Thursday, 7 July 2011

HOOKE'S LAW

MECHANICS - CASE STUDY

    Introduction


Cross Section of Test System

Specimen Testing Set Up
Problem Animation

Click to view movie (101k)
  Susan Placer, a new engineer at AAA Materials Testing Lab, is asked to find the Poisson's ratio for a new epoxy adhesive. The test method used at the lab compresses a 1 in cube specimen under compression. The test frame consists of a 1 in cube hole to place the sample. Then a load is place on the top surface (see figure). The total deflection of the top surface is monitored.
The test sample is a 1 in cube which is placed into a perfectly fitting frame. The frame is substantially more stiff than the sample and will not deflect during the test.

What is known:
  • Young modulus, E, is 2,000 ksi.
  • The frame is much stiffer than the specimen.
  • The cube dimension is 1 in on each side.
  • At 3,000 lbs, the cube deflects 0.001 in.
  • The sides are frictionless.
  • There are no gaps when sample is in the frame.
   
  Question

  What is the Poisson's ratio of the new epoxy adhesive? Also, what stresses are generated on the sides of the test frame?

     
    Approach

   
  • Since the frame is much stiffer than the specimen, the side deflections can be assumed equal to zero.
  • Stress in the x and y directions (horizontal sides of box) are not zero.
  • Use the three dimensional equations to set up the problem.
  • First, determine Poisson's ratio.
  • Second, determine the stress on each side using the Poisson's ratio

MECHANICS - THEORY

    One Dimensional Hooke's Law


1-D Hooke's Law
  Recall, Hooke's Law in one dimension (uniaxial loading), relates the normal stress and normal strain as
 
σ = Eε
 
The constant E is Young's modulus and represents the stiffness of the the material.
     
    Poisson's Ratio


Poisson's Effect
Click to view movie (21k)


Contraction in y- and z-direction
when Stressed in x-direction
  When analyzing more than one dimension, interaction between all directions needs to be considered. This is done through Poisson's ratio. Basically, Poisson's ratio is the amount of transverse contraction, or negative strain, when strained in a given direction. For a basic object pulled or strained in the x-direction, the Poisson's Ratio is defined as
 
-ν = εy /εx
 
Thus, when a member is pulled in the x-direction, there is a contraction strain in the y-direction (and z-direction). If it is pulled in the y-direction, then the contraction strain will be in the x-direction (and z-direction).
For a three dimensional object, Poisson's ratio will occur in equally in both perpendicular directions. If the load is in the x-direction, then strain in the y- and z-direction will be
     εy = εz = -νεx
     
    Two Dimensional Stress-Strain


Stress in Two Directions
  If a material is isotropic (homogenous in all directions, such as a solid metal) and is pulled in two directions, then due to Poisson's ratio, the overall normal strain will be the total of the two strains. For example, if there are normal stresses in both the x- and y-directions, then the total normal strain in the x-direction is
     εx total = εx due to σx + εx due to σy = σx /E - νσy/E
     εx = (σx - νσy) / E
Similarly, the normal strain in the y-direction would be
     εy = (σy - νσx) / E
     

Pure Shear Stress in a 2D plane
Click to view movie (29k)


Shear Angle due to Shear Stress
  However, Hooke's Law also relates shear strain and shear stress. If the shear stress and strain occurs in a plane then the stress and strain are related as
 
τ = G γ
 
or
     γ = τ/G = [2(1 + ν)/E] τ
where G is the shear modulus (a material property) and γ is the shear strain. The shear strain is defined as the angle (radians) caused by the shear stress as shown in the diagram at the left.
The shear modulus is related to Young modulus and Poisson's ratio,
 
G = E / 2(1 + ν)
 
Two dimensional stress-strain relationships are summarized in the table below.
     
