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CALCULATING BEAM DEFLECTIONS
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UNIVERSIDAD POLITÉCNICA SALESIANA
NOMBRE:FREDDY GUSTAVO CUCHIPARTE PASTUÑA
FECHA:03/01/20222
CALCULO DE DEFLEXIONES INSTANTANEAS Y A LARGO PLAZO
DATOS
Calcular las deflexiones inmediatas y a largo plazo a 3 meses y
a 5 años
≔
b 30 cm ≔
fc 210 ――
kgf
cm2
≔
fy 4200 ――
kgf
cm2
≔
h 55 cm =
―
h
b
1.833
≔
ωc 2400 ――
kgf
m3
≔
Es ⋅
2.1 106
――
kgf
cm2
≔
Φest 10 mm
≔
Ln 9 m (Cara a Cara de columna) ≔
rec 3 cm
≔
ΦL 20 mm
≔
As' =
⋅
⋅
2 π
⎛
⎜
⎝
――
ΦL
2
⎞
⎟
⎠
2
6.283 cm2
≔
d' =
+
+
rec Φest ――
ΦL
2
5 cm
≔
As =
⋅
⋅
4 π
⎛
⎜
⎝
――
ΦL
2
⎞
⎟
⎠
2
12.566 cm2
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≔
d =
-
-
-
h rec Φest ――
ΦL
2
50 cm
≔
qSCP 180 ――
kgf
m
(Sobrecarga Permanente)
≔
qL 450 ――
kgf
m
(Carga Viva)
≔
PorcentajeCS %
25
CALCULOS:
≔
qPP =
⋅
⋅
⋅
ωc b h 1 396 ――
kgf
m
Peso propio de la viga
≔
qD =
+
qPP qSCP 576 ――
kgf
m
carga muerta
≔
qCS =
+
qD ⋅
PorcentajeCS qL 688.5 ――
kgf
m
carga viva
≔
Ec =
⋅
15000
‾‾‾‾‾‾‾
⋅
fc ――
kgf
cm2
217370.651 ――
kgf
cm2
3. Creado con
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A) CALCULO DE LAS SOLICITACIONES
≔
MD =
―――
⋅
qD Ln2
8
5.832 ⋅
tonnef m
≔
ML =
―――
⋅
qL Ln2
8
4.556 ⋅
tonnef m ≔
MCS =
―――
⋅
qCS Ln2
8
6.971 ⋅
tonnef m
B) ALTURA MINIMA DE VIGAS
≔
hmin =
――
Ln
16
56.25 cm
=
if (
( ,
,
≥
h hmin “No verificar deflexiones” “Verificar Deflexiones”)
) “Verificar Deflexiones”
C) CARACTERISTICAS DE LA SECCION NO AGRIETADA
≔
Ig =
――
⋅
b h3
12
415937.5 cm4
≔
YCG =
―
h
2
27.5 cm
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D) CALCULO DE MOMENTO DE AGRIETAMIENTO
≔
fr =
⋅
2
‾‾‾‾‾‾‾
⋅
fc ――
kgf
cm2
28.983 ――
kgf
cm2
≔
MCR =
―――
⋅
fr Ig
-
h YCG
4.384 ⋅
tonnef m
D) CARACTERISTICAS DE LA SECCION AGRIETADA
≔
n =
――
Es
Ec
9.661 RELACION MODULAR
≔
At =
――――
⋅
(
( -
n 1)
) As'
2
27.209 cm2
≔
at =
――
At
ΦL
13.605 cm
≔
Ab =
――――
⋅
(
( -
n 1)
) As
2
54.418 cm2
≔
bt =
――
Ab
ΦL
27.209 cm
≔
YCG 26.8248 cm ≔
kd =
-
h YCG 28.175 cm
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≔
