Scribd
Upload
57 views1 page

Ejercicio 3

This document provides the steps to determine the stresses and deformations in the bars of a truss using the stiffness method of assembly. The truss has 5 degrees of freedom and is subjected to vertical loads of 80 klb and 50 klb. The cross-sectional area is 1 in^2 and modulus of elasticity is 30,000 klb/in^2. The document generates the transformation matrix [A] relating the displacements {D} to the bar elongations {d}.

Uploaded by

JOSE LUIS QUISPE GARAY
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as XLSX, PDF, TXT or read online on Scribd
57 views1 page

Ejercicio 3

This document provides the steps to determine the stresses and deformations in the bars of a truss using the stiffness method of assembly. The truss has 5 degrees of freedom and is subjected to vertical loads of 80 klb and 50 klb. The cross-sectional area is 1 in^2 and modulus of elasticity is 30,000 klb/in^2. The document generates the transformation matrix [A] relating the displacements {D} to the bar elongations {d}.

Uploaded by

JOSE LUIS QUISPE GARAY
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as XLSX, PDF, TXT or read online on Scribd
You are on page 1/ 1

I) Para la armadura mostrada determinar los esfuerzos y deformaciones en las barras, usando el metodo de rigidez por ensamblaje.

A =1 in2, E= 30000
klb/in2.
80 Klb

50 Klb
3
E= 30,000 Klb/in2
A= 1.00 in2
b1 b5
12 ft b3

1 b2 b4
2 4

20 Klb
16 ft 16 ft

SOLUCIÓN

1.- Definimos el sistema Q-D y q-d

D4
D3
3

d1
D2
i j
D1 4 D5 q-d
1
2
Q-D
5 GDL

2.- Generamos la matriz de transformacion de desplazamientos [A]

D1=1 Di =0 , i ≠1

0 b1
1 b2
b1 b5 C1 = 0 b3
b3 -1 b4
0 b5
b2 b4

D2=1 Di =0 , i ≠2

0 b1
0 b2
b1 b5
b3 C2 = -1 b3
0 b4
0 b5
b2 b4

D3=1 Di =0 , i ≠3

4/5 b1
0 b2
C3 = 0 b3
b1 b1
0 b4
b5
b3 b5 b3 - 4/5 b5

b2 b4
b2 b4

D4=1 Di =0 , i ≠4

3/5 b1
b1 b5
b1 b3 0 b2
b3 b5
C4 = 1 b3
b2 b4 0 b4
b2 b4 3/5 b5

D5=1 Di =0 , i ≠5

0 b1
b1 0 b2
b5 C5 = 0 b3
b3
1 b4
4/5 b5
b2 b4

D1=1 D2=1 D3=1 D4=1 D5=1


0 0 4/5 3/5 0 A1
1 0 0 0 0 A2
[A] = 0 -1 0 1 0 A3
-1 0 0 0 1 A4
0 0 - 4/5 3/5 4/5 A5

3.- Calculamos el [k'] matriz de rigidez de barra en el sistema local

E = 30000 Klb/in2 4,320,000 Klb/ft2 [𝑘′]=[𝐸𝐴/𝐿]


A= 1 in2 0.006944 ft2

barra 1: L1 = 20 ft barra 2: L2 = 16 ft

[k']1 = 1.50E+03 [k']2 = 1.88E+03

barra 3: L3 = 12 ft barra 4: L4 = 16 ft

[k'] 3 = 2.50E+03 [k']3 = 1.88E+03

barra 5: L5 = 20 ft

[k'] 3 = 1.50E+03

4. Transformamos el [k'] al sistema global [k]

[𝑘_𝑖 ]=[𝐴_𝑖 ]^𝑇∗[ 〖𝑘′ 〗 _𝑖 ]∗[𝐴_𝑖 ]

barra 1:
0
0
[A 1] T = 4/5 [k'1]*[A1] = 0 0 1.20E+03 9.00E+02 0
3/5
0

0 0 0 0 0
0 0 0 0 0
[k 1] = 0 0 9.60E+02 7.20E+02 0
0 0 7.20E+02 5.40E+02 0
0 0 0 0 0

barra 2:
1
0
[A 2] T = 0 [k'2 ]*[A2] = 1.88E+03 0 0 0 0
0
0

1.88E+03 0 0 0 0
0 0 0 0 0
[k2] = 0 0 0 0 0
0 0 0 0 0
0 0 0 0 0

barra 3:
0
-1
[A 3] T = 0 [k'3 ]*[A3] = 0 -2.50E+03 0 2.50E+03 0
1
0

0 0 0 0 0
0 2.50E+03 0 -2.50E+03 0
[k3] = 0 0 0 0 0
0 -2.50E+03 0 2.50E+03 0
0 0 0 0 0

barra 4:
-1
0
[A 4] T = 0 [k'4 ]*[A4] = -1.88E+03 0 0 0 1.88E+03
0
1

1.88E+03 0 0 0 -1.88E+03
0 0 0 0 0
[k4] = 0 0 0 0 0
0 0 0 0 0
-1.88E+03 0 0 0 1.88E+03

barra 5:
0
0
[A 5] T = - 4/5 [k'5 ]*[A5] = 0 0 -1.20E+03 9.00E+02 1.20E+03
3/5
4/5