2D Hooke's Law (Stress-Strain Relationship)
Compliance Format   Stiffness Format
 
or in matrix form

  or in matrix form
     
     
    Three Dimensional Stress-Strain


Stress Directions in 3D
(τxy = τyx, τyz = τzy, τxz = τzx)

  Just like 1D or 2D, Hooke's Law can also be applied to material undergoing three dimensional stress (triaxial loading). The development of 3D equations is similar to 2D, sum the total normal strain in one direction due to loads in all three directions. For the x-direction, this gives,
     εx total = εx due to σx + εx due to σy + εx due to σz
                  = σx /E - νσy/E - νσz/E
     εx = (σx - νσy - νσy) / E
Similarly, the other directions can also be determined. The final equations are summarized in the table below.
     
3D Hooke's Law (Stress-Strain Relationship)
Compliance Format   Stiffness Format
 


MECHANICS - CASE STUDY SOLUTION


1.0 inch Specimen Cube Inserted
into Test Frame

  A simple cube of material is inserted into a snug fitting test frame and then a 3,000 lb load is applied on the top surface. The sides of the test frame are stiff and will not deflect at the test load level. Also, the sides are smooth so there is no friction forces as the specimen is loaded.
The objective in testing the specimen is to determine the Poisson's ratio of the material. The Young's modulus, E, is known, 2,000 ksi. Also, the total deflection after loading is 0.001 inches.
   
    Determining Poisson's ratio


Cube Under Loading
  A good place to start solving this problem is the full 3D Hooke's law equations.
     
While these look complex, many terms are easily determined. The vertical stress can be calculated as
     σz = F/A = 3000 lb/(1 in)2 = 3,000 psi = 3.0 ksi
The other two horizontal stresses are unknown, but must be the same due to symmetric geometry and loading. Thus,
     σx = σy = p
     

Problem Diagram
  Next, the three strains can be determined. The vertical strain can be found since the deflection is known, giving
     εz = δ/L = (0.001 in) / (1.0 in) = 0.001
The other two strains are also known since the outside walls do not deflect (due to their high stiffness). Thus,
     εx = εy = 0.0
Also, Young's modulus is known, E = 2,000 ksi.
     

Graph of Function
f(ν) = (1-ν) / [(1+ν)(1-2ν)]
  Simplifying the 3D equations to two equations give,
     
Next, substituting specific values reduce these equations to
     
The negative signs represent compression for both strain and stress. There are now two unknowns, p and ν. The second equation can be rearranged to give,
     
This can be manipulated to give
     3ν2 + 0.5ν - 0.5 = 0
Using the quadratic equation, the positive root gives
     ν = 0.3333
This equation can also be solved numerically by graphical methods or by using numerical programs like MathCad, Mathematica, MathLab, or even Excel. The equation is graphed at the left which also gives
     ν = 0.3333
     
    Stresses



  The stresses in the x and y directions can now be determined,
  
  p = σx = σy = -1.5 ksi = 1,500 psi



MECHANICS - EXAMPLE

    Example


Plate with Applied Loads
and Displacement
  A 0.5 cm thick rectangular plate is pulled in tension by two loads in the x and y directions as shown. The total deflection in the x and y direction is 0.021 cm and 0.009 cm, respectively. What is the Poisson's ratio of the material? The z direction has no load and the deflection is not known.
   
    Solution


Stresses and Strains on Plate
  This problem involves loading from two directions, and thus requires at least the 2-D Hooke's Law. The 3-D Hooke's Law could be used, but since σz is zero, those equations will reduce to the 2-D equations. The equations are,
     εx = (σx - ν σy)/E
     εy = (σy - ν σx)/E
The strains and stresses in the x and y direction need to be calculated.
     σx = Px/Ax = 5/[(0.05)(0.005)] = 20 MPa
     σy = Py/Ay = 9/[(0.10)(0.005)] = 18 MPa
     εx = 0.021/10 = 0.0021 cm/cm
     εy = 0.009/5 = 0.0018 cm/cm
Substituting the stresses and strains into the 2-D Hooke's Law equations, gives
     0.0021 E = 20 - ν 18
     0.0018 E = 18 - ν 20
Solving for ν gives,

     ν = 0.1876