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
≔
T
1
⎛⎝t
1
⎞⎠
0
≔
T
2
⎛⎝t
2
⎞⎠
R
A
≔
M
1
⎛⎝t
1
⎞⎠
0
≔
M
2
⎛⎝t
2
⎞⎠
⋅
R
A
⎛⎝ −
t
2
2.4
⎞⎠
=
M
2
((5.4
))
150
⋅
=
T
2
((5.4
))
50
10
15
20
25
30
35
40
45
0
5
50
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
⎞⎠ ((
))
-140
-120
-100
-80
-60
-40
-20
-180
-160
0
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
⎞⎠ (( ⋅
))
3. Belka skręcana
≔
K
s
=
⋅
―
1
3
⎛
⎝
+
⋅
((0.7
))
3
((
+
20
40
))
⋅
((0.8
))
3
((
+
20
⋅
2 20
))
⎞
⎠
17.1
4
≔
E
210000
≔
ν
0.3
≔
E
1
=
―――
E
−
1
ν
2
230769.231
≔
G
=
―――
E
2 (( +
1
ν))
80769.231
≔
m
x
=
⋅
−
q 10
−1.5
≔
M
x
=
⋅
P 10
5
⋅
≔
α
=
‾‾‾‾‾‾
―――
⋅
G K
s
⋅
E
1
J
ω
0.316 ―
1
Układ równań
≔
Y
1
((ξ))
cosh ((ξ))
≔
Y
2
((ξ))
sinh ((ξ))
≔
Y
3
((ξ))
−
1
cosh ((ξ))
≔
Y
4
((ξ))
−
ξ
sinh ((ξ))
≔
Y
5
((ξ))
−
⋅
―
1
2
ξ
2
cosh ((ξ))
ξ
6
1 894
≔
ξ
=
⋅
α 6
1.894
Guess Values
Constraints
Solver
≔
M
x0
⋅
5
≔
Θ'
0
―
1
＝
0
+
−
+
⋅
⋅
Θ'
0
―
1
α
Y
2
((ξ))
⋅
⋅
M
x0
―――
1
⋅
⋅
G K
s
α
Y
4
((ξ))
⋅
⋅
M
x
―――
1
⋅
⋅
G K
s
α
Y
4
⎛
⎜⎝
−
ξ
⋅
⋅
α ―
4
10
6
m
⎞
⎟⎠
⋅
⋅
m
x
――――
1
⋅
⋅
G K
s
α
2
⎛
⎜⎝
−
Y
5
⎛
⎜⎝
−
ξ
⋅
⋅
α ―
7
10
6
m
⎞
⎟⎠
Y
5
((ξ))
⎞
⎟⎠
＝
0
+
−
+
⋅
Θ'
0
Y
1
((ξ))
⋅
⋅
M
x0
――
1
⋅
G K
s
Y
3
((ξ))
⋅
M
x
――
1
⋅
G K
s
Y
3
⎛
⎜⎝
−
ξ
⋅
⋅
α ―
4
10
6
m
⎞
⎟⎠
⋅
⋅
m
x
―――
1
⋅
⋅
G K
s
α
⎛
⎜⎝
−
Y
4
⎛
⎜⎝
−
ξ
⋅
⋅
α ―
7
10
6
m
⎞
⎟⎠
Y
4
((ξ))
⎞
⎟⎠
≔
=
⎛⎝
,
M
x0
Θ'
0
⎞⎠
−1094.474
0.001 ―
1
⎡
⎢
⎢
⎣
⎤
⎥
⎥
⎦
≔
M
x0
=
0
−1.094
⋅
≔
Θ'
0
=
1
0.001 ―
1
≔
x
,
‥
0
0.05
6
Siły przekrojowe
