n = 4 and (because of small cracks) no yield point.

Answer: TF(1 %) = 3.9 years

• 6. Suspension cables were randomly selected for tensile stressing-to-rupture testing. Using a linear ramp rate of R = dσ_Tensile/dt = 4 kpsi/min, the rupture data tended to obey a Weibull distribution with (σ_rupture)₆₃ = 250 kpsi and a Weibull slope of β = 12. Determine the expected time-to-failure for 1 % of the cables at an operational pressure of 2 kpsi. Assume a power-law acceleration factor with an exponent of n = 4 and (because of small defects) no yield point.

Answer: TF(1 %) = 854 years

• 7. With no yield point, Eq. (11.16) reduces to:

  $$ \mathrm{TF}\left(F\%\right)=\frac{1}{n+1}\left[\frac{\xi_{\mathrm{op}}}{R}\right]{\left[\frac{\xi_{\mathrm{bd}}\left(F\%\right)}{\xi_{\mathrm{op}}}\right]}^{n+1}, $$

  which is valid when the stress ξ is linearly ramped at a constant rate R (R = dξ/dt = constant) until failure occurs. Show that, if the stress is proportional to the power-law of some other parameter S,

  $$ \xi ={C}_o{S}^m, $$

  then the time-to-failure equation becomes:

  $$ \mathrm{TF}\left(F\%\right)=\frac{1}{n+1}\left[\frac{S_{\mathrm{op}}^m}{R_1}\right]{\left[\frac{S_{\mathrm{bd}}\left(F\%\right)}{S_{\mathrm{op}}}\right]}^{m\left(n+1\right)}, $$

  where the ramp rate R₁ is given by:

  $$ {R}_1={mS}^{m-1}\left(\frac{\mathrm{d}S}{\mathrm{d}t}\right)=\mathrm{constant}. $$

• 8. Using the results from Problem 7, metal storm shutters with small cracks were randomly selected for storm testing. The shutters were tested in a wind tunnel by ramping the wind speed S until the shutters failed. The stress σ in the shutters, due to the wind, is proportional to the square of the wind speed: σ = C₀S². Using a constant ramp rate of R₁ = 2S(dS/dt) = 10 (mph)²/min, the failure data tended to obey a Weibull distribution with (S)₆₃ = 100 mph and a Weibull slope of β = 10. Determine the expected time-to-failure for 1 % of the shutters with a nominal constant wind speed of 25 mph. Assume a power-law acceleration factor of at least n = 6, and because of the small cracks, no yield point.

Answer: TF(1 %) = 7.2 years