Engineering, 21.01.2021 22:20 ahmedeldyame

ASAE 1060 Steel wire (1 mm diameter) is coated with copper to form a composite with a diameter of 2mm. Use the following properties for parts a, b, and c of question 2: The elastic modulus of copper is 110 GPa The yield stress of the copper is 140 MPa The coefficient of thermal expansion of the copper is 17 times 10^-6/degree C. The elastic modulus of steel is 205 GPa The yield stress of the steel is 280 MPa The coefficient of thermal expansion of the copper is 10 times 10^-6/degree C Determine: a. The elastic modulus of the composite b. The maximum force that the composite will carry before either material yields c. The coefficient of thermal expansion of the composite material.

Answers: 2

Show answers

Answers

Answer from: preciousweeks11

a) $E_{m}$ = 133.75 Gpa

b) Fnet = 560 N

c) thermal expansion of the composite material = 14.31 $10^{-6 }$ / °C

Explanation:

Solution:

a) Elastic Modulus of the composite:

Area of steel wire = $\frac{\pi }{4}$ x ( $0.001^{2}$ ) = 0.8 x $10^{-6}$ $m^{2}$

Area of Copper wire = $\frac{\pi }{4}$ x ( $0.002^{2}$ ) - 0.8 x $10^{-6}$ $m^{2}$

Area of Copper wire = 2.4 x $10^{-6}$ $m^{2}$

Young's Modulus of Composite mixture:

$E_{m}$ = $F_{st}$ $E_{st}$ + $F_{Cu}$ $E_{Cu}$ Equation 1

here,

$F_{st}$ = Stress in Steel

$F_{Cu}$ = Stress in Copper.

We know that,

F = P/A

F is inversely proportional to Area, so if area is large, stress will less and vice versa. So, Take

Ratio for area of steel = $\frac{0.8. 10^{-6} }{(0.8 + 2.4) .10^{-6} }$

Ratio for area of steel = $\frac{0.8}{3.2 }$ = 0.25

Similarly, for Copper,

Ratio for area of copper = $\frac{2.4. 10^{-6} }{(0.8 + 2.4) .10^{-6} }$

Ratio for area of copper = $\frac{2.4 }{3.2}$ = 0.75

Put these values in equation 1:

$E_{m}$ = $F_{st}$ $E_{st}$ + $F_{Cu}$ $E_{Cu}$

$E_{m}$ = (0.25) $E_{st}$ + (0.75) $E_{Cu}$

We are given that,

$E_{st}$ = 205 Gpa

$E_{Cu}$ = 110 Gpa

So,

$E_{m}$ = (0.25) (205 Gpa) + (0.75) (110 GPa)

$E_{m}$ = 51.25GPa + 82.5 Gpa

Hence, the Elastic Modulus of the composite will be:

$E_{m}$ = 133.75 Gpa

b) maximum force:

Fnet = Fst + Fcu

We know that F = (Yield Stress x Area)

F = fst x Ast + fcu x Acu

And we are given that,

Yield stress of Steel = 280 Mpa

Yield stress of Copper = 140 Mpa

And,

Ast = 0.8 x $10^{-6}$ $m^{2}$

Acu = 2.4 x $10^{-6}$ $m^{2}$

Just plugging in the values, we get:

F = (280 Mpa) (0.8 x $10^{-6}$ $m^{2}$ ) + (140 Mpa) (2.4 x $10^{-6}$ $m^{2}$ )

F = 224 + 336

Fnet = 560 N ( because Mpa = $10^{6}$ N/ $m^{2}$ )

So, it means the composite will carry the maximum force of 560N

c) Coefficient of Thermal Expansion:

Strain on both material is same upon loading so,

(ΔL/L)st = (ΔL/L)cu

by thermal expansion equation:

( $\alpha .$ ΔT + $\frac{F}{A}$ $. \frac{1}{Est}$ ) = $\alpha .$ ΔT + $\frac{F}{A}$ $. \frac{1}{Ecu}$ )

Where $\alpha$ = Coefficient of Thermal expansion

Here, fst = -fcu = F

and ΔT = 1°

So,

Plugging in the values, we get.

( 10 x $10^{-6}$ x (1) + $\frac{F}{0.8.10^{-6} } . \frac{1}{205 . 10^{9} }$ ) = ( 17 x $10^{-6}$ x (1) + $\frac{-F}{2.4.10^{-6} } . \frac{1}{110 . 10^{9} }$ )

Solving for F, we get:

F = 0.71 N

Here,

fst = F = 0.71 N (Tension on Heating)

fcu = -F = 0.71 N ( Compression on Heating )

So, the combined thermal expansion of the composite material will be:

(ΔL/L)cu = ( 17 x $10^{-6}$ x (1°) + $\frac{-0.71}{2.4.10^{-6} } . \frac{1}{110 . 10^{9} }$ )

(ΔL/L)cu = ( 17 x $10^{-6}$ x (1°) - 2.69 x $10^{-6}$

combined thermal expansion of the composite material = 14.31 $10^{-6 }$ / °C

Answer from: Quest

answer: 500 wph.

explanation:

Answer from: Quest

answer:

6.3 is the answer

explanation:

because it is the ex

Another question on Engineering

Engineering, 03.07.2019 15:10

Heat is added to a piston-cylinder device filled with 2 kg of air to raise its temperature 400 c from an initial temperature of t1 27 cand pressure of pi 1 mpa. the process is isobaric process. find a)-the final pressure p2 b)-the heat transfer to the air.

Answers: 1

Answer

Engineering, 04.07.2019 03:10

What precautions should you take to prevent injuries when dealing with heavy loads?

Answers: 1

Answer

Engineering, 04.07.2019 18:10

Water in a partially filled large tank is to be supplied to the roof top, which is 8 m above the water level in the tank, through a 2.2-cm-internal-diameter pipe by maintaining a constant air pressure of 300 kpa (gage) in the tank. if the head loss in the piping is 2 m of water, determine the discharge rate of the supply of water to the roof top in liters per second.

Answers: 3

Answer

Engineering, 04.07.2019 18:20

Athin walled concentric tube exchanger is used to cool engine oil from 160°c to 60°c with water that is available at 25°c acting as a coolant. the oil and water flow rates are each at 2 kg/s, and the diameter of the inner tube is 0.5 m and the corresponding value of the overall heat transfer coefficient is 250 w/m2. oc. how long must the heat exchanger be to accomplish the desired cooling? cpwater=4.187 kj/kg-candcpengine el=2.035 kj/kg·°c, oil . 120]

Answers: 1

Answer

You know the right answer?

ASAE 1060 Steel wire (1 mm diameter) is coated with copper to form a composite with a diameter of 2m...

Questions