Lösungen - ifme.ovgu.deI_+II_+III... · Lösungen zur Aufgabensammlung Statik (Ausgabe 2001,2016)...
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Lösungen zur Technischen Mechanik - Statik - Ausgabe 2016
Transcript of Lösungen - ifme.ovgu.deI_+II_+III... · Lösungen zur Aufgabensammlung Statik (Ausgabe 2001,2016)...
Lösungen
zur
Technischen Mechanik
- Statik -
Ausgabe 2016
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Lösungen zur Aufgabensammlung Statik (Ausgabe 2001,2016) 2.1 Das zentrale ebene Kraftsystem Lösung 2.1.1
→: FRx = F1 cos 30° - F2 cos 45° - F3 cos 60° + F4 cos 30° = 286 N ↑ : FRy = F1 sin 30° + F2 sin 45° - F3 sin 60° - F4 sin 30° = - 276 N
F F F N
FF
FF
R Rx Ry
Ry
Rx
Rx
R
= + =
= = − = =
2 2 397
0 9657 0 7194tan cosR Rα α, , ,
αR = 316° Lösung 2.1.2
→: FRx = F1 cos 30° - F3 cos 60° - F4 = - 684 N ↑ : FRy = - F1 sin 30° - F2 - F3 sin 60° = - 2299 N
F F F N
FF
FF
R Rx Ry
Ry
Rx
Rx
R
= + =
= = = = −
2 2 2399
3 361 0 2851tan cosR Rα α, , , ,
αR = 253,4° Lösung 2.1.3
Ermittlung der αi aus tan iα =−−
y yx x
i
i
5
5 (unter Beachtung
des Lageplans) erforderlich! →: FRx = F1 cos 45° + F2 cos 80,6° - F3 cos 59° - F4 cos 11,3° = - 52,4 N ↑ : FRy = F1 sin 45° + F2 sin 80,6° + F3 sin 59° -F4 sin 11,3° = 86,5 N
F F F N
FF
FF
R Rx Ry
Ry
Rx
Rx
R
= + =
= = − = = −
2 2 101
1 65 0 518tan cosR Rα α, , , ,
αR = 121°
x
y
F1F2
F3
F4
α3
α2
α4
α1
x
y
F1F2F3
F4
30°60°
x
y
P1
P2
P3
P4
P5
F1
F2
F3
F4
5 5
5
5
Lösung 2.1.4
→: FHx + F1 + F2 cos 45° + F4 - F5 cos 30°- F6 cos 60° -F8 cos 45° = 0 ↑ : FHy + F2 sin 45° - F3 - F5 sin 30° - F6 sin 60° + F7 + F8 sin 45° = 0 FHx = 3360 N, FHy = - 674 N
F F F N
FF
FF
H Hx Hy
Hy
Hx
Hx
H
= + =
= = − = =
2 2 3427
0 2006 0 9804tan cosH Hα α, , , ,
αH = 348,7° Lösung 2.1.5
FRx = 0 : F1 sin α1 - F2 sin α2 = 0 ( 1 ) FRy = FR : F1 cos α1 + F2 cos α2 = FR ( 2 )
a.) aus ( 1 ) : sin sin 2α α α12
11
70005000
0 5 0 7 44 4= = ⋅ = = °FF
, , ,
aus ( 2 ) : F Ncos44,4R = ° °5000 +7000Ncos30 = 9635N b.) Elimination von α2 durch :
F FF F F
quadrieren und addierenR
1 2
1 2
sin sincos cos
1 2
1 2
α α
α α
=
− = −
: F F F F FR R12
12
222 0− ⋅ + − =cos 1α
[ ]F N F F N
FF
R11 12
12
2
7452 120812084000
0 5 0 151 8 7
= > =
= = ⋅ = = °
,
, , ,sin sin2 1 2α α α
c.)
( ) ( )
°=====
==+−
−==
+−=
60866,0sinsin2500sin
4330coscosF22cos220
cosF2
212
1212
11211
21
11
1
2
21
211
2112
αααα
αα
α∂∂
α
FFNFF
NFFFFF
FFFF
FFFFF
R
RRR
R
RR
Lösung 2.1.6
→: Fg1cosαg1 + Fg2cosαg2 + F1cosα1 - F2cosα2 - F3cosα3 + F4sinα4 = 0 ↑ : - Fg1sinαg1 + Fg2sinαg2 + F1sinα1 + F2sinα2 - F3sinα3 - F4cosα4 = 0 Lösung: Fg1 = -3694 N
Fg2 = 2658 N
x
y
F1
F2
F3F4
F5F6
F7
F8
45°
45°30°
30°
F1F2
FR
α1α2
x
y
x
y
Fg1
Fg2
αg1αg2
Lösung 2.1.7 FRx = F1cosα1 + F2cosα2 - F3sinα3 - F4cosα4 - F5cosα5 FRy = F1sinα1 - F2sinα2 - F3cosα3 - F4sinα4 + F5sinα5 FRx = 44,1N + 84,9N - 53,3N - 49,4N - 26,3N = 0 FRy = 121,2N - 49,0N - 53,3N - 28,5N + 9,6N = 0, d.h. FR = 0; Gleichgewicht Lösung 2.1.8
→: -FS1cosα1 + FS2cosα2 = 0 F FS
S2
1=cos
cos1
2
αα
↑ : FS1sinα1 + FS2sinα2 - mg = 0 FS1(sinα1 + tanα2 cosα1) = mg
cos
tan
α α
α α
1 1
2
2 2
2
44 5
0 8889 27 3
4 0626
0 677 34 1
= = = = °
=+ −
−= = = °
bl
a l bc b
,, ,
, , ,
Ergebnis: FS1 = 277N, FS2 = 297N Lösung 2.1.9
Es gilt: ( ) ( )F r h F r r hG⋅ − ≥ − −2 2 oder für die Grenzlage: F = FG cot α
( )cotα =
− −−
=−
−
r r hr h
also
F F rh hr hG
2 2
22
Lösung 2.1.10
Gleichgewicht für die ausgelenkte Lage: →: -FS sinϕ + FZ = 0 ↑ : FS cosϕ - FG = 0
Elimination von ϕ: [ ][ ]F F
F F
S Z
S G
sin
cos
ϕ
ϕ
2 2
2 2
=
=+ , also F F FS Z G= +2 2 =20590 N
ferner: ( )h l FF
FF F
G
S
G
Z G
= − = =+
12 2
cos cosϕ ϕ h l FF F
cmG
Z G
= −+
=1 75 5
2 2,
FS1FS2
FG = mgK
α1 α2
GrenzlinieF
FGFR r - h
( )r r h2 2− −
α
FG
FZ
FS ϕ
Lösung 2.1.11
← : FS1 cos α1 + FS2 cos α2 = FS cos αS ↑ : FAB = FS1 sin α1 + FS2 sin α2 + FS sin αS
tan sin cosα α α αS S S Shl
= = = = = °34
35
45
36 9,
( )F kN kNS = ⋅ + ⋅ =54
4 0 9962 8 0 9063 14 044, , , FAB = 12,156 kN
Lösung 2.1.12 Fall a: Gleichgewicht am freigeschnittenen Knoten K
→: -FS1 cosα - FS2 = 0 ↑ : FS1 sinα - FG = 0 FG = mg
ααα
-mgcot=cos;sin 121 SSS FFmgF −==
Fall b: Bilden von
r r rF F FR H G= + und Verschieben des Angriffspunktes von
rFR auf der Wir-
kungslinie nach K →: -FS1 cosα - FS2 - FH = 0 ↑ : FS1 sinα - FG = 0
( ) ( )HSHSS FFFFmgF ++−== ααα
mgcot-=cos;sin 121
Lösung 2.1.13
← : FS1 cos 45° - FN cos 60° - m2g = 0 ↑ : FS1 sin 45° + FN sin 60° - m1g = 0
( )
( ) ( ) NgmmFNgmmF
gmmFgmFF
gmFF
SN
N
NS
NS
4303132215
312
231
23
22
21
22
211
21
21
11
21
=++
==+−
=
−=+
−
=+
=−
B
FSFAB
FS1
FS2
αS
α1
α2
6m
8m
10m
αS
K
FG
FS1
FS2
α
mg
FH
FR
K
FS1
FS2mg
FH
FS1
FNm1g
m2g45°
30°
M
Lösung 2.1.14
← : FS1 cos α1 - FS2 cos α2 - F2 = 0 sinα1 + ↑ :- FS1 sin α1 - FS2 sin α2 - F1 = 0 cosα1 - FS2 (sinα1cos α2+ cos α1 sin α2)- F2 sinα1- F1 cosα1=0
F F F F F F
d h d h
S S22 1 1 1
1 2 1 21
2 2 1 2
1 2 1 2
1 1 2 22 81 3
65 1 2 81 7
58 7
= −++
=−+
= = ° = = °
sin coscos sin sin cos
sin coscos sin sin cos
tan tan
α αα α α α
α αα α α α
α α α α,,
; . . , ,,
; . . ,
FS2 = -18913N FS1 = 12312N 2.2 Das allgemeine ebene Kraftsystem Lösung 2.2.1
← : FS1 - FS2 cos 60° -FS3 = 0 ↑ : FS2 sin 60° - FG = 0 A : - FG a + FS3 a = 0
Ergebnis: F F N F F N F F NS G S G S G1 2 33 1
31577 2
31155 1000=
+= = = = =, ,
Lösung 2.2.2
↑: FS2 - FG - F2 = 0 A: FS3 2r + FG r + F2 2r - F1 r = 0 B: FS1 2r + FG r + F2 2r + F1 r = 0
Ergebnis: F F F F N F F F N F F F F NS
GS G S
G1
1 22 2 3
1 222
3500 2500 22
500= −+ +
= − = + = =− −
= −, ,
F1
F2
FS1 FS2
α1 α2
a
aFG
FS1
FS2
FS3
A S
F2
F1 FG
FS1 FS2
FS3
A
Sr
B
Lösung 2.2.3
FRx = ΣFix = F1cos60° + F2cos60° - F3cos60° + F4cos60° = 2Fcos60° = F FRy = ΣFiy = F1sin60° - F2sin60° - F3sin60° - F4sin60° = -2Fsin60° =−F 3 F F F F F F
FF
R Rx Ry
RRy
RxR
= + = + =
= = − ° = °
2 2 2 23 2
60 300tan tanα α
M M F r F r Fr
M M Fr
Fr
F r Fr
M M F r F r F r F r Fr
A iA
B iB
i
= = − − = −
= = − − + = −
= = − − − + = −
∑
∑
∑
2 3
1 3 4
0 0 1 2 3 4
3 3 3 3
23
23 3 3
23
23
23
23 2 3
Lösung 2.2.4 Schnittskizze:
A: F(r3 - r2sin60°) - FS1(r1 + r2) = 0 B: F(r3 + r2sin60°) - FS2(r1 + r2) = 0 C: Fr3 - FS3(r1 + r2) = 0 Lösung: FS1 = 8,9kN; FS2 = 55,1kN; FS3 = 32kN
Lösung 2.2.5
← : FS1 - FS2 cos 60° -FS3 cos 60°- Fcos 60° = 0 ↑ : FS2 sin 60°- FS3 sin 60°+ Fsin 60° - FG = 0
0 : - F a - FS3 a sin60°- FG .a2 = 0
Lösung:
F F F
F F F F
F F F F
S
SG
SG
1
2
3
23
116
2 33
1 87
23
1 45
= − = −
= −+ −
= −
= −+
= −
,
( ) ,
,
x
yF1
F2
F3F4
0A
B
r
r1r2
r3
A
B
CFS1
FS2
FS3
F
r2sin60°
r2sin60°60°
F
FGFS1
FS2
FS30
Lösung 2.2.6
← : FS2 cos 60° - FS3 cos 45° = 0 ↑ : FS1 + FS2 sin 60° + FS3 sin 45° - FG = 0 0 : 6FS2a sin60°+ 8FS2a cos60°-FG . 4a = 0 Lösung: FS1 = 0,406FG = 1623N FS2 = 0,435FG = 1740N FS3 = 0,308FG = 1230N
Lösung 2.2.7
← : F + FS1 cos 60° - FS3 cos 45° = 0 ↑ : - FS2 - FS1 sin 60° - FS3 sin 45° - FG = 0 0 : 2FS1a sin60° - FS1a cos60° + FG a = 0 Lösung: FS1 = - 812 N....Druckstab
FS2 = - 891 N....Druckstab FS3 = 840 N....Zugstab Lösung 2.2.8
← : -FS2 cos 30° -FS3 cos 45° = 0 ↑ : FS1 + FS2 sin 30°- FS3 sin 45° - FG = 0
0 : 2FS2a sin30°- 67
FG a = 0
Lösung: FS1 = - 0,59FG... Druckstab FS2 = 1,17FG... Zugstab
FS3 = -1,43FG... Druckstab Lösung 2.2.9
← : -FS3 cos 60° -FS2 cos 45° = 0 ↑ : -FS1 + FS3 sin 60°- FS2 sin 45° - mg = 0 0 : 6FS3a sin60° + 3FS3a cos60° - 3mga + M1 - M2 = 0 Lösung: FS1 = -160,5N... Druckstab FS2 = 32,3N... Zugstab FS3 = - 45,7N... Druckstab
FS1 FS2
0
FG
FS3
8a
S
0F
FG
FS1 FS2
FS3
S
0
FG
FS1
FS2
FS3
S
0
mgFS1 FS2
FS3
M1
M2
3a
3.1 Lagerreaktionen ebener Tragwerke Lösung 3.1.1
→ : FAH - F1 cos60° = 0 ↑ : FAV - F1 sin60° + FB - F2 = 0 A : -F1 sin60° . a + FB
. 2a - F2 . 3a = 0
Ergebnis:
F F F N
F F N
F F F N
B
AH
AV
= +
=
= =
= −
=
12
32
3 2365
12
1000
12
32
365
1 2
1
1 2
Lösung 3.1.2
→ : FAH - F cosα = 0 ↑ : -FAV - F sinα + FC = 0 A : -F sinα .
