问题描述:
反应谱工况下,程序输出的构件内力均为正值,该值代表其在地震作用下可能的最大值,那么它是如何与静力工况进行荷载组合的呢?
解答:
若反应谱工况下某构件的内力值为 X,实际内力的变化范围将在 -X 到 +X 之间。现举例来说明反应谱工况与静力工况的组合方式。
反应谱工况下的内力为:
重力工况下的内力为:
现定义仅包含反应谱工况的荷载组合Comb1,比例系数为1;同时包含重力工况和反应谱工况的荷载组合Comb2,两个工况的比例系数均为1。
1)对荷载组合Comb1,构件内力如下:
2)对荷载组合Comb2,构件内力如下:
-
Max:P =
(+500+100) = +600 kN;M2 =
(-50+200) = +150 kN-m;M3 = (+100+300) = +400 kN-m
-
Min:P = (+500-100) = +400 kN;M2 = (-50-200) = -250 kN-m;M3 =
(+100-300) = -200 kN-m
在进行设计内力组合时,为表示各内力分量间的对应关系,反应谱工况将采用如下所示的内力对:
-
P, Mx, My
-
P, Mx, -My
-
P, -Mx, My
-
P, -Mx, -My
-
-P, Mx, My
-
-P, Mx, -My
-
-P, -Mx, My
-
-P, -Mx, -My
3)对荷载组合Comb1,程序将给出如下所示的设计内力:
-
Comb1-1:P = +100 kN;M2 = +200 kN-m;M3 = +300 kN-m
-
Comb1-2:P = +100 kN;M2 = +200 kN-m;M3 = -300 kN-m
-
Comb1-3:P = +100 kN;M2 = -200 kN-m;M3 = +300 kN-m
-
Comb1-4:P = +100 kN;M2 = -200 kN-m;M3 = -300 kN-m
-
Comb1-5:P = -100 kN;M2 = +200 kN-m;M3 = +300 kN-m
-
Comb1-6:P = -100 kN;M2 = +200 kN-m;M3 = -300 kN-m
-
Comb1-7:P = -100 kN;M2 = -200 kN-m;M3 = +300 kN-m
-
Comb1-8:P = -100 kN;M2 = -200 kN-m;M3 = -300 kN-m
4)对荷载组合Comb2,程序将得到如下所示的设计内力:
-
Comb2-1: P = (+500+100) = +600 kN;M2 = (-50+200) = +150 kN-m;M3 = (+100+300) = +400 kN-m
-
Comb2-2: P = (+500+100) = +600 kN;M2 = (-50+200) = +150 kN-m;M3 = (+100-300) = -200 kN-m
-
Comb2-3: P = (+500+100) = +600 kN;M2 = (-50-200) = -250 kN-m;M3 = (+100+300) = +400 kN-m
-
Comb2-4: P = (+500+100) = +600 kN;M2 = (-50-200) = -250 kN-m;M3 = (+100-300) = -200 kN-m
-
Comb2-5: P = (+500-100) = +400 kN;M2 = (-50+200) = +150 kN-m;M3 = (+100+300) = +400 kN-m
-
Comb2-6: P = (+500-100) = +400 kN;M2 = (-50+200) = +150 kN-m;M3 = (+100-300) = -200 kN-m
-
Comb2-7: P = (+500-100) = +400 kN;M2 = (-50-200) = -250 kN-m;M3 = (+100+300) = +400 kN-m
-
Comb2-8: P = (+500-100) = +400 kN;M2 = (-50-200) = -250 kN-m;M3 = (+100-300) = -200 kN-m
故有反应谱工况参与的荷载组合都有8个子设计组合。
来源: CSI Knowledge Base
网址: https://wiki.csiamerica.com/display/kb/Load+combination+FAQ