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victus00 [196]
3 years ago
8

A 03-series cylindrical roller bearing with inner ring rotating is required for an application in which the life requirement is

40 kh at 520 rev/min. The application factor is 1.4. The radial load is 2600 lbf. The reliability goal is 0.90.
Required:
Determine the C10 value in kN for this application and design factor.
Engineering
1 answer:
-BARSIC- [3]3 years ago
7 0

Answer:

\mathbf{C_{10} = 137.611 \ kN}

Explanation:

From the information given:

Life requirement = 40 kh = 40 40 \times 10^{3} \ h

Speed (N) = 520 rev/min

Reliability goal (R_D) = 0.9

Radial load (F_D) = 2600 lbf

To find C10 value by using the formula:

C_{10}=F_D\times \pmatrix \dfrac{x_D}{x_o +(\theta-x_o) \bigg(In(\dfrac{1}{R_o}) \bigg)^{\dfrac{1}{b}}} \end {pmatrix} ^{^{^{\dfrac{1}{a}}

where;

x_D = \text{bearing life in million revolution} \\  \\ x_D = \dfrac{60 \times L_h \times N}{10^6} \\ \\ x_D = \dfrac{60 \times 40 \times 10^3 \times 520}{10^6}\\ \\ x_D = 1248 \text{ million revolutions}

\text{The cyclindrical roller bearing (a)}= \dfrac{10}{3}

The Weibull parameters include:

x_o = 0.02

(\theta - x_o) = 4.439

b= 1.483

∴

Using the above formula:

C_{10}=1.4\times 2600 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{1}{\dfrac{10}{3}}}

C_{10}=3640 \times \pmatrix \dfrac{1248}{0.02+(4.439) \bigg(In(\dfrac{1}{0.9}) \bigg)^{\dfrac{1}{1.483}}} \end {pmatrix} ^{^{^{\dfrac{3}{10}}

C_{10} = 3640 \times \bigg[\dfrac{1248}{0.9933481582}\bigg]^{\dfrac{3}{10}}

C_{10} = 30962.449 \ lbf

Recall that:

1 kN = 225 lbf

∴

C_{10} = \dfrac{30962.449}{225}

\mathbf{C_{10} = 137.611 \ kN}

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nirvana33 [79]

Answer:

Q(h=200)=0.35W

Q(h=3000)=5.25W

Explanation:

first part h=200W/Km^2

we must use the convection heat transfer equation for the chip

Q=hA(Ts-T∞)

h= convective coefficient=200W/m2 K

A=Base*Leght=5mmx5mm=25mm^2

Ts=temperature of the chip=85C

T∞=temperature of coolant=15C

Q=200x2.5x10^-5(85-15)=0.35W

Second part h=3000W/Km^2

Q=3000x2.5x10^-5(85-15)=5.25W

5 0
3 years ago
Two gage marks are placed exactly 250 mm apart on a 12-mm-diameter aluminum rod with E 5 73 GPa and an ultimate strength of 140
NNADVOKAT [17]

Answer:

81.76 N/mm² ( MPa), 1.71233

Explanation:

Modulus of elasticity = stress / strain

stress = modulus of elastic × strain

strain = ΔL / L = 250.28 mm - 250 mm / 250 mm = 0.00112

Modulus of elasticity E = 73 GPa = 73 × 10³ MPa where 1 MPa = 1 N/mm²

E = 73 × 10³N/mm²

stress =  73 × 10³N/mm²× 0.00112 = 81.76 N/mm² ( MPa)

b) Factor of safety = maximum allowable stress / induced stress = 140 MPa / 81.76 MPa = 1.71233

8 0
4 years ago
For this homework you are writing a main function, with 3 subfunctions. 2. The "main" function is to be named arm_prop a. Inputs
ra1l [238]

Answer:

MATLAB CODE:

clc;

clear all;

x=[1 2 3 4]; %Input vector of variable size

[gm,ra,hm]=arm_prop(x) %Main function calling , output values are hm, ra, hm and imput is x

%% Main function

function [gm,ra,hm]=arm_prop(x)

[gm]=geo_mean(x); %Calling Sub function 1

[ra]=rms_avg(x); %Calling Sub function 2

[hm]=har_mean(x); %Calling Sub function 3

 

%Sub function 2 =Rms average

function [gm]=geo_mean(x)    

n=numel(x); %calculating number of elemnts in vector x

temp=1; %Temporary value for multiplying each and every value of x

for i=1:1:n % For loop run for number of elements in x

temp=temp*x(i); %Multliplyng each and every element of x

end

gm=temp^(1/n); % Calculating geometric mean

end

%Sub function 2 =Rms average

function [ra]=rms_avg(x)

temp=0; %Temporary value for adding square of each and every value of x

N=numel(x); %calculating number of elemnts in vector x

for i=1:1:N % For loop run for number of elements in x

temp=temp+x(i)^2; %Adding square of each and every element of x

end

ra=(temp/N)^(1/2); % Calculating Rms average

end

%Sub function 3 =Harmonic mean

function [hm]=har_mean(x)

temp=0; %Temporary value for adding reciprocal of each and every value of x

N=numel(x); %calculating number of elemnts in vector x

for i=1:1:N % For loop run for number of elements in x

temp=temp+(1/x(i)); %Adding sreciprocal of each and every element of x

end

hm=N/temp; % Halculating Harmonic mean

end

end

Results:

gm =

2.2134

ra =

2.7386

hm =

1.9200

Here in one main function. 3 sub functions are written. Every sub function returns one output. As all 3 sub functions are inside main function, main function returns 3 output values gm, ra, hm.

Explanation:

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True or false : In improper integrals infinte intervals mean that both of the integration limits are should be infinity
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Explanation:

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