» Uniform Axial Member
In Chapter 3, we introduced an AXIAL MEMBER of length, L, subjected to a constant tensile force, P, acting through the centroid of its cross-sectional area, A. The following terms were then defined for the Axial Member:

e =

D L

;   s =

P A

;   s = Ee

Combining the above we get:

D = PL AE

» Discretely Varying Axial Member
To solve problems involving Discretely Varying Loads, Areas, or Modulus, break up the bar into lengths over which all the values of Force, Area and Modulus are constant. The total deflection can be obtained by treating each section of bar as a Uniform Axial Member:

Dtotal = Di = PiLi AiEi

» Continually Varying Axial Member
In many cases, the Loads, Areas, and/or Modulus of an Axial Member vary continuously. Reasons for these variations can include:

• Surface Friction Forces (i.e., dirt surrounding a pile, composite matrix on a fiber, etc.);
• Body Forces (i.e., the weight of a standing column);
• Varying Cross-sectional Area (i.e., a tapered concrete foundation or a tapered rod);

To solve problems involving Continually Varying Loads, Areas, or Modulus, the deflection can be determined by integrating over the length of the member:

Dtotal = L dD = L P(x) dx A(x) E(x)

» Displacement Method
In solving many Axial Member problems it is often useful to use the DISPLACEMENT METHOD. The 3 Steps of the Displacement Method are:

• Step1 - Compatibility. Define the Displacements (D) as required by the Kinematic and Geometric Boundary Conditions. (i.e., How must the parts move with respect to each other)
• Step2 - Hooke's Law. Solve for Stresses (s) as a function of Displacements (D), Length (L), and Young's Modulus (E):
  

  s = Ee = E

  D L

• Step3 - Equilibrium. The sum of the Internal Forces is equal to the sum of the External Forces: SF_internal = SF_external,     F = sA

Also note that the Stiffness of an Axial Member is given by:

K =

F D

=

EA L