≔
Θ ((ξ))
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
if
<
≤
0
ξ
⋅
⋅
α 6
―
4
10
‖
‖
‖
‖
+
+
⋅
⋅
Θ'
0
―
1
α
Y
2
((ξ))
⋅
⋅
M
x0
―――
1
⋅
⋅
G K
s
α
Y
4
((ξ))
⋅
⋅
m
x
――――
1
⋅
⋅
G K
s
α
2
⎛⎝
−
Y
5
((0))
Y
5
((ξ))⎞⎠
|
|
|
|
|
if
<
≤
⋅
⋅
α 6
―
4
10
ξ
⋅
⋅
α 6
―
7
10
‖
‖
‖
‖
−
+
+
⋅
⋅
Θ'
0
―
1
α
Y
2
((ξ))
⋅
⋅
M
x0
―――
1
⋅
⋅
G K
s
α
Y
4
((ξ))
⋅
⋅
m
x
――――
1
⋅
⋅
G K
s
α
2
⎛⎝
−
Y
5
((0))
Y
5
((ξ))⎞⎠
⋅
⋅
M
x
―――
1
⋅
⋅
G K
s
α
Y
4
⎛
⎜⎝
−
ξ
⋅
⋅
α ―
4
10
6
⎞
⎟⎠
|
|
|
|
|
if
<
≤
⋅
⋅
α 6
―
7
10
ξ
⋅
α 6
‖
‖
‖
‖
−
+
+
⋅
⋅
Θ'
0
―
1
α
Y
2
((ξ))
⋅
⋅
M
x0
―――
1
⋅
⋅
G K
s
α
Y
4
((ξ))
⋅
⋅
m
x
――――
1
⋅
⋅
G K
s
α
2
⎛
⎜⎝
−
Y
5
⎛
⎜⎝
−
ξ
⋅
⋅
α ―
7
10
6
⎞
⎟⎠
Y
5
((ξ))
⎞
⎟⎠
⋅
⋅
M
x
―――
1
⋅
⋅
G K
s
α
Y
4
⎛
⎜⎝
−
ξ
⋅
⋅
α ―
4
10
6
⎞
⎟⎠
≔
B ((ξ))
‖
‖
‖
|
|
|
|
|
if
<
≤
0
ξ
⋅
⋅
α 6
―
4
10
B (ξ)
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
if
<
≤
0
ξ
α 6
10
‖
‖
‖
‖
−
+
⋅
⋅
−
Θ'
0
――
⋅
G K
s
α
Y
2
((ξ))
⋅
⋅
M
x0
―
1
α
Y
2
((ξ))
⋅
⋅
m
x
――
1
α
2
⎛⎝
−
Y
3
((0))
Y
3
((ξ))⎞⎠
|
|
|
|
|
if
<
≤
⋅
⋅
α 6
―
4
10
ξ
⋅
⋅
α 6
―
7
10
‖
‖
‖
‖
−
−
+
⋅
⋅
−
Θ'
0
――
⋅
G K
s
α
Y
2
((ξ))
⋅
⋅
M
x0
―
1
α
Y
2
((ξ))
⋅
⋅
m
x
――
1
α
2
⎛⎝
−
Y
3
((0))
Y
3
((ξ))⎞⎠
⋅
⋅
M
x
―
1
α
Y
2
⎛
⎜⎝
−
ξ
⋅
⋅
α ―
4
10
6
⎞
⎟⎠
|
|
|
|
|
if
<
≤
⋅
⋅
α 6
―
7
10
ξ
⋅
α 6
‖
‖
‖
‖
−
−
+
⋅
⋅
−
Θ'
0
――
⋅
G K
s
α
Y
2
((ξ))
⋅
⋅
M
x0
―
1
α
Y
2
((ξ))
⋅
⋅
m
x
――
1
α
2
⎛
⎜⎝
−
Y
3
⎛
⎜⎝
−
ξ
⋅
⋅
α ―
7
10
6
⎞
⎟⎠
Y
3
((ξ))
⎞
⎟⎠
⋅
⋅
M
x
―
1
α
Y
2
⎛
⎜⎝
−
ξ
⋅
⋅
α ―
4
10
6
⎞
⎟⎠
=
B (( ⋅
α 5.4
))
−0.12084
⋅
2
≔
M
x.
