Page 204 - Shigley's Mechanical Engineering Design

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Deflection and Stiffness 179
For some problems even procedure 1 can be a task. Procedure 2 eliminates some
tricky geometric problems that would complicate procedure 1. We will describe the pro-
cedure for a beam problem.
Procedure 2
1 Write the equations of static equilibrium for the beam in terms of the applied loads
and unknown restraint reactions.
2 Write the deflection equation for the beam in terms of the applied loads and unknown
restraint reactions.
3 Apply boundary conditions to the deflection equation of step 2 consistent with the
restraints.
4 Solve the equations from steps 1 and 3.

EXAMPLE 4–15 The rods AD and CE shown in Fig. 4–17a each have a diameter of 10 mm. The second-
3
4
area moment of beam ABC is I = 62.5(10 ) mm . The modulus of elasticity of the
material used for the rods and beam is E = 200 GPa. The threads at the ends of the rods
are single-threaded with a pitch of 1.5 mm. The nuts are first snugly fit with bar ABC
horizontal. Next the nut at A is tightened one full turn. Determine the resulting tension
in each rod and the deflections of points A and C.
Solution There is a lot going on in this problem; a rod shortens, the rods stretch in tension, and
the beam bends. Let’s try the procedure!
1 The free-body diagram of the beam is shown in Fig. 4–17b. Summing forces, and
moments about B, gives
F B − F A − F C = 0 (1)
4F A − 3F C = 0 (2)
2 Using singularity functions, we find the moment equation for the beam is
1
M =−F A x + F B x − 0.2
where x is in meters. Integration yields
dy F A 2 F B 2
EI =− x + x − 0.2 + C 1
dx 2 2
F A 3 F B 3
EIy =− x + x − 0.2 + C 1 x + C 2 (3)
6 6
2
−9
4
9
The term EI = 200(10 ) 62.5(10 ) = 1.25(10 ) N · m .
Figure 4–17 200 150 F A 200 150 F C
Dimensions in mm. A B C A B C

x
F
600 B
800
(b) Free-body diagram of beam ABC
D
E
(a)