1
ESCUELA DE INGENIERIA CIVIL
ANALISIS ESTRUCTURAL
TEMA:ACTIVIDAD EN CLASES
Vigas y cargas puntuales “Método de Ecuaciones”
NOMBRE:ROSA GRANDA CASTILLO
AÑO:TERCERO “B”
DOCENTE:Ing. ALEXIS MARTÍNEZ
2011
2
Resolver la siguiente Viga.
Diagrama de Cortante
Diagrama de Momentos
∑ Fx=0
Ax=0
∑M=0
By . (6 )−152 ( 1
3.6)−5. ( 4 )=0
By=356
4 m 2 m
5 Tn52 Tn/m
∑ Fy=0
Ay+By=252
Donde:
Ay=203
152 Tn
203 Tn 35
6 Tn
52 h
x6
Si:526
=hx
∴ h=512
x
F=x .
512
x
2=
5 x2
24
5
356
203 Tn
1009
TRAMO I 0 ≤ x < 2
V=−356
−5 x2
24
M=35 x6
−5 x3
72Cuando:
x=0V=356;M=0
x=2V=5 ;M=1009
TRAMO II 2 ≤ x < 6
V=−356
−5−5 x2
24
M=35 x6
−5 ( x−2 )−5 x3
72Cuando:
x=2V=0 ;M=1009
x=6V=−203
;M=¿0
3
Resolver la siguiente Carga Puntual.
Donde:
W i=1
2 EI∫M2dx
W i=1
2 EI [∫0L8
(3Px )2dx+∫L8
L4
(2 Px+ PL8 )
2
dx+∫L4
L2
(Px+ 3PL8 )
2
dx ]W i=
12 EI {∫0
L8
9 P2x
2dx+∫
L8
L4
[4 P2x
2+2 (2Px )(PL8 )+ P2L2
64 ]dx+∫L4
L2
[P2x
2+2 (Px )( 3 PL8 )+ 9 P2 L2
64 ]dx }W i=
12 EI {∫0
L8
9 P2x
2dx+∫
L8
L4
[4 P2x
2+4 P2 Lx
8+P2L2
64 ] dx+∫L4
L2
[P2x
2+6 P2 Lx
8+
9P2L2
64 ] dx}W i=
12 EI {[3 P2 x3 ]L/8
0+[ 4 P2 x3
3+ P2Lx2
4+ P2 L2 x
64 ]L/ 4L/8
+[ P2 x3
3+3 P2 Lx2
8+ 9P2 L2 x
64 ]L/2L /4}
W i=1
2 EI {[ 3 P2L3
512 ]+[ 4 P2L3
192+ P2L3
64+P
2L3
256−4 P2 L3
1536− P2L3
256− P2L3
512 ]+[ P2L3
24+ 3 P2 L3
32+ 9P2L3
128−P2L3
192−3 P2 L3
128−9 P2L3
256 ]}
P P 2 P P P
L/8 L/8 L/8L/8L/4L/4
Ax=0Ay=3 PBy=3P
TRAMO I 0 ≤ x < L8
V=−3PM=3 Px
TRAMO II L8
≤ x < L4
V=−3P+P
M=3 Px−P(x− L8 )
M=2 Px+ PL8
TRAMO II L4
≤ x < L2
V=−3P+2P
M=3 Px−P(x− L8 )−P( x−L
4 )M=Px+ 3 PL
8
4
W i=1
2 EI {3 P2L3
512+ 49P2L3
1536+ 109 P2L3
768 }W i=
23 P2 L3
128EI
Condición:
W e=P2
(∂a+∂b+2∂c+∂d+∂e )
Ecuación de la parábola
y=A x2+Bx+C
Sistema de Ecuaciones
Donde: A=4∂
L2y B=0
Entonces:
y= 4∂
L2x2−∂
W e=P∂2 ( 7
16+ 3
4+2+ 3
4+ 7
16 )
W i=23 P2 L3
128EI
Igualamos W i=W e
23P2L3
128 EI=35
16P∂
x
y
-x
-d
(-L2 ; 0) (L2 ; 0)
(0 ; -d)0=A (−L
2 )2
+B(−L2 )+C
−∂=C
0=A ( L2 )2
+B (L2 )+C
0= AL2
4− BL
2−∂
0= AL2
4+BL
2−∂
0= AL2
4− BL
2−∂
0= AL2
4+BL
2−∂
AL2
2−2∂=0
Si: x= L4
03 L8
Es y= −3∂
4−∂
7 ∂16
5
∂=23 P L3
280 EI
Resolver la siguiente Viga.