((ξ))
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
if
<
≤
0
ξ
⋅
⋅
α 6
―
4
10
‖
‖
‖
−
M
x0
⋅
m
x
―
ξ
α
|
|
|
|
|
if
<
≤
⋅
⋅
α 6
―
4
10
ξ
⋅
⋅
α 6
―
7
10
‖
‖
‖
−
−
M
x0
⋅
m
x
―
ξ
α
M
x
|
|
|
|
|
if
<
≤
⋅
⋅
α 6
―
7
10
ξ
⋅
α 6
‖
‖
‖
−
−
M
x0
⋅
⋅
m
x
―
7
10
6
M
x
≔
M
s
((ξ))
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
if
<
≤
0
ξ
⋅
⋅
α 6
―
4
10
‖
‖
‖
+
+
⋅
⋅
⋅
Θ'
0
G K
s
Y
1
((ξ))
⋅
M
x0
Y
3
((ξ))
⋅
⋅
m
x
―
1
α
⎛⎝
−
Y
4
((0))
Y
4
((ξ))⎞⎠
|
|
|
|
|
if
<
≤
⋅
⋅
α 6
―
4
10
ξ
⋅
⋅
α 6
―
7
10
‖
‖
‖
−
+
+
⋅
⋅
⋅
Θ'
0
G K
s
Y
1
((ξ))
⋅
M
x0
Y
3
((ξ))
⋅
⋅
m
x
―
1
α
⎛⎝
−
Y
4
((0))
Y
4
((ξ))⎞⎠
⋅
M
x
Y
3
⎛
⎜⎝
−
ξ
⋅
⋅
α ―
4
10
6
⎞
⎟⎠
|
|
|
|
|
if
<
≤
⋅
⋅
α 6
―
7
10
ξ
⋅
α 6
‖
‖
‖
−
+
+
⋅
⋅
⋅
Θ'
0
G K
s
Y
1
((ξ))
⋅
M
x0
Y
3
((ξ))
⋅
⋅
m
x
―
1
α
⎛
⎜⎝
−
Y
4
⎛
⎜⎝
−
ξ
⋅
⋅
α ―
7
10
6
⎞
⎟⎠
Y
4
((ξ))
⎞
⎟⎠
⋅
M
x
Y
3
⎛
⎜⎝
−
ξ
⋅
⋅
α ―
4
10
6
⎞
⎟⎠
=
M
s
(( ⋅
α 5.4
))
−0.003506
⋅
≔
M
ω
((ξ))
‖
‖
‖
|
|
|
|
|
if
<
≤
0
ξ
⋅
⋅
α 6
―
4
10
M
ω
(ξ)
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
‖
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
if
<
≤
0
ξ
α 6
10
‖
‖
‖
+
+
⋅
⋅
⋅
−
Θ'
0
G K
s
Y
1
((ξ))
⋅
M
x0
Y
1
((ξ))
⋅
⋅
m
x
―
1
α
⎛⎝
−
Y
2
((0))
Y
2
((ξ))⎞⎠
|
|
|
|
|
if
<
≤
⋅
⋅
α 6
―
4
10
ξ
⋅
⋅
α 6
―
7
10
‖
‖
‖
−
+
+
⋅
⋅
⋅
−
Θ'
0
G K
s
Y
1
((ξ))
⋅
M
x0
Y
1
((ξ))
⋅
⋅
m
x
―
1
α
⎛⎝
−
Y
2
((0))
Y
2
((ξ))⎞⎠
⋅
M
x
Y
1
⎛
⎜⎝
−
ξ
⋅
⋅
α ―
4
10
6
⎞
⎟⎠
|
|
|
|
|
if
<
≤
⋅
⋅
α 6
―
7
10
ξ
⋅
α 6
‖
‖
‖
−
+
+
⋅
⋅
⋅
−
Θ'
0
G K
s
Y
1
((ξ))
⋅
M
x0
Y
1
((ξ))
⋅
⋅
m
x
―
1
α
⎛
⎜⎝
−
Y
2
⎛
⎜⎝
−
ξ
⋅
⋅
α ―
7
10
6
⎞
⎟⎠
Y
2
((ξ))
⎞
⎟⎠
⋅
M
x
Y
1
⎛
⎜⎝
−
ξ
⋅
⋅
α ―
4
10
6
⎞
⎟⎠
=
M
ω
(( ⋅
α 5.4
))
0.20903
⋅
Wykresy funkcji
0.002
0.003
0.004
0.005
0.006
