plz solve this stress strain question. some one plzzz?

Question

Q. A cylindrical bar of length ‘L₀’ and diameter ‘d₀’ is deformed using a force ‘F’. It should NEITHER experience a Plastic deformation NOR a reduction of diameter ‘Δd’ more than 7.5x10^(-3)mm. Which of the following material is suitable? Give a solution and reason to justify your answer.

length, L₀ =100 mm
diameter, d₀ = 10mm
change in diameter, Δd = 7.5x10-3 mm
Force, F= 27500 N

Material - Elasticity Modulus -- Yield strength --- Poisson Ratio
alloys - Giga Pascal -- Mega Pascal --- ratio
1. Alumin - 70 -- 200 --- 0.33
2. Brass -101 -- 300 --- 0.35
3. Steel - 207 -- 400 --- 0.27
4. Titanium - 107 -- 650 --- 0.36

Answer 1

1. Some usual definitions and relations
The stress in the bar is defined as quotient of force by cross sectional area
s = F/A = F/(pi/4*d²)
According to Hooke's law it is the product of elasticity modulus E and elongation e
s = E*e
Elongation is defined as:
e = ΔL/L₀
The contraction
c = Δd/d₀
is related to the elongation by the Poisson ratio p
c = -p*e
It reduces the cross sectional area to
A = pi/4*d₀²*(1+c)² = A₀*(1-p*e)²
Hence the stress is
s = F/(A₀*(1+c)²) = s₀/ (1- p*e)²
So stress rises with increasing elongation

2. To avoid tediuos calculations, lets neglect the stress rise for a moment, assume it constant at its lower limit
s ~ s₀= F/(pi/4*d₀²) = 350.14 MPa
That is greater than the yield strength of aluminum and brass. Hence there will surely occur a plastic deformation of both materials.
Next step is to estimate a lower limit of the contraction, which must hold the condition:
|c| < 7.5*10-4
From
s₀ = E*e = -E*c/p
=>
|c| = p*s₀/E
For the given modules you get:
steel |c| = 4.567*10-4 → ok
titanium |c| = 1.178*10-3 → too high
So only steel will (probably) fulfill the requirements of the problem.

3. The last step is to verify that steel fulfill the requirements by using the exact description:
s₀/ (1- p*e)² = E*e
or in terms of c
s₀/ (1+ c)² = E*c/p
=>
c*(1+c)² = p*s₀/ E
solve this cubic equation analytically or numerically to get
c = -4.571*10-4
That's ok (and just slightly greater than value calculated from 2).
The stress is
s = s₀/(1+c)² = 350.46 MPa
which is lower than the yield strength of 400Mpa.

Conclusion: steel is the right choice.

Answer 2

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Answer 3

I would have considered it, if you had spelled out the word Please and Plleeeeassee???

Don't be lazy, do it yourself.