Answer:
a)
= 133.75 Gpa
b) Fnet = 560 N
c) thermal expansion of the composite material = 14.31
/ °C
Explanation:
Solution:
a) Elastic Modulus of the composite:
Area of steel wire =
x (
) = 0.8 x
![m^{2}](https://tex.z-dn.net/?f=m%5E%7B2%7D)
Area of Copper wire =
x (
) - 0.8 x
![m^{2}](https://tex.z-dn.net/?f=m%5E%7B2%7D)
Area of Copper wire = 2.4 x
![m^{2}](https://tex.z-dn.net/?f=m%5E%7B2%7D)
Young's Modulus of Composite mixture:
= ![F_{st}](https://tex.z-dn.net/?f=F_%7Bst%7D)
+ ![F_{Cu}](https://tex.z-dn.net/?f=F_%7BCu%7D)
Equation 1
here,
= Stress in Steel
= 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} }](https://tex.z-dn.net/?f=%5Cfrac%7B0.8.%2010%5E%7B-6%7D%20%7D%7B%280.8%20%2B%202.4%29%20.10%5E%7B-6%7D%20%7D)
Ratio for area of steel =
= 0.25
Similarly, for Copper,
Ratio for area of copper = ![\frac{2.4. 10^{-6} }{(0.8 + 2.4) .10^{-6} }](https://tex.z-dn.net/?f=%5Cfrac%7B2.4.%2010%5E%7B-6%7D%20%7D%7B%280.8%20%2B%202.4%29%20.10%5E%7B-6%7D%20%7D)
Ratio for area of copper =
= 0.75
Put these values in equation 1:
= ![F_{st}](https://tex.z-dn.net/?f=F_%7Bst%7D)
+ ![F_{Cu}](https://tex.z-dn.net/?f=F_%7BCu%7D)
= (0.25)
+ (0.75)
We are given that,
= 205 Gpa
= 110 Gpa
So,
= (0.25) (205 Gpa) + (0.75) (110 GPa)
= 51.25GPa + 82.5 Gpa
Hence, the Elastic Modulus of the composite will be:
= 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
![m^{2}](https://tex.z-dn.net/?f=m%5E%7B2%7D)
Acu = 2.4 x
![m^{2}](https://tex.z-dn.net/?f=m%5E%7B2%7D)
Just plugging in the values, we get:
F = (280 Mpa) (0.8 x
) + (140 Mpa) (2.4 x
)
F = 224 + 336
Fnet = 560 N ( because Mpa =
N/
)
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:
(
ΔT + ![\frac{F}{A}](https://tex.z-dn.net/?f=%5Cfrac%7BF%7D%7BA%7D)
) =
ΔT + ![\frac{F}{A}](https://tex.z-dn.net/?f=%5Cfrac%7BF%7D%7BA%7D)
)
Where
= Coefficient of Thermal expansion
Here, fst = -fcu = F
and ΔT = 1°
So,
Plugging in the values, we get.
( 10 x
x (1) +
) = ( 17 x
x (1) +
)
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
x (1°) +
)
(ΔL/L)cu = ( 17 x
x (1°) - 2.69 x ![10^{-6}](https://tex.z-dn.net/?f=10%5E%7B-6%7D)
combined thermal expansion of the composite material = 14.31
/ °C