Tip 1 probleme partial MECANICA Problema 6
Să se determine torsorul sistemului de forţe din figură în raport cu punctul O. Să se determine sistemul echivalent cel mai simplu. Rezolvare
F6
F1
F2
F3
F4
F5
O C
DE
x
y
∑ ⋅−=⋅−°⋅⋅−°⋅⋅−⋅−°⋅⋅=⋅=
=−+=+=
−=°⋅−−°⋅−°⋅==°⋅+°⋅++°⋅−°⋅−=
mkNa,aFsinaFcosaFaFcosaFdFM
kN,),(,YXR
kN,sinFFsinFsinFY
kN,cosFcosFFcosFcosFX
iiO 78516023024304
451956103316
5610603060
331660303060
65432
2222
5621
54321
Sistemul echivalent cel mai simplu este dat de rezultanta R, aflată pe dreapta suport
ce întâlneşte axele în punctele de coordonate ),(),( BA yOBsiOxA .
ma,,
a,
R
Md
ma,,
a,
X
My
ma,,
a,
Y
Mx
O
O
B
O
A
6624519
7851
1733316
7851
945610
7851
===
=−−=−=
=−
−==
kNF
kNF
kNF
kN,F
kNFF
7
12
10
58
5
6
5
4
3
21
====
==
Ox
yM o
A
BR
d
Să se determine torsorul în O şi să se determine sistemul echivalent cel mai simplu pentru figura următoare.
y
xO
20 kN
15 kN 15 kN
10 kN
Ox
yMo
A
BR
d
Problema 10
Să se determine torsorul în O şi să se determine sistemul echivalent cel mai simplu pentru figura următoare.
y
xO
25 kN
15 kN
10 kN
15 kN
A
B
Rd
Rezolvare
kN44.2420sin1040sin2015Y
kN92.2020cos1040cos2015X
−=°⋅+°⋅−−==°⋅−°⋅+=
kN17.32R
96.1034)44.24(92.20YXR 2222
=⇒=−+=+=
mkN3.114
40cos5.12020sin410415315Mo
⋅−==°⋅⋅−°⋅⋅+⋅−⋅−=
m55.317.32
3.114
R
Md
m46.592.20
3.114
X
My
m68.444.24
3.114
Y
Mx
O
O
B
O
A
=−==
=−−=−=
=−−==
Rezolvare
kN34.191535sin2510Y
kN48.535cos2515X
=+°⋅+−==°⋅+−=
kN10.20R
07.40434.1948.5YXR2222
=⇒=+=+=
mkN52.54
35cos125310215515Mo
⋅==°⋅⋅−⋅−⋅+⋅=
m71.210.20
52.54
R
Md
m95.948.5
52.54
X
My
m82.234.19
52.54
Y
Mx
O
O
B
O
A
===
−=−=−=
===
Problema 11
Să se determine torsorul în O şi să se determine sistemul echivalent cel mai simplu pentru figura următoare.
y
x
10 kN
50 kN
30 kN
10 kN
O
Problema 12
Să se determine torsorul în O şi să se determine sistemul echivalent cel mai simplu pentru figura următoare.
x
y
50 kN
10 kN
15 kN
20 kN
O
Rezolvare
kN78.030sin5025sin1030Y
kN24.4430cos5025cos1010X
=°⋅−°⋅−=−=°⋅−°⋅+−=
kN25.44R
79.1957)78.0()24.44(YXR 2222
=⇒=+−=+=
mkN29.17
25cos5.2105.210
30sin55030cos5.1505.230Mo
⋅==°⋅⋅−⋅+
+°⋅⋅−°⋅⋅+⋅=
m39.025.44
29.17
R
Md
m39.024.44
29.17
X
My
m17.2278.0
29.17
Y
Mx
O
O
B
O
A
===
=−
−=−=
===
Rezolvare
kN15.30
25sin1030cos2035sin15Y
kN78.36
5025cos1030sin2035cos15X
−==°⋅−°⋅−°⋅−=
−==−°⋅−°⋅+°⋅=
kN56.47R79.2261
)15.30()78.36(YXR 2222
=⇒=
=−+−=+=
mkN32.123
25cos10225sin10230sin202
35cos15435sin152450Mo
⋅==°⋅⋅+°⋅⋅−°⋅⋅−
−°⋅⋅−°⋅⋅−⋅=
m59.256.47
32.123
R
Md
m35.378.36
32.123
X
My
m09.415.30
32.123
Y
Mx
O
O
B
O
A
===
=−
−=−=
−=−
==
Problema 13
Să se determine torsorul în O şi să se determine sistemul echivalent cel mai simplu pentru figura următoare.
