Ejercicios resueltos de análisis estructural

1. SEGUNDO EXÁMEN DE ANÁLISIS ESTRUCTURAL
Nombre: Ibarra Aliaga Katia M.
Tema 01: Calcular el grado de hiper estaticidad, los momentos en los extremos
en las barras, reacciones en los apoyos A, B, C, D, elaborar los diagramas de
cortante y momento flector de la viga que se muestra usando el método de
pendiente – deflexión.
Calcular el GHT y GL:
 𝐺𝐻𝑇 = 𝑁𝑅 − 𝑁𝐸𝐸 − 𝐶
𝐺𝐻𝑇 = 8 − 3 − 0
𝐺𝐻𝑇 = 5
 𝐺𝐿 = 3𝑁𝑁 − 𝑁𝑅
𝐺𝐿 = 3(4)− 8
𝑮𝑳 = 𝟒
1. Calculo de la rigidez de cada elemento(k)
𝐸 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑒
𝐾0 =
𝐼
𝐿0
=
𝐼
1.5
→ 𝟐𝑬𝑲𝟏 = 2𝐸
𝐼
1.5
=
2
1.5
= 𝟏. 𝟑𝟑
𝐾1 =
𝐼
𝐿1
=
2𝐼
4
→ 𝟐𝑬𝑲𝟐 = 2𝐸
2𝐼
4
=
4
4
= 𝟏
𝐾2 =
1.5𝐼
𝐿2
=
1.5𝐼
5
→ 𝟐𝑬𝑲𝟑 = 2𝐸
1.5𝐼
5
=
3
5
= 𝟎. 𝟔
𝐾3 =
𝐼
𝐿3
=
𝐼
4.5
→ 𝟐𝑬𝑲𝟒 = 2𝐸
𝐼
5
=
2
4.5
= 𝟎. 𝟒𝟒

2. 2. Calculo de los momentos de empotramiento
Tramo :
𝑴𝑽𝑶𝑳𝑨𝑫𝑰𝒁𝑶 = 2𝑇𝑛 𝑥 1.5𝑚 = 3𝑇𝑛. 𝑚
Tramo 1
Lado izquierdo:
𝑀𝐴𝐵 =
−𝑊𝐿2
12
=
−2 ∗ 42
12
= −2.67
Lado derecho:
𝑀𝐵𝐴 =
𝑊𝐿2
12
=
2 ∗ 42
12
= 2.67
Tramo 2:
𝑀𝐵𝐶 = 2𝐸𝐾2(2𝜃𝐵 + 𝜃𝐶 − 3𝜓) + 𝑀𝐸𝑃𝐵𝐶
𝑀𝐵𝐶 = 0.6(2𝜃𝐵 + 𝜃𝐶 ) + 0.48
𝑴𝑩𝑪 = 𝟏. 𝟐𝜽𝑩 + 𝟎. 𝟔𝜽𝑨 + 𝟎.𝟒𝟖

