USA: +1-585-535-1023

UK: +44-208-133-5697

AUS: +61-280-07-5697

Poisson’s Ratio


Poisson’s ratio is defined as the ratio of the lateral strain to the longitudinal strain. It is denoted as ll·


2.5.1 Uni-axial Stress System


If a bar is lengthened by axial tension, there will be a reduction in the transverse dimensions. Poisson in 1811 showed that the ratio of the unit deformations or strains in these directions is constant for stresses within the proportional limit.

As there is increase in dimension in z directions, there will be decrease in dimension

in the x and y directions.


Poisson’s ratio,            µ = lateral strain / longitudinal strain = – εx / εz = -εy / εz


Just as E is the property of material, µ is also a material property which can be

Determined experimentally

µ = – εxz

εx = – µ εz = – µ x σz / E

εy= – µ . σz / E

εz =  σz / E.


Equation (2.4) will give the strains in x, y and z directions due to loading on z direction.


2.5.2 Biaxial Stre ss System


Fig. 2.11 shows loading on two directions σx and σz. In this case for σ’z, there will be increase in dimensions in z direction but σx effects the lengthening through x direction and it tries to shorten through z direction.

εz =  σz / E. – µ . σx / E

εx =  σx / E. – µ . σz / E

εy =  σx/ E. – µ . σz / E


Equation (2.5) will give the strains in x, y, and z directions due to loading in x and z directions.


2.5.3 Triaxial Stress System


Here the effective strain is determined as the summation of direct effect and poisson’s effect.

+ Direct effect and -Poisson’s effect


εx =  σx / E. – µ. σy / E . – µ σz / E

εy =  σy / E. – µ . σx / E. – µ σz / E

εz =  σz/ E. – µ .  σx / E – µ σz / E

εx = 1/ E [ σx – µ (σy + σz)]

 εy  =1/ E [ σy – µ (σx + σz)]

εz =1/ E [ σz – µ (σz + σy)]


Equation (2.6) give us the general formula for a triaxial stress system


Here all the stresses are tensiles, if for a particular problem some of the stresses are compressive, adequate changes in signs should be made.

We can solve for crx, cr.v and crz from equations (2.6), Solution for ax and ay.for biaxial stresses

σx = ( εx + µεy) E / 1-µ2

for biaxial stresses

σy = (εy + µεx) E / 1-µ2