Page 202 - Shigley's Mechanical Engineering Design

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Deflection and Stiffness 177

EXAMPLE 4–14 The indeterminate beam 11 of Appendix Table A–9 is reproduced in Fig. 4–16.
Determine the reactions using procedure 1.
Solution The reactions are shown in Fig. 4–16b. Without R 2 the beam is a statically determinate
cantilever beam. Without M 1 the beam is a statically determinate simply supported
beam. In either case, the beam has only one redundant support. We will first solve this
problem using superposition, choosing R 2 as the redundant reaction. For the second
solution, we will use Castigliano’s theorem with M 1 as the redundant reaction.

Solution 1 1 Choose R 2 at B to be the redundant reaction.
2 Using static equilibrium equations solve for R 1 and M 1 in terms of F and R 2 . This
results in
Fl
R 1 = F − R 2 M 1 = − R 2 l (1)
2
3 Write the deflection equation for point B in terms of F and R 2 . Using superposition
of beam 1 of Table A–9 with F =−R 2 , and beam 2 of Table A–9 with a = l/2,
the deflection of B, at x = l, is

R 2 l 2 F(l/2) 2 l R 2 l 3 5Fl 3
δ B =− (l − 3l) + − 3l = − = 0 (2)
6EI 6EI 2 3EI 48EI
4 Equation (2) can be solved for R 2 directly. This yields

5F
Answer R 2 = (3)
16
Next, substituting R 2 into Eqs. (1) completes the solution, giving

11F 3Fl
Answer R 1 = M 1 = (4)
16 16
Note that the solution agrees with what is given for beam 11 in Table A–9.

Solution 2 1 Choose M 1 at O to be the redundant reaction.
2 Using static equilibrium equations solve for R 1 and R 2 in terms of F and M 1 . This
results in
F M 1 F M 1
R 1 = + R 2 = − (5)
2 l 2 l

Figure 4–16 y y
l F
F A B
l x
2 A B O
x
O M 1 R 1 ˆ x R 2
(a) (b)