Armaduras - Análisis Matricial

2016

ANÁLISIS MATRICIAL DE ESTRUCTURAS ARMADURAS

JIMMY DE LA CRUZ H. UNIVERSIDAD NACIONAL FEDERICO VILLARREAL JUNIO 2016

UNIVERSIDAD NACIONAL FEDERICO VILLARREAL FACULTAD DE INGENIERIA CIVIL

ANÁLISIS MATRICIAL DE ESTRUCTURAS

ARMADURAS Ecuacion de Rigidez de un elemento: La expresión matemática para el método matricial de rigidez es: ⎡⎣ Fi ⎤⎦ = ⎡⎣ Ki ⎤⎦ ⋅ ⎡⎣ Δi ⎤⎦ Donde: ⎡⎣ Fi ⎤⎦ = Vector de Fuerzas del elemento "i" ⎡⎣ Ki ⎤⎦ = Matriz de Rigidez del elemento "i" ⎡⎣ Δi ⎤⎦ = Vector de desplazamientos del elemento "i" Matriz de Rigidez en el eje Local de un Elemento:

Coeficientes correspondiente al 1 GDL y 2 GDL:

E⋅A p'i1 = ――⋅ u'i L E⋅A p'j1 = −――⋅ u'i L

−(E ⋅ A) p'i2 = ―――⋅ u'j L E⋅A p'j2 = ――⋅ u'j L

Elaborado por: Jimmy De La Cruz



Por superposición: E⋅A E⋅A p'i = ――⋅ u'i − ――⋅ u'j L L

−E ⋅ A E⋅A p'j = ――― ⋅ u'i + ――⋅ u'j L L

⎡ p'i ⎤ E ⋅ A ⎡ 1 −1 ⎤ ⎡ u'i ⎤ ⎢⎣ p' ⎥⎦ = ――⎢⎣ −1 1 ⎥⎦ ⋅ ⎢⎣ u' ⎥⎦ j j L

...(1)

[ p' ] = [ K' ] ⋅ [ u' ]

...(2)

E ⋅ A ⎡ 1 −1 ⎤ K'= ――⎢ L ⎣ −1 1 ⎥⎦

...(3)

Matriz de Transformación de desplazamientos:

ux'i = uxi ⋅ cos (ϕ) + uyi ⋅ sin (ϕ)

...(4)

ux'j = uxj ⋅ cos (ϕ) + uyj ⋅ sin (ϕ)

De (4),(5) ux'i = uxi ⋅ cos (ϕ) + uyi ⋅ sin (ϕ) ux'j = uxj ⋅ cos (ϕ) + uyj ⋅ sin (ϕ) ⎡ uxi ⎤ ⎡ ux'i ⎤ ⎡ cos (ϕ) sin (ϕ) 0 0 ⎤ ⎢ uyi ⎥ = ⋅⎢ ⎥ ⎢⎣ ux' ⎥⎦ ⎢⎣ 0 0 cos (ϕ) sin (ϕ) ⎥⎦ uxj i ⎢ ⎥ ⎣ uyj ⎦ [ u' ] = [ T ] ⋅ [ u ]

...(6)

⎡ cos (ϕ) sin (ϕ) 0 0 ⎤ T= ⎢ 0 cos (ϕ) sin (ϕ) ⎥⎦ ⎣ 0


...(5)



Rotación de Sistema de Referencia: Conocida la matriz de rigidez local de un elemento de una estructura "K'", para calcular la matriz de rigidez K en el sistema de coordenadas globales se realiza utilizando la matriz de rotación T

" [ T ] ". Para pasar de coordenadas locales a globales:

pxi = px'i ⋅ cos (ϕ)

...(7)

pxj = px'j ⋅ cos (ϕ)

...(9)

pyi = px'i ⋅ sin (ϕ)

...(8)

pyj = px'j ⋅ sin (ϕ)

...(10)

De (4),(5) ,(6) ⎡ pxi ⎤ ⎢ py ⎥ i ⎢ ⎥= pxj ⎢ ⎥ ⎣ pyj ⎦

y (7):

⎡ cos (ϕ) 0 ⎤ ⎢ sin (ϕ) 0 ⎥ ⎡ px'i ⎤ ⋅ ⎢ 0 cos (ϕ) ⎥ ⎢⎣ px'j ⎥⎦ ⎢⎣ 0 sin (ϕ) ⎥⎦ T

[ p ] = [ T ] ⋅ [ p' ] Reemplazando (2)y (6) T

[ p ] = [ T ] ⋅ [ K' ] ⋅ [ u' ]

...(11) en (11): [ u' ] = [ T ] ⋅ [ u ]

T

[ p ] = [ T ] ⋅ [ K' ] ⋅ [ T ] ⋅ [ u ] Se deduce que la matriz de rigidez en coordenadadas globales se obtiene realizando el siquiente triple producto matricial: T

[ K ] = [ T ] ⋅ [ K' ] ⋅ [ T ]


...(12)



T

Matriz de rotación ( T ): ⎡ cos (ϕ) 0 ⎤ ⎢ sin (ϕ) 0 ⎥ T T = ⎢ 0 cos (ϕ) ⎥ ⎢⎣ 0 sin (ϕ) ⎥⎦

⎡ cos (ϕ) sin (ϕ) 0 0 ⎤ T= ⎢ 0 cos (ϕ) sin (ϕ) ⎥⎦ ⎣ 0

Para pasar de coordenadas siguientes ecuaciónes:

globales

a

locales

se

utiliza

[ p' ] = [ T ] ⋅ [ p ] [ u' ] = [ T ] ⋅ [ u ] Error de Fabricación: ΔL = Variacíon de longitud debido al error de fabricación. Coordenadas Locales: E ⋅ A ⋅ ΔL px'i = ―――― L

py'i = 0

−E ⋅ A ⋅ ΔL p'xj = ―――― L

py'j = 0

⎡ E ⋅ A ⋅ ΔL ⎤ ―――― ⎥ ⎡ px'i ⎤ ⎢ L = ⎢⎣ px' ⎥⎦ ⎢ −E ⋅ A ⋅ ΔL ⎥ j ⎢ ―――― ⎥ L ⎣ ⎦ ⎡ E ⋅ A ⋅ ΔL ⎤ ⎥ ⎢ ―――― L f' = ⎢ −E ⋅ A ⋅ ΔL ⎥ ⎢ ―――― ⎥ L ⎣ ⎦

...(13)

Coordenadas Globales: Reemplazando en (11) T

[ f ] = [ T ] ⋅ [ f' ] ⎡ pxi ⎤ ⎢ py ⎥ i ⎥= ⎢ pxj ⎢ ⎥ ⎣ pyj ⎦

...(14)

E ⋅ A ⋅ ΔL ⎤ ⎡ cos (ϕ) 0 ⎤ ⎡ ―――― ⎥ ⎢ ⎢ sin (ϕ) ⎥ L 0 ⋅ ⎢ ⎢ 0 cos (ϕ) ⎥ −E ⋅ A ⋅ ΔL ⎥ ⎢ ―――― ⎥ ⎢⎣ 0 sin (ϕ) ⎥⎦ ⎣ L ⎦


las



Variación de Temperatura: α = Coeficiente de expansión térmica ΔL = Variacíon de longitud debido la variación de ΔT = Variación de temperatura Coordenadas Locales: ΔL = α ⋅ ΔT ⋅ L px'i = E ⋅ A ⋅ α ⋅ ΔT

py'i = 0

p'xj = −E ⋅ A ⋅ α ⋅ ΔT

py'j = 0

⎡ px'i ⎤ ⎡ E ⋅ A ⋅ α ⋅ ΔT ⎤ ⎢⎣ px' ⎥⎦ = ⎢⎣ −E ⋅ A ⋅ α ⋅ ΔT ⎥⎦ j

⎡ E ⋅ A ⋅ α ⋅ ΔT ⎤ f' = ⎢ ⎣ −E ⋅ A ⋅ α ⋅ ΔT ⎥⎦

...(15)

Coordenadas Globales: Reemplazando en (11) T

[ f ] = [ T ] ⋅ [ f' ] ⎡ pxi ⎤ ⎢ py ⎥ i ⎢ ⎥= pxj ⎢ ⎥ ⎣ pyj ⎦

...(16)

⎡ cos (ϕ) 0 ⎤ ⎢ sin (ϕ) 0 ⎥ ⎡ E ⋅ A ⋅ α ⋅ ΔT ⎤ ⋅ ⎢ 0 cos (ϕ) ⎥ ⎢⎣ −E ⋅ A ⋅ α ⋅ ΔT ⎥⎦ ⎢⎣ 0 sin (ϕ) ⎥⎦


temperatura.



Problema 01: Ensamblar la matriz de rigidez K, calcular las fuerzas en los elementos de la estructura. Datos: E ≔ 200 GPa A0 ≔ 100 mm

Solución: 1. Grados de Libertad:

2. Matriz de rigidez de cada elemento "Local":

EA ⎡ 1 −1 ⎤ K' = ―― ⋅ L ⎢⎣ −1 1 ⎥⎦


2



Elemento AB:

Aab ≔ 100 mm

Lab ≔ 14.142 m

2

⎡ 1.4142 −1.4142 ⎤ kN K'ab = ⎢ ―― ⎣ −1.4142 1.4142 ⎥⎦ mm Lbc ≔ 14.142 m

Elemento BC:

Abc ≔ 100 mm

⎡ 1.4142 −1.4142 ⎤ kN ―― K'bc = ⎢ ⎣ −1.4142 1.4142 ⎥⎦ mm 3. Matriz de rigidez de cada elemento "Global": ⎡ cos ⎛⎝ϕi⎞⎠ sin ⎛⎝ϕi⎞⎠ ⎤ 0 0 T = ⎢ 0 0 cos ⎛⎝ϕj⎠⎞ sin ⎛⎝ϕj⎞⎠ ⎥⎦ ⎣ Elemento AB:

ϕi ≔ 45°

ϕj ≔ 45°

⎤ ⎡ 0.707 0.707 0 0 Tab = ⎢ ⎣0 0 0.707 0.707 ⎥⎦ ⎡0⎤ ⎢0⎥ = ⎢ ⎥ 1 ⎢⎣ 2 ⎥⎦

T

Kab ≔ Tab ⋅ K'ab ⋅ Tab

GLab 0

0

2

1

0 ⎡ 0.707 0.707 −0.707 −0.707 ⎤ ⎢ 0.707 0.707 −0.707 −0.707 ⎥ kN 0 ―― Kab = ⎢ −0.707 −0.707 0.707 0.707 ⎥ mm 1 ⎢ ⎥ ⎣ −0.707 −0.707 0.707 0.707 ⎦ 2

Elemento BC:

ϕi ≔ −45°

ϕj ≔ −45°

⎡ 0.707 −0.707 0 ⎤ 0 Tbc = ⎢ ⎥ 0 0.707 −0.707 ⎦ ⎣0 T

Kbc ≔ Tbc ⋅ K'bc ⋅ Tbc 1

GLbc 2

0

0

⎡1⎤ ⎢2⎥ = ⎢ ⎥ 0 ⎢⎣ 0 ⎥⎦

⎡ 0.707 −0.707 −0.707 0.707 ⎤ 1 ⎢ −0.707 0.707 0.707 −0.707 ⎥ kN 2 ―― Kbc = ⎢ −0.707 0.707 0.707 −0.707 ⎥ mm 0 ⎢ ⎥ ⎣ 0.707 −0.707 −0.707 0.707 ⎦ 0


2



4. Matriz de rigidez de la estructura: 1

2

⎤ kN 1 ⎡ 1.414 0 ―― K=⎢ ⎣0 1.414 ⎥⎦ mm 2

⎡1⎤ GL = ⎢ ⎥ ⎣2⎦

5. Matriz de Fuerzas: ⎡ 0 ⎤ F≔⎢ kN ⎣ −100 ⎥⎦ 6. Desplazamientos: F = K⋅u

u≔K

−1

⎡ 0 ⎤ ⋅F=⎢ mm ⎣ −70.71 ⎥⎦

⎤ ⎡ 0 0 ⎥ ⎢ 0 0 uab ≔ ⎢ ⎥ mm 1 0 ⎢⎣ −70.71 ⎥⎦ 2

1 2

⎡ ⎤ 1 0 ⎢ −70.71 ⎥ 2 ubc ≔ ⎢ ⎥ mm 0 0 ⎢⎣ ⎥⎦ 0 0

7. Fuerzas en cada elemento: ⎡ 50 ⎤ ⎢ 50 ⎥ Fab ≔ Kab ⋅ uab = ⎢ kN −50 ⎥ ⎢⎣ −50 ⎥⎦

F'ab = −70.711 kN

⎡ 50 ⎤ ⎢ −50 ⎥ kN Fbc ≔ Kbc ⋅ ubc = ⎢ −50 ⎥ ⎢⎣ ⎥ 50 ⎦

F'bc = −70.711 kN

8. Reacciones en los apoyos: Rax ≔ 50 kN = 50 kN Ray ≔ 50 kN = 50 kN Rcx ≔ −50 kN = −50 kN Rcy ≔ 50 kN = 50 kN







EA ⎡ 1 −1 ⎤ K' = ―― ⋅ L ⎢⎣ −1 1 ⎥⎦


2


Elemento AB:


Lab ≔ 14.142 m

⎡ 1.4142 −1.4142 ⎤ kN ―― K'ab = ⎢ ⎣ −1.4142 1.4142 ⎥⎦ mm Lbc ≔ 14.142 m

Elemento BC:

⎡ 1.4142 −1.4142 ⎤ kN ―― K'bc = ⎢ ⎣ −1.4142 1.4142 ⎥⎦ mm Lbd ≔ 10.000 m

Elemento BD: ⎡ 2 −2 ⎤ kN ―― K'bd = ⎢ ⎣ −2 2 ⎥⎦ mm

3. Matriz de rigidez de cada elemento "Global": ⎡ cos ⎛⎝ϕi⎞⎠ sin ⎛⎝ϕi⎞⎠ ⎤ 0 0 T = ⎢ 0 0 cos ⎛⎝ϕj⎞⎠ sin ⎛⎝ϕj⎞⎠ ⎥⎦ ⎣ Elemento AB:

ϕi ≔ 45°

ϕj ≔ 45°

⎤ ⎡ 0.707 0.707 0 0 Tab = ⎢ ⎣0 0 0.707 0.707 ⎥⎦ T


GLab

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 1 ⎢⎣ 2 ⎥⎦

⎡ 0.707 0.707 −0.707 −0.707 ⎤ ⎢ 0.707 0.707 −0.707 −0.707 ⎥ kN ―― Kab = ⎢ −0.707 −0.707 0.707 0.707 ⎥ mm ⎢ ⎥ ⎣ −0.707 −0.707 0.707 0.707 ⎦



Elemento BC:

ϕi ≔ −45°


ϕj ≔ −45°

⎤ ⎡ 0.707 −0.707 0 0 Tbc = ⎢ ⎥ ⎣0 0 0.707 −0.707 ⎦ T

Kbc ≔ Tbc ⋅ K'bc ⋅ Tbc

GLbc

⎡1⎤ ⎢2⎥ = ⎢ ⎥ 0 ⎢⎣ 0 ⎥⎦

⎡ 0.707 −0.707 −0.707 0.707 ⎤ ⎢ −0.707 0.707 0.707 −0.707 ⎥ kN ―― Kbc = ⎢ −0.707 0.707 0.707 −0.707 ⎥ mm ⎢ ⎥ ⎣ 0.707 −0.707 −0.707 0.707 ⎦ Elemento BD:

ϕi ≔ −90°

ϕj ≔ −90°

⎡ 0 −1 0 0 ⎤ Tbd = ⎢ ⎣ 0 0 0 −1 ⎥⎦ T

Kbd ≔ Tbd ⋅ K'bd ⋅ Tbd

⎡0 0 ⎢0 2 Kbd = ⎢ 0 0 ⎢⎣ 0 −2

GLbd

⎡1⎤ ⎢2⎥ = ⎢ ⎥ 0 ⎢⎣ 0 ⎥⎦

0 0⎤ 0 −2 ⎥ kN ―― 0 0 ⎥ mm 0 2 ⎥⎦

4. Matriz de rigidez de la estructura: ⎡ 1.414 0 ⎤ kN ―― K=⎢ ⎣0 3.414 ⎥⎦ mm 5. Matriz de Fuerzas: ⎡ 0 ⎤ F≔⎢ kN ⎣ −100 ⎥⎦ 6. Desplazamientos: F = K⋅u u≔K

−1

⎡ 0 ⎤ ⋅F=⎢ mm ⎣ −29.289 ⎥⎦


⎡1⎤ GL = ⎢ ⎥ ⎣2⎦


⎤ ⎡ 0 ⎥ ⎢ 0 uab ≔ ⎢ ⎥ mm 0 ⎢⎣ −29.289 ⎥⎦


⎤ ⎡ 0 ⎢ −29.289 ⎥ ubc ≔ ⎢ ⎥ mm 0 ⎢⎣ ⎥⎦ 0

⎤ ⎡ 0 ⎢ −29.289 ⎥ ubd ≔ ⎢ ⎥ mm 0 ⎢⎣ ⎥⎦ 0

7. Fuerzas en cada elemento: ⎡ 20.711 ⎤ ⎢ 20.711 ⎥ Fab ≔ Kab ⋅ uab = ⎢ kN −20.711 ⎥ ⎢⎣ −20.711 ⎥⎦

F'ab = −29.289 kN

⎡ 20.711 ⎤ ⎢ −20.711 ⎥ kN Fbc ≔ Kbc ⋅ ubc = ⎢ −20.711 ⎥ ⎢⎣ ⎥ 20.711 ⎦

F'bc = −29.289 kN

⎤ ⎡ 0 ⎢ −58.578 ⎥ Fbd ≔ Kbd ⋅ ubd = ⎢ ⎥ kN 0 ⎢⎣ 58.578 ⎥⎦

F'bd = −58.578 kN

8. Reacciones en los apoyos: Rax ≔ 20.711 kN

Rcx ≔ −20.711 kN

Rdx ≔ 0 kN = 0 kN

Ray ≔ 20.711 kN

Rcy ≔ 20.711 kN

Rdx ≔ 58.578 kN







EA ⎡ 1 −1 ⎤ K' = ―― ⋅ L ⎢⎣ −1 1 ⎥⎦


2


Elemento AB:


Lab ≔ 1.414 m

⎡ 14.1443 −14.1443 ⎤ kN ―― K'ab = ⎢ ⎣ −14.1443 14.1443 ⎥⎦ mm Lbc ≔ 1.414 m

Elemento BC:

⎡ 14.1443 −14.1443 ⎤ kN ―― K'bc = ⎢ ⎣ −14.1443 14.1443 ⎥⎦ mm Lbd ≔ 1.000 m

Elemento BD:

⎡ 20 −20 ⎤ kN ―― K'bd = ⎢ ⎣ −20 20 ⎥⎦ mm 3. Matriz de rigidez de cada elemento "Global": ⎡ cos ⎛⎝ϕi⎞⎠ sin ⎛⎝ϕi⎞⎠ ⎤ 0 0 T = ⎢ 0 0 cos ⎛⎝ϕj⎞⎠ sin ⎛⎝ϕj⎞⎠ ⎥⎦ ⎣ Elemento AB:

