View
215
Download
0
Embed Size (px)
Citation preview
8/16/2019 Decr is Pincer Chas
1/11
Long. Elemento: 4.33 m
Long. De Separación: 7.00 m
Carga Muerta
Peso del canalón: 2 !g"m2
Peso de estructura met#lica: a !g"m2
Carga $i%a: 30 !g"m2
PD& '2()* +4.33+7&)2)2 !g&),2)2 -n& ),2) -n
Carga $i%a
Carga %i%a: 30 !g"m2
PL& 30'4.33* +7& 0.3 !g& 0.0 -n& 0.) -n
/allando las reacciones
Carga Muerta
7 (1.21)2
= R A= R L
2.89= R A= RB
Carga $i%a
7 (0.91)2
= R A= R L
3.19= R A= RB
Reacciones en las barras por la carga muerta:
udo 1
Σ F y=¿ 0
1.21= ABxSen(30,98 )
AB=2.35
8/16/2019 Decr is Pincer Chas
2/11
2.35 (cos (30,98 ) )= AC
AC =2.01
udo 1C
Σ F y=¿ 0
CB=0
Σ F X =0
AC =CE
2.01=CE
udo 1
Σ F y=0
BExSen (30,98 )=BAxSen (30,98 )+1.21
BE=4.70
Σ F X =¿ 0
BAxCos (30,98 )+BExCos (30,98)=BD
BD=6.04
udo 1E
Σ F y=0
EBxSen (30,98)= ED
ED=2.42
8/16/2019 Decr is Pincer Chas
3/11
Σ F X =0
EC + EG= EBxCos (30,98 )
EG=2.02
udo 1D
Σ F y=0
1.21+ DE= DGxSen (30,98 )
DG=7.05
Σ F X =0
DB+ DGxCos (30,98 )= DF
DF =12.98
udo 15
FG=1.21
Reacciones en las barras por la sobrecarga:
udo 1
Σ F y=0
T AB xSen (30,98)=5
T AB=9.71
Σ F X =0
8/16/2019 Decr is Pincer Chas
4/11
T AB xCos (30,98 )−T AC =0
9.71 xCos(30,78 )=T AC
T AC =8.33
udo 1
Σ F X =0
T AB xSen (30,98 )−T BE xSen (30.98 )=0
Σ F BE
=9.71
F X =0
T AB xCos (30.98)+T BE xCos (30,98)−T BD=0
T BD=16.65
udo 1E
Σ F y=0
9.71 xSen (30,98 )−T DE=0
Σ F X =0
T EG−8.33−9.71 xCos (30,98 )=0
T EG=16.65
udo 1D
Σ F y=0
5
−T DG xSen
(30,98
)=0
8/16/2019 Decr is Pincer Chas
5/11
F X =0
16.55−9.71 xCos (30,98 )−T DF =0
T DF =24.98
Reacciones en las barras por la carga viva:
udo 1
Σ F y=0
3.19−0.91−T AB xSen (30,98 )=0
T AB=4.43
Σ F X =0
T AC −T AB xCos (30,98 )=0
T AC =4.43 xCos (30,98 )
T AC =3.80
udo 1
Σ F y=0
T AB xSen (30,98)−0.91−T BE xSen (30,98)=0
4.43 xSen (30,98)−0.91=T BE xSen (30,98 )
T BE=2.66
Σ F X =0
T AB xCos (39,98)+T BE xCos (30,98)−T BD=0
T BD=6.08
8/16/2019 Decr is Pincer Chas
6/11
udo 1E
Σ F y=0
2.66 xSen (30,98)−T DE=0
T DE=1.37
Σ F X =0
T EG−3.80−2.66cos (30,98 )=0
T EG=6.08
udo 1D
Σ F y=0
1.37−0.91−T DG xSen (30,98 )=0
T DG=0.89
Σ F X =0
6.08+0.89 xCos (30,98 )−T DF =0
T DF =6.84
Long. Del elemento & 4.33
Long. De separaci6n & 7.00 m
Carga Muerta
Peso del canalón & 2 !g"m2
Peso de la estructura met#lica & ) !g"m
Carga $i%a & 30 !g"m2
PD&'20()*+4.33+7&)2)2
8/16/2019 Decr is Pincer Chas
7/11
PD&).2)
PL&30+4.33+7&0.3
PL&0.)
/allando uer8as internas con carga muerta
R A =1.21 x7
2
R A=4.24Tn
udo 1
Σ F y=0
4.24−1.21−T AB xSen (30,98 )=0
T AB=5.89
Σ F X =0
T AC +5.89 xCos (30,98)=0
T AC =5.05
udo 1C
T CE=T AC
T CE=5.05
udo 1
Σ F y=0
8/16/2019 Decr is Pincer Chas
8/11
5.89−1.21−T BE xSen (30,98)=0
T BE=9.09Tn
ΣF X =0
5.89 xCos (30,98)+9.09 xCos (30.98)−T BD=0
T BD=12.84Tn
udo 1E
Σ F y=0
9.09 xSen (30,98)−T DE=0
T DE=4.68Tn
Σ F X =0
T EG−9.09 xCos (30.98)−1.21=0
T EG=12.84Tn
udo 1D
Σ F y=0
4.68−T DG xSen (30,98 )−1.21=0
T DG=6.74Tn
Σ F X =0
12.84−6.74 xCos (30.98 )−T DF =0
T DF =18.62Tn
8/16/2019 Decr is Pincer Chas
9/11
8/16/2019 Decr is Pincer Chas
10/11
8/16/2019 Decr is Pincer Chas
11/11