1. Charakterystyki geometryczne:
≔
A
=
⋅
2 ((
+
⋅
0.8
((
+
+
10
10
10
))
⋅
0.7
((
+
20
10
))))
90
2
≔
S
y1
=
⋅
2 ((
−
⋅
0.7
((
+
⋅
20
10
⋅
⋅
0.5 20
20
))
⋅
0.8
((
+
⋅
⋅
0.5 10
10
⋅
10
10
))))
320
3
≔
z
0
=
――
S
y1
A
3.556
≔
J
z
⋅
2 ((
+
⋅
0.7
((
+
+
⋅
⋅
⋅
0.5 10
10
13.3333
⋅
⋅
⋅
0.5 20
10
16.6666
⋅
⋅
20
10
10
))
⋅
0.8
((
+
+
+
⋅
⋅
⋅
0.5 10
10
13.3333
⋅
⋅
⋅
0.5 20
10
16.6666
⋅
⋅
⋅
0.5 10
10
6.6666
⋅
10
10
≔
J
y1
=
⋅
2 ((
+
⋅
0.7
((
+
⋅
⋅
10
20
20
⋅
⋅
⋅
⋅
0.5 20
20
0.6666 20
))
⋅
0.8
((
+
⋅
⋅
⋅
0.5 10
10
6.6666
⋅
⋅
10
10
10
))))
11466.288
4
≔
J
y
=
−
J
y1
⋅
z
0
2
A
10328.51
4
Sprawdzenie
≔
J
y
=
⋅
2 ((
+
⋅
0.7
((
+
⋅
⋅
16.44
10
16.44
⋅
⋅
⋅
⋅
0.5 16.44
16.44
0.6666 16.44
))
⋅
0.8
((
+
+
⋅
⋅
⋅
⋅
0.5 13.56
13.56
0.6666 13.56
⋅
⋅
13.56
10
13.56
⋅
⋅
3.56
10
3.56
))))
10331.558
4
≔
S
y
=
⋅
2 ((
−
⋅
0.7
((
+
⋅
16.44
10
⋅
⋅
0.5 16.44
16.44
))
⋅
0.8
((
+
+
⋅
⋅
0.5 13.56
13.56
⋅
13.56
10
⋅
3.56
10
))))
−1.667
3
Obranie bieguna B i punktu K
≔
ω
1
0
≔
ω
2
=
−
ω
1
⋅
10
10
−100
2
≔
ω
3
=
+
ω
2
⋅
10
10
0
2
≔
ω
4
=
+
ω
1
⋅
10
20
200
2
≔
ω
5
=
−
ω
4
⋅
10
20
0
2
≔
z
B
−3.56
≔
J
ω.y
=
⋅
2 ⎛⎝
−
⋅
0.7
⎛⎝
+
⋅
⋅
⋅
0.5 200
2
10
13.3333
⋅
⋅
⋅
0.5 200
2
20
10
⎞⎠
⋅
0.8
⎛⎝
+
⋅
⋅
⋅
0.5 100
2
10
10
⋅
⋅
⋅
0.5 100
2
10
13.3333
⎞⎠⎞⎠ 27999.98
5
≔
z
A
=
――
J
ω.y
J
z
2.346
Obranie bieguna A i punktu K0
≔
ω
1
=
⋅
−
z
A
10
−23.464
2
≔
ω
2
=
−
ω
1
⋅
10
10
−123.464
2
=
−
176.54
⋅
―――――
+
176.54
46.93
10
3.3333
102.051
=
+
−23.46
⋅
⋅
―――――
+
23.46
176.54
20
0.3333 20
43.2
≔
ω
3
=
+
ω
2
⋅
10
⎛⎝
−
10
z
A
⎞⎠ −46.927
2
=
+
⋅
――
100
10
3.3333
23.46
56.793
10
20
176 536
2
≔
ω
4
=
+
ω
1
⋅
10
20
176.536
2
=
−
176.54
⋅
―――――
+
176.54
46.93
10
6.6666
27.561
=
+
−23.46
⋅
⋅
―――――
+
23.46
176.54
20
0.6666 20
109.86
=
+
⋅
10 6.6666
23.46
90.126
≔
ω
5
=
−
ω
4
⋅
10
⎛⎝ +
z
A
20
⎞⎠ −46.927
2
≔
