lösungen für aufgaben zur technischen mechanik – statik · lösungen für aufgaben zur...
TRANSCRIPT
Lösungen für Aufgaben zur Technischen Mechanik – Statik - Lösung 2.1.8
Lösung 2.2.3
Lösung 2.2.7
FS1FS2
FG = mgK
α1 α2
x
yF1
F2
F3F4
0A
B
r
S
0 F
FG
FS1 FS2
FS3
→: -FS1cosα1 + FS2cosα2 = 0 F FS
S2
1=cos
cos1
2
αα
↑ : FS1sinα1 + FS2sinα2 - mg = 0 FS1(sinα1 + tanα2 cosα1) = mg
°===−−+
=
°====
1,34677,06062,4tan
3,278889,05,4
4cos
2
22
2
11
αα
αα
bcbla
lb
Ergebnis: FS1 = 277N, FS2 = 297N
°=°−==
=+=+=
30060tantan
23 2222
RRx
RyR
RyRxR
FF
FFFFFF
αα
FRx = Σ Fix = F1cos60° + F2cos60° - F3cos60° + F4cos60° = 2Fcos60° = F FRy = Σ Fiy = F1sin60° - F2sin60° - F3sin60° - F4sin60° = -2Fsin60° =−F 3
← : 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
Ergebnisse: FS1 = - 812 N....Druckstab FS2 = - 891 N....Druckstab FS3 = 840 N....Zugstab
Lösung 3.1.2
Ergebnis: Lösung 3.1.12
Ergebnis: FB = 3,82F; FAH = - 1,09F; FAV = 2,7F 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
FC
FAH FAV F sinα
F cosα
3F FBcos30°
FAHFAV
4F
2F
FBsin30°
FAH FAV
2,5a 5qaF1
FGV
FGH
FGV
FGHF2
2a FB
I
II
MA
kNFFF
kNFosFF
kNFFF
AV
AH
C
1343sin
21
1521c
39343sin
23
===
===
===
α
α
α
→ : FAH - F cosα = 0 ↑ : -FAV - F sinα + FC = 0 ↶A : -F sinα . 3a + FC
. 2a = 0
→ : 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:
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 0 25
640 0 25
5 2440
3 2 252
8100
2 2 1 2
1 22
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: 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
FAV
FAH
MA FGVl FGHl
F1
F2
FB
FGHl
FGVl FGVr
FGHr
FGVrFGHr
FF
F
FB
FAH
FAV
III
III
3a 3a
2a
a a
l1 l2
γ α
α β
atanα
I
FAV
FS1FS2
α β
;;
;;
;;
221
2121
222
acFFcFaFM
acFFF
acFFF
acFFFFFFF
GVrA
GVlAV
BGHrGHlAH
⋅=⋅+⋅=
⋅+=⋅+=
⋅−====
: -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 3.2.25
→: 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 0 32
0 34sin
2 0 84
⋅ − ⋅ + ⋅ = = −
− ⋅ + ⋅ = =
↑ + = = −
= ° = −
sin
tan folgt = 63,4
αα
α α
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
II
F
FS1
FS4
FS3α
IIIFS6
FS5FS3
FS2
α γβ
FF
FB
FAH
FAV
FS3F
FB
FS1
FS2
αI
II
FBHFA
F1
F2
F3
FBV
FAz1
FL(z1)
FQ(z1)
M(z1)
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
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
FL(z2)a
F1 M(z2)
FQ(z2)FA
z2
FL(z3)a
F3M(z3)
FQ(z3)
FBVz3 a-z3FBH
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
-+
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⏐=⏐MB⏐ist nur für
negatives MA möglich.
