-m w-054x1J6xa270-66N 后413子0卫N a=08x09x270=61 Fa=0.08x0,429x÷2700=61.8 F,=0sx0429←x10=-61,4N 5=0x026x120=562N 鞋 。0sas（}2mo-2N 绿 eoaoe6 .=008T+0.0=008x0.4665+001-=0.04732 2G.-=270x5+2700x10+180x15=6750 =0-004723048=2,97 5=200t-a072k33048=2.9则 atal sesm sof th 9 [Example 4.6] One 3-story reinforce concrete building is located at site of the second group in category I1 . The stifiness of the beams is ¥ and two columns have the same cross-section . The weight of each floor, three mode shapes and corresponding periods are shown in Figure 4.3. The seismic zone of intensity for design is 7. Tg=0.3s. x=0.05. Draw the figures of the moment under the horizontal seismic actions by mode analysis method. G3= 2700kN T1=0.4665 T2=0.2086 1/3 T3=0.1348 2/3 1 G2= 2700kN G1= 1800kN 2/3 1 1/4 2/3 -1 -3/4 Participating factor 1.364 3 1 2700 3 2 1800 1 2700 3 1 2700 3 2 1800 1 2700 2 2 2 1＝ ＝ ÷ ø ö ç è æ ÷ + ´ ø ö ç è æ ´ + ´ ´ + ´ + ´ g Seismic actions 2700 66.3kN 3 1 F 0.054 1.364 11＝ ´ ´ ´ ＝ F 0.054 1.364 1 1800 132.6kN 2700 132.6kN 3 2 F 0.054 1.364 13 12 ＝ ＝ ＝ ＝ ´ ´ ´ ´ ´ ´ The seismic actions of mode 2 and mode 3 can be obtained in a similar way. Solution: Given: Mode 1, T1=0.4665 > Tg Seismic influence coefficient 0.08 0.054 0.4665 0.3 T T 0.9 max 0.9 1 g 1＝ ＝ ÷ ´ ＝ ø ö ç è æ ÷ ÷ ø ö ç ç è æ a a The moments of the frame can be drawn according to the actions above and we’ll finally get the figure of the moment after conbinating modes by SRSS method, shown in Figure 4.4 and Figure 4.5. 2700 61.8kN 3 2 F22＝0.08´ 0.429´ ´ ＝ F23＝0.08´ 0.429´ (－1)´1800＝－61.8kN F31＝0.08´ 0.26´1´ 2700＝56.2kN 2700 42.1kN 4 3 F 0.08 0.26 32＝ － ÷´ ＝－ ø ö ç è æ ´ ´ 1800 9.4kN 4 1 F33＝0.08´ 0.26´ ´ ＝ 2700 61.8kN 3 2 F21＝0.08´ 0.429 ´ ´ ＝ 414.4 -42.1 56.2 132.6 132.6 66.3 9.4 (c) 40.9 11.5 29.4 11.5 29.1 40.9 422.6 342.8 745.9 11.8 (a) 497.3 61.8 165.8 331.5 165.8 -61.8 61.8 (b) 77.3 77.3 414.4 331.5 77.3 77.3 77.3 77.3 11.8 40.9 29.1 29.4 183.3 183.3 526.1 342.8 422.6 342.8 183.3 342.8 526.1 765.4 497.3 Figure 4.4 Moments of Each Seismic Actions Figure 4.5 Moment of the Frame Structure after Modes Conbined [Example 4.7] Draw the figures of the moment under the horizontal seismic actions in Example 4.6 by base shear method. Equivalent values of the whole gravity loads Geq＝0.85´ (1800 + 2700 ´ 2)＝6120kN Seismic influence coefficient 0.08 0.054 0.4665 0.3 0.9 1＝ ÷ ´ ＝ ø ö ç è æ a Total seismic actions FEk＝a1Geq＝0.054 ´ 6120＝330.48kN Because T1 ＝0.4665＞1.4Tg＝1.4 X0.3=0.42 d n＝0.08T1 + 0.01＝0.08´0.4665+ 0.01＝0.04732 å= ´ + ´ + ´ n k 1 GkHk＝2700 5 2700 10 1800 15＝67500 (1 0.04732) 330.48 62.97kN 67500 2700 5 F1＝ ´ - ´ ＝ ´ (1 0.04732) 330.48 125.94kN 67500 2700 10 F2＝ ´ - ´ ＝ ´ (1 0.04732) 0.04732 330.48 141.58KN 67500 1800 15 F3 ´ = ú û ù ê ë é ´ - + ´ ＝ The moments of the frame according to the actions above is shown in Figure 4.6(a). According to Table 5.7 , top additional seismic action coefficient