MECHANICS  CASE STUDY SOLUTION

Beam Loading


The deflection at the beam tip can be determined by first finding
the moment equations and then integrating those equations. There are actually
two moment equations, one for each half, since the load and beam structure is
not continuous. Four boundary conditions will be needed, two for each beam section.





Moment Equation and Diagram

Moment and Shear Diagrams


One method to determine beam deflections is to integrate the moment equation.
This requires that the moment equation is known before starting the integration.
This can be done by cutting each beam section and developing a moment equation
as a function of the beam location, x (details for this process can be found
in the
Shear and Moment Diagrams section).
For the cantilever beam, there are two sections. The first one is from point
A to B. The second is from point B to C. Making a cut in the first sections and
solving for the moment and shear at the cut surface gives,
V_{1} = 0.48x N
M_{1} = 0.24x^{2} Nmm
The second cut in section 2 gives
V_{2} = 24.0 N
M_{2} = 24 (x  25) Nmm
= 24x +
600 Nmm
The equations are plotted at the left.






Beam Properties



The moment of inertia
for a rectangular cross section gives,
I_{1} = 12(3)^{3}/12 = 27.0 mm^{4}
I_{2} = 16(5)^{3}/12 = 166.7
mm^{4}
The material stiffness, E, is given as
E = 200 GPa = 200×10^{9} N/m^{2} (1 m/1000 mm)^{2}
= 200,000 N/mm^{2}






Integrating Moment Equations



The deflection of any beam can be found by integrating the basic moment differential
equation,
EIv´´ = M
However, each section must be integrated separately. Integrating section AB
twice gives,
Recall, v is the deflection and v´ is the slope of the beam. The constants
of integration, C_{1} and C_{2}_{}, must be determined from the boundary conditions (see
below).
Integrating the second beam section BC, gives






Boundary Conditions

Four Boundary Conditions 

There are four constants of integrating that need to be defined. This requires
four boundary conditions.
The first two conditions are due to the fixed joint at the right end.
This requires both the deflections, v, and the slope, v´, to be zero. These
are listed in the table at the left as conditions 1) and 2).
Another two conditions can be identified at the joint between beam sections
1 and 2. Since the beam is continuous, the beam deflection and slope on either
side of the joint must be equal. This gives the third and fourth condition, as
listed in the table.






Determining Constants



With the four boundary conditions defined, four equations can now be constructed
which will allow all four constants to be determined. Generally, boundary conditions
can be applied so that only one constant is present in a given equation. However,
sometimes two or three equations will need to be solved simultaneously.
Boundary Condition 2) v´_{2} = 0 at x = 100 mm
33.33×10^{6} v´ = 12x^{2} +
600x + C_{3}
33.33×10^{6} (0) = 12(100)^{2} + 600(100) +
C_{3}
C_{3} = 60,000 Nmm^{2}
Boundary Condition 1) v_{2} = 0 at x = 100 mm
33.33×10^{6} v = 4x^{3} + 300x^{2} +
C_{3}x + C_{4}
33.33×10^{6} (0) = 4(100)^{3} +
300(100)^{2}
+ 60,000(100)
+ C_{4}
C_{4} = 5.0×10^{6} Nmm^{3}
Boundary Condition 4) v´_{1} = v´_{2} at x
= 50 mm
C_{1} = 19,720 Nmm^{2}
Boundary Condition 3) v_{1} = v_{2} at x =
50 mm
C_{2} = 1,145,000 Nmm^{3}






Final Deflection Equations

Final Deflection Curve 

The final deflection equations for both beam sections are
v_{1} = 3.704×10^{9}x^{4} +
0.003652x  0.2120 mm
applies
for 0
≤ x ≤ 50
v_{2} = 1.2×10^{7}x^{3} +
9.0×10^{6}x^{2} + 0.0018x
 0.15 mm
applies
for 50 ≤ x ≤ 100
The maximum deflection at the tip (x = 0) is
v_{x=0} = 0.2120 mm