3a + FC . 2a = 0
Ergebnis:
F Fsin F kN
F Fcos F kN
F Fsin F kN
C
AH
AV
= = =
= = =
= = =
32
34
3 39
12
15
12
34
13
α
α
α
Lösung 3.1.3
→ : -FAH + FC cos45° = 0 ↑ : -FAV + FC sin45° - F = 0
A : FC sin45° . 2a - F . 3a = 0
Ergebnis: F F kN F F kN F F kNC AH AV= = = = = =32
2 106 32
75 12
25
FAH
FAV FB
F2F1 sin60°
F1 cos60°
FC
FAHFAV F sinα
F cosα
FFAHFAV FC sin45°
FC cos45°
Lösung 3.1.4
→ : FAH = 0 ↑ : FAV + FB - F - F - F - 3qa = 0
A : F . 2a +3qa . a2 - F . 2a + FB . 3a - F . 4a = 0
Ergebnis: F F F qa F F F F qa F FAH B AV= = − = = = + = =0 43
12
56
0 833 53
72
316
5 17, ,
Lösung 3.1.5
→ : FAH = 0 ↑ : FAV + FB - 2F - qa = 0
A : F a + qa . a2 + FB . 2a - 2F . 3a = 0
Ergebnis: F F F qa F qa kN F F qa kNB AV= − −
= − = = − + =12
6 12
52
14
21 12
54
15
Lösung 3.1.6
→ : FAH = 0 ↑ : FAV - F - qa = 0
A : F a + qa . a2 -MA = 0
Ergebnis: F F F qa N M Fa qa NmAH AV A= = + = = + =0 5000 12
52502
Lösung 3.1.7
→ : FAH = 0
↑ : FAV + FB - q0 .a2 - q0a = 0
A : q a a q a a F aB
002 3
23
2 0⋅ − ⋅ + ⋅ =
Ergebnis: F F q a F q aAH AV B= = =0 54
140 0
F FF3qa
FBFAH FAV
a2
2F
Faqa
FBFAH FAV
a2
Fqa
MA
FAH
FAVa2
23aq a0
2
q0 a
FBFAH FAV
a3
Lösung 3.1.8
→ : FBH - F = 0 ↑ : FA + FBV - 2q0a - 3q0a = 0
A : F a q a a q a a F aBV⋅ + ⋅ − ⋅ + ⋅ =2 3 32
2 43
4 00 0
Ergebnis: F F N F F q a N F F q a NBH AV BV= = = + = = − +
= −2002
13124
645 82
1124
145 80 0, ,
Lösung 3.1.9
→ : FBH - Fcos45° = 0 ↑ : FA + FBV - Fsin45° - 3qa = 0
B : 032
3345sin 0 =⋅−⋅+⋅°− aFaaqaF A
Ergebnis: NFqaFNFFNFqaF BVBHA 92143
22233535
223322
62
23
=+====−=
Lösung 3.1.10
→ : -FAH + 2q0a - F = 0 ↑ : FAV + FB - F- 2q0a = 0
A : − ⋅ − ⋅ + ⋅ + ⋅ =2 23
2 2 2 00 0q a a q a a F a F aB
Ergebnis:
F q a F F
F q a F F
F q a F q a F q a F F
B
AH
AV
= − =
= − =
= + − + = + =
53
23
2
2 53
13
2 73
0
0
0 0 0
Lösung 3.1.11
→ : - FAH + F3 = 0 ↑ : FAV + FB - F1 - F2 = 0 A : -F1 a - F2 3a + F3 a + FB 4a = 0
Ergebnis:
[ ] [ ]F F F F N F F N F F F F NB AH AV= + − = = = = + + =14
3 1000 4000 14
3 30001 2 3 3 1 2 3
32a
2q0a
FBV
FBHFA
43aF 3q0a
3qa
FBV32a FBHFA
Fsin45°Fcos45°
23a
2 0q a
2q0 a
FB
FAHFAV
a
F F
F3
FB
FAH
FAV
F2F1
Lösung 3.1.12
→ : FAH + 3F - FB sin30° = 0 ↑ : FAV + FB cos30° - 2F - 4F = 0 A : FB cos30° . 4a + FBsin30° . 2a - 2F . a - 3F . a - 4F . 3a = 0 Ergebnis: FB = 3,82F; FAH = - 1,09F; FAV = 2,7F
Lösung 3.1.13
→ : FAH - F1 = 0 FAH = F1 ↑ : FAV - F2 = 0 FAV = F2 A : -F1 a + F2 3a - M - MA = 0 MA = (3F2 -F1)a - M
Lösung 3.1.14
→ : -FAH + F1 + 2F1 = 0 FAH = 3F1 ↑ : FAV - 4F1 + 2F1 = 0 FAV = 2F1
A : F1 a -2F1 a - 4F1⋅a2 + MA +2F1 a = 0 MA = F1a
Lösung 3.1.15
Bestimmung von q(z): Ansatz: q(z) = A + Bz + Cz2
Bestimmung der Konstanten A, B, C: z = 0: q(0) = q0 A = q0
q’(0) = 0 B = 0
z = a: q(a) = 0 A + Ca2 = 0 C Aa
qa
= − = −202
3F FBcos30°
FAHFAV
4F
2F
FBsin30°
F1
FAH
FAV MA
F2
M
F1
FAH
FAVMA
2F1
2F1
4F1
a2
FR
zR
q(z)
FR
zR
FAH
MAFAV
q z q za
F q za
dz q z za
q a
zF
q za
zdzF
q z za q a
q a a
R
a a
RR
a
R
a
( ) = −
= −
= −
=
= ⋅ −
= ⋅ −
= ⋅ =
∫
∫
0
2
0
2
00
3
20
0
0
2
00
2 4
20 0
02
1
13
23
1 1 12 4
32 4
38
Lösung: → = ↑ = = ⋅ =: : :F F q a A M F zq a
AH AV A R R023 40
02
Lösung 3.1.16
→ : -FA + FBH = 0 ↑ : FBV - FGE - FG = 0 B : -FG . 3a -FGE . a + FA
. a = 0 Ergebnis: FA = FBH = FGE + 3 FG; FBV = FGE + FG
Lösung 3.1.17
↑ : F F m g m gA B+ − − =1 116
0
A : − ⋅ − ⋅ + ⋅ =m g a m g a FB1 13 16
8 9a 0
Ergebnis: F m g F m gB A= =1327
37541 1;
Lösung 3.1.18
↑ : FA + FB - FK - FT - FG = 0 A : FB 8a - FG 6a - FT 4a - FK 3a = 0
Ergebnis: F F F F F F F FB T K G A T K G= + + = + +12
38
34
12
58
14
;
↑ : FC + FD - FK - FG = 0 C : FD 3a - FG 6a - FK 3a = 0 Ergebnis: FD = FK + 2 FG FC = - FG
A
B
FA
FBH FBV
FG
FGE
FA FB
mgm1g
9a
6a
a
FA FB
FGFK
8a
6a4a
FT3a
FC FD
FGFK
3a
6a
Achsenbelastung der unteren Rolle:
( )
( )
F F F F F F
F F F Fa aa a
oder aa
F F
RH G RV G G G
R RH RV G
R G
= = − = − −
= + = −
=−−
= = ° = =
=
cos sin sin
sin
tan sin
α α α
α
α α α
1
2 16 26
45
38 6 441
0 6247
0 866
2 2
, ,
,
3.2 Scheibenverbindungen Lösung 3.2.1
Teil I: → : FAH - FGH = 0 ↑ : FAV + FB - FGV - F - 4qa = 0 A : -F . 2a -4qa . 2a + FB . 3a - FGV
. 4a = 0 Teil II: → : FGH = 0
↑ : FC + FGV - 2qa = 0 G : -2qa . a + FC . 2a = 0
Ergebnis: F F F qa N F F qa N
F F qa N F qa N
AH AV B
GH C GV
= = + = = + =
= = = = =
0 13
1667 23
4 5333
0 1000 1000
FS = FG
FS = FG
FG
FGFR
α
FAH FAV
2a F + 4qa
FB FGV
FGH
FGV
FGH2qa
a FC
I
II
Lösung 3.2.2
Teil I: → : FAH - FGH1 = 0 ↑ : FAV + FB - FGV1 - F1 - 2qa = 0 A : F1 . 2a -2qa . 7a + FB . 6a - FGV1
. 8a = 0 Teil II: → : FGH1 - FGH2 = 0 ↑ : FGV1 + FGV2 - 8qa = 0 G1 : -8qa . 4a + FGV2 . 8a = 0
Teil III: → : FGH2 = 0
↑ : FC + FD - FGV2 - F2 - 3qa = 0 D : F qa a F a F aGV C2 29a 3 152
6 2 0⋅ + ⋅ − ⋅ + ⋅ =
Ergebnis: FAH = 0 FB = 333 kN FGH1 = 0 FGH2 = 0 FC = 687,5 kN FAV = 116,7 kN FGV1 = 200 kN FGV2 = 200 kN FD = 262,5 kN Lösung 3.2.3
Teil I: → : FAH - FGH = 0 ↑ : FAV - FGV - F1 - 5qa = 0 A : MA - F1 . 3a - 5qa . 2,5a - FGV
. 5a = 0 Teil II: → : FGH = 0
↑ : FB + FGV - F2= 0 G : -F2 . 3a + FB . 5a = 0 Ergebnis: F F N F F F N F F F F qa N
M F a F a qa Nm
B GH GV AH AV
A
= = = = = = = + + =
= + + =
35
960 025
640 025
5 2440
3 2 252
8100
2 2 1 2
1 22
FAHFAV
4a
2qa
FB FGV1
FGH1
FGV1
FGH18qa
a
FC
I
IIFGV2
FGH2
FDFGV2
FGH23qa
1,5a
III
F1
F2
FAH FAV
2,5a 5qaF1
FGV
FGH
FGV
FGHF2
2a FB
I
II
MA
Lösung 3.2.4
→ : - FAH + F1 - FGH = 0 ↑ : FAV + FGV - F2 = 0 A : - F1 3a - F2 2a + FGH 6a + FGV 5a = 0 → : FGH - FBH = 0 ↑ : FBV - FGV - F3 = 0 B : F3 2,5a - FGH 6a + FGV 5a = 0 oder am Gesamtsystem: A : - F1 3a - F2 2a + FBV 10a - 7,5 F3 a = 0 ↑ : FAV + FBV - F2 - F3 = 0
Ergebnis: FAH = 6,3 kN; FAV = 5,65 kN; FGH = 5,7 kN; FGV = 4,35 kN; FBH = 5,7 kN; FBV = 9,35 kN Lösung 3.2.5
Teil I: → : FG1H - FG3H = 0 ↑ : FAV + FG3V - FG1V - 4qa = 0 A : FG3H 6a - 4qa 2a - FG1H 3a + FG3V 4a = 0 Teil II: → : FG3H - FG2H - FBH = 0 ↑ : FBV - FG3V - 4qa - FG2V = 0
B : 4qa 2a + FG2H 3a - FG3H 6a + FG3V 4a = 0 Teil III: → : - FG1H + FG2H = 0 ↑ : FG1V + FG2V - F1 - F2= 0 G1 : FG2V 8a - F1 2a - F2 6a = 0 Ergebnis: FAV = 28,5 kN, FBH = 0, FG1H = 10,7 kN, FG2H = 10,7 kN, FG3H = 10,7 kN FBV = 33,5 kN, FG1V = 12,5 kN, FG2V = 17,5 kN, FG3V = 0 Auflagerreaktionen am Gesamtsystem bestimmbar!
F1
F2
F3
FAHFAV
FGH
FGV
FGH FGV
FBH
FBV
FG1H
4qa 4qa
FG1V
FAV
FG3H
FG3V
FG3H
FG3V
FBH
FBV
FG2V
FG2HF1 F2FG1H
FG1V
FG2H
FG2V
I II
III
Lösung 3.2.6
Teil I: → : -FG1H - F1 + FG2H = 0 (1) ↑ : FG2V - FG1V = 0 (2) G2 : -F1 a - FG1H 2a + FG1V 3a = 0 (3) (Teil II:) → : -FG3H - FG2H + 4qa = 0 ↑ : FB + FG3V - FG2V = 0 G2 : 4qa 2a + FB 3,5a - FG3H 4a + FG3V 2a = 0
Teil III: → : FG1H + FG3H - FAH - F2 = 0 (4) ↑ : FG1V - FG3V + FAV = 0 (5) G3 : -FG1V 5a - FG1H 2a + F2 a – FAV 3,5a = 0 (6) Gesamtsystem: →: -FAH – 2F + 4qa = 0 FAH = 0 (7) ↑: FAV + FB = 0 (8) A: Fa + 3Fa + 5FBa – 8qa2 = 0 (9) FB = -FAV = 0 Ergebnis: FAH = 0, FB = 0, FG1H = - 0,5 kN, FG2H = 3,5 kN, FG3H = 4,5 kN FAV = 0, FG1V = 1 kN, FG2V = 1 kN, FG3V = 1 kN Lösung 3.2.7
→ : - FAH + FGH = 0 ↑ : FAV + FGV - F1 = 0 A : - F1 a - FGH a + FGV 2a + MA = 0 → : -FGH + F2 = 0 ↑ : FB - FGV - 1,5q0a = 0 B : -F2 a +3q0 a2 + FGV 3a = 0
Ergebnis: FAH = 3F; FAV = 3F; FB = 2F; FGH = 3F; FGV = - F; MA = 7Fa; Lösung 3.2.8
→ − =
↑ − + + =
− ⋅ + ⋅ + ⋅ =
:
:
:
F F
F F F F
G F a F a F a
GH D
B GV C D
B C D
12
2 0
12
2 0
4 12
2 2 12
2 6 0
FG1H
2a4qa
FG1V
FAV
FG2H
FG2V
FG3H
FG3VFAH
FB
FG2V
FG2H
F1
F2
FG1HFG1V
FG3H
FG3V
I II
III
FAV
FAHMA
FGV FGH
F1
F2
FB
FGH
FGV3
20q a
FC
FA
FGVFGH
FD
FFB
FGHFGV
→ − + =
↑ − + + =
⋅ − ⋅ =
:
:
:
F F
F F F
G F a F a
GH A
GV A
A
12
2 0
12
2 0
12
2 6 4 0
Ergebnis: FA = 0,942 F; FB = 1,225 F; FC = -1,558 F(Druckstab); FD = 0,942 F; FGH = 0,667 F; FGV = 0,333 F Lösung 3.2.9
→ : - FAH + FGHl = 0 ↑ : FAV - FGVl = 0 A : -FGVl a + MA = 0 → : - FGHl + FGHr = 0 ↑ : -FGVr + FGVl - F1 = 0 → : -FGHr + F2 = 0 ↑ : FB + FGVr = 0 G : F2 c + FB a = 0
Ergebnis:
;;
;;
;;
221
2121
222
acFFcFaFM
acFFF
acFFF
acFFFFFFF
GVrA
GVlAV
BGHrGHlAH
⋅=⋅+⋅=
⋅+=⋅+=
⋅−====
Lösung 3.2.10
→ : - FAH + FGH + 12
2F = 0
↑ : FAV - FGV - 12
2F = 0
A : - F a 2 - FGH a - FGV 2a + MA = 0 → : FGH = 0 ↑ : FB + FGV - 2q0a = 0 B : 2q0 a2 - FGV 2a = 0
Ergebnis (unter Verwendung von F qa= 2 ): FAH = qa; FAV = 2qa; FB = qa; FGH = 0; FGV = qa; MA = 4qa2 ;
FAV
FAH
MA FGVl FGHl
F1
F2
FB
FGHl
FGVl FGVr
FGHr
FGVrFGHr
FAV
FAHMA
FGV FGH
F
2qa
FB
FGH
FGV
a 2
a
Lösung 3.2.11
→ : FAH - FGH = 0 ↑ : FAV + FGV - F = 0 A : - F a + FGV 2a = 0
→ : FGH - FBH -q a0
2 = 0
↑ : FBV - FGV = 0
B : q a0
2
3 - FGH a + FGV a = 0
Ergebnis (unter Verwendung von q Fa0 = ):
F F F F F F F F F F F FAH AV BH BV GH GV= = = = = =
56 2
13 2
56 2
Lösung 3.2.12
→ : FAH - FS cos α = 0 ↑ : FAV + FC - FS sin α = 0 A : - FS sin α 9a + FC 12a = 0 → : FS cos α - FDH = 0 ↑ : FS sin α - FDV - F = 0 D : - F 12a + FS sin α 6a = 0 sin cosα α= = = =
63 5
25
33 5
15
aa
aa
Ergebnis: kNFFkNFF
kNFFkNFFkNFFkNFF
DVDH
CSAVAH
3;3
5,423;71,65;5,1
21;3
====
========
Lösung 3.2.13
→ : -FS - FCH = 0 ↑ : - FCV + F = 0 C : F . 200mm + FS . 350mm = 0 → : FCH - FDH = 0 ↑ : FDV + FCV - FE = 0 D : FCH . 100mm + FCV . 500mm - FE . 250mm= 0
Ergebnis: F F kN F F kN F F kN
F F kN F F kN F F kN
S CV CH
DH DV E
= − = − = = = =
= = = = = =
47
1 714 347
1 714
47
1 714 4335
3 685 7835
6 685
, ; ; ,
, ; , ; ,
FFAH
FAV
FGH
FGV
FGH FGV
FB H
FB V
q a0
223a
α
FAH
FAV
FS
FS
F
FC
FDHFDV
α
3a
6a3 5a
α
FFCH
B
CD
CE
FS
FCV
FEFDH
FDV FCH
FCV
200200150
250250
100
Lösung 3.2.14
→ : FCH - FGH - FS3 = 0 ↑ : FCV + FGV - qa = 0
G : - FS3 2a + FCH 4a + FCV 4a -qa ⋅a2 = 0
→ : FGH + FS1 12
2 = 0
↑ : FB - FGV - 5qa - FS2 - 12
2 1FS = 0
B : -5q a2 . 2,5 - FGV 5a - FS2 2a = 0
→ : FS3 - FS1 12
2 = 0
↑ : FS2 +12
2 1FS = 0
Ergebnis:
;75,34
15;75,34
15;3,524
15
;75,34
15;2;0;4
321 qaqaFqaqaFqaqaF
qaFqaqaF
qaFFqaF
SSS
GVGH
CVCHB
==−=−===
−=−=−=
===
Auflagerreaktionen am Gesamtsystem bestimmbar! Lösung 3.2.15
↑ : FDV - mg = 0 FDV = mg →: FDH - FA = 0 F mgDH =
45
D : F r mg rA ⋅ − ⋅ =7 5 6 0, F mgA =45
II: E : F r mg rBV ⋅ − ⋅ =2 7 0 FBV = 3,5 mg B : FScos 2r - mg 5r = 0α⋅ ⋅
F mg mgS = ⋅ =52
1 3 004cosα
,
→: F mgSsin - FBHα − = 0 FBH = 0,667mg
FGV
FGH
FB
qa
5qa
FS1 FS2
FS3
FCVFCH
FGH
FGV
FS3
mg
FA
FDH
FDV
7,5r6r
2r
3r3,61rα
FA
FDH
FDV
F
F
FF=mg
I
IIIII
mgmg
mg
A
B
G
D
A
BG
DC
FS
FBH
FBV FS
FBHFBV
EF
Bα α
Lösung 3.2.16
Gesamtsystem: → : FEH = 0 ↑ : FEV + FD -F = 0 E : F 3a – FD 4a = 0
kNFkNFF EVD 5,15,443
===
II : B : FCV 2a - F 3a = 0 F F kNCV = =32
9
C : FBV 2a - F a = 0 F F kNBV = =12
3
→ : FCH - FBH = 0 FCH = 0 I : A : 2FBH a + FEH 4a = 0 mit FEH = 0 wird FBH = 0 → : FBH + FEH - FAH = 0 FAH = 0 ↑ : FBV + FEV - FAV = 0 FAV = 4,5kN Lösung 3.2.17
Gesamtsystem: A F b F h F b
hF F kN
C F b F h F bh
F F kN
CH CH
AH AH
:
:
⋅ − ⋅ = = = =
⋅ − ⋅ = = = =
0 12
1
0 12
1
Teilsysteme:
kNFFaFaFC
kNFFabFaFbFA
BEADBEAD
BEBE
5,2202sinsin:
25,185
350
53:
=−==⋅+⋅
−=−=−==⋅⋅+⋅
αα
sin cosα α= =
35
45 II: ↑: ( )F F F F kN kNCV AD BE CV+ + = = − ⋅ = −
35
0 35
1 25 0 75, ,
Gesamtsystem: ↑: FAV + FCV - F = 0 FAV = F - FCV = 2,75 kN
2a
2a2a a
4a
A
B C
DE
B
A
C
FD
FAV
FAH
FAV
FAH
FB
V
FBHFB
VFBH
FEV
FEH
FCV FCH FCV
FCHF
I
II
III
F
FCH
FCV
FAH
FAV
b
a
h
aα
FAH
FAV
b
FCH
FCV
FAD FBE
I
II
Lösung 3.2.18
→ + = = −
↑ − = =