∑ Fx=0
Ax=0
∑M=0
−By . (1 )+6( 13
.6)−5. (Ay )=0
By=92
∑ Fy=0
Ay−By=6
Donde:-Ay - By = -6 5Ay+ By = 12
Ay=32
Si:26=hx
∴ h=13x
1 m
2 Tn/m
4 m 1 m
x
92 Tn3
2 Tn
x
x
6
Momento Max. Tramo 2
Diagrama de Cortante
Diagrama de Momentos
TRAMO I 0 ≤ x < 1
V= x2
6
M=−x3
18Cuando: x=0V=0 ;M=0
x=1V=16;M=−1
18
TRAMO II 1 ≤ x < 5
V= x2
6−3
2
M=−x3
18+ 3
2( x−1 )
Cuando:
x=1V=−43
;M=−118
x=5V=83;M=−17
18
TRAMO III 5 ≤ x < 6
V= x2
6−3
2−9
2
M=−x3
18+ 3
2( x−1 )+ 9
2(x−5 )
Cuando:
x=5V=−1116
;M=−1718
x=6V=0 ;M=0
0= x2
6−3
2⟹ x2=3
26∴ x=3
Mmax=32
18+ 3
2(3−1 )∴Mmax=
32
16
83
116
43
[V]+_
7
Resolver la siguiente Carga Puntual.
Donde:
Ax=0Ay=3 PBy=3P
TRAMO I 0 ≤ x < L5
V=−3PM=3 Px
TRAMO II L5
≤ x < 2L5
V=−3P+P
M=3 Px−P(x− L5 )
M=2 Px+ PL5
TRAMO II 2L5
≤ x < L2
V=−3P+P+2P
M=3 Px−P(x− L5 )−2 P(x−2 L
5 )M=PL
178
32
118
[M]+
_
2 P
L/5 L/5 L/5 L/5
P 2 P P
L/5
8
W i=1
2 EI∫M2dx
W i=1
2 EI [2∫0L5
(3 Px )2dx+2∫L5
2L5
(2 Px+ PL5 )
2
dx+2∫2L5
L2
(PL )2dx ]W i=
1EI {∫0
L5
9P2x
2dx+∫
L5
2 L5 [4 P2
x2+( 4 P2 xL
5 )+P2 L2
25 ]dx+∫2L5
L2
[P2L
2 ] dx }W i=
1EI {3P2 x3L/5
0+ 4 P2 x3
32 L/5L/5
+ 4 P2 xL2
102 L/5L/5
+ P2L2 x25
2L/5L/5
+P2 L2 x L/22L/5}
W i=1EI {3 P2L3
125+ 32P2 L3
375−4 P2L3
375+ 8 P2 L3
125−2P2L3
125+ P2 L3
125− P2L3
125+P
2L3
2−2 P2L3
5 }W i=
1EI {3 P2L3
125+ 28P2 L3
375+ 6 P2L3
125+P
2 L3
125+ 1P2L3
10 }W i=
191 P2L3
750 EI
Condición:
W e=P2
(∂1+2∂2+2∂3+∂4 )
Ecuación de la parábola
y=A x2+Bx+C
Sistema de Ecuaciones
Donde: A=4∂
L2y B=0
Entonces:
0=A (−L2 )
2
+B(−L2 )+C
−∂=C
0=A ( L2 )2
+B (L2 )+C
0= AL2
4− BL
2−∂
0= AL2
4+BL
2−∂
0= AL2
4− BL
2−∂
0= AL2
4+BL
2−∂
AL2
2−2∂=0
Si: x= L
100
3L10
Es y= −24∂
25−∂−16∂
25
-d
(-L2 ; 0) (L2 ; 0)
d1
d2 d3
d4
x
y
-x
9
y= 4∂
L2x2−∂
W e=P∂2 (24
25+ 16
25+ 16
25+ 24
25 )W e=
8 P∂5
W i=191 P2L3
750 EI
Igualamos W i=W e
191P2 L3
750 EI=8
5P ∂
∂=191 P L3
1200 EI
Resolver la siguiente viga.
