Elastic-Plastic Connections with Limited Capacity in both directions

The formulation of the problem in this case which provides the general elastic-plastic behaviour can be extracted from the formulation of previous sections. Noting that the artificial moment should be applied in either positive or negative direction, a general formula for evaluation of moment can be written as follows:

Mri= Mi + £ mj(X+ – X-) ,i = 1,…,n

j=1

in which in any time one of the artificial moments can be present. i. e. in the presence of either of Xf orX-, the other one will have zero value. This can be mathematically stated as: Xf XX-= 0. Substituting the general Mri from Eq.(10) into Eq.(3) for M+ – M – and using a combination of objective functions in Eq.(8) and Eq.(9) and the complementary condition of Xf X – = 0 yields the following problem for the general elastic-plastic behaviour of a nonlinear connection.

Minimize £ [Xf (Mi – Mm ) + X-(MfM:i – Mri) + X+X-]

i=1

Subject to Mm <Mi + ^ mj(X+ – Xj ) < M+mi, i =1,…,n

j=1

Example 1.

To illustrate the solution procedure, and capabilities of the method a one-bay, one-story portal frame as shown in Fig.(3-a) is considered. In this example the moment-curvature relation of connections of beam to columns are assumed to be elastic-plastic in both positive and negative directions. All members are assumed to have the similar cross-sections. It is assumed that the frame will be loaded laterally up to 40 KN. The Moment capacity of connections in each direction has been intentionally assumed to be 85% of the members’ plastic limit as shown in Fig (3-b).

Fig.(3): Example 1, a) Properties of a Portal Frame, b)M – relations for connections 2 and 3

To have a comparison tool in hand, first the non-linear analysis of the frame will be done in a historical manner and then it is compared to the results of proposed solution procedure. Assuming that the structure behaves linearly, it is analysed under total lateral load (40-H & 65-V KN). The result of linear analysis has been reported in the first row of Table 1.

Table 1: Results of analysis of Example 1

Loading

Mj

M2

Midspan

M3

M4

Total (Linear)

-23.892

0.961

41.456

-48.050

47.098

First Yield

-21.132

.850

36.668

-42.5

41.658

Load increment

-6.715

0.904

7.959

0

6.240

Total Nonlinear

-27.847

1.755

44.627

-42.5

47.898

Since value of M3 =-48.05 exceeds the plastic limit of the connection (42.5 KN. m), the structure will experience inelastic behaviour under this intensity of load. The values of moments in all joints/ connections at the threshold of inelastic behaviour can be obtained by scaling the initial results by 42.5 . The results have been reported at second row of Table1. In the next stage, for the rest of

48.050

loading (i. e. 4.620-H and 6.507-V), the frame model will contain a hinge at connection 3. The result of this load increment has been reported in the third row of the Table 1. The total moments due to nonlinear behaviour is the sum of second and third row of the Table and has been reported in the fourth row.

The analysis of the problem via proposed method requires analysis of the structure under unit moments that are applied in all elastic-plastic connections. The results of such analysis for unit loads at joints 2 and3 have been shown in Table 2.

Employing Eq. (10), the resultant moments at connections 2 and 3 can be written as follows:

Mr2 = 0.961 + (1 – 0.6823 )(X+ – X2- ) + (-0.5 + 0.5454)(X+ – X3- )

Mr3 =-48.050 + ( -0.5 + 0.5454)(X2+ – X2- ) + (1 – 0.6823 )(X3+ – X3- )

Now according to Eq. (11) a quadratic programming problem is established as follows:

2

Minimize ^[X+ (Mrt -(-42.5)) + X;(42.5-Mri) + X+X-]

i=1

Subject to Mr2 = 0.961 + 0.3177(X + – X2-) + 0.0454(X3+ – X3-) < 42.5 Mr2 = 0.961 + 0.3177(X2+ – X2-) + 0.0454(X3+ – X3-) >-42.5 Mr3 = -48.050 + 0.0454(X2+ – X2- ) + 0.3177(X3+ – X3- ) < 42.5 Mr3 =-48.050 + 0.0454(X2+ – X2- ) + 0.3177(X3+ – X3- ) >-42.5 Solution of this QP problem gives the following results:

X3+ = 17.46783 and X 2+ = X 2- = X 3- = 0

Table 2: Results of analysis of Example 1 for unit loads and Proposed Method.

Loading

M1

M2

Midspan

M3

M4

Unit moment at 2

+0.0458

1-0.6823

+0.1816

+0.5454-.5

-0.2264

Unit moment at 3

-0.2264

+0.5454-.5

+0.1816

1-0.6823

+0.0458

Total external loads

-23.892

0.961

41.456

-48.050

47.098

Unit moment at 3 x 17.46783

-3.955

0.793

3.172

5.55

0.800

Sum of the Two Above Rows

-27.847

1.754

44.628

-42.5

47.898

It remains to obtain internal forces in all sections using Eq. (10). The results of moment calculations have been shown in fifth and sixth rows of Table2. If the results of Table 1 and Table 2 are compared, it is observed that there is a very small difference between the two results which may have been because of numerical round off error in either of the algorithms/methods or both. As a conclusion, in this example it was shown that the proposed method is capable of solving the nonlinear problem and theoretically saying, it results in exact nonlinear solution.