→ − = = −
↑ = =
→ − + + = =
↑ − − − + = =
→ − − − = =
↑ + + = =
→ − +
:
:
:
:
:
:
:
:
:
F F F F
F F F F
F F F F
F F
F F F F
F F F F F F
F F F F F
F F F F F
F F
S S S
S S
S S S
S S
S S S S
S S S S S
S S AH AH
S S AV AV
S B
1 2 2
1 1
4 2 4
3 3
1 5 6 5
1 3 5 6 6
5 4
5 7
6
0 2
0 5
0 2
0 0
0 0
0 5
0 2
0
cos
sin
cos cos cos
sin sin sin
cos
sin
cos
α
α
α α α
α α α
α
α
α = =
↑ − − = = −
0 2
06 7 7
F F
F F F FB
S S S: sinα
Lösung unter Verwendung von: sin cosα α= =15
25
; ;
Anderer Lösungsweg: Erkannte Nullstäbe: Stab 3 und Stab 5, d.h. FS3 = 0 und FS5 = 0; dann gilt FS1 = FS6 und FS2 = FS4 Auflagerreaktionen auch am Gesamtsystem bestimmbar, Lösung 3.2.19
→: FS2 + FS1cosα = 0 ↑ : FS1sinα - F = 0
sin
cos
α
α
=
=
110310
Damit wird: F F F F FS S1 210 3 16 3= = = −, ;
: FS4 - FS1 - 2Fsinα = 0 : -2Fcosα - FS3 = 0 F F F F F F FS S3 4
610
1 9 10 210
3 79= − = − = + =, ; ,
I
FS1
FS2α
αα
α
α
IIIII
IV
V
FS3FS2 FS4
FS1
FS6
FS5FS3
FS6
FAHFAV
FS5
FS4
FBFS7
F
FS7
1
27
I
FS1F
FS2
α
3a
aα
2F
IIFS1
FS4
FS3
αα
α
→: FS6 - FS2 + FS5cosγ - FS3sinα= 0 ↑ : FS5sinγ + FS3cosα = 0 α = 18,43°; γ = 2α = 36,86° oder sin2 2sin cos 3
5; cos2 cos sin 4
52 2α α α α α α= = = − =
FFFFFFF S 65
12533;3
35
1092
sin2cos2F
sincosFF 6
2
S3S5 −=−−−==⋅==−=αα
γα
: FS8 - FS4 - FS5cosα - 2Fsinα = 0 : 2Fcosα + FS7 + FS5sinα = 0
F F F F
F F F F F
S
S
7
8
3110
2310
2 85
3 79 3310
2110
7 27
= − ⋅ − ⋅ = −
= + ⋅ + ⋅ =
, ;
, , ;
→: -FS6 + FS10 + FS9cosβ - FS7sinα= 0 ↑ : FS9sinβ + FS7cosα = 0
tan sin cosβ β β β= = = ° = =97
1 2857 52 1 9130
7130
, ; , ;
FS9 = 3,4F; FS10 = -9F
→: FB - FAH = 0 ↑ : FAV - 6F = 0 FAV = 6F A : F . 3a + 2F . 2a + 2F . a - FB . a = 0 FB = FAH = 9F
↑ : FAV + FS11 = 0 FS11 = -6F
IIIFS2
FS3
FS5
FS6
α
γ
αα
αα
α
2F
IVFS4
FS5 FS7
FS8
αβ
VFS6
FS7FS9
FS10
aβ
79
a
F
F2F2F FB
FAH
FAV
FAV
FAH
FS10
FS11
VI
Lösung 3.2.20 Gesamtsystem:
→: FB - FAH = 0 ↑ : FAV - F = 0 FAV = F A : F . a - FB . a = 0 FB = FAH = F
Stabkräfte: →: FS1 + FS2cos30° = 0 ↑ : -FS2sin30° - F = 0 FS2 = -2F = -40 kN;
kNFFFS 64,34332121 ==⋅=
→: - FS2cos30° + FS4cos45° + FS3cos30°= 0 ↑ : FS2sin30°-FS4sin45° + FS3sin30° = 0 Da sin45° = cos45° gilt, erhält man nach Addition der beiden Gleichungen
FS3(sin30°+cos30°) + FS2(sin30°-cos30°) = 0
F F kN F F F kNS S S S3 4 2 32 3 11 3
10 72 3045
35 86= −−
+= − =
°°
+ = −, ( ) ,sincos
→: FS5cos60° - FS4cos45° + FS6cos60°= 0 ↑ : FS5sin60°+FS4sin45° - FS6sin60° = 0 12
12
2 0 2 0
12
312
2 023
0
22
1 13
10 72 2 40
5 6 4 5 6 4
5 6 4 5 6 4
5 4 6 4 5
( )
( )
( ) ,
F F F F F F
F F F F F F
F F kN F F F kN
S S S S S S
S S S S S S
S S S S S
+ − = + − =
− + = − + =
= − = − = − = −
→: FB - FS5cos60° - FS1 - FS3cos30° = 0 ↑ : -FS5sin60° - FS7 - FS3cos60° = 0 F F F F kN
F F F kN
B S S S
S S S
= + + =
= − − =
1 3 5
7 5 3
12
3 12
20
12
3 12
14 6,
F
FB
FAH
FAV
I
FS1
F
FS2
30°
II
FS2 FS3
45°FS4
30°30°
FS4 FS5
FS630°
60°45°III
FS1FS3
FS5FS7
FB30°30°
30°IV
Lösung 3.2.21
→: FBH - FA = 0 ↑ : FBV - F = 0 FBV = F B : -F . 8a + FA . 8a = 0 FBH = FA = F
→: -FS1 - FS2cosα = 0 ↑ : -FS2sinα - F = 0
sin cos tan
sin
α α α
α
α
= = =
= − = − = −
= = =
35
45
34
53
33 3
43
26 7
2
1
; ;
,
,
F F F kN
F Fcot F kN
S
S
→: FS1 + FS3cosβ - FS4cosβ = 0 ↑ : -FS4sinβ - FS3sinβ = 0 FS3 = -FS4
FS1 - 2FS4cosβ = 0
coscos
ββ
= = = =213 2
133
24 0441F F F kNS
S , FS3 = -FS4 = -24,04kN
→: -FS5 - FS3cosβ - FS6cosγ + FS2cosα = 0 ↑ : -FS6sinγ + FS2sinα + FS3sinβ = 0
sin cos
sin
γ γ
γ
= =
=− −
= − = −
= − + + = =
25
15
5 44 7
43
23
13
6 7
6
5
;
,
,
F F F F kN
F F F F F kN
S
S
→: FS5 + FS4cosβ - FA = 0 ↑ : -FS7 + FS4sinβ = 0
FFA
FBVFBH
III
IIIIV
V
12
345
67
IFS1F
FS2
α
3a
4a
5a
α
FS4 FS3
FS1II
β
a 13
2a
3aβ
FS2
FS3
FS5
FS6
β αγ III
8a
4aγ
4 5 a
FA
FS4
FS7
FS5IV β
F F F F kN
F F F F F F kN Kontrolle
S S
A S S
7 4
5 4
133
313
20
13
23
20
= = ⋅ = =
= + = + = =
sin
cos
β
β ( )
Kontrolle: →: FBH + FS6cosγ = 0 FBH = - FS6cosγ = F = 20kN ↑ : FS7 + FS6sinγ + FBV = 0 FBV = -FS7 - FS6sinγ = -F + 2F = F = 20kN
Ritterscher Schnitt:
III: -4Fa + 3FS4acosβ + 2FS4asinβ = 0 FS4(3 cosβ + 2 sinβ) = 4F
F F F kNS44 136 6
13
13 24 04=+
= = ,
IV: 8Fa + 4FS6asinγ = 0 F F F kNS62 5 44 7= − = − = −
sinγ,
→: -FS5 - FS4cosβ - FS6cosγ = 0 F F F F kNS5 313 2
135 1
513
6 7= − ⋅ + ⋅ = = ,
Kontrolle: ↑ : -FS6sinγ - F - FS4sinβ = 0 2F - F - F = 0 Lösung 3.2.22
Symmetrie: FAH = 0; FAV = FB = 1,5F = 30kN
Zur Geometrie:
l a atan aa
l a
l a l a
1 1
2 1
3 46
23
113
6 73
23
33 7 311
15 25
97
52 15
= + = = =
= − =
= → = ° = → = °
= → = °
α α
α α β β
γ γ
; ;
, ,
,
tan
tan tan = al
tan =3al
1
2
→: FS1cosα + FS2cosβ = 0 ↑ : FAV + FS1sinα + FS2sinβ = 0 FS1= -91,4kN FS2 = 78,8kN
VFBV
FBH
FS7 FS6
γ
F
FS6
β
FS5
III
IIIIV
FS41
23
γ
FF
F
FB
FAH
FAV
III
III
3a 3a
2a
aa
l1 l2
γα
α
β
atanα
IFAV
FS1FS2
α β
: -FS1 + FS4 - Fsinα = 0 : -FS3 - Fcosα = 0 FS3 = -16,6kN FS4 = -80,4kN
→: FS6 -FS3sinα + FS5 cosγ - FS2cosβ = 0 ↑ : FS5sinγ + FS3cosα - FS2sinβ = 0 FS5 = 43,9kN FS6 = 40kN
Lösung 3.2.23
Aus Symmetriegründen: FAH = 0 FA = FB = 6F = 6 . 4,5kN = 27kN F = 4,5kN
→: FS1cosα + FS2 = 0 ↑ : FA + FS1sinα - F = 0 FS1= -50,3kN FS2 = 45kN
: -FS1 + FS4 - 2Fsinα = 0 : -FS3 - 2Fcosα = 0 FS3 = -8,05kN FS4 = -46,3kN
→: FS6 + FS5cosβ - FS2 - FS3sinα = 0 ↑ : FS3cosα + FS5sinβ = 0 FS5 = 9kN FS6 = 36kN
II
F
FS1
FS4
FS3α
IIIFS6
FS5FS3
FS2
α γβ
FA FB
F
2F2F 2F
Symm.
a 5FA
I
FS1
FS2
2a
a
α
α
F
II
2F
FS1
FS4
FS3α
FS2
FS3 FS5
FS6β
α
III
3a
4a5aβ
→: FS8 cosα - FS5cosβ - FS4 cosα + FS7sinα = 0 ↑ : -FS7cosα - FS5sinβ -2F - FS4sinα + FS8sinα= 0 FS7 = -12,07kN FS8 = -34,23kN
→: FS10 + FS9cosγ - FS6 - FS7sinα = 0 ↑ : FS7cosα + FS9sinγ = 0 FS9 = 11,38kN FS10 = 27kN
2F
IVFS4
FS5FS7
FS8
βα
α
FS6
FS7 FS9
FS10γ
α
V
a
3aa 10γ
Lösung 3.2.24
→ − =
↑ + − =
− ⋅ − ⋅ =
:
:
:
F F
F F F
A Fa F a F a
AH B
AV B
B B
12
2 0
12
2 0
9 12
2 92
12
2 32
3 0
Ergebnis: FB = 53,8kN; FAH = 38,04kN; FAV = -8,04kN
↑ − = = =
← + = = − = −
: ,
: ,
12
3 0 23
3 34 64
12
0 12
17 32
1 1
2 1 2 1
F F F F kN
F F F F kN
S S
S S S S
↓ : FS3cos30° + FS1cos30° = 0 FS3 = -FS1 = -34,64kN ← : FS4 + 0,5 FS3 - 0,5 FS1 = 0 FS4 = 0,5 .(FS1 - FS3) = 34,64kN
↓ : -FS3cos30° - FS5cos30° = 0 FS5 = -FS3 = 34,64kN ← : FS6 + 0,5 FS5 - 0,5 FS3 - FS2 = 0 FS6 = 0,5 .(FS3 - FS5) + FS2 = -51,96kN
↓ : FS7cos30° + FS5cos30° - 12
2FB = 0 FS7 = 9,3kN
← : 12
2FB + FS8 - 0,5 FS5 + 0,5 FS7 - FS4 = 0 FS8 = 9,24kN
↓ : -FS7cos30° - FS9cos30° = 0 FS9 = -FS7 = -9,3kN ← : -FS6 + 0,5 FS9 - 0,5 FS7 + FS10 = 0 FS10 = -42,66kN
↑: FS11cos30° + FAV = 0 FS11 = 9,3kN Kontrolle: →: FAH + 0,5 FS11 + FS10 = 0 erfüllt.
FB
F
FAH
FAV
45°32
3a
4,5a
I60°
F
FS1
FS2
FS1FS3
FS460°
30°
II
FS2
FS3
FS660°
30°
III
FS5
FS4
FS7
FS860° 60°
IV
FS5
FB
45°
FS10
FS7
FS6
60° 60°
V
FS9
FS11
60°FAH
VIIFAV
FS10
Lösung 3.2.25
→: F - FAH = 0 FAH = F ↑ : FB - F + FAV = 0 FAV = 0,25F A : -F . a - F . 2a + FB . 4a = 0 FB = 0,75 F
I F a F a F a F F
II F a F a F F
F F F F
aus F F
S B S
S B S
S B S
S
:
:
:
,
3 3
1 1
2 2
2
32
0 18
2 032
0 34sin
2 0 84
⋅ − ⋅ + ⋅ = = −
− ⋅ + ⋅ = =
↑ + = = −
= ° = −
sin
tan folgt = 63,4
αα
α α
FF
FB
FAH
FAV
FS3F
FB
FS1
FS2
αI
II
3.3 Schnittgrößen in Trägern und Systemen von Trägern Lösung 3.3.1
→ : FAH - F1 cos60° + F3 cos45° = 0 ↑ : FAV - F1 sin60° - F2 - F3 sin45° + FB = 0 A : - F1 sin60° . 2a - F2 . 4a- F3 sin45° . 7a + FB . 8a = 0 Ergebnis: FAH = -37,4 kN; FAV = 21,8 kN; FB = 49,3 kN
→ : FAH - F1 cos60° + FLC = 0 FLC = 42,4 kN ↑ : FAV - F1 sin60° - FQC = 0 FQC = 13,1 kN C : F1 sin60° . a - FAV . 3a + MC = 0 MC = 56,7 kNm Lösung 3.3.2
→ : - FBH + F2 cos60° + F3 cos60° = 0 ↑ : FA + F2 sin60° - F1 - F3 sin60° + FBV = 0 A : - F1
. a + F2 sin60° . 2a - F3 sin60° . 4a + FBV . 3a = 0 Ergebnis: FA = 756 N; FBH = 1000 N; FBV = 1244 N
0 1≤ ≤z a → : FL(z1) = 0; ↑ : FQ(z1) = FA = 756 N S1 : M(z1) = FA z1;
z1 = 0: M(0) = 0 z1 = a: M(a) = 756 Nm 0 2≤ ≤z a
→ : FL(z2) = 0; ↑ : FQ(z2) = FA - F1 = -1244 N S2 : M(z2) = FA (a + z2) - F1 z2; z2 = 0: M2(0) = 756 Nm z2 = a: M2(a) = -488 Nm
0 3≤ ≤z a
→ : FL(z3) = - FBH + F3 cos 60° = - 500 N; ↑ : FQ(z3) = - FBV + F3 sin 60° = -378 N; S3 : M(z3) = FBV (a - z3) - F3 sin 60° . (2a-z3); z3 = 0: M3(0) = - 488 Nm z3 = a: M3(a) = -866 Nm
FAH FAV
F1 F2 F3
FB
FAH FAV
2a a
F1
C
C FLC
FQC
MC
FBHFA
F1
F2
F3
FBV
FAz1
FL(z1)
FQ(z1)
M(z1)
FL(z2)a
F1 M(z2)
FQ(z2)FA
z2
FL(z3)a
F3M(z3)
FQ(z3)
FBVz3 a-z3FBH
0 4≤ ≤z a → : FL(z4) = F3 cos 60° = 500 N; ↑ : FQ(z4) = F3 sin 60° = 866 N; S4 : M(z4) = - F3 sin 60° . (a-z4); z4 = 0: M4(0) = - 866 Nm z4 = a: M4(a) = 0
M Nmmax = 866
Lösung 3.3.3
→ : FBH - F cos45° = 0 ↑ : FA - 3qa - F sin45° + FBV = 0 A : - 3qa . 1,5a - F sin45° . 4a + FBV . 3a = 0 Ergebnis: FA = 3322 N; FBH = 3535 N; FBV = 9213 N
0 31≤ ≤z a → : FL(z1) = 0; ↑ : FQ(z1) = FA - qz1 z1 = 0: FQ(0) = 3322N z1 = 3a: FQ(3a) = -5678N
S1 : M(z1) = FA z1 - 12 1
2qz ;
z1 = 0: M(0) = 0 z1 = 3a: M(3a) = -354 Nm Besondere Werte:
- Querkraft-Nullstelle: z F z z Fq
mQA
1 1 10 0 11: ( ) ,= = =
- Extremwert für M: ( )M z Nm1 184=
- 2. Nullstelle für M: $ : ($ ) $ ,z M z z Fq
mA1 1 10 2 0 22= = =
FL(z4)
F3M(z4) FQ(z4)
z4 a-z4
FBHFA
F1
F2
F3
FBV
FL in N+-
FQ in N
500
500
+ +
-
866
378
1244
756
M in Nm866
488
756
-+
FBHFA
Fcos45°3qa Fsin45°
FBV1,5a
FA z1
FL(z1)
FQ(z1)
M(z1)qz1
0 2≤ ≤z a → : FL(z2) = - F cos 45° = -3535 N; ↑ : FQ(z2) = F sin 45° = 3535 N; S2 : M(z2) = - F sin 45° . (a-z2); z2 = 0: M2(0) = - 354 Nm z2 = a: M2(a) = 0
M Nmmax = 354
Lösung 3.3.4
→ : - FBH + FA cos45° = 0 ↑ : F - 3qa + FA sin45° + FBV = 0 B : 3qa . 0,5a - FA sin45° . 2a + F . a = 0
Ergebnis: F qa F qa F qaA BV BH= = − =74
2 34
74
; ; ;
0 21≤ ≤z a
→ : FL(z1) = - FA cos45° = −74
qa ;
↑ : FQ(z1) = FA sin45° - qz1
z1 = 0: FQ(0) = 74
qa z1 = 2a: FQ(2a) =−14
qa
S1 : M(z1) = FA sin45° . z1 - 12 1
2qz ;
z1 = 0: M(0) = 0 z1 = 2a: M(2a) = 1,5 qa2
Besondere Werte:
- Querkraft-Nullstelle: z F z z aQ1 1 10 74
: ( ) = =
- Extremwert für M: ( )M z qa121 53= ,
FL(z2)
Fsin45°M(z2) FQ(z2)
z2 a-z2Fcos45°
z1 $z1
FBHFA
Fcos45°
3535
Fsin45°
FBV
FQ in N
-3535 FL in N
++-
5678
3322354
184+
-M in Nm
FBH F
FA cos45° 3qa
FA sin45°FBV1,5a
FAsin45° z1
FL(z1)FQ(z1)
M(z1)qz1
FAcos45°
0 2≤ ≤z a → : FL(z2) = 0 ↑ : FQ(z2) = - F + q(a - z2) z2 = 0: FQ(0) = - qa; z2 = a: FQ(a) = -2qa
S2 : M(z2) = F . (a-z2) - ( )12 2
2q a z− ;
z2 = 0: M2(0) = 32
2qa z2 = a: M2(a) = 0
M qamax ,=1 53 2
Lösung 3.3.5
→ : -FBH + F2 cos45° = 0 ↑ : - F1 - F2 sin45° + FBV = 0 B : - MB + F2 sin45° . 2a + F1 . 3a = 0 Ergebnis: MB = 5,82 Fa; FBH = 1,41 F; FBV = 2,41 F
0 1≤ ≤z a → : FL(z1) = 0; ↑ : FQ(z1) = - F1 = - F S1 : M(z1) = - F1 z1 = - F z1;
z1 = 0: M(0) = 0 z1 = a: M(a) = - Fa
0 22≤ ≤z a → : FL(z2) = - F2 cos 45° = - 1,41 F; ↑ : FQ(z2) = - F2 sin45° - F1 = - 2,41 F S2 : M(z2) = - F1 (a + z2) - F2 sin45° . z2;
FL(z2) F
M(z2) FQ(z2)
z2 a-z2
q(a-z2)
FBH F
FA cos45°
1,75 qa
FA sin45°FBV
1,75a
- FL
FQ
M
1,75qa0,25qa
qa 2qa
+-
+
1,53qa2 1,5qa2
F1F2
z1 z2 FBV
FBH
MB
α
F1
z1
FL(z1)
FQ(z1)
M(z 1)
FL(z2)aF1 M(z2)
FQ(z2)
F2
z2
α
z2 = 0: M2(0) = - Fa z2 = 2a: M2(2a) = - 5,82 Fa Mmax= 5,82 Fa
Lösung 3.3.6
A: M Fc ql ql c lA = − = −
12 2
2
M ist
positiv für c l
für c l
negativ für c lA
>
=
<
20
2
2
B: M Fc qlcB = =
MB ist positiv für beliebige c, bei c l≥
2 ist MB > MA, MA=MBist nur für
negatives MA möglich.
