Question

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.

Answer 1

a)
`E_(m)` = 133.75 Gpa

b) Fnet = 560 N

c) thermal expansion of the composite material = 14.31
`10^(-6 )` / °C

Step-by-step explanation:

Solution:

a) Elastic Modulus of the composite:

Area of steel wire =
`(\pi )/(4)` x (
`0.001^(2)`) = 0.8 x
`10^(-6)`
`m^(2)`

Area of Copper wire =
`(\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 =
`(0.8. 10^(-6) )/((0.8 + 2.4) .10^(-6) )`

Ratio for area of steel =
`(0.8)/(3.2 )` = 0.25

Similarly, for Copper,

Ratio for area of copper =
`(2.4. 10^(-6) )/((0.8 + 2.4) .10^(-6) )`

Ratio for area of copper =
`(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 +
`(F)/(A)`
`. (1)/(Est)`) =
`\alpha .`ΔT +
`(F)/(A)`
`. (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) +
`(F)/(0.8.10^(-6) ) . (1)/(205 . 10^(9) )` ) = ( 17 x
`10^(-6)` x (1) +
`(-F)/(2.4.10^(-6) ) . (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°) +
`(-0.71)/(2.4.10^(-6) ) . (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