MECHANICS  CASE STUDY SOLUTION

Beam Loading


The deflection at the tip of the equipment can be determined by integrating
the basic distributed load equation four times. There are actually
four equations, one for each half since the load and beam structure
differ in each section. A total of eight boundary conditions will be needed
to solve the four integration constants for both equations.




FreeBody Diagram

FreeBody Diagram


To help determine beam sections and boundary conditions, a freebody diagram
should be constructed. Each change in beam geometry and load requires a new beam
section and deflection equation. For this cantilever beam, there will be two
sections, one from point A to B and a second section from point B to C.
The uniform distributed load on the left part of the beam is
w = (0.04 N/mm^{2})(12 mm) = 0.48 N/mm
The actual values of the reactions do not need to be determined which is one
of the advantages of this method.






Beam Properties



The moment of inertia
using the equation 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 the LoadDeflection Equations



The deflection of any beam can be found by integrating the basic loaddeflection
differential equation,
EIv´´´´ = w(x)
for each beam section.
Section 1 (from point A to B)
The load function w_{1}(x) is the actual uniform distributed load of 0.48 N/mm
EIv_{1}´´´´ = 0.48 N/mm
EIv_{1}´´´ = 0.48x + C_{1} =>
V_{1}
EIv_{1}´´ = 0.24x^{2} + C_{1} x
+ C_{2} => M_{1}
EIv_{1}´ = 0.08x^{3} + C_{1} x^{2}/2
+ C_{2} x + C_{3}
EIv_{1} = 0.02x^{4} + C_{1} x^{3}/6
+ C_{2} x^{2}/2 + C_{3} x + C_{4}
Note that EIv´´´ is the shear and EIv´´ is the
moment. This will be needed when applying the boundary conditions.
Section 2 (from point B to C)
Section 2 is similar to section 1 except there is no uniform load. Thus, w_{2}(x)
is just 0.
EIv_{2}´´´´ =
0 N/mm
EIv_{2}´´´ = C_{5} => V_{2}
EIv_{2}´´ = C_{5} x
+ C_{6} => M_{2}
EIv_{2}´ = C_{5} x^{2}/2
+ C_{6} x + C_{7}
EIv_{2} = C_{5} x^{3}/6
+ C_{6} x^{2}/2 + C_{7} x + C_{8}






Boundary Conditions

Eight Boundary Conditions 

There are eight constants of integrating that need to be defined. This requires
eight 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).
The next two conditions are due to continuity 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.
Conditions 5 and 6 are from the free end which cannot have any shear or moment.
And finally, similar to slope and deflection, the shear and moment need
to be the same between beam sections 1 and 2. The shear is the the same since
there is no point load at the joint. Likewise, the moment is the same since there
is no applied point moment at the joint.






Determining Constants



With the eight boundary conditions defined, eight equations can now be constructed.
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 5) V_{1} = 0 at x = 0 mm
0 = 0.48(0) + C_{1}
C_{1} = 0
Boundary Condition 6) M_{1} = 0 at x = 0 mm
0 = 0.24(0) + C_{1} (0) + C_{2}
C_{2} = 0
Boundary Condition 7) V_{1} = V_{2} at x =
50 mm
0.48x + C_{1} = C_{5}
0.48(50) + 0 = C_{5}
C_{5} = 24 N
Boundary Condition 8) M_{1} = M_{2} at x =
50 mm
0.24x^{2} + C_{1}x + C_{2} =
C_{5}x + C_{6}
0.24(50)^{2} + 0 + 0 = 24(50) + C_{6}
600 = 1,200 + C_{6}
C_{6} = 600 Nmm
Boundary Condition 2) v´_{2} = 0 at x = 100 mm
33.33×10^{6} v´_{2} = 24x^{2}/2
+ 600x + C_{7}
33.33×10^{6} (0) = 12(100)^{2} +
600(100) + C_{7}
C_{7} = 60,000 Nmm^{2}
Boundary Condition 1) v_{2} = 0 at x = 100 mm
33.33×10^{6} v_{2} = C_{5}x^{3}/6
+ C_{6}x^{2}/2
+ C_{7}x + C_{8}
33.33×10^{6} (0) = 4(100)^{3} +
300(100)^{2}
+
60,000(100) + C_{8}
C_{8} = 5.0×10^{6} Nmm^{3}






Boundary Condition 4) v´_{1} = v´_{2} at x =
50 mm
C_{3} = 19,720 Nmm^{2}
Boundary Condition 3) v_{1} = v_{2} at x = 50 mm
C_{4} = 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  0.2119 mm
v_{2} = 1.2×10^{7}x^{3} +
9.0×10^{6}x^{2} + 0.0018x
 0.15 mm
The maximum deflection at the tip (x = 0) is
v_{x=0} = 0.2119 mm
This is the same (within rounding error) of the
previous solution.



