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Chapter 14

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.

fig14.7-2(a)

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:

\begin{matrix} \qquad \text{ Fig. 14.7-2} (b): & 6\lambda (4\phi )=4(15)(\phi ) & \longrightarrow \lambda =2.5 \\ \quad \quad (c): & 12\lambda (2\phi )=4(15)(\phi )& \longrightarrow \lambda =2.5 \\ \quad \quad (d): & 6\lambda (4\phi )=6(15)(\phi ) & \longrightarrow \lambda =3.75 \\ \text{ Fig. 14.7-2} (e): & 72\lambda \phi = 210\phi  & \longrightarrow \lambda =2.92 \\ \text{Joint  rotation,} & \frac{B \quad \qquad -2(15)(\phi )}{72\lambda \phi   \qquad =180\phi } \\ \text{Joint  rotation}, & \frac{D \quad \qquad -(15)(\phi )}{72\lambda \phi   \qquad =165\phi } &  \longrightarrow \lambda =2.29 \end{matrix}

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 2ɸ, 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,

Step 1:
The shear force for the member FG is calculated as (15 kN m + 15 kN m)/2 m = 15 kN.
Step 2:
The shear force at the end F of member DF is given by (27.48 kN – 15 kN) = 12.48 kN, where 27.48 kN is the known applied load and 15 kN is the shear force from Step 1. The bending moment at end D is therefore (12.48 kN) (2 m) – 15 kN m – 9.96 kN m
Step 3:
For member CD, the shear force is (15 kNm + 15 kNm)/(4m) = 7.5 kN.
Step 4:
The shear force at end C of member CB is given by the difference of the applied force and the shear for CD; that is 13.74 kN – 7.5 kN – 6.24 kN. The bending moment at end B is therefore (6.24 kN) (4 m) — 15 kNm = 9.96 kNm.
Step 5:
For member GH, the shear force is (15 kN m + 15 kN m) /(4 m) = 7.5 kN. The axial force is given by the shear force of member FG (see Step 1). The axial forces in member FG, DF and the force acting from the right of Joint D are all equal to the shear force in GH, namely 7.5 kN.
Steps 6

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.

Step 8:
For member 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.

fig14.7-2(abcd)
fig14-7.2e,f
fig14.7-3
fig14.7-4
fig14.7-5
fig14.7-6