## Chapter 14

## Q. 14.6.4

## Q. 14.6.4

Fig. 14.6-10(a) shows a fixed-end portal frame of uniform plastic moment of resistance M_{P}. Determine the minimum value of ** P** such that the loading shown would cause collapse.

## Step-by-Step

## Verified Solution

We can identify three mechanisms of collapse, as shown in Fig.14.6-10(b), (c) and (d):

**Beam mechanism (Fig. (b))**

The work equation is

or

\qquad P=2M_{p}/3a**Sides way mechanism** (Fig. (c))<

or

\qquad P=M_{p}/a**Combined mechanism** (Fig. (d)—see also Comment (2) at end of solution.)

or

\qquad P=3M_{p}/5aThe combined mechanism gives the lowest value of ** P**, that is the lowest upper bound value of the three. What we can say now is that neither the beam mechanism northe sidesway mechanism can be correct, but we cannot yet say that the combined mechanism is correct, because there may exist another mechanism which is more critical than that in Fig. (d).

As an exercise, the reader should carry out a statical check and verify that the bending moment diagram for the combined mechanism is as shown in Fig. (e)—*Hint*: see Comment (1) below. Fig. (e) shows that the portal frame in its collapse state in Fig.(d) satisfies the three conditions of mechanism, equilibrium, and yield. Hence, by the uniqueness theorem, the combined mechanism is the correct collapse mechanism and the corresponding value of ** P**, namely, P = 3M_{P}/5a is the correct answer.

COMMENTS

(1) If the reader has difficulties in drawing the bending moment diagram in Fig. 14.6-10(e), he should refer to the free-body diagram in Fig. (f):

*∑** Moments = 0 for member CD gives V = *2M_{P}/3a

*∑** Moments = 0 for member CDE as whole gives H = *2M_{P}/2a

** P** is, of course, 3M_{P}/5a as found earlier.

(2) In Fig. 14.6-10(d) the mechanism is referred to as a ‘combined’ mechanism, because it is obtained by combining the mechanisms in Figs, (b) and (c). Example 14.6-4 shows that it is possible to obtain a lower collapse load by combining two mechanisms to form a third mechanism. The combination of mechanisms has powerful applications in plastic analysis and will be dealt with in greater detail in Section 14.7.

(3) From Comment (2) above it can be seen that, although three mechanisms are shown in Fig.14.6-10(b), (c) and (d), only two are independent, the third being obtainable by combining the first two.

In algebra, a set of quantities Q_l, Q_2, Q_3 . . . Q_nare said to be linearly independent if none of them can be obtained from a linear combination of the others; that is if the linear equation

*\qquad c_1Q_1+c_2Q_2+ . . . c_nQ_n= 0*

can be satisfied identically only when the constants c_{1} c_{2}, . . . c_{n} are all zero. In plastic theory of structures, a group of mechanisms are called **independent mechanisms** if none of them can be obtained from a linear combination of the others. Thus the beam mechanism and the sidesway mechanism in Fig. 14.6-10 are independent of each other: the beam mechanism contains a plastic hinge at section C, which is not present in the sidesway mechanism, and hence the former mechanism cannot be obtained as a multiple of the latter. However, the three mechanisms taken together are not independent because each can be expressed as a linear combination of the other two. Specifically, in Fig. 14.6-10:

Combined mechanism = Beam mechanism + Sidesway mechanism

Any two of the three mechanisms can be regarded as the independent mechanisms.

(4) Of the mechanisms in Fig. 14.6-10, the sidesway and the combined mechanisms are **regular mechanisms **or complete mechanisms; they each contain R + 1 plastic hinges, where R is the number of redundancies of the framework. The beam mechanism is a **partial mechanism** or incomplete mechanism; it contains less than R + 1 plastic hinges. A regular mechanism is statically determinate; the complete bending moment distribution may be determined by statics from the applied loading and the known values of the moments at plastic hinge sections. A partial mechanism, on the other hand, is still statically indeterminate. In general, if a structure has ** R** redundancies, then the formation of

*R*plastic hinges will render it statically determinate, provided that these R hinges do not convert part of the structure into a (partial) mechanism; if they do convert part of the structure into a mechanism, then the structure remains statically indeterminate—that is, in the latter case, we cannot say that the formation of each plastic hinge reduces the degree of redundancy by one. Consider, for example, the structures in Fig. 14.6-11, each of which has 3 redundancies. In each of Fig. 14.6-11 (a), (b) and (c), the three plastic hinges convert the structure into a partial mechanism, and it is easy to see that the structure remains statically indeterminate. In Figs, (d), (e) and (f), the plastic hinges are so located that no mechanisms form, and again it is readily seen that all the structures are statically determinate. If an additional hinge now forms in any of the determinate structures in Figs, (d), (e) and (f), that structure will collapse as a regular mechanism.

(5) It can readily be shown that for a given structure the total number of independent mechanisms is given by

M = N – R (14.6-3)

where **R** is the number of redundancies and N is the number of **critical sections**. Critical sections (sometimes called **cardinal sections**) are possible locations for plastic hinges; thus for the portal frame in Fig. 14.6-10(a) the critical sections are ** A**,

**,**

*B***,**

*C***and**

*D***, which occur at the joints and at loading points. The bending moment diagram between two adjacent critical sections is a straight line; therefore a plastic hinge hinge cannot occur between critical sections (except in the particular case where the bending moment diagram between the sections is a rectangle of height M_p so that there is an infinite number of critical sections).**

*E*To derive Eqn 14.6-3 consider a structure having R redundancies and N critical sections. The complete bending moment diagram is defined by the moment values at the **N** critical sections. Each virtual work equation (such as Eqn 14.9-1 or Eqn 14.9-2) written for such a mechanism represents a condition of equilibrium between the external loads and the bending moments at some of the **N** critical sections. Hence, **M** independent mechanisms, providing **M** independent equations, will enable the determination of **M** of the N bending moments required to define the complete bending moment diagram, so that **N — M** bending moment values still remain unknown. Since the number of redundancies is known to be R, we must have **N — M = R**; hence Eqn14.6-3