- MA = MB oder − −⎛⎝⎜
⎞⎠⎟ = =ql c l qlc c l
2 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 =====
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− ;
Fql
l
B
FAVFAH
MA c
BMB
c
F
FAH FB3a
4qa2a
FAV
a
z1
FL(z1)
FQ(z1)
M(z1)qz1
FL(z2) FB
M(z2) FQ(z2)
z2 3a-z2
q(3a-z2)
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 3.3.13
→ : 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
-
--
+
+
FQ
M
10kN
16,7kN
13,3kN
5kNm
8,9kNm
FAV
F
MA
FAH
qa
FL(z1)
M(z1) FQ(z1)
z1
qz1
FL(z2)
z2
M(z2)
FQ(z2)
qa
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
⏐ Mmax ⏐ = 1,5 qa2
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) ⏐Mmax⏐I = ⏐Mmax⏐II
( )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= − = − + − = = − =( ) ,
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= = = = − = −;
FL(z3)
z3
M(z3)
FQ(z3)
qa
F
FL in qa FQ in M in qa2
- -
-
-- 1
1 1
1,5
0,5
zFL(z)
FQ(z)
M(z)
FA
qz
45°
45°45°
Fx
y
z
FS1
FS2
FS3
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
2 16
0 62
1 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 4.3.1
Nach dem Erstarrungsprinzip gilt:
Fu1 . r01 - Fu2 . r02 = 0 oder F F rru u2 1
01
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
a 5a 6
I
II F
a
2a
a
FS6
FS4
FS5
FS1 FS2FS3
FAz
FAxFAy
y Fu1
Fu2
Fa1
Fa2
Fr1
Fr2
FBx
zFBy
z1
Mx1My1
Mz1FL1FQx1FQy1
FAx FAy
FAz
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
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( ) ( )
z2
Mx2
My2
Mz2FL2FQx2
FQy2FAx FAy
FAz
l1
Fr1
Fa1
Fu1
z3
Mx3 My3
Mz3 FL3
FQx3FQy3 FBx
FBy
l3-z3
Lösung 5.4
[ ] [ ] [ ] [ ] [ ]i A mm x mm y mm x A mm y A mmi Si Si Si i Si i
2 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, ; , ;
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
S3A2
A1
A3
S2S1
x
y
xS
SyS
S1
S2
y y,
x
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 Ixx = (41,67 + 114,59)a4 = 156,26a4; Iyy = (15 + 41,25)a4 = 56,25a4 Ixy = (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 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
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
FFF
FFF
xFG
FH
FN
FS
A B
C
F
F
FNB
FNA
FHB
FHA
α
Lösungen für Aufgaben zur Technischen Mechanik – Festigkeitslehre - Lösung 2.16
→ =
↑ = − +
− − =
= −
: ( )
: ( )
: ( )( )
F
F F F qa
B F a qa F al f
BH
BV C S
S C
0 1
2 2
2 2 0 32 4
2
∆
Für die lineare Feder gilt: f FcC= und damit wird aus Gleichungen (3) und (4)
2 2
2 0
14
24
14
14
F F qaF lEA
Fc
F qaclEA
Fqa cl
EAclEA
F qaclEA
S C
S C
S C BV
− =
+ =
=+
= −+
=+
; ;
Lösung 2.22
↓ : FS1 + FS2 + FS3 + F = 0 ( 1 ) A : FS2 a + 2 FS3 a + F x = 0 ( 2 )
w F lEAi
Si i= ( 3 ), ( 4 ), ( 5 )
1.) w1 = w2 = w3 F aEA
F aEA
F aEA
F F F FS S SS S S S
1 2 32 1 3 1
32
2 23
12
= = = = ( 6 ), ( 7 )
( 6 ) und ( 7 ) in ( 1 ) : F F F F F FS S S S1 1 1 123
12
613
+ + = − = −
( 6 ) und ( 7 ) in ( 2 ) : 23
53
10131 1
1F a F a Fx x FF
a aS SS+ = − = − =
2.)
σ σ σ11
22
336
134
133
13= = − = = − = = −
FA
FA
FA
FA
FA
FA
S S S
B2qa FS
FC
f ∆l
x FA
FS1 FS2 FS3
3.)
( )w w w2 1 312
= + ( 6’ )
( 1 ) bis ( 5 ) und ( 6’) liefern FS1 + FS2 + FS3 = - F ( 1’ )
F a F a FaS S2 32 23
+ = − ( 2’ )
32
122 1 3F F FS S S= + ( 6’ )
FS1 + FS2 + FS3 = - F (I) FS1 -3 FS2 + 2FS3 = 0 (II) F F FS S2 32 23
+ = − (III)
I - II : 4 FS2 - FS3 = -F III + 2.(I - II) : 9 832F FS = −
F F F F F F
w FaEA
w FaEA
w FaEA
S S S2 3 1
2 3 1
827
527
1427
49
1027
1427
= − = − = −
= − = − = −
Lösung 4.1
→ =
↑ + − =
− − =
− =
= = = = −
= =
:
:
:
:
; ; ;;
F
F F ql
C F l ql M
G F l ql
F ql F F ql M qlF F ql
CH
B CV
B C
B
B CH CV C
GH GV
0
2 0
3 4 0
2 2 0
00
2
2
2
( )
0 2 00 0
12
1 2
1 2
1 1 2
1 1 12
2 2
≤ ≤ ≤ ≤
= =
= − = −
= − = −
z l z lF z F z
F z q l z F z ql
M z qlz qz M z qlz
L L
Q Q
( ) ( )
( ) ( )
( ) ( )
Aufgabe 6.2 (Statik): Ixx = 36,33h4
w1 w2 w3
2ql
FB
FB
2ql
FGV
FGH
MC
FCV
FCHz1 z2
12
2ql
ql
ql
ql2
-
-
+
+
FQ-Verlauf
M-Verlauf
Maximalwerte für y: Unterseite y = e1 = 3,25h Oberseite y = -e2 = -1,75h Spannungsverteilung an der Einspannstelle (Maximales Moment)
σ
σ
σ
zxx
z
z
y qlI
y
y e qlh
Druckspannung
y e qlh
Zugspannung
( )
( ) ,
( ) ,
= −
= = − ⇒
= − = ⇒
2
1
2
3
2
2
3
0 0894
0 0482
Lösung 4.8
σ
σ
( ) ( )( )
( ) ( ) ( )
( ) ( )
( )
z M zW z
M z F z W z bh z
h z h zl
W z bh zl
z F z
bh zl
xx
x
max
max
= = ⋅ =
= +⎛⎝⎜
⎞⎠⎟ = +⎛
⎝⎜⎞⎠⎟
=⋅
+⎛⎝⎜
⎞⎠⎟
2
002 2
02
2
6
1 26
1 2
6
1 2
Ort der max. Biegespannung entweder an der Einspannstelle (max. Moment) oder an der
Stelle, an der d zdz
σmax ( )= 0 ist.