y
xO
10 kN
15 kN
20 kN
25 kN
mkN96.125
52525cos105.125sin10535sin155.25.120Mo
⋅==⋅+°⋅⋅+°⋅⋅−°⋅⋅−⋅=
m8.422.26
96.125
R
Md
m62.522.23
96.125
X
My
m35.1017.12
96.125
Y
Mx
O
O
B
O
A
===
−=−=−=
===
Tip 2 Probleme partial MECANICA
Să se determine reacţiunile pentru următoarele structuri plane: Problema 1
Rezolvare
kN17.12
2525sin1035sin15Y
kN22.23
25cos102035cos15X
==+°⋅−°⋅−=
==°⋅−+°⋅=
kN22.26R28.687
)17.12()22.23(YXR 2222
=⇒=
=+=+=
( )
( )
∑
∑
∑
∑
=+⋅−⇔=
=⇒
⋅+⋅⋅=
⇔=⋅−⋅⋅−⋅
⇔=
=
⇒⋅−⋅+⋅⋅=
⇔=⋅−⋅⋅−⋅+⋅
⇔=
=⇒=−⇔=
0V8pV0V
:Verificare
kN87.41V8
354810V
03548p8V
0M
kN13.38V
8
65354810V
03548p6H8V
0M
kN5H05H0H
BA
BB
B
A
A
A
AA
B
AA
OK087.418013.38 =+−⇒
Problema 5
Problema 6
( )
( )
kN07.26V
7
3060305.122V
0305.3753205.1207V
0M
kN93.8V
7
305.12230V
0305.3755.1207V
0M
kN20H020H0H
B
B
B
A
A
A
A
B
BB
=⇒
−++=⇒
=−⋅⋅−⋅−⋅+⋅
⇔=
=⇒
−+−=⇒
=+⋅⋅−⋅+⋅
⇔=
=⇒=−⇔=
∑
∑
∑
Verificare:
( )OK03507.2693.8
075VV0VBA
=−+⇔
=⋅−+⇒=∑
( )
( )
kNm125.98M
0M30460cos2075.35.15
0M
kN10V
060cos20V
0V
kN82.24H
05.1560sin20H
0H
A
A
0
A
A
0
A
A
0
A
=⇒
=+−⋅⋅−⋅⋅−
⇔==
⇒=⋅−
⇔==⇒
=⋅−⋅−
⇔=
∑
∑
∑
Problema 7
Problema 8
( )
( )
kN42.8V
5.4
5.7375.30V
05.2325.25.435.4V
0M
kN08.5V
5.4
5.4375.3012V
025.25.435.13435.4V
0M
kN3H03H0H
B
B
B
A
A
A
A
B
AA
=
⇒+=⇒
=⋅−⋅⋅−⋅
⇔==
⇒++−=⇒
=⋅⋅−⋅−⋅+⋅
⇔=
=⇒=−⇔=
∑
∑
∑
Verificare:
( )OK05.1342.808.5
05.43VV0VBA
=−+⇔
=⋅−+⇒=∑
( )
( )kN25.5V
4
183V
036134V0M
kN75.0V4
