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MECHANICS - EXAMPLE

    Example


Test Specimens
 

To determine mechanical properties of a material, a torsion test and a tension test was done for two different diameter bar made from the same material as shown in the figures on the left.

During the torsion test, a maximum shear stress of 100 MPa was measured on a bar 50 mm diameter. The angle of twist was 2.5 degree over a length of 400 mm.

The tensile test was done with a bar of 20 mm diameter and 400 mm length. The specimen had an elongation of 0.5 mm when subjected to a tensile load of 40 kN.

Determine the Poisson's ratio of the material.

   
    Solution

   

Torsion test data:

     Diameter of bar, d1 = 2r1 = 50 mm
     Length, L1 = 400 mm
     Maximum shear stress, τ = 100 MPa
     θ = 2.5° = 0.04363 rad

Tension test data:

     Diameter of bar, d2 = 2r2 = 20 mm
     Length, L2 = 400 mm
     Tension, F= 40 kN
     Elongation, δL= 0.5 mm

To determine the poisson's ratio, it is necessary to know the modulus of elasticity and shear modulus of the material. These can be determined using the data from the two tests.

     

Torsion Test Details
 

For the tension test

     Strain,
     ε = Elongation / Length
       = δL / L2
       = 0.5 / 400
       = 0.00125

     Stress,
     σ = F / A
        = F / (π r22)
        = 40 / (π 0.012)
        = 127,300 kPa   
        = 127.3 MPa

     Modulus of Elasticity,
     E = σ / ε
        = 127.3 / 0.00125
        = 101,860 MPa
        = 101.9 GPa

     


Tension Test Details

 

For the torsion test

     Shear Modulus,
     G = L1 τ / θ r1
        = (400)(100) / (0.04363)(50/2)
        = 36,672 MPa
        = 36.67 GPa

Now for Poisson's ratio,

     G = E / 2(1 + ν)
     ν = (E / 2G) - 1
       = {101.9 / (2)(36.67)} - 1
       = 0.3894

Poisson's ratio, ν = 0.389

     
   
 
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