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

i mean i gues

explanation:

Other questions on the subject: Engineering

Engineering, 04.07.2019 18:10, koolgurl2003

Apump is used to circulate hot water in a home heating system. water enters the well-insulated pump operating at steady state at a rate of 0.42 gal/min. the inlet pressure and temperature are 14.7 lbf/in.2, and 180°f, respectively; at the exit the pressure is 60 lbf/in.2 the pump requires 1/15 hp of power input. water can be modeled as an incompressible substance with constant density of 60.58 lb/ft3 and constant specific heat of 1 btu/lb or. neglecting kinetic and potential energy effects, determine the temperature change, in °r, as the water flows through the pump.

Answers: 1

continue

Engineering, 04.07.2019 18:10, heidiburgos1own6c0

Fluids at rest possess no flow energy. a)- true b)- false

Answers: 3

continue

Engineering, 04.07.2019 18:10, abdirahmansoloman

Air is to be cooled in the evaporator section of a refrigerator by passing it over a bank of 0.8-cm-outer-diameter and 0.4-m-long tubes inside which the refrigerant is evaporating at -20°c. air approaches the tube bank in the normal direction at 0°c and 1 atm with a mean velocity of 4 m/s. the tubes are arranged in-line with longitudinal and transverse pitches of sl- st 1.5 cm. there are 30 rows in the flow direction with 15 tubes in each row. determine (a) the refrigeration capacity of this system and (b) pressure drop across the tube bank. evaluate the air properties at an assumed mean temperature of -5°c and 1 atm. is this a good assumption?

Answers: 1

continue

Engineering, 04.07.2019 18:10, bckyanne3

Afull journal bearing has a journal diameter of 27 mm, with a unilateral tolerance of -0.028 mm. the bushing bore has a diameter of 27.028 mm and a unilateral tolerance of 0.04 mm. the l/d ratio is 0.5. the load is 1.3 kn and the journal runs at 1200 rev/min. if the average viscosity is 50 mpa-s, find the minimum film thickness, the power loss, and the side flow for the minimum clearance assembly.

Answers: 1

continue

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...