习题课3 静定平面桁架的内力计算_659501715

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Fy1
2
1
22
(8)
S1 2FP 4FP A 3FP

2FP 1 C 2 FN a
Ⅰ a
解: 1) I-I右:
D B 3FP
a
∑F
S1
=0
a
复杂桁架
2 2 + 3FP ⋅ FN ⋅ 2 2 2 − 2 FP ⋅ =0 2 FN = − FP ( 压 )
2) 结点B:
∑ Fx = 0 ∑F
y
FxBC = FP FNBC = 2 FP
FP 2FP A FP
=
FP FP A FP
FP B FP
+
FP
D FP A B FP
6
(5)
S1 FP 2FP A FP

C D
Ⅱ S2 FP B FP a a 6根零杆
0 0 0 0 0 0
Ⅱ FN2 Ⅰ FN1 a a
由于荷载反对称,该桁架除下部水平链杆AB外, 其余杆件受力反对称,故 FNCD = 0 。 I-I右: ∑ FS1 = 0 II-II右:∑ FS 2
4 Ⅱ
E
解: 1) I-I右:
FP 4m
FP 4m
简单桁架
2 ∑ M D = 0 FN 1 = 3 FP (拉)
2) 结点E:
2 ∑ Fx = 0 Fx2 = 3 FP
2 ∑ Fx = 0 FN = − 3 FP (压) FN2
E FP
18
5 5 2 FP 3 FN 2 = Fx 2 = FP (拉 ) 4 6
∑F ∑F
y
y
4)或取I-I左:
∑F
FN 2 = FP (拉) 1 2 = 0 Fy1 = − FP FN 1 = − FP (压) 2 2
y
=0
= 0 Fy1 = −0.5FP
2 FN 1 = − FP (压) 2
d
(6)
联合桁架
B
1
3a
A a FP a
C FP FP a FP a a
解:作如图封闭截面,取ABC部分为隔离体:
3 4 Fy 5 = ⋅ FP = FP 4 3
2FP
4 Fx 5 = FP 3
FN 3 = 0
FP
D
2FP/3
19

(5)
复杂桁架
0 1.5FP 2 A FP 1 B FP FP 1.5FP d d
Ⅰ C d
20
C
解: 2)结点A: 3)结点B:
d
d
1)结点C:结构与荷载均对称,两斜杆轴力为零。
14
(3)
A 1 D a FP FP E
a
B a
C a/2
0
a/2
联合桁架
15
a/2
F
2
0
0
A -FP 1
FP

FP
E
a Ⅰ
-FP F 02 C a/2 =0
FP B FP -FP a
FN a/2 FN = 0 a
2 FP
解: 1)I-I右:
联合桁架
∑M
E
2)结点C: FN 2 = 0 3)结点F:
C
FNCA = 2 FP FNCB = 2 FP
24

(9)联合桁架
1 C
FP FP
a a
FP 3FP A D
a
解: I-I 左: ∑ M D = 0
3FP
a
2 Ⅰ a
B 3FP
∑M
C
=0
1 FN 1 = ⋅ 3FP a = FP (拉) 3a 1 FN 2 = (3FP ⋅ 3a − 3FP a ) = 2 FP (拉) 25 3a
∑M
A
=0
−1 ( FP ⋅ a + FP ⋅ 2a ) = −0.6 FP (压) FN 1 = 21 5a
(7) 联合桁架
C
a a
1 0 A FP Ⅰ FP a a FP
Fy1 =
解: I-I下:
Ⅰ a
B B FP
a
Fx1
FN 1
5
Leabharlann Baidu
∑MA = 0
1 2 (− FP ⋅ 2a ) = − FP 3a 3 5 2 5 (− FP ) = − FN 1 = FP (压) 2 3 3
9
二、用简捷方法求桁架指定杆轴力 (1)
E H A 1.5FP C FP a 2 1 F a/2 G D B FP FP a/2 a/2 a 简单桁架 1.5FP a/2
10
二、用简捷方法求桁架指定杆轴力 解: II-II左:
E H A 2

