Answer:
(b) Given the Weibull parameters of example 11-3, the factor by which the catalog rating must be increased if the reliability is to be increased from 0.9 to 0.99.
Equation 11-1: F*L^(1/3) = Constant
Weibull parameters of example 11-3: xo = 0.02 (theta-xo) = 4.439 b = 1.483
Explanation:
(a)The Catalog rating(C)
Bearing life:[tex]L_1 = L , L_2 = 2L[/tex]
Catalog rating: [tex]C_1 = C , C_2 = ? ,[/tex]
From given equation bearing life equation,
[tex]F\times\frac{1}{3} (L_1) = C_1 ...(1) \\\\ F\times\frac{1}{3} (L_2) =C_2...(2)[/tex]
we Dividing eqn (2) with (1)
[tex]\frac{C_2}{C_1} =\frac{1}{3} (\frac{L_2}{L_1})\\\\ C_2 = C*(\frac{2L}{L})\frac{1}{3} \\\\ C_2 = 1.26 C[/tex]
The Catalog rating increased by factor of 1.26
(b) Reliability Increase from 0.9 to 0.99
[tex]R_1 = 0.9 , R_2 = 0.99[/tex]
Now calculating life adjustment factor for both value of reliability from Weibull parametres
[tex]a_1 = x_o + (\theta - x_o){ ln(\frac{1}{R_1} ) }^{\frac{1}{b}}[/tex]
[tex]= 0.02 + 4.439{ ln(\frac{1}{0.9} ) }^{\frac{1}{1.483}} \\\\ = 0.02 + 4.439( 0.1044 )^{0.67}\\\\a_1 = 0.9968[/tex]
Similarly
[tex]a_2 = x_o + (\theta - x_o){ ln(\frac{1}{R_2} ) }^{\frac{1}{b} }\\\\ = 0.02 + 4.439{ ln(1/0.99) }^{\frac{1}{1.483} }\\\\ = 0.02 + 4.439( 0.0099 )^{0.67}\\\\a_2 = 0.2215[/tex]
Now calculating bearing life for each value
[tex]L_1 = a_1 * LL_1 = 0.9968LL_2 = a_2 * LL_2 = 0.2215L[/tex]
Now using given ball bearing life equation and dividing each other similar to previous problem
[tex]\frac{C_2}{C_1} = (\frac{L_2}{L_1} )^{\frac{1}{3} }\\\\ C_2 = C* (\frac{0.2215L }{0.9968L} )^{1/3}\\\\ C_2 = 0.61 C[/tex]
Catalog rating increased by factor of 0.61
