DINAMICA ESTRUCTURAL

1. CLASE DE
DINAMICA
REALIZADO POR:
ING. ROMEL VALENZUELA
ING. FERNANDO LEIVA
CLASE 13

2. Ejemplo:
Determine el esfuerzo normal máximo
= 20
= 218000
.
0.50x0.50m
Solución:
=
1
12
50 = 520,833.33
=
1
6
ℎ =
1
6
50 50 = 20,833.33
= + =
12
+
12
=
24
=
24 218,000 520,833.33
400
= 42,578.13
= =
20,000
9.81 ∗ 100
= 20.39 = 0.02039
= =
42,578.13
20.39
= 45.70
= 0
= 0

3. = =
.
= 0.14
∆ =
10
=
0.14
10
= 0.014
40 − 1.1 = 0
Calculando de la grafica de
fuerza externa se obtiene:
=
40
1.10
= 36.3636
= 40 − 36.3636
0 ≤ _ ≤ 1.10
=
∆
2
=
40 0.014
2 0.02039
= 0.1923
= 2 − ∆ − +
∆
= 2 − 45.70 0.014 0.19225 − 0 +
39.49 0.014
0.02039
= 0.68541

4. = 2 − ∆ − +
∆
= 2 − 45.70 0.014 0.68541 − 0.19225 +
39.49 0.014
0.02039
= 1.2727
= 2 − 45.70 0.014 1.2727 − 0.68541 +
39.49 0.014
0.02039
= 1.70887
= 2 − 45.70 0.014 1.70887 − 1.2727 +
39.49 0.014
0.02039
= 1.81043
= 2 − 45.70 0.014 1.81043 − 1.70887 +
39.49 0.014
0.02039
= 1.53093
= 1.81043

5. =
6
. =
6 218,000 520,833.33
400
(1.81043)
= 7709,195.41 ∗
= =
7709,195.41
20,833.33
= 370.04
Esfuerzo Dinámico
Esfuerzo Estático

6. Esfuerzo Normal MáximoEsfuerzo Estático
= 20,000
600
= 33.33
=
12
=
33.33 600
12
= 999,990 ∗
= =
999,990
20,833.33
= 47.99
= +
= 370.04 + 47.99
= 417.035