283V
012124134V0M
0H0H
BB
BA
AA
AB
A
=⇒+=⇒
=⋅⋅−−⋅⇔=
=⇒−+−=⇒
=⋅⋅+⋅⋅−+⋅⇔=
=⇔=
∑
∑
∑
Verificare:
( )OK0625.575.0
061VV0VBA
=−+⇔
=⋅−+⇒=∑
Problema 9
Problema 10
( )
( )
kN25V
3
60135V
012305.13303V
0M
kN125V
3
375V05.25303V
0M
0H0H
B
B
B
A
A
AA
B
B
=
⇒−=⇒
=⋅⋅+⋅⋅−⋅
⇔==
⇒=⇔=⋅⋅−⋅
⇔=
=⇔=
∑
∑
∑
Verificare:
( )OK015025125
0530VV0VBA
=−+⇔
=⋅−+⇒=∑
( )
( )
kN57.40V
5.3
41.263.791.89V
025sin5.22525cos5.32505.41.1205.3V
0M
kN09.4V
5.3
41.261.12V
025sin5.22555.01.1205.3V
0M
kN57.10H025sin25H0H
B
B
B
A
A
A
A
B
AA
=
⇒−+=⇒
=°⋅⋅+°⋅⋅−⋅⋅−⋅
⇔==
⇒+−=⇒
=°⋅⋅−⋅⋅+⋅
⇔=
=⇒=°⋅−⇔=
∑
∑
∑
Verificare:
( )OK066.222257.4009.4
025cos251.120VV0VBA
=−−+⇔
=°⋅−⋅−+⇒=∑
Problema 11
Problema 12
( )
( )
kN89,38V
5
47,194V
036156015sin2035V
0M
kN21,58V
5
06,291V
036156015sin20315cos2055V
0M
kN82,84H
061515sin20H0H
B
B
B
A
A
A
A
B
A
A
−=
⇒−=⇒
=⋅⋅+−°⋅⋅−⋅
⇔==
⇒=⇔
=⋅⋅−+°⋅⋅+°⋅⋅−⋅
⇔==
=⋅−°⋅+⇔=
∑
∑
∑
Verificare:
( )OK032,1989,3821,58
015cos20VV0VBA
=−−⇔
=°⋅−−⇒=∑
( )
( )
kN66,20V
6
02,124V
036156060sin5,12060cos6206V
0M
kN38V
6
94,227V
036156060sin5,42060cos6206V
0M
kN80H061560cos20H0H
B
B
B
A
A
A
A
B
AA
−=
⇒
−=⇒
=⋅⋅+−°⋅⋅−°⋅⋅−⋅
⇔==
⇒=⇒
=⋅⋅−+°⋅⋅−°⋅⋅+⋅
⇔=
=⇒=⋅−°⋅+⇔=
∑
∑
∑
Verificare:
( )OK034,1766,2038
060sin20VV0VBA
=−−⇔
=°⋅−−⇒=∑
Problema 13
10 kN/m 10 kN/m
20 kN
Problema 14
10 kN/m
20 kN
Probleme – Grinzi cu zabrele
Să se determine eforturile din barele următoarelor grinzi cu zăbrele: 1.