0.007
0.008
0.009
0
0.001
0.01
1.2
1.8
2.4
3
3.6
4.2
4.8
5.4
0
0.6
6
x ((
))
Θ (( ⋅
x α))
0
0.25
0.5
0.75
1
1.25
1.5
1.75
-0.5
-0.25
2
1.2
1.8
2.4
3
3.6
4.2
4.8
5.4
0
0.6
6
x ((
))
B (( ⋅
x α)) ⎛⎝
⋅
2
⎞⎠
1
1.5
2
2.5
100
150
200
250
-1.5
-1
-0.5
0
0.5
-2.5
-2
1.2
1.8
2.4
3
3.6
4.2
4.8
5.4
0
0.6
6
x ((
))
M
x.
(( ⋅
x α)) ((
⋅
))
-150
-100
-50
0
50
100
-250
-200
1.2
1.8
2.4
3
3.6
4.2
4.8
5.4
0
0.6
6
x ((
))
M
ω
(( ⋅
x α)) ((
⋅
))
-1.5
-1
-0.5
0
0.5
1
1.5
2
-2.5
-2
2.5
1.2
1.8
2.4
3
3.6
4.2
4.8
5.4
0
0.6
6
x ((
))
+
M
ω
(( ⋅
x α))
M
s
(( ⋅
x α)) ((
⋅
))
-4.5
-3
-1.5
0
1.5
3
4.5
6
-7.5
-6
7.5
1.2
1.8
2.4
3
3.6
4.2
4.8
5.4
0
0.6
6
x ((
))
M
s
(( ⋅
x α)) ((
⋅
))
4. Naprężenia normalne
≔
M
y
⋅
−31.718925
≔
M
z
⋅
150
≔
B
⋅
−0.12084
2
≔
J
y
10331.5582
4
≔
J
z
11933.3030
4
≔
J
ω
600906.7902
6
Punkt 1
≔
y
10
≔
z
−3.56
≔
ω
−23.46
2
y
10
=
⋅
――
M
y
J
y
z
10.93
=
⋅
−――
M
z
J
z
y
−125.7
=
⋅
―
B
J
ω
ω
0.47
≔
σ
x
=
――――→
+
−
⋅
――
M
y
J
y
z
⋅
――
M
z
J
z
y
⋅
―
B
J
ω
ω
,
explicit ALL
+
−
⋅
―――――――
⋅
−31.718925
10331.5582
cm
4
−3.56
cm
⋅
――――――
⋅
150
kN m
11933.3030
cm
4
10
cm
⋅
――――――
⋅
−0.12084
kN m
2
600906.7902
cm
6
−23.46
cm
2
−114.297
Punkt 2
≔
z
−13.56
≔
y
10
≔
ω
−123.46
2
=
⋅
――
M
y
J
y
z
41.63
=
⋅
−――
M
z
J
z
y
−125.7
=
⋅
―
B
J
ω
ω
2.48
≔
σ
x
=
――――→
+
−
⋅
――
M
y
J
y
z
⋅
――
M
z
J
z
y
⋅
―
B
J
ω
ω
,
explicit ALL
+
−
⋅
―――――――
⋅
−31.718925
10331.5582
cm
4
−13.56
cm
⋅
――――――
⋅
150
kN m
11933.3030
cm
4
10
cm
⋅
――――――
⋅
−0.12084
kN m
2
600906.7902
cm
6
−123.46
cm
2
−81.585
Punkt 3
≔
z
−13.56
≔
y
20
≔
ω
−46.93
2
=
⋅
――
M
y
J
y
z
41.63
=
⋅
−――
M
z
J
z
y
−251.4
=
⋅
―
B
J
ω
ω
0.94
≔
σ
x
=
――――→
+
−
⋅
――
M
y
J
y
z
⋅
――
M
z
J
z
y
⋅
