U
KŁADY STATYCZNIE NIEWYZNACZALNE
Politechnika Poznańska
Adam Łodygowski ®
1
O
BLICZANIE UKŁADÓW STATYCZNIE NIEWYZNACZALNYCH
METODĄ SIŁ
.
Zadana rama wygląda następująco:
Siły wewnętrzne od obciążenia zewnętrznego. Dobieram układ podstawowy w ten sposób
aby zachować symetrię:
Zapisuję układ równań kanonicznych:
=
∆
+
⋅
+
⋅
+
⋅
=
∆
+
⋅
+
⋅
+
⋅
=
∆
+
⋅
+
⋅
+
⋅
0
0
0
3
3
33
2
32
1
31
2
3
23
2
22
1
21
1
3
13
2
12
1
11
P
P
P
X
X
X
X
X
X
X
X
X
δ
δ
δ
δ
δ
δ
δ
δ
δ
∫
⋅
=
ds
EI
M
M
k
i
ik
δ
∫
⋅
=
∆
ds
EI
M
M
i
P
iP
U
KŁADY STATYCZNIE NIEWYZNACZALNE
Politechnika Poznańska
Adam Łodygowski ®
2
Rysuję wykresy momentów od poszczególnych sił jednostkowych:
M
1
M
2
M
3
U
KŁADY STATYCZNIE NIEWYZNACZALNE
Politechnika Poznańska
Adam Łodygowski ®
3
M
P
M
S
U
KŁADY STATYCZNIE NIEWYZNACZALNE
Politechnika Poznańska
Adam Łodygowski ®
4
Korzystając z metody Wereszczegina- Mohra całkowania iloczynu dwóch funkcji (w tym
jednej prostoliniowej) otrzymuje się:
[
]
[
]
0
504
10
48
1
12
3
2
6
12
2
1
1
6
6
6
2
2
1
6
3
2
6
10
2
2
1
2
1
0
3
2
23
2
2
22
1
2
21
=
⋅
=
+
=
=
⋅
⋅
⋅
⋅
⋅
+
⋅
⋅
⋅
⋅
+
⋅
⋅
⋅
⋅
⋅
⋅
⋅
=
⋅
=
=
⋅
=
∫
∫
∫
ds
EI
M
M
EI
EI
EI
EI
ds
EI
M
M
ds
EI
M
M
δ
δ
δ
[
]
+
=
=
⋅
+
⋅
⋅
⋅
⋅
+
⋅
+
⋅
⋅
⋅
⋅
⋅
⋅
+
⋅
⋅
⋅
⋅
⋅
⋅
=
⋅
=
=
⋅
=
+
=
⋅
=
∫
∫
∫
42
10
3
4
1
1
3
1
4
3
2
4
6
2
1
4
3
1
1
3
2
1
6
2
1
2
2
1
1
3
2
10
2
2
1
2
1
0
90
10
8
1
3
3
33
2
3
32
1
3
31
EI
EI
EI
ds
EI
M
M
ds
EI
M
M
EI
ds
EI
M
M
δ
δ
δ
[
]
+
−
=
⋅
⋅
⋅
⋅
+
⋅
+
⋅
⋅
⋅
⋅
+
⋅
+
⋅
⋅
⋅
⋅
⋅
⋅
−
−
⋅
⋅
⋅
⋅
⋅
⋅
+
⋅
⋅
⋅
⋅
⋅
⋅
⋅
−
=
⋅
=
∆
=
⋅
=
∆
+
−
=
⋅
⋅
⋅
⋅
+
⋅
⋅
⋅
−
−
⋅
⋅
⋅
⋅
⋅
⋅
+
⋅
⋅
⋅
⋅
⋅
⋅
⋅
−
=
⋅
=
∆
∫
∫
∫
1668
10
3
232
1
2
5
8
6
4
6
3
2
1
3
1
4
3
2
6
128
2
1
4
3
1
1
3
2
6
56
2
1
2
2
1
1
2
1
8
2
4
10
2
3
2
1
3
2
56
10
2
2
1
2
1
0
3744
10
464
1
6
8
12
4
12
3
2
6
56
12
1
6
2
1
8
2
4
10
2
3
2
6
3
2
56
10
2
2
1
2
1
2
2
3
3
2
2
2
2
1
1
EI
EI
EI
ds
EI
M
M
ds
EI
M
M
EI
EI
EI
ds
EI
M
M
P
P
P
P
P
P