d zdz
Fbh
zl
zl
zl
zl
zl
zl
zl
σmax ( )=
+⎛⎝⎜
⎞⎠⎟ − +⎛
⎝⎜⎞⎠⎟
+⎛⎝⎜
⎞⎠⎟
=
+⎛⎝⎜
⎞⎠⎟ − = =
6 1 2 4 1 2
1 20
1 2 4 0 12
02
2
4
( ) 20
max20max 3
243
2 bhFll
bhFll
==⎟⎠⎞
⎜⎝⎛ σσ
Die absolut größte Spannung beträgt σmax =34 0
2
Flbh
und tritt an der Stelle z l=
2 auf.
Lösung 4.18 Größte Beanspruchung an der Einspannstelle! Max. Spannung an einem Eckpunkt des Querschnittes, d.h.:
σzy
Zugsp.
Drucksp.
h(z)
z
h03h0
σmax(z)
z 0,5ll
σmax
mmbmmbFlh
hFlb
hFlb
hFlb
hbFl
bhFl
hblFM
bhlFM
MMx
My
M
gewzul
zul
zulzulzul
607,5824
1112
06246246
W
6W22
WWII
3
2
22
22max
2
yymax
2
xxmax
y
ymax
x
xmaxmax
yy
ymaxmax
xx
xmaxmax
==⎟⎟⎠
⎞⎜⎜⎝
⎛ ⋅+±=
=−⋅−⇒=+=
=⋅=
=⋅=
+=+=
σσ
σσσσ
σ
Nur das positive Vorzeichen ist physikalisch sinnvoll. Lösung 4.33
( ) ( )
F F F F
z a z a
M z F z Fz M z F a z F a z
B C
B C
= =
≤ ≤ ≤ ≤
= = = − = −
23
13
0 0 223
2 13
2
1 2
1 1 1 2 2 2
;
( ) ( )
EIv z Fz EIv z F a z
EIv z Fz C EIv z F a z C
EIv z Fz C z C EIv z F a z C z C
RB ÜB v z Cv z a C a C
v z a v z Fa C
′′ = − ′′ = − −
′ = − + ′ = − +
= − + + = − − + +
= = ⇒ =
= = ⇒ + =
= = = ⇒ − +
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
/ : ( )( )
( ) ( )
1 1 2 2
1 12
1 2 22
3
1 13
1 1 2 2 23
3 2 4
1 2
2 3 4
1 23
23
13
2
13
16
2
19
118
2
0 0 02 0 2 0
0 19 1
34
1 22
12
3
12
2 32
43
49
0 13
23
59
0 49
89
a Fa C
v z a v z Fa C Fa C
C Fa C C Fa C Fa
= − +
′ = = ′ = ⇒ − + = +
= = = − =
( ) ( )
v z FaEI
za
za
v z FaEI
za
za
za
v v z a v z FaEIF
( ) ( )
( ) ( )
1
31 1
3
2
32 2
22
3
1 2
3
95
188 4 6
0 49
= − ⎛⎝⎜⎞⎠⎟
⎡
⎣⎢
⎤
⎦⎥ = + − ⎛
⎝⎜⎞⎠⎟ + ⎛⎝⎜
⎞⎠⎟
⎡
⎣⎢
⎤
⎦⎥
= = = = =
z1 z2
F
FB FC
Lösung 4.45
Die Aufgabe ist einfach statisch unbestimmt. Gleichgewichtsbedingungen: ← − =
↑ + − =
⋅ − ⋅ =
:
::
F F
F F qaB F a F a
AH BH
AV BV
AV AH
0
2 00
( )
( )
( )
( )
M z F z qz M z q a z
EIv z qz F z EIv z q a z
EIv z qz F z C EIv z q a z C
EIv z qz F z C z C EIv z q a z C z C
AV
AV
AV
AV
( ) ( )
( ) ( )
( ) ( )
( ) ( )
1 1 12
2 22
1 12
1 2 22
1 13
12
1 2 23
3
1 14
13
1 1 2 2 24
3 2 4
12
12
12
12
16
12
16
124
16
124
= ⋅ − = − −
′′ = − ⋅ ′′ = −
′ = − ⋅ + ′ = − − +
