statika grede vjezbe
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-
Zadatak
Sile F1 i F2 nastaju djelovanjem ruku radnika na ruke elektrine builice .Zamijeniti ovaj sistem sila glavnim vektorom i glavnim momentom za taku O. Rezultate prikazati u vektorskoj obliku.
B
A
Glavni vektor jednak je vektorskom zbiru oba vektora:
N)1416(
421036
kjiF
kjkjiF
FF
O
O
iO
Najea komponenta sile je -14 N i njom se vri pritisak builice o dasku.
Glavni moment jednak je vektorskom zbiru momenata pojedinanih sila za taku O:
21
21
FrFrM
FOBFOAM
MM
BAO
O
OiO
Glavni vektor jednak je vektorskom zbiru momenata pojedinanih sila za taku O:
21
21
FrFrM
FOBFOAM
MM
BAO
O
OiO
4203,025,00
10363,0015,0
kjikjiMO
Nm)45,03,33,1(
4,045,03,39,0
kjiM
ikjiM
O
O
Komponente momenta MOx = 1,3 Nm i MOy = 3,3 Nm utiu na savijanje builice oko tih osa, dok komponenta MOz = - 0,45 Nm utie na uvijanje builice oko ose z.
-
Zadatak: Skup sila koji djeluje na kvadar prema slici redukovati na taku O (odrediti rezultantnu silu i moment). Zadano je:
1 2 3 4 120 NF F F F
0,3 ma
0,5 mb
0,7 mc
Rjeenje:
Geometrija kvadra
arctg cd
2 2d a b
2 2arctg c
a b
2 2
0,7arctg0,3 0,5
50,2
arctg ab
0,3arctg0,5
30,96
Komponente sile F1
1 1 cosxyF F
1 1 sinzF F
1 1
1 1
sin
cos sinx xy
x
F FF F
1 1
1 1
cos
cos cosy xy
y
F FF F
Glavni vektor rezultantna sila
1 4 1 4cos sinR xX F F F F
120 cos50,2 sin30,96 120RX
1 3 1 3cos cosR yY F F F F
120 cos50,2 cos30,96 120RY
1 2 1 2sinR zZ F F F F
120 sin50,2 120RZ
185,8615
99,516 N
N212, 4 N19
R
R
R
XYZ
159,516 185,869 212,194R R R RF X i Y j Z k i j k Y j Z k i j k159 516 185 869 212 194Y j Z kY j Z k 159 516 185 8696 185 869F X iX iii
2 2 2 2 2 2159,516 185,869 212,194324,066 N
R R R R
R
F X Y ZF
-
Ravan Oyz moment za osu x
2 1Ox yM F b F c
2 1 cos cosOxM F b F c
120 0,5 120 cos50,2 cos30,96 0,7106,109 Nm
Ox
Ox
MM
Ravan Oxz moment za osu y
0OyM
Ravan Oxy moment za osu z
4 1 3Oz yM F b F a F a
4 1 3cos cosOzM F b F a F a
120 0,5 120 cos50,2 cos30,96 0,3 120 0,3115 N,76 m1
Oz
Oz
MM
106,109 115,761O Ox Oy OzM M i M j M k i k i M j M k i k106 109 115 761i M j M kj M k 106 109
2 2 2 2 2106,109 115,76
157,034 Nm
1O Ox Oy Oz
O
M M M M
M
Zadatak Homogena ploa, oblika i dimenzija kao na slici, vezana je za vertikalnu osovinu u taki A pomou sfernog oslonca, a u taki B pomou cilindrinog oslonca. U taki D vezano je ue, iji je drugi kraj vezan za podlogu u taki E. Teina ploe je G=50 kN, a na plou djeluje vjetar w=10 kN/m2. Odrediti sve reakcije veza. Zadano je: BC=AD=2R, AB=3R, EA=R, R=1m.
R60
2R
1
,
5
R
1
,
5
R
R
x
y
z
A
E
D
C B
GW
T
Reakcije veza ploe Ploa e se osloboditi veza sa okolinom i njihov uticaj zamijenit e se reakcijama.
