-
7/25/2019 Ejercicioaplicacion
1/7
Ejercicio de aplicacin
Disear el eje de la figura.
DATOS
dB=2 ; rB=1
dC=4 ; {r} rsub {C} =2
=60
Acero 1040HR; Sy=30ksi y Su=50 ksi
n=2
SOLUCIN
Para la polea de 2 pulg
FBx z=2 30cos60 =30lb
FBx y=2 30 sen60 =51,96 lb
Entonces en plano xz
Diagrama de cuerpo libre
-
7/25/2019 Ejercicioaplicacion
2/7
Diagrama de Fuerzas
Diagrama de Momentos
+Fx=0
R1P1P2+R2=0
R1=R2+P1+P2(1)
+MA=0
P1 (60 )+P2(300)R2 (420 )=0
R2=
P1(60 )+P
2(300)
420(2)
Reemplzo FBx z
P1=30 lb! F
Cx z
=P2=100 lben2
-
7/25/2019 Ejercicioaplicacion
3/7
R2=(30 (60 )+100 (300))lb " pul#
420pul#
R2=75.71 lb
ReemplazoR2 en (1)
R1=(75.71+30+100) lb
R1=54.29 lb
Donde
MBxz=3257.14 lb" pul#
MCxz=9085.71lb " pul#
En el plano xy
Diagrama de Fuerzas
Diagrama de Momentos
-
7/25/2019 Ejercicioaplicacion
4/7
+Fx=0
R1P1+R2=0
R1=P1R2(3)
+ MA=0
P1 (60 )R2(420)=0
R2=P1 (60 )
420 (4)
Reemplzo FBx y=P1=51.96lben(4)
R2=51.96 (60 ) lb " pul#
420pul#
R2=7.42 lb
Reemplzo R2en (3)
R1=(51.967.42)lb
R1=44,54 lb
Donde
MBxy=2672.23 lb" pul#
Momentos Resultantes:
-
7/25/2019 Ejercicioaplicacion
5/7
MB=[ (MB )xy ]2+ [ (MB )xz]2
MB=(2672.23 )2+(3257.14)2lb "
MB=4213.048 lb"
MC= [ (MC)xy ]2+ [ (MC)xz]2
MC=(0)2+(9085.71)2 lb" pul#
MC=9085.71 lb" pul#
Determinacin del punto crtico
El punto C es el ms crtico.
Diseo del eje
! El es"uerzo e#ui$alente de %on Misses es&
($e% )& " M=$2
+3 '2
()
$=32M
( d3=
32 (9085.71) lb" pul#
( d3
(b)
'=16 )
( d3
)=FBx y r B=51.96 lb 1pul#=51.96 lb" pul#
16 (51.96 )lb " ( d
3(c)
'=
Dado n=2
($e% )& " M= Syn
-
7/25/2019 Ejercicioaplicacion
6/7
($e% )& " M=30 10
3lb" pul#
2
($e% )& " M=15000 lb "(d )
En (a)
($e% )& " M2=$2+3 '2(e)
Reemplazo (b) (c) (d) en (e)
150002=(32 (9085.71 )( d3 )
2
+3(16 (51.96 )( d3 )2
150002=
(32 (9085.71 ) )2+9(16 (51.96 ) )2
(2
d6
d6=
(32 (9085.71 ))2+9 (16 (51.96))2
(2(150002)
d= 638.0686 pul#
d=1.83pul#
!or la teora del es"uerzo cortante m#imo
n)*CM=0.5 Sy
'M+x
'M+x=0.5Sy
n)*CM
'M+x=(1
2$)
2
+'2
0.5(30103)2
=(
1
2(32 (9085.71 )
( d3
)
)
2
+
(
16 (51.96 )
( d3
)
2
-
7/25/2019 Ejercicioaplicacion
7/7
(0.5 (30 103 )
2 )2
=(12 ( 32 (9085.71)( d3 ))2
+(16 (51.96 )( d3 )2
75002=
(0.5(32) (9085.71 ))2+9 (16 (51.96 ))2
(2
d6
d6=
((0.5)(32) (9085.71 ))2+9 (16 (51.96 ))2
(2(75002)
d= 638.077pul#
d=1.35pul#
