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Problemas resueltos en clase beer
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Todos los problemas resueltos en clase de beer con anotaciones extras
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problemas,planificacion
EXAMER
FINAL
, FA
7
Il
B
Y ..1
= 5000KN
.
45
S
I
1
Poligono
5000kN -
> R
SoFB
=
En F + = = 5000 N
EFV = O -D
FA .
Sen 30 = FB ·
sen 45
FA = 366025kN
2ec
FB =
2588'19kN
EFH = O 2 incog
FA COS30 +
FB COS 45 5000
-
. =
NY
↑ FA
P = 50ON
Q = GSON 300( FB > X
-
D
400Q
GSON
V
500N
se puede hacer también con
Con ecuaciones de equilibrio
el poligono cerrado ,
EFu = 0
FA .
sen 50 = 500 + 650 .
cos 40
FA = 1302N
zec
EFH = 0 FB =
419-55N
zinc
500 650 FA
FA 50 FB
↓ 1408
COS
650 Sen 40
.
= + ·
X
500
650 48
2
S 50/c
FB
500
-950 =
; FB
FB
=
419'55N
X = 6527 FA-652'7 + 650 = 13027 N
, 20029 - Dx981 = 1962N
~
/u00 20
o17
TB
TA
V1962N
2FH = 0
TB Cos 20 sol
COS 40
.
=
TA .
TA = 212889N
zec
+ B =
173549N
2FV = 0 zinc .
1962 = TA - Sen 40 + TB .
sen 20
ry
Q
P = 300N
Q = 400N
go
S
(200 >x
&
D %
30
·
D
.
coS
38
O
13
(
D
↳
z
-P COS 30 sen 15
en X
MY
.
:
·
en 7 : P . sen 30 A
en z : p .
coS 30 .
cos 15
a COSC
. .
COSB
* = 300-COS -
Sen 15 , Sen 30 , cos 30 . cos 15) 2 x
B
R= D + Q =
Q
S
en X : Q .
cos 50 .
cos 20
R = (17437 45642 , , 1632) a-CosL
↳
I I I
en y z
: Q .
sen 50
RE módulo :
en z : -
Q- cos 50 ·
sen 20 RX
RY
>
-
Ir = 515'13N
Q = 400 (cos 50 . cos20 , sen 50 ,
-cos 50. Sen 20)
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