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Evgen [1.6K]
2 years ago
7

A cylindrical 1040 steel rod having a minimum tensile strength of 865 MPa (125,000 psi), a ductility of at least 10%EL, and a fi

nal diameter of 6.0 mm (0.25 in.) is desired. Some 7.94 mm (0.313 in.) diameter 1040 steel stock, which has been cold worked 20% is available. Describe the procedure you would follow to obtain this material. Assume that 1040 steel experiences cracking at 40%CW
Engineering
1 answer:
gtnhenbr [62]2 years ago
4 0

Answer:

procedure attached below

The material to be used will have a %Cw of 34.5%Cw which is < 40%Cw

Explanation:

<em>Given data:</em>

Minimum tensile strength = 865 MPa

Ductility = 10%EL

Desired Final diameter = 6.0 mm

20% cold worked 7.94 mm diameter 1040 steel stock

<u> Describe the procedure you would follow to obtain this material.</u>

assuming 1040 steel experiences cracking at 40%CW

<em>attached below is a detailed procedure of obtaining the material</em>

The material to be used will have a %Cw of 34.5%Cw which is < 40%Cw

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Answer:

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3 years ago
A number 12 copper wire has a diameter of 2.053 mm. Calculate the resistance of a 37.0 m long piece of such wire.
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Answer:

R=1923Ω

Explanation:

Resistivity(R) of copper wire at 20 degrees Celsius is 1.72x10^-8Ωm.

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3 years ago
A two-dimensional flow field described by
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Answer:

the answer is

Explanation:

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3 years ago
A 20-cm-long rod with a diameter of 0.250 cm is loaded with a 5500 N weight. If the diameter decreases to 0.210 cm, determine th
ss7ja [257]

Answer:

1561.84 MPa

Explanation:

L=20 cm

d1=0.21 cm

d2=0.25 cm

F=5500 N

a) σ= F/A1= 5000/(π/4×(0.0025)^2)= 1018.5916 MPa

lateral strain= Δd/d1= (0.0021-0.0025)/0.0025= -0.16

longitudinal strain (ε_l)= -lateral strain/ν = -(-0.16)/0.3

(assuming a poisson's ration of  0.3)

ε_l =0.16/0.3 = 0.5333

b) σ_true= σ(1+ ε_l)= 1018.5916( 1+0.5333)

σ_true = 1561.84 MPa

ε_true = ln( 1+ε_l)= ln(1+0.5333)

ε_true= 0.4274222

The engineering stress on the rod when it is loaded with a 5500 N weight is 1561.84 MPa.

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