Uncertainty Analysis of Poisson’s Ratio

Compare the exact distribution of Poisson's ratio with a first-order linear approximation

Given measurements of Young’s modulus

= (209.2 ± 1.2) kN/mm² and shear modulus

= (81.2 ± 1.5) kN/mm², assume symmetric triangular uncertainty distributions with 90% coverage probability. Determine the uncertainty of Poisson’s ratio,

ν=Y/(2G)−1

Store the measured values of Young's modulus and shear modulus and their uncertainties, expressed in kN/mm²:

In[1]:=

{μ,σ,μ,σ}=Quantity[{209.2,1.2,81.2,1.5},"Kilonewtons"/"Millimeters"^2]

Out[1]=



209.2

kN/

1.2

kN/

81.2

kN/

1.5

kN/



Compute the scaling factor that maps a symmetric triangular distribution on [−1,1] to a 90% coverage interval:

In[2]:=

=InverseCDFTriangularDistribution[{-1,1}],1+

2

Out[2]=

Define symmetric triangular probability distributions for Young's modulus and shear modulus with 90% coverage:

In[3]:=

y=TriangularDistribution[{μ-σ/,μ+σ/}]s=TriangularDistribution[{μ-σ/,μ+σ/}]

Out[3]=

QuantityDistributionTriangularDistribution[{207.445,210.955}],

kN/



Out[4]=

QuantityDistributionTriangularDistribution[{79.0063,83.3937}],

kN/



Verify that the distribution reproduces the stated 90% coverage interval:

In[5]:=

Probability[Abs[μ-]<σ,y]

Out[5]=

0.9

In[6]:=

Probability[Abs[μ-]<σ,s]

Out[6]=

0.9

Define the exact probability distribution of Poisson's ratio using the joint distribution of Y and G:

In[7]:=

pr=TransformedDistribution



2

-1,{y,s}

Out[7]=

TransformedDistribution-1+

x.1

2x.2

,{x.1TriangularDistribution[{207.445,210.955}],x.2TriangularDistribution[{79.0063,83.3937}]}

The exact distribution of Poisson's ratio is now defined. The following steps introduce a first-order linear approximation and compare it with the exact result.

Define a first-order Taylor expansion used to approximate the Poisson's ratio formula with a linear function:

In[8]:=

lin[e_,{x_,x0_},{y_,y0_}]:=Block{f=Function[{x,y},e]},f[x0,y0]+

(1,0)

[x0,y0](x-x0)+

(0,1)

[x0,y0](y-y0)

Approximate the result by a normal distribution with matching mean and standard deviation:

In[9]:=

lin[_]:=NormalDistribution[Mean[],StandardDeviation[]]

In[10]:=

prApprox=TransformedDistributionlin



2

-1,{,μ},{,μ},{lin[y],lin[s]}

Out[10]=

NormalDistribution[0.288177,0.0148769]

Find the mean ratio using the exact and the approximate distributions:

In[11]:=

{μpr,μprApprox}={Mean[pr],Mean[prApprox]}

Out[11]=

{0.288334,0.288177}

Find the standard deviation of the ratio using the exact and the approximate distributions:

In[12]:=

{σpr,σprApprox}={StandardDeviation[pr],StandardDeviation[prApprox]}

Out[12]=

{0.0148823,0.0148769}

Define the probability density function of Poisson's ratio from the exact distribution:

In[13]:=

pdf=PDF[pr,r];

Visualize the density function and compare it to the normal approximation:

In[14]:=

Plot[Evaluate@{PDF[pr,r],PDF[prApprox,r]},{r,μpr-4σpr,μpr+4σpr},PlotLegends{"Exact distribution","Normal approximation"},PlotRangeAll,ExclusionsNone]

Out[14]=

	Exact distribution
	Normal approximation

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Uncertainty Analysis of Poisson’s Ratio

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Uncertainty Analysis of Poisson’s Ratio

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