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## 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. ## 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 $M_{P} = 67.5 kN m$is the correct answer.  