ϕi ≔ 45°

ϕj ≔ 45°

⎡ 0.707 0.707 0 ⎤ 0 Tab = ⎢ ⎣0 0 0.707 0.707 ⎥⎦ T


GLab

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 1 ⎢⎣ 2 ⎥⎦

⎡ 7.072 7.072 −7.072 −7.072 ⎤ ⎢ 7.072 7.072 −7.072 −7.072 ⎥ kN ―― Kab = ⎢ −7.072 −7.072 7.072 7.072 ⎥ mm ⎢ ⎥ ⎣ −7.072 −7.072 7.072 7.072 ⎦



Elemento BC:

ϕi ≔ −45°


ϕj ≔ −45°

⎤ ⎡ 0.707 −0.707 0 0 Tbc = ⎢ ⎥ ⎣0 0 0.707 −0.707 ⎦

T


GLbc

⎡1⎤ ⎢2⎥ = ⎢ ⎥ 0 ⎢⎣ 0 ⎥⎦

⎡ 7.072 −7.072 −7.072 7.072 ⎤ ⎢ −7.072 7.072 7.072 −7.072 ⎥ kN Kbc = ⎢ ―― −7.072 7.072 7.072 −7.072 ⎥ mm ⎢ ⎥ ⎣ 7.072 −7.072 −7.072 7.072 ⎦ Elemento BD:

ϕi ≔ 0 °

ϕj ≔ 0 °

⎡1 0 0 0⎤ Tbd = ⎢ ⎣ 0 0 1 0 ⎦⎥ T

Kbd ≔ Tbd ⋅ K'bd ⋅ Tbd ⎡ 20 ⎢ 0 Kbd = ⎢ −20 ⎢⎣ 0

0 −20 0 ⎤ 0 0 0 ⎥ kN ―― 0 20 0 ⎥ mm ⎥ 0 0 0⎦

GLbd

⎡1⎤ ⎢2⎥ = ⎢ ⎥ 0 ⎢⎣ 0 ⎥⎦

4. Matriz de rigidez de la estructura: ⎡ 34.144 0 ⎤ kN ―― K=⎢ ⎥ ⎣ 0 14.144 ⎦ mm 5. Matriz de Fuerzas: ⎡ 100 ⎤ kN F≔⎢ ⎣ 0 ⎥⎦


⎡1⎤ GL = ⎢ ⎥ ⎣2⎦



6. Desplazamientos: F = K⋅u

u≔K

⎡ 0 ⎤ ⎢ 0 ⎥ uab ≔ ⎢ mm 2.929 ⎥ ⎢⎣ 0 ⎥⎦

−1

⎡ 2.929 ⎤ ⋅F=⎢ ⎥⎦ mm ⎣0 ⎡ 2.929 ⎤ ⎢ 0 ⎥ ubc ≔ ⎢ mm 0 ⎥ ⎢⎣ 0 ⎥⎦

⎡ 2.929 ⎤ ⎢ 0 ⎥ ubd ≔ ⎢ mm 0 ⎥ ⎢⎣ 0 ⎥⎦

7. Deformaciones en cada elemento: Ai = ⎡⎣ −cos ⎛⎝θi⎞⎠ −sin ⎛⎝θi⎞⎠ cos ⎛⎝θj⎞⎠ sin ⎛⎝θj⎞⎠ ⎤⎦ ui' = Ai ⋅ ui - ΔLi Elemento AB:

A = ⎡⎣ −0.707 −0.707 0.707 0.707 ⎤⎦

u'ab ≔ A ⋅ uab = 2.071 mm

Elemento BC:

A = ⎡⎣ −0.71 0.71 0.71 −0.71 ⎤⎦

u'bc ≔ A ⋅ ubc = −2.071 mm

Elemento BD:

A = ⎡⎣ −1 0 1 0 ⎤⎦

u'bc ≔ A ⋅ ubc = −2.929 mm

8. Fuerzas en cada elemento: ⎡ −20.714 ⎤ ⎢ −20.714 ⎥ Fab ≔ Kab ⋅ uab = ⎢ kN 20.714 ⎥ ⎢⎣ ⎥ 20.714 ⎦

F'ab = 29.294 kN

⎡ 20.714 ⎤ ⎢ −20.714 ⎥ kN Fbc ≔ Kbc ⋅ ubc = ⎢ −20.714 ⎥ ⎢⎣ ⎥⎦ 20.714

F'bc = −29.294 kN

⎡ 58.58 ⎤ ⎢ 0 ⎥ kN Fbd ≔ Kbd ⋅ ubd = ⎢ −58.58 ⎥ ⎢⎣ 0 ⎥⎦

F'bd = −58.58 kN




9. Reacciones en los apoyos: Rax ≔ −20.714 kN

Rcx ≔ −20.714 kN

Rdx ≔ −58.58 kN

Ray ≔ −20.714 kN

Rcy ≔ 20.714 kN

Rdy ≔ 0 kN = 0 kN







EA ⎡ 1 −1 ⎤ K' = ―― ⋅ L ⎢⎣ −1 1 ⎥⎦


2


Elemento AB:


Lab ≔ 4.472 m

⎡ 4.4723 −4.4723 ⎤ kN ―― K'ab = ⎢ ⎣ −4.4723 4.4723 ⎥⎦ mm Elemento BC:

Lbc ≔ 4.472 m

⎡ 4.4723 −4.4723 ⎤ kN ―― K'bc = ⎢ ⎣ −4.4723 4.4723 ⎥⎦ mm Elemento CD:

Lcd ≔ 4.472 m

⎡ 4.4723 −4.4723 ⎤ kN ―― K'cd = ⎢ ⎣ −4.4723 4.4723 ⎥⎦ mm Elemento BD:

Lbd ≔ 4.000 m

⎡ 5 −5 ⎤ kN ―― K'bd = ⎢ ⎣ −5 5 ⎥⎦ mm Elemento DE:

Lde ≔ 4.472 m

⎡ 4.4723 −4.4723 ⎤ kN ―― K'de = ⎢ ⎣ −4.4723 4.4723 ⎥⎦ mm 3. Matriz de rigidez de cada elemento "Global": Elemento AB:

ϕi ≔ 63.345°

ϕj ≔ 63.345°

⎡ 0.449 0.894 0 ⎤ 0 Tab = ⎢ ⎣0 0 0.449 0.894 ⎥⎦ T


GLab

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 1 ⎢⎣ 0 ⎥⎦

⎡ 0.9 1.793 −0.9 −1.793 ⎤ ⎢ 1.793 3.572 −1.793 −3.572 ⎥ kN ―― Kab = ⎢ −0.9 −1.793 0.9 1.793 ⎥ mm ⎢ ⎥ ⎣ −1.793 −3.572 1.793 3.572 ⎦



Elemento BC:

ϕi ≔ −63.345°


ϕj ≔ −63.345°

⎤ ⎡ 0.45 −0.89 0 0 Tbc = ⎢ ⎣0 0 0.45 −0.89 ⎥⎦ T


GLbc

⎡1⎤ ⎢0⎥ = ⎢ ⎥ 0 ⎢⎣ 0 ⎥⎦

⎡ 0.9 −1.793 −0.9 1.793 ⎤ ⎢ −1.793 3.572 1.793 −3.572 ⎥ kN ―― Kbc = ⎢ −0.9 1.793 0.9 −1.793 ⎥ mm ⎢ ⎥ ⎣ 1.793 −3.572 −1.793 3.572 ⎦ Elemento CD:

ϕi ≔ 63.345° ϕj ≔ 63.345°

⎤ ⎡ 0.449 0.894 0 0 Tcd = ⎢ ⎣0 0 0.449 0.894 ⎥⎦ T

Kcd ≔ Tcd ⋅ K'cd ⋅ Tcd

GLcd

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 2 ⎢⎣ 0 ⎥⎦

⎡ 0.9 1.793 −0.9 −1.793 ⎤ ⎢ 1.793 3.572 −1.793 −3.572 ⎥ kN Kcd = ⎢ ―― −0.9 −1.793 0.9 1.793 ⎥ mm ⎢ ⎥ ⎣ −1.793 −3.572 1.793 3.572 ⎦ Elemento BD:

ϕi ≔ 0 °

ϕj ≔ 0 °

⎡1 0 0 0⎤ Tbd = ⎢ ⎣ 0 0 1 0 ⎥⎦ T

Kbd ≔ Tbd ⋅ K'bd ⋅ Tbd ⎡ 5 ⎢ 0 Kbd = ⎢ −5 ⎢⎣ 0

0 −5 0 ⎤ 0 0 0 ⎥ kN ―― 0 5 0 ⎥ mm ⎥ 0 0 0⎦


GLbd

⎡1⎤ ⎢0⎥ = ⎢ ⎥ 2 ⎢⎣ 0 ⎥⎦


Elemento DE:

ϕi ≔ −63.345°


ϕj ≔ −63.345°

⎤ ⎡ 0.45 −0.89 0 0 Tde = ⎢ ⎣0 0 0.45 −0.89 ⎥⎦ T

Kde ≔ Tde ⋅ K'de ⋅ Tde

GLde

⎡2⎤ ⎢0⎥ = ⎢ ⎥ 0 ⎢⎣ 0 ⎥⎦

⎡ 0.9 −1.793 −0.9 1.793 ⎤ ⎢ −1.793 3.572 1.793 −3.572 ⎥ kN ―― Kde = ⎢ −0.9 1.793 0.9 −1.793 ⎥ mm ⎢ ⎥ ⎣ 1.793 −3.572 −1.793 3.572 ⎦ 4. Matriz de rigidez de la estructura: ⎡ 6.8 −5 ⎤ kN ―― K=⎢ 6.8 ⎥⎦ mm ⎣ −5

⎡1⎤ GL = ⎢ ⎥ ⎣2⎦

5. Matriz de Fuerzas: ⎡ 50 ⎤ F ≔ ⎢ ⎥ kN ⎣ 0 ⎦ 6. Desplazamientos: F = K⋅u

u≔K

⎤ ⎡ 0 ⎥ ⎢ 0 uab ≔ ⎢ mm 16.006 ⎥ ⎢⎣ ⎥⎦ 0 ⎡ 16.006 ⎤ ⎥ ⎢ 0 mm ubd ≔ ⎢ 11.769 ⎥ ⎢⎣ ⎥⎦ 0

−1

⎡ 16.006 ⎤ mm ⋅F=⎢ ⎣ 11.769 ⎥⎦ ⎡ 16.006 ⎤ ⎥ ⎢ 0 ubc ≔ ⎢ ⎥ mm 0 ⎢⎣ ⎥⎦ 0 ⎡ 11.769 ⎤ ⎥ ⎢ 0 ude ≔ ⎢ ⎥ mm 0 ⎢⎣ ⎥⎦ 0

⎤ ⎡ 0 ⎥ ⎢ 0 ucd ≔ ⎢ mm 11.769 ⎥ ⎢⎣ ⎥⎦ 0



F'ab = 32.113 kN



⎡ 14.407 ⎤ ⎢ −28.701 ⎥ kN Fbc ≔ Kbc ⋅ ubc = ⎢ −14.407 ⎥ ⎢⎣ ⎥ 28.701 ⎦

F'bc = −32.113 kN

⎡ −10.593 ⎤ ⎢ −21.103 ⎥ kN Fcd ≔ Kcd ⋅ ucd = ⎢ 10.593 ⎥ ⎢⎣ ⎥ 21.103 ⎦

F'cd = 23.613 kN

⎡ 21.185 ⎤ ⎢ 0 ⎥ Fbd ≔ Kbd ⋅ ubd = ⎢ kN −21.185 ⎥ ⎢⎣ 0 ⎥⎦

F'bd = −21.185 kN

⎡ 10.593 ⎤ ⎢ −21.103 ⎥ kN Fde ≔ Kde ⋅ ude = ⎢ −10.593 ⎥ ⎢⎣ ⎥ 21.103 ⎦

F'de = −23.613 kN

8. Reacciones en los apoyos: Rax ≔ −14.407 kN

Rcx ≔ −14.407 kN − 10.593 kN = −25 kN

Ray ≔ −28.701 kN

Rcy ≔ 28.701 kN − 21.103 kN = 7.6 kN

Rex ≔ −10.593 kN

Rby ≔ 28.701 kN + −28.701 kN + 0 kN = 0 kN

Rey ≔ 21.103 kN

Rdy ≔ 21.103 kN + 0 kN + −21.103 kN = 0 kN




Problema 05: Ensamblar la matriz de rigidez K y determinar las fuerzas en los elementos de la estructura. Datos: E ≔ 200 GPa A0 ≔ 30000 mm



EA ⎡ 1 −1 ⎤ K' = ―― ⋅ L ⎢⎣ −1 1 ⎥⎦


2


Elemento AB:

Lab ≔ 5.00 m


Aab ≔ 30000 mm

2

⎡ 1200 −1200 ⎤ kN K'ab = ⎢ ―― ⎣ −1200 1200 ⎥⎦ mm Lac ≔ 4.00 m

Elemento AC:

Aac ≔ 30000 mm

2

⎡ 1500 −1500 ⎤ kN ―― K'ac = ⎢ ⎣ −1500 1500 ⎥⎦ mm Lcb ≔ 3.00 m

Elemento CB:

Acb ≔ 30000 mm

2

⎡ 2000 −2000 ⎤ kN ―― K'cb = ⎢ ⎣ −2000 2000 ⎥⎦ mm 3. Matriz de rigidez de cada elemento "Global": Elemento AB:

ϕi ≔ 37°

ϕj ≔ 37°

⎡ 0.799 ⎢ 0.602 ⎤ ⎡ 0.799 0.602 0 T 0 Tab = ⎢ Tab = ⎢ ⎥ ⎣0 ⎦ 0 0.799 0.602 0 ⎢⎣ 0

⎤ 0 ⎥ 0 0.799 ⎥ 0.602 ⎥⎦

T

Kab ≔ Tab ⋅ K'ab ⋅ Tab ⎡ 765.382 576.757 −765.382 −576.757 ⎤ ⎡0⎤ ⎢ 576.757 434.618 −576.757 −434.618 ⎥ kN ⎢0⎥ ―― Kab = ⎢ GLab = ⎢ ⎥ −765.382 −576.757 765.382 576.757 ⎥ mm 1 ⎢ ⎥ ⎣ −576.757 −434.618 576.757 434.618 ⎦ ⎢⎣ 2 ⎥⎦



Elemento AC:

ϕi ≔ 0 °


ϕj ≔ −45°

⎡1 0 0 ⎤ 0 Tac = ⎢ ⎥ ⎣ 0 0 0.707 −0.707 ⎦ T

Kac ≔ Tac ⋅ K'ac ⋅ Tac ⎡ 1500 ⎢ 0 Kac = ⎢ −1060.66 ⎢ ⎣ 1060.66 Elemento CB:

0 −1060.66 1060.66 ⎤ ⎥ kN 0 0 0 ⎥ ―― 0 750 −750 mm ⎥ 0 −750 750 ⎦

ϕi ≔ 45°

GLac

ϕj ≔ 90°

⎡ 0.707 0.707 0 0 ⎤ Tcb = ⎢ ⎣0 0 0 1 ⎦⎥ T

GLcb

⎡3⎤ ⎢0⎥ = ⎢ ⎥ 1 ⎢⎣ 2 ⎥⎦

Kcb ≔ Tcb ⋅ K'cb ⋅ Tcb ⎡ 1000 1000 ⎢ 1000 1000 Kcb = ⎢ 0 0 ⎢ ⎣ −1414.214 −1414.214

0 −1414.214 ⎤ 0 −1414.214 ⎥ kN ⎥ ―― 0 0 mm ⎥ 0 2000 ⎦

4. Matriz de rigidez de la estructura:

⎡ 765.382 ⎤ 576.757 0 kN ⎢ K = 576.757 2434.618 −1414.214 ⎥ ―― ⎢⎣ 0 ⎥⎦ mm −1414.214 1750 5. Matriz de Fuerzas: ⎡ 30 ⎤ F ≔ ⎢ 0 ⎥ kN ⎢⎣ 0 ⎥⎦


⎡1⎤ GL = ⎢ 2 ⎥ ⎢⎣ 3 ⎥⎦

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 3 ⎢⎣ 0 ⎥⎦




u≔K

−1

⎡ 0.05907068 ⎤ ⋅ F = ⎢ −0.02637442 ⎥ mm ⎢⎣ −0.02131376 ⎥⎦

⎡ −0.02131376 ⎤ ⎤ ⎤ ⎡ ⎡ 0 0 ⎢ ⎥ ⎥ ⎥ ⎢ ⎢ 0 0 0 uab ≔ ⎢ mm u ≔ mm u ≔ ac cb ⎢ 0.05907068 ⎥ mm ⎢ −0.02131376 ⎥ 0.05907068 ⎥ ⎢⎣ −0.02637442 ⎥⎦ ⎢⎣ −0.02637442 ⎥⎦ ⎢⎣ ⎥⎦ 0 7. Fuerzas en cada elemento: ⎡ −30 ⎤ ⎢ −22.607 ⎥ Fab ≔ Kab ⋅ uab = ⎢ ⎥ kN 30 ⎢⎣ ⎥ 22.607 ⎦

F'ab = 37.564 kN

⎡ 22.607 ⎤ ⎥ ⎢ 0 kN Fac ≔ Kac ⋅ uac = ⎢ −15.985 ⎥ ⎢⎣ 15.985 ⎥⎦

F'ac = −22.607 kN

⎡ 15.985 ⎤ ⎢ 15.985 ⎥ Fcb ≔ Kcb ⋅ ucb = ⎢ ⎥ kN 0 ⎢⎣ −22.607 ⎥⎦

F'cb = −22.607 kN

8. Reacciones en los apoyos: Rax ≔ −30 kN + 22.607 kN = −7.39 kN Ray ≔ −22.607 kN Rcx ≔ −15.985 kN + 15.985 kN = 0 kN Rcy ≔ 15.985 kN + 15.985 kN = 31.97 kN






EA ⎡ 1 −1 ⎤ K' = ―― ⋅ L ⎣⎢ −1 1 ⎦⎥


2


Elemento AB:

Lab ≔ 3.5 m

⎡ 80 −80 ⎤ kN ―― K'ab = ⎢ ⎣ −80 80 ⎥⎦ mm Elemento BC:

Lbc ≔ 4.00 m

⎡ 70 −70 ⎤ kN ―― K'bc = ⎢ ⎣ −70 70 ⎥⎦ mm Elemento DC:

Ldc ≔ 3.50 m

⎡ 80 −80 ⎤ kN ―― K'dc = ⎢ ⎣ −80 80 ⎥⎦ mm Elemento AD:

Lad ≔ 4.00 m

⎡ 70 −70 ⎤ kN ―― K'ad = ⎢ ⎣ −70 70 ⎥⎦ mm Elemento AC:

Lac ≔ 5.315 m

⎡ 52.6811 −52.6811 ⎤ kN ―― K'ac = ⎢ ⎣ −52.6811 52.6811 ⎥⎦ mm Elemento BD:

Lbd ≔ 5.315 m

⎡ 52.6811 −52.6811 ⎤ kN ―― K'bd = ⎢ ⎣ −52.6811 52.6811 ⎥⎦ mm



3. Matriz de rigidez de cada elemento "Global": ϕi ≔ 90°

Elemento AB:

ϕj ≔ 90°

⎡0 1 0 0⎤ Tab = ⎢ ⎣ 0 0 0 1 ⎥⎦

GLab

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 1 ⎢⎣ 2 ⎥⎦

GLbc

⎡1⎤ ⎢2⎥ = ⎢ ⎥ 3 ⎢⎣ 4 ⎥⎦

T

Kab ≔ Tab ⋅ K'ab ⋅ Tab ⎡0 0 ⎢ 0 80 Kab = ⎢ 0 0 ⎢⎣ 0 −80

0 0⎤ 0 −80 ⎥ kN ―― 0 0 ⎥ mm 0 80 ⎥⎦

Elemento BC:

ϕi ≔ 0 °

ϕj ≔ 0 °

⎡1 0 0 0⎤ Tbc = ⎢ ⎣ 0 0 1 0 ⎥⎦ T

Kbc ≔ Tbc ⋅ K'bc ⋅ Tbc ⎡ 70 ⎢ 0 Kbc = ⎢ −70 ⎢⎣ 0

0 −70 0 ⎤ 0 0 0 ⎥ kN ―― 0 70 0 ⎥ mm ⎥ 0 0 0⎦

Elemento AD:

ϕi ≔ 0 °

⎡1 0 0 ⎤ 0 Tad = ⎢ ⎣ 0 0 0.707 −0.707 ⎥⎦


ϕj ≔ −45°


T

Kad ≔ Tad ⋅ K'ad ⋅ Tad ⎡ 70 ⎢ 0 Kad = ⎢ −49.497 ⎢ ⎣ 49.497 Elemento DC:

0 −49.497 49.497 ⎤ ⎥ kN 0 0 0 ⎥ ―― 0 35 −35 mm ⎥ 0 −35 35 ⎦ ϕi ≔ 45°

ϕj ≔ 90°

⎡ 0.707 0.707 0 0 ⎤ Tdc = ⎢ ⎣0 0 0 1 ⎦⎥

GLdc

T

⎡5⎤ ⎢0⎥ = ⎢ ⎥ 3 ⎢⎣ 4 ⎥⎦

Kdc ≔ Tdc ⋅ K'dc ⋅ Tdc

⎡ 40 40 ⎢ 40 40 Kdc = ⎢ 0 0 ⎢ ⎣ −56.569 −56.569 Elemento AC:

0 −56.569 ⎤ 0 −56.569 ⎥ kN ⎥ ―― 0 0 mm ⎥ 0 80 ⎦

ϕi ≔ 41.186°

⎤ ⎡ 0.753 0.659 0 0 Tac = ⎢ ⎣0 0 0.753 0.659 ⎥⎦ T

ϕj ≔ 41.186°

GLac

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 3 ⎢⎣ 4 ⎥⎦

Kac ≔ Tac ⋅ K'ac ⋅ Tac ⎡ 29.837 26.107 −29.837 −26.107 ⎤ ⎢ 26.107 22.844 −26.107 −22.844 ⎥ kN Kac = ⎢ ―― −29.837 −26.107 29.837 26.107 ⎥ mm ⎢ ⎥ ⎣ −26.107 −22.844 26.107 22.844 ⎦


GLad

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 5 ⎢⎣ 0 ⎥⎦


Elemento BD:

ϕi ≔ −41.186°

ϕj ≔ −86.186°

⎤ ⎡ 0.753 −0.659 0 0 Tbd = ⎢ ⎥ 0 0.067 −0.998 ⎦ ⎣0 T

Kbd ≔ Tbd ⋅ K'bd ⋅ Tbd ⎡ 29.837 −26.107 −2.637 39.559 ⎤ ⎡1⎤ ⎢ −26.107 22.844 2.308 −34.614 ⎥ kN ⎢2⎥ ―― Kbd = ⎢ GLbd = ⎢ ⎥ −2.637 2.308 0.233 −3.496 ⎥ mm 5 ⎢ ⎥ ⎣ 39.559 −34.614 −3.496 52.448 ⎦ ⎢⎣ 0 ⎥⎦ 4. Matriz de rigidez de la estructura: ⎡ 99.837 −26.107 −70 ⎡1⎤ 0 −2.637 ⎤ ⎢ −26.107 102.844 ⎢2⎥ 0 0 2.308 ⎥ kN ⎢ ⎥ GL = ⎢ 3 ⎥ ―― 0 99.837 26.107 0 K = −70 ⎢ ⎥ mm ⎢4⎥ 0 0 26.107 102.844 −56.569 ⎢ ⎥ ⎢⎣ ⎥⎦ 5 2.308 0 −56.569 75.233 ⎦ ⎣ −2.637 5. Matriz de Fuerzas: ⎡ 0 ⎤ ⎢ 0 ⎥ F ≔ ⎢ 30 ⎥ kN ⎢ 0 ⎥ ⎢⎣ ⎥⎦ 0 6. Desplazamientos:

F = K⋅u

⎡ 0.60524846 ⎤ ⎢ 0.15895649 ⎥ −1 u ≔ K ⋅ F = ⎢ 0.81286455 ⎥ mm ⎢ −0.33655426 ⎥ ⎢ ⎥ ⎣ −0.23671798 ⎦



⎤ ⎡ 0 ⎥ ⎢ 0 uab ≔ ⎢ mm 0.60524846 ⎥ ⎢⎣ 0.15895649 ⎥⎦

⎡ −0.23671798 ⎤ ⎡ 0.60524846 ⎤ ⎢ ⎥ ⎢ 0.15895649 ⎥ 0 mm u ≔ ubc ≔ ⎢ dc ⎢ 0.81286455 ⎥ mm 0.81286455 ⎥ ⎢⎣ −0.33655426 ⎥⎦ ⎢⎣ −0.33655426 ⎥⎦

⎤ ⎤ ⎡ ⎡ ⎡ 0.60524846 ⎤ 0 0 ⎥ ⎥ ⎢ ⎢ ⎢ 0.15895649 ⎥ 0 0 mm uac ≔ ⎢ mm ubd ≔ ⎢ mm uad ≔ ⎢ ⎥ ⎥ −0.23671798 0.81286455 −0.23671798 ⎥ ⎢⎣ ⎥⎦ ⎢⎣ −0.33655426 ⎥⎦ ⎢⎣ ⎥⎦ 0 0 7. Fuerzas en cada elemento: ⎤ ⎡ 0 ⎢ −12.717 ⎥ Fab ≔ Kab ⋅ uab = ⎢ ⎥ kN 0 ⎢⎣ 12.717 ⎥⎦

F'ab = 12.717 kN

⎡ −14.533 ⎤ ⎢ 0 ⎥ kN Fbc ≔ Kbc ⋅ ubc = ⎢ 14.533 ⎥ ⎢⎣ 0 ⎥⎦

F'bc = 14.533 kN

⎡ 9.57 ⎤ ⎢ 9.57 ⎥ Fdc ≔ Kdc ⋅ udc = ⎢ ⎥ kN 0 ⎢⎣ −13.534 ⎥⎦

F'dc = −13.534 kN

⎡ 11.717 ⎤ ⎥ ⎢ 0 Fad ≔ Kad ⋅ uad = ⎢ kN −8.285 ⎥ ⎢⎣ 8.285 ⎥⎦

F'ad = −11.717 kN

⎡ −15.467 ⎤ ⎢ −13.534 ⎥ kN Fac ≔ Kac ⋅ uac = ⎢ 15.467 ⎥ ⎢⎣ ⎥ 13.534 ⎦

F'ac = 20.552 kN

⎡ 14.533 ⎤ ⎢ −12.717 ⎥ kN Fbd ≔ Kbd ⋅ ubd = ⎢ −1.285 ⎥ ⎢⎣ ⎥ 19.268 ⎦

F'bd = −19.311 kN



8. Reacciones en los apoyos: Rax ≔ 0 kN + 11.717 kN + −15.467 kN = −3.75 kN Ray ≔ −12.717 kN + 0 kN + −13.534 kN = −26.3 kN Rdx ≔ 9.57 kN + −8.285 kN + −1.285 kN = 0 kN Rdy ≔ 9.57 kN + −8.285 kN + 19.268 kN = 20.55 kN




Problema 07: Ensamblar la matriz de rigidez K, calcular las fuerzas en los elementos de la estructura. Datos: E ≔ 210 GPa Aab ≔ 600 mm Abc ≔ 600 mm

2

2

Aac ≔ 848.528 mm



EA ⎡ 1 −1 ⎤ K' = ―― ⋅ L ⎣⎢ −1 1 ⎦⎥


2


Elemento AB:


Lab ≔ 1.00 m


Lbc ≔ 1.00 m

⎡ 126 −126 ⎤ kN ―― K'bc = ⎢ ⎣ −126 126 ⎥⎦ mm Lac ≔ 1.414 m

Elemento AC:

⎡ 126.019 −126.019 ⎤ kN ―― K'ac = ⎢ ⎣ −126.019 126.019 ⎥⎦ mm 3. Matriz de rigidez de cada elemento "Global": ϕi ≔ 90°

Elemento AB:

ϕj ≔ 90°

⎡0 1 0 0⎤ Tab = ⎢ ⎣ 0 0 0 1 ⎦⎥

GLab

T

Kab ≔ Tab ⋅ K'ab ⋅ Tab ⎡0 0 ⎢ 0 126 Kab = ⎢ 0 0 ⎢⎣ 0 −126

Elemento BC:

0 0⎤ 0 −126 ⎥ kN ―― 0 0 ⎥ mm 0 126 ⎥⎦

ϕi ≔ 0 °

⎡1 0 0 ⎤ 0 Tbc = ⎢ ⎥ ⎣ 0 0 0.707 −0.707 ⎦


ϕj ≔ −45°

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 1 ⎢⎣ 0 ⎥⎦



T

Kbc ≔ Tbc ⋅ K'bc ⋅ Tbc ⎡ 126 ⎢ 0 Kbc = ⎢ −89.095 ⎢ ⎣ 89.095 Elemento AC:

0 −89.095 89.095 ⎤ ⎡1⎤ ⎥ kN 0 0 0 ⎢0⎥ ―― ⎥ GLbc = ⎢ ⎥ 0 63 −63 mm 2 ⎥ 0 −63 63 ⎦ ⎢⎣ 0 ⎥⎦ ϕi ≔ 45°

ϕj ≔ 0 °

⎡ 0.707 0.707 0 0 ⎤ Tac = ⎢ ⎣0 0 1 0 ⎦⎥ T

Kac ≔ Tac ⋅ K'ac ⋅ Tac

⎡ 63.01 63.01 −89.109 0 ⎤ ⎢ 63.01 63.01 −89.109 0 ⎥ kN ―― Kac = ⎢ −89.109 −89.109 126.019 0 ⎥ mm ⎢ ⎥ ⎣ 0 0 0 0⎦

GLac

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 2 ⎢⎣ 0 ⎥⎦

4. Matriz de rigidez de la estructura: ⎡ 126 −89.095 ⎤ kN ―― K=⎢ ⎣ −89.095 189.019 ⎥⎦ mm

⎡1⎤ GL = ⎢ ⎥ ⎣2⎦

5. Matriz de Fuerzas: ⎡ 1000 ⎤ kN F≔⎢ ⎣ 0 ⎥⎦ 6. Desplazamientos: F = K⋅u

u≔K

−1

⎡ 11.9041633 ⎤ ⋅F=⎢ mm ⎣ 5.61111202 ⎥⎦

⎤ ⎡ 0 ⎥ ⎢ 0 mm uab ≔ ⎢ 11.9041633 ⎥ ⎢⎣ ⎥⎦ 0

⎡ 11.9041633 ⎤ ⎥ ⎢ 0 ubc ≔ ⎢ mm 5.61111202 ⎥ ⎢⎣ ⎥⎦ 0


⎤ ⎡ 0 ⎥ ⎢ 0 uac ≔ ⎢ mm 5.61111202 ⎥ ⎢⎣ ⎥⎦ 0



7. Fuerzas en cada elemento: ⎡0⎤ ⎢0⎥ Fab ≔ Kab ⋅ uab = ⎢ ⎥ kN 0 ⎢⎣ 0 ⎥⎦

F'ab = 0 kN

⎡ 1000 ⎤ ⎢ ⎥ 0 Fbc ≔ Kbc ⋅ ubc = ⎢ kN −707.107 ⎥ ⎢⎣ 707.107 ⎥⎦

F'bc = −1000 kN

⎡ −500 ⎤ ⎢ −500 ⎥ kN Fac ≔ Kac ⋅ uac = ⎢ 707.107 ⎥ ⎢⎣ ⎥⎦ 0

F'ac = 707.107 kN

8. Reacciones en los apoyos: Rax ≔ 0 kN + −500 kN = −500 kN Ray ≔ 0 kN + −500 kN = −500 kN Rbx ≔ 0 kN + 1000 kN − 1000 kN = 0 kN Rby ≔ 0 kN + 0 kN = 0 kN Rcx ≔ −707.107 kN + 707.107 kN = 0 kN Rcy ≔ 707.107 kN + 0 kN = 707.11 kN







EA ⎡ 1 −1 ⎤ K' = ―― ⋅ L ⎢⎣ −1 1 ⎥⎦


2


Elemento AB:

Lab ≔ 5.00 m


Lbc ≔ 4.00 m

⎡ 5 −5 ⎤ kN ―― K'bc = ⎢ ⎣ −5 5 ⎥⎦ mm Elemento CD:

Lcd ≔ 3.00 m

⎡ 6.6667 −6.6667 ⎤ kN K'cd = ⎢ ―― ⎣ −6.6667 6.6667 ⎥⎦ mm Elemento AE:

Lae ≔ 4.00 m

⎡ 5 −5 ⎤ kN ―― K'ae = ⎢ ⎣ −5 5 ⎥⎦ mm Elemento ED:

Led ≔ 4.00 m

⎡ 5 −5 ⎤ kN ―― K'ed = ⎢ ⎣ −5 5 ⎥⎦ mm Elemento BE:

Lbe ≔ 3.00 m

⎡ 6.6667 −6.6667 ⎤ kN ―― K'be = ⎢ ⎣ −6.6667 6.6667 ⎥⎦ mm Elemento EC:

Lec ≔ 5.00 m

⎡ 4 −4 ⎤ kN ―― K'ec = ⎢ ⎣ −4 4 ⎥⎦ mm Elemento BD:

Lbd ≔ 5.00 m

⎡ 4 −4 ⎤ kN ―― K'bd = ⎢ ⎣ −4 4 ⎥⎦ mm





3. Matriz de rigidez de cada elemento "Global": Elemento AB:

ϕi ≔ 36.87°

ϕj ≔ 36.87°

⎡ 0.8 0.6 0 0 ⎤ Tab = ⎢ ⎣0 0 0.8 0.6 ⎥⎦ T


GLab

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 3 ⎢⎣ 4 ⎥⎦

⎡ 2.56 1.92 −2.56 −1.92 ⎤ ⎢ 1.92 1.44 −1.92 −1.44 ⎥ kN Kab = ⎢ ―― −2.56 −1.92 2.56 1.92 ⎥ mm ⎢ ⎥ ⎣ −1.92 −1.44 1.92 1.44 ⎦

Elemento BC:

ϕj ≔ 0°

ϕi ≔ 0 °

⎡1 0 0 0⎤ Tbc = ⎢ ⎣ 0 0 1 0 ⎦⎥ T


0 −5 0 ⎤ 0 0 0 ⎥ kN ―― 0 5 0 ⎥ mm ⎥ 0 0 0⎦

Elemento CD:

ϕi ≔ −90°

GLbc

ϕj ≔ −90°

⎡ 0 −1 0 0 ⎤ Tcd = ⎢ ⎣ 0 0 0 −1 ⎥⎦ T


⎡0 0 ⎢ 0 6.667 Kcd = ⎢ 0 0 ⎢⎣ 0 −6.667

⎡3⎤ ⎢4⎥ = ⎢ ⎥ 5 ⎢⎣ 6 ⎥⎦

GLcd

⎡5⎤ ⎢6⎥ = ⎢ ⎥ 0 ⎢⎣ 0 ⎥⎦

⎤ 0 0 0 −6.667 ⎥ kN ⎥ ―― 0 0 mm 0 6.667 ⎥⎦



Elemento AE:

ϕi ≔ 0 °

ϕj ≔ 0 °

⎡1 0 0 0⎤ Tae = ⎢ ⎣ 0 0 1 0 ⎦⎥ T

Kae ≔ Tae ⋅ K'ae ⋅ Tae ⎡ 5 ⎢ 0 Kae = ⎢ −5 ⎢⎣ 0

0 −5 0 ⎤ 0 0 0 ⎥ kN ―― 0 5 0 ⎥ mm ⎥ 0 0 0⎦

Elemento ED:

GLae

ϕi ≔ 0 °

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 1 ⎢⎣ 2 ⎥⎦

ϕj ≔ 0°

⎡1 0 0 0⎤ Ted = ⎢ ⎣ 0 0 1 0 ⎥⎦ T

Ked ≔ Ted ⋅ K'ed ⋅ Ted ⎡ 5 ⎢ 0 Ked = ⎢ −5 ⎢⎣ 0

⎡1⎤ 0 −5 0 ⎤ ⎢2⎥ GLed = ⎢ ⎥ 0 0 0 ⎥ kN 0 ―― 0 5 0 ⎥ mm ⎢⎣ 0 ⎥⎦ ⎥ 0 0 0⎦

Elemento BE:

ϕi ≔ −90°

ϕj ≔ −90°

⎡ 0 −1 0 0 ⎤ Tbe = ⎢ ⎣ 0 0 0 −1 ⎥⎦ T

Kbe ≔ Tbe ⋅ K'be ⋅ Tbe ⎡0 0 ⎢ 0 6.667 Kbe = ⎢ 0 0 ⎢⎣ 0 −6.667

GLbe

⎡3⎤ ⎢4⎥ = ⎢ ⎥ 1 ⎢⎣ 2 ⎥⎦

⎤ 0 0 0 −6.667 ⎥ kN ⎥ ―― 0 0 mm 0 6.667 ⎥⎦




Elemento EC:

ϕi ≔ 36.87°

ϕj ≔ 36.87°

⎡ 0.8 0.6 0 0 ⎤ Tec = ⎢ ⎣0 0 0.8 0.6 ⎥⎦ T

Kec ≔ Tec ⋅ K'ec ⋅ Tec


GLec

⎡1⎤ ⎢2⎥ = ⎢ ⎥ 5 ⎢⎣ 6 ⎥⎦

⎡ 2.56 1.92 −2.56 −1.92 ⎤ ⎢ 1.92 1.44 −1.92 −1.44 ⎥ kN ―― Kec = ⎢ −2.56 −1.92 2.56 1.92 ⎥ mm ⎢ ⎥ ⎣ −1.92 −1.44 1.92 1.44 ⎦ Elemento BD:

