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Deflection and Stiffness 175
The integrals may be evaluated in a number of ways: by a program using Simpson’s
8
rule integration, by a program using a spreadsheet, or by mathematics software. Using
MathCad and checking the results with Excel gives the integrals as I 1 = 0.076 615,
I 2 =−0.000 159, and I 3 = 0.000 773. Substituting these into Eq. (1) gives

Answer δ = 0.077 23 in
Finite-element (FE) programs are also very accessible. Figure 4–14b shows a
simple half-model, using symmetry, of the press consisting of 216 plane-stress (2-D)
elements. Creating the model and analyzing it to obtain a solution took minutes.
Doubling the results from the FE analysis yielded δ = 0.07790 in, a less than 1 percent
variation from the results of the numerical integration.

4–10 Statically Indeterminate Problems
A system is overconstrained when it has more unknown support (reaction) forces and/or
moments than static equilibrium equations. Such a system is said to be statically indeter-
minate and the extra constraint supports are called redundant supports. In addition to the
static equilibrium equations, a deflection equation is required for each redundant support
reaction in order to obtain a solution. For example, consider a beam in bending with a wall
support on one end and a simple support on the other, such as beam 12 of Table A–9.
There are three support reactions and only two static equilibrium equations are available.
This beam has one redundant support. To solve for the three unknown support reactions
we use the two equilibrium equations and one additional deflection equation. For another
example, consider beam 15 of Table A–9. This beam has a wall on both ends, giving rise
to two redundant supports requiring two deflection equations in addition to the equations
from statics. The purpose of redundant supports is to provide safety and reduce deflection.
A simple example of a statically indeterminate problem is furnished by the nested
helical springs in Fig. 4–15a. When this assembly is loaded by the compressive force F,
it deforms through the distance δ. What is the compressive force in each spring?
Only one equation of static equilibrium can be written. It is

F = F − F 1 − F 2 = 0 (a)
which simply says that the total force F is resisted by a force F 1 in spring 1 plus the
force F 2 in spring 2. Since there are two unknowns and only one static equilibrium
equation, the system is statically indeterminate.
To write another equation, note the deformation relation in Fig. 4–15b. The two
springs have the same deformation. Thus, we obtain the second equation as
δ 1 = δ 2 = δ (b)

If we now substitute Eq. (4–2) in Eq. (b), we have
F 1 F 2
= (c)
k 1 k 2

8 See Case Study 4, p. 203, J. E. Shigley and C. R. Mischke, Mechanical Engineering Design, 6th ed.,
McGraw-Hill, New York, 2001.