Introduction | ||||||||||||||||||||||
| 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:
| |||||||||||||||||||||
Question | ||||||||||||||||||||||
What is the maximum shear strain in the area of OABC? | ||||||||||||||||||||||
Approach | ||||||||||||||||||||||
|
| 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 | |||||
| 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 εx´ or εy´ 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
| |||||
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
|
| 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.
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 | ||||||||||||||||||||||
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 = εx´ = 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 |
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° | ||
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° | ||
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 |