= − ⋅ + + = − + +
M z F zEIv z F z
EIv z F z C
EIv z F z C z C
BH
BH
BH
BH
( )( )
( )
( )
3 3
3 3
3 32
5
3 33
5 3 6
12
16
= − ⋅
′′ = ⋅
′ = ⋅ +
= ⋅ + +
RB/ÜB: 1 0 0 02 0 0 0
3 0 124
16
0 16
124
4 0 0 124
0 124
5 0 16
0 16
6 0 16
12
16
1 2
3 6
14 3
1 12 3
24
4 44
33
5 52
1 23 2
13
. ( ). ( )
. ( )
. ( )
. ( )
. ( ) ( )
v z Cv z C
v z a qa F a C a C F a qa
v z qa C C qa
v z a F a C a C F a
v z a v z qa F a C qa C
AV AV
BH BH
AV
= = ⇒ =
= = ⇒ =
= = ⇒ − + = = −
= = ⇒ + = = −
= = ⇒ + = = −
′ = = ′ = ⇒ − + = − + 3
3 22
53
3 32 37 0 1
216
13
16
. ( ) ( )′ = = ′ = ⇒ + = − + = +v z a v z F a C qa C C F a qaBH BH
Ergebnisse aus GGW und RB/ÜB:
aus 6. folgt mit den GGW F qa F F qaAV AH BH= ⇒ = =3
163
16 und F qaBV =
2916
FAH
FAV
FBH
FBV
z1 z2
z3
C
q
Momentenverlauf:
( )
( )
M z qz a z F z qa qz z a M z qa
M z q a z
M z qaz
Q( ) ( )
( )
( )
* * * *1 1 1 1 1 1 1
2
2 22
3 3
12
38
316
0 316
9512
123
16
= −⎛⎝⎜
⎞⎠⎟ = − = = =
= − −
= −
Verschiebung des Punktes C:
( )v v z aEI
C a C qaEIC = = = + =( )2 3 4
41 316
Lösung 4.50
FC = v(z2 = 0) . c
( )
( )
( )
( )
M z qz M z q a z F z
EIv z qz EIv z q a z F z
EIv z qz C EIv z q a z F z C
EIv z qz C z C EIv z q a z F z C z C
C
C
C
C
( ) ( )
( ) ( )
( ) ( )
( ) ( )
1 12
2 22
2
1 12
2 22
2
1 13
1 2 23
22
3
1 14
1 1 2 2 24
23
3 2 4
12
12
12
12
16
16
12
124
124
16
= − = − + +
′′ = ′′ = + −
′ = + ′ = + − +
= + + = + − + +
RB/ÜB:
1 0 12
43
2 0 23
13
2 32 3
2 44 3
. ( )
. ( )
′ = = ⇒ = −
= = ⇒ = −
v z a C F a qa
v z a C qa F a
C
C
3 0 12
43
4 0 2 56
1 2 1 32 3
1 2 2 4 14 3
. ( ) ( )
. ( ) ( )
′ = = ′ = ⇒ = = −
= = = ⇒ = − = −
v z a v z C C F a qa
v z a v z C C C a qa F a
C
C
516
2qa0,5qa2
316
2qa
M(z)
-9
5122qa
316 a
-
-
FC z1z2
MA
FA
D
F v z c qaEI
F aEI
c F qacEIa
c
F c qa
v v z CEI
qaEI
F aEI
v c qaEI
v c qaEI
qaEI
qaEI
CC
C
C
DC
D D
= = ⋅ = −⎛⎝⎜
⎞⎠⎟ ⋅ ⇒ =
+⎛⎝⎜
⎞⎠⎟
→ ∞ =
= = = = −
→ = →∞ = − =
( )
( )
( )
( ) ( )
2
4 3
3
12
4 3
4 4 4 4
0 1724 3
17
8 3
178
0 2 56
0 2 2 8548
1148
Lösung 5.6
( )14 4
100 100 25
216
1
100 1 100 1
11 100 16
151
22 2
2
3 4
4
.
% % %
.