2 T/m
4.501.50
1 2
1,5m
X2 T/m
h
10
ΣFx=0 ∑Fy=0 RAx=¿ 0 RAy+RBy-6=0
RAy+RBy =6 TN
Ray= 2,5 TN= 5/2 TNΣMA=0+←
RBy (6)-4, 50(1, 50+1/3*4, 50)−1,50( 23∗1,50) =0
RBy (6) = 4, 50*3+ 1, 50 RBy = 15/6 Tn =2, 5 Tn
TRAMO I 0 ≤ X < 1, 5
2
1,5= hX
h=2 X1,50
Si x=0 {V=−3,5M=0
Si x=1.5 { V=2M=4,5
V=−72
+ X h2
=0 M=72X− X2
1,50( X
3)=0
V=−72
+ X 2
1,50 M=7
2X− X3
4,50
4,5m
X2 T/m
h
11
TRAMO II 0 ≤ X1 < 4, 5
24,5
= hX
h=2 X4,50
Diagramas de cortante.
Si x=0 {V=2,5M=0
Si x=4,5 {V=−2M=4,5
Punto de inflexion
0=52− x2
4,52 x2−22,5=0x2=11,25x=3,354m
M máx=5,59 T-m
-2Tn
-72Tn
52Tn
0( V )+
- x=3.35
V=52−X1h
2=0
M=
52X1−
X12
4,50(X 1
3)=0
V=52−
X 12
4,50 M=5
2X1−
X13
13,50
12
Diagrama del Momento
Resolver la carga puntual
M=5PX M=5PX-2P (X-L/6) M=5PX-2P(X-L/6)-P(X-L/3)
Wi = 1
2EI¿
Wi = 1
2EI¿
Wi = 1
2EI¿
Wi = 1
2EI¿
Wi =
12EI [50 P2( L3
648 )+18 P2( L3
81− L3
648 )+4 P2L( L2
18− L2
72 )+2 P2 L2
9 ( L3 −L6 )+8P2( L3
24− L3
81 )+ 16 P2 L3
( L2
8− L2
18)+ 8
9P2L2( L
2− L
3)]
Wi = 1
2EI¿
Wi = 12EI ( 199 P2L3
162 )
0
92Tn.m
5.59Tn.m
( M )+
-
13
Wi = 199 P2 L3
3 2 4 EI
y=a x2+bx+c
y=a x2+bx+c
0=a(−L2
)2
+b (−L2 )+c−δ=a (0)+b(0)+c
−δ c=c y=a x2+bx+c
0=a( L2)
2
+b( L2 )+c
aL2
4+ bL
2+c=0 a (−L
2)
2
+b (−L2 )+c=0
aL2
4−bL
2+c=0
4 δcL2 ( L2
4 )+b ( L2 )−δ c=0
2aL2
4+2c=0
b=0
a L2
2+2c=0
a=4 δcL2
y=a x2+bx+c
y=4δ c
L2 x2−δc
14
W e=12(2Pδ a+Pδ b+4 Pδ c+Pδ d+2 Pδ e)
cuando x = -L/6 Si x= -L/3
δ b=4 δcL2 (−L
6)
2
−δ c δ a=4 δcL2 (−L
3)
2
−δ c
δ b=−89
δc δ a=−59
δc
cuando x = L/6
δ d=4δ c
L2 ( L6)
2
−δ c
δ b=−89
δc
Resolver la siguiente viga.
W e=12P(−5
9δc−
89δ c−δ c−
89δ c−
59δ c)
W e=1P2
(−35δ9
)
W e=(−35Pδ18
)
W e=W i
35Pδ18
=199 P2L3
324 EI
δ c=199P L3
630EI
ΣFx=0RAy=0
ΣMA=0 +←34 ( 2
3∗1.5)−27
4 ( 23∗4.5)+RBy (4.5 )=0
RBy (4.5 )=392
RBy= 392∗4.5
=133T
ΣFy=0
RAy+RBy=34+ 23
4
RAy+RBy=152T
RAy=152
−133
RAy=196T
1 Tn/m
3 Tn/m
RAy
1.50 4.50
1 2
15
TRAMO I 0 <= X < 1.5
TRAMO II 1.5 <= X < 6
11.5
= h1.5−X
h=(1.5−X )
1.5
V=1+ 1.5−X
1.52
∗X=
1.5+1.5−X1.52
∗X=
3−X1.52
∗X=3−X
3∗X
M=−[( 1.5−X1.5
∗X ) X2 +X (1−1.5−X
1.5 )2
∗( 23X )]
M=−[ 1.5−X3
∗X 2+
X ( 1.5−1.5−X1.5 )2
∗2
3X ]
Si x=0 {V=0M=0
Si x=1.5 { V=3/ 4M=−3/ 4
V= 34−19
6+
2(X−1.5)2
6=−29
12+(X−1.5)2
3
V=−34
(X−0.5 )+196
(X−1.5 )−(X−1.5)3
9
Si x=1.5 {V=−29 /12M=−3/4
Si x=6 {V=13 /3M=0
34.5
= hX−1.5
h=2(X−1.5)
3
PUNTO DE INFLEXION
0=−2912
+(X−1.5)2
2(X−1.5)2
2=29
12
(X−1.5)2=5812
16
Diagramas de momento y cortante realizados en Excel.