3. 𝑀𝐶𝐵 = 2𝐸𝐾2(2𝜃𝐶 + 𝜃𝐵 − 3𝜓) + 𝑀𝐸𝑃𝐶𝐵
𝑴𝑪𝑩 = 𝟎. 𝟔(𝟐𝜽𝑪 + 𝜽𝑩) − 𝟎. 𝟏𝟖
𝑴𝑪𝑩 = 𝟏. 𝟐𝜽𝑪 + 𝟎. 𝟔𝜽𝑩 − 𝟎. 𝟏𝟖
Tramo 3
𝑀𝐶𝐷 = 3𝐸𝐾3(𝜃𝐶 − 𝜓) + 𝑀𝐸𝑃𝐶𝐷 −
1
2
𝑀𝐸𝑃𝐷𝐶
𝑀𝐶𝐷 = 0.67(𝜃𝐶 − 0.017)− 4.22 −
1
2
∗ 4.22
𝑀𝐶𝐷 = 0.67𝜃𝐶 − 0.017 − 4.22 −
1
2
(4.22)
𝑴𝑪𝑫 = 𝟎.𝟔𝟕𝜽𝑪 − 𝟔. 𝟑𝟓
3. Momentos en los extremos
TRAMO 1:
MAB = 2𝐸𝐾1 ∗ (2𝜃𝐴 + 𝜃𝐵 − 3𝑅) ± 𝑀𝐸𝑃𝐴𝐵
MAB = 1 ∗ (2𝜃𝐴 + 𝜃𝐵 − 0) − 2.67
MAB = 𝟐𝜽𝑨 + 𝜽𝑩 − 𝟐.𝟔𝟕…………………………………. (1)
MBA = 2𝐸𝐾1 ∗ (2𝜃𝐵 + 𝜃𝐴 − 3𝑅) ± 𝑀𝐸𝑃𝐵𝐴
MBA = 1 ∗ (2𝜃𝐵 + 𝜃𝐴 − 0) + 2.67
MBA = 𝟐𝜽𝑩 + 𝜽𝑨 + 𝟐.𝟔𝟕…………………………………. (2)
TRAMO 2:
MBC = 2𝐸𝐾2 ∗ (2𝜃𝐵 + 𝜃𝐶 − 3𝑅) ± 𝑀𝐸𝑃𝐵𝐶
MBC = 0.6 ∗ (2𝜃𝐵 + 𝜃𝐶 − 0) + 0.48
MBC = 1.2𝜃𝐵 + 0.6𝜃𝐶 + 0.48…………………………………. (3)
MCB = 2𝐸𝐾2 ∗ (2𝜃𝐶 + 𝜃𝐵 − 3𝑅) ± 𝑀𝐸𝑃𝐶𝐵
MCB = 0.6 ∗ (2𝜃𝐶 + 𝜃𝐵 − 0) + 0.18
MCB = 1.2𝜃𝐶 + 0.6𝜃𝐵 + 0.18…………………………………. (4)
TRAMO 3:
MCD = 3𝐸𝐾3 ∗ (𝜃𝐶 + 𝜑) ± 𝑀𝐸𝑃𝐶𝐷 −
1
2
𝑀𝐸𝑃𝐷𝐶
MCD = 3 ∗ 0.22 ∗ (𝜃𝐶 − 0.017) − 4.22 −
1
2
4.22
MCD = 0.66𝜃𝐶 − 6.34 …………………………………. (5)

4. 4. Aplicamos sumatoria en los nudos
NODO A
NODO A = M voladizo + MAB = 0
NODO A = 3 + 2𝜃𝐴 + 𝜃𝐵 − 2.67 = 0
NODO A = 2𝜃𝐴 + 𝜃𝐵 = −0.33… …… …(𝟏)
NODO B
NODO B = MBA + MBC= 0
NODO B = 2𝜃𝐵 + 𝜃𝐴 + 2.67 + 1.2𝜃𝐵 + 0.6𝜃𝐶 + 0.48 = 0
NODO B = 𝜃𝐴 + 3.2𝜃𝐵 + 0.6𝜃𝐶 = −3.15… …… (𝟐)
NODO C
NODO C= MCB+ MCD= 0
NODO C= 1.2𝜃𝐶 + 0.6𝜃𝐵 + 0.18 + 0.66𝜃𝐶 − 6.34
NODO C= 0.6𝜃𝐵 + 1.86𝜃𝐶 = 6.16 … ……(𝟑)