% % %
m A l D l m A l D d l
m mm
dD
MW
MW
W D W W dD
WW d
D
v v h h
v h
v
vt
tvh
t
thtv th tv
h v
v
tv
th
= ⋅ ⋅ = ⋅ ⋅ = ⋅ ⋅ = − ⋅ ⋅
−⋅ = ⎛⎝⎜
⎞⎠⎟ ⋅ =
= = = = − ⎛⎝⎜⎞⎠⎟
⎛⎝⎜
⎞⎠⎟
−⎛⎝⎜
⎞⎠⎟ ⋅ = −
⎛
⎝⎜
⎞
⎠⎟ ⋅ =
− ⎛⎝⎜⎞⎠⎟
−
⎛
⎝
⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟⋅ = −
ρπ
ρ ρπ
ρ
τ τπ
τ ττ
⎛⎝⎜
⎞⎠⎟ ⋅ =
−⎛⎝⎜
⎞⎠⎟ ⋅ = −
⎛
⎝⎜
⎞
⎠⎟ ⋅ =
− ⎛⎝⎜⎞⎠⎟
−
⎛
⎝
⎜⎜⎜⎜
⎞
⎠
⎟⎟⎟⎟⋅ = −⎛
⎝⎜⎞⎠⎟ ⋅ =
100 6 7
3 100 1 100 1
11 100 16
151 100 6 74
% , %
. % % % % , %ϑ ϑϑ
h v
v
tv
th
II d
D
Lösung 7.11 Momente an der Einspannstelle: Biege- und Torsionsmoment erreichen an der Einspannstelle ihr Maximum.
( )( )
M F F a Nmm
M F F b Nmm
bmax
tmax
= − ⋅ = ⋅
= − ⋅ = ⋅
2 80 10
60 10
1 24
1 24
Spannungen: ( )σπmax =
⋅ ⋅−
M DD D
bmax a
a i
324 4 ( )τ
πmax =⋅ ⋅−
M DD D
tmax a
a i
164 4
Der erforderliche Innendurchmesser folgt aus der Beziehung σ στσ
σV zul4
2
1 3= +⎛⎝⎜
⎞⎠⎟ ≤max
max
max
( )σπ
πσ
zula bmax
i
tmax
bmax
i aa bmax
zul
tmax
bmax
D MD
MM
D DD M M
Mmm
≥−
+⎛
⎝⎜
⎞
⎠⎟
≤ − +⎛
⎝⎜
⎞
⎠⎟ =
321 3
2
321 3
242 3
4
2
4
2
4
Da4
,
Gewählt: Di = 42 mm
Spannungsnachweis: σ σVvorh zulN
mmN
mm= < =154 8 1602 2,
Lösung 7.13 Biege- und Torsionsmomente an der Einspannstelle:
FN = 0 M Fa M Fa M Fax y t= = =2 3 3
M M M Fa Fa Nmmres x y= + = + = = ⋅2 2 64 9 13 1 8028 10,
Spannungen: ( ) ( )
( )
σπ
τπ
σ σ τπ
max max
max max
=⋅
−=
⋅−
= + =−
+ ⎛⎝⎜
⎞⎠⎟
32 16
3 32 32
4 4 4 4
2 24 4
22
M dd d
M dd d
dd d
M M
res a
a i
t a
a i
Vmaxa
a ires
t
Erforderlicher Innendurchmesser: Aus σ σVmax zul≤ folgt
d d d M M mmi aa
zulres
t≤ − + ⎛⎝⎜
⎞⎠⎟ =4 2
2
432 3
294 9
π σ, digew = 94mm
σ σvorh zulN
mmN
mm= < =103 24 1202 2,
FF
3Fa3Fa
2Fa
Lösungen für Aufgaben zur Technischen Mechanik – Dynamik - Lösung 1.6
Bereich I x a x a t C x a t C t C
AB t x x C C
t t x t v x t s t va
s s a t m
Bereich II x x C x C t C
ÜB t t x v x s C v ms
C s vt m
t t x t v x t
A A A
AA
: && &
: & , ,
: &( ) , ( )
: && &
: & , ,
: &( ) , ( )
= = + = + +
= = = ⇒ = =
= = = ⇒ = = = =
= = = +
= = = ⇒ = = = − = −
= = =
12
1 2
1 2
1 1 1 1 1 1 12
3 3 4
1 1 3 4 1 1
2 2 2
12
0 0 0 0 0
20 12
160
0
16 160
( )
( )
( ) ( ) ( )
s vt s vt v t t s
Bereich III x a x a t C x a t C t C
ÜB t t x v x s C v a t C s vt a t
Endbedingung t t x x s S a t t v und
S a t v a t t s vt a t a t t v t t
B B B
B B
B
B B B B
2 2 1 1 2 1 1
52
5 6
2 2 5 2 6 2 2 22
3 3 3 2
32
2 3 2 2 22
3 22
3 2
12
12
0 0 1
12
12
12
= + − = − +
= = + = + +
= = = ⇒ = − = − +
= = = = ⇒ − + =
= + − + − + = − + −
: && &
: & ,
: & , ( )
( )
( ) ( ) ( )
+
= + ⇒ = +−
− = + + = =
= − = ⇒ = − − − − =
s
t t va
in t t S sv
va
s s t
t s s s S a t t v t t m
B Bges
B
2
2 3 3 11
2 2 3 22
3 2
2
2 12
20 23 8 51
51 16 35 12
400
( )
( )
Lösung 1.7 1 0
13
112
0 0 0 0 0 0 0
1
13
13
112
112
1
12
13
1 14
1 2
1 2
1 1 1 1 1 12
11
12 4
1 1 13
1 1 14
. :
( ) ( ) ( )
: ( ) ( )
: ( )
( ) ( )
Bereich t t
a t K t v t K t C s t K t C t C
t v s C C
t t a t a a K t K at
ms
v t K t ms
s t K t m
≤ ≤
= = + = + +
= = = ⇒ = =
= = ⇒ = = =
= = = =
212
13
13
1 23
112
112
12
23
14
2 23
43
12
1 2
1 1 3 12
3 4
1 2 1 1 1 1 1 3 3 3
2 1 1 1 4 4
2 2 2 1 2 3
2 2 1 2
.
