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Chapter 14

Q. 14.7.2

The frame in Fig. 14.7-7(a) is of uniform sections throughout, and supports the working loads as shown. If the collapse load factor is to be 2, determine the required plastic moment of resistance.

fig14.7-7a

Step-by-Step

Verified Solution

Number of critical sections N = 14 (one at each of joints A and F; one at each of loading points G and H; two at each of joints C and D; three at each of joints B and E).

Number of redundancies R = 6 (see Fig. 14.7-6)
From Eqn 14.6-3
M = N – R = 1 4 – 6 = 8

There are thus 8 independent mechanisms; these are the 4 in Fig. 14.7-7(b), (c), (d), (e) plus 4 joint rotations, at B, C, D and E.

The detailed calculations for the various combinations of mechanisms are left to the reader, but he should verify that the partial mechanism in Fig. 14.7-7(g) gives the highest value for the required plastic moment of resistance, namely

\qquad M_{p} = 67.5  kNm.

The partial mechanism of collapse is statically indeterminate, as explained in Comment (4) at the end of Example 14.6-4. However, if it is assumed that the bending moment at end E of the member ED has the (unknown) value m, then the bending moment diagram may be drawn as in Fig. 14.7-7(h). The reader should verify that, for 0 ≤ m ≤ 67.5 kNm the moment ordinates nowhere exceed 67.5 kNm. Hence, by the uniquencess theorem, it is concluded that M_{P} = 67.5  kN mis the correct answer.

COMMENTS
(1) In using the uniqueness theorem, it is not necessary to know the actual bending moment distribution; it is only necessary to show that a bending moment istribution can be found which satisfies the yield condition (see Proof in Section 14.9).
(2) The partial collapse mechanism in this example has only one redundancy and it is relatively easy to prove the existence of a bending moment distribution (such as that in Fig. 14.7-7(h)) which satisfies the yield condition. For complex frames, this could be a difficult problem, and the reader is reierred to specialist books1,3,6,7.

fig14.7-6
fig14.7-7