Zało
ż
enia:
EA
const
=
a
4
:=
h
2
:=
2
2
4
2
+
2
5
⋅
→
sin
α
2
2
5
⋅
:=
cos
α
4
2
5
⋅
(
)
:=
sin45
2
2
:=
cos45
2
2
:=
Układ podstawowy:
Reakcje do wykresy N
P
:
Σ
M10
V5
10
−
a
⋅
25 sin45
⋅
3
⋅
a
⋅
−
25 cos45
⋅
h
⋅
+
30 cos
α
⋅
4
⋅
a
+
5 a
⋅
48
5
⋅
5
25
2
⋅
4
−
2
−
→
:=
kN
V5 10.62741782
=
kN
Σ
Y
V10
10
25 sin45
⋅
+
30 cos
α
⋅
−
V5
+
(
)
25
2
⋅
4
12
5
⋅
5
−
8
+
→
:=
kN
V10
11.47227162
=
kN
Σ
X
H10
30 sin
α
⋅
25 cos45
⋅
+
25
2
⋅
2
6
5
⋅
+
→
:=
kN
H10
31.09407739
=
kN
1
Obliczenia sił do wykresy N
P
P10.1
V10
sin
α
5
25
2
⋅
4
12
5
⋅
5
−
8
+
⋅
→
:=
kN
P10.1 25.6527792
=
kN
P10.9
H10 P10.1 cos
α
⋅
−
54
5
⋅
5
16
−
→
:=
kN
P10.9 8.14953416
=
kN
P9.8
P10.9
54
5
⋅
5
16
−
→
:=
kN
P9.8
8.14953416
=
kN
P1.8
10
−
P10.1 sin
α
⋅
+
sin
α
5
12
5
⋅
5
25
2
⋅
4
−
2
+
⋅
−
→
:=
kN
P1.8
3.29209942
=
kN
P1.2
P10.1 cos
α
⋅
P1.8 cos
α
⋅
+
25
2
⋅
48
5
⋅
5
−
12
+
→
:=
kN
P1.2
25.88908648
=
kN
P2.8
P1.8 sin
α
⋅
25
2
⋅
4
12
5
⋅
5
−
2
−
→
:=
kN
P2.8
1.47227162
=
kN
kN
P8.7
P9.8 P1.8 cos
α
⋅
−
78
5
⋅
5
25
2
⋅
2
−
12
−
→
:=
P8.7
5.20499092
=
kN
P2.7
P2.8
sin
α
5
12
5
⋅
5
25
2
⋅
4
−
2
+
⋅
−
→
:=
kN
P2.7
3.29209942
=
kN
P2.3
P1.2 P2.7 cos
α
⋅
+
75
2
⋅
2
72
5
⋅
5
−
8
+
→
:=
kN
P2.3
28.83362971
=
kN
P3.4
P2.3 25 sin45
⋅
−
25
2
⋅
72
5
⋅
5
−
8
+
→
:=
kN
P3.4
11.15596018
=
kN
2
P6.5
V5
sin
α
5
25
2
⋅
4
48
5
⋅
5
−
2
+
⋅
−
→
:=
P6.5
23.76362867
=
kN
P4.5
P6.5 cos
α
⋅
96
5
⋅
5
25
2
⋅
2
−
4
−
→
:=
kN
P4.5
21.25483564
=
kN
P4.6
P6.5
−
sin
α
⋅
30 cos
α
⋅
+
25
2
⋅
4
12
5
⋅
5
+
2
+
→
:=
kN
P4.6
16.20539791
=
kN
P7.6
P6.5 cos
α
⋅
30 sin
α
⋅
+
126
5
⋅
5
25
2
⋅
2
−
4
−
→
:=
kN
P7.6
34.6712435
=
kN
P4.7
P4.6
sin
α
5
25
2
⋅
4
12
5
⋅
5
+
2
+
⋅
→
:=
kN
P4.7