( ) ( ) ( )
: ( ) ( )
( ) ( )
( )
( )
Bereich t t t
a t a v t a t C s t a t C t C
ÜB t t v t v t ms
ms
a t C ms
C C ms
s t s t m m m m C C m
t t v t a t C ms
ms
ms
s t a t
≤ ≤
= = + = + +
= = = ⇒ = + = + = −
= = ⇒ = − + =
= = + = − =
= 23 2 4 2 4
314
1112
+ + = − + =C t C m m m m
( )
mCsmC
CtCmmmmtsts
Csm
smCtKtK
sm
smtvtvttÜB
smtKK
sm
ttaKttKa
tKKtatt
tKKaatatt
CtCtKtKtsCtKtKtvtKKta
tttBereich
1219
38
634
1211
1211)()(
2621
34
34)()(:
31
00)(:
)(:61
21)(
21)()(
.3
65
6252322
552232223222
2332323
132331
33233
2321122
653
32
252
3232
32
=−=
+++−=⇒==
+−=++=⇒===
=−=−=−−
=⇒−=−
++===
−+===
+++=++=+=
≤≤
Lösung 2.1
ω ω
ω ω ω ω
ω ω ππ
π
W W
W G ZW K ZWG
KW
G
K
ZW G M KG
K
K
GM
K
G M
K
G M
K
G
D v vD
r r rr
rr
vD
r r rr
vD
rr
n rr
vn D
i
i rr
vn D
zz
⋅ = ⇒ =
⋅ = ⋅ ⇒ = ⋅ = ⋅
⋅ = ⋅ ⇒ ⋅ = ⋅⎛
⎝⎜
⎞
⎠⎟ = =
= = = =
22
2
2 2
0 18
22
,
Lösung 3.5
542422
322
2222
11111101
)(021)(
0)(
CtCtxCxx
CtCgttyCgtygy
CCtvtxvxHy
+===
++−=+−=−=
=+===
&&&
&&&
&
gHvvt
gHtvt
tvgtHtyty
vvtxtxttEB
CyCx
invCinvyvCvxtAB
EEE
EEEE
EEE
0
2
112,1
012
12
021
1221
32
52
2222
2422
2g
tang
tan02g
tan2
tan21)()(
cos)()(:
0000
sscoscos0:
−⎟⎟⎠
⎞⎜⎜⎝
⎛±=⇒=+−
⋅+−=⇒=
=⇒==
=⇒==⇒==⇒==⇒==
ααα
α
α
αααα
&
&
Zahlenwerte: v kmh
t s t sE E2 1 21200 3 67 55 1= = =, , ( , )
Lösung 3.13
( ) ( )J M J M t tt
t t J M t M Jt
J m D m d m e
&&
:
ϕ ω ω
ω
ωω
= − = −
= =
= = ⇒ =
= ⎛⎝⎜
⎞⎠⎟ − ⎛
⎝⎜⎞⎠⎟ +
⎧⎨⎩
⎫⎬⎭
0 0 0 0
0 0
1 1 0 1 01
1
1
2
2
2
22
0 0
12 2
6 12 2
( )[ ]( )[ ]
m a D m b d J aD bd d e
Mt
aD bd d e n
Zahlenwerte n s M Nm
1
2
2
24 2 2 2
01
1
4 2 2 21
1
11 1
0
4 4 326 8
326 8
2
4 77 286 5 0 3933
= = ⇒ = − +
= − + =
= ≡ =− −
ρπ
ρπ ρπ
ω ρπ ωπ
: , , ,min
m y2&&
m1
m2
m2g
x
y
m x2&&
J&&ϕ
ϕ
M0
J
Lösung 3.20
ZB x x x xr
J m r
F m g m x m g m xF m g m x m g m x
J F r F r J xr
m gr m xr m gr m xr
x g m m
m m m
S
S
S S
:
&& &&
&& &&
&&&&
&& &&
&&
3 1 2
22
1 1 1 1 1 1
2 3 3 3 3 3
2 2 1 3 3 1 1
1 3
1 2 3
12
