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Tasya [4]
3 years ago
5

A steel rectangular tube has outside dimensions of 150 mm x 50 mm and a wall thickness of 4 mm. State the inside dimensions, the

area of its cross section, and the weight of a piece 1.22 m long.
Engineering
1 answer:
lara [203]3 years ago
6 0

Answer:

 inside dimension  = 142 mm \times 42 mm

cross section area = 7.5\times 10^{-3} m^2

mass of 1.2 meter log steel  = 1.843\times 10^{-3} \rho

Explanation:

given data:

Outside dimension of steel rectangular = 150 mm\times 50mm

Thickness = 4 mm

Long = 1.22 m

inside dimension will be = (150- 8)mm \times ( 50-8)mm

                                         = 142 mm \times 42 mm

cross section area = 150\times 50 mm^2

                               = 7500\times 10^{-6} m^2

                               = 7.5\times 10^{-3} m^2

let the density be assumed as \rho

mass of 1.2 meter log steel will be = 1.2  \times (7.5\times 10^{-3} -  0.142\times 0.048)\times \rho

                                                       = 1.843\times 10^{-3} \rho

                                                       

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1. For ball bearings, determine: (a) The factor by which the catalog rating (C10) must be increased, if the life of a bearing un
ElenaW [278]

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:L_1 = L , L_2 = 2L

   Catalog rating: C_1 = C , C_2 = ? ,

From given equation bearing life equation,

F\times\frac{1}{3} (L_1)  = C_1   ...(1)   \\\\    F\times\frac{1}{3} (L_2)  =C_2...(2)

we Dividing eqn (2) with (1)

\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

The Catalog rating increased by factor of 1.26

(b) Reliability Increase from 0.9 to 0.99

R_1 = 0.9 , R_2 = 0.99

Now calculating life adjustment factor for both value of reliability from Weibull parametres

a_1 = x_o + (\theta - x_o){ ln(\frac{1}{R_1} ) }^{\frac{1}{b}}

= 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

Similarly

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

Now calculating bearing life for each value

L_1 = a_1 * LL_1 = 0.9968LL_2 = a_2 * LL_2 = 0.2215L

Now using given ball bearing life equation and dividing each other similar to previous problem

\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

Catalog rating increased by factor of 0.61

6 0
3 years ago
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