Diploma of Mechanical Engineering

Strain.doc Christchurch Polytechnic Page 1 S. E. Tomsett. Version 19-10-00 Institute of Technology Printed 25-Mar-09

Strain (and Stress) in two directions Firstly consider an element with a single applied stress. The strain on the element will be as shown in the following diagram.

The strain in the X-direction in found from x = E.x x = x /E The strain in the y-direction in found from y = .x or y = .x /E

Now consider an element with two applied stresses. We can calculate the final strain by adding the strain resulting from each stress.

Thus, the total strain in the X-direction is x = x /E - .y /E and, the total strain in the Y-direction is y = y /E - .x /E Strain gauges. If we record the strains in a material in two perpendicular directions, then we can solve for the stresses in those directions. For this, the equations are rearranged to,

Where x, y, and z and three perpendicular directions Problem 1 Given 1 = -100x10

-6 and 2 = 350x10-6, calculate the stresses in these

directions. E = 200 GPa v = 0.3 [1 = 1.1 MPa, 2 = 70.3 MPa] Problem 2 Given 1 = -1x10

-3 and 2 = 800x10-6, calculate the stresses in these directions. E

= 100 GPa, v = 0.4 [1 = -80.95 MPa, 2 = 47.6 MPa]

Note, The principle strain are the result of the principle stresses, and act in the same direction. Remember that the principle stresses are at 90 to each other, thus can be used in the above calculations.

x

x y

x

x = x /E y = .x /E y = .y /E

y =y /E

y

y

x

y

zyxx

E

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