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MECHANICS - EXAMPLE
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Beam with Loads |
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Example
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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.
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Solution
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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.
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Eight Boundary Conditions
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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.
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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) |
v´2(x=120) = 0 |
7) |
v´1(x=60) = v´2(x=60) |
8) |
M1(x=60) = M2(x=60) |
Boundary Condition List |
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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
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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
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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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