ϕi ≔ −36.87°

ϕj ≔ −36.87°

⎡ 0.8 −0.6 0 0 ⎤ Tbd = ⎢ 0 0.8 −0.6 ⎥⎦ ⎣0 T


GLec

⎡3⎤ ⎢4⎥ = ⎢ ⎥ 0 ⎢⎣ 0 ⎥⎦

⎡ 2.56 −1.92 −2.56 1.92 ⎤ ⎢ −1.92 1.44 1.92 −1.44 ⎥ kN ―― Kbd = ⎢ −2.56 1.92 2.56 −1.92 ⎥ mm ⎢ ⎥ ⎣ 1.92 −1.44 −1.92 1.44 ⎦ 4. Matriz de rigidez de la estructura: ⎡ 12.56 1.92 0 0 −2.56 −1.92 ⎤ ⎢ 1.92 8.107 0 −6.667 −1.92 −1.44 ⎥ ⎢ 0 ⎥ 0 10.12 0 −5 0 kN K=⎢ ⎥ ―― 0 −6.667 0 9.547 0 0 ⎢ ⎥ mm −2.56 −1.92 −5 0 7.56 1.92 ⎢ ⎥ 0 0 1.92 8.107 ⎦ ⎣ −1.92 −1.44 5. Matriz de Fuerzas: ⎡ 0 ⎤ ⎢ −100 ⎥ ⎢ 0 ⎥ F≔⎢ kN 0 ⎥ ⎢ ⎥ 50 ⎢ ⎥ ⎣ 0 ⎦


⎡1⎤ ⎢2⎥ ⎢3⎥ GL = ⎢ ⎥ 4 ⎢ ⎥ 5 ⎢ ⎥ ⎣6⎦


6. Desplazamientos:

F = K⋅u

⎡ 4.565 ⎤ ⎢ −32.7 ⎥ ⎢ 0.85 ⎥ −1 u≔K ⋅F=⎢ mm −22.835 ⎥ ⎢ ⎥ 1.721 ⎢ ⎥ ⎣ −5.135 ⎦


⎡1⎤ ⎢2⎥ ⎢3⎥ GL = ⎢ ⎥ 4 ⎢ ⎥ 5 ⎢ ⎥ ⎣6⎦

⎤ ⎡ 0 ⎥ ⎢ 0 uab = ⎢ mm 0.85 ⎥ ⎢⎣ −22.835 ⎥⎦

⎡ 0.85 ⎤ ⎢ −22.835 ⎥ ubc = ⎢ mm 1.721 ⎥ ⎢⎣ −5.135 ⎥⎦

⎡ 1.721 ⎤ ⎢ −5.135 ⎥ ucd = ⎢ ⎥ mm 0 ⎢⎣ 0 ⎥⎦

⎤ ⎡ 0 ⎥ ⎢ 0 mm uae = ⎢ 4.565 ⎥ ⎢⎣ −32.7 ⎥⎦

⎡ 4.565 ⎤ ⎢ −32.7 ⎥ ued = ⎢ ⎥ mm 0 ⎢⎣ 0 ⎥⎦

⎡ 0.85 ⎤ ⎢ −22.835 ⎥ ube = ⎢ mm 4.565 ⎥ ⎢⎣ −32.7 ⎥⎦

⎡ 4.565 ⎤ ⎢ −32.7 ⎥ mm uec = ⎢ 1.721 ⎥ ⎢⎣ −5.135 ⎥⎦

⎡ 0.85 ⎤ ⎢ −22.835 ⎥ ubd = ⎢ ⎥ mm 0 ⎢⎣ 0 ⎥⎦

7. Fuerzas en cada elemento: ⎡ 41.666 ⎤ ⎢ 31.25 ⎥ Fab ≔ Kab ⋅ uab = ⎢ kN −41.666 ⎥ ⎢⎣ −31.25 ⎥⎦

F'ab = −52.08 kN

⎡ −4.354 ⎤ ⎥ ⎢ 0 kN Fbc ≔ Kbc ⋅ ubc = ⎢ 4.354 ⎥ ⎢⎣ 0 ⎥⎦

F'bc = 4.35 kN

⎤ ⎡ 0 ⎢ −34.234 ⎥ Fcd ≔ Kcd ⋅ ucd = ⎢ ⎥ kN 0 ⎢⎣ 34.234 ⎥⎦

F'cd = −34.23 kN

⎡ −22.823 ⎤ ⎥ ⎢ 0 kN Fae ≔ Kae ⋅ uae = ⎢ 22.823 ⎥ ⎢⎣ 0 ⎥⎦

F'ae = 22.82 kN




⎡ 22.823 ⎤ ⎥ ⎢ 0 Fed ≔ Ked ⋅ ued = ⎢ kN −22.823 ⎥ ⎢⎣ 0 ⎥⎦

F'ed = −22.82 kN

⎤ ⎡ 0 ⎢ 65.766 ⎥ Fbe ≔ Kbe ⋅ ube = ⎢ ⎥ kN 0 ⎢⎣ −65.766 ⎥⎦

F'be = 65.77 kN

⎡ −45.646 ⎤ ⎢ −34.234 ⎥ Fec ≔ Kec ⋅ uec = ⎢ kN 45.646 ⎥ ⎢⎣ ⎥ 34.234 ⎦

F'ec = 57.06 kN

⎡ 46.021 ⎤ ⎢ −34.516 ⎥ kN Fbd ≔ Kbd ⋅ ubd = ⎢ −46.021 ⎥ ⎢⎣ ⎥ 34.516 ⎦

F'bd = −57.53 kN

8. Reacciones en los apoyos: Rax ≔ 41.666 kN − 22.823 kN = 18.84 kN Ray ≔ 31.25 kN + 0 kN = 31.25 kN Rdx ≔ 0 kN − 22.823 kN − 46.021 kN = −68.84 kN Rdy ≔ 34.234 kN + 0 kN + 34.516 kN = 68.75 kN







EA ⎡ 1 −1 ⎤ K' = ―― ⋅ L ⎢⎣ −1 1 ⎥⎦


2


Elemento BC:


Lbc ≔ 2.00 m

⎡ 150 −150 ⎤ kN ―― K'bc = ⎢ ⎣ −150 150 ⎥⎦ mm Elemento CD:

Lcd ≔ 2.00 m

⎡ 150 −150 ⎤ kN ―― K'cd = ⎢ ⎣ −150 150 ⎥⎦ mm Elemento AD:

Lad ≔ 2.00 m

⎡ 150 −150 ⎤ kN ―― K'ad = ⎢ ⎣ −150 150 ⎥⎦ mm Elemento DE:

Lde ≔ 2.00 m

⎡ 150 −150 ⎤ kN ―― K'de = ⎢ ⎣ −150 150 ⎥⎦ mm Elemento AC:

Lac ≔ 2 ⋅

2

‾‾ 2 m

⎡ 106.066 −106.066 ⎤ kN ―― K'ac = ⎢ ⎣ −106.066 106.066 ⎥⎦ mm

Elemento CE:

Lce ≔ 2 ⋅

2

‾‾ 2 m

⎡ 106.066 −106.066 ⎤ kN ―― K'ce = ⎢ ⎣ −106.066 106.066 ⎥⎦ mm 3. Matriz de rigidez de cada elemento "Global": Elemento BC:

ϕi ≔ 0 °

⎡1 0 0 0⎤ Tbc = ⎢ ⎣ 0 0 1 0 ⎦⎥


ϕj ≔ 0°


T


⎡0⎤ 0 −150 0 ⎤ ⎢0⎥ 0 0 0 ⎥ kN GLbc = ⎢ ⎥ 1 ―― 0 150 0 ⎥ mm ⎢⎣ 2 ⎥⎦ ⎥ 0 0 0⎦ ϕj ≔ −90°

ϕi ≔ −90°

Elemento CD:

⎡ 0 −1 0 0 ⎤ Tcd = ⎢ ⎣ 0 0 0 −1 ⎥⎦ T


⎡0 0 ⎢ 0 150 Kcd = ⎢ 0 0 ⎢⎣ 0 −150

GLcd

⎡1⎤ ⎢2⎥ = ⎢ ⎥ 3 ⎢⎣ 4 ⎥⎦

0 0⎤ 0 −150 ⎥ kN ―― 0 0 ⎥ mm 0 150 ⎥⎦

ϕi ≔ 0 °

Elemento AD:

ϕj ≔ 0 °

⎡1 0 0 0⎤ Tad = ⎢ ⎣ 0 0 1 0 ⎥⎦ T

Kad ≔ Tad ⋅ K'ad ⋅ Tad ⎡ 150 ⎢ 0 Kad = ⎢ −150 ⎢⎣ 0 Elemento DE:

⎡0⎤ 0 −150 0 ⎤ ⎢0⎥ GLad = ⎢ ⎥ 0 0 0 ⎥ kN 3 ―― 0 150 0 ⎥ mm ⎢⎣ 4 ⎥⎦ ⎥ 0 0 0⎦ ϕi ≔ 0 °


ϕj ≔ 0°




⎡1 0 0 0⎤ Tde = ⎢ ⎣ 0 0 1 0 ⎦⎥ T

Kde ≔ Tde ⋅ K'de ⋅ Tde ⎡ 150 ⎢ 0 Kde = ⎢ −150 ⎢⎣ 0 Elemento AC:

⎡3⎤ 0 −150 0 ⎤ ⎢4⎥ GLde = ⎢ ⎥ 0 0 0 ⎥ kN 5 ―― 0 150 0 ⎥ mm ⎢⎣ 6 ⎥⎦ ⎥ 0 0 0⎦ ϕi ≔ 45°

ϕj ≔ 45°

⎡ 0.707 0.707 0 ⎤ 0 Tac = ⎢ ⎣0 0 0.707 0.707 ⎥⎦

T


GLac

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 1 ⎢⎣ 2 ⎥⎦

⎡ 53.033 53.033 −53.033 −53.033 ⎤ ⎢ 53.033 53.033 −53.033 −53.033 ⎥ kN ―― Kac = ⎢ −53.033 −53.033 53.033 53.033 ⎥ mm ⎢ ⎥ ⎣ −53.033 −53.033 53.033 53.033 ⎦

Elemento CE:

ϕi ≔ −45°

ϕj ≔ −45°

⎡ 0.707 −0.707 0 ⎤ 0 Tce = ⎢ 0 0.707 −0.707 ⎥⎦ ⎣0 T

Kce ≔ Tce ⋅ K'ce ⋅ Tce

GLce

⎡1⎤ ⎢2⎥ = ⎢ ⎥ 5 ⎢⎣ 6 ⎥⎦

⎡ 53.033 −53.033 −53.033 53.033 ⎤ ⎢ −53.033 53.033 53.033 −53.033 ⎥ kN Kce = ⎢ ―― −53.033 53.033 53.033 −53.033 ⎥ mm ⎢ ⎥ ⎣ 53.033 −53.033 −53.033 53.033 ⎦




4. Matriz de rigidez de la estructura: ⎡ 256.066 0 0 0 −53.033 53.033 ⎤ ⎡1⎤ ⎢ 0 256.066 0 −150 53.033 −53.033 ⎥ ⎢2⎥ ⎢ 0 ⎥ 0 300 0 −150 0 kN ⎢3⎥ K=⎢ ⎥ ―― GL = ⎢ ⎥ 0 −150 0 150 0 0 mm ⎢ ⎥ 4 ⎥ ⎢ −53.033 53.033 −150 0 203.033 −53.033 ⎢ ⎥ 5 ⎢ ⎥ 0 0 −53.033 53.033 ⎦ ⎣ 53.033 −53.033 ⎣6⎦ 5. Matriz de Fuerzas: ⎡ 0 ⎤ ⎢ 0 ⎥ ⎢ 0 ⎥ kN F≔⎢ 0 ⎥ ⎥ ⎢ 0 ⎢ ⎥ ⎣ −30 ⎦ 6. Desplazamientos:

F = K⋅u

⎡1⎤ ⎢2⎥ ⎢3⎥ GL = ⎢ ⎥ 4 ⎢ ⎥ 5 ⎢ ⎥ ⎣6⎦

⎡ 0.4 ⎤ ⎢ −0.966 ⎥ ⎢ −0.2 ⎥ −1 u≔K ⋅F=⎢ ⎥ mm −0.966 ⎢ ⎥ ⎢ −0.4 ⎥ ⎣ −2.331 ⎦

⎤ ⎡ 0 ⎥ ⎢ 0 ubc = ⎢ mm 0.4 ⎥ ⎢⎣ −0.966 ⎥⎦

⎡ 0.4 ⎤ ⎢ −0.966 ⎥ mm ucd = ⎢ −0.2 ⎥ ⎢ ⎥ ⎣ −0.966 ⎦

⎡ 0 ⎤ ⎢ 0 ⎥ uad = ⎢ mm −0.2 ⎥ ⎢⎣ −0.966 ⎥⎦

⎡ −0.2 ⎤ ⎢ −0.966 ⎥ mm ude = ⎢ −0.4 ⎥ ⎢ ⎥ ⎣ −2.331 ⎦

⎤ ⎡ 0 ⎥ ⎢ 0 uac = ⎢ mm 0.4 ⎥ ⎢⎣ −0.966 ⎥⎦

⎡ 0.4 ⎤ ⎢ −0.966 ⎥ mm uce = ⎢ −0.4 ⎥ ⎢ ⎥ ⎣ −2.331 ⎦

7. Deformaciones en cada elemento: Ai = ⎡⎣ −cos ⎛⎝θi⎞⎠ −sin ⎛⎝θi⎞⎠ cos ⎛⎝θj⎞⎠ sin ⎛⎝θj⎞⎠ ⎤⎦ ui' = Ai ⋅ ui Elemento BC:

A = ⎡⎣ −1 0 1 0 ⎤⎦


u'bc ≔ A ⋅ ubc = 0.4 mm



Elemento CD:

A = ⎡⎣ 0 1 0 −1 ⎤⎦

u'cd ≔ A ⋅ ucd = 0 mm

Elemento AD:

A = ⎡⎣ −1 0 1 0 ⎤⎦

u'ad ≔ A ⋅ uad = −0.2 mm

Elemento DE:

A = ⎡⎣ −1 0 1 0 ⎤⎦

u'de ≔ A ⋅ ude = −0.2 mm

Elemento AC:

A = ⎡⎣ −0.71 −0.71 0.71 0.71 ⎤⎦

u'ac ≔ A ⋅ uac = −0.4 mm

Elemento CE:

A = ⎡⎣ −0.71 0.71 0.71 −0.71 ⎤⎦

u'ce ≔ A ⋅ uce = 0.4 mm

8. Fuerzas en cada elemento: ⎡ −60 ⎤ ⎢ 0⎥ Fbc ≔ Kbc ⋅ ubc = ⎢ kN 60 ⎥ ⎢⎣ 0 ⎥⎦

F'bc = 60 kN

⎡0⎤ ⎢0⎥ Fcd ≔ Kcd ⋅ ucd = ⎢ ⎥ kN 0 ⎢⎣ 0 ⎥⎦

F'cd = 0 kN

⎡ 30 ⎤ ⎢ 0⎥ kN Fad ≔ Kad ⋅ uad = ⎢ −30 ⎥ ⎢⎣ 0 ⎥⎦

F'ad = −30 kN

⎡ 30 ⎤ ⎢ 0⎥ kN Fde ≔ Kde ⋅ ude = ⎢ −30 ⎥ ⎢⎣ 0 ⎥⎦

F'de = −30 kN




⎡ 30 ⎤ ⎢ 30 ⎥ kN Fac ≔ Kac ⋅ uac = ⎢ −30 ⎥ ⎢⎣ −30 ⎥⎦

F'ac = −42.43 kN

⎡ −30 ⎤ ⎢ 30 ⎥ kN Fce ≔ Kce ⋅ uce = ⎢ 30 ⎥ ⎢⎣ −30 ⎥⎦

F'ce = 42.43 kN

9. Reacciones en los apoyos: Rax ≔ 30 kN + 30 kN = 60 kN Ray ≔ 30 kN Rbx ≔ −60 kN Rby ≔ 0 kN





2

Error de Fabricación: ΔLac ≔ −10 mm



EA ⎡ 1 −1 ⎤ K' = ―― ⋅ L ⎢⎣ −1 1 ⎥⎦



Elemento BC:


Lbc ≔ 4.00 m

⎡ 2 −2 ⎤ kN ―― K'bc = ⎢ ⎣ −2 2 ⎥⎦ mm Lcd ≔ 3.00 m

Elemento CD:

⎡ 2.6667 −2.6667 ⎤ kN ―― K'cd = ⎢ ⎣ −2.6667 2.6667 ⎥⎦ mm Lac ≔ 5.00 m

Elemento AC:

⎡ 1.6 −1.6 ⎤ kN ―― K'ac = ⎢ ⎣ −1.6 1.6 ⎥⎦ mm 3. Matriz de rigidez de cada elemento "Global": ⎡ cos ⎛⎝ϕi⎞⎠ sin ⎛⎝ϕi⎞⎠ ⎤ 0 0 Ti = ⎢ 0 0 cos ⎛⎝ϕj⎞⎠ sin ⎛⎝ϕj⎞⎠ ⎥⎦ ⎣ Elemento BC:

ϕj ≔ 0°

ϕi ≔ 0 °

⎡1 0 0 0⎤ Tbc = ⎢ ⎣ 0 0 1 0 ⎦⎥ T


0 −2 0 ⎤ 0 0 0 ⎥ kN ―― 0 2 0 ⎥ mm ⎥⎦ 0 0 0


GLbc

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 1 ⎢⎣ 2 ⎥⎦


Elemento CD:

ϕi ≔ −90°

ϕj ≔ −90°

⎡ 0 −1 0 0 ⎤ Tcd = ⎢ ⎣ 0 0 0 −1 ⎥⎦ GLcd

T


⎡0 0 ⎢ 0 2.667 Kcd = ⎢ 0 0 ⎢⎣ 0 −2.667

Elemento AC:


⎡1⎤ ⎢2⎥ = ⎢ ⎥ 0 ⎢⎣ 0 ⎥⎦

⎤ 0 0 0 −2.667 ⎥ kN ⎥ ―― 0 0 mm 0 2.667 ⎥⎦

ϕi ≔ 36.87°

ϕj ≔ 36.87°

⎡ 0.8 0.6 0 0 ⎤ Tac = ⎢ ⎣0 0 0.8 0.6 ⎥⎦

GLac

T


⎡0⎤ ⎢0⎥ = ⎢ ⎥ 1 ⎢⎣ 2 ⎥⎦

⎡ 1.024 0.768 −1.024 −0.768 ⎤ ⎢ 0.768 0.576 −0.768 −0.576 ⎥ kN Kac = ⎢ ―― −1.024 −0.768 1.024 0.768 ⎥ mm ⎢ ⎥ ⎣ −0.768 −0.576 0.768 0.576 ⎦

