Question 14.10.2: Using Eqn 14.10-26, obtain an upper bound on the shakedown l...
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)
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,
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.
