Deformation under load.   Since stress is force per unit area, and modulus of elasticity is stress divided by strain.  We can use the relationships to determine the deformation.  Lets see where the formula comes from.   Remember that strain is really deformation also, and is simply the change in length compared to the original length.  Therefore the following formula can be derived:
                  S = P / A;        E = S / e;      so  e = S / E    and      e = delta l / L

                 Therefore    E = (P / A) / (dl  / L)    or     E = (P x L) / (A x dl )

                 And,   dl   = (P x L) / (A x E)

Example Problem:  Find the change in length of a 24 inch aluminum rod 1/4 square rod supporting a 500 lb. load.

        Solution:   E for aluminum = 10, 000, 000 psi
                          A = .25 inches x .25 inches = .0625 in2
                         dl  = 500 lbs x 24 inches  /  (.0625 in2 x 10,000,000 lbs/in2)
                                   = 12,000 lb -inches  / (625,000 lbs)
                                   = .0192 inches
 
Temperature change and deformation.   Materials are also deformed by changes in temperature.  The rate of deformation as the temperature changes depends on the thermal coefficient  (alpha value) of the material.  The alpha value represents the change in length of a material per unit length per one degree change in temperature.   The alpha values of a few different materials is show below:
                                                      Alpha Values for Selected Materials
                                                          Note all values x 10-6    in/in/oF
                                          1020 Steel        6.5                             2014 Aluminum  12.8
                                          1030 Steel        6.3                             C3600 Brass      11.4
                                          CI145 Copper  9.9                             Concrete              6.0
                                          Wood (pine)     3.0                             Plate Glass           5.0
                                           Nylon (resin)  45.0                             ABS (resin)        53.0

          delta = alpha  x  L  x  delta temperature
               d = a  x  L  x dt
               d = ((in. / in.) / 0F)  x  in.  x  0F

Example Problem:   How much change in length would occur for a 50 foot length of railroad iron if the temperate at night is 40 degrees F and the temperature of the steel rail is raised to 120 degrees in the daytime?

          Solution:   a = .00000630;     L = 50 ft. x (12 in. / ft.) = 600 inches;  
                     dt = 120 - 40 = 80 degrees F
                           d  = .00000630 (in/in/oF)  x 600 in.  x 80 oF
                           d  = .3024 inches

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