100 kN50 kN 50 kN100 kN 100 kN
2.00 2.00 2.00 2.00
2.00
45°
100 kN50 kN 50 kN100 kN 100 kN
1
2
3
4
5
6
7
8
9
100 kN50 kN 50 kN100 kN 100 kN
V0
H0
V8
0
Rezolvare: Determinarea reacţiunilor cu ajutorul grinzii echivalente:
( )
( )
OKVVV
kNV
VM
kNV
VM
HH
04002002000503100500
2008
200400600400
021004100610085080
2008
200400600400
021004100610085080
00
80
8
80
0
08
0
=−+⇔=−×−−+⇒=
=+++=⇒
=×−×−×−×−×⇒=
=+++=⇒
=×−×−×−×−×⇒=
=⇒=
∑
∑
∑
∑
Metoda izolării nodurilor
1
50 kN
N10
N13
0
50
200
N02
N03
45°
N241502
N23
3N350
100
N340212.13
45° 45°
N572005
100
N54
4N46150
100 N47
45°
70.72
45°
Metoda secţiunilor
00
500500
13
1010
=⇒=
−=⇒=+⇒=
∑
∑NH
kNNNV
kNNcos.NNcosNH
kN.Nsin
NsinNV
15045172120450
1321245
150045200500
02020203
030303
=⇒°⋅=⇒=+°⋅⇒=
−=⇒
°−=⇒=°⋅−−⇒=
∑
∑
kNNNH
kNNV
15001500
00
2424
23
=⇒=+−⇒=
=⇒=
∑
∑
kNNNcossin
cos.H
kN.sin
NsinNsin.V
20004545
5045172120
727045
5004545132121000
3535
3434
−=⇒=+°⋅°
+°⋅⇒=
=°
=⇒=°⋅+°⋅−⇒=
∑
∑
kNNNH
kNNNV
20002000
10001000
5757
5454
−=⇒=+⇒=
−=⇒=+⇒=
∑
∑
kNNNcossin
cos.H
kN.sin
NsinNsin.V
15004545
504572701500
727045
500454572701000
4646
4747
=⇒=+°⋅°
+°⋅−−⇒=
=°
=⇒=°⋅+°⋅−⇒=
∑
∑
100 kN50 kN 50 kN100 kN 100 kN
2.00 2.00 2.00 2.00
2.00
45°
1
2
3
4
5
6
7
8
9
0
b
b
a
a
c c
d
d
e e
Secţ a-a
00
500500
13
1010
=⇒=
−=⇒=+⇒=
∑
∑NH
kNNNV
Secţ b-b
kNNN)M(
kN..
N.N)M(
1502
3000225022000
77212411
300041125022000
02023
03032
==⇒=⋅−⋅−⋅⇒=∑
−=−=⇒=⋅+⋅−⋅⇒=∑
Secţ c-c
kNNNH
NV
15001500
00
2424
23
=⇒=−⇒=
=⇒=
∑
∑
Secţ d-d
kN..
N.NN)M(
kNNN)M(
9270411
1000411225022000
2002
40002210045042000
3434352
35354
==⇒=⋅+⋅+⋅−⋅⇒=∑
−=−=⇒=⋅+⋅−⋅−⋅⇒=∑
Secţ. e-e
kNNNH
kNNV
20002000
1001000
5757
54
−=⇒=+⇒=
−=+⇒=
∑
∑
20 kN
40 kN
40 kN
20 kN
0
1
2 4
3
5
6 8
7
10
9
111.50 1.50 1.50 1.50 1.50 1.50
0.7
5 1.5
0 2.25
20 kN
011
40 kN 40 kN 20 kN
20 kN
0
40 kN 40 kN 20 kN
V11V0
H0
Rezolvare: Determinarea reacţiunilor cu ajutorul grinzii echivalente:
( )
( )
OK0120309002024020VV0V
kN309
6012090V