4. Matriz de rigidez de la estructura: ⎡ 3.024 0.768 ⎤ kN ―― K=⎢ ⎣ 0.768 3.243 ⎥⎦ mm

⎡1⎤ GL = ⎢ ⎥ ⎣2⎦

5. Fuerzas debido al error de fabricación ( fef ): Elemento AC:

ΔLac = −10 mm

⎡ E ⋅ A0 ⋅ ΔLac ⎤ ⎢ ――――⎥ Lac ⎥ f'ac ≔ ⎢ −E ⋅ A0 ⋅ ΔLac ⎥ ⎢ ――――― ⎢⎣ ⎥⎦ Lac


⎡ −16 ⎤ f'ac = ⎢ kN ⎣ 16 ⎥⎦


fac ≔ Tac ⋅ f'ac

⎡ −12.8 ⎤ ⎢ −9.6 ⎥ kN fac = ⎢ 12.8 ⎥ ⎢⎣ ⎥ 9.6 ⎦

fef = ∑ fi

⎡ 12.8 ⎤ kN fef ≔ ⎢ ⎣ 9.6 ⎥⎦

T


GLac

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 1 ⎢⎣ 2 ⎥⎦

6. Fuerzas concentradas: ⎡0⎤ fcon. ≔ ⎢ ⎥ kN ⎣0⎦ 7. Matriz de Fuerzas: ⎡ −12.8 ⎤ F=⎢ kN ⎣ −9.6 ⎥⎦

F ≔ fcon. − fef


u≔K

⎡ 0 ⎤ ⎢ 0 ⎥ mm ubc = ⎢ −3.704 ⎥ ⎢⎣ −2.083 ⎥⎦

−1

⎡ −3.704 ⎤ ⋅F=⎢ mm ⎣ −2.083 ⎥⎦ ⎡ −3.704 ⎤ ⎢ −2.083 ⎥ ucd = ⎢ ⎥ mm 0 ⎢⎣ ⎥⎦ 0

⎡1⎤ GL = ⎢ ⎥ ⎣2⎦ ⎡ 0 ⎤ ⎢ 0 ⎥ uac = ⎢ mm −3.704 ⎥ ⎢⎣ −2.083 ⎥⎦

9. Deformaciones en cada elemento: Ai = ⎡⎣ −cos ⎛⎝θi⎞⎠ −sin ⎛⎝θi⎞⎠ cos ⎛⎝θj⎞⎠ sin ⎛⎝θj⎞⎠ ⎤⎦ ui' = Ai ⋅ ui - ΔLi Elemento BC:

A = ⎡⎣ −1 0 1 0 ⎤⎦

u'bc ≔ A ⋅ ubc = −3.7 mm

Elemento CD:

A = ⎡⎣ 0 1 0 −1 ⎤⎦


u'cd ≔ A ⋅ ucd = −2.08 mm



Elemento AC:

A = ⎡⎣ −0.8 −0.6 0.8 0.6 ⎤⎦

u'ac ≔ A ⋅ uac − ΔLac = 5.79 mm

10. Fuerzas en cada elemento: ⎡ 7.41 ⎤ ⎢ 0 ⎥ Fbc ≔ Kbc ⋅ ubc = ⎢ kN −7.41 ⎥ ⎢⎣ 0 ⎥⎦

F'bc = −7.41 kN

⎤ ⎡ 0 ⎢ −5.56 ⎥ Fcd ≔ Kcd ⋅ ucd = ⎢ ⎥ kN 0 ⎢⎣ 5.56 ⎥⎦

F'cd = −5.56 kN

⎡ −7.41 ⎤ ⎢ −5.56 ⎥ kN Fac ≔ Kac ⋅ uac + fac = ⎢ 7.41 ⎥ ⎢⎣ ⎥ 5.56 ⎦

F'ac = 9.26 kN

11. Reacciones en los apoyos: Rbx ≔ 7.41 kN

Rax ≔ −7.41 kN

Rdx ≔ 0 kN

Rby ≔ 0 kN

Ray ≔ −5.56 kN

Rdy ≔ 5.56 kN





2

Variación de temperatura: ΔTac ≔ 150°F α ≔ 6.5 ⋅ 10



EA ⎡ 1 −1 ⎤ K' = ―― ⋅ L ⎢⎣ −1 1 ⎥⎦


−6

°F

−1


Elemento AB:


Lab ≔ 3.05 m

⎡ 32.7869 −32.7869 ⎤ kN ―― K'ab = ⎢ ⎣ −32.7869 32.7869 ⎥⎦ mm Elemento BC:

Lbc ≔ 3.05 m

⎡ 32.7869 −32.7869 ⎤ kN ―― K'bc = ⎢ ⎣ −32.7869 32.7869 ⎥⎦ mm Elemento DC:

Ldc ≔ 3.05 m

⎡ 32.7869 −32.7869 ⎤ kN K'dc = ⎢ ―― ⎣ −32.7869 32.7869 ⎥⎦ mm Elemento AD:

Lad ≔ 3.05 m

⎡ 32.7869 −32.7869 ⎤ kN ―― K'ad = ⎢ ⎣ −32.7869 32.7869 ⎥⎦ mm Elemento AC:

Lac ≔ 3.05 ⋅

2

‾‾ 2 m


Lbd ≔ 3.05 ⋅

2

‾‾ 2 m

⎡ 23.1838 −23.1838 ⎤ kN ―― K'bd = ⎢ ⎣ −23.1838 23.1838 ⎥⎦ mm 3. Matriz de rigidez de cada elemento "Global": ⎡ cos ⎛⎝ϕi⎞⎠ sin ⎛⎝ϕi⎞⎠ ⎤ 0 0 T = ⎢ 0 0 cos ⎛⎝ϕj⎞⎠ sin ⎛⎝ϕj⎞⎠ ⎥⎦ ⎣ Elemento AB: ⎡0 1 0 0⎤ Tab = ⎢ ⎣ 0 0 0 1 ⎦⎥

ϕj ≔ 90°

ϕi ≔ 90° T



⎡1⎤


⎡0 0 ⎢ 0 32.787 Kab = ⎢ 0 0 ⎢⎣ 0 −32.787 Elemento BC:

ϕi ≔ 0 °

⎡1 0 0 0⎤ Tbc = ⎢ ⎣ 0 0 1 0 ⎦⎥ ⎡ 32.787 ⎢ 0 Kbc = ⎢ −32.787 ⎢⎣ 0 Elemento DC:

⎡ 32.787 ⎢ 0 Kad = ⎢ −32.787 ⎢⎣ 0 Elemento AC:

GLbc

⎡3⎤ ⎢4⎥ = ⎢ ⎥ 0 ⎢⎣ 0 ⎥⎦

ϕj ≔ 0° T

0 −32.787 0 ⎤ 0 0 0 ⎥ kN ―― 0 32.787 0 ⎥ mm ⎥ 0 0 0⎦ ϕi ≔ 90°

ϕj ≔ 90° T


⎡0 0 ⎢ 0 32.787 Kdc = ⎢ 0 0 ⎢⎣ 0 −32.787

⎡1 0 0 0⎤ Tad = ⎢ ⎣ 0 0 1 0 ⎦⎥

GLab

⎡1⎤ ⎢2⎥ = ⎢ ⎥ 3 ⎢⎣ 4 ⎥⎦


⎡0 1 0 0⎤ Tdc = ⎢ ⎣ 0 0 0 1 ⎦⎥

Elemento AD:

⎤ 0 0 0 −32.787 ⎥ kN ⎥ ―― 0 0 mm 0 32.787 ⎥⎦


⎤ 0 0 0 −32.787 ⎥ kN ⎥ ―― 0 0 mm 0 32.787 ⎥⎦

ϕi ≔ 0 °

GLdc

⎡0⎤ ⎢5⎥ = ⎢ ⎥ 0 ⎢⎣ 0 ⎥⎦

GLad

⎡1⎤ ⎢2⎥ = ⎢ ⎥ 0 ⎢⎣ 5 ⎥⎦

ϕj ≔ 0 ° T

Kad ≔ Tad ⋅ K'ad ⋅ Tad 0 −32.787 0 ⎤ 0 0 0 ⎥ kN ―― 0 32.787 0 ⎥ mm ⎥ 0 0 0⎦ ϕi ≔ 45°

ϕj ≔ 45°

⎤ ⎡ 0.707 0.707 0 0 Tac = ⎢ ⎣0 0 0.707 0.707 ⎥⎦

T


⎡1⎤ ⎡ 11.592 11.592 −11.592 −11.592 ⎤ ⎢2⎥ ⎢ 11.592 11.592 −11.592 −11.592 ⎥ kN GLac = ⎢ ⎥ 0 Kac = ⎢ ―― −11.592 −11.592 11.592 11.592 ⎥ mm ⎢⎣ 0 ⎥⎦ ⎢ ⎥ ⎣ −11.592 −11.592 11.592 11.592 ⎦



Elemento BD:

ϕi ≔ −45°


ϕj ≔ −45°

⎤ ⎡ 0.707 −0.707 0 0 Tbd = ⎢ ⎥ ⎣0 0 0.707 −0.707 ⎦

T


⎡3⎤ ⎡ 11.592 −11.592 −11.592 11.592 ⎤ ⎢4⎥ ⎢ −11.592 11.592 11.592 −11.592 ⎥ kN GLbd = ⎢ ⎥ 0 ―― Kbd = ⎢ −11.592 11.592 11.592 −11.592 ⎥ mm ⎢⎣ 5 ⎥⎦ ⎢ ⎥ ⎣ 11.592 −11.592 −11.592 11.592 ⎦ 4. Matriz de rigidez de la estructura:

⎡ 253.409 ⎤ 66.192 0 0 0 ⎢ 66.192 253.409 ⎥ 0 −187.218 0 kip ⎢ K= 0 0 253.409 −66.192 66.192 ⎥ ―― ⎢ ⎥ 0 −187.218 −66.192 253.409 −66.192 in ⎢ ⎥ 0 66.192 −66.192 253.409 ⎦ ⎣ 0

⎡1⎤ ⎢2⎥ GL = ⎢ 3 ⎥ ⎢4⎥ ⎢⎣ ⎥⎦ 5

5. Fuerzas debido a la variacion de temperatura( fT ): Elemento AC:

ΔTac = 150

⎡ E ⋅ A0 ⋅ α ⋅ ΔTac ⎤ f'ac ≔ ⎢ ⎥ ⎣ −E ⋅ A0 ⋅ α ⋅ ΔTac ⎦

T

fac ≔ Tac ⋅ f'ac

°F

⎡ 97.5 ⎤ kN f'ac = ⎢ ⎣ −97.5 ⎥⎦ ⎡ 68.943 ⎤ ⎢ 68.943 ⎥ kN fac = ⎢ −68.943 ⎥ ⎢⎣ −68.943 ⎥⎦

GLac

⎡1⎤ ⎢2⎥ = ⎢ ⎥ 0 ⎢⎣ 0 ⎥⎦

fT = ∑ fi ⎡ 68.943 ⎤ ⎢ 68.943 ⎥ ⎥ kN fT ≔ ⎢ 0 ⎥ ⎢ 0 ⎢⎣ ⎥⎦ 0 6. Fuerzas concentradas: ⎡0⎤ ⎢0⎥ fcon. ≔ ⎢ 0 ⎥ kN ⎢0⎥ ⎢⎣ ⎥⎦ 0


⎡1⎤ ⎢2⎥ GL = ⎢ 3 ⎥ ⎢4⎥ ⎢⎣ ⎥⎦ 5



7. Matriz de Fuerzas: ⎡ −68.943 ⎤ ⎢ −68.943 ⎥ ⎥ kN F=⎢ 0 ⎢ ⎥ 0 ⎢ ⎥ ⎣ 0 ⎦

F ≔ fcon. − fT

⎡1⎤ ⎢2⎥ GL = ⎢ 3 ⎥ ⎢4⎥ ⎢⎣ ⎥⎦ 5

8. Desplazamientos:

F = K⋅u

⎡ −0.615885 ⎤ ⎢ −3.589638 ⎥ −1 u ≔ K ⋅ F = ⎢ −0.615885 ⎥ mm ⎢ ⎥ −2.973754 ⎢ ⎥ ⎣ −0.615885 ⎦

⎡ −0.615885 ⎤ ⎢ −3.589638 ⎥ mm uab = ⎢ −0.615885 ⎥ ⎢ ⎥ ⎣ −2.973754 ⎦

⎡ −0.615885 ⎤ ⎢ −2.973754 ⎥ ubc = ⎢ ⎥ mm 0 ⎢⎣ ⎥⎦ 0

⎡ 0 ⎤ ⎢ −0.615885 ⎥ udc = ⎢ ⎥ mm 0 ⎢⎣ 0 ⎥⎦

⎡ −0.615885 ⎤ ⎢ −3.589638 ⎥ uad = ⎢ ⎥ mm 0 ⎢ ⎥ ⎣ −0.615885 ⎦

⎡ −0.615885 ⎤ ⎢ −3.589638 ⎥ uac = ⎢ ⎥ mm 0 ⎢⎣ ⎥⎦ 0

⎡ −0.615885 ⎤ ⎢ −2.973754 ⎥ ubd = ⎢ ⎥ mm 0 ⎢ ⎥ ⎣ −0.615885 ⎦


A = ⎡⎣ 0 −1 0 1 ⎤⎦

u'bc ≔ A ⋅ ubc = 2.97 mm

Elemento BC:

A = ⎡⎣ −1 0 1 0 ⎤⎦

u'bc ≔ A ⋅ ubc = 0.62 mm

Elemento DC:

A = ⎡⎣ 0 −1 0 1 ⎤⎦


u'dc ≔ A ⋅ udc = 0.62 mm



Elemento AD:

A = ⎡⎣ 0 −1 0 1 ⎤⎦

u'ad ≔ A ⋅ uad = 2.97 mm

Elemento AC:

A = ⎡⎣ −0.71 −0.71 0.71 0.71 ⎤⎦

u'aC ≔ A ⋅ uac − α ⋅ ΔTac ⋅ Lac = −1.23 mm

Elemento BD:

A = ⎡⎣ −0.71 0.71 0.71 −0.71 ⎤⎦

u'bd ≔ A ⋅ ubd = −1.23 mm

9. Fuerzas en cada elemento: ⎤ ⎡ 0 ⎢ −20.193 ⎥ Fab ≔ Kab ⋅ uab = ⎢ ⎥ kN 0 ⎢⎣ 20.193 ⎥⎦

F'ab = 20.19 kN

⎡ −20.193 ⎤ ⎥ ⎢ 0 kN Fbc ≔ Kbc ⋅ ubc = ⎢ 20.193 ⎥ ⎢⎣ 0 ⎥⎦

F'bc = 20.19 kN

⎤ ⎡ 0 ⎢ −20.193 ⎥ Fdc ≔ Kdc ⋅ udc = ⎢ ⎥ kN 0 ⎢⎣ 20.193 ⎥⎦

F'dc = 20.19 kN

⎡ −20.193 ⎤ ⎥ ⎢ 0 kN Fad ≔ Kad ⋅ uad = ⎢ 20.193 ⎥ ⎢⎣ 0 ⎥⎦

F'ad = 20.19 kN

⎡ 20.193 ⎤ ⎢ 20.193 ⎥ kN Fac ≔ Kac ⋅ uac + fac = ⎢ −20.193 ⎥ ⎢⎣ −20.193 ⎥⎦

F'ac = −28.56 kN

⎡ 20.193 ⎤ ⎢ −20.193 ⎥ kN Fbd ≔ Kbd ⋅ ubd = ⎢ −20.193 ⎥ ⎢⎣ ⎥ 20.193 ⎦

F'bd = −28.56 kN



10. Reacciones en los apoyos: Rcx ≔ 20.193 kN + 0 kN − 20.193 kN = 0 kN Rcy ≔ 0 kN + 20.193 kN − 20.193 kN = 0 kN Rdx ≔ 0 kN + 20.193 kN − 20.193 kN = 0 kN Rdy ≔ −20.193 kN + 0 kN + 20.193 kN = 0 kN





Problema 12: Ensamblar la matriz de rigidez K y determinar las fuerzas en los elementos de la estructura. Datos: E ≔ 200 GPa A0 ≔ 300 mm



EA ⎡ 1 −1 ⎤ K' = ―― ⋅ L ⎣⎢ −1 1 ⎦⎥


2


Elemento AB:


Lab ≔ 3.50 m

⎡ 17.1429 −17.1429 ⎤ kN ―― K'ab = ⎢ ⎣ −17.1429 17.1429 ⎥⎦ mm Lac ≔ 3.85 m

Elemento AC:

⎡ 15.5844 −15.5844 ⎤ kN ―― K'ac = ⎢ ⎣ −15.5844 15.5844 ⎥⎦ mm Lcb ≔ 3.20 m

Elemento CB:

⎡ 18.75 −18.75 ⎤ kN ―― K'cb = ⎢ ⎣ −18.75 18.75 ⎥⎦ mm 3. Matriz de rigidez de cada elemento "Global": Elemento AB:

ϕi ≔ 38.66° ϕj ≔ 15°

⎡ 0.781 ⎢ 0.625 ⎡ 0.781 0.625 0 ⎤ T 0 Tab = ⎢ Tab = ⎢ ⎥ ⎣0 0 0.966 0.259 ⎦ 0 ⎢⎣ 0

⎤ 0 ⎥ 0 0.966 ⎥ 0.259 ⎥⎦

T

Kab ≔ Tab ⋅ K'ab ⋅ Tab ⎡ 10.453 8.362 −12.93 −3.465 ⎤ ⎢ 8.362 6.69 −10.344 −2.772 ⎥ kN ―― Kab = ⎢ −12.93 −10.344 15.995 4.286 ⎥ mm ⎢ ⎥ 4.286 1.148 ⎦ ⎣ −3.465 −2.772


GLab

⎡0⎤ ⎢3⎥ = ⎢ ⎥ 1 ⎢⎣ 2 ⎥⎦


Elemento AC:

ϕi ≔ 90°


ϕj ≔ 66.34°

⎡0 1 0 ⎤ 0 Tac = ⎢ ⎥ ⎣ 0 0 0.401 0.916 ⎦ T

Kac ≔ Tac ⋅ K'ac ⋅ Tac GLac

⎡0⎤ ⎢3⎥ = ⎢ ⎥ 0 ⎢⎣ 0 ⎥⎦

⎡0 ⎤ 0 0 0 ⎢ 0 15.584 −6.254 −14.274 ⎥ kN Kac = ⎢ ―― 0 −6.254 2.51 5.728 ⎥ mm ⎢ ⎥ ⎣ 0 −14.274 5.728 13.075 ⎦ Elemento CB:

ϕi ≔ −55°

ϕj ≔ −55°

⎤ ⎡ 0.574 −0.819 0 0 Tcb = ⎢ ⎥ 0 0.574 −0.819 ⎦ ⎣0 T

Kcb ≔ Tcb ⋅ K'cb ⋅ Tcb

GLcb

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 1 ⎢⎣ 2 ⎥⎦

⎡ 6.169 −8.81 −6.169 8.81 ⎤ ⎢ −8.81 12.581 8.81 −12.581 ⎥ kN ―― Kcb = ⎢ −6.169 8.81 6.169 −8.81 ⎥ mm ⎢ ⎥ 12.581 ⎦ ⎣ 8.81 −12.581 −8.81 4. Matriz de rigidez de la estructura:

⎡ 22.163 −4.524 −10.344 ⎤ kN K = ⎢ −4.524 13.73 −2.772 ⎥ ―― ⎢⎣ −10.344 −2.772 22.274 ⎥⎦ mm 5. Matriz de Fuerzas: ⎡ 100 ⎤ F ≔ ⎢ −100 ⎥ kN ⎢⎣ 0 ⎥⎦


⎡1⎤ GL = ⎢ 2 ⎥ ⎢⎣ 3 ⎥⎦




u≔K

−1

⎡ 3.81393069 ⎤ ⋅ F = ⎢ −5.81528101 ⎥ mm ⎢⎣ 1.04755967 ⎥⎦

⎤ ⎡ ⎤ ⎡ 0 0 ⎢ 1.04755967 ⎥ ⎢ 1.04755967 ⎥ uab ≔ ⎢ mm uac ≔ ⎢ ⎥ mm 3.81393069 ⎥ 0 ⎢⎣ −5.81528101 ⎥⎦ ⎢⎣ ⎥⎦ 0

⎤ ⎡ 0 ⎥ ⎢ 0 mm ucb ≔ ⎢ 3.81393069 ⎥ ⎢⎣ −5.81528101 ⎥⎦

7. Deformaciones en cada elemento: Ai = ⎡⎣ −cos ⎛⎝θi⎞⎠ −sin ⎛⎝θi⎞⎠ cos ⎛⎝θj⎞⎠ sin ⎛⎝θj⎞⎠ ⎤⎦ ui' = Ai ⋅ ui Elemento AB:

A = ⎡⎣ −0.781 −0.625 0.966 0.259 ⎤⎦

u'ab ≔ A ⋅ uab = 1.524 mm

Elemento AC: −17 −1 0.401 0.916 ⎤⎦ A = ⎡⎣ −6.123 ⋅ 10

u'ac ≔ A ⋅ uac = −1.048 mm

Elemento CB:

A = ⎡⎣ −0.574 0.819 0.574 −0.819 ⎤⎦

u'ab ≔ A ⋅ uab = 7.809 mm


F'ab = 26.134 kN

⎡ 0 ⎤ ⎢ 16.326 ⎥ kN Fac ≔ Kac ⋅ uac = ⎢ −6.552 ⎥ ⎢⎣ −14.953 ⎥⎦

F'ac = −16.326 kN



⎡ −74.757 ⎤ ⎢ 106.764 ⎥ kN Fcb ≔ Kcb ⋅ ucb = ⎢ 74.757 ⎥ ⎢⎣ −106.764 ⎥⎦

F'cb = 130.335 kN

9. Reacciones en los apoyos: Rax ≔ −20.407 kN + 0 kN = −20.41 kN Ray ≔ −16.326 kN + 16.326 kN = 0 kN Rcx ≔ −6.552 kN + −74.757 kN = −81.3 kN Rcy ≔ −14.953 kN + 106.764 kN = 91.81 kN





Problema 13: Ensamblar la matriz de rigidez K y determinar las fuerzas en los elementos de la estructura. Datos: E ≔ 200 GPa Aab ≔ 15000 mm Adc ≔ 15000 mm Adb ≔ 15000 mm

2

2

Abc ≔ 10000 mm Aad ≔ 10000 mm

2

2

2

Variación de temperatura: ΔTad ≔ 20 °C

α ≔ 1.2 ⋅ 10

Error de Fabricación: Solución:

ΔLdb ≔ −2 mm

1. Grados de Libertad:


EA ⎡ 1 −1 ⎤ K' = ―― ⋅ L ⎢⎣ −1 1 ⎥⎦


−5

°C

−1


Elemento AB:


Lab ≔ 10.00 m

⎡ 300 −300 ⎤ kN ―― K'ab = ⎢ ⎣ −300 300 ⎥⎦ mm Lbc ≔ 5.00 m

Elemento BC:

⎡ 400 −400 ⎤ kN ―― K'bc = ⎢ ⎣ −400 400 ⎥⎦ mm Ldc ≔ 10.00 m

Elemento DC:


Lad ≔ 5.00 m

⎡ 400 −400 ⎤ kN ―― K'ad = ⎢ ⎣ −400 400 ⎥⎦ mm Elemento DB:

Ldb ≔ 5 ⋅

2

‾‾ 3 m

⎡ 346.41 −346.41 ⎤ kN ―― K'db = ⎢ ⎣ −346.41 346.41 ⎥⎦ mm 3. Matriz de rigidez de cada elemento "Global": Elemento AB:

ϕi ≔ 60°

ϕj ≔ 60°

⎤ ⎡ 0.5 0.866 0 0 Tab = ⎢ ⎥ ⎣0 0 0.5 0.866 ⎦

⎡ 0.5 ⎢ 0.866 T Tab = ⎢ 0 ⎢⎣ 0

⎤ 0 ⎥ 0 0.5 ⎥ 0.866 ⎥⎦

T

Kab ≔ Tab ⋅ K'ab ⋅ Tab ⎡ 75 129.904 −75 −129.904 ⎤ ⎡0⎤ ⎢ 129.904 225 ⎥ kN −129.904 −225 ⎢0⎥ ―― Kab = ⎢ GLab = ⎢ ⎥ −75 −129.904 75 129.904 ⎥ mm 1 ⎢ ⎥ 129.904 225 ⎣ −129.904 −225 ⎦ ⎢⎣ 2 ⎥⎦



ϕi ≔ 0 °

Elemento BC:

ϕj ≔ 0 °

⎡1 0 0 0⎤ Tbc = ⎢ ⎣ 0 0 1 0 ⎥⎦ ⎡ 400 ⎢ 0 Kbc = ⎢ −400 ⎢⎣ 0


T


0 −400 0 ⎤ 0 0 0 ⎥ kN ―― 0 400 0 ⎥ mm ⎥ 0 0 0⎦ ϕi ≔ 60°

Elemento DC:

GLbc

⎡1⎤ ⎢2⎥ = ⎢ ⎥ 3 ⎢⎣ 4 ⎥⎦

ϕj ≔ 60°

⎤ ⎡ 0.5 0.866 0 0 Tdc = ⎢ ⎥ ⎣0 0 0.5 0.866 ⎦

T


⎡5⎤ ⎡ 75 129.904 −75 −129.904 ⎤ ⎢0⎥ ⎢ 129.904 225 ⎥ kN GLdc = ⎢ ⎥ −129.904 −225 3 ―― Kdc = ⎢ −75 −129.904 75 129.904 ⎥ mm ⎢⎣ 4 ⎥⎦ ⎢ ⎥ 129.904 225 ⎣ −129.904 −225 ⎦ ϕi ≔ 0 °

Elemento AD:

ϕj ≔ 0 °

⎡1 0 0 0⎤ Tad = ⎢ ⎣ 0 0 1 0 ⎥⎦ ⎡ 400 ⎢ 0 Kad = ⎢ −400 ⎢⎣ 0

T

Kad ≔ Tad ⋅ K'ad ⋅ Tad

0 −400 0 ⎤ 0 0 0 ⎥ kN ―― 0 400 0 ⎥ mm ⎥ 0 0 0⎦

Elemento DB:

ϕi ≔ 90°

⎡0 1 0 0⎤ Tdb = ⎢ ⎣ 0 0 0 1 ⎥⎦ ⎡0 0 ⎢ 0 346.41 Kdb = ⎢ 0 0 ⎢⎣ 0 −346.41

GLad

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 5 ⎢⎣ 0 ⎥⎦

GLdb

⎡5⎤ ⎢0⎥ = ⎢ ⎥ 1 ⎢⎣ 2 ⎥⎦

ϕj ≔ 90° T

Kdb ≔ Tdb ⋅ K'db ⋅ Tdb ⎤ 0 0 0 −346.41 ⎥ kN ⎥ ―― 0 0 mm 0 346.41 ⎥⎦





⎡ 475 ⎤ ⎡1⎤ 129.904 −400 0 0 ⎢ 129.904 571.41 ⎥ ⎢2⎥ 0 0 0 kN ⎢ ⎥ ―― GL = ⎢ 3 ⎥ 0 475 129.904 −75 K = −400 ⎢ ⎥ mm ⎢4⎥ 0 0 129.904 225 −129.904 ⎢ ⎥ ⎢⎣ ⎥⎦ 5 0 0 −75 −129.904 475 ⎣ ⎦ 5. Fuerzas debido al error de fabricación ( fef ): Elemento DB:

ΔLdb = −2 mm

⎡ E ⋅ Adb ⋅ ΔLdb ⎤ ⎢ ―――― ⎥ Ldb ⎥ f'db ≔ ⎢ ⎢ −E ⋅ Adb ⋅ ΔLdb ⎥ ――――― ⎢⎣ ⎥⎦ Ldb

⎡ −692.82 ⎤ f'db = ⎢ kN ⎣ 692.82 ⎥⎦

fdb ≔ Tdb ⋅ f'db

⎡ ⎤ 0 ⎢ −692.82 ⎥ fdb = ⎢ ⎥ kN 0 ⎢⎣ 692.82 ⎥⎦

fef = ∑ fi

⎤ ⎡ 0 ⎢ 692.82 ⎥ ⎥ kN fef ≔ ⎢ 0 ⎥ ⎢ 0 ⎢⎣ ⎥⎦ 0

T

GLdb

⎡5⎤ ⎢0⎥ = ⎢ ⎥ 1 ⎢⎣ 2 ⎥⎦

⎡1⎤ ⎢2⎥ GL = ⎢ 3 ⎥ ⎢4⎥ ⎢⎣ ⎥⎦ 5

6. Fuerzas debido a la variacion de temperatura( fT ) Elemento AD:

ΔTad = 20 °C

⎡ E ⋅ Aad ⋅ α ⋅ ΔTad ⎤ f'ad ≔ ⎢ ⎥ ⎣ −E ⋅ Aad ⋅ α ⋅ ΔTad ⎦

⎡ 480 ⎤ kN f'ad = ⎢ ⎣ −480 ⎥⎦

fad ≔ Tad ⋅ f'ad

⎡ 480 ⎤ ⎢ 0⎥ kN fad = ⎢ −480 ⎥ ⎢⎣ 0 ⎥⎦

fT = ∑ fi

⎡ 0 ⎤ ⎢ 0 ⎥ fT ≔ ⎢ 0 ⎥ kN ⎢ 0 ⎥ ⎢ ⎥ ⎣ −480 ⎦

T


GLad

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 5 ⎢⎣ 0 ⎥⎦

⎡1⎤ ⎢2⎥ GL = ⎢ 3 ⎥ ⎢4⎥ ⎢⎣ ⎥⎦ 5



7. Fuerzas concentradas: ⎡1⎤ ⎢2⎥ GL = ⎢ 3 ⎥ ⎢4⎥ ⎢⎣ ⎥⎦ 5

⎤ ⎡ 0 ⎥ ⎢ 0 fcon. ≔ ⎢ 282.85 ⎥ kN ⎢ −282.85 ⎥ ⎢ ⎥ ⎣ ⎦ 0 8. Matriz de Fuerzas:

F ≔ fcon. − fef − fT

⎡ ⎤ 0 ⎢ −692.82 ⎥ F = ⎢ 282.85 ⎥ kN ⎢ −282.85 ⎥ ⎢ ⎥ ⎣ 480 ⎦

⎡1⎤ ⎢2⎥ GL = ⎢ 3 ⎥ ⎢4⎥ ⎢⎣ ⎥⎦ 5

9. Desplazamientos: F = K⋅u ⎡ 13.276617 ⎤ ⎢ −4.230767 ⎥ −1 ⎥ mm u ≔ K ⋅ F = ⎢ 14.392 ⎢ −9.109224 ⎥ ⎢ ⎥ ⎣ 0.791741 ⎦

⎡1⎤ ⎢2⎥ GL = ⎢ 3 ⎥ ⎢4⎥ ⎢⎣ ⎥⎦ 5

⎤ ⎡ 0 ⎥ ⎢ 0 mm uab = ⎢ 13.27662 ⎥ ⎢⎣ −4.23077 ⎥⎦

⎡ 13.27662 ⎤ ⎢ −4.23077 ⎥ mm ubc = ⎢ 14.392 ⎥ ⎢⎣ −9.10922 ⎥⎦

⎤ ⎡0 ⎥ ⎢0 mm uad = ⎢ 0.79174 ⎥ ⎢⎣ 0 ⎥⎦

⎡ 0.79174 ⎤ ⎥ ⎢ 0 mm udb = ⎢ 13.27662 ⎥ ⎢⎣ −4.23077 ⎥⎦

⎡ 0.79174 ⎤ ⎥ ⎢ 0 udc = ⎢ mm 14.392 ⎥ ⎢⎣ −9.10922 ⎥⎦


A = ⎡⎣ −0.5 −0.866 0.5 0.866 ⎤⎦


u'ab ≔ A ⋅ uab = 2.974 mm



Elemento BC:

A = ⎡⎣ −1 0 1 0 ⎤⎦

u'bc ≔ A ⋅ ubc = 1.115 mm

Elemento DC:

A = ⎡⎣ −0.5 −0.866 0.5 0.866 ⎤⎦

u'dc ≔ A ⋅ udc = −1.089 mm

Elemento AD:

A = ⎡⎣ −1 0 1 0 ⎤⎦

u'ad ≔ A ⋅ uad − α ⋅ ΔTad ⋅ Lad = −0.408 mm

Elemento DB:

A = ⎡⎣ 0 −1 0 1 ⎤⎦

u'db ≔ A ⋅ udb − ΔLdb = −2.231 mm


F'ab = 892.307 kN

⎡ −446.154 ⎤ ⎢ ⎥ 0 kN Fbc ≔ Kbc ⋅ ubc = ⎢ 446.154 ⎥ ⎢⎣ ⎥⎦ 0

F'bc = 446.154 kN

⎡ 163.304 ⎤ ⎢ 282.85 ⎥ kN Fdc ≔ Kdc ⋅ udc = ⎢ −163.304 ⎥ ⎢⎣ −282.85 ⎥⎦

F'dc = −326.607 kN

⎡ 163.304 ⎤ ⎥ ⎢ 0 Fad ≔ Kad ⋅ uad + fad = ⎢ kN −163.304 ⎥ ⎢⎣ ⎥⎦ 0

F'ad = −163.304 kN



⎤ ⎡ 0 ⎢ 772.76 ⎥ Fdb ≔ Kdb ⋅ udb + fdb = ⎢ ⎥ kN 0 ⎢⎣ −772.76 ⎥⎦


F'db = −772.76 kN

12. Reacciones en los apoyos: Rax ≔ −446.154 kN + 163.304 kN = −282.85 kN Ray ≔ −772.761 kN + 0 kN = −772.76 kN Rdx ≔ 163.304 kN − 163.304 kN + 0 kN = 0 kN Rdy ≔ 282.85 kN + 0 kN + 772.76 kN = 1055.61 kN




Problema 14: Ensamblar la matriz de rigidez K y determinar las fuerzas en los elementos de la estructura. Datos: E ≔ 200 GPa Aab ≔ 1000 mm Aca ≔ 1000 mm

2

2

kN kr ≔ 2.5 ―― mm



EA ⎡ 1 −1 ⎤ K' = ―― ⋅ L ⎢⎣ −1 1 ⎥⎦


Acb ≔ 2000 mm

2


Elemento AB:


Lab ≔ 0.60 m

⎡ 333.3333 −333.3333 ⎤ kN ―― K'ab = ⎢ ⎣ −333.3333 333.3333 ⎥⎦ mm Lca ≔ 0.80 m

Elemento CA:

⎡ 250 −250 ⎤ kN ―― K'ca = ⎢ ⎣ −250 250 ⎥⎦ mm Lcb ≔ 1.00 m

Elemento CB:

⎡ 400 −400 ⎤ kN ―― K'cb = ⎢ ⎣ −400 400 ⎥⎦ mm kN kr = 2.5 ―― mm

Resorte:

⎡ 2.5 −2.5 ⎤ kN ―― K're = ⎢ ⎣ −2.5 2.5 ⎥⎦ mm 3. Matriz de rigidez de cada elemento "Global": ϕi ≔ 0 °

Elemento AB: ⎡1 0 0 0⎤ Tab = ⎢ ⎣ 0 0 1 0 ⎥⎦

ϕj ≔ 0 ° T


⎡ 333.333 ⎢ 0 Kab = ⎢ −333.333 ⎢⎣ 0

0 −333.333 0 ⎤ 0 0 0 ⎥ kN ―― 0 333.333 0 ⎥ mm ⎥ 0 0 0⎦

ϕi ≔ 90°

Elemento CA:

0 0⎤ 0 −250 ⎥ kN ―― 0 0 ⎥ mm 0 250 ⎥⎦


GLca

⎡3⎤ ⎢4⎥ = ⎢ ⎥ 0 ⎢⎣ 0 ⎥⎦

ϕj ≔ 90°

⎡0 1 0 0⎤ Tca = ⎢ ⎣ 0 0 0 1 ⎥⎦ ⎡0 0 ⎢ 0 250 Kca = ⎢ 0 0 ⎢⎣ 0 −250

GLab

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 1 ⎢⎣ 2 ⎥⎦

T

Kca ≔ Tca ⋅ K'ca ⋅ Tca


ϕi ≔ 53°

Elemento CB:


ϕj ≔ 53°

⎡ 0.602 0.799 0 ⎤ T 0 Kcb ≔ Tcb ⋅ K'cb ⋅ Tcb Tcb = ⎢ ⎥ ⎣0 0 0.602 0.799 ⎦ ⎡ 144.873 192.252 −144.873 −192.252 ⎤ ⎢ 192.252 255.127 −192.252 −255.127 ⎥ kN ―― Kcb = ⎢ −144.873 −192.252 144.873 192.252 ⎥ mm ⎢ ⎥ ⎣ −192.252 −255.127 192.252 255.127 ⎦ ϕ ≔ 0°

Resorte: ⎡1 0 0 0⎤ Tre = ⎢ ⎣ 0 0 1 0 ⎥⎦ ⎡ 2.5 ⎢ 0 Kre = ⎢ −2.5 ⎢⎣ 0

GLcb

⎡3⎤ ⎢4⎥ = ⎢ ⎥ 1 ⎢⎣ 2 ⎥⎦

T

Kre ≔ Tre ⋅ K're ⋅ Tre

0 −2.5 0 ⎤ 0 0 0 ⎥ kN ―― 0 2.5 0 ⎥ mm ⎥ 0 0 0⎦

GLre

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 3 ⎢⎣ 4 ⎥⎦


⎡ 478.206 192.252 −144.873 −192.252 ⎤ ⎢ 192.252 255.127 −192.252 −255.127 ⎥ kN K=⎢ ―― −144.873 −192.252 147.373 192.252 ⎥ mm ⎢ ⎥ ⎣ −192.252 −255.127 192.252 505.127 ⎦ 5. Matriz de Fuerzas: ⎡ 0 ⎤ ⎢ −4 ⎥ kN F≔⎢ 0 ⎥ ⎢⎣ 0 ⎥⎦