- MA = MB oder − −
= =ql c l qlc c l2 4
Lösung 3.3.7
→ : FAH = 0 ↑ : - 4qa + FAV + FB = 0 A : -4qa . a + FB . 3a = 0
Ergebnis: ;3,1334;7,26
38;0 kNqaFkNqaFF BAVAH =====
F1F2
z1 z2 FBV
FBH
MB
α
-
-
-
1,41FFL
FQ
F2,41F
FaM 5,82Fa
Fql
l
B
FAVFAH
MA c
BMB
cF
FAH FB3a
4qa2a
FAVa
0 1≤ ≤z a → : FL(z1) = 0 ↑ : FQ(z1) = - qz1 z1 = 0: FQ(0) = 0 z1 = a: FQ(a) = -qa = -10kN
S1 : M(z1) = - 12 1
2qz ;
z1 = 0: M(0) = 0 z1 = a: M(a) = - 0,5 qa2 = - 5kNm
0 32≤ ≤z a → : FL(z2) = 0 ↑ : FQ(z2) = - FB + q(3a - z2)
z2 = 0: FQ(0) = 53
qa = 16,7kN;
z2 = 3a: FQ(a) = −43
qa = - 13,3kN
S2 : M(z2) = FB . (3a-z2) - ( )12
3 22q a z− ;
z2 = 0: M2(0) = −12
2qa = -5kNm z2 = 3a: M2(3a) = 0
Besondere Werte:
- Querkraft-Nullstelle: mazzFz Q 66,1350)(: 222 ===
- Extremwert für M: ( )M z qa kNm228
98 9= = , = Mmax
Lösung 3.3.8
→ : FA - FS1 cos45° - F cos30° = 0 ↑ : FS2 + FS1 sin45° + F sin30° = 0 B : FS2
. 2a + F sin30° . 3a = 0 Ergebnis: FA = 1116N; FS1 = 354N; FS2 = -750N
0 1≤ ≤z a → : FL(z1) = -FA = -1116N; ↑ : FQ(z1) = 0
z1
FL(z1)FQ(z1)
M(z1)qz1
FL(z2) FB
M(z2) FQ(z2)
z2 3a-z2
q(3a-z2)
-
--
+
+
FQ
M
10kN
16,7kN
13,3kN
5kNm
8,9kNm
AFA
FS1 FS2
F
45°30°B C
FAz1
FL(z1)
FQ(z1)
M(z 1)
S1 : M(z1) = 0; 0 22≤ ≤z a
→ : FL(z2) = FS1 cos45° - FA = -F cos30° = - 866N; ↑ : FQ(z2) = FS1 sin45° = 250N S2 : M(z2) = FS1 sin45° . z2; z2 = 0: M2(0) = 0 z2 = 2a: M2(2a) = 500 Nm
0 3≤ ≤z a
→ : FL(z3) = - F cos 30° = - 866N; ↑ : FQ(z3) = - F sin 30° = -500N; S3 : M(z3) = F sin 30° . (a-z3); z3 = 0: M3(0) = 500 Nm z3 = a: M3(a) = 0
Mmax = 500 Nm
Berücksichtigung des Eigengewichtes:
Ermittlung von q aus m b h a q mga
b h g Nm
= ⋅ ⋅ ⋅ = = ⋅ ⋅ ⋅ =ρ ρ44
138
→ : FA - FS1 cos45° - F cos30° = 0 ↑ : FS2 + FS1 sin45° + F sin30° -4qa = 0 B : FS2
. 2a + F sin30° . 3a -4qa2 = 0 Ergebnis: FA = 1393N; FS1 = 745N; FS2 = -475N
0 1≤ ≤z a → : FL(z1) = - FA = - 1393N; ↑ : FQ(z1) = - qz1 ; FQ(0) = 0; FQ(a) = - qa = -138 N
S1 : M(z1) = −12 1
2qz ; M(0) = 0; M(a) = -69Nm
0 22≤ ≤z a → : FL(z2) = FS1 cos45° - FA = -F cos30° = - 866N; ↑ : FQ(z2) = FS1 sin45° - q(a+z2); ↑ FQ(0) = 389N; FQ(2a) = 113N
FL(z2)a
FS1 M(z2)
FQ(z2)FA z2
FL(z3) F
M(z 3)FQ(z 3)
z3 a-z3
-
+
-
+
FL
FQ
M
866N
500N
500Nm
250N
1116N
AFA
FS1
FS2
F
45°30°B C
4qa
FAz1
FL(z1)
FQ(z1)
M(z 1)qz1
FL(z2)a
FS1 M(z2)
FQ(z2)FA z2
q(a+z2)
S2 : M(z2) = FS1 sin45° . z2 - ( )12 2
2q a z+ ;
z2 = 0: M2(0) = - 69Nm z2 = 2a: M2(2a) = 430 Nm 0 3≤ ≤z a
→ : FL(z3) = - F cos 30° = - 866N; ↑ : FQ(z3) = - F sin 30° + q(a-z3); FQ(0) = -362N; FQ(a) = - 500N
S3 : M(z3) = F sin 30° . (a-z3) - ( )12 3
2q a z− ;
z3 = 0: M3(0) = 430 Nm z3 = a: M3(a) = 0 Mmax = 430 Nm
FL(z3)F
M(z3)FQ(z3)
z3 a-z3
q(a-z3)
-
+
-
-
FL
FQ
M
866N
500N
430Nm
362N
1393N
-113N389N
138N
69Nm+
Lösung 3.3.9
→ : FAH - 5qa = 0 ↑ : FAV - 2qa - 10qa + FB = 0 A : - 2qa . 2a - 10qa . 4a + FB . 5a = 0 Ergebnis: FB = 8,8qa; FAH = 5qa; FAV = 3,2qa
0 1≤ ≤z a → : FL(z1) = -FAH = - 5qa; ↑ : FQ(z1) = FAV = 3,2qa S1 : M(z1) = FAV z1;
z1 = 0: M(0) = 0 z1 = a: M(a) = 3,2qa2
0 22≤ ≤z a
→ : FL(z2) = -FAH = - 5qa; ↑ : FQ(z2) = FAV - qz2 FQ(0) = 3,2qa; FQ(2a) = 1,2qa
S2 : M(z2) = FA (a + z2) - 12 2
2qz ;
z2 = 0: M2(0) = 3,2qa2 z2 = 2a: M2(2a) = 7,6qa2
0 3≤ ≤z a
→ : FL(z3) = - 5qa; ↑ : FQ(z3) = - FB + 10qa = 1,2qa; S3 : M(z3) = FB (2a - z3) - 10qa . (a-z3); z3 = 0: M3(0) = 7,6qa2 z3 = a: M3(a) = 8,8qa2
0 4≤ ≤z a
→ : FL(z4) = 0 ↑ : FQ(z4) = - FB = - 8,8qa; S4 : M(z4) = FB . (a-z4); z4 = 0: M4(0) = 8,8qa2 z4 = a: M4(a) = 0
M qamax ,= 8 8 2
FAH
FB
2qa F
FAV
FAVz1
FL(z1)
FQ(z1)
M(z1)
FAH
FL(z2)aFAH
M(z2)
FQ(z2)FAV
z2
qz2
FL(z3)a
FM(z3) FQ(z3)
FBz3 a-z3
FL(z4) FB
M(z4) FQ(z4)
z4 a-z4
FL in qa-
FQ in qa
5
8,8
+
-
1,23,2
7,6
M in qa2
8,8
3,2+
Lösung 3.3.10
→ : - FAH = 0 ↑ : -F1 - 2q1a - F2 - 2q2a - F3+ FAV+ FB = 0 A : FB . 2a - 2q2a . 2a + F1 . a - F2 . a - F3 . 3a = 0 Ergebnis: FAH = 0; FAV = 75kN; FB = 105kN
Die Längskraft ist für alle Bereiche gleich Null.
0 1≤ ≤z a ↑ : FQ(z1) = -F1- qz1 z1 = 0: FQ(0) = -20kN; z1 = a: FQ(a) = -40kN
S1 : M(z1) = -F1 . z1 - 12 1
2qz ;
z1 = 0: M(0) = 0; z1 = a: M(a) = - 60kNm
0 2≤ ≤z a ↑ : FQ(z2) = FAV - F1 - q1(a + z2) z2 = 0: FQ(0) = 35kN; z2 = a: FQ(a) = 15kN
S2 : M(z2) = FAV. z2 - F1 . (a+z2) - ( )1
2 1 22q a z+ ;
z2 = 0: M2(0) = -60kNm; z2 = a: M2(a) = -10kNm 0 3≤ ≤z a
↑ : FQ(z3) = - FB + F3 + q2(2a-z3); FQ(0) = -10 kN; FQ(a) = -50 kN
S3 : M(z3) = FB (a - z3) - F3 . (2a-z3) - ( )12
22 32q a z− ;
z3 = 0: M3(0) = -10 kNm; z3 = a: M3(a) = -70 kNm 0 4≤ ≤z a
↑ : FQ(z4) = F3 + q2(a-z4) FQ(0) = 55 kN; FQ(a) = 15 kN
S4 : M(z4) = -F3 . (a-z4) - ( )12 2 4
2q a z− ;
z4 = 0: M4(0) = -70 kNm; z4 = a: M4(a) = 0
FAH
F1 F2
2q1a
FAV FB
2q2a
F3
z1
FL(z1)
FQ(z1)
M(z1)q1z1
F1
z2
FL(z2)
FQ(z2)
M(z2)q1(a+z2)F1
FAV
a
FL(z3)a
F3M(z3) FQ(z3)
FBz3 a-z3
q2(2a-z3)
FL(z4)
F3
M(z4)FQ(z4)
z4 a-z4
q2(a-z4)
Mmax = 70 kNm
Lösung 3.3.11
→ : FAH + F = 0 ↑ : FAV - 1,5 qa + FB = 0 A : - 3qa2 - F . a + FB . 2a = 0 Ergebnis: FAV = -0,5qa; FAH = -qa; FB = 2qa
0 21≤ ≤z a
q zqa
z d hq z z qz
a( ) , . .
( )1 1
1 1 12
3 2 6= ⋅
⋅=
→ : FL(z1) = -FAH = qa;
↓ : FQ(z1) = FAV - qz
a12
6
z1 = 0: FQ(0) = -0,5qa z1 = 2a: FQ(2a) = - 76
qa
S1 : M(z1) = FAV z1 - qz
az1
21
6 3⋅ ;
z1 = 0: M(0) = 0 z1 = 2a: M(2a) = - 139
2qa
0 2≤ ≤z a
q zqa
a z d hq z a z q a z
a( ) ( ), . .
( ) ( ) ( )2 2
2 2 22
32
22
26
= ⋅ +⋅ +
=+
→ : FL(z2) = -FAH = qa;
↑ : FQ(z2) = FAV + FB - q a z
a( )2
62
2+
z2 = 0: FQ(0) =56
qa z2 = a: FQ(a) = 0
S2 : M(z2) = FAV (2a+z2) + FB . z2 -
q a za
a z( ) ( )26
23
22
2+⋅
+;
FQ in kN
M in kNm-
+ +
- -1015
2040
35
50
5515
60 70
10
FAH
FAV
F1,5qa
FB
FAV z1
FL(z1)
FQ(z1)
M(z 1)
0,5q(z 1)z 1
FAH
FAV z2
FL(z2)
FQ(z2)
M(z2)0,5q(z2)(2a+z2)
FAH
FB
z2 = 0: M(0) = -139
2qa ; z2 = a: M(a) = - qa2
0 3≤ ≤z a ↓: FL(z3) = 0 ← : FQ(z3) = F S3 : M(z3) = -F(a-z3) = -qa(a-z3) z3 = 0: M(0) = - qa2 z3 = a: M(a) = 0
Mmax = 1,44 qa2
Lösung 3.3.12
→ : FAH - F = 0 ↑ : FAV + FB = 0 A : - M + F . a + FB . 4a = 0 Ergebnis: FAV = -5kN; FAH = 20kN; FB = 5kN
Es treten keine Längskräfte auf.
0 1≤ ≤z a ← : FQ(z1) = - F = -20kN S1 : M(z1) = -Fz1 z1 = 0: M(0) = 0 z1 = a: M(a) = -20kNm
0 22≤ ≤z a ↓ : FQ(z2) = FAV = -5kN S2 : M(z2) = FAV z2 - Fa z2 = 0: M(0) = -20 kNm z2 = 2a: M(2a) = -30kNm
0 23≤ ≤z a ↓ : FQ(z3) = -FB = -5kN
z3
FL(z3)
FQ(z3)
M(z3)
a-z3
F
FL in qa
1 +
FQ in qa
0,5
1,17
0,83
1
-+
+
m in qa2
--1,44
11
FAH
FAV
FM
FB
Fz1
FL(z 1)FQ(z1) M(z1)
FAH
FAV
FM(z2)
FQ(z2)z2
FL(z2)
FL(z3) FB
M(z3) FQ(z3)
z3 2a-z3
S3 : M(z3) = FB (2a- z3) z3 = 0: M(0) = 10 kNm z3 = 2a: M(2a) = 0
Mmax = 30 kNm
Lösung 3.3.13
→ : FAH + F = 0 ↑ : FAV - qa = 0 A : MA - F . a - qa . 0,5a = 0 Ergebnis: FAV = qa; FAH = -qa; MA = 1,5qa2
0 1≤ ≤z a ← : FL(z1) = 0 ↑ : FQ(z1) = qz1 FQ(0) = 0 FQ(a) = qa S1 : M(z1) = - 0,5q z1
2
z1 = 0: M(0) = 0 z1 = a: M(a) = - 0,5qa2
0 2≤ ≤z a ← : FQ(z2) = 0 ↓ : FL(z2)= - qa S2 : M(z2) = - 0,5qa2
0 3≤ ≤z a ← : FQ(z3) = F = qa ↓ : FL(z2)= - qa S2 : M(z2) = - 0,5qa2 - Fz3 z3 = 0: M(0) = - 0,5qa2 z3 = a: M(a) = - 1,5qa2
30
20
5
20
2010
--
+- -
FQ in kN
M in kNm
FAV
F
MA
FAH
qa
FL(z1)
M(z1) FQ(z1)
z1
qz1
FL(z2)
z2
M(z2)
FQ(z2)
qa
FL(z3)
z3
M(z 3)
FQ(z3)
qa
F
Mmax = 1,5 qa2
Lösung 3.3.14
→ : FAH - F = 0 ↑ : FAV + FB - 2qa = 0 A : M - F . a + FB . 3a - 2qa2 = 0 Ergebnis: FAV = qa; FAH = 2qa; FB = qa
0 21≤ ≤z a → : FL(z1) = -FAH = - 2qa; ↓ : FQ(z1) = FAV - qz1
z1 = 0: FQ(0) = qa z1 = 2a: FQ(2a) = - qa
S1 : M(z1) = FAV z1 - qz1
2
2 ;
z1 = 0: M(0) = 0 z1 = 2a: M(2a) = 0 Besondere Werte: - Querkraft-Nullstelle: z F z z aQ1 1 10: ( ) = =
- Extremwert für M: ( )M z qa121
2=
0 2≤ ≤z a → : FL(z2) = -FAH = - 2qa; ↑ : FQ(z2) = FAV - 2qa = -qa S2 : M(z2) = FAV (2a+z2) - 2qa . (a+z2); z2 = 0: M(0) = 0; z2 = a: M(a) = - qa2
0 3≤ ≤z a ↑ : FL(z3) = - FB = -qa; → : FQ(z3) = 0; S3 : M(z3) = 0;
FL in qa FQ in qa M in qa2
-+
-
-+
1
11
1,5
0,5
FAH FAV
F
M
FB
2qa
FAV z1
FL(z1)
FQ(z1)
M(z1)qz1
FAH
FAV z2
FL(z2)
FQ(z2)
M(z2)q2a
FAH
2a
FL(z3)M(z3) FQ(z3)
z3
FB
0 4≤ ≤z a ↑ : FL(z4) = - FB = -qa; → : FQ(z4) = F = 2qa; S4 : M(z4) = F . z4; z4 = 0: M(0) = 0 z4 = a: M(a) = 2qa2
0 5≤ ≤z a ↑ : FL(z5) = 0; → : FQ(z5) = 0; S3 : M(z5) = M = qa2;
Kontrolle des Momentengleichgewichts am Verzweigungspunkt C: C: M5(0) - M2(a) - M4(a) = 0 qa2 - 2qa2 + qa2 = 0 Gleichung ist erfüllt
M qamax = 2 2
FL(z4)M(z4) FQ(z4)
z4
FB
F
FL(z5)M(z5)
FQ(z5)z5
M
a-z5
CM2(a)
M4(a)
M5(0)
z1
2
-
1
+ 2
FQ in qa
2
FL in qa
-
++
-
1
1
1
0,5+
-
M in qa2
1
Lösung 3.3.15
→ : FAH + F1 + F3 = 0 ↑ : FAV - qa + F2 = 0 A : MA - F2 . a - qa . 0,5a + F1
. a + F3 . a = 0 Ergebnis: FAV = 2F1; FAH = - 3F1; MA = F1 . a
0 1≤ ≤z a →: FL(z1) = -F1 ↑ : FQ(z1) = F2 = 2F1 S1 : M(z1) = F2 . z1 ; z1 = 0: M(0) = 0; z1 = a: M(a) = 2F1a
0 2≤ ≤z a →: FL(z2) = F3 = 2F1 ↑ : FQ(z2) = q(a-z2) FQ(0) = qa = 4F1; FQ(a) = 0
S2 : M(z2) = - ( )12 2
2q a z− ;
z2 = 0: M2(0) = -2F1a; z2 = a: M2(a) = 0 0 3≤ ≤z a ↓ : FL(z3) = 2F1 ← : FQ(z3) = FAH =-3F1 S1 : M(z3) = FAH . z3 - MA ; z3 = 0: M(0) = -F1a; z3 = a: M(a) = -4F1a
Mmax = 4F1a
FAV
F1
MA
FAH qa
F2F3
z1
FL(z1)
FQ(z1)
M(z1)F2
F1
FL(z2) F3
M(z 2)FQ(z2)
z2 a-z2