YCG 26.8248 cm ≔
kd =
-
h YCG 28.175 cm
CALCULO CENTROIDAL E INERCIAL
CALCULO CENTROIDAL
≔
H1 2 cm
SEC AREA X Y A*X A*Y
1
≔
s1 =
⋅
b h 1650 cm2
≔
x 42.21 ≔
y 27.5 ≔
x1 =
⋅
s1 x 69646.5 cm2
≔
y1 =
⋅
s1 y 45375 cm2
2
≔
s2 =
⋅
at H1 27.209 cm2
≔
x 61.02 ≔
y 50 ≔
x2 =
⋅
s2 x 1660.298 cm2
≔
y2 =
⋅
s2 y 1360.454 cm2
3
≔
s3 =
⋅
bt H1 54.418 cm2
≔
x 70.82 ≔
y 5 ≔
x3 =
⋅
s3 x 3853.894 cm2
≔
y3 =
⋅
s3 y 272.091 cm2
4
≔
s4 =
⋅
at H1 27.209 cm2
≔
x 20.41 ≔
y 50 ≔
x4 =
⋅
s4 x 555.337 cm2
≔
y4 =
⋅
s4 y 1360.454 cm2
5
≔
s5 =
⋅
bt H1 54.418 cm2
≔
x 13.61 ≔
y 5 ≔
x5 =
⋅
s5 x 740.631 cm2
≔
y5 =
⋅
s5 y 272.091 cm2
≔
A =
+
+
+
+
s1 s2 s3 s4 s5 1813.254 cm2
≔
Σ =
+
+
+
+
x1 x2 x3 x4 x5 76456.66 cm2
≔
Ay =
+
+
+
+
y1 y2 y3 y4 y5 48640.089 cm2
≔
YCG =
――
Ay
A
26.825
≔
kd =
-
h ⎛
⎝ ⋅
YCG cm⎞
⎠ 28.175 cm
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CALCULO INERCIAL
≔
Ab 54.42 cm
SEC AREA Y A*Y Ix
1 ≔
s1 =
⋅
b ―
h
2
825 cm2
≔
Y1 14.43 ≔
y1 =
⋅
s1 Y1 11904.75 cm2
≔
Iy1 =
―――
⋅
b
⎛
⎜
⎝
―
h
2
⎞
⎟
⎠
3
12
51992.188 cm4
2 ≔
s2 =
⋅
at H1 27.209 cm2
≔
Y2 23.18 ≔
y2 =
⋅
s2 Y2 630.706 cm2
≔
Iy2 =
―――
⋅
at H13
12
9.07 cm4
3 ≔
s3 =
⋅
at H1 27.209 cm2
≔
Y3 23.18 ≔
y3 =
⋅
s3 Y3 630.706 cm2
≔
Iy3 =
―――
⋅
at H13
12
9.07 cm4
≔
A =
+
+
s1 s2 s3 879.418 cm2
≔
a =
+
+
y1 y2 y3 13166.163 cm2
≔
YCG =
―
a
A
14.971
SEC dy Steiner
1 ≔
d1 =
-
Y1 (
(YCG)
) -0.541 ≔
ST1 42150.795 cm4
2 ≔
d2 =
-
Y2 YCG 8.209 ≔
ST2 105377.49 cm4
3 ≔
d3 =
-
Y3 YCG 8.209 ≔
ST3 105377.49 cm4
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≔
ICr1 =
+
+
ST1 ST2 ST3 252905.775 cm4
SEC AREA Y A*Y Ix
1 ≔
s1 =
⋅
Ab H1 108.84 cm2
≔
Y1 21.83 ≔
y1 =
⋅
s1 Y1 2375.977 cm2
≔
Iy1 =
――――
⋅
Ab (
(H1)
)
3
12
36.28 cm4
≔
A =
s1 108.84 cm2
≔
a =
y1 2375.977 cm2
≔
YCG =
―
a
A
21.83
SEC dy St
1 ≔
d1 =
-
Y1 (
(YCG)
) 0 ≔
ST1 51878.198 cm4
≔
ICr2 =
ST1 51878.198 cm4
≔
Icr =
+
ICr1 ICr2 304783.973 cm4
≔
Icr =
+
252913.1137 cm4
51877.2484 cm4
304790.362 cm4
≔
Ix =
+
+
252913.1137 cm4
36.2787 cm4
⋅
108.8360 cm2
(
(21.8248 cm)
)
2
304790.362 cm4
f) MOMENTO DE INERCIA EFECTIVA