Determine el desplazamiento vertical en el Nodo “C” de la armadura mostrada:
TRABAJO VIRTUAL
Calculo de los momentos virtuales
≤
Px
D
4in
'
&
,
Epn
^
1
Calculo
detención
)
Dy
15ft
3in
'
4in
'
ZFY
-01^4
)
EIYA
-0Gt
EFX
-0
Ayitlookbb
3in
'
DY
-25-25=0
-201151-25140
)
-151×-0
Dx
+
AKO
,
a)
&
4in
'
B
4in
'
he
DY
Okbb
.
Dxitookllb
100
TAKO
v25
K
v15
K
,
bott
,
20ft
,
F-
=
29,000k
si
=
29,000kW
/
¡
n
'
trabajo
Virtual
:[
INIIMHLI
F-
(A)
tgd.HN
2
Nodo
"
c
"
d-
36.87°
Nodo
"
B
"
EFX
-0-4+1
EFY
.
-
Ott
"
"
Yy
"
"
"
"
"
≤
"
"
°
"
≤
"
°
"
"
"
"
±
"
°
"
"
"
"
"
"
"
$
ÍÍmÍ
FLESH
-
25
FEB
+
FCELOSL
"
iFBA}gÍ*
FBA
-
-33.33kt
"
FCE
"
"
,
,
Igp
,
FCE
=
25151136.87°
FCB
=
-416610536.87°
÷
"
"
25
FCE
_-
41.664lb
FCB
=
-33.334lb
Nodo
"
E
"
Nodo
"
A
"
TFEA
#
Efy
↑
-1
[
f-
✗
=D
→
1+1
FAD
^
Efy
-_
O
Rt
)
=µtµü_Ed
FBETFCF-sendtt-F-n.SN
"
FCECOSL
-
FEDTFEACOSL
i.FR#YgAF-FAD--FEAbenL
"
"
"
"
25
-141.665M
36.87°
ftp.A.ben36.870-04/.66l0SJ6.870=Fi=D-83.33cos36.87o
FEA
"
"
FAD
--
83.335cm
36.87°
'
-
FEA
"
"
"
FBE
+
FCE
50.00
tt-EAben36.tt
=D
f-
[
D=
100.00
Kbb
"
FEA
"
FEASCNL
FAD
_
-50.00k
db
FEASENLFCESENL
FEA
=
-83.334lb
×
D
4in
'
&
,
E-
1
Calculado
/
asturianos
:
^
Efy
-01^1-1
)
EFX
=D
Dy
15ft
3in
'
4in
'
DY
-1.00
DX
-
Ax
3in
'
EIYA
-
-0Gt
AX
--
8/3
,
a)
&
4in
'
B
4in
'
si
,
[
Dxtlb
)
-40111=0
,
bott
,
20ft
¡
1.00
D
#
813
2
Nodo
"
C
"
Nodo
"
B
"
EFY
-0
EFX
-0
RFCE
-
IFCESENL
EFY
-
O
EFX
-0
FBE
^
t-BE-OFBA-T-CB.tt/I-e-o-FcE5end-
1.00
FCB
--
FCECOSL
a-Yg-oo.si
BA
FBA
-1.334
FCELOS
?
1.00
FCESCM
36.870=1.00
5-
[
B=
1.667.10536870
FCB
FCE
-1.667
FCB
_-
1.334
Nodo
"
E
"
Nodo
"
A
"
FEASENL
^
TFEA
zfy
--
O
[
f-
✗
=D
EFY
=D
>
"
"
"
"
i.tt/#,-n--x.,Ygj-FcEcosdFBE-FCF-
sent-FEAbl.nl
FED
-
FEALOSLTFCECOSTN
FBA
FAD
-
FEA
Sent
"
0.00+1.66 7
sensato
.BZ?-FEA5en36.870FED--
1.66710536-871-1.6671053687
'
F-
Acosa
FAD
_-
1.6675113687°
FBÉ
"
"
pez
FEA
_-
1667
5-
F-
D=
2.667
"
FEA
"
FEASENL
FAD
-1.00
FCESCNL
