Consider the situation of Ex. 15.2 and 15.3, where a component made of alloy S-590 is subjected to a stress of 200 MPa at a temperature of 600°C. (a) Repeat the rupture life calculation of Ex. 15.3, using the Eq. 15.20 fit for PLM. (b) If the desired service life is 1.5 years, what are the safety

Question

Consider the situation of Ex. 15.2 and 15.3, where a component made of alloy S-590 is subjected to a stress of 200M Pa at a temperature of 600°C.

(a) Repeat the rupture life calculation of Ex. 15.3, using the Eq. 15.20 fit for [latex]P_{LM}[/latex].

[latex]\begin{gathered}\log t_r=\frac{P_{L M}}{T}-C \quad \quad (a)\ P_{L M}=b_0+b_1 x+b_2 x^2+b_3 x^3 \quad \quad (b)\end{gathered}[/latex] (15.20)

(b) If the desired service life is 1.5 years, what are the safety factors in life and in stress?

(c) What is the safety margin in temperature?

Accepted Answer

(a) From Table 15.3, the S-590 alloy constants for Eq. 15.20 are C = 17, [latex]b_{0} = 38,405 , b_{1} = −8206 , ext{ and } b_{2} = b_{3} = 0[/latex]. Noting that x = log σ, substituting these values gives [latex]P_{LM}[/latex] from Eq. 15.20(b), and then [latex]t_{r}[/latex] follows from Eq. 15.20(a) with substitution of T = 600°C + 273 = 873 K:

[latex]P_{L M}=b_0+b_1 x=38,405-8,206 \log (200 MPa )=19,523[/latex]

[latex]\log t_r=\frac{P_{L M}}{T}-C=\frac{19,523}{873 K }-17=5.363, \quad t_r=230,583 h =9608 ext { days }[/latex]

(b) The safety factor in life is calculated by comparing [latex]t_{r}[/latex] from (a) with the service life [latex]\hat{t}[/latex]:

[latex]\hat{t}=(1.5 ext { years })(365.25 ext { days } / ext { year })(24 h / ext { day })=13,149 h[/latex]

[latex]X_t=\frac{t_{r 2}}{\hat{t}}=\frac{230,583 h }{13,149 h }=17.5[/latex]

For the safety factor in stress, we need to determine the [latex]σ_{1}[/latex] that causes failure at [latex]t_{r} = \hat{t}[/latex], which depends on the value of [latex]P_{LM}[/latex] corresponding to [latex]t_{r} = \hat{t}[/latex] at temperature [latex]T = \hat{T}[/latex] . The needed [latex]P_{LM}[/latex] value can be calculated from Eq. 15.17:

[latex]P_{L M}=T\left(\log t_r+C ight)=\hat{T}(\log \hat{t}+C)=(873 K )[\log (13,149 h )+17]=18,437[/latex]

Substituting this [latex]P_{LM}[/latex] into Eq. 15.20(b) and solving gives [latex]σ = σ_{1}[/latex], and then Eq. 15.23 gives [latex]X_{σ}[/latex]

[latex]X_\sigma=\frac{\sigma_1}{\hat{\sigma}} \quad\left(t_r=\hat{t} ight)[/latex] (15.23)

[latex]P_{L M}=b_0+b_1 x, \quad 18,437=38,405-8,206 \log \sigma_1[/latex]

[latex]\log \sigma_1=2.433, \quad \sigma_1=271.2 MPa[/latex]

[latex]X_\sigma=\frac{\sigma_1}{\hat{\sigma}}=\frac{271.2}{200}=1.36[/latex]

(c) To obtain the temperature increase, [latex]\Delta T_f=T_f-\hat{T}[/latex] , that causes failure at [latex]\left(\sigma, t_r ight) = (\hat{\sigma}, \hat{t})[/latex], solve Eq. 15.17

[latex]P_{L M}=T\left(\log t_r+C ight)[/latex] (15.17)

for T . Then calculate [latex]T_{f}[/latex] by substituting [latex]t_r=\hat{t}[/latex] along with [latex]P_{LM}[/latex] from (a), as this value corresponds to [latex]\hat{\sigma}[/latex] . The result of these two operations is

[latex]T=\frac{P_{L M}}{\log t_r+C}, \quad T_f=\frac{19,523}{\log (13,149 h )+17}=924.4 K[/latex]

[latex]\Delta T_f=T_f-\hat{T}=924.4-873=51.4 K[/latex]

Comments Solving Eq. 15.20(b) for σ, as in (b) of this example, requires an iterative solution if this polynomial is of order 3 (or 5, etc.). But a closed-form calculation is possible in this case, due to [latex]b_{2} = b_{3} = 0[/latex] giving a linear relationship. Note that the safety factor in stress of [latex]X_{σ} = 1.36[/latex] is rather low. A higher value would require a safety factor in life even larger than the [latex]X_{t} = 17.5[/latex] that is calculated.

Question 15.5: Consider the situation of Ex. 15.2 and 15.3, where a compone...

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