Page 201 - Shigley's Mechanical Engineering Design

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176 Mechanical Engineering Design
Figure 4–15 F

k 1 k
2

(a)

F F
1 2

k k
1
2

(b)

Now we solve Eq. (c) for F 1 and substitute the result in Eq. (a). This gives

k 1 k 2 F
F − F 2 − F 2 = 0or F 2 = (d)
k 2 k 1 + k 2
Substituting F 2 into Eq. (c) gives F 1 = k 1 F/(k 1 + k 2 ) and so δ = δ 1 = δ 2 =
F/(k 1 + k 2 ). Thus, for two springs in parallel, the overall spring constant is
k = F/δ = k 1 + k 2 .
In the spring example, obtaining the necessary deformation equation was very
straightforward. However, for other situations, the deformation relations may not be as
easy. A more structured approach may be necessary. Here we will show two basic pro-
cedures for general statically indeterminate problems.

Procedure 1
1 Choose the redundant reaction(s). There may be alternative choices (See Ex-
ample 4–14).
2 Write the equations of static equilibrium for the remaining reactions in terms of the
applied loads and the redundant reaction(s) of step 1.
3 Write the deflection equation(s) for the point(s) at the locations of the
redundant reaction(s) of step 1 in terms of the applied loads and the redundant
reaction(s) of step 1. Normally the deflection(s) is (are) zero. If a redundant
reaction is a moment, the corresponding deflection equation is a rotational
deflection equation.
4 The equations from steps 2 and 3 can now be solved to determine the reactions.
In step 3 the deflection equations can be solved in any of the standard ways. Here
we will demonstrate the use of superposition and Castigliano’s theorem on a beam
problem.