―
B
J
ω
ω
,
explicit ALL
+
−
⋅
―――――――
⋅
−31.718925
10331.5582
cm
4
−13.56
cm
⋅
――――――
⋅
150
kN m
11933.3030
cm
4
20
cm
⋅
――――――
⋅
−0.12084
kN m
2
600906.7902
cm
6
−46.93
cm
2
−208.823
Punkt 4
≔
z
16.44
≔
y
10
≔
ω
176.54
2
=
⋅
――
M
y
J
y
z
−50.47
=
⋅
−――
M
z
J
z
y
−125.7
=
⋅
―
B
J
ω
ω
−3.55
≔
σ
x
=
――――→
+
−
⋅
――
M
y
J
y
z
⋅
――
M
z
J
z
y
⋅
―
B
J
ω
ω
,
explicit ALL
+
−
⋅
―――――――
⋅
−31.718925
10331.5582
cm
4
16.44
cm
⋅
――――――
⋅
150
kN m
11933.3030
cm
4
10
cm
⋅
――――――
⋅
−0.12084
kN m
2
600906.7902
cm
6
176.54
cm
2
−179.721
Punkt 5
≔
z
16.44
≔
y
20
≔
ω
−46.93
2
M
y
50 47
M
z
251 4
B
0 94
=
⋅
――
M
y
J
y
z
−50.47
=
⋅
−――
M
z
J
z
y
−251.4
=
⋅
―
B
J
ω
ω
0.94
≔
σ
x
=
――――→
+
−
⋅
――
M
y
J
y
z
⋅
――
M
z
J
z
y
⋅
―
B
J
ω
ω
,
explicit ALL
+
−
⋅
―――――――
⋅
−31.718925
10331.5582
cm
4
16.44
cm
⋅
――――――
⋅
150
kN m
11933.3030
cm
4
20
cm
⋅
――――――
⋅
−0.12084
kN m
2
600906.7902
cm
6
−46.93
cm
2
−300.926
Punkt 6
≔
z
−3.56
≔
y
−10
≔
ω
23.46
2
=
⋅
――
M
y
J
y
z
10.93
=
⋅
−――
M
z
J
z
y
125.7
=
⋅
―
B
J
ω
ω
−0.47
≔
σ
x
=
――――→
+
−
⋅
――
M
y
J
y
z
⋅
――
M
z
J
z
y
⋅
―
B
J
ω
ω
,
explicit ALL
+
−
⋅
―――――――
⋅
−31.718925
10331.5582
cm
4
−3.56
cm
⋅
――――――
⋅
150
kN m
11933.3030
cm
4
−10
cm
⋅
――――――
⋅
−0.12084
kN m
2
600906.7902
cm
6
23.46
cm
2
136.156
Punkt 7
≔
z
−13.56
≔
y
−10
≔
ω
123.46
2
=
⋅
――
M
y
J
y
z
41.63
=
⋅
−――
M
z
J
z
y
125.7
=
⋅
―
B
J
ω
ω
−2.48
≔
σ
x
=
――――→
+
−
⋅
――
M
y
J
y
z
⋅
――
M
z
J
z
y
⋅
―
B
J
ω
ω
,
explicit ALL
+
−
⋅
―――――――
⋅
−31.718925
10331.5582
cm
4
−13.56
cm
⋅
――――――
⋅
150
kN m
11933.3030
cm
4
−10
cm
⋅
――――――
⋅
−0.12084
kN m
2
600906.7902
cm
6
123.46
cm
2
164.846
Punkt 8
≔
z
−13.56
≔
y
−20
≔
ω
46.93
2
=
⋅
――
M
y
J
y
z
41.63
