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