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Lady_Fox [76]
3 years ago
13

A 4140 steel shaft, heat-treated to a minimum yield strength of 100 ksi, has a diameter of 1 7/16 in. The shaft rotates at 600 r

pm and transmits 40 hp through a gear. The hub upon which the gear is seated is 1.5 in wide. You have in stock 1137 OQT 1300 carbon steel that you are considering using as a key to seat the gear hub. What dimensions will the key need to be? Is this material appropriate for the application? If not, select a more appropriate material.

Engineering
2 answers:
MrMuchimi3 years ago
8 0

Answer:

The key dimensions; (4.54mm× 6.75mm×4.5mm)

Explanation

Check attachment

velikii [3]3 years ago
4 0
Answer:










Explanation:



4140-40 I’d pick wood




I hope this helps! :)
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Compare electromagnets and solenoids
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Create a separate function file fieldtovar.m that receives a single structure as an input and assigns each of the field values t
Soloha48 [4]

Answer:

Explanation gives the answer

Explanation:

% Using MATLAB,

% Matlab file : fieldtovar.m

function varargout = fieldtovar(S)

% function that accepts single structure as input, assigning each

% of the field values to user-defined variables

fields = fieldnames(S); % get the field names of the input structure

% check if number of user-defined variables and number of fields in

% structure are equal

if nargout == length(fields)

% if equal assign each value of structure to user-defined varable

for i=1:nargout

varargout{i} = getfield(S,fields{i});

end

else

% if not equal display an error message

error('The number of output variables does not equal the number of fields');

end

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%This brings an end to the program

4 0
3 years ago
A steel bolt has a modulus of 207 GPa. It holds two rigid plates together at a high temperature under conditions where the creep
VikaD [51]

Answer:

14.36((14MPa) approximately

Explanation:

In this question, we are asked to calculate the stress tightened in a bolt to a stress of 69MPa.

Please check attachment for complete solution and step by step explanation

7 0
3 years ago
A fatigue test was conducted in which the mean stress was 46.2 MPa and the stress amplitude was 219 MPa.
sleet_krkn [62]

Answer:

a)σ₁ = 265.2 MPa

b)σ₂ = -172.8 MPa

c)Stress\ ratio =-0.65

d)Range = 438 MPa

Explanation:

Given that

Mean stress ,σm= 46.2 MPa

Stress amplitude ,σa= 219 MPa

Lets take

Maximum stress level = σ₁

Minimum stress level =σ₂

The mean stress given as

\sigma_m=\dfrac{\sigma_1+\sigma_2}{2}

2\sigma_m={\sigma_1+\sigma_2}

2 x 46.2 =  σ₁ +  σ₂

 σ₁ +  σ₂ = 92.4 MPa    --------1

The amplitude stress given as

\sigma_a=\dfrac{\sigma_1-\sigma_2}{2}

2\sigma_a={\sigma_1-\sigma_2}

2 x 219 =  σ₁ -  σ₂

 σ₁ -  σ₂ = 438 MPa    --------2

By adding the above equation

2  σ₁ = 530.4

σ₁ = 265.2 MPa

-σ₂ = 438 -265.2 MPa

σ₂ = -172.8 MPa

Stress ratio

Stress\ ratio =\dfrac{\sigma_{min}}{\sigma_{max}}

Stress\ ratio =\dfrac{-172.8}{265.2}

Stress\ ratio =-0.65

Range = 265.2 MPa - ( -172.8 MPa)

Range = 438 MPa

8 0
3 years ago
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