=
⋅
−――
M
z
J
z
y
251.4
=
⋅
―
B
J
ω
ω
−0.94
≔
σ
x
=
――――→
+
−
⋅
――
M
y
J
y
z
⋅
――
M
z
J
z
y
⋅
―
B
J
ω
ω
,
explicit ALL
+
−
⋅
―――――――
⋅
−31.718925
10331.5582
cm
4
−13.56
cm
⋅
――――――
⋅
150
kN m
11933.3030
cm
4
−20
cm
⋅
――――――
⋅
−0.12084
kN m
2
600906.7902
cm
6
46.93
cm
2
292.084
Punkt 9
≔
z
16.44
≔
y
−10
≔
ω
−176.54
2
=
⋅
――
M
y
J
y
z
−50.47
=
⋅
−――
M
z
J
z
y
125.7
=
⋅
―
B
J
ω
ω
3.55
≔
σ
x
=
――――→
+
−
⋅
――
M
y
J
y
z
⋅
――
M
z
J
z
y
⋅
―
B
J
ω
ω
,
explicit ALL
+
−
⋅
―――――――
⋅
−31.718925
10331.5582
cm
4
16.44
cm
⋅
――――――
⋅
150
kN m
11933.3030
cm
4
−10
cm
⋅
――――――
⋅
−0.12084
kN m
2
600906.7902
cm
6
−176.54
cm
2
78.776
Punkt 10
≔
z
16.44
≔
y
−20
≔
ω
46.93
2
=
⋅
――
M
y
J
y
z
−50.47
=
⋅
−――
M
z
J
z
y
251.4
=
⋅
―
B
J
ω
ω
−0.94
≔
σ
x
=
――――→
+
−
⋅
――
M
y
J
y
z
⋅
――
M
z
J
z
y
⋅
―
B
J
ω
ω
,
explicit ALL
+
−
⋅
―――――――
⋅
−31.718925
10331.5582
cm
4
16.44
cm
⋅
――――――
⋅
150
kN m
11933.3030
cm
4
−20
cm
⋅
――――――
⋅
−0.12084
kN m
2
600906.7902
cm
6
46.93
cm
2
199.981
5. Naprężenia styczne
Momenty statyczne, Sy:
≔
S
y5
0
≔
S
y3
0
≔
S
y4
=
⋅
⋅
0.7
16.44
10
115.08
3
≔
S
y2
=
⋅
⋅
−0.8
13.56
10
−108.48
3
≔
S
y12
=
−
S
y2
⋅
⋅
0.8
―――――――
+
13.56
3.56
2
10
−176.96
3
≔
S
y14
=
−
+
S
y4
⋅
⋅
⋅
0.7
0.5 16.44
16.44
⋅
⋅
⋅
0.7
0.5 3.56
3.56
205.24
3
≔
S
y1
=
+
S
y14
S
y12
28.28
3
Momenty statyczne, Sz:
≔
S
z5
0
≔
S
y3
0
≔
S
z4
=
⋅
⋅
⋅
0.7
0.5 ((
+
20
10
)) 10
105
3
≔
S
z2
=
⋅
⋅
⋅
0.8
0.5 ((
+
20
10
)) 10
120
3
≔
S
z12
=
+
S
z2
⋅
⋅
0.8
10
10
200
3
≔
S
z14
=
+
S
z4
⋅
⋅
0.7
10
20
245
3
≔
S
z1
=
+
S
z14
S
z12
445
3
≔
S
z0
=
+
S
z1
⋅
⋅
⋅
0.8
0.5 10
10
485
3
Momenty statyczne, Sw:
≔
S
ω5
0
≔
S
ω3
0
≔
S
ω4
=
⋅
⋅
0.7
0.5 10
⎛⎝
−
176.54
2
46.93
2
