2 )}
节点1力矩平衡力方程 m2 m1 Q2l 0
m1 Q2l m2
m1
EJ l2
{6[(u2
u1 ) (v2
v1 )]
l ( 41
22 )}
二. 由单元刚度矩阵计算
{F e } [K e ]{ e } [K e ][T e ] { e }
0 0 0 0 u1 u1 v1
0 0
0 0
1 0
0 1
0 0
0 0
I
0 0 0 0 1 0
0 0 0 0 0 1
[
K
2 23
]
[T
2
]T
[
K
2 23
][T
2
]
[
K
2 23
]
若局部坐标与整体坐标方向一致, 则 [K e ] [K e ]
4. 求总刚度矩阵
节点号为i, j的单元, 其刚度矩 阵元素在总刚度矩阵中的位置
3i－2 3i－1 3i
k115 k215 k315
k116 k216 k316
k411 k511
k412 k512
k413 k513
k414 k514
k415 k515
k416 k516
k611 k612 k613 k614 k615 k616
2单元
kkk654222444
k425 k525 k625
k426 k526 k626
3.462 253.385
0
0 252
0 0 253.385
3.462 0
5.711 3.462 3.462
0 0 0 252 0
0 0 0 0 1.385
3.0000462uv00022
2.758
px3
6.25
5.208
py3 m3
3.462 0 0 0 0 0 0 0
u1
转换关系:
f ii
i
cos sin
0
sin cos
0
0 0 1
uvii
i
1
Fx1 x
1.轴向内力
N12
AE l
(2
1)
AE l
[cos (u2
u1) sin(v2
v1)]
AE [cos
l
sin
]uv22
u1 v1
AE [
l
]uv 22
u1 v1
2. 弯曲内力 y 向位移: 1节点: f1 sin u1 cos v1 u1 v1
3i－2 ＊＊＊
3i－1 ＊＊＊
3i
＊＊＊
3j－2 ＊＊＊ 3j－1 ＊＊＊
3j ＊＊＊
3j－2 3j－1 3j
＊＊＊ ＊＊＊ ＊＊＊
＊ ＊＊ ＊＊ ＊ ＊＊＊
单元刚度矩阵元素编号(相当于在总刚度矩阵中的位置)
1单元
kkk132111111
k112 k212 k312
k113 k213 k313
k114 k214 k314
y
1
b 2.5m a 2.5m
转换成整体座标:
故, ①单元的等效结点力:
0 2.0 {P121} 2202.5..05
{F e } [T ]{F e } {T }1 [T ]T
0 2 2 0
P121 T1 T 202.52.5 2022.5.5
1节点 2节点
②单元 N 2 N3 0
Al 2 0
J
0 0 u1 v1
12 6l
0
6l 2l 2
0
uuu2211vvv212
m2
0 0
12 6l 6l 2l 2
0 0
12 6l
6l 4l 2
2
矩阵相乘，求出各分量
N1
EA l
[u1
v1
u2
v2 ]
EA l
[
(u1
u2 )
(v1
v2 )]
N2
0 0 1 0 0 0
0 0 0 0 1 0
0 0 0 1 0 0
0 0 0 0 0 1
0 1 0 0 0 0
1 0 0 0 0 0
[T 1]T
0 0
0 0
1 0 0 0 0 0 1 0
0 0 0 1 0 0
0 0 0 0 0 1
[K112 ] [T 1]T [K112 ][T 1]
1.385 0 3.462 1.385 0 3.462
v1) 4l1
6(u2
v2 ) 2l2 }
EJ l2
{6[ ( u2
u1) (v2
v1)]
l ( 41
22 )}
m2
EJ l2
{6(u1
v1)
2l1
6(u2
v2 )
4l2 }
EJ l2
{6[ ( u2
u1) (v2
v1)]
l ( 21
42 )}
三 . 有非节点载荷时单元内力计算
252 0
0 252 0
0
0 1.385 3.462 0 1.385 3.462
0 3.462 11.542 0 3.462 5.771
252 0
0 252 0
0
0 1.385 3.462 0 1.385 3.462
0 3.462 5.771 0 3.462 11.542
单元刚度矩阵迭加成整体刚度矩阵
k427 k527 k627
k428 k528 k628
k429 k529 k629
kk872244
k725 k825
k726 k826
k727 k827
k728 k828
k729 k829
k924 k925 k926 k927 k928 k629
[
K e1 12
]
[K23e1]
1.385 0 3.462 1.385 0 3.462 0 252 0 0 252 0 3.462 0 11.542 3.462 0 5.771 1.385 0 3.462 1.385 0 3.462 0 252 0 0 252 0 3.462 0 5.771 3.462 0 11.542
2节点 3节点
总体载荷
2 0 2.5
0 0 0
px1
p y1 m1
2 px1
p y1 2.5 m1
2 0
0 6.25
1 2
3 4.25
2.5 5.208 0.05 2.758
0 0 0
0 6.25 5.208
px3
EJ l2
{6[(u2
u1) (v2
v1)] l(21
42 )}
Q2
EJ (12
f12 6l12 )
l3
EJ l3
{12[u(u1
u2 ) (v1
v2 )] 6l(1
2 )}
y方向平衡力方程 Q1 Q2 0
Q1 Q2
Q1
EJ l3
{12[u(u2
u1) (v2
v1)] 6l(1
0 6l 2l 2 0 6l 4l 2
252 0
0 252 0
0
0 1.385 3.462 0 1.385 3.462
[
K112
]
[
K
2 23
]
103
0 252
3.462 0
11.542 0
0 252
3.462 0
5.771 0
0 1.385 3.462 0 1.385 3.462
252 0 0 0 1.385 3.462
3.462 0 5.711 3.462 3.462 23.083 0 3.462 5.771
0 0 0 252 0 0 252 0 0
0 0 0 0 1.385 3.462 0 1.385 3.462
0 0 0 0 3.462 5.771 0 3.462 11.542
0 252 0 0 252 0
[
K121
]
103
3.462 1.385
0 0
11.542 3.462
3.462 1.385
0 0
5.771 3.462
0 252 0 0 252 0
3.462 0 5.771 3.462 0 11.542
②单元 0o , cos 1, sin 0
py3 m3
px3
6.25
5.208
py3 m3
6. 引入约束条件, 构成总体方程
2 px1 p y1
2.5 m1 3
4.25
1.385
0
3.462
103
1.385 0
0 252 0 0 252
3.462 0
11.541 3.462
0
1.385 0
[T 2 ]
0 0
0
0
0 0
0
0
0 0 1 0 0 0
0
0
0
0
0
0
0
0
0 0 000 1
1 0 0 0 0 0
0 1 0 0 0 0
0 0 1 0 0 0
0 0 0 1 0 0
0 0 0 0 1 0
0 0 0 0 0 1
I
1 0 0 0 0 0
0 1 0 0 0 0
[T 2 ]T
ql Q2 Q3 2 6.25kN
m2
m3
ql2 12
5.208 kN • m
转换成整体座标:
y
q 2.5kN / m