⎡1⎤ ⎢2⎥ GL = ⎢ ⎥ 3 ⎢⎣ 4 ⎥⎦

6. Desplazamientos: F = K⋅u ⎡ 0.00904 ⎤ ⎢ −0.94704 ⎥ −1 u≔K ⋅F=⎢ mm −1.20569 ⎥ ⎢ ⎥ ⎣ −0.016 ⎦


⎡1⎤ ⎢2⎥ GL = ⎢ ⎥ 3 ⎢⎣ 4 ⎥⎦

⎡1⎤ ⎢2⎥ GL = ⎢ ⎥ 3 ⎢⎣ 4 ⎥⎦


⎡ 0 ⎤ ⎢ 0 ⎥ mm uab = ⎢ 0.00904 ⎥ ⎢⎣ −0.94704 ⎥⎦


⎡ −1.20569 ⎤ ⎢ −0.016 ⎥ uca = ⎢ ⎥ mm 0 ⎢⎣ ⎥⎦ 0

⎡ −1.20569 ⎤ ⎢ −0.016 ⎥ mm ucb = ⎢ 0.00904 ⎥ ⎢ ⎥ ⎣ −0.94704 ⎦

⎡ 0 ⎤ ⎢ 0 ⎥ mm ure = ⎢ −1.20569 ⎥ ⎢⎣ −0.016 ⎥⎦ 7. Deformaciones en cada elemento: Ai = ⎡⎣ −cos ⎛⎝θi⎞⎠ −sin ⎛⎝θi⎞⎠ cos ⎛⎝θj⎞⎠ sin ⎛⎝θj⎞⎠ ⎤⎦ ui' = Ai ⋅ ui Elemento AB:

A = ⎡⎣ −1 0 1 0 ⎤⎦

u'ab ≔ A ⋅ uab = 0.009 mm

Elemento CA:

A = ⎡⎣ 0 −1 0 1 ⎤⎦

u'ca ≔ A ⋅ uca = 0.016 mm

Elemento CB:

A = ⎡⎣ −0.602 −0.799 0.602 0.799 ⎤⎦

u'cb ≔ A ⋅ ucb = −0.013 mm

Resorte:

A = ⎡⎣ −1 0 1 0 ⎤⎦

u're ≔ A ⋅ ure = −1.206 mm

8. Fuerzas en cada elemento: Elemento AB: ⎡ −3.014 ⎤ ⎥ ⎢ 0 kN Fab ≔ Kab ⋅ uab = ⎢ 3.014 ⎥ ⎢⎣ 0 ⎥⎦


F'ab = 3.014 kN



Elemento CA: ⎡ 0⎤ ⎢ −4 ⎥ Fca ≔ Kca ⋅ uca = ⎢ kN 0⎥ ⎢⎣ 4 ⎥⎦

F'ca = 4 kN

Elemento CB: ⎡ 3.014 ⎤ ⎢ 4 ⎥ kN Fcb ≔ Kcb ⋅ ucb = ⎢ −3.014 ⎥ ⎢⎣ −4 ⎥⎦

F'cb = −5.009 kN

Resorte: ⎡ 3.014 ⎤ ⎢ 0 ⎥ Fre ≔ Kre ⋅ ure = ⎢ kN −3.014 ⎥ ⎢⎣ 0 ⎥⎦ 9. Reacciones en los apoyos: Rax ≔ 3.014 kN + 0 kN = 3.01 kN Ray ≔ 0 kN + 4 kN = 4 kN


F're = −3.014 kN



Problema 15: Ensamblar la matriz de rigidez K y determinar las fuerzas en los elementos de la estructura. Datos: E ≔ 200 GPa Aab ≔ 500 mm

2

kN kr ≔ 2 ―― mm



EA ⎡ 1 −1 ⎤ K' = ―― ⋅ L ⎢⎣ −1 1 ⎥⎦

Elemento AB:

Lab ≔ 10.00 m

⎡ 10 −10 ⎤ kN ―― K'ab = ⎢ ⎣ −10 10 ⎥⎦ mm


Acb ≔ 500 mm

2


Elemento CB:


Lcb ≔ 5.00 m

⎡ 20 −20 ⎤ kN ―― K'cb = ⎢ ⎣ −20 20 ⎥⎦ mm kN kr = 2 ―― mm

Resorte: ⎡ 2 −2 ⎤ kN ―― K're = ⎢ ⎣ −2 2 ⎥⎦ mm


ϕi ≔ 0 °

ϕj ≔ 0 °

⎡1 0 0 0⎤ Tab = ⎢ ⎣ 0 0 1 0 ⎦⎥ ⎡ 10 ⎢ 0 Kab = ⎢ −10 ⎢⎣ 0

T


0 −10 0 ⎤ 0 0 0 ⎥ kN ―― 0 10 0 ⎥ mm ⎥ 0 0 0⎦

Elemento CB:

ϕi ≔ −45°

GLab

ϕj ≔ −45°

⎡ 0.707 −0.707 0 ⎤ 0 Tcb = ⎢ 0 0.707 −0.707 ⎥⎦ ⎣0

T


⎡ 10 −10 −10 10 ⎤ ⎢ −10 10 10 −10 ⎥ kN ―― Kcb = ⎢ −10 10 10 −10 ⎥ mm ⎢ ⎥ ⎣ 10 −10 −10 10 ⎦

GLcb

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 1 ⎢⎣ 2 ⎥⎦

GLre

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 1 ⎢⎣ 2 ⎥⎦

ϕ ≔ 90°

Resorte: ⎡0 1 0 0⎤ Tre = ⎢ ⎣ 0 0 0 1 ⎦⎥ ⎡0 0 ⎢0 2 Kre = ⎢ 0 0 ⎢⎣ 0 −2

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 1 ⎢⎣ 2 ⎥⎦

0 0⎤ 0 −2 ⎥ kN ―― 0 0 ⎥ mm 0 2 ⎥⎦


T




4. Matriz de rigidez de la estructura: ⎡ 20 −10 ⎤ kN ―― K=⎢ ⎣ −10 12 ⎥⎦ mm

⎡1⎤ GL = ⎢ ⎥ ⎣2⎦

5. Matriz de Fuerzas:

⎡1⎤ GL = ⎢ ⎥ ⎣2⎦

⎡ 0 ⎤ kN F≔⎢ ⎣ −50 ⎥⎦ 6. Desplazamientos: F = K⋅u

u≔K

−1

⎡1⎤ GL = ⎢ ⎥ ⎣2⎦

⎡ −3.57143 ⎤ ⋅F=⎢ mm ⎣ −7.14286 ⎥⎦

⎡ 0 ⎤ ⎢ 0 ⎥ mm uab = ⎢ −3.57143 ⎥ ⎢⎣ −7.14286 ⎥⎦

⎡ 0 ⎤ ⎢ 0 ⎥ ucb = ⎢ mm −3.57143 ⎥ ⎢⎣ −7.14286 ⎥⎦

⎡ 0 ⎤ ⎢ 0 ⎥ ure = ⎢ mm −3.57143 ⎥ ⎢⎣ −7.14286 ⎥⎦


A = ⎡⎣ −1 0 1 0 ⎤⎦

u'ab ≔ A ⋅ uab = −3.571 mm

Elemento CB:

A = ⎡⎣ −0.707 0.707 0.707 −0.707 ⎤⎦

u'cb ≔ A ⋅ ucb = 2.525 mm

Resorte:

A = ⎡⎣ 0 −1 0 1 ⎤⎦


u're ≔ A ⋅ ure = −7.143 mm



8. Fuerzas en cada elemento: Elemento AB: ⎡ 35.714 ⎤ ⎥ ⎢ 0 Fab ≔ Kab ⋅ uab = ⎢ kN −35.714 ⎥ ⎢⎣ 0 ⎥⎦

F'ab = −35.714 kN

Elemento CB: ⎡ −35.714 ⎤ ⎢ 35.714 ⎥ Fcb ≔ Kcb ⋅ ucb = ⎢ kN 35.714 ⎥ ⎢⎣ −35.714 ⎥⎦

F'cb = 50.508 kN

Resorte: ⎤ ⎡ 0 ⎢ 14.286 ⎥ Fre ≔ Kre ⋅ ure = ⎢ ⎥ kN 0 ⎢⎣ −14.286 ⎥⎦

F're = −14.286 kN

9. Reacciones en los apoyos: Rax ≔ 35.71 kN

Rcx ≔ −35.71 kN

Ray ≔ 0 kN

Rcy ≔ 35.71 kN




Problema 16: Ensamblar la matriz de rigidez K y determinar las fuerzas en los elementos de la estructura. Datos: E ≔ 200 GPa Aab ≔ 500 mm Lab ≔ 5.00 m kN kr ≔ 4 ―― mm



EA ⎡ 1 −1 ⎤ K' = ―― ⋅ L ⎢⎣ −1 1 ⎥⎦


2

Acb ≔ 500 mm Lcb ≔ 5.00 m

2


Elemento AB:


Lab = 5 m

⎡ 20 −20 ⎤ kN ―― K'ab = ⎢ ⎣ −20 20 ⎥⎦ mm Lcb = 5 m

Elemento CB:

⎡ 20 −20 ⎤ kN ―― K'cb = ⎢ ⎣ −20 20 ⎥⎦ mm kN kr = 4 ―― mm

Resorte: ⎡ 4 −4 ⎤ kN ―― K're = ⎢ ⎣ −4 4 ⎥⎦ mm


ϕi ≔ −60°

ϕj ≔ −60°

⎡ 0.5 −0.866 0 ⎤ 0 Tab = ⎢ 0 0.5 −0.866 ⎥⎦ ⎣0

T


⎡ 5 −8.66 −5 8.66 ⎤ ⎢ −8.66 15 ⎥ kN 8.66 −15 ―― Kab = ⎢ −5 8.66 5 −8.66 ⎥ mm ⎢ ⎥ −8.66 15 ⎣ 8.66 −15 ⎦ Elemento CB:

ϕi ≔ −120°

GLab

ϕj ≔ −120°

⎤ ⎡ −0.5 −0.866 0 0 Tcb = ⎢ 0 −0.5 −0.866 ⎥⎦ ⎣ 0

T


⎡ 5 8.66 −5 −8.66 ⎤ ⎢ 8.66 15 ⎥ kN −8.66 −15 ―― Kcb = ⎢ −5 −8.66 5 8.66 ⎥ mm ⎢ ⎥ 8.66 15 ⎣ −8.66 −15 ⎦ Resorte:

ϕ ≔ 90°

⎡0 1 0 0⎤ Tre = ⎢ ⎣ 0 0 0 1 ⎦⎥


⎡0⎤ ⎢0⎥ = ⎢ ⎥ 1 ⎢⎣ 2 ⎥⎦

T


GLcb

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 1 ⎢⎣ 2 ⎥⎦


⎡0 0 ⎢0 4 Kre = ⎢ 0 0 ⎢⎣ 0 −4


0 0⎤ 0 −4 ⎥ kN ―― 0 0 ⎥ mm 0 4 ⎥⎦

GLre

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 1 ⎢⎣ 2 ⎥⎦

4. Matriz de rigidez de la estructura: ⎡ 10 0 ⎤ kN K=⎢ ―― ⎣ 0 34 ⎥⎦ mm

⎡1⎤ GL = ⎢ ⎥ ⎣2⎦

5. Matriz de Fuerzas: ⎡ 0 ⎤ F≔⎢ kN ⎣ −100 ⎥⎦

⎡1⎤ GL = ⎢ ⎥ ⎣2⎦


u≔K

−1

⎡1⎤ GL = ⎢ ⎥ ⎣2⎦

⎡ 0 ⎤ mm ⋅F=⎢ ⎣ −2.94118 ⎥⎦

⎤ ⎡ 0 ⎥ ⎢ 0 uab = ⎢ ⎥ mm 0 ⎢⎣ −2.94118 ⎥⎦

⎤ ⎡ 0 ⎥ ⎢ 0 ucb = ⎢ ⎥ mm 0 ⎢⎣ −2.94118 ⎥⎦

⎤ ⎡ 0 ⎥ ⎢ 0 ure = ⎢ ⎥ mm 0 ⎢⎣ −2.94118 ⎥⎦


A = ⎡⎣ −0.5 0.866 0.5 −0.866 ⎤⎦

u'ab ≔ A ⋅ uab = 2.547 mm

Elemento CB:

A = ⎡⎣ 0.5 0.866 −0.5 −0.866 ⎤⎦


u'cb ≔ A ⋅ ucb = 2.547 mm



Resorte:

A = ⎡⎣ 0 −1 0 1 ⎤⎦

u're ≔ A ⋅ ure = −2.941 mm

11. Fuerzas en cada elemento: Elemento AB: ⎡ −25.471 ⎤ ⎢ 44.118 ⎥ Fab ≔ Kab ⋅ uab = ⎢ kN 25.471 ⎥ ⎢⎣ −44.118 ⎥⎦

F'ab = 50.943 kN

Elemento CB: ⎡ 25.471 ⎤ ⎢ 44.118 ⎥ Fcb ≔ Kcb ⋅ ucb = ⎢ kN −25.471 ⎥ ⎢⎣ −44.118 ⎥⎦

F'cb = 50.943 kN

Resorte: ⎤ ⎡ 0 ⎢ 11.765 ⎥ Fre ≔ Kre ⋅ ure = ⎢ ⎥ kN 0 ⎢⎣ −11.765 ⎥⎦ 8. Reacciones en los apoyos: Rax ≔ −25.47 kN Ray ≔ 44.12 kN Rcx ≔ 25.47 kN Rcy ≔ 44.12 kN


F're = −11.765 kN



Problema 17: Ensamblar la matriz de rigidez K y determinar las fuerzas en los elementos de la estructura. Datos: 6

E ≔ 30 ⋅ 10 psi A0 ≔ 2 in

2

Asentamiento del apoyo D: Δd ≔ −0.05 in



EA ⎡ 1 −1 ⎤ K' = ―― ⋅ L ⎢⎣ −1 1 ⎥⎦

Elemento AB:

Lab ≔ 4 ft

⎡ 15000 −15000 ⎤ kip ―― K'ab = ⎢ ⎣ −15000 15000 ⎥⎦ ft



Elemento BC:


Lbc ≔ 8 ft

⎡ 7500 −7500 ⎤ kip ―― K'bc = ⎢ ⎣ −7500 7500 ⎥⎦ ft Ldc ≔ 4 ft

Elemento DC:

⎡ 15000 −15000 ⎤ kip ―― K'dc = ⎢ ⎣ −15000 15000 ⎥⎦ ft Lad ≔ 8 ft

Elemento AD:

⎡ 7500 −7500 ⎤ kip ―― K'ad = ⎢ ⎣ −7500 7500 ⎥⎦ ft Elemento BD:

Lbd ≔ 8.944 ft

⎡ 6708.4079 −6708.4079 ⎤ kip ―― K'bd = ⎢ ⎣ −6708.4079 6708.4079 ⎥⎦ ft 3. Matriz de rigidez de cada elemento "Global": ϕi ≔ 90°

Elemento AB:

ϕj ≔ 90°

⎡0 1 0 0⎤ Tab = ⎢ ⎣ 0 0 0 1 ⎥⎦

T


⎡0 0 ⎢ 0 15000 Kab = ⎢ 0 0 ⎢⎣ 0 −15000

0 0⎤ 0 −15000 ⎥ kip ―― 0 0 ⎥ ft 0 15000 ⎥⎦

ϕi ≔ 0 °

Elemento BC:

ϕj ≔ 0°

⎡1 0 0 0⎤ Tbc = ⎢ ⎣ 0 0 1 0 ⎥⎦ ⎡ 7500 ⎢ 0 Kbc = ⎢ −7500 ⎢⎣ 0

0 −7500 0 ⎤ 0 0 0 ⎥ kip ―― 0 7500 0 ⎥ ft ⎥ 0 0 0⎦


GLab

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 2 ⎢⎣ 3 ⎥⎦

T


GLbc

⎡2⎤ ⎢3⎥ = ⎢ ⎥ 0 ⎢⎣ 0 ⎥⎦


ϕi ≔ 90°

Elemento DC:


ϕj ≔ 90°

⎡0 1 0 0⎤ Tdc = ⎢ ⎣ 0 0 0 1 ⎥⎦

T


⎡0 0 ⎢ 0 15000 Kdc = ⎢ 0 0 ⎢⎣ 0 −15000

0 0⎤ 0 −15000 ⎥ kip ―― 0 0 ⎥ ft 0 15000 ⎥⎦

ϕi ≔ 0 °

Elemento AD:

ϕj ≔ 0°

⎡1 0 0 0⎤ Tad = ⎢ ⎣ 0 0 1 0 ⎥⎦ ⎡ 7500 ⎢ 0 Kad = ⎢ −7500 ⎢⎣ 0

GLdc

T


0 −7500 0 ⎤ 0 0 0 ⎥ kip ―― 0 7500 0 ⎥ ft ⎥ 0 0 0⎦

Elemento BD:

⎡0⎤ ⎢1⎥ = ⎢ ⎥ 0 ⎢⎣ 0 ⎥⎦

GLad

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 0 ⎢⎣ 1 ⎥⎦

ϕi ≔ −26.565° ϕj ≔ −26.565°

⎤ ⎡ 0.894 −0.447 0 0 Tbd = ⎢ ⎥ 0 0.894 −0.447 ⎦ ⎣0

T


⎡2⎤ ⎡ 5366.731 −2683.36 −5366.731 2683.36 ⎤ ⎢3⎥ ⎢ −2683.36 1341.677 2683.36 −1341.677 ⎥ kip GLbd = ⎢ ⎥ 0 ―― Kbd = ⎢ −5366.731 2683.36 5366.731 −2683.36 ⎥ ft ⎢⎣ 1 ⎥⎦ ⎢ ⎥ 1341.677 ⎦ ⎣ 2683.36 −1341.677 −2683.36 4. Matriz de rigidez de la estructura:

⎡ 16341.677 2683.36 −1341.677 ⎤ kip K = ⎢ 2683.36 12866.731 −2683.36 ⎥ ―― ⎢⎣ −1341.677 −2683.36 16341.677 ⎥⎦ ft

⎡1⎤ GL = ⎢ 2 ⎥ ⎢⎣ 3 ⎥⎦

5. Matriz de Fuerzas: ⎡0⎤ F ≔ ⎢ 0 ⎥ kip ⎢⎣ 0 ⎥⎦


⎡1⎤ GL = ⎢ 2 ⎥ ⎢⎣ 3 ⎥⎦



6. Desplazamientos: F = K⋅u ⎡ 16341.677 2683.36 −1341.677 ⎤ ⎡ Δd ⎤ ⎡0⎤ ⎢ 0 ⎥ = ⎢ 2683.36 12866.731 −2683.36 ⎥ ⋅ ⎢ Δ2 ⎥ ⎢⎣ −1341.677 −2683.36 16341.677 ⎥⎦ ⎢ ⎥ ⎢⎣ 0 ⎥⎦ ⎣ Δ3 ⎦ ⎡ 12866.731 −2683.36 ⎤ ⎡ Δ2 ⎤ ⎡ 2683.36 ⎤ ⎡0⎤ = ⎢⎣ −2683.36 16341.677 ⎥⎦ ⋅ ⎢⎣ Δ ⎥⎦ + ⎢⎣ −1341.677 ⎥⎦ ⋅ Δd ⎢⎣ 0 ⎥⎦ 3