5.  APLICAMOS EL SISTEMADE ECUACIONES:
=2𝜃𝐴 + 𝜃𝐵 + 0𝜃𝐶 = −0.33
= 𝜃𝐴 + 3.2𝜃𝐵 + 0.6𝜃𝐶 = −3.15
=0𝜃𝐴 + 0.6𝜃𝐵 + 1.86𝜃𝐶 = 6.16
0.5997 -0.1995 0.0644 -0.33 𝜃𝐴 = 0.83
2 1 0 𝜃𝐴 -0.33 -0.1995 0.3990 -0.1287 -3.15 𝜃𝐵 = −1.98
1 3.2 0.6 𝜃𝐵 = -3.15 0.0644 -0.1287 0.5792 6.16 𝜃𝐶 = 3.95
0 0.6 1.86 𝜃𝐶 6.16
 REEMMPLAZAMOS EN LOS MOMENTOS
MAB = 2*0.83 - 1.98 - 2.67 = -3
MBA = 0.83 + 2*-1.98 + 2.67 = -0.46
MBC = 1.2* -1.98 + 0.6*3.95 + 0.48 = 0.47
MCB = 1.2*3.95 + 0.6 * -1.98 + 0.18 = 3.73
MCD = 0.66 * 3.95 – 6.34 = -3.73
________ _______ _________ ______
6.64 ↑ 2.41 ↑ 7.4 ↑ 4.8 ↑
 DFC

6.  DMF

8. 2.Calcular el grado de hiperestaticidad, los momentos en los extremos de barra, reacciones en los apoyos A, F, E del marco que se muestra, usando el método
pendiente – deflexión. Despreciar las deformaciones axiales yconsiderar como un marco con ladeo horizontal.
 GRADO DE HIPERESTATICIDAD
GH= 3B + R – 3N
B = 5
R = 9
N = 6
GH= 3*5+ 9 – 3*6= 6 …. VIGA HIPERESTÁTICA

9.  ECUACIONES BARRA
 ELEMENTO 1
K1 = I/L1 = 2I/5 2EK1 = 2E2I/5 = 0.8
R=
∆
𝐿
=
∆
5𝑚
= 0.2∆
MEPAB = -(WL^2) /12 = -(3*5^2)/12 = -6.25
MEPBA= 6.25
MAB = 2𝐸𝐾1 ∗ (2𝜃𝐴 + 𝜃𝐵 − 3𝑅) ± 𝑀𝐸𝑃𝐴𝐵
MAB = 0.8 ∗ (𝜃𝐵 − (3 ∗ 0.2∆)) − 6.25
MAB = 0.8 𝜃𝐵 − 0.48∆ − 6.25
MBA= 2𝐸𝐾1 ∗ (2𝜃𝐵 + 𝜃𝐴 − 3𝑅) ± 𝑀𝐸𝑃𝐵𝐴
MBA= 0.8 ∗ (2𝜃𝐵 − (3 ∗ 0.2∆))+ 6.25
MBA= 1.6𝜃𝐵 − 0.48∆ + 6.25
 ELEMENTO 2
K2 = I/5 2EK2 = 2EI/5 = 0.4

10. MEPBC= −
𝑃𝐿
8
=
5∗5
8
= −3.12
MEPBC= 3.12
MBC= 2𝐸𝐾2 ∗ (2𝜃𝐵 + 𝜃𝐶 − 3𝑅) ± 𝑀𝐸𝑃𝐵𝐶
MBC= 0.4 ∗ (2𝜃𝐵 + 𝜃𝐶 − 0) − 3.12
MBC= 0.8𝜃𝐵 + 0.4𝜃𝐶 − 3.12
MCB= 2𝐸𝐾2 ∗ (2𝜃𝐶 + 𝜃𝐵 − 3𝑅) ± 𝑀𝐸𝑃𝐶𝐵
MCB= 0.4 ∗ (2𝜃𝐶 + 𝜃𝐵 − 0) + 3.12
MCB= 0.8𝜃𝐶 + 0.4𝜃𝐵 + 3.12
 ELEMENTO 3
K3 = I/L3 = 2I/5 2EK3 = 2E2I/5 = 0.8
R=
∆
𝐿
=
∆
5𝑚
= 0.2∆
MEPCF = -(WL^2) /12 = -(3*5^2)/12 = -6.25
MEPFC= 6.25
MCF = 2𝐸𝐾3 ∗ (2𝜃𝐶 + 𝜃𝐹 − 3𝑅) ± 𝑀𝐸𝑃𝐶𝐹
MCF = 0.8 ∗ (2𝜃𝐶 + (3 ∗ 0.2∆)) − 6.25
MCF = 1.6 𝜃𝐶 + 0.48∆ − 6.25

