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I.
PROBLEMAS PROPUESTOS 4.1. Dibujar las líneas de influencia de las reacciones verticales en los apoyos B, E y G de la viga, así como también las líneas de influencia del momento C.
REACCIÓN B
+↺ ∑MD (izq) = 0
0 ≤ x ≤ 36
By (24) = 0
By (24) - 1 (36 - x) = 0 By =
+↺ ∑MD (der)= 0
By = 0
36−𝑋 24
Cuando: x=0
→
By = 1.5
x=12
→
By = 1
x=24
→
By = 0.5
x=36
→
By = 0
36 ≤ x ≤ 72
REACCIÓN E
+↺ G =0 +↺∑M ∑M = 0 0 0≤ x≤≤x36≤ 36 G
By (60) - 1 (72 - x) + Ey (24) = 0
By (60) - 1 (72 -36−𝑋 x) + Ey (24) = 0
Ey (24) = - ( )(60) + 72 – x 36−𝑋24 Ey (24) = - ( 𝑋−36)(60) + 72 – x E (24) = 24 (60) + 72 – x y
24
𝑋−36
Ey (24) =
5𝑋
Ey (24)24 =
+↺ ∑MG = 0
- 1 (72 - x) + By (60) + Ey (24)= 0 Ey =
(72−𝑥)
(60) + 72 – x
2
36 ≤ x ≤ 72
24
– 90 + 72 – x
5𝑋
Ey (24)Ey=(24) = –3𝑋90– 18 + 72 – x 2
Ey (24)Ey== Ey =
2
3𝑋 3𝑋 – 18
– 18
2
2 24
3𝑋 3𝑋 – 18 2 Ey =
Ey = Ey =
48
–
3
x=0
→
Ey = -0.75
4
x=12
→
Ey = 0
x=24
→
Ey = 0.75
x=36
→
Ey = 1.5
x=48
→
Ey = 1
x=60
→
Ey = 0.5
x=72
→
Ey = 0
24 3𝑋 − 36 Ey = 3 3𝑋 48
–
48
4
3𝑋 − 36 48
Cuando:
REACCIÓN G
+↑ ∑Fy = 0
0 ≤ x ≤ 36
By + Ey + Gy = 1 36−𝑋 24
Gy =
+ 7 4
3𝑋 48
−
72−𝑥
4
24
16
−
36−𝑋 24
Cuando: x=0
→
Gy = 0.25
x=12
→
Gy = 0
x=24
→
Gy = -0.25
x=36
→
Gy = -0.5
x=48
→
Gy = 0
x=60
→
Gy = 0.5
x=72
→
Gy = 1
36 ≤ x ≤ 72
By + Ey + Gy = 1
3
− + Gy = 1
𝑥
+↑ ∑Fy = 0
+ Gy = 1
Gy = 1 −
72−𝑥 24
MOMENTO C
+↺ ∑MC = 0 0 ≤ x ≤ 24 MC + 1 (24 - x) - By (12) = 0 MC = By (12) – 1 (24 – x) MC = MC =
36−𝑋 24 36−𝑋 2
Cuando: x=0
→
MC = -6
x=12
→
MC = 0
x=24
→
MC = 6
(12) – 1 (24 – x) – 24 + x
+↺ ∑MC = 0 24 ≤ x ≤ 36
Cuando:
MC - By (12) = 0
x=24
→
MC = 6
36−𝑋
x=36
→
MC = 0
MC =
24
MC =18 −
(12) 𝑋 2
Cuando:
+↺ ∑MC = 0 36 ≤ x ≤ 72
x=36
→
MC = 0
MC - By (12) = 0
x=48
→
MC = 0
MC = 0
x=60
→
MC = 0
x=72
→
MC = 0
5. CONCLUSIONES 6. RECOMENDACIONES 7. BIBLIOGRAFÍA
PROBLEMAS PROPUESTOS 4.1. Dibujar las líneas de influencia de las reacciones verticales en los apoyos B, E y G de la viga, así como también las líneas de influencia del momento C.
REACCIÓN B
+↺ ∑MD (izq) = 0
0 ≤ x ≤ 36
By (24) = 0
By (24) - 1 (36 - x) = 0 By =
+↺ ∑MD (der)= 0
By = 0
36−𝑋 24
Cuando: x=0
→
By = 1.5
x=12
→
By = 1
x=24
→
By = 0.5
x=36
→
By = 0
36 ≤ x ≤ 72
REACCIÓN E
+↺ G =0 +↺∑M ∑M = 0 0 0≤ x≤≤x36≤ 36 G
By (60) - 1 (72 - x) + Ey (24) = 0
By (60) - 1 (72 -36−𝑋 x) + Ey (24) = 0
Ey (24) = - ( )(60) + 72 – x 36−𝑋24 Ey (24) = - ( 𝑋−36)(60) + 72 – x E (24) = 24 (60) + 72 – x y
24
𝑋−36
Ey (24) =
5𝑋
Ey (24)24 =
+↺ ∑MG = 0
- 1 (72 - x) + By (60) + Ey (24)= 0 Ey =
(72−𝑥)
(60) + 72 – x
2
36 ≤ x ≤ 72
24
– 90 + 72 – x
5𝑋
Ey (24)Ey=(24) = –3𝑋90– 18 + 72 – x 2
Ey (24)Ey== Ey =
2
3𝑋 3𝑋 – 18
– 18
2
2 24
3𝑋 3𝑋 – 18 2 Ey =
Ey = Ey =
48
–
3
x=0
→
Ey = -0.75
4
x=12
→
Ey = 0
x=24
→
Ey = 0.75
x=36
→
Ey = 1.5
x=48
→
Ey = 1
x=60
→
Ey = 0.5
x=72
→
Ey = 0
24 3𝑋 − 36 Ey = 3 3𝑋 48
–
48
4
3𝑋 − 36 48
Cuando:
REACCIÓN G
+↑ ∑Fy = 0
0 ≤ x ≤ 36
By + Ey + Gy = 1 36−𝑋 24
Gy =
+ 7 4
3𝑋 48
−
72−𝑥
4
24
16
−
36−𝑋 24
Cuando: x=0
→
Gy = 0.25
x=12
→
Gy = 0
x=24
→
Gy = -0.25
x=36
→
Gy = -0.5
x=48
→
Gy = 0
x=60
→
Gy = 0.5
x=72
→
Gy = 1
36 ≤ x ≤ 72
By + Ey + Gy = 1
3
− + Gy = 1
𝑥
+↑ ∑Fy = 0
+ Gy = 1
Gy = 1 −
72−𝑥 24
MOMENTO C
+↺ ∑MC = 0 0 ≤ x ≤ 24 MC + 1 (24 - x) - By (12) = 0 MC = By (12) – 1 (24 – x) MC = MC =
36−𝑋 24 36−𝑋 2
Cuando: x=0
→
MC = -6
x=12
→
MC = 0
x=24
→
MC = 6
(12) – 1 (24 – x) – 24 + x
+↺ ∑MC = 0 24 ≤ x ≤ 36
Cuando:
MC - By (12) = 0
x=24
→
MC = 6
36−𝑋
x=36
→
MC = 0
MC =
24
MC =18 −
(12) 𝑋 2
Cuando:
+↺ ∑MC = 0 36 ≤ x ≤ 72
x=36
→
MC = 0
MC - By (12) = 0
x=48
→
MC = 0
MC = 0
x=60
→
MC = 0
x=72
→
MC = 0
5. CONCLUSIONES 6. RECOMENDACIONES 7. BIBLIOGRAFÍA
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