Wednesday, 6 July 2011

PLANE STRAIN

MECHANICS - CASE STUDY

    Introduction


Aircraft Fuselage Loading
  A bracket is expected to carry a large load and the designers need to know the maximum shear strain. The problem is there is no way to measure shear strain accurately. Also, they do not know what direction the maximum shear strain will be a maximum.
However, they do have a new laser instrument that can accurately measure the displacement between any two points on an object.
What is known:
  • Four points are spaced 1 cm apart to form a perfect square (before loading).
  • Laser measuring device determines the total distance between point O and the other three points after loading is:

        Distance (cm)
    Pt 1 Pt 2 Initial Final
    O A 1.0 1.00150
    O B 1.41421 1.41775
    O C 1.0 1.00200
     
    Question

    What is the maximum shear strain in the area of OABC?

     
    Approach

   
  • Calculate the normal strains between all points.
  • Set up a basic coordinate system with OC the x-axis and OA the y-axis.
  • Use the strain transformation equations to determine the shear strain in the x-y coordinate system.
  • Find the maximum shear strain


MECHANICS - THEORY


Strains at a point (Strain Element)

Shear Strain (radians)
  Strain Element

  Similar to stress analysis at a point, strain can also be rotated to different directions and thus maximum shear and normal strains can be found. Strain at a given point on a two dimensional object can be seen as a small strain element with two normal strains and one shear strain, as shown in the diagram.
The sign convention for strains at a point is similar to stresses at a point. Normal tension strain in both the x and y direction are assumed positive. The positive shear strain direction is shown in the diagram at the left. Shear stresses act on four sides of the stress element, causing a pinching or shear action. All shear stresses on all four sides are the same, thus
     γxy = γyx
Recall, the shear strain is actually defined as the angle of rotation or twist due to the shear stress. This angle is in radians and is shown at the left.
   
    Strain Rotation


Rotated Stress Element


Rotated Stress Element
  Stains at a point (strain element) can be rotated to give a new strain state at any particular angle. The rotation angle, θ, is assumed positive using the right hand rule (counter-clockwise in the x-y plane is positive). The new coordinate system is labeled as x' and y'. The new rotated strains are shown in the diagram at the left. The shear stresses, γx'y' and γy'x' are still equal.
As one may expect, the strain transformation equations are nearly identical to the stress transformation angles. The only difference is a factor of 2 for the shear strain. Other than that, the strain symbol, ε, can be simply substituted for σ to give the strain transformation equations (detail derivation is omitted). The final equations are,
 
 
     
    Principal Strains

    Generally, the largest normal strain is of most interest. This can be found by taking a derivative of either the ε or ε strain with respect to θ and equating it to zero. This will give the principal rotation angle, θp, that will produce the principal (maximum and minimum) strains. The resulting equations are
 
 
These strains, ε1 or ε2 are referred to as the principal shear strains.
     

Plot of Rotated Strain
  Maximum Shear Strain

  Similar to the principal strains, the maximum shear strain can be determine by taking a derivative of γx´y´ strain with respect to θ and equating it to zero. This gives
 



MECHANICS - CASE STUDY SOLUTION


Distance Between Points Before
and After Loading

  The distance between points A, B, C and O are measured using a laser device before and after the bracket is loaded. The results are listed in the table below.
    Distance (cm)
Pt 1 Pt 2 Initial Final
O A 1.0 1.00150
O B 1.41421 1.41775
O C 1.0 1.00200

These displacements can be used to find the normal strain in the direction of OA, OB, and OC. Then, the strains can be analyzed using the strain transformation equations. Finally, the maximum shear strain can be determined.
     
    Normal Strains


Normal Strain Between Points
  Since the distance before and after loading are known, the normal strains can be determined as
     
The εOA and εOC are in the direction of the y-axis and x-axis, respectively. Also, the strain εOB is in the direction of the new rotated coordinate x´. This gives
      εOA = εy = 0.00150 = 1,500 μ
      εOB = ε = 0.00250 = 2,500 μ
      εOC = εx = 0.00200 = 2,000 μ
Since strain is generally a very small number, the symbol μ "micro-strain" is commonly used to represent 10-6. This symbol does not represent units, and strain is still unit-less.
     
    Rotating Strains

     
    The stress element needs to be rotated 45o in the positive direction. Using the stress transformation equations, the stresses in the new x'-y' coordinate system will be,
    
     2,500 μ = 1,750 μ + 0 + γxy/2
     γxy = 750 μ = 0.000750
     
    Maximum Shear Strain

    The maximum shear strain can be found by
     
     γmax = 450.7μ = 0.0004507
The maximum shear strain will on a plane that is rotated θγ-max. That angle is,
     
     θγ-max = -16.85o


MECHANICS - EXAMPLE


Strain Element
  Example

  What is the largest principle strain for the strain element shown on the left?
   
  Solution

  The given strain components are,
       εx = 310 ×10-6 = 310 μ
     εy = -140 ×10-6 = -140 μ
     γxy = 260 ×10-6 = 260 μ
where m represents micro-strains.

     
    The principal strain equations could be used directly, but another way is to find the principal angles, and then use them to find the principal strains. The principle angles for this strain state are
     tan 2θp =  γxy / (εx - εy)
                = 260/(310 + 140) = 0.5778
     θp = (tan-1 0.5778) / 2
     θp = 15.0°, 105.0°
     

Rotated Strain Element, 15°
  Notice there are two possibilities. Both will give the same principal strains as will be shown below. For 15o, the two normal strains are,
     
     εx' = [310 + (-140)]/2 + [310 - (-140)]/2 cos 30
              + 260/2 sin 30
         = 85 + 194.9 + 65.0 = 344.9 μ
     εy' = [310 + (-140)]/2 - [310 - (-140)]/2 cos 30
             - 260/2 sin 30
         = 85 - 194.9 - 65 = -174.9 μ
The principal strains are,
     ε1 = εx' = 344.9 μ @ 15°
     ε2 = εy' = -174.9 μ @ 15°

     

Rotated Strain Element, 105°
  The other angle, 105°, will give the same results, but x' and y' will be reversed.      
     εx' = [310 + (-140)]/2 + [310 - (-140)]/2 cos 210
              + 260/2 sin 210
         = 85 - 194.9 - 65 = -174.9 μ
     εy' = [310 + (-140)]/2 - [310 - (-140)]/2 cos 210
             - 260/2 sin 210
         = 85 + 194.9 + 65.0 = 344.9 μ
The principal strains are,
     ε1 = εy' = 344.9 μ @ 105°
     ε2 = εx' = -174.9 μ @ 105°
As expected, the principal strain values are the same, the new coordinate, x' and y' are just reversed. But actually, the direction are the same sine this angle is 90 degree off from the previous angle