11. MFC= 2𝐸𝐾3 ∗ (2𝜃𝐹 + 𝜃𝐶 − 3𝑅) ± 𝑀𝐸𝑃𝐹𝐶
MFC= 0.8 ∗ (𝜃𝐶 + (3 ∗ 0.2∆)) + 6.25
MFC= 0.8𝜃𝐶 + 0.48∆ + 6.25
 ELEMENTO 4
K4 = I/5 2EK4 = 2EI/5 = 0.4
MEPCD= −
𝑃𝑎𝑏2
𝑙2
=
4∗3∗22
52
= −1.92
MEPDC=
𝑃𝑏𝑎2
𝑙2
=
4∗2∗32
52
= 2.88
MCD= 2𝐸𝐾4 ∗ (2𝜃𝐶 + 𝜃𝐷 − 3𝑅) ± 𝑀𝐸𝑃𝐵𝐶
MCD= 0.4 ∗ (2𝜃𝐶 + 𝜃𝐷 − 0) − 1.92
MCD= 0.8𝜃𝐶 + 0.4𝜃𝐷 − 1.92
MDC= 2𝐸𝐾4 ∗ (2𝜃𝐷 + 𝜃𝐶 − 3𝑅) ± 𝑀𝐸𝑃𝐶𝐵
MDC= 0.4 ∗ (2𝜃𝐷 + 𝜃𝐶 − 0) + 2.88
MDC= 0.8𝜃𝐷 + 0.4𝜃𝐶 + 2.88
 ELEMENTO 5
K5 = 2I/3 2EK3 = 2E2I/3 = 1.33
R=
∆
𝐿
=
∆
3𝑚
= 0.33∆

12. MEPDE= 0
MEPED= 0
MDE= 2𝐸𝐾5 ∗ (2𝜃𝐷 + 𝜃𝐸 − 3𝑅) ± 𝑀𝐸𝑃𝐷𝐸
MDE= 1.33 ∗ (2𝜃𝐷 − (3 ∗ 0.33∆))
MDE= 2.66 𝜃𝐷 − 1.32∆
MED= 2𝐸𝐾5 ∗ (2𝜃𝐸 + 𝜃𝐷 − 3𝑅) ± 𝑀𝐸𝑃𝐷𝐸
MED=1.33 ∗ (𝜃𝐷 − (3 ∗ 0.33∆))
MED= 1.33𝜃𝐷 − 1.32∆
 ECUACIONES DE NODO
MB= MBA+ MBC= 0
MB=1.6𝜃𝐵 − 0.48∆ + 6.25 + 0.8𝜃𝐵 + 0.4𝜃𝐶 − 3.12 = 0
MB= 2.4𝜃𝐵 + 0.4𝜃𝐶 − 0.48∆ = −3.13… …… …. (1)
MC= MCB + MCD+ MCF = 0
MC= 0.8𝜃𝐶 + 0.4𝜃𝐵 + 3.12 + 0.8𝜃𝐶 + 0.4𝜃𝐷 − 1.92 + 1.6 𝜃𝐶 + 0.48∆ − 6.25 = 0
MC=0.4𝜃𝐵 + 3.2 𝜃𝐶 + 0.4𝜃𝐷 + 0.48∆ = 5.05 … ……… .(2)

13. MD= MDC+ MDE= 0
MD= 0.8𝜃𝐷 + 0.4𝜃𝐶 + 2.88 + 2.66 𝜃𝐷 − 1.32∆ = 0
MD= 0.4𝜃𝐶 + 3.46𝜃𝐷 − 1.32∆ = −2.88 …… …… (3)
 ΣFH = HA + HF + HE + (3*5) – (3*5) = 0
HA + HF + HE = 0