Diagrama de Cortantes
X CORTANTE MOMENTO0 0,00 0,00
0,5 0,42 -0,111,5 0,75 -0,751,5 -2,42 -0,752 -2,33 0,443 -1,67 2,50
3,7 -0,80 3,384 -0,33 3,565 1,67 2,946 4,33 0,00
TRAMO I 0 <= X < 1.5
TRAMO II 1.5 <= X < 6
PUNTO DE INFLEXION
0=−2912
+(X−1.5)2
2(X−1.5)2
2=29
12
(X−1.5)2=5812
0
133 Tn
34Tn
2912Tn
( V )+
-
17
Diagrama de momentos
Resolver la carga puntual
34Tn.m
3.59Tn.m
( M )+
-
Wi= 12 EI∫M 2dx
We=12
(4 Pδ 1+3 Pδ2+3 Pδ3+4 Pδ 4 )M=7 Px
M=7 Px−4 P(x− L6 )=7 Px−4 Px+ 2
3PL=3 pX+ 2
3PL
18
Wi= 12 EI∫M 2dx
We=12
(4 Pδ 1+3 Pδ2+3 Pδ3+4 Pδ 4 )M=7 Px
M=7 Px−4 P(x− L6 )=7 Px−4 Px+ 2
3PL=3 pX+ 2
3PL
Wi= 1EI
¿
Wi= 1EI
[(49 P2 X3
3 )0
L6 +(9 P2 X
3
3+4 P2L
X2
3+ 4
9P2L2 X )L
6
L3 +( 25
9P2 L2 X)L
3
L2 ]
Wi= 1EI
[(49 P2 X3
3 )0
L6 +(3 P2 X3+2P2 L X2+ 4
9P2L2 X)L
6
L3 +(25
9P2L2 X )L
3
L2 ]
Wi= 1EI
[ 493P2( L6 )
3
+3 P2 X ( L3 )3
+2P2L X ( L3 )2
+ 49P2L2( L3 )−3 P2 X ( L6 )
3
−2 P2L X ( L6 )2
−49P2 L2(L
6)+ 25
9P2 L2 L
2−25
9P2L2 L
3]
Wi= 1EI
¿
Wi= 1EI ( 71
81P2L3)= 71
81EIP2L3
19
We=Wi
y= δ9−δ
12
(4 Pδ1+3Pδ 2+3 Pδ3+4 Pδ4 )= 7181EI
P2L3
12 (4 P
59δ+3P
89δ+3 P
89δ+4 P
59δ)= 71
81EIP2L3
Pδ2 ( 20
9+ 24
9+24
9+ 20
9 )= 7181EI
P2 L3
Pδ2 ( 88
9 )= 7181EI
P2 L3
Pδ (44 )= 719 EI
P2L3
δ= 71 P2L3
9∗44 EI=71 P2L3
396 EI
y (−L6 )( L6 )=4 δ
L2 ( L6)
2
−δ
y= δ9−δ
y=−89
δ (δ2 , δ 3)
y= 4δ
L2X2−δ
y (−L3 )( L3 )=4 δ
L2 ( L3)
2
−δ
y= 4δ9
−δ
y=−59
δ(δ1 , δ 4)
y=a x2+bx+c
(−L2
,0)
0=a(−L2
)2
+b (−L2
)+c
0=aL2
4−bL
2+c
(0 ,−δ)−δ=CC=−δ
( L2,0)
0=a( L2)
2
+b (L2)+c
0=aL2
4+ bL
2+c
0=aL2
4−bL
2+c
0=aL2
4+ bL
2+c
0=aL2
2+0+2c
a L2
4=−2c
a=4 δ
L2
( 4 δ
L2 ) L4 +b (L2 )=δ
δ+b=δb=0
Top Related