0
12
= = = =
=
= − = −
= + = +
+ − = = + + − +
=−
+ +
ϕ ϕ
ϕ
( )
( )
( )
T m x m x J x m m m
U m gx m gx gx m m
L T U x m m m gx m m
ddt
Lx
Lx
Lx
g m m Lx
x m m m ddt
Lx
x m m
= + + = + +⎛⎝⎜
⎞⎠⎟
= − + = − −
= − = + +⎛⎝⎜
⎞⎠⎟ + −
⎛⎝⎜
⎞⎠⎟− =
= − = + +⎛⎝⎜
⎞⎠⎟
⎛⎝⎜
⎞⎠⎟= +
12
12
12
12
12
12
12
0
12
12
1 12
3 32
22 2
1 2 3
1 1 3 3 1 3
21 2 3 1 3
1 3 1 2 3 1
& & & &
&
&
&&
&&&
ϕ
∂∂
∂∂
∂∂
∂∂
∂∂
( )
( )
( )
2 3
21 2 3 1 3
1 2 3 1 3
0
12
12
12
0
+⎛⎝⎜
⎞⎠⎟
+ = ⇒ + =
+ = + +⎛⎝⎜
⎞⎠⎟ − −
+ +⎛⎝⎜
⎞⎠⎟ − − =
m
T U konst ddt
T U
T U x m m m gx m m
xx m m m gx m m
.
&
&&& & Lösung 3.26
( )
( )
ϕ ϕ= = − = −⎛⎝⎜
⎞⎠⎟
= − −⎛⎝⎜
⎞⎠⎟
− − − = ⋅
+ + −⎛⎝⎜
⎞⎠⎟
⎧⎨⎩
⎫⎬⎭= −⎛
⎝⎜⎞⎠⎟
= + = =
xr
x R r Rr
x
F m g m Rr
x
F R r J xr
m x rr
x m Jr
m Rr
m g Rr
J J J J m r J m R
S
S
A B A A B B
21 2
1 1 2
22
2 2
2 222 1
2
1
22 2
1
1
0 1
1 1
2 12
12
&&
&&&&
&&
J&&ϕ2
x1 x3
ϕ2
FS1 FS2
m1g m3gm x3 3&&m x1 1&&
J2&&ϕ
m x2 2&&
x2
x1
FS
FS
m1g
11xm &&
FT
ϕ
FN
m2g
{ } { } { }{ }{ }
22
22
21
121
22
44
22
22222
244
2
222
11
1
2121
22
⎟⎠⎞
⎜⎝⎛ −
+
⎟⎠⎞
⎜⎝⎛ −
+=⎟
⎠⎞
⎜⎝⎛ −=
++
=
+=⇒+=+=
⋅=⋅=+=
rRr
J
rR
mm
gmxrRx
bRdrbRdrmJ
bRdrmbRdrmbRdrJ
bRmdrmmmm BABA
&&&&
πρπρπρ
ρπρπ
( ) ( )T m x m x J U m gx z B ddt
T U= + + = − + =12
02 22
1 12
22
1 1& & & . .ϕ
Lösung 3.28
( ) αα
αα
αα
αϕαϕ
ϕϕϕ
sin71sin
sin74sin2
21
21
0sinsin
0sin:0sin:
22
2121
11212222
2222
1111
2121
mggxxmF
gxmgmmmmx
mrJmrJmmm
gmxmrxJgmxm
rxJ
rgmrxmrFJmrgmrxmrFJm
rxxxx
S
S
S
=−+=
==⎟⎠⎞
⎜⎝⎛ +++
====
=−++−+
=⋅−+−=⋅−++
=====
&&&&
&&&&
&&&&
&&&&
&&&&
&&&&
Lösung 3.31
( )
( )
ZB sr
r r rr
sr
Masse m ms mg FScheibe J M F r
Scheibe J J F r F r
F ms mg F Mr
J rr
sr
J J sr
rr
M J rr
sr
msr mgr
S
A S
S S
S SA
A
:
: &&
: &&
/ : &&
&&&&
&& &&&&
ϕ ϕ ϕ ϕ
ϕ
ϕ
23
2 2 1 1 12
1 3
2
1 1 1 1
2 3 2 1 2 2 3
2 11
12
12
3
2 33
2
11
22
12
33 3
01 0
2 3 0
0
= = ⇒ =
+ − =
− + =
+ − + =
= + = −
+ − + + + =
( FS = 0 , da Riemenabtriebsseite ohne Last)
m x1 1&&