Sprawdzenie globalne delt:
+
⋅
⋅
=
+
+
+
+
+
+
+
+
=
+
⋅
⋅
=
⋅
⋅
⋅
⋅
⋅
+
⋅
+
⋅
⋅
⋅
⋅
+
⋅
+
⋅
⋅
⋅
⋅
⋅
+
+
⋅
+
⋅
⋅
⋅
⋅
+
⋅
+
⋅
⋅
⋅
⋅
⋅
+
⋅
⋅
⋅
⋅
⋅
⋅
=
=
∑∑
∫
∑∑
∫
942
10
3
340
1
942
10
3
340
1
12
3
2
6
12
2
1
1
4
3
1
1
3
2
6
1
2
1
1
3
1
4
3
2
6
4
2
1
2
1
13
3
1
16
3
2
16
6
2
1
16
3
1
13
3
2
13
6
2
1
2
1
13
3
2
10
2
13
2
1
1
33
32
31
23
22
21
13
12
11
2
2
EI
EI
EI
EI
EI
EI
ds
EI
M
ds
EI
M
i
k
ik
S
i
k
ik
S
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
δ
U
KŁADY STATYCZNIE NIEWYZNACZALNE
Politechnika Poznańska
Adam Łodygowski ®
5
+
⋅
⋅
−
=
∆
+
∆
+
∆
=
∆
+
⋅
⋅
−
=
⋅
⋅
⋅
⋅
+
⋅
⋅
⋅
⋅
⋅
⋅
−
−
⋅
⋅
⋅
⋅
+
⋅
+
⋅
⋅
⋅
⋅
+
⋅
+
⋅
⋅
⋅
⋅
⋅
−
−
⋅
⋅
⋅
⋅
+
⋅
+
⋅
⋅
⋅
⋅
+
⋅
+
⋅
⋅
⋅
⋅
⋅
−
−
⋅
⋅
⋅
⋅
⋅
⋅
+
⋅
⋅
⋅
⋅
⋅
⋅
−
=
⋅
∆
=
⋅
∑
∫
∑
∫
5412
10
3
1624
1
5412
10
3
1624
1
2
1
8
2
4
2
3
2
1
3
2
10
2
56
2
1
1
2
5
8
6
4
6
3
2
1
3
1
4
3
2
6
128
2
1
1
3
2
4
3
1
6
56
2
1
2
1
2
39
8
6
4
6
3
2
13
3
1
16
3
2
6
128
2
1
16
3
1
13
3
2
6
56
2
1
2
1
13
2
1
8
2
4
10
2
3
2
13
3
2
10
2
56
2
1
1
3
2
1
2
2
2
2
EI
EI
EI
EI
EI
EI
ds
EI
M
M
ds
EI
M
M
P
P
P
iP
S
P
iP
S
P
Mając dane wszystkie wielkości podstawiam je do układu równań i rozwiązuje go:
=
∆
+
⋅
+
⋅
+
⋅
=
∆
+
⋅
+
⋅
+
⋅
=
∆
+
⋅
+
⋅
+
⋅
0
0
0
3
3
33
2
32
1
31
2
3
23
2
22
1
21
1
3
13
2
12
1
11
P
P
P
X
X
X
X
X
X
X
X
X
δ
δ
δ
δ
δ
δ
δ
δ
δ
(
)
(
)
(
)
(
)
(
)
]
[
687978
,
27
0
]
[
489344
,
5
0
1668
10
3
232
42
10
3
4
0
90
10
8
0
0
0
504
10
48
0
0
3744
10
464
90
10
8
0
216
10
48
3
2
1
3
2
1
3
2
1
3
2
1
kN
X
X
kN
X
X
X
X
X
X
X
X
X
X
=
=
=
=
+
⋅
+
⋅
+
⋅
+
⋅
+
⋅
+
⋅
=
+
⋅
+
⋅
+
⋅
+
⋅
=
+
⋅
−
⋅
+
⋅
+
⋅
+
⋅
+
⋅
M
P
U
KŁADY STATYCZNIE NIEWYZNACZALNE
Politechnika Poznańska
Adam Łodygowski ®
6
T
P
N
P
Sprawdzenie kinematyczne:
M
P
U
KŁADY STATYCZNIE NIEWYZNACZALNE
Politechnika Poznańska
Adam Łodygowski ®
7
M
i
EI
EI
EI
u
ds
EI
M
M
u
i
i
n
i
031
,
0
6
8
6
4
6
3
2
6
6
2
687
,
15
623
,
4
1
4
623
,
4
40
2
1
3
8
2
4
40
3
2
1
2
2
=
⋅
⋅
⋅