R60
x
y
z
A D
C B
GW
S
zT
yT
XA
YAZA
XBYB
T
-
Teite ploe Da bi se postavila aktivna sila teine ploe i sila vjetra, potrebno je odrediti koordinate teita:
m5,1
i
iiT A
zAz
m797,0
232
342
232
2
2
RRR
RRRRRR
AyAyi
iiT
T
y
z
D
C B
zT
yT R
2R
1
,
5
R
1
,
5
R
y
z
A
T
1
2
T1 T2 yT1
yT24R 3
Sila vjetra:
kN3,442
62
2
wRRwAW
Ugao sa slike:
6,2621
2
RR
ADEAtg
R60
x
y
z
D
C B
GW
S
zT
yT
XA
YAZA
XBYB
R60
2R
1
,
5
R
1
,
5
R
R
x
y
z
A
E
GW
T
Statiki uslovi ravnotee ploe su: 1. 060sinsin WSXXX BA 2. 060coscos WSYYY BA 3. 0 GZZ A 4. 0360cos RYzWyGM BTTx 5. 0360sin RXzWM BTy 6. 02sin60sin RSyWM Tz
3R
Iz prethodnih 6 jednaina dobivaju se nepoznate reakcije oslonaca ploe:
Iz 3. kN50GZA
Iz 4. kN36,2460cos31
TTB zWyGRY
Iz 5. kN18,1960sin31
TB zWRX
Iz 6. kN19,3460sinsin21
TyWRS
Iz 2. kN79,3260coscos WSYY BA
Iz 1. kN85,360sinsin WSXX BA
Reakcije oslonaca A i B imaju intenzitete:
kN31
kN92,5922
222
BBB
AAAA
YXF
ZYXF
-
Homogena ploa ABCD teine G = 6 kN optereena je silama F = 9 kN u takama A i B prema slici. Na plou djeluje spreg sila u ravni ploe momenta M = 5 kNm. Ploa je oslonjena sa 6 tapova zanemarljivih teina prema slici. Odrediti reakcije tapova ako su dimenzije: a = 5 m, b = 5 m, c = 4 m, d = 2 m.
Rjeenje:
Ploa ABCD osloboena veza
2 2 2 2
5arctg arctg 51,3444arctg arctg 38,6655arctg arctg 455
5 5arctg arctg 74,212
accbba
a bd
= = =
= = =
= = = + += = =
-
i
F ( M
i
)
F
( A )
F
( B )
G M
1
S
2
S
3
S
4
S
5
S
6
S
i
X
F 0 0 /
1
s i nS 0 0
4
s i nS 0
6
s i n s i nS ( 1 )
i
Y
0 0 0 / 0 0
3
c o sS 0
6
s i n c o sS ( 2 )
i
Z
0 - F - G /
1
c o sS 2
S
3
s i nS 4
c o sS 5
S
6
c o sS ( 3 )
i
x
a a a / 2 / 0 0 0 0 0 a
i
y
0 b b / 2 / 0 0 b b b b
i
z
0 0 / 0 0 0 0 0 0
x
i i i i
M
y Z z Y
0
F b
2
b
G
0 0 0
3
s i nb S 4
c o sb S 5
S b
6
c o sb S ( 4 )
y
i i i i
M
z X x Z
0
F a
2
a
G
0 0 0 0 0 0
6
c o sa S ( 5 )
z
i i i i
M
x Y y X
0 0 0
M
0 0 0
4
s i nb S 0
6
6
s i n c o s
s i n s i n
a S
b S
( 6 )
J e d n a i n e r a v n o t e e n a o s n o v u t a b e l e :
F
1
s i nS 4
s i nS 6
s i n s i n 0S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ( 1 )
3
c o sS 6
s i n c o s 0S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ( 2 )
F G +
1
c o sS +2
S +
3
s i nS +4
c o sS +5
S +
6
c o sS = 0 . . . . . . . . . . . . . . . . . . . . . . . . . ( 3 )
3 4 5 6
s i n c o s c o s 0
2
b
F b G b S b S S b b S . . . . . . . . . . . . . . . . . . . . . ( 4 )
6
c o s 0
2
a
F a G a S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ( 5 )
4 6 6
s i n s i n c o s s i n s i n 0M b S a S b S . . . . . . . . . . . . . . . . . . . . . . . . . . . . ( 6 )
i z ( 5 ) :
6
2
4 4 , 0 9 1 k N
c o s
a
F a G
S
a
i z ( 2 ) :
6
3
s i n c o s
3 8 , 4 1 9 k N
c o s
S
S
i z ( 6 ) :
6 6
4
s i n c o s s i n s i n
1 , 2 8 1 k N
s i n
M a S b S
S
b
i z ( 1 ) :
4 6
1
s i n s i n s i n
2 8 , 1 7 4 k N
s i n
F S S
S
i z ( 4 ) :
3 4 6
5
s i n c o s c o s
2
2 4 , 8 k N
b
F b G b S b S b S
S
b
i z ( 3 ) :
2
S = F + G -
1
c o sS -3
s i nS -4
c o sS -5
S -
6
c o sS = 2 0 , 6 k N
-
Homogena ploa ABCD se odrava u ravnotei pomou sfernog zgloba, cilindrinog zgloba, te tapa zanemarljive teine, prema slici. Odrediti reakcije veza. Potrebni podaci su: G = 10 kN, F = 10 kN, a = 4 m, b = 4 m, c = 6 m
Rjeenje:
Ploa ABCD osloboena veza
6arctg arctg 56,3144arctg arctg 454
90 90 45 45
cbab
= = =
= = = = = =
-
iF F G AX AY AZ DX DY S
iX sinF 0 AX 0 0 DX 0 0 (1) iY cosF 0 0 AY 0 0 DY cosS (2) iZ 0 G 0 0 AZ 0 0 sinS (3) ix a a/2 0 0 0 0 0 a
iy b b/2 0 0 0 0 0 b
iz 0 c/2 0 0 0 c c c
x
i i i i
My Z z Y
= 0 2
b G 0 0 0 0 Dc Y sincosb Sc S
+ (4)
y
i i i i
Mz X x Z
= 0 2
a G 0 0 0 Dc X 0 sina S (5)
z
i i i i
Mx Y y X
=
cossin
a Fb F
+ 0 0 0 0 0 0 cosa S (6)
Jednaine ravnotee na osnovu tabele:
sinF + AX + DX = 0 ................................................................................... (1)
cosF + AY + DY cosS = 0 ........................................................................ (2)
G + AZ sinS =0 ...................................................................................... (3)
2b G Dc Y sin cosb S c S + =0 ......................................................... (4)
2a G + Dc X + sina S =0 ............................................................................ (5)
cos sina F b F + cosa S =0 ............................................................ (6)
iz(6): cos sincos
a F b FSa
+ = 25,495kNS =
iz (5): sin
2D
aa S GX
c
= 17,475 kNDX =
iz (4): sin cos
2D
b G b S c SY
c
+ = 3,333 kNDY =
iz (3): AZ =G + sinS 31,213 kNAZ = iz (2): AY = cosF DY cosS + 10,404kNAY = iz (1): AX = sinF DX 24,547 kNAX =
-
563. The cart supports the uniform crate having a mass of85 kg. Determine the vertical reactions on the three castersat A, B, and C. The caster at B is not shown. Neglect themass of the cart.
