clase de dinamica cls # 13
DESCRIPTION
DINAMICA ESTRUCTURALTRANSCRIPT
CLASE DE DINAMICA
REALIZADO POR:ING. ROMEL VALENZUELAING. FERNANDO LEIVA CLASE 13
Ejemplo:Determine el esfuerzo normal máximo
퐹푢푒푟푧푎 푒푥푡푒푟푛푎
푊 = 20푡표푛
퐸 = 218000푘푔푐푚
퐶표푙.0.50x0.50m
Solución:
퐼 =1
12 50 = 520,833.33 푐푚
푆 =16 푏ℎ =
16 50 50 = 20,833.33 푐푚
푘 = 푘 + 푘 =12퐸퐼퐿 +
12퐸퐼퐿 =
24퐸퐼퐿
푘 =24 218,000 520,833.33
400 = 42,578.13푘푔푐푚푠
푚 =푃푔 =
20,0009.81 ∗ 100 = 20.39 푘푔 = 0.02039 푡표푛
휔 =푘푚 =
42,578.1320.39 = 45.70 푠푒푔
푥 = 0
푣 = 0
푇 = =.
= 0.14 푠푒푔
∆푡 =푇
10 =0.1410 = 0.014 푠푒푔
40 − 1.1푚 = 0
Calculando de la grafica defuerza externa se obtiene:
푚 =40
1.10 = 36.3636
퐹 = 40 − 36.3636푡
0 ≤ 푡_푖 ≤ 1.10
푥 =퐹 ∆푡2푚 =
40 0.0142 0.02039 = 0.1923푐푚푠
푥 = 2 − 휔∆푡 푥 − 푥 +퐹 ∆푡푚
푥 = 2 − 45.70 0.014 0.19225 − 0 +39.49 0.014
0.02039 = 0.68541
푥 = 2 − 휔∆푡 푥 − 푥 +퐹 ∆푡푚
푥 = 2 − 45.70 0.014 0.68541 − 0.19225 +39.49 0.014
0.02039 = 1.2727
푥 = 2 − 45.70 0.014 1.2727 − 0.68541 +39.49 0.014
0.02039 = 1.70887
푥 = 2 − 45.70 0.014 1.70887 − 1.2727 +39.49 0.014
0.02039 = 1.81043
푥 = 2 − 45.70 0.014 1.81043 − 1.70887 +39.49 0.014
0.02039 = 1.53093
푥 = 1.81043푐푚푠
푀 =6퐸퐼퐿 푥 . =
6 218,000 520,833.33400 (1.81043)
푀 = 7709,195.41 푘푔푓 ∗ 푐푚
휎 =푀푆 =
7709,195.4120,833.33 = 370.04
푘푔푓푐푚
Esfuerzo Dinámico
Esfuerzo Estático
Esfuerzo Normal MáximoEsfuerzo Estático
푊 = 20,000푘푔푓600푐푚 = 33.33
푘푔푓푐푚
푀 = 푊퐿12 =
33.33 60012 = 999,990 푘푔푓 ∗ 푐푚
휎 =푀푆 =
999,99020,833.33 = 47.99
푘푔푓푐푚
휎 = 휎 + 휎
휎 = 370.04 + 47.99
휎 = 417.035푘푔푐푚