This example is the same problem as the case study previously solved. However, instead of using the principle of superposition, solve for point A deflection by integrating the load-deflection equation. The beam is a standard 4 x 2 structural steel tube. Assume Young 's modulus, E, is 29,000 ksi.

Recall, the load-deflection equation is

     EIv´´´´ = -w(x)

This equation can be integrated four times, but four boundary conditions are required to determine the integration constants.

Eight Boundary Conditions

The beam in this problem requires two sections due to the center support. This also means that there must be four boundary conditions. These conditions are listed in the diagram at the left.

Integrating the load-deflection equation for beam section one gives

For the second beam section, integration of the load-deflection equation gives,

Now each of the boundary conditions need to be applied which will produce eight equations. They are given below.

Boundary Condition 1)
     0 = 0 + 0 + 0 + C₃
     C₃ = 0

Boundary Condition 2)
     0 = 0 + C₁
     C₁ = 0

Boundary Condition 3)
     
     27×10⁶ = 1,800 C₂ + C₄

Boundary Condition 4)
     
     27×10⁶ = 36,000 C₅ + 1,800 C₆ + 60 C₇ + C₈

Boundary Condition 5)
     
     432×10⁶ = 288,000 C₅ + 7,200 C₆ + 120 C₇ + C₈

Boundary Condition 6)
     
     14.4×10⁶ = 7,200 C₅ + 120 C₆ + C₇

Boundary Condition 7)
     
     0 = -60 C₂ + 1,800 C₅ + 60 C₆ + C₇

Boundary Condition 8)
     
     0 = -C₂ + 60 C₅ + C₆

The five equations from boundary conditions 4 through 8 can be used to solve for the 5 unknowns, C₂, C₅, C₆, C₇ and C₈. These five equations are summarized in matrix form below.

It is difficult to solve these by hand but can easily solved with a matrix solver in most engineering calculators or with a spreadsheet. The solution for the unknowns are

     C₂= 39,000
     C₅= 5,400
     C₆= -285,000
     C₇ = 9,720,000
     C₈ = -237,600,000

Using boundary condition 3 gives

     C₄ = -43,200,000

All the integration constants are now known and can be used with the deflection equation. Young's Modulus is 29,000 ksi and the moment of inertia for the 4x2 steel tube is 5.32 in⁴ (from the Structural Shapes Appendix).

     EI v₁ = -wx⁴/24 + C₁ x³/6 + C₂ x²/2 + C₃ x + C₄

     (29,000,000)(5.32) v₁ = 0 + 0 + 0 + 0 - 43,200,000

      v₁ = 0.2800 in

This deflection at point A is the same as previously calculated using superposition.