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Evgen [1.6K]
3 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]3 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:

The code will be:

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3 years ago
A composite shaft with length L = 46 in is made by fitting an aluminum sleeve (Ga = 5 x 10^3 ksi) over a
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Answer:

Explanation:

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∅_{steel = ∅_B = 3.01 × 10⁻⁴ rad

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