05.1403405.4209V0M
kN909
90240300180V
05.4206405.7409209V0M
0H0H
110
11
110
0
011
0
=−+⇔=−×−−+⇒=
=++=⇒
=×−×−×−×⇒=
=+++=⇒
=×−×−×−×−×⇒=
=⇒=
∑
∑
∑
∑
894.0cos
447.0sin
=α=α
N01
N020
27°
20
90
N24140
2
N21
kN140N
894.06.156894.0NN0cosNN0H
kN6.156447.0
70N0sinN90200V
02
01020102
0110
=⇒
⋅=⋅−=⇒=α⋅+⇒=
−=−=⇒=α⋅++−⇒=
∑
∑
kN140N0N1400H
0N0V
2424
21
=⇒=+−⇒=
=⇒=
∑
∑
40 N13
156.6N14
1
140
N43
4
44.75
N46
40
3
111.8620
N35
N36
20
5
67.11 N57
N56
6100
56.5540
N67
N68
45°45°
7
N79
67.11
0 N78
45°
860
N89
N8-10
0
kN86.111N11.67N75.44)1(
kN75.44N94.89N2)2()1(
)2(6.156NN
0cosNcosNcos6.1560H
)1(11.67sin
30NN
0sinNsinNsin6.156400V
1313
1414
1314
1314
1314
1314
−=⇒=−−⇒−=⇒−=⇒+
−=+⇔
=α⋅+α⋅+α⋅⇒=
=α
=−⇔
=α⋅+α⋅−α⋅+−⇒=
∑
∑
kN100N0Ncos75.441400H
kN20447.075.44N0Nsin75.440V
4646
4343
=⇒−+α⋅+−⇒=
=⋅=⇒=+α⋅−⇒=
∑
∑
kN55.56N)11.67(632.014.14N)1(
kN11.67N100N894.0N447.010
100N894.0)N632.014.14(707.0
100N894.0N707.0
0cosN45sinNcos86.1110H
)1(N632.014.14N10N447.0N707.0
0sinN45cosNsin86.11120400V
3636
353535
3535
3536
3536
35363536
3536
−=⇒−×+−=⇒
−=⇒−=++−−=++−×⇒
⇒−=+⇔
=α⋅+°⋅+α⋅⇒=+−=⇒−=−⇔
=α⋅+°⋅−α⋅+−−⇒=
∑
∑
kN40N0sin11.67sin11.67N200V
kN11.67N0cosNcos11.670H
5656
5757
=⇒=α⋅+α⋅+−−⇒=
−=⇒=α⋅+α⋅⇒=
∑
∑
kN60N
0N045cos55.561000H
kN0N
045sinN45sin55.56400V
68
68
67
67
=⇒
=++°⋅+−⇒==⇒
=°⋅+°⋅−⇒=
∑
∑
kNN
sin.Nsin.V
kN.NcosNcos.H
0
0116711670
1167011670
78
78
7979
=⇒
=⋅+−⋅−⇒=
−=⇒=⋅+⋅⇒=
∑
∑αα
αα
kNNNH
kNNsinNV
600600
000
108108
8989
=⇒=+−⇒=
=⇒=⋅⇒=
−−∑
∑ α
10
N10-9
N10-1160
11
30
N11-9
N11-10
N01 N02 N21 N24 N13 N14 N43 N46 N36 N35 N56
-156.6 140 0 140 -111.86 -44.75 20 100 -56.55 -67.11 40
N57 N67 N68 N78 N79 N89 N8-10 N9-10 N10-11 N9-11
-67.11 0 60 0 -67.11 0 60 0 60 -67.11
20 kN
40 kN
40 kN
20 kN
0
1
2 4
3
5
6 8
7
10
9
111.5 1.5 1.5 1.5 1.5 1.5
0.8
1.5
2.3
a
a
b
b
c
c
d
d
e
e
f
f
Metoda sectiunilor Sect a-a
20 kNN01
N02
90
1
227°
kNNNH
NV
600600
00
11101110
109
=⇒=+−⇒=
=⇒=
−−
−
∑
∑
kNNNcos.H
kN.NsinNV