⎡ 12866.731 −2683.36 ⎤ kip K≔⎢ ―― ⎣ −2683.36 16341.677 ⎥⎦ ft

⎡ 2683.36 ⎤ kip ⎡0⎤ ―― F ≔ ⎢ ⎥ kip − ⎢ ⋅ Δd ⎣0⎦ ⎣ −1341.677 ⎥⎦ ft u≔K

−1

⎡2⎤ GL = ⎢ ⎥ ⎣3⎦

⎡ 0.00991 ⎤ in ⋅F=⎢ ⎣ −0.00248 ⎥⎦

⎤ ⎡ 0 ⎥ ⎢ 0 in uab = ⎢ 0.00991 ⎥ ⎢⎣ −0.00248 ⎥⎦

⎡ 0.00991 ⎤ ⎢ −0.00248 ⎥ ubc = ⎢ ⎥ in 0 ⎢⎣ 0 ⎥⎦

⎤ ⎡ 0 ⎥ ⎢ 0 uad = ⎢ ⎥ in 0 ⎢⎣ −0.05 ⎥⎦

⎡ 0.00991 ⎤ ⎢ −0.00248 ⎥ ubd = ⎢ ⎥ in 0 ⎢⎣ −0.05 ⎥⎦

⎡ 0 ⎤ ⎢ −0.05 ⎥ udc = ⎢ ⎥ in 0 ⎢⎣ 0 ⎥⎦


A = ⎡⎣ 0 −1 0 1 ⎤⎦


u'ab ≔ A ⋅ uab = −0.00248 in



Elemento BC:

A = ⎡⎣ −1 0 1 0 ⎤⎦

u'bc ≔ A ⋅ ubc = −0.00991 in

Elemento DC:

A = ⎡⎣ 0 −1 0 1 ⎤⎦

u'dc ≔ A ⋅ udc = 0.05 in

Elemento AD:

A = ⎡⎣ −1 0 1 0 ⎤⎦

u'ad ≔ A ⋅ uad = 0 in

Elemento BD:

A = ⎡⎣ −0.894 0.447 0.894 −0.447 ⎤⎦

u'bd ≔ A ⋅ ubd = 0.01239 in

8. Fuerzas en cada elemento: Elemento AB: ⎤ ⎡ 0 ⎢ 3.097 ⎥ Fab ≔ Kab ⋅ uab = ⎢ ⎥ kip 0 ⎢⎣ −3.097 ⎥⎦

F'ab = −3.097 kip

Elemento BC: ⎡ 6.194 ⎤ ⎢ 0 ⎥ kip Fbc ≔ Kbc ⋅ ubc = ⎢ −6.194 ⎥ ⎢⎣ 0 ⎥⎦

F'bc = −6.194 kip

Elemento DC: ⎡ 0 ⎤ ⎢ −62.5 ⎥ Fdc ≔ Kdc ⋅ udc = ⎢ kip 0 ⎥ ⎢⎣ 62.5 ⎥⎦


F'dc = 62.5 kip



Elemento AD: ⎡0⎤ ⎢0⎥ Fad ≔ Kad ⋅ uad = ⎢ ⎥ kip 0 ⎢⎣ 0 ⎥⎦

F'ad = 0 kip

Elemento BD: ⎡ −6.194 ⎤ ⎢ 3.097 ⎥ Fbd ≔ Kbd ⋅ ubd = ⎢ kip 6.194 ⎥ ⎢⎣ −3.097 ⎥⎦

F'bd = 6.925 kip

9. Reacciones en los apoyos: Rax ≔ 0 kip + 0 kip = 0 kip Ray ≔ 3.097 kip + 0 kip = 3.1 kip Rcx ≔ −6.194 kip + 0 kip = −6.19 kip Rcy ≔ 0 kip + 62.5 kip = 62.5 kip Rdx ≔ 0 kip + 0 kip + 6.194 kip = 6.19 kip Rdy ≔ −62.5 kip + 0 kip + −3.097 kip = −65.6 kip




Problema 18: Ensamblar la matriz de rigidez K y determinar las fuerzas en los elementos de la estructura. Datos: 2

E ≔ 200 GPa

A0 ≔ 500 mm

kN kr1 ≔ 1 ―― mm

kN kr2 ≔ 1.2 ―― mm

Asentamiento del apoyo C: Δc ≔ 5 mm



EA ⎡ 1 −1 ⎤ K' = ―― ⋅ L ⎢⎣ −1 1 ⎥⎦



Elemento AB:

Lab ≔ 4.00 m


Lbc ≔ 4.00 m

⎡ 25 −25 ⎤ kN ―― K'bc = ⎢ ⎣ −25 25 ⎥⎦ mm Elemento DC:

Ldc ≔ 4.00 m


Lad ≔ 4.00 m

⎡ 25 −25 ⎤ kN ―― K'ad = ⎢ ⎣ −25 25 ⎥⎦ mm Elemento AC:

Lac ≔ 4 ⋅

2

‾‾ 2 m


Lbd ≔ 4 ⋅

2

‾‾ 2 m

⎡ 17.6777 −17.6777 ⎤ kN ―― K'bd = ⎢ ⎣ −17.6777 17.6777 ⎥⎦ mm Resorte 1:

kN kr1 = 1 ―― mm

⎡ 1 −1 ⎤ kN ―― K're1 = ⎢ ⎣ −1 1 ⎥⎦ mm Resorte 2:

kN kr2 = 1.2 ―― mm

⎡ 1.2 −1.2 ⎤ kN ―― K're2 = ⎢ ⎣ −1.2 1.2 ⎥⎦ mm





3. Matriz de rigidez de cada elemento "Global": ϕi ≔ 90°

Elemento AB:

ϕj ≔ 90°

⎡0 1 0 0⎤ Tab = ⎢ ⎣ 0 0 0 1 ⎥⎦ ⎡0 0 ⎢ 0 25 Kab = ⎢ 0 0 ⎢⎣ 0 −25

T


0 0⎤ 0 −25 ⎥ kN ―― 0 0 ⎥ mm 0 25 ⎥⎦ ϕi ≔ 0 °

Elemento BC:

GLab

ϕj ≔ 0°

⎡1 0 0 0⎤ Tbc = ⎢ ⎣ 0 0 1 0 ⎥⎦ ⎡ 25 ⎢ 0 Kbc = ⎢ −25 ⎢⎣ 0

T


0 −25 0 ⎤ 0 0 0 ⎥ kN ―― 0 25 0 ⎥ mm ⎥ 0 0 0⎦ ϕi ≔ 90°

Elemento DC:

GLbc

T


0 0⎤ 0 −25 ⎥ kN ―― 0 0 ⎥ mm 0 25 ⎥⎦

Elemento AD:

ϕi ≔ 0 °

⎡1 0 0 0⎤ Tad = ⎢ ⎣ 0 0 1 0 ⎥⎦ ⎡ 25 ⎢ 0 Kad = ⎢ −25 ⎢⎣ 0

⎡0⎤ ⎢3⎥ = ⎢ ⎥ 2 ⎢⎣ 1 ⎥⎦

ϕj ≔ 90°

⎡0 1 0 0⎤ Tdc = ⎢ ⎣ 0 0 0 1 ⎥⎦ ⎡0 0 ⎢ 0 25 Kdc = ⎢ 0 0 ⎢⎣ 0 −25

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 0 ⎢⎣ 3 ⎥⎦

0 −25 0 ⎤ 0 0 0 ⎥ kN ―― 0 25 0 ⎥ mm ⎥ 0 0 0⎦


GLdc

⎡4⎤ ⎢5⎥ = ⎢ ⎥ 2 ⎢⎣ 1 ⎥⎦

ϕj ≔ 0° T


GLad

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 4 ⎢⎣ 5 ⎥⎦


Elemento AC:

ϕi ≔ 45°


ϕj ≔ 45°

⎡ 0.707 0.707 0 ⎤ 0 Tac = ⎢ ⎣0 0 0.707 0.707 ⎥⎦

T


⎡ 8.839 8.839 −8.839 −8.839 ⎤ ⎢ 8.839 8.839 −8.839 −8.839 ⎥ kN ―― Kac = ⎢ −8.839 −8.839 8.839 8.839 ⎥ mm ⎢ ⎥ ⎣ −8.839 −8.839 8.839 8.839 ⎦ Elemento BD:

ϕi ≔ −45°

GLac

ϕj ≔ −45°

⎡ 0.707 −0.707 0 ⎤ 0 Tbd = ⎢ ⎥ 0 0.707 −0.707 ⎦ ⎣0

T


⎡ 8.839 −8.839 −8.839 8.839 ⎤ ⎢ −8.839 8.839 8.839 −8.839 ⎥ kN ―― Kbd = ⎢ −8.839 8.839 8.839 −8.839 ⎥ mm ⎢ ⎥ ⎣ 8.839 −8.839 −8.839 8.839 ⎦ Resorte 1:

Resorte 2:

GLbd

⎡0⎤ ⎢3⎥ = ⎢ ⎥ 4 ⎢⎣ 5 ⎥⎦

ϕ ≔ 90°

⎡0 1 0 0⎤ Tre1 = ⎢ ⎣ 0 0 0 1 ⎥⎦ ⎡0 0 ⎢0 1 Kre1 = ⎢ 0 0 ⎢⎣ 0 −1

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 2 ⎢⎣ 1 ⎥⎦

T

Kre1 ≔ Tre1 ⋅ K're1 ⋅ Tre1

0 0⎤ 0 −1 ⎥ kN ―― 0 0 ⎥ mm 0 1 ⎥⎦

GLre1

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 4 ⎢⎣ 5 ⎥⎦

ϕ ≔ −135°

⎡ −0.707 −0.707 0 ⎤ 0 Tre2 = ⎢ 0 −0.707 −0.707 ⎥⎦ ⎣ 0 ⎡ 0.6 0.6 −0.6 −0.6 ⎤ ⎢ 0.6 0.6 −0.6 −0.6 ⎥ kN ―― Kre2 = ⎢ −0.6 −0.6 0.6 0.6 ⎥ mm ⎢ ⎥ ⎣ −0.6 −0.6 0.6 0.6 ⎦


T

Kre2 ≔ Tre2 ⋅ K're2 ⋅ Tre2

GLre2

⎡0⎤ ⎢0⎥ = ⎢ ⎥ 4 ⎢⎣ 5 ⎥⎦




⎡ 33.839 8.839 0 ⎤ 0 −25 ⎢ 8.839 33.839 0 ⎥ 0 0 kN ⎢ 0 0 33.839 8.839 −8.839 ⎥ ―― K= ⎢ ⎥ 0 0 8.839 34.439 −8.239 mm ⎢ ⎥ 0 −8.839 −8.239 35.439 ⎦ ⎣ −25

⎡1⎤ ⎢2⎥ GL = ⎢ 3 ⎥ ⎢4⎥ ⎢⎣ ⎥⎦ 5

5. Matriz de Fuerzas: ⎡ 0 ⎤ ⎢ 0 ⎥ F ≔ ⎢ 80 ⎥ kN ⎢ 50 ⎥ ⎢⎣ ⎥⎦ 0

⎡1⎤ ⎢2⎥ GL = ⎢ 3 ⎥ ⎢4⎥ ⎢⎣ ⎥⎦ 5

6. Desplazamientos: F = K⋅u ⎡ 0 ⎤ ⎢ 0 ⎥ ⎢ 80 ⎥ = ⎢ 50 ⎥ ⎢⎣ ⎥⎦ 0

⎡ 33.839 8.839 0 0 −25 ⎤ ⎡ Δc ⎤ ⎢ 8.839 33.839 ⎥ ⎢ Δ2 ⎥ 0 0 0 ⎢ ⎥ ⎢ 0 0 33.839 8.839 −8.839 ⎥ ⋅ Δ3 ⎢ ⎥ ⎢ ⎥ 0 0 8.839 34.439 −8.239 ⎢ Δ4 ⎥ ⎢ ⎥ 0 −8.839 −8.239 35.439 ⎦ ⎢⎣ Δ5 ⎥⎦ ⎣ −25

⎡ 0 ⎤ ⎢ 80 ⎥ ⎢ 50 ⎥ = ⎢⎣ 0 ⎥⎦

⎡ 33.839 ⎤ ⎡ Δ2 ⎤ ⎡ 8.839 ⎤ 0 0 0 ⎢ 0 33.839 8.839 −8.839 ⎥ ⎢ Δ3 ⎥ ⎢ 0 ⎥ ⋅⎢ ⎥+⎢ ⋅Δ ⎢ 0 8.839 34.439 −8.239 ⎥ Δ4 0 ⎥ c ⎢ ⎥ ⎢ ⎥ 0 −8.839 −8.239 35.439 ⎦ ⎣ Δ5 ⎦ ⎢⎣ −25 ⎥⎦ ⎣

⎡ 33.839 ⎤ 0 0 0 ⎢ 0 33.839 8.839 −8.839 ⎥ kN ―― K≔⎢ 0 8.839 34.439 −8.239 ⎥ mm ⎢ ⎥ 0 −8.839 −8.239 35.439 ⎦ ⎣ ⎡ −44.195 ⎤ ⎡ 8.839 ⎤ ⎡ 0 ⎤ ⎢ 80 ⎥ ⎢ 0 ⎥ kN ⎢ 80 ⎥ ⋅ Δc = ⎢ ―― F ≔ ⎢ ⎥ kN − ⎢ ⎥ ⎥ kN 0 50 50 mm ⎢⎣ −25 ⎥⎦ ⎢⎣ 125 ⎥⎦ ⎢⎣ 0 ⎥⎦

u≔K

−1

⎡ −1.30604 ⎤ ⎢ 3.13328 ⎥ mm ⋅F=⎢ 1.7773 ⎥ ⎢⎣ 4.72187 ⎥⎦


⎡2⎤ ⎢3⎥ GL = ⎢ ⎥ 4 ⎢⎣ 5 ⎥⎦



⎤ ⎡0 ⎥ ⎢0 uab = ⎢ ⎥ mm 0 ⎢⎣ 3.13328 ⎥⎦

⎤ ⎡ 0 ⎢ 3.13328 ⎥ mm ubc = ⎢ −1.30604 ⎥ ⎢⎣ 5 ⎥⎦

⎡ 1.7773 ⎤ ⎢ 4.72187 ⎥ udc = ⎢ mm −1.30604 ⎥ ⎢⎣ 5 ⎥⎦

⎤ ⎡0 ⎥ ⎢0 mm uad = ⎢ 1.7773 ⎥ ⎢⎣ 4.72187 ⎥⎦

⎤ ⎡ 0 ⎥ ⎢ 0 uac = ⎢ mm −1.30604 ⎥ ⎢⎣ 5 ⎥⎦

⎤ ⎡0 ⎢ 3.13328 ⎥ mm ubd = ⎢ 1.7773 ⎥ ⎢⎣ 4.72187 ⎥⎦

⎤ ⎡0 ⎥ ⎢0 ure1 = ⎢ mm 1.7773 ⎥ ⎢⎣ 4.72187 ⎥⎦

⎤ ⎡0 ⎥ ⎢0 ure2 = ⎢ mm 1.7773 ⎥ ⎢⎣ 4.72187 ⎥⎦


A = ⎡⎣ 0 −1 0 1 ⎤⎦

u'ab ≔ A ⋅ uab = 3.133 mm

Elemento BC:

A = ⎡⎣ −1 0 1 0 ⎤⎦

u'bc ≔ A ⋅ ubc = −1.306 mm

Elemento DC:

A = ⎡⎣ 0 −1 0 1 ⎤⎦

u'dc ≔ A ⋅ udc = 0.278 mm

Elemento AD:

A = ⎡⎣ −1 0 1 0 ⎤⎦

u'ad ≔ A ⋅ uad = 1.777 mm

Elemento AC:

A = ⎡⎣ −0.707 −0.707 0.707 0.707 ⎤⎦


u'ac ≔ A ⋅ uac = 2.612 mm



Elemento BD:

A = ⎡⎣ −0.707 0.707 0.707 −0.707 ⎤⎦

u'bd ≔ A ⋅ ubd = 0.133 mm

Resorte 1:

A = ⎡⎣ 0 −1 0 1 ⎤⎦

u're1 ≔ A ⋅ ure1 = 4.722 mm

Resorte 2:

A = ⎡⎣ 0.707 0.707 −0.707 −0.707 ⎤⎦

u're2 ≔ A ⋅ ure2 = −4.596 mm

8. Fuerzas en cada elemento: Elemento AB: ⎡ 0 ⎤ ⎢ −78.332 ⎥ Fab ≔ Kab ⋅ uab = ⎢ ⎥ kN 0 ⎢⎣ 78.332 ⎥⎦

F'ab = 78.332 kN

Elemento BC: ⎡ 32.651 ⎤ ⎥ ⎢ 0 Fbc ≔ Kbc ⋅ ubc = ⎢ kN −32.651 ⎥ ⎢⎣ 0 ⎥⎦

F'bc = −32.651 kN

Elemento DC: ⎤ ⎡ 0 ⎢ −6.953 ⎥ Fdc ≔ Kdc ⋅ udc = ⎢ ⎥ kN 0 ⎢⎣ 6.953 ⎥⎦

F'dc = 6.953 kN

Elemento AD: ⎡ −44.432 ⎤ ⎥ ⎢ 0 Fad ≔ Kad ⋅ uad = ⎢ kN 44.432 ⎥ ⎢⎣ 0 ⎥⎦


F'ad = 44.432 kN



Elemento AC: ⎡ −32.65 ⎤ ⎢ −32.65 ⎥ kN Fac ≔ Kac ⋅ uac = ⎢ 32.65 ⎥ ⎢⎣ ⎥ 32.65 ⎦

F'ac = 46.175 kN

Elemento BD: ⎡ −1.668 ⎤ ⎢ 1.668 ⎥ Fbd ≔ Kbd ⋅ ubd = ⎢ kN 1.668 ⎥ ⎢⎣ −1.668 ⎥⎦

F'bd = 2.359 kN

Resorte 1: ⎡ 0 ⎤ ⎢ −4.722 ⎥ Fre1 ≔ Kre1 ⋅ ure1 = ⎢ ⎥ kN 0 ⎢⎣ 4.722 ⎥⎦

F're1 = 4.722 kN

Resorte 2: ⎡ −3.899 ⎤ ⎢ −3.899 ⎥ kN Fre2 ≔ Kre2 ⋅ ure2 = ⎢ 3.899 ⎥ ⎢⎣ ⎥ 3.899 ⎦

F're2 = −5.515 kN

9. Reacciones en los apoyos: Rax ≔ 0 kN + −44.432 kN + −32.65 kN = −77.08 kN Ray ≔ −78.332 kN + 0 kN + −32.65 kN = −110.98 kN Rbx ≔ 0 kN + 32.651 kN + −1.668 kN = 30.98 kN Rby ≔ 78.332 kN + 0 kN + 1.668 kN = 80 kN Rcx ≔ −32.651 kN + 0 kN + 32.65 kN = 0 kN Rcy ≔ 0 kN + 6.953 kN + 32.65 kN = 39.6 kN


Armaduras - Análisis Matricial

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