Help to find force given a diameter reduction (Poissons Ratio)

In summary, the task at hand is to calculate the force required to reduce the diameter of a mild steel bar from 40mm to 39.99mm, given its dimensions and properties such as Young's Modulus and Poisson's ratio. After performing calculations, it was determined that the required force is 209.44 kN.
  • #1
LDC1972
65
0

Homework Statement



Mild steel bar 40 mm diameter and 100 mm long, subjected to a tensile force along its axis (through central length)
Youngs Modulus = 200GN m^-2
Poisson's ratio = 0.3

Calculate the force required to reduce the diameter to 39.99mm

(there is also a picture saying Use the x- y co-ordinate system above). I shall upload that momentarily if needed?

Homework Equations



Poissons ratio = Transverse strain = - εt / εl

Diameter = 0.04m
Lo = 0.1m
E = 200GN m^-2 = 200x10^9 Pascals (Pa)
V = 0.3

Transverse strain εt = (Reduction in Diameter - Original Diameter) / Original Diameter

Axial strain = Transverse strain / V

Δd = εtd

ε = εt / -V

ε = σ / E

σ = F/A


The Attempt at a Solution



So here's what I did and hoping it is right?

Transverse strain εt = (Reduction in Diameter - Original Diameter) / Original Diameter
(0.3999 - 0.4000) / 0.4000 = -0.00025 Transverse strain (εt)

Axial strain = Transverse strain / V
-0.00025 / 0.3 = .00083333333 Axial strain

I confirmed these by dividing Transverse strain by Axial strain = -0.3

Since decrease in diameter = transverse strain x Original Diameter then:
Δd = εtd = -0.00025 x .40 = 0.1mm which is correct.

As εt = -Vε Then:
-0.00025 = -0.3 x ε

ε = εt / -V
ε = 0.00083333333

ε = σ / E
σ is unknown ∴
σ = ε x E
σ = 16666666 Pa (stress value)

σ = F / A

Force = σ x A (Area)
= 16666666 Pa x 0.00125663706
= 20943.95019 Nm

I think here I can bring the figure down a thousand to 20.944 KNm?

Thank guys for ANY help! This took me ages!
 
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  • #2
1. What are the derived units for a Pascal?
2. Since when did force in SI become Newton-meters?
 
  • #3
SteamKing said:
1. What are the derived units for a Pascal?
2. Since when did force in SI become Newton-meters?

Oh, every time one of us Brit's ask a question regarding GNm, we get told to use the proper term GPa (I think that's the equivalent). So I swapped the GNm to Gpa. Yes just checked online and 1 Newtom metre = 1 Pascal.

Force SI should be Newtons? As in N or KN or MN but not NM?

Thanks!

I'm worried my force looks too low? .01mm should surely take more than my final figure?

Thanks

Lloyd
 
  • #4
SteamKing said:
1. What are the derived units for a Pascal?
2. Since when did force in SI become Newton-meters?

Just went through double checking figures as "felt" force too low to make a .01mm impact on diameter...

Found σ was missing a unit, one 6 short!

New final figure: 209439.51N
Or simplified to 209.43951KN
Or to 0.20943951 GN?
 
  • #5
LDC1972 said:
Oh, every time one of us Brit's ask a question regarding GNm, we get told to use the proper term GPa (I think that's the equivalent). So I swapped the GNm to Gpa. Yes just checked online and 1 Newtom metre = 1 Pascal.

Force SI should be Newtons? As in N or KN or MN but not NM?

Thanks!

I'm worried my force looks too low? .01mm should surely take more than my final figure?

Thanks

Lloyd

Not even close.

The pascal is the SI unit for pressure or stress, which implies a force divided by an area.

1 Pascal = 1 Newton / m^2
 
  • #6
F = 209.44 kN looks like a good result.
 
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Likes 1 person
  • #7
ignore this.
 
Last edited:

1. What is Poisson's Ratio?

Poisson's Ratio is a material property that describes the relationship between the lateral (transverse) strain and axial (longitudinal) strain of a material when it is subjected to a force.

2. How is Poisson's Ratio calculated?

Poisson's Ratio is calculated by dividing the lateral strain by the axial strain. It is represented by the Greek letter ν (nu) and has a range of -1 to 0.5, with most materials falling between 0 and 0.5.

3. What is the significance of Poisson's Ratio in a diameter reduction?

In a diameter reduction, Poisson's Ratio is important because it relates to the change in the material's volume. When a force is applied to a material, it will experience a decrease in diameter and an increase in length. Poisson's Ratio helps to determine the amount of lateral and axial strain that will occur in the material.

4. How does Poisson's Ratio affect the force required for a diameter reduction?

Poisson's Ratio can affect the force required for a diameter reduction because it is a measure of a material's resistance to deformation. A material with a higher Poisson's Ratio will require more force to achieve the same amount of diameter reduction compared to a material with a lower Poisson's Ratio.

5. Can Poisson's Ratio be changed?

No, Poisson's Ratio is a fundamental property of a material and cannot be changed. However, different materials have different Poisson's Ratios, so choosing the right material for a given application can help to optimize the force required for a diameter reduction.

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