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ziro4ka [17]
2 years ago
7

(a) A non-cold-worked 1040 steel cylindrical rod has an initial length of 100 mm and initial diameter of 7.50 mm. is to be defor

med using a tensile load of 18,000 N. It must not experience either plastic deformation or a diameter reduction of more than 1.5 x 10-2 mm. Would the 1040 steel be a possible candidate for this application? Justify your choice(s) using calculations. For the 1040steel: Elastic modulus is 205 GPa, Yield strength is 450 MPa, and Poisson’s ratio is 0.27. (b) If you are asked to perform a strain hardening process to increase the yield strength so the steel can be used in another application with larger force load. How much cold work would be required to reduce the diameter of the steel to 6.0 mm?
Engineering
1 answer:
serg [7]2 years ago
6 0

Answer:

A) 1040 steel is not a possible candidate for this application

B) 35.94%

Explanation:

Initial length = 100 mm =  0.1 m

Initial diameter ( d ) = 7.5 mm = 0.0075 m

Tensile load ( p ) = 18,000 N

Condition : The 1040 steel must not experience plastic deformation or a diameter reduction of more than 1.5 * 10^-5 m

<u>A) would the 1040 steel be a possible candidate for this application</u>

<em>Yield strength of 1040 steel < stress  ( in order to be a possible candidate )</em>

stress = p / A0 = ( 18000 ) / ( \frac{\pi }{4} ) * 0.0075^2

                      = 18,000 / (4.418 * 10^-5 )   =  407.424 MPa

Yield strength of 1040 steel = 450 MPa

stress = 407.424 MPa

∴ Yield strength ( 450 MPa ) > stress ( 407.424 MPa )  

Therefore 1040 steel is not a possible candidate for this application

<u>B) Determine How much cold work would be required to reduce the diameter of the steel to 6.0 mm</u>

Area1 = ( \frac{\pi }{4} ) ( 0.006 )^2 = 2.83 * 10^-5 m^2

therefore % of cold work done = ( A0 - A1 ) / A0  * 100 = 35.94%

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we have  

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7 0
2 years ago
A man weighs 145 lb on earth.Part ASpecify his mass in slugs.Express your answer to three significant figures and include the ap
adell [148]

Answer:

<em>a) 4.51 lbf-s^2/ft</em>

<em>b) 65.8 kg</em>

<em>c) 645 N</em>

<em>d) 23.8 lb</em>

<em>e) 65.8 kg</em>

<em></em>

Explanation:

Weight of the man on Earth = 145 lb

a) Mass in slug is...

32.174 pound = 1 slug

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x = 145/32.174 = <em>4.51 lbf-s^2/ft</em>

b) Mass in kg is...

2.205 pounds = 1 kg

145 pounds = x kg

x = 145/2.205 = <em>65.8 kg</em>

c) Weight in Newton = mg

where

m is mass in kg

g is acceleration due to gravity on Earth = 9.81 m/s^2

Weight in Newton = 65.8 x 9.81 = <em>645 N</em>

d) If on the moon with acceleration due to gravity of 5.30 ft/s^2,

1 m/s^2 = 3.2808 ft/s^2

x m/s^2 = 5.30 ft/s^2

x = 5.30/3.2808 = 1.6155 m/s^2

weight in Newton = mg = 65.8 x 1.6155 = 106

weight in pounds = 106/4.448 = <em>23.8 lb</em>

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mass on the moon = <em>65.8 kg</em>

3 0
3 years ago
Technician A says amperage cannot exist without both voltage and resistance. Technician B says if amperage is high, then you kno
Ivan

Answer:

Technician A

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Ohms law:  I= E/R so rest resistance must be present along with E/potential difference.  Even if just wire shorted together there is resistance but very little.

Tech B: Again ohms law.  Current flow is directly proportional to the voltage and inversely  proportional to R (resistance or impedance).

8 0
3 years ago
You don't know which insert you have, and the inserts are different sizes, meaning the amount needed for a 1:3 ratio is differen
VashaNatasha [74]

Answer:

Explanation:

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3 years ago
Calculate the wire pressure for a round copper bar with an original cross-sectional area of 12.56 mm2 to a 30% reduction of area
dybincka [34]

Answer:153.76 MPa

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\alpha =\frac{30}{2}=15^{\circ}

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B=\mu cot\left ( \aplha\right )=0.2985

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\sigma _{pressure}=350\left [\frac{1+0.2985}{0.2985}\right ]\left [ 1-\frac{8.792}{12.56}\right ]^{0.2985}

\sigma _{pressure}=153.76 MPa

8 0
3 years ago
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