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

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Discussion Overview

The discussion revolves around calculating the force required to reduce the diameter of a mild steel bar from 40 mm to 39.99 mm under tensile stress, utilizing concepts such as Young's modulus and Poisson's ratio. The context includes homework-related calculations and the application of material properties in a theoretical framework.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • The initial calculations presented by a participant involve determining transverse strain and axial strain based on the diameter reduction and Poisson's ratio.
  • Concerns are raised about the derived units for pressure and stress, particularly questioning the use of Newton-meters for force.
  • Another participant corrects the stress calculation, suggesting that the initial force value was too low and provides a revised force figure.
  • There is a discussion about the proper terminology for pressure units, with references to GPa and the confusion surrounding the use of GNm versus GPa.
  • One participant expresses uncertainty about the adequacy of the calculated force in relation to the expected impact on diameter.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of the calculated force and the appropriate units for force and pressure. There is no consensus on the final force value, as some participants question the initial calculations while others provide corrections.

Contextual Notes

Participants highlight potential errors in unit conversions and calculations, particularly regarding the stress value and its units. The discussion reflects uncertainty about the implications of the calculated force in relation to the physical changes in the material.

Who May Find This Useful

This discussion may be useful for students and professionals interested in material mechanics, particularly those dealing with tensile stress calculations and the application of Poisson's ratio in engineering contexts.

LDC1972
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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!
 
Physics news on Phys.org
1. What are the derived units for a Pascal?
2. Since when did force in SI become Newton-meters?
 
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
 
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?
 
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
 
F = 209.44 kN looks like a good result.
 
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ignore this.
 
Last edited:

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