q(a-z2)
z3
FL(z3)
FQ(z3) M(z3)
FAV
FAH MA
41
M in F1aFQ in F1
2
FL in F1
2
+-
+
3
2 +
-2
2
1
4+
--
Lösung 3.3.16
→: q0a + FAH - FB sin30° = 0 ↑ : FAV - F + FB cos30° = 0
A: FB sin30° . a + FB cos30° . a - 2Fa -q0a ⋅43
a =0
Ergebnis: FB = 13,177kN; FAH = 0,589kN; FAV = -6,412kN
0 21≤ ≤z a
q zq
az d h
q z z q za
( ) , . .( )
10
11 1 0 1
2
2 2 4= ⋅
⋅=
↑: FL(z1) = -FAV = 6412N;
→ : FQ(z1) = -FAH - q z
a0 1
2
4
z1 = 0: FQ(0) = -589N z1 = 2a: FQ(2a) = - 6589N
S1 : M(z1) = -FAH z1 - q z
az0 1
21
4 3⋅ ;
z1 = 0: M(0) = 0 z1 = 2a: M(2a) = - 5178Nm
0 2≤ ≤z a → : FL(z2) = -FAH - q0a = -6589N; ↑ : FQ(z2) = FAV = -6412N
S2 : M(z2) = -FAH . 2a + FAV . z2 -
23 0
2q a ;
z2 = 0: M(0) = -5178Nm; z2 = a: M(a) = - 11590Nm
0 3≤ ≤z a ↓: FL(z3) = -FBcos30° = -11412N ← : FQ(z3) = FBsin30° = 6589N S3 : M(z3) = FBsin30° . z3 z3 = 0: M(0) = 0 z3 = a: M(a) = 65989Nm
0 4≤ ≤z a ←: FL(z4) = 0 ↑ : FQ(z4) = F = 5000N S4 : M(z4) = - F . (a-z4) z4 = 0: M(0) = -5000Nm; z4 = a: M(a) = 0
FAH
F
FAV
FB30°
q0a
43
a
FAV
z1
FL(z1)
FQ(z1)M(z1)
0,5q(z1)z1
FAH
FAV
z2
FL(z2)
FQ(z2)
M(z2)
q0a
FAHFB
z3
FL(z3)FQ(z3)
M(z3)
FB
z4FL(z4)
FQ(z4)M(z4)
F
a-z4
Lösung 3.3.17
↑ : FB - FAV = 0 ← : FAH - F = 0 A : 2FB a - F a sin α = 0
F F F F Fsin FAH B AV= = = =12
14
α
0 ≤ ϕ ≤ 30° ϕ = ϕ1
F Fsin Fcos F F
F Fcos Fsin F F
M Fa Fasin M Fa
N N
Q Q
( ) ( )
( ) ( )
( ) ( ) ( )
ϕ ϕ ϕ ϕ ϕ ϕ
ϕ ϕ ϕ ϕ ϕ ϕ
ϕ ϕ ϕ ϕ ϕ ϕ
1 1 1 1 1 1
1 1 1 1 1 1
1 1 1 1 1 1
14
014
14
0 14
14
1 0 14
14
− − = = +
− + = = −
+ − − = = + −
sin cos
cos sin
cos sin cos
30 180 2°≤ ≤ ° = −ϕ ϕ π ϕ
( ) ( ) ( )ϕϕϕϕϕ
ϕϕϕϕϕ
ϕϕϕϕϕ
cos141cos1
41)(M0cos1
41)(M
sin41sin
41)(0sin
41)(
cos41cos
41)(0cos
41)(
22
22
22
+=−==−−
−=−==+
=−==+
FaFaFa
FFFFF
FFFFF
NN
0FN
+ - +
-+
FQ M
F14 F − 1
4 F
6,6-
6,4
11,4+
-
FL in kN FQ in kN0,6
6,46,6
6,6
5-
-
+
+
M in kNm
5,2
11,65
6,6-
-
+
Fa
FBFAH FAV
α
FN(ϕ1)
ϕ1ϕ1
ϕ1
FQ(ϕ1)
M(ϕ1)
F 14
F
M(ϕ)
14
F
FN(ϕ)
ϕ2ϕ
FQ(ϕ)
B
F F F F MF F F F M Fa
F F F F M
F F F
F
N Q
N Q
N Q
N Q
N
1 1
1 1
2 2
2 2
2
0 0 25 0 0 030 0 7165 30 0 741 30 0 467
30 0 2165 30 0 125 180 0
180 0 25 180 0
00 90
( ) , ( ) ( )( ) , ( ) , ( ) ,( ) , ( ) , ( )( ) , ( )
( )
° = ° = =
° = ° = ° =
° = ° = − ° =
° = − ° =
=
= = °
ϕ
ϕ ϕcos 2 2
Lösung 3.3.18
→: FBH - Fcos45° = 0 ↑ : FBV + FC - Fsin45° = 0 B: FC 2a - Fsin45° . a = 0 Ergebnis: FBH = 353,6N; FBV = FC = 176,8N
0 ≤ z ≤ a → : FN(z) = -FBH = -353,6N; ↑ : FQ(z) = FBV = 176,8N S : M(z) = FBV
. z ; z = 0: M(0) = 0; z = a: M(a) = 35,36Nm
0 ≤ ϕ ≤ 90° F F F F
F F F F
F acos F acos
N C N C
Q C Q C
C C
( ) ( )( ) ( )( ) ( )
ϕ ϕ ϕ ϕ
ϕ ϕ ϕ ϕ
ϕ ϕ ϕ ϕ
+ = = −
+ = = −
− = =
cos cos
sin sin
M M
0
0
0
F N F M NmF F N M
N Q
N Q
( ) , ( ) ( ) ,( ) ( ) , ( )0 176 8 0 0 0 35 3690 0 90 176 8 90 0° = − ° = =
° = ° = − ° =
FC
FBH
F45°
FBV
FBVz
FN(z)
FQ(z)
M(z)FBH
M(ϕ)FN(ϕ)
acosϕ
ϕ
FQ(ϕ)
FCϕ
-
+-
+
-
FN in N FQ in N M in Nm
353,6176,8
176,8
176,8
35,36
Lösung 3.3.19
→: FAH = 0 ↑ : FAV - 2qa = 0 A: MA - 2qa . 3a = 0 MA = 6qa2 FAV = 2qa
0 ≤ z1 ≤ 2a
→ : FL(z1) = 0; ↑ : FQ(z1) = -qz1; FQ(0) = 0 FQ(2a) = -2qa S : M(z1) = -qz1
. 0,5z1; M(0) = 0 M(2a) = -2qa2
0 ≤ ϕ ≤ 180°
( )
F qasin F qasinF qacos F qacos
qa a asin qa
L L
Q Q
( ) ( )( ) ( )
( ) ( ) ( )
ϕ ϕ ϕ ϕ
ϕ ϕ ϕ ϕ
ϕ ϕ ϕ ϕ
− = =
+ = = −
+ + = = − +
2 0 22 0 2
2 0 2 12M M sin
ϕ = 0°: FL(0°) = 0 FQ(0°) = -2qa M(0°) = -2qa2
ϕ = 90°: FL(90°) = 2qa FQ(90°) = 0 M(90°) = -4qa2
ϕ = 180°: FL(180°) = 0 FQ(180°) = 2qa M(180°) = -2qa2
0 ≤ z2 ≤ 4a
→ : FL(z2) = 0; ↑ : FQ(z2) = FAV = 2qa; S : M(z2) = MA -FAV
(4a- z2); M(0) = -2qa2 M(4a) = 6qa2
Mmax = 6qa2
2qa
FAH
FAV
M A
3a
z1
FL(z1)
FQ(z 1)
M(z1)qz1
M(ϕ)
FL(ϕ)
asinϕ
ϕ
FQ(ϕ)2qa
a
4a-z2FL(z1)
FQ(z1)
M(z1)MAFAH
FAV
2
FL in qa
FQ in qa
MA in qa2
2+
+
-2
+
-
6
2
24
Lösung 3.3.20
Ermittlung der Schnittreaktionen aus: F z q z dz
M z F z dz
Q
Q
( ) ( )
( ) ( )
= −
=
∫∫
Aus FQ sind die Lagerreaktionen bestimmbar. 0 ≤ z ≤ a Integration:
F z qa
az C
M z q aa
z C z C
Q ( )
( )
= +
=
+ +
0 1
0 1 2
22
22
ππ
ππ
cos
sin2
Bestimmung der Integrationskonstanten aus den Randbedingungen: z = 0: M(0) = 0 C2 = 0
z = a: M(a) = 0 Ca
q a1 0
21 2= − ⋅
π
F z qa
z
M z qa
z za
Q( )
( )
= −
= −
42 2
1
42
2 0
2 0
ππ π
ππ
a cos
a sin2
Auflagerreaktionen:
↑ : FBV = FQ(0) = −
42
12 0ππ q a = 0,23q0a
↑ : FC = -FQ(a) =42 0π
q a = 0,41q0a
Querkraftnullstelle: π π2 2
1 0cosa
z −
= z a= 0 56,
Maximales Moment: Mmax = M(z = 0,56a) = 0,085q0a2
Lösung 3.3.21
→: FAsinα - FBH = 0 ↑ : FAcosα - 2qa + FBV - F = 0 B: - FAsinα . 3a - FAcosα . 2a + 2qa2 -F . a = 0 Es ist: sin cosα α=
+= =
aa a2 24
15
25
Ergebnis: FA = 0,48qa; FBH = 0,21qa; FBV = 2,07qa
FBV z
q0
FC
q(z)FBH
FQ(0)
FBV
FQ(a)
FC
FA
2qa
FFBH
FBV
αα
a
0 51≤ ≤z a
( ) 211
2111
111
111
52cos
21)(
54ccos)(
52sincos)(
qzzFzqzFzM
qzFosqzFzF
qzqzzF
AA
AAQ
L
−=−=
−=⋅−=
−=⋅−=
α
αα
αα
z1 = 0: FL(0) = 0; FQ(0) = 0,48qa M(0) = 0 z a1 5= : FL( 5a ) = -0,89qa; FQ( 5a ) = -1,31qa M( 5a ) = -0,93qa2
Querkraftnullstelle: z Fq
aA1
54
0 6= = ,
Extremwert für M: M z qa( ) ,120 14=
Nullstelle für M: $ ,z Fq
aA1
52
1 2= =
0 ≤ z2 ≤ 2a
→ : FL(z2) = F - FBV = - 1,57qa; ↑ : FQ(z2) = FBH = 0,21qa; S : M(z2) = -FBH (2a-z2) - Fa; M(0) = -0,93qa2 M(2a) = -0,5qa2
0 ≤ z3 ≤ a → : FL(z3) = 0; ↑ : FQ(z3) = F = 0,5qa; S : M(z3) = -F(a-z3); M(0) = -Fa = -0,5qa2 M(a) = 0
|Mmax| = 0,93qa2
FA
z1
qz1cosα
α
z1cosα
FL(z 1)
M(z 1)
FQ(z1)
z2FL(z2)
FQ(z2)M(z2)
F2a-z2
FBVFBH
z3
FL(z3)
FQ(z3)M(z3) F
a-z3
FQ in qa
0,89
1,57
FL in qa
+
-
-
+
-+0,48
1,31
0,21 0,5
M in qa2
-
-
-+
0,93
0,5
0,140,6a
1,2a
Lösung 3.3.22
Alle horizontalen Kraftkomponenten sind Null! ↑: FAV - F + FG1V = 0 A: - F . a + FG1V . 2a = 0 ↑: -FG1V + FB - F + FG2V = 0 G2: F . a - FB . 2a + FG1V . 3a = 0 ↑: -FG2V + FC - F + FD = 0 D: F . a - FC . 2a + FG2V . 3a = 0
Ergebnis:
F F F
F F F F F F F F
F F F F
AH G H G H
AV B C D
G V G V
= = =
= = = =
= =
1 2
1 2
012
54
78
38
12
14
; ; ;
; ;
Ermittelt man die Schnittgrößen am getrennten System, sind 8 Schnitte erforderlich. Behandelt man die Gelenkkräfte als innere Kräfte, so kommt man mit 6 Bereichen aus. Alle Längskräfte sind Null! 0 ≤ z1 ≤ a
↑ : FQ(z1) = FAV = 0,5F; S : M(z1) = FAV . z1 M(0) = 0 M(a) = 0,5Fa
0 ≤ z2 ≤ 2a ↑ : FQ(z2) = FAV - F = - 0,5F; S : M(z2) = FAV(a+z2) - F . z2 M(0) = 0,5Fa; M(a) = 0(Gelenk); M(2a) = - 0,5Fa
0 ≤ z3 ≤ a ↑ : FQ(z3) = FAV - F + FB = 0,75F; S : M(z3) = FAV(3a+z3) - F(2a+z3) + FB . z3 M(0) = - 0,5Fa; M(a) = 0,25Fa
0 ≤ z4 ≤ 2a
FAH
FAV
FFG1H
FG1VF
FFG1V
FG1H
FB FG2V
FG2H
FG2V
FG2H
FC FD
z1
FL(z1)
FQ(z1)
M(z1)
FAV
z2FL(z2)
FQ(z2)
M(z2)
FAV
FG1
z3
FL(z3)
FQ(z3)
M(z3)
FAV
FG1
FB
z4
FL(z4)FQ(z4)
M(z4)
FAV
FG1
FB
F
↑ : FQ(z4) = FAV - 2F + FB = - 0,25F; S : M(z4) = FAV(4a+z4) - F(3a+z4) + FB(a+z4) - F . z4 M(0) = 0,25Fa; M(a) = 0(Gelenk); M(2a) = -0,25Fa 0 ≤ z5 ≤ a
↑ : FQ(z5) = F - FD = 0,625F; S : M(z5) = -F(a-z5) + FD(2a-z5); M(0) = - 0,25Fa M(a) = 0,375Fa
0 ≤ z6 ≤ a ↑ : FQ(z6) = -FD = -0,375F; S : M(z6) = FD(a-z6); M(0) = 0,375Fa M(a) = 0
+
-
+-
+-
+-
+-
+
FQ in F
M in Fa
0,5
0,50,25
0,750,37
0,62
0,5
0,5
0,25
0,25
0,37
Lösung 3.3.23
Aus Symmetriegründen im Teil I folgt F F qaA GV= =12
0 ≤ z ≤ l ↑ : FQ(z) = FA -qz
S : M(z) = FA . z - 12
2qz
M(0,5a) = 0,125qa2; M(a) = 0 M(l) = - 0,5ql(l-a) MmaxI = MmaxII
( )18
12
4 4 4 4 0 2 2 1 0 8282 2 2 2 2qa ql l a a l la a la l a l l= − = − + − = = − =( ) ,
z5
FL(z5)
FQ(z5)M(z5) F
a-z5 FD
z6
FL(z6)
FQ(z6)M(z6)
FDa-z6
zFL(z)
FQ(z)
M(z)
FA
qz
Lösung 3.3.24
→: FAH - FGH = 0 ↑: FAV - 3qa + FB - FGV = 0
A: -3qa ⋅32
a + FB . 2a - FGV
. 3a = 0
→: -FC + FGH = 0 ↑: - qa + FGV = 0
G: -qa ⋅12
a + FC . a = 0
Ergebnis: F qa F qa F qa F qa F qa F qaAH AV B C GH GV= = = = = =12
14
154
12
12
; ; ; ; ;
0 ≤ z1 ≤ 2a
→ : FL(z1) = -FAH = - 0,5qa
↑ : FQ(z1) = FAV -qz1 = q a z4 1−
FQ(0) = 0,25qa; FQ(0,25a) = 0; FQ(2a) = -1,75qa
S : M(z1) = FAV . z1 - 12 1
2qz = qz a z1
12 2−
M(0) = 0; M(0,25a) = 0,03qa2; M(2a) = -1,5qa2
0 ≤ z2 ≤ 2a
→ : FL(z2) = -FC = - 0,5qa ↑ : FQ(z2) = q (2a - z2) FQ(0) = 2qa; FQ(2a) = 0
S : M(z2) = FC . a - ( )12
2 22q a z−
M(0) = -1,5qa2; M(2a) = 0,5qa2
0 ≤ z3 ≤ a
↑ : FL(z3) = 0 →: FQ(z3) = -FC = - 0,5qa S : M(z3) = FC . (a - z3) M(0) = 0,5qa2; M(a) = 0
Mmax= 1,5 qa2
FAH FAV
3qa
FB
FGH
FGV
FGH
FGV
qaFC
z1FL(z1)
FQ(z1)
M(z1)
FAV
qz1
FAH
z2
FL(z2)
FQ(z2)M(z2)
2a-z2
q(2a-z2)FC
z3
FL(z3)
FQ(z3)
M(z3)
FCa-z3
FL in qa-0,5
FQ in qa-
+
-
+0,25
1,75
2
0,5
-+
+ M in qa2
0,03
1,5
0,5
0,25a
0,5a
Lösung 3.3.25
→: FAH - FGH + qa = 0 ↑: FAV - 2qa + FGV = 0
A: -2qa ⋅12
a - qa ⋅12
a + FGH . a +FGV
. a = 0
→: -FBH + FGH = 0 ↑: - FGV -2qa + FBV = 0
B: 2qa ⋅12
a + FGV . a - FGH
. a = 0
Ergebnis: F qa F F qa F qa F qa F qaAH BH GH AV BV GV= = = = = =
45
47
49
4 4; ; ;
0 ≤ z1 ≤ a
↑ : FL(z1) = -FAV = - 7
4qa
→ : FQ(z1) = -FAH -qz1 = -q a z4 1+
FQ(0) =- 0,25qa; FQ(a) = -1,25qa
S : M(z1) = -FAH . z1 - 12 1
2qz = -qz a z1
12 2+
M(0) = 0; M(a) = -0,75qa2
2qa
qa2qa
FAH
FAV
FGH
FGH
FGV
FGV
FBH
FBV
z1
FL(z1)
FQ(z1)M(z1)
FAV
qz1
FAH
0 ≤ z2 ≤ 2a → : FL(z2) = -FAH - qa = - 1,25qa
↑ : FQ(z2) = FAV -2qz2 = ( )q a z4
7 8 2−
FQ = 0 z a a278
0 875= = ,