1
F
a/2
∑M
C
G D
FN 1 ⋅ a + 1.5 FP ⋅ a = 0 FN 1 = −1.5 FP (压) 1.5FP
(10)
Ⅰ C
0.5FP
a
D A a FP B a
复杂桁架 Ⅰ a
1
解:
整体平衡:∑ M A = 0 I-I 上:
1 1 FxC = ⋅ FP a = FP (←) 2a 2
1 1 1 FN 1 = ⋅ FP a = FP (拉) 2a 2 4
∑M
D
=0
26
C Ⅰ Ⅰ Ⅱ FP FP FP a/2 a/2 a a
简单桁架
B
1.5FP
I-I下:
∑M
FN 2
D
=0 Fy 2 = −0.5 FP
Fy 2 a + FP a / 2 = 0
5
1
11
5 = −0.5 FP ⋅ = −1.118FP (压) 1
2
a/2
=0
(2)
3FP 1 2 A D d d B FP d
FyBC = FP
FNBC FP
FNBD B 3FP
23
= 0 FNBD = −4 FP
3)结点D:
2 ∑ FS 2 = 0 FN 2 − 2 ⋅ 4 FP = 0 FN 2 = 2 2 FP ( 拉 )
D FN2 S2 4FP
4)结点C:
FN1=2FP
∑F ∑F
x y
=0 =0
FNCA = FNCB FN 1 = 2 FP ( 拉 )
∑F
y
= 0 Fy1 = FP
FN 1 = 2 FP (拉)
16
a/2
0
D
(4)
A F C1 H 3 D FP 4m 2 E G
3m 3m
17
4 B FP 4m FP 4m
FP 4m
简单桁架
6.67FP A

F
1Ⅰ C H 3 2 FN DⅠ
G
3m
6.67FP B 4FP FP 4m FP 4m
13
3FP

1

3
2FP
d
FxA=2FP A

2 FN D d B Ⅰ FyB=2FP d d FP d C
FyA=FP
FyC=FP
整体平衡:
∑ M B = 0 FyA =
II-II左:
1 (3FP d + FP ⋅ 2d − 2 FP d − FP d ) = FP (↑) 2d
1 ∑ M D = 0 FN 1 = d (− FP ⋅ d ) = − FP (压) ∑ Fy = 0 FN 2 = FP − 3FP = −2 FP (压)
3m
6.67FP A

F
1Ⅰ C H 3 2 FN DⅠ
G
3m
6.67FP B 4FP FP 4m FP 4m
4 Ⅱ
E
FP 4m
FP 4m
3) II-II右: −1 ∑ M F = 0 FN 4 = 6 (4 FP + 8FP ) = −2 FP (压) FN5 4) 结点D: F
3m
N3
∑F
x
=0
联合桁架
3
2FP
d
C d
12
(2)
3FP

1

3
2FP
d
FxA=2FP A

2 FN D d B Ⅰ FyB=2FP d d
联合桁架
C FP d FyC=FP
FyA=FP
解: I-I右:
1) ∑ M C = 0 FN 3 d + FP d − 2 FP d = 0 FN 3 = FP (拉) 3) ∑ Fy = 0 FyC = FP (↑) 2) ∑ Fx = 0 FN = FP (拉)
习题课 3
静定平面桁架的内力计算
1
一、找出桁架的零杆 (1)
FP 0 0 0 0 0
0
0
0
8根零杆
2
(2)
0 0 0 0 FP 5根零杆 0
3
(3)
0 0
FP
FP FP 0 0
0
0
0 12根零杆
0 0 0 0 0 0 0 0 FP
4
(4)
FP
对称
A
12根零杆
5
(5)
C D a D FP a B a a FP
2 2 FN 1 + FP ⋅ − FP ⋅ 2 2 − 2 2 FN 2 + FP ⋅ + FP ⋅ =0 2 2 2 = 0 FN 1 = 0 2 2 = 0 FN 2 = 2 FP 2 7
(6)
FP FP 附属部分 0 0 0 0 0
0
6根零杆
8
(7)
0 0 FP 0 0 0 0 0
7根零杆
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