60011670
11670300
10111011
911911
=⇒=−⋅⇒=
−=⇒=+⋅⇒=
−−
−−
∑
∑α
α
kN6.156N05.1sinN5.1205.1900)M(
kN140N075.0N5.1205.1900)M(
01012
02021
−=⇒=⋅α⋅+⋅−⋅⇒=
=⇒=⋅−⋅−⋅⇒=
∑
∑
Sect b-b
20
40
0
1
290 N24
N14
N13
4
d1
3d2
3.00
27° d1
d2
3
40
11.68
1.68
341.1d2
d68.1
2
d68.1A
m341.1447.03sin3d
2
21
134
1
=⇒⋅=⋅=
=⋅=α⋅=
Sect c-c
20 kN
40 kN
40 kN
0
1
2 4
3
5
6N46
N36
N35
d3d4
45°
90
m2d2
d68.1
2
d12.2A
m59.1d25.2d2
4
43
356
3
22
3
=⇒×=×=
=⇒=⋅
( )( )( ) kN6.56N0dN25.21005.1403405.4205.4900M
kN5.67N0dN5.1403405.4205.4900M
kN100N05.1N5.1403203900M
363365
354356
46463
−=⇒=⋅−⋅−⋅−⋅−⋅−⋅⇔=
−=⇒=⋅+⋅−⋅−⋅−⋅⇔=
=⇒=⋅−⋅−⋅−⋅⇔=
∑
∑
∑
kN86.111N0341.1N5.1403203900)M(
kN140N075.0N5.1205.1900)M(
13134
24241
−=⇒=⋅+⋅−⋅−⋅⇒=
=⇒=⋅−⋅−⋅⇒=
∑
∑
( ) kN74.44N0dN5.1N5.1403203900M14214243
−=⇒=⋅+⋅−⋅−⋅−⋅⇒=∑
Sect d-d
8
7
10
9
30
6
5
N75
N76
N86
d5
d6
45°
m2dd
m59.1dd
64
53
====
( )( )( ) 0N0dN25.2605.4300M
kN5.67N0dN5.4300M
kN60N05.1N3300M
765765
756756
86867
=⇒=⋅+⋅+⋅−⇔=
−=⇒=⋅−⋅−⇔=
=⇒=⋅+⋅−⇔=
∑
∑
∑
Sect e-e
8
7
10
9
30
N97
N98
N10-8
d7
d8
m341.1dd
m341.1dd
28
17
====
( )( )( ) 0N0dN5.1603300M
kN11.67N0dN3300M
kN60N075.0N5.1300M
987987
978978
8108109
=⇒=⋅+⋅+⋅−⇔=
−=⇒=⋅−⋅−⇔=
=⇒=⋅+⋅−⇔=
∑
∑
∑ −−
Sect f-f
10
9
30
N11-9
N11-10
11
( )( ) kN11.67N0sin5.1N5.1300M
kN60N075.0N5.1300M
91191110
101101119
−=⇒=α⋅⋅−⋅−⇔=
=⇒=⋅+⋅−⇔=
−−
−−
∑
∑
Pentru simplificare, determinarea eforturilor din tiranti se face utilizand izolare de noduri.
20 kN 20 kN
80 kN 80 kN
100 kN
3 5 791
2
4
6
8
10
1.5
0
3.0
0
2.50 2.50 2.50 2.50
20 kN 20 kN80 kN80 kN 100 kN
20 kN 20 kN80 kN80 kN 100 kN
V1
H1
V9
50 kN
42°
17°
Rezolvare: Determinarea reacţiunilor cu ajutorul grinzii echivalente:
( )
( )
OK03005.1575.1420100280220VV0V
kN5.15710
75500800200V
05.1505.28051005.780102010V0M
kN5.14210
75500800200V
05.1505.28051005.780102010V0M
kN50H0H
91
9
91
1
19
1
=−+⇔=−×−×−+⇒=
=+++=⇒
=×−×−×−×−×−×⇒=
=−++=⇒
=×+×−×−×−×−×⇒=
=⇒=
∑
∑
∑
∑
957017
287017
743042
669042
.coscos