36.23637133
=
kN
P3.7
25 sin45
⋅
25
2
⋅
2
→
:=
kN
P3.7
17.67766953
=
kN
3
Reakcje do wykresy N
3
:
Σ
M10
V5
1 2
⋅
a
⋅
5 a
⋅
2
5
→
:=
Σ
M5
V10
1 3
⋅
a
⋅
5 a
⋅
3
5
→
:=
Obliczenia sił do wykresy N
3
P10.1
V10
sin
α
3
5
⋅
5
→
:=
P10.9
P10.1 cos
α
⋅
6
5
→
:=
P1.8
P10.1
3
5
⋅
5
→
:=
P1.2
P10.1 cos
α
⋅
P10.1 cos
α
⋅
+
12
5
→
:=
P2.8
P1.8 sin
α
⋅
3
5
→
:=
P2.7
3
5
−
1
+
sin
α
2
5
⋅
5
→
:=
P2.3
P1.2 P2.7 cos
α
⋅
−
8
5
→
:=
P3.7
P2.7 sin
α
⋅
2
5
→
:=
P3.6
P3.7
sin
α
2
5
⋅
5
→
:=
4
P6.5
P3.6
2
5
⋅
5
→
:=
P7.6
P3.6 cos
α
⋅
P6.5 cos
α
⋅
+
8
5
→
:=
P4.5
P6.5 cos
α
⋅
4
5
→
:=
P8.7
P7.6 P2.7 cos
α
⋅
+
12
5
→
:=
5
Układ równa
ń
i jego rozwi
ą
zanie:
δ
11 X1
⋅
δ
12 X2
⋅
+
δ
13 X3
⋅
+
∆
1P
+
0
=
δ
21 X1
⋅
δ
22 X2
⋅
+
δ
23 X3
⋅
+
∆
2P
+
0
=
δ
31 X1
⋅
δ
32 X2
⋅
+
δ
33 X3
⋅
+
∆
3P
+
0
=
X1
7.18372635
:=
kN
X2
12.15084239
:=
kN
X3
3.47097683
:=
kN
Rozwi
ą
zanie ko
ń
cowe
Reakcje:
V2
X3 3.47097683
=
:=
kN
Σ
M10
V5
10
−
a
⋅
25 sin45
⋅
3
⋅
a
⋅
−
25 cos45
⋅
h
⋅
+
30 cos
α
⋅
4
⋅
a
+
V2 2
⋅
a
+
5 a
⋅
12.01580855
=
:=
kN
Σ
Y
V10
10
25 sin45
⋅
+
30 cos
α
⋅
−
V5
+
V2
−
9.38968552
=
:=
kN
Σ
X
H10
30 sin
α
⋅
25 cos45
⋅
+
31.09407739
=
:=
kN
Sprawdzenie zgodno
ś
ci odkształce
ń
przez obliczenie przemieszczenia w
ę
zła 5 (pionowe)
V10
3
2
:=
V2
5
2
:=
P10.1
V10
sin
α
3
5
⋅
2
→
:=
P10.9
P10.1 cos
α
⋅
3
→
:=
6
P1.8
P10.1
3
5
⋅
2
→
:=
P1.2
P10.1 cos
α
⋅
P10.1 cos
α
⋅
+
6
→
:=
P2.8
P1.8 sin
α
⋅
3
2
→
:=
P8.7
P10.9 P1.8 cos
α
⋅
+
6
→
:=
P2.7
P2.8
−
V2
+
sin
α
5
→
:=
P2.3
P1.2 P2.7 cos
α
⋅
−
4
→
:=
P3.7
P2.7 sin
α
⋅
1
→
:=
P4.7
P3.7
sin
α
5
→
:=
P6.5
1
sin
α
5
→
:=
P7.6
P6.5 cos
α
⋅
2
→
:=
P4.5
P6.5 cos
α
⋅
2
→
:=
P2.3 P4.5
−
P4.7 cos
α
⋅
−
0
→
7