⎞⎠ 453.64
4
≔
S
ω2
=
⋅
⋅
−0.8
――――――――
+
46.93
2
123.46
2
2
10
−681.56
4
≔
S
ω12
=
−
S
ω2
⋅
⋅
0.8
――――――――
+
123.46
2
23.46
2
2
10
−1269.24
4
≔
S
ω14
=
+
S
ω4
⋅
⋅
⋅
0.5 0.7
20
⎛⎝
−
176.54
2
23.46
2
⎞⎠ 1525.2
4
≔
S
ω1
=
+
S
ω14
S
ω12
255.96
4
≔
S
ω0
=
−
S
ω1
⋅
⋅
⋅
0.5 0.8
23.46
2
10
162.12
4
Naprężenia styczne:
≔
T
z
−30.37388
≔
T
y
50
≔
M
ω
⋅
0.20903
≔
M
s
⋅
−0.003506
≔
τ
y2
=
――――
⋅
−
T
z
S
y2
⋅
J
y
0.8
−3.99
≔
τ
z2
=
――――
⋅
−
T
y
S
z2
⋅
J
z
0.8
−6.28
≔
τ
ω2
=
――――
⋅
−
M
ω
S
ω2
⋅
J
ω
0.8
0.3
≔
τ
y12
=
――――
⋅
−
T
z
S
y12
⋅
J
y
0.8
−6.5
≔
τ
z12
=
――――
⋅
−
T
y
S
z12
⋅
J
z
0.8
−10.47
≔
τ
ω12
=
――――
⋅
−
M
ω
S
ω12
⋅
J
ω
0.8
0.55
≔
τ
y4
=
――――
⋅
−
T
z
S
y4
⋅
J
y
0.7
4.83
≔
τ
z4
=
――――
⋅
−
T
y
S
z4
⋅
J
z
0.7
−6.28
≔
τ
ω4
=
――――
⋅
−
M
ω
S
ω4
⋅
J
ω
0.7
−0.23
≔
τ
y14
=
――――
⋅
−
T
z
S
y14
⋅
J
y
0.7
8.62
≔
τ
z14
=
――――
⋅
−
T
y
S
z14
⋅
J
z
0.7
−14.66
≔
τ
ω14
=
――――
⋅
−
M
ω
S
ω14
⋅
J
ω
0.7
−0.76
≔
τ
y1
=
――――
⋅
−
T
z
S
y1
⋅
J
y
0.8
1.04
≔
τ
z1
=
――――
⋅
−
T
y
S
z1
⋅
J
z
0.8
−23.31
≔
τ
ω1
=
――――
⋅
−
M
ω
S
ω1
⋅
J
ω
0.8
−0.11
≔
τ
z0
=
――――
⋅
−
T
y
S
z0
⋅
J
z
0.8
−25.4
≔
τ
ω0
=
――――
⋅
−
M
ω
S
ω0
⋅
J
ω
0.8
−0.07
Lewa strona
Prawa strona
≔
τ
2
=
+
+
τ
y2
τ
z2
τ
ω2
−9.98
≔
τ
2
=
+
+
−
τ
y2
τ
z2
τ
ω2
−2
≔
τ
12
=
+
+
τ
y12
τ
z12
τ
ω12
−16.43
≔
τ
12
=
+
+
−
τ
y12
τ
z12
τ
ω12
−3.42
≔
τ
4
=
+
+
τ
y4
τ
z4
τ
ω4
−1.68
≔
τ
4
=
+
+
−
τ
y4
τ
z4
τ
ω4
−11.34
≔
τ
14
=
+
+
τ
y14
τ
z14
τ
ω14
−6.8
≔
τ
14
=
+
+
−
τ
y14
τ
z14
τ
ω14
−24.04
22 38
24 46
≔
τ
1
=
+
+
τ
y1
τ
z1
τ
ω1
−22.38
≔
τ
1
=
+
+
−
τ
y1
τ
z1
τ
ω1
−24.46
≔
τ
0
=
+
τ
z0
τ
ω0
−25.47
≔
τ
0
=
+
τ
z0
τ
ω0
−25.47
Naprężenia w punkcie L
≔
S
yL
=
⋅
⋅
−0.8
13.56
5
−54.24
3
≔
S
zL
=
⋅
⋅
0.8
