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Solution

1. Calculate the G-factor with modification for column EF. Since the far end of the restraining girders is hinged, girder stiffness should be multiplied by 0.5 (see Table 31.2). Using the section properties in Example 31.1, we obtain

2. From the alignment chart in Figure 32.4b, K 1.22 is obtained.

### 31.5.2 Different Restraining Column End Conditions

To consider different far end conditions of restraining columns, the general effective length factor equations for column C2 (Figure 31.3) were derived by Duan and Chen (1988,1989,1996). By assuming that the far ends of columns C1 and C3 are hinged and using the slope deflection equation approach for the subassemblies shown in Figure 31.3, we obtain the following.

31.5.2.1 For a Braced Frame (Duan and Chen 1988)

\ga gAgB

where C and S are stability functions as defined by Equations 31.3 and 31.4; GA and GB are defined in Equations 31.7 and 31.8; and GAC1, GAC2, GBC2, and GBC3 are the stiffness ratios of columns at the A-th and B-th ends of the columns being considered, respectively. They are defined as

where ^ indicates the summation of all columns rigidly connected to the joint and lying in the plane in which buckling of column is being considered.

Although Equation 31.38 was derived for the special case in which the far ends of both column C1 and column C3 are hinged, this equation is also applicable if adjustment to GCi is made as follows: (1) if the far end of column Ci (C1 or C3) is fixed, then take GCi = 0 (except for GC2) and (2) if the far end of column Ci (C1 or C3) is rigidly connected, then take GCi = 0 and GC2 = 1.0. Therefore, Equation 31.38 can be specialized for the following conditions:

1. If the far ends of both column C1 and column C3 are fixed, we have GAC1 = GBC3 = 0 and Equation 31.38 reduces to

2. If the far end of column C1 is rigidly connected and the far end of column C3 is fixed, we have GAC2 = 1.0 and GAC1 = GBC3 = 0, and Equation 31.38 reduces to

3. If the far end of column C1 is rigidly connected and the far end of column C3 is hinged, we have GAC1 = 0 and GAC2 = 1.0, and Equation 31.38 reduces to

4. If the far end of column C1 is hinged and the far end of column C3 is fixed, we have GBC3 = 0, and Equation 31.38 reduces to

GbC / VGa gbj GaGb

5. If the far ends of both columns C1 and C3 are rigidly connected (i.e., the assumptions used in developing the alignment chart), we have GC2 = 1.0 and GCi = 0, and Equation 31.38 reduces to c2 - s2 + 2c(-^ + -H = 0

\ga gAgB

which can be rewritten in the form of Equation 31.5.

31.5.2.2 For an Unbraced Frame (Duan and Chen 1989, 1996)

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