SOLUCION:

F´c = 280 kg/cm2 fy= 36 kg/cm2

WD = 2 T/m WL = 3 T/m

Wu = 1.2(2) + 1.6(3) = 7 T/m

WDu = 1.2(2) = 2.4 T/m

TRAMO AB:

0 + 2MA (0 + 4.9) + MB (4.9) = 7¿¿9.8MA + 4.9MB = -205,886 ……………………..(1)

TRAMO ABC:

MA (4.9) + 2MB (4.9 + 7) + MC (7)= W ¿¿ + W ¿¿

4.9MA + 23.8MB + 7MC = -411.686 …………………………… (2)

TRAMO BC: (SEGUNDO TRAMO IMAGINARIO)

2MC (0 + 7) + MB (7) = 2.4¿¿

7MB + 14MC = -205,8 …………………………(3)

3.5MB + 7MC = -102.9

RESOLVIENDO: 1, 2 Y 3

9.8MA + 4.9MB + 0 = -205.886

(-2) 4.9MA + 23.8MB + 7MC = -411.686

9.8MA + 4.9MB + 0 = -205.886

9.8MA – 47.6MB - 14MC = 823.37

-42.7MB - 14MC = 617.484

AA B C

7 T/m

2.4 T/m

4.9 m 7 m

-42.7MB - 14MC = 617.484

7MB + 14MC = - 205.8

-35.7MB = 411.684

MB = -11.532

MC = -8.934

MA = -15.243

BARRA AB:

∑FY = 0

VA¨ + VB

´ = 34.3

∑MA = 0

15.243 + VB´ (4.9) = 34.3(2.45) + 11.532

VB´ = 16.40 VA

¨ = 17.9

TRAMO BC :

∑FY = 0

VB¨ + VC

´ = 16.8

A B

15.243 11.532

VB´VA

¨

4.9

34.37 T/m

VB ¨ VC

´

B C

11.532 8.93416.8

7

L1 = 0.7L2 4m ≤ L2 ≤ 8m

L1 = 4.9 m L2 = 7m

b = 30 m para L2 ≤ 5m

b = 40 m para 6m ≤ L2 ≤ 8m

f´c fy WD WL

4m ≤ L2 ≤ 5m 210 4200 1 2

6m ≤ L2 ≤ 7m 280 3600 2 3

L =8m 360 4200 3 4

a. dimensionar la viga para M+max.

b. Calcular As según corresponda para todos los M+ y M- N° 6 ≤ ᶲ ≤ N° 10

c. Diseñar por corte toda la viga N° 3 ≤ ᶲv ≤ N° 5

2. Trazar diagrama Trilineal

Seccion M+ max.

Seccion M- min.

AL1 L2

0.4