14.  ΣMB =
−𝑊𝐿2
2
+ 𝑀𝐴𝐵+ 𝑀𝐵𝐴 − 𝐻𝐴 ∗ 5 = 0
=
−3∗52
2
+ 0.8 𝜃𝐵 − 0.48∆ − 6.25 + 1.6𝜃𝐵 − 0.48∆ + 6.25 − 𝐻𝐴 ∗ 5 = 0
=−37.5 + 0.8 𝜃𝐵 − 0.48∆ − 6.25 + 1.6𝜃𝐵 − 0.48∆ + 6.25 = 𝐻𝐴 ∗ 5
HA =
2.4 𝜃𝐵−0.96∆
5
− 7.5
 ΣMC =
𝑊𝐿2
2
+ 𝑀𝐹𝐶 + 𝑀𝐶𝐹 − 𝐻𝐹 ∗ 5 = 0
=
3∗52
2
+ 0.8 𝜃𝐶 + 0.48∆ + 6.25 + 1.6𝜃𝐶 + 0.48∆ − 6.25 − 𝐻𝐹 ∗ 5 = 0
=37.5 + 0.8 𝜃𝐶 + 0.48∆ + 6.25 + 1.6𝜃𝐶 + 0.48∆ − 6.25 = 𝐻𝐹 ∗ 5
HF =
2.4 𝜃𝐶+0.96∆
5
+ 7.5
 ΣMD = 𝑀𝐷𝐸 + 𝑀𝐸𝐷 − 𝐻𝐸 ∗ 3 = 0
= 2.66 𝜃𝐷 − 1.32∆ + 1.33𝜃𝐷 − 1.32∆ = 𝐻𝐸 ∗ 3
HE =
3.99 𝜃𝐷−2.64∆
3
 HA + HF + HE = 0

15. 0 =
2.4 𝜃𝐵−0.96∆
5
− 7.5 +
2.4 𝜃𝐶+0.96∆
5
+ 7.5 +
3.99 𝜃𝐷−2.64∆
3
0 =
2.4 𝜃𝐵+2.4 𝜃𝐶
5
+
3.99 𝜃𝐷−2.64∆
3
0 = 0.48 𝜃𝐵 + 0.48 𝜃𝐶 + 1.33 𝜃𝐷 − 0.88∆ … … … … (4)
 SISTEMADE ECUACIONES
= 2.4𝜃𝐵 + 0.4𝜃𝐶 + 0𝜃𝐷 − 0.48∆ = −3.13
= 0.4𝜃𝐵 + 3.2 𝜃𝐶 + 0.4𝜃𝐷 + 0.48∆ = 5.05
= 0𝜃𝐵 + 0.4𝜃𝐶 + 3.46𝜃𝐷 − 1.32∆ = −2.88
= 0.48 𝜃𝐵 + 0.48 𝜃𝐶 + 1.33 𝜃𝐷 − 0.88∆ = 0
2.4 0.4 0 -0.48 𝜃𝐵 -3.13
0.4 3.2 0.4 0.48 𝜃𝐶 5.05
0 0.4 3.46 -1.32 𝜃𝐷 = -2.88
0.48 0.48 1.33 -0.88 ∆ 0

16.  REEMPLAZANDO EN MOMENTOS
MAB = (0.8 ∗ −2.53) − (0.48 ∗ −3.78) − 6.25 = −6.46
MBA= (1.6 ∗ −2.53) − (0.48 ∗ −3.78) + 6.25 = 4.02
MBC= (0.8 ∗ −2.53) + (0.4 ∗ 2.79) − 3.12 = −4.03
MCB= (0.8 ∗ 2.79) + (0.4 ∗ −2.53) + 3.12 = 4.34
MCF = (1.6 ∗ 2.79) + (0.48 ∗ −3.78) − 6.25 = −3.6
MFC= (0.8 ∗ 2.79) + (0.48 ∗ −3.78) + 6.25 = 6.67
MCD= (0.8 ∗ 2.79) + (0.4 ∗ −2.60) − 1.92 = −0.73
MDC= (0.8 ∗ −2.60) + (0.4 ∗ 2.79) + 2.88 = 1.92
MDE= (2.66 ∗ −2.60) − (1.32 ∗ −3.78) = −1.93
MED= (1.33 ∗ −2.60) − (1.32 ∗ −3.78) = 1.53