FQ(0) = 1,75qa; FQ(0,875a) = 0; FQ(2a) = -2,25qa
S : M(z2) = FAV . z2 - FAH . a - 12
222qa qz− = ( )− − +
q a az z4
3 7 422 2
2
M(0) = - 0,75qa2; M(0,875a) = 0,016qa2; M(2a) = -1,25qa2
0 ≤ z3 ≤ a
↑ : FL(z3) = - FBV = -2,25qa →: FQ(z3) = FBH = 1,25qa S : M(z3) =- FBH . (a - z3) = -1,25qa . (a - z3) M(0) = -1,25qa2; M(a) = 0
qa
FAH FAV
z2
FQ(z2)
FL(z 2)M(z 2)2qz2
z3
FL(z3)
FQ(z3)M(z 3)
FBH
a-z3
FBV
FL in qa
-
-
-
1,75
1,25 2,25
FQ in qa
0,25
1,25
1,75
0,875a 2,25
1,25
- +-
+
-- -
-
0,750,02
1,25
0,75a
M in qa2Mmax= 1,25qa2
Lösung 3.3.26
→: FAH - FGH + F = 0 ↑: FAV - FGV = 0 A: - F . 12a+ FGH
. 12a -FGV . 4a = 0
→: -FBH + FGH = 0 ↑: FGV -8qa + FBV = 0 B: 8qa ⋅4a - FGV
. 8a - FGH . 8a = 0
Ergebnis: F qa F F qa F qa F qa F qaAH BH GH AV BV GV= = = = = =
252
32
132
32
; ; ;
0 ≤ z1 ≤ 12a
↑ : FL(z1) = -FAV = - 3
2qa
→ : FQ(z1) = -FAH = -12
qa
S : M(z1) = -FAH . z1 = -12 1qaz
M(0) = 0; M(12a) = - 6qa2
0 ≤ z2 ≤ 4a
→ : FL(z2) = -FAH - F = - 2,5qa ↑ : FQ(z2) = FAV = 1,5qa
S : M(z2) = FAV . z2 - FAH . 12a = ( )− −32
4 2qa a z
M(0) = - 6qa2; M(4a) = 0
0 ≤ z3 ≤ 8a
→ : FL(z3) = -FGH = - 2,5qa
↑ : FQ(z3) = FGV -qz3 = q a z32 3−
FQ(0) = 1, 5qa; FQ(1,5a) = 0; FQ(8a) = -6,5qa
8qa
FAH
FAV
FGH
FGH
FGV
FGV
FBH
FBV
F
z1
FL(z1)
FQ(z1)M(z1)
FAVFAH
F
FAH
FAV
z2FQ(z2)
FL(z 2)M(z 2)
qz3
FGH
FGV
z3FQ(z3)
FL(z3)M(z3)
S : M(z3) = FGV . z3 - 12 3
2qz = ( )qz a z332
3 −
M(0) = 0; M(1,5a) = 1,125qa2; M(3a) = 0; M(8a) = -20qa2
0 ≤ z4 ≤ 8a
↑ : FL(z4) = - FBV = -6,5qa →: FQ(z4) = FBH = 2,5qa S : M(z4) =- FBH . (8a - z4) = -2,5qa . (8a – z4) M(0) = -20qa2; M(8a) = 0
z4
FL(z4)
FQ(z4)M(z 4)
FBH
8a-z4
FBV
FL in qa
-
-
-
6,5
2,5
1,5FQ in qa
-
++
-
2,5
6,51,5a
1,5
0,5
M in qa2
6
-
- -
-
3a20
+1,1
4. Das räumliche Kraftsystem 4.1 Das zentrale räumliche Kraftsystem Lösung 4.1.1
d c b cm
d a b cm
d a b c cmcd
bd
cd
dd
ad
bd
12 2
22 2
32 2 2
1 1
3
2
3
2 2
33 54
36 06
39 05
= + =
= + =
= + + =
= =
= =
= =
,
,
,
;
;
;
sin cos
sin cos
sin cos
α α
β β
γ γ
: ,
: ,
: ,
F F F F F dd
ad
F F ad
N
F F F F bd
F dd
bd
F bd
F bd
N
F F F F F F cd
F cd
N
Rx
Ry
Rz
= + = + ⋅ = + =
= + = + ⋅ = + =
= + + = + + =
1 4 1 42
3 21 4
3
2 4 21
42
3 22
14
3
3 2 4 3 21
43
253 65
409 36
304 68
cos sin
cos cos cos
sin sin
β γ
α β γ
α β
F N
Richtung FFFFFF
R
RRx
RR
RRy
RR
RRz
RR
=
= = = °
= = = °
= = = °
569 86
0 4451 63 5
0 7184 44 1
0 5347 57 7
,
: , ,
, ,
, ,
cos
cos
cos
α α
β β
γ γ
y
z
d1 c
bF2
α
β
γ
F2y
F2z
z
d2
cF4
F4z
F4xy
d3
y
x
F4xy a
b
d2
x
y
z
F1
F2
F3 F4Fz
Fxy ab
c
Lösung 4.1.2
x: FS2 sinα - FS1 sinα = 0 FS1 = FS2 y: FS2 cosα + FS1 cosα + FS3 cosβ = 0 z: -F + FS3 sinβ = 0
sin cos sin = 45
; cosα α β β= = = = = = =345
15
645
25
8100
6100
352 2 2 2
aa
aa
aa
aa
; ;
Ergebnis: F F F F F F FS S S1 2 323 516
54
= = − = − = =cotcos sin
βα β
;
Lösung 4.1.3
x: -FS2 sin45° + FS1 sin45° = 0 FS1 = FS2 y: -FS2 cos45° - FS1 cos45° - FS3 sin45° = 0 z: -F - FS3 cos45° = 0
Ergebnis: F F F N F F NS S S1 2 32
2707 2 1414= = = = − = −;
Lösung 4.1.4
31=cos
232=sin
53=cos
54=sin
β
β
α
α
51=cos
52=sin
215=cos
214=sin δδγγ
xy
z
α
βα
FFS1sinα
FS2sinαFS2cosα
FS1cosα
FS3sinβ
FS3cosβ
45°
45°45°
Fx
y
z
FS1
FS2
FS3
α
β γ
δFACcosγ sinδFADcosα
FABsinβ
FACsinγ
FADsinα
FAE
FACcosγcosδFABcosβ⋅ 1
2 2 FABcosβ⋅ 12 2
A
E B
A
A
E
E C
D
E
C
3a
4a5a
4a
a 2 a 5
4a
a
2a
F F
F F F F kN
AB AC
AB AC AB AC
cos12
cos sinβ γ δ⋅ = ⋅
= = =
2
16
2 221
1242
37 03,
( )
F F F
F F F
F F F kN
AD AB AC
AD AC AC
AD AC AD
cos cos cos cos
35
cos sin + cos
α β γ δ
γ δ δ
= ⋅ +
= =
= =
12
2
321
521
21 82,
F F F F
F F F F F kN
AD AB AC AE
AD AB AC AE AE
sin + sin + sin
+ 23
+ 421
α β γ =
= =45
2 69 825,
Lösung 4.1.5
Es ist: FW = FG Aus Symmetriegründen oder aus Summe aller Kräfte in x-Richtung folgt FS1 = FS2 x: FS1 sinα cosα - FS2 sinα
cosα = 0 FS1 = FS2 → y: -FS1cos2α - FS2cos2α + FS3cosα -FWcosα = 0 2FS1cosα - FS3 = -FG ↑ z: -FS1sinα - FS2sinα - FS3sinα - Fwsinα -FG = 0 2FS1sinα + FS3sinα = -FG(1+sinα)
( )( )
( )( )
F F F F kN
F F F kN
S S G G
S G G
1 2
3
1 22 1
2 1 33 3
3 155
11
2 33 3
0 155
= = −+
+= −
+= −
= −+ −
+= −
−= −
sinsin cos
cos sin sinsin cos
αα α
α α αα α
,
,
4.2 Das allgemeine räumliche Kraftsystem Lösung 4.2.1 r r r r
r
F F e F e F e
F F F F FF F F F F
F F F F F
F
R Rx x Ry y Rz z
Rx
Ry
Rz
R
= + +
= − = − =
= − + = − + =
= − = − =
=3 6
1 4
2 5
2 2 00
3 3 0
0
x
yz
αα
FS1cosα cosα
FS1sinα
FS1cosα
FS1
FS1cosα sinα x yz
αα
FG
FS3FWFS1
FS2
zyxR
Rz
Ry
Rx
zRzyRyxRxR
eFaeFaeFaMFaaFaFMFaaFaFM
FaaFaFM
eMeMeMM
rrrr
rrrr
35435
4
64
35
54
++−=
=⋅+⋅=
=⋅+⋅=−=⋅−⋅−=
++=
Lösung 4.2.2
M F b F lcosM F lsin Nm
M Mcos M Msin
M M M Nm
M F l F Ml
N
y
x
x y
x y
= +
= =
= = = = °
= + = = °− = °
= = =
1 2
2
2 2
3 3
38 97
1 5 56 3
70 29 90 33 7
39 05
α
α
β β β β
γ β
= 58,5Nm
tan =MM
y
x
,
, ,
, ,
,
Lösung 4.2.3
l h a l a
F Fhl
F
F F al
F
F F F
F F F
Cy C C
Cxz C C
Cx Cxz C
Cz Cxz C
2 2 234 6 34
0 394
34 0 92
334
0 4732
534
0 78864
= + =
= =
= =
= =
= =
,
,
,
,
,
sin = F
cos = F
sin = F
cos = F
C
C
Cxz
Cxz
α
α
β
β
AB
)6(053:
5,121)5(05
25:
394,032)4(032:
)3(0:)2(0:
)1(0:
=⋅−⋅
−=−==⋅−⋅−
===⋅+⋅−
=−
=+−+↓
=−+→
aFaF
kNFFaFaF
FFFaFaF
FFFFFF
FFF
BxBz
GByByG
CGCyCyG
CzBz
GCyByAy
CxBxAx
aus ( 4 ) : FC = 1,69 FG = 5,076 kN aus ( 3 ) : FBz = 0,78864 FC = 4 kN
aus ( 6 ) : F F kNBx Bz= =35
2 4,
aus ( 1 ) : FAx = 0
aus ( 2 ) : F F kNAy G= =16
0 5,
F1b
F2l
M
F3 .
30°
βγ x
y
xy
z
A
B
C
FG
FCFCxz
αβl h3a
5a
FCxz
AC
C
Lösung 4.2.4
:
:
:
− + ⋅ = =
− =
− ⋅ − ⋅ − ⋅ + = =
F y F a y a
F F
F a F a F a F x x a
Q Q S Q
S Q
S S S Q Q Q
4 0 43
3 0
2 5 0 83
)3(025324:)2(1202244:)1(024:
3212
23
23
2321
=⋅−⋅−⋅−⋅
==⋅−⋅
=⋅−++
aFaFaFapapaFapaaF
apFFF
SSS
SS
SSS
012)1(0548)3(24
122
2122
=++−
=−−=
SS
SSR
FFpaFFpapaF
221
22
211
2
312:)1(90436:)1()3(paFFpaFauspaFFpa
SSS
SS
=−=
==−+
Lösung 4.2.5 Aus Symmetriegründen folgt: FS4 = FS5 und FS2 = FS3
Knoten I:
↓ + = = = − = −
+ ⋅ = = −
=
: ,
:
F F F F F F
F F F F
F F
S S S
S S S S
S
216
06
21 2
2 56
25
0 2 63
2
5 5 4
6 5 6 5
6
Knoten II:
: ,
: ,
− + = = = =
↓ + = = = − = −
F F F F F F
F F F F F F
S S S S
S S S S S
6 1 1 6
1 2 2 3 1
12
0 2 2 2 2 8
22
2 22
0 12
1 4
Lösung 4.2.6
sin cos
sin cos
α α= =
°= ° =
15
25
45 45 12
2
;
x
y
FQFS
FS
FS
a
4a
a
4a2axQ
yQ
x
yFR
FS1
FS2FS3
a 5a 6
I
II F
a
2a
a
FS6
FS4
FS5
FS1 FS2FS3
A
α
FWFG
FS1FS2FS3 FS4
FS5FS6
2aa
ax y
z
x
y
x y
:
: ,
:
:
:
: ,
12
2 12
2 0
0 52
112
12
212
2 0
12
2 2 0
12
12
2 0
12
12
2 2 0 24
0 35
6 2 6 2
4 4
1 3 5 2 6 4
5 6
1 6
6 6 2
F F F F
F F F F F
F F F F F F F
A F a F a F a F a
A F a F a F a
A F a F a F F F F
S S S S
S W S W W
G S S S S S S
G W S S
G S S
W S S S W W
− = =
→ + = = − = −
↑ − − − − − − − =
− − − − =
− − − =
− − = = = − = −
cos
sin
α
α
Und damit: ( )F F F F F F FS G W S G S W1 5 314
2 12
34
= − − = − =; ;
Lösung 4.2.7 Vereinfachte Rechnung: Strahlensymmetrie zur Schwerelinie, d,h. FS1 = FS3 = FS5 = FSV und FS2 = FS4 = FS6 = FSD.
Gleichgewichtsbedingungen:
z F a M F Ma
z F F mg FMa
mg
aa l
la l
FMl
amg
FM a l
a
SD SD
SV SD SV
SV SD
:
:
36
3 0 23
3 3 02
3
23 3
23
2 2 2 2
2
2 2
2
sinsin
coscot
3
sin cos cot = la
αα
αα
α α α
⋅ + = = − ⋅
↑ − − − = = ⋅ −
=+
=+
= − = −+
Lösung 4.2.8
( )
→ − + = =
− ⋅ + ⋅ + ⋅ + ⋅ + =
=−
−
⋅ − ⋅ − ⋅ + = = + −
⋅ + ⋅ + ⋅ − ⋅ + =
= − + +−
:
:
:
:
F F F F
C F a F a Fa
Fa
M
FF F M
a
B F a F a F a M F F F Ma
C F a F a F a F a M
F F FM M
a
S y S y
S S x y z
Sy x z
S y z x S yz x
S S x z y
S x yx y
1 1
1 2
2
3 3
3 4
4
0
2 20
2
20
2
20
M
mg
S
FSVFSV
FSV
FSDFSD
FSD
α
α
α
a6
3
z
A B
CD
Fx Fy
Fz
MxMy
Mz
x
y
z
FS1
FS2
FS3 FS4
FS5
FS6
A F a F a F a M F F F M
a
A F a Fa
Fa
M FF F M
a
S x z y S xz y
S x y z Sx y z
:
:
− ⋅ + ⋅ + ⋅ + = = + +
⋅ − ⋅ − ⋅ + = =+
−
5 5
6 6
20
2
2 20
2
Lösung 4.2.9
( ) ( )
( )
AB x rsin r rcos r
AC l r r r
= = + + = +
= = + + = +
=+
++
= −
ϕ ϕ ϕ
ϕ ϕ
αϕ
αϕϕ
ϕϕ ϕ ϕ
2 2
2 2
2 1
2 1 3 2
13 2
2 13 2
12
1
(
(
(
cos )
cos ) cos
sin = rl cos
cos =cos )cos
sin2
cos sin = 1- cos2
°><°−>−+
>−+
+⋅=⋅=
=
−
=−
606001cos23cos
01cos23cos
cos23
lr-cos
1
0lr-cos
)1(0sin+cos:
00
0
0
ϕϕϕ
ϕϕϕ
ϕ
ϕ
αϕ
rM
rMF
rFM
rFrFM
S
S
SS
( )
( ) ( )
( ) :
( ) :
( ) :
( ) :
( ) :
2 01
3 1 0
4 0
5 1 2 02
1 11
6 2 012
0
0
→ − = =−
↑ + + + =
− + =
+ − + ⋅ = = −+ −
−
− + − ⋅ = = −
F F F Mr
F F F
F F F
F r F r F r F Mr
F F F r F F
Dz S Dz
Dy S Ey
Dx S Ex
S S Ey Ey
S S Ex Ex S
cos sin2
sincos 3 + 2cos
sin
cos cos2
sin sin3 + 2cos sin
cos 3 + 2cos
r cos sin2
r cos cos2
cos cos2
sin2
αϕ ϕ
ϕ ϕ
α
αϕ
α ϕϕ ϕ
ϕ ϕ
αϕ
αϕ
αϕ ϕ
( )( ) ( ):
( ) ( ):
3 12
5 12
1
4 12
6 12
− = − +
+ = +
rF F
rF F
Dy S
Dx S
sin + sin
cos2
sin2
cos
α ϕ
ϕ ϕα
A x B
l
FSFS
C
r
α
ϕ
FScosαA
Bx
ϕ2
2coscos ϕαSF
2sincos ϕαsF
FS
FDx
FDy
FDzFEx
FEy
FSsinα
rsinϕ
rcosϕ
M0
DE
D
D
Lösung 4.2.10 Symmetrie:
( )
( )
h b m l l a m
l l h b m
FF F
F mg lh b
kN
F mga
h bkN
Sp
S Sp
S
S
SS
− = = + =
= + − =
↑
− =
=⋅−
=
=−
=
11 18 25
21 31
22 0
22 375
0 669
2 2
2 2
,
,
::
,
( ),
sin - mg = 0cos sin
α
α β
kNmhFM
MkNmhFM
kNFalFFF
kNmgFF
kNFFF
Dz
DyDx
SDz
SDy
SDx
521
0303
2321coscos
226,121sin
3345.021sincos
=⋅=
==⋅=
==⋅==
===
===
βα
α
βα
Lösung 4.2.11
:
:
:
:
:
F F F F F N
F F F F F N
F a F h F Fah
N
