
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 loaddeflection equation.
The beam is a standard 4 x 2 structural steel tube. Assume Young 's modulus,
E, is 29,000 ksi.




Solution






Recall, the
loaddeflection 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 loaddeflection equation for beam section one gives
For the second beam section, integration of the loaddeflection equation gives,
Now each of the boundary conditions need to be applied which will produce
eight equations. They are given below.




No 
Condition 
1) 
v_{1}(x=0) = 0 
2) 
V_{1}(x=0) = 0 
3) 
v_{1}(x=60) = 0 
4) 
v_{2}(x=60) = 0 
5) 
v_{2}(x=120) = 0 
6) 
v´_{2}(x=120) = 0 
7) 
v´_{1}(x=60) = v´_{2}(x=60) 
8) 
M_{1}(x=60) = M_{2}(x=60) 
Boundary Condition List 

Boundary Condition 1)
0 = 0 + 0 + 0 + C_{3}
C_{3} = 0
Boundary Condition 2)
0 = 0 + C_{1}
C_{1} = 0
Boundary Condition 3)
27×10^{6} = 1,800 C_{2} + C_{4}
Boundary Condition 4)
27×10^{6} = 36,000 C_{5} + 1,800
C_{6} + 60
C_{7} +
C_{8}
Boundary Condition 5)
432×10^{6} = 288,000 C_{5} + 7,200
C_{6} + 120 C_{7} + C_{8}
Boundary Condition 6)
14.4×10^{6} = 7,200
C_{5} + 120 C_{6} + C_{7}
Boundary Condition 7)
0 = 60 C_{2} + 1,800
C_{5} +
60 C_{6} + C_{7}
Boundary Condition 8)
0 = C_{2} + 60 C_{5} + C_{6}






The five equations from boundary conditions 4 through 8 can be used to solve for the 5 unknowns, C_{2}, C_{5}, C_{6}, C_{7} and C_{8}. 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
C_{2}= 39,000
C_{5}= 5,400
C_{6}= 285,000
C_{7} = 9,720,000
C_{8} = 237,600,000
Using boundary condition 3 gives
C_{4} = 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 in^{4} (from the
Structural Shapes Appendix).
EI v_{1} = wx^{4}/24 + C_{1} x^{3}/6
+ C_{2} x^{2}/2 + C_{3} x + C_{4}
(29,000,000)(5.32) v_{1} = 0 + 0 + 0
+ 0  43,200,000
v_{1} = 0.2800 in
This deflection at point A is the same as
previously calculated using superposition.