0 0 0 0 0
0 0 0 0 0
[k5] = 0 0 9.60E+02 -7.20E+02 -9.60E+02
0 0 -7.20E+02 5.40E+02 7.20E+02
0 0 -9.60E+02 7.20E+02 9.60E+02

5.- Ensamblamos la matriz de rigidez [K]

[𝐾]=∑▒[𝑘_𝑖 ]

3.75E+03 0 0 0 -1.88E+03
0 2.50E+03 0 -2.50E+03 0
[K] = [k1]+[k2]+[k3]+[k4]+[k5] = 0 0 1.92E+03 0 -9.60E+02
0 -2.50E+03 0 3.58E+03 7.20E+02
-1.88E+03 0 -9.60E+02 7.20E+02 2.84E+03

6.- Calculo de {R}


R4 80 Klb
R4 R3
3 50 Klb
3 R3

b1 b5

R2 12 ft b3

1 R1 4 R5 R2
2 1 b2 2 R1 b4 4 R5
Q-D
20 Klb
16 ft 16 ft
Hacemos equilibrio en los nudos para determinar el vector {R}

Nudo 2:
Σ Fx = 0 ==> R1 = 0 Klb
Σ Fy = 0 ==> R2 = 20 Klb
Nudo 3:
Σ Fx = 0 ==> R3 = -50 Klb
Σ Fy = 0 ==> R4 = 80 Klb
Nudo 4:
Σ Fx = 0 ==> R5 = 0 Klb

Por lo tanto

R1 0 0 Klb
R2 20 -20 Klb
{R} = R3 = -50 ==> {Q} =- {R} = 50 Klb
R4 80 -80 Klb
R5 0 0 Klb

7.- Calculo de {D} de los desplazamientos en el sistema global


{𝑄}=[𝐾]∗{𝐷} {𝑄}=−{𝑅}
{𝐷}=[𝐾]^(−1)∗{𝑄}
1 2 3 4 5
0.00 3.75E+03 0 0 0 -1.88E+03 1
-20.00 0 2.50E+03 0 -2.50E+03 0 2
{Q} = 50.00 [K] = 0 0 1.92E+03 0 -9.60E+02 3
-80.00 0 -2.50E+03 0 3.58E+03 7.20E+02 4
0.00 -1.88E+03 0 -9.60E+02 7.20E+02 2.84E+03 5

5.33E-04 -3.56E-04 2.67E-04 -3.56E-04 5.33E-04


-3.56E-04 1.80E-03 -3.56E-04 1.40E-03 -7.11E-04
[𝐾]^(−1)= 2.67E-04 -3.56E-04 7.88E-04 -3.56E-04 5.33E-04
-3.56E-04 1.40E-03 -3.56E-04 1.40E-03 -7.11E-04
5.33E-04 -7.11E-04 5.33E-04 -7.11E-04 1.07E-03

Por lo tanto:
D1 4.89E-02 ft
D2 -1.66E-01 ft
{D} = D3 {D} = 7.49E-02 ft
D4 -1.58E-01 ft
D5 9.78E-02 ft

8.- Determinanos los esfuerzos y desplazamientos en el sistema local

Desplazaminetos

{𝑑_𝑖 }=[𝐴_𝑖 ]∗{𝐷}

barra 1 4.89E-02
{d1} = 0 0 4/5 3/5 0 -1.66E-01 -3.47E-02 ft
=
7.49E-02
-1.58E-01
9.78E-02

barra 2 4.89E-02
{d2} = 1 0 0 0 0 -1.66E-01 = 4.89E-02 ft
7.49E-02
-1.58E-01
9.78E-02

barra 3 4.89E-02
{d3} = 0 -1 0 1 0 -1.66E-01 = 8.00E-03 ft
7.49E-02
-1.58E-01
9.78E-02

barra 4 4.89E-02
{d4} = -1 0 0 0 1 -1.66E-01 = 4.89E-02 ft
7.49E-02
-1.58E-01
9.78E-02

barra 5 4.89E-02
{d5} = 0 0 - 4/5 3/5 4/5 -1.66E-01 = -7.64E-02 ft
7.49E-02
-1.58E-01
9.78E-02

Esfuerzos

{𝑞_𝑖 }={𝑞_𝑖 }𝑝+{𝑞_𝑖 }𝑐 donde:


{𝑞_𝑖 }𝑐=[ 〖𝑘′ 〗 _𝑖 ]∗{𝑑_𝑖 } NOTA: como no existen fuerzas en las barras debido el
{q}p = 0
Entonces: 0

{𝑞_𝑖 }={𝑞_𝑖 }𝑝+[ 〖𝑘′〗 _𝑖 ]∗{𝑑_𝑖 } {𝑞_𝑖 }=[ 〖𝑘′〗 _𝑖 ]∗{𝑑_𝑖 }

barra 1
{q1}= 1.50E+03 -3.47E-02 = -52.08 klb Comprensión

barra 2
{q2}= 1.88E+03 4.89E-02 = 91.67 klb Tracción

barra 3
{q3}= 2.50E+03 8.00E-03 = 20.00 klb Tracción

barra 4
{q4}= 1.88E+03 4.89E-02 = 91.67 klb Tracción

barra 5
{q5}= 1.50E+03 -7.64E-02 = -114.58 klb Comprensión

2
0
b1 b5
52.08 klb (C) .
b3 114.58 klb (C)
0
0
k
91.67 klb (T) l 91.67 klb (T)
b2 b b4
(
T
)

You might also like