Advanced Methods of Structural Analysis

12.2 Direct Method of Plastic Analysis 427

tension, and temperature changes; this is a fundamental difference between plastic
and elastic analysis. In the following sections, we will consider different methods of
determining plastic loads.

12.2 Direct Method of Plastic Analysis.....................................

The fundamental concept of plastic analysis of a structure may be clearly presented
using the direct method. Let us consider the structure shown in Fig.12.3a subjected
to loadPat pointK. The horizontal rod is absolutely rigid. All hangers have con-
stant stiffnessEA. The plastic analysis must be preceded by elastic analysis.

Elastic analysisThis analysis should be performed on the basis of any appropri-
ate method of analysis of statically indeterminate structures. Omitting this analysis,
which is familiar for reader and presents no difficulties, the distribution of internal
forces in members 1–4 of the structure is as follows (Fig.12.3b):

N 1 D0:4PI N 2 D0:3PI N 3 D0:2PI N 4 D0:1P

Plastic analysis
Step 1: Increasing of load leads to the appearance of the yield stresses. They are
reached in the most highly stressed member. In our case, this member is element 1.
LetN 1 become equal to limit load, i.e.N 1 DNy.SinceN 1 D0:4P, then it occurs

if external load would be equal toP D N0:4y D2:5Ny. For this loadP, the limit
tension will be reached in the first hanger.Internal forces in another members are
(Fig.12.3c)

N 2 D0:3PD0:32:5NyD0:75NyI N 3 D0:5NyI N 4 D0:25Ny

Step 2: If loadPwill be increased by valueP 2 ,thenN 1 DNyremains without
changes. It means that additional load will be distributed between three members
2, 3, and 4, i.e., the design diagram had been changed (Fig.12.3d). This structure
is once statically indeterminate. Elastic analysis of this structure due to loadP 2
leads to the following internal forces

N 2 D0:833P 2 I N 3 D0:333P 2 I N 4 D0:167P 2 :

As always, the most highly stressed member will reach the yield stress first. Since
first hanger is already in yield condition (and cannot resist any additional load), the
most highly stressed member due to loadP 2 is the second hanger. The total limit
load in this element equals

N 2 D0:75NyC0:833P 2 :

In this formula, the first term corresponds to initially applied loadP D2:5Ny
(Fig. 12.3c), while the second term corresponds to additional load P 2.