Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

462
STATICALLy INdETERMINATE
bEAMS
The roller reaction B y will be selected as the redundant. This reaction force is removed from
the beam, leaving a cantilever as the released beam. Note that the released beam is stable and
that it is statically determinate. Next, the deflection of the released beam at the location of the
redundant is analyzed for two loading cases. The first case consists of the cantilever beam with
applied load P, and the downward deflection v B at the location of the redundant is determined
(Figure 11.6c). The second case consists of the cantilever beam with only the redundant reaction
force B y , and the upward deflection v′ B caused by B y is determined (Figure 11.6d).
By the principle of superposition, the sum of these two loading cases (Figures 11.6c
and 11.6d) is equivalent to the actual beam (Figure 11.6a) if the sum of v B and v′ B equals
the actual beam deflection at B. The actual beam deflection at B is known beforehand: The
deflection must be zero, since the beam is supported by a roller at B. From this fact, a
geometry-of-deformation equation can be written for B in terms of the two loading cases:
vB + vB ′ = 0
(a)
The deflections v B and v′ B can be determined from equations given in the beam table found
in Appendix C:
v
B
3 3
5PL
BL y
=− and vB′ = (b)
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These deflection expressions are substituted into Equation (a) to produce an equation based
on the deflected geometry of the beam, but expressed in terms of the unknown reaction B y .
This compatibility equation can be solved for the value of the redundant:
3 3
5PL
BL y
5
− + = 0 ∴ By
= P
(c)
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16
Once the magnitude of B y has been determined, the remaining reactions can be found from
the equilibrium equations. The results are as follows:
11
3
A = 0 A = P M = PL
16
16
x y A (d)
The choice of redundant is arbitrary, provided that the primary beam remains stable.
Consider the previous propped cantilever beam (Figure 11.6a), which has four reactions
(Figure 11.6b). (For convenience, the beam of Figure 11.6a is reproduced in Figure 11.7a.)
Suppose that, instead of roller B, the moment reaction M A is chosen as the redundant, leaving
a simply supported span as the released beam. Removing M A allows the beam to rotate
freely at A; therefore, the rotation angle θ A must be determined for the released beam subjected
to the applied load P (Figure 11.7b). Next, the simple span is subjected to redundant
M A alone and the resulting rotation angle θ′ A is determined (Figure 11.7c).
Just as before, the sum of these two loading cases (Figures 11.7b and 11.7c) is equivalent
to the actual beam (Figure 11.7a), provided that the rotations produced by the two separate loading
cases add up to the actual beam rotation at A. Since the actual beam is fixed at A, the rotation
angle must be zero, a condition that leads to the following geometry-of-deformation equation:
θA + θA ′ = 0
(e)
Again from the beam table in Appendix C, the rotation angles for the two cases can be
expressed as
2
PL
MAL
θA
=− and θA′ = (f)
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