Poissson again - totally stumped as compressive column?

jeudi 1 août 2013

1. The problem statement, all variables and given/known data

Column supports a mass on its' top. So force is downwards.

Column properties:

Do = 50mm (outer dia)

Di = 40mm (inner dia)

E = 250 GNm^-2 (modulus of elasticity)

V = 0.33 (Poissons ratio)

2. Relevant equations

Poissons ratio = Transverse strain = - εt / εl

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

3. The attempt at a solution

Those that saw my recent post will realise I've worked out how to calculate axial strain / transverse strain etc.

But that was where strain was 'pulling' a member and thereby reducing its' diameter and increasing its' length.

This problem is the opposite! There are also some additional technical questions related to strain gauges and wheatstone bridges. But I'll discuss this with my tutor Monday when he is available...

In the mean time I want to crack the first part of the problem.

The question:

Sketch a graph of expected strain against applied load, for the range 0 to 500kg. Make load the horizontal axis and strain the vertical.

The column is fixed vertically and the mass is placed on top.

So intuitively the diameter will increase and the length will decrease.

My first question is do I apply the same formula as before in my last post? Or will being a column effect the required equations?

My previous workings:

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)

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So far I have done the following:

Calculated area by A = ∏/4 [(0.05m^2) - (0.04m^2)]

A = 7.0685834 x10^-4 m^2

E = 250 x10^9 Nm^-2

l = 0.2m

Force = mass x gravity

I calculated force at 50kg intervals to 500kg

So F = 490.5N @ 50kg

to F = 4905N @ 500kg

I then calculated stress as σ = F/A , using the 10 values of force I derived.

So σ = 693915.6 Pa @ 490.5N

to σ = 6939155.6 Pa @ 4905N

So I believe I have the load (force) worked out to plot already correctly?

And can use ε = σ / E to calculate strain for the vertical axis?

Many many thanks,

Lloyd