Question 14.10.3: Using the method of combination of mechanisms, determine the...
Using the method of combination of mechanisms, determine the shakedown load factor λ for the uniform frame in Fig. 14.10-6(a) if the load H can vary at random within the range 0 to 10λ kN and the load V can vary at random within the range 0 to 15λ kN. The plastic moment of resistance of each member is 50 kNm.
Three collapse mechanisms can be identified. These are the beam mechanism in Fig.14.10-6(b), the sidesway mechanism in Fig. (c) and the combined mechanism in Fig.(d).
The elastic moments due to the loads H and V are shown in Table 14.10-2,where the sign convention for bending moments is the same as that in Fig. 14.9-2;namely, a bending moment producing compression on that side of the member adjacent to the dotted line is designated a positive moment.
| Section | Load V | Load H | Combined Loading | ||||
| \mathscr{M}^{max}_{i} (1) |
\mathscr{M}^{min}_{i} (2) |
\mathscr{M}^{max}_{i} (3) |
\mathscr{M}^{min}_{i} (4) |
\mathscr{M}^{max}_{i} (5) |
\mathscr{M}^{min}_{i} (6) |
\mathscr{M}^{max}_{i} -\mathscr{M}^{min}_{i} (7) |
|
| A | 0 | -7.50 | 15.63 | 0 | 15.63 | -7.5 | 23.13 |
| B | 15.00 | 0 | 0 | -9.38 | 15.00 | -9.38 | 24.38 |
| C | 0 | -22.50 | 0 | 0 | 0 | -22.50 | 22.50 |
| D | 15.00 | 0 | 9.38 | 0 | 24.38 | 0 | 24.38 |
| E | 0 | -7.50 | 0 | -15.63 | 0 | -23.13 | 23.13 |
| Table. 14-10-2 | |||||||
Applying Eqn 14.10-26 to the beam mechanism in Fig..l4.10-6(b):
\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)
\qquad \lambda \left[\mathscr{M}^{max}_{B}\theta ^{+}_{B}-\mathscr{M}^{min}_{C}\theta ^{-}_{C}+\mathscr{M}^{max}_{D}\theta ^{+}_{D} \right] = 4M_{P}\thetaUsing the \mathscr{M} values in Table 14.10-2 and the θ values in Fig. 14.10-6(b).
\begin{matrix} \hspace{5em}\lambda \left[(15.00 M_{p})(\theta )- (-22.50 M_{P})(2\theta )(24.38)(\theta )\right] &=4(50)(\theta )\hspace{5em}\\ \hspace{20em} 84.38\lambda &= 200 \hspace{8em} \\ \hspace{22em} \lambda &= 2.37(an \ upper \ bound) \end{matrix}Applying Eqn 14.10-26 to the sidesway mechanism in Fig. 14.10-6(c).
\qquad \lambda \left[\mathscr{M}^{max}_{A}\theta ^{+}_{A}-\mathscr{M}^{min}_{B}\theta ^{-}_{B}+\mathscr{M}^{max}_{D}\theta ^{+}_{D} -\mathscr{M}^{min}_{E}\theta ^{-}_{E}\right] = 4M_{P}\theta \qquad \lambda \left[(15.63)(\theta )-(-9.38)(\theta )+(24.38)(\theta) -(-23.13)(\theta)\right] = 4(50)(\theta)\\ \hspace{5em} 72.52\lambda =200\\ \qquad \lambda =2.76(an \ upper \ bound)The method of combination of mechanisms previously used for static collapse is now applied to incremental collapse. The combined mechanism in Fig.14.10-6(d) is obtained by adding together Figs, (b) and (c), resulting in the cancellation of the plastic hinge at B. Referring to Fig. 14.10-6(b), the cancellation of the hinge B would mean removing the term \mathscr{M}^{max}_{B}\theta ^{+}_{B}, i.e. \mathscr{M}^{max}_{B}\left|\theta _{B}\right| , from the left-hand side of the incremental collapse equation, and removing M_{P}\left|\theta _{B}\right| from the right-hand side of that equation. Similarly, referring to the sidesway mechanism in Fig. 14.10-6(c), the cancellation of hinge B means the removal of \mathscr{M}^{min}_{B}\left|\theta _{B}\right|, rom the left-hand side of the incremental collapse equation andM_{P}\left|\theta _{B}\right| from the right-hand side. In other words,the incremental collapse equation for the combined mechanism is obtained by adding together the equations for the beam and sidesway mechanisms and then (I) removing from the left-hand side the quantity