.sinsin
.coscos
.sinsin
=°≈=°≈
=°≈=°≈
ββα
α
Metoda sectiunilor + metoda izolarii nodurilor
20 kN 20 kN
80 kN 80 kN
100 kN
3 5 791
2
4
6
8
1050 kN
42°
17°
a
a
b
b
c
c
d
d
Sect a-a
N13
N14
N24
1
2
20 kN50 kN
50
142.5
3
4
d1
d2
d3
20 kN50 kN
N24
N212
( )
kN24.161N
m67.1669.05.2sin5.2d
m15.2957.025.2cos25.2d
0dNdN5.1505.2205.25.1420M
kN44.169N
25.2N25.2505.25.1425.22075.0500)M(
kN08.52N
m44.1957.05.1cos5.1d
0dN5.1500)M(
14
3
2
3142243
13
134
24
1
1241
−=⇒
=⋅=α⋅==⋅=β⋅=
=⋅+⋅+⋅+⋅−⋅⇒=
=⇒
=⋅−⋅+⋅+⋅−⋅−⇒=−=⇒
=⋅=β⋅=
=⋅+⋅⇒=
∑
∑
∑
kN44.33N0sinNN200V212421
−=⇒=β⋅+−−⇒=∑
Sect b-b
N35
N45
N46
1
2
20 kN50 kN
50
142.5
3
4
d2
d3
80 kN
5
6
d5
170N34
N35
3 Sect c-c
N75
N85
N86
9
10
20 kN
157.5
7
8
d2
d3
80 kN
5
6
d5
( ) kN88.12N0dNdN5.2205.25.1570M
kN86.169N0dN5.28052055.1570)M(
kN78.152N025.2N5.2205.25.1570)M(
853852867
865865
75758
=⇒=⋅+⋅+⋅−⋅⇒=
−=⇒=⋅+⋅−⋅−⋅⇒=
=⇒=⋅−⋅−⋅⇒=
∑
∑
∑
( )
( )kN61.9N
0dNdN5.1505.2205.25.1420M
kN86.169N
0dN5.1505.28052055.1420M
kN44.169N
025.2N25.2505.25.1425.22075.0500)M(
45
4452463
46
5465
35
354
−=⇒
=⋅+⋅+⋅+⋅−⋅⇒=
−=⇒
=⋅+⋅+⋅−⋅−⋅⇒=
=⇒
=⋅−⋅+⋅+⋅−⋅−⇒=
∑
∑
∑
m87.2957.03cos3d
m67.1dd
m67.1669.05.2sin5.2d
m15.2957.025.2cos25.2d
5
34
3
2
=⋅=β⋅===
=⋅=α⋅==⋅=β⋅=
0N0V34
=⇒=∑
m87.2957.03cos3d
m67.1dd
m67.1669.05.2sin5.2d
m15.2957.025.2cos25.2d
5
34
3
2
=⋅=β⋅===
=⋅=α⋅==⋅=β⋅=
N68
100 kN
6N64
N65
Sect d-d
7
8
N97
N98
N10-8
9
10
20 kN
157.5
d1
d2
d3
20 kNN10-8
N10-9 10
N24 = -52.08 kN N13 = 169.44 kN N14 = -161.24 kN N21 = -33.44 kN N35 = 169.44 kN N46 = -169.86 kN N45 = -9.61 kN N34 = 0 N75 = 152.78 kN N86 = -168.86 kN N85 = 12.88 kN N65 = -100 kN N97 = 152.78 kN N8-10 = 0 N98 = -205.84 kN N9-10 = -20 kN N78 = 0
adev0cosNcosN0H
kN100N0100N0V
6864
6565
=β⋅+β⋅−⇒=∑
−=⇒=−−⇒=∑
( )kN84.205N
0dNdN5.2205.25.1570M
0N0dN0dN0)M(
kN78.152N
025.2N5.2205.25.1570)M(
98
39821087
10811085869
97
978
−=⇒
=⋅+⋅+⋅−⋅⇒=
=⇒=⋅⇒=⋅⇒=
=⇒
=⋅−⋅−⋅⇒=
−
−−
∑
∑
∑
kN20N0sinN20N0V109810109
−=⇒=β⋅+−−⇒=∑ −−−