## Chapter 14

## Q. 14.7.1

## Q. 14.7.1

The frame in Fig. 14.7-2(a) carries working loads of values indicated. If the plastic moment of resistance is 15 kNm for each member, determine the collapse load factor.

## Step-by-Step

## Verified Solution

The frame has 12 critical sections, as indicated by the short lines in Fig. 14.7-2(a). There are 6 redundancies (see Comment (1) at end). Therefore from Eqn 14.6-3, the number of independent mechanisms is

M = N – R = 1 2 – 6 = 6

These can be identified as

(1) the beam mechanism in Fig. 14.7-2(b)

(2) the beam mechanism in Fig. (c)

(3) the sidesway mechanism in Fig. (d) and

(4), (5) (6) the joint rotations at **B, D**, **G**.

The calculations can be set out systematically as follows:

Of course, we can apply a joint rotation at G to eliminate the column (beam) hinge, but then we end up with a beam (column) hinge rotation of 2*ɸ*, so that there is no reduction in plastic work.

The statical check in Fig. 14.7-3 shows that the bending moment nowhere exceeds 15 kNm. Hence the frame in its collapse state in Fig. 14.7-2(f) satisfies the three conditions of collapse, equilibrium, and yield. Therefore the load factor λ = 2.29 is the true collapse load factor.

The bending moments in Fig. 14.7-3 are based on the freebody diagrams in Fig. 14.7-4, in which the bold arrows indicate known forces and known moments, such as the applied load and the moments at plastic hinge positions. The circled numbers indicate the order in which the calculations are made: thus,

and 7:For Joint D, the moment on the left is the known plastic moment 15 kN m, that on the right is 9.96 kN m (from Step 2). Therefore, the moment acting at the bottom is (15-9.96)kN m = 5.04 kN m. Therefore, the bending moment at end D of member DE is also 5.04 kN m; this gives the shear force in DE as (5.04 kN m + 15 kN m)/(4 m) = 5.01 kN. Similarly, the shear force in the (very short) vertical member at Joint D is also 5.01 kN; hence the axial force in the short horizontal member on the left of Joint D is 5.01 kN m + 7.5 kN m = 12.51 kN. The shear forces in the horizontal members at Joint D are 7.5 kN and 12.48 kN, from Steps 3 and 2 respectively; herefore the axial force in the vertical member at that joint is 7.5 kN + 12.48 kN = 19.98 kN, which also gives the axial force in **DE**.

The axial forces in **CD** and **BC** are therefore 12.51 kN.

**AB**, the axial force is equal to the shear force in member BC (Step 4); the bending moment at end D is also obtained from Step 4 as 9.96 kNm. Hence the shear force in

**AB**is (15 kNm – 9.96 kN m)/(4 m) = 1.23 kN. As a check, the axial force in

**BC**plus the shear force in

**AB**is 12.51 kN + 1.23kN = 13.74 kN, which equals the applied horizontal force at Joint

**B**.

COMMENTS

(1) In using Eqn 14.6-3 to determine the number of independent mechanisms it was stated that the frame had 6 redundancies. One effective method of finding the number of redundancies of frameworks is to make vertical cuts as illustrated in Fig. 14.7-5; each cut releases three redundancies (a moment, a shear and an axial force), and since two cuts are required to render the structure statically determinate, we conclude that the frame has six redundancies.

Similarly, the frame in Fig. 14.7-6 also has six redundancies.

(2) The frame in Fig. 14.7-2(a) collapses as a regular mechanism. Regular mechanisms being statically determinate, it was possible to draw the complete bending diagram using the equations of statics only. Sometimes a frame may collapse as a patial mechanism, which by definition is statically indeterminate; in such a case, the bending moment diagram at collapse cannot be completely defined—see Example 14.7-2.