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Mrrafil [7]
4 years ago
6

Q7. A cylindrical rod of 1040 steel originally 15.2 mm (0.60 in.) in diameter is to be cold worked by drawing; the circular cros

s section will be maintained during deformation. A cold-worked tensile strength in excess of 840 MPa (122,000 psi) and a ductility of at least 12%EL are desired. Furthermore, the final diameter must be 10 mm (0.40 in.). Explain how this may be accomplished. (10 points)

Engineering
1 answer:
umka2103 [35]4 years ago
5 0

Answer:

11.2mm or 0.45in

Explanation:

The percent cold work, attendant tensile strength and ductility if drawing is carried out without interruption is given by the equation you will find in the attached file.

Please go through the attached file for a step by step solution to this question.

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Extra points!!!!! <br><br> List the general types of housing<br> (Example: condo)
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co-op, apartment, townhome, manor etc

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sergey [27]

Answer:   1.  Hand Plane

                 2. Rift Sawn

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2 years ago
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A balanced three-phase 208 V wye-connected source supplies a balanced three-phase wyeconnected load. If the line current IA is m
tatyana61 [14]

Answer:

6.004 Ω

Explanation:

For  a Y- connected system given that :

Line voltage, $V_L = 208 \ V$

Line current , $I_L=20\ A$

and specified that $V_L \ and \ I_L$ are in phase.

Hence the impedance will be pure resistive.

For Y-system

$V_L = \sqrt3 V_{ph}$

$V_{ph}$ = phase voltage

$V_{ph}$ $=\frac{V_L}{\sqrt3} = \frac{208}{\sqrt3}$

    = 120.08 V

Line current = Phase current

$I_L = I_{ph} = 20 \ A$

Now, $z_{ph} = \frac{V_{ph}}{I_{ph}}=\frac{120.08}{20}$

                          = 6.004 Ω

5 0
3 years ago
A full journal bearing has a shaft diameter of 3.000 in with a unilateral tolerance of -0.0004 in. The l/d ratio is unity. The b
sashaice [31]

Answer:

i) 154°F

ii) 0.0114 mm

iii) 0.075 btu/s

iv) 0.080 in^3/s

Explanation:

<u>i)Determine the average film Temperature </u>

( from Viscosity-temperature chart in US customary units for SAE10 )

at Temp =  154°F

absolute viscosity = 4.25 rev

and ΔT = 2 ( 154 - operating temp ) = 28°F

where : operating temp = 140°F as given in question

also from the chart applying Raimondi and Boyd boundary conditions

ΔT = 29°F  hence we can pick 154°F as the average film temperature

<u>ii) Calculate the minimum film thickness</u>

Cmin = Bore diameter - Journal shaft diameter / 2

         = 3.003 - 3 / 2 = 0.0015 in

Given that : h₀ / Cmin = 0.76

there h₀ = 0.0015 * 0.76 = 0.0114 mm

<u>iii)Determine the heat loss rate </u>

Also known as power loss ratio ( H ) = ( 2π*w*f*r*N ) / ( 778 *12 )

( 2π * 0.0132 * 675 * 1.5 * 10 ) / ( 9336 )

Heat loss rate = 0.075 btu/s

<u>iv)Calculate lubricant side-flow rate for minimum clearance assembly </u>

Side flow rate = 0.315 * Total volume flow rate

                       = 0.315 * ( 3.8 * 1.5 * 0.0015 * 10 * 3 )

                      = 0.080 in^3/s.

6 0
3 years ago
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