SOLUCION DE LA INTEGRAL DE DUHAMEL CON MATHCAD

1.-PROBLEMA 01: SOLUCION UTILIZANDO LA INTEGRAL DE DUHAMEL(CON MATHCAD PRIME 3.1)

k1 ≔ 15.65

m1 ≔ 2

π Tn1 ≔ 2 ⋅ ―― wn1

wn1 ≔ 88.46

Tn1 = 0.071

P1 ((t)) ≔ 15 ⋅ sin ((40 ⋅ t)) t

A1 ((t)) ≔ ⌠ ⌡ P1 ((x)) ⋅ cos ((wn1 ⋅ x)) d x 0 t

B1 ((t)) ≔ ⌠ ⌡ P1 ((x)) ⋅ sin ((wn1 ⋅ x)) d x 0

32.3 u1 ((t)) ≔ ―――⋅ ((A1 ((t)) ⋅ sin ((wn1 ⋅ t)) − B1 ((t)) ⋅ cos ((wn1 ⋅ t)))) m1 ⋅ wn1 f1 ((t)) ≔ k1 ⋅ u1 ((t)) d v1 ((t)) ≔ ―― u1 ((t)) dt t ≔ 0 , 0.01 ‥ 2

0.9 0.75 0.6 0.45 0.3 0.15 0 0

0.2

0.4

0.6

0.8

1

-0.15 -0.3 -0.45 -0.6 -0.75 -0.9

t

1.2

1.4

1.6

1.8

2

f1 ((t))

2.-PROBLEMA 02: SOLUCION UTILIZANDO LA INTEGRAL DE DUHAMEL(CON MATHCAD PRIME 3.1)

k2 ≔ 237.926 m2 ≔ 5

wn2 ≔ 6.8982

π Tn2 ≔ 2 ⋅ ―― wn2

Tn2 = 0.911

P2 ((t)) ≔ 16 ⋅ t − 4 t

A2 ((t)) ≔ ⌠ ⌡ P2 ((x)) ⋅ cos ((wn2 ⋅ x)) d x 0 t

B2 ((t)) ≔ ⌠ ⌡ P2 ((x)) ⋅ sin ((wn2 ⋅ x)) d x 0

32.3 u2 ((t)) ≔ ―――⋅ ((A2 ((t)) ⋅ sin ((wn2 ⋅ t)) − B2 ((t)) ⋅ cos ((wn2 ⋅ t)))) m2 ⋅ wn2 f2 ((t)) ≔ k2 ⋅ u2 ((t)) d v2 ((t)) ≔ ―― u2 ((t)) dt t ≔ 0 , 0.01 ‥ 3

1.5⋅10³ 1.35⋅10³ 1.2⋅10³ 1.05⋅10³ 900 750

f2 ((t))

600 450 300 150 0 0

0.3

0.6

0.9

1.2

1.5

-150

t

1.8

2.1

2.4

2.7

3

3.-PROBLEMA 03: SOLUCION UTILIZANDO LA INTEGRAL DE DUHAMEL(CON MATHCAD PRIME 3.1)

k3 ≔ 7714.51 m3 ≔ 2.5

π Tn3 ≔ 2 ⋅ ―― wn3

wn3 ≔ 55.55

Tn3 = 0.113

P3 ((t)) ≔ −2.5 ⋅ 0.7 ⋅ 9.8 t

A3 ((t)) ≔ ⌠ ⌡ P3 ((x)) ⋅ cos ((wn3 ⋅ x)) d x 0 t

B3 ((t)) ≔ ⌠ ⌡ P3 ((x)) ⋅ sin ((wn3 ⋅ x)) d x 0

32.3 u3 ((t)) ≔ ―――⋅ ((A3 ((t)) ⋅ sin ((wn3 ⋅ t)) − B3 ((t)) ⋅ cos ((wn3 ⋅ t)))) m3 ⋅ wn3 f3 ((t)) ≔ k3 ⋅ u3 ((t)) d v3 ((t)) ≔ ―― u3 ((t)) dt t ≔ 0 , 0.005 ‥ 1

50 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

-50 -150 -250 -350 -450 -550

f3 ((t))

-650 -750 -850 -950 -1.05⋅10³ -1.15⋅10³

t

a ≔ 0.7