Materials Tutorial 3 - Stress, Strain, and the Young Modulus

 

 

Stress and Strain

Wires obey Hooke's Law just like a spring.  This is because bonds between atoms stretch just like springs:

 

 

If we stretch a wire, the amount it stretches by depends on:

If we have two of the same material and length but of different thicknesses, it is clear that the thicker wire will stretch less for a given load.  We make this a fair test by using the term tensile stress which is defined as the tension per unit area normal to that area.  The term normal means at 90o to the area. 

 

We can also talk of the compression force per unit area, i.e. the pressure

 

Stress = Load (N) = F                                             

           area (m2)   A

 

In some text books you may see stress given the code s (sigma, a Greek letter 's'). Therefore:

 

You will have met the expression F/A before.  It is, of course, pressure, which implies a squashing force.  A stretching force gives an expression of the same kind.  Units are newtons per square metre (N/m2) or Pascals (Pa).

 

 1 Pa = 1 N m-2

 

You must always convert areas to square metres for this equation to work.  Remember that radii will often be given in mm or cm.  This is a common bear trap.

 

Question 1

Find the area of a wire of diameter 0.75 mm in m2

Answer

 

If we have a wire of the same material and the same diameter, it doesn’t take a genius to see that the wire will stretch more for a given load if it is longer.  To take this into account, we express the extension as a ratio of the original length.  We call this the tensile strain which we define as the extension per unit length.

 

                        Strain = extension (m)              

                                   original length (m)           

 

There are no units for strain; it’s just a number.  It can sometimes be expressed as a percentage. 

 

 

You will find that the same is true for when we compress a material.

 

Question 2

What is the strain of a 1.5 m wire that stretches by 2 mm if a load is applied?

Answer

 

 

Stress-Strain Curves

Stress-strain graphs are really a development of force-extension graphs, simply taking into account the factors needed to ensure a fair test.  A typical stress-strain graph looks like this:

We can describe the details of the graph as:

We can draw stress-strain graphs of materials that show other properties.

 

 

The Young Modulus

The Young Modulus is defined as the ratio of the tensile stress and the tensile strain.  

  So we can write:

            Young modulus = tensile stress

                                      tensile strain

  We know that

            tensile stress = force F

                                  area     A

  and that

            tensile strain =  ___extension __   = Dl

                                   original length        l

Young Modulus has the physics code E, so we can write:

which becomes:

Units for the Young Modulus are Pascals (Pa) or newtons per square metre (Nm-2).

The Young Modulus describes pulling forces

We can link the Young Modulus to a stress strain graph.

The Young Modulus is the gradient of the stress-strain graph for the region that obeys Hooke’s Law.  This is why we have the stress on the vertical axis when we would expect the stress to be on the horizontal axis.

The area under the stress strain graph is the strain energy per unit volume (joules per metre3).

Strain energy per unit volume = 1/2 stress x strain.

The units arise because stress is in N m-2 and strain is m m-1 (NOTE: This unit here is not "millimetres to the minus one", but metres per metre which mean no units).

        N m-2 x m m-1 = N m m-3.  N m is joules, hence Jm-3

Area is the strain energy per unit volume.  So we can write this equation:

Area = ½ × stress × strain

In Physics code:

The term E' is pronounced "E-prime" or "E-dashed" and is being used as the code for the elastic strain energy per unit volume.

Al is area × length = volume

 

Question 3

A wire made of a particular material is loaded with a load of 500 N.  The diameter of the wire is 1.0 mm.  The length of the wire is 2.5 m, and it stretches 8 mm when under load.  What is the Young Modulus of this material?

Answer

Question 4

 What is the elastic strain energy per unit volume for the wire in question 3?

Answer

 

Measuring Stress and Strain

We can measure the Young Modulus doing a simple experiment like this:

 

We need to measure the diameter of the wire using a micrometer.  A video tutorial in how to do this is in one of the links.  Then we measure the extension as we increase the load, recording the observations.  We need to convert the force into stress, and extension into strain.  From that we plot a stress-strain graph, using the gradient to calculate the Young Modulus.

 

The value we get for E is often rather low.  This is because the wire might suffer the following defects:

There are also many uncertainties.  The greatest uncertainty comes in measuring the diameter.