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## Q. 14.10.2

Using Eqn 14.10-26, obtain an upper bound on the shakedown load factor for the uniform two-span beam in Fig. 14.10-2, if the loads P and Q may each vary at random within the range 0 to $\lambda M_{P}L$.

$\qquad \lambda\left[\sum\mathscr{M}^{max}_{i}\theta^{+}_{i} – \sum{\mathscr{M}^{min}_{i}\theta^{-}_{i}} \right] =\sum{(M_{P})_i\left|\theta_{i} \right|}$                        (14.10-26)

## Verified Solution

Consider the mechanism in Fig. 14.10-5. The elastic moments in Table 14.10-1 are
relevant. Substituting appropriate values into Eqns 14.10-26,

$\begin{matrix} \lambda \left[(0.188 M_{p})(\theta )- (-0.203 M_{P})(2\theta )\right] &=M_{P}(\theta +2\theta )\\ whence \hspace{17em} \lambda &= 5.05\end{matrix}$

Ans. λ = 5.05 is an upper bound on the shakedown load factor.

 Section Due to P Due to Q Combined Loading $\mathscr{M}^{max}_{i}$ $\mathscr{M}^{min}_{i}$ $\mathscr{M}^{max}_{i}$ $\mathscr{M}^{min}_{i}$ $\mathscr{M}^{max}_{i}$ $\mathscr{M}^{min}_{i}$ B 0.094 $M_P$ 0 0.094 $M_P$ 0 0.188 $M_P$ 0 D 0 -0.203 $M_P$ 0.047 $M_P$ 0 0.047 $M_P$ -0.203 $M_P$ E 0.047 $M_P$ 0 0 -0.203 $M_P$ 0.047 $M_P$ -0.203 $M_P$ Table. 14-10-1

COMMENT For any arbitrary mechanism, Eqn 14.10-26 only gives an upper bound on the shakedown load factor. However, in this particular case it is seen that the mechanism in Fig. 14.10-5 is the correct incremental collapse mechanism; hence λ =5.05 is the correct shakedown load factor.