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MECHANICS - EXAMPLE


Beam with Loads
  Example

 

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.

   
  Solution

     

 

 

 

 

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.

     
No Condition
1) v1(x=0) = 0
2) V1(x=0) = 0
3) v1(x=60) = 0
4) v2(x=60) = 0
5) v2(x=120) = 0
6) 2(x=120) = 0
7) 1(x=60) = v´2(x=60)
8) M1(x=60) = M2(x=60)
Boundary Condition List
 

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.

     
   
 
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