BA
C0.2 m 0.5 m
0.6 m 0.35 m
0.1 m
0.4 m0.2 m
0.35 m
565. If and , determinethe tension developed in cables AB, CD, and EF. Neglectthe weight of the plate.
y = 1 mP = 6 kN, x = 0.75 m z
F
B
D
A
y
x
x y
E
C
P
2 m2 m
-
566. Determine the location x and y of the point ofapplication of force P so that the tension developed incables AB, CD, and EF is the same. Neglect the weight ofthe plate.
z
F
B
D
A
y
x
x y
E
C
P
2 m2 m
*568. Determine the magnitude of force F that must beexerted on the handle at C to hold the 75-kg crate in theposition shown. Also, determine the components of reactionat the thrust bearing A and smooth journal bearing B.
F0.1 m
0.2 m
0.5 m
0.6 m
0.1 m
z
xy
A
B
C
-
569. The shaft is supported by three smooth journalbearings at A, B, and C. Determine the components ofreaction at these bearings.
0.6 mx B
CA
z
0.9 m
0.6 m
0.9 m y0.9 m
0.9 m
0.9 m
900 N
500 N
450 N600 N
570. Determine the tension in cables BD and CD andthe x, y, z components of reaction at the ball-and-socketjoint at A.
z
y
x
C
B A
3 m
300 N
D
1 m
0.5 m
1.5 m
-
*572. Determine the components of reaction acting at thesmooth journal bearings A, B, and C.
0.6 m45
x y
C
z
B
A0.4 m
0.8 m
0.4 m
450 N
300 N m
573. Determine the force components acting on the ball-and-socket at A, the reaction at the roller B and the tensionon the cord CD needed for equilibrium of the quartercircular plate.
z
x
350 N
1 m
2 m
60
3 m200 N
200 N
yB
A
C
D
-
577. The plate has a weight of W with center of gravity atG. Determine the distance d along line GH where thevertical force P = 0.75W will cause the tension in wire CD tobecome zero.
z
F
B
D
A
H
y
xG
d
E
C
P
L2
L2
L2
L2
578. The plate has a weight of W with center of gravity atG. Determine the tension developed in wires AB, CD, andEF if the force P = 0.75W is applied at d = L/2.
z
F
B
D
A
H
y
xG
d
E
C
P
L2
L2
L2
L2
-
579. The boom is supported by a ball-and-socket joint at Aand a guy wire at B. If the 5-kN loads lie in a plane which isparallel to the xy plane, determine the x, y, z components ofreaction at A and the tension in the cable at B.
z
5 kN
5 kN
y
x
3 m
2 m
1.5 m
30
30
BA
585. The circular plate has a weight W and center ofgravity at its center. If it is supported by three vertical cordstied to its edge, determine the largest distance d from thecenter to where any vertical force P can be applied so as notto cause the force in any one of the cables to become zero.
A
d120
120120
Cr
P
B
-
586. Solve Prob. 585 if the plates weight W is neglected.
A
d120
120120
Cr
P
B
587. A uniform square table having a weight W and sidesa is supported by three vertical legs. Determine the smallestvertical force P that can be applied to its top that will causeit to tip over.
a/2a/2
a
-
*592. The shaft assembly is supported by two smoothjournal bearings A and B and a short link DC. If a couplemoment is applied to the shaft as shown, determine thecomponents of force reaction at the journal bearings and theforce in the link. The link lies in a plane parallel to the yzplane and the bearings are properly aligned on the shaft.
250 mm300 mm
400 mm250 N m
y
Ax
20
120 mm
30D
B
z
C
123456
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