A
B
C
2 kN
4 kN
6 kNm
4 kN/m
2m
2m
3m
3m
2m
4m
V
A
V
B
H
A
A
B
C
2 kN
4 kN
6 kNm
4 kN/m
2m
2m
3m
3m
2m
4m
V
A
=10 kN
V
B
=4 kN
H
A
=4 kN
REAKCJE 1
Zadanie 1.
Oblicz reakcje podpór układu z równań równowagi
1.
0
=
∑ x
0
]
[
4
=
+
kN
H
A
]
[
4 kN
H
A
−
=
2.
0
=
∑
A
M
0
]
[
10
]
[
6
]
[
6
]
[
4
]
[
5
,
5
])
[
3
]
/
[
4
(
]
[
2
]
[
2
=
⋅
+
+
⋅
+
⋅
⋅
−
⋅
−
m
V
kNm
m
kN
m
m
m
kN
m
kN
B
]
[
4 kN
V
B
=
3.
0
=
∑ y
0
]
[
3
]
/
[
4
]
[
2
=
+
⋅
−
−
B
A
V
m
m
kN
kN
V
0
]
[
4
]
[
3
]
/
[
4
]
[
2
=
+
⋅
−
−
kN
m
m
kN
kN
V
A
]
[
10 kN
V
A
=
Sprawdzenie:
=
⋅
+
+
⋅
⋅
+
⋅
+
⋅
−
⋅
−
=
∑
]
[
3
]
[
6
]
[
5
,
1
])
[
3
]
/
[
4
(
]
[
5
]
[
2
]
[
6
]
[
7
m
V
kNm
m
m
m
kN
m
kN
m
H
m
V
M
B
A
A
C
0
]
[
3
]
[
4
]
[
6
]
[
5
,
1
])
[
3
]
/
[
4
(
]
[
5
]
[
2
]
[
6
]
[
4
]
[
7
]
[
10
=
⋅
+
+
⋅
⋅
+
⋅
+
⋅
+
⋅
−
=
m
kN
kNm
m
m
m
kN
m
kN
m
kN
m
kN
0
=
∑
C
M
5 kNm
1 kN
2
4 kN/m
4
√
2 kN
4m
3m
2m
3m
2m
45°
A
B
C
P1
Zadanie 2.
Oblicz reakcje podpór układu z równań równowagi
Rysunek pomocniczy
5 kNm
1 kN
2 kN/m
4 kN/m
4
√
2 kN
4m
3m
2m
3m
2m
45°
A
B
C
P1
V
B
H
C
V
A
M
A
I
II
6 kN
4 kN
4 kN
4 kN
D
1.
0
=
∑ x
0
]
[
4
]
[
4
=
−
−
kN
H
kN
C
]
[
0 kN
H
C
=
2.
0
1
=
∑
I
P
M
0
]
[
1
]
[
4
]
[
2
]
[
5
]
[
1
]
[
3
]
[
2
]
[
6
=
⋅
−
⋅
−
⋅
−
⋅
+
⋅
−
m
kN
m
H
m
kN
m
V
m
kN
C
B
0
]
[
1
]
[
4
]
[
2
]
[
0
]
[
5
]
[
1
]
[
3
]
[
2
]
[
6
=
⋅
−
⋅
−
⋅
−
⋅
+
⋅
−
m
kN
m
kN
m
kN
m
V
m
kN
B
]
[
7 kN
V
B
=
3.
0
=
∑ y
0
]
[
1
]
[
6
]
[
4
=
−
+
−
+
−
kN
V
kN
kN
V
B
A
0
]
[
1
]
[
7
]
[
6
]
[
4
=
−
+
−
+
−
kN
kN
kN
kN
V
A
]
[
4 kN
V
A
=
4.
0
1
=
∑
II
P
M
0
]
[
5
]
[
4
]
[
4
]
[
4
=
+
⋅
−
⋅
+
kNm
m
kN
m
V
M
A
A
0
]
[
5
]
[
4
]
[
4
]
[
4
]
[
4
=
+
⋅
−
⋅
+
kNm
m
kN
m
kN
M
A
]
[
5 kNm
M
A
−
=
Sprawdzenie:
=
⋅
−
⋅
−
⋅
+
+
⋅
−
⋅
−
+
⋅
=
∑
]
[
1
]
[
3
]
[
1
]
[
1
]
[
5
]
[
6
]
[
4
]
[
1
]
[
4
]
[
6
m
H
m
kN
m
V
kNm
m
kN
m
kN
M
m
V
M
C
B
A
A
D
0
]
[
1
0
]
[
3
]
[
1
]
[
1
]
[
7
]
[
5
]
[
6
]
[
4
]
[
1
]
[
4
]
[
5
]
[
6
]
[
4
=
⋅
−
⋅
−
⋅
+
+
⋅
−
⋅
−
−
⋅
=
m
m
kN
m
kN
kNm
m
kN
m
kN
kNm
m
kN
0
=
∑
D
M
3m
2m
2m
2
√
2 kN
4 kNm
1 kNm
10 kN/m
45°
A
B
C
D
E
P1
2m
2m
Zadanie 3.
Oblicz reakcje podpór układu i siłę w pręcie
DE z równań równowagi
1.
0
1
=
∑
I
P
M
0
]
[
2
]
[
15
]
[
3
=
⋅
+
⋅
−
m
kN
m
V
A
]
[
10 kN
V
A
=
2.
0
1
=
∑
II
P
M
0
]
[
4
]
[
4
]
[
2
]
[
2
]
[
4
]
[
2
=
⋅
+
−
⋅
+
⋅
−
m
V
kNm
m
kN
m
kN
C
]
[
2 kN
V
C
=
3.
0
=
∑ y
0
]
[
2
]
[
15
=
−
+
−
+
kN
V
kN
V
V
C
B
A
0
]
[
2
]
[
2
]
[
15
]
[
10
=
−
+
−
+
kN
kN
kN
V
kN
B
]
[
5 kN
V
B
=
4.
0
=
∑
B
M
0
]
[
7
]
[
2
]
[
4
]
[
2
]
[
7
]
[
2
]
[
4
]
[
1
]
[
1
]
[
15
=
⋅
+
⋅
+
⋅
+
⋅
−
−
−
⋅
−
m
V
m
H
m
kN
m
kN
kNm
kNm
m
kN
C
C
0
]
[
7
]
[
2
]
[
2
]
[
4
]
[
2
]
[
7
]
[
2
]
[
4
]
[
1
]
[
1
]
[
15
=
⋅
+
⋅
+
⋅
+
⋅
−
−
−
⋅
−
m
kN
m
H
m
kN
m
kN
kNm
kNm
m
kN
C
]
[
6 kN
H
C
=
5.
0
=
∑ x
0
]
[
2
=
−
−
C
B
H
kN
H
]
[
8 kN
H
B
=
Rysunek pomocniczy
3m
2m
2m
2
√
2 kN
4 kNm
1 kNm
10 kN/m
45°
A
B
C
D
E
P1
2m
2m
I
IV
N
1
V
A
N
1
III
V
B
V
C
H
C
H
B
15 kN
2 kN
2 kN
II