F a F a F F N
F a F a F F N
AV BV G AV G
AH BH C AH C
G C C G
G BV BV G
BH C BH C
+ − = = =
+ − = = =
− ⋅ + ⋅ = = =
⋅ − ⋅ = = =
⋅ − ⋅ = = =
0 14
250
0 12
100
2 0 2 200
3 4 0 34
750
4 2 0 12
100
al EFSp
β
α
FScosαcosβ
FScosαcosβ
2FSsinα
2FScosαsinβ
h-blSp
lS
mgF
lSp
A DB
CFG
FAH
FC
FAV FBHFBV
A
A
AB
FSsinα
FScosαcosβ FScosαsinβ
FDx FDy
FDz
MDx
MDy
MDz
4.3 Schnittgrößen im räumlich belasteten Balken Lösung 4.3.1
Nach dem Erstarrungsprinzip gilt:
Fu1 . r01 - Fu2 . r02 = 0 oder F F rru u2 101
02
= ⋅
Fu1 = 4500N Fa1 = 1640N Fr1 = 1740N Fu2 = 10800N Fa2 = 3930N Fr2 = 4180N
Lagerreaktionen: FAx - Fr1 - Fu2 + FBx = 0 A: -Fu1l1 - Fr2(l1 + l2) + Fa2r02 + FBy(l1 + l2 + l3) = 0 ↑: FAy - Fr2 - Fu1 + FBy = 0 A: -Fr1l1 - Fu2(l1 + l2) - Fa1r01 + FBx(l1 + l2 + l3) = 0 →: FAz + Fa1 - Fa2 = 0 Ergebnis: FAx = 3660N FAy = 5580N FAz = 2290N FBx = 8880N FBy = 3100N 0 ≤ z1 ≤ l1
FL1 = - FAz = - 2290N FQx1 = - FAx = -3360N FQy1 = FAy = 5580N Mx1 = FAy . z1 Mx1(0) = 0 Mx1(l1) = 307Nm My1 = - FAx . z1 My1(0) = 0 My1(l1) = -201Nm Mz1 = 0
0 ≤ z2 ≤ l2
FL2 = - FAz - Fa1 = - 3930N FQx2 = - FAx + Fr1= -1920N FQy2 = FAy - Fu1 = 1080N Mx2 = FAy . (l1 + z2) -Fu1 . z2 Mx2(0) = 307Nm Mx2(l2) = 383Nm My2 = - FAx . (l1 + z2) + Fr1 . z2 - Fa1 . r01 My2(0) = -398Nm My2(l2) = -533Nm Mz2 = -Fu1 . r01 = -540Nm
0 ≤ z3 ≤ l3
FL3 = 0 FQx3 = FBx = 8880N FQy3 = - FBy = -3100N Mx3 = FBy . (l3 - z3) Mx3(0) = 186Nm Mx3(l3) = 0 My3 = - FBx . (l3 - z3) My3(0) = -533Nm My3(l3) = 0 Mz3 = 0
FAz
FAxFAy
y Fu1
Fu2
Fa1
Fa2
Fr1
Fr2
FBx
zFBy
z1
Mx1My1
Mz1FL1FQx1FQy1
FAx FAy
FAz
z2
Mx2
My2
Mz2FL2FQx2
FQy2FAx FAy
FAz
l1
Fr1
Fa1
Fu1
z3
Mx3 My3
Mz3 FL3FQx3
FQy3 FBx
FBy
l3-z3
2290N
3930N
-FL
y-z-Ebene x-z-Ebene
+-FQy
5580N1080N
3100N
Mx
+
-
307Nm 383Nm186Nm
540NmMz
-+
FQx
3660N 1920N
8880N
-201Nm398Nm
533Nm
My
Mbmax bei z2 = l2 = 70mm M M z l M z l Nm Nmx ybmax = = + = = + =2
22 2 2
22 2
2 2383 533 656( ) ( ) Lösung 4.3.2
FAx = F1 = 100N FAy = F2 + F4 = 300N FAz = F3 = 80N MAx = F2 . 7a + F4 . 3a = 130Nm MAy = F1 . 7a + F3 . 6a = 118Nm MAz = F2 . 6a = 60Nm
0 ≤ z1 ≤ 6a
FL1 = - F1 = - 100N FQx1 = - F3 = - 80N FQy1 = - F2 = -100N Mx1 = - F2 . z1 Mx1(0) = 0 Mx1(6a) = -60Nm My1 = - F3 . z1 My1(0) = 0 My1(6a) = - 48Nm Mz1 = 0
0 ≤ z2 ≤ 4a FL2 = F3 = 80N FQx2 = - F1 = -100N FQy2 = - F2 = -100N Mx2 = -F2 . z2 Mx2(0) = 0 Mx2(4a) = -40Nm My2 = - F3 . 6a - F1 . z2 My2(0) = -48Nm My2(4a) = -88Nm
F1
F2
F3 F4
FAxFAy
FAz
MAx
MAy MAz
F1
F2
F3
z1FQx1
FQy1
FL1
Mx1My1
Mz1
F1
F2
F3
z2FQx2
FQy2
FL2Mx2
My2
Mz2
Mz2 = F2 . 6a = 60Nm 0 ≤ z3 ≤ 3a
FL3 = F3 = 80N FQx3 = - F1 = -100N FQy3 = - F2 - F4 = -300N Mx3 = -F2 . (4a + z3) - F4
. z3 Mx3(0) = - 40Nm Mx3(3a) = -130Nm My3 = - F1 . (4a + z3) - F3 . 6a My3(0) = - 88Nm My3(3a) = -118Nm Mz3 = F2 . 6a = 60Nm
Lösung 4.3.3
FAx = F2 = 2qa FAy = F1 + 2qa = 3qa FAz = 0 MAx = F1 . 2a + 2qa . a = 4qa2
MAy = F2 . 2a = 4qa2
MAz = F1 . a = qa2
F1
F2
F3 F4
FQx3
FQy3
FL3
Mx3
My3 Mz3
z3
+
-
-
-FQy in N
FL in N
FQx in N
Mx in Nm
My in Nm
Mz in Nm
100
80100300
6040 130
+60
80100
-
48
118-
F1
F2 2qa
FAxFAy
FAz
MAx
MAy MAz
0 ≤ z1 ≤ a
FL1 = - F2 = - 2qa FQx1 = 0 FQy1 = - F1 = -qa Mx1 = - F1 . z1 Mx1(0) = 0 Mx1(a) = - qa2
My1 = 0 Mz1 = 0
0 ≤ z2 ≤ 2a
FL2 = 0 FQx2 = - F2 = - 2qa FQy2 = - F1 - qz2 FQy2(0) = - qa FQy2(2a) = - 3qa
Mx2 = -F1 . z2 -12 2
2qz Mx2(0) = 0 Mx2(2a) = -4qa2
My2 = - F2 . z2 My2(0) = 0 My2(2a) = -4qa2
Mz2 = F1 . a = qa2
F1
F2
z1FQx1
FQy1
FL1
Mx1My1
Mz1
F1
F2
qz2
z2FQx2
FQy2
FL2
Mx2My2
Mz2
-
-
-FQy in qa
FL in qa
FQx in qa
Mx in qa2
M y in qa 2
M z in qa2
2
13
1
4
+1
2-
4-
Lösung 4.3.4 0 ≤ z1 ≤ l
FL1 = F2 FQx1 = - F1 FQy1 = - F3 Mx1 = - F3 . z1 My1 = - F1 . z1 Mz1 = 0
0
2≤ ≤ϕ
π
: FLϕ = F1sinϕ + F2cosϕ : FQy2 = - F3 : FQr = - F1 cosϕ + F2 sinϕ
Mr = - F3 (Lcosϕ + Rsinϕ) My2 = - F1 (L + Rsinϕ) + F2 R(1 - cosϕ) Mϕ = F3 [Lsinϕ + R (1 - cosϕ)] Mr, My2 - Biegemomente Mϕ - Torsionsmoment
F1
F2 F3
z1 FQx1
FQy1 FL1
Mx1 My1
Mz1
F1
F2 F3
ϕ FQr
FQy
FLϕ
Mr
My Mϕ
ϕ ϕ
F1
F2
M
L
Lcosϕ Rsinϕ
ϕ
ϕ Rsinϕ Lsinϕ
R(1-cosϕ)
R(1-cosϕ)
Mϕ
5.Der Schwerpunkt Lösung 5.1
xx l
ly
y l
lS
Si ii
ii
S
Si ii
ii
= ==
=
=
=
∑
∑
∑
∑1
6
1
61
6
1
6; ;
[ ] [ ] [ ] [ ] [ ]i l mm x mm y mm x l mm y l mmi Si Si Si i Si i
2 2
1 20 10 10 200 200 2 10 20 15 200 150 3 30 35 20 1050 600 4 25 57,5 10 1437,5 250 5 65 32,5 0 2112,5 0 6 10 0 5 0 50 Σ 160 5000 1250
Damit xmmmm
mm ymmmm
mmS S= = = =5000160
31 21250
1607 8
2 2
, ; , ;
Lösung 5.2
1 2 3
4
5 6
x
y
xS
S yS
l1
l2
l3 l4
l5
A1 A2
A3 x
y
S2 S1
S3
[ ] [ ] [ ] [ ] [ ]i l a x a y a x l a y l ai Si Si Si i Si i2 2
1 28 0 0 0 0 2 9,42 2 1 18,84 9,42 3 2 -3 -1 -6 -2 4 3 -1,5 -2 -4,5 -6 5 3,61 -1,5 -1 -5,41 -3,61 Σ 46,03 2,93 -2,19
Damit x a a y a aSl Sl= = = − = −2 93
46 030 06 2 19
46 030 05,
,, ; ,
,, ;
[ ] [ ] [ ] [ ] [ ]i A a x a y a x A a y A ai Si Si Si i Si i2 3 3
1 48 0 0 0 0 2 -7,07 2 1 -14,14 -7,07 3 -3 -2 -1,333 6 4 Σ 37,93 -8,14 -3,07
Damit x a a y a aSF SF= − = − = − = −8 14
37 930 22 3 07
37 930 08,
,, ; ,
,, ;
Lösung 5.3 [ ] [ ] [ ] [ ] [ ]i l a x a y a x l a y l ai Si Si Si i Si i
2 2
1 3,14 1,27 1,27 4 4 2 3 0 3,5 0 10,5 3 5 1,5 3 7,5 15 4 1 3 0,5 3 0,5 5 1 2,5 0 2,5 0 Σ 13,14 17 30
l1
l2 l3
l4
l5
A1 A2
A3
x
y
S2 S1
S3
Damit x a y a aSl Sl= = = =17
13 141 29a 30
13 142 28
,, ;
,, ;
[ ] [ ] [ ] [ ] [ ]i A a x a y a x A a y A ai Si Si Si i Si i2 3 3
1 -3,142 0,849 0,849 -2,667 -2,667 2 3 1,5 0,5 4,5 1,5
3 6 1 2,333 6 14 x y rS S= =
43π
Σ 5,858 7,833 12,833
Damit x a a y aSF SF= = = =7 8335 858
1 34 12 8335 858
2 19a,,
, ; ,,
, ;
Lösung 5.4
[ ] [ ] [ ] [ ] [ ]i A mm x mm y mm x A mm y A mmi Si Si Si i Si i2 3 3
1 600 10 15 6000 9000 2 900 50 7,5 45000 6750 3 1100 90 27,5 99000 30250 Σ 2600 150000 46000
Damit x mm mm y mm mmSF SF= = = =150000
260057 5 46000
260017 7, ; , ;
SxS
yS
S3 A2
A1
A3
S2 S1 x
y
xS
S yS
Lösung 5.5
[ ] [ ] [ ] [ ] [ ]i l a x a y a x l a y l ai Si Si Si i Si i2 2
1 4 -2 -1,5 -8 -6 2 2,356 -4,955 -0,955 -11,674 -2,25 3 3,5 -3,75 0 -13,125 0 4 1,5 -2 0,75 -3 1,125 5 2 -1 1,5 -2 3 6 4,712 0,955 0 4,5 0 7 6,283 0 0 0 0 Σ 24,351 -33,299 -4,125
Damit x a a y a aSl Sl= − = − = − = −33 29924 351
1 37 4 12524 351
0 17,,
, ; ,,
, ;
[ ] [ ] [ ] [ ] [ ]i A a x a y a x A a y A ai Si Si Si i Si i2 3 3
1 3,534 0,637 0 2,25 0 2 6 -1 0 -6 0
3 3 -3 -0,75 -9 -2,25 x y rS S= =
43π
4 1,767 -4,637 -0,637 -8,194 -1,126 5 -3,142 0 0 0 0 Σ 11,159 -20,944 -3,376
Damit x a a y a aSF SF= − = − = − = −20 94411 159
1 88 3 37611 159
0 3,,
, ; ,,
, ;
l1 l2
l3 l4
l5 l6
l7
A1
A2 A3 A4
A5
SxS
yS
Lösung 5.6
[ ] [ ] [ ] [ ] [ ]i A mm x mm y mm x A mm y A mmi Si Si Si i Si i
2 3 3
1 400 10 10 4000 4000 2 -78,5 15,76 15,76 -1237,2 -1237,2 3 350 37,5 5 13125 1750 4 78,5 59,24 4,24 4650,3 332,8 5 1000 5 70 5000 70000 6 78,5 4,24 124,24 332,8 9752,8 Σ 1828,5 25870,9 84598,4
Damit x mm mm y mm mmSF SF= = = =258711828
14 1 845981828
46 3, ; , ;
Vergleich mit DIN-Angaben: x cm y cmSF SF= =1 45 4 65, ; , ; Lösung 5.7
xA
x A yA
y A A A A
x y A mm
x mm y mm A mm
xmm
mmmm y
mmmm
mm
S Sii
i S Sii
i
S S
S S
S S
= = = +
= = =
= − = − =
=− ⋅
= − =− ⋅
= −
= =∑ ∑1 1
0 0 2290
4 75 13004 13003590
1 475 13003590
27 2
1
2
1
2
1 2
1 1 12
2 2 22
3
2
3
2
,
,
,
, ; ,
Lösung 5.8
Es gilt: Die Statischen Momente der Fläche bezüglich der Achsen durch S sind Null.
S y A S x Ax Sii
i y Sii
i= = = == =
∑ ∑1
2
1
2
0 0
Symmetrie zur y-Achse, d.h. Sy = 0 ist erfüllt.
1
2 3 4
5
6
S
h
1
2
x
y
i Quadrat ya
h A a
i Dreieck y h A ah
S a h a ah h ah a
h a a Lösung
S
S
x
= = − =
= = − = −
= −
+ = − + =
= =
12
2 23 2
2 30 3 3
20
2 37 0 63
1 12
2 2
22
2 2
1
( ): ,
( ): ,
, ( , ( )scheidet aus), h 2
Lösung 5.9 Waagerechte Lage der Unterkante dann, wenn der Schwerpunkt S auf der Vertikalen durch A (y-Achse) liegt. Dann gilt Sy = 0.
acAufgabevergleiche
aaccbcbca
bcAcxDreiecki
bcaAcaxecki
S
S
63,0)8.5(
02330
62)(
2,
3:)(2
)(,2
:)Recht(1
2222
22
11
=
=+−=−−
=−==
−=−
==
Lösung 5.10
Äquivalenz:
( )
°=⇒==
=+⋅=⋅=
−=−=−°=
−=−°°=
9,4396154,0tan
0036,0
0026,0003,0321cos30
0025,0sin30+sin30
22
12
321
ϕϕS
S
GZSSGGZ
SGSG
SSG
xy
FFyxFMSFrF
rxFrFFrxF
ryrFrFrFyF
Lösung 5.11 Querschnitt symmetrisch zur x -Achse, d.h. yS = 0 .
xx A x A
A A
Normprofil x mm A mmStegblech x mm A mm
x mm
SS S
S
S
S
=++
− = − =
− = − =
= −
1 1 2 2
1 2
1 12
2 12
1 15 5 13502 52 5 500
25 5
: , ,: , ,
,
S 12
x
y
S M
Z1 2
3
x
y
6.Flächenträgheitsmomente Lösung 6.1
[ ] [ ] [ ] [ ] [ ] [ ]i A c y c y A c I c I c y A ci Si Si i xxi yyi Si i2 3 4 4 2 4
1 6 0 0 12 18 0
2 9 5 45 2434
34 225
Σ 15 45 2454
754 225
Damit y c c I c I c c I cS xx xx yy= = = +
= −
= =4515
3 2454
225 11454
135 6054
754
4 4 4 4; ; ;
Lösung 6.2
[ ] [ ] [ ] [ ] [ ] [ ]i A h y h y A h I h I h y A hi Si Si i xxi yyi Si i2 3 4 4 2 4
1 40 0,5 20 83,33 213,33 10 2 -24 0 0 -32 -72 0
Σ 16 20 51,33 141,33 10
y h h I I h IS xx xx yy= = = = − ⋅
= =2016
1 25 61 33h 61 33 2516
16 36 33h 141 33h4 4 4 4, ; , ; , , ; ,
Ixy = 0; I1 = Iyy; I2 = Ixx Lösung 6.3
Zerlegung der Sechskantfläche in vier gleiche Teilflächen. Diese werden weiter in zwei Dreiecke zerlegt.