⋅
−
⋅
⋅
+
⋅
+
⋅
⋅
⋅
+
⋅
⋅
⋅
⋅
−
⋅
=
⋅
=
∫
Dobieram odpowiedni przekrój dwuteowy:
43
,
96
5
,
19
1567
2
,
1
2
,
1
2
≥
≤
⋅
≤
⋅
W
cm
kN
W
kNcm
W
M
dop
σ
Dwuteownik 120:
[
]
cm
h
kNm
EI
cm
W
cm
I
0
,
12
4
,
672
7
,
54
328
2
3
4
=
=
=
=
U
KŁADY STATYCZNIE NIEWYZNACZALNE
Politechnika Poznańska
Adam Łodygowski ®
8
Siły wewnętrzne od osiadania podpór
.
Układ podstawowy przyjmuję podobnie jak w poprzednio:
( ) ( )
[
]
0
01
,
0
1
01
,
0
1
1
1
=
⋅
−
⋅
−
=
∆
=
∆
−
=
∆
∆
∆
∑
i
R
( ) ( )
(
)
[
]
004
,
0
012
,
0
2
01
,
0
1
01
,
0
1
2
2
=
⋅
−
⋅
+
⋅
−
=
∆
=
∆
−
=
∆
∆
∆
∑
i
R
U
KŁADY STATYCZNIE NIEWYZNACZALNE
Politechnika Poznańska
Adam Łodygowski ®
9
( )
( )
0
0
1
01
,
0
2
1
01
,
0
2
1
2
3
=
⋅
+
⋅
−
⋅
−
=
∆
=
∆
−
=
∆
∆
∆
∑
i
R
Delty wykorzystuję z obliczonego wcześniej układu podstawowego:
=
∆
+
⋅
+
⋅
+
⋅
=
∆
+
⋅
+
⋅
+
⋅
=
∆
+
⋅
+
⋅
+
⋅
∆
∆
∆
0
0
0
3
3
33
2
32
1
31
2
3
23
2
22
1
21
1
3
13
2
12
1
11
X
X
X
X
X
X
X
X
X
δ
δ
δ
δ
δ
δ
δ
δ
δ
(
)
(
)
( )
(
)
(
)
(
)
( )
[ ]
[ ]
[ ]
kN
X
kN
X
kN
X
EI
X
X
X
EI
X
X
X
EI
X
X
X
0
0041
,
0
0
0
0
42
10
3
4
0
90
10
8
0
004
,
0
0
504
10
48
0
0
0
90
10
8
0
216
10
48
3
2
1
3
2
1
3
2
1
3
2
1
=
−
=
=
=
⋅
+
⋅
+
⋅
+
⋅
+
⋅
+
⋅
=
⋅
+
⋅
+
⋅
+
⋅
+
⋅
=
⋅
+
⋅
+
⋅
+
⋅
+
⋅
+
⋅
M
∆
n
U
KŁADY STATYCZNIE NIEWYZNACZALNE
Politechnika Poznańska
Adam Łodygowski ®
10
Sprawdzenie:
[
]
[ ]
[ ]
m
m
V
EI
EI
EI
V
ds
EI
M
M
R
V
K
K
i
n
i
K
01
,
0
01000074
,
0
6
3
2
6
0492
,
0
2
1
1
6
6
0246
,
0
2
1
6
3
2
0246
,
0
40
2
1
1
012
,
0
1
1
≈
=
⋅
⋅
⋅
⋅
⋅
−
⋅
⋅
⋅
⋅
−
⋅
⋅
⋅
⋅
⋅
−
=
−
⋅
⋅
=
∆
+
⋅
∑
∫
∆
U
KŁADY STATYCZNIE NIEWYZNACZALNE
Politechnika Poznańska
Adam Łodygowski ®
11
Siły wewnętrzne od wpływu temperatur:
Schemat podstawowy przyjęto jak w poprzednim zadaniu:
[
]
C
t
kNm
EI
C
t
C
t
m
h
C
t
C
t
C
t
C
t
m
g
t
d
0
0
2
0
0
0
0
0
5
0
0
20
"
4
,
672
0
'
10
12
,
0
0
"
10
10
2
,
1
40
'
30
=
=
=
=
=
=
∆
−
=
⋅
=
=
∆
=
−
α
Delty od temperatur obliczam według wzoru:
∫
∫
+
∆
=
∆
ds
t
N
ds
h
t
M
t
i
t
i
it
0
α
α
M
1
N
1
4397893
,
0
10
12
12
,
0
40
12
6
10
12
12
,
0
40
6
40
2
1
2
6
6
0
−
=
⋅
⋅
⋅
⋅
+
⋅
⋅
⋅
⋅
⋅
⋅
−
=
+
∆
=
∆
−
−
∫
∫
ds
t
N
ds
h
t
M
t
i
t
i
it
α
α
U
KŁADY STATYCZNIE NIEWYZNACZALNE
Politechnika Poznańska
Adam Łodygowski ®
12
M
2
N
2
(
)
0
0
0
=
=
∆
=
+
∆
=
∆
∫
∫
t
i
symetria
ds
t
N
ds
h
t
M
t
i
t
i
it
α
α
U
KŁADY STATYCZNIE NIEWYZNACZALNE
Politechnika Poznańska
Adam Łodygowski ®
13
M
3
N
3
[
]
146738
,
0
20
10
2
,
1
6
1
10
12
12
,
0
40
6
2
5
2
10
12
12
,
0
40
1
40
2
1
2
5
6
6
0
−
=
⋅
⋅
⋅
⋅
−
⋅
⋅
⋅
⋅
⋅
−
⋅
⋅
⋅
⋅
⋅
⋅
−
=
+
∆
=
∆
−
−
−
∫
∫
ds
t
N
ds
h
t
M
t
i
t
i
it
α
α
Układ równań kanonicznych:
=
∆
+
⋅
+
⋅
+
⋅
=
∆
+
⋅
+
⋅
+
⋅
=
∆
+
⋅
+
⋅
+
⋅
0
0
0
3
3
33
2
32
1
31
2
3
23
2
22
1
21
1
3
13
2
12
1
11
t
t
t
X
X
X
X
X
X
X
X
X
δ
δ
δ
δ
δ
δ
δ
δ
δ
U
KŁADY STATYCZNIE NIEWYZNACZALNE
Politechnika Poznańska
Adam Łodygowski ®
14
Podstawiamy obliczone delty od wpływu temperatur:
(
)
(
)
(
)
(
)
( )
(
)
(
)
[ ]
[ ]
[ ]
kN
X
kN
X
kN
X
EI
X
X
X
EI
X
X
X
EI
X
X
X
5921
,
0
0
6184
,
0
0
146738
,
0
42
10
3
4
0
90
10
8
0
0
0
504
10
48
0
0
439789
,
0
90
10
8
0
216
10
48
3
2
1
3
2
1
3
2
1
3
2
1
=
=
=
=
⋅
−
⋅
+
⋅
+
⋅
+
⋅
+
⋅
=
⋅
+
⋅
+
⋅
+
⋅
+
⋅
=
⋅
−
⋅
+
⋅
+
⋅
+
⋅
+
⋅
Wykres końcowy od wpływu temperatury:
M
t
T
t
U
KŁADY STATYCZNIE NIEWYZNACZALNE
Politechnika Poznańska
Adam Łodygowski ®
15
T
t
M
i
U
KŁADY STATYCZNIE NIEWYZNACZALNE
Politechnika Poznańska
Adam Łodygowski ®
16
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Obliczam zadane przemieszczenie
Korzystam z twierdzenia redukcyjnego. Wykorzystuję końcowy wykres momentów dla
układu statycznie niewyznaczalnego i rysuję wykres momentów od przyłożonej jednostkowej
siły wirtualnej dla schematu zastępczego.
U
KŁADY STATYCZNIE NIEWYZNACZALNE
Politechnika Poznańska
Adam Łodygowski ®
17
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