\qquad \mathscr{M}^{max}_{B}- \mathscr{M}^{min}_{B}\left|\theta _{B}\right|which is the work term due to the range of elastic moments and (2) removing from the right-hand side the quantity
\qquad 2M_{P}\left|\theta _{B}\right|which is twice the plastic work at hinge B. Here 2M_{P} is 100 and, from Column 7 of Table 14. 10-2 \mathscr{M}^{max}_{B} – \mathscr{M}^{min}_{B}=24.38. We are now able to display the calculations as follows :
\qquad \begin{matrix}\text{Beam mechanism}: & 84.38\lambda=200 & \longrightarrow & \lambda =2.37 \\ \text{Sidesway mechanism}: & \underline{72.52\lambda =200} &\longrightarrow & \lambda =2.76 \\\text{ Adding}: & 156.90\lambda =400 \\ \text{ Cancel hinge B}: & \underline{-24.38\lambda -100} \\ \text{ Combined mechanism}: & 132.52\lambda =300 & \longrightarrow & \lambda= \underline{2.26}\end{matrix}In this example, it is seen that there are only three collapse mechanisms. Each of the three λ’s is an upper bound on the correct shakedown load factor, λ = 2.26 being the lowest upper bound is therefore the correct shakedown load factor.
COMMENTS
(a) For more complicated frames, it is sometimes difficult to be sure that all possible mechanisms have been investigated. To confirm that the answer obtained is correct, a statical check is necessary; methods are given in specialist texts3,6,7 .
b) The shakedown load factor resulting from the analysis of any assumed mechanism of collapse, whether or not it is the correct mechanism, can never exceed the corresponding static collapse load factor for the same assumed mechanism. The shakedown load factor determined from the correct incremental collapse mechanism may be anything between, say, 60%, to 90% of the static load factor determined from the correct static collapse mechanism, though for realistic structures and loading conditions the former is unlikely to be less than 75% of the latter6.
(c) Up to now it has been assumed that the structure is linear-elastic, so that the moment curvature relation is given by curve OAB in Fig. 14.10-1. It, referring to that figure, the moment curvature relation follows the more realistic curve OA‘B with an elastic range 2M_{Y} < 2M_{P} it is then necessary to add one further condition to Eqns 14.10-3 and 4:
\qquad \mathscr{M}^{max}_{i} – \mathscr{M}^{min}_{i}\leq 2M_{Y} (14.10-27)
which guards against the possible danger of alternating plasticity. Similarly, the additional condition
\qquad \lambda ( \mathscr{M}^{max}_{i} – \mathscr{M}^{min}_{i})\leq 2M_{Y} (14.10-28)
must be added to the Eqns 14.10-18 and 19 and to the Eqns 14.10-22 and 23. With
these modifications, the shakedown theorem and the lower-bound and upper-bound arguments would be valid irrespective of whether the moment -curvature curve is OAB or OA’B in Fig. 14.10-1. However, it should be noted that for practical structural frames and realistic loadings, alternating plasticity is rarely of critical importance3.
\qquad \mathscr{M}^{max}_{i} +\overline{m_{i}}\leq (M_{P})_{i} (14.10-3)
\qquad \mathscr{M}^{min}_{i} + \overline{m_{i}}\geq -(M_{P})_{i} (14.10-4)
\qquad \lambda \mathscr{M}^{max}_{i} +\overline{m_{i}}\leq (M_{P})_{i} (14.10-18)
\qquad \lambda \mathscr{M}^{min}_{i} + \overline{m_{i}}\geq -(M_{P})_{i} (14.10-19)
\qquad \quad \lambda \mathscr{M}^{max}_{i} +\overline{m_i}=(M_{P})_i \ if \ \theta _i \ is \ +ve (14.10-22)
\qquad \quad \lambda \mathscr{M}^{min}_{i} +\overline{m_i}=-(M_{P})_i \ if \ \theta _i \ is \ -ve (14.10-23)