Das Flächenträgheitsmoment des Kreises ist I Id
xxK yyK= =π 4
64
Wegen mehrfacher Symmetrie gilt: Ixx = Iyy und Ixx = IxxK - IxxS
Ibh
xxD =3
12
y y,
xS1
S2
S1 S2
y y,
x
x
y
x
y
b
h
h s
Is s s s
s
I I d s
xxS
xx yy
=
=
−
=
= = −
3
4 312
2 2 312
5 3144
645 3144
3 3
4
44π
Lösung 6.4
Für I20 entnimmt man DIN1025: A = 3,35 . 103mm3; Ixx = 21,4 . 106mm4; Iyy = 1,17 . 106mm4; h = 200mm; b = 300mm; c = 400mm
Gurte:
463
433
232 1033,53
12;1033,33
12;104 mmtcImmctImmctA yyxx ⋅==⋅==⋅=⋅=
[ ] [ ] [ ] [ ] [ ] [ ] [ ]i A mm x mm y mm I mm I mm y A mm x A mmi Si Si xxi yyi Si i Si i
2 4 4 2 4 2 4
1 3,35.103 150 0 21,4.106 1,17.106 0 75,38.106
2 4 .103 0 105 0,03.106 53,3.106 44,1.106 0 3 4 .103 0 -105 0,03.106 53,3.106 44,1.106 0 4 3,35.103 -150 0 21,4.106 1,17.106 0 75,38.106
Σ 14,7.103 42,87.106 109,0.106 88,2.106 150,8.106
Ixx = (42,87 + 88,2) .106mm4 = 131,1 .106mm4; Iyy = (109,0 + 150,8) .106mm4 = 259,8 .106mm4; Lösung 6.5 1. Differenz zweier Rechtecke
( )( ) ( )( ) ( ) ( )
( )( ) ( )( ) ( ) ( )
Ib h b h h h b mm
Ih b h b b b h mm
xx
yy
=+ +
−− −
= + + + = ⋅
=+ +
−− −
= + + + = ⋅
δ δ δ δ δ δ
δ δ δ δ δ δ
3 3 2 34 4
3 3 2 34 4
12 12 63b
63h 5 352 10
12 12 63h
63b 1 88 10
,
,
2. Dünnwandiger Träger, d.h. δ << h,b (Vernachlässigung von Größen, die in δ klein von höherer Ordnung sind)
x
y
s
h 0,5h
0,5s
1
2
3 4
( )
( )
I h b h h h mm
I b h b b b mm
xx
yy
= + ⋅
= + = ⋅
= + ⋅
= + = ⋅
212 2 6
3b 5 333 10
212 2 6
3h 1 867 10
3 2 24 4
3 2 24 4
δδ
δ
δδ
δ
,
,
Lösung 6.6
[ ] [ ] [ ] [ ] [ ]i A a x a y a x A a y A ai Si Si Si i Si i2 3 3
1 15 -1,5 2,5 -22,5 37,5 2 7,5 1 1,667 7,5 12,5 Σ 22,5 -15 50 x a a y a aS S= − = − = =
23
0 667 209
2 222, ; , ;
Fortsetzung der Tabelle: [ ] [ ] [ ] [ ] [ ] [ ]I a I a I a y A a x A a x y A axxi yyi xyi Si i Si i Si Si i
4 4 4 2 4 2 4 4
31,25 11,25 0 93,75 33,75 -56,25 10,42 3,75 3,125 20,84 7,5 12,5 41,67 15 3,125 114,59 41,25 -43,75 I xx = (41,67 + 114,59)a4 = 156,26a4; I yy = (15 + 41,25)a4 = 56,25a4 I xy = (3,125 + 43,75) = 46,875a4
I I y Axx xx S= − =2
(156,26 - 111,09)a4 = 45,17a4 I I x Ayy yy S= − =2
(56,25 - 10)a4 = 46,25a4 I I x y Axy xy S S= + = (46,875 - 33,317)a4 = 13,56a4
Hauptträgheitsmomente und Hauptachsen:
( ) ( ) ( )I a a1 22 2 4 445 71 0 54 13 56 45 71 13 57, ( , , , ) , ,= ± − + = ±
I1 = 59,28a4; I2 = 32,14a4
°==−
=−
= 1,4604056,156,13
17,4528,59tan 011
01 ϕϕxy
xx
III
Lösung 6.7 A = 13,0cm2; xS = 1,6538cm; yS = -2,6538cm; Ixx = 80,78cm4; Iyy = 38,78cm4; Ixy = -32,31cm4; I1 = 98,31cm4; I2 = 21,24cm4; ϕ01=-28,49°
Lösung 6.8 A = 18,0cm2; xS = 0; yS = 0; Ixx = 246cm4; Iyy = 61,5cm4; Ixy = -90cm4; I1 =282,63cm4; I2 = 24,87cm4; ϕ01=-22,15° Lösung 6.9
( )A D d mm cm
I I I I ID d d
cm
I I cm
p xx yy
= − = − ⋅ =
= = = = = − + ⋅
=
= =
π π π
π π π
2 24 2 2 2
1 2
4 42
24
1 24
46
4 410 6 10 73 83
2 2 2 232
632
354
923 43
461 72
,
,
,
Lösung 6.10 A = 530cm2; yS = 36,54cm; Ixx = I1 = 39,756 . 104cm4; Iyy = I2 = 30,71. 104cm4
Lösung 6.11 A = 105cm2; yS = 8,63cm; Ixx = I1 = 1579cm4; Iyy = I2 = 817,5cm4
7. Haftung und Reibung Lösung 7.1
:Fcosβ + FH - FGsinα = 0 (1) FHmax = µ0 FN (2) :FN + Fsinβ - FGcosα = 0 (3) FN aus (3) und (2) in (1): Fcosβ + µ0(FGcosα - Fsinβ) - FGsinα = 0
oder
Nin
FF G 6,426scoscossin0
0 =−−
=βµβαµα
Selbsthemmung: F = 0, d.h. sinα0 - µ0 cosα0 = 0 tanα0 = µ0 α0 = arctanµ0 = 11,3° Lösung 7.2
A: FS a cosα + 3F a cosα = 0 FS = -3F :FS cosα ± FH + FGsinα = 0 (1) FHmax = µ0 FN (2) :FN + FS sinα - FGcosα = 0 (3)
-3F cosα ± µ0 (FGcosα + 3F sinα) + FGsinα = 0
( ) ( ) NFFNFF GG 113sincos3
cossin293sincos3
cossinmin
0
0max
0
0 ==+
−==
−+
αµααµα
αµααµα
Lösung 7.3
C F a F a F F
B F x F a x
F F F
F x F a x x a x
G N N G
H G
Hmax N G
G G
:
:
⋅ − ⋅ = =
± ⋅ − −
=
= =
± ⋅ − −
= ± ⋅ − + =
20 1
2
20
12
12 2
0 12 2
0
0 0
0 0
µ µ
µ µ
x a x a mm x mm1 12 2 2
400 66700
±
= =±
≤ ≤µµ
FN FH
FG
F
α
β
FN FH
FG FS α
FS
F
A
α
α
α
xFG
FH
FN
FS
A B
C
Lösung 7.4
B: FNC . l - F . cosα . (l + x) = 0 C: ±FHB . l . cosα + FNB . l . sinα - F . x . cosα = 0 →: FNB ∓ FHC . cosα - FNC . sinα = 0 FHBmax = µ0 . FNB FHCmax = µ0 . FNC
( ) ( )tan
tantan
tanα µ
µ µ αα µ
µ µ α−
− +⋅ ≤ ≤
+− −
⋅02
02
0
02
02
01 2 1 2l x l
Lösung 7.5
( )
( )2;5,00123:45
0cotcot143
0scosls87cos
8:
0:0:
0201020
020
0max0max
−===−+°=
=−++
==
=⋅+⋅+⋅−⋅
=+−↑
=−−→
µµµµα
αµαµ
µµ
αααα
NBHBNAHA
HBNB
HBHA
NBNA
FFFF
inlFlFinFlFA
FFFFFF
Lösung 7.6
( )
::
:
− − + =
− + =
− ⋅ ⋅ −
+ ⋅ =
= =
=− +
F F FF F F
A Fl
a F b
F F F F
bl a b
NA G NB
HA G HB
G NB
HAmax NA HBmax NB
cossin
cos
tan
α
α
α
µ µ
µ α
00
20
2
0 0
0
Lösung 7.7
Beide Grenzlagen findet man schnell durch Konstruktion der Reibkegel. (Bild zeigt qualitativ die grafische Lösung) tanρ0 = µ0 = 0,5; ρ0 = 26,6° Mit Maßstab, z.B. 1m entspicht 2cm in der Zeichnung erhält man x1 = 0,2m ≤ x ≤ x2 = 4,2m
B
C
FNBFNC
F
FHCFHB
α
F
F
FNB
FNA
FHB
FHA
α
FG
FNAFHA
FNB FHB
α
x1
x2
α β
ρ0 ρ0
ρ0 ρ0
Lösung 7.8
: FN1sinα - FH1cosα - FN2 = 0 (1) : FN1cosα + FH1sinα - FH2 -FG1 = 0 (2) FH1max = µ01FN1 FH2max = µ02FN2 (3) : 2 FH2 - FG2 = 0 (4)
Aus (4) mit (3) : F FN
G2
2
022=
µ
In (1) und (2) eingesetzt, erhält man nach Elimination von FN1 die Lösung: ( )
( )( )
( )αµααµαµ
αµααµαµ
os
FFmitoderFFos
FGG
GG
G
csin2sincos
21
2csinsincos
01
0102
211201
20102
−+
=
=+−
+=
Lösung 7.9
Symmetrie!
( )
( )
( )
I F F F F
II F F F F
F F F F
III F a F d F b F cF acos dsin F b F c
F acos dsin b F c FF
c acot d bb
acot d c
G B B G
H G H G
Hmax N NG
B B N H
B N H
G G GG
:
:
:
,
+ = = − ⋅
− = =
= =
⋅ + ⋅ + ⋅ + ⋅ =
+ + ⋅ + ⋅ =
− + + + = ⋅
− − + =
=+ −
=
2 0 12
1
2 0 12
20
0
12 2 2
0
0
0 11
00
0
0
0
0
sinsin
cos sin
sin
αα
µµ
α α
α α
αα α
µµ
µ α
µα
Lösung 7.10
↑ − = = = =
→ − − = =
− ⋅ + ⋅ + ⋅ − ⋅ =
= ⋅ = =
:
:
:
F F F F F F F
F F F F F
B F h F b F c F c
b FF
h h mm
N N N N N Hmax N
N N N
N N N
N
1 2 1 2 0
0 00
0 0
0
0
02
2 20
2 48
µ
µ µµ
µ µ
µ
FG1
FG2
FN1 FN2
FH1
FH2
S
FN2FN2FH2
FH2
α
FB
FB FB FG
FC
I
II
III
FN FH
FG= mg FN FN FH FH
h b
F
FN1 FH1
FH2 FN2 c B
Lösung 7.11 → : FN2 - FH1= 0 ↑: FN1 + FH2 - FG = 0 : M- FH1 r - FH2 r = 0 Haftgrenzbedingung : FH1max = µ0 FN1 FH2max = µ0 FN2
FN2 - µ0 FN1 = 0 FN1 + µ0 FN2 - FG = 0 M - µ0 r(FN1 + FN2) = 0
( )F F F F M F rNG
NG
G102 2
0
02
0 0
021 1
11
=+
=+
=+
+µµ
µµ µ
µ
Lösung 7.12
p A F mg dF mgA
dA mg drl
M rdF Mmg
lrdr
M mgl
N N
N
l l
⋅ = = = =
− = − =
=
∫ ∫2 0 2 0
14
00
00
0
2 2
µ µ
µ
Lösung 7.13
( ) ( )
::
:
( )
( ) ( ) ( )
, , , ,
F F F F F FF F F
F a F b F a F F F ba
F F F F
F F F F ba
F
F ba
F
F F F F
A B G A G B
A A B B C
B C G B G C
A G B B B C
B B A C A G B A G B A C
C B A G B A A
C G C G
+ − = = −
+ − =
+ − ⋅ = = −
− + − =
− = − = − − −
+ −
= − +
= = =
00
20 1
2012
1 12
1 125 0 25 0 22 188 9
µ µ
µ µ
µ µ µ µ µ µ µ
µ µ µ µ µ
N
Lösung 7.14
kNmg
adF
mgFdFFa NN
35,221
02
:
0
0
=⋅=
==⋅−
µ
µ
Mr
FG
FN1
FN2
FH1FH2
r
drMµ0dFN
µ0dFN
b
a
FCFG
FA FB
µA FA µB FB
Fmg
FNµ0 FN
r
A
Lösung 7.15
MB = r ( FR1 + FR2 )
:
:
( )
( )
F r F r Frr
F F
F r F r Frr
F F
F F F F
F F F F
M Fr
N R R N
N R R N
R R
R R
B
1 1 1 1
2 2 2 2
1 1
2 2
2
2 0
2 0
1 2 21
1 2 21
41
− − = ⋅ =
− − + = ⋅ −
=
− = =−
+ = =+
=−
µµ
µµ
µ µµµ
µ µµµ
µµ
Lösung 7.16
Annahme: Stab sei dünn Gleitreibkraft FR = µFN
( )
→ + − + =
↑ − − − − =
⋅ − + =
→ − − − + =
↑ − + + − =
: ( )
: ( ): ( )
: ( )
: ( )
F F F F
F F F F FF b F a b
F F F F
F F F F
N N N N
N N N N G
N N
N N N
N N N GK
2 2 3 4
2 2 3 4
3 4
2 2 1
2 2 1
0 1
0 22 0 3
0 4
0 5
sin cos
cos sin
sin cos
cos sin
α µ α
α µ α µ µ
α µ α µ
α µ α
5 Gleichungen für 5 Unbekannte
Ergebnis: ( )[ ]
( )[ ] ( )F F F
a
a b a bGK G= + ⋅− + ⋅
− − − +µ
µ α µ α
µ µ α µ α
1 2
1 2 2
2
2 2
sin cos
cos sin
Lösung 7.17
Gleitreibkraft FR = µFN Darstellung der Verhältnisse für Heben:
( ) ( )( ) ( )
( )( ) 0sincos:
0cossin:0sincossincos:
0cossincossin:
32
32
21
21
=+−−↑
=−++→=−−+−↑
=+−+→
NN
NN
GNN
NN
FFFFF
FFFFF
βµβ
µβµββµβαµα
βµβαµα
FF
A
FN1 FN2
FR1 FR2
FN1 FN2FR1 FR2MB
ϕ
FG
F
FN4
µFN4
µFN3
µFN2
µFN1
µFN2
FN3
FN2
FN1
FGK
2
α
α
FG
µFN2
µFN3
µFN2
µFN1
α
α
β
βFN1 FN2
F
FN3
FN2
A
A
Ergebnis für Heben: ( )[ ][ ]
( ) ( ) ( )F F F NG= ⋅
+ +
+ + +=
1 - sin cos sin cos
1- sin cos
2
2
µ β µ β α µ α
µ α β µ α β
2
2124
Ergebnis für Senken: (Vorzeichen bei µ kehren sich um, nicht aber bei µ2)
( )[ ][ ]( ) ( ) ( )
′ = ⋅− −
+ − +′ =F F F NG
1- sin cos sin cos
1- sin cos
2
2
µ β µ β α µ α
µ α β µ α β
2
253
Lösung 7.18 Fest vertäutes Schiff: Haftung zwischen Seil und Poller.
Es gilt: F F eS S1 20= ⋅ µ α
ziehender Strang: FS1 = FS gezogener Strang: FS2 = FH Umschlingungswinkel: α = n . 2π F F e nS H
n= ⋅ = =µ π
µ π0 2
0
12
5 5ln FF
S
H
,
Lösung 7.19 Gleitreibung zwischen Seil und Rundholz
Es gilt: F F eS S1 2= ⋅ µα
Hochziehen: ziehender Strang: FS1 = FS gezogener Strang: FS2 = FG Umschlingungswinkel: α = 120°
F F e FS G G= =µ π2
3 1 52, Ablassen:: ziehender Strang: FS1 = FG gezogener Strang: FS2 = F’
S
F F e F F e FG S S G G= = =−' ' ,
µ π µ π23
23 0 66
1
2
FS
FH
FG
FSα
α F’S
12
FG
12
Lösung 7.20 Scheibe läuft unter feststehendem Bremsband durch.
Es gilt: F F eS S1 2= ⋅ µα (1)
( ) ( ) GGS
GSS
GSS
S
S
SGS
Ferl
raFFer
rF
FrreFFmit
FrrFFaus
FlaFaus
lFaFBrFrFrF
11
:)1(
:)2(
:)3(
)3(0:)2(0:0
112
122
112
2
2
1211
−⋅⋅
=−
=
−⋅=
−=
=
=⋅−⋅=⋅−⋅−⋅
µαµα
µα
F
FS1
FS2
FS2
FG
0
B
ziehend
gezogen
8. Das Prinzip der virtuellen Arbeit Lösung 8.1
( )
δ δϕ δϕ ϕ
ϕ δϕ ϕ
ϕ
W M Fl
M FlsinMFl
MFl
Gleichgew nur für MFl
möglich
= −
− = =
= ≤
0
00
0 0
0
1
sin = 0
sin
arcsin
*
* . .
Lösung 8.2
U m g ba
m g la
dUd
m ga
m gamm
= − +
− −
= − + = =
1 2
1 21
2
2
21
02
tancos
cossin
cossin2 2
αα
α ααα
α*
Grenzfälle: m1 > 2m2 kein Gleichgewicht möglich ( sin α*≤ 1 )
m1 = 2m2 απ* = =2
0für a
m1 = 0 α* = 0 Lösung 8.3
δ δ δ δ δα
α β δ αδα βδβ
β α β α
βδβ αδα δβαβ
δα
β β α
W F f F q q rf rcos lcos f rsin lsin
a lsin rsin rl
diff
rl
rl
rl
= − − =
= + = − −
= = =
= =
= − = −
1
2
1 1
(*)
.sin sin
cos cos coscos
cos sin sin2 2
(*) − + +
−
=
= +−
F r F rsin l rl
rl r
l
F Fsinr
1 2
1 2 2
10
11
δα αδα αα
α
δα
αα
α
sin cos
sin
rcossin
2
2
F
M0
ϕδϕ
ldϕlsinϕδϕ
m1gm2g
αa
l a−2
2cosαa
2cosα b
∇
f
δf
α
β
q
δq
l
r
a
Lösung 8.4
δ
δ δ δ δ
δ δ δ
WW F F F
F FS A B C
A B C S
=
= − − + =
= = =
03 0
2
Lösung 8.11
( )
( )
( )( )
( ) ( )
U F lcos F lcos F lcos c c
l F F c
U l F F c
U l F F c für Gleichgewi cht U
cl F F
cl F F
A
GS GS Gk T T
GS Gk T
GS Gk T
GS Gk T
T
GS Gk
T
GS Gk
= + + + +
= + +
′ = − + +
′′ = − + + ′ =
=+ +
=
12
32
2 12
12
2
2 52
2 5
2 5 0
52
52
2 2
2
ϕ ϕ ϕ ϕ ϕ
ϕ ϕ
ϕ ϕ
ϕ
ϕ ϕ
cos
sin
cos
sin
a.) A ≥ 1 : nur Gleichgewichtslage ϕ1 = 0 b.) A < 1 : es ex. weitere Gleichgewichtslagen ϕj ≠ 0, j = 1...N Stabilität:
( )( )( )
möglichinstabilauchalsstabsowohlAFFlUAb
LagestabcFFlUAa
jGkGSj
TGkGS
.cos2)(:1.).052)(:1.) 1
ϕϕϕ
−+=′′<>++−=′′>
Lösung 8.16
( )
δ δ αδα
ϕ δ ϕδϕ α ϕ δα δϕ
ϕ ϕ δϕ ϕ ϕ
W F x cx lcos x lsin
Flsin c cFl
T
TT
= − − =
= = − = =
− = =
02 2 2 2
2 4 0 2sin
FS
FS
F
3F
δC
δBδA
FGk
FGS
FGS
cT2ϕ
cTϕ
ϕ
Aϕ
ϕsin ϕ
x, δxϕ
MC
Lösung 8.17
( )U Fl
F l
U l F F lsin F F
Ul
F F lcosF
für labil
G Q
G Q Q G
G QG
= ⋅ − −
′ = − = =
′′ = − − = − < ≤ ≤
21
20 1
2
212
0 02
lsin cos
lcos cot
lsinl
sin
α α
α α α
α αα
απ ( )
Lösung 8.18
1.5
0
0.7167phi.
sin phi( )
34
π.0 phi0 0.79 1.57 2.36
0
0.5
1
1.5
FQ
l
α
l-lcosα∇
FG
stabil0U78,5instabil0U:0
cosNcm10313,92Ncm10225cosmglcrU
78,5,0
0,7167mglcrsin
0sinmglcrU
cosmglcr21
rfcosmglcf21U
2
1
22221
2
2
22
2
>′′°=<′′=
⋅⋅−⋅=⋅−=′′
°==
==
=⋅−=′
⋅+=
=⋅+=
:
,
ϕϕ
ϕϕ
ϕϕ
ϕϕϕ
ϕϕ
ϕϕ
ϕϕ
1,37 ≡ 78,5°