J
ω
⋅
2 ⎛⎝
+
⋅
0.7
⎛⎝
−
+
+
⋅
⋅
⋅
0.5 −46.93
2
10
27.56
2
⋅
⋅
⋅
0.5 176.54
2
10
102.05
2
⋅
⋅
⋅
0.5 176.54
2
20
109.86
2
⋅
⋅
⋅
0.5 23.46
2
20
43.2
2
⎞⎠
⋅
0.8
⎛⎝
+
⋅
⋅
⋅
0.5 23.46
2
10
56.79
2
⋅
0.5 123.46
2. Belka zginana
≔
q
15 ――
≔
P
50
Belka zginana w płaszczyźnie XZ
Metoda sił - belka statycznie niewyznaczalna
Guess Values
Constraints
Solver
≔
R
B
0
＝
−
⋅
⋅
q 4.2 m ―――
4.2
m
2
⋅
R
B
6
m
0
≔
=
⎛⎝R
B
⎞⎠ 22.05
Guess Values
Cons
traints
Solver
≔
R
A
0
＝
−
⋅
⋅
q 4.2 m ((
+
2.1
m
1.8
m))
⋅
R
A
6
m
0
≔
=
⎛⎝R
A
⎞⎠ 40.95
=
⋅
―
7
10
6
4.2
=
⋅
―
3
10
6
1.8
=
−
+
⋅
q 4.2
0
=
⋅ 1.8
39.69
⋅
≔
δ
10
+
+
⋅
⋅
⋅
⋅
0.5 39.69
1.8
⎛
⎜⎝
―――――
+
4.2
0.6
6
⎞
⎟⎠
⋅
⋅
⋅
⋅
0.5 39.69
4.2
⎛
⎜
⎜
⎝
――――
⋅
4.2
―
2
3
6
⎞
⎟
⎟
⎠
⋅
⋅
⋅
―
2
3
――――
⋅
q ((4.2
))
2
8
4.2
⎛
⎜⎝
―――
2.1
6
⎞
⎟⎠
≔
δ
11
⋅
⋅
⋅
0.5 1 6
―
2
3
≔
X
=
――
δ
10
δ
11
49.943
⋅
Wyznaczenie reakcji
alue
s
≔
R
A
0
ues
≔
R
B
0
Guess
V
Constraints
Solver
＝
+
−
⋅
⋅
q 4.2 m ―――
4.2
m
2
⋅
R
B
6
m
X
0
≔
=
⎛⎝R
B
⎞⎠ 30.374
Guess Val
u
Constraints
Solver
R
A
0
＝
−
−
⋅
⋅
q 4.2 m ((
+
2.1
m
1.8
m))
⋅
R
A
6
m
X
0
≔
=
⎛⎝R
A
⎞⎠ 32.626
Siły przekrojowe
≔
t
1
,
‥
0
0.01
4.2
≔
t
2
,
‥
4.2
4.21
6
=
⋅
―
9
10
6
5.4
≔
T
1
⎛⎝t
1
⎞⎠
− ⋅
q t
1
≔
T
2
⎛⎝t
2
⎞⎠
− ⋅
q 4.2
≔
M
1
⎛⎝t
1
⎞⎠
−
⋅
t
1
⋅
⋅
q t
1
―
t
1
2
≔
M
2
⎛⎝t
2
⎞⎠
−
⋅
t
2
⋅
⋅
q 4.2
⎛⎝ −
t
2
2.1
⎞⎠
=
M
2
((5.4
))
−31.719
⋅
=
T
2
((5.4
))
−30.37388
-19.5
-13
-6.5
0
6.5
13
19.5
26
32.5
-32.5
-26
39
1.2
1.8
2.4
3
3.6
4.2
4.8
5.4
0
0.6
6
t
1
(( ))
t
2
(( ))
T
1
⎛⎝t
1
⎞⎠ ((
))
T
2
⎛⎝t
2
⎞⎠ ((
))
-18
-9
0
9
18
27
36
45
-36
-27
54
1.2
1.8
2.4
3
3.6
4.2
4.8
5.4
0
0.6
6
t
1
(( ))
t
2
(( ))
−
M
1
⎛⎝t
1
⎞⎠ (( ⋅
))
−
M
2
⎛⎝t
2
⎞⎠ (( ⋅
))
Belka zginana w płaszczyźnie XY
≔
R
A
50
≔
M
=
⋅
⋅
50
―
6
10
6
180
⋅
≔
H
0
=
⋅
―
6
10
6
3.6
≔
t
1
,
‥
0
0.01
2.4
≔
t
2
,
‥
2.4
2.41
6
T ⎛t ⎞
0
T ⎛t ⎞
R