―――――
+
20
15
2
5
70
3
≔
S
ωL
=
⋅
−0.8
⎛
⎜
⎜
⎝
⋅
――――――――――――
+
――――――――
−
123.46
2
46.93
2
2
46.93
2
2
5
⎞
⎟
⎟
⎠
−170.39
4
≔
τ
yL
=
――――
⋅
−
T
z
S
yL
⋅
J
y
0.8
−1.99
≔
τ
zL
=
――――
⋅
−
T
y
S
zL
⋅
J
z
0.8
−3.67
≔
τ
ωL
=
――――
⋅
−
M
ω
S
ωL
⋅
J
ω
0.8
0.07
≔
τ
L
=
+
+
τ
yL
τ
zL
τ
ωL
−5.59
≔
z
−13.56
≔
y
15
≔
ω
=
−――――――――
+
123.46
2
46.93
2
2
−85.2
2
=
⋅
――
M
y
J
y
z
41.63
=
⋅
−――
M
z
J
z
y
−188.55
=
⋅
―
B
J
ω
ω
1.71
≔
σ
xL
=
――――→
+
−
⋅
――
M
y
J
y
z
⋅
――
M
z
J
z
y
⋅
―
B
J
ω
ω
,
explicit ALL
+
−
⋅
―――――――
⋅
−31.718925
10331.5582
cm
4
−13.56
cm
⋅
――――――
⋅
150
kN m
11933.3030
cm
4
15
cm
⋅
⋅
――――――
⋅
−0.12084
kN m
2
600906.7902
cm
6
−85.195
2
−145.204
...
≔
σ
zastL
=
‾‾‾‾‾‾‾‾‾‾
+
σ
xL
2
4
τ
L
2
145.63
Guess Values
tra
ints
≔
x
0
＝
+
−176.54
cm
2
⋅
―――――――――
⎛⎝
+
46.93
cm
2
176.54
cm
2
⎞⎠
x
0
Naprężenia w punkcie K
≔
S
yK
=
⋅
⋅
−0.7
16.44
5
−57.54
3
≔
S
zK
=
⋅
⋅
0.7
―――――
+
20
15
2
5
61.25
3
≔
S
ωK
=
⋅
⋅
−0.7
0.5 ⎛⎝
−
⋅
46.93
2
2.1
⋅
2.9
64.81
2
⎞⎠ 31.29
4
Co
n
Solver
10
cm
=
((x))
7.9
≔
τ
yK
=
――――
⋅
−
T
z
S
yK
⋅
J
y
0.7
−2.42
≔
τ
zK
=
――――
⋅
−
T
y
S
zK
⋅
J
z
0.7
−3.67
≔
τ
ωK
=
――――
⋅
−
M
ω
S
ωK
⋅
J
ω
0.7
−0.02
=
+
−176.54
2
⋅
―――――――――
⎛⎝
+
46.93
2
176.54
2
⎞⎠
10
5
−64.81
2
≔
τ
K
=
+
+
τ
yK
τ
zK
τ
ωK
−6.1
≔
z
16.44
≔
y
−15
≔
ω
−64.81
2
=
⋅
――
M
y
J
y
z
−50.47
=
⋅
−――
M
z
J
z
y
188.55
≔
σ
xK
=
――――→
+
−
⋅
――
M
y
J
y
z
⋅
――
M
z
J
z
y
⋅
―
B
J
ω
ω
,
explicit ALL
+
−
⋅
―――――――
⋅
−31.718925
10331.5582
cm
4
16.44
cm
⋅
――――――
⋅
150
kN m
11933.3030
cm
4
−15
cm
⋅
――――――
⋅
−0.12084
kN m
2
600906.7902
cm
6
−64.81
2
139.379
...
≔
σ
zastK
=
‾‾‾‾‾‾‾‾‾‾
+
σ
xK
2
4
τ
K
2
139.91
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