It Can Be Shown That . . .

Because the double beam apparatus has upper and lower beams that differ in composition will derive equations that will give the stiffness of one beam of the double beam apparatus. Once we know the relationships for one beam, we will then combine the stiffnesses of each beam to find the composite (or total) stiffness.

Starting with the relationship:

(1)     

where,

k is the stiffness (or spring constant) of the beam,

F is the applied force, and

d is the tip displacement.

In order to find the stiffness, we therefore need an equation that relates the tip displacement to an applied force.

To model a single beam of the double beam and end block apparatus, we include a moment, M0, that forces the slope at x=L to zero.

The constitutive equation that relates the material properties and the geometry of the beam to the local forces is the Euler-Bernoulli relation, which states for a beam with a constant cross-section and modulus the moment is related to the displacement by:

(1")     

where,

M is the local moment,

E is Young's Modulus of the material, and

I is the moment of inertia.

The fixed end of the beam imposes two boundary conditions, that are given by:

(2")     

and

(3")     

The end block clamping forces the following boundary condition which will let us solve for M0:

(4")     .

The local moment, M, is the sum of the load induced bending moment and the constant end moment (as given by the Balance of Angular Momentum) and is given by:

(5")     .

(Note that positive M gives tension in the bottom of the beam.)

By substituting Equation 5" into Equation 1" we have the following equation:

(6")     .

By integrating twice and substituting the boundary conditions, the tip displacement is given by:

(7")     

Then after substitution of Equation 7" into Equation 1 we have the stiffness for a single beam is given by:

(2)     .