J1 1&&ϕJ2 2&&ϕx1
x2
FS
FN1
FT1m1g
FT2
FN2m2g
rr
ϕ2
ϕ1 m x2 2&&
( )J J2 3 2+ &&ϕ
J1 1&&ϕ FS1
FS=0
FS2
mg s
ms&&
ϕ2ϕ1
r1
r2
r3
MA
( )
( )
KssKdssdssdssdKsoder
Kss
Kst
KtsKtsK
mr
JJrJ
rr
mgsMM
mgrrrMMsfürM
mr
JJrJ
rr
mgMrr
r
s
AA
AAA
A
221.3
212.2
0.1
2
2
23
322
1
1
2
3
2
min
2
31minmin
23
322
1
1
2
3
2
31
2
=====⇒=
===
++
+⎟⎟⎠
⎞⎜⎜⎝
⎛=⇒=
==⇒=
++
+⎟⎟⎠
⎞⎜⎜⎝
⎛
−=
&&&&&
&&&&
&&&
&&&&
Lösung 3.39
K
rrJ
rrJ
MMrrJ
rrJ
rJFrFJ
MrFJrrrr
UU
U
=
⎟⎟⎠
⎞⎜⎜⎝
⎛+
==−⎟⎟⎠
⎞⎜⎜⎝
⎛+
=⇒=−
=−+
=⇒=
2
12
1
21
22
12
1
212
22
2222
111
21
212211
0
0
0
ϕϕ
ϕϕ
ϕ
ϕϕϕϕ
&&&&
&&&&
&&
&&&&&&
( ) Nmbrbrt
rrnMbrrmJbrrmJ
nrrJ
rrJ
tMKttt
KtCCKtK
6,42222
2:
00)0(
22
212
11
2122
2
422
22
21
412
11
1
222
12
1
21
1
212221
2222
=+=====
=⎟⎟⎠
⎞⎜⎜⎝
⎛+=⇒===
==⇒=+==
ρππρπρ
πωωωωϕ
ϕϕϕϕ
&
&&&&&
Lösung 4.2
( )ϕϕϕ
ϕϕϕ
ϕϕ
ϕ
coscos2)(
cos)(21cos
)(21cos
0cos
0
20
222
101
2211
−=
+=
==
==+=+
grv
mgrmvmgr
mvTmgrU
TmgrUTUTU
( )
( ) ( ) NmgFmgmgvrmF
smgrv
SS 4,92833)(cos3cos2cos)()(
05,61cos2)(
02
0
=+=−=−=
=+=
πϕϕϕϕϕ
ϕπ
22ϕ&&J
11ϕ&&J
ϕ2 ϕ1
FU
M FN
rvm
2
U=0
(2)
mg
FS ϕ ϕ0
(1)
ϕ&&mr
Lösung 4.10
Energiebilanz im Rohr: mclvmvcl =⇒= 0
20
2
21
21
00000:
21
0
34
0102
432
3
211
=⇒==⇒==⇒==⇒==
++−=+−=−=
+===
CyhChyvCvxCxtAB
CtCgtyCgtygy
CtCxCxx
&
&
&&&
&&&
EB: chmgwlh
vwg
vwttvwywxtt EEE 2
0210
22
000 ==−⎟⎟
⎠
⎞⎜⎜⎝
⎛==⇒===
Zahlenwerte: l = 9,9 cm; v0 = 8,09 m/s Lösung 4.13
Bewegung auf der schiefen Ebene: αα sin221sin 1
21 glvmvmgl =⇒=
ββββ
ββω
ωϕ
cossin0cossin:
0sincos: 2
1
SNNS
NS
FmgFFFmg
FFmrrv
+==−+↓
=−−→
==&
cm
lr
wh
wwhÜberhöhungg
rfürF
FmgFmr
S
SS
6,20
sin21
tan1tansin:tan0
0sincos
sincos
2
*2
**
2**
2
=
⎟⎠⎞
⎜⎝⎛+
=
+=====
=⋅+
−−
α
βββωβββ
ββ
ββω
Lösung 4.14
21313
233
2113311
2112
22
2112211
)(221
21
221
21
vhhgvmvmghmvmghUTUT
vghvmvmvmghUTUT
+−=+=+⇒+=+
+==+⇒+=+
mg
x
y
mg β FS FN
mrω2 r
rvm
23 3
mg
FN
v3
( ) mrsmv
gvhhrr
gvrrrfürF
rvmmgF
rvmmgF
N
NN
917,295,62
0
0:
2
21
31*
23**
23
23
>=+−=>
=⇒==
−==+−↑